INTRODUCTION TO THE THEORY OF ENTIRE FUNCTIONS
This is Volume 56 in PURE AND APPLIED MATHEMATICS A series of Monographs and Textbooks Editors: PAUL A. SMITH AND SAMUEL EILENBERG A complete list of titles in this series appears at the end of this volume
INTRODUCTION TO THE THEORY OF ENTIRE FUNCTIONS A. S. B. HOLLAND Department of Mathematics University of Calgary Calgary, Canada
A C A D E M I C P R E S S New York and London A Subsidiary of Harcourt Brace Jovanovich, Publishers
1973
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Library of Congress Cataloging in Publication Data Holland, Anthony S B An introduction to the theory of entire functions. (Pure and applied mathematics; a series of monographs and textbooks) Bibliography: p. 1. Functions, Entire. I. Title. 11. Series. QA3.P8 vOI. [QA353.E5] 510’.8~ (515 ’.98] ‘72-88365 ISBN 0-1 2-352750-3
AMS (MOS)1970 Subject Classifications: 30-01,30-02, 30A62,30A66 PRINTED IN THE UNITED S T A W OF
AMERICA
CONTENTS
Preface
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
ix
Chapter I: A Study of the Maximum Modulus and Basic Theorems 1.1 1.2 1.3 1.4-1.8 1.9 1.10 1.11 1.12 1.13-1.14
1.15 1.16
The Nature of Singular Points . . . . . . . . . . . . . . . . . 1 Meromorphic Functions (Definition) . . . . . . . . . . . . . . . 4 Entire Functions (Definition) . . . . . . . . . . . . . . . . . . 5 Maximum and Minimum Modulus . . . . . . . . . . . . . . . 6 Order of Zeros . . . . . . . . . . . . . . . . . . . . . . . . 13 Algebraic Entire Functions . . . . . . . . . . . . . . . . . . . 14 Rate of Increase of Maximum Modulus and Definition of Order . . 15 The Disjunction of Zeros of a Nonconstant Entire Function . . . . 16 Fundamental Properties of the Complex Number System: Elementary Theorems on Zeros of Entire Functions . . . . . . . . . . . . 17 Hadamard's Three-Circle Theorem and Convexity . . . . . . . . . 19 Infinite Products . . . . . . . . . . . . . . . . . . . . . . . . 22
Chapter 11: The Expansion of Functions and Picard Theorems 2.1 2.2-2.3 2.4 2.5 2.6 2.1
Residues . . . . . . . . . . . . . . . . . . . . . . . . . . . Expansion of a Meromorphic Function . . . . . . . . . . . . . Expansion of an Entire Function . . . . . . . . . . . . . . . . Rouche's Theorem . . . . . . . . . . . . . . . . . . . . . . . Hurwitz's Theorem . . . . . . . . . . . . . . . . . . . . . . . Picard Theorems for Functions of Finite Order . . . . . . . . . . V
26 27 29 30 30 31
vi
Contents
Chapter 111: Theorems Concerning the Modulus of a Function and Its Zeros 3.1 3.2 3.3 3.4 3.5 3.6 3.1
Inequalities for B { f ( z ) } . . . . . . . . . . . . . . . . . . . . . Poisson's Integral Formula . . . . . . . . . . . . . . . . . . . Jensen's Theorem . . . . . . . . . . . . . . . . . . . . . . . The Poisson-Jensen Formula . . . . . . . . . . . . . . . . . . Carleman's Theorem . . . . . . . . . . . . . . . . . . . . . . Schwarz's Lemma . . . . . . . . . . . . . . . . . . . . . . . A Theorem of Bore1 and CarathCodory . . . . . . . . . . . . . .
40 42 43 41 48 52 53
Chapter IV: Infinite Product Representation : Order and Type 4.1 4.2 4.3 4.4 4.54.6 4.1 4.8 4.9 4.10 4.1 1 4.12413 4.14
Weierstrass Factorization Theorem . . . . . . . . . . . . . . . . Order of an Entire Function . . . . . . . . . . . . . . . . . . Type of an Entire Function . . . . . . . . . . . . . . . . . . . Growth of f(z) in Unbounded Subdomains of the Plane . . . . . . Enumerative Function n ( r ) . . . . . . . . . . . . . . . . . . . Exponent of Convergence . . . . . . . . . . . . . . . . . . . . Genus of a Canonical Product . . . . . . . . . . . . . . . . . . Hadamard's Factorization Theorem . . . . . . . . . . . . . . . Order and Exponent of Convergence . . . . . . . . . . . . . . . Genus of an Entire Function . . . . . . . . . . . . . . . . . . Order and Type of an Entire Function Defined by Power Series . . On an Entire Function of an Entire Function (G . P6lya) . . . . .
56 59 61 62 63 65 66 68 71 14 74 80
Chapter V : Standard Functions and Characterization Theorems 5.1 5.2 5.3 5.4 5.5 5.6 5.1 5.8 5.9-5.10
The Gamma Function . . . . . . . . . . . . . . . . . . . . . 83 Analytic Continuation of T ( z ). . . . . . . . . . . . . . . . . . 88 Conjugate Points . . . . . . . . . . . . . . . . . . . . . . . . 92 Bessel's Function . . . . . . . . . . . . . . . . . . . . . . . . 92 The Function F&) = exp(-ta) cos z t dr (a > 1) . . . . . . . 93 Order of the Derived Function . . . . . . . . . . . . . . . . . 95 Laguerre's Theorem . . . . . . . . . . . . . . . . . . . . . . 96 Convex Sets and Lucas's Theorem . . . . . . . . . . . . . . . . 91 Mittag-Leffler Theorem . . . . . . . . . . . . . . . . . . . . . 100
Jr
Chapter VI: Functions with Real and/or Negative Zeros: Minimum Modulus I and Sequences of Functions 6.1 6.2
Functions with Real Zeros Only The Minimum Modulus m ( r ) .
. . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . .
109 115
vii
Contents 6.3 6.4 6.5
Sequences of Functions . . . . . . . . . . . . . . . . . . . . . Vitali’s Convergence Theorem . . . . . . . . . . . . . . . . . Montel’s Theorem . . . . . . . . . . . . . . . . . . . . . . .
.
118 121 122
Theorems of PhragmCn and Lindelof . . . . . . . . . . . . . . . The Indicator Function h(0) . . . . . . . . . . . . . . . . . . Behavior of m ( r ) . . . . . . . . . . . . . . . . . . . . . . . .
124 129 133
Chapter VII: Theorems of PhragmCn and Lindelof: Minimum Modulus I1 7.1-7.7 7.8 7.9
Chapter VIII: Theorems of Bore]. Schottky. Picard. and Landau: Asymptotic Values 8.1 8.2 8.3-8.5 8.6 8.7 8.8 8.9 8.10 8.11 8.12
a-Points of an Entire Function . . . . . . . . . . . . . . . . . Borel’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . Exceptional-B Values . . . . . . . . . . . . . . . . . . . . . . Exceptional-P Values . . . . . . . . . . . . . . . . . . . . . . Schottky’s Theorem . . . . . . . . . . . . . . . . . . . . . . Picard’s First Theorem . . . . . . . . . . . . . . . . . . . . . Landau’s Theorem . . . . . . . . . . . . . . . . . . . . . . . Picard’s Second Theorem . . . . . . . . . . . . . . . . . . . . Asymptotic Values . . . . . . . . . . . . . . . . . . . . . . . Contiguous Paths . . . . . . . . . . . . . . . . . . . . . . .
145 146 147 148 149 155 155 157 159 161
Chapter 1X: Elementary Nevanlinna Theory 9.1 9.2 9.3 9.4 9.5 9.6-9.7
Enumerative Functions: N ( r . a ) . m(r. a ) . . . . . . . . . . . . . . The Nevanlinna Characteristic T(r) . . . . . . . . . . . . . . . A Bound for m(r. a ) on 1 a I = 1 . . . . . . . . . . . . . . . . Order of a Meroniorphic Function . . . . . . . . . . . . . . . Factorization of a Meromorphic Function . . . . . . . . . . . . The Ahlfors-Shimizu Characteristic To( r ) . . . . . . . . . . . . .
Appendix . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Suggestions for Further Reading
. . . . . . . . . . . . . . . . . .
163 168 170 174 175 176
182
185
Bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
186
INDEX . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
219
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PREFACE The purpose of this monograph is to acquaint the reader with some of the basic analysis and theorems central to a study of functions analytic in the entire finite complex plane. The study of these entire functions, a subclass of the family of meromorphic functions, has been assumed by some to be somewhat out of vogue in the present decade, but judging from the results of various specialist conferences, e.g., La Jolla, 1966 [“Entire Functions and Related Parts of Analysis” (Proc. Symp. Pure Math., 1 Zrh), AMS, 19681,this assumption is not particularly valid. This short monograph will display some of the intrinsic beauty of the subject which is far from “exhausted,” and array in a sequential form theorems central to the study of entire functions. Since the chapters are developed in a logical order, for those not particularly familiar with the basic function theory it is advisable not to skip sections. An appendix has been added listing a few of the more important theorems and definitions. An attempt has been made to prove results fully and to use the phrase “it can be shown,” as seldom as possible. Anyone with a good first course in complex function theory should have little difficulty in reading all chapters. A considerable amount of work has been accomplished, and indeed is still being done, in the subjects of distribution of zeros of polynomials, Hilbert spaces of entire functions, iteration of functions, etc. It is partly with this in mind that this survey is written, since most of what follows is basic to an understanding of not only the geometry of zeros, but the growth and behavior of the maximum modulus, the minimum modulus, the apoints, etc. ix
X
Preface
Very few of the results are new, in the sense that one could not find them in function theory texts, however the claim to originality of this survey is its continuity, its amplification of somewhat recondite theorems and its grouping of results scattered throughout a wide spectrum of texts, articles, and lecture notes. For some of the latest advances in entire function theory in particular, the reader is referred to an extensive bibliography which, by virtue of the scope of the subject, is by no means comprehensive. I should like to add that a further purpose of this monograph is to prepare the reader for study of papers which have expanded the frontiers of the subject and, in particular, for study of such erudite works as “Meromorphic Functions,” by W. K. Hayman, Oxford University Press (Clarendon), 1964, and “Entire Functions” by R. Boas, Academic Press, 1954. I have reproduced very little of the material in Dr. Boas’ book, which among other things, deals at length with functions of exponential order: most of the present survey deals with functions of arbitrary order. The difference between a lemma and a theorem is always debatable, and in this survey a lemma is used to mean a result of lesser importance than a theorem, but normally needed for the theorem which it precedes. Proper names appended to theorems only mean that the name is that by which the theorem is normally known, and is not meant to imply that the theorem is exclusively the discovery of that particular person. In attempting to develop a style and arrangement of material, I have looked to the works of Dr. E. C. Titchmarsh, whose several very readable texts are well known to all analysts. I have had several occasions to use material, in particular from “The Theory of Functions” by E. C . Titchmarsh, second edition, 1939, for which I wish to acknowledge the kind permission of the Oxford University Press (Clarendon). I wish also to thank Dr. G. Pdya for permission to include his result on an entire function of an entire function, and also to acknowledge the very considerable assistance given me by my colleague, the late Dr. H. D. Ursell.
INTRODUCTION ,TO THE THEORY OF ENTIRE FUNCTIONS
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CHAPTER I
A STUDY OF THE MAXIMUM MODULUS AND BASIC THEOREMS
We commence our study of entire functions by acquainting ourselves with some of the basic properties of analytic functions. Elementary analysis such as properties of Laurent and Taylor series expansion of functions, Cauchy’s theorem, Cauchy’s formulas and logarithms of complex functions has been assumed. Although there are several ways to treat the concept of an entire function, the whole treatment has been kept classical. Many far-reaching ramifications of some of the later theorems can be explored with techniques of modern function theory, for example, the use of elliptic modular functions in Picard’s theorems, or the structure of Hilbert spaces of entire functions. However, the exploration of these techniques has been left to the reader who will find many references to them in the bibliography. An initial start is now made by examining the behavior of aalytic functions at singular points. 1.1
THE NATURE OF SINGULAR POINTS
A singular point of a complex function is a place where the function ceases to be analytic (see Appendix). The point zo is said to be an isolated singularity for a single-valued function f(z) if 36 > 0, such that f(z) is analytic in 0 < I z - zo I < 6 but not in I z - z,, I < 6. 1
2
I. A Study of the Maximum Modulus and Basic Theorems
Most singularities we encounter are singularities of functions such as l/(z - a) at z = a or I/ez at z = 0. However, a function defined to be
clearly has a singularity at the origin. Note that z = 0 is not an isolated singularity of the multivalued function f(z) = z1I2. In the first place, z1I2 does not possess a derivative at z = 0 and furthet , there is no neighborhood of z = 0 in which the function is single-valued. Thus the requirement that f(z) be single-valued excludes branch points. To analyze the behavior of a function at an isolated singularity we use a Laurent series. Let f(z) have an isolated singularity at z = z,. Then we know thatf(z) can be expanded as a Laurent series in a sufficiently small circle about z, (except at z, itself). Let the Laurent series be m
f(z) =
C
n=-m
an(z - z J n
(see Appendix). We divide isolated singularities into three classes.
I. Let f(z) be bounded in the neighborhood of z,. Assuming the Laurent series for f(z) to exist, 3 M > 0, such that
Cauchy's inequalities show that 1 a, 1 5 M / r n for r arbitrarily small. Thus all coefficients with negative index vanish. Hence, except for z = z,,f(z) is represented by a Taylor series f(z)
= a,
+ al(z - z,)
-f
..
*
+ a,(z
- ZJ
$-
'
.* .
The Taylor series defines an analytic function throughout the interior of
a circle of convergence about zo and the series coincides withf(z) at all of these points except z = z,. If we redefine f(z) at z, by f(zo) = a,, then the function is analytic in I z - zo 1 5 r . This singularity is due to a break in the continuity of f(z). Thus to make the function analytic at z,, it suffices to redefine f(z) at zo so as to make f(z) continuous. The singularities of this type are called removable. 11. Letf(z) be unbounded in the neighborhood of a point z,. The Laurent expansion off in the deleted neighborhood of zo contains negative in-
3
1.1 The Nature of Singular Points
dex terms. Thus
where 1
Do
g ( z - ZO) =
n =O
an(z - z,,)~
and
The series I?( I /(z - z,,)) is called the principal part off(z). If the principal part has a finite number of terms, i.e., a,!0, n > k > 0, and a_k # 0, then 7
and we say that f(z) has a pole of order k at z = z,,. If the principal part has an infinite number of non zero terms, we say that 2,) is an essential singularity of f ( z ) , e.g., ellZ has an essential singularity at z = 0. The following theorem illustrates the nature of a function with an isolated essential singularity. 1.1.1 Theorem (Weierstrass). I n the neighborhood of an isolated essential singularity, an analytic function comes arbitrarily close to every complex value. (Picard's theorem states that in every neighborhood of an essential singularity, an analytic function takes on every finite complex value with one possible exception). Proof. Let f ( z ) have an essential singularity at z,,. Let f ( z ) be bounded away from the value a in a neighborhood of z,,, i.e., let I f ( z ) - a ] be bounded away from zero. Then
is bounded and analytic in a deleted neighborhood of z,,, i.e., O
Thus g ( z ) has at most a removable singularity at z,,. for otherwise
would have only a pole at z,,.
Further, g(zo) # 0,
Thus if g ( z ) approaches a nonzero limit as
4
I. A. Study of the Maximum Modulus and Basic Theorems
z + z,, then f(z) has a removable singularity at z , , which conflicts with the assumption. Thusf(z) is not bounded away from a value a in the neighborhood ofz,. 0 We call a point of accumulation of poles in a domain wheref(z) is otherwise analytic an essential singularity. 111. Let z,, be an essential singularity off(z). Letf(z) be bounded below in the neighborhood of z,. That is, 3 M > 0 such that
Then l/f(z) is analytic and z, is a point of accumulation of zeros of l/f(z) in its domain of analyticity. Hence 1 /f(z) is identically zero which is clearly impossible. Thus an accumulation point z,, of poles does not behave like a pole. Singularities at infinity may be discussed similarly, since if f(z) is analytic in IzI > R , O I R <-, then F(w) =.f(l/w) is analytic for
and the three cases follow. Further, an infinity of separate poles can have an accumulation point at infinity.
1.2 Definition. A meromorphic function has no singularities in the finite plane except poles. A meromorphic function is rational meromorphic if z = 00 is a regular point or a pole and transcendental meromorphic if z = 00 is an essential singularity. One simple characterization of a meromorphic function is as follows. 1.2.1 Theorem. A rational meromorphic function must be a rational function (i.e., quotient of two polynomials). Proof. Let f(z) be a rational meromorphic function. Let the points zo, zl, . . ., z, (one of which may be the point at 00) include all the poles. The number of poles is finite, otherwise they would possess a point of accumulation in the whole plane and the function would possess an essential singularity.
1.3 Entire Functions (Definition)
5
Let
Po(--)z -1 z,
Y
. . ' PI,(--)z Y
1
- z,
denote the principal parts of the Laurent expansion of f(z) about zo, . . . , z,, , respectively. Then
is analytic in the whole z-plane including the point at 00. Thus it is uniformly bounded and by Liouville's theorem it may be shown to be constant. Hence f ( z >=
i:
"=o
A+)z
1 -
z,
+ c.
However, the py's are polynomials in the respective arguments, therefore f(z) is the quotient of two polynomials, i.e.,
and the equation gives the decomposition of f(z) into its partial fractions where the z, are the zeros of Q(z). 0
1.3 Definition. An entire function is a function analytic in the entire finite complex plane. Thus an entire function may be represented by an everywhere convergent power series a,zn. Entire functions are special cases of analytic functions. There are three ways for an entire function to behave at infinity:
x;=o
i. f(z) can have a regular point at 00. Then we will show by Liouville's theoremf(z) = K . ii. f(z) can have a pole of order k 2 1 at 00. Then f(z) is a polynomial. iii. f(z) can have an essential singular point at 00. Then f(z) is a transcendental efitire function. The sum, difference and product of a finite number of entire functions are entire functions, and the quotient of two entire functions is an entire function (provided that the denominator is nowhere zero). An entire function of an entire function is an entire function.
6
I. A Study of the Maximum Modulus and Basic Theorems
We consider the behavior of one of the simplest of entire functions e'. Since
ez . e-'
=
1
for all finite z,
ez has no zeros in the finite plane. Since
then
has infinitely many roots. NOTE,log z is not an entire function (see z since
=
0, for example). However,
+ i arg ez = log ex + i(y + 2nn) = x + iy + i2nn = z + 2nni, log ez represents an infinite family of entire functions, viz. z, z + 2ni, . . . . log ez = log I ez I
1.3.1 Theorem. If f(z) is an entire function which does not vanish at any point, thenjlz) = egiz)where g(z) is an entire function. Proof. Since logf(z) represents an infinite family of entire functions differing from each other by integral multiples of 2ni, denote any one of these by g(z) and we have that log,f(z) = g(z)
+ 2nni.
Thus f(z)
= elogf(z) = eg(z)+ztlni = e u i z ) .
(2ni is the period of the exponential function.)
0
1.4 MAXIMUM MODULUS
We now study the maximum modulus M ( r ) = max If(z) I of a function f(z) analytic on the disk 1 z I 5 r. The maximum principle illustrated by the maximum modulus theorem, a theorem central to all analytic function theory, demonstrates the remarkable property of these functions, viz., that if a functionf(z) is analytic in an open connected set K and is not constant in K , then f(z) has no maximum value in K . Further, if f(z) is continuous in
1.4-1.8
I
Maximum and Minimum Modulus
a closed connected set D and analytic in the interior of D,then I,f(z) I attains its maximum value on the boundary of D. The problem of finding the exact position where a function attains its maximum absolute value can offer considerable difficulty. We expect the maximum modulus of an entire function to be on the boundary of the domain D since (as we shall show in Section 6.3) M ( r ) increases with r, however, this is certainly not evident for all analytic functions. We prove the maximum modulus theorem for a function analytic in and continuous on the boundary of a domain D. Considerf(z) to be analytic in a domain D,including the origin. Write m
f(z)
=
C
I z I < R.
anzfl,
,=O
For r
< R, m
m
~ f ( z 12> =f(z>
f(z) =
1 a,rmeime . n=o C cinrne-ine.
m=O
Since both series are absolutely convergent (within the original circle of convergence), they may be multiplied, the resulting series being uniformly convergent for 0 5 0 5 2n and for any given r < R ( I z I in general cannot approach R ) . 1.4.1 Lemma. Iff(z) is analytic for
Iz
-a
I < R, R > 0, write
M ( r ) = max I f ( z ) l ,
O ( r < R.
Iz-al=r
Then If(a)I 5 M(r), 0 5 r
if and only if f(z)
< R,
and
= constant =
If(z) I = M(r), 0
M(r)eia(a real) and
1 z - a 1 < R.
Proof. Since we may write m
f(z)
C
= U,(Z n=o
- a)n,
I z - u I < R,
we have
[valid at r
=0
since
1 a, l2 = I f ( a ) 12]. Thus m
C I an 12rzn5 (M(r)}' n=O
I. A Study of the Maximum Modulus and Basic Theorems
8 and
I a, l2 5 { M ( r ) } 2 ,i.e., I a, I 5 M ( r ) . Also, I a,
l2 + I a, I2r2+ .
since
+ I a,
5 {M(r)}2
I a, I < M ( r ) , 0 < r < R, unless a, = 0, n = 1, 2, . . . whence f(z) is a constant = M ( r ) , 0 < r < R . Thus eitherf(z) is constant for I z - a I < R or I f(z) I takes values greater than I f(a) I in every neighborhood of z = a. Alternatively, if f(z) is analytic and not constant in a domain D, then I f ( z ) I cannot take a maximum value at any point a E D. 1.4.2 The Maximum Modulus Theorem. If f(z) is analytic in a bounded domain D and continuous on 6 and M = maxZecIf(z) 1, where C is the boundary of D, then in D. If(z) I 5 M Further,
I f(z) I < M in
D except in the case where f(z) is a constant.
Proof. a. Clearly, if f(z) = K (constant) Vz, z E D, then since f is continuous on 6 , f(z) = K Vz, z E d, and hence f(z) = K Vz, z E C so that I KI = M and If(z)I = M Vz, Z E 6. b. Assumef(z) is not constant in D. Write K = maxIf(z)I. Z€D
Since 6 is compact, K is finite and If(z) in d. By the previous lemma, If(z)
I # K,
I attains
the value K at least once
VZ = a E D
and I f(z) I > I f ( a ) I in every neighborhood of a. Thus the maximum value K of I f ( z ) I on 6, is attained at some z E C . Accordingly K = M and If(z) I < M in D. 0
1.5 The Minimum Principle. Iff(z) is analytic in D and is never zero there, then‘ [,f(z)]-’ is analytic in D. Since the minimum of I f ( z ) I, i.e., m(r), is attained at the same points as the maximum of I f ( z ) I-l, it follows from the maximum principle that If(z) I cannot attain its minimum in the interior of D. Further, since the inability of a function to attain its maximum at an interior point is a “local” property, the maximum principle is valid for analytic functions not single-valued in a multiconnected domain.
1.41.8
9
Maximum and Minimum Modulus
Thus we have the minimum principle, viz., that I f ( z ) I attains its minimum value on the boundary of a region D and that if its minimum value is found inside D, then f is necessarily a constant. With the maximum modulus theorem at our disposal, we now consider M ( r ) first from the point of view of its relation to the coefficients in the power series for f ( z ) , later in its growth as compared to the modulus of a standard function like eL.Later still, we compare the growth of the maximum modulus to the growth of the minimum modulus. It transpires that in many cases the growth of the maximum modulus is not a particularly suitable object with which to study the more delicate behavior of entire functions. A new object called the Nevanlinna characterisric is introduced to which a final section of the survey will be devoted. 1.6 MAXIMUM MODULUS OF COEFFICIENTS OF A POWER SERIES
Let f(z)=ao
+ a,z + ... + apzP + ...,
I zI 5 r < R,
and hence f(Z)/ZP = aoz-p
+ u1z-P+1 +
+ up + - - .
* * *
*
Integrating with respect to q5 from 0 to 272,
1
2n
2n
uoz-P d$
f ( z ) / z P dq5 =
0
+ . . + j2'updq5 +
0
-
.
a
.
0
(We can integrate the infinite series, since for variable $ and fixed r the series converges uniformly.) Now
1 =s 2n
0
zm dq5
2n 0
Thus 2nap =
rm(cosmq5
+ i sin m$) dq5 = rm[ sinmm$ ~
- -i
S," f ( z ) / z pdq5 and 1
a =2n
s
2n
0
f(z)/zPd+
for p
= 0,
1, 2, . . .
.
Let M ( r ) denote the maximum modulus o f . f ( z )on a closed disk of radius r, center z = 0, i.e., M ( r ) = max,,,,, I f ( z ) 1. Since
10
with
I. A Study of the Maximum Modulus and Basic Theorems
IzI
=
r, If(z) I 5 M ( r ) , we have
I up I 5 M(r)/rp
(Cauchy's inequality).
Three theorems follow immediately. 1.6.1 Theorem. A function which is analytic and bounded for all finite z is constant.
Proof. If f(z) is analytic for all finite z, the Taylor series f(z) = C2='=o a,zrL converges for all z . If I f ( z ) 1 I M , we have I u , ~] 5 Mr-ll for all n and r. Letting r + 00, I a, I 5 0 for n > 0. Thus a,] = 0, n > 0, and
f @ ) = a,.
0
1.6.2 Theorem. Iff(z) is analytic for all finite z and if as I z
I
-
00
thenf(z) is a polynomial of degree less than or equal to k.
I a, I 5 M ( r ) P .
Proof. By Cauchy's inequality
I a n I 5 O(rk-n)
-+
Since M ( r ) = O ( r k )
for n > k, r
0,
--f
00
and a, = 0
for n > k
Thus f ( z ) is a polynomial of degree less than or equal to k . 1.6.3 Theorem. (Liouville). If an entire function is not an absolute constant, then M ( r ) + 00,r + 00.
+
+
+
Proof. Let f(z) = a, a,z . . a,zn . . be an entire function. Since I a, I 5 M ( r ) / r nfor M ( r ) the maximum modulus of f(z) on 1 z I 5 r and since M ( r ) is a nondecreasing function of r, either 3 a constant C such that I
M ( r ) 5 C,
C > 0,
M(r)-+oo,
r-m.
or
Assume the existence of C, then
I U, I 5 C/rn
for r
> 0, n = 0,
1,
... .
1.4-1.8
11
Maximum and Minimum Modulus
For n 2 I , C / r n+ 0, r 00. Since the left-hand side is independent of r, we have I an 1 I 0, i.e., a,l = 0, n 2 I , and thus f(z) = a,. Iff(z) $ constant, then M ( r ) is not bounded, and since it is nondecreasing, M ( r ) 00, r + w . c] --+
--+
1.7 MAXIMUM MODULUS OF A POLYNOMIAL
Consider a polynomial = a,
+ a,z + - + a,,z", *
*
an # 0, n 2 1 ,
Clearly, as r + 00
Thus for
E
> 0, in particular,
Hence for r > To(&) and
E
< I , 3rO(&)such that for r > ro(e),
Iz I I r,
Similarly, for a point z, on
IzI
=
r
12
I. A Study of the Maximum Modulus and Basic Theorems
and for r > To(&)
I P(Z0) I 2 I a, I r n ( l - E l . Thus for arbitrary
E
I a, I
> 0 and sufficiently large r r"(1 - E ) I I P ( z ) I 5 I a, I r n ( l -1-
IzI 5r
This inequality is true at a point in the disk attains its maximum M ( r ) , i.e.,
M ( r ) i I a,, I rn(l
+
E).
at which
I P(z) I
r > To(&).
E),
Also since the value of 1 P(z) I at a point zo on M(r), thus M ( r ) 2 I a, I rn(l - E ) and 1 -E I M ( r ) / a, I rn 5 1
Iz I =r
+
cannot exceed
E.
Since E is arbitrarily small, we have lim M ( r ) / a , I r n = 1, r+m
i.e., the maximum modulus of a polynomial of degree n is asymptotically equal to the modulus of the highest degree term of the polynomial. 1.8 MAXIMUM MODULUS OF AN ENTIRE FUNCTION
We define M ( r ; f ) to be the maximum modulus of f(z) in a closed disk I z I 5 r, e.g.,
I cos z 1 .
M ( r ; cos z ) = max It1
Since cosz= 1
--+2! 22
*
'
a
I Z I (00,
,
then lcoszlI 1
iZi2 +-+-+ 2!
1 2 1 4
4!
*
*
a
.
Hence
r2 r4 ~ c o s z ~ ~ l + - + - + 2! 4! and
-..=coshr
13
1.9 Order of Zeros
However, at z
and (el Thus
=
ir on the circle,
+ c r ) / 2coincides with the maximum modulus of cos z in M ( r ; cos z) = (e7 + e-')/2.
I z 1 5 r.
Similarly M ( r ; sin z)
=
and
(er - ecr)/2
M ( r ; eZ) = er.
1.8.1 Theorem. If an entire function is not a polynomial, its maximum modulus increases faster than the maximum modulus of any polynomial, i.e., lim M ( r ; P ) / M ( r ; f )= 0, r+m
where P(z) is the polynomial and f(z) is the entire function.
+
+
+
+
Proof. Write f(z) = a, a,z . a,zn , and let P ( z ) be of degree m. Denote the coefficient of the highest degree term of the polynomial by 6 , .f 0 and for E, > 0 , M ( r ; P ) 5 bmrm(l
+
EJ,
a
r > r,.
+
Consider f(z): choose a term of degree p 2 rn 1 (ap # 0). From Cauchy's inequality, I up I 5 M ( r ; f ) / r p and we have M(r; f )
2 rp I up I 2 I up I rm+l.
Choose r > 1 , then for r > ro ,
+
M ( r ; P) < I bm I rm(l E,) M(r;f) I up I rm+l Thus limr+cx,M ( r ; P ) / M ( r ;f ) = 0 .
I bm l(1
+
~ 0 )
Iaplr
0
1.9 ORDER OF THE ZEROS OF A NONCONSTANT ENTIRE FUNCTION
If f(z)
+ a,z + . + a,zn + . . , we expand about z = a to obtain + c,(z - a>n + - . f(z) = c, + cl(z - a ) +
= a,
*
* *
If a is a zero of f ( z ) , then c, = 0. It may be that cl, c z , etc. are also zero. (Not all ci = 0 since we would havef(z) = constant.)
I. A Study of the Maximum Modulus and Basic Theorems
14
Thus 3k, the smallest number for which ck # 0. The rank k of this coefis called the order of the zero. ficient or exponent in the expression (z Thus if the order of a zero a off(z) is k,
and f(z)
=
+ Ck+,(Z
(z - U ) k [ C k
-
a)
+.
'
I.
Note the series converges for arbitrary z. The series represents an entire function andf(z) = (z - u)~C$(Z), where C$(z) is an entire function of which z = a is not a zero. 1.10 Definition. An analytic functionf(z) is said to be algebraic if it satisfies an equation
Po(z>
+ P,(Z)f(Z) + Pz(z>.f'(z) + - + P,,(z)fYz) * *
=
0,
for all z in some given domain, and where Pi(z) are polynomials (of arbitrary finite degree) for i = 0, . . . , n, n 2 1, and P , ( z ) 3 0. We recall the definition of a transcendental number to be a (real) number which is not a root of any algebraic equation with rational coefficients and of finite degree. 1.10.1 Theorem. If an entire function is not a polynomial, it is nonalgebraic.
Proof. Let f(z) be a transcendental entire function, and suppose that it satisfies the polynomial equation of degree n, P"(Z)
-I- . . *
+ P,,(z)f"(z)
0.
Consider a sequence of disks, centered at the origin, radius r = I , 2, 3, . . . . Let zk be a point in the disk of radius k , at which If(z) I attains a local maximum, i.e.,
I f(zr.1 I
=
M(k; f ) .
