Introduction to Real Analysis (2nd Edition)

Introduction to Real Analysis

(*, ) Manfred Stoll

English reprint copyright ®2004 by Pearson Education North Asia Limited and China Machine Press. Original English Language title: Introduction to Real Analysis, Second Edition by Manfred Stoll EISBN 0-321-04625-0 Copyright ®2001 by Addison Wesley Longman, Inc. All right reserved. Published by arrangement with the original publisher, Pearson Education. Inc. , publishing as Addison Wesley Longman, Inc. 1 L EI1

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Contents

PREFACE x

TO THE STUDENT xv

i.

J

The Real Number System

i

1.1 Sets and Operations on Sets 2 1.2 Functions 6 1.3 Mathematical Induction 15 1.4 The Least Upper Bound Property 20 1.5 Consequences of the Least upper Bound Property 28 1.6 Binary and Ternary Expansions 30 1.7 Countable and Uncountable Sets 34

Notes 43 Miscellaneous Exercises 44 Supplemental Reading 46

21 Sequences of Real Numbers 2.1 Convergent Sequences 48

2.2 Limit Theorems

53

2.3 Monotone Sequences 60

2.4 Subsequences and the Bolzano-Weierstrass Theorem 67 2.5 Limit Superior and Inferior of a Sequence 73 2.6 Cauchy Sequences 80

2.7 Series of Real Numbers 86 Notes 90 Miscellaneous Exercises 90 Supplemental Reading 92

vi

47

vii

Contents

3 1 Structure of Point Sets 3.1 Open and Closed Sets 3.2 Compact Sets 101 3.3 The Cantor Set 107

93

93

Notes 110 Miscellaneous Exercises 111 Supplemental Reading 113

Q Limits and Continuity 116 4.2 Continuous Functions 130 4.3 Uniform Continuity 144 4.4 Monotone Functions and Discontinuities

115

4.1 Limit of a Function

148

Notes 162 Miscellaneous Exercises 162 Supplemental Reading 163

5 j Differentiation 5.1 The Derivative 166 5.2 The Mean Value Theorem 5.3 L'Hospital's Rule 190 5.4 Newton's Method 197

165 176

Notes 203 Miscellaneous Exercises 204 Supplemental Reading 205

The Riemann and

6 Riemann-Stieltjes Integral

207

6.1 The Riemann Integral 208 6.2 Properties of the Riemann Integral 223 6.3 Fundamental Theorem of Calculus 23A 6.4 Improper Riemann Integrals 239 6.5 The Riemann-Stieltjes Integral 245 6.6 Numerical Methods 260 6.7 Proof of Lebesgue's Theorem 272 Notes 276 Miscellaneous Exercises 277 Supplemental Reading 278

7J Series of Real Numbers 7.1 Convergence Tests 280 7.2 The Dirichlet Test 294

279

viii

Contents

7.3 Absolute and Conditional Convergence 7.4 Square Summable Sequences 306

299

Notes

313 Miscellaneous Exercises 314 Supplemental Reading 315

$

Sequences and Series of Functions 8.1 Pointwise Convergence and Interchange of Limits

317 318

8.2 Uniform Convergence 323 8.3 Uniform Convergence and Continuity 330 8.4 Uniform Convergence and Integration 337 8.5 Uniform Convergence and Differentiation 339 8.6 The Weierstrass Approximation Theorem 346 8.7 Power Serves Expansions 353 8.8 The Gamma Function 372 Notes 377 Miscellaneous Exercises 377 Supplemental Reading 378

j ..

q r Orthogonal Functions and Fourier Series

379

9.1 Orthogonal Functions 380 9.2 Completeness and Parseval's Equality 390 9.3 Trigonometric and Fourier Series 394 9.4 Convergence in the Mean of Fourier Series 404 9.5 Pointwise Convergence of Fourier Series 415 Notes 426 Miscellaneous Exercises 428 Supplemental Reading 428

Lebesgue Measure and Integration 10.1 Introduction to Measure 430 10.2 Measure of Open Sets; Compact Sets 432 10.3 Inner and Outer Measure; Measurable Sets 444 10.4 Properties of Measurable Sets 449 10.5 Measurable Functions 455 10.6 The Lebesgue Integral of a Bounded Function 462 10.7 The General Lebesgue Integral 473 10.8 Square Integrable Functions 484 Notes

491

Miscellaneous Exercises

Supplemental Reading

492 493

429

ix

Contents

Logic and Proofs

APPENDIX:

A.1 Propositions and Connectives 496 A.2 Rules of Inference 500 A.3 Mathematical Proofs 507 A.4 Use of Quantifiers 515 Supplemental Reading 521

Bibliography 522 Hints and Solutions to Selected Exercises

Notation Index Index

545

543

523

495

Preface

The subject of real analysis is one of the fundamental areas of mathematics, and is the foundation for the study of many advanced topics, not only in mathematics, but also in engineering and the physical sciences. A thorough understanding of the concepts of real analysis has also become increasingly important for the study of advanced topics in economics and the social sciences. Topics such as Fourier series, measure theory, and integration are fundamental in mathematics and physics as well as engineering, economics, and many other areas. Due to the increased importance of real analysis in many diverse subject areas, the typical first semester course on this subject has a varied student enrollment in terms of both ability and motivation. From my own experience, the audience typically includes mathematics majors, for whom this course represents the only rigorous treatment of analysis in their collegiate career, and students who plan to pursue graduate study in mathematics. In addition, there are mathematics education majors who need a strong background in analysis in preparation for teaching high school calculus. Occasionally, the enrollment includes graduate students in economics, engineering, physics, and other areas, who need a thorough treatment of analysis in preparation for additional graduate study either in mathematics or their own subject area. In an ideal situation, it would be desirable to offer separate courses for each of these categories of students. Unfortunately, staffing and enrollment usually make such choices impossible. In the preparation of the text there were several goals I had in mind. The first was to write a text suitable for a one-year sequence in real analysis at the junior or senior level, providing a rigorous and comprehensive treatment of the theoretical concepts of analysis. The topics chosen for inclusion are based on my experience in teaching graduate courses in mathematics, and reflect what I feel are minimal requirements for successful graduate study. I get to the least upper bound property as quickly as possible, and emphasize this important property in the text. For this reason, the algebraic properties of the rational and real number systems are treated very informally, and the construction of the real number system from the rational numbers is included only as a miscellaneous exercise. I have attempted to keep the proofs as concise as possible, and to x

Preface

xi

let the subject matter progress in a natural manner. Topics or sections that are not specifically required in subsequent chapters are indicated by a footnote. My second goal was to make the text understandable to the typical student enrolled in the course, taking into consideration the variations in abilities, background. and motivation. For this reason, Chapters 1 through 6 have been written with the intent to be accessible to the average student, while at the same time challenging the more talented student through the exercises. The basic topological concepts of open, closed, and compact sets, as well as limits of sequences and functions, are introduced for the real num-

bers only. However, the proofs of many of the theorems, especially those involving topological concepts, are presented in a manner that permits easy extensions to more abstract settings. These chapters also include a large number of examples and more routine and computational exercises. Chapters 7 through 10 assume that the students have achieved some level of expertise in the subject. In these chapters, function spaces are introduced and studied in greater detail. The theorems, examples, and exercises require greater sophistication and mathematical maturity for full understanding. From my own experiences, these are not unrealistic expectations.

The book contains most of the standard topics one would expect to find in an introductory text on real analysis-limits of sequences, limits of functions, continuity, differentiation, integration, series, sequences and series of functions, and power series. These topics are basic to the study of real analysis and are included in most texts at this level. In addition, I have included a number of topics that are not always included in comparable texts. For instance, Chapter 6 contains a section on the Riemann-Stieltjes integral, and a section on numerical methods. Chapter 7 also includes a section on square summable sequences and a brief introduction to nonmed linear spaces. Both of these concepts appear again in later chapters of the text. In Chapter 8, to prove the Weierstrass approximation theorem, I use the method of approximate identities. This exposes the student to a very important technique in analysis that is used again in the chapter on Fourier series. The study of Fourier series, and the representation of functions in terms of series of orthogonal functions, has become increasingly important in many diverse areas. The inclusion of Fourier series in the text allows the student to gain some exposure to this important subject, without the necessity of taking a full semester course on partial differential equations. In the final chapter I have also included a detailed treatment of Lebesgue measure and the Lebesgue integral. The approach to measure theory follows the original method of Lebesgue, using inner and outer measure. This provides an intuitive and leisurely approach to this very important topic. The exercises at the end of each section are intended to reinforce the concepts of the section and to help the students gain experience in developing their own proofs. Although the text contains some routine and computational problems, many of the exercises are designed to make the students think about the basic concepts of analysis, and to challenge their creativity and logical thinking. Solutions and hints to selected exercises are included at the end of the text. These. problems are marked by an asterisk (*). At the end of each chapter I have also included a section of notes on the chapter, miscellaneous exercises, and a supplemental reading list. The notes in many cases pro-

xii

Preface vide historical comments on the development of the subject, or discuss topics not included in the chapter. The miscellaneous exercises are intended to extend the subject matter of the text or to cover topics that, although important, are not covered in the chapter itself. The supplemental reading list provides references to topics that relate to the subject under discussion. Some of the references provide historical information; others provide alternative solutions of results or interesting related problems. Most of the articles appear ih the American Mathematical Monthly or Mathematics Magazine. and should be easily accessible for students' reference. To cover all the chapters in a one-year sequence is perhaps overly ambitious. However, from my own experience in teaching the course, with a judicious choice of topics it is possible to cover most of the text in two semesters. A one-semester course should at a minimum include all or most of the first five chapters, and part or all of Chapter 6 or Chapter 7. The latter chapter can be taught independently of Chapter 6; the only dependence on Chapter 6 is the integral test, and this can be covered without a theoretical treatment of Riemann integration. The remaining topics should be more than sufficient for a full second semester. The only formal prerequisite for reading the text is a standard three- or four-semester sequence in calculus. Even though an occasional talented student has completed one semester of this course during their sophomore year. some mathematical maturity is expected, and the average student might be advised to take the course during their junior or senior year.

Features New to the Second Edition In content, the second edition remains primarily unchanged from the first. The subject of real analysis has not changed significantly since publication of the first edition. In this edition I have incorporated many of the valuable suggestions from reviewers, instructors, and students. Some new topics have been included, and the presentation of others has been revised. New examples and revised explanations appear throughout this edition of the text. The second edition also contains additional illustrations and expanded problem sets. The problem sets in all sections of the first six chapters have been expanded to include more routine and computational problems. The challenging problems are still there. With the addition of more routine problems, instructors using this text will have greater flexibility in the assignment of exercises. The supplemental reading lists have all been updated to include relevant articles that have appeared since 1996. Two of the more substantive changes are the inclusion of a proof of Lebesgue's theorem in Chapter 6, and the addition of an appendix on logic and proofs. In the first edition, Lebesgue's theorem was stated in Section 6.1 and then proved in Chapter 10. At the recommendation of my colleague Anton Schep, I have included a self-contained proof of Lebesgue's theorem as a separate section in Chapter 6. The proof is based on notes that he has used to supplement the text. In the proof, as in the statement of the theorem, the only reference to measure theory is the definition of a set of measure zero. With this change it is now possible not only to state but also to prove this important theorem without first having to develop the theory of Lebesgue measure and integration. The greatest difficulty facing many students taking a course in real analysis is the ability to write and to understand proofs. Most have never had a course in mathemati-

Preface

xiii

cal logic. For this reason I have included a brief appendix on logic and proofs. The appendix is not intended to replace a formal course in logic; it is only intended to introduce the rules of logic that students need to know in order to better understand proofs. These rules are also crucial in helping students develop the ability to write their own proofs. The various methods of proof are discussed in detail, and examples of each method are included and analyzed. The appendix also includes a section on the use of quantifiers. with special emphasis on the proper negation of quantified sentences. The

appendix itself is independent of the text: however, references to it are included throughout the first several chapters of the text. The appendix can be included as part of the course, or assigned as independent reading.

LL

Acknowledgments I would like to thank the students at South Carolina who have learned this material from

me, or my colleagues, from preliminary versions of this text. Your criticisms, comments, and suggestions were appreciated. I am also indebted to those colleagues, especially the late Jeong Yang, who agreed to use the manuscript in their courses. Special thanks are also due to the reviewers who examined the manuscript for the first edition and provided constructive criticisms and suggestions for its improvement: Joel Anderson, Pennsylvania State University; Bogdam Baishanski, Ohio State University; Robert Brown, University of Kansas; Donald Edmondson, University of Texas at Austin; Kevin Grasse, University of Oklahoma; Harvey Greenwald, California Polytechnic State University; Adam Helfer, University of Missouri, Columbia: Jan Kucera. Washington State University; Thomas Reidel, University of Louisville; Joel Robin, University of Wisconsin, Madison; Stuart Robinson, Cleveland State University; Dan Shea, University of Wisconsin, Madison; Richard B. Sher, University of North Car-

olina: Thomas Smith, Manhattan College. Your careful reading of the manuscript helped to turn the preliminary drafts into a polished text. 1 would also like to thank Carolyn Lee-Davis and the staff at Addison Wesley

Longman for their assistance in the preparation of the second edition, and the reviewers for this edition for their comments and recommendations: William Barnier, Sonoma State University; Rene Barrientos, Miami Dade Community College; Denine Burkett, Lock Haven University; Steve Deckelman, University of Wisconsin; Lyn

Geisler, Randolph-Macon College; Constant Goutziers, State University of New York; Christopher Heil, Georgia Institute of Technology; William Stout. Salve Regina University. Special thanks go to my colleague, George McNulty, for his careful reading of the appendix. His constructive criticisms and suggestions were appreciated. I am also grateful to the readers who informed me of errors in the first edition, and to the instructors who conveyed to me some of the difficulties encountered while using

the book as a text. Hopefully all of the errors and shortcomings of the first edition have been corrected. Finally, I would especially like to thank my wife, Mary Lee, without whose encouragement this project might never have been completed. Manfred Stoll

To the Student

The difference between a course on calculus and a course on real analysis is analogous to the difference in the approach to the subject prior to the nineteenth century and since that time. Most of the topics in calculus were developed in the late seventeenth and eighteenth centuries by such prominent mathematicians as Newton, Leibniz, Bernoulli, Euler, and many others. Newton and Leibniz developed the differential and integral calculus; their successors extended and applied the theory to many problems in mathematics and the physical sciences. They had phenomenal insight into the problems, and

were extremely proficient and ingenious in deriving complex formulas. What they lacked, however, were the tools to place the subject on a rigorous mathematical foundation. This did not occur until the nineteenth century with the contributions of Cauchy, Bolzano, Weierstrass, Cantor, and many others.

In calculus, the emphasis is primarily on developing expertise in computational techniques and applications. In real analysis, you will be expected to understand the concepts and to develop the ability to prove results using the definitions and previous theorems. Understanding the concept of a limit, and proving results about limits, will be significantly more important than computing limits. To accomplish this, it is essential that all definitions and statements of theorems be learned precisely. Most of the proofs of the theorems and solutions of the problems are logical consequences of the definitions and previous results; some, however, do require ingenuity and creativity. The text contains numerous examples and counterexamples to illustrate the particular topics under discussion. These are included to show why certain hypotheses are required, and to help develop a more thorough understanding of the subject. It is crucial that you not only learn what is true, but that you also have sufficient counterexamples

at your disposal. I have included hints and answers to selected exercises at the end of the text; these are indicated by an asterisk (*). For some of the problems I have provided complete details; for others I have provided only brief hints, leaving the details to you. As always, you are encouraged to first attempt the exercises, and to look at the hints or solutions only after repeated attempts have been unsuccessful. xc

To the Student

xV

At the end of each chapter I have included a supplemental reading list. The journal articles or books are all related to the topics in the chapter. Some provide historical information or extensions of the topics to more general settings; others provide alternative solutions of results in the text. or solutions of interesting related problems. All of the articles should be accessible in your library. They are included to encourage you to develop the habit of looking into the mathematical literature. An excellent source for additional historical information and biographies of famous mathematicians is the MacTutor History of Mathematics archive at the University of St. Andrews, Scotland. The URL of their webpage is http://www-history.mcs.standrews.as.uk/ On reading the text you will inevitably encounter topics, formulas, or examples that may appear too technical and difficult to comprehend. Skip them for the moment; there will be plenty for you to understand in what follows. Upon later reading the section, you may be surprised that it is not nearly as difficult as previously imagined. Concepts that initially appear difficult become clearer once you develop a greater understanding of the subject. It is important to keep in mind that many of the examples and topics that appear difficult to you were most likely just as difficult to the mathematicians of the era in which they first appeared. The material in the text is self-contained and independent of calculus. I do not use any results from calculus in the definitions and development of the subject matter. Occasionally, however, in the examples and exercises I do assume knowledge of the elementary functions and of notation and concepts that should have been encountered elsewhere. These concepts will be defined carefully at the appropriate place in the text. Manfred Stoll

The Real Number System 1.1

Sets and Operations on Sets

1.2 Functions

1.3 Mathematical Induction 1.4 The Least Upper Bound Property 1.5 Consequences of the Least Upper Bound Property 1.6 Binary and Ternary Expansions

1.7 Countable and Uncountable Sets

The key to understanding many of the fundamental concepts of calculus, such as limits, continuity, and the integral, is the least upper bound property of the real number system It As we all know, the rational number system contains gaps. For example. there does

not exist a rational number r such that r2 = 2, i.e., f is irrational. The fact that the rational numbers do contain gaps makes them inadequate for any meaningful discussion of the above concepts. The standard argument used in proving that the equation r2 = 2 does not have a solution in the rational numbers goes as follows: Suppose that there exists a rational number r such that r2 = 2. Write r = m/n where in, it are integers that are not both even. Thus m' = 2n2. Therefore in2 is even, and hence in itself must be even. But m2, and hence also 2n2, are both divisible by 4. Therefore n2 is even, and as a consequence n is also even. This, however, contradicts our assumption that not both in and n are even. The method of proof used in this example is proof by contradiction; namely, we assume the negation of the conclusion and arrive at a logical contradiction. A discussion of the various methods of proof is included in Section A.3 of the Appendix. The above argument shows that there does not exist a rational number r such that r2 = 2. This argument was known to Pythagoras (around 500 B.C.), and even the Greek mathematicians of this era noted that the straight line contains many more points than the rational numbers. It was not until the nineteenth century, however, when mathematicians became concerned with putting calculus on a firm mathematical footing, that the development of the real number system was accomplished. The construction of the real number system is attributed to Richard Dedekind (1831-1916) and Georg Cantor 0

2

Chapter I

The Real Number System

(1845-1917), both of whom published their results independently in 1872. Dedekind's aim was the construction of a number system. with the same completeness as the real line, using only the basic postulates of the integers and the principles of set theory. Instead of constructing the real numbers, we will assume their existence and examine the least upper bound property. As we will see, this property is the key to many basic facts about the real numbers that are usually taken for granted in the study of calculus. In Chapter 1 we will assume a basic understanding of the concept of a set and also of both the rational and real number systems. In Section 1.4 we will briefly review the algebraic and order properties of both the rational and real number systems and discuss the least upper bound property. By example we will show that this property fails for the rational numbers. In the subsequent two sections we will prove several elementary consequences of the least upper bound property. In Section 1.7 we define the notion of a countable set and consider some of the basic properties of countable sets. Among the key results of this section are that the rational numbers are countable, whereas the real numbers are not.

Sets and Operations on Sets Sets are constantly encountered in mathematics. One speaks of sets of points, collections of real numbers, and families of functions. A set is conceived simply as a collection of definable objects. The words set, collection, and family are all synonymous. The notation .r e A means that x is an element of the set A; the notation x e A means that x is not an element of the set A. The set containing no elements is called the empty set and will be denoted by 0.

A set can be described by listing its elements, usually within braces { }. For example,

A = {-1,2.5.4} describes the set consisting of the numbers -1, 2, 4, and 5. More generally, a set A may be defined as the collection of all elements x in some larger collection satisfying a given property. Thus the notation

A = {x : P(x)} defines A to be the set of all objects x having the property P(x). This is usually read as "A equals the set of all elements x such that P(x)." For example, if x ranges over all real numbers, the set A defined by

A = {x:1 <x<5} is the set of all real numbers that lie between 1 and 5. For this example, 3.75 E A. whereas 5 E A. We will also use the notation A = {x E X : P(x)} to indicate that only those x that are elements of X are being considered. Some basic sets that we will encounter throughout the text are the following:

N = the set of natural numbers or positive Integers = {1. 2, 3, ...}, Z = the set of all integers = {. . ., -2, -1, 0, 1, 2... .},

1.1

Sets and Operations on Sets

3

O = the set of rational numbers = { p/q : p, q E Z, q # 0}. and IlR = the set of real numbers. In addition, we will occasionally encounter the set 10, 1, 2, 3, .. .} of nonnegative integers.

Real numbers that are not rational numbers are called irrational numbers. Since

many fractions can represent the same rational number, two rational numbers r, = p,/q, and r, = p2/q2 are equal if and only if q, P2 = ptg2. Set-theoretically the rational numbers can be defined as sets of ordered pairs of integers (m, n), n * 0, where two ordered pairs (p,, q,) and (p2, q2) are said to be equivalent (represent the same rational number) if ptq, = p2q,. We assume that the reader is familiar with the algebraic operations of addition and multiplication of rational numbers. A set A is a subset of a set B or is contained in B, denoted A C B. if every element of A is an element of B. The set A is a proper subset of B, denoted A C# B if A is a subset of B, but there is an element of B that is not an element of A. Two sets A and

B are equal, denoted A = B, if A C B and B C A. By definition, the empty set (6 is a subset of every set.

Set Operations There are a number of elementary operations that may be performed on sets. If A and B are sets, the union of A and B, denoted A U B, is the set of all elements that belong either to A or to B or to both A and B. Symbolically,

AUB={x:xEA or xEB}. The intersection of A and B, denoted A fl B, is the set of elements that belong to both A and B; that is,

Af1B={x:xEA and xEB}. Two sets A and B are disjoint if A fl B = ¢. The relative complement of A in B, denoted B \A, is the set of all elements that are in B but not in A. In set notation,

B\ A= {x : x E B and x q-A}. If the set A is a subset of some fixed set X. then X \ A is usually referred to as the com-

plement of A and is denoted by A`. These basic set operations are illustrated in Figure 1.1 with the shaded areas representing A U B, A fl B, and B \ A, respectively.

AUB

AnB Figure 1.1

BMA

4

Chapter 1

The Real Number System

There are several elementary set theoretic identities that will be encountered throughout the text. We state some of these in the following theorem; others are given in the exercises.

1.1.1

THEOREM If A. B, and C are sets, then

(a) An(BUC)=(AnB)u(Anc), (b) AU(BnC)=(AUB)n(AUC), (c) C \ (A U B) = (C \ A) n (C \ B),

(d) c \ (A n B) = (C \ A) U (C\ B). The identities (a) and (b) are referred to as the distributive laws, whereas (c) and (d) are De Morgan's laws. If A and B are subsets of a set X, then De Morgan's laws can also be expressed as

(AUB)` =A`nB`,

(A nB)` =A`UB`.

A more general version of both the distributive laws and De Morgan's laws will be stated in Theorems 1.7.12 and 1.7.13. Proof. We will provide the proof of (a) to illustrate the method used in proving these results. The proofs of (b)-(d) are relegated to the exercises (Exercise 7). To prove the identity in (a) we must prove that

An(BUC)C(AnB)U(AnC)

(AnB)U(AnC)CAn(BUC). Suppose xEAn(BUC).Then xEAand xEBUC.Since xEBUC,xEBor xEC.IfxEB,then xEAnBandtherefore n C x (A n C). This then proves that xE x An(BUC) c (AnB)u(Anc). and

To complete the proof, it still has to be shown that (A n B) U (A n C) C An (B U C), thereby proving equality. If x E (A n B) U (An c). then by definition

xEAnBor xER An C,or both. But if xEAnB,then xEE Aand xEB.Since xEBwealso have xEBUC.Therefore xEAn(BUC).Similarly, ifxEAnC,

then xEAn(BUC). Q

If A is any set, the set of all subsets of A is denoted by 91(A). The set 9'(A) is some-

times referred to as the power set of A. For example, if A = {I, 2}, then

91(A) = {4, {l}, (2), {1, 2}}. In this example, the set A has 2 elements and 91(A) has 4 or 22 elements, the elements in this instance being the subsets of A. If we take a set with 3 elements, then by listing the subsets of A it is easily seen that there are exactly 23 subsets of A (Exercise 8). On the basis of these two examples we are inclined to conjecture that if A contains n elements, then @'(A) contains 2" elements.

1.1

Sets and Operations on Sets

S

We now prove that this is indeed the case. We form subsets B of A by deciding for each element of A whether to include it in B or to leave it out. Thus for each element of A there are exactly two possible choices. Since A has n elements, there are exactly 2" possible decisions, each decision corresponding to a subset of A. Finally, if A and B are two sets, the Cartesian product of A and B, denoted A X B. is defined as the set of all ordered pairs' (a. b), where the first component a is from A and the second component b is from B, i.e.,

A X B= {(a, b) : a E A, b E B}. For example, if A = (1, 2} and B = {-1, 2, 4}, then

A X B = ((l, -1), (1, 2), (1, 4), (2, -l),(2,2),(2,4)). The Cartesian product of R with R is usually denoted by R2 and is referred to as the euclidean plane. If A and B are subsets of R, then A X B is a subset of R2. The case where A and B are intervals is illustrated in Figure 1.2.

B

b

a A

Figure 1.2

The Cartesian Product A x B

EXERCISES 1.1 1. LetA = {-1,0,1,2},B = {-2,3}, and C = (-2,0,1,5). a. Find each of the following: (A U B), (B U C), (A n B), (B n C), (A n C). A n (B u C), A \ B. C \ B.

A\ (BUC).

1. A set theoretic definition of ordered pair can be given as follows: (a, b) - ((a), {a, b}}. With this definition two ordered pairs (a, b) and (c, d) are equal if and only if a = c and b = d (Miscellaneous Exercise 1).

6

Chapter 1

The Real Number System

b. Find each of the following: (A X B), (C x B), (A x B) n (C X B). (A n c) x B. c. On the basis of your answer in (b), what might you conjecture about (A n c) X B for arbitrary sets A. B. C°

2. Let A={xER:-1<-x:5 5}.B={xE18:0Sx:53),C=(.rElB:2sx:5 4). *a. Find each of the following: AnB.An7L,BnC,AUB,BUC. *b. Find each of the following: A X B. A X C, (A X B) U (A X C). c. Sketch the sets (A X B) U (A X C) and A X (B U C). d. On the basis of your answer in (c), what might you conjecture about A X (B U C) for arbitrary sets A, B. C.' 3. If A and B are sets, prove that

a. Anm=d., AU46 =A. b. AfA=A. c. AnB=BnA, AUB=BUA.

AUA=A.

4. Prove the following associate laws for the set operations U and n:

*a. An(BnC)= (AnB)nc.

b. AU(BUC) _ (AU8)UC.

5. If A C B. prove that

a. AnB=A.

b. AUB=B.

6. If A is a subset of X. prove that

a. A U A` = X, c. (A`)` = A.

b. A n A` = ¢,

7. If A. B. and C are sets, prove that

*a. AU(Bnc) _ (AUB)n(AUC),

b. C\(AUB) _ (C\A)n(c\B).

c. C\(A n B) = (C\A) U (C\B). 8. *Verify that a set with three elements has eight subsets. 9. If A and B are subsets of a set X. prove that A \ B = A n B. 10. If A and B are any sets, show that A n B and A \ B are disjoint and that

A = (AnB)U(A\B). 11. Prove that A X (B, U B,) = (A X B,) U (A X B,)12. Suppose A, C are subsets of X. and B, D are subsets of

Y.

Prove that

(AxB)n(CxD)=(Anc)x(BnD).

1.21

Functions We begin this section with the fundamental concept of a function. In many texts, a func-

tion or a mapping f from a set A to a set B is described as a rule that assigns to each element x E A a unique element y E B. This is generally expressed by writing y = f(x) to denote the value of the function f at x. The difficulty with this "definition" is that the terms "rule" and "assigns" are vague and difficult to define. Consequently we will define "function" strictly in terms of sets, using the notation and concepts introduced in the preceding section. The motivation for the following definition is to think of the graph of a function; namely the set of ordered pairs (x, v) where .v is given by the "rule" that defines the function.

Functions

1.2

1.2.1

7

DEFINITION Let A and B be any two sets. A function f from A into B is a subset of A X B with the property that each x E A is the first component of precisely one ordered pair (x, y) E f ; that is, for every x E A there exists Y E B such that (x, y) E f, and if (x, y) and (x, y') are elements of f, then y = y'. The set A is called the domain off, denoted Dom f The range of f, denoted Range f, is defined by

Range

f= (yEB:(x,y)Ef forsomnexEA).

If Range f = B, then the function f is said to be onto B. (See Figure 1.3.)

AxB I

1-

- - -f- -

(x.y)

I

A

Figure 1.3 A Function as a Graph

If f is a function from A to B and (x, y) E f, then the element y is called the value of the function f at x and we write

y = f (x)

or

f : x -+ y.

We also use the notation f : A -± B to indicate that f is a function from (or on) A into (or to) B, or that f maps A to B or is a mapping of A to B. If we think of a function f : A -> B as mapping an element x E A to an element y = f (x) in B, then this is often represented by a diagram as in Figure 1.4. If f : A -* Q8, then f is said to be a realvalued function on A. A function f from A to B is not just any subset of A X B. The key phrase in Defini-

tion 1.2.1 is that "each x E A is the first component of precisely one ordered pair (x, y) E f." To better understand the notion of function we consider several examples.

8

Chapter I

The Real Number System

Figure 1.4 A Function as a Mapping

1.2.2

EXAMPLES

(a) Let A = 1-3, -2, -1, 0, 11 and B = Z. Consider the subset f of A X B given by

f = {(-3. 2). (-2. -2). (-1.4). (0. -6), (1.4)}. Since each x E A belongs to precisely one ordered pair (x. y) E f. f is a function from

A into B with Range f = (-6. -2. 2.4). Figure 1.5 indicates what f does to each element of A. Even though the element 4 in B is the second component of the two distinct ordered pairs (- 1, 4) and (1, 4). this does not contradict the definition of function. -3

2

-2 --2

0'--6 4

-1

4

1

Figure 1.5

(b) Let A and B be as in (a) and consider g defined by

g = ((-3,2).(-2.4).(-2.I).(-l.4),(0,5). (1, I)). Since (-2.4) and (-2, 1) are two elements of g with the same first component, g is not a function from A to B.

(c) In this example we let A = B = R. and let h be defined by

h=((x,v)ERXR:y=x'+2). This function is described by the equation), = x2 + 2. The standard way of expressing this function is as

h(x)_X''+2,

Domh=R.

This specifies both the equation defining the function and the domain of the function. For this example, Range It = {y E R : y ? 2}.' 2. Since x' z 0 for all x E R. the range of It is a subset of (v E R :

? 2). To obtain equality. we rey quire that for every e > 2 there exists an .r E R such that x' + 2 = v. The existence of such a y will follow as a consequence of Example 1.4.8.

1.2

Functions

9

(d) Let A be any nonempty set and let i = {(.r, x) : x E Al. Then i is a function from A onto A whose value at each x E A is x; i.e., i(r) = x. The function i is called the identity function on A.

(e) Let A and B be two nonempty sets and consider the projection function p from A X B to A defined by p = {((a, b), a) : (a, b) E A X B).

In this example Dom p = A X B. Since p : (a, b) -* a, we denote this simply by p(a, b) = a. For example, if A = 11, 2, 3) and B = { -1, 11, then A X B = {(1, -1).

(1, 1), (2, -1), (2,1), (3, -1), (3, 1)} and p(l, -1) = 1,p(1, 1) = l,p(2, -1) = 2. etc.

As was indicated in (c), if our function h is given by an equation such as y = x' + 2, we will simply write h(x) = x'' + 2, Dom h = Fl, to denote the function h. It should be emphasized, however, that an equation such as h(x) = x2 + 2 by itself does not define a function; the domain of h must also be specified. Thus

h(x) = x' + 2,

Dom h = I8,

and

g(x)=x2+2,

Domg={xEIB:-1 :5 x:5 2},

define two different functions.

Image and Inverse Image 1.2.3

DEFINITION

Lei f be a function from A into B. If E C A, then f(E), the Image of

E under f, is defined by

f(E) = {f(x) : x E E}. If H C B, the inverse Image of H, denoted f -'(H), is defined by

f-'(H)=(xEA:f(x)EH). If H = {y}, we will write f -'(y) instead of f ''({y}). Thus for Y E B.

f"'(y) _ {x E A : f(x) = y}. It is important to keep in mind that for E C A, f(E) denotes a subset of B, while for H C B, f-' (H) describes a subset of the domain A. It should be clear that f(A) = Range f, and that'f is onto B if and only if f(A) = B. To illustrate the notions of image and inverse image of a set we consider the following examples.

10

Chapter 1

1.2.4

The Real Number System

EXAMPLES

(a) As in Example 1.2.2, let A = {-3, -2,-1.0, 1},B = 71, and f : A -* 2 the function given by

f = ((-3,2),(-2. -2),(-l,4).(0. -6),(1.4)1. Consider the subset E _ {-1, 0, 1) of A. Then

f(E)

{f(-1), f(0). f(l)}

{-6,4}.

If H = (0, 1, 2, 3, 4}, then

f-'(H)= {xEA: f(x)EH}= {-3. -1,1}. Since both f(- 1) and f (l) are equal to 4, f '(4)

-1, 1). On the other hand, since

(x, 0) 0 f for any x E A, f -'(0) _ 46. (b) Consider the function g : 7 -+ Z given by g (x) = x 2, and let E

-1, -2,

-3,...). Then g(E) = {(-n)2 : n E NI} _ {1,4,9,. ..}. On the other hand,

g-'(g(E)) = 71\ {0}. For this example, E C# g-'(g(E)).

(c) Let h be the function defined by h(x) = 2x + 3, Dom h = R. If E _ {x E R :

-1 < x s 2}, then

h(E) = {2x+3:-1 5 xs2} _ {yER: l

y:5 7).

For the set E we also have

h-'(E) ={xER:2x+3EE}= {x:_2

s x s - 1} 2

For each y E R, x E h-' (y) if and only if 2x + 3 = y, which upon solving for x gives

x = 1(y - 3). Thus for each y E R. h-'(y) _ {2 (y - 3)}. Since h-'(y) #

for each

y E R, the function h maps R onto R. Suppose f is a function from A into B. If A, and A2 are subsets of A with A, C A2, then from the definition of the image of a set it follows immediately that f(A1) C f(A2). Likewise, if B, C B2 C B, then f -' (B,) C f -' (B2). The operations of finding the image or inverse image of a set usually also preserve the basic set operations of union, intersection, and complementation. There are two important exceptions, which are presented in part (b) of the next theorem and in Exercise 9.

1.2.5 THEOREM Let f be a function from A into B. If A, and A2 are subsets of A. then (a) f(Ai U A2) = f(A,) Uf(A2) (b) f(A, n A2) C f(A,) nf(A2)

1.2

Functions

11

Proof. To prove (a), let y be an element of f(A, U A,). Then Y = f(x) for some x in A, U A2. Thus x E A, or x E A,. Suppose x E A,. Then y = f(x) E f(A,). Similarly, if x E A2, Y E f (A2). Therefore y E f (A,) U f (A2). Thus

f(A, U A2) C f(A,) U f(A,). Since it is clear that f(A,) and f(A2) are subsets of f(A, U A,), the reverse inclusion also holds, thereby proving equality. Since f(A, n A2) is a subset of both f(Al) and f(A2), the relation stated in (b) is also true.

To. see that equality need not hold in (b), consider the function g(x) = x2 Dom g = 7L, of Example 1.2.4(b). If A, = (0, -1, -2, -3, ...), and A2 = (0. I, 2, 3, .

.

. } , then f(Al) = f(A2) _ {0, 1, 4, 9, .

.

.}, but A, n A, = {0}. Thus

f(A, fl A2) = f({0}) = {0} # f(A,) n f(A2) _ (0, 1, 4, 9, ...).

1.2.6

THEOREM Let f be a function from A to B. If B, and B2 are subsets of B, then

(a) f -'(B, U B2) = f -'(B,) U f-'(B2), (b) f -'(B, n B2) = f -'(B,) n f-'(B2),

(c) f-'(B\B,) = A\f-'(B,). Proof. To illustrate the method of proof we will prove (a), leaving the proofs of (b) and (c) to the exercises (Exercise 8). Suppose x is an element of f -'(BI U B2). Then by

the definition of the inverse image of a set, f (x) E B, U B2. Thus f (x) E B, or f(x) E B2. If f(x) E B,, then x is in f '(81), and thus x E f -' (B,) U f (B,). The same argument also applies if f(x) E B2. In either case we have f -'(BI U B2)

C f-(B1)Uf-'(B2)

To prove the reverse containment, we first note that both B, and B2 are subsets of

B, U B2. Therefore both f -' (B,) and f - (B2) are subsets of f `(BI U B2), and thus f -' (B,) U f -' (B2) C f (B, U B2). This proves the reverse containment thereby giving equality.

Inverse Function 1.2.7

DEFINITION A function f from A into B is said to be one-to-one if whenever x, * x2, then f(x,) * f(x2). Alternatively, a function f is one-to-one if whenever (x,, y) and (x2, y) are elements

off then x, = x2. From the definition it follows that f is one-to-one if and only if f `(y) consists of at most one element of A for every y E B. If f is onto B. then f -'(y) # rb for every y E B. Thus if f is one-to-one and onto B, then f `(y) consists of exactly one element x E A and

g={(y,x)EBXA: f(x)=y} defines a function from B to A. This leads to the following definition.

12

Chapter 1

1.2.8

The Real Number System

DEFINITION

If f is a one-to-one function from A onto B. let

f -' = {(y, x) E B X A: f(x) = y}. The function f from B onto A is called the inverse function of f. Furthermore. for each y E B.

x = f -'(y) if and only if f(x) = y. There is a subtle point that needs to be clarified. If f is any function from A to B. then f''(y) (technically f -'({y})) is defined for any y E B as the set of points x in A such that f(x) = y. However, if f is a one-to-one function of A onto B, then f -'(y) denotes the value of the inverse function f -' at y E B. Thus it makes sense to write f -'(y) = x whenever (y. x) E f '. Also, if f is a one-to-one function of A into B. then

f'' defined by f -' = {(y, x) : y E Range f and f(x) = y} is a function from Range f onto A. (See Figure 1.6.)

Figure 1.6 The Inverse Function

1.2.9

EXAMPLES

(a) Let h be the function of Example 1.2.4(c); that is, h(x) = 2x + 3, Dom h = R. The function h is clearly one-to-one and onto R with

x=h-'(v)=2(y-3).

Domh-' - R.

(b) Consider the function f defined by the equation y = 2. If we take for the domain of f all of R, then f is not a one-to-one function. However, if we let

Domf=A={xER:xa0}, then f becomes a one-to-one mapping of A into A. To see that f is one-to-one, let x,, x2 E A with x, # x2. Suppose x, < x2. Then x1 < x2; that is, f(x1) # f(x2). Tberofore f is one-to-one. To show that f is onto A. we need to show that for each y E A. y > 0, there exists a positive real number x such that

x2=y.

1.2

Functions

13

Intuitively we know that such an x exists; namely, the square root of y. However. a rigorous proof of the existence of such an x will require the least upper bound property of the real numbers. In Example 1.4.8 we will prove that for each y > 0 there exists a unique positive real number x such that X' = y. The number x is called

the square root of y and is denoted by \. Thus the inverse function of f is given by

Domf ' _ {y E R:y :0}.

f-'(y) _ Composition of Functions

Suppose f is a function from A to B and g is a function from B to C. If a E A. then f(a) is an element of B, the domain of g. Consequently we can apply the function g to f(a) to obtain the element g(f(a)) in C. This process, illustrated in Figure 1.7. gives a new function h which maps a E A to g(f(a)) in C.

C

h(a) =g(f(a))

Figure 1.7

1.2.10

Composition of g with f

DEFINITION If f is a function from A to B and g is a function from B to C, then the function g of: A -i C. defined by

gof= {(x.z)EA X C:z=g(f(x))} is called the composition of g with f

(f

If f is a one-to-one function from A into B, then it can be shown that o f) (x) = x for all x E A and that (f o f ') (y) = y for ally E Range f (Exercise

10). This is illustrated in (b) of the following example.

1.2.11

EXAMPLES

(a) If f (x) = v -+x with Dom f = {x E R : x z - I } and g(x) = x', Dom g = R. then

(g ofXx) = g(f(x)) = (

)'' = I + x,

Dom(gof)= {x ER:xz -1}.

14

Chapter 1

The Real Number System

Even though the equation (g of)(x) = 1 + x is defined for all real numbers x, the domain of the composite function g o f is still only the set {x E R : x,2: -1 }. For this example, since Range g C Dom f, we can also find f o g; namely,

(f o g)(x) =.f(g(x)) =

1 + x2,

(b) For the function fin (a), the inverse function f

f-'(y)=y2- 1, Dom f'' =

is given by

Range f = {y E R :Y z 0}.

Thus for x E Dom f,

(f

Dom f o g = R.

f)(x) = f _' U W) = (f (x))2 - I =

(\/j)2 - 1 = X.

and for y > 0,

(fof-')(Y) =f(f+I(Y)) =

= Y.

(

R EXERCISES 1.2 1. Let A = {-1, 0, 1, 2) and B = N. Which of the following subsets of A X B is a function from A into B? Explain your answer.

b. g = {(-1,2),(0,7).(1,3),(2,7)}

a. f= {(-l.2).(0.3).(2.5)} -1)}

c. h =

d. k = {(x, y) : y = 2r + 3, x E A}

2. *a. Let A = {(x, y) E R X R : x2 + y2 = 11. Is A a function? Explain your answer. b. Let B = {(x, y) E A : y z 0}. Is B a function? Explain your answer.

3. Let f : N -* N be the function defined by f(n) = 2n - 1. Find f(E) and f -'(E) for each of the following subsets E of N.

c. N b. {1,3,5.7} *a. {1,2,3,4} 4 . Let f= {(x,y):xE R,y =x3 + 1}. *a. Let A = {x : -1 <- x 5 2}. Find f(A) and f '(A). b. Show that f is a one-to-one function of R onto R. *c. Find the inverse function f - 1.

S. Let f, g mapping Z into Z be given by f(x) = x + 3 and g(x) = 2x. Find (f o g)(N) and (g of)(N). 6. For each of the following real-valued functions, find the range of the function f and determine whether f is oneto-one. If f is one-to-one, find the inverse function and specify the domain of f-'.

*a. f(x)3x-2, Domf=R *c. 1(x) = 1

e. Ax) _

1

+

s Dom f= {x E R : O s x < l} I,

Dom f= {x E R : -1 5 x s 11

*g. f(x. y) = x. Dom f = R2

b. f(x)=5x+4. Domf=R d. Ax) = I X X Dom f= {x E R : 0 < x < 1) 1. Ax) = sin x,

Dom f = {x E R : 0!5 x :s r)

1.3

Mathematical Induction

15

7. Let A = {t E R : 0:5 t < 21r} and B = R2. and let f : A -+ B be defined by J(t) = (co. r. sin t). *a. What is the range of f? *b. Find f ' of each of the following points: (1, 0), (0, -1), ;-, ), (0, 1). c. Is the function f one-to-one? 8. Prove parts (b) and (c) of Theorem 1.2.6.

9. Letf:A -+ BandletFCA. a. Prove that f(A) \f(F) C f(A \ F). b. Give an example for which f(A) \f(F) * f(A \ F). 10. Let f be a one-to-one function from A into B. Show that' o(f f)(x) = x for all x E A and that (f of for ally E Range f. 11. *Let f : A -* B and g : B -> A be functions satisfying (g o f)(x) = x for all x E A. Show that f is a one-to-one function. Must f be onto B?

12. If f : A -s B and g : B -+ C are one-to-one functions, show that (g of)

1.3

f -' o g' on Range (g of).

Mathematical Induction Throughout the text we will on occasion need to prove a statement, identity, or inequality involving the positive integer n. As an example, consider the following identity. For each n E N,

r+r2+

+r"= r -

r"+'

Mathematical induction is a very useful tool in establishing that such an identity is valid for all positive integers n.

1.3.1

THEOREM (Principle of Mathematical Induction) For each n E N, let P(n) be a statement about the positive integer n. If

(a) P(1) is true, and (b) P(k + 1) is true whenever P(k) is true, then P(n) is true for all n E N. The proof of this theorem depends on the fact that the positive integers are wellordered; namely, every nonempty subset of N has a smallest element. This statement is usually taken as a postulate or axiom for the positive integers; we do so in this text. Since it will be used on several other occasions, we state it both for completeness and emphasis.

1.3.2

WELL-ORDERING PRINCIPLE

Every nonempty subset of N has a smallest element.

The well-ordering principle can be restated as follows: If A C F%, A # d,, then there exists n E A such that n :5 k for all k E A. To prove Theorem 1.3.1 we will use the method of proof by contradiction. Most theorems involve showing that the statement P implies the statement Q; namely, if P is

16

Chapter 1

The Real Number System

true, then Q is true. In a proof by contradiction one assumes that P is true and Q is false. and then shows that these two assumptions lead to a logical contradiction; namely, show that some statement R is both true and false. Further details on the method of proof by contradiction are provided in Section A.3 of the Appendix.

Proof of Theorem 1.3.1. Assume that the hypotheses of Theorem 1.3.1 are true, but that the conclusion is false; that is, there exists a positive integer n such that the statement P(n) is false. Let

A = {k E N : P(k) is false}. By our assumption, the set A is nonempty. Thus by the well-ordering principle A has a smallest element k,,. Since P(1) is true, k > 1. Also, since ko is the smallest element of A. P(k,, - 1) is true. But then by hypothesis (b), P(k,,) is also true. Thus k, a A. This,

however, is a contradiction. Consequently, P(n) must be true for all n E N. Q

1.3.3

EXAMPLES We now provide two examples to illustrate the method of proof by mathematical induction. The first example provides a proof of the identity in the introduction to the section. An alternative method of proof will be requested in the exercises (Exercise 7).

(a) To use mathematical induction, we let our statement P(n), n E N, be as follows:

+r"= r - r"+i . r # 1. I -r

r+ When n = I we have

r(1 -r) _ r-r2 I - r , provided r * 1. r = (1 - r) Thus the identity is valid for n = 1. Assume P(k) is true for k i' I; i.e.,

+r

r+

r - rA+I

r # 1.

I-r

We must now show that the statement P(k + 1) is true; that is.

r-

r+

r(A+ p+ i

I -r

'

r # 1.

But

r+

+ rk+i = r +

+ rk +

rk+1,

which by the induction hypothesis r - rk+i

r - rA+i + (1 - r)rk+l

I-r

I - r

r - rk+'

1-r

Mathematical induction

1.3

17

Thus the identity is valid for k + 1, and hence by the principle of mathematical induction for all n E N.

(b) For our second example, we use mathematical induction to prove Bernoulli's in-

equality. If h > -1, then

(I +h)">1 + nh for alln EN. When n = 1, (1 + h)' = I + h. Thus since equality holds, the inequality is certainly valid. Assume that the inequality is true when n = k, k -- 1. Then for is = k + 1,

(I + h)'-I = (I + h)'"(1 + h), which by the induction hypothesis and the fact that (1 + h) > 0

?(1 +kh)(1 +h)= I +(k+ 1)h+kh= 1 + (k + 1)h. Therefore the inequality holds for n = k + 1, and thus by the principle of mathematical induction for all n E N. Although the statement of Theorem 1.3.1 starts with n = 1, the result is still true if we start with any integer n,, E Z. The modified principle of mathematical induction is as follows: If for each n E 77, n ? n,,, P(n) is a statement about the integer n satisfying .

(a)

is true, and

(b') P(k + 1) is true whenever P(k) is true. k ? no, then P(n) is true for all n E 77, n ? n,. The proof of this follows from Theorem 1.3.1 by simply setting

Q(n)=P(n+n,- I),

n E N,

which is now a statement about the positive integer n.

Remark. In the principle of mathematical induction, the hypothesis that P(I) be true is essential. For example, consider the statement P(n):

n+i=n, nEN. This is clearly false! However, if we assume that P(k) is true, then we also obtain that P(k + 1) is true. Thus it is absolutely essential that P(n,) be true for at least one fixed value of n,.

There is a second version of the principle of mathematical induction that is also quite useful.

1.3.4

THEOREM (Second Principle of Mathematical Induction) For each n E N. let P(n) be a statement about the positive integer n. If

(a) P(1) is true, and (b) for k > 1, P(k) is true whenever P(j) is true for all positive integers j < k, then P(n) is true for all n E N.

18

chapter 1

The Real Number System

Proof.

Exercise 3.

Q

Mathematical induction is also used in the recursive definition of functions defined for the positive integers. In this procedure, we give an initial value of the function fat

n = 1. then assuming that f has been defined for all integers k = 1. .... n, the value off at n + I is given in terms of the values off at k, k 5 n. This is illustrated by the following examples.

1.3.5

EXAMPLES

(a) Suppose f : N -' N is defined by f(1) = t and f(n + 1) = nf(n). n E N. The values off for n = 1, 2, 3.4 are given as follows: 3.2.1.

f(1) = 1, f(2) = lf(1) = 1, f(3) = 2f(2) = 2.1, f(4) = 3f(3) =

Thus we conjecture that f(n) (n - 1)!, where 0. is defined to be equal to one, and for n E N, n! (read n factorial) is defined as

The conjecture is certainly true when n = 1. Thus assume that it is true for n = k, 1; that is, f(k) = (k - 1)!. Then for n = k + 1, k

f(k + 1) = kf(k), which by the induction hypothesis

k(k - 1)! = k! Therefore the identity holds for it = (k + 1), and thus by the principle of mathematical induction, for all it E N.

(b) For our second example, consider the function f : N - R defined by f (l) = 0, f(2) = 3, and for n ? 2 by f (n) = (1, +i) f (n - 2). Computing the values of f for n = 3, 4, 5, and 6, we have

f(3) = 0,

f(4) = 5,

f(5) = 0.

f(6) =

From these values we conjecture that 0,

f(n) =

1

tn

if n is odd, if it is even.

To prove our conjecture we will use the second principle of mathematical induction. Our conjecture is certainly true for it = 1, 2. Suppose n > 2, and suppose our conjec-

ture holds for all k < n. If n is odd, then so is (n - 2), and thus by the induction hypothesis f(n - 2) = 0. Therefore f(n) = 0. On the other hand, if n is even, so is (n - 2). Thus by the induction hypothesis f(n - 2) = ;,?-y. Therefore n

f(n)

-(n1) n+l

f(

n-2)

-(n-11

1

=

I

n+1)n-1 n+l

.

IN

1.3

19

Mathematical Induction

EXERCISES 1.3 1. Use mathematical induction to prove that each of the following identities are valid for all n E N. n(n + 1)

+n=

1+2+3+

a.

.

+n2=

12+22+

*b.

2

n(n + 1)(2n + 1) 6

1 +3+5+ +(2n- 1)=n2

d. 13+23+

+n3=[Zn(n+l)]2

+2"=2(2"- 1)

e. 2+22+23+ *f For x, y E Il,

x"+1 -y",i =(x-y)(x"+x"-iy+...+y")

g' 1(2)+2(3)+

n+1 n

n(n+1)

+

2. Use mathematical induction to establish the following inequalities.

b. 2" > n2 for all n E N. n ? 5 d. 13 + 23 + + n3 < n4 for all n E 1N, n

*a. 2" > n for all n E N *c. n! > 2" f o r all n E !N, n L - 4 e. 13 + 23 +

2

+ n3 < £n4 for all n E !N, n >- 3

3. Prove Theorem 1.3.4.

4. *Let f : N - N be defined by f (l) = 5, f(2) = 13, and for n z 2, f(n) = 2f(n - 2) + f(n - 1). Prove that

f(n)=3.2"+(-1)"foralln E Ni. 5. For each of the following functions f with domain N, determine a formula for f(n) and use mathematical induction to prove your conclusion.

a. f(1)=2,andforn> 1,f(n)=(n-1)f(n-1)-n+1. *b. f(l)=1,1(2)=4, andforn>2,f(n)=2f(n-1)-f(n-2)+2. c. f (l) = 1, and for n > 1, f (n) = * d f(1) = I f(2) = 0 and '

(n

3n 1) f (n - 1).

for n > 2 f(n) _

- 2) - f(n n(n-1)*

C. For a,, a2 E R arbitrary, l et f ( l ) = a,, f(2 ) = a2, and for n

> 2 , f (n)

f. For a 1, a2 E R arbitrary, let f (l) = a,, f (2) = a2, and for n > 2,f(n) _ 6. Let f: N! -*NI be defined by f(1) = 1, f(2) = 2, and

f(n + 2) =(n + 1) + f(n)). Use Theorem 1.3.4 to prove that 1 s f(n)s 2 for all n e N. 7. *Prove that

r+r2+

+r"= r - r

,

r#

1

without using mathematical induction.

nEN

An - 2) n(n - 1)'

(__4)f(n - 2).

20

8.

Chapter)

The Real Number System

Use mathematical induction to prove the arithmetic geometric mean inequality. If a,, a,.. nonnegative real numbers, then ala, ,

.

-a. S (

a, + a2 + '

-

.

. a,,, n E N. are

- +a.

with equality if and only if a, = a, =

1.4

.

= a".

The Least Upper Bound Property In this section we will consider the concept of the least upper bound of a set and introduce the least upper bound or supremum property of the real numbers R. Prior to introducing these new ideas we briefly review the algebraic and order properties of 0 and R. Both the rational numbers 0 and the real numbers R are algebraic systems known as fields. The key fact about a field is that it is a set F with two operations, addition (+) and multiplication (), that satisfy the following axioms:

1. If a,bE F.then a+bE 2. The operations are commutative; that is, for all a, b E F

a+b=b+a

and

3. The operations are associative; that is. for all a, b, c E F

a + (b + c) _ (a + b) + c

and

a (b c) = (a b) c.

4. There exists an element 0 E F such that a + 0 = a for every a E F. 5. Every a E F has an additive inverse; that is, there exists an element -a in F such that

a+(-a)=0. 6. There exists an element 1 E F with 1 # 0 such that a 1 = a for all a E F. 7. Every a E F with a * 0 has a multiplicative inverse; that is, there exists an element a' in F such that 1.

8. The operation of multiplication is distributive over addition; that is, for all

a, b, c E F, The element 0 is called the zero of F and the element I is called the unit of F. For a # 0, the element a-' is customarily written as or 1/a. Similarly, we write a - b in-

stead of a + (-b), ab instead of a b, and alb or instead of a b-. The real numbers R contain a subset P known as the positive real numbers satisfying the following:

1.4

The Least Upper Bound Property

21

(01) If a, b E P, then a+ b E P and a -b E P. (02) If a E R then one and only one of the following hold:

aEP,

-aEP,

a =0.

Properties (01) and (02) are called the order properties of R. Any field F with a nonempty subset satisfying (01) and (02) is called an ordered field. For the real numbers we assume the existence of a positive set P. For the rational numbers 0, the set of pos-

itive rational numbers is given by P n 0, which can be proved to be equal to

{p/q:p,gE7L,4 *O,pgE1i}. Let a, b be elements of R. If a - b is positive, i.e., a - b E P. then we write a > b or b < a. In particular, the notation a > 0 (or 0 < a) means that a is a positive

element. Also, a s b (or b ? a) if a < b or a = b. The following useful results are immediate consequences of the order properties and the axioms for addition and multiplication. Let a, b, c be elements of R.

(a) If a > b, then a + c > b + c. (b) If a > b and c > 0, then ac > be. (c) If a > b and c < 0, then ac < be. (d) If a * 0, then a2 > 0. (e) If a > 0, then 1/a > 0; if a < 0, then 1/a < 0. To illustrate the method of proof, we provide the proof of (b). Suppose a > b; i.e., a - b is positive. If c is positive, then by (01), (a - b)c is positive. By the distributive law,

(a - b)c = ac - be. Therefore ac - be is positive; that is, ac > be. The proofs of the other results are left as exercises.

Upper Bound of a Set We now turn our attention to the most important topic of this chapter; namely, the least upper bound or supremum property of R. In Example 1.4.5(c) we will show that this property fails for the rational numbers Q. First, however, we define the concept of an upper bound of a set.

1A.1

DEFINITION A subset E of R is bounded above if there exists,0 E R such that x 5 /3 for every x E E. Such a (3 is called an upper bound of E. The concepts bounded below and lower bound are defined similarly. A set E is bounded if E is bounded both above and below. We now consider several examples to illustrate these concepts.

1A.2

EXAMPLES

(a) Let A = {0, 12, 3, s, ...} = { l is bounded below by any real number r

n = 1, 2, 3, ...} (see Figure l.8). Clearly A 0 and above by any real numbers ? 1.

22

Chapter 1

The Real Number System

o

2

1

.1

a

a 4s

2 Figure 1.8

(b) N = {1, 2, 3, ...}. This set is bounded below; e.g., I is a lower bound. Our intuition tells us that N is not bounded above. It is obvious that there is no positive integer n such that j n for all j E N. However, what is not so obvious is that there is no real number .6 such that j s /3 for all j E N. In fact, given /3 E R, the proof of the existence of a positive integer n > /3 will require the least upper bound property of R (Theorem 1.5.1).

(c) B = {r E 0 : r > 0 and r2 < 2}. Again it is clear that 0 is a lower bound for B. and that B is bounded above; e.g., 2 is an upper bound for B. What is not so obvious, however, is that B has no maximum. By the maximum or largest element of B we mean an element a E B such that p a for all p E B. Suppose p E B. Define the rational number q by

2-p2

_

2p+2

q-p+(p+2) p+2 With q as defined, a simple computation gives

q2-= 2

z_ (p+2)2

Since p2 < 2, q > p and q2 < 2. Thus B has no largest element. Similarly, the set

has no minimum or smallest element. Intuitively, the largest element of B would satisfy p2 = 2. However, as was shown in the introduction, there is no rational number p for

which p2 = 2.

Least Upper Bound of a Set 1A.3

Let E be a nonempty subset of R that is bounded above. An element a E R is called the least upper bound or supremum of E if DEFINITION

(1) a is an upper bound of E, and (ii) if /3 E R satisfies /3 < a, then 13 is not an upper bound of E. Condition (ii) is equivalent to a /3 for all upper bounds J3 of E. Also by (ii), the least upper bound of a set is unique. If the set E has a least upper bound, we write

a = sup E to denote that a is the supremum or least upper bound of E. The greatest lower bound or ifmum of a nonempty set E is defined similarly, and if it exists, is denoted by inf E.

1.4

The Least Upper Bound Property

23

There is one important fact about the supremum of a set that will be used repeatedly throughout the text. Due to its importance we state it as a theorem.

1 AA THEOREM Let A be a nonempty subset of H that is bounded above. An upper bound or of A is the supremum of A if and only if for every 13 < a, there exists an element x E A such that

13 <x5a. Proof. Suppose a = sup A. If 13 < a, then 13 is not an upper bound of A. Thus there exists an element x in A such that x > B. On the other hand, since a is an upper bound

ofA,x: a. Conversely, if a is an upper bound of A satisfying the stated condition, then every

(3 < a is not an upper bound of A. Thus a = sup A. Q

1.4.5

EXAMPLES In the following examples, let's consider again the three sets of the previous examples. (a) As in Example 1.4.2(a), let A = {0, Z, 3, ;, ...}. Since 0 is a lower bound of A and

0 E A, infA = 0. We now prove that sup A = 1. Since I - I'-1 < I for all n = 1, 2, ... , 1 is an upper bound. To show that I = sup A we need to show that if 13 E R with $ < 1, then f3 is not an upper bound of A. Clearly if 0 0, then $ is not an upper bound of A. Suppose, as in Figure 1.9, 0 < 13 < 1. Then our intuition tells us that there exists an integer no such that

n

1

1

1

1 E A, and thus 13 is not an upper bound. Therefore sup A = 1. The existence n of such an integer no will follow from Theorem 1.5.1. In this example, inf A E A but sup A e A.

(b) For the set N, inf N = 1. Since N is not bounded above, N does not have an upper bound in R. (c) In this example we prove that the supremum of the set

B={rEQ:r>0 and

r2 <2},

if it exists, is not an element of 0. Suppose a = sup B exists and is in 0. Since a is rational, a2 0 2. Thus a2 < 2 or a2 > 2. But if a 6 B, then since B contains no largest

24

Chapter 1

The Real Number System

element, there exists q E B such that q > a. This contradicts the fact that a is an upper bound of B. Similarly, if a2 > 2, then there exists a q < a such that q2 > 2. But then q is an upper bound of R, which is a contradiction of property (ii) of Definition 1.4.3. The least upper bound of B in N is V2 (Section 1.5, Exercise 9), which we know is not rational.

Least Upper Bound Property of R The following property, also referred to as the completeness property of R. distinguishes the real numbers from the rational numbers and forms the foundation for many of the results in real analysis.

1A.6 SUPREMUM OR LEAST UPPER BOUND PROPERTY OF N Even nonempn subset of P that is bounded above has a supremum in R. For our later convenience we restate the supremum property of P as the infimum property of R.

1.4.7

INFIMUM OR GREATEST LOWER BOUND PROPERTY OF P Ever nonempn subset of H that is bounded below has an infimum in B. Although stated here as a property, which we will assume as a basic axiom about

P, the least upper bound property of P is really a theorem due to both Cantor and Dedekind, both of whom published their results independently in 1872. Dedekind, in the paper "Stetigkeit and irrationale Zahlen" (Continuity and irrational numbers), used algebraic techniques now known as the method of Dedekind cuts to construct the real number system P from the rational numbers O. He proved that the system R contained a natural subset of positive elements satisfying the order axioms (01) and (02), and furthermore, that P also satisfied the least upper bound property. The books by Burrill and t by Spooner and Mentzger cited in the Supplemental Readings are devoted to number systems. Both texts contain Dedekind's construction of R. Cantor, on the other hand,

constructed P from 0 using Cauchy sequences. In the miscellaneous exercises of Chapter 2 we will provide some of the key steps of this construction.

1.4.8

EXAMPLE In this example we show that for every positive real number y > 0, there exists a unique positive real number a such that a2 = y; i.e., a = v. The uniqueness of a was established in Example 1.2.9(b). We only prove the result for y > 1, leaving the case 0 < y s I to the exercises (Exercise 6). Let

C={xEH:x>0 and x2 1, 1 E C and thus C is nonempty. Also since y > 1, y2 > y, and thus y is an upper bound of C. Hence by the least upper bound property, C has a supremum in P. Let a = sup C. We now prove that a2 = y. To accomplish this we show that the assumptions a2 < Y and a2 > v lead to contradictions. Thus a2 = y.

1.4

The Least Upper Bound Property

25

Define the real number $3 by

_ y(a + 1) C2-_ /3=a+(a+r) a+r

(1)

Then y(! - I)(a2

)

P'- - y =

(2)

(a + V)2

If a2 < y, then by (1) $3 > a, and by (2) $32 < v. This contradicts that a is an upper bound for C. On the other hand, if Cr' > y, then by (1),6 < a and by (2), j32 > Y. Thus if x E R with x ? $3, then x2 > y. Therefore $3 is an upper bound of C. This contradicts

that a is the least upper bound of C. Since S defined by (I) may not be rational, the same proof will not work for the set B of Example 1.4.2(c). However, using Theorem 1.5.2 of the following section, it is possible to also prove that sup B = 12. For convenience, we extend the definition of supremum and infimum of a subset E of N to include the case where E is not necessarily bounded above or below.

1.4.9

DEFINITION

If E is a nonempty subset of R, we set

sup E = oo

if E is not bounded above, and inf E _ - oo if E is not bounded below.

For the empty set ¢, every element of i is, an upper bound of 46. For this reason the supremum of the empty set 44 is taken to be -oo. Similarly, inf d = cc. Also, for the symbols -oo and oo we adopt the convention that -oo < x < cc for every x E R.

Intervals Using the order properties of R, we can define certain subsets of R known as intervals.

1.4.10

DEFINITION

For a, b E R, an-55 b, the open interval (a, b) is defined as

(a,b)={xE68:a<x
In addition, we also have the half-open (half-closed) intervals

[a,b)={xEJl:a<x
(a,oo)={xER:a <x
26

Chapter 1

The Real Number System

with analogous definitions for (-cc, b) and (-oo, b). The intervals (a. 00), (-oo, b) and (-oo, oo) = R are also referred to as open intervals, whereas the intervals [a, oo) and (-oo, b) are called closed intervals. (See Figure 1.10.) (a. b) b

a

[a, b] a

b

(a, b)

Figure 1.10 Types of Intervals

In this definition, when b = a, (a, a) = 0 and [a, a] = {a}. Although the empty set 0 and the singleton {a} do not fit our intuitive definition of an interval, we will include them as the degenerate case of open and closed intervals, respectively. It should

be noted that the intervals of the form (a, b), (a, b], [a, b), and [a, b] with a, b E R. a s b, are all bounded subsets of R. An alternative way of defining intervals without use of the adjectives open and closed is as follows.

1.4.11

DEFINITION A subset J of R is an interval if whenever x, y E J with x < y, then every t satisfying x < t < y is in J. This definition also allows the possibility that J is empty or a singleton. One can show that every set J satisfying the definition is one of the intervals defined in Definition 1.4.10 (Exercise 21).

EXERCISES 1.4 1. Use the axioms for addition and multiplication to prove the following: If a E R, then -a. 0. b. a.

c. -(-a) = a. 2. Let a, b E It Prove the following. b. If ab - 0, then either a = 0 or b = 0. a. If a # 0, then I/a # 0. 3. Let a, b, c E R. Prove the following.

a. If a > b, then a + c > b + c. c. If a * 0, then a2 > 0. 4. *If a, b E IR, prove that ab

b. If a > b and c < 0, then ac < bc. d. If a > 0 then I/a > 0, and if a < 0 then 1/a < 0.

](a2 + b2).

1.4

The Least Upper Bound Property

27

5. Find the supremum and infimum of each of the following sets.

*a.A1,ll1,...}= 248 1l

'c. C

1J

nEll} '):E N ll

{n cosna:nEN}

"e.E{2nn:nEt`11 "g. G{(1

nEN)

b.Bsinn 22 :nEN} ll

J

d. D

{cos ni :nEN}

f. F

{(-lr-!:nGN} 1(

h. H

+(-lY')":nEN}

j.J={x2:-2<x<2}

i.l={xER:xz<4}

6. If 0 < y < 1, prove that there exists a unique positive real number x such that x2 = y. 7. Prove that there exists a positive real number x such that x3 = 2. 8. Let A be a nonempty subset of R. If a = sup A is finite, show that for each e > 0, there is an a E A such that

a - e < a < a. 9. Let E be a nonempty subset of R that is bounded above, and set U = jig E R that sup E = inf U.

is an upper bound of E}. Prove

10. Let A be a nonempty subset of R and let -A = (-x: x E A). Prove that infA = -sup(-A). 11. Use the least upper bound property of R to prove that every nonempty subset of R that is bounded below has an infimum.

12. If A and B are nonempty subsets of R with A C B, prove that

inf B s infA s sup A <_ sup B. 13. Suppose that A and B are bounded subsets of R. Prove that A U B is bounded and that sup(A U B) = sup{sup A, sup B}. 14. For A. B, subsets of R, define

A+B=(a+b:a6A,bEB)

and

A = (-1, 2, 4, 7) and B = {-2, -1, 1), find A + B and A

B.

'b. If A and B are nonempty and bounded above, prove that

sup(A + B) = sup A + sup B. c. If A and B are nonempty subsets of the positive real numbers that are bounded above, prove that

B) = (sup A)(sup B). d. Give an example of two nonempty bounded sets A and B for which sup(A B) # (sup A)(sup B). sup(A

15. Let f, g be real-valued functions defined on a nonempty set X satisfying Range f and Range g are bounded subsets of R. Prove each of the following.

*a. sup{f(x) + g(x) . X E X) s sup{f(x) : x e X} + sup{g(x) : x E X} b. inf{f(x) : x E X} + inf{g(x) : x E X} s inf{f(x) + g(x) : x r= X} c. If f(x) : g(x) for all x E X. then sup{f(x) : x E X} s sup{f(x) : x e X}. 16. Let X = Y = [0, 1 ] and let f : X X Y -+ R be defined by f (x, y) = 3x + 2y. a. For each x E X, find F(x) = sup{f(x, y) y E Y}; then find sup{F(x) : x E X}. b. For each y E Y, find G(y) = sup{ f(x, y) : x E X}; then find sup(G(y) : y E Y). c. Find sup{f(x, y) : (x, y) E X X Y}. Compare your answer with your answers in parts (a) and (b).

28

Chapter 1

The Real Number System

17. Perform the computations of Exercise 16 with X = [-1, 11. Y = [0, 2], and f(x, v) = 3x - 2y. 18. Let X, Y be nonempty sets, and let f be a nonnegative real-valued function defined on X X Y. For each x E X and y E Y, define G(y) = sup{f(x. v) : x E X}. F(x) = sup{f(x, y) : y E Y}, Prove that

sup{F(x) : x E X} = sup{G(v) : y E Y} = sup{f(x, y): (x, v) E X X Y}. 19. Let X, Y be nonempty sets and let f : X X Y -+ R be a function with bounded range. Let F(x) = sup{f(x, y) : y E Y} and H(y) = inf{f(x, y) : x E X}. Prove that inf{F(x) : x E X}. sup{H(y) : y E Y} 20. Let X = Y = [0, I ]. Perform the computations of Exercise 19 for each of the following functions f(x, y).

b. f(x, y) =

a. f(x, v) = 3x + 2r

1,

x=y,

x*v

21. Let J be a subset of R that has the following property: if x, y E J with x < y, then t E J for every t satisfying x < t < Y. Prove that J is an interval as defined in Definition 1.4.10.

1.5 1

Consequences of the Least Upper Bound Property In this section we look at a number of elementary properties of the real numbers, which

in more elementary courses are usually taken for granted. As we will see, however, these are all actually consequences of the least upper bound property of the real numbers.

1.5.1

THEOREM (Archimedian Property) tive integer n such that

If x, y E Ll and x > 0, then there exists a posi-

nx > Y.

Proof.

If y <- 0, then the result is true for all n. Thus assume that y > 0. We will

again use the method of proof by contradiction. Let

A={nx:nEN}. If the result is false, that is, there does not exist an n E N such that nx > y. then nx y for all n E N. Thus y is an upper bound for A. Thus since A * 46, A has a least upper

bound in R. Let a = sup A. Since x > 0, a - x < a. Therefore a - x is not an upper bound and thus there exists an element of A, say mx, such that

a - x<mx. But then a < (m + 1)x, which contradicts the fact that a is an upper bound of A. Therefore, there exists a positive integer n such that nx > y.

Remark. One way in which the previous result is often used is as follows: given e > 0, there exists a positive integer n such that noe > 1. As a consequence, 1

- < E n

for all integers n, n ? no.

1.5

Consequences of the Least Upper Bound Property

29

1.5.2 THEOREM If x, .v E l and x < v, then there exists r E O such that

x 0, by Theorem 1.5.1 there exists an integer n > 0 such that

n(y - x) > 1

or

ny > I + nx.

Again by Theorem 1.5.1, {k E N : k > nx} is nonempty. Thus by the well-ordering principle, there exists m E N such that

m - 1 :5 nx<m. Therefore

nx<m<_I+nx
x< n 0, then the result is obvious. Finally, if x < y < 0, then by the pre-

ceding calculations there exists r E 0 such that -y < r < -x; i.e., x < -r < y. The conclusion of Theorem 1.5.2 is often expressed by the statement that the rational numbers are dense in the real numbers; that is, between any two real numbers there exists a rational number. A precise definition of "dense" is given in Definition 3.1.12. Another consequence of the least upper bound property is the following theorem concerning the existence of nth roots.

1.5.3

THEOREM For every real number x > 0 and every positive integer n, there exists a unique positive real number y so that y" = x. The number y is written as lcx or x'1" and is called the nth root of x. The uniqueness of y is obvious. Since the existence of y will be an immediate consequence of the intermediate value theorem (Theorem 4.2.11), we omit the details of the proof in the text. A sketch of the proof of Theorem 1.5.3, using the least upper bound property, is included in the miscellaneous exercises. It should be emphasized that the proof of Theorem 4.2.11 also depends on the least upper bound property.

1.5.4

COROLLARY

If a, b are positive real numbers, and n is a positive integer, then (ab)Un = al/nb'l".

Proof. Set a = a'"" and R =

P". Then ab=a"/3"

Thus by uniqueness, af3 = (ab)'t".

30

Chapter 1

The Real Number System

0 EXERCISES 1.5 1. *If r and s are positive rational numbers, prove directly (without using the supremum property) that there exists

an n E N such that nr > s. 2. Given any x E R, prove that there exists a unique n E Z such that n - 1 5 x < n. 3. If r * 0 is a rational number and x is an irrational number, prove that r + x and rt are irrational. 4. Prove directly (without using Theorem 1.5.1) that between any two rational numbers there exists a rational number. 5.

If x. y E R with x < y, show that x < ty + (I - t)x < y for all t, 0 < t < 1.

6. *a. Prove that between any two rational numbers there exists an irrational number. *b. Prove that between any two real numbers there exists an irrational number.

7. If x > 0, show that there exists n E N such that 1/2" < x. 8. Let x, y E R with x < y. If u E P with u > 0. show that there exists a rational number r such that x < ru < Y.

9. Let B = {r E 0 : r > 0 and r2 < 2} and a = sup B. Prove that a2 = 2.

1.6

Binary and Ternary Expansions3 In our standard base 10 number system we use the integers {0, 1.. . . , 9} to represent real numbers. Base 10, however, is not the only possible base. In base 3 (ternary) we use only the integers {0, 1, 2}, and in base 2 (binary) we use only {0, I }. In this section we will show how the least upper bound property may be used to prove the existence of the expansion of a real number x. 0 < x 5 1, in a given base. For purposes of illustration, and for later use, we will use base 2 and 3, which are commonly referred to as the binary and ternary expansions, respectively.

In base 10, given the decimal .1021, what we really mean is the real number given by

-+ 1

10

0

2

1

107+70+70-

However, in base 3, the expansion .1021 represents 1

0

2

1

3+32+3;+34, which is the ternary expansion of 34/81. For a given x, 0 < x s 1, the ternary or base 3 expansion of x is defined inductively as follows. 3. This section can be omined on first reading. The ternary expansion of a real number is only required in Section 3.3 and can be covered at that point.

1.6

1.6.1

DEFINITION

Binary and Ternary Expansions

31

Let n, E {0, 1, 2} be the largest integer such that n 3 < X.

Having chosen n1, ... , nk, let nk+ 1 E {0, 1, 2} be the largest integer such that n, 3

The expression .n,n2n3

+n2+. . .+nk+nk+1 <X. 32

3k+'

3k

is called the ternary expansion of x.

If we set

E={3 +.+3k:k=1,2, ..}, then E * 0 and E is bounded above by x. As we will shortly see, sup E = x. In terms of series, which will be covered in detail later, we have 00 n

k.

X =

k=1 3

The binary expansion of x, or the expansion of x to any other base, is defined similarly. For the binary expansion, the integer nk at each step is chosen as the largest integer in {0, 1).

1.6.2

EXAMPLES

(a) We now use Definition 1.6.1 to obtain the ternary expansion of 3. As the first step. we must choose n, as the largest integer in {0, 1, 2} such that < 3'

3

This inequality fails for n, = 1, 2. Thus n, = 0. To find n2, we choose the largest integer n2 E {0, 1, 2) such that

3+32<3' which is satisfied by n2 = 2. To find n3 we must have

0+ 3

2

+ n3 <

32

1

3.

33

It is left as an exercise to show that this is satisfied for n3 = 0, 1, 2. Thus we take n3 = 2. At this stage we conjecture that nk = 2 for all k >_ 2, and that 1

3 = .02222

(base 3).

32

Chapter t

The Real Number System

To see that this indeed is the case, we use the fact that for the geometric series.

rk=(11

O
r),

k=o

Thus

- _2

.0222...=

2

I

2

k

9 kap 3

k=, 3k

1

1

i

9 1

3

In (c) we will illustrate how mathematical induction may also be used to prove such a result. The above ternary expansion is not unique. The number ;also has the finite expansion I 3

=

(base 3).

.1000.

We will discuss this in greater detail at,the end of this section.

(b) The binary expansion of is given by 3 = .010101

(base 2).

This expansion can be obtained using the definition and induction (Exercise 3). Alternatively, using the geometric series we have X

r_I 2

1 .

I"

I

I

I

4 = 3.

4

-U

(c) The ternary expansion of 1 is given by 1

=.1111

.

2

(base 3).

We now show in detail how this expansion is derived. We will use the second principle

of mathematical induction to prove that nk = I for all k E K. Since < ; and i > n, = 1. Thus, the result is true for k = 1. Let k > I and assume that n; = I for

I

all j < k. By definition nk is the largest integer in {0, 1, 2) such that

1+

3

+

3k-1

+nk<1 (3)

2

3`

Using the identity

r+r2+

+r"=

r - r"+i 1-r'

r # 1, (Example 1.3.3(a))

with n = k - 1, we obtain 3+

+

3kli = '

2

`

1 - 3k1_

1

1.6

Binary and Ternary Expansion

33

Substituting into equation (3) and multiplying by 2 gives 2;tk

1-

3k. +

< I.

This inequality is true if nk = 0 or I and is false if nk = 2. Since nk is to be chosen as the largest integer of {0, 1, 2} for which (3) holds, nk = 1. Thus by Theorem 1.3.4.

n; = 1 for all k E N.

1.6.3

THEOREM 1.6.1, let

Let x E U8 with 0 < x <- 1. and with {n} as defined in Definition (n1

nk

3

3'

Then sup E = x.

Proof. Let a = sup E. Since x is an upper bound, a s x. Suppose a < r. Let k be the smallest positive integer such that 1

3k

<.r - a

.<x.

a+

or

3k

Since a = sup E,

But n1

3

I +...+3k+;k= 3 +... i irk

n1

i+ nk +

nj.._ 1

3k

<

x.

3k

If nk = 0 or 1, this contradicts the choice of nk. If nk = 2. then we have 11 3

+

.

.

.

+

nk_ r + 1 3k_1

<

x.

If any n;, 1 s j <_ k - l is 0 or 1, we have a contradiction to the choice of n;. If all the ni = 2, then we obtain

3+3<xsl,

2

1

which is also a contradiction. The expansion

x = .nln,n3 is finite or terminating if there exists an integer m E tk such that nk. = 0 for all k > m. Otherwise, the expansion is infinite or nonterminating. The expansion of a real num-

34

Chapter I

The Real Number System

ber in a given base is not always unique; when x has a finite expansion, it has an infinite expansion as well. For the ternary expansion, when

x=3m,

aEN with O
and 3 does not divide a, then x has a finite expansion of the form

+ +3m, a,"E{1,2}.

x= 3

or an infinite expansion of the form

x=

a, 3

+-

0 +3m+

2

k k=m+i 3*

when a,"=1,

k,

when am = 2.

or

x=3+

+

+ I

k=m +1 3

© EXERCISES 1.6 1. Find the ternary expansion of each of the following. *a. i b. i3 2. Find the real number determined by each of the following finite or infinite ternary expansions. In the case of an infinite expansion, use the geometric series as in Example 1.6.2 (c) to determine your answer.

c..1010 b. .00222 e..001001001 f. .121212 *d..101010 3. *Find the binary expansion of 3. Use induction to prove the result. 4. Find both the finite and infinite binary expansions of 6. 5. If 0 < x s 1, prove that the infinite ternary expansion of x is unique. 6. If x = a/2m. with a E N odd, and 0 < a < 2m, show that x has a finite binary expansion of the form ni al 0 x= + + a'", + + 1 and an infinite expansion x= a.m = 1, 2k. 2'" 2 2 2m k.m+i *a. .0022

+7

1.71 Countable and Uncountable Sets In discussing sets, we all have an intuitive understanding of what it means for a set to be finite or infinite, and what it means for two finite sets to be of the same size; that is, to have the same number of elements. For example, the sets A = {2, 7, 11, 211 and B = {7, 3, 19, 32} both have the same number of elements; namely, four. We discover

1.7

Countable and Uncountable Sets

35

this by counting the number of elements in each of the two sets. Alternatively, we can reach the same conclusion without counting, by simply pairing up the elements:

7+2 3+7 19 + 11 32 + 21

For infinite sets, the concept of two sets being of the same size or having the same number of elements is vague. For example, let S denote the squares of the positive integers; namely, S = {12, 22, 32, .

. J.

Then on one hand, S is a proper subset of the positive integers N, yet as Galileo (1564-1642) observed, the sets N and S can be placed into a one-to-one correspondence as follows:

1+12 2+22 3+32

This example caused Galileo, and many subsequent mathematicians, to conclude that the standard notion of "size of a set" did not apply to infinite sets. Cantor, on the other hand, realized that the concept of one-to-one correspondence raised many interesting questions about the theory of infinite sets. In this section we take a closer look at infinite sets and what it means for an infinite set to be countable. We begin by defining the concept of equivalence of sets.

1.7.1

DEFINITION Two sets A and B are said to be equivalent (or to have the same cardinaiity), denoted A -- B, if there exists a one-to-one function of A onto B.

The notion of equivalence of sets satisfies the following:

(I) A -A. (reflexive) (ii) If A B, then B --A. (symmetric) (iii) If A - B and B -- C, then A ^- C. (transitive)

1.7.2

For each positive integer n, let N. = {1, 2, ... , n}. As in Section 1.1, N denotes the set of all positive integers. If A is a set, we say: DEFINITION

(a) A is finite if A N. for some n, or if A = t¢. (b) A is infinite if A is not finite. (c) A is countable if A N.

36

Chapter 1

The Real Number System

(d) A is uncountable if A is neither finite nor countable. (e) A is at most countable if A is finite or countable. The reader might want to ponder Exercise 21 at this point. Countable sets are often called denumerable or enumerable sets. It should be pointed out that some textbooks, when using the term countable, include the possibility that the set is finite.

1.7.3

EXAMPLES 112 ,

22,32, J. Then the function g(n) = n2 is a one-to-one mapping of N onto S. Thus S - N, and S is countable. (b) For our second example, we show that 7L -- N. To see this, consider the function

(a) As before, let S =

f: N -+ Z defined by n

.f(n)

(n

(n even),

2 2 1),

(n odd).

Figure 1.11 illustrates what the mapping f does to the first few integers. It is left as an exercise (Exercise 1) to show that this function is a one-to-one mapping of N onto Z. Thus 7L -- N, and the set Z is also countable.

Figure 1.11

As another illustration of countable sets consider the following theorem.

1.7.4 THEOREM N x N is countable. Proof. For this example, it is easier to construct a one-to-one mapping of N X N onto N. Such a function is given by

f(m, n) = 2-'(2n - 1). It is left as an exercise (Exercise 3) to show that f as defined is a one-to-one mapping

ofNXNontoN. 0 One of our goals in this section is to show that the set Q of rational numbers is countable.

1.7

Countable and Uncountable Sets

37

Sequences 1.7.5

DEFINITION

If A is a set, by a sequence in A we nean a function f from N into A.

For each n E N, let x = f(n). Then x is called the nth term of the sequence f. For notational convenience, sequences are denoted by or just rather than the function f Note, however, the distinction between which denotes the sequence, and {x : n = 1, 2, . . .}, which denotes the range of the sequence. For ex-

ample, {1 - (-1)"} denotes the sequence f where

f(n) = X. = 1 - (-1)". On the other hand, {x : n = 1, 2, ...} = {0, 2}. By definition, if A is a countable set, then there exists a one-to-one function f from N onto A. Thus

A = Rangef = {x,,:n = 1,2,...}. The sequence f is called an enumeration of the set A; i.e., A = {x : n = 1, 2, .. .} with x * x,R whenever n * m. This ability to enumerate the elements of a countable set plays a key role in the proof of the following theorem.

1.7.6

THEOREM Every infinite subset of a countable set is countable. Proof.

Let A be a countable set and let Ix.: n = 1, 2, ...} bean enumeration of A.

Suppose E is an infinite subset of A. Then each x E E is of the form xk for some k E N. We inductively construct a function f : N -) E as follows: Let n, be the smallest pos-

itive integer such that x.. E E. Such an integer exists by the well-ordering principle.

Having chosen n1..... nk - ,, let nk be the smallest integer greater than nk _ , such that

E E\

Set f(k) = x,,. Since E is infinite, f is defined on N. If m > k, then nk and thus x,,. Therefore f is one-to-one. The function f is onto E since if x E E, then x = x, for some j. By construction, nk = j for some k. and thus f(k) = x.

1.7.7 THEOREM If f maps N onto A. then A is at most countable. Proof. If A is finite, the result is certainly true. Suppose A is infinite. Since f maps N onto A, each a E A is of the form f (n) for some n E N. For each a E A, by the wellordering principle

f-'({a}) = {n E N J (n) = a} has a smallest integer, which we denote by n,,. Consider the mapping a -a nd of A into N. If a * b, then since f is a function, n. * nb. Also, since A is infinite, {n, : a E A) is an infinite subset of N: Thus the mapping a--+ n, is a one-to-one mapping of A onto an infinite subset of N. Therefore by-Theorem 1.7.6, A is countable.

Indexed Families of Sets In Section 1.1 we defined the union and intersection of two sets. We now extend these definitions to larger collections of sets. Recall that if X is a set, 91(X) denotes the set of all subsets of X.

38

Chapter 1

1.7.8

The Real Number System

DEFINITION Let A and X be nonempty sets. An indexed family of subsets of X with index set A is a function from A into 9(X).

If f : A -+ 13(A), then for each a E A, we let Ea = f(a). As for sequences, we denote this function by {Ea}aE,%. If A = N, then {En}nEN is called a sequence of subsets of X. In this instance, we adopt the more conventional notation {En} 1 to denote {En}nEN

1.7.9

EXAMPLES The following are all examples of indexed families of sets.

(a) The sequence {N.} 1, where N. = {1, 2.... , n}, is a sequence of subsets of N. Then {1n} 1 is a sequence of sub(b) For each n E N, set 1n = {x (E R : 0 < x < sets of N.

(c) For each x, 0 < x < 1, let

Ex= {rE0:0t=- r<x}. Then {E;},,E(0.1)is an indexed family of subsets of 0. In this example, the open interval (0, 1) is our index set.

1.7.10

DEFINITION Suppose {Ea}aEA is an indexed family of subsets of X. The union of the family of sets (Ea}aEA is defined to be

6A E. = {x E X : x r= Ea for some a E A). The intersection of the family of sets {Ea}aEA is defined as

n E. = {x E X: x E Ea for all a E A}. If A = N we use the notation 00

00

fl En U E. and n-1

u E.

instead of

n=1

nEN

and

fl En,

nEN

respectively. Also, if A = NO then k

nEN, E.

is denoted by

U En,

n=1

with an analogous definition for the intersection. Occasionally, when the index set A is fixed in the discussion, we will use the shorthand notation U. E. or !1a E. rather than

aU E. 1.7.11

or

f A Ea.

EXAMPLES We now consider the union and intersection of the families of sets given in the previous example.

1.7

Countable and Uncountable Sets

39

(a) With N. = {1, 2, ... , n}, we have 00

00

nN,, = {1},

and

n=1

UN,, = N. n-i

Since I is the only element that is in N. for all n, fl,, Ni,, = {1}. For the union, since N. C Ni for all n, U. Nn C Ni. On the other hand, if n E N, then n E NI,,, and as a consequence, N C U,, N,,, which proves equality. (b) As in the previous example, for n E NI, let In = {x E R:0 < x < .1}. We first show that

00

n in = (Q.

Suppose the conclusion is false. Then there exists x E H such that x E In for all n; i.e.,

0<x< 1,n

forallnEN.

This, however, contradicts Theorem 1.5.1, which guarantees the existence of a positive integer n such that nx > 1. For the union, since In C 11 for all n ? 1, 00

={xE08:0<x< 1}.

R=1

(c) We leave it as an exercise (Exercise 9) to show that if E, is defined as in Example

1.7.9(c); i.e., E, = {r E O : 0 5 r < x}, then

sEn E,={0}

and

1}.

As for a finite number of sets, we also have analogues of the distributive laws and De Morgan's laws for arbitrary unions and intersections.

1.7.12

THEOREM (Distributive Laws)

(a)

En( U Ea) = aU (E n E

1.7.13

If Ea, a E A. and E are subsets of a set X, then

U (fl Ea)

n (E U E.).

*eA

THEOREM (De Morgan's Laws) If {EQ},EA is a family of subsets of X, then

(a)

= aEA fl Ea,

UA EQ

aE

c

(b) *GA n Ea = aEA u E:The proofs of both of these theorems are left to the exercises. The following theorem is an analogue of Theorems 1.25 and 1.2.6.

40

Chapter 1

1.7.14

The Real Number System

THEOREM Lei f be a function from X into Y. (a) If {E,},EAis a famil of subsets of X, then

f (,U E0) = U f(E ) f(n E,) C n f(E,). (b) If {B,}OEA is a family of subsets of Y. then

P

f

(U

B,)

= U f -'(B.).

'(n B.) = n f-,(B,.). GA

A

Proof. (a) To illustrate the method of proof, we will prove that f (n. E,) C fl, f (E,). The proof that f (U, E,) = U f(E,) is left to the exercises. Suppose y E f(fl, E,.). Then by definition of the image of a set, Y = f (.Y) for some

x E fl, E,. But then x E E, for all a E A, and thus f (x) E f (E.) for all a E A. There-

fore y E fl, f(E,). (b) Again we will only prove one of the two identities. The other is left to the exercises.

In this instance we will show that f -' (U, B,) = U. f Let B = U, B,. Since B. C B for all a e A, we have f `(B,) C f `(B) for all a e A. Thus U. f "' (B,) C f -' ( U,, B.). In the other direction, suppose x E f -'(U,, B,). Then by the definition of the inverse image of a set, f (x) E U. B,. Thus f (x) E B,,, for some index a,, E A.

But then x E f -' (B,), and hence x E U, f '(B.). This proves equality. U

The Countability of 0 The countability of the set of rational numbers Q will. follow as a corollary of the following theorem.

1.7.15

THEOREM If (E,)'".1 is a sequence of countable sets and x.

S = n-I U then S is countable.

Proof.

Since E is countable for each n E N. we can write

E = {x,,,k:k= 1,2,...). Since E, is an infinite subset of S, the set S itself is infinite. Consider the function

h: N X N - Shy h(n, k) = x,k.

1.7

Countable and Uncountable Sets

41

.The function h, although not necessarily one-to-one, is a mapping of N X N onto S. Thus since N X N - N, there exists a mapping of N onto S. Hence by Theorem 1.7.7 the set S is countable.

1.7.16

COROLLARY

CQ is countable.

Proof. For each m E N, let Em=

Then E. is countable, and since 0 =

{iE}. Ua_, Eby

Theorem

1.7.15 the set 0 is count-

able.

The UncountabUity of 08 In November 1873, in a letter to Dedekind, Cantor asked whether the set R itself was countable. A month later he answered his own question by proving that 01 was not countable. We now prove, using Cantor's elegant "diagonal" argument, that the closed interval [0, 11 is uncountable,. and thus 08 itself is uncountable. For the proof we will use the fact that as in Section 1.6, every x E [0, 11 has a decimal expansion of the form x = n ,n, with n; E {0, 1, 2, . .. , 9}. As for the binary and ternary expansions, the decimal expansion is not necessarily unique. Certain numbers such as ,'-o have two expansions; namely. 10

.100

and

10 = .0999

This, however, will not be crucial in the proof of the following theorem.

1.7.17

THEOREM The closed interval [0, 11 is uncountable.

Proof. Since there are infinitely many rational numbers in [0, 1 ], the set is not finite. To prove that it is uncountable, we only need to show that it is not countable. To accomplish this, we will prove that every countable subset of [0, 1} is a proper subset of [0, 1 ]. Thus [0, 1 ] cannot be countable. Let E = {xn : n = 1, 2, .. .} be a countable subset of [0, 1]. Then each xn has a decimal expansion xn = xn.lxn.2xn,3 .

.

.

where for each k E N, x,.k E 10, 1, .. .. 9}. We now define a new number Y = .yly2Y3 ..

as follows: If x, ,, c 5. define yn = 6; if xn ? 6, define yn = 3. Then y E [0, 1]. and since yn * 0 or 9, y is not one of the real numbers with two decimal expansions. Also. since for each n E N. yn # xn,n, we have y * xn for any n. Therefore Y E E; i.e., E is a proper subset of [0, 11. Another example of an uncountable set is given in the following theorem.

42

Chapter)

1.7.18

The Real Number System

THEOREM If A is the set of all sequences whose elements are 0 or 1, then A is uncountable.

Remark. The set A is the set of all functions f from N into {0, I }. Thus a sequence f E A if and only if f(n) = O or I for all n.

Proof. As in the previous theorem, we will also prove that every countable subset of A is a proper subset of A, and thus A cannot be countable. Let E be a countable subset of A and let {s : n = 1 , 2, .. .} be an enumeration of the set E. For each n, s. is a sequence of 0's and l's. We construct a new sequence s as follows: for each k E N, let

s(k) = I - sk(k). Thus if sk(k) = 0, s(k) = 1, and if sk(k) = 1, s(k) = 0. Thus s e A. Since for all k E N

s(k) # sk(k), we haves* s for any n E N. Therefore s a E; i.e., Ec A. 0 In the previous two theorems we proved that the closed interval [0, 1 ] and the set of all sequences of 0's and I's are both uncountable sets. One may ask, are these two sets equivalent? Considering the fact that every real number x E [0,1 ] has a binary expansion, which is really a sequence of 0's and I's, one would expect that the answer is yes. This indeed is the case (Miscellaneous Exercise 7).

EXERCISES 1.7 1. a. Prove that the function f of Example 1.7.3(b) is a one-to-one function of N onto Z. b. Find a one-to-one function of Z onto N. 2. *Let 0 denote the set of odd positive integers. Prove that 0 -- N. 3. Prove that the function f of Theorem 1.7.4 is a one-to-one function of N X N onto N.

4. *a. If a. b e R with a < b, prove that (a, b) - (0, 1). b. Prove that (0, 1) -- (0, oo). 5. Suppose X. Y, Z are sets. If X - Y and Y- Z, prove that X

Z.

6. *a. If A ^- X and B Y, prove that (A X B) (X X Y). b. If A and B are countable sets, prove that A X B is countable. 7. If X - Y, prove that 9(X) - 9(Y). , A. for each of the following sequences of sets & Find U , A. and

fl°

nEN nEN

x51nEN,n; 2

e.A ={xER:

b.A.={xCE R:-1,<x<1}. nEN x:5 l-,-,), nEN nEN

9. For each x E (0, 1), let Ex = {r E Q : 0 s r < x}. Prove that

n E,={0}

xE(0,I)

and

10. Prove Theorem 1.7.12. 11. Prove Theorem 1.7.13.

U E,={r E Q:0sr< 1).

Notes

43

12. Let f be a function from X into Y.

a. If {Ea}eA is a family of subsets of X, prove that f(Ua Ea) ='U. f(E*). b. If {Ba}aEA is a family of subsets of Y, prove that f -' (laBa) = fla f -'(B.). 13. a. If A is a countable subset of an uncountable set X. prove that X \ A is uncountable. b. Prove that the set of irrational numbers is uncountable. 14. Suppose f is a function from X into Y. If the range of f is uncountable, prove that X is uncountable. 15. a. For each n E N, prove that the collection of all polynomials in x of degree less than or equal to n with rational coefficients is countable.

b. Prove that the set of all polynomials in x with rational coefficients is countable.

16. Prove that the set of all intervals with rational end points is countable.

17. Let A be a nonempty bounded subset of R and let a = sup A. If a a A, prove that for every e > 0. the interval (a - e, a) contains infinitely many points of A. 18. *a. Prove that (0, 1) - (0, 1 ]. (This problem is not easy!) b. Prove that (0, 1) -- [0. 1 ]. 19. *A real number a is algebraic if there exists a polynomial p(x) with integer coefficients such that p(a) = 0. Prove that the set of algebraic numbers is countable. 20. Prove that any infinite set contains a countable subset. 21. Prove that a set is infinite if and only if it is equivalent to a proper subset of itself. 22. *Prove that any function from a set A to the set 9(A) of all subsets of A is not onto 9(A). 23. *Prove that [0, 1] x [0, 1 ] - [0, 1].

NOTES The most important concept of this chapter is the least up-

per bound property of the real numbers. This property will be fundamental in the development of the underlying theory of calculus. In the present chapter we have already seen its application in proving the Archimedian property

(Theorem 1.5.1). For the rational number system this property can be proved directly. For the real number system it was originally assumed as an axiom by Archimedes (287-212 B.c.). Cantor, however, proved that the Archimedian property was no axiom, but a proposition derivable from the least upper bound property. In subsequent chapters the least upper bound property will occur in proofs of theorems, either directly or indirectly, with regular frequency. It will play a crucial role in the characterization of the compact subsets of R and in the study of sequences of real numbers. It will also be required in the proof of the intermediate value theorem for continuous functions. One of the corollaries of this theorem is Theorem 1.5.3 on the existence of nth roots of positive real numbers. Many other results in the text will de-

pend on previous theorems that require the least upper bound property in their proofs.

The emphasis on the least upper bound property is not meant to overshadow the importance of the concepts of countable and uncountable sets. The fact that the rational numbers are countable, and thus can be enumerated, will be used on several occasions in the construction of examples. In all the examples and exercises. every infinite subset of R turns out to be either countable or equivalent to [0, 1 ]. Cantor also made this observation, and it led him to ask whether this result was true for every infinite subset of R. Cantor was never able to answer this question; nor has anyone else. The assertion that every infinite subset of R is either countable or equivalent to [0, 11 is known as the continuum hypothesis. In 1938 Kurt Godel proved that the continuum hypothesis is consistent with the standard axioms of set theory; that is. Godel showed that the continuum hypothesis cannot be disproved on the basis of the standard axioms of set theory. On the other hand, in 1963 Paul Cohen showed that the continuum hy-

44

Chapter 1

The Real Number System

pothesis is undecidable on the basis of the current axioms of set theory. Cantor's creation of the theory of infinite sets was

motivated to a great extent by problems arising in the study of convergence of Fourier series. We will discuss some of these problems in greater detail in Chapter 9. The notion of one-to-one. correspondence allowed Cantor to define the "power" or what is now called the "cardinality" of a set. For a given set A, the cardinal number or power of A is denoted by A. Two sets A and B are said to have

is equivalent to some subset of B. We also write A < 8 provided A s B and A # B. C)early, N < [0, 1 ]. Also. Exercise 22 of the previous section proves that for any set A, A < Using equivalence of sets, it follows that if

A s B and B C, then A s C. This naturally leads to the following question: If A and 8 are sets such that

ample 0 = 0, N, = n, N = K. and [0. 1) = c.

A c B and B < A. is A = B? The answer is yes, and the result is known as the Schroder-Bernstein theorem. Cantor did more than just introduce cardinal numbers; he also defined addition, multiplication, and exponentiation of cardinal numbers. Cantor's original work on cardinal numbers and the theory of infinite sets can be found in his monograph listed in the Supplemental Readings. A more modern treatment. including a proof of

An ordering can be defined on cardinal numbers as follows: If A and B are sets, then A B provided that A

the Schrdder-Bemstein theorem, is available in the text by Halmos.

the same cardinality, denoted A = B, provided that A is equivalent to B. Some commonly encountered sets have special symbols attached to their cardinal number. For ex-

MISCELLANEOUS EXERCISES 1. Let A and B be nonempty sets. For a E A, b E B. define the ordered pair (a. b) by (a, b) _ {{a}, {a, b}}.. Prove that two ordered pairs (a, b) and (c, d) are equal if and only if a = c and b = d. The following-two exercises are detailed and lengthy. The first is a sketch of the proof of Theorem 1.5.3. The second shows how the least upper bound property may be used to define the exponential function b', b > 1.

2. LetE= {t E R:t> 0 and t' <x). a. Show that E # ¢ by showing that x/(x + 1) E E. b. Show that I. + x is an upper bound of E. Let y = sup E. The remaining parts of the exercise are to show that y' = x. This will be accomplished by showing that y" < x and y" > x lead to contradictions, leaving y" = x. To accomplish this, the following inequality will prove useful. Suppose 0 < a < b, then

b"-a"=(b-a)(b"-'+ab" +... '+ a`)
x-y n(v + I)

Use (*) to show that. y + h E E. d. Show that the assumption y" > x also leads to 'a contradiction of the definition of y as follows: Set

k=yy"_x Show that if tat y-k,then t44;E. 3. Fix b > 1. a. Suppose m, n, p, q are integers with n > 0 and q > 0. If m/n = p/q, prove that

(hl" =

-Thus if r is rational, b' is well defined.

'

Miscellaneous Exercises

45

b. If r, s are rational, prove that b"' = b' b'. c. If x E R, let B(x) = {b': r E 0. t x}. Prove that b' = sup B(r) when r E 0. Thus it now makes sense to define b' = sup B(x) when x E R. d. Prove that b"i" = bxb'' for all real numbers x,

The following two exercises provide a detailed development of the field of complex numbers.

4. Definition. A complex number is an ordered pair (a. b) of real numbers. If ;, = (a, b) and w _ (c, d), we write z = w if and only if a = c and b = d. For complex number z and w we define addition and multiplication as follows:

z+w=(a+ c, b + d) (or - bd, ad + bc). The set of ordered pairs (a, b) of real numbers with the above operations of addition and multiplication is denoted by C. a. Find elements 0 and 1 in C such that 0 + z = z and lz = z for all z E C. zw

b. Show that if z = (a, b), then -z - (-a,-b) is the additive inverse of z. c. For z E C with z * 0, find the multiplicative inverse z-'. d. Prove that the set of complex numbers C with addition and multiplication as defined is a field.

e. Set i = (0, 1). Show that i2 = -1. f. Show that every complex number z can be written as z = a + bi where a. b E R. The real numbers a and b are called the real part and the imaginary part of z, respectively. We write a = Re(z) and b = Im(z). g. Prove that C is not an ordered field. 5. Definition. If z = a + bi E C. then the complex number i = a - bi is called the conjugate of z. The absolute a2 + value of z, denoted IzI, is defined by IzI = a. Prove each of the following. (1)

i

z+

(ii) zw = z N (iii) z + i = 2 Re(z),

z - z = 2i Im(z)

(iv) zz = 1z12

b. Prove each of the following. (1)

IZI = IzI

(ii) Izwl = IzIIwI (iii) IRe(z)I IzI, IIm(z)I s IzI (iv) Iz + w12 = Izl2 + Iwl2 + 2 Re(zw) (v) Iz - w12 = 1z12 + Iwl2 - 2 Re(zw) (vi) Iz + wl IZI + Iw'I

The following result, known as the Schroder-Bernstein theorem, is nontrivial, but very- important. It is included as an exercise to motivate further thought and additional studies. A proof of the result can be found in the text b) Halmos listed in the Supplemental Reading.

6. Let X and Y be infinite sets. If X is equivalent to a subset of Y. and Y is equivalent to a subset of X. prove that X is equivalent to Y. 7. As in Theorem 1.7.18, let A denote the set of all sequences of 0's and I's. Use the previous result to prove that

A-[0,1].

46

Chapter i

The Real Number System

SUPPLEMENTAL READING Buck, R. C., "Mathematical induction and recursive definition:' Amer. Math. Monthly 70 (1963),128-135. Burrill, Claude W., Foundations of Real Numbers. McGraw-Hill, Inc., New York, 1967. Cantor, Georg, Contributions to the Founding of the Theory of Transfrnite Numbers (translated by Philip E. B. Jourdain), Open Court Publ. Co., Chicago and London, 1915. Dauben, Joseph W., Georg Cantor: His Mathematics and Philosophy of the Infinite, Princeton University Press, Princeton, N.J., 1979.

GOdel, Kurt, "What is Cantor's continuum problem?" Amer. Math. Monthly 54 (1947), 515-525. Halmos, Paul. Naive Set Theory, Springer-Verlag. New York. Heidelberg. Berlin, 1974.

Richman, F. "Is 0.999... = 0- Math. Mag. 72 (1999), 396-400. Shrader-Frechette. M., "Complementary rational numbers," Math. Mag. 51(1978), 90-98. Spooner, George and Mentzer, Richard, Introduction to Number Systems, Prentice-Hall, Inc., Englewood Cliffs, N.J., 1968.

Sequences of Real Numbers 2.1 Convergent Sequences

2.2 Limit Theorems 2.3 Monotone Sequences

2.4 Subsequences and the Bolzano-Weierstrass Theorem

2.5 Limit Superior and Inferior of a Sequence 2.6 Cauchy Sequences

2.7 Series of Real Numbers

In our study of sequences of real numbers we encounter our first serious introduction to the limit process. The notion of convergence of a sequence dates back to the early nineteenth century and the work of Bolzano (1817) and Cauchy (1821). Some of the concepts and results included in this chapter have undoubtedly been encountered previously in the study of calculus. Our presentation, however, will be considerably more rigorous-emphasizing proofs rather than computations. We begin the chapter by introducing the notion of convergence of a sequence of real numbers and by proving the standard limit theorems for sequences normally encountered in calculus. In Section 2.3 we will use the least upper bound property of 1q to prove that every bounded monotone sequence of real numbers converges in R. The study of subsequences and subsequential limits will be the topic of Section 2.4. In this section we also prove the well-known result of Bolzano and Weierstrass that every bounded sequence of real numbers has a convergent subsequence. This result will then be used to provide a short proof of the fact that every Cauchy sequence of real numbers converges. Although the study of series of real numbers is the main topic of Chapter 7, some knowledge of series will be required in the construction of certain examples in Chapters 4 and 6. For this reason, we include a brief introduction to series as the last section of this chapter. Even though our emphasis in this chapter is on sequences of real numbers, in subsequent chapters we will also encounter sequences of functions and convergence of sequences in normed linear spaces. A good understanding of sequences of real numbers will prove very helpful in providing insight into properties of sequences in more general settings. 47

48

Chaprer2

Sequences of Real Numbers

2.11 Convergent Sequences Before we begin our study of sequences we first introduce the absolute value of a real number.

2.1.1

DEFINITION

For a real number x. the absolute value of x, denoted ix . is de-

fined by V.

1x1 =

ifx > 0,

-.r, ifx - 0.

For example, 141 = 4 and 1-5I = 5. From the definition, Ixl ? 0 for all x E R. and 1x1 = 0 if and only if x = 0. This last statement follows from the fact that if x * 0. then -x * 0 and thus 1x1 > 0. The following theorem, the proof of which is left to the exercises (Exercise 1), summarizes several well-known properties of absolute value.

2.1.2

THEOREM

(a) I-xI = Ix1 for all x E R. (b) lxvI = I.xI IyI for all x. Y E R.

(c) IxI = \ for all x E R. (d) If r > 0. then lxl < r if and only if -r < x < r. (e) - Ixl s x < Ixl for all x E R. The following inequality is very important and will be used frequently throughout the text.

2.1.3

THEOREM (Triangle Inequality)

For all.r, y E R, we have

Ix +v1l< lx1+h'I. Proof. The triangle inequality is easily proved as follows: For x,.%- E R.

0:!:_ (x+v)-=x-+2xv+vIxl' + 21x1 hvi + lyl' = (Ixl + w;)=. Thus by Theorem 2.1.2(c),

Ix+yl=

(x+y)':S

(L,I+Iv1)'-1xl+i1.

As a consequence of the triangle inequality, we obtain the following two useful inequalities. 2.11A

COROLLARY

For all x, v, E IL we have

(a) Ix - yl s Ix - zl + j:.-yl,and (b) IIx I - ly11

Ix - yl.

2.1

Convergent Sequences

49

Proof. We provide the proof of (a), leaving the proof of (b) as an exercise (Exercise 2). If x, y, z E 08, then by the triangle inequality,

Ix - yl =

I(x-z)+(z- y)I S Ix - zI + Iz - yI.

I]

The following example illustrates how properties of absolute value can be used to solve inequalities.

2.1.5

EXAMPLE Determine the set of all real numbers x that satisfy the inequality 12x + 41 < 8. By Theorem 2.1.2(d), 12x + 41 < 8 if and only if -8 < 2x + 4 < 8. or equivalently, -12 < 2x < 4. Thus the given inequality is satisfied by a real number

x if and only if -6 < x < 2. Geometrically, 1x1 represents the distance from x to the origin 0. More generally, for x, y E 1i8, the euclidean distance d(x, y) between x and y is defined by

d(x,y)= Ix - yl For example, d(-1, 3) = I-1 - (3)1 = 1-41 = 4, and d(5, -2) = 15 - (-2)1 = 7. The distance .d, may be regarded as a function on R X R which satisfies the following properties: d(x, y) _,- 0, d(x, y) = 0 if and only if x = y, d(x, y) = d(y, x), and d(x, y) s d(x, z) + d(z, y)

for all x, y, z E R. This last inequality, also referred to as the triangle inequality, follows from Corollary 2.1.4(a).

Neighborhood of a Point The study of the convergence of a sequence or the limit of a function requires the notion of one real number being "close to" another. Since the euclidean distance between two points a and x is given by d(a, x) = j a - xl, saying that x is "close to" a is equivalent to saying that the distance lx - aI between them is "small" A convenient method for expressing this idea is with the concept of an e-neighborhood of a point. This concept will prove useful not only in the study of the limit of a sequence but also in our study of the limit of a function and the structure of point sets in R.

2.1.6

DEFINITION

Let p E R and let e > 0. The set

N1(p)_{xER:Ix -pl <e} is called an e-neighborhood of the point p. Whenever we use the term neighborhood, we will always mean an e-neighborhood

with e > 0. By Theorem 2.1.2(d), for fixed p E R and E > 0,

N,(p)={x:p-E<x
50

Chapter2

Sequences of Real Numbers

p-E

p+E

p

Figure 2.1

N.(p)

Convergence of a Sequence Recall from Definition 1.7.5 that a sequence in R is a function f :N -i R. For each n E N, p = f(n) is called the nth term of the sequence f, and for convenience, the sequence f is denoted by I or simply

2.1.7

DEFINITION A sequence {p.)' t in R is said to converge if there exists a point p E R such that for every e > 0, there exists a positive integer no such that p,, E Ni(p) p is the limit for all n a n,. If this is the case, we say that { of the sequence (p.), and we write

if {

yip" = p or p" _+p. does not converge, then {p,,) is said to diverge.' In the definition, the statement p E N,(p) for all n at n, is equivalent to

1p - p l < e for all n ? n,. As a general rule, the integer n,, will depend on the given e. This will be illustrated in the following examples.

2.1.8

EXAMPLES

(a) For our first example we show that the sequence t converges to 0 in R. The proof of this is the remark following Theorem 1.5.1; namely, given e > 0, there exists a positive integer n, such that n,e > 1. Thus for all n _> no,

<e.

H_o

Therefore 1im,, = 0. In this example, the integer n, must be chosen so that n "o

n, > lie.

defined by p = p for all n E N is called the constant (b) If p E R, the sequence sequence p. Since Jp - pl = 0 for all n E 101, we have lim p = p. (e) Consider the sequence

1E,0_ 2n + 2

I.

We will show nthat **0

lim 3n+2=3. 1. The negation of the definition of convergence of a sequence is included in Section A.4 of the Appendix.

2.1

Convergent Sequences

51

Since

2n+ l

2

3n + 2

3

l

1

3(3n + 2) < 9n'

given e > 0, choose n E N such that n > 1/9e. Then for all n ? no, I

2n+ 1

2

3n + 2

3

< e.

Thus the given sequence converges to 3.

(d) The sequence { Ip.+11 - (-

diverges in R. To prove this, we first note that for = 2 for all n. Suppose p -+ p for some p E R. Let this sequence, j ph 0 < e < 1. Then by the definition of convergence, there exists an integer n, such that

Ip - pj < efor all n?n0.But if n-a no,then

2=1pn-pn+11 _,; 1p0-pI+Ip-p,,+ij< 2e<2. This, however, is a contradiction. Thus our assumption that the sequence converges is false; i.e., the sequence diverges.

(e) As our final example we prove that li(Vn+ I - Vn) = 0. We first note that

( n+1-y) (Vn + I +V) 1

( n+1+\)

I

n+1+Vn1

2Vn-

Given e > 0, we want to choose no E NI such that 1/(2\) < e for all n y n,. This is easily verified to be the case if n, E N is chosen so that n, a 1/4E2. With this choice of n, we now have I n + 1 - VnI < e for all it at n,.

2.1.9

DEFINITION A sequence { in R is said to be bounded if there exists a positive constant M such that M for all it E N. This definition is equivalent to saying that the range {p : it E N} of the sequence {p,} is a bounded subset of R.

2.1.10

THEOREM

(a) If a sequence in ir'8 converges, then its limit is unique. (b) Every convergent sequence in R is bounded.

Proof.

(a) To prove (a) we use the method of proof by contradiction. Thus we assume that the sequence converges but that its limit is not unique. So suppose the se-

52

Sequences of Real Numbers

Chapter2

quence {p"} converges to two distinct points p,q E R. Let e = 11p - q1. Since p" - * p, there exists an integer n, such that I p" - p I < e for all n ? n1. Also, since p" - q, there exists an integer n2 such that I p. - qI < e for all is ? n2. Thus if n max{ni, n2}, by the triangle inequality (Corollary 2.1.4(a)), 2

Ip - qI s Ip" - pi + Ipn - qI < 2e=-glp-qI The inequality Ip - qI <'Ip - qI however gives a contradiction. Thus the limit must be unique. (b) Let {p"} be a convergent sequence in 11 that converges top E R. Take e = 1.

For this e, there exists an integer na such that I p" - pI < 1 for all n > no. Thus by Corollary 2.1.4(b), I p" I < I p I + 1 for all n > no. Let

M = max{Ip,l.

.Ip".I. Ipl + 1}.

Then Ip"I s M for all n E N. Therefore the sequence is bounded.

2.1.11

O

EXAMPLES

(a) According to the previous theorem, every convergent sequence is bounded. The converse, however, is false. The sequence (I - (-1)"}"= I is bounded, but by Example 2.1.8(d), the sequence does not converge. The sequence is bounded since Ip"I =

II -(-1)"1:5 2forallnEN. (b) The sequence {n(- I r} is not bounded in R, and thus cannot converge.

EXERCISES 2.1 1. Prove Theorem 2.1.2.

2. *Prove Corollary 2.1.4(b). + X.1 Sri I + 3. Prove that for xi...._x. E R, Ix, + + Ix" I 4. If x, y E R, show that (x + yl = Ixl + lyl if and only if xy ? 0. 5. If a, b e R, prove that labs s 21(a2 + b2). 6. Determine all x E R that satisfy each of the following inequalities.

"a.13x-21<-I1

b. Ix2-4I<5

d. Ix-li
'c.lxl+Ix-11<3

7. Determine and sketch the set of ordered pairs (x, y) in R X R that satisfies the following.

b.IxI:IyI C.Ixyl!5 2 a. 4x1=II 8. For each of the following sequences, prove, using an e, no argument that the sequence converges to the given limit p; that is, given e > 0, determine n such that I p" - pI < e for all n ;' no. b2n+5 _ 1 n2+ I _ 3 13n ++5 P 'a. c 2n2 2\ 2n 7 ' 2 6n-3 p 3 'P

j1-(rJ,p=1 l n (

/1

1

M'p=0

f. In

1

+R

1

J},p=2.

2.2 Limit Theorems

+*c. !

53

9. Show that each of the following sequences diverges in R.

*a {n(1 + (-1)")}

b.

ll

d. j .sin

2

e.

10. If b > 0, prove that lim

1

n-- 1 + nb

11. *a. If b > 1, prove that lim

l(+)nl

sin}

J

= 0.

b = 0.

b. If 0 <- b < I, prove that nix lim b" = 0. 12. *Let {a,} be a sequence in Il withnix lim a" = a. Prove that lim a, 13. Let {a.} be a sequence in R withn-+oc lim a, = a. Prove that lim n xq3, = a3.

14. *If an a 0 for all n and lim an = a, prove that lim \ _ V. 15. Prove that if {a,} converges to a, then {la,I} converges to ]a]. Is the converse true?

16. Let {a.} be a sequence in P with lim an = a. If a > 0, prove that there exists n,, E N such that an > 0 for all

nano. 17. Let {a,} be a sequence in P satisfying Ia. - a.. I ? c for some c > 0 and all n E N. Prove that the sequence {a,} diverges.

2.2

Limit Theorems In this section we will emphasize some of the important properties of sequences of real numbers and investigate the limits of several basic sequences that are frequently encountered in the study of analysis. Our first result involves algebraic operations on convergent sequences.

2.2.1

THEOREM

If {a.} and

are convergent sequences of real numbers with

lima.=a

n- oo

and

limb,=b,

n- 00

then

lira (a, + b,) = a + b, and (a) n_00

(b) ltm a,b, = ab. .-000

L. = b. (c) Furthermore, if a * 0, and a. * O for all n, then lint n--an

a

Proof. The proof of (a) is left to the exercises (Exercise 1). To prove (b), we add and subtract the term a,b to obtain

la,b. - abl = I(a,b, - a,b) + (a,b - ab)l

c la,llb. - bl + Iblla. - al.

54

Chapter2

Sequences of Real Numbers

Since {an} converges, by Theorem 2.1.10(b), {an} is bounded. Thus there exists a constant M > 0 such that Ia.I <- M for all n. Therefore,

lanbn - abl s Mlbn - bl + IbI Ian - at. Let e > 0 be given. Since an -+a, there exists a positive integer n, such that

In - al < n

E

2(lbl + 1)

for all n a n1. Also, since bn -+ b, there exists a positive integer n_ such that

Ibn-bt<2mf for all n at n2. Thus if n a max{n1, n2},

abl < M(2M

+ IbIC2(IbI+ 1)) < E.

Therefore lira anbn = ab. To prove (c) it suffices to show thatA00 lira I/an = 1/a. The result (c) then follows from (b). Since a # 0 and an a, there exists a positive integer n, such that

Ian - al <

21

21aI

for all n a n,. Also, since

IaISIa - and+lal

<2ial+lanl

for n a no, we have

Ianl z Ilai for all n a n,. Therefore,

Ia - a.I

1

Ian

a

Ianl Iai

I

2

<

Ian - al.

Iaj2

Let e > 0 be given. Since an -+a, we can choose an integer n, a n, so that

Ian - al < E 2for all n a n1. Therefore,

a_aI<E I

2.2 Limit Theorems

55

for all n >_ n1, and as a consequence, I

--

lim n-+oo a, - a'

2.2.2

COROLLARY

for any c E R,

If {a,,} is a convergent sequence of real numbers with lim an = a, then n"°`

(a) lim (an + c) = a + c, and (b) Inn can = ca. n-ow Proof. If we define the sequence {c,} by c, = c for all n E N, then the conclusions follow by (a) and (b) of the previous theorem.

2.2.3

THEOREM Let {a,,} and {b,,} be sequences of real numbers. If {b,,} is bounded and

lima,, = 0, then lim anbn = 0.

n-.oo

Proof. Exercise 3. Remark. Since the sequence {b,} may not converge, Theorem 2.2.1(c) does not apply. The fact that the sequence {b,,} is bounded is crucial. For example, consider the sequences {,1,} and {3n}.

2.2.4

THEOREM Suppose {a,,}, {b,,}, and {c,,} are sequences of real numbers for which there exists n, E N such that

an
n-+oo

limb,,=L.

n--.W

Proof. Exercise 4. This result, commonly called the "squeeze theorem," is very useful in applications. Quite often to show that a given sequence {a,,} in R converges to a, we will first prove that

la, - al S Mb, for some positive constant M and a nonnegative sequence {b,,} with lim b, = 0. (See, for example, the proof of Theorem 2.2.1(c).) At this point we can uselheorem 2.2.4

to conclude that the sequence {(a,, - a1} converges to zero, or equivalently, that lim an = a. now

56

Chapter 2

Sequences of Real Numbers

This can also be proved directly. SinceM-OG lim b" = 0, given e > 0. there exists n,, E IN such that 0 S b" < e/(M + 1) for all /n ? no. Thus for n ? n,,,

Ia"-aI5Mb"<M`M+ 1)<E. Therefore lim a" = a.

Some Special Sequences We next consider some special sequences of real numbers that occur frequently in the study of analysis. For the proof of Theorem 2.2.6 we require the following result.

2.2.5

THEOREM (Binomial Theorem)

For a E l8, n E N,

(1+a)"= k±(k)a&= (0)

+(n)a+(2)a2+

+(n)a"

where n

k

_

n!

k!(W:: k)!

is the binomial coefficient. As in Section 1.3, for n E N, n! (read n factorial) is defined by

2-1, with the usual convention that 0! = 1. Since the binomial theorem is ancillary to our main topic of discussion, we leave the proof, using mathematical induction, to the exercises (Exercise 13). An alternative proof using Taylor ser ies will be provided in Section 8.7.

2.2.6

THEOREM

(a) If p > 0, then limo

0.

(b) If p > 0, then lim

= 1.

(c) lim " n = 1. (d) If p > I and a is real, then

lint no

- = 0. P

(e) If IPI < 1, then lim p" = 0.

(f) For all p E R, lim nt = 0. Proof. The proofs of (a) and (b) are left to the exercises (Exercise 5). The proof of (a) is straightforward and the proof of (b) (for p > 1) is similar to the proof

2.2 Limit Theorems

57

of (c). For the proof of (c), let x" _ '"In-n - 1. Since x" is positive, by the binomial theorem

.(nl

n = (1 + x")"

1)

n(n2-

/f xn =

for all n ? 2. Therefore, xn 5 2/(n - 1) for all n ? 2, and as a consequence,

Thus by (a) and Theorem 2.2.4, lim x" = 0, from which the result follows. " (d) Let k be a positive integer so that k > a. Since p > 1, write p = (I + q), with q > 0. By the binomial theorem, for n > 2k,

P"=0 +q)"> kn qk=

qk.

k!

Since k < in, n - k + 1 > in + 1 > in. Therefore,

n(n- 1)' '(n-k+ 1)

nk

kt

2kk!'

and as a consequence, n°

05

k-"

c MnF"

The result now follows by part (a) and Theorem 2.2.4.

(e) Write p as p = ±1/q, where q > 1. Then, IP"I = IPA" =

4"

which by (d) (with a = 0) converges to 0 as n -- oo.

(f) Fix k E N such that k > dpi. For n > k, n!

IPr < klk-t) IPI " (k - 1)! ( k }

- n!

Since ip(/k < 1, the result follows by (e).

2.2.7

,

EXAMPLES We now provide several examples to illustrate the previous theorems.

(a) As in Example 2.1.8(c) consider the sequence'

f 2n + 1

in + 2

2n+1_n(2+'-)

2+-',

3n + 2

3+

} We write

58

Chapter 2

Sequences of Real Numbers

1

Since lim = lam 2 = 0, by Corollary 2.2.2(a), n-"o n n-.oo n

lim2 + 1n I = 2

1im3 + ? I = 3.

and

n--"O(

R"00(

n

Therefore by Theorem 2.2.1,

oo\2+nJ3+?3 )=2-1=2. /

(2+n?)-_1 3+n

1

1

3

(b) Consider the sequence

12N/n1. We first note that

+7

05

-

(-1)"

I2\+7

2\ I

1

Thus by Theorem 2.2.4 and Theorem 2.2.6(a) with p = 2, On

0.

"-Cc 2 n+

3l

n

(c) For our next example we consider the sequence

+n}

3'

As in (a), we first want

to factor out the dominant power in both the numerator and the denominator. By Theorem 2.2.6(d), lim n°/p" = 0 for any a E R and p > 1. This simply states that p" (p > 1) grows taster than any power of n. Therefore the dominant terms in the numerator and denominator are 2" and 3", respectively. Thus

2"+n3 _ 2"(1 +29') _( 2 (1 +Y) 3n + n2

3n(1 + 3,)

3

y

(I +

By Theorem 2.2.1 and Theorem 2.2.6(d),

"'

1+

1.

11+F 3

Finally, since lim ()" = 0 (Theorem 2.2.6(e)), we have n-4W

3

n-. 2n +n2=0. 3"

(d) As our final example we consider the sequence {n((1 + n)'2 - 1)}. Before we can evaluate the limit of this sequence we must first simplify the nth term of the sequence. This is accomplished as follows:

1)=n((l }!)2

- 1)=n((n+1)2- 1)

2n - 1 _ -2n2 - n (n + 1)2

F n+ l)2

2.2 Limit Theorems

59

Now we can factor out an n2 from both the numerator and the denominator. This gives

-2 -

(1 + i)2.

X,

"

Using the limit theorems we now conclude that lim x" = -2.

EXERCISES 2.2 1. Prove Theorem 2.2.1(a).

2. Let {a"} and {b"} be sequences of real numbers. a. If {a"} and {a" + b"} both converge, prove that the sequence {b"} converges.

b. Suppose b" * 0 for all n E N. If {b"} and

both converge, prove that the sequence {a"} also converges.

3. Prove Theorem 2.2.3. 4. Prove Theorem 2.2.4. 5.

a. If p > 0, prove that lim

1= v 0.

n

*b. If p > 0, prove that lim

= 1.

6. Find the limit of each of the following sequences.

(3n2+2n+l)°° 5n2 - 2n + 3 ".. ( Sc. 5

n

{\

°°

n

11+"

+a -

g.

n2+n-n}.,

L{

- i)E

h. {(2" + 3)""r..,

7. For each of the following sequences, determine whether the given sequence converges or diverges. If the sequence converges, find its limit; if it diverges, explain why.

I + (-1)"1°0 ((

n

1

-sin n

"-1

d.

Sc. 1n2\2n+3)2 } 00 a8n3 +51ao

L

e. fV9W 8. -Prove that "lim R cos n

2

-} 2

" l 2"+n2J.i

In cos nhr

2n+3

00

}

= 0.

9. Let {x"} be a sequence in R with x" - 0. and x" * 0 for all n. Prove that lim x" sin

= 0.

z

60

Chapter2

Sequences of Real Numbers

10. Let {a"} be a sequence of positive real numbers such that lim .-. a"

=L.

*a. If L < 1, prove that the sequence {a"} converges and that lim a = 0. RtiOG b. If L > 1, prove that the sequence {a"} is unbounded. c. Give an example of a convergent sequence {a"} of positive real numbers for which L = 1.

d. Give an example of a divergent sequence {a"} of positive real numbers for which L = 1. 11. Use the previous exercise to determine convergence or divergence of each of the following sequences. *a. {n2a"},0 < a < 1

b.

C'

{l n! }.0
d.

e.

{-}.aER.p>

{}.o<
}

I

12. *Suppose Sao lim (a" - 1)/(a" + 1) = 0. Prove that lim a" = 1. n--.oo I !s k s n, prove that 13. a. For n E N,

(k)+\k- I )

(n k

l)*b.

Use mathematical induction to prove the binomial theorem (Theorem 2.2.5).

14. Let {ak}k i be a sequence in R. For each n E N, define

a,+ +a"

s"

n

*a. Prove that if li-m ak = a, then lim s" = a. b. Give an example of a sequence {ak} which diverges, but for which {s") converges.

2.3' Monotone Sequences In this section we will briefly consider monotone sequences of real numbers. As we will see, one of the advantages of such sequences is that they will either converge

in R or diverge to oo or -oo.

2.3.1

DEFINITION A sequence {a"},°_ 1 of real numbers is said to be:

(a) monotone Increasing (or nondecreasing) if a" s a,, + t for all n E N; (b) monotone decreasing (or nonincreasing) if a" ? a"+ 1 for all n E N; (c) monotone if it is either monotone increasing or monotone decreasing. A sequence {a"} is strictly increasing if a" < a"+, for all n. Strictly decreasing is defined similarly.

2.3

Monotone Sequences

61

As a general rule, bounded sequences need not converge; e.g., {I - (-1)"}. For monotone sequences, however, we have the following convergence result.

2.3.2

THEOREM If {a,

is monotone and bounded, then {a"},` , converges.

Proof Suppose (an} is monotone increasing. Set

E=fan: n=1,2,...}. Then E # ¢ and is bounded above. Let a = sup E. We now show that

lim a" = a. n-ti00

Let e > 0 be given. Since a - E is not an upper bound of E, there exists a positive integer n,, such that

a - E
a - e < an<- a forallnatn0. Thus an E NE (a) for all n ? no and therefore lim a" = a. ano-I

anu anyN

a-t

a

Figure 2.2

Nested Intervals Property As an application of the previous theorem we prove the following result, usually referred to as the nested intervals property. The term nested comes from the fact that the sequence {I"} of intervals satisfies 1" D In+1 for all it E N (see Figure 2.3).

2.3.3

COROLLARY (Nested Intervals Pro)erty)

a,

a2

an

a,.«,

If {In}

b,,., b

Figure 2.3

bounded intervals with In D In+, for all n E N, then 00

nI ,,*o.

,

is a sequence of closed and

b2

b,

62

Chapter2

Sequences of Real Numbers

Proof. Suppose In = [an, bn), a", b" E R, an 5 bn. Since 1n 3 /".," for all m ? 0, an!5 an+m:s bnt,,,<_b,,, for all n, m E N. Thus the sequence {a"} is monotone increasing and bounded above by every b" m E N. Thus by the previous theorem, a = lim an exists with a s b", for all m E N. Therefore a E In, for all m E N and thus "~'° 00

a E nm-l/,,,, which proves the result. Q

Remark. Similarly, if b = lim bthen b E In for all n, and thus [a, b] C

1 1n.

In fact, one can show that equality holds (Exercise 1).

2.3A

EXAMPLES

(a) Our first example shows that the conclusion of Corollary 2.3.3 is false if the intervals 1" are not closed. As in Example 1.7.11(b), for each n E N set 1" = (0, ). Then 1,, 3 1n+, for all n, but 00

n 1" = o. The conclusion of Corollary 2.3.3 may also be false if the intervals 1" are unbounded (Exercise 2).

(b) Consider the sequence {j?'}. , with 0 < p < 1. Even though Theorem 2.2.6(e) applies, we use the results of this section to prove thatn-WO lim p" = 0. For n E N set s" = p". Since p > 0, s" > 0 for all n E N. Thus {sn} is bounded below. Also, "+'

The last inequality follows since s,, > 0 and 0 < p < 1. Thus the sequence {s"} is monotone decreasing, bounded below, and hence by Theorem 2.3.2, is convergent. Let

s = lim s". Then n-O oW

s= lim sn+, = n-+oo limp"+' =p lims"=ps. n-"O Therefore s = ps or s(1 - p) = 0. But p # I and thus we must have s = 0. (c) Let a, = 1 and for n ? I set an+i = 6(2a,, + 5). The first three terms of the sequence {a"} are a, = 1, a2 = 6, and a3 = y. Thus we suspect that the sequence {a"} is monotone increasing and bounded above by 2. Since a, < 2, if we assume that a" <_ 2, then

a"+,= I(2a,, +5)56(4+9)<2.

2.3

Monotone Sequences

63

Thus by mathematical induction a, s 2 for all n E N. Likewise, since a, < a2, if our induction hypothesis is an < then the result is true for n = 1. and an+I =

(tan + 5) <

6

6(2an+I + 5) = an+2.

Therefore the sequence {an} is monotone increasing, bounded above, and thus is convergent. Let a = lim an. Then nr00

a

lima..,,

R-KO

aim

6 (tan + 5) =

(2a + 5).

6

Solving the equation a = '-6(2a + 5) for a gives a = ;.

(d) Let a, = 1, and for n ? 1, set

To investigate the convergence of the

sequence {an}. we will establish by induction that

a.
1

for all n r= N. When n = 1, we have

1=a,
I S ak+t = V Lak < V 2ak+, = ak+2, and

ak+2= VL[lk+, < V4=2. Thus the sequence {an} is monotone increasing, bounded above by 2, and hence by Theorem 2.3.2 is convergent. It is possible to prove directly that

sup{an:n= 1, 2, ...}=2. However, if we let a = lim an, then

a = lim an+, = lim n-"0 w_00

= \1521

The last equality follows by Exercise 14 of Section 2.1. Therefore a is a solution of

a2 = 2a, which since a a 1, implies a = 2. N

Euler's Number e 2.3.5

EXAMPLE In this example we consider in detail the very important sequence {tn}= 1, where for each n E N,

tn='1+nJn. We will show that the sequence {t,} is monotone increasing and bounded above, and thus has a limit. The standard notation for this limit is e (in honor of Leonhard Euler); i.e.,

I+ e= lim n-

11n n

64

Chapter 2

Sequences of Real Numbers

By the binomial theorem

t"=C1+n)" I

+n

n+

n(n - 1) 1 1.2

n(n - 1) 1.2

+

n2+

1

n

1

(1)

n"

For k = 1, . .. , n, the (k + 1)-st term on the right side is

n(n- 1). . .(n-k+ 1)

1

1.2...k

nt'

which is equal to

n)...(1 -k n'

n)(1

(2)

in the same way, we obtain n + 2 terms, and for k = 1, 2.. if we expand the (k + 1)-st term is

1

2

1

.

