Mathematical Analysis- m
'MEASURE THEORY Space, Hilbert Space, Spectral Theory) [For Honours and Post-Graduate Students]
(Including Banach
By
KESHA WA PRASAD GUPTA B. Sc. HO;IS., M.Sc. Department of IIfathematic!>, P.P.N. College, Kanpur.
KRISHN A Prakashan Mandir 11.
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Meerut-250001 (U. P.) INDIA.
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MEASURE THEORY Thirteenth E dition 1994-95 All rights reserved by the author.
Price Rs. 55'00 only. Published by KRISHNA Prakashan Mandir, 11, Shivaji Road, Meerut-l (U.P.) India and Printed at Fine Art Press, Meerut.
PREFACE
How to define the measure of a set in a pure Mathematics is a modern concept and it is needless to state that there is hardly a book by any Indian author on this topic. of writing such a book under the
Hence the necessity
caption,
'The Theory of
Lebesgue Measure and Integration' was taken because of the huge and unbearable cost of foreign books on the subject. The subject being very modern in
nature, all needed defini
tions have been given at the beginning of the chapters, and though apparently the volume looks
very thin but it
covers the entire
course as laid down in various Indian Universities. It is needless to emphasise that the style adopted in the book is lucid, cleat, easy and clearly understandable to
the students. A
good many solved and unsolved examples have been given in every chapter so that students may have enough practice in the subject. I am grateful to Dr. (Mrs.) P. Srivastava, matics,
B. H. U.,
Dr. S. N. Lal
and
Reader in Mathe
Sri Ram Bechan Ram
Lecturers in Maths., B. H. U., for their valuable guidance. also thankful to my friends and well
I am
wishers Dr. M. P. Jaiswal,
Prof. R. G. Gupta, S. P. S. Bhadauria for their valuable sugges tions.
Thanks are also due to the printers and publishers
who
took pains in bringing out this work. This book may have some defects and
I shall thankfully
receive constructive suggestions for its improvement from all those interested in studying the subject. Kanpur
K. P. GUPTA PREFACE TO THE THIRTEENTH EDITION
The present edition has been thoroughly revised and enlarged. This edition solves the purpose of Measure Theory and Functional Analysis (Modern Analysis) both. All the chapters have also gone
Substantial changes.
An attempt has been made to remove any
type of mistakes of previous edition. K. P. GUPTA
'l'f''f'!ff'flm'l fircrivr~ I fet!' CltTT"~ ~~ ~~~ II f~ f~~U( fiffet~~ fifU~ I f~q"~mn~m ~s~ II
We adure you the guardian of the south-east quarter and Ruler of the whole uni~'erse, eternal bliss personified, the omnipresent and all pervading Brahma manifest in the form of Vedas. We worship Lord Shiva shining in His own glory, devoid of material attributes, undifferentiated, desire less, all pervading consciousness, having nothing to wrap about hifl1se/f except ether.
- Authors and Publishers
Contents Chapters
1. 2. 3. 4. 5. 6.
Basic concepts of set Basic set operations Functions and sequences Bounded. derived, open and closed sets on the real line Countability of sets Measure and outer measure Ring of sets, a-ring of sets Algebra of sets a-algebra of sets. Finitely additive Translation invaJ iant Postulates for all ideal measure function Measure space Completeness of measure function Caratheodory's postulate for outer measun: Measurable set Problems related to measure functions Problems related to ring of sets, a-algebra
7.
Lebesgue measure of a set Measure of open and closed intervals Measure of rectangle and a parallelopiped Exterior and Interior measure Some Theorems and solved problems Equivalence of Lebesgue-exterior measure and Caratheodory outer measure Fundamental Theorems Limiting sets Covering in the sense of Vitali Vitali's covering Theorem Sets of the Type Fa and Gil Borel measurable set Unsolved problems
8.
Measurable functions Measurable functions. Almo,l cverywerc Equivalent functions Characteristic function Simple function. Limit supelior and limit inferior Some Theorems and solved problems Lebesgue measurable functions Borel mensurability of functions Little Wood's three principles
Pages
1 6 11 16 21 46 46
47 47 49 49 50 50 50 51 51 63
73 73 74 74 75 95 99 100 106 107 110 110 III
115 lI5 115 116 117 118 136
136
( ,i ) Chapters 9. The Lebesgue integral of a function Lebesgue integral of a function Lebesgue integral of an unbounded function First mean value Theorem Countable additivity property of the integral
10. Theorem on convergence of sequences of measurable Functions Convergence in measure F. Reisz's Theorem D. F. Egor's Theorem Lebesgue bounded convergence theorem Lebesgue's dominated convergence theorem Beppo-Levis's Theorem Faton's Lemma
11. Absolute continuous functions. Indefinite integral and Differentiation Continuous function, Absolute continuous function Indefinite integral. Dift'erentiable Increasing and decreasing functions Functions of bounded variation Lebesgue point Problems related to functions of bounded variation Problems related to absolute continuous functions Problems related to indefinite integral Fundamental theorem of integral calculus Problcms related to Lebesguc point of a function
12.
V-Space Conjugate number, LP-space Norm of an elemcnt of LP-space Convergence sequence. Cauchy sequence Completeness of Ll'-space, Banach space Some theorems Holder's inequality Minkowski's inequality Schwarz's inequality LP-space is a normcd linear space Reisz Fisher theorem Some theorems
13.
Further theorems on Lebesgue Integration Integration by parts Second mean value theorem; Relation between Lebesgue integral and Stieltjes integral Cumulative distribution function Lebesgue-St ieitjes integral
Pages
140 140 144 147 150
176 176 178 179 181 183 185 186
194 194 195 195 195 196 197 205 214 219 220
225 225 225 226 226 226 229 231 232 237 238 241
246 246 248 250 254 255
(
vii
)
Chapters 14. The Weierstrass approximation theorem and semicontinuous function Bernstein polynomial S. N. Bernstein's theorem Weierstrass approximation theorem Semi-continuous function Theorem on semi-continuous functions
15. Signed measure
Pages
256 256 256 260 261 262
265
Signed measure. Positive and negative sets 266 Theorems on positive and negative sets 267 Hahn decomposition theorem 272 Singular measures 273 Jordon decomposition. Absolutely continuous measure function 274 Radon Nikodym theorem 274 Lebesgue decomposition theorem 277
16. Product measure
280
17. Fourier series
284
Periodic function. Trigonometric series 284 Finite discontinuity 284 Fourier series and Fourier Coefficients. Even and odd functions 285 Parsevel's identity for Fourier series " 292 Riemann--Lebesgue Theorem 294 Dirichlet's integral 295 Summation of series by arithmetic means 297 Fejer's integral 297 Summability of Fourier series. Fejer Theorem 298
18. Banach space
301
19. Hilbert Space
352
20. Finite Dimensional Spectral Theory
416
SYMBOLS AND THEIR MEANING Symbol 3 =>
iff & ¥ E
rf; :::>
c o N R R+ Q Def.
Q+ s.t. U
n
e.g. w.r.t. an~a
!no (S)
me (8) mi (8) In
(8)
L-integrable R-integrable E (f(x) > a)
t
f(x) dx
a.e.
Meaning there exist or there exists implies is implied by implies and is implied by if and only if and for every belongs to does not belong to is a superset of ·is a subset of null set Set of natural numbers Set of real numbers set of positive real numbers set of rational numbers Definition set of positive rational numbers such that Union intersection is cardinally equivalent to sequence consisting of points a" a2. a3, a4, ... sequence consisting of points at, a2, 03, ... for example with respect to the sequence converges in limit to a outer measure of a set S Exterior measure of a set S Interior measure of a set measure of a set S Integrablelin the sense of Lebesgue Integrable in the sense of Riemann {x E E : f(x) > a}
Lebesgue integral orr(x) over E
AI
almost everywhere complement of A
Tx
T(x)
1 Basic Concepts of Set The basic idea of a set is the origin point in the study of Modern Algebra. 1'0. Der. A set is a collection of well distinct objects. The objects of a set are called the elements or members of that set and their membership is defined by certain conditions. The elements of a set can be anything: mango, pen, sun, moon, river etc. Examples. (i) The set of interior points of a circle of unit radius with centre at the origin. (ii) The set of pen, ihkpot, pencil. (iii) The set of Ram, Mohan, Shyam, Bhole Shanker. (iv) The students of Banaras Hindu University. (v) Collection of letters a, b, c, d, e forms a set. Remark. (i) Set and aggregate both have the same meaning. (ii) The elements of a set must be distinguished from one another. 1'1. Notation. S~ts are usually denoted by capital letters A, D, C, D, E, P, Q, X, ... and their elements are denoted by corresponding small letters
a, b, c, d, e, p, q, x, ... It is not necessary that the elements of a set A are denoted by a. We can also write the symbol b for the general element of the set A. If a is an element of a set A, then this fact is denoted by the symbol a E A. The symbol E is used for anyone of the following phrases: (i) belong to. (ii) belongs to. (Iii) is in. (iv) are in. "belongs to", determined from the context "a E A" is read as "a belongs to A". If a is not a member of A, then we shall write a rt. A
2
BASIC CONCEPfS OF SET
It is customary to put a vertical line "I" or. an oblique "I" througt. a symbol to indicate the opposite meaning of the symbol. If we write "Let x E X", then the meaning of this is "Let x be an element of X". Remark. (i) It should be noted that order is not preserved in case of set, where as the Older is necessarily preserved in case of a sequence. That is to say, each of the sets {I, 2, 3}, {3, 2, I}, {t, 3, 2} denotes the same set. (ii) The repetition of an element does not change the nature of a set, i.e., each of the sets {I, 2, 2, 3}, {I, 3, }, 2}, {I, 2, 2,3, 3, 3} {t, 2, 3} denotes the same set. 1'2. Set of sets. It may happen that the elements of a set are sets themselves. A set consisting of a number of sets is called set of sets. Set of set,; is also expr.:ss·ed by saying "family of sets" or "class of sets". In order to make a distinction between a set and a family of sets, we use bold letters. A, D, C, P, ...... in case or family of sets. For example {A, B, C, D, E} is a family of sets. 1'3. Subset. If every element of A is also an element of a set B, then A is called a subset of B. This relationship between A and B is denoled by the sFmbo I A C R and is read as "A is a subset of B". or "A is contained in R". "Let A C !l" is read as .. Let A be a suhset of B".
Symbolically A is called a sub,et of B iff any a E A ~ a E B. Examples. (i) {a, {b}, {c}, {b, c} are subsets of the set la, b, c}. (ii) Evidently any a E A ~ a E A. Hence, by def. of subset, A C A for every set A. 1'4. Super set. A C B is also expressed by writing R::> A and is read as "B contains A" or "B is a superset of A". 1'5. Equality of sets. Two sets A and B are said to be equal if every elemeut of A belongs to B and every element of R :l1ongs to A. This relationship between A and B is denoted by the symbol A= B.
B.4.SIC CONCEPTS OF SE r
3
Sy,mbolically A=B iff any x E A ~ x E B. Alternately A C B, B C A {:> A=B. If A=B is not true, then we write A::;eB. 1'6. Proper subset. A is called a proper subset of B if A c B and A::;eB and this relationship is denoted by the symbol ..4 ~ B and is read as "A is a proper subset of B". From this definition it follows that every element of If is an element of Band B contains at least an element which does not belong to A. Examples. (i) {a, b} is a proper subset of la, b, c}. (ii) Since A C A and A =A and therefore A is not a nroper subset of A, i.e., A k A is not true for every set A 1'7. Finite set. A set consisting of a finite number of elements is called a finite set. Example. {a, b}, {I, i, 3}, {a1' a2, aa, ...... , a.} are aU finite sets. 1'8. Infinite set. A set consisting of an infinite number of elements is called an infinite set e.g., N, Q, R, Z, ... are all infinite sets. 1-9. Null set. A set containing no element is called a null set and is denoted by symbol.p. It is also called empty set or void set. Examples. (i) {x: x:;t:x1=~. For x::;ex is not possible for any element x. (ii) {x E R: X2 < O}=.fo. For the square of any real quantity ;;iii 0 and so x· < 0 is impossible for any x E R. Remark. There will be only one empty set and so we shall always speak of the empty set instead of an empty set. 1'10. Power set. The family consisting of all subsets of a set is called the power set ot'that set and is denoted by P, e.g. {B: B C A} is the power set of A and so we write P={B: B C A}.
Evidently the value of P varies with the value of A, i.e., P depends upon A and so we denore the power set of A by the symbol P (A). Finally, P (A)={B : B C AJ.
4
BASIC CONCEPTS OF SET
e.g. it A={l, 2, 3}, then
P (A)={ {I}. {2}, {3}, {I, 2}, {2, 3}, {3, I}, {I, 2, :}, q, }
Ex. 1. Prove that, C A for every set A. Solution. Since, does not contain any element and therefore the condition any x E ' " x E A is fulfilled and hence the result follows. Ex. 2. If a set A contains n elements, then P (A) contains 2" elements. Solution.
The total number of elements of P (A)
=nCO+nC1 ~-"C2+···+nc" =(1 + I)", by Binomial expansion
=2". 1'11. Universal set. If all the sets under consideration are subsets of a fixed set, then this fixed set is called universal set or universe of discourse and is denoted by the symbol X or by U. Singleton set. A setjconsisting of only one element is called a singleton set e.g. each 6f the following sets is a singleton set {a}, {b}, {I}, {2}. 1'12. Indexed set and Index set. Let At be a non-empty set for each r in a set 6. In this ca~e the sets AI' A 2 , 04 3 , ...... , A. are called indexed sets and the sets 6.={l, 2,3, ...... , n} is called irdex set. Here the suffix r E II of Ar is called all index. Such a family of sel." is denoted by {Ar : r E 6.} or {Ar} r E 6. Example. Let A 1 ={a, b, c}, A 2 =:c, d, /, m}, A3={m, n, q}, Ac={p, q, r, sl, A.;=(a, I, d, f}, 6={1, 2, 3, 4, 5}.
Here we find that 3 a non-emp y set At "f r E 6.. Hence 6 is called an index set and the sets AI' A!, A3 , AI' A; are c:.llied indexed sets. 1'13. Heriditary property. A nOll-empty family A={Ar} of sets is said to be heriditar y if Ar C A., A, E A => Ar E A. 1'14. Definition. A family {A.} of sets i5 sa id to be p.·rwise disjoint if
Ar n
..4.=,
¥
r, sEll S.t. r=l=s.
BASIC CONCEP rs OF ~ ET
Problems Set 1. Prove the following: (i) A C t/J => A=t/J (ii) A c B, B C C ,. A C C. 2. Are the sets ~, to} and {t/J} different? 3. If A={l, a}, then find P (A). 4. State whether each of the following statements is correct or incorrect. Here S is any set and S::;6~. (i) S E P (S), (ii) S C P (S), (iii) {S} E P (S). Answer.
3. 4.
{t/J, A, {I}, (i) correct.
{a} }
(ii) incorrect.
(Iii) incorrect.
2 Bas ic Set 0 perations Introduction. From arithmetic we know that(0 a+b is defined as the sum of the members a and b. (ii) a-b is defined as the difference of a and b. (iii) ab is defined as the product of a and b. (iv) ~ is defined as division of a and b, provided b#O in the last-case. In a similar fashion we will define the operations of union, intersection, difference, product in case of sets. Union. The union of two sets A and B, defined by A U B, is defined as the set of those elements which either belong to A or to B. Symbolically, A U B={X : x € A or x € B} A U B is usually read as 'A union B'. The following have the same meaning: union, sum, logical sum, join. The union of a finite number of sets A10 AI.· .. · .. , A,. is denoted by fl
Al U A2 U ............... U A,.
or by U Ar
,=1
CD
Similarly Al U A. U .......... U A .. U ...... =
U A, •
•
Also U Ar={x: x E Ar for some value of r, I ~ r ~ n}. '=1
In this diagram the shaded portions represents A U B. Such type of diagram is called Vend diagram. Examples. (i) P the set of points which lie on the line AB and Q that of the line CD, then P U, Q is the set of those points which lie either on AB or CD.
~IC SET OPlRATION3
If A={l, 2}, B={a, b, c, I}, then A U B={l, 2, a, b, c, I}, {2, a, b, c, I}={ I, 2, a, b, c}. Remarks. (i) Some writers write A+B in place of A U B. (ii) The meaning of "for some element a E A" or "for at least one element a in A" is the same. Intersection. The intersection of two sets A and B, denoted by A n B, is defined as the set containing those elements which " belong to A and B both. Symbolically, A n B={x : x E A and x E B} A n Breads" A intersection B". The intersection of a finite number of sets AI' A2 ,· ........ A.. is denoted by (ii)
,. or by n Ar • r-l
n
By def.
n Ar ={x : x E Ar
r=l
"f r S.t.
1
~ r ~ n}.
Examples. (i) If A={a, b, c}, B={l, ?, a}, then A n B={a}. (ii) If A={l, 2, 3}, B ={c, d, e}, then An B=t/I. In the diagram the shaded portion represent A n B. Remarks. (i) Some writers write A.B for A n B. (ii) The following words have the same meaning: (iii) The phrases ''for every r" and ''for all values of r" both have the same meaning. (iv) It is ea-;y to prove that ,A n B=B n A AU B=B U A (A U B) U C=A u (B U Cj (A n B) n C=A n (B n C) (v) From the definition it is quite obvious that and 'X E A n B ~ x E A and any x E A n B <.;> x E B so that A nBC A, A nBC B.
8
BASIC SET OPERA TJONS
(vi)
Evidently any a E A => a E A U B. and any a E B ~ a E A u B. so that A C A U B, B C A u B. Observe the difference between (v) and (vi). Disjoint sets. Any two sets A and B are said to be disjoint sets if their intersection is ,p. More precisely A and B are disjoint sets if A n B=t/I, e.g. {l, 2}, {a, b} are disjoint sets. Difference or sets. The difference of a set A w.r.t. a set Bis a set which contains only those elements of A which do not belong to B and is denoted by the symbol A-B. Symbolically A-B={x: x E A and x f1. B}. Similarly we define B-A={x : x E B and x ft. A}. Example. (i) If A={l, 2, 3, a, b}, B={a, b, c, d}, then A-B={l, 2, 3}, B-A={c, d}. Complement of a set. The complement of a set A w.r.t. the universal set X is defined as the set X -A and is denoted by X'. Symbolically, A'=X-A={X : x E X and x It is also clear that any x E A' ~ X ft A. Remarks. (i) X'=rjl For X--X=rP (ii) ,p'=X. For ",'=X-,p=X (iii) A U A'=X, A n A'=c!> (iv) A-B=A n B', B-A=B n A' For A-B={x: x E A, x ft B} ={x : x E A, x E B'} =A n B' Similarly B-A=B n A' (v)
ft
A}
(A,),=A
=> x E A or x E B x rt. A U B => x rt. A and x (/. E, Symmetric of difference sets. The symmetric difference of two sets A and E, denoted by A 6 B, is defined as A A R=(A--B) U (B-A)
(vi)
x E A U B
9
8ASIC SET OPERATIONS
Example.
. If A={l, 2, 3, 4}, B={3, 4, 5, 6} then A-B={I, 2}, B-A={5, 6}, A 6 B=(A.-B) U (B-A)={1,2} U {5,6}
={l, 2, 5, 6}. Distributive law. The union of sets is distributive w.r.t. the intersection of sets and conversely, I.e. (A U B) n C=(A n C) u (11 n C) (A n B) u C=(A u C) n (11 u C). De-Morgan's law. Let AIX be a non-empty set for each IX in an index set 6.. Then (i)
(
u AIJ. )' = ( () ~IX' ) (I.E 6 «ELl
( () ArJ. )' = U Aac' (I.E 6 IXE6 For proof refer to author' book on "Set theory and Related Topics." Ordered pair. An element of the form (a, b) is called an ordered pair. The elements a, b in the ordered pair (a, b) are called the first element and the second element respectively. Equality of ordered pairs. Let (a, b) and (c, d) bc any two ordered pairs. We define (a, b)=(c, d) iff a=c, b=d i.e. (a, b)=(c, d) <=> a=c, b=d. Product of sets. Let A and B be any two given sets. The product of A, B is defined as the set consisting of all ordered pairs (a, b) where a E A, b E B and is denoted by the symbol A x B. Symbolically, AxB={(a, b): a E A, b E B} A X 11 is read as "A cross .B" . The product set A x B is sometimes called Direct product or Cartesian product. The reason for the name Cartesian product is that the set A X B can be plotted in a Cartesian plane. Similarly BXA={(b, a) : b E B. a E A} (ii)
rr
J'
_ cJ ___ ~{(L
bl
_____.fb n ..,.. .----- A
-
------::::
)(
10
BASIC SET OPERATIONS
Evidently A x B::j:.B xA. If either B or A equal to ~, then AxB=t/J. Example. If A={I, 2}, B={a, b, c}, then AxB={(l, a), (I, b), (I, c), (2, a), (2, b), (2, c)} AxA={I, 2}x{l, 2} ={( I, J), (I, 2), (2, 1), (2, 2)}. From this example it is clear that if A contains 2 elements and B contains 3 elements, then A x B contains 2 X 3 = 6 elements. Generalising this result, we find that if A contains m elements and B contains n elements, then A X B contains m x n elements. This fact implies that A x B is an infinite set if either A or B or both are infinite sets. Product sets in general. The product set of the sets, A, B, C is denoted by A X B x C and is defined as A·xBxC={(a, b, c) ; a E A, b E B, c E C}. Generalising this result to a finite number of sets, we have A1 xA2 x .. x An={(al' a2 , ••• , an) : ar E Ar ¥ r, s. t. l~r~n} Examples. (i) The product R x R [or simply R2J is defined as Euclidean plane (of two dimensions). (ii) The product set R x R x R [or simply R3J is defined as Euclidean space of three dimensions.
3 Functions and Sequences 3'0. Function. Suppose X' and Yare two given sets. By some given rule, if each element x E X corresponds to a unique element Y E Y, then that rule is called a map of X into Y and is denoted by f. We write f: X... Y. This is read as "f maps X into Y" or . 'I is a mapping of X into Y". Mapping sometimes called operation or map or correspondence or function.
Let an element y E Y be corresponded by an element x EX, then Y is called the image of x and is denoted by f (x). Thus y=/(x). "/(x) is read a'S 1 of x" Here x is defined as the pre-image of y. The set X is defined as the domain of the ftinction/ and Y is called co-domain of j, I(X) is cabed the range or the image set off. It is also expressed as /(X)={/(x) : x E X} Evidently /(X) C Y. Remark. From the definition of a map it is clea.r that a map ./: X_ Y is said to be well-defined if (i) any element x EX=> f(x) E Y (ii) any element x EX=> a unique element/(x) € Y. In other words 'no element of X can have more than one I-image, i.e., given any x E X, we can not find two or more than two elements Yl' Y2 E Y such that Yl-/(X), Y2 f(x). (iii) Two or more than two elements of X may have the same/ -image in Y. i.e., if Xl. x 2 • Xa, , .. E X, then, Examples. gram:
/(x1)=/(X z) =/(Xa) = ... (i) Let a map /: A-B be defined by the dia-
12
" FUNCtIoNS' AN[j SEQUENCES
f
8
Here a E A ~ 3 1,2 E B such that 1 =!(a)=2. This ~ ! is not well-defined. (iii) The map ! A~B defined by the following diagram is well defined.
Aj~~f _"; f. ------..-= 10--b
•
I
domain of f {a, b, c} range of 1={2, I} Co-domain of/=(2, 1, -I). 3'1. Onto and into mappings Letf: X---,>-Ybe a map. I is called an 'into' map if 3 at least one element in Y which does not have a pre-image in X. i.e.,fis an 'into' map if/(X) is a proper subset of Y. i.e., if!(X) C Y. / is said to be an 'oIllO' map if !(X)= Y, i.e., if evcry element in Y has a pre-image in X. 3-2. One-one and many-one mappings. Let: X --;. Y be a map. / is called a one-olle map if Here
/(X1)=(X 2) ! Xl> X 2 01'
E X ~ XI=X~,
equivalently
Xl :;CX2 : Xl> X 2 E X ~ f(xd:;tf(x 2} if different elements In X have different images in Y. . / is called a many-ol/e l1lap if dilfercnt clement s in X hav~ the same i-image in Y i.e., if
i.e.
j(x})=f(x") : Xl, x 2 E X ~ X'*X 2'
13
pUNCflONS AND SEQUENCES
Examples. gram:
(i)
Let f: A ~B be a map defined by the dia-
e j
Here
f(A)={f(x) : x E A} ={x, y, z}=B.
Also /is one-one. Finally, f is one-one onto map. (i) The map f: A~ B defined by the follow ing diagram is many-one into map. fj
A
3·3 Real niued map. A map is said to be real valued map if a set is mapped into the set R by it. That is to say, any map f: A ~R is called a rea' valued map. 3·4. Set Function. A function is called a set Junction if its domain consists of a family of sets e.g., the map f: {A. : i E N}-+N defined by the formula
is a set fu nctio n. 3·S. Real Valued Set Functi.n A function is called real valued set function if its domain is a family of sc!s and its co· domain is the set of real numbers R. e.g.,
a map f:
{Ar
t
r E L;..
-?
R
is real valued set function and this map is said to be finitely additive if
14
FUNCTIONS AND SEQUENCES
I (
U Ar
,=1
)= UI ( ,=1
Ar )."
The map I is said to be couatable additive if
I (
~1
Ar
)= r~ I (
Ar )-
3'6. Extended real valued set function. An extended real valued set function is a real valued set function which also takes either of the value -{-
00, -- 00.
_[-00,00] such that f(Ar}=+oo or for some value of r. It may be noted that 00 and -00 both are not real numbers. i.e.,
a map
I: {
Ar}r E 6.
-00
Step function. We divide the interval [a, b] by means of points Xl' X 2, ...... , X" s.t. a=xo < Xl < X 2 <, ...... , < x,,=b. A function/s.t./(x)=c, for X'-l < x <x, is called step function where c, are constants ;=1, 2, ...... , n. 3'7. Sequence'. By a sequence we mean a map I whose domain is N, That is to sap, a may I: N _ A is called a sequence. The value of the function I at n E N is denoted by f or The set {I(n) : nE N} is defined as the range of the sequeace. If this sequence is in X, then the sequence is said to be in X• . This X may be R or the set of complex numbers or any other set of numbers. 3 8. Convergent sequence. A sequence < aft > is said to converge to a if given E > 0, 3 no E N S.t. ¥ n ? no => I a-a" I < ~ A sequence < a" > is said to be convergent if given 6> 0,3 no ~ N S.t. ¥ m ~ 110 =:> I am-am+fl I < ~ ¥ pEN. 3'9. Bounded sequence. A sequence < an > is said to be bounded if 3 a number M > such that I a" I < M ¥ n I::. N. It is easy to prove that Convergence => boundedlles~. 3'10. Metric spa.:e. Let X7:-tP be any given space. Let X, y, z E X be arbitrary
°
Fl.JNt::TIONS AND
~EQUENCfS
15
A function P: XxX .... R having (he properties listed below (i) p (x, y) ;;;. 0 (ii) P (x, y)=O iff X= y (iii) p (x, y)=P (y, x) (iv) P (x, y)..Lp (y, z) ~ P C<, z) (triangle inequality) is called a distance functioll or a metric for X. 3'11. Definition. A sequence < an > is said to be strictly monotonic increasing if an < an+! ¥ n E N. A sequence < an > is said to be strictly monotonic decreasino... if an+! < a.,. >.f n E N. A sequence' < an > is said to be monotonic increasing or monotonic non-clccrcasing if an ,,;;; an+! ¥ 11 E N. A s~quence < an > is said to be monoton'c decreasing or monotonic non-increasing if an+! ~ an ¥ 11 E N.
A sequence < an > is said to be monotonic or monotone if it is either monotonic increasing or monotonic decreasing. Notice that a sequence will be either monotonic increasing or monotonic decreasing. 3'12. Axiom of Choice. Cartesian product of a family of non-empty sets is non-empty. Axiom of choice may also be stat·~d as "there exists a choice fu nction for any non-empty fami Iy of nOll-empty sets".
4 Bounded Sets, Derived Sets, Open Sets and Closed Sets on the Real Line Rea) Jine. The elements of the set R can be represented by means of points on a straight line. The origin of this straight line is taken to represent the number o. The real numbers x > 0 are represented on the right of lero, where as the real numbers x< 0 are represented on the left of o. Each point of this straight line represents a unique real number and each real number determines a unique point on the line. For this reason the set R is referred to as the real line or real axis. This is why, we use the words point and number conversely. Remarks. Throughout this chapter we shall take R as the universal set provided nothing is contradictory stated. O.,en and closed intervals. Let a, b E R be arbitrary. We write [s, b]={x E R : a <; x ~ b}, (a, b)={x E R: a < x < b}, and can these sets of real numbers by the names closed interval and op~"inferval res pecti ve Iy. Also we write [a, b)={x : a ~ x < b} (a, b]={x : a < x ~ b} and call each of these sets 'of real numbers by the name semi-open interval or semi-clo ..ed interval or half-open interval or half closed interval. Neigbbourbood of a point. Let ~ > 0 be any real number. Let Xu be any point on the real line. Then the set {x E R: I x -xo I < ~} is defined as thJ. 6-neighbourhood of the point Xo. Thus, . -neighbourhood of the point Xo ={x E R: I x -Xo I < ~}={x E R : Xo --4I < x < XO+Ii}
17
SETS ON TtiE REAL LINE
=(XO-E,
xo+·)·
If A is a neighbourhood of x, then A-{x} is called a deleted neighbourhood of x, i.e., the set (X-f, x+.)-{x} is called deleted .-neighbourhood of x. Example. The set of real numbers x satisfying the condition 3 < x < 7 is 2-neighbourhood of the point x=s. The idea of open set and closed set is connected with the idea of neighbourhood. Open set. A set A c R is caned an open set if given x e: A, 3 f > 0 such that the 6-neighbourhood of x is a ·subs~t of A.
From this it is clear that every 6 -neighbourhood is a.n open set but an open set is not necessarily an • - neighbourhood. (i) The set of all real numbers x such that 2 is a I-neighbourhood of the point Xo= 1 as well as this set is an open set. Example.
o< x <
(ii) The set pf all real numbers x such that 0 < x < 5 is an open set but not an ,- neighbourhood for any t > O. Continuity. A real valued function f is said to be continuous at the point Xo if given E > 0, 3 8 '> 0 such that I f(x)-/(xo) I < E whenever I x-x, I < 8. If a function is continuous at each point x for which it is defined, then it is called a continuous function. Bounded linear set : (i) Let A C R b! any given s~t. Then A is c!l.lIed linear set. (ii) A real number b is called an upper bouocl for the set A if x~byxEA
(iii) An upper bound b of A is called the lU'1t upper bound Or supremum of the sct A if b ~ y for every upper bound y of A. In brief. the supremum of A is written a~ sup (A). Also the lea'>t upper bound is written as. I. u. b. Example. b is the sl!premuln fllr all the reat numbers in (a, h] and h+x is an upper bound for the set [a, hJ ¥ X e: R+. (iv) A set A is called bounded abuvf: or bounded on tbe ri&bt if it has an upper bound.
18
SETS ON THE REAL LINE
Examples. (I) The set of all re3J numbers in [a, b] is bounded above. (2) The set {x E R : x < 5} is bounded above. (3) The set (X E Q+ : Xl < 5) is bound:!d above and its ¢ Q.. Hence supreupper bounds are 6, 7, 8, 1l/2, ... since mum of the set does not exist. (4) The set {I, I, i, t, ...... ,} is bounded above. (v) A set A is said to be unbou .ded on the right if it is not bounded on the right e.g. the set lx E R : x < 5/. (vi) A real number a is called a lower bound for the set A if
vS
aWi;x¥xEA
A lower bound m of A is called greatest lower bound or talBlum of A if a ~ m for every lower bound a of A. In short the in{~mum of.4 is written as inf (.4) and greatest lower bound is Wrttten as g. 1. b. A set A is called bounded on the left or bounded below if it has a lowo:r b.)und. An interval which has at least one end point as 00 or - «. is an unbounded interval. Examples. (1) The set {x E R : x > 5}, is bounded below. (2) The set N={l. 2, 3, 4, ...... } is bounded below. (3) The set {x E Q+ : x ~ 5} is b.Junded on the left (i.e. bounded below) and its greatest lower bound is 5. (4) The set of real numbers [a, b] is bounded b.!low and its infimum is a. (vii) The set A is said to be unbaun:lej o:t tile lert if it is not bounded on the left e.g. the set {x E R : x < 5}. (viii) The set A is called bcnm!led if it is bounded both below and above e.g. the set of all real numbers in [a, b] and (a, b). Another definition. The set A is said to be bounded if 3 a numb!r m. > 0 E R such tbt I x I ~ m ¥ x E A. l .. EVidently A c [-m, mj=a finitei:ltervaL So we also define that the set A ii bound ~d if it is a SU~l:::t of a finite interval. (ix) The_ set A is caJled unbo1in~ed if it is not bounded. Eumples. 0) A finite set is alw.lYs bounded. (2) lhe set A={l, i, l, 1, ... } is bounded. For A C TO, 1 . (3) An infinite set mayor may not b! bounded. (4) Let A={ 5, 6, 7, 8, 9} be a subset of the set N of natural ftUl!lbers. Then 5 is lower bound of A as 5 ~ x "rI- x E A. Similarly I, 2, 3, 4 are also lower bounds of A. 5 is the greatest
19
SETS ON THE REAL LINE
lower bound of A. 9 is the upper bound of A as x ~ 9 ¥ x E A Similarly 10, I J, 12, ... etc. are also upper bounds of A, wh.:reas 9 is the least upper bound of A. (x) If A~tP, then we ado;:,! th~ convention, sup A= -- 00, inf (A)= +00. Limit point and derived set. A point x is said to be a limit point of a set A if every nej!h~ bourhood G containing x contains a point of A other than x, I e., x is said to be a limit point of A if for every neighbourhood G with x E G, (G-{x})
n
A#t/J.
The set of all limit points of a set A is called the derived set 0/ A and is denoted by D(A). Evidently x E A => x E D(A) or x fI. D(A). e.g. the limit point of the set {J, j, 1, t, ... } is O. Condensation point. A point x is said to be a condensation point of a set A if every neighbourhood containing x contains an infinite number of points of A. In accordance with this definition a finite set has no condensation point. Closed set. A set A is said to bl! closed if every limiting point of A belongs to the set A itself. Symbolically, a set A is said fo be closed if D(A) C A.
Examples.
(I)
{O,
1,1, I, 1, ...... }=A is a
closed set.
For
D(AJ={O} C A. (2) A closed interval is a closed set.
For D«(a, bJ)=(a, b] C la, bJ. (3) Every real number is a limit point so that ration'll limit points belongs to the set Q. This => D(R)=R C R D (Q)=R
20
SETS ON THE REAL LINE
Examples. (1) R is an open set. (2) An open interval is an open set. (3) Q is neither open nor closed. R.emarks. (i) An arbitrary union of open sets is open. (ii) The intersection of a finite number of open sets is open, whereas arbitrary intersection of open sets mayor may not be open. (iii) An arbitrary intersection of closed sets is closed. (iv) A finite union of closed sets is closed whereas an arbitrary union of closed sets mayor may not be closed.
5 Countability of Sets Introduction. A question may arise on any two given sets that they have the same number of elements or not. The answer can be found by counting the number of elements of the set if the sets are finite. On the other hand if the selS are infinite sets, then the answer depends upon how one defines two sets to have the same number of elements. 5'0. Definition. A set A is said to be cardinally equivalent to a set B if 3 (at least one) one-one map from A onto B. This fact is denoted by the symbol A-B. This fact is also expressed by saying that A is numerically equivalent to B. (ii) A is equivalent to B. (iii) A is equipollent to B. (iv) A and B have the same power.
(i)
Examples (i) Any two singleton sets are equivalent (ii) Let N={l, 2, 3, ... }, E={2, 4, 6, ... }.
Define a map!: N-+E by the!ormulal(n)-2n. Evidently I is one-one and onto. Hence N--E. This example shows that the set N is equivalent to a proper subset of itself. This is a striking property of infinite sets. Let A={J. 2. 3, ... }. Ba:::{alt a., a..... }. Then Ar-B under the mapping n ~a,..
(iii)
(iv) Let A=[2, 6], B=[a, bJ, where [2, 6] denotes tlv! set 01 all real numbers in the closed inter,al [2, 6]. Similar meanillg lor [a, b].
Then [2, 6]-[0, b] under the mapping
x-a+ b-a 4-
(x-2).
22
COUNTABILITY OF SETS
Remark. In order to find the mapping from [2, 6] to [a, b], firstly find the equation of a straight line joining the point (2, a) to (6, b) by the formula Y-Y1=)'2- Yl (X-Xl) X 2 -X1
and then put y-f(x).
Theorem l. To prove that the relation A,...,B ill the family of sets is an equivalence relation. Proof. The given relation is (i) reflexive: A-A for every set A. For the identity map /04 : A-+A given by onto. (ii) Symmetric: A-B => B---A.
fA
(x) .... x is one-one
onto
For
A-B => 3 a one-one map f: A ~--+ B.
=> f- 1 : B :$
onto ---+
A is also one-one.
B-A.
(iii) transitive: A-B, B-C:o> A--C. onto
For A-B, B-C .. 3 one-one mapsf: A ---+ B onto
and g : B - - .. C. onto
=- gf: A - - - C is also one-one, .. A--C. given relation satisfies all the conditions of an equivalence relation. Hence the result !"t. Cardinal numbers. [Kanpur 71, 70; Utkal 5] The relation A-B in the family of sets is an equivalence relation. Hence the rela.tion decomposes the family of sets into di!;joint equivalence classes. Every equivalence class defines a unique cardinal number. In other words the cardinal number of a class of equivalence sets is any representative of the class. The cardinal number of a set is sometimes called its power or potenc~'. If P is any finite set consisting of p elements, then its cardinal number is defined as p. The cardinal number of ~ is defined as O. A cardinal number corresponding to a finite set is called a finite cardinal number. A cardinal number corresponding to an
rt means that the
COUN'fABILIrY OF
~ETS
23
infinite set is called a transfinite cardinal number. It follows from these definitions that all the transfinite cardinal numbers a"C 'greater than any fi!1ite cardinal number. Notation. The cardinal numbers of the sets P, Q, R, S, ........ . arc denoted by the corresponding small letters P. q. r, s, ........ . Also the cardinal number of a set A is denoted by I A I or card A. Then, by definition / P 1= card P=p, I Q /=card Q=q. The cardinal numbers of the set N and the set of all real numbers [0, l~ are denoted by a and c respectively. Then . / N 1=8, I [0, l] /=c. The symbol Xo (read as aleph nUll) is also used to denote I N /. This symbol was originally used by Cantor. If P""Q then by definition of cardinal number, 1 P/·=IQI· 5'2. Sum of cardinal numbers. Let P and Q be any two sets with cardinality p and q S.t. their intersection is empty i e., I PI=p.1 Q I=q, P () Q=; Then the sum of p and q is defined as p+q=1 P u QI· More generally, let P". be a cardinal number corresponding to the set p~ ¥ CIt in an index set 1::,.. Further suppo'se that. . Prx. ("\ P~=,p >tf Of, ~ € 6 s.t. 1X:;t=~. We define the sum of cardinal numbers as E p",= U P", J -EDCltEI::,. • Examples. (I) 1/ A={l, 2}, B={5, 6, 7), then 1A 1=2, I B 1=3 and A () B=~ so that 1A U B 1==1 A 1+1 B 1- 2+3-5. Also A u B={I, 2, 5, 6, 7} and hence 1 A U .B 1-=5. This is a verification of out definition 5'2.
(ii)
Then
...
Let
A={a, b, c}, B:oo{a, p, q, r, s}.
A U B=(a, b, c} U {a, p, q, r, s}. ={a, b, c, p, q, r, s}. 1A
u
B
1=7.
24
COUNfABILITY OF SETS
Evidently 1A 1=3, 1B 1=5 and A n B::;t:~. :. By definition 1A UBI =I=- I A 1+ I B I i.e., 7 ;::3+ 5. This example also verifies our definition of sum of cardinal numbers and it also shows that An B=; is a necessary condition for defining the rule 1AuB 1=1 A 1+1 B I. S·3. Prodlict of cardinal numbers Let P and Q be any sets s.t. I PI =p, 1 Q I=q. We define the product of cardinal numbers p and q as pxq= I PxQ I. More generally, let Pa. be a non-empty set for each Gt in an index set A and let pa.=1 Pa.. Then we define the product of cardinal numbers as npa.=j x Pa.I. where ilpa. stands for the Pl.P2' Ua· .... and xPrJ. stands for the cartesian product. Example. Let P={a, b}. Q={a. J, 2, 3}. Then I P 1=2, I Q 1=4, pnQ,ptP. By definition, 1 P x Q I =2 x 4=8. Now I Px Q 1=1 {(a, a), (a, I), (a, 2), (a, 3), (b, a), (b, 1), (b. 2), (b, 3)} I
=8. This verifies our definition of product of cardinal numbers. !§·4. Definition. A set A is called an infinite set if A is equivalent to a proper subset of itself. In the language of notation, a set A is called an infinite set if A -.I to a proper subset of A e.g. N, Q, R, C etc. 5'5. Definition. A set A is called a finite set if it is equivalent to {I, 2, 3, ... ,n} for some value of n E N. S·6. A set A is called a denumerable set if A ..... N. e.g. tal' aa, a8,·· ... } is denumerable, since this set -- N under the map an .... n. Moreover this is a sequence and therefore a sequence is a/ways denumJrable.
A denumerable set is sometimes called enumerable or countably infinite .
•: A.....,N ~ 1A 1=1 N I => I A 1=8. Also A.-..J N ~ A is denumerable. Hence a denumerable set can also be defined as : A set A is denumerable if I A 1=8.
25
COUNTABILITY OF SETS
We shall always use this very definition in future 5'7. Definition. Countable sets. A set is called (:ount4ble or atmost countable if it is finite or denumerable. Example. (i) A={I, 2, 3, 4}. Then A is finite so that, by definition, A is countable. (ii) Let A={a1 • a2 • aa,·"} so that A is denumerable. Hence by definition A is countable. Here observe tbe difference between coulitable and countably infinite. 5 8. Definition. A set A is ca.lled an uncountable set if A is an infinite set and A is not cardinally equivalent to N. e.g. Rand C are uncountable sets. The following have the same meaning: Non countable, non-denumerable, non-enumerable, uncountable. 5'9. Definition. A set A is said to have the power of contil1um if A r- [0, 1]. Also we say that the cardinality of A is (:. Theorem 2. To prove that
I AxB 1=1 B 1+1
B 1+1 B I.. ·to Proof. Let A and B be any two sets. AxB={(x, y): x E A, Y E B} U
=
{x, y) : )' E B}
xEA Then AxB
I
I A I terms.
1=\
u {(x, y) : y E B} x E A
... (1)
For a fixed x E A. consider the map .f: 8~{(x. y) : )' E B} given by .f (y)= (x, y)
~ )'
E B.
Evidently.f is one-one onto. Then B '" {(x: y) : y E B} This ~ 1 B 1 =1 {(x, y) : y E B} I. In this event (') shows that I AxB 1=1 B 1+1 B 1+· .. to
1A 1terms.
Q.E.D. Verification of the theorem 2. Let A={l, 2}, 8-(1,2, 3}. Then
1 A 1=2, I B 1=3 I AxB 1=2.3=6
A >
26
COUNTABILITV OF SETS
If we take
A1={(I, I), (I, 2), (I, 3)}, B1 =t(2, I), (2, 2), (2, 3)}. then A x B=.A I U B t • Al () Bl =;. that 1AxB 1=1 Al u Bl lie. 2.3=3+3. Remark. Every property of addition and multiplication of natural numbers is not true for cardinal numbers in general. e.g. a+a=a does not imply a=O a.a=8= La does not imply a= 1. This proves that Cancellation law is not true for the operpations of addition and multiplication of cardinal numbers. Ex. t. If fI. be OIly cardinal number, then fI. E;; 1 A 1 ~ 01 => 1 A 1 -«. Solution. Let« be any cardinal number. Let II E;;: 1 A 1 <; fl.. To prove that 1 A 1 =cx. Let 01=1 B I Then we have to prove that is eq.uivalent to proving that A-B. « EO;;
I AI
.. :0
IAI
E;at=>
I B I .s;; I A I B- to a subset of A or B-A IAI
E;
I A I -I B I.
which
... (1)
IBI
=> A- to a subset of B or A-B. ...(2) The statements (I) and (2) taken together imply the required result. Ex. 2. If () and {J are cardinal numbers such that fI. ~ f3 and {J ~ ar, prove that at = 11. [Banaras 1969] Solution. Let A and B be two sets s.t. 1 A 1 =a, 1 B 1 =-~. Also let at ~ {J. (J ~ ac. To prove that Cl""'~. we have to prove that: A I - 1 B I. GlC;;B=> IAI ~ IBI lO A -- to a subset of B or A-B ... (1) ~~Cl"IBI~IA: ~ B- to a subset of A or B-A. . .. (2) The statements (I) and (2) imply the required result. Theorem 3. A set A is countably infinite iff A can be put In the form of the sequ[!nce tOI' O 2 , tl3 • .• 1 of tlte distinc: elements. Proof. A set A is countably infinite iff 3 a one-one onto map f: N ~ A i.e. iff (i) m.;f:n and m, n E N ~ f (m)-:;cf (n). (ii) Given any element b E A, 3 an e1em'!nt a C N s.t. f(a) b i.e., iff f[N]=A and f is one-one. 111
27
COUNT ABILI j Y OF SETS
or iff given any element n € N. 3 Of! € A lI.t.j(n):::a n andf is one-one. Above means A={al. a2 • aa •... }' where a10 02' 0a.· .. are distinct elements. Theorem 3ft. Every subset of a cvuntable set is coufttable. [Meerut 1987, 88] Proof. Let A be a countable set. Then A is either finite or enumerable. (I) When A is finite. Then every subset of A is also finite and so is countable. (ii) When A is eoumerable. Then A can be written as ... (1) A={x1, XI''''} Let B be a subset of A. (i) If B=;. then evidently B is countable. (ii) If BtF"', then B is expressible as B={x. l , x n2 ,· .. }, where XIII E A. Evidently B is enumerable, by virtue of (l). Theorem 4. The set of all real numbers iI' the closed illferval [0. 1] is not denumerable. [Banaras 1971 ; Kallpur 68] Proof. Let A denote the set of an real numb.:rs in tne eto.ed interval lO, I J. SymbolicallY we write
A={x E R : 0 ~ X ~ l}=[O, 1). To prove that A is uncollntable. Suppose not. Then A is couhtable. This implies that every element of A must appear in the sequence x"' .. .of distinct elements. i.e., B={xlo x 2 , xa, .. ·} ... {I) Write decimal expansion of those xl's as follows: x 1=0'xu Xl 2 Xu xu .... · X 1m . .. .. X10 X 2 , Xa,· ......
X~=O.X21 X 22 X 23 X~, ...... x:.,. ..... .
X.,,=0·Xm1 XtII~
X"'a X""
..•..•
XIII"",
where XII E {O, 1. 2, 3, 4, S, 6, 7, 8, 9} ¥ i and j and each decimal contains an infinite number of non-zero elements. We can write 1 and i in two ways as given below: J 1=1'0000... ... ... . 1 1-0'5000 ....... ..
28
COUNTABILIrY OF SETS
J t =0'99999 ...... =0'9 } 1 1=0'49999 ...... =0'49
and
... (2) But in the present case we shall write the decimal representation of the elements A in the form (2). Construct a real number ~=O, ~l> ~a ... ~". ... s.t. Xu = 5, write ~l =4 if and if Xu :f.:5, write fl = 5 in general if x",,,,=5, write ,,,,=4 and if x",,,,;t:5, write f",=5. In either case, it is clear that x".".;t:'''' ¥ m. . .. (3) Since ~1It=4 or 5 ¥ m E [0, 1], i.e., ~ E A ... (4)
ea.
:. e
x",=~ ~ O.X"'1 X.I xma ...... =O'~l ~l! ~a""" => xmm=~", T m. Contrary to (3). :. x",:;;c~ ¥ m.
e
In this event (I) shows that ~ A. Contrary to (4). Hence the required result follows. Q.E.D. Theorem S. To prove that the set R is uncountable. [Ufkal 6S ; Puajab 62 ; Vikram 61] Proof. Firstly we shall show that I R/=c. We know that cardinality of the set of real numbers in the open interval (0. b) is c, so that I (a, b) /=c ¥ 0, b E R s.t. a < b. This .. \ ~ )i=c. ...(1) The set R is the set of all real numbers lying in the open interval (-00, (0), i.e., R=(-O, co).
(-r'
Define a mapj:
(-~, ~ )~
(-00,00) by the formula
/(x)=tan x. Evidently j is well defined. Also it is easy to verify that! is one-one onto. :.
This
i.e. This
(-~, ~)-
I(
(-00, (0).
~ -~, ~ )=1 (-co,
00)
c=/ R I.
with (I).
~
in
a~ordance
R is uncountable,
I· Q.E.D.
29
COUNTABILITY OF SETS
Ex. 3. To ~ow that lor every real number x, the real numbers in lhe semi-open interval [x, x+ I) lorm an uncountable set. Solution. Let x E R be arbitrary. Define a map I: [x, x+I)_[O, 1] by the formula.!(X)=X-x, Evidently l(x)=x-x=O l(x+l)=x+l-x=1 :. I is well defined. f(XJ=/(X2 ) ; Xl> X 2 E rx, x+l) => X1-X=XI-x This => I is one-one. I is a continuous map => I is an 'onto' map. :. [x, X+I) '" [0, 1) This => I [x, x+l) 1=1 [0, I) ,. But the set of all real numbers in [0, 1) form an uncountable set and hence the set of all real numbers [x, x+ I). Proved. Theorem 6. (i) II Ai is countable infinite set, then fI
U A. is countable infinite and hence deduce that n.a =a. 1=1
-
(ii)
If Ai
is countably infinite set lor i = 1,2, 3, ... ...... then
U Ai is counfably infinite and hence deduce that
8+8+8+ ...... +to 8 terms=a. [Meerut 1972; Kurukshetra 69] Or Union 01 countable collection 01 cOll'Jtable sets is countable. [Meerut 1987, 86] Proof. (i) Let A. be a countably infinite set. Let
A=U" A • • =1
To prove that A is countably infinite. Let the elements of A; be displayed as Al : au a12 Q I8 ...... 0 1....... . ,42 :
au
A8 : 0al
at~
a 21 .... .. 0 2" ..... .
aaJ 0 83 , ..... as" ......
.
Write B={au , 012'''' ... a 1" .. • ; 0 21 , a22' ...... ; 0"1' a2,,, .. ·}· The S(~t B considered as the set of distinct elements is coun:. , B I =8. ...( 1) tably infinite.
30
COl!NTABILlTY OF SElS
n A}= r/> for i;f=j, then evidently I A I = I B, If Ai n A, t:.d> for i :;t:j, th'.!n A is equipollent If A,
=a.
with some subThis:? I A I ~ ! B I -'-~a or I A I E;;;. a ... (2) Al C A and 1 Al I =8 ... (3) :. a= I Al I ~ I A I or a ~ I A I Combin:ng (2) and (3), we get II ..; I A I ~ a, which:? I A I = a. In either ca<;e I A I =a This proves the required result. Dedu<:tion. Further assume that A, n A,=,p for i#j and ;,j=l, 2, ...... , n. Then by what we have established it follows that I A I =8.
set of B. But
or
IU At I=a,
or
a+a, ...... + to n terms=a, or IZ.a=a.
1'=1
(ii)
or ;
1=1
I A, I =8.
..
Let AJ be a countably infinite set, and A=- U
1_1
A.
To prove that A is countably infinite. Let the elements of A, be displayed as follows all : a2l Aa : 03l
Al :
0 12 013·· .... 01.... • .. .
A~
0 22 0 28 ...... a ....... ..
032
a33 · ..... 0." .... ..
First we shall assume that A, n A/=I/I for i::;t:j. Divide the elements of A into blocks s.L mIll block contain m elements and at} wili be in mtll block iff; +j=m + l. Within the m'" block the second sufii'< j of Oi! increase from I to m. Thus A can be expressed as
A={ou ; 0 21 , 0 12 ; 0 31 , 022' 013 ; ... ; 0"1>"" 0 1,,; .. }. Clearl.t' A is expressible in the form of a sequence of distinct elements and hence denumerable so that ! 11 =a. Secondly we assume that A, n AFi=t/J for i .,cj. In this case A is equipollent with some subset of N' so that ! A I ~ I N I or I A I ~ a. . .. (4) But Al C A and I Al I ~ 2.
31
COUNT ABILITY OF SETS
so that
a= I Al IE;;/ A ' or a ~ I A / This => / A / =a, by (4) and (5).
... (5)
This proves the required result. Deduction. Further assume that B. n A,=~ for i.:l=j. Then from what ha'i been done it follows that
I A 1 -a::>·
IUA.I=a => ; I 1
'~1
=.
/ A. 1
=> a+a+ ... +... to a terms=a This completes the proof. . Theorem 7. To prove that N X N is countable, or to prove that a2 =a.a=a. Proof. To prove a.a=a is the same as to prove I NxN 1= 1 N ,. The elements of N x N may be displayed a'i follows: (1.1)
(1,2)
(2, I)
(2. 2)
(1,3) ...... (I,m) ... (2, 3) ...... (2. m) ...
(n, I)
(n, 2)
(n. 3) ...... (n, m) ...
If we write then
(i)
.
At={(i. 1). (I. 2). (i,3) ....... } •
NxN= u A.
\ ,-I
(ii) A. nA,=q. for i#j (iii) A, is a countably infinite "V i .
..
These facts lead to the conclu,ion that U
Ai is
countabl~
infinite i.e., N )( N is countably infinite and hence NxN ...... N . This => / NxN /=/ N I Ex. 4. Prove I P (A) 1=2 / A / ;f A is any finite set. Solution. Let A be any finite set with cardinality n. P (A) is the family of all subsets of A. / P (A) / =·co+ ..C'+ ..C3+ ........('" =(1 + 1)"=2"=2 I A / :. I peA) / =2 I A I . 1 heorem 8. (I) If A, is non-enumerable set ¥ .. e: :01.
32
COUNTABILIlY OF SETS
-
then U AI is non-enumerable and hence deduce that ~ /
1=1
(U)
c+c+c+ ...... to a terms=c. If Ai is non-enumerable set for 1 <; i ~ n.
It
U AI
1=1
is non-enumerable and hence deduce that
Proof.
.
c+c+c+ ...... to n terms=c. (i) Let AI be a non-enumerable set ¥ i E N so that
we can write I A, I
=c
¥ i E N.
..
Let A=U A•. 1=1
To prove that A is non-enumerabie. FOr proving this we must show that 1A I=c. 1A, I=c shows that the set A, is numerically equivalent to the set of real numbers in the interval [0, I). That is to say, AI-[O. i) Similarly A2
r- [ } .
Aa"'" [
i.e. A 1 -[0,
I).
A2
1
}+~) I
1
1
-I},}) Aa -
A __ ..
1)
2+22' 2+2i+V
[},
~)
[1- 2,,-1 .1_ ' 1- 2" !.).
If we a~sume that A, n A, =~ for i :;t:.j, then above results taken together imply that
U A • ..., U[1- 2-L ,1-~) 2
).1
or
A-lO, I)
or
1=1
or
1A 1=1 [0, 1
A
J)
\=c
I=c.
If we assume that A, nA,:;i:t/> for i;#:j, then 00
u A. is equivalent with some subset of L0, J). 0-1
33
COUNfABILlTY OF SETS
IU I~ ,-1
Ai
c or 1 A
1
~
C
AICA and 1 AII=c. 1 ~ 1 A, or I A ~ c. 1 A 1 ~ c, 1 A 1 ~ c => c ~ / A / ~ c => I A 1=c. In either case,' [ A I=c. Deduction. Further assume that AinA,=t/> for i#j.
But
c=I Al
We have seen that!u Ai I=c, this => i-I
or
E
iEN
/ Ai /=c
c+c+c+ ...... to a terms=c. (ii) Let Ai be a non-enumerable set for I ~ i ~ n. Then we can write I Ai I=c for I E;;; i ~ n.
.
To prove that U Ai is non-enumerable. ;=1
1 Al I=c implies that Al is cardinally equivalent with the set of real numbers in the semi-open interval [a}, O 2 ) where a10 a2 E R and a1 < a2• This is expressed by writing AI-[Olo 4.!). Here we shall suppose that Or < Or+!. where ar, Or+! E R for r=l, 2,3, ... , n. Thus we have
A~ .......
[0[, O2 ), A2 ,.." [a 2 • 0 3), A3 '" [oa,' a4 ), ... An ,..., [a... 0"+1)' If we assume that· Ai n A,=q, for i#j, then the above state-
ments taken together imply that
This =>
i~l Ai \=1
[01' 0"1-1)
If we assume that Ai
n
..
u
;=1
Ai
-.J
[01' Ontl).·
=c =>, /;91 I=c.
1
Ar;t=~
Ai
for ;;j::j, then
n
U Ai is equivalent with some sub,et of [ai' o"tl) so that i=l
.. (1)
But
Al
" C U 4.=1
Ai
and
1Al I=c.
34
COUNTABILlTY OF SETS
:.
c=
1 Al I
~ li~
By (I) and (2), r
:.
At
I
or c
~ \ ~lAt
... (2)
UAi I=c,
J=1
In either case
I UA, I'==c. t·ol
From this the required result follows. Deduction. (i) Further assume that Ai n As-¢> for I#}.
Then
I UAt I=c => IUAi I=c, ... 1
/==1
=> c+c+ ... +c+ ... to n terms=c, This completes the proof. (ii) To prove that ~+~=at for every finite cardinal a. [Banaras 1971J Solution. Let A and B be two disjoint infinite sets s.t.
I A 1=«-1 B I.
Then A U B is also an infinite set and so I A U B A
n B=~ => I A => III =
U B
+1&.
1=1 A 1+1
B
1=1&
1
Proved. Theorem 9. If afinlte set 0/ elements is added to an enumerable set, the reSUlting set is also enun!rable. Or to prove thlt n+a=l, n being any finite cardinal number. Proof. Let A be an enumerable set. Let B be a finite set with cardinality n. Then 1A I=a, I B I==n. Let A n B =.p. To prove that B u A i'1 enumerable. For this we must show that I B U A / =1 N I.
or
:x
n+a=a Write B={blt b2 , ba.... , bn1 A, being an enumerable set, is expressible in the form of a sequence. So we can write / A={an~l' an+2. anH. an+, ... } Define a map f: N-B U -A by the formula /(r)=l br ~f 1 <;; r <;; n 1Qr If r ~ n Evidently r is one-one onto. Hence N "'" B U A. This..=>_1 N /=/ B u A/=> a=n+a. Proved.
3S
CC::>UNTAB(LITY OF ,SETS
Theorem 10. cardinal number.
Prove n+«=~ 'V n E N,
at
being any infinite
Or If a finite set of elements is added to an infinite set, the power of the set is unaffected. Proof. Let A be any infinite set with power IX. Let C be any finite set with cardinal number n. Let A n C=rP. We have to prove that 1A U C 1=1 A I. For proving this we prove that cx+n=lX. A is an infinite set => 3 B C A s.t. I B I==a. Write A-B=P. Then AU C=B U PuC or A U C=B U CUP. . .. (1) B U C is enumerable, being a finite union of a finite set and an enumerable set. B U C -." N. Now 1B I=a => B,..., N, => N '" B. B U C ,.., N, N '" B ~ B U C '" B. Also, P '" P (by reflexivity). Now, B U C r- Band P '" P. Therefore B U CUP'" B U P which => I B lJ CUP 1= 1B U P 1
=> 1B U CUP 1=1 A I => 1A U C 1=1 A 1 => cx+n=« => n+GI=«.
:
A-B=P
on using (I)
Hence the proposition. Theorem 11. Prove a+«=GI. CIC being any transfinite cardinal number. Or, If an enumerable set is added to an infinite set, the power of the infinite set is unaffected. Proof. Let A be any infinite set with cardinality" so that I A 1=«. Let A n N=.p. We have to prove that I A U N 1=1 A I. If we show that «+a=2, ,the result will follow. A is an infinite set => 3 B C A s.t. 1 B 1=a .. We have A U N=(A-B) U B U N=(A-B) U (B U N) or A U N=(A-B) U (B UN). . .. (1) BUN, being a finite union of countably infinite sets, is countably infinite. :. BUN'""" N. But B -- N. .: B is enumerable. The relation M ,..., N. where M and N are any two sets, is an equivalence relation in the family of sets.
36
COUNT ABILIfV OF SETS
:. B --' N => N -." B BUN ,.., N, N I"W B => BUN - B. Now BUN - Band A-B - A-B. Combi ning these tw:> we have, (A--B) U (B U N) ,.... (A-B) U B Using (1), we get A UN,.. , A. This => I A U N 1==1 A ,.
(by symme try) (by transiti vity) (by reflexivity)
=> lZ+a=~. Q.E.D. Ex. 5. If we subtract an enumerable set from a non-enumer-
able so:t, then the remaining set is non-enumerable. [Agra 19661 Proof. Let A be a non-en umerab le set. Let B be enumerab1e set. Let A-- B=P. We claim P is non-en umerab le. Suppos e the contrar y. Then P is an enume rable set. Moreo ver A =:B U P. A, b~ing a finite union of counta ble infinite sets, is an cnum~rable set_ A co:ttra.diction. For A is nonenum~rable. Hence the require d result follows . Ex. 6. If an enumerable set is subtracted from all enumerable set, the rema;n;ng set .will be eltllmerab Ie. Proof. Let A a'1d B b:>th be enume rable sets. To prove that A - B is enum;! rable. Let A -B=P . We have to prove that Pis counta bly infinite. Supp0'ie the contrar y. Then P i'i non-en umerab le. A=B U P. A is non-enumerable, being a finite union ,of a non-en umerab le and an ~numerable set. A contrad iction. For A is enumerable. Hence the require d result follows. Ex. 7. If (I. ;s any transfinite card;n~l nu-nber, then a Proof. Let A h~ any infillite set with cardina lity It, i.e., 1 A I ~7a:. To prove thrtt a <; '%. A is infinite set => 3 B cAs t. l B I=a B C A => I B I ~ 1 A I => a ~ at. Ex. 8. Prove that the set of all rationals is countable. (Punjab 1969; Kurukshetra 69; Kanpur 68) Proof. A rationa l numbe r is real numb~r expressible in the form ~ where m and n are integers and n:;t:O. n We claim Q is counta ble.
COUNfABILITY OF SETS
37
Define a map I: Q+-+N X N by the formula
1(: )=(m, n) where m and n are positive integers prime to each other. Evidently
1 is one-one. Q+ is equipollent with some subset of N x N. 'Q+ I ~ I NxN I = I N I =a I Q+ I ~ a. . .. (1) Since N C Q+, :. I N I ~ I Q+ I, or 8 ~ I Q+ I ... (2) From (1) and (2), 8 ~ I Q+ I ~ a, \\hich ~ I Q+ I =a, Hence
or
Q+ is cardinally equivalent to Q- under the map T!_ -+~~ q
q
:. I Q+ I = I Q- I =8 i Q I = I Q+ U Q- u to} I =8+n+l =(a+a)+1 :. associative law holds. =a+l=a :. I Q I =8, which:::> Q is enumerable and hence countable.
Ix. 9.
Prove that the et 01 integers is countable. Proof. To prove that Z is countable, it is enough to prove that I Z I =8.
Write
Z+={l, 2, 3, ...... } Z-={-I,-2,-·3, ... }. Then Z= Z+ U Z- U {O}. Also Z+, Z-, to} arc disjoint sets :. I Z I = I Z+ i + I Z- I + : to} I = I Z+ I + I Z- I + I to} I Z+ N under the mapping n -+ n Z- ,.... N under the mapping -n -- n so that I Z+ I = I N I , I Z- I = I N I I Z+ I = I Z- I = I N I =8. or Substituting these values in (I), we get I Z i =8+8+1=(a+a)+I=a+:=8.
... (l)
"oJ
:. : Z I =a. Ex. to. Prove thai a < c. Solution. To prove that 8 < c. We know I N 1 - a, R i =c, NCR. NCR;} ! N I ~ R i :) 8
. .. (2)
Q.E.D.
I
~ C.
38
COUNTABt-LlTY OF SETS
N is not cardinally equivalent to R under any mappi·ng. :. a:;t:c. Now a:;t:c, a ~ c ~ a < c Theorem 12. (Cantor's theorem) : To prove that I A I A,.., peA) so that 3 one-one map onto
I:
A
---l-
P (A)
Take B=={x E A : x ff. f (x)} x E A .. I (x) E P (A), by (3) .
.. f
... (3) ... (4)
(x) C A, since P (A) is the family of all subsets of A.
Also B C A, according to (4). Thus, Be A,/(x) C A for any x E A. This . 3 CIt E A. s.t.f(II.)=B. CIt e: A. => 11. e: B or tx ft. B. Consider the case in which 11. e: B. . .. (5) Now, /It e: A. ell e: B => ell ft. I (a.), according to (4) . • ell ft. B, Contrary to (5). .: I (tx)=B. :. The possibility Ie e: B is ruled out. Consider the second possibility in which tx fF. B. .. .(6) Now, • e: A. => ell ft. B ... II ¢ I (tx) .: I (II)-B. . . II e: B according to (4). Contrary to (6). :. The possibility CIt (/:. B is also ruled out.
COUIIITABILITY OF SETS
39
It amounts to saying ~ E A does not imply at E B or at B. Again we get a contradiction. It means that our initial assumption is wrong. This means that (2) is true. Remark. If I A I =n, (finite cardinal number) then the last theorem implies that n < 2n.
rt.
[Kanpur 1910; Banaras 69] ":
I
P (A)
I
=2 1 A '=2 n •
Ex. 11. Let X be any set. Let A (X) be the family of all the cizaracteristic fUllctions of X. 7 hell prove that the power set of X is equil'alent to A (X). . Solution. Let A (X) denotc the family of all the characteristic functions of a sct X. To prove that A (X) f"oJ P (X). Let B C X be arbitrary. Let X8 denote the characteristic function of B relative to X. Then X B: X ... {I, O} s.t. X B (x)=
JI if x E B
to if x
(/:. B
Evidently X B is an arbitrary element of A (X) and B is an arbitrary element of P (X). Defined a mapf: A (X) - P (X) by requiring thatf(XB)=B. This shows that f maps the characteristic func~ion of B relative to X into the set B. Evidently f is onc-one onto. Ex. 12.
A (X) '" P (X). Q.E.D. Prove that the set of algebraic numbers is countable. [Punjab 65; Kanpur 72; Bhagalpur 68]
Proof.
First we shall show that the set of polynomials P (x)=aO+a1x+a2 x 2 + '" +amxm with integral coefficients is enumerable. Let m be a fixed positive integer. Let Pmn qenote a polynomial of degree m in which I ao I + I a1 I a2 I + I + ... + I am I =n, ... (1) where n is any positive integer. Then Pm2 denotes a polynomial of degree m with coefficients s.t. I 00 I + I a1 I I a2 I +... + I am I = 2.
+,
40
COUNTABILITY OF SETS
Write Pm= U {Pm,. : n EN}. Then P. is a countable union of polynomials Pm. of fixed degree m in which the coefficients are subjected to the condition (1). Since N is enumerable and therefore Pm is enumerable. Write
P=u
{Pm: mEN}
P is enumerable, being a countable union of enumerable sets. Evidently P is the set of aU polynomials with integral coefficients. Secondly we shall show that the sct of algebraic numbers is countably infinite. Write
00
P'= U {p" (x) : p,. (x)=O} II-I
'
where P.. (x) stands for AJlQiynomial of degree n with in,tegral coefficients. From what has been done it follows that P' is enumerable. An algebraic number is defined as the root a polynomial P (x)=O with integral coefficients. Take A,,={x : x is a, solution of p,. (x)=O}. The A" is finite. SilWe a polynomial of degree n has at most n roots. A= U {Aa : EN}. Define Evidently A is the set of algebraic numbers. A being a countable union of countable sets, is countable. Since A is not finite and hence A is enumerable. This completes the proof Ex. 13. To prove that the set 0.( a/l irrationals in any interval is non~enumerable. [Raj. 1965; Agra M.Sc. Statistics 65J ,Proof. We know that the set of aU real numbers (i.e. the set of all rationals and irrationals) in any interval is uncountable. We also know that the set of all rationals in .any interval is enumerable. Then the complement of the set of rationals relative to the set of real numbers in any interval is uncountable, i.e., the set of irrationals in that interval is uncountable. For complement of an enumerable set w.r.t. a non-enumerable set is non-enumerable. Ex, 14. Show that the set of a/l transcendental numbl?rs ill alH' interval is non-enumera~/e. . (Punjab 1965; Agra \i.Sc, Statistics 65J
41
COUNTABlLln OF SETS
Proof. We know that the set of all real numbers (i.e. the set of all algebraic numbers and transcendental numbers) in any in· terval is non·enumerable. We also know that the set of all algebraic numbers in any interval is enumerable. Then the complement of the set of all algebraic numbers in any interval relative to the sct of aU real numbers in that interval is uncountable, i.e., the set of alI transccndentals ·in that interval is uncountable. For the complement of enumerable set reia'jve to a non-enumerable set is non-enumerable. Ex. 15. Two enumerable sets are equivalent. Proof. Let A and B ~e both enumerable sets.
To prove that By hypothesis,
A -- B. A r-- N, B "-' N.
Since the relation A """"' B is an equivalence relation in the family of sets. (by symmetry) B -- N ~ N ""'"' B. By transitivity A,... N, N '""" B => A -- B. Theorem 11. Prove that C.c=c. fBanaras 1967] Proof. Let A be the set of all real numbers in the closed interval [0, 1]. In symbol A=[O, IJ. Then I A I =c. To prove that c.c=c, it is enoguh to show that
IAXA/=IAI· Write B={(O, x) : x E A}. Define a map I: A---'>- B by the requirement
1 (x) = (0,
..,. x E A
x)
Evidently 1 is one-one onto so that A -- B. Evidently
:.
I A i '-= / B I B c A X A. IB I ~ IAXA I I A I ~ I AxA I
.
... (1)
... (2), on using (I> Let x, yEA be arbitrary. Then x and y can be uniquelY written as x=0·X1X S X3 ••• y=O·yly2Y.··.··.··. in the form of infinite decimals which do not end wit:. zer~. That is to say, this infillite decimal contains non-7ero digit'>. We write i=0·4999, .... =049 instead of t=0'500
or
42
COUNTABILITY OF SETS·
Define a map /1
: A x A~A by
the formula
It (x, Y)=O·X 1Y\X2Y2XaYs .. ·· .. Evidently /1 is one-One. This => lAxAI~IAI
Combining this with (2), we get IA:~[AxAI~IA[
or
IAxA!=IAj.
Find the power 0/ an aggregate
Ex. 16. M
2m
'
0/ numbers
M and m being positive and integral.
Solutio~.
Let
A={~
Q.E.D. gil'en by
[Agra 1964J
: M, mEN}
To determined the power of A.
AM={~
Write
:mE
N}
yo
ME
N
Elements of AM can be displayed as follows:
Al : ~
1
}a ... }........
2 A2 : 2i
2
2
3 A3 : 2i
.2i
22
23
3
3
28
... 22ft ...... ."
n
23
Obviously (i) AT is enumerable (ii) Ar n Ar'=r/J
...
¥ ¥
3
2ft ...... n
...
2ft ....
rE N. r, r' E N s.t. r-:pr',
(iii) A= U A r • r--l
Being an enumerable union of enumerable sets, A is enumerable and hence its cardinal number is a i.e., the power of the given set is a. Ans. Ex. 17. Show that the set 0/ points in the closed interval [2, 4] and in the open-interval (J, 2) are cardinali} equivalent. [Banaras 1970J Solution. Consider the map
---r
/: [2, 41- [I, 2] s.t. f(x) = ( ,:-2 ) + 1.
f is continuous => / is onto.
43
COUNTABJLlTY OF SETS
I'.() f( Xl) =J~X2
::>
~12-~+I-_X22-·2+1
=> X 1 =X 2 •
This says that! is one-one. Thus f is one-one onto. Hence [2,41 '" [I, 2J. This . .. (1 ) => I r2, 4J I = I [l, 2J I . ... (2) But I [1, 2J I = I (1,2) I . Combining (I)' and (2), we get the required result. Theorem 12. (Dedekind~peirce). El'er)" infinite set is equivalent to its proper subset. • or Every infinite set can be put into a one-one correspondence wilh a proper subset of itself. Proof. Let S be an infinite set. Then S contains a denumerable subset, say, Write S*=S-{al, a z, ... } and Sl=S-{a1} Then SI is a proper subset of S. Also Sand S1 both are infinite sets. Define a map f S-+SI S.t. f (a,.)=a'+ 1 for n= I, 2, 3, ... and /(s)=s ¥ s E S*. Evidently / is one-one onto map Hence S is equivalent to SI' It follows that S is equivalent to its proper subset. Note. This fact was used by Dedekind to define infinite sets as follows. A set is iJ~rinite iff it call be put hlto a one-one correspondence with a proper subset of itself. Theorem 13. (GaliJeo's Paradox). Any denumerable set can be put into a one-olle correspondence with a proper subset of itself.
Proof. Let A be a denumerable set. Then A is expressible as A={at. a3 , aa,· .. }. Write B=A-{al}={a2 , aa, ... } Define a map f : A-+B s.t. f(a,.)=o"+1 forn=l, 2, 3, ... Evidently f is one-one and onto map. Also B is a proper subset of A. Hence the theorem.
44
COUNTABILl1Y OF SETS
Theorem 14. Every nOll-empty open set 011 real line is the countable union of nan-empty disjoint open intervals. [Meerut 1988,86] Or A bounded set of lion-overlapping intervals is countable. Proof. (i) Let G C R be an open set and x E G be arbitrary. Then 3 an open interval I", with centre x such that x E I., C G. Let C. denote union of all such open intervals. Then C.,= u tI",CG}. Evidently COl is the largest open interval containing x such that x E C., C G. Moreover, each x EGis contained in /'" and each / .. is contained in G. The suggests that G= U C"'. We have either CIII n C,=t/> or C., n CY=l=~, ¥ x, y E G. Consider the case CIII n Cy::j::.cp. Then C", U Cy is an open interval containing both x and y such that C.. C (CIII U Cy) C G, C, C (C", u C,) C G. By definition of Ce; and C y , we have (Co: u Cy) c C.. ' (CIII U Cy) C Cy. Hence C",=C~ U Cy and Cy=CIII U ell, i.e. C.. =Cy Thus, '" x, y E G, we have either Cm=C, or CIII n C,=r/I, Also G=U Cm• This declares tl-at G is the union of pair-wise disjoint nonempty intervals. It remains to prove that the intervals C", form a countable collection. (m Let G={G,.} be a collection of open subsets of R such that (i) G,.:/ t/J. 'V n, Oi) G.. n Gm=t/> for Il ,em. We know that each G •• contains an infinite number of rational and irrational numbers. Let K=lK,,}, where K" is the set of all rational numb;!rs in Gn. D::fine a map f: K - G such thatf(k,,)=G n • . .since for each k .. E K, there is a unique open set G.. E G, md conversely. :. f is one-one onto. Therefore K -- G. But K considered as the set of rational numbers U KA , is .::ountable. Since K C Q. Hence G is countable. Theorem 15. ~r {G,.} is all)' cdlectioll ~ non-empty disjoinl open sets ofR, then prOl'e that (G,,} is c(Juntable. {Kanpur 1979, 771 Proof. For proof, sec (ii) part (Of '[ h.:orcm 14 Problem 18. SIIolI' ,IIat a cOllllfahle set is a Boret set.
COUNTABILITY OF SETS
45
Solution. Let A be a countable set so that A is expressible as A={a1 , a2 , a~, ... } {ar}={x: x=ar }
= U ~ x : ar ~ x < an + ~}
"=1 ( n =Countable inters!?Clion of closed sets. {or} Also A= U r EN These facts prove that A can be obtained by the formation of countable union and intersection of closed sets and open sets and hence A is a Borel set. 5'10. Continum Hypothesis. It states that there exists no set having its cardinal number between a and c. But this hypothesis does not assert that there exists no set with cardinal numb.;r greater than c. In this connection Cantor had proved that the pow.:!r set peA) of any set A has cardinal number greater than the cardinality of A. Consequently the cardinality of P (R) is greater than c.
6 Measure And Outer Measure 6'0. Def. Boolean ring. (or ring of sets). [Gujarat 70] A non-empty family A of subsets of X is called a ring 'of sets or B-ring if any A, B E A ~ A-B E A, A U BE A. That is to say, ring of sets is closed under the formation of unions and differences. Remark. (1) A n B=.A.-(A-B), A t:. B=(A--B) U (B - A), where A t:. B stands for the symmetric difference of A and B. These equations dec/are that A n B E A, A t:. B E A if A is a ring of sets. (2) ", E A. <·For .=A~A. (d. DeC. (1-ring of sets. [Gujarat 70J A non-empty family A of subsets of X is called a (1- ring of sets if it is closed under the formation of differences and countab'e unions, i.e., if A, B E A ~ A - B E A 00
and
A, E A
~
U Ai E A. ~~1
Remark.
00
""
eo
n Ai=A- U (A-A.) where A= U Ai. i=1
,=1
i=1
This equality says that (1- ring is also closed under the formation of countable intersection. Examples of ring. (i) P (X), the power set of X, is a-ring where X is any set. ring.
(ii) If X is any set, then class of all finite subsets of X is a This class is (1- ring if X is finite.
47
MEASURE AND OUTER MEASURE
62.' Def. Algebra of sets (or Boolean Algebra or Field). A family A of subsets of X is called an algebra of sets if (I) A E A => A' E A where A'=X-A (2) A, B E A .. A U B E A. 6'3. Def. a-algebra of sets (or a-field). A non-empty family A of subsets of X is called a-algebra of sets or a-Boolean Algebra of Borel Field if (i) A E A => A' E A GO
(ii) Ai E A =>
U At E A.
Remark. (1). Each algebra (a-algebra) A is a ring (a-ring) because if A and B are in A, then A-BEA, since A--B=(A'u B)'. (2) A ring A of subsets of X is an algebra in X iff X E A. For if A is an algebra, then X = A' U A where A E A so that X E A. Conversely if X E A, then A' E A ¥ A E A. For A'=X -A so that A is an algebra.
6'4. Der. Semi-ring. A family A of subsets of X is called a semi-ring if (1) tP E A and (2) A, B E A=> A-B E A (3) AI E A S.t A, n A1=f# for i-:f::j .. 3 A E A GO
6·S. Det. Monotone class. A non-empty class of sets A is called a mO:lOtone class if for every monotonic sequence < AI > of sets in A, we have lim An E A. Examples If A.,. C A.,. pi, then is an increasing sequence. 00
For increasing sequellce, we write
t >. If A= u
n-l
we denote it as An t A. If An+-! C An, then
<
An
>
An, then
is decreasing
GO
sequence and we write <;A,.
~
> and if n An=A, then we write ft=J
An.j, A If M is a monotone class, then AE M for every sequence
in M. Any a-ring is a monoton~ class. A monotone class is a a-ring iff it is a ring. , 6'6. DefinItion. Complete lattice. The set R.=R U {-co, ~}=[-oo, co]
48
MEASURE AND OUTER MEASURE
is called Extended set of real numbers. Briefly R. is called complete lattice. 6'7. Definition. Set Function. Refer to D!finition (3'4), Chapter 3, on p~ge 13. 6·S. Definition. Extended real valued set function. Refer to Definition (3'6), Chapter 3, on page 14. 6'9. Definition A fU!1ction (set function) is said to be finite function (finite set function) if all its functional values are finite. Def. Additive Class. [Kanpur Statistics 75] Let Q be a space and A be a cla~s of subsets of ~:I. A is said to be finitely additive class of sets if (i) cf, E A (ii) A, B E: A ~ A-B E: A, A U BE: A. This definition is parallel to the definition of ring of sets. Hence we can say that every ring of sets is finitely ajditive class and vice-versa. Def. Completely Additive Cla..s. (Kmpur Statistics 77, 75) Let Q be a space. The class A is said to be a complete ly additive class of sets if (i) r/J E A (ii) A E A '* A' E: A (iii) If is any sequence in A, then U An E: A. All these condition<; prove that Q E A. Hence every completely additive class might b~ a field or a-field and vice-versa. DeC. Completely Additive Set Function. [Kanpur Statistics 77, 75] An extended real valued set function I-' is said to be completely additiv~ set function if I-' : A....,. [ - 'X> , <Xl] s. t. (i) A is completely additive class. (ii) I-' (cp)=O. (iii) Ther\.. e; of disjoint sets in A S.t. I-' (
~l Ai
)= igl
I-' (Ai).
Def. Continuou.. Function. Let J.I. be a measure function on the ring A. Then J.I. is additive, positive and monotone. Now if < E" > is an incre:lsing sequence, then is also increasing sequence. Moreov!r if E" l E and En, £ E: A, and ",(£,,) t tl-{E).
49
MEASURE AND OUTER MEASURE
We express this condition by using that ~ is continuous frell! below. Similarly ~ is continuous from above if En ~ E. "10. Definition. Let X be a set. Let A be a a-ring of seu which are subsets of X• .i.et m be an extended real valued set function defined on tke g-ring A. (1) m is called non-negative if m (A) ;;;;, 0 V E A. (2) m is called additive if ¥ A, B E A s.t. A n B=1> => m (A U B)=m (A)+m (B). m is called finitely additive if A. E A for i=l, 2, ... , n S.t. A. n AI=1> for i=l=j =>
(3)
m( (4)
UA,
1=1
)= i:
1=1
m (A,)
m is called Countably additive if
A, E A ¥ i E N s.t. A,nAI=1> for i:;f;j (5)
~ m( UA,)=;m (A,) i=l
i=l
m is monotone if ¥ A, B E A s.t. A C B => m (A) EO;; m (B)
m is subtractive if A, BEA s.t. A :J B, m(A) is finite => m(A-B)=m(A)-m(B) (7) In is Countably subadditive if
(6)
A, E A ViE N =>
In (
'~1 A.) ~ '~l m (Ai)
m is translation invariant if A E A, x EX=> m (A+x)=m (A), where A+x={y+x : yEA} (9) m is regular if AEA, E > 0 => 3 G, FE A S.t. m (G)=E~m(A) ~ m(F)+" 6'11. Postulates for an ideal measure function. A measure is an extended real valued set function m defined on g·ring A of subsets of X s.t. (I) meA) ~ 0 ¥ A E A. In particular m (1))=0. (2) A, B E A s.t. A C B ~ m (A) ~ m (B) (3) A, B E A s.t. A n B=1> => meA U B)=m(A)+m(B). (4) For an intervall, m(l)=l (/)=length of the intervall. (8)
so
MEASURE
(5) If
OT,!!.ER MEASURB
is a sequence of pai rwi se d i sjoi n t sets in A, then
m
(6) m
AND
(u ) i=1
A, =
E '=1
m (A,)
is translation invariant, i.e.,
where
-
m (A+x)=m (A) A+x={y+x: yEA}.
Definition.
6'12.
Measurable space.
y.
x€X,
[Kanpur 1969]
subsets of X and m is m eas ure on A, then the tripl e (X, A, m) is called a measurable sp ace. If A is
a
algebra of
If m assumes only the values 0 and 00, then the m ea sure space (X, A, m) is called degenerate. Any set A E A is called a measurable set ( relative to the
m).
measure
Definition. Let
6'13.
m be a measure function on a a-ring A
of subsets of X. (1) Completeness
of me�sure function. Th e measure function is said to be complete if B e A s.t. m(B)=O, A C B =- A � A i.e., if all the subsets of a set of me asu re zero are measurable. In this case (X, A, m) is call'Jd a comple te measure space. m
(2)
An y set A€ A i9 said to have finite measure if m(A)< IX..,
(3)
The measure of any set A e A is said to be C7-finite if n EN> s.t.
3 a sequence < A" E R: 00
,
(ii)
m (A,,)
(i)
A C U A..
(4)
The measure fu nct io n
(5)
The measure function
11=1
< 00 \j n e N.
m is cal led finite or a-finite accor
ding as the measure of every set is finite or a-finite. a-finite)
(i) (ii)
m
is called totaUy finite (totally
if X e A i.e. A is algebra of sets. m(X) is finit,e, (�r a-finite).
6''(4. Herid �ta.ry froperty. A n:1n-empty class S={Si} of sets is said to. have. heriditary if SI C S,. S� e S => SI E S. 6'15.
Caratbeedory's postulate for outer measure. [Kolbapur 70; Banaras 66,65] o
Let an extended real valued set fu nction mo defined on 'U
51
MERSURE AND OU fER MEASURE
heriditarya-ring A={A.} have the following properties: (i) mo(A,) ~ 0 and mo(Ar)=O if A,=p (ii) Ar C A. <=> mo(A,) ~ mo(A.) (iii) mo (u Ar) 1:mo(Ar) r
(iv) mo (A, u A.)=mo (A,)+mo (A.) provided A, n A,=rfo. These properties are called Caratheodory postulates for outer measure.
6'16.
Definition.
Measurable set. [Banaras 71; Poona 70; Kolhapur 70] Let m~ denote an outer measure function defined on a heriditary a --ring A={Ai}. Any set A E A is said to bc measurable w.r.t. the given outer measure function mo if m~(T)=mo(A n T)+m.(A' n T) V TeA or equivalently, mo(T) ;;;?; mo(A n n+miA' n T). If we say that l.I. set A is measurable, then we shall write mo(A) =m(A). Remark. FO'r any set E, E=A E+A' E. In accordance with Definition (6' 15), this gives nlo(E)=mo(A n E+A' n E). <: mo (A n E) +mo(A' n E). 1I10(E) ~ t11 o(A n E)+mo(A' n E).
n
i.e.
n
To prove mo(E)'-~mo(A
() E)+mo(A'
n
E),
it is enough to provc that 1I1o{E)?-lIl o(A
n E)+mo(A' n E).
6'17. Elementary set. A plane set is said to be an elementary set if it is expressible as a finite union of pairwise disjoiat rectangles. Problems related to measure functions. Theorem I. A a -additive set function is a continuous set function. [Kanpur Statistics 77J Proof. Let p. be a a -additive set function be any sequence of sets in the class A on which the a-additive set function is defined, where A is an algebra. (i) Let be an increasing sequence in the class A so that Al C A2 C A3 C ......
52
MEASURE AND OUTER MIlASURE
-
Let B1 =A1 , Dn=A .. -An_1 for n
Then U An E A. ,,=1
_
>
1.
00
Then D. n Dm=rP for n#m and U D,,= U An, _1
Ak =
,,=1
k
~
n
Dn and so I-' (A k )= E I-' (Dn)
w=1
"=1
... (1)
or
I-' (lim An) = Lim
n .. oo
I-'(A,,)
This => II- is continuous from below. (ii) Let be a descreasing sequence, then At :::> A2 ::> A3 :::> A, :l...... Then AI-An C A t -An+1 and A 1 -
co
co
.. =1
.. =1
U· A,.= U (At-An) .
Hence, by part (i), I-' (
Al-n~l An
)= =
"~1 (A
I-' [
Lim n~oo
=p.{A1 )
or or or
"'(AI)
--p.( U An n=1
I-' (
U An
,,=1
I-' Lim
n-oo
1 ·- An) ]
I-'{AI-An), by (1) ...
Lim Il-+oo
)=I,(A1 ) - Lim
n-+oo
)=
Lim
n-+ 00
(,1)= Lim nn
n _;
00
I-'(A..) I-'(An)
p.(An)
I-'
(A ). ..
This ~ II- is continuous from above. Hence po is a continuous function.
S:f
MEASURE AND OUTBR MEASURE
Theorem 2. Let X be a space 0/ at least two points and X. For each A C X, we define A (A)=JO ~f ~ 11 If xoeA then prove that ~ is an outer meQsure. [Kanpur Statistics 1976] Proof. Let A, Be X be arbitrary. Then (i) ,...(A);il O. This follows from the definition of p. In particular ~(.p)=0 as Xo l/:. ~ Q' ",(.p)=0. (ii) A C 8 => ",(A) ~ ",(8). For A C B => (l).xo EB if Xu e A (2).xo ft B if x. ft A ... I'(B)=l=,...(A) if Xo E A & 1'(8)=0=I'(A) if Xu ft. A => I'(A)=I'(11) if Xo E A or A. Again if x. e: B, then Xo e: A or Xo ,. A. Hence if Xu e: B, then ",(A)= 1 or O. This ... ",(A) ~ I'(E). For 0 < 1. Finally A C B ~ I'(A) ~ p.(E). (ii) To prove ",(A U 8)=I'(B)+I'(B) if A n B=.p. We have Xu ¢ A U R or Xo E:: A U B x. ,. A U B => Xu ~ A and x. ~ B. Conscquenrly I'(A U B)=O => ",(A)=O=",(.8). :. ,..(A U 8)=I>'(A)+I'(B). In the second case, Xu e A U .B => Xo E A and Xo ¢ B or Xu e: B and Xu ¢. A,for A n B=rf.. :. I'(A U B)=1 => ,.(A)= 1,1'(8)=0 or I'(B) = I, 1'(04)=0• . This => I'(A u B)=,...{A)+I'(B). Hence in both cases, (iii) is proved {iv) To prove p.( U A .. ) C;;; E 1'(14,,), where A" C X.
Xo
e
x,,.
x.,.
"
"
We have x. ¢. U A. or Xo E U A.. If Xo ~ U A", then x. ~ An ¥ n :. If 1'( U A.. )=O, then ,,(...4,,)=0 ¥ II so that
..
i: ~(A,,)=O+O+O+ ... +0=0 :.
p,(U A .. )=1: ",(A,,). "
II
In the second case if Xo E U A", then x. one value of If or Xo E A" V-
ft.
e
A
n.
for at least
MEASURE AND OUTER MEASURE
i.e.,
p{A
no
)= 1
or p{An)= 1 ¥ n.
Hence if p( U An)= I, then
.
E p{An )=O+1+0+ ... +0=1 n 00
l.: (J.(An) = 1 +l-f-} + ...
or
+ 1=00.
