Introduction to the Theory of Entire Functions

INTRODUCTION TO THE THEORY OF ENTIRE FUNCTIONS This is Volume 56 in PURE AND APPLIED MATHEMATICS A series of Monograp...

Author: Anthony S.B. Holland

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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

COPYRIGHT (D 1973. BY ACADEMIC PRESS, INC. ALL RIGHTS RESERVED. NO PART OF THIS PUBLICATION MAY BE REPRODUCED OR TRANSMITTED IN ANY FORM OR BY ANY MEANS, ELECTRONJC OR MECHANICAL, INCLUDING PHOTOCOPY, RECORDING, OR ANY INFORMATION STORAGE AND RETRIEVAL SYSTEM, WITHOUT PERMISSION IN WRITING FROM THE PUBLISHER.

ACADEMIC PRESS, INC.

111 Fifth Avenue, New York, New’ York 10003

United Kingdom Edition published by ACADEMIC PRESS, INC. (LONDON) LTD. 24/28 Oval Road. London NWI

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

This Page Intentionally Left Blank

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 R , O I R =

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


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

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

 ro(e),

Iz I I r,

Similarly, for a point z, on

IzI

=

r

12


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 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.)


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.


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


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


[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


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 , )


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 }
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


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


2.7.3 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


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


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


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


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


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


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


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 #JI

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
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.


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


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


+

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


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 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 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 )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,,


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.


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,
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


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 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


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


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


[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


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


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


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


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,

-


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)


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

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 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


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


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


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


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@,

102


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 ) )= 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


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

~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


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


+

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


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


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


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 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


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.


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 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 # 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 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,


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.


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


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


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


< 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-,
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


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


thus the inequality is true if the suffixes 1 and 2 are interchanged. Hence

where Kl depends on

M(e) <
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


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


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


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

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


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 1, i.e.,

- r cos 8 > 0,

r cos 0

< 0,

cos 8 < 0,

742

< 8 < 3n/2

172


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


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.


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


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)

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AHLFORS, L. V.,and GRUNSKY, H. (1) “Uber die Blochsche Konstante.” Math. Z. 42 (1937)671-673. AHLFORS, L., and HEINS,M. (1) “Questions of regularity connected with the Phragmtn-Lindelof principle.” Ann. Math. (2)50 (1949) 341-346. 0

AHLFORS,L. V., and SARIO, L. (1) “Riemann SurQces.” Princeton Univ. Press, Princeton, New Jersey (1960). ALANDER, M. (1) “Sur les fonctions entitres non-rklles.” Ark. Mat., Astron. Fysik 18 (1924) 1-9. AL-KATKFI,W. (1) “On the asymptotic values and paths of certain integral and meromorphic functions.” Proc. London Mafh. SOC.(3) 16 (1966) 599634. AMIRA,B. (1) “Sur un thtorkme de M. Wiman dans la thtorie des fonctions entitres.” Math. Z. 22 (1925)206-221.

ARIMA,K. (1) “On maximum modulus of integral functions.” J. Math. SOC.Japan 4 (1952),62-66. BAKER, I. N. (1) “Sets of non-normality in iteration theory.” J. London Math. SOC.40 (1965)499-502. (2) “The distribution of fixpoints of entire functions.” Proc. London Math. SOC.(3) 16 (1966)493-506.

(3) “Zusammensetzungen ganzer Funktionen.” Math. Z. 69 (1958) 121-163. (4) “The existence of fixpoints of entire functions.” Math. Z. 73 (1960) 280-284. (5) “Repulsive Fixpoints of Entire Functions.” Math. Z. 104 (1968) 252-256. (6) “Permutable entire functions.” Math. Z. 79 (1962)243-249.

188 (7) (8) (9)

Bibliography “On some results of A. Rknyi and C. Rknyi concerning periodic entire functions.” Acta Sci. Mafh. 27 (1966) 197-200. “Entire functions with linearly distributed values.” Math. Z. 86 (1964) 263-267. “Fixpoints of polynomials and rational functions.” J . London Math. SOC.39 (1964)

615-622. (10) “Fixpoints and iterates of entire functions.” Mafh. Z. 71 (1959) 146153. (1 1) “Some entire functions with fixpoints of every order.” J. Ausfr. Math. SOC.(to ap-

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BAKER,I. N., and GROSS,F. (1) “On Factorizing Entire Functions.” Proc. London Mafh. SOC.(3) 18 (1968) 69-76. BELLMAN, R. (1) “A generalization of a Zygmund-Bernstein theorem.” Duke Mafh. J. 10 (1943) 649451.