.k

n+

(1

1)(

1)...(I

n+

- n + 1),

which is greater than the corresponding term in (2). Thus t < r,, we also obtain 4

+

+

1+ +1.2 1.2.3 2

1

+r---

1+1+2+22+

+ r'-' _ (1 - r")/(1 - r), r # 1,

which by the identity I + r +

=1+

+

1

for all n. From (1)

I

1 -g


..

!-

. = 3 ..

i -g

Thus {t"} is bounded above by 3, and we can apply Theorem 2.3.2. Since t" s 3 for all n e N we also have that e s 3. To five decimal places, e = 2.71828 . The number e is the base of the natural logarithm function which will be defined in Example 6.3.5 as a definite integral.

Infinite Limits If a monotone increasing sequence {a"} is bounded above, then by Theorem 2.3.2 the .sequence converges. If the sequence {a'} is not bounded above, then for each posi-

tive real number M there exists n E N such that a" ? M for all n ? n,,. Since the real number M can be taken to be arbitrarily large, this is usually expressed by saying that the sequence {a"} diverges to oo. We make this concept precise, not only for monotone sequences, but for any sequence of real numbers, with the following definition.

23 Monotone Sequences

2.3.6

65

Let {a,j be a sequence of real numbers. We say that {an} approaches infinity, or that {an} diverges to oc, denoted a,, -+ oo, if for every positive real number M, there exists an integer n,, E N such that DEFINITION

a > M for all n ? n,,. We will also use the notation lim an = oo to denote that a - oc as n -+ oc. The n"X concept of a,, - -oo is defined similarly.

2.3.7

THEOREM n--)I oo.

If {a"} is monotone increasing and not bounded above. then a,, -> oc as

As a consequence of Theorems 2.3.2 and 2.3.7, every monotone increasing sequence {an} either converges to a real number (if the sequence is bounded above) or diverges to oo. In either case,

lim a,, = sup{an : n E N}.

Remarks. Although the definition of diverging to infinity is included in this section on monotone sequences, this should not give the impression that Definition 2.3.6 is applicable only to such sequences. In the following we give an example of a sequence that diverges to infinity but is not monotone. Also, it is important to remember that when we say that a sequence converges, we mean that it converges to a real number.

2.3.8

EXAMPLE Consider the sequence {n(2 + (- I )")}. If n is even, then n(2 + (- I Y') = 3n; if n is odd, then n(2 + (-1)") = n. In either case,

n(2+(-1)")ten, and thus the sequence diverges to oo. The sequence, however, is clearly not monotone.

EXERCISES 2.3 1. Let 1n = [an, x

n E N, be closed and bounded intervals satisfying I,, D I,,. i for all n. Prove that

n 1n = [a, b], "=1

where a = sup{an : n E N} and b = inf{b : n e NJ. 2. 'Show by example that the conclusion of Corollary 2.3.3 is false if the intervals 1,, with 1. D 1_ are not bounded. 3. Show that each of the following sequences are monotone. Find a lower or upper bound if it exists; find the limit if you can.

t

It

b.

{ a+If)(a- J-)1,wherea> + cos-

d. {sn}, where sn = cos'

2

1

2+

+ cos'

1117

66

Chapter2

Sequences of Real Numbers

4. Define the sequence {a,,} as follows: a, ', and a,r, = V + Ct". a. Show that a" s 2 for all n. b. Show that the sequence (a") is monotone increasing. c. Find lim a". 5. *Let a, > 1, and for n E N, define a"_, = 2 - I/a". Show that the sequence {a") is monotone and bounded. Find n-+x lim a". 6. Let 0 < a < 1. Set t, = 2, and for n E N, set:,, = 2 - alt,. Show that the sequence {t") is monotone and bounded. Find lim t". 7. For each of the following, prove that the sequence (a") converges and find the limit.

n

b. a"+, _ V, a, = 3

a. a"+, = 1(2a" + 5), a, = 2

*c.a"+ti2a, a,=1

*e. a"_, = 3a" - 2, a, = 4 8. Set x, = a. where a > 0 and let x"

d.a"+,=V2.. + 3, a4

f. an+, = 3, a, = I

x" + (1/x"). Determine if the sequence {x"} converges or diverges.

9. Let a > /0. Choose x, > Va. For n = 1, 2, 3, .... define xn+1 =

21X,+SJ.

\

*a. Show that the sequence {x"} is monotone and bounded.

b. Prove that lim x" = %/a-. c. Prove that 0 c x" - NF. s (x. 10. In Exercise 9, let a = 3 and x, = 2. Use part (c) to find x" such that Ix" - \/31 < 10-5. 11. Let A be a nonempty subset of R that is bounded above and let a = sup A. Show that there exists a monotone increasing sequence {a"} in A such that a = I m a". Can the sequence (a"} be chosen to be strictly increasing? 12. Use Example 2.3.5 to find the limit of each of the following sequences.

*a. lI *c.

1

+

nb. d.

1

+ 2n/3n}

j(1

- n) +

'

jl l

n

Sn= +2+

1

1

+n

Show that {s,} is monotone increasing but not bounded above.

14. For each n E N, let I + 72+

s"=

I

+

n

Show that the sequence {s"} is monotone increasing but not bounded above.

15. *For each n E N. let I S"= i2+22+.+n2. 1

1

Show that the sequence {s"} is monotone increasing and bounded above by 2.

16. Let 0 < b < 1. For each n E N, let s" = I + b + b2 + creasing and bounded above. Find lim s".

+ b". Prove that the sequence {sue} is monotone in-

2.4

Subsequences and the Bolzano-Weierstrass Theorem

67

17. Show that each of the following sequences diverges to 00. n

n

18. *Which of the sequences in the previous exercise are monotone? Explain your answer.

19. If a -+oc and (b^} converges in R, prove that (a, +

diverges to oc.

20. If a, > 0 for all n E h and lim a^ = 0, prove that I la, -+ oc. 21. Suppose a, > a, > 0. For n ? 2, set a^ _ i = ; (a, + a^ 1). Prove that a. {au, ;} is monotone decreasing.

b. {a2k) is monotone increasing, and

c. {a^} converges. 22. Let {s,} be a bounded sequence of real numbers. For each n E N. let a^ and b^ be defined as follows:

a, = inf{sk : k ? n}, b" = sup{sk : k ? n}. a. Prove that the sequences {a^} and (b,,) are monotone and bounded. b. Prove that lim a^ = lim b" if and only if the sequence {s^} converges. nix nix 23. *In Theorem 2.3.2 we used the supremum property of R to prove that every bounded monotone sequence converges. Prove that the converse is also true; namely, if every bounded monotone sequence in l8 converges. then every nonempty subset of R that is bounded above has a supremum in R. 24. *Use the nested intervals property to prove that [0, 1 ] is uncountable.

2.4

Subsequences and the Bolzano-Weierstrass Theorem In this section we will consider subsequences and subsequential limits of a given sequence of real numbers. One of the key results of the section is that every bounded sequence of real numbers has a convergent subsequence. This result, known as the sequential version of the Bolzano-Weierstrass theorem, is one of the fundamental results of real analysis.

2.4.1

Given a sequence { p^} in OB, consider a sequence {nk}k 1 of positive . Then the sequence { p^,}k , is called a subsequence of the sequence { p"}. DEFINITION

integers such that n, < n, < n3 <

If the sequence {p,,} converges, its limit is called a subsequential limit of the sequence { p,}. Specifically, a point p E 18 is a subsequential limit of the sequence { p"} if there exists a subsequence {p,,,} of {p,,} that converges to p. Also, we say that oo is a subsequential limit of { p"} if there exists a subsequence { p,, } so that p^, -+oo as

k -+ oo. Similarly for -no.

2.4.2

EXAMPLES

(a) Consider the sequence {(I - (-1)")}. If n is even, then a, = 0, and if n is odd, then a = 2. Thus 0 and 2 are subsequential limits of the given sequence. That these are the only two subsequential limits is left to the exercises (Exercise 1).

68

Chapter 2

Sequences of Real Numbers

(b) As our second example, consider the sequence {(-1 )" + }. Both I and - I are subsequential limits. If n is even, i.e., n = 2k, then

an = au = l +

2k'

which converges to t. On the other hand, if n is odd, i.e., n = 2k + 1, then 1

a,, = a2k+ i = -1 +

2k + 1'

which converges to -1. This shows that -1 and I are subsequential limits. Suppose {a",} is any subsequence of {a"}. If the sequence Ink) contains an infinite number of both odd and even integers, then the subsequence {a",} cannot converge. (Why?) On the other hand, if all but a finite number of the nk are even, then {a",} converges to 1. Similarly, if all but a finite number of the nk are odd, then {a",} converges to -1. Thus -1 and I are the only subsequential limits of {a"}.

If n is even, then n(1 + (-I )") = 2n, (c) Consider the sequence {n(1 + whereas if n is odd, n(1 + (-1)") = 0. Thus 0 and oo are two subsequential limits of the sequence. The same argument as in (b) proves that these are the only two subsequential limits. Our first result assures us that for convergent sequences, every subsequence also converges to the same limit.

2.4.3

THEOREM Let (p,,) be a sequence in R. If {p.1 converges to p, then every subsequence of { p"} also converges to p.

Proof. Let {p",} be a subsequence of {p"}, and let e > 0 be given. Since p"-+p, there exists a positive integer n such that 1p. - pl < e for all n ? n0. Since {nk} is strictly increasing, nk ? no for all k ? n". Therefore,

Ipn, - pI < e

forallk? n0;i.e.,p",-+p. 2.4A

EXAMPLES

(a) In our first example we illustrate how Theorem 2.4.3 may be used to compute the

limit of a sequence. Consider the sequence {p"} where 0 < p < 1. By Example 2.3.4(b), the sequence {p"} is monotone decreasing, bounded below, and hence converges. Let

a = lim p". By Theorem 2.4.3 the subsequence {pen} also converges to a. But p2" = (p")2, and thus

a = lim Thus a2 = a. Since 0

p2"

= lim (p")2 = a2.

a < 1, we must have a = 0.

2.4

4

Subsequences and the Bolzano-Weierstrass'fheorem

69

(b) In our second example we show how the previous theorem may be used to pros e divergence of a given sequence. Consider the sequence { sin n07r}, where 8 is a ra-

tional number with 0 < 8 < 1. Write 0 = alb, with a, b E N and b ? 2. When n = kb, k E N, then sin n8ir = sin kair = 0. Therefore 0 is a subsequential limit of the sequence. On the other hand, if n = 2kb + 1, k E N, then

sin n0ar = sin (2kb + l) 6 ar = sin 12kalr + b ir) = cos(2kaar) sin b ar = sin b ar.

Since 0 < alb < 1, sin bar * 0. Thus sin bar is another distinct subsequential limit of {sin nBar}. Hence as a consequence of Theorem 2.4.3, the sequence { sin n07r} diverges. The result is still true if 0 is irrational. The proof however is much more difficult.

Limit Point of a Set In order to state the Bolzano-Weierstrass theorem we need to introduce the concept of a limit point of a set. This concept is fundamental in the study of analysis and will occur again when we discuss the structure of point sets and limits of functions.

2.4.5

DEFINITION

Let E be a subset of R.

(a) A point p E R is a limit point of E if every e-neighborhood N,(p) of p conrains a point q E E with q* p. (b) A point p E E that is not a limit point of E is called an isolated point of E.

Remarks. In the definition of limit point it is not required that p is a point of E. Also, a point p E E is an isolated point of E if there exists an e > 0 such that

N,(p)flE={p}. 2A.6

EXAMPLES

(a) E _ (a, b), a < b. Every point p, a < p < b, is a limit point of E. This follows from the fact that for any e > 0, there exists a point x E (a, b) satisfying p < x < p + e. These, however, are not the only limit points. Both a and b are also limit points of E, but they do not belong to E. (b) E _ {,'-, : n = 1, 2, ...}. Each n is an isolated point of E. If a is chosen so that

0<E<

I

n(n + 1)

_

I

n+I

-I n'

then N,(,-,) n E _ Hence no point of E is a limit point of E. However, 0 is a limit point of E which does not belong to E. To see that 0 is a limit point, given

70

Chapter2

Sequences of Real Numbers

e > 0, choose n E IN so that 1- < e. Such a choice of n is possible by Theorem 1.5.1. Then

1 EN,(0)flE, n

and thus 0 is a limit point of E.

(c) Let E =o fl [0, 1 ]. If p (4 [0, 1 ], then p is not a limit point of E. For if p > 1. then for e = 1(p - 1) we have k(p) fl E = 46. Likewise when p < 0. On the other hand, every p E [0, 1] is a limit point of E. Let e > 0 be given. Suppose first that 0 s p < 1. Then by Theorem 1.5.2 there exists r E Q such that

p 0, N,(p) contains a point r E E with r * p. The same argument also proves that every point of R is a limit point of Q.

2.4.7

THEOREM Let E be a subset of 08. (a) If p is a limit point of E, then every neighborhood of p contains infinitely many points of E. (b) If p is a limit point of E, then there exists a sequence { E N, such that

p

Proof.2 (a) Suppose there exists a neighborhood Na of p which contains only finitely q, # p. Let many points of E, say q1, ... ,

emin{lp-q1J... >lp-q.11. Then N,(p) fl E contains at most p. Thus p is not a limit point of E. (b) We construct the sequence in E as follows: Since p is a limit point of E, for each positive integer n, by the definition of limit point, there exists p E E with p # p such that I p - p I < ,1-,. The sequence {p.) clearly satisfies p --.>p. L)

2.4.8

COROLLARY A finite set has no limit points.

2.4.9

EXAMPLE

In this example we illustrate part (b) of the previous theorem. Since is a limit point of Q (Example 2.4.6(c)), by Theorem 2.4.7(b) there exists a sequence of rational numbers with r -> N/-2. Such a sequence, however, need not be unique.

If r V, then the same is also true for the sequence {r + ,1-,}.

2. The proof of (a) is an example of a proof by contraposition. In this method of proof we assume the nepation of the conclusion and prove the negation of the hypothesis.

2.4

Subsequences and the Bolzano-Weierstrass Theorem

71

Bolzano-Weierstrass Theorem We are now ready to state and prove the Bolzano-Weierstrass theorem. This important result was originally proved by Bernhard Bolzano (1781-1848) and modified slightly in the 1860s by Karl Weierstrass (1815-1897). The theorem is one of the fundamental results of real analysis. Its importance will become evident in subsequent sections. Although not immediately obvious from the proof, the least upper

bound property plays a crucial role in the proof of the theorem. The least upper bound property was used in proving the nested intervals property (2.3.3), and this result is crucial in the proof.

2A.10 THEOREM (Bolzano-Weierstrass) Every bounded infinite subset of R has a limit point.

Proof. Let S be a bounded infinite subset of R. Since S is bounded, there exist a, b E Il with a < b such that S C 11 = [a, b]. Divide 11 into two closed subintervals

b1 fa+b

a,a

2

Jl

2

b] J

each of length (b - a)/2. At least one of these, call it 12, must contain infinitely many points of S. If not, then S would be the union of two finite sets and thus be a finite set itself. Repeating this process, we obtain a sequence {1"} of closed and bounded intervals satisfying

(a) [a, b]=11DIZD.. .D1"D..., (b) length of I" = (b - a)/2"-, and (c) for each n, in fl s is infinite.

By Corollary 2.3.3, lono- 11, * 0. Let x e fly 1 In. It remains to be shown that x is a limit point of S. If e > 0 is given, choose n E N so that (b - a)/2"-1 < e. If y E I", then

ly - xj S length I" < e. Therefore 1" C N.(x). Since In fl S is infinite, there exists y E S fl N1(x) with y * x. Since e > 0 was arbitrary, x is a limit point of S. The conclusion of the Bolzano-Weierstrass theorem may fail if either hypothesis is removed. By Corollary 2.4.8 a finite set has no limit point. On the other hand, the set N is an infinite unbounded subset of II with no limit points. The following corollary is often called the sequential version of the Bolzano-Weierstrass theorem.

2A.11

COROLLARY (Bolzano-Weierstrass) gent subsequence.

Every bounded sequence in R has a conver-

72

Chapter 2

Sequences of Real Numbers

Proof.3

Let {p,,} be a bounded sequence in R, and let S = { p" : it = 1, 2, . is finite, then there exists a point p E S and a sequence Ink.) with n, < n_ < that

.

.}. If S such

Pn,=Pn_=.=P

The subsequence {pn,} obviously converges to p. If S is infinite, then by the Bolzano-Weierstrass theorem, S has a limit point p E R.

Choose n1 so that I p - pn,I < 1 . Having chosen n1, ... , nkchoose an integer nk > nk _ , so that

IP - Pn,I < kI Such an integer nk exists since every neighborhood of p contains infinitely many points of S. The sequence {p",}k , is a subsequence of (pn) that by construction converges to p. An argument similar to the one used in the previous Corollary may be used to prove the following.

2.4.12

THEOREM Let { pn} be a sequence in R. If p is a limit point of { p" : n E K), then such that p,,, -+p as k -+ co. there exists a subsequence {p.,) of {

Proof.

Exercise It.

As an application of Theorem 2.4.12 we consider the following example.

2.4.13

EXAMPLE Since the set E = a fl [0, 1 ] is countable, there exists an enumeration {rn : n E NJ of E. The sequence {rn}W , is called an enumeration of the rational numbers

in (0,1 ]. By Example 2.4.6(c), every p E [0, 1 ] is a limit point of {r" : n = 1, 2, ...}. Thus if p E [0, 1 ], there exists a subsequence {rn } of {r,} such that rn, -+p. The sequence {rn} has the property that every p e [0, 1) is a subsequential limit of the sequence. This sequence also provides an example of a sequence for which the set of subsequential limits of the sequence is uncountable.

EXERCISES 2.4 1. a. Prove that 0 and 2 .are the only subsequential limits of the sequence (1 - (-1n b. Prove that 0 and oo are the only subsequenual limits of the sequence {n(1 + 2. a. Construct a sequence {sn} for which the subsequential limits are {-oo, -2. 1}. b. Construct a sequence {s"} for which the set of subsequential limits of the sequence is countable. 3. The proof of Corollary 2.4.11 is an example of a proof by cases (see Appendix A.3). If S = { p, : n E101}. then there are two cases; namely. S is finite or S is infinite. For each case we prove the existence of a convergem subsequence of {p"}.

2.5

Limit Superior and Inferior of a Sequence

73

-

3. Find all the subsequential limits of each of the following sequences.

*a. s sin 2

b. s cos Slln

c.

Sll-I)" +12sin

l

*e.

-I"

I--

2

l 1

d.

l

2

sin?

f {(1.5

n

4. Use Example 2.3.5 to find the limit of each of the following sequences. Justify your answer. *a.

b. .CI

I \ + 3n1J

+

j

c_

2nI

1(I + n/nj

= 1.

5. Suppose p > 1. Use the method of Example 2.4.4 to show that lim 6. For n E Nl, set p" = n'I".

a. Prove that I < p, , < p" for all n ? 3. b. Let p = limp". Use the fact that the subsequence {p,,,} of {p,,} also converges to p to conclude that p = 1. 7. Determine the limit points and the isolated points of each of the following sets. (( "

*a. S1ll

n

c. (0. l) U {2}

:nEN!

b.

-1)"+!:nE!i } 1

d. N

f. Qfl(0,1) e. Q8\Q 8. Let A be a nonempty subset of U8 that is bounded above and let a = sup A. If a E A. prove that a is a limit point of A.

9. Let { p"} be a bounded sequence of real numbers and let p E R be such that every convergent subsequence of converges to p. Prove that the sequence {p") converges to p. 10. a. Construct a subset of 68 with exactly two limit points. b. Find an infinite subset of R with no limit points. c. Construct a countable subset of U8 with countably many limit points. d. Find a countable subset of R with uncountably many limit points. 11. Prove Theorem 2.4.12. 12. Prove that every sequence in R has a monotone subsequence. 13. Use the Bolzano-Weierstrass theorem to prove the nested intervals property (Corollary 2.3.3). 14. *Use the nested intervals property to prove the least upper bound property of R. 15. Prove that every uncountable subset of R has a limit point in R.

25 Limit Superior and Inferior of a Sequence In this section we define the limit superior and limit inferior of a sequence of real numbers. These two limit operations are important because unlike the limit of a sequence. the limit superior and limit inferior of a sequence always exist. The concept of the limit superior and limit inferior will also be important in our study of both series of real numbers and power series.

74

Chapter 2

Sequences of Real Numbers

Let {s"} be a sequence in R. For each k E N, we define ak and bk as follows:

ak = inf{s" : n ? k}, bk = sup{s" : n ? k}. Recall that for a nonempty subset E of R, sup E is the least upper bound of E if E is bounded above, and co otherwise. From the definition, ak s bk for all k. Furthermore, the sequences {ak} and {bk} satisfy the following:

ak < ak.l and bk'- bk+l (3) for all k. To prove (3), let Ek = {s" : n zz k}. Then Ek+, C E.. Therefore, if bk = sup Ek, s" s bk for all n ? k. In particular

s" : bk for all n ? k + 1. Therefore bk+, = sup Ek+, s bk. A similar argument will show that the sequence {ak} is nondecreasing. As a consequence of (3), the sequence {ak} is monotone increasing and the sequence {bk} is monotone decreasing. Thus by Theorems 2.3.2 and 2.3.7, these two sequences always have limits in i8 U {-oo, oo}.

2.5.1

Let {s"} be a sequence in R. The limit superior of {s"}, denoted lim s" "~" or lim s", is defined as DEFINITION

ligoo s"

urn bk A sup{S" : n > k}.

k

The limit inferior of {s"}, denoted lim s" or lim s", is defined as "-00

lim s" = lim ak = sup inf{s" : n at k}. k-oo

nToo

kEN

We now give several examples for which we will compute the limit inferior and limit superior. As will be evident, these computations are very tedious. An easier method will be given in Theorem 2.5.7.

2.5.2

EXAMPLES

(a) {1 + (-1. Let s,, = 1 + (-1)". Then s" = 2 if n is even, 0 otherwise. Thus ak = 0 for all k and bk = 2 for all k. Therefore lim s" = 2

and

lims"=0.

(b) {n(1 + (-,. In this example, s"=n(1 +(-1)")={0 2.

ifnisodd. if n is even.

Set Ek = {s,,: n = k}. Then if k is odd Ek = {0, 2(k + 1), 0, 2(k + 3),. ..} Ek = (2k, 0, 2(k + 2), 0. 2(k + 4),. ..} if k is even.

2.5

Limit Superior and Inferior of a Sequence

75

Therefore ak = inf Ek = 0, and bk = sup Ek = oo. Thus

limsn=0

and

n-+x

ii-m- S, = 00. n-x

(c) {(-1)"+,,,}°,. Setsn= (-1)"+1.Then

-1+n

n odd,

1+

n even.

n

The first few terms of the sequence {sn} are illustrated in Figure 2.4.

$5 S3

$3

0 5

I

3

S( 'i4

52

75

3

64

2

Figure 2.4

To compute the limit superior and inferior of the sequence {sn}, we set Ek = {sn : n ? k}. If k is even, then

Ek1+k,-1+k+1' I+kI

.

1.

Therefore, for k even,

bk = sup Ek = 1 + k

ak = inf Ek = - I.

and

Similarly, fork odd, 1

bk=supEk= 1+ k+ l

and

ak=infEk= -I.

As a consequence,

1 + k,

a. _ -1

k even,

for all k. and b. _ 1+

1

k

k odd. 1

Thus

lim

and

limsn=-1. or-_+W

Our first theorem provides an (e, n0) characterization of the limit superior. An analogous characterization for the limit inferior is given in Theorem 2.5.4.

76

Chapter 2

2.5.3

Sequences of Real Numbers

THEOREM Let {sn}' , be a sequence in R.

(a) Suppose lim sn E R. Then /3 = lim s if and only if for all e > 0,

(i) there exists n E N such that s < 0 + e for all n ? n0, and (ii) given n E N, there exists k E N with k ? n such that sk > /3 - e. (b) lim s = oo if and only if given M and n E N, there exists k E N with n such that sk ? M.

(c) n-00 lim sn = -oo if and only if sn -+ -oo as n -4 oo. Remark. The statement "s,, < (3 + e for all n a

means that sn < /3 + e for all but finitely many n. On the other hand, the statement "given n, there exists k E N with

k ? n such that sk > /3 - e" means that sn > /3 - e for infinitely many indices n. 2.5A THEOREM Let {sn} be a sequence in H.

(a) Suppose urn sn t R. Then a = lim sn if and only if for all E > 0, n+oc

n-4ou

(i) there exists no E N such that s > a - e for all n ? no, and (ii) given n E N, there exists k E Ni with k ? n such that sk < a + e. (b) lim sn = -oo if and only if given M and n E N, there exists k E N with i_-"_Q

k ? n such that sk <_ M. (c) lim s = oo if and only if s,, - oo as n --> oo.

Proof of Theorem 2.5.3. We will only prove (a). The proofs of (b) and (c) are left to the exercises (Exercise 5).

(a) Suppose /3 = lim S. = rlim bk where bk = sup{ s,, : n ? k}.

Let e > 0 be given. Since lim bk = (3, there exists a positive integer n,, such that bk < /3 + e for all k >t n0. Since sn s bk for all n ? k, k

s,,
ing, bk ? /3 for all k. In particular, bn ? /3. By the definition of b,,, however, given e > 0, there exists an integer k z n such that

sk> b,, - e? /3 -e, which proves (ii).

Conversely, assume that (i) and (ii) hold. Let e > 0 be given. By (i) there exists

n. E N such that s < /3 + e for all n a na. Therefore,

b,,,= sup{s,,:n?n0}:/3+e.

2.5

Limit Superior and Inferior of a Sequence

77

Since the sequence {bn} is monotone decreasing. bn <- 6 + E for all n ? n,,. Thus

Since c > 0 was arbitrary, urn sn S /3. Suppose $3' = lim sn < (3. Choose e > 0 such that $3' < 6 - 2E. But then there exists n such that

sn<13'+E<$3-E foralln?n0, which contradicts (ii). Thus Tim sn = /3.

p

To illustrate the previous two theorems, consider the sequence sn = (- I)" + 1/ of Example 2.5.2(c). For this sequence, lim s" = 1 and lim s,, _ -1. Given e > 0, then

s"< I + E for all n E N with n >_ e. Since the odd terms get close to -1, we can never have the existence of an integer n such that sn > 1 - E for all n ? no. On the other hand, given any n E N, there exists an even integer k ? n such that Sk > I - E. An immediate consequence of the previous two theorems is as follows.

2.5.5

COROLLARY

Jim sn = lim s" if and only if Jim s" exists in R U {-oo, oo}.

n-+00

n-M

nix

Proof.. Suppose lim s" = lim sn = a E R. Let e > 0 be given. By (a) of the previous two theorems, there exist positive integers ni and n2, such that

s" < a + e for all n ? n,, and sn > a - E for all n ? n,. Thus if n,, = max{nt, n2},

a - e < sn
2.5.6 THEOREM Let {a"} and {b"} be bounded sequences in R. Then lim a,, + Jim bn S lim (an + bn) <_ lim a, + lim b"

n-.oo

n-too

n--.x

w

<- nyOG lim (an + bn) S ntiR lim an + lim b,,.

Proof.

Exercise 6. U

The following theorem relates the limit superior and inferior of a sequence to the subsequential limits of the sequence, and is in fact very useful for finding Tim- S. and lin s of a sequence {sn}.

2.5.7

THEOREM Let {sn}' 1 be a sequence in IR, and let

E = the set ofsubsequential limits of {sn} in R U {-oo. oo}.

78

Chapter 2

Sequences of Real Numbers

n:

Then lim s, and lim s" are in E and "x-100

(a) lim s" = sup E. and (b) lim s" = inf E. "-400

Proof. Let s = lim s". Suppose s E R. To show that s E E, we show the existence of a subsequence {s"j of {s"} which converges to s. Take e = 1. Let n, be the smallest integer such that

s - I <S". < S + 1. Such an integer exists by (i) and (ii) of Theorem 2.5.3(a). Suppose the integers nl < n2 < .. < nt have been chosen. Take e = 1/(k + 1). Let n1 be the smallest integer greater than nk such that

$ -k+1<S",. <S+k+1 1

1

Again, such an integer exists by (i) and (ii) of Theorem 2.5.3(a). Then (sn j is a subse-

quence of {s"} which clearly converges to s. Therefore s E E. The case s = no is treated similarly. Ifs = -oo, then by (c) of Theorem 2.5.3, sn -i -oo as n -* no. Since s E E, s < sup E. It remains to be shown that sup E = s. If s = no we are done. Otherwise, suppose sup E = S > s. Suppose S # no. Then there exists a E E such that

sass.

Since a E R, we can choose e > 0 such that s + e < a - e. For this e, there exists n0 E N such that s" < s + e for all n z n0. Hence there can exist only finitely many k such that

(sk - al < e. Consequently no subsequence of {s"} can converge to a. This contradiction shows that

sup E = s. The case S = no is treated similarly. Q 2.5.8

EXAMPLES In the following examples we use Theorem 2.5.7 to compute lim s" and lim s" for each of the given sequences {s"}.

(a) Let s,, = (- I)" + 1. By Example 2.4.2(b), the set of subsequential limits of {sn} is {-1, I}. Thus by the previous theorem,

lim sn = -1

and

lim s" = 1.

(b) Let sn = n(1 + (-1)"). By Example 2.4.2(c) the subsequential limits of {sn} are 0 and no. Therefore,

lim s" = 0

and

lim s" = no.

2.5

Limit Superior and Inferior of a Sequence

79

(c) Let s" = sin" If n is even, i.e., n = 2k, then s,t = sin kir = 0. On the other hand, if n is odd, i.e., it = 2k + 1, then s,+, = sin(2k + 1)121 _ (-1)r. Hence the set of subsequential limits of the sequence {s"} is {-1, 0, 1}. As a conse.

quence,

lim s" = -1

and

Urn s" = 1.

EXERCISES 2.5 1. Find the limit inferior and limit superior of each of the following sequences.

b. j(1 +(-Ir)sin *d I - 2(-1)"n f 3n+2

}

*a.

n + (-1)"n2)

n2+1 e.

1

ll

+ cos nir) l 11+n(I n

g. {cos n9w},

f, {[ 1.5 + (- I)"J"}

9EO

2. Let {a"} be a sequence in R. If ii-m- la"I = 0. prove that lim a" = 0. 3. *Let Jr.) be an enumeration of the rational numbers in (0, 1). Find lint r" and urn r".

4. Let {s"} be a sequence in R. If s E H satisfies that for every e > 0, there exists no E N such that s" < s + e for

all n ? n prove that liras" s s. 5. a. Prove Theorem 2.5.3(b). b. Prove Theorem 2.5.3(c). 6. Let {a"} and {b"} be bounded sequences in R.

*a. Prove that lim a" + limb"slim (a" + b") <_ Iim a" + limb,,. b. Prove that lim (-a") = -Fun a.. c. Given an example to show that equality need not hold in (a).

7. a. If a" and b" are positive for all n, prove that lira not of the form 0 oo.

(Tim a")(Iim b"), provided the product on the right is

b. Need equality hold in (a)?

8. *Let s, = 0. For n E N, n > 1, let s" be defined by S2m =

S2m-I

2

1

,

S2,"+I = 2 + s2,,,.

Find Iim s" and lim s".

9. Let a" > 0 for all n. Prove that Tim " a"

lim

a='. a"

10. *Suppose (a"}, {b"} are sequences of nonnegative real numbers with limo b, = b * 0 and lim a" = a. Prove that

lima"b"=ab.

n- =

80

Chapter2

2.6.

Sequences of Real Numbers

Cauchy Sequences In order to apply the definition to prove that a given sequence 1 1),, } converges, we must For this reason, theorems that provide sufficient know the limit of the sequence

conditions for convergence, such as Theorem 2.3.2. are particularly useful. The drawback to Theorem 2.3.2 is that it applies only to monotone sequences of real numbers. In this section we consider another criterion that. for sequences in R. is sufficient to ensure convergence of the sequence.

2.6.1

DEFINITION A sequence {p,),',-, in R is a Cauchy sequence if for every r: > 0. there exists a positive integer it,, such that

IP,.-P.,I <E for all integers n, m ? n,,.

Remark. In Definition 2.6.1, the criterion I p - p,,,l < e for all integers n. m ? it,, is equivalent to Inn-4 - P,.I < E

for all n ? it,, and all k E N. Thus if {

is a Cauchy sequence in R.

PnI = 0

(4)

for every k E N. The converse, however, is false: namely, if { p,,} is a sequence in 6a that

0 for every k E N. this does not imply that'the sequence satisfies lim n-.x is a Cauchy sequence (Exercise 4). The hypothesis (4) only implies that for each E for all k E N, given E > 0, there exists a positive integer n such that Ip,,.a n at n,,, 2.6.2

THEOREM

(a) Every convergent sequence in R is a Cauchy sequence. (b) Every Cauchy sequence is bounded.

Proof.

(a) Suppose that {p,,} converges to p E R. Let e > 0 be given. Then for the given e, there exists a positive integer n,, such that

In. - PI <

l e

for all n ? n,,. Thus by the triangle inequality, if it, m a n,,, Ip..

P1 c!pn

PI+IP

pn.I
(b) Take e = 1. By the definition of Cauchy sequence, there exists n E N such

2.6

Cauchy Sequences

81

that I p,, - p0I < I for all n, m ? no. Therefore, with m = nby Corollary 2.1.4(b).

I PnI { 1 + IP I for all n ? n,,. Let

M = max{1 + Ipnj, Ipil,.

.

., Ipn _,I}.

Then for all n, I pnI :SM. Thus {pn} is bounded. 0

2.6.3

THEOREM If { pn} is a Cauchy sequence in l that has a convergent subsequence, then the sequence { converges.

Proof. Suppose { p,, } is a convergent subsequence of { pn} with lim pn, = p. Let e > 0 be given. Since {pn} is Cauchy, there exists an integer ni such that

IPA - PmI <2e foralln,m?n,. Since p,, -pp, for the given e, there exists an integer k, such that

IPn, - PI <2e forallkaki. Let no = max{k,, n,}, and choose nk such that k at no. Then nk a n,. Thus if is >_ no, by the triangle inequality,

IPn - pi < IPn - Pn,I + IPn, - PI < 2e + 2e = e. Therefore lim p,, = p, which proves the result. 0

2.6.4

THEOREM Every Cauchy sequence of real numbers converges.

Proof. Let {p,,} be a Cauchy sequence in R. By Theorem 2.6.2, the sequence { pn} is bounded. Thus by Corollary 2.4.11, the sequence {pn} has a convergent subsequence.

The result now follows by Theorem 2.6.3. U

Remark The statement that "every Cauchy sequence in R converges" is often expressed by saying that R is complete. Since the proof-of Theorem 2.6.4 used the Bolzano-Weierstrass theorem, the completeness of R ultimately depends on the least upper bound property of R. Conversely, if we assume completeness of K, then we can prove that R satisfies the least upper bound property (Exercise 12). For this reason, the least upper bound or supremum property of R is often called the completeness property of R. 2.6.5

EXAMPLES

(a) For our first example we consider the sequence {sn} where for n E N,

1+

2+. .+'2. .

2

n

82

Chapter2

Sequences of Real Numbers

For k E N, (n+l)z+...+(n+k)z

(n n

-n+ 1) +

+(n+k-

n+k/]

1

n + k

In the above we have used the inequality

1

1

n+m-1

(n+m)2

n+m

valid for all n, m E N. Since the sequence {.1} converges, it is a Cauchy sequence. Thus given e > 0, there exists n, E Nl such that I1, kI < e for all n > no and all k E N. Thus IS. +k - S. I < e for all n ? no and all k E N. Therefore the sequence {s"} is a Cauchy sequence. "n

(b) In this example we illustrate how the concept of a Cauchy sequence may be used to prove convergence of a given sequence. Additional applications will be given in the exercises. Let a,, a2 be arbitrary real numbers with a1 # a2. For n > 3, define an inductively by an = I(an-1 + an-2)

(S)

Our first goal is to show that the sequence {a"} is Cauchy. We first note that

an+1-an=-2(an-an-1) As a consequence, for n ? 2, 1

2"

an+1 - an

(a2 - a1).

(6)

This last statement is most easily verified by induction (Exercise 5). For m ? 1, consider Ian+," - a,, 1. By the triangle inequality,

m:Ian+k+I - an+k) S I -1

Ian+'n - anI

7,(an+k+I

-0

which by (6) I

I

1

Ia2 - a]I 7, 2n+k-f By Example 1.3.3(a) rk

kl

m

1

2"-21a2 - all 7, 2k.

r-r'* 1-r

r # 1.

(7)

2.6 Cauchy Sequences

83

Thus with r = 2, M

2 - 2)m+I

1

k=12k

1-Z

l

= 1 - - < 1. 2m

Therefore, 1

Ian+m - and :S-

22 Ia2 -

2

all

(S)

for all n >- 2 and m E N. Let e > 0 be given. Choose no such that lag - a, I/2"-2 < e for all n ? no. Then by (8), Ian+m - and < E

for all m E N, n ? nn. This, however, is just another way of stating that

Ian - am < e for all m, n ano. Therefore the sequence {an) is a Cauchy sequence in R, and thus by Theorem 2.6.4.

a = lima" n- 00 exists in R.

Can we find the limit a here? If we follow the approach in Example 2.3.4(c), by taking the limit of both sides of equation (5), we only get a = a. To find the value of a, let us observe that

an+1 - a, = (an+, - an) + (an - a"-,) + ... + (a2 - a1) n

Y, (ak+ I - ak),

k=1

then use (6) to get I

k-1

(a2-al)I(-2) (

n

The last equality follows from formula (7). Since an+, ing the limit of both sides we obtain 2

a -a, = (a2 - a,)

or

and (-3f -+0, upon tak-

a=a,+3(a2-a,).

Contractive Sequences One of the key properties of the sequence {an} of the previous example was that Ian+, - and

< Ia,, - a.-,I 2

84

Chapter 2

Sequences of Real Numbers

for all n ? 2. This property was used to show that the sequence {a"} was a Cauchy sequence and thus converged. Such sequences are commonly referred to as contractive sequences. We make this precise in the following definition.

2.6.6

DEFINITION A sequence { p"} in R is contractive if there exists a real number b, 0 < b < 1, such that IPn+I - P"I t bIP" - Pn

forallnEN,na2. If {p.) is a contractive sequence, then an argument similar to the one used in the previous example shows that

IPn+I - P"I `- b"-'IP2 - P1I for all n >_ 1, and that IP"+m-P.1

b" Cb"-'IP2-PII(l+b+. +b'"-I)< 1-bIP,-PII

for all n, m E N. As a consequence, every contractive sequence is a Cauchy sequence. Therefore, every contractive sequence in R converges to a point in R. We summarize this in the following theorem.

2.6.7

THEOREM Every contractive sequence in R converges in R. Furthermore. if the sequence {p.1 is contractive and p = lim p", then

(a) IP - P"I

l_b

I(b)

P-P"::=:':r'

where 0 < b < 1 is the constant in Definition 2.6.6.

Proof. We leave the details of the proof to the exercises (Exercise 7). 0

2.6.8

EXAMPLE Suppose we are given that the polynomial p(x) = x2 - 3x + I has exactly one zero in the open interval (0, 1). If c r= (0, 1) is such that p(c) = 0, then c = (c2 + 1). We start with cl E (0, 1) arbitrary, and for n a- I we set

i

1

,

c"+1=3(c;,+I). Since cl E (0, 1) we have c2 E (0, 1), and by induction, c" E (0, 1) for all n E N. To prove that the sequence {c"} converges we prove that it is contractive. For n > 2 we have

Ic"+I-c"I=I3(c+1)-3(c,,-1+1)I

2.6

Cauchy Sequences

85

= 3I(Cn - C.-1)(c + Cn-1)I 2

:5

3ICn - Cn-1I

Thus the sequence is contractive with constant b = 3. If c = nlimc then c max. (c2 + 1) or p(c) = 0. Suppose we begin with c1 = .5 and we wish to determine the value of n such that Ic - Cl < 10-3. By Theorem 2.6.7 (a) it suffices to determine n such that bn-t

(b - 1)

IC2 - ct I <

By computation c2 = aZ. Thus since b = 3, we want to determine n such that 4(3)"-i

12

3

2

< 10-1.

Solving the above inequality for n - I gives

(n - 1) >

- In 4 - In 2 = 13.62 (to two decimal places).

3 In 10

n3

Thus the choice of n = 15 is sufficient to ensure accuracy to within 10-3. This estimate on n however is very crude. In fact, after six steps, part (a) of Theorem 2.6.7 only gives .Ic - c6I < 3(3)SIc2 - c1I = .032922. The actual approximation however is much better than this. By computation we find that c3 = .391204, c4 = .384347, cs = .382574.

and c6 = .382121. By part (b) of Theorem 2.6.7, Ic - c6I

21c6 - csl = 000906,

and this is less than 10-3.

0 EXERCISES 2.6 1. If

and are Cauchy sequences in R, prove (without using Theorem 2.6.4) that {a + also Cauchy sequences.

and

are

2. For each of the following, determine whether the given sequence is a Cauchy sequence.

+l}

b. {(-I)"}

-a. In in

adI . 1(1)'n 1n2+3

ll((

{.1 + (-l)nn2} 2n2 +3

3. For n E 4 let S.

+21+31++n1.

Prove that is a Cauchy sequence. defined for n = 1, 2, 3, ... by 4. Consider the sequence s=1+2+:..+1

(n+(-I}"}

c.

n

l

86

Chapter 2

Sequences of Real Numbers

*a. Show that {sn} is not a Cauchy sequence.

b. Even though {sn} is not a Cauchy sequence, show that I'M Isn.k - s,, = 0 for all k E N. n+x

5. Use mathematical induction to prove identity (6). 6. Let {an} be the sequence of Example 2.6.5. a. Use mathematical induction to prove thatf

aa, , = 22I i (a, +a, ) +

I

(a, + 2a.,)I

I

b. Use the result of (a) to find lim an. 7. Prove Theorem 2.6.7.

8. 'Let a, > 0, and for n ? 2, define an = (2 + a, -

Prove that {an} is contractive, and find lint nixan.

s (c.2 + 2). 9. Let c, E (0, 1) be arbitrary, and for n E N set cn _ a. Show that the sequence {c,} is contractive. b. Let c = lim en. Show that c is a solution of x' - 5x + 2 = 0. 10-3. c. Let c, = Z. Using the result of Theorem 2.6.7(b), determine the value of n such that Ic 10. Consider the polynomial p(x) = x3 + 5x - 1. It can be shown that p(x) has exactly one root in the open interval

(0. 1). Let a, E (0, 1) be arbitrary, and for n ? 1, set a _ ;(I - a,3,). a. Prove that the sequence {an} is contractive.

b. Show that if a = lim an, then p(a) = 0. c. Let a, = Z. Using the result of Theorem 2.6.7(b), determine the value of n such that la - an! < 10-'. 11. Let a, # a, be real numbers, and let 0 < b < 1. For n ? 3, set an = ban _, + (l - b)an -,. a. Show that the sequence (an) is contractive.

*b. Find nom lima,. 12. Prove that if every Cauchy sequence in R converges, then every nonempty subset of R that is bounded above has a supremum.

_2.77

Series of Real Numbers In this section we will give a brief introduction to series of real numbers. Some knowledge of series, especially series with nonnegative terms, will be required in Chapter 4. The topic of series in general, including various convergence tests, alternating series, etc., will be treated in much greater detail in Chapter 7. We begin with some preliminary notation. If {an}, is a sequence in N and if p, q E N with p q, set

Tlak=ap+ap,,+

+av.

k=p

2.7.1

Let {an}'°_, be a sequence of real numbers. Let {s,,}., be the sequence where for each n E N, sn = 7-k=, ak. The sequence is called obtained from an infinite series, or series, and is denoted either as DEFINITION

2.7 Series of Real Numbers

+ an +

or as a1 + a2 +

ak

87

k-I

For each n E N, s" is called the nth partial sum of the series and a" is called the nth term of the series. The series 7,I ak converges if and only if the sequence (s"} of nth partial sums converges in R. if im s" = s, then s is called the sum of the series, and we write 00

s = I a, k=1

If the sequence {s"} diverges, then the series joko. I ak is said to diverge.

2.7.2

EXAMPLES

(a) For (rl < 1, consider the geometric series 00

lrrk

k=1

For n(=- N1,

s"=

k-I

rk=r+r2+ +r

Thus

(1 -r)s"=s"-rs"=r-r"+', and as a consequence, S" _

r - r"+

I-r

Since Iri < 1, by Theorem 2.2.6(e),"-.00 lim r" = 0. Therefore lira s" = r/(1 - r), and thus '~°° °O

_

r

A rk For Iri

n!
1 the series =1 rk diverges (Exercise 3).

(b) Consider the series Moo I ak, where for each k E N, ak = (k Then

1

k

n

s" _ j ak k-I

=(1

- 2)+(2

=1- n+1 1

Thus lim s" = 1 and hence Jk I ak = 1.

3)+...+(n

n+ 1}

1)

88

Chapter 2

Sequences of Real Numbers

(c) Consider Y,(-1 )k. Then

sn = I (k=1

1)k

0,

-

_

1,

if n is even. if n is odd.

Thus since {sn} diverges, the series diverges.

The Cauchy Criterion The following criterion, which provides necessary and sufficient conditions for the convergence of a series, was formulated by Augustin-Louis Cauchy (1789-1857) in 1821.

2.7.3

THEOREM (Cauchy Criterion) The series Y,k I ak converges if and only if given e > 0, there exists a positive integer n,,. such that < e

for all in > n 2,- n0.

Proof. Since k_n+I akI

the result is an immediate consequence of Theorem 2.6.2 and Theorem 2.6.4. O

Remark The previous theorem simply states that the series I ak converges if and only if the sequence {s,,} of nth partial sums is a Cauchy sequence.

2.7.4

In this example we show that the series 7" 1 k diverges. We accomplish this by showing that the sequence (sn) of partial sums is not a Cauchy sequence. Consider EXAMPLE

stn-sn=n+t.+.

+2n'

nEN.

There are exactly n terms in the sum on the right, and each term is greater than or equal to 1/2n. Therefore, (s2.

-sn ? n

)= 2n

J

1'

2

The sequence {sn} therefore fails to be a Cauchy sequence and thus the series diverges. The divergence of this series appears to have been first established by Nicole Oresme (1323?-1382) using a method of proof similar to that suggested in the solution of Exercise 13 of Section 2.3.

2.7

2.7.5

Series of Real Numbers

89

COROLLARY If ET , ak converges, then Jim ak = 0.

Proof. Since ak = sk - sk_ i, this is an immediate consequence of the Cauchy criterion.

Remark. The condition lim at = 0 is not sufficient for the convergence of I ak. For example, the series 1 diverges, yet slim k = 0. -oc k

2.7.6

THEOREM S u p p o s e ak at 0 f o r all k E N. T h e n E is bounded above.

Proof. Since ak a 0 for all k, the sequence orem 2.3.2, the sequence

is monotone increasing. Thus by Theconverges if and only if it is bounded above.

EXERCISES 2.7 1. *'Using the inequality 1

I

1

1

k2

k(k - l)

k-I

k

prove that the series 71

k-ikz

converges.

2. Prove that the series X,,

k., k2+k

converges.

3. If Irl _- 1, show that the series 7,k'., rk diverges. 4. Prove that the series

I

converges. (See Exercise 3 of Section 2.6.)

k 0 k!

5. *Suppose ak >_ 0 for all k. Prove that if B ak converges, then

k.,

6. If Jk , ak and 1:T=, bk both converge, prove each of the following.

a Yk , cak converges for all c E R. b. 7,'k., (a,, + bk) converges.

7. If I' , (ak + bk) converges, does this imply that the series

a, +b, +a2+b2+ converges?

8. Suppose bk?ak?0for all kE N. a. If 10ko. , bk converges, prove that J:e , ak converges. b. If 7Q1. , ak diverges, prove that 7,x I bk diverges.

9. Consider the series T, k,,, p E R. k

a. Prove that the series diverges for all p 5 1. b. Prove that the series converges for all p > 1.

, at converges if and only if

k

converges.

90

Chapter2

Sequences of Real Numbers

NOTES This chapter provided our first serious introduction to the limit process. In subsequent chapters we will encounter limits of functions, the derivative, and the integral, all of which are further examples of the limit process. Of the many results proved in this chapter, it is difficult to select one or two for special emphasis. They are all important! Many of them will be encountered again-either directly or indirectly-throughout the text. Some of the concepts and results of this chapter have certainly been encountered previously; others undoubtedly are new. Two concepts that may not have been previously encountered are limit point of a set and the limit superior (inferior) of a sequence of real numbers. The notion of limit point of a set is one of the fundamental concepts of analysis. We will encounter limit points again when we characterize the closed subsets of R. The notion of limit point will also be crucial in the definition of the limit of a function. The results of Theorem 2.4.7, although elementary, are very useful. The fact that every limit point of a set is the limit of a sequence of distinct points in the set will be exploited in several instances in subsequent chapters. The primary importance of the limit superior and inferior of a sequence is that these two limit operations always exist in R U {-oo, oo}. As we will see in Chapter 7, this will allow us to present the correct statements of

the root and ratio test for convergence of a series. The limit superior will also be required to define the radius of convergence of a power series. There will be other instances in the text where these two limit operations will be encountered.

In this chapter we have proved several important

consequences of the least upper bound property of R. The least upper bound property was used to prove that every bounded monotone sequence converges. This result was subsequently used to prove the nested intervals property. which in turn was used to prove the Bolzano-Weierstrass theorem. By Exercise 23 of Section 2.3 and Exercises 13 and 14 of Section 2.4. each of these imply the least upper bound property of R. Another property of the real numbers that is equivalent to the least upper bound property is the completeness property of R; namely, every Cauchy sequence of real numbers converges. Other consequences of the least upper bound property will be encountered in subsequent chapters. Cauchy sequences were originally studied by Cantor

in the middle of the nineteenth century. He referred to them as fundamental sequences and used them in his construction of the real number system R (see Miscellaneous Exercises 4-11). The main reason that these sequences are

attributed to Cauchy, rather than Cantor, is because his 1821 criterion for convergence of a series (Theorem 2.7.3) is equivalent to the statement that the sequence of partial sums is a Cauchy sequence. The fact that Cauchy was a more prominent mathematician than Cantor may also have been a factor. In later chapters we will encounter examples of spaces of functions that have defined on them a function, called a norm, having properties analogous to those of the absolute value function. For such spaces it will also be possible to define both convergence of a sequence and the notion of Cauchy sequence. Many of these spaces will also have the property that they are complete; that is, every Cauchy sequence in the space converges.

MISCELLANEOUS EXERCISES The first three exercises involve the concept of an infinite product. Let {ak} be a sequence of nonzero real numbers. F o r e a c h is = 1, 2, ... , define

pn = flak = a, . a2 .

.

. an.

k=1

If p = lim pn exists, then p is the infinite product of the sequence {ak}k 1, and we write TM1

p=1 ak. k-1 If the limit does not exist, then the infinite product is said to diverge. Some authors require that p * 0. We will not make this requirement; rather we will specify p # 0 if this hypothesis is required in a result.

Miscellaneous Exercises

91

1. Determine whether each of the following infinite products converge. If it converges, find the infinite product.

c.frlll-

a. H(- I)k

2. If II' 1 ak = p with p * 0, prove that lim ak = 1. 3. If a ? 0 for all n E N, prove that (1 + ak) converges if and only if

ak converges. k-1

k-1

To prove the result, establish the following inequality:

a,+ +a,,

(1+a,) (1+a°):5 e°'` '°

CONSTRUCTION OF THE REAL NUMBERS In the following exercises we outline the construction of the real number system from the rational number system using Cantor's method of Cauchy sequences.

in 0 is Cauchy if for every r f-= Q, r > 0, there exists a positive integer n° such that Ia - a,,' < r for all n, m z n°. A sequence in 0 is called a null sequence if for every r (=- 0, r > 0, there exists a positive integer n° such that Ia I < r for all n z n°. Two Cauchy sequences provided (a° in Q are said to be equivalent, denoted is a null sequence. and Let Q denote the set of rational numbers. A sequence

4. Let

and

be Cauchy sequences in Q. Prove the following.

a. {a°) --

c. If

and

e. If {a°) - {cn} and {b°}

then {a°}

then {a +

b. If

{b.), then {b°} -

d. If

{b°}, then

{c +

and

{c,d°}.

Given a Cauchy sequence in Q. let denote the set of all Cauchy sequences in Q equivalent to The set is called the equivalence class determined by S. Given two Cauchy sequences

and (b,,) in 0, prove that

provided

and

n [{b°}] = 0 otherwise. Let 9t denote the set of equivalence classes of Cauchy sequences in Q. We denote the elements of 9t by lower-

case Greek letters a, 6, y, .... Thus if a E 9t, a = [{a°}] for some Cauchy sequence in Q. The sequence is called a representative of the equivalence class a. Suppose a = and P = [{b,,)]. Define -a, a + j9, and a - 6 as follows:

-a =

One needs to show that these operations are well defined; that is, independent of the representative of the equivalence class. For example, to prove that -a is well defined, we suppose that are two representatives and of a; i.e., {a°} -- {b°}. But by 4(d), Therefore, [{-a.)) = [(-b°)]. This shows that -a is well defined.

6. Prove that the operations + and are well defined on 9t. For each p E Q, let { p} denote the constant sequence p. and set ap = [{ p}]. Also, we set 0 = [{0}], t, = [{l}]. As we will see, the element 0 will be the zero of 9t and k will be the unit of R. A Cauchy sequence in 0 belongs to 0 if and only if b -> 0. Similarly, E i if and only if (a - 1) -+ 0. The following problem provides us with the multiplicative inverse of a # 0.

92

Chapter2

Sequences of Real Numbers

7. If a # 0, prove that there exists

E a such that a, * 0 for all n E N, and that {

is a Cauchy sequence.

Define a-' _ S. Prove that 9t with operations + and

is a field.

We now proceed to define an order relation on 9t. A Cauchy sequence

in Q is positive if there exists r E Q, r > 0. and no E N such that a > r for all n >_ no. Let 91 be defined by 91 _ is a positive Cauchy sequence}. 9. Prove that the set 'satisfies the order properties (01) and (02) of Section 1.4. 10. Show that the mapping p -> a,, is a one-to-one mapping of Q into % which satisfies

ap+a, =ap,q for all p, q E Q. Furthermore, if p > 0, then a,, E

'.

11. Prove that every nonempty subset of 9t which is bounded above has a least upper bound in 9t. The above exercises prove that 9t is an ordered field that satisfies the least upper bound property One can show that any two complete ordered fields are in fact isomorphic, that is, there exists a one-to-one mapping of one onto the other that preserves the operations of addition, multiplication, and the order properties. Thus 9t is isomorphic to the real numbers R.

SUPPLEMENTAL READING Aguirre, J. A. F., "A note on Cauchy sequences," Math. Mag. 68 (1995), 296-297. Bell, H. B., "Proof of a fundamental theorem on sequences;' Amer. Math. Monthly 71 (1964), 665-666. Goffman. C., 'Completeness of the real numbers;' Math. Mag. 47 (1974), I-8. Newman, Donald J. and Parsons, T. D., "On monotone

subsequences;' Amer. Math. Monthly 95 (1988),

44-45. Staib, J. H. and Demos, M. S., "On the limit points of the sequence (sin n 1:' Math. Mag. 40 (1967), 210-213. Wenner, B. R., 'The uncountability of the reals;' Amer Math. Monthly 76 (1969), 679-680.

3

Structure of Point Sets 3.1 Open and Closed Sets

3.2 Compact Sets 3.3 The Cantor Set

In this chapter we introduce some of the basic concepts fundamental to the study of limits and continuity, and study the structure of point sets in R. The branch of mathematics conceded with the study of these topics-not only for the real numbers but also for more general sets-is known as topology. Modern point set topology dates back to the early part of this century; its roots, however, date back to the 1850s and 1860s and the studies of Bolzano, Cantor, and Weierstrass on sets of real numbers. Many important mathematical concepts depend on the concept of a limit point of a set and the limit process, and one of the primary goals of topology is to provide an appropriate setting for the study of these concepts.

Although we restrict our study to the topology of the real line, all of the concepts encountered in this chapter can be defined in the more general setting of metric spaces. A thorough understanding of these topics on the real line will prove invaluable when they are encountered again in more abstract settings. On first reading, the concepts introduced in this chapter may seem difficult and challenging. With perseverance, however, understanding will follow.

3 ,1

Open and Closed Sets In the previous two chapters we used the terms open and closed in describing intervals in R. The purpose of this section is to give a precise meaning to the adjectives open and closed, not only for intervals, but also for arbitrary subsets of R. Before defining what we mean by an open set, we first define the concept of an interior point of a set. 93

94

Chaprer3

3.1.1

Structure of Point Sets

Let E be a subset of R. A point p E E is called an interior point of E if there exists an e > 0 such that N,(p) C E. The set of interior points of E is denoted by Int(E), and is called the interior of E. DEFINITION

Recall that for p E R and e > 0, the e-neighborhood N,(p) of p is defined as

{xER:Ix - pl < e}. 3.1.2

EXAMPLES

(a) Let E = (a, b] with a < b. Every p satisfying a < p < b is an interior point of E. If e is chosen such that

0 < e <min{Ip-al,lb-pl}, then N,(p) C E. The point b, however, is not an interior point. For every e > 0, N,(b) = (b - e, b + e)contains points that are not in E. Any x satisfying b < x < b + e is not in E. This is illustrated in Figure 3.1. For this example, Int(E) = (a, b). NN(P)

N,(b)

P

b

a

Figure 3.1

(b) Let E denote the set of irrational real numbers, i.e., E = R \ Q. If p E E, then by Theorem 1.5.2, for every e > 0 there exists r E Q fl N,(p). Thus N,(p) always contains a point of R not in E. Therefore no point of E is an interior point of E; i.e., Int(E) = 0. Using the fact that between any two real numbers there exists an irrational number (Exercise 6, Section 1.5), a similar argument also proves that Int(Q) _ ¢.

Open and Closed Sets Using the notion of an interior point, we now define what we mean by an open set.

3.1.3

DEFINITION

(a) A subset 0 of R is open if every point of 0 is an interior point of 0. (b) A subset F of P is dosed if F` = R \ F is open.

Remark. From the definition of an interior point it should be clear that a set 0 C R is open if and only if for every p E 0 there exists an e > 0 (depending on p) so that N,(p) C 0. In Theorem 3.1.9 we will provide a characterization of closed sets in terms of limit points.

3.1.4

EXAMPLES

(a) The entire set N is open. For any p E N and e > 0, N,(p) C R. Since R is open, by definition the empty set 0 is closed. However, the empty set is also open. Since 4)

3.1

Open and Closed Sets

95

contains no points at all, Definition 3.1.3(a) is vacuously satisfied. Consequently R is also closed.

(b) Every e-neighborhood is open. Suppose p E I8 and E > 0. If q E NE(p), then

p - qI < e. Choose S so that 0 < S s e - p - qI. If x E N5(q), then

Ix-PI:5 IP - qI+Ix - qI < IP - qi + S<_ IP-qi +e - IP - qi = e. Therefore N6(q) C NE(p) (see Figure 3.2). Thus q is an interior point of N1(p). Since q E NE(p) was arbitrary, NE(p) is open. Ne(p)

-N6(q) p-E

p

q

p+E

Figure 3.2

(c) Let E = (a, b], a < b, be as in Example 3.1.2(a). Since the point b E E is not an interior point of E. the set E is not open. The complement of E is given by

E`=(-oo,a]U(b,oo). An argument similar to the one given in Example 3.1.2(a) shows that a is not an interior point of E`. Thus E` is not open and hence by definition E is not closed. Hence E is neither open nor closed.

(d) Let F = [a, b], a < b. Then F` _ (-oo, a) U (b, oo), and this set is open. This can be proved directly, but also follows as a consequence of Theorem 3.1.6(a) below. Thus F is closed.

(e) Consider the set 0. Since no point of 0 is an interior point of Q (Example 3.1.2(b)), the set Q is not open. Also, C is not closed.

The use of the adjective open in describing the intervals (a, b), (a, on), (-cc, b) and (-oo, oo) is justified by the following theorem.

3.1.5

THEOREM Every open interval in R is an open subset of It

Proof. Exercise 1. 3.1.6

THEOREM

(a) For any collection {Oa},,EA of open subsets of R, n o. is open. &FA

(b) F o r any f i n i t e collection {O,, ... ,

of open subsets of R. fl O; is open.

96

Chapter 3

Structure of Point Sets

Proof. The proof of (a) is left as an exercise (Exercise 6). (b) If f l o, = 46, we are done. Otherwise, suppose

pE0=.n 0;. Then p E 0, for all i = 1, . .. , n. Since 0; is open, there exists an e; > 0 such that NE,(p) C 0..

Let e = min{e,, ... , e,}. Then e > 0 and NE(p) C 0,

for all i. Therefore

NE(p) C 0; i.e., p is an interior point of 0. Since p E 0 was arbitrary, 0 is open. C) As a consequence of the previous theorem every closed interval [a. b], a, b E R with a s b, is a closed subset of R. Since H \ [a, b] = (-oo, a) U (b, oo) is the union of two open intervals, by the previous theorem P \ [a, b] is an open subset of R. Thus [a, b] is a closed set. For closed subsets of R we have the following analogue of the previous result.

3.1.7

THEOREM

(a) For any collection {Fa}aEA of closed subsets of P. n F. is closed.

(b) For any finite collection IF,, ... ,

of closed subsets of R. U F is closed j°

Proof. The proofs of (a) and (b) follow from the previous theorem and De Morgan's laws:

(")c

EA

A

j=1

j .1

Remark. The fact that the intersection of a finite number of open sets is open is due to the fact that the minimum of a finite number of positive numbers is positive. This guarantees the existence of an e > 0 such that the e-neighborhood of p is contained in the intersection. For an infinite number of open sets, the choice of a positive e may no longer be possible. This is illustrated by the following two examples.

3.1.8

EXAMPLES We now provide two examples to show that part (b) of Theorem 3.1.6 is, in general, false for a countable collection of open sets. Likewise, part (b) of Theorem 3.1.7 is, in general, also false for an arbitrary union of closed sets (Exercise 6b).

(a) For each n = 1, 2, ..

.

, let 0 =

(-,1-,,,-,). Then each O is open, but

00

n1 O = {0}, which is not open.

3.1

Open and Closed Sets

97

(b) Alternatively, if we let G,, = (0, 1 + ), n = 1, 2.... , then again each G. is open, but x

n n-I

G,, = (0. 1 ],

which is neither open nor closed.

The following theorem provides a characterization of the closed subsets of R. Before stating the theorem we recall the definition of a limit point (Definition 2.4.5). For

E C R, a point p E R is a limit point of E if for every e> 0. (N,(p)1 { p}) n E t-

3.1.9

4,.

THEOREM A subset F of R is closed if and only if F contains all its limit points.

Proof. Suppose F is closed. Then by definition F` is open and thus for every p E F' there exists e > 0 such that N, (p) C F'; that is, Ne(p) fl F = 0. Consequently. no point of F` is a limit point of F. Therefore F must contain all its limit points. Conversely, let F be a subset of R that contains all its limit points. To show F is closed we must show F'' is open. Let p E V. Since F contains all its limit points. p is

not a limit point of F. Thus there exists an e > 0 such that N,(p) n F = 44. Hence N,(p) C F' and p is an interior point of V. Since p E F` was arbitrary, F` is open and therefore F is closed.

Closure of a Set 3.1.10

DEFINITION

If E is a subset of R, let E' denote the set of limit points of E. The

closure of E, denoted E, is defined as

E=EUE'. 3.1.11

THEOREM If E is a subset of R, then (a) E is closed. (b) E = E if and only if E is closed. (c) E C F for every closed set F C R such that E C F.

Proof. (a) To show that E is closed, we must show that k is open. Let p E E`. Then p 44 E.and p is not a limit point of E. Thus there exists an e > 0 such that

N,(p)flE_¢. We complete the proof by showing that NE(p) fl E' is also empty and thus N,(p) fl E =

Therefore N,(p) C E`; i.e., p is an interior point of E`.

Suppose N,(p) fl E' * 4,. Let q E NE(p) n E', and choose S > 0 such that Na(q) C N,(p). Since q E E, q is a limit point of E and thus N8(q) fl E * 46. But this

implies that N. (p) n E * tb, which is a contradiction. Therefore N,(p) n E' = 46, which proves the result.

_ (b) If E = E, then E is closed. Conversely, if E is closed, then E' C E and thus _ E. (c) If E C F and F is closed, then E' C F. Thus E C F.

98

Chapter3

3.1.12

Structure of Point Sets

DEFINITION A subset D of H is dense in R if D = H. The rationals 0 are dense in R. By Example 2.4.6(c), every point of H is a limit point of Q. Hence 0 = H. This explains the comment following Theorem 1.5.2. The rationals are not only dense; they are also countable. Countable dense subsets play a very important role in analysis. They allow us to approximate arbitrary elements in a set by elements chosen from a countable subset of H. Since the rationals are dense in R,

given any p E 11 and e > 0, there exists r E 0 such that Ip - rI < e. Additional examples of this will occur elsewhere in the text.

Characterization of the Open Subsets of R' If {In} is any finite or countable collection of open intervals, then by Theorem 3.1.6, U = U I is an open subset of N. Conversely, every open subset of H can be expressed as a finite or countable union of open intervals (see Exercise 17). However, a much stronger result is true. We now prove that every open set can be expressed as a finite or countable union of pairwise disjoint open intervals. A collection {!n} of subsets of R is pairwise disjoint if 1, fl Im = 0 whenever n # m.

3.1.13

THEOREM If U is an open subset of R, then there exists a finite or countable collection {1n} ofpairwise disjoint open intervals such that

U=UI.. Proof.

Let x E U. Since U is open, there exists an e > 0 such that

(X-E,x+E)C U. In particular (s, x) and [x, t) are subsets of U for some s < x and some t > x. Define r, and lx as follows:

r, = sup{t : t > x and [x, t) C U}, lx = inf{s : s < x and (s, x] C U}.

and

Then x < r, < oo and -oo s 1, < x. For each x E U, let 1, = (t r,). Then (a) 1, C U, and (b) if x, y E U, then either Ix = I, or 1, fl /,. = 4). The proofs of (a) and (b) are left as exercises (Exercise 16).

To complete the proof, we let 2 = {I, : x E U}. For each interval I E 9, choose rt E Q such that rt E I. If !, J E 2 are distinct intervals, then rt * rj. Therefore the mapping I - rt is a one-to-one mapping of T into O. Thus the collection Y is at most countable and therefore can be enumerated as {I;}iEA, where A is either a finite subset of N, or A = N. Clearly

U=iE Ii, 1. This topic can be omitted upon first reading of the text. The structure of open sets will only be required in Chapter 10 when defining the measure of an open subset of R.

3.1

Open and Closed Sets

99

and by (b), if n * j, then 1 f11, = 4). Thus the collection {1%EA is pairwise disjoint.

Relatively Open and Closed Sets One of the reasons for studying topological concepts is to enable us to study properties of continuous functions. In most instances, the domain of a function is not all of R, but rather a proper subset of R, as is the case with f(x) = NA for which Dom f = [0, oc). When discussing a particular function we will always restrict our attention to the do-

main of the function rather than all of R. With this in mind we make the following definition.

3.1.14

DEFINITION

Let X be a subset of R.

(a) A subset U of X is open In (or open relative to) X if for every p E U, there exists e > 0 such that NN(p) fl x C U. (b) A subset C of X is closed in (or closed relative to) X if X \ C is open in X.

3.1.15

EXAMPLE

Let X = [0, oo) and let U = (0, 1). Then U is not open in R but is open

in X. (Why?) The following theorem, the proof of which is left as an exercise (Exercise 21), provides a simple characterization of what it means for a set to be open or closed in X. 3.1.16

THEOREM Let X be a subset of R.

(a) A subset U of X is open in X if and only if U = X fl O for some open subset

OofR. (b) A subset C of X is closed in X if and only if C = X fl F for some closed subset F of R.

Connected Sets2 Our final topic in this section involves the notion of a "connected set." The idea of connectedness is just one more of the many mathematical concepts that have their roots in Cantor's studies on the structure of subsets of R. When we use the term "connected subset' of R, intuitively we are inclined to think of an interval as opposed to sets such as the positive integers N or (0, 1) U {2}. We make this precise with the following definition.

3.1.17

DEFINITION A subset A of R is connected if there do not exist two disjoint open sets U and V such that

(a) AlU*4)andAflV*e(,,and (b) (An u)U(Af1V)=A. 2. this concept, though important and used implicitly in several instances in the text, will not be required specifically in subsequent chapters except in a few exercises. Thus the topic of connectedne ss can be omitted upon first reading of the text.

100

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The definition for a connected set differs from most definitions in that it defines connectedness by negation; i.e., defining what it means for a set not to be connected. According to the definition, a set A is not connected if there exist disjoint open sets U and V satisfying both (a) and (b). As an example of a subset of R that is not connected, consider the set of positive integers N. If we let U = (1, Z) and V = (, co). then U and V are disjoint open subsets of 68 with u n Nl

l}

and

v n N = {2, 3, ...}

that also satisfy (U n N) U (V n N) = N. That the interval (a, b) is connected is a consequence of the following theoreth, the proof of which is left to the exercises (Exercise 25).

3.1.18 THEOREM A subset of 118 is connected if and only if it is an interval.

EXERCISES 3.1 1. Prove Theorem 3.1.5. 2. *Show that every finite subset of I8 is closed. 3. Show that the intervals (-oo, a) and [a, oo) are closed subsets of R.

4. For the following subsets E of R, fill in the chart.

E

Int(E)

E'

Isol. pts. of E

E

Open?

Closed?

(0, 1) U {2}

(a, b)

(a,b] [a, b]

[o, 1]na 5. a. Let F be a closed subset of R and let { p,} be a sequence in F which converges to p E R. Prove that p E F. b. Show by example that the conclusion is false if F is not closed. 6. *a. Prove Theorem 3.1.6(a).

b. Give an example of a countable collection of closed subsets of R such that U.=, F. is not closed. 7. Let A, B be subsets of R. b. Show that Int(A n B) = Int(A) n Int(B). a. If A C B, show that Int(A) C Int(B). c. Is Int(A U B) = Int(A) U Int(B)? 8. Let E be a subset of R.

*a. Prove that Int(E) is open. b. Prove that E is open if and only if E = Int(E). c. If G C E and G is open, prove that G C Int(E).

3.2

Compact Sets

101

9. Let A. B be subsets of R.

*a. Show that (A U B) = A U B. b. Show that (A fl B) C A fl B. c. Give an example for which the containment in part (b) is proper. 10. Prove that the set of limit points of a set is closed. 11. Let E C R. A point p E 18 is a boundary point of E if for every e > 0, N,(p) contains both points of E and points of E. Find the boundary points of each of the following sets. d. Q c. N *a. (a, b) b. E = {,'-, : n E N} 12. a. Prove that a set E C B8 is open if and only if E does not contain any of its boundary points. b. Prove that a set E C 18 is closed if and only if E contains all its boundary points. 13. *Prove that the set of irrational numbers is dense in R.

in D with lim p = p. 15. Let Do = {0, 11, and for each n E N, let D" = {a/2" : a E N, u is odd, 0 < a < 2"}. Let D = U `-0 D". Prose 14. *If D is dense in 18, prove that for every p E IB there exists a sequence that D is a countable dense subset of [0, I ].

16. Prove statements (a) and (b) of Theorem 3.1.13. 17. *Prove that there exists a countable collection 9 of open intervals such that if U is an open subset of E8 and

p E U, there exists / E I with p E /" C U. 18. Let X = (0, oo). For each of the following subsets of X determine whether the given set is open in X. closed in X. or neither. *c. (0, 1 ] U (2, 3) d. (0, 1 ] U (21 e. (1, : n E NJ b . (0, 1 ) *a. (0, 1 ]

19. For each of the following subsets of 0, determine whether the set is open in 0, closed in Q. both open and closed in 0, or neither.

a.A={pE0:I
b.B=(pE0:2
c.N

20. *If U is an open subset of 68, prove that E C U is open in U if and only if E is an open subset of R. 21. Prove Theorem 3.1.16. 22. Let X be a subset of R. Prove each of the following.

a. If {Ua},EA is any collection of subsets of X with U. open in X for all a E A. then

Ua is open in X.

b. If { U,, ... , Ua} is any finite collection of subsets of X with each U; open in X. then f;'= , U, is open in X. 23. For each of the following, use the definition to prove that the given set is not connected.

*a. (0, 1) U {2}

b.

n= 1.2,...}

c. (pE0:p>0andI
24. *a. If A is connected, prove that A is connected. b. Give an example of a set A such that A is not connected, but A is connected. 25. *Prove Theorem 3.1.18.

Compact Sets In this section we introduce the concept of a "compact set" and provide a characterization of the compact subsets of R. The notion of a compact set is very important in the study of analysis, and many significant results in the text will depend on the fact that every closed and bounded interval in F8 is compact. The modern definition of a compact

102

Chapter3

Structure of Point Sets set given in Definition 3.2.3 dates back to the second half of the nineteenth century and the studies of Heine and Borel on compact subsets of R.

3.2.1

DEFINITION

Let E C It A collection {O,},.EA of open subsets of R is an open

cover of E if

E C QU 0.. An alternative definition is as follows: The collection {O,}REA of open sets is an open cover of E if for each p E E, there exists an index a E A such that p E 0,.

3.2.2

EXAMPLES

(a) Let E = (0, 1) and 0n = (0, 1 - ), n = 2,3..... Then

2

is an open

cover of E. To see this, suppose x E E. Then since x < 1, there exists an integer n such

that x < 1 - R. Thus x E 0,,, and as a consequence, all

E C n=2 U 0,,. which proves the assertion. In fact, since 0 C E for each n. we have E = U. 't2 0n.

(b) Let F = ;0, oo) and for each n E I,1, let U,, _ (-1, n). Then {Un}nENis an open cover of F.

(c) If E is any nonempty subset of R, then for every e > 0. the collection {N,,(x): x E E} is an open cover of E.

3.2.3

DEFINITION A subset K o f U8 is compact i f every o p e n cover o f K has a f i n i t e subcover o f K; that is, i f {0,}a,EA is an o p e n cover o f K , then there exists a,,. .. , a E A

such that R

K C U0,. Remark To prove (using the definition) that a given set K is compact, we must prove

that f o r m= open cover {O,} of K, there exists a finite subcollection 0,,, ... , 0.. whose union covers K. This is illustrated in (a) of the following examples. It is W sufficient to find an open cover of K for which a finite subcollection also covers K. Such an open cover always exists. (Why?) On the other hand, to show that a set E is not compact, it suffices to find (construct) an open cover {O,} of E for which the union of no finite subcollection contains E. This is illustrated in Examples 3.2.4 (b) and (c).

3.2A EXAMPLES (a) Every finite set is compact. Suppose E = {p,, ... , pJ is a finite subset of N and {0,},EA is an open cover of E. Then for each j, j = 1, ,.. , n, there exists a; E A such that pi E 0,; But then {O,}7. is a finite subcollection which covers E.

3.2

Compact Sets

103

(b) The open interval (0, 1) is not compact. For the open cover {O"} 2 of (0, 1) in Example 3.2.2(a), no finite subcollection can cover (0, 1). Suppose on the contrary that a finite number, say O . . . , O",, covers (0, 1). Let N = max{n . . . , Pik}. Then N

k

(0.1)000"CUO"= 0,1-N, which is a contradiction.

(c) The closed set F = [0, oo) is not compact. For the open cover CU. of F, no finite subcollection can cover F. If there exist a finite number of sets in al1 which cover F, then there exists N E N such that F C (-1, N). (Why?) This, however, is a contradiction.

Properties of Compact Sets Before we provide a characterization of the compact subsets of U8, we first prove several properties of compact sets that require only the definition of compactness. As a consequence, all three of the following theorems are true in more general settings; e.g. in n-dimensional space R", and in general metric spaces. (For a discussion of metric spaces, see the notes at the end of the chapter.)

3.2.5

THEOREM

(a) Every compact subset of IR is closed and bounded. (b) Every closed subset of a compact set is compact.

Proof. (a) Let K be a compact subset of R. To show that K is bounded, consider the open cover {(-k, k)}5EN of 6$, and hence also of K. Since K is compact, there exist

k,, ... , k" E IN such that

KCU If N = max{k,, ... , k"}, then K is a subset of (-N, N). Thus K is bounded. To show that K is closed, we need to show that K` is open. Let p E K` be arbitrary.

For each q E K, choose eq > 0 such that

N,,(q)nN.,(p) _ 4, Any eq satisfying 0 < eq < z p - ql will work. Then {N,(q)}qE,, is an open cover of K. Since K is compact, there exists q,, ... , q" such that

K C U N. (qj). j=1

Let e = min{eqj : j = 1, ..

.

qj

, n}, which is positive. Then

N,(p)f N,gj(g) = 0 for all j = 1.. ... n. Thus N,(p) fl K = 0; i.e., N,(p) C K. Therefore p is an interior point of K` and thus K` is open.

104

Chapter 3

Structure of Point Sets

(b) Let F be a closed subset of the compact set K and let {4".},,E.a be an open cover of F. Then {Oa}aEA U {F`}

is an open cover of K. Since K is compact. a finite number of these will cover K. and hence also F. As a consequence of the previous theorem, the open interval (0. 1) is not compact since it is not closed. Likewise, the set A = {n : n E N} is not compact since again A is not closed. By Example 2.4.6(b), the point 0 is a limit point of A but 0 0 A. Also, the closed set [0, oo) is not compact since it is not bounded. In Theorem 3 .2.9 we will prove the converse of Theorem 3.2.5(a); namely, every closed and bounded subset of l is compact.

Remark. In proving that the compact set K was closed, compactness allowed us to select a finite subcover from the constructed open cover of K. Finiteness then assured that the e as defined was positive. This method of first constructing an open cover possessing certain properties and then using compactness to assure the existence of a finite subcover will be used on other occasions in the text.

3.2.6 THEOREM If S is an infinite subset of a compact set K. then S has a limit point in K. Proof.3 If no point of K is a limit point of S, then for each q E K, there exists a neighborhood Nq of q so that Nq contains at most one point of S, namely q if q E S. Since S is infinite, no finite subcollection of {Nq}qEK can cover S, and consequently no finite subcollection of {Nq}qEK can cover K. This is a contradiction.

Remark. Since every compact subset of R is closed and bounded, the previous theorem can also be proved using the Bolzano-Weicrstrass theorem (2.4.10) (see Exercise 6). Moreover, as we will see in Theorem 3.2.9, if K is a subset of R having the property that every infinite subset of K has a limit point in K, then K is compact. The property that "every infinite subset of K has a limit point in K" is usually referred to as the

Bolzano-Welerstrass property. Another useful consequence of compactness is the following analogue of the nested intervals property.

3.2.7

THEOREM If

K

t

is a sequence of nonemply compact subsets of R with

K. + t for all n, then

K=

fIK,,

is nonempty and compact. 3. The proof of Theorem 3.2.6 is again a proof by contradiction. In the proof we assume that the hypothesis "S is an infinite subset of K and K is compact" is true, and that the conclusion "S has a limit point in K" is false.

3.2

Compact Sets

105

Proof. We first show that fl , K # ¢. Let O = K. By Theorem 3.2.5, K. is closed and thus O is open. Furthermore, 00

00

n K. = 45 if and only if

Ul O = R.

is an open cover of R, and thus also of K1. But K, then < nk such that is compact. Therefore there exists n, <

Thus if fl. , K.

k

U 0,,J.. K, C j_

fl K,,, _ ¢. This, however, is a contradiction, since the inBut then K, fl K,,, fl tersection is equal to K,,,, which by hypothesis is nonempty. Thus K= fl K # 40. By Theorem 3.1.7, K is closed, and hence by Theorem 3.2.5(b), K is compact. Q

Characterization of the Compact Subsets of R We now turn to our goal of providing a characterization of the compact subsets of the

real line R. The first of the following two results is attributed to Eduard Heine (1821-1881) and Emile Borel (1871-1956); the second is due to Bernhard Bolzano (1781-1848) and Karl Weierstrass (1815-1897). The two theorems rank very high among the many important advances in the foundations of analysis during the nineteenth century. The importance of these results will become evident in later chapters. As is to be expected, the least upper bound property of u8 will play a crucial role in the proofs of these theorems.

3.2.8

THEOREM (Heine-Borel)

Proof.

Every closed and bounded interval [a, b) is compact.

Let °1L = {U,},,, be an open cover of [a, b] and let

E = {r e [a, b] : [a, r] is covered by a finite number of the sets U,}. The set E is bounded above by b, and since a E U. for some a E A, E is nonempty. Thus by the least upper bound property, the supremum of E exists in R. Let c = sup E. Since b is an upper bound of E, c < b. We first show that c E E; i.e., [a, c] is covered by a finite number of sets in 91. Since c E [a, b], c E Up for some (3 E A. Since Up is open, there exists e > 0 such that (c - e, c + e) C Up. Furthermore, since c - e is not an upper bound of E, there exists r E E such that c - e < r c. But then [a, r] is covered by a finite number, s a y U.,,, ... , U , , of sets in °U. But then the finite collection (U.,, ... , U,, , Up} covers [a, c]. Therefore, c E E. To conclude the proof, we show that c = b. Suppose c < b. If we choose s < b such that c < s < c 4- e, then the collection {U,,, ... , U,,, Up} also covers [a. s]. Thus s E E, which contradicts that c = sup E. Hence c must equal b. Q An alternative proof of the Heine-Borel theorem using the nested interval property (2.3.3) is outlined in the exercises (Exercise 11). The statement of the Heine-Borel theorem was initially due to Heine, a student of Weierstrass, who used the result implicitly in

106

Chapter3

Structure of Point Sets

the 1870s in his studies on continuous functions. The theorem was proved by Borel in 1894 for the case where the open cover was countable. For an arbitrary open cover the result was finally proved in 1904 by Henri Lebesgue (1875-1941). Using the Heine-Borel theorem, we now prove the following characterization of the compact subsets of R.

3.2.9

THEOREM (Heine-Borel-Bolzano-Weierstrass) Let K be a subset of R. Then the following are equivalent:

(a) K is closed and bounded. (b) K is compact. (c) Every infinite subset of K has a limit point in K.

(b). Since K is bounded, there exists a positive constant M so that K C [-M, M]. Since [-M, M] is compact and K is closed, by Theorem 3.2.5(b), K is

Proof. (a) compact. (b)

(c). This is Theorem 3.2.6.

(c) = (a). Suppose the set K is not bounded. Then for every n E N there exists p E K with p * p. for n * in such that Ip.i > n for all n. Then (p,,: n = 1, 2,

...)

is an infinite subset of R with no limit point in R. and hence none in K, which is a contradiction. Thus K is bounded. Let p be a limit point of K. By definition of limit point, for each n E N there exist

p E K with p * p such that

IN - A <

1n

Then S = {p,,: n = 1, 2, ... } is an infinite subset of K, and p is a limit point of S. To complete the proof we must show that p is the only limit point of S, and hence by hypothesis must be in K; i.e., K is closed.

Suppose q E QR with q # p. Let e = iI p - 9I, and choose N E N such that 1/N < e. Then for all n a N,

Ip - qI < Ipp - ql+Ipn - pI < Ipp - qI + n
1p.-qI > Ip-qI -e=2Ip-qI. Thus N1(q) contains at most finitely many p, and as a consequence q cannot be a limit point of S. Therefore, no q E R with q * p is a limit point of S.

Several additional consequences of compactness are given in Exercises 5 and 6. Another is the following theorem.

3.2.10

THEOREM Let K be a nonempty compact subset of R. Then every sequence in K has a convergent subsequence that converges to a point in K.

Proof.

Exercise 7.

3.3

The Cantor Set

107

D EXERCISES 3.2 1.

Let A = {;,:n = 1, 2,.

.

}.

.

a. Show that the set A is not compact by constructing an open cover of A that does not have a finite subcoser of A.

b. Prove directly (using the definition) that K = A U {0} is compact.

2. Suppose is a convergent sequence in R with lim p = p. Prove, using the definition. that the set A = {p} U {p" n E N} is a compact subset of W. :

3. Show that (0, 1 ) is not compact by constructing an open cover of (0. 1' that does not have a finite subcover.

4. Suppose A and B are compact subsets of R. 'a. Prove (using only the definition) that A U B is compact. b. Prove that A (1 B is compact. 5. 'Let K be a nonempty compact subset of R. Prove that sup K and inf K exist and are in K. 6. Let K be a compact subset of R. Use the Bolzano-Weierstrass theorem (2.4.10) to prove that every infinite subset of K has a limit point in K. 7. Prove Theorem 3.2.10.

8. Prove that a compact set has only a finite number of isolated points. 9. Let K be a compact subset of R and p E R \ K. a. Prove that there exist points a, h E K such that

Ja - p, = inf{lx - pI

:

.r E K}

and

jb - p = sup{jx - pI

:

x E K}.

b. Are either of the above still true if K is only closed'?

10. a. Prove that the intersection of an arbitrary collection of compact sets is compact. b. Find a countable collection {K,} . , of compact subsets of R such that U -, K, is not compact. 11. This exercise outlines an alternative proof of the Heine-Borel theorem. Suppose [a. h: is not compact. Then there exists an open cover 1t = of [a, h] such that no finite subcollection of 1l covers 'a. b'.We now proceed to show that this leads to a contradiction. Divide [a, b] into two closed subintervals [a. "- * n] and [" t. b'. each of length (b - a)/2. At least one of these, call it h, cannot be covered by a finite number of the U,. Repeating this process, obtain a sequence of closed and bounded intervals satisfying

(a) (a, b] D /, D 1, D

D/D

(b) length of 1 = (b - a)/2", and

(e) for each n, 1" is not covered by a finite number of the U,,. Now use Corollary 2.3.3 to obtain a contradiction.

3.3 1

The Cantor Set In this section we will construct a compact subset of [0, 1 ], known as the Cantor set. that has a number of interesting properties. This set is constructed by induction, the first two stages of which are illustrated in Figure 3.3. Let P. = [0, 11. From PO remove the middle third open interval (3, 3). This leaves two disjoint closed intervals

J,.= [o.'']

and

J,,, =

(

2I

1

108

Chapter 3

Structure of Point Sets

J 1.2

f1rY

`iu}

VF1}

r 2

3

9

9

zs

JZ ,

3

7

8

9

9

Figure 3.3

Set P, = J,,, U J,,2. From each of J,., and J1,2 remove the middle third open intervals 1

C 32,

2

7 8 32,

and

32

32

of length 1/32 This leaves 22 disjoint closed intervals J2.1' J2.2, J2,3, J24 of length 1/32; namely, 7

326

L

32' 3 J

L

' 3 ,'

32' 3 ,

Set P2 = J2,, U J2.2 U J2,3 U J2.4. In Figure 3.3, the shaded intervals indicate the open intervals that are removed at each stage of the construction. We continue this process inductively. At the nth step, each P. is the union of 2" disjoint closed intervals each of length 1/3"; i.e.,

r P. = U J..j, where for each j, J",; is a closed interval of the form zj z!+ 1 Jn.j = 3n' 3" Since each P, is a finite union of closed intervals, P. is closed and bounded, hence compact. Furthermore, since P0 D P, D P2 D , by Theorem 3.2.7,

a P= t5

P"

is is a nonempty compact subset of (0, 1 ]. The set P is called the Cantor ternary set. We now consider some properties of the set P. PROPERTY I

P is compact and nonempry.

PROPERTY 2 P contains all the end points of the closed intervals {J",r}, n = 1, 2, . .

, k = 1,2,.

PROPERTY 3

.

,2".

Every point of P is a limit point of P.

3.3

The Cantor Set

109

Proof. Let p E P and let e > 0 be given. Choose in E I 1 such that 1/3< e. Since p E P P E J,".k for some k, 1 <_ k s 2"'. But Jm. k =

Since the length of

xk L L 1 3,,,

3m

= 1/3'" < E, Jm.k C N,(p). Thus both end points of Jm., are in

P fl Nf(p), and at least one of these is distinct from p. 0 PROPERTY 4

The sum of the lengths of the intervals removed is 1.

Proof. At step 1, we removed one interval of length 1/3 . At the second step, we removed two intervals of length 1/32. At the nth step, to obtain P", we removed 2" -' intervals of length 1/3". Thus the sum of the lengths of the intervals removed

'

=3+232+...2"-13"+..

x

2'?3" t

31

1x 2 3

3

O

3

As a consequence of Property 4,

PROPERTY 5 P contains no intervals. be the ternary expansion of x. As we indicated For x E [0, 1 let x = .nln2n3 in Section 1.6, this expansion is unique except when

x=

3m,

aEN

0 < a < 3m,

with

where 3 does not divide a. In this case, x has two expansions: a finite expansion

x=3+

+ 3m,

a. E {1, 2},

and an infinite expansion. If am = 2, we will use the finite expansion. If am = 1. we will use the infinite expansion

x=-+ a1

3

++ 0

3m

2 co 71 3k k.m+l

With this convention we have PROPERTY then

6 if for each x E [0,1 J, x = .nln2n3

is the ternary expansion of x,

x E P if and only if nk E {0, 2}.

110

Chapter 3

Structure of Point Sets

Proof.

Exercise 2.

As a consequence of Property 6 and Theorem 1.7.18,

PROPERTY 7 P is uncountable. For each n, the set P. has only a finite number of end points. As a consequence, the set of points of P that are end points of some open interval removed in the construction is countable. Since P is uncountable, P contains points other than end points. By Exercise 1(a) of Section 1.6, the ternary expansion of; is 1

4

Thus

;

= .020202

E P. but 1 is not an end point of any closed interval J,",k.

Remark. By Property 4, the sum of the lengths of the intervals removed is 1. This seems to imply that P is in some sense very "small." On the other hand, by Property 7. P is uncountable, which seems to imply that P is "large:'

EXERCISES 3.3 1. Determine whether ; is in the Cantor set. 2. Prove Property 6 of the Cantor set. 3. Let 0 < a < 1. Construct a closed subset F of [0, 1 ] in a manner similar to the construction of the Cantor set such that the sum of all the open intervals removed is equal to a. 4. *Prove that the Cantor set P is equivalent to [0, I ].

NOTES Without a doubt, the most important concept of this chap-

ter is compactness. The fact that every open cover of a compact set has a finite subcover will be crucial in the study of continuous functions, especially uniform continuity. As we will see in many instances, the applications of compactness depend on the ability to choose a finite subcover from a particular open cover. A good example of

this is the proof of Theorem 3.2.5. Other instances will occur later in the text. Since compactness is the most important concept, Theorems 3.2.8 and 3.2.9 are the two most important results. In the Heine-Borel theorem we proved that every closed and bounded interval is compact, whereas in the Heine-Borel-Bolzano-Weierstrass theorem, we character-

set of K has a limit point in K. The converse of this result is also true, not only in R (Theorem 3.2.9), but also in the more general setting of metric spaces. A proof of this important result is outlined in the miscellaneous exercises. The notion of a metric and a metric space is motivated by geometrical considerations in R and n-dimensional euclidean space R". Many of the results involving the structure of point sets in R depend on the fact that the euclidean distance d(x, y) = Ix - yI between the points x

and y in R satisfy the following properties: d(x. y) z 0, d(x, y) = 0 if and only if x = y, d(x, y) = d(y. x), and d(x, y) s d(x. z) + d(z, y)

for all x, y, z E R. Using these properties as a motivation.

ized the compact subsets of R. In Theorem 3.2.6 we

the notion of distance can be extended to sets other than

proved that if K is a compact set, then every infinite sub-

P as follows.

Miscellaneous Exercises

111

(b) d(x, y) = 0 if and only if x = y,

of a metric. It allows one to prove that every e-neighborhood of a point is an open set. Also, combined with the property that d(p, q) = 0 if and only if p = q, it guarantees that the limit of a convergent sequence in X is unique. With the exception of Theorems 3.1.13 and 3.1.18, all of

(c) d(x, y) = d(y, x),

the results of Section 3.1 are still valid for any metric

DEFINITION Let X be a nonempty set. A real-valued function d defined on X X X satisfying

(a) d(x, y) ? O for all x, y E X.

(d) d(x, y) s d(x, z) + d(z, y) for all x, y, z E X,

is called a metric or distance function on X. The set X with metric d is called a metric space, and is denoted by (X, d).

Inequality (d) is called the triangle inequality. If d is a metric on a set X, then as in Definition 2.1.6, if p E X and e > 0, the e-neighborhood N,(p) of p is defined as

N,(p) = {x E X : d(p,x) < e).

space (X, d). In fact, the proofs follow verbatim the proofs

given in the text. The proofs of Theorems 3.1.13 and 3.1.18, however, require properties of the real number system and thus cannot be expected to be valid in an arbitrary metric space. As for the real line, if (X, d) is a metric space, a subset K of X is compact if every open cover of K has a finite subcover of K. Again, since the proofs of Theorems 3.2.5. 3.2.6, and 3.2.7 only require the definition of compactness. these results are again valid for an arbitrary metric space. In particular, if K is a compact subset of (X, d) then K is closed

Since the notions of limit point of a set, limit of a se-

and bounded. In this setting, a subset E of X is bounded if

quence, and interior point of a set are all defined in terms of e-neighborhoods, all of these concepts can be defined verbatim for an arbitrary metric space. As in R, if (X, d) is a metric space, a subset 0 of X is open in (X, d) if and only if every point of 0 is an interior point of O. It is important to note, however, that different metrics on a given set result in different c-neighborhoods, and thus a set that is open with respect to one metric may not be open with respect to a different metric. If X is any nonempty subset

there exists a positive constant M and p e X such that d(p, x) M for all x E E. The converse, however, is false. For example, if we set p(x, y) _ Ix - yI/(1 + Ix - v ). then p is a metric on R satisfying p(x, y) < I for all x, y E R (see Miscellaneous Exercise 4). In particular.

of lR, then d defined on X X X by d(x, y) = Ix - yI is a metric on X. In this setting, the concept of a subset U of X being relatively open in X is consistent with U being an open subset of X. The triangle inequality (d) is crucial in the definition

[0. oo) is a closed and bounded subset of (R, p) that is not compact. Thus the Heine-Borel-Bolzano-Weierstrass theorem is not valid in an arbitrary metric space. This is to be expected since the least upper bound property of R was used in part of the proof. However, there is one important part of this theorem that is still valid; namely, if a subset K of a metric space (X, d) satisfies the Bolzano-Weierstrass property, then K is compact. The proof of this very important result is outlined in Miscellaneous Exercise 7.

MISCELLANEOUS EXERCISES The first two exercises involve the geometric and euclidean metric structure of IB". For n > 2, R" _ {(xt, .. , x") xi E R, i = 1, .. , n). For p = (pt, . q") in I2" and , p"), q = (qi,

c E R. define p + q = (pt + qt, .. , p" + q"), and cp = (cpt..... cp.). Also. let 0 = (0, ... , 0). For p, q E R", the inner product of p and q, denoted (p, q), is definedas

(p,q)=ptq.+"+p"q"

1. Prove each of the following. For p, q. r E P8",

a. (p, p) a 0 with equality if and only if p = 0. b. (p, q) _ (q, p).

112

Chapter3

Structure of Point Sets

c. (ap + bq, r) = a (p, r) + b (q, r) for all a, b E d. 1(p, q)I z (

(9

This last inequality is usually called the Cauchy-Schwarz inequality. As a hint on how to prove part (d), for A E R, expand (p - Aq, p - Aq) and then choose A appropriately. Note that by part (a). (p - Aq, p - Aq) z 0 for

allAER. 2. For p = (Pi.... ,p") E R', set lIP112 = the euclidean length of the vector p.

)P)P

=\p

a. Use the result of Exercise 1(d) to prove that Ilp + g112

.

The quantity IIPII: is called the norm or

IIPI2 + I1g112 for all p, q E R".

b. Using the result of part (a), prove that d2(p, q) = lip - q4' is a metric on R". c. In R2 sketch the 1-neighborhood of 0.

3. For p=(pi.. .,p.,)and d, (p,q)_

in R", set

1p:-q,1.

a. Prove that d, is a metric on R".

b. In R2 sketch the 1-neighborhood of 0.

c. Suppose that {pk} is a sequence in R", where for each k E N. pk = (pk,,, ... , pk."). Prove that the sequence

{pk} converges top = (p ... , p") if and only if lim pk.; = p; for all i = 1, ... , n. d Prove that a sequence {pk} in R" converges to p E R' with respect to the metric d, if and only if it converges to p with respect to the metric d2. Specifically, prove that lim d,(pk, p) = 0 if and only if Alimd2(pk, p) = 0.

4. a. If (X, d) is a metric space, prove that p(x, y) = d(x, v)/(I + d(x, y)) is also a metric on X. b. Prove that a subset U of X is open in (X, d) if and only if it is open in (X, p). 5. If E is an uncountable subset of R. prove that some point of E is a limit point of E. (Hint: Use Exercise 17 of Section 3. 1).

6. Let {D"} be a countable collection of dense open subsets of R. Prove that (1

, O" is dense in R.

The following exercise is designed to prove the converse of Theorem 3.2.6; namely, if K is a subset of a metric space (X, d) having the properly that every infinite subset of K has a limit point in K. then K is compact.

7. Let K be a subset of a metric space (X. d) that has the property that every infinite subset of K has a limit point in K.

a. Prove that there exists a countable subset D of K which is dense in K. (Hint: Fix n E N. Let p, E K be arbitrary. Choose p2 E K, if possible, such that d(p,, P2) ? Suppose pl.. . . , pi have been chosen. Choose pj+,, if possible, such that d(pi, pj ) >_ in for all i = 1, ... , j. Use the assumption about K to prove that this process must terminate after a finite number of steps. Let 91 denote this finite collection of points, and let D = U "eN P". Prove that D is countable and dense in K.)

b. Let D be as in (a), and let U be an open subset of X such that u n K * 0. Prove that there exists p E D and n E N such that N,t"(p) C U. c. Using the result of (b), prove that for every open cover qt. of K, there exists a finite or countable collection { U")" C °U, such that K C U Un.

d. Prove that every countable open cover of K has a finite subcover. (Hint: If is a countable open cover of K, for each n E N. let W. = U, U,. Prove that K C W. for some n E N. Assume that the result is false, and obtain an infinite subset of K with no limit point in K, which is contradiction.)

Supplemental Reading

113

SUPPLEMENTAL READING Asic, M. D. and Adamovic, D. D., "Limit points of sequences in metric spaces," Amer. Math. Monthly 77 (1970), 613-616. Corazza, P., "Introduction to metric-preserving functions:' Amer. Math. Monthly 106 (1999). 309-323. Dubeau, Francis, "Cauchy-Bunyakowski-Schwarz inequality revisited," Amer. Math. Monthly 99 (1990), 419-421. Espelie, M. S. and Joseph, J. I-, "Compact subsets of the Sorgenfrey line," Math. Mag. 49 (1976), 250-251. Fleron, Julian F., "A note on the history of the Cantor set and Cantor function;' Math. Mag. 67 (1994),136-140.

Geissinger, Ladner, "Pythagoras and the CauchySchwarz inequality," Amer. Math. Monthly 83 (1976), 40-41. Kaplansky, Irving, Set Theory and Metric Spaces. Chelsea Publ. Co., New York, 1977. Kraft, R. L., "A golden Cantor set." Amer. Math. Monthly 105 (1998), 718-725. Labarre, Jr., A. E., "Structure theorem for open sets of real numbers;' Amer. Math. Monthly 72 (1965), 1114.

Nathanson, M. B., "Round metric spaces," Amer. Math. Monthly 82 (1975), 738-741.

Limits and Continuity 4.1 Limit of a Function 4.2 Continuous Functions

4.3 Uniform Continuity 4.4 Monotone Functions and Discontinuities

The concept of limit dates back to the late seventeenth century and the work of Isaac Newton (1642-1727) and Gottfried Leibniz (1646-1716). Both of these mathematicians are given historical credit for inventing the differential and integral calculus. Although the idea of limit occurs in Newton's work Philosophia Naturalis Principia Mathematica of 1687, he never expressed the concept algebraically; rather, he used the phrase "ultimate ratios of evanescent quantities" to describe the limit process involved in computing the derivatives of functions. The subject of limits lacked mathematical rigor until 1821 when Augustin-Louis Cauchy (1789-1857) published his Cours d'Analyse in which he offered the following definition of limit: "If the successive values attributed to the same variable approach indefinitely a fixed value, such that finally they dijf°er from it by as little as desired, this latter is called the limit of all the others." Even this statement does not resemble the modem delta-epsilon version of limit given in Section 4.1. Although Cauchy gave a strictly verbal definition of limit, he did use epsilons, deltas, and inequalities in his proofs. For this reason, Cauchy is credited for putting calculus on the rigorous basis we are familiar with today. Based on the previous study of calculus, the student should have an intuitive notion of what it means for a function to be continuous. This most likely compares to how mathematicians of the eighteenth century perceived a continuous function; namely, one that can be expressed by a single formula or equation involving a variable x. Mathematicians of this period certainly accepted functions that failed to be continuous at a finite number of points. However, even they might have difficulty envisaging a function 115

116

Chapter 4

Limits and Continuity

that is continuous at every irrational number and discontinuous at every rational number in its domain. Such a function is given in Example 4.2.2(g). An example of an increasing function having the same properties will also be given in Section 4 of this chapter.

4.11 Limit of a Function The basic idea underlying the concept of the limit of a function f at a point p is to study the behavior off at points close to, but not equal to, p. We illustrate this with the following simple examples. Suppose that the velocity v (ft /sec) of a falling object is given

as a function v = v(t) of time t. If the object hits the ground in t = 2 seconds, then v(2) = 0. Thus to find the velocity at the time of impact, we investigate the behavior of v(t) as t approaches 2, but is not equal to 2. Neglecting air resistance, the function v(t) is given as follows: (t)

32t, 05t<2, 0,

t ? 2.

Our intuition should convince us that v(t) approaches -64 ft/sec as t approaches 2. and that this is the velocity upon impact. As another example, consider the function f(x) = x sin x * 0. Here the function f is not defined at x = 0. Thus to investigate the behavior off at 0 we need to consider the values f(x) for x close to, but not equal to 0. Since I

V(x) I = jx sin x 5 kxi

for all x * 0, our intuition again should tell us that f(x) approaches 0 as x approaches 0. This is indeed the case as we will see in Example 4.1.10(c). We now make this idea of f(x) approaching a value L as x approaches a point p precise. To make the definition meaningful, we must require that the point p be a limit point of the domain of the function f.

4.1.1

DEFINITION

Let E be a subset of 68 and f a real-valued function with domain E.

Suppose that p is a limit point of E. The function f has a limit at p if there exists a number L E Q8 such that given any e > 0, there exists a 8 > O for which

1f(x)-LI <e for all points x E E satisfying 0 < Ix - pI < S. If this is the case, we write 1 m f(x) = L

or f(x) - L

as x -> p.

The definition of the limit of a function can also be stated in terms of e and 6 neigh-

borhoods as follows: If E C R, f : E --- R, and p is a limit point of E. then lim f (x) = L XP

4.1

Limit of a Function

117

if and only if given e > 0, there exists a S > 0 such that

f(x) E N,(L) for all x E Efl (Na(p)\{p}). This is illustrated graphically in Figure 4.1.

L+E

Lt

Figure 4.1

lim f(x) = L x-p

Remarks (a) In the definition of limit, the choice of S for a given e may depend not only on e and the function, but also on the point p. This will be illustrated in Example 4.1.2(g).

(b) If p is not a limit point of E, then for S sufficiently small, there do not exist any x E E so that 0 < Ix - pI < S. Thus if p is an isolated point of E, the concept of the limit of a function at p has no meaning. (c) In the definition of limit, it is not required that p E E, only that p is a limit point of E. Even if p E E, and f has a limit at p, we may very well have that

lf(X) *f(p). This will be the case in Example 4.1.2(c).

(d) Let E C R and p a limit point of E. To show that a given function f does tlt have a limit at p, we must show that for every L E R, there exists an e > 0, such that for

every S > 0, there exists an x E E with 0 < Ix - pI < S, for which

Lf(x)-Llae. We will illustrate this in Example 4.1.2(e).

118

Chapter 4

4.1.2

Limits and Continuity

EXAMPLES

(a) Let E be a nonempty subset of 1R and let f, g, and h be functions on E defined by f (x) = c (c E 18), g(x) = x, and h(x) = x2, respectively. If p is a limit point of E. then

lim f(x) = c,

lim g (x) = p.

lim h(x) = p2.

X-P

These limits are also expressed as limc = c, limx = p, and limx'' =

p2.

Even though we may feel that these limits are obvious, theypstill have to be proved. We illustrate the method of proof by using the definition to prove that lim h(x) = p2. The proofs of the other two limits are left to the exercises (Exercise 2). Fpor x E E,

Ih(x) - p2I = Ix2 - p2I = Ix - pIIx + pi If

(IkI + Ipj)Ix - pI

pl < 1,then kI < IpI + 1. Hence forallxEEwithIx - pl < 1, Ih(x) - p2I < (2lpl + l)Ix - pI.

This last term will be less than e provided Ix - pl < e/(21 pl + 1). Thus given e > 0, we choose 8 = min{ 1, E/(2 1 pI + 1)}. With this choice of 8, if x E E with 0 < Ix - pl < S,we first of all have Ix - pl < 1, and therefore also

Ih(x) - p2I < (2lpl + 1)Ix - pl < (2IPI + 1) (2IPIE+

1)

= e.

Thus lim x2 = p2. X" p

(b) For x * 2, let f(x) be defined by f (X) = X2 -4

x-2 The domain of f is E = (-oo, 2) U (2, oo), and 2 is clearly a limit point of E. We now show that lim f (x) = 4. For x # 2,

If(x)-4I = Ix-2 -4I = Ix+2-4I = Ix-2I. Thus given e > 0, the choice S = e works in the definition. (c) Consider the following variation of (b). Let g be defined on R by

x2 -4 g(x) = x - 2 2,

x#2,

x2.

For this example, 2 is a point in the domain of g, and it is still the case that lim g(x) = 4. However, the limit does not equal g(2) = 2. The graph of g is given in Figure 4.2.

(d) Let E = (-1, 0) U (0, oo). For x e E, let h(x) be defined by

h(x)=

x+1-1 x

4.1

Limit of a Function

g(x), x s 2

2

Figure 4.2

We claim that l

Graph of g

h(x) = Z. This result is obtained as follows: For x # 0,

x+1-ill x+1+1

x+1-1 x

777-1 + 1/

x

_

x

1

x+I+1

x( x+1+1)

From this last term we now conjecture that h(x) -+ as x --+0. By the above, h(x) - 12

(

+1

121

-

x +X+ l + 11)

I2( I

x+1)(1+Vx- -+I) _ 2(V, +1+1)2

-x

I -I2(V+

IxI

2( x+1+1)2 For x E Ewe have (

x + 1 + 1)2 > 1, and thus

Ih(x)I <

2I

Given e > 0, let S = e. ThenI for all x e E with 0 < jxI < S, Ih(x)

and thus

21 < ZI < 2 < E,

liim h(x)

(e) Let f be defined on R as follows:

f(x) - 10, x E O, x le a.

119

120

Chapter 4

Limits and Continuity

We will show that for this function, lim f(x) fails to exist for every p E R. Fix p E R. Let L E R and let

e = max( IL - 11, ILI).

Suppose e = IL - 11. By Theorem 1.5.2. for any S > 0, there exists an x E 0 such

that 0 < Ip - xI < S. For such an x,

If(x)-LI=11-LI=elf e = ILI, then by Exercise 6, Section 1.5, for any 8 > 0, there exists an irrational

number x with 0 < Ix - p l < S. Again, for such an x, V (x) - LI = e. Thus with

e as defined, for any 8 > 0, there exists an x with 0 < Ix - pI < S such that V (x) - LI ? e. Since this works for every L E 6L lim f(x) does not exist.

(f) Let f : 68 -* l be defined by 0' x E Q, AX) - IX, x Q. Then lim f (x) = 0. Since L f (x)1 jx i for all x, given e > 0. any S. 0 < 8 e, will work in the definition of the limit. A modification of the argument given in (e) shows

that for any p 0 0, iim f (x) does not exist. An alternative proof will be provided in Example 4.1.5(b).

(g) Our final example shows dramatically how the choice of S will generally depend not only on e, but also on the point p. Let E _ (0, oo) and let f : E - R be defined by

f(x)

I

We will prove that for p E. (0, oo), 1 liml = p

-+p x

If x > p/2, then

1-11 x

=

Ix-PI

2

<-.

P XP Ix - PI, Therefore, given e > 0, let S = min{p/2, pee/2}. Then if 0 < Ix - pI < S. x > p/2,

and

x -P < p2Sse. The S as defined depends on both p and e. This suggests that any S that works for a given p and a must depend on both p and e. Suppose on the contrary that for a given

e > 0, the choice of 8 is independent of p E (0, oo). Then with e = 1, there exists a S > 0 such that

zI

p1<1

4.1

Limit of a Function

121

for all x, p E (0, oo) with 0 < Ix - p I < S. Since any smaller S must also work, we can

assume that 0 < S < 1. But now if we take p = 18 and x = 8, then 0 < Ix - pI < S, and thus I f(x) - f(p) I < 1. However,

If(x) -1(P)I

(s

8I

S

> 1.

This contradiction proves that the choice of S must depend on both p and e. 0

Sequential Criterion for Limits Our first theorem allows us to reduce the question of the existence of the limit of a function to one concerning the existence of limits of sequences. As we will see, this result will be very useful in subsequent proofs, and also in showing that a given function does not have a limit at a point p.

4.1.3

THEOREM Let E be a subset of l1, p a limit point of E, and f a real-valued function defined on E. Then lim f (x) = L if and only if X-+p

lint f (pn) = L

n-+oc

for every sequence {pn} in E, with pn # p for all n, and lim p = p.

Remark. Since p is a limit point of E, Theorem 2.4.7 guarantees the existence of a sequence { pn} in E with pn # p for all n E N and pn -- p.

Proof. Suppose xlp iim f (x) = L. Let {pn} be any sequence in E with pn * p for all n and pn -+ p. Let E > 0 be given. Since x-.p limf(x) = L, there exists a S > 0 such that V (x) (x)

- L I < e for all'x E E, 0 < Ix - p I < S.

(1)

Since lint pn = p, for the above S, there exists a positive integer no such that m

0 < Ipn - P I < S for all n ? no. Thus if n ? n by (1), l f (pn) - LI < e. Therefore, lint f (pn) = L. R-00

Conversely, suppose f(pn) -4 L for every sequence { pn} in E with pn * p for all a and pn -+ p. Suppose lim f (x) . # L. Then there exists an e > 0 such that for every x-Px E Ewith 0 0, there exists an

n r= N, take S = ,-,. Then for each n, there exists p,, E E such that

0
and

If(Pn)-LIae.

Thus pn -+ p, but {f(pn)) does not converge to L. This contradiction proves the re-

sult. p

The following uniqueness result is an immediate consequence of the previous theorem.

122

Chapter 4

4.1.4

Limits and Continuity

If f has o limit at p, then it is unique. Theorem 4.1.3 is often applied to show that a limit does not exist. If one can find a sequence {p"} with p" -- p, such that {f(p")} does not converge, then lim f(x) does not exist. Alternatively, if one can find two sequences {p"} and {r"} both converging to p, but for which COROLLARY

litn .f(p,,) * iim f(rM),

-00

1W

then again 1 m f(x) does not exist. We illustrate this with the following two examples.

4.1.5

EXAMPLES

(a) Let E = (0, oo) and f(x) = sin (11x), x E E. We use the previous theorem to show that lim sin x-.0

1

X

does not exist. Let p" = zi + 1 r;. Then

sin(2n + 1)2 = (-1)". Thus ii f (p,) does not exist, and consequently by Theorem 4.1.3. liim f (x) also does

"7

not exist. The graph of f(x) = sin(1/x) is given in Figure 4.3.

n

1

2rr

V

Figure 4.3 Graph of f(x) = sin(l/x), x > 0

4.1

limit of a Function

123

(b) As in Example 4.1.2(f) let J0, X E Q, AX) = lx, x qE Q. Suppose p E R, p * 0. Since Q is dense in R, there exists a sequence C Q with pn * p for all n E N such that pn -> p. Hence, lim 0. On the other hand. since R \ Q is also dense in R, there exists a sequence " *00{qn} of irrational numbers with

qn -? p. But then lim n-+oo

lim q = p. Thus since p * 0, by Theorem 4.1.3,

n- 00

lim f (x) does not exist.

Limit Theorems 4.1.6 THEOREM Suppose E C R, f. g : E -+ P. and p is a limit point of E. If limf(x) = A

and

lim g(x) = B,

x-4p

then

(a) xlim [f(x) + g(x)] = A + B, (b) 1 m f(x)g(x) = AB, and

(c) lim

x

= B, provided B * 0.

Proof. For (a) and (b) apply Theorem 4.1.3 and Theorem 2.2.1. We leave the details to the exercises (Exercise 11). Proof of (c): By (b) it suffices to show that 1

1

lim g(x)

=B

We first show that since B # 0, g(x) * 0 for all x sufficiently close to p, x * p. Take e = IBI/2. Then by the definition of limit, there exists a St > 0 such that

Ig(x) - BI <

11 2

for all x E E, 0 < Ix - pI < SI. By Corollary 2.1.4, Ig(x) - BI > Ilg(x)I - IBII. Thus

forallxEE,0< Ix - pI < Si. We can now apply Theorem 4.1.3 and the corresponding result for sequences of Theorem 2.2.1. Let { pn} be any sequence in E with pn -+ p and pn * p for all n. For

124

Chapter 4

Limits and Continuity

the above S1, there exists an no E Rl such that 0 < jp" - pl < S1 for all n ? no. Thus g(p") * 0 for all n ? n0. Therefore by Theorem 2.2.1(c), lim

1

"°° g(p")

=

1

B.

Since this holds for every sequence p" -> p, by Theorem 4.1.3,

lmg(x) =

1 B.

L1

The proofs of the following two theorems are easy consequences of Theorem 4.1.3 and the corresponding theorems for sequences (2.2.3 and 2.2.4). First, however, we give the following definition.

4.1.7

DEFINITION A real-valued function f defined on a set E is bounded on E if there exists a constant M such that V(x)l t M for all x E E.

4.1.8

THEOREM Suppose E C R, p is a limit point of E, and f, g are real-valued functions on E. If g is bounded on E and lim f (x) = 0, then t-.p

lim f(x)g(x) = 0.

Proof. Exercise 12.

4.1.9

THEOREM Suppose E C R, p is a limit point of E, and f, g, h are functions from E into R satisfying g (x)

If lire g (x) =lim h(x) = L, Proof,

f (x) : h (x) for all x E E. then

1 m f (x) = L.

Exercise 13.

We now provide several examples to illustrate the previous theorems.

4.1.10

EXAMPLES

(a) By Example 4.1.2(a), lim x = c. Thus, using mathematical induction and Theorem 4.1.6(b), x.+c lim x" = c" for all n E N. If p(x) is a polynomial !unction of degree n, that is,

.+a,x+ao, where n is a n o n n e g a t i v e i n t e g e r and a 0 ,

. .. , a" E R with a" * 0, then a repeated application of Theorem 4.1.6(a) gives lim p(x) = p(c). (b) Consider lim

x-.-2

x'+2x2-2x-4 x2 - 4

4A

Limit of a Function

125

By part (a), xlim2 (x' + 2x2 - 2x - 4) = 0 and dint (x2 - 4) = 0. Since the denominator has limit zero, Theorem 4.1.6(c) does not apply. In this example, however, for

x # -2, (x+2)(x2-2)

x3+2x2-2x-4 x2-4

(x+2)(x-2)

x' -2

x-2

Since slim (x - 2) = -4, which is nonzero, we can now apply Theorem 4.1.6(c) to conclude tat lim

x3+2x2-2x-4= lim x2-21 x--2x-2 2 x2-4

(c) Let E = R \ {0}, and let f : E -r R be defined by f (x) = x sin

x.

Since I sin (l /x) I <- I for all x E i8, x # 0, and Ii m x = 0, by Theorem 4.1.8

lim x sin 1 = 0. x

x-w

The graph of f(x) = x sin(1/x) is given in Figure 4.4.

-0.2

Figure 4.4 Graph of f(x) = xsin(1/x), x # 0

(d) Let E = (0, oo) and let f be defined on E by

f(t) _

sin t

We now prove that

lim

r-0

-smt= 1. 1

126

Chapter 4

Limits and Continuity As we will see in the next chapter, this limit will be crucial in computing the derivative of the sine function. From Figure 4.5, we have

area(LOPQ) < area(sector OPR) < area(DOSR).

Figure 4.5

In terms of t, this gives

2sintcost<2t<2tan:. Therefore,

sin t

cost <

<

1

cost

t

Using the fact that lim cos t = 1 (Exercise 5), by Theorem 4.1.9 we obtain li m

sin t 1

= I.

Limits at Infinity Up to this point we have only considered limits at points p E N. We now extend the definition to include limits at oo or -oo. The definition of the limit at oo is very similar to

lim f(n) where f : N , R; that is, f is a sequence in R.

4.1.11

DEFINITION

Let f be a real-valued function such that Domf tl (a, oo) * ¢ for

every a E P. The function f has a limit at oo if there exists a number L E R such that given e > 0, there exists a real number M for which

lf(x) - LI < e

4.1

Limit of a Function

127

for all x E Dom f fl (M, oo). (See Figure 4.6.) If this is the case, we write lim (x) = L. X-*00f

Similarly, if Dom f fl (-oo, b) * O for every b E R,

lim f (x) = L

X--00

if and only if given e > 0, there exists a real number M such that

If(x) - LI < e for all x E Dom f fl (-oo, m). The hypothesis that Dom f fl (a, oo) * 0 for every a E R is equivalent to saying that the domain of the function f is not bounded above. If Dom f = N, then the above definition gives the definition for the limit of a sequence. Readers should convince themselves that all theorems up to this point involving limits at a point p E R are still valid if p is replaced by o0 or -oo.

L+E L

L-E

Figure 4.6

4.1.12

lim f(x) = L

x-t=

EXAMPLES

(a) As our first example, consider the function f(x) = (sin x)/x defined on (0, oo). Since Isin xl

1,

U(x)I

x

for all x E (0, oo). Let e > 0 be given. Then with M = 1/e, I f (x)I < e Therefore, 1

(sin x)/x = 0.

for all x > M.

128

Limits and Continuity

Chapter 4

(b) For our second example consider f(x) = x sin irx. If we set p" = (n + ;). n E N. then

f(p") = (n + !)sin(n + ;)ir = (-1)"(n + 2). Thus the sequence {f(p")}n°_ I is unbounded, and as a consequence, lim x sin zrx does not exist. x

EXERCISES 4.1 1. Use the definition to establish each of the following limits. b6 fim2 (3x + 5) = -1 *a. lim (2x - 7) = -3

*c. I m

_

x

l+ x

I

d.

2

i

e..-- I x + 1 - 3

lim 2x2-3x-4=

x-+-i

L lim

x3-2x-4 = 5 x2-4

2

2. Use the definition to establish each of the following limits.

a. lim c=C x-v

b. lim x = p

C. lim x3 = p3

d. lim x-0 x" = p", n E RI

x-+v

-v

I

p>0

v

x +o

x

2Vp

, p>0

3. For each of the following, determine whether the indicated limit exists in R. Justify your answer!

*a x-.° limWx-

b.x-.i limx2x +- I

*C. lim cos-

d. lim

I

X-O

e. lim

x

f. lim

14

4. *Define f : (-1, 1)

cos x

)2 (x + lx

1

R by

f(x)=x2-x-2

x+l Determine the limit L of f at -1 and prove, using e and S, that f has limit L at -1. 5. *a Using Figure 4.5, prove that I sin hl -- jhI for all It E R. b. Using the trigonometric identity I - cos It = 2 sin 2 2, prove that (i) limo cos h = I.

00lim

1 - cosh h

0.

6. Let E C R, p a limit point of E, and f : E -* R. Suppose there exist a constant M > 0 and L E R such that [f(x) - LI c Mix - pl for all x E E. Prove that lim f(x) = L. X-P

4.1

Limit of a Function

129

7. Suppose f: E -i R, p is a limit point of E, and limf(x) = L. x-p *a. Prove that lint If(x)I = ILI x-o

= VL.

b. If, in addition, f(x) >- 0 for all x E E, prove that lim c. Prove that lim (f(x))" = L" for each n E N.

8. Use the limit theorems, examples, and previous exercises to find each of the following limits. State which theorems, examples, or exercises are used in each case.

.a, littt 5x2 + 3x - 2

b. lim

se lint

d.

x-1

_-'-+

m 2x + 5

x-.-t

x3 _ x2 + 2 x+1 Ix + 13n

xm2

x + 2)

f lim-

x-4

I

-

.r(

'1 1

Ix+21

s 9. 'Suppose f : (a, b) -+ R, p E [a, b], and + m f (x) > 0. Prove that there exists a 8 > 0 such that f (x) > 0 for all

x E (a, b) with 0 < Ix - pi < S. 10. Suppose E C R, p is a limit point of E, and f : E -+ R. Prove that if f has a limit at p, then there exists a positive constant M and a S > 0, such that [f(x)I s M for all x E E, 0 < Ix - pi < 6. 11. a. Prove Theorem 4.1.6(a). b. Prove Theorem 4.1.6(b).

12. 'Prove Theorem.4.1.8. 13. Prove Theorem 4.1.9.

14. Let f, g be real-valued functions defined on E C R and let p be a limit point of E.

a. If limf(x) and lim (f(x) + g(x)) exist, prove that lim g(x) exists. xrp x-p x-p b. If lim f(x) and lim (f(x)g(x)) exist, does it follow that lim g(x) exists? x-ap X-P X-P

15. Let E be a nonempty subset of R and let p be a limit point of E. Suppose f is a bounded real-valued function on E having the property that iiin f(x) does not exist. Prove that there exist sequences (p.) and {q"} in E with

lim p" = lim q" = p such that lim f(p") and lim f(q") exist, but are not equal. .-CO "-W

16. *Let f be a real-valued function defined on (a, oo) for some a > 0. Define g on lim f (x) = L if and only if lint g(t) = L.

by g(t)

17. Investigate the limits at oo of each of the following functions defined on (0, oo).

a. f(x) =

c. f(x) =

3x2

2+ +x

!

V-47+ 1

e. f( x) =

X

b. f (x) =

1 + x2

d. f(x) - 2x + 3 x+l V x - 2x AX) =

2NA + 3x h. f(x) = x sin R be such that lim x f (x) = L where L E R. Prove that lim f (x) = 0.

g. f (x) = x cos x

18. Let f : (a, oo)

1

Prove that

130

Chapter4

Limits and Continuity

19. Let f : R -> R satisfy f(x + y) = f(x) + f(y) for all.r. v E R. If lim f(.r) exists, prove that a. lim f(x) = 0, and b. lim f(x) exists for every p E R. =-v

4.2j Continuous Functions The notion of continuity dates back to Leonhard Euler (1707-1783). To Euler, a continuous curve (function) was one that could be expressed by a single formula or equation of the variable x. If the definition of the curve was made up of several parts, it was called discontinuous. This definition was sufficient to convey the concept of continuity in Euler's time as mathematicians were primarily concerned with elementary functions; namely, functions built up from the trigonometric and exponential functions, and inverses of these functions, using algebraic operations and composition. The more modern version of continuity is credited to Bolzano (1817) and Cauchy (1821). Both men were motivated to provide a clear and precise definition of continuity in order to prove the intermediate value theorem (Theorem 4.2.11). Cauchy's definition of continuity was as follows: "The function f (x) will be, between two assigned val-

ues of the variable x, a continuous function of this variable if for each value of x between these limits, the numerical value [i.e., absolute value] of the difference f(x + a) - f(x) decreases indefinitely with a'. Even this definition appears strange in comparison with the more modem definition in use today. Both Bolzano and Cauchy were concerned with continuity on an interval, rather than continuity at a point.

4.2.1

Let E be a subset of U8 and f a real-valued function with domain E. The function f is continuous at a point p E E, if for every e > 0, there exists a S > 0 such that DEFINITION

IA(x) -f(P)I < e for all x E E with Ix - pI < S. The function f is continuous on E if and only if f is continuous at every point p E E. This definition can be rephrased as follows: A function f : E - IR is continuous at

p E E if and only if given e > 0, there exists a S > 0 such that

f(x) E NE(f(p)) for all x E N5(p) fl E. This is illustrated in Figure 4.7.

Remarks (a) If p E E is a limit point of E, then f is continuous at p if and only if

Lmf(x) =f(p).

X-P 1. Cauchy Course d'Analyse, p. 43.

4.2 Continuous Functions

131

Figure 4.7

Also, as a consequence of Theorem 4.1.3, f is continuous at p if and only if

f(p) for every sequence

in E with p, -+ p.

(b) If p E E is an isolated point, then every function f on E is continuous at p. This follows immediately from the fact that for an isolated point p of E, there exists a S > 0 such that N8(p) (1 E = { p}. We now consider several of the functions given in previous examples, and also some new examples.

4.2.2

EXAMPLES

(a) Let g be defined as in Example 4.1.2(c), i.e.,

1xZ-4 g(x)= 2,

x-2.

x#2,

x=2.

At the point p = 2, lim g(x) = 4 # g(2). Thus g is not continuous at p = 2. However, if we redefine g at p = 2 so that g(2) = 4, then this function is now continuous at

p=2.

(b1 Let f be as defined in Example 4 . 1 . 2(f ): i . e., f (x) =

{0. x E O, x x Q

132

Chapter4

Lindta and Continuity

Since

limf(x) = 0 = f(0), f is continuous at p = 0. On the other hand, since limf(x) -P fails to exist for every p * 0, f is discontinuous at every p E Il, p * 0. (e) The function f defined by

-

11, xE0,

f(x)

x VE Q,

0,

of Example 4.1.2(e) is discontinuous at every p E R. (d) As in Example 4.1.2(g), the function f (x) = 1/x is continuous at every p E (O,x). Thus, f is continuous on (0, a).

(e) Let f be.defined by

x = 0,

0, f (X)

x sin

-x

,

x:# 0.

By Example 4.1.10(c),

li mf(x) = 0 = f(0). Thus f is continuous at x = 0. (f) In this example we show that f (x) = sin x is continuous on R. Let x, y E R. Then

If(s) -.f(x)l = Isiny - sinxl = 2 cos (y + x) sin 2(y - x) I 2

{2lsinI(y-x+ ).

By Exercise 5 of the previous section, I sin h I < Ih 1. Therefore,

If(y) -f(x)I S Iy - xl, from which it follows that f is continuous on R. (g) We now consider a function on (0, 1) that is discontinuous at every rational num-

ber in (0, 1) and continuous at every irrational number in (0, 1). For x E (0, 1) define

0, AX)

= in1,

if x is irrational,

if x is rational with x =

m in lowest terms. n

The graph of f, at least for a few rational numbers, is given in Figure 4.8. To establish our claim we will show that

lira f(x) = 0 X-P

4.2

Continuous Functions

133

for every p E (0, 1). As a consequence, since f(p) = 0 for every irrational number p E (0, 1),f is continuous at every irrational number. Also, since f(p) # 0 when p r= 0 fl (0, 1), f is discontinuous at every rational number in (0, 1).

2

4

8 16

-

1

16

I

3

I

5

3

7

8

16

4

16

8

16

1

9

S

11

3

13

'7

U

2

16

8

16

4

16

8

16

Figure 4.8

Fix p E (0, 1) and let e > 0 be given. To prove that lim f (x) = 0 we need to show X __*P

that there exists a 8> 0 such that

lf(x)I < E for all x E Na(p) 11(0, 1), x * p. This is certainly the case for any irrational number x.

On the other hand, if x is rational with x = m/n (in lowest terms), then f (x) = I/n. Choose n0 E 1\J such that 1/n,, < E. There exist only a finite number of rational numbers m/n (in lowest terms) in (0, 1) with denominator less than no. Denote these by

rl, ... , rk, and let

8=min{Ir1-pI:i= 1,.. ,k,r,*p}. (Note: Since p may be a rational number and thus possibly equal to r, for some i = 1, ... , k, we take the minimum of fir, - pI) only for those i for which r, # p). Thus 8 > 0, and if r E 0 fl Na(p) fl (0, 1), r * p, with r = m/n in lowest terms, then n ? n0. Therefore, if(r)I

< E.

Thus I f (x) I < e for all x E Na(p) fl (0, 1), x # p. If f and g are real-valued functions defined on a set E, we define the sum f + g, the difference f - g, the product fg, and the absolute value lfl off on E as follows: For x E E.

(f + g) (x) =1(x) + 8(x).

(f - 8)(x) = f(x) - 8(x),

134

Chapter 4

Limits and Continuity

(fg) (x) = f(x)g(x), Ifl(x) = V(X)1-

Furthermore, if g(x) * 0 for all x E E, we define the quotient fig by

\g/(x)

g(x)*

More generally, if f and g are real-valued functions defined on a set E. the quotient f/g can always be defined on E, = {x E E : g(x) * 0}. As an application of Theorem 4.1.6 we prove that continuity is preserved under the algebraic operations defined above. The proof that Ifl is continuous whenever f is continuous is left as an exercise (Exercise 6).

4.2.3

THEOREM

If E C R and f, g : E -> 68 are continuous at p E E, then

(a) f + g and f - g are continuous at p, and (b) f g is continuous at p. (c) If g(x) * 0 for all x E E, then fig is continuous at p.

Proof. If p is an isolated point of E, then the result is true since every function on E is continuous at p. If p is a limit point of E, then the conclusions follow from Theorem

4.1.6. Q Composition of Continuous Functions In the following theorem we prove that continuity is also preserved under composition of functions.

4.2.4 THEOREM Let A, B C l18 and let f : A -+ R and g : B - l be functions such that Range f C B. If f is continuous at p E A and g is continuous at f(p), then h = g e f is continuous at p.

Proof.

Let e > O be given. Since g is continuous at f (p), there exists a S, > 0 such that

lg(y) - g(f(p))I < e for ally E B fl NN,(f(p)).

(2)

Since f is continuous at p, for this S,, there exists a S > 0 such that

Lf(x)-f(p)I <S, for all xEAf1N8(p). Thus, if x E A with Ix - pl < S, by (2)

lh(x) - h(p)I = Ig(f(x)) - g(f(p))I < e. Therefore h is continuous at p.

4.2.5

EXAMPLES

(a) If p is a polynomial function on R, then by Example 4.1.10(a), lim p(x) = p(c) for every c E R. Thus every polynomial function p is continuous on i$.`~`

4.2

Continuous Functions

135

(b) Suppose p and q are polynomials on P and E _ {x E R : q(x) = 01. Then by Theorem 4.2.3 the rational function r defined on H \ E by r(x)=p(-,

q(x)

xER\E,

is continuous on H \ E.

(c) By Example 4.2.2(f), f(x) = sin x is continuous on P. Hence if p is a polynomial function on R, by Theorem 4.2.4, (f o p)(x) = sin(p(x)) is also continuous on R.

Topological Characterization of Continuity Before considering several consequences of continuity, we provide a strictly topological characterization of a continuous function. In more abstract courses this is often taken as the definition of continuity.

4.2.6

THEOREM Let E be a subset of P and let f be a real-valued function on E. Then f is continuous on E if and only if f -'(V) is open in E for every open subset V of R.