"""I
This or
~ p
(U An)=E P{An) p (U An)<E p.(An) ~ p(U An)~Ep(An)'
From (i), (ii), (iii) and (iv), it follows that p is an outer measure function. Theorem 3. An outer Ineasure is monotonic and a - sub additive. [Kanpur M.Sc. Statistics 1976; Banaras 69] Proof. Let A be a a --ring of sets which are subsets of X. Let (X. A, mol be a (outer) measure space. (i) To prove that mo is monotonic. For this we have to show that lno(A) IE;; lno(B) ~ mo(C) ::;:;; me(D) ~ ...... if A C Be C cD ..... . Consider the case in which A C B. Notice that B=(B -A) U A, (B-A) n A-t/l. This::> mo (B)=lno (B-A)+mo (A) ~ mo (A). For Ina (B-A) ;;;;;, 0 ::> mo (B) ~ mo (A) ~ Ino (A) ~ Ino (B). Similarly B C C, C C D . ::> Ino (B) ~ mo (C). mo (C) ~ Ino (D).
Combining these results mo (A) ~ 111" (8) ~ mo (C) if
E; Ino (D)
<;; ......
ACBCCCDC ...... (ii) To prove that 1110 is a --subadditive. For this we have to show that
mo
(U
1=1
AI) :s:;;
.E mo (A.), where <
.=1
All
>
is a sequence of sets in A.
Write
Bn=An-
8-1
U i=l
Then
A,
55
MEASURE AND OUTER MEASURE 00
00
This shows that => (1) U E,= U At 1=1
i=l
n BI=~ for i~j.
(2) E, (3) B.
(2)
=> mo
(,~
=> mo
(U A,)== 'f
c
Et)= ,~ mo (D,) 1110
(E,) by (I)
~}; 1110
(A,) by (3)
i_I
t=l 00
,=1
mo
A.. ¥ n.
(u AI) ~ rr: • --1
1=1
mo (A,) .
Theorem 4.
If T be a subset of the set of real numbers s.l. and if p.(A)=m. (A n T) 'V A C R, then show that p. is an outer measure (Caratheodory's outer measure).
0< ma (T) < Proof.
00
Let A, B, C, T C R be arbitrary, but Tis fixed and
o< Also we define
Ii
mo (T)
<
00.
real valued set function p.(A)=mo (A
n
I'
as follows:
T).
To prove that p. is outer measure. (1)
p.(A)
For mo (A (2) For
~
n
O. T);;' 0 .. I-'(A)=mo (A * p.(Al ~ O.
n T) ;;. 0
f£(~)=O.
A.=~ ... A () T=rp'" Ino (A n T)="1o (rp)=O ~ I-'(A)=mo (A n T):o=O ~ p.(A):za:O.
(3) ACE * peA) ~ p.(B) For ACE ... A nrc B n T * mo (A T) ~ mo CE T) c> p.(A) ~ p.(B). (4) I'(A U B)=p.(A)+p.(B) if A n E=rp For A n B= if> ~ (A n T) n (B n T)=,p mo [(A n T) U(B n T)]==mo (A n T>+mo (8 Q p.(A)+p.(B)=mo [(A U D) n TJ=p{A CJ B)
n
*
c> I'(A U B)=p(A)+I.L(B)
n
n
'f)
56
MEASURE AND OU'fER
(5)
MEA~U1U!
t,) <; E I'(An)
1'( U A "
/I
For I'(U An)=mo [(U An)
n
T)J=lIlo [U (Au
" "
~
..•
n
T)]
n
X mo (A
"
n T)=
X I'(An) ..
1'( U An} E;; E p(A..)•
"
"
From (1), (2), (3), (4) and (5), it follows that I' is outer measure. Problems related measurable sets. Theorem 5. A set is outer measurable iff A' is outer measurable. Proof. Let a set A be outer measurable so that for any set T, mo (T)=mo (T n A)+mo (T n A') =mo (T () A')+mo (T n A) =mo (T n A')+mo [T n (A')']. This proves that A' is outer measurable. Conversely suppose that A' is outer measurable so that for any set T,
mo (T)=mo (t =mo tT -mo (T
n A')+mo [T n
n n
(A')'] A')+mo (T n A) A)+mu (T n A').
This .. A is outer measurable.
Proved.
Theorem 6. The union of two outer (mo) measurable sets is outer measurable. [Meerut 1972; Banaras 65J Proof. Let SI, S2 be any two outer measurable sets. To prove that SI U Sa is outer measurable .. Taking T as any set and applying the criterian of measurabi. iity of SI mo (T)=m o (T n Sl)+mO (T n SI')· ... (1) Taking T n SI' for the set T and applying the criterian of neasurability of S., we get mo (T n Sl'),=mO (T n SI' n S2)+mO (T n SI' n S~'). Using the fact that (SI n S~)' -SI' U S~'. we get
mo (T
n S/)=rna (T n S/ n S2)+m
O
IT
n (St
U S3}·J. .. (2)
MEASURE AND OUTER MEASURE
Lastly taking T n (SI u S2) for the set T and applying the criterian of IlIeaslilrability to SI. we get n70 [Tn (SIU S2)]=m O [Tn(SIU S~) ns\] +1»0
[Tn(Sl uSII)
ns1']
Making use of the fact that (SI U S2) n 8 1 =SI (SI US2)nS1 ' =(SI n S l')U (S2nSl') =¢U(S2 nSl')
-S2 nSl' we have mo [Tn(SI U S2)]=n70 (TnS1)+m O (TnS2 nS1') using (2), we get mo [Tn (SI u S2)]=mO(Tn SI)+mO (Tn SI')]
-mo [Tn(SlUSlI)'] using (1), we get mo [Tn(SlUS2 )]=mO (T)-m o [TU(SIU S2)'] i.e .• mo [Tn(SIUS~)]+mo [Tn (SlU S2),J :::atno{T). This proves that SI U S2 is outer measurable. Proved. Problem 1. The intersection oj two outer measurable sets is outer measurable. Solation. Let AI' A2 be any two outer measurable sets. To prove that Al nA 2 is outer measurable. AI' A2 are outer measurable sets. =;> A/, A2 ' arc outer measurable sets. => AI' U A/ is an outer measurable set. ~ (AI' U A 2 ')' is an outer measurable set. ~ (AI')' n (All') is outer measurable. => Al n A~ is outer measurable. Hence the result. Probl(m 2. ~r SI' S2 are our;'/, measurable sets and S2 CS1' thell show I haf SI- S2 is outer measurable set. Solution. Let SI and S2 b~ outer measurable sets and let S2 CSI
To prove that SI-S2 is outer measurable. S1> S2 are outer measurable sets => SI. S2' are outer measurable => $, ns/ is outer measurable ... (*) Sln..s~'=Sln(X -SJ =SlnX- SlnS2=Sl-SZ
•
58
MEASURE AND OUTER MEASURB
:. SI- S2 is outer measurable, by (:It). Problem 3. If SI and S2 are measurable sets and if S2 n81 -cp, then
m (S1US2)=m (SI)+m (S2)'
Solution.
Let S10 S2 be measurable sets so that mo (Sl)=m (SI)' mo (S2)=m (S2)' and mo (S) US2 )=m (SIUSa), [Prove this as in Th. S] Also suppose that SI n Sa=CP' To prove that m (SluSS)=m (81)+m (Sa)' By the postulates of Caratheodory outer measure, mo (SI U 8.)=mo (SI) +mo (Sa), In accordance with initial assumption, this takes the form m (SlU S2)=m (Sl)+m (Sa). Alternative method. Taking S1US2 for the set Tand applying the criterian of measurability to S1> we get mo (SIU Sa)=mo [(SIU Sa) nSl ]+mO [SI'() (SIU S2)] =mO(SI)+mO [(SI' nSl )u (St' nSz)] =mO(SI)+m O [cpU S2n(¥-Sl)] =m o (Sl)+mO (S2) j For Sl()Sa=rp -> S2eX-SI =mo lSl)+mO (S2) 1 => S2n(X - SI)==8. In accordance with initial assumption this takes the form m (SlUS2)'=nI (SJ+m ('sa). Theorem 7. If the outer measure of a set is zero, then the set is measurable. [Banaras 1971] Proof. Let A be any set such that mo (A)=O To prove that A is outer measurable. For any set T, T=AnT+A'nT.
By Caratheodory postulates for outer measure,
n
or
mo (T)=mo (An T,A' T) t;;; mo ~AnT)+mo (TUA') nlo (T) E>;; mo (A T) +mo (TnA') ': TnAC:A mo (TnA) ~ nlo (A)-O " mo (TnA) <; 0 But mo (TnA) ;JP 0 :. 0 ~ mo (TnA) ~ 0 Consequently /)/" (TnA)-O
n
.,.(1)
.
" (2)
59
MEASURE AND OUTER MEASURE
Again :.
TnA'cT Ino (TnA') ~ mo (T)
using (2), mo (TIJA')+mo (TnA) ~ Ino (TIJA')+mo (TnA) ~
or
Ino
nlo
(T)+O (T).
",(3)
Combining (I) and (3), we get Ino
(T)=m o (TnA' )+1110 (Tn A).
This proves thnt A is outer measurable. Problem 4. Show that every subset of a set of measure zero is measurable and its measure is equal to zero. Solution. Let A be a set of measure zero so that A is measurable and In (A)o~O. Let BCA be arbitrary. To prove that B is mcasur"blc and In (B)=O. BCA, m(A)=O => m(B) ~ m(A)=-O => m(B)
But
.s;;; 0
~ 0 ¥ B. m(B) ~ 0, !nCB) ~ 0 => m{B)=I.J.
m( 8)
Since any set of measure Zero is measurable and hence B is measurable. Proved.
Problem~.
Let A be a mo measurable subset of B,
< <'''', A C B, then mo [WnBnA']=mo (WnB) - Ino (WIJA) Solution. A is mo measurable => mo (T)=11I0 (TnA)+mo (Tn A ) ¥ T Taking T= Wf"lB, we get mo (WnB)=mo (WnBnA) tmo (WnBIJA') mo (WnB) - n1 0 (WnA)=mo (Wn.Bf"lA') ACB -=> AnB=AJ. If W is any set with mo (W)
or
r:r.;
Problem 6. !f A is mo-measllrable subset with and B is allY set cOl/tail/jllg A, then 1110
n1 0
(A)
<
00
(BnA )=nlg (B) - mo (A).
Sohlfiou.
01"
then
Suppose ACE and A is outer measurable so that mo (T)=mo (TIJA)+m o lTnA') ¥ S. Taking T=B, we get mo (R)=/n o (Bf'lA)+m o (BIJA') 1110 (B)-mo (A)=1110 (RnA') as ACB ~ A f'lB=A. Problem 7. Let A be a measurable set. If B is any set, /I/O)
lA U R)+I11(j (AnB)=-lilo (A)+m o (B).
(Meerut 1972; Kanpur 70J
60
MEASURE AND OUTER MEASURE
Solution.
Suppose A is measurable so that m. (1)=mo (TnA) + mo (Tn A')
Taking T=B and T=AUB, we get two equations mo (O)=mo (BnA)+m o (BnA').
...(1) ..(2)
"'0 (AUB)=mo [(AUB)nA]+m o [(AUB)nA'].
Evidently (A U B) n A=A, and
n A'=(A nA')u(BnA')=~ U (BnA')=BnA' Putting these values in (2), mo(AuB)=mo(A)+mo(BnA') ... (3) (1)-(3) gives mo(B)-mo(AUB)=mo(BnA)-mo(A) or mo(B)+mo(A)=mo(AuB)+mo(BnA). Der. Let (X, d) be a metric space and 14* be an outer measure on P(X) S.t. p.*(AUB)=p.*(A)+p.*(B) whenever d\A, B) > O. Then 14ft. is cailed outer measure. Theorem 8. To show that an open set in a metric space is measurable with respect to any outer measure. tBanaras 1965J Proof. Let (.Y, d) be a metric space and G be an open subset of X. To show that G is outer measurable w.r.t. metric d. Let ECGCX and G'=X-G. (AUB)
EO!'-~{XEE: d(x, G') ~~}follo\\s that < E.. > is a monotonic
lJcfiue
From this it hence Lim t .... exists. Also Lim E.. =E. For allY set TC X.
... (1)
sequence
T=TnG-t-TnG'.
...(ft.)
By Caratheodory' postulates for outer measure. i.e.
ma(T)=m\l(TnG-t-1'nG') ~ mo(TnG)+mo(TnG') moU') '- mo( . nG)+mo(TnG')
Tn
Sct Then
... (2)
G=e
T=,EU(TnG'):JE.. U(TnG'), by (1) (110(T) ~ 1110 [Ly, U (TnG')] Evidently d (En, TnG') > O. :.
This follows frum our construction (1). Then 1110 lE.. U (TnG')]=m o (En)+m o (TnG') Making n ..... oo.
... (3)
61
MEASURE AND OllTER MEASURE
Lim rn [E" U (T
n
G')]= Lim rno (E.. )+mo (T
n-+oo
n G')
=mo (E)+rno (T n G') =lno (T n G)+mo (T n G')
... (4)
Making ll-oo in (3) and using (4), mo (T) ;;;;, rno (T n G)+rno (T n G'). Combining this with (2), we get rno (T)'=rn o (T n G)+mo (T n G'). This => G is outer measurable. Proved. Theorem 9. If < E,. > is monotonic non-increasing sp.quence of measurable sets (outer measuralyle sets), then the limit set ('~
E=
n Ek is a measurable (outer measllrable) set and for every T
k~l
ojfinite outer measure, rno (T Proof.
(i)
n
E)=
Lim
n-co
rno (T
n
En).
To "how that Lim rno (T n-+oo
Lim rno (T
n-+ 00
n
En)=m o (T
n
n
E.. ) exists and
E).
L':!t T be a set of finite measure. Since El ::J E2 ::J £3 ::J ...... It follows that T n £1 ::J Tn E2 :J T n £3 ~ ..... . Therefore 1110 (T n EI ) ~ Ino (T n E2 ) ~ Ino (T n E3)";;;.··. Consequently < 1110 (TE,,): /lEN> is monotonic non-increasing sequence. In brief we shall denote Al n A2 n Aa n ... , by AIA2Aa ..... Similarly the set A U B U C is denoted by A+B+C. Also every member of < rno (TEn) : n EN> is non-negative. Hence this sequence has a limit so that Lim
n--.oo
1110
(TEn) exists_
00
~n~C~=>EC~=>TnEcTn&
"=1
=> TE
C TEn
Lim
=> rno(TE)
=> n--".>oo rno (TE)
n_---
~
< n-+oo rno (TEn)
Lim · T ak mg rno (TEn)=},., we get rno (TE) 'IV
rno (TE,,)
Lim
< .\
••
·.(1)
62
MEASURE AND OUTER MEASURE
The set T can be broken as T=TE+TE1' + E1TE2 ' +E2 TEa'+·· .. ··+&.-1 TE..' ...... (*) Making use of Caratheodory postulates for outer measure, Ino (T) ~ mo (TE)+m o (TE1')+m O (EITE2')-i-mo (F2TEs') ...... +mo (£"-1 TE,,') ... (2) Since each En is outer measurable, mo (T)=nto (TE.. )+mo (TEn') Ino (T)=rn o (TE1)+mO (TEl')
... (3) Taking E'II-l T for the set T and applying the condition of meac;urability to En, we get mo (EII - 1 T)=rn O (E.._1TE,,)+mo (En- 1 TE.. ·) or rno (E.- 1 T)=rn O (EnT)+rn o (En_lTEn') [For En-I::> En => En n En_1 =EnJ. or rno (E"-lTE,/)=rno (TEn- 1)-rn o (TEn). ...(4) Writing (2) with the help of (3) :md (4) rno (T) ~ 1110 (TE)+[rn o (T)-mo (TE 1)J+[rnO (TEl)-rno(rE~)] +[nto (TE2)--mO (TEa)]+· .... · =[mo (TEn-I) -rno (TEn)] =m o (TE)+rno (T)-mo (TEn). Taking limit as n-'t-oo, mo (T) ~ mo (TE)+m o (T)- ,\ ... (5) or ,\ ~ rno (TE).
Combining (I) and (5), we get A=rno (TE) Lim i.e. lila (TE,,)=mo (TE). n-'?oo (ii)
To show that E
co=
n
,,=1
En is measurabl:::.
For any set T C .Y, we have T=TE+TE' ... (6) From which we get rno (T) ~ mo (TE)+rn o (TE'). ...(7) From (6), TE'=T-TE using this in (*), we get TE' =T' E1 ' + El TE2 ' +~ TEa' + ..... + E..~1 TEn' This gives mo (TE):::;;; mo (TE1')+mo (E1TE/)+m o (E2 TE3')+ ...... + mo (t~'-lT£,.. )
63
MEASU!,E AND OUTER MEASURE
Writing this with the help of (3) and (4), ~ [mo (T)-mo (TE 1 )]+[mO(TEl)-(mO (TE2)]
mo (TE')
+[mo (TE2)-rno (TEs)]+···+ [rno (TE"-I)--m O (TE,.)l =mo (T)- rno (TEn)
Making
n--l>-rrJ,
mo (TE') or
rno (T)
m (T) -- Lim rno (TEn)
n-oo
rn (TE') + Lim rno (TEn).
n-oo
Lim mo (TE,,)=m o erE)
But or
~
~
Ino
(T). ~ rno (TE'H-1I10 (TE)
Combining (7) and (8), we get rno (T)=mo (TE')+mo (TE) ~ E is outer measurable. Problems related to ring of sets, a--algebra. Problem 8. A Boolean ring B containing X (whole space) ;s an algebra Solution. Let B be a Boolean ring s.t. X E B. To prove that B is an algebra we have to prove that ¥ A eX=> A' E B A, X E B => X --A=A' E B, by def. of ring. Problem 9. Intersection of two B-rings is a B-ring. [Gujarat 1970]
This
Solution. Let 8 1 and B2 be two Brings. To prove that 8 1 n B~ is a B-ring, we have to prove that A, B E Bl n B~ => A-B, A U BE Bl n B2
n B2 A, B E Bl and A, B E 8 2 => A -- B, A U B E 8 1 and A -B, A U B E B2 ~ A-B, A U BE 8 1 n B 2 • A, B E Bl ~
Problem ]0. EI'ery algebra is a ring. Solution. Let A be an algebra. To prove A is a ring, we have to prove A, B E A=> (1) A -B E A. (2) A U Be A. (2) follows from the definition of algebra. A, B E A ~ A', B'. BE A, by def. of algebra. ~ A' U B E A ~ (A U B), E A ~ AN n B'=A n B'=A-B e A
64
MEASURE AND OUTER MEASURE
Theorem
to.
A mOll%ne ring j..s a a-ring.
Proof. Let A be a monotone ring. Let Ai E A ¥ i. By assumption A is closed under the formation of finite unions and n
so Al , Al U A 2 •
••••.• ,
U Ai etc. belong to A.
1=1
Write B,,-=.· U" Ai' Then Bn E A. Also
>
<
Bn
~
A is a a-ring.
is an increll.-
t=1 00
00
"=1
n=1
sing sequence and U B,.= U An. A is a monotone cIa&s => lim Bn E A 00
~
U An E A n=1
co
00
For lim Bn = U B .. =.JJ An. n=1
n=1
Theorem 11. A a field is a monotone and monotone field is a a-field. Proof. Let A be a a-field and < An > be a monotonic sequence in A. (i) If < A,.
>
00
is increasing sequence, then lim A.. = U A.. EA, n=l
from the definition of a-field so that A is a monotone class. (ii) If is decreasing sequence, then
r
00
1m An=" A,.EA, n-,; 00 1=1
by the property of a-field, Hence, in either case, a-field is a monotone class. Second part. Conversely let M be a monotone field. C-onsider a sequence < Bn > in M. Let &.= U BIr. if < Bn > is non-decreasing sequence. and
k<.n F,,= n BIr. if < k<.n
BII
>
is non-increasing sequence
lim M} because M IS . a monotone cIass. lim En Fn E E M
65
MEASURE AND OUTER MEASURE
lim Bn
t=
u
11;;;::1
lim Btl
J.
=
n
11~1
..
Bn=U [
B,.=
n ..
l
U
k~n
8/:]= U En= lim E.. n
n Bk]= n F,.=lim F..
k~n
"
Now we can say that lim Bn t , lim B,. t E M ~
00
U
BII
tE
'Y;
M,
n
Bn ~ E M
=> M is a a-field. Problem 10. Show that a ring is closed under formation of symmetric difference and illlersections. [Gujarat 1970] Solution. Recall that a non-empty family A of subsets of X is called a ring of sets if A, BE A ~ A--B E A, A U BE A We want to show that (i) A, B E A => A n B E A Oi) A, B E A => A L':::. B E A where 6. denotes the symmetric difference of two sets. (i) A, BE A :::> A E A, A -B E A => A - (A -- B) E A => A n BE A For A---(A-B)=A n B. (ii) A, BE A :::> A-B, B-A e A => (A-B) U (B--A) E A => A 6 B=(.4-B) U (B-A) E A => A L':::. BE A. Theorem 12. If mo is an outer measure, then the class A of !no measurable sets is a-algebra (or a-jield). [Kanpur Statistics 1977] Proof.
Let A be the class mo - measurable sets.
Let A, B E A be arbitrary, then A, B are outer mea<1urable sets. To prove that A is a a - algebra of sets, we have to show that (i) (Ii)
(iii)
A' E A A U BE A jf < A, > js a sequence of sets in A, then
66
MEASURE AND OUTER MEASURE
(1) We have supposed that A is an outer measurable set so tha' A' is outer measurable. (2) We have seen that if A and B are outer measurable sets, then A U B is an outer measurable set. (3)
Let
<
Ai
>
be a sequence of outer measurable sets. n
00
Also let A= U Ai, B,.= U Ai. ;=1
i-I
To prove that A is outer measurable. By (2), A1> A2 are outer measurable sets => Al U A 2) is outer measurable => (AI U A2 ) U .42 is outer measurable, by the same reasoning. Generalising this, we can say that any finite union of outer measurable sets is outer measurable a.nd so Bft is outer measurable. Consequently for any T, rno (T)=rno (T n Bn)+rn o (T n B,.') ... (1) 00
..
A= U Ai:J U A.=Bn i-I
i=1
This => A :J Bn => 8.' :J A' => A' C B,.' => Tn A' C Tn Bn' => rno (T n A') ~ rno (T n Bn') Now (I) becomes rno (T)=me (TnB,.)+rn o (TnB,.') ~ rno (TnB,.)+rn o (TnA') or rno (T) ~ rno (T n B .. )+rno (T n A') ... (2)
•
Bn= U A. :J An. i=1
This => B.. :;) A.
n
8n=A...
Thus An n B.. =A.. Similarly Bn n A.'=Bn_ 1 • Since An is outer measurable and hence rno (T)=rno (T n A.)+rno (T nAn'). Taking T n B. for T, this becomes rno (T n B,,)=rno (T n B" n A,.) +rno (T n Bn n A.;) =rno (T n An)+rn o (T n B .. _1 ) or rno (T n B,,)=rno (T n A,,)+rno (T n B n_ 1 ) n=2, 3, 4, ...... , n -1, n in the equation (3), Putting rno (T n B 2 )=rnO (T n A 2 )+ rno (T n B l ) rno (T n Bs)=rno (1' n Aa)+rno (T n B 2 ) rno (T n B.)=rno (T n A4 )+rnO (T n Ba)
...
(3)
67
MEASURE AND OUTER MEASURE
mo (T mo (T
n n
B.'-!)=m o (T n An-l)+mO (T n B .. - 2) B,,)=mo (T n AnH-mo (T n Bn- I )
Upon addition, we obtain n
E mo (T
,-2
This
~
n
" L
B,.)=
mo (T
r~2
fIlo (T
n
n
,,~
A,)+ E mo (T r-l
n
to
D,.)= E Ino (T () A,)+nl o (T
Dr)
n
Bt )
r~-2
fa
= ,=1 E rno (T n A,).
n
nlo (T
:.
.
D,,)= E rno (T ,=1
n
For Dl=A).
A,).
. .. (*)
Using this in equat!on (2), n
rno (1')
~
r:
mo (T
.=1
A,)+mo (T
U Ai => T n A= Tn ('U<=1 Ai
A=
j~t
z> rno
(T n
00
==> }; nlo (T ,=1
or
n
A)=rno
n Ar)
n
... (4)
A')
)= U .=1
(1'
n A,)
(~1 (T n A,») ~! rno (T n A.)
~ rno
(T
n
A).
Using this in (4) rno (T) ;;?! rno (T n A)+rno (T n A'). But T=T n A+T n A' mo (T)=mo (T n A+T n A') ~ mo (T n A)+rno (T n A') mo (T) ~ rno (T n A)+mo (T il A'). Combining (5) and (6), rno (T)=fn o ~ (T n A)+mo (T () A')
. .. (5)
... (6)
00
This ==> A is outer measurable ==> A= U A, E A. 1=1
Theorem 13. (i) (ij)
Let (X, A, m) be measurable space. A, D E A s.t. A C B => m (A) ~ m (D) If < An > is a sequence of sets in A, then
Then
68
MEASURE AND OUTER MEASURE
m(
.91 A. ) ~.~ 2' (A.).
[Indore 1978J
Proof. Let (X, A, m) be a measurable space. (i) Let A, B E A s.t. A C B. To prove m (A) .s;;; m (B). Notice that B=(B-A) U A, (B-A) n A=rp This ~ m (B)=m (B-A)+m (A) ~ m (A) Also m (B- A) ~ O. ~ m (B) ;;0. m (A) ==- m (A) ~ m (B). (ii) Let < An > be a sequence of sets in A. "-I
Write
Bn=AII - - U A. '=1
Then Bl=Alo B 2 =A 2 -A1 , Bs=Aa-(AI U A2),······ This shows that (1) B. n BI=t/I for i::j:j 00
00
(2) U A,= U 1=1
Bn C A" ¥ n.
(3) (1)
B~
;=1
)= ,~
==- m
(~1 B,
=> m
[~1 Ai ]=;~
m (BI) m (B , ), by (2).
00
~
L m(A.), by (3). '=1
=> m
[U
1=1
AI]
~
;
1=1
m (A,),
Problem 11. Let A be an algebra of subsets of X and let < An > be a seq.Jeuce of sets in A. Theil prol'e that there is a sequence < Bn > of sets in A slIch that Bn n Bm=!,~for nz-:j:.11 alld 00
00
1=1
;=1
U B , = U A,.
[Kanper Statistics 1976; Gujrat 1970J Let A be an algebra of subsets of X so that A, B ~ A => A U B E A
Solution. (i)
(ii)
A' E A.
MEASURE AND OUTER MEASURE
69
(iii)
To prove A-B E A A, B E A => A, B' E A => A n B' E A => A () B'=A-B E A. Let < A" > be a sequence of sets in A. To prove that 3 a sequence < Bn > of sets in A. s.t.
B,.
n Bm=,p for m:;t;n and
00
00
U B.= U Ai.
>=1
i=1
By assumption, Ai E A for i= 1, 2, 3, 4, ...... 11-1
Write
Bn=A,. -
U Ai' 1=1
Then . 0 0 0 0
This shows that (I) U Bi= U Ai i=1
i=l
(2) B .. n Bm=,p for m=l=n. (3) Bn C All "I n. => Al U A2 E A ~ Al U A2 U Aa E A.
(i)
"-1
Generalising this, we see that U Ai E A. i=1
n-l
U Ai E A, An E A
~
"-1
An -- U Ai E A, by (iii)
<=1
;=1
=> Bn E A => < B" > is a sequence of sets in A. Also we have seen that < B" > has the properties (I) and (2).
Hence Proved. Problem 12. If I' is completely a-additive set function on a class E and if < En t > is any sequence of sets from the class E, thell lim (I' £11)=1' (lim E,,). [Kanpur Statistics 1975. 76J Solution. < Eit > is non-decreasing sequence and its limit 00
will be U Eft.
This increasing sequence is converted into sequ-
ence of disjoint sets as : Write Eo=rb, Bn =EII -En _ 1 for n=l, 2, 3, ... 00
00
"=1
II-I
Then we have U B.. = U E,,=lim En
70
MEASURE AND OUTER MEASURE
... ...
go
lim &=
u
n=1
00
En= U (E.. -Bn-l) 10=1
Problem 13. Let", be a measure on an algebra A. (i) If An C A"+I' An e A. 1 EO; n < <X1 00 and ~ U An E A, thell p.
ai)
(
If A" :::> A"+l' A" '" (A 1)
'" (
<
()()
)
U A"
=
~
lim n_~
E A, 1 ~ n
<
(A,,).
p. 00,
00
00 and
n
n=1
~1 An )= n~":'"
A" E A, then
(An).
p.
Solution. Refer theorem 1. Problem 14. If be a sequence of disjoint measurable sets. then for any set A, mo [ A
Solution.
n
(U )]=.f Ei
'=1
.=1
mo -(A U Ei )
[Meerut 1987J
Prove equation (*) as in Theorem 12 that n
mo (r () Bn)= E
!no
(T () A,)
r=l
or
mo (T
n
It
Bn)= E !no (T
n
E,)
r=l
where
00
It
0=1
1=1
A= U Eh Bn = U E,
(I.e. replace A, by E, in theorem 12)
... (4)
MEASURE AND OciTER MEASURE
71
Replacing T by A in (4), we get
mo [An
C9tEi ) J=I~ mo(A nE.)
Exercises 1. Show that the additive set function defined over a ring can not assume both the values +00 and -00. [Poona 1970] 2. What is a measure space? [Kanpur 1969] 3. State Caratheodory's criterian of measurability of a set. [Banaras 1966] 4. State Caratheodory's criterian of measurability and prove the union of two measurable sets is measurable. LBanaras 1965] 5. (a) Define a a-filed and a monotone class of sets, show t4at a monontone filed is a a-filed. (b) Define continuty of a set function, and show that a a-additive set function is continuous. (c) If fL is a measure function, defined on a class E of sets. If E is any sequence of sets from E for which p (E..) <00, then p. (lim ~ub 1:...) :> lim sup pE,.. [Kanpur Statistics 1977] 6. such that Ee Lim sup I.. , and E in < co, where i" is the length of lit. [Banaras 65] 9. Let po be a countably additive non-negative measure defined on an algebra of subsets of a set X such that p.(X) < 00. Show that if {A" : E N} is a contracting sequence of p.-measurable subsets of X, then
po (
n~NA" ]=;~mN p (A,,).
f2
MEASURE AND OUTER MEASURE
Also give a counter example to show that the result is not valid if f4 (X)=OC!>. [Banaras 64] Hint. Refer problem 13. 10. Write a short note on the problem of measure in Euclidean space. [Banaras 64] 11. of sets in A such that B" n B... =rp for m#:n co
00
and
U Bi= U Ai i=l
1=1
[Gujrat 70] tw~ B-rings (Boolean rings) is a B-ring. [Gujrat 70] 13. The union of two rno measurable sets is mo measurable. [Meerut 72] 14. (a) EXplain Caratheodory's concept of an outer measure po* on a set X. Define p.*-measurubility of a subset A of X. If p.*(A}=O. show that A is measurable. (b) If is a contracting sequence of po-measurable sets and po(A I ) < 00, prove that f' (lim A.. )=lim f4 (Aft) [Banara~ 71] (Give an example to show that the finiteness condition is essential). [Banaras 71J 12.
Intersecti:m
of
7 Lebesgue Measure of a Set Introduction. The concept of measure is an extension of concept of length. Measure of an interval in R is usually defined as its length, the measure of a polygon in R2 is usually its area and so on.
7·0.
Measure of an open set and a closed set. An open set is an enumerable union of non-overlapping open intervals. The measure of an open set is defined as the limiting sum of the lengths of its intervals. If the set G is contained in an interval (a, b), then the measure of G, denoted by m (G), has the property that m (G)
~
b-a.
We know that the complement of a closed set w.r.t. an open set is open. Let a closed set F be contained in an open interval (a, b). We define the measure of F as follows: m(F)=b--a --m(F'), where F'=(a, b)-F. 7·1. Measure of an open interval. The length of any open interval G is defined as its length and is denoted by the symbol meG). Examples: (i) m [(2, 5)]=5-2=3 (ii) m[(a, b)]=b-a. Evidently meG) > o. 7·2. Measure of a closed interval. Let La, bJ be the smallest closed interval containing a closed set F. Then we define m(F)=b-a-m(F'). F' being the complement of F w.r.t. the interval. To prove that the measure of a closed interval is its length. Let A=[a, b]. Then A is a closed interval and hence a closed set. [0, b] is the st:Jlallest closl1d interval containing A.
74
LEBESGUE MEASURE OF A SEr
Complement of A relative to [a, b]=A'=cp. For A=[a. b]. By definition, m(A)=b-a-m(A')=b-a-m(q,)=b-a-O =b-a i.e. m([a. b])=h-a. 7'3. Measure of rectangle. The area of an open rectangle R (a < x < b, c < y -< d) i.e. (b-a) (d-c) is defined as the measure of R. Thus m(R)=(b-a) (d-c). The area of a closed rectangle R (a ~ x ~ h, c ~ y ~ d), i.e. (b -a) (d -c) is defined as the measure of R. Thus m(R)=(b-a) (d-c). 7'4. Measure of a parallelopiped. The volume of an open parallelopied V (a < x < b, c < y < d, 1< z < m) is defined as the measure of V. Thus m(V)=(b-a) (d-c) (m -I). Similarly the measure of a closed parallelopiped V' (a ~ x ~ b, c ~ y ~ d, I ~ z ~ m)
is defined as its volume. m{V')=(b--a) Cd-c) (m-/). Thus 7'5. Exterior and interior measure. (Lebesgue measurable set) [Kan~ur 75. 73. 68] The Lebesgue exterior measure (or simply exterior measure) of any set A, denoted by m,(A), is denned as follows: m,(A)=inf. {meG) : G :J A, G is open} ..• (1) From this it follows that meG) > me(A). Hence given. > 0, 3 an open set G :J A s.t. m(G) < me{A)+,. . .. (2) If the set A is contained in an interval (a. b), then
o ~ m,,(A) ~ b-a. Also and
Al C A2 =:> m.(A 1 ) <;; m.(A 2) m.(,) =0.
The Lebesgue Interior measure (or simply interior measure of a set A contained in an interval (a, b), denoted by m.(A), defined as follows: m,(A)=sup {m(F) : F is closed, F C A}. It can also be expressed as [The equation (2) is of vital importance for further study].
LEBESGUE MEASURE OF A SET
75
m. (A)=b-a-m, (A'). . .. (2) From the definition it is clear that mi (A) ~ 0 and mi ( (a +', b -() " .(2) and m (Ol)+m (0 2) ~ m (0) ... (3) (2) => m (0) ~ m [(a+., b-e)]=b-e-(a+e) => m (G) ~ b-a-2E. . .. (4) By (1), m (Ol)+m (0 2) < m. (A)+111. (A')+2e. Using (3), m (G) ~ m (Gl)+m (G 2 ) < m. (A)+m, (A')+2e or m (0) < m. (A)+m_ (A')+2 •. Using (4), we get or b-a-2f. < m, (A)+m, (A')+26 or b-a m, (A') < m. (,.,)+4& or m, (A)=b- a-m. (A{) < m. (A)+4. or m. (A) < m, (A>+4e. Since e is arbitrary and hence making E~O, we get the required result m. (A) ~ m. (A). Theorem 2. A sei .A is measurable (i.e. Lebesgue measurable) iff its complement A' is measurable. Proof. Let a set A be contained in an interval (a, b). Then we know that ... (1) m. (A)=b-a-m. (A').
76
LEBESGUE MEASURE OF
~
SET
Let A be measurable so that mi (A)=m, (A)=m (A). . .. (2) To prove that A' is measurable, we have to show that m, (A') =mi (A'). . .. (3) Writing (1) with the help of (2), m (A)=b-a-m, (A') or m. (A')=b-a-m (A). ...(4) (1) is true ¥ A. Replacing A by A' in (1), we obtain mi (A')=b-a-m, (A). For (A,),=A. Using (2), we get mt (A')=b-a-m (A). ...(5) Since, R.H.S. of (4)=R.H.S. of (5) Hence, L.H.S. of (4)=L.H.S. of (5) i.e. m, (A')=tni (A'). This => A' is measurable. (ii) Let A' be measurable so that ...(6) me (A')=mi (A/)=m (A'). To prove that A is measurable By virtue of (I), we have tnt (A)=b-a-m, (A') and mi (A')=b-a-m, (A). In view of (6), these equations become mi (A)=b-a-m (A') ... (7) m (A')=b-a-m. (A). ...(8) Equating the two values of m (A') b-a-m. (A)=b--a-m, (A). mi (A)=m. (A). or This => A is measurable. Proved. Note. Let a measurable set A be contained in [a, b). Then In; (A)=b-a-m. (A') m (A)=b-a-m (A') or m (A)+m (A')=b-a. or For A is measurable => A' is measurable. Theorem 2. (a) Evay bounded open set and bounded closed set are measurable. lMeerut 1987, 86] Proof. Let G be a bounded open set. Form G by means of a finite number of disjoint closed intervals. Then the sum of the lengths of these intervals is equal to the length (measure) of G. Hence m, (G)=m (G) ... (1) (i)
77
LEBESGUE MEASURE OF A SET
Thus G can be obtained in an open set consisting of G itself, and contains a closed set s.t. difference of measures of closed set and open set is arbitrarily small. Consequently m. (G)=m (G)=m. (G)
This ~ G is measurable. Since complement of a measurable set is measurable. Also complement of open set is closed set. This ~ Gf is measurable and G is closed. Theorem 3. . Let A and B be any two sets such that A e B. Then (i) mi (A) ~ m. (B) (U) m, (A) ~ m. (B). Proof. Let A e B. (i) To prove that mi (A) ~ m; (B). By definition of interior measure, Ini (A)=sup {m (F) : F is closed, Fe A} lni (B)=sup {m (F) : F is closed, FeB}. Let S be the set of numbers consistjng of measures of all closed subsets of A. Similarly we define the set T for B. Then m; (A)=sup S, mi (B)=sup T. . .. (1) F is any closed set s.t. F e A => F CAe B. For A C B. f
~
Fe B
:;. F is closed subset of B.
This shows that any m (F) E S ~ m (F) E T. Hence SeT. Consequently sup S ~ sup T. For least upper bound, i.e. supremum of a subset of any set can not exceed the least upper bound of the set itself. sup S ~ sup T =? mi (A) <; nil. (B), by virtue of (I). (i) To prove that m, (A) ~ m, (B). By definition of exterior measure, m, (A)=inf {m (G) : G is open, G ::J A} m. (B)=inf {m (G) : G is open, G ::J BJ. Let S be the set of numbers consisting of measures of all open supersets of A. Similarly \\'e define the set T for B. Then m. (A)=inf S, me (B)=inf T. . .. () G is any open set S.t. G ::J B =? G ::J B :> A. For B :> A. => G :J A. ~ G is open super set of A. This prove, the fact that and III (G) E T =-> 1J1 (G) E S.
78
LEBESGUE
MEA~UPE
OF A SET
Hence T C S. Consequently inf S <;; inf T. For greatest lower bound of a set can not exceed greatest lower bound of a subset of that set inf S <;; inf 1 => m. (A) ~ m~ (B), according to (2). Theorem 4. El and E% are measurable (in the sense of Lebesgue), then El U E2 is measurable. [Meerut 1988, 86; Kanpur M. Sc. P. 79] Proof. LetE1: be a me:1surable set for k= 1,2 and
,r
E=£1 U E2 •
To prove that E is measurable. Consider an arbitrary num bel' f > O. EI: is a meastJrable <;et => 3 a closed set F1: and an open set G" S.t. F; C cI: C G"o m (Gd-m (Fd
< T"
Write F=F1 U F2 , G='G 1 U G2 • Then F is a closed set and G is an open set so that F and G are measurable sets. F1, C Ek C GI •
:::>
2
2
2
/:~'l
1:=1
le·,1
U Fl. CUE/;. C U
G,:.
=>FCEcG => FeE, E c G => mi (F) <;; 11'1, (E), mil (E) ~ 11'1 (G) ~ 11'1 (F) ~ 11'1 .. (E) ~ me (E) ~ 11'1 (G) [For 11'1, (F)=m (F), m. (G)=m (G) and ml (A) < mil (A) v A] => m. (E)-mt (E) ~ m (G)-- m(F). ...(1) F'=A finite intersection of open sets
G -F=G n =An open set=A measurable set m (G-F)=m (G)-·m (F).
...(2)
Similarly m (G/<-F,,)=m (G,,)-m (F,,). 2
Notice that G-F C 1: (G1.- -F1)
"-1
This => m (G - F)
~
I
1: m (G,.-- F;-) 1<=1 2
.. m(G)-m (F) ~ E [m (G/.)-m (Fl')] k-l
<
•
-2 .2=.
LEBESGUE MEASURE OF A SEl ~ m (G) -In (F) < Ii => m. (E)-mi (E) ~ m (G)-m (F) <:
::>
m. (E)-mi (E)
<
6,
by (1)
e.
Making E-+O me (£) -mi (E) ~ O. For E is arbitrary. This => me (E) ~ mi (E). But m. (E) ~ m, (E) is true ¥ E. Hence m. (E)=m, (E). This ~ E is measurable. Problem t. If X and Yare measurable sets, then X n Y i~ measurable. What can you say about X U Y. [Kanpur 1975, 791 Solution. For the sake of convenience we write E1 =X, £2= Y (i)
To prove X U Y=E1 U E2 is measurable. (prove this as in theorem 4)
(ii) X, Yare measurable => X' Y' are measurable, by theorem 2. => X' U y' is measurable by (i) => (X n Y)'=X' u Y'is measurable.
=> X n Y is measurable. [For A is measurable ¢> A' is me.asurableJ. Theorem 5. If G1 and G2 are any two open sets in[ a, b], then, m (G1 )+111 (G 2 )=m (G 1 U G2 )+m (G 1 n G2 ) i.e., m (G 1 )+m (Gz)=m (G 1 +G2 )+m (G 1G 2 ) [Kanpur 1973] Proof. Here m(Gl) = length of interval in G1 The length of the interval which is common in G1 n G2 is equal 10 the length of the interval in G1 n G2 =G1G2• On the other hand the lengths of common intervals in m (G1 ) +m G2 occurs twice. Consequently m(G1 )+m(G2 )=m(G I +G 2 ) +m(G1G2). Theorem 6. If El and E2 are subsets o/[a, bJ, then m.(El)+m.(Ez) ~ m.(E1 U E2 )+I1I.(E1 n £2) alld mi(£1)+mi(E2 } ~ mi(EI U E2)-miCE, n £2) Or To prove that (i) m*(E1 )+m*(£2) ~ m*(E1 U t~)+m*(I:;l (ii) m*(£!)+m*(£2) ~ m*(£1 U £2)+m*(EI
E
>
n
E2 )
n t z)
[Banaras 1968J Proof. Let El and £~ be subsets vi closed interval [a, bJ. Fix O. Choose open sets G1 and G2 s.t. El C G1> Ez C G2 ,
~o
LEBESGUE MEASURE OE A SET
m(Gt ) < m.(E1 )+E/2 and m(G 2 ) < m'(Es)+(E/2). Then m(G1)+m(G2 ) < m.,(E1 )+m..(E2 )+f But m(G1)+mtG2)=m(Gl U G 2 )+m(Gl n G2 ) Therefore meG} u G2 )+m(Gl n G 2 ) < m.(E1)+m,,(E2 )+, But Gl U G 2 and Gl n G2 are sets containing El U £2 and £1 n E2 respectively. Then the last gives m,,(EI U E2)+m. (E) n E2 ) < m. (E1)+m.(E2)+" Since" is arbitrary and hence making E-+O. We obtain the first required result namely ... (1) miEI U E2)+m.(E I n E 2 ) ~ m.\E1)+m.(E2) This is true I.f E 1 , E2 C [a, b]. Replacing Ez, E2 by E 1 ', E2 ' respectively in (1), we obtain m/E/ U E2')"" m"(£l' n £2') ~ m.(£t')+m,(E2 ') ••• (2) where E1'=[a, b]-El' £2'=[a, b]-E2' Applying De-Morgan's law in (2). m~[(EI n E 2 )']+me [(El U E 2 )'] ~ m.(E/)+me(E/). By definition of interior measure, this gives [b-a-mi (E1 n E2)J+[b-a-mi (El U E2 )] E;;; [b-a-mi(E1)]+[b-a- m;(E2 )] Simplifying this we get mi(El n E2 )+mi(El U E2) ~ mi(El)+m.(Es). This proves the second required result. For mi(A)=b-a-nle(A') ¥ A c [a, b] and x < y => -x > - y. Proved. Theorem 7. If El and £2 are measurable subsets of [a, b] then prove that E2)' m (E1 )+ m (E2)=m (El U Ez)+m (El [Meerut 19S8, 87; Kanpur 79, 88; Kolhapur 70; Banaras 69] Solution. Let El and E2 be measurable subsets of[a, b) so that m.(E1)=m,(E1)=mr E1 ) } m.(E2 )=mi(Ez)=m(E2 ) ... (*) Firstly we shall prove that (i) m"(£1)+m,,(E2)~m.(EI U E 2 )+m.(E'1 n £2) (ii) mi(E1) .. mi(B2)e.l11i(F."t U E2)+mi(E1 n E2)· Prove this as in Theorem 6. Wr:ting (i) and (ii) with the help of (*) (i)' II1(E 1 )+m(E2 ) ~ m(£1 lJ E2 )+m (E 1 n £2) (ii)' IJl (E1 )+I11(E2 ) ~ m(EI U EzH-m(E l E2 ).
n
n
81
LEBESGUE MEASURE OF A SET
For finite union and intersection of measurable sets are meac;urable so that El U E2 and El n E2 both as measurable and so m. (El U Ea)=m, (El U Ea)=m (El U E2 ) me (El n E 2)=m, (El E2)=m (E1 nEs). Combining (i)' and (ii)' we get the required result. Problem. If El and E9 are measurable subsets 0/[0, b], prove that El U E 2, El n E2 and E1 -& are measurable and show also that (i) m(h.;)+m(E2)=m(El n E2 )+m (El U E2 ) (ii) m(E1 -E2 )=m (E1 )-m(E2) if E z C El [Kanpur M.Sc. 1979] Solution. (i) Prove as in theorem 7. (ii) Since difference of measurable sets is measurable :. E1 -E2 measurable set
n
(E1 -E2 ) n E2 =rp m [(E1 - E2 ) U E2J= m (E1 -E2 )+m(E2) or m(E1 )=m (E1 - E2)+m(E2) or m(El) - m(E2 )=m (E1 -E2) Problem 2. If A, B, 'C, are p.-measurable sets, show that p.(A)+p.(B)+ p.(C)=p. (A U B U C)+ p. (B n C) +p. (C n A)+p. (A n B)-p. (A n B n C). [Banarsas III, 1970] Solation. Here we shall use the following lemma: Lemma. If E1 , E2 are measurable subsets of [a, b], then m(E1 )+m(E2 )=m (El U E2 )+m (E1 n E2 )· Now we come to the proof of the problem. Let A, B, C be measurable subsets of [a, b]. Since finite union and intersection of measurable sets are measurable and hence A n B n C, A U B U C, A U B, A n C etc. are all measurable. Taking E]=A U 8, E2 =C in the lemma, we get m (A U B)+m (C)=m (A U B U C)+m[(A U B) n C]
:.
... (1)
Again, by the lemma. m (A)+ m(B)=m (A U B)+m (A n B) >r m (A)+m (B)-m(A n B)=m (A U B). Hence m (A U B) n C] =m[(A n C) u (B n C)]
... (2)
82
LEBESGUE MEASURE OF A SET
==m (A
n C)+m (B n C)-m (A n C n B n C).
[This follows from (2)] =m (A n C)+m (B n C)-m (A n B n C) or m[(A U B) n C]=m (A n C)+m (B nC)-m (A n B n C) . ... (3) Writing (1) with the help of (2) and (3) m (A)+m (B)-m (A n B)+m (C) =m (A U B U C)+m (A n C)+m (8 n C)-m(AnBn<:) or m (A)+m (B)+m (C)=m (A U B U C)+m (B n C) +m (C n A)+m (A n B)-m (A n B nC). If A, B, C are ,,-measurable sets, then the las~ gives the required result. " (A)+ " (B)+" (C) (.4 U B U C)+" (B
=="
n C)+I' (C n .4)+" (A n B) - " (A n B n C).
Problem 3. Let E 1 , E2 , ... be sequence of sets of refl/ numbers, prove that m* (E1 +E2+"~) ~ m* (E1)+m* (E2 )+ ... where m*(E) is the Lebe.gue outer measure. If in addition the sets E, are pairwise disjoint, i.e., no two of them have common poinl8 then show that m* (E1 +E2 + ... ) ~ m* (E1 )+m.(E2 )+ ... where m* (E) denotes the inner measure of E. [Kanpur 1972] SolutioD. We write m. (E) in place of m* (E) and m, (E) in place of m* (E). To prove that (i)
m. (
E Er
r-l
)
if E1 , E 2 , ... are pairwise disjoint. To prove (i). Given. > 0, 3 an open set Gn S.t. E. C G,., m (G n ) < m. (E;.)
E
+ Tn
LEBESGUE MEASURE OF A SET
83
An arbitrary union of open sets is open '1nd therefore -"
]; G, is an open set s. t.
'~1
... (1)
This follows from the fact that En C G", ¥ n. (1) =>
m.( UE, )-E;;; m, (U G, .)=m(u G r~1
r=1
'=1
00
r )'"
1: m(G,)
r_l
~ m. (,! E. ) ~ !,m (G,) <.!. [ m. (E.l+; ] ~ m(! ( r~l Fr
)
<
,~ mo (E,)+I! ..
Since c is arbitrary and hence making 1-0, we get
m,( gl
E, ) <;
tl m~
(E,)
This is also expressed as
m, (
!1 Er ) <; r~lm. (E,l
To prove (ii). Suppose E1 , E2 , ... ••• are pairwise disjoint so that en n En=t/> for n#=m. Consequently m. (Em n En)=O=m. (E". n En) if m;#:n. We know that m. (E1)+m. (~) E:; m, (El U EA)+m, (El In accordance with (2), this becomes
m. (E1)+m, (£2)
~ mj (£1 U
E2 )
... (*)
. .. (2)
n
£2)'
84
LEBESGUIi MEASURE OF A
sn
By induction, it follows that
m,
(Ql
Er);.
!1
... (3}
mi (Er)
n being a positive integer. 00
E/idently
and so m, or
(u
,_1
m,
U
,=-1
,.
Er :::> U E., r=-l
Er);;. m,
(gl
Er )
(u
,-1
Er ')
~i
'1'=1
m, (E,.), by (3)
~ !lm, (E,).
Making n__ , we get m,
(Ql Er ') ~ £1 m, (Er)·
This is also expressed as m,
(X E,.) ~ ~ m, '1'-1
(E,).
'1'=1
This completes the proof. Problem. Let < E, > be a sequence of measurable sets. Then prove that
m(~l' E, )
<;
,!
m (Ei ).
[Meerut 1999. 87]
Solution. E. is measurable set => m (E,)=m. (E,)
... (2)
00
and
U is measurable and so
'1'-1
... (3)
Now prove equation (*) of the above problem (3), I.e.,
m.
(u
'1'-1
Er) <;
E m. (E,). '1'-1
Putting (2) and (3) in (*), we get the required result.
. .. (*)
8S
LBBIISGUB MEASURE OF A SET
Problem 4. If M is a measurable set, then prove that for any set E. [Kanpur 1972] m, (E)=m. (EM)+m. (E-EM). Sollltion. We know that (l~ -me (E 1 +£2) ~ m. (E1)+m. (E2) (2) m. (E1 )+m. (E2) ~ m. (E1 +E2)+m. (E1E2) By virtue of (I), m. (E) ~ m. (E-EM)+m. (EM). . .. (3) By (2), me (E)+m. (M) ~ m. (EM)+m. (E+M) ;;;;. m. (EM)+m. (E-EM+M) [as E+M ::> E-EM+M] ~ m. (EM)+m. (E-EM)+m, (M) [as m. (E1 +E2) ~ m. (E1)+m, (EJ if El n E 2=c;6]. But m. (M)=m. (M)=m (M) as M is measurable m. (E)+m (M) ;;;;. m. (E4l)+m. (E-EM)+m (M) or m. (E),;;;;' m. (EM)+m. (E-EM). ...(4) Combining (3) and (4), m. (E)=m. (£ -EM)+1Y.. (EM). Proved. Theorem 8. Prove that a set E of real numbers is measurable iffgiven e > 0, there exists ~, el' e2' S.t. E=t+e1-e2, and m. (el ) < e, m. (e 2) < Ii and E is the sum of a finite number of intervals. [Kanpur 1973] Proof I. Let E be a measurable set and E C R. To prove t'ha' E=~+el-e2 where m. (el) < t, m. (e 2) and is the sum of a finite numb!r of intervals. By assumption 3 open'set G s.t. G ::> E and m (G) < m (E)+ •.
e
Write and ~ is a union of enough intervals of G.
.. _(1) ...(2)
Also take
el=G-~
and
<-
. .. (3)
... (4) m. (el ) < t. m. (e 2 )=m (G --E)=m (G)-m (E) < c, by (1) :. m. (e2) < e. [Since G --E is measurable] (2)-(3) => e2-el=G-E--(G-~)=~--E
=> E=~+el-e2' From (4), (5) and (6), the required result follows.
... (6)
86
LEBE~GUE
MEASURE OF A SET
II. Let E=1;+e 1 -e2 with m, (e 1 ) < e, m. (e 2) < e, E C R. To prove that E is a .measurable set. Let E C [a, bJ. Since E=(!;+e1 )-eZ' Hence m. (E)=m, (f+e 1 )-m. (e a) ~ m. (1;+e l ) ~ m. (f)+m, (e l ) < m (~)+4! or m. (E) < m (~)+E. ; .. (7) E=~+11-12 ~ [a, bJ---E=[a, bJ-f+/2 - /1 ~ E =~'+1.,-11::;' m. (E')=m. (~'+/2)-m. (11) o m. (E') < m, (~'+/2)-e < m. (1;'+/2) ~ m.(~')+m (/2) => m. (E') < m (f)+E as E is measurable and so is 1;'. o [a, bJ-m. (E') > [a, b]--m (~')-4! ~ mi (E) > m (~) -41 as m, (E)=[a, b]-me (E') and-_ - m. (E)-4 < m < m, (E)+e, by (7). => m, (E)-fi < mi (E)+fi ~ m. (E) ~ m, (E) on making E~O. But m. (F) ~ m, (E). Hence m. (E)=m, (E). This::> E is a measurable set. Problem 5. Is the Lebesgue measure of the union of two mea[Kanpur 1975] surable sets equal to the sum of their measures? Solution. Since m (A lJ B)=m (A)+m (B) if A n B=r/>. Ans. Yes, if their intersection is null set. Problem 6. Construct a non-dense perfect set in the interval '[Kanpur 1968] [0, IJ whose measure is i. Solution. Consider the closed interval [0, IJ. Let ,\ E [O~ 1]. Delete a portion of length I" from the interval [0, I] s.t.
m
11 +2/2 +22 / 2 + ... +2,,-1 1,,=>.-- n~-2 Stage 1st 2nd 3rd
Number of intervals dropped 1 2
22
nth 2"-1 2,,-1 I" The sum of the lengths of the intervals dropped upto nth stage =/1 +2/2 +22 / 3 +", +2"-1 I" =.\- n~2' by assumption.
The limiting sum of the dropped intervals
LEBESGUE MEASURE OF A SE r
= nl:oo
l
A-
87
n~21=~-0=A'
Let A denote the set of end points of the interval {III} with their limiting points, then A is a non-dense perfect set. Also m(A)=m [0, I]-length of dropped intervals=l-A. If we take A=i, then m (A) = l-i=i. The set A formed as above is a non-dense perfect set with measure 1. Theorem 9. 'Any set A is measurable iff an open set G containing A and a closed set H contained in A can be so determined that I G I - I H I < 6, where I G I stands for length of open intervals in G. ' Proof. Definition of exterior measure => given. > 0, we can find an open set G :J A s.t. I G I < m, (A)+e/2. ...(1) Similarly we can find a closed set H S.t. m. (A) < I HI +e/2 or - I HI < (E/2)-m, (A) ... (2) (1)+(2) gives I G I - I HI < e+m. (A)-m. (A). ...(3) I. Let the set A be measurable so that m.(A)=mi (A). To prove 1G 1- 1if I < e. ...(4) Now (3) => I G 1- I HI < e+O. Hence the result. II. Let (4) be true. To prove A a'l measurable. The definitions of exterior and interior measure show that m. (A) ~ 1 Gland mi (A) ~ I HI. SO that m. (A)-m; (A) < 1(;1 - I HI < e, by (4). But E is arbitrary and so making e~O, m. (A)-m; (A) 0 or m. (A) < m, (A). But m. (A) ~ n1i (A) is true ¥ A. Hence m. (A)=m; (A) so that A is measurable. Problem 9. A subset E of [a, b) is measurablt> ~ffgiven e>O, :I open sets G1 , G2 S.t. G1 :J E, O2 => E', m (G1 n G2 ) < e. Solution. Let E C [a, b]. The definition of exterior measure sllg~ests that 3 op(!n sets G1JE, OJ :J E' s.t. m(G1 ) <me (E)+6/2, m (G 2) < m~ {£,)+./2. This => m (G1)+m (G~) < me (E)+m. (E )+e. But m (OI)+m {G 2 )=m (G l U G2 )+m (01 n O2 ), Hence m (G 1 U G2 )+m (G 1 n G2 ) < m. (E)+m, (£')+e .. (I)
<
88
LEBESGUE MEASURE OF A SET
G1 J E, G2 ::> E' ~ G1 UG 2 ::> EUE'=[a, b] => G1 UG 2 ::> [a, b] => G1 UG 2 =[a, b] as G1 U G. c [a, b] => m (G 1 U G2 )=b-a.
Now (1) => b-a+m (G 1 n G2) < m. (E)+m, (£')+_ ... (2) I. Let E be measurable so that mil (E)=mi (E). . .. (3) To prove
m (G 1
n
G2 )
<
6.
We know that mi (E)=b-a-m. (E') m. (E)+m ll (E')=b-a, by (3). Now (2) becomes b- a+m (01 n G2 ) < b-a+c. This ~ m (G 1 n G2 ) < IS. II. Let m (G1 n Gs), < E To prove E is measurable. G1 ::> E => m. (E) ~ m (Gl ). Similarly m. (E') ~ m (G 2 ) Hence mll(E)+m.(E');:;,:;; m(G1 ) +m(G2)=m(G1 U G2)+m(G1 nG2) or mil (E)+mll (E') ~ b-a+6. This ~ m. (E) ~ b-a-m. (E'H'E=m. (E)+ •. Making £_0, we get m. (E) ~ m, (E). But m. (E) ;;;;, m, (E). This => m. (E)=mi (E) s> E is measurable. Theorem 11). If A ill a countable set then prove that m.(A)=O. Proof. Let A be an enumerable (countable) set. To prove m, (A)=O. By assumption, A is expressible as A={xft : n EN}. or
Let each point x" be enclosed by an open interval of length in such a manner that these intervals form a family of disjoint open intervals. The measure of this family of sets is
_/2
ft
-
E
(J)
..1!.1 _/2"= I -(t) =f. This => A can be contained in an open interval of length _. This Q m, (A) EO; _. But II! is arbitrary and hence making 5 .... 0, we get mil (A) But m. (A) ;;iii 0 ¥ A. Hence m. (A)=O.
~
O.
Corollary l. Since the set Q of rational numbers is countable and so m. (Q)=O. Also m. (N)=O=m. (I). Corollary 2. Since the length of the interval (0, 1) or [0, 1] is 1 so that m. «0, ]»=m ([0, l])=l:;t:O.
89
LEBESGUE MEASURE OF A SET
Hence [0, 1] is not countable. Corollary 3. The converse of Theorem lO is not always true. Example. Outer measure of Cantor's set is zero but it is uncountable set. Also Cantor's set is itself measurable. Hence any set with outer measure different from zero is uncountable. Theorem 11. lfme (A)=O, then A is measurable. Or, to prove that a linear set 0" Lebesgue exterior measure zero is Lebesgue measurable. [Banaras 1966] Proof. Let m. (A)=O. To prove A is measurable. We know that m. (A) ;:;at mi (A) ~ 0 Y A. This => 0 ;;;, m, (A) ;:;at 0 => rni (A)=O. Thus me (A)=O=m, (A) or m. (A)=m. (A), this => A is measurable and its measure is zero. Ex. An enumerable set is measurable and its measure is zero. [Kanpur 1976 ; Nagpur 64 ; Banaras 69] Solution. Prove as Theorem lO that me (A)=O. This => A is measurable and its measure is zero. Problem 7. The difference of two measurable sets is measur[Meerut 1986J able. Solution, Let El and E2 be measurable sets. To prove that EI-E2 is a measurable set EI-E2=El
n
E 2 '·
E2 is measurable => Ez' is measurable ~ El n E 2 ' is measurable ~ EI-E2 is measurable. Problem 8. If a measurable set G1 contains another measurable set G2 , then show that G1 -G2 is measurable alld its measure is 1I1(G1 ) -m(G2 )· Solution G~ is measurable => G2 ' is measurable => G1 n G2 ' is measurable => G1 -G2 =G 1 n G?' is measurable Remains to prove that m (G 1 - G 2 )=m (G 1 )-m (G 2 ). Since (G 1 -G2 ) n G2 =rp. Hence 111 [(G 1 -G2 )+G1 J=m (G 1 -G 2 )+m (G 2 ) or In (G1)=m (G 1 -GZ)+m (G 2 ) or In (G 1 }-m (G 2 )=m (G 1 -G2 ). Problem 9. A singleton set is a meas!lrable set and its measure is zero.
90
LEBESGUE MEASURE OF A SET
Proof. Let A={a} be a singleton set. To prove that A is a measurable set. Since every singleton set is a closed set and a closed set is a measurable set. Hence A is measurable. Furthermore A can be regarded as closed interval [a, a] which contains only the point a. By definition of measure m ([a, a])=a -a=O. m (A)=O. ' Problem 10. Sh~w that the Lebesgue measure of the following set is zero {x E R : 0 <: x < 1 and x has a decimal expansion not using [Banaras III, 1970] the digit 7}. An alternate statement. Prove that the measure of the set of points in the interval (0, 1) representing numbers whose expansion as infinite decimals do not contain some particular digit [Agra 1973, 65] (say 7) is zero. Solution. Divioe the open interval (0, 1) into 10 equal parts. Delete the interior of the 8th part. Again subdivide the remaining 9 intervals, each in 10 equal parts al\d delete the interior of 8th parts each of them. This process will be continued indefinitely. The points, which remain, constiute the required set. We denote this set by E. Then E={x E R : 0 < x < I and x has a decimal expansion not using the digit 7}. (0, l)-E is the complementary set E'. The set E' consists of an enumerable number of disjoint open intervals. Here E' is an open set. An open set is always measurable whose measure is the sum of measures of intervals which form the open set. The number of intervals deleted and their lengths can be written as Length of each No. of intervals No. of subdivision 'deleted deleted interval 1st 2nd 3rd nth
1
]/10
9
1{102 1{103
92
1{1O"
91
LBBSGUE MEASURE OF A SBT
1 9/102 1 9/102 10 = 10 +[='9710=10 + -J(Hf=10=1 m (E')=1 m (EU E')=m (0, 1)= 1-
°
or m (E)+m (E')=1. For EnE'=q, or m (E)=l-m (E')=I--1=0 or m (E)-=O. Therefore measure of the required set is zero. Remark. The end points of the deleted intervals with their limiting points from a non-dense perfect set. Problem 11. Let x be a number in the interval (0, 1) written in the scale of 10 as x=0.x1x.xs ... x" ... where none of the x's is 5. Find the content (or measure) of the set so defined Solution. Divide the interval (0, 1) into 10 equal parts. Delete the interior of the tth part. Again subdivide the remaining 9 intervals each in 10 equal parts and delete the interior of 6th part in each of them. Tt,.is process will be continued indefinitely. The points, which remain, constitute the required set. We denote this set by E. Then E={x€R: O<x<1 and x has a decimal expansion not using the digit 5}. «( , 1)-£ is complementary ~et E'. The set £' consists of an enumerable number of disjoint open intervals. Here E' is an open set. An open set is always measurable whose measure is the sum of measure of intervals which form the open set. The number of intervals deleted and their lengths can be written as No. of subdivision No. of intervals Length of each deleted interval deleted. 15t 1 1/10 2nd 9 1/102 ~rd 9CI 1/103
11th
]/10" 1
9
92
S,,-l
m (E')=IO+ 102 +103+"'+ 10"-+'"
1 9/102 _1 9/102 =10+1-9/10-10+./10 =1 m (E')=I.
92
LW3E GUE MEASURE OF A SEf
m (EuE')=m (0, 1)=1-0=1,
EnE'=~
m (E)+m (E')= 1
or m (E)=l-m (E')=1-1=0 Measure of the required set is zero. Problem 12. Find the measure of the set {x.} in' (0, 1) where x,. expressed as an infinite decimal does not contain the digits 5 and 7. Solution. Divide the interval (0, 1) into equal parts. Delete the interiors of 6th and 8th parts. Again subdivide the remaining intervals, each in 10 equal parts and delete the interiors of 6th and 8th parts in each of them. This process will be continued indefinitely. The points, which remain, constitute the required set. We denote this set by E. Then E={xER : O<x
No. of intervals deleted 2 2-8 2'8 2
Length of each deleted interval 1/10 1/102 1/103
1/10" m (E
,
2
2-8
2-8 2 _
2-8,,-1
)=[6+ IOZ+)]3 +-_.+ ~ +-_. 2/10
2/10
=1~871()=2/10 =1
or
m (E')=1. But m (EUE')=m (0,1)=1-0, and E~E'=rp m (E)+m (E')=I-O=1 m(E)=l-m(E')=I-l=O
Measure of the required set is O.
Ans.
LEBESGUE MEASUEE OF A SET
93
76. Definition. A relation which holds except in a set of measure zero is said to hold almost everywhere. The phrase "almost everywhere" is, in short, written as 'a.e'. Remark. The phrase "almost everywhere" is a particular case of "everwhere." 7'7.
Cantor's tennary set. [Vikram 1969; Banaras 66, 64; Meerut 1990, 88, 86] 1r-~F1~-rt::.=::::::=='1 0121
M
2!!
iii
3
9
9
Take the closed interval [0, 1). Divide the closed interval [0, 1) into three equal parts. Take away the interior of the middle part, i.e., the open interval (1/3, 2/3). This is the first stage of our construction. Again subdivide the remaining two closed intervals [0, 1J and [2/3, 1], each in three equal parts. Delete the interiors of the middle parts i.e. the open intervals
(3~" 3~ ) and ( ~
,
~
)-
This is the second stage of our construction. Similarly at ptll I stage, 2p - 1 open intervals are removed of length 3:11 each. The points, which remain, constitute Cantor's tennary set. We denote this set by E-[O, I] - E is the complementary set £' w.r.t. [0, 1] and consists of an enumerable mutually disjoint open intervals .. (i) Hencl;! E' is an open set. An open set is always measurable whose measure in the sum measures of intervals which form the set £' 1
2
2J
2"-1
m(E')=3+ji+ 33+······Jri + ...
=i [l +2/3 +(2/3)2+ ... ] 1
1
=3' 1-2/3 =1. •
m(E')=I
m (E)+m (£')=1 or m (£)=1-1=0. This zero is the measure of the set E. [Kanpur 1989; Meerut 88, 86; BaDaras 6"]
94
LEBESGUE MEASURE OF A SET
(ii) Since g' is open and hence E is closed. ~ D (E) C E. This Also here we get ECD (E). so that E=D(E). This ~ E is a perfect set. (iii) Let the points of E be represented by infinite decimals
(in the scale of 3). Let O.olo2aa ... in the scale of 3) be a point of E. S' 0 a1 a2 + a3 a. Ince .a,02aa···=3+F 33-+"'+3"'" Since we are expanding decimals in the scale of 3. Hence the a1, a2> 0a, ... can not take the value 3 or greater than 3. Hence the
only possibility is that 0 1 , a2 , aa, ... can take the values 0, 1,2. At the first step all the points with a1 = 1 are removed. Similarly second step removes all the remaining points with 02= 1. Similarly at nth stage all the points with 0" as 1 are taken away. It means that a1 , a 2 , ... can not take the value 1. Thus we see that if o a1a2a3' .. is an element of E, then a.. =O, 2 ¥ n E N. The points 1/3 and 2/3 are taken to be represented by 00222 ... and 0'2000 ... (iv) To prove tbat the set E is uncountable. [Meerut 1990, 86] Suppose not. Then the points of E form a countable set. It means that every point in E must appear in the sequence. of distinct points. i.e., E={x...: n E NJ. Writing decimal expansion of these x/s,
. .. (1)
Xl =O'Xll X 12 XU" ,Xltn
X2=0'X 21 X 22 X 23 · "X1'"
X,,=O'X"l X"2 X"3" . X"", . ..
X,p=O or 2 ¥ rand ¥ h. where each Construct real number
~=0'fl~2~3"
s.t. if
x ...",=O, write
~",=2
. .. (2)
95
LEBESGUE MEASURE OF A SET
x".".==2, write ~=O. In either case x..¢~". ¥ m. This => x",¢~ ¥ m =>~ it E ... (3) according to (I). ": f",=O or 2 ¥ m. This => t e: E ... (4), according to (2). Evidently (4) contradicts (3). Hence the required result follows. Tbeorem 12. Show that the Lebesgue exterior measure ;s a [Banaras 1967] Caratheodory outer measure. Proor. To show that Lebesgue exterior measure is Caratheodory outer measure, it is enough to show that
and if
(i)
m. (E) ;;. O.
(ii) El C E z => m, (E1 ) <; m. (E2 ). (iii) For any finite or infinite sequence of s.ets EIO E 2 ...... , E,........
.
m. (u E k ) <; E m, (Ek )
"
(iv) m. {El U Ez)==m. (El)+m. (E2) jf E'l. n E.==",. By definition, m. (E)=inf {m (G) : G is open, G :J E}. (i) From this it is quite clear that m. (E) ;;. O. For m (G) ;;. O. (ii) Let El C E 2• To prove m. (E1) <; m. (EJ. For proof refer Theorem 3 (ii), Page 76. (iii) To establish the result (iii), it is enough to show that m" (E)
< ~. m.(E 00
k ).
00
where E= U E 1,.
1'-1
t-l
Case 1. When the series Em. (EI:) is divergent. Since every covering of E is a covering of E l , E 2 , : and therefore m. (E) ~ 00. Case 2. When the series E m. (EI:) is convergent. Let the /I
family {ItI : j= I, 2, ...... } of open intervals be open covering of Ek y. k. Given E > 0,
Here the totality of these open intervals is at most enumerable. Evidently this is open covering of E.
96
LEBESGUE MEASURE 'OF A SET
=
00
E m, (Ek
k-l
)+_.
00
Thus m, (E) ~ ~ m. (E.,)+E. k~l
Making E-+O, we get the required result m. (E)
~
00
E m. (E,,) k=l
Let El n E 2 =q, so that d (EI , E 2 ) > O. d being the metric on X. Let d (EIO E 2 )=3 > o. Also let E=E1 U E 2 • To prove that m. (E)=m. (El)+m. (E2 ). Let {I., : k=l, 2, ...... } be an open covering to E S.t. d {I,,) < a, d (lk) represents the diameter of I". Case I. When either m, (El ) or m2 (E2 ) is finite. Let T={Ik' : k= 1,2, ..... } be a sub-family of {II: : kEN} (iv)
00
s.t. U /,.' covers E1 • /.=1
Similarly suppose that S={I/: k=l, 2, ...... } is a sub-family 00
of(l,,: k=l, 2, ... } s.t. U
I/:- covers E 2 , where
1.:_1
k E N}={I/, 12' la'" .. ... , 11' , / 2', 13- , ...... }. Obviously Tis an open covering of E. and has no points in E2 • Similarly S is an open covering of E2 and has no points in E1• Then
{lk :
,
m.(El)+m,(E2)~
=
00
~ m /:=1
00
00
E m(llc')+ 1J m(ll:) "~l
11-1
00
(llC' U [,:)= 1: m (l/l:)=m. (E) k=I'
m. (E1 )+me (E2 ) ~ m. (E). By (iii), just proved, m. (E1)+m, (E2 ) ;;.. m, (E1 U E2 )=m. (E) i.e. m. (E1)+m. ;;. m. (E). Combining (3) and (4), we get the required result.
i.e.,
...(4)
97
LEBESGUE MEASURE OF A SET
Theorem 13. The uniOll of a finite number of measurable sets is measurable. Proof. Let E1 , E2 •••••. , En be measurable sets and let II
E=U E". k=1
To prove that E is measurable. Consider an arbitrary fi > O.. E~ is a measurable set => 3 a closed set P,: and an open set G" S.t. F" C E" eGa,
m (G,..}-m (F,.) n
Write
< nF.
... (1)
n
F=u F,,, G= U G. k=1
k=1
Being a finite union of closed sets, F is closed. Similar arguments prove that G is an open set. ,.
II
R
F" C E,.. C G" => U F" C U Ek C U G" "=1 k~l "=1
=> Fe E C G. => FeE, E c G. => m, (F) ~ mt (E), m. (E) ~ m. (G) (Refer Theorem 3) => m(F)=mi(F) ~ mi (E), m. (E) ~ m. (G)=m(G) [For Closed sets and open sets both ore measurable sets] => m(F) ~ m, (E) ~ m. (E) ~ m (G) => m. (E)-m, (E) ~ m (G) -11'1 (F). . .. (2)'. G-F=G n F'. For G n F'=Grt(X-F)=GnX--GnF=G-F. =A finite intersection of open sets =An open set=A measurable set G={G-F) U F and (G -F) n F=r/>. :. m (G)=m (G-F)+m (F). From which, m (G-F)=m (G)-m (F) Similarly, III (G,. -F, )=m (G,) In (Ft ). Notice that G
n
Feu (G,. -FJ. This => k-l
III
"
(G -f) ~ E k=l
III
(G k - Fd
98
LEBESGUE MF.ASURE OF A SET
~
m (G)-m (F) .;;; E• [m (Gt)-m (Ft )] 11=1
.. m (G)-m (F) <:
•
<
-.n=., by (1)
<
fi
n
E.
Using this in (2), or
m_ (E)-m, (E) C;; m (G)-m (F) m. (E)-m, (E) < •.
But. is arbitrary. Hence it can be made as small as we please. Making ..... 0 in the last inequality. m,(E)-m,(E) E;; O. This ~ m,CE) .;;; ml(E) But m_(E) ~ m,(E) is always true. Combining the last two, m.(E)=mi(E) This proves that E is a measurable set. Tbeorem 14. The intersection of a finite number of measurable sets is measurable. Proof. Let E1• Ea ....... ' En be measurable sets. To prove that
u"
E1, is measurable.
11=1
We know that: The union of finite number of measurable sets is measurable. Let all the sets E/: be contained in an interval fl. Then Et'=complement of Ek w.r.t. fl. By De-Morgan's law
UE,,' = [ .=1 U Ek]'
11=1
Ele is measurable .. its complement E'/: is measurable. ~
u"
11=1
e~'
is measurable.
~ [ "'=1 U E1:]' is measurable. " is measurable. onE" "-1
... (1)
99
LEBESGUE MEASURE OF A SET
Theorem 15. First Funclamental Theorem. are disjoint measurable sets and
If E1 ,
E2, •••
E=E1 +E2•••••• 00
then E is measurable and m (E)= E m (EI»' 1>=1
[Meerut 1990; Kanpur M.Sc. P .. l98S; Banaras 68] Proof. Let E J , E 2 , ...... be disjoint measurable sets so that m. (Ek)=m, (Ek)-m (E k ) yo k E, n EI=cfi for i='l=j.
Let all the sets be contained in an interval (a, b] then
or
... (1)
•
•
r~l
r=l
Let S,.= E 1-1" then m(S.)= Em (E,). For ErnE.=cfi for r='l=s 00
Then S" C 1: Er=E,
i.e., S. C E so that E' C Sfa'.
r=1
E' C S,,'
~
m. (E') ~ m. (S,.')=m (S.') => m. (E') ~ m (S,/) [For S. is measurable~ by theorem 13 and so S,,' is measurable]. => m. (E') ~ m (S,,')=b-a-m (Sn) =b-a- "E m (E,,) r=1
~
m. (E')
~
" m (Er) b-a- E r=1
n
=> E m (Er) r=1
~
b-a-m. (E')=m, (E),
100
LEBESGUL MEASUF.E OF A !'oET
:0
• m (E) m, (E) ;;. E 7'=1
Making n-+oo; we obtain m. (E)
~
00
E m (Er) 7'=1
=> m, (E) ;;;;.
E m (Er) ~ m. (E) ... (3), On using (I)
7'=1
.. m, (E) ;;. m. (E). m, (E) ~ m. (E) is always true. :. m. (E)=m. (E). This ., E is measurable and But
.. m (E)
~
00
E m (Ek ) ~ m (E), by (3). ~_1
This concludes the problem. Theorem 16. Second Fundamental Theorem.
If E 1 , E 2 ,···are
00
meosuroh/e sets, then the set U Er is measurable. 7'=1
Proof. E=
00
U E~ is a measurable set. k_l
Let
By De-Morgan's law,
F'=[ nE,; ]= UE t'
~=l
k-l
Ek is measurable
::>
its complement Ek ' is measurable. 00
=> 1: E,r' is measurable 1:=1
.. F' is measurable => F is measurable.
Lialtinl Sets. Theorem 17.
Suppose that
<
E"
>
is a mOl/otonic increasing
101
LEBESGUE MEASURE OF A SET
~equence,
i.e., each set of the sequence contains the following one. 00
If E is their limiting sum, i.e. E= U £.,., then 11=1
(E) Lim E} m \. = n_oom ( n.
[Gujrat 70]
An Alternate Statement. Define a set function on a ring R. ~ on a ring R is countable additive and there exist A and a sequence < .An > e R such that A, C .If'+! and
If a set function 00
.4.= U ...4., prove that 9> (An) ~ rP (A) as n~a::>. ra=1
[Kanpur M.Sc. P. 1910] Proof. Let < E" > be a monotonic increasing sequence of measurable sets, so that 00
El C E2 C £3 C···· .. Also let E= U £1" lII=1
An enumerable union of measurable sets is measurable. Also difference of two measurable sets is a measurable set. Hence E and E,+l-E,. are measurable sets. Notice that
£=E1 +(E2 - E1 )+( E3 - £2)+'" +(E,-E,-I)+ ..... . 00
£=E1
i.e.
+ ,=1 ~ (E"+l-E,,)
is a disjoint union of measurable sets. Therefore 00
m (E)=m (El) + ,,_1 ~ m (Er+l-Er)
... (J)
" (m (Er+l)-m (E,)] m (E)=m (El )+ 1:
or
,=1
or
... (2)" .
For
00
2-' [m (Er+J--m (Er)]
,.=1
-=Lim [{m (E?)-1J1 (E1)}+{m (Ea)-m (E2 )} n-+oo
+ ... +{m (En)-m (Ell-1m
102
LEBESGUE
MEASU~E
OF A. SET
Lim m (B)= n_oom (En).
From (2),
Ex. Show that if < Eft > is sequence of measurable sets all contained in a set offinite measure and Lim En exists, then Lim Eft is measurable and m (Lim E,,)=Lim m (Eft). [Banaras 1967J Soilltion. Let < En > be a sequence of measurable sets. Then this sequence will be either monotonic increasing or monotonic decreasing. To be particular at this stage, we further suppose that the sequence is monotonic increasing so that E1 C E2 C E8
c ..... .
00
Let E= U Ek • Now prove as in the abOYe theorem 17 that .10-1
m
,lim En )=lim \n_oo n_oo
In
(E). /I
Theorem 18. If E 1 , Ez, ...... are measurable sets and 00
E= r_1 n E~ and if El ::> E2 :) ... , then Lim m (Eft), z.e. . m (Lim Eta ) = Lim m (En). m (E)= n~;rJ n_ 00 n_oo Or Let p. be a countably additive non-negative measure defined on a a-algebra of subsets of a set X such that I' (X) < 00. ~lrow that if {An: n E N} is a contracting sequence of I'-measurable subsets of X, then I'
[
n ] Lim n E N An == nE N
J.I. (An).
[Banaras 1964J Proof. Let < Eft > be a contracting (monotonic decreasing) sequence of measurable sets so that ~I ::> E2 ::> Ea ::> ...... Also let
00
E= n En.
,,-I
Tken E=
Lim
"_00 E.
103
LEBESGUE MEASURE OE A SET
i.e.
... (1)
is a disjoint union of measurable sets. Since an enumerable union and intersection of measurable sets are measurable. Also difference of two measurable sets is measurable. Now (1) is expressible as 00
m (E1)=m (E)+ l: m (Er-Er+1 ).
...(2)
r=1 00
m (E1)=m (E)+ E [m (Er)-m (Er +J )]
or
r=l
"-1
or
E [m (Er)-m (Er+l)]
r=l
or
... (3)
= n-+oo Lim [m (El)-m
)-riz (E.)1 + ... T[m (E"-l)-m (E
(Eq)]+[m (E2 ~
rI ) ]
= Lim
(3)
[m (E
) -
m (En)]
1 n-.oo can also be written as
Lim m (E1)=m (E)+m (E1)- n'+oo m (En)
or
m (E)=
Lim n~OO
III
(En).
Theorem 19. The set of all measurable (i.e. Lebesgue measurable) sets is a--algebra (or a-field). Proof Let A be the set of all measurable sets. Let A, B E A be arbitrary so that A and B are measurable sets.
To prove that A is a -a-algebra, we have to show that (i)
A' E A
104
LEBESGUE MEASURE OF A SET
(f'i) AU.:; .2 A (iii) if < En > is a sequence of measurable sets in A 00
u
then
E, E A.
~=1
Let all the sets belongiilg to A be contained in an interval (a, b). (i) Let A be measurable so that m. (A)=m, (A)=m (A) .... (1) To prove that the complement A' of A is measurable. ...(2) We know that m, (A)=b-a-m. (A'). This is true for every A. Replacing A by A', m, (A/)=b-a-m. (A) ... (3) writing (2) and (:S) with the help of (I). m (A)=b-a-m. (A') mi (A')=b-a-m (A) m. (A/)=b-a-m (A)=m, (A') From which, or m. (A')=m, (A'). This => A' is measurable. (ii) Let E 1 , E2 E A so that El and £2 are measurable sets and E=E1 U E 2• let To prove that E E A, we have to show that E is measurable. Consider an arbitrary number c > O. Ek is a measurable set => :I a closed set F" and an open set Gle e s.t. F.,. C E" C Gk , m (G,,)-m (FTe)
E
< T'
Write Then F is a closed set and G is an open set so that F and G are measurable sets, 2
FTe C Ei C G" => U Fit .=1
~
o
I
I
e i=1 u Ek e II_I u
G7r,
Fe Ee G.
=> FeE, E e G• .. m, (F) ~ m, (E), m, (E) m, (E) ~ m. (6) ~ m (G)
~ In
(G)
m (F) ~ [For m, (F)=m (F), m. (G)=m (G) and m, (A) <; m. (A) ¥ A] => m. (E)-m. (E) ~ m (G)-m (F). •.. (4)
LEEr~Gt'E
lOS
MEASURE OF A SEt
This
2
~ m
(G-- F) E;', l: m (G,,-Fre ) k=l
=> m (G) -m (F)
~
k:l 2
[11'1 (Gre)- 11'1 (Fk ) ]
e
< 2.2=.
=> m (G)- m (F) < , => m. (E)-m. (E) ~ m (G)-m (F)
=> m, (E)-m. (E) <
<
I.
by (4)
E.
MakjngE~O,
This But
(iii) Let
m. (E)-I1I. (£) ~ O. For I is arbitrary. => m. (E) ~ In. (E). m. (E) ~ m. (E) is true for any set E => m. (E)=m. (E) => E is measurable.
<
En
>
be a sequence of measurable sets and let 00
E=
l- En .
.. =1
To prove that £ is a measurable set. For the sake of convenience we suppose Ek nE/:'=0 for k:l=k'. Here the operation E stands for u. We know that
or or -
m. (E) Let
00
~ ~ .=1
m (E,,).
. .. (5)
•
S.. = 1.' Ek • 1c~1
El and E. are measurable sets. El U E. is measurable by (ii), => S.. is measurable
106
LEBI:SGUe MEASURB OF A SET ~
its complement S ..' is also measurable.
m (S..)=m (
.
i.e.
1:
Jt~l
EII)=
i:
k=1
m (Ek )
m (S..)= 1: m (Et ).
. .. (6)
11=1
S.. = 1:" EkCE
i.e.
Sn C E so that E' C S,.'.
1=1
E' C S ..'
~ m~ ~ ~
==-
(E')
~
(E)
~
me (S,.')=m (S,/) m~ (E') .s;; m (S,,')=b-a-m (S.. ) m. (E') ~ b-a-m (Sn) m (5..) .s;; b-a-m. (E')=m. (E) m (S.. ) ~ mi (E) II
~ nt.
m (S..)= }; m (EA:) .=1
~ 111.
(E)
"
~ E 11=1
m (Ek ).
.Making n-'?oo, .
m. (E)
.
~ .
00
L: m (Ek )
1:=1
;;;;.
me (E), on using (5).
mi (E) ~ m~ (E). But m. (E) ~ m. (E) is true for any set. Consequently m. (E)=m. (E). This ~ E is measurable. This concludes the problem. 7'8. Definition. Convering in the sense of Vitali. [Banaras IV 70] Let E ·be a set and M be family of closed intervals such that none of them is a singleton set. The set E is said to be covered by the family M in the .sense of Vitali if I.rf x E E and ¥ ~ > 0 ~ 3 a closed interval I E M s.t.XE/,
or
In
(I)
<
6.
That is to say, E is said to be covered oy the family M in the sense of Vitali if every point of the set E is contained in some small closed intervals belong to M.
107
LEBESGUE MEASURE OF A SET
Theorem 20. Vitali's Covering Theorem. ~f a bounded set E is covered by a family M of closed intervals in the sense of Vitali then it is possible to find a countable (finite or enumerable subfamily of closed intervals {/k} of M such that l",n/.. =0 for m=l=n and m. [E-E I,:J=O. I,
[Punjab 1967; Banaras IV 70J Proof. Let E be a bounded set which is covered by a family M of closed intervals in the sense of Vitali so that x E E, e > 0 ~ 3 a closed interval I!. E M s.t. m (/10) < e ...... (*). To prove that 3 a pairwise disjoint sub-family {/k} of M with the property m. [E - E Ie] = 0, Ie
where the symbol E stands for union. Choose an interval (a, b) S.t. E C (a, b). Form a family Mo of closed intervals from M s.t. M o={I,: EM: II, C (a, b)}. From the construction it is clear that Mo C AI and every member of Mo is contained in (a, b). Also Mo covers £ in the sense of Vitali. Pick an interval II E M. If E ell' then the theorem is proved. For m. (£-/1 )=0, by virtue of (*). In the contrary case pick another interval 12 E Mo s.t. 12n/I=~ If E c 12 , then the theorem follows. For
m. ( E -
1:
k-l
I"
)=0,
by virtue of (*).
III the contrary case, pick an interval 13 E Mo S.t. I". n 1.. =~ for m=l=n and tn, n= I, 2, 3. Repeating this process Il times, we have a family consisting of pairwise disjoint closed intervals 110 12 , ....... I".
" Ir, the proof is complete. If £ C E ,=1 In the contrary case, E
so that
108
LEBESGUE MEASURE OF A SET
.(1)
,. In this case, Sl!t F,.= E I r , G,,=(a, h)--F,.. r~l
Now we consider those intervals of M which are subsets of the open set 0,.. By virtue of (1) such intervals exist and their length ~ m «a,
b».
Let k .. denote the least upper bound of the lengths of these intervals and let 1,.+1 be one of them for which k m (1.+1) >
i
Of course k" > O. For no interval of the family is a singleton set Therefore I ..+1 nI.. =rp for r=l, 2, ......... , n-l. If the process of constructing the intervals II' 12 ........... ends after a finite numl'er of steps, then the theorem is proved. If this process docs not come to an end after a finite number of steps, then we are left with a sequence < I. > of pairwise disjoint closed intervals. Now we shall show that this sequence also satisfies the given condition, i.e.,
me
[E- ~
k=l
h ]=0.
. .. (2)
For proving this, we construct a closed interval D/: ¥ k s.t. (i) the length of Dk is 5 times the length of I,,, i.e. m (D 1,)=5m (lk)' (ii) the middle point of Dk coincides with the middle point of II;. By initial assumption, every element of the sequence is contained in (a, b) so that
III (1~1 Consequently III
I.
)
~ 111 «a. b».
(~l Die )=sm( tl
lit; )
< In> ...(3)
~ m «a, b»
=finite number.
LEBESGUE MEASURE OF A SET
... m( ; D.. ) 7<-1
<
109 00 •
-
In order to prove (2), we shall prove that
ELet
E I .. C E Die ¥ i.
E G" Gi is open
. .. (4)
7.:=,
x E E - E I", then x e= G, •
X
-
-
.=1
1t=1
¥
i.,
=> i a closed interval I E Mo s.t. x E Ie Gi •
. .. (5) However, it is impossible that ICC ... This follows from the fact that m (I) ~ k tt ~ 2m (I'Hl) ¥ II. This is impossible. For m (1,,)-+0, by (3). Consequently the relation (5) is not satisfied for some nand hence the relation I n F..=I=. ... (6) holds. Next we assume that n is a least positive integer s.t. (6) holds. Therefore n > i.
For 1 n Fi=rP and Fl C F2 C Fa··· From the definition of n, it follows that 1 n F..- 1 =;· We can now deduce two results from this. The first is . .. (7) I n I.=I=if. and the second j., , C G"- l . .. (8) and m (I) ~ k"-1 < 2m (I.. ). From (7) and (8), we find that I C DII •
-
As a result of which ICE D/: so that x E Iccd
-
X Dk
1 ~t
Hence (2) follows. Remark. This proof was originally given by the famous mathematician Banach. Under the hypothesis of the prl!vi<.'us Theorem 19, we can prove the following theorem: Theorem 21. For every, > 0, 3 a fillire sub-srstem '1' 12 "" I .. of pairwise di.~;oil/l elored il/tena/s of the system :\,1 S.t.
110'
LABESGUE MEASURE OF
m.
[E - i: It J< t=1
~
SET
E.