BERNSTEIN, S. “Sur une proprittk des fonctions entitres.” C. R. Acad. Sci. Paris 176 (1923) 16031605.

‘‘LeGons sur les propriktks extremales et la meilleure approximation des fonctions analytiques d‘une variable rkelle.” Gauthier-Villars, Paris, 1926. “Sur la meilleure approximation de I x JP par des polynbmes de degrks trts klevks.” Izv. Akad. Nauk SSSR Ser. Mat. (1938), 169-190 (Russian and French). “Sur la borne supkrieure du module de la dkrivke d’une fonction de degrk fini.” Compt. Rend. Dokl. Acad. Sci. URSS (N.S.) 51 (1946) 567-568. “Sur la meilleure approximation sur tout I’axe reel des fonctions continues par des fonctions entikres de degrk fini, I, 11, 111, IV, V.” Compt. Rend. Dokl. Acad. Sci. URSS (N.S.) 51 (1946) 331-334, 487490; 52 (1946) 563-566; 54 (1946) 103-108, 476-478.

(10)

(11) (12) (13)

(14)

“Dtmonstration nouvelle et gknkralisation de quelques formules de la meilleure approximation.” Compt. Rend. Dokl. Acad. Sci. URSS (N.S.) 54 (1946) 663-664. “On limiting relations among constants of the theory of best approximation.” Dokl. Akad. Nauk SSSR (N.S.) 57 (1947) 3-5 (Russian). “On properties of homogeneous functional classes.” Dokl. Akad. Nauk SSSR (N.S.) 57 (1947) 11 1-114 (Russian). “Limit laws of the theory of best approximation.” Dokl. Akad. Nauk SSSR (N.S.) 58 (1947) 525-528 (Russian). “A second note on homogeneous functional classes.” Dokl. Akad, Nauk SSSR (N.S.) 59 (1948) 1379-1384 (Russian). “A generalization of an inequality of S. B. Stechkin to entire functions of finite degree..” Dokl. Akad. Nauk SSSR (N.S.) 60 (1948) 1487-1490 (Russian). “The extension of properties of trigonometric polynomials to entire functions of finite degree.” Izv. Akad. Nauk SSSR. Ser. Mat. 12 (1948) 421-444 (Russian). “Remarks on my paper “The extension of properties of trigonometric polynomials to entire functions of finite degree.” Izv. Akad. Nauk SSSR. Ser. Mat. 12 (1948) 571-573 (Russian). “On additive majorants of finite growth.” Dokl. Akad. Nauk SSSR ( N . S . ) 66 (1949) 545-548 (Russian).

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Introduction to the Theory of Entire Functions

Introduction to the theory of entire functions

Lectures on entire functions

Entire and meromorphic functions

Introduction to Arithmetic Theory of Automorphic Functions

Introduction to Arithmetic Theory of Automorphic Functions

Lectures on Entire Functions

Lectures on entire functions

Introduction to the theory of algebraic functions of one variable

Introduction to The Theory of Functions of a Complex Variable

Introduction to the Theory of Algebraic Functions of One Variable

Distribution of zeros of entire functions

Entire and Meromorphic Functions (Universitext)

Entire and Meromorphic Functions (Universitext)

Entire and Meromorphic Functions (Universitext)

Introduction to the arithmetic theory of automorphic functions

An introduction to the theory of real functions

An Introduction to the Theory of Real Functions

Introduction to the arithmetic theory of automorphic functions

An Introduction to the Theory of Stationary Random Functions

Introduction to the Theory of Algebraic Numbers and Functions

An Introduction to the Theory of Real Functions

Introduction to the Arithmetic Theory of Automorphic Functions

An introduction to the theory of stationary random functions

An Introduction to the Theory of Real Functions

An Introduction To The Theory Of Multiply Periodic Functions

Introduction to the arithmetic theory of automorphic functions

Introduction to Bessel Functions

Introduction to G-functions

Introduction to Bessel Functions

The theory of functions

Introduction to the Theory of Entire Functions