Proof. Recall (Definition 3.1.14) that a set U C E is open in E if for every p E U there exists a 8 > 0 such that Na(p) fl E C U. Suppose f is continuous on E and V is an open subset of R. If f ''(V) we are done. Suppose p E f -(V). Then f(p) E V. Since V is open, there exists e > 0 such

that N1(f(p)) C V. Since f is continuous at p, there exists a S > 0 such that f(x) E NE(f(p)) for all x E Na(p) fl E; i.e., Na(p) fl E C f-'(V). Since p E f -t(V) was arbitrary, f -'(V) is open in E. Conversely, suppose f -'(V) is open in E for every open subset V of R. Let p E E and let e > 0 be given. Then by hypothesis f -'(N,(f(p))) is open in E. Thus there exists a S > 0 such that

E fl Na(p) C f -'(Nf(f(p))); that is, f(x) E N.(f (p)) for all x E Na(p) fl E. Therefore f is continuous at p. O 4.2.7

EXAMPLES

(a) We illustrate the previous theorem for the function f(x) = \, Dom f = [0, oo). Suppose first that V is an open interval (a, b) with a < b. Then

f-,(V)=

1

¢,

b < 0,

[O,b2),

as0
(a2, b2),

0 < a.

Clearly 46 and (a2, b2) are open subsets of H and hence also of [0, oo). Although [0, b2) is not open in R,

[0, b2) = (-b2, b2) fl [0, oo).

Thus by Theorem 3.1.16, [0, b2) is open in [0, oo). If V is an arbitrary open subset of R, then by Theorem 3.1.13 (or Exercise 17, Section 3.1) V = U 1,,, where is a fi-

136

Chapter 4

Limits and Continuity

(Theorem nite or countable collection of open intervals. Since f -'(V) = U f is open in [0, oo), f -'(V) is open in [0, oo). Therefore 1.7.14) and each f f(x) _ N/x is continuous on [0, oo).

That f -'(V) is open in 10, x) for every open set V can also be proved as follows: Let p E f"'(V) be arbitrary. Since f(p) E V and V is open, there exists an open in-

terval I with f(p) E I C V. But then p E f -'(I) C f -(V). By the above, f -'(I) is open in [0, cc). Thus there exists a 8 > 0 such that N8(p) fl [0, x) C f -'(I) C f - '(V ). Since p E f -'(V) was arbitrary, f -'(V) is open in (0, oo). (b) In this example we show that if f : E -> R is continuous on E and V C E is open in E, then f(V) is not necessarily open in Range f. Consider the function f : R - R given by

f(x)

x2,

- {3-2x,

x < 1,

x> 1.

Then f is continuous on R and Range f = R (Exercise 10). However, f((-I, 1)) [0, 1), and this set is not open in R. (See Figure 4.9.)

I

Figure 4.9 Graph of f(x) _

X2

13-2x,

xx> I.

>I

Continuity and Compactness We now consider several consequences of continuity. In our first result we prove that the continuous image of a compact set is compact. In the proof of the theorem we use only continuity and the definition of a compact set. An alternative proof using the Heine-Borel-Bolzano-Weierstrass theorem (Theorem 3.2.9) is suggested in the exercises (Exercise 25).

4.2.8 THEOREM If K is a compact subset of P and if f : K -.> H is continuous on K, then f (K) is compact.

4.2

Continuous Functions

137

Proof. Let { V,},EA bean open cover of f(K). Since f is continuous on Kj -'(V.) is open in K for every a E A. By Theorem 3.1.16, for each a there exists an open subset U. of l such that

f -'(V,) = K n U,. We claim that {U,}0EA is an open cover of K. If p E K, then f(p) Ef(K) and thus f(p) E V. for some a E A. But then p is in j-'(V.) and hence also in U,. Since each U. is also open, the collection { U«}.eA is an open cover of K. Since K is compact, there

exists a,,.

.

.

, an E A such that

K C ,U U.. I

Therefore, n

n

K =U (u.fl K) = Utf j- -'(V,), j=t and by Theorem 1.7.14(a), n

f(K) = Uf(r'(V0,)) Since f (f -'(V.,)) C V,, f (K) C U; _ IV.,. Thus f (K) is compact. As a corollary of the previous theorem we obtain the following generalization of the usual maximum-minimum theorem encountered in calculus.

4.2.9

COROLLARY Let K be a compact subset of R and let f : K -+ R be continuous. Then there exist p, q E K such that

f(q)<_f(x)sf(p) forallxEK. Proof. Let M = sup{f(x) : x e K}. By the previous theorem f(K) is compact. Thus, since f(K) is bounded, M < oo. Also, since f(K) is closed, M E f(K). Thus there exists p E K such that f(p) = M. Similarly for m = inf{f(x) : x E K}. We now provide examples to show that the result is false if K C 68 is not compact: that is, not both closed and bounded.

4.2.10

EXAMPLES

(a) Suppose E is a subset of R. If E is unbounded, consider

f(x) =

x2

I +x" Then f is continuous on E, sup { f (x) : x E E} = 1, but f(x) < I for all x e E. To see that the supremum is 1, we note that since E is unbounded, there exists a sequence {xn} in E such that x2. --* oo as n -+ oo. But then

lim f(zn) = lim

R-00

n-xl + X 1

1.

138

Chapter 4

Limits and Continuity

Thus if 0 < (3 < 1, there exists an integer n such that f (x.) > p. Hence sup{ f (x) : x E E} = 1. The graph of f with E = (0, oo) is given in Figure 4.10.

I

4 1

2

3

E=(0.-)

Figure 4.10 Graph of f(x) =

4

x2

+z2

5

6

x E [0,oo)

(b) If E is not closed, let x be a limit point of E that is not in E. Then 1

8(x) _ .

1 + (x - x,)2

is continuous on E, with g(x) < 1 for all x e E. A similar argument shows that sup{g(x) : x E E} = 1. This is illustrated in Figure 4.11 with E = (0, 1) and xo = 1.

Intermediate Value Theorem The following theorem is attributed to both Bolzano and Cauchy. Cauchy, however, im-

plicitly assumed the completeness of R in his proof, whereas the proof by Bolzano (given below) uses the least upper bound property. An alternative proof is outlined in the miscellaneous exercises.

4.2.11

THEOREM (intermediate Value Theorem) Let f: [a, b) -> R be continuous. Suppose f(a) < f(b). If y is a number satisfying

f(a) < y < f(b), then there exists c E (a, b) such that f(c) = y.

4.2 Continuous Functions

139

I

0.5

E=[0, I)

0

Figure 4.11

Graph of g(x) =

I

I+

(x

- 1)2, x E (0, 1)

The statement and conclusion of the intermediate value theorem is illustrated in Figure 4.12. The theorem simply states that if f is continuous on [a, b] with f(a) < f(b), and y E 68 satisfies that f(a) < y < f(b), then the graph off crosses the line y = y at least once. Proof.

Let A = {x E (a, b) : f(x) 5 y}. The set A * 45 since a E A, and A is

bounded above by b. Thus by the least upper bound property, A has a supremum in R. Let c = sup A. Since b is an upper bound, c s b. We now show that f(c) = y. Since c = sup A, either c E A or c is a limit point of A. If c e A, then f(c) < y. If c is a limit point of A, then by Theorem 2.4.7 there exists a sequence in A such that x -* c. Since x. E A, f 5 y. Since f is continuous,

f(c) = lim f(x,,) <_ y Thus in either case, f(c) s y. Suppose f(c) < y. Then e = 2(y - f(c)) > 0. Since f is continuous at c, for this e there exists a S > 0 such that

f(c) - e < f(x) < f(c) + e

for all x E

fl [a, b].

Since f(c) < y, c # b. Thus (c, b] fl N5(c) # 0. But for any x E (c, b] with c < x <

c + S,

1(x)
140

Chapter 4

Limits and Continuity

Figure 4.12 Intermediate Value Theorem

But then x E A and x > c, which contradicts c = sup A. Therefore f (c) = y. The intermediate value theorem is one of the fundamental theorems of calculus. Simply stated, the theorem implies that if I is an interval and f : 1 R is continuous. then f(!) is an interval. Due to the importance of this result we state it as a corollary.

4.2.12

COROLLARY If I C R is an interval and f :I -> R is continuous on 1, then f(1) is an interval.

Proof. Let s, t e f(I) with s < t, and let a, b E I with a * b be such that f(a) = s and f(b) = t. Suppose y satisfies s < y < t. If a < b, then since f is continuous on [a, b], by the intermediate value theorem there exists c E (a, b) such that f(c) = y. Thus y E f(1). A similar argument also holds if a > b. There is an alternative way to state the previous corollary using the terminology of

connected sets. If I is a connected subset of R and f : I-+ R is continuous on I, then f(!) is connected. This result can be proved using only properties of continuous functions and the definition of a connected set (Exercise 28). The corollary now follows from the fact that a subset of R is connected if and only if it is an interval (Theorem 3.1.18). The proof of Theorem 3.1.18, however, also requires the least upper bound property of R. Consequently, the supremum property of the real numbers cannot be avoided in proving Corollary 4.2.12. The following two corollaries are additional applications of the intermediate value theorem. Our first result is the proof of Theorem 1.5.3.

4.2.13

COROLLARY For every real number y > 0 and every positive integer n, there exists a unique positive real number y so that y° = y.

4.2

Continuous Functions

141

Proof. That y is unique is clear. Let f(x) = x", which by Example 4.2.5(a) is continuous on R. Let a = 0 and b = y + 1. Since (y + 1)" > y, f satisfies the hypothesis of Theorem 4.2.11. Thus there exists y, 0 < y < y + 1, such that

f(y) = y" = y

4.2.14

COROLLARY

If f : [0, 1 ] -* [0, 1 ] is continuous, then there exists y E [0, 1 ] such

that Ay) = y.

Proof. Let g(x) = f(x) - x. Then g(0) = f(0) ? 0 and g(1) = f(1) - I there exists y E [0, 11 such that g(y) = 0; i.e., f(y) = y. Q 4.2.15

0. Thus

EXAMPLES

(a) In the proof of Theorem 4.2.11 continuity of the function f was required. The following example shows that the converse of Theorem 4.2.11 is false; that is, if a function f satisfies the intermediate value property on an interval [a, b], this does not imply that f is continuous on [a, b]. Let f be defined on [0, 2] as follows,

0<xs?,

- -1,

x = 0.

1

f(x) =

x

or,

Then f(0) _ -1, f(r) = 1, and for every y, -1 < y < 1, there exists an x E (0, such that f(x) = y. However, the function f is not continuous at x = 0 (see Figure 4.3).

(b) In this example we show that the conclusion of the intermediate value theorem is false if the interval [a, b] of real numbers is replaced by an interval of rational numbers.

Let E _ {x E 0 : 0 5 x < 2}, and let f (x) = x2. Then f is continuous on E with f(O) < 2 < f(2). However, there does not exist r E E such that f(r) = 2. 0

EXERCISES 4.2 1. For each of the following, determine whether the given function is continuous at the indicated point x".

2x2-5x-3 x03, f(x)

o

x 6, 3

1- cos x `c. g(x) =

x

x= 3,

x# 0,

x = 0, 2. Let f : R -' R be defined by 0,

f(x) _

8x,

2x2 + 8,

at x" = 3

at x° = 0

when x is rational, when x is irrational.

a. Prove, using e and 8, that f is continuous at 2. b. Is f continuous at 1? Justify your answer.

h(x)

d. k(x)

Vc-2 x#4,

_

x

-

J

4

4

x2,

1 4 - x,

x

x 5 2,

x>2.

4

at x" = 4

at x, =.2

142

Limits and Continuity

Chapter 4

3. Let f : R -r R be defined by

x E Q,

x2,

f(x) __

x+2, x0Q.

Finds all points (if any) where f is continuous. 4. *Prove (without using Example 4.2.7) that f(x) = V is continuous on [0, oo). 5. Define f : (0, 1 ] - R by

f

x+l x

a. Justify that f is continuous on (0, 1]. b, Can one define f(0) so that f is continuous on [0, 1]? 6. Let E C R, and suppose f : E --+ R is continuous at p e E. *a. Prove that Ifl is continuous at p. Is the converse true?

b. Set g(x) =, x E E. Prove that g is continuous at p. 7. *Let E C R and suppose f : E-*R is continuous on E. Prove that f" defined by f'(x) _ (f(x))" is continuous on E for each n E N. 8. a. Prove that f (x) = cos x is continuous on R. b. If E C R and f : E -+ R is continuous on E, prove that g(x) = cos (fix)) is continuous on E. 9. For each of the following equations, determine the largest subset E of R such that the given equation defines a continuous function on E. In each case state which theorems or examples are used to show that the function is continuous on E.

*a. AX) =

x

4x

4)5

d. k(x) =cos

*c.

h(x)

b. g(x) = sinz

sin x

x2 +

10. Prove that

f(x) _

x2,

x s 1,

{ 32x, x> 1,

is continuous on R and that Range f = R. 11. As in Example 4.2.7(a), use Theorem 4.2.6 to prove that each of the following functions is continuous on the given domain.

a. f(x) =

1, X

Dom f = (0, oo)

b, g(x) = x2, Dom g = R

12. Suppose E is a subset of R and f, g : E -+ R are continuous at p E E. Prove that each of the functions defined below is continuous at p. *a. max{f, g}(x) = max{f(x), g(x)}, x E E b. min{f, g}(x) = min{f(x), g(x)}, x E E

c. f'(x) = max{f(x), 0), x E E 13. Suppose f : [a, b] -* R and g : [b, c] - R are continuous on [a, b] and [b. c], respectively. Leth : [a. c] -+R be defined by

h(x) _ {f

(x a s x s b,

g(x).

b < x s c.

Prove that h is continuous on [a, c] if and only if f(b) = g(b).

4.2

Continuous Functions

143

14. Use the intermediate value theorem to prove that every polynomial of odd degree has at least one real zero.

15. Prove that there exists x E (0, z) such that cos x = x. 16. *Suppose f : [ -1, 1 ] -+ R is continuous and satisfies f(- I) = f(l). Prove that there exists y E [0, 17 such that

f(Y) = f(Y - 1) 17. Suppose f : [0, 1 ] --? R is continuous and satisfies f(0) = f(l). Prove that there exists y E (0, ;] such that ICY) = f(y + D. 18. *Let E C R and let f : E -a R be continuous. Let F = {x E E : f(x) = 0}. Prove that F is closed in E. Is F necessarily closed in R? 19. Suppose f : (0, 1) -- R is continuous and satisfies f(r) = 0 for each rational number r E (0, 1). Prove that

f(x) = 0 for all x E (0, 1). f(x) + f(y) for all x, y E R. a. If f is continuous at some p E R, prove that f is continuous at every point of R. *b. If f is continuous on R, prove that f(x) = cx for all x E R, where c = f(1). 21. Let E C R and let f be a real-valued function on E that is continuous at p E E. If f(p) > 0, prove that there exists an a > 0 and a S > 0 such that f(x) 2- a for all x e N$(p) fl E. 22. *Let f : E.- R be continuous at p E E. Prove that there exists a positive constant M and S > 0 such that

20. Let f be a real-valued function on R satisfying f(x + y)

lf(x)I sMforallxEEflN6(p). 23. Let f : (0, 1) --> R be defined by

f(x) _

-

0, In,

if x is irrational, if x is rational with x=m/n in lowest terms.

a. Prove that f is unbounded on every open interval I C (0, 1). b. Use part (a) and the previous exercise to conclude that f is discontinuous at every point of (0, 1). 24. Suppose E is a subset of R and f, g : E -- R are continuous on E. Show that {x E E : f(x) > g(x)} is open in E. 25. *Let K be a compact subset of R and let f : K -a R be continuous on K. Prove that f(K) is compact by showing that f(K) is closed and bounded.

26. Let E C R and let f be a real-valued function on E. Prove that f is continuous on E if and only iff-'(F) is closed in E for every closed subset F of R. 27. Let A, B C R and let f : A -- ). R and g : B -a R be functions such that Range f C B.

a. If V C R, prove that (g -f)-'(V) = f-'(g-'(V)). b. If f and g are continuous on A and B respectively, use Theorem 4.2.6 to prove that g of is continuous on A. 28. Suppose 1 is a connected subset of R and f : I--> R is continuous on I. Prove, using only the properties of continuity and the definition of connected set, that f(1) is connected.

29. *Let K C R be compact and let f be a real-valued function on K. Suppose that for each x E K there exists e, > 0 such that f is bounded on NE,(x) n K. Prove that f is bounded on K.

30. Let A C R. For p E R, the distance from p to the set A, denoted d(p, A), is defined by d(p, A) = inf{Ip - xJ : x E A}. a. Prove that d(p, A) = 0 if and only if p E A. b. For x, y E R, prove that Id(x, A) - d(y, A) Jx - y1. e. Prove that the function x -+ d(x, A) is continuous on R. d. If A, B are disjoint closed subsets of R, prove that d(x, A) AX) = d(x, A) + d(x, B) (x)

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Limits and Continuity

is a continuous function on R satisfying 0 s f(x) 0, x E A.

1 for all x E R. and

f(x) - 11, xEB.

31. Let f be a continuous real-valued function on R satisfying f(0) = I and f(x + v) = f(x)f(v) for all x.y E R. Prove that f(x) = a` for some a E R. a > 0.

Uniform Continuity In the previous section we discussed continuity of a function at a point and on a set. By Definition 4.2.1, a function f : E -i R is continuous on E if for each p E E, given any

e > 0, there exists a S > 0 such that Lf(x) - f(p)I < e for all x E E fl N5(p). In general, for a given e > 0, the choice of S that works depends not only on a and the function f, but also on the point p. This was illustrated in Example 4.1.2(g) for the function f(x) = 1/x, x E (0, oo). Functions for which a choice of S independent of p is possible are given a special name.

4.3.1

DEFINITION

Let E C P and f : E - R. The function f is uniformly continuous

on E if given e > 0, there exists a S > 0 such that 1f(x) - fCv)I < e

for all x,vEEwith x-v1 <S. The key point in the definition of uniform continuity is that the choice of S must depend only on e, the function f and the set E; it has to be independent of any x E E. To illustrate this, we consider the following examples.

4.3.2

EXAMPLES

(a) If E is a bounded subset of R, then f(x) = .r` is uniformly continuous on E. Since E is bounded, there exists a positive constant C so that 1x1 C for all x E E. If x, y E E, then

Lf(x) - f(v)I = 1x2 - y2I = Ix + yI Ix - v1

(Ix1 + yI)Ix - yI 2CIx - v1. Let e > 0 be given. Take S = e/2C. If x, y E E with Ix - yI < S, then by the above

Lf(x) -f(y)1 s 2CIx - yI < 2C8 = Therefore, f is uniformly continuous on E. In this example, the choice of S depends both on e and the set E. In the exercises you will be asked to show that this result is false if the set E is an unbounded interval.

(b) Let f(x) = sinx. As in Example 4.2.2(f),

Lf(y) -f(x)I < Iy - xI for all x, y E R. Consequently, f is uniformly continuous on R.

4.3

Uniform Continuity

145

(c) In this example we show that the function f(x) = I /.r, x E (0. oo) is r= uniformly continuous on (0, oo). Suppose, on the contrary, that f is uniformly continuous on (0, oo). Then, as in Example 4.1.2(g), if we take e = 1, there exists a S > 0 such that If(x) - f(Y) I

X- y < I

for all x, y E (0,-oo) with Ix - yl < S. Choose no E Rl such that 1/n < S, and for n E f i set x = 1/n. Then for all n ? no, X. - X.+1 = I

n-n+1 <

n

<

but

If(x.) -f(xn+i)I = In - (n + 1)I = 1. This however contradicts that If(x) - f(y) I < 1 for all x, y E (0. oo) with Ix - yl < S. Thus f is not uniformly continuous on (0, oo). The function f(x) = l/x, however, is uniformly continuous on (a, oo) for every fixed a > 0. Suppose x, y E [a, oo). Then

[f(x) - f(y)I = ly xyx l 5 a21x - yl. Hence, given e > 0, if we choose S such that 0 < S < a2e, then as a consequence of the above inequality, if(x) - f(y)I < e for all x, y E [a. oo) with Ix - yi < S.

Lipschitz Functions Both of the functions in Example 4.3.2(a) and (b), and the function f(x) = 1/x with Dom f = [a, oo), a > 0, are examples of an extensive class of functions. If E C R, a function f : E -+ R satisfies a Lipschitz condition on E if there exists a positive constant M such that [1(x) - f(Y) I < Mix - Y I for all x, y E E. Functions satisfying the above inequality are usually referred to as Lipschitz functions. As we will see in the next chapter, functions for which the derivative is bounded are Lipschitz functions.. As a consequence of the following theorem, every Lipschitz function is uniformly continuous. However, not every uniformly continuous

function is a Lipschitz function. For example, the function f(x) = V is uniformly continuous on (0, oo), but f does not satisfy a Lipschitz condition on [0, oo) (see Exercise 5).

4.3.3

THEOREM Suppose E C R 'and f : E - R. If there exists a positive constant M such that

V(x) - f(y)I <- MIx - yI for all x, y E E, then f is uniformly continuous on E.

146

Chapter 4

Limits and Continuity

Proof. Exercise 1. O

Uniform Continuity Theorem If the function f does not satisfy a Lipschitz condition on E, then determining whether f is uniformly continuous on E is much more difficult. The following theorem provides a sufficient condition on the set E such that every continuous real-valued function on E is uniformly continuous.

4.3.4 THEOREM If K C R is compact and f : K -+ R is continuous on K, then f is uniformly continuous on K.

Let e > 0 be given. Since f is continuous, for each p E K, there exists a 6P > 0 such that

Proof.

(3)

If(X) - f(p) I < 2 for all x E K fl N2s,(p)

The collection {Nso(p)},eK is an open cover of K . Since K is compact, a finite number of these will cover K. Thus there exist a finite number of points P i . . . . . p in K such that

K C U Ns,(pi). i=[ Let

S = min{SP, : i = 1,

, n).

Then S > 0. Suppose x, y E K with Ix - yi < S. Since x E K, x E N,&(pi) for some i. Furthermore, since k - yI < S s Sp, x, y E Nzs,.(Pi)

Thus by the triangle inequality and inequality (3),

LAX) - f(y)I C If(x) - f(pi)I + U(pa) - f(y)I < 2 + 2 = e.

4.3.5

COROLLARY A continuous real-valued function on a closed and bounded interval [a, b] is uniformly continuous. The definition of uniform continuity and the proof of Corollary 4.3.5 appeared in a paper by Eduard Heine in 1872.

4.3.6

EXAMPLE In this example, we show that the properties closed and bounded are both required in Corollary 4.3.5. The interval [0, oo) is closed, but not bounded. The function f(x) = x2 is continuous on (0, oo), but not uniformly continuous on [0, co) (Exercise 2). On the other hand, the interval (0, 1) is bounded, but not closed. The function f(x) = 1/x is continuous on (0, 1), but is not uniformly continuous on (0, 1).

4.3

Uniform Continuity

147

EXERCISES 4.3 1. Prove Theorem 4.3.3. 2. Show that the following functions are not uniformly continuous on the given domain.

*a. AX) = x2, Dom f = [0, oo)

b, g(x) =

I.

Dom g = (0, oo)

c. h(x) = sin x,

Dom h = (0, co)

3. Prove that each of the following functions is uniformly continuous on the indicated set.

*a. AX)

1 + x , x E [0. oo)

c. h(x) = x2

xER

+

e. e(x) = x + 1, x E (0, oo)

b. g(x) _ .r'-,

x E hi

d. k(x) = cos x, x E R

*1 f(x) =

sinx,

x E (0, I)

4. Show that each of the following functions is a Lipschitz function.

*a. f(x) =

,

Dom f = [a, oo), a > 0

c. h(x) = sin x, Dom h = [a, oo), a > 0

b. g(x) = x2 + 1, Dom g = (0. oo)

d. p(x) a polynomial, Dom p = [ -a, a], a > 0

5. *a. Show that f(x) _ \ satisfies a Lipschitz condition on [a, oo), a > 0. b. Prove that V is uniformly continuous on (0, oo). c. Show that f does not satisfy a Lipschitz condition on (0, oo). 6. Suppose E C R and f, g are Lipschitz functions on E. a. Prove that f + g is a Lipschitz function on E. b. If in addition f and g are bounded on E, or the set E is compact, prove that fg is a Lipschitz function on E.

7. Suppose E C R and f. g are uniformly continuous real-valued functions on E. a. Prove that f + g is uniformly continuous on E. *b. If, in addition, f and g are bounded, prove that fg is uniformly continuous on E. c. Is part (b) still true if only one of the two functions is bounded? is a Cauchy sequence in E, prove that { f(x )} is a 8. Suppose E C R and f : E - R is uniformly continuous. If Cauchy sequence.

9. Let f : (a, b) -+ R be uniformly continuous on (a, b). Use the previous exercise to show that f can be defined at a and b such that f is continuous on [a, b]. 10. Suppose that E is a bounded subset of R and f : E - R is uniformly continuous on E. Prove that f is bounded on E.

11. Suppose -oo s a < c < b <_ oo, and suppose f : (a, b) - R is continuous on (a. b). a. If f is uniformly continuous on (a, c) and also on (c, b), prove that f is uniformly continuous on (a, b). b. Show by example that the conclusion in part (a) may be false if f is not continuous on (a, b).

12. Let a E R. Suppose f is a continuous real-valued function on [a, oo) satisfying lim f(x) = L. where L E R. Prove that

a, f is bounded on [a, oo), and b. f is uniformly continuous on (a, oo).

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Chapter 4

Limits and Continuity

13. Let E C R. A function f : E E is contractive if there exists a constant b, 0 < b < I. such that If(x) - f(y)I b Ix - yI. *a. If E is closed, and f : E -* E is contractive, prove that there exists a unique point x E E such that (Such a point xo is called a fixed point of f.)

x,,.

b. Let E _ (0,;]. Show that f(x) = r' is contractive on E. but that I 'does not have a fixed point in E.

14. A function f : Il -> R is periodic if there exists p E R such that f(x + p) = f(x) for all x E R. Prove that a continuous periodic function on R is bounded and uniformly continuous on R.

44.4

Monotone Functions and Discontinuities In this section we take a closer look at both limits and continuity for real-valued functions defined on an interval 1 C R. More specifically, however, we will be interested in classifying the types of discontinuities that such a function may have. We will also investigate properties of monotone functions defined on an interval I. These functions will play a crucial role in Chapter 6 on Riemann-Stieltjes integration. First, however, we begin with the right and left limits of a real-valued function defined on a subset E or R.

Right and Left Limits 4.4.1

Let E C I8 and let f be a real-valued function defined on E. Suppose p is a limit point of E fl (p, oo). The function f has a right limit at p if there exists a number L E IB such that given any e > 0, there exists a S > 0 for which DEFINITION

Lf (x) - LI < e for all x E E satisb,ing p < x < p + S. The right limit of f, if it exists, is denoted by f (p +), and we write lim f(x). A p+) = x lim- Ax) = x-.p x>p

Similarly, if p is a limit point of En( - oo, p), the left limit off at p, if it exists, is denoted by f( p- ), and we write

f(p-) = lim.f(x) = limf(x). x
The hypothesis that p is a limit point of E fl (p, oo) guarantees that for every

S > 0, E n (p, p + S) * 4i. If E is an open interval (a, b), -oo < a < b s oo, then any p satisfying a s p < b is a limit point of E fl (p, oo). Similarly, if -oo <- a < b < oo, then any p satisfying a < p s b is a limit point of (-oo, p) fl E. If 1 is any interval with Int(!) # 0, and f : I -> R, then f has a limit at p E lnt(I) if and only if

(a) f (p+) and f (p-) both exist, and

(b) f(p+) =f(p-). The hypothesis that p E Int(!) guarantees that p is a limit point of both (-oo, p) fl 1 and I fl (p, oo). If p is the left endpoint of the interval 1, then the right limit of f at p

4.4 Monotone Functions and Discontinuities

149

coincides with the limit off at p. The analogous statement is also true if p is the right endpoint of I. We also define right and left continuity of a function at a point p as follows.

4A.2

Let E C R and let f be a real-valued function on E. The function f is right continuous (left continuous) at p E E if for any e > 0, there exists a S > 0 DEFINITION

such that

If(x) -f(p)I < e

for all xEEwith p<_x 0 such that E fl (p, p + S) = 4,. Thus if f': E -+ R and e > 0 is arbitrary, then If (x) - f (p)l < e for all x E. E fl ['p, p + S). Thus every f : E -> R is right continuous at p. In particular, if E is a closed interval [a, b], then every f : [a, b] -+ R is right continuous at b. Also, f is left continuous at b if and only if f is continuous at p. The following theorem, the proof of which is left to the exercises, is an immediate consequence of the definitions.

4A.3 THEOREM A function f : (a, b) -+R is right continuous at p E (a, b) if and only if f (p+) exists and equals f (p). Similarly, f is left continuous at p if and only if f (p- ) exists and equals f (p).

Proof.

Exercise 1. Q

Types of Discontinuities By the previous theorem a function f is continuous at p E (a, b) if and only if

(a) f (p+) and f (p-) both exist, and

(b) f(p+) =f(p-) =f(p). A real-valued function f defined on an interval I can fail to be continuous at a point p E 1 (the closure of I) for several reasons. One possibility is that lim f(x) exists but xZ either does not equal f (p), or f is not defined at p. Such a function can easily be made continuous at p by either defining or redefining f at p as follows: lim f(x). f(p) = X-P

For this reason, such a discontinuity is called a removable discontinuity. For example, the function

-4

g(x)

x-2'

x #.2,

2,

x = 2,

of Example 4.2.2(a) is not continuous at 2 since

1 m g(x) = 4 * g(2).

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Chapter 4

Limits and Continuity

By redefining g such that g(2) = 4. the resulting function is then continuous at 2. Another example is given by f (x) = x sin(] /x), x E (0, oo), which is not defined at 0. If we define f on [0, oo) by

x=0

0,

f(x)

x sin

i , x > 0,

x

then by Example 4.2.2(e), f is now continuous at 0.

Another possibility is that f(p+) and f(p-) both exist, but are not equal. This type of discontinuity is called a jump discontinuity. (See Figure 4.13.)

4AA DEFINITION Let f be a real-valued function defined on an interval I. The function f has a jump discontinuity at p E Int(1) if f(p+) and f (p-) both exist, but f is not continuous at p. If p E I is the left (right) endpoint of 1, then f has a jump discontinu-

ity at p if f (p+) (f( p- )) exists, but f is not continuous at p.

Ap+)

AP-)

p

Figure 4.13 Jump Discontinuity off at p

Jump discontinuities are also referred to as simple discontinuities, or discontinuities of the first kind. All other discontinuities are said to be of second kind. If f( p+ ) and f (p-) both exist, but f is not continuous at p, then either

(a) f(p+) # f(p-), or

(b) f(p+) =f(p-) #f(p)

4.4

Monotone Functions and Discontinuities

151

In case (a) f has a jump discontinuity at p, whereas in case (b) the discontinuity is removable. All discontinuities for which f (p+) or f (p-) does not exist are discontinuities of the second kind.

4.4.5

EXAMPLES

(a) Let f be defined by x,

0<x-1,

f(x)= 3-x2, x> 1.

The graph off is given in Figure 4.14. If x < 1, then f(x) = x. Therefore,

f(I-)= lint f(x) = limx= 1 =f(1). X +F X-1 Likewise, the right limit off at 1 is

2

1

Figure 4.14

f(1+)=Xlimf(x)=lim(3-x2)=2. Therefore, f (I -) = f (l) = 1, and f (I +) = 2. Thus f is left continuous at 1, but not continuous. Since both right and left limits exist at 1, but are not equal, the function f has a jump discontinuity at 1.

(b) Let [x] denote the greatest Integer function; that is, for each x, [x] = largest integer n that is less than or equal to x. For example, [2.9] = 2,[3.1) = 3, and [

1.5] = -2. The graph of y = [x] is given in Figure 4.15. It is clear that for each n E Z,

lim [x] = n - 1

and

lim. [x] = n.

Thus f has a jump discontinuity at each n E Z. Also, since f (n) = [n] = n, f (x) = [x]

is right continuous at each integer. Finally, since f is constant on each interval (n - 1, n), n E Z, f is continuous at every x E R\7L.

152

Chapter4

Limits and Continuity

34-

24-

I +

-1

2

1

3

00 -1 Figure 4.15

Graph of [x]

(c) Let f be defined on R by if x<_0,

0,

f(x) =

sin

X

if x > 0.

Then f(0-) = 0, but f(0+) does not exist. Thus the discontinuity is of second kind. (d) Consider the function g : R -+ R defined by g(x) = sin (21rx[x]).

For x E (n, n + 1), n E Z, x[x] = nx, and thus g(x) is continuous on every interval (n, n + 1), n E Z. On the other hand, for n E 71,

lim sin(2ax[x]) = sin(21rn2) = 0, sn+

and

lim sin(2irx[x]) = sin(2an(n - 1)) = 0. Since g(n) = sin(21rn2) = 0, g is also continuous at each n E Z. Thus g is a bounded continuous function on R. The function g, however, is not uniformly continuous on R

(Exercise 7). The graph of g for x E (-4.4) is given in Figure 4.16. 0

Monotone Functions 4.4.6

DEFINITION

Let f be a real-valued function defined on an interval I.

(a) f is monotone increasing (increasing, nondecreasing) on I if f(x) < f(y) for all x, y E I with x < y.

4.4

AAIA

I

A

Monotone Functions and Discontinuities

153

A

AA

IV

TV-11-

Figure 4.16

A

V

Graph of g(x) = sin(2ax[x]), x E (-4,4)

(b) f is monotone decreasing (decreasing, nonincreasing) on I if f (x) at f (y)

for all x, y E I with x < y. (c) f is monotone on 1 if f is monotone increasing on I or monotone decreasing on I.

A function f is strictly increasing on I if f (x) < f (y) for all x, y E I with x < y. The concept of strictly decreasing is defined similarly. Also, f is strictly monotone on I if f is strictly increasing on I or strictly decreasing on I. Our main result for monotone functions is as follows.

4A.7 THEOREM Let I C R be an open interval and let f : I -+ H be monotone increasing on I. Then f (p+) and f (p-) exists for every p E I and

inff(x). supf(x) =f(p-) :5f(p):5 f(p+) = p<x x
Furthermore, if p,q E I with p < q, then f(p +) 5 f(q-). Although we stated the theorem for monotone increasing functions, a similar statement is also valid for monotone decreasing functions.

Proof. i+ix p E I. Since f is increasing on 1, {f(x) : x < p, x E I} is bounded above by f (p). Let

A = sup{ f (x) : x < p, x e I}. Then A

f (p). We now show that

lim f (x) = A.

X- +P

The proof of this is similar to the proof of Theorem 2.3.2. Let e > 0 be given. Since A is the least upper bound of {f (x) : x < p}, there exists xo < p such that

A-e
Lf(x) - AI < e x+p

f(x) = A. Similarly,

f(p) f(p+) = inf{f (x) : p < x, x E l}.

154

Chapter 4

Limits and Continuity

Finally, suppose p < q. Then

f(p+)=inf{f(x):p<x,xE1}sinf{f(x):p<x
4.4.8

sup { f (x) : x < q, x E l } = f (q - ).

COROLLARY If f is monotone on an open interval 1, then the set of discontinuities

off is at most countable.

Proof. Let E

p E 1: f is discontinuous at p}. Suppose f is monotone increasing

on I. Then

p E E if and only if f (p-) < f(p+). For each p E E, choose r1, E 0 such that

f(p-) < r, <.f(p+). If p < q, then f(p+) s f(q-). Therefore, if p. q E E, r, * r,, and thus the function p -), ro is a one-to-one map of E into Q. Therefore E is equivalent to a subset of 0 and thus is at most countable.

Construction of Monotone Functions with Prescribed Discontinuities We now proceed to show that given any finite or countable subset A of (a, b), there exists a monotone increasing function f on [a, b] that is discontinuous at each x E A and continuous on [a, b] \A. We first illustrate how this is accomplished for the case when a finite subset of (a, b). To facilitate this construction, we define A = {a,, ... , the unit jump function Ion R as follows.

4.4.9

DEFINITION

The unit jump function 1: R -> R is defined by

1(x) = 10.

1,

when x < 0, when x ? 0.

The function I is right continuous at 0 with 1(0+) = 1(0) = 1 and 1(0-) = 0. For

aER,set la(x) = I(x - a) =

0 1,

when x < a, when x ? a.

Then la has a unit jump at a and is right continuous at a (see Figure 4.17). Suppose {c,, ... , are positive real numbers. Define f on [a, b] by

f(x) _

ck1(x - ak). k- l

The reader should verify that the function f is

(a) monotone increasing on [a, b],

(b) continuous on [a, b]\{a,, a2, ... , aj,

4.4

Monotone Functions and Discontinuities

155

I

a

Figure 4.17

Graph of 1(x - a)

(c) right continuous at each ak, k = 1, 2, ... , n, and (d) discontinuous at each ak with f(ak+) - flak-) = ck for all k = 1. 2, ... , n. of (a, b) is not surprising. That such a function exists f o r any finite subset {a,, ... , However, that such a function exists for any countable subset A of (a, b) may take some convincing, especially if one takes A to be dense in [a, b]; e.g., the rational numbers in (a, b).

4A.10 THEOREM Let a, b e R and let

be a countable subset of (a, b). Let be any sequence of positive real numbers such that 7,0,0. , c converges. Then there exists a monotone increasing function f on [a, b) such that c,,, (a) f(a) = 0 and f(b) _ (b) f is continuous on [a, b]\ {x : n = 1, 2, ...},

for all n, i.e., f is right continuous at all x,,, and (d) f is discontinuous at each x with (c)

f(xJ Proof.

For each x E [a, b], define 00

AX) _ 7, c 1(x =t

Since 0 s c I(x -

x E [a, b], we have 00

Ck <

Ckl(X - Xk) S

s (X) = k-t

Ck.

k=t

Thus for each x E [a, b], the sequence {s (x)} of partial sums is monotone increasing and bounded above, and hence by Theorem 2.7.6 converges. Since

n= 1,2,.. .

156

Chapter 4

Limits and Continuity

for all x, y with x < y, f is monotone increasing on (a, b). Furthermore, since x > a for all n, 1(a 0 for all n. Therefore f(a) = 0. Also, since 1(b I for all n,

f(b) = 7, C. k-l

This proves (a).

We now prove (b). Fix p E [a, b), p # x for any n. Let E = {x : n E N}. There are two cases to consider.

(1) Suppose p is not a limit point of E. If this is the case, there exists a S > 0 such

that NB(p)flE= ¢. Then

I(x-xk)=1(p-xk) forallxE(p-6,p+6) and all k = 1, 2, .... Thus f is constant on (p - 8, p + 8) and hence continuous. (ii) Suppose p is a limit point of E. Let e > 0 be given. Since the series 7, 1 Ck converges, by the Cauchy criterion there exists a positive integer N such that Ck < E.

k-N+1

Choose S such that

0 < S < min{1p -

n = 1, 2, ... , N}.

With this choice of 8, if xk e N6(p) fl E, we have k > N. Suppose p < x < p + S. Then

1(p-xk) =I(x-xk) forallk= 1,2,...,N. Furthermore, for any x > p, we always have

021(x-xk)-1(p-xk): 1 forallk6N. Therefore, for p < x < p + S, 00

0 t f(x) - f(P) 5

'0 k-N+l

Ck(I(x - xk) - 1(P - xk)) 5.7, Ck < E. k=N+1

Thus f is right continuous at p. Similarly, f is left continuous at p, and therefore f is continuous at p.

For the proof of (c), fix an x E E. If x is an isolated point of E, then as above, for all y, x < y < x + S. Thus f(x ). Suppose x is a limit point of E. Let e > 0

there exists a S > 0 such that En (x,,, x + S) = ¢. Therefore, f(y) = be given. Again, choose a positive integer N so that

I 00

kN+l

Ck < E.

4.4

Monotone Functions and Discontinuities

157

As in (b), there exists a S > 0 such that if xk. E (x,,, x,, + S) fl E, then k > N. Thus a 7, ck < e for all y E (x,, x + S). 0 s f(y) k=N+l

f is right continuous at each x,,. Therefore, For the proof of (d), suppose y < x,,. Again, if x is an isolated point of E, there ex-

ists a S > 0 such that (x - S, for all y <

n E = 4. Therefore, for all k * n,

xn,

I(xn-xn)=1(0)= 1. Therefore,

f(xn) - f(y) = cn

for all y, x - S < y < x,,.

N+I ck < e. For

Suppose x is a limit point of E. Given c > 0, choose N such that

fl E then k > N. Then for all

this N, choose S > 0 such that if xk E (x - S,

<x,,

yE[a,b] with C.

f(xn) - f(y) < C. + 7, ck < C. + e. k-N+1

Therefore, f(xn) - f(xn-) = Cn O

4A.11

EXAMPLES

( a ) T a k e c = 2', x = 1 = 1/(n + 1), n = 1, 2, ... , and (a, b) = (0, 1). As in Theorem 4.4.10, let 00

Ax) _ 71

n-1

< I. If 0 < x <

In this example, the sequence {xn} satisfies 0 < xt < x2 < 0 for all n. Thus xt = 1, then I(x -

f(x)=0, xE [0'2). If x1

x < x2 = 3, then !(x - x1) = 1 and !(x - xk) = 0 for all k =- 2. Therefore,

f(x)=c1=2,

X

If x2 5 x < x3 = 4, then I(x - xk) = 1 for k = 1,2 and I(x - xk) = 0 fork a 3. Therefore, AX)

+c2=

1

2+

34,

xE

and so forth. The graph off is depicted in Figure 4.18.

J23. 34 ,

158

Chapter 4

Limits and Continuity

7

0-0

8

.-o

4

2

S

0

4

0 I

2

3

4

2

3

4

S

Figure 4.18

(b) Let c" = 2-", and let {x"} be an enumeration of the rationals in (0, 1). Theorem 4.4.10 guarantees the existence of a nondecreasing function on [0. lj, which is discontinuous at each rational number in (0, 1). and continuous at every irrational number in (0, 1).

(c) If in the proof of Theorem 4.4.10, we take to be a countable subset of R and choose {c"}"EN (c" > 0) such that Z'n'o- , c" = 1, then we obtain a nondecreasing, real-valued function f on R satisfying lie f(x) = 0 and lim f(x) = 1. (See Exercise 20.) Such a function is called a distribution function on d. Snch functions arise naturally in probability theory.

Inverse Functions Suppose f is a strictly increasing real-valued function on an interval I. Let x, y E I with

x 0 y. If x < y, then since f is strictly increasing. f(x) < f(y). Similarly, if x > y then f(x) > f(y). Thus f(x) # f(y) for any jr, y E I with x # y. Therefore, f is one-to-one, and consequently has an inverse function f -' defined on f(1). In the following theorem we prove that if f is continuous on 1, then f -' is also continuous on f(1).

4.4.12 THEOREM Let I C R be an interval and let f : I-+ R be strictly monotone and continuous on !. Then f -' is strictly monotone and continuous on J = f(1).

Proof. Without loss of generality, we consider the case where f is strictly increasing on I. Since f is continuous, by Corollary 4.2.12, f(1) = J is an interval. Furthermore, since f is strictly increasing on I, f is a one-to-one function from I onto J. Hence f is a one-to-one function from J onto I. Suppose y,, y2 E J with y, < y2. Then there exist distinct points x,, x2 E I such that f(x;) = y;, i = 1, 2. Since f is strictly increasing, we have x, < x2. Thus f -'(y,) < f -' (y2); i.e., f ` is strictly increasing.

4.4

Monotone Functions and Discontinuities

159

It remains to be shown that f ` is continuous on J. We first show that f `' is left continuous at each y, E J for which (-oo, yo) fl J * 0. This last assumption means only that y, is not the left endpoint of J. Let xa E I be such that f(xo) = y0. Then (-oo, x,) n l * ¢. Let e > 0 be given. Without loss of generality, we can assume that e is sufficiently small so that x, - e E I. Since f is continuous and strictly increasing,

f((xo - e, xo]) = (f(xo - e),f(xo)) = (yo - 5, yo] where S > 0 is given by S = f(x,) - f(x, - e). Thus since f-' is strictly increasing,

x, - E= f


for all y E (y, - 8, y,) (see Figure 4.19). Hence,

1f -'(yo) - f -'(y)l < e

for ally E (y0 - s, yo]

Figure 4.19

Therefore, f "' is left continuous at y,. A similar argument also proves that f -' is right continuous at each y, E J that is not the right endpoint of J. Thus f -' is continuous at

each y, E J. For a strictly increasing function f on an open interval I, an alternative proof of the continuity of the inverse function f `' is suggested in Exercise 14.

4.4.13

EXAMPLE The function f(x) = x2 is continuous and strictly increasing on I = [0, oo)

with J = f(I) = [0, oo). Thus the inverse function f `'(y) = N /y is continuous on (0, oo). As a consequence, the function g(x) = N/x is continuous on (0, oo). Applying the same argument to f (x) = x" shows that the function g(x) = 'Clx is strictly increas-

ing and continuous on (0, oo). 0

160

Chapter 4

Limits and Continuity

Remark. In the statement of Theorem 4.4.12 we assumed that f was strictly monotone and continuous on the interval 1. The fact that f is either strictly increasing or strictly decreasing on I implies that f is one-to-one on the interval 1. Conversely, if f is one-to-one and continuous on an interval 1. then as a consequence of the intermediate value theorem the function f is strictly monotone on I (Exercise 15). This, however, is false if either f is not continuous on the interval 1, or if Dom f is not an interval. (See Exercises 16, 17, and 18.)

EXERCISES 4.4 1. Prove Theorem 4.4.3. 2. For each of the following functions f defined on R \ {0}, find ,4 m_ f(x) and -Ii m, f(x), provided the limits exist.

a. f(x) = Ixl

*b. f(x) _

Ix[x].

X < 0,

[X]

X>0

X

C. f(x) = [I - x2]

d. f(x) = [x2 - 1]

e. f(x) _ [ X1,

*L f(x) = x[ x

3. For each of the functions f in Exercise 2, determine whether f has a removable discontinuity, a jump discontinuity, or a discontinuity of second kind. at x = 0. If f has a removable discontinuity at 0. specify how f(0) should be defined so that f is continuous at 0.

4. *Investigate continuity of g(x) = (x - 2) [x) at x = 2. 5. Let f(x) = x - [x]. Discuss continuity of f. Sketch the graph of f. 6. *Let f be defined by f(x) = _ {x x2[x}, + b,

x < -2. x > -2.

Determine the value of b such that f has a removable discontinuity at x = -2. 7. *Prove that g(x) = sin(2rrx[x]) is not uniformly continuous on R. 8. Prove that the function f of Example 4.4.11(a) is continuous at x = 1.

9. Let E C Fl and let f be a real-valued function on E. Suppose p E R is a limit point of E fl (p, oo). Prove that lim f(x) = L if and only if 1 m f(p") = L for every sequence {p"} in E with p" > p for all n E N and p" --> p.

X-P 10. Let f be a real-valued function on (a, b].

a. If f is continuous on (a, b} and lim f(x) exists. prove that f is uniformly continuous on (a, b].

b. If f is uniformly continuous on (a, b), prove that lint f(x) exists. 11. Let f : [0, 2] - R be defined by

f(x) -

X,

.'

a-

05x 1,

1 < x 2. Show that f and f -t are strictly increasing and find f([0, 2]). Are f and f 1 + x2,

respective domains?

12. *If m E 7L, n E N, prove that f(x) = x'"1" is continuous on (0, oo).

continuous at every point of their

4.4

Monotone Functions and Discontinuities

161

13. a. If f and g are monotone increasing functions on an interval I. prove that f + g is monotone increasing on I. b. If in addition f and g are positive on 1, prove that fg is monotone increasing on I. c. Show by example that the conclusion in part (b) may be false if f and g are not both positive on I.

14. Let I C R be an open interval and let f : I -+ R be strictly increasing and continuous on I.

a. If U C I is open, prove that f(U) is open. b. Use (a) and Theorem 4.2.6 to prove that f'' is continuous on f(I). 13. Let I C R be an interval and let f be a one-to-one continuous real-valued function on 1. Prove that f is strictly monotone on I. 16. Let f : [0,1 ) -' R be defined by

AX)

2x,

Osx<2.

3-2x,

?5xsl.

a. Sketch the graph of f. b. Show that f is one-to-one on [0, 1 ], but not strictly monotone on [0, 1 ].

c. Show that f([0.. I ]) = [0, 2). d. Find f-' (y), y E [0, 2]. and show that f-' is not continuous at y, = 1. 17. Let E = [0, 11 U (2, 3), and for x E E set

f(x) -

x2,

Q :!:-: x < 1,

I 4-x, 25x<3.

a. Sketch the graph of f. b. Show that f is one-to-one and continuous on E.

e. Show that f(E) _ (0, 2]. d. Find f-' (y), y g [0, 2], and show that f`' is not continuous at y, = 1. 18. Let f : [0, 1] -+ R be defined by x,

ll-x,

f(x) - S

x E dQ,

Prove that f is one-to-one, that f([0, 1 ]) _ [0, 1], but that f is not monotone on any interval 1 C [0, 11.

19. Let ! C R be an interval and let f : I -+ R be monotone increasing. For p E Int(I), the jump of f at p, denoted

Jf(p), is defined by Jf(p) = f(p+) - f(p -) If p is the left endpoint, set Jf( p) = f( p +) - f(p), and if p is the right endpoint, set Jf(p) = f(p) - f(p -). a. Prove that f is continuous at p E I if and only if Jf(p) = 0. b. If p e Int(I), prove that Jf(p) = inf{ f(y) - f(x) : x < p < y, x, y E I}. 20. Let be a countable subset of R and a sequence of positive real numbers satisfying E c = 1. Let

f:R -iRbe defined by

f(x) _

C. !(x -

Prove that lint f(x) = 0 and X__ lim f(x) = 1.

162

Chapter4

Limits and Continuity

NOTES The limit of a function at a point is one of the fundamental tools of analysis. Not only is it crucial to continuity, but also to many subsequent topics in the text. The limit process will occur over and over again. We will encounter it in the next chapter in the definition of the derivative and again in the chapters on integration, series, etc. Another very important concept that will be encountered often in the text is uniform continuity. Uniform continuity is important in that given e > 0, it guarantees the

a unique positive real number y such that y" = x. Even though the existence of nth roots can be proved without the intermediate value theorem, any such proof is simply the statement that the function f(x) = x" satisfies the intermediate value property on [0, a] for every a > 0. Other applications of the intermediate value theorem will occur elsewhere in the text.

The proof of the intermediate value theorem de-

on [a, b], a, b E R, is Riemann integrable on [a, b].

pended on the fact that the connected subsets of R are the intervals (Theorem 3.1.18) and that the continuous image of a connected set is connected (Exercise 28 of Section 4.2). Assuming these two results, the intermediate value theorem is an immediate consequence as follows: Sup-

Other applications of uniform continuity will occur in

pose f is continuous on [a, b]. Let I = f([a, b]). Then I

many other theorems and in the exercises. One of the most important results of this chapter is the intermediate value theorem (Theorem 4.2.11). The intermediate value theorem has already been used in Corollary 4.2.13 to prove the existence of nth roots; namely, for

is connected and thus must be an interval. Thus if y satis-

existence of a S > 0 such that If(x) - f(y) I < e for all x, y E Dom f with Ix - yI < 6. In Chapter 6 we will use this to prove that every continuous real-valued function

every positive real number x and n E 101, there exists

fies f(a) < -y < f(b), then y E I, and hence there exists c E [a, b] such that f(c) = y. That the continuous image of a connected set is connected follows from the definition. However, the proof that the connected subsets of R are the intervals requires the least upper bound property.

MISCELLANEOUS EXERCISES 1. Let f be a continuous real-valued function on [a, b] with f(a) < 0 < f(b). Let c, = 2(a + b). If f(c,) > 0, let c, = } (a + c,). If f(cl) < 0, let c2 = 1(c, + b). Continue this process inductively to obtain a sequence {c"} in (a, b) which converges to a point c E (a, b) for which f(c) = 0. 2. Let E C R, p a limit point of E, and f a real-valued function defined on E. The limit superior of f at p, denoted 1 m f(x), is defined by 1 m f(x) = anfo sup{ f(x) : x E (N8( p) \ { p}) fl E}.

Similarly, the limit inferior off at p, denoted lim f(x), is defined by =FP !,f f (x) = ssu$ inf{}(x) : x e (N8(p) \ { p}) fl }.

x-P

Prove each of the following:

a. lim f (x) <_ L if and only if given e > 0, there exists a S > 0 such that f(x) < L + e for all x E E,

0 0 and 8 > 0, there exists x e E with 0 < I x - p I < S such that J-P

f(x)>L-e.

c. If I m f(x) = L, then for any sequence {xn} in E with x" # p for all n E N. and limx" = P, lim f(x") 5 L. d. There exists a sequence {x"} in E with x" # p for all n E N, such that and limx" = P, lim f(xn) = lim f(x). xp 3. Let X C R and f a real-valued function on X. For p E X, the oscillation of fat p, denoted w(f; p), is defined as w(f; p) = inf sup{ If(x) - f(y) I : x, y e Na(p) fl X}. 8>0

Supplemental Reading

163

Prove each of the following:

a. The function f is continuous at p if and only if w(f: p) = 0. b. For every s E i8, the set {x E X : w(f; x) < s} is open in X. c. The set {x E X : f is continuous at x} is the intersection of at most countably many sets that are open in X. 4. Find w(f; x) for the functions f of Example 4.1.2(e) and Example 4.2.2(g). The following set of exercises involves the Cantor ternary function. Let P denote the Cantor ternary set of Section 3.3. For each x E (0, 1], let x = .aia2a3 denote the ternary expansion of x. Define N as follows: oo,

N

if a # I for all is E fJ,

min{n : a = 11, otherwise.

Define b = 1a for is < N, and bN = 1, if N is finite. (Note: b E {0,1} for all n.) 5. If x E (0, 1) has two ternary expansions, show that

"b

T,

n-i

2

is independent of the expansion of x.

I

The Cantor ternary function f on [0, 1 ] is defined as follows: f(0) = 0, and if x E (0. 1 ] with ternary expansion x = .aja2a3 . , set b

f(x)= _12

where N and b are defined as above. 6. Prove each of the following:

a. f is monotone increasing on [0, 1]. b. f is constant on each interval in the complement of the Cantor set in [0, 1 ]. c. f is continuous on [0, 1 ].

d. f(P) = [0, 1]. e. Sketch the graph of f.

SUPPLEMENTAL READING Bryant, J., Kuzmanovich, J. and Pavlichenkov, A., "Functions with compact preimages of compact sets:' Math. Mag. 70 (1997), 362-364. Bumcrot, R. and Sheingorn, M., "Variations on continuity: Sets of infinite limits,' Math. Mag. 47 (1974), 41-43. Cauchy, A. L., Cours d'Analyse, Paris, 1821, in Oeuvres compldtes d'Augustin Cauchy, series 2, vol. 3, Gauthier-Villars, Paris, 1899. Fleron, Julian F., "A note on the history of the Cantor set and Cantor function;' Math. Mag. 67 (1994), 136-140.

Grabinger, Judith V., "Who gave you the epsilon? Cauchy and the origins of rigorous calculus;' Amer. Math. Monthly 90 (1983), 185-194. Martelli, M., Dang, M. and Seph, T.. "Defining chaos;' Math. Mag. 71 (1999), 112-122. Snipes, Ray F., "Is every continuous function uniformly continuous?" Math. Mag. 57 (1994),169-173. Straffin, Jr., Philip. D., "Periodic points of continuous functions;' Math. Mag. 51 (1978), 99-105. Velleman, D. J., "Characterizing continuity," Amer. Math. Monthly 104 (1997), 318-322.

Differentiation 5.1 The Derivative

52 The Mean Value Theorem 5.3 L'Hospital's Rule

5.4 Newton's Method

The development of differential and integral calculus by Isaac Newton (1642-1727) and Gottfried Wilhelm Leibniz (1646-1716) in the mid-seventeenth century constitutes one of the great advances in mathematics. In the two years following his degree from Cambridge in 1664, Newton invented the method of fluxions (derivatives) and fluents (integrals) to solve problems in physics involving velocity and motion. During the same period, he also discovered the laws of universal gravitation and made significant contributions to the study of optics. Leibniz, on the other hand, whose contributions came

ten years later, was led to the invention of calculus through the study of tangents to curves and the problem of area. The first published account of Newton's calculus appeared in his 1687 treatise Philosophia Naturalis Principia Mathematica. Unfortunately, however, much of Newton's work on calculus did not appear until 1737, ten years after his death, in a work entitled Methodus fuxionum et serierum infinitorum. Mathematicians prior to the time of Newton and Leibniz knew how to compute tangents to specific curves and velocities in particular situations. They also knew how to compute areas under elementary curves. What distinguished the work of Newton and Leibniz from that of their predecessors was that they realized that the problems of finding the tangent to a curve and the area under a curve were inversely related. More importantly, they also developed the notation and a set of techniques (a calculus) to solve these problems for arbitrary functions, whether algebraic or transcendental. In Newton's presentation of his infinitesimal calculus, he looked upon y as a flowing quantity, or fluent, of which the quantity v was the fluxion or rate of change. Newton's notation is still in use in physics and differential geometry, whereas every student of calculus learns the 165

166

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Differentiation

d (for difference) and f (for sum) notation of Leibniz to denote differentiation and integration. Many of the basic rules and formulas of the differential calculus were developed by these two remarkable mathematicians. In the paper A New Method for Maxima and Minima, and also for Tangents, which is not Obstructed by Irrational Quantities, published in 1684, Leibniz gave correct rules for differentiation of sums, products. quotients, powers, and roots. In addition to his many contributions to the subject. Leibniz

also disseminated his results in publications and correspondence with colleagues throughout Europe. Newton and Leibniz, with their invention of the calculus, had created a tool of such novel subtlety that its utility was proved for over 150 years before its limitations forced mathematicians to clarify its foundations. The rigorous formulation of the derivative did not occur until 1821 when Cauchy provided a formal definition of limit. This helped to place the theory on a firm mathematical footing. Cauchy's contributions to the rigorous development of calculus will be evident in both this and subsequent chapters. In this chapter we develop the theory of differentiation based on the definition of Cauchy, with special emphasis on the mean value theorem and consequences thereof. The first section presents the standard results concerning derivatives of functions obtained by means of algebraic operations and composition. In the examples and exercises we will derive the derivatives of some of the basic algebraic and trigonometric functions. However, throughout the chapter we will assume that the reader is already familiar with standard techniques of differentiation and some of its applications. Therefore, we will concentrate on the mathematical concepts of the derivative. emphasizing many of its more subtle properties.

5.1

The Derivative In an elementary calculus course, the derivative is usually introduced by considering the problem of the tangent line to a curve or finding the velocity of an object moving in a

straight line. Suppose y = f(x) is a real-valued function defined on an interval [a, b]. Fix p E [a, b]. For x E [a, b], x # p, the quantity Q(x) = f(x)

- f(p)

x-p represents the slope of the straight line (secant line) joining the points (p,f(p)) and (x,f(x)) on the graph off (see Figure 5.1). The function Q(x) is defined for all values of x E [a, b], x # p. The limit of Q(x) as x approaches p, provided this limit exists, is defined as the slope of the tangent line to the curve y = f(x) at the point (p,f(p)). A similar type of limit occurs if we consider the problem of defining the velocity of a moving object. Suppose that an object is moving in a straight line and that its distance s from a fixed point P is given as a function of t. namely, s = s(t). If t,, is fixed. then the average velocity over the time interval from to to t, t # to, is defined as

s(t) -

t-r

5.1

The Derivative

167

Figure 5.1

The limit of this quantity as t approaches to, again provided that the limit exists, is taken as the definition of the velocity of the object at time ta. Both of the previous examples involve identical limits; namely,

limf (x) - A p)

and

x-p

lira 1-.1,

s(t) - s(t°)

t - t,

These limits,. if they exist, are called the derivatives of the functions f and s at p and t respectively. The term derivative comes from the French fonction derivee.

5.1.1

DEFINITION Let I C R be an interval and let f be a real-valued function with domain I. For fixed p E 1, the derivative off at p. denoted f'(p), is defined to be

-f(p) Pp) = limf(x) x-,P x-p provided the limit exists. If f(p) is defined at a point p E 1, we say that f is differentiable at p. If the derivative f is defined at every point of a set E C 1, we say that f is differentiable on E.

If p is an interior point of I, then p + h E I for all h sufficiently small. If we set x = p + h, h # 0, then the definition of the derivative of f at p can be expressed as

f'(P) =

lt_mf(P

+ hh' - f(P)

provided the limit exists. This formulation of the derivative is sometimes easier to use.

In the definition of the derivative we do not exclude the possibility that p is an endpoint of I. If p E I is the left endpoint of 1, then P(P) = Rimp.

f(x) - f(p) =

x-p

ti!

f(p + h) - f(p) h

provided, of course, that the limit exists. The analogous formula also holds if p E I is the right endpoint of L In analogy with the right and left limit of a function, we also define the right and left derivative of a function.

168

Chapter5

5.1.2

Differentiation DEFINITION Let I C I be an interval and let f be a real-valued function with domain I. If p E I is such that I fl (p, oo) # 0, then the right derivative off at p, denoted f' (p), is defined as

f+(P) = lim h-4O

f(P + h) - f(P) h

provided the limit exists. Similarly, if p E I satisfies (-oo, p) fl I # 4,, then the left derivative off at p, denoted f'_(p), is given by

f'(P)=hli3t-

f(P + h) - f(P) h

provided the limit exists.

Remarks (a) If p E Int(/), then f'(p) exists if and only if both f"+(p) and f_(p) exist and are equal. On the other hand, if p E I is the left (right) endpoint of I, then f'(p) exists if and only if f+(p) (f_(p)) exists. In this case, f'(p) = f+ (P) (.f' (P)) The reader should note the distinction between f+(p) and f'(p+). The first denotes the right derivative off at p, whereas the latter is the right limit of the derivative; i.e.,

f'(P+) = lim f'(-). Here, of course, we are assuming that f' is defined for all x E (p, p + 8) for some

8>0.

(b) If f is a differentiable function on an interval 1, we will also occasionally use Leibniz's notation d dx

f(x),

"& ,

or

'*,

to denote the derivative of y = f(x). (c) If f is differentiable on an interval I, then the derivative f'(x) is itself a function on I. Therefore, we can consider the existence of the derivative of the function f' at a point p E I. If the function f' has a derivative at a point p E 1, we refer to this quantity as the second derivative off at p, which we denote f"(p). Thus

f"(P) =

hmf'(P + h) -f'(P) h

In a similar fashion we can define the third derivative off at p, denoted f'"(p) or f(')(p).

In general, for n E N, ft"t(p) denotes the nth derivative off at p. In order to discuss the existence of the nth derivative off at p, we require the existence of the (n - 1)st derivative off on an interval containing p.

The Derivative

5.1

5.1.3

169

EXAMPLES

(a) In the exercises (Exercise 2) you will be asked to prove that if f(x) = x", n E Z, then f'(x) = nx"- ' for all x E IB (x * 0 if n is negative). For the function f(x) = xz, the result is obtained as follows:

fi(x) =

,y9 (x +

z -

h

h) = 2x.

=

A similar computation shows that f"(x) = 2.

(b) Consider f(x) = N /x, x > 0. We first note that for h * 0,

f(x+h)-f(x)

-vrx h

=

( x+h-V)( x+h+V) h

x+h+Vx)

(

x+h+;7x Since h x -+h = ix, we have

f'(x)=h+

1

1

=

x + h + vx

2Vx

(c) Consider f(x) = sin x. From the identity

sin(x + h) = sin x cos h + cos x sin h, we obtain

sin(x + h) - sin x h

r cos

= sin x l

h-I

h

I+ cos x f

sin h h

By Example 4.1.10(d) and Exercise 5, Section 4.1, sin h lmi=l h

and

lim

cos h - 1 = 0.

Therefore,

f'(x).

sin(x + h) - sin x h

LIn

= sin x lim i

cos h - 11

= cos X.

+ cos x l

J

In Exercise 3 you will be asked to prove that

d (cos x) = -sin x.

[ sin h J

170

Chapters

Differentiation (d) Let f be defined by

x ? 0,

x

.f(x) = IxI =

-x, x<0.

Then

_

h

.f+(0) = lim

h

and

o h= 1,

h

ti

ii

01

= lim _h

Thus f (0) and f'_(0) both exist, but are unequal. Therefore f'(0) does not exist. (e) In this example let g(x) = x312 with Dom g = [0, oo). Then for p = 0, x3/2

8'(0) = 8+(0) =I-,Q limX- = Jim V = 0. .Ho Thus g is differentiable at 0 with g'(0) = 0. (f) Let f be defined by

x x#0, xsin, x

f(x) =

For x # 0, f'(x) _ -x cos X + sin I (Exercise 8). When x = 0,

f'(0) _ h m

h f(h)

- f0

=him sin h,

which by Example 4.1.5(a) does not exist. Therefore, f'(0) does not exist.

II

(g) Consider the following variation of (f). Let

g(x) =

x2 sin x 0,

,

x # 0,

x0.

This example is very important! In Exercise 9 you will be asked to show that g'(x) exists for all x E R with g'(0) = 0, but that the derivative g' is not continuous at 0.

When Cauchy gave the rigorous definition of the derivative, he assumed that the given function was continuous on its domain. As a consequence of the following theorem this requirement is not necessary.

5.1.4 THEOREM If 1 C R is an interval and f : I -+ R is differentiable at p E 1, then f is continuous at p.

Proof.

Fort # p,

f(t) -f(P) =

(f(t` _ P(P))(t

- P)

5.1

The Derivative

171

Since

limi(t) -f(p) '-+p t-p exists and equals f'(p), by Theorem 4.1.6(b),

lim(f(t) - f(p)) = lim _P r-+p

(f(t) - f(P)1Jlim(r - P) = Pp) 0 = 0.

t-p

r-+p

Therefore, lm f(t) = f(p) and thus f is continuous at p. In the above, if p is an endpoint, then the limits are either the right or left limit at p, whichever is appropriate.

Remark In both Examples 5.1.3(d) and (f), the given function is continuous at 0, but n o t d i ff e r e n t i a b l e a t 0. Given a f i n it e number of points, say p i, ... , pn, it is easy to construct a function f which is continuous but not differentiable at pI, .. . , p,,. For example, n

f(x)= k-I I Ix-Pk1 has the desired properties. In 1861, Weierstrass constructed a function f which is continuous at every point of 11 but nowhere differentiable. When published in 1874, this example astounded the mathematical community. Prior to this time mathematicians generally believed that continuous functions were differentiable (except perhaps at a finite number of points). In Example 8.5.3 we will consider the function of Weierstrass in detail.

Derivatives of Sums, Products, and Quotients We now derive the formulas for the derivative of sums, products, and quotients of functions. These rules were discovered by Leibniz in 1675.

5.1.5 THEOREM Suppose f, g are real-valued functions defined on an interval L 1f f and g are differentiable at x E 1, then f + g,fg, and fig (if g(x) # 0) are differentiable at x and

(a) (f + 8)'(x) = fl(X) + g'(x),

(b) (J)'(x) = f'(x)g(x) +f(x)g'(x) (c) ()'(x) = fl(x)8(x) -

f2x)g'(x)'

provided g(x) # 0.

(g(x))

Proof. The proof of (a) is left as an exercise (Exercise 4). For the proof of (b), by adding and subtracting the term f (x + h)g(x), we have for h * 0,

(fg)(x + h) - (fg)(x)

= f(x + h)(Ax

+ h) - g(x)

+ (f(x + h) - f(x)lAX) h

J

172

Chapters

Differentiation

By Theorem 5.1.4, since f is differentiable at r, i a f(x + h) = f(x). Thus since each of the limits exist, by Theorem 4.1.6,

(fg)'(x) = lim ti0

(.fg)(x + h) - Ug)(x) h

/

=ii mf(x+h)ii

mig(x+hh-g(x)

+\h)

- f(x)

+ g(x) him/f(x l

h

= f(x)g'(x) + g(x)f'(x). To prove (c), we first prove that (1/g)'(x) = -g'(x)/[g(x)]2, provided g(x) * 0. The result then follows by writing f/g as f (1/g) and applying the product formula (b). Suppose g(x) * 0. By continuity of g at x, there exists a S > 0 such that g(x + h) * 0 for all h, I h i < S. Thus for h sufficiently small and nonzero, 1

1

g(x + h)

g(x)

-

g(x + h) - g(x)'

h

1

) g(x)g(x + h)

h

Again, using the fact that lim g(x + h) = g(x), by Theorem 4.1.6, 1

1ll'

C-/ 8 (x)

1

g(x + h) __ lim h-0

g(x)

h

(g(x + h) - g(x)1

_ -lim 11-0 \

) lim n-o g(x)g(x + h)

h

-g'(x) = g2(x) ,

The Chain Rule The previous theorem allows us to compute the derivatives of sums, products, and quotients of differentiable functions. The chain rule, on the other hand, allows us to compute the derivative of a function obtained from the composition of two or more differentiable functions. Prior to stating and proving the result, we introduce some useful notation. Suppose f is differentiable at x E 1. For t E 1, t # x, set

Q(t) = f(t) - f(x) t - x

Then by Definition 5.1.1, Q(t) --+f(x) as tax. If we let u(t) = Q(t) - f'(x). then u(t) -- 0 as t -+ x. Therefore, if f is differentiable at x, fort * x,

f(t) - f(x) = (t - x)[f'(x) + u(t)] where u(t) -+0 as By setting u(x) = 0, the above identity is valid for all r E I.

(1)

5.1

5.1.6

The Derivative

173

THEOREM (Chain Rule) Suppose f is a real-valued function defined on an interval I and g is a real-valued function defined on some interval J such that Range f C J. If f is differentiable at x E I and g is differentiable at f(x), then h = g of is differentiable at x and

h'(x) = g'(f(x))f'(x) Proof. Let y = f(x). Then by equation (1),

f(t) - f(x) _ (t - x)[f'(x) + u(t)I.

(2)

g(s) - g(y) = (s - y)[g'(y) + v(s)],

(3)

where t E I, s E J, and u(t) -> 0 as t -> x and v(s) -+0 as s -+y. Let s = f (t). Since f is continuous at x, s --* y as t -> x. By identity (3) and then (2),

h(t) - h(x) = g(f(t)) - g(f(x)) [f(t) - f(x)I [g'(y) + v(s)] = (t - x)[f'(x) + u(t)][g'(v) + v(f(t))]. Therefore, for t * x,

h(t)- hx_ t-x

) = [f'(x) + u(t)N91(y) + v(f(t))).

Since v(f(t)) and u(t) both have limit 0 as t -+ x, lim

h(t) - h(x) =

r(X) g'(y) = g'(f(x))f'(x). O

t-x

To illustrate the previous theorem we consider the following examples.

5.1.7

EXAMPLES

(a) By Example 5.1.3(c) the function f (x) = sin x is differentiable on R. Hence if g :1 * R is differentiable on the interval 1, h(x) = If o g)(x) = sin g(x) is differentiable on I with

h'(x) = f'(g(x)) g'(x) = g'(x)cos g(x). In particular, if g(x) = 1/x2, Dom g = (0, oo), then by Theorem 5.1.5(c),

- 2x

g'(x) _ (x2)2 =

2 x3,

x E (0, oo).

Therefore,

d dx

2 -=-cossmx2 x2 .

1

1

2.

(b) By Exercise 2,

dx"=nx"-' forallnEfol.

Differentiation

Chapters

174

Thus if f : I -+ R is differentiable on the interval 1, then by the chain rule g(x) = [f(x) ]1.

n E N, is differentiable on I with g'(x) = n[f(x)]"-'f'(x). This formula can also be obtained from Theorem 5.1.5(b) using mathematical induction.

EXERCISES 5.1 1. Use the definition to find the derivative of each of the following functions.

*a. f(x)=x;, x E R

b. g(x) =

*c. h(x) = 1 X, x # 0

d. k(x) =

x+2, x> -2 x > -2

l '

*e. f (x)

x* -1

x+1

f. g(x) = xZ + 1' x E R

2. *Prove that for all integers n,

dx" = nx"-' (x * 0 if n is negative). 3. *a. Prove that

d (cos x) = -sin x. b. Find the derivative of tan x = sin x/cos x. 4. Prove Theorem 5.1.5(a). S. For each of the following, determine whether the given function is differentiable at the indicated point x°. Justify your answer!

*a. f(x) = xIxl,

at X. = 0

b.

x E 0'

f(x) = lo0, ,

*c. g(x) = (x - 2)[x], at x° = 2 e.

x

f()=

d. h(x) =

x00,

atx° = 0

x + 2, x E [ -2, oo), at x° _ -2

sin X, xEQ.

X. =0 x00. atx

x,

6. Let f(x) = lxI3. Compute f'(x), f"(x), and show that f"'(0) does not exist. 7. Determine where each of the following functions from R to R is differentiable and find the derivative.

*a. 1(x) = x(x]

b.g(x)=Ix-21+Ix+tI *c. h(x) = I sin x I

{x2[]. 0,

:::"

8. Use the product rule, quotient rule, and chain rule to find the derivative of each of the following functions.

a. f (z) = x sin

x*0

c. f(x)= x+1Y

b. f(x) = (cos (sin x) ), n, m E h!

d. f(x)=x°(2+sink), x * 0

5.1

The Derivative

175

9. Let g be defined by

g(x) =

x2 sin1, x0, x

a. Prove that g is differentiable at 0 and that g'(0) = 0. b, Show that g'(x) is not continuous at 0. 10. Let f be defined by f(x) = fx2 + 2, x s 2,

ax+b, x> 2.

a. For what values of a and b is f continuous at 2? b. For what values of a and b is f differentiable at 2? 11. Let f be defined by

x < -l, x3+1. -lsx:5 2, ax + b,

f(x) =

cx+d,

x> 2.

Determine the constants a, b, c, and d such that f is differentiable on R.

12. Assume there exists a function L : (0, oo) -+ R satisfying L'(x) = l/x for all x E (0, oo). Find the derivative of each of the following functions.

a. f(x) = L(2x + 1), x > -2

b.g(x)=L(x2), x#0 *c. h(x) = [L.(x)]', x > 0 d. k(x) = L(L(x)), x E {x > 0 : L(x) > 0} 13. For b real, let f be defined by

f(x) =

1x° sin 1-,

x > 0,

0,

x50.

x

Prove the following:

a. f is continuous at 0 if and only if b > 0. b. f is differentiable at 0 if and only if b > I. c. f is continuous at 0 if and only if b > 2. mfx

14. a. If f is differentiable at x, prove that li

,+h- fx, - h )

b. If Ii f (x

+h)-f(xo-h) 2h

=fI M

exists, is f differentiable at x0?

15. If f : (a, b) -+ R is differentiable at p E (a, b), prove that f'(p) = lim n[f(p + ,l,) - f(p)]. Show by example that the existence of the limit of the sequence {n[ f(p + f(p)j) does not imply the existence of f'(p). 16. Leibniz's Rule: Suppose f and g have nth order derivatives on (a, b). Prove that (nk) f (k)(X)e - kkX).

(fg)t11(x) _ 4-0

176

Chapter5

Differentiation

5.21 The Mean Value Theorem In this section we will prove the mean value theorem and give several consequences of this important result. Even though the proof itself is elementary, the theorem is one of the most useful results of analysis. Its importance is based on the fact that it allows us to relate the values of a function to values of its derivative. We begin the section with a discussion of local maxima and minima.

Local Maxima and Minima 5.2.1

DEFINITION Suppose E C R and f is a real-valued function with domain E. The function f has a local maximum at a point p E E if there exists a 8 > 0 such that f (x) s f(p)for all x E E fl Na(p)- The function f has an absolute maximum at p E E iff(x) 5 f(p) for all x E E.

Similarly, f has a local minimum at a point q E E if there exists a 8 > 0 such that

f(x) ? f(q) for all x E E fl %(q) and f has an absolute minimum at q E E if f (x) ? f (q) for all x E E.

Remark. As a consequence of Corollary 4.2.9. every continuous real-valued function defined on a compact subset K of I has an absolute maximum and minimum on K.

The function f, illustrated in Figure 5.2, has a local maximum at a, p,, and p, and a local minimum at and b. The points (p4,f(p4)) and (p,.f(p,)) are absolute maxima and absolute minima, respectively.

Figure 5.2

5.2

The Mean Value Theorem

177

The following theorem gives the relationship between local maxima or minima of a function defined on an interval and the values of its derivative.

5.2.2

THEOREM Let fbe a real-valued function defined on an interval 1, and supposef has either a local minimum or local maximum at p E Int(l). 1f f is differentiable at p, then

f(P) = 0. Proof. If f is differentiable at p E Int(1), then f'- (p) and f; (p) both exist and are equal. Suppose f has a local maximum at p. Then there exists a S > 0 such that f (t) f (p) for all t E I with I t - p I < S. In particular, if p < t < p + S, t E 1, then

f(t) - f(p) s 0. t-P Thus f+(P)

0. Similarly, if p - S < t < p,

f(t) - f(P)

t-p

0,

and therefore f'_(p)' 0. Finally, since f+ (p) = f'_ (p) = f'(p), we have f(p) = 0. The proof of the case wheref has a local minimum at p is similar. O As a consequence of the previous theorem we have the following corollary.

5.2.3 COROLLARY Lei f be a continuous real-valued function on [a, b]. 1f f has a local maximum or minimum at p E (a, b), then either the derivative off at p does not exist,

or f(p) = 0. Remark. The conclusion of Theorem 5.2.2 is not valid if p E I is an endpoint of the interval. For example, if f :[a, b] -- R has a local maximum at a, and if f is differentiole at a, then we can only conclude that f'(n) = f+ (a) 5 0. This is illustrated in the following examples.

5.2.4

EXAMPLES

(a) The function

f(x)= x- lZ , 05x2, has a local maximum at p = 0 and p = 2, and an absolute minimum at q = ;. By com-

putation, we have f'(0) _ -1, f' (2) = 3, and f (2) = 0. The graph of f is given in Figure 5.3.

(b) The function f (x) _ x 1, x E [ -1, 11, has an absolute minimum at p = 0. However, by Example 5.1.3(d) the derivative does not exist at p = 0.

Rolle's Theorem Prior to stating and proving the mean value theorem, we first state and prove the following theorem credited to Michel Rolle (1652-1719).

178

Chapter 5

Differentiation

24J 21

Il

4

U

Figure 5.3

5.2.5

Graph of f(x) _ (x - 1)2, 0 5 x 5 2

THEOREM (Rolle's Theorem) Suppose f is a continuous real-valued function on

[a, b] with f(a) = f(b), and that f is differentiable on (a, b). Then there exists c E (a, b) such that f (c) = 0. Since the derivative off at c gives the slope of the tangent line at (c, f (c)), a geometric interpretation of Rolle's theorem is that if f satisfies the hypothesis of the theorem, then there exists at least one value of c E (a, b) for which the tangent line to the graph off is horizontal. For the function f depicted in Figure 5.4, there are exactly two such points.

Proof. If f is constant on [a, b], then f'(x) = 0 for all x E [a, b]. Thus, we assume that f is not constant. Since the closed interval [a, b] is compact, by Corollary 4.2.9,f has a maximum and a minimum on [a. b]. If f(t) > f(a) for some t, then f has a maximum at some c E (a, b). Thus by Theorem 5.2.2, f'(c) = 0. If f(t) < f(a) for some t, then f has a minimum at some c E (a, b), and thus again f'(c) = 0.

Remarks (a) Continuity off on [a, b] is required in the proof of Rolle's theorem. The function

Ix, 05x< 1,

f(x)= 0, x=1

is differentiable on (0, 1) and satisfies f(0) = f(1) = 0; yet f' (x) * 0 for all x E (0, 1). The function f fails to be continuous at 1. (b) For Rolle's theorem, differentiability of f at a and b is not required. For example,

the function f(x) = - xZ, x e [-2, 2], satisfies the hypothesis of Rolle's theorem, yet the derivative does not exist at -2 and 2. For x E (-2, 2), fi(x) _

V4

- X2

and the conclusion of Rolle's theorem is satisfied with c = 0.

5.2 The Mean Value Theorem

Figure 5.4

179

Rolle's Theorem

The Mean Value Theorem As a consequence of Rolle's theorem we obtain the mean value theorem. This result is usually attributed to Joseph Lagrange (1736-1813).

5.2.6

THEOREM (Mean Value Theorem) If f : [a, b] -+ R is continuous on [a, b] and differentiable on (a, b), then there exists c E (a, b) such that

f(b) - f(a) =f(c)(b - a). Graphically, the mean value theorem states that there exists at least one point c E (a, b) such that the slope of the tangent line to the graph of the function f is equal to the slope of the straight line passing through (a, f (a)) and (b, f (b)). For the function of Figure 5.5, there are two such values of c, namely c, and c2.

Proof. Consider the function g defined on [a. b] by

g(x) = f(x) - f(a) -

{f(b)b

- a(a),(x - a).

Then g is continuous on [a, b], differentiable on (a, b), with g(a) = g(b). Thus by Rolle's theorem there exists c E (a, b) such that g'(c) = 0. But

g'(x) =f'(x) -

f(b) - f(a)

for all x E (a, b). Taking x = c gives f(c) = now follows.

b - a

f (b)

- a(a), from which the conclusion

180

Chapters

Differentiation

Figure 5.5

Mean Value Theorem

The mean value theorem is one of the fundamental results of differential calculus. Its importance lies in the fact that,it enables us to obtain information about a function f from its derivative f'. In Example 5.2.7 we will illustrate how the mean value theorem can be used to derive inequalities. Other applications will be given later in this section and in the exercises. It will also be used in many other instances in the text.

5.2.7

EXAMPLE In this example we illustrate how the mean value theorem may be used in proving elementary inequalities. We will use it to prove that x 1 + x

s ln(1 + x) < x for all x > - 1,

where In x denotes the natural logarithm function on (0, oo). This function is defined and considered in detail in Example 6.3.5 of the next chapter. There it is proved that the

derivative of In x is 1/x. Let f(x) = ln(1 + x), x E (-1, oo). Then f(0) = 0. If x > 0, then by the mean value theorem, there exists c e (0, x) such that In(1 + x) = f(x) - f(0) = f(c)x.

But f'(c) _ (1 + c)-' and (I + x)-' < (1 + c)-t < 1 for all c E (0, x). Therefore, x 1

+x
and as a consequence,

1

x +x`ln(1 +x)<x for all xz0.

5.2 The Mean Value Theorem

181

Now suppose -1 < x < 0. Then again by the mean value theorem there exists c E (x. 0) such that ln(1 + x) = f(x) - f(0) = 1 + c,

But since x < c < 0, 1 < (1 + c)-' < (1 + x)-', and since x is negative 1 + x

< ln(1 + x) < .r.

Hence the desired inequality holds for all x > -1, with equality if and only if x = 0.

U The following theorem, attributed to Cauchy, is a useful generalization of the mean value theorem.

5.2.8

THEOREM (Cauchy Mean Value Theorem) If f, g are continuous real-valued functions on [a, b] that are differentiable on (a, b), then there exists c E (a, b) such that [f(b) - f(a)]g'(c) _ [g(b) - 8(a)] .`'(c)

. Proof.

Let

h(x) = [f(b) -f(a)]g(x) - [8(b) - g(a)]f(x). Then h is continuous on [a, b], differentiable on (a, b) with h(a) = f(b)g(a) - f(a)g(b) = h(b).

Thus by Rolle's theorem, there exists c E (a, b) such that h'(c) = 0, which gives the

result. U The geometric interpretation of the Cauchy mean value theorem is very similar to that of the mean value theorem. If g'(x) * 0 for all'x E (a, b), then g(a) 0 g(b) and the conclusion of Theorem 5.2.8 can be written as

f(b) - f(a) _ f (c) g(b) - g(a) guppose.x = g(t), y = f(t), a s t s b, is a parametric representation of a curve C in the plane. As t moves along the interval [a, b], the point (x, y) moves along C from the

point P = (g(a),f(a)) to Q = (&)J (b)). The slope of the line joining P to Q is given

by [f(b) - f(a)]/[g(b) - g(a)] (see Figure 5.6). On the other hand, the quantity f'(t)/g'(t) is. the slope of the. curve C at the point (g(r), f(t)). Thus one meaning of Theorem:5.2.8 is that there must be a point on the curve C where the slope of the curve is the same as the slope of the line joining P to Q.

182

Chapters

Differentiation

Q=(g(b). f(b))

P=(g(a), f(a))

Figure 5.6

Applications of the Mean Value Theorem We now give several consequences of the mean value theorem. Additional applications are also given in the exercises. In the following, I will denote an arbitrary interval in R.

5.2.9 THEOREM Suppose f : 1--+ R is differentiable on the interval L (a) If f'(x) ? O for all x E I, then f is monotone increasing on 1. (b) If f (x) > O for all x E I, then f is strictly increasing on I. (c) If f'(x) 5 O for all x E 1, then f is monotone decreasing on L (d) If f'(x) < O for all x E 1, then f is strictly decreasing on I. (e) If f'(x) = O for all x E I, then f is constant on I.

Proof.

Suppose x1, x2 E I with x, < x2. By the mean value theorem applied to f on

[x1, x2].

f(x2) -f(xl) =f'(C)(x2 - xl) for some c E (x1, x2). If f(c) ? 0, then f(x2) ? f(x1). Thus, if f'(x) a 0 for all x E 1, we have f(x2) ? f(xl) for all xj, x2 E I with xi < x2. Thus f is monotone increasing on I. The other results follow similarly. O

Remark It needs to be emphasized that if the derivative of a function f is positive at a point c, then this does not imply that f is increasing on an interval containing c. The

function f of Exercise 18 satisfies f'(0) = 1, but f'(x) assumes both negative and positive values in every neighborhood of 0. Thus f is not monotone on any interval containing 0. If f'(c) > 0, the only conclusion that can be reached is that there exists

a 8 > 0 such that f(x) < f(c) for all x E (c - S, c) and f(x) > f(c) for all x E (c, c + 8) (Exercise 15). This, however, does not mean that f is increasing on

(c - 8, c + 8). However, if f'(c) > 0 and f' is continuous at c, then there exists a 8 > 0 such that f'(x) > 0 for all x E (c - 8, c + 8). Thus f is increasing on

(c-8,c+ 8).

5.2

The Mean Value Theorem

183

Theorem 5.2.9 is often used to determine maxima and minima of functions as follows: Suppose f is a real-valued continuous function on (a, b), and c E (a, b) is such that f'(c) = 0 or f(c) does not exist. Suppose f is differentiable on (a, c) and (c, b). If f'(x) < 0 for all x E (a, c) and f'(x) > 0 for all x E (c, b), then by Theorem 5.2.9. f is decreasing on (a, c) and increasing on (c, b). As a consequence, one concludes that f has a local minimum at c. This method is usually referred to as the first derivative test for local maxima or minima. The natural inclination is to think that the converse is also true; namely, if f has a local minimum at c, then f is decreasing to the left of c and increasing to the right of c. As the following example shows, however, this is false!

5.2.10

EXAMPLE

Let f be defined by

f(x) = {x4(2

x 0,

,

x = 0.

The function f has an absolute minimum at x = 0; however, f'(x) has both negative and positive values in every neighborhood of 0. The details are left as an exercise (Exercise 19). The graph of f'(x) = 4x3(2 + sin 1/x) - x2 cos l/x, x * 0, for x in a neighborhood of zero is given in Figure 5.7.

Figure 5.7

Graph of f'(x) = 4x3(2 + sin j)

- x2 cos , x * 0

The following theorem, besides being useful in computing right or left derivatives at a point, also states that the derivative (if it exists everywhere on an interval) can only have discontinuities of the second kind.

5.2.11

THEOREM Suppose f :[a, b) -+ R is continuous on [a, b) and differentiable on (a, b). If lim f'(x) exists, then f+ (a) exists and x-+a

f, (a) = 1 m, f'(x). Xa

184

Chapter5

Differentiation

Proof. Let L = lim' f'(x), which is assumed to exist. Given e > 0, there exists a x-ra S > 0 such that

If'(x)-LI <e for all x, a < x < a + S. Suppose 0 < h < 8 is such that a + h < b. Since f is continuous on [a, a + h] and differentiable on (a, a + h), by the mean value theorem f(a + h) - f(a) = f'(C,,)h for some Ch E (a, a + h). Therefore,

f(a + h) - f(a) _ LI h

= If'(Sh) - LI < e

for all h, 0 < h < S. Thus f; (a) = L. Q 5.2.12

EXAMPLES

(a) To illustrate the previous theorem, consider the function

f(x) ° For x < 1, f'(x)

Jx2 + 1, x < 1,

3-x, x> 1.

2x, which has a left limit of 2 at x = I. Thus by the theorem,

f'_(1) = lim 2x = 2. Similarly,

f+'(1) = lim (-2x) = -2. (b) The converse of Theorem 5.2.11 is false. The function Ix2 sin 1

$(x) =

0,

x

x # 0, x = 0,

of Example 5.1.3(g) has the property that g'(0) exists but lidg'(x) does not.

Intermediate Value Theorem for Derivatives Our second important result of this section, credited to Jean Gaston Darboux (18421917), is the intermediate value theorem for derivatives. The remarkable aspect of this theorem is that the hypothesis does not require continuity of the derivative. If the derivative were. continuous, then the result would follow from Theorem 4.2.11 applied to f

5.2.13

THEOREM (Intermediate Value Theorem for Derivatives) Suppose I C R is an interval and f : I - P is differentiable on I. Then given a, b in I with a < b and a real number A between [(a) and f'(b), there exists c E (a, b) such that f(c) = A.

Proof, Define g by g(x) = f(x) - Ax. Then g is differentiable on I with g'(x) _

f'(x) - A.

5.2 The Mean Value Theorem

185

Suppose f'(a) < A < f'(b). Then g'(a) < 0 and g'(b) > 0. As in the remark following Theorem 5.2.9. since g'(a) < 0, there exists an x, > a such that g(.r,) < g(a)-

Also, since g'(b) > 0, there exists an x, < b such that g(x.) < g(b). As a consequence, g has an absolute minimum at some point c E (a, b). But then

g'(c) = f'(c) - A = 0,

i.e., f'(c) = A. Q The previous theorem is often used in calculus to determine where a function is increasing or decreasing. Suppose it has been determined that the derivative f is zero at c, and c2 with c, < c2, and that f'(x) * 0 for all x E (c,, c2). Then by the previous theorem, it suffices to check the sign of the derivative at a single point in the interval (cr, c2) to determine whether f is positive or negative on the whole interval (c c,). Theorem 5.2.9 then allows us to determine whether f is increasing or decreasing on (cj, c2).

Inverse Function Theorem We conclude this section with the following version of the inverse function theorem.

5.2.14 THEOREM (Inverse Function Theorem) Suppose ! C R is an interval and f : I -+ R is differentiable on I with f'(x) * O for all x E I. Then f is one-to-one on I, the inverse

function f` is continuous and differentiable on J = f(!) with 1

(f-')'(f(x)) = f'(x) for all x E 1.

Proof. Since f'(x) * 0 for all x E 1, by Theorem 5.2.13, f is either positive or negative on 1. Assume that f'(x) > 0 for all x E I. Then by Theorem 5.2.9,f is strictly increasing on I and by Theorem 4.4.12, f -' is continuous on J = f(I). It remains to be shown that f -' is differentiable on J. Let y, E J, and let {y.) be any sequence in J with yn y0, and y 0 y for all n. For each n, there exists x E 1 such that y,,. Since f'' is continuous, x -+x0 = fHence

f1'

xn - x

f - `Cy0)

Y. - y0

- - oof(xx) - f(x0)

=f (xo)

Since this holds for any sequence definition of the derivative,

with y, -r y,,, y * y,,, by Theorem 4.1.3 and the

(f TO = rl(xo)' v

186

Chapter 5

Differentiation

Remark, The hypothesis that f'(x) # 0 for all x E I is crucial. For example, the function f(x) = x3 is strictly increasing on [ -1, 1 ] with f'(0) = 0. The inverse func. tion f -'(y) = y'13, however, is not differentiable at y = 0.

5.2.15

EXAMPLES

(a) As an application of the previous theorem, we show that f (x) = x'1", x E (0, oo), n E N, is differentiable on (0, oo) with n

for all x E (0, oo). Consider the function g(x) = x", n E N, Dom g = (0, oo). Then g'(x) = nx"-' and g'(x) > 0 for all x E (0, oo). By the previous theorem, g-' is differentiable on J = g((0, oo)) _ (0, oo) with

(g-')'(g(x)) =

g'(x)

=

1-1

If we set y = g(x) = x", then x = y"" and

(g-')'(y) = n(y )"-1 = n

yI/"-t.

Since f = g' the desired result follows. (b) As in Example 5.2.7, let L(x) = In x denote the natural logarithm function on (0, co). Since L'(x) = l/x is strictly positive on (0, oo), the function L is one-to-one, the inverse function L-' is continuous on Q8 = Range L, and by Theorem 5.2.14,

If we set E = L', then E'(L(x)) = x, or E'(y) = E(y) where y = L(x). The function E(x), x e 08, is called the natural exponential function on } and is usually denoted by e', where e is Euler's number of Example 2.3.5. The exponential function E(x) is considered in greater detail in Example 8.7.20.

(c) In this example we consider the inverse function of g(x) = cos x, x E [0, ir]. Since g'(x) = -sin x is strictly negative for x E (0, ir), the function g is strictly decreasing on [0, a] with g([0, a]) _ [- I, 1 ]. Therefore its inverse function g- 1, which we denote by Arccos, exists on [ -1, 1 ]. Thus for y E [ -1, 1 ], x = Arccos y if and only if y = cos x. Finally, since g'(x) 0 for x E (0, ir), by the inverse function theorem,

(8-,)'(g(x)) = g'(x)

-1 -sinx = N/l

z

'

or since y = cos x,

-l dy

Arccos y =

The graphs of both cos x, x E [0, ir], and Arccos x, x E [ -1, 13, are given in Figure 5.8.

5.2 The Mean Value Theorem

Figure 5.8

187

Graphs of cos x, x E [0, ar], and Arccos x, x E [-1, 11

EXERCISES 5.2 1. For each of the following functions, determine the interval(s) where the function is increasing or decreasing, and find all local maxima and minima.

*a f(x)=x3+Gx-5, xER

b.g(x)=4x-x°, xER

c. h(x) = 1+x2x2, x e R

d. k(x) _ V-

e.l(x)=x+7, x#0 2. Let f (x) _

2

x,

xz0

f f(x)x-b, a*b,x*b

(x - a,)2, where a,, a2, ... , a are constants. Find the value of x where f is a minimum.

3. As in Example 5.2.7, use the mean value theorem to establish each of the following inequalities. a.

1 +xs1+Ix, x> -I

b.exZI+x, xER

C., (I +x)a? I + ax, x> -1,a> I 1. For a E N, this inequality was proved by mathematical induction in Example 1.3.3(b). In this exercise, and in Exercise 4(b), you may assume that for a E R.

d x° = ax°-

188

Differentiation

Chapter 5

4. Prove each of the following inequalities.

a.all. -b""<(a-b)'/", a>b>0,nEE N,na: 2 *b.a°b'-°:S aa+(1 -a)b, a,b>0,0
*a. If f"(c) > 0, prove that f has a local minimum at c. b. If f"(c) < 0, prove that f has a local maximum at c. c. Show by examples that no conclusion can be made if f"(c) = 0. 6. Suppose f : (a, b) -> R satisfies If(x) - f (y)I <- Mix - yI" for some a > I and all x, y E (a, b). Prove that f is a constant function on (a, b). 7. 17ind a polynomial P(x) of degree less than or equal to 2 with P(2) = 0 such that the function

x s 1. x > 1.

x2,

f(x}

P(x),

is differentiable at x = 1. 8. Let g be defined by

g(x)- f2sinx+cos2x, xs0, axe+bx+c, x>0. Determine the constants a, b, and c such that g'(0) and g"(0) exist.

9. a. Suppose f is differentiable on an interval I. Prove that f is bounded on I if and only if there exists a constant

M such that If(x) - f(y)I s MIx - yI for all x, y E I. b. Prove that I sin x - sin y I < I r - y l for all x, y E R.

c. Prove that I N/x - V I c 2'-- Ix - yI for all x, y E (a. oo), a > 0. 10. a. Show that tan x > x for x E (0, ). b. Set

f(x) -

sinx x 1,

x E (0,;

x=0.

Show that f is strictly decreasing on [0, i ]. c. Using the result of pan (b), prove that x < sin x <- x for all x E [0, F].

11. Give an example of a uniformly continuous function on [0, 1 ] that is differentiable on (0, 1) but for which f is not bounded on (0, 1).

12. ea Suppose f'(x) exists for all x E (a. b). Let c E (a. b). Show that there exists a sequence {x"} in (a, b) with x" * c and x" -+ c such that f'(x") - +f'(c). b. Does f'(x,) -+f'(c) for every sequence {x"} with x" -* c? 13. Let f, g : [0, oo) - R be continuous on (0, oo) and differentiable on (0, oo). If f(0) = g(0) and f'(x) a g'(x) for all x e (0, oo), prove that f(x) z g(x) for all x E (0, oo). .

14. A differentiable function f : [a, b] - R is uniformly differentiable on [a. b] if for every E > 0 there exists a S > 0 such that

f(t) - f(x)

t-x

<E

5.2 The Mean Value Theorem

189

for all r, x E (a, b] with 0 < It - xl < S. Show that f is uniformly differentiable on (a. b] if and only if f is continuous on [a, b]. 15. *Suppose f : [a, b] -> R with f. (a) > 0. Prove that there exists a S > 0 such that f (x) > f(a) for all

x, a<x
16. Prove that the equation x3 - 3x + b = 0 has at most one root in the interval [ -1. 1 ]. M. 'Suppose gas differentiable on (a. b) with I g'(x) I s M for all x E (a. b). Prove that there exists an e > 0 such that the function f(x) = x + eg(x) is one-to-one on (a, b). 18. Let

I x+2x2 sin

f(x)

X,

x#0,

x=0.. 0, a. Show that f'(0) = 1.b. Prove that f'(x) assumes both positive and negative values in every neighborhood of 0.

19. Let f be defined by .

f(x) =

x'(2r.,+ all) X :A 0, ..

x=0.

0,

a. Show that f has an absolute minimum at x = 0. b. Show that f'(x) assumes both negative and positive values in every neighborhood of 0.

20. Let f(x)=x2,g(x)=x3,xE[-1,l]. a. Find c E (-1, 1) such that the conclusion of Theorem 5.2.8 holds. b. Show that there does not exist any c E (-1, 1) for which

Al) -f(-1) f(c) g(1) - g(-1) g'(c)' 21. For r E 0 and x > 0, let f(x) = x'. Prove that f'(x) = rx'-'. 22. Suppose L : (0, oo)

R is a differentiable function satisfying L'(x) = 1/x with L(1) = 0. Prove each of the fol-

lowing.

a. L(ab) = L(a) + L(b) for all a, b E (0, oo)

b, L(I/b) = -L(b), b > 0 c. L(b') = rL(b), b > 0, r E R d. L(e) = 1, where e is Euler's number of Example 2.3.5

e. Range L = R 23. Let g(x) - tan x, -f < x < . a. Show that g is one-to-one on (-f, i) with Range g = R. 'b. Let Arctan x, x E R, denote the inverse function of g. Use Theorem 5.2.14 to prove that

d

An:tan x

+ x2 .

24. a. , Show that f(x) = sin x is one-to-one on [ -i, f] with f([ - z, ]) _ [ - I,.1 J. z

b. For x E (- I, 1]. let Arcsin x denote the inverse function off. Show that Arcsin x is differentiable on (-1, 1). and find the derivative of Arcsin x.

190

Chapter 5

Differentiation

25. Let f : (0, oo) -> R be differentiable on (0, oo) and suppose that lim f'(x) = L. f (x + h f(x) a. Show that for any h > 0, lim = L. h

b. Show that lim f

5.3

-

x

)) = L.

L'Hospital's Rule As another application of the mean value theorem, we now prove ('Hospital's rule for

evaluating limits. Although the theorem is named after the Marquis de l'Hospital (1661-1704), it should be called Bernoulli's rule. The story is that in 1691, l'Hospital asked Johann Bernoulli (1667-1748) to provide, for a fee, lectures on the new subject of calculus. L'Hospital subsequently incorporated these lectures into the first calculus text, L'Analyse des infiniment petis (Analysis of infinitely small quantities), published in 1696. The initial version (stated without the use of limits) of what is now known as l'Hospital's rule first appeared in this text.

Infinite Limits Since l'Hospital's rule allows for infinite limits, we provide the following definitions.

5.3.1

DEFINITION Let f be a real-valued function defined on a subset E of R and let p be a limit point of E. We say that f tends to oo, or diverges to oo, as x approaches p. denoted

Jim f (x) = 00, X-+p

if for every M E R, there exists S > 0 such that

f(x) > M for all x E E with 0 < Ix - pI < S. Similarly,

lim f(x) _ -00, if for every M E R, there exists a S > 0 such that

f(x) < M for all x E E with 0 < Ix - pI < S. For f defined on an appropriate subset E of P. it is also possible to define each of the following limits:

lint, f (x) =too,

X+p

lim f (x) =too,

X-+p

Jim f (x) =too,

X- 00

Jim f (x) =too.

XH-x

Since these definitions are similar to Definitions 4.1.11 and 4.4.1 they are left to the exercises (Exercise 1).

5.3

!Hospital's Rule

191

Remark. Since we now allow the possibility of a function having infinite limits, it needs to be emphasized that when we say that a function f has a limit at p e I3 (or at ± oo), we mean a finite limit.

L'Hospital's Rule L'Hospital's rule is useful for evaluating limits of the form gx(x)

lim

where either (a) lim f(x) = lim g(x) = 0 or (b) f and g tend to ± oo as x -+p. If (a) holds, then 1 m W 419W) is usually referred to as indeterminate of form 0/0, whereas in (b) the limit is referred to as indeterminate of form oo/oo. The reason that (a) and (b) are indeterminate is that previous methods may no longer apply. In (a), if either lim f (x) or lim g(x) is nonzero, then previous methods discussed in x-ip

Section 4.1 apply. For example, if both f and g have limits at p and lim g(x) * 0, then p by Theorem 4.1.6(c),

f(x) lim -_ x-*p 8(x)

lim f(x) lim g(x)

On the other hand, if lim f (x) = A * 0 and g(x) > 0 with lim g(x) = 0, then as X-P x-ip, f(x)/g(x) tends to `Z oo if A > 0, and to -oo if A < 0 (Exercise 5). However, if lim f(x) = lim g(x) = 0, then unless the quotient f(x)/g(x) can somehow be simplified, previous methods may no longer be applicable.

5.3.2

THEOREM (L'Hospital's Rule) Suppose f, g are real-valued differentiable functions

on (a, b), with g'(x) # O for all x E (a, b), where -oo lim

f,x)

x-.a g 'W

a < b s oo. Suppose

= L, where L E R U {-oo, oo).

if

(a) lim f(x) = lim g(x) = 0, or x-sa* (b) lint g(x) = ±oo, then lim

f ((x()

x-.a' g(x)

Remark. The analogous result where x

= L.

b- is obviously also true. A more elementary version of 1'Hospital's rule, which relies only on the definition of the derivative, is

192

ChaprerS

Differentiation

given in Exercise 2. Also, Exercise 7 provides examples of two functions f and g satisfying (a) for which lim (f(x)/g(x)) exists but lim (f'(x)/g'(x)) does not exist. x-a x-a

Proof. (a) Suppose (a) holds. We first prove the case where a is finite. Let

be a

sequence in (a, b) with x -+a and x * a for all n. Since we want to apply the generalized mean value theorem to f and g on the interval [a, x,), we need both f and g continuous at a. This is accomplished by setting f(a) = g(a) = 0.

Then by hypothesis (a), f and g are continuous at a. Thus by the generalized mean value theorem, for each n E NI there exists c between a and /x such that p

[f(xa) - J

g(a))f (cn),

or f(xn)

_ f'(Ca)

g(xa)

g'(Cn)

Note, since g'(x) # 0 for all x E (a, b), g(x,) # g(a) for all n. As n --I- oc, c -> a'. Thus by Theorem 4.1.3 and the hypothesis,

'_°° AX-) =

l-

f' (x)

= L.

m g'(x) -

with x tea', the result follows. Suppose a = -oo. To handle this case, we make the substitution x = -1/t. Then as t -> 0+, x -> -oo. Define the functions (p(t) and *(t) on (0, c) for some c > 0 by Since the above holds for every sequence

ap(t) = f (- )

and

*(t) = g(- ! ).

We leave it as an exercise (Exercise 3) to verify that t-0 +G r)

x

lim

g(X)L,

and that

1m P(t) = lim iy(t) = 0. Thus by the above, t

xli.m- g(x) = 1-'0

00)

= L.

(b) Suppose I m, g(x) = oo. The case in which g(x) -+ -oo is treated similarly. Rather than treating the finite case and infinite case separately, we provide a proof that works for both.

5.3

CHospital's Rule

193

Suppose first that -oo s L < oo, and /3 E R satisfies /3 > L. Choose r such that

L < r < /3. Since lim f, (x)

r-.a g,(x)

< r,

there exists c, E (a, b) such that

r for all C, a < < c1.

g'(C)

Fix a y, a < y < c1. Since g(x) -* oo as x -+a+, there exists a c2, a < c2 < y, such that g(x) > g(y) and g(x) > 0 for all x, a < x < c2. Let x E (a, c2) be arbitrary. Then by the generalized mean value theorem, there exists E (x, y) such that

f(x) -AY) = f'(C) < r. g(x) - g(Y)

(v)

g'(C)

Multiplying inequality (4) by (g(x) - g(y))/g(x), which is positive, we obtain

f(x) - f(Y) < g(x)

r( l--

g(Y)1 g(x)

or

g(x)

<

g()

+ rl l -

(5) g(x

for all x, a < x < c2. Now for fixed y, since g(x) -+ oo, xm

g-) =

hm.

= 0. 8X)

Therefore,

1'm_IgX)+rl I L<

Thus there exists c3, a < c3

--ii =r
c2, such that

/

- +r(l for all x, a < x < c3. Thus by inequality (5),

f(x) x AX)

< /3 for all x, a < x < c3.

(6)

If L = -oo, then for any /3 E Q8, there exists c3 such that inequality (6) holds for all x, a < x < c3. Thus by definition,

f(x) 'tea' g(x )

-oo.

194

Chapter5

Differentiation

If L is finite, then given e > 0, by taking /3 = L + e, there exists c3 such that gXx))

x


for all x, a < x < c3.

(7)

Suppose -oo < L <_ oo. Let a E R, a < L be arbitrary. Then an argument similar to the above gives the existence of c3 (=- (a, b) such that

f(x) > a for all x, a

<x
.

g(X)

If L = oo, then the above implies that limf(x)

= oo.

,/

x-+a' g(x)

On the other hand, if L is finite, taking a = L - e gives the existence of c3 such that f(x)

x 8()

>L-e forallx , a<x
Combining this with inequality (7) proves that lim_

f(x)

x-*a 8(x)

= L. Q

Remarks (a) The proof of case (a) could have been done similarly to that of (b), treating the case where a is finite and -oo simultaneously. I chose not to do so since making the substi-

tution x = -I It is a useful technique, reducing problems involving limits at -oo to right limits at 0. Conversely, limits at 0 can be transformed to limits at too with the substitution x = 1/t. These new limits are in many instances easier to evaluate than the original. This is illustrated in Example 5.3.4(c).

(b) In hypothesis (b), we only required that lim g(x) = too. If lira f (x) is finite, then x-,a' x-a it immediately follows that f (X) lint-=0,

x+a 8(x)

and l'Hospital's rule is not required (Exercise 4). Thus in practice, hypothesis (b) of 1' Hospital's rule is used only if both f and g have infinite limits.

For convenience we stated and proved l'Hospital's rule in terms of right limits. Since the analogous results for left limits are also true, combining the two results gives the following corollary.

5.3.3

COROLLARY (L'Hospital's Rule)

Suppose f, g are real-valued differentiable functions

on (a, p) U (p, b) with g'(x) * Oforall x E (a, p) U (p, b), where -oo Suppose

lim' (x) = L, where L E R U {-oo, oo}. x-ip g'(x)

a < b < oo.

5.3

L'Hospital's Rule

195

If

f(x) = lim g(x) = 0, or (a) limf(x) (b) lim g(x)

00,

then

f(x)

limg(x) = L.

5.3.4

EXAMPLES

(a) Consider lira x-.0

ln(1+x X

where In is the natural logarithm function on (0, oo). This limit is indeterminate of form

0/0. With f(x) = In(1 + x) and g(x) = x,

f'x x

.O g'(x)_

10

1+x=1.

Thus by 1'Hospital's rule,

ln(1 + x)

lira A

X

= 1.

Although l'Hospital's rule provides an easy method for evaluating this limit, the result can also be obtained by using previous techniques. In Example 5.2.7 we proved that +xsln(1+x)5x

forallx> -1. Thus 1

:5

1 +x

ln(1+x) X

for all x > 0. Thus by Theorem 4.1.9,

ln(1 + x) x

11>o

-

1.

(b) In this example we consider 1 - cos x

O

xz

This is indeterminate of form 0/0. If we apply l'Hospital's rule we obtain 1

stn x 2x

196

Differentiation

Chapter5

which is again indeterminate of form 0/0. However, applying ('Hospital's rule one more time gives lim

x-O

cosx 2

I

=

2

Therefore,

lim

1-cosx

xti0

I

2.

x2

(e) Consider e-six

lim

x-.0-

X

Since lim a-11x = 0, the above limit is indeterminate of form 0/0. If we apply I'Hospix0' tal's rule, we obtain

0X e- tix

and this limit is more complicated than the original limit. However, if we let t = 1/x, then

e-11X

t

-,0 X = lim e

.

This limit is indeterminate of form oo/oo. By l'Hospital's rule, lim

t -= lim -=0. ,tea e' 1

roc er

Therefore, lim a-llx/x = 0.

EXERCISES 5.3 1. Provide definitions for each of the following limits.

a. A-P limf(x)=oo

b. limf(x)=oo x-4ao

2. *Suppose f g are differentiable on (a, b), xa E (a, b) and g'(xa) * 0. If f (xa) = g(xa) = 0, prove that

f(x)

f'(xa)

X _X, g(x)

g,(x0)

lim

(Hint: Apply the definition of the derivative.)

3. Let h(x) be defined on (-oo, b). Show that there exists a c > 0 such that rp(t) = h(-;) is defined on (0, c), and that l-.-aa lim h(x) = lim op(t). 1H04 4. 'Let f, g be real-valued functions defined on (a, b). If lim f(x) exists in R andx-a' lim g(x) = oo, prove that lint f(x)/g(x) = 0. xa'

5.4

Newton's Method

197

5. Suppose f, g are real-valued functions on (a, b) satisfying lim, f(x) = A * 0, lim* g(x) = 0. and g(x) > 0 for all x E (a, b). If A > 0, prove that lim f(x)/g(x) = oc. If A <0, prove that lim. f(x)/g(x) _ -oc. 6. Use I'Hospital's rule and any of the differentiation formulas from calculus to find each of the following limits. In the following, In x, x > 0, denotes the natural logarithm function. 1

a. lim

x5 +2x-3

b. ,Imi

2x3 - x2 - I

-x

c. lim

In x

d. lira _.0

e lim_ox° In x where a > 0 g. lim (i.

lim

=-o'

(In x)°

e` .r

,

f. lim ,-.o

h. lim Ay

where a > 0

x`+2x-3 2x' - x' - I I - cos 2Asin x sin x

(In x)" y

.r

,

p, q E fl

sin .rJ

7. Let f(x) = x2 sin(1/x) and g(x) = sin x. Show that lim

f( x)

exists but that lim

fg'(-x(x)

) does not exist.

box)

p(x), p(X) where p and q are polynomials of degree it and in. respectively. 8. Investigate lim and lim . q(x) x_'-« q(x) 9. Let f (x) = (sin x)/x for x * 0, and f (0) = I . a. Show that f'(0) exists, and determine its value. b. Show that f"(0) exists, and determine its value.

10. Let f be defined on R by, f(x) =

5.4

e_

for x # 0. and f(0) = 0. Prove that ft"'(0) = 0 for all n = I. 2. .

Newton's Method2 In this section we consider the iterative method, commonly known as Newton's method.

for finding approximations to the solutions of the equation f (x) = 0. Although the method is named after Newton, it is actually due to Joseph Raphson (1648-1715). and in many texts the method is referred to as the Newton-Raphson method. Newton did derive an iterative method for finding the roots of a cubic equation; his method, however. is not the one used in the procedure named after him. Suppose f is a continuous function on [a, b] satisfying f(a)f(b) < 0. Then f is of opposite sign at the end points a and b and thus by the intermediate value theorem (Theorem 4.2.11), there exists at least one value c E (a, b) for which f (c) = 0. If, in addition, f is differentiable on (a. b) with f'(x) * 0 for all x E (a, b), then f is either strictly increasing or decreasing on [a, b]. In this case the value c is unique; that is, there is exactly one point where the graph off crosses the x-axis.

2. The topics of this section are not required in subsequent chapters.

198

Chapter5

Differentiation

An elementary approach to finding a numerical approximation to the value c is the method of bisection. For this method, differentiability off is not required. To illustrate

the method, suppose f (as in Figure 5.9) satisfies f(a) < 0 < f(b). Let

c,=2(a+b). If f(c,) = 0, we are done. If f(c1) * 0, then c belongs to one of the two intervals (a, c1) or (c1, b), and thus Ic1 - ci < I' (b - a). Suppose (as in Figure 5.9) f(c1) > 0. Then c E (a, c1), and in this case we set co = a, and C2 = 2 (CO + CO-

If f(c2) = 0, we are done. If not, then suppose (as in Figure 5.9) f(c2) < 0. Then c E (c2, c1), and as above, we set C3 = I (Cl + C2).

c1,C2,...,cn,n?2

I//

C. = lCn-1 + C)

Figure 5.9

have been determined with , n - 2. If by happenstance f (cn) = 0, then we have obtained the exact value. If f(cn_ 1)f(cn) < 0, then c lies between cn_ 1 and c,, and we define

In general, suppose

for some j = 0, .

.

.

cnr1 = -2 2 (C. + Cn_1),

5.4

199

0, then c lies between c and c,, and in this case.

On the other hand, if we define

This gives us a sequence

Newton's Method

which satisfies

:2I(b-a). Thus lim c = c and by continuity, f(c) = 0. Although this method provides a sequence of numbers that converges to the zero off. it has the disadvantage that the convergence is rather slow. An alternative method, due to Raphson, uses tangent lines to the curve to find successive points c approximating the zero off. As we will see, this method will often converge much more rapidly to the solution.

As above, assume that f is differentiable on [a, b] with f (a) f (b) < 0 and f'(x) # 0 for all x E [a, b]. Let ct be an initial guess to the value c. The line tangent to the graph off at (c,, f (c,)) has an equation given by

Y =f(ct) +f(ct)(x - CO. Since f (c,) # 0, the line crosses the x-axis at a point that we denote by c2 (Figure 5.10). Thus

0 =f(ci) +f'(ct)(c2 - Cl), that, upon solving for c2, gives C2 = C, -

Figure 5.10

f(CI)

f'(CI)

Newton's Method

200

Chapter5

Differentiation

We now replace the point c, by the second estimate c2 to obtain c3, and so forth. Inductively, we obtain a sequence (cn) given by the formula c

-----

,

c"

n = 1,2...

(8)

in which c, is an initial guess to the solution f(c) = 0. As we will see, under suitable hypothesis, the sequence {c"} will converge very rapidly to a solution of the equation f(x) = 0. Before we prove the main result, we illustrate the above with an example.

5.4.1

EXAMPLE Let a > 0 and consider the function

f(x)=x2-a. If a > 1, then fhas exactly one zero on [0, a], namely \. If 0 < a < 1. then the zero off lies in [0, 1). Let c, be an initial guess to Va. Then by formula (8), for n ? 1,

cl. - a Cn+1 - Cn

2c"

l

a

- 2(C" + Cn).

This is exactly the sequence of Exercise 9 of Section 2.3, where the reader was ask to prove that the sequence converges to %/a-. With a = 2, taking c, = 1.4 as an initia guess yields c2 = 1.4142857, c3 = 1.4142135,

which is already correct to at least seven decimal places.

5.4.2

THEOREM Let f be a real-valued function on [a, b] which is twice differentiable on

[a, b]. Suppose that f(a)f(b) < 0 and that there exist constants m and M such that If'(x)I m > 0 and [f"(x) 1 s M for all x E [a, b]. Then there exists a subinterval I of [a, b] containing a zero c off such that {for any c, E 1, the sequence {c"} defined by J (Cn) T(c.), nEN, Cn+l=cn-f,(c),

is in 1, and n-co lim c" _. c. Furthermore,

C"+, - cl S

ICn - W.

(9)

Prior to proving Theorem 5.4.2 we first state and prove the following lemma. The result is in fact a special case of Taylor's theorem (8.7.16), which will be discussed in Chapter 8.

5.4.3

LEMMA Suppose f : [a, b] -a R is such that f and f' are continuous on [a, b] and f"(x) exists for all x E (a, b). Let x" E [a, b). Then for any x E [a, b], there exists a real number between x, and x such that

f(x) =f(xo) +f'(xe)(x - xo) + 2f"(C)(x - xo)2.

5.4

Proof.

Newton's Method

201

For x E [a, b], let a E R be determined by

f(x) = f(x,) +

CO -

Define g on [a, b] by

x,) - a(t - x,)2.

g(t) = f(t) - f(x.) -

If x = x, then the conclusion is true with C = x". Assume that x > x". Then g is continuous and differentiable on [x x] with g(x") = g(x) = 0. Thus by Rolle's theorem there exists c E (x0, x) such that g'(c) = 0. But

g'(t) = fi(t) -f'(x,,) - 2a(t - x,). By hypothesis, g' is continuous on [x c], differentiable on (x c), and satisfies g'(x,) _ g'(c) = 0. Thus by Rolle's theorem again, there exists C E (x c) such that 0. But

g"(t) = f"(t) - 2a. Therefore, a = if"(C)

Proof of Theorem 5.4.2. Since f(a) f(b) < 0 and f'(x) # 0 for all x E [a. b], f has exactly one zero c in the interval (a, b). Let x, E [a. b] be arbitrary. By Lemma 5.4.3 there exists a point C between c and x, such that

0 = f(c) = f (xo) + f'(xo)(c - x,) + f"(C)(c - x,)2, 2

or

- Ax.) = f' (xo)(c - x0) + 2 f "(3'xc - x,)2.

(10)

If x, is defined by X1 = X.

f(x.) - f,(xo),

then by equation (10).

X, =x,+(c- x,)+ 2 f,"(0 f )(c-x,)2. 1

0

Therefore, "

IX, - cl = 2

x,

Ic - x,12 s

Ic - x,12.

(11)

Choose s>0sothat 8<2m/M andI=(c-S,c+S] C [a,b].Ifc"E1,then lc - c"I < S. If c"+, is defined by formula (8), then by formula (11),

c"+,-c1<-

M

82<S.

Therefore, c"+, E I. Thus if the initial choice c, E I, then c" E I f o r all n = 2, 3, ... .

202

Chaprer5

Differentiation

It remains to be shown that 1 m c = c. If ct E 1, then by induction

ICn+I - CI <

2

n

-

I.

But by our choice of S. (M/2m) S < 1, and as a consequence cn -*c. Q

Remarks (a) For a given function f satisfying the hypothesis of the theorem, the constants M and m, and thus S can be determined. To determine the interval I, one can use the method of bisection to find an approximation xn to c satisfying Ix,, - x - I I < S. If cl is taken to be xn, then Icl - ci < S. In practice, however, one usually makes a judicious guess for c1 and proceeds with the computations.

(b) Let en = c - cn be the error in approximating c, and let K = M/2m. Then inequality (9) can be expressed as ien+1i C Kleni2.

Consequently, if I e,,I < 10', then I en + I I < K 10-2. Thus, except for the constant factor K, the accuracy actually doubles at each step. For this reason. Newton's method is usually referred to as a second order or quadratic method. (c) Even though Newton's method is very efficient, there are a number of things that can go wrong if c1 is poorly chosen. For example, in Figure 5.11, the initial choice of c1 gives a c2 outside the interval, and the subsequent cn tend to -cc. Such a function is given by f (x) = x/(x2 + 1). In Figure 5.12, the initial choice of c1 causes the subsequent values to oscillate between c1 and c2. A function having this property is given by g(x) = x - J Taking c1 = 1 gives c2 = -1, c3 = 1, etc. For this reason, the initial choice of c1 for many functions has to be sufficiently close to c in order to be sure that the method works.

(cl.ftc1))

(c2. f(c2))

Figure 5.11

Notes

203

Figure 5.12

EXERCISES 5.4 1. *For a > 0, apply Newton's method to f(x) = x3 - a to obtain a sequence of a.

that converges to the cube root

2. Use Newton's method to find approximations to the roots, accurate to six decimal places, of the given functions on the interval [0, 1 ].

*a. f(x) = x3 - 3x + 1

b. f(x) = 3x3 - 5x2 + I

c. f(x)=x`-3x3+1 3. Use Newton's method to approximate the real zeros of f(x) = x' - 4x - 3 accurate to four decimal places. 4. Show that f(x) = In x - x + 3, x E (0, oo), has two real zeros. Use Newton's method to approximate them accurate to four decimal places. 5. Suppose g : [a, b] -i, R is differentiable on an interval [a, b] and suppose that there exists a constant c, 0 < c < I, such that Ig'(x)I s c for all x E [a, b]. Prove, that there exists a unique point x, E [a, b] such that g(xo) = x,. (See Exercise 13, Section 4.3.) 6. Let f : [a, b] -* R be differentiable on [a, b] with f (a) < 0 < f(b). Suppose there exist constants m and M such that 0 < m <_ f'(x) s M for all x E [a, b]. Let c, E [a, b] be arbitrary, and define

C-1 = f(c,,) M Prove that the sequence

converges to the unique zero off on [a, b]. (Hint: Consider the function

g(x) = x - f(x)/M.)

NOTES Without question the most significant result of this chapter is the mean value theorem. The simplicity of its proof disguises the importance and usefulness of the result. The theorem allows us to obtain information about the funcdon from its derivative. This has many applications, as

was illustrated by the subsequent theorems and exercises. Additional applications will be encountered throughout the text. Although the mean value theorem is attributed to Lagrange, his proof, which appeared about 1772, was based

M

Chapter 5

Differentiation

on the false assumption that every function could be expanded in a power series. Cauchy, in his 1823 text Rfsumf des Lecons donnees a L'Ecole Royale Polytechnique sur le Calcul Infinitesimal, used the modern definition of the derivative to provide a proof of the mean value theorem. His statement and proof of the theorem, however, differ from the version detailed here in that he assumed continuity of the derivative. What Cauchy actually proved

was that if f is continuous on [a, b], then the quantity {f(b) - f(a)}/{b - a) lies between the minimum and maximum values of f on [a, b]. (See Miscellaneous Exercise 4.) Then by the intermediate value theorem (Theorem 4.2.1 1) applied to the continuous function f, there exists c E (a, b) such that

f(b) - f(a) = f'(c)(b - a). It is worth noting that our proof of the mean value theorem depends ultimately on the completeness property of R (through Rolle's theorem and Corollary 4.2.9). The mean value theorem can justifiably be called the fundamental theorem of differential calculus. It allowed the development of rigorous proofs of many results that were previously taken as fact or "proved" from geometric constructions. Although the modern proof of l'Hospital's rule uses the mean value theorem, it should be remembered that the original version for calculating the limit of a quotient where both the numerator and denominator become zero first appeared in 1696, over 70 years before Lagrange's proof of the mean value theorem. The original version was stated and "proved" in a purely geometric manner without reference to limits. For further details, including a history of calculus, the reader is referred to the text by Katz listed in the Bibliography.

The Bernoulli brothers, Jakob (1654-1705) and Johann (1667-1748), were among the first mathematicians in Europe to use the new techniques of Newton and Leibniz in the study of curves and related physical problems. Among these were finding the equations of the catenary

and isochrone.` Both brothers also contributed to the study of differential equations by solving the Bernoulli equation: y' + P(x)v = Q(x)y". Through their numerous publications and correspondence with other mathematicians, the Bernoulli brothers helped to establish the utility of the new calculus. The first text on differential calculus by ('Hospital also contributed significantly to popularizing the subject. Leonhard Euler (1707-1783), one of the most prolific mathematicians in history, contributed significantly to establishing calculus as an independent science. Even though the calculus of exponential and logarithmic functions was basically developed by Johann Bernoulli, it was Euler's expositions in the eighteenth century that brought these topics into the mainstream of mathematics. Much of what we know today about the exponential, logarithmic. and trigonometric functions is due to Euler. He was also among the first mathematicians to define the concept of a function. However, to Euler-as to the other mathematicians of that period-a function was one that had a power series expansion. It is important to note that most mathematicians of the eighteenth century. including Euler, were primarily concerned with computations needed for the ap, plications of calculus; proofs did not gain prominence until the nineteenth century. For this reason, numerous re-

sults of that era that were assumed to be true were subsequently proved to be true only under more restrictive conditions.

MISCELLANEOUS EXERCISES 1. Let f : (a, b) -> R and suppose f" exists at x E (a, b). Prove that

f"(z,)

limf(x,+h)+f(x0-h) -2f(x") h2

hn0

Give an example where this limit exists at x,, but f" does not exist at x,.

3. The catenary problem involves finding the equation of a freely hanging cable, whereas the isochrone problem involves finding the equation of a curve along which an object would fall with uniform vertical velocity.

Supplemental Reading

205

2. Let f be a real-valued differentiable function on R. a. If there exists a constant b < I such that I f'(x) I < b for all x E Ca, prove that f has a fixed point in R. (See Exercise 13, Section 4.3.) b. Show that the function f(x) = x + (1 + e`)-' satisfies I f'(x)l < I for all x E R but that f has no fixed point in R.

3. A function f is convex (or concave up) on the interval (a. b) if for any x, y E (a, b), and 0 < t < 1,

f(tx + (I -r)y)<-tf(x)+(I -t)f(y). a. If f is convex on (a. b). prove that f is continuous on (a, b). b. If f is convex on (a, b), prove that f+(p) and f _(p) exist for every p E (a, b). Show by example that a convex function on (a, b) need not be differentiable on (a, b). c. Suppose f"(x) exists for all x E (a, b). Prove that f is convex on (a, b) if and only if f"(x) at 0 for all

x e (a, b). 4. Suppose fis differentiable on [a, b] and that f is continuous on [a, bl.Without using the mean value theorem, prove that

min{f'(x):x e [a, b]} sf(bb

Q(a)

- max{f (x) :x E [a, b]}.

5. (T. M. Flett) If f is differentiable on [a, b] and f'(a) = f'(b), prove that there exists ( E (a, b) such that

f(C) - f(a) = C - a

R is differentiable at c E (a, b). If (s") and {t"} are sequences in (a, b) with s,, < c < t" and limoo (t" - s") = 0, prove that

6. Suppose f: (a, b)

lim

f(s") =

t"-s"

SUPPLEMENTAL READING Baxley, J. V. and Hayashi, E. K., "Indeterminate forms of exponential type," Amer. Math. Monthly 85 (1978), 484-486. Cajori, Florian, "Historical note on the Newton-Raphson method of approximation;' Amer. Math. Monthly 18 (1911), 29-33. Corless, R. M., "Variations on a theme of Newton;' Math. Mag. 71 (1998), 34-41. Flett, T. M., "A mean value theorem," Math. Gaz. 42 (1958), 38-39. Hall, W. S. and Newell, M. L., 'The mean value theorem for vector valued functions;' Math. Mag. 52 (1979), 157-158. Hartig, Donald, "L'Hospital's rule via integration;' Amer. Math. Monthly 98 (1991). 156-157. Langlois, W. E. and Holder, L. 1., "The relation of

f (a) to f'(a+)," Math. Mag. 39 (1966), 112-120. Katznelson, Y. and Stromberg, K., "Everywhere differentiable, nowhere monotone function;' Amer. Math. Monthly 81(1974). 349-354. Miller, A. D. and Vyborny, R., "Some remarks on functions with one-sided derivatives;' Amer. Math. Monthly 93 (1986),471-475.

Rosenholtz, "Mere is no differentiable metric for W." Amer. Math. Monthly 86 (1979), 585-586. Rotando, L. M. and Korn, H., "The indeterminate form 0°," Math. Mag. 50 (1977),41-42. Thurston, H. A., "On the definition of the tangent line;' Amer. Math. Monthly 71 (1964), 1099-1103. Tong, J. and Braza, P. A., "A converse of the mean-value theorem;' Amer. Math. Monthly 104 (1997). 939-942.

The Riemann and Riemann-Stieltjes Integral 6.1 The Riemann integral 6.2 Properties of the Riemann Integral 6.3 Fundamental Theorem of Calculus 6.4 Improper Riemann Integrals 6.5 The Riemann-Stieltjes integral

6.6 Numerical Methods 6.7 Proof of Lebesgue's Theorem

When Newton and Leibniz developed the calculus, both considered integration as the inverse operation of differentiation. For example, in the De analysi. t Newton proved

that the area under the curve y = a''(m/n # -1) is given by an

t

m+n by using his differential calculus to prove that if A(x) represents the area from 0 to x then A'(x) = ax'"/". Even though Leibniz arrived at the concept of the integral by using sums to compute the area, integration itself was always the inverse operation of differentiation. Throughout the eighteenth century, the definite integral of a function f(x) on [a, b], denoted fa f(x)dx, was defined as F(b) - F(a) where F was any function whose derivative was f (x). This remained the definition of the definite integral until the 1820s. The modem approach to integration is again due to Cauchy, who was the first mathematician to construct a theory of integration based on approximating the area under the curve. Euler had previously used sums of the form J'=, f(xk _ ,)(xk - x,-,) to approximate the integral of a function f(x) in situations where the function F(x) could not be computed. Cauchy, however, used limits of such sums to develop a theory of integration that was independent of the differential calculus. One of the difficulties with 1. "The Mathematical Works of Isaac Newton;' edited by D. T. Whiteside, Johnson Reprint Corporation. Ness York, 1964.

208

Chapter 6

The Riemann and Riemann-Stieltjes Integral

Cauchy's definition of the integral was that it was very restrictive: only functions that were continuous, or continuous except at a finite number of points, were proved to be integrable. However, one of Cauchy's key achievements was that using his definition, he was able to prove the fundamental theorem of calculus; specifically. if f is continu-

ous on [a, bi, then there exists a function F on [a, hl such that F'(x) = f(x) for all x E [a, b]. The modern definition of integration was developed in 1853 by Georg Bernhard Riemann (1826-1866). Riemann was led to the development of the integral by trying to characterize which functions were integrable according to Cauchy's definition. In the process, he modified Cauchy's definition and developed the theory of integration that bears his name. One of his achievements was providing necessary and sufficient conditions for a real-valued bounded function to be integrable. In Section 6.1 we develop the theory of the Riemann integral using the approach of Jean Gaston Darboux (1842-1917). In this section we also include the statement of Lebesgue's theorem, which provides necessary and sufficient conditions that a bounded real-valued function defined on

a closed and bounded interval be Riemann integrable. A self-contained proof of Lebesgue's theorem will be given in Section 6.7. The equivalence of the Riemann and Darboux approach will be proved in Section 6.2. In Section 6.5 we will consider the more general Riemann-Stieltjes integral that will give meaning to the following types of integrals: +

f1

f J '(x) d[xor

f(x) dx', J0

f(x) da(x), ,

11

where a is a monotone increasing function on [a, b]. These types of integrals were developed by Thomas-Jean Stieltjes (1856-1894) and arise in many applications in both mathematics and physics. The theory itself involves only minor modifications in the definition of the Riemann integral; the consequences, however, are far-reaching. The Riemann-Stieltjes integral permits the expression of many seemingly diverse results as a single formula.

6.1

The Riemann Integral There are traditionally two approaches to the theory of the Riemann integral: the original method of Riemann and the method introduced by Darboux in 1875 using lower and upper sums. I have chosen the latter approach to define the Riemann integral because of its easy adaptability to the Riemann-Stieltjes integral. We will, however, consider both methods and show that they are, in fact, equivalent.

Upper and Lower Sums Let [a, b], a < b, be a given closed and bounded interval in R. By a partition 91 of [a, b] we mean a finite set of points 91 = {xo, x1..

a=xo<x,<

.. ,

<xb.

such that

6.1

The Riemann Integral

209

There is no requirement that the points x, be equally spaced. For each i = 1. 2.... , n. set

Ax, = x, - x;_,, which is equal to the length of the interval [x,x,].