7'9. Definition. Set of the type Fa. A set E is said to be of the type Fa if it is expressible as a union of an enumerable number of closed sets Flu i.e. GO
E= U Ft. k=l
7'10. Definition. Set of the type G,. A set E i'i said to be of the type Ga if it is expressible as an intersection of an enumerable number of open sets G/;> i.e. 00
E=
n G". k=1
Evidently complement of Fa is G6 and conversely. 7'11. Definition. Borel Set. A set E is called a Borel set if it can be obtained from closed and open sots by using a finite or an enumerable number of union and intersection operations. Example. The sets of the type Fa and G6 are Borel sets. 7'12. Definition. Borel measorable. A bounded Borel set is said to be Borel measurable. Most writers refer to all Borel sets as being Borel measurable. Theorem 22.' The sets of the type Fa and G. are measurable sets. Proof. Let E be a set of the type Fa. Then E is expressible as
-
E= EFt.
where FI: is a closed set.
/1=1
Fir. is a closed, set ~ Ft is a measurable set. Also an enumerable union of measurable sets is measurable. Hence
00
U F;" i.e.
E is measurable
"=1
(ii)
as
Let E be a set of the type Ga so that E can be expressed
111
LEBESGUE MEASURE OF A SET
Gk is a open set * Glo is a measurable set. Also an enumerable intersection of measurable sets is measur-
able.
00
Hence
. U G1" i.e.
E is measurable.
110=1
A Borel measurable set is Lebesgue measurable . .£Meerut 1987, 86J Proof. T~et a set E be Borel measurable so that E is a bounded Borel set. To prove that E is Lebesgue measurable. E is expressible as E=(U Gk ) n (n Fk ) Tbeorem 23.
k
k
where Gk is an open set and Flo is a closed set. E is bounded ~ all Gk and aU F_ are bounded. Also G1, is open and F" is closed. ~ Gk and F1, both are Lebesgue measur~ble sets. ~ U G1" n Fr, are Lebesgue measurable sets. k
~
k
For a countable union or intersection of measurable sets is measurable. (U Gk ) n (n Flo) is also Lebesgue measurable Ie
Ie
by the same reasoning. =:>
E is Lebesgue measurable.
Exercises 1. Define a set function on a ring R. If a set function.p on a ring R is countably additive and there exist A and a sequence < An > E R such that Ar C Ar+1 and 00
A= U
A,., prove that .p (A .. ) -+ c/> (A) as n_oo •
..=1
[Kanpur M. See P. 1980] 2. If E1 and E2 are measurable subsets of [a, b], prove that E1 U E2 , El n £2 and E1-E2 are measurable and show also that . (i) m (E1)+m (E2 )=m (E1 U E 2 )+m (E1 n E 2 ) (ii) m (E1}-m (E2 )=m (E1 -E2 ) if E2 c E1 • [Kaapur M. Sc. P. 1979] 3. Define the Leb~sgue measure of an open set of real numbers. Prove that an enumerable set is measurable and its measure is zero. Show further that a continuous function defined [Kanpur 19731 in a cl03ed interval is measurable.
LEBE~GUE
112
MEASURE OF A SET
4. Define outer and inner measures. If 1-'''' is an outer measure on Q and M is the class of all ",·-measurable sets, then show that M ill a _a-field and fL'" is a measure on M. [Kanpur Statistics 1977] S. Define measurable sets in the sense of Labesgue giving two examples. If X and Yare measurable sets, show that Xn Y is measur e. able. What can you say about X U Y? LKanpur 1975] 6. (i) Give an example of a set which is not measurable in the Lebesgue sense. Justify your example. Oi) Is the Lebesgue measure of the union of two measurable sets equal to the sum of their measures? [Kanpur 1974] 7. (a) Define the Lebesgue measure of an open set of real numbers. If 0 1 and O2 are any two open sets then prove that m(01+ 0 ;l)+m(Oh 02)=m(01)+m(02)' where m(O) dendtes the Lebes'!ue measure of 0 1 , (b) Prove that a set E of real numbers is measurable iff, given E > 0, EI set~, el' e2 s.t. E=~+el-e2 and ~ is the sum ofa finite number of intervals, and m. (eJ < E, me (e2) < c. [Kanpur 1973] 8. (a) Let E 1 , E 2, ... , be a sequence of sets of real numbers, Prove that m* (E1 +E2 + .. ,) ~ m* (E1)+IJ-.* (E2 )+, .. , where m'" (E) is the Lebesgue outer measure. If in addition the sets E, are pairwise disjoint, i.e., no two of them have common points, then prove tha t m* (E1 +E2 +, .. ) ~ m", (E1)+m", (E2)-+-:" where m* (E) denotes the jnner measure of E. (b) If M is measurable set, then prove that for any set E, m'" (E)=m""(EM)+m'" (E-£.\4). lKanpur 1972) 9. Define Lebesgue measure of a set. Construct a non-dense perfect set in the interval (0, 1) whose measure is 1/2. lKanpur 1968J 10.
A of R.
Define the Lebesgue exterior measure m"'(A) of a subset For any sequence < A.. > of subset of R, prove that m"'( U An) ~ I. 11/* (A .. ). [Banaras Ill, 1971J ,
113
LeBESGUE MEASURE OF A SET
11.
If A, B, Care I'-measurable sets, show that peA) +I'( B) +1-'(C) =,..( A U B U e) +1'(Bn e) + l'(en.4) (a)
+1'(.4 n B)-I-'(AnBnC).
(b) zero:
Show that the Lebesgue measure of the following set is
{x E R : 0
<
x
<
1 and x has a decimal expansion not
using the digit 7}.
[Banaras III, 1970]
Define measures of open and closed sets and prove that m(E1 )+m(E;)=m(E)UE2 H-m(E1 nE2) where E) and E t are both open or both closed sets. Show that a countable set is measurable and is of measure zero. 12.
Prove that an outer measure is monotonic and cr-subadditive. [Danans 1%9] 13. Ii)
(ii)
Prove that m*(E1 uEz)+m*(E1nE2 ) ~ m*(E1)+m*(Ez) m.(EI UE2 )+m.(EI nE2 ) ~ m*(EI )+m.(E2). 00
.-1
Show that if E 1 • E2 ... are measurable and disjoint, then U Ell is measurable and m
(U
t=I
Ek)=
Um(E
k ).
Ie-I
[Danaras 1968]
14 Construct a set which is not measurable in the sense of Lebesgue. [Danaras 1968] 15. (a) Prove that Lebes~ue exterior measure is a Caratheodory outer measure. (b) Show that if < E.. > is a sequence of measurable sets all contained in a set of finite measure and lim Eft exists, then lim E,. is measurable and m (lim E,,)=lim In (E.. ) where the limit on the right hand side of tile equltlity exists. [Banaras 1%7] 16. (a) State Caratheodory's criterion of measurability of a set. Prove that a linear set of Lebesgue exterior measure zero is Lebesgue measurable. (b) DGscribe the construction of Cantor's ternary set and find i.ts Lebesgue measure. (Baoaras 1966]
114
LBBESGUE MEAWRE OF A SET
17. (a) Establish the equivalence of Caratheodory's and Lebesgue Criterion of Lebesgue measurability of a bounded subset of the real line. (b) Show that the Lebesgue measure of the Cantor subset of the unit intt:rval is zero. [Banaras 1964]
18. If,.,. is a measure on a ring Rand < E,. > is an increasing sequence of sets in R for which lim E.. E R, prove that
.
,.,. (lim E.. )=lim ,.,. (En). ,.
n
[Gujrat 1970J
8 Measurable Functions 8'0.
Definition.
Measurable function. [Poona 1970; Meerut 71; Kolhapur 69; Kanpur 1983,74,73, M.Sc. P. 75,76; Gujarat 70] An extended real valued function/. defined over a measurable set E, is said to be measurable (in the sense of Lebesgue) if the set {x E E :/(x) > al ...(1) is measurable for every extended real number a. The set {x E E: / (x) > a} is denoted by the symbol E(/> a). Remark (1) The condition (1) of the above definition can be replaced by any of the three conditions of theorem 4, page 118. Remark (2) The measure of the set E (/ > a) may be finite or infinite. Remark (3) If E=R, then the set E (/ > a) becomes an open set. 8'1. Definition. Almost everywhere. A relation, which holds except ID a set of ,measure zero, is said to hold almost everywhere. That is to say, a property P is said to hold almost everywhere on the set E if 3 Eo C E S.t. (i) the property P holds 'V x e E-Eo (ii) m'(E.,'=O. The following have the same meaning (i) almost everywhere on the set E (ii) for almost all points of E. The phrase almost everywhere, is denoted by the symbol a. e. or p. p. S·2. Definition. Equivalent functions. Two functions/and g defined on the same set E, are said to be eqUivalent if m [E (f:,cg)]=O. That is to say, functions / and g defined on the same set E, are said to be equivalent if 3 A, BeE S.t. E=A U B,J=g on A,J:;t:g on B, m (B)=O.
116
MEASURABLE FUNCTIONS
.·3.
De6aition. Characteristic function. [Meerut 1990] Let A be a subset of a set E. The characteristic function KA of A is defined as : I if x E A KA(X)= { 0 if x E E. . A
The function Kix) is measurable iff A is measurable. Hence existence of a non-measurable set implies the existence of a nonmeasurable function. The function K,.(x) is also called indicator flUlCtJo. of A . ..... Deftnition Simple Function. [Kanpur Statistics 1977, 75] A function is said to be simple function jf the range of the function is finite. For example, characterist;c function. , A real valued function ~ is called simple if it is measurable and assumes only a finite number of values. If ~ is simple and has finite number of values Ill> 112, ..... «ta, then", is expressible a'l
.
~(x)=
E _, KAi '=1
where A.={x : ~(X)="I}' The sum, product alld difference of two simple functions are simple. 8'5.
Limit Superior aad Limit Inferior Let < Ar : r EN> be a sequence of sets.
The supremum
00
of this sequence, denoted by set U A, and the limit superior of r~l
__
__
,,",,00
this sequence, denoted by Lim A" is defined as lim A.= n U A... n=1 tn._II
The iufimum of the sequence, is defined as the set
00
n
A.. and
..=1
lilllit ioferior of the sequence is denoted by Lim A•• and is defined as
Lim A.. =
00
00
a-I
_::::IB
unA...
If Lim A.. = Lim A". then we sar that the limit of the sequeace exists and the common value is dC!noted by lim A... Thus Lim A.=Lim A,,=Lim All if Lim A .. exists.
117
MEASURABLE FUNCTIONS
Monotone Sequence. A sequence of sets is said to be non-decreasing (expanding) if A,. C A,,+! V n and is denoted by < A. t >. A sequence of sets is said to be non-increasing (contracting) if A.. +! C A,. ¥ n and is denoted by . In either case the sequence is said to be a monotonic sequence.
Theorem. Every monotone sequence of sets is convergent. Proof. Let be a non-decreasing sequence, then
By def. lim inf A.. = lim sup A.. =
n01
U
n=1
[n Am]= UA,. Ift=..
[In~" A.
... (1)
11=1
J
... (2) 00
But we see that for any non-decreasing sequence, U Aft is in,,=11
dependent of k, so we may take k= I. Lim sup A,,=
lienee, by (2),
n [u J= U R=l
m-.l
A",
m~l
A".=
U A...
n-l
00
Hence, lim sup An= U A.. =lim inf A.. .. =1 00
or lim sup A.. =lim inf A .. = lim A .. =
U An. n=1
Hence monotonic non-decreasin~ sequence is convergent. Similarly we can also prove that monotonic non-increasing sequence is convergent. Theorem 1. Let f be a measurable function defined over a 00
measurable set E,: ¥ kEN and E = U E/... 1<=1
Then f is measurable
Ol'er
E
Proof. Let a be any real number. Let (be a function over a measurable set Ek V kEN so that Ek (f > a) is measurab;~ ¥ kEN.
mea~urable
118
MEASURABLE FUNCTIONS
Let To prove that f is measurable over E, we have to show that > a) is measurable. Being an enumerable union of measurable sets, E is measurable.
E (f
E (f> a)=(
UE
1:
)
(f:> a)=
k=l
U [E
t
(f
>
a)]
k=l
=An enumerable union of measurable sets =A measurable set. a) is a measurable set.
E (f > Theorem 2. Iff is a constant function over a measurable set E, then f is measurable over E.
0r
If f(x)=c (constant) for evpry vaiue of x E E, then f is measurable over E, E being a measurable set.
ProoC. Letfbe a constant function defined over a measurable set E so thatf(x)=c ¥ x E E. Let a be any real number. To prove thatfis measuTable over E, we have to show that E (f > a) is a measurable set. Evidently E (f > a)=JE .if c > a l4> If c ~ a In both cases, E (f > a) is measurable set. For E and tP both are mea<;urable sets. Theorem 3. Iff is a measurable function over a measurable set E, then f is also me.uurable over any set ACE, A being a measurable set.
ProoC. Let a be any real number. Let f be a measurable function defined. over a measurble set E so that E (f > a) is a measurable set. Let ACE be any arbitrary measurable set. To prove thatfis measurable over A, it is enough to show that A (f > a) is a measurable set. A (f > a)=[ E (f E > a)] n A = Intersection of two measurable sets =Measurable set. :. A (f > a) is a measurable set. . Lsbesgue measurable Cunctions. Theorem 4. To gi.'e four equivalent definition for the Lebesgue measure of a real valued function.
Prove their equivalence.
(Kanpur 1968; JSanara~ 69, 66]
119
MEASURABLE FUNCTIONS
Proof. Let f be an extended real valued function defined over a measurable set E. Thenf is Lebesgue measurable (or simply measurable) function jff anyone of the following is measurable: (ii) E(f~o), (i) E (f > 0), (iii) E (f < 0), (iv) E (f ~ a), where 0 is any real number and E (f > a) stands for the set (x E E :f(x)
>
o}.
Similarly we define the other three sets. To prove that these four sta.tements are equivalent. (i) ~ (ii).
,For
E(f~ a)= ~1
[(
Ef> a-
~)]
an arbitrary intersection of measura.ble sets is measurable. (ii)
~
(i).
For £
(I >
0)=
~1 E( f ~ a+ ~ [
)]
and an arbitrary union of measurable sets is measurable. (ii) <> (iii).
For E (f < a)=[E (I ~ a)J' and a set is measurable iff its complement is. Here complement of every set is taken w.r.t. the set E. (iii) ~ (iv). For
£(/~ a)=
n£(f< ~+~) n
n=1
and an arbi~rary intersection of measurable sets is measurable. (iv) ~ (iii). For
E(f < a)= ~1 [E ( f ~ a -
! )]
and an arbitrary uni"n of measurable sets is measurable. This completes the proof. 00
,
Remark. 1.1. {XEE :f(x);;tGt}= U {x E E :f(x) > «-(lin)} n_1
2.
E(/<<<)=£-[£(/><<)]
120
MEASUR~BLE
3.
E(f~ IX)=
00
n
{x E E:f(x)
,,=1
4.
<
PUNC nONS
cx+(ljn)}
E(f> «)=E-[E(f< «)] 00
n
5. {x E £ :/(x)=oo}
{x E E: f(x)
>
n}
11=1 00
6. {x E E:f(x)=-oo}= n {x E E:f(x) ,
_1
<
-n}.
Remark 2. An extended real valued function f is said to be measurable (Lebesgue measurable) if its domain is measurable and if it satisfies one of the first four statements. Remark 3. / is said to be Lebesgue measurable over E if (v) E (/=a) is measurable for every extended real number a. To prove that the first four statements imply the statement (v). (ii) and (iv). (v) ¥ a E R. For £(/=a)=[E(f< a)] n [(Ef;;;' a)] and a finite intersection of measurable sets is measurable. (ii)
~
(v) for a=oo. 00
For
E(/=oo)=
n
[£(/~
,,=1
Similarly (iv) => (v) for a= Finally, (ii) and (iv) This proves the
~
n)].
00.
(v) for every extended real number a.
r~quired
result.
Theorem 5 Let/ be a measurable function defined orer a measurable set E. Let c be any real number. Show that
c/,/+c,
I /I,P, }- (if/
vanishes nowhere
Oil
E) are measur-
able functions. [Meerut 1990, 88; Kanpur 85, 71; Gujrat 70J Proof. Letf be a measurable function defined over a measurable set E. Let c be any real constant.
>
< a) both are measurable set 'tf a E: R. (i) To prove that cf is measurable over E. [Mi:erut 1972J If c=O, then cf is constant and hence measurable. Consider the case in which c:;t:O.
/ is measurable over E
~
E (f
a) and E (f
121
MEASURABLE FUNC I IONS
£(
I > :) if e > 0
E(
I < :)
E (el> a)= {
if e
< O.
Both the sets on R.H.S. are measurable. Hence E (ef > a) is measurable and so is el. Therefore el is measurable VeE R. (ii) To prove that I f I is measurable. [Indore 78J E ( III < a)=[E (I < a) U [E (j > -a)] if a ~ O. Also a finite union of two measurable sets is a measurable set. E is given to be measurable. Therefore the measurability of E ( II I < a) follows. [Kaopur 83J Consequently III is measurable over E. For I x I < a => -a < x < a. (iii) To prove that c+1 is a measurable function. E(c+l> a)=E(/> a--e) =E (I> b), on setting b=a-c:, = measu rable set. Hence the result (iii) follows. (iv) To prove that 12 is a measurable function. E (f2 < a)=E ( III < ya) if a ~ 0 But E( If I < va)=[£(/< va)] u [E(f> -va)J. : . .E(f2 < a)=[E(/< v'a)J U [EU> -ya)] if a ~ O. Hence the measurability o~~~onseqUentlY (2 is measurable over F. . -_·;i ,'- '0 . (v) Let / vanish no\\ here on E so that f (x):;i:O V x E E. . H ence fIeXists. E ( / > 0) ifa=O if a;;;.. 0 fE(/>O)]n[E(/ a --= [[E(f< O)J n lE(/< I/a]) U [E(/>O)] ifa
measurable. Remark.
)
f
Hence E ( ; > a ) is measurable and so is ;
.
By the theorem, just proved cl is measurable function VeE R :;. ~I is measurable function, on taking c= - I.
.r is measurable =>
122
MEA SURA BLE FUNCTIONS
Theorem 6. Let f and g be measurable functions defined over a measurable set E. Show that f+g, f - g, fg ore measurable fune. tions over E. [Meerut 1986 ; Banaras 69, 67 ; Kanpur 79, 77, 71] Proof. Let f and g be measurable functions defined over a measurable set E. To prove that f+g, f-g, fg are measurable over E. Firstly we shall establish the following lemma: Lemma. If f and g are measurable functions defined over a measurable set E, then the set E (f > g) is measurable. [Kanpur 76, 7S, 72] The proof of the lemma starts. Let f> g, then by the corollary to the axiom of Archimedes there exists a rational number r S.t. f> r > g. Therefore E (f > g)= U [[E (f > r)] n [E (g < r)]]
reQ =An enumerable union of measurable sets =measurable set Hence the lemma. For (i) Q is an enumerable set (ii) fand g are measurable => E (f > r) and E (g <'r) are measurable set. (iii) Intersection of t\\O measur.able sets is a measurable set. Now the proof of the main theorem starts, g is measurable :> cg is measurable >.f c E R =- a +cg is measurable >.f a, c E R => a-g is measurable on taking c= -1. I. To prove that f+g is measurable over E. [Gujarat'70] Let a be any real number. E(f+g > a)=E( f> a-g) = measurable set, by the lemma just proved. For f and a -g both are measurable functions over E. Thus E (f+g > a) is a measurable set and hence f+g is a measurable function. II. To prove that f-g is measurable. f and g are measurable over E => f and - g are measurable over E q f-g is measurable over E . •: f+ (-g) =f-g.
123
MEASURABLE FUNCTIONS
III. To prove that fg is measurable over E. f and g are measurable functions over E => f +g, f - g are measurable functions over E => (f+g)2, (f-g)! are measurable functions over E . ~ (ftg)2_(f _g)2 is a measurable function over E =>
!
[(f - g)2-(f-g)2] is a measurable function over E
=> fg is a measurable function over E. Alternate method. To prove that f+g is measurable. Since for any real number «, 3 a rational number r s.t. f(x) < r < t1.-g (x) whenever f(x)+g (x) < t1. {x E E: f(x)+g (x) < «} =U ,. [(x E E: f(x) < r} n {x E E: g (xi < «-r}l. This union is taken over a set of rational numbers. It follows that f+g is measurable. Similarly we can prove that f - g is also measurable. Theorem 7. If f and g are measurable functions defined over a measurable set E and if g vanishes nowhere on the set E, tlren the quotient function
.~
is measurable Ol'er E.
[Kanpur 1979, 77J
Proof. Let f and g be measurable functions defined over a measurable set E. Also let g vanish nowhere on E so that g (x):;eO '" x E E. Then
I.
~ g
exists.
To prove that fg is measurable. lGujarat 1970J f and g are measurable functions => f+g. f--g both are measurable functions ~ (f +g)2 and (f - g)2 both are measurable functions => (f+g)L·(f g)2 is a measurable function => 1 [(f+g)Z-(f-g)2] is a measurable function fg is a measurable function.
=-
II. To prove that..! is measurable over E. g
Let a be any real number.
124
MEASURABLE FUNCTIONS
E(g>O) (E (g > 0)] () (E (g
>
[E (g
O)]U[(E (g
ifa=O
< I(a)] if a>O < O)]()[E Ig < I/o)]]
ifa
mea~urable:
>
Hence E ( ;
a ) is measurable in every case.
Therefore) is measurable over E. g
III.
To prove that f
Evidently
g
is measurable over 1:;.
Lg =f. J.-. , ~g g
is measurable function, by II
= product of two measurable functions. =measurahle function, by J.
Hence the result (Ill). Pr0l1e that the space of measurable functions is closed under the usual operations of Arithmetic. [Bbagalpur 1968J Theorem 8.
Proof. The fundamental operations of Arithmetic are: addition. subtraction, multiplication, division. Let f, g be arbitrary members of the space of measurable functions. If we show thatf+g, f-g, fg and f
g
(if g vanishes nowhere
on E) are mea~urahle functions, the result wili follow. Of course the functions f and g are measurable over a measurable set E. Here write the proof of Theorems 6 and 7. Ex. 1. !If and g are real valued measurable functions on a measurable space X, prr l'e that the f U g, f () g are measurable [Gujarat 1970J functions. Solution. Let c he any real constant. Letl and g be real
va.lued measurable functions defined on a measurable set X so that X ( f > c) and X (g > c) are measurable sets. (i) To prove that f U g is measurable. X (f U g> c)=[X (f > c)] U [X (g > c)] =finite union of measurable sets =measurable set.
125
MEASURARLE FUNenONS
This proves the result (i). (ii) To show that f n g is measurable X (
rn
g
>
c)=[X (f
>
c)] () [X (g
>
e))
=finite intersection of mea!>urable sets =measurable set. Thi~
=> the result (ii).
Ex 1. Show that the set of all measurable functions form a vector space over R. [KaDpur 1914]
Solutio.. Let E be a m~asurable set. Let V be the set of all measuruble functions defined over E. Then Y={f:f: E+R is a measurable function}. To prove that Y (R) is a vector space. Firstly we shall prove the following lemmas. I.emma 1. Iff is measurable, then af is measurable, a being any real number. Also let c be any real number.
if a
>
0
if a
<
0
Both the sets on the R.H.S. are measurable. Hence E (af> c) is measurable so that af is measurable. Lemma 2. Iff and g are measurable, then f+g is also measurable E(f>g)= U [E (! > r > g)], Q=set or rational numbers reQ U
[E (f
>
r)]
n
[E (g
<
r))
rEQ =An enumerable union of measurable sets =measurable sd. g is me:\surable => ag is measurable => -g is measurable => c - g is measurable. Now E {f+g > a)=E ( r < a -g)=measutable set For E ( f> g) is measurable. Now we come to the main problem. Let!. g, hEY be arbitrary. Then a, b, c E R, R being the set of real numbers. (i) (Y, +) is abelian group l. !+g E Y, by Lemma 2. 2. 3 0 E V s.t.f+O=O+f=/.
126
MEASUI
Here 0 is zero function so that 0 (X)=O >,f x. It means that 0 is a special case of a constant function and so it is measurable. Also (/+0) (x)-f(x) + 0 (x) =1 (x) +0=1 (x), this. 1+0-1. 3. -f E V s.t. -1+1=0, by Lemma 1. 4. f+g=g+/as (R, +) is an abelian group. 5. 1+ (g+h)=(f+g)+h as (R, +) is an abelian group. (ii) al E V, by Lemma 1. 6. (a+b)/=a/+b f 7. a (/¢-g)=a/+ag. 8. J.I=f. (6), (7). (8) follow from the fact that (R, +, .) is a field. From what has been shown it follows that V (R) is a vector space. Theorem 9. Let < /" > be a sequence 01 measurable /unctions defined over a measurable set E. Then sup {f1'/2'/3 ... } and inl { J;. h. la ....... } are measurable over E. [Kanpur 1986, 76] Proof. Let < In > be a sequence of measurable functions defined over a measurable set E so that E ( I .. > a) is a measurable set ¥ n, a being any real number. Write g(x)=sup {/,,(x) : n=l, 2, 3, ...... } h (x) =inf f/n(x): n=l, 2, 3, ...... } hex) can be defined in terms of supremum as follows: h(x)=-sup {-f,,(x): n=l, 2, 3, ...... }. To prove that g and h are measurable over E, it is enough to show that g is measurable over E. E (g
>
c
a)= U [E (fn ,.z:;l
=
""U
.=1
> a»)
{x E E :f,. (x)
>
a}
= An enumerable union of measurable sets =measurable set Theorem 10. Let < III > be a sequence of mea.mrabe functions defined over a measurable set E. Show that
sup {J;'/2""'f,.},
in/{fh!2"'" !.},
Lim!", Lim!n, are measurable Gver E.
Henc~ show that Lim /t. is
measurablover E if Lim!n exists. [Kanpur 1986, 76, 72, 70; Banaras 197., 68, 65]
127
MEASURABLE FUNCTIONS
Proof. Let:< In > be a sequence of measurable functions defined over a measurable set E. Step I. To prove that sup {Ir : 1 EO; r ~ n} and inf { J.. : ] :s:;;; r < n} are measurable over E. Define M (x)=sup {/r (x) : 1 EO; r < n} =
U" r-l
IT (x)
< <
and m (x)=inf {/r (x) : 1 r n} m (x) can also be defined in terms of supremum as follows:
<
m (x)=-sup {-Ir (x): 1 ~ r n}. If we show that M (x) is measurable over E, the re&ult will
follow. For proving this, it is enough to prove that E (M measurable set, a being any real number. In is a measurable function over E => E (f,.
>
>
a) is a
a) is a measurable set V n.
Evidently E[M> a)= U" E[(fr
>
a)]
=A finite union of measurable sets =A measurable set. This proves that E (M > a) is a measurable set.
Step II.
To prove that
(Tm
In and Lim fn are measurable
functions over the set E. D.!fine Mk (x)=sup { f" (x)},
rBanaras 1965]
n;Pk 111,. (X)=
inf {I" (x)}. n;pk
Then Lim fn (x)=
inf {Mk (x)} k~l
Lim I .. (x)= sup {mk (x)}. -k ~ 1 By case (i), M,. (x) and mk(x) both are measurable functions ~ dcli.ned over E V kEN and ag lin, by case (i),
128
ME,\SUR,\BLE FUNCTIONS
inf {Mk(X)}, k;;;tl
SUp {mk(x)} k:?1
are measurable over E, i.e. lim/.. , Lim
f. both ate measurable
over E. Step. III.
Let Lim / .. exist: [Meerut 1971; Kolhapur 70]
To show that Lim /. is measurable over E. By case (ii), Llm"j., Lim/.. both are measurable over E. By hypothesis, Limf.=Lim /.=Limf,•. From what has been done, it follows that Lim /. is measur· able over E. Theorem 11. A continuous/unction defined over a. measurable set E is measurable. [Kanpur 76; Meerut 71.; Kolbapur 69] Or, Show that a continuous function defined in a closed interval is measurable. [Kanpar 1987J Proof. Let / be a continuous function defined over a mea· surable set E. To prove that / is a measurable function over E. Let a· be any real number. We claim E (f;;;' a) is a closed set. A=E (/ ~ a). . .. (1) Let To prove that A is closed, it is enough to prove that D (A) C A. . .. (2) Let Xo E D(A) be arbitrary, then Xo is a limit point of A so that 3 a sequence <x,.> whose elements x" E A lim s.t. n-+oo x,.=xo· / is continuous at xo, X,.-X o ~ /(x") ~ /(x o). This follows from Heine's definition of continuity
. .. (3)
By (t), x" E A => ft x,.) ~ a => lim j (X,.) ~ a n-+oo => /(x.) ~ a, according to (3), => Xd E A, by (I).
129
MIlASURABLE FUNCTIONS
Finally, any This
Xo
~ Xo E A. D(A) c A => A is closed ~ A is measurabl. => E( I ~ a) is measurable
E D(A)
~
=> I is a measurable function over the set E. Theorem II (a). Let I be a function delined on a measurable set E. Then f is measurable iff lor any open set G C R, 1-1 (G) is a measurable set. Proof. Suppose a function f is defined on a measurable set E. Let G C R be an open set, where R=set of real numbers. I. Suppose I is measurable on E. Aim. /-1 (G) is a measurable set. Since any open subset G of R is expressible as a countable union of disjoint open intervals. Hence G is expressible as 00
G= U In, where 1.. = (a,,, bit) n=1
/-1 (G)= U {x E E: I(x)
.
E
, ft }
But/(x) E I,,=(a n , b.. ) ~ /(x)E (a .. , b.) ~ an aft) n E ( / < bn )]
,.
I
is measurable => E ( /
... (1)
<
an) and E
(I <
b..)
both are measurable sets. Also intersection of measurable sets is measurable. Now R.H.S. of (I) is measurable. Hence 1-1 (G) is measurable. II. Conversely suppose that/-1 (G) is measurable. Aim. f is measurable. Take G=(a, 00), where a > O. Then I-I (G)=(x E E : I(x) E (a, oo)} ~{x E E: a a) or I-I (G)=E (/ > a) Also 1-1 (G) is measurable. Hence E (f > a) is measurable. Consequently lis measurable on E.
128
MEASURABLE FUNCTIONS
are measurable over E, i.e. Limf.. , Lim f,. both ate measurable over E. Let Lim f .. exist:
Step. III.
[Meerut 1971; Kolhapur 70] To show that Lim f. is measurable over E. By case (ii), Llin-f., Limf.. both are measurable over E. By hypothesis, Limf.=Lim f.=Limf, •. From what has been done, it follows that Lim able over E.
f. is measur-
Theorem 11. A continuous function defined over a measurable set E is measurable. [Kanpur 76; Meerut 72; Kolbapur 69] Or, Show that a continuous function defined in a closed interval is [Kanpar 1987j measurable. Proof. Let f be a continuous function defined over a measurable set E. To prove that f is a measurable function over E. Let a be any real number. We claim E (f ~ a) is a closed set. A=E (f ~ a). ...(1) Let To prove that A is closed, it is enough to prove that D (A) C A. ...(2) Let Xo E D(A) be arbitrary, then Xo is a limit point of A so that 3 a sequence <x.. > .whose elements x .. E A Jim s.t. n-+IX) x .. =xo•
f is continuous at x o, x .. -Xo ~ f(x,,) ~ /(x o)'
... (3)
This follows from Heine's definition of continuity By (1), x ..
e
A => Ax..)
~ a => lim
n-+oo
j (x .. )
~a
=> /(x.) ~ a, according to (3), => x" E A, by (I).
129
MEASURABLE FUNCTIONS
E D(A) => Xo E A. => D(A) c A => A is closed => A is measurabl. => E( f ~ a) is measurable => / is a measurable function over the set E. Theorem II (a). Let f be a function defined on a measurable set E. Then f is measurable iff for any open set G C R, f- 1 (G) is a measurable set. Proof. Suppose a function f is defined on a measurable set E. Let G C R be an open set, where R=set of real numbers. I. Suppose / is measurable on E. Aim. /-1 (G) is a measurable set. Since any open subset G of R is expressible as a countable union of disjoint open intervals. Hence G is expressible as Finally, any This
Xo
00
G= U I", where 1,,= (a,,, b,.) ,.=1
..
f- 1 (G)= U {x E E: f(x) E ,ft } Butf(x) E I,.=(a", h,,) => f(x)E (a,., btl)
=> an < lex) < b..
:. f- 1
(G)= U [E (/ > a.. )
"
f is measurable => E ( f
n
E (f
<
bn )]
... (1)
<
an) and E (f
< b,.)
both are measurable sets. Also intersection of measurable sets is measurable. Now R.H.S. of (I) is measurable. Hence f- I (G) is measurable. II. Conversely suppose thatf-I (G) is measurable. Aim. f is measurable. Take G=(a, (0), where a > O. Then /-1 (G)={x E E :I(x) E (a, oo)} --={x E E: a < f{x) < oo} =£«(>a)
or [-1 (G)=E (/ > a) Also f- 1 (G) is measurable. Hence E ( / > a) is measurable. Consequently / is measurable on E.
130
MIlO\SURABLIL FUNCTIONS
Remark. The converse of Theorem 11 is not ture. That is to say, a measurable function ne.ed not be continuous. Example. Consider a function f: R ... {O, I} defined by { lifO ~ x <; : f(x)= 0 otherwise. Evidently I is measurable but not continuous. The point x=O is a point of discontinuity. Here R=set of reals. Theorem 12. If < f. > is a sequence of measurable functions [Meerut 1986] then show that Limf. is measurable. 0, To prove thqt the limit 01 a convergent sequenr.e 01 meas.urable lunctions is measurable. [K8n.,ur M. Se. P. 1986) Proof. Let < Ita' be a sequence of measurable functions. To prove that Lim f. is measurable. t. < I. > is a monotonic sequence and hence it is either monotonic increa'!ling or monotonic decreasing. If < /.1 > is a monotonic increasing sequence, then Lim f .. = sup {f.}. n ~ 1 If < I. > is a monotonic decreasing sequence, then ·Limffl= inf {lit}. n~1
Hence if we prove that sup {/1,f•• ...... } and inf{/l,h, ... · .. } are measurable. then the problem will be p-roved. Let g(x)=sup {fIJ., ...... } ... inf {h. fs •...... }. 00
Then £(g > a)= U [£(f.
>
a)]=An enumerable union of
measurable sets. Hence E(g > a) is mea'iurable and so g is measurable. hex) is expressible as h(x)=--sup {-I.(x} : n=l, 2, ...... }. This .. h(x) is also mea'!lurable as g is measurable. This completes the proof. Ex. Let < I. > be a sequ:mce 01 real valued measurable "=1
functions. If lim
n-+-IX)
Solution.
1.=1 eXists,
then show that I is also measurable.
[Kanpur 1975] Here write the proof of above Theorem 12.
MEASURABLF. FUNCTIONS
131
Tbeor,em 13. III is a meafurable function, defined on a measurable set E and if g and f are equivalent functions, then g is a measurable function on E. An Alternate Statement. Iff is a measurable function and if f=g almost every where, then g i~ measurable. Proof. Let E be a measurab'.e set and let a be any real number. Also let fbe a measurable function on E so that E (f > a)
is a measuPabte set. Further suppose that f and g are equivalent functions on E so that f=g a.e.· on E ... (1) To prove that g is measurable over E, we have to show that E (g > a) is a measurable set. (I) c> 3 ACE S.t. m (A)=O, f(x)=g (x) V- x E E-A=B (say). ~ f:#g on A. f=g on B, m (A)=O, A n B=0, E=A U B. m (A) =0 ~ A is measurable. E and A are measurable sets => B=E -A is measurable. f is measurable on E, BeE is measurable ~ f is measurable over B. Refer theorem 3, page 118. => A (f > a) is a measurable set => B (g > a) is a measurable set. For f=g on B c> B (f > a)=B (g > a) A (g > a)={x E A: g{x) > a} C A. Also m(A)=O. This c> A (g > a) is measurable. For every subset of a set of measure zero is measurable. E (g > a)=[A (g > a)] U [B (g > a)]. For E=AuB. = union of two measurable sets. =measurable set. :. E (g > a) is a measurable set. Ex. Answer the following questions in < Yes' or 'No'. Is a function equal to a measurable function almost everywhere itself measurable? [Kaapur 1974] Solution. Yes. (Refer Theorem 13). Theorem 14. To show thar the characteristic function of a set E is measurable iff E is a measurable set. [1\1eerut 1990, 88; Kolbapur 69; Banaras 69; Gujarat 10]
132
MEASURABLE FUNCTIONS
Or The set E and its characteristic function both are measurable or both non-measurable. Proof. Let E be a subset of a measurable set A. The characteristic function KII of E is defined as {I
KE(x)= 0
if x E E if· x E A-E
We shall show that (i) if KE is measurable, then E is measurable. (ii) if E is measurable, then KE ~s measurable. (i) Let KE be measurable on A so that A (K F. > 0) is a measurable set. . But E=A (KI> > 0). Hence E is measurable. (ii) Let E be measurable and a be any real number. Then 0 if a ~ 1 A (KE > a)= I E If 0 ~ a < I lAir a < 0 Every set on the R.H.S. is measurable.
r
Hence A(KE>a) is measurable and so KE is measurable on A Theorem IS. Iff and g be measurable real valued functions defined on X and F tf(x), g(x)]=h(x), x E X be real and continuous function on the Euclidean plane RI, show that h is measurable. LKanpilr M.Sc. 1984, 81; Kanpur 69J Proof. Let a be any real number. Let , G={(u, v) : F (u, 11) > a}. Then G is an open subset of R2 !oo that G is expressible as 00
G= U In, R~l
where <./,,> is a sequence of open interval,> s t. /,,=:(11, }.) : an < II < bn , en < I' <: dll}. Evidently {x EX: f(x) E (a", b,.)}
X: an a.} n {x E X :f~x) < b,,} =An intersection or two measurable sets. [For f is measurable] = measurable set.
={x
E
MEASURABLE FUNCTIONS
:. {X EX: I (X) E (a". bA)} is a measurable set. Similarly measurability of g over X implies that {x EX: g(x) E (cn , d.)} is a measurable set. Then it follows that {x EX: F(f(x). g(x» E T.\ {x E X :f(x) E (a", b..)}n{x EX: g{x) E (c.. , d,.)} is measurable set. Now =(x EX: hex) > a}={x EX: F (/(x), g(x» E G} =
-
U {x € X: F (/(x), g(x» E f .. }
11-1
=measurable set. Consequently h is measurable on X. Ex. 3. A/unction f is measurable iff the set {x :/(x) < r} is [Kolbapur 1970J measurable /~r every rational number r. Solutiou. Suppose a ,function / is defined over a measurable set E. Let r be an arbitrary rational number. (i) Suppose / is measurable over £ so that E (I < c) is a measurable set, c being any real number. To prove that {x : f(x) < r} is measurable. E (I < c) is measurable y. c E R ::> in particular, E (f < c) is measurable ¥ CEQ. For QCR. ~ £ (f < r) is measurable for r E Q. ~ {x: /(x) < r)={x E E: f(x) < r} is measurable. ~ {x :/(x) < r} is measurable. (ii) Conversely suppose that {x : f(x) < r} is measurable. To prove that f is measurable. Observe that £(/< c)={x E E:f(x) < c} = U {x E E: f{x} < r < c} rE.Q
U {x E E ;f(x) < r, r E Q} r
134
MEASURABbE FUNCTIONS
Solution. Let G be an open set and g be a measurable function. Let A={x : g(x) E G}. . .. (1) To prove A is a measurable set. G is an open set ::;. G is expressible in the form G=U·/.. , where 1.. =(0", b.. ).
..
Hence A={x : g(x) E G}=U {x: g(x) E I .. }
..
.4= U [{x: g(x) E (a... b..)}= U {x:
.. .. = U [{x: g(x) > a,,}n{x : g(x) < ..
ot
0 ..
<
g(x)
<
b..}
bIt}]
=enumerable union of measurable sets =measurable set. Hence the result (1). {x :f(g(x» < a}={x : g(x) E G} ... (2) where G={y : fey) < a} f is continuous => G={y : fey) < a} is open => {x : g(x) E G} is measurable, by (1) => fog is measurable, by (2). Theorem 16. (E. Botel). Suppose a measurable function f defined on a closed interval [a, b], is finite almost everywhere on the sume inrerval. For all numbers. a • • > 0, there exists a continuousfunction", [a, b] such that m [(E If-; I ~ a)] < E. [Paojab 1969) Proof. Letfbe a measurable function defined on [a, b] S.t.
f is finite almost everywhere ~ [a,
b].
Case I. Letfbe bounded. Then 3 a number k > 0 s.t. If(x} I < k ... (1) ¥ X E [a. b]. Let a, • > 0 be arbitrary numbers. Fixing C1, 6 by ~hoosir,~ a positive integer m so large that (kIm) <: a. Write
aDd
E",=E
where r=l-m. 2-m, ...... m-I -1 . ) m (
m
k ~f < k •
Evidently E,nEI J~ for i¢j. . where i,j=l-m, 2-m, ....... m.
llS
MEASURABLE FUNCTIONS
[a, b]=
.
1: E,
,_l-m
Select a closed interval F, C E, s.t. m (E,) < m (Er )+(./2m). Take Then m ([0, b])-m (F) < •. Depne a function ,p on F by writing
... (2)
rk ,p (x)=¥ x E F" r=l-m, 2-m, ...... m. m Then ,p is constant on each closed set Fr. J
Also Conseqvently
F,()Fr'=0 for r¢r'.
,p is continuous on Fr.
I ~ (x) !=I~ I=~k ~ :. I ~ (x) I ~ k.
k.
Finally, :f x E F r , then
I/(X)-~ (x) \=1 k-:'(=~I m-rl ~ ~ < d. . Ilex) -~ (x) I <
0
""
x E F,.
Now we shall state a lemma without proof. Lemma. Let F be a closed set contained in the closed interval [a, b]. If the function ,p (x) is defined and continuous on the set F, then it is possible to find out a function r/J(x} on [a, b) s.t. (i) 1\1 (x) is continuous (iiI 1\1 (x)=,p (x) V x E F (iii) max. I 1\1 (x) !=max. 1,p (x) I. By the lemma, just stated, it is possible to find a fUnction I\I(x) defined on [a, b) having all the properties stated in the lemma. Moreover
! ~ a) c 1/-1\1 I;;;;. 0)] ~ m [E ('/_.1\1) ~ 0] < E. E (1/-'"
Hence i.e.
m [E(
[a, b]--F. m ([a, bJ-F)
<
t,
by (2)
Thus we have proved the theorem for a bounded functlon/. Case II. Let/be unbounded. Let us suppose that /(x) is unbounded. Then we can find out a bounded function g such that m.[(E/#g)]
<
i·
MEASURABLE FUNCTIONS
136
Applying Case I, just proved, to the bounded function g, we are able to find a continuous function cjI s.t.
m [ECI
g-cjll ;;;;. 11)] <
f'
From this we can deduce that E ( If-cjll ;? 11) C E (f:Fg)+E ( I g-cjll ~ 0). This => m [E ( If-cjll ;;;;. 0)] ~ m [E (f:Fg») +m [E( I g-.p E
I~
0)]
E
<2"+2"=.' m[E(If-cjll ;;;;'0)] < •. The function .p satisfie'i all the requirements. Hence the theorem is also true in case of unbounded function. 8'6. Borel measurability fllnctions. A functionf is said to be Borel measurable if the set {x :/(x) < a} ¥ a E R is a Borel set. (I) Ev~ry Borel measurable function is (Lebesgue) measurable function. F,o'r every Borel measurable set is a Lebesgue measurable set,. (2) If I is Borel measurable function and A is Bord set, then 1-1 (A) is Borel measurable. (3) The product of two Borel measurable functions is a Borel measurable function. 8'7. Little Wood's three principles. In connection with the theory of functions of a real variable, J.E. Little Wood says. "The extent of knowledge required is nothing like so great as is sometimes supposed. There are three principles, roughly expressible as follows: 1. Every (measurable) set is nearly a finite union of intervals; 2. every (measurable) function is nearly continuous; 3. every convergent sequence of (measurable) functions is nearly uniformly continuous." Problem. Show that the function I defilled on R by (x+5, x < -1 I(x)= -< 2 , -1 ~ x < 0
Lxi , x ~ 0 is a measurable function. Soilltion. Let a be any real number. R (I ~ a) .. x+5 ~ a => 'x ~ a-5.
Then a
< 0 and
]37
MEASUREBLE FUNCTIONS
:. .R. (I ~ a)=[ -00, a-5]. I n view of this, we get
a)=j(-OO, a-5]- 5U] ·{OJ
if a < 0 if a=O (-00, a--5] U [O,ya] if 0 < a < 2 (-oc, a-5] U [-1, ya] !f 2 =E;; a ~ 4 (-00, Va], If4:::"" a Since each set on R.H.S. is measurable.
R (/~
(
:. R
-
00,
(I ~ a) is measurable
I is measurable,
for every real value of a. Hence
. EXERCISES
1. 2.
3.
IfI is a measurable function, then
[Kanpur 83J Let I and g be measurable real valued functions defined on X and F be real and continuous on R, if h (x)=F (/(x), g (x», prove that" is measurable. [Kanpur M.Sc. P. 1981] If/ and g are measurable functions on [a, b], prove that / +g, / - g, /g are also measurable functions. If g (x).,i:O
~
when a
4.
5.
6.
II I is also measurable.
x
~
b, prove that
f-. g
is also measurable.
[Kanpur M.Sc. P. 1979] If I, g are measurable functions on [a, b], show that functions l+g,lg are measurable. If g(x).;JCO (a ~ x ~ b), show that Ilg is also measurable. lKanpur 1977J Define mea~urable function. If / and g are finite and measurable functions defined on a set E, prove that E (/ > g) is measurable. Show Ihnt the functions/+g and/g are measurable. lK' is a sequence of measurable functIOns on [a, bJ such that the sequence is bounded for evt;ry x in [a, bJ, then prove that the functions sup "
lim/.. are all measurable on [a, b]. 7.
/r.,
inf /n, lim )", n
[Kanpur 1976]
Define measurable functions. Show that the sum and product of n (;;;;. 2) measurable functions are also measurable. Let be a sequence of real valued measurable functions. If -lim
In .../
n-oo
exists, then f is also measurable. lKanp.u 1975J
138
MEASURABLE FUNCTIONS
8.
Define 'measurable functions'. Show that the set of aU measurable functions form a vector space over R. Show also that the limit of pointwise convergent sequence of measurable functions is measurable. [Kanpur 1974]
9.
Define' measurable function'. If I and g are finite and measurable functions defined on a measurable set E, prove that the set E (I> g) is measurable. Show further that the functions I+g andfg are measurable. [Kanpur 1973J
10.
Define measurable function. If is a ~equence of measurable functions, prove that lim
n_oo
I ..
and
lim
- - I.. n-+oo
are measurable. Show also that a continuous function defined in a closed interval in measurable. [Kanpur 1972] 11.
12.
Define a measurable function. Prove that if I(x) and g(x) are finite and measurable, then the functions I (x)±g(x) and /(x).g(x) are measurable. [Kanpur 1911] (a) If g : R_R is continuous and I: R_R is measurable, show that gol is measurable. (b) Let be a sequence of Lebesgue measurable subsets of R such that
00 00
and
13.
14.
B=
n
U A...
1"=1 fI=k
Show that B is measurable and that p.(B)=O. [Banaras III 1970] Let I be an extended real valued func~ion whose domain is measurable. Prove the,equivalence of the following statements: (i) For eaeh real number a, the set {x : f(x»a} is measurable. (ii) For each real number a, the set {x : f(x)~a} is measurable. Deduce that if/and g are measurable functions, then/+g and Ig are also measurable. Show that characteristic function of a selt E is measurable iff E is a measurable set. [Banaras 69J Construct a set which is not measurable in the sense of Lebesgue.
ME \SURABLE FUNCTIONS
139
< In > is a sequence of measurable functions, (Banaras 68J then -Lim/n , Lim I,. are measurable. Prove that if the functions I and g are finite measurable, then Show that if
15. 16. 17.
18.
19.
20.
21.
l+g,1 -g andfg are mea'iurable functions. [Banaras 67J If < fn > is a sequeGce of measurable functions show that Lim sup In, Lim inff" are measurable. [Banaras 67] (a) Define a measurable function on a set X, and show that the characteristic function of a measuraple set is 3t. measurable function. (b) If I and g are real valued measurable functions on a measurable space X, prove thatf+g,fg,f U g and! n g are measurable. [Gujrat 70J (a) Iff and g be measurable functions and c be a constant, prove that functions cf and f+ g arc measurable. (b) A real valued function wluch is continuous in an open interval is measurable. lMeerut 1972] When is a function said to be measurable? Prove that the limit of a convergent sequence of measurable functions is measurable. [Meerut 71] Prove that the following condition is necessary and sufiiclent that the function be measurable: tire set {x: I(x) ~ r} is measurable for every rational number r. [Kolbapur 69J A functionf is measurable ifr the set {x :f(x) < r} is measurable for every rational number. [Kolhapur 70J
9 The Lebesgue Integral of a Function Introduction. Lebesgue's definition of an integral is more general in comparison to Riemann's. It enables us to integrate those functions for which Riemann's method fails. That is to say, if a function is integrable in the sense of Riemann in an integral (0, b), then it is also ~ebesgue integrable over (a, b) but not conversely, i.e. if a function is Lebesgue integrable over an interval (a, b), then it is not necessarily Riemann integrable over (a, b). The phrase "Integrable in the sense of Lebesgue" is written as "L-integrable." From now onwards, we adopt the following conventions:
r f(x) dx=Lebesgue integral of/ex) over E.
,8
(R)
J E
lex) dx=Riemann integral of/ex) over E.
9'0.
To define the Lebesgue Integral of a function. [Meerut 1987; Kanpur 86,72, 70J First Method. Sets in two dimensional world are ref.:rred to as plane sets. In two dimensional world, an interval is taken to be a rectangle (in particular a space) and the area of the rect:lOgle represent'> the measure of the rectangle. Let lex) be an txtended real valued function defined over a measurable set E. We define Q U: E)=-, {(x, y) : x E E, 0 ~ y ~ Ax)}. Do (/, E)={(x, y) : x E I::, 0 ~ y < /lx)}. The function/(x) is said to be Lebesgue \ Integrable over E if {J (/, E) is measurable and the Lebesgue integral of lex) over E is dd'ined as the plane measure of D (/, E), i.e.
l,. /(x) dx=m [.0 (/, E)].
141
THE LEBESGUE INTEGRAL OF A FU:-ICTION
Second Method. Case (I). When f(x) is bounded over E. [Kanpur 1983,32, 77, 76,70 ; Indore 78] Let I(x) be a bounded measurable function defined over a measurable set E. We can take 0:
E= U el:,
f'k
n
ek'=.p for k:j:k'.
10=0
Since I(x) is measurable over E and hence the sets measurable. Therefore, we have
el:
are
11-1
m(E)= , meek)' 10=0
11-1
Define
11-1
S= L: YI< m(e,.), S= l} YAH m(eIJ· 7:=0
k=O
Then sand S are respectively called Lower and upper Lebesgue sums corresponding to this mode of sub-division. Now we shall prove a lemma. Lemma. To prove that s <; S for any mode of sub-division. Let jlo and So be Lebesgue sums corresponding a to particular mode of sub-division. We shall first show that if we add a new point y of sub-division and again compute sums sand S, then So ~ s, S E;;; So i.e. lower sum does not decrease and upper sum does not increase, when new sub-division points are added.
Let YI <; Y ~ YHI' We see that for k=t=i the sets ei; and semi-open intervals [Yb YI:H) come into new method of subdivision as well as in old.
y -Y
The set e, is replaced by two disjoint sets ePI and ePI, where
Hence
e/(II=£ (Yi ~ I(x) < y) e,(2)=E (y ~ /(x) < Yl+l)' 'l1(e,)=m (l';(J')+m (l'/2».
142
THE LEbESGUE INTEGRAl. OF A FUNClION
The semi-open intervals [)\, )',.1) is replaced by two disjoint and [y, Yi+I)' semi-open intervals [Y;,
y,
Thus, i-I
n-l
s= I' YI; m(ek)+Y' m(ep»+y.m(e/ 2»+
1: Yk.m(e,,) 1:-i+1
k=O '-1
~ 1: YI .. m(e;,)+Yt.m(eP»+Ytm(ePi)+ k=O
n-l
1:
y" meek)
k_i+l
1\-1
= E Yk.m(e,J=so beO
s ~ So I e., So ~ s. By making parallel arguments we can show that S ~ So This proves the result (I). Going back to the actual lemma continued above, proceed as follows to prove it.
we
Let (Sh Sl) and (S2' S2) be Lebesgue sums corresponding to any two modes of sub-division of the segment [ae, tl), say I and II. Set up a method III, the method III consists of division points of the method I and II. Here we also suppose that the method II consists of division points in addition to the method I. They by what we have established, it follows that SI ~ S2' SI ~ Sa, S2 ~
Sa, S2 E;;; SI' Sa EO; SI' Sa"':::;; S2' SI S2 ~ Sa, Sa ~ S2 ~ SI
<
so that .:.(2) where Sa and Sa are Lebesgue sums corresponding to the method III. Since Sa ~ Sa. In this ev~nt (2) takes the form
.. (3)
$1 ~ S2 ~ Sa ~ Sa ~ S2 ~ SI' From this the required result follows, i.e., S ~ 5 for any mode of sub-division.
Q.E.D. (lemma) For (3) it is quite clear that the sequence <.s/<> is bounded abovc and hcnce sup. {s,,} exists k~l
r--
-1--1--1-1-1-1"--1-'
o
~
~
~
U
V
~
~
~
143
THE LEBESGUE INTlGRRL OF A FUNCTION
It is also clear trom (3) that the sequence bounded below and hence info {S.~} exists.
< S" : kEN>
is
k~l
Write
u=sup {.flo}, V=inf {Sk} k~l k ~ 1 Then U ~ St, V eo;; SI. We claim U V. Suppose not. Then U> V.
<
For any.
>
0, we have U
I! - "2 <
810
V'
'2' E
~
S l'
Now,
i.e.
8 1 -S1
>
By assumption, U> V so that we can take U-V> 2,. Then (4) reduces to 8 1 -S1 > • Making • -+ 0, A contradiction. Hence our initial assumption is wrong. This ~ U~ V. Thus S1 <; U ~ V ~ S1 From which
i.e,
0< V-U
... (4)
U-V-E.
8 1 - S1 ~
For
SI
O.
~ SI·
"-1 < k=O E (Y"+1-YI,).m(ek)'
Taking A=max (YkH -Yk), we get
o~
"-1
V-U ~ E A.m(ek) k=O
o ~ v-u ~·A,m(E).
or
... (5)
Since .\ can be made arbitrarily small and therefore making A-+O in (5), we get
o eo;; i.c.
V---U V=U
~
0
(in limit) Definition. The common value of the members U and V is called Lebesgue integral of ftx) over E, i e.
t
!(x) dx=common value of the memb:!rs U and V.
144
THE LEBESGUE INTEGRAL OF A FUNCTION
If we take E=(o, b), then the following symbol is used:
J:
f(x) ix=common value of raembers U and V.
Thus we see that a bounded mea.urable function is Lebesgueintegrable. . Case II. When the function f(x) is unbounded. [Kanpur 76, 75, 73, 71, 70 ; Banaras 67, 71] Let f(x) be an unbounded measurable function defined over a measurable set E. Suppose, without loosing generality, that I(x) ~ O. For if the result is applicable to positive functions, it is equally applicable to negative functions and therefore it is so by addition in general case. Let mEN .be arbitrary. Define a function [/(x)].. as follows: when f(x) ~ m when /(x) > m. Now we shall show that [/(x)] .. is bounded and measurable over E. Evidently [f(x)Jm=min {I(x), m}, rough upper bound of [f(x)Jm=m rough lower bound of [f(x)]m=O. This ,hows that [f(x)]", is bounded over E. Let a be any real positive number. [/(x)]m=
{
E{[f(x)Jm
f(x) m
>
a}={ E[f~ >
0]
!~: ~:
f(x) is measurable over E implies the measurability of E [/(x»o],
o
is also a measurable set. From what has been done, it follows that E {[f(x)Jm > a} is measurable and hence [J~x)]m is measurable over E. The fUIlction [f(x)]m being bounded and mt'l\slIrable, is L-integrable over E. Notice that [f(x)], ~ [ftx)\+l ¥ r E N and therefore, by Theorem 4, •
J E
[f(x)J, dx
~
J E
[f(x)]r+l dx .
This proves that the sequence
f(x)1, dx : r
EN>
145
THE LEBESGUE INTEGRAL OF A FUNCTION
is monotonic non-decreasing and hence Lim m·..o,.oo
J [/(x)Jm dx exists I>
and we write
f/
f
(x) dx= Lim [f(x)]", dx. . .. (1) m-..o,.oo E If the second member of (I) is finite, then we say that function/ (x) is L-infegrable or summable over E. Thus we see that every non-negetIve unbounded measurable function is L-integrable but only functions baving finite integrals are called summable. E
In order to define the integral of any unbounded measurable function we proceed as follows: /(x) ~ 0 for x E El j(x) < 0 for x eEl' Let E=...oEI U E2 , then El n El=0. By countable additivity property of the integral, i.e. Theorem 3.
and
f
I>
l (x)]", dx~oJ
~
r/(x)]", dX+[ ~ [J~x)]", dx.
...(2)
Making m-+oo, we get
I/ 6
If
J
£1
(x) dX=J
El
lex)
/(x) dx=oo,
J
E,
dX+[ / (x) dx. £2
/(x) dx=
-00,
then
f f(x) dx=(l-oo)+(-oo). whi~h is meaningless. The integral of a function has a meaning if and only if one of the integrals on the R. H.S. of (2) is finite. Definition. Let f be a measurable function on [a, bJ. If both fiand/- are L-integrable on (a, b), then we say that/is L-integrable. In this case we write / E L (a, b). Thcorme I. ~r tile ordinate set Q (/, E) is measurable then / (x) is measllrable over the measurable sel, E. Proof. Let D E) be Lebesgue-measurable, E being a measurable set. To prove that (x) is measurable over E, is is enough to prove that E (j(x) >0) is measurable, a being any real positive number. mEN be arbitrary. Let Deline t.. . -tx E E: r,"<) > a}=E (f.> a)
«(.
r
146
THE LEBESGUE INTEGRAL OF A FUNCTION
Ea+l/m={X E E :f(x) > a
+1}=E ( f > a+k)
1=( a, a+k)
and Evidently and therefore
Ea X I ::J EOH/.,. X I. Also EJ X I ~ 0 n (R" X /) ::> E"+llm X I ... (1) where Rn=Rx Rx ...... (n factors). o is known to be measurable. R" is always measurable. An interval is always measurable and hene e I is measurable. From -this it follows that Q n (Rn x I) is measurable so that m. [.on (R"x/)]=m e [f.'n (R"xln..=m LOn (Rnxl)) ... (2) From (1) m, (Ea X /) ;;;. m; [SJ n (R" x I)] and m.[Dn (R"x J)] ~ m. (Eo+l/rn X I). Making use of (2), we get m. (Eo x I) ~ m [Un (R" x I)] and m [D n (RI> x l)] ~ m, (E,,+1/m X I). Combining the last two inequalities, we have ~
m, (EI> x I) ~ m [D n (R"E I)) ;;;. m. (E"tl,,,, x /) From which
m, (Fax I)
i.e. i.e. or
~
m. (E"aH;mx/) (I) ? ·m. (EO+1Im) X m. (I) m, (Eo) X m (I) ~ me (EaH,m) X m (I) m, (Eo) ~ m. (EaH/m)·
tn.. (Eo) Xm,
Making m-+-oo and observing that
Lim",-+o:> m, (Ea-tl/"')= m. (Lim,..-+- 00 E O +1I"') we get, m. (Eo) ~ me (Eo). But me (Eo) ~ m, (E.) is true lor any set E". Combining (3) and (4), we get m; (Ed)=m e (Eo). This proves that Eo is measurable, i.e., Hence proved. E (f > a) is measurable. Ex.1. The Lebesguf! integral of a bounded measurable tionf(x) defined over a measarable set E is the common limit upper and lower Lebesgue sums s alld S. Sol. To prove that
...(3} . .. (4)
Juncof the
THE LEBE("GUE INTEGRAL OF A FUNCTION
IE f
147
[in the limit]
(x) dx=s=S.
We know that s
Lf
~
(x) dx
~
S. ~
Taking limit as max. (YH1- y;.) s
tf ff
~
~
(x) dx
S=s
[in limit]
(x) dx=S=s
E
If a bounded measurable
First Mean Value Theorem 2. tion f satisfies the inequality II
then
<,;
rJI..m (E)
~
0, we get
I
fI Ii
(x) <,;
~
(x) dx
~ ~.m
func-
on the measurable set E, (E).
Proof. let I be a bounded measurable function defined over a measurable set E. Let I (x) satisfy the inequality.
<, I (x)
lit
To prove that a..m(E) <,;
~ ~
tf
¥ X E E.
(x) dx <;
~.m (E).
Let n E N be such that IX
<'/(x) <;; fJ
takes the form 1
IX-if
Taking
at
-.!....=a, n
1
(x) <,~+;;- on the set E.
()+!...=b, we get It
a < f (x) < b l,f X E E. Divide the interval [a, bJ by means of points (l=Y. < Yl < Y2 < ...... < y.=b. Let e,,=E (Yk ~/(x) < YH1)'
Evidently
"-I
E = U e", e,:
n
ek = 0 for k::j:k'.
~-o
Since I (x) is measurable over E and hence the measurability of the sets ek • A-I
'J.'hen m (E)= 1: m ~e.. ). k=l
)48
THE LEBESGUE Il'iITEGRAL OF A FUNCTION
a <; Yk ~ h on the set e k , then a.m (el') ~ Yk m (e.) ~ h.m (et). Summing the result f6r the whole set E, we get
If
"-1
"-1
k_\
1;=0
"-I
1: a m (ek) EO; 1: Yt m (ell) I!i',; 1.' b m (ek) i~O
." "-I
a m (E) <; 1: Ytm (ei)
I.e.
~
h m (E)
10=0
Making max. (Yk+l-h) a m (E)
(IX-~)m(E)
or
~
~
0, we get
J (x) dx .;; E'
E;; tJ(X)dX
h m (Lj
<;;(fH-,n m(E)
Making n-+oo, we get the required result. IX m (E)
~
Ie J (x) dx < fJ.m (E).
Ex. 2. If the Junction J is constant on the measurable set E, say,f (x)=c, then
J .f
(-¥) dx=c. m (E).
E
Sol. Let.f be real valued function defined over a measurable set E s.t . .f (x)=c (constant) V x E E. To prove that
I
E
f (x) dx=c.m (E).
f(x)=c 'V x E E ~ c <;. f(x) ~ c V x E E. Using first mean value Theorem (2), c.m (E)
~
This
t
~
t
f(x) dx
< c.m, (E).
f(x) dx=c.m (E).
Ex. 3. Let f be a bounded measurable functioll defilled o\'er a measurable set E S.t. m (£)=0. SholV that
ff Ii
(x) dx=O.
Sol. Let f be a boundc~ measurable function defined over a measurable set E s.t. m (£)=0. We claim
In f(;)
dx=O.
149
THE LEBESGUE INTEGRAL Of A FUNCTION
We can take at <;; f(x) ~ f) on the set E. Using first mean value Theorem, we get
«.m (E) EO;;
Is f(x) dx ~ p.m
(E)
Using the fact that m(E)=O, we get
o <; IE
L.
or
f(x) dx=O f(x) dx=O.
This completes the proof. Ex. 4. Let f be a bounded measurable function defined over a measurable set E s.t. (i) f(x) ~ 0 (ii)
IE
f(x) dx=O.
Prove that f(x) =0 almost every~here in E, i.e., f(x) is equiva(Kanpur 1989; Bbagalpur 68) lent to zero function on E. Sol. f be a bounded measurable function defined over mt:asurable set E s.t. (i) f (x) ;;;J 0 ¥ x E E
IE
(ii)
f(x) dx=O.
To prove that f(x)=O Let Eo=E (f(x)=O),
E.=E
(n~i < f(x)
almost everywhere in E.
<;;
~)
a.;. x
e
N.
M being an upper bound of/(x) in E. Aim. f(x)=O ¥ x on E. and m (E-Eo)=O. 00
Then E= U E., EknE,,'=(lj for k::j:k'. "",-0
Siuce f is measurable over E and hence the measurability of the sets En. it means that ",(I) M
on the set E. ¥ n E N. We have nt - .
All' n+
m(En)
<
IE {(x) dx ~ I tI
B
f
dx=~O
150
THE LBBESGUE INTEGRAL OF FUNCTION
M n+ l·m(E,,) ~ 0
Finally,
V nEN
m(£,,) ~ 0 vnEN But m(E,,) ;ils 0 V neN is true for any set Erl • Combining the last two inequalities, we get m(E,,)=O V nE N
or
Thus =>
00
E
,,=1
m(En) =0.
Making use of (I), we get m(E - Eo) ;. O. Thus we have shown that f(x) = 0 V x E Eo C E s.t. m (E-Eo)=O. This proves that f(x)=O a.e. (almost everywhere) on the set E. Hence the result. Remark 1. Show that iff is integrable and
J:
ftt) dt=O
V x E [a, b],
/-0 a.e. in
then
(a, b).
(Meerut 1972 ; Banaras 68) Hint. Write the so¥ution of Ex. 4, Page 149. Remark 2. Let f(x~= -1 V x E (0, i), f(x) = 1 ¥ x E (i, 1). Evidently f{x) is defined over (0, I)
J f(x) dx= _J1 /2 f(x) dX+Jl l
o
0
__ _ J1 /2 dx+Jl o
f(x) dx
1/2
dx
1/2
=-i+i=O. Thus,
J:
f(x)=o,
but f(x)-#O for any x E (0, I). This proves that the integral of the function vanishes although the function is not equivalent to zero. Contrary to the result, just proved. Theorem 3. CountabiJity additive property of tbe integral (in ease of a bounded function). Let f be a bounded measurable function defined over a measurable set E and let E be the union of pairwise disjoil1t sets Et. Show
151
THE LEBESGUE INTEGRAL OF FUNCTION
J [(x)dx= E fE [(x) dx,
that
"=1
£.
'"
00
E= U Ek S.t. Ek nE,c'=0 [or k¥=k'.
where
11:=1
[Indore 78] Proof. Let[ be a bounded measurable function defined over a measurable set E. Let 00
E U Ek s.t, E",nE,,'={lj for ki=k'. "=1
To show that
f [(x) dx= E IE f(x) dx. E
k_l
k
We can take « < [(x) < fJ on the set E, interval [oe, f.l] by means of points Cl=Yo
and introduce the sets
Divide the closed
Yl < Y2 < ... < y .. =/l e,,'=E (y" ~ [(x) < YA:+l)'
<
"-I
Evidently E= U e" s.t.
ek
n e,,'=0 for
k=l:-k'.
l=O
Since [ is 'measurable and hence the measurability of the fI-l
sets e/c, so that m (E)= E m (e",). k=O
Let
E=E1 UE2 , E1 nE2 =0. e,PI=E1 (Yk ~ [(x) < YkH), e.(2)=E2 (Y" [(x) < YII:+l)' E 1 nE2 =0 :0> ek(llnek(2)=0.
<
Then We have "-1
I; k=O
m(e/t)=m (ek(l)+m (e,,(2».
Making max. CYk+l
Ie
"-1
"-I
k=O
k=O
Y" m (e,,) = E Yk.m (e,(l1)+ E Yrc.m (e..!I'). -Yk)~O,
J
we have
f(x) dx= El f(x) dx+
JE2 [(x) dx.
This shows that the result is true in case E= E 1 UE2 s.t. El()E2~-0.
" .(1)
152
THB LEBESGUE INTEGRAL OF A HJNCTION
Let the result be valid in case II-I
U
E=
E~
S.t. E_nEk'=r/> for k#;k'.
J
f(x) dx= II-I 1:
1t=1
Then we have
£
f
E f(x) dx.
11:=1
k
Consider the case, when
..
E= U Ek
S.t.
n El '=-=t/> for
Et
A:~I
expH~ssed
Hence E can also be
E= ( "-1 U
E.~
)
k::j:.k'.
as
E...
U
k=1
Then by what we have establised it follows that
fB
I
f{x) dx= f(x) dx+
IE. f(x) dx
II-I
U
Ell
I,~I
=
If IE ,=1
= /'-1 E This shows that
" E= U
t~e
E.~;
~
IE'
II
f(x) dx+JE
n
f(X)
dx
f(x) dx.
required result is true in case S.t.
Ek()E/=t/I for k:;t=k'.
k~1
This means that the required rc:su/t is true for k= J, 2, 3, ... n. By mathematical induction, it can be shown that the result is true for any possible value of k. Consider the case in which 00
E= U E1• S.t. E.nEk '= for kl=k'. 1._1
Here E can also be expressed as
153
THF LlHlI!SGUE INTEGRAL OF A FUNCTION
..
Taking Rn+t=
U
*=,. j-l
Fie. we get E=
(..U E" ) U RD+!. ,
k--l
From what has been done it follows that
L
fl(X) Jx+JR
rex) dx
I
or
,
ire:
dx~ - ':=1
,f(x)
t;;
lit-!
J(x) dx
f(x) dx+JR
eJ:
. .. (2)
f(x) dx. fltl
Since f(x) is bounded on the set E and therefore we can take Cl ~ lex) ~ (J on the set R II +1' On using the first mean value theorem, we get
~
•• m (R .. t l )
JR
f(x) dx '
~
(J.m (R n+1)'
..• (3)
00
k,#k'.
for /,
~."
11 00
111 ~ R"/-l)
lim III
n-+oo
(R", 1)=
1:
"-"tl
11m
111
(F.)
....
2'
n-+oo "
III
(L',J
lit I
-,0 (hy general principle of convergence) lim til
1/->-00
(H" II)=~()' II-+~""
Taking limit as
lim
c(.O ~
11-00
)illl
() -<'" -::::,
fI :. 00
lim
or
f/--:>CXi
Making
J
~.
1/-+ 00
~R "II'
lex) lix , {J.O
J '.
I(.\) , /'-<"') .\ "'" t ,
R
I . "'1
R
"H
./(x) dx
0,
... (5)
in (2) and observing (5), we get
"f
I(x) dx - Ii III 1/
".(4)
in lJ) and observing (4), we get
~
~;
00 Ie 1
r. J
L'
"_I "'k
/:' f( \) dx+O .~
/(x) dx.
Q.E.D.
154
THE LEBESGUE INTLGRAL OF A FUl'CTlON
Ex. If A and B are disjoint measurable subsets 0/ [a, b] and if/is bounded function Lebesgue integrable on [a, b], then (Meerut 1912) Sol.
For the sake of convenience, we take, E1=A, E2 =B, E1 U E 2 =E. Then E1> E2 C [a, b] and E1 n E2 =<$. Now we have to show that
IE [dx+ fE/dx + fEz [dx. Now write the proof of the above theorem 3, Page 150, upto equation (1). Theorem 3b. If/is a bounded [unction in L [a, b], and if a < c < b, then/e L [a, c],/e L [c, b]
and
(Kanpur MSc. P. 1978) Proof. Suppose / is a bounded function in [a, b], then A={x E [a, b] : [(x) > ex} is a measurable set. Let E1={x E [a, c] : lex) > Cl}=[a, c) n A =Intersection of measurable sets. =measurable set Hence £1 is measurable so that / is measurable [a, c]. Then f is bounded and measurable on [a, c] so that f is Lebesgue integrable on [a, c], i.e., f E L [a, c). Similarly we can prove that [E L [c, b]. Remains to prove that
Lf=JE/+JE2f where £=(a, b), E.=(a, c), £2=(C, b). Then E=E, U E 2 , E1 n E2 =1>· Let ek(l)=E1 (h ~/IX) < YHl), =£2 (Yk ~ [(x)
e/c(2 1
El
Then This ~
n
<
Yk+l)
E2=", => ek(l) n e,p)=cp. m (e/:)=m (e1P»+m(e k (2».
155
THE LEBESGUE INTEGRAL OF A FUNCTION
Making max. (YTe+l - Yk)--+O, we have
f
j(x) dx=!
11
E~
dx+J
f(x)
E2
f(x) dx.
Theorem 4. Let f and g be any two bounded mcasnrable functions definerJ on a measurable ~et E. SholV that (i) (ii)
(iii)
J" t I
~
f(x) dx
L
L
[f(x)+g(x)] dx-=
CJ(x) dx=c
I>
if f{x) ~
g (x) dx
f{x) dx+
g.(X)
L, g(x) dx (Bhagalpur 66)
I
f(x) dx, c being any constant
E
Proof. Let I and g be any two bounded measurable functions, defined over a measurable set E. We can take. on the set E. at
<
< '" <
Yz
ek=E (Yt <; /(x}
<
Yk+l)'
·1-1
Then
Yn={3
n e,:=q, for k=/;k'.
E= U ek s.t. et "=0
Since I(x) is measurable over £ and hence the sets measurable, so that
e~
are
"-1
m (£)= E m (e,). k~O
Step (i).
~
Let f(x)
g(x).
To prove that
f
f(x) dx
Clearly
Y_
< I(x)
E
f
~
~
[;
g(x) dx.
g(x)
From which y" ~ g\x) on the set elt. By first mean value theorem 2, y,:.m
(e~)
<
f
elt
g(x) dx.
Summing the result for the whole set E,
¥
X
E
ell.
156
THE LEBESGUE IN I EGEAL OF A FUNCTION
Using countable additivity property of the integral, we have
Making max. (YHI--Yk)-)oO, we get
f
E
Step
n.
(x) dx c::;;;
I
E
g(x) dx.
To prove that
Ie [f(x)+g(x)] dx=L f(x) dx+t g(x)dx. (Kanpur 73, M.Sc. P. 75) Let s be the lower Lebesgue sum relative to f(x) with the scale Yo, YI, y~, ... ,y.. Let s' be the lower Lebesgue sum relative to f(x)+a with the scale yo+a, YI+a, Y2+a, ... ,y,.+a, a is any real positiv lumber, ..
I
~')'k.11l
s=
(Ct)
" ·0
ra-l
S·
=-. ~' (y*+a) m (Ck) /:'-"'11
A-I
-= J.'
"-I
)'k.1II (t'k)
-I-a }.'
11/
(c,,)
I..-U
"
1
I.
II
}.. rlc.1II (cJ I a.1II (1:")
-=:-s \-a.11I (1:').
Making max. (Yk+
f
1
.l't)-·"O, we get
(f-j a) tis -""
E
J
b
tJx + (1.111
(J~')
... ( 1)
By cOllntubh: additivity propl'rty of the integral,
I
I:
( (-I g) dx' -
", f U+g) t
1.;
"
-II
,
(
k
lix
..
,
(~)
157
THE LEBESGUE INT'flGRAL OF A FUNCTION
~ ~: Ie
k
(Yk+g) dx.
In accordance with (I), this gives
L
~ ~~:
(I+g) dx
(fek g dx I h 111 (e/;) ]
L. (f-f-g) dx ~ Ie g dx+ J: y".m (e
Finally
If we replace Yk hy
In
Yld-I
(/+,1;'). dx
<
l .).
...
(3)
on the R.H.S. of (3), we get
L
,I;'
dx+
:~:J Yk+l m
(e/J.
...(4)
Combining (3) and (4), we get
+
n-l
E
YHI
m(el..)·
k~O
Making max. (YHI - yd-+ 0, we get
f
I:.
gdx+J
This
~
J
f (/+g) dx ~J gd~+I d'(=j . dx+f dx. ~
Idx
E E l - .
(f+g)
f
E
fdx.
g
.. .(5)
~
I:.
By mathematical induction, it can he shown that
J(j~ +/~+ ... +1;.) (h=-= I
~
fl dx
+J
1-.
f~ dx
j .•.
+L ("
dx
where < fn(x) : 11 EN> is a sequence of bounded measurable functions defined over E. I Kanpur M.Sc. P. 1985] Step (iii).
Let c he any constant.
To prove that
L
c(.\) dx=c
L
/(x) dx.
lIndorc 1978]
158
THE LEBESGUE INTEGRAL OF A FUNCTION
Let s be lower Lebesgue sum relative to [(x) with the scale Yo, YI' Y2' ... , Yn. Let c > O. Let s' be the lower Lebesgue sum relative to cf (x) with the scale CYo, ('YI' CY2' ... , cy". Evi
s' = E cy,.. m(eJ;) k=O "-1
=c X Yk m (ed=cs k=O
i.e.,
s'=cs. Making max (h-u- Yk)-70, we get
f
E
cf dx=c
ff
dx.
[';
If c=O. the result is obvious. Ex. If f and g are bounded measurable functions offinite measure prove that .
t
(af+bg)=a
Ie f+b t
01/
a set E.
g.
[Kolbapur 69 ; Banaras 69] , Solution. Let.f and g be bounded measurable functions defined over a set E of finite measure. First we shall establish the following results: (i) (ii)
f
E
t
[f{x) +g (x)] dx = cf (x) dx=c
S f(x) dx+ B
t
g(x) dx
Jlex) dx, c bemg any constant.
For proof. see Theurem 4, just proved. Our main aim is to show that
t
(af+bg)=a
SE
/+h
t
g
a and b being constants. f and g are bounded measurable functions => af+bg is also bounded measurable. Now
Ie [a/(x)-1-hg(x)] d~=In alex) dx+Lbg(x) dx, by (i) =0
IE
f{x) dx+b
IE
g(x) dx, by (ii).
Proved.
159
THE LEBESGUE INTEGRAL OF A FUNCTION
Ex. Iff and g are integrable (boundC!d or unbounded) functions over a measurable set E (of finite or infinite measure), then prove that f+g is also integrabla over E and that
L
f+g=t
f+fE
g.
[Kanpur M.Sc. 7.3, M.Sc. P. 75] Solution. Theorem 4.
Here prove the equation
(5)
of the
above
Theorem 5. If two bounded measurable functions f and g are equivalent on a measurable set E, i.e., if f=g a.e. Of! E,
then they have the same integral, i.e.,
t
f(x) dx=
Is g(x) dx.
or, The bounded measurable functions which are equal almost everywhere have the same integral. Also prove that the converse of this theorem is not true. Proof. Letf(x) and g(x) be bound.:::d measurable functions defined over a measurable set E. Let f(x)=g(x) a.e. on the set E. To prove that
J fix) dx=J ,.
E
g(X) dx.
Our assumptiol1 implies that f(x)=g(x) ¥ x E E except in a set of measure zero. This fact is expressed by saying that f(x)=g(x) ¥ x E £1 C £ s t. m (£ -E1 )=0. Clearly E=(8-El) U E1 , (E-E1) () El=0. By countable additivity property of the integral,
/,. (f-g) dX=!(E_E1 ) (f-g) dx+JEl (f-g) dx. f(x)=g(x)
¥
. .. (1)
x E El => IEl (f-g) dx=JEl :f-I) dx.
... (2)
160
THE LEBFSGUE INTEGRAL OF A FUNe flON
m(E- £1)=0
~ [(£-£1)
C[-g) dx=O.
. .. (3)
Writing (I) with the help of (2) and (3), we get
or
t t f/
or
IE
«(-g) dx=
(f--g) dx=O
01
F.
or
dx+/
6
-g dx=O
f dx-J£ g dx=O
L,fdx-J,.gdX.
The converse of this theorem is not true. i e., if
i.-e., if
f t ([ £
.f(x) dx=
IE
g(x) dx, then f(x):;t=g (x) a.e. on the set £
.g) dx=O, then (x)-g(x):;i:O a.e. on E,
i.e., in particular if
ff
dx=O, then /(x):;t:O a.e. on E.
E
We need only give an example in favour of this statement. Define a function (in the closed interval r--I, I] st. when x ~ 0 when x < o.
/(X)={-1 +1
Then
[1 (x) dx= [1 (x) dx+ I: .f(x) dx
II
or
/(x) dx=jO
-1
Finally
dx+fl -dx
-I
fl
0
=[ II -( x I =1-1=0. (dx)=O
-1
(x):;eO a.e. in the integral [--I , J 1.
but
Thus we see that the integral of (vani<;hes but the function is not equivalent to zero.
Theorem 6.
What is Lchesguc integral (~/f(x) in
Show that if (R)
f:
I(x) dx ('xi.fts, so docs
01/
interval?
161
iRE. LEBESGUE INTEGRAL OF A FU"IC;nON
(L)
1: 1 (x) dx (and they arc equal), but not conversely.
[Bannras 65; Kolhapur 1 0 ; Mt'erut 90, 88, 81; Kanpur90, 87] Proof I. For the proof of the fir"t part of the theorem, refer first method of Theorem 1. To prove that if (R)
J: f
I:
f
(x) dr. exists, then (x) dx exists and
f:f(X) (R)
1: f
dx=(R)
~f
exists
J:f(X) dx.
is continuous and bounded in
[a, bJ
f is continuous => f is measurable. Now f is bOllnded and measurable and so it is L-integrable over [a, bJ. Hence (L)
f:f(X) dx exists.
Divide the interval [a. bJ by me.lnS of points Yo, YI' Y2'·· .. ·····• Y .. s.t. a=yo< Yl
Lower Riemann
"-I
L m\'(Yk+l - rk)
WIU=
k -0 "-I
Upper Riemann sum = ].' Mk(htl'- y,J k~O
rex)
i;;. R·int,;grahle over [a, b]. Hence in the limiting case whch /'t1~Yb both Riemann sums tend to a definite limit "R)
f~ f(x)
dx.
L-intcgral of /{y) over
la, b],-
J:f(X) dx
=
'EI jY1d1 f(x) dx
k~O
Yk
the integral is additivt:,
162
THE LEBESGUE INTEGRAL OF A FUNCTION
Using first mean value theorem,
I
< ell f(x) dx < M~.m(ek) mil (Yl<+l-Yk) < JYk+lf(x) dx <; M" (YHI -·Yk)· Yk m/:.m(ek)
i.e.,
Summing this for the whole interval [a, b], we get
Making max. (Yk+l (R) I:f(X) dx
~
which
~
Yk)~O,
we get
/:f(X) dx
<; (R)
I:
f(x) dx
l:f(X) dx=(R) I:f(X) dx.
II. The converse of this tbeorem is not true,
,·,e., if (R)
I:
I:
f(x) dx
exists, then
it
is
not
necessary that
f(...;) dx exists.
We need only give an example of a function which is integrable in the sense of Lebestue but not in the sense of Riemann!. [Meerut 88; Kanpur M.Sc. P. 1982] Letf\x) be a function of x defined over an interval [0, 1] s.t. f(x) = 1 when x is irrational and /(x)=O when x is rational. [Kanpur 74; Banaras 69] Clearly f(x) is discontinuous. Points of discontinuity form measurable sets. Every sub·inteual of (0, 1) contains rationals as well as irrationals. For all values of r and for all modes of sub-division, M,=l, m,=O . • -1
.-1
r=O
r:::::110
Lower Riemann sum= l. m, (y,+:-y,)= }; O.(Y'+l-Y,)=O. ,
a-l
.-1
,-0
,=0
Upper Riemann sum= 1: M'.(Y'H -y,)= E 1.(Y,+1 -y,)=1. Upper Riemann sum~Lower Riemann sum. This => f (x) is not R-integrable over (0, I).
163
THE LEBESGUE INfEGRAL OF A FUNCTION
This
~
(R)
J:
f{x) dx does not exist.
Let El and E2 be the sets of points of irrationals and rationals in [0, 1] respectively. Let E=E1 UEa• Evidently
El n EI =
flJ .
DU, E)=Q(f, E1>+U(f, E~).
.
D( f, E 1 )= {(x, Y) : x EEl' 0 <;; Y <; Ax)} ={(x, Y) : x EEl' 0 ~ Y ~ I} ={x: x E E1}x{y: Y € [0, I]} m DU, E1)=1 X 1=1.
Again
D( I. E2 )=f(x, y} : x E E2 ,O " Y ~ f(x)} ={(x, y): x E..E~, 0 ~ Y t;;; O}=E.x{O} m D(f, E2)=m(E2)xm({O})=m(E2 )xO=O.
tf(X) dx=mU(f, E)=mU(f, E1)+m!l (f, E.)=I+O=1. This proves that f(x) is L.integrable over [0, 1], i.e.
f:
f (x) j;e exists.
Finally, tf(X) dx exists but (R)
J:
f(x) dx does not exist.
Ex. Prove that a function integrable over a closed interval I=[a, b) in the sense of Riemann is also integrable over I in the sense of the Lebesgue. Discuss the integrability in the Lebesgue sense of the fwzction f over [0, I] defined by if X is irrational I { f(x)= 0 if x is rational. (Banaras 1969] Solution. Prove this as in Theorem 6. Ex. Let a function f be defined on the interval (0, I) as follows:
I(X)={~
X X
is irrational is Irrational.
Is I integrable ill Riemann unse? Is If integrable in the lKanpur 74J Lebesgue sense '1 Solution. Here write I I part of Theorem 6. Here notice that U ([, E1)={(x, y) : x EEl, 0 ~ y ~ I(x)} ={x: x € E 1}x {y: 0 ~ .v ~ O} =E1xlO}.
164
THE LEBESGUE INTEGRAL OF A Fl1NCTION
Hence mQ(/, E1 )=m (Et).m({O})=O as m({O})=O, m[Q(J, E2 )]=m(E2).m([O, 1])=1 x 1=.::1.
Ex, 5.
>
Let p
0, q> 0, then show that
J:;:,-:q dX=~ "-/+q +p~fq --p-l-)q +, ..
(i)
and deduce that (it)
LOg2~-'-1--}+~-}.+ _,
(iii)
4 =1-- 3 +5---=r+ ,.,
~
J
1
I
Solution. We have x P- 1( 1+XO)-l = x P -) (1 _.. xtl.!- x'q _x3Q +,.\4Q -
.,)
00
= XP-l
L' (x2I1Q ._ X(2,,+1)Q)
'.=0
=
, 00
1:
[X2"~l-p-l_X(2"+1)G-t-P-ll
,,=0
Integration vie Ids the .result .. /
Jo ~V-l 1T x
dx= ~ [
1
q
.. =0
X2 ..
q~::, _ _ _~~~~~q+p
11
(2 + I) q +p ~(j=O
'J.llq r P
1 +____ - __
J l
III = ____ P p',-q 1'+2:.1 XoP-l
o 1 +Xfl
1 ! I dt=-·- .--- -- + .-~p
p+q
p +i..q
p+3q
-.-1. -+.,. .v +3'{
.L I
'"
, .. (l)
Hence the result (i). Put p=q = I and observe that
J1 1' . dx= [ o I
+x
log (! +x) 11 I
log 2= I·· i
we get
=
Jo
+ t -- t +'"
Jog 2, Hence the rc,ult (ii).
Taking p= I, 1j=2 in (I) a.nd obscuing that
}11 I Q
+X2
dx= [ tan- J x
J1
r.: n ='0j',
165
THE LEBESGUE INTEGRAL OF A FUNCTION
we get Henc~
the result (iii).
sin x . b . .~. P rove that th e fiunctlOll IS /lot Le esgue integr-
Ex., 6
x
able over [0, 00[.
[Banarlls 1971]
T 0 Sh ow t hat th e f unctIon . sin-x.IS not L -mtegra ' bIe. · So1Ut 100.
x
-x
If' we show t hat sin.x.IS not summabl e over [0,00[, we can conclude the result.
For this we must show that
J
I
OO
o
sin x , 00.
!-I- - - !
X
Consider the integral
I"'"
sin ~ dx=
o
X
Jr.
£ r- 1
I~}..!l
X
J dx
X
(r-ll"
{y+
r~l
~
" J" I
1.:
r~l
n
== 2'
r~ 1
Lim n_oo
0
I -
~.,
rIT.
I siny I d)'=
0
x
0
00
I if)'
rn
In-lsin~.J dx ~ ~
1 L~~~J x
i.e.,
sin...ll+(r--l)1t)
1t
E !r
2 1.:.. _.
r·' I
rIt
=~x :JO=co
.-1
7t
dx= 00.
o
Ex. .Give an example of a function which is not integrable in the sens~ of Lebesgue. [Kanpur 74, 75] Solution.
Ex. 7.
Write the solution of the above solved Ex. 6.
Prove thai the function f(x)=-ddx
(X2 sin x·~)=2X sin~-~ cos ~ xx x
is not L-integrable over [0, 1]. Solution.
I'.
•
Let J(x)=2x sm
1 2 Xi-x
[Banaras 67; Kanpur 88] I
cos,Xi'
166
THE LEBESGUE INTEGRAL OF A FUNCTION
The function 2x sin
x~ is bounded and continuous over
[0, 1]. Continuity of 2x sin
~ implies that it is measurable over
[0, 1.]. Being bounded and measurable over [0, I]. 2x sin
~
x
is
L-integrable over [0, 1]. To show thatf(x) is not L-integrable over [0, 1], it suffices to show that -
~
x
cos!... is not summable
x
over [0, J]. For ~ llis we must show that
J:~\ cos ~I dx=~. a,,={(2n+l)rc}-I/1, b,={(2n- U rc}-I/.,
Let
a •
E;
x
<.: btl ~ !a" ~ !.. ;;.!.. ~ ..!.. ~ .!... ~-x b,. b.'" Xli a,,1
, .. (2n-A) rc
1 x <; (2n+i>
~ "I
1t
~. 1 \ 1 ,.,. I cos Xl ;;t 2"
I
It I dx=,l; ~ 1cOS"2
Va x Q
~
t./JbOnn -21x
ft_l
X
I
J
,,-I
Ibnan -1 IcOS"l1 II dx X
X
• b,,· 1 · 6n+ 1 dx= E llog -2=4- 1: log 6--1=00 .....1 an n=1 n-o
00
Jll Icos 1I dx=oo. -
oX
"21 X
I
This proves the required result. Ex. Answer the following questions in 'Yes' or 'No'. (vii) Do the Lebesgue integrable functions over (0, 1) form a vector space 'l [Kanpur 74] (viii) Is a function equal to Q measurable function almost [Kanpur 74] everywhere itself measurable? (Ix) Is the Lebesgue measure of the union of two measurable [Kanpur 74J sets equal to the sum of their measureJ ? (x) Is a function, which is differentiable over (0, 1), Lebesgue integrable over (0, 1) '1 [Kanpur 74]
167
THE LEBE5GU 3 INTEGRAL OF A FUNCTION
(xi)
Is a continuous function over (0, I) Lebesgue integrable? [Kanpur 75] Solution. (vii) yes, (viii) yes, (ix) no, (x) yes, (xi) yes. Ex. 8. Show that if f(X)=!