Since limk+mM ( k ; f ) = 00, then I f(zk) I is unbounded as k we may assume f(zk) # 0, k 2 1. Set z = zk and thus
-
00
and thus
1.11 Rate of Increase of Maximum Modulus and Definition of Order
15
If Pn(zk)is a polynomial of degree greater than or equal to one, lim I Pn(zk)I
(since
= 00
I zk I
k+m
-
00
as k
-
00).
Thus we have that
and
Since limn.+, M ( k ; ,f)
= 00, the
right-hand side tends to zero. Thus
implying P,(zk) 0. However P,,(zk) cannot approach zero since Pn is either a polynomial of degree greater than or equal to one or a nonzero constant. Therefore, by contradiction f is nonalgebraic. 17 --f
We now touch on one of the methods by which we characterize an entire function, that is, the “order,” which compares the rate of growth of the maximum modulus with the growth of the modulus of a simple entire function, namely e’. Evgrafovt devotes much space to this concept. Several definitions of order, all of which are equivalent, will be presented as they are required. 1.11 RATE OF INCREASE OF THE MAXIMUM MODULUS
Since
1. M(r; e Z ) = e“, 2. M(r; expzk) = exprk,
3. then
M(r; exp e Z ) = exp e“,
log M(r; e*)
=
r,
log log (M(r; ez) = log r,
log M(r; exp z k )
= rk,
log log M(r; exp zk) = k log r,
log M(r; exp ez)
= e7,
log log M(r; exp e*) = r.
t M. A. Evgrafov, “Asymptotic Estimates and Entire Functions.” Gordon and Breach, New York, 1961.
I. A Study of the Maximum Modulus and Basic Theorems
16
These observations prompt us to formulate the following definition of the growth of an entire function.
1.11.1 Definition. The order Q of an entire function f(z) not identically zero, is defined as the superior limit of the ratio of log log M ( r ; f) to log log M(r; e") as r + 00, i.e., Q =
e.g., iff(z)
= cos
lim log log M ( r ; f )
r+m
log log M ( r ; e")
log log M ( r ; f ) logr '
-
= lim r+m
z
M(r; cos z ) =
e'
+
= er
.
1
2
+ e-2r 2
and log log M(r; cos z) log r
log lr[ I
+
r-1
(log
+; - 2 r ) ] l 9
log r
therefore
r+m
log r
=
1.
For arbitrary entire functions, order is somewhat difficult to calculate exactly. However, more often than not an estimate of the magnitude of this expression is good enough. 1.12 Theorem. Iff(z) is an entire function not identically zero, then for all z = z, in the plane, 3 a disk centered at z, in whichf(z) has no zeros except possibly z = z, itself.
Proof. Suppose f(zo) # 0. Then I f(zo) I is a positive number. Since f(z) is continuous (an entire function, therefore differentiable), 3 a disk centered at z,, such that If(z) -f(zo) 1 < E for all E > 0. Thus
Taking
E =
If(z,) 1, we have
1.13-1.14
Complex Number System: Theorems on Zeros of Entire Functions
17
and hence I f(z) I # 0. Thus forf(z,) # 0, 3 a disk centered at zo containing no zero of f(z). Further, suppose f(zo) = 0 and k is the order of the zero z = z,. Then f(z) = (z - Z ~ > ~ # ( Z where ), #(z) is an entire function and #(zo) # 0. We have just proved 3 a disk centered at z , , in which #(z) # 0, hence in that disk f(z) has no zeros other than 2,. 0 For the next few theorems we need one or two fundamental properties of the complex number system. We assume the following property.
1.13 The Nested Interval Property. Given a nested sequence of closed (bounded) regions in the complex plane, viz., II , I,, I s , . . . , such that Z,,+l E I, (n = 1, 2, . . .), then 3 at least one point common to all the regions. And we prove the following theorem. 1.13.1 The Bolzano-Weierstrass Theorem. Every bounded infinite set S in R2 or C has a point of accumulation.
Proof. We must find a point P such that every neighborhood of P contains infinitely many points of S. Since S is bounded, we may assume that it lies completely in a square KO of side 1. We then divide KO into four equal squares by intersecting lines. One of these smaller squares must contain infinitely many points of S, for otherwise S would be finite. Denote such a square by Kl and divide K, into four squares obtaining K, that contains infinitely many points of S. We repeat the process continuously, obtaining a sequence of squares (all closed and bounded) with KO 2 Kl
2
K,
2
2
K,
2 * * .
each of which contains infinitely many points of S. By the nested set property there must be a point P that lies in all K,. We want to show that this point P is a point of accumulation of S. Let U be any neighborhood of P, a disk with center P and radius E . We note from our construction that the length of the sides of K, is 112". Hence when n is sufficiently large, such that 112" < ~ 1 2 the , square K, must lie completely in U . Hence U contains infinitely many points of S. 0
1.13.2 Theorem. An entire function f(z) not identically zero cannot have infinitely many zeros in a disk of finite radius.
1 z I 5 r . Then by the Bolzano-Weierstrass theorem there exists a point zo on the disk which is a point of accumulation of the set of zeros off(z). Therefore in an arbi-
Proof. Suppose f(z) has infinitely many zeros in the disk
I. A Study of the Maximum Modulus and Basic Theorems
18
trary disk centered at z, , there are infinitely many zeros off(z) contradicting Theorem 1.12.
NOTE.An entire function may have an infinite set of zeros in the whole complex plane, e.g., sinz
=0
for z
= nn,
n
= 0,
1,
...
1.13.3 Theorem. If the values of two entire functionsf(z) and g(z) coincide at infinitely many points belonging to a disk K of finite radius, then the functions are identically equal, i.e., f(z) = g(z). Proof. Sincefand g are entire functions, the diRerence is an entire function, i.e., f(z> - g ( 4 = 4J(Z>? where #(z) vanishes at all points for whichf(z) = g(z). If 4(z) $0, we have a contradiction from the previous theorem. Thus +(z) = 0, i.e.,f(z) = g(z) Vz. In particular, if the entire functionf(z) takes the same value A at an infinite number of points on a disk K, thenf(z) = A . 0
1.13.4 Theorem. The equation a,
+ u,z +
*
’.
+ u,zn = 0,
n 2 1, a,
# 0,
has at least one complex root.
Proof. (By contradiction). Let P,(z)
= a,
+ a,z +
* * *
+ u,zn
and assume P,(z) # 0 in the complex plane, i.e., f(z) = l/P,(z) is an entire function (quotient of two entire functions). Also, f(z) is nonconstant since P,(z) -+ 00 as z + 00. We have then that since f is a nonconstant entire function lim M ( r ; f) = 00. 7 m ’
However,f-t 0 as 1 z
1 --+ 00 and the contradiction proves the theorem.
We have the immediate corollary that a polynomial of degree n has exactly n roots by writing P,(Z)
= a,
+ a,z + - - + unzn = (z - Z1)Pn-*(Z) *
1.15
19
Hadamard's Three-Circle Theorem and Convexity
where z , is the (complex) root whose existence was established and where Pll-l(z) is a polynomial of degree n - 1.
1.14 We can regard a transcendental entire function f ( z ) = a,
+ a,z +
* *
.
+ qlzn+
* *
.
as a sort of polynomial of infinite degree. If the analogy is valid,
a,
+ a,z +
f
*
+ a,,z" -+ .
1
.
=
0
must have infinitely many solutions. Clearly this is not true generally. For example, ~ + - + Z- + . . . +2 T2 + . . . = O
I!
Zn
2!
n.
or ez = 0 has no solutions for finite z. To improve the analogy, consider a,
+ a,z + . . . t- a,?z"
=
A,
where A is an arbitrary complex number. This polynomial equation of degree n has n roots. Thus with P,,(z) = A , where P,,(z) is a polynomial of degree I I having roots equal in number to the degree, let us consider, for example,
which has infinitely many roots for A # 0. Thus the number of roots in a sense equals the degree. Similarly, by writing cos z = A , A # 0, the equation has infinitely many solutions which we see by writing ,iz
cos z
=
+ ,-iz
2
,
etc.
This concept will be generalized later when we discuss Picard-like theorems.
1.15 Hadamard's Three-Circle Theorem, If f ( z ) is analytic and singlevalued in Q < I z I < R and continuous on I z I = Q , I z 1 = R and if M ( r ) denotes the maximum of l f ( z ) I on 1 z 1 = r, p < r < R, then log M ( r ) is a convex function of log r, i.e., for p < rl < r2 < r3 < R , log r3 - log r2 log M r 2 ) 5 log M r , ) log r3 - log rl
+
log
M(r3)
log r2 - log rl log r3 - logr , (1)
20
I. A Study of the Maximum Modulus and Basic Theorems
[We recall that a function is convex (downward) in the following sense namely, y = #(x) is convex if the curve #(x) between x1y xz is always below the chord joining (xl, 4(x1)) ( x Z , #(xZ)), i.e.,
Equation (1) is equivalent to
ProoJ Let # ( z )
= zjif(z), where 3, is to be a determined constant (real). Since the function z"f(z) is not in general single-valued in rl 5 1 z 1 5 r3, we cut the annulus along the negative part of the real axis obtaining a domain in which the principal branch of this function is analytic. The maximum modulus of this branch of the function in the cut annulus is attained on the boundary of the domain. Since 1 is real, all the branches of z"f(z)have the same modulus. By considering a branch of the function analytic in that part of the annulus for which n/2 5 arg z 5 3n/2 it is clear that the principal value cannot attain its maximum modulus on the cut and therefore must attain it on one of the boundary circles of the annulus. Thus the maximum of I # ( z ) 1 occurs on one of the bounding circles, i.e.,
I (PG)I 5 max{r,"M(r,), Hence on
r31Mr3)).
(1)
I z 1 = r2, I f(z> I 5 max{rlArr;aWrl),r3+idM(r3)}
(2)
We choose I such that rlAM(rl)= rSAM(r3)and 3, =
-{log Wr3)/~(rl)}/{b37r3/r1>.
With this I in (2), m - 2 )
and
I (rz/rl)-Awl)
M(r2)1ogPm1 < - (rz/rl)log{iM(t3)/.~(ri)) M(r1)log[rs/rl]
and taking logarithms,
5 M ( r l ) l o ~ ~ r 3 i ~ a 1 M ( r , ) l ~ ~ I r a ~0 ri). M(rz)10g(73/rl) Clearly equality is achieved when # ( z ) is constant, i.e., j ( z ) is of the form Czu for some real u.
I . 15 Hadamard's Three-Circle Theorem and Convexity
21
X
FIGURE 1
A sufficient condition for the convexity of $(x) is that +"(x) #'(x) is nondecreasing (Fig. 1 ). Thus for xI < x < x2,
2 0, i.e.,
integration of which gives the previous definition of convexity. Alternatively, putting x = &(x, xz) we obtain another more general definition of convexity, viz.,
+
+ Sx,)
4{&x1
5
ac4cx1>
+ +(xz)l.
Note that Zl
and (x2 - X ) ~ ' ( X )5
s"
$'(t) d t .
2
Actually a lot more can be said about convex functions which are of particular interest in the theory of Harmonic functions. We state one or two results which are not difficult to prove.+
I. A functionf(x) continuous in an open interval I is convex if and only if for every pair of points xl, x2 E I and every pair of nonnegative numbers A ,p 2 , p 1 p 2 > 0, we have
+
f( P l X l
PI
.+PZXZ ) I
+Pz
Pl.f(X1)
P1
+ + Pz
Pzf(X2)-
This clearly reduces to our previous definition if p 1 = p 2 = I . See S. Saks and A. Zygmund, "Analytic Functions," Monographie Maternatyczne Vol. 28. Polska Akademia Nauk, Warsaw, 1965.
22
I. A Study of the Maximum Modulus and Basic Theorems
11. If g(x) is continuous in a closed interval [a, b ] andf(x) is convex in an open interval containing all the values of g(x), then
1.16 INFINITE PRODUCTS
In order to examine the behavior of entire functions, it is necessary to express them in as many different ways as possible. We will review properties of infinite products with a view to expressing entire functions in this form. Later we will study infinite series representation in some detail. Clearly some entire functions, the polynomials, have only finite product representation. A study of convergence and divergence of infinite products will now be undertaken. Consider ( I al)(1 az) containing an infinite number of factors, i.e., RE, (1 a%),where no ai = - 1. Then let us write
+
+
+
-
+
1.16.1 Definition. The infinite product converges if there exists a finite L (not zero) such that lim P, = L . n-rw
If P,? 0, the product is said to diverge to zero. It is necessary that a, 0 since P,,= P,-l a,,Pn-, , but clearly not suficient since (l/n)) diverges yet a, --+ 0. --+
n:&(1 +
+
1.16.2 Theorem. If a, 2 0 for all n, then verge or diverge together.
n:=,(l+ a,)
and C =:,
--f
a,? con-
Proof. We note that P, is a nondecreasing function of n (since each term is greater than or equal to one). Thus P,, either converges or approaches 00. We have
+
and if P,, is bounded, then Cy=,ai is bounded. Also, if exp(C%, ai}is bounded, and thus P,, is bounded.
If a,, 5 0 for all n, write a,
=
C:=l
ai is bounded,
-b,, and consider n:==,(l - bn).
23
1.16 Infinite Products
1.16.3 Theorem. If b,] 2 0, b,, # I for all n, and - b,) converges.
nr=l(l
xT=lb , converges,
then
Proof. Since x7=lb, converges, 3 N such that
and b,
< 1 for n 2 N , Thus
and
>
and P,{P,-l is decreasing [since for n N , (1 - bff) < I ] and has a positive lower limit. Therefore P,8/Pn-l tends to a positive limit, and since P,-, # 0, P , converges. 0
1.16.4 Theorem. If 0 5 b, < 1 for all n and n2==l(l - b,f)diverges to zero.
xrTlb,
b
diverges, then
Proof. If 0 5 b < I , 1 - b 5 e-b. Thus ( I - bl)(l - b,)
..
(1 - b,]) 5 e-bl-ba-..*-bn
and the right-hand side approaches zero, hence the result.
0
+
1.16.5 Definition. The product n;=,(l a,,) is said to be absolutely convergent if 1 / a, I ) converges. Thus a necessary and sufficient condition that the product should be absolutely convergent is that x2=lI a, I should converge.
+
1.16.5.1 Lemma. An absolutely convergent product is convergent. Proof. As before, write
n n
p,=
m-1
(1 + a , )
I. A Study of the Maximum Modulus and Basic Theorems
24
nr==,(l +
Now if 1 a, I) converges, q, approaches a limit and Cr=l(qn- qn-d converges, and by the comparison theorem Cr=l(Pn- P n d 1 )converges, i.e., Pn approaches a (nonzero) limit. 1
-
zr=l
We see that this limit cannot be zero, since if 1 a, I converges and 1, the series 1 a,/(l a,) 1 converges. Thus the product
+ a,
zrXl
+
approaches a limit, but this product is l/Pn, thus limn+ooP, # 0. 1.16.6 Theorem. A necessary and sufficient condition for the convergence of the infinite product n,"=l(l a,), a, complex, is the convergence of Cr'=l log(1 a,) where each log has its principal value.
+
+
Proof. Write
s, =
c log(l + n
a,).
7=1
To establish sufficiency, we have that Pn = exp(S,,) and since the exponential function is continuous, S , + S which implies P, eS. To establish necessity, we have that --f
S,
=
log P,,
+ 2q,ni,
q,, an integer.
Since the principal value of the log of the product is not necessarily the sum of the principal values of its factors, q, is not necessarily zero. We show that q, is constant for all n > N and from this necessity follows. Write u,, Pn the principal values of the arguments of (1 a,) and P n , respectively. If the infinite product converges, a, -+ 0 (since a, 3 0) and
+
1.16 Infinite Products
25
However, since qn is an integer, this implies qn = q for all sufficiently large n. Thus if Pn tends to the finite nonzero limit P as n -+ 00, it follows that
Sn
--f
and the condition is necessary.
1.16.7 Theorem. The product a region where the series C:].
log P
+ 2974
0
n:==, [ 1 + U,,(z)]is uniformly convergent in
I Ui7(z)I converges uniformly. Proof. Let M be an upper bound of the sum Cgl I UJi{z)I in the considered region. Then
( I -1I U,(z) /}(I
+ I U,(z) I }
(1
+ I U n ( z )I } <
elU1(z)+-*+Un(z)l
- e.‘ <
Let
Then
hence C:=2(Pn(z) - P,,_l(z)}converges uniformly since m
m
and Cr=lI U i l ( z )I converges uniformly. Thus P, approaches a limit and the absolutely convergent product converges. A considerable number of practical examples should be worked through in the standard texts, since there are many theorems or lemmas which could be cited, most of which are special cases of the previous theorems.+
+ See in particular, the examples in E. C. Titchmarsh, “The Theory of Functions.” Oxford Univ. Press, London and New York, 1939.
CHAPTER II
THE EXPANSION OF FUNCTIONS AND PICARD THEOREMS
As a preamble to the expansion of a meromorphic and an entire function,
we recall the following. 2.1 RESIDUES
Let f(z) have a pole of order k at z = z,. Write
and
Thus
dk-1 [(z - z,)kf(z)]= (k - l)!a.-l dzk-l (k I)! al(z - zo)2 ... ,
+
+
+
2!
dk-1
26
+ - k!al !o ( z
- z,)
27
2.2-2.3 Expansion of a Meromorphic Function
and
2.2
EXPANSION OF A MEROMORPHIC FUNCTION
Since a meromorphic function is analytic in a region except possibly at a finite number of poles, we have the following theorem. 2.2.1 Theorem. Let f ( z ) be a function whose only singularities, except at
infinity, are poles. Suppose all poles are simple. Let them be a,, a,, . . . and be ordered such that 0 < I a, I 5 I a, I 5 and let the residues at the poles be 6 , , b,, . . . . Suppose there is a sequence of contours C,,, such that C,, includes a, , a,, . . . , a,, , but no other poles. Let the minimum distance Rn of C, from the origin approach 00 with n, while Ln the length of C,>is O ( R , ) and such that on C , , ,f ( z ) = o(R,). This last condition is satisfied if, for example, f ( z ) is bounded on all C,, Then
-
.
for all z except poles. Proof. Consider
I=
1
___
2ni
f ( w ) dw c, w ( w - z )
where z E C,,. The integrand has poles at arn,0, and w bmlam(a, - z ) , - f(O)/z,f ( z ) l z , respectively
==
z, with residues
and
by the conditions imposed. Thus
Note, that the series converges uniformly inside any closed contour such that all poles are inside. [7
28
11.' The Expansion of Functions and Picard Theorems
2.3 POLES AND ZEROS OF A MEROMORPHIC FUNCTION
2.3.1 Theorem. If f ( z ) is analytic in and on a closed contour C apart from a finite number of poles and if f(z) # 0, z E C, then
where N is the number of zeros in C (zero of order m counted m times) and P is the number of poles in C (pole of order m counted m times).
Proof. Let z
=a
be a zero of order m. Then in the neighborhood of z f ( z >= ( z
- a>"&>
and analytic in the neighborhood of z
= a,
for g ( z ) # 0 = a.
Thus
The last term is analytic at z = a, therefore f ' ( z ) / f ( z )has a simple pole at z = a, with residue m. The sum of the residues at the zeros of f(z) is N (the number of zeros). Similarly by writing -m for m, the sum of the residues at the poles of f ( z ) is -P (number of poles). Iff(z) is an entire function, the number of zeros
2.3.2. Corollary. If # ( z ) is analytic in and on C and if f ( z ) has zeros at a l , . . . , a p and poles at b,, . . . , b,, multiplicity being counted as before,
for if z z
=
=a
is a zero of order m, we have in the neighborhood of - a)mg(z) where g ( z ) # 0 and analytic. Thus
a, f ( z ) = ( z
The last term is analytic at z = a, therefore the left-hand side has a simple pole at z = a with residue m$(a). Applying the previous theorem we have the result.
29
2.4 Expansion of an Entire Function
2.4 EXPANSION OF AN ENTIRE FUNCTION AS AN INFINITE PRODUCT
Supposef(z) has simple zeros at the points a,, a 2 , . . . , a,. In the neighborhood of a , , f(z) = (z - a,)g(z), where g(z) is analytic and nonzero. Thus f‘(z> f(z)
1
-
z
- an
+-g‘g(z)(z)
with g’(z)/g(z) being analytic at a,, . Hence f’(z)/f(z) has a simple pole at z = a, with residue 1. Suppose f’(z)/f(z) is a function of the type considered in the expansion of a meromorphic function f(z). Then P(z) = f’(z)/f(z) has poles at a , , a 2 , . . . , a, and
Clearly P(0) = f’(O)/f(O) and b,, = 1 for all n. Thus
Integrating from 0 to z along a path not passing through a pole, we have logf(z) - logf(0)
=
+ f {log(z -
zf’(o) f(0)
n=l
where the value of the logarithms depend upon the path. Taking exponentials,
As an example, consider f(z)
sin z
= -Z
fi’ (1 -$)expnn,
Z
n=--m
or m
sinz=z
(1 n-1
-=). Z2
The next theorem called Rouchk‘s theorem follows somewhat naturally.
30
11. The Expansion of Functions and Picard Theorems
2.5 Theorem. Iff(z) and g(z) are analytic in and on a closed contour C, and if I g(z) I < If(z) 1 on C, then f(z) and f(z) g(z) have the same number of zeros inside C.
+
Proof. Let 4 ( z ) = g(z)/f(z). Then I +(z) I < 1 on C. Note that neither f(z) norf(z) g(z) has a zero on C. If the number of zeros off(z) g(z) inside C is denoted by N ‘ , then
+
+
=N+-
2ni
4’{1 - 4
+
+2
- . . . } dz.
Also,
thus by uniform convergence of the series we have N ‘
=
N.
0
The following theorem is now a consequence.
2.6 A Theorem of Hurwitz. Let f;,(z) be a sequence of functions analytic in a region D bounded by a simple closed contour. Let fn(z) tend to f(z) uniformly. Assume f(z) $0, and let zo be an interior point of D.Then z, is a zero off(z) if and only if z, is a limit point of zeros ofj;(z), n = 1, 2, . . . and points which are zeros for infinite n are counted as limit points. Proof. Choose e small and such that the circle I z - zo I = e is in D and contains or has on it no zero off(z) except possibly z,. Then If(z) I has a positive lower bound on the circle, i.e., If@) I L m > 0. Having fixed
e and m, choose N
If,(z) -f(z)
I <m
so large that
for n L N , z E { I z
Since
f ( z ) =f(z)
+ {fn(z> -f(z>),
-
zo I
=
e}.
31
2.7 Picard Theorems for Functions of Finite Order
Jl(z) has the same number of zeros in the circle asf(z). This follows from Rouchts theorem because f(z) and f,,(z) -f(z) are analytic in and on a closed contour C in a region D and If,(z) -f(z) I < If(z) I. Thus if f(zO) = 0, then f,,(z) has exactly one zero in C for all n 2 N so that z0 is a limit point of zeros off,(z). Also, iff(z,) # 0, thenf,(z,,) # 0 in C. 0
2.7 The next group of theorems although elementary in nature, examine more carefully the relation between order and the number of roots of an equation of the form f(z) = AP(z) where f(z) is entire, A is a constant, and P(z) is a nonzero polynomial. The only deficiency in the theorems is that they do not deal with functions of infinite order, a ramification which will be dealt with later on in the survey. 2.7.1 Picard Theorems. It will be useful to establish first of all, a few lemmas concerning inequalities for coefficients of power series. . . . anzn . . Write Consider the entire functionf(z) = a, a,z
+ + a,,= an + iBn
and z
= r(cos
6
+ i sin 0),
r
+
+ -
2 0,
Then m
f(z)
=
C
(orn
n=o
+ i,Bn)rn(cos8 + i sin 0 ) n .
If we write f(z> = U(r, 0)
+ Wr, 0
then m
U(r, 0)
= a0
+C
(or, cos n0 - B,, sin nO)rn.
(1)
n=1
Fixing r and integrating with respect to 0 from 0 to 2n, we have Llg =
1 J 2 n U(r, 0) do. 2n
0
To compute orpr p 2 1 multiply ( I ) by cosp0 and integrate with respect to 0 from 0 to 2n. Thus U(r, 0) cos p6 d8 = aprp
32
11. The Expansion of Functions and Picard Theorems
from which a
=-j nrp
Pp
=-
2n
V ( r , e) COspe dB
p = 1,2, ..
and similarly nrp
Also,
j2nV(r,0 ) sinpBd0,
p
=
1, 2, . . .
.
12n
2a0 fuprp = U ( r , e)(i f cospe) dB n o
and 2a0
j
Bprp = - 2n V ( r , e ) ( l f sinp0) do. n o
Note that the factors multiplying U(r, 0 ) are now nonnegative. Let p ( r ) = maxosB52nU(r, 0 ) on the circle of radius r. Then
j2’ f cos pe) d0
2u, f uprp 5 ’(‘) (1 n o
= 2p (r).
2.7.2 Lemma. If U(r, 0 ) the real part of an entire function f ( z ) satisfies U(r, 0 ) 5 p ( r ) 5 Crs,
S > 0,
for all r > N , then f(z) is a polynomial of degree not exceeding n = [S], i.e., the integer part of 6.
Proof. Since
I up I
1% I 5
and
I /-II,5 2 { p ( r ) - uo}/rP,we
2 (Cr6 - uo) ,p
and
IBPI
+
5
have that
2(Crs - uo) rp
If p > [a], since p is an integer, p 2 [S] 1 > 6 hence u p , r - + o o , and up iBP = ap = 0 for p > [d] = n. 0
+
PP
-
0,
33
2.7 Picard Theorems for Functions of Finite Order
2.7.3 Lemma. I f f ( z ) is a transcendental entire function and P ( z ) and Q ( z )
+
are polynomials of degree m, n, respectively and if P ( z ) 0, then the order el of P ( z ) f ( z ) Q ( z ) coincides with the order e off(z), i.e.,
+
el = e. Proof. Writing
M ( r ) = max I f ( z ) I2153
we confine ourselves to values of the functions at points on the circle I z I = r. Let a,z*, b,zn denote, respectively, the leading terms in the polynomials P(z) and Q(z). Taking E = & in the inequalities for polynomials previously evaluated, we assert for I z I = r > r o , that
However, at z1 (on the same circle) at which 1 P(z)f(z) its maximum M l ( r ) , we have Mi(r)
=
+ Q(z) I attains
I P(zilf(zi) + Q(zi) I I 8 I a, I rrnM(r)+ 8 I bn I rn*
Thus
Sincef(z) is a transcendental entire function it has been proved that M ( r ) increases faster than the maximum modulus of any polynomial and hence faster than any power of r. Therefore each { . } + 1 as r 00 and hence for r sufficiently large, the right-hand < 2 and the left hand { } > Q , i.e.,
-
{ - . a }
i I a, I rmM(r)L MI@)5 3 I a, I rnM(r).
34
11. The Expansion of Functions and Picard Theorems
We now require
- log log M(r) lim r+m log r
Y
i.e.,
and each {. . . } -+ 1 as r + 00. Hence the right-hand { } > 4, and left-hand {
---
9
.} < 2
and the
Q log M ( r ) 5 log M,(r) I 2 log M(r). Taking logarithms again,
- log log M(r) lim
,-+m
consequently [P(z)fOl.
logr
log log M l ( r ) 5lim logr
,+m
e = el and
~
lim ,-+-
log log M ( r ) logr '
the orders off(z) and P(z)f(z)
+ Q(z) are equal
0
2.7.4 Lemma. The order of an entire function
where P ( z ) , Q(z), and g(z) are polynomials and P(z) $0, is equal to the degree of g(z). Proof. From the previous lemma, the order der of $(.) = e@). We need to show that
- log log Ml(r) lim r+.n
where max,,,,,
logr
e off(z)
coincides with the or-
=n
1 $(z) I = M,(r) and n is the degree of g(z). Set
and z
= r(cos
0
+ i sin 0).
35
2.7 Picard Theorems for Functions of Finite Order
By hypothesis, en = 1 c,
I # 0,
thus
goekrk(cos n
g(z) = =
ak
n
+ i sin Cr,)(cos kO + i sin kO) + ke) -/-
Qkrk(COS(ai,
i S i l l ( a k $-
k0))
Z O
and
and
We require now log rnaxlzI,,I $(z)
hence for
E
I.
For fixed r and 0 5 O 5 2n
in (0, 1) and r > r(E),
and Let zo be a point on I z I are n such points). Then 1% Mi(r) 2 1%
=
r such that cos(a,,
I d(zo) I = @ J n
+
2 enrn
-
+ no) = 1 (actually there
n-1
@krk cos(ffk Z O
+ ke).
zo n-1
@krk
1 + en
> enrn(l - E ) Thus
ep(l
- E)
< log M l ( r ) < enrn(l + E )
and -
lirn
r+m
...
r for r > r(E).
log log M , ( r ) logr
= n.
0
en
36
11. The Expansion of Functions and Picard Theorems
2.7.5 Lemma. If g(z) is an entire function and if the order of the function f(z) = eg(z)is finite, then g(z) is a polynomial and hence the order off(z) is an integer.
Pvoof. If z
=
r (cos 0
+ i sin 0) and 1% I f(z) I
U(r, 0) is the real part of g ( z ) , then =
w, 0).
Let
max If(z)
I = M(r)
and
max
osean
IzI-r
U(r, 0)
= p(r).
Then log M ( r ) = p ( r ) . Suppose 6 is the order off(z). Then - log
lim
r+m
Given
E
log M ( r ) log r
= 6.
> 0, 34.9) > 1 such that for r > r(E) log log M ( r ) log r
< 8 + E
i.e., log log M ( r ) < log(r6+")
or log M ( r ) < rd+E. Thus from (I), p ( r ) < r6+"
for r > r(E).
From Lemma 2.7.2 it follows that g ( z ) is a polynomial the degree n of which satisfies n 5 [S E ] and E is arbitrarily small. Thus n 5 [6] and (by Lemma 2.7.4) the order of f ( z ) coincides with n, i.e., 6 is integral. 0
+
2.7.6 Theorem. Let f ( z ) denote a transcendental entire function, the order 6 of which is finite but nonintegral. Then if P(z) is a nonzero polynomial, the equation f(z)=
has infinitely many roots for every complex number A (no exceptions).
Proof. Suppose 3 A
=
A, for which f ( z ) = A,P(z) has only finitely many
37
2.7 Picard Theorems for Functions of Finite Order
roots if any at all. Then the entire functionf(z) - A,P(z) has only finitely many zeros. Therefore by a previous theorem we may write f(z) - A,P(z)
=
Q(z)eg(z)
where Q(z) is a nonzero polynomial (equal to I iff - AoP has no zeros). Further g(z) is an entire function, and we have f(z)
=
A,P(z) -t Q(z)eg'.'.
Thus the order 6 off(z) coincides with the order of eg(z)which, being finite by a previous lemma, is integral. The contradiction proves the theorem. 0 2.7.7 Theorem. Iff(z) is a transcendental entire function of finite integral order n and if P(z) is a nonzero polynomial, then
f(z) = AP(z) has infinitely many roots for every complex number A , with the possible exception of one value.
Proof. Suppose there exists at least two values a, b at whichf(z) = A P ( z ) has only finitely many roots. Then f(z) - aP(z) and f(z) - bP(z) have only finitely many zeros. Thus f(z) - aP(z)
=
Q,(z)e"'z)
(1 1
f(z) - bP(z)
=
Qz(z)eez(z)
(2)
and
where Ql(z) and Qz(z) are nonzero polynomials and gl(z) and gz(z) are entire functions. Hence the orders of egi(.) and egz(z)coincide with the order n off(z). We also conclude that g,(z) and g,(z) are polynomials of degree n. Hence n 2 1 since for n = 0, g,(z) and gz(z) would be constants and f(z) would be a polynomial but not a transcendental entire function. Subtracting (1) from (2), Q1(z)eQ1(')- Qz(z)eQa(-') = (b - a)P(z)
=
R(z),
where R(z) is a nonzero polynomial since b # a and by hypothesis P(z) # 0. We need to show that this equation is not possible for Q,(z), Q,(z) and R ( z ) nonzero polynomials and gl(z), gz(z) polynomials of degree greater than or equal to 1. Differentiating (Qi'
+ Qigi')e@~ (Qz' + Qzgz')eQa -
=
R'
38
11. The Expansion of Functions and Picard Theorems
also =
Ql egl - Q ,
R.