Suppose f is a bounded real-valued function on [a, b]. Given a partition of [a, b], for each i = 1.2.. ... n. let

9 = {xo, x......

in, = inf{f(t):x,_, <_ t <_ x,}.

M, = sup{f(t):x, _, 5 t s x,}. Since f is bounded, by the least upper bound property, the quantities m, and M, exist in R. If f is a continuous function on [a, b], then by Corollary 4.2.9. for each i there exist

points t s; E [x,_,.x,] such that M, = f(t,) and m, = f(sj. The upper sum °U(9, f) for the partition 9 and function f is defined by

1110"D = Y M;ox,. Similarly, the lower sum Y(9, f) is defined by n

20"D = I,8=1MAX, Since m, <_ M, for all i = 1, .

.

.

, n, we always have

s for any partition 91 of [a, b]. The upper sum for a nonnegative continuous function f is illustrated in Figure 6.1. In this case °lt(2P, f) represents the circumscribed rectangular approximation to the area under the graph off. Similarly, the lower sum represents the inscribed rectangular approximation to the area under the graph off.

Figure 6.1

Upper Sum U(9. j)

210

Chapter6

The Rlemann and Riemann-Stleltjes Integral

Upper and Lower Integrals If the function f satisfies m : f(t) <- M for all t e [a, b], then

m(b - a) c

--5 M(b - a)

(1)

for any partition 9 of [a, b]. To see that inequality (1) holds, let iI' = {x°, ... , any partition of [a, b]. Since M; <- M for all i = 1, ... , n, 61491, f)

M4Axi `-

M(X,

be

- xi -1) = M(b - a).

Similarly, -W(9, f) z m(b - a). Thus the set {°G.(91, f) : 91 is a partition of [a, b]} is bounded above and below, as is the set {2(9', f)}.

6.1.1

Let f be abounded real-valued function on the closed and bounded interval [a, b]. The upper and lower Integrals off, denoted fQ f and fa f respectively, are defined by DEFINITION

b f = inf{°U,(9, f) : 91 is a partition of [a, b]), rb

f = sup{.T(91, f) : 9' is a partition of [a, b]}.

Ja

Since the sets {° L (91j )j )} and {2(9,f)} are nonempty and bounded, the lower and upper integrals of a bounded function f : [a, b] -+ R always exist. Our first goal is to prove that f; f s f. f for any bounded real-valued function f on [a, b]. To this end, we make the f!1owing definition.

6.1.2

DEFINITION A partition 9* of [a, b) is a refinement of 9 if 9' C 9*. A refinement of a given partition 9 is obtained by adding additional points to 91. If 911 and 9'2 are two partitions of [a, b], then 9, U 9'2 is a refinement of both 91, and 912.

6.1.3

LEMMA If 91* is a refinement of 9, then

2(91,f) 2(91*,f) s w.(91*,f) < °u(91,f). Proof. Suppose 9' _ {x°, x1, ... , xa} and 91* = 9 U {x*}, where x* * x; for any j = 0, 1, ... , n. Then there exists an index k such that xk_ < x* < xk. Let 1

Mk = sup{f(t) : t E [Xk_ 1, x*]}, Mk' = sup{f(t) : t E [x*, Xk]}.

Since f(t) s Mk for all t e [xk_,, xk], we have that f(t) s Mk for all t E (xk_ 1, x*] and also for all t E [x*, xk]. Thus both Mk and Mk are less than or equal to Mk. Now k-1

n

M;Ax, + MM(X* - Xk-1) + Mk(xk - x*) +

1149'*,f) _ 1=1

MMAzr l-k-I

6.1

The Riemann Integral

211

But Mk(X*

- Xk-1) + Mi(xk - X*)

MkAXA.

Therefore,

The proof for the lower sum is similar. If P* contains k more points than 91, we need only repeat the above argument k times to obtain the result.

6.1.4 THEOREM Let f be a bounded real-valued function on [a, b]. Then

t

jaf :S:

f

Proof. Given any two partitions 9, 9 of [a, b].

2(91,f) S 2(9 U 9.,f) S 1(9 U 9.,f) t Thus 2(91,f) < °ll.(a f) for any partitions 9, 9.. Hence

Ji= up X(O,f) S Gll.(a,f) for any partition a. Taking the infimum over 9, gives the result. Q

The Riemann Integral If f : [a, b] -+ R is bounded, then the lower and upper integrals off on [a, b] always exist and satisfy fo f < f° f. As we will shortly see, there is a large family of functions for which equality holds: such functions are said to be integrable.

6.1.5

Let f be a bounded real-valued function on the closed and bounded interval [a, b]. If DEFINITION

then f is said to be Riemann integrable or integrable on [a. b]. The common value is denoted by Jb b Jf(x)dx,

or a

f

a

and is called the Riemann Integral or integral off over [a, b]. We denote by 9t[a, b] the set of Riemann integrable functions on [a, b]. In addition, if f E 9t[a, b], we define Jbaf

-Jabs

212

Chapter6

The Riemann and Riemann-Stieltjes Integral

If f : [a, b) -> R satisfies m s f(t)

M for all t e [a, b], then by inequality (1),

m(b-a): Jbf

Ihf <_M(b-a).

a

a

If, in addition, f E 91[a, b], then

m(b-a)<- Jf b sM(b-a). a

In particular, if f(x) ? 0 for all x E [a, b], then fa f >_ 0. If f E 9t[a, b] is nonnegative, then the quantity fa f represents the area of the region bounded above by the graph

y = f(x), below by the x-axis, and by the lines x = a and x = b.

6.1.6

EXAMPLES

(a) The following function, attributed to Dirichlet, is the canonical example of a function that is not Riemann integrable on a closed interval. Let f be defined by

f(x) =

x E Q,

11,

0, x0Q.

Suppose a < b. If 9 = {xo, ... , x} is any partition of [a, b], then m; = 0 and M; = 1 for all i = 1, ... , n. Thus

2(9,.f) = 0

°1L(9,f) = (b - a)

and

for any partition 9 of [a, b]. Therefore, fb

f=0

J f=(b-a),

and

1a

a

and as a consequence, f is not Riernann integrable on [a, b].

(b) Let f : [0, 1 ] -> l be defined by

0:5 x < 1

In this example we prove that f is integrable on [0, 1] 9 = {x°, x,, .

with f, f = 2. Let

.. , x,} be a partition of [0, 1 ], and let k E {1, ... , n} be such that

xk_, < I - xk (See Figure 6.2.) If mi and M; denote the infimum and supremum off on [xi _ 1, x;] respectively, then

- f0, f

m'

1,

i = 1,.. ., k,

i=k+I,..

,n,

and

M;

-

f0, 1,

i = 1,..

i=k,..

k - I, n.

6.1

The Riemann Integral

213

i

I

Xk

xn=1

2

Figure 6.2

Thus n

x(91, f) =

i-k+t

Axi = (1 - xk),

and

ft

a(L(P,f) = I Oxi = (I - xk-1) i-k

Since 1 - xk s 2 < 1 - xk_ 1, 2(91,f) s 2 <- °lt(91, f) for all partitions 91 of [0. 1 ]. Thus fo f s 2 s fo f. Hence if f is integrable on [0, 11, then fo f = 2. We conclude by proving that f is indeed integrable on [0, 1). Since I - xk _ , = (1 - xk) + (xk - xk _ , ), we have

`'ll(`',f) = 2(9',f) + (Xk - xk-1) Let e > 0 be arbitrary. If 9 is any partition of [0, 1) with i xi < e for all i, then nu(9P,f) < 2(91,f) + e. Thus I

f f - ifk49-1f)s°uP1,f) r1

s 2011f) +

e c

sup2(`9-,f)+e51 f+e, 0

where the infimum and supremum are taken over all partitions a of [0, 1). Since E > 0 was arbitrary, we have j' If = fo f. Thereforef is integrable on [0, 11 with f, f = 2.

214

Chapter 6

The Riemann and Riemann-Stieltjes Integral

(c) We now provide another example to illustrate how tedious even a trivial integral can be if one relies only on the definition of the integral. Luckily, the fundamental theorem

of calculus (Theorem 6.3.2) will allow us to avoid such tedious computations. Let f(x) = x, x E [a, b]. For the purpose of illustration we take a > 0 (Figure 6.3). Interpreting the integral as the area under the curve, we intuitively see that

tbx

=

-a)(b+a)=(b2-a22).

This is obtained from the formula for the area of a parallelogram. Let 91 be any partition of [a, b]. Since f(x) = x is increasing on [a, b], {x0, x,, ... ,

m;=f(x;_,)=x;_,

M;=f(x;)=x;.

and

Figure 6.3

Therefore,

4911f)

x; _ , Ox;

and

949, f)

x; Ax;-

For each index i,

x;-1 < 2(xi_, +x) <_x;. If we multiply this by Ox; = x; - x;_, and sum from i = 1 to n, we obtain

l (xi

-

_')

°d(g1,f)

But ;_, 2

(x?-xi_i)= 2 (xn-4o)= 2 (b2-a2).

6.1

The Riemann Integral

21S

Finally, if we take the supremum and infimum over all partitions 9' of [a, b], we have J6

xdxs2(b2-a2):5 f xdx. a

This in itself does not prove that f(x) = x is integrable on [a, b] with Jb

xdx = 2 (b2 - a2). a

To prove that f is integrable on [a, b], we note that for any partition 9' of [a, b], fb

0c a

rb

xdx - J x dx <_ alt(9i, f) - T(91,f) = I (xi - xi -1)Axi i-1 a

< (1ma5 Axi) (b - a).

Let e > 0 be given. If we choose a partition 9' such that Axi <. a/(b - a + 1) for all i, then fax dx - fax dx < e. Since this holds for every e > 0, we conclude that

fa x dx = fax dx. Thus f(x) = x is Riemann integrable on [a, b] with fax dx = 2 (b2 - a2).

(d) Consider the function f(x) = x2, x E [0, 1]. For n E Ni, let 9' be the partition {0, ,'- 2, ... , 1}. Since f is increasing on [0, 1 ], its infimum and supremum on each

interval [Y,11 are attained at the left and right endpoint respectively, with m, = (i - 1)2/n2 and Mi = i2/n2. Since Axi = 1/n for all i, and

2(9,,,f)°n3[12+22+.+(n- 1)2],

n

Using the identity 12 + 22 +

+ m2 = 6m(m + lX2m + 1) (Exercise 1, Section

1.3), we have

491.1f) = 6 C1

n/

(2 - n

and

1UP1,,,f) = 6 ( 1 -1

n)(2+1).

Thus sup.T(9D,,, f) = 1 and inf °lL(91,,, f) = 3. Since the collection 191.: n e N} is a subset of the set of all partitions of [0, 1 ], 1

3

= sup 201., f) :5

`-f( ',f) = Jx2 dx,

and

0

3 = inf °lt(9,,, f) f °lt( P, f) = J.x2dx. 0

Therefore f(x) = x2 is integrable on [0, 1 ] with fo x2 dx =

i

.

216

Chapter6

The Rietnann and Riemann-Stieltles Integral

Riemann's Criterion For Integrability The following theorem, the original version of which is attributed to Riemann, provides necessary and sufficient conditions for the existence of the Riemann integral.

6.1.7

THEOREM Abounded real-valued function f is Riemann integrable on [a, b] if and only if for every E > 0, there exists a partition 9 of [a. b) such that E. (2) RL(9',f) Furthermore, if 9 is a partition of [a, b] for which inequality (2) holds, then the in-

equality also holds for all refinements of 9.

Proof.

Suppose inequality (2) holds for a given e > 0. Then

0C I f- Jf_
a

a

Thug f is integrable on [a, b]. Conversely, suppose f is integrable on [a, b]. Let e > 0 be given. Then there exist partitions 9, and 92 of [a, b] such that

jbf °lL(92,f) -

<2

and

tbf - 491'-f) < 2

Let 9 = 91 U 92. Then

jh 0149V) 90149V)

L(92,f) < u

f + 2 < `-P(('1Pi,f) + E

E.

Therefore, °U.(9, f) - %(9,f) < e, which proves inequality (2). If D. is any refinement of 9, then by Lemma 6.1.3,

0 <_ qt(L,f) - 2(a,f) :9 6u(P,f) - e(91,f) < E. Thus inequality (2) is also valid for any refinement 2 of 9. Q

Integrability of Continuous and Monotone Functions As an application of the previous theorem, we prove that every continuous teal-valued function and every monotone function on [a, b] is Riemann integrable on [a. b]. As we will see, both of these results will also follow from Lebesgue's theorem (Theorem 6.1.13).

6.1.8 THEOREM Lei f be a real-valued function on [a, b]. (a) If f is continuous on [a, b), then f is Riemann integrable on [a, b]. (b) If f is monotone on [a, b], then f is Riemann integrable on [a, b].

Proof. (a) Let e > 0 be given. Choose nl > 0 such that (b - a)rl < E. Since f is continuous on [a, b], by Theorem 4.3.4,f is uniformly continuous on [a, b]. Thus there

exists a S > 0 such that

[f(x) - f(t) I < 77

(3)

6.1

The Riemann Integral

217

for all x, t E [a, b] with Ix - tl < S. Choose a partition 91 of [a, b] such that Ox; < S for all i = 1, 2, ... , n. Then by inequality (3),

Mi - misrl for all i = 1, 2, ... , n. Therefore,

61(P,f)-201,f)=

(M,-mi),%xi:S71

Ax,=rl(b-a)<e.

Thus by Theorem 6.1.7,f is integrable on [a, b].

(b) Suppose f is monotone increasing on [a, b]. For n E h1, set h = (b - a)/n. is a partition of . , n, set xi = a + ih. Then 9' = { x o , x,, ... , [a, b] that satisfies Axi = h for all i = 1, . . . , n. Since f is monotone increasing on

Also f o r i = 0, 1 , ..

[a, b], mi = f(xi_ i) and Mi = f(x,). Therefore,

[f(x;) -f(xi-1)]Ax,

NOV) - `^e(911f) = =h

(f (xi) - f (xi- 01

(b - a) n

[f(b) - f(a)].

Given e > 0, choose n E N such that

(b - a) n

[f(b) - f(a)] < E.

For this n and corresponding partition 9, OIL.(91, f) - 2(91j) < e. Thus f is integrable on [a, b].

The Composition Theorem We next prove that the composition iP -f, of a continuous function gyp, with a Riemann integrable function f, is again Riemann integrable. As an application of Lebesgue's theorem we will present a much shorter proof of this result later in the section.

6.1.9 THEOREM Let f be a bounded Riemann integrable function on [a, b] with Range f C [c, d]. If ip is continuous on [c, d], then tp -f is Riemann integrable on [a, b].

Proof. Since 9 is continuous on the closed and bounded interval [c, d], ip is bounded and uniformly continuous on [c, d]. Let K = sup{IQp(t)I: t E [c, d]}, and let e > 0 be

given. Set e' = e/(b - a + 2K). Since ip is uniformly continuous on [c, d]. there exists 8, 0 < S < e', such that (4) 19(s) - (P(01 < e' for all s, t E [c, d] with Is - tI < S. Furthermore, since f E 9%[a, b], by Theorem 6.1.7 there exists a partition 9 ' = {xo, . . . . . x}} of [a, b] such that

%(9',f) - _y(9', f) < S2.

218

Chapter6

The Rlemann and Riemann-Stieltjes Integral

To complete the proof we will show that (5)

°1491, (P of) - 201, (P of) C E.

By Theorem 6.1.7 it then follows that (p of E Jt(a, b]. For each k = 1, 2, . . . , n, let ink and Mk denote the infimum and supremum off on [xk_ 1, xk). Also, set

mA = inf{cp(f(t)) : t E [x,-,, xJ) M,*t = sup{p(f(t)):1 E [xk_I,xk]}.

and

We partiton the set 11, 2, ... , n} into disjoint sets A and B as follows:

A={k:Mk-mk<6}

and

B={k:Mk-mka6}.

We first note that I f(t) - f(s)l 5 Mk - ink for all s, t E [xk_,, xk]. Hence if k E A. then by inequality (4),

IcP(f(t)) - V(f(s))j < e' for all s, t e [xk_,, xk]. But Mk

- m4 = sup{cp(f(t)) -'p(f(s)): s, t E [Xk_1,xk]}.

lift*

- mk Therefore, Mk* -mk < 2K. Thus

s e' for all k E A. On the other hand, if k E B. then

11491, (P -f) - 2(9"P of) _

(Mk - mk )AX* + kEA

tea

(Mk - mk)AX&

5 E' 7, Axk + 2K T, AXk kEB

kEA

5 e'(b - a) + 2K Y, Axk. kEB

But for k E B, 6 s Mk - Mk. Therefore, 7' (Mk - mk)I&Xk s s (043V)

Oxk 5 kEB

S kEB

- 49V)) < S < e'.

Hence by the above,

'k(91, V o f) - £('P, qp o f) 5 e'(b - a) + 2KE' = e. This establishes inequality (5), and thus q) of E cJt[a, b]. U As a consequence of the previous theorem, if f is Riemann integrable on [a, b], then so are the functions If I and f2 . For emphasis we state this as a corollary.

6.1.10 COROLLARY If f E 9t[a, b], then If ! and f 2 are Riemann integrable on [a, b]. A natural question to ask is whether the composition of two Riemann integrable functions is Riemann integrable. In Example 6.1.14(b) we will show that the answer is an emphatic no!

6.1

The Riemann integral

219

Lebesgue's Theorem In Theorem 6.1.8(a) we proved that every continuous function on [a, b] is Riemann integrable on [a, b]. By Exercise 16, this is also true for every bounded function on [a, b] that is continuous except at a finite number of points. On the other hand, as a consequence of Theorem 6.1.8(b), every monotone function on [a, bJ is Riemann integrable. is an enumeration of the rational numbers in [0, 1 ] and Hence, for example, if c > 0 are such that T, c, converges, then by Theorem 4.4.10, 1

00

f(x) = 71 C. 1(x n=1

is monotone increasing on [0, 1 ], and thus is Riemann integrable on [0, 11. By Theorem 4.4.10, the function f is continuous at every irrational number and discontinuous at every, rational number in [0, 1]. We now state the beautiful result of Lebesgue that provides necessary and sufficient conditions that a bounded real-valued function on [a, b) be Riemann integrable. To properly state Lebesgue's result we need to introduce the idea of a set of measure zero. The concept of measure of a set will be treated in detail in Chapter 10. The basic idea is that the measure of an interval is its length. This is then used to define what we mean by measurable set and the measure of a measurable set. At this point we only need to know what it means for a set to have measure zero.

6 1.11

DEFINITION A subset E of P has measure zero if given any e > 0, there exists a finite or countable collection {1n)n of open intervals such that

ECUJ

and

e, n

where l(1n) denotes the length of the interval In.

6.1.12

EXAMPLES

(a) Every finite set E has measure zero. Suppose E = {x1,

. .

. , xN} is a finite subset of

R. For each n = 1, 2, ... , N, as in Figure 6.4 (with N = 6), let

/

E

Nxn+2N, .

Then N

N

ECU /

> l(!n) = e.

and

n=1

n=1

Therefore, E has measure zero. it

14

12

(0) (0) Xt

X4

X2

16

l5

X3 X6

XS

13

Figure 6.4

220

Chapter 6

The Riemann and Riemann-Stleltjea Integral

(b) Every countable subset of R has measure zero. Suppose E = {x,}00 I is a countable

subset of R. Let f > 0 be given. For each n E N. let E l

E

Since xn E 1 for all n, E C U. 11,,. Thus since 1(I4) = E12n, 00

j l(In) = E

w=1

2n = E. n=1

As an example, the set Q of rational numbers has measure zero.

(c) The Cantor set Pin [0, 1] has measure zero (Exercise 21). We now state the following theorem of Henri Lebesgue, the proof of which will be given in Section 6.7. This result appeared in 1902 and provides the most succinct form of necessary and sufficient conditions for Riemann integrability.

6.1.13 THEOREM (Lebesgue) A bounded real-valued function f on [a. b] is Riemann integrable if and only if the set of discontinuities off has measure zero.

Remark. If f is continuous on [a, b], then clearly f satisfies the hypothesis of Theorem 6.1.13 and thus is Riemann integrable. If f is a bounded function that is continuous except at a finite number of points, then by Example 6.1.12(a) the set of discontinuities off has measure zero. Hence f E 9t[a, b]. If f is monotone on [a, b], then by Corollary 4.4.8, the set of discontinuities off is at most countable, and thus by Example 6.1.12(b).

has measure zero. Hence again f E 9t[a, b]. As an application of Lebesgue's theorem we give the following short proof of Theorem 6.1.9.

Proof of Theorem 6.1.9 Using Lebesguevs Theorem. As in Theorem 6.1.9, suppose f E 9t[a, b] with Range f C [c, d], and suppose rp : [c, d] -, R is continuous. Let

E = {x E (a, b): f is not continuous at x} and F = {x E [a, b): rp of is not continuous at x}. By Theorem 4.2.4, F C E. Since f is Riemann integrable on (a, b], the set E has measure zero, and as a consequence so does the set F. Therefore. V of E 9t[a, b]. Q

6.1.14

EXAMPLES

(a) As in Example 4.2.2(g), let f be defined on [0, 1] by

11, x=0, f(x) =

0, 1n,

if x is irrational,

if x = n in lowest terms, x * 0.

6.1

The Riemann Integral

221

Since f is continuous except at the rational numbers, which have measure zero, f is Riemann integrable on [0, 1 ]. Furthermore, since $(P, f) = 0 for all partitions 9 of [0, 1 ],

Jf(x)dx = 0. (b) Let f be the Riemann integrable function on [0, 1 ] given in (a), and let g : [0, 1 ] - R be defined by

0, x0,

g(x) __ 11, x E (0, 1].

Since g is continuous except at 0, g E %[0, 1 ]. But for x E [0, 1 ],

(g of)(x) = J 1, if x is rational,

0, if x is irrational.

By Example 6.1.6(a), g of 6E Jt[O, 1).

EXERCISES 6.1 1. Let f (x) = 1 - x2, x E [ -1, 2]. Find Sf(9, f) and qt(9J) for each of the following partitions of [ -1, 2].

a. 9 = {-1,0, 1,2) b.9={-1,-12,12,1,2,2} 2. Show that each of the following functions is Riemann integrable on [0, 2], and use the definition to find f02 I

*a f(x) =

2, 15 x 5 2 '

b f (x)

1,

0, sx < 1

3,

25 x<2

-2,

sx 52

2 3. Show that each of the following functions are Riemann integrable on [a, b] and find f; f.

0,

*a. f(x) = c

(c a constant)

b. f(x) =

2, 1,

a s x < c, x = c.

c<xsb

4. Use one of the methods of Examples 6.1.6 to find fo f for each of the following monotone functions f. *a. f(x) = [3x], where (x) is the greatest integer function b. f(x) = x[2x]

c. f(x) - 2x + I

d. f(x) = 1 - x2

5. Prove that f. *f = 0 for any real-valued function f on [a. a]. 6 . *If f , g E 3t(a, b) with f ( x ) 5 g ( x ) f o r all x E [a, b], p r o v e that f f s f g.

7. a. Suppose f is continuous on (a. b] with f(x) z 0 for all x E [a, b]. If f. f = 0. prove that f(x) = 0 for all x r= [a. b]. b. Show by example that the conclusion may be false if f is not continuous.

222

Chapter 6

The Rlemann and Riemann-Stieltjes integral

& *Let f : [(0, 11 - R be defined by

xEa, f()= {0, X. xEQ Compute fo f and fo f. Is f integrable on [0, 11?

9. Suppose f is a nonnegative Riemann integrable function on [a, b] satisfying f(r) = 0 for all r E Q fl [a, b]. Prove that f; f = 0. 10. *Use the method of Example 6.1.6(c) to show that

Ja

xdx= 3(b3-a3)(0:a
11. Suppose f is monotone increasing on [a, b]. For n E N, set It = (b - a)/n. L e t 9n = {xo, x,, ... , xn} where for

each k = 0,.

.

., n,xk = a + kh.

a. Prove that

f

0 s Wt(9n,f) -

(b

a)

[f(b) -f(a)]

b. Prove that f; f = lim bU(9n, f). 12. Use the previous exercise to evaluate the following integrals. b (I (I c. J x3 dx d. x3 dx x dx b. J (3x - 2) dx *a. I

J0

-2

0

1a

13. Let f be a bounded function on [a, b]. Suppose there exists a sequence {9n} of partitions of [a, b) such that lim SC(9n, f) = lim QU (9n, f) = L (L E 18).

ntiPO

n-wo

Prove that f is Riemann integrable on [a, b] with fa f = L.

14. a. If f e 9t[a, b], prove directly (without using Theorems 6.1.9 or 6.1.13) that If I E 9t[a, b].

b. If Ill E 9t[a, b], is f e 9t[a, b]? 15. *a. If f e 91[a, b], prove directly that f2 E 9t[a, b].

b. Give an example of a bounded function f on [a. b] for which f2 E 9t[a, b], but f e 9t[a, b]. 16. *Suppose f is a bounded real-valued function on [a, b] that has only a finite number of discontinuities. Prove directly that f is Riemann integrable on [a, b].

17. a. If E has measure zero, prove that every subset of E has measure zero. b. If E,, E2 have measure zero, prove that E, U E2 has measure zero. ao

c. If each En, n = 1, 2, ... , has measure zero, prove that U E. has measure zero. n=I 18. Use Theorem 6.1.13 to prove that if f, g r= 9t[a, b], then f + g E 9t[a, b]. 19. Prove directly that the function f of Example 6.1.14(a) is Riemann integrable on (a, b). 20. Let f : [a, b] -+ R be continuous. Suppose that for every Riemann integrable function g : (a, b] -+ R the product fg is Riemann integrable and f; fg = 0. Prove that f(x) = 0 for all x E [a, b].

21. Prove that the Cantor set P in [0. 1) has measure zero.

6.2

S.2

Properties of the Riemann Integral

223

Properties of the Riemann Integral In this section we derive some basic properties of the Riemann integral. As in the pre-

vious section, [a, b], a < b, will be a closed and bounded interval in R, and dtra, b; denotes the set of Riemann integrable functions on [a, b]. 6.2.1

THEOREM Let f, g E aAt[a, b]. Then rh

b

(b

f g, with jci = c j f, and

(f + g) = J f +

(a) f + g E 1.[a, b] with Ja

b

(b) cf E It [a, b] for all c E

a

a

a

a

(c) fg E Ft[a. b]. Proof. (a) The integrability of f + g is actually a consequence of Theorem 6.1.13. However, in establishing the formula for the integral of (f + g), we will also obtain the be a partition of [a, b]. integrability of f + gas a consequence. Let 91 = { x 0 . . . . . For each i = 1, . .. , n, let

M;(f) = sup{f(t):tE [x,_I,x,]}, M,(g) = sup{g(t) :I E [x,- i, x,]}.

Then f(t) + g(t) s M,(f) + M, (g) for all t E [x,_ x,], and thus M,(f) + M,(g). sup{f(t) + g(t) : t E [x1_ 1,x1]} Therefore, for all partitions 9' of [a, b],

U49,f + g) c 'U(QP,f) + g). Let e > 0 be given. Since f, g E 9t[a, b], there exist partitions 91f and 91, of [a, b] such that b

°lt(9f,.f) <

jf+ a

2

Jg±.

and

a

Let 9 = al'fU 2P8. Since 91 is a refinement of both al'f and 9914, b

au(D,f + g) < 1u(9'f,f) + att(9g, g) <

j f+ f g + e. b

a

Therefore, b

f

b

(f+g) < f+

fb

ag+e.

Since the above holds for all e > 0, b

jb(f+ g) _- jaf+ f bg

a

224

Chapter6

The Riemann and Riemann-Stieltjes Integral

A similar argument proves that

jn(f + g) 2: J f +

f g.

Thus the lower integral of (f + g) is equal to the upper integral of (f + g), and as a con-

sequence, (f + g) E 9t[a, b] with

fb(f+g)=

ff+ fhg.

(b) The proof of (b) is left as an exercise (Exercise 1). (c) To prove (c), we first note that

fg =

4[(f+g)2- (f-g)2].

By part (a) the functions (f + g) and (f - g) are integrable on [a, b]. and by Corollary 6.1.10, (f + g)'- and (f - g)2 are integrable. Hence by parts (a) and (b),fg is integrable on [a, b].

6.2.2

THEOREM 1f f E 9t(a, b), then I f I E 9t[a, b] with I

f E 9t[a, b], by Corollary 6.1.10, I f (E 9t[a, b]. Let c = ±1 be such that I 5"f I = c fa f . Since cf(x)

I f(# for all x E [a, b], °11(9, cf) <_ °U(9, If I)

for any partition 9 of [a, b]. Therefore, since both cf and If I are integrable, fa cf f.' I f I. Combining. the above we have

ff1 =cff6.2.3

jhj<

li I

THEOREM Let f be a bounded real-valued function on [a, b], and suppose a < c < b. Then f E 9t[a, b] if and only if f e 9t[a, c] and f E 9t[c, b]. if this is the case, then

Jf=faf+ff

(6)

Proof.

Let c E R satisfy a < c < b. We first prove that if f is a bounded real-valued function on [a, b], then

ff= faf+ ff

(7)

Suppose 9, and 92 are partitions of [a, c] and [c, b], respectively. Then 9 = 91 U 92

6.2 Properties of the ltiemann Integral

225

is a partition of [a, b] with c E 9. Conversely, if 9 is any partition of 'a, b with c E 91, then 99 = 91, U 912 where 91, and 912 are partitions of [a, c] and [c, b]. respectively. For such a partition °.P,

°tt(P,f) _ °u(P1,f) + aLt(92,f) >

Ji + Ji. a

a

If 9. is any partition of [a, b], then 91 = 9. U {c} is a refinement of 9. containing c. Therefore, b

c

'149-1f) ' 014911f) ' J f

+ J

a

Takin$ the infimum over all partitions a of [a, b] gives

f

f

c

'

f-jaf+If

To prove the reverse inequality, let e > 0 be given. Then there exist partitions 91, and 91, of [a, c] and [c, b], respectively such that

%(91,.f)<

(f+2

fb

and

E

f + 2*

°lt(P2, f) < J

a

. Let 91 = 9', U 9'2. Then

J f S °u(oy,f) = °u(9'1,f) + gt(9D2, f) < J`f + J f+ a

a

c

Since e > 0 was arbitrary,

t f:5 fa'+

Jf c

which when combined with the previous inequality proves equation (7). A similar argument also proves

f f=

+ f 'ff f" c If f is integrable on [a, c] and [c, b], then by equations (7) and (8), a

.

(8)

a

b

af+jf=lfslf=Jf+Jf

J

Therefore, f E Jt[a, b] and identity (6) holds. Conversely, if f E 9t[a, b], then by equations (7) and (8),

Jf+if=lf+If

226

Chapter 6

The Riemann and Riemann-Stleltjes Integral

Since the lower integral off is always less than or equal to the upper integral off, the above holds if and only if

a

ff

f-ff

11

and

u

b

Hence f E k[a, c] and f E dt [c, b]. Q

Riemann's Definition of the Integral We close this section by comparing the approach of Darboux with the original method of Riemann.

6.2.4

DEFINITION

Let f be abounded real-valued function on [a. b] and let 9

{xo. xl,....x} be a partition of [a, b]. For each i = 1, 2, ... , n. choose t, E [xi _ 1, x,]. The sum

f(p.f) = ,if(oaxi I

is called a Riemann sum off cith respect to the partition 91 and the points {t,}.

Figure 6.5 A Riemann Sum f(91. f) of f

In the Riemann approach to integration, one defines the integral of a bounded realvalued function f as the limit of the Riemann sums off. Since 9(91,f) depends not only on the partition °,D = {x0, xi,... , but also on the points t; E [x;_ 1, xi], we first

6.2 Properties of the Riemann Integral

227

need to clarify what we mean by the limit of the Riemann sums &'(9, f ). For a partition

_ {x0, x1.... , xn} of [a, b]. set

max{oxi: i = 1,2,...,n}.

I1911

The quantity Il9 11 is called the norm or the mesh of the partition 9.

6.25

DEFINITION

Let f be a bounded real-valued function on [a, b]. Then

lim -Y(9, f) = 119)1-'o

if given c > 0, there exists a S > 0 such that

l f (oaxi - I < E

(9)

i=n[

for all partitions 9 of [a, bl with 11911 < S, and all choices of ti E [x, _ I, xi].

6.2.6 THEOREM Lei f be a bounded real-valued function on [a, b]. If lim SP(9, f) = I, 11911-'0

then f E 9t[a, b] and fa f = 1. Conversely, if f E 9t[a, b], then 111im0 &(91,f) exists and

el-

rb

lim 11911-.0

x(9,.1) = J f. a

Proof. Suppose , la tprt0 P(9, f) = I. Let e > 0 be given, and let S > 0 be such that inequality ( 9 ) holds f o r all partitions 9 = {x0, xl, ... , xn} of [a, b) with 11211 < S, and all ti E [xi_ 1, xi]. By the definition of Mi, for each i = 1, ... , n, there exists i E [xi_ 1, xi] such that f M, - e. Thus n

OU(911f) =

<

i=1

M4xi f(OAXi + E //

i=1

i.l

AX,

< 1+ E + e(b -a]= 1+ E[ 1 + b -a]. Similarly T(21,f) > I - E[ 1 + b - a]. Therefore, °U.(9, f) - 2(91,f) < 2E[ 1 + b - a]. Thus as a consequence of Theorem 6.1.7, f E 9t[a, b) with fa f = I. Conversely, suppose f r= Jt[a, b]. Let M > 0 be such that I f(x)I s M for all x E [a, b]. Let e > 0 be given. Since f E 9t[a, b], by Theorem 6.1.7 there exists a partition . of [a, b] such that Jb

e <,f)

U(,.f)< f f+E.

228

Chapter 6

The Rlemann and Riemann-Stieltjes Integral

Suppose % = {xa, . . . , x,,,}. Let S = e/NM, and let 9 = ( ) , 0 ,...,y . ) be any partition of [a, b] with 11 9 II < S. As in the definition of the integral, let

M;=sup{f(x):xE[x;_1,x;]}, i= I,.

.

.. N.

Consider any interval [yk_ 1, yk], k = 1, . .. , n. This interval may or may not contain

points x. E a. Since 9 . contains N + I points, there are at most N - I intervals [y,-,, yk] which contain an x, E a, i # 0, N. Suppose, as in Figure 6.6, {.x t, ... , xj+m} C [Yk- 1, Yk] [f xx * yk_ 1, set M k = sup{ f (x) : x E [yk_,, xj]}. Similarly, if xl+m * set Mk = sup{f(x) : x E [x, +m, y'k]}. Yk - 1

yk,

Y4

X1

xj+1

x142

X1

Figure 6.6

Let tk E [yk_ 1, Yk] be arbitrary. Since 1(t) - f(s)I s 2M for all t, s E [yk_,, yk], 2M + M1 s, s = 1, . .. , m, 2M + Mk, i = 1, 2.

f(tk) f(tk)

and

Therefore, m

f(tk)AYk =f(rk)(x,

Yk-1) + Y f(rk)OXI+, + f(1k)O'k - x,+m) $=1

< 2M.Yk + Mk(x, - Yk-1) +

M,+,'&x,+, + M4(Yk - X,-m)

M s=1

< 2M6 + 01L(9k, f ), where 2 - P , = { y k -, x,,

-

, x;+., yk} is a partition of [yk- 1, Yk]. If the interval [Yk-,. yk]

contains no x; E a, i * 0, N. we simply let 9k = {yk_ 1, yk}. Let 9' = U. Then 9' is a partition of [a, b] that is also a refinement of 2. Since at most N - I intervals [Yk-1, Yk] contain a point of 9, other than xo and xN, n

f(9j) _

n

f(tk)DYk < 2M(N - 1)8 +

6u(9k,f) k-1

k=1

b

<2e+04(9',f)<2e+Ru(a,f)<3e+

f. a

A similar argument proves that

`9'(P,f) > jfb - 3e. a

herefore,

f(P,f) -

JiI < 3e.

f.

6.2

229

Properties of The Riemann Integral

Since this holds for any partition QP _ {yo, ... , y,,} of [a, b] with 11 1,1P11 < a and any k = 1, . . . , n, choice of tk e rb O

6.2.7

EXAMPLE In this example we will use the method of Riemann sums to evaluate f; x dx. Since f(x) = x is Riemann integrable on [a, b], rb x dx = lim 9(91, f). Ja

Since the limit exists for any t, E [x1_,, x,], we can take t; = (x;_, + x;)/2. With this choice of t,, 1,

fyfy /!11

"(.7,f)

R

R

T

1 7,(x,_, + xj)(x, - x,_,) 2j_j

2

2

2

(x, - x?2 (b - Q ), 1

i -11

which proves the result.

EXERCISES 6.2 1. Prove Theorem 6.2.1(b). 2. a. Use the method of Riemann sums to evaluate f; x2 d, b. Use the method of Riemann sums to evaluate fa x" dx, n E N, n >_ 3.

3. a. Let fbe a real,valued function on [a, b] such that f(x) = 0 for all x # c, , ... , cR. Prove that f E A[a, b] with f; f = 0. *b. Let f, g E.9t[a, b] be such that f(x) = g(x) for all but a finite number of points in ra, b]. Prove that

faf=fog. c. Is the result of part (a) still true if f (x) = 0 for all but countably many points in [a, b]?

4. Let f E 9t[-a, a], a > 0. Prove each of the following. a. If f is even (i.e., f (-x) = f(x) for all x (E [ -a, a]), then f f = 25f.

b. Iff is odd (i.e.,f(-x) _ -f(x) for all x E [-a, a]), then f f = 0. 5. *Let f be a bounded real-valued function on [a, b] such that f E k[c, b] for every c, a < c < b. Prove that

fE

f; f= lira, f b f

6. Let f be continuo6s on [0, 1]. Prove that

f(jf(x)dx.

lim

7. Use the previous exercise to evaluate each of the following limits. (You may use any applicable methods from calculus to evaluate the definite integrals.) R

1 a. lim R-"OnYk k'

R

b. lim Y

i n2

R

k

k2

c. lim

n

R-rook=, n2 + k2

230

8.

Chapter 6

The Riemann and Riemann-Stieltjes Integral

AS in Example 4.4.11(a), let f be the monotone function on [0, 1) defined by

f(x) _

2" 1(x -

x = n/(n + 1), n E N. Find fo f(x) dx. Leave your answer in the form of an infinite series. 9. Suppose f E 9t[a, b] and c E R. Define f, on [a - c. b - c] by f.(x) = f(x + c). Prove that

ffEdt[a-c,b-c]with

Jf(x)ds = I f(x) dx. 6

10. Let f. g, h be bounded real-valued functions on [a. b] satisfying f(x) s g(x) S h(x) for all x E [a. b]. If

f. h E Il[a. b] with f. f = fu h = 1, prove that g E 'lt[a. b] with ff g = 1.

6.

I

Fundamental Theorem of Calculus In this section we will prove two well-known and very important results. Collectively they are commonly referred to as the fundamental theorem of calculus. To Newton and Leibniz, integration was the inverse operation of differentiation. In Leibniz's notation, this result would simply be expressed as f f'(x) dx = f (x), where the symbol f (denoting sum) was Leibniz's notation for the integral. This notation is still used in most calculus texts to denote the indefinite integral of a function. Since the modern definition of the integral is based on either Riemann or Darboux sums, we now need to prove that integration is indeed the inverse operation of differentiation. Both versions of the fundamental theorem of calculus presented here are essentially due to Cauchy, who proved the results for continuous functions. As was illustrated in Examples 6.1.6 and 6.2.7, the computation of the integral of a function, using either Darboux's or Riemann's definition, can be extremely tedious. For nontrivial functions, these computations are in most instances impossible. The first version of the fundamental theorem of calculus provides a major tool for the evaluation of Riemann integrals. We begin with the following definition.

6.3.1

Let f be a real-valued function on an interval L A function F on I is called an antiderivative off on I if F'(x) = f(x) for all x E 1. DEFINITION

Remark. An antiderivative, if it exists, is not unique. If F is an antiderivative off, then so is F + C for any constant C. Conversely, if F and G are antiderivatives off, then

F'(x) - G'(x) = 0 for all x E [a, b]. Thus by Theorem 5.2.9, G(x) = F(x) + C for some constant C.

6.3.2

THEOREM (Fundamental Theorem of Calculus) antiderivative off on [a, b], then

if f E R[a, b] and if F is

J f(x) dx = F(b) - F(a).2 2. The quantity F(b) - F(u} is usually denoted by F(x)l;.

an

6.3 Fundamental Theorem of Calculus

231

Let 9 = (x0, x 1 . . . . . xa} be any partition of [a, b]. If F is an antiderivative off, then by the mean value theorem, for each i = 1, ... , n, there exists t, E (x,_ x,) Proof.

such that

F(x,) - F(x;_,) = f(t,)Ox,. Therefore, a

a

FIf(y)Ox, _

[F(xi) - F(x;_1)] = F(b) - F(a).

Since a

i=1

we have

f f (x) A 5 F(b) - F(a) < J f (x) dx. Ja

a

Thus if f is Riemann integrable on [a, b],

Ibf(x)dx=F(b)-F(a). U a

Remark. The above version of the fundamental theorem of calculus is considerably stronger than the version given in most elementary calculus texts in that it does not require continuity of the function f. We only need that f is integrable and that it

has an antiderivative on [a, b]. We will illustrate this in part (b) of the following example.

6.3.3

EXAMPLES

(a) If f (x) = x", n E N, then F(x) _

+

a,bER,

1

X"11 is an antiderivative off. Thus for any

fb

x" dx = a

I n+l

(bm+1 - an+1).

(b) Consider the function F on (0, 11 defined by

F(x) =

xZ sin

X,

x # 0,

A straightforward computation gives

-cos lx + 2x sin 1x , x # 0 ' f(x) - F'(x) = I 0,

x=0.

232

Chapter6

The Riemann and Rlemann-Stieltjes Integral

Then f is bounded on [0, 1] and continuous everywhere except at x = 0. Thus f E 9t[0, 1 ], and by the previous theorem, i

J

f=F(1)-F(0)=sin1.

(c) Let f be defined on [0, 21 by f(x)_

1, 0 5x< 1, x- 1, 1 5x 52,

and for x E [0, 2] let F(x) be defined by rr

F(x) =

J0

'f(r) dt.

If 0 s x S 1, then F(x) = fo 1 dt = x. On the other hand, if 1 < x s 2, then

F(x)= 1 1dt+ 1(t- 1)dr= o

Thus

0sxs1,

x,

F(x)

2x2-x+Z.

i

2x2-x+2, 1 <x<-2.

The graphs off and F are given in Figure 6.7. Even though f is not continuous at x = 1, the function F is continuous everywhere on [0, 2]. This in fact is always the case, as will be proved in Theorem 6.3.4.

1

The following version of the fundamental theorem of calculus, also attributed to Cauchy, proves that for continuous functions, integration is the inverse operation of differentiation.

6.3.4

THEOREM (Fundamental Theorem of Calculus) [a, b] by

F(x) =

Ja

'f(t) dt.

Let f E 9t[a, b]. Define F on

6.3 Fundamental Theorem of Calculus

233

Then F is continuous on [a, b]. Furthermore, if f is continuous at a point c E [a, b:, then F is differentiable at c and F'(c) = f(c).

Proof. The proof that F is continuous is left as an exercise (Exercise 1). Suppose f is continuous at c E [a, b). We will show that F' (c) = f (c). Let h > 0. Then by Theorem 6.2.3, c+h

F(c + h) - F(c)

J

c+h

c

f(t) dt -

f(t) dt = {

f

o

f(t) A

c

Therefore,

F(c +

h) - F(c) - f(c) h

C

c+h

f(t)

hJ

-

1

h

f (C)

dt -

f" [f(t) - f(c)) dt.

Let E > 0 be given. Since f is continuous at c, there exists a S > 0 such that

NO - f(c) I < E f o r all t,

I t - c I < S. Therefore, if 0 < h < S,

F(c + h) - F(c) - f(c)

Jc+h

I

If(t) - f(c)I dt

h

c

h

Jc +h


Edt=E. c

Thus

m

F(c + h)' - F(c) =

hli-,o

h

Ac);

i.e., F+' (c) = f (c). Similarly, if f is continuous at c E (a, b], F'_ (c) = f (c), which proves the result. Q

Remarks (a) If f is continuous on [a, b], then an antiderivative of f always exists; namely, the function F given by

F(x) =

-f(t) dt, a 5 x :S b.

Since f is continuous, F'(x) = f(x) for all x E [a, b]. As a consequence, we obtain the following more elementary version of Theorem 6.3.2 normally encountered in the

study of calculus: If f is continuous on [a, b] and G is any antiderivative of f, then

f f = G(b) - G(a).

234

Chapter6

The Riemann and Rlemann-Stieltjes Integral

(b) Integrability of a function f on [a, b] does not imply the existence of an antiderivative off. For example, if f is monotone increasing on [a, b] and F(x) = f` f, then for

any c E (a, b), F' (c) = f (c+) and F_(c) = f(c-) (Exercise 14). Thus if f is not continuous at c, the derivative of F does not exist at c.

The Natural Logarithm Function As our first application of the fundamental theorem of calculus, we use the result to define the natural logarithm function In x.

6.3.5

EXAMPLE For x > 0, let L(x) be defined by

dt.

L(z) = i

Since f (t) = 1/t is continuous on (0, oo), by Theorem 6.3.4, L(x) satisfies L'(x) = l/x for all x > 0. Furthermore, since L'(x) > 0 for all x E (0, oo), L is strictly increasing on (0, oo). We now prove that the function L(x) satisfies the usual properties of a logarithm function; namely,

(a) L(ab) = L(a) + L(b) for all a, b > 0, (b) L(e) _ -L(b), b > 0, and (c) L(b') = rL(b), b > 0. r E R. To prove (a), consider the function L(ax), x > 0. By the chain rule (Theorem 5.1.6),

d L(ax) = dX

.a=X=L'(x). ax

Thus by Theorem 5.2.9, L(ax) = L(x) + C for some constant C. From the definition of L we have L(l) = 0. Therefore, L(a) = L(1) + C = C.

Hence L(ax) = L(a) + L(x) for all x > 0, which proves (a). The proof of (b) proceeds analogously. It is worth noting that for the proof of (a) and (b) we only used the fact

that L'(x) = l/x and L(1) = 0. To prove (c), if n E N, then by (a) L(b") = nL(b). Also by (b),

L(b-") = L\\b/"/

=

nL(b)

-nL(b).

Therefore, L(b") = nL(b) for all n E Z. Consider L('6) where n E N. Since nL(-6) = L(b), L(-6) _ L L(b). Therefore, L(b') = rL(b) for all r E 0. Since L is continuous, the above holds for all r E R.

6.3

Fundamental Theorem of Calculus

235

Our final step will be to prove that L(e) = 1, where e is Euler's number of Example 2.3.5. To accomplish this we use the definition to compute the derivative of L at 1. Since L'(1) exists,

1 =L'(1)= ]im n--,ao

L(l +

L(1) i

1mnL(l+nl

=

= lim L(t I +

!T) = L(e). n

The last equality follows by the continuity of L and the definition of e. Therefore, L(e) = 1 and the function L(x) is the logarithm function to the base e. This function is usually denoted by log, x or In x, and is called the natural logarithm function.

Consequences of the Fundamental Theorem of Calculus We now prove several other consequences of Theorem 6.3.4. Our first result is the mean value theorem for integrals.

6.3.6 THEOREM (Mean Value Theorem for Integrals) Let f be a continuous real-valued function on [a, b]. Then there exists c e [a, b] such that b

f

f =f(c)(b - a).

a

Proof.

Let F(x) = fa f. Since f is continuous on [a, b], F'(x) = f(x) for all x E [a, b]. Thus by the mean value theorem (Theorem 5.2.6), there exists c E [a, b] such that

Jbf = F(b) - F(a) = F'(c)(b - a) =f(c)(b - a). 0 a

An alternative proof of the above can also be based on the intermediate value theorem using the continuity of f. This alternative method will be used in the proof of the analogous result for the Riemann-Stieltjes integral.

6.3.7

THEOREM (Integration by Parts Formula) Let f, g be differentiable functions on

[a, b] with f, g' E 9t[a, b]. Then Jb fg' = f(b)g(b) - f(a)g(a) -

gf'.

'Proof. Since f, g are differentiable on [a, b], they are continuous and thus also integrable on [a, b]. Therefore by Theorem 6.2.1(c), fg' and gf' are integrable on [a, b]. Since

(fgY = gf' + fg',

236

Chapter 6

The Riemann and Riemann-Stieltjes Integral

the function (fg)' E 9t[a, b]. By the fundamental theorem of calculus (Theorem 6.3.2), b b Vg), b

= J gf + J fg',

f(b)g(b) - f(a)g(a) = J

a

a

a

from which the result follows. 0

6.3.8

THEOREM (Change of Variable Theorem) Let tp be differentiable on [a, b] with rp' E &[a, b]. If f is continuous on I = rp([a, b]), then h

v(b)

f(x) dx.

f((p(t))w'(t) de a

e(a)

Proof. Since tP is continuous, I = rp([a, b]) is a closed and bounded interval. Also, since f a rp is continuous and tp' E 9t[a. b], by Theorem 6.2.1(c), (f o W)tp' E 9t[a, b]. If I = r([a, b]) is a single point, then tp is constant on [a. b]. In this case rp'(t) = 0 for all t and both integrals above are zero. Otherwise, for x E I define

F(x) = J

f(s) ds.

(a) Since f is continuous, F'(x) = f(x) for all x E 1. By the chain rule,

d F(1P(t)) = F'((p(t))1P'(t) =f(tp(t))w'(t)

for all t E [a, b]. Therefore by Theorem 6.3.2, Jbf((p(t))tp'(t) dt = F((p(b))

F(,p(a)) = J b)f(s) ds. (a)

a

Remark Another version of the change of variable theorem is given in Exercise 10.

6.3.9

EXAMPLE To illustrate the change of variable theorem, consider fo t/(1 + t2) di. If we let (,p(t) = 1 + t2 and f(x) = 1/x, then Joe

1 + t2 which, by Theorem 6.3.8

dt = 2

_Js

102f(`p(t))(p'(t) dt

xdx=21n5.

2

EXERCISES 6.3 1. *Let f E 9t[a, b]. For x E [a. b], set F(x) = f' f. Prove that F is continuous on [a, b].

6.3 Fundamental Theorem of Calculus

237

2. For x E [0, 1 ], find F(x) = fo f(t) dt for each of the following functions f defined on [0, 11. In each case verify that F is continuous on [0, 11. and that F'(x) = f(x) at all points where f is continuous. 1,

a. f(x)=xz-3x+5

0 s x <2

*b. f(x) =

-2, 1 5 x <-1 d. f(x) = x[3x]

c. f(x) = x - [3x]

3. Let f(t) be defined by

f(t)

0st<1, {b'i2, 15t<-2, 0 s x 5 2.

and let F(x) be defined by F(x) = fo f(t) dt,

a. Find F(x). b. For what value of bin the definition off is F(x) differentiable for all x E [0, 2]?

4. Let f be a continuous real-valued function on [a, b] and define Hon [a, b] by H(x) = f; f. Find H'(x). 3. Find F'(x) where F is defined on [0, 1 ] as follows. 1

a. F(x) _

1+t

0

*b. F(x) = fcosf2dr

dt

t

e. F(x) _

1 + t3 dt

d. F(x) =

JIQ) dr, where f is continuous on [0, 1 ]. 0

6. *Let L(x) be defined as in Example 6.3.5. Prove that g1/x) _ -L(x). 7. Suppose f: [a, b] -+R is continuous and g, h : [c, d] -+ [a, b] are differentiable. For x E [c. d] define (AX)

H(x) =

f(t) dt.

1 Mx)

Find H'(x). & Let f : R -+ R be continuous. For a > 0 define g on R by

g(x) = J

f (t) dt.

a a

Show that g is differentiable and find g'(x).

9. *Let f be a continuous real-valued function on [a, b], g E 31[a, b] with g(x) 2z 0 for all x E [a. b]. Prove that there exists c E [a, b] such that

Jef(x)g(x)dx =-f(c) o

J°g(x)dx. a

10. Prove the following change of variables formula: Let rp be a real-valued differentiable function on [a, b] with (x) # 0 fof all x E [a, b]. Let 0 be the inverse function of rp on I = tp([a. b]). If f : 1-+ R is continuous on 1, then (o(b) f(t)4x'(t)dt. . f(rp(x))dx = a

)v(a)

The Riemann and RIemann-Stieltjes Integral

Chapter6

238

11. Evaluate each of the following integrals. Justify each step.

a

2 In x

+b

dx

X

I

(I

c. J

vx

dx

x

d.

x In x d x

0

J

t

at + b dr.

a, b> 0

0 I

e, 10 12.

(4 iI

1/X

1+V

f

j4

dx

dx I

x

Suppose f is continuous on [0, 1 ]. Prove that lim J0

'f

(x") dx = f (O).

13. Suppose f : (a. b] -> R is continuous. Let M = max{jf(x)j: x E [a, b]}. Show that

/b tim

I If(x)r dx

= M.

14. Let f be a monotone increasing function on [a, b] and let F(x) = fo f(t) dt. Prove that F' (c) = f(c+) and

F_(c) = f(c-) for every c E (a, b). 15. Use Theorem 4.4.10 and the previous exercise to construct a continuous, increasing function F on [0, 1) that is differentiable at every irrational number in [0, 11, and not differentiable at any rational number in (0, 1 ].

16. Cauchy-Schwarc Inequality for Integrals: Let f, g E &(a, b]. Prove that

(jb J,bf(x)g(x) dx I 2 5

f 2(x) dx/

(Lb g2(x) dl/

(Hint: For a, 0 E R consider fa (af - /3g)2.) 17. The Exponential Function: As in Example 6.3.5, let L : (0, oo) - R be defined by

f!df.

L(x) I

t

a. Show that L is strictly increasing on (0, oo) with Range L = R. b. Let E : R -i (0, oo) denote the inverse function of L. Use Theorem 5.2.14 to prove that E'(x) = E(x) for an x E R. (The function E is called the natural exponential function, and is often denoted by exp(x).)

c. Prove that E(x + y) = E(x)E(y) for all x, y E R. d. Prove that E(x) = e` for all x E R. where e is Euler's number, and e` is as defined in Miscellaneous Exercise 3 of Chapter 1.

6.4

d.4

Improper Riemann Integrals

239

Improper Riemann Integrals In the definition of the Riemann integral of a real-valued function f. we required that f be a bounded function defined on a closed and bounded interval [a, b. If these two hypotheses are not satisfied, then the preceding theory does not apply, and we have to modify the definition. In this section we will briefly consider the changes required if the function j is unbounded at some point of its domain, or if the interval on which f is defined is itself unbounded.

Unbounded Functions on Finite Intervals If f E It[a. b] with If (x)I s M for all x E [a. b], then by Theorem 6.2.3, f E A [c, b] for every c E (a, b), and

f of

- hf I

=

f I < Ja` U1 < M(c - a).

IJ

As a consequence, if f E lt[a, b], then f E k[c, b] for every c E (a, b) and e

limf

n

f=Jf. a

C

a bounded real-valued function defined only on (a, b] with f E R[c, b] for every c E (a, b). If we define f(a) = 0, then f is a bounded function on [a. b) satisfying f E 1t[c, b] for every c E (a. b). By Exercise 5. Section 6.2. f E %[a, b] with lim

C

a- J f

o

f.

f= a

Furthermore, by Exercise 3 of Section 6.2, the answer is independent of how we define

f(a). Using the above as motivation, we extend the definition of the integral to include the case in which f is unbounded at an endpoint. This extension of the integral is also credited to Cauchy.

6.4.1

Let f be a real-valued function on (a, b) such that f E dt[c, b] for even c E (a, b). The improper Riemann integral off on (a, b], denoted f,' f, is defined to be DEFINITION

b

b

Ja

limf f, Ca' f

provided the limit exists. If the limit exists. then the improper integral is said to be convergent. Otherwise, the improper integral is said to be divergent.

240

Chapter6

TheRlemann and Riemann-Stleltjes Integral

A similar definition can be given if f is defined on (a, b) and is unbounded at b. If f is unbounded at a point p, a < p < b, we consider the improper integrals off on the intervals [a, p) and (p, b]. If each of the improper integrals exists, then we define the improper integral off on [a, b] to be the sum b

i,

af+Jf.

l 6.4.2

EXAMPLES

(a) Consider the function f(x) = 1/x on (0, 1 ]. This function is clearly unbounded at 0. Since f is continuous on (0, 1 ],f e Jt[c, 1 ] for every c E (0, 1). By Example 6.3.5. 1

X

dx=Inc.

To evaluate lim In c we consider In r", where 0 < r < I and n E N. Since In r" _ C- Oo

n In r and In r < 0,

lim In r" = limn In r = -oo. Therefore, since r" _+0 as n -+ oo and In x is monotone increasing, lirm In c = -oo. Thus the improper integral of 1/x on (0, 1) diverges. (b) For our second example we consider the improper integral of L(x) = In x on (0, 1 ]. Since L(x) is continuous on (0, 1), L E 9t[c, 1 ] for every c E (0, 1). Consider ff' In x dx. If we take g(x) = x, then by the integration by parts formula, j In x dx =

L(x)g'(x) dx J

= L(l)g(l) - L(c)g(c) - J L'(x)g(x) dx

=-cInc-1+c. By the substitution c = 1/t, t > 0, and l'Hospital's rule,

lira c In c = lim

1-.0o

-1n t t

= -lim 1 = 0. t-00 t

Therefore, lim I In x dx = -1. C

c-+O'

Hence the improper integral of In x converges on (0, 1 ) with

I 0

lnxdx = -1.

6.4

Improper Riemann Integrals

241

(c) Consider the function f defined on [ -1, 1 ] by

-1 sx<-0.

0,

f(x)=

1,

x

0<xsl.

Here f is unbounded at 0. For this example, the improper integral off over [ -1, 1' fails to exist because the improper integral off over (0, 1 ] does not exist. (d) There are some significant differences between the Riemann integral and the im-

proper Riemann integral. For example, if f E 9t[a, bJ, then by Theorem 6.2.1. f2 E 9t[a, b]. This however is false for the improper Riemann integral! For example. if f (x) = 1/Vx, x E (0,1 ], then

Jldx ,l° x

z

Thus the improper integral off converges on (0, 1]. However, f2(x) = t/x, and the improper integral of 1/x on (0, 1 ] diverges. Also, if f E 9t[a, b], then [f I E 9t[a, b]. This .again is false for improper Riemann integrals. In Exercise 4, we provide an example of a function f for which the improper integral off converges, but the improper integral of L f I diverges.

Infinite intervals We now turn our attention to functions defined on infinite intervals.

6A.3

DEFINITION

Lei f be a real-valued function on [a, co) that is Riemann integrable on

[a, c) for every c > a. The Improper Rlemann integral off on [a. oo), denoted f, f. is defined to be

j= lim J f, C-400 n

provided the limit exists. If the limit exists, then the improper integral is said to be convergent. Otherwise, the improper integral is said to be divergent.

If f is a real-valued function defined on (-oo, bJ satisfying f E 9t[c, b] for every c < b, then the improper integral off on (-oo, b] is defined as

f

f bf,

provided the limit exists. If f is defined on (-oo, oo), then the improper integral off is defined as

xf+I

30f,

242

Chapter 6

The Riemann and Riemann-Stieltjes Integral

for some fixed p E R, provided that the improper integrals off on (-oo, p] and [ p, oo) are both convergent. For a function f defined on (-oo, oo), care must be exercised in computing the improper integral off. It is incorrect to compute

For example, if f (x) = x, then lim 1 x dx = c--tlim 1 (c2 oo 2 -c

C-.0o

- (-c)2) = 0.

However, the improper integral off on (-oo, oo) is divergent since

fc

JOO

f

lim c-.oo

1

x dx = lim - c2 = oo. c-.oo 2

Remark. If f is nonnegative on (a, oo) with f E Jt[a, c] for every c > a, then f.f is a monotone increasing function of c on [a, oo). Thus lim f. f exists either as a real number or diverges to no. For this reason, if f (x)

0 for C

all x E [a, oc), we use the notation rb

Jf(x)dx < oo

or

J

f(x) dx = 00

to denote that the improper integral off on [a, oo) converges or diverges to no respectively.

6.4.4

EXAMPLES

(a) Let f (x) = 1/x2, x E [ 1, oo). Since f is continuous on [ 1, co), f e Jt[ 1, c) for everyc > 1. Therefore,

Ji=l

c-+ao

Ix

imf'Idx=liml-1+1)=1.

c-.oo\

C

Thus the improper integral converges to the value 1. (b) In this example we consider the function f (x) = (sin x)/x, x E [v, oo). Sincef is

continuous, f E 9t[a, c] for every c > v. This function has the property that the improper integral off on [a, o0) converges, but the improper integral of VI diverges. The proof of the convergence of the improper integral off is found in Exercise 7. Here we will show that

IfI = r

Isin xl

=oo.

6.4

Improper Riemann Integrals

243

For n E N, consider sin xI J

J

km1

x

a

((k "i° Isin xl dx. x ka

dx =

Since the integrand is nonnegative, (k +3/4h.

(k+I)" {sin xf

>

x

J(k+1/4hr

I

>_

(k

t )ir

x

\/2. Also,

On the interval [(k + 1)1r, (k + ;)1r], sin xl

x

Isin

hr,+4)JrJ. forallxErlk+4(k / \\

/

(k+3/4)n Isin x

x

Ia 22(k+1)rr

ll

ll

k+ 1' and as a consequence, "+

f

sin xl

x

1,/2 4

dx

k_ik+l

By Example 2.7.4, the series 7,k 1 k diverges. Therefore,

(°° !sin J A

X

dx = ".4 lim

("+I)vIsinxI ]r

x

dx = oo.

As the previous example shows, the convergence of the improper integral off does imply the convergence of the improper integral of If 1. If f is a real-valued function on [a, oo) such that f e ,[a, c] for every c > a and the improper integral of If I converges on (a, oo), then f is said to be absolutely integrable on [a, oo). An analogous definition can be given for unbounded functions on a finite interval. We leave it as an exercise to prove that if f is absolutely integrable on [a, oo), then the improper integral off also converges on [a, oo) (Exercise 5). We conclude this section with the following useful comparison test for improper 'integrals.

244

Chapter 6

6.4.5

The Riemann and Rlemann-Stleltjes Integral

THEOREM (Comparison T e s t )

Let g : [a, oo) -+ R be a nonnegative f u n c t i o n satis-

fying g E 9t[a, c] f o r every c > a and f g(x)dx < oo. If f : [a. oo) --+R satisfies (a) f E Jt[a, c] for every c > a, and (b) I f (x) 5 g(x) for all x E [a, oo), then the improper integral off on [a, oo) converges, and

J

:5 J g(x)dx.

(x) dx

Proof. The proof is left to the exercises (Exercise 6).

N EXERCISES 6.4 1. For each of the following functions f defined on (0, 1), determine whether the improper integral off converges. If it converges, find fo f.

'a f(x) = Xp> 0 < p < I

1>

*d. f(x) = x In x

f(x) _

c. f(x) =

x

e. f(x) = (I + x) I

Inxx

f. f(x) = tan

+ x)

(Zx)

2. For each of the following, determine whether the improper integral converges or diverges. If it converges, evaluate the integral. *a.

fee*dx

b.

°O In.r

X ax

d.

Jx_'dx. p > I

C.

x

i

o

°D

dx

2

f'xZ+,dx h.

g.

dx

°G

*e. E x l n x

2

f(x2+ 1)pdx. P> I

x(l n x)P"

P> 1

f IV

I.

(x2+ 1)(x+

1)dx

3. For each of the following, determine the values of p and q for which the improper integral converges. *a. f I/2xpjlnxjpdx

c.

b.

0

Jaxp[ln(l +x)]°dx 0

2

4. Petfbedefined on(0, 1]by

f(x) =

±(X2 sin

X2)- 2x sin X2 -

cos X2.

Show that the improper Riemann integral off converges on (0, I ], but that the improper integral of VI diverges on (0,1 1.

5. *If f is absolutely integrable on [a. oo) and integrable on [a, c] for every c > a, prove that the improper Riemann

integral off on [a, oo) converges.

6. Prove Theorem 6.4.5.

6.5 The Riemann-Stieltjes Integral

245

x E [1r,oo). +a. Show that the improper integral of If I converges on [1r, oo).

7. Let f(x) _ (cos

*b. Use integration by parts on [ir, c], c > ir, to show that J

sin x dx exists. x

8. Show that fox-' sin x dx converges for all p, 0 < p < 2.

9. Forx>0,set

The function r is called the Gamma function. a. Show that the improper integral converges for all x > 0. b. Use integration by parts to show that r(x + I) = xr(x), x > 0.

c. Show that r(1) = I. d. For n r= Rl, prove that r(n + 1) = W.

The Riemann-Stieltjes Integral In this section we consider the Riemann-Stieltjes integral, which, as we will see, is an extension of the Riemann integral. To motivate the Riemann-Stieltjes integral we consider the following example from physics involving the moment of inertia.

6.5.1

EXAMPLE Consider n-masses, each of mass m;, i = 1, ... , n, located along the x-axis at distances r, from the origin with 0 < rt < < r (Figure 6.8). The moment of inertia 1, about an axis through the origin at right angles to the system of masses, is given by

1 = ; r?m;.

0

HIM

Figure 6.8

3. Since the results of this section are not specifically required in subsequent chapters, this topic can be omitted on first reading of the text.

246

Chapters

The Riemann and Riemann-Stieltjes Integral

On the other hand, if we have a wire of length I along the x-axis with one end at the origin, then the moment of inertia 1 is given by ! = jxp(x)dx.

where for each x E [0, 1). p(x) denotes the cross-sectional density at x.

Although these two problems are totally different, the first being discrete and the second continuous, the Riemann-Stieltjes integral will allow us to express both of these

formulas as a single integral. In the definition of the Riemann integral, we used the length Axi of the ith interval to define the upper and lower Riemann sums of a bounded function f. The only difference between the Riemann and Riemann-Stieltjes integral is that we replace Axi by

Aai = a(xi) - a(x; _ 1), where a is a nondecreasing function on [a, b]. Taking a(x) = x will give the usual Riemann integral. Although the modification in the definition is only minor, the consequences are far-reaching. Not only will we obtain a more extensive theory of integration, but also an integral that has broad applications in the mathematical sciences.

Definition of the Riemann-Stieltjes Integral Let a be a monotone increasing function on [a, b], and let f be a bounded real-valued function on [a, b]. For each partition 9 = {xo, x1 , .. . , of [a, b], set Aai = a(xi) - a(x;_ 1),

i=

..

, n.

Since a is monotone increasing, Aai ? 0 for all i. As in Section 6.1, let m, = inf{ f(t) : t E [x,_ i, xi]},

M, = sup{f(t):t E [xi_1,xi]}. As for the Riemann integral, the upper Riemann-Stieltjes sum off with respect to a and the partition 91, denoted all.(9, f, a), is defined by

kt(9, f, a) _

r=

M;Aa,.

Similarly, the lower Riemann-Stleltjes sum off with respect to a and the partition 9, denoted 2(9, f, a), is defined by -Y(9, f, a) _

miAai. i=1

Since m; s M, and Aai ? 0, we always have 2(91,f a) < all(9, f, a). Furthermore, if m s f(x) s M for all x E [a, b], then m[a(b) - a(a)] <_ Z(9, f, a) 5 Rt(9, f, a) < M[a(b) - a(a))

(10)

6.5

The Riemann-Stieltjes Integral

247

for all partitions 9 of [a, b]. Let 96 be any partition of (a, b). Since Mi s M for all i and

Aai ? 0, n

Mla1 = M

Mi4ai -.5

Lai = M[a(b) - a(a)).

i=1

i=1

Thus °U,(9), f, a) <_ M[a(b) - a(a)]. The other inequality follows similarly. In the above we have used the fact that n

I Aai = (a(xl) - a(x0)) + (a(x2) - a(xl)) + .

+ (a(xn) - a(xn-1))

i-

= a(xn) - a(xo) = a(b) - a(a)In analogy with the Riemann integral, the upper and lower Riemann-Stleltjes integrals off with respect to a over [a, b], denoted fQ f da and f; f da respectively, are defined by b

f da = inf{ott(91, f, a) : 9 is a partition of [a, b]},

Jfda = sup{f(9, f, a) : 9' is a partition of [a, b]}, n

By inequality (10) the set {qL(91, f, a) : 9' is a partition of [a, b]} is bounded below, and thus the upper integral off with respect to a exists as a real number. Similarly, the lower

sums are bounded above, and thus the supremum defining the lower integral is also finite. As for the Riemann integral, our first step is to prove that the lower integral is less than or equal to the upper integral.

6.5.2

THEOREM Let f be abounded real-valued function on [a, b], and a a monotone increasing function on [a, b]. Then

Jfda. e Jabfda:5 Proof. As in the proof of Lemma 6.1.3, if 9'* is a refinement of the partition 9, then X(91,f a) <_ X(91*,f, a) :5 gL(96*,f a) c ah(y,f, a). Thus if 91, 2, are any two partitions of [a, b],

f, a) s 2(9 U a j a) :S %(91 U ll, f, a) < 61L(,, f, a). Therefore T(91, f, a) <_ 6tL(9;, f, a) for any partitions 9, 9.. Hence

Jfda = sp96,f,a)

a)

a

for any partition 9.. Taking the infimum over a gives the result. O

248

Chapters

6.5.3

The Riemann and Riemann-Stieltjes Integral

Let f be a bounded real-valued function on [a, bl, and a a monotone increasing function on [a, b). If DEFINITION

h

h

fda = Jfda. f

then f is said to be Riemann-Stieltjes integrable or integrable with respect to a on [a, b). The common value is denoted by h

(h

fda

or

f (x) da (x),

! a

a

and is called the Riemann-Stieltjes integral off with respect to a. As was indicated previously, the special case a(x) = x gives the usual Riemann integral on [a, b).

6.5.4

EXAMPLES

(a) Fix a < c 5 b. As in Definition 4.4.9, let 1,.(x) = I (x - c) be the unit jump function at c defined by

= 1`(x)

f0, x
t 1,

x?c.

We now prove the following: If f is a bounded real-valued function on [a, b] that is con-

tinuous at c, a < c < b, then f is integrable with respect to 1, and

'f(x)

h

dl (x - c) = f (c).

f dl,. =

1

Ja

a

For convenience, we set a(x) = 1,.(x), which is clearly monotone increasing on [a, b). Let 91 = {xo, x1..... x,} be any partition of [a, b]. Since a < c b, there exists an

index k, 1 5 k 5 n, such that Xk- I < C 5 xk.

Then

Da* = a(xk) - a(xk_1) = I - 0 = 1,

and

Aa;=0, foralli#k. Therefore,

olt(91, f, a) = Mk&ak = Mk = sup{f(t) : xk_, <_ t 5 xk}, Z(91, f, a) = mk&ak = Mk = inf{f(t) : xk_ 15 t 5 xk}.

Since f is continuous at c, given e > 0, there exists a S > 0 such that

f(c) - e < f(t)
and

6.5

The Riemann-Stieltjes Integral

249

f o r all t E [a, b] with I t - cI < S. If 91 is any partition of (a, b] with x, - xx_, < S for all j, then AC) - E S Mk c Mk :5 AC) + E. Therefore, f (c) - E S Y(91, f , a) : 5U(91, f, a) <_ f(c) + e. As a consequence.

f(c) - E

b

f dc, _ I b f da

Jn

f (c) + E.

o

Since e > 0 was arbitrary, the upper and lower integrals off are equal, and thus f is integrable with respect to a on [a, b] with Jb

f da = f (c).

(b) The function f(x) _

1

x E

0, x6EQ, is not integrable with respect to any nonconstant monotone increasing function a.

Suppose a is monotone increasing on [a, b], a < b, with a(a) * a(b). If 91 = (xo, x,, ... , is any partition of [a, b], then m; = 0 and M; = 1 for all i = 1, ... , n. f, a) = 0 and Therefore, i a; = a(b) - a(a).

Rt(91,f, a)

Thus f is not integrable with respect to a.

Remark. It should be noted that in Example 6.5.4(a) only left continuity off at c is required (Exercise 2(a)).

The following theorem, which is the analogue of Theorem 6.1.7 for the Riemann integral, provides necessary and sufficient conditions for the existence of the RiemannStieltjes integral. The proof of the theorem follows the proof of Theorem 6.1.7 verbatim and thus is omitted.

6.5.5

THEOREM Let a 'be a monotone increasing function on [a, b]. A bounded realvaluedfunction f is Riemann-Stieltjes integrable with respect to a on [a, b] if and only if for every e > 0; there exists a partition 9 of [a, b] such that

a) -

a) < E.

Furthermore, if 9 is a partition of [a, b] for which the above holds, then the inequality also holds for all refinements of P.

We now use the previous theorem to prove the following analogue of Theorem 6.1.8. Except for some minor differences, the two proofs are very similar.

250

Chapter 6

The Riemann and Riemann-Stieltjes Integral

6.5.6 THEOREM Let f be a real-valued function on [a, b) and a a monotone increasing function on [a, b). (a) If f is continuous on [a, b), then f is integrable with respect to a on [a, b]. (b) 1f f is monotone on [a, b] and a is continuous on [a, b], then f is integrable with respect to a on [a, b].

Proof. (a) The proof of (a) is identical to the proof of Theorem 6.1.8(a) except that given e > 0, we choose 71 > 0 such that

[a(b) - a(a)),q < e. The remainder of the proof now follows verbatim the proof of Theorem 6.1.8(a). of [a, b] (b) For any positive integer n, choose a partition 91 = {x°, x,.... , such that

Aa; = a(x;)

- a(x;_,) = n [a(b) - a(a)].

Since a is continuous, such a choice is possible by the intermediate value theorem. Assumef is monotone increasing on [a, b]. Then M; = f(xi) and m; = f(x,_1). Therefore,

1!t(91,f a) - 2(91,f, a) _ i (Ax) - f(x+-,)]Dal

i-t

[a(b) - a(a)] n

U(x;) - Ax,-0] i=1

[a(b) - a(a))

n

U(b) - f(a)).

Given e > 0, choose n E N such that

[a(b) - a(a)] n

[f(b) - f(a)) < e.

For this n and corresponding partition 91, °ll(?, f, a) -

f, a) < e, which proves

the result. Q Remark. In part (b) above, the result may be false if a is not continuous. For example, if a < c < b, then the monotone function 1, is not integrable with respect to 1, on [a, b] (Exercise 2b). The article by Kenneth Ross, listed in the Supplemental Reading section, suggests an alternative approach to the Riemann-Stieltjes integral that eliminates this difficulty.

Properties of the Riemann-Stieltjes Integral We next consider several basic properties of the Riemann-Stieltjes integral. For rotational convenience, we make the following definition.

6.5

The Riemann-Stieltjes Integral

251

6.5.7

For a given monotone increasing function a on [a, bJ, 9R(a) denotes the set of bounded real-valued functions f on [a. b] that are Riemann-Stieltjes integrable with respect to a.

6.5.8

THEOREM

DEFINITION

(a) If f, g E gi(a), then f + g and cf are in %(a) for every c E R and Jb

b

b

fda + f

(f + g)da = 1a

a

b

g da,

a

b

cf da = c a

I

fda.

(b) If f E 9t(a;), i =Jfd(ai 1, 2, then f E 9R(a1 + a2) and

+ a 2) = Jfdai + Jfda2. (c) If f E °Jt(a) and a < c < b, then f is integrable with respect to a on [a, c] and [c, b] with b

a

b

fda= f.cf da+ Jcfda.

f

(d) If f, g E t(a) with f (x) <_ g(x) for all x E [a, b), then rb

rb

Ja

Ja

f da

g da.

(e) If I f(x)I 5 M on [a, b] and f E 9R(a), then If I E 9t(a) and b Ja

f

f da < blf I da

M[a(b) - a(a)].

a

Proof. We provide the proofs of (b) and (e). The proofs of (a), (c), and (d), along with other properties of the Riemann-Stieltjes integral. are left to the exercises. (b) Since f E %(a;), given e > 0, there exists a partition 9D,, i = 1, 2, such that

-U.(gli, f, ai) - X(`l...f a.) < Z Let 9' = 91, U 912. Since 91 is a refinement of both 91, and 912, inequality (11) is still valid for the partition 91. Thus since

A(a1 + a2); = A(a,)1 + A(a2)i

forall i = 1,...,n,

252

Chapter 6

The Riemann and Riemann-Stleltjes Integral

61L(@1 j, al + a2) - £(9, J, al + a2)

= °lL.(9,f a,) - `-e(91,f, a,) + °1L(J,f a2) - `-e(g,f a2) e

E

< 2 + 2 = 6. Therefore by Theorem 6.5.5, f E 9t(al + a2). Furthermore, for any partition 91 of [a, b),

2(91j, a, + a2) _ ZP,f al) + £(9j a2) t((b

rb

Jfda,+Jfda2 a

a

a, + a2).

< °d(91,f, a,) + °tt(9a,f, a2)

Thus since f E 9t(al + a2),

f

b

b

f d(a, + a2) =

f da, +

I

b

f f dal.

(e) Suppose f E 9t(a) and 91 _ {x°, x,. ... . xa} is a partition of [a, b]. For each

i= 1,. ..,n,let

M; = sup{f(t): t E [xi_,,xi]}, M* = sup{I f(t)I:t e [xi_,,xi]}, mi = inf{f(t): t e [xi_,,x.]}, m* = inf{I f(t)I : t E [xi_,,xi]}. If t, x E [xi_,, xi], then

II f(t) - If(x)II

1f(t) -f(x)I <_Mi - mi.

Thus M*i - m * 5 Mi - mi, for all i = 1, . .. , n, and as a consequence, °u(91, If 1 , a) - ,x(91, I f I, a) s tlL(91, f, a) - X(Ol, f, a).

Therefore by Theorem 6.5.5, If I E 9t(a). Choose c = t l such that c f f da -e 0. Then

JQfda =cJafda= J cfdas jbIfIdcr

MJ da

= M [a(b) - a(a)]. As for the Riemann integral, we also have the following mean value theorem and integration by parts formula for the Riemann-Stieltjes integral.

6.5.9

THEOREM (Mean Value Theorem) Let f be a continuous real-valued function on [a, b], and a a monotone increasing function on [a, b]. Then there exists c E [a, b] such that

J

f da = f(c) [a(b) - a(a)J-

6.5

The Riemann-Stieltjes Integral

253

Proof. Let m and M denote the minimum and maximum off on [a. b} respectively. Then by Theorem 6.5.8(d), b

m[a(b) - a(a)] s

f fda <_ M[a(b) - a(a)]. a

If a(b) - a(a) = 0, then any c E [a, b] will work. If a(b) - a(a) # 0, then by the intermediate value theorem there exists c E [a, b) such that b

f (c) = a(b) -1 a(a) J. fda. which proves the result.

6.5.10 THEOREM (Integration by Parts Formula) Suppose a and Pare monotone increasing functions on [a, b]. Then a E 91.(/3) if and only if fl E 9t(a). If this is the case, J b

a d$ = a(b)S(b) - a(a)9(a) - J

a

b/3

da.

a

Proof. By Exercise 9, for any partition 91 of [a, b], and

1u(91, a, (3) = a(b)(3(b) - a(a)f(a) - 2(91, (3, a),

`-x(91, a, 0) = a(b)j3(b) - a(a)f(a)

111(91, (3, a).

Therefore, 11[(91, a, /3) - 201, a, /3) = 1401, /3, a) - 49. is, a).

From this identity it immediately follows by Theorem 6.5.5 that a E 9t(t) if and only if /3 E M(a). Furthermore, if /3 E 9t(a), then given e > 0, there exists a partition 9' of [a. b] such that W(91, /3, a) >

Jb!3da_e. a

Hence, Jb/3

Jba d p

1U,(9', a, $) < a(b)f3(b) - a(a)$(a) -

a

da + E. a

Since the above holds for any E > 0, b

f

b

a d( < a(b)R(b) - a(a)$(a) - J (3da.

A similar argument using the lower sum proves the reverse inequality.

254

chapter 6

The Riemann and Riemann-Stieltjes Integral We conclude this section with two results that represent the extremes encountered in Riemann-Stieltjes integration. As in Example 6.5.4(a), let I(x - c) be the unit jump are nonnegfunction at c E R. Suppose {s"},N,- I is a finite subset of (a, b] and ative real numbers. Define the monotone increasing function a on [a, b] by N

a(X) _ I cn I (X - Sn). n-1

N 1bN

If f is continuous on [a, b], then by Example 6.5.4(a) and Theorem 6.5.8(b), b

fa

f da =

c"

1

n=1

a

I Cnf (sn) f(x) dl (x - sn) = "'1

(12)

Suppose {s"}: I is a countable subset of (a, b] and {cn}: , is a sequence of nonnegative real numbers for which E cn converges. As in Theorem 4.4.10, define a on [a, b] by

x

CO) _ I, c" 1(x - sn).

(13)

"=I

Since 0 : I (x - sn) <_ 1 for all n, the series in equation (13) converges for every x E [a, b], and a is a monotone increasing function on [a, b]. For such a function awe have the following theorem.

6.5.11

THEOREM Let f be a continuous real-valued function on [a, b], and let a be the monotone function defined by equation (13). Then f E Jt(a) and oo+

rb

( f da = G c"f(s") n=1

fa

Proof.

Since f is continuous on [a, b], f E lt(a) (Theorem 6.5.6(a)). Let e > 0 be

given. Choose a positive integer N such that 00

I C. < e. n-N+1

Define II and 02 as follows:

/

/31(x)

N

nil

Cn1(x - Sn),

/332(x)

Cn1(x - sn). n=N+I

Then a = 6, + $2, and by identity (12), N

{(b

J f d$l = I Cnf (sn) fa

n=1

Let M = max{I f(x)I : x (=- [a, b]). Then by Theorem 6.5.8(b) and (e),

1 Ja

f da -n7, cnf(sn)

I

=

J

.fdf321 c M [$2(b) - /32(a)]

6.5

The Riemann-Stieltjes Integral

255

W

:5 Mlc,,<Me, n=x+i

which proves the result.

At the other extreme, if the monotone function a is also differentiable, then we have the following result.

6.5.12

THEOREM Suppose f E 9t[a, b], and a is a monotone increasing differentiable function on [a, b] with a' E 9t[a, b]. Then f E &(a) and J f da = J f(x)a'(x) dx. a

a

Proof. Since both f and a' are Riemann integrable on [a, b], by Theorem 6.2.1(c), fa' E 9t[a, b]. Let e > 0 be given. Since a' E 01[a, b], by Theorem 6.1.7 there exists a partition 91 of (a, b) such that

°U.(9, a') - Y(), a') < e.

(14)

Let °. 2. _ {xo, ... , xn} be any refinement of 91. As in Theorem 6.2.6, for each i = 1, ... , n, we can choose s, E [xi_,, x,] such that f(s,)Aa; + e.

° L(°9 f, a) <

(15)

By the mean value theorem, for each i = 1, ... , n, there exists t; E [xi _,, x,] such that

Dai = a(x;) - a(x,_1) = a'(ti)Axi. Therefore, n

a

I f(s,)Aar = Yf(sr)a'(y)Ax,. i=i

(16)

i=1

Let M = sup{ f(x)I :x E [a, b]}, and for i = 1, ... , n, let in, and M; denote the infimtun and supremum respectively of a' over the interval [x_ ,, x,). Then

la'(si)-a'(t;)l:!-- Mi - m, f o r all i = 1, ... , n. Therefore,

i

f(si)a'(ti),xi - In As;)a'(sAXi`

i-l

1

i- 1f(si)I1a'(1i) - a'(sMAxi `M(Mi - mi)ixi = M(%(a, a') - -W(9, a')) < Me.

The last inequality follows by inequality (14) since .. is a refinement of 91. Therefore, n

i-i

f(s;)a'(ti)Ax; S

i- f(si)a'(s;)Ax, + Me <_ °U.(9, fa') + Me.

256

Chapter 6

The Riemann and Rlemann-Stieltjes Integral

Thus by equations (15) and (16).

'lt.(a, f, a) < 0U.(a. fa') + (M + 1)e.

Since fa' E 9t[a, b], there exists a partition ). of [a, b]. which is a refinement of 91. such that b

G(L(2, fa') < J fa' + E. G

Thus b

rh

f da <

t

fa' + (M + 2)e.

Since this holds for any E > 0, fQ f da <_ f' fa'. A similar argument using lower sums proves the reverse inequality. Thus f E J1(a) and h

b

J fda = I fa'. a

1a

We now give several examples to illustrate the previous two theorems.

6.5.13

EXAMPLES

(a) First, we illustrate the finite version of Theorem 6.5.11; namely, identity (12). Consider

j exd(x].

Forx E [0, 2],

[x]=1(x- 1)+1(x-2). Therefore since ex is continuous, it is integrable with respect to [x], and by identity (12),

+ e2.

J exd[x] =

(b) To illustrate Theorem 6.5.12, if a(x) = x2 on [0, 1), then for any Riemann integrable function f on [0, 1 ],

ff(x)dx2 = 2 J fix) x dx. 0

0

In particular, if f(x) = sin lrx, then 1 f sin Trx dr' = 2 0

f x sin irx dx, 0

The Riemann-Stieltjes Integral

6.5

257

which by the integration by parts formula (Theorem 6.3.7)

+ n J cos irx dx

X cos Trx I

0

0

I

_ -- ITcos V +

sin Vx U

?r

(c) As another illustration of Theorem 6.5.12, 13[x] 0

2 J ;[x]e2x dx 0 3

= 21 e - dx + 21 2e2' dx = 2e6 - e° - e2. 2

1

The Riemann-Stieltjes integral has important applications in a variety of areas. including physics and probability theory. It allows us to express diverse formulas as a single expression. To illustrate this, we again consider the moment of inertia problem of Example 6.5.1.

6.5.14

EXAMPLE

In this example we will show that both formulas of Example 6.5.1 can be

expressed as I

I=

x dm(x), 0

where m(x) denotes the mass of the wire or system from 0 to x. Clearly the function m(x) is nondecreasing, and thus since x2 is continuous, the above integral exists. In the first case, since 0 < rt < r2 < < r,,, and the masses m; are located at r, (see Figure 6.8), m(x) is given by m(x)

m; l (x - r;).

Thus by Theorem 6.5.11, r

x2dm(x) _

r?m;.

Io

On the other hand, if m(x) is differentiable and m'(x) is Riemann integrable, then by Theorem 6.5.12, Jx2 dm(x) = Jx2m'(x) dx. 0

258

Chapter 6

The Riemann and Riemann-Stieltjes Integral

It only remains to be shown that m'(x) is the density. The density p(x) of a wire (mass per unit length) is defined as the limit of the average density. In the interval [x, x + Ax, the average density is

m(x + Ax) - m(x) Ax

Thus if m(x) is differentiable, p(x) = m'(x).

Rlemann-Stieltjes Sums We conclude this section with a few remarks concerning Riemann-Stieltjes sums. As in

Definition 6.2.4, let f be a bounded real-valued function on [a, b], a a monotone in-

creasing function on [a, b], and 9' _ {xo, ... , i = 1, 2, . , n, choosey E [x,_,, x;]. Then

a partition of [a, b]. For each

F(`.'P, f a) = Ef(t,)Aa, is called a Riemann-Stieltjes sum off with respect to a and the partition 9. A natural question is whether the analogue of Theorem 6.2.6 is valid for the Riemann-Stieltjes integral. Unfortunately, only part of the Theorem holds. If lim MPH-+0 W(9, f, a) = I, then f e 911(a) on [a, b] and f. f da = I (Exercise 11). The converse, as the following example will illustrate, is false! However, if f and a satisfy either of the hypotheses of Theorem 6.5.6, then 1b

lim 9 f da = u9n-+O

f, a).

a

The result where f is continuous and a is monotone increasing is given in Exercise 10.

6.5.15

EXAMPLE Let f and a be defined on [0, 21 as follows:

0, 0S x11,

f(x) a(x)

_

1,

1<x

0,

0 5 x < 1, 1 c x < 2.

11,

2,

Since f is left continuous at 1, by Example 6.5.4(a), fo fda = f(1) = 0. Suppose = {xo, x,, ... , is any partition of [0, 2] with 11Z 91. Let k be such that xk_, < I < xk. Then Aak = 1 and Aa; = 0 for all i # k. If tk E (1,xk],then f, a) = 1. On the other hand, if tk = 1, then f(°,P, f, a) = 0. As a consequence, n

does not exist.

lim 9'(91,f, a)

6.5 The Riemann-Stieltjes Integral

259

EXERCISES 6.5 1. *Evaluate f ' f (x) da(x) wheref is bounded on (-1. 11 and continuous at 0, and a is given by

-1, x<0, a(x) =

0, 1,

x = 0, x > 0.

2. a. In Example 6.5.4(a), prove that if the function f is left continuous at c, then f is integrable with respect to 1, on [a, b]. b. Show that 1, is not integrable with respect to 1,

3. Let a be nondecreasing on (a, b). Suppose f is bounded on [a. b] and integrable with respect to a on [a. b]. For

X E [a, b] set F(x) = f. f da. *a. Prove that IF(x) - F(y)l s Mla(x) - a(y)l for some positive constant M and all x. y E [a, b]. b. Prove that if a is continuous at x E [a, b), then F is also continuous at x0.

4. a. Prove Theorem 6.5.8(a). b. Prove Theorem 6.5.8(c). c. Prove Theorem 6.5.8(d). 5. Use the theorems from the text to compute each of the following integrals. W/2

x d(sin x)

*a.

b. J 3[x] dx2

J(o

1([x] + x) da(x)

d.

3

*c. J x2 d[x] o

n

where a(x) - x2 + e'

I

4

ex d[x]

e.

I

f. 1 (x - [x]) dx3

1

sin ax d[4x]

g. J 0

I

6. Verify the integration by parts formula with parts (b) and (c) of the previous exercise.

7. Find fu fda, where f is continuous on [0, 1 ] and CO)

Leave your answer in the form of an infinite series.

& Let a be as in Exercise 7. Evaluate each of the following integrals. Leave your answers in the form of an infinite series. Ix da(x)

*a.

b. I 'a(x) dx

Jo

Jo

9. Suppose a and j6 are monotone increasing on [a, b], and 91 is a partition of [a, b]. Prove that

11L(91, a, P) = a(b)$(b) - a(a)3(a) -

(3, a).

10. Prove that if f is a continuous real-valued function on [a, bland a is monotone increasing on [a, b], then a

imoY(P, f, a) exists and ii it,o pp, f, a) = f; f da.

260

Chapter 6

The Riemann and Riemann-Stleltjes Integral

11. Let f be a bounded real-valued function on [a, h] and a a monotone increasing function on (a. b]. Prove the following: If Arn0Y(91. f, a) = 1, then f E 9t(a) and f ."f da = 1. 12. *a. Let a be a monotone increasing function on [a, b]. If f E R(a), prove that j2 E Jt(a).

b. If f, g E llt(a), prove that f g E Jt(a). 9t(a) on [a, b]. 14. Suppose f is a nonnegative continuous function on [a, b], and a is nondecreasing on [a, b]. Define the function R on [a, b] by 6(x) = fa f da. If g is continuous on [a, b], prove that 13. If f E Jt(a) on [a, b] with Range f C [c, d], and rp is continuous on [c, d], prove that V of E

h 1

b

g df = J fg da. 4,

Numerical Methods4 In this section, we will take a brief look at some elementary numerical methods that are useful in obtaining approximations to Riemann integrals. Even though the fundamental theorem of calculus provides an easy method for evaluating definite integrals, it is useful only if we can find an antiderivative of the function being integrated. To illustrate this, in Example 6.3.9 we showed that z

dr = 2 In 5. JO f 1 + 1--

This, however, is not particularly useful if we do not know the value of In 5. By Example 6.3.5,

ln5 = J

t

dr.

To obtain an approximation to In 5, we can choose from several available methods for obtaining numerical approximations to the definite integral.

Approximations Using Riemann Sums Since upper and lower sums are in general difficult to evaluate, Darboux's definition of the Riemann integral is not particularly useful in obtaining numerical approximations. The most elementary numerical method is the use of Riemann sums. One of the first mathematicians to use numerical methods was Euler, who considered sums of the form

I J(x

k-1

..

t)(Xk - Xk-1)

4. The topics of this section may be omitted on first reading.

6.6 Numerical Methods

261

as an approximation to the integral. This is nothing but the Riemann sum for the partiof [a, b] with tk = xk_, for all k. tion 91 = {x0, x,, . . . , In using Riemann sums to approximate the integral of f, it is convenient to take

equally spaced partitions. Let n E N, and set h = (b - a)/n. Define

xo=a,

x,=a+h,

Thus if 9 = {x0, x,, ... , i = 1,

(17)

with xi as defined, we always have

.x; = h for all

, n, and n

Y,f(ti)Axi = h Yf(t,), i-t

i_,

where for each i = 1, ... , n, t; E [xi_,,xi). If we take t; = xi_ 1 for all i, then we obtain Euler's formula above. Similarly, we could take t; to be the right endpoint xi of the interval [xi_ 1, xi]. Another choice of t; would be the midpoint; i.e., t; = (xi_, + x;)/2. For monotone functions, it is intuitively obvious that the midpoint gives a better approximation than either the right or left endpoint to the integral of f over the interval [xi_ 1, xi](see Figure 6.9). With xi as defined by equation (17) and t; = (xi _, + x,)/2, the above formula becomes

M (f) is called the nth midpoint \`approximation to the integral off over [a, b].

Regardless of the method used, it is important to be able to estimate the error between the true value and the approximate value. If f E 9t[a, b], then for any partition

Figure 6.9

Midpoint Approximation

262

Chapter 6

The Rlemann and Riemann-Stieltjes Integral of [a, b]. we always have b

f f(x) dx - Y(P''f)

t(P'f) - 20'f)

(19)

1

for any choice of tk E [xk _ 1. xk ]. Inequality (19) follows from the fact that both the Rie-

mann sum of f and the integral of f lie between the lower and upper sum of f. If f is monotone increasing on [a. b], then by the proof of Theorem 6.1.8(b),

GU(91rt,f) - T(91,,,f) = h[f(b) -f(a)]. Thus by inequality (19), b

If

f(x)dx - h7,f(t;)` :5 hIf(b) -f(a)I i-

(20)

for any choice of t; E [xi _ 1, x;]. Although we proved inequality (20) for a monotone increasing function, the inequality is also valid for a monotone decreasing function. Thus for monotone functions, inequality (20) provides an estimate of the error between the true value and the approximate value.

6.6.1

EXAMPLE The function f(x) = 1/x is decreasing on [ 1, 5]. Thus by inequality (20),

for n E Id and h = (b - a)/n,

1f(ti)I :Sn15i

11n5

5

1

for any choice of t; E [xi _ 1, x;]. Thus with n = 8, the error is less than

0.4; n = 160 only guarantees an error of less than slo = 0.02. We would be required to take very large values of n to be guaranteed a sufficiently small error.

With n=8,h=Z andxi=I+i,i=0,1,...,8,andt,=xi_j=(1+i)/2

(the left endpoint), 8

9(Ps'f) =

1

1 +i

=2+3+4+5+6+7 I

8+9

= 1.8290 (to four decimal places).

To four decimal places, In 5 = 1.6094. Thus the error is 0.2196-less than the predicted error of 0.4. If we use the midpoint approximation (18) (again with n = 8), then t; = 1 + (i - i) i' which upon simplification equals (3 + 2i)/4. Thus 1

$

4

M8(f)=2j3+2i

6.6

Numerical Methods

=2\5+7+9+ 111+ 13+ 15+ 17+

263

19!

= 1.5998 (to four decimal places). Using the midpoint approximation, the error is less than 0.01. This is considerably better than predicted by inequality (20). As we will shortly see, this improved accuracy is not a coincidence.

Inequality (20) provides an error estimate for monotone functions. For arbitrary real-valued functions, under the additional hypothesis that the derivative is bounded, we have the following theorem.

6.6.2 THEOREM Lei f be a real-valued differentiable function on [a, b] for which f'(x) is bounded on [a, b]. Then for any partition 01 of [a, b],

jf(x)dx

-

II91IIM(b - a),

a

where M = sup{If'(x)I : x E [a, b]}.

Proof. Suppose 01 = {xp, x,, .

.

.

, xj is a partition of [a, b]. Then

qt(o,f) -

= G (Mi - mi)Ozi. i=t

Since f is continuous on [a, b], for each i there exist si, si E [xi_,, xi] such that

Mi - mi = f(si) - f(si) By the mean value theorem, f(si) - f(s;) = f'(ti)(si - s:) for some ti between si and s;. Therefore, since If'(y)I Isi - s'i) s M11 91 II, n

NOV) - 491,f) S MII91II

i-,

1 xi = II01IIM(b - a).

The result follows by inequality (19).

If f satisfies the hypothesis of the theorem, h = (b - a)/n, n E N, and the x; are defined by equation (17), then 1a

f(x)dx - hY f(ti)

M(b - a)h

for any choice of ti E [x,_ 1, xi]. A slight improvement of this inequality is given in the exercises (Exercise 6). If denotes the error between the true value and the approximate value, that is,

En() =

f (x) dx Ja

f ),

264

Chapter 6

The Riemann and Riemann-Stleltjes Integral

then the above inequality can simply be written as IE"(f)I s Ch, where C is a fixed constant depending only on f and the interval [a, b]. Since the term "h" occurs to the first power, this method is commonly referred to as a first order method. In Example 6.6.1 we saw that the midpoint rule provided much greater accuracy than using the left endpoint. This is not a coincidence as the following theorem proves.

6.6.3

THEOREM Iff is twice differentiable on [a, b] and f"(x) is bounded on [a, b], then f (X) I

Ja

- M"(f)

I

<-

b2'

M( 24 a)

where M = sup{I f"(x)I: x e [a, b]}.

Remark Since the error between the true value and the approximate value involves the term h2, the midpoint approximation is a second order method.

Proof. To prove the result, we first prove that for each i = I, ... , n, xXf(x)dx -

hf(a+( i - ')h)I

4h'

(21)

For t E [0, 2], consider the function gi(t) =

j

.+1

c,

f(x)dx - 2tf(c;), r

where ci = (x, _ I + xi)/2. Then

gi(2)=

f(x)dx - hf

(a +(i-

h)).

Since f is continuous, by Theorem 6.3.4

g;(t) = f(ci + t) + f(ci - t) - 2 f'(c), g1(t) = f'(ci + t) - f'(ci - t).

and

By the mean value theorem,

f'(c, + t) - f'(c; - t) = f"(C)2t for some C E (ci - t, ci + t). Since If"(l;)1

M, we obtain

Ig;(t)I <- 2Mt

for all t E [0, J. Since g;(0) = 0, by the fundamental theorem of calculus,

jIg7(x)Idr2Mfxdr1.rMt2. W(t)I = [fg7(x)dx I S 0

0

0

6.6

Numerical Methods

265

Also, since gi(0) = 0, again by the fundamental theorem of calculus, nr

J

j

8'(t)drl

nr

j8;(t)I dt

Mr2) = 24hj

0

0

which proves inequality (21). Therefore,

f(x)dx-M"(t)I

=

I

j f(x)dx-hf1a+li-2)h)

j8i(2)I jt,

82 24Mh3n = M(b - a) h

24

2.

h

The Trapezoidal Rule We now consider another common second order approximation method known as the trapezoidal rule. In using Riemann sums, regardless of the choice of the points t, we used rectangles to approximate the integral of the function f. In this method we will replace rectangles by trapezoids. Let f be a Riemann integrable function on [a, b] and let . , x"} be a partition of [a, b]. As in Figure 6.10, for each interval [xi_,,x;], the area of the trapezoid formed by the points

91 = {x0, xr, .

.

(x;, f(xi)), (xi, 0) is given by

2[J(xi-I)

+f(xi)]

x.

xi

Figure 6.10

Summing these up gives

7, [f(xi-i) +f(x) ]Axi 2;_,

266

Chapter 6

The Riemann and Rlemann-Stleltjes Integral

as an approximation to the integral off on [a. b]. If, as previously, we set h = (b - a)In

and x; = a + ih, i = 0, ... , n, then the above sum becomes

-I h h" 2 7, [f(a + (i - 1)h) + f(a + ih)J = 2 {f(a) + 2 f(a + A) + f(b),. If f is Riemann integrable on [a, b] and n E N, the quantity TT(f) defined by n-I

h

Tn(f) = 2 [f(a) + 2 E f(a + ih) + f(b)

J,

(22)

where h = (b - a)/n, is called the nth trapezoidal approximation to the integral off on [a, b). If we set yi = f(a + ih), then T"(f) can be expressed as h l TV) =2 Yo+22yi+Y.J. The following theorem, under suitable hypothesis on f, provides an estimate of the error of the trapezoidal approximation.

6.6A THEOREM 1f f is twice differentiable on [a, bJ and f"(x) is bounded on [a, b], then b

l

f(x) dx - TT(f)

M(b - a) h2 12

where M = sup{jf"(x)j : x E [a, b]}. The above error estimate can also be expressed as follows: If f satisfies the hypothesis of the theorem, then for n E N, rb I

J AX)

dx - TT(f) I

- M(b - a)3 I 12

(23)

n2

In this form it is possible to determine the value of n required to guarantee predetermined accuracy.

Proof. The proof of this theorem is very similar to the proof of Theorem 6.6.3. As a consequence, we leave most of the details to the exercises (Exercise 5). The first step is

to prove that for each i = 1,...,n, I

f(x)dx-2[f(xi_l)+f(x;)]1 <

.,-1

12

h3.

To accomplish this, consider the function ,+r

{('x `

Jx;_i

f(x)dx

2[f(x,-1) +f(xi-1 + t))

6.6

Numerical Methods

267

fort E [0, h]. Then g;(0) = 0 and g,(h) =

jf(x)dx - 2 [f(x;-1) +f(x,)]

By computing g;(t) and g"(t) one obtains from the hypothesis on f that Ig"(t)I S 1W. Since g,(0) = g;(0) = 0, applying the fundamental theorem of calculus twice we obtain, as in Theorem 6.6.3, that Jg,(h)I s ,2Mh3. The remainder of the proof is identical to Theorem 6.6.3.

Simpson's Rule The trapezoidal approximation T (f) amounts to approximating the function f with a which passes through the points {(x;, f(x;))}"=0 (see piecewise linear function Figure 6.11). Our intuition should convince us that one way to obtain a better approximation to the integral off over [a, b] is to use smoother curves. This is exactly what is done in Simpson's rule, which uses parabolas to approximate the integral. To use quadratic approximations we will need to use three successive points of the partition of [a, b].This is because three points are required to uniquely determine a parabola.

Figure 6.11

Trapezoidal Approximation

Prior to deriving Simpson's rule, we first establish the following formula: If p(x) = Ax2 + Bx + C is the quadratic function passing through the points (0, yo), (h, y, ), (2h, Y2), then ( 2k J p(x) 0

dx =

h

s [Yo + 4Yi + Y2]

(24)

One way to derive this formula would be to first determine the coefficients A, B, C so that p(O) = yo, p(h) = yi, p(2h) = y2, and then integrate p(x). This, however, is unnecessary. By integrating first,

268

Chapter 6

The Riemann and Riemann-Stieltjes Integral

J

up(x) dx = 3 A(2h)3 + 2 B(2h)2 + C(2h) 0

=

[8Ah2 + 6Bh + 6C) 3

=

3

[p(O) + 4p(h) + p(2h)] =

3

[Yo + 4Y. + Y2].

Let f E 9t[a, b] and let n E N be even. Set h = (b - a)/n. On each of the inter-

vals [a + 2(i - 1)h, a + 2ih], i = 1, ...

we approximate the integral off by the integral of the quadratic function that agrees with fat the points

Yo = f(a + 2(i - 1)h),

y' = f(a + (2i - 1)h),

y2 = f (a + 2ih).

By identity (24) this gives

[f(a + 2(i -- 1)h) + 4f(a + (2i - l)h) + f(a + 2ih)] 3 as an approximation to the integral off over the interval [a + 2(i - 1)h, a + 2ih].

Summing these terms from i = 1 to i gives

3[f(a)+4f(a+h)+2f(a+2h) +4f(a+3h)+ +4f(a+(n-1)h)+f(b)] as an approximation to the integral off over (a, b]. The quantity S (f) is called the nth Simpson approximation to the integral off over [a, b]. If we set y; = f(a + ih), then is given by

SM(f)= 3[Yo+4Yi

(25)

+2Y2+4Y3+.

The following theorem, again under suitable restrictions on f, provides an error estimate for Simpson's rule.

6.6.5 THEOREM If f is four times differentiable on [a, b] and f t4i(x) is bounded on [a, b], .

then for n E N even, e

f(x)dx a

Sr(f)1

< M(b - a) ISO

h4,

where M = sup{jf(4)(x)j : x E [a, b]}.

Remarks. Since the error term involves h4, Simpson's rule is a fourth order method. Also, if f(x) is a polynomial of degree less than or equal to 3, then f(4i(x) = 0 and thus

f f (x) dx = S&). 0

6.6

Numerical Methods

269

If f satisfies the hypothesis of the theorem and n E RI is even, then the above error estimate can be expressed as rb

1, f (x) dx - S"(f) I `-

M(b - a)5

I

180

n4

(26)

Proof. The proof of the theorem proceeds in an analogous manner to the proofs of the previous two results. We first prove that for i = 1.... , ;, I

tc, f(x) dx - 3 [ f(c. - h) + 4 f(c.) + f(c. + h)] - _hs,

(27)

ti

where c; = a + (2i - 1)h. To accomplish this, define g, on [0. h] by

dx- 3 [f(c+ - t) + 4f(cr) +f(cr + t)).

gr(t)

To prove inequality (27), we are required to show that Ig,(h)I <- Mh5/90. Upon computing the successive derivatives of g, we obtain

gi(t)=3[f'(c;-t)-f'(c,+t)]+3[f(ci-t)+f(cr+t)]-4f(ci).

g7(t)=3[ f"(c,--f"(c;+t)]+3[

t)+f'(c;+t)],

and

gr' (t) = 3 [f"'(C, - t) - f"'(c; + r)].

By the mean value theorem, f"(c, - t)

t) =

for some

in

(c; - t, c; + t). Therefore. I8,""(t) l s

Mt g.

3

As in the previous two theorems, since g;(0) = g;(0) = g"(0) = 0, upon three integration we have Ig;(h)l s Mh590. This proves inequality (27). Finally, n/2 1a

nr_

f(x)dx - S"(f)I = Igi(x)I Mh5 n

90(2) .6.6.6

EXAMPLE

lgr(x)l

M(b - a) 180

h

In this example we will use the trapezoidal rule and Simpson's rule with

n = 8 to obtain approximations to In 5. For f(x) = x-', f"(x) = 2x'3, and f(4kx) _ 24x-'. Therefore,

sup{If"(x)I: x E [1, 5]} = 2

and

sup{Ift4I(x)I: x E [1, 5]) = 24.

270

Chapter 6

The Riemann and Riemann-Stieltjes Integral

By Theorem 6.6.4 with h = Z. the error E8 in the trapezoidal approximation satisfies 6 IEa(f)I

212(-21 Y

= 0.16666

=

On the other hand, the error Es in using Simpson's rule is guaranteed to satisfy a

IEsl)1

2180

30 =

2

which is considerably better. With x, = 1 + ih and y; = f(x;), i = 0, 1, 2

1

Yo=1,Yi=3,

2

..

.

0.03333

, 8, 2

1

1

2

1

Y6=4,Y7=9,Ye5

Therefore,

1

Ts(f)=4 Yo+2ly,+Ys i=1

I

4+3+1+5+3+7+2+9+5 = 1.6290 (to four decimal places). Since In 5 = 1.6094 to four decimal places, the error is less than 0.02, well within the tolerance predicted by the theory. With Simpson's rule, 1

Ss(f)= 6[rrYo+4Yi +2Yz+4Y3+2Ya+4Ys+2Ye+4Y7 +Ya]

=I[I+3+1+8+3+g+2+9+I = 1.6108 (to four decimal places). With Simpson's rule, the actual error is less than 0.0014. We can also use the results of the theory to determine how large n must be chosen

to guarantee predetermined accuracy. For the trapezoidal rule, since M = 2 and (b - a) = 4, by inequality (23) 3

IEn(f)It212

n2'

Thus to obtain accuracy to within 0.001, we need 3

n2 > 212 10;, which is accomplished with n ? 104. On the other hand, using Simpson's rule, 1<

24.45 1 180

n4.

6.6 Numerical Methods

271

Thus to obtain accuracy to within 0.001, we need n E Rl even with

24.4' nS

>

180

103.

This will be satisfied with n ? 20.

EXERCISES 6.6 1. a. Use the midpoint rule, trapezoidal rule, and Simpson's rule to approximate In 2 = f, 2(1/x) dx with n = 4. For each method, determine the estimated error and compare your answer to In 2 = 0.69315 (to five decimal places).

b. Repeat part (a) with n = 8. c. For each of the three methods (midpoint rule, trapezoidal rule, Simpson's rule) determine how large a must be chosen to assure accuracy in the approximation of In 2 to within 0.0001.

2. a. Using the fact that ' dt

it

f01+t

4

use the midpoint rule, trapezoidal rule, and Simpson's rule to approximate ,r/4 with n = 4. For each method, determine the estimated error.

b. Repeat part (a) with n = 8. c. For each of the three methods determine how large n must be chosen to obtain an approximation of n accurate to within 0.0001. 3. Use Simpson's rule to obtain approximations of each of the following integrals accurate to at least four decimal places. i

2

a. 1a

1 + aP

"b.

dx

10

1 + 2 dx

c. to sin(x2) dx

4. Determine how large a must be chosen so that the trapezoidal approximation T. approximates 5 a-'2dx with an error less than 10-6. 5. Fill in the details of the proof of Theorem 6.6.4. 6. Let f be a differentiable function on [a, b] with f(x) bounded on [a, b]. Let n E N. Prove that e

I J f(x) dx - h

f(a + ih) I S

M(b - a) 2

h,

where h = (b - a)/n and M = sup{j f'(x)h : x E [a, b)). T"(f )). 7. a. Show that Ta,(f) = i b. Show that S5,(f) _ JM"(f) + 3 8. Prove the following variation of Theorem 6.6.4: If f is twice differentiable on [a. b] and f" is continuous on [a. b]. then there exists a point c E [a, b] such that b

f(x) dx =

(b - a)h2 12

f"(c).

272

Chapter 6

The Rlemann and Riemann-Stieltjes integral

9. a. Use the previous exercise to show that ,O,Z < TA - In 2 < 384

b. Using part (a) and Exercise 1(b) show that .6915 < In 2 < .6938.

Proof of Lebesgue's Theorem In this final section we present a self-contained proof of Lebesgue's characterization of the Riemann integrable functions on [a, b]. Recall that a subset E of R has measure zero if given any e > 0, there exists a finite or countable collection {In} of open intervals with E C U. I, and En l(ln) < oo, where I (In) denotes the length of the interval I. We begin with several preparatory lemmas.

6.7.1

LEMMA A finite or countable union of sets of measure zero has measure zero.

Proof. We will prove the lemma for the case of a countable union of sets of measure zero. The result for a finite union is an immediate consequence. Suppose {En}FEN is a countable collection of sets that measure zero. Set E = U E and let e > 0 be given. Since each set E. is a set of measure zero, for each n E N there exists a finite or countable collection {1n.k}k of open intervals such that E.C and kl(I,,.k) < e/2". Since we can always take In, k to be the empty set, there is no loss of generality in assuming that the collection {I , k}k is countable. Then {In, k}n. k is again a countable collection of open intervals with E C U,,, k In, k.

Since N X N is countable, there exists a one-to-one function f of N onto N X N.

For each m E N, set J. =

Then {J,n}T=, is a countable collection of open intervals with EC Un, J,n. Since f is one-to-one, for each N E N, the set F N = f({ 1, ... , N)) is a finite subset of N X N. Hence there exist positive integers N, and KI such that for all (n, k) E FN we have I S n < N, and 1 <_ k <_ K1. Hence N

11 (J.) =

m=1

(n.k)EP,,

I(In.k) c Y, I(In.k) Isn
IsksK,

But N,++

I (In, k) = t

nsN,

15ksK,

Thus

6.7.2

K,

N,

x

N,

G I I (In. k) c 7.1 I (In. k) < Y n=1k=1 n=Ik=I

l (J,n) < e. Therefore E has measure zero.

n< e.

n-12

0

LEMMA Suppose f is a nonnegative Riemann integrable function on [a, b] with f, ,"'f = 0. Then {x E [a, b] : f (x) > 0} has measure zero.

Proof. We first prove that for each c > 0, the set EE _ {x E [a, b]: f(x) ? c} has measure zero. Let e > 0 be given. Since fo f = 0, there exists a partition 9 = {x0, x1, , x,,} of [a, b) with °lt(9, f) < ce, where

..

6.7

Proof of Lebesgue's Theorem

273

and Mi = sup{ f (x) :x E [xi _ 1, xi] }. Let I= {i : E,0 (xi _ 1, x, ] # ¢}. If i E 1, then Mi at c. Hence

CE > V(P,f) 2t I MiAxi

CY1AX,

idl

tEl

Thus J:1Ei1([x,_,, xi]) _ 7,iEi Axi < e. Finally, since EC Uiel [xi_ 1, xi], it follows that Ec has measure zero.

To conclude the proof we note that {x E [a. b] : f(x) > 0} = U' 1 E. where for each n E (\J,

En={xe[a.b]:f(x)!}. By the above, each E. has measure zero. The conclusion now follows by Lemma 6.7. 1.

J

6.7.3

DEFINITION If E is a subset of IL the characteristic function of E, denoted Xe, is the function defined by

XE(x) _

1,

x E E,

0,

x 44 E.

As in Section 6.2, if 9 = {xo, xl, ... , xn} is a partition of [a, b], the norm 119111 of the partition 9 is defined by 119I1 = max Axi. If f is a bounded real-valued function on [a, b], 15i5n

the infimum and supremum off on [xi_ 1, xi] are denoted by mi and Mi respectively; i.e..

m; = inf{f(x) : x E [xi_ 1.xi]}

and

Mi = sup{f(x) : x E [xi_ . xi]}.

The lower function L f and the upper function Uf for f and the partition P are defined respectively by Lf(x)

Uf (x) =

i

and

i=1

Mi X(x, _,. x,)(x)-

The graphs of the lower function L f and upper function Uf off are depicted in Figure 6.12. Since mi s f(x) c Mi for all x E [xi_ 1, xi),

L1(x) s f(x) s Uf(x) for all x E [a, b). Also, since L f and Uf are continuous except at a finite number of points, they are Riemann integrable on [a, b] (Exercise 1) with

J Lf=T(9P,f) a

and

Ja Uf=a13,(2,f)

274

Chapter 6

The Riemann and Riemann-Stieltjes Integral

Figure 6.12 Graphs of the Lower and Upper Functions off

THEOREM 6.1.13 (Lebesgue) Abounded real-valued function f on [a, b] is Riemann integrable if and only if the set of discontinuities off has measure zero.

Proof.

1b1

Assume first that f is Riemann integrable on [a, b]. Then for each n E N, 1/n, such that

there exists a partition 91, of [a, b] with

0:S

and

0 5 O) (91,,,f) - f-

f
(28)

a

Since 2(91.,f) t 2(91,f) and V(91, f) s 61L(91, f) for any refinement 91 of a,,, the partitions 9P,,, n = 1, 2, .. . , can be chosen so that SPn+I is a refinement of 91.. For each n E N, let L,, and U respectively be the lower and upper functions off corresponding to the partition P,,. Then L,(x) s f(x) <_ U,(x) for all x E [a, b) and

= 49., f)

bLn 1a

and

f

bUn

=

(29)

a

Since 91n+ l is a refinement of p,,, the functions L. and U. satisfy La(x) _< L,,, i(x) and Un+ 1(x) s U,(x) for all x E [a. b]. Define L and U on [a, b] by

L(x) = lim L,(x)

and

U(x) = lim U,(x).

Then L. (x) 5 L(x) 5 f(x) 5 U(x) s U,(x)

for all x E [a, b). Hence 16Ln a

JbL< a

bL: Jbf 1a

a

rbUC bUC JbUn Ja

a

a

6.7

Proof of Lebesgue's Theorem

275

But by equations (28) and (29). h

!,

lim J lim

h

I

L = Jim f n-.x

u

f.

a

n

Hence L and U are Riemann integrable on [a, b] with b

I L = JbU= fbf Since U(x) - L(x) ? 0 for all x E [a, b], by Lemma 6.7.2, {x E [a, b] : U(x) # L(x)} has measure zero. Furthermore, since each 91 has measure zero, by Lemma 6.7.1, the set

E = {x E [a, b] : U(x) # L(x)} U (U 9'n) also has measure zero. We conclude by showing that f is continuous on [a, b]\E. Fix x, E [a, b]\ E, and let e > 0 be given. Since L(x,,) = U(x,), there exists an integer k E N such that Uk (x,) - Lk(x,) < e. Also, since xa E 9'k, the functions Uk and Lk are constant in a neighborhood of x0. Hence there exists a S > 0 such that 0 < Uk (x,,) - Lk (x) = Lk (x) - Lk (xo) < E

for all x E [a, b] with Ix -

S. Finally, since Lk(x) <- f(x)

Uk(x) for all

X E [a, b), - E < Lk (x) - Uk (xo) C f (x) - f (x,,) < Uk (x) - Lk (xo) < E

for all x with Ix - x0I < S. Therefore f is continuous at x,,. Conversely, suppose f is continuous on [a, b]\E, where E is a set of measure zero

and a < b. Let M > 0 be such that I f(x)I

M for all x E [a, b], and let e > 0 be

given. Since the set E has measure zero, there exists a finite or countable collection

of open intervals such that E C U I and E 1(1,,) < E/4M. Also, since f is continuous

on [a, b]\E, for each x E [a, b] \ E there exists an open interval J, such that 1f (z) - f (y) I < E/2(b - a) for all y, z e J, fl [a, b]. The collection U {J, : x E [a, b]\ E} is an open cover of [a, b]. Thus by compactness, a finite number, say {Ik}x=, and {JX}; , also cover [a, b]. Let

9' = {a = to, t,, ..

.

, tN = b}

be the partition of [a, b] determined by those endpoints of Ik, k = 1. ... , n, and J,,, j = 1, ... , m, that are in [a, b]. For each j, 1 s j c N, the interval (t,_,, t,) is contained in some Ik or J,,. Let J = {j: (t, _ 1, t,) C Ik for some k}. Also, for each j E {1, ...,NJ, let m, and M; denote the infimum and supremum off on [t,_,. t;] respectively.Then M, - m, <- 2M for all j E J, and M, - m, < E/2(b - a) for all j 44 J. Thus

R(9P,f)-X(P,.f)= I(Mi-mi)Oti+

52M:01 + ,EJ

f

1CJ

I(M!-mi)At,

E c1t!

2(b - a)1EJ

Chapter 6

276

The Rlemann and Riemann-Stieltjes Integral

E

:5 2M7, 1 (1k)+2(b-a

N

4r;

E E 52M\4M)+2(b-a)(b-a)=E.

Hence by Theorem 6.1.7 the function f is Riemann integrable on [a, b].

1]

W EXERCISES 6.7 1. Let f be a bounded real-valued function on [a, b] and 91 a partition of [a, b]. Prove that the lower function Lt and upper function Uf off for the partition ' are Riemann integrable on [a, b] with rb

jb Lf = Y(9,f) 0

= W495,f)

and 1a

2. Let P be the Cantor set in [0, 1 ]. Show that Xr E %[(), 1) and find f.' kAX)dx.

3. Let f E R[a, b], and for x E [a, b], set F(x) = fo f(t) dt. Prove that there exists a subset E of [a, b] of measure zero such that F'(x) = f(x) for all x E [a, b] \ E. 4. *Suppose f E %[a, b) and g is a bounded real-valued function on [a. b). If {x E [a, b) : g(x) # f(x)} has measure zero, is g Riemann integrable on [a, b)?

NOTES The fundamental theorem of calculus is without question the key theorem of calculus; it relates the Cauchy-Riemann the-

ory of integration to differentiation. For Newton, Leibniz, and their successors, integration was the inverse operation of differentiation. When Cauchy, however, defined integration independent of differentiation, the fundamental theorem of calculus became a necessity. It was crucial in proving that Cauchy's integral was the inverse of differentiation, thereby providing both a convenient tool for the evaluation of definite integrals and proving that every continuous real-valued function defined on an interval I has an antiderivative on /. Although we stated the fundamental theorem of calculus as two separate theorems, for continuous functions they are the same. Specifically, if f is a continuous real-valued function on [a, b] and F is an antiderivative off, then for

any x E [a, b],

F(x) = Fa) + 11f(t) dt. Conversely, if f is continuous and F is defined as above.

then F(x) = f(x).

Lebesgue's theorem is one of the most beautiful results of analysis. It provides very concise necessary and sufficient conditions for Riemann integrability of a bounded real-valued function f. Although Riemann was the first to provide such conditions, his result lacked the simplicity and elegance of Lebesgue's theorem. Unfortunately for Riemann, the concept of measure had not yet been developed when he stated and proved his result. In

Chapter 10 we will develop the theory of measure and Lebesgue's extension of the Riemann integral. The need for numerical methods for the evaluation of definite integrals was recognized as early as the eighteenth century. Euler and Thomas Simpson (1710-1761), among others, used numerical techniques to approximate the definite integral in problems where an antiderivative could not be found. Even the error estimates developed in Section 6.6 date back to that era. With the availability of efficient calculators and high-speed computers, numerical methods have increased in importance in the past few decades. This has led to the development of very sophisticated numerical algorithms for the evaluation of definite integrals.

Miscellaneous Exercises

Although Newton and Leibniz are credited with inventing the differential and integral calculus, many mathematicians prior to their time knew formulas for computing tangents and areas in particular instances. Archimedes (around 200 a.c.), in his treatise Quadrature of the Parabola, used the method of exhaustion by inscribed triangles to derive a formula for the area under a parabolic segment. By the mid-1640s, Pierre de Fermat (1601-1665) had determined the formulas for the area

under any curve of the form y = x' (k * -1), and for finding tangents to such curves. Isaac Barrow (16301677), a professor of gegmetry at Cambridge, developed an algebraic procedure that is virtually identical to the

277

differential calculus for finding tangents to a curve in his 1670 treatise Lectiones geometriae. However, all these early methods were developed using geometric arguments. Newton and Leibniz developed the concepts, the notation. and the algorithms for making these computations for arbitrary functions. Most importantly, however, both men

realized the inverse nature of the problems of tangents and areas. For these reasons they are credited with the development of the differential and integral calculus. Further information on the historical development of calculus may be found in the text by Katz listed in the Bibliography and the article by Rosenthal listed at the end of the chapter.

MISCELLANEOUS EXERCISES A real-valued function f on [a, b] is a step function if there exist a finite number of disjoint intervals (Ii}n- 1 with U Il = [a, b) such that f is constant on each of the intervals lt, j = 1, ... , n.

1. a. If f is a step function on [a, b), prove that f E 9t[a, b) with b -

a

f=7,

'-1

where ci is the value off on I;.

b , Iff E 9t[a, b] and e > 0 is given, prove that there exists a step function h on [a, b] such that

Ifaf - fahI< e. c. If h is a step function on [a, b] and e > 0 is given, prove that there exists a continuous function g on [a, b]

such thatIfah - f°gI <e.

d. If f E 9t[a, b] and e > 0 is given, prove that there exists a continuous function g on [a, b] such that

I fa f - fa gl <e. Let f be a real-valued function defined on (0, oo). The laplace transform off, denoted T(f ), is the function defined by

1(f)(s) =1 e "f(t) dt,whenever the improper Riemann integral exists. 0

2. Let f be defined on (0, oo). Prove that there exists a e R U {-oo, oo) such that f(f)(s) is defined for all s e (a, oo), and that the integral defining Z(f)(s) diverges for all s E (-oo, a). (Hint: First show that if `d;(f)(sa) exists for some sa, then T(f)(s) exists for all s > se.) 3. Suppose f E 9t[0, c] for every c > 0, and that there exists a positive constant C, and a E R. such that I f(r) 1 s Ce- for all t ? 0. Prove that T(f)(s) exists for all s > a.

278

Chapter6

The Riemann and Rlemann-Stieltjes Integral

4. Compute the Laplace transform of each of the following functions. In each case, specify the interval on which `af(f)(s) is defined

a. At) = 1 b. f (r) c. f(t) = cos wt

d. f(t) = sin tot e. f(t) = t", n E N (Use induction.) I. f(t) = I (t - c), where J(t - c) is the unit jump function at t = c g. f(r) = t°, a > - I (See Exercise 9, Section 6.4.) 5. Suppose f is differentiable on [0, oo) and a E R is such that `.f(f)(s) exists for all s > a. If lim e-"f(t) = 0 for all

s > a, prove that Y(f')(s) = s .(f)(s) - f(0).

® SUPPLEMENTAL READING Bagby, R. J., "A convergence of limits," Math. Mag. 71

(1998),270-277. Bao-lin, Z., "A note on the mean-value theorem for integrals." Amer. Math. Monthly 104 (1997), 561-562. Bartle, R. G., "Return to the Riemann integral;' Amer. Math. Monthly 103 (1996), 625-632. Botsko, M. W., "An elementary proof that a bounded a.e. continuous function is Riemann integrable," Amer. Math. Monthly 95 (1988), 249-252. Bullock. G. L., "A geometric interpretation of the Riemann-Stieltjes integral:' Amer. Math. Monthly 95 (1988), 448-455. Goel, S. K. and Rodriguez, D. M., "A note on evaluating limits using Riemann sums:' Math. Mag. 60 (1987), 225-228. Grabinger, J. V., "Was Newton's calculus a dead end? The Continental influence of Maclaurin's Treatise of Fluxions," Amer. Math. Monthly 104 (1997), 393-410. Hartman, F. and Sprows, D., "Oscillating sawtooth functions:' Math. Mag. 68 (1995), 211-213. Jacobson, B., "On the mean-value theorem for integrals:' Amer. Math. Monthly 89 (1982), 300-301.

Jones, W. R. and Landau, M. D., "One-sided limits and integrability," Math. Mag. 45 (1972), 19-21. Klippert. J.. "On the right-hand derivative of a certain integral function;' Amer. Math. Monthly 98 (1991). 751-752. Kristensen, E., Poulsen, E. T. and Reich, E., "A characterization of Riemann integrabiliry"Amer. Math. Monthly 69 (1962), 498-505. Quadling, Douglas, "a in my eye," Math. Gaz. 77

(1993),314-321. Rickey, V. F. and Tuchinsky, P. M., "An application of geography to mathematics: History of the integral of the secant," Math. Mag. 53 (1980), 162-166. Rosenthal, Arthur, "The history of calculus:' Amer. Math. Monthly 58 (1951), 75-86. Ross, K., "Another approach to Riemann-Stieltjes integrals:' Amer. Math. Monthly 87 (1980).660-662. Stein, S. K., "The error of the trapezoidal method for a concave curve:' Amer. Math. Monthly 83 (1976)

643-645. Williams, K. S., "Note on 5

44 (1971),9-11.

(sin xlx)dx," Math. Mag.

71 Series of Real Numbers 7.1 Convergence Tests

7.2 The Dirichlet Test

7.3 Absolute and Conditional Convergence 7.4 Square Summable Sequences

Although the study of series has a long history in mathematics,' the modem definition of convergence dates back only to the beginning of the nineteenth century. In 1821, Cauchy, in his text Cours d'Analyse, used his definition of limit to provide the first formal definition of convergence of a series in terms of convergence of the sequence of partial sums. The Cauchy criterion (Theorem 2.7.3) was the first significant result to provide necessary and sufficient conditions for convergence of a series. Cauchy not only stated and proved the result, he also applied his result to prove convergence and divergence of given series. Many of the early convergence tests, such as the root and ratio test, are due to him. Cauchy, with his formal development of series, placed the subject matter on a rigorous mathematical foundation. In this chapter we will continue our study of series of real numbers. Our primary emphasis in Section 7.1 will be on deriving several tests that are useful in determining the convergence or divergence of a given series. In Section 7.3, we will study the concepts of absolute convergence, conditional convergence, and rearrangements of series. One of the key results of this section is that every rearrangement of an absolutely convergent series not only converges, but converges to the same sum. As we will also see, this fails dramatically if the series converges but fails to converge absolutely. In Section 7.4, we give a brief introduction to the topic of square summable sequences. These play an important role in the study of Fourier series. One of the main results of this section will be the Cauchy-Schwarz inequality for series. The section also contains a brief introduction to normed linear spaces. 1. See the notes at the end of the chapter.

279

280

Chapter 7

Series of Real Numbers

Convergence Tests In Section 2.7 we provided a very brief introduction to the subject of infinite series. In the study of infinite series it is very useful to have tests available to help determine whether a given series converges or diverges. For example, Corollary 2.7.5 is very useful in determining divergence of a series. If the sequence {ak} does not converge to zero, then the series 7- ak diverges. On the other hand, if lim ak = 0, then nothing can be ascertained concerning convergence or divergence of the series I ak. In this section we will state and prove several useful results that can be used to establish convergence or divergence of a given series. Additional tests for convergence will be given in the exercises and subsequent sections. With the exception of Theorem 7.1.1, all of our results in this section will be stated for series of nonnegative terms. I will As in Definition 2.7.1, given an infinite series Ja I ak of real numbers, denote the associated sequence of partial sums defined by n

sn _ Y, ak. k=1

The series 7, ak converges if and only if the sequence of nth partial sums converges in R. Furthermore, iffl-00 lim s = s, then s is called the sum of the series, and we write

M

I ak = S. k=1

If the sequence

diverges, then the series I ak is said to diverge. Furthermore, if

lim s = oo (or -oo), then we write 7, ak = oo (-oo) to denote that the series diverges to 00 (or - oo). If ak at 0 for all k, then by Theorem 2.7.6 the series 7, ak converges if and only if lim s < oo. Thus for series of nonnegative terms we adopt the notation cc

I ak < 00 k=1

to denote that the series converges.

Remarks (a) Although we generally index a series by the positive integers N, it is sometimes more convenient to start with k = 0 or with k = ka for some integer ko. In this case, the resulting series are denoted as

I ak,

k=0

T, ak

k=k

7.1

Convergence Tests

Also, from the Cauchy criterion (Theorem 2.7.3) it is clear that and only if 7, a k., ak converges for some, and hence every, k,, E N.

281

I ak converges if

(b) Given any sequence {sn}.'= 1 of real numbers, we can always find a series I ak whose nth partial sum is sn. If we set a1 = s1 and ak = sk - sk_ 1, k > 1, then n

T. ak = Si + (SZ - S1) + .

.

+ (Sn - S;_,) = Sn.

k=1

7.1.1

THEOREM If Yoko= , ak = a and 7.0ko- I bk = A, then

a

(a) Y, cak = ca, for any c E R, and k-1 DO

(b) F; (ak+bk)=a+$3. k-1

Proof. The proof of (a) is similar to (b) and thus is omitted. To prove (b), for each n E N, let n

n

Sn =

ak

and

to =

bk.

k-1

k-1

Since the series converge to a and $3 respectively, tim s = a and lint t = $3. Therefore, by Theorem 2.2.1, lim(sn + tn) = a + $3. But n

n

ak + T, bk = 7, (ak + bk).

S. + to = k=1

k-1

k-1

Therefore, sn + to is the nth partial sum of the - series Z(ak + bk). Since the sequence {sn + t,,} converges to a + (3, OD

I (ak+bk)=a+A

k-1

Comparison Test One of the most important convergence tests is the comparison test. Though very elementary, it provides one of the most useful tools in determining convergence or divergence of a series. It is useful both in applications and theory. Several of the proofs of subsequent theorems rely on this test. In applications, by comparing the terms of a given series with the terms of a series for which convergence or divergence is known, we are then able to determine whether the given series converges or diverges.