(0
< x
E:;; 1), f(0)=19.
thenf is not integrable on [0, I]. [Kanpur 1979, 77] Solution. Let n be a natural number. Evidently f is an unbounded function. Recall that [f{x)] f{x) !fO ~f(x) ~ n n
Here we have
=1
n
[f(x)].=! if!
x
[f(x)]
n
>
Iff(x)
~x
n.
<1
< n => X-1 ~ n ::;. n-1 ~ X
n
[f(X)]n=n ifO\.< x
<~
[f(O)].= 19. The value of[ f(x)]n at x=o does not affect our calculations. / I [f]" dx=Jl t1 n dx+Jl ~ dx
I
o
lin X
0
=[ nx lIn +[ log 1:n X
Lim n-+oo
fl 0
1 n
--log -+I=I+log n [f]. dx= I + Lim log n. n-+ 00
Since the value of R.H.S. does not exist, hence f is not Lebesgue integrable over [0, I]. Ex. 9. Show that the function f defined by t
f(x)=x l /3 'O
<
< x
I,
and f(O) =0 is Lebesgue integrable over [0, 1]. Solution. Let n be a real number. Recall that (X)] f(x) !f 0 ~ f(x) ~ n
[Ji
"
={
n
Iff(x)
>
n
Here we have
[rex)] =-h. x n~ ~ x ~ 1 n
168
THE LEBESGUE INTEGRAL OF A FUNCTION
l
~ n~
For /(x)
-}-p <; n <
[ j(X) ] It =ll if 0
~,3 ~ X
;;>
]
< I/'~'l
x
[j(O) ] .. =0
1[f]
dX=JIJ;
1
o
3
~3
In x
II
3 ( X 2j3 =2
dX+jl/n3
)
Il
dx
0
I
1
ljn 3
+n
( ) l/n 3
x
0
__} (1--!..)+~- 23 !.n
-2 Lim
n-+oo
This
~
I
JI [I]
n a-
n2
Z
dx=~.
2 is Lebesgue integrable over [0, 11-
Remark.
0
,.
Although (R)
J: I(x) dX.=~.
Theorem 1. Suppose I is measurable on a !l/f!a:,urable set E. To prove thai f is integrable iff II I is integrable and that
ISEfl ~ LIll·
[Kanpur 73J
Proof t Let f be a measurable function on II. measurable set E. Suppof>e f is Lebesgue integrable over E. Then its positive and negative parts 1+ and /_ are also Lebesgue integrable over E. :But 1/1=/++1-. This'means that If!' is l.ebesgue integrabl.: over E. II. Letf be a measurable function on a measurable set E S.t.
I!I is Lebesgue Since
0
ir.tegrabJe over E, then
f
£
00.
B
~f+(x) ~ If(x)
Hence t.f+(x) dx or
f 1f I <
I+(x) dx
~
<
t
00,
IV
I/(x)
I dx <
00
that:;::> . f+ is Lebesgue integrable:
Similarly we can prove that 1- is But f=l... -1-·
x E E.
Lebe~gue
integrable.
Hence f is Lebesgue integrabk on E.
16'1
1 HE LEBESGUE INTEGRAL OF A fUNCTION
Ifl. Letf(x) ~O on £1 and.1~x) <0 on E~. Theil E 1 n£2=0 and E=£1 U E 2 • By countable properly of tho;: integrai
J Ifi dx=fE If I dX+J~- If I dx E
and J,fdx=I.fldx+J £1
E
This
I
.~!
1
~
£2
fdX=Jr
If
iJ
E1
...
(1)
dx I/ldx-Jo.lfl 1 °2
:
I I r olfidx+1 E I/i d '; "'''\E E ldxi<:l I ,1:. 1 I, 2 i J 1
J
1(1+ £-111= ~
JEli
O'j
:::>IJIE fll ~ Jill. E This completes the proof. Theorem 8, Let I be a bnunded measurable lunction deji'ned
If
I'
Thl'.I1 , E f \ ~ E I fl. [Meerut 1972, 71; Kolhapur 70; Banaras 71j Proof. Hert' write III Part of the about Theorem 7. Theorem 9a. Let f and g be unhounded measurable fUllctiuns defined over measurable set E. Show that
over a measurable set £.
(I) JEfdX=
1!1 fEkldx
0
E=I~l E .. Sot,
where
n £/=0 for k::j=k'. [Meerut 1975, 72, 71J Proof. Let f and g be unbounded measurable fUllctions defined over a Tll('asurable set £. Suppost', without losing generality, that/(x) ~ 0, g (X) ~ o. r or if the result is applicable to positive functions, it is equally applicable to negative ';unctions therefore it is so by addit;on in general case. Let mEN be arbitrary. Define [!tx)]m and [g (x)] ... as follows: Ek
when/ex) ~ when/ex) >
[!(X)]m={f(X)
and then we define
J E
m (f)
dx= Lim
Similarly define
f
E
g (x) dx= Lim
m·,.oo
m~oo
f
E
I
1:-
[f]
0
(g (xli", dx.
In
dx.
til
111
170
THE LEBESGUE INTEGRAL of A "UNCTION
Step
(i)
00
Let E= E E. s.t. Ek n It=l
To prove that
J
jJ
f{x) dx=
E
1:=1
Ek'={lJ for k�k'.
IE f(x)
dx.
"
[f (x)]", is bounded and measurable over fore, the integral of[f(x)]", is additive, i.e .• Since
I Making In-+«>
E
E
rf(x)]", dx
.=1
IE
E f(x) dx ;a
and there
[f(x)]", dx
[f{x)�111 dx
II i: � 1<-1 and then n .... «>.
J
J.:o�k
E
we get
'£1 IEk f(x)
. .. (1)
dx.
Again
IE [f(x)]", dx= ,,! lEk [f(x)]111
dx �
; 17;< f(x) dx Lk
.... 1
[f(x)]..
.: i.e.,
Making
m
...
! JEt f{x)
I
E [/(x)�.. dx��
OCt we
get
I
f E dx <
t1 IE" f dx.
Combining (1) and (2), we ge t
lEI Similar Problem.
fix.
dx=
k�J
the requird result. Ek
f dx.
sets
and
I"
is the measure.
measurable and non-negative on X and for A E M,
Solution.
. ..
( 3)
Let X be measurable space in which M is
the O-ring 01 measurable define. t/>(A)
...(2)
=
L
f dl", then 1> is countably additive.
Prove as above that
fEI dl'== E, I 1
Suppose f is
Ele
/"1"
(Kanpur 1986]
...(4)
THE LEBESGUb INTEGRAL OF A FUNCTION
But
:.
L
~(A)= I if> (E)=
dll, E=
I~l r/>(E
k)
or
~l E
1~ Ek
k=
rp (
171
k~ Ek )=~! ¢(E
k)
This proves that cP is countably additive. Theorem 9b. Let I, g be non-negatIve valued functions on la, b]. If I, gEL [a, b], thenl+g E L [a, b] and
Ie (/+g)=JE I+/ E g. [Kan{}ur 1972, 73; M. Sc. P. 75] Proof.
To prove that
IE (/+g) dx=fE Idx+JE g d.". Let 1\1 (x)=/(x)+g (x). Now we have to show that
IE
q, (x)
dX=/Ef(X) dx+fE g(x)dx.
Case I. When I(x) ? 0, g (x) ~ 0 so that Firstly we shall show that
[,p (x)]", Let
1 (x)
~ l/(x)1",+[g (x)]", ~
EO;; m, g (x)
4 (x)
~ O.
[1\1 (x)hlll.
< m ; then
. .. (3)
[J(x)]",=/(x), [g (x)]",=g (x) [1\1 (x)]m EO;; '" (x) =f(x)+g (x) =l f (x)]",+[g (x)] ..
i.e.
[1/1 (x)]",
< [f(x)] ... +lg (x)]"..
If either! (x) or g (x) is greater than m, then
4 (x) >
...(4) m so
that
r
[1/' (x).)",=m < f(x)]",+[g (x)lm [I), (x)]", < [f(x ]",+ [g (x)]"' From (4) and (5), it follows that [1jI (x)]", [hx)]",+[g (x)]", is always true. Exactly in a similar way, we can show that [!(X)]m+[g (x)]", ~ [tjI (X)]2"" Combining (6) and (7), we get the result (3). Integrating (3).and observing that
i.e.
<
...(5) ... (6)
... (7)
172
THE LEBESGUE INTJ:GRAL or A rU?-:CTION
J {~f(x )]... +!g we
~et IE!Y
IE [1(x)1". dx+ JE [g (.0:) 1m dx, IE [f(X)]", dx+ IE [g (x)]", dx
(X)Jm} dx=
E
~
(x)Jm dx
< JE [t.jJ(x)h'n dx Maldng m-7OC,
IE 4{x) dx ~ fE/ex) dx+ IE g (x) dx ~ IE y; (x) dx or
L~ '/1 (X)
I
dx= Efpq dx+t g (x) dx.
When j(x) ~ O. g (x) < 0, q, (x) ~ O. Ij.(x)-H - g(x»=f(x). Now the resuIt follows from case 1, since-g(x) > 0. Case III. When [(x) ~ 0, g(x) < 0, tI:(x) < O. Then (-1/1 (x»+f(x)==( --g (x». E v~ry function in this equation is > O. Now the result follows from caSt: I. R.:mark. Iff is an unbounded measurable function defined ova (l lllc~surable set E, then it can be easily shown that Case U.
Then
(i) (ii)
(iii)
iJ~J(X)
dx
f. (I (x)
dx=c
J 1:.
If ((x)
!~ t:
~
0,
I
Ij(x) I dx
[Kanpur J991J
r
.J(X) dx
,L
l:.J(x) dx=O, thenf(x)=O
almost everywhere i fI E. (i.) Ifj{x)= go (x) a e. on the set E, then
f
Ef(X) dx=
f
£ g(x)
dx.
I::XtRCJS[S 1. Iff (x) i~ bounded IDca:;urable fundion on [a, b], show thatJ(x) is L.:-bc:sgu..: integrable over [a, b]. [Klinpur M.Sc. P. 1983, 82J 2. Given an example of a functIOn which is integrable in the sense of Lebe~gue but nol in tbe ~ense of Riemann. I Kanpur M.Sc. P. 1982J 3. Iff is a bounded function in L [a, b] and if a < c < b,
prove that JE L La, C],fE L [e, b]
and
J:
f'- [f+
J:
f·
LKanpur M.Sc. P. 1978j
J73
THE LEBESGUE INfEGRt\L OF A. fUNCTI0N
4. Iff is a bounded measurable function on [a, b], show that f is L -i ntegrable on [a, b j. [Kanpur 1977, 76] 5. Explain how the Lebesgue intt:;gral of a bClunded function is extended to unbounded functions. Show that if x ~ I
!(x)=-,-ll/x,O <
1
0, x=O
then f is not integrable in the Lebesgue sense over [0,
11.
~Kanpur J976]
6. Define the Lebesgue integral of ~n unbounded measura· ble function over a measurable set. Gjve an exa.mple of a function which is not integrable in the Lebesgue sense. lKanpur 1975J 7. Iff and g are integrable functions over a measur::..blc set E, th';n prove that f+g is also integrablt: over E and tha.t
J (f-rg)=J E
E
f+f E g
lKallpur M.Sc. Pre. 7S, :"I.1.Sc. Flnal 73] 8. Give an example of a function which is not Lebesgue integrable. Justify four answer. [Kanpur 197'4] 9. Define the Lebesgue integral of atl unbounded measurable function over!l measurable sec E. Prove that f is integraLle iff If I is integrable and that
:if z: f~f z: d r x-oSill · L e t ji(X ) ~= dxL X21
l'
In
If:
. I (0 , )I . t hcopen Intcrva
SI ,1OW
thilt f is not integrable in the Lebesgue sense over (0, I). lKanpur 1913J 10. (a) Define the Lebesgue integral of a bounded mcasurabk function f in a bounded mea.surahle se~ E. proving the existence of the integral. (b) rftwo fllnctionsj:lnd g are Lehesgue integrable over a ~et E, then prove that f i-g is also integrable and
t U+g)=L.r+t g·
(c) Let f be integrahle over a set E and suppose tl:at £ is the union (sum) of a count&ble family of pairwise disjoint measurable sets E.; i.e., E=-=l::d-E2+ ..... . Prove that
f /=1' JE r
n
.. "
r.
~ Kaupllr
1972!
174
THE LEBESGUE INTEGRAL OF A FUNCTION
11 Explain how the Lebesgue integral of a bounded function is extended to unb:>unded functions. Prove that the function
l(x)=2x sin
~-~x cos.!.... x
x
is not integrable in the Lebesgue's sense over (0, 1). [Kanpur 1971] 12. Prove that a bounded measurable function on [a, b] is Lebesgue integrable over (a, b]. [Kanpnr 1970] 13. Iff is a non-negative measurable function defined over a set E of measure > 0 and
tldX=O then/(x) is a e. zero. [Bhagalpur 1968] 14. D.!fine the Lebesgue integral of an unbounded function. on [a, bJ. Show that if
f(x)=~ (0 <
x
~
1)'/(0)=19,
then/is not integrable on [0, 1]. [Kanpur 1970] IS. Define the Lebesgue integral of a function I, with [Kanpur 1969] respect to the measure over a given set. 16. Explain the full definition of the Lebesgue integral of a ",-measurable function/: X~R. Point out the case in which the integral is not defined. Define L' (X, ,.,.). [Banaras 1971] Hint. Case II of § 9-0. 17. Let (X~ M, ,.,.) be a measure space and I: X-+R be ,,-measurable and non-negative. Prove that
I"
fdp., ME M
is countably additive on M. [Banaras 1971] Hint. Here write Theorem 9a Page 166. 18. Decide whether or not each of the following two functions is in L' (R, m), and give reasons for your decisions: . sin oX (1) 1(0)= I, I(x)=----"- for x::;cO
x
(ii) g is even and
J2-,,-1 for x E [2k, 2k+1l g(x}=1_2-k - 1 for x E [2k+ I, 2k+2] where kEN. I [Ban~ras 1971] Hint. For first part, see Ex. 6, Page 16~. 19. Show that I: [0, 1]~R is Riemann integrable iff the discontinuities off form a set of Lebesgue meflsure zero. [Banaras 1970]
THE LEBeSGUE INTEGRH OF A FUNCTION
20.
175
Iff=g a.e., show that
L
f=,t g f d.nd g being bounded measurable functions on a set E of finite measure. [Banaras 1969] 21. Prove that if f(x) =0 at every point of Cantor's ten nary set andf(x)=p in each of the complementary intervals of length 3-1', then
J: f<x) dx exists in the Lebesgue sense and is equal to 3. [Banaras 1968] 22. Define Lebesgue integral of an unbounded function. Show that the function
f(x)=~ ~) dx (X2 sin X2 is not integrable in the sen!.e of Lebesgue over (0, 1). [Banaras 1967] 23. Give an example of a bounded real function on [0, I] which is L'!besgue integrable but not Riemann integrable. Also give an example of a function f on [0, 00] such that f is not Lebesgue integrable over (0, 00] though the improper Riemann integral off exists and is finite. [Banaras 1966] 24. Write a note comparing Riemann's definition of an integral with Lebesgue's. [Banaras 1965] 25. Define the Lebesgue integrable of a measurable nonnegative function f over a measurable set E. [Meerut 1970] 26. Iff is bounded and integrable in the Riemann sense in a closed interval [a. b) then prove that f is measurable and integrable in the Lebesgue sense and the Riemann integral off over [a, b] is equal to the Lebesgue integral over [a, b]. [Meerut 1970] 27. Iffis Lcb,:<;gue integrable on [a, b], and
J:
/(1) dt=O V x E [a, b],
prove thatf(t)=O almost everywhere in [a, b].
[Ml'erut 1972]
10 Theorems on Convergence of Seq uences of Measurable Functions Introduction. Convergence in measure is essentially weaker than convergence almost everywhere. There exist sequences of measurable (even continuous) functions which are divergent at every point, but which are convergent in measure. 10'0. Der. Convergence in mean. A sequence < f" > of Lebesgue integrable functions is said to converge in mean to a function I if
Lim n-+oo
J 1/.. --/1 E
dx=O.
[Indore 1978) 10'1. Definition. Convergence in measure. Let < In > be a sequence of measurable function defined over a measurable set E. Let/be a measurable functicn defined over E S.t. (i) /(x) < 00 a.e. on the set E. (ii)
Lim
n-+oo
m[E(I/n-fl
~E)]=O
¥
E
>
O.
Then the sequence < f" > is said to converge in measure to the function! 1O· 2. Definition. Pointwise convergence. Let < fn > be a sequence of measurable function defined over a measurable set E. If there exists a measurable functionf over E S.t. Lim (,,(x) = lex) 'V x E E, n--+ CI,
then we say that the sequence
<
fn
>
converges
pointwj',~ to
fon E. 1Q. 3. Definition. Convergence almost c\'erywhere. [Indore 1Q78! Let ,/ In / be a sequence of measurable functions defined
177
THEOREMS ON CONVERGENCE OF SEQUENCES
over a measurable set E. If 3 a measurable function f over E and (i) m (A)=O
a lIet A S.t.
lim I.n (X)=J\x ,rr) \f X E E-A n-+oo (i.e., converges pointwise to f on E-A) then we say that the sequence < f .. > converges to I almost everywhere on E. 10'4. Definition. Uniform Convergence. A sequencl:: < I .. > of measurable functions defined over a measurable set E is said to converge uniformly almost everywhere to a functionf if :I ACE s.t. m(A)=O and given E > 0, 3 no E N s.t. ¥ n ;;;::: no and V x E E--A => If..(x)-/(x) 1< e. Theorem 1. Let < fn > be a sequence of i1;ltegrable functions which converge zn mean to a lunction f, then f ..-I in measare. [Indore ,1970] Proof. Let < f,. > be a sequence of integrable functions defined over a measurable set E. Let En=E (1/.. -1 I ~ 8) where 8 > O. Then 1/,,(x)-/(x) I ~ 8 y X E En.
(1'\')
This => fEn I/n(x)-j(X) 1 dx Since/R~{in
lim n-+oo
... (1)
m(En).8
mean on E and hence
I
Ifn(x)-(x)
E. C E => fEn (2) and (3)
~
I dx=O
IIn-I I ~
... (2)
IE
11..- II
... (3)
~ IE" I/n-/I=O as n-+OO.
In this event (I) But m(En)
~
0,
=> m(En }.& ~
V Ell'
0
::;>
Hence
m(E,,) EO;; 0 as n-H>O.
lim n--jooo
m(En) =0.
This => m [E (/I. -II ~ 8)]=Oas 11_00 Y a> O. => 1,,-1 is measure. Theorem 2. Let < f,. > be a sequence of measurable lunctions
lim a measurable set E an d I,'f' n_oo thell show that I is measurable 011 E. Oil
Proof.
I.• (x) =J,I') (X a.e.
on E .
Let A={x E E: lim In (x):;i: I(x)}. Then b} def. n-+oo of almost everywhere, m(A)=O. Define, g..(x)=/n(x) if E A and g(x)=O if x E A.
178
THEOREMS ON CONVERGENCE OF SEQUENCES
This => lim gn =lim/n if x ,. A ~ g(x)=/(x) if x ,. A and g(x)=O if x E A. The above facts prove that < g,. > converges pointwise to g on E. Also gn is mea'iurable V n. Hence limit function, i.e., g is mea'iurable. Consequently I i'i measurable. Theorem 3. (F. Riesz). Let <: f,. > be a sequence ollunctions which converges in m'!asure to the lunction Ion a measurable jet E. Then there exists a subsequence which also converges to the functionl almost everywhere. [Meerut 1986j Proof. Let be a monotonic decreasing sequence of positive terms sach that lim Uf! =0. Let En;: be a convergent series 00
of positive terms so that E nk=O. 1:=1
< In >
converges if} measure to I
lim n
-+00
... ( I)
It means that corresponding to the seq!lence < Uk > we can construct a st:quence < n. > satisfying the condition (1), is a sequence of < I .. >. Now we want to show that lim 1.1«x)=flx) a.e. on E. k-.oO .. (2)
Write
From this it follows that < A" > is a monotonic decreasing sequence of m:!a~urable sets. This declares that lim m(A,,)=m(B) .. (3). (Refer Theorem 18. Chapter 7, page 102). 00
It follows from (I) and (2) that m(A .. ) \
.
< 2;,,=nn/c.
Consequently
lim m(An) =0 .. (4) as En" is convergent. This => m(B)=O, by (3). Let y E E -B b~ arbitrary. Then y f!. B y ,. B => y t Ano for some natural number no => y If. E( lin:, II ~ Ok) 'V- k ~ nO
=> IIn~(") -/lY) ! < lim => k_oo
Ok
1,.··(y)=/(Y)
,?L
k ~ no
V Y
E E-B.
179
THEOREMS ON CONVERGIlNCE OF SEQUENCES
Also
m(B)=O.
lim
/ ..t=/
Hence k -+00
a.e.
on E.
Theorem 4. (D.F. Egorff). If a sequence 0/ measurable/unctions is convergent almost everywhere on a measurable set E, 0/ finite measure, then/or every 6>0,3 a measnrable set EoeE s.t. (i) m(Eo)< 8, (ii) the sequence is uniformly convergent on E---Eo· [Meerut 1988,90] Proof. Let < f. > be a sequence of measurable functions defined on a measurable set E s.t. < fra > converges almost everywhere to a finite measurable function / on E. Let lim/..{x)=/(x) V x E A. Then, by def. of a.e., m (E-..4)=O. Since every function is measurable on a set of measure zero and
E::' =
Ii [E (Ift{X) f,.-+/ on E
:. s.t. m
~
-f(x)
I
.lim m
n~ao
> 0, 3 positive (E-E~.. ) < e/2m Given«
(11m) )] and E"'=
]1 E:'.
(E - E"')=O. ..
integer no depending on .. and m
v
Define Eo=
U (E --E:
m=l
•••
)
g
Being countable union of measurable sets, Also Eo C E m (Eo)=m [
(1)
!?u is
measurable.
U (E-E"0m )] ~ E m (E-E"') < 110
'"= 1
E,
by (1).
m=1
This ¢ m(Eo)< 6. This completes the first part of the promblem E-Eo=EThus
l.,f
U (OE -Ernno )=E n
"'=1
I
(u
E"') ... =1 no
n ~ rio and x E E -Eo ~ x E E~
=> If.. (x) - f(x) I < 11m. Hence < f" > converges uniformly t%n £-Eo· Theorem 5. (D.F. Egorff). If a sequence of measurable functions converges to a/inite limit almost ef'erywhere ill a measurable
180
THEOREMS ON CONVERGENCE OF SEQUENCES
set E, then, given 8, we can lind a set 01 measure greater than m (E) -/l in which the sequence converges uniformly. Proof. Let < In > be a sequence of measurable functions defined over a measurable set E. Let this sequence converge to a finite measurable function! alntost everywhere on E. Let 8 > O. To prove that :i ~ measurable set ACE S.t. (i) meA) > m (E)-8 Oi) < In > converges uniformly on A. Consider a convergent series ~7)k of positive terms and a sequence < an > of positive terms converging to zero s.t. 171 > 172 > 173 > ........ . 00
An (17)= E E (1/k--f1 :? 0),0> O.
Write
k='TI
Then m (An (17)1=0 as n- oo • ...(1) For In-f. By virtue of (J). corresponding to every positive integer k, we can find another positive inte~er 7)k s.t. m [(ARk) (0)] < Tlk. Now we cho.))e a positive integer
ko
00
E
S.t.
7).
<
8
i=ko vo
Set B= . 2-' An, ("I)' Then m (8) < ~. l=ko Write A=E-·B. Then E=A U B, A or
n
B=0.
rn (£)=111 {A)+m (8) m (A)=m (£) -m (B)
> m (E) -
8. For m (B) < 8 ~ -m (8) > -8. :. m (A) > m (£)-8. Rem'lins to ShOW that Itt (x)-"/{x) uniformly on A. Let. > 0 be arbitrary. Then we can find an integer k S.t. k ;;;;t ko, 0" < c ailY x E A ~ y ~ B. For A=E--B. ~ x
f'I.
Ani
(a,)
~ .Y f'I. E ( Ilk ~ /It(x) fix) ~ /ft{x) - /Ix)
I I ~ ai) I < 0, < ~ !< E ¥ k
for
k;;;-:",
k ~ ni ~ n" x E A
~ In "'/uniformly on A. Let g be a function s.t./and g are equivalent on £ so that
m
[E ( 1--1= g)]=O.
. .. (2)
181
THEOREMS ON CONVERGENCE OF SEQUENCES
To prove that the sequence < In the function g. For this we have to show that
..
I.e.
>
converges in measure to
I/,.-g I ~ 6)J=0. . .. (3) Iln-g I ~ E) C E(f~g)+E( lin-II;;'.)
[E (
In
Evidently E( This => m [E ( I I .. -g I ';J 6)J ~ m [E (l;i=g)]+m [E ( Using (1) and (2), we get m [E ( 1 /..-g I ~ 6)J ~ 0+0=0 m [E (
I/n-g I ~
Il,.-I I ~ .)].
6)J ~ O.
,... (4)
But measure is a non-negative quantity so that m [E( Iln-g I;;" 6)] ~ O. . .. (5) Combining (4) and (5), we get the required equation (3). Proved. Theorem 6. Lebesgue bounded convergence theorem. Let < In > be a sequence 01 bounded measurable lunctions defined over a measurable set E s.t. I/.. (x) I < M >.f n E Nand>.; x E E. Let < /., > converge ill measure to a measurable junction I on the set E, M being a positive cOllstant. Then
Lim
n~oo
f
E
j,,(x) dx=J
E
j{x) dx.
[Meerut 1986; Kolhapur 76 ; Kanpur 85, 70, 73) . Proof. Let < /.. > be a ~equence of bounded measurable functions defined over a measurable set E S.t. Iln(x) 1 < M>,f n E Nand >,f x E E. ...(1) Then In is integrable over E >.; n E N, M being a positive constant. Let < In > converge in measure to a measurable function Ion E so that Lim m[E( If"
IH'OO
II ~E)]=O,
... (2)
for every. > O. To prove that Lim
1l~ 00
f
E
In (x)
dx=J !(x) dx. E
From (I) and (2), I/(x) I < M. Suppose 8 > 0 is arbitrary and set A.=E( lin-II ;;" 0), Bn=E( 1/.. -/1 < 0). Then E=A" U B.. and An n B,,=(lj.
"
.(3)
. .. (4)
182
THEOREMS ON GONVERGENCE OF SEQUENCES
The definition of convergence in measure ensures that Lim m (An)-O.
n-oo
. .. (5)
This follows from (2). By countable additivity property of the integral
t
lin-II dX=JAn 1/11-/1 dX+JBn lin-II dx.
We have lin-II <: 8 ¥ Using first mean value theorem.
or
IBn I/.. ~/I dx <
8.m
IBn I In-I I dx <
3.m (E).
Take an arbitrary
E
8.m (£)
(Btl)
. .. (6)
E BII •
X
E;;; 3.m (E)
•• (7)
> 0 and choose 8 such that e < "2
In this event (7) takes the form
IB.. l/n-/1 dx < ; Now 8 is fixed
lin-II or
~
... (8)
II.. I + III < M+M=2M lin-II < 2M.
Using first mean value theorem, we get
fl/n-/I dx <
2M.m
(An)
... (9)
(5) =>
Given.
>
0, :I no E N S.t. n ~
110
=>
I m (A .. ) I < 4~
~ m (A .. )
:.
• < 4M
m (An)
~
... (10)
O.
From (9) and (10),
I I fn -I I dx < ; ~ t 1/.. -1 I dx < ; + ; =~ A..
¥
n
110
... (11)
Writing (6) with the help of (8) and (II), we get
or
L, I I,,-{ I dx <
6
¥ n
~ 110
¥
n ;;. no
183
THEORFMS ON CONVERGENCE OF SEQUENCES
It It
or This
(fn-/) dx
~
Lim
n~oo
or
Lim n-+oo
Lim
or
I. t I< J (J. I
(f.. -f) dx E;;
n-+oo
fi
I>
I>
/!n-f I dx
"'I n
<•
;a, no·
-f) dx=O
If
fn (x) dx-
J /.. (x) dx= I>
J I
f(x) dx=O
I>
I>
f{x) dx. Hence the result.
Ex. State and prove the theorem of bounded convergence for Lebesgue integrals. Show that this theorem is not true lor Riemann [Kanpur 1973] integrals. Tbeorem 7. Tbe Dominated Convergence Theorem. Let < fn : n EN> be a sequence of measurable functions defined over a measurable set E S.t. I fn (x) I < '" (x) yo x E E and "'I n E N where", is integrable over E and the sequence < fn > converges in measure to a measurable I on E.
Then
J f{x) dx= n-oo Lim I I>
fi
f. (x) dx.
[Kanpur 1991, 89, 87; Punjab 69; Banaras 65; Meerut 88, 90J Proof. Let < In > be a sequence of measurable functions defined over a measurable set E. Let", be integrable over E s.t. Ifn (x) I <.CiJ (x) "'I x E E and "'I n E N. . .. (1) This ~ I. (x) is bounded V- n. Also fn (x) is given to be measurable. :. < In > is a sequence of bounded measurable functi~ns and hence these are L-integrable. Let the sequence < In> converge in mea!>ure to a mcac;urable function I on E so that Lim m [E n-+oo
(I f,n -f I
To prove that Lim n-+oo
~
&)J=O
J In (x) dx= I B
I>
... (2)
for every e
/(x) dx.
> o.
184
THEOREMS ON CONVPRGENCE OF SEQUENCf:S
From (1) and (2), I/(X) 1 < IjI (x) ... (3) Moreover, tfo(x) is integrable over E. Then (3) shows that I(x) is integrable over E. It means that (In - f) is integrable over E. Let ~ > 0 be arbitrary and suppose that An=E (i In-I I ~ 8), Bn=E (I In -II < S). Then E=An U Bn, An n Bn=0. The definition (10-1) of convergence in measure ensures that Lim
n~oo
meA..)=0
. .. (4)
By countable additivity property of the integral,
t;
lin-II dX=!An I/.. --fl dX+!BIJ
IJ~-/I
dx
... (5)
We have I f,,-fl < 8 ¥ X E Bn. Using first mean value theorem, we get
fBn
I In-I I dx < ~.m(B.. ) ~
JI
Bn I .. -II dx
i.e.,
>
Take e
<
8.m(E)
a.m(E)
... (6)
0 and choose 8 such that 8.11I(E)
E
< "2
Then (6) reduces t.
f
Bn 1 In - I I dx <
~
.. ,(7)
Now 8 is fixed,
I f~--/I
or
~
1fn 1+11 I < q,+1\i=2tjJ
lin-II <
~I\I V X E E.
On integrating, we get
IAn I In-f I dx <
-.
fAn q,(x) dx
... (8)
1f..I
-:
This
~
~ IJAn I\i(x) dx \= JA~ tjJ(x) dx.
In this event (8) takes the form JA..
(4) ~ Given
'I}
>
I I .. -f I dx < i
IJAn IjI(X) dx r
0,3 no E N S.t. n ~ no =>
I meAn) I <
... (9) TJ
185
THEOREMS ON CONVERGENCE OF SEQUENCES
=> •
m(A,..} < 1/ In (An) ~ 0
Making use of absolute continuity of the integral, we have IJA" 4(x) dx
I<
;
[For this refer theorem (11) of chapter (11) page 210] whenver m(AII) < lJ for n ;? no' Using this in (9), we get
fAn I f" - f
I dx < ;
... (10)
Writing (5) with the help of (7) and (10), we get
t
I f,,-f I dx < ~+; =~
L,
or
Ifn-fl dx
II It
(fn-f) dx
or
Lim
or
00
Lim
11-+00
or
Lim
i~
(fn-f) dx:
This => n~
<
Il-'J>o 00
I (f~-t) I I E'
.
€
for
t <
11
~
110
1.1;,-fl dx E
¥
n
~
<
e
no
dx=O
f;.(x) dx--J f(x) dx=O
f,
E
E
/;,(x) dx=J f(x) dx. E
This concludes the proof. Q.E.D. Ex. State and prove Lebesgue's dominated convergence theorem. [Banaras 1965, 67. IV 70] Beppo-Levi's Theorem 8. l.et < In > be a non-decreasing sequence of integrable jimctions defined over a measurable set E. L e1
Lim j'.. b e 'lI1tegra ble n -+ (Xl
J'
E
ol'er~.
r
tlell
'T'I ~
Lim Lim I ..(x) dx. J,,(x) dx= n -+ CX) E "' I: tl-'J>o 00 [Meerut 1988, 87; Banaras 69, IV 70; Kanpur 1'14,82,69] Proof. Let < in > be a non-decreasing sequence of integrable functions defined over a measurable set E.
L et
Lim f," b e 'mtegra bl e over E.
11-+00
186
THEOREMS ON CONVERGENCE OF SEQUENCES
J
f
Lim I .. (x) dx. To prove that Lim In(x) dx= n->oo B B n_oo Since .< In > is a non-decreasing sequence and hence
11 ~/2
~/3 ~
..... .
This ~ 11 ~ In ¥ n ~ 1.. - 11 ~ 0 ~ I\In ~ 0 ¥ n, on taking ~ .. =/n-fl' Moreover < In > is a sequence of integrable functions implies that <,p.. > is a sequence of integrable functions. Finally is a sequence of non-negative integrable functions. Applying this to the Lebesgue's bounded convergence theorem, lim
f
n .... oo B or
lim
n ..... oo
or
lim
11_00
dX=
I
lim
Bn-oo
J (/. -fJ dx=I B
B
.1. 'f'n
dx-
E
E
dx=
I ..(x) dx=.
E
B
dX
lim (f.. -/l) dx
n-oo
f In J /1 f n-+oo r I n_oo' n_OO,E Lim
or
.f. '!In
lim In dx-
J 11 d'( E
lim (,,(x) dx.
Hence the result. Remark. The theorem 8 is also known as "Lebesgue's monotone convergence theorem". Ex. State and prove Lebesgue monotone convergence theorem. [Meerut 1988; Kanpur M.Sc. P. 81] 9. Fatou's lemma. Let < f" > be a sequence of non-negative integrable lunctions delined over a measurable set E S.t. Lim.Iniff,,,= ( a.e. on F .
(l')
n-+cc
I
Lim. n_oo ml I,,(x) dx <
(ii)
Then
I
~
J
Lim.
n-+oo 1111 E (n(X) dx. [!\leet'uf 198.8, 87; Kanpur 81; Bhagalpur 68; Banaras 64; InQore 78] Proof. Let < fn > be a sequence of non-negative integrable fu nctions defined over a measurable set E s. t. E
(J')
(x) d"
00.
.
~
f = Lim.In f f,n a e on Il-+ 00 .
E
187
THEOREMS ON CONVERGENCE OF SEQUENCES
(ii)
Lim infI. fn(x) dx
n-+-::t:>
Ii
r
To prove that
<
co.
~
Lim infI fn(x) dx. n_CO £ Define gl"(X)= inf {fn(x)}=inf {fn(x) : n ~ k} f(x) dx
£
n~k ~
Then gn(x) and therefore
t
fn(x) ¥ n g .. (x) dx
~
\ £ fn(x) dx.
Consequently,
I
gn(X) dx ~ Lim infI fn(x) dx Lim ... (1) n-co F. n-+ co Ii Since < gn > is an increasing sequence of non-negative integrable functions and hence by Beppo-Levi's Theorem 8,
Lim
n_cx>
i.e.
I
Ii
gn(X) dx=
=J
I
,r(X) dx
E
F.
Lim gn(x) dx
n-+ 00
=1 n-oo Lim inf
Lim gn(X) dx= n-+co Ii Using this in (1), we get
J
I
~
£
f(x) dx
J f(x) dx. E
Lim inf
n-oo
fn(x) dx
I
fn(x) dx.
£
Hence the result. Ex. State and prOl'e Fatou's lemmaIor integration. [8bagaJpur 1969 ; 8anaras 64] Problem 1. State the l.ebesgue's dominated convergence theorem and use if to e~'aluate the following integral: Lim f,,(x) dx n-'>- 00
J1
0
n3 / 2 x
' where fn(x) = 1 +n2x~' 0 ~ x ~ 1, n= 1,2, 3,... ...
,
[Kanpur 1976, 74, 72] Solution. For the statement, see Theorem 7. 1 n3 / 2 X2 1 ---:,. -=·I.(x) say 2x J• ~ x ,fn(x)=-. x l+n '1'. ,
Then fn(X)
~ Ij.(x).
Also 4(;\) is integrable in [0, 11·
188
THEOREMS ON CONVERGENCE OF SEQUENCES
Hence, by Lebesgue dominated convergence theorem, lim
Jlo fn(x) dx= =
-=
r
lim fn(x) dx
... (1)
·0
I: 11:: (r+~~2- ) 1: n~~ (In)( .~ :X
dx
)
2
112
=Jlo O. (·~2) dx=fl 0 dx=O. U+x
ADS.
0
Problem 2.
> 0, prove that
If rI.
11-.lim+00 J"0(1- nx- )"
X'- 1
dx=
10 00
e-rxCl -
I
dx
where the integrals are taken in the Lebesgue sense. Let fn(x) = ( 1-:
Solution. Then fn(x)
.p
~
0
0.
XCI-I.
0
lim 11_00
f,.(x) dx
J" (1--::')" x"- dx= Joo en
lim
l
Il_ex>
0
0
W
X~-1 dx.
Show that the theorem of bounded cotlv.!rgence
Problem 3.
applies to .(,.(x) = 1 Solution.
>
Hence, by Lebesgue dominated
J" f,.(x) dx= JOO
lIm
IX
X)" =e-X.
n r} (x) is Lebesgue integrable. convergence theorem, 11-+ ex>
Il-oo
x a- 1 •
(x) V n where Ij!=e-ID
Notice that lim (1-
i.e.
r
[Kanpur 1973]
::2X2for °< x < J.
lKaDpur 1972]
nx
fn(x) = 1 +n 2x 2
1
=-1--
'
I
'~---1---------2
nx +I1X [\/(nx) -V'(IIX)] Take M=i. Then I f,,(x) I ~ M. Hence, bv Lebesgue convergence theorem, lim lim f,,(x) dx. f,,(x) dx=
Jl
11
Il-+ ex>
L.H.S. of (J)=lim
'-.
0
0 11-+00
JIo 1+11\ : IlX
1 log (l +n2x2), dx=lim -2 1/
~
t
+2
..• (1)
(form~) 00
189
THEOREM<; ON CONVERGENCE OF SEQUENCES
Uf(I +~~:~].2nx2 =Iim ~- =0
=Iim R.H.S. Cf(I)=J:lim
1+
2
112X2
CI;;X2) dx=I:n~:[M(i7n~+X2)JdX
=[1o (0). (0" x -.l~ , 21 dx=fl 0 dx=O. 0
:. L.H.S.=O==R.H.S. This:;. L.H S,=R.H.S. This verifies Lebesgue bounded convergence theorem. Theorem 10.' Let f be a non-negative fLlllction which is integrable over a measurable E. Then given £>0, 3 '>0 s.t. V AcE with
m (A) < S, we have
Jf < c.
[Meerut 1986]
A
Proof. Let f ~0 defined on a measurable set E. Let ACE S.t. m (A) < S
... (1) ... (2)
When f is bounded on E.
Case I.
> 0 S.t. I f This ~ lex) ~ M Then 3 M
(x)
I~
asf~
...(3)
M
0
Using first mean value theorem, we get
f
.A
f
< M.m (A)
<
M 3,
by (2)
J .f < M.S
or
... (4)
A
Given
E
> 0, we determinc i
15
Then (4)
give~
< M
or
Lf <
e
1)
s.L.
M3
<
E.
Case II. Whcn f is unbounded on E. We define a measurable function .I;, on £ as follows:
r
')={
. " C\
f(x) Il
if.f(x)
if/ex) >
II
Observe that .I;. (x) ~ f"tl (x) 0 /~ (x) 11. It means that < f .. (x) > is a sequence of bounded measurable function defined on E. We write
and
<
<
190
THEOREMS ON CONVERGENCE OF
~E9UENCES
J I=Lim f f,. Here we have In
I
~
In ~ LI
and therefore
J.4
or
L
(I-.f..)
This
~
given.
>
~
0
,.,(6)
0, EI no E N s.t. n
It (I-In) \<
L(I-i < f.4 f ~
or
no )
Since
.rno ~ no and so
Using (2), Taking
JI .4
no
<
~
no.
E'
L(f-In») <.'
or
...(5)
n~oo
E
vn~no
E'
llo
... (7) nil meA)
llo·8
< .-, we get
llo~
f f <~. L[(r-fnJ+InJ -< .4
(7) +(8) gives
...(8)
no
E'+."=e, say
LI<E
or,
Thus in both cases we have proved that
L/< Problem 4,
by then
E.
a
and
II -1 <: < 0 I: (0, Ixo , 0 < x ~ 1 ji) \X =l o , x=o
ver~fy Beppo Levi' s th~orem. Solution. For each n, we define In : (0, 1] -> R S.t.
f,,(x) ={xa
o
!f lin ~ x ~
If 0 ~
X
I
< 1/11
1] -. R be defined
191
THEOREMS ON CONVERGENCE OF SEQUENCES
observe that
< In >
In-I
is an increasing sequence of functions S.t.
I: In= a~l (1- n~+I)
and
Io/n I
or
~
1
a+l
"f
nE
"f
nEN
N
By Monotone convergence theorem,
J: I
dx exists and
Io/=n--:r:> [1o/n=a.f.-T Lt
I
I
Note: fis 1,InbQunded and it is not R-integrable over [0, 1]
alth~ugh (R)
I:
I
exists.
EXERCISES 1. State and prove Lebesgue's monotone convergence theorem. [Kanpur M.Sc. P .. 1984,82] 2. If E E M and < In > is a sequence of non-negative measurable functions such that lim inff,.(x)=/(x), x E E, prow that
n-+XJ
_
f
E
Idp.
~
lim infI I" dr"
n-+ co
E
[Kanpur M.Sc. P. 1981]
3. State and prove Lebesgue's dominated convergence [Kanpur 1983] theorem, and use it to evaluate the following integral: lim 1 J j~{x) dx, where
n_ co
0
n3 / 2 x In(x)= l+n t \-2' 0 ~ x ~ t, n=l, 2, 3, [Kanpur 1976] 4. State the Lebesgue's dominated convergence theolem and usc it to evaluate the following integral: lim n3 '2 x (,,(x) dx where In (X) = 1 + 2 -2' 0 ~ X ~ 1.
fl
11-+ co
0-
n x
[Kanpur 1974] 5. (a) State and prove the theorem of bounded convergence for Lebesgue integrals. Show that this theorem is not true for Riemann integrals.
192
THEOREMS ON CONVERGENCE OF SEQUENCES
(b)
If ~ > 0, prove that
lim
t n(l_~)n xa-1 dX=J"'"
e-1JIj X,,-l dx, n 0 where the integrals are taken in the Lebesgue sense. [Kanpur 1973] 6. (a) Let < ,f,. > be a sequence of Lebesgue integrable functions in a set E such that 11-+00
1.
fn(x) ~ 0 for x E E and
almost everywhere in E.
lim n-;..oo
Prove that
J E
f
fn(x)=f(x)
f
~ n-+oo lim fn. E > for which strict
Give an example of a sequence < fn inequality holds in the above result. Hint. See Theorem 9, Page 186. (b) State and prove Lebesgue's theorerp of dominated convergence. Show that the theorem of bounded convergence f, (x)
applies to
nx =1+n20 and the theorem of dominated cqnvergence to g,. (X)=11 3 / 2 x/(1 +112X2) for 0 E;;; x ~ 1. [Kanpur 1972] 7. Let < fn > be a sequence of measurable functions defined on a set E of finite measure, and suppose that there is a real number M such that I f,,(x) I .;:; ; M for all n and all x. lim . If f(x) = fn(x) for each x In E, then prove that n
11-00
I
E
f- lim
-11_ 00
J J, E
n'
[Kanpur 1970]
8. Define the Lebesgue-integral of a functionf, with respect to the measure Over a given set. Let < fn > be a sequence of measurable functions defined on the p.-rneasurable space E wch that ~ fl (x) ~ f2 (x) E;;; ... ... (x E E). lim . If f,,(x)-;..f(x), then show that
°
11-;" 00
f fn dp. J.f d/-', (n.-~CXj). -?
l:.
Hint. Refer Theorem 9. 9. If f: R ... R is Lebesgue summable and for n=l, 2, 3, ....... .. E.={x E R : If~x) I > n}
[Kanpur 1969]
193
THEOREMS ON CONVERGENCE OF SEQUENCES 00
~'
show that the series
p. (En) converges.
[Banaras III 701
n=1
10. State and prove Beppo-Levi's theorem. [Banaras IV 70] 11. State and prove Lebesgue's dominated ~onvergence theorem. [Banaras 1965, IV 70) 12. Let < In > be a sequence of real valued measurable functions on (-00, 00) and -let 00
1/" (x)
1:
I=~ (x) E
V (-00, 00).
n=1
Prove that
f: L~)
In
(x)
dX=n~
S: I.
(x) dx
provided that either side of the equality exists. [Banaras IV 70] Hint. r/> (x) E V (-00, (0) => ~ (x) is measurablo and ~ (x) is L-intep·able on ( - 00, (0). 13. Let < I .. > be a monotonic non-decreasing sequence of integrable functions over E and let prove that
f 1=
lim
J
Lim
n-»oo
In i,
1ft.
[Banaras 69] 14. State and prove the theorem of dominated convergence. [Banaras 67] 15. Let fbe a (finite) rea) valued Lebesgue measurable function on [0, 1] and let E .. ={x E [0, I]: n-l <:'f(x) < n} for n=O, ±I, ±2, ...... , show that E
1f 1
o
dp. <:
0<",
n-+oo
iff
00
1.' _00
B
I nip. (Eft) <
00.
[Baaaras 66]
11 Absolute Continuous Functions, Indefinite Integral and Differentiation 11'0. Continuous function. A real vatu.ed function [(x) is said to be continuous at given
> 0,38 > 0 s.t. I f(x) -f(xo} i
f
<
Xo
if
Ii
whenever I x'-:"'xo I < 8. [Baoaras 65,67, IV 70] 11'1. Absolute continuous function. [Meerut 87 ; Kolhapur 70] Let f be a finite real vilued function defined over a closed interval [a. b] s.t. given
f
>
0, 3 8
>
0 s.t./1' [f(b r )'- f(a r )] , r=1
whenever E (br r-l 01
<
-
ar )
bi E;;;;
< O2
I<
E
8, where
<
b2
~
...
«;;
an
< bn •
Without altering the sense of definition, we replace the condition
Ir~l
[f(br} --[(or)] to
X
r=1
I<
E
by
th~ stronger condition.
I f(b r) -flor) I <
e.
Then the function f(x) is said to be absolutely continuous in the interval [a, b). An absolutely continuous function is continuous. Since we can take the above sum consisting of only one term.
ABSOLUTE CONTINUOUS FONCTIONS, INDEFINITE INTEGRAU
195
11'2. Indefinite Integral. The fundamental theorem of integral calculus states that integration and differentiation are reverse processes of one another. Let f(x) be inte~rable over [a, b] and let F(X)=[ f(t) dt+c
¥ x E [a, b]. where c is any constant. Then the function F(x) is defined as an indefinite integral (or simply integral of f(x». The term indefinite refers to the variable x-the upper limit of the integral. 11'3. Differentiable. If f(x) is a continuous function, then F(X)=[ f(t) dt+F(a)
is said to be differentiable, F(a) being any finite constani.
it '4. Def. Monotonic functions. Suppose a function f is defined on a closed intervai
[a, b].
Let x. y E [a, b]. f is said to be strictly increasing function if x > y => f(x) > f(y) f is said to be monotonic non-decreasing if
x > y => f(x) ;;;;J f(y). f is said to be monotonic non-increasing if x > Y => flX) ~ Ji.y). f is said to be strictly decrea<;ing function if x > y => /(x) < Jiy).
Increasing and decreac;ing functions are called monotonic functions. Monotonic functions are a/ways regarded as finite valued functions.
11'5.
Definition: Function of bounded Variation. [Indore 1978 ; Banaras 67] A real valued functionf is said to be of bounded variation in a closed interval [a, b] iff it can be expressed as f(x)=¢ (x) -1\I(x) '>of x E [a, b) where -;6(x) and ~(x) are monotonic functions. --.... Alternatively a function of bounded variation is defined as follows: Suppose a real valued function is defined on a closed interval [a, b]. The interval [a, b] is divided by means of points a=xo < Xl < x 2 < ...... <: x .. =b Define
"-1 v= k=O 1: I !(xk+l)-flXJ;.) I·
196
ABSOLUTE CONTINUOUS FUNCTIONS, INDEFINITE INTEGRALS
The least upper bound of the set of all possible sums V is ,..
•
b
called the total variation of I and is denoted by V ( I) o
i.e.,
b
b
V (f)=sup V. o
If V=finite, then 11
I is called a function of finite variation or function of bounded variation. We also say that I has linite variation in [a, b]. 11'6. Der. I..ipschitz condition. A function I is said to satisfy Lip~hitz condition if :I positive constant M s.t. I/(x)--I(y) I ~ M I x-y i. 11'7. Del. Lebesgue point. If lim
h_O
1 J,"+I' II", I I(t)-I(x) I dt=O,
then x is said to be a
Lebesgue point of the function l(t). U·8. Fundamental Theorem or Integral CalcuJus. [Meerut 1990] If I<x) is continuous and if F(X)=J"" l(t) dt+F(a)
... (1)
then the theorem states that F'(x)=/(x). . .. (2) That is to say differentiation and integra~jon are reverse processes. By elementary theory of fntegral Calculus, (2) is proved as follows: To prove (2). From (1),
F(x+h)-.F(X)=I~" I(t) dt+F(a)=
f:
=
J
tll
I(t) dt+
1:
u:
l(t) dt+F(a) ]
l(t) d(
",+II
e
l(t) dt.
J.+Ii
F(x+h)-F(x) _ I(x) = l.. l(t) dt-/(x) -.!... ("+" dt h h ,. h J.. 1 f",+11 = II Ie [f(t) -/(x)] dt
ABSOLU rE CONTINUOUS FUNCtIONS, INDEFINITE INTEGRALS
!F(X+!)-F(X) -f(x)
\=m' \
[+"[f(t)--/(X)] dtl
~1-hIIJ:t"lf(t)-f(X} I dt. Since f(x) is continuous and therefore given, > 0, 3 3 > 0 s.t. I f(t) - f(x) ! < e whenever I t-x I Using this in (3), we get IF(x+hl=-F(X) -f(x}
or
197
\< I ~ I J:
IF(X+~)-F(X) -f(x} \ <
e,
t
"
IhI<
E.dt,
... (3) <
~.
8
I h I < ~.
Making h-,;O and consequently .......0, lim F(x+h)-F(x) _ /(x) =0 a.e. h_O h
or or
lim F (x+h)-F(x) h-,;O I,
lex) a.e.
dF (x) (ix-=f(x) a.e. F'(x}=/(x) a.e.
Hence the result. Problems related to functions of bounded variation. Theorem 1. A monotonic/unction on [a, b] has finite variation on [a, b]. That is to say, a monotonic function defined on closed interval [a, b] is of bounded variation. [Indore 1978] Proof. It is enough to prove the theorem for an increasing function. Let f be an increasing function defined on the interval [a, b] so thatf(x,} <; f(xr+l) for Xr <; xr+l' Divide the closed interval [a, bJ by means of points a=xo < Xl < XI < ...... < xn=h. or
For f(x r +1 ) -/(xr ) ;;:a, 0
+
[ft x ..) - f\X"-l)] --f(xn} -f(xv} = /(b) --/(a) = finite number. [For f is monotonic;? ftb) and f(a) are finite numbers] =[ f(x l ) -/(xo)] +[ f(x z) - ftXl)]+'" ...
Thus V=finite quantity. This proves that the total variation- is cOJ!stant and is inde-
198
ABSOLUTE CONTINUOUS FUNCTIONS, INDEFlNITE INTEGRALS
pendent of the mode of sub-division. Hence f is of boullded variation. . Theorem 2. Iff is of bouneed variation on [a, b], then V=P+N and P-N=j(b)--f(Q), where V, P, N, respectively denote total, positive and negative vari:ltions off on [a, b]. Proof. Let II-I
V= X I f(xr+t)-f(x r ) r=l
I.
Hence the closed interval [a, b] i:i
divi~ed
by means of points
a=xo < Xl < Xl < ...... < xn=b. Let p be the sum of those differences f(xr+.)-f(x,) which are positive. -n that the sum of those diff~rences which are negative. Evidently / f(b)-f(a)=p -no v=p+n, From which we get v+f(b)-f(a)=2p v-f(b)+f(a)=2n ... (1) v=2p+ f(a) ---/(b) i.e. v=2n+ f(b)-f(a) ... (2) Set P=sup p, N=sup n, V=sup v where we take the suprema over all possible sub-division of [a, b). Taking suprema in (1) and (2), V=2P+f(a)--/(b) .,.(3) ... (4) V 2N+f~b)--fta) upon addition, 2V=2P+2N or V=P+N (3) ... (4) gives 0=2 (P- N)+2 [f(a)- f(b)] or f(b)-fta)=P-N. Theorem 3. To prove that a function f is of bounded variation if and only if it is expressible as a difference (If two monotonic functions both non-increasing or both non-decreasing. '. [Meerut 1988; Banaras 69] Proof. Let a function f be defined and finite on a closed interval [a, b] so that f (a) and.f (b) are finite numbers. We shall show that , (i) iff is of bounded variation, then it is representable as a difference of two monotonic increasing functions.
ABSOLUTE CONTINUOUS FUNCTIONS, INDEFINITE INTEGRALS
199:
(ii) iff=g-h, where g and h both are monotonic increasing functions, then f is of bounded variation. (iii) Let f be of bounded variation. Divide the interval [a, b] by means of points
a=xO<x1 <x2 <···<xn =b.
Let
V=
!: /
f(Xr+J-f(X r )/
Let p be the sum of those differences f(x r +1)-- f(xr) which are positive, - n that the sum of those differences which are negative. Evidently v=p+n, feb) - f(a)=p-n. From which we get v+f(b)-f(a)=2p v-f(b)+f(a)=2n v=2p+f(a)- feb) ] i.e. v=2n+ feb) - f(a) ... (1) Set P=sup p, N=sup 11, V=sup v, where the suprema is taken over all pos~ ible subdivisions of [0, b]. Taking suprema in (I), we obtain V-2P+f(a)-flb) ... (2) V=2N+ftb)--f(a) ... (3) Further we suppose that Vex), P(x), N(x) respectively denote total variation, posItive and negative variations off in the interval [a, x], where x :E;; b. With the help of (2) and (3), V=2P(x)+f(a)-f(x) ... (4) V=2N(x)+!(x) --f(a) ... (5) (4)- (5) gives 0=2 [P(x)-Nlx)]+2 [f(a)-f(x)] .f(x)=P(x) - N(x)+ f(a). Taking f(a)+P(x)=P'(x), we get f(x)=P'(x)-N(x). . .. (6) It is easy to verify that P(x) and N(x) both are monotonic inc(c:lsing functions. P(x) is an increasing function implies that P(x) is also increasing function. Now the required result at once follows from (6). (ii) Letf=g-h, where g and h both are increasing functions. For any mode of sub-divison of [a, b],
or
200
ABSOLUTE CONTINUOUS FUNCTIONS, INDEFINITE INTEGRALS -1
V (/)= E I j(x'+l)-/(x,.) ,.=0
I
a=xo<x1 <x2 <.·.<xn =b
where
I I(Xr -tl)-/(x,,)I=1 = ~
[g(x~+1)-h(xr+J] - [g(xr)- hex,)] I I [g(X r+1)-g(x,.)]-[h(xr+l)-h(x,)] i I g(X'+I)-g(X,) 1+ Ih(x,+1)-Iz(X,.) I
=[g(x,+1)-g(x,)] + [h(X"+l) -h(x,.)]. For g and h both are monotonic increasing
functions.
...
n-l
E If(xr+J-I(x,,)
r-O
I <;;
11-1
E
r~O
[g(X'+l)-g (x,)]
=g(b) - g(a) +h(b) - h(a) =a finite number. [For 1 is finite ,valued ~ g and h both are finite valued]. V (f) <; a finite number. This. 1 is of boundld variation. Theorem 4. The su~, difference and product of two lunctions 1 and g 01bounded variations are lunctions 01 bounded variations.
...
Hence show that L is 01 bounded variation if g . I g (x) I ;;;at 0 > 0 ¥ x. [Kanpur M.Sc. P. 1988] Proof. Let/and g be functions of bounded variations on the closed interval [a, b]. Divide this interval by means of points
a=xO<x1 <xl<'" <xn=b eo
Denote the total variation of1 in [a, b] by V (/). Then, by a
assumption eo
V (I)
~
a
(g)
< finite number}
<
finite number
(i) Letl+g=s. To prove that s is ~f bounded variation in [a, b].
•.• (1)
201
ABSOLlJ. E CONTINUOUS FUNCTIONS, INDEFINIE INTEGRALS
I S(XHl)-S(X~) 1=1 [ftXHl)+g(X'+l)]-[!(X/c)+g(X k )] I· =1 [!(X,:+1)-!(Xk)]+[g(X,,+1) -g(X~)] I ~!!(XHl)-!(Xk) ft-l
ft-l
k=O
k~O
1: I S(Sk+l)- S(S/.) 1 ~
:.
1+1 g(Xk+I)-g(X
I!(X k +1)-!(Xk)
};
k) ,
I
71-1
J: I g(XHl) - g(X,,)
+
;:=0 b
or
V (s) a b
V (S)
or
a
<
This
~ S
(ii)
Let d=!-g.
I
V a
b
(! )+
V (g) a
by virtue of (I).
00,
is of bounded variation in [a, b].
d(x"+1)-d(Xk)
...
b
:<
I = I [f(hl·l)-g(XHl)]-[!(x .. )-g(x/.:)l I = I [!(XHl)-!(X/.:)]-[g(x/.: 41 )-g(xt)l I ~ 1 !(Xhl) - !(x.. ) I + I g(XHl)- g(Xi) I·
"-1
}; 1 d(XHl) --d(x,,)
k=D
+ b
V (d) ~
or
a
a
a
This
~
V (d) u
b
V
b
(!)+ .,
(g}
<
n-l
1:
,,~O
00,
I g(x,:+1) -
g(x/.:)
I
by virtue of (1).
n
< ex.
=> d is of bounded variation in [a, b]. (iii) Let p(x)=!(x) g(x).
To prove that the function p is of bounded variation in [a, Write A=sup { I lex) I : x E [a, b]} B=sup { 1 g(x) I : x E [a, b]}. I P(XHl) - p(xd 1
= I f(XH1) g( Xl:+l) - !(x,t) g(x,J I
bJ.
202
ABSOLUTE CONTINUOUS FUNC110NS INDEFlNI1E INTEGRALS
=1 [(XI<+I) g(XHl)-[(Xk) g(XHl)+[(Xk) g(XT'+l)-f(Xk) -[(XT,)]+[(Xk) [g(Xk+l)-g(X,,)] I
I
g(Xk)
=1 g(X kH ) [[(XkH) ..;;;;1 g(XkH) ~ B
:.
I . I [(XHl)--[(X T,)
1+1[(Xk) I . I g(XHJ-g(Xk) A I g(Xk+l)-g(X k) I
1/(XT,+l)-/(Xk) 1+
"-1
ft-l
10=0
k=O
E I p (Xk+l)- p(Xk) I ~ BEl [(Xk+l)-[(Xk) I
b
b
b
G
a
<
Y (p) ~ B V ([)+A V (g)
or
I
a
b
This => V (p)
<
00,
by virtue of (I).
qo
a
=> p is of bounded variation in [a, b]. (iv) Let g(x) ~ (1 > 0 1.1 x E (a, b]
h=!. g
Set
To prove that h is of bounded variation in [a, b]. I 11=- , g(x) ~ g
0
>
I
0 1
=> h(x)=-( gx) ~ (1
¥ x E [a, b]
>
0
y. x E [a, b].
h (x k) /'=1' _ I_ - _1_ I, h(xl'+l). g(XHl) g (x,)
(=/' g(Xk)-g(Xkl-l) I' g(x) g(XHl)
~ ;o •I g(xk ) -g(XHl) I :.
or
I
"-1 E h(x',H) -hex,,') 110=0
b
a
C1
b
or
,b
-Y (h) ~ '2 Y (g)
V (h)
<
a
I
~ 2I
<
f1
1:
n-l k=O
I
g(x;;+t) -h(x,J
I
00
00.
Il
Accordingly h is of bounded variation. By case (iii),
Jh=J.
!..= g gJ
is of bounded variafion in [a ,
b] •
ABSOLUTE CONTINUOUS FUNCTIONS, INDEFINITE INIEGRALS
203
Theorem 5. Let f be a function of bounded variation in [a, b]. Iff is continuous at X=X o, then the function 0-
(x)=V(f)
'It
a
is €ontinuous at x=xo' Proof. Let a < Xo means of points Xo
<
Xl
<
b. Divide the closed interval [xo, b] by
< xs< .... ····.< x,.=b.
Vfe can find an • >0 s.t. b
V
b
>
. .. ( I)
V (f)-.
"
Xo f is continuous at x=xo ::;. I f(XI)-f(xo)
I<
e, u,henever
I Xl-xu 1< 8.
--...
From (1),
n-l
= 1,=1 E : RXHI-/(Xk) I +2E= b
Thus But
b
V(/)+2e
b
V(n
Xl
b
b
Xl
V (f)=V(f)+ V (f) Xo
...
Xl
or
Xl
b
V (f)+ V (/> Xo
V (f)
«
b
< V (/)+2£
2e ... (2)
Xa Xl
Xo
1t (XI )-1t (Xo) = V (/)- V (/) II
II
204
AESOLUTE CONTINUOUS FUNCTIONS JNL LFINITE INTEGRALS
XJ
b
"
C
= V (f). For V= V+ V so a
c
tha~
a
II
c
II
V-V=V a
a
Xo
or
or
n(xl ) -n(xo) < 26 I ~(Xl) - 7t(xo) I < 2e whenever
I Xl -xo I < 8 Generalising this, I n(xk)-7t(XO) I <2E whenever I Xk-XO I < 8. This => n(x) is continuous at x=xo. Tberoem 6. An integral is afunction of bounded l'ariation. An Alternate Statement. If f(x) is a function of integrable on
[a, b] then prove that the function F(x) =
J: 1
(u) du,
has finite
variation on [a. b). [Indore 1978J Proof. Let F(x) be an indefinite integral of f (x) defined over (a, b). To prove that F(x) is a function of bounded variation in the open interval (a, b). -Our assumption implies thatj(K) is integrable over (a, b) and F(x) = [
fIt) df+F(o) ¥ x E (0, b),
... (1)
F(a) being any finite constant.
Any function/(x) i:; expressible in the form /(x)=f+(x)-f-(x), (+(x =
where
.
J/
»)
x) ¥
0
X
... (2)
~ 0
vx
"t x < a. 0 V x ~ 0 Since f (x) is integ'able over (a, b) and therefore integrating (2). we get
f-(x)=I
and
-f (x)
1-
[
/(t) dt=
J:
/+(r) dt--J: I-(t) dt V x E (a, b)
Using (I). we get F(X)-F(a)=[ J+(f) dt-"-i:f-(t) dl "t x E (u, b)
or
F(X)=F(!l)+J:f+(t) dt-[f-(t) dt
ABSOLUTE CONflNUOUS FUNCTIONS, INDEFINITE INTEGRALS
Taking
1:
205
f+(t) dt=FI(x)
and
J:
we get
F(x)=F(a)+FI(x)-F2(X)
f-(t) dt=F2 (x), ... (3) V- x
E (a, b).
Since the integrand of the integrals
J:
f+(/) dl and
1:
f-(t) dl
is non-negative and is therefore non-decreasing function of x, i.e., FI(x) and FI(x) are montonic non·decreasing functions of x. Consequently F(a)+FI(x)=Fa(x) (say) and F2(x) are monotonic non-decreasing functions of x. Then (3) becomes F(x)=F3(X)-Flx) 'V x E (a, b). This shows that F(x) is a function of bounded variation in (a, b).
Problems related to absolute continuous (unctions. Theorem 7. If I (x) and g(x) are absolutely continuous functions, then their sum, difference and product are also absolutely continuous (unctions. Further if g(x) does not vanish for any x, then the quotient
B(X» gx
is a/so absolutely continuous.
Proof Letf(x) and g(x) be absolutely continuous functions over the closed interval ra, b] so that given • > 0, 3 8 > 0 s.t •. n
X I fib,·)-f\aJ
I < e and
"=1
whenever
n
k
I g(bk)-g(a,,) I <
E
11=1
"
l,'
(b;;-aJ
< 8
~I
V- aI' bI , a2> b 2, ...... , an, b,. s.t. a 1 < bI ~ a~ < b2 ~ ...... ~ an < b,.. Step (i). To prove that /(x)±g(x) is absolutely continuous over [a, b]. n
E I f(b,:)±g(b,J -l f(akLf::g(ak)] 11=-1
I
ABSOLUTE CONTINUOUS FUNCTIONS, IN~EFINiTE INTBJRALS
206
Finally
From t~ t~required result follows. Step (ii). To prove that f(x) g(x) is absolutely continuous over [a, b). n
1: If(b l.) g(b,)-f(ai') g(ak)
I
1:".1
<
n
n
I . I g(blc) -g(ak) I + 1: I g(ak) I I !(b~) -feat) It
1: I RbI:)
lc=]
t-l
We know that absolute continuity => continuity ~ bounded ness ~ f(x) and g(x) are bounded in [a, b] => f(x) Ml' g(x) M 2 • Of x E [a, b], Ml and M2 are upper bounds of f(x) and g(x) respectively in [a, b].
<
<
In this event (I) takes the form
Setting n
E 1:~1
E
(I
Ml
I + I M21
)=~', we get
I Rbi) g(b,J -flak) g(a/;) I < .'.
From this the required result follows. Step (iii). Let g(x) vanish now here in [d, b]. so that :I a > 0 S.t.
I g(x) I ;:.
0
>of x E [a, b].
To prove that !Ix) is absolutely continuous over [a, b]. gt X)
207
ABSOLUTE CONTINUOUS FUNCTIONS, INDEFINlTE INl EGRALS
i: 1_1___I _1 =
i; I g(bt)-g(at) I
1
k=1
g(ilk )
g(ak)
g(b k ) (a k ) I
k=l'
<=a2
Setting ~=«;', we get a
n 1 I 1 \ < ~l g(b k ) - i(a~)
e'.
This proves that _(1 is absolutely continuou over [a, bJ. g X)
By step (ii), /(x). _1_ = .
[a,
~x)
f(Jf~ is absolutely continuous over
~~
J
bJ. This concludes the problem. Ex. Iff, g : [0, 1]-+ k are absolutely continuous functions,
prove that f
+g and fg are ,absolutely continuous functions.
[Banaras IV 1970] Solution. For solution refer to proof of Theorem 7, steps (i) and (ii). Theorem 8. Let f(x} be tln absolutely continuous function on the closed interval [a, b]. Let afunction F(y) satisfy the Lipschitz condition in the ugment [A, B} C [a, b). The composite function F[f(x») is absolutely continuous in [A, B]. Proof. Let /(x) be an absolutely continuous function on the closed interval [a, bI then/{x) is a~solutely continuous on [A, BJ era, b] so that ¥
I
E f(bk)-/(a~)
11=1
¥
numbers
01'
E
> 0, 3 S > 0 s.t.
J<
6
...
(1)
bl , ...... , an, b n S.t.
A=al
and
1.' (blt-ak)
-
_Also let F(y) satisfy Lipschitz condition in [4, B) so that 3 s.t. I F(Blt)-F(Alt) I :( 'rJ I B,,-A. ,. Then
It~: IF(,f(blt )
-F(/{ak» \
:(\~l
\f(bk)-f(ak )
'I
> 0
I
<'rJe, on using (l)
208
ABSOLUTE CONTINUOUS FUNCTIONS, INDEFINITE INTEGRALS
Taking'rj
€=fi',
we obtain
.£11 F(f(b1c»-F(f(ak» 1< 15', wheneverTc~
(bTc-aTc)
for a finite sequence of pairwise disjoint intervals
<
< 8
(all, bk)
>.
This proves that F(f(x» is absolutely continuous in [A, B]. Theorem 9. Every absolutely continuous function is of bounded [Meerut 1988, 90] variation. Proof. Letf be an absolutely continuous function on a closed interval [a, b] so that we can select a 3 > 0 S.t.
~
1c~1
\ f(h k) - ftat) \ ,
<
1 whenever
E (bk-a1c) < 8
110=1
for aU numbers a1> bI , a 2, bz, ...... a,., b,. where a=a1 < bi ~ (12 < b2 ~ ...... ~ an < bn=b. Again divide the closed interval [a, b] by means of points a=co -< C1 < C2 < ..... c =b no in no parts s.t. CkH -CI: < 8. Consequently for any subdivision of [CTc,
E I f(XHl)-f(x.)
•
I <;;;
1 where
Xt+1>
CkH]
XI E
[CTc, CI:+1]
V~~+1 (f) ~ 1.
i.e.
It follows that
f ...... +V C"o Vba (f )=V C1 (f)+V c. ()+ Co
~
C"O_l
(f)
1+1+ ...... =n o
~!
i.e.
C1
( f)
<
00.
Consequently f is of bounded variation. Theorem 10. If a function f is absolutely continuous is an interval and iff' (x)=O almost everywhere, thenfis constant. [Meerut 1986; Banaras 71, 68; Punjab 65, 68; Kol~apur 70] Proof. Suppose f is absolutely continuous in an interval [a, b] and
f'(x)=O
a.e. on [a, b].
ABSOLUTE CONTINUOUS FUNCTIONS, INDEFINITE INTEGRALS
209
To prove thatfis cW)nstant in [a, b], we have to show that f(c) /(a) for any c E [., b]. Let E be a set of those points for which f'(x) =0. Set E~[a, c]. any x E E => !'(x)=O
=> 3
7],
h
0 s.t. If(x+~~ - f(x) I < 'I
>
=> If(x+h)-f(x)
I<
'I h.
Hence corresponding to each x E E, 3 a small closed interval [x, x+hJ contained in [a, c] s.t. If(x+h)-f(x) I < 7] h. . .. (1) It means that the intervals [x, x+h] cover E in Vitali's sense. Hence we can find a finite number of pairwise disjoint closed intervals lk s.t. /l=(X I, YI], i 2=[X2, Yz].···:·· .. ·, I .. =[x... Yn] which cover all of E except for a set of measure zero less than a. where a is pre-assigned number> Qcorrespor.Jing to each E in the definition of absolutely continuity of f If we suppose X k ~ Xk+l' then a=yo <;; Xl < Yt xl! < Y2 X .. < y" X ..+1=C and
k~O I
<
(Xk+ 1
-Ylr) I <
I),
l~O
<...... <
k(x
-<
m ) - /(Yk) 1 <
E.
For f is absolutely continuous. Notice that
,~
k
(y,.) --I(x,)
I" .:~: <
i.e.
iI/f(Y/.:) -f(x k )
Now I/(C)-f(a j .
I<
7j
(y, - x,). ,,,,,oni;ng:o (I).
(b-a)
'I (b -a)
... (2)
1=1 ;,: [f(x +1)-f(Yl:)l+ E[f(Yk)-f(xt)]1 k=G
k
k..o
",~ If{X.H) -/(y.) 1+ !. kLY·l-f{X.) I < .+'Ij (b-a)
210
ABSOLUTE CONTINUOUS FUNCTIONS, INDEFINITE--INTEGRALS
:. I/(c)-J(a) I < 6+17 (b-a). But 6 and 'TJ are arbitrary. Hence they can be made arbitrary small as we please
:. IJ(c) --J(a) 1=0 ~
Proved. J(c)=Jta). Theorem 11. Let a Junction J be integrable over a measurable set E and let < E It > be a seqilence oj subsets oj E such that Lim m(E,,)=O
k-+oo
equivoimtly
f
Lim E.J(X) dx=O k-H~)
Then or
....
IIE~J(X) dX! <
E,
k
~
ko, whenel'er m(E,,)<
o.
(This property of integral is known as absolute continuity of the integral). Proof. Let /(x) be.,tategrable over a measurable set E and let <EIt> be a sequence of subJets of E S.t.
Urn
k -+f>o m(Ek)=O.
,Lim f To prove that k_do '£ J<x) dx=O. ;
Ie
Case I. ~henJ(x) is Throughout the discussion we shall suppose that
E" C E ¥ kEN. Slncej(x) is bounded over E and hence we can take
« ~ J(x) ~ (3 on the set Ell.
Using the first mean value theorem, we get
Making k-+oo and observing (1) we get
f Lim f o ~ k-+oo E"
Lim «.0 ~ k-oo EIcJ(x) ~x ~ ~.O or This
J(x) dx ~ 0
~ k~:fE".f(X) dx=O
• •. (1)
211
ABSOLUTE CONTINUOUS FUNCTIONS, INDEFINITE INTEGRALS
Case II. When f(x) is unbounded. Suppose, without losing generality, thatf(x) ~ O. For if the result is applicable to positive functions, it is equally applicable to negative functions and therefore it is so by addition in general case. Let mEN be arbitrary.
Define [f(x)]", as follows: when /(x) ~ m whenf(x) > m.
[/(x)]",= {f{X)
m
Then [.r(x)]", is L-integrable and we define
J Ii
.r(x) dx= Lim
m-.oo
Observe that [.r(x)] ...
f
Ii
L.r(X)].n dx.
. .. (2)
< f(x)
and therefore JEk [f(x)]", dx <; fE./(X) dx
f
or
E" {f(x) -[ f(x)]m} dx
This
O.
... (3)
Q
given
E'
> 0, 3 ng E N S.t. m
lJEk {f(x)-[f(x)Jm} dX\
~
~
<
~ ".
Ii'
1 < ,f
J
E. [ f(x)-[ f{x)]", dx
according to (3,1
=- JEJf(x)-[ f{X)ln.l tlx < Ii',
... (4)
in particular_ Since
< n•. m(E
lex) ]". dx giv 0, 3 kg E
and therefore JEJ (1) ~
...(5)
k ).
N s.t. k ~ k.
=- I m(E,.) I < 3 .. M(Et ) -:
In view of (6), (S) becomes JEJ f(x)
1.
dx
~
n. m(E,.)
<
,...&
<
3 ... (6)
m(E'J.):::O= O.
212
ABSOLUTE CONTINUOUS FUNCTIONS, INDEFINITE INTEGRALS
or
<
JEt [ f(x) ]no dx
Taking noo <
~',
no 8 for
k~ko·
... (7)
we get
JEt [f(x)] no
dX < ~"
~
for k
... (8)
ko
Adding (4) and (8), we get fEk {JiX)-[ f(x)
lnJ dx+IEt [f(X) ]nodx < ,'+,'.
Setting e=E'+e, we get JEtf(X) dx .. I(x)
~
o.
:. JEk If(x) I dx < e Now, or
IIEk/(X) dX\
I
JEkf(X) dx
This
=>
< II
I<
E
for k
/r o•
_.. (9)
~ JEk If(x) I dx <
for k
I
~
k~: E~ f(x)
~
kJ whenever
6
m(E~) <
dx=O.
8
Q.E.D. ,
Theorem 12. An integral is a continuous fUllction. Proof. Let F(x) be an indefinite integral of f{x) defined over the open interval (a, b) so thatf(x) is integrable over (a, b) and F(x) = [itt) dt+F(a) ... (*) ¥ x
e
(a, b)
F(a) being any finite constant.
Firstly we shall establish a lemma: Let f(x} be integrable over a measurable set E and Lim m(Ek)=O. let <En'> be a sequencf! ofsub,ets of E S.t. k Lemma.
·~oo
Then
Lim k-+OQ
fEft
f(x) dx""=-O·
The pro"r of the lem n 1 start~. For proofreferto Theorem 11. Q.E.D. (lemma) Going back to the actual theorem, continued above, we pwtced as follows to prove it.
.213
ABWLUTE CONTINUOUS FUNCTIONS, INDEFINiTE INlEGRALS
Let x E (a, b) be arbitrary. Let Xl' X~ E (x-i), x+3). Set E1<:=[x1 , xz]. Then m(Ek)=m ([Xl> x 2]) < 3. By the lemma, just proved,
IfE f(/) dl I < I,
f!
for k ~ k. whenever meek) < 3.
In present case this takes the form
if;:
fl/) dll
<•
<
3.
I<
8.
whenever m ([Xl> XI))
In view of (*) this becomes
I F(x2)-FtXI) I <
e
whenever
I Xa-XI I <
8.
Finally, given
• > 0,38> 0 s.t.
I F(X2)-F(XI) I < •
whenever I XZ-xi This proves that F(x) is a continuous function.
Theorem 13. An indefinite integral is an absolutely cOfl/inuous [B.n.V. 1967] function. Proor. Let F(x) be an indefinite Lebe~gue integral of f(x) defined over [a, b] so that F(x) =
and
r:
fu) dt + F(a)
... (.)
¥ x E [a, b] over [a, b]. F(a) being any constant To prove that F(x) is absolutely continuous over [a, b]. Now we prove a lemma.
lex) is integrable
Lemma.
Let f(x) be integrable over a measurable set E and lim let < £1'> be a sequence of subsets of E S.t. k _00 m(E1<;)=O.
I
Then kLim E 1: f(x) dx=O. _00 The proof of the lemma starts. The proof of this lemma is given in Theorem 11. Going back to the actual theorem continued above, we proceed as follows to prove it. By the lemma, just proved
214
AESOLUTE CONTINUOUS FUNCTIONS, INDEFINITE INTEGRALS
\fEk f(x) dx I<
G,
k ~ ko. whenever m(E~) < (j. Having this in mind, let us prove the the,orem. Let aI' hI, a2 , b2 , . " ••• an, b.be 2n real numbers s.t. 0 1 < b1 <.; O 2 < b2 ~ ...... ~ a. < bn • "
Let m(Ek )= E m(b" • _1
1: fb,a. f(t) dt
Then \
0=1
I~
i.e.
.
0_1
n
0,)= ~ (b.-a.) '~1
<
<
S, where
l)
>
O.
E
[F(bi)-F(Ot)]/<E
according to (*) We have the final result as given
6
> 0,38 >
0
s·t.1 i
.=1
[F(b.)- F(a,)]
1<
e
n
whenever
1,'
i_t
(bi-o,)
<
~
for all values of 01> b1 , a 2 , b2 • ...... , an, b•• s.t. a1 < b1 ~ a 2 < b 2 ~ ...... ...;; On < 5... This proves that F(x) is absolutely continuous function. Problems related to indefinite integral. Theorem 14. Every absolutely continuous function is an indefinite integral of its own derivative. [Banaras 1967] Or To' show that F(x) is an indefinite integral if F(x) is absolutely continuous. [Banaras 1965] Proof. Let f(x) be an absolutely continuous function ill a closed interval [a, b] so that
f'
(x) and
J: f'
(I) dt
exists finitely ¥ x E [a, bJ. Let F(x) be an indefinite integral of F(x)=/(a) +
To prove that
F{x)~/(x).
1: f
rex), so that
(1) dr, x E [0, b]
... (1)
215
ABSOLUTE CONTINUOUS FUNCTIONS, INDEFINITE INTEGRALS
By Theorem 13, F(x) is absolutely continuous in [a, b]. Also (I) ~ F' (x)=1' (x) a.e. From which
fx [F(x)-f(x)]=O.
Upon integration, F(x)--f(x)=c (const.) Tak,in g x=a in (1), F(a) =f(a) + i.e. or
J:
F(a)=f(a) F(x)-f(x)=O
... (2)
I'(t) dt=f(a)
for x=a.
Subjecting (2) to this condition, we obtain O=c. Then (2) is reduced to F(x) - f(x)=O a.e. i.e. . F(x)=f(x) a.e. In this event (l) proves that f(x) is indefinite integral of its own derivative. Theorem 15. A necessary and sufficient condition that a function should be an indefinite integral is that it should be absolutely continuous. [Meerut 1986 ; Punjab 66; Banaras 67, 71J Proof.
(i) Let F(x) be an indefinite Lebesgue integral of
(x) defined over [a, b] so that
F(X)=[ [(I) dt+f(a)
... (*)
. ¥ X E. [a, b], and f(x) is integrable over [a, b],
F(a) being any constant. To prove that F(x) is absolutely continuous over [a,
b1.
For proof refer to Theorem 13. (ii) Let f(x) be an absolutely continuous function in a closed interval [a, b] so that III
I' (x) and exist finitely
J" f' (f) dt
V x E [a, b).
Let F(x) be an indefinite integral of I (x) so that F(x)=f(a)+J"'"
f'
(f) dt
where x € [a, b].
216
ABSOLUTE CONTINUOUS FUNCTIONS, INDEFINITE INTEGRALS
To prove that F(x) f(x). For proof refer to Theorem 14. Theorem 16. Let F(x) be an indefinite integral of a bounded measurable function f(x). Then F' (x)=ftx) a.e. An alternate statement. If f is a bounded and measurable function on [a, b] and Ftx) =
1:
f(t) dt+F(a)
then prove that F' (x) .... f(x) a.e. in [a, b]. [Kanpur 199O,Meerut 1988,87, 86 ; Punjab 67, 65 ; Banaras 67] Proor. Let F(x) be an indefinite integrvl of a bounded measurable function f(x) so that f(x) is integrable and F(X)=[ f(t) dt+F(a),
... ( I)
F(a) being any finite constant. To prove that F'(x)=f(x) a.e. By hypothesis f(t) is bounded and therefore we can write I f(t) I <; "'I, where M is an upper boudd of flt). From (1),
! U:+Tl == ! [J:+~
F(x+hl-·F(X)=
[(t) dt-:-F(a)-
J:
/(t) dl- F\a) ]
f(l) dt ]
Taking modulus of both sides
J:H f(t) dr i
IF(X+h~-F(X)[=m I ~
1 JIII+I> ThT I f(tJ I . I dt I OJ
AI
~ThT
JIII+I> 01
AI
Idt/=ThT·1hl=M.
:. IF(X+~-.F(X)1 ~ M F(x+h)-F(x) I . and h -+ F (x) a.e.
... (2) as.h~O.
Let h take a se.quence of values tending to 0, i.e. It-.O as n~oo.
217
ABSOLUTE CONTINUOUS FUNCTIONS, INDEFiNITE INTEGRALS
Let
C
E [a,
bJ. By Lebesgue bounded convergence theorem,
X= J'1m fao I'1m [F(X+h>-F(X)]d h
or
J:
F'(x) dx=lim (} =Iim
J:
a
F(x+h)-F(x)d ----- x
h
F(x+h) dx- ~
u- [::
=lim ~
JC
U:H
F(I) dt+}
F(t) dt+
. as
jb..
J:
J:
F(x) dx
1
F(x) dx
on putting x +hr=- t
J:-Ih F(t) dt ]
F(x) dx= Ib F~t) dt ,a
rl J~h F(I) dt-ll 1 J~h a F(t) dt
=lim lll.
=
1:
1
]
=F(c)-F(a) I(x) dx, by (1).
[See Remark written, just below] or
Ja" [F' (x)-/(x)] dx=O
'"
c E [a, b]
This => F'(x)--J(x)=O ~ F'(x)=/(x) a.e. Remark, If we write g'(t)=Ft1). then
li~ ~ [+'- F(I)
df=lim }
1:+'- [~l get)
I [
=Iim /, g (x)
] dt
Jc-c+'-
C _ ' ( ) - 1:'( ) -I' - 1m j?'(c-l-h)-gi II - )--g c -C' C ,
Theorem 17. III: [a, b] .... R is all increojillg lUi cticn (nondecreasing). then I I : [a. b] .... R is L-integrable and
I:
I '(x) dx
~I
Proof. Consider a sequence tions I. : [a, bJ~R defined as
(b)-I (a).
[Punjab 65, 67; Banaras 69, 71J of non-negative func-
< In >
In(x)=n [ I (x+}) ---/(x)
1'"
x E [a, b].
218
ABSOLUTE CONTINUOUS FUNCTIONS, INDEfiNITE IN1EGKALS
Since I ..(x) is an increasing function and hence it is L-integrable. The sequence < In > converges to I (x) a.e. I
f'(x) =
i.e.
Lim n-+oo
.
I .. (x) a.e.
Applying Fatou's lemma. we get
I
b
..
f'(x) dx
~
Lim inf
n-+ 00
fib
I ..(x) dX}
... (1)
Q
Extend the definition of the function I by setting l(x)-4(b) If b <: x ~ b+l Since the integral
.J: I .. (x) dx is
monotonic function of n and therefore it can
be taken in the sense of Riemann. Lim inf
n-+oo
1 b
Consequently
I,.(x) dx
0
[1: I(x+~)-/(X)] dx =n~: inf n [J: I lx+~ ) dx--J: I(x) dx ] =n~: =
infn
Lim inf II n-+ 00
[f~-tl/" Itt) dt--JI I\x) dx ].
Ja -1'"
a
. x+-=t 1 . t h erst fi on puttIng m integral n Lim = ,,-+ 00
inf II [jt+l/" IlX) dx+
Lim == n_oo
ipf n
Ia
0+1."
= Lim inf n n-+oo
I(x) dx
1
li
[fD+l/" I(x) dx+ Ia I(x) dx ] b a+l/" [Jb+l/n
[1
b
I\x) dx - 1'1+1/" I(x) dx ]
]
/J
Lim.mf n - [/\b)-'Aa)] = n-J>oo n by remark of Theorem 16. P. 217. =/tb) - I( '1). Finally, Lim inf
n-+
00
Jil /J
I,,(x)
dx=/tb)-/~a).
..
ABSOLUTE CONTINUOUS FUNCTIONS, INDEFINITE INTEGRALS
219
Using this in (I), we obtain
J:
rex) dx :S;" f(b)--f(a).
Proved. Theorem 18. Fundamental theorem of integral calculus : If f is an integrable function its indefinite integral F(x) has almost everywhere a finite differential coefficient or Iff is an integrable function on [a, bJ and if F(x) = then
1:
F(t) dt+F(a).
for almost everywhere on la, b]. [Banaras 1967] Letfbe an integrable function on [a, b] and let
F'(x)=f(x)
Proof.
F(X)=jlDa f(t) dt+F(a),
... ( I)
1'(0) being any constant.
To prove that F (x)=f(x) a.e. 011 [a, b). Without loss of generality, we may suppose that f(x) ~ 0 tv x. Let m be a positive integer and define a function [f (Xj]". as follows
If(X)Jm={f~X) i.ff(x) ~ m m Iff~x) > m Evidently [f(x)l", ~f(x) in every case Ax) - [f~x)]m ~ O. so that Consequently
1: l f(l)-[/(I)].,.] dt=G",(x)
__ (2)
is an increasing function of x. As a result of which Gm{x) has non-negative differential coefficient. Since [fV)]". is bounded and measurable and so
~x By (2),
1:
[f(t)Jm dt=[f(x)]". ... (3), by theorem 16.
J: f(t) dt=G".(x) +J: [f(l)],,, dt, F(x)-F(a)":'Gm(x)+ J: [f(t)l",
dl, by (I).
i.e.
Differentiating w.r.t. x, F'(x)--O=G'.,.(x)+[f(x)]m, by (3) ~ [f(x)].,. as G'".(x);;;;r o. :. FJx) ~ [f\x)]m a.e. But m i, arbitrary and hence m~ OJ, give~ ':F'(x)'~ f(Jc}
... (4)
220
ABSOLUIE CONTINUOUS FUNCTIONS, INDEFINITE INTEGRALS
1: J:
But Also
f(x) dx=F(b)-F(a), by (1) r(x) dx=F(b)-F(a).
The last two equations show that
J: I:
or
r(x) dx=
J:
f(x) dx
[r(x)-f(x)] dx=O.
Also F'(x)-f(x) ~ 0 a.e. by (4). Then we must have F'(x) - f(x)=O a.e. or F'(x}-f(x)
a.e.
Problems related to Lebesgue point of a function. Theorem 19. Let x be a Lebesgue point of afunctionf(t). Then show that the indefinite integral F(x) =f(a)
1:
+
f(t) dt
is aijj'erelltiable at cach point x and r(x)=f(x).
Let x be a Lebesgue point of the functionf(t), so
Proof. that
Lim I J",+h h-+O 11", I f(t)-f(x) Also let
F(x)=fw)+J"'o /(t) dt
To prove that Since
I dt=O
... (1)
... (2)
F'(x) =f(x).
F (x) =Lim F(x+h) -F(x) h ~O It
Hence if we show that Lim \ F(x+h~ -F(x) _ (x)
11--+ }
"
.
1=0,
... (3)
the result will follow. From (2). F(x+h)--F(x)= ",+A a f(f) dt- I'"0 /(t) dt
J
=.j
1·+" /\t) dt
",+A j. t) dt+ JOII f(t) dt= "'
221
ABSOLUTE CONTINUOUS FUNCTIONS, INDEFINITE INTEGRALS
:. !I~±h)h =- ~i~) = h~ I"'+"
f(t) dt
:II
•••
J,"+II f(x) dt=h-1 f(x)"J,"+h
1 Also 11:11 or
1 JGI+" II 81 f(x)
or
1 JGI+" f(x)=h 81 f(x) dt
I dt=h f(x)
(4)
[1'"+" t '" =f(x)
dt=f(x) . (5)
Subtracting (5) from (4),
IF (X+~-F (x) -- f(x) \=1 ~f:+h [[(t) - f(x)ldt \ 1 ~ h-I., 'f(t)-f(x)! dt. £+TI
•
Making h-+O and using (I), we get
I
I
~im F(x+h~--F(x) -f(x) ~ 0
,1-+0
But
Lim
h
~
... (6)
I~(x+h)-F(x) -f(x) I ~ 0 hi?'
... (7)
h ...O,
Since modulus of any quantity is always non-negative. Combining (6) and (7), we get the required result. Theorem 20. Every point of continuity of a summable function f(t) is a Lebesgue point off(t). Proof. Letf(t) be sum mabie over the closed interval [a, b]. Also letftt) be continuous. at t=xl • To prove that Xl is a Lebesgue point of the functionf(t). Recall thatftt) is summable over [a, b] if
L b
•
f(t) dt=a finite quantity
f(t) is continuous at f=X l implies given e > 0, there exists 8 > 0 such that 'f(t) -/(.'"1) , < t whenever I t From which we get
I
Xl+h
Xl
or
1
II
\
I fll)--/tXl) I dt < •
fXl+h Xl
I f(l) -f(x!) I dl <
- Xl
jX2+h
I<
3.
dt=f,h
Xl E
... (2)
whenever I h
I< 3
ABSOLUTE CONTINUOUS FUNCTIONS, INDEFINITE INTEGRALS
I i.e.