Regarding these as a system of two equations in the unknowns eel and the determinant of the system d ( z ) is given by Qi(Qz’ = -QzQi’ =
+ Qgz‘)
+ QiQz‘
- QdQi‘
eg:,
+ Qigi’)
- QiQzki’
- gz’).
We must now show that d ( z ) $0. If we assume that A ( z ) = 0, divide by -Q,Q, to obtain el’
Qz’
Qz
Qi
+ (8,
- g,)‘
= 0.
Integrating this equation we have Qi log -
Q,
+ g , - g, = constant = C,
Thus
However, by dividing Eqs. ( I ) and (2) we obtain
Q @1-9n 1
=
Q2
f-aP f-bP‘
~
Thus
f-aP f-bP
--
-c
and
f ( z ) ( l - C ) = P(z)(a - bC). Since b # a , we have C # 1 and hence
which contradicts the hypothesis, forf(z) is a transcendental entire function and P ( z ) is a polynomial. Thus d ( z ) $ 0. Solving for egl and eh, we obtain
2.7 Picard Theorems for Functions of Finite Order
39
However, these equations contain a contradiction since the left-hand side is a transcendental entire function and the function on the right-hand side is a rational function, and thus a polynomial since it is entire. We can say [for P(z) = I ] thatf(z) takes on all values an infinite number of times with one possible exception. 0 The previous two theorems are special cases of the following theorem. 2.7.8 Theorem. Let Q(z) be a transcendental meromorphic function. For every complex number A (finite or infinite) with two possible exceptions, Q(z) = A has infinitely many roots.
Proof. If Q ( z ) is an entire function, it does not become infinite at any z. Thus A = 00 is an exceptional value for every entire function. Consequently, on the basis of the previous theorem there may exist at most one finite exceptional value. If f(z) is a transcendental entire function and P(z) is a nonzero polynomial, the meromorphic function Q(z) = f(z)/P(z)becomes infinite only at a finite number of points, namely the zeros of P(z). Thus 00 is an exceptional value, and further, one other finite exceptional value is possible. Thus f ( z ) / P ( z )= A or f(z) = A P ( z ) has infinitely many roots for every A except possibly a single finite value of A . Letf(z) and g(z) denote two transcendental entire functions whose ratio is not rational. Hence f(z)/g(z) is meromorphic. Thus f(z)/g(z) = A has infinitely many roots with two possible exceptions. 0
CHAPTER I I I
THEOREMS CONCERNING THE MODULUS OF A FUNCTION AND ITS ZEROS
It will now be expedient to study the real part of an analytic function. Several theorems transpire, the most important of which is Jensen's theorem which relates the modulus of a function to its zeros. The Poisson-Jensen theorem is somewhat of a starting point for Nevanlinna theory which is in effect a much more delicate analysis of the behavior of meromorphic functions. We examine a theorem similar to Cauchy's inequality but involving the upper bound of B{f(z)) on 1 z I = r rather than M(r). Let A(r) be the upper bound of the real part off(z) on 1 z [ = r.
3.1 Theorem. Iff(z)
=
C z 0a,,zn, then
I a, I r n 5 max(4A(r), 0) - 2 . 9 { f ( O ) 3 for all n > 0 and r.
Proof. Let z
= reie and 03
f(z)
=
C
a,zn
=
+
U(r, Q) iV(r, 0)
n-0
40
41
3.1 Inequalities for 9{ f(z)}
and let a, = an
+ ib,,. Then (u,cos no - 8, sin n8)rn.
The series converges uniformly with respect to 8. Thus we may multiply by cosn0 or sinno and integrate term by term. Thus
J
2n
U(r, e) cos ne dB
=
0
St'
u,rn cos2 no d8 = nu,rn.
Similarly
ST'
U(r, 0 ) sin nf3 de
=
-q!lnrn,
n > 0;
also,
J:"
U(r, 0 ) de
Hence a,rn
=
(a,
1 + i@,)rn= n
= 2nao.
1
2n
o
V ( r , 8)e-in0dB,
n >0
and
Thus
+
Note, I u [ u = 0 for u < 0. Hence if A ( r ) < 0, the right-hand side of (1) becomes zero and if A(r) 2 0, the right-hand side does not exceed
Thus
1 a, I rn 5 max{4A(r), O } - 2.92{f(o)}, 0 This may be improved as follows. Write a,
= a,
+ i&,
=
1 a, 1 eien;
42
111. Theorems Concerning the Modulus of a Function and its Zeros
then
Thus
3.2 POISSON'S INTEGRAL FORMULA
3.2.1 Theorem. Let f ( z ) be analytic in a region D including J z 1 5 R and let U(r, 8) be 9 { f ( z ) } .Then for 0 5 r < R U ( r , 0)
=
-
R2 - r2
R2- 2Rr cos(8 - #J)+ r2 U ( R , 4) d$.
[Similarly for V(r, 8) = . 7 { f ( z ) } .]
Proof. Let M
f(z)=
C (a, + ipl,)rneinO, n=O
r 5 R.
We have co
U(r, 8) =
C (a,cos n8 - p, sin n8)rn, n=O
3.3 Jensen's Theorem
43
Inversion of the summation and integral process is justified by uniform convergence. To sum
f cos n(O - # > ( r / R P ,
n-
1
use m
m
C=
C cos nu
iS= i
and
xn
n-1
2 sin nu . xn n-1
Thus C=
x cos n - x2 1 - 2xcosu x2
+
and adding & we obtain Poisson's formula. 3.3 Jensen's Theorem. Letf(z) be analytic for I z I < R. Supposef(0) # 0 and let r l , r 2 , . . . , r,i, . . . be the moduli of the zeros of f(z) in I z I < R arranged as a nondecreasing sequence. If r, 5 r 5 r,l+l, then
where a zero of order p is counted p times. (This formula connects the modulus of a function with the moduli of the zeros.) Proof: First write the formula in another way. Let n ( x ) denote the number of zeros off@) for I z I 5 x . Then if r, I r 5 r,+, ,
1%
rn r1r2 . rn
.
n
=n
log r
-
1 log rm Til-1
=
'2m(log r m f l
-
m=1
log r,)
+ n(1og r - log rn)
44
111. Theorems Concerning the Modulus of a Function and its Zeros
NOTE.For r m 5 x < r,n+l, m
=
n ( x ) and n
=
n ( x ) for rn 5 x
< r . Thus
{Check:
Thus we require to prove
Clearly both sides of the formula are equal for r Iff(z) has no zero on I z 1 = r,
= 0.
We clearly cannot divide by r, integrate with respect to r, and take real parts, by virtue of infinities of the integrand. I n an interval between the moduli of two zeros r, , rnfl each side of Jensen's formula has a continuous derivative. We examine the assertion of the theorem. The derivative of the left-hand side d n dr ... r The derivative of the right-hand side is 1
2rr
x S,
d dr
- {log If(re")
I } dB = ;n
J:
:
- {logf(reiO) + logf(re-")} d0
This result equals n ( r ) / r = n / r by ( 1 ) also.
45
3.3 Jensen’s Theorem
Hence the two derivatives are equal in any such interval. Hence the two sides of Jensen’s formula differ by a constant in any such interval. Since both sides of Jensen’s formula are equal for r = 0, the constant is 0 (in that particular interval). Therefore it is sufficient to prove that each side (of Jensen’s formula) is continuous when r passes through a value r R .This is clearly true for the left-hand side. We consider the right-hand side. It is sufficient to assume that there is one zero of modulus r,? and amplitude zero. Thus z,, = rri and f(z) = (z,, - z ) # ( z ) , where #(z> is analytic and nonzero in the neighborhood of z = z,, . Consequently log I f ( z ) 1
=
log I rn - reio I
+ log I #J
I
where y ( r , 0 ) # 0 and continuous in the neighborhood of z = z,. Since we are considering a neighborhood of z = z,, i.e., r = r , , let r/rn < 2 and 10 I < n. Then
and therefore
2r =1--cosf3+ ‘n
Further, in the neighborhood of the real zero z = r, , we require to consider a range of integration about 8 = 0, viz [- 6, 61 as 6 -+ 0. Thus
r log I 1 - -eioI de rn
By writing
I 41,I <
r log I 1 - -eie 1 d€J rn
46
111. Theorems Concerning the Modulus of a Function and its Zeros
Also, for 0 < 8
< 3212, we have sin 8 > 26/32. Thus log sin 8 = log I sin 8 I > B
+ log 8,
(since sin 8 is positive) and
I log I sin 8 11 < I B + log 8 I I c + 1 log 8 I
(log I sin 8 j
< 0).
Hence (see Fig. I )
a
= 2 ~ 6 - - 2 J iogede,
&+o
e
= 2A6
-0
- 26 log 6
+ 26,
E
log E + 0,
since 6 l o g 8 + 0 .
Thus a contribution to the integral in the neighborhood of zero is arbitrarily small; thus the whole integral is continuous and Jensen’s theorem is established. The theorem may be extended to functions with poles as well as zeros. Let f(z) satisfy the same conditions of the original theorem and let it have zeros a, , . . . ,a,,, and poles 6 , , . . . , b, with moduli less than or equal
3.4 The Poisson-Jensen Formula
41
to r. Then
for let
Since g(z) has zeros a , , . . . , a,,f(O)
= g(O),
and r"h(0)
log
1 b,b2 . - b, I
= log
rn
I b,b2
. * *
b, 1
1 --
2n
s
log I h(reie) I d0
(2)
The result follows by subtracting (2) from (1). 3.4 THE POISSON-JENSEN FORMULA
Let f(z) have zeros at a, , a 2 , . . . , a , and poles at b, , b, , . . . , b, , inside the circle I z 1 5 R and letf(z) be analytic elsewhere inside and on the circle. Then
This contains Poisson's and Jensen's formulas as special cases. For no zeros or poles, we have Poisson's formula for the real part of log f ( z ) , viz., log I f(reie) I
For r
= 0,
=-
R2 - r 2 R2 - 2Rr cos(r3 - 4)
+ r 2 log If(Rei+) 1 dq5.
111. Theorems Concerning the Modulus of a Function and its Zeros
48
I. Let f(z)
=z
log [ reee- a I
- a, 1 a ] < R. We require to prove that
=-
R2 - r2 R2 - 2Rr cos(0 - 4)
fr
+ r2 log I Rei@- a I d4
R2 - r2 R2-2Rrcos(0-4)$r2
log I Rei$- a I d+;
and this is true by applying Poisson’s formula to the real part of log(R - (iiz/R)) which is analytic for 1 z 1 5 R. Since U(r, 0) corresponds to log 1 R - (lireie/R) 1, U(R, 4) corresponds to log I R - dei$ 1, a = a i,8, and
+
+ (,8 cos 4 - a sin c ) ) ~ ) , log 1 Rei@- a 1 = 8 log{(R cos 4 - a ) z + (R sin 4 - ,8)2),
U(R,
4) = 4 log{(R
-a
cos 4 - ,8 sin q5)2
thus log I R - iiei@1
= log
I Rei$ - a I.
11. If f(z) = l/(z - b), the Poisson-Jensen formula is equivalent to Poisson’s formula for the real part of log(R - (6z/R)). 111. Iff(z) is analytic with no poles or zeros in I z 1 5 R, the formula is Poisson’s formula for the real part of logf(z).
We now add all these cases to obtain the Poisson-Jensen formula. 3.5 A theorem is included now, which has considerable application in
gap and density theorems. Jensen’s theorem applies to a circular region and the following theorem is similar but applies to a half-plane.
3.5.1 Carleman’s Theorem. Let f(z) be analytic for
and let it have zeros r,eiOl, r2eiez. . ., r,,eien, inside the contour consisting arg z 5 n/2 and the parts of semicircles I z 1 = e, 1 z 1 = R, - z / 2 I of the imaginary axis joining them. Let f(z) have no zeros on the contour.
3.5 Carleman’s Theorem
49
Then
Where O(1) denotes a function of R-tm.
e, R
which for fixed p is bounded as
Proof. Consider
FIGURE 2
3 C
D
taken around the contour C : ABCDEF (Fig. 2 ) starting at z = ip and a fixed determination of the log starting at z = ie and thereafter varying continuously, JCQ is bounded. On the negative imaginary axis z = - iy and the contribution to I is
On the large semicircle, z
=
Reie and we obtain
+
since eiO cia = 2 cos 0. Along the positive imaginary axis, the contribution to I becomes
Adding and taking the real part, we obtain the right-hand side of Carleman’s
50
111. Theorems Concerning the Modulus of a Function and its Zeros
theorem, viz.,
Also, integrating I by parts, we obtain
In moving around the contour starting at z = ie and initial value of the log, the log increases by 2nni finishing at z = ie. Hence
where n is the number of zeros off’in the contour. Thus the integrated term becomes purely imaginary, viz.,
By the corollary to Theorem 2.3.1 we have
hence the result.
0
NOTE.In the proof of Carleman’s theorem, logf(z) is not a single-valued analytic function in the region and we do not apply Cauchy’s theorem to any function involving log$ (If we did, we should have to make our region
51
3.5 Carleman’s Theorem
r,
simply-connected by a canal as in Fig. 3, joining all zeros o f f to the boundary C. We simply choose one of the possible values of logf at i= ie, follow it around C, and use it in
The first part shows that there is no ambiguity in . 9 { I c , - y } in spite of the ambiguity in log .f and therefore in I . Then @ { I c - y } gives the main term on the right-hand side of the theorem.
It also remarks that once the branch of logfat ie is chosen, I y , and so also 9 { I Y } , is bounded. (This is not “as e + 0” but “ as R 00.’’ In fact e is not varied.) Thus.9{IY}is the O(1) term. Integration by parts then replaces I by “integrated terms” and the integral --f
To this new integral the theory of residues is applied. The proof that a { I , , } is O( I ) is via I,, = o( 1). For this it is necessary that we fix a value of logfaround the whole of y and that no part of y uses a logf with some large addition, Znni. So we fix a value of the logf at ie or at some fixed point iy, above i p , and deduce logffrom there onward counterclockwise. “Fix” means determine independently of R . To fix at - ie would have the same effect, since
+
However, to fix partway along y, e.g. at e, would fail to prove that the unknown correction is O(1). Both IY and the “integrated terms” would give contributions of order n and we would have to prove that they cancel
52
111. Theorems Concerning the Modulus of a Function and its Zeros
+
out. Actually 0(1)= A ( B / R 2 ) ,where A , B are independent of R. For applications of this theorem, see Titchmarsht and Levinson.$ For example, it is now a very simple matter to show that iff(z) is analytic and bounded in the right half-plane and if the zeros in the right half-plane are r,eiol, r2eiea, . . . , then the series
5%
p=1
converges.
rp
We continue with several theorems giving bounds for the absolute value of a function in terms of various parameters. It is possible in some cases to “tighten-up’’ the theorems, to improve bounds, or to generalize, however, since we only require the following forms of the theorems we leave it to the reader to consult the literature for further improvements. 3.6 Theorem. (Schwarz’s lemma). If f ( z ) is analytic for uous for I z 1 I R, and f ( 0 ) = 0, then
I f(reis) I I rM/R,
1 z 1 < R, contin-
0 5 r 5 R,
where M = maxlrl-R I f ( z ) 1. Equality only occurs when f ( z ) = zMeiv/R, and y is a real constant. Proof. Let
Then g ( z ) is analytic for 0 < I z I < R and continuous for 0 5 1 z I 5 R. Thus max I g ( z ) I occurs either at z = 0 or I z 1 = R.
+ E. C. Titchrnarsh, “The Theory of Functions.” Oxford Univ. Press, London and New York, 1939. N. Levinson, “Gap and Density Theorems’’ (AMS Colloq. Publ.), Amer. Math. SOC., Providence, Rhode Island, 1940 (reprinted 1963).
53
3.7 Theorem of Bore1 and Carathdodory
equality holding when g(z) is a constant Me'v/R. Therefore
I f ( r e f 0 )I 5 r M / R equality holding only when f(z)
=
for r 5 R,
zMeiY/R.
0
3.7 A THEOREM OF BOREL AND CARATmODORY
The following result enables us to deduce an upper bound for the modulus of a function on I z 1 = r , from bounds for its real or imaginary parts on a larger concentric circle I z I = R.
3.7.1 Theorem. Let f(z) be analytic for I z 1 5 R and let M ( r ) and A ( r ) denote the max I f(z) 1 and max 9{ f(z) 1 on I z I = r . Then for 0 < r < R ,
Proof: The result is clearly true forf(z) = constant. For f(z) nonconstant suppose f(0)= 0. Then A(R) > A ( 0 ) = 0. Since by the Poisson-Jensen formula
and if we let A(R)
=
maxO U(R, #), then U(R,
4) 5 A ( R ) , V4
and
Thus U ( 0 ) = A ( R ) implies U(R, 4) = A(R), V$ and A ( 0 ) 5 U ( 0 ) I A(R) unless the function is a constant. Let
Then #(z) is analytic for J z 1 5 R since the real part of the denominator iv, does not vanish. Now $(O) = 0 and iff(z) = u
+
54
111. Theorems Concerning the Modulus of a Function and its Zeros
Also, Schwarz's lemma gives
I #(.I I I rlR since
I 4(z) I is bounded by 1. Thus
+
(use I R R# I 2 R - R I # I). Thus we obtain the result for f(0) = 0. If f(0) # 0, apply the result to f ( z ) -f(O) and
since f(0) may be negative. Adding I f( 0 ) I to both sides, we have that
We also show that
since
where C is the circle
which ensures C is inside 1 z
1=R
(Fig. 4). The previous theorem gives
3.7 Theorem of Bore1 and CarathCodory
FIGURE 4
55
CHAPTER IV
INFINITE PRODUCT REPRESENTATION: ORDER AND TYPE
A study will now be made of the infinite product representation of an entire function. Several distinctions between functions with an infinite number of zeros and functions with a finite number of zeros will emerge. The concept of “order” will be dealt with in more detail, leading to a further object for studying these functions, namely “type.” We start with a well-known and very illuminating theorem of Weierstrass. 4.1 THE WEIERSTRASS FACTORIZATION THEOREM
Since an entire function is an analytic function with no singularities except at 00, consider a polynomial f(z) with zeros at zl,z2,. . , z,. The polynomial can then be factorized as
.
Thus given any finite set of points {zl, z2, . . . , z,~},there always exists an entire function with zeros zl, z2, . . . , 2,. Consider now an infinite sequence ofpointsz,, z2, . . . ,z , , . . . such that0 < I z1 15 I z2 I 5 . . . 5 I z, I 5 and whose sole limit point in 00, i.e., limn+oo1 z, 1 = 00.
-
56
57
4.1 Weierstrass Factorization Theorem
We must construct an entire function which vanishes at these points and nowhere alse - (z/z,)) unfortunately may diverge}. This is, in effect, accomplished by Weierstrass's factorization theorem, which demonstrates the construction and existence of such a function. We prove a form of the theorem stated as follows:
{n:=l(l
4.1.1 Theorem. If ( z , ~ is } an arbitrary sequence of complex numbers different from zero and whose sole limit point is 00 and if m is a nonnegative
integer, then 3 an entire function G ( z ) having roots at the points zl, z2 , . . . (and these points only) and a root of multiplicity m at the point zero. Further, G ( z ) can be defined by the absolutely uniformly convergent product
n
G ( z ) = eg(z)Zm
m
eQu(z/zn),
where
eg(z)
n-1
is an arbitrary entire function and Q,(z) is a polynomial such that QU(z)= z
4-+ 2
+y . ZU
22
*
*
a
The nonnegative integer v has the property that the series converges uniformly in the whole plane.
C:=E"=I I / I z, lufl
Proof. Consider (1 - (z/zn))eQv(Z/Zn) where Q J z ) is a polynomial of degree v. This is an entire function which vanishes for z = z,~.Since
where
I z/z, I < 1,
(1 -
e)
i Z
Z'!
exp ~ , , ( z / z , )= exp - - - -z, 2Zn2
nr=l(l
. . . + Q.($))
We are required to determine v such that - (z/z,))eQv(*/Zn) is absolutely and uniformly convergent for I z I < R and arbitrarily large R . Choose R > 1 and u such that 0 < a < I . Then 3 a positive integer q such that I zq I 5 R / a and I z q f l I > R/u. Thus we see that the partial
58
IV. Infinite Product Representation Order and Type
n:=,(
product 1 - (z/z,))eQv(E’Zn) is trivially an entire function. Consider the remainder - (z/zn))eQv(z’rn) with I z I 5 R. For n > q, I z, I > R/a, i.e., I z / z , I < a < I . Wenowestimateeachfactor ( 1 U,(z)> in the product for n > q
n&+l(l
+
Thus
since em - 1 5 mem. Hence
Two cases arise:
I. either 3 a positive integer p such that 11. there does not exist such an integer.
CASEI. Take Y
=p - 1
C2=l1/ I z ,
Ip
< 00, or
and RP
I Un(4 I < p I znl
&l-a *
since 1 z I 5 R, and C(n)I Un(z)I < 00 for I z I 5 R. Thus, for l-IEZ+l(l U,(z>)converges absolutely and uniformly.
+
1 z I 5 R,
4.2
Order of an Entire Function
CASE11. Take v
=n
-
59
1 so that
I U,,(z)1 < 00 with the same result as before. Thus by the root test - ( z / ~ , ~ ) ) e ~ ~is( ~analytic ’ ~ n ) in 1 z 1 5 R, and since R is arThus nF=l(l bitrary and v # v ( R ) the product represents an entire function. If, further, z = 0 is a zero of order m of C(z), then G ( z ) / z m C l ( zhas ) no zeros and equals eg(z),say, where G,(z) is the above infinite product, and
Here g ( z ) is an arbitrary entire function.
0
If G ( z ) is subject to further restrictions, it should be possible to say more about g(z). The factor is clearly inserted to produce convergence of the infinite product. The factors
are called “primary factors” and the product formed with these factors is called the “canonical product,” provided that v is the smallest integer for converges. which Ctn)I / 1 z,, 4.2
ORDER OF AN ENTIRE FUNCTION
We define order of an entire function such thatfis of finite order, if there exists a constant 1 such that 1 f(z) I < exp r A for I z I = r > r,, and for nonconstant f of finite order, A > 0. If the inequality is true for a certain 1,then it is true for A‘ > A, thus there exists an infinity of 1’s > 0 satisfying this inequality. The lower bound of these A’s is called the “order off.” Denote this lower bound by e. Then given E > 0, 3,such that
If@)
I < expre+“
for
I z I = r > r,.
This implies
M ( r ) = max I f ( z ) 1 < exp re+e 111-r
for r
> r,,
IV. Infinite Product Representation Order and Type
60
while M ( r ) > exp re-e for an infinite number of r + log log M ( r ) log r
<e +E
-+
00.
log log M ( r ) log r
and
Thus
> @ - E
for an infinite number of r + -t 00, hence
- log log M ( r ) e = lim logr
7+M
1z I =r
Alternatively, f(z) is of finite order, if 3 A > 0 such that as If(z) Thusf(z) is of order
I
=
-
00
0(exP r A > .
e if
for every E > 0 but not for any negative E . The constant implied in 0 depends in general on E , otherwise E could be replaced by zero in the formula. A few theorems will now be proved using the above expression for the order. 4.2.1 Theorem. If f i ( z ) and J2(z)are entire functions of order el and p 2 , respectively, and if el < e2, then the order of F(z) = fl(z) fi(z) is equal
+
to
ez.
Proof: We suppose e2 is finite. Since Wr;
h +J2)I W r ; fd + M ( r ; fi) < exp re^+^ f exp r e a t e < 2 exp rezfs for r > r,,(E),
+
we have that e I p2 E and hence e 5 ez. On the other hand 3 a sequence of numbers r, -+ 00 such that M(r,?;fz)> exp Thus M ( r n ;fi + f 2 ) 2 exp r2a-E - exp ril+E
=
exp r p { I
-
exp ( r i ~ +-~r,$?-E)}
> 4 exp
+
provided E is so small that el E < 0%- E and n is sufficiently large. Thus e 2 ez and the order of the sum, fi f2, equals e2. 0
+
NOTE.The proof also applies without significant changes when p2 = 00. It should be noted that the theorem is sometimes false when el = e2.
61
4.3 Type of an Entire Function
Taking fl(z) = e: and fi(z) = -e2, that, if el 5 e2, then p 5 p a .
el = e2 = 1
and
e = 0.
We can say
4.2.2 Theorem. If J;(z) and f2(z) are entire functions of order el and p 2 , respectively, where e, 5 p2, then the order e of F(z) -,fi(z)h(z) is such that 0 5 p a . Proof. Given
Thus
Q
E
> :
5 p2 -I-
ic
0 and r sufficiently large, we have
and hence 0 I p2.
0
4.2.3 Theorem. Iff(z) is an entire function of order e and P ( z ) is a nonzero polynomial, then the product.f(z)P(z) is of order 0. If the quotientf(z)/P(z) is an entire function, then it also is of order e. Proof. From the previous theorem the order off(z)P(z) does not exceed e. Since I P(z) I > I for z sufficiently large, for these values of z, I.f(z)P(z) I 2 I,f(z) I. Thus the order off(z)P(z) is not less than e, which proves the first part of the theorem. We note also that if f(z)/P(z) is entire, then its order is the same as P(z)f(z)/P(z) = f ( z ) and the second part of the theorem is proved. 0 4.3
TYPE
Entire functions may be further subdivided as follows. Given an entire function of order g (finite), suppose 3 a positive k such that M ( r ) < exp kr@ for r > R,. Thenf(z) is said to be of finite type. The greatest lower bound cr = inf k 2 0 of the k’s for which M ( r ) < exp kre holds (starting from some sufficiently large r ) , is called the rype off(z), e.g., ez has order e = 1 and type cr = I .
4.3.1 Definition. f(z) is called normal, mean, or finite type if 0 < cr < 00 and f(z) is called minimum type if cr = 0. f(z) is called maximum fype or infinite fype if cr = 00 (i.e., M ( r ) exceeds exp kre for arbitrary large r ) . As a consequence, we have the following result.
IV. Infinite Product Representation Order and Type
62
4.3.2 Theorem. The type u of an entire function of order
Q,
0 < Q < 00,
is given by the formula - log a = lim r+m
Proof. Since u = inf k, given
E
M ( r ) < exp(u
M(r) re
> 0, ~ R ( E>) 0 such that
+ &)re
for r > R ( E ) .
Also, there is a sequence { r , } such that
0 < rl < - . -< r, <
and
Thus log M(r)/re < u
+e
M(r,) > exp(a - E)r,Q.
for r > R ( E )
while log M(rn)/r,; > u - E
for arbitrary large r,.
This is precisely what is meant by u = hr-+log M(r)/rQ.
0
EXAMPLE. (er - 1)/2
5 M ( r ) = max I sin z 1 5 (el 121-r
and sin z is of order
Q =
1 and type u
+ 1)/2
== 1.
Note that the maximum modulus M ( r ) describes the growth of an entire function in the neighborhood of the point at infinity but gives no information about the behavior of f(z) in various unbounded subdomains. Consider for example ez (order 1, type I )
4.4
In the closed angle -(n/2)
and thus
@-+GO
as r - m .
+
E
I 8 5 ( 4 2 ) - E,
E
> 0,
However, in every closed angle
4.5-4.6
63
Enumerative Function n(r)
and ez+O as r - m . Thus 3 two open angles each of 7c radians, viz., the right half-plane 9 ( z ) > 0 and the left half-plane *(z) < 0, such that limr+m e’ = 00 in every angle in 9 ( z ) > 0 while lim,+m e’ = 0 in every angle in 9 ( z ) < 0. 4.5 There are several types of “enumerative” functions, some of which will be studied in the section dealing with the so-called elementary Nevanlinna theory. The simplest of these is the function n ( r ) , viz., the number of zeros off(z) in J z I 5 r. We now give several theorems dealing with n(r). 4.5.1 Theorem. Iff(z) is an entire function of order e < 00 and has a n infinity of zeros withf(0) # 0, then given E > 0, 3 R, such that for R 2 R,,
where n ( R ) denotes the number of zeros off(z) in Proof. Iff(z) is analytic in in I z 1 5 R / 3 , then for
1 z I 5 R and
a,, u 2 ,
1 z I 5 R. . . . , a, are zeros off(z)
M/2n, where If(z) 1 5 M for we have I g(z) I I 5 R / 3 , p = 1, . .., n. Thus (for I Z J = R )
1 z/ap 123
and
By the maximum modulus theorem 5 M/2” and
If, further, f(z) is of order result follows. 0 4.5.2 Corollary.
I. n ( R ) = O(Re+&).For
e, then
1
1 - z/up J
I z I = R and
ap J
2 2.
1 g ( 0 ) I 5 M/2”. Therefore
for r > r,, M ( r ) < exp re+&and the
64
IV. Infinite Product Representation Order and Type
and --log If(0)
I < R:+u'
say, then for R 2 R,
and n ( R ) = O(Re+8). 11. Since n(R) denotes the number of zeros for which I a, I 5 R, then n ( R ) is a nondecreasing function of R which is constant in intervals. It is zero for R < I a, 1 iff(0) is not zero. By virtue of Jensen's formula,
Sincef(z) is of order
e, i.e.,
J f ( R e i e 1) < k, exp Re+s, E > 0, then
log 1 f(Reie) I
< kRe+e.
Thus
Since n(R) is nondecreasing,
jr%
dx 2 n(R)
rR* R
= n ( R ) log 2
X
and
Thus, roughly, the higher the order of a function the more zeros it may have in a given region.
e < 00 and rl , r 2 , . . . are the moduli of its infinite number of zeros, then CCnf l/rRa< 00 for a > e.
4.6 Theorem. If f(z) is of order
Proof. Let B be a number between a and e, i.e., e Putting r = r,, we have
n < Ar,p
and
< < a. Then n(r) < Ara.
rZa < A,n-a'p.
4.1 Exponent of Convergence
65
Since alp > 1,
NOTE.Clearly the result is trivial for a finite number of zeros. 4.7 Definition. The lower bound of the positive numbers a for which &,) l/rna is convergent, is called the exponent of convergence of the zeros and is denoted by el. Formally, the empty set has el = 0 and if the series diverges for all positive CL,then el L- 00. Thus
and
We have proved that Q, 5 e since it may be possible to find numbers less than e for which the series converges. 4.7.1 Lemma. The number defined by the equations
log n - log n(r) el = lim -- lim n+m
logrn
logr
r+m
has the property: if log n el (finite) = lim n+m
then
log rn
- log n(r)
or lim
1% r
r+m
el is the exponent of convergence of the zeros of f(z), E > O
and E
> 0.
Proof. The limit implies that
for n
>N
and
for n > N
i.e.,
66
IV. Infinite Product Representation Order and Type
From inequality (2), n > r$L-€ and thus l / n < l / r p but since
Using inequality (I), let
E'
> 0 and let
n < r@+e,
E =
4 E ' , then 3 N such that
Vn > N.
Define
and thus 6 > 0. Hence
'dn > N
The series &) l/r$ may either converge or diverge. For example, take r, = n or r, = n(log n)z. If the zeros off(z) are finite or nil, el = 0. Thus el > 0 implies 3 infinitely many zeros.
NOTE.We can have el < e, e.g., iff@) = eZ, e = 1 but since there are no zeros of ez then el = 0. Let f(z) be an entire function of order e < 00, f(0)# 0,f(z,) = 0, n = 1, 2, . . . . Then 3 an integer p 1 such that C,"=l1/ I z, IP+l < 00. By the previous theorem any integer exceeding e will serve as p 1.
+
+
4.8 Definition. The smallest integer p for which Cgp,, I / I z, IP+l < 00, is called the genus (rank) of the canonical product. The genus of the general entire function
67
4.8 Genus of a Canonical Product
will be defined later. Sometimes the two will coincide. Thus by Weierstrass’s theorem,
where Y = p. (We were previously looking for Y such that Cr=lI z/zn I v + l would converge.) If the z,’s are finite, define p = 0 and the product as IX=:=,( 1 - (z/zn)).
