Complex Function Theory

COMPLEX FUNCTION T HEOR Y Pure and Applied Mathematics A Series of Monographs and Textbooks Edited by Paul A. Smith a...

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COMPLEX FUNCTION T HEOR Y

Pure and Applied Mathematics A Series of Monographs and Textbooks Edited by Paul A. Smith and Samuel Ellenberg

Columbia University, New York

1 : ARNOLD SOMMERFELD. Partial Differential Equations in Physics. 1949 (Lectures on Theoretical Physics, Volume VI) 2 : REINHOLD BAER.Linear Algebra and Projective Geometry. 1952 3 : HERBERT BUSEMANN A N D PAUL KELLY.Projective Geometry and Projective Metrics. 1953 4 : STEFANBERGMAN A N D M. SCHIFFER. Kernel Functions and Elliptic Differential Equations in Mathematical Physics. 1953 5 : RALPHPHILIP BOAS,JR. Entire Functions. 1954 BUSEMANN. The Geometry of Geodesics. 1955 6 : HERBERT 7 : CLAUDECHEVALLEY. Fundamental Concepts of Algebra. 1956 8 : SZE-TSENHu. Homotopy Theory. 1959 9 : A. M. OSTROWSKI. Solution of Equations and Systems of Equations. Second Edition. 1966 Foundations of Modern Analysis. 1960 10: J. DIEUDONNB. 11 : S. I. GOLDBERG. Curvature and Homology. 1962 HELGASON. Differential Geometry and Symmetric Spaces. 1962 12 : SIGURDUR Introduction to the Theory of Integration. 1963 13: T. H. HILDEBRANDT. ABHYANKAR. Local Analytic Geometry. 1964 14: SHREERAM L. BISHOPA N D RICHARD J. CRITTENDEN. Geometry of Manifolds. 1964 15 : RICHARD 16: STEVEN A. GAAL.Point Set Topology. 1964 17: BARRYMITCHELL.Theory of Categories. 1965 18: ANTHONYP. MORSE.A Theory of Sets. 1965

Pure and Applied Mathematics A Series of Monographs and Textbooks

19 : GUSTAVE CHOQUET. Topology. 1966 A N D I. R. SHAFAREVICH. Number Theory. 1966 20: Z. I. BOREVICH 21 : JosB LUISMASSERA A N I ) J U A NJORGE ScrrAI'FER. Linear Differential Equations and Function Spaces. 1966 22 : RICHARD D. SCHAFER. An Introduction to Nonassociative Algebras. 1966 23: MARTINEICIJIXR. Introduction to the Theory of Algebraic Numbers and Functions. 1966 24 : SHREERAM A R H Y AKNA I L Resolution of Singularities of Embedded Algebraic Surfaces. 1966 TREVES. Topological Vector Spaces, Distributions, and Kernels. 1967 25 : FRANCOIS 26: PETER D. L A Xand R A L P H S. PHILLIPS.Scattering Theory. 1967 27: OYSTEINORE.The Four Color Problem. 1967 28 : MAURICEHEINS.Complex Function Theory. 1968

in trrparaiion: R. M. BLUMENTHAI, A N D R. K. GETOOR. Markov Processes and Potential Theory. HANSFREUDENTHAL A N D H. DE VRIES.Linear Lie Groups.

This Page Intentionally Left Blank

COMPLEX FUNCTION THEORY Maurice Heins DEPARTMENT OF MATHEMATICS UNIVERSITY OF ILLINOIS

URBANA, ILLINOIS

1968

A C A D E M I C P R E S S New York and London

COPYRIGHT 0 1968, BY ACADEMIC PRESSINC. ALL RIGHTS RESERVED. NO PART OF THIS BOOK MAY BE REPRODUCED IN ANY FORM, BY PHOTOSTAT, MICROFILM, OR ANY OTHER MEANS, WITHOUT WRITEN PERMISSION FROM THE PUBLISHERS.

ACADEMIC PRESS INC.

1 I 1 Fifth Avenue, New York, New York 10003

United Kingdom Edition published by ACADEMIC PRESS I N C . (LONDON) LTD. Berkeley Square House, London W. 1

LIBRARY OF CONGRESS CATALOG

CARD

NUMBER: 68-16518

PRINTED IN THE UNITED STATES OF AMERICA

PREFACE

24 nueeiv

pue&iv

This book is in two parts. The first is intended to serve as a basis for a first course on complex function theory for both undergraduate and graduate students. It is not expected that the whole of Part One will be “ covered ” in a semester. Experience and realism would say that the larger part of the first seven chapters is manageable.Part One can constitute a semester course when an adequate level of maturity is available. Chapter IX, the final chapter of Part One, culminates in the proof of the prime number theorem of Hadamard and de la VallCe Poussin-a particularly handsome application of topics treated in earlier chapters. I dare say that a syllabus consisting of the material presented in Part One forms a reasonable and, in fact, mathematically appealing program that can serve as a basis for the requirements in complex function theory for all doctoral candidates in mathematics at present. Part Two seeks to supplement and round out the exposition for students who are to use complex function theory seriously in their professional work. It should serve as a foundation for subsequent courses in the area of complex analysis on such topics as the Nevanlinna theory, Riemann surfaces, several variable theory, and differential equations. It may be used for a proseminar in analysis for first- or second-semester graduate students; or taken in conjunction with chs. VlII and IX it could serve as the basis of a second-semester course on complex function theory in which active student participation is a predominant note. I have envisaged the role of Part Two as twofold: to increase the student’s experience with the ideas and methods of Part One and to exploit the applicability of these methods.

Viii

Preface

It is my intention to be specific about material from other disciplines which will be used in this book. In particular, this will be the case for the theorems of elementary real analysis, including the differential calculus. Precisely here more than anywhere else the instructor of a first-semester course on complex function theory encounters his greatest tribulations. The scope of elementary courses on real analysis is so varied that frequently students of excellent ability come to their first course on complex function theory with a less than rudimentary grasp of the basic and most elementary theorems of the one and several variable real differential calculus. I have tried not t o presume on the reader’s background for material from other fields. Exceptions are: the representation of a positive integer as a product of primes in Ch. VIII for which see [14]; improper integrals in Chapters V, VII, IX, XIV for which see [32a]; and standard information from the elements of linear algebra in Chs. XII, XVIII for which see [58]. It is reasonable to suppose that the reader will be familiar with these questions. The student should ultimately recognize that four principal methods dominate complex function theory, methods closely associated with the founders of the subject: the power series approach, the complex integral approach embracing the Cauchy theory in its full range, the approach based on the connection with the theory of harmonic functions, and the mapping approach. The first method has its roots in the work of the eighteenth century and was developed with grandeur by Weierstrass. The last two methods find their sources in the work of Gauss and Riemann. At the beginning of the study of complex function theory the Cauchy theory stands out because of its dazzling virtuosity. The status of complex function theory has changed greatly during this century. At the beginning it had many triumphs. Today they are fewer.It is also characteristic that complex function theory does not appear in splendid isolation but rather along with other branches of mathematics as a component interrelating with other components of an organic whole. The proud vaunt of “ rein funktionentheoretisch ” belongs to an era that is past. At the stage of the student’s development when he encounters complex function theory, he will almost inevitably be subjected to bouts of rigor, to “rectitude with exactitude.” It is in the nature of things and doubtless it would be a mistake to put aside such aspects of mathematical training. However, I hope the student will seek a light touch. I wish to acknowledge the influence of the books of Carathkodory [21] and. of Saks and Zygmund [108]. The book of Saks and Zygmund is a model for exposition of the highest standard which is free of refinement profitless at this level. Special thanks are due to Professor F. M. Stewart for having communicated to me earlier his treatment of the Cauchy theorem.

Preface

ix

There remains for me to express my thanks to Professors Eilenberg and Smith for their long-standing invitation to write a volume on complex function theory for their series, to the University of Illinois for affording me the opportunity to write this book, and to Academic Press for its helpful cooperation.

January 1968 Urbana, Illinois

MAURICEHEINS


CONTENTS

vii

PREFACE

PART ONE

Chapter I. The Real Field Prefatory Remarks 1. Sets 2. Groups, Rings, Fields

3. Ordered Fields The Real Number System Real Sequences, Monotone Sequences Limit Superior, Limit Inferior Cauchy Sequences

4. 5. 6. 7.

3 3 5 7 9 12 15 16

Chapter II. The Complex Field. Limits Prefatory Remarks 1. The Complex Field

2. Geometry of the Complex Plane 3. Limits 4. Infinite Series 5. Power Series

17 18 21 24 27 29

Contents

Xii

Chapter III. Topological and Metric Spaces 1. 2. 3. 4. 5. 6.

Prefatory Remarks The Notion of a Topological Space Compactness Connectedness Arc Metric Space Compactification

38 39 43 46 49 50 53

Chapter IV. Complex Differential Analysis Prefatory Remarks 1. Complex-Valued Functions of One Real Variable 2. Complex-Valued Functions of One Complex Variable 3. Power Series 4. Further Properties and Applications of the Exponential Function, Trigonometric Functions 5. Fourier Aspects of Power Series 6. Functions of Two Real Variables 7. The Cauchy-Riemann Equation(s) 8. Harmonic Functions 9. The Theorem of Gauss on the Zeros of the Derivative of a Polynomial

57 58 62 63 69 75 79 85 87 91

Chapter V. Cauchy Theory Prefatory Remarks 1. Line Integrals 2. The Simplicia1Cauchy-Goursat Theorem 3. The Cauchy-Goursat Theorem for a Singular Cell and Some 4. 5. 6. 7. 8. 9.

Consequences Simple Connectivity, Primitives Analytic and Continuous Logarithms and Powers Certain Integrals Representing Analytic Functions Some Sufficient Conditions for Analyticity Runge’s Theorem Mapping Properties of Analytic Functions

94 95 104 109 124 127 131 139 140 143

Chapter VI. Laurent Expansion. Meromorphic Functions Prefatory Remarks 1. Laurent Expansion

2. Isolated Singularities of Analytic Functions 3. Residue 4. Meromorphic Functions

150 151 158 164 167

Confenfs

xiii

Chapter VII. Further Applications of the Cauchy Theory Prefatory Remarks 1. The Residue Theorem 2. Application of the Residue Theorem to the Calculation of Integrals 3. Application of the Method of Cauchy

182 183 186 191

Chapter VIII. The Zeros and Poles of Meromorphic Functions Prefatory Remarks The Theorems of Mittag-Leffler and Weierstrass The Isomorphism Theorem of Bers Infinite Products Sums and Products 5. The Jensen Formula 6. Entire Functions of Finite Order I. Bounded Analytic Functions

1. 2. 3. 4.

1 94 196 200 204 210 21 3 216 224

Chapter IX. The Gamma and Zeta Functions. Prime Number Theorem Prefatory Remarks 1. The Gamma Function 2. The Riemann Zeta Function and Its Functional Equation 3. The Theorem of Hadamard and de la Vallke Poussin on the Zeros of the Riemann Zeta Function 4. The Prime Number Theorem of Hadamard and de la Vallb Poussin

229 230 231 242 243

PART TWO Some General Observations

251

Chapter X. Supplementary Developments Concerning C Prefatory Remarks 1 . Other Constructions of C 2. Characterizations of Absolute Value

3. Geometry in C

253 253 254 254

Chapter XI. Complex Differential Coefficients Prefatory Remarks 1. Complex Differential Coefficients

2. Conformality 3. Complex Differential Coefficientsand Harmonic Functions

260 260 262 264

Contents

XiV

Chapter XII. Topics in the Theory of Power Series 1. 2. 3. 4. 5. 6.

Prefatory Remarks Formal Aspects Sums, Cauchy Multiplication Holomorphic Functions of Two Complex Variables Cauchy’s Majorant Calculus Vector-Valued Functions Analytic Continuation

266 261 268 269 213 218 286

Chapter XIII. Harmonic and Subharmonic Functions Prefatory Remarks 1. Poisson Integral and Applications 2. Weierstrass Approximation Theorem and the Dirichlet Problem for an Annulus 3. Principles of Schwarz and Carleman 4. Subharmonic Functions

294 295

300 302 304

Chapter XIV. Complements to the Cauchy Theory Prefatory Remarks 1. Gaussian Sums 2. Doubly Periodic Functions 3. The Weierstrass Preparation Theorem 4. Runge’s Theorem on Polynomial Approximation 5. Integrals of Cauchy Type, Cauchy Principal Value 6. The Cauchy Integral Theorem for Weierstrass Classes

310 31 1 313 314 317 319 321

Chapter XV. Mobius Transformations Prefatory Remarks 1 . Classification of Miibius Transformations

325 325

Chapter XVI. The Modular Function X. The Picard Theorems Prefatory Remarks 1. The Modular Function h 2. The Big Picard Theorem

329 330 338

Chapter X W . Some Questions Concerning Univalent Analytic Functions Prefatory Remarks 1. The Riemann Mapping Theorem (plane Case)

340 341

Contents

2. 3. 4. 5.

Caratheodory’s Theorem on Variable Regions Application of the Schwarz Reflexion Principle Schwarz-Christoffel Functions Regions with Simple Polygonal Frontiers

xv 343 345 347 351

Chapter X W I . Riemann Surfaces Prefatory Remarks Riemann Surfaces in the Sense of Weyl-Radb Weierstrass Classes, Analytic Entities Compact Analytic Entities and Riemann Surfaces The Problem of Riemann and Klein 5. Distinguishing Meromorphic Functions 6. Existence Theorem for Auxiliary Harmonic Functions 7. Field Isomorphism and Conformal Equivalence of Compact Riemann Surfaces

354 355 358 368 379 380 383

REFERENCES

394

NOTATION

400

INDEX

407

1. 2. 3. 4.

389


PART ONE


BB Chapter I @@

T H E REAL F I E L D

PREFATORY REMARKS It is taken for granted that the user of this book has already studied elementary real analysis. Desiderata for the purposes of complex function theory are : the properties of the real number system, limit processes, sequences, continuity, the Riemann-Stieltjes integral, and the basic theorems of the differential calculus in one and several variables. However, there is considerable variation of opinion concerning what elementary real analysis consists of. Also, the actual amount of material treated in any given class is apt to be very variable even in the presence of a fairly standard syllabus. A period of adjustment between the reader and the book is probably inevitable. In this chapter we give a succinct summary of material that should by and large be well known to the student. The familiar parts should be read for review, attention being given to the exercises. It is my experience that the notions of limit superior and limit inferior are much less commonly known by students than instructors might think. Since these notions are of considerable use in complex function theory, they merit prompt attention. For readers wishing supplementary material we suggest Ch. 1 of [133], Chs. 1 and 2 of [36], and Chs. 1, 2, and 3 of [107]. 1. SETS

The usual material of elementary set theory is so current that we may certainly take it for granted. Thus, without explanation we use as known the 3

4

1. The Real Field

notations a E A , A c B, A u B, A n B. We also use 8 for the empty set and A - B for the set of elements in A but not in B. For the set specified by a sentential function P(x), the set of x such that P(x) holds, we write { x : P(x)} and often the simpler { P ( x ) } . The universe of x in question will be clear from context. We also use curly brackets to indicate singletons and finite sets whose elements are exhibited explicitly. Thus, { a } denotes the set whose sole member is a and {a, b} denotes the set whose members consist exactly of Q and b. We take for granted the notion of an ordered pair (a, b) [following N. Wiener and Sierpitiski we may define (a, b) as { { a } ,{ a , b } } ]and that it has the property that (a, b) = (c, d ) if and only if a = c and h = d.7 A relation is a set of ordered pairs. Given sets A and B, by the Cartesian product of A and B, denoted A x B, is meant the set of ordered pairs (a, b) satisfying a € A , b E B. A function is a relation satisfying the condition that distinct ordered pairs belonging to it have distinct first components. Given a function f and (a, 6) ~ f , we denote b by f ( a ) . The set of the first components of the ordered pairs of a function is called the domain of the function, the second components the tralues, the set of second components the image. A function f is said to be constant provided that f ( x ) =f ( y ) for all x, y in the domain off. When the context renders the matter clear, we shall denote a constant f by the same notation as that used for its unique value when f # 8. Given a function f with domain A and image contained in B, we say that .f maps A into B and denote this assertion by f:A --f B.$ If the image is B itself, we say that f maps A onto B. Given a function f and C , a subset of the domain o f f , by the restriction off to C, denoted f I C, is meant the set of ordered pairs belonging to f whose first components belong to C. Clearly, f I C is a function. By the image of C with respect t o f , denotedf(C), is meant the image off I C . We shall sometimes denote f ( C ) by {f ( x ) : x E C } . Given f : A -+ B, D c B, by thepreimage of D with respect tof, denoted f - ‘ ( D ) , ismeant { x :f ( x ) E D } . A note of caution should be made for the case of numerical-valued functions: it is necessary to distinguish preimage from image with respect to the reciprocal of the given function. Suitable care will be taken notationally.