282

Chapter 7

7.1.2

Series of Real Numbers

THEOREM (Comparison Test) Suppose 7, ak and 7, bk are two given series of nonnegative real numbers satisfying

0 < at

Mbk

for some positive constant M and all integers k a kn, for some fixed k E N. 00

(a) If

bk < oo, then Y, ak < oo k=1

k-1

W

x

(b)IfY, at = oo, then 7, bk = oo. k=1

k-1

Proof. Suppose the terms {ak} and {bk} satisfy at 5 Mbk for all k > ko and some positive constant M. Then for n > m ? ko, n

n

k=m+I

k=m+I

05 1 akSM 7,

bk.

Suppose I bk converges. Then given e > 0, by the Cauchy criterion (2.7.3) there exists an integer n,, ? k such that km+1 bk < M

for all n > m ? n,. Thus 0 <_ L+k=m+I at < e for all n > mat n0. Hence by the Cauchy criterion E at converges. On the other hand, if I at diverges, then 1 bk must also diverge. As a corollary of the previous theorem we also have the following version of the comparison test.

7.1.3

COROLLARY (Limit Comparison Test) of positive real numbers.

Suppose E at and E bk are two given series

(a) If liimW bn = L with 0 < L < oo, then 7. at converges if and only if E bk converges.

(b) If limb = 0 and B bk converges, then I at converges.

-oo

Proof. The proof, the details of which are left to the exercises (Exercise 6), follows immediately from the definition of the limit and the comparison test.

Remark. If Jim an/bn = 0 and I at converges, then nothing can be concluded about the convergence of the series 7, bk. In 7.1.4(d) we provide an example of a divergent series 7, bk and a convergent series E at for which lim 0. On the other hand, n-+oo

7.1

Convergence Tests

283

in Exercise 23, given a convergent series E ak with ak > 0, you will be asked to construct a convergent series I bk with bk > 0 such that lim a"Jb, = 0.

7.1.4

EXAMPLES

(a) As an application of the comparison test, consider the series °O

k

ksi 3

Thus we We will compare the given series with the convergent geometric series 7, wish to show that there exists ko E Fk1 such that k/3k s 1/2k for all k a k0. By Theorem 2.2.6(d), k

k 3 = 0.

k

Thus by taking e = 1, there exists an integer ko such that k(3)k <_ 1 for all k ? k,. As a consequence,

3k s 2k

for all k ? k0.

Since the geometric series E('-i)k converges (Example 2.7.2(a)), by the comparison test, the given series also converges. A similar argument can be used to prove that I k"/ak converges for any n E 7L and a E R with a > 1 (Exercise 2(1)). This is accomplished

by comparing the given series with 7, b-k, where 1 < b < a. Thus by Theorem 7.1.1, p(k)/ak converges for any polynomial p and any a E R with a > I (Exercise 11). (b) As our second example, consider the series °°

+1 zk3+ 1

To use the comparison test, we first have to determine what series we want to compare with. Since

k+ 1 _

2k +1

1

1 +k

k 2+k

and

kltm.

1 k-1j,

2+k

2

we will compare the given series with the series Y, 1/k. This series is known to diverge (Example 2.7.4). If we take e = 4%52, then we can conclude that there exists ko E N such that

1+k

11vF2

2+k:2- e

4

for all k at k0. Thus there exists a positive constant M and k, E N such that

k+1 I 2k3+1aMk

284

Chapter 7

Series of Real Numbers

for all k ? ko. Since the series E 1/k diverges, by the comparison test the given series also diverges. (c) The divergence of the series

k+ 1

co

1

2k'+l

can also be obtained by the limit comparison test. Comparing the terms of the given series to the terms of the series Y, 1/k we have

?kT

lint -"o

1+k = 2+k -2

__

1

k-r+moo

k

Thus since the series 11/k diverges, by the limit comparison test the series °O

k+1

I 'V 2k' + 1 also diverges. 2_k

(d) Let ak =

and bk = 1/k. Then 7, ak converges, 7, bk diverges, and

'-

b,,

aoo

2" = 0.

Thus if rim " = 0, convergence of the series 7, ak does not imply convergence of the

b

series

bk.

Integral Test

Our second major convergence test is the integral test. Recall from Section 6.4, if f is a real-valued function defined on [a, oo) with f E %[a, c] for every c > a, then the improper Riemann integral off is defined by

dx = lint

jf(x) dx, a

u

provided the limit exists. If f(x) at 0, we use the notation 00

f(x) dx < oo a

f f (x) dx = oo \ o 00

to denote that the improper integral off converges (diverges).

7.1.5

THEOREM (Integral Test)

Let {ak} I be a decreasing sequence of nonnegative real numbers, and let f be a nonnegative monotone decreasing function on( 1, oo) satisfying f (k) = akfor all k E N. Then

F, ak < 00 if and only if

k-t

Jf(x) dx < coo.

7.1

Convergence Tests

285

Proof. Since f is monotone on [ 1, oo), by Theorem 6.1.8 it is Riemann integrable on [ 1, c] for everyc > 1. Let n E (\i, n ? 2, and consider the partition `2P = { 1, 2, .... n } of [ 1 . n]. Since f is decreasing, f o r each k = 2, 3, ... , n (see Figure 7.1),

sup{f(t) : t E [k - 1, k]} = f(k - 1) = ak_ I, inf{f(t) t E [k - 1, k]} = f(k) = at.

Figure 7.1

and

Integral test

Therefore,

k-2

.f(x)dx C

ak = X(9j)

k-1

I

ak

from which the result follows.

7.1.6

EXAMPLES

(a) As our first application of the integral test, we consider the p-series

When p = 1 this series is referred to as the harmonic series. If p

0, then {k-°} does

not converge to zero, and thus by Corollary 2.7.5, the series diverges. Suppose

p > 0, p

1. Let f(x) = xo, which is decreasing on [ 1, oo). Then `

1

Jx-Pdr=---___[i

p-1

--J 1

1

c

Therefore, rco JI

x-odx=lim `~00

JI

1

p > 1,

00,

p < 1.

x-°dx= p- 1

286

Chapter 7

Series of Real Numbers

Thus by the integral test, the series diverges for p < I and converges for p > 1. When p = 1, then by Example 6.3.5,

f i

= Inc. `Xdx

Since lim In x X-W

lim+ In t = oo (Example 6.4.2(a)), by the integral test the series 1-.0 also diverges for p = I. Summarizing our results, we find °O+

G kr

f converges, diverges,

if p > 1, if p

1.

(b) As our second application of the integral test, we consider the series

I 00

1

k..,- kInk*

Let f(x) = (x In x)'', x E [2, oo). Since

(1+lnx) f'(x)

(x in x)2

f'(x) < 0 for all x > 1. Thus f is monotone decreasing on [2, oo). But

lim J c

I

C- 2 x In x

dx = lim ln(ln c) - ln(ln 2) = oo.

Thus by the integral test the given series diverges.

(c) As our final example, we consider the series

'° Ink

7 kp. pEIR. We first consider the case p = 1. Since f(x) = (In x)/x is decreasing on [e, oo) and

`Inx lira - dx = lim0100 (in c)2 - 1 = oo, X

C

by the integral test the series I'k*-2 (In k)/k diverges.

Suppose now that p > 1. Write p = q + r, where q > 1 and r > 0. By I'Hospital's rule M On x)/x' = 0. Thus there exists k E N such that (In k)/k' < 1 for all k ? ko. As a consequence,

Ink _ 1 Ink kq k' kr

1

kq

7.1

Convergence Tests

287

for all k ? k0. Since q > I the series I 1/kq converges. Hence by the comparison test, the series I (In k)/kP also converges. Finally, if p < 1, then again by the comparison test, the divergence of I 1/k' implies that 7, (In k)/kP also diverges. This follows from the fact that (In k)/k" ? (In 2)/k" for all k E N, k ? 2. Summarizing our results we find

ink k__Z

kP

f converges, diverges,

if p > 1, if p 5 1.

Ratio and Root Tests We now consider the well-known ratio and root tests. Although useful in determining the convergence or divergence of certain types of series, both of these tests are also very important in the study of power series. The ratio test, attributed to Jean d'Alembert (1717-1783), is particularly applicable to series involving factorials. The root test, attributed to Cauchy, will be used in the next chapter to define the interval of convergence of a power series. Because of the close similarity between these two tests we state them together. Prior to stating the ratio and root tests we recall the definition of the limit inferior (lim) and limit superior ( lim) of a sequence of real numbers (Definition 2.5.1). If {sn}

is a sequence in R, then

lim s = sup inf{s : n ? k} = lim inf{s : n a k}, k-+0o k6N

lim s,

kE

sup{s : n

and

k} = slim sup(s : n !- k).

By Theorem 2.5.7, if E denotes the set of subsequential limits of {sn} in OI U {-oo, oo}, then

!Lm s = inf E

n _CG

and

lim s,, = sup E.

n

In particular, if the sequence {sn} converges (or diverges to either - oo or oo), then Urn S. = lira S. = lim S.

7.1.7

THEOREM (Ratio Test) Let I ak be a series of positive terms, and let

R = lim ak'+ k-+oo ak

and

r = lim

00

(a) If R < 1, then I, ak < 00k-1 00

(b) If r > 1, then I ak = oo. k-1

(c) If r 5 1 5 R, then the test is inconclusive.

k-+oo ak

a"-+'.

288

Chapter 7

Series of Real Numbers

7.1.8 THEOREM (Root Test) Let 7, ak be a series of nonnegative terms, and let k

a = lim k-0oo

ak

(a) If a < 1, then 7, 00 ak < oo. k-I 00

(b) If a > 1, then

', ak = oo. k=1

(c) If a = 1, then the test is inconclusive. Before proving the theorems we give several examples that illustrate these two convergence tests.

7.1.9

EXAMPLES

(a) Consider the p-series °O

1

A kp, p E R.

k- I

limakk1=lim(1+k1 =

kroo a

1.

kyoo \\

Also, by Theorem 2.2.6,

r urn

oo

k1

k-*w

P

Ok)

1.

Therefore, for this series,

r=R=a=1 for any p E R; thus both tests are inconclusive. The series diverges, however, if p s 1, and converges for p > 1.

(b) As our second example, we consider pk

kl,

0
Here ak = pk/k!. Thus

p 4+I +l

lim

ak+1

k-+oo ak

pk

= lim k-boo

1

(k + 1)!

.

= p lim

I k-+= k + 1

= 0.

7.1

Convergence Tests

289

By the ratio test the series converges for all p, 0 < p < oo. As a consequence of Corollary 2.7.5, we also obtain k

limp =0

k-.oo k!

for all p E R. In this example, the presence of k! makes the root test difficult to use.

(c) Consider I a where

an

if n = 2k,

Zk,

_

l

ifn=2k+1.

3k.

Here, 00

1+2+3+22+32+

-i

By computation, 2

(3)

a.+,

1

121, n=2k+l. '1/3k

a"

For the subsequence of {a,,

with even n,

lim

k- oo

a2k+I a2k

lk 21

= lim k-,oo 3

= 0.

For the subsequence with odd n,

lim

a2k+2

k-+oo a2k+ t

1

2

liM

/3\k

k-G21 2 '

00.

Thus {0, oo} is the set of subsequential limits of Consequently, r = 0 and R = oo. Therefore, the ratio test is inconclusive. On the other hand, 1

1 M )1/(2k+ t),

n= 2k,

n=2k+1.

By Theorem 2.2.6(b), urno (V)v(2k+ q = 1. Thus the sequence {

} has two sub-

sequential limits; namely, 1/V and 1/V/3. Therefore, a = 1/N/2, and the series converges by the root test.

290

Chapter 7

Series of Real Numbers

Proof of Ratio Test. Suppose lim

n-.m

a= a,,

= R < 1.

Choose c such that R < c < 1. Then by Theorem 2.5.3(a), there exists a positive integer n, such that an

±an1 s c

for all n ? no.

In particular, ane, 1 s can . By induction on m, a,, ,,,, < C'"a,, Therefore, writing

n=no+m,mz0,

an <_ Mc"

for all n at no,

where M = a,. /c"°. Thus since 0 < c < 1, by Example 2.7.2(a), the series E c" converges. Therefore, E an also converges by the comparison test. Suppose

lima"}=r> I.

nToo an

Again choose c so that r > c > 1. As above, there exists a positive integer n, such that a,, z Mc" for some constant M and all n _> n,. But since c > 1,1 c" = oo, and thus by

the comparison test, B an = oo. Q Proof of Root Test. Let

a = lim
" an < c for all n a n,. But then a,, s c" for all n ? n and E an < oo by the comparison test.

If a > 1, then

> I for infinitely many n. Thus a" > 1 for infinitely many

n, and as a consequence, {a,} does not converge to zero. Hence by Corollary 2.7.5, the

series B a,, diverges. Q Example 7.1.9(c) provides an example of a series for which the ratio test is inconclusive, but the root test worked. Thus it appears that the root test is a stronger test; a fact which is confirmed by the following theorem.

7.1.10

THEOREM Let {an} be a sequence of positive numbers. Then

lim aa,

n-rm an

lim

=oo

lim

s lim a-` n-+oo an

7.1

Convergence Tests

291

Remark As a consequence of the theorem, if a series E ak converges by virtue of the ratio test, i.e., R < 1, then we also have a < 1. Similarly, if 7. ak diverges by virtue of the ratio test, i.e., r > 1, then we also have a > 1. Thus if the ratio test proves convergence or divergence of the series, so does the root test. The converse, however, is false (as indicated above).

Proof. Let F,ma-+i

R

n-+oo a"

If R = oo, then there is nothing to prove. Thus assume that R < oo, and let P > R be arbitrary. Then there exists a positive integer n, such that an an

s0

for all n ? n,.

As in the proof of the ratio test, this gives an s MI" for all n ? n where M = a,.19" Hence

"a"<_0u for all n=?n,. = 1, we have

Since lim

M-000

llm "a:s p.

n-,00

s R, which proves the result. The inequality for the limit inferior is proved similarly. Q

Since J3 > R was arbitrary, Tim

7.1.11

EXAMPLE

If ak = k!, then lim(ak+I/ak) = oo. Thus as a consequence of the pre-

vious theorem,

lim 'N_! = oo. S

k-+oo

EXERCISES 7.1 1. If a and b are positive real numbers, prove that W

I

Al (ak + b)o converges if p > I and diverges if p c 1. 2. Test each of the following series for convergence: sA,

2

k

k' +I

M

b.

kt,

_

k2+2k-1

01+

k2

C. G2k

Chapter 7

292

Series of Real Numbers

00

3k

k'e_k

d. F,

*e.

k1

k-1

k+ 1 - f

(kl)2

*9.

*L

(2k)!

(/ - 1)k

J.

1.

k.

k.

k_1

-j, a> 1, nEZ

k-1

*m.

sin

k-1

(1k)

p>0

p>0

0.

ykln(l+k k=1

3. For each of the following, determine all values of p E R for which the given geometric series converges, and find the sum of the series.

*a 7, (sin 0

b.

3

(l ± p)

c.

(p * 1)

k-11

4. Suppose ak a 0 for all k E N and I a,, < oo. For each of the following, either prove that the given series converges, or provide an example for which the series diverges. 00

00

00

a,2,

4; 1 + ak

c. k-I

k-1

M e.

*d. 7, kak

V k ak

f.

k-1

t-1

g. I a,,,, where (a j is a subsequence of 5. *Determine all values of p and q for which the following series converges:

I

1

k4(ln k)P'

(Hint: Consider the three cases q > 1, q = 1, q < 1.) 6. *Prove Corollary 7.1.3. 7. If I ak converges and B bk = oo, prove that E (ak + bk) = oo. is a sequence in R with a > 0 for all n E N. For each k E 101 set 8. Suppose bk =

1

k

- ± a,,. k n=I

Prove that Jk , bk diverges. 9. Suppose that the series E ak converges and {n;} is a strictly increasing sequence of positive integers. Define the sequence {bk} as follows:

b,=a,+

+a

,

bk=a.,_,+,+...+a,,:

Prove that E bk converges and that $k , bk = k I ak. (This exercise proves that if the series E ak converges, then any series obtained from Eat by inserting parentheses also converges to the same sum. The following exercise shows that removing parentheses may lead to difficulties.)

7.1

Convergence Tests

293

10. Give an example of a series F. ak such that Ek , (a2k _, + a2k) converges, but E ak diverges. 11. Prove that p(k) k-1

converges for any polynomial p and a > 1. 12. *Suppose that the series E ak of positive real numbers converges by virtue of the root or ratio test. Show that the series Yt , k"ak converges for all is E N. 13. *Show that the series 1

1

+23+32++... I

1

converges, but that both the ratio and root tests are inconclusive.

14. Apply the root and ratio tests to the series I ak where 2k.

when k is even,

2t+2'

when k is odd.

at

0 for all k E N. Prove that the series Ik , ak converges if and only if some subsequence {sj of the

15. Suppose ak sequence

of partial sums converges.

16. *Cauchy Condensation Test: Suppose that a, z a2 z a3 a

? 0. Use the previous exercise to prove that

P 2kag converges.

7, R , ak converges if and only if

17. Use the Cauchy condensation test to show that $4 , 1/n^ converges for all p > 1, and diverges for all p,

0
1

*a. k_' 2 k Ink

b.

O0

00+

1

k_'2 k(In k)"

(P > I)

e' L k(ln kiln Ink)

19. *For k E N, let ck be defined by

Prove that {ck} is a monotone decreasing sequence of positive numbers that is bounded below. The limit c of the sequence is called Euler's constant. Show that c is approximately 0.577.

20. Raabe's Test: Let ak > 0 for all k r= N. Prove the following.

a. If ak+,/ak -- I - r/k for some r > 1 and all k a k k, E Id, then T. ak < oo. (Hint: Show that

(k - 1)ak - kak1 a (r - I)ak for all k ? k,.) b. If ak+,/ak z I - 1/k for all k k k0 E N, then I ak = oo. (Hint: Show that {kak,1} is monotone increasing fork z k,.) 21. *lf p, q > 0, show that the series

(p+ (q+

1)(p+2)... (p+k)

1)(q+2)...(q+k)

converges for q > p + I and diverges for q 5 p + 1.

294

Chapter 7

Series of Real Numbers

22. For p > 0 consider the series L

(2k

I'+I2.4Ir+...+(1.2 4.

+.. 1)

1

a. Show that the ratio test fails for this series. b. Use Raabe's test to show that the series converges for p > 2, diverges for p < 2, and that the test is inconclusive when p = 2. c. Prove that the series diverges for p = 2. 23. Let {an} be a sequence of positive real numbers.

*a. Suppose I ak converges. Construct a convergent series 71 bk with bk > 0 such that lim akbk = 0. kroc b. Suppose I ak diverges. Construct a divergent series E bk with bk > 0 such thatk m bb/ak = 0.

The Dirichlet Test In this section we prove the Dirichlet convergence test, named after Peter Lejeune Dirichlet (1805-1859), and apply it to both alternating series and trigonometric series. The key to the Dirichlet test is the following summation by parts formula of Neils Abel (1802-1829). This formula is the analogue for series of the integration by parts formula.

7.2.1

THEOREM (Abel Partial Summation Formula) Let {ak} and {bk} be sequences of real numbers. Set

Ao=0

A'=

and

n Iak

ifn?1.

k-I

Then if15p5q, 4-1

q

Y, akbk = E Akbk - bk+ 1) + Agbq - Ap-1b .

k=p

k=p

Proof. Since ak = A k - Ak_ 1, q

q

I akbk = I (Ak - Ak-1) bk

k=p

k=p

q-1

4

Akbk+1

Akbk k-p

k-p

k=p-I

Akbk - bk+1) + Agbq - Ap-1b

.

O

As an application of the partial summation formula we prove the following theorem of Dirichlet. Another application is given in Exercise 1.

7.2 The Dirichlet Test

7.2.2

295

THEOREM (Dirichlet Test) Suppose {ak} and {bk} are sequences of real numbers satisfying the following:

(a) the partial sums A. _ Jk=, ak form a bounded sequence.

(b) b, a b2 a b3 a

? O, and

(c) k-occ liimbk=0. Then Jko' , akbk converges.

Proof. Since {A,} is a bounded sequence, we can choose M > 0 such that IA. 1 s M for all n. Also, since b -+ 0, given e > 0, there exists a positive integer n, such that b < e/2M for all n z n,. Thus, if n, s p s q, by the partial summation formula, q- I

k.p akbkl =

L , Ak(bk - bk+1) + Aqbq - Ap_Ibp

k-p 1

qj IAkI(bk - bk+1) + I AqI bq + IAp-1lbp k-p

Cq.P -1

M7, (bk-bk+1)+bq+bp

2Mbp<E. Hence by the Cauchy criterion (Theorem 2.7.3), Jk , akbk converges.

0

Alternating Series Our first application of the Dirichlet test is to alternating series. An alternating series is a series of the form Y, (-1)k bk or I(-1)k+ 1bk, with bk > 0 for all k

7.2.3

THEOREM (Alternating Series Test) If {bk} is a sequence of real numbers satisfying

(a) b1?b2z (b)

k

0, and

bk = 0,

OD

then I (-1)k+lbkconverges. k-1

Proof. Let ak = (-1)k+ 1. Then IA. I < 1 for all n, and the Dirichlet test applies. O Remark The hypothesis (a) that {bk} is decreasing is required. If we only assume that bk a 0 and Jim bk = 0, then the conclusion may be false (Exercise 2). For an alternating series satisfying the hypothesis of Theorem 7.2.3, we can actually do better than just prove convergence. The following theorem provides an estimate to the sum of the series. on the rate of convergence of the partial sums

296

Chapter?

Series of Real Numbers

7.2.4 THEOREM Consider the series

7,(-1}k+Ibk, where the sequence {bk} satisfies the

hypothesis of Theorem 7.2.3. Let 00

(-1)k+Ibk

Sn =

s = 7, (-1)k+1bk

and

k=1

k=I

T h e n I s - s.1 < b.+ 1 for all n E N.

Proof. Consider the sequence {stn}. Since 2n

sz, = 7, (_ 1)k+Ibk = (b1 - b2) + ... + (b,,_ I k-1

is monotone increasing. Similarly and (bk_ 1 - bk) >_ 0 for all k, the sequence {s2i+1} is monotone decreasing. Since {sq} converges to s, so do the subsequences

{s2,,} and {s2,+I}. Therefore, s2, s s <- s2,, +I for all n E N. As a consequence,

Is-SkI:5 ISk+I-SkI=bk+I for all kEN.

7.2.5

E)

EXAMPLE Since the sequence fl/(2k - l)} decreases to zero, by Theorem 7.2.3, the series 00 (-1)k+I k- I

2k - l

converges. As we will see in Chapter 9 (Example 9.5.7), this series converges to 7r/4. Thus by Theorem 7.2.4, if s is the nth partial sum of the series,

4 - s" 12n + 1 for all n E N. Although this can be used to obtain an approximation to ar, the convergence is very slow. We would have to take n = 50 to be guaranteed accuracy to two decimal places.

Trigonometric Series Our next application of the Dirichlet test is to the convergence of trigonometric series. These types of series will be studied in much greater detail in Chapter 9.

7.2.6 THEOREM (Trigonometric Series) Suppose {bk} is a sequence of real numbers satisfying bI ? b2 ?

? 0 and kim bk = 0. Then

(a) Y,'I bk sin kt converges for all t E R, and (b) Yk I bk cos ki converges for all t E P. except perhaps t = 2pir, p E Z.

The Dirichlet Test

7.2

297

Proof. To prove the result, we require the following two identities: For t * 2pir,

pE1,

cos Zt - cos(n + 2)t

71si n kt = k=1

2 sin i

(1) tt

sin(n + !)t - sin i t

n

Ylcos kt = k=,

2

2 sin gt

(2)

2

We will prove identity (1), leaving identity (2) for the exercises (Exercise 4). Set A. = Ink= I sin kt. Using the trigonometric identity sin A sin B = 2 (cos (A - B) - cos(A + B)) we obtain

(sm)An= 2

k=t

sin

i t sin kt 2

=2kCosk2t-cosk+2

ItJ

=2fcos2t-cosn+21IJ. Thus for t # 2pir, p E 7L, An =

Cos i t - cos(n + ;)t 2sin 2t

Therefore,

lCos Its + jcos(n + 21 sin Zt{

Z)t(

sin 2tl '

which is finite provided t * 2pir, p e Z. Consequently by the Dirichlet test, the series in (a) converges for all t # 2 pir, p E Z. If t = 2prr, p E Z. then sin kt = 0 for all k. Thus the series in (a) converges for all t E R. The proof of the convergence of the series in (b) is similar. However in this case, when

t = 2pir, cos kt = 1 for all k E N, and the given series may or may not converge. Q

7.2.7

EXAMPLE By Theorem 7.2.6, the series

k-i k

Coskt

298

Series of Real Numbers

Chapter 7

converges for all t * 2pir, p E Z. When t = 2prr, p e Z, then cos kt = I for all k and the series E k diverges. On the other hand, the series 00

k-I

1

- zcos ki k

converges for all t E R.

EXERCISES 7.2 2.

*Abel's Test: Prove that if ak converges, and {bk} is monotone and bounded, then E akbk converges. bk = 0. *Show by example that the hypothesis of Theorem 7.2.3 cannot be replaced by bk z 0 and ,,

3.

If I ak converges, does E ak always converge?

4.

*Prove that

1.

k:1 S.

cos kt =

sin(n + 2)t - sin 21 ,

t * 2p7r, p G Z.

2 sin 2 t

Test each of the following series for convergence.

a, (-1)k+I' p>0 k'

kal

kk

00

* d.

(k + 1) k

k-1 si

6.

kkt rER,>0

00 (-1)klnk

b. k-2 7

k

!. k-1D -1) *h.

*G kk

k+1

k Cs kt

(k + I)k+1

*f

(-Ir Mklnk sink k_Z In k

IE R ,p > 0

Given that

w (-i)k+1 k.l

k2

jr2 12'

determine how large n E N must be chosen so that I nrz/12 - s^ I < 10-', where s^ is the nth partial sum of the 7.

series. If p and q are positive real numbers, show that 00

(ln k)a

converges. &

*Suppose that E ak converges. Prove that

lkak=0.

lim 0-00 n k.1

7.3

9.

Absolute and Conditional Convergence

+

As in Exercise 19 of the previous section. let ck = I + ; +

b = I - l2 +

1 3

- ... _ 1 = 2n

299

- In k. Set

(-1)k+1 k=1

k

Show that lim b = In 2. (Hint: b, = c_ - c + In 2.)

7.3

Absolute and Conditional Convergence In this section we introduce the concept of absolute convergence of a series. As we will see in this and subsequent sections of the text, the notion of absolute convergence is very important in the study of series. We begin with the definition of absolute and conditional convergence.

7.3.1

DEFINITION A series ak of real numbers is said to be absolutely convergent (or converges absolutely) if 1 ak I converges. The series is said to be conditionally convergent if it is convergent but not absolutely convergent. We illustrate these two definitions with the following examples.

7.3.2

EXAMPLES

(a) Since the sequence {l/k} decreases to zero, by Theorem 7.2.3 the series (-1)k+ 1/k converges. However,

° (-I)k+I II k

k=1

xI I

=

Ik=oo.

k-1

(-1)+'/k is conditionally convergent. (b) Consider the series _ (-1)k+1/k2. By Theorem 7.2.3 the alternating series converges. Furthermore, since 7, 1/k2 < oo, the series is absolutely convergent. U Thus the series

Our first result for absolutely convergent series is as follows.

7.3.3

THEOREM

If Z ak converges absolutely, then F_ ak converges and

k=1

akl

k-1 lakl.

Proof. Suppose Zak converges absolutely; i.e., `jak1 < oo. By the triangle inequality, for 1 < p s q,

7,ak) < y IakI k-p

k-p

300

Chapter 7

Series of Real Numbers

Thus by the Cauchy criterion (Theorem 2.7.3) the series 2 , ak converges. Finally. with p = I in the above,

kI ak l = 1im l

t a,

:!s:

l

I

t

lakl :9

kY"

IakI. Q

Remark, To test a series I ak for absolute convergence we can apply any of the appropriate convergence tests of Section 7.1 to the series 7, Iakl. There is, however, one important fact which needs to be emphasized. If the series the ratio or root test, i.e.,

r = lim Ian+ > 1 w_-_

Iak I diverges by virtue of

a = lim » Ianl > 1, n-wo

or

Ia.I

then not only does 71 1akl diverge, but I ak also diverges. To see this, suppose a > 1. Then as in the proof of the root test, IakI > I for infinitely many k. Hence the sequence {ak} does not converge to zero, and thus by Corol-

lary 2.7.5, 1 ak diverges. Similarly, if r > 1 and if I < c < r, then as in the proof of Theorem 7.1.7(b), there exists a positive integer n and constant M such that

Ianl y Mc" for all n at no. Thus again, since c > 1, (an} does not converge to zero, and the series 71 ak diverges. We summarize this as follows.

7.3A THEOREM (Ratio and Root Test) Let I ak be a series of real numbers, and let

aslim

.

Also, if ak # O for all k E N, let

R= Tim k-*oo

a, ak

r = lim

and

k+m

Iak+, I ak

(a) If a < I or R < 1, then the series I ak is absolutely convergent. (b) If a > 1 or r > 1, then the series 7, ak is divergent. (c) If a = 1 or r 5 1 <_ R, then the test is inconclusive.

7.3.5

EXAMPLE To illustrate the previous theorem, we consider the series Z. -I P(k)o. where p is a polynomial and b E R with I b I < 1. In this example, ak = p(k)b` and thus

p(k + 1)

ak+, ak

Ib1 I

Since kim p(k + 1)/p(k) = 1 (verify this), tio test the series is absolutely convergent.

,im

p(k)

(

Iak+,/a,I = IbI < 1. Thus by the ra-

7.3

Absolute and Conditional Convergence

301

Rearrangements of Series. We next take up the topic of rearrangements of series. Loosely speaking, a series I a,. is a "rearrangement" of the series 7- ak if all the terms in the original series 7, a4 appear exactly once in the series 7, ak, but not necessarily in the same order. For example, the series

l

+

1z+

+

22+

3-

l2+ 5

4-

7-

k= k

The following provides a formal definition of this concept.

7.3.6

DEFINITION A series 7, ak is a rearrangement of the series

ak if there exists a one-to-one function j from N onto N such that ak = ate? for all k E N.

A natural question to ask is: "If the series 7, ak converges and I ak is a rearrangement of I ak, does the series ak' converge? If so, does it converge to the same sum?" As we will see, the answer to both of these questions is no! 7.3.7' EXAMPLE Consider the series 1

k-l

which converges, but not absolutely. Consider also the following series, which is a rearrangement of the above:

+3-I+5+I-4+I+11-6 +... 2

9

7

s=

k k=1

As in the proof of Theorem 7.2.4, s <

s<s3

for all n E N. In particular,

=1-23 +

6'

Let s,, denote the nth partial sum of the series in (3). Then

1-3+4k-1 1

3'3-

R 71

k -I 4k

_

1) 2k)

(3)

302

Chapter 7

Series of Real Numbers

Since

+

I

4k - 3

1

1

4k - 1

2k

-

8k-3 2k(4k - 1)(4k - 3)'

we have

_

I

1

0<4k-3+4k-1

1

1

for some constant M. Thus the sequence {sin} is strictly increasing, and by the comparison test converges. Let s' = lim sin. Since n-,00

1

A. +I -'S3n+ 4n + I

and

1

1

S3n+2=S3,,+4n+

1 +4n+3'

the sequences {sin+I} and {s',+2} also converge to s'. Therefore, lim sn = s'. Thus the series (3) also converges. However, since

6=Si<Sg<Sy< s' = lim s,, > 6. Thus the series (3) does not converge to the same sum as the original n-i00 series. This, as we will see, is because the given series does not converge absolutely.

7.3.8 THEOREM If the series E ak converges absolutely, then every rearrangement of 7. ak converges to the same sum.

Proof.

Let 7, ak be a rearrangement of I ak. Since exists a positive integer N such that

lakl < oo, given e > 0, there

lakl <E

(4)

k-n

for all m z n a N. Suppose ak = atk), where j is a one-to-one function of N onto N. Choose an integer p (p ? N) such that {1, 2, .

.. , N} C {j(1),j(2), ... ,j(p)}.

Such a p exists since the function j is one-to-one and onto N. Let a

Sn

k-I

n

ak

and

s;, _

k-I

Ifnap, n

Sn - sn = k-!

rn

ak - L./ a1(kN k-I

ak.

7.3

Absolute and Conditional Convergence

303

By the choice of p, the numbers a,, ... , aN appear in both sums and consequently will cancel. Thus the only terms remaining will have index k or j(k) greater than or equal to N. Thus, by inequality (4),

Is.-S;,I 52E for all n a p. Therefore, lim s;, = lim sn, and thus the rearrangement converges to the n-o0

n- 00

same sum as the original series. p

For conditionally convergent series we have the following very interesting result of Riemann.

7.3.9

THEOREM Let ak be a conditionally convergent series of real numbers. Suppose a E R. Then there exists a rearrangement I ak of 7, ak which converges to a. Before proving the theorem, we illustrate the method of proof with the alternating series 00

1

k-I

This series converges, but fails to converge absolutely. The positive terms of this series

are Pk = 1/(2k - 1), k E N; and the absolute values of the negative terms are Qk = 1/2k, k E N. Since 7, 1 = oo, we also have 7, Pk = 7, Qk = oo. Suppose for purposes of illustration a = 1.5. Our first step is to add "just enough" positive terms to exceed a. More precisely, we let m, be the smallest integer such that P, + + P,,, > a. For a = 1.5. m1 = 3; i.e., 1 + 3 + 3 > 1.5, whereas 1 + 3 < 1.5. Our next step is to go in the other direction; namely, we let n 1 be the smallest integer such that

PI+...+pm,-Q,-. . .-Qn
Proof. Without loss of generality, we assume that ak * 0 for all k. Let pk and qk be defined by

Pk = 2 (IakI + ak)

and

qk = 2 (IakI - ak).

Then pk - qk = ak and Pk + qk = I ak I I. Furthermore, if ak > 0, then qk = 0 and Pk = ak ; if ak < 0, then Pk = 0 and qk = IakI. We first prove that the series I Pk and 7, qk both diverge. Since 00

00

I(Pk+qk)= k-I I IakI, k-I they cannot both converge. Also, since n

n

n

I ak = 7, Pk - 7, qk k-I k=I k-I

304

Chapter 7

Series of Real Numbers

the convergence of one implies the convergence of the other. Thus they both must diverge.

Let P1, P;, P3,. .. denote the positive terms of 7, ak in the original order, and let Q1, Q2, Q3, ... be the absolute values of the negative terms, also in the original order. The series Pk and 7, Qk differ from I pk and I qk only by zero terms, and thus are also divergent. We will inductively construct sequences {mk} and {n*} such that the series

p1 + ... + Pm, - QI - . . . - Qn, + P.'+1 + + Pm, - Qn..1 - Q,, + . . .

(S)

has the desired property. Clearly series (5) is a rearrangement of the original series. Let m1 be the smallest integer such that

XI=PI+. ..+P,n'>a. Such an m1 exists since 7, Pk = oo. Similarly, let n1 be the smallest integer such that =X1-Q1-.

Y1

. .-Qn,
Suppose {m1, ... , mk} and {n1, . . . , nk} have been chosen. Let mk+ 1 and nk+ I be the smallest integers greater than mk and nk respectively, such that Xk+I = Yk + Pm,+l + . .. + Pm,., > a,

Yk+I -Xk+1 -Qn,+I --

and

. - Qn,.,
Such integers always exist due to the divergence of the series I Pk and I Qk. Since mk+1 was chosen to be the smallest integer for which the above holds,

Xk+I - Pm, 5a. Therefore,

0<Xk+I -a-15 P.,., Similarly,

0
lim Xk = k-.oo lim Yk = a.

kroo

Let S. be the nth partial sum of the series (5). If the last term of Sn is a Pn, then there exists a k such that

Yk<Sn:!:-: Xi+I If the last term of Sn is -Qn, then there exists a k such that Yk+1 E S. < Xk+1.

In either case, we obtain 1 m Sn = a, which proves the result.

7.3

Absolute and Conditional Convergence

305

Remark. By a variation of the above proof one can show that if the series 7. ak is conditionally convergent, then given a, 6 with

-oo
lim S = a n1

lim S. = /3,

and

m

Atia

where S. is the nth partial sum of E ak (Exercise 13).

EXERCISES 7.3 1. Prove the following:

a. If lim k°ak = A for some p > I, then F, ak converges absolutely.

b. If iimn k ak = A * 0, then Y, ak diverges. k

c. If kim k ak = 0, then the test is inbonclusive.

2. 'Suppose I ak < oo and

Mk < oo. Prove that the series V, ak bk converges absolutely.

3. a. Prove that if I ak converges and I bk converges absolutely, then 7, ak bk converges. b. Show by example that the conclusion may be false if one of the two series does not converge absolutely. is monotone decreasing with lim b = 0. If {a,} is a sequence in R satisfying 4. Suppose that the sequence for all n E N, prove that ak converges absolutely. bekak converges for every choice of ek E { -1. I }. 5. a. If 7, ak converges absolutely, prove that b. If Y_ ek ak converges for every choice of ek E { -1. 1 }, prove that 7, ak converges absolutely. 6. Test each of the following series for absolute and conditional convergence. I

(_I)k+l = 71

k-1

b.

Nrk

d.k-a-I(-I)kln(lnk) In k (-1)k+lkk

9. k-1 1 (k +

1)1`

I

(-1)k

G N/k In(In k) {-I)k

k)P

'e. k

z k (In

sG k-I

h.

= (-I)k+Ikk

P>0

00

,(-1+1 sin

()

k-1

kv

k-1 (k + 1)

=

(-1)k

I.

I k

,2kt+(_l)k

7. Test the series 1;k 1 pk/k°, p E R, for absolute and conditional convergence. 8. 'Show that the series I + - 3 + a + s - 1 + diverges.

9. Determine whether the series I - z - i + + s

a

i +s+s

o

ill

+

converges or diverges.

10. 'Prove that the series Ik 1 (sin k)/k is conditionally convergent. 11. Prove that if 7, ak is conditionally convergent, then there exists a rearrangement 7, ak of

ak which diverges to c.

12. If ak a 0 for all k E N, and B ak = co, prove that 7, ak = oo for any rearrangement ak of I ak. 13. Suppose the series I ak is conditionally convergent. Given a,)9 with -oo s a <- rA s oo, show that there exists a rearrangement I ak of ak such that lim S. = a and lim S. where S. is the nth partial sum of 7, ak.

ft

14. Suppose every rearrangement of the series

M-00

ak converges. Prove that

ak converges absolutely.

306

Chapter 7

7.41

Series of Real Numbers

Square Summable Sequences2 In this section we introduce the set l2 of square summable sequences of real numbers and derive several useful inequalities. This set occurs naturally in the study of Fourier series.

7.4.1

DEFINITION A sequence {ak}k 1 of real numbers is said to be in 12, or to be square

summable, if 0C

Iak
For {ak} E I2 set II {ak} II 2 =

Xak.

The set 12 is called the space of square summable sequences, and the quantity II{ak}II2 is called the norm of the sequence {ak}.

Remark. Since a sequence {ak} in R is by definition a function a from N into R with ak = a(k), it is sometimes convenient to think of 12 as the set of all functions a: N - R for which

F

II a Ill =

7.4.2

I a(k)I2 < C.O. 1

EXAMPLES

(a) For the sequence { l/k}x 1,

kk2. H{'}1122

Since this is ap-series with p = 2. the series I 1/k2 converges and thus { 1/k} E 12. On the other hand, since

l %Ilk

II2

k-11k - '

we have { 1 / Vk} 4E l2.

2. The topic of square summable sequences, although important in the study of Fourier series, is not specifically required in Chapter 9 and thus can be omitted on first reading. The concept of a nonmed linear space occurs on several occasions in the discussion of subsequent topics in the text. At that point, the reader can study this topic more carefully.

7.4

Square Summable Sequences

307

(b) For fixed q, 0 < q < oc, consider the sequence { I /k v }. Then

{ klq

k'

22

By Example 7.1.9(a) this series converges for all q > ; and diverges for all q s ;-Thus { 1/k°} E 12 if and only if q > ;

Cauchy-Schwarz Inequality Our main goal in this section is to prove the Cauchy-Schwarz inequality for sequences in l2. First, however, we prove the finite version of this inequality.

7.4.3

If n E NI, and at,.,., an and b1..... bn

THEOREM (Cauchy-Schwarz Inequality) are real numbers, then n+

n

k=1

Proof.

Iakbkl `-

, bk

Fk=

k=1

Let A E R and consider

k=1

k=1

` Iakbkl + A2 , bk. n

n

0 ` 7,(Iakl - AIbkl)2 = 1 ak - 2A

k-1

k.1

The above can be written as 0 <_ A - 2AC + A2B,

where A = 7,k=, ak, C = 1k-1 Iakbkl. and B = 2;k=1 bk. If B = 0, then bk = 0 for all k = 1, ... , n, and the Cauchy-Schwarz inequality certainly holds. If B 0, we take k = C/B which gives B2

OVA or C2 <_ AB; that is,

k

lak

5

()(n)

Taking the square root of both sides gives the desired result. 0 As a consequence of the previous result we have the following corollary.

7A.4 COROLLARY (Cauchy-Schwarz Inequality) If {ak}, {bk} E 12, then 7,i 1 akbk is absolutely convergent and

k-1

lakbkl s II{ak}112II{bk}II2.

308

Chapter 7

Series of Real Numbers

Proof.

For each n E N, by the previous theorem Iakbkl

II{ak}II2II{bk}II2-

bk

ak

k-1

k-l

k-I

Letting n --> oo gives the desired result.

Minkowski's Inequality Our next result shows that the norm 11 112 satisfies the "triangle inequality" on 12.

7A.5 THEOREM (Minkowski's Inequality) If {ak} and {bk} are in 12. then {ak + bk} E 12 and II{ak + bk}II2 c II{ak}1I2 + II{bk}II2-

Proof. By hypothesis, each of the series 7, a2 and I bi' converge. Also, by Corollary 7.4.4, the series E ak bk converges absolutely. Since

(ak + bk)2 = ak + 2akbk + bk <_ ak + 2lakbkl + Mk, we have -0

00

II{ak + bk}IIi = 2 (ak + bk)2 5 II{ak}IIz + 2 7, Iakbkl + II{bk}IIi k-1

k=1

which by the Cauchy-Schwarz inequality, II{ak}IIi + 211{ak}II2II{bk}II2 + II{bk}II2 S

= (11{ak}II2 + 110J12)2-

Taking the square root of both sides gives the desired result.

Although not specifically stated, Minkowski's inequality is also true for finite . , sums. In particular, if n E N and a,, . .. , a,,,'bi, are real numbers, then

..

j(ak+bk)2S.Jj a,2k + k-1

bk.

(6)

k=1

I

In the following theorem we summarize some of the properties of the norm 112 on 12. These are very similar to the properties of the absolute value function on R. As for the absolute value, the inequality II

11{a4 +

bk}112

< I1{ak}II2 +

II{bk}II2

is also referred to as the triangle inequality for 12.

7A.6 THEOREM The norm 11 16 on 122 satisfies the following properties: (a) ll{ak}112 ? 0/oral! {ak} E 12. (b) II{ak}1I2 = 0 if and only if ak = O for all k E N.

7.4 Square Summable Sequences

309

(c) If {ak} E 12 and c E R, then {cak) E 12 and II{cak}II2 = IcI (d) If {ak}, {bk} E l2, then {ak + bk} E 12 and II{ak + bk}IL II{ak)II2 + II{bk}II2. II{ak}II2.

Proof. The results (a) and (b) are obvious from the definition, and (d) is just a restatement of Minkowski's inequality. The verification of (c) is left as an exercise (Exercise 5). 0

Euclidean Space U8" The finite versions of both the Cauchy-Schwarz and Minkowski inequalities can be used to provide some useful results about n-dimensional real euclidean space R". For n E N, R" consists of all n-tuples (a,, a2, ... , aR) of real numbers. When n = 1, R' is just the real line R. For notational convenience, we denote the above n-tuple by a; i.e..

a = (a,, a2, ... , a"). The point a is also referred to as the vector a with components a,, k = 1, 2, ... , n. Two points a and b of R" are equal if and only if ak = bk for all k = 1, 2, ... , n. If a = (a,, ... , a"), b = (b,,. . . , b") E R^ and c E R, we define a + b and ca by

a+b=(at+b,,...,a,+b"), ca = (ca,, .

.

. , ca").

For two vectors a and b in IR", the Inner product of a and b, denoted (a. b). is defined by

(a, b)

akbk.

kai

Also, for a E l8" we define the norm or length of the vector a by ak.

118112= \/(a, a)=

As a consequence of the Cauchy-Schwarz inequality, I(a, b)I s IIaI6IIbII2

for all a, b E W. The eucildean distance d2(a, b) between a and b is defined as d2(a, b) = IIa - N12 = V 7, (ak - bk)2. n k=,

When n = 1, d2(a, b) simply becomes Ia - bI . By Minkowski's inequality (6) we have

d2(a, b) s d2(a, c) + d2(c, b)

for all a b. c E R".

Nonmed Linear Spaces The space 12 as well as Q8" are both examples of vector spaces over It For completeness we include the definition of a vector space.

310

Chapter 7

7.4.7

Series of Real Numbers

DEFINITION A set X with two operations " + ", vector addition, and " ", scalar multiplication, satisfying

x+ y E X for all x, y E X,

and

c'xEX forallxEX,cER is a vector space over R if the following are satisfied:

(a) x + y = y + x. (commutative law) (b) x + (y + z) = (x + y) + z. (associative law) (c) There is a unique element in X called the zero element, denoted 0, such that

x+0=x forallxEX. (d) For each x E X, there exists a unique element -x E X such that

x+(-x)=0. (e)

+a'y. (g) (a+b)'x=a'x+b'x. (h) I'x=x.

(f)

It is clear that R" with + and ' defined by

a+b=(a,+b,,. . .,an+bn), c' a = (cab.

.

.,Can)

is a vector space over R. Similarly, if addition and scalar multiplication of sequences {ak}, {bk} in 12 are defined by

{ak} + {bk} = {ak + bk}, c E R, c {ak) = {cak}, then it is easily shown that 12 is a vector space over R. The fact that {ak + bk} and {cak} are in 12 is part of the statement of Theorems 7.4.5 and 7.4.6. The zero element 0 in l2

is the sequence {ak} where ak = 0 for all k E N. In addition to being vector spaces, the spaces R" and 12 are also examples of normed linear spaces. The concepts of a "normed linear space" and a "norm" are defined as follows:

7.4.8

DEFINITION

Let X be a vector space of R. A function !III : X -+ R satisfying

(a) llxll ? Ofor all x G X, (b) IIxIl = 0 if and only if x = 0,

(c) llcxll = lcl llxll for all c E R, x E X, and

(d) llx+yll!5 llxll+IIYIIforallx,yEX is called a norm on X. The pair (X, 11 11) is called a normed linear space.

Inequality (d) is called the triangle inequality for the norm 11 11. By Theorem 7.4.6, is a nonmed linear space. Additional examples of normed linear spaces will occur elsewhere in the text and the exercises. (12, II Il2)

7.4

If (X, 11

11)

Square Summable Sequences

311

is a normed linear space, the distance d(x. y) between two points

x, y E X is defined as d(x, y) = II x - y II. From the definition of the norm II 1. it immediately follows that

(a) d(x, y) ? 0 for all x, y E X. (b) d(x,y) = 0 if and only if x = y, (c) d(x, y) = d(y, x), and (d) d(x, y) s d(x, z) + d(z. y) for all x, y, z e X. Thus d is a metric on the vector space X. As in Definition 2.1.6, fore > 0. p E X. the e-neighborhood Nf(p) is defined as

For example, in 1I2. with norm II II2 an e-neighborhood of a = (a,, a2) is the set of

points x = (x,, x,) satisfying (x, - a,)2 + (x, - a2)2 < e-, that is. an open disc with center a and radius a (See Figure 7.2). In 1183, an e-neighborhood of a is simply an open ball with center a and radius e. X2

a,

a,

Figure 7.2

An e-Neighborhood of a = (a,, a2) E R2

Since the notions of "convergence of a sequence" (2.1.7) and "limit point of a set" (2.4.5) are defined in terms of e-neighborhoods, both of these concepts have analogous definitions in a normed linear space (X, 11 11). Thus for example, a sequence in X

converges to x E X, if given e > 0, there exists n,, E IO' such that IIx - xfl < e for all n E N. n ? no. If this is the case, we say that x,, converges in the norm II II to x. This type of convergence, referred to as norm convergence, will be encountered again later in the text. As a general rule, many of the results involving sequences of real numbers are still valid in the setting of a normed linear space. This is especially true for those theorems whose proofs relied only on properties of the absolute value. Great care, however, must

312

Series of Real Numbers

Chapter 7

be exercised with theorems that rely on the supremum property of R. For example. the Bolzano-Weierstrass theorem fails for (12, 11 112) (Exercise 7). On the other hand, the Bolzano-Weierstrass theorem still holds in (R". Ii IIz) (Exercise 6).

6 EXERCISES 7.4 1. Determine which of the following sequences are in 12. sa.

(

1

Ink k=z

b.

]

"

c I In k

J !,in kj"

d.

t Vk Ink I k- 2

Vk k-z 2. Determine all values of p E R such that the given sequence is in 12.

{k"Ik

d.

k

k-1

Vk-(Ink)P k.:

3. If {a,.} E 12, prove that tli m ak = 0.

, at/k converges absolutely. 4. *If {ak} E 12, prove that 5. If {ak} E 12 and c E R. prove that {cak} E 12 and II{cak}IIz = Iclll{ak}IIz

6. a. Suppose that (%)-A., is a sequence in R", where f o r each k E N, pk = ( p , , . . . . . pk.,,) Prove that the sequence { pk} converges top = (p,, ... , p") in the norm 11112 if and only if m pk..; = p, for all i = 1..... n.

k-x

b. Prove that every bounded sequence in R" has a convergent subsequence.

7. For each n E N. let e" be the sequence in 12 defined by

e"(k)=

(0, kn, 1.

k=n.

Show that II e" - emit = V if n # m. (Remark: The sequence {e" } is a bounded sequence in 12 with no convergent subsequence. Thus the Bolzano-Weierstrass theorem (2.4.11) fails in 12.) 8. Show that equality holds in the Cauchy-Schwarz inequality if and only if for all k E N. h, = cak for some c E R. 9. For a, b E R", let (a. b) denote the inner product of a and b. Prove each of the following.

a. (a,a)?0 b. (a, a) = 0 if and only if a = (0.. ... 0) c. (a, b + c) _ (a, b) + (a, c)

d. (a, b) _ (b, a) e. (a. b) = 12 (IIAV2 2 + Vblli - IIa - 15111). 10. *For a, b E R", use the law of cosines to prove that (a. b) = Ilal2llbf 2cos 0, where B is the angle between the vectors a and b. 11. Let (X, II V) be a normed linear space. Prove that I lIxII - Ilyll l 5 Ox - yll for all x. y E X. 12. Let 1' denote the set of all sequences {ak} satisfying H{ak}ll, _ Ik , lakl < oo. a. Prove that (1', II II,) is a normed linear space. b. Prove that 1' C 12. 13. Determine all values of p E R for which each of the sequences in Exercise 2 is in 1'. 14. Let X be a nonempty set, and let S(X) denote the set of all bounded real-valued functions on X. For f E 31(X), set Ilf L = sup{I f(x) I : x r= X}. Prove that (11(X), 11 IL) is a nonmed linear space.

Notes

313

NOTES The geometric series is perhaps one of the most important series in analysis. It forms the basis for the proof of the ratio and root test and thus also for the study of convergence of power series. In the seventeenth and eighteenth centuries, the convergence of many series was established by comparison with the geometric series. The geometric series dates back to Euclid in the third century B.C. The formula for a finite sum of a geometric progression appeared in Euclid's Elements, and Archimedes. in his treatise Quadrature of the Parabola, indirectly used the geomet-

gence. An excellent illustration of these eighteenth-century techniques is Euler's derivation of the sum of the series

ric series to find the area under a parabolic arc,3 Even though Greek mathematicians knew how to sum a finite geometric progression, they used reducto ad absurdum

As power series were used more frequently to approximate mathematical quantities, the emphasis turned to deriving precise error estimates in approximating a function by a finite sum of the terms of its power series. In 1768. d'Alembert made a careful examination of the binomial

arguments to avoid dealing with infinite quantities. Infinite series first appeared in the middle ages. In the fourteenth century, Nicole Oresme (1323?-1382) of Italy provided a geometric proof to the effect that the infinite series 1

2

n 2+22+

2"

had sum equal to 2, a result now quite easy to prove using power series. He was also the first to prove that the harmonic series diverged, whereas the Italian mathematician Pietro Mengoli (1624-1686) was the first to show that the

sum of the alternating harmonic series I - i + ; - o + is In 2. The results of Oresme and Mengoli, as well as others of this era, were stated verbally, the infinite sum notation did not appear until the seventeenth century.

With the development of the calculus in the midseventeenth century, the emphasis shifted to the. study of power series expansions of functions and computations involving power series. The expansion of functions in power series was, for Newton and his successors, an indispensable tool. In his treatise, Of the Method of Fluxions and Infinite series,s Newton includes a discussion of infinite series techniques for the solution of both algebraic and differential equations. Both the geometric series and the binomial series were employed in many of his computations.

During the seventeenth and eighteenth centuries, mathematicians operated with power series in the same way

as with polynomials; often ignoring questions of conver-

1I

I

+3-+.. I

The convergence of this series had been established earlier by Johann Bernoulli. Using results from the theory of equations and applying them to the power series expansion of sin x, Euler was able to show that the sum of the given series was 7r2/6.'

series (I + r)", p E Q, discovered earlier by Newton.6 D'Alembert computed the value of n for which the absolute value of the ratio of successive terms is less than 1, thereby ensuring that the successive terms decrease. More importantly, d'Alembert computed the bounds on the error made

in approximating (I + r)" by a finite number of terms of this series. His argument relied on a term-by-term comparison with a geometric series, similar to the proof used in es-

tablishing the ratio test. D'Alembert's result can easily be converted into a proof of the convergence of the series.

Computations analogous to those of d'Alembert were also undertaken by Lagrange in his book, Theorie des Foncrions Analvtiques, published in 1797. The remainder term for Taylor's formula (to be discussed in the next chapter) first appeared in this text. Cauchy was undoubtedly influenced by the works of d'Alembert and Lagrange in his definition of convergence of a series. However, unlike the results of d'Alembert, which applied only to the binomial series, and those of Lagrange for Taylor series, Cauchy's definition of convergence was entirely general: applicable to any series of real numbers. Using his definition of convergence. Cauchy proved that the statement now known as the Cauchy criterion was necessary for the convergence of a series of real numbers. This was also known earlier to Bolzano. However, without the completeness property of the real number system, neither Bolzano nor Cauchy was able to prove that the Cauchy criterion was also sufficient for convergence of the series.

3. See p. 105 of the text by Katz.

4. The Mathematical Works of Isaac Newton, edited by D. T. Whiteside, Johnson Reprint Corporation. New York. 1964.

5. For details, see the article by J. Grabiner. 6. "REf exions sur les suites et sur les ravines imaginaires;' Opuscules Methematiques, vol. 5 (1768). pp. 171-215.

314

Chapter 7

Series of Real Numbers

MISCELLANEOUS EXERCISES 1.

Given two series Ek oak and Boko.o bk, set c, = I , _ o ak b _ k, n = 0, 1.2.. . ..The series !.O c is called the

Cauchy product of E ak and I bk. a. If I ak andd I bk converge absolutely, prove that E c converges absolutely and that

-oc"-\koak)(kobk).

b. Let ak = bk = (- I)k/, k = 0, 1, 2. ... Prove that the Cauchy product of F. ak and B bk diverges. c. Prove that the result of part (a) is still true if only one of the two series converge absolutely: the other series must still converge. 2.

Let X be a nonempty set. If f is a real-valued function on X, define II f 11, = sup S Y,

.eF

I jr (x) I

: F is a finite subset of X} .

In the above, the supremum is taken over all finite subsets F of X. Denote by l'(X) the set of all real-valued functions f on X for which II f 11, < 00.

a. Suppose X is infinite. If jr E l'(X), prove that {x E X:f(x) # 0} is at most countable.

b. If {x E X:f(x) # 0} _ (x,: n E A}, where A C N, prove that 11f11, _ Z.EA I f(x.)I. c. Prove that (1'(X), II II,)is a normed linear space. 3.

For f E °t[a, b], set (f(x))2dx]1/2

IIf112= I I a. Prove that for f, g E R[a, b], f. I jr (x)g(x) I dx s 11 f 11211 g I1r

b. Forf, g E JL[a, b], prove that 11f + g 112 s 11f112 + 118112

c. Prove that 11 16 defines a norm on %[a, b], the space of continuous real-valued functions on [a, b).

d. is11 I6 anorm on Jt(a,b]? 4.

b] denote the vector space of continuous real-valued functions on (a, b). For As in the previous exercise, let f E 19[a, b], set IIf1I. = max{ If(x) I : x r= [a, b]}. II. is a norm on %(a, b]. b. Prove that a sequence {f.) in 4[a, b] converges to f e 14[a, b] if and only if given e > 0. there exists n. E N such that I f (x) - f (x) I < e for all x E [a, b] and all n E N, n a n,.

a. Prove that 11

DEFINITION (a) Let (X,11 11) be a normed linear space, and let E C X. A function f : E -+ R is continuous at p E E If given

e > 0, there exists a S > 0 such that If(x) - f(p) I < e for all x E E with IIx - pIi < S. (b) Let X be a vector space over It A function f : X -+ R is linear if f(ar + by) = af(x) + bf(y) for all x, y E X and a, b E R. 5. Let (X, II II)be a normed linear space and let f : X - R be a linear function. Prove that the following are equivalent.

a. f is continuous at some p E X. b. f is continuous at 0.

Supplemental Reading

315

c. There exists a positive constant M such that If(x)1 <- MIIxiifor all x E X. d. There exists a positive constant M such that I f(x) - f(y)I <_ Mix - ylifor all x, y E X. e. f is uniformly continuous on X. X

6. For fixed b E 12, define 1' : 12 -+ R by r(a) = (a, b) = I a(k)b(k). Prove that r is a continuous linear function on l k=1

7. As in Exercise 3, let b] denote the vector space of continuous real-valued functions on [a. b] with norm l N2. For fixed g E 9t[a, b], prove that r defined by r(f) = fa f(x)g(x) dx is a continuous linear function on `g[a, b].

SUPPLEMENTAL READING Behforooz, G. H.. "Thinning out the harmonic series;' Math. Mag. 68 (1995), 289-293. Boas, R. P., "Estimating remainders:' Math. Mag. 51 (1978), 83-89. Boas, R. R, "Partial sums of infinite series, and how they grow," Amer. Math. Monthly 84 (1977), 237-258. Cowen, C. C., Davidson. K. R., and Kaufman, R. P., "Rearranging the alternating harmonic series," Amer. Math. Monthly 87 (1980), 817-819. Efthimiou, C. J., "Finding exact values for infinite series:' Math. Mag. 72 (1999), 45-51. Goar, M., "Olivier and Abel on series convergence: An episode from early 19th century analysis:' Math. Mag. 72 (1999), 347-355. Grabinger, Judith V., The Origins of Cauchy's Rigorous Calculus, The MIT Press, Cambridge, Massachusetts, 1981.

Jungck, G., "An alternative to the integral test;' Math. Mag. 56 (1983), 232-235. Katz, V. J "Ideas of Calculus in Islam and India," Math. Mag. 68 (1995),163-174.

Kortam, R. A., "Simple proofs for 7 2 k_I k 00

sin x

=x

k.1

=I 6

and

x2

l - a f ; 'Marh. Mag. 69 (1996),

kir'

122-125.

Maher, P., "Jensen's inequality by differentiation:' Math. Gaz. 73 (1989). 139-140. Tolsted, E., "An elementary derivation of the Cauchy, Holder, and Minkowski inequalities from Young's inequality," Math. Mag. 37 (1964), 2-12. Young, R. M., "Euler's constant:' Math. Gazette 75 (1991), 187-190.

pl Sequences and Series

J of Functions

8.1 Pointwise Convergence and Interchange of Limits

8.2 Uniform Convergence 8.3 Uniform Convergence and Continuity 8.4 Uniform Convergence and Integration 8.5 Uniform Convergence and Differentiation 8.6 The Weierstrass Approximation Theorem 8.7 Power Series Expansions

8.8 The Gamma Function

In this chapter we will study convergence properties of sequences and series of realvalued functions defined on a set E. In most instances E will be a subset of R. Since we are dealing with sequences and series of functions, there naturally arise questions involving preservation of continuity, differentiability, and integrability. Specifically. is the limit function of a convergent sequence of continuous, differentiable, or integrable functions again continuous, differentiable, or integrable? We will discuss these questions in detail in Section 8.1 and show by examples that the answer to all of these questions is generally no. Convergence by itself is not sufficient for preservation of either continuity, differentiability, or integrability. Additional hypotheses will be required. In the 1850s Weierstrass made a careful distinction between convergence of a sequence or series of numbers and that of a sequence or series of functions. His concept

of uniform convergence is the additional hypothesis required for the preservation of continuity and integrability. It was also Weierstrass who constructed a continuous but nowhere differentiable function, and who proved that every continuous real-valued function on a closed and bounded interval can be uniformly approximated by a polynomial. As prominently as Cauchy is associated with the study of sequences and series of numbers, Weierstrass is likewise associated with the study of sequences and series of functions. For his many contributions to the subject area, Weierstrass is often referred to as the father of modern analysis. The study of sequences and series of functions has its origins in the study of power series representation of functions. The power series of In (I + x) was known to Nicolaus Mercator (1620-1687) by 1668, and the power series for many of the transcendental 317

318

Chapter 8

Sequences and Series of Functions

functions such as Arctan x, Arcsin x, among others, were discovered around 1670 by James Gregory (1625-1683). All of these series were obtained without any reference to calculus. Newton's first discoveries, dating back to the early months of 1665. resulted from his ability to express functions in terms of power series. His treatise on calculus, published posthumously in 1737, was appropriately entitled A treatise of the method of fiuxions and infinite series. Among his many accomplishments. Newton derived the power series expansion of (1 + x)" using algebraic techniques. This series and the geometric series were crucial in many of his computations. Newton also displayed the power of his calculus by deriving the power series expansion of ln(I + x) using termby-term integration of the expansion of 1/(1 + x). The mathematicians Colin Maclaurin (1698-1746) and Brooks Taylor (1685-1731) are prominent for being the first mathematicians to use the new calculus in determining the coefficients in the power series expansion of a function.

8.1' Pointwise Convergence and Interchange of Limits In this section we consider a number of questions involving sequences and series of functions and interchange of limits. Some of these questions were incorrectly believed to be

true by many mathematicians prior to the nineteenth century. Even Cauchy in his text Cours d'Anahse "proved" a theorem to the effect that the limit of a convergent sequence of continuous functions was again continuous. As we will shortly see, this result is false! To begin our study of sequences and series of functions we first define what we mean by pointwise convergence of a sequence of functions. If E is a nonempty set and if for each n E N, f is a real-valued function on E, then we say that { is a sequence of real-valued functions on E. For each x E E, such a sequence gives rise to the sequence of real numbers, which may or may not converge. If the sequence { converges for all x E E, then the sequence is said to converge pointwise on E, and by the uniqueness of the limit,

f(X) = lim L(x) defines a function f from E into R. We summarize this in the following definition.

8.1.1

Let { f } , be a sequence of real-valued functions defined on a set E. The sequence if.) converges pointwise on E if { converges for every x E E. If this is the case, then f defined by DEFINITION

f(x) = I i m

x E E.

defines a function on F. The fiaiction f is called the limit of the sequence If,).

In terms of e and n, the sequence {f,) converges pointwise (of if for each x E E, given e > 0, there exists a positive integer n,, = e) such that

If,,(x) - f(x)I < e

8.1 Pointwise Convergence and Interchange of Limits

319

for all n ? na. The expression no = na(x, e) indicates that the positive integer n, may depend both on e and x E E. If as above, {fa}°a° is a sequence of real-valued functions on a nonempty set E, then with the sequence {fa} we can associate the sequence {S0} of nth partial sums, where for each n E N, S. is the real-valued function on E defined by 1

Sa(x) = (A + ... + fa)(x) = I fk(x) k=1 The sequence is called a series of functions on E denoted by I* j fk or simply E fk. The series 1 fk converges pointwise on E if for each x E E the sequence {Sa(x)} of partial sums converges; that is, the series Z. I fk(x) converges for each x E E. If the sequence {Sa} converges pointwise to the function S on E, then S is called the sum of the series E fk and we write S = E. , fk, or if we wish to emphasize the variable x, 00

S(x) = 7, A(x), x E E. k=1

Suppose fa : [a, b] -+ R for all n e Ni, and f(x)

each x E [a, b].

Among the questions we want to consider are the following:

(a) if each fa is continuous at p E [a, b], is the function f continuous at p? Recall that the function f is continuous at p if and only if

limf(t) = f(p). Since f (x) = lim fa(x), what we are really asking is does

lim{ lim ff(t) = lim lim fa(t) r-4p

M-600(1-+p

(b) If for each n E N the function Ja is differentiable at p E [a, b], is f differenSable at p? If so, does

f(p) =

lim f,(p)? n

(c) If for each n E N the function fa is Riemann integrable on [a, b], is f Riemann integrable? If so, does rb a

jb

f= a--.f

f.?

a

We now provide a number of examples to show that the answer to all of these questions is generally no.

320

Chapter 8

8.12

Sequences and Series of Functions

EXAMPLES

(a) Let E = [0, 1 ], and for each x E E, n E N. let f"(x) = x". Clearly each f is continuous on E. The graphs of f, f,-, f4, f6, and f,i, are given in Figure 8.1. Since fn(1) = I 1. If 0 :5 x < 1, then by Theorem 2.2.6(e), lim fn(x) = 0. Therefor all n, Jim n-r x fore.

u m f,(x) = f(x) = S 0,

0 ` i < 1,

The function f, however, is not continuous on [0, 1 ].

Figure 8.1

Graphs of f,, f2,f4,ff,f10

a

(b) Consider the sequence {fk}j 0 defined by ,

f(x)=(1+

xE68.

For each k = 0, 1 , 2, .. . , fk is continuous on R. Consider the series each x E R is given by x l k x (1 fk(x)= + x2 k=0 L=O

0 fi

which for

Pointwise Convergence and interchange of limits

8.1

321

We now show that this series converges for all x E I and also find its sum f. If x = 0. then fk(O) = 0 for all k, and thus f(0) = 0. If x * 0, then 1/(1 + x2) < 1, and hence by Example 2.7.2(a), X2

71

(-

= 1 + xz = f(x).

k=O

Therefore, 0,

f(x)

x = 0,

1 +xz, x#0,

which again, is not continuous on R.

(c) Let {xk} be an enumeration of the rational numbers in [0, 1]. For each n E N, define fn as follows:

fn(x) =

if x = xk, 1 <_ k 5 n, otherwise.

J 0, 1,

Since each fn is continuous except at xi, ... , xn, fn is Riemann integrable on [0, 1 ] with fn fn(x) dx = 1. On the other hand, if x is rational,

f(x) = s10,1, if x is irrational,

lim

which by Example 6.1.6(a) is not Riemann integrable on (0, 11.

(d) For x E [0, 1], n E N, let fn(x) = nx (I - x2)n. Since each fn is continuous, fn is Riemann integrable on [0, 1 J. If 0 < x < 1, then 0 < 1 - x'' < 1, and thus by Theorem 2.2.6,

lim nx(1 - X'-)n = 0.

nyoc

0, we have

Finally, since

f(x) = 1 m L(x) = 0 for all x E [0, 1 J. Thus f is also Riemann integrable on (0, 1 ]. On the other hand,

dx = n 10

1

x (1 - xZ)r dx =

0

1

n

2n+1

Therefore.

lim

n-ioc

f'

frt(x) dx = 2 # 0 =

0

I

f (x) dx.

J

11

(e) As our final example, consider fn(x) = (sin nx)/n, x E R. Since I sin nx -5 1 for all x E=- R and n E N,

f (x) = lim fn(x) = 0 n-+oo

for all x E R.

322

Chapcer8

Sequences and Series of Functions

Therefore, f'(x) = 0 for all x. On the other hand,

f;'(x) = cos nx.

In particular f;(0) = 1 so that lim f',(0) = 1 # f'(0). This example shows that in general dx

lyltto fn(X)) # llm f n(x)

Additional examples are given in the exercises.

EXERCISES 8.1 1. Find the pointwise limits of each of the following sequences of functions on the given set. sin nx

*a. 11 _, }, xE(0,oo)

b.

*c. {(cos x)Z"}, x E R

d. {nxe-"},

11+nx

X E f 0. oo)

.

x E R

2. For n E N, define f, : N -+ R by f, (m) = n/(m + n). Prove that HIM

lim f,(m))

# liml!m,fn(m)).

3. Consider the sequence {f.) with it nZx,

f(x)

2n-n2x, 0,

2, defined on [0, 1 ] by

05x5 1,n 1I

n

<xs 2n

?<x<_1. it

a. Sketch the graph of f, for it = 2, 3, and 4. b. Prove that lim f,(x) = 0 for each x E [0, 1 ].

c. Show that fo fn(x) dx = I for all n = 2, 3.... . 4. Let g (x) = e-n`/n, x E 10, oo), it E N. Find lim g,(x) and lim g;,(x). ny0o n-+oo 5. Let f,(x) _ (x/n)e- `1n, X E [0.00).

a. Show that tim fn(x) = 0 for all x E [0, oo). b. Given e > 0, does there exist an integer it, E N such that If,(x) I < e for all x E (0, oo) and all it z it,,. (Hint: Determine the maximum of fn on [0, oo).) c. Answer the same question as in pan (b) for x E [0, a], a > 0. 6. *If a,, m ? 0, it, m E N, prove that - cc cc cc L an.,,, 71 71 an.,, = n=Im=1

n=1

with the convention that if one of the sums is finite, so is the other, and equality holds. Conversely, if one is infinite, so is the other.

8.2 Uniform Convergence

8.2

323

Uniform Convergence All of the examples in the previous section show that pointwise convergence by itself is not sufficient to allow the interchange of limit operations; additional hypotheses are required. It was Weierstrass who, in the 1850s, realized what additional assumptions were needed to ensure that the limit function of a convergent sequence of continuous functions was again continuous. Recall from Definition 8.1.1, a sequence { f,} of real-valued functions defined on a set E converges pointwise to a function f on E if for each x E E, given c- > 0, there exists a positive integer n = n,,(x, e) such that

Nx) - L WI < E for all n ? n0. The key here is that the choice of the integer n,, may depend not only on e. but also on x E E. If this dependence on x can be removed, then we have the following.

8.2.1

DEFINITION A sequence of real-valued functions {f} defined on a set E converges uniformly to f on E. if for every e > 0, there exists a positive integer n such that

If,(x) - f(x)1 < E for all x E E and all n ? n,,. Similarly, a series 7-4'- i fr of real-valued functions converges uniformly on a set E if and only if the sequence of partial sums converges uniformly on E.

The inequality in the definition can also be expressed as

f(x)- E
y = f(x) - e and y = f(x) + E. This is illustrated in Figure 8.2 with E = [a, b]. 8.2.2

EXAMPLES

(a) For x E [0, 1 ], n E t\i, let f"(x) = x". By Example 8.1.2(a), the sequence {f.1 converges pointwise to the function

0sx< I,

f(x) _ {°, x= 1.

We now show that the convergence is not uniform. If the convergence were uniform, then given E > 0, there would exist a positive integer n, such that I "(j) - f(x) 1 < e for all n >_ n" and all x E [0, 1 ]. In particular,

x<e

for all x E [0, 1).

324

Chapters

Sequences and Series of Functions

a

b

Figure 8.2

This, however, is a contradiction if e < 1. Even though the convergence is not uniform on (0, 1 ], the sequence does converge uniformly to 0 on [0, a] for every a, 0 < a < 1. This follows immediately from the fact that for x E [0, a], LAW I = I x' I -S a". (b) Consider the series 00 k71

{ixe-k'' - (k -

0:9 x < 00.

Since the series is a telescoping series, the nth partial sum S"(x) is given by

S"(x) = nxe"";. It is easily shown (Exercise Id, Section 8.1) that

S(x) = lim S"(x) = 0 for all x E (0, oo). M- 00 We now show that the convergence is not uniform. Suppose that the sequence {S"} converges uniformly to 0 on [0, oo). Then if we take e = 1, there exists a positive integer n0 such that

IS.(x) - S(x) l = S"(x) = nxe -"'_ < I

for all n a n, and x E [0, oo). However, for each n E N, by the first derivative test S. has a maximum at x, = with M. = max S"(x)

F2e_

This term, however, is greater than 1 for n =' 6. Thus the convergence is not uniform. The graphs of S4, Ss, and S16 are given in Figure 8.3. Since the maximum of each S. moves along the x-axis as n -* oo, such functions are often referred to as "slidinghump" functions. For this example, M. -+ oo as n -+ oo.

8.2 Uniform Convergence

Figure 8.3

325

Graphs of S Se, S16

The Cauchy Criterion Our first criterion for uniform convergence is the Cauchy criterion. The statement of this result is very similar to the definition of Cauchy sequence.

8.2.3

THEOREM (Cauchy Criterion) A sequence of real-valued functions defined on a set E converges uniformly on E if and only if for every e > 0, there exists an integer no E NI such that (1)

Ifn(x) - fm(x)I < E

for all x E E and all n, m ? n0.

Proof. If (f.} converges uniformly to f on E, then the proof that inequality (1) holds is similar to the proof that every convergent sequence is Cauchy. Conversely, suppose that the sequence if.) satisfies inequality (1). Then for each x E E, the sequence { f (x)} is a Cauchy sequence in R, and hence converges (Theorem 2.6.4). Therefore,

f(x) = lim .fi(x) exists for every x E E.

We now show that the sequence {f,,} converges uniformly to f on E. Let e > 0 be given. By hypothesis, there exists n, E NI such that inequality (1) holds for all x E E and all n, m a no. Fix an m >_ n0. Then f

LAX) - fm(x)I = ha Wx) - fm(x)I

E

for all x E E. Since the above holds for all m ? n the sequence {f.} converges uni-

formly to f on E. Q The analogous result for series is as follows.

326

Chap{er 8

Sequences and Series of Functions

8.2A COROLLARY The series lx I ft of real-valued functions on E converges uniformly on E if and only if given e > 0. there exists a positive integer n,,, such that kn 171-fx)

nsl

I

<E

for all x E E and all integers m > n ? no. Proof. The proof of the corollary follows by applying the previous theorem to the partial sums S"(x) of the series fk(x). Q

8.2.5 THEOREM Suppose the sequence (f.} of real-valued functions on the set E converges pointwise to f on E. For each n E N, set M.

p

sup

I fn(x) - ! (x) I I.

Then I f.1 converges uniformly to f on E if and only if lim M. = 0. n-oo

Proof. Exercise 1. Q 8.2.6

EXAMPLES

(a) To illustrate the previous theorem we consider the sequence S"(x) = nxe-"XI,

n = 1, 2, ..

.

of Example 8.2.2(b). For this sequence, lim S"(x) = 0 for all x, 0 s x < oo. However,

M. = sup S"(x)

n

xE[0,oo)

which diverges to oo. Thus the convergence is not uniform on (0, oo). However, the sequence {S"} does converge uniformly to the zero function on [a, oo) for every a > 0 (Exercise 6).

(b) Consider the sequence { f} of Example 8.2-2(a) given by f"(x) = x", x E [0,1 ]. This sequence converges pointwise to the function f (x) = 0, 0 < x < 1, and f (l) = 1. Since

Jx", 0x :S x < i, I.f"(x) -f(x)I = l0, x= = 1, we have

Mn = SU

:E[0, Ij

Ifnx) -f(x)I = 1.

Thus since {M"} does not converge to zero, the sequence { f"} does not converge uniformly to f on [0, 1 ]. On the other hand, if 0 < a < I is fixed, then Mn xE[ Dal

I fn(x)I = a".

8.2 Uniform Convergence

327

Since lim an = 0, by Theorem 8.2.5 the sequence { fn} converges to zero uniformly on n-400

[0, a] for every fixed a, 0 < a < 1.

The Welerstrass M-Test The following theorem of Weierstrass provides a very useful test for uniform convergence of a series of functions.

8.2.7

THEOREM (Weierstrass M-Test) Suppose { fk} is a sequence of rral-valued functions defined on a set E, and {Mk} is a sequence of real numbers satisfying k(x) L 00

Mk for all x E E and k E F.

00

If Y, Mk < oo, then 7, k(x) converges uniformly and absolutely on E. k-1

k-1

Pro of. Let S. (x) = Zk.I fk(x). Then for n > m, n

ISn(x) - sm(x)l = Ik

n

n

IJi(X) < ± Mk

fk(x)

m+l

k-m+I

k=m+1

Uniform convergence now follows by the Cauchy criterion. That 7, I fk(x) I also converges is clear.

8.2.8

EXAMPLES

.(a) If Zak converges absolutely, then since Iak cos kxl s IakI for all x E R, by the Weierstrass M-test, the series I ak cos kx converges uniformly on R. Similarly for the series I ak sin kx. In particular, the series

7 k-1

cos kx

00

kr

k=1

sin kx

k°

p>

1

converge uniformly on R.

(b) Consider the series 7,I (x12)k. This is a geometric series that converges for all

x E R satisfying I x I< 2. If 0< a < 2 and I x s a, then (2)k. \2)k I

Since a/2 < 1 the series 7, (a/2)k converges. Thus by the Weierstrass M-test, the se-

ries I', (x/2)k converges uniformly on [-a, a] for any a, 0 < a < 2. The series, however, does not converge uniformly on (-2, 2) (Exercise 11). Although the Weierstrass M-test automatically implies absolute convergence, the following example shows that uniform convergence as a general rule does not imply absolute convergence.

Chapter 8

328

8.2.9

Sequences and Series of Functions

EXAMPLE Consider the series 00

T, (-1)k+1

k-1

k

0 - x c 1.

For each k E N, set ak(x) = xk/k. For x E [0, 11, we have

a,(x) z a2(x) at ... ? 0

and

li-m ak(x) = 0.

Thus by Theorem 7.2.3, the series I (-I)k "ak(x) converges for all x E [0, 1]. Let

(-1Y+'ak(x).

S(x) _ k=1

If S (x) is the nth partial sum of the series, then by Theorem 7.2.4

IS(x) -

1(x) s n + 1

for all x E (0. 1 ].

converges uniformly to S on [0, 1 ]. However, the given series does not conThus verge absolutely when x = 1. The converse is also false; absolute convergence need not imply uniform conver-

gence! As an example, consider the series Y,1 x2(1 + x2)-k of Example 8.1.2(b). Since all the terms are nonnegative, the series converges absolutely to 1

f (x) =

0, x = 0, l + x2, x # 0,

on R. However, as a consequence of Corollary 8.3.2 of the next section, since f is not continuous at 0, the convergence cannot be uniform on any interval containing 0. The series I (- 1)t+ txk/k, x E [0, 1]. also provides an example of a series that converges uniformly on [0, 1 ] but for which the Weierstrass M-Test fails.

EXERCISES 8.2 1. Prove Theorem 8.2.5. a. If (f.) and converge uniformly on a set E, prove that {f. + converges uniformly on E. +b. If (f.1 and converge uniformly on a set E, and there exist constants M and N such that I f.(x)I <- M and 1g.(x)l s N for all n E N and all x r= E, prove that { converges uniformly on E. c. Find examples of sequences f f.) and that converge uniformly on a set E, but for which { f does not converge uniformly on E. 3. Show that if (f.1 converges uniformly on (a, b) and (f. (a)} and { ff (b)} converge, then (f.) converges uniformly on (a, b]. 4. Let f .(x) = nx(1 - x2r, 0 5 x 5 1. Show that does not converge uniformly to 0 on (0, 1 2.

5. Let

x"'

0 s x 5 I.

a. Show that (f.} converges uniformly to 0 on [0, a] for any a, 0 < a < 1. b. Does I f.1 converge uniformly on [0, fl?

8.2 Uniform Convergence

329

converges uniformly to 0 on [a. no) for every a > 0.

6. Show that the sequence {nxe 7. For each n E N. set

x x + - sin nx, x r= R. n

Show that the sequence { converges uniformly to f(x) = x for all x E (-a, a], a > 0. Does (f.1 converge uniformly to f on R? 8. Show that each of the following series converge uniformly on the indicated interval.

; k-

+ xk-1

00

b, i e-k`xt, 0 5 x< oo

0 5 x < oc

,

Ca.

k-1 00

"c. Y, k-1

1 5 x < on

d.

00

(-l)k*1

k-,

k+x

,

0 5 x< on

9. Test each.of the following series for uniform convergence on the indicated interval. sin 2kx x E G8 (2k + 1 ?/2 '

00

'a.

1

*c'

1

(kx+2(k+1)x+2

k

2k+1

k=O

e.

(xi<_1

xk

k(In k)2 '

x5I

05x51 00

(-1)k+lx2k+l

00

d.

00

b F,

//\

Xsinikol, p> 1.JxIs2

10. Show that each of the following series converge uniformly on (a, oo) for any a > 0, but do not converge uniformly on (0, oo). I

M

'a. 2

00

I + k2x

I

b. I ii-+;

11. Show that the series E, , (x/2)k does not converge uniformly on (-2, 2). 12. If Jk o ak converges absolutely, prove that I-k o akxk converges uniformly on [ -1. I ]. 13. If I .oak converges, prove that Iku akxk converges uniformly on [0, 1]. 14. Let {ck) be a sequence of real numbers satisfying 7, Ickl < on, and let {xk} be a countable subset of (a, b]. Prove that the' series lk , ck !(x - xk) converges uniformly on [a, b]. Here ! is the unit jump function defined in 4.4.9. 15. Dirichlet'17est for Uniform Convergence: Suppose {)k) and {gk} are sequences of functions on a set E satisfying Ik., gk(x) are uniformly bounded on E; i.e., there exists M > 0 such that (a) the partial sums

<Mforalln E NlandxEE, (b) k(x) >- fk I(x) > 0 for all k E hi and x E E, and

(c) jim fk(x) = 0 uniformly on E. Prove that I fk(x)gk(x) converges uniformly on E. 16. Prove that

. sin kx . t os kz k=1

kP

(p > 0)

k-1

converge uniformly on any closed interval that does not contain an integer multiple of 21r.

330

Sequences and Series of Functions

Chapter 8

17. Define a sequence of functions If.) on [0, I ] by

if

f(x)= 0,

2"

ii < x `

2"

elsewhere.

Prove that 1

1 fn(x) converges uniformly on [0, 1 ], but that the Weierstrass M-test fails.

18. 'Let F0 be a bounded Riemann integrable function on [0. I ]. For n E 101, define F .(x) on [0, 1 ] by F (x) = fo F.- 1(t) dt. Prove that 7,; o Fk(x) converges uniformly on (0. 1 ].

8.31 Uniform Convergence and Continuity In this section we will prove that the limit of a uniformly convergent sequence of con-

tinuous functions is again continuous. Prior to proving this result, we first prove a stronger result that will have additional applications later.

8.3.1

is a sequence of real-valued functions that converges uniTHEOREM Suppose formly to a function f on a subset E of R. Let p be a limit point of E, and suppose that

for each n E N, lim fn(x) = A,,. Then the sequence {An} converges and

lim f(x) = lim A,,.

x-.p

n-+co

Remark, The last statement can be rewritten as/ lim I lim

X_p \n-iW

I lim /I = lim n-.00 \X- p

It should be noted that p is not required to be a point of E; only a limit point of E.

Let e > 0 be given. Since the sequence I f.1 converges uniformly to f on E. there exists a positive integer n such that

Proof.

Ifn(x) - fm(x)I < E

(2)

for all n, m _a n and all x E E. Since inequality (2) holds for all x E E. letting x -+p gives

IAn - Amt s e

for all n, m ? no.

Thus {An} is a Cauchy sequence in R, which as a consequence of Theorem 2.6.4 con-

verges. Let A= nti00 limA,

8.3 Uniform Convergence and Continuity

331

It remains to be shown that lim f(x) = A. Again, let e > 0 be given. First, by the uniform convergence of the sequence { fn(x)} and the convergence of the sequence there exists a positive integer m such that

1f(X) - fm(x)I <

3

for all x E E, and that

IA - Aml < 3. Since lim fm(x) = Am, there exists a S > 0 such that

lfm(x) - Anl < 3 for all x E E, 0
If(x) - At < If(x) -fm(x)I + Ifm(x) - Aml +

IAm -

Al

< e + I fm(x) - A, I

Thus if x E E with 0 < Ix - pi < 8, If(x) - Al < E; i.e., lim f (x) = A.

8.3.2

[]

COROLLARY Let E be a subset of R.

(a) If {

is a sequence of continuous real-valued functions on E. and if {f.)

converges uniformly to f on E, then f is continuous on E. (b) If { fn} is a sequence of continuous real-valued functions on E, and if I converges uniformly on E, then

,f

S(x) _ 71 fn(x) M=1

is continuous on E.

Proof. (a) If p E E is an isolated point, then f is automatically continuous at p. If p E E is a limit point of E, then since f is continuous for each n E N, Jim

MX) = MP).

Thus by the previous theorem,

m f(x) = urn fn(P) = f(p). n-.oo

s

Therefore, f is continuous at p.

332

Chapter 8

Sequences and Series of Functions

For the proof of (b), let fk(X)

SS(x) = krl

converges uniformly to S on E,

Then for each n E N, S. is continuous on E. Since by part (a) S is also continuous on E.

8.3.3

EXAMPLE The sequence {xn},.,, x E [0, 1 ], of Example 8.1.2(a) does not converge uniformly on [0, 1 ] since the limit function

}.(x) =

J0,

x < 1,

o

1, x= 1,

is not continuous on [0, 11. Likewise, the series 00 k

(

1

__

,1 + x2 Jk

0, x = 0. 1 + x2, X # 0,

of Example 8.1.2(b) cannot converge uniformly on any interval containing 0, since the

sum of the series is not continuous at 0. 0

Dini's Theorem' A natural question to ask is whether the converse of Corollary 8.3.2 is true. Namely, if f and f. are continuous for all it and f -+f pointwise, is the convergence necessarily uniform? The following example shows that this need not be the case. However, in Theorem 8.3.5 we will prove that with the additional assumption that the sequence { f (x)} is monotone for all x, the convergence is indeed uniform.

8.3.4

EXAMPLE As in Example 8.2.2(b), for each n e N, let S (x) = nxe-""s, x E [0. 1 ].

Then S is continuous on (0, 1 ] for each n, and lim SS(x) = S(x) = 0, which is also "-OOD

continuous. However, since

max Osssl

S (x) = n ,

by Theorem 8,2.5 the convergence cannot be uniform.

8.3.5. THEOREM (Dins) Suppose K is a compact subset of R and (f.} is a sequence of continuous real-valued functions on K satisfying the following:

(a) {f,,) converges pointwise on K to a continuous function f, and

(b) f (x) ? f,,. 1(x) for all x E K and n E N. Then

converges uniformly to f on K.

1. This topic is not required in subsequent sections and thus can be omitted on first reading.

8.3 Uniform Convergence and Continuity

333

Proof. For each n E N let g = f - f. Then g is continuous on K, g (x) gn+i(x) ? 0 for all x E K and all n E N, and lim g (x) = 0 for each x E K. Let e > 0 be given. For each n E Nl, let p be a limit point of K,,. By Theorem 2.4.7 there exK ists a sequence {xk} in K,, that converges to p. Since g is continuous and g (xk) >- a for

allkEN, gn(P) = t m ga(xk) y E. .Therefore, p E K,,. Thus K,, is closed, and as a consequence of Theorem 3.2.5, also compact. Furthermore, since g (x) >_ g..#) for all x E K, K,,. i C K,, foralln. Finally, since g,,(x) -+ 0 for each x E K, 00

n-i However, by Theorem 3.2.7 this can only be the case if K,,, = 4 for some n E N. Thus

for all n ? no,

for allxEK. Therefore the sequence {g,} converges uniformly to 0 on K. Q

8.3.6

EXAMPLE We now provide an example to show that compactness is required. For each n E N, set I

nx+ V

0<x
Then { f,(x)} monotonically decreases to f (x) = 0 foreach x E (0, 1). However, since linof,,(x) = 1, by Theorem 8.3.1 the convergence cannot be uniform.

The Space `C[q b12 We conclude this section with a brief discussion of the space `t;[a, b) of all continuous real-valued functions on the closed and bounded interval [a, b]. If f, g E 1.9[a, b] and c E R, then by Theorem 4.2.3 the functions f + g and cf are also continuous on [a, b]. Thus c4[a, b] is a vector, space over R where for the zero element we take the constant function 0; that is, the function given by f(x) = 0 for all x E [a, b]. We define a norm on %[a, b] as follows. 2. This topic can be omitted on first reading. The concept of a complete normed linear space (Definition 8.3.10) is only required in Section 10.8.

334

Chapter 8

8.3.7

Sequences and Series of Functions

DEFINITION

For each f E'g[a, b], set

IIf(I. = max{l f(.r)I : x E [a, b]}. The quantity II f M is called the uniform norm off on [a, b]. b] is left to the exercises (Exercise 12). We now That II 11. is indeed a norm on take a closer look at uniform convergence of a sequence of continuous real-valued functions and introduce the concept of convergence in norm. We first note that a sequence of continuous real-valued functions on [a, b] is nothing but a sequence in the set %[a, b]. Suppose that the sequence { in `P[a, b] converges uniformly to f on (a. b].

Then by Corollary 8.3.2, the function f E'F[a, b]. By the definition of uniform convergence, given c > 0, there exists a positive integer n,, such that I fn(x) - Ax) I < E

for all x E [a, b] and all n ? n,,. This is illustrated in Figure 8.2. Since f - f is continuous, if n z no, by Corollary 4.2.9 IIf,, - fllu = max{ I f0(x) - f(x)l : x E [a, b]} < e. Therefore, 11f. - f Il u < e for all n ? no. Conversely, suppose f, f E 'P[a, b] satisfy the following: For each e > 0 there ex-

ists a positive integer n,, such that N f - f j. < e for all n ? n0. But then If(x) - f (x)I f on [a, b]. This < E for all x E [a, b] and all n ? n0; i.e., { proves the following theorem.

8.3.8 THEOREM A sequence {

f E %(a, b) if and only if given e > 0, there exists na E N such that IIf - f,11. < e for all n ? no. Using the above as motivation, we define convergence in a normed linear space as follows.

8.3.9

DEFINITION

Let (X, 11

11) be a normed linear space. A sequence

in X converges

in norm if there exists x E X such that for every e > 0, there exists a positive integer ne such that IIx - x.11 < e for all n ? no. If this is the case, we say that converges in norm to x, and denote this by x

i

x as n -+ oa.

in X converges in norm to x E X From the definition it is clear that a sequence if and only if lim 11x - x11 = 0. Also, as in the proof of Theorem 2.1.10, if converges in norm "&n its limit is unique. Using the norm it is also possible to define what we mean by a Cauchy sequence in a normed linear space.

8.3.10

DEFINITION

(a) A sequence {x} in a normed linear space (X, ll h) is a Cauchy sequence if for every e > 0, there exists a positive integer n, such that Ilxn - xmll < E

for all integers n, m ? n,,. (b) A normed linear space (X,11 II) is complete if every Cauchy sequence in X converges in norm to an element of X.

8.3 Uniform Convergence and Continuity

335

As for sequences of real numbers, every sequence {x,,} in X that converges in norm to x E X is a Cauchy sequence. In Theorem 2.6.4 we proved that the normed linear space (R,I I) is complete. The following theorem proves that (%[a, b], II ll.) is also complete.

8.3.11

THEOREM The normed linear space (`P[a, b], II II.) is complete.

Proof. Let {fn} be a Cauchy sequence in %[a, b]; i.e., given e > 0. there exists a positive integer no such that Ilfn - fmll. < E for all n,m >- no. But then Ifn(x) - fm(x)I

Il fn - fmllu < E

for all x E [a, b] and all n, m a n,. Thus by Theorem 8.2.3 and Corollary 8.3.2, the sequence { fn} converges uniformly to a continuous function f on [a, b]. Finally. since the convergence is uniform, given e > 0, there exists an integer no such that

I fn(x) - f(x)l < E for all x E [a, b] and n >- n,. As a consequence, we have

1Ifn -

converges to f in the norm 11

f 1j. < E for all n a no. Therefore, the sequence { fn}

L. Q

Contraction Mappings In Exercise 13 of Section 4.3 we defined the notion of a contractive function on a subset E of R. We now extend this to normed linear spaces.

8.3.12

DEFINITION

Let (X, N II) be a nonmed linear space. A mapping (function) T : X -+X

is called a contraction mapping (function) if there exists a constant c, 0 < c < 1. such that

IIT(x) - T(y)iI < cllx - yll for all x, y E X. Clearly every contraction mapping on X is continuous, in fact uniformly continuous on X. As in Exercise 13 of Section 4.3, we now prove that if T is a contraction mapping on a complete normed linear space (X, II 1). then T has a unique fixed point in X.

8.3.13

THEOREM Let (X, II II) be a complete normed linear space and let T : X -+ X be a contraction mapping. Then there exists a unique point x E X such that T(x) = x.

Proof. Suppose T : X - X satisfies 11T (x) - T(y)p <- cll x - yq for fixed

c,

0 < c < 1. and all x, y e X. We now define a sequence {xn} in X as follows: Let x, E X be arbitrary. For n E N set xn = T(xn_,). That is, x, = T(x,), x2 = T(xi), etc. Since

IIxn+1 - xnll = lIT(xn) - T(xn-1)II

cllxn - xn-111,

the sequence {xn} is a contractive sequence in X (see Definition 2.6.6). An argument similar to the one used for a contractive sequence in Section 2.6 shows that Ilxn+m - xnll C

C. 1

C

Ilx, - xoll

336

Chapter 8

Sequences and Series of Functions

for all n, m E N. The details are left as an exercise (Exercise 10). Since c" -* 0. the sequence {x"} is a Cauchy sequence in X. By completeness, the sequence {x"} converges (in the norm) to some x E X. But by continuity of the mapping T,

x = lim x"+t = lim T(x") = T(x); i.e., T(x) = x. Suppose y E X also satisfies T(v) = y. But then

Ily - Al = IIT(y) - T(x)ll

cliv - xll.

Since 0 < c < 1, the above can be true if and only if Ilv - xii = 0; that is. v = x. Thus x is unique.

EXERCISES 8.3 1. *Show that the series I'U x (1 - x)0 cannot converge uniformly for 0 s x 5 1. 2. For n E N. let f .(x) = x"/(I + x"), x (5 [0, 1 ]. Prove that the sequence f f.) does not converge uniformly on [0, 1]. 3. Give an example of a sequence f f.) of functions on [0, 1]. such that each f" is not continuous at any point of [0, 11, but for which the sequence {f.) converges uniformly to a continuous function f on [0. 11.

4. *Suppose that f is uniformly continuous on R. For each n E N. set f"(x) = f(x + Prove that the sequence ff.) converges uniformly to f on R. S. Let (f.1 be a sequence of continuous real-valued functions that converges uniformly to a function f on a set E C R. Prove that liim f"(x") = f(x) for every sequence {x"} C E such that x" -*x E E. 6. *Let E C R and let D be a dense subset of E. If (f.} is a sequence of continuous real-valued functions on E. and if if.) converges uniformly on D. prove that (f.) converges uniformly on E. (Recall that D is dense in E if every point of E is either a point of D or a limit point of D.) 7. Find a sequence {f"} in IC[0, I ] with II f.11. = I such that no subsequence of (f,) converges (in norm) in %[0. C. 8. Suppose {f.1 is a sequence of continuous functions on [a, h] that converges uniformly on [a, b]. For each x E [a, b], set g(x) = sup{f" (x)}. a. Prove that g is continuous on [a, b].

b. Show by example that the conclusion may be false if the sequence {f"} converges only pointwise on [a. b].

9. For each n E N and x E It, set f,(x) = (1 + )". Use Dini's theorem to prove that the sequence {f"} converges uniformly to e' on [a, b] for any fixed a, b E R. 10. Let (X, II II) be a normed linear space and let T : X -* X be a contraction mapping with constant c, 0 < c < 1. If {x,} is the sequence in X as defined in the proof of Theorem 8.3.13, prove that

11x... - x,, lI S (c"/(1 - c)) II x, - x.11 for all n, m E N. 11. Define T : [0, 1 ] -a %[0, 1 ] by (Tto)(x) = fo cp(t) dt, a. Prove that I (T2V)(x)I s z X211011..

0 S x S 1, v E %[0, 1 ], and set T2 = ToT.

b. Show that T2 is a contraction mapping on `P[0, I ] and thus has a fixed point in `P[0. 1 ]. c. Prove that T has a fixed point in `4[0, 1 ].

12. Prove that

b], f Q is a normed linear space.

13. Prove that (12, Ii 02) is a complete normed linear space.

8.4 Uniform Convergence and Integration

-8.4

337

Uniform Convergence and Integration In this section we will prove that uniform convergence of a sequence If.) of Riemann integrable functions is again sufficient for the limit function f to be Riemann integrable, and for convergence of the definite integrals of fn to the definite integral off The analogous result for the Riemann-Stieltjes integral is left to the exercises (Exercise 2).

8.4.1

THEOREM Suppose fn E 9t[a, b] for all n E N. and suppose that the sequence {fn} converges uniformly to f on [a, b]. Then f E gt[a, b] and Jb

b

f

f(x) dx = li m

fn(x)dx.

n-M

a

a

Proof. For each n E lkU, set

en =sEnlax,[!x) -f(x)I. a. bj V. Since fn --).f uniformly on [a, b]. by Theorem 8.2.5, n-x lim e. = 0. Also, for all X E [Q, b],

fn(x) - En 5 f(x) t fn(x) + En. Hence

rb

(fn-En) :

fb

b

f ff:

Vn+En) a

Therefore, b

0-

f-

rb

f<2En[b-a].

Ja

fa

Since en -> 0, f E &[a, b]. Also by inequality (3), rh

rh

f(x) dx -

Ja

fn(x)dxI G En[b - Q],

Ja

and thus b

hm

b

fn(x) dx =

j f(x) dx. a

fa

8.4.2

COROLLARY

/f fk E k[a, b] for all k E N, and if 00

f(x)

k-l

fk(x),

x E [a, b],

(3)

338

Chapter 8

Sequences and Series of Functions

where the series converges uniformly on [a, b], then f E Jt[a, b] and b Ja

m

b

f A (x) dx. k.1 a

f(x) dx = 7,

Proof. Apply the previous theorem to S"(x) = 1,"t. I fk(x), which by Theorem 6.2.1 is integrable for each n E N. O Although uniform convergence is sufficient for the conclusion of Theorem 8.4.1, it is not necessary. For example, if f"(x) = x", x E [0, 1 ], then {f"} converges pointwise, but not uniformly, to the function

f(x) =

J0, 1,

0 : 5x<1 , x = 1.

The function f E 9t[0, 1 ] and

°n+

1

=0= f f(x) dx.

In Section 10.6, using results from the Lebesgue theory of integration, we will be able to prove a stronger convergence result that does not require uniform convergence of the sequence If.}. However, it does require that the limit function f is Riemann integrable. For completeness we include a statement of that result at this point.

8.4.3

THEOREM (Bounded Convergence Theorem) Suppose f and fn, n E N, are Riemann integrable functions on (a, b] with lim f"(x) = f(x) for all x E [a, b]. Suppose n-.oo

also that there exists a positive constant M such that [fn(x) I < M for all x E [a, b] and

allnE N. Then f(x) dx.

f"(x) dx =

lim

n-iao

ab

Jab

It is easily checked that the sequence If.} on (0, 1], where for each n E N fn(x) = x", satisfies the hypothesis of the prwtvious theorem. Also, since the limit func-

tion f is continuous except at x = 1, f E J.[0, 1].

EXERCISES 8.4 1. *If I I ak l < oo, prove that

,

f (akx*)dx= I 7 0

ak

k+1

8.5 Uniform Convergence and Differentiation

339

2. Let a be a monotone increasing function on [a, b]. Suppose f, E A(a) on [a, b) for all n E R;. and suppose that the sequence converges uniformly to f on :a. b,. Then f E .R(o) and (h

(

f f da = lim f. f da. nx/(l + nx), x E [0, I ] . Show that the sequence { f } converges pointwise. but not uniformly, to an integrable function f on i 0, 1 ]. and that lim f,' f;,(x) dx = fol f(.r) dr.

3. For each n E N. let

4. *If f is Riemann integrable on [0, 11. use the bounded convergence theorem to prove that Iim f r f(x) dr = 0.

5. Let (f,} be a sequence in :7i[a, b] that converges uniformly to f E .R[a, b'. For n E ) set F,(x) = f' f , and let F(x) = f f, r E [a, b]. Prove that converges uniformly to F on [a, bl. R is continuous. Prove that Iim f f(r") dr = f (0). 6. Suppose f : [0, I ] 7. *Let {r"} be an enumeration of the rational numbers in [0, 11, and let f : [0, 11 -- R be defined by Y

I

f(x) = I Y2I (x - ri). where / is the unit jump function of Definition 4.4.9. Prove that f E Jt [0. I 8. Define g on R by g(x) = x - [x]. where [x] denotes the greatest integer function. Prove that the function K(nr)

f(x)

is Riemann integrable on [0, 1 ] and find f f(x) dx. This function was given by Riemann as an example of a function that is not integrable according to Cauchy's definition.) 9. *Let g E R[a, b] and let {f} be a sequence of Riemann integrable functions on [a, b] that converges uniformly to f on [n, b]. Prove that Iim f ,".ff, g = f b fg.

10. *Let g be a nonnegative real-valued function on ;0. oc) for which fo g(x) dr is finite. Suppose if,) is a sequence of real-valued functions on (0, oo) satisfying f;, E dt[0, c] for every c > 0 and ! f (x) <_ g(x) for all x E [0. oo) and n E N. If the sequence (f.1 converges uniformly to f on [0. c] for every c > 0, prove that liym fo .f (x) dx = fo .f(.r) Jr.

11. For

1fII

= f,°1f(x)Idx.

a. Prove that (%[a, b], II J,) is a normed linear space.

*b. Show by example that the normed linear space (E[a. b], III ) is not complete.

i

Uniform Convergence and Differentiation In this section we consider the question of interchange of limits and differentiation. Example 8.1.2(e) shows that even if the sequence (f J converges uniformly to f, this is not sufficient for convergence of the sequence ff.'} of derivatives. Example 8.5.3 will further demonstrate dramatically the failure of the interchange of limits and differentiation. There we will give an example of a series, each of whose terms has derivatives of all orders, that converges uniformly to a continuous function f. but for which f fails to exist at every point of R. Clearly, uniform convergence of the sequence (f.1 is not sufficient. What is required is uniform convergence of the sequence If.}.

340

Chapters

8.5.1

Sequences and Series of Functions

THEOREM Suppose

is a sequence of differentiable functions on [a, b]. If

(a) {f,,) converges uniformly on [a, b], and (b) { f (x,)} converges for some x E [a, b],

then If.} converges uniformly to a function f on [a, b], with

f'(x) = Jim f;,(x).

Remarks (a) Convergence of { f (x,)} at some x, E [a, b] is required. For example, if we let need not converge for any x E [a. b]. f (x) + n, then g"(x) = f"(x), but In Exercise I you will be asked to show that uniform convergence of (f.'} is also required; pointwise convergence is not sufficient.

(b) If, in addition to the hypotheses, we assume that f" is continuous on [a, b], then a much shorter and easier proof can be provided using the fundamental theorem of calculus. Since f" is continuous, by Theorem 6.3.2 f(x) = fn(xo) +

Jf.(t) dt

I.X

for all x e [a, b]. The result can now be proved using Corollary 8.3.2 and Theorem 8.4.1. The details are left to the exercises (Exercise 2).

Proof.

Let e > 0 be given. Since { f (x,)} converges and If.') converges uniformly, time exists no E RI such that fn(x,,) - fm(xo) I < 2 for all n, m z n

(4)

and

I f"(t) - f",(t) I < 2(b

a

a) for all t r= [a, b] and all n, m ? n,.

(5)

Apply the mean value theorem to the functions f - fm with n, m ? n, fixed. Then for x, y E [a, b], there exists t between x and y such that

I(ff(X) - f,,(X)) - (My) - fm&))I = I [f"(t) - f;'(t)](x - A. Thus by inequality (5),

I(f (X) - fm(X)) - (AY) - fm&))I

2(b

a)I X - yl < 2

(6)

Take y = x, in inequality (6). Then by inequalities (4) and (6). for all x E [a, b] and

n, ma n

If,(x) - fm(X) I

I MW - fm(x)) - (ff(x0) - fm(X0))I + I f.(xo) - fm(X0) I

<2E +2E =E.

8.5 Uniform Convergence and Differentiation

341

Hence by Theorem 8.2.3, the sequence (f,) converges uniformly on [a. b). Let

f(x) = .limff(x) It remains to be shown that f is differentiable, and that

f'(x) = hm f;,(x) for all x E [a, b]. Fix p E [a, b], and fort # p, t E [a, b], define

A() = f(t) - f(P)

8n(t) = fn(t) - fe(P)'

t- p

t-p

Then g(t)

each t e [a, b], t * p, and for each n 1 m g.(t)

= fe(P)

Let E = [a, b] \ {p}. Take y = p in inequality (6). Then for all t E E. a gm(t) I <_ 2(b for all n, m z no.

a)

Therefore,

converges uniformly to g on E. Hence by Theorem 8.3.1,

f'(P) _r-P lim g(t) =f.'(p) O 8.5.2

EXAMPLE To illustrate the previous theorem, consider the series k-i

.

2

Since 12-k sin kxl <_ 2-'for all x E R, by the Weierstrass M-test this series converges uniformly to a function S on R. For n E N, let

S, (x) =k,` 7 Then

S.'(X)

sin kx 2

k cos kx k=1

2k

Since E k2 -k converges, by the Weierstrass M-test the sequence {S,,} converges uniformly on R. Thus by Theorem 8.5.1,

S'(x) = lim S; ,(x) _

k cos kx 2k

A Continuous Nowhere Differentiable Function We conclude this section with the following example of Weierstrass of a continuous function that is nowhere differentiable. When this example first appeared in 1874, it astounded the mathematical community.

342

Chapter8

Sequences and Series of Functions

8.5.3 EXAMPLE Consider the function f on R defined by f(x) =

00 Cos akirx k-0

,

2k

(7)

where a is an odd positive integer satisfying

a>31r+2. Since cos ak9rx

1

2k,

2k

the series (7) converges uniformly on R, and hence f is continuous. The graphs of the partial sums (with a = 13) Si(x) = cos(wx), S2(x) = Si(x) + 1 cos(13ax), and S3(x) _ S2(x) + 2, cos(1321rx) are illustrated in Figure 8.4. 2

2 S2

S 1

r

-1

Figure 8.4

We prove that f is nowhere differentiable by showing that for each x E R, there ex-

ists a sequence h -* 0 such that

f(x + hn) - f(x) 1im I 11i00

= 00.

hn

For n E N, set Cos ak?rx

A-0

RR(x) =

0

k=n

2'

cos aklrx

2k

Then for h > 0,

f(x+h)-f(x)-SR(x+h)-SR(x)+RR(x+h)-RR(x) It

It

It

8.5 Uniform Convergence and Differentiation

343

Our first step will be to estimate

S"(x + h) - S"(x) h

from above. By the mean value theorem, Cos ak?r(X + h) - cos ak7rx

k

- = -a it sln[a 1r(x + Q,

h

for some , 0 < < h. Since y 15 aklr, I -ak it sm[aklr(x + b)]

we obtain

S"(x + h) - S"(x) It

I`2kI :5 IT

cos akvr(x + h) - cos akirx h

"-' (a)*

__

k-o 2

(8)

n[l - (2)"] 1- 2

a

27r


a"x = k" + B",

where k" is an integer, and -2 5 B" < . Set

h" -

1-B" am

Since -'-z s B" < '-z, we have 2 >- 1 - B" > '-i. Therefore,

2 a" :

< 2a".

(9)

h"

Fork? n, akir(X + h") = a-"a"ir(x + h") ak_"ir(a"x + =

ak "Tr(k" + 1).

Since ak-" is odd and k" is an integer,

cos[akir(x + h")] = cos[ak-"ir(k" + 1)] Also, akirx = ak-"a"Trx = ak-"ir(k,, + B"). Using the trigonometric identity

cos(A + B) = cos A cos B - sin A sin B,

344

Chapter 8

Sequences and Series of Functions

and the fact that sin(k"ak-" ir) = 0, we have cos aklrx = cos(a "k"1r) cos(ak "S"1r)

= (-I )k- cos(a-"S"1r). Therefore,

cos a kir(x

+ h") - Cos akirx

(- 1)k- cos(ak -"S"1r) + cos(a k- "5.1r)).

As a consequence, 1 R"(x + h") - R"(x) I

-

°° I cos aklr(x + h - cos a`1rx k-1 2k h"

I

h,,

. (- I)t(. h"

k=" 1

=h"

[ I + cos ak - "S"1r, I 2k

I + Cos ak "S"1r

°°

1

1 - Cos 8"1r

h"

2"

2k

k-n

Since -Z s S" < Z, cos 8"1r 2t 0. Therefore, by inequality (9) and the above,

R"(x+h,,)-R"(x)

"

3(21".

1

(10)

Using the reverse triangle inequality, I a + b I ? I a I - I b I, we have

f (x + h") - f (x) I

R& + h") - R"(x) (- I S"& x + h") - S"(x)

h"

h"

h"

E

which by inequalities (8) and (10),

>2 a 3 l2

21r

a

a-2 2

2

(

\2J"[3 Since i > 1, we obtain f(x + h") - f(x) h

f - 0o

as n -+ oo,

n

provided a is an odd positive integer satisfying a 21r 2 3

>0;

i.e., a > 31r + 2. Since 1r < 3.15, we need a z 13. N

345

8.5 Uniform Convergence and Differentiation

Remark. The above proof is based on the proof of a more general result given in the text by E. Hewitt and K. Stromberg. There it is proved (Theorem 17.7) that

)

f( x) = '

°C cos akrrx bk

k=0

has the desired property if a is an odd positive integer, and b is any real number with

b > I satisfying

6> 1+3ir. The above function was carefully examined by G. H. Hardy [Trans. Amer. Math. Soc., 17, 301-325 (1916)] who proved that the above f has the stated properties provided

1 :<.b s a. These are by no means the only examples of such functions. A slightly easier construction of a continuous function that is.nowhere differentiable is given in Exercise 7.

EXERCISES 8.5 1. For n E N, set f(x) =

that the sequence that on

[0, 1

f

that

on [0,

converges

converge to

[a, b]

on [a, b] for

a on

converges

for on

and

f

that

n

of

the

and that f'(x) = li m

for all

xE

a of real numbers satisfying I kjak I < oo. Show that the series Yk o akxk converges unio formly to a function f on jx] 5 1 and that f'(x) = joko-, kakxk- ' for all x, ] x 1 s 1. 4. *Let If.) be a sequence of differentiable real-valued functions on (a, b) that converges pointwise to a function f on (a, b). Suppose the sequence {f;,} converges uniformly on every compact subset of (a, b). Prove that f is differentiable on (a, b) and that f'(x) = lim f;,(x) for all x E (a, b). S. State and prove an analogue of Theorem 8.5.1 for a series of functions 7, fk(x). 6. Show that each of the following series converge on the indicated interval and that the derivative of the sum can be obtained by term-by-term differentiation of the series.

--1i,

x E (O' OD)

b.

M

C. 2xk, ]x]
a 0o

d.

x E (0, oo)

xk

k, x E (-oo, oo)

k=0

7. This exercise provides another construction of a continuous function f on R which is nowhere differentiable. Set g(x) = x {, = 15 x is 1, and extend g to R to be periodic of period 2 by setting g(x + 2) = g(x). Define f on

R by

f(x) _ k=o(31 g(4kx). 4

346

Chapter 8

Sequences and Series of Functions

a. Prove that f is continuous on R. b. Fix xo E R and m E N. Set 8," Z4-'", where the sign is chosen so that no integer lies between 4'"x" and 4'"(xe + 8,"). Show that g(4"(x" + 8,.)) 0 for all n > m. c. Show that

f (x° +

s) - f (X.) > 2 (3"' + 1). m

Since 8," -+ 0 as m - oo, it now follows that f'(.t") does not exist.

8.6

The Weierstrass Approximation Theorem In this section we will prove the following well-known theorem of Weierstrass.

8.6.1

THEOREM (Weierstrass) If f is a continuous real-valued function on [a, b], then given e > 0, there exists a polynominal P such that 11(x) - P(x) I < E

for all x E [a, b]. An equivalent version, and what we will actually prove, is the following: 1f f is a continuous real-valued function on [a, b], then there exists a sequence {P"} of polynomialf such that

f (x) = lim P"(x) uniformly on [a, b]. n- oc Before we prove Theorem 8.6.1, we state and prove a more fundamental result that will also have applications later. Prior to doing so, we need the following definitions.

8.6.2

DEFINITION A real-valued function f on R is periodic with period p if

f (x + p) = f (x) for all x r= R. The canonical examples of periodic functions are the functions sin x and cos x, both of which are periodic of period 2ir. The graph of a periodic function of period p is illustrated in Figure 8.5. The graphs of a periodic function of period p on any two successive intervals of length p are identical. It is clear that if f is periodic of period p, then

f(x+kp) =f(x) forallkEZ. Another useful property of periodic functions follows.

8.6.3

THEOREM 1f f is periodic of period p and Riemann integrable on [0, p), then f is Rie-

mann integrable on [a, a + p] for every a E R. and +P

f° c

Proof.

Exercise 2.

f(x) dx = J r f (x) dx. J0

8.6 The Weierstrass Approximation Theorem

Figure 8.5

347

Graph of a Periodic Function of Period p

Approximate Identities 8.6A

DEFINITION A sequence {Qn} of nonnegative Riemann iruegrable functions on [-a, a] satisfying a QQ(t) dt = 1, and

(a) J

(b) lim J n-roo

dt = 0 for every S > 0,

{dsiro

is called an approximate Identity on [ -a, a]. An approximate identity {Qn} is sometimes also referred to as a Dirac sequence.

Remark. In (b), by the integral over the set {8 s I t 1} we mean the integral over the two intervals [-a,-8] and [S, a]. As we will see in Theorem 8.6.5, and again in Chapter 9, approximate identities play a very important role in analysis. An elementary example of such a sequence Qnn=°° 1 is as follows:

QQ (t)

=

n

1

2'

n

0,

1

Sxs n'

n

It is easily shown that the sequence {Q,,} is an approximate identity on [ -1, 1 ] (Exercise 3). Other examples will be encountered in the proof of Theorem 8.6.1 and in the exercises, and still others when we study Fourier series.

As a general rule, the Q. are usually taken to be even functions: QQ(-x) = Qn(x). T h e f a c t that the integrals over the set {t : f t I

i.e., S} become small as

348

Chapter 8 - Sequences and Series of Functions

n --), oo seems to suggest that in some sense the functions themselves become small as n becomes large. On the other hand, since the integrals over [-a, a] are always 1, by property (b) of Definition 8.6.4 (8

dt = 1

lim J a

for every 8 > 0. This seems to indicate that the functions are concentrated near 0 and must become very large near 0 (see Exercise 6). The graphs of the first few functions are given in Figure 8.6. Q1, Q2, and Q3 of a typical approximate identity

-t

0

-0.5

Figure 8.6

8.6.5

0.5

Graphs of Q Q2, Q3

THEOREM- Let bean approximate identity on [-I, I], and let f be abounded real-valued periodic function on R of period 2 with f E 9t[ -1, 1 ]. For n e N, x E R, define

S (x) = J l f (x +

dt.

(11)

1f f is continuous at x E R, then lima Sn(x) = f(x).

Furthermore, if f is continuous on [ -1, 1 ], then lim S (x) = f (x)

uniformly on R.

Proof. We first note that since f is periodic and integrable on [-1. 1). f is integrable on every finite subinterval of R. Thus the integral in equation (11) is defined for all x E R. Also, since f is bounded, there exists a constant M > 0 such that L f (x)I s M for all x E R.

8.6 The Weierstrass Approximation Theorem

349

Suppose first that f is continuous at x E R. By property (a) of Definition 8.6.4.

f(x) =

Jf(x)Q0(t) dt.

Therefore,

IS.(x) -f(x)I = I f (f(x + t) -f(x)]Qn(t)dt '

f

I f(x + t) - f(x)I Q,(t)dt.

(12)

Let e > 0 be given. Since f is continuous at x, there exists a 8 > 0 such that

lf(X+t)-f(X)I < 2 for all t, I t I < S. Therefore,

J

Lf (x + r) - f (x) I Q. (t) dt < a

2

5

t

a

Q,,(t) dt

Q0(t) dt = 2.

J

(13)

2 On the other hand,

J{851:I} If(x + t) - f(x)I Q,(t)dt

2M J

Q.(t)tit.

{85I'I}

Since {Qn) is an approximate identity , by property (b) there exists no E N such that

J{8s111}

Qn(t) dt <

4M

for all n a no. Thus if n a no,

f(f(x + t) - f(x)I Qn(t) dt < 2. {8slr1}

Therefore, by equation (12) and the above,

IS. (X) -f(x)I s Ja If(x + t) -f(x)IQ0(t)dt If(X + t) - f(X)I Qn(t) dt

+ {asIr1} E

E

<2+2=E for all n z n,. Thus lim S0(x) = f(x). n-Ow

350

Chapter 8

Sequences and Series of Functions

Suppose f is continuous on [ -1. 1]. Since f is periodic, this implies that f (-1) = f (I). By Theorem 4.3.4, f is uniformly continuous on [ -1, 11, and hence by periodicity, also on R (Exercise 1). Thus given e > 0, there exists a 8 > 0 such that

I f(x + t) - f(x)I <

2

for all x E R and all t, I t l < 6. As a consequence, inequality (13) holds for all x E R. Therefore, as above, there exists n E N such that I

f(x) I < E

for all x E R and all n ? no. This proves that the sequence

converges uniformly to

f on R. Q Proof of the Weierstrass Approximation Theorem. We now use Theorem 8.6.5 to prove the Weierstrass approximation theorem (Theorem 8.6.1). Suppose f is a continuous real-valued function on [a, b]. By making a change of variable, i.e.,

g(x) = f((b - a)x + a), x E [0, 1 ], we can assume that f is continuous on [0, 1 ]. Also, if we let

g(x)=f(x)-f(0)-x[f(l)-f(0)], xE[0,11, then g(0) = g(l) = 0 and g(x) - f(x) is a polynomial. If we can approximate g by a polynomial Q and set

P(x) = Q(x) + [f(x) - 8(x)] then P is also a polynomial with I f(x) - P(x) I = Ig(x) - Q(x)I. Therefore, without loss of generality, we can assume that f is defined on [0,1 ] satisfying

f(0) = f(1) = 0. Extend f to [ -1, 1 ] by defining f (x) = 0 for all x E [ -1, 0). Then f is continuous on [ -1, 1]. Finally, we extend f to all of R by defining

f (x) = f (x - 2k), k E Z, where k E Z is chosen so that x - 2k E (-1, 1 ] (see Figure 8.7). Then f is continuous and periodic of period 2 on all of R, and thus satisfies the hypothesis of Theorem 8.6.5.

f -2

-1

o

Figure 8.7

2

8.6 The Weierstrass Approximation Theorem

351

Our next step is to find an approximate identity {Q"} on [ -1, 1 ] such that the corresponding functions S. of Theorem 8.6.5 defined by equation (11) are polynomials. To accomplish this we let

Q,(t) = C"(1 - t2)",

where c > 0 is chosen such that Q"(t) tit = 1. 12,

Thus the sequence {Q"} satisfies property (a) of Definition 8.6.4. To show that it also satisfies (b) we need an estimate on the magnitude of c". Since I

I

1=c"J(l -t2rdt=2c"J(l i

-t2rdt

0

?2c"J

(1-t2)"dt

o

2c,, o

(1-nt2)dt=2c" (

1

-

1

37 /

4c"

3\fn we obtain

In the above we have used the inequality (1 - t2)" z 1 - nt2 valid for all t E [0, 1] (Example 1.3.3(b)). Finally, for any 5, 0 < S < 1,

Q"(t)=c"(1-t2rsV(1-52r for all t,SsIt 1<-1. Thus since 0 < (1 - 82) < 1, by Theorem 2.2.6(d), limo"(t) = 0 uniformly in S <_

1t1

<- 1. Therefore,

lim

Q"(t) tit = 0.

"-'OD J{8 5 1 fl)

For x E [0, 1 ], set P"(x) = j 1 f (x + t) Q"(t) tit.

352

Chapters

Sequences and Series of Functions

This is the function S"(x) of Theorem 8.6.5, restricted to x E [0, 1]. Let x E [0, 1 Since f(t) = 0 for t E [ -1.0] U [ 1, 2],

f(x f(x +

Jt

+ t)QQ(t) dt,

dr c

which by the change of variables s = t + x gives P"(x) = f 'f(s)Q. (s - x) ds.

Therefore, P"(x), for x E [0, 1 ], is a polynomial of degree less than or equal to 2n. As a consequence of Theorem 8.6.5,

lim P"(x) = f(x)

n-M

uniformly on [0, 1).

thereby proving the result. O Remark. The above proof of the Weierstrass approximation theorem is a variation of a proof found in the text by Walter Rudin listed in the Bibliography.

EXERCISES 8.6 1. If f : lR - R is periodic of period 2 and continuous on [-1, 11, prove that f is uniformly continuous on R. 2. *Prove Theorem 8.6.3. 3. Forn E N, define Q. on [ -1, 1 ] as follows: n

Q"(x) _

2 0,

-ln cx

ln,

1 <1xI sl.

Show that {Q"} is an approximate identity on [ -1, 1 ].

4. FornEN,setQ"(x)=c"(1 - IxI)",x E[-1,1]. a. Determine c" > 0 so that f', Q"(r) dr = 1. b. Prove that with the above choice of c", the sequence {Q"} is an approximate identity on [ -1. 1 ].

c. Sketch the graph of Q"(x) for n = 2, 4, 8. S. Forn E N, set Q"(x) = c,, x l e -', x E [ -1, 1 ]. a. Determine c" > 0 so that f', Q"(r) dr = 1.

b. Prove that with the above choice of c", the sequence {Q"} is an approximate identity on [ 6. *If {Q"} is an approximate identity on [ -1, 1 ], prove that Fim-

1, 1 ].

x E [ -8. S)) = oo for every S > 0.

7. Let f be a continuous real-valued function on [0, 1). Prove that given e > 0, there exists a polynomial P with rational coefficients such that 1f(x) - P(x) I < e for all x E [0, 1 ].

8.7

Power Series Expansions

353

8. Suppose f is a continuous real-valued function on [0, 11 satisfying

f f(x)x" dx = 0 for all n = 0. 1. 2.... 0

Prove that f(x) = 0 for all x E [0, 1 ]. (Hint: First show that fo f(x)P(x) dx = 0 for every polynomial P. then use the Weierstrass theorem to show that fo f2(x) dx = 0.)

8J

Power Series Expansions In this section we turn our attention to the study of power series and the representation of functions by means of power series. Because of their special nature, power series possess certain properties that are not valid for series of functions. We begin with the following definition.

8.7.1

DEFINITION the form

Let {ak}k_o be a sequence of real numbers, and let c E R. A series of

00

I ak(X - c)k = a0 + a1(x - c) + a,(x - c)2 + aj(x - c)? + k=0

is called a power series in (x - c). When c = 0, the series is called a power series in x. The numbers at are called the coefficients of the power series. Even though the study of representation of functions by means of power series dates back to the mid-seventeenth century, the rigorous study of convergence is much more recent. Certainly Newton and his successors were concerned with questions involving the convergence of a power series to its defining function. It was Cauchy, however, who, with his formal development of series, brought mathematical rigor to the subject. As an application of his root and ratio test, Cauchy was among the first to use these tests to determine the interval of convergence of a power series. This is accomplished as follows: Consider a power series 7, ak(x - c)k. Applying the root test to this series gives

km where a = lim k-.oo

Ix - cl A

Iakl

= Ix - cla.

m

k

I ak [

. Thus by Theorem 7.3.4, the series converges absolutely if

alx - cl < 1, and divergesifalx - cl > 1.Ifa=0,then alx-cl < I for all x E R. If 0 < a < oo, then

aix - cl < I 8.7.2

DEFINITION defined by

if and only if

Ix - cl <

Given a power series 7- ak(x - c)k, the radius of convergence R is

R = litn. 1

If lim

°

Ia

oo we take R = 0, and if lim 'Ml q, = 0 we set R = oo.

354

chapter8

Sequences and Series of Functions

When R = 0, the power series E ak(x - c)k converges only for x = c. On the other hand, if R = oo, then the power series converges for all x E R.

Remark If ak # 0 for all k and lim I ak+, I / I ak I exists, then by Theorem 7.1.10 the radius of convergence of E ak(x - c)k is also given by I

R

lak+1I

= lim Iaj k-,oo

This formulation is particularly useful if the coefficients involve factorials.

8.7.3

THEOREM Given a power series 5:k'. O ak(x - c)k with radius of convergence R, 0 < R 5 oo, then the series (a) converges absolutely for all x with Ix - cI < R, and (b) diverges for all x with I x - c I > R. (c) Furthermore, if 0 < p < R, then the series converges uniformly for all x with

Ix - ci :S P. Proof. Statements (a) and (b) were proved in the discussion preceding the statement of the theorem. Suppose 0 < p < R. Choose /3 such that p < 6 < R. Since

= RI < I,

{ ak I <

for all k at no.

lim

k

k

k-roo

Hence for k a n, and I x- c i 5 p, lak(x - c)kj

Iakjpk <

(Q/k.

But (p//3) < I and thus 7,(p/j3)k < oo. Therefore, by the Weierstrass M-test, the se-

ries converges uniformly on Ix - c) s p. 0 The previous theorem provides no suggestion as to what happens when }x - c l = R. As the following examples (with c = 0) illustrate, the series may either converge or diverge when I x I = R.

8.7.4

EXAMPLES

(a) The series 00

xk

71 k-1

has radius of convergence R = 1. This series diverges at both x = I and -1.

8.7

Power Series Expansions

355

(b) The series J , xk/k also has radius of convergence R = 1. In this case, when x = 1 the series diverges; whereas when x = -1, the series is an alternating series which converges by Theorem 7.2.3.

(c) Consider the series 7,k'-, xk/k2. Again the radius of convergence is R = 1. In this example the series converges at both x = I and - 1.

(d) Consider the series I + 2x + 32x2 + 23x3 + 34x4

a,,xk k=0

where ak

__

3k,

if k is even,

2' if k is odd.

Hence Tim V1 akl = 3, and therefore, R = 3. The series diverges at both x = 3 and

X = -3. (e) Finally, consider the series 7, k!xk. Here ak = k!, and a541 lim= kim(k + 1) = oo.

k--.oo ak

Thus by Theorem 7.1.10, verges only for x = 0.

k /l ak

- oo and R = 0. Therefore, the power series con-

Abel's Theorem Suppose we are given a power series Y, ak(x - c)k with radius of convergence R > 0. By setting

AX) = Y,00 ak(x - c)k,

(14)

k-0

we obtain a function that is defined for all x, f x - cI < R. Functions that are defined in terms of a power series (as in series (14)) are usually referred to as ral.aoalytic

functions. Fix p with 0 < p < R. Since the series converges uniformly to f on

Ix - cI p, by Corollary 8.3.2, f is continuous on Ix - cI <- p. Since this holds for all p < R. the function f is continuous on Ix - c I < R. If the series (14) also converges at an endpoint, say at x = c + R, then f is continuous not only in (c - R, c + R) but also at x = c + R. This follows from the theorem of Abel, which is shown next. For convenience, we take c = 0 and R = 1.

8.7.5 THEOREM (Abel's Theorem) Suppose f(x) = Y-k o akxk has radius of convergence R = 1, and that 7_' o ak converges. Then 00

f(x) = Y, ak. k=0

356

Chapter 8

Sequences and Series of Functions

Set s_I = 0, and f o r n = 0, 1, 2, ... let s = 7,1.0 ak. Then by the partial summation formula (7.2.1), Proof.

k-0

Sk(Xk - xk+I) + Sin

akXk =

/

k=0

n-I

(1 - X) I Skxk + k-0

Since the sequence {sn} converges, if we let n -a oo, then for all x. Ix I < 1, x+

f (X) _ (1 - X) L, SkXk k-0

Lets = lim s,,, and let e > 0 be given. Choose n,, E N such that I s - sn I < 1 e for all n Z no. Since 00

(1 -x)7, xk= 1, IXI < A-0

we have for all x, 0 < x < 1,

If(x) - SI = 0 - X)kI(Sk - S)xkl s (1 - X)k

ISk - SIX'

(I-X)Sk-SI +E(1-x),xk k=0 2

k=n,+I

s(1-x)M+2 where M = Yk,olsk - sI. If we now choose 8 > 0 such that I - S < x < I implies that (1 - x) M < e, then if(x) - s l < e for all x, 1 - 8 < x < I. Thus

lim f (x) = s. Q 8.7.6

i

EXAMPLE To illustrate Abel's theorem, consider the series Icko-o (-1)ktk. This series has radius of convergence R = 1. Furthermore, the series converges to f (t) = 1/(1 + t) for all t, I t I < 1. Since the convergence is uniform on I t I s I x I where I x I < 1. by Corollary 8.4.2, dt

ln(1 + x) = 0

°°

7(-1)J = k.0

+

0

(-1)t+l xk

00

(-1)kxk+I = k.ok + 1 k.I

0

for all x, I x I < 1. The series

tk d

(-1I k-1

k

xk

k

8.7 Power Series Expansions

357

has radius of convergence R = 1, and also converges when x = 1. Thus by Abel's theorem,

001

n2=

1

1-2+3-4

k

Differentiation of Power Series Suppose the power series Yw 0 ak(x - c)k has radius of convergence R > 0. If we differentiate the series term-by-term we obtain the new power series,

0

00

2 kak(x - c)k-1 = G (k + 1)ak+1(x - c)k. k-l k-0

(a5)

The obvious question is what is the radius of convergence of the differentiated series

(15)? Furthermore, if f is defined by f(x) = I' oak(x - c)k, lx - cI < R, does the series (15) converge to f'(x)? The answers to both of these questions are provided by the following theorem.

8.7.7 THEOREM Suppose 7,- o ak(x - c)k has radius of convergence R > 0, and

Iak(x-c)k, Ix-cl
k-0 Then 00

(a) 2 kak(x - c)k-1 has radius of convergence R, and k.1

ao

(b) f'(x) = 7, kak(x - C)k-'for all x, Ix - cl < R. k-l

Proof

For convenience we take c = 0. Consider the differentiated series 7, kalxk -1.

By Theorem 2.2.6, lim \ = 1, and for x * 0, lxI

_

k k I

fi-m k

k-1 = W. . By Exercise 10 of Section 2.5, for x * 0, k IakIkIXIk-1

=lxliklakl. Therefore, if R = oo, the differentiated series (15) converges for all x, and if 0 < R < oo,

the differentiated series converges for all x, IxI < R, and diverges for all x, Ixl > R. Thus the radius of convergence of Joko- 1 kakxk-1 is also R.

358

Chapter 8

Sequences and Series of Functions

Furthermore, for any p, 0 < p < R, by Theorem 8.7.3 the series 7, kakxk-' converges uniformly for all x, I x 1 s p. Thus by Theorem 8.5.1, the series (15), obtained by term-by-term differentiation, converges to f'(x), i.e., OQ

f'(x) = Y, kakxk-'

for all x, I xI < R.

k=1

8.7.8 COROLLARY Suppose Ioko- o ak(x - c)k has radius of convergence R > 0, and x f (x) = Y, ak(x - c)k,

I x - c I < R.

k=o

Then f has derivatives of all orders in Ix - c I < R, and for each n E IN,

ft")(x) _

x ',k(k - 1)

k-n

(k - n + 1)ak(x - c)k-".

(16)

In particular, ft")(c) = n! a,,.

(17)

Proof. The result is obtained by successively applying the previous theorem to f, f', f", etc. Equation (17) follows by setting x = c in equation (16).

8.7.9

DEFINITION A real-valued function f defined on an open interval I is said to be infinitely differentiable on I if f(x) exists on I for all n E N. The set of infinitely differentiable functions on an open interval I is denoted by C'(1).