Combining the last two inequalities. Lim
I.!-.JXI+h
h.-+-O 'h
I f(t)-f(x l ) I dt
Xl
1=0
Lim ~ IXI+h I f(t)--f(x l ) I dt=O h-+O h Xl showing thereby Xl is a ~ebesgue point off(t). EXERCISES 1. What is an absolutely continuous function? If I is abs~. lutely continuous on [a, b] and f '(x)=O a.e. prove that I is constant. [Banaras 1971; Meerut 72] 2. (a) Let f be an increasing real valued function on [a, b]. Show that f is Lebesgue measurable and that
i.e.
J:
f' (x) dx
~ feb) -f(a)
Letf: [a, b] _ R be such that I ./(y) - f\x) I <;; k (y-x) whenever a ~ X < Y ~ b. Prove that (b)
J:
f' (x) dx=/(b) -/(0) .
[Baaaras 1971] 3. Prove that if f is absolutely continuous, then f'(x) exists almo~t everywhere. [Baoaras 1971J 4.
Show that the function f defined on lO, 1] by Ii x>={X cos (1tX/2) for 0 < ~ EI; 1 (0 for x=o
is continuous but not of bounded variation on [0, 1]. [Baaaras M.Sc. Final III 1970]
ABSOLUl E CONTINUOUS fUNCTIONS, INDEFINITE INTEGRALS
223
5.
Define 'absolute continuity' of a function I: [0, I ]-+R. If/, g: [0, l]-+R are absolutely continuous, prove that/+g and Ig are absolutely continuons [Banaras M.Sc. Final IV 1970] 6. Prove that a monotone function is differentiable 'almost everywhere'. [Banaras M.Sc. Final IV 1970] 7. Let [a, b] and [c, d] be compact intervals of the real line. Let/b~ a bou.nded summable real valued function on [c, d].
f: .r
~ x ~ d) prove that Foi/J is .absolutely continuous, .p being an absoulte continuous function on [a, b] into [c, 4J. Defining
F(x) =
(U)
du
(c
[Banaras M~Sc. Final IV 1970] Hint. Refer Theorem 8, Page 207. 8. Prove that if.r (x) is absolutely continuous in an interval and I (x)=O almost everywhere, thenl (x) is constant. Let E be a measurable set and let E" be the linear set which is the-section of E by the ordinate x from the origin. Then show that E~ is measurable for almost all x and " (E)=
f
m
(E~) dx,
where'f' denotes the plane measure. [Banaras 1968] 9. Define (a) function of bounded variation and (b) absolutely continuous function. Prove that a necessary and suffi;;ient condition that a function should be an indefinite integral is that it should be absolutely continuous. 10.
Show that if F(x)=F(a)
+ J:.r (t) dt
[Banaras 1967] then F(x)=I(x) almost everywhere. Hint. See Theorem 16, Page 216 ll. Define absolute continuity for a real function of a real variable. Show that F (x) is an indefinite integral if F is absolutely continuous. [Bansras 1965]
n. State, without proofs, the following theorems: and explain all the concepts occurring in the statements: (i) Fubini's Theorem, [Banaras 1964} (ii) Radon = Nikodym Theorem.
224
ABSOLUTE CONTINUOUS FUNCTIONS, INDEFINITE IN rEGRALS
13. Explain what do you understand by f(x} is an absolutely continuous function on an interval (a, b]. If f is integrable on [a, hI, show that F(x) =
I:
f(t} dt, a
< x< h
is absolutely continuous on a given interval (a, b]. Give an example of a function which is continuous but not absolutely continuous on a given interval. [Kanpur 1970] 14. State and prove Vitali's covering theorem and use it to prove that an increasing real valued function on the interval [a, b1 is differentiable almost everywhere and that
r.
f'(x) dx
~ f(b)-f (a).
[Punjab 1967]
12 LP-Spaces 12'0. Conjugate number. Let p, q be any two positive real numbers S.t. (i)
P
>
(ii)
I
~ +.l= 1.
P q Then q is called conjugate to p. If p=2, then q=2, so that 2 is self-conjugate number. 12'1. V-Space. By LV (a, b) we mean a class of functions I(x) S.t. (i) J<x) is meltslIrable over (a, b). (ii) / f /P is L-integrable over (a, b) for P > 0, i.e.
J:
I f II' dx <
00
for P > O.
Sometimes we denote the class of such functions by the symbol LP in case the mention of interval is not necessary. Then [1 or simply L denote<; the class of measurable function I(x) which are also L-integrable. Examples. (i) If I(x), g(x) E LV (a, h), then j(x)±g(x) E U (a, b), where (a, b) is finite or infinite. (ii) If I(x) E U' (a, b), I(x) E U (a, b), then I(x) E U (a, b) if p < r < q. (iii) Every bounded function bebngs to Lv (a, b) >.f p, where (a, b) is a finite interval. (iv) The spaces LV (a, b) and Lv (a, b) are said to be conjugate to each ot her if
~ + ~q = I.p
12'2. Norm of an clement of LP-Spaee. Let /(x) E /.1' (a, b) be arbitr~ry. The norm of (V':), denoted by /I f lip, is defined as follows:
ilfl!~-(J: Ifl pdPf
1 •
226 If p=J, then we write
11/111=11/11=1:
III dx,
where (a, b) is finite or infinite interval. 12'3. Defi.ition. 'Let I(x), g(x) E LP (a, b) be arbitrary. d (I, g)=11 I -g II,,· We define Then d is called distance function or metric on L'P (a, b), d (f, g) denotes the distance between the functions/(x) and g(x). 12'4. Convergent sequence. Let < In > be a sequence of functions belonging to £Pspace. The sequence is said to converge in mean with index p, if given II> 0,3 no E N s.t. m, n ~ no => 111m (x)-In (x) lip < • or if
f
I I,,(x)-/m(x) IP
dx~O as m-+oo and n~oo.
In this ca'le the sequence is called meanlundamental sequence
Or Cauchy sequence. The sequ~nce is said to converge in mean with index p to a function I(x) if given I! > 03 no. s.t. n ~ no => If/-I,. lip < 6 or if
II
eN
1-/n'IP
dx~(} as n_oo.
12·S. Definition. Cauch; sequence. Let < lro > be a sequence of functions belonging to a L'P space, This sequence is said to be a Fundamental sequence or Cauchy sequence if given II> 0, 3, no E N s.t. m, n ;iii no :0> iif.. -/.... \lp < •. Later on we shall show that every convergent sequence is a. Cauchy sequence. (Refer Theorem 13, Page 241). 11'6. Definition. Completeness of L'P·space. An LP-space is said to be complete if every Cauchy sequence in tne space is convergent at some point of the space,' i.e. for every Cauchy sequence < I .. > in the space, there is an element lin the space s.t. /.,-+1. A complete normed linear space is called Banach space. Theorem I. II IE Lp and g ~ I, then g E V,. Proof. Let IE Lp (a, b), so that (i) I is measurable over (a, b), (ii) I/I P is integrable i.e.,
Also let
J: I/P'dx <
g(X) <; Ji.x).
00.
221
[!>·SPACES
To prove that g E U' (a, b), it suffices to show that (iii) g is measurable over (a, b) (iv) I g 121 is integrable over (a, b), i.e.,
I:
!g
II' dx <
00.
Let at be any real positive number. {x E (a, b) : g(x) > at}={X E (a, b) :
< g(x)} (a, b) : at < g(x) <; I(x)} (a, b) : CIt < I(x)} CIt
={x E ={x E =measurable set, according to (i) Finally, {x E (a, b) : g(x) > ot}=rneasurable set. This => g(x) is measurable over (a, b). Hence the result (iii). •• g(x) ~ I(x)
. J:gig ;" or
)t
If)
dx ~ •
If 121 dx <
00,
by (ii)
" I II' dx < J This proves the result (iv). g
00.
Theorem 2. lIfE L'P (a, b), p > 1 ; then IE L (a, b). Proof. Let f E L" (a, b), p > I. To prove that IE L (a, b), it is enough to prove that (i) {is measurable over (a, b) (ii) 1 I I is integrable over (a, b).
By hypothesis, (iii) I is measurable over (a. b). (iv) I 1/1' is integrable over (a, b). Let E=(a, b). Clearly (iii) => 0). Let A=E( III ~ I), B=E( Then E=A U B, A n B=0.
III <
1).
By countable additivity property of the integral,
L, !I I ~x LII I +L, II I dx
=
I ft x ) I < 1
'ff
dx
... (1)
x E B.
Using first mean value theorem. ,
B
I/(x)
I dx < m(B)=a finite quantity.
I:his proves that I/(x)
I is integrable over B.
\
. .• (2)
228 .:
f(x)
~
1
¥ x E A.
:. / f / ~ / f III on the set Integrating, we get
LIf I
i.e.
~
dx
LI Fill
dx
<
A. For p
00,
>
1.
according to (iv)
LlfldX
This proves that I f I is inte1rable over A. . .. (3) From (I), (2) and (3), the result (ii) follows. Theorem 3. If fE 0', g E L'P, thenf+g E P. [Meerut 1988] Proof. Let f, g E LP (a, b) so that (i) f, g are measurable over (a, b) (ii) I f ill, I g III are integrable over (a, b), i.e.
I:
Ifll> dx
<
J: I gill
co,
dx
<
co.
To prove that f+g E P (a, b), it suffices to prove that (iii) f+g is measurable over (a, b). (iv) If+g III is integrable over (a, b), i.e.,
J: If+g IP
Evidently (i) Let
Let Then
~
dx
<
00.
(iii). £=(a, b).
A=E( If I '" A n B=0,
/ g I),
B=E(
if I > I g I)·
E=A U B.
By countable additivity property of the integral
L. I
fi-g ill
I f +g II'
~~ ~
d~=-:'L I f+g ill dx+
t
( I f I + I g I )Il ( I g I + / g / )1l=2 P
/
/f+g /11 dx g
ip
... (1)
on the set A
Integrating, we get
LI f+g III
L
i.e.
.Ix
2p
LI
Kill dx
< 2 p x oo=oc, according to (ii)
If+gjI'dx
Again
~ <
00
/ f + g /1' ~ ( I j I + I g / )7' <" 21' I r I'. on the set B.
... (2)
LP·SPACES
229
Consequently
I.
l/+g
I" dx < 2"
Is
If I" dx
< 22'xoo=00,
Js lf+gl2'dX
i.e.
by (ii) . .. (3)
In view of (2) and (3) : (I) shows that
t
I/+g 12'
<
or
00
J:
If+g
I" dx <
00.
Hence the result (iv). Tbeorem 4. Holder's inequalities. Let lEU (a, b), g E L l. Let U (a, b) and L'l (a, b) be conjugate to each other. Show thatfg E L (a, b) and II/g 1/
real numbers such that and ifIE LP,
I I -+-= I
p ,q
g E L'l,
f.g E V and
then
J/fg I ~ (Jlfl fill (II g I Clr P
q
dl:EJiJ
with equality only when a.e. [Banaras i969] Proof. Letf E LP (a, b), g E U (a, b), p > l. Let £1' (a, b) and La (a, b) be conjugate to one another 1 I so that - + - = I. p
To prove that (i)
q
Ig E
L (a, b).
(ii) 1/ fg II ~ /I f 11,,·11 g II" 0 < at < I. Consider the function .p(x)=x«- (XX for O<x< 00. It is ea~y to see that ljJ(x) is maximum for X= I. that ljJ(x) ~4(1) for O<xO, B>O, we get
Let
i..)« -at (~) /' ) -a. (B B ~,
or
or
A«Bl-Ql-rI. A AI1.Bl-«
~ ~
or
A«B-z-«
(l-':'rI.) B Aat+B (I-rl.)
It means
~_./ 1-.
'B~
... (1)
230
B
UJ-SPACES
This inequality has been proved under the condition A > 0, 0, but (1) also holds when A and B or both equal to zero.
>
Setting
ac=!... p
>
so that p
Therefore
c:
I.
1 ac+-= 1, q
at
<
I)
1 I-cx=q
or
Then (1) beeomes AlfP Bl/v
~
A
B -+q
... (2)
""'p
If either f(x) or g(x) is equivalent to zero, then the theorem is trivial. Excluding this trivial case we assert that both the
integral I
J: If I" dx and J: 1g Iv dx
are stric.t1y positive. meaning. •
Then each of the following functions has a
f(x)
g(x)
II;' G(x)= Wg~
Wnte F(x) = II f
If we set Al/"=I F(x) then (2) reduces to
I, Bl1a=1 G(x) I,
I F(x) G(x) I ~
If(x!1'+1 G(x) Iv p q
... (3)
Integrating it,
J"I F(x) G(x) I dx ~ p~ J" I F(x) IV dx+l.q J" I G(x) I" dx / o
a
=!...~" "Ifl"dx p
1
a
or
blgllldx
J: If
J:
J:II
a
(I
II
or
+C
J If 11> dx J Jig I" dx
a
=PIG I f or
a
111 dx
Iv dx
1 F(x)
1
+q
I:
\g Iv dx
II
J 1g Iq dx P
G(x) I dx
a
f II" II iY; . - - ..;
II
~
1
q
<1
\/(x) g(x) 1 dx
IIfg
1
=-+-=1
]
IIfllpII, g IIv.
This proves the second required result. f E L" (a, b), g E La (0, b) implies
.. (4)
...(5)
231
UJ-SPACES
(iii) (iv)
I and g are measurable over (a, b). II 121 and I g Iq are integrable over (a, I lip dx < 00, 1 g Iq dx < 00,
J:
U: II
i.e.
J:
121 dx fP < 00,
U: I
b) i.e.
g Iq dx ) l/q < 00
" I II, < 00, II g IIq < 00. Using this in (5), we find that Il/g II < 00 This => fg is integrable over (a, b). (iii) => I(x) g(x) is measurable over (a, b). The last two statements taken together imply the result (i). Theorem 5. Minkowski's inequality. Iff. g E LP (a, b) where p> 1, then i.e.
11/+g I,
~ II
111,+11 g 1121'
(Burdblwan 1990,Meerut 1988, 87 ; Banaras 70, 68; Indore 78J Proof. Let/, g E £P (a, b), p > 1. To show that 0) I+g E Lp (a, b) (ii)
11/+g II,
< 111/1.+11 gil,·
Let q be a number conjugate to p, then
L +!. = 1. q
p
I, g E £P (a, b) .,. I+g E LP (a, b) Refer Theorem 3. g E LP (a, b) => I+g E LV (a, b) • (/+g)plq E Lq (a, b). Now 1 E £P (a, b), (f+g)Plq E U (a, b).
and J,
On applying Holder's inequality to these functions, we get
I: or
If 1·1 J~ g I plq dx ~
(J: III" dx rIP U: 'f+g
f:lfl.lf+glp,qdX.~ O:I/PJdx
I(P/q)q
f'p u: I/+g Iv
dx )"'1
dxf/'l ... (1)
Since /, g E LV (a, b) and therefore interchanging f and g in the last inequality, we get
J: I g I·I[+g
Iv/ q
dx ~
(J: I g Iv dxf'V U: If+g IV dx/,q
... (2)
Adding (I) and (2), we get
J: If ,.,
[+ g IV;q dx
+J: I g 1·lf+g Iv/q dx
232
OJ-SPACES
~
[(J"OI/IJ>dx )1/11+ (Jbo,gl1l dX )1/11] ((bJ.I/+gI1ldx )1111 ... (3)
..• p~+~=l :• q So tong
l+I!.._p q .
as I/+g I'=I/+g 1.I/+g
111«
1
~ (1/1+lg 1).I/+g 111111 =1/1.I/+g 1"}I1+1 g '.I/+g Ipla.
Integrating, we get
J: II+gIPdx~ J: 1/1.I/+gllllqdx+J: Igl·l/fglPlqdx J: I I+g P' dx " [0: I I I" dx rIP +0: 1 I" dx tv J(J: if+g dx r On using (3), we get
Dividing by
0: or
(J: 11+
g
Ip dx
til
ll
IP
g
and observing that
1 1 1--=-, we get q ,p
I/+g!, dx
r'p~(J: 1/111 dx Y'' +0: I
g
I" dx
f'P
II/+g II, ~ 11/11,,+ II gil,,· Ex. State and prove Minkowski's inequality lor V'-spaces. [Banaras M.Sc. Final IV 1970] Theorem 6. Schwarz's inequality. II f, g E L2 (a, b); then Ig eLand IIlg 112 ~ 11/1/2 II g 112' Proof. Suppose I, g E LI (a, b) so that (i) f and g are measurable over (a, b) Oi) 1I 12, I g 12 are integrable over (a, b), a.e.,
J: I j(x) dx < f: g(x) dx < J: I j(x) I~ dx+ I: 1g(x) Ii dx < 12
00,
I
This =>
To prove that Ig E L(a, b) (iv) IIlg II ~ 11/112' II g 112 (i) => fg is measurable over (a, b)
12
00.
00.
. .. (1)
(iii)
... (2)
233
O'-SPACES
Suppose x E (a, b) is arbitrary. Then obviously ( I f(x) /-1 g(x) I )2 ~ O. This gives 2 I f(x) 1.1 g(x) I Et;; I f(x) 12+1 g(x) 12
or
n
f(x) g(x) I dx
~!
By virtue of (1), this
U: ~
'f(x) 12 dx+
I:,
J: I
g(x)
f(x) g(x) I dx <
/2 dx ]
00
This => 'f(x) g(x) , is integrable over (a, b) From (2) and (3), the result (iii) follows. Let at be any real number. Then (II If'
or or
at 2
oe 2
2 +211 I fl· If 1
~ 0
I g I + I g 12
J: If 12 dx+2:x J: Ifg I dx+j:
where p=J:
I.flla dx,
If p=O, i.e, if
0
I g 12 dx ~ 0
J:
q=2
If /2
J:
Ifg
... (4)
I dx,
~x=o then
r=J:
Ig
12 dx
If 12 =0
/(x)=O a.e. on (a, b) In this case both the sides of the inequality namely
I:
are Zero.
Ifg
I dx
~
U: 'f 12 fa 0: I dx
or
Set
f1.=~q
q2
q2
in (4). Then ~!:+
Putting the values of p, q, r we get
2
g
/2 dx
t2
Excluding this trivial case we suppose that p:;toO.
----+r ~ 0 or -. q2+4pr 4p 2 p ' q ~ 2pl'2 r1/2.
or
or
~
a 2p+aq+r ~ 0
or
or
+ I g I )2
... (3)
(
~
;pq) q+r ;.. 0
0
c?'
i
J: IlgldX~2(J: Ifl t U: Igl~dx fa J: I I ~ (1: If t2 0: / t2 2
fg
dx
dx
2
12 dx
g 12 dx
This is also expressed by writing IIjg II ~ II I liz II gl!z.
Hence the result (iy).
.(5)
234 Remark. This theorem IS also expressed by saying that: Let f and g be measurable functions on [a, b]. If f and g are square integrable, then fg E L [a, b] and
Solution.
1:
Then
[Banaras 1971] First write the proof of the previous Theorem 6. 1 fg I
~(J:
dx
If 12 dx
r 0: /2
I
g
12 dx r~
Since If 12-P and het'lce the last gives
J:
I/g
1
~
dx
IJ: fg dx I~ I:
But
Ex. t.
IIJ .. fg dx b
lJo1f(x)
Solution. we get or
I L
I~ [j1 0
Proved .
(J
I]. show that
I f(x) 12 dx
)1/2 .
[Banaras 1971]
If f.g E V (0, 1], then by Schwarz's inequality, IIfg II ~ 11/.1'2 II g 112
1: I fg 1dx ~ U:
J: ! f I [For
or
by (6)
[
If 12
dx
Taking g(x) = 1 V x, we get
or
... (6)
I fg I dx
{J
IflE L2 [0, dx
dxtlO: g2 dXY'2
[ jbf = dx ]I/B[Jb.. 154 dx ]1/1 , ~ b12 dx )1/2[Jb g2 dx ]1/2 . ~
...
(J:r
dx
I:
~
[1: II
Igl2 dx
IJ: f dx I ~ J:
12 dx
rIll: I
I' d.~J/I
r /2
=(t dX=1]
I f I dx
~
U: I
If: I~ U: I f r'2. f dx
g
f
II dx Tla
12 dx
Ex. 2. Iff and g are Jquare integrable in the Lehesge sense, then prove that 1+ g is also square integrable in the Lebesgue sense Qnd [Meer ut 1971J
235
f and
Solution. sense so that
f,
g are square integrable in the Lebesgue g E L2 [a, b].
(I +g)
To prove
E L2 [a, b]
li/+g 112 ~ 11l112+11 gilM
and
By Schwarz's inequality, f, geL2[a, b] ~ fgE L[a, b] ..• (1) Alsof, gEU[a, b) ~ p, g2EL[a, b] ... (2) (1) and (2) :::> P+g2+2/gEL[a, b] ~ (/+g)2EL[a, b] ~ l+gE Uta, b]. This proves the first required result I1+ g 12 ~ ( I f I + I g 1)2
= 1/1 2 + IgI2+2Ifl·lgl·
= I/I!+/g
. J: I f+g 12 ~ J:
12
+21lg
If !I+
1
C
I
g i2+2
1: Ilg I .
By Schwarz's inequality, this gives
f: I
f+ g
12
J: I11 + J: I
~
1
+2[1:
g
12
T/2 U:
If 12
I g 12
J/
2
=[0: I t2 +0: I \2 fa r (J: I/+gI2t2~(j: 1/12 f2+(J: I t2 fl2
g
Taking positive square root
g,a
II 1+ g 112
or
~
'I f
II~+II
g 112·
Tbeorme 7. Let < fn > be a St quence of integrable functiuns which converge in mean to a IUllction f 1 hell the sequence converges in mea,mre to f Proof. See theorem I, chapter 10, page 177. Theorme 8. Let I (x) ~ 0 be an integrable lunction a e. 011 a measurable set E. Then! E
I
(x) dx=O iff f (x)=O a.e. on E.
The proof of this theorem is left .as an exercise for students,
236
LP-SPACES
Therome 9. Show that (U', d) is a metric space. [Kolhapur 1968J Proof.
In, f,,,
Let
We define d
(In,
E LP (a, b) be arbitrary.
f,,,)= II In-1m
111'=0: lin-I", Ip dx t
P
We have to prove that (£1', d) is a metric space. (i)
:.
i.e. (ii) d
(I,
...
I in
-1m (x) I ~ 0 I In f.,. I p dx II/n--Im II p=d (In.!m) (x)
0:
t
g)=O <=> II
~
;;;.
0
O.
I -g II p=O
*S: I I-g <=>
P
121 dx=O
11- g 121 =0
a.e., By Theorem 8. a.e., <=> I =g a.e. • d (.f. g)=O ifl=g a.e. (iii) d(/, g)=d(g,/) For II--gi = Ig-II (iv) IIf-g II (f--h)+(h-g) II" where!, g, hEU'. On applying Minkowski's inequality, we get IIf-gll,,= II (f-h)+(It--g) II" E;;; II f-/H,,+ II h-g lip =d (f, h)+d (h, g)
-=- 1/- g 1 =0 11
11,,=
<
Finally, d (j, g) d (I. h)+d (h, g). From what has been done the required result follows. Theorem 10. An U'-space is a linear space. Proof. Let I, gE V' (a, b) and cE R be arbitr:!.!'y. To prove that D'-space is a linear space, we have to show that (1)
f. gEU' (a,
(2)
fE LP (a, b) and cER ::} cfE LV (a, b)
b) => f+ gE L (a, b)
For proving (I), See theorem ..>. (3) LeI fED' (a, b) and cER. Theil fis measurable over (a; b)
J: I
(4)
flP dx<
... (4)
00
=> cfis measurable over
(a,
b). For c is a constant.
237
J: I clip
dx=1 ell'
J:
I/IP dx
< ! c Ip
fb L I ({ Ip dx <
i.e.
00=00,
bv (5)
00
Also cl is shown to be measurable over (a, b). These statements imply that c/E LP (a, b). Hence the result. Theorem n. An LP-space is a normed linear space. [Meerut 1990 ; Kolhapur 70] Proof. Throughout the discussion we shall suppose that I, g E LP (a, b) and c E R. To prove that U' (a, b) is a normed linear space, we have to prove that l. I+g E U (a, b) 2. cl E DJ (a, b) 3. II/il" ~ 0 In particular nfllp-=O iff/=O a.e. 4. IIclll,,=1 cl.ll/i" 5. U+Kilp~lI/llv+lIglip I. To prove /. g E V' (a, b) ~ (, K arc measurable functions on (a, b)
(f: I/I tl' < (f:
and ~
I
~~g
I gIPf'J'
=,
P
<:
00
is measurable on (a, b)
(J: !I+g \" fP < 00
and
f+g E LP (a, b)
Q
II. IE L"
(a, /1)
(f"I 11') < ~ dis mca<;urable and (J: I Cfi/'fJP « .., =;.
f is measurable and
b
\
1/1'
L:f
00
=> .IE V' (0, b).
III. Since II
i
d ()
and
Next we want to prllv.!
~o (
r
~
I f'!/yiJ> ;;;..
II f ii,,=O
.:>
0; i.e., lI,n,
~ ()
l=u.
!i I ! p=O. 0 and II f'I ,,-~O, then (=0. (Refer solved prohlem 4, ch'lptcr 9, page 149).
It i" obvious that 1=--00 ~
Again if
J:
238
L1'-SPACES
IV. V.
!:
II cIW=O: I cliP dXf'P =1 c I ( Illp dXY'1' =1 c 1·llfl'"
For proving this, see Theorem 5, page 231. This concludes the problem.
Theorem 12. (Riesz.Fischer Theorem). I. The necessary and sufficient condition lor a sequence < j" "> ollunctions In E LP (a, b) to converge in mean to a lunction I E LP (a, b) is that "/m--/.. 111' .... 0 as m ..... oo , n-+oo. [Banaras 1971] Or An U-space is complete space. [Burdhw8n 1990. Meerut 1988] Proof. Let <: In > be a sequence of functions belonging to L" (a, b). Let this sequence converge in mean to a function f E Lp (a, b) so that given. < 0, 3 no E N s.t. m ~ 110 => II/- /m II,. < • ... (1) To prove that given
E
>
For m, n
0, i ~
Lim
m-+oo n-+oo 110
"In -1m 11,,=0, it
E N s.t. m. n ~ no =>
is enough to show
\I 1m --In III' <
E.
no, we have
I'ln- 1m d,,=lIln-I+I-lm II..
~"f,.-I"II+"I-I...
III',
by Minkowski's inequality.
<6+e=2e. Fmally, 111m -In jI,,<2e
V- m, n~no From this the required result follows. II. Conversely suppose that <.f~ > is a sequence of func-
tions belonging to U(a, b) s.t. given e>O, i noEN s.t. m, n~l1o => Ii J.n -In 11,,<. ... (2) To prove that the givcn sequence convcrges in mean to a function fE LP (a, b). I Set e=2k and no=nk. Then
1
III.,. -In III' < 2k
¥
m, n ~ nk'
Or we can take 1/ fm--fn" I, < e for m>nk' we have 111
... (3)
239 1
-=1
2"
00
E II g,_+I-gl·lIp
i.e.
k=l
<
1.
This proves that the series E II gHl- g"
II"
is convergent.
k
... (4)
with the conventipn that g(x)=oo if the series on r.h.s. of (5) is divergent.
(1:
I g I" dx
r" =mL~: U: [
I gl
I
+k~11 gHl -gk I
Making use of Minkowski's inequality, we get
IIg",,~
r
rl"
Lim (ligl/lp+XIIgHI-gkllp]
m~oo
=
k-l
00
II gll'p+ . ,,=1 Ell g'+l-g"
II"
< II gl /I,,+I=A finite quantity. II g 1/" < A finite quantity.
i.e.
gE LP (a, b). This proves that Let E be the set of those points x for which g(x)=oo. D.:fine l(x)=O, xEE x$Ebut xE(a, b)
and If x(/. E, then m. I(x)= Lim [ gl + E (g,
m
~oo
k-l
H
--g,J
]
= Lim g", () x .. m-+oo Lim
1 (x)= m ....oo g",(x), Also 1 (x)=O,
XE (a,
b) s.t. x$. E·
XEE s.t. xE (a, b).
It follows that I(x)=
Lim In ... ex:.
g",(x) a.e on (a, b)
•••
(5)
240 . ThIS
Lim
~
111-+ 0 0
If -- g",
1=0
a,e. in the set (a, b).
00
Evidently g", (X)=gl+ l: (gHl--g,.) k=l
I g",
00
I ~ I gll+ L
k=l 00
< I gl I + E
k=l
:.
I gIll I
I gHl-- gk I
I gHl-glc I=g.
~ g
>of
m.
Making m~oo and observing (5), we get
I ~ If I + I g", I ~ If-g", I ~ 2g.
Also If-g",
I f I < g.
gt-g=2g.
Thus we have shown that 3 g(x) E LP (a, b) s.t. I f- gm I ~ 2g ¥ m and
Lim m-+oo
I .f- g", 1-0 -
a.e·.
. (a,
In
b)
On applying Lebesgue's Dominated Convergence Theorem, we get
Lim
111-+ 00
or or or or
f~
If-g",llIdx=f b Lim
Q
m~:
U:
a
m-? 00
I f--gm I"
Lim m-+oo
,;~r:
"f . gIll
I" dx= Jba
O.dx=O
dx J/l'=O
1/,,=0 '
II f- -/;.",
Lim II.f' --J,In
111-+00
If -gm
II" - O.
II,,-~' 0 .
(,"'" (x)=j~", (X) (·n u"ing (3).
This proves the required result. Remark. This theorem can 'also be stated in the foIlowing manner. (i) The LII-Space is complete. [Meerut 1988, 87; Punjab 691 (ii) The LII-Space is a Banach spa!.". (iii) A convergence sequence <: fn . ~ in Lt'-space ha~. a limit in LII-Space. (iv) Every Cauchy sequence "(II> in the f.!·-Spacc converges to a limit function r E: [}'-Spacc.
24( Ex. The LP-spaces are complete. [Meerut 88, 87; Punjab 69] Solutio.. To prove the problem, we have to prove every Cauchy sequence < fn > in LP (a, b) is convergent. Then given E > 0.3 no E N s,t. m, n ~ no => IIf", --fn II" < E. Now write proof of" II part of Theorem 12 . • Theorem 13. Suppose < f .. > is a sequence of functions belonging to an LP-space. 1/ this sequence is convergent, then it is a Cauchy sequence. [Meerut 1990] Of A sequence 0/ functions belonging to an LP-space is a Cauchy sequence if this sequence has its limit in the space. Proof. Let the sequence < /n > of functions belong to an £1'space S.t. this is a convergent sequence so that it must converge to a functionf E LP-space (say) i.e., 3f E £P (a, b) S.t.
Lim fn=f
a.e.
n~oo
To prove that (1) ~ given E
>.;f /I
IIfm
~ no:;:>
11/- f"
-In II ,,=[1 (." -/+f f"
II
,< 11/", -/11 + 11/-/","
,
~
..( I)
< {" > is a Cauchy sequence. > 0, there exists a positive integer II
E
II
< '2
no S.t.
¥ m, n ~ nil
p
hy Minkowsk:'s inequality
p.
<:2+2=£' 11/."-/,, Ii " < E ' " In, n ~ no wt"dch proves that the sequl:nce < fll > is a Cauchy sequence. Ex. If < /.. > is a Cauchy sequence, then it has a limit. Hint Refer to first part of Theorem 12, the result£ at ollce follows. Theorem 14. A sequence <. LP-space has at most olle limit.
/~
:>
0/ functions
belonging to an
Prooi. If po~slhle let /n~/, rr.~g. Then the identity .f g=/ /n +f,.-g gives ii/ g II ~ :: / -Iii Ii + II I .. - g n . This follows from Minkowski in~quality. Since -.-/:;:> !If,,-/il =0 as n~oo Ift-+g ~ II/.. -g II =0 as ,,--,00. Now (I) shows that IIf- g II ~ 0
r.,
. .. (1)
242
IIf-g II
But
~
0 is always true.
Combining the last two,
IIf-gil=o a.e. => f==g
This
Proved.
Approximation by continuous functions. Theorem 15. Iff is a bounded measurable function on [a, b] thenf.r a given ~ > 0, a a continuous function g on [a, b] s.t.
IIf-g 112 < •. Proof. Suppose f is a bounded closed interval [a, b] and let
mea~urable
function over the
F(X)=S: f(t)dt,x E [a;b]
... (1)
... (2)
Then F' (x)=f(x) a.e. on [a, b]. Let M be an upper bound of I/(x) 1in [a, b]. Then.
I f(x) I ~ M V x
E [a, b].
If x, x+h E [a, b], then F(x+h) -F(x)=
[+11 f(t) dt -
1:
f(t) dt
[+II f(t) dt+ 1: f(/) dl= [+II J(t) dl F(x+h)-F(x)= j f(t) dt =
Gl+1I
:.
01
... (3)
From which, 1 F(x+h)--F(x)
i=/ (+TIf(t) dt I ~ J:+h If(l) J dt=Mh. ~
I dt
GI+II 01
M
1 F(xth)-f~x) 1 ~ Mh.
< 8 and Mh < El , we get I F(x+h)-F(x) I < tl' whenever
Taking h
h
< 8.
This => F(x) is continuous in [a, b]. For each natural number n, we define Gn(x)=n
r
Gl+(l/"'
J
,
/(t) dt, x E [a, b]
.•• (4)
V'-SPACES
243 = n[ F
(X +~) - F( x) ]
G,.(x)=n [ F(X+!
)-F(X)]
... (5)
Also F(x) is continuous in [a, b]. Consequently G.(x) is continuous in [a, b]. From (5),
(X+n!-.) --F(x)
/ F Lim
Lim ~
n-+oo u,,(x)= n-+oo - - - lIn - - -
_ Lim F(x+h) -F(.~'Le=F'(X) = r.(x) by (2)
-h-+O
h
Lim
,
Gn(x)=/(x) a.e. on [a, b]
n-?oo
From (4),
J~
I G.. (x) I = I n !
~
...
j"'t1'" .f(t) dl \ ~ n 1.,+1/11
",+l/..
11M J
1/(1)
11M dt=- =M.
n
x
I G,,(x) I ~ M I G,,-/12 ~ (I ~
¥ n E N and V x E [a, bJ
Gn 4M2,
I + III y.
)2 ~ (M+M)2=4M2
n.
By virtue of (6), Lim n-~
0Ci
G..(x)-/(x)=O
or
Lim
n->oo
- 0 I Gn-11 2 .
Hence, by Lebesgue's bounded corivergence theorem, Lim
J: I G.. -/12 :.
r
x=
Lim n ... oo This
1: Lim I G.. -/12 dx= J:
Lim
n-+oo
(fb
fb I Gil-liZ dx=O CI
I Gra -/12 dx )1 12=0.
CI
'* Lim II G.. -/1I2=0 '* given > 0, 3 positive integer no S.t. fi
V n ~
110
I dt
(;C
(l:
\
G•• -/12
... (6)
=> II/-G.. 11
~
<
11
0 dx=O
244 ~ II 1-- Gnu II :I < • ~ II/-g II 2 < f, Gnu=g·
Thus 3 a continuous function 1 IIHl/no
g(x)=-
/(1) dt. x E (a, b)
no • II/-gll:a<.
s.t. This completes the proof.
lI-
.>0.
EXERCISES
1.
(b) IfI e L2 [0, 1], show that
II:/(X) dx I ~ u: I/(x) 12 dXr~ [Banaras 1971)
2. (a) Let < I.> E LI [0, 1] and IE L2 [0, 1] such that In-./in the n1!an. Show that there exists a subsequence
<1",,>
of
< I.> such that I (b)
Let
I, g
h(x)oo:
J~oo
Stat~
(b)
I, g
J:
I
/(x - y) g(y) dy ( - 00
and
pro\'~
< x<
J:
00).
[Banaras IV 1970]
Minkowski's
inequality,
for
E V [0, J] and define
l(t) dl, G(x)=
Prove that
a.e. on [0, 1].
he V (-00, co).
3. (a> D'-Spaces.
F(xl=
-+
E V (-co, 00) and define
Prove that
Let
nk
J:
Fg dX=[ FG
get) dt where x E [0, 1].
t --I: IG
dx
and cst:lhli~h that the integral .. appearing i \ the formula do cxi<;t and are finite [Baoaras IV 19701 4. Show that (i) x-I E D' for p (ii)
~
< 2 in (0, 1). -I (1+log I x I r- I ill V, in (0,
(0, 00). for a.ny value of p other than 2.
(0) but nm in L'. in
IBaullr.... 1,iil
245
O'·SPACES
5. LetIJt.!2' ...... E £P. Then prove that a necessary and sufficient condition that
11/,.-/11,,-+0
as as
n-:;oo, ·is that 1If..-/m m and n-+oo, where
II
11,,_0
~ /lp= (11 rp(x) Ip dX) 1/1'
Show also that the limit function I is uniquely determined by the sequence of functions , except for a set of measure zero, and that there exists a sequence < f > such that lim f f nr
almost everywhere.
nr
[Banaras 1967J
13 Further Theorems on Lebesgue Integration Integration by parts. The formula of integration by parts in the Lebf sgue theory is the same as the ordinary one i.e., J:fg dx=[(f fdx ) g
J: -J: [~~. f 1 fdx
Theorem 1. If f(x) is an indefinite integral and indefinite integral of g(x) , then
J:
lex) g(x)
dx=-~[ f(x)
G (x)
dx.
if G(x) is al1
J: -
J:f'(X) G(x) dx.
Proof. We know that the product of two absolutely continuous functions is an abt;olutely continuoas function. Let G(x) be an indefinite inte:ral of g(x) for x E [n, bJ so that G(x)=
l
1:
get) dt + C ... (l), C being constant
is an indefinite integral (1)
~
f is absolutely continuous.
=> G(x) is absolutely continuous on [a, b] and g(x) is Lebesgue integrable in [a, b].
Since f and G both are absolutely continuous and hence f(x) G(x) is absolutely continuous. Consequently f(x) G(x) is Lebesgue integrable. (Refer to Theorem 14, Chapter 11, page 214).
Hence
1:
f(x) G(x) dx exists.
J(t) is indefinite integral ~ f(t) is absolutely continuous => J(t) is indefinite integral of its derivative
ret).
=> f'(t) is integrable => f'(t) G(t) is integrable.
.. I: r
(t) G(t) dt exists.
2~7
fUkTHBlt THEOREMS ON LEBESGUE INtEGRA nON
Thus we have shown that the integrals
J:
f(t) G(t) dt and
From (1),
I:
1:
['(t) G(t) dt both exist.
G' (x)=g(x)
;x [f(x) G(x)J dX=[ f(x) G(x)
But
~x [f(X) G(x)
l=f'
1:
a.e. on [a. bJ. ... (2)
(x) G(x)+/(x) G'(x)
=f'(x) G(x)+f(x) g(x) .
. Using this in (2),
J: [f~(x)
G(x)+f(x) g(x)] dX=[ f(X)G(X)l
r:
1:
or
J:
or
J:[(X) g(x) dX=[ f(x) G(x) ]: - J:f'(X) G(x)dx.
f'(x) G(x) dx+
[(x) g(x) dX=[ f(x) G(x)
Ex. Let f, g both integrable over [0, I]. Also let f. g both be measurable over [0, I] and F(x) =
J:
f(t) dt, G(x) =
Prove that
1:
J:
g(t) dt, where xE [0, J].
Fg dX=[ FG
J: -I:
fG dx
and established that the integrals appearing in the formula do exist and are finite. [Banaras IV 1970J Solution.
1: ~x (FG] dX=[ FG 1: But
~x
... (1)
[ FG ]=F'G+FG'=--fG+Fg.
:. ~x [FG ]=fG+Fg. For F(x) =
1:
...(2)
f(t) :;. F (x)=[(x) a.e.
and G(x)= [ get) dt => G'(x)=g(x) a.e.
From (I) and (2), or
r
J:
(fG+Fg) dX=[ FG
J:
fG dX+Jl Fg dX=[FG]l
,0
0
0
248
r
or
FURTHER THEOREMS ON LEBESGUE INTEGRATION
Fg dX=[FG
r_ Jl
fG dx.
... (3) Complete the proof of existence of the integra! appearing in (3) by the help of Theorem 1. I Theorem 2. First mean value theorem. Suppose f .(.t) i~ bozJzded and g(x) ;;a 0 in a measurable set E and are inleKra.le over E; then 3 a number" between the bounds ofI(x) in E s.l. ,
0
.0
0
I I(x) g(x) dX=A IE g(x) dx. Ii
Proof. Since f(x) is bounded in E and so it will have its g.l.b. (greatest lower bound) and l.u.b. (least upper bound) Gt and (3 respectively. Then IX ~ I(x) E; (3. This => a g(x) ~ f(x) g(x) ~ f3.g(x) as g ~ 0 -
IX
f Ii g(x) dx
~
f Ii f(x) g(x) dx
~ (3 IE
g(x) dx
J
[asf, g are integrable => fg is integrable]
Jrg
f~.fg. dX=A ..
=>
where q. Corollalt'y.
dx
... (1)
~ i\ ~
To prove t f g
13.
dx=/(~)
Ie
g dx.
f is integrable
=> f is continuous => 3 x, y E E S.t. f(x)=rz,J(Y)=fJ and/(x) takes every value between a and (J
=> 3 , E E s.t. f(~)=A where
:. t
fg
dx=f(~)
t
IX
~
).
~
p.
g dx, by (I).
Theorem 3. Second mean value theore... If f( x) is integrable over (a, b), and g(x) is positive, bounded, and non-increasing, then f/(X) g(x) dx=g (a+O)f:f(X) dx where
~
is some number between a and b.
Fix s > 0 S.t. e < g (a+O) -g (b -0). This => 3 a point Xl s.t. Proof.
g(a+O)-g(~)
< e
LEBE~GUE
fURTHER TREOREMS ON
249
INTEGRATION
Similarly there are points x 2 , xa," ... , x" s.t. g(xr - 1 +O)-g(x)
<
f,
~
E
so long as g(x'_l+O)-g(b-O)
(x r - 1 < x (x > x r ),
>
E.
<
x.)
Otherwise we take x,,=b.
Thus the point b is reached after a finite number of steps, since the variation of g(x) in each interval (Xr-h x.) is at least E. Write ifJ(x)=g(x.+O) in each interval Xr ~ x < Xr+I' Then o :::;;; Iji(x) -g(x) < f!, except possibly at the points
I: ~(x)
f(x) dx=
J:
F(x) =
Suppose
::+1
l~l g(Xr+O) J
f(x) dx.
f(T) dt.
... (1) Also suppose that In and M are lower and upper bounds of F(x) respectively. Then, by virtue of Abel's lemma, mg (a+O)
But
~
J:
Iji(x) f(x) dx
\J: ifJ(x) f(x) dx- J: g(x) f(x) dx I~ Ii-+
Mg(a+O).
f:
. .. (2)
'/(x) , dx.
as E_O.
!Jb
I a
ljif dX-Jb fg dx ',1=0 a
Lim
or
.-+0
Making
IS
°
This tends to Hence Limo
~
~-)-O
Jb ~f dx= a
Lim JO fg dx
e~ 0
... (3)
a
in (2) and observing (3), we get
Jag(X)f(x) dx
~ Mg(o+O)
Jba g(x) fIx) dx
~ M.
iJ
lIIg(a+O) ~
or
In :::;;;
J
g(a+O)
But an indefinite integral is a continuous function and hence F(x) is a continuous functi on. Consequently F(x) takes every value between In and M so that at x=~, we have
1 F(e)= g a+O)
J"' g(x) f(x) a
dx
... (4)
250
FURTHER. THEOREMS ON LIiBESGUE INTE.GRATlON
But F(x) =
1:
J'o
f:
or
f(t) dt
... (5)
I Jba g(X) leX) dx f(t) dt=g(Q+O)
f(x) g(x) dx=g (a +0)
J:
by (4) & (5).
f(x) dx ..
where a
J:
f(x) g(x) dx=g(b-O)
r:
f(x) dx,
where a<~
"-1
sum 11= L f(Ek) [g(XHl) -g(xk)]. 1=0
If the sum a tends to a finite limit I as y=max (Xt+t-X1J .... 0, independent of both the method of subdivision and the choice of the point ~k' The limit / is called the stieltjes integral of tbe functionf w.r.t. the function g and is denoted by
.f:
(x) dg(x)
or
by (S)
I:
lex) dg(x),
where S denotes that the integral is taken in the sense of Stieltjes. From the definition of Stieltjes integral it is clear that Riemann integral is a special case of Stieltjes integral, obtained by putting g(x)=x. Theorem S. Relation between L~!lesglJe 'integral and Stieltjes integral. If f is a continous function and g is absulutely continuous function both 011 [a, b], ,hell
251
fURTHER THEOREMS ON LEBESGUE INTEGRATION
(S)
Ib
j<x) dg(x)==(L)
a
Jb f(x)
g'(x) dx.
~
Proof. Evidently both the integrals exist. Divide the closed interval [a, b] by means of points
a=xJ <xl <x 2 < ... . ... <xn=b. Choose a point fie in the interval (x., x1'+l)'
"-1
E f('C,d [g(Xi.:-tI)- g(Xk))'
Form the sum
(1=
Write
'\=max. (XHI-Xk).
k=O
. Lim a= (S) Th(;n, by definition .\-~O
1°aI(x) dg(x)
... (1)
g(x) is absolutely continuous implies
(L)
fXkH g'(x) dx=g(X k11 )-g(Xk)
x.
a-(L)
= =
1:
f(x) g'(x) dx
k:/(ek) [g(Xk-tI)-g(x_)l- i~ :n (~)
~
E~f(~k). (L)
k=o
fX x::+1 f(x) g' (x) dx
fXk +1 g'(x) dx- "i,1 (L) Xk k~O
I
Xk+1 f(X) g'(x) dx Xk
If we denote the oscillation of .r(x) in (x", Xk~ 1) by the last becomes 'I
\=\ n,e (L)
a--(L) JO (x) g' (x) dx v
1e=0
~ ~; I(L) e:f-
t Irk
JXk+1 Wk X"
Wk,
g'(x} dx
then
I
g'(x) dx \
=< "-1 I~O 11'". (L) JXx:-1-1 I g'(x) I dx ~ "-I lJ «.(L) k=l'
IX k+l x"
I g'C~) I dx, where ot=malC.
(w.).
252
FURTHER THEOREMS ON LHIESGUE INTEGRATION
=0:
J:
(L)
I g'(X) I dx.
Finally,
I a-(L) J:f(X) g'(x) dx
Making
.\~O,
consequently
~~~ Ia-(L) But
Lim .\~O
~
at~O, we
I~ 0
Ia--(L) Jb,/(x) g'(x) dx
/ ;;.;;t 0
(L)
Lim 1 a=(L)
JO
(1-
"1.-+0
=> (S)
J:
Theorem 6.
..
1:
I:
I g'(x) I dx
obtain
f{x) g'(x) dx
~~~ I
Therefore This
1:
i E;; ~ (L)
f(x) g'(x) dx /=0
f(x) g'(x) dx
f(x) dg (x)=(L)
J:
fiX) g'(x) dx by (I).
Proved. Helly's second theorem. Let f(x) be a continuous
function defined 011 the interval (a, b) and < gn(x) > be a sequence of functions which converges to a finite fUllctkm g(X) at erery point of [a, b]. If
V/) (gn) a
<
kl
'V n, then
Lim n~ 00
Jb f(x) IJ
dg .. (x)=
Jb f(x) dg (x). Q
Proof. Divide the interval [a, b] by means of points xo, X.. s.t. a=xo < Xl < ...... < xn=b. b n_ 1 [ [h Then!" (gm) = to g ... (Xk+l)-g ... (xd· But ~ (gn)
<
kl
Xl"',
"'f
n.
.. ,(I)
... (2)
as lim gm=g. b
or
V (g) II
~
kl
... (3)
253
fURl HER THEOREMS ON LEBESGUE INTEGRATION
J f(X) dg (X) = }..' b
fX t-ll
11-1
k=o
a
J~ f(X) dg (X) = "-I l..' • I:H
X,:
k=O
X"
[ I(x) -I(x,,) +/(x,,) ] dg (X).
Since I is continuous and hence continuous in [x •• X::l-11· SO we can take l/(x) ·j(Xk) i < (e/3k 1) \I XE[X •• Xlc+l] ... (4)
~ i;1 JXk Ll I I(x) - I(Xk) I . I dg (x) I 1c=0 Xk
:. Ilfb
I dg
~ 3-~~
. kl +l'
II
a
or
1;:-11
i f(Xk) I . I dg (x) I
II: r I~ ; + :~: ! dl!
f(Xk)
J
by (2)
Ig(XkU) -g (Xk)
This can al<;o be expressed as
\r rdg I< ~-3 + '~:l , f(x~)
g(Xkll) --g (Xt)
&
a
k=O
,
I
... (5)
I
where O<8~1. Similarly we can ~et
:',b f I
a
dg",
I E;;; 68' i' + "-I ;,; i f(Xt l
Since
Lim m.'>oo
Henl'e 3
I
\ i
I.~O :
i where
i gm (X'·H)
... (6)
g." =- g
po~iti\'c
in:egcr
170
\ .'~l f(."(t) [g,., (Xk.I)- g.,. (Xk)] k~O
'i.t.
III ~ "0
-"if I(Xt) tc~o
:>
(g(X;,
1
t>
g(x)
In this event (5) and (6) taken together imply that
lI~ I a f dgm
faa r dg < I
Ii
1;I < ~I-J
254
FVRrHER THEOREMS ON l.EBESGUE INTEGRATION
Thi<;::>
~
I
b
Lim
In_CO
Lim
n_oo
a
Jba
I(x) dgm (X)=Jb I(x) dg (x) a
I{x) dgn (X)=J b I(x) dg (x). a
7. Definition. Cumulative distribution function. Let B be a family of Borel sets, each of which is a subset of R. A measure IIdefined 011 B is called Borel measure on the real line. If II- is finite Borel measure, therl we associate to it a function F known as its cumulath'c distribution function by setting ... ( I) F(x)=f1o (-00, x] ¥ x E R It is easy to verify that F is a real valued monotone increasing function Also ~, (a, b]=F(b)-F(a). ...(2) F(b)-F(o)=p (-00, D]_.p (-00, aJ=f' (a, b]. For Since
n (a, b+~] n
(a,
b]= ,,=1
Hence it (a, b]= Lim f' ( a, .
n-+oo
b+J..] n
[Refer to Theorem 17, Chapter 7, page 100] Consequently F(b)= Lim F (b+~)=F(b+). n_co n This proves, by definition, that F(x) is continuous on the right. Furthermore p(b}= Lim p (b- ~,b ] n_oo n =
n~:
[
F(b)
-F (b- !)],
=F(b) -
by (2)
!.)
Lim F (b=F(b)-F(b-). n_oo n If p{b} were zero, i.e., Borel measure of a singleton set {b} is zero, then F(b)=F(b- ,. This shows that F(x) is continuous on the left. have shown that F(b)=F(b+ )=F(b-). This shows that F(x) is ccntinuous if p{b} =0 00
Since
n (- ,-nl=0,
n~l
m(0)=0.
Finally we
255
FURTHER THEOREMS ON LEBESGUE INTEGRATION
Therefore
Lim
n -+--00
F(n)=O
or
Lim
x-+-oo
F(x) =0.
Thus we have shown that 'if po is Borel measure on the real line, then the cumulative distribution function F is a monotone increasing bounded function which is continuous on the right hand and Lim F(x) =0.
x-.-oo Definition. Lebesgue-Stieltjeg integral. Let f be a non-negative Borel measurable function and F a monotone increasing function which is continuous on the right. Then the Lebesgue-Stieltjes integral off w.r.t. F is defined as f
f
dF=f f dpo,
where po is a finite Borel measure and F its cumulative distribution function. : Let f be any Borel measurable function, then we define f+(X)~-{ f(x) if f(x) ~ 0 o if f(x) < 0 J 0 if f(x) ~ 0 f-(x) = 1 -f(x) if f(x) <. 0 Then f(x)=f+ (x) -f- (X). In this case J f dF= J f+ dF- Jf- dF ... (3) The fuuction f is said to be integrable in the sense of Lebesgue-StieItjes if both the integrals on r.h.s. of (3) are finite.
14 The Weierstrass Approxi mation Theorems and Sem i-continuous Functions 14'0. Definition. A Polynomial function (or poiynomial) is a function P which is expressible as P(x)=OO+OlX+ 02XZ+ .....
+On X,. ,
where n is some non-negative integer and 0 ,°1, 02' .... .. 0" are constants. 14'1. Definition. Let f b\! a finite function defined on the closed interval rO, ;J. The polynomial
°
Bn (x)=
l: Cx'
k~o I..
(l_-X),,-k j
i~ called Bern.,tein polynomial of degree
(" . k
I~
('!...) n
n for function f.
Here
. I coe ffi' t h ~ B'Inomla Clcnt k' , n ! k -, . _
•
(n- ).
Theorem t. (S.N. Bernstein's Theorem). If a jUlIction f(x) is continuous all the segment lO, I], thell Bn (x) -+ f(x) uniformly lI'.r.l. x as n~oo. Proof. The following proof is due to Bernstein. Here we shaH make use of Bcrnste in polynomial Bn (x) defined as Bn (x)= £" 1,0
TO C
Xl:
• (I-x)n-!. f
(
-k)
... (1 )
11
1
(.~ =.
II! t - - _.- - • k:(n-k)! The polynomial B" (x) is calIed Bernstein polynomial of degr\!e !1 for .I~x). n Lemma. To prole 1: C (k IlXf.xk.(i - X)"-I: ~ 4'
where
k
"...
~
257
SEMI CONTINUOUS FUNCTIONS
Let a, b be arbitrary rOO,1 numbers. By Binomial theorem. ..
n
]J C ak bn-k=(a+b)n <=0 k
... (2)
Differentiating partially this w.e.t. a we obtain "
n
E C.ka"-l.bn-k=n (a+b)lt-l 1:=0 k
On simplification this gives n
"
... (3)
E C.k.ak.b"-"=na (a+b) ..-l. k=o"
Differentiating: this w.r.t. a. , n
It
E C k 2.a H b'H=n (a+b),,-l+n (If-I) a (a+b)n-%
or
k~o
k
.
"
.
E C.k 2a"b"-k=na (a+b),,-l+ n (n--1) a2 (a+b),,-t
... (4)
k=O k
Throught the discussion we shall suppose that x E [0, I]. Set a=x, b=l-x so that a+h=l. Then (2), (3) and (4) yield respectively n
n
E C k~o
k
It
n
x' (1- x)n-k= 1
~' C.k.xk k=o k n
.. (5)
(l_x)n-k=nx
... (5')
'N
E C.k 2 .xk. (I--x),,-I:=nx+n (n -1) x1!=nx [I +n:c-x]
~r
k=o
k
"
..
E C.k 2 .x!'.(I-x)"-k =nx (l+nx-x) k=o"
Multiplying (5), (5'), (5") by n 2x 2 • then adding, "
~'
-
2nx, 1 respt'ch vely
It
C (k-nx)%.xk.(l-x)"-k=nx(l-x) "=0 " Since (2x-I)1 ~ 0 or 4x1-4x ~ 1 or x (I-x) or nx (L -x) E;; (n/4). Now (6) is reduced to
,l.Od
... (6) ~
1/4
258
SEMI CONTINUOUS FUNCTIONS
..0 . It
It
X (k-nx)'I. ext (l-x)"-·
~
-
n 4
... (7)
Hence the lemma. Since fis coDtinuou~ and therefore I f(x) - fey) I < « given , > 0, 3 8 > 0 s.t. whenever I x-y I < ~; x, y e (0, I). Write M=ma'C. { l1(x) I : x E [0, I]}.
Multiplying (5) by f(x) and then subtracting (I) from this we obtain ¥ n E N, fix)--BIt (x)=
i: l- f(x)-f (~)] ex· (l-x),,-t n
.~
II
=£' +L'.
where £' is the sum over those values of k s.t.l
! -x !<~,
... (8) where
I
E" is the sum over those values of k s.t./ ~ -x ~ 8. .n Divide 'all the ntlmbers k=O. 1.2, ... ·...• n into two parts A and B according to the rules . k
and
E
k E
Aul: --xl < B if I: -x I~
8 8.
It means that A and B correspond to ];' and £. respectively. If k E A, then /f
I E' I
=/
"~ <
...
(-! )--f(x) , <
E' [f(X) - f ( : )]
];'
t.
~ x. (l-x)"-t
If(~)-f (,n!!-.-)II C x· (I - x),,-ar • •
t.E' C JK" (l--x),,-I:=c, by (5) .
•
I E' 1<<< .k E B ~ 'n Jk --x II ~ 8 ~
... (9) f
k - nx/
-lis
~ 1
259
SEMI CONTINUOUS FUNCTIONS
(k-nx)2
=> -
n2~' ~
1. For k, n, x all are real.
If k E B, then
I );"1 ~ E" If(X)--f ( : ~ EW [ I f{x) I
~ x~ (l-x),,-t
) \
+k (~ )IJ f
~ 2ME" (k-nx)2 "" 2.2
x' (I -x)"'"
c· XIt (1- X)ft-I: J
nOli
Finally
I E· I <; 2~2-'
... (10)
Writing (8) with the help of (9) and (10), we get
M
I f(x)-B .. (x) I < E+ 2nF' Choose no E N S.t. no Then
II
~ no
M
> 2.8;-'
=-
n ~n ~
=>
6
>
... (11)
M 2n/)2
"D
M > 271f. => M
=> 2nJiI-
M =- 2n~i-+E <
<
n
>
M
2E81
E
2E.
Now (II) becomes ... (12) I.f<x)- B .. (X) I, < 2E B.. (x) cOl1vergl!:s uniformly to I(x) w.r.t. x E [0, 1) as n-HIO. Proved. Problem. If a function f(x) is continuous on [0, 1], then (or every _ > 0, there is a polynomial P (x) s.t. I /(x)-P(x) f < • yo iC E (0, 1].
Solution. Here write proof of Theorem 1. In the end of the proof put B,.(x)=P(x).
260
SEMI CONTINUOUS FUNCTIONS
Problem 2. For ellery continuous/unction/(x) on (0, I), there is a sequence 0/ polynomials < Pn(x» S.t. < Pn{x) > converges uni/crmly to J(x) w.,..t. x E LO, 1] as n~tX). Solution. Here write proof of theorem I. In the end of the proof put B,.(x)=Pn(x).
.. C.x~.(t _x)n-k (k- nx)1 ~ (n/4).
Problem 3. Prove that E
1>:=0 I>:
Solution. Here write the proof of equation (7) Theorem 1. Now we shall prove a very important theorem known as Weierstrass approximation theorem of real analysis. Theorem 2. (Weierstrass Theorem). Let lex) be a continuous /ullction defined on the closed interval [a, b J. For every 6 > 0, there exists a polynomial P(x) S.t., I/(x) -P(x)
1<
6
Y x E
fa
b].
,Thi.' theorem can aho be rephrased as : For each continuous Junction, there exists a sequence COnl'erges uniformly toftx) w.r.t. X E fl, bl as n~oo. Proof. Firstly we shall establish the following lemma: Lemma. For every continuous functionJ(x) defined on the closed interval [0, I], there exists a sequence of polynomials < P .. (x) : n EN> s.t. < p,.(x) > converges uniformly to lex) w.r.t. x E [0, I] as n~OO. The proof of the lemma starts [For proof refer to Theorem I]. This completes the proof of the lemma. Going back to the proof of the actual theorem, suppose that / is a continuous function on [a, b]. If [a, b]=[O, I), then theorem is proved. Now we suppose (a, b]:;e[O, 1]. Define g(y)=/[a+(b-a) y] Then g(O) =/(a) , g(l)=/tb). Also/is continuc.us on La, b]. This leads to the conclusion that g(y) is continuous on (0. 1]. Hence, by the lemma, just proved, :I a polynomial Q(y) s.t. I g(y)-Q(y) ! < 6, ..... (1), 0 ~ y ~ 1. If x E [a, bJ a.nd y E [0, I], then set' x a y= -_._-- so that boa
'61
SEMI CONTiNUOUS FUNCTIONS
g(y)~g (:=:)=1 [a+(b-a) (~=:)]=f(X). Then we have j!(X)-Q
(:=:) I<
The polynomial Q
. e, a ~ x
~ b, by (l2).
(~=:) .sati!>fies our requirements.
Theory and problems related to semi-continuous runctio~s. 14'2. Definition. Limit superior and Limit inferior. Let a function f be defined on a set E and let Xu be a limit point of E. Consider an arbitrary real number ~ > O. Write M,,(xu)=sup {[(x)}, m.,(xo)= inf {f(x)}. x E (xo 8, x o +3) n E. The quantity M~(xo) does not increase and M~(xo) does not decrease with decreasing 8. Hence the (finite or infinite) limits Lim Lim M(xo) = 8_0 M.,(x o), m(xo)=8_0 m.,(xo) exi!,t. M(xo) is called limit superior and m(xu) is called limit inferior. We also write Lim M(xo) =
x-+xo
[(x),
If Xo E E, then we have the relation m(x,,) E;;, I(xo> C; M(xo)· Notice that the function may not be defined at Xo itself. 14'3. Definition. Semi· continuous function A function I, defined on a closed intervallO, Ij, is said to be lower semi·continuous or upper semi-continuous at the point Xo according as Lim - - (x)-/(xo)
X-Xo'
or
Lim x-+xu
r
.
J(X)=j x I 0
Iff is continuous at the point x~, then it will be both upper and lower continuous. F·or I"IS contmuous at Xo => Ji1,x)= Lim
X-Xo
ji(x)= Lim r- -- J{x)
X--'o
262'
SEMI CONTINUOUS FUN'!TIONS
Conversely if1 is lower and upper both continuous at the point Xo then 1 is continuous at XII provided 1 is a finite function. Then lower semi-continuity of a fmlction at a point Xo is equivalent to upper semi-continuity of the function - I at the same point xo' 14'4. Definition. A function I is said to be lower semi continuous on the segment [a, b) if it is defined and lower semi continuous at each point of this seginent. In a similar fashion we definf upper semi-continuity of a function on a segmen~ [a, b]. Theorem 3. Let a lunction I be defined in a closed interval [a, b] and let Xo E [a, b]. Also let < x" > be an arbitrary sequence 01 points in [a, bJ s.t. < x" > converges to xI) Then a necessary anc1 sufficient condition that 1 be lower semi continuous at Xo is that Lim I(x.) , n_oo I(x,,) Proof. Lim X,,=X o => 3 a subsequence < X n-+oo n" Lim Lim k-oo l(x/1,) = /1.-+ oc I(x,,)
;
kEN> s.t.
... (1)
Making use of the result : The limit superior of the function I at a point Xu is the greatest of the numbers which are limits of the sequence of the for~ l(x1),f(X 2),f(xa), ... where x" E [a, hI and x,,~Xo We see that Lim I(x) ~ Lim /(x ) n_ CI) 11/: x ... Xo . . Lim Lim Usmg (1), thIS becomes '.-- f(x) ::;;;; - . (x,,) fI-+CIJ
" - Xo
Let f be lower semi-continuous at ;r 0 so that Lim f(xo) = ":::':;Yo f(x). Lim To prove thatf(xo) <; n-+ch I(x .. ).
... (2)
(i)
In view of (3), (2) is reduced to I(:xo) (Ii)
Convi:rsely suppose that/(xo)
. .. (3)
~ .!:i' ' .1 I(x,,) n-+ 00 Lim
~ - - IlXn ) n~ 0Cl
... (4)
263
SEMI CONIINUOUS FUNCTIONS
To prove that f is lower semi continuous at Xe. Lim f(x) E;; Lim
From (2),
x~xo
n~oo
f(x ff,)
x .. -xo ~ f(x,.)- f(xo)
But
Lim - - /(x) ~ f(xo) X-Xo Combining (4) and (5)
:.
This
:a>
f
..• (5)
Urn- ji(X ) = ji( xo)' X-Xo is lower semi continuous at Xo.
Theorem 4. Let a function f be gefilled on a closed interval [a, b], Xo E [a, b] and f(x o) >- 00. A necessary and sufficient condition that I be lower semi continuous at x, is that 'to every I(xo} > A, there corresponds a 0 > 0 such that I(x) > A provided I x-xo I < b, X E [a, b]. Proof. [a, b] and
(i)
where
Xo
Suppose a functionf is defined on a closed interval E [a, bJ. Also suppose thatf(xo} > - co.
Let f be lower semi continuous so that Lim f(x o) = - - f(x) 3 8> 0 s.t. )C -+Xo Lim Lim ' f(x)=m(xo) = --~ m.l (x..) x ~Xo x-)oxo
..• ( I)
... (2)
m.,(xo)=inf {I(x) : x E (xo-8, xo+o) n [a, b]}. Also let f(x o' > A. To prove that 3 XE [a, b] s.t.f(x) > A whenever I x -Xo I < O. From (J) and (2), it follows that 3 x E (xo-'S, x o+8)nla, b]s.t. m~(xo) > A .•. (3) Since f(x} ~ ma(xo) for I x - Xo I < 8. On using (3),j(x) > A for I x-xo I < 8. Conversely, suppose that/(xo) > A. Now we can find a 8 > 0 corresponding to this A s.t. ma(xo) ~ A, which in turn implies that m(xo) ;;;;. A.
with
Increasing A and taking the limit as we obtain m(xo) d: I(xo)' m(xo) is always true.
A - /(xo) ,
But
f(x o) ~
264
SEMI CONTINUOUS FUNCTIONS
Combining the last two inequalities, we get Lim f(x o}= m(xo) = f(x o}· X-+Xo
This proves that/ex) is lower semi
contin~ous
at
Xo·
Remark. If (x.) is a finite number, then the above theorem can also be restated in the following form: A necessary and sufficient condition that /tx) be lower semicontinuous at the point Xo is that to every to> 0, 3 a number ~>O s.t. f(x o) - E < f(x) provided I x-xo I < Ii and x E (a, b]. The theorem gives a relation between semi continuity and continuity. Theorem S. Suppose that the function f and g are defined on the segment [a, b] and also suppOle that f and g both are lower semi continuous at x=rxo. If the sumi+g is defined on [a, bJ, then it is lower ~emi continuous at xO. Solutiod. We may suppose thatf(xo) +g(xo) > - 00, hut such of the numbersf(xo} and g(x o) is distinct from -d:). Consider an arbitrary number A s.t. /(x..,}+g(x o) > A. Then there exist numbers Band C s.t. f(x o} > B, g(xll ) > C so that B+C > A. Also f(x) and g(x) both are semi continuous at xu. Therefore a a number ~ > 0 s.t. !(x»B, g(x»C where xe[a, b) and I x-xo I < 8. lRefer to Theorem 4J For this x, j(x) > B, g(x) > C .. f(x)+g(x) > B+C > A => f(x)+g(x) > A. This proves that f{x) + g(x) is lower semi continuous at xo. (Refer to Theorem 4] Theorem 6. A necessary and sufficient condition that a fUIlClioll f defined on the segment F- [a, b] be lower semi cofltinuous 011 E is that the set E (f ~ c:) be closed for arbitary real c. Proof. Let a function/be defined on the segment £=(a, b1. Also suppose that c is any real number. (i) Let f be lower semi continuous on E. Let A ~E (f ~ c). To ~rove that 4 is closed.
SEMI
CONrI~UOLJ~,
265
HIN('noNS
If the points of sequence <x,,> belong to A and if x,,~xo' then lower sl'mi continuity off implies that f (xo ) ~ -Lim.) --- J~xn . [Refer to Theorem 3] I1~OO
Xn
E A
~
f
(x,,) E;
Lim
~n~oo'
C
r( ) X,.
=> ,(xo)
~
=> f (xu)
~ C
c'
..... C
Lim --. f(xn}
I1~OO
~ C
=> xoEA.
Thus x,,-Xo E A. This => limit point of the set {xn: 11=1, 2, 3, ...... } C A belongs to A => A is closed. (ii) Conversely suppo<;e that A is a closed s~t. To prove that f is lower semi continuous on E. · B= Lim W rIte - Ii(x). x-:>xo
If possible, let us suppose that f (xo) > B. Our assumption implies that 3 a sequence
<
Xn'>
. .. ( I) in E S.t.
X"-+Xo'
Ifwe suppose that c is any number lying between B andJ(xo). then B
f
i.e. This =>
f
(xo)
<;;
X~~o f{x)
is semi continuous at
XO'
15 Singed Measure Introduction. We have seen in the preceding chapter that a measure is a non-negative function. If a measu.re is allowed to take both positive and negative values, then this idea leads to the consideration of signed measure which is defined as follows: IS·9. Definition Signed measure. [Kanpur Statics 1976] Let A be a a-algebra of sub-sets of X. An extended real valued set function
v: A-+[ - co, co] is called a signed measure on a measurable space (X, A) if it satisfies the following conditions: (i) v takes atmost one of the values - 00 or + 00. (ii) l'(0)=O (iii) v is countably additive, i.e., v(
U A,
1-1
)= r:
'=1
v(A,)
for any scquen~e < A, > of disjoint measurable sets. Froin this definition it is clear that a measure is a particular ca'ie of a signtd measure. That is to say, every measure on A is also a signed measure. The converse of this is not true in general, i.e., signed measur.! is not a measure in general. A singed measure v is said to be finite if v (A) < co \f A E A. 15'1. Definition. Positive and negative sets. A subset E of X is said to be positive relative to a signed measure v defined on a measurable l-pace (X. A) if (i) E is measurable. (ii) ¥ ACE s.t. A is measurable =:t v (A) ~ O. Notice that every measurable sub!;et of a positive set is a positive set. A subset E of X is said to be a negative set relative to a signed measure v defined on a measurable space (X. A) if.
267
SIGNED MEASURE
E is measurable. v(A) ~ O. (ii) y AC E S.t. A is measurable A set E( C X) is said to be a null set relative to a !>igned measure v if E b both positive and negative. 15'2. Distinction between a set of measure zero and a null set. A measurable set E is a set of measure zero iff every measurable subset of it has v measure Zero. The measure of every null set is zero. A set of measure zero may be a union of 1wo measurable sets whose measures are not zero but are negative of each other. Similarly we can explain difference between a set of posith'e measure and a positive set. Note that 0, the null set, is a positive sub:>et of X. Theorem 1. A union of any coul/fable collection of positive subsets of X is positive. Proof. Let <. An> be a sequence of positive sets in X. Let (i)
*
00
A= U Ai.
Also let B be any measurable subset of A.
i=l
To prove that A is a positive set. Write Bn=B nAn' n A'''-ln .... n A'J V nEN Here dashes denote complement 0 respectIve sets w.r.t. X, i.e., A,,'=X -·A n . We know that if a set is measurable then its complement is also measurabk. (\ Iso a countable inters~ction of measurable sets is measurable. These facts lead to the conclusion that Bn is a measurable 'iub~et of tIll' positive set All and hence v(B.) ;;;;a 0, by definition of po~itlve set. From the con:.tIUction of B .. it is clear that sets Btl are disjoint. Moreover 00
Of)
B,.= U Bn. ,.=J
Hence
v(B)= U v(R,.) ,_1
~
O.
Thus we have shown that A is a measurable set. For An is a positive set => An is a measurable set
(1)
LJ
A,. is measurable => A is measurable .
.. =1
(2)
¥
BCA s.t. B is measurable set => v,B) ;;;;a O.
268
SIGNED MIlASURIl
By definition, this proves that A is a positive set Similarly we can prove the following theorem:
Theorem 2. EI'ery subset of a 11egatire set is negative set and a countable union of negati1'e sets is a negative set. Theorem 3. If E is a measurable set with !inite negative measure, i.e. if - 00 < v(£)
Evidently Hencc or I"(E1 )
E=(E-E) u £1
and
(E- £1)
n
E1=0.
V(E)=I'(E -£1) +1'(£1) 1'( E-E1):=· v(E) - v(El~
Sim:e \.( E) is finite and hence (l) dcclarc~ that v(E both are finite.
Since I'(E) (2) shows that
<0.
... (1) ... (2)
E 1 ) and
Hence v(£) is a negative finite number. Then
r(E- E 1 )<0. If E _. El is a negative set then we may take A=E--E1 • Otherwise E-El contains subsets ofpositivc measure. Let /1 2 be a smallest positive number S.t. there is a measurable set E~cE·· E1 with the property that I'(El »(I,'1/2) E=[E --FlU E z1u(E1 UEt
),
(E·-EtU E2 )n(l;;lUE2 )=0.
\.(E) = I'(E - FlU E 2 )+ I"(Etl+ I'(E.,) =--o\'(E- El U £2) I'(E 1 U E~) 1'(E-E1 u E.!)=I'(E)-I'(E1)-l'iEt )
+
or
For I'(E) of pO'iitive interers and a sequence < E" > of distinct me~surable ~ets s.t.
A of E S.t. I'(A)
!- <
n.
v(Ek ) <:
0')
1.69
SIGNeD MEASURE
In the latter case, suppose that
A=( E- U Ei)
.. (3)
,_~J
A .. u.. ual, from this equation we can deduce that
~1
v(E)=v(A)+v ( 00
=v(A)+ E
Ei )
l'(E~)
i==l
>v(A)+ ... (4)
Since v assumes atmost one of values - co, and co, and veE) is finite, therefore, (4) says that v( A) is finite and the series 00
E
1'~1
From (4),
1 . - IS convergent.
n_
~'(A)
00
E
"~l
n,.
=finite negative-finite negative =a finite negative number.
...
(A)
(3) says that A is a measurable set.
For an enumerable u1}ion of m.!llsurable set~ is measurable, and difference of two measurable sets is measurable. It remains to show tlrat A is a negative set. ror this let B C A be any arbitrary measurable set. 00
;-1
BCA=E- U EkCE- U Eu ~~1
t=l
i.e.
i-I
BCE- U £1' k~l
270
SIGNED MEASURE
The choice of positive integer shows that v(8)
~
I -11,-1 "'if i E N.
... (5)
This is true Making ni--'>- 00 in (5), we get vCR) O.
<
Thus we have shown that A is a measurable set s.t. (v) A and "'if RCA S.t. R is a measurable set => vCR) ~ O. This proves that A is a negative set wIth v(A)
<0
This concludes the problem.
Tbeorem 4. If E i~ a measurable set oj finite positive measure, i.e.,O < v(E) :I a measurable set El C E and a smallest positive .
1
IOteger n 1 S.t. v( E1 ) < -- --- . n1
Evidently E=(E-El) U El and (E-E1) Hence v(£)=v(E-EIH-v(EI) or
n
v(E-El)=v(E)- v(El)'
E1 =0· ... (1) ... (2)
Since veE) is finite and hence (1) declares that veE -EJ and veE!) both are finite. Given that veE) >0. Hence vee) is positive. Also, by assumption, v(EI) is negative. Hence (2) shows that vee-Ell is finite. If E -El is a positive set, then we may take A=E--E1 • Otherwise E - £1 contains subsets of negative measure. Let n2 be a smallest positive integer S.t. there is a measurable set E 2 CE--£1 with the property that I'(E2)
< -- n}- . 2
(E-El U E l ) n (E1 U E2 )=0 and E=(E-·El U E~) U (E1 U E2 ). This => v(E)=v(E-E~ U E~)+v(E1 U E'IJ => v(E)-v(El)-V (E2 )=v (E-El ~ E2 ) => v(E-·El U £2) < o. For vee) > 0 and veE,) < 0 so that -veEr) > 0 for r= 1,2.
271
SIGNED MEASURE
This proves that E-El U E2 is a set of positive measure. If E-El U E2 i<; a positive set, then we may take A=E-El U E 2• Otherwise we repeat above process. Continuing this process, we shall get either a positive subset > 0 or a sequence < nt : kEN> of positive integers and a sequence < E:, > of disjoint measurable sets s.t.
A of E S.t. v(A)
-00
<
< - n"-I
v(E!.) .
In the later case, suppose that 00
... (3)
E~.
A=E- U
As usual, from this equation we can obtain ... (4) 00
•
\'(E, < v(A) _ E
or
1
nlc
~,=I
Since v assumes atmost one of values - 00 and 00, and veE) is. finite. Therefore, (4) says that veAl is finite and the series 00
E
Fr.>m (4),
1 .
n~
-
IS
II-I
v( .4)
>
v( £)
convergent.
+
00
l,' t=l
n"
=Sum of two finite positIve numbers =positive number. Hence veAl > O. (3) says that A is a measurable set. For an enumerable unior. of measurable sets is measurable, and difference of two measurable sets is measurable. Remains to show that A is a positive set. For ,this Jet B C A be arbitrary S.t. B is measurable. Then or
00
i-I
.-1
.=1
U E~ C E -
B C A = EI-I
BeE - U E t 1:=1
U Et
272
SIGNED MEASUR,E
The choice of positive integers shows that I v(B) ~ - .
••• (5)
11, -1
This is true viE N. Making ni-~ 00 in (5), we get
v(B) ~ O. Thus we have seen that A is a measurable set S.t. veAl and "t B C A S.t. B is measurable ~ veAl ~ O.
> ()
A is a positive 'iet with veAl > O. This concludes the problem. Theorem 5. (Hahn Decomposition Theorem). Suppose r is a signed measure on a measurable space (X, A). Then :I a positil' . set P and a negatii'e set Qs.t. P n Q=0, X=P U Q A being a-a/~ebra of subsets of X. rBurdbw an 199.0, Punjab 1967•. 65 j Proof. Let A be a-algebra of subsets of X. Let v be a signed measure on a measurable space (X, A). Since v assumes atmost one of the values - co and + 00. Without loss of generality we can suppose that v does not take - co. Consider the family B of all negative subsets of X and let Con~equcntly
A=inf [I'(E) : E E B}. . B s.t. Lim v( En) Th en:J a sequence <En> In
n--* CO
\
-n.
B is a family of negative sets
::;. <Eft> is a
~equence
of negative sets
oc
=> U E t is a negative set, by i=l
Theorem 2.
=> Q is a negative set on taking 00
Q= U E,. ;=1
17hus Q is a negative subset of X. Then, acc.Jrding to (1), Next we consider the subset Q-E.. of Q. Q=(Q- En) U En
I'(Q) ~.\.
<
or
v(Q)=v(Q-En}+v(En ) v(En) v(Q) <; v(Em)
V n E N.
En E B V n E N. In view of (I). these two facts prove that v(Q) ~ .\.
273
SIGNED MEASURE
Thus we have shown that v(Q) ~ >.. and v(Q) ~ .\. This ~ v(Q)=1. ::>.\>-00 Next our aim is to show that P=X-Q is a positive subset ofX. Suppose the contrary. Then P is not positive and so P is negative. Hence by definition. for every measurable set E C p. vee) < O. Now E is a measurable subset of X with negative measure. Making use of the result : If E is a measurable set of finite negative measure. i.e., - 0 0 < veE) < 0, then E contains a set A with yeA) < 0, we obtain a negative set ACE s.t. YeA) < O. Since A and Q both are disjoint negative subsets of X and their union A U Q is also negative. Refer to Theorem 2. Consequently v(A U Q) ~ .\, by virtue of (1). But ,\ ~ v(A U Q)'= v(A) + v(Q) = v(A)+ >... or .\ ..; v(A)+.\ or veAl ;;. O. Contrary to the fact that YeA) < O. Hence our assumption is wrong, i.e.," P is not positive" is wrong. Therefore P is positive. Thus P=X - Q is positive and Q is negative. :. X=p U Q, P n Q=0· A decomposition of a measurable space X into t~ subsets P and Q s.t. X=P U Q, P n Q=0, where P and Q are positive and negative sets respectively relative to. the signed measure v, is called Hahn decomposition for the signed measure v. Also P and Q are called positive and negative components of X. Remark. The Hahn decomposition is not unique. If {Po Q} is a Hahn decomposition for v, then we may define two signed measures ~+ and v- s.t. y=v+-v- by setting v+(E)=v(E n P) v-(E)=-v(E Q).
n
IS·3. Singular measures. Two measures VI and V2 are said to be mutually singular (in symbol "I .L v2 ) if 3 a measurable set A C Xs.t. v1(A)=0=Vi(X -A).
For example, the measures v" and mutuaIly singular.
v-, defined
as above, are
274
SIGNED MEASURE
15'4. Definition. Jordan decomposition. [Punjab 67,65] Let v be a signed measure on a measurable space (X, A). By Jordan decomposition of v, we mean a pair (VI' Va) of mutually singular measures VI and VB satisfying the condition V=V1-V I • For example a pair of measures (v+, v-) defined as above, is a Jordan decomposition of v. 15'5. Definition. Absolutely continuous measure function. Let v and ~ be measure functions defined on a space (X, A). The measure v is said to be absolutely continuous w.r.t. p. iff p.(A)=O ¥ A E A ~ v(A)=O and is denoted by V ~ ".. In case ~ is a-finite, the converse is also true. Theorem 6. (Radon Nikodyn Theorem). Let (X, A,,...) be a a-!inite· measure space. LeI v'be a measure defined on A s.l. v is absolutely. continuous w.r.t. p.. TlJen there exists a non-negative measurable function f s.t. V(E)=isfdf' ¥ Ee A. The function f is uniqne in the sense that if g is any measurable functioll with this property, then g(x)-=f(x) almost everywhere in X w.r t. ".. {BuJ'dbw,an,1990;,Kanpur Statistics 1977, 76; Punjab 66, 65; Banaras 64] Proof. To establis,h the existence of the functionf, let us assume that p. is finite. ~ is finite *v -fJ'" is a signed measure on A for each rational number «.
Let T denote the set of all non-negative rational number. For every t e T, let (Ph Q,) be a Hahn decomposilion for the signed measure V-Ip.. If at ;;. t, then Q, is negative for v -11.".. For if E is a measurable subset of Q" then v(E)-Glf' (E) <; v(E)-t. For E is measurable subset of Q" at ;;. t =- v(E)-fJp.(E) ~ v(E) -lp.(E) ~ 0 since - I I <; -to 0:::> v(E)-atp.(E) ~ 0 ~ (v-Gl~) (E) EO; O. Similarly P, is positivt: for each signed measure V-att if at~t. Write P= n {P, : t e T}. Being an intersection of countable collection of measurable sets, Pis mea-surable. Let E be any mea-surable subset of P. Then
275
SIGNED MEASURE
E
c
p=n P, => EC()P,CP, Y 1 e T ~ ECP, Y 1 E T .. (v-tl') E :> O. For P, is positive => v(E)-tjJ(E) :> 0 ~ veE) ;;;. II'(E) => t I' (E) <; veE) Y t E T ~ I'(E)=O or v(E)- 00.
Hence I'(E) > 0 => v(E)=oo. Now we define a non-negative function I: X-[-oo, CX)] by writing f(x)=inf{t e T: x E Q,} Y x'E Z. Here we adopt the usual convention that the infimum of any empty collection of real numbers is 00. Therefore I(x) = 0 Y x E P. To prove that / is measurable. Let r E Rand q E Q be arbitrary. Suppose S.= U {Q, : y t < q} = Union of measurable set =measurable set. Now definition of I declares that
e Q s.t. q > r} =intersection of measurable sets . measurable set. Thus {x EX: /(x) c; r} is measurable and so I is measurable. Next our aim is to show that {x E X :/(x) Et;; r}=() {Sf: q
V(E)-t.ldl' Y EE A.
Let us now suppose that f'(Enp)-O. If I'(E()P) > 0, then v(Enp)=O.
I" /
I
I
Also djA ;;;. E() P djA= co. This => "I dl'= 00. [For /(x)=O Y x E P, hence/: E ... [ -00, 00] is not integrable] Hence
V(E)=ff/df'
This establishes the equality in case I' (E n P) > O. Absolute continuity of v w.r.t. I' implies that v(Enp)-O. Select a positive integer n and set .
Et:={X
E E:
~~1 < lex) <; ~
J.
276
SIGNED MEASURE
For each rational number q
>
!:., we have n
E1c C{x EX :f(x) ~ n}CS.= n{Q,: y t < q}. Also Q, is a negative set for the signed measure v-ql' and Elt i~ a sub~et of Q,. Therefore E" is a negative set. (Refer Theorem 2). This => (v-ql') (Et ) ~ 0 => v(Ek ) ~ qp.(Ek ) k
=> v(Ek) ~ -n '" (E,,). On the other hand, since
.... (1)
E" C { x E X :f<.x)
and hence [ v -- (k~ This gives
k--ll
> 11 f c
Pet-t,/ft
1) '" ] (E,,) ~ O. For P(k-t,/ft is positive,
v(Ek )
~
(
k ~ 1) 1'(Ek ).
• ••
(2)
From (I) and (2),
k-l) ",(E (-n
Ie )
~ v(E/,) ~
k Ii
",(E~).
. .. (J)
Since E is a disjoint u"ion of sets En P and E". Hence ",(Enp)='=v(Enp). It follows from coulltahle additive prop .• y of integral that
ff Ii
d",=
~ froLt f
11=0
d"" v(E) =
E v (Et )
k-l
By definition of E •• k-l -
k
~f(x)~-on
n
E".
"
Applying first mean va.lue theorem,
n (k-l)
f
IA(E,,) ~ Etf(X) dx ~ k n",(E".).
From (3) and (4), ( k--
-n° 1) ",(E
lt)
E/: f dx - v(Ek) ~ J
k
n y(E.). .
Summing this over k. -
~ I' (E) ~
n
I
l>
f(x) dx-v(E) <;
Since n is arbitrary and hence making 0=
f£
f(x) dx-v(E)
~
0,
~ ",(E). -
n
11 ..... 00,
. .. (4)
. 277
SIGNED MEASURE
f
or
£
f(x) dx=v{E).
This completes the eXistence proof. Second part. To prove the almost uniqueness of the function
J:
X-+[ -00, 00]. Let g: X .[ -<X>, co] be any measurable function satisfying
the condition For each n
l'(E)==
e:
f,. g
d". ¥ E
e:
A.
N, let
A .. ={x E X :J(x)-J(x) ~ J/n} E A and Bn={x e: x: g(x)-/(x) ~ lIn} E A. By definition of An, I(x) -g(x) ;;t JIn on A...
Applying first mean value theorem, we get fA .. (I-g) d".
or
~ ~ I£(A
fA .. Id".-fAn gdfl
II )
~ ~"'(A.)
v(A..)-v(A,,) ~ (lIn) fl(A.) (lIn) ".(A .. ) ~ 0 or I£(An) ~ O.
or or
But I£(A,.) ;;;;. 0 is always true :. ".(An) =0. Similarly we also have I'(B,.)=O. Let
C={x
e:
-
X : J(x) #g(x)} = U (A,.UB .. ). n_l
."
Then p(C)=1: p(A,.)+}J ,,(B..)==1: 0+1: 0=0.
"
..
Then p(C)=O. This => f-g a.e., on X w.r.t. p. Exactly in a similar way the theorem can be proved in case ~ is a - infinite. Remark. The functionJ: X-+[ -00,00], defined as above, is called Radon Nikodom derivativ~ of the measure v w.r.t. 1£. In symbol,
f= dv. dp.
Theorem 7. (Lebesgue Decomposition Theorem). [Punjab 69, 66. 65] Lft (X, A. ".) be a a-finite measure space end v a.-finite measure defined on A. Then 3 two uniquely determined measllres \I. and 1'1 s.t. V= VO+V1 • Va 1. "., VI « ",.
278
SIGNED MEASURE
Proof. >'=I-'+v. p. and v are G-finite implies that>. is a-finite.
" < ~ and v
Evidently
>..
p.
By Radon Nikodym theorem, we are able to find non-negative functions/, g: X~[ -00, 00] s.t. P(E)=J fdA, v(E)=f g d>. "I E E A.
'E
Let A={x EX: f(x) Then X
E
>
OJ, B={x EX: f(x)=O}.
··~UB, A()B=0.
Furthermore p.(B)=LfdA=O:
Defir ~ two functions Vo, VI: A-.[oo, 00] by requiring that Vo (E)=v (EnB), VI (E)=v (EnA) ¥ EEA. Then Vo and VI are measures on A and satisfy the condition V=VO+Vl'
v. (A)=v(AnB)=v (0)=0, or Vo (A)=O. Thus p(B)=O=vo (A)=vo (X -B) i.e. p.(B)=O==vo (X -B) This ~ Vo is mutually singular to p. ;;:> vol-I-'· To show that VI ~ p.. For this let EEA be arbitrary s.t. I-'(E)=O.
Then
Isf dA=p(E)=O
fE/ d)..=O.
or
Also /(x)
~0
a.e. on E relative to ~. on A()E and hence VI (E)==v (EnA) by definition of VJ ~" (EnA)=O. :. VI (E) ~ O. But VI (E) ~ o. Combining these two results, VI (E)=O. Thus we have shown that ,.(£)=0 =>
¥
xE E.
This ~ f=O Since f>O
Th~
To prove the uniqueness of
VI
(E)=-O
.~<~ Vo
and
VI'
To show that v. and VI are unique, let vo' and VI' be measures s.t. 1I=v. +Vl' and has the same property as that of the measures "0 aru1"vl respectively. Then v=VO+v1 and V= vo' +Vl' are the two Lebesgue de.::ompositions ofv. Then Vo -VO'=Vl'-V1, Agaip
279
SIGNED MEASURE
v1' -VI is absolutely continuous and Vo- vo' is singular relative to We have vo"'v.', V1==V1'. Hence Vo and VI are unique. , Problem 1. Let A and B be two measurable sets and V a signed measure s.t. A C B and I V (B) I < co. Then show that I v (A) I < co. . Solutiou. Since B==(B-A)+A, (B-A) n A=ti. Hence v (8)-=v (B-A)+v (A). Hence if v (B-A) and v (A) both are w, then v (D) is w. If v(B-A) or veAl is GO, then again v(B) is co. Given I v (B) I
EXERCISES
1.
2. 3.
State and prove Hahn decomposition theorem and the lQl"dan Decomposition theorem far signed measures. [PunJab 1970] State and prove Lebesgue decomposition theorem. [Punjab 1969, 66] State and prove Radon Nikodym Theorem. [Punjah 19661
16 Product Measure Introduction.
Recall that Xx Y={(x. y) : xE X, yE Y}. Xx Y is called cartesian product of X and Y.
16'0. Definition.
If ACX, Be Y, then
AxBCXx Y. A x B is defined as rectangle and the components A, Bare known as its sides.