Composition. Given f: A --f B and g : B -+ C . By the composition of g with o f , is meant the set of ordered pairs (x, gLf(x)]), x E A . It is a map of A into C .

f, denoted g

Znuerse. Given that f is a function. The set g of ordered pairs (b, a), (a, b) ~ f is, not necessarily a function.7 In fact, g is a function if and only if Similarly we take for granted the notions of ordered triples (a,b, c) and ordered quadruples (a,b, c, d ) . We shall also allow “f: A -+ B” to read “f mapping A into B.” Ti g is the inverse of the relation f.

2. Groups, Rings, Fields

5

,f is univalent [or in other words (1, l)], that is, iff ( x ) =f ( y ) implies x = y . In this case g is termed the function inverse t o f , or more simply the inverse off.

It will be denoted byf-’. It is to be noted that in the case of a univalent map of a set A onto a set B, the preimage of C( c B ) is the image of C with respect to f -1. Union and intersection of families of sets. Given a set A , by ‘$(A) we denote the set whose elements are the subsets of A , the set of parts of A . Given cp: A + V ( A ) , by Uleh cp(1) is meant the set of x belonging to cp(1)for some 1 E A and by cp(1) is meant the set of x belonging to ~ ( 1for ) all , IE A. Commonly cp(1) is designated indicially by a notation such as X , .

nAEh

Unions of functions. It may occur that each cp(1) is a function, say fA. It f, .For the purposes of complex makes sense set-theoretically to consider UAEA function theory the operation of taking unions of functions is a particularly convenient means of constructing functions from “ pieces.” For this reason the answer to the following question is wanted. When is the union a function? We leave for the reader the treatment of this question in Ex. 1.1 below.

EXERCISES 1.1. In the situation just considered a necessary and sufficient condition that

UAeAfi be a function is that whenever a belongs to the domains of fi and fu, then

fXa) =f,(a).

1.2. The map L of A onto itself given by L ( X ) = x, x E A , is termed the identity map of A onto itself. Show that i f f is a univalent map of A onto B, then f 0 f = L and f 0 f -’ is the identity map of B onto itself.

A matter of notation. It frequently occurs that we specify a function f by declaring its domain A and the element f ( x ) that f assigns to x E A . One speaks of the function f “assigning f ( x ) to x E A” and more briefly (and inexactly) of “ the function f ( x ) with domain A,” “the function f(x), x E A,” or simply “ the function f (x),” such licenses being taken when the facts are clear from context.

2. GROUPS, RINGS, FIELDS

We consider a set G( # 0) and a map * of C x G into G and we denote the image of (a, b) with respect to * by ab or a * b. Such a map is termed a binary operation on G. The composite (G, .) is termed a group provided that (1) the associative law (ab)c = a(bc), a, b, c E C , holds, and (2) the equations ax = b and y a = 6, a, b E G, admit solutions in G. It belongs to the very first lines of

I. The Real Field

6

group theory to show that the solutions are unique and further that there exists a unique element e (neutral element, identity) such that ae = ea = a, a E G . The equations ax = e and ya = e have a common solution. It is termed the inverse of a and is denoted a - ' . It is easily seen that given a E G the map taking x into ax is a univalent map of G onto itself. The same holds for the map taking x into xa. The term " group " will be applied to G (the underlying set) as well as to (G, The elements of G are consequently called elements of the group. The tenor of this remark remains in force for all structures (algebraic and topological) which we shall meet. If, in addition, the group operation is commutative, that is, ab = ba, a, b E G, then the group is termed abelian (after the Norwegian mathematician N. H. Abel). In the case of abelian groups the additive notation is frequently (but not invariably) used, the group operation being written " + " rather than "*," the neutral element being termed zero and denoted 0, the solution of a + x = b being denoted b - a, and the inverse of a being denoted - a . Given a group (G, .), a nonempty subset H of G is termed a subgroup provided that ( H , I H x H) is a group. This is the case if and only if ab-' E H whenever a, b E H . If A c G, the intersection of all subgroups of G which contain A is itself a subgroup of G. It is termed the subgroup of G generated by A . Groups of various kinds will appear in our study of complex function theory. One situation of interest is the following. Let A be a set and let G denote the set of univalent maps of A onto itself (automorphisms of A ) . We define the operation to be composition. Then (C, is a group, the group of automorphisms o f A . In concrete situations we frequently encounter subgroups of a group of automorphisms which are of particular interest. On several occasions in this book we shall meet such subgroups. Given A(# 8) and binary operations + and * on A, we say that the composite (A, +, is a ring provided that ( A , +) is an abelian group and that, in addition, (1) * is associative, and (2) the distributive laws a(b + c) = ab + ac, (b + c)a = ba + ca, a, b, c E A , hold. If, in addition, * is commutative, A is termed a commutative ring. It is an easy consequence of the distributive laws that Oa = a0 = 0, a E A . The definitions for subring and the subring generated by a given subset of a ring parallel those for groups and need not be stated explicitly. The same observation holds for the following paragraph. A commutative ring A is termed afield provided that A contains more than one element and that the equation ax = b, a( # 0), b E A , admits a solution (necessarily unique) in A. The solution is denoted b/a. A field A contains a unique element 1( # 0) satisfying la = a, a E A . It is called the identity (element) of A . If a, h E A , then ah = 0 if and only if at least one of the elements a, b is 0 . If a E A and a # 0, then the unique solution of ax = 1 is not 0. It is termed the reciprocal of a and is denoted by a-l as well as by I / a . a).

0)

e )

7

3. Ordered Fields

We note that A - ( 0 ) together with * restricted to the Cartesian product of A - ( 0 ) with itself is an abelian group.

3. ORDERED FIELDS Given a field A , P( c A ) is said to be a positive part of A provided that (I) P x P is mapped into P by both and and (2) the sets {0}, P, and {x: - x E P} are disjoint and have A as their union. By an ordered$eld is meant a composite ( A , +, *, P) where ( A , - ) is a field and P is a positive part of A . We are concerned for the remainder of this section with an ordered field A . An element p is termed positive provided that p E P; an element n is termed negatioe provided that -n E P. With the aid of the distributive law it is seen that -(ab) = a( -b) and ab = (-a)( - b). It follows readily that if p is positive and n,, n, are negative, then pn, is negative and n,n2 is positive. Noting that 1 = (1)(1) = (- 1)( - l), we see that 1 is positive.

+

+,

EXERCISE 3.1. a2

Prove the following property of an ordered field: Given elements a and b, then

+ b2 is positive or 0. It is 0 if and only if a

= b = 0.

Inequality. Given a, b E A , we say that a is less than b, denoted a < b, provided that b - a E P. We say that a is greater than 6, denoted a > b, provided that b < a.t It is elementary that the law of trichotomy holds: Given a, b E A , then one and only one of the following is valid: a = b, a < b, b < a. We write “ a I b” to denote “ a b or a = b.” The elementary rules of inequalities follow readily by use of the definitions just put down and the notion of a positive part. We give some of the principal rules in the exercises that follow. It is to be understood that a, b, c, d E A , p is positive, n is negative.

EXERCISES 3.2. a > 0 if and only if a is positive. a < 0 if and only if a is negative. 3.3. If a < b, then a T c < b c. 3.4. If a < b and c < d, then (I c < b d.

+

+

+

t It would be better to use terms like “precede ” and need to break with convention.

‘I

follow,” but there is no serious

a

I. The Real Field

3.5. What is the corresponding conclusion when a I b and c 2 d? 3.6. If a 2 0, b 2 0, and a b = 0, then a = b = 0. 3.7. If a < b and b < c, then a < c. What is the corresponding conclusion when a b and b < c ? Show that if a I b, b I c , a = c, then a = b. 3.8. If a < b, then pa nb. 3.9. If 0 a < b and 0 < c < d, then 0 I ac < bd. Consider on the basis of your

+

experience what possibilities may occur in the real number system when the conditions 0 I a and 0 I c are dropped. 3.10. If 0 < a < 6 , then 0 < b-’ < u - l . 3.11. Show that a 2 + b2 2 2ab. When does the case of equality occur? The use of inequalities is extremely frequent in analysis and it is well to secure a firm control early. The cautionary note of Ex. 3.9 is to be heeded. For an authoritative systematic account of inequalities we cite [50]. We continue our review of notions that arise in the study of an ordered field and consider X c A. An element a E A is termed an upper bound of X provided that a 2 x , x E X ; a is termed a lower bound of X provided that a 5 x , x E X. The set X is said to be bounded above provided that there exists a E A which is an upper bound of X . A corresponding definition is made with “ above ” and “ upper ” replaced by “ below ” and “ lower,” respectively. We say that X is bounded provided that Xis both bounded above and bounded below. There is at most one element in X which is an upper bound of X. When such an element exists, it is termed the maximum of X and is denoted max X. A corresponding assertion holds with “ lower bound ” replacing “ upper bound ” and we arrive at the notion of the minimum of a set (min X). Given X , we consider the set Y of upper bounds of X and the set 2 of lower bounds of X.When Y has a minimum, we term min Y the supremum of X and denote it sup X ; when Z has a maximum, we term max 2 the injimum of X and denote it inf X. It is to be noted that the terms “least upper bound” (1.u.b.) and “ greatest lower bound ” (g.1.b.) are employed commonly in English-speaking countries for “ supremum ” and “ infimum,” respectively. If X possesses a maximum, then max X = sup X. Similarly, if X possesses a minimum, then min X = inf X . To say that sup X = + a signifies that X is not bounded above and to say that inf X = - co signifies that X is not bounded below. These formal notations will be seen to go into meaningful assertions when X is a subset of the real number system and one considers the extended real line ($6, Ch. 111). It is to be emphasized that, given a E A, X c A, when we write “ sup X = a,” it is implicit that the above Y does have a minimum. Cognate remarks are to be in force when we deal with notions such as “limit ” and “ limit superior.” The set {a, b } has a maximum and a minimum as follows at once from the alternatives: a I b, a > b. From the point of view of analysis it is convenient to introduce a’, the positive part of a, defined as max{a, 0); a - , the negative part of a, defined as min{a,O}; lal, the absolute value of a,

4. The Real Number System

9

defined as max{a, - a } . It is also convenient to introduce A" = { x 2 0} and A- = {x L O } .

EXERCISES 3.12. Verify the following: u=u++a-,

lal=a+ - u - ,

(U+b)+IU++b+,

+

-lul a. Then q = mln serves. 4.12. If a E R,then

4.10. 4.11.

a= sup{qEQ,qIa}=sup{qEQ,q a) E R. Let f be a continuous map of [a, b] into R; that is, it is supposed that for each x E [a, 61 and each positive 7 there exists 6 positive such that If(y) -f(x)l < 7 when ly - X I < 6. Suppose that f ( a ) < c < f ( b ) . Then there exists d E] a, b[ such that f(d)= c. Prove this well-known theorem of elementary analysis in the following way. Introduce {f(x) 5 c} and show that d = sup{f(x) 5 c } is actually max{f(x) = c}. 4.14. Given n EN, and a k , k =0, ..., n, satisfying at ER+,x ; a t > O . Show with the aid of Ex. 4.13 that there is a unique positive real number x satisfying X " + ' - x t a k x ' = O . Hint: Consider 1 - x " o k X k - ( n + l ) f o r x > O . Apply this result to show that if a E R + and n is a positive natural number, then there exists a unique x E R + such that x" = a . We use the notations a''" and (6 when n = 2) for the unique x in question. 4.15. Let f:R +R, not constant, satisfy 4.13.

vt/a

Show that f is the identity map of R onto itself. Hint: Show that f ( x ) = x on N,Z, Q successively and exploit the fact that if a > 0, then b2 = a for some b E R to conclude that if x < y , then f ( x ) < f ( y ) . p h e result of the exercise says that the only isomorphic map of R into itself is the identity map.] 4.16. Determine the mapsf: R -tR satisfying: ( l ) f ( x y ) = f ( x ) + f ( y ) , x , y :.R, (2) f is continuous at some a E R;that is, for each positive 7 there exists 6 positive such that If(x) -f(a)l < r] when Ix - a1 < 6. 4.17. Suppose that A is a field and that P I and Pzare positive parts rendering A a complete ordered field. Show that PI =PI.

+

5. REAL SEQUENCES, MONOTONE SEQUENCES The sequences in this and the following two sections are assumed t o have domain N and take values in R. We suppose that the reader has encountered

5. Real Sequences, Monotone Sequences

13

the usual elementary material on real sequences, that is, sequences with image in R. Thus, we take for granted familiarity with the definition of limit of a sequence which follows: Given a real sequence (a,,)and a E R, we say that (a,,) has limit a provided that for each positive q there exists m E N such that la - a,,l < q for n 2 m. In addition, we assume familiarity with such matters as uniqueness of limit; that is, if (a,,) has limit /3 also, then a = /3; the usual notation a,, (also lim a,,); the definitions for sums, (a,,) (b,,)= (a,, + b,,), and products (a,,)(b,,)= (a,,b,,)endowing the set of real sequences with the structure of a commutative ring; and the associated limit theorems for sums, differences, products, and quotients. Possibly less familiar but no less immediate is the result that if (a,,)has limit a, then every subsequence of (a,,) has limit a, which follows from the definition of limit and the fact that if v : N --f N and is increasing, then v(n) 2 n, n E N. The notion of a limit may be viewed in the following manner. Let 2 denote a ) where (a,,) is a real sequence, a E R, and the set of ordered pairs ((a,JnPN, (a,,) has limit a. Then 2 is afunction and lim a,, is just 2((a,,)).It might be well to note that implicit in taking for granted that “ lim a,, ” is meaningful is the assumption that (a,) belongs to the domain of 2,that is, that for some a E R the sequence (a,,) has limit a. These remarks hold elsewhere in the study of limits. Corresponding observations hold for limit superior and limit inferior ($6, Ch. I). In this section we consider the existence of limits of monotone sequences. A real sequence (a,) is said to be increasing provided that a,,, < a,, whenever m < n. It is said to be never decreasing provided that a,,, I a,, whenever m < n. Corresponding definitions are made for decreasing (a,,, > a,, whenever m < n) and never increasing (a, 2 a,, whenever m < n). We write “(a,,)?” to denote “ (a,,) never decreasing ” and “ (a,,)l” to denote “ (a,,) never increasing.” A sequence is termed monotone provided that it is either never decreasing or never increasing. We consider (a,,)?. There are two possibilities to consider. One is that {a,,: n E N } is not bounded above. Here for each b E R there exists m such that a,,, > b and consequently a,, > b for n 2 m. Hence a,, = + oo.t The other possibility is that {a,,: n E N} is bounded above. By the completeness property of R,the set {a,,:n E N } has a supremum a. Given positive q, a - q is not an upper bound of {a,,: n E N } . Hence a,,, > a - q for some m E N and consequently

+

a - q b, n 2 m .

=

+ co,”

I. The Real Field

14

Hence a monotone never decreasing real sequence either has limit 4- co or else has limit 0: for some a E R.t Obviously, a corresponding statement holds for (a")l.

EXERCISES 5.1. Following A. Hurwitz [59; pp. 706-7211 we may establish the existence of the square root of a positive real number on the basis of the theorem just treated concerning the limit of a monotone sequence and, in fact, conclude the existence of an increasing function L on P satisfying

L(XY) = L(x)

+ Ldv),

x, y

E

p.

(5.1)

Given a > 1, we consider the sequence (a")satisfying uo = a ,

anti = (a.

+ a;Ia)/2,

n E N.

Show that (a,) is decreasing and that a = lim a. > 1 and az = a . Conclude the existence of a positive square root for positive a. Continuing with L,(x) = 2n(x1/2n- l), x > 0, show that for each positive x # 1 the sequence (Ln(x)) is decreasing and has a finite limit and that L,(l) = 0. Let L(x) = limn+ L..(x). Show by studying L.(xy) - [L.(x) L.(y)] that L satisfies Eq. (5.1). Show that L is never decreasing, that is, L ( x ) 2 L(y) for x < y, and conclude with the aid of Eq. (5.1) that L is continuous at 1 and thereupon on P.[Cona > 1.1 Next, using the fact that (1 2-n)zn< e, n E N, but sider (L(U*'~")), lim(1 2-n )2 n= e, e being lim[l (n l)-l](ntl), conclude

+

+

+ +

L[(1

+

+ 2-")'"] m(n), la - arl < 2-(”+’)}, n E N, and show that b. = am,”) serves. 6.3. Given (an)bounded, that is, {an:n E N} is bounded, then there exists a subsequence possessing a limit in R. 6.4. Given a G R and (a”).Then lim a, = a if and only if lim sup.,, a, = a and lim inf.,, a. = a. State and prove the corresponding result for lim a, = co.