As a consequence of Corollary 8.7.8, if 7, ak(x - c)k has radius of convergence

R > 0 and if f is defined by f(x) = 7-'k-, ak(x - c)k for Ix - cI < R, then the function f is infinitely differentiable on (c - R, c + R) and its nth derivative is given by equation (16). We illustrate this with the following example.

8.7.10

EXAMPLE For Ixi < 1, 00

xk.

1-x Thus by the previous corollary, 00

1

(1 - x)2 = 2

(1 - x)3

kz k-1 =

kYa

co

1(

k

+ l)x k ,

00

=

00

k( k

k

-2

- 1)x k-2 =

k

',( k +

2)( k

-0

and for arbitrary n E N,

(n-1)!(1 -xY'

J(k+n-1 )...(k+ 1)xk.

+ 1)xk ,

8.7 Power Series Expansions

359

Uniqueness Theorem for Power Series The following uniqueness result for power series is another consequence of Corollary 8.7.8.

8.7.11

ak(x - c)k and 7, bk(x - c)k are two power series which converge for all x, Ix - c < R, for some R > 0. Then COROLLARY Suppose

00

00

'x - ci < R,

G ak(x - c)k = T, bk(x - c)k, k-0

k-0

i f and only i f ak = bk f o r all k = 0, 1, 2, ... .

Proof. Clearly, if ak = bk for all k, then the two power series are equal and converge to the same function. Conversely, set

AX) = T,00 ak(x -

c)k

00

g(x) = Y, bk(x - c

and

k=0

k-0

If f (x) = g(x) for all x, x - c I < R, then r)(x) = gt") (x) for all n = 0, 1, 2.. . . and all x, I x - c I < R. In particular, f")(c) = gO)(c) for all n = 0. 1, 2, .... Thus by equation (17), a" = b" for all n.

Representation of a Function by a Power Series Up to this point we have shown that if a function f is defined by a power series, that is 00

Ix - c) < R,

f(x) _ Y, ak(x - c)k, k-0

with radius of convergence R > 0, then by Corollary 8.7.8, f is infinitely differentiable on (c - R, c + R) and the coefficients ak are given by ak = f clo(t)/k!. We now consider the converse question. Given an infinitely differentiable function on an open interval I and c E 1, canf be expressed as a power series in a neighborhood of the point c? Specifically, does there exist an e > 0 such that 00

f (x) = G; ak(x - c)k k-0

f o r all x, I x - c I < e, with ak = f ()(c)/k! f o r all k = 0, 1, 2, ... ? The following example from Cauchy shows that this is not always possible.

8.7.12

EXAMPLE

Let f be defined on R by

x=0.

0,

Since X

m

e" = liim a-`' = 0, f is continuous at 0. For x * 0, .

x3

360

Chapter8

Sequences and Series of Functions

When x = 0, we have f (h) f'(0) = lim h-*0

-h A o) = lim h-0

e

= lim ' '

h

t. = 0. e'

The last step follows from 1'Hospital's rule. Thus,

X, e '/'', x 0 0,

.f'(x) _

x = 0.

0,

By induction, it follows as above, that for each n E N, P(s)e_'fix,

f(n)(x) = 1

x # 0,

x0,

0,

(

where P is a polynomial of degree 3n. The details are left to the exercises (Exercise 15).

Thus the function f is infinitely differentiable on R. If there exists R > 0 such that f(x) = J o akxk for all x, IxI < R, then ak = 0 for all k. As a consequence, f cannot be presented by a power series that converges to f in a neighborhood of 0.

Taylor Polynomials and Taylor Series We now consider the problem of representing a function f in terms of a power series in greater detail. Newton derived the power series expansion of many of the elementary functions by algebraic techniques or term-by-term integration. For example, the series expansion of 1/(1 + x) can easily be obtained by long division, which upon term-byterm integration gives the power series expansion of In (1 + x). Maclaurin and Taylor were among the first mathematicians to use Newton's calculus in determining the coefficients in the power series expansion of a function. Both realized that if a function f (x) had a power series expansion ak(x - c)k, then the coefficients ak had to be given by

f(c)/k!. 8.7.13

DEFINITION Let f be a real-valued function defined on an open interval 1, and let c E I and n E N. Suppose f W(x) exists for all x E I. The polynomial (k) (c)

r.( C)(x) _ kmo

f kf

(x - c)k

is called the Taylor polynomial of order n off at the point c. If f is infinitely differentiable on 1, the series

= fk)(c) k.0 k!

is called the Taylor series off at c.

(x - c)k

8.7 Power Series Expansions

361

For the special case c = 0, the Taylor series of a function f is often referred to as the Maclaurin series. The first three Taylor polynomials, To, T1, T2, are given specifically by T0(f, c)(x) = f(c),

T1(, c)(x) = f(c) + f'(c)(x - c),

f 2r) (x - c)2.

T2(f, c)(x) = f(c) + f'(c)(x - c) +

The Taylor polynomial T,(f, c) is the linear approximation to f at c; that is, the equation of the straight line passing through (c, f(c)) with slope f'(c). In general, the Taylor polynomial T. of f is a polynomial of degree less than or equal to n that satisfies T (k)(f c)(c) = f tk)(c),

for all k = 0, 1, ... , n. Since ft">(c) might possibly be zero, T. (as the next example shows) could very well be a polynomial of degree strictly less than n.

8.7.14

EXAMPLES In the following examples we compute the Taylor series of several functions. At this stage nothing is implied about the convergence of the series to the function.

(a) Let f (x) = sin x and take c = Z. Then

f(2) = sin 2 = 1,

f'(2)=cos2=0, f.("(2) = -sin 2 = -1, J 3)(2) = -Cos Z = 0. Thus 2

T3(f2)(x)=1-2t x-2/, which is a polynomial of degree 2. In general, if n is odd, even, P 20(f) = (-1)k. Therefore, if n is even,

n12 (-I r

T ( f , 2)(x) = T.

1(J,

2)(x) = 0 (2k) \x -

The Taylor expansion of f(x) = sin x about c = 2 is given by

00 (-1)k

()!

IX

ir

2k

2)

(b) For the function f(x) = e-"X', by Example 8.7.12

0)(x) = 0 for all n E N.

0, and if n = 2k is

-2\zk

362

Chapter 8

Sequences and Series of Functions

Thus the Taylor series of f at c = 0 converges for all x E l ; namely, to the zero function. It, however, does not converge to f. (c) In many instances, the Taylor expansion of a given function can be computed from a known series. As an example, we find the Taylor series expansion of f(x) = 1/x about

c = 2. This could be done by computing the derivatives off and evaluating them at c = 2. However, it would still remain to be shown that the given series converges to f (x). An easier method is as follows: We first write 1

1

x

2-(2-x)

_l

1

(2

2

For lwl < 1,

rwk. °O

1-w

k_o

Setting w = (2 - x)/2, we have

(2 - xk =

1

2k O

X

0O

(-1)k

71 2k+1 k.O

2k

(x - 2)k

for all x, I x - 21 < 2. By uniqueness, the given series must be the Taylor series of f(x) = 1/x. In this instance, the power series also converges to the function f(x) for all

x satisfying Ix - 21 < 2.

U 11

Remainder Estimates

To investigate when the Taylor series of a function f converges to f(x), we consider

R (x) =

c)(x) = f (x) -

c)(x).

The function R. is called the remainder or error function between jr and

(18) c).

Clearly,

f (x) = Jim TT(f, c)(x) if and only if

lim

c)(x) = 0.

n

Since the Taylor polynomial T. is the nth partial sum of the Taylor series of f, the Taylor series converges to f at a point x if and only if nliini this fact, we state it as a theorem.

8.7.1 5

c)(x) = 0. To emphasize

THEOREM Suppose f is an infinitely differentiable real-valued function on the open interval I and c E I. Then for x E 1,

AX) = if and only if limn

c)(x) = 0.

fck)(c) (x - c) , k

00

k-O

k. E

8.7

Power Series Expansions

363

The formula

f(x) =f(c) +f'(c)(x - c) +

+f2(x - c)2 +

f n() (x - c)" + R"(f, x)(x)

is known as Taylor's formula with remainder. We now proceed to derive several formulas for the remainder term R. These can be used to show convergence of T. to f.

Lagrange Form of the Remainder Our first result, attributed to Joseph Lagrange (1736-1813), is called the Lagrange form of the remainder. This result, sometimes also referred to as Taylor's theorem, was previously proved for the special case n = 2 in Lemma 5.4.3.

8.7.16

THEOREM Suppose f is a real-valued function on an open interval 1. c E I and n E N. If f (n+ ')(t) exists for every t E 1, then for any x E I, there exists a C between x and c such that Rn(x) = Rn(.f, c)(x) _

+ 1) (x - C)n+l

(19)

Remark. Continuity of f("+) is not required.

Proof. Fix x E 1, and let M be defined by f(X) = T"(f, c)(x) + M(X - c)n+ To prove the result, we need to show that (n + 1)! M = f("+ ')(C) for some tween x and c. To accomplish this, set

be-

g(t) = f(t) - TTJ, c)(t) - M(t - On +, = R"(t) - M(t - c)n+1 First, since T. is a polynomial of degree less than or equal to n, g(n+Ikt) = f(n+t)(t) - (n + 1)! M.

Also, since Tne)(f, c)(c) = f (t)(c), k = 0, 1, ... , n,

g(c) = g'(c) = ... = g(")(c) = 0. For convenience, let's assume x > c. By the choice of M, g(x) = 0. By the mean value theorem applied to g on the interval [c, x], there exists x,, c < x, < x, such that

0 = g(x) - g(c) = g'(xi)(x - c). Thus g'(x,) = 0. Since g'(c) = 0, by the mean value theorem applied to g' on the interval [c, x, ], g"(x2) = 0 for some x2, c < x2 < x,. Continuing in this manner, we obtain a point x" satisfying c < xn < x, such that g(")(xn) = 0. Applying the mean value

364

Chapter 8

Sequences and Series of Functions

theorem once more to the function g(n) on the interval [c, xn], we obtain the existence of

a t E (c, xn) such that 0 = 8(n)(Xn) - g(n)(c) = Jn+ l)( )(X - C).

0; i.e., f(n+1)(C) - (n + 1)! M = 0, for some C between x and c. 0

Thus

In Example 8.7.20 we will give several examples to show how the remainder estimates may be used to prove convergence of the Taylor series to its defining function. In the following example we show how the previous theorem may be used to derive simple estimates and inequalities.

8.7.17

EXAMPLES

(a) In this example we use Theorem 8.7.16 with n = 2 to approximate f (x) =

1 + x,

x > -1. With c = 0 we find that f(0) = 11

f'(0) =

f"(0) =

,

-4

.

2 Therefore, T2(f, 0)(x) = 1 + 2 x - e x, and thus 1

1 +x = 1+2x-8x2+ RZV, 0x ) ( ). 1

By formula (19), R2(f, 0)(x) =

f

(3)

L)

X3

3!

= 16 I (1 + )-SR x3 -

for some C between 0 and x. If x > 0, then

> 0, and thus (1 + C)"s2 < 1. There-

fore, we have

V1 -+x - T2(f, 0) (x) I <

16

x3

for any x > 0. If we let x = 0.4, then TZ(f, 0)(.4) = 1.18, and by the above, 1.4 - 1.181 < 0.004, so that two-decimal-place accuracy is assured. In fact, to five decimal places 1.4 = 1.18322. I

(b) The error estimates can also be used to derive inequalities. As in the previous example, 1

-1/1+1= 1 + 2 x - 8x2 + R2(f, 0) (x). For x > 0 we have 0 < R2(f, 0)(x) < 16x3. Thus

I+2x-8x2< VT +x<1+2x-gx2+ I x3 16

for. all x > 0.

8.7

Power Series Expansions

365

Integral Form of the Remainder Another formula for R.V, c) is given by the following integral form of the remainder. This, however, does require the additional hypothesis that the (n + l )st derivative is Riemann integrable.

8.7.18 THEOREM Suppose f is a real-valued function on an open interval 1, c E I and n I-, N. If f ("+'kt) exists for every t E 1 and is Riemann integrable on every closed and bounded subinterval of I, then

R"(x) = R"U c)(x) = n! { `f("+°(r)(x - t)"dt,

x E 1.

(20)

C

Proof. The result is proved by induction on n. Suppose n = 1. Then

R1(x) = f(x) - f(c) - f'(c)(x - c), which by the fundamental theorem of calculus

f

= l " f'(t) dt - f'(c) dt =

J

[f'(t) - f '(c)) dt.

From the integration by parts formula (Theorem 6.3.7) with

u(t) = f'(t) - f'(c), v'(t) = 1, u'(t) = f"(t), v(t) _ (t - jr), we obtain j CX JC

[f'(t) - f'(c)) dt = [f'(t) - f'(c))(r - x) I; -

f"(t)(t - x) dt

C

Is

(x - t) f"(i) dt.

To complete the proof, we assume that the result holds for n = k, and prove that this implies the result for n = k + 1. Thus assume Rk(x) is given by equation (20). Then Rk+ I(x) = f(x) - Tk+ IV, c)(x) p(k+ (c)

f(x) - Tk(f c)(x) - (k + 1)! (x - c)k+i f(k+I)(

=Rk(x)- (k+ 1)i) (x - cr+i kt 1

js (X

- t)kf (k+')(t) dt - k' f (k+')(c) J I (x - t)k dt k

(k+I)

(k+')

366

Chapter a

Sequences and Series of Functions

As for the case n = 1, we again use integration by parts with u(t) = f (k+'1(t) - f(k+'1(c)

v(t) =

and

k

+

(x -

1)k-I,

which upon simplification, gives

Rk+i(x) =

1

J

(k + 1)! J.

s

(x - t)k''f(k+'>(t)dt.

Cauchy's Form for the Remainder Under the additional assumption of continuity of f the remainder as follows.

8.7.19

we obtain Cauchy's form for

COROLLARY Let f be a real-valued function on an open interval 1, c E 1 and n E N. If f("+1) is continuous on I, then for each x E 1, there exists a C between c and x such that Rn(x) = R,(f, c)(x) =

P +14) (X - cxx - i)". n!

(21)

Proof

Since f("+')(t)(x - t)" is continuous on the interval from c to x, by the mean value theorem for integrals (Theorem 6.3.6), there exists a C between c and x such that

f fn+I)(:)(x - t/ dt = (x - c) f("+'kC)(x l - C)". C

The result now follows by equation (20).

We now compute the Taylor series for several elementary functions, and use the previous formulas for the remainder to show that the series converges to the function.

8.7.20

EXAMPLES

(a) As our first example, we prove the binomial theorem (Theorem 2.2.5). For n E N

let f(x) = (1 + x)", x E R. Since f is a polynomial of degree n, if k > n then f (k)(x) = 0 for all x E R. Therefore, by Theorem 8.7.16,

f(x) _

f (k)(0) = n!/(n - k)! f o r k = 0, 1, ... , n. Therefore, n!

k=a k!(n - k)!

8.7 Power Series Expansions

367

The series expansion of (1 + x)° for a E R with a < 0 is given in Theorem 8.8.4, whereas the expansion for a > 0 is given in Exercise 7 of the next section. For rational numbers a, the expansion of (1 + x)° was known to Newton as early as 1664.

(b) Let f (x) = sin x with c = 0. Then f ("i(x)

=

(

n = 2k + 1,

1)k cos x,

(-1)ksinx,

n = 2k.

Thus f(")(0) = 0 for all even n E N, and f ii(0) = (-1)'`, whenever n = 2k + 1, k = 0, 1, 2, .... Therefore, the Taylor series off at c = 0 is given by 00

(-1)k

k=O

(2k + 1)!

x2k+i

To show convergence of the series to sin x we consider the remainder term Rn(x). By Theorem 8.7.16, for each x E R there exists a C such that f(n+ 1 (SS)

R"{x) = (n + 1)! x

n+l

Since Ii (" ")(x) 1 -5 1 for all x, we have Ixln+1

Rn(x),

(n + 1)!

By Theorem 2.2.6(f), lim jxI"+'/(n + 1)! = 0 for any x E R. As a consequence. lim Rn(x) = 0 for all xnE+1R, and thus 00

sinx = 7 k_o

(-+) (2k + 1)`

x +', x E R.

The sine function, as well as the cosine function, can be defined strictly in terms of power series. For further details, see Miscellaneous Exercise 3. (c) As our third example, we derive the Taylor series for f (x) = ln(1 + x), where, as in Example 6.3.5,

lnx=fix Idr, x>0, denotes the natural logarithm function on (0, oo). Then f(0) = In (1) = 0, and by the fundamental theorem of calculus, f '(x) = 1/(1 + x). Thus for n = 1, 2,

... ,

f("i(x) _ (-1)n+i (n - 1)!

0+x)"

In particular, f W(0) _ (-1)"+ 1(n - 1)!, and the Taylor series off at 0 becomes

. (-1)n+I 7' x n., n

368

Chapter 8

Sequences and Series of Functions

Although we have already proved that this series converges to In (1 + x) for all

x, -1 < x

1 (Example 8.7.6), we will prove this again to illustrate the use of the remainder formulas. Suppose first that 0 < x <- 1. By Theorem 8.7.16, (_ 1)"+zx"+l

R (x) = R If 0)(x1 =

(n + 1)(l + C)"''

for some C, 0 < < x. In this case, (1 + C) > 1, and thus IR"(x)

I

n+ l x

"+l<

n+ 1

for all x, 0 < x s 1. Therefore, 1im R"(x) = 0 for all x e [0, 1 ].

n-Oc

We next consider the more difficult case -1 < x < 0. By the Cauchy form of the

remainder, if - I < x < 0, there exists ', x s s 0, such that ft"+l'(C)(x

R"(x) = n!

- C)"x = (-

l)"+2 (Ix+

)rx

n+l'

Therefore, IR"(x)I

(II+I0 [(l

+

)]"'

Consider the function ap(t) defined on [x, 0] by x

(P(t) = 1 + t

Then cp'(t) = (I + x)/(1 + t)2, which is positive on [x, 0] provided x > -1. Therefore,

q,(0) = -x = IxI

(P(t)

for all t, x s t <_ 0. Thus IR"(x)I `-

(1

I+I0lxln.

Since IxI < 1, lim R"(x) = 0. Therefore, the Taylor series converges to In (1 + x) for

all x,-1 <x -_ 1;i.e.,

ln(l+x)=

00

(-I)n+l

n=l

n

x",

-l<xsl.

8.7

Power Series Expansions

369

The Taylor expansion of In(l + x) was first obtained in 1668 by Nicolaus Mercator. Newton shortly afterward obtained the same expansion by term-by-term integration of the series expansion of 1/(1 + x) (see Example 8.7.6). (d) As our final example, we consider the natural exponential function E(x) = exp x that is defined as the inverse function of the natural logarithm function L(x) = In x. The

domain of L is (0, oo) with range (-oo, oo). Since L'(x) = 1/x is strictly positive on (0, oo), L is a strictly increasing function on (0, oo). The inverse function E(x) is defined by

y = E(x)

if and only if x = In y.

By the inverse function theorem (Theorem 5.2.14), E is differentiable on R with

E'(x) = E'(L(y)) = L'(y) = y = E(x). Thus E'(x) = E(x) for all x E R. Since In ab = In a + In b, it immediately follows that

E(x + y) = E(x)E(y)

and

E(-x) =

E(x)

Also, E(O) = 1, and by Example 6.3.5, E(1) = e where e is Euler's number defined in Example 2.3.5. Therefore, E(n) = e" for every integer n. If r = m/n, m, n E 1 with

n * 0, then E(nr) = E(m) = e. But E(nr) _ (E(r)r. Therefore, if r E C4,

E(r) = e'. For arbitrary x E R we define e" by e` = E(x). This definition of e' is consistent with the definition given in Miscellaneous Exercise 3 of Chapter 1 (See Exercise 16). Since E(")(x) = E(x) for all x E R, E(")(0) = E(0) = 1. Thus the Taylor series expansion of E(x) about c = 0 is given by 00

'X k-0 k 1

It is left as an exercise (Exercise 11) to show that this series converges to e' for all

x E R. There is a more subtle question involving power series representation of functions that we have not touched upon. The question concerns the following: How is the radius of convergence R of the Taylor series off related to the function f? The full answer to this question requires a knowledge of complex analysis and thus is beyond the scope of this text. However, we will illustrate the question and provide a hint of the answer with

the following examples. If f(x) = 1/(1 + x), then the Taylor expansion off about x = 0 is given by 1. (- I )kxk, k=0

Chapter 8

370

Sequences and Series of Functions

which has radius of convergence R = 1. This is expected since the given function f is not defined at x = -1, and thus the series could not have radius of convergence R > 1. If it did, this would imply that f would then have a finite limit at x = -1. On the other hand, the function g(x) = 1 /(1 + x2) is infinitely differentiable on all of R. However, the Taylor series expansion of g about c = 0 is given by 00

,(-I)kX2k, k=0 which again has only radius of convergence R = 1. The reason for this is that even

though g is well-behaved on R, if we extend g to the complex plane C by 1

g(z) = 1 + z2 then g is not defined when z2 = -1; i.e., z = ±i, where i is the complex number that satisfies i2 = -1.

EXERCISES 8.7 1. Find the radius of convergence of each of the following power series. 00

3k

a. k-1 k

D

*b.

Xk

x+12*

k=0 4 00

*d.

when k is even,

1

C.

I,, akxk where ak - 2" . k'0 2k +2

71

1 - 1 xk

00

/

1

k

kt

f. j k1 2) k1

when k is odd

2. For each of the following, determine all values of x for which the given series converges. a.

k,

k-0 X

(X * 0)

3kxk

tb G k- 1 2(1

x)"k

(X # 1)

C.

I

l+x k 1-x

(x

#1)

3. Using the power series expansion of 1/(1 - x) and its derivatives, And 00

+a. T', kxk, k-1

IXI < 1

b. I k2xk, IXI < t k_1

00

nkk

c. k-1

4. a. Use Theorem 8.7.16 to show that 1i - (I + 1 x - 9 x2) I < 81x3 for all x < 0. 'Mi, and provide an estimate of the error. b. Use the above inequality to approximate 5. Determine how large n must be chosen so that I sin x - T (sin, 0)(x)l < .001 for all x, IXI s 1. 6. Use the Taylor series and remainder estimate of Example 8.7.20(c) to compute In 1.2 accurate to four decimal places.

8.7

Power Series Expansions

371

7. Suppose f(x) = Ik o ak(x - c)t has radius of convergence R > 0. For Ix - cl < R, set F(x) = f' f(r) dr. Prove that 00

F(x) =

a, k+1(x-c)k+i,

k

Ix - cl < R.

8. *a. Use the previous exercise and the fact that

Arctan x =

1 + x2

to obtain the Taylor series expansion of Arctan x about c - 0.

b Use pact (a) to obtain a series expansion for ir. *c. How large must n be chosen so that the nth partial sum of the series in part (b) provides an approximation of it correct to four decimal places? 9. Use Exercise 8 and Abel's theorem to prove that

(-1

a

k=o2k+1

4'

10. Find constants ao, a,, a2, a3, a4 such that

x4+3x2-2x+5=a4(x- 1)4+a3(x- 1)3+a2(x- 1)2+a,(x- 1)+ao. 11. Prove that the Taylor series of e` (with c = 0) converges toe` for all x E R. 12. Using any applicable method, find the Taylor series of each of the following functions at the indicated point, and specify the interval on which the series converges to the function. *b. f (x) = In x, c = 1 a. &) = cos x, c = 0

c. f(x) = In1 l

± \),

c=0

x

*d. f(x) = (I - x)-". c = 0

*e. f(x) = Aresinx, c = 0 g. f(x) = V X-, c = 1

L f(x) =

x 1

x2,

c=0

h. f(x) = Arctanx, c = 1

13. Suppose f (x) _= Ik o akxk, I xl < R, where R > 0. Prove the following. a. f (x) is even if and only if ak = 0 for all odd k. b. M) is odd if and only if ak = 0 for all even k. 14. Suppose f(x) = 7.i oakxk, I x I < R1, and g(x) _ Zm o bk xk, I x I < R2. Prove that

f(x)g(x) _ Jk ockxk, lxi < min{R,, R2}, where ck = Yi-oaibk-i 15. Let f: R -> R be defined by f(x) = e "" for x * 0, and f(0) = 0. Prove that for each n E N,

Mx)

P()e-V'

x * 0,

0,

x = 0,

where P is a polynomial of degree 3n.

16. Suppose b > 1. For x E R define b(x) = E(x In b), where E is the natural exponential function. a. Prove that b(r) _- b' for all r E Q.

b. For x E R, prove that b(x) = sup(b' : r E 0, r < x}.

372

Chapter 8

Sequences and Series of Functions

8.81 The Gamma Function We close this chapter with a brief discussion of the Beta and Gamma functions, both of which are attributed to Euler. The Gamma function is closely related to factorials, and arises in many areas of mathematics. The origin, history, and development of the Gamma function are described very nicely in the article by Philip Davis listed in the supplemental reading. Our primary application of the Gamma function will be in the Taylor

expansion of (1 - x)-, where a > 0 is arbitrary. 8.8.1

DEFINITION

For 0 < x < oo, the Gamma function r(x) is defined by ]F(x) =

Jte1d:.

(22)

When 0 < x < 1, the integral in equation (22) is an improper integral not only at oo, but also at 0. The convergence of the improper integral defining f(x), x > 0, was given as an exercise (Exercise 9) in Section 6.4. The graph of r(x) for 0 < x < 5 is given in Figure 8.8. The following properties of the Gamma function show that it is closely related to factorials.

6

I

2

3

Figure 8.8 Graph of f(x), 0 < x s 5

8.8.2

THEOREM

(a) For each x, 0 < x < oo, r(x + 1) = x r(x). (b) For n E ICU, r(n + I) = n!.

Proof. Let 0 < c < R < oo. We apply integration by parts to 1 R

t'e-' dt.

8.8

The Gamma Function

373

With u = tx and v' = e-', R

+xJR

tx -Ie-'di

txe-'dt= -r`e-'

R

Rx

e

Since im; cxe-` = 0 and m R*e-R = 0, taking the appropriate limits in the above R yields

r(x + 1) = J txe-'dt = x

0

0

0

1x-le-`dt = xI'(x).

This proves (a). For the proof of (b) we first note that

I'(1) =

e-' dt = 1. J000

Thus by induction, r(n + 1) = W.

8.8.3

EXAMPLE Since the value of I'(!) occurs frequently, we now show that I'('-2) = . By definition,

r

J = Joy t- 'ne-' dt.

2 With the substitution t = s2.

0

To complete the result, we need to evaluate the so-called probability integral f0 e-" ds. This can be accomplished by the following trick using the change of variables theorem from multivariable calculus. Consider the double integral

J=

Joo f 0

e-xl-''dxdy.

0

By changing to polar coordinates

x = r cos 9, y = r sin 9,

with 0< r < oo, 0 E (0, z ), WRe `rdrd9 J= 100 0 10 a (O0 = ?J 0

e_,: rdr4. IT

374

Chapter 8

Sequences and Series of Functions

On the other hand,

J=

0o (oo

J 0

>o

J

e-:= a-)" dx dy =

J 0

e-x:

2

dx

0

Therefore, f0c e-'2 dx =

2

0

from which the result follows.

The Binomial Series As an application of the Gamma function, we will derive the power series expansion of

f(x) = (1 - x)-°, where a > 0 is real. The coefficients of this expansion are expressed very nicely in terms of the Gamma function. By Example 8.7.10, for n E N, 00

(1-x)-n= 1

(n - 1)!

k.0

k!

which in terms of the Gamma function, gives

_

1

(1 - xr

r(k+n)

1

1'(n) k=o

k!

k

x

We will now prove that this formula is still valid for all a E R with a > 0.

8.8.4

For a > 0,

THEOREM (Binomial Series)

_

1

x)°

1

oc

r(n + a)xn,

IxI

n!

< 1.

Proof. We first show that the radius of convergence of the series is R = 1. Set an = 17(n + a)/n!. Then

an+,r(n+1+a)

a

(n + 1)!

n!

T(n + a)*

But by Theorem 8.8.2, I'(n + 1 + a) = (n + a)I'(n + a). Therefore,

a^+, =limn + a = 1. lim n-"o a -oo n + 1 and as a consequence of Theorem 7.1.10, we have R = 1.

8.8 The Gamma Function

375

To show that the series actually converges to (1 - x)-a, we set

00 r(n + a)

1

fa(xW =

r(a)

n!

x",

IxI < I.

Since a power series can be differentiated term-by-term, 1

fv(x) = r(a)

I- nr(n + a) n-1

x X.

n!

Multiplying by (I - x) gives 1

'

(l - x)fa(x) = r( a) Y' _

-

nr(n + a) n!

(1 - x)x rt-'

a r(n + a) n_i (n - 1)! x

1

r(a) I

`F(n + I + a)

1

F(a) o

-°

nr(n + a) n1

n-I

nF(n + a) 1

L

n!

n!

x

X.

But r(n + I + a) - nr(n + a) = ar(n + a). Therefore, (1 - x)fa(x) = afa(x) As a consequence,

d [(1 - x)afa(x)] = -a(1 - X)a-,fa(x) + (1 - x)af,(x)

_ -a(1 - 40-If (X) + all - x)°-'fa(x) = 0. Therefore, (1 - x)a fa(x) is equal to a constant for all x, I x I < 1. But f.(0) = 1. Thus (1 - x)afa(x) = 1; that is,

fa(x) = (1 - x)-a, which proves the result.

0

The Beta Function There are a number of important integrals that can be expressed in terms of the Gamma function. Some of these, which can be obtained by a change of variables, are given in the exercises. There is one integral, however, that is very important and thus we state it as a theorem. Since the proof is nontrivial and would take us too far astray, we state the result without proof. For a proof of the theorem, see Theorem 8.20 in the text by Rudin.

8.8.5

THEOREM For x > 0, y > 0,

J' t'`-'(i - ty-' dt = o

r(x)F(y) F(x + y).

376

Sequences and Series of Functions

Chapter 8

The function r(x)r(y)

B(x' A

T(x + y)' x, y > 0'

is called the Beta function.

EXERCISES 8.8 1. 'a. Compute r(z), r('2222).

b. Prove that for n E N,

CIn+2)= (2n)! n . 2. By making a change of variable, prove that

r(x)= f0(In-

I

dt, 0<x
3. Evaluate each of the following definite integrals.

*a

I edt

f

dt, n E N. t 4. By making the change of variable t = sin 2 u in Theorem 8.8.5, prove that 0

(Ar Jo

b.

0(

(sin u)2"`'(cos u)z `' du =

1 r(n)r(m) n, m > 0. 2 r(n + m)'

5. Evaluate each of the following integrals.

a

f+.R

(sin x)2" dx,

nEN

J0

b.

f

n!1

(sin x)'' *' dx, n E N

0

6. Use the binomial series and term-by-term integration to find the power series expansion of

Arcsinx =

f

(1 - t') -'/2 dt.

0

7. Let a > 0. Set (g) = 1 and fork E N, set

(a

a(a-1)(a-2)

\k/

(a-k+l)

k!

Note, if m e N, i

k

k! (m - k)!'

k 5 m,

and (k) = 0 fork > m.

a. Prove that the series I:-k-0(k) x' converges uniformly and absolutely for x E H, 11. b. Prove that

xt = (1 + x)°, x E [-1, 1 ].

Miscellaneous Exercises

377

NOTES Without question the most important concept of this chapter is that of uniform convergence of a sequence or series

of functions. It is the additional hypothesis required in proving that the limit function of a sequence of continuous or integrable functions is again continuous or integrable. As was shown by numerous examples, pointwise convergence is not sufficient. For differentiation, uniform

convergence of {f"} is not sufficient; uniform convergence of the sequence of derivatives (f' ) is also required. The example of Weierstrass (Example 8.5.3) is interesting for several reasons. First, it provides an example of a continuous function which is nowhere differentiable on R. Furthermore, it provides an example of a sequence of infinitely differentiable functions that converges uniformly on R, but for which the limit function is nowhere differentiable. Exercise 7 of Section 8.5 provides another construction of a continuous function f that is nowhere differ-

entiable. Although this construction is much easier, the partial sums of the series defining the function f are themselves not differentiable everywhere. Thus it is not so surprising that f itself is not differentiable anywhere on It The proof of the Weierstrass approximation theorem presented in the text is only one of the many proofs available. A constructive proof by S. N. Bernstein using the socalled Bernstein polynomials can be found on page 107 of the text by Natanson listed in the Bibliography. The proof in the text, using approximate identities, was chosen because the technique involved is very important in analysis

and will be encountered later in the text. In Theorem 9.4.5 we will prove a variation of the Weierstrass approximation theorem. At that point we will show that every continuous real-valued function on [-vr, w] with f(-ir) = f(r) can be uniformly approximated to within a given e > 0 by a finite sum of a trigonometric series.

MISCELLANEOUS EXERCISES 1. Using Miscellaneous Exercise I of Chapter 6 and the Weierstrass approximation theorem, prove the following: If f E 9t.[a, b] and e > 0 is given, then there exists a polynomial P such that

tk PI <

e.

£1_ 2. DefinefonRby

f(x) =

{cex(i_:-). 0,

Ixi < I. IxI 5 1.

where exp(x) = ex, and c > 0 is chosen so that f f(x) dx = 1. For A > 0, set fx(x) = i f(Ax). a. Prove that fa E C(R) for all A > 0. b. Prove that fa(x) = 0 for all x E R, I x I z A, and that f ,, fx (x) dx = 1. c. Prove that for every 8 > 0, flasl1l} fa(r) dr = 0. 3. In this exercise we show how the trigonometric functions may be defined by means of power series. Define the functions S and C on R by S(x)

(-1)k x2k,1 k=o (2k + I)i

_W

_

C (x)

(-1)k u

k-o (2k)! x

a. Show that the power series defining S and C converge for all x E R.

b. Show that S'(x) = C(x) and C'(x) = -S(x), x E R. c. Show that S"(x) = -S(x) and C"(x) = -C(x).

378

Chapter8

Sequences and Series of Functions

d. Show that if f : R --)- R satisfies f "(x) = -f(x) with f(O) = 0, f'(0) = 1, then f (x) = S(x) for all x E R. e. If f: R - R satisfies f"(x) = 1(x), prove that there exist constants c,, c2, such that f(x) = c,S(x) + c,C(x). L Show that (S(x))2 + (C(x))2 = 1. (Hint: Consider the function f(x) _ (S(x))2 + (C(x))2.) g. Show that C(x + y) = C(x) C(y) - S(x) S(y) and S(x + y) = S(x) C(y) + C(x) S(y) for all x, y E R.

SUPPLEMENTAL READING Andrushkiw, J. W., "A note on multiple series of positive terms," Amer. Math. Monthly 68 (1961), 253-258. Billingsley, P., "Van der Waerden's continuous nowhere differentiable function," Amer. Math. Monthly 89 (1982), 691. Blank, A. A., "A simple example of a Weierstrass function;' Amer. Math. Monthly 73 (1966), 515-519. Boas, Jr., R. P., "Partial sums of infinite series and how they grow," Amer. Math. Monthly 84 (1977), 237-258. Boas, Jr., R. P. and Pollard, H., "Continuous analogues of series;' Amer. Mark Monthly 80 (1973),18-25. Cunningham, Jr., F., "faking limits under the integral sign;' Math. Mag. 40 (1967), 179-186. Davis, P. J., "Leonhard Euler's integral: A historical profile of the Gamma function;' Amer. Math. Monthly 66 (1959), 849-869. French, A. P., " he integral definition of the logarithm and the logarithmic series;' Amer. Math. Monthly 85 (1978), 580-582. Kestleman, H., "Riemann integration of limit functions;' Amer. Math. Monthly 77 (1970),182-187. Lewin, J. W., "Some applications of the bounded convergence theorem for an introductory course in analysis," Amer. Math. Monthly 94 (1987), 988-993.

Mathf, P., "Approximation of Holder continuous functions by Bernstein polynomials;' Amer. Math. Monthly 106 (1999), 568-725. Miller, K. S., "Derivatives of non-integer order," Math. Mag. 68 (1995), 183-192.

Minassian, D. P. and Gaisser, J. W.. A simple Weierstrass function," Amer. Math. Monthly 91 (1984). 254-256. Patin, J. M., "A very short proof of Stirling's formula," Amer. Math. Monthly % (1989),41-42. Roy, Ranjan, "The discovery of the series formula for rr by Leibniz, Gregory and Nilakantha," Math. Mag. 63 (1990),291-306. Sagan, H., "An elementary proof that Schoenberg's space filling curve is nowhere differentiable" Math. Mag. 65 (1992), 125-128. Schenkman, Eugene, "The Weierstrass approximation theorem;' Amer. Math. Monthly 79 (1972), 65-66. Weinstock, Robert, "Elementary evaluations of dr. Jo eos x2 dx, and Jo sin x2 dx; 'Amer. Ja a Math. Monthly 97 (1990), 39-42.

Orthogonal Functions y J and Fourier Series 9.1 Orthogonal Functions 9.2 Completeness and Parseval's Equality

9.3 Trigonometric and Fourier Series

9.4 Convergence in the Mean of Fourier Series 9.5 Pointwise Convergence of Fourier Series

In this chapter we consider the problem of expressing a real-valued periodic function of period 2a in terms of a trigonometric series 00

2 ao + Y, (a cos nx + b sin nx), n-l where the an and bn are real numbers. As we will see, such series afford much greater generality in the type of functions that can be represented as opposed to Taylor series. The study of trigonometric series has its origins in the monumental work of Joseph Fourier (1768-1830) on heat conduction in solids. His 1807 presentation to the French Academy introduced a whole new subject area in mathematics while at the same time providing very useful techniques for solving physical problems. Fourier's work is the source of all modern methods in mathematical physics involving boundary value problems and has been a source of new ideas in mathematical analysis for the past two centuries. To see how greatly mathematics has been influenced by the studies of Fourier, one only needs to look at the two-volume work Trigonometric Series by A. Zygmund (Cambridge University Press, 1968). In addition to trigonometric series. Fourier's original method of separation of variables leads naturally to the study of orthogonal functions and the representation of functions in terms of a series of orthogonal functions. All of these have many applications in mathematical physics and engineering. Fourier initially claimed and tried to show, with no success, that the Fourier series expansion of a function actually represented the function. Although his claim is false. 379

380

Chapter 9

Orthogonal Functions and Fourier Series

in view of the eighteenth-century concept of a function this was not an unrealistic expectation. Fourier's claim had an immediate impact on nineteenth-century mathematics. It caused mathematicians to reconsider the definition of "function." The question of what type of function has a Fourier series expansion also led Riemann to the development of the theory of the integral and the notion of an integrable function. The first substantial progress on the convergence of a Fourier series to its defining function is due to Dirichlet in 1829. Instead of trying to prove, like Fourier, that the Fourier series always converges to its defining function, Dirichlet considered the more restrictive problem of finding sufficient conditions on the function f for which the Fourier series converges pointwise to the function. In the first section, we provide a brief introduction to the theory of orthogonal functions and to the concept of approximation in the mean. In Section 9.2 we also introduce the notion of a complete sequence of orthogonal functions and show that this is equivalent to convergence in the mean of the sequence of partial sums of the Fourier series to its defining function. The proof of the completeness of the trigonometric system { 1, sin nx, cos nx}' , will be presented in Section 9.4. In this section we also prove Fejdr's theorem on the uniform approximation of a continuous function by the nth partial sum of a trigonometric series. In the final section, we present Dirichlet's contributions to the pointwise convergence problem.

9.1

Orthogonal Functions In this section we provide a brief introduction to orthogonal functions and the question of representing a function by means of a series of orthogonal functions. Although these topics have their origins in the study of partial differential equations and boundary value problems,' they are closely related to concepts normally encountered in the study of vector spaces. If X is a vector space over R (see Definition 7.4.7), a function (,) : X X X -+ R is

an Inner product on X if

(a) (x, x) ? 0 for all x E X, (b) (x, x) = 0 if and only if x = 0, (c) (x, y) (y, x) for all x, y E X, and (d) (ax + by, z) = a(x, z) + b(y, z) for all x, y, z E X and a, b E R. In R", the usual inner product is given by

(a, b) = I, aab1 i=1

f o r a = (a,, ... , a") and b = (b,, ... , b") in R". If () is an inner product on X, then two nonzero vectors x, y E X are orthogonal if (x, y) = 0. The term "orthogonal" is syn-

1. For a detailed treatment of this subject see the texts by Berg and McGregor or Weinberger listed in the Bibliography.

9.1

Orthogonal Functions

381

onymous with "perpendicular;" and comes from geometric considerations in R". Two nonzero vectors a and b in IB" are orthogonal if and only if they are mutually perpendicular; that is, the angle 9 between the two vectors a and b is z or 90° (see Exercise 10, Section 7.4).

In the study of analysis we typically encounter vector spaces whose elements are functions. For example, in previous sections we have shown that the space 12 of square

summable sequences and the space 11[a, b] of continuous real-valued functions on [a, b] are vector spaces over R. With the usual rules of addition and scalar multiplication, 9t[a, b], the set of Riemann integrable functions on (a, b), is also a vector space over R. If for f, g E 9t[a, b] we define

U, g) = f f(x)g(x) dx, b

I

then it is easily shown that (,) satisfies (a), (c), and (d) of the definition of an inner product. It does not, however, satisfy (b). If a < b and ci, . . . , c" are a finite number of points in [a, b], then the function 0,

x # c,,

f(x) = 11 , x=ci, is in 9t[a, b] satisfying (f, f) = 0, but f is not the zero function. Thus technically () is not an inner product on 9t[a, b]; a minor difficulty which can easily be overcome by defining two Riemann integrable functions f and g to be equal if f(x) = g(x) for all x E [a, b] except on a set of measure zero. This will be explored in greater detail in Chapter 10. Alternatively, if we restrict ourselves to the subset b] of 9t[a, b], then (f, g) as defined above is an inner product on b] (Exercise 11).

Orthogonal Functions We now define orthogonality with respect to the above inner product on 9t[a, b].

9.1.1

DEFINITION A finite or countable collection of Riemann integrable functions {¢n} on [a, b] satisfying fa 40.' * 0 is orthogonal on [a, b] if

&(X)4m(X) dx = 0 for all n * m.

\Wn+ m) = 1a

9.1.2

EXAMPLES

(a) For our first. example, we consider the two, functions ¢(x) = I and 4i(x) = x, x E [ -1, 1]. Since f f O(x)*(x) dx = f t X dx = 0, the functions 46 and are orthogonal on the interval [-1, 11.

382

Chapter 9

Orthogonal Functions and Fourier Series

(b) In this example we show that the sequence of functions (sin nx}. , is orthogonal on [-ir, ar]. By the trigonometric identity sin A sin B = 2[cos(A - B) - cos(A + B)],

for n # m, Jsin nx sin mx dx =

[cos(n - m)x - cos(n + m)x] dx

J

2

u

_ i sm(n - m)x 2

_

sin(n + m)x

(n+m)

(n - m)

0.

For future reference, when n = m,

(I - cos 2nx) dx

sine nx dx = 2 -a

X-

sin nXlI'

= Jr.

-p

2\

(c) As our final example, we consider the collection 11, sin ¶, cos T )"., on the interval [ -L, L] where L > 0. As in (b), if n * m, then 1L

sinnlrx L sin m7rx L dx = 0.

L

Thus the collection (sin ¶ ) is orthogonal on [ -L, U. Also, by the trigonometric identities cos A cos B = I [cos(A - B) + cos(A + B)]

sin A cos B = 2[sin(A - B) + sin(A + B)], we have for n # m, (L

f

1 L cos

narx L

mirx

(L

cos L dx = 1

L

sin

nirx L

m7rx dx

cos L

= 0.

Thus the functions in the collection {cos z} are all orthogonal on [-L, L] as are the functions sin T and cos L" for all n, m E N with n * m. For m = n,

f sin nUx cos L

dx=

2n

sin' "'

ILL

= 0.

9.1

Orthogonal Functions

383

This last identity shows that the functions sin Of and cos "LF are also orthogonal on [-L. L] for all n E N. Finally, since

Jsindxr fcos for all n E N, the constant function 1 is orthogonal to sin "lr` and cos " for all n E N. In this example we also have L

f

Lsin

2 nWX

L

2 ri?lX

L

dx = J

Co. .L dx = L.

If in Example 9.1.2(b) we define ¢"(x) _ j sin nx, then the sequence {4"(x)}R , satisfies

fA

4)"(x)¢m(x) dx =

{0, 1,

when n # in. when n = m.

Such a sequence of orthogonal functions is given a special name.

9.1.3

DEFINITION A finite or countable collection of Riemann integrable functions {di"}

is orthonormal on [a, b] if Jb

d."(x)O,"(x) dx = a

10,

1;

when n * m, when n = m.

Given a collection {¢"} of orthogonal functions on (a, b], we can always construct a family {fir"} of orthonormal functions on [a, b] by setting

where c" is defined by b

c. = J ¢^(x) dx. a

Approximation in the Mean Let {O"} be a finite or countable family of orthogonal functions defined on an interval [a, b]. F o r each N E N a n d c, ... , cN E R, consider the Nth partial sum N

SN(x)

_

c4,,(x)

(1)

A natural question is, given a real-valued function f on [a, b], how must the coefficients c" be chosen so that SN gives the best approximation to f on [a, b]? In the Weierstrass

384

Chapter 9

Orthogonal Functions and Fourier Series

approximation theorem we have already encountered one form of approximation; namely, uniform approximation or approximation in the uniform norm. However, for the study of orthogonal functions there is another type of norm approximation that turns out to be more useful. If X is a vector space over R with inner product (,), then there is a natural norm on X associated with this inner product. If for x E X we define

IIxII=

,

II is a norm is left to the exercises (Exercise 12). The crucial step in proving the triangle inequality for II II is the following version of the Cauchy-Schwarz inequality: For all x, y E X.

then II II is a norm on X as defined in Definition 7.4.8. The details that II

I(x,Y)I -s 11x11 11Y11.

The proof of this inequality follows verbatim the proof of Theorem 7.4.3. For the vector space R[a. b] with inner product (f, g) = fa f(x)g(x) dx, the norm of a function f, denoted I I f 112, is given by

111112 = If

f

E &[a, b], the problem to consider is, how must the constants c be chosen in order to minimize the quantity

Ilf- SNII2 = f [f(x) - SN(x)]2dx? a

This type of norm approximation is referred to as approximation in the mean or least squares approximation. The following theorem specifies the choice of (ca) so that SN provides the best approximation to f in the mean.

9.1A THEOREM Let f E gt[a, b] and let

be a finite or countable collection of orthogonal functions on [a, b]. For N E N, let SN be defined by equation (1). Then the quantity b

[f(x) - SN(x)]2 dx Ja

is minimal if and only if fa f(x)-OR(x) dx cn

n = I , 2, ... N.

(2)

fa 02(x) dx

Furthermore, for this choice of c,,, rb

[.f(x) - SN(x)]2 dx =

jf2(x) dx -

N

b

c2 I

eel

a

.0,22(x) dx.

(3)

9.1

Orthogonal Functions

385

Prior to proving the result, we give the following alternative statement of the previous theorem.

9.1.5

COROLLARY Let f E Ii(a, b] and let SN(x) = I:- I caaya(x) where the ca are dean E R, then

fined by equation (2). If TN(x) = 7.N_ 1 b

f

b [f(x) - TN(x)]2 dx,

[f (x) - SN(X)]2 dx

J

f o r any choice o f aa, n = 1, 2, ... , N. For fixed N E N,

Proof of Theorem 9.1.4. b

0 5 J [f(x) - S,,(x)]2 dx a

b

f 2(x) dx - 2 J bf (x)SN (x) dx + J SN(x) dx.

J

a

By linearity of the integral (Theorem 6.2.1), 1bf N (x)SN(x) dx

bf (x)4in(X) dx.

cn

Ja

n°I

a

Also, hS2N(x)dx

=

bSN(x\ 'C4Y'n(x) f dx =

J

/

J

nI

c,J hSN(x)4n(x) dx. a

But b Ja

N

SN(X)wn(X) = I ck I b 0k(X)0.(x) dx, a

which by orthogonality, b

0(x) dX.

Cn

I Therefore, b

N= c . N

jS(x)dx

2

n=1

+(

l Wn(X) dx. JJJ

Upon substituting into equation (4) we obtain

0 G i [f(x) - SN(X)]2 dx b

N C.

bf2(X) Ja

dX - 2

nil

Cn

bf(X)fn(X) 1a

dx + jnil

6 1a

'On(x) dx.

(4)

386

Chapter 9

Orthogonal Functions and Fourier Series

which upon completing the square N

Ia

f 2(X) dx +

-I aJ

b

n(x)

Ic

b

L

f. b

2J

L./ f bo-12 6 2

fa n

n=1

The coefficients cn occur only in the middle term. Since this term is nonnegative, the right side is a minimum if and only if

cn = a

n

With this choice of cn, we also obtain formula (3) upon substitution.

9.1.6

U

EXAMPLE As was previously shown, the functions 01(x) = 1 and ¢2(x) = x are orthogonal on [ -1, 11. Let f (x) = x3 + 1. Then c '

f',f(x)01(x) = l J (x3 + 1) dx = 1. 2 I f!14i(x)dx

and

f_',f(x)d2(x)dx C2 =

0z(x)

=-

(' I3

(x,

+ x)dx =

2 -,

3 .

5

Therefore, S2(x) = I + sx is the best approximation in the mean to f(x) = 1 + x3 on [-I, 1]. The graphs off and S2 are given in Figure 9.1.

-1

-0.5 Figure 9.1

0

0.5

Graphs off and $2

1

9.1

9.1.7

Orthogonal Functions

387

I be a sequence of orthogonal functions on [a, b] and let DEFINITION Let f E %[a, b]. For each n E 1, the number f,,f(x)4n(x) dx

(5)

fa 4.'(x) dx

is called the Fourier coefficient off with respect to the system

I

The series

00

is called the Fourier series off. This is denoted by

=1 00

AX) -- I cn r (x)

(6)

n-1

in the Remark. The notation "--" in formula (6) means only that the coefficients series are given by formula (5). Nothing is implied about convergence of the series'.

9.1.8

EXAMPLE In Example 9.1.2(b) it was shown that the sequence of functions {sin nx},1 is orthogonal on [ -Tr, ir]. Since

n = 1, 2, ..

,

if f E Jt[ -ar, Tr], the Fourier coefficients c, n = 1. 2_ ... , off with respect to the orthogonal system {sin nx} are given by

c = ir-

f (x) sin nx dx,

and the Fourier series off becomes f (X)

=I

c sin nx.

As indicated above, nothing is implied about convergence. Even converge, it need not converge to the function f. Since the terms of the series are odd functions of x, the series, if it converges, defines an odd function on [ -ir, vr]. Thus unless f itself is odd, the series could not converge to f. For example, if f(x) = 1, then _ sinnxdx= - mIr cosnx

-if f

1r 1

n

a

- =0.

In this case, the series converges for all x. but clearly not to f (x) = 1.

Bessel's Inequality For each N E N, let SN(x) denote the Nth partial sum of the Fourier series off, i.e.. N

SN(x) = 71 CIAn(x), n=1

388

Chapter 9

Orthogonal Functions and Fourier Series

where the cn are the Fourier coefficients off with respect to the sequence thogonal functions on [a, b]. Then by identity (3) of Theorem 9.1.4.

f

of or-

b

f2(x)dx-

N r0.

cnJ

Therefore, fb

bf 2(X)dx.

0.2(x)dx <

cn

J

a

a

Since this holds for every N E J. by letting N -+ oo we obtain the following inequality.

9.1.9

THEOREM (Bessel's Inequality) If f E 9t[a, b] and {cn}R , are the Fourier coefficients off with respect to the sequence of orthogonal functions {tbn}". ,, then ao

n=I

,!,

cn fa V'n(x) dx :5

rb f 2(x) dx. J

a

In Example 9.1.8 with f(x) = 1, f % f 2(x) dx = 2rr, and cn = 0 for all n = 1, 2, .... Thus it is clear that equality need not hold in Bessel's inequality. How-

ever, there is one consequence of Theorem 9.1.9 that will prove useful later.

9.1.10

is a sequence of orthogonal functions on [a, b]. If

COROLLARY Suppose

f E 9t[a, b], then lim

n-oo

f f (x)fn(x) dx = 0. a

Ia (6.2(x) dx

Proof.

Since f E 9t[a, b]. fa f2(x) dx is finite. Thus by Bessel's inequality, the series c. f° d, converges. As a consequence, Jim

b

46.(x) dx = 0,

cn

n-+oo

a

and thus,

fboa f(x)On(x) dx = 0. (x)

a

-n

b

faV

2(

n(x) dx

EXERCISES 9.1 1. *Let f(x) = sin rrx, ¢,(x) = 1, and 42(x) = x. Find c, and c2 so that S2(x) = c,4,(x) + c2O2(x) gives the best approximation in the mean to f on [ -1, 1 ].

9.1

389

Orthogonal Functions

2. a. Show that the polynomials Po(x) = 1, PI(x) = x. and P2(x) = 12x2 - 1 are orthogonal on [ -1, l ].

b. Let (0, AX)

-1 <_ x < 0,

0<-xs 1.

1,

Find the constants co, c,, c2, such that S2(x) = cnPo(x) + c^(x) + c2P2(x) gives the best approximation in the

mean to f on [-1, 1]. 3. +a. Let 00(x) = 1, ¢t,(x) = x - a,, 42(x) = x2 - a2x - a3. Determine the constants a,, a2, and a3, so that {40, m,, 42} are orthogonal on [0, I ]. b. Find the polynomial of degree less than or equal to 2 that best approximates f(x) = sin rrx in the mean on

[0,1]. 4. Let {4nr.- t be a sequence of orthogonal functions on [a, b]. For f, g E J1[a, b] with f - Man4n and g show that for a, jS E R,

Ybn4n.

(af + Pg) - 7, (aan + Pb.) .. n" 1

a. Show that the sequence {sin is orthogonal on [0, err]. b. For f E 4t[0, in, show that the Fourier series off with respect to the sequence (sin nx) is given by

S.

m

b. sin nx n=1

where b" = ? f (x) sin nx dx. Ir o

'c. Find the Fourier series of f(x) = x on [0, ir] in terms of the orthogonal sequence {sin nx}. a. Show that the sequence {1, cos nx}' , is orthogonal on [0, ir].

6.

b. For f e %[0, +r], show that the Fourier series off with respect to the sequence {l, cos ax} is given by I

2

,

°O+

ao + !

an cos n x where

2

'

ao = - f (x)dx and a" 7r

nom,

Q

7r

f (x) cos nx d x , n = 1, 2, .. o

'c. Find the Fourier series of f(x) - x on [0, zr] in terms of the orthogonal sequence 11. cos nx}.

7. If f E Jt[0, in], prove that lim fo f (x) sin nx dx = lim fo f (x) cos nx dx = 0. n-"D n-+00

7p

8. Let {4n} be a sequence of orthogonal functions on [a, b]. If the series I an4n(x) converges uniformly to a func-

n"t

tion f(x) on [a, b], prove that for each n e N, a" is the Fourier coefficient of f.

9. Let {an} be a sequence in (0, 1) satisfying 0 < an+, < an < I for all n E N. Define 4" on [0, 1 ] by

05x
0.

an+I

4,(x) =

-2(x - an) an - an+,

,

1(an

x

an+,

an),

+ an)

an<x:5 1.

0,

Show that {4n} is orthogonal in 9t[0, l] and compute II4nII2 for each n E N. 10. Let Po(x) = 1, and for n E N let .

P"(x)

x2)",

x E [-1, 1].

2"n! de (1 The polynomials P. are called the Legendre polynomials on [ -1, 1 ].

390

Chapter 9

Orthogonal Functions and Fourier Series

a. Find P1, P2, and P3. b. Show that the sequence {Pn}moo is orthogonal on [-1, 1 J. (Hint: Use repeated integration by parts.)

11. For f.g E T[a, h] define (f, g) = fa f(x)g(x) dx. Prove that () is an inner product on 12. Let X be a vector space over l with inner product (,). Prove each of the following.

b].

*a. I (x. y) l s 11 x I I 11 Y 11 for all x, y E X. b.

11

11

is a norm on X.

13. Cauchy-Schwarz Inequality: Use the previous exercise to prove that if f, g E 3t[a. b], then J hf (x) g(x) dx 12 < (t"12

g2(x)

(x) d`) ( jah

dx).

14. Let X be a vector space over IIi with inner product Q. If {y,, ... , yn} are nonzero orthogonal vectors in X and + cyn) II is a minimum if and only if x E X. prove that the quantity II x - (c,y, + _ (x. Y;)

`'

IIY;Ii-

for all i = 1.... , n. (Hint: Imitate the proof of Theorem 9.1.4.)

9.2

Completeness and Parseval's Equality In this section we look for necessary and sufficient conditions on the sequence {0n} of orthogonal functions on [a, b] for which equality holds in Bessel's inequality. To accomplish this it will be useful to introduce the notion of convergence in the mean.

9.2.1

DEFINITION A sequence

of Riemann integrable function on [a, b] converges

in the mean to f E 9i[a, b] if h

lim

f

[f(x) -

If we consider cJ&.[a, b] as a normed linear space with norm

f rh 111112 =

f(x)dx

LJ

then convergence in the mean is nothing but convergence in norm as defined in Definition 8.3.9. Thus a sequence in 91t[a, b] converges to f E &[a, b] in the mean if and only if lim 11f - f 112 = 0. Convergence in the mean is sometimes also referred to as mean-square convergence. It is natural to ask how convergence in the mean is related to pointwise or uniform convergence. Our first theorem proves that uniform convergence implies convergence in the mean. As should be expected, pointwise convergence is not sufficient (Exer-

9.2

Completeness and Parseval's Equality

391

cise 2). In the other direction, we will show in Example 9.2.3 that convergence in the mean does not imply pointwise convergence, and thus certainly not uniform convergence. There we construct a sequence (f.1 of Riemann integrable functions on [0, 1 ] such that II f 11 2 -+ 0, but for which { f (x)} fails to converge for any

xE[0,1). 9.2.2

THEOREM I f f , f , , , n = 1, 2, . .. , are Riemann integrable on [a, b], and (f.) converges uniformly to f on [a, b], then converges in the mean to f on [a, b].

Woof.

Since the proof of this result is similar to the proof of Theorem 8.4.1, we leave it as an exercise (Exercise 1).

9.2.3

In this example we construct a sequence I f.1 on [0, 1] that converges to zero in the mean, but for which { f (x)} does not converge for any x E [0, 1 ]. This seEXAMPLE

quence is constructed as follows: For each n E N, write n = 2k + j where k = 0, 1, 2, ... , and 0 s j < 2k For example, 1 = 21+0,2 = 2' +0,3 = 2' + 1, etc. Define ff on [0, 1 ] by

jksxsj+

1,

0, of

1

k

2

2

otherwise.

The first four functions ff, f2, f3, and f4 are given as follows: fi(x) = 1, and

f

0:5 z:S 2,

1,

fi(x)- {0,I, .fa(x) _

0<x<2,

isx<

1,

0 :5

:5

41

0, 1<xsl.

For each n e K f. E 3R[0, 1 ] with

J 0

1 dr = 2k

fn(X) dx = 1

J/2'

Thus lim Jt; f n(x) dx = 0. On the other hand, if x E [0, 1 ], then the sequence { f (x)} n +00an infinite number of 0's and l's, and thus does not converge. contains In the following theorem we prove that convergence in the mean of the partial sums of the Fourier series is equivalent to equality in Bessel's inequality.

392

Chapter 9

9.2.4

Orthogonal Functions and Fourier Series

THEOREM Let {Wn}: I be a sequence of orthogonal functions on [a, b',. Then the following are equivalent: (a) For every f E 9t[a, b],

jb - SN(X)]2 dx = 0,

limo

where SN is the Nth partial sum of the Fourier series off. (b) For every f E 9t[a, b],

x

b

cn 1

n-I

b

4n(x) dx = J f2(x) dx.

(Parseval's Equality)

a

u

where the cn are the Fourier coefficients off. N

Proof. Suppose SN(x) _

n-1

cnd n(x) is the Nth partial sum of the Fourier series off.

Then by Theorem 9.1.4, b

b

Ja

[f (x) - SN(x) ]2 dx =

b

J f2(x) dx -

Cn

n-1

a

J 4' (x) dx. a

From this it follows immediately that {SN} converges in the mean to f if and only if Par-

seval's equality holds. 0

9.2.5

DEFINITION A sequence {¢n}O I of orthogonal functions on [a, b] is said to be complete if for every f E 9t[a, b], ao

I Cn J 0n(x) dx = J f2(x) dx.

n-I

u

a

As a consequence of the previous theorem, the orthogonal sequence {0n} is complete on [a, b] if and only if for every f E 9t(a, b], the sequence {SN} of the partial sums of the Fourier series off converges in the mean to f. We now prove some additional consequences of completeness of an orthogonal sequence.

9.2.6

THEOREM If the sequence {¢n}f j of orthogonal functions on (a, b] is complete, and if f is a continuous real-valued function on [a, b] satisfying b

J f(x)4n(x) dx = 0 for all n = 1, 2, .. a

then f (x) = O for all x e [a, b].

,

9.2

Completeness and Parseval's Equality

393

Proof. The hypothesis implies that the Fourier coefficients c off are zero for all n E N. Thus by Parseval's equality, fb

f2(x) dx = 0.

Ja

Since f2 is continuous and nonnegative, by Exercise 7, Section 6.1, this holds if and only if f 2(x) = 0 for all x E [a, b]. Thus f (x) = 0 for all x E [a, b]. E) There is a converse to Theorem 9.2.6. Since the proof of the converse requires a knowledge of the Lebesgue integral, we only state the result. A sketch of the proof is provided in the miscellaneous exercises (Exercise 3) of Chapter 10.

9.2.7

If {4n}' , is a sequence of orthogonal functions on [a, b] for which the only real-valued continuous function f on [a, b] satisfying THEOREM

b

J f (x)4n(x) dx = 0 for all n = 1, 2, . a

is complete.

is the zero function, then the system

For the orthogonal system {sin nx}M-, on [ -vr, vr] and f (x) = 1, we have c = 0 for all n = 1, 2, .... Thus as a consequence of Theorem 9.2.6, the orthogonal system {sin nx} is not complete on [ - vr, vr]. However, as we will see in Exercise 3 of Section 9.4, this system will be complete on [0, vr]. Another consequence of completeness is the following: Suppose {46,} is complete on [a, b] and f, g are continuous real-valued functions on [a, b] satisfying Jbf(x)On(x) dx =

Jg(x)4a(x) dx a

a

for all n = 1, 2, ... , then f(x) = g(x) for all x E [a, b]. The above assumption simply means that f and g have the same Fourier coefficients. To prove the result, apply Theorem 9.2.6 to h(x) = f(x) - g(x).

9.2.8 THEOREM Suppose f (x)

is complete on [a, b], f, g E °k[a, b] with 7, C.-On(x)

n-t

and

g(x)

Y, b,,4 (x) n-I

Then b

f Proof. Exercise 5. 0

b

oo

f(x)g(x) dx = 7, cnbn J n-I

a

46.1(x) dx.

394

Orthogonal Functions and Fourier Series

Chapter 9

El EXERCISES 9.2 1. Prove Theorem 9.2.2.

2. For n E f0i, define the function f on [0, I ] by

f() =

v 0<x< !, n

elsewhere.

0,

converges to 0 pointwise but not in the mean.

Show that

3. Consider the orthogonal system

I

1, cos

nnx L

mrr '°

, sin

L

_, on [ -L, L].

a. Show that if f E 9t[ -L, L], then the Fourier series off with respect to the above orthogonal system is given by

a cosnrx L+b

I

f(x)

2 ao +

nirx

L

a,, =

L

J f(x)cos nLx dx, n = 0. 1, 2, .... and b _

f(x)sin'

J

dx, n = 1.2.... .

L

L

b. Show that Bessel's inequality becomes L

a-', +

1

(a.' + b;) s 1

f2(x) dx. L

2

4. *a. Assuming that the orthogonal system {sin nx} is complete on [0, in, show that Parseval's inequality becomes

Jnf(x)sinnxdx. b2 _ 1 f 'f 2(x) dx, where b = ,,=

0

*b. Use Parseval's equality and the indicated function to find the sum of the given series.

(1) 7,

I

(2k

1)2'

f(x) = I

(ii) kI

ji,

f(x) = x

S. *Prove Theorem 9.2.8.

6. *Show by example that continuity off is required in Theorem 9.2.6.

9.3 1

Trigonometric and Fourier Series In Section 9.1 we introduced Fourier series with respect to any system of orthogonal functions on [a, b]. In this section we will emphasize the trigonometric system 1, cos

nlrx, L

--

sin

nTT x

L

_,'

9.3

Trigonometric and Fourier Series

395

which by Example 9.1.2(c) is orthogonal on [-L, L]. For convenience we will take L = Tr. Any series of the form

2Ao + I (A cos nx + B. sin nx), n=1

where the A. and B,, are real numbers, is called a trigonometric series. For example, the series 00 sin nx

and

..2 In n

I

cos nx

MMI

are both examples of trigonometric series. Since the coefficients and

are nonnegative and decrease to zero, by Theorem 7.2.6 the first series converges for all

x E R, whereas the second converges for all x E R except x = 2pir, p E Z.

Fourier Series For the orthogonal system {1, cos nx, sin nx} si on [-1r, 7r] we have 4

ff

sin 2 nx dx =

-v

n

f-n

cost nxdx = ir,

and

I

12 dx = 21r.

1 tr

Thus by Definition 9.1.7, the Fourier coefficients of a function f E 9t[ -a, ir] with respect to the orthogonal system are defined as follows.

9.3.1

DEFINITION

Let f E R[-7r, 1r]. The Fourier coefficients off with respect to the

orthogonal system (1, cos nx, sin nx} are defined by

Jf(x) dr, -ir

ao =

an

b b.

"

1

,f (x) cos nx dx,

"

1

J

f (x) sin nx dx,

n = 1, 2-

n = 1, 2, ... .

n

Also, the Fourier series off is given by 1

AX) - 2ao +

°O

,

(a cos nx + b sin nx).

3%

Chapter 9

Orthogonal Functions and Fourier Series

Remark. For the constant function 0o = 1, since f T, 020 = 21r, the term ao should be defined as za f', f (x) dx, according to Definition 9.1.7. However, for notational convenience it is easier to define ao as J°a f(x) dx and to include the constant 1 in the definition of the Fourier series. For the orthogonal system ( 1, cos nx, sin nx), by Exercise 3(b) of the previous section, Bessel's inequality 9.1.9 becomes 00

I ao +

-

(a + b2) <-L

n-1

a

-a

V(x))2 dx

(Bessel's Inequality)

Thus for f E &[a, b] the sequences {a"} and {b,,} of Fourier coefficients off are square summable sequences. In Theorem 10.8.7 we will prove that if {an} and {b"} are square summable sequences, then there exists a Lebesgue integrable function f such that {a"} and {b"} are the Fourier coefficients off.

Remark on Notation. In subsequent sections we will primarily be interested in realvalued functions defined on all of N that are periodic of period 27r. As a consequence, rather than defining our functions on [ -vr, vr], in the examples and exercises, we only

define the function f on [-vr, vr) with the convention that f(vr) = f(-r). This allows us to extend f to all of R as a 2vr periodic function according to the following definition.

9.3.2

DEFINITION

9.3.3

EXAMPLES

For a real-valued function f defined on( - ir, vr), the periodic extension (of period 21r) off to R is obtained by defining f (x) = f(x - 2kir). where k E 1 is such that x - 2kvr E [-vr, vr).

-trsx<0, tf0, 0 <_x
(a) Let f (x) Then

ao=-faf(x)dx=1aldx= 11

and for n = 1, 2, .. a"

_

b" =

1

I

1yr

a

1,

0

,

fir cos nx dx =

n 1

sin nx dx = I

0

= 0,

sin nx

0

cos nx -I nvr

Iff 0 IT

0

I

-[1 - cosnvr] = n[1 - (-1)"]. n7r

In the above we have used the fact that cos nvr = (-1)". Thus the Fourier series off is given by

f(x) ^-

1

2

00

+ n

-l

na[1 - (-1)"]sinnx = 2 + yr T 2k + 1 sin(2k - I)x. k=0

9.3 Trigonometric and Fourier Series

397

If SN(x) denotes the Nth partial sum of the Fourier series, then S1 and S., are given by S1(x) =

2

+ 2 sin x

SAX) = 2 + 2 [sin x + 3 sin 3x ir

and

7r

The graphs of f, S1. S3, S5, and S15 are given in Figure 9.2. S1

Figure 9.2

S3 S5 S15

Graphs off, S1. S. S5, S15

(b) Let f (x) = x. Before we compute the Fourier coefficients, we will make several observations that simplify this task. Recall that a function g(x) is even on [-a, a] if

g(-x) = g(x) for all x, and g(x) is odd if g(-x) = -g(x) for all x. By Exercise 4 of Section 6.2, if g(x) is even on (-a, a], then fa

g(x)dx = 21 ag(x)dx, a

0

whereas if g(x) is odd, a

! -a

g(x) dx = 0.

The functions sin nx are all odd, whereas cos nx are even for all n. Therefore, since

f (x) = x is odd, x cos nx is odd and x sin nx is even. Thus. a" = 0 for all n = 0,

1,2,...,and

b=

2

f x sin nx dx, a

which by an integration by parts

_

2

I n cosnxlr+ -IAcosnxdx n L

a

0

=-ncosnar=n(-1)"+i

398

Chapter 9

Orthogonal Functions and Fourier Series Therefore, x, (-1)n+1

x -- 2

It

n=1

sin nx.

Riemann-Lebesgue Lemma There is one additional result from the general theory that will be needed later. For the orthogonal system { 1, cos nx, sin nx}, Corollary 9.1.10 is as follows.

93.4 THEOREM /f f E lim

n- 00

then

lim jf(x) cos nxdx = n-x

Jf(x) sin nxdx = 0.

Theorem 9.3.4 is commonly referred to as the Riemann-Lebesgue lemma. The following example shows that integrability of the function f is required.

9.3.5

EXAMPLE

Let f be defined on [-1T, Ir) as follows:

f(x)

0,

-irSx50,

1,

0 < x < vr.

Ix Then

(°

A sin nx

f (x) sin nxdr =

dx =

o

A

na sin x x o

Hence,

lim J-"" f(x) sin nxdx = n-row lim

n-.00

f n" sinxdx = 0

fO0sinx dx x 0

2

Is a Trigonometric Series a Fourier Series? Since every Fourier series is a trigonometric series, an obvious question to ask is whether every trigonometric series is a Fourier series. More specifically, given a trigonometric series 00

2Ao + 7, (An cos nx + B. sin nx), n-1

2. The value of a/2 for the improper integral is most easily obtained by contour integration and the theory of residues of complex analysis. A real variables approach that computes the value of this integral is given in the article by K. S. Williams listed in the supplemental readings.

9.3

Trigonometric and Fourier Series

399

converging to zero, does there exist a function f on [ -nr, a; such with and that the coefficients A,, and B,, are given by Definition 9.3.1? As we will see, the answer is no! First, however, in the positive direction, we prove the following.

9.3.6

THEOREM If the trigonometric series ;Ao + 7-(A cos nx + B. sin nx) converges uniformly on [ -ir, a], then it is the Fourier series of a continuous real-valued function on [ -?r, Tr].

Proof. For n E Nl, let S (x) = 1 Ao +

(Ak cos kx + Bk sin kx).

2

k-l

Since the series converges uniformly on [ -ir,1r], and S. is continuous for each n,

f (X) = urn S (x) is a continuous function on [-ir, vr]. Form E N, consider

Jf(x) cos mxdx = L1( n-+w lim S (x)) cos mxdx = lim

A

cos mxdx.

J

Since for each m, the sequence {S (x) cos mx} converges uniformly to f(x) cos mx on [-ir, 1r], the above interchange of limits and integration is valid by Theorem 8.4.1. If

n > m, then fSn(x)cosmx = 2Ao J

Ak r cos kx cos mxdx

cos mxdx + k-1

i,

a

n

+

Bk

sin kx cos mxdx.

k-1

Thus by orthogonality, S (x) cos mxdx = AmIr.

Letting n -> oo gives

A. =

91

r

f (X) cos mx dx.

The analogous formula also holds for Bm, and thus the given series is the Fourier series of f.

Remarks (a) If 2 I Ak I and E I Bk I both converge, then by the Weierstrass M-test, the series 00

2 Ao + kY, (AA cos kx + Bk sin kx)

400

Chapter9

Orthogonal Functions and Fourier Series

converges uniformly on R, and thus is the Fourier series of a continuous function on [ -ir, ir]. Convergence of the series I IAk I and 7, I Bk I is not necessary, however, for uniform convergence of the trigonometric series. For example, the trigonometric series

1 sinn Innxn °`

converges uniformly on 11 (Exercise 12), yet x

1

= 00.

=2n In n

(b) In 1903, Lebesgue proved the following stronger version of Theorem 9.3.6: If AX) = 2Ao + l ° t(Ak cos kx + Bk sin kx) for all x E (-1r, a), and if f is continuous (in fact, measurable), then Ak and Bk are the Fourier coefficients of the function f.3 (See also Miscellaneous Exercise 1.)

We now turn to the negative results. Consider the series

0 sin nx a2 Inn

(n

which by Theorem 7.2.6, converges for all x E R. However, there is no Riemann integrable function f on [ -7r, or] such that

Inn =

b = If * f(x) sin nxdx. n

If such a function f exists, then by Bessel's inequality we would have °,

00

n=2

hi

=2

(Inn)'

'

f2(x)dx. n

jr, so is f2, and thus the integral is finite. But since f is Riemann integrable on [-jr], On the other hand, °C

n-2 (ln n)2

00,

which gives a contradiction.

The above argument only shows that the series (7) is not obtained by means of Definition 9.3.1 from a Riemann integrable function. There remains, however, the ques-

tion of whether this is still the case if we extend our definition to allow the class of Lebesgue integrable functions to be introduced in Chapter 10. As we will see in Section 10.8, the answer to this is still no! 3. "Sur les series trigonometric:' Annales Scienriftques de I'Ecole Normale Supfrieure. (3) 20 (1903). 453-485.

9.3 Trigonometric and Fourier Series

401

Fourier Sine and Cosine Series We close this section with a brief discussion of Fourier sine and cosine series. As we have seen in Exercise 5 of Section 9.1, the sequence {sin nx}" I is orthogonal on [0, Tr]. Also, by Exercise 6 of Section 9.1, the same is true of the sequence {1, cos nx} -1. For f E 9t[0, in, the Fourier series with respect to each of these two orthogonal systems are called the Fourier sine and cosine series off respectively. Since A

sin2 nx dx =

I

cost nx dx =

IT

0

0

the formulas of Definition 9.1.7 give the following.

9.3.7

For f E 9t[0, a], the Fourier sine series off is given by

DEFINITION

00

7, b sin nx,

f(x)

n-I

where

rA

bn = ir J f (x) sin nxdx, n = 1, 2, .. 0

are the Fourier sine coefficients of f Similarly, the Fourier cosine series of f E 9t[0, 7r] is given by

AX) -

1

a0 +

00

a, cos nx,

n= where 110 =

A

2

f(x) dx

and

0

2

a=a

f(x) cos nxdx,

n = 1, 2, ..

0

are the Fourier cosine coefficients off. There is a simple connection between Fourier series and the Fourier sine and co-

sine series. As in Example 9.3.3(b), we first note that if f is an even function on [-ar, vr], then by Exercise 1, 1

AX)

2

00

a0 + I a cos nx, where a _

jf(x)cosnxdx.

n-l

0

Similarly, if f is an odd function on [ -Tr, ir;, then 00

f (x) -r

b,, sin nx, n=1

where b =

A

2 77

I

0

f (x) sin nxdx.

402

Chapter 9

Orthogonal Functions and Fourier Series

depend only on the values of the function f on and Thus the coefficients [0, ar]. Conversely, given a function f on [0, ir), the following definition extends f both as an even and an odd function to the interval (-i, ir).

9.3.8

DEFINITION

Let f be a real-valued function defined on [0, ir). The even extension of

f to (-it, it), denoted f is the function defined by fe(x) =

f(-x). -a < x < 0, 0 s x < V.

f (x),

Similarly, the odd extension off to (-a, a), denoted f0, is the function defined by

1-f(-X). -T < x < 0, f,(x) =

x = 0, 0 < x < V.

0,

P X),

It is easily seen that the functions f,(x) and fo(x) are even and odd respectively on (-1r, it), and agree with the given function f on (0, ar). For the function f given by the graph in Figure 9.3, the graph of the even and odd extensions off are given in Figures

9.4(a) and (b) respectively. From the above discussion it is easily seen that if f E 9t[0, ir], then the Fourier sine series off is equal to the Fourier series of f and the Fourier cosine series off is equal to the Fourier series of f, (Exercise 2).

A

Figure 9.3

A

-A

Figure 9.4

(a) Even Extension off

(b) Odd Extension off

9.3 Trigonometric and Fourier Series

403

Remark, In the definition of the even and odd extension we only assumed that f was defined on [0, a), and then defined f, and f, on (-a, Ir). If Air-) = xlim_ f(x) exists, then for the even extension we define f,(- v) = f(zr-). Thus f, is now defined on [-ir, ir) and hence can be extended to all of TlR as a 2v-periodic function. For the

odd extension f we set f0(-w) = f(ir-), thereby defining f on [-1r, a).

EXERCISES 9.3 1. Let f E 9t(-a, ar). Prove the following. 'a. If f is even on [ -a, in, then the Fourier series off is given by

fv f (x) cos nxdx, it = 0, 1, 2, .

Ax) ^- 1 as + Y , a,, cos nx, where a

n-I 2 o b. If f is odd on [ -7, a], then the Fourier series off is given by OD

f (x) ^- 7, by sin nx,

where b _ ?

f (x) sin nxdx,

n = 1, 2, ..

N-1

2. Let f E 9t[0,ir]. Prove the following. a. The Fourier sine series off is equal to the Fourier series of the odd extension f, off. b. The Fourier cosine series off is equal to the Fourier series of the even extension f, off. 3. Find the Fourier series of each of the following functions f on [ -a, 1r):

a f(x)-{

1, 1,

-ar s x < 0,

0,

b f(x)-{x,

0sx
'c. f(x) = lx(

-IT

X :S 0,

0<x<ar

d. f(x) = x2

e. f(x) = I + x

0,

-rr <_ x < -2,

1,

0

0,

2

x< x<

2, in

4. On the interval [-2A, 2Tr) sketch the graph of the 2w-periodic extension of each of the functions in Exercise 3. S. Find the Fourier sine and cosine series on [0, ar) of each of the following functions.

b. f(x) = x

'a. f(x) = I

c. f(x) = x2

'd. f(x) _ {0'

0 is 2

f(x)-{-1, 0sx<2, 1,

25 x
6. Let h be defined on [0, in) as follows: cx,

h(X) =

OsxsT2-

0 -4 2<x
where c > 0 is a constant.

f. f(x) = e-'

x
404

Chapter 9

Orthogonal Functions and Fourier Series

a. Sketch the graph of h on [0, or). b. Sketch the graph of the even extension h, on (- 7r, 9r).

c. Find the Fourier series of h, d. Sketch the graph of the odd extension h on (- IT. 7r). e. Find the Fourier series of h,,. 7. a. Find the Fourier series of f (x) = sin x on [ - a, a]. b. Find the Fourier cosine series of f(x) = sin x on [0, zr).

8. *Suppose f, f are continuous on [-ir, Ir]. and f" E 91[ -7r, or]. Also, suppose that f(-v) = f(ir) and f '(-vr) = f'(ar). Prove that the Fourier series off converges uniformly to f on [ -or, zr]. (Hint: Apply integration by parts to show that Jail and Ib,l are less than C kfor some positive constant C.) 9. Show that the trigonometric series

sinnx n=i

n

converges for all x E IR but is not the Fourier series of a Riemann integrable function on [ -ir, zr]. 10. If f is absolutely integrable on [ -it, zr] (see Section 6.4), prove that n

n

lint J

f(x) cos nxdx = lim

f(x) sin nxdx = 0. J

n

11. Using the previous exercise, show that each of the following hold.

a. lim J In x sin nxdx = nix lim n-,x I n

b. lim J n xa sin nxdx = lim n-.oc 0

In x cos nxdx = 0 1o

x° cos nxdx = 0 for all a > -1 10

12. Suppose an ? 0 for all n E N, and {nan} is monotone decreasing with lim nan = 0. Prove that J. ,an sin nx converges uniformly. (Hint: Write I an sin nx as I (nan) (" n') and use the fact that the partial sums of ";"' are uniformly bounded. See also Exercise 15 of Section 8.2.)

9.4

Convergence in the Mean of Fourier Series Let f be a real-valued function defined on [ -1r, ir), and extend f to all of R to be periodic of period 21T. Throughout this section we will assume that f e 9t[-v, ir]. Let

AX) ^ -ao + 2

(an cos nx + bn sin nx) n=1

be the Fourier Series off, and for each n E N, let n

S,(x) =

2

ao +

(ak cos kx + b* sin kx)

be the nth partial sum of the Fourier series.

9.4 Convergence in the Mean of Fourier Series

405

Our goal in this section will be to prove that if f E Jt[-a, irJ, then the sequence converges in the mean to the function f on [ - ir, or]. By Theorem 9.2.4, this is equivalent to completeness of the trigonometric system (1, cos nx, sin nx}. To investigate mean-square convergence, and also pointwise convergence in the next section, it will be useful to obtain an integral expression for S..

The Dirichlet Kernel 9.4.1.

THEOREM Let f E 9t[-Tr, or]. Then for each n E N and x E R, Sn(x) - V

JA

f(t)D,.(x - t) dt.

A

where D. is the Dirichlet kernel, given by

sin (n + j)t 2

t#2pir,pEZ,

2sin Z

D (t) = 1 + 7, Cos kt =

n+2,

k-t

1

t=2pir.

Proof. By the definition of the Fourier coefficients ak and bk (Definition 9.3.1), S (x) = 1ao + 7, (ak cos kx + bk sin kx)

k-t

2

f

2 f (r) dt + ' (-'-j"f(r) cos kt dt) cos kx

',

+ k7, (1 .1

f(t) sin kt dt sin kx

ir J

It

jA At) 12 +

=

'r j

A

1

1

"

2

k-1

f (t) - + A

cos k(x - t) J dt.

In the last step we have used the trigonometric identity

cos k(x - t) = cos kx cos kt + sin kx sin ki.

Set D (s) = z + eke I cos ks. Then by the above, t) dt.

J A

406

Chapter 9

Orthogonal Functions and Fourier Series

To conclude the proof, it remains to derive the formula for

If s = 2pir, p E Z.

then

2

ka,

Y'I=n+1. 2

By identity (2) of Theorem 7.2.6, for s # 2pa,

1 + sin(n + j)s - sin Z _ sin(n + Z)s 2 sin 2

2

which establishes the result.

2 sin 2

L3

If the sequence I were an approximate identity, convergence results would follow easily from Theorem 8.6.5. Unfortunately, the functions D are neither nonnegative nor do they satisfy (b) of Definition 8.6.4. They do, however, still satisfy

I

JDn(t) dt = 1,

(8)

A

a fact which will prove useful later. To see that equation (8) holds, it suffices to integrate the sum defining term by term. The only nonzero term will be the integral involving Z, and thus

Ltt=Lt= The graphs of D,, D3, and D5 are illustrated in Figure 9.5.

4T

Figure 9.5 Graphs of D,, D3, DS

.

9.4 Convergence in the Mean of Fourier Series

407

The Fejer Kernel To prove mean-square convergence, it turns out to be more useful to first consider the arithmetic means of the partial sums S. For each n E N. set

n+1

o"(x) =

By Exercise 14 of Section 2.2, if limS"(x) = S(x), then we also have lim o"(x) = S(x).

However, it is possible that the sequence {S"(x)} diverges for a particularx, whereas the sequence {o"(x)} may converge.

9.4.2

LEMMA

For n E N, let S"(x) = iao + Y,k.1(ak cos kx + bk sin kx). Then

(a) o"(x) =

Zao

+ J71 ( 1 - n + 1) (aj cos jx + bj sin jx)

and A

(b)

J if

[1x) - S"(x)]2 dx <

j

[1(x) - on(x)]2 dx.

Proof. The proof of (a) is left as an exercise (Exercise 1). Since o" is itself the partial sum of a trigonometric series, the result (b) follows by Corollary 9.1.5. Our next step is to express o"(x) as an integral analogous to that of S"(x) in the previous theorem.

9A.3 THEOREM Let f E Jt[-ir, zr]. Then (for each n E N and x E R,

o"(x) _ ! J f(t)F"(X - t) dt, where F. is the Fejer kernel, given by

F"(t) =

I

r pk(t) n+lk,.o

sin(n + I )L'

1

2(n + 1)[

sin 2

n+1 2

2

2p7r,

t# t - 2pzr.

'

408

Chapter 9

Orthogonal Functions and Fourier Series

The graphs of F1, F3, and F5 are illustrated in Figure 9.6. 3

I

Figure 9.6

Graphs of F,. F3. Fs

Proof. By Theorem 9.4.1, fA

s) ds,

where D. is the Dirichlet kernel. Therefore, f(s)Fn(x - s) ds, where

7, Dk(t).

1

n+ 1 k=0

If t = 2pir, p E Z, then Dk(2pvr) = k + , and thus

F 2or) = "(P

n+I2) =(n+1) 1

A0( k +

t

2

If t * 2p7r, p E Z, then 1

IT

FF(t)

k+ t = 2(n + 1) sin 71 sin 2

2(n + 1) sine 2' k=0

sin

sin + k +

2

2)t.

9.4 Convergence in the Mean of Fourier Series

409

By the identity sin A sin B = Z(cos(B - A) - cos(B + A)],

r

r

k-0

sin -sin k +

i2t

2

1

t=-

2k-0

(cos kt - cos(k + 1)t)

(1 - cos(n + 1)t) = sin2(n + 1) 2

=

2 Therefore, for t * 2pir, sin(n + 1)_

I

2(n + 1)L

sin 2

2

O

We now prove the following properties of the Fejer kernel.

9.4.4 THEOREM (a) F. is periodic of period 21r with F (t) i? 0 for all t. (c)

dt = 1.

IV

1

0 uniformly for all t, 8 s 1 tt c ir.

(d) For 0 < 8 < ir, 1im Proof. (a) Clearly sin{ (n

1)

Also, since

(t + 2zr)) = sin( (n + 1) 2) cos(n + 1)ir = (- l)"'sin(n + 1) 2,

2 and sin 12(t + 21r) = -sin 2, substituting into the formula for F. gives F(t). Therefore, F. is periodic of period 21r. The proof of (b) is obvious. (c) Since f DD(t) dt = 1, we have A

n

if F"(t) dt = tr

n + 1 ka0 ;

Dk(t) dt = 1.

To prove (d), we first note that sine 2 --- sine I for all t, 0 < 8

Itt

ir. Also,

since Isin(n + 1);I s 1 for all t,

F (t) <

1

2a

2(n + 1) sin 1

for all t, 8 :s Itt

Therefore, for any S, 0 < 8 < -tr, 1im F (t) = 0 uniformly on 6 s 1 tl <_ ir

410

Chapter 9

Orthogonal Functions and Fouler Series

As a consequence of (b), (c), and (d), the sequence

FF(t)},°_, is an approximate

identity on [- or, or]. If f is a real-valued function on R that is periodic of period 2or, then

on(X) _ -

t)dt,

I

J n

which by the change of variable s = t - x,

=I

I

w-zf(s + x)FF(s) ds.

Since the function s -+ f(s + x)F (s) is periodic of period 21r, by Theorem 8.6.3 f (s + x)FF(s) ds.

on(x) I

n

Thus if f is continuous on [-or, a] with f(-1r) = f(or), by Theorem 8.6.5,

limo,(x) = f(x) uniformly on [ -7r, 1T). We summarize this in the following theorem of L. Fejer (1880-1959).

9.4.5

THEOREM (Fej4r) If f is a continuous real-valued function on [-or, or] with

f (-or) = f (7r), then f (x)

uniformly on [ -ot, or].

9A.6

COROLLARY If f is a continuous real-valued function on[-1r, or] with f(-v) = f(or), then n lim

[f(x) - SS(x)]Z dx = 0.

_n

Proof. The proof of the Corollary is an immediate consequence of Lemma 9.4.2(b).

Remark. There is a similarity between Theorem 9.4.5 and the Weierstrass approximation theorem. An alternative way of expressing. Theorem 9.4.5 is as follows: Let! be

9.4 Convergence in the Mean of Fourier Series

411

continuous on [ -vr, vr] with f(-v) = f(ir). Given e > 0, there exists a trigonometric polynomial n

T (x) _

Ao + 2

k-I

(Ak cos kx + Bk sin kr)

such that

I1(x) for all x E

vr, vr]. The function complex form it can be rewritten as

E

is called a trigonometric polynomial since in

T.(x) = L;

cke'kx,

k--

where by De Moivre's formula e' = cos kx + i sin kx.

Convergence in the Mean Corollary 9.4.6 only proves convergence in the mean for the case where f is a continuous real-valued function on [ -vr, n]. We ate now ready to prove the main result of this section.

9A.7 THEOREM /f f E 9t[-vr, vr], then Jim f [f(x) - S,(x)]2 dx = 0, where S. is the nth partial sum of the Fourier series off. For the proof of the theorem we require the following lemma.

9A.8

LEMMA, Let f E 3t[a, b] with I f(x)I s M for all x E [a, b). Then given e > 0, there exists a continuous function g on [a, b] with g(a) = g(b) and I g(x)I <- M for all

x E [a, b], such that

f

b

If(x)

g(x)I dx < e.

a

Before proving the lemma, we will first use the result to prove Theorem 9.4.7, then consider some consequences of this. theorem.

Proof of Theorem 9.4.7. Suppose I f(x)I <_ M with M > 0, and let e > 0 be given. By the lemma, there exists a continuous function g on [-n. vr] with g(-vr) = g(vr) and Ig(x)I 5 M for all x E [-vr, vr] such that A

f

If(x) - g(x)I dx < A

80

(9)

412

Chapter 9

Orthogonal Functions and Fourier Series

Let S (g)(x) be the function defined by

S0(gxx) =

aJ

g(t)D0(x - t) dt,

where D. is the Dirichlet kernel. Then since g is continuous, by Corollary 9.4.6 the sequence {S (g)} converges in the mean to g on [ - ir, it ]. Thus there exists n,, E N such that J

for all n

A [g(x) - S0(g)(x)]2 dx < 4

no. Consider f ,[f(x) - S (g)(x)]2 dx. Since

f(x) - S0(gxx)l c I1(x)

g(x)l + 1g(X) - S.(g)(x)l,

we have If(x) - S0(g)(x)12 -_ 2[jf(x) - g(X)12 + lg(x) -

But by inequality (9),

11(x)-g(x)I2dx52MIf(x)-g(x)Idx<4 Therefore, for all n ? n,,, [f(x) - Sn(g)(x)]2 dx < 2 + 2

A

[g(x) - S0(g)(x)]2 dx < E.

1

However, for each n E N, S (g) is the nth partial sum of a trigonometric series. Thus if S is the nth partial sum of the Fourier series off, by Corollary 9.1.5

[f(x) - S0(x)]2 dx s 1 A [.f(x) - S0(gxx)}2 dx. Thus, given E > 0, there exists an integer n0, such that

[f(x) -

E

for all n > n,; i.e., {S.) converges in the mean to f on [ -rr,1r).

0

As a consequence of Theorem 9.2.4 and Theorem 9.4.7 we have the following corollary.

9.4.9

COROLLARY (Parseval's Equality)

1f f E %(-ir, ir], then

00

2 ao +

(a,2, + b.) I,

I

fr f 2(X) dx. A

9.4 Convergence in the Mean of Fourier Series

9.4.10

413

EXAMPLE By Example 9.3.3(b), the Fourier series of f(x) = x is given by

- 2(-1)".i X_

sin nx. n

Here an = 0 for all n, and b = 2(-1 r "/n. Thus by Parseval's equality, 00

4

2 _n X2dx=-a2,

n2

3

1

which gives

= .,

1r2

6

_n2

0

Proof of Lemma 9.4.8. Let f E Jt[a, b] with if (x)I s M for all x E [a, b]. Since f,(x) = f(x) + M is nonnegative, if we can prove the result for the function f,, the result also follows for the function f. Thus we can assume that f satisfies 0 <- f(x) s M

for all xE[a,b]. Let e > 0 be given. Since f is Riemann integrable on [a, b], there exists a partition of [a, b] such that

_ {xo, x,, ... ,

0s

b jf(x) dx - f(9,f) < 2'

where m;Ox;,

and m; = inf { f(t) : t E [x;_,, x;]}, i = 1 , 2, ... , n. For each i = 1, 2... , n, define the functions h; on [a, b] by

hi(x) =

mi,

x..1 <_ x < X"

0,

elsewhere,

and let h(x) =

"_, h,{x) (see Figure 9.7). The function h is called a step function on [a, b]. Since his continuous except at a finite number of points, h is Riemann integrable and rb

h(x) dx

n

m;Ox; = Y(J', f).

J Therefore,

0s I[f(x)-h(x)]dx<2. b

a

414

Chapter9

Orthogonal Functions and Fourier Series

h

m3 + M5 t M4 -4-

M2 t mi t

0

mb + I

I

X1

X2

I

xp=a

I

X3

i

X4.

I

I

X5

x6=b

Figure 9.7 Graph of the Step Function h

Also, since m; = inf{f(t) : t E [x;_ 1, x;]}, 0:5 h(x) s f(x) for all x E [a, b]. By tatting slightly shorter intervals and connecting the endpoints with straight line segments. we leave it as an exercise (Exercise 9) to show that there exists a continuous function g on [a, b] with g(a) = g(b) = 0, 0 s g(x) h(x), such that Jb

[h(x) - g(x)]dx < 2. a

Then 0 is g(x) s f(x), and b

b

0

J

b

[f(x) - g(x)] dx = J [f(x) - h(x)] dx + J [h(x) - g(x)] dx a

a

a


2. *Using the Fourier series of f(x) = x2 and Parseval's equality, find

1/n4.

3. *Prove that the orthogonal systems {sin nx} -I and {1, cos nx}; , are both complete on [0, a]. sin x, sin i, ...} is complete on [0, ir]. 4. Show that the set {sin 5. Let f E 9t[ -ar, ir], and let S. denote the nth partial sum of the Fourier series of f. a. Using the Cauchy-Schwarz inequality for integrals (Exercise 13, Section 9.1), prove that JIR

(Sa(x) - f(x))dxl

29rt

Jp IS.(x) - f(x)I3dx

9.5

Pointwise Convergence of Fourier Series

415

b. Prove that A

lim

(!I

S.(x)dr =

f(x)dx. J

6. *Use the previous exercise to prove the following: Suppose f E 1t[ -7r, a] with

AX) - ao +

(a cos nx + b sin nx).

For x E [-7r, ar], set F(x) = f

af

(t) dt - 1

Then F is continuous on [

v. ar] with F(-7r) = F(n). and

for all x E [-ir, ir], F(x) =

2

a,ir +

(-fl sin nx - b" (cos nx - cos nIr)). n

n

where the convergence is uniform on [-a, Tr]. 7. By integrating the Fourier series of f (x) = x (Example 9.3.3(b)), find the Fourier series of g(x) = x2. 8. Suppose f is a real-valued periodic function of period 2a on R with f E 1t[ -n, 1r], and .r E R is such Z(f(xo-) + f(xo+)]. both exist. Prove that limo that f(x,-) and

9. Complete the proof of Lemma 9.4.8; namely, given the step function h on (a. b] and e > 0. prove that there exists a continuous function g on [a, b] with 0 < g(x) <- h(x), g(a) = g(b) = 0, and fa [h(x) - g(x)] dx < e. 10. "Let f E &(a, b]. Given e > 0. prove that there exists a polynomial p(x) such that fa If(x) - p(x) I dx < e.

9.

I Pointwise Convergence of Fourier Series We now turn our attention to the question of pointwise convergence of a Fourier series to its defining function. It is known that even if the function is continuous, this is not always possible; additional hypothesis on the function f will be required. As was indicated in the introduction, Dirichlet was the first to find sufficient conditions on the function f so that the Fourier series off converges to f. For the proof of his result we need several preliminary lemmas.

9.5.1

LEMMA Let f be a periodic real-valued function on P of period 2Tr with f E R[ -Tr, 7r]. Fix x E R. Then for any real number A,

f(x - t)

v[f(x + t)

A=2J 0

2

-

1

d4

where D. is the Dirichlet kernel and S. is the nth partial sum of the Fourier series off.

Proof.

By Theorem 9.4.1, S. (x)

J_/(tfl(X - t) dt,

416

Chapter 9

Orthogonal Functions and Fourier Series

which by the change of variables = x - t. (-A

+A

ds =

- f (x - s)D.(s) ds.

1

x+A

A

Since both f and D are periodic of period 2a, by Theorem 8.6.3 I

x4A

7r

[

A

1

- f (x - s)D.(s) ds =

J

V

A

J f (x - s)D.(s) ds. A

Therefore,

f (x - s)D(s) ds

S (x) _ J 0

In the first integral set u = -s. Then since Jo

D (u), 1

f(x -

ds = -- J of(x

A

+ u)DD(u) du

A

=

1IT

fn

f(x + s)D.(s) ds. 0

Therefore,

S,(x) _ i1 A [f(x + s) + f(x - s))D.(s) ds. 0

Finally, since fo

ds.= 1,

Z (1f(x+s) f(x- s) 2

Jo

which is the desired identity. U 9.5.2

LEMMA

Ifg E

0Jt[0, ir), then

lim f

,g(t)

sin(n +

f t dt = 0.

2 Extend g to [-or, or) by defining g(x) = 0 for all x, -ors x < 0. Since 0

Proof. sin(n + J)t = cos i sin nt + sin i cos nt, we have

JA g(t) sinl n + 2) t dt = J Ag(t) sinl n +

t dt

2)

o

A

rA

=J

gi(t) sin nt dt + A

g2(t) cos ,it dt, J -A

9.5

Pointwise Convergence of Fourier Series

417

where g,(t) = g(t) cos 1 and g2(t) = g(t) sin 2. Since g is Riemann integrable on [-ar, ir;, so are both g, and $2. Thus by Theorem 9.3.4, A

( lim 1

nom=

A

g,(i) sin nt dt = nlim 00

from which the result follows.

g2(t) cos nt dt = 0, -A

U

Dirichlet's Theorem Before we state and prove Dirichlet's theorem, we briefly review some notation introduced in Chapter 4. For a real-valued function f defined in a neighborhood of a given point x the right and left limits off at x0, denoted f (x,+) and f (x,-) respectively, are defined by

f(x,+) = lim f(x)

f(x,-) = lim_ f(x),

and

provided, of course, that the limits exist.

9.5.3

Let f be a real-valued periodic function on R of period 2ir with f E Jt[-ir,1r]. Suppose x, E R is such that

THEOREM (Dirichiet)

(a) f (x,+) and f (x,-) both exist, and (b) there exists a constant M and a S > 0 such that

If(x,+ t) - f(x,+)1 <- Mt

l f(x, - t) - f(x,-)I s Mt

and

for all t,0
provided, of course, that the inequalities (10) hold.

Proof. Set A =

2[f(x,+) + f(x,-)]. By Lemma 9.5.1,

2IV[f(x°+t)

f(x°-t)

2

0

!I A[(

(xo + t) -f(xo+))

0

lsin(n + ;)t

+ (f(x. - t) - f(xo-))1

2 sin i

(A

_

J 0

dt

A

)tdt +

g,(t) sin n +

2

I

to

g2(t) sin (n + 2)tdt.

(10)

418

Chapter9

Orthogonal Functions and Fourier Series where

=f(x + t) -f(xo+)

gi(t)

and

2sin1

g2(t)

=f(x,, - t) -f(x.-) 2sin f

To prove the result, it suffices to show that the functions g, and g2 are Riemann integrable on [0, Tr]. If this is the case, then by Lemma 9.5.2.

lim1gi(t)sinln+2ftdt=0, i = 1,2.

\

U

Therefore,

A = 2[f(xo+) +f(xo-)]To finish the proof we still have to show that g, E R[0, ar], i = 1, 2. We will prove

the result for the function g,, the proof for g2 being similar. Set h(t) = f(x, + t) f (x,+ ). Since f E Jt[ -vr, 1r] and is periodic of period 2ir, h is Riemann integrable on [0, a]. Also, (sin Z)-' is continuous on (0, v] and thus Riemann integrable on [c. 1r] for

any c, 0 < c < ir. Therefore,

gilt) = 2 h(t) sin is Riemann integrable on [c, 17] for any c, 0 < c < a. Let S > 0 be as in hypothesis (b). Then for 0 < t < S, using inequality (10), I

g

i

(t)I =

If(; + t) - f(x,+)I t

12 stn 21

If(xo + t) -f(x,+)I Itl

Si

SMlsinll.

Since

lirax-

X-,0' sin x

there exists a constant C such that tt2 Z

sC

sin 2t

for all t, 0 < t s ir. Therefore, g, is bounded on (0, S), and hence also on (0, w]. Thus since g, is Riemann integrable on [c, ir] for any c, 0 < c < r, by Exercise 5 of Section 6.2, g, is Riemann integrable on (0, w].

Piecewise Continuous Functions Dirichlet's theorem required that f(x,-) and f(x,+) both exist at x0, and that f satisfy the inequalities (10) at the point x0. The existence of both f(x,-) and f(x,+) means that either f is continuous at x or in the terminology of Section 4.4, has a removable

9.5 Pointwise Convergence of Fourier Series

419

or jump discontinuity at x0. Functions that have only simple discontinuities, and at most a finite number on an interval [a, b], are said to be piecewise continuous on [a, b . We make this precise with the following definition.

9.5A DEFINITION A real-valued function f is piecewise continuous on [a, b] if there exist finitely many points a = x < x, <

-

< x = b such that

(a) f is continuous on (xi_,, x;) f o r all i = 1, 2, . .. , n, and

(b) for each i = 0, 1, 2, ... , n, f(x;+) and f(x;-) exist as finite limits. For i = 0 and n, we of course only require the existence of the right and left limit respectively. Also, we do not require that f be defined at xa, x1.... , The graph of a typical piecewise continuous function is illustrated in Figure. 9.8.

xf

b

Figure 9.8

In addition to piecewise continuous functions, it will also be useful to consider functions for which the derivative is piecewise continuous. If f is piecewise continuous on [a, b], then the derivative f' is piecewise continuous on [a. b), if there exist a finite

number of points a = xa < x, <

. < x - b such that

(a) f(x) exists and is continuous on each interval (xi _ 1, xi), i = 1, 2,. .. , n, and

(b) for each i = 0, 1, 2, ... , n, the quantities f'(x;+) and f'(x;-) both exist as finite limits. Again, for the endpoints a and b we obviously only require the

existence of f (a+) and f'(b-). 9.5.5

EXAMPLES

(a) Consider the function

f(x)-

x,

xZ+2,

0<x<1, I <x<2.

Since f is continuous on (0, 1) and (1, 2) and f(0+), f(1-), f(1 +), and f(2-) all exist, f is piecewise continuous on [0, 2]. Also, since f,(x)-{1,

2x,

0<x<1, 1 <x<2,

and f'(0+), f'(I-),f'(1 + ), and f'(2 -)all exist, f' is also piecewise continuous on [0, 2].

420

Chapter 9

Orthogonal Functions and Fourier Series

(b) As our second example, consider the function

x=0,

0,

f(x)

xI sin 1, X

O < x <- 1.

The function f is both continuous and differentiable on [0, 1 ] with

f

x=0,

0,

'x

2x sin 1

x

cos 1,

x

0 < x < 1.

However, since f'(0+) does not exist, f' is not piecewise continuous on [0, 1].

Returning to Dirichlet's theorem, suppose f is such that both f'(x,-) and f'(x,+) exist at the point x0. If f' is piecewise continuous on [ -Tr, in, then there exists x, > x such that f and f' are continuous on (x x,). If f is not continuous at x redefine f at x, as f (x,+ ). Then f (redefined if necessary) is continuous on [x, x,), and thus by Theorem 5.2.11, li

Ax. + o

Ax" +) t) I

As a consequence, there exists a constant M and a S > 0 such that

If(xo + t) - f(x,+)I <- Mr for all t, 0 < t < S. Similarly, the existence of f'(x,-) implies the existence of a constant M and a S > 0 such that

If(x - t) -f(x,,-)I <_ Mr for all t, 0 < t < S. Thus f satisfies the hypothesis of Dirichiet's theorem. On combining the above discussion with Theorem 9.5.3 we obtain the following corollary.

9.5.6

COROLLARY Suppose f is a real-valued periodic function on R of period 2a. If f and f' are piecewise continuous on [ -Tr, Tr], then

xS.,(x) = 2[f(x+) +f(x-)] for all x E R. 9.5.7

EXAMPLES

(a) Consider the function f defined on [-ir, ar) by

-ITsx<-

-2.

.

2'

2,

2 <x<ar.

9.5

Pointwise Convergence of Fourier Series

421

Extend f to all of U8 as a periodic function of period 2a. The function f is then continuous except at x = 2 it + krr, k e Z. At each discontinuity x,,,

I[f(x"+) + f(xo-)] = 2 The graph of 2[f(x+) + f(x-)] on the interval [-27r, 21r] is given in Figure 9.9.

-O -IT

-3°

-2A

2

O-3

0

-°

°

OA

2

'-

3°

2v

2

Graph of 1[f(x+) + f(x-)]

Figure 9.9

To discuss convergence of the Fourier series off we first note that f is piecewise

continuous on [-1r, ar] and differentiable on (-1r, -2), (--, 1), and (Z, zr) with f'(x) = 0 in the respective intervals. Thus f' is also piecewise continuous on [ -ir, ir].

As a consequence of Corollary 9.5.6, the Fourier series of f converges to 2[f(x+) +f(x-)] for all x E 08. The Fourier series off is obtained as follows: 1

ao=- J

a2

3dx=3, °/2

(2

a°

6 3cosnxdx=sin n 2,n=1,2, ..

n/2

and since f is an even function, b = 0 for all n E Ni. Therefore, 3

f (x)

2

+

nit

6

Gnir- sin 2 cos nx. n_l

If n is even, sin!' = 0. If n is odd, i.e., n = 2k + 1,

sin(2k + 1)

sin(ka + 2)

=

2

Thus

= (-1)&.

sin

2 I)k

AX)

+6

ka

I cos(2k + 1) x.

2k

2

When x = 0, the series converges to f(O) = 3. As a consequence 3

3

6

`

°° 0

(-1)t

2k+l'

422

Chapter 9

Orthogonal Functions and Fourier Series which upon simplification gives

(-1)k 00 2k + 1 k-o

- rr 4*

(b) Dirichlet's theorem can also be applied to the Fourier sine and cosine series of a real-valued function f defined on [0,1r). As an example, consider the cosine series of f(x) = x on (0, 7r). By Exercise 5(b) of Section 9.3, the cosine series of f(x) = x is given by

x^

I

4 00

2

k-I (2k

2 cos(2k - I)x.

1

1)'

Since the even extension f, off to [-a, Tr] is given by f,(x) x 1, the cosine series of x is the Fourier series of I x I on [ -ir, 7r]. Since both f, and f; are piecewise continuous on [-ir, ir], the Fourier series converges to the 21r-periodic extension of I x for all x E R. This function is illustrated in Figure 9.10.

_p

-2n

IT

2n

By Dirichlet's theorem, the Fourier series converges to I x I for all x E [-ir, rr]. Taking x = 0 gives 0 00

or

1

I (2k - 1)2

_

7r -

.

ir

2

4

00

1

k-l ( 2k- 1)2' )

.

8

Remarks. Although the inequalities (10) of Theorem 9.5.3 are sufficient, they are by no means necessary. There are variations of Dirichlet's theorem that provide sufficient conditions on f to guarantee convergence of the series to 1 [ f (xo+) + f (xo-) ]. For example, the inequalities of Theorem 9.5.3 can be replaced by the following:

If(xo + t) -f(xo+)1 < Mt° and

If(xo - t) -.f(xo-)I c Mt°

(11)

9.5

Pointwise Convergence of Fourier Series

423

for all t, 0 < t < 8, and some a, 0 < a 5 1.1f f satisfies the above at x0, then the conclusion of Dirichlet's theorem is still valid. An even more general condition comes from

Ulisse Dini (1845-1918). who proved that if f E k[ -zr, zr] satisfies

o8lf(xo + t) - f(x"+)I t

J

dt <

and

oo

ff&If(x"-t)-.f(x.-)Idt
to

for some S > 0, then the Fourier series off converges to [f(x"+) + Both of the above hold if f satisfies inequalities (11) at x0.

at x".

2

Differentiation of Fourier Series Our final result of this section involves the derivative of a Fourier series. Consider the function f (x) = x. Since both f and f are continuous on [ - ir,1r], the Fourier series of f converges to f for all x E (-,r, ir). Therefore, by Example 9.3.3(b),

2(-1)n'' x=

sin nx,

x E (-ii', Tr).

n

n=1

The differentiated series 00

2(-1)n+tcosnx

n=1

does not converge since its nth term fails to approach zero as n -> oo. The 21r-periodic extension off has discontinuities at x = ±(2k - 1) it, k E Ni. It turns out that continuity of the periodic function is important for differentiability of the Fourier series. Sufficient conditions are given by the following theorem.

9.5.8

THEOREM Let f be a continuous function on [-ir, zr] with f(-ir) = f(v), and let f be piecewise continuous on [-ir,,r]. If

Ax) = 1 ao + ',00(an cos nx + bn sin nx), x E [ -ir, it ], 2

n=t

is the Fourier series off then at each x E (-1r, ar) where f"(x) exists, 00

f'(x) =

(-nan sin nx + nbn cos nx).

(12)

Remark. At a point x where f"(x) does not exist but f"(x-) and f" (x+) both exist, the series (12).converges to 2[f'(x-) + f'(x+)].

424

Chapter9

Orthogonal Functions and Fourier Series

Proof.

Suppose x E (-vr, ar) is such that f"(x) exists. Since fis continuous at x. by Dirichlet's theorem 1

f' (x) =

2

00

as + I (a" cos nx + P. sin nx).

Since f is continuous with f(-7T) = f(ir), and f E Jt.[-ar, zr], as =

f'(t)dt =

2

[f(l) - f(-a)] = 0.

2

J

Also, by integration by parts,

a" =

1

- f'(t) cos nt dt 2ar

= 2[ f(ar) cos nar - f(-v) cos(-nir)] + tar 1 A f(t) sin nt dt = nb".

Similarly, /3" = -na,,, which proves the result. O

9.5.9

EXAMPLE Consider the function f(x) = x2. Since f is even on (-ar, ar), 00 i

f (x) = 2 ao +

a" cos nx.

Also, since f satisfies the hypothesis of Theorem 9.5.8, 00

f'(x) = Zr = 7, (-na" sin nx). n.1

On the other hand, by Example 9.3.3(b),

4(-1)"+'

2x = "_I 7, 00

n

sin nx.

Therefore, a" = 4(-1)"/n2 for n = 1, 2, 3, . ... To find ao we use the definition. This gives ao = 2ar2/3, and thus .2f

x2 = 3

2)"

+ 4 001 ( "_1

n

cos nx,

x E [-ar, 71.

Remarks. In closing this section, it should be mentioned that there exist continuous functions f for which the Fourier series of f fails to converge at a given point. This was first shown by P. du Bois Reymond. Other examples were subsequently constructed by L. Fej6r and Lebesgue. The example of Fej6r can be found on p. 416 of the text by

9.5

Pointwise Convergence of Fourier Series

425

E. C. Titchmarsh. Given any countable set E in (-7r, IT), it is possible to construct a continuous function f on [-vr, 7r] such that the Fourier series off diverges on E and converges on (-1r, 1r) \ E.4 In fact, it is possible to construct a continuous function on [-7r, 7r] whose Fourier series diverges on an uncountable subset of (-a, rr).5 The existence of functions having such pathological behavior is due to the fact that for the Dirichlet kernel D0, (lr

lim I

n- 00

I Dn(1)I Lit = oc.

0

The details are left to the miscellaneous exercises. For all of the above examples, the Fourier series off converges to f except on a set of measure zero. This is the case for the Fourier series of every Riemann integrable function f on [-ir, a]. In 1926, Kolmogorov showed that there exist Lebesgue integrable functions whose Fourier series diverge everywhere.6 The biggest question about the convergence of Fourier series was asked by Lusin: If a function f satisfies the hypothesis that f2 is Lebesgue integrable on [-ir, ir], does the Fourier series off converge to f, except perhaps on a set of measure zero? This problem remained unanswered for 50 years. The first proof that this was indeed the case was provided by L. Carleson in 1966.'

EXERCISES 9.5 1. For each of the functions of Exercise 3, Section 9.3, sketch the graph of the function on the interval [ - 21r. 2v to which the Fourier series converges. 2. ;Use the Fourier series of f(x) = x2 to find

I 00

CO (-1)M+1 n22 Y,

and

n-t

1

n2.

n=1

3. Using the Fourier cosine series of f(x) = sin x on [0, ar) (Exercise 7, Section 9.3), find each of the following sums:

;a

sb

1

4rt2

(-1)" Ls+t 4n2 - 1

I

4. a. Find the Fourier series of f(x) = (jr - IxI)2 on [-1r,1r]. b. On the interval [,-2r, 21r] sketch the graph of the function to which the series in part (a) converges. c. Use the results of part (a) to find the sums of the following series: 00

G 12 ,

CO

a 4. See Chapter VIII of the text by A. Zygmund. 5. Ibid. 6. "Une s6rie de Fourier-Lebesgue divergente partout;' Compte Rendus, 183 (1926), 1327-1328. 7. "On convergence and growth of partial sums of Fourier Series" Acta Math., 116 (1966) 135-157.

426

Orthogonal Functions and Fourier Series

Chapter 9

S. a. Show that for -it < x < it,

e' =

eII

e-'[, - 2(cos; (-1 nx - n sin nx) nz+ '

2

2a

b. Using part (a) find W 1-11.-I

G z

W

1

and

z

6. *a. Find the Fourier cosine series of e' on [0, ar).

*b. On the interval [-21r, 21r], sketch the graph of the function to which the series in part (a) converges. 7. Let f be a continuous function on [-ar, it] with f(-ir) = f(ir). If, in addition, f' is piecewise continuous on [ -zr, a), prove that the Fourier series off converges uniformly to f on [ -tr, it]. & Suppose f E 9R[-ir,1r], and x, E (-it, ar) is such that f(x,-) and f(x,+) exist, and that l

f(x,

f(xo+) .

+ t)t

l

f(xo - t)t f(x.- )

both exist as finite limits. Prove that f satisfies the hypothesis of Dirichlet's theorem (9.5.3).

9. Suppose f E 9t[-zr, ir] and x, E (-a,.it) is such that f(x,, ) and f(x,+) both exist If f satisfies the inequalities (11) at x, for some a, 0 < a { 1, prove that the. Fourier series off at x, converges to x[f(x,+) + f(x,-)].

NOTES There is no doubt about the significance of Fourier's contributions to the areas of mathematical physics and applied mathematics; one only needs to consult a text on partial differential equations. The methods he developed relating to the theory of heat conduction are applicable to a large class of physical phenomena, including acoustics, elasticity, optics, and the theory of electrical networks, among others. Fourier's work, however, is even more significant in that it inaugurated a new area of mathematics. The study of Fourier series led to the development of the fundamental concepts and methods of what is now called real analysis. The study of the concept of a function by Dirichlet and others was directly linked to their inter-

est in Fourier series. The study of Fourier series by Riemann led to his development of the Riemann integral. He was concerned with the question of finding sufficient conditions for the existence of the integrals that gave the Fourier coefficients of a function f, that is,

a.= -

f(x) cos nx dx -II

and

The quest for an understanding of what types of functions possessed Fourier series is partly responsible for the development of Lebesgue's theory of integration. The need for a

more extensive theory of integration was illustrated by Lebesgue in 1903 in the paper Sur les series trigonometric.° In it he constructed an example of a function that is not Rie-

mann integrable but that is representable everywhere by its Fourier series. Such a function is f(x) _ -1a12 sin j1, whose Fourier series is I. ,(cos kx)/k. This series converges everywere to the function f on [ - 7r, 1r), but since f is

unbounded, it is not Riemann integrable on [ -ar, zr]. It is, however, integrable in the sense of Lebesgue. The article by Alan Gluchoff provides an excellent exposition on the influence of trigonometric series to the theories of integration of Cauchy, Riemann, and Lebesgue. There are several important topics that we have not

touched upon in this chapter. One of these is the Gibbs phenomenon, named after Josiah Gibbs (1839-1903). To explain this phenomenon we consider the Fourier se-

ries of f(x) = 0, x E [ -a, 0), and f(x) = 1, x E [0, or) of Example 9.3.3 By Dirichlet's theorem we have

II

b=

1

f(x) sin nx dx. -II

8. Annales Scient (1903), 453-485.

s de PEcole Normale Supdrieure, (3) 20

Votes

0, -ar<x<0.

2+-1(

I

1)

sin(2k-1)x= 2 , 1,

of (0,7r] given by v , = j k. j = 0. 1, 2, ... , k. and

x0,

ti _;(Vi-,+vy),

0 < x < 7r.

it sin

2i

it 2k

l

3w

+ 3 sin 2k

A careful examination of the graphs (see Figure 9.2) of the partial sums S3, Ss, and S,s shows that near 0 each of the functions S,, i = 3, 5, 15, has an absolute maximum at a point t;, where the t; get closer to 0, but S,{t;) is bounded away from 1. We now consider the behavior of the partial sums S.

1

+ 2k

-I

(2k - 1)a

I

=

2k

k

R(r,) ;.

r

n

stz xx

(X

dr.

0

To approximate the integral we use the Taylor series expansion of sin x. This gives

sin 3x

x

3

sin x

+ (2k

+

sin

Therefore.

more closely. For n - (2k - 1), k E hl, S7k_,(x) = 2 + 2I sin x +

427

1) sin(2k - 1)x

x4

x2

x6

xt 13t+st7t+9!-..

for all x E R. Therefore,

and

+ cos(2k - 1)x].

Sy,_,(x) = 2[cosx + cos3x +

ir

1 (" sin x x

0

dt= Ir-3,31+S. St Tr3

1

n-7

If we multiply SSt_,(x) by sin x and use the identity

ir9

-7.T++9 91sinxcosjx = 2 [sin(j + I)x - sin(j - I)x],

,R2

=1

we obtain

,T4

3.3!+5.5!

,F6

7.7!

sin x Syk_,(x) = 1 sin kx.

- 1 - .54831 + .16235 From this it now follows that S2k_ t(x) has relative maxima

and minima at the points

- .02725 + .00291 .59 (to two decimal places).

2kx = ± 9r, ± 21r, .. , ± 2(k - 1)1r.

The notation'*-" denotes approximately equal to. There-

fore, 1.09. Even though ultimo S. (x) = 1 These points are equally spaced in (-,it, a). Consider the for all x E (0, Ir), if we approach zero along the points xt, points xk = ar/(2k), at which each S3k_, has an (absolute) then S7k_, (xk) overshoots the value I by approximately maximum with

.09; i.e., kim LSy,_,(xk) - f(xt)l - 0.09. If the Fourier

Su_,(xk) = 2+

I sin

series converges uniformly, this cannot happen. The above behavior, known as the Gibbs phenome-

sin 2k

2k+

3

+

+

I

Isin

(2k - 1)7r 2k

To find kiimaoS2k_,(xk), we write the above sum as a Riemann sum of the function g(x) = (sin x)/x. This function is Riemann integrable on [0, a]. With the partition '

non, is due to the fact that f has a jump discontinuity at x0 = 0, and is typical of the behavior of the Fourier series of a piecewise continuous function at a jump discontinuity. Furthermore, if f and f are piecewise continuous on [ -a, 1r], then the amount of overshoot at a discontinuity x0, due to the Gibbs phenomenon, is approximately equal to 0.09[f(x0+) The article by Shelupsky in

Chapter 9

428

Orthogonal Functions and Fourier Series

the supplemental readings provides a very nice discussion of this phenomenon. Another important question involves the uniqueness of the representation of a function by a trigonometric series. Specifically, if

1(x) _ 7, (Ak cos kx + Bk sin kx) k-0

a finite number of x in [-rr, rr]. This then led Cantor to consider the uniqueness problem for infinite subsets of

(Ck cos kx + Dk sin kr)

[ - rr, rr]; specifically, if E C [ - ir, rr] is infinite and

k-0

for all x E [ -a, 'r], must Ak = Ck and Bk = Dk for all k = 0, 1, 2, . .? Alternatively, if t

- Ao + T, (Ak cos kx + Bk sin kx) = 0 2 k-l

rr, rr]. where S. is the nth partial sum of the series. In 1870 Eduard Heine proved that if the sequence (S.} converges uniformly to 0 on [ -ir, rr] then AA =Bk = 0 for all k. This is Theorem 9.3.6. The general uniqueness problem was solved by Cantor in the early 1870's. Furthermore, he also proved uniqueness if equation (13) holds for all but

for all x E

(13)

for all x E [ -rr, rr], must Ak = Bk = 0 for all k = 0, 0 I.2, ... ? By equation (13) we mean that N-.OU

equation (13) holds for all x E [ -rr, a) \ E, must Ak = Bk = 0 for all k? Since point set theory was undeveloped at this time, Cantor devoted much of his time and effort to studying point subsets of R. For a thorough discus-

sion of the uniqueness problem, refer to the article by Marshall Ash listed in the supplemental readings, or to the text by A. Zygmund listed in the Bibliography.

MISCELLANEOUS EXERCISES 1. Suppose f is continuous on [ -rr, rr) and f(x) = i A0 + 7,k i (At cos kr + Bk sin kr). Let S. be the nth partial sum of the series. If there exists a positive constant M such that IS.(x)I < M for alI x E [ -rr. rr] and n e N. prove that Ak and Bk are the Fourier coefficients off.

2. Prove that lim j ID.(t)I dt = oo, where D, is the Dirichlet kernel. (See Example 6.4.4(b).)

3. Let f(x) = -In12 sin J. Show that the Fourier series of f is given by

1(cos kx)lk (Note: Since f is unbounded

at x = 0, the integrals defining ak and bk have to be interpreted as improper integrals.)

SUPPLEMENTAL READING Ash, Marshall J., "Uniqueness of representation by trigonometric series:' Amer. Math. Monthly % (1989),873-885. Askey, R. and Haimo. D. T., "Similarities between Fourier series and power series,' Amer. Math. Monthly 103 (19%), 297-304. Gluchoff, Alan D., "Trigonometric series and theories of integration;' Math. Mag. 67 (1994). 3-20. Gonzalez-Velasco. Enrique A., "Connections in mathematical analysis: The case of Fourier series;' Amer. Math. Monthly 99 (1992), 427-441. Halmos, R. "Fourier series;' Amer. Math. Monthly 85 (1978), 33-34.

Lanczos, Cornelius, Discourse on Fourier Series. Hafner Pubi. Co., New York, 1966. Mulcahy, C., "Plotting and scheming with wavelets;' Math. Mag. 69 (1996), 323-343. Shelupsky, D., "Derivation of the Gibbs phenomenon;' Amer. Math. Monthly 87 (1980). 210-212. Shepp, L. A. and Kruskal, J. B., "Computerized tomography: The new medical X-ray technology;' Amer Math. Monthly 85 (1978), 420-439. Simon, Barry, "Uniform convergence of Fourier series;' Amer. Math. Monthly 67 (1969), 55-56. Williams, K. S., "Note on f o (sin x/x) dx," Math. Mag. 44 (1971),9-11.

1 O Lebesgue Measure and Integration 10.1 Introduction to Measure 10.2 Measure of Open Sets; Compact Sets 10.3 Inner and Outer Measure; Measurable Sets 10.4 Properties of Measurable Sets

10.5 Measurable Functions 10.6 The Lebesgue Integral of a Bounded Function 10.7 The General Lebesgue Integral

10.8 Square Integrable Functions

The concept of measure plays a very important role in the theory of real analysis. On the real line, the idea of measure generalizes the length of an interval, in the plane, the area of a rectangle, and so forth. It allows us to talk about the measure of a set in the same way that we talk about the length of an interval. The development of the Riemann integral of a bounded function on a closed and bounded interval [a, b depended very much on the fact that we partitioned [a, b] into intervals. The notion of measure and measurable set will play a prominent role in the development of the Lebesgue integral in that we will partition [a, b] not into intervals, but instead into pairwise disjoint measurable sets. The theory of measure is credited to Henri Lebesgue (1875-1941) who, in his famous 1902 thesis, defined measure of subsets of the line and the plane, as well as the Lebesgue integral of a nonnegative function. Like Riemann, Lebesgue was led to the development of his theory of integration while searching for sufficient conditions on a function f for which the integrals defining the Fourier coefficients off exist. In this chapter we will develop the theory of Lebesgue measure of subsets of I following the original approach of Lebesgue using inner and outer measure. Although this approach is somewhat more tedious than the modern approach developed by Constantin Carathdodory (1873-1950), it has the advantage of being more intuitive and easier to visualize. In the first section we will illustrate the need for the concept of measure of a set and measurable function by considering an alternative approach to integration developed by Lebesgue in 1928. Although this ultimately will not be how we define the 429

430

Chapter 10

Lebesgue Measure and Integration

Lebesgue integral, the approach is instructive in emphasizing the concepts required for the development of the Lebesgue theory of integration. In Section 10.2 we use the fact that every open subset of R can be expressed as a finite or countable union of disjoint open intervals to define the measure of open sets, and then of compact sets. These are then used to define the inner and outer measure of subsets of R. A bounded subset of R is then said to be measurable if these two quantities are the same. In Section 10.6 we develop the theory of the Lebesgue integral of a bounded real-

valued function using upper and lower sums as in the development of the Riemann integral. However, rather than using point partitions of the interval, we will use measurable partitions. The key result of the section is that a bounded real-valued function on [a, b] is Lebesgue integrable if and only if it is measurable. As we will see, the class of Lebesgue integrable functions contains the class of Riemann integrable functions, and for Riemann integrable functions, the two integrals coincide. One of the advantages of the Lebesgue theory of integration involves the interchange of limits of sequences of functions and integration. We will prove several very important and useful convergence theorems, including the well-known bounded convergence theorem and Lebesgue's dominated convergence theorem.

10.1

Introduction to Measure In Definition 6.1.1 1 we defined what it means for a subset of lR to have measure zero. In this chapter we will consider the concept of measure of a set in greater detail. When

introducing a new concept it is very natural to ask both "why," and "does it lead to something useful?" Both of these questions were answered by Lebesgue in 1903 when he exhibited a trigonometric series that converged everywhere to a nonnegative function f that was not Riemann integrable. ' The function f, however, is integrable according to Lebesgue's definition and the trigonometric series is the Fourier series off. In this section we will illustrate why it is necessary to consider the concept of measure of a set by considering an alternative approach to integration. As we will see later in the chapter, this approach leads to greater generality in the types of functions that can be integrated. In addition, Lebesgue's theory of integration allows us to prove interchange of limit and integration theorems without requiring uniform convergence of the sequence of functions. Let f be a bounded function on [a, b]. In developing the theory of the Riemann integral we partitioned the interval [a, b] and defined the upper and lower sums off corresponding to the partition. Lebesgue's approach to integration is to partition the range of the function, rather than the domain. For purposes of illustration, suppose f is nonnegative with Range f C (0, /3). Let n E N, and partition [0, /3) into n disjoint subintervals

1U-1)nn

j = 1, 2,

, n.

See Figure 10.1 with n = 8. 1. Annales Scientifiques de 1'i`cole Normale Superieure. (3) 20 (1903). 453-485.

10.1

Introduction to Measure

431

Y 4

1$

f

1

----- -----------------

S:

j

----I

{

11

6 {

1

1

I

{

a

x

x1

;2 -3 -'14 '1 b

Figure 10.1

For each j = 1, 2, ... , n, we let Ej _

{x e [a,b]:(j-1) n <_ f(x) < ja} n

In Figure 10.1,

E4=

{xE [a, b] : 3 8 s f(x) < 4 8 }

= [a, x1) U (xz, x3) U (xs, xs]

For a set such as E4, a finite union of disjoint intervals, it is reasonable to define the measure of E,, denoted m(E,), as the sum of the length of the intervals, i.e.,

m(E4) = (x1 - a) + (x3 - x2) + (x5 - X,). Assuming that we can do this for each of the sets Ep a lower approximation to the area under the graph off is given by

f, and an upper approximation is given by

i- j8m(Ei) Taking limits as n -), oo, assuming that they exist and are equal, would in fact, provide another approach to integration. That this -indeed is the case for a large class of functions, including the Riemann integrable functions, will be proved in Section 10.6. For nice functions, namely those for which the sets E, are finite unions of intervals, the above is perfectly reasonable. However, suppose our function f is defined on [0, 1: by

f(x) = J0, x E a fl [0, 1], x,

elsewhere.

432

Chapter 10

Lebesgue Measure and Integration As above, for each j = 1, 2, . .. , n, let

E= {xE[O.l]:-1f(x)
I

x irrational :

n

< x < in

Here the sets Ej are no longer unions of intervals, so what is meant by the measure of the set is by no means obvious. Our goal in the next two sections is to define a function A on a family .44 of subsets of I8, called the measurable sets. The collection M will contain all the intervals and A will satisfy the following:

(a) for an interval J, A(J) = length of J, (b) A(E + x) = A(E) for all E E At and x E I8, where E + x = {y + x:y E E}. and (c) A(U Ek) _ A(Ek) for any finite or countable family {Ek} of pairwise disjoint sets in .At.

10.2

Measure of Open Sets; Compact Sets We begin our discussion of measure theory by first defining the measure of open and compact subsets of R.

10.2.1

DEFINITION length of J.

If J is an interval, we define the measure of J, denoted m(J), to be the

Thus if J is (a, b), (a, b], [a. b), or [a, b], a, b E I8, then

m(J)=b-a. If J is R, (a, oo), [a, oo), (-oo, b) or (-oo, b], we set m(J) = oo. In dealing with the symbols oo and -oo, it is customary to adopt the following conventions:

(a) If x is real, then x + oo = oo, x - oo = -oo. (b) If x > 0 then x oo = oo, x (-oo) -oo. (c) If x < 0 then x - oo = - oo, x (- oo) = oo. (d) Also, oo + co = oo, -oo -oo = =oo, oo (too) = ±oo,

-00*(±oo)=zoo. The symbols oo - oo and -oo + oo are undefined, but we shall adopt the arbitrary convention that 0 oo = 0.

10.2

Measure of Open Sets; Compact Sets

433

Measure of Open Sets Our first step will be to extend the set function m to the open subsets of R. Since this extension relies on Theorem 3.1.13, we restate that result at this point.

THEOREM 3.1.13 If U is an open subset of R, then there exists a finite or countable collection {I } of pairwise disjoint open intervals such that

U=U In. R

Recall, the family {1"} is pairwise disjoint if and only if I" f l 1,n = 40 whenever n * m.

10.2.2

If U is an open subset of R with U = Unln, where {1"} is a finite or countable collection of pairwise disjoint open intervals, we define the measure of U, denoted m(U), by DEFINITION

m(I,).

m(U)

Remarks (a) For the empty set 0, we set m(¢) = 0. (b) The sum defining m(U) may be either finite or infinite. If any of the intervals are of infinite length, then m(U) = no. On the other hand, if

U=UI,,, where the 1n are pairwise disjoint bounded open intervals, we may still have 00

m(U) = I m(1") = 00, n-1

due to the divergence of the series to no. Since m(I,) ? 0 for all n, the sequence of partial sums is monotone increasing and thus will either converge to a real number or diverge to no.

10.2.3

EXAMPLES

(a) Foreach n= 1,2,. ..,set

In=(n

-

2",n+Zn

Then 11=(1 -2,1+1),12=(2-1,2+1),etc. Since n+2-"< (n+1)-2''"+ 4 2 4 for all n e N, the collection {I"}" 1 is pairwise disjoint. Let U = U' 0- 11n. Then 1

m(U) = G m(I4) _ n-1

n 11

2-2 =

°D

n-0

1

1

2.