Theorem 1. If (X, A) and (Y, B) be two measurable spaces and if A E A, .dE B, then the rectangle A X B is a measurable set. 16'1. Definition. Let (X, A) and (Y, B) be two spaces. Let xEX, YEY and ECXx Y. Write E.,={y : (x, y)E E}. E,,={x : (.~, Y)EE}. Then E. is called section of E determined by x. section is determined by an element x and therefore X-section. Every section of a measurable set is Therefore E.. , E, are measurable sets.
mca'.>urable
Since this Ez is called measurable.
16'2. Definition. Let (X, A) and (Y, B) be two measurable spaces. Let xEX,yEY, EeXxY be ar!'itrary. Letfbe measurable function on E. Write f.(y)=f(x, y), fy(x)-/(x, y).
Thenf,(x) and f ..(y) are functions over the sets Eyand E.. respectively. J. is called X-se~tion of J. Since every section of a
281
PRODUCT MEASURE
measurable function is measurable and !herefore j.. andly are measurable functions. 16'3. Definition. Product measure. [Punjab 1967J Suppose (Y, A, 1') and (X, B, v) are a-finite measure spaces. Let EcXx Y. Consider a real valued set function 7>, i.e. 1t : Xx Y-R 1t(E)=J v(E.,) dJlo=f f,(E,,) dv
l'.t.
and
1t(A X B)=p.(A).v(B)
Then
1t
where AEA, BEB. is a measure function and is referred to as product of
Jlo and v.
Also we write :t=Jlo xv. 16'4. Definition. Double Integral. IfI is Lebesgue integrable over a measurable set E, then
IE JI(x, y) dx dy is called iterated integral and
f
f(x, y) d7C is called double integral.
If A is a measurable subset of E, then we define
fJ 6
l(x, y) dx dy=
fJ I(x, y) XA(X, y) dx dy.
2. Fubini's Theorem. Suppose E=[a, b] x [c, dJ. Also suppo'ie that I is lebesgue integrable over E. Then
tI
lex, y) dx dy=
1: U: I(x, y) dy] dx
J fI(x, y) d~ dy= J: [J: I(x. y) 1dy dx
6
(Banaras IV 1970, 67; Puajitb 67, 66J Ex. I. State Fubini's theurem and explain why
J: 0: (;~;;a
dX} dy#
J: U: (~:~;:)2
dy}dX.
[Banaras 1970, 67J
where (x, y)
E..~
[0, 11.
282
PRODUCT MEASURE
..· ...·
..· Thus we see that
J: {J:[(X, y) dX} dy=-~#: . . J: U: J: {J:/(X, dX} J: {J:
[(x, y) dy } dx
or
y)
dy#
I(x, y) dy} dx.
The reason for this is that [ is not Lebesgue integrable Over [0, 1] x [0, 1].
EXERCISES
1. State Fubini's theorem. Discuss its applicability to the function .
f(x, y)
x2_y2
(X2+y2)2'
X2+yl#O, Xc [0, 11. y6 [0, 1].
[Banaras 1967J
283
PRODUCT MEASURE
2.
State Fubini's theorem and explain why y Jl{Jl (X2+y2)2 o
X2_
0
lll
dx
}
dy=j:.
Jl {II 0
X II _ 0 (Xl:
y
2 +y2)2
dy
}
dx
[Banaras IV 1970] 3. Define ptoduct measure. State and prove Fubini's [Punjab 1967] theorem. 4. Develop the ~oncept of product measure to the state needed to prove Fuhini's theorem and prove the latter theorem. [Punjab 1966]
17 Fourier Series Introduction. 00
+
A trigometrical series is a series of the form
...
1J (On cos nx+bn sin nx)
... (1)
tI~l
where the coefficients 0 0 , 01 • •••. .. 0'1, b1 , b2 , ... bn are independ~nt of x. The problem of representing a given series by a series of the form (1) was first discovered by Fourier in a problem of the conduction of heat. 17'0. Periodic function. A function f(x) is said to be periodic if f(x +1) =f(x) V x. If I is the smallest positive number s.t. f\x+/)=flX) ¥ x, then the fUnction /(x) is called periodic function with period I. Examples: (i) Each of the functions sin x, con x, sec x.' cosec x is a periodic function wIth period.2n. (ii) tan x, cot x both are periodic functions with period n. (iii) sin nx, cos nx are periodic functions with period 2r./n.
( 2n) .
For sin n x+jJ
=SIJl
nx
'V
x.
(iv) The sum of a number cf periodic functions is
II.
periodic
00
function e.g. 1: (an cos nx+ bn sin nx) is a periodic function with ...1
period 21t. 17'1. Finite discontinuit). A function f(x) is said to be continuous at Xo if Lim Lim " .... 0 f(xo+h}= h~O f(xo-h)=/(xo}· "If both the limits/(xu -0) and/(xo+O) exist, but are not equal, then we say thatf(x) has a finite discontinuity at XU' The F,Jurier series off(x) converges to i [/(xo+O)+f(xo-O)].
285
FOURIER SERIES
17'2. Even and odd functions. A functionf(x) is said to be an even function if f( -X)-f(x) ¥ x. For examole ~os x, Xl, x 3 sin x, x 2 cos x all are even functions. A functionf(x) is said to be an odd function if f( ---x}= - f(x)
For example sin x,
X2
x.
1.;1
sin x, x 3 cos x all are odd functions.
The geometric characteristic of the graph of an even function f(x) is that the graph is symmetric W.I.t. y-axis and for the graph of an odd function, we have symmetry about the origin. L[
17'3. Ortht)gonal functions. The functions fl' f2,'" ..... .in TC] are said to be orthogonal w.r.t. the intervall-1t, 'It] if
-ft,
l. I·
f".(x)f.(x)
d~ =0 for m::j;n.
The functions arc said to be orthonormal if -w
f".(x)f.. {x) dx=
. I Th e functIOn -
1°
if m:;l:n )'f m=n
1
sin nx cos nx are orthonormal functions. - - ----
V2TC' V'" ' y1t
17·4. Trigono'metric polynomial.
Tn(X)::"~2°-+
1·
"=0
A polynomial of the form
(a.. cos nx+b" sin IlX)
is called trigonometrical polynomial: 17·5. Fourier series. Suppose a fU,Detion f is Lebesgue integrable over the interval [c, c+2lt] and a
f(x)=
f
II
+,,':1 [an
COS
nx+b. sin nx]
is lIniformly convergent in [c, c+21tl. I
0 .. -=-,,7t
Ie+2tr J(x) cos nx dx, r
... (1)
Then
J
J c+2" f(x) sin nx dx, b.=1t
c
..
(n=O, 1,2, ... ). The series (1) is called Fourier series of !(x).
The constants a.. , bn are called Fourier constants 17'6. We shall use the following result in the subsequent articles.
286
(I')
FOURIER SERIES
Let m and n be any two integers. cos nx ]C+21t c sin nx dx= ---n- c: -0 ifn:;t:O·
fe-fa"
[
f
Similarly (ii)
0CHtf
cos nx dx=O if n:;t:O.
J:t~. sin mx sin nx dx=i
=i
J:+
21r
[CdS
(m-n) x-cos (m+n)x]dx
[Sin m-n (m-n)x
sin (m+n)x
m-t-n
]c+lI
1t
c
=0 if m:;en.
(',,',') fC+2. c:
cos mx cos nx dx=i JC+2. c [cos (m+n)~+cos (m-n)x]d.:t
.
=0 ifm:;t:n.
jC+21t sin
(iv)
Similarly
1
C+2.
•
2
nx dx= i
1<+2# ,
IC+21t
cos! nx dx= f .
f
=1t
(v)
f"+
C 2Ir
(I-cos 2nx) dx=rc if n:;t:O. '
sin mx cos nx dx=i
if n#O.
.+rtf
C
(l +cos 2nx) dx
[sin (m+n)x+sin (m -n)x] dx
=0 ifm#n. Here c takes any real value. Theorem 1.
a If a functionf(x) = 2°
00 • +n-l E (an cos nx+bll sm nx)
is Lebesgue integrable and uniformly convergent in [c, c+21t]. Then a.. =I- fC+~. lex) cos nx dx, bn= ~
Proof.
"
(n=O, 1, 2, 3 ...... )
a Let f(x) = 2°
fC+
2•
"
f(x) sin nx dx,
+ ..001:_1 (a.. cos nx+b.. sin nx)
... (1)
In'egrating (I), '
1
C+:.
c
a
f(x} dx=f·[c+27t-c] 00 + :'1 a.. [ Jc+2tr cos ItX dx+ bn 1C+I.] 0 sin nx dx C
=7[ao (all other integrals vanish)
287
FOURIER SERIES
t
or
00=1t
r"+I. f(x) dx
... (2)
0
Again, for any fixed positive integer m, we mUltiply (1) by cos mx and integrate. Then
I
C+II.
II
[(x) cos mx dx=
0')
TEa"
t le+I" cos
Q
II
niX
dx
[0+2. cos mx cos nx• dx+ 00E b" JO+1. cos mxsin nX dx
_1
fI-l
0
/I
Again the int~grals on the right vanish except one, namely
J cos 1/+2. Hence 1 II+2tr
1/
/0+2. cos
lex) cos mxdx==a",.
II
...
mx cos mx dx, (n=m)·
I Ic+ttr
0",= -
n
D
l
mx dX=rt am
/(x) cos mx dx •
. .. (3)
Similarly, if we multiply (I) by sin mx and integrate, then we can easily obhin bm = I-
1 0
rt
r
+1 • !(x) sin mx dx .
0
1 O+2• flx) , By (2) and (3). am =cos mx dx, (m=O, 1.2, ... ) 1t
•
This completes the proof. Deduction: (i) Taking c=O, we get I Q,.=ft
J2' !(x) cos nx dx, for n=O, ], 2, 3,
1 JlItr lex) sm . nx dx, for n=l, 2, b,.=ft • . (ii)
o'
0
3....
Taking c= -71, we get
f' ~ I'
a.. =1 b,.=
1t
_"
'/I:
_"
"x) cos nx dx for n=O, 1,2, 3, ... f(.t) sin nx dx, for n=r I, 2, 3, ..
Furthermore
00
f(x) = (0./2) + E (r" cos nx+h,. sin nx) ... 1
288
FOURIER SERIES
= 2~ ["
fey)
~v+() In)'!l
C.
fey)
COS
ny
COS
nx
+sin ny sin nx) or f(x)=
2~ [ . f(y) dy+(l/lr) ,,~ /:,.
~v
fey) cos n(x-y) dy,
where Remark. Remember the result of deduction. Remark. If f(x) is given, then we can compute the coeffi· cients associated with f(x) with the help of Theorem 1 and thus Fourier series of functionf(x) is determined .. Now the question arises that the Fourier series will converge and that will represent f(x) or not. A very wide class of functions possess Fourier series and that do converge and represent f(x). If a function defined in the interval [c, c+2n:] possesses Taylor series expansion, then it will certainly possess a Fourier series representing it. If a function does not posses Taylor series expansion, then it will not possess a Fourier series representing it. Theorem 2. If a function I(x) ;s uniformly convergent in the interval [ -I, I] and
f(X)=~+E [
a.=+ (,
then
where Proof.
a" cos
(~;X) +b" "in (n;x) ]
f(x) co's (n;x) dx, bn =
+
flex) sin (n;x) dx,
(n=O, 1,2, ......... ). The proof is exactly similar to Theorem I.
Remark. If I(x) is de.fined in the interval (c, c+21t)
I(x)=~ +"~1 [0,. cos (n;x) -tb.. sin (n~x)]
and if then
0 ..
+
=+1:+2" f(x) cos (n;x).dX' b,.= ]:+•• f(x) sin (n;x) dx
where n=O, 1,2,3, ..... . Deductions. For half range series: (a)
00 fIC If f(x)=:r+ .~1 o. cos (nnx) -,- ,when I(x) E L [0, /)
289
FOURIER SERIES
then
all=~ 1: f (X) cos (n;x) (b)
Iff(x) =
,.~ h"
dx, (n=O, 1,2.......... )
sin (n;x) andf(x) E L [0, /],
b"=7 /:f(X) sin (n~x) dX, (n=l, 2,3•....... )
then
Remember the above results.
':1
"" (a,. cos nx+b,. sin nx), a If S.,.(x)=/+
Tbeorem 3.
Sm(X)=! (" fU+x) sin (~-tJ2!. dt TtJ-... 2sm(tf2) where f(x) has a period of length 2~. Proof. Tt(a" cos nx+b,. sin nx) then
=cos nx 1:"./(U) cos nu du+sin nx [" feu) sin nu du
=J:." feu) Also Hence
ao=!
7t
J"
cos n(u-x) du
feu) du
-ow
m a S",(X)=2....!!+ E (a" cos nx+b,. sin nx)
,.=1
=2I
"It
I"
_"
]J"
=-
"It
Tt
feu)
-"
=!J" _w
. . I'"
feu) du+ E -1
f(u)
.. =1 "It
[)-2+
f(u) cos n(u-x) du
_II'
III cos n(u -x) 1:
,.=1
]
du
[Sin2 (~+l) (U-X)] du I (u-x) SIn
Putting u-x=t, we get S",(x)=! ("-.. sin ~m+t) t .f(t+x} dt "It L,,-o: 2 sm (6t) Since the length of interval [ -1S -x, r.-x] is 2x and hence We can replace this interval by another interval of length 2"1t namely [ ~7t, "It]. Thus 1 sin (m+l) t S",(x)=;t _., 2 sin (it) f(t+x) dt
J"
Hence the result.
290
FOURIER SERIES
Find Fourier series for f{x) = X2 , 0
Problem 1. Solution.
X
<
21f.
f(x)=~-t:.~o (a .. cos nx+bn sin nx)
a.=-I ICH. f{x) rr:
<
0
11
cos nx dx==rr:
2•
... (1)
f(x) cos nx dx,
0
on taking c=O
=!rr: J~"0 X2 co.. nx dx=~rr: l{~ sin nx}2. - ~ J211 non 0 = __ ~ [{_ X mt
or
C~}2. n
0
a.. =~ if n,rO. n I 1 fll1l ao=f(x)dx=-
f2.
lt
b.. =-1
r.: 0
0
X2
x sin nx dX]
+J2" ~ dxJ= nrr: 4: cos 2nlt tl
0
1 87;3 8 dX=:'-3-=3-
fl.l!.
•
J2" xl! sin nx dx=-1 [{ ---xl! cos IlX}2.
lt~
lt
II
0
+ii2J2. 0 1 [ - 4n =-
rr:
n
I/X
dx ]
l cos 2nrc+"22 ( x sin nx )2. 2 1 sin nx dx1 n no --2
2•
0
~_ (
=! [ __ 4rr:2 _ n
It
x cos
n2
__
~
!) (cos nx)2"J= __ 4Tt. 11
n
0
Putting the values in (1), l ) =8r.-2+.. ~,[4"2 cos nx-4rr:.] Ji\x sm IIX 6
tI-l
n
n
Problem 2. Expandf(x)=x, 0 sine series, (b) cosine series. Solution. where
b,,=;
(aJ
Let f(x)
/:f(X) "in
=
!l b,.
("_;X) dx.
<
x
sin
<
2 in a half range (a)
(~;x)
Here 1=2.
211 . (Ilrr:X) "rr:~) ·n:f. 2 l2 -2 dx-= [f. -x (co> 2
b"=2 • X
SIR
JI
Ans.
2 • cos (n7tx) +mc T dx ]
.•. (1)
FOURIER SERIES
291
4 cos nlt+-.2 2 (. = [ -sm -n1tX)2] = -4- (-1)". nrc nrc nrc 2, nrc Hence (I) becomes
- -4 (- I)" sm . (nrcx) 1: -2 "=1 nrc
X=
a I(x)=f+
:1a" 01)
cos (nrcx) -,-
... (2)
Here 1=2.
(b)
Let
a,,=
f- t I(x) cos (n~x) dX=~ f: I(x) cos (n;x) dx
=
.
r x cos (nltx) dx= (2X sin n1tX)2 _~ II sin ~ dIC 2 nit 2 . nrc, 2
J8
4 ( nrcx)! 4 =0+ T2 cos -2 =""""iZ
nlt
=-iz [( -I)" --I] jf n Tt
:.
an =
nit
4)
nrc-I]
n:;t:O.
~.~ jf n= 1, 3,
nit
r
LCOS
5, 7, ...... and a,.=O if n=2, 4 .... .
=1- J: J{x) dx=~ r: x dx=2.
ao
01)
nltx
4
Now (2) becomes, f(x) = I +}; 22 [( -I)" -I] cos -2 "=1 n 7t
f(x)=l-~[~ cos (r.X)+~ cos (3nX) n: P 2.:z 2
or
+~ cos(5~X)+ ...... ] Problem 3. cosine series. . Solution.
where
Expand f(x)==sin x, 0
We have
I=!Jit rc 0
n: ,
x
in a Fourier
,
sin x cos
I'IX
[sin (I-n)
dx.
Here /=rc.
dx
... (1)
x+sin (l+n) xl dx
=_! [COS (I-II) X + cos (n+ 1) rc
< x
a ':-'1 a. cos (nrex) I(x)=t+ -,-
a..=~ f~ Itx) cos (";X, a.. _3
<
1-n
n+1
xr
if n:;Cl.
292
FOURIER SERIES
-~[l~n {COS (l~n) 7t-l}+ n~l {COS (n+l) 1t-I}] 2'
(nil-I)
By (I),
7t
[J +cos mc] if n;;i:J.
J" sin x cos x dx=-1t1 J" sin 2x dx =l [-cos;x 1: =0.
a1=-2 1t
0
'lo=~J'Ir lto
0
sin x
dx=~1t (-cos x)'Ir0 =~1t by (I).
Putting the values in (I). 4 aD 1 f{x)=2-+ 0 - E (2 1) 1t
,,~l
n -
7t
[1 +cos nlt] cos nx
. 2 4[ 2 1 -l cos2x+ 1_ cos4x+ 1_ cos6x+ ... ] or smx=;-; L 4ll 1 62 1 17'7. The L2-theory of Fourier series. A functionf is said to be square integrable over [0, 27t] when / is measurable and
J:tr /2 dx <
00.
In this case we also write f2 E L [0, 21t] or
fELl [0, 2n']. The space of square integrable functions is denoted by the symbol L2 [0, 2n]. Theorem 3.
Parsevel's identity for Fourier series.
Let the Fourier series '!2J!+
E(a" cos nx+b" sin nx)
"=1
of f(x) converge uniformly to f{x) at every point of the interval (0, 21t) and Iet/{x} E L (0, 21t). Then
Proof.
a
Letf(x)=t+
,,:'1 (a.. cos nx tb.. sin "x) 00
Multiplying (1) by f<x) and then integrating. we get
._.(1)
FOURIER SERIES
l°·W
293
a
[f(X)]2 dX=-2°
J2W f(x}
00
J:~ f' x) cos nx dx
dx+ E [ a.
..~
+b.
J:-It X) sin nx dx J
Making use of the fact that 1 a ll =n
we get
J2. lex) cos nx dx, b.=J2W f(x) sin nx dx, 0
0
J:o
.
[[(x)]· dx-=(ao/2).7Wo+ 00 E [t7o"l+nb,,']. 11-1
Problem 4.
To prove tltat for all positive integers m.
J'
a,1 + (a"lI+b,,2) ~ ~I [/(x)]1 dx 2 n-l -I where all and bll are the Fourier coef/icientl of I(x). Ilx) being piecewise continuous in (-I, I).
E
Solution.
We
have/(x)=~+
!1 [a"
cos (n;x)
+b" sin (~~)]
where
a,,=~
f, f(x)
cos
b"=rIJ'_,I(x) sm. (mcx) --r'fake
S,"(x)-~- +
1
(n7~ ) dx
... (1)
... (2)
dx.
!1 [all cos (y)+b
n sin
{n~x)] ... (3)
Evidently [[_-S",]2
[,JI dx+ t,SIII
~ o. Hence JI
f/ Sil
-I
[f-SrnlH dx
~ O.
dx ;;;a. 2 dx ... (4) Multiplying (3) by f and integrating, we get in virtue of (2), as
or
2
294
fOURIER SERIES
... (5)
or Squaring (3) and then integrating,
I~,
S.- dx=!!!i.2/+ !ix t, [0,,2 +2 Etil
,,-1
cos2
(n~x) +b..2sin (~~~ 2
I' [
. all an cos (nnx -,- ) +ao b" Sill
-I
)]
dx
\nnt-) J dx
... (6)
and
For
fl
cos
f,
cos 2
r~X) dX=1
(n~x ) dx==O= f~, sin \ n~x
f, [1
+cos
) dx
(2'~"~)] dx=~I·-O""1 eh
Putting the value from (5), (6) in (4) and simplifying, we obtain ~
or
1 -I
J' -I
P dx;;. a2° + /1=1 L' (a,,2 +htJ2) 2
Iii
... (7)
Hence the result. Deduction. Taking lim it as m .. 00 in (7). we get lkssch.
inequality namely
Tlleorem 4. Riemann-LebEsgue theorem. !If E:: L [-1t, nj and if < an > and < bn > are FOl/ri('r c()efficip.llt.~ of lJ functior lim lim f , then n-?oo 0 11 =0, n-+oo b,,-=O.
295
FOURIER SERIES
Proof. Fix Ii > O. Define an unbounded measurable function g on [ -n, 1t] s.t.
J:"
Then g E 1..2
1 f{x)--g(x) I
dx
j!.
<
. .. (1)
n] so that
[ -1t, <:lO
, E (A ..=+Bn 2 )
<
... (2)
00
».=1
J"_11' B,,=! J" -'It
A,,=1
where
g(x) cos nx dx
7t
g (x) sin If X dx
1t
(2)
:=>
E A,,2
<
00
::>
Lim A,,=O n-+oo
::> 3 no E N s.t. I An 1 < e/2 ¥ n ~ no·
••• (3)
For any n, lan-An
=11
I
(II'
1-11'
1t
~- 7t~J"_" Thus
I a,,--A.. ! <
If--g
rf(x)-g(x)] cos nx dx I dx
(6/2)
a"=(a.• -A.. )+A,,
... (4) E
e
I a" I <. I a" --An I + I A .. I < T+2"
=E It 11
~"o
[This follows from (3) and (4)].
This
Lim b =0. n- 00 It Theorem 5. Establish Dirichlet's integt;,al i.e., iff E L (0, 2,,) and if S" deflates the nth partial slim of the Fourier series ofivc). Exactly in a similar fashion we can show
thell 1 S", (x)~==-2
7t
J'It sill (n+l) u ---. - '(-I .')- [nf(x+u)+f(x-u)] duo 0
Sill
nU
[Banars 1961, 63] Proof.
U
m
Sm (x)= ~-+ 1: (a" cos -
"~l
flX-t
btl sin
I1X).
. _.( I)
296
FOURIER SERIES
Q,.=!KJo(2" f(t)
1
nt dt, b,.=! 2• f(t} sin nt dt n 0 Putting these values in (1),
with
1t
Sm=--K1 Ill.. f(t) 0
COS
[ i+
m1:
(cos nt cos nx+ sin nt sin nx) ] dt
n-l
r.Sm=J:' f(t} [~ +n~l cos n (t-x) ]dt
or
Put 1t
t-x=u. Then
Sm= J2"-« f(x+u) -~
[I-2+
m ~
cos nu ] dll
n-l
. "! sin (m+i) u Smce j+ 1: cos nu= - 2 . (i) ,.~l SIn u
Dm(u), say.
Hem:e 1t,Sm=f_r"II-1I f(xtU) Dm (ll) du.
. .. (2)
Evidently Dm (u) is a periodic function of period 2/1 and feu)
is given to be periodic fUI)ction of period 21t. Hence the interval [-x, 21t-x] can be replaced by another interval of length 2 .. namely [0, 21tJ. Thus Tt
Sm= J•2" l(x+u) Dm (u) du=
I" 0
f(x+u) Dm (u) du
+J:" f(x+ u) Dm (u) du Putting u= -t in the second integral, 71:
Sm=I' f(u+x) Dm (u) du+ o
-=
1:
f(u+x) Dm (u) du+
r-
2
..
J-1I"
r:
flx--t) Dm (-t) (-dt)
I(x -I) D", (t) dt as Dm (/)=Dm (- 1)
=[ f(x+u) Dm (u) du+[ I(x-t) Dm
. . 1: or
SIII=l
1t
(I)
dt,
by the
same reasoning discussed above [/(x+IIJ+/(x-U) Dm (1I) du
J*' [f(x+u)+f(x-u)] 0
Dm (u) du
297
FOURIER SERIES
or
S,,= ~ 11:
J" [!(x+u)+!(x-u)] Dn (u) duo 0
17'7. Summation of series by arithmetic means. [Burdbwan 1990] Let E an be a series of real numbers with partial sums
If l: an is not convergent, i.e., if lim S" does not tend to a limit, then it is sometimes possible to associate with the series a 'sum' in a less direct way. This simplest method is known as 'Summation by arithmetic means.' . W rIte a,,= Sl +S2+n "+S,, - .
Then a .. is the arithmetic mean of the partial sums. If S" -+ S, then also (1,,- S. But an may tend to a limit even though lim Sf! does not exist. A series for which a" tend to It limit, is said to be summable by arithmetic means, or by (C, 1). Hence C stands for Cesaro's fur first order. Examples (i) E( - 1)" is summable (C, 1) to the sum i. (ii) I+O-I+I+O-l+ ............ is !>ummablc (C, I) to the sum 2/3. Theorem 6. Fejer's integral. Establish rh,' illtegral
-_ 1
0,. ~2--
nn
10
sin 2 (.nll) --'~(l
Sin
211)
, • lflX+U)+!(X-U)] duo .
Proof. Let S,. denote the 11th partial sum of the Fourier series of f. Then
S,,=! R
Wh ere
j"
l.!(x+u) +/(x-u)j Dn
(U)
dx.
0
~~
l. D,. (U)=2+
~
f)bl
... (3)
sin(11+.) U cos mu .... -,-- - -2 Sill (iU)
[Prove this as in Theorem 5]. Write then, by (3), a .. _1-. nn
J" [/(x+u)+f(x-II)) [ 0
l·
",_1
Em (u) ]dU
298
1:
nut
1
Dm (u)=
m~l
m",l
sin ('!z+i)!~ = sin2 (/~'!.!.3l. 2 sm (iu) 2 Sll1~ (~U)
. 1 sin 2 (nuI2) Taking Kn (u) =2. 2( ) ,t
r1n=~ f1t n
or
r1n
Sill
tu
,
we get
£!(x+u)+f(x-u)] Kn (u) du
... (4~
0
=-21-
Jao [ f(x+u)+f(x-u) ] sin 2(nuI2) 12 duo 0 SID (U ) 2
n!;
This is known as Fejer's integral. Remark. It was Fejer who found that the operation of' "summation by arithmetic means" is applicable to the Fourier, series
178.
Sumtnability of Fourier series. Here we shall discuss the case, for what values of x the Fourier seri,!s of f e L [- Jr, Jr] is (C, 1) summable to f(x) ? i.e., lim for what x e [-Jr, n], a.. (x)=f(x). n-.oo 1
Here
all (x)=n (1n
(x)=.!n
,.
Sm (x). We have se~n that
E m=l
I'"a
[!(x+t)+f(x- I)] K,. (t) dt
... (4)
[This is equation (4) of Theorem 6] tn a gpecial case if l= I, then Sn(x)=1 V Il so that all (x)=-=(nln) = 1. Now (4) gives 2 ft 2 Kn(t)dt or /lx)=f(;'(.)K.. (t)dt (5)
1=-J 1t
jft
no
0
1
.. ·
U +fl .\ -- t) -·f(x) Kn (t) df (4) - (5) gives c. (x) l(x);..+..;2 J1t0 ({(X+ '--~2'--
If 2 then -;-
lim lim. a" (X)=/(x), i.e., n .... oo n-oo [a .. (x)-j(x)]=O
JIt [!(X+l) -I) --J~x) ----2+f(x '... --0
Theorem 7.
~CL+ :2
£
.. _1
1
K.. (1)=0 as Il~OO.
1// is continuous and Ie L t --tt, ttl. then
(a .. cos nx+bn
Sin
I/x)=f(x), (c, I).
FOURIER SERII:.. S·
Proof.
299 1
We have seen that 1 ~ 'It [f(x+t)+I(x-t)] Dn (t) dt Sn (x)=0
1t
a,.
(x)=~
; Sm(X)=-~ Jft n m~l nr. " 1
[f(x+t)+/(x-~t)]
J
·n
11
.>Jut - X Dm (/)-- E n 111_1 n 111-1
sin
(m+~) t
2 f>in ( "2
2n sin 2
1t
) =Kn (I)
~ '2 )
11
J l/(x+t)-t-/(x--l)] Kn
(t) dt
... (I)
0
1=:= 1,
In a special case if 011
E
Dm (t) ] dl m=l"
-~--'---t-"
= sin:! (n+.n/ 1 Hence u,. (x)= -
[
then S .. (x)= 1 ¥
n so
that
(X)= (njn) = 1. Then (1) ~
2
11
It
0
2
10"
1=-j
or
I(x)=
1:
Kn(t)dt
... (2)
I(x) K" (f) dl
... (3)
(1)-(3) gives
I
CI"
(x)-fvc)
< !- lir 1:
:.
0
I"'" ! 11: [J(x+tH f(x .-. t)-2/(.t)] K .. (I) dr\
[I /(.t+t)-/(x) 1+1 (X--'I) -f(x) I ] I K" (t) I dl ... ( 4', f is continuO\JI; ~ I l(x1) - f(xz) I -< 6 whenever I X 1 -X2 I <; s.
I CI"
I<
(x) -f(x)
I on-f I
or
1
~ It
<. 24 1T
Here ! Xl - x 2 I < Also
0
<
8 <:
J" 0
I K .. (1) I dt
a ~ I x±t-x l < ~ 0 '" t < o. 7:.
From (2) and (5), 1 0 ,. -II
or
ft J0 «+1) I Kn (t) I dl
2e
1C
0~
.
It I<
(5)
S
=.
< 1C 2' I an (x) -/(x) ! < e ¥ nand 0 < t -< 8.
... (6)
300
FOtJRIER SERIES
If t
;;;a 8, then Kn
· In\!J"
••
~
0
t
(I) " 211 sinl -(8j2Y
[}i(x+t)-/(x)+!(x-t)-!(x)]KtI{t)dt
II
i 1: [f(x+t)-/{x} I+I/(x-t)-f(x) IJ· ~2n~ si!2 (8/2) dl Now (4) declares that IOn (x)-f(x) I < '1
.By (6) and (7), lim
n-+oo
Ott
(x)-/(x),
... (7)
(C,
1).
18 Banach Space 18'1. Vector Space or (Linear Space). Let (F,+, .) be a field. Then a non-empty set V together with two binary operations vector addition and sca lar multiplication is called vector space o¥er tne field F if the following conditions are satisfied: . '(i) (V, +) is an abelian group, i.e.,
+ ),
(1) V is closed w.r.t. ( i.e., • 11- U, vE V=> u+v E V.
(2) A
u+O=O+u=u 'V- U E V. (4) Given any uE V,:I a unique vector -u E V S.t. u+( --u)=( -u)+u=O. (5) Addition is commutative, i.e.
u+v=v+u 11- u, VE V. (ii) 'I E V, a E F => au E V (scalar multipltication law). This law must satisfy the following conditions: (6) a(u+v)=au+av. (7) (a+b) ll=au+bu. (8) (ab) u=a (bu) (9) For the unity element I EF lu=u where II, V, W E Vand a, b E F. Instead of saying "V is a vector space over a field F", we always say "V(F) is a vector space." The elements of V are called vectors and clements of Fare called scalars. When F=R the field of real numbers, then VCR) is called real vectors space. Similaqy V(C) is called cpmplex vector space.
302
BANACH SPACE
18'2. Subspace. Let V(F) be a vector space. A non-empty subset Wof V is called a subspace of V if W itself is a vector space over F w.r 1. the operations of vector addition and scalar multiplication in V. The spaces to} and V(F) are called trivial subspaces or improper sub<;paces of V, others are said to be DOD-trivial or proper subspaces of V. 18'3. Linear sum or sum. Let U and W be subspaces of a vector space V(f). The sum of U and W is denoted by U+ Wand is defined as U+W={u+w: U E U, w E W}. 18'4. Direct sum. A vector space V (F) is said to be the direct sum of its subspaces U ahd W if every v E V is uniquely v=u+w where U E U, w E W. expressible as We denote direct sum of U and W by UEB W. Evidently V=UEBW';' V=U+W. Note that V=UEBW ~ V=U+W, Un W={O}. In this case U and Ware called complementary subspaces of V.
If U n W={Ol, then U and Ware called disjoint spaces. 18'5. Quotient space. Let W be a subspace of a vector space V (F). Let v E V be arbitrary. Then v+ W={v+w : w E W} is called a coset of Win V. Denote the collection of all cosets of W in V by VI W. Then VIW={v+W: v e V} V/W is a vector space over F w.r.t. vector addition and scalar mullipiication as follows: (u+ W)+(v+ W)=(u+v)+ w a(u+W)=au i-W \f u, v E V and a E F. This vector space V; IV is called Quotient space or Factor space of V by W. 18'6. Basis or Hamel basis. Any subset S of a vector space V{F) is said to be basis for V if S is linearly independent and vectors of S generate V i.e. L(S)= V. The idea of base or basis is due to German mathematiciar Hamel and so basis is also called Hamel basis. There exist more than one basis for the same vector spac1 V{F) but the humber of elements in each basis is the same. The number of elements in a basis of a finite dimension
303
BANACH SPACE
vector space is called dimension of V. In brief dimension V is denoted by dim V. 1S·7. Linear transformation. Any map T: X(K)-ioY(K) is called linear transformation or linear map or homomorphism if T(Oll+bl')=aTu+bTv V- a, b E K and ll, v E V. A linear map T: X(K)~X(K) is called linellr operation on X. A linear map T: X(K)-ioK is called linear funcHonal on X. A linear map T: X(K)~Y(K) is called isomorphism if T is one-one and onto. The sec of all linear maps from X into X is denoted by L(X, Y) and is a vector space over K w.r.t. vector addition and scalar multiplication. i.e.,
(Tl +T2) (1I)=T1u+T2u where
(aTl ) (u)=aT1u 1I E X.
V T 1• T2 E L(X, Y) ¥ a E K
A special case of L(X, Y), i.e., when Y= R. The clements of L(X, R) ar~ called linear functionals and is denoted by X*.
IS·8. Partially orllered set. A relation ~ defined in a set A with prorerties li~,ted bleow (i) refl ;xive, x ~ x (ii) antisymmetric. x ~ y, y ~ x '"" x=y (iii) transitive, x ~ y, y ~ Z:::? X ~ z ¥ x, y, Z E A is called a partial order in A. In this case the set A is called a partially ordered set and is denoted by (A, ~) or simply by A. Any two elements x, y in a partially ordered set A are said to be e()mpar~ble if either x ~ )' or y ~ x. If ev..:ry pair of elements of a partially ordered set is comparLlble, then A is called chain or linearly ordcrcil or totally ordered set. ]8'9. Minimal and maximal clement... Let (A, ~) be a 'p
x ~ a for every comparable element x E A. An clement b E A is called minimal element of A jf h ~ x • for every comparable element x in A.
304
BANACH SPACE
18'10. Supremum. Infimum. See def. in Chapter 4. 18'11. Zorn's Lemma. If S is a partially ordered set in which every totally ordered subset has an upper bound, then S has a maximal element. 18'12. Normed Jinear space. Let V(K) be a vet or space. For any u E V, we define a function (norm function) II II: V~R s.t. ¥ x, Y E V, we have (n I ) II x II ~ 0, (n 2 ) II x II =0 if x=O (na) 1\ (XX II = I ex I . II x II , O l E K (n 4 ) II x+y II <;; II x II + /I y /I • The pair (V, II II) is called normed linear space. Here K=C or R. The real number II x II is called norm of vector x. 18'13. Bounded map. Let X and Y be normed linear spaces. A linear map T: X-Y is said to bounded if there is an M s.t. /I Tx II ~ M II x !I ¥ x E: X, M > O. Theorem 1. Let N be a normed linear space. Let d be J function on N defined as d(x, y)= 1\ x-y 1\ ¥ x, YEN. Then dis a metric on N Proof. (I) x, YEN => x-y E N. Since II x-y II ;a. 0, by (n I ). Hence d(x, y) ~ O. (2) d(x, Y)= II X-Y II = II (-1) (y-x) II = I (-1) I .11 y-X II = II y-x U =d(y, x) :. d(x, y)=d(y, x) (3) d(x, z)= /I x-zlI=lI{x-y)+(y-z) II ~ II x-y II lIy-zll· by (114) =d(x, y)+d (y, z) or d(x, z) ~ d(x, y)+d(y, z). Thus d is a metric on N. Remark. Thus we have seen that a normed linear space can be treated as a metric space. Thus we have given a topological structure to normed linear space. Hence we can define closed sphere, limiting point etc. in a normed linear space. 18'14. Banach space. [Kanpur 1984]. Any sequence <Xn> in a normcd linear space (N, II II) is said tv converge to Xo E N, i.e., xn-+xo if given E > 0, 3 positive integer no S.t. n ;;? no ~ II Xn - Xo II < E. It follows that lim 0 'f n-+oo II xn-xo II = I xn~xo· A sequence <XII> in N is called Caucby sequeD~e if given • > 0, 11 positive integer 110 s.t. m, n ;;> no ~ II X",-x,. II < E.
+
305
BANACH SPACE
A normed linear space is said to be complete if every Cauchy sequence in N is convergent in N. A normed linear space which is complete as a metric space, is called a Banach Space or B space. [Kanpur M.Sc. F. 1983]
Examples. (i) The vector space C [0, I] with the norm II x II =max { I XCI) /, t E [0, I]} is a Banch space. (ii) The vector sp8.G,e C. of convergent sequence wiih the II x II =sup { I x-I : n=I, 2... } [Kanpur 1989] norm is a Banch space (iii) The vector space 11" p ~ I of sequ,ences for which
.!
I x.. \p <
00
with the norm II x II
=[ !)
x. 11'
J/P
is a Banach space.: Theorem 2. To show that (LP, d) is metric space. [KoJhapur 1968] Proof. See Theorem 9, chapter Il. IS·1S. Continuity. Let Nand M be norilled linear spaces. A map f: N~ M is said to be continuous at Xo E N iff given any • > 0, a 8 > 0 s.t. II X-~o II < 8 => IIf(x)-f(x.) II < t. ) The map f is said to be continuous at every point of N iff it is continuous at each point of N. We also say that f is continuous at Xo if for every sequence < x" > E N converging to xo. the sequence < f(x~) ~ E M conver~ea tof(xo) E M, i.e., if x" .. x o =:;,.'f(x..) -+ I(xo). IS·16. Bounded map. A linear map T: N-M is said to ~ bounded map if a a real number k ~ 0 s.t. II T (x) II <; k II x II 1.;1 x E N. 18·17. Isomorphism. Let Nand M be normed linear spaces and T: N~M be a linear map. Then T is called topological isomorphism if (i) T is one-one and onto (ii) T and T-l both are continuous maps. IS· IS. Isometric isomorphism. L~t Nand M be Donned linear spaces. A linear map T: N-M is called isometric isomorphism if T is on e-one 11 T(x) II = II x U ¥ x E N. and
306
BANA-CH SPACE
18'19. Grapb of a function. [Kanpur 1980) Let/: X- Y be map. Then graph ofa function lis defined as a set consisting of elements of the from (x. /(x» where x EX. Thu,> graph/={(x,f(x» : x F: X} C Xx Y as/Ix) E Y.
Symbolically we write graphj=/" 18'20. Closec! linear map. Let N and lV' be normed linear spaces. A linear map T: N • N' is said to be closed map if
< x n > eN. xn .... x.
T(x.)-y
~
x E N, where y= T(x). 18'21. Some elem~ntary defil1itions. (iJ Let (X, d) be'a metric space and A C X. A point Xo E X is called limit point or
limiting point or cluster point or accumulation point if every open sphere S~ (xo) contains points of A other than xo, i.e., (Sr (xo)-{xol) n A~ fl) S,(xo) = open sphere of radius r and centre Xo ={x EX: d(x. Xo) < r}. (ii) The set of all limit points of A is denoted by D (A) and is called derived set of A. (iii) A sequence < Xn > in a metric space (X. d) is called a Cauchy sequeuce if given c > 0, 3 positive integer no S.t. m, n ~
110 ~
d(x,., xm)
<
f.
A Cauchy sequence of real numbers is a convergent sequence. (iv) A point x of a metric space (X. d) is called an interior point of a set A C Y if 3 ope 1 sphere S, (x) s.t. S, (x) C A. ' The set of all interior points of A is denoted by A' or by set (AO). AU is always an open set AO C A. Also A is open do A=Ao. (v) Closure Let (X, dJ be a metric space and A C X. Then closure of A is denoted by Aand i" defined a3 A = n {F: F J A, F is closed, F C X} A is closed and A C A 'A is closed -<* A=A (vi) Adherent or contract point. A point x E X is called adherent point of A if every neighbourhood of x contains a point of A. Thus it is clear that if x is adherent point of A, then (i) x is a limit point or (ii) x is not a limit point of A but x E A.
BANACH SPACE
307
(vii) If x is not a limit point of A, then x is called isolated point of A. (viii) Den' e set. A set A is called somewhere dense set if int (::t)=(At:;t:0 i.e. if closure of A contains some open sets. Dense Subset. Let X be a normed linear space. A subset A of X is called dense subset of X if every Cauchy sequence in A converges to a point in ]{. 4 is called nowhere dense set or non-dense if (.4t=0. The space X is said to be Separable if 3 ACX st. A is countable s.t. A=X. (iv) Baire's Category Theory. Let (X, d) be a metric space and A C X. The set A is called of first. category if it can be expressed as a countable union of non-dense sets. The set A is called of the second category if it is not of the first category. Baire's Category Theorem. A complete space is a set of second category. Result. A closed subspace of a complete metric space is complete. 18'22. Jl"unctional Conjugate Space. Let N be a normed linear space over the field K of '>calars, where K=C or R represents respectiveiy the set of complex numbers or real numbers. Then any continuous linear mapf:N~K is called functional. The set of all such functionals is denoted by N*. The set N* is called conjugate space (or adjoint space or dual space) of N. The set (N*)*=N** is called second conjugate space of N. [Kanpur M.Sc. F. 1983] Convergence. Let Nand N' be norm~d linear spaces. Let B(N, N/) denote normed linear space of all bounded linear maps Tn: N ... N'. (i) Strong Convergence. The :.equence of B(N, N') s
is said to converge strongly to T, written a; Tn - - + T if ¥ u E N, Tn (u)~T(u) i.e., if given f > 0, :I no E N s.t. II Tn (u) - T (u) 1\ < Ii ¥ n ~ no· (ii) Uniform Convergence. The sequence Tn E B(N, N/) is said to converge uniformly t~ T if given fi > 0,:1 no EN s.t.1\ Tn--TII <_ ¥ n;;;;t no·
308
BANo\CH SPACE
Weak Convergence. < T" > E B (N, N') is said to . w converge weakly, written as T .. -~ T if ¢[T. (u)] _ I,6[T (u)] y 1,6 E N*, dual space of Nand ¥ u E N. It can be proved easily that uniform convergence 0 strong convergence .. weak conveJgence. 18'23. Projection. Def. A projection on a Banach space B is a linear map S.t. Ea=E and E is continuous map. Tbeorem 1. Let N be a normeo linear space and x, yEN. Then I II x II - II y II I ~ II x-y n . [Kanpur M.Sc. F. 1987.89] Proof. Evidently !I xII = II (x-y)+yU ~ Ilx-yll + Ilyll . as II u + v II EO; II u II + II v II or IIxll-lIy!I~lIx-YH ... (1) Interchanging x and y in (I) II y II -- II x n<; II }' -x II = II ( - J) (x - y) II =1-11 .lIx-YII·= nX-YIi or II y n - ux II <: II x- y II as UIIU II = I GI 1 • II u II or - ( II x II - "y II ) ~ II x -- y II or IIxll-IIYJl~--lIx-YIl ... (2) By (t)and (2), -llx-JyIl ~ IIxll-IIYII E;; Ux-YII or III x II - II ~ 11\ <; II x - y II [For -4 ~ a ~ 4 ~ I a 1 <; 4] Theore~ 2. Let N be a normed linear space witil norm II II. Then the mBp/: N-+R S.I.f(x)= II x II is continuous, ,hal is, the [Kanpur 1987; Meerut 87j norm II lion N is a continuous map. Proof. Let <XfI > be a sequence in N s.t. lim n_ 'Xl x.=xo ... (1 , Given/(x)= II x II ¥ x E N. Then I/(x,,)-/(xu) 1 = I II x,,11 - II Xo II I ~ II X,,- Xu II , by Theorem I. -+0 as n-+ oo , by (I). :. I I(x,,) -/(xo} 1 <: 0 as n __ «> But I I(x,,) -/(xo) Il( 0 is always true. :. 1Ilx.)- Ilxo} I = 0 as n .. 00. (iii)
This
lim /(XfI)--f(XO) n_ou By (1) and (2), we see that x,,· ... x• .. I(x,.)-+/txo) 0
... 12)
309
BANACH SPACE
This, by definition, proves that! is a continuous map. Lemma. If a ;;, 0, b ~ 0 and
J I -+-=I,I
then
ab ~
p
q
aP b -+. "" p q fl
Proof. If 0=0 or b=O, the lemma is obvious. So Jet us suppose that a > 0, b :> O. Define !(t)=(I-.l)+At-t A for 0 Then !'(t)=>'-A!).-l=A (l-t A- 1 ) Also 0 < A < I ;> I-A> 0
••• (1)
< A<
1.
... (2)
f'(t) = -A (A-1) t).-2 .f'(t)='\ (I >.) tA-I. ...(2a) /'(t)=O gives>. (1- 1).-])=0 or t=l as A < 1
or
Now At t=I.0 <,\ < I,J:(t)=positive. by (2a). This proves that the function has minimum value at t= 1. Consequently !(i) ~!(I)=O. or !(t}?;J0 or l-A+At-I).;;;'O .. (3) Write ,\== IIp. This is justifiable step as Jfp < 1 and also
A<1. Then (3) gives 1--
I t -+-
P P
~+~ P
or
_tl/'P -.
q
Put
?;J 0
_t 1 /P
0
as
,
Then Il 1 'Pb-_ _ + _a_ _ (apb-fl)l/p ;;;;t 0
I=apb-
q
p
Multiplying both sides by bf , btl aP -+-ahtl-f{P ?;J 0 q p
[ But
.!..+!...=IQ·i+l=q:> q- [=1
p
q
p
bq
or
a -+-ab?;J 0 q p
or
ab
P
P
btl -+_. "'" p q ~
aP
This completes the proof of the lemma.
J
I 1 -+-=1 q
p
310
BANACH SPACI:
Theorem 3. It states that if X=(Xt> X2 , Y., ... , y,,), 1 < p < 00, then
" '_1.'1 I x,y, I ~
or
•••••• ,
x,,) and Y===(Yl'
II x II" II Y I/ll
where the scalars x" Yi are real or complex. Proor. If x=o or y=O, then the inequality is obvious. So let us suppose that x¢O, ncO. Firstly we shall prove the following l e m m a . ' (H ,re state and prove the above lemma). Now we come to the proof of the main theorem. By the lemma, q &.. , . a,p+b. aw,~
I x. I
Put a.=1Xjj;'
.= -IIYI u.-I .
I x. 1·1 y. I II x II. Ii Y Ii.
Then
11
E
or
b
I x.y, I
/-1
1\ X
But n x
-q
p
y,
.... !.. ~ '<:
n
<;
II" II Y IIIl
E ~
P
+!_ ..!.1!.t
II x I\.p q
P
I x, Ip
i_I
II Y "
+!-.
E
Hv"
I Yi III
i_1
q TJ~-n;"-
II x I"P
II.=-{ '~ll X,IPF'P .. II x 111'21= i~ IXII"
In view of this the last gives,
or It states that if x-(Xtt P < eo, then
Theorem 4. Minkowskl's Ineqoality. XI' ... , x,,),
Y=(Yh Ya, ... ,
I: x+Y
)'n)
and I
<
II" < II x 1,+11 Y ill'
311
BANACH SPACE
i.e.,
" I' }1/" [ ,2:'" I x,+y, Ip ]1/f1 ~~ { E" I X, jP }lll> { Ely, '-1 .=1 '-1
whue Xl, Y. are scalars (real or complex). Proor. Step I. State and prove the Theorem 3.
Step II. II x+y or
11,=
..
{ 1:" I x.+y, II' }I/II
II x+y 11,11= '_E1 I x'+Y.
1_1
, I x,+y, I . I x,+y, IP-l
IP=E
Applying Step I to R.H.S., \I x+y Il.p ~ p; I x, IP}I/p {E
•
I x.+y, l(p/v).9}l/'. • +{E I y.lfI}I/P C1: I x,+y,1
,
(11/9)
"J l /1
~ II X /I, {II x+y /I,P}'j<1+11 y III' {II x+y II.'P/II =(11 x 1/,,+11 y /I,,) (II x+y 1/.)1'/11 by nx+y /I.P/Q and noting that
:. II x+y Dividing
.
11"i>
we <>et P - q~= 1 as 1+E-=p, q -
:. II x+y III' ~ II X /1,+11 y II,,· Problem 1. ShoH! that the /inear spaces R" and en of all ntuples x=(xtt X 2 • .... x .. ) of real and complex numbers are Banach spaces under the norm [Meerut 1908] (These spaces are called Euclidean and unitary spaces respectively). Solution. (i) !I x " ~ O. For each I x, I ~ 0 (ii) II x 11=0 ~ x=O.
312
BANACH SPACE
For II x
1/=0
<:> {
~ 1 xd 2 }i/l =0 {;:> E
t_l '
1 Xi 12=:0
i
<:> Xt=O ¥ i <:> X-(X 1, Xli' ... , x,,)=CJ. (iii) II x+ y II .;; " x II + II y II • This follows from Minkowsi inequality for p=2. Hence CIt and R" are normed lineal" spaces. These spaces are metric spaces w.r.t. the metric d defined by d (x, Y)=II x _.y II To prove the Completeness of en (or R"). Suppose < x" > is a Cauchy sequence in Crt or ~". Since each x" is an n tuple of complex {or real) numbers and hence
x m=(Xm l'
X mz , .. ,
x m.. ).
This -> Xm~ is kth co-ordinate of X m . By definition of Cauchy sequence given E > 0,3 no € N s.t. m, I ~ no => II X m - XI II < 6
=> II
Xm - XI
112
<
.
t
2 -> E 1 Xm,-Xu 12 >=1
<
.2
=> I Xml-X" pI < ~2 ¥ i ~ 1 Xm,--XU I < 6 ¥ i. This proves that the sequence < Xmi> is a Cauchy sequence \f i. But C (or R) is complete. Hence every sequence < Xml > . z, In . C (R) or so t h at Lim Xml=Z' ¥ l.. converges to a pomt m-.C>o
Consequently every Cauchy sequence < x,,> converges to a point z=(zt> Z2' .•• , z .. ) E en (or R"I. Hence Cn and R" are complete spaces and consequently they are Banach spaces. Similar Problem. Define a normed linear space. Prove /fl the linear space of aU sequences x={x1 , X 2 , .•• , x" ... }, of scalars such that 00
E _1
I Xn IV <
00
with norm defined by
is a complete normed linear ~pacL. [Kanpur 1988, 85J Tbeorem 5. Let M be a closed linear subspace of a nornled linear space N. If norm of a coset x + M in the quotient space is defined as "x+M II=inf{" x+m II: m E M} then NIM is a normed linear space. Further if N is a Banach space, then so is NIM. [Kanpur M.Sc. F. 198Z, 81, 79 ; Meerut 82 ; Buodelkbaod 83, 82 ; Madras 79; Amritsar 82J
313
BANACH SPACE
Proo[ Let x, yEN and m, nt' E M be arbitrary. We know that NjM is a linear space w.r.t. the operations
(x+M)+(y+M)=(:c+ y)+M and ~ (x+M)=lXx+M where IX is a scalar. M is the zero element of Nj M.
(I) ... (2)
...(3)
Aho
x+M=M <=> x E M. ...(4) Now we verify all the postulates for a norm. (i) II x+MII ~ O. For" x+m II ~ 0 and every set of non-negative real numbers is bounded below and so inf { II x+m II: m E M} exists and is non-negative. (ii) To prove II x+M II =0 <=> x+ M=M=zero element of NIM. Observe thatx+M=M => x E M. This suggests that IIx+MI\=inf{l!x-lmll:x,mE M~ =inf { II y II : y=x+m E M} =inf { II y II : y E M)=O. [Being l)ubspace, M cnnt~ ins zero vector whose norm is real number 0]. ...(5) Finally x+ M=M => II x+M /1=0. Conversely suppose that II x+M 11=0. :. inf{1I x+m II: Tn E MII=-O. This => 3 a sequence <mn> E M where n= I, 2, .. S.t. Lim "x+m n II =0. n-+ 00 Lim 171,,= -x. Therefore 11-+00
But M is closed and <mn> is a convergent sequence in M converging to -x. -x E M. This => x E Mas M i~ slI:bspace Thus" x+M II =0 => X EM=> x +-M=M. .. (6) From (5) and (6), the result (ii) follows. (iii) To prove II (x+M)+(y+M) II ~ 11 x+ Mil + II y+M II. By (\), II (x+M)+ m, m' E M as Mis subspace] =inf { II (x+m)+(y+m') II : m, 1/1' E M} ~inf { II x+m II + II y+m' II : m, m' c: ,\1} [For II x+ y II ~ U.'( II + II y II in NJ
314
BANACH SPACE
=inf {il x+m Ii: m E M} +inf {II y+m' II: m' E MJ
=/1
x+M II
+ II y+M
/.
Hence the result (iii). {iv) To prove liCIt (x+M) II = I or. I • /I x+M \I. II IX (x+ M) II = inf { II or. (Xi m) : m E M} =inf { I IX I . II x+m II : m E M} [For IIClxu=lat/.l/xll = I III I . inf {II x+m II: m E M} = I CltI·ll x-r M II· From what has been done it follQws that N/M is a norm~d linear space. Further suppose that N is complete (or Banach space). To prove that N/M is complete (or Banach space). Here we make use of the fact that a Cau:hy sequence is convergent iff it has a cOll·vergent sub~equence Let <s,,+M> be aCauchy:\cquence in NIM. Takee=l, by def of Cauchy sequence. given E= i > 0 3 ni e N s.t. I, m ~ "1 =- II (s,+MI- (s.n+M) II < E=i => iI (s +M)-(s",+M) II < •. "1
Write
S
=x1• Then !I nl
ni
(Xl +M)-(sm+M) II
Similarly for 1=(1)2, 3 a positive integer n2 + M)-(sm+M) II < Ul for m > n 2 •
S.t. II (s
II.,
Set sn 2 ;"'''2' then
11
(x 2 +M)-(s... +M) II
< (nl .
Gene"aling this concept, we have positive integers " 12 > n l s.t. II (x:. t-M) (s",+ M) II < (f)" where x,.=s and 111 > nk • • 11k
n.> ... >
> na >
Finally we have obtained a subsequence <x.. +M> of 5"quence <s•• + M> in N{M 5 t Hence II (XHl +M) -(Xk+M) II < (J)k. . .. (7) Remains to show that this subsequence convt:rges to a point in N!M We chooo;e a vector YI E x I M. Af.er choosing Yl' we select
+
Y2 E
x 2 +M s.t.
,
\I )'1-Y2 II < J. We select Y8 E xa+M 5.t. II )'2- Y3 H < (i)I. Continuing in this manner, we get a sequence in N S.t. \I y,,- Y"+t II < n)" for n= I, 2, 3,... ...(8) This assumption is possible due to (7).
315
bANACH SPACE
We claim < l'n> is a Cauchy s~quence. Let ~ > 0 and choose a po~itive integer I
2,,0--1 <
e S.t.
n, m
11~ E
N so large that
~ no
nYm-Y .. /1=11 (Y"'--Y"'+I)-!-()'m+l-Ym+2)+-··+Cv .. l-Y") II ~ II y".- Ym+l iI ... II Y"'+l- )'mH !i+ ... + 1Y"-l-)' .. 1I < 0)'" H!)"Hl+··.+(i)"-l, by (8), < (t)"'·t(Hm+1+ .. ·+0)n-L t- ...... _ (V'" _ I I -1-1~2m-=t < iilO=l< e < y .. > is a Cauchy sequence. But N is complete. lim lim :. 3 yEN S.t. y,.= y. Consequently x,,= y. 11_<'..0 ll--'?OO
Hence x ..+M ·y+M E NIM. Thus < sn+M > is a Cauchy sequence in NIM v.hich has a CO'lvergent subsequence converging to a point y+M E NIM. Consequently the Cauch sequence <5 n +lvl> converges to a point in Normed linear space NIM, meaning thereby NIM is complete Consequently NIM is complete. Problem 2. Let C(X) dCllote the linear .~pace of all bounded continuolls scalar l'alued functions defined 0" a topological space X. Show that C(X) is a Banach spact! under the nurm i~fll =sup {If(x) I : x EX}, whuefE (X). [hombay 1976; Meerut 79] Solution. Let c(x)={f: f is l. continuous and bounded map from Xto X} where X is a topological space. Let.t~ gEe (X) are arbitrary. Define ilfli =Sup { If\x/ I : x EX}. It is easy to verify that C(X) is a linear space over operations defined by (f+g) (x)=f(x)+g (x) (vector addition) «(1f) (x)=~f (x) (scalar multiplication) To show that OX) is a normed linear space. (i) Ilfll ~ O. For I fix) I ;;a 0 >,f x (ii) IIfll =0 => 1=0 (zero function) For II III =0 ~ sup { I f,x) I : x E X}=O <:> II(x) I ~- 0 V x. <:> f(x)=O
V x <:> f=
.
3I 6
BANACH SPA(E
+ IIgll For II/+g II =sup { I (!+g) (x) I : x E X} =sup { ! lex) +g(x) I : x E X} ..;; sup { I f(x) I + I g(x) I : x E X} For I a+b I ~ I a I + I b I =sup { If(x) I: x E X}+sup {I g(x) I :x E X} = II f II + II g ". Hence the result (iii). (iv) II "fll = IIX I .lIfll· For II afil =sup { I (IX!) (x) I : x E X} =sup { I a.f(x) I : x E X} =,sup { I Cl I . I fix: I : x E X} == I Cl I .sup i I f{x) I : x E X} (iii) II/+gl1 ~ \1/11
=
IIX I
.lIfll·
Hence C(X) is a normed linear space. Also C(X) is a metric space w.r.t. induced metric d s.t. d (f, g)= IIf-g" =sup { I /(x)-g(x) I : x EX}. Remains to prove that C(X) is complete as a metric space. Let be a Cauchy sequence in C(X). Then each fn : X -x is a continuous map. By def. of Cauchy sl'quence. ~ d (f, .. f,,,) < • given < > 0; 3 nil E N ~.t. tn, n ~ .:> IIf.. -f", II < E ~ sup { If,,(x)-fn, (X) I: x E X} < • ~ Ifn(xJ-fm(x) I ~ sup I !fn(.A)-f",(x) I: x E X}<_ -:> I f,.(x)- fm(x) I < E ¥ X E X. This proves that is a uniformly convergent sequence of continuous functions.
"0
Hence <[II> must COl verge to a continuous Illap f on X. Accordingly CLX) is a complete normed linear space and hence a Banach space, Problem 2a Pro!'e that th~ conjugate spact! 01 a normed linear I Kanpur M.Sc. F. 1981] space is a lJanach spare. Solution. Let N be a normed linear space over that field K of scalars. Here K =C or C. Also Rand C both are Banach spaces. By def. N*, the conjugate space of N is the set of linear and continuous functions from N into K, i.e., N*={f:f N-K is a linear continuous map}. Replacing C(X) by N* in the above Problem 2, we get the proof of present problem.
317
BANACH SPACE
Problem 2b. If Nand N' are normed linear spaces, then the set B (N, N') of al/ continuous (or bounded) /inepr transformations of Minto N' is itself a normed linear space w.r.t. pointwise linear operations and the norm is defined by IITII=sup{IIT(x)II:lIxll~ I}.
FurTher space.
if N' is a Banach space, then B (N, N'),is also a Banach [Jiwaji 1983; Kanpur 86, 91; Meerut 88, 87]
Solution. To prove that B (N, N) is a linear space. Let S be the set of linear transformations from a normed linear space N into a normed linear space N'. Let T 1 , T2 E S and a, J3 E K and x E N. Then S is a linear space w.r.t. the linear operations (T1 +T2 ) (x)=Tt (x)+T2 (x)
(GtT1) (x) =«T] (x)
and
Here K is a field of scalars. Let T], T2 E B (N, N'), then Tl and 1'2 are continnous linea.r m'tps from N into N'. Then 3 kl' k2 > 0 s.t. il T](x) II ~ kl II x II , II T2(x: II ~ k2 II x II [For continuity do boundcdness] Tl and T2 are linear mars => (1.T1 +.81'2 is a linear map.
" (a.T1 +.BT,) (x) "
= II
a.T1 (x)+fiTs (x) II
+ II () T2 (x) II I Gt I . II T1(x) I! + I .B I . " T 2(x) II I Gt I .kL " x Ii + I Gt I .k? \I x II
<; II «Tl (x) II =
~
or
"(GtT1 +()T2 ) (x)" ~ [ This => crT1 +()T2 is bounded =>
I a. I kl+ I ~ I .k~]
II x II
GtTI +~ T2 is continuous map
=> CltT1 +fiT2 E B (N, N'J. E B (N, N') => «T1 +.8T2 E B (N, N'). This proves that B (N. N') is a subspace of a linear space S. Thu~ T 1 , 1~
Hence B (N, N') is a linear space. II. B (N, N) is a normed linear space. Let T 1 • T2 E B (N. N') and x E N S.t. II x II
<
I.
(i) :I T) " ~ O. For \I TI II =sup { 1/ T1(x)!t: II x II ~ J} and " T1(:c) /I ~ 0, by def of norm.
318
BANACH SPACE
II Tl II =0 ~ Tt=O II Tl II -0 <'> SUp { II T1(x) II : II X /I ~ I}=O
(ii)
<'> <::>
II T1(x) T1=0
1/
=0
¥
X
<'>
T1(x)=O
¥ X
+ " T2 "
(iii)
II Tl + T2 II ~ II Tl II
For
"T1 +T2 11 =sup {" T t (x)+T2 (x) II: II X II ~.1} ~ sup { 0 Tlx) II 1/ T 2(x) II : II x jI ~ I} [For I a+b I ~ I a i Ib I] =sup { !i T1(x) " : " x II I} +sup { 1\ T 2(x) n : II x It ~ I}
+
<
+
="
Tlli + II T211 (iv) "rxTt /I = I at I . II Tl " For "rxTt " =sup { II (ceTl) (x) 11 : II x II ~ 1} . = , at I .sup { II TJ(x) II : II x II <;; l}= I II I . \I Tl n From what has been done, it follows that B (N, N ) is a normed linear space. fII. Remains to prove that B (N, N') is a complete space, where lV'· is a Banach space and so it is complete as a metric space. Let < 7~ > b~ a Cauchy sequence in B (N, N,). Then Tn E B (N, N') Tn N-+N' is a continuous linear may "f n. By def. of Cauchy sequence, II T,. - Tm ,,-+0 as n, m-+oo ... (1) For every x E N, T..(x) E N' ¥ n
*
:. <
=
"(T.. --T,,,)(x) " ~" T.. -Tm II. !! x II -+0 as m, n-+ oo , by (1) T,,(x) > is a Cauchy sequence in N' which is complete.
Also" T..(x)-Tm(x) Ii
Hence:l T(x) E ,v' s.t.
lim T..(x) = lex).
n-+oo
...
This defines a map T: N-+N'. Now we shall show that T is continuous and linear. lim x, YEN, then T(lIx+Py)= T.. (otX+~y) IJ-+OO
=atT.(x)+~T..(y)
:.
as Tn is iinear =zT(x)+,9T(y), by (2). T is linear.
(2)
If
319
BANACH SPACE
I a-b I ~ I a Hence I (II Tn II - 1\
Since
or
I
ITm
Ib I 1\) I ~ II
T .. -··T,,, II -
0 as m, n-+ oo , by (I)
II - II Tm II I -0 as m, n -+00. This ~ < II Tn II > i .. a Cauchy sequence of real numbers => < II Tn II > is convergent sequence ~ < II Tn !I > is a bounded sequence => 3 K > 0 S.t. Ii Tn II ~ K y. n. .. (3) 1\ Tn
Consequently, II T(x) II = 111im T,,(x) :1 =lim II T"(x) II ~ lim II Tn II . II X 1\ ~ k II X 1\ by (3) or II T(x) II K. 1\ X II This => T is bounded => T is continuou~. As < T,,(x) > is a convergent sequence with limit T(x) and so by def. II Tn(x)- T(x) 11 < f l.,f n ~ no ... (4) II T-T" II =sup. (II (T-T,,) (X\ II: II x II ~ I} < 6, by (4) Thus II T - Tn II < 6 "I n ~ no
<
This =>
lim T.. =T n-+-oo ... (5) T is linear and continuous ~ TE B (N, N') lim , => T.. EB(N, N). by (5) n-+oo Thus B(N. N') is complete and ~o that it is a Banach space. Problem 3. Let N be a IIOIl-zero lIormed linear space and M={x : XE Nand" x II = I}. Then prm'e that N is a Banach space ~ M is complete. [Kanpur 1991,89, 83 ; Meerut 79) Solution. Let N be a Banach space. To prove thar M is complete. N is a Banach space => N is complete as a metric space. Let < x" _ be a Cauchy sequence in M, then II x .. II =1 'rll ..• (1) < x" > is a Cauchy iicquence in MeN => < x .. > is a Cauchy sequence in N. Also N is complete => 3 x E N S.t. X" "x. But norm is a continuous function. :. II x" II -+ II x II lim or II oX II = n ... oo II x" II = I , by (1)
320
BANACH SPACE
II x!l =1, XEN By def. of M, thIs ~ XE IH. This Cauchy sequence < x .. > in M converges to x E M. Hence M is complete. Conversely. Suppose that M is complete. To prove that N is a Banach ~pace. Since N is a normed linear space and so it is enough to prove that N is complete. Let < x .. > be a Cauchy sequence in N. Then, by def.
or
lim
m, n-oo II X,,-X m II =0 But I II x .. II - II Xm II I ~ II X,,-X m II _ lim m, n-+oo III x" II - II Xm 111=0.
... (1) O~
by (1).
This:? < :I x" iI > is a Cauchy sequence. Thus < II X .. II > is a Cauchy sequence of real numbers. Also R is complete. Hence 3 « e R s.t. lim n_oo II x" II =ot ... (2) . y,,=-x" W fIte II x.1I
"f
II Y.. - Ym II = 1\ II;: II - \I =
~ "'"
... )
~: II Ii
x" lI.xx"') (Xm Xm) III II (1ix..T!.. 1I + IiX:f - II Xm \I I nx .. -Xm II + nx~JUL~~ II -/I Xm 1111 II x.. II
~ II X"-X,,, II "'" II x .. II
or
(3\
n
+ 11
D x" II • II Xm
X,,-X m
II x,,11
II 11 by theorem 1 '
II ~ 211 X"-X m II • Ym '<::: II x" It Making n, m-+oo and noting (I) and (2), 11 Y .. -Ym II ~ 0 as m, n~ao. But II Y.. -Ym n ~ O. Hence. lim m, n_oo II Y.. -Ym II =0. II
.V..
-
This:?
<
y ..
> is a Cauchy sequence in
II )'.. ~lln
~: II 11= ::~: : = 1
or
~ve
get
N
II y .. II = 1.
'" (4)
BANACH SPACE
321
=-
This y .. E M. Finally, < Yn > is a Cauchy sequence in M which is given lim to be complete. Henee 3 y E M s.t. y,.=y. Then
n-oo
y,.~ y
=-
x,.
\I x,. II -
Y ~
x,.
-+ IX}"
by (2)
=- nlim x,.=acy .... oo Y E M,
IX
is a scalar, M is linear space
~
••• (S) aty E MCN=-a.yEN.
Now (5) proves that the Cauchy sequence < x. > in N converges to N at a point exy E N, so that N is complete and so N is a Banach space. Theorem 6. Let Nand N' be normed line'lr spaces and T a linear transformation of N into N ' . Then the following conditions are equivalent to one another: (i) T is continuous (ii) T is continuous at the origin [i.e., xn-+O .. T(x,,)-+O] (iii) T is bounded [i.e., 3 a real number k ~ 0 s.I. nT(x) II ~ k II x II \f x E N] (iv) If S={x : II x II -I} is closed sphere in N, then its image T(S) is bounded set in N'. [Meerut 1988; Kanpur89, 85; Jiwaji 83) Proof I. (i) ~ (ii).
Firstly we prove (i) ~ (ii). Let T: N-+N' be a continuous map and sequence in N s.t.
lim
n-+oo By continuity of T,
<
XII
> be a
xn=O.
xn-+O ~ T(x ft )-+ T(O). But T(O)=O for any linear map :? T(xlI)-+O.
This proves that T is continuous at origin. Now we claim (ii) ~ (i). Let T be continuous at the origin. To prove (i). . N S.t. lim X,,=X E N • L et < X,,:> b e a sequence 10 n_oo Then
x,.~x
=> x.-x-+O:? T(x,.-x)-+T(O)=O. [For T is continuous at origin]
322
BANACH SPACE;
=> T(x,,) T(x)--+O =>
T(x,.) .... T(x)
Thus X,,-+x ~ T(x .. ) , T(x). By def. this proves that T is continuous. II. To prove that (ii) ~ (iii). To prove (ii) => (iii). Let T be continuous at x=O. To prove (iii). Suppose not. Then T is not bounded. Then for each positive integer n, we can find vector x_EN S.t. II T(x.) II > n II x" II ... (1)
· T a k log Yn= IIx"
n x" II x.. Also II y .. II = n II x,.
II
• •• as
' we get
'
UT(y,,) II
III'I = nII \Ixx....II
1\
> 1.
...(2)
=,;'1
lim II y" II =0 n-+oo
.. (3)
~=O, CX>
As norm is continuous function and so lim n-+oo
II T(y,.) II > I,
lim n-+oo
But
by (2).
Finally, y ..-+O and T(y .. ) does not tend to zero as n-HLJ. A contradiction. For T is continuous at x=O. Hence (iii) is true. To prove (iii) => (ii). Let l' b! bound\!d so that :I a real number k ;;;;t 0 S.t. II T(x} II ~ k II x 1\ V x E N. ...(4) Lim Let < x .. > be a. sequen<::e in N S.t. n ...oo x.. =O E N. This => II x .. II -+ 0 as n "00. For norm is a continuJus map. By (4), \I T(x,,) II ~ k \I x,./I. Using (5), we find that \I T(x .. ) \I _ 0 as n -+00. This ~ T(x .. )_O as n -+00. This shows that T is continuous at the origin.
...... (5)
BANACH SPACE
323
III. (iii) <:> (iv). To prove (iii) => (iv). Let (iii) be true, i.e., 3 k
II T(x) (6) => II T(x) II
1/ ,
~ 0 S.t. k II x II V x € N
... (6) . .. (7)
~ k II x II 'VX ESC N. But XES => T(x) E T(S). Hence (7) proves that T(S) is b.lunded.
To prove (iv) => (iii). Let (iv) be true, then 3 k ~ 0 s.t. 1\ T(x) II ~ k II x II 'V x € S. . .. (8) If x=O, then T(x)=O so that (8) is true for x=o. If x*O, then put
Y=1fXj so that
I/y II
II x 1/ ==fiTi=I.
Consequently yES and so by (8), or or
. .. (9)
x
1\
T(y) 11 E:;; k II y
n
~ II ) 11, k j II/ ~ II 11=k : ~ :: =k
" T ( II
T(x) 1/ ~ k 1/ x II 'V x~O. By (9) and (10), II T(x) n , k II x II ¥ X E N. This => T is bounded. Hence (iii). 1\
. .. (10)
This completes the proof. Problem 3a. Let X and Y be normed linear spaces. Show that a linear trans/ormation T on X into Y is bounded iff T is continuous at a point Xo E X. [Kanpur 1979] Solution I. Let T: X -+ Y be a linear map s.t. T is bounded. To prove that T is continuous. By assumption, 3 k>O s.t. II T(x) II E:;;k II x II ¥ X E X. . .•(1) Let < x .. > be a sequence in X s.t. lim Xn=X EN. <x,
n--? 00
As norm is continuous function and so Lim n-oo II x" 11=11 xii. This
=>
Lim n~oo
II x .. -x II =0.
By (1), II T(x. -x) II ~ k nx.-x II. Making n--? 00 and using (2), we get
••• (2)
324
BANACH SPACE
But Lim
Finally,
n"'OO Lim
or
n-400
Thus, Xn-4X =>
II T(xn-x) II II T(x.. -x) II
~
II T(x .. -x)
1=0
0 as n-..oo. 0 y. n.
~
T(xn)=T(x).
Lim 1J~00
T(xn)= T(x).
By def. this proves that T is a continuou~ map. Let T be a continuous linear map. To prove that T is bounded. Suppose not. Then T is not bounded so that for every positive integer n, we can find a vector x .. E X s.t. II T(x .. ) II > n II Xn II I or n \I Xn II II T(x n ) II > I. .. (3)
II.
Write .Also
)'n=
II
-~~--, then (3) ~ !I T(Yn) II > I
n II
)'n
XII
I=
h
y. n
... (4)
II n IIx..x .. II !\ I= n IIII x"x" IIII =n1
Lim I 11)'.. 1/=··---=0. n .. oo 00 . . Lim Yn= 0. A s norm IS contmuous,
. .. (5)
n~oo
> I, according to (4). n-4 co Finally, Y1l-'0 => T(y .. } does not tend to zero as n-4OO. By (4), Lim T(y .. )
A contradiction. For Tis continuou'l at x =0. Consequently T is bounded. Problem 4. Suppose that a Banach space B is the direct sum of the linear suhspaces M and N, so that B=MffiN and let z=x+y be the unique expression of a vector z in 8 as the SUm of vectors x and y in M and N. Shf)w that a new norm can be defined on the linear space B by II z II' = II x II + II y II and verify that it is actually a norm. If 8' symbolizes the linear space equipped with this norm, prOl'e that B' is a Banach -space if M and N are closed in R [Meerut 1988] Solution. Note that Band B are same sets. (B, /I II) is a Banach space. Let B= M(fJ V and let M and N be closed as a
325
BANACH SPACE
metric space in D. Then z=x+y with x e M, yeN is unique representation of z. Let II Z 1'= II x II II Y n ... (1) To prove tltat (B', II 11') is a Banach space, where D' =B.
+
Let
Z, Zl' Z2
E D', then
z=x+Y, ZI==XI+Ylt z.=X2+Y2 where x, Xl' X 2 E M and y, Yl> Ya E N. Then XI +X2 EM, YI+Ya eN 'is M and N are subspaces. Let CIt be a scalar.
II Z 11 ';;:' O. For II X n .11 Y H ~O ~ II Z II' ~ 0, by (1) II Z 11'=0 ~ z=O. For 11 zl\'=O <:> II X II + II Y II =0 ~ II x II =0= II Y II ~ x=O=y <=> z=x+y=O+O=O. (iii) II ZI+Z2 11'=11 (XI+ yJ+(x2+y.) 1I'=:I(xl+xJ+(YI+YJII' = II Xl +X 2 II + II Yt + Y211 , by (I) ~ II XIII + II XI II + II Yl II + II Yz \I = ( II X~ II + II YI II ) +( II XI II + /I Y2 II ) = II Zl II' + II Z. II'
(i)
(ii)
or
IIzI+z211'~lIzllI'+lIz2il'
(iv) \I rl.Z 11'= I at 1·11 Z II . For CXZ=I1(X+Y)=llx+atY :. II atZ 1\ ' = II atX II + II atY II =
I 11 I .( II X II
= I at I . II X II + I at I . II Y II + II Y (1)= I II I . II Z \I '
(v) From what has been done it follows that (D', II II') is a normed linear space and is also metric !.pace w.r t. the induced metric d defined by d(Zl' zz)= II Zl-ZZ II '. To prove that D' is complete. M and N are closed subspaces of complete space ~ M and N are complete, Let be a Cauchy sequence in 8'.
z.. =X.. +Yn with x" E M. Yn E N
Then
Zn Zm=(Xn +Yn)-(x":+ y",) =(xn-x m )+ (y"-y,,,). By def. of Cauchy sequence, :I no E N s.t.
m, n This
~
~
"0
II Z"-z,,, It '
<
...(2)
I.
+
II X"-X,,, II II y,,-y ... 11 < E, by (1) and (2) ::> II XII-X", 1\ <" ,/2 and II Yn - Ym II <6/2 .
0
¥
Hence
<
XII
>
and
<
y.
>
n, m ;:;, no
are Cauchy sequences in M and
326
BANACH SPACe
N respectively which are given to be complete. Hence 3 p E M and q E N s.t. X,,-+p E M, y,.-+q E N This=> x,,+y,,-+p+q => z-+p+q E B'. Thus every Cauchy sequence < z,. > in B' converges to p+q e B'. Hence B' is complete. Consequently it is a Banach space. Problem 5. If M is a closed linear subspace of a normed linear Ipace Nand T : N- NI M is a natural map S.t. T(x)=x+M. Show that T is continuous and II T II ~ I. [K'anpur 1994488, 87; Meerut 80; Amritsar 82] Solation. Let M be a closed subspace of normed linear space N, then NIM is a normed linear space with the norm defined by ... (1) II x+M II =inf { II x+m /I : m E M} NIM is a linear space w.r.t. the operations ... (2) (x+M)+(y+M)a::(x+y)+M ... (3) a(x+M)=ax+M wnere x, yEN and a. b ~re scalars. by def. of T T(ax+by)=(ax+bvY+M, by (2) -=(ax+ .M')+(by+ M) ==a(x+M)+b(y+M) by (3) -aT (xH bT (y). This => T is linear. n T(x) II = IIx+MII cooinf{ IIx+m II: m € M} <;; II x+mll Thus II T(x) II ~ II x+m II ¥ m E M. ...(4) In particular for m=O, we get ... (5) II T(x) II ~ II or II T(x) II ~ I. II x II "I x E N By def. this proves that T is bounded and so T is linear, by problem (30.). Second part. " T" = sup { II T(x) II : x E N, II x II ~ I} ~ sup { 1\ x II: n x II < I}, by (5) =1. or IITII~ I ..
x"
Problem 6. Let N(K) and N' (K) be normed linear spaces. Suppose that T: N-+N' is linear map. Theil ker (T), null space of T. is linear manifold and ker (T) is closed if T is conti"uous.
BANACH SPACE
Soultion. By def. ker (T)==N(T)={x eN: T(x)-O} Observe that 0 E Nand T(O)=O :. 0 E ker (T). Let x, y E ker (T) and a, b E K. then T (x)=fl= T(y) T (ax+by)=aT(x)+bT(y) as T is Iienar =aO+bO=O. :. a~+by E ker (T). Finally, x. y € ker (T) => ax+ by € ker (T).
327 ... (1) . .. (2)
...(3)
By (2) and \3). it follows that ker (T) is linear manifold. Sceond Part. Let T be continuous map. By def. of inverse of an element, T-l[{O}]=ker (1 ) N' is a nonned linear space and is a metric space w.r.t. induced d(x', y')= II x'-y'li 'of x', y' E N' metric d s.t. In every metric space !>ingleton set is a closed set and so {OJ is a closed set in N'. Also inverse image of a' closed set under a continuous map is a closed set. Hence T-l [{O}] is closed and therefore ker (T) is closed. Porblem 7. If T Is cotinuous linea, transformation on a normed linear space N Into normed lmear shace N' and if M is its null space. show that T induces a natural linear map T': NIM_N' and II T' II-II Til [Meerut 198Z] Solution. Suppose T: N ... N' is a continuous linear map. where Nand N' are normed linear spaces ovt:r the same field K. Let ker (T)=M then M is a closed linear subspace of N. , (Refer Problem 6). Hence NIM is a normed linear subspace w.r.t. the norm defined by
II x+MII =inf{" x+m II: me M}. We define a map T' : NIM~N' st.
...(1)
T' (x+M)=T(x) ... (2) Let x+M, y+M E NIM and a, be K. T' [a(x+MJ+b (y+M)]=T'[(ax+M)+(by+M)1 = T'[(ax +by)+M]= T (ax+by) =aT( x) +bT(y) as T is linear =aT'(x+M)+bT' (y+M) by def. of T'. This proves that T' is linear. By (1).11 x+M II -inf { II x+m II : In E M} ~ II x+m II •
328
BANACH SPACE
In pArticular for m=O E M. II x+M II ~ " x II V x E N (3) " T' II =sup { II T' (x+M) II : II x+M II ~ I ¥ x EN}. Using def. of T', we get II T' II =sup { II T(x) II : II x+ Mil <; 1 V x E N} or 1/ T' II =sup t " T(x) II: II x+m II ~ 1, x E N, m E M} ... (4) x E N, m EM=> x, mEN as MeN ~ x+m E N II T 1/ =sup { II T(x) Ij : 11 x II .;; I; x E N} =sup{IIT(x+m) u:nx+mll~ 1,xE N,mE M}. But T(x+m)=T(x)+T(m)=T(x)+O=T(x as m EM=> T(m)=O /I Til =sup { II T(x) II: II x+m II .;; 1, m EM}. . .. (5) By (4) and (5), II T' II = II TO. Theorem 7. Let Nand N' be normed linear spaces and let T: N-N' be any linear map. If N is finite dimensional then Tis ~ontinuous map (or bounded). [Kanpur 1989; Meerut 76, 73] Solution. Let dim N=n and {elf e2 , •• , en} be a basis for N. Then and x E N is uniqu¢lY expressible as
..
X= 1: at e" where aI' a2,00., a. are scalars. 1-1
It can be proved that II x II norm is called the zeroth norm. ,.
II T(x) II
=
0=
max I at I is a norm of N. This
II
II E a, T(e,) II
~ E
i-l
i_l
I a,
I . II
Te, II
,. ~ 1:
1_1
II x II o· II Te, II
..
Since the basis is fixed and so 1: U Te, II is a positive constant . • _1
II
Let k==l:
i-I
1/ Tel
II . Then /I T(x) II
~ k II x H 0
This => T is bounded => T is continuous. (Refer Problem 3a). Theorem 8. (Hahn-Banach Theorem). Let M be a linear sub8pace 0/ a normed linear space N. LeI f be a linear functional on
BANACH SPACE
M. Thenf can be extended to a function F defined on the whole spare N s.t. II FII = ilfll. [Meerut 1988,82; Kanpur 82, 78, 79; Indore 79J Proof. We first prove the following lemma. Lemma. Let M be a linear subspace of normed linear space Nand Xo E Nand Xo $. M. Letf be a functional on M. If Mo=(MU{xo})={x-t-ClXo : x E M, 1& is real} is the linear subspace generated by M and Xo. thenf can be extended to a functionalfo defined on Mo such that lifo II = IIfll . [Kanpur M.Sc. F 19°41 Proof of the lemma. I. Let N be a real normed linear space. fo : Mo-R. . .. (1) Define a map s.t. flu)=fo(x+lltXu)=/(x)+cxro ... (2) where " is an arbitrary but fixed real number, R is the set of real numbers. u, v E Mo are expressible as u=x+cxxo, v=y+~xo M. Let a, b E R=set of reals. au+bv=a (x+Clxo)+b (yt f3xo)=(ax+by)+xo (ocx+b[:i) x, y ~ M and a, b e R => ax+hy E M [For M is subspace] 10 (au+bl')=/o[(ax+by)+xo (acx+bf3)]
.vhere x, y
e
=/(ax+by)+(acx+h~) '0' according to (2). =aj(x)+bj(y)+(acx+ hfJ) '0' asf is linear. =0 [/(x)+cx'o]+b [/(Y)+~'ol =afo (x+lltXo)+b/o (y+~xo), by (2) =0/0 (uH blo (v).
This proves that 10 is linear. linear functional on Mo and
10
Now (I) proves that 10 is a
fo(x)=/(x) ¥ x E M, by putting cx=O in (2). is an eXlension off.
Hence
II. Remains to prove that II 10 II = UI \I. 11/011 =sup {I.fo(x) I: x EMu, II x II ~ I} ~sup ( I/o(x) I : x E M, II x II ~ I} as Mo ::> M =sup { If(x) I : x E M, II x II <. IJ as 10=1 on M
= Ilf II· This:::> 11/0 II ;a 11//1· III. Now our problem is to show that suitable value of
'0.
If 10 " .;;;;
... (3) II f II under
330
BANACH SPACE
Let
E M, then
Xl' X 2
/(Xl)-/~X2)=/(Xl-.'C·J ~ 1/(Xl -X2) I ~ lI/n·1I s;;: II/II • II (Xl +XO)-(X2+XO) II
xl -x2 11
II . Ii [ II Xl +Xo II + II X 2 + Xo II ] II/II . II XI+Xo II + II/II. II X2+Xo II -/(x~) - 11/11.11 X 2 +Xo II ~-/(XI)+ IIlll.11 Xl+X o /I.
< II / =
or
lt is true Y Xl' XI This proves that
e
M.
sup { /(y)-lIl 11,11 y+xo II} ~ inf {-/(y)+11 / 11·11 y+xo II}· yEM yeM Choosing a real number s.t.
'0
sup {-/
y=~ in (4) and noting that at
/(y)=/
-!f(x)-~
If 11.
>
I x' 1/(x). we, get ,;)-=«
If 11.1 l~+x. II ~ro ~-!." /\x)+I1/ /.·\1:,11~ +Xu II Ot' 1
.••
(5)
0, then (5) gives
I - [--f(x)lit
11/11.11 x+:xxo II] ~
ro
~
I
-ot [-/(x)+II/1i .11 x+ClXo II J.
This ~-/(.'()- II/II. II u /I ~ IJro <;. /\x)+ 11/11.11 u II and
=> -r.f(x)+IJ'o· EO. II/II. II u II /(x)+Qru ~ u II.
III"."
Taking the or or
sec~nd
inequality,
/(x)+aro ~ IIfil . II u II /o(x +orXo) ~ II/II ." u II /o(u) ~ II/II . II u II. Similarly if
c,(
...(6)
< 0, then by taking first inequality in (5), we
get (6). Replacing u by - 11 in (6). -/o(u) E;;; O/II.II-ull-II/II.II ull -/.(u) .;;; II/II. II u II. By (6) and (7),
... (7)
BANACH SPACE
331
... (8) Ilo{ u I ~ II I II . II U II =sup ( I/o(u) I : U E M o, II u II < I}. .. (9) Using (8), 11/0 II ~ 11/11. By (3) and (9), we get II fo II = II I II . The same result can be proved "hen N is a complex norm(d linear space. Proolol the main theorem. Let P denote the set of all ordered pairs (/~, M 1 ) where f), is an extension of I to the subspace
nfo II
MI ::J M and II IA II =111 . We define a partial order ~ 0 .. P by setting
(II, M I ) and/'A=I. on M A•
c
~ (/~, M~) iff M" M,. f, M) E P and so P
Since (
=F- 0. Let
Q={(j" Mt}}
be a chain in P, then Q has an upper bound (g, U M,) where g(X)=Ji(x) ¥ x E M,. Here U M, is a subspace of N. By Zorn's lemma. 3 a maximal element (F, H) in P. It can be shown that H=N. Here we have used Zorn's lemma which states that, "Every non-empty partially ordered set in Which each chain has an upper bound has a maximal element". By the lemma, \I ,. II = 11/11.
Theorem 9. Generalized Hahn-Banach Theorem. 011 X .J t.
Let X be a
real vector space, p a real valued function
p(x+y) .-;. b(x)+p(y) p(ax)=ap(x) for a ;,il 0 and Y a subspace of X. {f f is linear Oil Y alld f(y) ~ p(y) V Y ~ Y, thtll there exists a linear function F on X s.l. F(J')=j(y) F(x) e;;;; p(x) on X.
011
Y and
lKanpur 79, 78; Meerut 89, Indor 79) Proof. T he main problem is to extend f to g defined 011 Y1= Y U {xo} where Xo ~ Y but XI) E X so that Ily)=g(y) ¥ Y E Y. This can be easily done. The existence of F will be proved by the help of Zorn's lemma. Let E be the family of all extensions g of f defined a~ ahow. Define partial order relation ~ on E as follo",s : gl EO;; g~ if g2 is an extension of gl
332
BANACH SPACE
i.e., if g\(X)=g2(X) and doma n of gl C domain of g2 where gl' g2 E E. Then (E. ~) is a partially ordelcd set. Let C be a chain (Totally ordered subset) in E so that C C E. Define a linear functional T S.t. (i) Domain of T=union of domains of g l.rf g E C. (ii) If x E domain of T, then we define T(x)=g(x) .•. (I) where gEe and x E domain of g. From the constructions of E, C, T it is clear T is a linear extension off and T(x)=g(x)=flX) ~ p(x) T(x) ~ p(x) on its domain ... (2) so that TEE and it is an upper bound of C. By Zorn's lemma, E has a maximal element F. Consequently F is an extension of f and its domain is X and F(x)=f(x) on Y, by (1) and F(x) ~ p(x) on X, by (2).
Theroem 10. If N is a normed linear space and Xo is non-zero vector in N, then 3 fUllctional fo in N* s.t. fo (xo)= II XU II and II fo II = I. [Meeurt 79, 76] Proof. Supposc N is a normed linear space over the field K of scalars. Let Xu E N s.t. xo~O. Wirte M={axo: a E K}. Then M is a linear subspace of N and IS generated by Xo' Define a map f: M-+ K S.t. f( x o)= II XU II· (I) LeT II, v E M, then u=a;xo• I'=~x•• \\here Of, fJ E K. Let P. q E K. pu-+-q~'=p /Xxo+q 8xo=(px+qP) X o_ Pr1.+qf3 E J(. By (1),f(pu-1 qv)=(~+q~) II Xo II=pcr; IIXo ,I +q{J /I Xo II =pf (7X..)+ qf(f3xo)= pf(u>+qftv). This proves that f is linear (X
I feu) 1= I Of III
or
111= IIX I . 11 Xo I = II ( Xo \1= II u 1\ < \I 2u II I < 2 I u II. ..(2) By def., this provcs that f is bounded a~d ht'nce continuous XO
If(u)
by Problem 3'(a). II! II=sup { I feu) I : U E M, /I U II ~ I} =sup { 1/ U II : U E M, II t.I II <" I}= I or IIfil=1. Also!(xo)=HxoilontakingOt=l.