EXAMPLES. If zn = n, then p = 1, (C(,) I / I z, l2 < w); if z, = 0, and if z1 = 4 log 2, z, = log n, n 2 2 3 no finite p.
=
en, then
p
4.8.1
Summarizing: For
el < c,
m
m
but
since p is the smallest integer for which the preceding equation holds. Thus, if el is not an integer, p = [el]. If el is an integer, then either
or
4.8.2
Note that
z;=l
1 / I z, lei. It is useful to subdivide but we have no information for entire functions into two further classes depending upon whether the function f(z) has zeros such that CnQ)=l 1 / r 2 converges or diverges, but we shall not pursue this subject of convergence or divergence class any further.
We have two cases to consider.
68 CASEI. p
IV. Infinite Product Representation Order and Type
4-1 = el,
CASE11. p = el. Hence p 4 el, but
pIe14e
and
el 5 e, thus p 4 e l l p + l
4.9 Hadamard’s Factorization Theorem. If f ( z ) is an entire function of order e with zeros z l , z 2 , . . . , [ f ( O ) # 01, then f ( z ) = eQ‘Z)P(z) where P(z) is the canonical product formed with the zeros off(z) and Q ( z ) is a polynomial of degree not greater than e. (The canonical product of course includes the exponential convergence producing factor which may be unity.)
Proof. We have that since f ( z ) is an entire function, f ( z ) = ~ ( O ) P ( Z ) ~ Q ( ~ ) , where P ( z ) is a product of primary factors and Q ( z ) is an entire function. We require to prove that Q ( z ) is a polynomial. Let v = [el. Thus p 5 v. Taking logs and differentiating v 1 times, we obtain
+
{Note that
To show that Q ( z ) is a polynomial of degree at most v, we require to show that Q(’+’)(z) = 0. Let
n,,,,,,(l
- (z/zrZ))-l Since gn(z) is entire, ( f ( 0 ) # 0, f ( z ) is entire and cancels with factors in f ( z ) } , then I g R ( z )I = 0(exp(2R)etE) also for 1 z 1 < 2R (by the maximum modulus theorem). Let hR(z)= logg,(z), the logarithms being determined for hR(0)= 0. Then hR(z) is analytic for Iz II R since gR(Z) # 0 in I z 1 5 R and 9 { h R ( z ) }< KRQ“. (We have absorbed 2Q+ein K.) The real part may be negative but cannot be - 00 in J z l I R .
69
4.9 Hadamard’s Factorization Theorem
By a previous result (Theorem 3.7.1 ),
and we have
then
Thus
-
+
1 > e. Terms O(RQ+e-v-l) and so also for I z I < 4 R. Since v = [el. v 0 as R 00, provided E is small. Also, since Cg1I z, I-”-’ converges, + 0 as R 00 and C l r , l > R I z, I-”-’ becomes terms O(Clznl,R1 z, in effect the remainder term for R sufficiently large. Since Q(”+”(z)is independent of R it must be zero, and the theorem follows. [7
-
--f
What we have shown is that f ( z ) = eOCz)P(z)where Q(z) is a polynomial of degree v 5 Q and ~ ( z=)
fi ( I z)e x p ( 5 + . .
n-1
- Zn
where p is the smallest integer for which
1 +P ($)’),
CF=l I / I z,
IP+’
< 00.
As an example of Weierstrass’s theorem and Hadamard’s theorem we express sin n z as an infinite product. The zeros are z = +n and all are simple. Arrange the zeros in a sequence 0, +1, - 1 , +2, -2, . . . .
I. We consider
70
IV. Infinite Product Representation Order and Type
[the polynomial Z
Qv($)
has v and v
=p = 1
+
= Z,
such that 1/ I zn lp+’ converges and p is integral, p = 1, and QY(z/zn)= z/n].Then (1) becomes ZI-II;P=~(~- (z2/n2))and
fi (1 -(zt/nz>).
n-1
Taking the logarithmic derivative
n cos n z sin n z
+ 71 + c
22
O3
=d(Z)
n-l
___
z2 - n 2 ‘
If we use the fact that ncotnz =-
1 z
+c n=l
22 ~
z 2 - n2 ’
then g’(z) = 0 and g(z) = constant. Since sin n z / n z + 1 as z
-+
0, then
eo(Z) = n
and
n ( I -(z2/n2)). m
sin n z = n z
n-1
11. By Hadamard’s theorem, since the order of sin nz is e = 1, then Q ( z ) is a polynomial of degree less than or equal to p = 1 and hence Q ( z ) = A Bz. Since sin n z eQ(2)= z n ; , (1 - ( Z W )
+
and
n ( I - (z2/n2)). m
sin n z
=
xz
n-1
With Hadamard’s theorem, we are now in a position to prove the following result.
71
4.10 Order and Exponent of Convergence
4.9.1 Theorem. Iff(z) is an entire function of order e and g ( z ) is an entire function of order e' 5 e and if the zeros of g ( z ) are all zeros off(z), then H ( z ) = f ( z ) / g ( z ) is of order e, at most. Proof. Writing PI@), Pz(z) to be the canonical product of f ( z ) and g(z), respectively, we have,
f ( z ) = Pl(z)eQ1(')
and
g ( z ) = P2(z)eQa(r),
Q1,Q, being appropriate polynomials. Thus H ( z ) = p(Z)eQi(Z)-Qa(z) where P ( z ) = Pl(z)/P,(z) is the canonical product formed from the zeros of P l ( z ) which are not zeros of P2(z). Since the exponent of convergence of a sequence is not increased by removing some of the terms, the exponent of convergence and hence the order of P(z), does not exceed e. Further, Q l ( z ) - Q 2 ( z )is a polynomial of degree not exceeding e, thus the order of H ( z ) = f ( z ) / g ( z ) is of order e, at most. 0
4.10 Theorem. The order of a canonical product equals the exponent of convergence of its zeros. Proof. Since for any entire function el 5 e, we require to prove that e 5 el for a canonical product. Let the zeros be z, , z z , . . . , and k be a constant greater than 1. Let P ( z ) be the canonical product, and we have
For
Cz,since I z 1 = r a n d I zn I > kr, I z/zn 1 < 1 and
Thus llog(1
-t)
Z
exp(-+Zn
1 P
... +-
(3) I
IV. Infinite Product Representation Order and Type
Ifp
+1
=
el,
+
+
(we recall that p 5 el 5 p 1). If p 1 > el converges), and E is small enough, then
+
E
(recall that CEl r;'Ql+s'
since
In C1, I z/z, small. Since
and
I 2 I / k . Note that I z/z, I can be large but cannot be
4.10 Order and Exponent of Convergence
73
where K depends on k only. Thus,
since (i) (ii)
1 z, I are bounded, C I z, I - Q ~ - ~ is a finite series, lznl c k t
Thus log I P ( z ) I = O{ I z
p+e}
and
I P ( z ) 1 = O{exp r e l + e } from which we conclude that the order of P(z), viz., and since el 5 e, el = e. 0
e is such that e 5 el
A particularly useful result is the following lemma.
e is not an integer, el = e. Proof. In any case el 5 e. Suppose el < e. Then P(z) is of order el, i.e., P(z) is of order less than e. If Q(z) is of degree q, eQ(') is of order q 5 e but q < e since q is integral and e is not. Hencef(z) is the product of two functions each of order less than e. Thus,f(z) is of order less than e which contradicts the hypothesis that f(z) is an entire function of order e. 0 4.10.1 Lemma. If
A consequence is that a function of nonintegral order must have an infinity of zeros. (Since if the number of zeros is finite, el = 0 = e.) Also, if the order is not integral, the function is dominated by P(z), whereas if the order is integral, P(z) may reduce to a polynomial or a constant and the order depends entirely on the factor eQcz). 4.10.2
In any case, since P(z) is of order
el and
e = m a x h el>.
e*(z)is of order q, then
74
IV. Infinite Product Representation Order and Type
4.11 Definition. The genus of the entire functionf(z) is the greater of the two integers p and q and is therefore an integer. Since p 5 p and q 5 Q, the genus does not exceed the order.
EXAMPLE 1. For the function m
s i n z = z n (I-=),
22
n=l
(actuallyn,"==,z( 1 & (z/nn))efz/n) the order of eQcz)is q = 0, e Q W n ) = &In, and p = 1 . The genus is max(p, q ) = 1. = 1 since the series C,"=ll/(nn)1+8 and the order
< 00,
E
> 0,
e = max(q, el) = 1. Hence the genus is 1 and the order is I .
EXAMPLE 2. For the function
the order of eQ(z)is q = 0. The genus is max(p, q ) = 0 since e Q W z n ) = 1, The order is max(0, 1) = 1 since el = 1 for C;=z l/n(log n)z < 00. We need to establish that for r > l
only.
(We can use for example, Gauss's test for infinite series.) Hence the genus is 0. And the order is 1. 4.12 If we have the power series representation of an entire function, we can calculate order and type fairly simply as shall be illustrated by the next two theorems. In order to study more sophisticated functions we will need Stirling's approximation for the gamma function. Both the approximation and the gamma function will be studied a little later. 4.12.1 Theorem. A necessary and sufficient condition that m
should be an entire function of order p, is that
4.12-4.13
Order and Type of an Entire Function Defined by Power Series
75
Proof. We use the fact that C& I a,zn I does not differ much from its greatest term, and that I f ( z ) I lies between the two. Let
where ,u is zero, positive, or 00. Then for every log
(&) > (p
-
8)
n log n
E
>0
for n
> no,
i.e., I a, 1 < r n ( g - e ) . If ,u > 0, Cp==o a,zn converges for all z so that f(z) is an entire function. If p is finite,
Let denote the part of the last series for which n 5 (2r)1/14-eand let be the remainder
Cz
and e 5 1/(,LA - 8 ) . Making E + 0, e 5 1 /,u. If ,u = 00, the same argument with an arbitrarily large ,u shows that e = 0. On the other hand, given E , 3 a sequence of values of n for which
Take r = ( 2 n ) p f e and solving for n,
76
IV. Infinite Product Representation Order and Type
Since Cauchy’s inequality gives M ( r ) 2 I a, I rn, then for a sequence of values of r tending to 00 M ( r ) > exp{Ar’/P+“},
+
thus e >_ l/(p tion, its order e
E)
=
and for l/y or
E
--f
0, e 2 l/p, i.e., if f(z) is an entire func-
Further, if p = 0, thenf(z) is of infinite order. Letf(z) be a function of finite order e. Then a, -+ 0 and p is nonegative and the argument has shown that y = l/e. NOTE.Iff(z) is entire and if
thenf(z) is of infinite order, since the limit as n being entire.
+ 00
can be 0 withoutf(z)
A similar theorem for the type follows from the following lemma: 4.13 Lemma. Letf(z) have a Taylor series expansion EgoU,Z’~. Suppose 3 numbers p > 0, A > 0 and an integer N = N ( p , A) > 0, such that I a, I < (epA/n>””‘ for all n > N . Then f(z) is an entire function and given any E
> 0 there is a number R
=
M ( r ) < exp{(R
Proof. Since 1 a,
Thus
vm
+
R ( E )> 0 such that
+
&)rli]
for all r > R.
1 < (epA/n)n’/‘,
0, n
-+
00,
and f(z) is entire. Further,
if n > no = no(r) = {2/1epAr/L}.Choosing R1 = R ’ ( p , A) > 1 and so large that n,,(r) > N , if r > R’, then
provided n > no.
4.12-4.13
Order and Type of an Entire Function Defined by Power Series
77
We now deduce an upper bound for M ( r ) .
However,
and max
I a, I rn 5 max 1 an I rn < max X+1 c n
N+lsnsn,
(e,,ul/n)"'p rn
N+lcn
5 max (epA/n)"'p rn 15n
= exp
The maximum is achieved for n
(W) .
= ,uW, thus
I an I rn < exp (W) .
max N+lcnsno
Hence if r > R',
M ( r ) < rN
N
C I an I n=o
+ (no - N ) exp(lrp) + 1
N
=
r A V
C I a, I + ( 2 ~ e p W- N ) exp(lrp) + 1 n=O
= exp(lrp){ 2pepirP
-
N
+ r." exp(-Arp)
c N
n=o
1 an I + exp(-lr#)
Given any E > 0, 3 a number R = R ( E )> R' such that the expression in brackets is less than exp Er/' provided r > R. Hence M ( r ) < exp{(l
+ e)r@}
for all r > R.
0
78
IV. Infinite Product Representation Order and Type
4.13.1 Theorem. Iff(z) is an entire function of finite order and type a, then 1 -
a=-
ee
e (0 < e < a)
lim n la,
n+m
(1)
Proof. Suppose u is finite. Then given any k > u, 3 a number R = R ( k ) > 0 such that M ( r ) < exp kre for r > R. According to Cauchy's inequality,
The minimum value of exp (kre)/rnoccurs for r
I a, I < (eek/n),/e
if n > N and r
=
(n/ke)'/e, thus
=
(n/ke)'/'=' > R ( k ) .
Rewriting, 1
k >-n ee
I a,
(eln
Therefore
Since k is an arbitrary number exceeding cr,
where the right-hand side is clearly finite. Now let k' be any number exceeding the right-hand side of (I). Then 3 a number N = N(k') > 0 such that I a, I < (epk'/n)"'e for all n > N. Applying the lemma with A = k' and p = e, given any E , 3 a number R = R ( E )> 0 [not to be confused with R(A)], such that
M ( r ) < exp{(kl
+ &)re)
for all r > R.
Thus a 5 k1 and because of the choice of k',
Hence the result. Also, if the right-hand side of ( I ) is finite so is u and if (3 is infinite, so is the right-hand of (I). 0
4.12413 Order and Type of an Entire Function Defined by Power Series
79
EXAMPLE 1. The function
c m
f(z)=
(e@fJ/n)n’Qzn
n=l
is of order
e and type u.
EXAMPLE 2. Since
characterizes an entire function of order zero, any function with coefficients I a, I = I / n n I E n where (&,} is a sequence of positive numbers-converging to zero is of order zero. For example,
has
e
=
0 (examine log(l/ln/Ja,J)).
EXAMPLE 3. The condition
(same as
together with
characterizes an entire function of infinite order, e.g., consider I a, 1 = I / n n s n {en} a sequence of positive numbers converging to zero slowly enough that lim E, log n = 00 n+m
(since we require (1/=1) The sequence E, = l/(log n)l-a (n
-
= 1,
00).
2, . . .) meets these requirements
80 if 0 < 6
< 1 (since E,
-
IV. Infinite Product Representation Order and Type
0 but limn+,
E,
log n
-
m).
Thus the series
represents an entire function of infinite order. 4.14 We terminate the chapter with a paper of G . P6lya which is quoted almost verbatim. There is not a great deal which can be done to improve the explanation or substance. In order to prove the result to follow we need a theorem of H. Bohrt which says, 4.14.1 “Let e be a number such that 0 < e < I , and let w function analytic for I z I 5 1 and satisfying
+(O)
= 0,
max
=
# ( z ) be any
I # ( z ) 1 = 1.
kI=e
Let r4 denote the radius of the largest circle 1 w 1 = r,+ whose points all represent values taken by # ( z ) in the circular domain I z I 5 1. Then r4 is not less than C, C = C(e) being a positive number depending upon e.” With this theorem, we can now prove the following (see P6lyat). 4.14.2 Theorem. Suppose thatf(z), g(z), h(z) are entire functions connected by the relation
f(4 = g { h ( z ) ) .
(2)
Suppose further that h(0) = 0.
Let F(r), G(r), H ( r ) denote the maximum moduli off(z), g(z), h(z), respectively, in the circle I z 1 5 r. Then there is a definite number c, greater than 0 and less than 1, independent of g(z), h ( z ) , and r, and such that
“uber einen Satz von Edmund Landau,” Scripta Univ. Hierosolymitanarum 1 No.2 (1923) 1-5. t G. Pblya, “On an integral function of an integral function,” J . London Math. SOC.1 (1926), 12-15; J. E. Littlewood, “Lectures on the Theory of Functions,” pp. 225-227. Oxford Univ. Press, London and New York, 1944.
81
4.14 An Entire Function of an Entire Function
We could substitute any positive fraction for 4 provided c is replaced by some other suitable constant. The opposite inequality
m)5 G { H ( r ) l is an immediate consequence of the definition. Proof. To fix our ideas let us take theorem of Bohr to the function
e = 4, put
C(4) = c, and apply the
which satisfies the conditions (1). We see that the function w = h(z) maps the circular domain 1 z I 5 r on a Riemann surface extended over the w-plane whose various sheets cover the whole length of a certain circle of center w = 0 and of radius R , which is not less than cH(4r). Suppose that w,, is a point on the circle I w 1 = R, such that
I g(wJ I
=
G( I wo 1)
Then there is at least one point z, inside
=
G(R).
I z I 5 r,
such that
It follows that
We now prove the main result. 4.14.3 Theorem. If g(z) and h(z) are entire functions and g { ( h ( z ) ) is an
entire function of finite order, then there are only two possible cases: either (a) the internal function h(z) is a polynomial and the external function g(z) is of finite order, or else (b) the internal function h ( z ) is not a polynomial but a function of finite order, and the external function g(z) is of zero order. Proof. The case where g(z) or h(z) is a constant is of no interest and will be excluded. Considering, if necessary, h ( z ) - h(0) instead of h(z), and g{w h(0)) instead of g(w), we can and shall assume that (3) is true. Then we have, adopting the notation (2), the inequality (4). Observe that F(r), G(r), H ( r ) are increasing functions. We may express the hypothesis thatf(z) is of finite order by the inequality
+
F(r) < Aexpra
(5)
IV. Infinite Product Representation Order and Type
82
Put h(z) = alz
and assume I a ,
+ a2z2+ - . - + a,zm +
* * *
I > 0. We have
and by virtue of (4)-(6)¶ G(c 1 a , 1 2-,rm) 5 G { c H ( + r ) )5 F(r) < A exp ra C(c I a, I 2 - 9 ) 5 A exp ralm. That is to say, the order of g(z) does not exceed a/m. If h(z) is not a polynomial¶ m can be chosen arbitrarily large and in this case the order of g(z) is zero. In any case there is an inequality for g(z), analogous to (6), let us say ( I b, I > 0, n L 1). G(r) 2 I b, I rn Combining this with (4) and ( 5 ) , we obtain
Thus the order of h(z) is not greater than a. The chief point being settled by Theorem 4.14.2 is that there is naturally no difficulty in finding closer relations between the orders of magnitude of F(r), G ( r ) , and H ( r ) . The case (b) of Theorem 4.14.3 is actually possible. Put g(w) = 1
+ 2-’w + P w 2 + 2 - W + - .. ;
h(z) = e?
The entire function g{h(z)) = 1
+ 2-’e2 + 2-4e2r + .
(7)
is the “upper half” of a theta series. The zeros and the order of magnitude of the whole theta series being perfectly known, we conclude on general principles that the function (7) is of the second order. We can easily obtain more precise information by direct calculation. Let M ( r ) denote the maximum modulus and n(r) the number of the zeros of the function (7) in the circle 1 z 1 5 r. Then we have lim r-a log M ( r ) = 2 lim r-%(r) r+m
r+m
=
1 4log2 *
-
CHAPTER V
STANDARD FUNCTIONS AND CHARACTERIZATION THEOREMS
The gamma function is now studied in some detail, firstly with a view to consolidating some of the theorems we have developed and secondly, to illustrate the concept of analytic continuation. For further details, e.g., integrals involving T ( z ) , power series, relation to the zeta function, etc., the reader is referred to Whittaker and Watson.+ 5.1 THE GAMMA FUNCTION AND ITS PROPERTIES
Define
Jo+ e-tt2-1 dt, m
~ ( x=)
x real.
5.1.1 The integral converges at the upper limit since for all x, tX-le-l
Then t E (0,
si e - W 1 dt
=
t-2tX+le- - O(t-2),
t -+a.
does not converge for x < 0 if 0 < 6
11, et
< 1 since for
<e.
E. T. Whittaker and G. N. Watson, “A Course of Modern Analysis.” Cambridge Univ. Press, London and New York, 1962. 83
V. Standard Functions and Characterization Theorems
84
Thus e-t > e-l and
j1tX-le-t dt > j tX-le-' dt 1
6
6
(1 - S"}
-
which diverges as S -+ 0 for x < 0. Also, for x > 0, since e-t < 1 for t > 0
j'
tx-le-t dt
<
6
j'
tX-' dt
6
which remains bounded. 5.1.2 The integral converges uniformly for 0
< a 5 x 5 6, for
independently of x. Hence the integral represents a continuous function for x > 0. If z is complex, Jomtz-le-t dt is again uniformly convergent over any finite region in which . 9 ( z ) 2 a > 0, for if z = x iy 5.1.3
+
I tZ-' I
=
P-'
and we use 5.1.2. Hence r ( z ) is analytic for 9 ( z ) > 0. 5.1.4
For x > 1, integration by parts gives
r ( x ) = (x - I)r(x- I), and
T ( n ) = (n - I)! 5.1.5
r(O+) = +
00,
F(x) >
for positive integral n.
for
1'
tX-'eutdt
O+
tx-1 dt
=
1 ex
--+
00,
as x -+ O+.
5.1 The Gamma Function
85
Also lim x r ( x ) = 1, x+o+
for x r ( x ) = r ( x
+ l ) , and since r ( x ) is continuous, lim T ( x + l ) = T ( l ) = 1 . x+o+
5.1.6 For x > 0, y > 0, r(x)r(y) r ( x+y)
So
P-1
U3
=
(1
+ ty+v dt .
Since T ( x ) r b ) = Jm
dt
tx-1e-t
0
Jm
sv-le-8 ds
for x
> 0,
y
> 0,
0
put s = tv and
Letting u = t ( l
+ v)
Inversion of the integrals is justified, since the individual integrals converge uniformly for x 2 E > 0 and y 2 E > 0. Sometimes r ( x ) r ( y ) / I ‘ ( x y ) is called the Beta function B ( x , y ) which equals tX-l(l - t)v-’ dt by a suitable transformation.
+
Jt
5.1.7 Putting x
=y =
f, v
= tan2 8,
we obtain
(r($)I2 = 2 Sn”dO = n. 0
Since T(1)> 0, we obtain
s
a,
r(x)r(l -x) =
0
r ( f=) 6. Also, putting y = 1 - x ,
u-z
-du = l+u
n
sin(1 - x)n ’
O<x
86
V. Standard Functions and Characterization Theorems
and
by contour integration. Thus 7c
T(X)T(l- x) = sin x x
for O < x < l
5.1.8 Asymptotic Behavior of T(x): Stirling's Formula. Consider T(x),
where x is an integer, say n, then F(n) = ( n - l)!. We have
c log
n-1
log(n - l)! =
j
v+ll2
also
log t dr =
,-1/2
s
112
{log(Y
+ t ) + log@ - t ) >dt
{log Y2
+ log(1 -
0
=
J
112
0
= log Y
s
n-112
=
Y,
v=l
+ c,,
log t dt -
112
t2/v2))
cv= O(l/Y2). n-1
C
C,
v-1
- 4)- (n - 4)- B log s (n - 4) logn - n c o(l),
= (n - B) log(n =
dt
+ +
+ s - c c, -k m
0(
1)
V-1
where C is a constant. Before establishing the nonintegral case, we prove the following lemma. 5.1.8.1 Lemma. For n large
T ( n ) / F ( n+ a ) = U P . Proof. lim nu r(a)I'(n)/T(u n +w
Transform t
+ n) = lim nu n+m
=
I'
u/n. Then the right-hand side is
F ( 1 -
0
r)n-l
dt.
87
5.1 The Gamma Function
Thus lim n‘r(n)/r(a
+ n) = 1
n+m
and r ( n ) / r ( n
+ a) = n4
for large n.
0
+
-
If x is not an integer, let x = n a where n is integral and 0 < a From the previous lemma we have that since r ( n a ) r(n)nu, log P(x)
= log
r(n
+
+ a) = log r ( n ) + a log n + o(1) =
< 1.
(n-~)logn-n+C’+alogn+o(l)
+ + C’
(x - a - 4) log(x - a) - x a a log(x - a) o(1) = (x - &)log x - x c o(1).
=
+
+
+ +
+
To evaluate C, consider r(2x)r(3) = 2*-lr(x)r(x &), a recurrence relation obtained by considering, for example, B(x, x). Taking logs, (2x - 8) log 2x - 2x =
+ c + o(1) + l
(2x-l)log2+ -x -
o g 6
(X-~)logx-x+CCo(l)+xlog(x+4)
& + c + o(1).
Thus
+ 2x log x - 4 log2x - 2x + c + o(1) + log1/R = 2xlog2 - log2 + x l o g x - hlogx - 2x + 2 c - 4 + x l o g x - 1 + - + 4 1 ) ( :x)
2 x log 2
and l o g 6 + & log 2 = c - 4 giving C
= log
6+ o(1).
1 +x - + o(1) 2x
=
Hence
+l
o g 6
and for x not a negative integer, since &(l)= 1
+ o(l),
log r ( x )
=
c + o(l),
(x - 4) log x - x
+ o(1)
V. Standard Functions and Characterization Theorems
88
Stirling’s formula for complex z can be shown,t the result holding uniformly for - 7c 8 5 arg z 5 7c 6, 6 > 0. The negative real axis is excluded since r ( z ) has an infinite number of poles on it.
+
5.2
+
ANALYTIC CONTINUATION OF
r(z)
For 9 ( z ) > 0, r ( z ) = JTe-tfz-l dt is an analytic function. We require now to extend analytically into the rest of the complex z-plane. Consider Z(z)
=
J, e+(-
5‘)L-I
where C i s the real axis from m to 6 > 0, I E the real axis from 6 to m (Fig. 1). Define (-
I
I
d5,
=
6 in the positive sense, and and choose
E)r-l = e(z-*)log:-t)
C-p‘anr
C
FIGURE 1
log(- 6) to be real for t = -8. The integral converges uniformly in any finite region of the z-plane (since the integral depends upon C and the circle does not pass through the origin, i.e., 8 is fixed.) Thus I ( z ) is analytic for all finite values of z. To evaluate Z(z), set 5 = pi+. Then log(-
E)
=
log e
+ i ( # - n)
on the contour (so as to make log(- [) real for 5 = -6). The inetgrals on the portions of C, consisting of (00, 6) and (8, m) give
=
- 2i sin nz
J,
e-Qez-’dp.
t See E. C. Titchmarsh, “The Theory of Functions.” Oxford Univ. Press, London and New York, 1939.
5.2 Analytic Continuation of r ( z )
Also, on the circle
89
1 5' I = 8,
1 6 I = 8 gives
and the integral around the circle
as 8 - 0
O ( @ ) = o(1)
if x > 0 (562-'
:J d#). Letting 6
I(z)
.--f
0 we obtain
I
e-Qpz--'de,
= - 2i
sin nz
= - 2i
sin nz I'(z).
*(z)
>0
Since Z(z) is analytic for all finite z, iiZ(z)cscnz is analytic except possibly for poles of csc nz. Further, iiZ(z) csc nz
=
I'(z),
g ( z ) > 0.
Hence iiZ(z) csc nz is the analytic continuation of r ( z ) in the entire zplane. Since the poles of csc nz are z = 0, f l , f 2 , . . . and F(z) is analytic at z = 1, 2, . . . , the only possible poles of iiZ(z) csc nz are z = 0, -1, -2, . . . . These are actually poles of T(z), for if z is one of these numbers, say -n, then (--,y-l is single-valued in C and Z(z) can be evaluated directly by Cauchy's theorem 2ni (n!
l)n+n+l
- 2ni =n!
*
Thus Z(-n) = -2ni/n! and the poles of csc nz at z = 0, -n are actually poles ofI'(z). ( I ' ( - n ) = Z(-n) 1/(-2i sin nn).)The residue at z = -n is
-
- 2ni z+-n
(actually lim Z(z) i csc nz . (z 2
z+-n
The formula I'(x)F(l - x )
+ n) = I ( -
=n
")
n) i lim 2 z+-n sinnz
.
csc nx and others, can now be justified
V. Standard Functions and Characterization Theorems
90
for complex values. Thus I'(z)I'(l - z ) = n csc nz for all nonintegral z (since the left-hand side is n csc nz for 0 < z < 1 and T(z)has an analytic continuation to the whole plane). Thus l/I'(z) is an entire function (since poles of T(l - z ) are canceled by zeros of sin nz or 1
1
T(Z)
?L
-=-
1 Z(1 - z ) - -i sin nz -i Z(l - z) 2 sinn(1 - z ) 2n
and 7(1 - z) is analytic everywhere in the finite plane). 5.2.1 To prove 1/ T ( z )is of order 1.
Proof. Since l/F(z)
= 1(1
- z)/(-2in), consider
i.e.,
I(1 - z)
=
I,
(- w)-"e-Odw.
Take C the unit circle together with the real axis from 1 to twice. On the circle, - o = ei@,- ?L 5 4 5 n and
00
described
The integral around the circle is O(em), { I e-O I 5 e}. On the rest of the contour, - w = te", t > 1, 4 = kn.Therefore for this part of the integral
If n
=
[r],
+
+
(since n < r I ) , and (r 2)'+2 = O(exp ~ l + by ~ )taking logs and observing that terms O(r log r ) = O(rl+s), E > 0. Also, all terms including O(em)are O(exp rl+a), thus e 5 1 . Since e 2 el = I [observing the poles of r ( z ) ]we conclude that e = 1. 0
91
5.2 Analytic Continuation of T ( z )
We are now in a position to develop an expansion for l/I'(z),
Since
r(1)= 1, lim L+O
and 1
viz.
=
B, b
=
1
--
zI'(z)
-
lim L+O
1
I'(z
+ 1) = I
0. Putting z = 1,
Taking logs,
+ log n=1 n m
0=a and Q
=
- (log2 - 1) + log-
{
(
2
) + ...}
--
2
1 ... + --logn n
n+m
(For the existence of the limit we can use, for example, the integral test for series.) Thus
Further, since
5.2.2
Also, according to Gauss,
It follows easily now that F(z established.
+ 1) = zI'(z).
Similar formulas can be easily
92hh
V. Standard Functions and Characterization Theorems
5.3 CONJUGATE POINTS
The gamma function assumes conjugate values at conjugate points, a property which facilitates calculation of absolute values. Thus, for example,
I I'(4
+ it) l 2 = I'(4 + it)F(t - i t )
and since
I'(s)F(l
- s) = n/sin
ns
I I'(4 + i t ) la = n/sin n(4 + i t ) = n/cos int and
We can similarly establish the asymptotic behavior of large y, showing that
I F(x + iy) I for
for finite x. One or two examples now follow which illustrate, in particular, the use of previous theorems and lemmas in the calculation of order. For a particularly practical function, we consider the following. 5.4 BESSEL'S FUNCTION Jv(z)
For
Y
a nonnegative integer Jv(z) =
(- I)* (z/2)v+2n n! I'(Y n 1) -
+ +
Thus Z-~J~(Z) is an entire function since the series converges absolutely for all z. To calculate the order:
thus
=
lim
n+w
n log 2
+ log(n/2)! + log n log n
F(Y
+ ( 4 2 )+ 1 )
5.5
Function Fa(z) =
J'$exp(--ra)
cos zt dr
(a > 1 )
93
Since log r ( n ) = (n -
Hence
4) log n - n + C + o(1)
Z - ~ / ~ J ~ (is~ an / Y entire )
for C constant,
function of order
4 [for
z-,J,(z) = O(exp 1 z I1iC),
-
changing z l/t gives the result]. Thus ~ - ~ ' ~ J , , ( has f i ) an infinity of zeros and further, J,(z) hasan infinity of zeros (for if J , ( f i ) = 0 for z = a, J,(Z> = o for z = l/ a ).
5.5 The f o l l o w i n g example requires a little more care. We require to calculate the order of
F,(z) = J
exp(- t a )cos zt dt
(a > 1).
0
For a > 1 the integral converges uniformly for all'finite z, thus Fa(z) is an entire function. Also,
where inversion is justified by convergence of
( 2 -)(2n)! ,.ZntZn
Jmexp(0
Now
Thus
ta)
n-0
dt
=
Jmexp(0
tm) cosh
rt d t .