+

t In 96, Ch. 111, we shall see that the notational devices employed above may be transformed in a unified way into meaningful assertions in terms of the compactification of R to be introduced there.

I. The Real Field

16

7. CAUCHY SEQUENCES It is readily verified that if lim a,, = u E R,then the following condition, the so-called Cauchy condition, is fulfilled :

For each positive q there exists m E N such that la, - a41 < q when p , 4 2 m . (C)

-=

It suffices to take m such that la,, - QI q/2 when n 2 m. For then la, aql I lap - Q I la - a,l < q when p , q 2 m. The converse holds. It states an important property of R.A Cauchy sequence is one that satisfies the condition (C). We have:

-

+

Every Cauchy sequence possesses a limit (in R). A particularly simple proof may be given with the aid of the concepts of limit superior and limit inferior. Let (a,) be a Cauchy sequence. Invoking (C) for q = 1 we see that (a,,) is bounded. Let u p = sup{a,+,,: n E N } and let p, = inf{u,+,: n E N}. The sequences (a,,) and (p.) are bounded and monotone and hence possess limits in R.Given positive q we see that from a,+, - aq < q, p , q 2 m and n E N, we have a, - a4I q, p , q 2 m,and hence up - p, I q for p 2 m. Hence lim ap = lim p,. We denote the common value by a. We assert that lim a, = a. Indeed, given positive q, for some p we have

+ q,

a- q

< p, I a, < u

u-q

< a,, < u + q, n 2 p .

and hence

BB Chapter I.. @@

T H E COMPLEX F I E L D . L I M I T S

PREFATORY REMARKS It is a commonplace that R is not aIgebraicaIIy closed; that is, it is not true that every polynomial of positive degree with real coefficients takes the value 0 at some x E R. A most simple example is x2 + 1. Since x2 2 0 for x real, x2 + 1 2 1 > 0 for x real and hence x2 + 1 does not take the value 0 for any x E R. Complex domain algebra and analysis begin with the search for a satisfactory field to operate in. The search is conditioned by the imposed demands. We might, for example, seek a field containing R as a subfield and at the same time algebraically closed. We might ask for ostensibly less, say, a field C containing R as a subfield and an element i satisfying i2 + 1 = 0 such that C is generated by R u {i}. This demand is a reasonable one and the consequences of the choice are fruitful. In particular, we shall see in $2, Ch. I11 (“fundamental” theorem of algebra) that C is algebraically closed. In fact, C is “ the ” minimal algebraically closed extension of R. If F is an algebraically closed field containing R, then there exists i E F satisfying i z + 1 = 0 and the subfield of F generated by R u {i} fulfills the conditions imposed on C and, trivially, is contained in F. We shall see in the next section that the C are algebraically equivalent.? In $2 the geometric language of the complex field is introduced. Limit questions, infinite series, and power series are considered in the remaining sections of the chapter. The power series is part of the stock-in-trade of

t Hence the minimality property holds up to an isomorphism. 17

II. The Complex Field. Limits

18

complex function theory and will be encountered frequently throughout the book.

1.

THE COMPLEX FIELD

We consider a field C satisfying: (1) R is a subfield; (2) there exists i E C such that i 2 + 1 = 0; (3) C is generated by R u { i } . There are several easy remarks to be made. Given a, b E R, then a + bi = 0 if and only if a = b = 0. The sufficiency is trivial. From a + bi = 0 we obtain

a2 + b2 = (a + bi)(a - bi) = 0, whence a = b = 0. Suppose that a + bi # 0. Then from

+

(a + b i a’) ( b2 e ) = 1, we see that

(a + bi)-’ =a2 + b2 U

+-a2( -+b )b2 i’

It is now concluded that the set {a + bi: a, b E R} is a subfield of C since differences and quotients of elements of the form a + bi, a, b E R, are also elements of the same form. If a E R, a = a + Oi; i = 0 + li. Hence C = {a bi: a, b E R}.Given o! E C, there is a unique (a, b) E R x R such that a = a bi. For if a bi= c + di, a, b, c, d E R , then (a - c ) + (b - d)i= 0, whence the assertion. The rules for addition and multiplication,

+ +

+

(a

+ bi) + (c + d i ) = (a + c) + ( b + d ) i , (a + bi)(c + di) = (ac - bd) + (ad + bc)i,

(1.1)

are obvious and are stated only for reference. We term C the complexjie1d.t It is to be observed that the C are algebraically equivalent in the sense of isomorphism. That is, if K also fulfills the imposed conditions, there exists a univalent map f of C onto K satisfying f(x y ) =f(x) +f(y) andf(xy) =f(x)f(y), x, y E C.$ To see this, let j E K satisfyj’ 1 = 0. A mapfis given by

+

+

{(a

+ bi, a + bj): a, b E R},

t Again, a traditional definite article. A mapfof C onto K fulfilling these conditions is termed an isomorphism of C onto K. The definition is, of course, applicable to more general algebraic structures.

1. The Complex Field

19

as is easily verified. Note that it is harmless here to use the same symbols “ + ” and relative to both C and K. ‘‘e”

EXERCISES 1.1. Verify that f s o specified does have the stated properties. 1.2. Determine the isomorphismsfof C onto itself satisfyingf(x)

= x,

x E R.

From an axiomatic point of view we are on as good ground here as in the study of R.However, just as one constructs R from more primitive structures, so here we may establish the existence of a C constructively. The construction given below in Ex. 1.3 is frequently used as the starting point of the study of the complex field. It will be seen that the construction is motivated by the preceding axiomatic considerations.

EXERCISE

+

+

1.3. Let K = R x R; let + map ((a, b), (c, d ) ) onto (a c, b d ) ; let * map ((a,b), (c, d ) ) onto (ac - bd, ad + bc). Show that (K, .) is a field, that its zero is (0, O), its identity is (1, 0), and that [(0, 1)]’+ ( 1 , O ) = (0,O). Show further that R, = {(a, 0 ) : a E R} is a subfield that is a complete ordered field with positive part = {(a,0): a > 0}, and that K is generated by R I u {(0, l)}.

+,

We have seen (Ex. 4.15, $4, Ch. I) that from an algebraic point of view R is fairly simple. The only isomorphism of R onto itself is the identity map. We remark without proof that the set of isomorphisms of C onto itself is huge-a curious fact not needed for the purposes of classical analysis, but one which does show us that C is not without complication from an algebraic point of view.? Real and imaginary part, conjugate, absolute value. We introduce certain useful maps of C into R and C. Here a, b E R. (1) Re, the map taking a + bi into a. We term the image of z E C the realpart of z. (Notational remark: We write Re z for Re(z) when it is convenient.) (2) Im, the map taking a + bi into b. We term the image of z the imaginary part of z. Here, Im z for Im(z). (3) -, the map taking a + bi into a - bi. We term the image of z the conjugate of z and denote it 5. (4) I 1, the map taking a bi into J(u2 + b2). We term the image of z the absolute value of z and denote it IzI. The elements of the form ai, a E R, are termed pure imaginary. We want to learn a number of the basic rules for calculating with Re, Im, -, I I. The more obvious are stated without proof.

+

t To establish the remark

one turns to transcendental field extensions (cf. (133, 1391).

II. The Complex Field. Limb

20

(1) Re and Im are linear over R and each maps C onto R. Re is a projection of C onto R. In other words, given z, w E C, a E R, we have:

+ w) = Re z + Re w, Im(z + w) = Im z + Im w,

Re(a z) = a Re z;

Re(z

Re(C)

Im(a z) = a Im z;

= Im(C) = R;

Re(Re z ) = Re z.

(2) The conjugate map is an isomorphism of C onto itself. We have: -z = Z + Z, ZW = (Z)(W). Further, 2 = z if and only if z E R. Theconjugate map is an involution, that is, Z = z, z E C.

(3) Re z = (z (4)

1212

+ 2)/2,

Im z = (z - Z)/2i.

= zz.

The following observation is to be made: for elements of R the definition of absolute value is consistent with that given earlier. In fact, given a E R, a = a + Oi, so that the present definition of la1 yields J.', and this is precisely the absolute value of a in the sense of real analysis. We recall the convention on J in R. Two important rules pertain to the absolute value of a sum and the absolute value of a product. They are Iz + wl I

IZI

+ IWI,

(1.2)

For (1.3) we may proceed very simply as follows: lzwl2 = (zw)(zw) = 1212 l W l Z = (121 Iwl)'.

Hence lzwl = IzI IwI since x2 defines an increasing function on R . ' we note that Iz

For (1.2)

+ w12 = l z l z + lwlz + 2 Re(zF) I

1Zl2

= (14

+

lWl2

+ 2 lziq

+ lw1>2.

Hence Iz + wI I IzI + Iwl. We note that equality holds if and only if zW 2 0.

EXERCISES 1.4. 1.5.

Let U = { u E C, IuI

= l}.

Show that U is a subgroup of {C - {0}, }.

ZW >_ 0 if and only if there exist h, p 2 0 but not both zero such that hz = pw.

2. Geometry of the Complex Plane

21

+

IIm zI 5 2I'' Izl. When does equality occur on either side? 1.6. IzI I/Re zI 1.7. IIm(zk)I 5 k IIm zI k a positive integer. 1.8. ( Z wI ' IZ - wI' = 2(1~1' 1~1'). Under what conditions 1.9. Given zk E c, k = 1, . . .,n. Show that 1x1 I 2

+ +

+

does equality hold? 1.10. Show that if z"+' + C;ar zk= 0, n E N,CF,lak\ > 0, then IzI does not exceed the unique positive root of x"+' - C; laklx' = 0 which in turn is not greater than max{z; lakl, 1) (cf. Ex. 4.14, Ch. I). Hint: From z"+' = - C"0ak zk, conclude IZI"+'

cF,lakl I Z l k .

+

1.11. Given ak real, k = 0,...,n 1, satisfying a. 2 al 2 * . . 2 a. + I > 0. Show that if Z;+' akzk = 0, then IzI 2 1 (Enestrom). Hint: Apply the Abel identity to

C:''

at,?'.

0

i=a-bi

FIG. 2.1

2. GEOMETRY OF THE COMPLEX PLANE The map 6 :R x R + C taking (a, b) into a + bi is a univalent map of R x R onto C. We term 6 the canonical map of R x R onto C.It is to be noted that absolute value in C permits us to introduce a notion of distance in C. That is, given z , w E C,we define the distance from z to w as Iw - z1 and we see that distance so defined is a metric, for it satisfies Iw - zI = 0 if and only if w = z, Iz - W I = Iw - zI, Iw - zI I Iu - zI Iw - u1.t Given

+

t The concept of a metric will be considered in Ch. 111.


22

(a, b), (c, d) E R x R, we see that

ICT(C,d ) - O ( U , b)l

= J[(c

-

+ (d - b)’]

(2.1)

The relation (2.1) asserts that CT preserves distance. Using CT and the coordinatization of the Euclidean plane induced by the introduction of a rectangular coordinate frame we coordinatize the Euclidean plane via C.We shall term such a coordinatization of the Euclidean plane Gaussian or complex. We note that R corresponds to the first axis of the frame which we term the real axis and the set {ai: a E R} corresponds to the second axis, which we term the imaginary axis. The complex number 0 corresponds to the origin. Re z is the coordinate of the orthogonal projection on the real axis of the point with complex coordinate z.Corresponding geometric interpretations for (Im z)i, Z, and IzI are left to the reader. While the preceding paragraph serves to relate C with 2-dimensional Euclidean geometry, for the purposes of complex function theory we may as well operate directly in C.Thus we call C itself the Gaussian plane and introduce all the geometric concepts to be employed in terms of the identity map of C onto itself and the operations and special maps (Re, Im, -, I I) already introduced. Thus R is the real axis and {ai: a E R} is the imaginary axis. We are dealing with a realization of 2-dimensional Euclidean geometry. Starting this way, the reader need not concern himself with the refinementsinherent in an axiomatic account of 2-dimensional Euclidean geometry. We shall see that thanks to the canonical map o we obtain a “faithful” copy of the conventional 2-dimensional coordinate geometry. We introduce several notions and notations that will be constantly employed. By a straight line is meant a set of the form { a + bt: t E R}, where a, b E C,b # 0. Given a, b E C , by the segment [a, b] is meant the set ((1 - t)a tb: 0 5 t I 1). It should be verified that, when a, b E R, b 2 a, the present definition is consistent with that of $3, Ch. I. Given a, b E C , b # 0, by the ray associated with (a, b) is meant the set ( a + bt: t E R’}.

+

EXERCISES 2.1. Given a, b( # a) E C. There exists a unique line containing a and 6. In fact, it is ((1 - t)a rb: t E R}. What does this say about segments? 2.2. A ray admits a unique representation of the form

+

+ ut: t E R+},a E C, IuI = 1. Given a, b E C, b # 0. Let L(a, b) = {a + bt: t E R}. Determine a necessary {a

2.3. , L(a1, bl) n L ( a 2 ,bz) = 0, and sufficient condition for L(u1, bl) = L ( u z , b ~ ) for where the (at, br) are admitted (a, b).

2. Geometry of the Complex Plane

23

Given a E C , r E , 'R we denote by C(a; r ) the set { [ z - a1 = r } . It is the circle (or circumference) with center a, radius r. For positive r we denote by A(a; r ) the set {lz - a1 < r } . It is the circular disk (alias open ball) with center a, radius r.

EXERCISE 2.4.

If A(u; r ) = A@; s), then a = b and r

= s.

Convexity. We say that E ( c C ) is convex provided that [a, b] c E, whenever a, b E E. Convex sets have many interesting properties. We shall encounter the notion of convexity in the theorem of Gauss concerning the roots of the derivative of a polynomial (99, Ch. IV). Given A c C, there is a smallest (in the sense of set inclusion) convex set containing A . In fact, it is the intersection of all convex sets containing A . It is termed the conuex hull of A .

EXERCISES The following sets are convex: C; A(a; r ) ; [a, b ] ; {a < Re z I b ) , a < b ; [c, d ] ) ,a < 6, c < d. 2.6. Given A convex; n a positive integer; ak E A , pk 20,k = 1, . . . , n ; c: pk = 1. Then c: pkak E A . Hint: Proof by induction. 2.7. Let A be a given subset of C; let n be a positive integer. Let B. denote the set of sums of the form c1 pkak, where ak E A , pk 2 0, k = 1, ...,n, and 2 pk = 1. Show that the convex hull of A is U k e N Bk+,.Show that if A is the finite set {a,, . . . , an},then the convex hull of A is B.. 2.5.

o([a, b] x

Linear maps of C into C . Given a, b E C , we consider the linear function I with domain C given by

I(z) = az

+ b.

(2.2)

We put aside the trivial case where a = 0 and seek to interpret geometrically the action of 1. When is 1 an isometry? In other words, when does 1 preserve distance? From IKW)

- Wl

= la1 Iw -

4,

(2.3)

we see that 1 is an isometry if and only if la1 = 1 . If a = 1, we term 1 the translation through 6. We then note that if, in particular, b # 0, the line L(c, b) is mapped onto itself by 1. If la1 = 1 but a # 1, we term 1 a rotation.


24

When 1 is a rotation and c = ac

+ by then

I(z) - c = a(z - c).

(2.4)

Clearly I preserves distance from c. If a > 0 and a # 1, from (2.4) taken in the present context, we see that rays associated with (c, d ) , d~ C - { 0 } ,are mapped onto themselves by 1. Again from (2.4) we see that distances from c are multiplied by a. A linear map I for which a > 0 and there exists a $xed point, that is, a point c such that l(c) = c, is called a homothety. It is immediate that a linear map of C into C which is not constant is a translation, a rotation, a homothety, or the composition of a homothety and a rotation. It suffices to consider the case where a E C - R’. Let c denote the fixed point of 1. Let m(z) = c + (a/lal)(z- c); let n(z) = c la1 ( z - c). We have I = n 0 m. Theconversestatement isimmediate.