2n 12

434

Chapter 10

Lebesgue Measure and Integration

The set U is an example of an unbounded open set with finite measure. (b) Let J n = (n, n n = 1, 2, ... , and let V = U' I J,. Then '° 1

00

m(V) n-

m(J,) _ I - = oo. n-In

(c) As in Section 3.3, let U denote the complement of the Cantor set in [0, 1]. Since U is the union of the open intervals that have been removed, by Property 4 of the Cantor set, m(U)= 1. We now state and prove several results concerning the measure of open sets.

10.2.4 THEOREM If U and V are open subsets of U8 with U C V. then m(U) 5 m(V).

Proof. The statement of the theorem appears to be so obvious that no proof seems to be required. However, it is important to keep in mind how the measure of an open set is defined. Suppose

U=UI, n

and

V=UJm,

where {In}n and {Jm}m are finite or countable collections of pairwise disjoint open in-

tervals. Since U C V, each interval 4 C J. for some m. For each m, let

Since the collection {Jm}m is pairwise disjoint, so is the collection {Nm}m, and

U=U1.=UnU I,,. Therefore,

m(U) _mnEN, .

m(I.).

But by Exercise 1,

I m(I4) c m(Jm), FE N.

from which the result follows.

Cl

Remark. As a consequence of the previous theorem, if U is an open subset of (a, b), a, b E R, then

m(U)sb-a. Thus every bounded open set has finite measure.

10.2

10.2.5

Measure of Open Sets; Compact Sets

435

THEOREM If U is an open subset of R, then kim m(Uk),

m(U)

00

where for each k E 1N, U. = U fl (-k, k).

Proof. For each k, Uk is open, with Uk C Uk+1 C U

for all k E N. By Theorem 10.2.4, the sequence {m(Uk)} is monotone increasing with m(Uk) : m(U) for all k. Therefore, rlim m(Uk) 5 m(U).

oo

(1)

If U is bounded, then there exists ko E N such that

Ufl(-k,k)=U for all k a ko. Hence m(Uk) = m(U) for all k k and thus equality holds in equation (1). Suppose that U is an unbounded open subset of R with

U=UIi, n

where {In} is a finite or countable collection of pairwise disjoint open intervals. If m(I,) = oo for some n, then m(U) = oo, and for that n, either In = R or In is an interval of the form (-oo, an)

or

(an, 00)

for some a E R. Suppose 1n = (an, oo). Choose k, E N such that k, a lad. Then for all k at k0,

In fl (-k, k) = (an, k), and thus

oo = 1im m(1, (1(-k, k)) s Item m(Uk) s m(U). Therefore equality holds in equation (1). The other two cases follow similarly. Suppose m(In) < oo for all n. Since U is unbounded, the collection {In} must be infinite. If the collection were finite, then since each interval has finite length, each in-

terval is bounded, and as a consequence U must also be bounded. Let a E R with a < m(U). Since

0

G n-1

m(I4) = m(U) > a,

there exists a positive integer N such that N

> m(I4) > a. M-1

436

Chapter 10

Lebesgue Measure and Integration

Let V = UN 11,,. Then V is a bounded open set, and thus by the above.

m(V) =slim m(V fl (-k, k)). Since m(V) > a, there exists k E N such that

m(V n(- k, k)) > a for all k a k0. But v fl (-k, k) C Uk for all k E N. Hence by Theorem 10.2.4,

tn(V fl (-k, k)) s m(Uk) and as a consequence

m(Uk) > a for all k ? ko. If m(U) = oo, then since a < m(U) was arbitrary, we have m(Uk) -+ oo as k -> no.

If m(U.) < no, then given e > 0, take a = m(U) - e. By the above, there exists k E N such that m(U) - e < m(Uk) s m(U) for all k an k,,. Therefore, lim m(Uk) = m(U).

Remark. In proving results about open sets, the previous theorem allows us to first prove the result for the case where U is a bounded open set, and then to use the limit process to extend the result to the unbounded case. Our next goal is to prove the following.

10.2.6

THEOREM

If { Un}n is a finite or countable collection of open subsets of R, then

tn(U U.)

t

m(U.). n

For the proof of the theorem we require the following lemma.

10.2.7

LEMMA If {In}R i is a finite collection of bounded open intervals, then N

N

m( U In I

M(In)-

n=1

n=1

It should be noted that the collection {In} is not assumed to be pairwise disjoint. The lemma is most easily proved by resorting to the theory of the Riemann integral.

10.2.8

DEFINITION If E is a subset of fib, the characteristic function of E, denoted XE, is the function defined by

XE(x) =

1,

x E E,

0,

x (4 E.

10.2

Measure of Open Sets; Compact Sets

437

Suppose 1 is a bounded open interval. Choose a, b E R such that I C [a, b]. Since Xf is continuous on [a, b] except at the two endpoints of 1, x, E k[a, b] with Jb XI(x) dx = m(1).

It should be clear that this is independent of the interval [a, b] containing 1. If U is an open subset of [a, b] with m

U= UJk, k=1 where the {Jk} are pairwise disjoint open intervals, then m

Xu(x) = I Xj,(x) k-I

and thus n

m(U) =7,, m(Jk) _I: k=I

k=1

Ja

n

Xj,(x) dx = J Xu(x) dx.

(2)

a

P r o o f of lemma 10.2.7. Let {la} 1 be a finite collection of bounded open intervals and let U = U =11,,. Choose a, b E I8 such that U C [a, b]. Then Xu is continuous except at a finite number of points, and N

Xu(x) t 7, X4(x). n=1

Therefore by identity (2) and the above discussion, e

m(U) = Jb Xu(x) dx <_ a

Xl.(x) dx = Ja

n=I

n=I

Jb Xr,(x) a

_ , m(In) O I

Proof of Theorem 10.2.6.

Since the result for a finite collection follows obviously from that of a countable collection, we suppose {Un}'° 1 is a countable collection of open sets and 00

U= UIU,,. Since U is open, U = U. J,n where {Jm}m is a finite or countable collection of pairwise disjoint open intervals. Also, for each n,

U. = U In, k. where for each n, {I,, k}k is a finite or countable collection of pairwise disjoint open intervals.

438

Chapter 10

Lebesgue Measure and Integration

We assume first that U is bounded with m(U) < oo. Let e > 0 be given. If the collection {Jm}m is infinite, then since F'm m(Jm) < oo, there exists a positive integer N such that 00

I m(J,) < e.

m-N+1

Thus N

m(U) < Y, m(J,) + E. M-1

If the collection {Jm}m is finite, the previous step is not necessary. Consider the collection {Jm}m. 1. For each rn = 1, 2, ... , N. let K. be an open interval such that K. C Jm

m(Jm) <

and

N

Then N

N

M(U) < Y, m(J,,) + e < I m(K,) + 2e.

(3)

m-I

M=1

Let A = U v=1 Km. Since each k. is closed and bounded, so is the set A. Thus A is compact, and since

AC U=nU1( the collection {I,,, k} is an open cover of A. Hence by compactness there exists a finite k,,, i = 1, ... , J. j = 1, . . . , mi, such that number

ACUIn,.k i.!

Since the intervals {Km} are pairwise disjoint

mU aJ

Y, m(K.) = m(U Km) m=I

m =1

which by Lemma 10.2.7 !

ij

m;

MV., k,) = 17, m(1 i-I !-I o0

.1

m(U,,) s

`-

n=I

!=1

Combining this with inequality (3) gives

m(U) <

2e.

n-I

k, )

10.2

Measure of Open Sets; Compact Sets

439

Since e > 0 was arbitrary, the result follows for the case where U is bounded. If U is unbounded, then for each k E N,

x

00

m(Ufl(-k,k))<_y m(u,, n(-k. k))52 m(U,). n=1

n=1

The result now follows by Theorem 10.2.5. Q

10.2.9

THEOREM

If U and V are open subsets of R, then

m(U) + m(V) = m(u U V) + m(u n v). Proof. (a) If both U and V are unions of a finite number of bounded open intervals. then so are u n V and U U V. Thus the functions XU,

Xv,

Xuuv,

Xunv

are all Riemann integrable on some interval [a, b] with

Xu(X) + Xv(x) = Xuuv(x) + Xunv(x) for all x E [a, b]. Therefore by identity (2),

m(U) + m(V) = m(u U V) + m(u n v). (b) For the general case, suppose 00

U= U In n-I

and

V = U Jn, n=1

where the collections {In} and {J.} consist of pairwise disjoint open intervals respectively. If one of m(U) or m(V) is oo, then by Theorem 10.2.4, m(U U V) = oo and the conclusion holds. Thus we can assume that both m(U) and m(V) are finite. Let e > 0 be given. Choose N E N such that 00

00

Y, m(I,) < e

m(JJ) < C.

and n-N+1

n=N+I

Let U * and U ** be defined by U

N n- I In,

00

U##

n N U+ I

In.

Also let V* and V** be defined analogously. Then

m(U) = m(U*) + m(U**)

and

m(V) = m(V*) + m(V**).

Since m(U **) < e and m(V **) < e,

m(U) + m(V) < m(U*) + m(V*) + 2e. Since the sets U* and V* are finite unions of open intervals, by part (a)

m(U*) + m(V*) = m(U* U V*) + m(U* n v*),

440

Chapter to

Lebesgue Measure and Integration

which by Theorem 10.2.4

sm(UUV)+m(UnV). The last inequality follows since both U * U V * and u * n V * are subsets of U U V and u n V respectively. Hence

m(U)+m(V) <m(UUV)+m(UnV)+2e. Since e > 0 was arbitrary, we have

m(U)+mm <_m(UUV)+m(UnV). We now proceed to prove the reverse inequality. We first note that

U U V = (U * U V *) U (U ** U V **), and as a consequence,

m(UUV)sm(U*UV*)+2e. Also by the distributive law,

un v= (u- u u**)n(v* u v**) = (U*nV*)U(U**nV*)U(U*nV**)U(U**nV**) C(U*nV*)UU**UV**. Therefore,

m(U n v) < m(U* n v*) + 2e. Combining the above gives

m(UUV)+m(unv) < m(u* u v*) + m(u* n v*) + 4e, which since U * and V * are finite unions of intervals

= m(U*) + m(V *) + 4e s m(U) + m(V) + 4e. Again since e > 0 was arbitrary, this proves the reverse inequality.

Measure of Compact Sets We now define the measure of a compact subset of R. If K is a compact subset of R and U is any bounded open set containing K, then

U=KU(U\K). Using the fact that U.\ K is also open and bounded, and thus has finite measure, we define the measure of K as follows.

10.2.10

DEFINITION

Let K be a compact subset of R. The measure of K, denoted m(K), is

defined to be

m(K) = m(U) - m(U \ K), where U is any bounded open subset of R containing K.

10.2

Measure of Open Sets: Compact Sets

441

We first show that the definition of m(K) is independent of the choice of U.

10.2.11

THEOREM If K is compact, then m(K) is well defined.

Proof.

Suppose U and V are any two bounded open sets containing K. Then by Theorem 10.2.9,

m(U) + m(V\ K) = m(U U (V \ K)) + m(U n (V\ K))

=m(UUV)+m((UnV)\K). In the above we have used the fact that

UU(V\K)= UUV

and

Un(V\K)=(UnV)\K.

Similarly,

m(U \ K) + m(v) = m(U U v) + m((U n V) \ K). Therefore,

m(U) + m(V \ K) = m(V) + m(U \ K). Since all the terms are finite,

m(U) - m(U \ K) = m(V) - m(V \ K). Thus the definition of m(K) is independent of the choice of U. i.e., m(K) is well defined.

O 10.2.12

EXAMPLES

(a) In our first example we show that for a closed and bounded interval 1,a, b), a, b E R, Definition 10.2.10 is consistent with Definition 10.2.1; namely,

m([a, b]) = b - a.

LetU=(a-E,b+ e),e>0.Then U\[a,b] = (a - e,a)U(b,b + E). Therefore,

m([a, bj) = m(U) - m(U \[a, b])

=(b-a)+2e-2e=b-a. (b) If K = {xi, ... ,

with x; E R, then m(K) = 0. Choose 8 > 0 such that the in-

tervals

1j =(x,-S,xj +8), j=1,2,.. are pairwise disjoint, and let U = U'.. 1;. Then

U \ K = 0(x,-6,x;)U(xx,x;+S). Thus m(U) = m(U \ K); i.e., m(K) = 0. U

n,

442

Chapter 10

Lebesgue Measure and Integration

10.2.13

THEOREM

(a) If K is compact and U is open with K C U, then m(K) c m(U). (b) If K, and K2 are compact with K, C K,, then m(K,) s m(K2).

Proof. The proof of (a) is an immediate consequence of the definition. For the proof of (b), if U is a bounded open set containing K2, then since U \ K2 C U \ K,,

m(K,) = m(U) - m(U\K,) <- m(U) - m(U\K2) = m(K2). O If I is an open interval and a, b E R, then i fl [a, b] is an interval and thus my fl [a, b]) is defined with

m(I fl [a, b]) = m(1 fl (a, b)). We extend this to open subsets of R as follows.

10.2.14

If U is an open subset of R with U = U,, 1,,, where is a finite or countable collection of pairwise disjoint open intervals, and a, b E R, we define DEFINITION

m(U fl [a, b])

m(1 fl [a, b)).

Since my. fl [a, b)) = m(1 fl (a, b)) for all n, we have

m(U fl [a, b]) = m(U fl (a, b)). Remark. What the above definition really defines is the measure of relatively open subsets of [a, b) (see Definition 3.1.14). By Theorem 3.1.16, a subset G of [a, b] is open in [a, b] if and only if there exists an open subset U of R such that G = U n [a, b]. Since the set U is not unique, we leave it as an exercise (Exercise 4) to show that if U. V are open subsets of ltd with

u fl [a, b] = V fl [a, b], then m(U fl [a, b]) = m(V fl [a, b]).

10.2.15

THEOREM If U is an open subset of lib and a, b e I8, then

m(U fl [a, b]) + m(U` fl [a, b]) = b - a. Recall from Chapter 1 that If = R \ U = (x E H : x 44 U). If U is open, then U` is closed and thus U` fl [a, b) is a compact subset of [a, b].

Proof. Suppose V D [a, b] is open. Let K = U` fl [a, b]. Then since V D K. m(K) = m(V) - m(V \ K). But

V \ K = V fl (U` fl (a, b])" = (V fl U) U (V fl [a, b]`) D V fl U. Therefore, m(V fl U) <- m(V \ K). Since U fl [a, b) C U) v,

m(U fl [a, b]) + m(K) < m(U fl v) + m(V) - m(V \ K) s m(V).

10.2

Measure of Open Sets; Compact Sets

443

Given e > 0, take V = (a - e, b + e). Then

m(Un[a,b])+m(U`n[a,b]):r= b-a+2e. Since e > 0 is arbitrary, this proves

m(Un[a,b])+m(U`n[a,b])<--b-a. To prove the reverse inequality, let 1, = [a + e, b - e], where 0 < e < 2(b - a). Then m(U n [a, b])+ m(u° n [a, b]) >- m(U n (a, b)) + m(U` n 4). Since (a, b) is an open set containing U` n /E,

m(u` n 4) = b - a - m((a, b) \ (u° n f)) But

m((a, b) \ (U` n 4)) = m(((a, b) n U) U ((a, b) n /;)) = m(((a, b) Cl U) U (a, a - e) + (b - e, b)), which by Theorem 10.2.6 <- m(U Cl (a, b)) +

Therefore,

m(Un[a,b])+m(U`n[a,b])?b-a-2e. Since e > 0 was arbitrary, the reverse inequality follows. Q

EXERCISES 10.2 1. If

is a finite or countable collection of disjoint open intervals with U 1 C (a, b), prove that

m(4)

m((a, b)).

2. *If U * io is an open subset of R, prove that m(U) > 0. 3. 'Let P denote the Cantor set in [0, 1 ]. Prove that m(P) = 0. 4. Suppose U, V are open subsets of R. a, b E R with U n [a, b] = V fl [a, b]. Prove that m(U n [a, b]) = m(V n [a, b]). 5. If A. B are subsets of R. prove that XAnB(x) = XA(x)XB(x),

XAUB(x) = XA(x) + XB(x) - XAnB(x),

XA'(x) = 1 - XA(x) 6. *If K, and Kz are disjoint compact subsets of R, prove that m(K, U K2) = m(K,) + m(K2).

444

Chapter io

10.3

Lebesgue Measure and Integration

Inner and Outer Measure; Measurable Sets Our goal in this section is to define a function A on a large family l1 of subsets of R. called the measurable sets, which agrees with the function m on the open and compact subsets of R. We begin with the definition of inner and outer measure of a set.

10.3.1

DEFINITION Let E be a subset of H. The Lebesgue outer measure of E, denoted A*(E), is defined by

A*(E) = inf{m(U): U is open with E C U}. The Lebesgue inner measure of E, denoted A*(E), is defined by A*(E) = sup{m(K) : K is compact with K C E}.

10.3.2

THEOREM

(a) For any subset E of R, 0 < A*(E) t A*(E). (b) If EI and E2 are subsets of a8 with El C E,, then A*(EI)

A*(E,)

A*(EI) < A*(E,).

and

Proof. (a) If K is compact and U is open with K C E C U, then 0 5 m(K) If we fix K, then m(K) over all such U gives

m(U).

m(U) for all open sets U containing E. Taking the infimum 0

tn(K)

A*(E).

Taking the supremum over all compact subsets K of E proves (a). The proof of (b) is

similar and is left as an exercise (Exercise 7). 0 10.3.3

EXAMPLES

(a) If E is any countable subset of R. then

A*(E) = A*(E) = 0. Suppose E = {xn}" I. Let E > 0 be arbitrary. For each n, let e

In = Xn - 2n, Xn +

2n

and set U = U' I In. Then U is open with E C U. By Theorem 10.2.6,

z m(U)

n=1

°°

m(In)

E

F- T = 2e. n=1

Therefore, A*(E) < 2e. Since e > 0 was arbitrary, A*(E) = 0. As a consequence, we also have A*(E) = 0.

10.3

Inner and Outer Measure; Measurable Sets

445

(b) if I is any bounded interval, then A*(1) = A*(/) = m(1).

Suppose I = (a, b) with a, b E R. Since I itself is open,

A*(I) <_ m(1) = b - a.

On the other hand, if 0 < E < (b - a), then [a + $, b - fl ] is a compact subset of !. and as a consequence

b -a - e=ml [ a+2,b-2Jj_A*(1). Therefore,

b-a-ESA*(1)SA*(1)Sb-a. Since e > 0 was arbitrary, equality holds. A similar argument proves that if 1 is any closed and bounded interval, then

A*(I) = A*(!) = m(I). As a consequence of Theorem 10.3.2(b), the result holds for any bounded interval 1.

(c) For any open set U,

A*(U) = A*(U) = m(U). By definition, A*(U) = m(U). But m(U) _ collection of open intervals with U =

where {I,} is a pairwise disjoint

Suppose a E R satisfies a < A*(U). Since m(U) > a, there exist a finite number of intervals 11, ... ,1N such that $N t m(1) > a. For each j, choose a closed and bounded interval Ji C I, such that

j tm(J,) > a. Let K = U t t J. Then K is a compact subset of U and thus A*(U) > m(K). Finally, since the intervals {J% I are pairwise disjoint, by Exercise 6 of Section 10.2,

m(JJ) > a.

m(K) _ J=1

Therefore A*(U) > a. If A*(U) = oo, then by the above A*(U) > a for every a E R; that is, A*(U) = oo. On the other hand, if A*(U) is finite, take a = A*(U) - E, where e > 0 is arbitrary. But then A*(U) ? A* (U) > A*(U) - e for every e > 0. From this it now follows that A*(U) = A*(U) = m(U).

Measurable Sets In both of the previous examples, the inner and outer measures of the sets are equal. As we shall see, all subsets of R built out of open sets or closed sets by countable unions, intersections, and complementation will have this property. This includes most sets encountered in practice. In fact, the explicit construction of a set whose inner and outer measures are different requires use of an axiom from set theory, the Axiom of Choice,

446

Chapter 1 0

Lebesgue Measure and Integration which we have not discussed. The construction of such a set is outlined in the miscellaneous exercises.

10.3A

DEFINITION

(a) A bounded subset E of ff is said to be LAbesgue measurable or measurable if

A*(E) = A*(E).

If this is the case, then the measure of E, denoted A(E), is defined as

A(E) = A*(E) = k*(E). (b) An unbounded set E is measurable if E fl [a, b] is measurable for every closed and bounded interval [a, b]. If this is the case, we define A(E) = klint A(E fl [-k, kJ).

Remarks (a) If E is unbounded and E fl l is measurable for every closed and bounded interval 1, then by Theorem 10.3.2 the sequence {A(E rl [-k, k])} t is nondecreasing, and as a consequence A(E) = lim A(E fl [ -k, k]) exists.

(b) There is no discrepancy between the two parts of the definition. We will shortly prove in Theorem 10.4.1 that if E is a bounded measurable set, then E fl 1 is measurable for every interval 1. Conversely, if E is a bounded set for which

A*(E fl [a, b]) = A*(E fl (a, b]) for all a, b E R, then by choosing a and b sufficiently large such that E C [a. b], we have A*(E) = A*(E). The two separate definitions are required due to the existence of unbounded nonmeasurable sets E for which A* (E) = A*(E) = oo. An example of such a set will be given in Exercise 5 of Section 10.4.

10.3.5 THEOREM Every set E of outer measure zero is measurable with A(E) = 0. Proof. Suppose E C R with A*(E) = 0. Then for any closed and bounded interval 1. A*(E fl I)

A*(E) = 0.

Thus A* (En I) = A*(E fl I) = 0, and hence E fl I is measurable for every closed and

bounded interval I. Since A(E fl [ -k, k]) = 0 for every k E N, A(E) = 0. 0 As a consequence of the previous theorem and Example 10.3.3(a), every countable set E is measurable with A(E) = 0. In particular, Q is measurable with A(Q) = 0. Another consequence of Theorem 10.3.5 is that every subset of a set of measure zero is measurable.

10.3

10.3.6

THEOREM

Inner and Outer Measure; Measurable Sets

447

Every interval I is measurable with A(1) = m(l).

By Example 10.3.3(b), if 1 is a bounded interval, A*(l) = A*(l) = m(l). Thus I is measurable with A(l) = m(1). On the other hand, if 1 is unbounded, then I fl [a, b] is a bounded interval for every a, b E R, and thus measurable. In this case.

Proof.

A(/)

10.3.7

THEOREM

klym

A(1 fl [-k, k]) = iim m(1 fl [-k, k]) = oo.

For any a, b E R and E C R, A*(E fl [a, b]) + A* (E` fl [a, b]) = b - a.

Proof. Let U.be any open subset of R with E fl [a, b] C U. Then U` fl [a, b] is compact with U` fl (a, b] C E` fl [a, b]. Therefore, m(U) + A*(E` fl [a, b]) ? m(U fl [a, b]) + m(U` fl [a, b]) = b - a. The last equality follows by Theorem 10.2.15. Taking the infimum over all open sets U containing E` fl [a, b] gives

A*(E fl [a, b)) + A*(E` fl [a, b]) ? b - a. To prove the reverse inequality, let K be a compact subset of E` fl [a, b]. Then K`

is open with K` fl [a, b] D E fl [a, b]. Therefore, A*(E fl [a, b]) + m(K fl [a, b]) <_ m(Kc Cl [a, b]) + m(K fl [a, b])

= b - a. The last equality again follows by Theorem 10.2.15 with U = K. Thus taking the supremum over all compact subsets K of E` Cl [a, b] gives

A*(E fl [a, b]) + A*(E` fl [a, b]) s b - a, which combined with the above, proves the result.

10.3.8 LEMMA For any subset E of R, A*(E) = kiimn A*(E fl [-k, k]).

Proof. Since the proof is similar to that of Theorem 10.2.5 we leave it as an exercise (Exercise 8).

10.3.9

THEOREM Suppose E,, E2 are subsets of R. Then (a) A*(E1 U E2) + A*(E, Cl E2) <_ A*(E1) + A*(E2), and (b) A*(E1 U E2) + A*(E, fl E2) ? A*(E1) + A*(E2).

Proof. (a) If A*(E;) = oo for some i, i = 1, 2, then inequality (a) certainly holds. Thus suppose A*(E;) < oo for i = 1, 2. Let e > 0 be given. By the definition of outer measure, for each i, we can choose an open set U. containing E, such that m(U;) < A*(E;) + 2

448

Chapter io

Lebesgue Measure and Integration

Therefore,

e + A*(E,) + A*(E,) > m(U,) + m(U,), which by Theorem 10.2.9

=m(U,UU2)+m(U,nU2) ? A*(E, U E2) + A*(E, n E2). The last inequality follows from the definition of outer measure. Since e > 0 was arbitrary, inequality (a) follows. (b) Let a, b E Il8 be arbitrary. By (a) applied to [a, b] n E;, we have

A*([a, b] n E,) + A*([a, b] n El) ? A*([a, b] n (E, U El)) + A*([a, b] n (E, n E;)) = A*([a, b] n (E, n E2)`) + A*([a, b] n (E, U E,)`). But by Theorem 10.3.7, for any E C R,

A*([a, b] n E`) = (b - a) - A*(E n [a, b)). Therefore,

A*(E, n [a, b]) + A*(E2 n [a, b]) <_ A*([a, b] n (E, n E2)) + A*([a, b] n (E, U E,)). For each k E N. set 4 = [ -k, k]. By the above, A*(E, n Ik) + A* (E2 n Ik) 5 A*((E, n E,) n 1k) +A*((E, U E2) n 1k) s A*(E, n E2) + A*(E, U E2). The result now follows by Lemma 10.3.8.

EXERCISES 10.3 1.

a. If E C R. a, b E IR, prove that A*(E n (a. b)) = A*(E n [a, b]) and A*(E n (a, b)) = A*(E n [a. b]). *b. If E C R, prove that A*(E + x) = A*(E) and A*(E + x) = A*(E) for every x E R, where

E+x=(a+ x: aEE).

2. Prove that every subset of a set of measure zero is measurable. 3. *Let E, C R with A*(E,) = 0. If E2 is a measurable subset of R, prove that E, n E. and E, U E2 are measurable. 4. Let E _ [0, 1 ] \ Q. Prove that E is measurable and A(E) = 1. 5. Let P denote the Cantor set in [0, 11. Prove that A*(P) = 0. of open sets with E C U. for all n E N such that 6. *If E C R, prove that there exists a sequence

-A*(E) = A*( n U,). 7. Prove Theorem 10.3.2(b). & *Prove Lemma 10.3.8. 9. a. Prove that every open set U is measurable with A(U) = m(U). b. Prove that every compact set K is measurable with A(K) = m(K). c. Prove that every closed set is measurable.

10.4

10.4

Properties of Measurable Sets

449

Properties of Measurable Sets In this section we will study some of the basic properties of measurable sets. Our first result proves that the union and intersection of two measurable sets are again measurable.

10.4.1

THEOREM

If E, and E, are measurable subsets of R, then

E,flE,

and

E,UE,

are measurable with

A(E,) + A(E,) = A(E, U E,) + A(E, fl E,).

Proof. (a) We consider first the case where both E, and E, are bounded measurable sets, in which case, E, fl E, and E, U E, are also bounded. Since A(E) = A*(EE) = A*(E)

i = 1, 2,

by Theorem 10.3.9,

A(E,) + A(E,)

A*(E, U E,) + A*(E, fl E,) A*(E, U E2) + A*(E, fl E,) : A(E,) + A(E2).

Therefore,

A*(E, U E,) + A*(E, fl E,) = A*(E, U E,) + A*(E, fl E,), and as a consequence,

A*(E, U E,) - A*(E, U E2) = A*(E, fl E,) - A*(E, fl E,). But for any bounded set E, A*(E) - A*(E) ? 0. Thus equality can hold in the above if and only if both sides are zero; namely,

A*(E, U E,) = A*(E, U E,)

and

A*(E, fl E,) = A*(E, fl E,).

Therefore, E, fl E, and E, U E, are measurable, with

A(E,) + A(E,) = A(E, U E,) + A(E, fl E,). (b) Suppose one or both of the measurable sets E, and E, are unbounded. Let 1= [a, b] with a. b E R. If both E, and E, are unbounded, then E, n/ and E, fl 1 are measurable by definition. If one of the two sets, say E,, is bounded, then by part (a), E, fl l is measurable. Thus in both cases, E, n l and E, fl 1 are bounded measurable sets. But then

(E, f1 E,) fl I

and

(E, U E,) f11

are measurable for every closed and bounded interval I with

A(E, fl I) + A(E, fl 1) = A((E, fl E2) fl I) + A((E, U E2) fl 1).

(4)

Since E, U E2 is unbounded, E, U E, is measurable by definition. Also, if E, fl E. is

unbounded, then it is measurable by definition. On the other hand, if E, fl E, is bounded, choose I such that E, fl E2 C 1. In this case,

(E, fl E2)f11=E,flE2.

450

Chapter 10

Lebesgue Measure and Integration

Therefore, El n E2 is measurable. Finally, to prove that A(E1) + A(E,) = A(E1 U E2) + A(E, n E2),

take Ik = [ - k, k] in equation (4) and let k -+ oo. Q

Remark. As a consequence of the previous theorem, if E is a bounded measurable subset of R, then £111 is measurable for every bounded interval 1. Also, if El, are measurable sets, then by induction

l Ek

and

... , E.

U Ek

are measurable.

10.4.2

THEOREM A subset E of R is measurable if and only if

A*(E n [a, b]) + A*(E`' n [a, b]) < b - a for every a, b E R.

Proof.

Let E C R. Interchanging the roles of E and E` in Theorem 10.3.7 gives

A*(E n [a, bJ) + A*(E` n (a, b]) = b - a for any a, b E R. If E is measurable, then E n [a, b] is measurable for every a, b E R. and thus

A*(E n [a, b]) = A*(E n [a, b]). Thus if E is measurable, A*(E n [a, b]) + A*(E` n [a, b)) = b - a. Conversely, suppose E satisfies A*(E n [a, b]) + A*(E` n [a, b])

b - a for

every a, b E R. Since we always have

A*(E n [a, b]) + A*(E` n [a, b]) = b - a, we obtain

A*(E n [a, b)) - A*(E n [a, b]) 5 0.

Since A*(E n [a, b]) c A*(E n [a, b]), the above can hold if and only if A*(E Cl [a, b]) = A*(E n [a, b]).

Thus E n [a, b] is measurable for every a, b E R. Hence E is measurable. Q

10.4.3

COROLLARY A set E is measurable if and only if E` is measurable.

Proof. This is an immediate consequence of the fact that E satisfies Theorem 10.4.2 if and only if E' satisfies Theorem 10.4.2. Q Our next goal is to show that the union and intersection of a countable collection of measurable sets is again measurable. For the proof of this result we require the following theorem.

10.4

Properties of Measurable Sets

451

10.4A THEOREM

(a) If

is a sequence of subsets of 18, then

A*(U En) s n=l ; A*(En) (b) If {En}O,O- , is a sequence of pairwise disjoint subsets of R, then

k(

U'

n=1

E- ?I R-I

Proof. (a) If :, A*(En) = oo, then the conclusion in (a) is certainly true. Thus we assume that Y,, A*(EE) < coo. Let e > 0 be given. For each n E N. there exists an open set U. with En C Un such that m(UU) < A*(EE) + 2n

Let U = U00 , Un. Then U is an open subset of R with E = U" , E. C U. Thus A*(E) c m(U Un), which by Theorem 10.2.6

nil

m(U,) <

n=1

(A*(En) + 2n)

00

_

A*(E.) + e. M=I

Since this holds for all e > 0, the result follows. (b) Suppose the sets En, n = 1 , 2, ... , are pairwise disjoint. Since E, fl E2 = 46, by Theorem 10.3.9(b), A*(E,) + A*(E2) S A*(E, U E2). By induction,

kF, A*(EE) 5 A*(U

k-I

E) <_ A"(r1 Ek t.

Since the above holds for all n E N, letting n --* cc gives the /desired result.

10A.5 THEOREM Let {En}., be a sequence of measurable sets. Then 00

U En

00

and

nE

are measurable with (a)

A(U En) 5 n-1 A(E )

O

452

Chapter lo

Lebesgue Measure and Integration

I f in addition, the s e t s ER, n = 1 , 2, . .. , are pairwise disjoint, then A(nUx En) =

(b)

,r A(En) R=l

Proof. Let E = Un°_ I E. Without loss of generality we can assume that E (and hence each of the sets ER) is bounded. If not, we consider

En[a,b] = RU (Enn(a,b]) for a, b E R. Set A, = El, and for each n E N, n ? 2, set An = ER \

(U Ek) = k=I

(n

EE)c

k-1

Since finite unions, intersections, and complements of measurable sets are again measurable, An is measurable for each n E N. Furthermore, the sets An, n E N, are pairwise disjoint with «;

U An = E. Thus by Theorem 10.4.4(a) and (b),

a 00 T, A(An) g A*(E) . A*(E) < 7, A(An) n=1

n=I

Therefore, A*(E) = A*(E), and since E is bounded, E is measurable. Furthermore, by Theorem 10.4.4,

a A(E) 5 7, A(E,,), n

with equality if the sets En, n = 1, 2. .. , are pairwise disjoint. By Corollary 10.4.3, E'0 is measurable for each n. Thus by the above, U',-, E' . is measurable. Since M

fEn= n=1

U00

n-1

the intersection is also measurable. Q 10.4.6

THEOREM

(a) If {En}' 1 is a sequence of measurable sets with E, C E2 C AI U En J = lim A(E0). N=1

noc

, then

10.4

Properties of Measurable Sets

(b) If {E.)' I is a sequence of measurable sets with E, A(EI) < oo, then

E, D

453

and

/

ill (l En = lim A(E,). n-+x n

I

Proof. (a) Let E = U w I En. By the previous theorem, E is measurable. If A(Ef) = 00 for some k, then A(E,,) = oo for all n z k and A(E) = oo, thus proving the result. Hence we assume that A(E,,) < oo for all n. Set Eo = ¢, and for n E N, let A. = E.\En Then each A. is measurable, the collection {An} is pairwise disjoint, and 00

UAn=E. n-1 Thus by Theorem 10.4.5(b), N

A(E) =

Nun T, A(En\ En_ I)

A(An)

n-1

n=1

But E. = (En \ En _ I) U E. _ 1. Since the sets E. \ En _ 1 and En _ , are disjoint.

A(En) = A(EE\ En_1) + A(E._I). Therefore, N

N

I A(En\En-1) _ 2 [A(En) - A(En-1)] = A(EN) - A(Eo) = A(EN),

n-1

n-1

from which the result now follows. (b) Let E= no,*- I En. Again by the previous theorem. E is measurable. Since M

EI = E U U (En \ En+ I), n-I

which is a union of pairwise disjoint measurable sets, by Theorem 10.4.5(b),

A(E,) = A(E) +

n-I N

= A(E) + lim T, [A(E.) - A(En+1)] n=1

= A(E) + A(EI) - slim A(EN+I) Since A(EI) < oo, A(E) = N

A(EN+ 1).

The Sigma-Algebra of Measurable Sets Let .At denote the collection of measurable subsets of R. By Theorem 10.3.6, every interval I is in .AA, with A(I) = m(I). Since every open set U can be expressed as a finite or

454

Chapter 10

Lebesgue Measure and Integration

countable union of pairwise disjoint open intervals, by Theorem 10.4.5 every open set U is measurable with A(U) = m(U).

Since the complement of every measurable set is measurable, At also contains all the closed subsets of R. In particular, every compact set K is measurable with A(K) = m(K), where m(K) is as defined in Definition 10.2.10. These by no means exhaust the measurable sets. Any set obtained from a countable union or intersection of open sets or closed sets, or of sets obtained in this manner, is again measurable. The collection .,il. is very large. To illustrate just how large, we use the fact that the Cantor set P has measure zero. Thus any subset of the Cantor set has outer measure zero, and as a consequence of Theorem 10.3.5, is measurable. By Property 6 of the Can-

tor set (Section 3.3), P has the same cardinality (Definition 1.7.1) as the set of all sequences of 0's and l's. By Miscellaneous Exercise 7 of Chapter 1, the set of all sequences of 0's and l's is equivalent to [0, 1 ], and this set has the same cardinality as all of R. Thus the set of all subsets of P is equivalent (not equal) to the set of all subsets of R. As a consequence, in the terminology of equivalence of sets, .At has the same cardinality as the set of all subsets of P. However, nonmeasurable subsets of P do exist. The construction of such a set will be outlined in the miscellaneous exercises. We conclude this section by summarizing some of the properties of M.

10.4.7

THEOREM

(a) IfEEA,then E`E.M.. (b) 0, H E A. ( e ) I f E E .M., n = 1 , 2,

.

, t h en en

00

00

UEnEAM,

M-1

and

n-1

(d) Every interval I E .At with A(I) = m(I). (e) If E E At, then E + x E .At for all x E P with

A(E + x) = A(E).

Proof.

The result (a) is Corollary 10.4.3, whereas (b) follows from Theorem 10.3.6 and Corollary 10.4.3. The statement (c) is Theorem 10.4.5, and (d) is Theorem 10.3.6. The proof of (e) follows from Exercise lb of the previous section.

Any collection sd of subsets of a set X satisfying (a), (b), and (c) of the previous of sets. The o denotes that the collectheorem is called a sigma-algebra tion s4 is closed under countable unions.

Remark An alternative approach to the theory of measure is credited to Constantin Carathdodory. In this approach a subset E of P is said to be measurable if

A*(E fl T) + A*(E` fl T) = A*(T)

(5)

10.5

Measurable Functions

455

for every subset T of R. Since T = (E fl T) U (E` fl T ). by Theorem 10.3.9 one always has A*(T) s A*(E fl T) + A*(E, fl T ). Thus E satisfies equation (5) if and only if

A*(E fl T) + A*(E, fl T) 55: A*(T). The advantage to this approach is that it does not require the concept of inner measure. and it includes both unbounded and bounded sets simultaneously. If a subset E of Il8 satisfies equation (5), taking T = [a, b], a, b E R, gives

A*(E fl [a, b]) + A*(E` fl [a, b]) = A*([a, b]) = b - a. Thus E satisfies Theorem 10.4.2 and hence is measurable. In Exercise 6. the reader will be asked to prove that if E is measurable as defined in the text, then E satisfies equation (5) for every subset T of Ft

D EXERCISES 10.4 1. Find a sequence OAI fl

of measurable sets with E, D E, D

such that

E I + lim n-'

2. *If E is a measurable subset of I8, prove that given e > 0, there exists an open set U D E and a closed set F C E such that A(U \ E) < e and A(E \ F) < e. 3. Let E be a bounded subset of R. Prove that E is measurable if and only if given e > 0. there exists an open set U and a compact set K with K C E C U such that A(U \ K) < e.

4. *If E E, are measurable subsets of [0. I], and if A(E1) = 1. prove that A(E, n E,) = A(E2). 5. Suppose E, is a nonmeasurable subset of [0. I ]. Set E = E, U (l, oo). Prove that E is nonmeasurable but that A*(E) = A*(E) = oo. 6. *Prove that a subset E of R is measurable if and only if A*(E fl T) + A*(E` fl T) = A*(T) for every subset T of R.

10.5

Measurable Functions In our discussion of Lebesgue's approach to integration, we defined the sets

E) ={xE [a.b]:(j-1)2n -_ f(x) where f : [a, b] -+ (0, 9) is a bounded real-valued function. As we saw in Section 10.1, in order to define the integral off by partitioning the range, it is necessary that the sets E;, j = 1, ... , n, be measurable. Thus we make the following definition.

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Chapter 1 o

10.5.1

Lebesgue Measure and Integration

DEFINITION Let f be a real-valued function defined on (a, b]. The function f is said to be measurable if for every s E 18, the set

{xE[a,b]:f(x)>s} is measurable. More generally, if E is a measurable subset of I8, a function f : E - R is measurable if

{xEE:f(x)>s} is measurable for every s e R. Since f -'((s, oo)) = {x : f(x) > s}, f is measurable if and only if f -'((s, oo)) is a measurable set for every s E R. We illustrate the idea of a measurable function with the following examples.

10.5.2

EXAMPLES

(a) Let A be a measurable subset of I8 and let XA denote the characteristic function of A. Then

{X: XA(x) > s} =

R, A,

0:S s<1,

0,

s ? 1.

s < 0,

Since each of the sets ¢, A, and l are measurable. XA is a measurable function on R. (b) Let f : [0, 1) -+ I8 be defined by

f(x) = J0, x e Q n [0, ii, x,

x E [0, 1 ] \ Q.

Then 1

[0, 1],

ifs < 0,

{x E [0, 1] : f(x) > s} = Q ` fl (s, l), if 0 <_ s < 1, ¢, if s?1. Again, since each of the sets is a measurable subset of f8, f is measurable.

Properties of Measurable Functions We now consider some properties of measurable functions. Our first result provides several equivalent conditions for measurability.

10.5.3 THEOREM Let f be a real-valued function defined on a measurable set E. Then f is measurable if and only if any of the following hold:

(a) {x : f (x) > s} is measurable for every s E R. (b) {x : f(x) ? s} is measurable for every s E R.

10.5

Measurable Functions

457

(c) {x : f (x) < s} is measurable for every s E R. (d) {x : f (x) < s} is measurable for every s E R.

Proof. The set of (d) is the complement of the set in (a). Thus by Corollary 10.4.3. one is measurable if and only if the other is. Similarly for the sets of (b) and (c). Thus it suffices to prove that (a) is equivalent to (b). Suppose (a) holds. For each n E N, let

En = {x : f (x) > s - ,,}. By (a), E. is measurable for all n E N. But {x : AX) a s} = 1

; Ea,

which is measurable by Theorem 10.4.5. Conversely, since 00

{x:f(x) > s} =R=1U{x:f(x)?s+n}, I

if (b) holds, then by Theorem 10.4.5, (a) also holds. 0 10.5A THEOREM Suppose f, g are measurable real-valued functions defined on a measurable set E. Then

(a) f + c and cf are measurable for every c E R, (b) f + g is measurable, (e) fg is measurable, and (d) 11g is measurable provided g(x) # O for all x E E.

Proof. The proof of (a) is straightforward and is omitted. (b) Lets E R. Then f(x) + g(x) > s if and only if f(x) > s - g(x). If x E E is such that f(x) > s - g(x), then there exists r E 0 such that

f(x) > r > s - g(x). Let {r,,},°,°__1 be an enumeration of Q. Then

{x : f(x) + g(x) > s} = U ({x : f(x) > ra} fl {x: ra > s - g(x)}). Since f and g are measurable functions,

{x : f(x) > ra}

and

{x : ra > s - g(x)}

are measurable sets for every n E Ni. Thus their intersection and the resulting union is also measurable. Therefore, f + g is measurable. To prove (c) we first show that f2 is measurable. If s < 0, then

{x E E : f 2(x) > s} = E, which is measurable. Assume s ? 0. Then

{x : f 2(x) > s} = {x : f (x) > \6} U {x : f (x) < -N/S-1-

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But each of these two sets are measurable. Thus their union is measurable. Since

f8 = 4[(f + 8)Z - (f - 8)2]1 the function fg is measurable. The proof of (d) is left as an exercise (Exercise 5).

10.5.5

THEOREM Every continuous real-valued function on [a, b] is measurable.

Proof. Exercise 7.

A Property Holding Almost Everywhere A very important concept in the study of measure theory involves the idea of a property being true for all x except for a set of measure zero. This idea was previously encountered in the statement and proof of Lebesgue's theorem in Chapter 6; namely, a bounded real-valued function if on [a, b] is Riemann integrable if and only if {x : f is not continuous at x) has measure zero. An equivalent formulation is that f is continuous except on a set of measure zero. In this section we will encounter several other properties that are assumed to hold except on sets of measure zero.

10.5.6

DEFINITION A property P is said to hold almost everywhere (abbreviated a.e.) if the set of points where P does not hold has measure zero, i.e., A({x : P does not hold}) = 0.

Remark The assertion that a set is of measure zero includes the assertion that it is measurable. This, however, is not necessary. If instead we only require that A*({x : P does not hold}) = 0, then by Theorem 10.3.5 the set {x: P does not hold) is, in fact, measurable.

We will illustrate the concept of a property holding almost everywhere by means of the following examples.

10.5.7

EXAMPLES

(a) Suppose f and g are real-valued functions defined on [a, b]. The functions f and g are said to be equal almost everywhere, denoted f = g a.e., if

{x E [a, b] :f (x)

g(x))

has measure zero. For example, if g(x) = I for all x E [0, 11 and

AX) = 11, l0,

x E [0, 1] \ 0,

x E [0, ii fl a,

then {x E [0, 1] : f (x) # g(x)) = 0, 1] (l a, which has measure zero. Therefore, f = g a.e.

(b) In Theorem 10.5.4 we proved that if g is a real-valued measurable function on [a, b] with g(x) # 0 for all x E [a, b], then 11g is also measurable on [a, b]. Suppose we replace the hypothesis g(x) # 0 for all x E [a, b] with g # 0 a.e.; that is, the set

E={xE[a,b]:g(x)=0}

10.5

Measurable Functions

459

has measure zero. If we now define f by g(x),

f(x)

1,

x E [a, b] \ E,

xEE,

then f(x) # 0 for all x E [a, b] and f(x) = g(x) except for x E E, which has measure zero. Thus f = g a.e. on [a, b]. As a consequence of our next theorem, the function f will also be measurable on [a, b]. (c) A real-valued function f on [a, b] is continuous almost everywhere if {x E [a, b] : f is not continuous at x} has measure zero. As in Example 6.1.14, consider the function f on [0,1 ] defined by

1(x) =

if x is irrational,

0,

m.

1

if x = - in lowest terms, x # 0. n

n

As was shown in Example 4.2.2(g), the function f is continuous at every irrational number in [0, 1], and discontinuous at every rational number in [0, 1 ]. Therefore, A({x E [0, 1 ] : f is not continuous at x}) = Ap fl [0, 11) = 0.

Thus f is continuous a.e. on [0, 1].

(d) Let f and f, n = 1, 2, ... , be real-valued functions defined on [a, b]. The sequence {f.) is said to converge almost everywhere to f, denoted f --+f a.e., if {x E [a, b] : {f. (x)} does not converge to f (x)}

has measure zero. To illustrate this, consider the sequence (f,,} defined in Example 8.1.2(c) as follows: Let {xk} be an enumeration of 0 n [0, 1 ]. For each n E N, define f, on [0, 11 by AX) __

0, 1,

if x = xk, 1 <_ k:5 n, otherwise.

Then

hm f(x)

0. 1,

x is rational, if x is irrational.

Thus {x E [0, 1 ] : { f (x)} does not converge to I) = an to, 1 ], which has measure zero. Hence f -a 1 a.e. on [0, 1). One of the key results needed in the sequel is the following.

10.5.8

THEOREM Suppose f and g are real-valued functions defined on a measurable set A. !f f is measurable and g = f a.e., then g is measurable on A. Pr oof. Let E = {x E A:g(x) # f(x)}. Then A(E) = 0. Let B = A\ E. Since E is measurable, so is B. Also on B, g(x) = f(x). Fix s E R. Then

{xEA:g(x)>s}={xEB:g(x)>s}UIx EE:g(x)>s} = {x E B:f(x) > s} U {x E E: g(x) > s}.

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Since E, = {x E E: g(x) > s} is a subset of E and A (E) = 0, the set E, is measurable. Also, since f is measurable, {x E B : f(x) > s} is measurable. Therefore, {x: g(x) > s}, and thus g, is measurable.

10.5.9 THEOREM Let f f.}:: , be a sequence of real-valued measurable functions defined on a measurable set A such that (f (x)}' , is bounded for every x E A. Let rp (x) = sup{ f" (x) : n E N)

and

Vi(x) = inf { fn (x) : n E NJ.

Then 9 and ifr are measurable on A.

Proof. The result follows by Theorem 10.4.5, and the fact that for every s E R.

{x:(p(x) > s) = U00 {x: f"(x) > s}

and

n-i

10.5.10

a (x: 4,(x) < s} = n-i U {x: f"(x) < s}.

be a sequence of real-valued measurable functions defined COROLLARY Let on a measurable set A, and let f be a real-valued function on A. /f f" -+f a.e. on A, then f is measurable on A.

Proof. Let E = {x: { f"(x)} does not converge tof(x)}. By hypothesis A(E) = 0. Set gn(x) =

{f"ax),

xEA\E, x E E.

Then g" = fn a.e., and thus is measurable. Also, lim gn(x) = g(x) exists for all x E A. But

g(x) = lim g" (x) = lim gn (x) = inf sup{ fk(x) : k at n}. n-oo

n-"o

n

By the previous theorem, each of the functions

F.(x) = sup{fk(x):k at n}

and

g(x) = inf{F,(x):n E N}

are measurable on A. Finally, since f = g a.e., f itself is measurable. Suppose { fn} is a sequence of measurable functions on [a, b] such that fn --).f a.e.

Then by definition there exists a subset E of [a, b] such that A([a, b]\E) = 0, and slim fn (x) = f (x) for all x E E. Exercise 15 provides a significant strengthening of this result. There you will be asked to prove that given e > 0, there exists a measurable set E C [a, b], such that A([a, b] \ E) < e, and {fn} converges uniformly to f on E. This result is known as Egorov's theorem.

10.5

Measurable Functions

461

EXERCISES 10.5 1. *Let f be defined on [0, 1) by

x=0,

0,

f(x)=

x, :.<'Prove

directly that f is measurable.

2. Let f be a real-valued function on a measurable set A with finite range, i.e., Range f = {ai, ... , aq}. a1 E R. Prove that f is measurable if and only if f -I (a;) is measurable for all j = 1. ... , n. 3. Let A be a measurable subset of R, and let f : A - R be measurable. Prove that for each n E N, the function j defined by

f (x) =

ff(x),

l n,

if If(x)I s n. if )f(x)I > n,

is measurable on A.

4. If f is measurable on [a. b], prove that f + c and cf are measurable on [a, b] for every c E R. 5, *If g is measurable on [a, b] with g(x) * 0 for all x E [a, b], prove that 1/g is measurable on [a, b]. 6. Let f be a real-valued function on [a. b]. Prove that f is measurable if and only if f `(U) is a measurable subset of [a, b] for every open subset U of R. 7. Prove that every continuous real-valued function f on [a, b] is measurable. 8, Let A be a measurable subset of R. If g :A -+ R is measurable and f : R -* R is continuous, prove that f o g is measurable.

9. If f is monotone on [a, b] and g: R - [a, b] is measurable, prove that f e g is measurable. 10. Let E be a measurable subset of R, and let f be a measurable function on E. Define the functions f ' and f on E

as follows: f'(x) = max{f(x), 01, f -(x) = max{-f(x), 01. *a. Prove directly that f' and f- are nonnegative measurable functions on E with f(x) = f' (x) - f- (x).

b. Prove that If(x)I = f'(x) + f-(x) and that If I is measurable. *c. If f is a real-valued function on [a, b] such that I f I is measurable on (a, b), is f measurable on [a. b]?

d. If f(x) = + cos x, x E (0, 21r), find f' (x) and f-(x). 11. Let A be a measurable subset of R and let {f.) be a sequence of measurable functions on A. Let E = {x E A : { f (x)}, .. i converges}. Prove that E is measurable.

12. Let f be a bounded measurable function on [0, 1] and let (f.1 be defined on [0. I ] by f .(x) = x"f(x). Prove that each j is measurable and that (f.} converges to 0 almost everywhere. 13. Let {xk} be an enumeration of the rational numbers in [0, I J. For each n E N let f(x)=

I1, if x=xk,l5k<-n, 0,

otherwise.

Show that f is measurable for each n E N. and that (f.1 converges to 0 almost everywhere on [0, I ].

14. *If f is differentiable on [a, b], prove that f' is measurable on [a, b].

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15. Egorov's Theorem: Let {f} be a sequence of measurable functions on [a. b] such that f. --+f a.e. on [a. b]. Given e > 0. prove that there exists a measurable subset E of [a, h] such that A([a. b] \E) < e and {f} converges uniformly to f on E. 16. Construct a sequence (J,} of measurable functions on [0. 1 ] such that { f (x)} converges for each x E [0, 1 j but that (f.) does not converge uniformly on any measurable set E C [0. I ] with A([0, I ] \E) = 0.

1

10.61 The Lebesgue Integral of a Bounded Function There are many different approaches to the development of the Lebesgue integral. One is the method outlined in the first section of this chapter that is further pursued in Exercise 1. The drawback to this approach is that it a priori assumes that f is a measurable function. The approach that we will follow is patterned on the Darboux approach to the Riemann integral.

10.6.1

DEFINITION Let E be a measurable subset of R. A measurable partition of E is a of pairwise disjoint measurable subsets of E such that finite collection P = {E1, . .. ,

IDEA.=E.

k=1

Suppose 91 = {x,,, x1, ... , xn} is a point partition of [a, b] as considered in Chap-

ter 6. Set E1 = [x,,, x,], and f o r k = 2, .. . , n, set Ek = (xk_ 1, xk]. Then the collection 91 = {E,, . . . , En} is a measurable partition of [a, b]. A measurable partition of [a, b], however, need not consist of intervals. For example, if E, = [a, b] n0 and E2 = [a, b] \E1, then {E1, E,} is a measurable partition of [a, b]. As for the Riemann integral, if f is a bounded real-valued function on [a, b], and 01 = {E1, . .. , En} is a measurable partition of [a, b], we define the lower and upper Lebesgue sums off with respect to 91, denoted L(9), f) and ql.L(91, f) respectively, by n

n

YLO' f) = 7, mkA(Ek)

7, Mk A(EI),

and

k=I

k-1

where mk = inf{ f (x) : x E Ek}

and

Mk = sup { f (x) : x E Ek}.

Clearly, 510, f) : OUL(O, f) for every measurable partition P of [a, b]. As for the Riemann integral, we could now define the upper and lower Lebesgue integrals off, and then define a function to be Lebesgue integrable if and only if these two quantities are equal. The following theorem shows, however, that this is unnecessary.

10.6.2

THEOREM Let f be a bounded real-valued function on [a, b]. Then sUUp Y.(2, f) = inf °11L(9., f ),

where the inftmum and supremum are taken over all measurable partitions 0 and [a, b], if and only if f is measurable on [a, b].

of

The Lebesgue Integral of a Bounded Function

10.6

463

Remark. Although the previous theorem is stated for a closed interval [a, b], the result is also true for f defined on any measurable subset A of l with A (A) < cc. As a consequence of the previous theorem, which we will shortly prove, we make the following definition.

10.6.3

If f is a bounded real-valued measurable function on [a, b], the Lebesgue Integral off over [a, b], denoted f(a bj f dA (or f. f dA), is defined by DEFINITION

t

f dA = sup `.EL(9, f), b)

where the supreemum is taken over all measurable partitions of [a, b]. If A is a measurable subset of [a, b], the Lebesgue integral off over A, denoted fA f dA, is defined by

f dA =

fXA dA, J (a.bj

A

Remarks (a) In defining fA f dA, it was implicitly assumed that f is defined on all of [a, b]. If f is only defined on the measurable set A, A C [a, b], then fXA can still be defined on all of [a, b] in the obvious manner; namely,

(fxA)(x) _

{f(x)' x

A.

Alternatively, if f is a bounded measurable function defined on a measurable subset A of [a, b), we could define

jfdA = sup ZL (a, f ), A

where the supremum is taken over all measurable partitions a of A. However, if . _ {E,, ... , E} is any measurable partition of A, then

9' = {E,,... ,E,,, [a, b] 1A} is a measurable partition of [a, b] with fL(Lf) = 2L(9,fXA) Conversely, if 9 = {El, . . . , Ea} is any measurable partition of [a, b], then 9. = {E, fl A, ... , E. ('I A} is a measurable partition of A for which 21, (Q,f) _ 2L(91,fXA). Thus the two definitions for fA f dA give the same value.

(b) To distinguish between the Lebesgue and Riemann integral of a bounded realvalued function f on (a, b), the Riemann integral off, if it exists, will be denoted by

f bf(x) a

dx.

464

Chapter to

Lebesgue Measure and Integration

If in the Lebesgue integral we wish to emphasize the variable x, we will write f.''d f(x) dA(x) to denote the Lebesgue integral off. The two different notations should cause no confusion. In fact, in Corollary 10.6.8 we will prove that every Riemann integrable function on [a, b] is also Lebesgue integrable, and the two integrals are equal. Prior to proving Theorem 10.6.2 we first need the analogue of Theorem 6.1.4.

10.6.4

Let E be a measurable set and let l? be a measurable partition of E. A measurable partition . of E is a refinement of 9 if every set in a is a subset of some DEFINITION set in 91.

and 9 _

A useful fact about refinements is the following: If 9 = {E1,..., {A,, ... , Am} are measurable partitions of E, then the collection {E ; fl

is a measurable partition of E that is a refinement of both 31 and 9.

10.6.5

LEMMA If 91,.q are measurable partitions of [a, b] such that 2 is a refinement of 91, then

-YL(a,f) c qLL(a.f) As a consequence, scup `.TL(91,f)

Proof. The proof of the lemma is almost verbatim the proof of Lemma 6.1.3 and Theorem 6.1.4, and thus is left to the exercises (Exercise 2).

10.6.6

EXAMPLE

In this example, we calculate the Lebesgue integral of what is commonly

called a simple function on [a, b]. A simple function on [a, b] is a measurable realvalued function on [a, b] that assumes only a finite number of values.

Suppose s is a simple function on [a, b] with Range s = {a,, .. a, * of whenever i 0 j. For each i = I, . . . , n, set

.

,

where

E; = {x E [a, b] : s(x) = a;} = s-'({ail). Since s is measurable, each E; is a measurable set, and n++

S(x) _ !r a;XE,(x)

(6)

Furthermore, since ai * a i if i * j, the sets E i = 1, ... , n, are pairwise disjoint with U'. I E; = [a, b]. Equation (6) is called the canonical representation of s. If all the sets E; are intervals, then s is a step function on [a, b]. We will now show that every simple function s is Lebesgue integrable on [a, b] and compute the Lebesgue integral of s.

10.6 The Lebesgue Integral of a Bounded Function

10.6.7

465

LEMMA If s is a simple function on [a, b] with n

s = 7,, arXE, where {E;}"= I are pairwise disjoint measurable subsets of [a, b] with U"= I E, = [a, b], then s is Lebesgue integrable on [a. b] with

aA(E,).

s dA

J(u, hl

j=1

Proof. To show that s is Lebesgue integrable on [a, b], we will prove that s) = inf 61L.L(9., s),

sup a

.4

where the supremum and infmum are taken over all measurable partitions 9. of [a, b].

Since 9' = {E1..... ER} is a measurable partition of [a, b] and s(x) = a; for all x E E. aiA(E;).

-WL(9', s) = eILL(91, s) _

But then n

n

I ai A(Ej) !-__ sup $L(9., s) < inf altl(9 s) :5:- 1 ai A(Ei). a

-1

a

;-1

Therefore, s is Lebesgue integrable on [a, b] with f[, bis dA = I°=1 a;A(E;). 0

Remark Suppose f is a bounded real-valued measurable function on [a, b]. If 91 = {E ,, ... , En} is a measurable partition of [a, b], set (x) = E mkXE,(x)

Ij!(x) = I MkXE,(x),

and

k=1

(7)

k=1

where Mk = inf {f(x) : x E Ek} and Mk = sup{ f (x) : x E Ek}. Then tD and alt are simple functions on [a, b] with O(x) f(x) s *(x) for all x E [a, b]. Furthermore, by Lemma 10.6.7

J[a. bJ

apdA = k-1

mkA(Ea) = 20',.f),

and

MkA(EE) = °LL(9,.f)J[a. b] OdA = Y, k-1

Thus if f is a bounded real-valued measurable function on [a. b],

fdA = sup

a, b;

f

,p dA, u. hJ

where the supremum is taken over all simple functions ap on [a, b] satisfying V(x)

for all x E [a, b].

f(x)

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Chapter 10

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Proof of Theorem 10.6.2. Suppose f is a measurable function on [a. b) with m f (x) < M for all x E [a, b]. Let (3 = M - m, and for n E N, partition (m, M) into n subintervals of l e n g t h 13/n. F o r each k = 1, ... , n, set

Ek= {xE [a,b]:m+(k-1 n -5f(X) Then 91. = {E1, ... , E,,} is a measurable partition of [a, b). Also, if mk = inf{f(x) : x E Ek} and Mk = sup{ f (x) : x E Ek}, then

mk?m+(k-1)Pn

Mksm+kR. n

and

Therefore,

0 s innf °ltL(°.P, f) - suup Et(L,f) s °ut(9n,f) - XL(On,f)

L (Mk -

mk)A(Ek)

k-1 n

\\

[(m

= - J A(Ek) =

A(Ek)

n

(b - a).

Since n E N is arbitrary, by letting n -+ oo we obtain

suup L('Q, f) = inf VL(9,f)

(8)

Conversely, suppose equation (8) holds. By taking a common refinement if necessary, for each n E N, there exists a measurable partition !Pn of [a. b] such that °1fL(9n,f) < innfG(LL(°,f) + -TL(O'n,f)

,

and

5P

Since equality holds in equation (8),

'(LL(9'n,f) - 20n,f) < n For the partition p,,, let cpn and On be simple functions on [a, b] as defined by equation (7), satisfying (p,,(x) s f(x) s 11,,(x) for all x 6 [a, b], and

J(a. b]

(p,, dA = 21fl'n, f ),

1(a. b)

'Yn dA = °ltt{OImf)

Define (p and 11 on [a, b] by v(x) = sup,, (p,,(x) and 11(x) = infra 10.5.9, the functions (p and 11 are measurable functions on [a, b], with (p(x) s AX) < (G(x)

for all x E [a, b].

By Theorem

10.6 The Lebesgue Integral of a Bounded Function

467

To complete the proof we will show that V = 41 a.e. on [a, b]. Then as a consequence of Theorem 10.5.8, the function f will be measurable on [a, b]. Let E _ {x E [a, b] : (p(x) < i/i(x)}, and for each k E Nl, let

Ix E [a, b] : V(x) < +l'(x) - k

Ek =

for all n E lJ For n, k E N. let

Then E = Uk , Ek. If x E Ek, then v,,(x) < k

An. k = 1 x : 4p,(x) < 1P .(X) -

If x E A

, k,

then 4#.(x) -

`1.

k

Consider the simple function

S.(x) = (4r,(x)

Vn(X))XA..,(x).

Suppose the measurable partition 91 is given by {B,, .... BN}. Then N

s (x) _ 71 (M, - mi)Xe nA,..,(x),

i-I

where Mi and mi denote the supremum and infimum off, respectively, over Bi. The collection

9 = {Bi flA,,,k}" 1 U {[a, b] \A,,,k} E Bi fl A,, k}, then mj > 1/k is a measurable partition of [a, b]. If m* = for all j = 1, ... , N. Also, since s (x) = 0 for all x E [a, b) \ A., k, m j A(B, n A,,, k) >

L(9-, S.) _ i=1

1 A(B f l A,,. k) = k i.,

A(A. k)

k

On the other hand, N

i=I

(Mi - mi)A(Bi n A,,. k)

N+

t G(M1 - mi)A(Bi) _

,(91.,f) <

XT,

n

1=I

Combining the above two inequalities gives A(A,,. k) < k/n for all k, n E N. Since Ek C A,,, k for all n; for each k,

A(Ek) < n

for all n E N. Therefore, A(Ek) = 0. Finally, since m

A(E) <_ 1 A(Ek), k-1

we have A(E) = 0. Thus V = 4 a.e. on [a. b), which proves the result.

468

Chapter io

Lebesgue Measure and Integration

Comparison with the Riemann Integral The definition of the Lebesgue integral is very similar to that of the Riemann integral, except that in the Lebesgue theory we use measurable partitions rather than point partitions. If 91 = {xo, xt, . . . , is a partition of [a, b], then 9'* _ {[x0, x1 ]} U {(xk_ 1, Xk]}R= 2

is a measurable partition of [a, b]. Furthermore, if f is a bounded real-valued function on [a, b), then

2(9.f) < 2L(?*,f)

qt(QP.f) > qtL(?*.f)

and

Therefore, the lower Riemann integral off satisfies h

i f = sup{J'(P, f) : 91 is a partition of (a, b]} s sup{.TL(a, f) : I), is a measurable partition of [a, b]}. Similarly, for the upper Riemann integral off we have

J f ?.inf {°ltt(a, f) : 1 is a measurable partition of [a, b)}. If f is Riemann integrable on [a, b], then the upper and lower Riemann integrals off are equal, and thus J46f

(x) dx 1:-: sup `.eL(a, f)

ml 04L(2, f)

s

la bf

(x) dx,

where the supremum and infimum are taken over all measurable partitions a of [a, b]. As a consequence of Theorem 10.6.2, this proves the following result.

10.6.8

COROLLARY 1f f is Riemann integrable on [a, b], then f is Lebesgue integrable on [a, b], and h

(

fdA =

+

f Ax) dx. u

The converse, however, is false! This is illustrated by the following example.

10.6.9

EXAMPLE Let E = [0, I ] \ 0, and set

f(x) = XE(X) = {0,

when x is irrational, when x is rational.

By Example 6.1.6(a) the function f is not Riemann integrable. On the other hand, since f is a simple function, f is Lebesgue integrable, and by Lemma 10.6.7,

f dA = A(E) = 1.

Jlo. l i

10.6

The Lebesgue Integral of a Bounded Function

469

Properties of the Lebesgue Integral for Bounded Functions The following theorem summarizes some basic properties of the Lebesgue integral for bounded functions.

10.6.10 THEOREM Suppose f, g are bounded real-valued measurable functions on [a. b). Then (a) for all a,,8 E R,

(af+ 6g)dA=a f fdA+0 f a b]

gdA.

' a b]

i b]

(b) If A,. A2 are disjoint measurable subsets of [a, b], then

JA,UA2

fdA= JfdA+1 fdA. A,

A,

fdA ? J

(c) If f a g a.e. on [a, b], then

I(a. b]

(d) Iff=ga.e.on[a,b],then J fdA=

fdA < Jla, b]

f

gdA.

(a, b]

(a. b]

(e)

gdA.

[a, b]

If 1A. J(a, b]

Proof. Since the proof of (a) is similar to the proof of the corresponding result for the Riemann integral we leave it as an exercise (Exercise 4). For the proof of (b), by definition

f

fdA =

A,UA2

f

f%A,UA, A.

ia.b]

Since A, fl A2 = fXA,UA, = fXA, + fXA2, and the result now follows by (a). (c) Consider the function h(x) = f(x) - g(x). By hypothesis h ? 0 a.e. on [a, b i. Let

E, = {x:h(x) >_ 0}

and

E2 = [a, b]\ E,.

Consider the measurable partition 9' = {Et, E2} of [a, b]. Then

h A ? XL(91, f) = m,A(E,) + m2A(E,). a. b]

470

Chapter 10

Lebesgue Measure and Integration

Since h(x) z 0 for all x E E,, m, = inf{h(x):x E E,} > 0. On the other hand, since h ? 0 a.e., A(E2) = 0. Therefore, fa h dA ? 0. The result now follows by (a). The result (d) is an immediate consequence of (c), and (e) is left for the exercises. The measurability of f I follows from Exercise 8 or 10 of the previous section. Q

Bounded Convergence Theorem One of the main advantages of the Lebesgue theory of integration involves the interchange of limits. If if.} is a sequence of Riemann integrable functions on [a, b] such that fn(x) converges to a function f(x) for all x E [a, b], then there is no guarantee that f is Riemann integrable on [a, b]. An example of such a sequence was given in Example 8.1.2(c). For the Lebesgue integral, however, we have the following very useful result.

10.6.11

is a sequence of realvalued measurable functions on (a. b) for which there exists a positive constant M such

THEOREM (Bounded Convergence Theorem) Suppose that I fa(x)I

M for all n E N. and all x E [a, b]. If liim fn(x) = f (x) a.e. on [a, b],

then f is Lebesgue integrable on [a, b] and

f dA = lim

1

[a. bj

n-4ooJ

fn dA.

la. bj

Remark. Although we state and prove the bounded convergence theorem for a closed and bounded interval [a, b], the conclusion is still valid if the sequence {fn) is defined on a bounded measurable set A. The necessary modifications to the proof are left to the exercises.

Proof. Since fn --> f a.e., f is measurable by Corollary 10.5.10, and thus Lebesgue integrable. Let

E = {x e (a, b): fn(x) does not converge to f(x)). Define the functions g and g,,, n E N, on [a, b] as follows:

gn(x) = {f(x), 0,

x E (a, b) \ E,

xEE,

and

g(x) _

f(x),

x E [a, b] \ E.

xEE.

0,

Since A(E) = 0, gn = fn a.e. and g = f a.e. Therefore, r6 Ja

b

b

"dA =

fndA 1a

and

b

Ja gdA= I fdA.

Furthermore, gn(x) -+g(x) for all x E [a, b]. Let e > 0 be given. Form E N. set

E. _ {x E [a, b]:Ig(x) - gn(x)I < e forall n >_ m}.

10.6

Then E, C EZ C

The Lebesgue Integral of a Bounded Function

471

with U° , E. = [a, b]. Therefore, a fly Em = 45.

Here Em = [a, b] \ Em. Thus by Theorem 10.4.6, lim A(E,) = 0. Choose m E N such e for all7n*w m and all x E Em. Therefore, that A(Em) < e. Then I g(x) -

f6fdA- jbfdAl =

I

bgdA

IQ

-

f bgndAl

-'5

1

Ig

gn)dA

fa. bJ

jt

=

- g d + t1& - gdA

< e A(E,) + 2M A(Ec) < [b - a + 2M]. Since e > 0 was arbitrary, we have lire fla

b]

f dA = fla b, f dA. Q

Combining the bounded convergence theorem with Corollary 10.6.8, we obtain the bounded convergence theorem for Riemann integrable functions previously stated in Chapter 8. The theorem does require the additional hypothesis that the limit function f is Riemann integrable.

THEOREM 8.4.3

with lim f(x) such that [

Let f and f,,, n E N, be Riemann integrable functions on [a, b] all x E [a, b]. Suppose there exists a positive constant M M for all x E [a, b] and all n E N. Then lim fl-

10.6.12

f (x) dx. I

a

a

EXAMPLES

(a) In the first example, we show that the conclusion of the bounded convergence theorem is false if the sequence f f,,) is not bounded; that is, there does not exist a finite constant M such that I f (x)l <_ M for all n E N, and all x E [a, b]. For each n E N, define f on [0, 11 by

(x) = Mx)

f Sl

n, 0<x51n 0,

otherwise.

Then { is a sequence of measurable functions on [0, 1] that is not bounded, but that satisfies

1im f (x) = f (x) = 0 for all x E [0, 1 ].

472

Chapter 10

Lebesgue Measure and Integration

However, f, f. dA = nA((0, ]) = 1. Thus fdA.

f dA = 1 vk 0=

lim

n-fM

110,1:

J [0.11

(b) As our second example, we consider the sequence { of Example 9.2.3. For each n E N, write n = 2k +j. where k = 0, 1, 2, .. . , and 0:!:-: _ j < 2k. Define f on [0, 1 ] by

j 2k

.f "(x) 0,

j+1 `- x `-

2k

otherwise.

The first few of these are as follows: f, = %,o_ :, f,- _ X[o,1/41

X;0.1/: .f3 - X.1/2. l: f4

.. For each n E N, f. E Jt[0, 1 ] with 1

f

J0

1

dx = 2k

Thus lim fo f (x) dx = 0. On the other hand, if x E [0, 1], then the sequence { f (x)} contains an infinite number of 0's and 1's, and thus does not converge.

EXERCISES 10.6 1. *Let f be a bounded real-valued measurable function on [a, b) with m < f(x) < M for all x E [a, b]. Set

S=M-m.FornENandj=1....,n,letE1={xE[a,b]:m+(j-1)9,<_f(x)<m+jj}.The mJ , (m + (j -1))A(E,). Prove that

Lebesgue sums for f are defined by lint

fdA.

2. Prove Lemma 10.6.5.

3. Let f be a bounded measurable function on ;a, b]. If A is a measurable subset of (a, b) with k(A) = 0. prove that

4. a. Prove Theorem 10.6.10(a). b. Prove Theorem 10.6.10(e). S. *Let f be a nonnegative bounded measurable function on [a, b]. If E, F are measurable subsets of (a, b) with E C F. prove that fE f dA s f,,f dA. 6. Let f be a bounded measurable function on [a, b]. For each c > 0, prove that

A({x E [a, b]: I f(x)I > c}) s 1

1 fl dA.

Lbl 7.

*Let f be a nonnegative bounded measurable function on [a, b] satisfying fta bl fdA = 0. Use the previous exercise to prove that f = 0 a.e. on [a, b].

10.7

The General Lebesgue Integral

473

8. Fundamental Theorem of Calculus for the Lebesgue Integral: if f is differentiable on ia. bj and f' is bounded on [a, h], then f' is Lebesgue integrable, and f,,.,, f' dA = f(b) - f (a). 9. Let.f be a bounded measurable function on [0. 11. Show that x"f(x) is measurable on [0. 11 for n = I. 2.... . and find lint

f be a bounded measurable function on [a, b] such that f [ < I a.e. on [a, b]. Prove that lim f

f " dA = 0.

11. Let

be a sequence of nonnegative measurable functions on [a. h' satisfying /;,(.r) - M for all x E :a. b' and n E N. If f f,) converges to f a.e. on [a. b], prove that

lint f

f e f- A = f

fe-f dA.

" b:

12. *If f is a bounded real-valued measurable function on [a. bl, prove that there exists a sequence {s"} of simple

functions on [a, b] such that lint s"(x) = f(r) uniformly on [a, b]. 13. Modify the proof of the bounded convergence theorem when the interval [a. b] is replaced by a bounded measur-

able set A.

14. Use Egorov's theorem (Exercise 15. Section 10.51 to provide an alternative proof of the bounded convergence theorem. 15. *If p is a simple function on (a. h] and e > 0 is given, prove that there is a step function b on :a. h_ such that

c(x) = h(x) except on a set of measure less than e. 16. Let S"(x) = A0 + Yi-i (Ax cos kx + BA sin kx). If , S"(x) I s M for all .r E [ - ar. a] and n E N. and f(x) = lint S"(x) exists a.e. on [-a, a], prove that f is measurable and that the A4 and B5 are the Fourier coefficients off.

7

10.71

The General Lebesgue Integral in this section we extend the definition of the Lebesgue integral to include both the cases where the function f is unbounded, and where the domain of integration is unbounded. We will then prove the well-known results of Fatou and Lebesgue on the interchange of limits and integration. We first consider the extension of the Lebesgue integral to nonnegative measurable functions.

The Lebesgue Integral of a Nonnegative Measurable Function Suppose first that A is a bounded measurable subset of R. and that f is a nonnegative measurable function defined on A. For each n E N, consider the function f" defined on A by

f(x)

n} =

jf(t:), iff(x) n , I. n. if f(x) > n.

474

Chapter 10

Lebesgue Measure and Integration

Then (fn) is a sequence of nonnegative bounded measurable functions defined on A. with l fn(x) = f (x) for all x E A. Furthermore, if m > n, then MX) 5 fm(x) 5 f(x)

for all x E A, and thus the sequence

JAfn

dAK

is monotone increasing, and converges either to a real number or diverges to oo. This leads us to the following definition.

10.7.1

DEFINITION

(a) Let f be a nonnegative measurable function defined on a bounded measurable subset A of R. The Lebesgue integral off over A, denoted fA f dt, is defused by

A

A

A

(b) If A is an unbounded measurable subset of R and f is a nonnegative measurable function on A, the Lebesgue Integral off over A, denoted fA f dA, is defined by

Jfr,J A

-+oc

fdA.

Afl[-n, n]

In part (b) of the definition, the sequence f dAl }nEN

fArj -n, n]

J

is also monotone increasing, and thus converges either to a nonnegative real number or diverges to oo. In the definition we do not exclude the possible value of oo for the integral off. If the integral is finite, however, we make the following definition.

10.7.2

DEFINITION A nonnegative measurable function f defined on a measurable subset A of R is said to be (Lebesgue) Integrable on A if fA f dA < oo.

Remark If A is either a finite or infinite interval with endpoints a, b E R U (-00, oo), then the integral of a nonnegative measurable function f on A is also denoted by f; f A.

The General Lebesgue Integral

10.7

10.7.3

475

EXAMPLES

(a) For our first example, we consider the function f(x) = I// defined on (0,1). Then for each n E IN,

0<x<

n,

fn(x) = min{f(x), n} =

1 n-

1

Therefore,

ndx+

f fndA= 1 0

0

1

l

dx=

n

J

n

1 n

As a consequence, I

f dA = liml 2 - 1) = 2.

f dA = lim 1

The answer in this example corresponds to the improper Riemann integral of the func tion f. This will always be the case for nonnegative functions for which the improper Riemann integral exists (Exercise 18).

(b) Let g(x) = 1/x, 0 < x K 1. For the function g,

min{g(x), n} =

n, 0<x< 1,n 1

-<x< 1. n 1

x' Therefore, I/n

1

J min{g, n} dA = I 0

t

n dx +

0

1

J - dx = 1 + Inn. 1/fix

Thus

Jgdlt=lim(1 +Op

+ In n) = oo.

Since the Lebesgue integral of g is infinite, g is not integrable on (0, 11.

(c) As our final example, consider f (x) = 1.2 defined on A = (1, oc). In this example,

for n='2,Afl[-n,n] =(1,n),and

jfdA = jx2dx = I - 1.n

476

Chapter 10

Lebesgue Measure and Integration

Therefore,

I` fdA=lim1-n=1. I

Thus f is integrable on (1, oo).

The following theorem summarizes some of the basic properties of the Lebesgue integral of nonnegative measurable functions. Integrability of the functions f and g is not required.

10.7.4 THEOREM Let f, g be nonnegative measurable functions defined on a measurable set A. Then

(a)

j(f+g)dA= jfdA+

jgdA

and

tcfdcJA1A

for all c > 0. (b) If A,, A2 are disjoint measurable rsubsets of A, then

fdl +

fdA = JA, UA.

1A,

(c) If f s g a.e. on A, then jfdA

j

fdA.

A,

I g dA, with equality if f = g a.e. on A.

A

A

Proof. We will indicate the method of proof by proving part of (a). The remaining proofs are left to the exercises (Exercise 2). Suppose first that the set A is bounded. Let

h = f + g. Since min{ f (x) + g(x), n} 5 min{ f (x), n) + min{g(x), n} <_ min{ f (z) + g(x), 2n},

we have h < f + g <_ h2R for all n E N. As a consequence, fAh- dAs

!g,, dA _ Ih2.dA. A

A

A

Suppose f, g are integrable on A. Then

mHf,, dA + fgndA= ) 1m A

A

jf dA + Aliimm f g dA = JfdA+ JIdA. A

A

A

Therefore, since f

lim h dA = lim J h2i dA = j(f + g) dA,

n-.Wf

A

A

A

A

10.7

The General Lebesgue Integral

477

the result follows. If one or both of the sequences { fA fn dA}, { f, gn dA} diverges to x,

then so does their sum. In this case, we obtain fA(f + g)dA = oo. If A is unbounded. then by the above for each n E N,

j

fdA +

(f + g)dA = J

gdA, JAnl-n.r,

An;-n.._

Ant-n.n)

Q

and the result again follows by letting n -+ co.

As a consequence of the previous theorem, if f and g are nonnegative integrable functions on the measurable set A, then so is f + g and cf for every c > 0. Furthermore, if f <_ g a.e. and g is integrable, then so is f.

Fatou's Lemma Our first major convergence theorem for integrals of nonnegative measurable functions is the following result of Fatou.

10.7.5

THEOREM (Fatou's Lemma) If {fn} is a sequence of nonnegative measurable functions on a measurable set A, and slim fn (x) = f(x) a.e. on A, then 1 f dA <_ lim JfndA.

Proof. Suppose first that the set A is bounded. For each k E N, let and

hn(x) = min{ fn(x), k}

h(x) = min{ f (x), k}.

Then for each k E N, the sequence {hn} converges a.e. to It on A. Since I hn(x) 1 s k for all x E A, by the bounded convergence theorem,

J min{ f, k) dA = lim f min{ fn, k} dA s lim Ji

°'A

A

A

Since the above holds for each k e N,

f f dA = lim J min{ f, k} dA s lim J f, dA. A

noc A

k-ooo A

If A is unbounded, then by the above for each k e N.

fAn(-k. kJ

f dA s lim W__-00

f

fn dA c lim fn dA.

Letting k -+oo will give the desired result. Q

W__00

f

A

478

Chapter 10

10.7.6

Lebesgue Measure and Integration

EXAMPLE In this example we show that equality need not hold in Fatou's lemma. Consider the sequence ( fn} on [0, 1 ] of Example 10.6.12(a). For each n E N, fn(x) = n

if 0 < x s 1, and fn(x) = 0 elsewhere. This sequence satisfies

0=

f (lim lim J f dA. nix fn) dA < I =n-.oo 0, l1

f 0, I j

Remark. Fatou's lemma is often used to prove that the limit function f of a convergent sequence of nonnegative Lebesgue integrable functions is Lebesgue integrable. For if lim fA fn dA < oo and if fn -+f a.e. on A with fn ? 0 a.e. for all n, then by Fatou's lemma, fA f dA < oo. Thus f is integrable on A.

The General Lebesgue Integral We now turn our attention to the case where f is an arbitrary real-valued measurable function defined on a measurable subset A of R. As in Exercise 10 of Section 10.5, we define the functions f' and f " on A as follows:

f'(x) = max{f(x),0},f-(x) = max{ f(x),0}. If f (x) > 0, then f +(x) = f (x) and f -(x) = 0. On the other hand, if f (x) < 0, then f''(x) = 0 and f- (x) = -f(x). If f is measurable on A. then f+ and f- are nonnegative measurable functions on A with

f(x) =f'(x) -f-(x)

and

lf(x)I =f`(x) + f (x)

for all x E A. the integral off over A by Our natural inclination is to definejf+dA

jfdA

- jrdA.

=

A

A

A

The only problem with this definition is that it is possible that fAf' dA = fA f- dA = co, giving the undefined oo - oo in the above. However, if we assume that both f' and fare integrable on A, then the above definition makes sense. Furthermore, if f is measur-

able, and f' and f- are both integrable on A, then I f j is also integrable on A. Conversely, if f is measurable and f I is integrable on A, then since f' s If I and f - s if 1, by Theorem 10.7.4(c) both f' and f - are integrable on A. Therefore, we make the following definition.

10.7.7

DEFINITION Let f be a measurable real-valued function defined on a measurable subset A of R. The function f is said to be (Lebesgue) integrable on A if If I is integrable

on A. The set of Lebesgue integrable functions on A is denoted by 2(A). For f E 2(A), the Lebesgue integral off on A is defined by

f f dA = A

f f+ dA - jf- dA. A

A

10.7

The General Lebesgue Integral

479

Remark. The set Z(A) of Lebesgue integrable functions on A is often also denoted by -T'(A).

The definition of the general Lebesgue integral is consistent with our definition of the Lebesgue integral of a bounded function on [a, b'. If f is a bounded real-valued

measurable function on [a, b] then so are the functions f' and f-. Let E, = {x: f(x) a 0} and E2 = {x: f(x) < 0}. Then E, and E2 are disjoint measurable subsets

of [a, b] with E, U E2 = [a, b]. Furthermore, AE, = f" and fXE, = -f -. By Theorem 10.6.10(b) f

jfdA = j fdA + jfdA fb

f bb

b

(b

dA = j f + dA - 1 f - dA.

fXE, dA + a

a

a

a

Remark For a nonnegative measurable function, the Lebesgue integral and the improper Riemann integral are the same, provided of course that the latter exists (Exercise 18). This, however, is false for functions that are not nonnegative. For example, consider the function f(x) = (sinx)/x, x E [7r,oo), of Example 6.4.4(b). By Exercise 7 of Section 6.4, the improper integral off exists on [-rr, oo). However, as was shown in Example 6.4.4,

j

tn

IfIdA= lintj

"I

sinxl x

dx=00.

it

Thus f is not Lebesgue integrable on [ir,oo). Another such example for a finite interval is given in Exercise 23. The crucial fact to remember is that a measurable function f is Lebesgue integrable on a measurable set A if and only if IA If I dA < oo.

Our first result is the following extension of Theorem 10.7.4 to the class of integrable functions.

10.7.8

THEOREM Suppose f and g are Lebesgue integrable functions on the measurable set A. Then

(a) f + g a nd cf, c E R, are integrable on A with

j(f +g )dA= jf dA f g dA A

+

A

A

and

f cfdA=cjfdA. A

(b) If f 5 g a.e. on A, then

jfdA 5 A

j g dA, with equality if! = g a.e. A

A

480

Chapter i0

Lebesgue Measure and Integration

(c) If A, and A, are disjoint measurable subsets of A, then

= J fdA+ 1 fdA.

JAU! Proof.

A;

A,

(a) The proof that cf is integrable and that fA cf dA = cfA fdA follows im-

mediately from the definition. Before proving the result about the sum, we first note that the definition of the integral off on A is independent of the decomposition f = f - - f -.

Suppose that f = f, - f2 where f, and f, are nonnegative integrable functions on A. Then

f'+f

=f-+ f,,

and thus by Theorem 10.7.4,

jf+dA

+

jfdA = JA1A+

J f,dA.

Since all the integrals are finite,

J fdA = A

jfdA_ f f-dA = ffidA_ f f2 A. A

A

A

A

If f and g are integrable on A, then by definition so are the functions f' + g` and

f + g-. Since f + g = (f+ + g+) - (f- + g-), by the above

j(f+ g)dA =

A

f (f++g+)dA- j(f-

g)dA,

A

A

which by Theorem 10.7.4

=

ji dA - ji- dA + jg dA A

A

=

f fdA +

A

f g' dA A

jgdA. A

A

(b) If f s g a.e., then g

f ?0 a.e. Therefore by part (a),

O-fJ(g)dA= jgdA - ffdA. A

A

A

from which the result follows. The proof of (c) also follows from (a) and the fact that since A, n At = b, fXA,UA, -- f XA, +.fXA;

10.7

The General Lebesgue Integral

481

Lebesgue's Dominated Convergence Theorem Our second major convergence result is the following theorem of Lebesgue.

10.7.9

THEOREM (Lebesgue's Dominated Convergence Theorem) Let {fn} be a sequence of measurable functions defined on a measurable set A such that Jim fn (x) = f(x) exists a.e. on A. Suppose there exists a nonnegative integrable fiinction g on A such that If. (x) I g(x) a.e. on A. Then f is integrable on A and

jfdA

I'mJAfndA.

=

Proof. Since g is integrable on A, and I fn 15 g a.e. on A, by Theorem 10.7.4 each fn is also integrable on A. Also, by Corollary 10.5.10, the function f is measurable on A. Furthermore, by Fatou's lemma,

AA

jIfId!jIfnIdA s

jgdA < oo. A

Thus f is also integrable on A. .By redefining all the f,,, n E N. on a set of measure zero if necessary, we can without loss of generality assume that If. (x) I <_ g(x) for all x E A. Consider the sequence (g + fn}nEN of nonnegative measurable functions on A. By Fatou's lemma,

J (g + /) dA A

j(g+fn)dA

=

I

=

f gdA +

A

A

lim

f fn A.

nix A

Therefore,

f f dA <_ lim f n~x A A

A.

S imilarly, by applying Fatou's lemma to the sequence {g quence of nonnegative functions on A,

f (g - f) dA s lim j (g - fn) dA = A

A

j g dA + Jiim j (-fn)dA. A

But lim fA ( fn)dA = -lim fA f A. Therefore. n-oc n-*=

f f dA > n-.x lim f fn dA. A

which is again a se-

A

Combining the two inequalities gives the desired result. O

A

482

Lebesgue Measure and Integration

Chapter 10

Remark. The hypothesis that there exists an integrable function g satisfying I fn I

!f- g

a.e. is required in the proof in order to subtract f g dA in the above inequalities. This is not possible if f gdA = oo. As the following example shows, if such a function g does not exist, then the conclusion may be false.

10.7.10

EXAMPLE As in Example 10.6.12(a) consider the sequence { fn} on [0, 1] defined by

In, fn(x) =

0 <x<

1

n' 10, elsewhere. For the sequence { fn} we have lim fn (x) = 0 for all x E [0,1 ] but f o If,, dA = 1 for

all n. We now show that any measurable function g satisfying g(i) ? fn (x) for all X F: [0,I] and n E N satisfies f o 11gdA = oo. Since g(x) _' fn(x) f o r all x E (0,I] we have g(x) ? n for all x E (n:11 ,1-, ]. Since the collection {(kk + 11, k]}k=t of intervals is

pairwise disjoint, by Theorem 10.7.4, n

gdAa

gdA

J

k=1 ft I

[0 11

k+I) 1-1

,kA\I k+1 1 k1J/

k=1

Since the series 7 k+lt diverges, we have f(o 1)gdA = 00.

EXERCISES 10.7 1. Let f be a nonnegative measurable function on a measurable set A. If f4 f dA = 0. prove that f = 0 a.e. on A.

2. *a. Prove Theorem 10.7.4(b). b. Prove Theorem 10.7.4(c).

3. Let A be a measurable subset of R. a. If f is integrable on A and g is bounded and measurable on A, prove thatfg is integrable on A. b. If f and g are integrable on A, is the function fg integrable on A?

4. *Let f,(x) = x-°, x E (0, 1). Prove that f, is integrable on (0, 1) for all p, 0 < p < 1, and that 1

1 --p' 1 p.

(Q,, J(-' 1)

5. Define f on [ 1, oo) by

f(x) _

{ V, 0,

if x E [n, n + l ln'"), n = 1, 2, ..

,

otherwise.

Show that f E X([ 1, oo)) but f2

`.P([ 1, oo)).

6. Let P denote the Cantor set of Section 3.3. Define f on [0, I ] as follows: f(x) = 0 for every x E P. and f(x) = k for each x in the open interval of length 1/3k on [0, 1] 1 P. Prove that f is integrable on [0, 1 ] and that fo fdA = 3. 7. *Let f be a nonnegative integrable function on [a, b]. For each n E N, let E. = {x: n s f(x) < it + 1 }. Prove that

7, , nA(E,) < oo. 8. Evaluate( each of the following limits:

(I -

a, lim J (0. 1)

dx

b. lim n-.oo

(1 - x/n)nex"2 dx 1

L. W

10.7 The General Lebesgue Integral

483

9. Let f be an integrable function on (a, b]. Given e > 0, prove that there exists a bounded measurable function g on [a, b] such that fla bl If - g I dA < e. 10. *Suppose f is integrable on a measurable set A with A (A) = oo. Given e > 0, prove that there exists a bounded measurable set E C A such that LIE I f I dA < E. 11. Show by example that Fatou's lemma is false if the functions f,,, n E N, are not nonnegative. 12. Show by example that the bounded convergence theorem is false for a measurable set A with A (A) = oo. 13. *Let f be a Lebesgue integrable function on a measurable set A. Prove that given e > 0, there exists a 8 > 0 such that fE If I dA < e for all measurable subsets E of A with A (E) < S. 14. Let f be amr integrable function on (a, b), where -oo <_ a < b s oo. Define F on (a, b) by

F(x) = 4 z f dA, x E (a, b). J,

a. Prove that F is continuous on (a, b). b. If f is continuous at x, E (a, b), prove that F is differentiable at x, with F'(x,) = f(x,). 15. If f E `.e([0, I ]), show that for every t E R the function x - sin (rf(x)) E `.e([0, 1]), and that g(r) = flo it sin(tf(x))dA(x) is a differentiable function oft E R. Find g'(t). 16. If f G X(R), prove that lim JR f(x) sin nx dA = 0. (Hint: First prove the result for step functions.)

17. If f e T(R), prove that ,lim JR If(x + h) - f(x) I dA(x) = 0. 18. Let f be a nonnegative measurable function on (a, b] satisfying f E %[c, b] for every c, a < c < b. Prove that fa fdA = Clinn+f" f(x) dx. 19. *Monotone Convergence Theorem: Let I f.1 be a monotone increasing sequence of nonnegative measurable functions on a measurable set A. Prove that fA (lim dA = qim jA f,, dA. 20. Show that the monotone convergence theorem is false for a monotone decreasing sequence of measurable functions. 21. *a. Let f be a nonnegative measurable function on a measurable set A, and let {A,} be a sequence of pairwise disjoint measurable subsets of A with U. A. = A. Prove that

= 7, jfdA. 00

n"I A, b, Prove that the conclusion of part (a) is still valid for arbitrary f e T(A). 22. Let { f,} be a sequence of measurable functions on (a, b) satisfying I f,(x)I s g(x) a.e., where g is integrable on [a, b]. If liimf,(x) = f(x) exist a.e. on [a, b], and h is any bounded measurable function on [a, b], prove that (b

fh dA = lim f f,h dA. 1 111

a

23. *As in Exercise 4, Section 6.4, let f be defined on (0, 1) by

f(x)

=±(xZ sin z) I

Prove that f is not L.ebesgue integrable on (0, 1).

24. Let f be a nonnegative measurable function on [a, b]. For each t

mf(t) = A({x E [a, b] : f(x) > t}). a. Prove that m,(t) is monotone decreasing on (0, oo).

b. Prove that fa fdA = fo m1(t) dt.

0, let

484

Chapter 1 o

10.8

Lebesgue Measure and Integration

Square Integrable Functions In analogy with the space 12 of square summable sequences, we define the space 222 of

square integrable functions as follows.

10.8.1

Let A be a measurable subset of R. We denote by T2(A) the set of all measurable functions f on A for which I f I2 is integrable on A. For f E f2(A), set DEFINITION

IIfII2=\JaIfI2dA/ The quantity IIf II2 is called the 2-norm or norm off. Clearly. IIf II2 >_ 0, and from the definition it follows that if f E _T22(A) and c E 18, then cf E Y2(A) with II cfII2 =

IcIIIf1I2. We will shortly prove that if f, g E `.P2(A), then f + g E Y2(A) with Ill + 8 112 < IVII2 + 119 112- Thus .T2(A) is a vector space over R.

If f E 22(A) satisfies IIf II2 = 0, then by Exercise 1, Section 10.7, 1 f 122 = 0 a.e., and

thus f = 0 a.e. on A. This does not mean that f (x) = 0 for all x E A; only that f = 0 except on a set of measure zero. Thus II 112 satisfies all the properties of a norm except for IIf II2 = 0.if and only if f = 0. To get around this difficulty we will consider any two functions f and g in X2 (A) for which f = g a.e. as representing the same function. Formally, we define two measurable functions f and g on A to be equivalent if f = g a.e. In this way it is possible to define 22(A) as the set of equivalence classes of square integrable functions on A. Rather than proceeding in this formal fashion, we will take the customary approach of simply saying that two functions in f22 are equal if and only if they are equal almost everywhere. With this definition Y2(A) is a normed linear space.

10.8.2

EXAMPLES

(a) For our first example, let f (x) = 1/Vc, x E (0, 1). By Exercise 4 of the previous section, f is integrable on (0, 1) with 1(0 1)f dA = 2. Since f 2(x) = 1/x, by Example 10.7.3(b), f2 is not integrable on (0, 1) and thus f `12((0, 1)). On the other hand, if g(x) = x- 'I, then g2(x) = x-21-', which by Exercise 4 of the previous section is integrable. Thus g E `12((0, 1)) with IIBII2

x 2/'dx=3.

J 0

(b) Consider the function f (x) = 1/x for x E [ 1, oo). For any n E N, n ? 2,

J IfI2dA = J x2dx = rI L i

- 1J. n

Thus IIfII2 =

hfI2dA = nix lim

I

J°IfI2dA = I.

I

Therefore, f E 12([ 1, oo)) with 11f112 = 1. It is easily shown that f E1([ 1, x)).

10.8

Square Integrable Functions

485

Cauchy-Schwarz Inequality Our first result will be the analogue of the Cauchy-Schwarz inequality for 12. The following inequality is sometimes also referred to as Holder's inequality.

10.8.3

THEOREM (Cauchy-Sehwarz Inequality) Let A be a measurable subset of R. If f, g E 5f2(A), then fg is integrable on A with

j

1181 dA s 111 1I211g112

A

Proof. By Theorem 10.5.4, the productfg is measurable, and for any x E A, we have I f(x)g(x) 1:S

1 (I f(x) 12 + Ig(x)12). 2

Since by hypothesis f, g E `.E2(A), the function I fgI is integrable on A, and thusfg is integrable on A. As in the proof of Theorem 7.4.3, for y E R,

0s

j lfgI dA.

f (IfI -yIBI)2dA=IIf A

(9)

A

If II9II2 = 0, then by Exercise 1 of Section 10.7, g = 0 a.e. on A. As a consequence, fAl fgl dA = 0, and the conclusion holds. If I19112 * 0, set y = (fAlfgldA)/Ilgll2. With y as defined, equation (9) becomes 0

(fAIfgI dA)2 111112 IIg112

and thus (fA I fg I dA)2

11f 112118112, which proves the result. Q

Our next result is Minkowski's inequality for the space X22. Since the proof of this is identical to the proof of Theorem 7.4.5, we leave the details to the exercises.

10.8.4 THEOREM (Minkowski's Inequality) Let A be a measurable subset of R. If f, g E 2(A), then f+ g E `.C2(A) with 111+g112s111112+118112

Proof. Exercise 4. Q

The Normed Linear Space ,22([a, b]) If A is a measurable subset of R, the norm II Ikon 2(A) satisfies the following properties: (a) Of 112 ? 0 for any f E _T2(A).

(b) 111112 = 0 if and only if f = 0 a.e. on A. (c) IIcf 112 = I c 1111 112 for all f E W2(A) and c E R. (d) IV + 9 112

111112 + IIgh12 for all f, g E e2(A).

486

Chapter 10

Lebesgue Measure and Integration

Properties (a) and (c) follow from the definition, and (d) is Minkowski's inequality. With the convention that two functions f and g are equal if and only if f = g a.e.. T2(A) is a normed linear space. By Definition 8.3.9, a sequence {/n} in ce2 converges in norm to a function f in `.EZ if and only if Jim Of - fn112 = 0. Example 10.6.12(b) shows that convergence in Y2 does

not imply pointwise convergence of the sequence. As in Chapter 9, convergence in S2 is usually called convergence in the mean. We now prove that T22([a. b]) is a complete normed linear space.

10.8.5

THEOREM The normed linear space (.T2([a, b]), 11 A2) is complete.

Remark. Although we state and prove the result for a closed and bounded interval. the same method of proof will work for T2(A) where A is any measurable subset of R. Before proving the theorem, we first state and prove the following lemma.

10.8.6

LEMMA Let A be a measurable subset of R. Suppose {f.1 is a monotone increasing sequence of nonnegative measurable functions on A satisfying for all n E IN, A

is finite a.e. on A.

and for some finite constant C. Then f (x) = lim

Proof. Since { f (x)} is monotone increasing for each x E A, the sequence either converges to a real number or diverges to oo. Let f(x) = liymeOfn(x). and let

E_ {xEA:f(x) = oo}. We will prove that A(E) = 0. For each k E N, let

EE= {xEA:f(x)>k}. Then Ek D Ek+, for all k E tN with lkE Ek = E. For fixed k E IN, set An, k= {x E A: fn(x) > k}, n E N. Then A..k C An+I.k With U. EN A..k = Ek. Thus by Theorem 10.4.6(a),

A(Ek) = lim A(An.k) But

A(A..k)= j 1dA5 1 j .fndA IjfndA!Ck k A..,

A..,

kA

Therefore, A(Ek) s C/k for all k E N. Since A(E)) < oo, by Theorem 10.4.6(b),

A(E)= limA(EE)=0. Therefore, f is finite a.e. on A. 0

10.8

Square Integrable Functions

487

Proof of Theorem 10.8.5. Let {fn} be a Cauchy sequence in 22. Then given e > 0, there exists no E Ni such that Ilfn - fn II2 < e for all n, m ? no. For each k E NI,

let nk be the smallest integer such that Ilfn - fn 112 < 1/2k for all m, n ? nk. Then n1cn2<...

CnkG...,and

ll2 <

IIJn,.,

2k.

For each k E N, set

gk = Ifn,l + Ifn_ -fn,I + .. ' +

IA.,

-Jn.I.

By Minkowski's inequality, gk dA = 118k112

\

[a.b]

llfn,ll2 +

Ilfn

j-1

(Ilf.,112 +

2k)2

(Ilfn,ll2 + 1)2.

Thus the sequence {gk} satisfies the hypothesis of Lemma 10.8.6. Therefore, urn k is finite a.e. on [a, b]. But then lim gk is also finite a.e. on [a, b]. As a consequence, the k-+oo series 00

If, (x)I + Y, If.,.,(x) -fn,(x)I j-1

converges a.e. on [a, b], and therefore so does the series 00

fn,(x) + Y,(fn,.,(x) -fn,(x)) j=1

But the kth partial sum of this series is Therefore, the sequence {fn,}kEh converges a.e. on [a, b]. Let E denote the set of x E [a, b] for which this sequence converges. Then A([a, b) \ E) = 0. Define

f(),x=-

J

lim fn,(x)' 0,

x E E, otherwise.

Then {f,} converges to f ae. on. [a, b]. It remains to be shown that f E 22, and that { fn} converges to f in 2 2. Since

IfnIf

..I

k

+

p

j=1

Ifnf., -fn,l = 8k

by Fatou's lemma

IfI2dA5 {im gkdA
k-ico a b]

Thus f r= 22. Finally, since

f(x) - f., (x) = (f0,.,(x) - f., (x)) a.e.,

488

Chapter 10

Lebesgue Measure and Integration

by Fatou's lemma again, 2

iab]

(a, b]

(f n'.' - f., 11 < \2

Therefore, II f - fn,112 < 1/2k for all k E N. Thus the subsequence { f,,,}kEN converges to f in the norm of _y2. Finally, by the triangle inequality,

Of -f,112 < Of -fn,ll2 + Ilfn, -fnll2 < 2k + IIfn, -fnll2From this it now follows that the original sequence { fn} also converges to f in the norm

of 22.

Fourier Series We close this chapter by making a few observations about Fourier series and the space

22([-ir, ir]). As in Definition 9.3.1, if f is Lebesgue integrable on [-a, ir]. the Fourier coefficients of f with respect to the orthogonal system {1, cos nx, sin nx}., are given by A

an =

I

f(x) cos nx dx,

n = 0, 1, 2,.

f(x) sin nx dx,

n = 1, 2,

.

,

J

bn =

1

Since f is Lebesgue integrable, the functions f(x) sin nx and f(x) cos nx are measurable on [-zr, -rr], and in absolute value less than or equal to I f(x)I. Thus the functions

f(x) cos nx and f(x) sin nx are all integrable on [ -ir, a]. The same method used in proving Bessel's inequality for Riemann integrable functions proves the following (Exercise 8): If f E :E2([-ar, ir]) and {ak} and {bk} are the Fourier coefficients off, then 00

Za0+kY,

(ak+bk)s

i

7r

f2dA. I

J

(Bessel's Inequality)

A

Thus the sequences {ak}k o and {bk}' , are square summable. We now use completeness of the space 22([-ir, it)) to prove the converse.

10.8.7

THEOREM If {ak}k=o and {bk}k , are any sequences of real numbers satisfying 1

2 ap +

00

(ak + bk) < oo,

k=I

then there exists f E `.E2([-tr, rr]) whose Fourier coefficients are precisely {ak} and {bk}.

10.8

Square Integrable Functions

489

Proof. For each n E t\!, set n

S,(x) = 1 ao + I (ak cos kx + bk sin kx). 2

k-1

Since each S. is continuous, Sn is square integrable on [ -ir, a]. If m < n, then A

11Sn - S'11'2 = J n

2 (ak cos kx + bk sin kr)dx,

Ik=m+ 7, I

which by orthogonality

= jr j, (a + b2E). Since the series converges, the sequence {Sn] is a Cauchy sequence in `.C2([-Tr, a i). Thus by Theorem 10.8.5, there exists a function f E X2([-zr, ar]) such that Sn converges to f in 22; i.e., lim 11f - S,,92 = 0. If n > m, then A

A

and

Sn(x) cos mx dx = -cram

S. (x) sin mx dx = 1rbm.

-A

A Therefore,

Jf(x) COS mx dx - as A

V (x) - S,(x)) cos mx dx.

1

--

1

1 A

which by the Cauchy-Schwarz inequality 11f- S.11211COS 'nX 112 =

Ilf- S,112-

Since this holds for all n > m, letting n --> oo gives am -

1

jf(x)cosmxdr. -A

A similar argument proves that the b,n are the sine coefficients off.

O

Is Every Trigonometric Series a Fourier Series? In Section 9.3 we showed that the series sin nx

Inn even though it converges for all x E R, is not the Fourier series of a Riemann integrable

function on [-a, ir]. Since 7, (in n ) -2 = oo, by Bessel's inequality neither is it the Fourier series of a square integrable function. This, however, does not rule out the possibility that it is the Fourier series of a Lebesgue integrable function. The following

490

Chapter 10

Lebesgue Measure and Integration

classical result is very useful in answering this question. Since the proof of the theorem is beyond the level of this text, we state the result without proof.2

10.8.8

THEOREM

(a) If b > 0 for all n and 7,0,0-1 b,/n = oc, then 00

b, sin nx is not the Fourier series of a Lebesgue integrable function. (b) If {a,} is a sequence of nonnegative real numbers with lim a = 0. satisfying a, c 1 (a, _, + a,, 1), then the series 00

7, a,cosnx

n-t

is the Fourier series of a nonnegative Lebesgue integrable function on [ -v, it 1.

Since the sequence { 1/(ln n)} satisfies hypothesis (a), the series

00 sin nx

n.2 Inn is not the Fourier series of any integrable function on it, a]. However, it is interesting to note that the sequence {1/(ln n)} also satisfies hypothesis (b) (Exercise 13), and thus the series cos nx 71 Inn ,-2 is the Fourier series of a nonnegative Lebesgue integrable function.

EXERCISES 10.8 1. *For x E (0, 1) let fo (x) = x'°, p > 0. Determine all values of p such that f, E Z2((0, I )). 2. For each n E N, let 1, = (n, n + =,). For a given sequence {c,} of real numbers, define f on 11. oo) by f(x) = 7,W , c, Xtn (x). Show that f E 22([ 1. oo)) if and only if

00c2

a < oo.

n.l n

3. Find an example of a real-valued function f on (0, oo) such that f2 is integrable on (0, oo). but If V° is not in-

tegrable on (0, oo) for any p, 0 < p < oo, p * 2. (Hint: Consider the function 8(x)

l

x(1 + llnxl)2

2. A proof of the result can be found in Chapter V of the text by Zygmund.

Notes

491

4. *Prove Theorem 10.8.4.

S. Let A be a measurable subset of R with A(A) < oo. If f E f2(A), prove that f E Y(A) with f If I dA < tlf 112 (A(A))1/2 A

6. Let { fa} be a sequence in 22([a. b]). Suppose {fa} converges in 22 to f E T2 and {fa} converges a.e. to some measurable function g. Prove that f = g a.e. on [a, b].

7. 'Let If.) be a sequence in 22(A) that converges in cy2 to a function f r= lim

a-0D J A

fag dA =

If g E 222(A). prove that

jfdA. A

& If f E T2([-ir, ar]) and {ak} and {bk} are the Fourier coefficients of f, prove that I

2

ao+ I(at+b})s-ar-° f2dA. O°

1

k°1

9. Which of the following trigonometric series are Fourier series of an 22 function? a.

- cos nx

b

sin nx

C.

cos nx 2

VnInn

10. Suppose E is a measurable subset of (-a, ir) with A(E) > 0. Prove that for each S > 0, there exist at most finitely many integers n such that sin nx z 8 for all x E E. 11. Let f r= 12([a, b]). Prove that given e > 0, there exists a continuous function g on [a, b] such that 11f - g 16 < e. (Hint First prove that there exists a simple function having the desired properties, and then use Exercise 15, Section 10.6, and Lemma 9.4.8.) 12. For f E ?2([-w, ar]) let {ak} and {bk} be the Fourier coefficients off. a. Prove Parseval's equality: 1

2ao+

00

(qk' +b,t)=

f2dA. A

b. If at = bk = 0 for all k, prove that f = 0 a.e. 13. *Show that the sequence (I /(In n)} satisfies the hypothesis of Theorem 10.8.8(b).

NOTES Lebesgue's development of the theory of measure and integration was one of the great mathematical achievements

of the twentieth century. His proof that every bounded measurable function is Lebesgue integrable was based on the new idea of partitioning the range of a function, rather

than its domain. Lebesgue's theory of integration also permitted him to provide necessary and sufficient conditions for Riemann integrability of a bounded function f. In addition to the fact that the Lebesgue integral enlarged the family of integrable functions, its power results from the ease with which it handles the interchange of limits and integration. For the Riemann integral, uniform

convergence of the sequence (f.) is required. Otherwise, the limit function may not be Riemann integrable. On the other hand, if { fa} is a sequence of measurable functions on [a, b], then its pointwise limit f is also measurable.

Hence if f is also bounded, then f is integrable. The bounded convergence theorem is notable for its simplicity of hypotheses and proof. It only requires that (f.) be uniformly bounded and converge a.e. on [a, b]. This is sufficient to ensure that Urn

n

-.m

f, dA = a, b]

o. bJ

(lim .x f.) A.

492

Chapter 10

Lebesgue Measure and Integration

With the additional hypothesis that the pointwise limit f is Riemann integrable, the bounded convergence theorem is also applicable to a sequence if,) of Riemann integrable functions. The bounded convergence theorem and the dominated convergence theorem are the tools required to prove the fundamental theorem of calculus for the Lebesgue integral.

THEOREM A 1ff is differentiable and f is bounded on [a, b], then f is Lebesgue integrable, and

fa f'dA = f(b) - f(a). The proof of this result was requested in Exercise 8 of

Section 10.6. It follows simply by applying the bounded defined by convergence theorem to the sequence g (x) = n[f(x + '-,) - f(x)]. Since f is differentiable, the converges pointwise to f. Also, by the mean sequence is uniformly bounded on value theorem the sequence [a, b]. This then establishes the analogue of Theorem 6.3.2 for the Lebesgue integral. If instead of bounded, one assumes that f is Lebesgue integrable, then the result also follows by Lebesgue's dominated convergence theorem.

The Riemann theory of integration allows us to prove that if f E gt[a, b], and F is defined by F(x) f.'f(t) dt, then F'(x) = f(x) at any x e [a, h] at which f is continuous. Since f E 9l[a, b] if and only if f is continuous a.e. on [a, b], we have that F'(x) = f(x) a.e. on [a, b]. Although not proved in the text, this result is still valid for the Lebesgue integral. THEOREM B Let f be Lebesgue integrable on (a, b), and define F(x) = f, f dA. Then F'(x) = f(x) a. e. (a, b).

f

By writing f = f * - f -, it suffices to assume that

0. and thus F is monotone increasing on [a, b). It is a fact independent of integration that every monotone

function is differentiable a.e. on [a, b).3 A slight generalization of Theorem A then gives that

F'dA = F(x) - F(a) = J fdA, or that f.[ F' - f] dA = 0 for all x E [a. b]. As a consequence of Miscellaneous Exercise 5 we have F' = f a.e. Newton and Leibniz realized the inverse relationship of differentiation and integration. The above two versions of the fundamental theorem of calculus provide a rigorous formulation of this inverse relationship for a large class of functions. The Lebesgue theory of integration also provides the proper setting for the study of Fourier series. The bounded convergence theorem was used by Lebesgue to prove uniqueness of a Fourier series. If

f(x)=2Ao+ k-l

(Ak cos kx + 8k sin kx).

for all x E [ -it. zr], then f, being the pointwise limit of a sequence of continuous functions, is automatically mea-

surable on [-rr, a]. If f is also bounded, then f is integrable. If the sequence of partial sums of the trigonometric series is uniformly bounded, then by the bounded convergence theorem, the trigonometric series is the Fourier series off. In his 1903 paper Sur les series trigonometric,' Lebesgue showed that uniform boundedness of the partial sums may be removed; that boundedness of the function f itself was sufficient. This result was extended in 1912 by de la VallEe-Poussins to the case where the function f is integrable on (- rr, a ). See the article by Alan Gluchoff for an overview of how trigonometric series has influenced the various theories of integration.

MISCELLANEOUS EXERCISES 1. Let A be a measurable subset normed linear space.

of R. For f E 2(A) set lI f

11,

= f,, I f I dA. Prove that (.T(A ), (I (I,) is a complete

3. See page 208 of the text by Natanson. 4. Annales Scientifiques de 116cole Notmale Supdrieure. (3) 20 (1903).453-485. 5. "Sur l'unicitd du developpement trigonomEtrique;' Bull de t'Acad. Royale de Belgique (1912).702-718-. see also Chapter 9 of the text by Zygmund.

Supplemental Reading

493

2. Let A be a measurable subset of R, and let f, f, n = 1, 2, ... be measurable functions on A. The sequence If.) is said to converge in measure to f if for every 6 > 0. lim A({x E a-m

f(x)l

6}) = 0.

a. If

converges in the norm of .Y(A) to f E `.E(A). prove that converges in is a sequence in :E(A), and measure to f. b. Find a sequence ff.) of measurable functions on a measurable set A that converges to a function fin measure. but does not converge to f in norm.

c. If A(A) is finite and (f.} is a sequence of measurable functions that converges in measure to a measurable function f, prove that there exists a subsequence { fk} of {f} such that fk -+f a.e. on A. d. Show that the sequence { f } of Example 10.6.12(b) converges to 0 in measure.

e. Find a sequence {f.) of measurable functions on a measurable set A such that f -a 0 everywhere on A but { does not converge to 0 in measure. be a sequence of orthogonal functions on [a, b] having the property that the zero function is the only con3. Let tinuous real-valued function f satisfying fa fkp, dA = 0 for all n E N. Prove that the system is complete. 2([a, b]) satisfies Ja' (Hint: First use the hypothesis to prove that if f dA = 0 for all n E N. then f = 0 a.e. Next use completeness of `. b2 to prove that Parseval's equality holds for every f E `f ([a, b]).)

4. Construction of a Noumeasurable Set: For each x E [-12, Z], define the set K(x) by

K(x)={yE[-12,Z]:y-xE0}. a. Prove that for any x, y E [ -s, ;], either K(x) n K(v) = 4 or K(.r) = KO,). (Note: K(x) = K(y) does not imply that x = y; it only implies that x - y is rational.) Consider the family & = {K(x):x E [ -12,121} of disjoint subsets of J. Choose one point from each distinct set in this family and let A denote the set of points selected. The ability to choose such a point from each of the disjoint sets requires an axiom from set theory known as the axiom of choice. Further information about this very important axiom can be found in the text by Halmos. Let rk, k = 0, 1, 2, ... be an enumeration of the rational numbers in [ -1. I ], with r1 = 0. For each

k=0,1,2..., setAkA+rk.

b. Show that the collection {Ak} is pairwise disjoint with [- i, ] CU

u Ak C [- _, ;

c. Use the above to show that A*(A) = 0 and A*(A) > 0, thus proving that A is nonmeasurable. S. Suppose f is Lebesgue integrable on [a, b]. If f. f dA = 0 for every x E [a, b], prove that f = 0 a.e.

SUPPLEMENTAL READING Botts, Truman, "Probability theory and the Lebesgue integral," Math. Mag. 42 (1969). 105-1 I1. Burkill, H., `"The periods of a periodic function;' Math Mag. 47 (1974),206-210. Darst, R. B., "Some Cantor sets and Cantor functions;' Math. Mag. 45 (1972). 2-7. Dressler, R. E. and Stromberg, K. R., '"The Tonelli integral," Amer. Math. Monthly 81 (1974), 67-68. Gluchoff, Alan D.. 'Trigonometric series and theories of integration;' Math. Mag. 67 (1994), 3-20.

Katznelson, Y. and Stromberg, K., "Everywhere differentiable, nowhere monotone function;' Amer. Math. Monthly 81(1974), 349-354. Kraft, R. L., "What's the difference between Cantor sets?" Amer. Math. Monthly 101 (1994). 640-650. Ladder, G., "A simple introduction to integral equations," Math. Mag. 69 (1996), 172-181. Maligranda, L., "A simple proof of the Holder and Minkowski inequality," Amer. Math. Monthly 102 (1995), 256-259.

494

Chapter 10

Lebesgue Measure and Integration

McShane, E. J., "A unified theory of integration;' Amer. Math. Monthly 80 (1973), 349-359. Priestly, W. M., "Sets thick and thin;' Amer. Math. Monthly 83 (1976), 648-650.

Varberg, Dale E., "On absolutely continuous functions. Amer. Math. Monrhl' 72 (1965). 831-841. Wade. W. R., 'The bounded convergence theorem;' Amer. Math. Monthly 81 (1974). 387-389.

Appendix: Logic and Proofs A.1 Propositions and Connectives

A.2 Rules of Inference

A.3 Mathematical Proofs A.4 Use of Quantifiers

In searching for new mathematics, mathematicians use their imagination and ingenuity to make a conjecture about new theories based on observations of particular cases or phenomena. This approach is called inductive reasoning. Natural or social scientists may test the conjecture by making further observations. If the results are incompatible with the conjecture, then scientists must usually reject or modify the theory. Mathematicians, on the other hand, attempt to change the conjecture into a theorem by proving (or disproving) that the conjecture follows logically from the accepted axioms, the

hypothesis, and proved theorems of other mathematicians. In constructing a proof mathematicians are restricted to the use of deductive reasoning-the use of logic to draw conclusions based on hypothesis and statements accepted as true. In developing new mathematics, mathematicians have the advantage of not being restricted to the study of observable phenomena or even axioms that have been accepted as true. Euclid's fifth postulate (the parallel axiom) had been accepted as true for over 2200 years until the early nineteenth century when Janbs Bolyai (1802-1860), Carl Friederich Gauss (1777-1855), and Nicolai Lobatchewsky (1793-1856) challenged this axiom. The acceptance of alternatives to the parallel postulate led to the development of new and interesting non-euclidean geometries, many of which are still very important today. Deductive reasoning provides mathematics with an efficient means of both exposition and organization of its subject matter. In what follows we will attempt to explain this formal, logical side of mathematics. Formal logic will not indicate the steps necessary to prove a conjecture; it only provides a system of rules and procedures used to 495

496

Appendix

Logic and Proofs

decide whether a given statement follows from some given set of statements. Intuition. insight, and knowledge are still required to formulate the necessary steps in a proof.

A.1 1

Propositions and Connectives The English language, as most languages, allows for a variety of sentences: interrogatory sentences (Will I pass this course?), exclamatory sentences (I got an A:). and sentences that are either true or false. In the language of logic the latter are called propositions.

A.1.1

DEFINITION A proposition is a sentence or statement that is either true (T) or false (F). Some examples of propositions are

(a) "4 is an even number."

(b) "\ is rational.' (c) "The author of this text was born in Frankfurt. Germany." We certainly accept that (a) has the value T, whereas in the introduction to Chapter I it is proved that (b) has the value F. For (c), however, even though you may not know whether the statement is true or false, the statement does have either the value T or F. The above are all examples of simple propositions; that is, propositions that consist of a single statement. Simple propositions can be combined using the logical connectives "and;' "or," and "not" to form compound propositions.

A.1.2

DEFINITION

Given two propositions P and Q,

(I) The negation of P. denoted - P, is the proposition "not P." The proposition P is true exactly when P is false. (ii) The conjunction of P and Q, denoted P A Q, is the proposition "P and Q." The proposition P A Q is true exactly when both P and Q are true. (iii) The disjunction of P and Q, denoted P V Q, is the proposition "P or Q." The proposition P V Q is true exactly when at least one of P or Q is true. The truth values of the above connectives can be summarized very nicely by means of the following truth table. P

Q

-P

PAQ

T T

T

F F

T

F F

T

F F

T T

F F F

PVQ T T T F

The connectives "and," "or;" and "not" are illustrated by the following examples. Using (a), (b), and (c) above we have

-- (b) "V is not rational." (a) A (b) "4 is an even number and V is rational,"

A.1

Propositions and Connectives

497

(a) V (c) "4 is an even number or the author of this text was born in Frankfurt, Germany."

Since the statement "V is rational" is false, the proposition (a) A (b) is also false. On the other hand, since (a) is true. (a) V (c) is true regardless of where the author was born.

In logic the use of the word "or" is the inclusive or, meaning "one or the other, or both" as opposed to the exclusive or, meaning "one or the other, but not both" For propositions P and Q the exclusive or can be expressed in terms of A, V, and by (P V Q) A -- (P A Q) (see Exercise 2). In addition to and, other common connectives used in English are but, while, and although. Each of these has the same meaning as and, and is represented symbolically by A. For example, if P is "It is cold" and Q is "It is snowing;" then the statement "It is cold, but it is not snowing" would be expressed symbolically by P A - Q. In addition to the connectives A, V, and -, there are two other connectives that are commonly used in logic; namely, the conditional and biconditional.

A.1.3

DEFINITION

Given two propositions P and Q,

(i) The conditional sentence P => Q (read "P implies Q") is the proposition "If P, then Q." The statement P Q is a true statement unless P is true and Q is false, in which case it is a false statement. (ii) The biconditional sentence P * Q is the proposition "P if and only if Q." The sentence P q Q is true exactly when P and Q have the same truth values, otherwise it is false. The symbol " " is referred to as the implication or conditional symbol, whereas t* is the biconditional symbol. In "P ' Q," the proposition P is the hypothesis or an. tecedent and Q is the conclusion or consequent. The truth values for P ' Q and P a Q are given in the following table.

P T

Q

P=*Q

PaQ

T

T

T

T

F

F

F F

T

T T

F F

.

F

T

In the truth table for P = Q, the only line where P Q is true and P is true is the first line, where Q is also true. Thus the conditional statement P Q is often also expressed by saying that P is a sufficient condition for Q (if P is true then Q follows), or that Q is a necessary condition for P (P cannot be true unless Q is true). That Q is a necessary condition for P is sometimes also expressed by the phrase "P. only if Q" Since the biconditional P,4* Q is true exactly when P and Q have the same truth values, this is often also verbally expressed by "P is necessary and sufficient for Q:' To illustrate the truth values assigned in the conditional sentence, let us consider the following example. Your professor agrees "If you earn an A on the final, I will assign you an A for the course."

498

Appendix

Logic and Proofs

Here the antecedent P is "You earn an A on the final" and the consequent Q is "He assigns you an A for the course" The only case in which you have reason to be angry (the sentence is false) is when P is true and Q is false. If both P and Q are false, you may not be happy, but you have no cause to be angry with your professor. On the other hand, if P is false and Q is true, you certainly will not be angry. Closely related to the conditional sentence are the converse and the contrapositive of P =*. Q.

A.l.4 DEFINITION For propositions P and Q, the converse of P contrapositive of P

. Q is Q =o. P, and the

. Q is --Q' --P.

The truth values for each of these is given in the following table. P T T F F

Q

P=#. Q

Q .P

T F T

T

T T

F

T T

F

--Q=:I, --P T F

T T

F

T

A propositional formula is an expression involving finitely many logical connectives

(such as n, V, --, =, and e') and variables (such as P, Q, R, etc.). For example,

(P A (Q V R)) V -R is a propositional formula; it becomes a proposition when the letters P, Q, and R represent propositions and thus have either the truth value T or F. Two propositional formulas are equivalent if and only if they have the same truth values for all assignments of truth values to the simple propositions making up the propositional formulas. Simply stated, two propositional formulas are equivalent if and only if they have the same truth table. For example, the formulas P . Q and -P V Q are equivalent. This is verified by the following truth table.

P

Q

P=Q

-PvQ

T T

T

T

T

F

F

F

F F

T

T T

T T

F

Also, from the table for the contrapositive we see that -Q

P

A.1.5

-P is equivalent to

Q. To emphasize this we state it as a theorem.

THEOREM The proposition P

Q is equivalent to its contrapositive -Q = -P.

Exercise I contains several pairs of equivalent propositional formulas. These are important and should be memorized. Some propositional formulas have the property that they are always true regardless of the assignment of T or F to the simple propositions making up the formula. For example, it is easily verified by a truth table that each of the propositional formulas

PV - P

and

(P A Q) r* P

A.1

499

Propositions and Connectives

have the value T for any assignment of T or F to P and Q. Such propositions are called tautologies. A tautology is a propositional formula that is true for every assignment of

truth value to its components. A contradiction is the negation of a tautology. Thus (P V -P) is a contradiction. By means of a truth table it is easily verified that -- (P V -P) is equivalent to -P A P, which simply states that "P" and "not P" cannot both be true simultaneously. This is sometimes referred to as the law of the excluded middle. Exercise 3 contains several basic tautologies that will be very useful in our discussion on rules of inference. From the definitions of equivalence of propositional formulas and the biconditional it should be clear that two propositional formulas P and Q are equivalent if and only if P q Q is a tautology. To emphasize its importance we state it as a theorem.

A.1.6 THEOREM Two propositional formulas P and Q are equivalent if and only if P q Q is a tautology.

To illustrate this we consider the equivalent propositional formulas P -P V Q. The truth table for (P Q) a (-P V Q) is as follows:

Thus (P

Q and

(P * Q) q (--P V Q)

P

Q

T T F F

T

T

F T F

F

T T

T T T T

T F T T

Q) q (-P V Q) is a tautology.

, EXERCISES Al

1. For propositions P. Q, and R, verify by means of a truth table that each of the following pairs of propositional formulas are equivalent.

a. ^(-P) and P c. PA (QVR)and(PA Q)V (PAR) e. ^-(P A Q) and -P V --Q.

b. P A Q and Q A P; P V Q and Q V P d. P V (Q A R)and(P V Q) A (Q V R) f. ^-(P V Q) and ^-P A --Q

S. P4* Qand(P-r Q) A (Q=P) Parts (c) and (d) above are referred to as the distributive laws for A and V, whereas parts (e) and (f) are De Morgan's laws for A and V. 2. By means of a truth table show that (P V Q) A -- (P A Q) is true if either P or Q is true, and false otherwise. 3. Prove that each of the following is a tautology a. P v -P (Excluded Middle) b -(-P) q P (Double Negative) c. P a [-P (R A --R)] (Contradiction)

d. (P A Q) - P e. P (P V Q)

(Conjunctive Simplification) (Disjunction)

500

Appendix

Logic and Proofs

f . [PA(P=* Q)] 'Q g. [(P Q) A -- Q] ' -P h. [(P= Q) A (Q =-,, R)] a (P

(ModusPonens) (Modus Tollcns) R)

(Transitivity)

4. For each of the following determine whether it is a tautology, a contradiction, or neither.

a.PA --P

b.P''PA(PvQ)

c. (P A Q) A(-' PV - Q)

d. -(PAQ)''(P '- Q)

e. IQ A (P=* Q)] 'P

f. [P=* (Q A R)]

'[(Q=R) V (R 'P))

g S. In this exercise no knowledge about sequences is required. Let M, B, and C denote the following statements. is monotone." M: "The sequence B: "The sequence C: "The sequence

is bounded." converges:'

Express each of the following sentences symbolically, using the convention that "divergent' is the negation of "convergent" and "unbounded" is the negation of "bounded" (In each of the statements the tern sequence refers to a sequence of real numbers). a. The sequence

is monotone and bounded.

b. A convergent sequence is bounded.

c. A sequence converges, only if it is bounded.

d. In order that a sequence is bounded, it is necessary that it converges. diverges, then it is unbounded. e. If the sequence 6. Provide an appropriate negation of each of the sentences in Exercise 5. Express your answer first in symbolic form and then in English.

A.2J Rules of Inference Given a propositional formula R, the method of truth tables provides a reliable means to determine whether R is a tautology. This method can even be turned into a computer program which would accept any formula R as an input and would determine whether R is a tautology. However, checking formulas of even modest length, say twenty symbols, turns out to consume inordinant amounts of time. What is worse, this situation cannot be improved substantially by clever computer programming. It can be proven, using the techniques of computational complexity-a branch of mathematical computer science-that the problem of determining whether a propositional formula is a tautology is'intractable in the sense that any program for its solution will place insurmountable demands on computational resources. Where computational methods fail, mathematical ingenuity can still succeed. Confronted with a propositional formula R it may be possible to offer a proof that it is a tautology. Roughly speaking, a proof of R is a sequence of sentences or propositional formulas, the last one being R, such that each of the sentences in the sequence is either an axiom, a hypothesis, or a statement that follows from the previous sentences in the se-

A.2

Rules of Inference

501

quence by some principle of logical inference. To make this unambiguous. we have to specify these principles of logical inference. A principle of logical inference may have some premises (like the previous sentences referred to above) and a conclusion.

The Form of a Rule of Inference: From P,,.

.

.

, P, one can infer Q.

Symbolically this can be expressed as P,

P1....,P. :. Q,

or as P.

Q,

Here the formulas P,, ... , P are the premises and Q is the conclusion. (The symbol :. is used in mathematics to denote therefore.) We even allow the case when no premises are present. In that case, Q can be regarded as an axiom. The most important thing about a rule of inference is that it should be logically valid; that is,

P, A .

.

A P. = Q (or just Q if no premises are given)

should be a tautology. For example, to show that P, P Q :. Q is logically valid, it suffices to show that (P A (P Q)) Q is a. tautology. This is easily verified by means of a truth table as follows.

P A (P=Q)=:;- Q

P

Q

P=* Q

T T

T

T

T

F

F

F F

T

T T

F F F

F

T T T T

Since it becomes impractical to always verify by means of a truth table whether Pt A A P. Q is a tautology, we utilize tautologies to construct rules of inference. We can take any tautology of the form above and convert it into a rile of inference. Of course, we would gain nothing if we allowed ourselves to have a rule of inference for every tautology. We still have to use truth tables to verify that each rule of inference comes from a tautology. Fortunately, a handful of fairly simple tautologies is all that is needed. Each of the following tautologies can be verified by the method of truth tables without too much effort

PV -P PVQ [P A (P . Q)] P

Q

[(P = Q) A --- Q) =! -- P

(Excluded Middle) (Disjunction) (Modus Ponens) (Modus Tollens)

502

Appendix

Logic and Proofs

P A Q . P (or Q) [(P Q) A (Q R)} ' (P

R)

(Conjunctive Simplification) (Transitivity)

From these tautologies we obtain the rules of inference listed below. Some additional useful and important tautologies are as follows.

--(--P) q P P V Q a Q V P, P n Q a Q n P

Pv(QVR)a(PVQ)VR P A (Q A R) q (P A Q) A R Q --(P n Q)

-(P v Q) q -P A

Q

[P A (Q V R)] q [(P A Q) V (P A R)]

[Pv(QAR)]e*[(PVQ)A(PVR)] j P=*Qa-Q=*--P Pa[--P==O-(R A-R)]

(Double Negative) (Commutative Laws) (Associative Laws)

(De Morgan's Laws) (Distributive Laws)

(Contrapositive) (Contradiction)

One of the most fundamental rules of inference, in fact usually taken as an axiom, is the following.

Rule of the Excluded Middle: One can infer P V

P, for any statement P.

This rule simply states that the propositional formula P V --P can be inferred, for any proposition P. For example, if an argument involves a real number x, one can assert at

any time that "Either x = 0 or x # 0."

Rule of Conjunction: From P and Q, one can infer P A Q.

Rule of Disjunction: From P, one can infer P v Q.

Modus Ponens Rule or Rule of Detachment: From P and P

Q, one can infer Q.

Modus Tollens Rule or Rule of Contrapositive Inference: From -Q and P = Q, one can infer -P.

A.2

Rules of Inference

503

Rule of Conjunctive Simplification: From P A Q, one can infer both P and Q.

Rule of Transitive Inference or Hypothetical Syllogism: From P

Q and Q

. R, one

can infer P

. R.

The rule of conjunction, although not listed as a tautology, simply asserts that P, Q :. P A Q is logically valid; that is, P A Q P A Q is a tautology. Likewise, the rule of disjunction follows from the fact that P = P V Q is also a tautology. The

modus ponens rule is again a verbal statement of the modus ponens tautology [P A (P ' Q)] Q. The implication P Q by itself, even if known to be true, infers Q is also true when both P and Q are false. Hownothing about Q. The implication P ever, if both P and P =*- Q are true, then Q must also be true. The rule of conjunctive simplification is a restatement of the corresponding tautol-

ogy. However, it also follows as a consequence of the modus ponens rule. Since (P A Q) P (or Q) is a tautology, by the rules of conjunction and modus ponens

(P A Q) A [(P A Q)

P]

P (or Q).