94
and a,
V. Standard Functions and Characterization Theorems = 0,
log-
n odd. For n even, 1
I a, I
= logn! - log
= nlog
n n - -1ogn a
+ O(n)
and
Hence e
= a/(. -
1). For a
=
2,
F2(z)= G e x p ( - z2/4), showing directly that
e = 2.
5.5.1 The convergence of this integral is not particularly obvious and we study it a little further. If w = u iu,
+
Thus if u is large and positive, cos w
25
8 ciw
I cos w I = 4 e'.
and
If u is large and negative,
I cos w I
-
4 cU.
In the integral
Fe(z)=
J:
exp(-
2")
cos zt dt,
the factor coszt may be large in absolute value, in fact it will be large unless z is (exactly) real. For z = x iy, y > 0, t real,
+
and
95
5.6 Order of the Derived Function
However I coszt 1 is no bigger than this, thus except for a factor I cos zf I 5 etlzl and for a > 1,
I exp(--ta)
4,
cos zt 1 5 exp(t 1 z I - ta)5 exp(- t)
+
for all t for which ta-l > 1 I z 1. Hence the integral is uniformly absolutely convergent in any bounded range of z. For 01 = 1, I z 1 < I ,
tending to poles at z = f i , thus the integral converges only between the lines y = hi. We now consider the derivativef’(z) of an entire function and the following theorem is not unexpected. 5.6 Theorem. The derived entire function
f ‘(z) is of the same order (and
type) asf(z>. Proof. Let M ’ ( r ) = max,,,,, f’(z). Then
since
and by taking the integral along the straight line we have ~ ( r5)
J’“IY’W I df + I ~ ( 0I)I rM’(r) + I ~ ( o I. )
However, f’(z) where C is
I w -z 1
=
I f’(4I = M ’ ( r ) We choose R
= 2r
J
=-
2ni
R -r (1z
( wf (w ) z ) ~dw =r
< R ) and choosing z such that
M‘(r) 5 M ( R ) / ( R - r ) . and
V. Standard Functions and Characterization Theorems
96
Since
e =lim r +m
log log M ( r ) log r
the result follows. Note log M(r) is a convex function of log r and either log M(r)/log r < constant A or + 00; therefore
If log M(r)/log r -+00, the second term tends to zero, and if log M(r)/log r A > 1, the second terms tends to zero also. Similarly for the left-hand side of the inequality. -+
The “type” is discussed similarly. The next theorem is of considerable consequence. Extensions and ramifications may be found in papers due to M. Marden. 5.7 Theorem (Laguerre). If f(z) is an entire function, real for real z, of order e < 2 and with real zeros, then the zeros of f ’ ( z ) are also real. Further, the zeros off’@) are separated from each other by zeros of f(z).
Proof. The hypotheses give
where k is zero or positive and c, a, zl, . . . , zn are all real. Taking logs and differentiating we obtain
If z
=x
+ iy, then =0
Thus f‘(z)
=0
for y
=0
on the real axis only. Also
only.
97
5.8 Convex Sets and Lucas’s Theorem
is real and negative for real z. Thusf’(z)/f(z) decreases steadily as z increases through real values from z, to z,,, and it cannot vanish more than once between z, and z,+, (there can be no inflections). Clearly it changes sign (sincef = 0 at z, , z,+, andfdoes not change sign between z, , z,+, but f’does (Fig. 2)). Consequently it vanishes exactly once in the interval and the theorem is proved. 0
5.7.1 We can now deduce that if the zeros off ‘(z) are z:, z i , . . . , then the series 1/ I z, I‘, Zzp=, I / I z,’ I u converge or diverge together. Thus the zeros off’(z) have the same exponent of convergence as those off(z). It can be shown thatf(z) andf’(z) have the same genus. The only case to consider is clearly when e = 1 then the genus is 0 or 1. Since f ’(2) has the same order asf(z) and has real zeros only, applying the same theorem to f’(z), f ” ( z ) has real zeros only and so on for f”’(z), etc. We may extend the proof to functions of order 2 but of genus I . + It is simple to see that the theorem is not true for functions of genus 2. Let
xr-l
f(z)
=z
exp z2,
f ’(z) = ( 2 2
+ 1 ) exp z2
then the zeros off are real and the zeros off ’ are complex. Alternatively, let f(z) = (z2 - 4) exp(z2/3), f’(z) = #z(zz - 1) exp(z2/3) and the zeros o f f ‘ are real but not separated by the zeros off. It is convenient at this stage to include Lucas’s theorem on the zeros of a polynomial and its derivative. Extensions which have been deduced in various papers, require the techniques which have now been studied. 5.8
CONVEX SETS
We introduce the notion of convexity, and establish a theorem relating to the distribution of zeros of a polynomial and its derivative. + See E. C. Titchmarsh, “The Theory of Functions.” Oxford Univ. Press, London and New York, 1939.
98
V. Standard Functions and Characterization Theorems
Definition. A set R is convex whenever z1 and z, are two points of R, and the points [z I z = z1 a ( z 2 - z , ) ] ,0 5 01 5 1, also belong to R. Clearly a convex set must be arcwise connected, but need not be a domain (an open arcwise connected set.) The interior of a circle is a convex domain. Also, the subset of S: [z I z2 - 1 I 5 1, z f 01 (lemniscate) lying to the right or left of the half-plane is a convex region. Consider the interior of a triangle A with vertices at z , , z , , z,. The interior is the point set,
5.8.1
+
4 = [z I z = alzl + aazz + a3z3; a, L 0,
3
a, L 0, a, 2 0;
1
C a, = I .
7-1
Similarly, 17 is a convex polygon if 17 is the boundary of a convex domain. If a polygon I7 has vertices z, , z, , . . . ,z, , the interior 17i is such that
- . . +u,z,;
zIz=a,z,+
a,>0;
1-1
z=, a,z,, and a, 2 0, C",, a,
5.8.2 To prove: If z = the convex hull of the z,'s.
I
...; fa,= 1 =
1, then z lies in
Proof (by induction). For the case n = 2, if z' is between z1 and z, dividing the line joining z , and z, in the ratio 1: 1 - 1, 0 < 1 < 1 (Fig. 3), then vectorially
ozl = G' + El oz, = oz' + fi, oz' = +{ozl + oz, + ZZ' + Fz'}
and
+ Az,: C coefficients = i. Conversely, if z' = alzl + a,z,; a, + a, = 1 ; a,, a, 2 0, then z is in the convex hull of z , , z , . Let z = alzl + . . . + a,z,; C:=, a, = 1 ; a? 2 0. z'
=
(1 - l ) z l
5.8
99
Convex Sets and Lucas’s Theorem
We assume 0
< a, < I , and
z = (1 -a,) { ___ l - a n zl+...+-
an-1
1 - a,
.,-I}
+ anzn.
Write z’l
-- a, 1 - a, Z 1 +
+- 1an-1 - a,
*..
Zn-1.
Since a A + ... +-=an-1
1 - a,
1 - a,
1,
+
then z” is in the convex hull of zl, . . . , z,-~. Now z = (1 - an)z” a,z, and since a,, 1 - a, 2 0 and {a, (1 - a,)} = 1, then z is in the convex hull of z” and z,, i.e., zl, . . . , z,. 0
+
The converse proposition is similarly demonstrated. 5.8.3 Lucas’s Theorem. The zeros of the derivative P ’ ( z ) of a polynomial P(z), are contained within the convex hull of the zeros of P(z).
Proof. Let P(z) have zeros zl,z2, . . . , z,. Let 17 be the least convex polygon containing these zeros. We show that P‘(z) cannot vanish anywhere in the exterior of 17. Since P(z) = ( z - zl)(z - z2) . ( z - z,), then
- -
P(z)
dz
If P’(zo) = 0,then
Thus
and I)
n
V. Standard Functions and Characterization Theormes
100
Since zo =
f a7z,
with
f: a,
=
1, a, 2 0,
7-1
7-1
we deduce that zo lies within the convex hull of the z;s.
0
Some extensions of Lucas’s theorem to entire functions can be found in a paper by M. Marden.+ In this paper it is illustrated how Lucas’s theorem is not true for arbitrary entire functions, e.g., 5.8.4
has only one zero z
=
-1,
f’(z)
but the derivative = (1
+ z + z2) exp(z2/2)
has two zeros; z1 = e2niI3 and z2 -- e4nif3 and Lucas’s theorem clearly does not hold. The theorem does extend to functions whose order e is such that OI@
is the most general meromorphic function which coincides with Mo(z) in its poles and corresponding principal parts.
Proof. If Mo(z) and M(z) are two meromorphic functions which coincide in their poles and corresponding principal parts, their difference Mo(z) M ( z ) is evidently an entire function, say G(z) and M(z) = Mo(z) G(z). 0
+
For example, cot z and 2i/(eaii”- 1) are two meromorphic functions which coincide in poles and corresponding principal part, thus they differ + “On the zeros of the derivative of an entire function.” Amer. Math. Monthly 25, No. 8 (Oct. 1968). s W. K. Hayman, “Meromorphic Functions,” Oxford Mathematical Monograph. Oxford Univ. Press (Clarendon) 1964. London and New York, 1964.
5.9-5.10
101
Mittag-Leffler Theorem
only by an additive entire function. The poles of cot z are the zeros of sin z = 0, i.e., z = nn,n = 0, & l , . . . . The poles of 2i/(e2iz- 1) are the roots of e2iz = 1 = ezfini, i.e., z = nn, n = 0, i l , . . . . The difference is i, an entire function. 5.9.1 Lemma. Suppose f ( z ) is analytic in the entire finite complex plane except for poles z l , z 2 , . . . , zk. Let P(z, z,) be the principal part o f f ( z ) at z = z,, r = 1, . . . , k. Then there exists a function # ( z ) , analytic in the entire finite complex plane such that
Moreover, # ( z ) has the same principal part at z
= 00
asf(z).
Proof. Consider
Clearly g ( z ) is analytic in the entire finite complex plane except at points zl,. . . ,z k , where it has poles. Since for s # r, P(z, z,) is analytic at z = z,, the principal part of g ( z ) at z == z, is P(z, z,). Hence g ( z ) has the same poles and principal parts at z l , . . . ,zk as f ( z ) and is analytic everwhere else. Consider # ( z ) = f ( z ) - g(z), z # z r , r = 1, 2, . . . . This function is analytic in the entire complex plane except for possible poles at z l , z 2 , . . . , zk . Since g ( z ) and f(z) have the. same poles and principal parts, the difference # ( z ) is such that the principal part a t each pole is zero. Thus for
and # ( z ) has removable singularities at z l , . . . , zk. Thus # ( z ) is analytic. Since g ( z ) + 0, z 00, g ( z ) has a removable singularity at z = 00. Thus # ( z ) has the same principal part at z = 00 as f ( z ) . Writing --f
f ( z ) = g(z) the lemma is established. [7
+ #(z>,
102
V. Standard Functions and Characterization Theorems
5.10 CHARACTERIZATION OF A MEROMORPHIC FUNCTION
A rational meromorphic function is determined to within an additive constant by the behavior at its poles. We want to find how completely an arbitrary meromorphic function is determined by the character of its poles. Weierstrass’s theorem gives an infinite product representation of an entire function. We now prove the Mittag-Leffler theorem which represents a meromorphic function by an infinite decomposition into partial fractions. Since a meromorphic function may be characterized by the nature of its poles only to within an added entire function and since the question has been answered for meromorphic functions with a finite number of poles, the only other possibility to answer is when 00 is an accumulation point of poles. Clearly no finite point is an accumulation point of poles ( I l f = 0). 5.10.1 Mittag-Leffler Theorem. Let z,, z l , . . . , z,, . . . be any sequence of distinct points tending to 00. Suppose to each z, there is associated a polynomial P , ( l / ( z - z,)) in the variable l / ( z - zn). It is possible to find a meromorphic function f ( z ) having poles at the points z, but no other points, and with corresponding principal parts P,(l/(z- z,)). Then f ( z ) may be represented in the form
+ v=oc m
f(z)=
P”(l/(Z -Z J ) -9”(4)
where qy(z) are polynomials and w ( z ) is an entire function of z. €‘roo/: Unlike the finite case we must ensure that the given representation converges. Suppose the sequence (2,) is ordered. Thus I zo 1 5 1 z1 1 I * . since the only point of accumulation is infinity. Possibly zo = 0, but all other points differ from zero. We suppose that zo#O and the function P , ( l / ( z - z”)) being analytic everywhere except at z,, must itself be analytic at the origin. Thus P, has a Taylor expansion at the origin given by
.
P”(I / ( z - z,))
= at’
+ayz +apz + -
*
The radius of convergence is clearly 1 z, I. The series converges uniformly in the circle C,: I z 1 I 4 1 z , 1. Thus in C,, P,(l/(z - z”)) can be approximated by a finite sum as closely as we please. In particular, for q J z ) = at’ a p z * +-&)zk,
+
+- -
I P ” ( W - .,I)
- 9 , ( 4 I < 1/2‘
throughout C,. The series ~ ~ , , ( P , ( l / (z 2,))
- 9,(z)) converges to the
5.9-5.10
Mittag-Le&r
103
Theorem
desired meromorphic function in every circle about the origin since any such circle can be contained within one of the Cis. In C , , CL20(P,(l/(z- z,)) - qn(z)} is well behaved. It is an analytic function with no singularities but the prescribed poles. Thus
(zp=,)
is analytic in C, and dominated by C&(1/2”). The series then converges uniformly in C,. Thus since a uniformly convergent series of analytic functions converges to an analytic function, the second part of the series introduces no new singularities into C,. Thus the theorem is proved (if zo = 0, we add on Po(l/z)). 0 5.10.2 In general, the polynomials q,(z) will not be uniformly bounded (to ensure convergence). However, in special circumstances all the qv(z) may be chosen of the same finite degree. It is sufficient to take the degree k, of the polynomial q,(z) [i.e., the sum of the first k, terms of the power series for P,( 1 / (z - z,))] so large, that having chosen an arbitrary R > 0, the terms I P,(l/(z - 2,)) - q,(z) I of the series for all 1 z I 5 R and large v, remain less than the terms of a convergent series of positive terms.
EXAMPLE. The convergence producing terms q,(z) are not always necessary. If, e.g., the points 0, 1, 4, . . ., v2, . . . are to be poles of order one with respective principal parts l/(z - v2), then
mythe
is a solution. Note that for R > 0 and m > v =m 1 with I z I 5 R, converges uniforinly because
+
1 I/(z
- v’)
1 5 I/(va - R ) < I / ( v *
-
series from
4 v2) = 2/v2.
Similarly if the function has only simple poles
P”(1/(z - Z”>)= a,/(z - 2,) , then the q,(z) may all be chosen of degree n if the series Ego1 a, I/ I z, converges. Since
In+’
104
V. Standard Functions and Characterization Theorems
choosing q,(z) a polynomial of degree n, viz,
converges, it assures the convergence of Thus if EgoI a, I/ 1 z, &,)((U,/(Z - z,) - q,(z)). This case is the one that arises in most applications. Although the Mittag-Leffler theorem can be used to expand a function with simple poles into partial fractions, the entire function o(z) has still to be determined. This special case has a more direct approach. Let C be any simple closed curve not passing through any pole of f(z). If z is a regular point inside C, then
i.e.,
where res{f(z,)/(z, - z ) } ,means residue o f f ( t ) / ( t- z) at 5 = z,, the sum taken over all singularities z, off(z) in C.
NOTE.f(z) is the residue off(t)/(t
- z)
at the pole
t = z.
Since f(z) is assumed to have simple poles,
and
Since the poles are isolated 3 a sequence of closed curves C, such that C , 3 C, 3 C, , each avoiding all the poles of f(z) and 2 C, such that the distance of C, from the origin tends to infinity with n. If for some such sequence,
5.9-5.10
105
Mittag-Leffler Theorem
denoting poles in the annulus between Cn-l, C, by zp), the series
converges and gives the decomposition of the function into partial fractions. As an example of the theorem and the above method we consider n cot nz. 5.10.3 n c o t nz. The poles are the zeros of sin nz viz., poles of order 1 at z = 0, f l , f2, , . . . The residue at z = n is
lim (z-n)
a
-n
2-m
cos n z = 1. sin nz
Thus the principal parts 1 ~
pv( z - z,
)
1 1 -Z - -z - z, zv z,2
---
..
for Y = 1, 2, 3, . . . . We may take the degree k, of the polynomials q,(z) to be zero, hence q,(z) = - l/z, and = Y,
,,z = 0,
and for large
z2,
= -Y,
I z, 1, viz., I z, [
=Y
> 4R,
Therefore R
IZvlIZ,--Rl
4
<-.-R lZ,l
31zvI
2R
<m
and hence I P, - q, I is less then the terms of a convergent series of positive terms. Thus 1 f ( 2 ) = n cot 7zz = o(z) 3- z
+ c {W
,=I
1
- z,
+ L} z,
I06
V. Standard Functions and Characterization Theorems
or rewriting,
Further, o ( z ) is an entire function and the sum is uniformly and absolutely convergent in any bounded region at positive distance from the integers or more precisely, on every set of the form
For this reason the terms of the series can be reordered to give = 44 S(d, where
+
IC
cot nz
(Note that Cz,l/(z - n) is not convergent.) The problem of determining w(z) can sometimes be difficult. If we know that m
n-1
taking logs and differentiating, we have 1
zcotnz=--+
c za22- n2
____
n-1
showing that o ( z ) = 0. However, this is a rather circular argument and we consider the following. S ( z ) is periodic with period 1, since S(Z) - S ( z
1 I + 1) = -z z+l
Also, S(z) is uniformly convergent in the region
I~(IR,
{z=x+iy1
I z - n ~ r d > ~n ,E Z } .
However, it is not convergent for unbounded y since if any E is given satisfying 0 < E < 4 and if an N = N(E)could then be found such that
-
22
<E
for all z,
Mittag-Leffler Theorem
5.9-5.10
putting z
= 2iN
107
we would have
4N
'
n=N+l
n2
2N
+ 4N2 2 n J + l
4N 1 -4 N 2 + 4N2 2
Then n cot nz is periodic with period 1 and so o ( z ) must be also periodic with period 1. Further
x cot nz = n
cos nx cosh n y - i sin nx sinh ny sin nx cosh ny i cos nx sinh ny
+
e-imz
-fn-=Fin
as y - f c o
ie-%*
+
and thus .?t cot n z is bounded as I y I + 00. Similarly I S(z) 1 is bounded as y I -+ +co, since using 1 z2 - n2 l2 - (n2 y 2 - l)a2 0 or observing that I f 1 2
+
-s{f},
and
for y l 2 .
-
f 03. Thus o ( z ) Note also that S(iy) = -in coth ny -+ in as y is periodic with period 1 and bounded as 1 y I + 00, hence it is bounded in (z = x iy I 0 5 x 5 1 } and so is bounded in the whole plane. However, it is an entire function, thus it is a constant. Since w ( - z ) = - w ( z ) , we have that w ( z ) = 0 for all z [or examine ~ ( i y ) ]Hence .
+
ncotnz=-+
c -.
1
22
z2 - n2
z
Alternatively, let I,
=
's
2ni
R,
n cot xz dz. z -6
V. Standard Functions and Characterization Theorems
108
+
+
Let& bethesquare I S ( z ) I = n Q, I ~ ( z )I = n Q, n the vertical sides passing between the poles at n, n 1 1 n cot 762 I, = dz 4ni { J R n z - t
+
s
Rn
+
= 0,
fl,. . . ,
n cot n(- z ) - z - t
- 1 JR,,
- 1
-4ni
1
J
R,
JRnn
ncotnz
z-t
" ncot nzdz.
dzl
1 .(bound for I n cot nz I on R,) 5-
4n
-+O
and n cot nt
as n - m = S O).
. (n + 2 1 "- I 5 l2 8(0++) Q)2
CHAPTER VI
FUNCTIONS WITH REAL AND/OR NEGATIVE ZEROS: MINIMUM MODULUS I AND SEQUENCES OF FUNCTIONS
6.1 FUNCTIONS WITH REAL ZEROS ONLY
Some important complex functions have real zeros only, e.g., l / r ( z ) has no complex zeros. However, it can be very difficult sometimes to decide whether the zeros are real or not.+ The question can be decided sometimes. 6.1.1 Theorem. Letf(z) be a polynomial with
) an entire function of genus 0 or 1, all of whose zeros are real. Let ~ ( o J be all zeros being real and negative. Then
has all its zeros real, with as many positive, zero, and negative zeros as
f(d. t Cf. E. C. Titchmarsh, “The Theory of Functions,” Section 8.6, p. 268. Oxford Univ. Press, London and New York, 1939.
109
110
VI. Zeros: Minimum Modulus I and Sequences of Functions
Proof. Let
Consider
=
(1
+
$1
f(z)
(where
19 = z -
Then gl(z) has as many zeros at z = 0 as f ( z ) since g, (0)= f(0). Putting h ( z ) = zmlf(z),a1 > 0 , the zeros o f h are z = 0 and the zeros o f f . Hence, if h ( z ) vanishes, e.g., at z = b, , b, [zeros o f f ( z ) J ,then h'(z,)
=0
for b,
< z1< b, (Rolle's theorem).
Thus gl(z) has the same number o f positive and negative zeros as doesf. [Note g,(z) is a polynomial of degree p and so is f(z).] Since a, > 0 and f i s a polynomial, zalf-+ 0 as z --+ 0, hence there is another zero z,' between 0 and z,. Thus the operator
(1 + -
3
:1)
(I+--
cannot decrease the number of real positive zeros (it may increase them). The same result holds for
where
for then
6.1 Functions with Real Zeros Only
Transforming z
= z‘
exp k,, k,
g,(z‘expk,) = a,
and agn(z’exp k,)
= aa,
111
-
=k
zgl l/av,we have
+ aI$,(l)z’
+ aa14,(l)z’
exp
exp k -
= C,(Z).
(
1- + -
3
Y
* *
Thus where
*
-
?)}( 1 + $). . .(1 + $) an
Thus @,(w) -+ 4 ( w ) uniformly in any finite region, since 4 is an entire function, and by Hurwitz’s theorem, the zeros of g(z) are limits of zeros of C,(z). Then g,(z) has at least as many positive and negative zeros asf(z), also, g(z) has as many zeros at z = 0 as f(z) since if g(z) has a zero of order r ( ( p ) then a,, a,, . . . , a,.-, = 0. Thus f(z) has a zero of order r. (Note this has nothing to do with Rolle’s theorem.) Hence the theorem is proved. 0 A direct consequence is as follows. 6.1.2 Theorem. Suppose that 4 ( w ) satisfies the conditions of the previous
theorem and that f(z) is an entire function of the form m
,
a and z,
all positive.
n-1
Let f(z)
= CEO anzn. Then 00
g (2) =
C n-0
a n 4 (n>zn
is an entire function all of whose zeros are real and negative.
112
VI. Zeros: Minimum Modulus I and Sequences of Functions
+
PruuJ Since (1 x ) r z 5 1, x 2 0, g ( z ) is an entire function. Thus 1 #(n) I 5 I a I eknand lim
5 lim
n+m
n+w
. I a Il’nek = o
since f ( z ) is an entire function. Hence the g ( z ) series converges everywhere. Let
a polynomial of degree 2p. All zeros off,,(z) are real and negative. Thus by the previous theorem so are the zeros of
We show that g p ( z )-+ g ( z ) uniformly in any finite region. Since lim f p ( z )= f ( z ) P+W
then afl,Q-+ an as p
-
00
as
for fixed n. Also
(I
+ :)2
1pt,
I an,p1 5 1 a, I for all n, p since
(l+$)P=l+IIZ+
-
We can choose N so large that the second term is less than E [since g ( z ) is entire]. Then for this fixed N the first term tends to zero and g p ( z ) g ( z ) . As in the previous theorem, the result follows from Hurwitz’s theorem. 0
113
6.1 Functions with Real Zeros Only
EXAMPLE1. Forf(z) = 8,if 4(0) satisfies the conditions of the theorem, then F ( z ) = x2=0[4(n)/n!]zn is an entire function with all zeros real and negative. EXAMPLE 2. For 1
+(w) =
T(w + Y
+ 1) '
v>
- 1,
this is an entire function of genus 1 with zeros at w - Y - 2, . . . all real and negative. Thus the zeros of
=
- Y - 1,
are all real and negative and the zeros of J , ( z ) are all real. [Take z = - a. Then 2 i G = - 2a1Ia which is real.]
6.1.3 Functions with Real Negative Zeros. If all the zeros of a function are real and negative, the modulus of the function is related to the distribution of the zeros in a very simple way. Let f ( z ) be such a function with order e < 1. Then
If z is real and greater than 0, then logf(z)
=
2 log( 1 + z) 2 n{ log( I + 5) log( 1 + -
fl-1
=
Zn
-
n-i
(The first integral step depends upon the zflbeing real, greater than 0, and ordered increasingly.)
-
6.1.4 Theorem. If as t + 00, n ( t ) AP(A > 0 ) (excludes e = 0 except for f a polynomial), then log f ( x ) n l x e csc ne, x + 00, where all the zeros offare real and negative,f(O) = 1,fis entire, and of order 0 < e < 1. 'V
114
VI. Zeros: Minimum Modulus I and Sequences of Functions
ProoJ: We have
(A - E ) t e < n ( t ) < (A
+ &)re for t >>to(&).Thus
since
which is independent of x. Also, the integral converges at the lower limit since n ( t ) = 0 for t < zl.Transforming t = xu in the second integral, we obtain
+ J,l+udu = xQ(A+ &)n csc ne, . m
XQ(A
Ue-1
0 <e
A similar result holds for A - E and the theorem follows.
Generally, log f(reie)
-
< 1.
0
-
eieenA csc ne re
for fixed 8 E (- n,n),log f ( z )denoting the branch which is real on the positive real axis. In fact the expression for logf(z) as an integral for real z, holds by analytic continuation for - n < arg z < n (excludes real negative axis), and as before log f(rei0)
- J reie
Turning the line of integration to t
as before.
I t @ dr. t(r& t)
= ueie,
+
115
6.2 The Minimum Modulus m(r)
-
-
6.1.5 Conversely it can be shown that if as x 00 though real values logf(x) nilxe csc ize, (0 < e < 1, il > 0) then n(r) Are. This theorem is more difficult than the previous one. A clear proof is given by Boas,+ and a lengthy proof is given by Titchmarsh.t Several extensions and further results concerning functions with real and/or negative zeros may be found in Boas’s book.+ 6.2
.--f
THE MINIMUM MODULUS
Let m(r) denote the minimum modulus of If(z) I on 1 z I = r. We cannot expect m(r) to behave as simply as M(r) since it vanishes whenever r is the modulus of a zero off(z). Except in the immediate neighborhood of these exceptional points, a lower limit can be set for m(r). Generally m(r) -+O in somewhat the same way as l/M(r). Note m(r) can be quite large for certain functions. Consider a canonical product P(z) of order e, with zeros ZI, z2, . . . , z,, . . . . 6.2.1 Theorem. About each zero z, ( I z, I > l), we describe a circle of radius l/rna, h > e. Then in the region excluded from these circles, 1 P ( z ) 1 > exp(- re+e), r > ro(E), E > 0.
Proof. By Theorem 4.10
L
c
rnskr
log11 -
since
= O(reie)
R. Boas, “Entire Functions.” Academic Press, New York, 1954.
t E. C. Titchmarsh, “On integral functions with real negative zeros,” 26 (1927) 185-200. Proc. London Math. SOC.(2).
116
VI. Zeros: Minimum Modulus I and Sequences of Functions
and
=
O(rQ++').
+
+
We have assumed that e E - p > 0 and that p 1 - e - E L 0. This 1 but we then only need C,,)r;Q= & , r p l < 00 and fails if e = p in this case we replace E by 0 throughout. Thus
+
Further, since C(,,)l/r?< 00, the sum of the radii of the circles is finite (radius R, = 1/rnh).Thus 3 circles centered at the origin and of arbitrarily large radius which lie entirely in the excluded region.
FIGURE 1
NOTE.The disk centered at z, and with radius r i h in Fig. 1 is confined to the annulus r, - r i h < 1 z I < r, rgh. This meets the real axis (or any ray from the origin) in an interval of length 2rih. These intervals are to be excluded. The intervals may overlap but the total length of the excluded intervals is less than or equal to Cc=.,,2r;h = S a n d S < 00. If we now consider
+
R
R+StI
FIGURE 2
+ +
any range of r of length greater than S, say, R 5 r 5 R S 1 (Fig. 2), then the excluded intervals of r in this range have total length less than or equal to S so they do not cover the whole range. Thus 3 r in this range not excluded and r L R is arbitrarily large. If z lies outside every circle with cen-
6.2 The Minimum Modulus m(r)
117
ter z, and radius rLh7since r, 5 kr, then
Hence
(Note re > A log r.) Also,
Since for r, 5 1,
I z/z, 1 > 2,
and
1% I> Z
log - - 1
0.
Thus log I P ( z ) I > - ref2&- O(re+&) and
I P(z) 1 > exp(- re+&). 0 6.2.2 The result now follows that iffis a function of order
m ( r ) # o(exp(- re+&))
for any
E
e, then
> 0.
It was conjectured that for functions of finite order, m ( r ) f o{M(r)}-I-&, > 0, however, this was disproved by Hayman.+ For f(z) = P(z)eQfZ), where Q(z) is a polynomial of degree q 5 e, E
I~
X QP( z ) I
2 exp(- Arq) 2 exp(- Are)
for large r.
t W. K. Hayman, “The minimum modulus of large integral functions,” Proc. London Math. SOC. (3) 2 (1952) 469-512.
1 I8
VI. Zeros: Minimum Modulus I and Sequences of Functions
Also, by the previous result [with
P(41,
1 P ( z ) 1 > exp(-
el
the order of the canonical product
re+’) 2 exp(- re+€‘).
and m ( r ) 2 exp(- retE), E suitably adjusted. 6.3 THEOREMS ON SEQUENCES OF FUNCTIONS
We commence this section with the well-known Heine-Bore1 theorem. The theorems that follow all deal with the concept of compactness. We present the results without further amplification since they are somewhat sequential and quite self-explanatory.
6.3.1 Definition. A set or region is compact if it is both bounded and closed, e.g., ( I ) the set of all complex numbers z, such that 1 z 1 5 k (constant) is a compact set. However, (2) the set of all real numbers in I = (0, 1) is not compact since it is bounded but not closed. 6.3.1.1 Theorem. (Heine-Borel). Let S be a compact set. Suppose there is a family {G,} of open sets such that each point of S is contained in at least one of the G , . Then there exists a finite subfamily {G,,} ( j = I , 2, . . . , n ) of {G,} such that every point of S is in at least one of the G a j . is no finite subfamily { G a j } j, = I , 2, . . . , n of the G, such that every point of S is in at least one of the G,,.) Since S is a bounded set, it is contained in some closed square Q whose sides have length k . Subdivide Q into four closed congruent squares, the length of whose sides is k/2. Then there must be at least one of these, say Q , , such that no finite subfamily of {C,} covers S n Q, (that is the part of S contained in PI). Subdividing Q1into four closed congruent squares with sides of length k/2,, for at least one of these squares, denoted by Q,, there is no finite subfamily of {G,} which covers S n Q2. Continuing this process we obtain an infinite nested sequence of closed squares Q 13 Q1 2 . such that diameter (Q,) + 0 and no finite subfamily of {G,} covers S n Q,. There is a point zo common to a11 the squares Q , . This point is in S and hence contained in one of the sets of {G,}, say G a p . Since Gap is an open set, there exists an E > 0 such that all z satisfying
€‘roo$ Suppose that no finite subfamily of {C,} covers S. (That is, there
119
6.3 Sequences of Functions
1 z - z, I < E are contained in G Q p .Furthermore, since zo is contained in all the Q, and diameter (Q,) 0, it follows that for n sufficiently large, I z - z, I < E for every z in Q, and thus Q, is contained in G a p .Hence GQPcovers S n Q,. This contradicts the condition that no finite subfamily of {G,} covers the part of S contained in any Q,. Thus the theorem is established. 0 -+
6.3.2 Lemma. If (i) { f , ( z ) }is a sequence of functions each analytic in a
domain D, (ii) f,(z) -+ f(z) uniformly as n + 00, in every compact region R c D, (iii) f ( z ) is not constant in D and (iv) f(z) = a at some point z, E D,then all but a finite number of the functions .f,(z) take the value a in D.