+

EXERCISES 2.8. The linear maps of C into C with coefficient a # 0 map C univalently onto itself and constitute a group when taken together with composition. Show that the following are subgroups: the set of translations, the set of homotheties having a given fixed point, the set of rotations having a given fixed point together with the identity. What about the set of all homotheties, or all rotations together with the identity map? 2.9. When f: X X , we define f., the nth iterate of f, by fo = 1, the identity map,f.+, =f 0 f., n EN.Show that in the case of I, In(z)= z nb if a = 1, while if a # 1, &) = c an(z- c) where c is the fixed point of 1. --f

+

+

3. LIMITS We are concerned with sequences (a,,):N -,C. The definitions for the notions of a limit and of a Cauchy sequence are mere paraphrases of those given in elementary real analysis with absolute value now being taken in the sense of C.With this agreed, given (a,) and a E C, the statement “(a,,)has limit a” needs no gloss nor do the notations a,,,” “lim a,,.” We say that (a,,)possesses limit co, limn+ma,, = 03, provided that lanl = + co. A set in C is termed bounded provided that it is contained in A(0; r ) for some positive r. A function, in particular a sequence, with image in C is termed bounded provided that its image is. A sequence (a,,) is said to have aJnite limit provided that it has limit a for some a E C. Certain routine verifications are readily carried out: (1) a sequence has at most one finite limit; (2) the usual rules for sum, difference, product, and quotient hold; (3) if a sequence has limit a E C (respectively co), then so does

3. Limits

25

every subsequence; (4) a sequence possessing a finite limit is bounded; a sequence possessing limit co is unbounded; ( 5 ) for real sequences the notions of finite limit and Cauchy sequence agree with those given in real analysis; (6) given (a,) and u E C, lim a, = u if and only if lim Re a, = Re u and lim Im a, = Im u ; (7) (a,) possesses a finite limit if and only if (a,) is a Cauchy sequence. It is well to point out a geometric interpretation of a limit: lim a, = u E C if and only if for each positive q it is the case that { a m + n : n E N )c A ( ~ ;V ) for some m E N , or equivalently that { n : a,

$+

A(a; $1

is bounded. We note further that if lim a, where e(n) = sup{ la,+k - a1 : k E N}.

= C((E C),then

( ~ ( n ) ) lim l , ~ ( n= ) 0,

EXERCISES 3.1. Check assertions (1)-(7). A link between the real and complex situations is furnished by the inequality ]Real, IIm a1 I la1 I IRe a / IIm al. This observation is useful for (6) and (7). 3.2. Given lim a. = 0. Let 6. = (ZEak)/(n+ 1). Show that lim b. = 0. Hint: For fixed m E N write

+

and estimate the size of 6.. 3.3. Given lim a. = a E C. Let c. = a. - a so that lim c. = 0. Taking the notation of Ex. 3.2. in the present context show that lim b. = a. This result may be stated: if (a")possesses

a finite limit, then the sequence whose nth term is the arithmetic mean

of a. , . . . , a. possesses the same limit (Cesaro). Show by the example of (( - l ) n ) n E ~

that there exist sequences (a") not possessing a limit for which the corresponding (6.) do possess finite limits. Sequences of functions. It should be emphasized that when we are dealing with limit problems for sequences of functions, the nature of the common domain of the functions in question is not important initially. It is only when we are interested in whether or not such properties as continuity are preserved under limit operations that the nature of the domain enters our considerations; obviously we need a medium where the property in question makes sense. Let there be given a set X and functionsf, f,, n E N , mapping X into C. We say that (f,)tendspointwise to f or that f is the pointwise limit o f ( f , ) provided

26

11. The Complex Field. Limits

that lim f,(x) =f (x) for each x E X. We say that (f,) tends uniformly to f or that f is the uniform limit of cf,) provided that for each positive q there exists m E N such that

If(x) - fn(x)l < 4 for x E X , n 2 m. We see that iff is the uniform limit of for n sufficiently large the set

(3.1)

(f,)and X # 0,then

{If(x) - fn+k(X)I: X E X ,k ~ N 1 is bounded, # 0.Let c(n) denote the supremum of this set when n islarge. For some m E N, c(m n) = 0. The converse is immediate. Given Y c X , we say that f, tends pointwise (resp., uniformly) to f on Y provided thatf, I Y tends pointwise (resp., uniformly) to f I Y.

+

EXERCISES If(f.) tends uniformly tof, then (f.) tends tofpointwise. Show by example that the converse is false. 3.5. Suppose thatf, tends uniformly tofand g . tends uniformly tog, g and g , mapping X into C. Show that (f. g,) tends uniformly t o f + g but that (f.g.) need not tend uniformly to a complex-valued function with domain X. It is to be observed that the following definitions are in force. Let A denote a ring, let f, g : X +A . Then byff g is meant the map taking x intof(x) g(x) and by f g is meant the map taking x into f ( x ) g ( x ) . With addition and multiplication so defined the set off: X +. A is a ring, commutative when A is. Given a E A, we define af as the map taking x into a f ( x ) . [By license the af notation will apply when OL E C - (0)and A is replaced by the extended complex plane (56, Ch. 111), af being given as in the preceding sentence by the pointwise product, or when a E C and f is meromorphic ($4, Ch. VI) on an open set 0, afin this case being construed as the product in the sense of meromorphic functions of the constant with domain 0 taking the value a and the function5 Also, here and elsewhere the obvious conventions prevail when X = 0; to be sure, this situation is not of great concern to us.] To show failure for the product let X = 10, 1 1 , f . ( x ) = g.(x) = x - ' (n I ) - ' . Show that if in addition f and g are bounded, then f . g n tends uniformly to f g . Show that if lf.(x)l, If(x)l>rn > 0, then (h,) tends uniformly to h where 3.4.

+

+

e,

+ +

h,(x) = Lf.(x)l-', h(x) = Lf(x)l-'.

Cuuchy sequences. Givenf,: X 4 C , n E N. We say that (f,) is a pointwise Cuuchy sequence provided that (f.(x)) is a Cauchy sequence for each x E X. Such a sequence possesses a unique pointwise limit. We say that (f,) is a uniform Cuuchy sequence provided that for each positive rj there exists m E N such that

4. Infinite Series

27

for x E X , p and q 2 m. We see that if (f,)is a uniform Cauchy sequence and X # 8, then for n large the set {Ifp(x) - f&)I: x

E

x,P, 4 2 n)

is bounded, # 8. Let w(n) denote its supremum when n is large. For some m E N, limn+,,, o ( m + n) = 0. The converse is immediate. Given Y c X , we say that (f,)is pointwise (resp., uniformly) Cauchy on Y provided that (f,I Y ) is pointwise (resp., uniformly) Cauchy. It is clear that a uniform Cauchy sequence is a pointwise Cauchy sequence and hence that it possesses a unique pointwise limit, say$ There can be no other candidate for the uniform limit.

EXERCISE 3.6. Show thatfis in fact the uniform limit.

4. INFINITE SERIES

An infinite series is a sequence associated with a given sequence of a certain kind in a special way that we specify presently. Indeed, given (a,): N -,G where G is an abelian group (in the first applications G is C considered additively or the set of complex-valued functions with a given domain X ) , by the infinite series associated with (a,) is meant the sequence (s,) where

c n

s, =

z$

a, *

0

conr

We denote (s,) by a, . s, is termed the nth partial sum of a, . For the sake of convenience we define ak as a,+, when q E Z and (a,) is a sequence mapping { k E Z, k 2 q} into G and refer other subsequent ; a, to the corresponding use of use (to denote sum) of the notation 1 a,+,. Of course, we can equally well define x7 ak as (s,),~,, where s, = a,, in the first instance, and for subsequent limit questions refer directly to sequences with domain { k E Z, k 2 q } using the appropriate definition, To continue, we shall want to be more specific about G. We consider the ak is convergent provided that numerical case (G = C) first. We say that (s,) possesses a finite limit, and in this case we denote limn+ms, by a,-a concession to tradition since, in spite of the double use of the notation, there is no danger of confusion when the notation is taken in context. We term

c," conr

x;

c$

I$


28

c;

1; 1;

limn+ms, the sum of ak (now a sequence). We say that ak is divergent provided that (s,) does not possess a finite limit. We say that ak converges absolutely provided that lakl is convergent.? We next consider the case where the a, are functions mapping a given set X into C. We introduce {x: ak(x) converges}. This set is termed the convergenceset of ak.Iftheconvergence set is Xitself, then, trivially, ak possesses a pointwise limit. I n this case we say that C; ak convergespointwise (on X ) and we term the pointwise limit in question the (pointwise) sum oj akand denote it by ak. We say that ak converges absolutely (on X ) provided that lakl converges pointwise.$ We say that ak converges uniformly provided that it possesses a uniform limit. The limit in question is called the uniform sum of ak . Given Y c X,we say that ak converges (akI Y ) converges (converges (converges uniformly) on Y provided that uniformly). In the respective cases, the limit (uniform limit) in question is ak o n Y. Absolute convergence on Y is called the sum (uniform sum) of correspondingly defined with " absolutely " replacing " uniformly." K . Knopp's Infinite Series [68] is to be recommended as a n excellent account of the classical theory of infinite series and power series. The following exercises serve for reviewing and for correlating the present material and the reader's earlier experience with infinite series.

1:

1;

12

1;

c?

1;

1;

1;

1; 1;

1;

EXERCISES In Exs. 4.1-4.5 it is understood that ak E C. 4.1. If C," ak is convergent, then lim at = 0. 4.2. If ak 2 0, k E N,then either c,"ak is convergent or lirn sk = + 03. 4.3. If 0 I bk < a k , k EN, and c,"ak converges, then c,"bk converges and

c,"bk c,"ax. 4.4. If c$ak converges absolutely, then c,"ak converges. Prove this theorem in the following two ways: (a) by showing that (s,) is a Cauchy sequence, (b) by considering the following series:

4.5. Given c; at absolutely convergent, c,"bk divergent, bk positive, k E N.Show that lim inf., lan/b,,l= 0. 4.6. Given k E N, let 6 ( z ) = zk, z E C. Determine the convergence set of c$& (geometric series). Hint: Recall the identity (cz z k ) ( l - z ) = 1 - z"+'. 4.7. Given ak > 0, k E N. If lirn SUpk-. ak+l/ak < 1, then c,"ak is convergent. If oIJ

t Cautionary note: In spite of the sound of it, the fact that a series converges when it converges absolutely requires proof! 3 We denote by If1 the composition of the absolute value function and f: X+ C.

5. Power Series

29

+

lim inf,, -, a k + ,/ax> 1 (or lim ak + I / & = m), then Zc,"ak is divergent. Here lim inf a, + 1 , h k is construed to be in R. 4.8. The Weierstrass M-test. Given M r 2 0, ji: X + C,k E N. Suppose that (a) cco"M, is convergent, (b)< I&(x)l Mk, x E X , k E N.Then c2&converges absolutely and converges uniformly. Hint: For uniformity show that c2 & is uniform Cauchy. [This is the celebrated criterion for absolute and uniform convergence due to Weierstrass. " M" for '' majorant." The M-test is very useful in practice.] 4.9. Given 0 < r < 1. Show that Zgfk of Ex. 4.6 cnnverges absolutely and uniformly on { IzJ< r } .

5. POWER SERIES

This section is our first serious encounter with complex function theory. We are concerned with a special instance of infinite series of functions. We are given (a,) : N -+ C. We let X = C and introduce f,(z) = a, z", z E C. The series fk is termed the power series with kth coeficient 0,. We denote both this power series and its sum at a point z of its convergence set by akzk.t Power series play an important role in complex function theory, their long history reaching back to the eighteenth century. They furnish methods of investigation of great value in the theory of differential equations, as we shall see later. It should be kept in mind that i n the study of power series the special case a, = 1 (geometric series) is very important in that it is an instrument of comparison about which we have very complete information. Let us turn to the question of the nature of the convergence set of a power series. It is easy to construct power series with convergence set (0) as well as power series with convergence set C. It suffices to take a, = n ! for a power series of the first kind and a, = ( n ! ) - ' for a power series of the second kind.$ Suppose now that we are concerned with a power series a k z kfor which neither of these extreme situations arises. There exist c, d E C - (0) such that ak ck converges and ak d' diverges. As a first step in our study we shall prove the following lemma.

c$

1;

1;

1;

12

Lemma 5.1: Let 0 < r < IcI. Then verges uniformly on { 12) I r } .

1; a k z kconverges absolutezy and con-

PROOF: ( a k C k ) is bounded. Let M be a positive number such that E N. Then for IzI I r we have

lakckl 5 M , k

t Given

c,"

z E C, of course &zk denotes as well the infinite series associated with The meaning of the notation should be clear in any given case from the context. The utterly trivial a. = 0 would do for a series of the second kind, but we wish to point out a nontrivial instance. (a.2").

11. The Complex Field. Limits

30

The lemma follows by virtue of the Weierstrass M-test and the convergence properties of the geometric series. Now let E = {Izl: a k z kconverges}. We note that IcI E E. Further, Id1 is an upper bound of E, for by the lemma if IzI > Id(, a k z k diverges. Hence E possesses a positive supremum p 2 IcI. If IzI p, then there exists w, IzI < IwI 5 p, such that a k w kconverges. Hence in the present case we see that the convergence set of a k z k contains A(0; p ) and is contained in {lzl 5 PI. We are led to the following theorem.

1;

1;

-=

c;

Theorem 5.1: The convergence set of the power series 1; a k z ksatisjies one of the following mutually exclusive conditions: (a) it is {0}, (b) it is C,(c) there exists positive p (necessarily unique) such that A(0; p ) is contained in the convergence set and { IzI < p } contains if. In case (b) 1; a k z f converges absolutely and converges uniformly on { IzI < r } for each positive r; in case (c) this is true for r, 0 < r < p.

EXERCISE Furnish the details of the proof which are not contained in the preceding discussion.

5.1.

According as (a), (b), (c) holds, we say that the radius of convergence o f a k z k is 0, co,p. In case (c) we term C(0; p ) the circumference of convergence. The geometric series has radius of convergence 1. Its convergence set is A(0; 1). It is easy to give examples of power series with radius of convergence 1 which converge on C(0; l), in fact, even uniformly on { IzI < I}. It suffices to take a, = (n 1)-’. If we take a, = (- l)”(n + l)-’, we have a power series with radius of convergence 1 which converges for z = 1 and diverges for z = - 1. In general, the study in the third case of the points of C(0; p ) at which ak zk converges and of the nature of the sum function is delicate and comp1icated.f The sum is, however, well behaved in A(0; p ) for case (c) and in C for case (b) as we shall see shortly. Before we turn to this question, we shall continue our study of the radius of convergence by showing how it may be determined in terms of (a,,). The solution is due to Cauchy and Hadamard. Into it two themes enter: the geometric series and the notion of limit superior. Suppose that a k z k has radius of convergence 0. Given r positive,

+

+

1;

1;

t For some classical examples, cf. [70].

5. Power Series

31

{k: la,+ I I r k f l > 1) is unbounded. Otherwise for some m E N we should have

c$

k 2 m, and the radius of convergence of a k Z k would not be 0. Hence l a k + l l r k f> l 1, and, consequently, k + v l a k + l l > r-' holds for an unbounded set of k and hence SUP{k+vlak+ll:k € N }

= +O0.

That is, lim SUPk-tmk + v l a k + l I = +a.

+

I$

Suppose that a,, zk has radius of convergence 00. Given r positive, ak rk is convergent, there exists m E N such that la,, I r k + l I 1 and since consequently k + v l a k + I l I r-I for k 2 m. Hence, since r is an arbitrary positive number, limk+mk + v l a k + , l = 0. Suppose that a k Z k has positive radius of convergence p. Let O < s 1) is bounded since akSk is convergent, and { k : l a k + l l t k + '> l } is unbounded as we see on appeal to the reasoning used in the case of radius of convergence 0-here we should be led akzk on A(0; t). From the first assertion we see that to the convergence of (k+vlak+l1) is bounded. Suppose that lim S U P k - t m k + v l a k + l l = ci. From the first assertion ci I s-' and from the second CL 2 r - I . Given the arbitrariness of s and t, we see that

c:

I$

1:

x$

lim SUPk-tmk+vlak+ll =p-'. It is now easy to see by the disjunction of the above conclusions that the following theorem holds. Theorem 5.2 (Cauchy-Hadamard): has radius of convergence +co. a k z k has radius of convergence where ci is a positive number, then

I$

c:

1:

Iflimk-tmk + v l a k + l l = 0, then akZk I f Iim S U P k - t m k + v l a k + l l = +a,then 0. If (k+vlak+ll) has limit superior ci, akzk has radius of convergence c i - ' .

If we anticipate the language of $6, Ch. 111, where we shall have the extended real line available, we may put Theorem 5.2 in the tidier form: the radius of convergence of a k Z k is l/(lim S U P k - t r n k + v l a k + l l ) .

EXERCISES Let r be positive. If { k : ' + v l a k + > r ) is unbounded, then either the radius of convergence is zero, or else it is positive but no greater than r - ' .

5.2.