The modus tollens rule follows from the modus tollens tautology. It is also a simple Q is equivalent consequence of the contrapositive law and modus ponens. Since P to -Q -P, it is easily verified by a truth table that the formula -Q A (P Q) is

equivalent to -Q A (-Q

-P). Thus by the modus ponens rule --P can be

inferred. The above argument illustrates the use of the replacement rule. In the formula

-Q A (P = Q). the expressions -Q and P =:;- Q are subformulas of the original formula. The new formula -Q A (-Q =*- --P) was obtained from the original formula by replacing the subformula P Q by its equivalent formula -Q = -P. The resulting formula is then equivalent to the original. As a rule of inference, this is stated as follows.

Simple Replacement:

If P is a subformula of a formula P and P q Q', then from P one can infer any formula Q that results from replacing an occurence of P in P with Q'.

There are several additional rules that are worthy of mention. The justification of these rules is left to the exercises.

504

Appendix

Logic and Proofs

Disjunctive Syllogism: From P V Q and --P, one can infer Q.

Rule of Inference by Cases: From P

Q and R

Q. one can infer (P V R)

Q.

Rules of Biconditional Inference: From P From P

Q and Q P, one can infer P q Q. Q, one can infer P . Q and Q . P.

A more detailed discussion of the propositional proof system would have taken us

too far astray from our main goal; namely, to provide basic rules of inference with which to construct valid arguments. A formal discussion of the propositional proof system can be found in the text by Bums listed in the Supplemental Reading section. In that text the author proves that every tautology, and only tautologies, can be derived from the listed rules of inference. T o p r o v e the validity of an inference PL, ... , P. .. Q. we simply have to verify that using the rules of inference, we can infer the conclusion Q from the given premises

Pr, ... , P. Each statement in the proof should be either a premise or an axiom, or should follow from previous statements by one of the accepted rules of inference. Devising a proof is a feat of mathematical ingenuity. However, once a purported proof is in hand, it can be easily checked step by step for validity. We illustrate the use of the rules of inference with the following examples.

All EXAMPLES (a) As our first example we consider the following verbal argument. If John is a Democrat, he associates with Democrats. But John does not associate with Democrats. Therefore, John is not a Democrat. This argument can be written in symbolic form as P=:0. Q -P.

By the rule of contrapositive inference the argument is valid, whatever the truth or falsity of the statements in it may be. (b) As our second example we consider the following.

(a) P (b) P = Q

A.2

Rules of Inference

505

(c) P=(QTR) By the modus ponens rule, from (a) and (b) we can infer Q. Likewise from (a) and (c) R, we can infer R. Thus the inferwe can also infer Q R. But now from Q and Q ence is logically valid. In symbolic form, the above proof can be written as follows:

3.

P P P

4.

Q

5.

Q

..

R

1.

2.

(premise) (premise)

Q

(Q

R)

(premise)

(modus ponens I & 2) (modus ponens l & 3) (modus ponens 4 & 5)

'R

(c) For our final example we consider the following argument concerning Cauchy sequences. For the purposes of illustration we need to know nothing about sequences, with the exception that they are usually denoted as and that divergence is the negation of convergence. diverges. The sequence A bounded sequence has a convergent subsequence. Every Cauchy sequence that has a convergent subsequence converges. Therefore, the sequence is not Cauchy.

In this particular example the conclusion happens to be true, but the argument is not valid. To see this we will write the argument in symbolic form. Let P, Q, R, and S denote the following statements

P: Q: R: S:

"The sequence {p,} is Cauchy:' 'The sequence {p} converges:' "The sequence is bounded:' 'The sequence has a convergent subsequence."

In symbolic form the above argument is expressed as follows:

(a) - Q (b) R=* S (c) (P A S)

Q

By (a) and (c) and the modus tollens rule we can infer -(P A S). But by De Morgan's

law, -(P A S) is equivalent to the statement -Pv - S. However, from -P V -- S we can infer neither --P nor -S. If we knew that S (i.e., -(--5)) was true, then by the rule of disjunctive syllogism we could infer -P. Unfortunately, from the given premises nothing can be inferred about S. If S has the truth value F, then -P V -S is true regardless of the truth value of P. This allows us to obtain an assignment of truth values that make the premises true and the conclusion false. Thus the argument is not valid.

506

Appendix

Logic and Proofs

If the statement R had been included as a premise, then the resulting inference would be valid. As is often the case in proofs written by the beginning student, statement (b) of the proof is totally extraneous.

EXERCISES A.2 1. Construct a truth table for [(P' Q)A -.QJ

-P to verify the validity of the argument in Example A.2.1(a).

2. a. Justify the rule of disjunctive syllogism. b. Justify the rule of inference by cases. 3. Justify the rules of biconditional inference. In Exercises 4-12 use the rules of inference or a truth table to test the validity of each of the following.

4.

If L, is 11 to L2 and L2 is 11 to L3, then L, is 11 to L,.

L,istltoL2L2 is to L3.

L, is to L3If m is even, then 2 divides m.

5.

If 2 divides m, then 4 divides m2. If m is even and 4 divides m2, then m2 is even. m2 is even. 6.

SQ

7.

S

--P V -S -.P 8.

R

SaQ (RVS)*Q Q

P

9.

PAR RCS (R A S)mo,Q Q

P

10.

R

PAS (RAS) .Q .. 11.

Q

P

Q

R*S P V R

QVS

A.3

12.

The sequence The sequence

Mathematical Proofs

507

diverges. is bounded.

A bounded sequence has a convergent subsequence. Every Cauchy sequence that has a convergent subsequence converges.

The sequence

A.3

is not Cauchy.

Mathematical Proofs In mathematics a proof is a logically valid deduction of a theorem from the premises of the theorem, the axioms, or previously proved statements or theorems. The truth of any statement in a proof should be traceable back to some initial set of axioms or

postulates that are assumed true. A proof should not be just a string of symbols. Every step in a proof should express a complete sentence, including the justification of the step. In this section we look at several methods that are commonly used to prove a theorem. Most theorems in mathematics are stated in the form "If P, then Q"; that is. Q. Any theorem stated as a biconditional sentence "P, if and only if Q" is proved P by first proving P Q, and then Q ' P.

Direct Proof Q" is the direct proof; namely, we asThe most straightforward type of proof of "P sume the hypothesis P and use the axioms, computations, or other theorems and the rules of logic to infer Q.

Direct Proof of P= Q Proof.

Assume P.

Therefore Q. Thus P

Q. t

We illustrate the method of direct proof and the use of the rules of inference in justifying the validity of an argument with the following examples.

A.3.1

EXAMPLES

(a) In Section 2.6 of the text the following theorem about Cauchy sequences is proved.

Theorem 2.6.4. Every Cauchy sequence of real numbers converges.

is used to mark the end of a proof. Some authors prefer to use QED. which is an I In the text the symbol abbreviation of the Latin "quod eras demonstrandum." meaning "which was to be proved."

508

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Let P and Q denote the following statements respectively. P: Q:

is a Cauchy sequence of real numbers"

'The sequence 'The sequence

converges."

The theorem to be proved is "If P, then Q"; that is, "If { is a Cauchy sequence of real numbers, then the sequence converges:' Within Section 2.6 and in previous sections the following related theorems are proved.

Thl "Every Cauchy sequence is bounded:' Th2 "Every bounded sequence of real numbers has a convergent subsequence." Th3 "If (p.} is a Cauchy sequence of real numbers that has a convergent converges" subsequence, then the sequence is bounded" and "The sequence Let R and S denote the statements "The sequence has a convergent subsequence" respectively. Using P, Q, R, and 5, theorems Th l, Th2, and Th3 can be written symbolically as

Thl

P

Th2 Th3

RCS, (PAS)

R, Q.

From Th 1 and Th2, by the transitive rule we can infer P ' S. Thus from our assumption P. by modus ponens we can infer S. Hence by the rule of conjunctive inference we can infer P A S. But now by Th3 and the modus ponens rule we have

{(P A S) A [(P A S)*Q]}z*Q. It is important to again emphasize that the fact that (P A S) Q is true does not allow us to infer anything about Q. It is also required that P A S must be true. In symbolic form, the above proof can be written as follows: 1.

P

(hypothesis)

2.

PAR

(Thl)

3.

RCS

(Th2)

4.

P

5.

S

(transitive rule) (modus ponens)

6. 7.

PAS

.

.S

(PAS) = Q Q

(conjunctive inference) (Th3) (modus ponens)

The above provides a very,methodical argument illustrating the validity of the implication P Q. For a short proof this works very nicely. However, there is not a single proof in this text that is written in such detail using symbolic logic to proceed from P to Q. Mathematical proofs should be written in complete sentences, including justifications. The truth of any statement in the proof must follow from the initial hypothesis, the axioms, or previously proved theorems. A typical proof of the theorem, using the

A.3

Mathematical Proofs

509

facts introduced above, would proceed as follows. The comments in parenthesis are not part of the proof; they are included as explanations of the statements. be a Cauchy sequence of real numbers. (This asserts the truth of the hyProof. Let is bounded. (This asserts the implication pothesis P.) By Theorem 2.6.2 the sequence P R.) Thus by Corollary 2.4.1 the sequence 1p.1 has a convergent subsequence. (This is the implication R'' S, which by the transitive rule gives P * S. In the next step we invoke Theorem 2.6.3; namely, that P A S =:O- Q.) The result now follows by Theorem 2.6.3. Q

A better and more careful way, especially for the novice, to express the last senhas a convergent subsetence would be as follows: "Since the Cauchy sequence converges." quence, by Theorem 2.6.3 the sequence (b) For our second example we prove the following statement. "If m is an even integer, then m2 is divisible by 4:' Let P be the statement "The integer in is an even integer," and Q the statement "The integer m2 is divisible by 4." Thus we wish to prove that P Q. Note: If m and n are integers, we say that m divides n, or n is divisible by in, if n = km for some integer k.

Proof. (Assume P.) Suppose m is an even integer. Then m = 2k for some integer k. (Here we use the definition of an even integer.) Then m2 = 4k2. Thus m2 is divisible by

4. (The conclusion Q.) U

Proof by Contraposition Since the implication P prove the implication P

Q is equivalent to its contrapositive -Q Q by assuming --Q and deriving

P, we can

P. Such a proof is called

a contrapositive proof or proof by contraposition. 10,

Contrapositive Proof of P

Proof. Assume -Q.

.Q

Conclude - P by means of a direct proof. Thus

^-Q' --P. ThereforeP =:o-Q. Q

We illustrate the method of proof by contraposition with the following elementary example.

A.3.2

EXAMPLE Let n be an integer. If n2 is even, then n is even.

Proof. Suppose n is not even

Then n is odd. (Here, and below, we use the fact that an integer is odd if and only if it can be expressed as 2k + I for some integer k.) Thus n = 2k + 1 for some integer k. But then

n2=(2k+ 1)2=4k2+4k+ I =2(2k2+2k)+ 1.

510

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Logic and Proofs

(In the above we have used the rules concerning algebraic operations on the integers. i.e., proved theorems.) Since m = 2k2 + 2k is an integer, we have

n2=2m+1 for some integer m. Thus n2 is odd (-P).

U

Indirect Proof or Proof by Contradiction Our next method of proof is the indirect proof or proof by contradiction. A proof by contradiction makes use of the tautology

.[--P=(RA-R)].

P

The statement RA - R is a contradiction. It is worthy of mention that R does not appear anywhere on the left side of the tautology. Thus any proposition R that will do the job will suffice. As opposed to our previous two methods of proof, proof by contradiction can be applied to any statement "P." Direct proofs and contrapositive proofs only apply to implications "P =;,- Q."

Proof of P by Contradiction Proof. Assume -P. Therefore R.

A.3.3

Therefore -R. Thus P is true.

EXAMPLE To illustrate the method of proof by contradiction we consider the folis irrational. This proof is due to Pythagoras, and is the lowing classical proof that first known proof using contradiction. The statement P to be proved is as follows:

P "V is irrational." Proof. (We assume - P; that is,

is not irrational, i.e., rational.) Suppose that is rational. Then V2 = m/n, where m and n are integers, with m and n not both even. (The sentence "m and n not both even" will be our statement R.) But then m2 = 2n2. Therefore m2 is even. Thus by Example A.3.2 the integer m is even. Since m is even, by Example A.3.l(b) the integer m2 is divisible by 4. Since m2 = 2n2, 2n2 is divisible by 4. Thus n2 is even, and by Example A.3.2 the integer n is also even. Thus m and n are both

even. (This is our statement -- R. The negation of "not both are even" is "both are even.") This is a contradiction. Thus V is irrational. U

P

The method of proof by contradiction can also be used to prove the implication Q. Using the law of contradiction we have that (P = Q) .

[- (P =* Q) = (R A - R) ].

A.3

Mathematical Proofs

511

Thus to prove "P Q" by contradiction we must prove that -(P = Q) implies a contradiction R A - R. Since P =:: Q a -P V Q, by De Morgan's law and double negation

-(P=* Q)q(PA-Q). Q by contradiction we must show that the assumption PA ^ Q logically implies a contradiction RA R, for some appropriate statement R.

Thus in the proof of P

Proof of P= Q by Contradiction Proof. Suppose PA

Q.

Therefore -R. Thus P

Therefore R.

Q

O

A.3.4

EXAMPLES

(a) To illustrate the method of proof by contradiction we will prove the following theorem about the integers.

Theorem. If a, b, c are integers satisfying a2 + b2 = c2, then a or b is an even integer.

Let P denote the statement "a, b, c are integers satisfying a2 + b2 = c2" and Q denote the statement "a or b is an even integer." In the proof by contradiction we assume P A -- Q. Since a and b are assumed to be integers, from the assumption PA -- Q we conclude that a and b are odd integers. Thus we can write a = 2m + 1 and b = 2n + 1, where m and n are integers. But then

c2=a2+b2=(2m+ 1)2+(2n+ 1)2, which upon simplification gives c2 = 4k + 2 for some integer k. Thus c2 is an even integer. But then by Example A.3.2 the integer c itself is even. Hence c = 2p for some integer p. This gives

c2=4p2=4k+2

or

(p2-k)=1.

p2 - k is an integer. (This is our statement R.) On the other hand, 1 is not an integer. (This is our statement -R.) From our assumption Since p and k are integers,

PA _ Q we derived a contradiction R A - R. (b) For our second example we prove the theorem of Example A.3.2 by contradiction. For an integer n, let P and Q denote "n2 is even" and "n is even" respectively. To prove P Q we assume P and -Q; that is, n2 is even, and n is odd. Since n is odd, n = 2k + I for some integer k. Therefore,

n2 = (2k + 1)2 = 2(2k2 + 2k) + 1. Therefore n2 is odd (-P). This is a contradiction (PA

P).

512

Appendix

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In the previous example we assumed PA - Q and showed that -Q leads to -P. Thus P itself plays the role of R in the method of proof by contradiction. This is cerQ and derives -P without using P. tainly permissible. However, if one assumes PA then the proof is in fact a proof by contraposition, rather than a proof by contradiction. A WORD OF CAUTION! It has been my experience that students have a tendency to

overuse-and even misuse-proofs by contradiction. Quite often, in proving P

Q,

the student will assume PA -Q and derive Q thereby obtaining the contradiction Q and -Q. Inevitably, if the proof is correct, buried in the details is a direct proof of P Q. Similarly, if the assumption PA -- Q leads to -P then the proof is in all likelihood a proof by contraposition. This was the case in part (b) of the previous example. Another problem with indirect proofs or proofs by contraposition is that they involve the negation of statements. This is not always easy in analysis, especially if the statements themselves are complicated and involve one or more quantifiers. (See Examples A.4.4 in the next section.) Before attempting an indirect proof or even a proof by contraposition, the student is advised to first attempt to find a direct proof. A direct proof often has the advantage of being more constructive. If the statement involves the existence of a certain object, a direct proof may in fact provide a method for constructing the given object.

Proof by Cases The method of proof by cases is based on the rule of inference by cases. Thus to prove that (P V Q) R it suffices to prove that P R and that Q - R. We illustrate this with the following examples.

A.3.5

EXAMPLES

(a) For our first example we prove the following: "If n is a positive integer, then

n2 + n + l is odd:' If n is a positive integer, then n can be either even or odd. Thus if P and Q represent the statements "n is an even positive integer" and "n is an odd positive integer" respectively, then the statement we wish to prove is (P V Q) R, where R represents the

statement "n2 + n + 1 is odd:' By the rule of inference by cases it suffices to prove

P

R and Q =* R. The details are left to the exercises (Exercise 6).

'(b) For our second illustration of the method of proof by cases we consider the following theorem.

Theorem.

There are two irrational numbers a and b such that ab is rational.

is rational. If this is the case we take a = b = V. Second First case: case: N/r2`r' is irrational. In this case we take a = \/'2-"2 and b = V2-. So

Proof.

ab=(V "i)V2

_

2(`/-2)'=

2=2,

which is rational. Therefore, there must be irrational numbers a and b so that ab is rational.

A.3

Mathematical Proofs

513

In the proof, we let R denote the statement of the theorem and P the statement " V-2 % 2 is rational." In Case 1 we have P R, and in Case 2 we have -P = R. Thus by the rule of inference by cases we have (P V -- P) R. But P V -- P is true regardless of P. Hence R follows by the modus ponens rule. This example not only illustrates the method of proof by cases, it also illustrates how P V -- P can always be asserted in a proof.

Counterexamples Some conjectures in mathematics are simply statements that are either true or false. For example, "V is irrational:' Other conjectures are general in that they assert something about a whole class of objects. For example,

"Every Cauchy sequence of real numbers converges:' If n is an even integer, then n2 is even." The first makes an assertion about all Cauchy sequences, whereas the second makes an assertion about all even integers. Both of these statements are true. If we are presented with a statement about a class of objects, then such a statement is true if and only if it is true for every object in the class. Thus to conclude that such a statement is false it suffices to exhibit one object in the class for which the statement is not true. Such an object is called a counterexample. To illustrate this we consider the following conjecture:

"If n is a positive integer, then n2 - n + 5 is prime." With a little bit of thought, most students will immediately conclude that this conjecture is false. If we check for n = 1, 2, 3, 4, then n2 - n + 5 becomes 5, 7, 11, and 17, which are indeed all prime. However, when n = 5, n2 - n + 5 is equal to 25, which is certainly not prime. Since we have exhibited an object (n = 5) for which the hypothesis P is true but the conclusion Q is false, the implication P Q is false. The previous example was very elementary, and to find a counterexample was not very difficult. This, however, is not always the case. As an example, consider the following conjecture. "Every continuous function on a closed and bounded interval is differentiable except perhaps at a finite number of points:'

Most students who have completed a basic calculus sequence might be inclined to believe that this conjecture is true. Certainly, on the basis of most examples encountered in calculus such a conjecture seems reasonable. In fact, many mathematicians through the mid-nineteenth century accepted this, or a variation of it, as true. It was not until 1874, when Weierstrass constructed an example of a continuous function that was nowhere differentiable (see Section 8.5), that the above conjecture was proved false.

Helpful Hints The most common complaint heard from students when asked to prove a theorem is "I don't know where to start" Unfortunately, there are no easy rules that can be used to tell someone how to prove a theorem. Many of the problems and theorems in the exer-

514

Appendix

Logic and Proofs

cises follow from the definitions or from previous theorems. Some, however, require insight and creativity, and for these the students must devise their own arguments. Some helpful hints in constructing proofs follow.

(1) Make a list of all hypotheses and of what you want to prove. Do not ignore any of the hypotheses. As a general rule they are all required. If you have not used all the hypotheses then most likely your proof is incorrect. (2) Refresh your memory with the pertinent definitions. If necessary, write them out. This will help you to memorize them and also to understand them. It is very important that you know and understand all the definitions. They are the foundations upon which the theory is built.

(3) Search for theorems that have similar hypotheses or similar conclusions. Proved theorems are not just results; they are also tools that enable you to develop the theory further. Suppose you are given P and R as hypotheses and are asked to prove Q. If you have a theorem that P S and you can prove that (R A S) Q, then you have your desired proof. An alternative approach is to work backward; that is. to find a theorem that has the same conclusion Q with given hypothesis S, and see if you can prove S from the given hypotheses P and R. (4) Learn the statements and the proofs of theorems. When I teach real analysis. I always require students to memorize the statements of all the theorems and the proofs of selected theorems. Contrary to student beliefs, this is not done to torture them. The statements of the theorems are the tools used in proving other theorems; the proofs provide useful techniques that may be used elsewhere. They also provide a good model of how to write a correct proof. Some common errors committed by students include the following. (1) Using theorems for which all the hypotheses are not satisfied. (2) Making extra assumptions beyond those given in the statement of the problem or theorem. (3) When asked to prove something, for example all continuous functions f, the theorem is proved for a particular function such as f (x) = x-. Even though this is incorrect, by attempting to prove the theorem for a special case, the student may in fact gain insight on how to prove the result for the general case.

R EXERCISES A.3 1. As in Example A.3.l(a), write the proof of Example A.3.I(b) in symbolic form. 2. Same as the previous exercise for Example A.3.2. 3. Construct proofs of each of the following statements about the positive integers. Suppose k. m, and n are positive integers.

a. If m and n are odd, then mn is odd. c. If m2 is odd, then m is odd.

e. If n is odd, then n2 + 1 is even.

b. If m is odd, then m, is odd. d. If k divides m and m divides n, then k divides n.

A.4

4. Consider the following statement. "If n is a positive integer, then a. a direct proof, b. a proof by contradiction. 5. Prove that

>

Use of Quantifiers

515

Prove the statement by pros iding

is irrational by contradiction.

6. Complete the details of the proof in Example A.3.5. 7. Prove that the equation a2 = 4b + 3 has no integer solutions. S. Provide a counterexample to each of the following statements.

a. If n is a positive integer, then n2 - n + 41 is prime. b. If n is a positive integer, then n! < 2'. c. If n is a positive integer, then n' < (n + I)". d. Every continuous function is differentiable.

_AAJ Use of Quantifiers2 We have already touched on the notion of quantifiers while discussing counterexamples in the previous section. There we discussed the difference between sentences that are simply statements that are either true or false, and sentences that make assertions about a collection of objects. In this section we make the latter more precise.

Quantified Sentences The sentence "x'- = 4" is not a proposition as it is neither true nor false. If we replace x by specific values then the statement "x 2 = 4" becomes a proposition. For example, the sentence is true for x = 2 and false for x = 3. Likewise, the sentence "x is a rational number" is neither true nor false until x is replaced by a specific quantity. In the sentences "x 2 = 4" and "x is a rational number," the 'x' is called a variable and the sentences themselves are called formulas or open sentences in the variable x. Specifically, a formula (in logic) is a statement containing one or more variables which becomes a proposition when the variables are replaced by particular objects. W e will use the notation P(x) to denote that P is a formula in the variable x. Likewise, a formula in the varixk will be denoted by P(xt, .. . , xk). For example, "xi = x2 + x3' is a ables x 1 ,. formula in the variables xi, x2, x3. Before the truth of a formula P(x) can be determined we must specify what objects are available for discussion. This is called the universe U for P(x). For example. for the formula x2 = 4 an appropriate choice for the universe U may be either the set of positive integers N, the set of all integers 71, or even the set of real numbers R. It is not enough, however, to just specify the universe: For example, in the formula x - y = x. a meaning must also be given to " " and If we are discussing 2 X 2 matrices. then in addition to specifying the universe C' as the set of 2 X 2 matrices, we must also

2 Since this section requires some basic knowledge of the terminology of sets it is best postponed until Section 1.1 has been read.

516

Appendix

Logic and Proofs

define matrix multiplication and equality of matrices. For logical considerations, we must also require that the universe be nonempty. For a formula P(x) with specified universe U, the truth set of P(x) is the set of all x E U such that P(x) is a true proposition. In the notation of sets, the truth set is simply

{x E U: P(x)}. This is read as "the set of x in U such that P(x)." For example, {x E N : x2 = 4} = {2}, whereas {x E TL : x2 = 4} = {-2, 2}. Consider the two formulas P(x) and Q(x) given by "x2 = 4" and "(x + 1)2 = x2 + 2x + 1" respectively. If we take as our universe the set of real numbers R, the truth set for P(x) is nonempty. This is expressed by saying that there exists an x E 1R such that P(x) is true. On the other hand, the truth set for Q(x) is all of R, and this is ex-

pressed by saying that Q(x) is true for all x E R. We make this precise with the following definition.

AA.1

DEFINITION Suppose P(x) is a formula in the variable x with universe U. (1) The sentence (Vx) P(x) is read 'for all x, P(x)" and is true precisely when {x E U : P(x)} = U. The symbol V is called the universal quantifier. (ii) The sentence (3x) P(x) is read "there exists x such that P(x)" and is true precisely when {x E U : P(x)} # 4. The symbol 3 is called the existential quantifier.3

The expressions (Vx)P(x) and (3x)P(x) are called quantified sentences. The phrase "for every" is synonymous with "for all" If we wish to emphasize the universe U we write (Vx E U)P(x) and (3x E U)P(x). These are read as "for all x in U. P(x)and "there exists x in U, such that P(x)" respectively. For example, with the formulas

(x+ 1)2=x2+2x+landx2=4wehave (`dxE R)[(x+ 1)2 = x2+ 2x+ I],

(3xCt8)(x2=4). Most mathematicians avoid using V and 3 in publications. In fact, with the exception of this section, they are used nowhere else in the text. Expressing mathematical statements in quantified form is first of all not always easy, and second makes for awkward reading. Quantifiers, however, are crucial when it comes to negating complicated statements.

AA.2

EXAMPLES

(a) As our first example we express the following theorem of the text as a quantified sentence.

Theorem 2.6.4. Every Cauchy sequence of real numbers converges.

Clearly, the quantifier to be used is V. For our universe we take U to be the set of all sequences of real numbers, and let C(x) and Q(x) denote the open sentences "x is a Cauchy sequence" and "x converges" respectively. Consider first the sentence

3 The symbol 3! is often used to denote the existence of a unique x for which P(x) is true.

A.4

Use of Quantifiers

517

(b'x)[C(x) A Q(x)]. This sentence would be translated as follows: "For all sequences x, x is a Cauchy sequence and x converges:' This sentence, however, is the same as "every sequence is a convergent Cauchy sequence:' and this is clearly not the intent of the original statement. is a Cauchy seIn other texts, Theorem 2.6.4 is sometimes expressed as "if quence of real numbers, then {an} converges" This statement can also be rewritten as "For all sequences {an} of real numbers, if {an} is a Cauchy sequence, then {an} converges:' This version, although somewhat awkward grammatically, is now easily written in symbolic form as (Vx E U)[C(x) ' Q(x)l. In general, a sentence of the form "All P(x) are Q(x)" is expressed in symbolic form as (Vx)[P(x) Q(x)]. (b) As another example consider the statement "Some bounded sequences converge."

As in (a) let Q(x) be the sentence "x converges" and B(x) be the sentence "x is a bounded sequence." Since "some" is taken to mean at least one, the proper quantifier is 3. However, should the statement be expressed symbolically as (3x E U)[B(x) Q(x)]

or as (3x E U)[B(x) A Q(x)]? The first would be interpreted as "There exists a sequence x, such that if x is bounded, then x converges." This clearly is not the intent of the sentence. It does not ensure the existence of a bounded sequence that converges. The second, (3x)[B(x) A Q(x)l, reads "There exists a sequence x, such that x is bounded and x converges," and this is the correct interpretation. In general, the statement "Some P(x) are Q(x)" is expressed symbolically as (3x)[P(x) A Q(x)]. (c) Most expressions in mathematics require the use of many quantifiers. To illustrate this, consider the definition of convergence of a sequence of real numbers as given in Section 2.1 of the text.

DefWtion 2.1.7. A sequence {pn} in Il8 is said to converge if there exists a point

p E R such that for every e > 0, there exists a positive integer n such that

Ipn - pI < efor all n?no. This definition uses the quantifiers V and 3, not only once, but several times. We have

"3p E R," "3no E N," "(Ve)(e > 0)," and "(Vn)(n Z n.)." To write this statement in symbolic form, we begin with (3p E R)[' ]. Consider the sentence "for every e > 0, there exists ..:' This phrase, properly stated, should be expressed as "for all e E 13, if e > 0, then ...,"which in symbolic form would be written as (Ve)[(e > 0) . (. )].

This leaves us with the final phrase "there exists a positive integer n, such that Ipn - pI < e for all n n,,. This would be written as (3no E N) ("Ipn - p] < e for all n ? no"). The statement "Ipn - pi < e for all n ? no," again if properly stated, would read as "for all n E N, if n z n,,, then Ipn - pI < e," or in symbolic form, (`tin E N)((n a I p, - p I < e). Combining all of the above finally gives "The sequence { pn} in R is said to converge" if

(3p e R)[(`de){(e > 0) ' [(3no E NI){(Vn E N)((n a no)

Ipn - pI < e)}]}].

4 In mathematics the term "some" is taken to mean at least one. This differs from the colloquial interpretation where "some' is occasionally taken to mean two or more.

518

Appendix

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This can also be expressed as

n = tp - pl < e)],

(3p)(Ve)[e > 0

where the universe for each of the quantifiers is understood.

Negation of Quantified Sentences The next thing we want to consider is how to negate a quantified statement. First, however, we need to define what it means for two quantified sentences to be equivalent. Recall that two propositional formulas P and Q are equivalent if and only if P a Q is a tautology. Suppose P(x) and Q(x) are two formulas with nonempty universe U. Then P(x) and Q(x) are equivalent in U if and only if P(x) and Q(x) have the same truth value for all x E U; that is, (Vx E U)(P(x) #' Q(x)). Two quantified sentences P(x) and Q(x) are

equivalent if and only if they are equivalent in every universe. For example. since P Q is equivalent to P V Q, if P(x) and Q(x) are formulas or open sentences in x with universe U, then

(Vx E U)[(P(x)

Q(x)) a (-- P(x) V Q(x))).

Thus the quantified sentences (Vx)(P(x) alent.

Q(x)) and (Vx)(- P(x) V Q(x)) are equiv-

Consider now the quantified sentences (Vx)P(x) and (3x)P(x). The sentence -- (Vx)P(x) is true in a given universe U if and only if (Vx)P(x) is false: that is, if and

only if the truth set {x E U: P(x)} is not equal to U. But this is true if and only if {x E U : -- P(x)} is nonempty,5 i.e., (3x) P(x). Since this argument holds for any universe U, the quantified statement (Vx)P(x) is equivalent to (3x) -- P(x). A similar argument also proves that (3x)P(x) is equivalent to (Vx) - P(x). To emphasize these we state them as a theorem.

A.4.3 THEOREM Suppose P(x) is a formula in the variable x. Then (a) -(Vx)P(x) is equivalent to (3x) -P(x). (b) -(3x)P(x) is equivalent to (Vx) - P(x). In the following examples we find the negation of each of the quantified sentences of Examples A.4.2.

A.4.4

EXAMPLES

(a) As in Example A.4.2(a) we consider the statement "Every Cauchy sequence converges.' Using the same notation, in symbolic form this sentence was expressed as (Vx)[C(x) .Q(x)]. The negation of the statement becomes (3x) - [C(x) Q(x)J. Q(x)] is equivalent to (- C(x) V Q(x)), and by De Morgan's law the Now ( C(x) Q(x)] is latter is equivalent to C(x) A -- Q(x). Thus the negation of (Vx E U)[C(x) (3x E U)[C(x) A -- Q(x)]. This last statement would be read as "Mere exists a Cauchy

5 Here it is required that the universe U itself is not empty.

A.4

Use of Quantifiers

519

sequence in R that diverges." (Note: This .statement, however, is false in the real number system R, but true in the rational number system 0.)

(b) For our next example consider the negation of the statement "Some bounded sequences converge" In symbolic form this was expressed as (3x)[ B(x) A Q(x)] (see Example A.4.2(b)). The negation of this statement becomes (Vx) -- [B(x) A Q(x)j. which Q(x)]. But - B(x) V -- Q(x) is by De Morgan's law is equivalent to (Vx)[- B(x) V equivalent to B(x) -- Q(x). Hence the negation of (3x)[B(x) A Q(x)] is equivalent to of real (dx)[B(x) - Q(x)]. This last statement would read as "For all sequences is bounded, then diverges," or more simply as 'All bounded senumbers, if quences in R diverge:' (c) As our final example we undertake the negation of the definition of convergence of a sequence of real numbers. In symbolic form, a sequence {pn} in R is said to converge if

Ip,, - pi <

(3p (=- l)[(`de){(e > 0)' [(3n E N){(`dn E N)((n >_

We proceed to negate this sentence step by step. First, the negation of (3p E R)[ . } ] we have (`de){P(e)w becomes (Vp E 111) -- [ ]. Now inside the bracket [.

Q(e, no, n)}, where P(e) denotes "e > 0" and Q(e, no, n) denotes "(3n0 E N) {(V n E N)((n ? nn) a 1p. pi < e)}." But (Ve){P(e) Q(e, n,,, n)} is equivalent to (3e){P(e) A -- Q(e, no, n)}. It is left as an exercise (Exercise 7) to show that the negation of Q(e, no, n) becomes

(Vn,, E N){(3n E N)[(n ? n,) A (IPn - pi Combining all the above gives us the following. "A sequence verge" if

e)]}.

in R is said to di-

(Vp E R)(3e){(e > 0) A (Vn E N)[(3n E N)(n =' no) A (Ip - pi ? e)]}, or

(Vp E lR)(3e > 0)(Vn,, E N){(3n ? n,)IPp - pI ' e). Translating everything into English gives `A sequence

in 18 is said to diverge if for

all p E R, there exists e > 0 such that for all n, E N, there exists n ? n,, such that IPn - PI e.

0 EXERCISES A.4 1. Express each of the following sentences in symbolic form. Specify an appropriate universe for each. a. All men are mortal.

b. Not all mortals are men. C. Some isoceles triangles are equilateral triangles.

d. Some triangles are isoceles triangles and some are equilateral triangles. e. Some triangles are isoceles and equilateral triangles.

520

Appendix

Logic and Proofs

f. Not all isoceles triangles are equilateral triangles. g. Between any two distinct real numbers there is a rational number. 2. In the following, let the universe U be the set of all sequences of real numbers, and let M(x), B(x), and C(x) denote the sentences "x is monotone, "x is bounded,' and "x converges" respectively. Express each of the following statements in symbolic form using the quantifiers 3 and V. a. All convergent sequences are bounded.

b. There exists an unbounded monotone sequence. C. Some monotone sequences are unbounded. d. Every bounded monotone sequence converges. e. Not all bounded sequences converge. L Every divergent sequence is unbounded. 3. Determine which of the following quantified sentences are equivalent.

a. (Vx)[P(x) A Q(x)] and [(Vx)P(x) A (Vx)Q(x)] b. (Vx)[P(x) V Q(x)] and [(Vx)P(x) V (Vx)Q(x)] (Vx)Q(x) c. (Vx)[P(x) Q(x)) and (Vx)P(x) d. (Vx)(Vy)P(x. y) and (Vy)(Vx)P(x, y) e. (3x)[(3Y)P(x, y)] and (3Y)[(3x)P(x, y)] 1. (3x)[(VY)P(x, y)] and (Vy)[(3x)P(x, y)]

4. Find the negation of each of the following quantified sentences. a. (Vx)[P(x) V Q(x)]

b. (Vx)[P(x) V -- Q(x)] e. (3x)[P(x)' . Q(x)] d. (3x)[(P(x)* Q(x)) A (R(x)=* Q(x))] e. (Vx)[{P(x) A (P(x) z* Q(x))}

f. (Vx)[(3Y)(P(x, y)

' Q(x)]

' Q(x, Y))]

5. Find the negation of each of the following quantified sentences.

a. (Vx E R)(2x > x) b. (Vx E R)[(x > 0) .(2x > x)) C. (3x

IE

NX5x + 11 = 3x + 14)

d. (Vx E R)(3yE l8)(x + y = 0) e. (Ve E R)[e > 0 ' (3nE I@!)(1, < e)] 6. Find the negation of each of the statements in Exercise 2. Express your answer first in symbolic form and then in English.

7. Show that the negation of (3n,)((Vn)((n ? n,) p, - pI < e)) is (Vn,)(3n)((n 3 n,) A (Ip, - pI a e)]. 8. In the following, f is a real-valued function defined on an open interval (a, b) and p is a point in (a, b). The definition of the limit of the function f at p is as follows: "The function f has a limit at p if there exists a number LE R such that given any e > 0, there exists a 8 > 0 for which [f(x) - LI < e for all xE (a, b) with

0<]x-pI <S."

a. Express the definition of the limit of a function in symbolic form. b. Find the negation of the definition of the limit of a function.

Supplemental Reading

521

SUPPLEMENTAL READING Burris, S. N., Logic for Mathematics and Computer Science, Prentice Hall, New Jersey, 1998. Lemmon, E. J., Beginning Logic, Chapman & Hall/CRC, Boca Raton, Florida, 1998.

Smith, D., Eggen, M., and St. Andre, R.. A Transition to Advanced Mathematics, Brooks/Cole. Pacific Grove, CA, 1997.

Bibliography

Berg, P. W. & McGregor, J. L., Elementary Partial Differential Equations, Holden-Day. Oakland, CA, 1966. Hewitt, E. & Stromberg, K., Real and Abstract Analysis, Springer-Verlag, New York, 1965. Katz, Victor J., A History of Mathematics, 2d ed., Addison-Wesley, Reading, MA, 1998. Natanson, I. P., Theory of Functions of a Real Variable, vol. 1, Frederick Ungar Publ. Co., New York. 1964. Rudin, W., Principles of Mathematical Analysis, McGraw-Hill, Inc., New York, 1976. Titchmarch, E. C., The Theory of Functions, Oxford University Press, 1939. Weinberger, H. F., Partial Differential Equations, John Wiley & Sons, New York. 1965. Zygmund, A., Trigonometric Series, vol. I & 11, Cambridge University Press, London. 1968.

522

Hints and Solutions to Selected Exercises

S CHAPTER 1 EXERCISES 1.1 page 5

2.(a)AfB=B,Af17L={-1,0,1,2,3,4,5},Bf1C={x:2:5x!53} (b)AXB={(x,y):-l<x<_5, O!5 y<_3} 4. (a) Suppose xEAfl(Bf1C).ThenxEAandxEBflC.SincexEBf1C. xEBandxEC. Thus x E A fl B and x E C. Therefore, x E (A f7 B) fl C. This proves An (an c) c (A fl B) fl C. The reverse containment is proved similarly. 7. (a) Let x E A U (B fl C). Then x E A or x E s fl C. If x E A. then x E A U B x E A U C, i.e., x E (A U B) fl (A U C). Thus A U (B n C C (A U B) fl (A U C). The reverse containment is proved similarly. 8. If A = { 1, 2, 3} then .(A) = (d,. 11). {2}, (3), (1, 2), {1, 3), (2, 3), Al.

EXERCISES 1.2 page 14 2. (a) No! The ordered pairs (0, 1) and (0, - I) are both elements of A. This, however, contradicts the definition of function. 3. (a) f((1, 2, 3, 4)) = { 1. 3, 5, 71 and f '({1, 2, 3, 4}) = 1, 2). 4. (a) f(A) = (y : 0:5 Y S 9);

f-'(A) = {x : x' + I E A) = {x:-'5' s x s 1 } (c) f-'(x) = j x - 1, x r= R. 6. (a) Range f - R, f is oneto-one, and x = f-'(y) = 3(y + 2). (c) Range f = {y : I s y < oo}, f is one-to-one, and x = (i - I)/y. (g) Range f = l8, f is not one-to-one: If y, # y2. then (x, y,) # (x. y2) for any x, yet f(x, y,) = f(x, y2).

7. (a) Range f = {(x, y) E R X R : x2 + y2 = I } (b) f -'((1, 0)) = 0, f-'((0, -1)) = f 11. Assume f is not one-to-one and show that this leads to a contradiction.

EXERCISES 13 page 19 1. (b) For n = 1, 1 = l2. Assume the result is true for n = k. Then for n = k + 1, I +3+ . +(2k- 1)+(2(k+ 1) - 1)=k2+(2k+ 1)=(k+ 1)2. (f) When n = 1, .r' - y2 = (x - y)(x + y) and equality holds. For n = k + 1, write xk+2 xk-2 - Xyk-' + XY`-' - Yk+ = X(xk'' - k-')+ (x - y)yk+', and now apply the induction hypothesis. 2. (a) The result is true for n = 1. Assume that fork E N. 2k > k. Then by the induction hypothesis, 2k-' = 2k 2 > k 2 = k + k ? k + 1. (c) For n = 4,4! = 24 > 16 = 24. Thus the inequality is true when n = 4. Assume that n! > 2" for some n z 4. Then (n + I)! = (n + I )n! > (n + 1)2" > 2 2" = 2"-'. Thus by the modified principle of mathematical induction the inequality holds for all n E N. n ? 4. 4. For n E N let P(n) be the statement f(n) = 3 2" + (-1)". Then P(n) is true for n = 1, 2. Fork ? 3, assume that P(j) is true for all 523

524

Hints and Solutions to Selected Exercises

j E N1, j < k. Use the fact that f(k) = 2f(k - 2) + f(k - 1) and the induction hypothesis to show that P(k) is true.

5. (b) f(n) = n2 (d) f(n) = 0 if n is even, and f(n) _ (- 1)t"-'n/n! if n is odd. 7. For each n E N let S" = r + r2 + + r". Then S. - rS" = r - r"+', from which the result follows. 8. Hint: Let A = '(a, + + a") and write a" xA for some x a 0. Use the induction hypothesis to prove that (a, a" '') s x'1t"+')A. Thus by the second principle of mathematical induction, the result holds for all n E N.

Now use Bernoulli's inequality to prove that xj'j"'') s (n + x)/(n + 1). From this it now follows that xii("+')A 5 n + x A = n + 1

1

n + 1

(a,+.+a"+a"+,).

EXERCISES 1A page 26 4. Consider (a - b)2. S. (a) inf A = 0, sup A = I (e) inf C = -oo, sup C = 00 (e) inf E = 1, sup E = 3 (g) inf C = 0. sup C = no (i) inf I = - 2, sup ! = 2 14. (b) Since A and B are nonempty and bounded above, a = supA and is = sup B both exist in R. Since a = supA we have a s a for all a E A. Similarly b :5 0 for all b E B. Therefore, a + b 5 a + $ for all a E A, b E B. Thus a + /3 is an upper bound for A + B, and thus

y=sup(A+B)5a+J3. To prove the reverse inequality, we first note that since y is an upper bound for A + B, a + b s y forall a E A. b r= B. Let b E B be arbitrary, but fixed. Then a c y - b for all a E A. Thus y - b is an upper bound for A and hence a <_ y - b. Since this holds for all b E B, we also have that b s y - a for all b E B. Thus S s a; i.e., a + / <- y. An alternative way to prove that y = a + /3 is by contradiction. Suppose y < a + P. Set e = a + /3 - y. By assumption e > 0. Now consider a - Ie and 0 - ?e, and use Theorem 1.4.4to obtain a contradiction. 15. (a) Let a = sup{f(x) : x E X}, )3 = sup{g(x) : x E X}. Since the range off and g am bounded, a and P are finite with f(x) + g(x) a + 6 for every x E X. Therefore, a + 0 is an upper bound for {f(x) + g(x) : x E X}. Thus sup{f(x) + g(x) : x E X} s a + /3. 16. (a) F(x) = 3x + 2. sup{F(x):x E [0, 1)} = 5 20. (a) F(x) = 3x + 2 and H(y) = 2y. Thus sup{H(y) : y E [0, 111 = 2 and inf{F(x) : x E [0, 1 ]} = 2.

EXERCISES 15 page 30 1. First prove that if p and q are positive integers, then there exists n E N such that np > q. If p > q then n = 1 works. If p s q, consider (q + I )p. 6. (a) Use the fact V2/2 is irrational. (b) Use Theorem 1.5.2 and (a).

EXERCISES 1.6 page 34 2. (a) .0022 = + 51 + i + 1..0202020 (d) .101010

_ +;+ +5 +,+

= 27 + 81 =

_i

00

1

ar=13

1

1 ,] _-

3. .0101010

9

EXERCISES 1.7 page 42 2. Let f : 101-s 0 be defined by f (n) = 2n - 1. 4. (a) g(x) = a + x(b - a) is a one-to-one mapping of (0, 1) quo (a, b). 6. (a) Since A - X. there exists a one-to-one function It from A onto X. Similarly, there exists a one-to-one function g from B onto Y. To prove the result, show that F : A X B --X x Y defined by F(a, b) = (h(a), g(b)) is one-to-one and onto. 8. (a) U A" = R, f1 A. = {x : - I < x < 1) (c) U A. _ (-1, 2), fl A. _ [0, 1 18. (a) Consider the function on (0, 1) that for each n E N, n ? 2, maps to "=_ ,, and is the identity mapping + a,x + ao, consider the height h of the polynomial defined by elsewhere. 19. For a polynomial p(x) = ax" +

h=n+Iaol+lat[+...+Ia"I Prove that there are only a finite number of polynomials with integer coefficients of a given height It, and therefore only a finite number of algebraic numbers arising from polynomials of a given height It. 22. If f is a function from A -+ '(A), show that f is not onto by considering the set {x E A : x 6E f(x)}. 23. For a, b E [0,1 ] with decimal expansion a = a, a2 and b = b, b2 , consider the function

f : [0 1) x [0, 1 ) - [0, 1) by f(a, b) = a, b, a2b2 ... .

Hints and Solutions to Selected Exercises

525

CHAPTER 2 EXERCISES 2.1 page 52 2. Ix I = Ix - y + y 1 5 Ix - y I + I y 1. Therefore, I x I- I y 15 I x - y 1. Interchanging x and y gives I y i- Ix 1 5 I y- x I= I x- y I Now use the definition of II x I- I y I I. 6. (a) -3:5,x:5 13/3 (c) -1 <x < 2 & (a) Let a" = (3n + 5)/(2n + 7). Then I a. - 21 = 11/2(2n + 7) < 3/n. Given e > 0, choose n E N such that n, a 3/e. < -. Then for all n a n I a. - 2 I < e. Therefore, lim a" = 2(c) Set a" = (n2 + l )/2n2. Then I a. - III = Given e > 0, choose n, E N such that no a 1/2e. Then for all n a n", I a. - 121 < e.Thus lim a" = 2. 9. (a) If it is even, then n(1 + (-1)") = 2n. Thus the sequence is unbounded and diverges in R. (c) When n is even, i.e., n = 2k, then sin i = sinker = 0. On the other hand, when it is odd, i.e., n = 2k + 1, then sin If = sin(2k + 1) = (-1)k. Thus sin 9 assumes the values -1, 0, and 1 for infinitely many values of n. Therefore the sequence diverges. 11. (a) Write b = 1 + a where a > 0. By Example 1.3.3(b), b" ? I + no. Now use the previous problem. 12. First show that I a2. - a21 5 (I a" I + I a I) I a" - a I. Now use the fact that since {a"} converges, there exists a positive constant M such that I a" 15 M for all n e N. 14. Consider a = 0 and a > 0

separately. For a > 0, V - V = (a" - a)/(V + Va-). EXERCISES 2.2 page 59

S. (b) If p > 1, let x" = -Vp - 1. Apply the inequality of Example 1.3.3(b) to (I + x")". 6. (a) Is 7. (a) Converges to 1. 1

n2 + 1

n2 \2n +

2

3

(c) -1 (e) 2

(c) Since

(1 + 1/n2)2

(2 + 3/n)2'

by use of the limit theorems the sequence converges to 4.

8. Use the fact that I cos x 1 5 1 for all x E R. 10. (a) Suppose lim a"+ ,/a" = L < 1. Choose e > 0 such that L + e < 1. For this a there exists n. E 101 such that a"+ ,/a" < L + e for all n a n,. From this one obtains that for i > no, 0 < a" <_ (L + M(L + e)", where M = a",(L + Since (L + e) < 1, by Theorem 2.2.6(e). slim (L + e)" = 0. The result now follows by Theorem 2.2.4. 11. (a) With a" = n2a", 0 < a < L = Iimoa". i/a" = a lim (I + n)2 = a. Thus since L = a < 1, the sequence converges. 12. Set x" (a" - 1)/(a" + 1), and solve for a". 13. (b) Verify the result for it = 1. Assume the result holds for it = k. For n = k + 1, (1 + a)(k+1) _ (1 + a)(1 + a)k, which by the induction hypothesis

+ k

=(1 +a) Li

I=a

ja

k

±i

i=o

k

k+1

a'+

71 i-t

Using (a) show that the above is equal to

that one can assume a = 0.

k

(- _ 1

).v.

Ck + I

i

I {a'

j

14. (a) By considering the sequence {ak - a} show first

EXERCISES 2.3 page 65 2. Take I, _ [n oo). 3. (a) Set x" _ ( n2 + I)/n. Then x.2= 1 + , > 1 +

" ; = x.'+,. Thus x2. > x,2,+, > 1. From this it now follows that x" > x"+1 > 1 for all n E N. and that lim x" = 1. (c) Since a > 1, a"+' = a a" > a". Thus {a"} is monotone increasing. Let or = sup{a" : n E N!}. If a < oo, then a = lim a"+ = a lim a" = as > a. This, however, is a contradiction. Thus a" -4 oo. S. Use mathematical induction to show that a" > 1 for all n (=- N. From the inequality 2ab s a2 + 62, a, b >_ 0, we have 2a" 5 a.2 + I or a"+, = 2 - 1/a" s a". Therefore, {a"} is monotone decreasing. Finally, if a = Jim a", then

1=2- 1.a a = woo lint a"+,="moo limit- a") Therefore, a = 1. 7. (c) By induction, the sequence {a"} is monotone increasing and bounded above by 3. If a = Jim a", then a2 = lim a"+, = lim(2a" + 3) = 2a + 3. Thus a is a solution of a2 - 2a - 3 = 0. The two solutions

526

Hints and Solutions to Selected Exercises

are - I and 3. Since a must be positive, we have a = 3. (e) lim a" = 2. 9. (a) Use the inequality ab !< i (a2 + b2) to prove that x"., > \ for all n. To show that {x"} is monotone decreasing, consider x",, - x" and simplify. 12. (a) e2 (e) e312 13. To show that {s"} is unbounded, show that s2. > I + "I. Hint: First show it for n = 1. 2. and 3; then use mathematical induction to prove the result for all n E N. 15. Hint: Fork = 2. t < k,k ! l) = k i - ;. 17. (b) n + n(n - 1) (d) Hint: Write a = (1 + b) with b > 0. Now use the binomial theorem to show that a"In° ? cn for some positive constant c and all n sufficiently large. 18. (c) The sequence is not monotone increasing: If x" = n + (-1)'Vn-, then xu,, < x2,,. 23. This problem is somewhat tricky. It is not sufficient to just choose a monotone increasing sequence in the set; one also has to guarantee that the sequence converges to the least upper bound of the set. Let E be a nonempty subset of R that is bounded above. Let aqt denote the set of upper bounds of E. Since E * 0, we can choose an element x, E E. Also, since E is bounded above, OIL * 46. Choose /3, E V. Let a, = 12 (x, + 6, ), and consider the two intervals [x,, a, ] and (a,,19, ). Since x, E E, one or both of these intervals have nonempty intersection with E. If (a,, /3,] fl E * 0. choose x2 E E such that a, < x2 s S,. In this case, set $2 = /3,. If (a,, /3, ] fl E _ 0, choose x2 E E such that x, < x2 a,, and set /32 = a,. In this case, $2 E OU.. Proceeding inductively, construct two monotone sequences {)"} and {/3"} such that (a) {x"} C E with x" 5 x"+, for all n, (b) 0" - x" <- 2-"+ 1(b, - x1). Assuming that every bounded monotone {/3"} C OIL with ,6. ? /3"+, for all n, and (c) 0 sequence converges, let A = Jim /i". By (c) we also have /3 = lim x". It only remains to be shown that /3 = sup E. 24. Suppose A = {x" : n E N} is a countable subset of [0, 1 ]. To show that A C [0, l ] proceed as follows: At least one of the three closed intervals [0, 13], [13.23). [j, 1 ] does not contain x,. Call it!,. Divide 1, into three closed intervals of length 1/32. At least one of these, say 12, does not contain x2.

EXERCISES 2.4 page 72 3. (a) {-1, 0. 1} (e) {1} 4. (a) e2. Set p" = (1 + i )3n. By Example 2.3.5, Jim p" = e. The answer now follows since (I + 3 )6i = p.2. 7. (a) The set has only one limit point; namely { 11. Every point of the set is an isolated point. 14. See the solution of Exercise 23 of Section 2.3.

EXERCISES 2.5 page 79 1. (a) -oo, oo (c) -1, 1 (d) -3, j 3. 0, 1 6. (a) Let a = lim a" and y = I i(a" + b"). Since the sequences are all bounded, a, y E R. Let e > 0 be given. Then by Theorems 2.5.3 and 2.5.4 there exists n° E N such that

a" > a - e/2 for all n ? n° and (a" + b") < y + e/2 for all n ? n°. Therefore. b" < y + e/2 - a" < y - a + e. From this it now follows that limb" s y - a + e. Since e > 0 was arbitrary, we have limb" 5 y - a; i.e.. lim a" + Fim b" s lira (a" + Q. The other inequality is proved similarly. {s2.,} and {s2i,,+,}.

8. 1. 1. Hint: Consider the subsequences 10. By Theorem 2.5.7 there exists a subsequence {a",} of {a"} such that a", --+a. Since {b"}

converges to b, a,,,b,,, -+ ab. Therefore, ab is a subsequential limit of inequality follows similarly. The fact that b * 0 is crucial.

Thus Tim

z ab. The reverse

EXERCISES 2.6 page 85 2. (a) The sequence {(n + 1)In} converges and thus is Cauchy. (d) Convergent sequence; thus Cauchy 4. (a) Show that stn - s" Z. & For n 3, I a"+, - a" I < Ia, - a"_, I. Therefore, {a"} is contractive. If a = Iim a,,, then

;

0 < a < 1 and is a solution of a2 + 2a - 1 = 0. (a"_, - a") _ (b - 1)"-'(a2 - a,). Therefore, n-1

n

a.+l - al =

11. (b) Since (a,,., - a") = (b - lxa" - a"_,), by induction

1(b-1)"

(ak+, - a*) _ (a2 - a,) Y, (b - 1)k = (a2 - a,) 2k-S

b

.

Letting n - oo gives a = a, + i! b (a2 - a,).

EXERCISES 2.7 page 89

1. Lets"=

Sincek sk

-1,k>2,forna2,s"{ 1 + Ik=2(k

l - i)=2-2.

Therefore, {s"} is bounded above and hence converges by Theorem 2.7.6. 5. Use the inequality ab s 2 (a2 + b2), a, b ? 0. 9. (b) Since the sequence (s"} of partial sums is monotone increasing, it suffices to show that some

Hints and Solutions to Selected Exercises

+

subsequence is bounded. Consider the subsequence {s,,) where nk = 2k - I. and prove that s,,. < I + a + where a = 2-tP-1l. For example, when k = 2, nk = 3 and

s3=1+2P+

3P

527

=1+( +]1<1+P=1+2v -,-' P

0 CHAPTER 3 EXERCISES 3.1 page 100 2. By Corollary 2.4.8 a finite set has no limit points. Now apply Theorem 3.1.9. This can also be proved directly by showing that the complement of a finite set is the finite union of open intervals. 6. (a) Let p E 0 = U,O0 be arbitrary. Then p E 0, for some a E A. Since 0, is open, there exists e > 0 such that N,(p) C 0,: But then N,(p) C 0; i.e., p is an interior point of 0. 8. (a) Let p E Int(E) be arbitrary. Since p is an interior point of E, there exists an e > 0 such that N,(p) C E. To show that N,(p) C Int(E) it remains to be shown that every q E N,(p) is an interior point of E. 9. (a) Since A U B is a subset of A U B (which is closed), by Theorem 3.1.11(c), A U B C A U B. The reverse containment follows analogously. 11. (a) (a, b) 13. Use Exercise 6(b) of Section 1.5. 14. If p E D, set p, = p for all n E N. If p is a limit point of D, use Theorem 2.4.7. 17. Let {r,}'_, bean enumeration of 0, and {t:,}- , an enumeration of the positive rational numbers. Take ' = {N,lr,) : j, n E N). 18. (a) Closed in X. (c) Neither open nor closed in X. 20. Suppose U C R is open, and suppose E C U is open in U. Let p E E be arbitrary. Use the fact that E is open in U and that U is open to show that there exists an e > 0 such that N,(p) C E. Thus p is an interior point of E. Since p E E was arbitrary, E is open in R. The converse is obvious.

23. (a) Let U = (0, 1) and V = (;, ;). 24. (a) Use the fact that A = A U A', and if U open satisfies u n A' # 0 then U fl A # 4,. 25. Let A be a nonempty subset of R. If A contains at least two points and is not an interval, then there exist r, s E A with r < s and r E P with r < r < s. but r E A. The open sets U = (-no, r) and V = (r, oo) will prove that A is not connected. Therefore, every connected set is an interval. Conversely, suppose A is an interval and A is not connected. Then there exist disjoint open sets U and V with A fl U 0 ¢.. A fl V * 4,, and A C U U V. Suppose a E A fl U and b E A fl V. By Theorem 3.1.13 applied to U and V, there exist disjoint open intervals f and J such that a E I and b E J. Suppose a < b and J = (r, s). Show that t 14 U U V butt E A. This contradiction proves that A is connected.

EXERCISES 3.2 page 107 1. (b) Use the fact that 0 is a limit point of A. 4. (a) Let ° t = { U.J. be an open cover of A U B. Then 11 is also an open cover of A and B respectively. Now use the compactness of A and B to obtain a finite subcover of A U B. 5. Since K is compact, by Theorem 3.2.9, K is closed and bounded. Let a = sup K. This exists since K is nonempty and bounded. Now use the fact that K is closed to show that a E K.

EXERCISES 3.3 page 110 4. If x E P with x = a, a2, a, E (0, 2), set b, = )a,. Consider the function x --+.b,b,

CHAPTER 4 EXERCISES 4.1 page 128 1. (a) If f(x) = 2x - 7, L = -3, then 1f(x) - L 1 = 21x - 21. Given e > 0 take S = e/2. Then for all x with Ix - 21 < 8, 1 f (x) - L 1 = 2 ; x - 21 < 2S = e. (c) If f(x) = k/(1 + x), then

< 2 lx- 1 21x+1111 j(x) 21 for all x > 0. Hence given e > 0, choose S = min {2e, I ). With this choice of 8, x > 0 and thus 11(x) - I < i S s e. 3. (a) The limit does not exist. For x > 0, x/I x 1 = 1, whereas for x < 0, x/I x 1 (c) The limit does not exist. Consider the sequence { I /nzr), which has limit 0 as n -+ no.

Figure 4.5, for 0 < t < -rr/2, sin t = length of PQ < length of arc PR = t.

4. lim1 f (x)

7. (a) Use the inequality

3

5. (a) By

528

Hints and Solutions to Selected Exercises

IIf(x)I - ILII s If(x) - LI (see 2.1.4). (c) Use induction on n. S. (a) 0 (c)

(e) J (g) 2. Note: 9. Let L = lim f(x). By hypothesis L > 0. Take e = 1 L and use the definition of limit. 12. Since x-r g is bounded on E, there exists a positive constant M such that Ig(x) I s M for all x E E. Thus 1f(x) g(x) 15 MI f(x)'. for all x E E. Now use the fact that lim f(x) = 0. 16. Suppose lim f(x) = L. Let e > 0 be given. By definition there exists M > 0 such that I1(x) - L I < e for all X-M x E (a, oo) with x > M. Let S = 1/M. Then for all t E (0, 1/a) with t < S, 1 It E (a, oc) and I It > M. Therefore. IS(r) - LI = If(',) - LI < e. The proof that limo g(r) = L implies iim f(x) = L is similar. 17. (a) 22 (c) 2 -M (e) 2 (g) Limit does not exist. For all x > ;, cos ; > 1. Thus x cos x > 1 x and x cos ; - oo as x - oo. 2(

).

'v

EXERCISES 4.2 page 141 1. (c) Since I - cos x = 2 sin2(x/2), for x # 0, g(x)

sin2(x/2). Now use the fact that I sin t I s 11

4.Ifp>0.

1.

If(x) - f(P)I = If - VI = Ix - PI/(V + V P - ) < LIx - PI Let e > 0 be given. Set S = m i n { p , V p e } . Then I x - p I < S implies that 11(x) - f(p)I < e. Therefore, f is

continuous at p. If p = 0, set S = e'-. Alternatively use Theorem 4.1.3 and Exercise 14 of Section 2.1. 6. (a) See Exercise 7a of Section 4.1. 7. Use Theorem 4.2.4, and the fact that x" is continuous on R for all

n E N. 9. (a) R \ (-2.0,2). (c) R. 12. (a) Use the fact that max {f(x), g(x)} =12(1(x) + g(x) + I f(x) - g(x) 1). 16. Consider g(x) = f(x) - f(x - 1), xE(O, I J. 18. Let p E E be a limit point of F. Use Theorem 2.4.7 and continuity off to show that p E F. 20. (b) By induction, f(nx) = nf(x) for all n E h) and x E R. In particular, f (n) = en where c = Al). Also, c = Al) = An -.) = nf(.1). Therefore, Al) = cln. Since f is continuous, letting n -ioo gives f(0) = 0. From this it now follows that f(-x) = f(x) for all x E R. Thus f(n) = cn for all n E Z and f(r) = cr for all r E 0 (write r = m/n, m, n E Z). Finally, by continuity Ax) = cx for all x E R. 22. Take e = 1. Then for this choice of a there exists a 8 > 0 such that 11(x) - f(P)I s 1 for all x E Na(p) fl E. Show that this implies that If(x)I c (If(P)I + 1) for all x E Na(p) () E. 25. Theorem 2.4.7 and Theorem 3.2.10 should prove helpful. 29. By hypothesis, for each x E K there exists e, > 0 and M, > 0 such that I Ay) 1 5 M, for all y E N1(x) fl K. The collection {N,,(x)},Ex is an open cover of K. Now use compactness of K to show that there exists a positive constant M such that I f(y)I S M for all y E K.

EXERCISES 4.3 page 147 2. (a) Suppose f(x) = x2 is uniformly continuous on [0, oo). Then with e = 1, there exists a S > 0 such that J Ax) - f C Y ) I < I f o r all x, y E [0, oo) satisfying I x - y I < 6. Set x" = n and y" = it + 1. If n E IOl is such that no6 > 1, then Iy" - x" I = I_11 < S for all n a n,,. But I f&.) -1(x")1 = 2+3, z 2 for all n. This is a contradiction!

3. (a) For all x, y E [0, oo),

If(x)-f(y)I =

l+x

l+y

(I+x)(ly+Y)
Thus given e > 0, the choice 6 = e will work. (f) Set g(x) = sin x/x, x E (0. 1 ] and g(0) = 1. Then g is continuous on [0, 1 ], and thus by Theorem 4.3.4 uniformly continuous on [0, 1 ]. From this it now follows that f is uniformly con

tinuous on (0, 1). 4. (a) Show that If(x) - f(y)I 5

1 x - yI for all x, y E [a, oo). a3

5. (a) For x, y E [a, 00),

Imo- VY- I = Ix-yI/(V + VY- )52:rIx-yl. 7. (b) Suppose I f I and I g I are bounded by M, and M2, respectively. Then

I f(x)g(x) - f(y)g(y) I

I AX) I I g(x) - g(y) I + I g(y) I I f (x) - f(y) I

5 MI I g(x) - g(y) I + M2I f(x) - f(y) I

Hints and Solutions to Selected Exercises

13. (a) Let x, E E be arbitrary. For it = 1 set x,,.

Now use the uniform continuity off and g. the sequence {x"} is contractive.

EXERCISES 4.4 page 160 <x 2. (b) m. f(x) = ,lito. f(x) = 0 (f) lin f(x) = I. Hint: For bounded near x,, = 2. 6. b = 14 7. For n E N set x = n and y = it +

6

'[

529

= f(x.). Show that

] = n. 4. Use the fact that [.r; is

and show that I f (x.) - f(;)1 = sin 2. 12. If it E N, g(x) = x" is continuous and strictly increasing on (0, cc) with Range g = (0, oo). Therefore by Theorem 4.4.12, its inverse function g-'(x) = x"" is also continuous on (0, oo). From this it now follows that f(x) is continuous on (0, oo). 14. (a) Suppose first that U = (a, b) C 1. Then since f is strictly increasing and continuous where is a finite or on 1. f((a, b)) = (f(a), f(b)), which is open. For an arbitrary open set U C 1, write U = countable collection of open intervals, and use Theorems 1.7.14 and 3.1.6.

CHAPTER 5 EXERCISES S.1 page 174 1. (a) For f(x) = x3. f(x + h) - f(x) f '(X) = lim ilm h (c) For h(x) = 1/x, x # 0,

0

h '(.x) = .1im

.,

h

,=

(x + h)3 - x'

_

h

=

n-0 x(x + h)

lim (3x2 + 3xh + h2) = 3x.

x'

(e)Forf(x)=x+1'x#-1'f(x)=ni0(x+1)(x+h+I)

(x+l

),.

2. If n E N, by the binomial theorem (x+h)"-x'= F,1n)hkx"-'l

k-t k

2(n)

-I+h k

22

k

hr-'x"-4.

Dividing by h and taking the limit as h -- 0 proves the result for n E N. If n E 71 is negative, write x" = I /xf, m E N. and use Theorem 5.1.5(c). 3. (a) Since cos x = sin(x + ), by the chain rule cos x = cos(x + i (c) No. 7. (a) f'(x) exists for all x E 12\71. -sin x. Alternatively, use the definition of the derivative. S. (a) Yes Forx E (k, k + 1), k E Z,f(x) = x[x] = kr. Thus f'(x) = k for all x E (k, k+1), k E Z. (c) The function h is differentiable at all x where sin x # 0. For x E (2kw, (2k + I ytr), k E 71, h'(x) = cos x. For x E ((2k - I )w, 2kwr), k e Z, h'(x) -cos x. 9. (b) For x f 0, g'(x) = 2x sin - cos Since lynx cos does not exist. ii g'(x) does

0

12. (a) f'(x) = 2/(2x + 1) (c) h'(x) = 3[L(x)]2/x equals 0 if and only if (b - 1) > 0; i.e, b > I. not exist.

13. (b) f;(0) = hliw, ht''-'t sin ,which exists and

EXERCISES 5.2 page 187 1. (a) Increasing on R. (c) Decreasing on (-oo, 0) and increasing on (0, oo), with an absolute minimum at x = 0. 4. (a) Show that the function f (x) = xv" - (x - 1)'r" is decreasing on the interval 1 <- x :5 alb. (b) Set f(x) = ax - x°, x =' 0. Prove that f (x) ? f (l) to obtain x° s ax + (1 - a), x ? 0. Now take x = alb. 5. (a) If f"(c) > 0. then there exists a S > 0 such that f'(x)/(x - c) > 0 for all x. I x - c I < S. Therefore, f (x) < 0 on (c - S, c) and f (x) > 0 on (c, c + S). Thus f has a local minimum at c. 7. Since P(2) = 0 we can assume that P(x) = a(x - 2)2 + b(x - 2). Now use the fact that P must satisfy P(I) = I and P'(I) = 2 to determine a and b.

12. (a) Let t, -c. Since f'(c) exists, lim (f(r") - f(c))/(t - c) = f'(c). Now apply the mean value theorem.

15. Since f+(a) = lira (f(x) - f(a))/(x - a;70, there exists a S > 0 such that (f(x) - f(a))/(x - a) > 0 for all x, a < x < a + S. 17. Hint: Consider f (x). 22. (a) For fixed a > 0 consider f(x) = L(ax), x E (0, x). (c) By (a) and (b), L(b") = nL(b) for all n E Z and It E (0, oo). But then L(b) = L((b"")") = it L(V"). From this it

530

Hints and Solutions to Selected Exercises

now follows that L(b') = rL(b) for all r E Q. Now use the continuity of L to prove that L(b") = xL(b) for all x E R. where If = sup{b' : r E Q. r : x}. 23. (b) Since tan(Arctan x) = x. by Theorem 5.2.14 and the chain rule, tan(Arctan x) = (sec(Arctan x))(' Arctan x) = I. The result now follows from the identity sec2(Arctan x) = 1 + x2. To prove this, consider the right triangle with sides of length 1. 1 x (, 1 + x2 respectively.

EXERCISES 5.3 page 196

f(x) -f(x,')/g(x) - g(xo) 2f(x) . = x - X. x - X. g(x)

Now use Theorem 4.1.6(c) and the definition of the derivative.

4. Use the fact that since ,ilitf(x) exists, f(x) is bounded on (a, a + 8) for some S > 0. 1 Inx x3+2x-3 = lint 5x4+2 = 7 (e) By L'Hospital's rule. lim - = =lim= -x = 0. 6. (a) lim +-'= x !- 2x3 - x2 - 1 =-+I 6x2 - 2x 4"

(e) Make the substitution x = 1/t.

EXERCISES 5.4 page 203 (2c; + ar)/3c2, 2. (a) f(0) = I and f (1) _ -1. Therefore. 1. Let c, > 0 be arbitrary. By Newton's method c", f has a zero on the interval (0, 1 ]. With c, = 0.5. c2 = 0.33333333, f(c2) = .037037037, c3 = 0.34722222, f (c,3) = 0.000195587, c4 _ .34729635,f(c4) = 0.000000015.

CHAPTER 6 EXERCISES 6.1 page 221

(a) f(x)

_

1, 0Sx< 1, Let 91= xa { x j, ,, ... , x"} be any partition of [0, 2] and let k E {l.... , n} be such

2, 1 s x <- 2. that x4 _, < I s x4. Then W(9D, f) = 4 - 3x4 and gt(9P, f) = 4 - 3x4_ ,. Thus 4k(9', f) - W(9', f) = 3(x4 - x4-,). By Theorem 6.1.7 it now follows that f is Riemann integrable on (0, 2]. Also, since x,,- I < I S x4, 4914) 5 1 < 14911f) for any partition P. Hence f f = 1. Alternatively, consider the partition 91 = {0, c, 1, 2} where 0 < c < I is arbitrary. For this partition, 2(9j) = 1 and 411.(9', f) = 4 - 3c. The result now follows as above. 3. (a) If 91 = {xo, x,, .... x"} is a partition of [a, b], then inf{ f(t) : t E [x;_ x;]} = sup{f(t) : t E [x;_,, xj]} = c. Therefore, 20,J) = %4911f) = jt c(x; - x,-,) - c(b - a). 4. (a) Since f(x) _ [3x] is monotone increasing on (0, I ], f is Riemann integrable on (0, 1]. For n 2:4 consider the partition P. _ {0, 1- t 1 }. For this partition 2(91, f) = I and 9t(9>", f) = I + I. From this it now follows that Jo [ 3x] dx = 1. 6. Let 91 = {xj,, x,, .... ..x.) be a partition of [a, b). Since f(x) <- g(x) for all x E [a, b), sup{f(t) : t E [x,_,, x,]) :s sup{g(t) : t E [xj_, x,]) for all i_= 1.... , n. As a consequence, alt(9P, f) s '9t(9', g) for all partitions 9 of (a, b). Ta ag the infimum over 9' gives f f s " j g. The result now follows from the fact that f, g r= 9t[a, b]. 8. f f - 0, fo f = 3. 10. Since a z 0, and M, = x}. Therefore. f is increasing on [a, b]. Thus if 91 = {xo, x,, ... , x"} is a partition of [a, b], "

JI

.T(9,e) = j zj-. £1

`L(',f) _

and

ice!

x Axj.

Now show that x2_ i Axj :53(x3 - x3_,) s x,2Ax4. From this it will now follow that 2(9, f) s 3(b3 - a3) s Since f is continuous, f E 9t(a, b) with f; f = 3(b3 - a3). 12. (a) With 91" =.-O.: k = 0, 1, ... , n} and f(x) = x,

VW)= " (k)l

(=;k;

k.

By Exercise la, Section 1.3, Zk_, k = 3n(n + 1). Therefore, fo x dx = him "V = 3.

15. (a) Since f is

bounded on [a, b], If(x)1 s M for all x E [a, b]. Let 9 - {xo, x...... x"} be any partition of [a, b). For each i let

Hints and Solutions to Selected Exercises

531

M(f2) and m,(f2) denote the supremum and infimum of f'- respectively over [x,_,, x,] with an analogous definition for M,(f) and m,(f). Let e > 0 be given. Then for each i there exists s t, E [x, _ x,] such that M,(f2) f2(t,) -/ `Z e. Therefore,

0:5 We) - m,V 2) <.f2(si) - f2(,,) + e 5 I As,) +f(ti) I I f(s,) - 1(ri) I + e 5 2M[M,(f) - m,(f)] + e Since this holds for any e > 0, M,(f2) - m,(f2) <_ 2M[M,(f) - m,(f)]. Now use Theorem 6.1.7. 16. Assume that f is continuous on [a, b] except perhaps at a orb, or both. Since f is bounded, there exists M > 0 such that I f(x) I <_ M for all x E [a, b]. Let e > 0 be given. Choose y,, y2, a < y, < y2 < b so that y, - a < e/8M and b - Y2 < s/8M. Then

[sup{f(t):t E [a, y,]} - inf{f(r):r E [a.y,]}](y, - a) 5 2M(y, - a) < 41 e. Similarly for the interval [y2, b]. Since f is continuous on [y,, y2] there exists a partition 91 of : y y2] such that ql(g, f) - $(9', f) < 2 E. Let 9* = 01 U (a, b). Then @1* is a partition of [a. b]. and by the above. 'tt(91*, f) 2(91*. f) < E. Thus by Theorem 6.1.7, f E Jt[a, b]. Suppose f is continuous on [a, b] except at a finite number of . < c" 5 b. Apply the above to each of the intervals [a, c, ], [Cl- c2], .. points c,, c2, ... , c with a s C1 < c2 < [c", b] to obtain a partition of [a, b] for which Theorem 6.1.7 holds.

.

.

EXERCISES 6.2 page 229 2. (a) Take t, = 3(x; + x,x,_, + x;_,). 3. (a) If c, * a, set co = a. Similarly, set c"., = b if c * b. First prove that f E gt[c,_ c,], i = 1, 2, ... , n + I with f .,f = 0. Now use Theorem 6.2.3. (b) Set h = f - g. 5. Since f is bounded, I f(x)I s M for all x E [a, b] for some M > 0. Use equations (7), (8), and the fact that f E 01 [c, b] for every c E (a, b), to prove that (b

(b

('c

O S J f- J f_ J f-

rc

f<- 2M(c - a). J

Use this to show that f E i [a, bJ. 7. (a) lim

-n3

series defining f becomes a finite sum on [0, c]. Evaluate

k2 =

Ju

x2 dx = -. 3

8. For each fixed c, 0 < c < 1, the

and use Exercise 5.

EXERCISES 6.3 page 236 1. Sincef is bounded, I f(x)I <- M for all x E [a, b]. If x, y E [a, b] with x < y, then

IF(y)-F(x)I = JH Thus F satisfies a Lipschitz condition on [a, b] and hence is uniformly continuous.

2. (b) F(x) = x for 0 5 x :s

F(x) = i - 2x for I < x 5 1.

2:

S. (b) F'(x) = cosx2 6. By the chain rule, -dO-.L(() = -1 = j[-L(x)]-Therefore, L(F) = -L(x) + C for some constant C. Taking x = I shows that C = 0. 9. Let m and M denote the minimum and maximum off on [a. b], respectively. Since g(x) z 0 for all x, mg(x) s f(x)g(x) 5 Mg(x). If fQ g > 0, then from the previous inequality one obtains

Jb1g/Jbg m a

SM.

a

Nowapply the intermediate value theorem to f. If f, g = 0, use Theorem 6.2.2 to prove that fa fg = 0 and thus the conclusion holds for any c E [a, b]. 11. (a) With rp(x) = In x. by Theorem 6.3.8 f '; dx = fo" 2 x dx = 2(In 2)2. i (b) Use integration by parts. (e) 2 In 2 - 1. Use Exercise 10 with (p (x) _ V. and f(t) = t/(1 + t).

S32

Hints and Solutions to Selected Exercises

EXERCISES 6.4 page 244 1. (a) Since 0 < p < 1, jox-adz = oJim+ f'x-vdx = lim j[ 1 - c'-'] _

,

A Note: If p s 0, then x_a is (d) Converges, with jox in x dx 2. (a) Converges, with fo e x dx = Jim fo e x dx = Jim 1 - e ` = 1. (e) f2 b dx = In (In c) - In (ln 2), which diverges to coo as c -+ oo. (i) Use partial fractions to obtain

continuous on [0. 1], and if p z 1, then the improper integral diverges.

Thus Jo

+2 _1

x

(

Jo (x2+1)(x+1)

1dx

+ l)( xz+

(

)

c+11+2Arctanc-2.

4

ar

2.

3. (a) For p > -1, the improper integral converges for all q E R. and

for p < -1, the improper integral diverges for all q E R. When p = -1. the improper integral converges for all q < -1, and diverges for all q ;2! -1. 5. Use the fact that 0 s I f(x) I - f(x) s 21 f(x) I 7 . (a) I;{ Idx 5 j;i dx = ; - Thus Jim J,,l fI < oo. (b) By integration by parts. s - j: r ZZ dx. By (a) and Exercise 5. Wn j*'X2 dx exists. Also, Win = 0.7herefore. !; dx = C-00 Jim f dx exists. 9. (a) To show convergence of the improper integral, consider the integrals of t"' a-' over

-

C-W

the two intervals (0, 1 ) and [ 1, oo). If jr ? 1, then t'-'e-'is continuous on [0, 1). and thus the integral over [0, 1 ]

clearly exists. If 0 < x < 1, then since a-' s 1 fort E (0, 1 ],

Os

f t'-'e'dts I

e

C

Thus 1 to + f'tx-'e-'dt < on. For tit 1, use l'Hospital's rule to show that there exists a t, ? 1 such that

tx''

07 for all t 2t t,. Thus tx' a-' s e'"2 for all t Z t, and as a consequence, Pm. IR tx-'' tit < on. Thus the

improper integral defining r (x) converges for all x > 0.

EXERCISES 6.5 page 259 1. 2f(0) 3. (a) See the hint to Exercise 1, Section 6.3. 5. (a) f - 1. Use Theorem 6.5.10. (C) 12 + 22 + 32. For x r= [0, 3], [x] = 1(x - 1) + 1(x - 2) + I(x - 3), now use formula (12). & (a) jx , ii2. 12. (a) As in the solution of Exercise 15a of Section 6.1, if 9 = {xo, x1, .... x"} is a partition of (a, b), Mi(f2) - m,(j2) s 2M[M,(f) - m,(f)] where M > 0 is such that I f(x)I s M for all x r= [a, b]. From this it now follows that

0 s ft (9,f2, a) - 2(9, f2, a) s 2M[olt(9, f, a) - T(9,f, a)]. Now apply Theorem 6.5.5.

EXERCISES 6.6 page 271

.

2. (a) With n = 4, h = .25. Set x, _ .25i, y, = f(x,), i = 0, 1.2.3.4. Then yo = 1.00000, y, = 0.94118, Y2 = 0.80000, y3 = 0.64000, y4 = 0.50000. Therefore,

T4(f) =

S4(f) =

25 3

Yo + 2Yi + 2Y2 + 2Y3 + Y4] = 0.782795.

[Yo + 4y, + 2y2 + 4Y3 + Y4] = 0.785393.

By computation f"(t) - 2(3t2 - 1)/(1 + t)3. Using the first derivative test, f(t) has a local maximum of at t = 1. Therefore, I f (t) I s((I for all t E [0, 1]. Thus by equation (23) with n = 4, I4_T4(f)1s12\2)42=0.0026042.

Hints and Solutions to Selected Exercises

533

Since 4 = 0.7853982 (to seven decimal places), I , - T4(f) I = 0.0026032. 3. (b) By computation, f (41(X) 3(4x2 - 1)(1 + X2)-712. By the first derivative test, the function f (") has a maximum on [0, 2] at x = v5/2. Thus ft4>(x) I s 6/ (1 -+41Y < 1. If we choose n (even) so that fo f - S (f)j < 10-3, then we will be guaranteed accuracy to four decimal places. By inequality (26) with M = 1, we need to choose n so that Ik'o -L < 10-3, or

n4 > 17,778. The value n = 12 will work. This value of n will guarantee that E12(f) < 0.0000086. Compare your answer with the exact answer of V + 2ln(2 + V5).

EXERCISES 6.7 page 276 4. No. Consider g = Xc on [0, 11 and let f be the zero function.

CHAPTER 7 EXERCISES 7.1 page 291 2. (a) Diverges (c) Converges (e) Diverges (g) Converges by the ratio test (I) Converges (m) Converges for p > 1; diverges for 0 < p s 1 3. (a) Converges to 1 /(1 - sin p) for all p E R for which sin p I < 1; that is, for all p * (2k + 1)j, k E Z 4. (b) Since 7, ak converges, lim ak = 0. Thus there exists k, E N such that 0 s ak s 1 for all k -- k,. But then 0 5 ak s ak for all k a k and 7, ak converges by the comparison test. (d) Take ak = I/k2. 5. The series diverges for all q < 1, p E R, and converges for all q > 1, p E R. If q = 1, the series diverges for L, where 0 < L < oo. Take e = IL. For this e, there p s 1 and converges for p > 1. 6. Suppose exist k, E N such that #L 5 ak/bk s }L for all k z k,. The result now follows by the comparison test. 12. The proof Vk- = 1 for all n r= Z. 13. The given series is the sum of the two series u++ses the fact that lim(LPY' + + a,,, and tk = a1 + 2a1 + Ijt and Mk t ?Jp, each of which converges. 16. let s, = a, + a2 + `k 1 , show that if n < 22, then s s tk, and if is > 2k, 2ka2-. By writing s, = ak + (a2 + a3) + (a4 - as + a6 + a7) + then s Z I'tk. From these two inequalities it now follows that F ak < oo if and only if I 2ka2, < oo. 1& (a) Diverges. If ak - 1/(k In k), then 2ka2. = 1/(k In 2). 19. Use Example 5.2.7 to show that ck - ck+ 1 a 0 for all k. Thus {ck} is monotone decreasing. Use the definition of In k and the method of proof of the integral test to show that ck =' 0 for all k. 21. Write ak+I/ak - I - xk/k where xk = (q - p)(k/(q + k + 1)). 22. (c) When p = 2,

k-fI.3...(2k- 1)12

2.4... 2k )

112 Tk r =11`1-2,J zl-2k ( 1)a

Now use the fact that M9(1 + h)I"k = e. 23. (a) Set s, = L4., ak and let s = lim s,,. Consider the series E bk

where b, = (1< - W-771 ) and fork z 2, bk = (Vs_, - V). EXERCISES 7.2 page 298 1. If {b,} is monotone increasing to b. consider $00- , (b - bk)ak. coos kt, then 1 even. 4. If D. = E[SiO(k 1 sin

r

1

2

I

2

2

2. Take bk = 1/k fork odd, and bk = i/k2 for k

\

1

sin ?tJ.

5. (a) Converges (c) Converges (d) Diverges; kim , , = * 0 (f) Converges (h) Converges for all t * 2nsr, n E Z. If t - 2nar, then the series converges for p > 1 and diverges for 0 < p s 1. & Use the partial summation formula to prove that n

n-1

n

I kak = nA, - Y, Ak, where Ak = Y, al-

k-1 k-1 Now use Exercise 14 of Section 2.2.

k-I

534

Hints and Solutions to Selected Exercises

EXERCISES 7.3 page 305 2. Use the inequality lab 1 s 2 (a2 + b2), a, b E R. 6. (a) Converges conditionally (c) Converges absolutely for p > I and conditionally for 0 < p <- 1 (e) Converges absolutely for p > 1, and conditionally for 0 < p s I S. Fast note that n-I 1 I

S. - -

3k + I + 3k + 2

-

I

3k +

3)'

Now show that Si,, -+oo as n --, oo. 10. By Theorem 7.2.6 the series converges. To show that S ' t = oo, show that for any three consecutive integers, at least one satisfies I sin k I

EXERCISES 7.4 page 312 1. (a) 11{l/(In k)}112 = Moko-2 1/(In k)2, which diverges (Exercise 5, Section 7.1) 2. (a) Ip I < 1 (c) p z 2 4. Since { 1/k} E 12, the result follows by the Cauchy-Schwarz inequality. 10. If we interpret the vectors a and b as

forming two sides of a triangle, with the third side given by b - a, then by the law of cosines, I(b - a(I2 = Ilbll2 + 11a112 - 211 a 112 11 b 112 cos 0. Now apply Exercise 9e.

CHAPTER 8 EXERCISES 8.1 page 322 nx _ 0, x = 0, 1. (a) lim 3. (c)

rf n = j

10

1, x>0

(c) lira (cos x)2n __

"-.o,

0, 1,

x * k,r, k E Z,

x=kar,kEZ

2/

iM

(2n - n2x) dx = i + = 1 S. (a) If x = 0, fn(0) = 0 for all n e N. If x > 0,

n2x d x + f

0

1/n

then 0 < fn(x) < x/n, f r o m which the result follows. (b) F o r each n E N& (x) has a maximum of e ' at x = n.

6. Use the fact that forN,MEN,

N(M

"-I

M-1

M an. m

M-1

M

N

R-t

an. m

00

M-1 R=t

00

a, m

00

(

an. m 11-1

The above inequalities hold since a0, m > 0 for all n, m E N. Now first let M - oo, and then N -b oo, to obtain an. m

an. m

n=1

mil

m=1

nil

The same argument also proves the reverse inequality.

EXERCISES 8.2 page 328 2. (b) Suppose {f,} and {g,} converge uniformly to f and g respectively on E. Then I f"(x)gn(x) - f(x)g(x)1 I g"(x) I I f"(x) - f(x) I+ I f"(x) I I g" (x) - g(x) 1. By hypothesis, I S. (x) I s N for all x E E. n E N. Also, since I f"(x) I s M for all /x E E, n E! /N, I f(x)1 fm M for all x E E. Therefore, Ifn(x)8nx) - f(x)8(x)i

N1 MX) - f(x)I + Ml 8.(x) - 8(x)I

Now use the definition of uniform convergence of { fn} and {g,} to show that given e > 0, there exists no E N such that I fn (x) g" (x) - f(x)g (x) ( < e for all x E E and n a n0. 4. Find Mn = max{ fn (x) : x E [0, 111, and show that M. -+ oo. S. (a) For x E [0, a], I f"(x)1 s a". If 0 < a < 1, then lim a" = 0. Thus given e > 0, there exists n, E N so that d` < e for all n at n,; that is I f"(x) I < e for all x e[ -w O, a], n z n,. Therefore, f f.) converges uniformly to 0 on [0, a] whenever a < 1.

Hints and Solutions to Selected Exercises

8. (a)

ii +

< co. the series 2

+ converges uniformly by the Weierstrass k' j? (c) For x ,- 1, k2e-k: s k2(1/e)'t. Since 1/e < 1. the series 7, k2(1/e)k converges.

M-test.

9. (a)

for all x E It Since .1

535

X

2

I I

(2k +))3n

(2k

1

1)3/2

5 C R for all x E R. Since 7

uniformly for all x E R by the Weierstrass M-test.

(c) Hint: Let S,(x) =

< co, the given series converges 1

ku(kz + 2

1

- (k + 1)x + 2)

10. (a) For x ? a > 0. 1 + k2x ? ak2. Thus since 1/k2 < no. by the Weierstrass M-test, the given series converges uniformly on (a, oo) for every a > 0. To show that it does not converge uniformly on (0. oo), consider M for all x E [0, 1 J. Show (S, - 5_t)(1/n2), where S. is the nth partial sum of the series. 1& Suppose I Fa(x) for all x E [0, 1], n E N. Now use the Weierstrass M-test.

that I F,(x) I < M n.

EXERCISES 8.3 page 336 1. Show that

kLx(1 - xyt = {?'

Thus by Corollary 8.3.2, the convergence cannot be uniform on

[0, 1 ]. 4. Since f is uniformly continuous on R, given e > 0, there exists a S > O such that I f(x) - f(y) I < e for all x, y E R, Ix - yI < S. Choose n, E N such that 1/n, < S. Then for all n ? n I f(x) - f,(x) I < e for all x E It 6. Let e > 0 be given. Since (f.1 converges uniformly on D. there exists n, E N such that I f,(x) - f(x) I < e for

all x E D. n, m > n,. Use continuity of the functions and the fact that D is dense in E to prove that

If(y)-fm(y)I Sefor all yCE E,n,in >n,. EXERCISES 8.4 page 338 1. By the Weierstrass M-test and the hypothesis on {ak}, the series akxk converges uniformly on [0, 1). Now apply Corollary 8.4.2. 4. Since f E R[0, I ],f is bounded on (0, 1 ], i.e., I f (x) 15 M for all x e [0, 1 ]. Now apply the bounded convergence theorem to g,(x) - x"f(x), which converges pointwise to g(x) = 0.0 5 x < 1, and f(1) when x = 1. 7. For each k E N the function 2-kl(x - rk) is Riemann integrable on [0, 1] with fa 2-k 1(X - rk)dx = 2-k(1 - rk). By the Weierstrass M-test the series converges uniformly on [0, 1). Thus f E 31.[0, I) with fo f = TOO 2-k(l - rk). 9. By Theorem 6.2.1, f g E 9t[a, b] for all n (=- N. Show that { f g} converges uniformly to fg on [a, b] and apply Theorem 8.4.1. 10. Since If(x) 15 g(x) for all x E [0, oo). n E N, the same is true for I f(x) I. By Exercise 5, Section 6.4, it now follows that the improper integrals of f n e N. and f converge. Since fog < on,

show that given e > 0, there exists c E R, c > 0, so that f,'g < Ie. Now show that

If f-Jf.1 5Jlf-fa[+2Js. to f on [0, c] to finish the proof. 11. (b) To show that (19[a, b), n I6) is not Use the uniform convergence of complete, it suffices to find a sequence If.) of continuous functions that converges in the norm II 1i to a Riemann integrable function f that is not continuous.

EXERCISES 8.5 page 345 2. By the fundamental theorem of calculus, f,(x) = f,(x,) + fx'f,(t) dt for all x r= [a, b]. If { f )converges uniformly tog on [a, b], use Theorems 6.3.4 and 8.4.1 to prove that {f} converges uniformly to a function f on [a, b) with f (x) = g(x) for all x E [a, b]. 4. Let x E [a, b] be arbitrary, and choose c, d such that a < c < x < d < b. Now apply Theorem 8.5.1 to the sequence {f,} on [c, d], to obtain that f is differentiable at x

with f (x) =

f;(x).

536

Hints and Solutions to Selected Exercises

kx)_2

6. (a) Use the comparison test to show that the given series converges for all x > 0. Let S(x) _ 7,k , (1 + and S (x) = Mk=, (1 + kx)-2. Then S.(x) = -228k= I k(1 + kx) -3. Use the Weierstrass M-test and the comparison test to show that the sequences {S (x)} and IS' (x)) converge uniformly on [a. oo) for every a > 0. Thus by Theorem 8.5.1. 00

S'(x) = limS., (x) _ -2 ;k(1 +

kx)_3

k-I

for all x E (a, oo). Since this holds for every a > 0, the result holds for all x E (0, oo).

EXERCISES 8.6 page 352 2. Let 91 _ {xo, x ... , be a partition of [a. a + p]. Set y1 = x, - a. Then 91* = {y y ... , of [0, p]. If t E [x;_,, x;], then r = s + p for some s E [y; _,, y;]. Since f is periodic of period p,

is a partition

and as a consequence,

f(t) = f(s + p) = f(s). Therefore, sup{f(t) : t E [x,x,]} = sup{f(s) : s E xlL(91, f) = q.L(9", f). From this it now follows that: J.+P a

v

f=f'f

The proof for the lower integral is similar. Thus f E 9L[0,p] if and only if f E 9t[o,a + p].

4. (a) c = }(n + 1) 6. Set A,(8) = sup{Q (x) : x E [-S, S]). Then 0 < 8, < S2 implies

A (S2).

Suppose tim A (S,) < oo for some 81 > 0. Then there exists a finite constant C. and n, E N such that A (S) s C for is an approximate identity. all n ? n 0 < S 5 81. Use this fact to obtain a contradiction to the hypothesis that

EXERCISES 8.7 page 370

1. (b) R = 2 (d) R = e 2. (b) By the root test the series converges absolutely for all x, - 2 < x < t and 1 (- l)kx2k, Ixl < 1. Use the diverges for all other x E R. 3 (a) x/(1 - x)2 S. (a) + x2 = 1 - (-x2) = 1

A-D

previous exercise and the fact that Arctan x = fo(l + t2)", dt to find the Taylor series expansion of Aretanx at c = 0. (c) Use Theorem 7.2.4. 12. (b) By Example 8.7.20(c),

(-

kk+I

(x- l)k,

lnx=ln(l +(x - 1)) = which converges for all x, 0 < x 5 2. (d) By computation, f O) (x) _ Taylor series expansion of (I - x) -'/' is given by

1+

00 1.?2 .

(k-2xk

2

k=I

(k - })(t - x)-(k' 1R). Therefore, the

kl

For -1 < x s 0, use Theorem 8.7.16 to show that

(n + 1)! Use convergence of the series 13

(n + )

IxIA.I. IxI < 1, (n + l)! to conclude that lim R0(x) = 0, - i < x 5 0. If 0 < x < 1, use Corollary 8.7.19 to show that

7

1.3...(n+ n!

fxx (1-x)341-0)

Hints and Solutions to Selected Exercises

537

for some {', 0 < C < x. Now use the method of Example 8.7.20(c) to show that lim R"(x) = 0 for all x, 0 < x < 1. Thus the series converges to (I - x)-'R for all x, I x I < I. (e) Use the fact that ' Aresin x =

1

1 - t2

1

dt, Ix I < 1.

EXERCISES 8.8 page 376

1. (a) r(2) = r(3 + 1) _ ]r(i) = }V;i 3. (a) foe-'r'ldt = r(2) = WW S. (a)

ran

la

i r(n + 21)x(21)

(sinx)z'dx = 2

r(n + 1)

CHAPTER 9 EXERCISES 9.1 page 388

1. j' 0i = j- 1 = 2 and j

022 = j x2 = ;. Therefore, c, _ If 1, sin wx dx = 0 and c2 = 2 j' 1 x sin ax dx Thus by Theorem 9.1.4. S2(x) _ ;x gives the best approximation in the mean to sinir x on [ -1, 1 ].

1, a2 = 1. a3 = -6 S. (c) b" = : fo x sin nx dx - - w cos mr = 2,(-1)"''. Therefore,

3. (a) a, f(x) ^' 2

k)k+ sin kx. k=1

6. (c) x - 2 - a E00 ( k-O

+ l)2 cos(2k + 1)x 12. (a) As in the proof of Theorem 7.4.3, for A E R,

O :s II x - kyll2 = Kxil2 - 2A(x,y) + A=IIy112. If y * 0, take k = (x,y)11IYII2 to derive the inequality.

EXERCISES 9.2 page 394 4. (a) For the orthogonal system {sin nx}.° , on [0, w], f sin2nx dx = 16. Therefore, b" = w f f(x) sin nx dx. Thus Parseval's equality for the orthogonal system (sin nx) becomes CC

M-1

fIr 0

2

2

(b) (1) g ,

S. Use Parseval's equality and the fact that fg = i((f +'g)2 - f2 - g2] 6 that is identically zero except at a finite number of points will satisfy ja f dx = 0. (11)

6. Any function

EXERCISES 9.3 page 403 1. (a) If f is even on [ -zr,vr], then f (x) sin nx is odd and 1(x) cos nx is even. Thus b = O for all n = 1, 2... , and

;jof(x)cosnxdx,n=0,1,2.... 3. (a) f(x) -(c) I xI ^-

5. (a) 1,

2

1,

(1

(-1)k) k

IT

- ,rr F, (2k + 1)2

sin kx =

a k-0 2k + l

sin(2k + 1)x

cos(2k + 1)x. (e) 1 + x -- I + 2 1

1 4 ao sin(2k + 1)x a k<0 2k + 1

k=0 00

(d)

2

1

(- k

sin kx

k

a k.o 2k + I

cos(2k + 1)x, -?

x 1 I (-1)" - cos

a"=o n LL

I2

sin nx

538

Hints and Solutions to Selected Exercises

6. (c) Since h, is even, the Fourier series of h, is the cosine series of h. Therefore.

(aR

o

a0 =

l

i

r*

x dx + I (ir - x) dx = itc, iran

(a/2

a" =

x cos nx dx +

-J

and

a

(in - x)cos nx dx =

Z l 2cos

a/2

0

c-

Thus h,(x) ^-

4

c

a k-,

Z + (- I)"- ' - I

L

cos 2kx. & By integration by parts, bk = -

k2

!

Z1

irk

a

, f'(.x)sin kx dx.

Since f" E fit[-ir, ir], it is bounded on [-ir, ir]; i.e.,I f" (x) I <- M. Therefore, I bk I S 2M/k2. Similarly for ak. Thus by the Weierstrass M-test, the Fourier series off converges uniformly on [-ir, ir).

EXERCISES 9A page 414 a

2. 90 3. To show that {sin nx}.° , is complete on [0,ir] it suffices to show that Parseval's equality holds for every f E 9t[0,ir]; i.e., 0°

f2(x)dz.

b,2,

^' 1

o

To accomplish this, let f^ denote the odd extension off to [ -ir,ir]. Since f" is odd, a^ = 0 for all n = 0, 1, 2. . is complete on [-a, ir], and b^ = A fo f(x)sin nxdx. Since the orthogonal system {I,cos nx,sin

^-1

a

a

pp

? f2(x)dx. b2, =7fI J fo(x)dx = 7r 0 a

6. Let S^(t) = lao + 1k-1(ak cos kt + bk sin kt). If x E [-ir, irthen

` f(t) dt - jx S^(t) dt s J If (1) - S^(t) I dt. 1

1

J

s

a

a

k-i\1

Thus by Exercise 5a and Theorem 9.4.7, lim f'. S^(t) dt = f', f(t) dt, with the convergence being uniform on

[-ir,ir]. But

S^(t) dt = 2ao(x + ir) + ± I JJa

from which the result follows.

k

sin kx - kl (cos kx - cos kir)

.

10. Use Lemma 9.4.8 and the Weierstrass approximation theorem.

EXERCISES 9.5 page 425 2.

12

6

3. (a) 1 (b) 1 2

2

1r

4

6. (a)

I) air

+

((_1)ke"*

IT

_

a2 + k2

I)

coskx

(b) On [-ir,a),theseries

converges to el'i, and thus to the 2ir-periodic extension of e'l on alI of R.

CHAPTER 10 EXERCISES 10.2 page 443 2. Since U is open and nonempty, there exists x E U and r > 0 such that (x - r, x + r) C U. Thus by Theorem 10.2.4, m(U) z m((x - r, x + r)) = 2r > 0. 3. First show that m(P^) = 2m(P^_1) for all n E N. From this it now follows that m(P) s (3)' for all n E N.

Hints and Solutions to Selected Exercises

539

6. First show that there exist disjoint bounded open sets U1, U2 with U, 1) K, and U2 J K2. Then

m(K,UK2)=m(U,U(.2)-m((U,UU2)\(K,UK2)).But(U,UU2)\(K,UK2)=(U,\K,)U(U2\K2).Now use Theorem 10.2.9.

EXERCISES 10.3 page 448 1. (b) First show that if U is any open set, then U + x is open and m(U + x) = m(U). Use this and the definition to prove that A*(E + x) = A*(E). If K is compact and U is a bounded open set containing K. show that (U + x) \ (K + x) = (U \ K) + x. Use this to show that m(K + x) = m(K) and A*(E + x) = A*(E). 3. Since E, fl E2 C E, and A*(E,) = 0, A*(E, fl E2) = 0. Thus by Theorem 10.3.5, E, fl E2 is measurable. For E, U E2 apply Theorem 10.3.9. 6. If A*(E) < oo, then for each k E N there exists an open set Uk with UA D E such that m(Uk) < A*(E) + k. Now use the fact that E C fl U" C Uk for all k E N. S. Set Ek = E fl (-k, k), k E N. Then {A*(Ek)} is monotone increasing with A*(Ek) <- A*(E) for all k E N. Let a = 51im A*(Ek). Suppose a < A*(E). Choose jS E R such that a < P < A*(E). By definition there exists a compact set K with K C E such that m(K) > )3. Use this to show that there exists k G N such that A*(Et) > ,B for all k ? k, which is a contradiction-

EXERCISES 10A page 455 2. If E is bounded, the result follows from the definition of A*(E) and A*(E), and Theorem 10.4.5(b) (for a finite

union). If E is unbounded, let E. = En(-n. n). Given e > 0, choose U. open such that E C U. and A(U" \ E) < e/2". Let U = U U,;. Show that U \ E C U (U" \ E"). Now use Theorem 10.4.5 to show that A(U \ E) < e. To obtain a closed set F C E satisfying A(E \ F) < e, apply the result for open sets to E. 4. First show that A(E, U E2) = 1; then use Theorem 10.4.1. 6. If E satisfies A*(E fl T) + A*(E` fl T) = A*(T) for every T C R, then E satisfies Theorem 10.4.2 and thus is measurable. Conversely, suppose E is measurable and T C R. If A*(T) - oo, the result is true. Assume A*(T) < oo. Let e > 0 be arbitrary. Then there exists an open set U D T such that A(U) < A*(T) + e. Since E and U are measurable, E fl U and E` n U are disjoint measurable sets with (E fl U) U (E` fl U) =U. Furthermore, E fl U D E fl T and E` fl U D E` fl T.Thus by Theorem 10.3.9,

A*(T) s A*(E fl T) + A*(E` fl T) S A(E fl U) + A(E` fl u) = A(U) < A*(T) + e. Since the above holds for every e > 0, we have A*(E fl 7) + A*(E` fl T) = A*(T).

EXERCISES 10.5 page 461

1. {x : f(x) > c} a

[0,1],

if c < 0.

(0, 1 ],

(0, 1) U {1},

if OS c< 1, if l s c < 2.

(0, 1).

if 2 s c.

> c} = {x : g(x) > 0} fl ix: g(x) < 1}. Since g is measurable, each of the sets {g(x) > 01 5. If c > 0, then {x : and {g(x) < <} is measurable. Thus their intersection is measurable. The case c < 0 is treated similarly. I0. (a) If

c z 0, then {x : f'(x) > c} _ {x : f(x) > c}, and if c < 0. then {x : f'(x) > c} = E. Since each of the sets {f(x) > c} and E are measurable, f' is measurable. (c) Not in general. If E is a nonmeasurable set, consider the function that is 1 on E and -1 on E`. 14. Since f is differentiable on [a, b], f'(x) - lim n(f(x + 1) - f(x) for all x E [a, b]. For each n E N, g"(x) = n[f(x + ,'-,) - f(x)] is measurable (justify). Thus by Corollary 10.5.10. the function f is measurable. 15. First show that given e, d > 0, there exists a measurable set E C [a. b] and n, E N such that A([a, b] E) < e and I f"(x) - &) I < S for all x E E and n ? no. To accomplish this, for each k E N consider Ak = {x : I f"(x) - f(x) I < S

for all n z k}.

Now show that lim A(Ak) = 0. Here Ak = (a, b] \Ak. Complete the proof of Egorov's theorem as follows: By the

above, for each k E N, there exists a measurable set Ek and integer nk such that A(Ek) < e/2k and I f(x) - f,(x) I < for all x E Ek and n z nk. The set E= fl Ek will have the desired properties.

540

Hints and Solutions to Selected Exercises

EXERCISES 10.6 page 472 1. For each n E N, let rp, = (m + (j -1)@ )XE. Then gyp, is a simple function on [a, b] with f. q,. dA = S,(f). Furthermore, for each x E [a, b], 0 s f(x) - (p,(x) s $/n. Therefore, liimrp,(x) = f(x) for all x E [a, b]. Now apply the bounded convergence theorem. S. By Theorem 10.6.10(b), fF f fE f dA + fE. f dA >- fE f dA. 7. For each n E N, let E. = {x : f(x) > 11. Then U E, = {x : f(x) > 0). Use the previous exercise to show A(E) = 0. Now use Theorem 10.4.5. 12. The function rp, defined in the solution to Exercise 1 satisfies I f(x) - (p,(x) I < $/n for all x E [a, b]. Thus .{9.1 converges uniformly to f on (a, b]. 15. Suppose first that rp = XA where A is a measurable subset of [a, b]. By Exercise 2, Section 10.4, there exists an open set U ) A such that A(U \ A) < e/2. Use the set U to show that there exists a finite number of disjoint closed intervals {J,}.N=, such that V = U J. C U and A(U \ V) < e/2. Let h = -X ,. j.. Then his a step function on [a, b] and {x : h(x) * rp(x)} C (U 1 V) U (U \ A). If rp = JJ.,a;XAJ, where the Al are disjoint measurable subsets of [a, b), approximate each XA by a step function hi that dAp=

agrees with XA, except on a set of measure less than a/n.

EXERCISES 10.7 page 482 2. (a) Assume first that A, and A2 are bounded measurable sets. For each n E N. set f, = min{f, n}. By Theorem 10.6.10(b),

r

(

f,dA= J f,dA+

1A,UA,

f,dA.

1A,

A,

Since each of the sequences { fA f,} ,, i = 1, 2 are monotone increasing, they converge either to a finite number, or diverge to oo. In either case,

/ = +4-+O j of dA = i(

J

JAUA

f,dA + J f,dA) A,

A,

JfdA

j.fdA. A

A.

If either A, or A2 is unbounded, consider the integral off over (A, U A2) n[ - n, n], and use the above. 4. For vv

f(x) = x -P, x E (0,1), f,(x) = min{f(x), n} =

x'v

n- o <_ x < I Therefore,

I

ff,dA= 0

Jn 1pndx+ o

n,,

x"Pdx n

I

.4

I

1

I

p

In<

,1vvvJ

I

` pLI

J.

rs<1PPYP

Since (1 - p) > 0, lire f. f. dA = 1/(l -p).

7.05

nA(E,)5

JfdA..JfdAIbfdA
o

10. Let A. = An[ -n, n). By definition, fA f dA = lim fA. f dA. Since f is integrable, given e > 0, there exists is E N such that O :C- fA fdA - fA, fdA < e. By Theorem 10.7.4(b), fA f dA - fA f dA = fA%A. f dA. Thus E = A. is the desired set. 13. Let f, = min{ I f I. n}. Then 0 < f, s f I. Since f is integrable, by Lebesgue's dominated convergence theorem,

jIfIdA.

lim f, dA 1A

A

Therefore, given e > 0, there exists n E N such that fA(I f I - f,)dA < e/2. In particular, if E is any measurable subset of A.

JIf I dA < jfdA+. E

Hints and Solutions to Selected Exercises

541

But if A(E) < oo, fEf dA : nA(E). Hence choose S > 0 so that nS < e/2. 19. Since {f"} is monotone increasing on A, f(x) = lim f"(x) exists, either as a finite number or as oc, for every x E A. By Fatou's lemma, fA f dA s lim fA f dA. On the other hand, since f s f for all n E N. lim fAf" dA t5 fA f dA. Combining the two inequalities proves the result.

21. (a) Use the monotone convergence theorem. 23. Hint: I f(x)I ? (I/x)Icot I/r=1 - 2x z x ' - 2x on each of the intervals ((2n + 3)1r)-? s x <- ((2n - 3)rr)-'.

EXERCISES 10.8 page 490

1. 0 < p < i Of +gIL=

4. Since If + g 12 s 2(1 f 12 + I g I').f + g E 22(A). Assume II f + g 112 ;E 0. Then

jii+gi2:5 jtf+llfi+jIf+IIi. A

A

which by the Cauchy-Schwarz inequality <- 11 f+ g Jz lI f lL + 11f+ g 16IIg II,. The result follows upon simplification. 7. Use the Cauchy-Schwarz inequality. 12. Let f(x) = I/ln x, x E (1, oo). Since f"(x) > 0 for all x E (1, ac),

f is convex on (1, oo) (see Miscellaneous Exercise 3, Chapter 5). Therefore f(lx + ,'-y) 5 1/(x) + ;-f(y) for all x, y E (1, oo). Since n = 1(n - 1) + 11(n + I) the inequality follows.

Notation Index

closed interval, 25 equivalent sets, 35 indexed family of sets. 38 sequence of sets. 38

CONNECTIVES AND QUANTIFIERS

A V

V

conjunction, 496 disjunction, 496 negation, 496 conditional or implication, 497 biconditional. 497 for all, 516

3

there exists, 516

SETS AND SET OPERATIONS N

positive integers, 2

N.

[I,... ,n),35

Z

R

integers, 2 rational numbers, 3 real numbers, 3

46

n-dimensional euclidean space, 309 empty set, 2

0

9(A). AL

E

0 AUB A f1 B

ACB

AB AFB A`

(a, b)

AXB inf A sup A

(a, b) (a, bl, [a, b)

all subsets of A, 4 sigma algebra of measurable sets, 453 element of, 2 not an element of, 2 union. 3 intersection, 3 subset, 3 proper subset, 3 relative complement, 3 complement, 3 ordered pair, 5 cartesian product, 5 infimum of A. 22 supremum of A, 22 open interval. 25 half open intervals, 25

U.

E.- UE. =I

union. 38

m

(l e, flEe

intersection. 38

N.(P)

s-neighborhood. 49

Int(E) E' E

interior of E, 94 set of limit points of E. 97 closure of E. 97

FUNCTIONS f: A --i B

Domf Range f f (E)

f -'(H)

Pg

function from A to B. 7 domain of a function. 7 range of a function. 7 image of a set. 9 inverse image of a set. 9

f'

composition of functions, 13 inverse function, 12

fg

sum or difference of functions. 133 quotient of functions. 134

fig

product of functions, 133

ftg f*, f lim f (x)

li " f (x) lim f (x). lim f (X) (0 (f. p)

f'

positive. negative parts of function. 461 limit of a function, 116 limit at infinity. 126 right, left limit at p. 148

limit superior, inferior. 162

oscillation off 162

f+.fl

derivative off 167 right, left derivative. 168

ft^1

nth derivative. 168

Notation Index

544

B(.r. )-)

Xe

D. E(r), e`. exp.t f(.r) [.r} 1(x) K.

Beta function. 376 characteristic function. 273, 436 Dirichlet kernel. 405

exponential function. 238 Gamma function. 372

Y(f)

greatest integer function. 151 unit jump function, 154 Fej6r kernel, 406 Laplace transform. 277

!_(x), In x c)(x). R.(-Y)

natural logarithm function. 234 remainder or error function. 362

Simpson's approximation. 268 measure of an interval. 432 measure of an open set, 433 measure of a compact set. 440 inner measure of E. 444 outer measure of F. 444 Lebesgue measure of E. 446 Lebesgue integral. 463

nt(J )

,n(U) nt (K)

A,(E) A(£)

f, fdA.

SEQUENCES AND SERIES sequence. 37

lim a FUNCTION SPACES `G [a, h]

C`(!) di[a, b) !R(a) 12

[(at)

2

continuous functions on [it. b], 333 infinitely differentiable functions on !, 358 Riemann integrable functions. 212 Riemann-Stieltjes integrable functions. 251

lim

square summable sequences, 306 l2 norm, 306 norm. 310 euclidean norm, 309 inner product, 309

1-1

II f II

uniform norm. 334

Ii f I)

2' norm. 492 inner product in JtIa. bl, 381 Lebesgue integrable functions. 478 square integrable functions. 484 f2 norm. 484

..,f g. E(A). Y'(A) 12(A) II f II2

MEASURE AND INTEGRATION 91

IIphI

f(p.f)

partition of (a, 6), 208 norm of a partition, 227 lower, upper sum. 209 Riemann sum, 226

f°ff,f

lower, upper integral. 210

fo1 f,f(x)dx

Riemann integral, 211

fa f(x)dv

improper Riemann integral. 241

'f(3,f,a),

lower, upper Riemann-Stieltjes sums. 246

3[(.9.f. a) f, "f da. f,, f da

lower, upper. Riemann-Stieltjes integral, 247

f; f da

Riemann-Stieltjes integral, 248 midpoint approximation. 261 trapezoidal approximation. 266

T,(f)

limit of a sequence. 50

a

convergence of I lira it,,

flat c =.

to a. 50 limit superior, inferior. 74 infinite product. 90

a,

nth partial sum. 87

tr,

I al

series of real numbers, 87 sequence of functions. 318

series of functions, 319 norm convergence. 334

GREEK ALPHABET alpha beta chi delta epsilon eta gamma iota

kappa

lambda mu no omega phi Pi

psi sigma tau theta upsilon xi zeta

Index

A

Biconditional, 497

Abel, Niels, 294 Abel's partial summation formula. 294 test, 298 theorem. 355 Absolute convergence, of series. 299

Absolute maximum (minimum), 176 Absolute value, 45, 48 Absolutely integrable, 243 Algebraic number, 43 Algebraic properties of Q, R, 20 Alternating series test, 295 Almost everywhere. 458 Antecedent, 497 Antiderivative, 230 Approaches (tends to) infinity, 65, 190 Approximate identity, 347 Approximation in the mean. 383

sentence, 497

Binary expansion, 31 Binomial theorem, 56. 366 series, 374 Bisection, method of, 198

Archimedes. 43, 277, 313 Archimedean property. 28 Arithmetic-geometric mean

inequality, 20

Bolzano-Weierstrass property, 104

Bolzano-Weierstrass theorem:

Change of variable theorem. 236. 237

Bolzano. Bernhard. 47.71. 105, 130

for infinite sets, 71 for sequences, 71

Borel, Emile, 105 Bound: lower, 21 upper, 21

Bounded above (below), 21 function, 124 sequence, 51 set, 21

Bounded convergence theorem, 338,470

C Cantor, Georg. 1, 24, 35, 43, 90,

99,428

Associative, 20,310, 502 Axiom of Choice, 493

Cantor ternary set, 108 function, 163 Caratheodory, Constantin. 429, 454

B

Cardinality. 35.44

Barrow. Issac, 277 Bernoulli. Jakob. 204 Bernoulli. Johann, 190, 204, 313

Cartesian product, 5 Cauchy, Augustin-Louis. 47, 88, 115. 130, 166, 204, 207, 279, 313 Cauchy condensation test, 293 convergence criterion, 88. 325 mean value theorem. 181

Bernoulli's inequality, 17 Bessel's inequality, 388, 396, 488 Beta function, 376

384

for integrals, 238,485 for series, 307 Chain rule. 173

Approximation using Riemann sums, 260

product, 314 sequence. 80.91. 334 Cauchy form of the remainder. 366 Cauchy-Schwarz inequality. 112, 307.

Characteristic function. 273.436 Closed interval. 25

in, 99 set, 94 Closure of a set. 97 Coefficients: of Fourier series. 387.395.488 of power series. 353 Collection. 2

Commutative. 20. 310.502 Compactness, preservation of, 136 Compact set. 102, 111 measure of. 440 Comparison test: for improper integrals. 244 for series. 282 Complement of set. 3

Complete nonmed linear space. 334 Complete orthogonal sequence. 392 Completeness property of R, 24.81 Complex number, 45

absolute value of. 45 conjugate of. 45 imaginary part of. 45 real part of, 45

545

546

Index

Composition of functions, 13 continuity of, 134 differentiability of. 173

integrability of. 217 Conclusion, 497 Conditional, 497

0

Equivalent functions. 484

D'Alembert, Jean, 287, 313 Darboux, Jean Gaston. 208 Decimal expansion, 41 Decreasing function, 153

propositions. 498 sequences. 91 sets, 35

Error function, 362 Error in numerical approximation. 263,

sequence, 60

convergence, 299

Dedekind, Richard. 2. 24

sentence, 497

De Morgan's laws, 4, 39, 502

282

Conjunction. 496 Connected set. 99 preservation of, 140, 143 Consequent. 497 Constant sequence, 50

Dense set. 29, 98

Contained in, 3

Derivative of sums, products, and quotients. 171 Differentiable function. 167

Continuous: almost everywhere, 459

Denumerable, set, 36

space, 111. 309

Derivative, 167

Euler, Leonhard. 63. 130. 204 Euler's constant, 293

left, right, 168 second, 168

at a point. 130, 314

Differentiation of Fourier series, 423

inverse theorem, 158 on a set, 130

of power series, 357 Dini's theorem, 332 Dirac sequence. 347

uniformly, 144 Continuous function, 130

integrability of, 216, 250 Continuum hypothesis. 43 Contractive function, 148, 335 sequence,84

Contradiction, 499, 502 proof by , 15. 510.511 Contrapositive. 498, 502

conditional . 299 of Fourier series, 415, 417, 420

of improper integrals, 239. 241 in the mean, 390,404, 411, 486

in measure. 493 in norm. 311. 334. 390

of series. 87, 280 of series of functions. 319 uniform, 323 Converse, 498 Convex functions, 205 Cosine coefficient, 401 series. 401

Countable set, 36

329

jump, 150 removable, 149

of second kind, 150 simple, 150 Disjoint sets. 3

Disjunction, 496 Distance, 49, 111, 311 from a point to a set. 143

Distributive, 20 law(s), 4, 39, 502 Divergence:

pointwise. 318 radius of. 353 of sequence. 50

of sequence of functions, 318

,

Dirichlet's theorem , 417 Discontinuities: of first kind, 150

proof , 509

of improper integrals, 239, 241

of sequence, 50 of series. 87, 280

Diverges to infinity, 65. 280 Domain, of function, 7

Fatou's lemma, 477 FejEr , L.. 410, 424

Fej r kernel . 407 FejEr's, theorem . 410 Field, 20 Fermat . Pierre de, 277 Finite expansion , 33 set, 35 First derivative test , 183 First order method, 264 Fixed point, 148

Formula (in logic), 515 Fourier, Joseph. 379.426

Fourier coefficients, 387, 395.488 Fourier cosine series, 401 sine series, 401 Fourier series, 387, 395

differentiation of. 423 integration of, 415 pointwise convergence of. 415.417. 420

Fourth order method. 268 Function(s), 7 Beta, 376

E

bounded,124

e-neighborhood, 49, 311

Cantor ternary, 163 characteristic. 273, 436 composition of. 13 continuous, 130. 314 contractive, 148. 335

Egorov's theorem, 462

Element of a set. 2 Empty set, 2

Counterexample, 513

Enumerable set, 36 Enumeration of a set, 37 Equal almost everywhere, 458

Cover, open, 102

Equivalence class, 91

Countability of Q, 41

function, 229. 397 Existential quantifier, 516 Exponential function, 186, 238, 369

Factorial, 18, 56 Family, 2

Dirichlet, Peter Lejeune, 294. 380.415 Dirichlet kernel , 405 Dirichlet test . 295 for uniform convergence

number(e). 63 Even extension, 402

F

Direct proof, 507

Convergence:

absolute, 299 almost everywhere , 459

Euclidean distance, 49, 309 length (norm), 112, 309

convex, 205 decreasing, 153

derivative of, 167

Index

difference of. 133 differentiable, 167

domain of, 7 even. 229. 397 exponential. 186. 238, 369 Gamma. 245. 372 graph of. 6 greatest integer, 151 identity, 9 increasing, 152

infinitely differentiable, 358 integrable. 211. 248. 474, 478 inverse. 12

jump of. 161

limit inferior of, 162 limit of. 116 limit superior of. 162 linear. 314

Lipschitz. 145 measurable, 456 monotone. 153 natural logarithm, 234 nondecreasing (increasing), 152 odd, 229, 397 one-to-one, I I onto, 7 orthogonal, 381 orthonormal. 383 oscillation of, 162 periodic, 148, 346 periodic extension of, 396

piecewise continuous, 419 polynomial, 124 product of. 133 projection. 9 quotient of, 134 range of. 7 rational. 135 real analytic. 355 real-valued. 7 sequence of, 318 series of, 319 simple. 464

square integrable, 484 step. 277. 413 strictly decreasing (increasing), 153

sum of. 133 uniformly continuous, 144 unit jump. 154 value of, 7 Fundamental theorem of calculus, 230, 232 for Lebesgue integral. 473

Q

Integrable:

Gamma function, 245, 372 Geometric series, 87 Gibbs phenomenon, 426 Graph, of function. 6 Greatest integer function, 151 Greatest lower bound, 22 property of R. 24 Gregory, James, 318

absolutely. 243 Lebesgue. 474. 478 Riemann, 211 Riemann-Stieltjes, 249 Integral: improper, 239. 241 Lebesgue. 463, 474.478 lower (upper). 210. 247

547

Riemann, 212

H Half-closed (open) interval, 25 Harmonic series, 285 Heine. Eduard, 105, 428 Heine-Borel-Bolzano-Weierstrass theorem, 106 Heine-Bore) theorem, 105 Hypothesis, 497

Riemann-Stieltjes. 248 Integral form of the remainder. 365 Integral test, for series, 284 Integration by parts. 235. 253

Interchange theorem: relating to continuity. 331 relating to differentiation. 340 relating to integration, 337, 338.470.

477,481.483 1

Identity function. 9 Image (of a set), 9 Improper integrals, 239, 241 convergent, 239, 241 divergent. 239. 241 Increasing function, 152 sequence, 60 Index set, 38 Indexed family, 38 Indirect proof, 510 Induction, mathematical, 15 Inequalities: arithmetic-geometric mean, 20 Bernoulli, 17 Bessel, 388, 396.488 Cauchy-Schwarz, 112,238. 307, 384, 485 Minkowski. 308, 485 triangle, 48, 111, 308, 310 Infimum, 22

property of R. 24 Infinite:

relating to limits. 330 Interior. 94 point, 94 of a set, 94 Intermediate value theorem, 138 for derivatives, 184 Intersection of sets. 3. 38 Interval(s), 25, 26 closed, 25 half-closed (open), 25 infinite, 25 measure of, 432 nested. 61 open, 25 Inverse function. 12

continuity of, 158 differentiability of. 185 Inverse function theorem. 185 Inverse image (of set), 9 Irrational number, 3 Isolated point. 69

J

expansion, 33 interval, 25

Jump discontinuity. 150

limit, 65, 190

L

product, 90

Lagrange, Joseph, 179, 203, 313, 363 Lagrange. form of the remainder, 363 Laplace transform. 277 Largest element. 22 Least square approximation, 384 Least upper bound. 22

series, 86 set, 35

Infinitely differentiable, 358 Inner measure, 444 Inner product, I11. 309, 380 Integers, 2 Integrability theorems. 216. 220. 249. 250, 274, 462

property of R. 24 Lebesgue. Henri, 106, 219. 276. 426. 429, 492

548

Index

Lebesgue integral, 463. 474, 478 of nonnegative function. 474 properties of. 469.476, 479 Lebesgue measure. 446 sums. 472 Lebesgue's dominated convergence theorem. 481 Lebesguc's theorem, 220, 274 Left continuous, 149 derivative. 168

limit, 148 Legendre polynomials, 389 Leibniz, Gottfried. 115, 165. 207. 276 Leibniz's rule. 175

:Hospital, Marquis de. 190 L' l lospital's rule. 191, 194 Limit: of a function. 116 inferior, 74, 162 infinite. 65. 190

at infinity. 126, 190 left (right). 148 of a sequence, 50 of a sequence of functions, 318 subsequential, 67 superior. 74. 162

Limit comparison test, 282 Limit of sum, product. quotient: of functions. 123 of sequences, 53 Limit point, 69, 97 Linear approximation. 361 Lipschitz condition. 145 function, 145 Local maximum (minimum), 176 Logarithm, natural, 234 Logically valid, 501 Lower bound, 21 integral. 210, 247 sum, 209, 246,462

M Maclaurin. Colin. 318 Maclaurin series, 361 Mapping. See Function Mathematical induction, 15, 17 Maximum: absolute, 176 local, 176

inner, 444 of interval. 432 of measurable set. 446 of open set. 433 outer. 444 zero. 219

Measurable function, 456 partition, 462 set. 446

Mengoli. Pietro, 313 Mercator. Nicolaus. 317 Mesh, of partition, 227 Metric. 111. 311 space, I I I Midpoint approximation, 261 Minimum: absolute. 176 local, 176

Minkowski's inequality, 308, 485 Modified principle of mathematical induction, 17 Monotone convergence theorem, 483 Monotone function. 152 integrability of, 216. 250

Monotone sequence. 60 convergence of. 61

N nth derivative, 168 nth partial sum, 87. 280, 319 nth root, 29

nth term. 37, 50. 87 Natural exponential function. 186, 238 logarithm function, 234 Natural numbers, 2 Negation, 496 of quantified sentence. 518 Neighborhood. 49. 311 Nested intervals property. 61

Newton, lssac, 115. 165. 197, 207, 276. 313. 318 Newton's method. 197 Nondecreasing (increasing): function, 152 sequence.60

Nonmeasurdble set. 493 Nonnegative integers. 3 Nonterminating expansion, 33 Norm. 310

Maximum element, 22

of integrable function. 492

Mean-square convergence. 390 Mean value theorem, 179, 181

of square integrable function, 484 of square summable sequence, 306 uniform. 334

for integrals, 235.252 Measure: of compact set, 440

of vector, 112, 309

Norm convergence. 311, 334. 390

Norm of partition. 227 Normed linear space. 310 Nowhere differentiable function. 342 Null sequence. 91

Number(s): algebraic, 43 complex, 45 irrational. 3 natural. 2 rational. 3 real. 3

0 Odd extension. 402

function. 229. 397 One-to-one function, II Onto function. 7 Open: cover. 102 in, 99

interval, 25 Open set. 94, 111 measure of. 433

Order properties of S2, 21 Ordered field, 21 Ordered pair, 5 equality of. 5 Oresme. Nicole. 88, 313 Orthogonal. 380 functions. 381 Orthonormal functions. 383 Oscillation of a function. 162 Outer measure. 444

P p-series. 285 Pairwise disjoint, 98, 433 Parseval's equality. 392, 412. 491 Partial sum. 87, 280. 319 Partition. 208 measurable. 462 norm (mesh) of. 227 refinement of, 210, 464 Periodic extension, 396 function, 148. 346 Piecewise continuous function, 419

Point: interior, 94 isolated. 69

limit, 69 Pointwise convergence. 318. 319 of Fourier series. 415. 417. 420

Polynomial. 124 degree of, 124 Positive integers, 2 rational numbers. 21

Index

real numbers. 20 sequence, 92 Power series, 353 differentiation of, 357 integration of, 371

radius of convergence of. 353 uniform convergence of. 354 uniqueness of. 359 Power set, 4 Preservation of compactness, 136 of connectedness, 140, 143 Principle of mathematical induction, 15 Projection function. 9 Proof, 500 by cases. 512 by contradiction, 15, 510, 511 by contraposition, 509 Proper subset, 3

Proposition, 496 compound, 496 negation of. 496 simple, 496 Propositional formula, 498 equivalent, 498

Removable discontinuity, 149 Riemann, Georg Bernhard, 208, 276 Riemann's criterion for integrability. 216 Riemann integral, 211 properties of. 223 sum, 226 Riemann-Lebesgue lemma. 398 Riemann-Stieltjes integral, 248 properties of, 250 sum, 258

Right continuous, 149 derivative, 168 limit. 148 Rolle, Michel, 177 Rolle's theorem. 178 Root test, 288, 300 Rules of inference: biconditional, 504 by cases, 504 conjunction, 502 conjunctive simplification. 503 contrapositive inference. 502 disjunction, 502 disjunctive syllogism, 504 detachment, 502 excluded middle, 502 hypothetical syllogism, 503

Q Quadratic method, 202 Quantified sentence, 516

negation of, 518 Quantifiers: existential. 516 universal, 516

R

modus ponens. 502 modus tollens. 502 simple replacement. 503 transitive inference, 503

$

Range, of function, 7 Raphson. Joseph. 197 Ratio test, 287, 300

Secant line, 166 Second derivative, 168 test, 188 Second order method, 202. 264 Second principle of mathematical induction, 17

Rational number, 3

Sequence(s), 37.50

Raabe's test, 293 Radius of convergence, 353

function, 135

Real analytic function, 355 Real numbers, 3 order properties of, 21 Real-valued function, 7 Rearrangement of series, 301 Rearrangement theorem, 302 Recursive, 18 Refinement, of partition, 210,464 Relative complement, 3 Relatively closed (open), 99 Remainder in Taylor's theorem, 362 Cauchy form, 366 integral form, 365 Lagrange form, 363

bounded, S I

Cauchy, 80. 91, 334 constant. 50 contractive. 84 convergent. 50 decreasing. 60 Dirac, 347 divergent. 50 divergent to infinity. 65 equivalent. 91 increasing. 60 limit inferior of, 74 limit of, 50 limit superior of, 74 monotone, 60

nth term of. 37. 50 nondecreasing (increasing). 60 null. 91 positive, 92 of sets, 38 square summable, 306 subsequence of. 67 Sequence of functions. 318

limit of, 318 uniform convergence of. 323 Sequential criterion for limits, 121 Series. 86. 280

absolutely convergent. 299 alternating. 295 conditionally convergent. 299 convergent. 87. 280 divergent. 87. 280 Fourier. 387, 395 of functions, 319 geometric. 87 harmonic. 285 Maclaurin. 361 nth partial sum. 87. 280 nth term. 87 power, 353 rearrangements of. 301 sum of. 87, 280 Taylor, 360

trigonometric, 296. 395 uniform convergence of, 323 Set(s). 2 at most countable. 36 bounded, 21 Cantor ternary. 108 Cartesian product of, 5 closed. 94 closed in. 99 closure of, 97 compact. 102. 1 1 1

complement of. 3 connected, 99 contained in. 3 countable. 35 dense. 98 denumerable. 36

disjoint. 3 element of. 2 empty. 2 enumerable, 36

enumeration of. 37 equality of, 3 equivalent, 35 finite. 35 image of, 9 indexed family of. 38

549

550

Index

infinite. 35 interior of. 94 interior point of. 94 intersection of. 3, 38 inverse image of, 9 isolated point of. 69 limit point of, 69. 97 measurable, 446 nonmcasurable. 493 open. 94. 111 open in. 99

pairwise disjoint, 98.433 power. 4 relative complement of. 3

sequence of, 38 uncountable, 36 union of. 3. 38

lower. 209. 246. 462 partial, 87. 280.319

Riemann. 226 Riemann-Stieli jes. 258 of a series, 87, 280. 319 upper, 209. 246.462 Supremum. 22 property of R. 24

Tangent line. 166 Tautology. 499 Taylor. Brooks. 318 Taylor polynomial. 360 series, 360 Taylor's theorem. 362 Terminating expansion. 33

Ternary expansion .

Simple discontinuity. 150 Simple function , 464 canonical representation of. 464 integral of, 465 Simpson, Thomas. 276 Simpson's approximation, 268 rule, 267 Sine coefficient. 401

function. 163 Tests for convergence of series: alternating series. 295 Cauchy condensation. 293 Cauchy criterion. 88. 325 comparison. 282 Dirichlet, 295. 329 integral. 284 limit comparison. 282

series. 401

Strictly decreasing (increasing): function. 153 sequence. 60

Subsequence. 67

Subsequential limit. 67

Subset. 3 Sum:

Lebesgue. 472

Uncountable set. 36 Uncountability of R. 41 Uniform convergence. 323 of power series. 354 of trigonometric series. 399.410 Uniform continuity theorem. 146 Uniform norm. 334

Uniformly bounded. 329

T

Sigma algebra. 454

Square integrable function. 484 summable sequence, 306 Square root. 13 Step function . 277 . 413 Stie)tjes . Thomas-Jean . 208

U

31

Raabe, 293

ratio, root, 287. 300 of trigonometric series. 296 Weierstrass M-test . 327 Tests for local maximum or minimum: first derivative, 183 second derivative , 188 Trapezodial approximation . 266 rule, 265 Triangle inequality. 48, 111. Mg, 310 Trigonometric polynomial. 411 series. 296, 395 Truth set. 516

continuous, 144 differentiable. 188 Union of sets, 3.38 Uniqueness theorem for power series. 359

Unit jump function. 154 Universal quantilier. 516 Universe. 515 Upper bound. 21

integral. 210. 247 sum. 209. 246. 462

V Value, of function. 7 Vector(s). 112. 309 norm (of), 112. 309 orthogonal, 380 Vector space. 310

W Weierstrass approximation theorem. 346

M-test, 327 Weierstrass , Karl . 71 . 93. 105 . 171 , 317 . 341 . 377

Well-ordering principle .

Z Zero measure. 219

15