NOTE.The result is not necessarily true if (iii) is omitted, e.g., f,(z) = z/n (n = 1, 2, . . .), f ( z ) = 0. Then f,(z) -+f(z) uniformly in every compact region as n for all z E D.
-+
03.
However, if D does not contain z
= 0,
then f,(z) # 0
Proof. Clearly f(z) is analytic in D. Take e > 0, such that z E D when I z - z, I 5 e (Fig. 3) and f(z) # a for 0 < I z - z, I I e [possible by (iii)]. Let R be a compact region I z - z, I 5 e and C the circle
I z - zo 1 = e. Write m = limzGcIf(z) - a I, then m > 0 since C is a compact set. Since f,(z) -+f(z) uniformly in R as n -+ 00, 3 v such that I f , ( z ) -f(z) I < m for n > v and all z E R, hence If,(z) - f ( z ) I < If(z) - a I
for n > v and all z E C.
Also iff(z) is analytic in D, by (i) and (ii), so aref,(z) -f(z) Hence by Roucht’s theorem, if n > v,
andf(z) - a.
and f(z) - a have the same number of zeros within C. Thus if n
> v,
120
VI. Zeros: Minimum Modulus I and Sequences of Functions
f,(z) = a at some point z within C and hence at some point z E D (all but afinite number comes from n > Y). 0
6.3.3 Definition. If a function w = f ( z ) is single-valued in a region M and its inverse z = #(o)is single-valued in N (the set of points o corresponding to all possible points z of M), the transformation o = f ( z ) is one-one or biuniform in M and f(z) is univalent or schlicht in M, i.e., under a one-one transformation o = f(z), any two distinct points of M map onto two distinct points of N, i.e., two points cannot “merge.” 6.3.3.1 Theorem. If (i) { f , ( z ) } is a sequence of functions analytic in a domain D, (ii) f , ( z ) -*f(z), n + 00, uniformly in every compact region R c D,(iii)f(z) is not constant in D,(iv) eachf,(z) is univalent in D,then f(z) is univalent in D. Proof. Suppose f ( z ) is not univalent in D. Then 3 points z, , z2 in
D such that f(zl) = f(z,) = k, for example. Let V, , V2 be neighborhoods of z , , z 2 , respectively (Fig. 4), such that V, c D, V, c D, and V , n V2 = 4. By the lemma, with D replaced by V,, 3v1 such that f,(z) takes the value k at a point of V , for all Y, < n, similarly 3v2 such that for n > v 2 , f,(z) takes the value k at a point of V 2 . Hence if n > max(v,, y2), f,(z) = k at a point both in V, and V2, contradicting (iv). Hencef(z) is univalent. 0
FIGURE 4
6.3.4 Definition. A set E is said to be dense in a set A, if A G E, i.e., if every neighborhood of any point z E A contains at least one point of E.
EXAMPLE. The set E of rational points of (0, 1) is dense in A A
=
=
[0, I ] since
E.
6.3.5 Definition. A sequence of functions { f , ( z ) } is said to be uniformly bounded in a set S, if 3 M greater than zero such that If,(z) I < M for n = l , 2 , ... a n d a l l z E S . 6.3.5.1 Theorem. If (i) { f , ( z ) } is a sequence of functions analytic in a
121
6.4 Vitali's Convergence Theorem
domain D, (ii) {fn(z)} is uniformly bounded in every compact region R c D, then {fn'(z)} is uniformly bounded in every such R. Proof. By the Heine-Bore1 theorem, it is sufficient to prove the result for a compact circular region S c D (Fig. 5 ) . Suppose a E D and
0 < 6 < @(a,bd(D)).
I
Consider the compact region S : { z 1 z - a 1 5 S}, then S c D. Take r such that 6 < r < @(a,bd(D)). Let C be the circle I z - a I = r and write
FIGURE 5
I
= { z I z - a I 5 r } , then S, is a compact subset of D. Since {fn(z)} is uniformly bounded in S , , 3 M such that 1 fn(z) 1 < M for n = 1,2, . . . , and V z , z E S,. Let E be any point of S. Then
S,
For z E C,
since the distance
)[(',fI
1 z - l 1 2Ir -6 I. I 5 Ma
Hence for all
r / ( r - 6)2
and {f,'(z)} is uniformly bounded in S.
ES
0
6.4 Vitali's Convergence Theorem. Let {fn(z)} be a sequence of functions
analytic in a region I). Let If,(z) I 5 M for all n, z E D. Letf,(z) tend to a limit as n -,00, at a set of points having a limit point (i.e., point of accumulation) in D.Then f,(z) tends uniformly to a limit in a region bounded by a contour interior to D, the limit being an analytic function of z.
VI. Zeros: Minimum Modulus I and Sequences of Functions
122
Proof. It is sufficient to take D a circle with limit point the center. Thus in
the general case uniform convergence is proved in a circle with center the limit point interior to D. Repeating the process with any point of this circle, by analytic continuation, the domain of uniform convergence is extended to any region bounded by a contour in D. We may take the limit point to be the origin. Let the radius of the circle D, be R and let fn(4
= a0,n
+ a1,nz +
* * *-
Thus
IAtW - f n ( O ) I 5 If,(Z) I Also, fn(z) - f , ( O ) Schwarz’s lemma
is zero at z
= 0.
+ If(0)
II2M.
Further, since fn(z) is analytic, by
Let z’ (#O) be a point where the sequence converges.
ThCn z’ may be chosen so that the first term is arbitrarily small (by hypothesis). Also, sincef,(z’) approaches a limit, choosing n large the second term is arbitrarily small for all positive m. Hencef,(O), i.e., u,,, tends to a limit, say a,. Consider
This tends to a limit at z’ since ao,n tends to a limit. For I z I = R, I g,(z) I 5 2 M / R [since a0,%=fn(0)]. Thus it is also true for I z 1 < R (maximum modulus theorem). Consequently g,(z) satisfies the same conditions asf,(z) and q n-+a,. Similarly av,n+ a,, Vv, The convergence is R - E , by Cauchy’s uniform (with respect to v and z), since for I z I I inequalities I av,,,I 5 MIRVand since every term tends to a limit, the sum tends to a limit uniformly (Weierstrass M-test), for I z I 5 R - E. 0 6.5 Montel’s Theorem. Letf(z) be a function analytic in the half-strip S: a < x < 6, y > 0. If f(z) is bounded in S and f(z) I as y + 00 for a --f
6.5
I23
Montel’s Theorem
certain fixed value 6 of x, a < 6 < b, thenf(z) + I uniformly on every line x = xo in S. Further, f(z) + 1 uniformly for a 6 5 x 5 b - 6, 6 > 0.
+
+
Proof. Consider a sequence of functions f , ( z ) = f ( z in), n = 0, 1, . . . , in the rectangle R: a < x < b, 0 < y < 2. Then f , ( z ) + I at every point of the line x = 5 (by hypothesis), since as y -00, f(z) behaves like in), n 00. By Vitali’s theorem, f,(z) I uniformly in a region inf(z terior to R, e.g., a 6 < x < b - 6, & 5 y I $. This is because (i) f,(z) is uniformly bounded in R by hypothesis, (ii) f , ( z ) is analytic for all n. { f(z in) is a change of origin} and (iii) x = 6 is a set of points each of which is a limit point of points of convergence in R . Thus f , (2) + 1 uniformly on every line x = xo in R. 0
+
--f
-
+
+
+
NOTE.Using f(z in), n = 0, 1, . . . , means that whereas the function is analytic in the rectangle 0 < y < 2, a 5 x 5 b, the whole strip y + 00 becomes the region of analyticity. Then f(z in2) would not do, since 3 gaps in the strip in which no information is forthcoming.
+
6.5.1 By a conformal transformation z = i log w, the strip in the z-plane becomes an angle in the w-plane and the theorem states: “If # ( w ) is bounded in an angle a < arg w < ,l? and if # ( w ) + I as w 00 along any line arg w = constant between a,/?,then # ( w ) -+ I uniformly in any angle --+
a
+ 6 5 a r g w s p - 6,
6
> 0.
CHAPTER VIl
THEOREMS OF PHRAGMEN AND LINDEL6F: MINIMUM MODULUS 11
7.1 THEOREMS OF PHRAGM@N AND LINDELOF
These theorems are important extensions of the maximum modulus theorem and are useful in a more delicate discussion of entire functions bounded in specific directions. A monograph of considerable extent has been written on the subject of entire functions with particular reference to results of PhragmCn and Lindeliif.+The following theorems are a cross section of the simplest theorems contained therein. 7.2 Theorem. Let C be a simple closed contour and letf(z) be analytic in and on Cexcept at one point P E C (Fig. 1). Let If(z) 1 5 M on C except at P. Suppose 3 a function w ( z ) analytic, nonzero, and whose absolute value I w I 5 1 in the region D bounded by C. Suppose further, that w ( z ) is such that if E > 0, 3 a system of curves arbitrarily near to P and connecting the two sides of C around P, on which
I
{ w ( z ) } ”.f(z) I
(i.e., on y, see Fig. l),
5M
then Jf(.z) 1 5 M at all points in D. Viz. M. L. Cartwright, “Integral Functions.” Cambridge Univ. Press, London and New York, 1962. 124
7.1-7.7
Theorems of Phragmh and Lindelaf
125
Proof. Consider the function F(z) = (~(z))~.f(z)which is analytic in D. By hypothesis, if zo is in D, 3 a curve surrounding zo on which
Thus
and taking
E
3
0, lf(zo) 1 I M.Hence the result.
0
The exceptional point P may be replaced by any finite number or indeed an infinite number of points, provided that the functions w(z) corresponding to them with suitable properties may be found. Instead of starting with the previous theorem, it is usually simpler to start with a special auxiliary function adapted to the region considered. In practice, the exceptional point P is always at infinity. The previous theorem gives results about the behavior of a function in the neighborhood of an essential singularity. By a preliminary transformation the exceptional point can be placed at infinity.
7.3 Fundamental Theorem. Let f ( z ) be an analytic function of z = reie, regular in a region D between rays making an angle n / a at the origin and on the straight lines themselves. Suppose that 1 f(z) 1 I M on the lines and as r - m ,
f(z)
= O(exp rfl)
where
B
(uniformly).
Then If(z) 1 5 M throughout D. Proof. Suppose without loss of generality that the rays are 8 Let F(z) = exp(- EZY)~(Z), B < y < a
and fixed E > 0. Then
=
f n/2a.
126
VII. Theorems of Phragmen and Lindelof: Minimum Modulus I1
On the lines 8 = f n/2a, cos 70 > 0 since y < a, thus on these lines I I5 If@) I 5 M (Fig. 2). Also, on the arc I 8 1 I n / 2 a , I z I = R, and we have
m
and the right-hand side approaches 0 as R 00. Hence for R sufficiently large, I F ( z ) I I M on this arc also. By the maximum modulus theorem, I F(z) I L M throughout the interior of the region 1 0 1 5 n/2a, r 2 R and since R is arbitrarily large, throughout D. Further, by (l), I f ( z ) 1 < M exp(sry) in D and making E -+ 0 the result stated follows. --f
7.3.1 The straight lines may be replaced by curves approaching 00. Notice the relationship between the angle in the theorem and the order of f(z) at infinity. The wider the angle 8, the smaller the order off@) must be for the theorem to be true (since @ < a). In the next theorem, the order is not small enough for the previous proof to apply. 7.4 Theorem. The conclusion of the previous theorem still holds if f ( z ) = O(exp 6ra) for every positive 6 (uniformly in the angle).
Proof. As before, take - n/2a 5 0 5 n/2a. Let F ( z ) = exp(- &za)f(z), E fixed (temporarily). Taking 6 < E (e.g., e/2), then on the real axis
I.&>
I
=
O(exp dxa)
and
1 F ( x ) I = exp(- exa)O(exp 6 x a )
-
0
as x -+
00.
127
7.1-7.7 Theorems of Phragmkn and Lindeliif
On the real axis the full benefit of the factor exp(- E Z ~ ) is felt. Hence 1 F(x) I 5 M' for all real x. Considering F(z) in the angle n/2a, write M" = max(M, M'). Then I F(z) I 5 M" on the sides of the angle, and I F(z) I = O(exp dr") uniformly. Since I exp(- Era cos 8a) I < exp Era we have I F ( z ) 1 = O(exp ro) with any B such that a < @ < 2a. Thus the previous theorem applies in each of the half-angles and gives 1 F(z) I 5 M" in the full angle. Suppose if possible that
M' = sup z>o
1 F(x) 1 > M
so that M" = M' > M. The supremum is not approximated as x + 00 since F(x) + 0, x --* 00, and 1 F(0) I 5 M since 0 lies on the sides of the angle. Thus the supremum (1) is only approximated inside the real positive axis and is therefore attained at some point xo. Hence F(z) is analytic in the domain, I F(z) I 5 M' = M" throughout the domain, yet I F(xo) 1 = M' for an xo inside. By the maximum modulus theorem, F ( z ) is constant = M' in the domain, contradicting the fact that I F(z) 1 5 M < M' on the sides. Hence the supposition is not possible. Therefore M' 5 M,M" = M, and 1 F(z) I 5 M throughout the angle. Thus If@) 1 5 M I exp(ez') I ; now making E 0, If(z) I 5 M . 0
-
7.5 Theorem. Iff(z) + a as z-00 along two straight lines and f(z) is analytic and bounded in the angle between them, then f ( z ) + a uniformly in the whole angle.
ProoJ We may assume that the limit a is zero. We may also assume that the angle between the two lines is less than n, since the general case may be reduced by a substitution of the form z =: wk. Let the lines be 8 = f O', 8' < 4 2 . Let
and
Now If(z) I 5 M, say, everywhere, and
If(z) I < E Let J
= rlM/E.
for r > rl(E), 8
=
f 8'.
Then
I F(z) I < r M / J I E
(provided r I r l )
128
VII. Theorems of PhragmCn and LindelBf: Minimum Modulus I1
and
Thus by the main PhragmCn-Lindelof theorem region. Hence
1 F(z) I 5 E
in the whole
-
7.6 Theorem. If f ( z ) - + a as z-00 along a straight line, f ( z ) b as z 00 along another straight line, and f(z) is analytic and bounded in the angle between, then (1) a = b and (2) f(z) -+ Q uniformly in the angle. ---+
-
ProoJ Let f ( z ) a along 0 = u and f(z) + b along 8 = ,6 where u < B. The function {f(z) - $ ( a b)}a is analytic and bounded in the angle and tends to )(a - b)2 on each straight line, hence it tends to this limit uniformly in the angle, i.e.,
+
{f(z) - $(a + b)I2 - i(Q- b)2 = { f ( z >- Q>{f(Z) - b } tends uniformly to zero. Thus to any
I
E
corresponds an arc on which
{f@) - Q>{f(Z)- b> I5 E *
At every point of this arc, either I f ( z ) - a I I 1/or If(z) - b I I 1/y (or both). We may suppose that the former inequality holds at 8 = u and the latter holds at 8 = /?.Let 8, be the supremum of values of 8 for which the former holds. Then 8, is a limit of points where the former holds and is either a point where the latter holds or a limit of such points. Sincef(z) is continuous, both inequalities hold at 0,,. Taking z to be this point,
Making E -+ 0, it follows that a uniformly. 0
= b,
7.7 THE PHRAG-N-LINDELOF
thus by the previous theorem f(z) -+ a
THEOREM FOR OTHER REGIONS
The angle of the previous theorem may be transformed into other regions, e.g., into a strip. Consider the fundamental theorem applied to a region 2 1, I e I I n/2ff. Put s = i log z,
f ( z ) = +(s),
s =0
+ it.
129
7.8 The Indicator Function h(0)
Lines arg z = f n/2a approach parallel lines a = f n/2a and t = log I z I since - ia t = log I z I i8. Hence if I #(s) 1 5 M on the upper half of the two parallel lines and on the segment joining them (real axis), while #(a i t ) = O(exp eet), e < a (since 1 z 1 = el), in the strip between them, then I #(s) I I :M throughout the strip.
+
+
+
7.8
THE PHRAGIb@N-LINDEL~F FUNCTION
Previous theorems have considered the way in which a function behaves as z + 00 in different directions. A more systematic study of this question is now made. Consider f ( z ) = exp((u
+ ib)ze)
and we have I f ( z ) I = exp(r@(acos e8 - b
sin P O ) ) .
The behavior of log If(z) I depends firstly on re which is independent of 8. Behavior in different directions is determined by a factor
h ( 8 ) = u cos e8
-b
sin
=
log I f ( z ) [/re.
Although this is a special case (Titchmarsh points out), the general case is not so different from it as may be expected. Suppose that f ( z ) is analytic for a < 8 < B and 1 z I 2 r,, also that f ( z ) is of order e in the angle, i.e.,
lim r+m
uniformly in 8 for all positive general as
log If(reie)I re+e E
=o
but not for any negative
E.
Define h ( 8 ) in
where V ( r ) depends on a function considered. One chooses V ( r ) such that h ( 8 ) is finite and not identically zero: we choose the simplest case V ( r ) = rQ [we could choose re(1og r)P(log log r)Q.. - I.+ 7.8.1 Theorem. Let
t See M. L. Cartwright, “Integral Functions,” p. 41. Cambridge Univ. Press, London and New York, 1962.
130
VII. Theorems of Phragmh and Lindeldf: Minimum Modulus I1
Let
h(0,) I h,
h(0,) I h,.
and
Let H ( 0 ) be the function of the form a cos e0 ues h, , hz at e,, 8,. Then h(e) s , H m
+ b sin ee which takes val-
el I eI 0,.
Proof. Observe that H(0) =
+
h, sin ~ ( 0, 0 ) h, sin e(0 - 0,) sin ~ ( 0, 0,)
However, this expression is not required here. Let
Ha@) = a d cos e0
+ bd sin ee
thus for r sufficiently large,
thus h(0) is the best (least) number such that If(reie) I I exp(rQ(h(0)
+ S})
for every 8 > 0 and V r > ro(8, 0).
Thus If(reiel)
I
=
O(exp(rQ(h,
+ 6)))
since h(0,) 5 h, . A similar result holds for F(re"1) and by the fundamental theorem F ( z ) is bounded in the angle (0, , 0,) uniformly.
NOTE.In order to conclude F(z) = 0(1) in the angle, we need I F(z) I = O(exp r@)in the angle for some < n/(0, - 0,). Now n/(0, - 0,) > e
131
7.8 The Indicator Function h(8)
+
by hypothesis, putting nl(f3,- 0,) = e p we obtain p 1 F ( z ) I = O(exp rp) for some /3 < e p. By hypothesis
+
> 0 and we need
0 (exp rQ+pI3),
=
O(exp kre) = O(exp re+pI3)
+
- log I f(reie)I I lim H8(8)re (bounded) h ( 8 ) = r+m lim re r+m re
= H8(8)
7.8.2 Characteristic Behavior of h ( 8 ) and H(f3). We have a function analytic in the angle 5 arg z 5 8 , . For each 8 = arg z in this range define h ( 8 ) fromf, i.e.,
- log I f(reie) I
h ( 8 ) = lim
I +m
re
so that h ( 8 ) is the best number (least) such that If(reie) I 5 exp(re{h(8)+&}) for every E > 0 and for all r > T o ( & ) = r0(&,8 ) . Then H ( 8 ) is a smooth function of the form H ( 8 ) = a cos e8
+ b sin e8 = c cos(e8 + a)
and is in fact the particular one which agrees with h ( 8 ) at el, and 8 = O2 (Fig. 3). The previous theorem says h ( 8 ) 6 H ( 8 ) . The curve of h(8), is “convex” in a plane geometry in which the “straight lines” are the graphs of such H ( 8 ) functions.
132
VII. Theorems of PhragmCn and Lindelof: Minimum Modulus I1
7.8.3 If
-+
+
-+
If1
= o(exp r ~ f e )
If I
= O(exp r B + s )
lim - yp+r) log'f' - o
for all q > 0,
r+m
q = 2 ~ for , example. The first and last statements are the same, thus all are equivalent. The order e is the least number for which this is true (3 a least number). For each of these statements, the numbers e form a righthand class. A second condition is needed (least or false for E < 0), to make e the critical value. Thus the definition for e is equivalent to for all 1 > e but no 1 < e.
f = O(expr2)
--
7.8.4 From the previous theorem we may have that one or both of he h(8,), h(8,) approach --. The conclusion is then that h ( 8 ) = for 8, < 8 < 8,. The same proof applies with one or both the h l , h i s being arbitrarily large and negative. Thus h ( 8 ) = --bo for some 8 only if h ( 8 ) = - 00 for all 8 in the given angle, in which case f ( z ) = 0. 7.8.5 Theorem. If h(0i) and h(8,) are finite for O1 5 8 5 8, and Os - 8, < n/e and if' H ( 8 ) is the sinusoid such that h(8,) 5 H ( 8 , ) with h(%) 2 H@d,then W,)2 H(8,).
> - 00 by the last statement in 7.8.4. If 3 a positive 6 such that h(8,) I H(8,) - 6, consider
Proof. First, h(8,)
HB(e)= H ( e ) - 6 sin(0 - el) csc(e, - el). Then
fw,) H(fA), =
m 8 2 ) < H(8,),
and by the previous theorem,
which is a contradiction.
0
fw,)
=W . 3 )-
6
133
7.9 Behavior of m(r)
7.8.6 Theorem. If for 8, h(8,) sin ~ ( 8 , 8,)
< 8, < 8,, 8,
-
8, < n/e, 8,
-
8, < n/e, then
+ h(8,) sin e(8, - 8,) + h(8,) sin ~ ( 8 ,
-
8,) 2 0.
Proof. For any H ( 8 ) ,
Choosing H ( 8 ) such that H(8,) = h(8,) and H(8,) = h(0,) then by the previous theorem h(8,) 2 H(8,). Substitution in (1) gives the result. 0
7.8.7 Theorem. The function h(8) is continuous in any interval where it is finite. Proof. Let h ( 8 ) be finite in 8, 5 8 5 8, and let 8, < 8, < 8,. Let H1,,(8) be the sinusoid which takes the values h(8,), h(8,) at 8,, 8,; define H2,,(8) similarly. Then by theorems 7.8.1; 7.8.5, h ( 8 ) 5 H1,,(8), 8, 5 8 5 8,, and h ( 8 ) L H2,3(8), since h(8) = H2,3(8)at 8 = 8,, 8,. Thus outside this range, in particular, 8, 5 8 5 8,, h ( 8 ) exceeds or equals H2,,(8) by a previous theorem. Thus H2,3(8)
I h(e) I ~,,,(e),
0, I e 5 0,
and similarly, H,,,(o) I h(e) I H2,3(8),
0, I 8 i 0,.
Hence in whichever of these intervals 8 lies
The extreme terms tend to limits as 8- 8, [the same since H1,,(8,) = H2,,(8,)] and I h ( 8 ) - h(8,) I < K 1 8 - O2 I for k independent of 8. Hence h ( 8 ) is continuous at 8, and continuous everywhere in the interval. Also it has a left-hand and right-hand derivative at 8,. 0 7.9 MINIMUM MODULUS
We now study m ( r ) in more detail. We show in particular that for functions of small order not only can m ( r ) be large, but for a large proportion of r.
134
VII. Theorems of Phragmkn and Lindelof: Minimum Modulus I1
7.9.1 Theorem. If e < 4, there is a sequence of values of r tending to infinity through which m(r) -+ 00. Proof. There is no line arg z = constant on whichf(z) is bounded, since the
whole plane bounded by this line forms an angle 2n < n/e if e < 4. Hence if f(z) is bounded on this line, it is bounded everywhere and so reduces to a constant. Let
and
(i.e.,
# has real negative zeros). Then since 1 1 - (z/zn) I 2 1 1 -( 2 q ( r ) = I #(- r ) I
and
r/rn)I,
Mf(r> 5 #
Thus min,r,-7I f(z) I 2 I #(- r) I. Also #(- r) is unbounded since #(z) is an entire function of the same order as f(z). Then #(- r ) is unbounded in particular along the negative real axis as r -+ 03. Hence it is possible to find values of r, tending to 03 along which m ( r ) is unbounded. 0 We prove a lemma now which enables us to establish a far more precise result. 7.9.2 Lemma. Given that
and f is of order
e, 0 < e < I, then for e < a < 1,
and
Proof. dx
7.9 Behavior of
m(r)
135
provided (2) The sum in (2) is, on writing Q = 9 ( s ) ,
which is less than condition,
00
if
(T
> el, in particular for 0 < e < < 1. On that (T
dt
=
n s sin ~ t s
-
by residues, or
f
0
log(1
+ t ) dr =
r1-(f
tl+S
---1
s*
1- s
+
log(1 fl+8 t ) dt =
-ta
2 1 2(2-s)
log t
J:t1-8{
+ . . .) dt + ...
+ log( 1 + +)} dt
Combining the two results we obtain
- -18'
2 (-
n-1
1)n.
2 ~
na - sa
It --
s sin ICS *
136
VII. Theorems of Phragmkn and Lindeldf: Minimum Modulus I1
Similarly
-1
dx
X
rn
1-tldt
provided that
and that
S,I-.-l log 1 1 -
t
I df
=
II;
(4)
s tan ns
Condition (3) reduces to
The integral is finite if 0 < u < 1, while the series converges if u > el i.e., e ( s ) > el as before. To prove (4),
s: s,
t+-' log 1 1 - r I dt
t-8-1
=
-1 1 n-1 n(n -s)'
I 1 - tl
~
c n(nl +
1 " log I 1 - t I dt = - sa
n-1
11 - t l
=t
s)
=
1 - t, 0 5
'
- 1 = t(l - l/t)¶ t > 1.
15
1,
7.9 Behavior of m(r)
137
and
-
s{sz1 I
1
O0
2s
+
n s tan ns
. o
The more precise result now is the following theorem.
7.9.3 Theorem (Pdya). If 0 < e < 1, there are arbitrarily large values of r for which m ( r ) > {M(r))cOsnQ-e. Proof. Defining f ( z ) and # ( z ) as in theorem 7.9.1, we take c = 1, k = 0. It is sufficient to prove the theorem for # ( z ) . If z’ is a point where l f ( z ’ ) I = m(r), then
Thus if the theorem is true for #(z), since arbitrarily large r (by hypothesis), then m ( r ) M ( r ) 2 I # ( r ) I1+CoanQ--e
I #(-
r)
I > {#(r))cOsnQ-=for
- { M(r)}l+cosne--e. >
Thus forf(z), m ( r ) > {M(r))C09”e-€,r >> 1. We now establish the theorem for + ( z ) . If the theorem is false for #(z), 3 positive constants E , a such that a 2 1, and log I #(- x ) I < (cos ne - E ) log # ( x ) ,
For 0 < e
vx
> a.
< 9 ( s ) < 1 by the previous lemma,
Thus
=
- I’{cos ns log # ( x ) - log I #(- x> I Ix-’-l dx 0
138
VII. Theorems of Phragmtn and Lindellif: Minimum Modulus I1
at least in
e < 9 ( s ) < 1 where the last two integrals exist. Now -
s:
{COS
n~log 4 ( x ) - log
1 4 (-
X)
1 )x-"l
dx
is regular at every point in [0, a] except zero. At zero, +(XI'=
#(O)
+ x(b'(0) +
* * -*
Thus {cos 7cs log $ ( x ) - log I #(- x )
-
I}x-8-1-
A(c0s ns
+
l)x-8
where log #(x) Ax and A = #'(O) = l / r n . Hence for x in the neighborhood of zero, converges only if 9 ( s ) < 1. [The condition e < a ( s ) is required for convergence of the integral at m.] Consequently ' .}x-'-* dx converges and defines a function of s regular in 0 < *(s) < 1. We know that F(s), which is given in a narrower strip e < 9 ( s ) < 1 by J,", is in fact regular (i.e., analytically continuable) in the wider strip 0 < 9 ( s ) < 1. Taking
so
c{.
41w = (cos ne - E ) log 4&) = log +(XI, y(s) = cos 7cs - cos 7ce
- 1%
+
I 9(-
x)
I,
E,
then
and #l(x) > 0, Vx real).
> a (by hypothesis), also, #z(x) > 0. Figure 4 graphs
~ ( s )(s
NOTE.We use s complex since we must know that F(s) has a circular disk of regularity with center so > e and enclosing s' = so - h. We use in effect the Taylor series about so, with displacement - h. In (I) of Fig. 5,
1.
m
F(s) =
In (I) and (II), F(s) is regular and is
=
-
r. 0
-
s," only.
7.9 Behavior of m(r)
139
n
FIGURE 5
I
0‘
‘P
I
The underlying idea now, is that since F(s) defines a regular function for s real and greater than e and since F(s) is in fact regular right down to s = 0, it ought to hold also at points s = s’ < e, particularly as in the range of s being considered, y(s) is positive so that there can be no cancellation of positive and negative contributions in (#1 y #z)x-8-1. If F(s) were regular at s’, i.e.,
+ -
we should obtain m
E
[ & ( x ) x - ~ ” - ~ ~<xy(s’) J[m#2(x)xJ’-1 Ja a
dx
since then cos zs’> cos ne (s’ real) and
whence &) rLs’ < 00 contrary to the definition of e. However, we cannot simply apply Jr{#l(x) y ( ~ ) # ~ ( x ) } x - dx ~ - ~at s’. We have a function F(s) defined in the whole strip 0 < g ( s ) < I but we only know that it satisfies J,”{C$~ y#z}x-s-l d.v = F(s) for s in a narrower strip e < 9 ( s ) < I . Choose so real and greater than e but so near to e that 3 a circle centered at so with e inside it and which lies entirely in the strip 0 < 9 ( s ) < 1 in which we know that F(s) is regular (Fig. 6 ) . Further, we suppose that
+
+
0 < s,
-
Q
< h < k < 1 < min(so, 1 - so).
On the circle centered at so and radius I, F(s) is regular and therefore bounded, say I F(s) I 5 A on I s - so I = 1. Hence
I F‘*’(s,)/n!I 5 A/In
(Cauchy’s inequality).
140
VII. Theorems of Phragmtn and Lindelof: Minimum Modulus I1
FIGURE 6
,.
Thus with sr = so - h (sr < p), D = d/ds, > 0, we have
F(s') = echDF(s0)
where
I Un I 5 Mn = hnA/ln.Then
c Mn
(n)
and
&)
(Taylor expansion about so)
A = 1
- (h/l)
= B,
Un converges uniformly. Note that
where 1 Un,,,I I I Un I 5 M,. Thus we have that the series &Un verges to F(s') and
c Un,m+ c Un
(n)
= F(s')
as m
con-
-00.
fn)
There is no difficulty in principle in evaluating D'n'F(~,J= P n )(so)
+
by differentiating Jr{Ol y#z]x-8-1 dx under the sign of integration. Thus we have
The difficulty arises when we try to justify taking the summation under the sign of integration. To meet this difficulty, P6lya (whose method we are
141
7.9 Behavior of m(r)
following essentially) considers the finite object (1 - (hD/m))minstead of e-hD. As m -00, the operator (1 - (hD/m))mtends to e-hDand by the above analysis (1 - (hD/m))”F(s0) F(s’) as m 3 0 0 . For each m, &Un,m is finite and therefore the summation is valid under the sign of integration. We now consider (1 - (hD/m))mF(so).