32


What is the radius of convergencewhen ak = kk,k 2 ,k - l , k - k ?Let bk = 1 when some m E N and otherwise let bk = 0. What is the radius of convergence when ar =kkbk, k'br, k-'bk, k-*bk? 5.4. The convergence sets of E," akzkand c.2 a m + t z km, E N, are the same. 5.5. What is the radius of convergence when at = sin k ? [We presuppose for the sake of the exercise properties of the trigonometric functions which will be officially introduced in Ch. IV. The reader should check to see whether any properties of the sine function are invoked which do not come under the arithmetic treatment of Ch. IV. The irrationality of T is not to be assumed. Hint: Show that 5.3.

k

= m2 for

lim supr- lsin k l > 0 using the addition theorem for the sine and the fact that sin 1 # 0.1 5.6. What can be said about the radii of convergence of C," (Re &)zk and

C:o"(Im ak)zkwhen the radius of convergence of C," akzk is known? 5.7. The radii of convergence of c,"akZk and C," bkzk are known. Investigate the radius of convergence of C," ckzk for the following situations: cI = ak + b k , ck = ak b k , ck = Uk/bk.In the last case bk# 0, k E N. Construct examples to test the precision of your statements.

The reader should recall from elementary real analysis the theorem concerning the continuity of the uniform limit of a sequence of continuous functions. Actually, this theorem is a special case of a general theorem considered in Ex. 5.3, c h . 111, which subsumes the case of complex-valued functions. For the present we note that the definition of continuity for complex-valued functions with domain in C is merely a paraphrase of that used in the real case and that the uniform limit property holds. Suppose that the radius of convergence of ak zkis not 0.We see by Theorem 5.1 that f, in case (b) the sum and in case (c) the sum on A(0; p), is continuous. Many other properties off may be inferred. We continue by noting that if { k : ak # 0} # 8, then there is a least member m.Now

1:

1;

where g has the same context relative to am+kzk as doesf for ak zk.By virtue of the fact that g(0) # 0 and g is continuous at 0, we conclude that there exists r positive such that g(z) # 0 for all z E A(0; r ) . t Hence f(z) # 0, 0 < Iz(< r. We conclude: rfinf{ IzI :f(z) = 0, z # 0} = 0,then ak = 0,k E N. The hypothesis is fulfilled, in particular, iff(z) = 0, IzI < r, for some positive

t Note that there exists r positive such that Ig(0) -&)I such z, g(z) # 0.

< 1g(O)( for all z E A(0; r). For

5. Power Series

33

r or all the more so iff(z) = 0 for all z in the domain off. The converse is trivial. There is an important corollary, the identity theorem for power series.

c:

Suppose that b k z k converge, IzI < r, where r is positive, and a k z k and have sums on A(0; r ) which are equal. Then ak = b , , k E N. It suffices t o consider 1 ; (ak - bk)zk.It would, of course, be enough to require that

Note that the corollary yields the unique representation of a polynomial a k z k .Here { k : ak # 0} is bounded. function in the form We now give some exercises whose object will be to provide experience in reasoning with power series.

c$

EXERCISES Given(ak):N+Csatisfyingal = l,C,"klakl I l.Thenf(z) = ~ $ a r z kIzI , < 1, is univalent. Hint : Let z, w( # z ) E A(0; l), consider f ( w ) -f(z) and represent it as a product ( w - z)A and obtain a lower estimate for 14 .1 which renders the fact that A # 0 clear. Deduce the following corollary : Let ukzk have radius of convergence not zero and let a , # 0. There exists r positive such that the series has a univalent sum on A(0; r ) . Hint: Consider (u1r)-l (axrk)zk for small positive r. 5.9. We consider akZkconverging for IzI < r , r positive, and denote the sum on A(0; r ) by f. Let u E C(0; 1) and satisfy u"+ # 1, n E N. Let IzJ< r and introduce A.(z) = ( n I ) - ' %f(ukz). Show that A.(z) = a o =f(O). Hint: It is convenient to represent A. as the sum of a power series on A(0; r ) and to control the mth partial sum and the difference of A. and the mth partial sum. Conclude that laol 2 M(s) where 0 < s < r and If(z)l 2 M(s), Izl = s. By considering f(z) z-', j E N, show that ( ~ sJ~ 1.

a's of the identity. Sketch the set E for r

We now consider two theorems which yield as a conclusion the hypothesis of the Abel theorem, that is, the convergence of a series ak with complex terms when certain information is available on the allied power series a k Z k and its sum f on [0, I [ . We suppose that C$ a k z k converges on A(0; 1) and that f I [O, l[possesses afinite limit a at 1. Does ak converge? (- l)kzk.The theorem that we shall prove is due to Not always. Example: Tauber. It says: If the condition lim(ka,) = 0 is added, then I $ ak = a. Far more sophisticated results are known such as the celebrated Tauberian theorem of Littlewood [cf. 701. Their study would carry us outside of our program.

c$

ak . The proof is based upon a PROOF OF TAUBER'S THEOREM:Let s, = skillful choice of a sequence (x.)?, 0 < x, < 1, n E N, lim x, = 1 which insures that limlf(x,) - s,] = 0. To that end, we note that if 0 < x c 1 , then n

m

so that

I (n

+ 1) w(0) (1 - x) + w(nn ++l 1) (1 - x)-l ~

(5.3)

where

w(n) = SUp{(n + k)

lUn+kl:

k

E N}.

We suppose that w(n) > 0, n E N. Otherwise the theorem would be obvious. Our choice of (x,) is motivated by the following elementary fact: If A and B are positive, then the minimum value of Ax Bx-', x > 0, is equal to

+

36

II. The Complex Field. Limb

2Jz and is attained at from the identity

Jmand only at JG.This assertion follows

Returning to the proof we take

and conclude that The theorem follows. We obtain the theorem of Fejtr when we replace in the theorem of Tauber the condition lim(kak) = 0 by the condition klakI2converges. This condition may appear special at a first glance. It does have an interpretation in terms of area which is developed in the geometric theory of functions [43]. The proof of the FejCr theorem proceeds along lines similar to those of the Tauber theorem but calls for the use of the important Cauchy inequality [50] (not to be confused with the Cauchy inequalities), which it will be opportune to treat here thanks to (5.4): Given a k ,bk E R, k = 1, . . . , n. We have

(2 .)'(2 2 ak

ak')(

(5.5)

bk2)

(Cauchy inequality). Equality holds in (5.5) i f and only i f there exists (1,p), l,p real not both zero, such that l a k + p bk = 0, k = 1, . . . ,n. PROOF: We put aside the trivial cases: ak = 0, all k ; bk = 0, all k . In the remaining case, at all events (5.6) so that x > 0.

Hence taking the minimum value on the left we obtain

5. Power Series

37

and replacing bk by - b k , a corresponding inequality. The two so obtained inequalities taken together yield (5.5). Equality in (5.5) yields, when we retrace our steps, that (5.6) is an equality for some x or else (5.6) with b, replaced by -bk is. The stated proportionality follows. Equality in ( 5 3 , when the stated (A, p) exists, is readily verified. The following corollary is easy to see: Given (ak),(bk):N + R such that ak2and bkZconverge. Then akbk converges absolutely and

cz

Equality holds in (5.7) Aak /lbk = 0, k E N.

+

if and only fi

there exists

(A,p) as above such that

EXERCISES Prove Fejer’s theorem. Hint: Introduce r ( n ) = [%’ k lak12]1’2.We put aside the trivial case r ( n ) = 0, n large. With 0 < x < 1, show, using the Cauchy inequality and its corollary, that 5.17.

If (4Show that a sequence (x.) may be chosen advantageously (as in the proof of the Tauber theorem) by taking

when n 2 1. Construct (ak)for which the hypotheses of the Tauber theorem but not of the Fejer theorem are fulfilled and vice versa.

5.18.

BB Chapter 111 GG

TOPOLOGICAL AND M E T R I C SPACES

PREFATORY REMARKS When we introduce topological and metric spaces in a first book on complex function theory, we may appear to be insisting on an excessive amount of generality. Of the many justifications a principal one is that in complex analysis we encounter again and again certain general ideas that are particularly important in achieving global, as distinguished from local, results ; central among these ideas are compactness and connectedness. They serve as binding agents. The reader will have many occasions to see these two ideas put to work. Since no extra effort is needed to introduce them in a general setting and since it is sound pedagogy to have general principles wanted for many parts of modern mathematics thoroughly assimilated early, no further defense is needed for the level of generality employed in this chapter. The object is not, of course, to teach general topology while " standing on one foot," but rather to present a useful uade mecum for the remainder of this book. For a detailed authoritative modern account of general topology cf. [65]. There is another side to the topological requirements for the worker in certain areas of one variable complex function theory; it is the topology of the plane and, more generally, of two-dimensional manifolds. The demands for a specialized knowledge of such material do not appear in the early 38

1. The Notion of a Topological Space

39

chapters of our subject. We shall not have to concern ourselves anywhere in the present book with such topics as the Jordan curve theorem.? A very readable treatment of the theory of metric spaces and the topology of the plane is that of Newman [92]. For the student who has taken a standard undergraduate course on general topology this chapter will be little more than a review. Experience of the present time indicates that only a fraction of a beginning class in complex function theory has had a prior course on general topology.$ The remainder of such a class, having had a course on elementary real analysis, should find it a simple matter to become familiar with the rudiments of general topology and of the theory of metric spaces, as presented here, save possibly for the notion of connectedness which the newcomer may find is not directly referable to his earlier experience. However, it is not more “ difficult’’ than many concepts already encountered. By the study of examples in R and C the beginner should convince himself that the concept is “ reasonable ” and by the study of the role it plays that it is important. A model for the modern exposition of general topology is the treatise of Bourbaki (cf., in particular, [19]). Its influence will be strongly felt in this chapter. 1. THE NOTION OF A TOPOLOGICAL SPACE

The student is familiar with the following notion of an open subset of R : O( c R) is said to be open provided that for each a E 0 there exists r positive such that ]a - r, a r [ c 0. It is to be observed that the definition distinguishes a family of subsets of R, those that are open, and that the following useful properties hold : (a) given an arbitrary family @ of open parts of R, UOEm 0 is itself open; (b) if Ot and 0,are open parts of R, sois 0, n 0,; (c) 0 and R are open; (d) given a, b E R, a # b, there exist open Ot and 0, such that a~ Ot, b E O , , Ot n 0,= 8. The verifications are routine. It is not hard to sense that the stated properties may be thought of as applying to the members of a distinguished family of subsets of a given set, reference to R being eliminated. Granting on the basis of experience that the properties (a)-(d) inclusive are primordial in the study of questions involving open subsets of R, we have reasonable grounds to suppose that one can isolate useful general working principles by dealing with the question in an abstract setting.

+

t To the interested reader we recommend F. Riesz’ beautiful proof of the Jordan curve theorem which is fully accessible by the end of $5, Ch. V of this book. For an account in English cf. [53]. 3 Perhaps the preceding sentence will soon be obsolete!

40

III. Topological and Metric Spaces

We start with a set X . Let T c ‘ p ( X ) . We say that T is a topology on X provided that the counterparts of (a), (b), and (c) hold for the subsets of X which are members of T, that is, provided that T satisfies: (a) if 0 c T, then UOE0 0 E T ; (b) given O,,0, E T, then O1n 0, E T; (c) 0,X E T. Tis said to be a Hausdorff topology provided that in addition T satisfies: (d) given a, b E X , a # b , there exist 01, 0,E Tsuch that a E 01,b e O,, 0 , n 0, = 8. The composite ( X , T ) of a set X together with a topology T on X is termed a topological space; it is termed a Hausdorff (topological) space provided T is a Hausdorff topology. The members of T are termed open sets. We speak loosely of X as a topological space. The elements of Xare termed thepoints of x. Given S c ‘ p ( X ) ,the intersection of the topologies containing S is itself a topology, T ( S ) ,and it contains S. It is the smallest topology containing S (that is, it is contained in every topology containing S ) . T(S) is termed the topology on X generated by S.

EXERCISES 1.1. The topology on R given in the first paragraph of this section ( T = family of open sets with “open” as defined there) is generated by the family of intervals ]a,b[, a, b E R, a < b. Every open set is the union of a family of intervals. We term this topology the standard topology or more simply the topology on R. Of course there are other topologies on R [for example, ‘p(R)], however we shall use only the standard topology. 1.2. By the standard topology (or more simply the topology) on C is meant the topology generated by the family of circular disks A(a; r), a E C, r positive. Show that it is a Hausdorff topology and that A( c C ) is open in the sense of this topology if and only if for each a E A there exists r positive such that A(a; r ) c A (equivalently, if and only if A is the union of a family of disks). We shall use only the standard topology on C. 1.3. Given a E R, the following subsets of C are open: {Re z >a}, {Re z < a}, {Im z > a}, {Im z < a}, { IzI > a}.

Continuing with a topological space ( X , T ) we introduce several notions that occur commonly. A set F c X is termed closed provided that X - F is open. Given x E X , a set V c X is said to be a neighborhood of x provided that there exists 0 E T such that x E 0 c V. We note that A(a; r ) and { Iz - a1 I r } , r positive, are neighborhoods of a. We note also that old usage referred only to special kinds of sets as neighborhoods (intervals, disks, . ..). The modern usage is that of the definition. Given a set A c X and a E X , we say that a adheres to A (or a is a contact point of A ) provided that V n A # 0 for each neighborhood V of a. By the adherence or closure of A , denoted 2, is meant the set of points that adhere to A . By the frontier of A , denoted fr A , is meant A n X - A . We say that A clusters at a provided that a E A - { a } .

1. The Notion of a Topological Space

41

EXERCISES Given A c X , k is closed. A is closed if and only if A = A. If A is open, fr A = k - A. In C we have: A@; r ) = { ) z - a1 I r ) , fr A(a; r ) = C(a; r ) .

1.4. 1.5. 1.6.

Given topological spaces X and Y, each with its own topology. Let .f: X -, Y and x E X . We say that f is continuous at x provided that for each neighborhood V off (x) there exists a neighborhood U of x such thatf( U )c V, or, equivalently, f ( y ) E V , y E U.

EXERCISE 1.7. Show that f i s continuous at x if and only i f f -‘(V) is a neighborhood of x whenever V is a neighborhood off(x).

We say that f is continuous on X (or that f is a continuous map of X into Y ) provided that f is continuous at each point of X.?

EXERCISES 1.8. fis continuous on X if and only if for each 0 open in Y (that is, a member of the topology on Y ) f - ’ ( O )is open in X . 1.9. Composition. Suppose that X , Y,Zare topological spaces,f: X + Y,g: Y + Z , fis continuous at x E X , g is continuous atf(x). Then g ofis continuous at x.

Relatioe topology. Given a topological space ( X , T ) and A c X . The family of sets 0 n A , 0 E T, constitutes a topology TA on A , which is a Hausdorff topology when T is. It is termed the relative topology on A . Note that the standard topology on R is the relative topology on R in the sense of C. If now f : A + Y where Y is a topological space, we say that f is continuous at a E A (resp., on A ) provided thatfis continuous at a (on A ) in the sense of T, . The consistency is obvious when A = X .

EXERCISES 1.10. Check that for the case A c X = R, Y = R, the definition of continuity just given is equivalent to the usual “ e , 6 definition. The same holds if R is replaced by C for either X or Y. ”

t

“

Onto ” replaces “ into” when it is known that Y = f ( X ) .

42


1.11. Given X, Yare topological spaces, A c X,f:A + Y. Thenfis continuous on A if and only iff: A + f ( A ) is continuous on A in the sense of the relative topologies of A and f(A).

Homeomorphism. Given topological spaces (XI,T,) and ( X , , T,) we say that f is a homeomorphism of X , onto X , provided that f is a univalent map of X , onto X , such that the map with domain T, taking a set 0 intof(0) has image T, . It is to be noticed that iffis a homeomorphism of X , onto X , , then bothfandf-' are continuous on their domains by virtue of Ex. 1.8 of this section. The converse is also true: iffis a univalent map of X , onto X , such thatfandf- ' are continuous on their respective domains, then f i s a homeomorphism of X , onto X , . Indeed, f(0) E T, for 0 E T, and f-'(R) E TI for R E.T,, so that the map taking 0 intof(0) maps T, onto T2 . Topology is concerned with properties preserved under homeomorphisms.

EXERCISE 1.12. Show that adherence and frontier are preserved under homeomorphism, that is, if f i s a homeomorphism, then f ( A )=f(A),f(fr A ) = frlf(A)].