#'
333
BAN '.CH SPACE
r:
Finally, M~K s t. f «(1X o)=(1. II Xo II is continuous linear tran~formation, / Hence.! is a functional on A! with the property that .!(Xo)=
II
Xo
/I, IIfll =1
.. ,(3)
By Hahn-Banach theorem, the function f can b;! extended to a functio:lal fo on N S.t. "/0 II = II f II ... ( 4) By (3) and (4) lifo II =1. ... (~) fo is an extension off ~ /O=fon A! => fu(xo)=/(x o) as Xo E M. Using (3), /o(xo)= II Xo II ... (6) Finally, 3 fo E N-It< S.t. /o(x) = II XO II and II fo II = I. This completes the theorem. Remark. x,t:y => x-y#O => II x y II =0 => If(x-y) I = II x-y II ~O, by (2) => f(x--y).:;t:O => f(x)*fIY)· This proves the following result: if x.:;t:y- and x, YEN, then 3 fo E N if, s.t. fu(x)-~J~(Y). Theorem 11. If M is a c/oJed sub.)pace of a normed linear space Nand Xo fI. M, then 3 functional/o ill N :' s.t. foe M)= 0 and fo(.'(o)=r-0. [Kanpur 1980; Meerut 80] Proof I. Let M be a closed linear subspace of a normed linear space N, then N/A! is a normed linear space w.r.t. the norm detlned by II x+M!1 =inf { II x+m II: x E N} Xo fI. A! => xo+Af-:/'M "'" xo+M is not a zero vector of N/M. Hence,3 a functional f E (NI M)* S.t. f(xo+M)= II xo+M II ... (\) and IIfll =1 ... (2) (Refer Theorem 10) II. Define a map T: N-.N/M s.t. T(x)=x+M ••• (3) It is easy to see that T IS linear. For T (ax+by)=(ax+by)+M'=(ax+M)+(by+M) =a (x+M)+b (y+M)=a T(x)+b T(y) II T(x) /I = II x+M II =inf { II x+m II: mE M} E;; II x+m II or II T(x) " ~ " x+m " '
334
BANACH SPACE
By def., this proves that T is bounded. Hence T is continuous, by Problem 3 (a). III. Since T: N+N/M./: N/M-+-K. This ~ f T : N ~K. Now we define ... (4) fo(x)=( fT) (x) "" x EN. Composite map of two linear maps is a linear map. Also composite map of two continuous maps is a continuous map. Consequently fo is continuous linear functional on N. Hence fo E N* .;.(5) Also fim)=(fT) (m)=f(T(m)]=f(m+M)= II m+M II =0.
[For
M=O' is a zero vartor of N/M and mE M => m+M=M=O']. :. fo(m)=O "f m EM This ~ fo(M)=O !o(xo) =( fT) (xo)-l fT(xo)] =f (xo+M)= II xo+M 11*0, [For xo+M=I=M i.e., xo+m.,t:zero vector of N/M] or fo(x o)=1=0
.. (6) ... (7)
by (1) ... (8)
From (5), (7) and (8), the required result follows. Problem 8. Let M be a closed linear subspace of a normed linear sp:zce N, and let x" fI. M. If d =d (Xo, M), then 3 functional fo E N* S.t. fo(M)=O, fo(xo) = I, 11.1;11 =-l/d [Meerut 1988, 87,86; Kanpur 80; Delhi 81] Solution. Suppose M is a closed linear subspace of a normed linear space Nand Xo fI. M. Write Mo=(M u {x,}) ={x /-cxxo : « is any real number}. Then Mo is a linear subspace generated by M and {xo}. Xo fI. M ~ any y E Mo has a unique representation y=x+cxxo ... (1) fOl' some real number Cl. Let d=distance of x. from M.
Then
V x E M ... (2) d=inf { "x-xo II: x E M} ~ II x-xo II Xo ft. M ~ x=l=xo y. x EM=> II x -- Xo II > V x e M ;::.d>O
°
d
~ II x - Xu II =>
d.!... ;;;.
1 II x -Xo II
335
BANACH SPACE
/
Ii x -xu II
Define a map I:
~
} ¥ x Mo-..K s.t. jiy)= (
E
M.
. .. (l)
... (4)
where y is given by (I). Since representation of y is unique and so map I is well defined l(xo)=(O-l-lx~}=1 or I(xo) = I mE M => Itm)=/(m +o.xo) =0 ~ I(M)=O. Thus we hav' proved that j(M) = O,f(xo) = 1 ... (5) I/(y)
1= I x 1=
ac I II Y II II y"
I ct I . II y
II
x+atXo II
II
I at I . II y II
l7ITxIOl+xo \I
II y II with _::.. to M IIxo-(-x/ex)ll Ct
~ i! ~J! , according to (3) or
I fiy) I EO; II y IL
... (6)
d
Il/il=sup{
Ily>1 :yE
~ sup {~
M o.llyll<: I}
: y E Mo. II y II E;; 1 }
This saggests that II I I= I jd. Finally, we have shown that f is a linear functional on Mo S.t. !.M)=O,f(xo)=I, 11/11 =I/d. Hence, by Hahn-Banach theorem, the function I can be extended to a functional/o on the whole space N S.t. II 10 II = II I II lifo iI = 11/11 and fill II =Ijd ~ 11/0 II =Ijd. 10 is exten~ion of I=> fo =/on Mo => fo~=1 on Mo :J M => fo(M)=/(M). But j(M)=O => lo(M)=O Also J.,(xo)=.!i xo)' But I,x o)= 1. Hence .f~(xo)= I. This completes the proof. Definition. Let Nand N' be normed linear spaces with the same scalars. Then a one-one linear map T of N into N' S.t. II T(x) II = II x II ¥ X E N, is called an i'Jomlltric isomorphism of N into N', If such an isometric isomorphism cxi,ts, we say that N is isometrically isomorphic to N.
336
BANACH SPACE
Theorem 12. Natural imbe«!ding of N into N-. Let N be a normed linear space. Thm each vector x in N induces a lunctional Frr; on N*' d lined by FflJ ( I )=i(x) V lEN*' s.t. 1\ FflJ II = II x If • Further the map J: N-+N'I<* s.t. J(x)=F. V x E N delines an isometric isomorphism 01 N into N*''*'. [Kanpur 199.0; 87; Meerut 86] Proof. Let x E N and IE N*, then I: N_K is a linear continuous map. . .. (1) Let F.,: N*'-Kbe a map s.t. FflJ(/)=/(x). I. To prove that FflJ is functional we have to show that FflJ is linear and continuous. LetJ, g E Nit< and a, b E K be arbitrary, where K is a field of scalars. FflJ (ol+bg)=(al+bg) (x), by def. of FflJ =al(x)+bg(x)=a Fz (f) b Fz(g)
+
Thl:' => F" is a linear map ! FaI(/) 1= I/(x) , E;;; 11/11.11 x II Taking K= II x II =constant for the map F." we get 'Frr;
(I)'
~ KII/II
By def. this proves that F", is bounded and hence continuous. Thus F", is linear and continuous and so F:r; is a functional on Nil<. Consequently F:r; E (N'*')*=N'*'*. II. II F:r; II =sup UFaI(f) , : 11/11 ~ I} . =sup { 'I(x) I : 11/11 ~ I}, by def. of F", <sup {II/II . II x II: IlIIi ~ I} as '/lx) , ~ IIltlll x II ~IIXII
II F", II ~ II x II ¥ x E N III. Let x be any non-Zl!ro vector. Then:l a functional g E N'*' s.t. g(x)= II x II ... (3) and If g II = 1 (Refer Theorem 10) Also g(X)=F",(g), by (1). Now using (3), F",(g)= If x II ==g(x). This proves that 1\ x /I =g(x) ~ sup { , g(x) I : gENII<, II g II ~ I} = sup { I F",(g) I : g EN"", II g II ~ I} or
or
=IIF",/I /lxll';;;IIF",1I
..(4)
BANACH SPACE
337
"x
II Fill I! = II ••• (5) To show that the map J : N + N •• s.t. J(x)=F. l.rf x E N ..• (6) is an isomorphism. Fc.+&fI) (f)-/(ax+by)=af(x)+bf(y) =a FfIJ(f}+b F,I (f)=(a F.·+b F,) (f). Using (5), J(ax+bYJ=a J(x)+b J(y}. . .. (7) This .. J is linear. By (5) and (6), II J(x} " = " F. II = II x Ii or II J(x) II = II x II ... (8) This proves that J preserves the norm. (8) =- II J (x-y) n= "x-y" ~ /I J(x)-J(y}II=lIx-y 1/ ... (9) J(x)=J(y) .. J(xJ"-J(y)=O => J{x-y) .... O .. II x--y II =0, by (9) .. x=y. :. J is one-one map. ...{l0) From (7), (8) and (10). it follows that J is an isometric isomorphism of N into Nit •• This completes the proof. Deduction. Since J is an isometric isomorphism of N into NIt:* and therefore we may regard N as a part of N·. without changing any of its structure a~ a normed linear space. We write N C N**. Hence the map J is called natural imbedding (or canonical map) of N into N·*. [Kanpur 1978] Not~. If J is onto as well, then Nand Nit'!< are isometric. At this moment N=N*·, the sign of equality in the same sense of isomorphism under the map J.- Then the map J is called reflexive map and also N is called Norm reflexive or simply reflexive. Lemma 13. Let Band B' be Banach spaces and T a continuous linear map of B onto B'. Then the image of each open sphere centered on the origin in B contains an open sphert' centered on the origill in B'. [Kanpur 1978; Meerut 77] Pro..r I. Let Sr and S,' denote respectively the open spheres centered at the origin and radius r in Band B' respectively. Then Sr={xeB: II x" < r}. Sr'={xeB : II x 1/ < r} Observe that By (2) and (4y
IV.
Sr={
XE
B:
til- < I}.
Sr={rYEB: \I y 1/
<
Takip.g x=ry.
l}=r {yeB : " y II
<
I}
338
BANACH SPACE
or This
~
or
S,=r SJ T (S,)=nr Sl)=r T(Sl) T(Sr)=r T(Sl)
... (1)
as T is a linear map. Aim.
T(S,) = image of Sr contains some S,'
II. First we try to show that T(Sd contain'! Sr'. For each positive integer n, consider open sphere S.. in B. Then
As T :,B -+B' is onto and linear and so B'=T(B)=T [
~l S.. J= "~1 T(S.. )
00
or
B'=
u
T(S.. ).
... (2)
.. =1
Being Banach space, B' is complete as a metric space and so it is of second category, by Baire's category theorem [Refer Definition 18'21]. In this event (2) proves that the set T(Sm) for some value m of n is not non·dense set in B'. int
[T(')'",)]~I/I
T(Sm) has an interior point, say, Yo s.t. Yo E T(S.,.).
It is ea,y to see that the map f: B'-B' s.t·[.y)=y-yo is a homomorphism. Then [T(Sm)) Yo has the origin as an interior point Now we assert that T(Sm)-Yo C T(S2fR). Let y E T(Sm)-Yo be arbitrary. then i x E Sm s.t. y=T(x)-yo Again Yo E T(S... ) => 3 Xo E Sm S.t. Yo = T(xo) Now by (4), y=T(x)-T(xo)=T(x-xo) where x, Xo E Sm. This => " x -xo II
=>
T(x -xo) E T(S.II") T(S2"')' by (5). E T(Sm)-Yo ~ y E T(S_)
=>:1, E
Thus
i
< 2m=diameter of open sphere Sm
jc-x~ E S2fR ~
... (3)
... (4) ... (5)
BA~ACH
:.
or
339
SPACE
T(Sm)-Yu C T(S2fIl)=2m T(St), by (1) T(S.. )-y. C 2m T(Sl)
This => [T(S... ) - Y.] C [2mT(Sl)]=2m T(S1) ~
.•. (6)
[T(S.. )]-yo C 2m [f(SI)]
From (3) and (6), it follows that the origin is also an interior
S:
point of T(Sl)' Hence:l E > 0 s.t. c T(Sl)' We complete the proof by showing that So' C T(S.). Let yes: be arbitrary, then I~ Y II < E, by (7). ~
Y E T(Sl), this
... (7) . .. (8)
Y is an adherent point of T(Sl)
~ 3 Yl E nSl) s.t.1l Y-Yt II
< E/2.
By (7), Yt E r(St) ~ 3 Xl E Sl s.t. Yl==T(x 1 ),
1/ ~
a<
1/2'.
Again we observe from (7) that S'(012) C [T(SU2)j Also II Y-Yl II < ./],. This ~ l'-Yl e S"/2 c (T(St/.)]. Proceeding a above, 3 Y2 E T(Sl/2) s.t.
II (y - YJ) -Y2 II and
<
II XI
(E/2!) where Y.= T{X2) II < i·
Continuing in this way, we obtain a sequence s.t.
II x,. II
and
<
X,,= T(x,,).
in B ... (9)
2"-1
11 Y--(Yl+Y2+···+Y.) n <
. where
< x. >
I
.. Write s,,= X
E
2W
..• (10)
Xl.
t-l
For n > m, we have II s"-s,,, or
II s,,-s... H
n=11 t_l' i: x,- '-E 1 Xi II
=11 i_m+l ~, Xt II ~ II Xm+t 1\ + II x.+.11 + ... + II x. n
I
1 (h' h '. . G P ) + 2-+1 1+"'+2,,-1 W IC IS In "
<
~
=
i,; (l--~):: _1__ .-!-=0 1 2-2,,-m-l 1
1-2"
as m,
n~oo.
340
BANACH SPACE
<S,,> is a Cauchy sequence in a complete
Consequently
space B and so 3 x E B s.t.
II
Lim n~oo
s,,=X.
s" /I = II i~l x,1I ~ II Xl II 1
+ ... + II x" II
I-U)"
I
< I +i+ P +"'+ 2,i="1 =-1-.-=2 or
[I-~l <
2
II s" II <:: 2 II x II = II lim sro!l ==;lim II This
~
Sn
II <2<3.
xESa
x=lim
Sn ~
"
x=lim 1:
4_1
00
~
n
Xi
-= E Xi i=l
(1)
T(x)= E T(Xj)= E y, as T is continuous i=1
'=1
-> T(x =Iim (Y1+Y2+ .. +y,,)=y, by (10). Thus, y=T(x), 1\ X:I < 3 and so YET(Sd)' Fin!lliy, we h1.Ve shown that YES.' ~ yEO T(Sa) So' C T Sa)'
Of
This completes the proof. Theorem 14.
The open mapping theorem.
If 8 and 0' are Banach spa::es and f T is a continuous linear map of B ihto B', t"en T is an npen map. [Meerut 1987; Delhi 81 ; Buodelkhand 82; Kanpur 79] Proof. Let Band B' be Banach spaces over the same field K of scalars. Let T: B~B' be a continuou,> linear map. Aim. T is a'1 open map. Let G be an open set in 8. Then we have to show that 1'(G) is an open set. Let y ~ T(G) , then 3 xEG s.t Y= T(r). G is open, XEG ~ 3 an op~n sphere
sex,
sex. r) in
B s.t.
r) C G ... (1) S(x, r) stands for open sphere centered at x and radius r, S,
~ANACH
SPACE
341
/
,tands for open sphere centered at the origin and radius r. Then
Sex, ,)=x+S,. becomes x+S, c G.
Now (I) By the Lemma 13, 3 open sphere S' S'
or
y+S'
C T(S,)
'1 C
...(2)
'1
in B' S.t.
y+T(S,)
'1
or or
(\) -> T(x+ S,) C T(G) Q 7'(x) + T(Sr) C T(G) y+ T(Sr) C T G) Also, by (3), y+ S' c y+ T\S,) C T(G) y+S'
or
or
... (3)
'1 '1
C T(G)
) C T(G). y, '1 Thus we have shown that S'(
any y
e
This
T(G) ~ 3 open sphere S ( ~
y,
'1) in B' S.t. S'(,).-, '1) C T(G).
T(G) is open set.
Theorem 15. A one to olle t:ontinuous linea, map of Banach .fpace B onto a Banoch space B' is a homeomorphism. In particular, if a one-one linear map T of a Banach space onto Itself is continuous, then its inverse T-J is automatically continuous. [Meerat 1976J Proof. Let Band B' be Banach spaces, Let T: B-B' be continuous linear map s.t, T is one-one onto ma.p. To prove that T is a homeomorphism, we have to prove that (i) T i., one-one onto (ii) T is continuous (iii) T-t is continuous. Our allsumption :> (i) and (ii). Remains to prove (iii). To prove (iii), it is enough to prove that T is an open map. Prove it as in Theorem 14. Second part follows atonce from first part if we put B' =B. Remark. This theorem is sometimes known as Banach theorem.
342
BANACH SPACE
Theorem 16. Suppose P Is a projection on a Banach .space B lind M and N are its range and nulL spaces respectively. Then M and N are closed linear sub3pace3 of B st. B=M(J)N. Proof. Suppose P is a projection on a Banach space B, then PI=P and P :.B-+B is a continuous map. Let M and N be range and null space of P, then M={x e B: P(x)=x}={x e B: P(x)=/lx)} or ... (1) M={x E B : (P-/) (x)=O} ... (2) N={x E B: Px=O} N=p-l {O}, M=(P-l)-l (0) Evidently B is Banach space ~ B is metric space ~ every singleton set is closed :> {OJ is closed set => p-l {OJ is closed as P is continuous => N is closed. Beip.g identity map, I is continuous map and so p-/ is con· tinuous. Consequently M is closed any x E M n N ~ x E M, x EN=> Px=x, !*x=O ·.'x=O :. M n N={O} ... (3) any z E 11 => z=P(z)+t-P(z)=p(z)+(I-P) (z) Taking m=P(z). n=(/-P) (2'), we get z=m+n P(m)=PP(z)=PI(i)= f(z)=nt P(n)=P(I-P)(z)=(PI-PI)(z)=(P-P)(z)=O (z)=O Thus P(m)=m, P(n) =0. this => m Eo: M, n e N.
... (4)
Now (4) => B=M+N. By (3) and (5), B=M(fJN. . .. (5) Theorem 17. Let B be a Banach space and supp03e that M and N are two closed linear subspaces oj B 3.1. B=M$N. Ijz=x+y is the unique representation oj a vector in B as a SUm of vectors in M and N, then the map P defined P(z)=x is a projection on B whose range space and null space are M and N. [Meerut 1970j Proof. Our3Ssumption. P(z)=P(x+y)=x where x E M, y E Nand M={x e B: Px=x}:=arange space N={x E B: Px=O}=nulI space
343
BANACH SPACE
P2(Z) = P(PZ) = PX=X, =P(Z). :.
P2=P.
by def. of M. by def. of P It can be shown that P is a linear map.
Remains to prove that P is continuous. Let D' denote the linear space B equipped with new norm" /I' defined· by 1/ Z 11'==11 x 11+/1 y /I. Then B' is a Banach space (Refer Problem 2). Also II P(z) 11=11 x II ~ II x 11+11 y 1,=11 z I/' or \I P(z) " <. II z '" This => P is bounded and is continuous. Remains to prove that Band B' have the !lame topology. For this let T: D-+.D' be an identity map S.t. T(z)=z. Then II T(z) 1/=11 z 11=11 x+y \I ~ II x 1/ + lIy" = II zl/'. or II Ttz) II ~ II z ;1' This => T is bounded => T is continuous. Also T is one-one onto map. Finally Tis. one-one Onto and continuous map. This => T is a homeomorphism (Refer Th. 15). Hence Band B' have the same topology. The proof is now complete. Theorem 18. The closed graph theorem Let Band B' be Banach spaces. Let T: B- B' be a linear map. Then, T is continuous iff its graph is closed. [Kanpnr 1988. 84; Meerut 82, 88] Proof. Let Band B' be Banach spa(es. Let T: B-+B' be a linear map. Then the graph of T is TG={(x, T(x» : x E B} C B x B'. Recall that a set A is closed if A=A where A=closure of A. I\,lso A C A is true V- A. I. Let T be continuou<;. To prove that TG is closed. ·Let (x, y) E fa be arbitrary. (x, Y) E TG= Tr::r U D(TG) =- (x, y) E TG or (x, y) E D(TG ) => (x, y) is a limit point of TG ,.. :I se'luence < x .. , T(x .. ) > E TG s.t. lim n-+oo (x", T(x,,» == (x, y) ... (1) But x,,-x => T(x.. ) - T(x) ... (2) as T is continuous. Abo T(x,.) -+ y •.• (3), by (I).
344
BANACH SPACE
Combining (2) and (3), we get T(x)=y. :.
(x, y)=(x, T(x» E To
*
Thus (x, y) E To (x, y) E TG :. T C To. But To C T G is always true. Hence T c= To, this ~ To is closed.
II. Let Tu be closed. To prove that T is continuous. Band B' are Banach spaces => Band B' are complete spaces => BxB' is complt!te.
Also closed subspace of a complete space is a complete space. Hence To is a complete space. Define a map I: To-.B s.t.f[(x, T(x))]=x Evidently I is one-one onto and linear map. Also or
II/«x, T(x»] 11=// x II ~ II x 1/+/1 T(x) 11=1/ (x, Tlx») II 1I/[(x, T(x))] II ~ I " (x, T(x)) "
This => lis bounded => I is continuous map; Then I is homeomorphism. (Refer Theorem 15) 1 Consequently 1- : B-.1'o is continuous and so it is bounded. (Refer Problem 3a).
II T(x) \I C;; " T(x)
1/+/1 x 1/=1\ (x,
<; ".f- " • " x II "T(x) II .; "f- 1 II • " x II Also f- 1 is bounded and so T continuous.
T(x» 1\ =i//-1 (x) II
1
or
i~
bounded and therefore Tis
Theorem 9. The uniform bounded ness theorem (Banach Steinhaus Principle). Let B be u Banach space and N Q normed linear space. If {T,} Is u non-empty set of continuous linear transformation B_N s.t. {T.(x)} is bcunded subset 01 N ¥ x ~ B, then { " T, "} Is a bounded set 01 numbers, i.e. {Ti} is a bounded [Kanpur 1988,82] .fubset ofD (B, N). Proof. T,: B .. N is a linear map and so T. (x) E N ¥ x E B. For every positive integer, n, we define F,.-{x : x E B, " T, (x) " ~ n "" i} ... ( I) S, (x)=open sphere with centre x and radius r S" (xl-closed sphere with centre x and radius r x E F,. II T, (x) " ~ n, T. (x) E N
'*
345
9ANACH SPACE ~ Tt (x) ~
...
x E
E Sn [0]
n
Tt-I
~ x
E T,-l {S" [OJ} ¥ I
{S" [O]}
i
Fn=
n
T,-l {Sn [OJ}
.. (2)
t
We know that S. [0] is closed set in Nand Ti : n. . . N is confinuous => T.-l {Sft (0]1 is closed. Being an arbitrary intersection of clo~t:d sets, F,. is closed according to (2) .. ...(3)
-
Also
B= U Fn
".(4)
according to (2).
n is Banach,.
B is complete .. B is of second category, by Baire's Theorem. [Refer Def. 18'21 (ix)] ~ B is not of first category. => one of the F.. 's, say Fm is dcnlle set.
=> int
(F",):;t:{21
[But Fm i<; clo~ed => F",=Fr..J .. int (F.,,)::I=0 => (F",t:r=0 3 Xu E (F",t => Xu i<; an interior point of £;" => 3 closed sphere Sr [xu] s.t. s, [xo] C F,,, where S, [xo] s.t.
=-
If
/I y II fiib I, and ;=y, then
II ~ 1'1-11 y Ii ~ I
~ r, this 0 Z € Sr [0] z+x" E Sf' Ixo] C Fm .. z+xo E F", ~ II T, (z+xo) II .;;; m
or
"
z II
~
... Also or
or
Also(Fm)OC Fm ¥ Fn> F. ... this => II Tt (x o) II ~ m II T, (y) /1=1/ T, (z/r) 1/= Ifr II T, (z) 1\ =1/r II T, [(z+x o) xo] II II T, (y) II=I/r [II T, (z+xo)-T, (xu) II <.1/r [ U T, (z+xo) 11+11 T, (xu) II J <'1/r [m+m]=2m/r, by (4) and (5). II T, (y) U<; 2m/r,
... (4)
xoE: (Fmt. XO E
... (5)
346
BANACH SPACE
Thus II T. II=sup { II T. (y) II : II y II ~ I} ~ 2m/r II T, II (: 2m/r ¥ r. Thus the set { II Ti II } is a bounded set of numbers. Remark. This theorem is also known as Banach-Steinhaus theorem. Theorem 20. A non-empty subset X of a normed linear space N is bounded <;!>f(X) Is bounded set o/nun/bersfor eachfin N*. [Kanpur 1981, 80; Meerut 88, 80] Proof. Let X be a non· empty subst:t of a normed linear space N. Letf E be arbitrary, then/: N~K is a continuous linear may, K is a field of s~alars. f is continuouli do / is bounded. (Refer Problem 3a) .¢> a> 0 S.t. IIf(x) II ~ a II x II ... (1) ¥ x E N I. Let X be bounded, then 3 b :> 0 s.t. II x II b V x E X. . .. (2) To prove that/tX) is bounded. By (1) and (2). I f(x) I , ab y x € X This ~ /(X) is bounded set. II. Letf(X) be bounded To prove that X is bounded For convenience, we write X = {Xi} Then flX) is bounded Q {/(x,)} is bounded. ..(2) Detine a map Fx, : N*~K S.t. Fx. (/)=/(Xt) 1 hen Fx. is a continuous linear map (Refer theorem 12). This ~ Ex, E (N*)*=N** Now. by (2). {Fx,(f)}. i.e. {f(xd} is bounded ¥ f N* Also N* is compl!;'t. (Refer Problem 2a) Applying uOllform boundedness theorem 19, we find that { II Fx, II } is bounded subset of N-it<*. But Ii Fx, Ii = II Xi II (See cquatlOn~ of Theorem (2) { II x, } is bounded set. But {x.}=X. This => X is bounded ~et
or
,..*
<
e
Conjugate of an operator. lKanpor 1983, 81J Let N be a Dormed linear space ov'er the field K of scalars. Let T: N~Nbe a continuous linear map. LetfE N*. Then T: N~N,f: N~K This.. / T : N~K ~ /T e N* Definition.
18'24.
347
BANACH SPACE
Define a map T* : N* ... N* s.t. T*' (/}=IT V lEN· This ~ [T~ (/)1 (X)=( IT) (x) T~' is defined as conjugate or adjoint of T T*islinear. ForifO!,8 E Kand/,gE N*,then [T* (<
{.II r~x~!1
-sup {
:X:;CO}
... (2) by (I)
1f~1~~)U : IIf!l=l, x~o}
But If(T(x)] 1=1 (T·(f») (x) I=li pel) Ii . II x II Using this in (2), 1\ T lI=sup {II T·
II T II = II T" 1/. This equation is expressible as :
the map T: ~ T*' is a norm prcser.ving mapping of B(N) into BUV· ). [K aopur 1983] Problem 10. Let 1 be an operator on a Banach Space B. Show ,hal
and
T has inv!'rse T-' <=> conjugate r* of T has (T,i'<)-l=(T-l)lt,.
all
inverse (T·r 1
th~t
Solution.
T has an inverse T-l <=> TT-l=T-1T=1 ~ (Tr-l)"'=(T-1T)*=/~
<=> (T-l)* T*=T* (T-l)*_1 <:>
inverse of T-it is (T-l)t, For AB=BA=I ~ A=B-l
<:>
(T·)-l:a=(T-·l).,
('If
B-A-·I
348
BANACH SPACE
Problem. If {x ..}. {y,,} are sequences in N and {a,,}. {,sIll are scular sequences such that x,,~x, y .. ~y and Cl,,~ ot, ~ ....... ~, then "nx.+ P"r...... u+J:y. [Kanpur J989, 86] Solution. II o"x" + p..y,,-(u+py) II
-II Of"X.-tlt"x+""x-u+ P")'"-fJ,,y+fJ,,y-~y II =11 Cln (X,,-X) +X (ot"-Il)+P,, (Yn-Y)+y (~n-~) II ~ I a .. I •
Ii x .. -
-10.1. 0 +0.1
X
X
II +lorn - ex I • " X II + I ~n I· II y" - y
11+/ P 1.0+11 y
1/.0
,,+ /I Y
II • I ~,,- #
I
=0 as n_oo Lt
... 1.
(arnx..
n-+oo
+ !J"Yn)=a.x+fJy
Exercises la) Define a normed linear space. Prove that I. 1he linear space of all sequence X= < Xl' XI' ... > of scalars s.t.
-I J::
.. -1
II
Xn
I" <
00
r
X
with norm defined by
1 r: I x" I" ~ L.... j
I!,= ~
1
1/ p
,(I
~P<
00)
is a complete Hormed iinear space. (b) If M is a ciol>t:d linear subspace of normed linear space N, and if onto
T: N --+ N/M defined by 7(x)=x+M. show that T is continuous linear transforma 2.
3.
4.
1ion. for which !I T Ii E:: 1. lKanpur 1988] (a) State and prove the Banach-Steinhaus PrlOciple of 'Uniform Boundedness". (b) Prove that a normed linear space N is separable if Its conjugate space N* is separab!c. [Kanpur 19l18] (a) Define weak convergence and strong convergence in a "'ormed linear space. Prove that strong convergence implies weak convergence, but the converse is Iflot tlue in generaJ. (b) State and prove the closed graph theorem. [Kanpur 1988] (a) Define a Normed linear space (NLS). Prove the follow· mJ in a NLS (1'1, II 11):
BANACH SPACE
(i)
349
x /l-IiYi!I"lIx-Yll. < x" > be a sequence in N s.t. x" ....... x. then
,"
(iiI If
i: x;. I: - !I x " (iii) If <x.>, are sequences in Nand <7,,>. are scalar <;equences s t. x;.~x. Yn->Y. at" -+Of, ~" ... {3, then ot"x .• +~ y ..-Gfx+fJy. (b)· Prove that the linear sp'lce of all bounded sequences x=·<x•• Xl' .. > of scalars with norm degncd by II X II =sup ! x I is a Banach spllce. [Kaopur 1989, 87] (a) Explain tl!e natural imbedding of a NLS N in its second conjugate N** and show that it is an isometric isomorphio;m. (b) If M ic; a closed linear sub-space of a NLS N. and If T is the natural mapping of N onto NI M defined by T(x) =x +M. show that T is a continuous linear transforma.tion for which II Til <= I. [Kanpu... 1987J (a) Define weak convergence and strong convergence in a NLS. Prove that in a finite dimensional space the notation of weak and strong convergenc.! are equivalent. (b) Let N be a non-zero NLS. Prove that N is a Banach space? {x : II x II = I} is complete. lKanpur 1987J Prove that the linear space C [0, 1J of all continuous real functions defined on the closed unit interval [0. I] is a normed linear space (NLS) w r.t. the norm as an elementi defined by
..
5.
6.
7.
"(II=C R.
9.
liU)ldr
but not a Banach space. ·lKanp.... 199.0,86] Let Nand N' be normed linear spac:!s with the same scalars. and Bf N, N') the set of all bounded linear operators of N into N. Prove that B(N, N') is it~elfa NIS w.r.t. the pointwise linea .. operations and the norm defined by II Til =sup f " Tx II: II x n ~ I}. Further, if N' h a Banach space. then B (N, N) is all>o a Banach space. (Kanpur 1986] (a> Define a normed linear spa.ce (NLS). Prove /" the linear space of all sequences
x={xIo x 2 .···.x., ... } 00
of sca1a-rs such that 1:
R:=1
I x" Ip <
00
350
with norm defined by II x
BANACH SPACE
11.= rEI
1
x"
I" 1'" 1
f
11=1
is a complete Normed linear space. (b) Let Nand N' be normed linear spaces and T a linear transformations of N into N . Prove that the following conditions on T are all equivalent : (i) T is continuous (ii> T is continuous at the origin (iii) There exists a real number k
~
0
s.t. " Tx /I <;k " x II ¥ x E N. (iv) lf S is a closed unit sphere in N, then its image T(S) is bounded in N'. [Kanpur 1985] 10. (a) Define a Banach space. Let M be a linear subspace of a normed linear space N and let F be a functional defined on M. If Xo is a vector not in M and if Mo=M+{xo} is a linear subspace spanned by M and Xu then prove that F can be extended to a functional Fo defined on Mo such that IIF.,II= II F II. (b) Describe the natural imbedding of N into N** where N is a normed linear space and show that it is an isometric isomorphism. [Kanpur 1984] 11. State and prove the closed graph theorem. [Kanpur 1984, 83] 12. (a> Define a Banach space. Let N be a non-zero normed linear space. Prove that N is a Banach space <:> {x : II x /I = I} is complete. (b) Define the conjugate space of a normed linear space. If N is a normed linear S1'lace, then prove that the closed unit sphere S* in N* is a compact Hausdorff space in the weak* topology. [Kanpur 1983] 13. (a) Define a normed linear space. Prove that in a norm.:d linear space 1 IIxll-lIyll I <;;lIx-yli and use it to show that norm is a continuous function. (b) State and Prove Hahn-Banach theorem. [Kanpur M.Sc. F. 1982] 14. Let N be a normed linear space. Describe the natural imbedding of N into N** and show that is an isometric isomorphism.
BANACH SPACE
35(
(b) State and prove the uniform boundedness theorem. [Kanpur M.Sc. F. 1982] 15. (a) Prove that if M is a closed linear subspace of a normed linear space X, and the norm of a coset X + M in the quotient space XI M is defined by II X+MI! =inf{ II x+m 1/: m EM} then XIM is a normed linear space. (b) Prove that the conjugate space of a normed linear space is a Banach space. [Kanpur M.Sc. F. 1981] 16. Define the conjugate of a bounded linear operator Ton a normed linear space X into itself. Prove in detail that a non-empty subset A of a normed linear spac:! X is bounded if and only if J(A) is bound~d for every JE X*. [Kaopur M.Sc. F. 1981] t 7. (a) Prove that if M is a closed linear subspace of a nor~ med linear space Nand Xo is a vector not in M, then there exists an.!;' in N* such that !o(M) =0 and !1I(XO).t:O. (b) Prove that a normed linear space N is separable if its [Kanpur M.Sc. F. 1980] conjugate space N* is. 18. (a) Prove that the closed unit ball S* in the conjugate N* of a normed linear space N is a compact Hausdorff space in the weak* topology. (b) Explain the concept of the graph of an operator. State [Kaopur M.S.c. 1980] and prove the closed Graph Theorem. 19. (a) State and prove the uniform boundec ness theorem. (b) Prove that a non-empty subset X of a normed linear space N i5 bounded if and only if f(X) is a bounded set of num[Kapour M.S.c. F. 1980] bers for each f in N*.
19 Hilbert Space 19'1. Inner product space. Let V be a vector space (linear space) over a field K (real or complex). Let a, b E K and u, v. WE V be arbitrary. Then < ,> is called an inner product on V if the following are satisfied:
< < < <
(11) (12) (/a)
u, ,
>
au+bv,
E K IV
>=a < u,
IV
>+b < v,
IV
>
u, v > = , the complex conjugate of
u, u > ;;t 0 and < u, u >=0 iff u=O. The vector space V with an inner product is called an illner product space. A real inner product space is called a Euclidean space, ard a complex inner product space is called Ii unitary space. (/4)
19'2.
Hibert space
[Kanpur M.Sc. F. 1984, 83]
A Hilbert space is a Banach space in which norm is defined from a scalar valued function called an inner product.
A veetor space together with inner product is called inner product space. An inner product space is a normed vector space. If this normed vector space is complete under the norm obtained from the inner product, then it is called Hilbert Space. That is to say, a complex Banach space H(K) together with complex function < x, y >, called inner product is called Hilbert Space if the following conditions are satisfied:
> E K (ii) < x, y> = < y, x > (iii) < aX+J1y, Z > =ot < x, z > +fJ < y, z > (iv) < x, x >= II x W V- x, y, Z E H and ex, fJ (i)
< x,
y
E K,
353
HILBERT SPACE
Example. Consider the Banach space I." consisting of all n-tuples of complex numbers with the norm of a vector x=(xl> xI> "" x,,) defined by II x II
=( ._1.E I
<
y
X,
Xi /2)1/2
> = E"
and
X, }i,
i=1
where Y=< )'1' Y2, ...• y" >, Z=< complex numbers. Evidently
ZIt Z2' ... , ZIt
> ;
Xi,
Yi,
at,
fl
ar~
< x, Y > E C
(i)
=(Xl Vl)+(X 2 y~)+ ... +(x" )i,,) n
=X1Yl+XIlY2+"'+X"
y,,= 0=1 1: y, x,=
x>
(iii) OIx+flY=1I (Xl> ... , x,,)+~ (Yl' ... , y,,)
=(oeX1 +/lYh "', u,,+/ly,,)
<
ax+fly,
Z
>=
..
=a: (iv)
""
E (oex,+fly,) Zt=1X E Xi z,+fl E Y. Zf
.=1
1=1
<
x, z >+fl
f\
< x, x > = 1=1 E
'"
Xi
<
y, z
X.= E / Xi t::cl
i-I
>
/1= II X III
Hence 1'1" is a Hilbert space. [Kanpur 1983] 19'3. Adjoint operator. Let T be a linear operator on a Hilbert space, then its adjoint denoted by T*, is a unique operator on H satisfying the condition "f u, v E H. < Tu, v >=
To prove that T* is linear. Let u, v, w E H and a, b be scalars. < u T,'" (av+bw) >=< Tu, av+bw >, by def. of T* =< Tu, av >+< Tu, bw > =0 < Tu, v >+b < Tu, w >
354
HILBERT SPACE
==a (u, T* v)+/j (u, T*' w) =(u, aTf:. v)+(u, bT"" w) _=(u, aT"" v+bT* w) This is true ¥ u E H. Hence the last gives T* (av+bw)=aT* v+bT"" w This proves that T* is linear. The set of all linear operators on the Hilbert space H is denoted by B (H) It can be shown that B (H) is a Banach space.. 19-... Definitions of different types of oper.tors. Let T be a linear operator on a Hilbert space H. Then (0 Tis DOrmal if TT-»=T""r (ii) T is unitary or orthogonal if T*T=TT""=/.. (iii) T is self adjoi nt or Hermitian if r = T* • (iv) Tis positive if (Tu, u) ~ 0 ~ u E Hand T=T*'. 1"5. Convex set Let H(K).be a real or complex Hiibert space. A non-empi) subset S of H is said to be convex if x, YES ~ (I-Gl) X +ay E S where IJI is rea) and 0 ~ Cl ~ 1. If we take
2=i, tben
x.
yES ..
xt
y E S.
Tbeorem. A closed s.ubset of a complete space is complete. 19'6. Conjugate space H *. Let H be Hilbert space~ A continuous linear map f: H~C is called linear fuuctional or more precisely functional on H. The set L (H, C) of all functionals on H is denoted by H* and is caned cODjugate space of H. 19·7. Continuity. . An operator T on a Hilbert space H is said to be continuous at Xo e H if given any sequence (xIt ) e H s.t. X"-+X o E H ~ TXIt -+ Txo in H. 19'8. Ortbogonality. Let u, v be arbitrary elements of a Hilbert space H. Also let W, U be subspaces of H. (i) Definition. The vectors u, v are said to be ortbogonal if (u, v)=O. Result 1. The relation of orthogonality is symmetric. For u..Lv ~ (u, v)=O => (v, u)=(u, v)=O=O. v.Lu u.L v means "u is orthogonal to v". Result 2. The zero vector is orthogonal to every vector. For (0, u)=(O U, u)=O (u, u)=O. . Result 3. PytbagoreaD tlaeorem. Iff is orthogonal to g, then
II/+gI1 2 =lIfIP' +ilgll l •
Proof. lis orthogonal to g => (r. g'==O, (g,/)=O
... (1)
HILBERT SPACE
355
n/+g 111=< I+g.l+g) = (.f. I+g) +(g.l+ g) =(/.1)+<1. g)+(g./)+
=11 1111+11 g Ill.
by (i). (ii) Definition. Any vector u is said to be orthogonal to W. i.e. u .1 W jf (u, w)=O y w e w. (iii) Definition. The spaces U ~nd Ware said to be orthogonal i.e. U.l W if (u, w)=O ¥ u e U and w e w. (iv) Definition. The orthogonal complement of U, denoted by UJ., is defined as the set of vectors v e H s.t. v is orthogonal to every u E U. Symbolically. UJ.={x E H: (x. u)=O yo u E U}. 199. Orthogonal set. Let {u • ...• u.. 1 be a subset of a Hilbert space H s.t. each u;;cO. The set {u,} is said to be orthogonal set if the vectors are pairwise orthogonal, i.e. if (Ui, u/)=O for i:;ej The set {ud is said to ,be orthonormal set 'If
( Ut,
)"
Ul =Oi}=
{I0
i=j h=j
19'10. Complete orthonormal set. An orthonormal set is said to be complete if it is not contained in any orthonormal set. 19'11. Definition. Let T be a linear operator on a Hilbert space H. then (i I T preserves norm'i if II T(a) II = II ex H "t at E H In this case Tis also called isometry. (ii) T preserves inner product if (Tat, 713)=(11, ~> yo at, ~ E H 'F' ; .. called isomorphism if T is one-one onto and T preserves inner prouu,",,~, 19'11. Perpendicular projection. Let T be a linear operator on H. Then T is said to be projection 011 U along W if TfV)= T(u+w)=u with u+w=v E H=U~W, u E U. we W. A linear operator Ton H is said to be perpendicular proJection on U along UJ. if H=UffiUJ. and Tv=u where ,=u+w, vE H, uE U. weUJ.. Remark. (i) Since UJ. is uniquely determined by U and hence we only say that T is a perpendicular projection on (j instead of sa"ing 'T' is a perpendicular projection on U along UJ. (ii) Range of T= U. Null space T UJ..
356
HILBBRT SPACE
19'13. Parallelogram law. If ac, fJ are two vectors, then
"1I+fJ 112 +11 «-fJ 111=2 [II «IP'+II fJ In
Theorem 1. (Some properties of a Hilbert space). In a Hilbert space H, prove that (i) (a.x-fJy, Z)=II (x, z) -~ (y, z) (ii) (x, fJy+yz)=13 (x, y)+.y (x, z) (iii) (x, 0)=0 and (0, x)=O ¥ x, y, Z E H a"d II, {J are scalars. Proof. (i) ( ..x-{Jy, z)=(<<x+( -~) y, z)
(ii)
(x,
=11
(x, z)+( -~) (y, z)
=/l
(x, z)-{J (y, z).
{Jy+)'z}=(~y+)'z,
x)={J (y, x)+" (z, x)
=13 (y, x)+y (.x)
r
=13 (x, y)+y (x, z). Put y=O and so (x, ~y)=j:1 (x, y) where as (~x, y)={J (x, y). (iii) (x, O)=(x, 00)=0' (x, 0)=0 (x, 0)=0 (0, x)=(OO, x)=O (0, x)=O. Theorem 2. (Schwarz inequality). If x, yare two vectors in ~ Hilbert space H, then I (x, y) I ~ II x" II y II· [Kanpur 19l9~9'O,188, 86, Meeruti 69) Proof. Evidently inequality is valid when x=o or y=O. Let us suppose that x,cO and (y. x) x z=y- 11 x W • ...(1) / (y,x)x '\. Then (z, x)="y - II x II_ ,x/ (y,x) =(y, x)- IfX1f (x, x)=O.
For
(x, x)=11
:. Now ·;>r
or
X
112
so that (x, z)=O. " z II ~ 0 or (z, z) ;;;. 0 /y_ (y, x)~ z';;;. 0 " IIxW' /
(z, x)=O'
(y, z) -
~r~ ~I?
(x, z)
~0
... (2j
357
HILBERT SPACE
Using the result (2), we get / (y, x> x",,-· (y, z) ~ 0 or ""- y, y- II X 112 / ~ 0 (y, x)
(y, Y)-"if"Xli2 (y, x)
or
~
0
II Y \12 \I x 111- I (y, x) III ~ 0 II y II 2 II X III ~ I <x, y) I I I (x, y) I ~ II x II • 1\ y II • Theorem 3. In a Hilbert space the inner product is jointly continuous, i.e., Xn~X, yn~y => <xn, yn)~<x, y). [Kanpur 1988, 90J Proof. I (xn, Yn) -(x, y) I = I (x .. , Y.. )-(xn, y)+(x.. , y) -(x,y) I = I (Xn, y .. -y)+(x.. -x, y) I ~ I (x .. , Yn-y)I+l<x .. -x, y) I
or Gr or
as
I a,+tJ I ~ I a, 1+1 fJ I
~lIxnll.lly,.-yll+llx.. -xlI.IlYIl
... (1) by Schwan. inequality.
Now
Xn~X,
Yn-+y
when
n-+oo.
Hence II xn-x II - 0, \I Yn-y II Now (l) gives But Hence
I <Xn,yn)~(X,y) I ~ 0 I (x n , Y .. ) -(x, y) I ~ 0 Lim n~oo
I (xn, Yn) -
~
0 as n-+oo.
for n~oo. is always ture.
(x, y) \=0 or
Lim (Xn' y .. )=(x, y) n-+oo If x, yare two vectors
Problem lao (Law of parallelogram). of a Hilbert space H, then I x+y 11 2 + \I x-y 112=2 [II x 112+ II yin and 4(x, y)=11 x+y :I2-1! x- y Il ll +i II x+iy II L i II x- iy III. [KanIJur 19118] Solution. Ii x+y
112=(x+y, x +y)
=(x, x+Y)+(Y. x+ y) =(x, x)+(x, y)+(y, x)+(y, y) II x+y 112=1: x 112+11 y iI 2 +(x, y)+(y, x). Also II x-y 11 2 =(x-y, x -y)=<x, x-y) -
0,
. .. {l)
...(2)
358
HILBERT SPACE
1/ X+y 1/2+ II x--y 1!1=2 [ 1/ X 112+11 y 11 (1) - (2) gives Second Part. II X+y 112_ II x-y 112=2 [(x, y)+(y, x)] ... (3) Replacing y by iy, we get II x+i y \12_ II x-I y 111=2 (x, iy) -(iy, x)] =2 [T(x, y)+i (y, x)]
"x +iy 11:1- II x- iy 1/2=-2; [(x, y) - (y, x)]
or,
as ;=-;
Multiplying both sides by I.
. .. (4) i" x+iy 112-i II x iy 11"=2 [<x, y)-(y, x)] (3) + (4) gives II x+y 111- II x-y 112+i 1/ x+iy 112-1 II x-iy 112 =4 (x, y). Problem lb. If B is a Complex Banach space whose norm obeys the Parallelogram law, and ifinner product is defined on
Bby 4 <x, y)= II x+y II 2_ II x-y 111+; II x+iy III-I" x-iy 1111 then B is a Hilbert space. [Kanpur 1986; Meerut 77J Solution. let
Let
x, y. z €
B
be arbitrary and
ex
scalar.
Also
(1) II x+y 112+ II x-y 111=2 [ II x 112+ II y 112] 4(x, y)= II x+y II 2_ II x-y 112+i II x+ iy 1\2-i 1\ x-iy 1\11 . • .
... (2)
To Prove B is a Hilbert space, we have to prove that (i)
(x, x) = !I x II
11
(ii) <x, y)=(y, x) (iii) (x+y, z)=(x, z)+(y, z) (iv) (ex x, y)=1& (x, y). Here note that the Properties (iii) and (iv) are equivalent to
the single property
(l&x+[3y, z)=ot (x, z)+11 (y, z) in (2), Put y=X 4 (x, x)= 1/ 2x IPI- II 0 II 2+; II x (1 +i) II 2-i II x (I-i) II and I a I II u II Using the fact that II a u II (0
=
=\I'2= l l-i I 11+1 1=\1'(1)>+12) we get 4 <x, x)=4 II x I/I+i 2 II x /!I-i 2 II x /12=4
:I
359
HILBERT SPACE
or
(x, x)= (ii) "
1\
xIII.
Hence the result (i).
Til = II 1 II . Since norm of any number is a real number
and hence II x+iy 11= II x+ly II . In view of this, we take the conjugate of both sides in (2); 4 (x, y)=:\ x+y 1/ 1 -- 1\ x'-y 111-1 1\ x+iy 1\2+1 II x-Iy III ... (3)
Interchanging x and y in ,(2), 4 (y, x)= 1\ y+x 112_ II y-x 111+1 II y+lx 112_1 II y-Ix using the fact that II y+lx /1 2 = II I (x-iy) /1 2 = 1 i 1111 x-Iy III =lIx-iylll and II y-ix 11== II -I (x+iy) 1/1= I -I Pili x+iylll= II x+ly 1/ 1 II y-x 111= II (-I) (x-y) nl = I -111/1 x-y 111= II x-y 1/ 2 we get 4 (y, x)= II x+y 111- II x- y 1I11-t ill x-iy 112-1 /I x+iy II·
"I
... (4)
Comparing (3) and (4), we find tllat 4 (x, y)=4 (y, x) or (x, y)=(y, x) . Hence the result (ii). (iii) Aim (u+v, w)=(u, w)+(v, w) By (2),
4
111- II u-will+i II u+iw Ir'-i II u-iw /I' ... (5)
and 4
Adding (5) and (6), we get 4[(u, w)+(v, w)]={ II u+ w 11 1 + II v+w III} -{ II u-wll l + II v -w liZ} +1 { II u+lw /11+ II v+iw III} -I { II u~iw 112+ II v-iw III} ... (7) By (2),
4(u+v, w):;: II (u+v)+w "1- II (u+v)-w 11 1+/11 u+v+iw III -I II u+v-iw /1 1 ... (8) By (I), II (uf v)+w II '= II (u+w)+v 112 =2 [II u+w 11 1 + II v II I J- I' u tw-v" t =2 [ II u+w 11 1 + II v 11 2J- II u+(w~ v) ~ I again by (I), =2 [ II u+w II 1+ II v II'J -{2I1u 11'+211 w-v jl2 - 1/ u-(W_V)!!2} =2 [lIu+wli l + I v 112 -II u III-II w-v 112] +lIu+v-wll Z
360 or
H-II:BERT SPACE
"u+V+W 111- II U+v-w 112=2 [ II U+W 11 2+ II v 112_ 1\ U 112 - II W-V II 2 ••• (9) Interchanging u and v, we get " U+V+w \1 2 _ II u+v-w 111=2 [II v+w It+ II u 112_ II V 112 - II w-u 121 ... (10)
Adding (9) and (10) and then dividing the whole equation by 2, n u+v+w 112_ II u+v-w 111= II U+w 11 2 + II v+w 112 - II w-v II 2_ II w- U II 2 ... (11) Replacing w by iw and the mUltiplying by i on both sides, ill u+v+lw 111-/ II u+v-iw 112 =i [ I' u+iw 112+ II v+iw 112_ II iw-v 112_ II iw- U112] But II iw-v II = II v-iw II
:.
ill u+v+iw III-i II u+v-iw 112 =i [II u+iw 111+ II v+iw 112_ II v-iw 112 - II u-iw 112] ... (12) Adding (11) and (12) and then using (8)
4(u+v, w)= n u+w I: 1+ II v+w 12_ II v-w" 2_- I u--w 112 +i [ II u+iw 1 2+ II v+iw 112 __ II v-iw 112_ II u-iw n 2J Comparing this with (7), we find that or or
or or
4(u+v, w)=4[(u, w)+(v,w)] (u+v, w)=(u, w)+(v, w). (x+y, z)=(x, z)+(y, z)
.. _(13) Hence the result (iii). (iv) We establish this result in several steps. (a) When lit is a positive integer. Put x=y in (13), (2x, z)=(x, z}+(x, z)
(2x, z)=2(x, z) (2x, y)=2(x, y) Thus the result (iv) is true for :%=2. Let it be true for a.=n (Positive integer), then (nx, y)=n(x, y) «n+l) x, y)=(nx+x, y)=(nx, y>+(x, y), by (iii) =n(x, y) +(x, y)=(n+ I) (x, y) :. (iv) is true for a.=n+ I if it was true for lIt=n. Hence (iv)
is true for all positive integer a.. ~ (b) When Gt is a negative integer. Replacing x by -x in'(2), 4( -x, y)= II -x+y 112_ II -x- y 11 2 +; U -x+ly 112 - i II -x-iy 112 = II x- Y 1\2_ II x+y 1\2+; II x-iy II 2_; II x+iy II 2
361
HILBERT SPACE
or
Comparing this with (2), 4( -x, y)= -4(x, y) (-x, y)=-(x, y) Let «= - {J, where fJ is positive integer. «xx, y)=(-~x, y)=-(fJx, y), by (14) = -fJ(x, y), by (a)=«(x, y) (c) When« is a rational number.
... (14)
Let «=~ where p and q are integers and q:;t.O. q
/p
«lX, Y)='i. x, y
"/ P (X) /x y /, , /=, q ,Y"/=P ""q' by (a) and (b)
/x
"-
(aX, y)=P" q' y /
or
... (15)
X
or
-=z, then x=qz q (qz, y)=q(z, y) by (a) and (b). /x y "(x, y)=q(z, y) or 1 (x, y)=""ii' /.
q
Putting this in (15), (ax, y)=~ (x, y)=«(x, y). q
Similarly it can be proved that (<<x, y)=cx(x, y) when « is real. (d) When« is a complex number. Replacing x by Ix in (2), 4(ix, y)= II ix+ y iI2-- II ix- y !12+1 II Ix+ly \l 2-i II i'x-iy 1,1 2 " x+y hi = I i i2 11 x-iy 112_ I i 12 lI.x+iy 112+i I i 1 -; I i III \I x - Y li2 = II x-iy 1:2_ II x+iy 111 +1 II x+y 112 -i II x --y 112 Comparing this with (2), 4(ix, y)=4i(x, y) or (ix, y)=i(x, y) Let «=CX1+;«2 where Cltl and 01 2 are real.
.:.(16)
360
or
HILBERT SPACE
/I U+V+w 111- II u+v-w 111=2 [ " U+W 11
+ II v 112_ II U 112 - II W-V II 2 ••• (9) 2
Interchanging u and v, we get II u+V+w 112_ II u+v-w 11.=2 [II v+w /1+ II u 112_ II V 112 - II w- u / 2] ... (10) Adding (9) and (10) and then dividing the whole equation by 2, nu+v+w 112_ II u+v-w II Z= II u+w 111+ II v+w III - II w-v II 2_ II w- u II 2 ..• (11) Replacing w by iw and the multiplying by i on both sides, i" u+v+iw" I-I II u+v-iw II 2 =i [II u+iw" 2+ II v+iw 111- "iw-v 112_ II iw-u III] But II iw - v II = II v- iw II :. i" u+v+iw III-i II u+v-iw II 2 =i [ II u+iw 111+ II v+iw 112- II v-iw 112 - II u-iw 112] ... (12) Adding (11) and (12) and then using (8) 4(u+v, w)= II u+w I: 1+ II v+w 11- II v-w 112~ I u-·w 112 +i [ II u+iw 1 1+ II v+iw II 2__ II v-iw 112_ II u-iw n 2] Comparing this with (7), we find that 4(u+v, w)=4[(u, w)+(v,w)] (u+v, w)=(u, w)+(v, w). or ... (13) (x+y, z)=(x, z)+(y, z) or Hence the result (iii). (iv) We establish this result in several steps. (a) When at is a positive integer. Put x=y in (13), (2x, z)=(x, z)+(x, z) or (2x, z)=2(x, z) or (2x, y)=2(x, y) Thus the result (iv) is true for :1=2. Let it be true for rx.=n (Positive integer), then (nx, y)=n(x, y) «n+l) x, y)=(nx+x, y)=(nx, y)+(x, y), by (iii) =n(x,y)+(x, y)=(n+l) (x, y) :. (iv) is true for rx.=n+ I if it was true for at=n. Hence (iv) is true for all positive integer at. . (b) When at is a negative integer. Replacing x by -x in ·(2), 4( -x, y)= II -x+y 112_ II -x- Y II I +i U-x+iy 112 -ill-x-iYIl2 = II x-y 1\2_ II x+y 11 2 +i II x-iy II I-i II x+iy 112
361
HILBERT SPACE
Comparing this with (2), 4( -x, y)=-4(x, y) or Let
(c)
(-x, y)= -(x, y) fJ is positive integer. (cxx, y)=<-~x, y)=-(fJx, y), by (14) = -fJ(x, y), by (a)=cx(x, y)
ot= -
.. ,(14)
p, where
When
ot
is a rational number.
Let IX=~ where p and q are integers and q:;t:.O. q
/p
(lXX, y)=,q. x, y
, , / Pi' (X) y"/=P "i' /x y /, ' /=" by (a) and (b)
/x (aX, y)=P" (i' y "/
or
,., (15)
x
or
-=z, then x=qz q (qz, y)=q(z, y) by (a) and (b). 1 /x" (x, y)=q(z, y) or (x, Y)='ii' y / '
q
Putting this in (15), (ax, y)=~ (x, Y)=IX(X, y). q Similarly it can be proved that (CltX, y)=IX(X, y) when CIt is real. (d) When IX is a complex number. Replacing x by Ix in (2), 4(ix, y)= II ix+y iI2-- II ix-y 112+i II ix+iy 112-i II ix-iy 1,1 = / j j211 x-iy 1/ 2 _ / i 1211.x+iy 112+i / ; /2 nx+y h2 -; / i \2 II x-y li2 = II x-iy 1: 2- II x+iy 111+1 II x+y \12_; II x-·y 1/ 2 Comparing this with (2), 4(ix, y)=4i(x, y) or (ix, y)=i(x, y) ,:.(16) Let IX=a 1+;(1(2 where Cil:1 and (X2 are real. «(Xx, y)=«CX1+;:J2) x, y)=(CZ1X+icx2 x, y) =(CXIX, y)+(icx 2x, y)~ by (iii) =otl(X, y)+Clt2(ix, y), by (c) =a1(x, y)+iot9(X, y), by (16) =(Clt1+iClt2) (x, y)=cx(x, y) Hence (<<x, Y)=IX(X, y) for every scalar CIt. Hence B is a Hilbert space.
362
a E
HILBBR'r ~PACE
Theorem 4. An inner product space is a "ormed vector space. Proof. Let V(K) be inner product space. Let u, v E V and K be arbitrary. We define II u II=v(u, u). But (u, u) is defined on V. :. II u II is defined on V. (i) (u, u) ~ 0 and (u, u)=O iff u=o. Consequently, II u II ~ 0 and II u II =0 ifu=O. (ii) II au ii=/ a! I! u i/. For I! ou !1 2 =(ou, ou)=a (u, ou)=aii (u, u) = I a 12 I! u II·. Taking square root, I! au I! = I a III u II.
I! u+ v II ~ II u il + I! v II I! u+v I/I=(U+V, u+v)=(u, u+v)+(v, u+v) =(u, u)+(u, v)+(v, u)+(v, v)
(iii)
.:
= I! = II ~ II ~ I!
U ~/2+ u //2+ u //2+ u 112+
I! v 112+(u, 1')+(u, v) I! v 111+2 real part (u, v) II v //2+2 I (u, v) I II v 112+211 u /III v I! , by Schwarz's inequality =( I! u I! + II v I! ). II u+v II ~ II u II + I! v :1, .: norm of any vector ~ O. Thus the norm defined on an inner product space satisfies all the conditions of a normed vector space and hence it is a normed vector space. Theorem 5. A closed convex subset C of a Hilbert space H contains a unique vector of smallest norm. [Kanpur M.Sc. 88, 84, 83; Meerut 70] Proof. Write a= inf {II x II : x E C}. Then there exists Lim sequence (x .. ) E C s.t. n .. oo II x" II =a so that
...
II x" II -. 0 as II -+ C1J • Let x"' x'" belong to the sequence (x,,) . bs fH ~ X-..2 + X.. IS a convex su et 0 - .E C Hence, by def. of a, a <; this =>
II
X .. +Xm 1/ 2 ~ 4a 2
I, x",t Xm II, or
- II
X ..
+Xm
UI ~ -4a2•
HILBERT
SPAC~
363
By parallelogram law,
II x .. +x'" III+I! x"-x,,, 112=2 [11
x"
Il2+11 ~'" tll)
II X"-X,,, li 2 = -Ii X,,+X'" 112+2 [/I x" 112+11 x.lI l ]
or
~ -4a2
+2 [II Xn 112+11
x'" 112].
Making m, n-«>,
x'" Ij2 ~ - 40 2 + 2 [a 2 +a2 ] = 0 as n 00. Lim ' Hence U X"-x,,, 11=0. m, n-?oo This => (x,,) is a Cauchy sequence. Since H is complete and C is closed subset of H. Hence C is complete so that every Cauchy sequence in C converges at some point in C so that X,,-?Xo E C, say.
Ii x" -
Now II Hence
Xo Xo
Ii=Illim x" I =lim II Xn I\=a. is a vector in C with the smallest norm a.
Uniqueness of xO' If possible let xo. Yo E C s.t. II Xo lI=a, "Yo I =a. To show that xo=Yo' C is conveX ::. X,,;Y o E C => a
I
~ 1 ~oiYo\
i
<
or or
=> 4al II Xo+Yo 112. By parallelogram law, II Xo-)'o 1/ 2+ II xo+)'o 1'2= 2 [II Xo 1,8+ II Yo \1 2]=2 [all f al] II Xo- Yo 112=4a 2 -11 Xo+ ,1'0 112 ~ 4all -4a2=0 II xo-Yo 112 ~ O. But I xo-Yo 1\ ~ O.
Hence II Xo- Yo 1\ =0, this => Xo= Yo· Tbeorem 6. Let M be a closed linear subspace of a Hilbert space H. Let x 11. M but x E H. Let d be the distance from x to M. Then there exists unique Yo e M S.t. I x-Yo ll=d. [Meerut 19 8; Kanpur M.Sc. F. 79; Banaras 71] Proof,
By def. of distance d=d (x, M)=inf{1I x Y I: Y E M} Then:l sequence (y,,) in M s.t. 1\ x-y" II-?d. Consider tv. 0 vectors y", y", of sequence (y,,). . a l'mear su bspace -..... y,,+Y M IS - 2 -... E M . By def. of d, d
~
.\Ix-
(y"+y,,,) 2
I\
364 or or
or
HILBERT SPACE
4d 2 ~ II 2x-(Yn+ Ym) liS ... (1) 1/ (x- Yn)-l--(X-Ym) 1\2 ~ 4d 2 By parallelogram law, II (X-Yn)+(X-Ym) 1\2+11 (X-Yn)--(X-Y... ) WI =2 [1/ X-Yn ./1 2 +11 X-Ym 11 2J =2 [d2+d21~_tl.s m, n_oo II Yn-Y'" 1\2=4d2-112x-(Yn+x",) 1\ as m, n.OO ~ 4d2 -4d2 =0 by (1) But \I Yn - Ym 112 ~ 0 ¥ n, m. Hence \I Yn-y ... l1= 0 as m, n-+oo This => (Yn> is a Cauchy sequence in M. Being a closed subset of a complete space, M ill complete. Hence (Yn) converges to some point, say)'o e M
:.
lim Yn=Yo x-Yo 11=11 x-lim Yn 11=\1 \im (x-Yn) \I = lim Hence Yo is a vector in M S.t. II x- Yo lI=d.
\I
Uniqueness of Yo. If possible, let 3 Yo, Yl II x--Ylll=d, \I X-Yo \I=d. To show that Yl=YO' Bydef.ofd,
e
\I
X-Yn /:=d
M s.t.
...(2)
d~llx-YltYOil
112x-(Yl+YO) 112 ~ 4d2 • .._(3) By parallelogram law, \I (x- Yo)+(x- Yl) 1/2+ II {x-Yo)-(x-yJ III =2 [Ii X-YtlII2+11 (x-yJ III] =2 [d 2 +d1 j, by (2) or \I Yx -Yo 112=4d~-112x-{Yx+Yo) Ipi ~ 4d2 -- 4d 2 =0, by (3) or II (Yx - Yo) 112 ~ O. -But II )'c Yo II ;;;a O. Hence II YI-JO 11=0, this => 31=Yo' Theorem 7. If M is a proper closed subspace of a Hilbert space H, then 3 zo;t 0 E H S.t. Zo 1. M. fKanpur M.Sc. F. 19.3, 89 ; Meerut 88J Proof. M is a proper subspace of H => :I x E H but x ft M. Let d be the distance from x to M, then d=,d{x, M)=inf {II x-Y II: Y E M}.
or
365
HILBERT SPACE
Since M is closed and so 3 unique vector Yo E M s.t. II x-Yo lI=d (Refer theorem 6). Write zo=x-Yo, then x, Yll E H ::;. X-Yo=Zo E H. Also II Zo lI=d. Cliam. Zo .1.. M. Let Y E M and ac be a real number. Yo. Y EM!:> Yo+acy E Mas M is subspace => dl <; " x--(Yo+cxy) Ill, by def. of d => dB <; II Zo-cxy 111 as ZII=X-Yo or (Zo-«Y, Zo-«Y) ~ d2 or (Zo. Zo-GlY) -II (Y. Zo-IXY) ;;;:;: d2 or (Zo. ::0)-1£ (Zo' Y)-IX [(y, Zo)-ct (y, y)] ;;. dB [ac is real => GC=ii!] or II z(' IIIl-cx (zo, y)-cx (y, zu)+Cl 2 11 Y hll ~ dB or or or
n zo 112- ac [(zo, Y)+(Zo, y)]+«211 y il2
~ dB
d2 - 01.2 Re (zo, y)+acll II y 112 ;at dB [Re=real part] -2« Re (zo, y)+ml! II y 1/ 2 ;at 0 for every at. Hence this proves that Re (zo. y)=O Replacing y by iy, 1m (zo. y)=O Accordingly, (zo, y)=O "rI y E M :.
zo.L M.
Theorem 8. Let W be a non-empty sllbset of a Hilbert space H. Then Wi is!l closed linear subspace of H. [Meerut 1970] Proof. By def. Wi={X Ell: (x, y)=O y YEW}. Let x, y E Wi be arbitrary and two scalars GC, fl. Also let u E W be arbitrary. Then (x, u)=O=(y, u) Then (a.x+{:Jy, u)=rx. (x, u)+~ (y, u)=rx..O+{:J.O=O :. IXx+{:Jy E Wi, showing thereby Wi is a subspace of H. Remains to prove that Wi is a closed subset of H. For this we must show that Xu is any limiting point of Wi => Xo E Wi Now Xo is any limiting point of Wi => 3 sequence (x.. ) in Wi S.t. lim x,,=xo YEW, XII E WL => (x,,, y) =0 for every y
366
HILBERT SPACE
lim (x n , y)=O => (lim x n , y)=O
n-+oo
[For inner. product is continuous map] ::>
o Theorem 9.
(xu, y)=O. Also yEW E Wol => Wol is closed.
Xo
If M is a linear subspace of a Hilbert space H,
show that M is closed <:> M =MJ.J.. [Kanpur 1987] MJ.J.=fx e H: (x, y)=O Y y E Mol) I. Let M=Molol. To prove M is closed. By theorem 8, Mol is closed subspace of H so that (Mol)J·=Molol is again closed subspace
Proof.
But M=Molol. Hence M is closed.
II. Let Mbe closed. To prove M=Molol. Mol={x E H: (x, y)=O
Y y E M}
and
x EM=> (x, y)=O ¥ y E Mol :::> x E Ml.l. :. Me Moll.. ...(t' Now suppose the relation (1) is proper so that M;f:.MolJ.. Now M is a closed proper subspace of a Hilbert space MJ.J.. Hence, 3 a non-zero vector Za E MJ.ol s.t. zo..L.M.
(Refer Theorem) zo..L.M ~ z" E Mol. Thus Zo E Mol and Zo E MJ.ol so that Zo E MJ. n Molol ={O} This ... zo=O. A contradiction. For zo;f:.O. Hence M=MJ.J.. Theorem 10. Let S, SI' S2 be sub$eta of a Hilbert space H. Th~" (i) {O}J.=H. (ii) HJ.=={O}, (iii) sns.1.={O}, (Iv) S. C S. :::> S20l C SI J.. Proof. (i) {O}ol={x H: (0, x)=O}=H For (0, x)=O ~ x e H
e
(ii)
Hl={X E H: (x, y)=O ¥ y E H} = set of all' tbose vectors of H which are pel pendicular to all vectors of H inclUJ
(iii)
Any x E Snsol => XES, x E Sol .. (x, x)=O .. x=O
(iv) Let SI C S. Any x E S~ol ::> x ..L. S2 => (x, y)=O
¥ y E S2
367
HILBERT SPACE
• in particular (x, y)=O x J. S1 .. x e S1.l :. any x e StJ. .. x e S1J. This S1. C S1.l. Theorem 11. If M and N are closed linear subspaees of a Hilbert space H st. M LN, then M +N is also closed. (Meerut 1987,81; Kaupar M.Sc:. F'. 80] Proof. Let z be a limit point of M + N. If we show that zeM+N, the result will be proved. z is a limit point of M +N ~ 3 sequence (Zn) in M +N s.t. lim Zn=Z MiN=> MnN={O}. ForSnS.1={O} • M+N=M(BN . • a unique elements x"e M, Yn => )v s.t. Z. = x,. +Yn. z",-zn=(x..+y.,.)-(Xn+Yn)=(x.-X")+Cv,,, y,,) Also x", --x.. E M, YIlt -Yne N, M J. N This => (Xm -x..)L(Y.-YII)' Hence, by Pythagorean theorem, II (x", -x,,)+(Ym-Yn) 1I1:.~11 x.-x.. I1 I +/l y .. -y.. /II or "zm--z.. I1I~/! X,,,-x .. HII+l! y .. -y.. /III (zoo) is a convergent sequence => 1/ z.,.-z.I/_ 0 as m~oo • • ~oo. ~
:. or
Nx", -XII 111+// y.-y.. III _ 0 II x," -XII 1/=0=// Y",-Yn" as.m, n-+oo.
~ (X,.)€ M, (y,,) e N are Cauchy sequences . • 3 XE M, ye N s.t. lim XII=X, lim YII= Y => z=lim z.. =lim (x,,+y,,)=xt-ye M+N • z€ M+N.. For M and N are closed subspaces of a Hilbert space H, fe., complete space. Hence M and N are also complete Hence every Cauchy sequence in M er N is convergent to some point in M or N. Also x€ M, YEN. x+yeM+N. Theorem 11. If M ;s a closed linear subspace of a Hilbert space H, then H-M(f)MJ.. . [Kaupar M.Sc. F. 1984 ; Meerut 82, 87]
This
368
HILBERT SPACE
Proof. Let M be a closed subspace of a Hilbert space H. Then ML is also a called subspace of H (Refer Theorem 8). Also we know that M.LnM={O}. [Refer Theorem 10 (iii)]. Hence if we show that H=M+M.L, the result will be proved. For H=M(f)M.L ~ H=M +M.L, MnMl.={O}. Evidently M +M.Lc H (I). Remains to prove HCM+M1. suppose not. Then M+M1. is a proper closed subspace of H. Hence 3 zo,cOe Hs.t. zo.LM+M1.. (Refer Theorem 7) so that zo.f.M and zo.LM1.. This
~ ZoE
=>
M1. and zoEMl..L <> ZoE M1.nM1.1.={O} zo=O. A contradiction. For zo#O.
Hence HCM+M1. ... (2). H=M+M.L.
Combining (1) and (2), we get
Problem 2.
If S is a non-empty subset of a Hilbert space, then S1.=S1.1.1.. [Kanpur 1978]
Solution.
(i) By def.,
any
xe S => x.LS1. => xeS.L.L S1.CS1.1.
or (ii)
... (1)
Replacing S by S1. in (I), we get S.L C(S.L ).L.L
or
S.L CS.L.L.L
... (2)
But ACB => B.LCA.L, by Theorem 10. In view of this (I) gives or
(Sl..i.).LCS.L S.L.L.L C S.L
... (3)
Combining (2) and (3), S1.=S.L.L.L. Problem 3. If S is a non-empty subset of a Hi/bert space H, SIIOW that S.i.L is the closure of the set of all linear combinations 0/ the vectors in S, i.e., [Meerut 1976] Solution. We know that SCS.L.L ... (1) (Refer Problem 2). It means that S.L.L is a closed subspace of a Hilbert space H S.t. SCS.l.L. By def. of closure, [S] is the smallest closed subspace of H containting S. .
369
HILBERT SPACE
It follows that .[S] C SJ..1. We have But ACB
~
... (1)
S C [S1. BJ. C A.1. ... (2), by Theorem 10.
:. [S]J. C SJ.. Again applying (2), SHe [S1 J.J. [S] is
... (3) ~
a closed subspa.ce
... (4) (Refer Theorem 9)
(S1=[S)J..1.
By (3) and (4), SHe [SJ
... (5)
Combining (I) and (5), we get [S]=S.1..L. Theorem 13. (Bessel inequality). Let {Xl> x 2 , ... Xn} be ajinite orthonormal set in a Hilhert space H. If xE H, then
I
.E
(x, x,)
• ~l
Proof.
12~ /I x II!.
[Kanpur M.Sc. F. 1984, 87; Meerut 88] By definition of orthogonal set, (x,. xi)=8
t
JI '=10
if i=j if i,ej
..
V/rite y=x- E (x, Xi) X,. Evidently YEH. i-I fI
/I Y 1I 1=(y, Y)=(X- E (X, Xi) Xi, y)
'_1
fI
=(x, y)- E (X, x,) (XI, y) 1=1
=" /
X, X -
fI " ~ (x, X,) X, / -
't~l
..
=(x, x) - ]: (x,
E.. (X, Xt)
~_1
(XI'
Xi) (X, XI)
(_1
- E(X, XI) [ (" XI, X- 1-1 i: (x, x,) XI ~I
fFor
(x,
IXv)=a (x, y)]
>J
y)
370
HILBERT SPACE
+ E" (X,
X,) (X, X,)
~_1
" =11 X 111- ,-I l: I (X, X.> Since II y or
Ipi ;;;;t 0 and so \I x iiI - l:" I (x, ~_1
" I (X,
II X 112 ;;;;t l: .
Theorem independent.
r'
14~
XI)
II
;iii 0
Proved.
X,) :-
1-1
An orthogonal set of non-zero veclf!rs is linearly [Kanpur M.Sc. P. 1977,75,70, M.Sc. 'F: 12J
Proof. Let S={Xl> x z, .... x,,} be an orthogonal set of noozero Vf'ctors in a Hilbert space H. To prove S is linearly independent. Syppose l:" a,x, =0
... (1) where a, are scalars.
I-I
From which, (
i
'-'-I
l:" a, <X,. ,-1
a~. ~,. xr/'=
x,)=o
By a'i~umption, (Xt, x,)=O
¥ i except i =r.
HILBERT SPACE
371
Hence ar (xr, xr)=O, or a,,=O for r= 1,2, ... , II. Now (l) declares that S is linearly independent. Theorem 15. A" orthonormal set of vectors is linearly independent. Proof. Let S={x1 • x 2 • •• , x .. } be an orthonormal set in a Hilbert space H. To prove that S is linearly independent .
a
..
Let E a,x.=O :I-I
... (1) where a, are scalars.
~ a~ x" Xr/" =(0, x,) ,,=1
Then .(
.
~ 0,
.=1
(x,. x,)=O
~ a, 8',=0. But ~',=O
or
.=1
"I i except
or
i=r
0,8',=0 or ,ar.l:...,O or a,=O. This => a;=O for i-'I, 2, ... , n. Now (I) proves that S is linearly independent. Theorem 16. (Gram Schmidt orthogonali.&ation process). Let S={U., u." .. , u~} be a linearly independent set (or basis) in a Hilbert space H. Then there exists orthonormal set of vectors Sl={Wl> W2 , ... , W,.} S.t. L(S)=L(SI)' [Meerat 19701 Proof. Let S={u1 , ... , u..} be a L.I. set,in a Hilbert space H. To construct an orthonormal set SI={Wl> .. " w.. } S.t. L(S)=L(Sl)' L.I. means 'Linearly independent'. L.D. means 'Linearly dependent .. Now we shall construct an orthogonormal set Sl-{K!t, 'wa, ... ,w,.} in H by the help of elements of S. The main idea behind this construction is that each w, is a linear combination of vectors belonging to S. S is basis .. Sis L.l. ~ u,:;cO for 1 .so;; i .so;; n. Set
V1=Ul>
vlI=O ..
=>
U2
(u 2 , VI) lit· Va=U.- II Villa •
is a scalar multiple of VI =Ul
{UIl, u1} i6 L.D.
972
HfLBERT SPACE
A contradi ct i on. For {u�. u,} is L.I. Since every subset of L.I. set is L.I. Consequently v,:;co. (Ul• VI) VI "( Vi' VI >/ V =" U2- IIVI lit • , / " / (UII• UI) III _
-,,-u.--U�' <
> (u2, UI) U2, Ul - II (Ul• UI) UI lit =0. For (u .. uI)= 1/ Ul )1'. Hence V2 is orthogonal to VI and therefore {VI' V2} is ortho· gonal set. =
Set
� (ua• v.) Vc va=ua - '_1 '" II VI IIii =
Ua
(ua. VI)
(ua. V,) V9
- rrv;rra - -I
V3:;i:O. For va=O implies that {UtI U�, U3) is L.D. which is impossi ble. For {ul• Ul, ua} is L.I.
(ua• VI) VI (ua• V2) Vz V " I/ II Vt II' - Tv;fI" < a'VI) -
/ '
Now
s.t. i:/:j and I,
j= I, 2, 3.
Thus we have constructed an orthogonal set {VI' v� • of k (where k < n) vectors s.t.
Then (v" VI)-O s.t. i;t;j and i, j= 1,2, . . . , k. Ie
Set Vk+l=UHl- � -
'-1
(UH1' Vi) II Vi III�
VI
.
Then Vt+I:;i:O. For Vk+l=O implies that L.D. A con+radiction. For {ul .. . . U1:+1} is L.I.
l
. . .• Vio
{ul• U2, .. ·, Ilk+1}
373
HILBERT SPACE
If 1 <j E; k. then (VH1> VJ
)
:.
=
(
UII+1. V,
)
'
v,) (VI, VI) 1/ VI " • •
(UHIt
[For (v,. ",)=0 s.t. l#j and l,j=l. 2.... , k] =(Uk+1' VI)-(UII+1> 11/)=0 . :. (v'+l v,)=O. Consequently (VI. v/)=O s.t. l::j:j and 1<;1. j~k+ I. Our aim is now completed by induction. Thus we have cons.. tructed an orthogonal set {VI' vz•..• , v.. } of n vectors. Put Then {WI. w2 •• •• , w..} is an orthogonal set in which every vector is unit vector. meaning thereby Sl:=::twi. • w.. } is orthoe normal S&;t of vectors in V\f'). Since an orthonormal set is L.I. Hence Sl is L.I. By the above. expressio~. L[{u l .••• u}]=L[{w l • W ••.••• wr }]. Hence by induction. the required construction is obtained. Problems related to operators. Theorem 17. Every non-zero Hilbert space contains a complete orthonormal set. Proof.
Let H be a non-zerQ Hilbert space. 3 x E: H s.t. *~O.
Then
Normalize the vector x by wtitiDg Xl . II ~".
Then {Xl} is
clearly an orthonormal set in H. Thus every non-zero Hilbert space contains orthonormal set. Consider the collection or all possib!e orthonormal sets in H. By Zorn's lemma. this collection hilS a maximal number M. We assert that M is complete. Let us suppose y.:;;e.O s.t. y4..M. Set y Yl=lfYil'
374
HILBERl' SPACE
Then MU {Yt} is also an orthonormal set s.t. M U {yil ::l M;
contradicting the maximaJity of M as M is complete Finally Y..LM ~ y=o :. M is complete orthonormal set. Theorem 18. Let H be a Hilbert space. and leI {eel be an orthonormal set in H. Then the following conditions are all equivalent to one another (1) {e,} is complete (2) x..L{e,} ~ x=o (3) if x is, an arbitrary v(!ctor in H, then x=E (x, e,) e, I
(4)
if x is an arbitrary vector in H, then
,
n x III=E I (x, e,) 12.(Kanpur 1990,8!; Meerut 82, 79] Solution. (1)
~
'
(1). It is given that the orthonormal set {e,1
is compl~te. Suppose that xJ. {e,}. To prove that x=O, Suppose not.
Then' ¢O. Write e=" ~ II' Then e is a unit vector s.t.
e..L{e,} so that {ec, e} is ~mhonormal set which properly contains It means that {e,l is not complete orthonormal set. A contradiction. Hence x=O Finally x..L{e.} ~ x=O. (2). (3). It is given that x..L,e} => x=O.
{e,l.
Let x
E H be
arbitrary • To prove that x::.£E <x, e,). I
Take Y7x-E (x, e,) e,. I
Then '
(y, el)=(x-E (x, e,) e" e,) I
-(x, el)-E (x, e,) (~" el) I
=(x, e/)-(x" e,> (eh el>. For I, i=j (e" el)= { 0, i=t=j -(x, e,)-(x, el)=O
375
HILBERT SPACE
This .. y .i el .. Y .1 {e,l 01> x-:E (x, e,) e,=O
... y=O, by assumption.
I
o x=:E (x. e.) e,. I
(3)
0
(4). It is given that
x=:E (x, e,) e,. I
To prove that II x II I=:E I (x. e,) I • By as.)umption, II x II 2=<X. x)=(:E (x, e,) elt x) I
=E (x, l',) (eh x)=E (x, e.) (x. l',) I
I
=E I (x, e.) (4) => (1).
Ii
as z'z ..; I i
1'1.
It is given that
II x II '=E I (x, l',) II I
II
x
e: H.
To prove that {e,} is complete. Suppose not. Then {e,}. is not complete so that it is a proper subset of ali or'tho~ormal set {e, e}. Consequently eJ. {e.} 80 that (e. e,)=O
By assumption. II or
II
i.
en ·=E I (e. e,) I ·=0 I
or 1=0. This contradicts the fact that e is a u.nit vector. Hence {e,} is complete. Problem. Let E={e,} be an orthogonal set in a Hilbert space H. Show that B is compete iff E I (x. e,) I 1= II x UI [Meerct 1987] lEI . Solution I. Here write (4)·... (I) of the above Theorem 18. II. Given {e,} is complete. Aim. II x II ·=E I (x. e,) II nell =0.
I
Let
y=x-E (x, e,) e, I
... (1)
HIL~ERT SPACE
376
Tllen prove as in above Theorem 18 that (y, e1)=O. This .. y J.e1 But {e.} is complete. Consequently y=O. Now (1) gives x=X (x, e,) e,
...
II x 1I
1
•
=(x,
x)=(}; (x, e,) ej, x) i I
,
==X (x, e,) (e,. x) .
=X (x, e,) (x, e.)=E I (x, 1',) I
I
II.
•
Problem 4. Prove that Hilbert space is finite dimensional every complete orthonormal set is basis. Solution. Let H be a Hilbert space over a field K of scalars. Let S={e,} be a complete, orthonormal set in H. . .. (1) Then S is linearly independent (Refer 'theorem 15) For any x € H, we have ... (2) x=X (l(, ec) e"
*
,
by Theorem 18. Case I. Let dim H=n=finite To prove that S is basis (2) .. S generates H. . ... (3) By (1) and (3), it follows that S is a basis, dim H=n also proves that number of vectors in S must be n. Case 11. Let S be a'basis. To prove that dim J/=finite. We claim that S is a finite set. Suppose S is infinite. Then 3 A caSu. A ill denumerable. Take
A-{e1, ea,··., eft .... }. QO
Consider that series X
It-I
1
i IItl I == nl
as
e
GO
-i1: Uft. n _-1 ! eft I =
1
Then ¥
n.
All<> X ~ is a convergent series (with p-2 > I).
317
HILBERT SPACE
Hence, by Weierstrass M-test, the series X e~ is uniformly n convergent and so converges to some 'vector x E H so that ... (4) AI~
S is basis for H. Hence we can write
x=aAeA+ ... +a~e~.
... (5)
Letj be a positive integer having value different from the values of indices A, ... , 1-'. By (5), (x, e,)=a). (eA> e,) +a,. (e,., e,) =a~.O+ ... +a,..O, by orthogonal properlY
+...
or
(x, eJ)=O
.. (6)
By (4), (x. e,)=( E e:, eJ '/ , .. =1 n
or
t
(x , eJ)=P .
By (6) and (7),
-:z1 =-=0'
J Hence S is finite. H=finite.
... (7) or
~=o J
which is not possible.
Also S is given to be basis.
This .. dim
Remark. The above problem proves that: an infinite orthonormal set in a Hilbert space cannot be a basis. Problem 5. Prove thaf any two complete orthonc rmal sets in a Hilbert spare H halle the same cardinal number. ~olution. Let S] and S2 be 1\\ 0 complete orthonormal seta in a Hilbert space H. Case I. Let S}={el' e2 • ... , e .. } be finite. S} is complete orthonormal set
.. S} is basis of H.
(Refer Problem 4)
• di:n H=n S2 is complete orthonormal se't in l-J ~ S2 is basis of H
378
HILBERT SPACE
But dim H=11 (already proved) => S, has n vectors. Thus SI and S2 both have 11 vectors and so their cardinal number is the same. Case n. When SI is infinite Let x E S2 be arbitrary and S2(X)=(Y : y E S2 and (y, x)=;eO} ..• (1) Then S2(X) C S2 and S.(x) is countable. Let z E S9 be arbitrary, then by'Parseval's identity, we have II z :1'= E I (z. x) II XESt
as SI is complete orthonormal set. But z E S2 • z is unit vector => II z 11=1 :. )= E I (z, x) II xeSI
This => iI x E SI S t. (z, x)=;eO => z E S2(X) for some x E S10 by (I) Finally. This suggest that S2=
U St(X) xeSI
... (2)
card (SI) = 11 1 , card (S2) =n 2• (2) => card S2=card [ U S2(X)] <; card SI Let
=> n 2 ~
XESI
n
... (3) interchanging the roles of SI and S2' we get . .,(4) E;; n2• By (3) and (4), we fet "1="2 . ot ca'rd Sls=cud S2' Problem 6. Jf ani I be :!ero and identity operator3 on a Hilbert'space H. Ihen 0*-'0, J*=I. l,
"1
°
Hence show that if T is non-singular ope;'alor on H, then T* is also non-sil1gular and in this case (T*)-l=(T-l)*. Solution. Let x, y E H. By def., O(X)FO ¥ x and {(x)=x "
379
HILBERT SPACE
Therefore from the uniqueness of adjoint operator, we have O(x)=O*(x) or 0=0*. . .. (3) (/(x), y)=(x, y) II. . .. (4) (f*(x), y)=(x, I(y» = (x, y) By (3) and (4), (/(x), y)=(l*(x), y). => I(x)=l~(x)
This
=> 1=1*.
Uf. tet T be non-singular so that T is invertible,. then This
==>
=>
TT-l=T-IT=f. (,rr- 1 )*=(T-IT)*=:I* (T-l)* T*=T*(T-l)*=I. For (AB).::B*A* (T*)-I=(T-l)*
" (5)
For B*A*'=A'li:B~=1 => (A*r 1 =B;'t. (5) proves that T* is invertible and hence T* is non-singular. Theorem 19. The sc(f adjoint operators in B(H) form a closed real linear subspace of B(H) and therefore a rcal Banach space which contailts the identity trallsformation. [Kaopur 1985, 80 ; Meerut 76 j Proof. Let B(H) denote the set of all operators on H, then B(B) is a Banach space. Let S be the set of self adjoint operators on H. Then . .. (1) S={T E B(H) : T*=T}. Let ex and f3 be real numbers. then ex=iX, S=~.
Let At.
. .. (2)
A~
E S be arbitrary, then Al*=Al' A!*=A2' (exAI +f3A2)*= iiAl *+ "fJ,42 * =«A 1 +fJA 2 •
... (3) by (2) and (3)
This => 02A t + (3A z E S, by (1) Thus At. A2 e: S and ex, fJ are real numbers => exAt +IJA2 e: S. This proves that S is real linear subspace of B(B). Remains to prove that S is closed. Let (A .. ) be sequence of operators in S s.t. A..-A, then A,,*=A...