-
h5 m
[Transform x
= e‘
and consider (1
+ (hE/m) - (hD/m))m- 1 .]
where
It will be sufficient for our purpose to establish that at s = so,
Now
I y(”(s) I 5
. nfi
and 5 must be positive for we require
Thus
5 2h(en’2- 1) since h is certainly less than provided
4. Consequently I E 1 < 9h and E 2 - &,u(s) 9h I Bytoso).
(1)
Now, by hypothesis, y(e) = E is positive and (1) shows that we want so
142
VII. Theorems of Phragmtn and Lindelof: Minimum Modulus LI
so near to e that y is still positive there. Moreover we require so - e < h (i.e., s‘ < e). Thus we require 18(s0 - e) < y(so). This is true at so = e and by continuity of y, for all so sufficiently near (1) can be satisfied. The case that concerns us is so > e whence
Y ( $ 0 ) > Y (el - 4 s o - el since y(e) - y(so) = E - cos nso
+ cos ne -
E
n n e) sin (so - e) 2 2 n 5 2 sin - (so - e) < n(so- e ) 2 = 2 sin - (so
+ .
because so - e is very small. Thus it is sufficient to choose s,, so that
(there are other conditions such as so - e arrange that E 2 - #y(s) and hence
< 1 - so). Now we can indeed
As m --+ 00, everything on the right-hand side is positive and increases while the left-hand side is bounded by B, in fact it tends to F(so - h) = F(s’). Hence B 2 ~wx-ao-leAC#
dz(x)y (so) dx
a
and since 6 = log x > 0 ~ ~ - 8 ’ - 1 # z (dx x ) < 2B/y(so)= C.
In particular, the integral on the left-hand side converges. Hence, apparently converges for a value of the exponent less than e. This contradiction shows that the original supposition was untrue and the theorem is proved. 0
7.9
143
Behavior of m( r)
For functions of order 1 and exponential type, i.e., functions f ( z ) 0( & I r ’ ) we have the following theorem :
=
7.9.4 Theorem. If f ( z ) = O(ek?),there are arbitrarily large values of r for
which m ( r ) > e-(k+c)r.
Proof. Let
and
Sincef(z) = O(eklz”>, n(r) = 0 ( r ) , thus 3 constant k such that llr, < k/n since n(rn) = n = O(rn).Thus
(cf. product for sin 0). LA
- log I #(reie) I
h(0) = lim
r-b
2/1
Then h(0) Ink, VO. Since I # ( z ) I 2 I c la if 9 ( z ) 4 0, we see that h(B) is finite for - n/2 5 0 5 n/2. Thus by a previous theorem if h ( 8 ) is finite somewhere, it is finite everywhere, Also h ( - 0 ) = h(0) and with 0 < 0 < n
e,
e,
= - e,
= 0,
e,
=
e,
=
4,
we have
e
h(B) sin - - h(0) sin 0 2
e + h(8) sin >0 2 -
from Theorem 7.8.6 hence
Since
I #(z) I
I c la for z real and positive
144
VII. Theorems of Phragmen and Lindelbf: Minimum Modulus I1
Therefore h(0) 2 0 for - ?c 5 0 5 z [from (I)] and since h(0) is continuous, h(n) 2 0. Thus
from the definition of h(0). For such r,
whence m ( r ) > e-(k+e)r. 0 A considerable amount of work has been done on the minimum modulus of functions of both finite and infinite order. For example, Hayman+ has shown that for functions of infinite order, m ( r ) > M ( r ) - Alogloglogm(r) for some arbitrarily large r provided A > e 2 / z More recently, Kjellberg’ has proved a much more delicate result than Theorem 7.9.3. He shows that for each non constant entire function f ( z ) and for each 1 satisfying 0 < 1 < 1, the following holds: Either
(1) log m(r) > cos 7d log M ( r )
for certain arbitrarily large values of r, or if (1) is not fulfilled, the limit
exists and is positive or is infinite. This is, of course, a considerable improvement on the similar theorem we have already proved.
+ W. K. Hayman, “The minimum modulus of large integral functions,” Proc. London Math. SOC.(3) 2 (1952) 469-512. t B. Kjellberg, “A theorem on the minimum modulus of entire functions,” Math.
Scand. 12 (1963) 5-11.
CHAPTER VIII
THEOREMS OF BOREL, SCHOTTKY, PICARD, AND LANDAU: ASYMPTOTIC VALUES
8.1 THE +POINTS OF AN ENTIRE FUNCTION AND EXCEPTIONAL VALUES
Rather than discussing the zeros of a function, we now consider the distribution of points where the function takes any value a, the a-points. The following result has already been obtained: “Iff(z) is of finite order e where e is not an integer, thenf(z) has an infinity of zeros and the exponent of convergence of the zeros is e.” Clearlyf(z) - a is also of order e where a is any constant. Hencef(z) has an infinity of a-points and their exponent of convergence is e, i.e., their density is roughly the same for all values of a. Similarly for functions of zero order, such a function has an infinity of zeros unless it reduces to a polynomial. We can obtain similar results for functions of positive integral order by using Hadamard’s factorization theorem, but this tells us nothing about functions of infinite order. We shall show, using methods which are independent of concepts of order, Picard’s theorem, viz., “An entire function which is not constant takes every value with one possible exception, an infinite number of times.” Picard’s proof used elliptic modular functions. A further proof is now given depending upon Schottky’s theorem. As Titchmarsh points out, the main feature of 145
146
VIII. Theorems of Borel, Schottky, Picard, and Landau: Asymptotic Values
Picard’s theorem is that it admits the possibility of there being one exceptional value. The value may actually exist, e.g., ez # 0. A value with this property is called exceptional-P. There is another sense in which a value may be exceptional. A function may take the value a but only at points which have a convergence exponent less than e. For example f ( z ) = ez C
has e
=
O S G
1, but the zeros are
z,
=
((2n
+ 1)~/2}~,
n = 0, f 1 ,
...
Thus the exponent of convergence is 4. A value with this property is called exceptional-B, i.e., in the sense o f Borel. A value which is exceptional-P is necessarily exceptional-B. For entire functions of positive integral order, Picard’s theorem is a consequence of the following theorem of Borel which shows that 3 not merely at most one value exceptionalP, but that 3 at most one value exceptional-B. Alternative proofs of Picard’s theorem can be given depending upon theorems of Bloch.t 8.2 Borel’s Theorem. If the order of the entire function f ( z ) is a positive
integer, then the exponent of convergence of the a-points o f f ( z ) is equal to the order, with one possible exceptional value of a.
Proof. Suppose 3 two exceptional values a and b. Then f ( z ) - a = zkleQl(Z)Pl(z)
(1)
and f(z)-6
= zkneQ~(z)Pz(z)
where Q,(z) and Qa(z) are polynomials of degree p and P l ( z ) and Pg(z) are canonical products of order less than e. Note the order of f ( z ) - a equals e which is max(e, el) and el the order of Pl is assumed less than e since a is an exceptional value. Subtracting,
b - a = ZkleQl(z)pl(Z)- ZheBdz)Pz(z)
(3 1
and &pl(z)eQl(Z)-Qa(z) = zhP,(z)
+ (b - =)e-Qdz).
(4)
t See P. Dienes, “The Taylor Series,” Chapter VIII. Oxford Univ. Press, London and New York, 1931.
8.3-8.5
147
Exceptional-B Values
Since Q2(z) is of degree e, the right-hand side is of order e. Thus the lefthand side is of order @ and Q,(z) - Q2(z) is of degree @ since Pl(z) is of order less than e. Differentiating (3) we obtain,
+ k,zk1-lP, + zklPl’)eQI= (zkrP2Q2’+ k2~ka-1P2+ zkaP2’)eQa.
(zklPIQ1’
The order of Pi’ is the same as Pi (see Theorem 5.6) and is less than e, hence the coefficient of eQ1is of order less than e and similarly, so is that of eQa. Since the brackets are entire functions we may write them as zk3P3eQ3 and zk4P4eQ4,respectively, where Q3 and Q4 are polynomials of degree e - 1 at most (since e is an integer and the order of each bracket is less than e). Also, P3 and P4 are canonical products, thus = zk4p4eQz+Q4, zksP,eQ~+Qs
the two sides having the same zeros k, Qi
+
Q3
=
Q2
+
Q4
or
=
k, and P3 = P4. Thus
QI
-
Q2
=
Q4
-
Q3
and Q4 - Q3 is of degree not exceeding @ - 1, whereas Qr - Q2 was of degree e. The contradiction proves the theorem. 0
8.3 We proceed with the theory of exceptional values by studying meromorphic functions rather than entire functions. The exponent of convergence of the poles of a meromorphic functionf(z) is defined in the same way as for zeros of an entire function. If the exponent of convergence of the poles is finite, their canonical product H(z) is of finite order. We define the order @ off(z) as the larger of the orders of H ( z ) and G(z) where G(z) = z”H(z)f(z), m being the order of the pole at the origin (cf., Theorem 4.9.lj. If the expon.ent of convergence of the poles is infinite, the order of f(z) is infinite. If the exponent of convergence of the poles is less than the order of f(z), then 00 is an exceptional-B value. Any three exceptional values, a, b, c, can be reduced to 0, 1, 00 by putting
and the order of F(z) is equal to the order off(z). Iff(z) is an entire function, it is usual to confine the discussion of exceptional values to finite exceptional values. We extend Borel’s theorem a little, by the following theorem.
148
VIII. Theorems of Borel, Schottky, Picard, and Landau: Asymptotic Values
8.4 Theorem. Meromorphic functions of finite nonintegral order have at most one exceptional-B 'value. In particular, entire functions of finite nonintegral order have no exceptional-B values.
Proof. Suppose that the order e off(z) is not an integer and suppose that the values a and c are exceptional-B (implying that e > 0). We may suppose that a = 0 and c = 03 and write f(z) = G(z)/H(z) where the factor due to a pole at the origin has been absorbed in H(z). Since the exponents of convergence of the zeros and poles off(z) are each less than e, the order of H(z) is less than e and the order of G(z) is less than e, unless e = q. However, q is an integer and e is not, and further, e is the larger of the orders of G and H so we have a contradiction. Thus, either a or c is not an exceptional-B value. Since an entire function is a special case of a meromorphic function for which 03 is exceptional-B, the proof is complete. 0 8.5 Theorem. A meromorphic function of finite positive integral order has
at most two exceptional-B values. Proof. Suppose that f(z) is meromorphic and of order e, where e is a positive integer. Suppose further, that the function has three values exceptionalB, which we may take as 0, 1, and 00. The maximum of their exponents of convergence is el < e. Thus
where P, P,, and H are of order nomials of degree e and
el at most. Thus Q and Q1must be poly-
P(z)eQ(z)- P1(z)eQ1(z) = H(z).
(1)
If we consider the behavior of the functions as I z I -+ 00, we see that the terms of highest degree in Q and Q, must be the same, a,z@,say. MultipIying by exp(- a,zQ), we have the left-hand side of (1) of order
e - 1) < e and the right-hand side of order not less than e. Hence we have a contradiction which proves the theorem. 0 max(e1,
8.6 FUNCTIONS OF ZERO ORDER
The definition of Borel exceptional values fail for functions of zero order, however, similar methods apply. If f ( z ) is meromorphic and f(z) - a has only a finite number of zeros, a is a Picard exceptional value (i.e., excep-
8.7
149
Schottky’s Theorem
tional-P). Iff(z) has only a finite number of poles, We now have the following theorem.
00
is exceptional-P.
8.6.1 Theorem. A meromorphic function of zero order which is not a rational function has at most one value exceptional-P and an entire function of zero order which is not a polynomial has no finite values exceptional-P.
Proof. Let f(z) be a meromorphic function of zero order, with two values exceptional-P, these values as before may be 0 and 00. Thenf(z) = P(z)/H(z) where P and H are polynomials and thusf(z) is rational. 0 Further theorems concerning exceptional values can be studied from the point of view of “Lines of Julia.” 8.7 The main theorems which follow, viz., Picard’s “little” and “great” theorems, depend upon a result due to Schottky. This theorem has been considerably expanded in order to illustrate the analysis required. Although it appears that somewhat long-winded theorems and lemmas are needed, all the mathematical apparatus is elementary in nature. A proof of Picard’s theorem using modular functions is considerably more recondite than that presented here.t 8.7.1 Lemma. Let $ ( r ) be a real function of r, for 0 5 r 5 R, and let 0I $(r) I M for 0 < r 5 R,. Further, suppose 3 a constant C such that
Then 3 a constant A such that $ ( r ) < A C 2 / ( R ,- r)4,
0 < r < R,
Proof. The actual form of the result is not particularly important. What is important is that it depends only on r, R,,and C and not on M. We have, $(r)
< m % 3 / ( r l - r)’
t See in particular M. L. Cartwright, “Integral Functions,” Chapter VII. Cambridge Univ. Press, London and New York, 1962. t M. E. Picard, “Sur une proprikte des fonctions entihres,” C. R. Acud. Sci. Paris 88
(1879) 1024-1027.
150 for r
VIII. Theorems of Borel, Schottky, Picard, and Landau: Asymptotic Values
< r, I R,. Thus for 0 < r < r2 < r, 5 R,, +(r,) < CM1/a/(r,- r,),
and therefore
Further,
for 0
< r < r3 < r, < r, I R,. Generally,
for 0
< r, < rn-, <
+ +
< rl 5 R,. We now put
+
As n -+ 00, 1 1 - * . n/2,-' converges, 1 and M1/2n 1, and the result follows. 0
+ 1/2 + . . - + 1/2,-'
.-+
2
--+
8.7.2 Theorem (Schottky). Iff(z) is analytic and does not take either of the values 0 or 1 for I z 15 R , , then for I z 15 R < R,,
where K depends on f ( 0 ) only. For all functions which satisfy the given conditions and are such that 6 < I f ( 0 ) 1 < 1/6, 1 1 -f(O) 1 > 6, K depends on 6 only.
Proof. The actual form of the upper bound is not particularly required. This could be considerably improved if necessary. The importance lies in the fact that it depends on f(0) in the manner stated and on R/R,. Let gl(z) = log{f(z)}, g,(z) = log(1 -f(z)} where each log has its principal value at z = 0. Thus gl(z) and g2(z) are regular for I z I 5 R, (by hypothesis). Let M l ( r ) and M , ( r ) be the maxima of I gl(z) I and
151
8.7 Schottky’s Theorem
1 g2(z) 1, respectively, on I z I = r < R. Let M ( r ) = max{M,(r), M,(r)}. Let Bl(r) = - min .9{gl(z)} = max log 1/ 1 f(z) I Itl-r
121-7
[since for u real, max(- u ) = - min(u)]. We now apply CarathCodory’s theorem to - gl(z)and obtain
There are now two possibilities: either B,(r) 5 1, in which case inequality (1) is a result of the required type, or B,(r) > 1 in which case 3 a point z‘ on I z I = r where If(z’) 1 is small. However, if If(z’) I is small, g,(z) is (apart from a term 2nni) approximately equal to -f ( z ’ )[expand log(1 -f)]. Thus applying Carathtodory’s theorem to log g, , we have on the left-hand side MI (not log Ml as expected) and on the right-hand side log M,. If M , < k, then9(g2} < k and If I < 1 ek, V r and we have an equivalent inequality. If M,(r) + 00 as r 4 R,, it suffices to take log M , = O(dK) as M, -+ 00. Suppose B,(r) > 1 and let z’ be a point where
+
then
Therefore 3 an integer n such that
hence
Let h(z) = log{g,(z) - 2nni) where the log has its principal value at z = 0. Then h ( z ) is analytic for I z 1 2 R, sincef(z) # 0, thus g2(z) # 2nni.
I52
VIII. Theorems of Borel, Schottky, Picard, and Landau: Asymptotic Values
Carathkodory’s theorem then gives
Now
I h(z) I = I log{g,(z)
- 2nni) I =
I log l/[g,(z)
-
2nni]
I
2 log I l/[gz(z) - 2nniI I hence max
I h(z) I 2 log I 1/[8,(z) - 2nnil 1
121=7
and in particular at z’. Therefore the left-hand side is greater than or equal to log l/[I.f(z’> I
+ If(z’) 1, + * . * I > 1%
{1/2 If(z’> I >
and log 1 l/[g,(z’) - 2nni] 1 > B,(r) - log 2. On the right-hand side, since
+
has its principal value, then g2(0) = (real part) i(imaginary part) where 1 Imag 1 5 n. Thus I Imag {g,(O) - 2nni) I 2 n and I g,(O) - 2nni I 2 I Imag{gz(0) - 2nni) 1 2 n > 1. Since
153
and
+ 11 + I log for r 5 R,
(since e < r). We may interchange g, and g, in the whole argument because we can interchange f and 1 -f: 1
-ff
{
1 if f f 0
0
if f f l ,
154
VIII. Theorems of Borel, Schottky, Picard, and Landau: Asymptotic Values
thus the inequality is true if the suffixes 1 and 2 are interchanged. Hence
where Kl depends on
M(e) < <
I gl(0) 1 and I g2(0)I only. Take r = &(e+ R ) and
32Rla ( R - @)a {log M ( R ) R12
( R - el2
+ Kl},
since R
+ e < 2R < 2R1, etc.
K 2 d m , since log M ( R ) = O ( d m ) .
Thus using the lemma, M(,o) < K3R14/(R- Q
) and ~
we have
I f ( z ) I IeM(')< exp{KR,4/(R - r)", when K is constant and note r < e. Because K depends on I g,(O) I only, the last part of theorem is true also. 0
I gl(0) I
and
but if either MI or M2 were bounded, we would immediately obtain bounded.
If I
FURTHER NOTESTO SCHOTTKY'S THEOREM. If M ( R ) = 0, log W R ) # O(dM(R)),
155
8.9 Landau’s Theorem
There is nothing to consider unless M -+ 00, when log M = O(.t/M), and this bound for log M is taken in order to apply the lemma. We have the following statement: (1) If f is regular (analytic) in I z I 5 R , and f # 0, 1 by hypothesis, in I z 1 I R , , then If(z) I < B(f(O), I z I) in I z I 5 R , . The bound obtained approaches 00 as I z I + R, [not, of course, the actual value of I f l , but the bound obtained using this meagre information of the value off(0)l. The bound could be improved, but it would still approach 00 as I z I R , . Since we assume f is analytic in I z I 5 R 1 , f(z) is bounded there since any continuous function is bounded as such, but there is not a uniform bound for all f satisfying (1) and with the chosen value off(0). However, there is a uniform bound for all of them in any smaller circle. The assumption thatf(z) does not assume the values 0 or 1 was made only to fix attention, the essence of the theorem remains unchanged if we assume thatf(z) does not take any two finite values a and B. It is sufficient then to apply the theorem to the function G(z) = (f(z) - a ) / @ - a ) not taking the values 0 or 1.
-
8.8 Picard’s First Theorem. An entire function which is not constant takes every value, with one possible exception, at least once.
Proof. Suppose that f(z) does not take either of the values a or b (a f b). Then g(z) = (f(z) - a ) / @ - a) does not take either of the values 0 or 1. Thus by Schottky’s theorem,
Taking R ,
= 2R,
I g(z) I < C. Hence g(z) is constant. 0
A geperalization of Picard’s theorem is as follows. 8.9 Landau’s Theorem. If a is any number and any number not equal to 0, then there is a number R = R ( a , B) such that every function f(z) analytic for
=a
+ pz +
Q2Z2
+ agz3 +
* * *
,
1 z I 5 R , takes in this circle one of the values 0 or
1.
Proof. We may suppose a f O or 1 otherwise the result is immediate [since if a = O , f ( O ) = 0 and if a = l,f(O) = I]. Iff(z) does not take either of the values 0 or I , then by Schottky’s theorem,f(z) < K ( a ) for I z I 5 R / 2
156 because a
VIII. Theorems of Borel, Schottky, Picard, and Landau: Asymptotic Values = f(0).
Thus
Therefore what is said is, iff # 0, 1, then the radius of the circle R is less than or equal to a constant. Taking a larger circle, f must take one of the values 0 or 1, i.e., 3R(u, j3) such that Vf of the above form take one of the values 0 or 1 in this circle. The argument uses in effect, contrapositive logic. 0 Schottky's theorem saying that, the hypothesis that an analytic function (in I z 1 5 R ) does not assume two values, e.g., 0 or 1, is already a strong limitation on the behavior of the function. Landau's theorem confirms this fact from another direction.
8.9.1 The theorem can be stated in a different way: For every a, f 0, if the function f ( z ) = a 12 uzz2 . - is analytic in 1 z I 5 R and does not assume the values 0 and 1 in it, then R 5 R(a, 1).Further, since a function has a t least one singular point on the circle of convergence we can say the following.
+ +
+
+ +
+ -
8.9.2 For every a and ,b f 0, the function f(z) = a ,bz u2z2 must assume the value 0 or 1 in the closed circle I z I 5 R(a, 8) or the series has a singular point in this circle. The assumption
Bf
0 may be replaced by a more general condition.
+
8.9.3 Theorem. Let us suppose that A f 0 and that f ( r ) = a Iz' ulflzl+' - is analytic in 1 z 1 5 R and does not assume the values 0 or 1 there. Then R I R(a, A, l), i.e., R depends on 01, A, 1 only.
+
+
+
a
Proof. As before,
thus
8.10
157
Picard’s Second Theorem
Papers by Jenkinst deal with the connection between a and p. Every entire function which is not constant, may be written in the form f(z) = a lz’ az+lzr+l . Since the radius of convergence is then infinite, the inequality R 5 R(a, A, I ) is not satisfied. Thus f ( z ) must assume at least one of the values 0 or 1. Therefore Landau’s theorem includes Picard’s ‘‘little’’ theorem in a stronger form, for it gives an estimate of the radius R of a circle (centered at origin) in whichf(z) certainly assumes either the value 0 or 1.
+
+
+ .
8.10 Up to this point, Picard’s theorem has been stated in terms of entire functions, i.e., functions with an essential singularity at infinity. A corresponding theorem holds for any function with an isolated essential singularity. We recall that iff(z) is analytic throughout a neighborhood of a point a, i.e., for I z - a I < R, excepting the point a, then a is called an isolated essential singulariry of the function. Further, if f(z) is analytic in the annulus R‘ I :I z - a I 5 R , then f(z) may be expanded in a series of positive and negative powers of (z - a), converging at all points of the annulus. Consider the principal part C:=lb,(~ - a)-,. Iff(z) has a pole at z = a, then f(z) I .+ 00 as z -+ a, for
.
As z -+ a, { - . } + I b, I, thus the whole expression approaches 00, i.e., if f(z) = 0( I z - u 1 -k) as 1 z - a I 0, the singularity is at most a pole of order k. -+
8.10.1 Picard’s Second Theorem. In the neighborhood of an isolated essential singularity, a single-valued function takes every value with one possible exception, an infinite number of times; or, if f(z) is analytic for 0 < I z - zo I < e and 3 two unequal numbers a, b such that f(z) f a and f(z) # b for I z - zo 1 < e, then z, is not an essential singularity. Proof. We may suppose that z, = 0, p = 1, a = 0, and b = 1. We prove 3 a sequence of circles I z, I = r, where r, -+ 0 and on whichf(z) is bounded. t J. A. Jenkins, “On explicit bounds in Schottky’s theorem,” Cunud. J. Math. 7 (1955) 7 6 8 2 ; “On explicit bounds in Landau’s theorem,” ibid. 8 (1956) 423425.
158
VIII. Theorems of Borel, Schottky, Picard, and Landau: Asymptotic Values
This then precludes the existence of a singularity at z = 0 except possibly a removable one, for if f(z) = O( I z - a I-k) as I z - a I + 0, the singularity is at most a pole of order k and in particular if f(z) = O(l), there is no singularity. Using Weierstrass’s theorem which states that in the neighborhood of an isolated essential singularity a function comes arbitrarily close to any given value an infinity of times, there is a sequence of > 1z,1-+0 and points zl,z,, ... such that lzll > Iz21 >
If(z,)
-2
I < 4.
Thus it follows that if all the points are inside 1 z I = 1, Schottky’s theorem enables us to construct a sequence of circles with these points as centers and in whichf(z) is bounded. These circles do not include the origin (since f is not analytic at the origin), however, this is only because Schottky’s theorem was proved for the convex curves, viz., circles. The apparent difficulty is removed by making a conformal transformation which replaces the circles by elongated curves which though they exclude the origin, pass around it and overlap on the far side (Fig. 1). Let z = ew and consider the
half-strip in the w-plane u < 0, - n 5 u 5 n: this corresponds to the interior of I z I = 1. Let w, = log z,,
-n
< c7{w,} I n,
Thus
9 { w n } = log I zn I + - 00. Let f(z)
= g(w) =
{ f ( e w ) } . We apply Schottky’s theorem to
h(w’) = g(wn
+ w ’ ) =f(exp(w, + w’)). +
If n is large enough, z, -+ 0 and exp(w, w’) ern 0. Choose I w’ I 5 477, then h(w’) is analytic if n is large enough. (Since if w = w, w’, I w - w, I = I w’ I 5 4n and the circle centered at w, is sufficiently removed from the origin if the radius is less than or equal to 4n as in Fig. 2. Also z = e W = exp(w, w’), 0 < I z I < 1). Further, h(w’) does not take the values 0 or 1. Thus I h(w’) I < k = k{h(O)} for I w’ I 5 2n. Moreover,
+
-+
-+
+
8.11 Asymptotic Values
159
FIGURE2
@
-
0’
= f ( z , ) , satisfying 1 f(z,) - 2 1 < 3, I z, I 0. We can replace the right-hand side by an absolute constant. Consequently
W )= g(w,)
and in particular for u = SP(w,}, - n 5 u 5 n. Thus If(z) Izl=lz,l. 0
1 < A,
8.11 ASYMPTOTIC VALUES
If as 1 z 1 +00 along some continuous path y, f ( z ) + a, then a is an asymptotic value of f(z) and y is a path of finite determination. If as I z 1 -+ 00 along y, I f ( z ) I + 00, then infinity is an asymptotic value of f(z) and y is a path of infinite determination. If I f ( z ) I is bounded on y but f(z) does not tend to a limit, then y is a path offinite indetermination. Iff(z) is meromorphic for I z I > r, the definitions also hold. Zero is an asymptotic value of P since ez -+ 0 as I z I -+ 00 along the negative real axis. Consider 1,”exp(- t Q )dt, q a positive integer. This function has q asymptotic values exp(2niklq) Jwexp(- t Q )dt
k
= 0,
1, . . . , q - 1 as z-00
0
along the lines arg z gration t = re2nik’g,
= 2nk/q.
dt
For transform (rotate) the line of intee2niklq dr
and
J:
exp(-
tq)
dt + exp(2nik/q) = exp(2nik/q)
J: J:
exp(- rQ. eZnik)dr
as
I z I + 00
exp(- rg) dr.
8.11.1 Theorem. Every function with an isolated essential singularity at and which is not constant, has 00 as an asymptotic value (or has a
00,
path of infinite determination).
160
VIII. Theorems of Borel, Schottky, Picard, and Landau: Asymptotic Values
Proof. By Laurent's theorem, such a function is of the formf(z) -t g(z) where f ( z ) is an entire function and g(z) tends uniformly to a limit as I z I + 00. We consider thus entire functions only. The maximum modulus M ( r ) of a nonconstant entire function has the property that M ( r ) 00 steadily. Consider an indefinitely increasing sequence of numbers X , = M(r,), X , = M(r,), . . . . Clearly there is a point outside I z I = r , at which If(z) I > X,. The set of points where If(z) I > XI constitutes the interior of one or more regions bounded by curves on which I f ( z ) I = X,. These regions must be exterior to 1 z I = r , . Let one such region be D,. Then D, must extend to 00 (Fig. 3) otherwise we should have a finite re--+
gion with I f ( z ) I = X , on the boundary and If(z) I > X , inside, contrary to the maximum modulus theorem. Also, f(z) is unbounded in D, otherwise the Phragmtn-Lindeloff principle would show that I f(z) I 5 X , at all points inside D,. The Phragmtn-Lindeloff argument of 'Theorem 7.2, applies with P at infinity and w = rl/z. Thus 3 a point of D, at which I f ( z ) I > X , and consequently a domain D, interior to D, such that If(z) I > X , at all points of D,.We repeat the argument with X,, . . . . Hence 3 a sequence of infinite regions D,, D,,. . . each interior to the preceding one and such that If(z) I > X , in D, and If(z) I = X,,, on the boundary. We take a point on the boundary of D, and join it to a point on the boundary of D, by a continuous curve lying in D,,and repeat from D, to D,, and so on. Thus a continuous curve is obtained along which If(z) I +w.
0
161
8.12 Contiguous Paths
8.11.2 Theorem. If an entire function does not take the value a (i.e., if a is exceptional-P), then a is an asymptotic value.
Proof: If f(z) is an entire function and f(z) # a, then I /(f(z) - a ) is an entire function and thus by the previous theorem has the asymptotic value 0 0 ; consequently f(z) + a. 0 NOTE.The argument in Theorem 7.6 shows that if an entire function has asymptotic values on two curves and is bounded between them, then these asymptotic values must be the same. Asymptotic values not so connected should be considered distinct, whether they are equal or not. 8.12 CONTIGUOUS PATHS
If as I z I + 00, f(z) -+ a on two nonintersecting paths y1 and y2 starting from the same point z o , then y1 u y2 determines two infinite domains. Iff(z) + a uniformly in the closure of one of these domains, then y1 and yz are defined to be contiguous paths of finite determination and the closure of the domain concerned is a tract of determination. A similar definition holds for contiguous paths of infinite determination, or finite indetermination. For all other cases, paths of the same kind are defined to be noncontiguous.
Clearly, if f ( z ) + a as I z I + 00 along y1 and f(z) b as I z I +00 along y z , either a = b or If(z) I is unbounded between y1 and y z . ---f
8.12.1 Theorem. Between any two noncontiguous paths of finite determination and between any two noncontiguous paths of finite indetermination, there is a path of infinite determination.
Proof. Suppose f(z) is bounded in one of the domains bounded by y1 u y2 and that f(z) ---* a as ,I z I + 00 along y l and f ( z ) b along y 2 . By Theorem 7.6 we have that f(z) - + a uniformly in that domain. Thus y1 and y2 are contiguous paths and we have a contradiction. Similarly, using Theorem 7.2, I f ( z ) I is unbounded in the case of paths of finite indetermination. The method of Theorem 8.1 1.1 now gives the result. 0 --+
8.12.2 Denjoy conjectured that an entire function of finite order p can have at most 2p different asymptotic values (proved by Ahlfors).t We can see, however, that there can be at most 2p straight lines from 0 to cu along t L. V. Ahlfors, “Uber die asyrnptotischen Werte der meromorphen Functionen endlicher Ordnung,” Acta. Acud. Abo. Math. Phys. 6 (1932) 3-8.
162
VIII. Theorems of Borel, Schottky, Picard, and Landau: Asymptotic Values
which a function of order Q has distinct asymptotic values, since by Theorem 7.3 the angle between two such lines must be at least n/e. (N.B. 2Q * n/Q= 2n).
8.12.3 For further results concerning asymptotic values, the reader is directed to a thesis by Al-Katifi.+
W. Al-Katifi, “On the asymptotic values and paths of certain integral and meromorphic Functions,” Imperial College, London (1963) (also in Proc. London Math. SOC.(3) 16 (1966)599634.
C H A P T E R IX
ELEMENTARY NEVANLINNA THEORY
9.1 MEROMORPHIC FUNCTIONS
It will be convenient to use the letters M. F. to mean “Meromorphic Functions” by W. K. Hayman.+ The theory depends largely on the general Jensen formula, viz., for a function with zeros a,, az,. . . a,, and poles b l , . . . ,bn with hoduli not exceeding r and arranged with noncbtrreasing moduli,
Suppose that in the neighborhood of the origin f(z) k is any integer. Apply Jensen’s formula to z-kf(z).
-
czk
+ . . , where
W. K. Hayman, “Meromorphic Functions,” Oxford Mathematical Monograph. Oxford Univ. Press (Clarendon), London and New York. 1964. 163
164
IX. Elementary Nevanlinna Theory
Let n(r, 0) be the number of zeros off(z) in I z 1 5 r . If k > 0, v
= n(x, 0 ) -
k for I a, I L x < I a,+, I
(has k zeros at the origin), hence
(= N(r, l/f), M.F. with f(z) such that k = 0). If n(r, w) is the number of poles of f(z) in I z I 5 r, then by similar reasoning
log
I
rn b,b,**.b,
n(x’ m,
I
(= N ( r , f ) , M.F.).
dx
If k < 0, it appears in the second integral instead of the first. Let N(r, a) =
n(x’
- n(o’a)dx X
0
+ n(0,a ) log r
[M.F. N ( r , f ) corresponds to our N ( r , w), since n(0, w) = 01. We obtain, N ( r , 0 ) - N ( r , m) = 2n where n(0, 0)
=
jr log I f(re’ie)I
d0 - log I c
k. Now write
log+ a = max(1og a, 0),
IX
> 0.