Limit. Given topological spaces X and Y, the latter being Hausdorff. Let - {a} and let f:A + Y where A is X - {a} or X. By the Hausdorff property there is at most one element b E Y such that f I ( X - { a } )u (a, b) is continuous at a. When such a b exists, we say that f has limit b at a and denote b by limo$ Once more we note that our conventions on the employment of the limit notation are to prevail; implicit in its use without qualification is the existence of a limit. aEX

Limit of a sequence. Given (6,) : N -, Y, there is at most one element b E Y such that for each neighborhood Y of b there exists m E N such that 6, E V, n 2 m. When such b exists, we say that (b,) has limit b and denote b as in the earlier situations, by lim b, .t T,) and ( X , , T,) we may Product topology. Given topological spaces (X,, endow X , x X , with a useful topology as follows. Let T denote the family of unions of sets of the form 0, x O,, (O,, 0,) E T, x T2 . It is readily verified that T is a topology on XI x X , . T is called the product fopology of X , x X,. When X , and X , are Hausdorff, so is X , x X,. Further the projection map carrying (x,,x2)into x, is a continuous map of XI x X , onto

t We shall see in 86, this chapter, that the notion of the limit of a sequence may be subsumed under that of a map once we introduce the extended real line.

2. Compactness

43

X,, k = 1,2, when X , and X , are not empty (and vacuously when XI and X , are empty). EXERCISE 1.13. Verify the assertions of the preceding paragraph.

2. COMPACTNESS

Given a set X , Q, c q ( X ) is said to be a cooering of X provided that X = UEE* E. Y is said to be a Q, subcovering provided that Y c Q, and Y is a covering of X . When ( X , T ) is a topological space, @ is said to be an open couering of X provided that @ is a T subcovering. A topological space Xis said to be compact provided that Xis a Hausdorff space and that the Heine-Borel-Lebesgue property holds: f o r each open covering R of X , there exists ajinite R subcovering. The value of the notion of compactness is to be found in its fundamental properties and in the fact that a number of important spaces are compact. A set A contained in a Hausdorff space X is said to be compact provided that A is a compact space in the sense of its relative topology. This definition is consistent when A = X . It is easy to see that A c X i s compact f u n d only i f wheneuer R c T has the property A c UOER 0 , there exists @finite c SZ such that A c Uoco 0. A second simple fact concerning compact sets is: a compact set is closed. Indeed, suppose that A is a compact set in X . We show that X - A is open. We suppose that A # 0, X , for otherwise the assertion would be obvious. Let b E X - A and note that by the Hausdorff property for each U E A there exist open sets ~ ( a )O(a) , of X satisfying a € w(a), b € O(a), o(a) n O(a) = 8. By the compactness of A there exists a finite set {w(a,), . . . , o(a,,)} such that A c UlSks,, ~ ( a , ) .Now bE

n

O(a,) c X - A

lsk b}), b > fl, would constitute an open covering of X and a second application of compactness yields that X = f - ' ( { x > b } ) for some b > /?, so that inff(X) 2 6. Contradiction. Corresponding statements hold for upper semicontinuity for which the definitions are given by reversing the inequalities in the above definitions.

EXERCISES 2.3. LetnEN,akEC, k = O ,..., n . L e t p ( ~ ) = z " + ~ + ~ ~ a ~ z ~ , z € C . S h o w t h a t (PI has a minimum. Hint: Usingp(z) =zn+'[l +Za k Z k - ( " + l ) J show that for some positive r we have Ip(z)l 2 IzIn+'/22 Ip(O)l, IzI 2 r . 2.4. Apply the method of Hurwitz (Ex. 5.9, Ch. 11) to f(z) = I%bkzk12, considering h ( n l)-' C ; f ( u k z ) ,to show that if IEZ bkzkII M , IzI = r , then ZZ J b k l z r Z5kM 2and thereupon conclude the corresponding inequality for a power bkzk,namely if IZ," brzklI M , IzI = r, then c,"\bklZr2'I M 2 ,it being series assumed that the radius of convergence is +a or is positive but greater than r . Conclude that if M = lbol and r is positive, then b k = 0, k 2 1. 2.5. Returning to the situation of Ex. 2.3 show that if Ip(c)l = rnin{Ip(z)I:z E C } , then p(c) = 0, thereby establishing the " Fundamental " Theorem of Algebra. Hint: Supposep(c) # 0 and apply Ex.2.4 to 1/p(z c), z small. Prove as a corollary: If Z$ akz' has radius of convergence co and { k : ak # O } is unbounded, then the sum of C," akrkon C is not representable as the quotient of

+

x,"

+

+

polynomials.

3. CONNECTEDNESS We continue with a topological space X not necessarily Hausdorff. By a splitting of X is meant a pair {01, 0,) where (1) 0,, 0,are open and nonempty, (2) 0, n 0,= 8,and (3) O1 v 0,= X . We say that X is connected provided that there does not exist a splitting of X . Given A c X, we say that A is connected provided that A is a connected topological space in the sense of its relative topology. The importance of the notion will unfold. A premonition may be had by observing that if we are concerned with a sentential function P(x), x E X , where Xis a connected topological space and we know that the sets {P(x)} and {not P ( x ) } are both open, then one is empty, so that either P(x) holds for all x E X or else P(x) holds for no x E X.This observation yields a method of proof, termed the standard connectedness argument (cf. Theorem 4.2, this chapter). By a region we understand a nonempty open connected subset of a topological space. By a continuum we understand a

3. Connectedness

47

closed connected subset of a topological space which contains more than one point. We show

Theorem 3.1: Girlen f : X + Y where X and Y are topological spaces, X is connected, f is continuous on X. Then f ( X ) is connected. PROOF: We assume, as we may, that f ( X ) = Y. Suppose that Y is not connected and that {01,O,} is a splitting of Y. Then { , f ' ( O , ) ,f - ' ( 0 2 is ) }a splitting of X . Contradiction. Connected subsets of R. From the point of view of classical analysis it is essential to have precise information on the connected subsets of R. As we shall see, it is not hard to give a complete answer. The situation in C is far more complicated. Suppose then that A c R is connected and that a, b E A , a < b. Suppose that c E ]a, b[ n (R - A ) . Then { A n {x < c } , A n {x > c } } is a splitting of A . This is impossible. Hence [a, b] c A . If sup A = + co, then {x 2 c } c A , c E A . Similarly, if inf A = - co, then {x 5 c } c A , c E A . If A is bounded above and is not empty, then { c 5 x < sup A } c A , c E A , a corresponding statement holding for A bounded below. On considering the various possibilities we see that a nonempty connected subset of R must be a set of one of the following kinds: R, { a < x}, { a 5 x}, { a > x}, { a 2 x}, {a}, ]a, b [ , [a, b[, ]a, b ] , [a,b ] , where a, b E R, a < b. However, as yet we do not know whether, apart from { a } ,any of the listed sets is connected, the connectedness of the singleton being trivial. The following general theorem will permit us to reduce the question to the case of a segment and we shall be able to conclude that all the listed sets are connected.

Theorem 3.2: Let X be a topological space. Let A , c X , 1 E A, where each A , is connected and A , # 0. Then U l E A , is connected.

nAEA

PROOF: We suppose, as we may, that A # 8, the case where A = 0 being A , = X . Suppose that X is not connected and let trivial, and that UIEA {O,, 0,) be a splitting, the indexing being such that 0, contains some point A , , Since A , is connected, A , n 0, = 0,1E A, and hence 0, = 0. of Contradiction. The theorem follows. We return to the listed subsets of R. We note that

nAEA

R

=

U [-(n n€N

+ l), n + I ] ,

and ]a,b[ =

U [a + (n + m ) - ' , b - (n + m ) - ' ] EN

48


where m is a sufficiently large positive integer. The representation of the remaining sets as unions of segments having a nonempty intersection is readily supplied and is left to the reader. There remains to show that a segment is connected. Suppose that [a, b] is not connected and that {0,,O,} is a splitting such that a E 0,. Let c = sup{x: [a, XI c

O,}.

Clearly, a < c Ib. Suppose that c c 6. If c E 0,, then [c - h, c + h] c 0, for sufficiently small positive h, and [a, c + h] c 0,for such h. This is not possible. Hence c E 0, and consequently [c - h, c h] c 0, for some small positive h. For such h we have [a, x ] 0, for c - h < x and hence the supremum property of c is contradicted. Hence c = b, [a, b[ c O,,{ b } = 0 2 . This is not possible, since 0, contains more than one point. We conclude that a segment is connected. The connected subsets of R are now under control.

+

+

EXERCISES 3.1. If A is a connected subset of a topological space X and A = B c A, then B is connected. 3.2. Given a, b E R, a < b. Suppose that f is a continuous map of [a, b] into R. Show that f ([a, b]) is a segment. Generalize. 3.3. Show that f of Ex. 3.2 is univalent if and only if it is increasing ( f ( x )f (y), a I 3.4. Suppose that8 E + R where E is a connected subset of R,f is univalent and is continuous on E. Show that f is a homeomorphism of E onto f (E)and that, if 0 is an open subset of R which is contained in E , f ( O ) is an open subset of R. [We shall see that the maps one encounters in complex function theory, subject to obvious restrictions, have the property that they carry open subsets of C in their domain into open subsets of C. The present exercise is intended to depict a prototype in the one real variable theory.]

Components. Given a topological space X . If x E X , there is a unique maximal connected set, C(x), containing x (maximal in the sense that it contains all connected sets in X of which x is an element). It suffices by virtue of Theorem 3.2 to take C(x) as the union of all connected sets of X of which x is an element. By Ex. 3.1, C(x) is a closed subset of X . We term C(x) the component of X containing x. Given E c X , we say that E is a component of X provided that there exists x E X such that E = C(x). Given x , y E X , then either C(x) = C(y) or C(x) n C(y) = 0. For if C(x) n C(y) # 0, then C(x) u C(y)is connected and C(x)and C(y)must both be equal to C(x) u C(y) and hence to each other. The notion of component is applied to A c X , A being taken with its relative topology.

4. Arc

49

In the special case of a topological space X,each point of which is contained in a connected neighborhood, each component of X is open. The components are then regions. This is the situation for an open subset of C as we shall learn in the next section.

4. ARC

We define an arc of a topological space X as a continuous map y of a segment [a, b ] , a < b, of R into X ;?(a)is termed the initialpoint o f y and y(b) is termed the terminal point ofy. The term “path is also used for “ arc.” The map y need not be univalent. Much license is customary in connection with the terminology pertaining to arcs and the term “arc” is applied to what should be called ”image of arc.” We shall use y to denote also the image of the map y. The meaning will be clear from context, but the distinction is an essential one. Given arcs y1 and y, , each having domain [0, 11, the terminal point of yI being the initial point of y, , we define y1 y z as the arc given by y ( t ) = y1(2t),0 I t I +,= yz(2t - l), 4 I t I 1 (product arc). The space X i s termed arcwise connected provided that given p , q E X there exists an arc with initial point p and terminal point q. The following useful theorem is very easy to show. ”

-

Theorem 4.1 : An arcwise connected topological space is connected.

PROOF:Suppose that {01,0,} is a splitting of the space. Let p E 01, q E 0, , and let y denote an arc with initial point p and terminal point q. Then {y-’(O1), y-’(Oz)} is a splitting of the domain of y. However a segment is connected. Contradiction. For regions in C we have a converse. Theorem 4.2: A region in C is arcwise connected.

PROOF: The argument that follows will be our first encounter with the standard connectedness argument. Let A be a region in C and let p E A . Let B denote the set of q in A for which there exists an arc y lying in A of whichp is the initial point and q the terminal point. We show that B and A - B are both open. Since B # 8, it will follow that B = A . We note that if + b E A(a;r), then [a, b] c A(a; r ) . Let ab denote the map with domain [0, 11 taking t into a(l - t ) + bt. Suppose now that q E B. Let A(q; r ) c A. Let y be an arc in A with initial point p , terminal point q, and domain LO, I]. Then -+ y * qs is an arc in A, s E A(q;r). Hence A(q; r ) c B. Consequently, B is open.


50

Now if q E A - B and s E A(q; r ) c A , then s E A - B. For if s E B, there exists an arc y in A with initial point p , terminal point s, domain [0, 11. Then + y sq is an arc in A , so that q E B. Contradiction. We conclude that A - B is open. Hence B = A . Sincep is an arbitrary point of A , the theorem follows.

EXERCISE 4.1.

Given A

c C, A

convex. Show that A is arcwise connected. If A(a; r ) c B c A@; r ) ,

then B is convex. Establish the final remark of 53, this chapter.

5. METRIC SPACE Here we are concerned with a set on which a “reasonable” notion of distance is defined. Specifically, by a metric space is meant a set X together with a map p : X x X - t R + satisfying the following conditions: (1) p(x,y ) = 0 if and only if x = y ; (2) given x, y E X , we have p(x, y ) = p(y, x ) (symmetry); (3) given x, y , z E X we have p(x, z ) I p(x, y ) + p ( y , z) (triangle inequality). We term p(x,y ) the (p-) distance between x andy and p a metric on X . Important cases for us are R and C with p(x,y ) = Ix - yl. The p so specified is termed “standard.” By the ball with center a E X , radius r E P, is meant {p(a,x ) r } . It is denoted B(a; r). Metric spaces serve as a convenient medium for treating uniform limits and uniform continuity. The degree of abstraction is not anything like the ultimate but is very adequate for our purposes. However, before turning to these questions, we first consider a topology on Xassociated with p , and (quite independently) the notion of a complete metric space. We ask for the topology T on a metric space X generated by the family of balls. It is termed the topology on X induced by p . We show that T consists of the unions of families of balls. To establish this assertion we note that if c E B(a; r ) n B(b; s), then B(c; t ) c B(a; r ) n B(b; s) when

-=

t = min{r - p(a, c), s - p(b, c)}. Consequently, a point belonging to A , n A , , where A l and A , are unions of balls, is the center of a ball contained in A , n A , , Hence A l n A 2 is a union of balls. It is now immediate that the family of unions of balls is a topology on X and hence being contained in T is T. It now follows that a subset A of X i s open ifand only iffor each a E A there exists rpositive such that B(a; r ) c A.

5. Metric Space

51

The topology T is a Hausdorff topology. Given a, b E X , a # b, let r = p(a, b)/2. By the triangle inequality B(a; r ) n B(b; r ) = 0.

EXERCISE

+

Consider the maps of C x C into C taking (z, w) into z w and taking (z, w) into zw. Show that they are continuous. Show that the map of C - {0} into itself taking z into z-' is continuous. Conclude the standard continuity and limit theorems for sums, products, and quotients of complex-valued functions and sequences [cf. 46, this chapter, for the remarks concerning sequences]. 5.1.

Completeness of a metric space. We need hardly note that the term "complete" has been employed before. It is not our object to overturn established usage. The context should make the meaning clear. A metric space Xis said to be complete provided that every Cauchy sequence possesses a limit, a sequence (a,,):N + X being termed Cauchy provided that, as one would expect, for each positive q there exists m E N such that p(a,, a4)< q when p , q 2 m. Clearly, R and C are complete in the sense of their standard metric. Sequences of maps. Let X be a given set and let ( Y , p ) be a metric space. Let Jf,:X + Y, n E N. The definitions of 93, Ch. I1 concerning pointwise and uniform limits and pointwise and uniform Cauchy sequences are to be paraphrased with p(a, b) replacing Ib - a1 (and that concerning ~ ( nwhen ) X # 0). The notion of completeness plays the role of guarantor: when Y is a complete metric space, a pointwise (resp., uniform) Cauchy sequence possesses a pointwise (resp., uniform) limit.

EXERCISES 5.2. Prove the last assertion.

5.3. Let X be a topological space, a E X.It is supposed that (f.) tends uniformly to fon X and that eachf. is continuous at a. Show that fis continuous at Q. Hint: Para-

phrase the usual argument of elementary real analysis, starting with f Lf(Q),f

( 4 l If Lf(a), fn(a)l+ f[f"(QX fn(x>l+ P [ f . ( X k fWl I 2 44 + f Lf"(4, f X 4 1

for n large. We shall have frequent occasion to consider the case where X is a Hausdorff space (often an open set in C) and to treat sequences (f,,) satisfying

Ill. Topological and Metric Spaces

52

either (1) for each compact set A c X the sequence (fn I A) possesses a uniform limit or (2) for each compact A c X the sequence (f,I A) is a uniform Cauchy sequence. Following the terminology of Valiron we say that (f,)possesses a uniform limit I N X provided that (I) holds and that (f,)is a uniform Cauchy sequence IN X provided that (2) holds. Uniform continuity. Suppose that (X, p ) and (Y, p ) are metric spaces. Let f: X + Y. The notion of uniform continuity paraphrases the classical version. We say that f is uniformly continuous provided that for each positive q there exists a positive 6 such that PLm),f(Y)l < 9 when p(x, y ) < 6. In the nontrivial case where X # 0, there is a very elegant means of expressing the notion of uniform continuity due to de la VallCe Poussin. For 6 2 0 we consider { P C f ( x ) , f ( Y ) : P(X, Y ) l < 61.