. HILBERT SPACE
380
To show that A is self adjoint
II A -A* n = II A-A .. +A.. -A ..*+A ..*-A* II II A.--A.. II + II A .. -A,,* II + II A,,*--A* II Using the fact that A,,=A,,* and II Til == II -T II we get
<
II A-A* II
+"
E; !I A-An II (A-A .. )* II :::::211 A A.. II. For 11 T* II = II Til ¥ T. Making n~ 00 and notin!! that lim A~=A, we get II A -A'* II =0 or A=A* or A E S. Thus A,,--+A E S. This '0 S is closed. Thus S is a closed and real linear subspace of a Banaeh space B(H). Consequently S is itself a real Banach space. Also 1*=1, this ~ IE S. :. S contains identity transformation. Problem 7. If Al and A2 are self adjoint operators on H, then their product Al 4~ is self adjoint <¢' A1A2=A2.41 A1 and A2 commute. [Bundelkhand 1981] Solution. Let A1 and A~ be self adjoint operators on a Hilbert spac~ H. then Al"'=Att A2*=A~. A!l\O (Al.4~)*=A/ A1*=A2A1=AtA2 iff A1.42=.42.41 (A 1 4 2 )*=A1 A 2 iff A1 A 2 ==.4 2 A 1 • This proves that A1A2 is self adjoint <: AI and .42 commute. Theorem 20. The real Banach space of all self adjoint operators on a Hilbert space H is a partiolly ordered set whose linear structure and order structure ore related by the following properties: (0) ff A) ~ A2 Ihe'l A 1+.4 <; .4 2 +.4 ¥ A (b) A] ~ Ai and IX ~ O. then tlAl ~ IIA 2 • [Meerut 1976] Proof. Suppose that B is real Banach space of all self adjoint operators on a Hilbert space H. Let TJ> T2 .1 T3 E B and x E H be arbitrary. We define ' T. ~ T2 <=> (T}x, x) E;
*
HILBERT SPACE
381
For (TIX, x)=(T1x, x)
T 1= Tt TJ ~ T 1• (ii) Symmetric. Tl <;; TI , T'I. ~ T 1 .,. 1"1 = Ts For Tl <;; T2 • T9 ~ Tl ~ (TJx, x) <;; (T2X. x) and (TaX. x) EO;. (TIX, x) .,. (TIX, x)=(T2x, x) => T 1 =T2• (iii) Transitive. Tl C;; T~. T2 ~ Ta => Tl ~ Ta For Tl <; T2 , T2 T. ~ (Tlx, x) ~ (T2X, x) and (TaX. x) (TaX, x) $ (Tlx, x) <;; (T2X, x) <;; (TaX, x) ~ (T1x, x) <;; (T"dx, x) => Tl ~ T3 The above facts prove that (B. ~) is a partially ordered set. Second part. (i) Tl < T2 => 1"1 + T ~ Ta +T For T1 ~ T2 ::> (T1X, x) ~ (TaX, x) ~ (T1X, x")+(Tx, x) .;;; (T2x, x)+(Tx, x) ~ «T1 + T) (x) x)_ .;;; 2 + T) (x), x) ~ Tl + T ~ T 2 + T, by (I) (ji) Tl ~ T2 and at ~O ~ otTl < otT2 For Tl ~ T2 ~ (TIX, x> ~ (T\!x, ~ 0: (Tlx, x) ~ ot (TaX, x) o 1'1) (X), X) ~ ««T2) (.x), oX) Y. ~ 0 => «01: TI ) (X), X) ~ «OtT2 ) (X). X) => OtTl ~ acT2 \I IX ~ 0, by (1). This completes the proof. Theorem 21. If A is positive operator on H, then I+A is nonsingular. In particular, I+T* T and 1+ TT* are non-singular for arbi:rary operator T on H. [Meerut 1987] Proor. Let A be positive operator on a Hilbert space, then A=A* ... (1) and (Ax, x) ~ 0 ... (2) Let x,y€ H. (i) To show that (/ +A) (x)=O => X=O ¥ X € H (I+A) (x)=O ::> 1{;"(;)+Ax=O ~ x+Ax=O • Ax=-x ~ (Ax, x)=< -x, x)= - II x /II => (AX, x)= -/I x /II.,. -II X 112 ~ 0, by (2) .. IIxdl~O $
~
<
<
«T x>
«a
a
382
HILBERT SPACE
II X 112 ;;;;;e 0 is always true :. "x /12=0 or II x II =0 or x=O (ii) To show that I+A is one-one. (I +A) (x)=(I+A) (y) => (I+A) (x- y)=O => x-y=O, by (i) => x=y But
I+A is one-one. To show that I+A is onto. Let M be the range space of 1+ A. Then M={(l+A) (x) : x e HI Now we claim that if is closed. " (I+A) (x) 11:r= III (x)+A (x) 112= II x+.4 (x) II! :.
(iiI)
... (3)
=(x+Ax, x+Ax) =(x, x+Ax)+(Ax, x+Ax) =[(x, x)+(x, Ax)]+[(Ax, x)+(Ax. Ax)] = " x1l2 + II A¥ W'+(Ax, x)+<x, Ax) = II x 112 + II Ax II 2+2 (Ax, x) [For (x, Ax)=(A'*x, x)=(Ax, x) as A*=A] = II x 112+ II Ax 11 2 +2 (Ax, x) ;;;;;e II x II a [For (Ax, x) ;;;;;e 0, by (2)]
II (l+A) (x) /1 2 " x II ~ " (1+A) (x) II ... (4) Let «(1+ T) (x .. » be a Cauchy sequence in M. Let m, n be any two positive integers I! x " 2<
or or
II Xm -x.. "
< II (1+14) (xm--X,.) "by (4)
= /I (1+A) (x m ) -(l+A) (x.. ) II _I) as m, n .. ex> This follows from the definition of Cauchy sequence. This => (x .. ) is a Cauchy sequence in H. But H is complete and so 3 x E H S.t. lim x .. =x and so (I+A) (X) E M lim (I+A) (x .. )=(i+A) (lim x .. )=(1+A) (x) E M, by (3) Thi~ => (1+A) (xn)-(l+A) (X) E M. Thus the Cauchy sequence «/+A) (x n » in M converges to (J+A) (x) e M. Consequently M is complete. But every complete subspace of a complete space is closed. This means that M is closed. Next we claim that M=H. Suppose not. Then M#-H.
HILBERT SPACE
It means that M is a proper closed suhspace of H. non·z:!ro vector x .. E H S.t. Xo.l.M. (Refer Theorem 7) Xo E H ~(I+A) (xo) E M, by (3)
or or or
and
383 Then EI
Also xo.LM. These two statements prove that Xo 1 (l +A) (x o), this => (U+A) (xo), xo>=O (xo+Axo, xo)=O (x6 • xo>+(Axo, xo)=O .. (5) II x 1j2+(Axo• xo)=O Also II Xo liz ~ 0 and (Axo. xo> ~ ()', by (2). Now (5) => II Xo iI =O= xo=O, in particular. A contradiction. M=H. ronsequently T+A is onto map. From what has been done it follows that 1+.4 is one-one onto map. [Meerut 1987J This ~ l+A is isomorphism => I+A is,non singql!J.r..
Second Part. Let T be an arbitrary operator. on H. Then TP and T*T both are positive operators. :. It follows from the first part that 1+ TT* and I+- T* T both are non-singular maps. Theorem 22. An operator Ton H is unitary ¢> it is an isometric isomorphism of I{onto itself. [Meerut 1974] Pj\)~f. Let T be an operator on a Hilbert space H. Let x,yE H. I. Let T be unitary. Aim. T is an isometric isomorphism. T is unitary => TT*=T*T=l => T is one-one onto S.t. f*= T-l => T is isomorphism. (Tx, Ty)=(x, (f* F) (y»=(x, I(y»)=(x, y) Taking x=y, (Tx, Tx)=(x, x) or I Tx 112=11 Xii" or 1\' !x 11 =:1 xU This proves that T is isc~etric. II. Let T be isometric and iSOl~osphism.
384
HILBERT SPACE
To prove that T is unitary. T is isomorphism => T is one-one and ORtO. =:. 1'-1 exists. Tis isometric =- 1I Tx '1=1\ x /I ~ II Tx III=I! X 112 => (Tx, Tx)=(x, XI => «T-*TI (x), x)=(l(x), x) 0:> «T:tT-I) (x), x)=O .:> T*'T-/=O => T*T=1 => T is unitary. Remark. The following problem can also be proved with the help of the above theorem. If T is an operator on a Hilbert space H, then the following are equivalent to one another. (i) T*T=I (ii) (Tx, Ty)=(x, y) iii) 1\ Tx II = II x II. Problems 8. If H is a finite dimensio'lal Hilbert space. show that every isometric isomorphism of H into itself is unitary. Solution. H is finite dimensioal space => T is onto map. • h' N ow T: H - onto --+ H is isometric Isomorp Ism. Then T is one-one onto so that 11 exists. This follows from the fact that T is isomorphism. . T is isometric ~ II Tx 1l=11 x II ~ II Tx 111=/1 X 11 2 00> (Tx, Tx)=(x, x) => «T*T) (x), x)=(/(x), x) ::. «T*T-I) (x), x)=O => T*T-l=O ~ T*T=I Similarly we can prove that T*'T=I. Thus T*T=TT*=I, showing thereby T is unitary. Problem 9. Show that an operator T on a Hi/bert shace H is unitary .;:> (T(ecH is a complete orthonormal set whenever {e,l is. [Meerut 1986] Solution. Let lei} be a complete orthonormal set in a Hilbert space Hand T be an operator on H. Then for i=j <et, e'>=~H= 0 for i::j:j I. Let The unitary, then TT*==T*T=I.
, {I
HILBI!RT SPACE
385
Aim. {Tef} is complete orthonormal set (Tx, Ty)=<x, (T*T) (y»=(x, I (y»=(x, y) This proves that (Te" Te;)=(e" e,)=Il" This ~ {T(e,)} is orthonormal set in H. Remains to prove that {Te.} is complete xl {Tei} 0> (x, Te.)=O ~ (T* x, ef)=O ~ T*x 1. ret} and {e,l is complete. ~ T*x=O (Refer Theorem 18) ~ TT*x=TO=O ~ J(x)=O ~ x=O. Now x 1. {Ted ~ x=O. This proves that {Tei} is complete orthonormal set. Theorem 18). II.
(Refer
Let {Tee} be complete orthonormal set in a Hilbert space
H. To prove that T is unitary. First we claim that T is isometry i.e .• II Tx /i=/i x I! '" x E H. But II T(O) 11=/1 0 1/ Hence II Tx 11=11 x I) for x=o. Let
x~O,
then { M; II
By assumption, {eel is orthonormal
~
. .. (1 )
f is an orthonormal set in H.
{T(e,)} is orthonormal.
Hence { T ( 11; II )} is orthonormal set
J T(X)} lliXiI
or
This
.
IS
orthonormal set.
~ \ l~~~i \!= 1 .. ~
11 T(x)
II T(x)
iI=1I xII
I!= 1/ x If for x:;t:O
... (2)
By (1) and (2), 11 T(x) 11=11 x II Itf x ... (3) .'. T preserves norms or, T is isometric. Let M be the range of the map T : H-+H, then T(H)= M, then M is a subspace of H. We claim M is closed. Let y be the limit point of M, then 3 a sequence T(x.. in M
< »
386
HILBERT SPACE
s.t. T(xn)-+y so that (T(x,,» is convergent. Consequently (T(xn» is a Cauchy sequence. . .. (4) By virtue of (3),
II Xn -x", 112= II T(xn -XIII) 112
=/1
T(x,,)-T(x ...) W! as Tis a linear. -0 as m, n-. 00, by virtue of (4) II X,,-X m II -+0 as m, n-+oo. This (x,,) in II is a Cauchy sequence.
=-
But H is complete. Hence (x,,) converges to a point x E H. Hence T(x) E M. Now y==' lim
n~co
T(x,,)= T (Um
X,.)
all T is continuous
=T(x) EM y EM
or
y is a limit point of M. y E M. This proves that M is a closed subspace of a complete space H. As a result of which M is complete. Now we shall show that :.
T(H)=M=H. Suppose not. space of H.
Then M #-H so that M is a proper closed sub-
By theorem 7. 3 non-zero vector Yo e H S.t. Yo 1. M. Now Yo -L M, M is complete. J'0=0 (Refer Th. 18) A contradiction. :. M-H. This. -T: H-+ H is onto ~p. Also T is isometric & T is one-one. by virtue of (3). :. T is isometric and isomorphism of II onto H. Hence T is unitary. (Refer Theorem 22). Theorem 23. Riesz representltiea tlaeerem for c,ontinuous linear junctionais on a Hilbert space. Let H be a Hilbert Space and let / E H* be arbitrary. Then there exists a unique ve:-tor J' E H s.I./=/". i.e./(x)=(x, y) Y x E H. [Kanpur M.Sc. F.II', 13; .....y Measure Theory 66; Meerut 88, Pr&of. Firstly we shall show that if such y exists, i.e., f(x)-(x, y) ... (1) then :t is unique. If possible let y be not unique and so 3 y' with the property that f(x) = (x, y'). Also ftx)==(.~, y). Then
"'J
387
HILBERT SPACE
(x, y)=(x, y') or (x, y-y')=O V- x E H. Taking x=y-y', then (y-y', y-y')=O. This => y-y'=O => y=y', showing thereby y is unique. Step II. To prove the existence of y occurring in the equation I(x)=(x, y).
Ifl=0 (zero functional), then choosing y=O, theorem immediately follows. So we suppose
1#0. Write M={x E H :/(x)=O}. Then M is a null space of/so that M is a closed subspace of H. Consequently 3 yo,t:O E H s.t. Yo .1 M (Refer Theorem 7). Then we shall show that y=rxy.. will serve our purpose for suitable value of scalar ex. ... (2) Yo 1. M => (x, Yo)=O ¥ x E M Now if x E M, then (x, y)=(x, cY.)=« (x, Yo> =i (i}-I, , ,", (l~. ~ Also I(x) =0, by def. of M :. l(x)=O=(x, y> or f(x)~(x, y) so that (I) is satisfied ¥ x E M and ¥ II.
01
To find the condition for (I) for X=Yo, l(yo)=(Yo, ~Yo)=« (Yo, yo)=li II Yo 1/ 1 - f(yo) f(yo) «=il1o lit or IX=II Yo ,11 Thus the vector /(Yo) y=CltYo= 11-'-,I'Yu Yo !,
... (3}
... (4}
will satisfy (I) X E Mand X=y•. Call this fact by the name (*). Step III. To prove/(x}=(x, y) "f x, Y E H. Any x E H can be expressed in the form x=m+lly, for SOme scalar 11 and vector m E M. :. /(x)=/(m+~yo)-/(m}+Il/(Y.) as / is linear =(m. y) +~ (Yo, y), by (*). =(m+.Byu. y)=(x, y) when y is given by (4). ¥
388
HILBERT SPACE
, _f(Yof
FinaHy )0-11 Yo 11~'Yo is a unique vector in H satisfying the condition flx)=(x y) Hence the theorem. Theorem 24. Let H be a Hi/bert space. Let T* be the adjoint of operator Ton Hand QI a sell/or. Also let S he an operator on H. Then (i) (T+S)"*=T*+S*, (it) {rtT)*=dT*, (iii) (TS)*~-=S Til: (iv) (T*)*=1'. [Meerut 1988; Kaopur M Sc. F. LiBear Algebra 76] Proof. Let u, v E H be arbitrary. (i) «T+S) (U), v)=(T(u)+S(u). v) =(T(u), v>+ (SlU), v)",--(u, T*(v»)+(u, S*(v)/ =(u, (T*) (V)+S*(v»=(u. (T*+S*) (v». But «T+S) (u), (v»=(u, (T+ S)iI':(v» :. (u, (T+S)*(v» ~ (u, (TJi<+S'At) (v». Uniqueness of adjoint implies (T+S)*=T·+S· lGuru Nanall. 82; Meerut 72] (ti) «aT) (u), v)=(aT(u). v)=a (T(u), v) =a (u, T-~(v»= (TS)*=S*T*. [Kaopur 1984; Meerut 72] (iv) ': (T*(u), v)=(u, T(v» Also (T*(u), v) =(u, (T·)* (v» .: (u, T(v) ;(u, (T·)* (v». This ~ T=(T*)*. [Meerut 1970] Theorem 25. Let T be a linear operator on a Hilbert space H. Then
(I) (il)
IIT!/-= II T* 1\
II T 112=11 Tr* II. [Kanpor 84; Gurullaoak 82: Meerut 88, 87J
HILBERT 5:PACE
Proof.
Let x, y € /I be 3rbitrary.
111= = as T**=T II TT*y II· I y II, by Schwarz's lemma. ~ II Til·!i T*r II· ily II Dividing by !I T*y Ii, we get /I T*y 1\ ~ II T \1-/1 yll ~ \I T II if II y !I < 1 :1 T*y 1/ <; II T I ¥ Y s.t. II y /I ~ 1 sup {IIT*y II: II y II ~ I} ~ sup II TII=II Til (i)
II
389
T*y ~
or or
UT*lIoE;IITIi Replacing T by T*, II T \I ~
or
II r*'II· Combining (1) and (2), II T 11=11 T* II· (ii) Since 11 ~T II ~ !I S iI·1/ T /I. Hence II IT* II ~ II Til II T* 1/= I T Ipl, by (3) II IT* II ~ II T III II Tx IPI= =
... (1)
... (2) ...(3) •.• (4)
..; Ii T*Tx 11·11 x \I, by Schwarz's lemma. " II T*T II . II x II . II x II
= II T*T II lI:x II 2
J ~ 1/ T*T II if II x II <; 1 II Tx II J CO;;; II T--T II if II x II ~ I. Taking supremum of both sides and noting that sup \I T*TII = nT*TII ,we get sup { II Tx II 2: II x II Et: I} ~ II T*TII or [sup {II Tx II: II x II Et; 1}]2 '" II T*TI~ or II T II 2 ~ II T*T II ... (5) Combining (4) and (5), we get II T 111= II T*T H• Similar Problem. Prove that the mal' T-+ T* is a norm pre[Xaopar 1983] serving mapping oj 8(N} into 8(N·) Solution. Here prove that II Til=- \I T* II • Theorem 26. 1/ NI> N:. are normal operators on a Hilbert space H with the property that either commutes with the adjoint 0/ the other, then NI+N. line! NI NI are normaloperatf!rs. [Meerut 1979; BUndelkband 83; (Kanpur 199,o,Kaopur M Sc. F. Linear Algebra 77] Proof. By assumption NI N 1*=NI *N). N2* N2=Na NI* ... (1) (N} N2*=N2* Ni> N I • N2=N2 N I "') ... (2), (i) To prove that N I +\N2 is normal
or
390
HILBERT SPACE
(N1+Nz) (Nt +N2 )*=(Nt +N2 ) (Nt*+N2*') =NtNI*+NtNz*+N2NI*+NzNz* Also (N1+N2 )* (N1 +N2 )=(NI*+Nz*) (NI+NJ
... (3)
=~*~+~*~+~*~+~*~
=NI Nt*+NI * N 2 +N2* Nt+Nz N z*, by (I) =N1 Nt*+N. Nt*+NI N z*+N2 N2*' by (2) -(N1 +N2) (NI +N2 )*, by (3)
This. 1I1 +N2 is normal. (ii) To prove Nt Nz is normal. (NI N,) (Nt Nz)*=(NI N z) (Nz* NI*)=Nt (Nz N'J.*) NI* -NI (N2* N z) N t *, by (!) =(NI N 2 *) (N2 Nl*)= (N2* Nt) (Nt * N z), by (2) =N2* (Nt N I · ) N2=N2* (NI* N I ) N 2 • by (I) =(N2* N I *) (NI N.)=(Nt N~)· (NI N~) :. (NI N2 ) (NI Nz)*=(Nl N,,)· (Nl N2 ) • . This => Nt Nz is normal. Problem 27. Let T. be a linear operator on complex Hilbert Ipace V. .
T=O ~
-0 Y II € V. [BundelkhluMll983; Kanpur M.Sc. Final SO, 9",86] Proof. Let Tbe a In1ear operator on a complex Hilbert space V(F) and let II e: V be ,arbitrary. Then
Suppose To prove
T=O.
(i)
(Ii)
=O
=<0"(<<), «>=<0, «>=0. Suppose =0 Y gr..
. To prove T~O. (1) => (T(II+~), rt+~=O. {3 E
...(1)
v.
=> (T(cx)+T({3), «+IJ)=O «+f3>++(T([3), Cl)+(T(~}, f3>-0 => O+(T(<<), ~>+(T(fSkll)+O=O. by (I) ~ (T(<<). f3)+ (T(P), Cl)=O.
=> (T(<<), Q
This is true V lit, f3 E V. Replacing ~ by i{3 in (2) (T(ll), if3)-t
. ... (2)
HILBERT
SPAC~
191
7 (T(Il), f3)+i (T(8), at)=O -i (T(ac), /3)+i (T(P), at)=O or -(T(<<), ~)+(T(IJ), P)-O For I~O or (T(~), P) -·(T(~), at)=O. : .. (3) (2)+(3) gives (T(a), p)=O ¥ II, IJ E V. Taking ~= T(at), we get (T(Il), T('1)=O ¥ at e V. This ~ T(CIt)=O y. a. e V [For (u, u) = 0:0- u=O]
or or
=:>
T=fl.
Remark
In the above theorem, we have used complex numbers to prove the result a.nd hence this theorem may fail to be true for real inner product space. Theorem 28. Let T be a symmetric (self adjoint) linear transformation on Hilbert V(F). Show that (T(<<) ,«)=0 ~ T=O. [lanpur M.Sc F.1980J Proof. Let T be a symmetric linear transformation on Hilbert space V(F) so that T=T*. Let at, P e V be arbitrary.
Let To prove (i)
Now (li)
T=f). (T(Il), Gr.)=0. (T(at), Gr.)=(O(at), 11)=(0,
Let
To prove
at)==O.
(T(at), 4) -0.
... ( I)
T=O
(T( a.+ P), «+/I)-(T(4) + T(/I),
at+/I)
=(T(at), «+/I>+(T(/I), «+~) =(T(at), at)+(T(at), /I> (T(I1) , II)
+
+(T(~),
or
or or
F=R. i.e.
P),
By virtue of (I). this becomes O=O+ + +O +=O < T(at) , (J>+<~, T*(ct»=O < Tt«), /3>+ =0 ... (2) For T*=T Ca.,e I. When V(F) is a real inner product space so that In this case (at, fj)=(fJ. at)=(fJ, at) for (fj, at) E R. =<~, at>.
392
HILBERT SPACE
Using this in (2), +=O 2 =0 or -O =0, on taking {J=T(rt.).
or or
=> T(IX)=O
This
¥
IX
E V
A
o T=O. Case II. When V(F) is a complex Hilbert space then F=C. Replacing fJ by ;{J in (2), we ¥et + =0. This =>
7+i <~, T(Ot) > =0
=> i[-+ -+<{J, T(<<) > =0 => -<<<, Tlex»=O. Adding (2) and (3), =0 V (j E V. => =0, on taking {J=T{fI.)
This =>
T(IX)=O
¥
CIt
. .. (3)
E V=> T=O".
In either case 1'= O. . Theorem 29. Let T be a linear trans/ormation on Hilbert space V(F). Show that
T-O <:> =0 V IX, P e V. Proof. Suppose T is a linear operator on a Hilbert space V(F). Suppose at, fJ E V. (i) Suppose T=O. To prove that (T(at), (j>=0. (T(ex), (it)
Let
11>=<0(2), {J)=(O, r;>=o.
V ex, (J e V.
To prove T=O. In particular (I) => (T(2o), T(at»=O, where .. T(Ot)=O
V
at
e
. .. (1)
~=T(Gt)
Y
• T=41
Theorem 30. Suppose T is self-adjoint operator on Hilbert space V(K).
393
HILBERT SPAC8
Then S*TS is self adjoint for el'ery linear operator Son V. Further if S is invertible and S*TS is self-adjoint, then T is also selfadjoint. [Kanpur M.Sc. F. Linear Algebra 1976J Proof. Let T and S be arbitrary linear operators on a Hilbert space V(K). Step (i). Let l' be self-adjoint so that T = T*. To prove that S*TS is self-adjoint (S· TS)* = S*T* S, *= S*P"S= S*TS (S'1',S)*=S*TS. From this the required result follows. Step (ii). Let·S be invertible so that S-1 and (S*)-1 exist. S iTS is self-adjoint. => (S* n)* = s* TS. ~ S*T'*'S';'~=S*TS ~-S*'T*S=S*TS
S*T*(SS-I)=S*T(5S-}) (S*)-1 S*T*I=(S*)-1 S*TI ~ IT*I=ITI ;> T*=" T::> l' is self-adjoint. Theorem 31. Every linear operaTor T 011 a complex Hilbert space is uniquely expressible as 1'= T} + iT2 , where Tl alld T2 are Hermitian (self adjoint) operators alld i= v( -1) fKanpur 1991;86] ::>
::>
Proof. space.
Let T be a linear operator on a complex Hilbert
Evidently
-1'*)
1
T~=~i (1'-1'*), we
get
T=i (T+T*)+;
Taking T)=1 (1'+T*),
Gi
(T
T~, T 1 +iT2 • Remains to prove that Tl and T2 are Hermitian operators
on V. T/I<=1 (T+T-.\O)*=' (T*+T)~' (1'+1'*)=1'1' I 1 I T 2*=-i{ (T--T*)*=-2i (T*-T)=2; (T-P)=1'z• :. Tl·~Tl' T'1,*=T2This => T) and T2 are Hermitian. Uniqueness. If possible, let l' =P+ iQ be anol her representation where P*=P, Q*=Q. Then r*=(p+iQ)*=P"'+'iQ*=P-iQ
394
HlLBEIlT
N
T=P+iQ l:)Ti-l'*=p T
ow T"*'=P-iQf
2
~PACE
T*' =Q
'2;
~ T 1 =P, T2 =Q· Hence the representation T= T1 + iT2 is unique. Theorem 32. A necessary and sufficient. condition that Q linear operator T on a (omplex Hilbert Space V (unitary space) be selfadjoint (or Hermitian) is that be real for every «. [Delhi 1981; Kanpur M.Sc. F. 79] Proof. Let T be a linear operator on complex Hilbert space
V(F). (i)
Let T be self-adjoint so that T=T*. To prove that is real for every. E V. <'T(<<), cr)=(<<, T·; «(7.»=(<<, Tt.»=(T~CIt), «)
or
(T{CIt), (I.)= is real ¥ at E V. (ii) Let < T(CI), at> be real ¥ cx E V so that E R ¥ at E V. To prove that T=T*.
ThIs
~
(I)
. .. (1)
.. (T(<<), a)=(T«(I.), «)=(<<, T(a» =(T*(at), «) => (T(<<), Q[)= ~
(T(CIt), a)-
.. (T(IX)-T-!«(a), cr)=O ~ «(T~T*)
(a), «)=0
=> T-T.==O T=T*. Theorem 33. Let U be a linear operator on a Hilbert space V( K).
Then the following are equh·alent.
U*=U-1, UU*=U*V==I, (3) =(u, ).) ¥ lI, Y E Y, i.e., V preserves inner products (4) IIV(u)!I='/UI!¥UE v l Kanpur M Sc. Final Linear Algebra 1976] Proof. Let U be a linear operator on a Hilbert space V(K). Let lI, )' E V be arbitrary. We shall show that (1) (2)
(I) (I)
0:>
(2) ~ (3) ~ (4) •
~l).
~ (1).
U·=U-1". VU=*UU- 1, U.U=U-1U
HILBERT SPACE:.
395
=> UU*=I. U*U=! => UU*=U*U=I. (2) => (3).
UU*=", U·V=I => < (U*V) (u), v>=
.: I(u) = "
~
=> = . (3) => (4) .. " l' (V(u), U(u»=(u, u) => /I U(U) 112=:! u /1 2 => II U(U) !/=/I u II. (4)
=-
(1)
II V(u) Ii=il u!!
~ ii U(U) 1:2=11 u !12 => ::0- «U*U) (U), U>== «U*V-/) (U), U> =0 =>
U*U-I=O
u":;
By Th. 21.
~ U*U=I+O=1 ~ U*UU-l=IU-1 => U*/=U-1 .. U*=U-1. Corollary.
U is unitary ¢o II U(u -VI II = II u-v II ..;:. distance between the vectors u and V =distance between U(/I) and U(v) <* U is an isometry.
Ex. 10. Suppose T is unitary (orthogonal operator on a Hilbert shace Vand W is a T-im·ariant. Then _.how that W.l. Is also T·invariant. Solution. Suppose T is a unitary operator on a Hilbert space V(F) so that TT*=T*T=I. Let W be T-invariallt so that We v any I\' E W => T~ 1\') E W. To prove that Wi jc; T-invariant. it is enouguh to show that any WE Wi=> T(lr) e Wi.
Let
It'
E
Wand u E Wi be arbitrary; then E W.
=O and T(W)=Wl Tis u ,itrary, < T,u), 1\'\ >~ = <
T~u), T(II'»
= =0
:'196
HILBERT SPACE
Thus, < T~u). wl > = 0 ¥ wl E W. This * T(u) 1. W => T(u) E W J.. :. any u E WJ. => T(u) e WJ.
Proved.
Theorem 34. (a) let T be a linear operator on a complex lIilbert space V. Theil T is normal - 11 T*(u) 11= II T(u) II vuE V. [Deihl 1980; Bundelkhand 81; Meeurt 80J Proved. Suppose 7 is a linear operator on a complex Hilbert
space V. (i) T is normal => TT*= T*T. «rr*) (u), u)=«1 *T)(u), u) .. (T(T*(u», u)=(1fr(T(u)J, u) => (T*(u), T*(u»==(T(u), T**(u» => " T*(u) lill=(T(u), T(u»=I: T.(u) 1111 => II T*(u) 11=11 T(u) II· (U) Again 11"'(u) Ii.eli T(II) II. II T*(u) 112=11 T(u) III
=> (T"(u), T*(u»=(T(u), T(u» => «TT*) (u), II)=«T*T) (u), u) u>=O. Also V is wtnplex Hilbert space => 1T*- T*T=O (Refer Theorem ':J.7) => TT*=T*T => «TT*~ T"'T) (u),
Proved, T is normal Theort:m 34. (b). If T is a normal orerafor on Hi/bert space H, thm II TlII= 1/ T 11 2 • Proof Let U E /l be arbitrary. Then T is a normal operator => I: Tu I/=I! T*u /I .. (1), by Theorem 34a. II 12 11== SIJP iii 1'u !I : 1 Ii I ~ J} =sup (i T(Tu) 1/ : I II I! ~ I}. using (J) "..:sup {II T*(TII) Ii : iI u ii ~ I} =t;UP {I! (T*7) (u' Ii : i! Ii I! ~ I} = :! T*T 1i=I!T III, by Theorem 25. Theorem 35. Suppose T is a linear operator on a Hilbert space. Then T Is normal iff its real and imaginary parts commute. [Delhi 1981; Kanpur M.Sc. F. Linear Algebra 67, 77J Proof. We know that every linear opetator T on a Hilbert space is expressible as T=A+iB, where A=i (T+T*), =>
B=~ (T--T*)
are self-adjoint operators so that
A*=-·~.·f
B=B*.
397
HILBERT SPACE
T*=(A+iB}*=A*+TB*~=A iB. TT*,,,-(A+iB) (A-iB)~"'A2+ B2 +j (BA·-AB), T*T=(A.-iB) (A+iB)c--=A2+B2+i (AB-Bo4). T i" normal ~ TT*~-~ T*T --. A2 +-B2+i (BA -AB)'==A2+B~+i (AB-BA) e> BA - AB:-·=AB-BA ¢ 2AB=2BA <=> AB=BA <=> A and B commute ¢¢> real and im:lginary part') of I commute. Ex 11. Show that the set of all unitary operators on a Hilbert space V(K) is group under the composition of operators. [Kanpur M.Sc. F. 1985, 82; Meerut 69, 70] Solution. Let G denote the set of all unitary operators on a Hilbert space V(K). Let T, S, FE G be arbitrary so that T, S, F
are unitary operators on V(K). Consequently
, Tr*=T*T=I, SS*=$"*S=I. Firstly we shall show that TS is a unitary operator. (TS) (TS)*=(TS) (S*TI(:)=T(SS*) T*'=TIT* =TT*-=/ (TS)· (TS)=(S*T*) (TS)==S* (T*T) S=S*}S =S*S=/ (TS)* (TS)=(TS) (TS)*=I. This
~
1"S is a unitary operator. E G.
=> TS
(i) T. S E G -> TS E G. (if) 3 identity element I (identity operator)
Tl=IT=T. 1/*= 1=1*/ (iii) Every element T E G has its inverse For TT*=T*T=I ~ T*=T-1. Also T-I (T-1)*=T* (T*r 1
-= G S.t.
For
=1.
Similarly
(T-1)* (T-l)=i. Thus (T-1) (T-1)*=I=(T-1)*, (T-1) This::;. T-1 is a unitary operator ::;. T-l E G. (iv) (TS) F= T(SF). For [(TS) FJ (u)=(TS) [FlU)), uE V =
T lS(F(u)}]
j1
E G.
398
HILBERT SPACE
= T [(SF)(/I)] =[T(SF)](u) [(TS)F] (u)=[T(SF)](u) ¥ u e V. This => (TS) F=T(SF). The results (i), (ii), (iii) and (iv) taken together imply that G is a group w.r.t. operation of product of operators. E~. 12. 1fT is an arbitrary linear operator on afinite dimensional complex Hilbert space and if a and b are complex numbers such that, a 1=1 b 1=1, then aT+bT* is normal. {Meerut 1971; Kanpur 88] Solution. Let I a 1= I b I= I so that laj2=/bI 2 =1 or aa= bjj = 1 (aT+bT")*={aT* +bT*)*=aT""'+ bT (aT+bT*)*{a r +bT*)=(aT""'+bT) (aT+bT*) =aaT*T+ab (T*)2+baT2+bbTT* =T*T+ab (T*)2+abT2+TT* ... (1) (aT+hT*) (aT+bT*)*=(aT+bT*) (aT*+bT) =aaTI*+abT2+ba (T*)2+bbT*T =TT*+abT2+ab {T*)2+T*T =T·q+ab (T"",)2+al)TI + TT.. ...(2) Equating (1) to (2), we get (aT +bT*) (aT+hP)*=(aT +bT*)* (aT+bT*). This ~ aT +hT'iI< is a normal operator. Ex:. 13. Let T he a normal operator on a Hilbert space and fa polynomial with complex coefficients. Then the operator f(T) is normal. [Kanpur 1987J Letf=ao+a1x+a2x 3 + ... +a",x"', then /(T)=ool+a 1T+a 2 TJ+ ... +o",T"', (T*)A=T*T* ... T* (k factors) =(TT... T k factors)* =(Tk)*. Hence (T*)k=(Tk)* [f(T)]*=(aol+a1 T+ ... + a", T"')* =dol+d1T*+ ... +d", (1"")* [J(T)]*=dll+o1T*+ ... +a... (T*)",. Tis norm:!1 ~ TT*= T*T => T(T*f=T[T*(T*,)t-l]=(TT*) (T*YH =(T*'T) (T*) ..-1
Solution
or
... (1)
... (2)
.. (3)
399
HILBERT SPACE
=T*[T(T*)·-IJ= ...
= TX<[Tiit1 (T*)k-S] ~
T(T'*)"=(Ti'/<)2
[T(T;\~)a-2]
=(T:Jt:)t T T(T*)"=(T*" T. Similarly we can show that T"T*= T:Jt:7". By virtue of (3) and (4). it is easy to show that ~
.. (4)
[f(T)) [f(T»)*=[f(T)]* [f(1')]· f( T) is a normal operator. DeC. Projection If P is a project:on with range space M and null space N, then This
=-
M={x E H; Px=x}, N={x E H; Px=O}. Here we also say that P is a projeJtion on M along N. Also H=M~N.
A projection P on a Hilbert space H is called perpendicular projection if M and N are orthogonal. In this case N .L M and so N=Ml or M==;Nl.. Thus a perpendicular projection P on M is a projl!ction on AI along Ml. rn the theory of Hilbert space, we shall consider only tho~e projections which are perpendicular. Hence we omit the repeated use of the word perpendicular. Thus by a projection P on n, we shall always mean a perpfndicular projection on H. An operator P on a Hilbert space H is a projection iff P2=P,
p=p* Also Px=x ¥ x E M. Important result. (i) Px =X \rI x E M (ii) II Px II ~ Ii x \I (iii) pa=p*=p (iv) (Px, x)= II Px III. Theorem 36. P is a projection on a closed linear subspace M
on H ~ I-P is a projecTion on Mi. Proof. Let P be projection on a closed linear subspace M of a Hilhert space H. Aim. I-P is a projection on Mi. We have pZ=P*=P, Px=x >.,f x E M. (1_ p)I=/S+P2_IP-P/=J+P_P_1' =I-P (/-P)*=I*-P*=l-P Let N be range space of-I :-P.
400
HILBIlRT SPACE
any x EN.¢> (1-- P) (x)=x ¢> J(x) -P(x)=x <=> x-P(x)=x ~ --P(x)=O ~ P(x)=O ~ x is in the null space of P. ~ x E MJ.
.•.
N=MJ. •
Finally, we have proved that (l-P)2=I-P, (/-P)*=l-P, (I_·P) (x)=x ¥ x E Ml.. These facts prove that 1- P is a projection on MJ.. Conversely. Suppose that J-P is projection on MJ., where M is a closed linear subspace of H so that MJ.J.=M. It follows from the first part that l-(l-P) is a projection on (M.l) 1 =MJ.J. i e., P is a projection on M. Problem 14. If P is a projection on a closed linear subspace Mof H, then x E M ~ P(x)=x 4> " P(x) /1=/1 x II Solution. P is a projeetion on M ~ p2=P*=P I. Let Px=x where x E H. x E H ;... Px E M as M is range space => x E M as Px=x. Thus Px=x ~ x E M Conversely suppose that x E M: To prove Px=x. or
Let Px=y, then PPx=Py or Py=P 2x=Px Px=Py or Px-Py=O or P(x-y)=O This
~
x-y is in null space of P x-y E MJ. => 3 z E Ml. S.t. x--y=z ~ x=y+z. Px EM:$> Y E M as Px=y Now x=y+z with y E M, z E Ml. where H=MffiMJ.. Also x=x+O where x E M (given) and 0 E MJ.. Thus x+O=x=y+z Also x has unique representation. Comparing both sides, z=O, x=y. But y E M and so x E M. Thus x EM=> Px=x Finally x E M <:> Px=x. II. Aim. Px=x <:> " Px 1/=11 x" Obviously Px=x => /I Px !/=Il X" ...(1) ~
401
HILBERT SPACE
Conversely suppose that II Px
li=1/ x 1/.
••• (2)
To prove that Px=x. Evidently II Px+(I-P) x 11=11 x II. ...(3) But Px E M and (I-P) (x) E Ml.. This ~ Px .l (I-P) (x). By Pyttlagoras theorem. I Px+(l-P) (x) 111=11 Px 111+11 (I-P) (x) III using (2) and (3), II x 112=11 x 111+ ,,(/-P) (x) liZ or II (I-P) (x) !I =0 or "x-P(x) li=O or x-P(x}=O or Px=x. . .. (4) Finally x=Px <:> II Px 11= II x II, by (I) and (4). Theorem 37. (Projection Theorem). q P is a projection on a Hi/bert space H with range space M and null space N, then M .L N <:> P is self adjoint [Meerut 1973; Kanpur 84, 81] and in this case N=Ml.. Proof. Let P be a projection on a Hilbert space H, i.e., P is a projection on M along N so that H=MffiN, and z E H is uniquely expressible as z=x+y with x E M, YEN. Also Pz= P(x+y) =x. I Let M I. N, then (x, y)=O=(y, x). Now (Pz, 'z)=(x, z)=(x, x+y)=(x, x)+<x, y) =(x, x)+O=I! x Ill. . .. (1) Also (P*z, z)=(z, Pz)=(z, x)==(x+y, x) =(x, x)+(y, x)=(x, x)+O=il x II·. ... (2) Equating (I) to (2), (Pz, z)=(P*z, z) or «P -P*) z, z)=O yo z e H. This 0 P-P*=O ~ p=p* 0 P is self adjoint. U. Let P be self adjoint. To prove M .l N. x E M~ Px=x, By def. of M. y E No Py=O, by def. of N. (x, y)=(Px, y)=(x, p·~y)=<x, Py)
HILBERT SPACE
402
=(x, 0)=0 (X, y)=O x E M, YEN.
:.
Also This
=> M .L N.
... (1) Let M.l N. To prove N= Ml. x E N ~ x..LM ~ x E M.l. any . .. (2) /. Ne Mi. . .. (3) Suppos e (I) is not true, then N~M.1.. By (2) and (3), N is a proper subset of Mi. ~ N is a closed subspac~ of Hilbert space M.l. {For N is null space ~ N is closed] zo-:cO E Ml. S.t. zo...LN. ~ MJ.. => zoA..M. Also zo.lN. E Zo Now
III.
a
Hence we can conclu de zo.lH as H=M( JjN. This is possibl e only if zo=O. A contrad ication .
N=M.1.. Problem 15. If T is an arbitra ry linear operator 011 afinite dimensional comple x Hilbert space and if a al1d b are complex numbers such that I a 1=1 b I, /hen aT+b. * is normal. (Meeru t 1971]' Solution. Let I a / :-~ I b / so that I a 12~1 b /2 or ad=bli . (a r +bT-iF)* =dT*+ b(T*)* =dT*+ bT (aT+b T*)* (aT+b T* =(dr* +iiT) (aT+b T*) ... (1) =daT* T+dh (T*)2 +b'1r2 +bbTT * liT) (dT*+ T*) ~aT+bT*) (aT+b T*)*= (aT+b =aiiTr *+abT 2+bd (T*)2+ bbT*T . Using ad = bfj, in first and last term, (aT+b T*) (aT+bT *)*=bb TT-'F+ afjT2+ bd(T*) 2 . .. (2) +adr* T. get we (2), to Equati ng (I) (aT+b T*) (aT+b P')*=( aT-t-b T*)* (aT+bT~). Thi'l ~ aT+br *' is a normal opera.t or. Proble m 16. A closed linear subspace M of a Hi/bert space H is invariant under an operator T <:> MJ. is invariant under T". Solutio n I. Let M be invaria nt under T. To prove that M.l. Hence
is invaria nt under T'*'. Let
XE MJ..,
yE M, then (x, Y) =0.
HIL8ERT SPACE
403
By def. of adjoint, (T*'x, y)=(x, Ty)=(x, Yl)=O where Yl=TyEM. Thus (Tx:'F, y)=O. Also yEM. Hence this declares that T*'xeMJ.. Thu,; xEMl. ~ T*xEMJ.. :. MJ. is T*-invariant.
II. Let M be a closed linear subspace of H so that MJ.J.=M. Let Ml. be' invariant under T*. Aim. M is invariant under T. It follows from the first part that (Ml.)J. is invariant under (T*)* i.e. Ml.l. is invariant under T** i.e., M is invariant under T. Theorem 38. If P is a p:rojection on closed linear subspace M of a HUbert space Hand T is a linear operator on H. then , (Meerut 1987. 86) M reduces T <:> TP=PT. Proof. M is a closed subspace of H => H=M(f)Ml.. Suppose P is a projection on M along Ml., Pz=x, Px=x, Py=O then ... (1) where any zE H is uniquely expressible as z=x+y with xEM, yEMl.. I. Let M reduce T so that M and Ml. both are invariant under T. To prove TP=PT. By assumption, Tx E M, TyE MJ.. . .• (2) (PT) (z)=(PT) {x+y)=P [Tx+Ty]. Take TX=X1EM. Ty=y,EMl.. (PT) (z)=P(x1+Yt)=:=PX1+PYl=X1+O=X1 ••• (3), by (1) .. (4) (TP) (z) = T(fz) = Txd.x 1 by (1). Equating (3) to (4), (PT) (z)=( TP) (z). This => PT= TP. II. Conversely let TP=PT. To prove M reduces T, i.e., to prove M and Ml. both are invariant under T.
404
HILBERT SPACE
P (Tx)=(PT) (x)=(TP) (x)=T(Px)=Tx, by (I) P (Tx)=Tx, this::;. Tx E M P (Ty)=(PT) (y)=(TP) (y)=T(Py)=T(O)=O, by (I). P (Ty)=O, this TYEMJ. ••
xEM => TxEM
YEMJ. => TyEMJ..
Hence M reduces T. Theorem 39. Let El and E2 be perpendicular projections on U1 a71d U2 respectively, where U 1 and Ua are closed subspaces of a Hi/bert space V(K).· Then sholl' that El and £2 are orthogonal <=> U1 and U2 are orthogonal. E 1E 2 =0 or E2 E1 =0 <=> U}.1U2• [Kanpur M.Sc. 81, 78} Proof. Suppose E} and E2 are perpendicular projections on U1 and U2 respectively which are subspaces of a Hilbert space V so that E}~=El=El* E22=E2=E~*
'
"IE VI L 1:.~(U2) ="2 ~ "2 E U2 Let E1 and Ea be orthogonal so that E1(U1)
Step (i)
}
=U1
<=>
r
E1E~=O, E2 E1=0.
.. (I) ... (2) ...(3)
Let "IE VI' U2 E U 2 be arbitrary. To prove that VI and V,. are orthogonal, it is enough to show that ".,)=0.
<"1'
(u1o
ul E VI ~ E1(U1)=U1 }, by (2). u.E V 2 => E 2 ("-4)=U 2 u2 )=(E1(U 1 ), £2(U2 )=«E2 * E 1 ) (u 1 ), u2 ) =«E2E1) (U1), u2), by (I).
=(0 (u 1 ), u~), by !3) =(0, u2 )=0. Step (ii) Let VI and V 2 be orthogonal so that vd.U2 => U1 =U.J. ... (4) To prove that E1 and E2 are orthogonal, it is enough to show
that
H1LBER T SPACE
40S
Let U 1 E VI be arbitrary, then UtE U2 .L, by (4) "lEUI :::> £1(U1)=U1 :::> E1(u1)eUt =U2.L => £1(u 1)EU2.L => E1(U1)¢U2 For U2 nU2J.={0}. => £2(£I(U1»=0 For u¢U2 => E2(u)=0 => E 2E 1(U1 )=0 ¥ ul E U1
=> E2E 1 =0 => (£2£.)*=(0)*=0 :::> El*E2*=0,. E1 £2=O, by (1). Pr,oved. Theorem 40. If P is a projection on a closed linear subspace M of H, then M is invariant under an operator T ~ TP=PTP. Proof. Let P be the projection on a closed subspace M of H so that I'X=X ¥ xEM ... (1) ze H =>' PZE M. and ... (2) I. Let M be invariant under an operator T, then xEM => TxeM ... (3) Aim. TP=PTP ZEH:::> PzEM => T(PZ)EM, by (3) => r(pz)=pz ... (4) Further (PTP)(z)=aP [r(pz)J=P(Pz), by (4) _p2Z =PZ as PI=P -T(Pz), again by (4). =(TP) ,z) :. PTP=TP. PTP=TP n. Let .. ,(5) To prove that M is invariant under T. Let xeM, then Px=x. Now (5) gives (PTP) (x)=(TP) (x) or (PT) (Px)=T(Px) or (PT) (x)= Tx or p(Tx)e::. Tx This :> TXEM, according to (I). Thus xe M => TXE M This proves that Mis T-invariant. Theorem 41. If Pl' P 2• .. , P.,. are projections on closed linear sub,\p~lces M 1 , M 2 ,··., Mn of H, thell
406
HILBERT SPACE II
p= ~ P, is projection ~ P,'s are pairwise orthogonal
,-,
and in this case P is a projection on M = M J + M2 + ... + M ". lKanpur 1981, 78J Proof. Let P, be projection on closed linear subspaces of M, of H, then ... (1) P,I=P,=P,· and Pi (X,)=x, ¥ x, E Mt
Also let P*=(
or I.
P=PI +P2 + ... +P,,= "~ P, i_I
~ P,=P '£ _1 p,)* = '=£1 Pt*= .-1
P*=P Let P,'s be pairwise orthogonal so that
p,p,=O for i::pj Ai.... P is a projection. Then
P'==(
£ P,
=~
... (3)
)1 =~
P.2+2
~
i::pj P.+O=P, by (1) and (3)
,i-l
i
PiP,
Finally, J'2=P, p=p* This pr~es that P is projection. 11. Let P be projection. To prove that PIs are pairwise orthogonal. If T is any projection and zE H, then (Tz, z)=(Tlz, z)=(TTz, z)=(Tz, T*z) =(Tz, Tz)=1i Tz iiThus T is projection ~
nx 112=\1 P,(x) IP' ~ 1_1 ~" or
... (2)
... (4)
II P,(x) 11 2 = E" (PI x, x), by (4)
'_I
==(P1 x, x)+ ... +
407
HILBf:RT SPACE
From this we conclude that the sign of equality must hold throu~hout the above calculation. Hence to
II P,x
.
W= 1-1 ~
II P,(x) 1/2
~
II PJ(x) ill=O ¥ j s.t. j=l=i if j::p.i => x E Null space of P1 => x € M,L ifj¢:i => x J..MJ ¥ j s.t. j#i where x € M, This proves that M,.1. MJ for i:j:i
This
= Pj{X)=O
This ~ p,P,=O for i#}. Ill. Finally we have to prove that to
P is projection on M = ~ M, J-1
For this we have to prove that R(P)=M. any x E R(P) => Px=x
~ ( ~ P,
) (x)=x
x=P1X+P 2 X+ .. +p"x But PIX e Ml • PzX e Ma etc. Therefore x E M1+M.+ ... +M,,=M Thus x E R(P) => x € M o
.'!. R(P) C M ... {S) We have proved that II py 11=11 x II ¥ X € M, and each M, is contained in the range of P and so M is also contained in the range of P.
:. Me R(P) By (5) and (6). R(P)=M This completes the proof.
Problem 17.
... (6)
If P and Q are prujections on closed linear sub-
SpUCf:S M wnd N of H, 'pnJl'e that PQ is projection ... PQ=QP
n N. [Kanpur 1981, 78; Meerut 88.87] Solution. Suppose P and Q are projections on closed linear subspaces M and N respectively. Then P2=P=P*, Q2=Q=Q*. ...(1) In this case, show that PQ is projection on M
408
HILBERT SPAce
Px=x ¥ x E M Ilnd Qx=x ¥ x E N ... (2) Step I. Let PQ be a projection. Aim. PQ=QP. By assumption, (PQ)*=PQ or Q~p·=PQ Using (I), QP=PQ. Step II. Let PQ=QP ... (3) To prove that PQ is projection. (PQ)2:a(PQ) (PQ) = (PQ) (QP), by (3) =p(Q2) P=PQP, by (I) =P(QP)=P(PQ), by (3) =P"Q=-PQ, by (I) or (PQ)I=PQ ... (4) (PQ)*=Q*P*=QP, by (I) =PQ, by (3) or (PQ)*=PQ .. (5) By (4) and (5), PQ is projection. Step III. To show that PQ is projection on Mn-N For this we must show that R(PQ)=MnN and x EM n N ~ x € M, x € N ~ Px=-x, Qx=x => (PQ) (x)==P(Qx)=Px=x ~ (PQ) (x)=x ~ x € R(PQ) :. M nNe R(PQ) ... (6) • J Agam any y € R(I-Q) =-> (PQ) (y)=y ... (7) => P((PQ) (y)]==Py ~ (P2Q) (y)=Py =-> (PQ) (y)=Py, by (I) Usjn~ (7), we get Py=y, this => Y E M Using PQ=QP in (7), (QP) (y)=y ... (8) This ~ Q[(QP)y]=Qy ~ (Q2P) (y)=Qy ~ (QP) (y)=Qy ~ Qy=y, by (8) ~ Y € N. Thus y E R(PQ) =-> y E M, YEN ~ y E M n N :. R(PQ) C M n N ... (9) By (6) and (9), R(PQ)=M n N
This proves the required result. Problem 18. .If P is projection on (i) P is positive operator, i.e., P (ii) O~P~I (iii) II Px II E; II x II ¥ X e H. (iv) II P II ~ 1.
Solution. P is H.
proj~ction
on a
Q
Hilbert apace H, then 0
~
subspa~e
M of a Hilbert space
409
HILBERT SPACfl
=> PX=p "". and To prove (i)
X
p2=P*=P Let x E H
E M
... (1) ... (2)
=== = .... i/ Px !I~ ~ 0 Now ~ 0 and PI=P By def. this ~ P is positive operator 01> P;;' 0 ... (3) To pro"e (ii) P is projection on H ~ /- P is projection on H (Refer Theorem 36) ~ I-P ~ 0, by (i) => I ~ P .. (4) By (3) and (4), 0 E;.; P <; I To prOle (illl Let x E H. If R(P)=M, then R(I-P)=Ml., by Theorem (36). Evidently Px E M and (I-P) (XfE M~ Thi~
=> PXJ...(I-P) (x)
By Pythagoras theorem, " Px+(I-P) (x) 1!2= II Px 1i1+1i (I· x) (x) 1111 or I x 1i 2 =1t Px 112+1i x--Px !l'! ~ " Px j:~ or II x 112 ~ II Px liZ or II Px II ~11 x II To prO\'e (i\') By def. " P ii-sup { II Px II: /I x II <;; I} But II Px II ~ !! x /I , by (iii) part II P II ~ su p { II x II : II x \I ..; I} ~ I or II P II ~ 1. Problem 19. If Pond Q are projcctiultJ on closed linear subspaces M and N of a Hilbert space H, thell 5how that tile folluwing are equivalent: (I) P ~ Q. (Ii) II P,'{ II ~ II Qx II 'V x e H (iii) MeN. (ill) QP=P. (v) fQ=P. [Meerut 1980J Solution. In order to prove the Problem, we shall prove that (i) => (ii), (ii) => (iii), (iii) => (iv) , (iv)~ (V), (v) => 0)
===-:.'«Px, Px> =lil'x ii~ ... (1) v- x E II.
or
410
HILBERT SPACE ~
(ii) P ~ Q ~ Px ~ Qx ~ < Px, x > <; < Qx, x > => \I Px 112 II QX 11 2, by (I) => II Px II ~ /I QX " (ii) '"" (iii) '>.f x EM==- Px=x ~ II Px 112=/1 X 112 -::> II X 1/=/1 Px II EO;; II QX H=> II X II II QX II ... (2) But II QX" E; " X II ... (3) by Problem 18. Bv (2) and (3), II QX 1:=1/ X II This => Qx=x, by Problem 14. =>xEN 1 hus any ~ EM=> X E N : . MeN. (iii) => (iv). It is given that MeN. Px E M=range space of P. But MeN. :. Px E N and so Q(Px) E N. This preves that Q(Px)=Px. For yEN => Qy=y or (QP) (x)=Px ¥ X E H :. QP=P. (iv) ~ (v). It is given that QP=P. This ::;> (QP)*=P* => P'~Q*=P* => PQ= P. For P and Q are projections. (v) => (i). It is given that PQ=P. Let x E H. Then by (I), =11 Px 112=il (PQ)X 112 as PQ=P. =11 P(Qx) 1P' <" II Qx 1/2= , by (I). [For II Py II II y II ¥ yand '>.f P] or . This => Q. Problem 20. ~f P and Q are projections QII closed linear subspaces M a11d N of a Hilbert ~pace H, then prove that Q -P is a projection => P ~ Q. [Meerut 1974] Also prove that Q--P is Projection on NnMJ.. Solution. Let Q-P be a projection To prove that P ~ Q. Since every proje(;tJOn on a Hilbert space is a positive operator and so Q-P ~ 0, by problem 18 or Q ~ P, .or P ~ Q. Conversely. Suppose that P E;; Q. Aim. Q-P is a projection. P is projection => pI = p* =-' P In view of this, we have (i)
<
<
P"
<
<
411
HILBERT SPACE
(Q -P)*=Q*-P*=Q-P (Q_P)2=(Q _ P) (Q_P)=Q2+p2_PQ_QP =Q+P-PQ-QP=Q+P-P-P=Q-~
For P ~ Q => PQ=P, QP=P, by problem 19. Finally, (Q_P)2=Q_P, (Q-P)*=Q--P. This proves that Q-P is a projection. Second Part. To show that Q- P is a projection on NnM.1.. For this we must prove that R (Q-p)=NnM.1. and x E R (Q-P) _.. (Q-P) (x)=x ... (1) => Q (Q--P) (X)= Qx "'" (QI_QP) (x)=Qx. But Qll=Q; QP=PQ=P => (Q-P) (x)=Qx => x=Qx, by (1) => xERtQ) => xEN Again by (I), P(Q-P) (x)=Px => (PQ_P2) (x)=Px => (P-P) (x)=Px :;> O=Px "'" x E null space of P => x C M.1.. Finally, x E R (Q-P) .. x E N, x E M.1. => x E NnM.1.. :. R (Q-P) C NnM.1.. . .. (2) Again any y E NnMJ.. "'" YEN, Y E M.1. .. Qy=y, Py=O .. (Q-P) (y)=y-O=y => y e R (Q-P). :. NnM.1. C R (Q-P). ...(3) By (2) and (3), NnM.1.=R(Q-P). This completes the proof. Theorem 42. If N is a normal operator on a Hilbert space H, then II N 1/ 2 = II N2 II. lMeerut 19801 Proof. Let x E H be arbitrary. N is normal':> II Nx 1/=11 N*x II ... (1), by TIl 26. Replacing x by Nx, we get II NNx 1/=/1 N*Nx II or II N2x II = II N* Nx II .. (2) . By def. II NI II =sup t II N2x II : II x II ~ I} =sup { \I N*Nx II: II x II .;;; I}, by (2) = IIN*NII or II NI II = II N* N n using the fact that II N 111= II N*N II is true for any operator N, we get
II NI \I
= II Nlil.
Pro"ed:.
412
HILB£;(;. T SPACE
Exercises 1. (a) Define a Hilbert space. Prove the following in a Hilbert space H : (i) I <x, y> I .;;. II x II • H y II Oi) XII- X. Yn-+)' => (X n , YII) ~ (x, y) where <XII>, are sequences in H and x, Y E H. (b) Prove that a closed convex subset C of a Hilbert space H contains a unique vector of smallest norm. IKanpur 1988] 2. Define normal and unitary operators on a Hilbert space H. If T is an arbitrary operator on H and if IX and ~ are scalars S.t. i IX I = I fJ :, show that «.T+fJT-~ is normal. [Kanpur 1988] 3. (a) Define a Hilbert space. Prove the following in a Hilbert space H : (i) Jf M is a linear subspace of the Hilbert space H, show that M is closed <* M=Ml..i. (ii) If {P,} is an ortho normal set in H, then ~ I < x, et > il II X liz, x E H
<
i
(b) Define normal and unitary operators on a Hilbert space H. Let T be a normal operator on a Hilbert !>pace il and p(x) a polvnomial with complex coefficients. Prove that p[TJ is a normal operator. [Kanpur 1987] 4. Define a positive operator r on a Hilbert space H. Show that it hHs on unIque positive sequare root. lKan, ur 19871 S (a) Define a normed linear spaces (NLS). Prove the following5 in NLS (N, II II). (i) / [I x II - II Y ii/ ~ II x-J II (ii) lf {Xn} be a sequence in N, such tllat "',,-X, then II Xn II ... II x {Iii) If {'''n}, {Y .. } are sequences in Nand {IX n }. {(3.. } are scal~r sequences such that x n- x, J,,'-~Y and Cl" .... CIt. Bn-B. then ex"x,,+(;")'n-:- x+r~.r. [Kanpur 1989) (b) Prove that the linear spa:e C lO, 11 ot all contilluoUS real functions defined 011 the dosed unit interval [0, 1] is a normed linear space (NLS) w.r.t. the norm oran element/defined by
.
11/11
=
but not a Banach space.
f: 11(t) I
dt, [Kanpur 19861
HILBERT SPACE
413
6. (a) Define a Hilbert space. If B is a Banach space whose norms obeys the parallelogram law and if an inner product is delin d on B by
4(x,
y)=iJ
x+y
l/a_11 x-y :il+i" x+;y 111-; II x-Iy II!
then prove that B is a Hilbert space. (b) If x and yare any two elements of a Hilbert space H, prove that
I (x, y) I ~ II x II . II y
"
Explain when does equality hold. [Kanpur 1986] 1. (a> Show that an operator T on a Hilbert space H can uniquely he expressed as T= T1+iTjj where Tl and T2 are self adjoint operators H. (b) If T is an operator on a Hilbert space H for which (Tx. x)=O for all x, then show that T=O. (c) Prove that every Hilbert space i<; reflexive. [Kaopur 1986J 8. (a) Define a Hilbert space. Prove the following in Hilbert space H: (i) I (x, y) I < "x II • II y II (ii> Xn-X, Yn-y ~ (x .. , Yn)-(X, y) where !xnl, {Y .. } are sequences in H and x, Y E H. (b) Let H be a Hilbert space and {e,,} be an orthonormal set in H. Prove that the following conditions are all equivalent to one another: (i) fed is complete (ii) x..L {ei} x=O (iii) x E H => x=l; (x, e,:) e, (iv) x E H => II x i I=l; I (x, e,) 12 LAaopur 1985] 9. (a> Define self adjoint operator on a Hilbert space. Prove that the self adjoint oper!ltors in 8{H) from a closed real linear subsp!lce of B(H) which c.:mtains the identity transformation. (b) Define normal and unitary operators on a Hilbett space H. Prov~ that unitary opera.tors on H form a group. (Kanpur 1985] 10. Define a Hilbert space H and prove that a closed convex subset C of H contains a unique vector of smallest norm. [Kaopur 1984) 11. (a) If M is a closed linear subspace of a Hilbert space H, then prove that H=M(£)Ml., stating all the theorems that are used in the proof. (b) State and prove Bessel's inequality. [Kaaput 1984] U. (a> Define the adjoint operator r-~ r* on B (H) and prove that
=-
414
HILBERT SPACE
(Tt T2 ):'«=T2*' Tt '1: II T* II 17 Til (iii) II T* Til = II Till. (i)
(ii)
=
(b) If P is a projection on a Hilbert space H with range M and null space N then prove that M J. N <=> P is self adjoint. [Kanpor 19K4] 13. (a) S:ate and prove the closed graph theorem. (b) Define the conjugate T* of an operator T on a normed linear space N and prove that the mapping T-+T*' is a norm pre[Kanpor 1983] serving of B (.TIl) into B (N*'). 14. (a) Define a Hilbert space and prove that a closed convex subset C of a Hilbert space H contains a unique vector of smallest norm. (b) If M is a proper closed linear subspace of a Hilbert space H. then prove that there exists a non-zero vector Zo in H s.t. zoLM lKaDpar 1983] IS. (a) Let H be a Hilbert space and/be an arbitrary functional in H*. Prove that there exists a unique vector yin H such /(X)=(x. y) Y x E H. (b) Define the adjoint operation T-T~ on B (H) and prove that it is one-one onto as a mapping of B (H) into itself. [Kanpar 1983] 16. (a) If {ej} is an orthonormal set in a Hilbert space B, then show that 1:: I {x. e,} 12 ~ II X 1!2 for every vector x in H. (b) Prove that a Hilbert space H is separable <:> every orthonormal set in H is countable. [Kanpor M.Sc. F. 198Z] 17. (a) Lei: B be an ordered b:lsis of a Hilbert space H. Describle how using this fixed basis B a matrix [T] can be assigned to each operator Ton H. Such that: (i) the mapping T-+[T] is a one-one mapping of the set of all operators on H onto the set of all matrices. (ii) algebraic operations can be defined on the set of all matrices in such a manner that the mapping T-+[T] preserves the algebraic structure of B(B). (b) Show that the dimension of B( H} is fll, where n=dim H. [Kanpor M.Sc. F. 1981] 18. Prove that a closed convex subset lof a Hilbert space contains a unique element of smallest norm. [Kanpur M.Sc. F. 1981]
HILBERT SPACE
415
J9. (a) Prove that P lo P20 •• P n arc projections on closed linear sub;paces M i , M 2 , .••. , M .. of a Hilbert space H, then P=PI+Pa+ ... + 'n isa projection ifand only if P,'S are pairwise orthogonal. When this is the ca'>' , show further that P IS the projection on M = MI + M2+'" + M ... (b) If P and Q are projections on closed linear subspaces M and N of a Hilbert space H, prove that PQ is a projection if and onlv if PQ=QP. In this case, show that PQ is a projection on M n N. [Kanpur M.Sc. F. 1981, 78] 20. (a) If P is projection on H with range M and null space N, then show that MJ.. N <> P is self adjoint and in this case N=ML. (b) Show that the unitary operators on H form a group. [Kanpur M.Sc. F. 1982] 21. Prove that a Hilbert space H is separable if and only if every orthonormal set in H is countable. [Kanpor M.Sc. F. 19R11 (a) If M is a proper closed linear subspace of a Hilbert space H, prove that there exists a non-zero vector Zo in H, such that zoJ.M. (b) Prove that if M and N are closed linear subspaces of a Hilbert space H such that M .lN, then M +N is also closed. [Kaupur M.Sc. 1980, 78] 22. (a) Let H be a Hilbert space. Prove that the self adjoint operators in B(H) form a closed linear subspace of 8(H). (b) If an operator T on a Hilbert space /J is such that (Tx, x)=O for all x in H, show that T=O. [Kanpur M.Sc: F. 1980, 79J "0
20 Finite Dimensional Spectral Theory 20'1. Der. Let T be a linear operator on Hilbert space H(K). Any scalar I. E K is called an eign ~alue of T if 3 non-zero vector v E H s.t. TV=AV. Such a vector v is called eign vector of T corresponding to the eigen value ". The following have the same meaning: proper vector, eigen vector, characteristic vector, latent vector. . Similarly the following have the same meaning: latent value, elgn value, proper value, characteristic root, spectral value. The set of eign values of an operator T is called the spectrum of T and is denoted by aCT). Let M denote the set of all veetors of H. s.t. Tv='\v i.e. M={v E H: Tv="v} Then M is called eign space of~. [Kanpur 1988, 85] 20'2. Spectral Solution. Let T be a linear operator on a Hilbert space H. If:l distinct ~ and non-zero pairwise orthogonal complex numbers projections PI' Pz, "', P .. S.t .
"10 "I' .. ..
... (1 )
and
;=1
Then the expression (I) is called spectral solution for T. 20'3. Invariant. Let T be a linear operator T on a Hilbert space H. A closed subset M of H is said to be T-invarlant or invariant j under T if any x EM=> Tx E M. 20'4. Reducibility. Let T be linear operator on a Hilbert ~pace H. If M is a closed subspace of H, then T is said to be reduced by M if M and MJ. both are invariant under T. If M is invariant under T, then we also say sometimes that T is reduced by M or M reduces T. Theorem I. Let A be an eigen value 0/ an operator T on a
FINdE DIMeNSIONAL SPECTRAL THEORY
417
Hilbert Space H. LeI M be the eigen space of A, then M is a closed subspace of H. Proof. We have M={x E H: Tx=Ax}. Let a, b be scalars. x, y E M ~ Tx=Ax, Ty=Ay => T (ax+by)=aTx+bTy as T is linear ==aAx+bAy=A (ax+by) => T (ax+by)=A (ax+by). ax+by E H ~ ax+by EM. This proves that M is a subspace M={x E H: Tx=Ax=Mx}={x E H: (T-M)(x)=O} =Null space of linear map T-M =c1osed space. For null space of linear map is always closed. Hence the result. Theorem 2. The eigen space of a linear operator T on a Hilbert space H is invariant under T. Proof. Let M be an eigen space of T corresponding to eigen Then M ={x E H : Tx=Ax} ;\ E K, x EM=> Ax E M. by def. of subspace ~ Tx EM, by def. of M .. x E M~ Tx E M. Hence M is invariant under T. Theorem 3. The eigen values of a self adjoint operator are a/l real. , [Kanpur M.Sc. 1974, 71] Proof. Let S be a self adjoint operator on a Hilbert space H (K) Let Tx=;\x, then A is an eigen value of T relative to eigen vector x. To prove A is real. ;\ <x, x)= ;\-):" =0 => ,\=~ =:. ,\ is real. Theorem 4. If'x, yare eigen vectors of a self adjoint operator T on a Hilbert space H relative to different eigen values of T, [Kaapur M.Sc. P. 1971, M.Sc. F. 90] then x .L' y. Proof. Let T be self adjoint operator on a Hilbert space H (K). Let Tx=Ax, Ty=IJ.Y where Ai-II-' value A.
I
418
FINITE DIMEN5lONAL SPECTRAL THEORY
The prove T is self adjoint (Refer Theorem 3) ,\ (x, Y)=(AX, y)=( Tx, y) =(x, T*y) =(x, Ty) =(x, I-'y)
or
=p (x, Y)=/L (x, y)
(A-I') (x, y) =0
But Hence Tbeorem 5. Then
'\:F 1-'. (x, y)=O or x.1y. LeI T be a normal operator on a Hilhert space H.
Tu=,\u ~ T*U=AU, u E H [Kanpur M.Sc. F. Linear Algebra 1978, 76] Or Every vector of a normal operator T is also characteristic vector of T*. [Meerut 1988J Proof. Let T be a normal operator on a Hilbert space H, then TT* = T*T. The result will be proved in two steps. (i) Let x E H and a scalar II TtIC 1I1=(Tx, Tx) =(K, T* (Tx,)=('(, (T*T) (x» = (x, (Tf*) (x» (T*x, T*x)=11 T*x 112 This ~ \I Tx 11=11 T·x II. •.. (1) (ji) To prove T -aI is normal (T-al)*= T*-(a/)* =T*-aI·=T*-iiJ (T-al)(T-al)*={T-al) (T*-tii) =TT*.:aTI-aIT*+auI2
-a r
aT* + aul (T-:lI)*(T-all=(T*-aJ) (T-a/) = IT*
.. (2)
=T*T-aT*I-uIT+aaI2 =T*T-aT*-ar+aii, Equating (2) to (3) and noting Tr* -=T*T, we get (T-al) (T-a/)*=(T-aI,* (T-aI). This ~ T -aI is normal. (iii) Now we can replace T by T-AI in (I). Hence II (T -AI) (u) II = II (T -Al)* (u) II or
II Tu -AU II
= IIlT*-M I (u)
or
II Tu-,\u II
=
II T*u-fu II
II
... (3)
FINITE DIMENSIONAL SPECTRAL THEORY
419
This declares that II Tu-Au \I =0 <=> • T*II-AU H =0
Tu =A.u ~ T*u=Xu. Theorem 6 If T is a normal operator on a Hilbert space H. then eigen veclors of T belonging to distinct eigen values are [Kanpur 1988; Meerut 88] orthogonal. er
Proof. Suppose T is a normal operator on a Hilbert space H(K). Let
Tx=oU, Ty=p.y To prove
where p.-:FA.
xl.y
Tx=).x, Ty=p.y => T*x=Ax, T*y=j;)., by Theorem 5. ). (~, y)=(Ax, y)=(Tx, y)=(x, T*y)=('IC, p-y) =p. (x, y)
or (.\- p.) (x, y) =0 giving y)=O, this => xl.y. Remark. The above problem can also be expressed as follows: If Ml and M. are eigen spaces of a normal operator T relative to two ei!!en values ).10 ).2, then Ml and M. are orthogonal. Theorem 7. Supp',se T is a normal operator on a Hilbert space H(K). Let M be an eigen space of .\, where)' is an eigen value of T. Then M and Ml. both are invariant under T. [Kanpur M.Sc. F. Topology 1977, Linear Algebra 78] Proof. We have M={x E H : Tx=).x} ). E K, x EM=> Tx=).x.'\x E M as M is a subspace => Tx E M :. M is invariant under T. T is normal => Tr*-T*T T [T* x] =(17'*) (x) -(T*T)(x)-T*(Tx) = T*(.\)C) =).T*x .or TlT*x]=A (T*x).
<x
t
420
FIN1TE DI\fENSIONAl SPECTRAL THEORY
By def. of, M, this => T*x E M. Thus ¥ x EM=> , *x E M. This proves M is invariant under T*. Now M is invariant under T* => M 1. is invariant under (T*)* = T (Refer remark written just below this theorem) ~
M1. is invariant under T.
Hence the result. Remark. LeI M be invariant under any linear operator T so that
x EM=> Tx EM. To prove M1. is invariant undt-r T*. Let x E M1., Y E M. then (x. y)=O. By def. of adjoin{. (T*x. ?)=(x. Ty)=(x. YJ)=O ¥
where
Yl=Ty E M [For x E M1., YI ~ M => (x. YJ)=O]
Thus
(T*(x), y)=O
Also Y EM. Hence tbis declares that T* x E M L. Thus
•
M.l.
X
EM1. => T*x E M1..
ts T*-invariant.
Tbeorem 8. (Spectral Theorem for normal operator). Let T be an operator on a finite dimensional Hilbert space H. Let "1> be distinct eigen values of T, let MiJ M 2 • .... Mil be their eigen spaces; and let Pl. Pt , .... Pn be the projections 011 these eigen spaces. Then the following statemen!s are equivalent to one another:
"a..... ""
M/s are pairwise orthogonal and .'pan H. (ii) P;'s are pairwise orthogonal, (/)
n
n
I=E. PI, T=E "IPI, I-I
(iii) Tis numal.
Proof.
1_1
[Meerut 1988 ; Kanpur M.Sc. F. 83. 79. 77!
(i) => (ii). i e .• suppose (i) is true.
To prove (ii). By assumption.
H=M 1(£; M 2 (fJ ... (fJMn •
421
FINITE DIMENSIONAL SPECTRAL THEORY
Hence any x E H is uniquely expressible as n
X= E XI with XI E M, n-l
¥
I.
P,'S are projections on M,'s and Mt's are pairwise orthogonal, ~
P;'s are pairwi!>e orthogonal
... (1)
=> P;Pj=O=PjPI for l::I=j. P; is projection on M; => PI(X)=XI
... (2)
n
This => 1= ~ PI
... (3)
I-I
Tx=T (
i
XI )' =
I-I
i
TXI as T is linear
'-1
n
~
=
XI E MI
A;xl as
=>
TXI=}.lx,
.. (4)
From (I), l3) and (4), the required result follows. To prove (ii) => (iii), i.e., suppose (ii) is true. To prove (iii).
f AIP,r
Tt=[
=
i: ~I PI*= 1: -"I PI
as PI is projection => P,=P,*=P;I,
7' A;P) (
Tr*=( (ii)
:.
f
fx,P/)-f
... (5)
AlxjPIP,.
=> Pt's are pairwise orthogonal => P,P,=O for ;::I=j. TT*=E '\/5:; P;P,=E 1'\1 \2 P,2=E I '\; II Pi by (5).
Thus
;
;
TT*=E t
I Ai \2 Pi'
I
422
FINITE DIMENSIONAL SPE'TltAL tHEORY
T*T=E I A, \' PI'
,
Similarly
Hence TT*=T*T, showing thereby by T is normal. To prove (iii) => (i), i e., suppose T is normal To prove (i). T is a normal operator on Hilbert space Hand M/s are eigen spaces of T. => M/s are pairwise orthogonal, (by Th. 6). ...(6) • P,'S are pairwise orthogonal projections on M/s, (Refer Theorem 39 of Chapter 19) From (6), the first part of the r~sult is established. Write \ Then M is a closed subspace of H associated with projection n
"
P= 2' P, s.t. P,P,=P;P,=O for i¥=j.
'-1
.
T is a normal operator on H, M, is eigen space of T => M, reduces T 101 i, by Theorem 7. Also P, is a projection on M, • TP,= PIT of i, by Theorem 9.
TP=T ~ P,=E I
or
,
TP,=~ P,T=(~ ,
I
P,) T=PT
TP=PT, also T
IS operator on M. Hence M reduces T, by Theorem 38, Chapter 19. Hence M and MJ. both are invariant under T. Let T' be the restriction of T to MJ.. Then T' : ML-+Ml. S.t.
T'(x)=T(x)
¥
x E MJ..
... (1)
Let x be an eigen vector of T' relative to eigen value A, then T' (X)=Ax. ...(2) But T' is defined on MJ., (2) => x E M_L => T(x)=T' (x), by (I) using (2), we get T (X)=Ax => x is an eigen vector of T. :. Every eigen vector of T' is also an eigen vector of T. But T has no eigeo vector in MJ. as all the eigeo vectors of T are in M. Also M n MJ.={Ol. Consequently T' : ¥J.~M.J. is an operator with no eigen :. M~={O}. vectors and so no eigeo values.
FINITE DIlY.ENSlO AL SPECtRAL lHEORY
This
Q
=>
423
H=M as M(£;Ml.:::aH M,'s span H.
" Also H=M(£;Ml.=M= EM,. /-1
This Qo M,'s span H. Problem J. Let T be a normal operator on a finite dimensional Hilbert space and let p be a polynomial. Show that if A is an eigen value ot T, then p~A) is an eigen vahle of p(n. [Meerut J987] Solution. Let T be a no, mal operator on a finite, dimensional Hilbert space H. Let Tu=Au, i.e., A is an eigen value of T. I. To prove that PlT) is a normal operator. Prove this as in Ex. ) 3, page 398. II. To prove that peA) is an eigen value of p(T). p(x) =ao +a,x +' a.x l + ... +amxm PlT)=Qol+a1T+al T1 + ... + a,,,Tm .. (5) Tu=Au => T'u= T. Tu=T.(Au)=A(Tu)=A.Au => T 2u=A2 u Generalising tbis, ~e get T"u=A"u. In view of this (5) gives plT)u=aolu +a1Tu+ollT2u+ ... +arnT"'u = 0ou+a1Au+ a.;\'u+ +arnAmu =(oll+alA+a.;\2+ ... +amAm)u _p(A)u or, p( T)u=p(A)u. This => p(;\) is an eigen value of p(T). Problem 2. Let T be a normal operator on a Hilbert space
H. Let {A1' A2 • .... An} be the set of eigen values of T. Suppose M1-={x E H: ,Tx=A,x}. Prove that M,'s span H. [Meerut 1986] Hint. Here prove (iii) => (i) of Theorem 8. Problem 3. Let T be 0 linear operator on a jinlte-dimen.siona/ complex Hilbert .~pace V(K). Then T is normal iff the adjoint T* is a polynomial in T. [Meerut 1982, 79) Solution. Step (i). Suppose T is normal. To prove tbat 1* is polynomial in T. .
424
FINIfE DIMENSIONAL &PECTRAL THfOHY
Since T is normal and hence T is expressible as the spectral n
solution namely T= 1: c/Ej where c/ E K. i-I
and E/ is perpendicular pairwise orthogonal projections so that E j *=Ej =E/2. T is normal => TT*=T*T
f;
T*==(
CjE/
/-1
)* =
f; CiEi*=
i-I
i: CiEI
i-I
n
T* =
or
}; CiEi'
/~1
We know that each E/ is a polynomial in T. Consequently T* is a polynomial in T. Step (ii). Let T* be a polynomial in T so that T*=/(T}. To prove that T is normal.
TT*=T/(T}, T*T=!
... (1)
be a spectral resolution of T, where P;'s are pairwise non-zere orthogonal projections S.t. ..,(2)
and "/s are distinct complex numbers Step t. W ~ shall show that '\;'s are eigen values of T. There exists a non-zero vector x in the range of Pi S.t. PiX =X. TX=(
,f;
J-I
AiP,) X=
f
1-1
>'JPjPI (x)
using the fact that PiP 1= 0 if i ¥.j, we get TX=>.;P;2 (X}=-.\i PiX as Pil=p/ =.\/x or Tx=.\x
425
FINITE DIMENSIONAL SP CfRAL THEORY
This proves that A/
Step II.
'S an eigen \'alue of T
Write A ={Ah A2 ,
••••
¥ i.
A,,).
Aim. All the eigen values of T are precisely members of T. If possible, let A be an arbitrary eigen value cf T s.t. Tx=Ax.
Then
or
Tx=A lex)
(
"
£ AiPI )
i-I
X=A (
i
i-I
P; ) x
II
E AiPI(X)=A E PI(x)
or
I-I
i-I
};" A; p)p/ (X)=A E" PjP; (x)
or
'-1
1-1
or
>.) Pl (X)=A P/ (x)
or or
(A}-A) x=O.
A,P, X=AP, (x) ••
But
or
AjX='\X
x~O.
l-ience every eigen value of T is a member of A.
>',=A.
Thus we have proved that in the spectral resolution (I) of T, the scalars J..;'s are dIstinct eigen values of T.
Step III. Remains to prove the uniqueness of the solution (i). til
Let T= E
.-1
I)(.;QI ••• (3)
be another spectral resolution of T, where
ext's are distinct eigen values of T and Qt's are pairwise orthogonal projections.
We can also write (3) as
= E AI 2P,"= 1: A,2P; f
,
or
T2= 1:" A,2P,'
This:;>
TI= 1:
'-I n
1_1
A/l
Pi
... (5) ... (6)
426
FINn E DIMENSIONAL SPiCTRAL THEORY
Let get) by any polynomial \\ ith complex coffident in the complex variable I. Taking Irnear combination, we get n
g(T)= E g ('\i) PI /=1
~
p; is a polynomial s.t. Pi ('\,)= {
if i=j if i =1)
.. (7)
Taking p; in place of g,
Thus p,(T)=Pj • If we take po (t)= -<1-'\1) (I - '\2) ..• (I+'\n) I
(,\/-,\d (,\;-,\2) ••• (,\/_oA n )
then p; sat isfies (7). If we apply this discussion for Q/s then we shall get Q/ ""' p,(T) for each i. Therefore p/= Q; for each i. Hence the two spectral resolutions of T are the same. PlOblem 4. If T is an operator on a finite dimensional Hilbert .Ipace If, then prove the following statements. (0) V is singular ¢> 0 E (J (T), lKanpur 1984,81, 7S] i.e. TOis s ngular (/fO is the spectrum ofT.
(b) (c)
If T is non-~ingular, then A E (J (T) ::;.. ,\-1 E
(J
E (T-l)
[Kanpur 1984]
If A is r.onosingular, then (] (ATA-1)= (J(T)
If A E (] (Tl, and if p is any polynomial, then p (A) E a [p I T»). [Meerut 1982, 80] Hence (J (T) denote~ tire spectrum of T, i.e. the set of all eigen values of T. Solution la)o T is a sipgular operator ~ 3 x E H S.t. Tx=O where x.;iO ¢> Tx=Ox ¢> 0 is eigen value of T (d)
¢>
(b)
0E
(J
(T).
T is non-singular, " E
(J
(T)
:::. ,\ is eigen value of non-singular operator T ~ A=FO eigen value of non·~ingular T
427
FIN I 'E DIMf:!I"SIONAL SPECTR \L THEORY
~
3 x E If S.t x#O and Tx=Ax X=T-l (Ax)=AT-l (x) :;.. A-I X=T-l (x) => A-I E a (T-l).
~
Let S=ATA-], Then S-Al= ATA-I-A1AA-I=ATA-I_A(AlA·-I) =A [T--AT] A-I det(S-.\l)=det A det (T-Al).det A--I =det A. det A-I. det (T - AI) =det l.det (T -AI) as det A det A-l=det AA-l = I det (T- AI) or det (S~,u)=det (T-A1) :. det (S-,u)==-O ,'", det (T-Al)=O (c)
.. (1)
But det (S-M) -=- 0 ," A E a (S). :.
By (1), A E a (S) <=> A E a (T)
or or
a (8)=a (T) a (ATA-l)=a (T).
(d) (1)
Let ,\ E a (T), then 3 non-zero vector x S.t. Tx=Ax T 2x= T (Ax)=ATx=A AX=,\2x ~ 1'2x =A'x ~ T (Px)=T (AlIX)=.VTx => T3 X =A 2 . Ax ~ T3 X =A 3X.
.•. (2)
:;;10
Finally, TX=AX, T2 X=A 2 )(., T8X=AIX. Generalising, we get TnX=A ..X. Let p (1)=aO+alt-~ a2(2+ ... + 0 ,,1" be a polynomial in t of degree n \\-here an¥O. Then p tTl x=(ool +a1 T+aaP+ ... +anP) x ==aOx+a 1 Ax+a2 A2x+ ... +a.A n x = (au+olA+02A2+ ... +anA") x=p (A) x or p (T) xc=p (A),X with r,,=AX. This proves that
,\ E
(J
(T) => P (A) E
(J
[p (T»).
Problem 5. Let T be any operator 011 a fillite dimem;onal Hilbert space Hand N 0 normal operator on H. Shmv that if T commutes with N. then T alw commutes with N*. Solution Let T commute with normal operator N, then NN*=N*N ... (J) and TN",-NT .. (2) To prove tbat TN*=N*T I. We claim TN'''=NmT ... (3)
428
FINITB DIMENSIONAL SPECTRAL THlORY
The result is true for k = 1, by (2). Let (J) be true for k =m, TN'" = Nn'T. ...(4) Now TNtn+l=(TN"') N=(N"'T) N, by (4) =Nm (TN, =Nm(NT) = Nnl-tlT or TNm+l=Nm+l T which proves that the result is true for n=m+ 1 if it is true for n=m. Hence tbe result (3) is true by Induction. Let p (t)=a O+a l t+a.t 2+ ... +ant" be aby polynomial with complex coefficients. II. Tp (N)=T [ao/+alN+azNI+ ... +anNn] =a"T + alTN +a~TN2+ ... +anTNn =aolT +alNT+a2 N2T+ ... +anNnT, by (3) =(oo/+a l N+a 2N2+ ... + a"N") T or Tp (N)=p (N) T. ...(5) then
III.
Let N = 1:" Ai Pi be the I>pectral resolutIOn of the normal ;-J
operator N, where PI is projection and so that p/2=PI*=p/.
Then N* = (I' A/PI)· = 1: )./p/* =1: I
or
~/P/
I
N*=}; ~jPI'
... (6)
i
But each operator Pi is a polynomial in N. Hence, by virtue of (5), N* commutes with T. Problem 6 Let T be a normal operator on a finite dimens;onal Hilbert space H with lipearllnl AI. A2, ... , Am. Then prove the/ollowing statements (a) Tis Sl:'lj adjoint ~. each '\; is real, (b) T is positive "c- each ~ 0, (c) T is unitary «> I A; I= I ¥ i. c[Kanpur 1989, 91; Meerut 82J
"I
Ii
Solution.
Let T= I' "iP,
... (1)
1=1
be the spectral resolution of T, where AI. "', An are eigen values of. l' and PI, P 2 , ,Pn are non-zero pairwise orthogonal projections. This,* P?=P;*=P; ¥ i and P;P,=O for i~j. Here we also ha\e I'" P;=T. 1-1
...(2)
4
FINITE DIMENSIONAL SPECTRAL THEORY
T*=(E A;P;}*=E
).IP'"
-.c=E )./p;
/
... (
T"'=E ~Pi (a> let T be self adjoint, then T = T"'. Putting the values from (I) and (3) in (4), we get E (i;P;) = E AlP"
or
... (
i
or
P;=O.
E ().;-A/) i
or or or
Multiplying both sides by P j and nothing that p,p) ='1 ¥ i except i=j, we get C~/-Ai) P/'=O (~J-'\~) P/=O ("i,-A;) Pj=O or '3:,-A j =O ,\j= 1./.
This ==- Aj is real ¥ j. Conversely Suppose that A; is real l.tJ i, then A/= )./. . •. (5 Aim. T is self a~joint. By (3) and (5), T'" =E AiP/, By (I) and (6), T"'=T, showing thereby T is self adjoint. (b) Let x E H be arbitrary. ==<E Ai Pi x, E Pi x> J
=2-' E .\/ i
i
= E A/ /' j
• {Pil But P; Pj=PjP/= 0 Usin~
< p/
<
x, Pj
X
> ... (7.
Pi'" p/ x, x>
for &' lor
this in (7), we get < Tx. x >=E A/ I
i =- j "
':1=J
• (Pj =PJ ).
< P,2
X, X
>
== =\1 PI x 112. Using this in (8), <. Tx, x>=E A; \I Pi x \1 2 •
But
,
Let T be positive, then < TJI, x> ;;:: Aim. AI ~ 0 ¥i. By (9) and (10), E A. 1 PIX il 2 ~ O.
o.
,
Let us suppose that x is in the range of Ph then p.x=x, Plx=O ¥ i except i= 1.
. .. (9) .•• (10) ••• (11)
4JO
FINITE DlMLN310NAL SPECTRAL THEORY
Then (II) gives Al ~ O. Similarly we can prove A2 ~ O. Gelleralising, we get Ai ~ 0 ¥ i. Conversely suppose that Ai ;;;:. 0 ¥ i. Aim. T is positive. By (9) and (12), Th is ::> T is positive. (c)
~
... (12)
O.
TT· = (.r Ai P;) (E)./ PJ) i
= E
I' }
or
AI ~'J PI Pj=E Ai ii P I2 I
TT*=E I AI 12 PI.
... (13)
I
T is unitary
-¢>
TT*=I
-¢>
E I
~
I '\1 \2 P I =
E Pi>
by (2) aDd (13) ..
E (IA/12-1) PI =,0. I
Multiplyi"g both sides by PI and noting that PiP} =0 ¥ i except i =--j. we get (I
",\2_]) PJ=O
I A} \2-1=0 {:> IAj 1=]
vj. Exercises 1. (a) Define eigen value. eigen vector, eigen space and spectral resolution of an operator T on a Hilbert space H. Prove that. if A be an eigen value of an operator T, on a finite dimensional vector space H. then I A I ~ II Til. [Kanpur 1988J 2. (a I Define a Banach Algebra. Show that the set G of all regular elements in a Blnach Algebra is an open set. Further show that the mapping x-*x- 1 is continuous on G. [Kanpur 1989J (b) For an element x in a Banach Algebra A, prove that a (x) is non-empty. [Kanpur 1988] 3. Define spectrum and spectral resolution of an operator T on a non-zero Hilbert space. Prove that eigen values of an operator T constitutes a non-empty finite subset of the complex plane and the number of points in thi~ set does not exceed the [Kanpur 1987] dimension of the space. 4. (a) Define a Banach Algebra A. Also define regular and singular points in A. Prove that every elem:nt x for which II x . 1 II < ] is regular and its inverse is given by the formula -¢>
00
x- 1 =1+E (I-x)". ....1
[Kanpur 1989]
FI"ITE
DIMEN~JON"L
SPECTRAL THEORY
431
(b) Define spectral radius rex) of an element x of a Banach Algebra. Prove that
lim
n-+oo \I x' I lin . [Kaopur 1987] 5. (a) Define eigen vector, eigen value, eigen space, spectrum and spectral resolution of an operator T on a non-zero Hilbert splce H. Prove that jf A be an eigen. value of a operator T on a finite dimensional Hilbert space H, then I A I ~ II Til. (b) Let T be a normal operator 00 a Hilbert space H with spectrum {At. A2 , ••• , Am}. Use the spectral resoJution of T to prove th ~ following statements. (i) T is ~elf adjoint -= each Al is real. (ii) T is positive ¢> Ai ;?- 0 for each i, (iii) T is unitary I A/ I= I for each i. lKaopur 1986] 6. (a) Define eigen value, eigen vector, eigen space and spectral resolution of an operator T on a Hilbert space H. (b) Prove that the spectral resolution of a normal operator on a finite dimensional non-zero Hilbert space is unique. [Kanpur 1985] 7. (a) Prove that two matrices in A. are silT ilar';:> they are the matrics of a single operator H relative to different bases. (b) D~fine the spectrum (} (T) of an operator T on Hand prove that 0) T is singular.;:> 0 E a (T). (ii) If T is non-singular, then A E a (T) ¢> >,-1 E a (TI). {Kaopur 198'] 8. (a) Define regular and singular elements of a Banach Algebra A Prove that every element x for which II X -I II < I is reg'Jlar and the inverse of such an element is giver) by the formula r (.x) =
-=
X-I
"" = 1 + E (I-x)n. n-l
(b\ For an element x in the Banach Algebra A, prove that a (xl is non·empty. [Kanpur 1984] [Kanpur 198J] 9. State and prove the spectral Theorem. 10. (a) If S denotes the set of singular element of a Banach Algebra and Z denotes the set of all topologkal divisors of zero then prove that (i) Z C S. (ii) the boundary of S is a subset of Z. (b) Prove the following: (i) If r is an element of R then 1 -r is regular.
432
FINITE DIMENSIONAL SPECTRAL THEORY
(ii) If t -xr is regular then 1 -rx is also regular. . [Kaopur 1983] 11. Let G and S denote the set of regular elements and the set of singular elements respectively in a Banach Algebra A. Prove the following: (a) G is an open set. (b) The mapping X~X-I of G into Gis homcomorpliism of G onto itself. (c) The boundary of S is a subset of the set of all topological divisiors of zero. [Kanpur M.Sc. F. 19821 12.