Thus log a [since if a
=
1 a
= log+ a - log+ -
l/h, h > 1, log a
< 0,
loga
=0-
log+ a log+ h
=
max(1og a, 0) = 0,
= -
[Similarly for a > 1, log+(l/a) = 0.1 Let
log h
= log(l/h)].
I
(1 1
9.1
Enumerative Functions: N(r, a), m(r, a )
165
and
m(r, GO) = m(r, f )
=
2n
log+ I f(reiO) I do.
Then (1) may be written as
since
We now apply this formula to f(z) - a where a is any number. If f ( z ) - a = c,zk . . in the neighborhood of the origin,
+
noting that the term N ( r , m) is unaltered since poles off@) - a are the same for each a. {With a = 0 this corresponds to the M.F. formula
We can of course verify this by the following simple calculation
and # =f - a. Applying (2) to
4, we have
and
1
=-
2n
1
2n
0
log+ If(reiO)- a I d0
= m(r,f-
a).
166
IX. Elementary Nevanlinna Theory
Thus
m(r, a)- m(r, 0)
1 h
=-
N(r, 0) =
n(xyO)
2n
o
- n(o’ X
-
0
- m(r, a )
= m ( r , f - a)
log I f(reie) - a I do. dx
+ n ( 0 , O ) log r,
applied to f. Then n ( x , 0) n(x, a), since the number of zeros of f in 1 z 1 5 x becomes the number of a-points of f (or zeros of f - a) in l z 1 -< x, and n(0, 0) + k, since f(z) - a = c,zk . . . Also, N ( r , m) applied to f becomes N(r, co) applied to f- a. Thus
+
a
We observe further, that
depending upon whether
or (iv)
If1
<1
and
J a l( 1 .
Verifying, for example, the first result: if If1 2 1 and I a I 2 1, then I f 1 I a I L I a I and I a I I f 1 L Ifl. Thus I f 1 I a I 5 2 I f . a I. Hence
+
log{ I f 1
+ I a I} I log+ I f 1 + log+ I a I + log2.
Each case should be verified, e.g., if log{ If1
I f l 2 1, I a I < 1, then
+ I a I > i log2 + log If1 = log
2
+ log+ Ifl,
(4)
167
9.1 Enumerative Functions: N(r, a) m(r, a )
and log+ I a I = 0. Since log'
(for if log+ If-
If-
If-
a I 5 log+ If1
aI5
If1 + I a I,
+ log+ I a I + log 2
[from Eq. (4)]
a 1 > 1, log+ Ifa 1 = log Ifa 1 and if Ifa 15 1, a I = 0 which is certainly less than the right-hand side). Also,
If-
168
IX. Elementary Nevanlinna Theory
Thus if f ( z ) is a meromorphic function and not a constant, the value of the sum m(r, a ) N(r, a ) , for two given values of a, differ by a bounded function of r. Since all the sums are to this extent equivalent, we can represent them all by the one with a = 00. Thus putting
+
{corresponding to m ( r , f )
+ N ( r , f ) , M.F.}
for all a,
where $(r, a ) is (for each a ) bounded as r .-+ 00, and T(r) is called the characteristic function off (z). Then Eq. ( 5 ) is called Nevanlinna's first fundamental theorem. 9.2 Theorem. T(r) is an increasing convex function of log r.
Proof. We apply Jensen's formula to f(z) - ei2 (A real). For a function q5(z), with $(O) = c # 0 and $ ( z ) analytic in I z [ 5 R, we have that N ( r , 0)- N(r, 00)
=-Jrlog
2n
1 $(reie) I dB - log I c I.
Substituting in this formula, we note that N ( r , m) is the same whether we use f ( z ) or f ( z ) - e". n(x, 0 ) in the formula for N ( r , 0) becomes n(x, eiA)for f(z) - e'iA,whereas n(0, 0) applied to f(z) becomes n(0, eiA) applied to f ( z ) - eta. Thus for the function f(z) - eu we have,
provided f ( 0 ) f eiA.Further, for any value of a, 1
2n
nJolog 1 eie - a I dB by applying Jensen's theorem tof(z)
=z
=
- a,
dB
log+ 1 a
1
r = 1. Note that where rn I rI r,,,.
9.2 The Nevanlinna Characteristic T,(r)
169
I f a s 1 , l o g + l a l =Oand
If a > 1, log+ I a I
=
loga, and
since if a > I , and r l , r 2 , . . . , r,, is the product of the moduli of the zeros not exceeding one, the denominator of the left-hand side log is just the number 1, i.e., 3 no zeros of z - a whose moduli are less than or equal to one. In both cases log I eie - a I d0
I 1.
= log+ a
We now multiply ( 6 ) by 1/2nand integrate with respect to 1 over [0, 2n], i.e.,
Thus
1:"{&-
{ N ( r , eiA)- N ( r , c o ) }d1 = 1 2n
Sanlog I eia-f(reie) I dA} d0 0
= m(r, 00) - log+ If(0)
I.
Hence N ( r , ein) dA
Now for any a
+ log+ If(0) 1,
0
(8)
170
IX. Elementary Nevanlinna Theory
and
Therefore
which is nonnegative and nondecreasing. Thus N(r, a ) is an increasing convex function of r (since the second derivative is positive). Thus T ( r )has the same property d4
and the integrand is nonnegative and increasing).
0
9.3 We also establish a bound for m(r, a ) on the circle 1 a the result (3) with a = eie, we have
m(r, eie)
+ N(r, eie) = m(r,f
- eie)
+ N(r,
I = 1. Using
1 f ( 0 ) - eie 1.
GO)
- log
GO)
+ m ( r , f - eie)
Hence m(r, ese)
+ N(r, eie) + log I f ( 0 ) - edej = N(r, =
T(r)
+m(r,f-
eie) - m(r,co).
Thus
+
T ( r ) = m(r, eie) N ( r , eie) where I G ( 0 ) I I log 2
+ log 1 f ( 0 ) - eie I + G ( 0 ) ,
since
1 m(r, f - eie) - m(r, m) 1 5 log+ I eio 1
+ log 2 = log 2.
We now integrate both sides with respect to 6, from 0 to 27-2 and using (8), T ( r )d0
= T ( r )= -
27L
':I
m(r, eie) d0
+ T ( r ) - log+ If(0) I
9.3 A Bound for m(r,a) on I a I = 1
171
Thus we have 1 0=2n
an 0
m (r, ,ae) d8 - log+ If(0) 1
1 + log+ If ( 0 ) I + 2n
rn
G ( 8 ) dB
0
and 1 -
2n
2n
0
m(r, eie) d8
=
1 2n
--
2n
G ( 8 ) d8 5 log 2 .
0
Consequently m(r, a ) is bounded on the average in the circle I a I = 1 and a corresponding result holds on any other circle. Thus N(r, a ) is a convex increasing function of log r and so is T ( r ) , however m(r, a ) need be neither increasing nor convex in general. For example, if f(z) = z/(l - z2), then If(z) I < 1 for I z 1 < 4 and I z 1 > 2. Thus m ( r , f ) = 0 for r 5 4 or r 2 2, however, f(& 1) = 00, thus m ( 1 , f ) > 0. We note that N(r, a ) measures the number of times the function f(z) takes the value a. Since the largest contribution to m(r, a ) comes from arcs wheref(z) is nearly equal to a, m(r, a ) measures in a sense the intensity of the approximation off(z) to a (or as in M.F., thq average smallness in a sense o f f - a on I z I = R ) . We could describe m(r, a ) N(r, a ) as the total affinity of the functionf(z) for the value a. For a given function, certain valuks may be exceptional, i.e., in the sense that the function does not take these values.,The above theorem shows that there can be no exceptional values in the sense that the total affinity of the function for every value is the same, apart from bounded functions of r [since m(r, a ) N(r, a ) = T(r) 41. We can show that actually, the term N(r, a ) predominates. A few examples will help to consolidate the theory so far.
+
+
+
1. Consider P' which does not take the value 0 or EXAMPLE ues may be regarded as limiting values as z + & 00. Then n(r, 0 ) = 0, N(r, 00)
= 0,
r m(r, 0) = -, 7d
00.
These val-
since n ( x , 0) = 0 and n ( 0 , O ) = 0. since ez has no poles in the finite plane. since log+ I e-L I
=
log+ earcose= log e-rcOse
for e-rcose> 1, i.e.,
- r cos 8 > 0,
r cos 0
< 0,
cos 8 < 0,
742
< 8 < 3n/2
172
IX. Elementary Nevanlinna Theory
and
and m(r, m) = r / n by the same argument. For a # 0 or
and
4 is bounded.
00,
Hence
EXAMPLE2. Let f ( z ) be a rational function equal to P ( z ) / Q ( z ) , with P ( z ) of degree p and Q ( z ) of degree
Y,
P and Q having no common factor.
a. Suppose p > v. Then m(r, a) = O(1). In either case of log+, the integral is bounded, thus
I Q / ( P - aQ) 1 = O(ry-”). N(r, a ) = p log r
+ OU),
since dN(r, a)/dr = n(r, a ) / r . And the number of times P / Q assumes the value a is the number of times P - aQ = 0, which being a polynomial of degree p has p zeros. Thus dN/dr = p / r and N = p log r constant.
+
m(r, 00)
in
=-
since log O(r”-”)
-
j:’log+l$l
de
log cr”-”.
N(r, m) = v log r
+ O(l),
since the number of poles off in a large enough circle is the degree of Q, viz., v.
9.3 A Bound for m(r, a ) on
1a I=1
173
b. For ,u < v and a # 0,
m(r, a ) = O(1) N ( r , a)
since
=v
log r
i.e., the number of zeros of P
1
1 (P/Q) - a I
+ O(1)
=
O(r”-”)= O(1).
since n(r, a) = v,
- aQ is the degree of Q . For a = 0,
since n(r, 0) = degree of P. c. For ,u = Y, consider the leading coefficient (xb) in P and Q . Then if Q f . a,/b,, P since - - a
m(r, a) = O(1)
Q
a0
=60
+0
Then
is bounded for r -00
and
Jr,
etc. is bounded.
N(r, a) = ,u log
+ W),
since again P - aQ is of degree ,u = v. If a = ao/b, ( p = v) we have
and
where a is the degree of boP - aoQ. Therefore
since the number of times f = P/Q takes the value ao/b, is the number of zeros of boP - aoQ. In all cases T ( r ) = O(log r ) .
174
IX. Elementary Nevanlinna Theory
9.4 ORDER OF A MEROMORPHIC FUNCTION
9.4.1 Definition. The meromorphic function f ( z ) is said to be of order
- log T ( r ) lim
r+o
so that T ( r ) = O(re+#)for all
E
log r
e, if
=e
> 0 but not for E < 0.
We show that this agrees with the definition of order in the case of an entire function. 9.4.2 Theorem. Iff(z) is an entire function,
T ( r ) 5 log+ M ( r ) 5 R + r T(R) R-r
for
o < r < R.
Proof. Since an entire function has no poles, N ( r , m) = 0. Thus T ( r ) = m(r, a). The left-hand inequality is thus
log+ If(re”) I d8 5 log+ max If(reie) I which is clearly true since log+max I f 1 2 log+ Ifl. Also, by the Poisson-Jensen formula, log I f(reie)
I =2n
I”
R2 - r2 R8 - 2Rr cos(8 - 4) + r2 log If(Rei+)I
4
and since I R2 - ii,re$e 1 > I R(reie - a,) I because r < R and R > I a, each term after the integral is negative. Note also, that R2 - 2R cos(0 - 4)
I,
+ r2 2 ( R - r)2.
Thus taking 8 so that the left-hand side is a maximum,
t; -
log M ( r ) 5 -
s r l o g If(Rei+)I d+ 5 4- T ( R ) . R-r
Thus the right-hand side of the inequality is true. Note, (R > 1 and T ( R ) 2 0. 0
+ r ) / ( R- r )
175
9.5 Factorization of a Meromorphic Function
If we take R = 2r, the identity of the two definitions of order of an entire function is clear. since T ( r ) 5 log+ M ( r ) 5 3T(2r)
and log log+ M ( r ) < log T(2r) log T ( r ) I log r logr - log r 9.4.3 Now let rn(a)be the moduli of the zeros off(z) - a, r n ( m )the moduli of the poles off(z), and we have the following results. Iff(z) is of order e, then for every a
(i) m(r, a) = O(re+&), (ii) N ( r , a)
=
O(re+&),
(iii) n(r, a) = O(re+&)
and
We observe that since m(r, a ) = T ( r ) [and m(r, a ) ] 4: 0, that m(r, a) I T(r)
+ O(1)
and
C I/{rn(a)}e+e< 03.
+ O(1) - N(r, a) N(r, a) 5 T ( r )
and N(r, a)
+ O(1).
Thus (i) and (ii) follow (for T ( r ) I log+ M ( r ) = O(1og exp re+e)}, and since n(r, a) = r dN(r, a)/dr, (iii) follows. Also, f - a is a meromorphic function iff = P/Q and the zeros off - a are the zeros of P - aQ which is an entire function. Thus
as previously proved for entire functions. More precise results can be obtained.+ 9.5 FACTORIZATION OF A MEROMORPHIC FUNCTION
Let f(z) be a meromorphic function of order e, with zeros an and poles bn [f(O) # 01. Then 3 integers p 1 and p z not exceeding e, such that
t
See papers by R. Nevanlinna listed in the Bibliography.
IX. Elementary Nevanlinna Theory
176 and
are convergent for all values of z. The function fl
(z) = f(z)Pz(2)
is an entire function. Consider
fi(4: w , fl)
=
(since log+fP,5 log+f
m(r7 00,
fl)5 m(r,
00,
f) +
d r 7 0 0 7
P2)
+ log+ Pz)and
T(r,fl) 5 U r , f )
+ W ,Pz)
+ 0 (re+&)
= 0 (re+&)
[Pzis an entire function and we have proved that T ( r )I tog+ M ( r ) ] . Thusfl(z) is of order e at most. Hence fl(z)= e Q c r ) P l ( zwhere ) Q is a polynomial of degree not exceeding e. Thus we have proved that f(z) = eQ(2)P1 (z)/Pz (z), an extension of Hadamard’s theorem to meromorphic functions. 9.6 THE AHLFORS-SHIMIZU CHARACTERISTIC
This is now a second formulation of the first fundamental theorem. 9.6.1 Lemma. Let D be a bounded domain, bounded by a finite system of analytic curves y. Letf(z) be analytic in D and on y and let G ( R ) be twice
continuously differentiable (C E Cz)on the set of values R assumed by
f(z)in D and on y. Then
where g(R) = G”(W
+ R1 G’(R),
s denotes arc length along y and d/an is differentiation along the normal to y out of D.
Proof. Consider Green’s formula, the two-dimensional analog of the divergence theorem. The divergence theorem states
JJJ div F dV JJ
F-ndS
=
V
D
177
9.6-9.7 The Ahlfors-Shimizu Characteristic T,(r)
If F = F(x, y), we choose a cylinder V generated by a line segment of unit length parallel to the z-axis, its lower end describing a contour C . The triple integral now reduces by way of z-integration to div F dS, where D is the closed domain in the xy-plane bounded by C. The double integral over the curved surface of V reduces to the line integral around the contour C (after z-integration). Clearly, the integrals over the ends cancel since n is positive on top and negative on the bottom. Since we now have
Is,
ss
V .FdS=
D
JcF
nds,
choose F = V G and
and then
Taking G = G ( If]), we evaluate V 2 G ( Ifl) supposing first that f # 0 in D. We have that v = log I f / is harmonic in D. Put I f 1 = eu and G ( If I) = G(ev),then
a
av ax
-{G(e")} = e"G'(e")ax
3
and a2
-{G(e")} = e2"G" aY2
Adding and using the fact that v is harmonic,
Writing ev = R
=
I f ( z ) I and noting that
+ e"G'(eu)?.aY
d2v
IX. Elementary Nevanlinna Theory
178 we have
The result now follows. Iffhas zeros in D, we exclude them by small circles over which the contribution of J (aG/dn) ds is negligible since by hypothesis BG(R)/BRis bounded near R = 0. 0 We apply the lemma with G(R) = l o g d m and thus
Let f(z) be a meromorphic function in I z I I r and suppose f has no poles on I z I = r. Exclude poles 6, of multiplicity k, in 1 z I < r, by small circles of radius e (Fig. 1). On such a circle, f ( z ) = a-k,/(z - 6,)" * * . Thus
+ -
f(z)
where 4 is analytic in and on
= U/(Z -
~Jk+W
I z - 6 , I = e. Hence
I f(z)I = (1 /ek.) I 4 I where C, is an upper bound for log dl since
C,/ek,
4 since 4 is analytic. Also,
+ eei'9 12-
1%
+ eei9 I
If@,
I f 1 3 1 on the little circle. logdl
and
+ If@,
-
+ If@,
+ eeiw)
12
=
1
k, log -
e
+ o(I)
179
9.6-9.7 The Ahlfors-Shirnizu Characteristic T&)
since the outward normal to D is directed into the little circle. Hence in J,, aG(R)/dnds, a pole of multiplicity k , gives rise to a term (k,/e) 2ne = 2nk,. Thus
-
since d/an becomes a/dr + d/dr, ds becomes r do and since &,k, becomes n ( r , f ) the number of poles off in I z 1 < r (poles of order p counted p times). Hence 1 r 2n
~
d dr
s
an
logd 1
+ I f(reie)la dB + n(r, f)
Now call the right-hand side of the previous identity A(r). We divide by r and integrate from 0 to r and obtain
{using M.F.,
+
Consider the transformation W = (1 i i w ) / ( w - a), w the resulting transformation W = F(z). Write
Iw-al
k(w, a ) = 2/(1
+ I a l”(1 + I w 12)
=f ( z )
= k(a, w )
and call
180
1X. Elementary Nevanlinna Theory
for w, a both finite and
k(a, m) =
1
= k ( m , a).
dn7-P-
Thus k ( w , a) I 1 always. Now dW dz
+ I a 12)
(1
(W
- a)'
dw dz
and 1 1+IWl'
Thus
Writing
and "o(',
1 2n
a) = -
so
2n
log k{f(re"), a )
do
we apply (9) to F(z)
Clearly,
= N ( r , a) since f ( z ) - a has a zero of order p , at a pole of order p of F ( z ) , for F(z) = (1 cTf(z))/(f(z)- a). Thus the first fundamental theorem is obtained in a form due to Ahlfors and Shimizu, viz., the following theorem.
+
181
9.69.7 The Ahlfors-Shimizu Characteristic To@)
9.6.2 Theorem. If f ( z ) is meromorphic in I z 1 < R where 0 then for every a finite or infinite, and 0 < r < R , T o@)=
s:
A ( t ) / t dt
= N(r, a )
< R < 00,
+ mo(r, a ) - m,(O,a )
provided f ( 0 ) # a. Proof. Since 1 mo(r, a ) = 276
s
2n
log41
0
+ I F ( z ) l2 d0
and
mo(O,a ) = l o g 4 1
+ I ~ ( 0 1)2 ,
the proof follows for we have applied the previous result (9) to F(z). 9.7
0
We note further, that since
Thus To(r)and T ( r )differ by a bounded term. They may be used interchangeably in most applications. Then To(r) is called the Ahlfors-Shimizu characteristic and T ( r ) is called the Nevanlinna characteristic.
APPENDIX
We list several definitions, theorems, and observations to complement the introductory chapters and leave it to the reader to consult the standard texts for most of the usual introductory definitions and properties of point sets in the complex plane.
Definition. A nonempty open subset A of the complex plane is connected if and only if for every pair of points a, b in A there exists a polygon in A that joins a to b. Definition. A domain is an open connected set. Definition. A domain D is said to be simply connected if every simple closed curve in D contains only points of D in its interior (i.e., D has no “holes” in its interior). Definition. Let f ( z ) be a single-valued function defined in a domain D of the complex plane. Let zo be any fixed point in D. Thenf(z) is said to have a derivative at the point zo if the number defined by
exists, is finite, and is independent of how z + zo.
Definition. f(z) is analytic (holomorphic) at z 182
= zo
if and only if it is dif-
183
Appendix
ferentiable at every point in some neighborhood of zo. Thus f ( z ) is analytic in a domain D if and only if f ( z ) is analytic at every point of D.
Theorem. A necessary and sufficient condition for a function f ( z ) = u(x, y ) iu(x, y ) to be analytic in a domain D is that the four partial derivatives u,, uY, u,, uY exist, are continuous, and satisfy the Cauchy-Riemann conditions
+
u,
= UY,
uY
= - u,
at each point of D.
Definition. Given a multivalued function w = f ( z ) defined in a domain D in the complex plane, let Dobe a domain contained in D.Then by a branch of w in Do,we mean a single-valued function w, = g ( z ) defined in Do, such that for any z in D o , g ( z ) is one of the values of f ( z ) . Definition. C is a continuous arc if C is the set of all ordered pairs ( x , y) such that x = f ( t ) and y = g ( t ) where a 5 t 5 b and f(t) and g ( f ) are continuous. Definition. C is a contour if C is a continuous arc, f ( t ) and g ( t ) have sectionally continuous first derivatives andf(t) and g ( t ) are not zero for the same t.
Definition. C is a simple closed curve if (1) C is a continuous simple arc and ( 2 ) for a 5 i 5 B, z ( a ) = z w ) where z = z ( t ) = f(t) ig(t).
+
Theorem (Jordan curve theorem), Let C be a simple closed curve. Then C divides the plane into two disjoint domains D, and D,. Further, C is the boundary of D, as well as D,.Alternatively, the points not on C form two disjoint domains. Every point of C is an accumulation point of D, as well as of D, . Fina'lly, D,and D,have no boundary points other than the points of c. Theorem. Every power series C&,u,(z - zo)n has a radius of convergence R such that when 0 < R < 00 the series converges absolutely for I z - zo I < R and diverges for 1 z - zo 1 > R . When R = 0, the series
184
Appendix
converges only for z = zo and when R The number R is given by
1 -=lim R
= 00,
the series converges for all z.
qm.
n+m
The series may or may not converge on the “circle of convergence,” (z-zOI = R.
Definition. A sequence of functions {Sn(z)} defined on a set R is said to conuerge unijbrmly to a function S ( z ) if, for every E > 0, there exists a positive integer N ( E )depending upon E only, such that for all points z in R
I S,(z)
- S(z)
I <E
for all n
> N(E).
Theorem. Let the power series CLoa,(z - zo)n have a nonzero radius of convergence R. For any circle P, center zo and radius r < R, the power series C&,a,(z - z0)n converges uniformly within and on r. Theorem (Taylor series). Let f(z) be analytic within a circle C, = R. Then at each point z interior to C,
I z - zo I
The series converges and hasf(z) as its sum function.
Theorem (Laurent series). Let S be the region bounded by the concentric circles C, and C, with center zo and radii rl and r 2 , respectively, rl < r 2 . Letf(z) be analytic within S and on C, and C,. Then at each point z in the interior of S, f ( z ) can be represented by a convergent series of positive and negative powers of (z - zo),
where
a,
=-
n = 0 , 1,2,
...
and bn = -
n = 1,2,
...
the integral along C, and C, being taken in the positive direction.
SUGGESTIONS FOR FURTHER READING
Boas, R. P., Jr. “Entire Functions.” Academic Press, New York, 1954. Branges, L. de, “Hilbert Spaces of Entire Functions.” Prentice-Hall, Englewood Cliffs, New Jersey, 1968. Cartwright, M. L. “Integral Functions.” Cambridge Univ. Press, London and New York, 1962. Gross, F., ed., Proc. NRL Conf Classical Function Theory. Math. Res. Center, Naval Res. Lab., Washington, D. C., 1970. Hayman, W. K. “Meromorphic Functions.” Oxford Univ. Press (Clarendon), London and New York, 1964. Markushevich, A. I. “Entire Functions.” American Elsevier, New York, 1966. Tsuji, M. “Potential Theory in Modern Function Theory.” Maruzen, Tokyo, 1959. Valiron, G. “Theory of Integral Functions.” Chelsea, New York, 1949. Levin, B. Ja. Distribution of Zeros of Entire Functions, A.M.S. Translations of Math. Mono. No. 5, 1964.
185
BIBLIOGRAPHY
The following moderately expanded bibliography covers not only the subjects touched on in this monograph, but extensions of theorems proved herein. The subject of entire functions is far too wide to be able to produce an exhaustive set of references, however an attempt has been made to give a reasonably comprehensive cross section of papers and books relating to the subject.
S. AGMON, (1) “Functions of exponential type in an angle and singularities of Taylor series.” Trans. Amer. Math. SOC.70 (1951) 492-508.
AHIEZER, N. 1. (AKHYESER, AKHIEZER, ACHYESER, etc.) (1) “Sur les fonctions entikres d‘ordre entier.” Rend. Circ. Mat. Palerrno (I) 51 (1927) 390-393. (2)
“On some properties of integral transcendental functions of exponential type.” Izv. Akud. Nauk SSSR Ser. Math. 10 (1946) 411-428 (Russian; English summary).
(3) “Lectures on the Theory of Approximation.” Ogiz, Moscow and Leningrad, 1947 (Russian); (translation: Ungar, New York, 1956).
“On the theory of entire functions of finite degree.” Dokl. Akud. Nauk SSSR (N.S.) 63 (1948) 475-478 (Russian). (5) “On the interpolation of entire transcendental functions of finite degree.” Dokl. Akad. Nauk SSSR (N.S.) 65 (1949) 781-784 (Russian). (6) “The work of Academician S. N. Bernstein on the constructive theory of functions.” Usp. Matem. Nauk (N.S.) 6, No. 1 (41) (1951) 3-67 (Russian). (7) “On entire transcendental functions of finite degree having a majorant on a sequence of real points.” Izv. Akud. Nauk SSSR. Ser. Mat. 16 (1952) 353-364 (Russian). (4)
186
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INDEX A
Cauchy-Riemann equations, 183 Circle of convergence, 184 Closed curve, 183 Compact set, 118 Connected set, 182 Contiguous paths, 161 Continuous arc, 183 Contour, 183 Convex function, 20, 21, 22 Convex hull, 99 Convex set, 9 6 9 7
Ahlfors, L. V., 161 Ahlfors-Shimizu characteristic, 176-18 1 Ahlfors-Shimizu theorem, 180-181 Algebraic function, 14 Al-Katifi, W., 162 Analytic function, 82-1 83 a-points, 145-146 A ( r ) , 179 Asymptotic values, 159- 162
B
D
Bessel’s function, 92-93 Beta function, 85 Bloch, A., 146 Boas, R., 115 Bohr, H., 80 Bolzano-Weierstrass theorem, 17 Borel and Carathkodory, theorem of, 53-55 Borel exceptional value, 146
Dense set, 120 Denjoy, A., 161 Domain, 182 simply connected, 182
E Entire function, 5 a-points, 145-146 asymptotic values, 159-162 of an entire function, 81-82 exponent of convergence of, 65-66, 7173 genus of, 74 infinite product expension of, 29 with real zeros only, 109-115
C Canonical product, 59 Carleman’s theorem, 48-52 application of, 52 Cartwright, M. L., 124, 129, 149 Cauchy inequality, 10 21 9
220
Index
Entire functions (continued) transcendental, 5 type, see Type of entire function Evgrafov, M. A., 15 Exponent of convergence, 65, 66, 71-73, see also Entire function
F Functions of infinite order, 144 with real negative zeros, 113-115 with real zeros only, 109, 113-115 of zero order, 148-149 Fundamental theorem of algebra (1.13.4), 18 G
Gamma function, 83-92 analytic continuation of, 88-90 asymptotic behaviour of, 86-88 conjugate values of, 92 Gauss representation of, 91 infinite product representation of l / r ( z ) , 91 order of I / ~ ( z ) , 90 Genus (rank) of canonical product, 66-67 H
Hadamard’s factorization theorem, 68-69 Hadamard’s three-circle theorem, 19-20 Hayman, W. K., 117, 144, 163 Heine-Bore1 theorem, 118-1 19 Hurwitz, theorem of, 30-31, 111, 112 I
Infinite product, 22-25 absolute convergence of, 23 convergence of, 22 divergence of, 22 uniform convergence of, 25 J
Jenkins, J. A., 157 Jensen’s theorem (formula), 43-48, 163 Jordan arc, 183 Jordan curve theorem, 183 Julia, lines of, 149
K
Kjellberg, B., 144 k(w, a), k ( w , a), 179-180
L Laguerre, theorem of, 96-97 Landau, theorem of, 155-1 56 Laurant’s theorem (series), 184 Levinson, N., 52 Liouville, theorem of, 10 log+a, 164 Lucas, theorem of, 97, 99-100
M Marden, M., 96, 100 Maximum modulus, 6 of coefficients of power series, 9 of entire function, 12 growth of, see Order in unbounded subdomain, 62 of polynomial, 1 1 Maximum modulus theorem, 8 Meromorphic function, 4 characterization of, 102 expansion of, 27 factorization of, 175-176 of finite nonintegral order, 148 of finite positive integral order, 148 order of, 174 poles of, 28, 101 rational, 4 transcendental, 4, 39 of zero order, 148-149 zeros of, 28 Minimum modulus, 115-1 18, 133-144 Minimum modulus principle, 8-9 Mittag-Leffler theorem, 102-105 Montel’s theorem, 122-123 m(r, a), 164, 170 m ( r , f ) , 165 m(r, m), 165 mdr, a), 180 mdr, -), 180
N Nested interval property, 17 Nevanlinna, R., 175
22 1
Index characteristic function T(r), 9, 168, 181 convexity of T(r), 168-170 first fundamental theorem, 168 n(r), 63-64 n(r, a), 164-175 n(r, m), 164 N(r, a), 164-175 N ( r , f ) , 164 N(r, W), 164 N(r, m), 164 0 Order, 15-16 of Bessel function, 92-93 of canonical product, 70 of derived entire function, 95-96 of entire function, 15-16, 33-39, 59-61, 73-76 of l / r ( z ) , 90 of meromorphic function, 174175 of poles, 3, 157 of power series representation of entire function, 74-76 of product and sum of entire functions, 6Cb61 of quotient of entire functions, 71 of zeros, 13-14
R Residues, 26-27 RouchC’s theorem, 30
S Schottky’s theorem, 150-155 Schwarz’s lemma, 52-53 Sequences of functions, 118-123 Simple arc (Jordan arc), 183 Singularities essential, 3 4 , 157 isolated, 1, 157 removable, 2 Stirling’s formula, 86-88
T Taylor’s theorem, 184 Titchmarch, E. C., 52, 115, 145 T(r), 168-181 Tu(r), 180-181 Transcendental number, 14 TY Pe of entire function, 61-62, 78-79 finite, normal or mean, 61 infinite or maximum, 61 of power series, 78
P Path of finite determination, 159 of finite indetermination, 159 of infinite determination, 159 Phragmtn-Lindelof function h(O), 129-1 33 characteristic behavior of, 131 continuity of, 133 Phragmtn-Lindelof theorems, 124-129, 159-1 60 Picard, M. E., 3 exceptional value, 146, 161 great (second) theorem, 149, 157-159 little (first) theorem, 149, 155 theorems, 31-39 Poisson’s integral formula, 42-43 Poisson-Jensen formula, 4748, 174 Pblya, G., 80, 137 Primary factors, 59 Principal part, 3
U Uniform convergence of sequence, 184 Uniformly bounded set, 120 Univalent (Schlicht) function, 120
V Vitali’s convergence theorem, 121-122
W Watson, G. N., 83 Weierstrass, K., 3 Weierstrass factorization theorem, 56-59 Whittaker, E. T., 83 Z
Zeros, order of, 14
Pure and Applied Mathematics A Series of Monographs and Textbooks Editors Paul A. Smith and Samuel Ellenberg
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