(5.1)

When this set is bounded, we denote its supremum by @,(a). The set of 6 for which (5.1) is bounded is a connected subset of R + containing 0. The function wf assigning to each such 6 the value ~ ~ (is 6termed ) the modulus ofcontinuity off. Clearly of is never decreasing, that is, w,(a) I w,(b) for a, b in the domain of w, satisfying a < b. The relation between uniform continuity and modulus of continuity is given in Ex. 5.4.

EXERCISES 5.4. fis uniformly continuous if and only if the domain of w f does not reduce to 0 and w f is continuous at 0. It is supposed that X # 0. 5.5. Determinewfin thefollowingcases:X = [0, l ] , f ( x )= x; X = R + , f ( x )= l / x ; X=[O, l],f(x)=x’; X=R+,f(x)=x’. 5.6. Given f: [0, 11 Y where Y is a metric space. If limo 6-’ ~ ~ ( = 6 0,) then f is

constant. 5.7. Given f: X + Y and g: Y + Z are both uniformly continuous, X # 0. Show 6 W,[W,(~)]. ) that with h = g 0 f for 6 small we have ~ ~ ( 2

Important because of its applications is

Theorem 5.1: L e t f : X + Y where X is a compact metric space, Y is a metric space, and f is continuous on X. Then f is uniformly continuous. PROOF: We put aside the trivial case X = 8. Let q be positive. For each x E X let 6(x) be positive and such that p [ f ( x ) , f ( y ) ]< q/2 when

6. Compactijcation

53

p(x, y ) < 2 6 ( x ) , p and p being the metrics of X and Y, respectively. The

family of balls B [ x ;S(x)],x E X,is an open covering of X . By the compactness of X there exists a finite subfamily

with union X. Given y , z such that Pb,

2)

< 6 = min{Kx,),

*

9

*,

6(xn)),

we see that y E B [ x j ;S(x,)] for s o m e j E (1, n), say k. By the triangleinequality p(xk >

z,

p(xk Y ) + p ( Y , z , < 6(x/c), 3

so that p[f(Xk),f(Z)]< q/2. Since p [ f ( x k ) , f ( Y ) ]< q/2, we conclude that p [ f ( y ) , f ( z ) ]< q. The uniform continuity off is established.

EXERCISES 5.8. Let A denote a compact subset of C, A # 0, and F a closed subset of C, F # 0. Then the function f given by f ( z , w ) = Iz - wI, (z, w) E A x F, has a mini-

mum. Show that this is not true when “compact” is replaced by “closed.” 5.9. A compact metric space X is sequentially compact in the sense that given a sequence (x$ of points of X , there exists a subsequence (xV(“Jpossessing a limit in X. Prove this assertion by establishing the existence of a point a E X such that for

each positive r the set { n : x. E

B(a; r ) )

is unbounded. Prove Theorem 5.1 with the aid of the sequential compactness property. 5.10. Let ( X , p ) and ( Y , u) be metric spaces. Show that

m1,Y l ) ,

( X I . , Y2)l = P b I ,

xz)

+

O(Y1, Y 2 )

defines a metric on X x Y which induces the product topology on X x Y. Show that if X and Y are compact and f is a continuous map of X x Y into a metric space Z , thenfis uniformly continuous in the sense of the metric T . It is to be noted that this result extends to products XI x - .. x X. of compact metric spaces XI,

..., X ” .

6. COMPACTIFICATION Given a topological space A’, a topological space Y will be termed a compactijcation of X provided that (1) Y is a compact topological space, (2) X c Y, (3) the given topology of X i s its relative topology with respect t o Y. It is clear that if X admits a compactification Y, then X is also a compactification, closure being taken in the sense of Y. A necessary condition

54

I l l . Topological and Metric Spaces

for a topological space X to admit a compactification in which it is open is that X be Hausdorff and that for each x E X there exists a compact neighborhood V (neighborhood in the sense of the topology on X so that V c X). The converse is true. We put aside the trivial case where Xis compact and show: If X is a noncompact Hausdorfl space with the property that every point possesses a compact neighborhood, then there exists a compactijication consisting of X and a point oo $ X. Such a compactification is termed an Alexandroff (or one-point) compactification. To establish the assertion, we take as a clue the observation that if Y is a compactification of Xand A is a compact subset of X, then Y - A is open. We propose Y = X u {a}, where 00 is an element not in X and as a candidate topology for Y the union of the topology on X and the collection of sets Y - A, A compact c X. It serves.

EXERCISE 6.1. Verify the last assertion. Show also that any two Alexandroff compactifications of X are homeomorphic.

It is appropriate to ask what this means for C and complex function theory. To be sure, we are not dealing with the indispensable, but rather the convenient and the illuminating. As we see at once, C admits an Alexandroff compactification. We denote the adjoined element by 00 (infinity) and the compactification by 2‘. We term 2‘ the extended complex plane. Limit statements for l/z now have a concrete meaning. Circumlocutions go into precise statements. We define 1/0 (also denoted 0 - I ) to be 00 and 1/oo (also denoted oo-l) to be 0. We now see that the map of into itself taking z into l/z is a homeomorphism of 2‘ onto itself. Hence, iff is a continuous map of a topological space X into 2‘,then so isg of, where g(z) = l/z, z E We shall see later that it is very convenient to allow functions taking the value 00, especially in the study of meromorphic functions. In general, iff is a map with image in 2‘,we denote 9 o f by I/$

e

c.

EXERCISE 6.2. Relate for (a”):N + C the statement “lim a. “lim a. = 03 ” in the sense of the topology on

e.

= 03

”

of 93, Ch. I1 and

We introduce by agreement various operations involving co. Thus when a # 0, we define aoo and a/O to be 0 0 ; when a E C, we define a/co to be 0

6. Compucti/cution

+

55

and co + a , a co, co - a , and a - 03 to be 00. Thereby f given by az (resp., a/z) is continuous on when a E C - { 0 } , and f given by a + z is continuous on when a E C.

c

c

The extended real line. For the purposes of real analysis we wish to distinguish the “ large positive ” and the large negative.” To that end we introduce two distinct elements, neither in R, designated -co and +a, respectively. We define R as R u { - 00, + 0 0 ) and proceed by first defining an order < on fi which agrees with that on R for elements in R and thereupon introducing a topology on R rendering R a compactification of R. Given a, b E R, we say that a < b if and only if one of the following holds: (1) a = -a,b # -a, (2) U E R and either b = +co or b E R and b - a is positive. We propose as a topology T for R the collection of unions of sets of the following kinds: ]a, b[, { - 00 I x < a}, { a < x I co}, where a, b E R, a c b. T serves. “

+

EXERCISE 6.3. Verify the last assertion.

We term fi the extended real line. We introduce “ ] ” and “ [ ” as for R but now with reference to 8. Every subset of R has a supremum and an infimum defined in terms of the order on R. It will be convenient to write sup, f for the supremum of the image off I A wheref: X --t R and A c X , similar agreements holding for inf, f and, when meaningful, for maxAf and min,f. We define ( + co) a and a (+ co) to be 00 when a > - 0 0 ; ( - 00) a and a + ( - c o ) to be -co when a < +a;(+co)a and a(+co) to be +co when a > 0 and -co when a < 0; (-00)a and a(-00) to be -a when a > O and +co when a < 0 . We define I-col, I+col, and 1001 to be +a. Finally, we extend the definition of a+ as max{a, 0 } and a- as min{a, 0 ) to a E R. Given f, g : X -+ R, by “f < g ” is meant “ f ( x ) < g(x), x E X ”; I, > , and 2 are similarly defined.

+

+

+

+

EXERCISES Show that if A c R and A is nonempty and bounded above, then sup A in the sense of R is the same as sup A in the sense of R, and that a similar observation holds relative to the notion of infimum. 6.5. Subsume the definition of the limit of a sequence u_nder that of a function using the relative topology of N considered as a subset of R.

6.4.


56

6.6. Paraphrase the developments leading to the notions of limit superior (inferior) for (a”):N +R, noting that here “ sup ” and “ inf” exist. Compare for (a”):N +R

the R-theory and the R-theory. Return to the other material of Chs. I and I1 in which the symbols + a, and - m were used and reinterpret the developments in question in terms of the language of 3 (in particular, the Cauchy-Hadamard theorem, framing the appropriate conventions for the reciprocal on [0, a]). 6.8. Let a, b E C. Let 0 I rI < r 2 Ico and let 0 Is1 < s1 2 co. Show that if 6.7.

+

It:rl

then (a, rl, r 2 ) = (6, sl,

+

+

< Iz - a1 < r 2 } = { z :sl < Iz - bl < s2},

SZ).

+

6.9. Given (a”):N + C , let b, = la,tll-’/(“tl)if anti # 0, and let 6. = 03 if a,, = 0, n E N. Develop the Cauchy-Hadamard theorem in the following form: the radius of convergence of Co“akzkis lim inf,, b,, giving a proof starting with the

,

sequence (h). This is a convenient moment t o furnish a gloss for modes of expression such as (1) “ P ( x )when x is (sufficiently) near a,” (2) “ P ( x ) when x is small,” (3) “ P ( x ) when x is large.” We shall understand for (1) “there exists a neighborhood U of a such that P ( x ) holds when x E U.”For ‘‘x is small” we construe ‘‘x is near 0” and for ‘‘x is large” we construe “ l / x is small,” reference being made to x in a subset of 2‘. For “ P ( x ) when x is near but different from a” we shall understand the gloss for ( I ) with “ x E U ” replaced by “ x E U, x # a.” When we say “ f ( x ) is (arbitrarily) near b when x is (sufficiently) near a,” we mean t h a t f u {(a,b)} is a function continuous at a. When we say “ f ( x ) is near b when x is near but different from a,” we mean that the assertion glossed in the preceding sentence holds with.f replaced by the restriction off to its domain less {a}. “Near u” and “near b” in these assertions will be replaced on occasion by “ large,” “ small,” the first meaning “ near 00,” the second ‘‘near 0.”

BB Chapter IV

@@

COMPLEX D I F F E R E N T I A L ANALYSIS

PREFATORY REMARKS We are now ready to undertake a study of the differential calculus in the complex domain. Some of the initial parts of the program are little more than paraphrase and transplantation of the one real variable theory. However, to reveal our pedagogic hand, the time devoted to review, extension, and comparison is a useful period of synthesis for many students. Often some of the most basic ideas of real differential analysis become alive for them at just this period of development. After these preliminaries we shall continue by studying the differentiability of functions represented by power series (53) and shall find that we have at our disposal a method for obtaining solutions of a number of important differential equations-a method whose power the mathematicians of the eighteenth century found intoxicating: try to find a solution admitting a power series representation (method of undetermined coefficients for power series). The way is opened for the study of that function the worker in complex function theory would most reluctantly surrender, the exponential function (on C).The moment is at hand to prepare a number of fundamental working tools : the polar representation of complex numbers, logarithms, arguments, powers. The apparatus so introduced allows us to treat early the important Poisson integral representation of the real part of an analytic function (95). In the present chapter we take for granted familiarity with the elementary 51

IV. Complex Differential Anaijsis

58

properties of the Riemann integral. However, our wish to keep the exposition essentially “ self-contained ” will not be violated for the Riemann-Stieltjes integral will be treated in Ch. V. Thus we shall account for the material concerning the Riemann integral used in the present chapter. An important question is the relation between differential analysis involving one complex variable and differential analysis involving two real variables ($7), in other words, the link between the differential analysis ofJ a function of a complex variable, and the function of two real variables obtained from f with the mediation of the canonical map B of Ch. 11. We arrive at the Cauchy-Riemann equation(s) ($7) and the connection with harmonic functions ($8). To pave the way for $97 and 8 it will be sound pedagogically to review some aspects of two real variable theory. We shall do so in $6. The Gauss theorem on the zeros of the derivative of a polynomial is treated in $9. Its statement fits nicely in the realm of one complex variable differential analysis. Its proof brings into play the notion of the logarithmic derivative, the multiplicative decomposition of a polynomial, and the attractive notion of convexity.

1. COMPLEX-VALUED FUNCTIONS OF ONE REAL VARIABLE We consider fi V + C where V is a neighborhood (in the sense of the R-topology) of a E R, and recall that f is said to possess a derivative at a provided that If(x) - f ( a ) ] / ( x - a) has a limit at a (of course, it is to be emphasized, in C). The notation “ Lf(x) -f (a)]/(x- a) ” intends a function with domain V - { a } . When f possesses a derivative at a, we term x-a

the derivative off at a and denote it by f ’(a). We say “f ’(a) exists ” for “f possesses a derivative at a.” When we assert “f ’(a) = A ” where A E C,we mean that f possesses a derivative at a and the derivative off at a is A. The notion of derivative may be given an elegant and useful equivalent definition due to CarathCodory: f possesses a derivative at a provided that there exists g : V + C, continuous at a, such that f ( x ) -!(a) = g(x) (x - a),

x E V.

EXERCISES 1.1.

Iffhas a derivative at a, thenfis continuous at a.

(1.1)

1. Complex- Valued Functions of One Real Variable

59

1.2. Show that there is at most one g satisfying the imposed conditions. When f ’ ( a )exists, f ’ ( a ) = g(a). 1.3. u = Refand D = Imfboth possess derivatives at a if and only if fdoes. In this case f ’ ( a ) = u’(a) i ~ ’ ( a )[Here . “ Ref ” for “ Re 0 f ” and “ Im f ” for “ Im 0 f-”1

+

How incisive Carathtodory’s formulation is may be seen by applying it to the proof of the chain rule for the derivative of a composite function: Given f : V -+ U , g : U + C, where V is a neighborhood of a E R , U is a neighborhood of b =f (a) E R , both V and U being neighborhoods in the sense of the R-topology, f ‘(a)and g’(b) exist. Let h = g of. Then h’(a) = g’(b)f ‘(a). Here is a proof using the formulation of Carathtodory. We have

f(x)

- b = V(X) ( X - a),

x E V,

where cp: V + R and is continuous at a, and g(x) - g(b) = $(XI ( x - b),

where I): U

-+

x E u,

C and is continuous at b. Hence

= [I) o f ( x ) l V ( x )( x - a),

x E V.

The proof is completed by noting that (I) f ) q is continuous at a and that its value at a is g’(b)f ’(a). When Vis not assumed to be a.neighborhood of a but is assumed to contain some segment [a, b], a < b, we may consider the existence of a limit at a for [f + ( x ) - f + ( a ) ] / ( x- a), where f + =f I ( V n [a, +a[). When it exists, it is termed the right-handed derivatiue o j f at a and is denoted f+’(a).Analogous remarks hold for the left-handed derivative, the notationf-’(a) now being used. 0

Sign of derivative. We note that i f F : X + R , where X is a topological space, and F is continuous at a E X and F(a) > 0, then there exists a neighborhood V of a such that F(x) > 0, x E V. Next we introduce the signfunction, sg: C -+ C, defined by sgo=o;

sgz=:,

I4

ZZO,

(1.2)

and note that sg is multiplicative: sg(zw) = (sg z)(sg w). Let us now turn to f possessing a derivative at a, f real valued, f ’ ( a ) # 0. From (1.1) we have

in some neighborhood V , of a. This is the starting point of important considerations of the elementary differential calculus. From (1.3) follows: y

60

I V. Complex Differential Analysis

f ( a ) 2 f ( x )for x in some neighborhood of a, that is, iff has a relative maximum at a, or iff has a relative minimum at a, then f ' ( a ) = 0. We now see that the theorem of Rolle and the mean-value theorem of the differential calculus are easily concluded thanks t o the observation just made and the theorem of Weierstrass concerning the existence of a maximum and a minimum of a continuous real-valued function. The mean-value theorem states: Let f:[a, b] + R, - 03 < a < b < co. Suppose that f is continuous on [a, b] and that f '(x) exists for all x E ]a, b[. Then there exists c E ]a, b [ such that

+

EXERCISES Treat the special case where f ( a ) =f(b) (theorem of Rolle) by showing the existence of c E ]a, b[ at whichfattains either its maximum or its minimum. 1.5. Establish the mean-value theorem by applying the theorem of Ex. 1.4 to g defined by 1.4.

1.6. Show that if the hypotheses of the mean-value theorem are fulfilled and

f ' ( x ) > 0 for all x E ]a, b[, then f is increasing on [a, b], that is, f ( x ) < f ( y ) when aI x 0 and thatf, its sum on A(0; p), satisfiesf (0) = 1 and f '(z) =f (z) when (zI < r, for some r E 10, p ] (and hence, a posteriori, when Iz( c p). Using the condition f (0) = 1 , Theorem 3.1 of the present section, and the uniqueness theorem for power series we obtain

1;

a,

= 1,

+ I)ak+

= ak ,

(3.3) t The function exp satisfies (3.2), the terms involved being taken in the sense of R,and exp(0) = 1.

(k

1

k E N.

3. Power Series

67

Hence there is at most one such power series, for necessarily ak = (k!)-' if there is a solution. The uniquely available candidate fulfills the imposed conditions. First, (k!)-'zk has radius of convergence co. Let E denote its sum on C. Since ( ( k ! ) - ' ) is the solution of the system (3.3), we see that E(0) = 1 and that E' = E. We agree to term E the exponential function (on C) and proceed to show that

+

2:

EIR = exp.

(3.4)

Once this has been accomplished we shall not hesitate to denote E by exp and to speak of it simply as the exponenfiulfunction. We shall denote E ( z ) by both the traditional exp(z) and the ancient ez. A justification for the latter notation is to be found in the addition theorem and the conventions made on powers (cf. the discussion at the end of $4, this chapter). To avoid difficulties we use ez scrupulously as E(z); in case of need, one should furnish a gloss. We first show that E satisfies the addition theorem:

E(a

+ b) = E(a) E(b),

a, b E

C.

(3.5)

The following elegant proof based on the differential equation (3.2) was given essentially by CarathCodory [21, vol. I, p. 2331. Letf(z) = E(z) E(a + b - z), z E C. Then f ' ( z ) = 0, z E C. Hence f ( a ) =f(O) and (3.5) follows. Taking b = -a in (3.5), we see that E(a) # 0. We conclude that E(a) > 0 for a E R. It is elementary to see by the binomial theorem applied to n+ 1

that

II-

m

whence E( 1) = e. Since, in addition, E I R satisfies the addition theorem and is continuous, E I R = exp. We introduce A, the primitive of (1 + z)-' with domain A(0; l), satisfying L(0) = 0. We considerf= exp A and note that 0

and hencef(z) (1 + z)-' has derivative equal to 0 at each point of A(0; l), so that f(z) (1 + z ) - =f(O) = 1. Hence the useful identity exp[A(z)] = 1 + z ,

IzJ < 1.

(3.6)

IV. Complex Differentia.!Analysis

68

By the addition theorem for the exponential function,

+ z)”,

-= 1, N.Taking z = (n + 2)-’ and replacing n by n + 2, we obtain a second exp[n

A(z)]

= (1

IzI

for n E proof that exp(1) = e on considering the limit as n tends to that limo z-’ A(z) = X(0) = 1.

00.

It is to be noted

EXERCISES 3.8. Establish the addition theorem for the exponential function using the Cauchy multiplication rule for power series. 3.9. Abel’s binomial series. Let a E C. Consider the differential equation (1

+ z)y’ = ay.

(3.7)

Seek power series with radius of convergence p > 0 having sum S, on A(0; p ) satisfying S.(O) = 1 and (3.7). Show that there is exactly one such series, that p = co

+

when a E N and p = 1 when a E C - N, and that the kth coefficient is recursively by

(0“) The power series Note that

=”

(k:

1)

=

a

(k)

a-k k+l’

(3

kEN.

z‘ is termed the binomial series associated with parameter a.

(i)is tOhe usual binomial coefficient of elementary algebra for

a E N.We

have an example of formal extension. Show by considering [S. exp 0 (-aA)l’(z),

IzI < 1,

= exp[a

Izl< 1.

that S.(z)

A(z)],

+

In particular, when a = n - I , n a positive integer, we have [S,,.(z)~ = 1 z, that is, S,,. furnishes in power series form an analytic nth root of the function 1 z with domain A(0; 1). We shall see in the next section in what sense S.(z)is to beinterpreted as an ath power of (1 z), IzI < 1. Show, arguing from the differential equation (3.7), that S,,#(z) = S.(z) SB(z), IzI < 1. Give a second proof by first establishing the identity

+

+

using the elementary binomial theorem for a, /3 E N and properties of polynomials and thereupon Cauchy multiplication.

4. Further Properties and Applications of the Exponential Function

69

3.10. Sl,z.The binomial series, a = 4, is particularly interesting. It plays a useful role in approximation theory in the real domain. Show that Sllz(x)= d-, -

1 < x < 1 . From S I l 2 ( - x ) > 0 , O < x < 1 , conclude that

and hence that

:(Y)

__

1

I)(;‘

is convergent

zk converges uniformly and absolutely on A(0; 1). The sum

on [ - 1 , 11 i s d l + x . Show, noting that - 1 I x z - 1 I O f o r -1 S x l l , that 1x1 may be approximated uniformly on [- 1 , 1 1 by a polynomial. 3.11. Let f ( z ) = XY akzk, Iz1< r, where r > 0. Let F(z) = C,” lak+llzk and let M = lim, F1]0, r[. It is supposed that M < +a.Let g : A(0; s) +A(O; r), s > 0, and satisfy f o g ( z ) = z , IzI <s. Show that /g(z)/2 Izl/M. Apply to the following situation: n a positive integer, r = 2l/”- 1 , s = 1 , f ( z ) = ( 1 z)” - 1 , g ( z ) Sl,.(Z) - 1. 3.12. Seek a power series with radius of convergence p > 0 whose sum w on A(0; p ) satisfies w(0) = 1 and z w”(z) w’(z) z w(z) = 0, IzI < p. Show that there is exactly one such series and p = co. The differential equation is Bessef‘s equation of’ order 0, the solution in question, customarily denoted Jo , is the Bessel function of order 0. Carry out the same program when w is to satisfy w(0) = 1 and

+

+

z(l

-

+

+

z ) w”(z) [c - (a

+

+ b + l ) z ] w’(z) - ab w(z) = 0

where a, b, c E C, - c $ N (hypergeometric equation of Gauss). Show that p 2 1 and that the kth coefficient of the power series in question is k- 1

k- 1

Show that if b = c = 1, then w(z) = L O ( - z ) . In this exercise the power series approach to linear differential equations is treated ad hoc. In Ch. XI1 we shall see how the majorant method of Cauchy permits us to put the power series approach to differential equations on a systematic basis.

4. FURTHER PROPERTIES AND APPLICATIONS OF THE EXPONENTIAL FUNCTION, TRIGONOMETRIC FUNCTIONS

In this section we treat mapping properties of the exponential function. Further we introduce the sine and cosine arithmetically, that is, in terms of analysis without appeal to geometric considerations. In addition, the notions of logarithm, argument, root, and power will be studied. The latter concepts are elementary, but in practice care is required for their correct use. From the power series representation of the exponential function we

IV. complex Diflerential Anabsis

70

see that exp(ii) = exp(a) and consequently lexp(a)l = exp(Re a). Hence lexp(ia)l = 1 if and only if a E R. We define sine and cosine on C arithmetically by cos z = (e“

+e-92,

sin z = (e“ - e-’“)/2i.

(4.1)

+

Hence exp(iz) = cos z i sin z and exp( - iz) = cos z - i sin z, the former equality being the formula of Euler. The function cis with domain C defined by cis@)= exp(iz) is convenient.? By virtue of the addition theorem for the exponential function we have the de Moivre formula cis(a

+ b) = (cis a)(cis b),

a, b E C.

Taken together with the formula of Euler it yields the standard addition theorems for cos(a b) and sin(a b). We now consider the action of cis I R. We have from (4.1) cos‘ = -sin, sin’ = cos. Consider cos(3). We have

+

+

Hence

< 1 - (%)(*)< 0. Here the estimate is made by considering the zeroth partial sum. On the other hand cos(0) = 1. Hence cos x = 0 for some x E ]0,3[. Let 4 2 be the smallest positive root of cos x = 0. From cos x > 0, 0 < x < 4 2 , it follows that sin is increasing on [0, n/2] and hence sin x > 0, 0 < x I 4 2 , and sin(n/2) = 1.T From cos’ = -sin, we conclude that cos is decreasing on [O,n/2]. We see that cis is univalent on [0, n/2]. Note that cis(n) = [ci~(n/2)]~ = - 1 and that cis(2n) = 1. On using the identity (cis t)[cis(n - t ) ] = - 1, we see that cos is decreasing on [n/2,n]. Consequently cis is univalent on [0, n]. Further the image of [0, n] with respect to cis is C(0; 1) n {Imz 2 O}. From (cis t)[cis(2n - t)] = 1, we see that cis maps [n,2x1 univalently onto C(0; 1) n {Im z SO} and hence that cis maps [0,2n[ univalently onto C(0; 1). It is now easy to see that ez = 1 if and only if z is of the form 2nin, n E Z. For

t “cis ” = “cos + i sin.”

2 Gloss: when we say that a functionfhas a certain property on a subset E of its domain,

we mean thatf I E has the stated property-as “fincreases on E” or “fis univalent on E.”

4. Further Properties and Applications of the Exponential Function

71

if eit = 1, then t E R, and if m is the maximum of the integers not exceeding t/2n, we have 2nm I t < 2n(m 1). From this inequality and cis(t - 2nm) = 1 we see that t = 2nm. The converse is obvious. Hence we conclude that e" = eb if and only if b = a 2nin for some n E Z.

+

+

The image of the exponentialfunction. We know that 0 is omitted. Suppose that a # 0. Then a/lal = cis a for some a, 0 I a 2n. Hence

-=

+ ai). {log JaJ+ ( a + 2nn)i:

a = exp(1og la1

The preimage of { a } is exp(C) = C - (0).

n E Z } . We see that

EXERCISES exp maps {O I Im z < 2n) univalently onto C - {O}. The image of {Im z = a} with respect to exp is a ray less the point 0 and the image of {Re z = a) is C(0;eg), a E R.What are the images with respect to exp of {O I Im z I v/2},of {a < Re z m > O , a < x < b ,

then V [ l / f ] < m - ’ V[f].

1. Line Integrals

97

*(f) I f f ’ is continuous on [a, b], then

W lI (max,,,ol If’’l)@ - 4. Hint: For (d) the identity BD - A C = ( B - A ) D + A ( D - C ) is useful. Riemann-Stieltjes Integral. We proceed with a, b as above and C and form sums given in terms off, 9, a partitioning P of [a, b] whose associated sequence is ( x k ) ! , and a finite sequence u = ( u k ) t - satisfying x k I uk I X k + l , k = 0, .. . , m - 1. Such a u will be termed P-admitted. We introduce

f,g: [a, b]

--f

m- 1

s ( p ; u, =

C f c u k ) b(Xk+l) 0

-g(xk)l

and consider the behavior of S ( P ; u) for IPI small. We say that f possesses a Riemann-Stieltjes integral with respect to g provided that there exists A E C such that for each positive q there exists a positive 6 such that

I V ; u) -4 W Mx)).

When g(x) = x, a I x I b, we term the integral simply the Riemann integral off, denoting it p

x

(or Ja>(x)dx)*

In general, givenf, g as above and that a I a I p 5 b, by p d g

we understand 0 when a = p and the integral o f f I [a, p] with respect to g I [a, p] when a < p, assuming that it exists. Since we are concerned with developing the notion of the RiemannStieltjes integral for our specific needs in complex function theory, we turn directly to the important classical case where f is continuous on [a, b] and V[gJ < +a and show

V. Cauchy Theory

98

Theorem 1.1: I f f is continuous on [a, b] and V [ g ]< Riemann-Stieltjes integral with respect to g and

+ co, then f possesses a

PROOF: We introduce w , the modulus of continuity off, and proceed to compare the S ( P ; u ) . Suppose that Q is a rejinement of P and that t is Q-admitted. Following the cue given by (1.1) and using the v(k) notation, we see that m-1 v(k+l)-l

where (yk)$m)is the sequence associated with Q. Hence IS(Q;t ) - S ( P ; u>ls 4 P I )

US].

Let P and Q now be arbitrary partitionings of [a, b]. Using R = P u Q and z that is R-admitted we see that IS(Q; t ) - S ( P ; u)l I IS(Q; t ) - S ( R ; z)l

+ IS(R; z ) - S ( P ; u)l

I Cw(lQI) + 4 l P I ) I V b l .

(1.2)

Now consider

P , = { a + 2-"k(b - a ) ; k = 0, ..., 2 " } ,

n E N,

and u, = ( a

+ 2-"k(b - a));"-l.

Let S, = S(P,; u,,). We see that IS, - S,+kl IU[2-"(b - a ) ] v [ g ] ,

k E N.

Hence (S,) is a Cauchy sequence. Let A = limn- S, . Using (1.2), we see that for arbitrary P and P-admitted u we have J S ( P ;u ) - A1 5 IS(P; u ) - S,I

+ IS, - A1

I (4lPI) + wC2-"(b - a ) ] } VCgI + IS, whence, taking the limit as n + 03, we have

I V ; u ) - 4 I 4 l P I ) VCSl.

A19

(1.3) The inequality (1.3) shows that f possesses an integral with respect to g and brings to light very vividly the roles played by the continuity off and the

1. Line Integrals

99

finiteness of V [ g ] .The inequality of the theorem follows at once from (1.3). T h e theorem is established.

EXERCISES 1.3. In this exercisef(with or without subscript) denotes a complex-valued function with domain [a, b], continuous on [a, b]; g (with or without subscript) denotes a complex-valued function with domain [a, b], satisfying V [ g ]< $-a; k E C. Show the following. (a) If a < c < b, then

laf b

+

ds = J " f 4 a

1f b

ds.

C

(b) We have

Ja

Ja

Ja

.b

[Bilinearity.] (c) If (f.) tends to f uniformly on [a, 61, then b

,litjafn ds

Jb f b

=

ds.

(d) Let C(a) = 0, C ( x ) = V [ y I [a, XI], a < x I b. Then G is never-decreasing o n [a, 61 (that is, C(x) I G ( y ) when x < y ; "never-increasing'' similarly) and

When g(x) = x , a I xI b, then

jab If I dG

1 If1 b

=

dx.

(e) Suppose that p is a never-decreasing continuous map of [A, B] onto [a, 61, A, B E R. Then

Jl" f B

=I

b

0

p 4 s p) O

fds.

a

(f) Given a partitioning P of [a, b] with associated sequence (x& and u that is Padmitted. Then

100

V. Cauchy Theory

exists and is equal to

This last assertion holds also when g is continuous on [a, b]. Hint: In both cases compare S(P;u) for (gf’,x ) and for (9,f’). 1.4. Withfcontinuous on [a, 61, the function F given by

F(x) = rf(0 dt,

a I x I b,

a

is a primitive off. Hence if G is a primitive off,

j a b f ( x )dx = C(6) - G(a). 1.5. Review the earlier uses of the Riemann integral in this book. Formulate explicitly the properties used and justify them in the framework of the present treatment of the integral. 1.6. Let X be a compact metric space and let f be a continuous map of X x [a, b] into C. Let g be as initially given in Ex. 1.3. Show that the function

Iab f d’(0 (x,

1. Line Integrals

101

with domain X is uniformly approximated by sums of the form m- 1

4 0

for /PI small; that is, show that given E positive, there exists a positive 6 such that when IPI < 6, the difference of sum and integral has absolute value < E, x E X. Conclude that the integral defines a continuous function on X(cf. Ex. 5.10, Ch. 111). If X = A(c; r), 0 < r < m, and if

+

uniformly on [u, b] (that is, given positive E, there exists a positive 6 such that the difference of F ( t ) and the difference quotient on the left is in modulus less than E when 0 < Iz - < 6 for all f E [a, b1), then

CI

lab

F ( d dg(t)

is the derivative at c of

The variation o f g when g’ is continuous on [a,b ] . We shall establish under the stated assumption the useful formula

PROOF: From B

g(p) - g(u) =

g’ d x ,

a Ict 5 B 5 b,

Ja

using (d) and (a) of

Ex. 1.3, this section, we see that v ( g ; P ) IJabls‘ldx,

and hence

VCsl 5

Is’l d x . Jab

On the other hand, with (xk);lldenoting the sequence associated with have

P

we

V. Cuuchy Theory

102

Hence, using the standard inequalities for the absolute values of sums and differences, we have

Consequently,

5'

Ig'(xk)l

(xk+ 1

+ o(lpl)( b - a),

- xk)

0

where o now denotes the modulus of continuity of 9'. We conclude

Jabls'l d x 5 VCSl. The asserted equality follows.

EXERCISES 1.7. y(t) = cis t on [0, a], a > 0. We have L [ y ]= a. 1.8. Let y(0) = 0, y(t) = t it' sin(?r/t), 0 1, but is not when 0 < a 5 1.

+

Line integrals. Let y be a rectifiable path in C with domain [a, b] and let where E contains the image of y. Suppose that f 0 y is a continuous map of [a, b] into C . By the line integral off along y, denoted

f: E-,

e,

S,fd., is meant

p

0

Y)dY*

Letf: A --t C where A is a region in C and f is continuous on A. Let y be a rectifiable path in A with domain [a,b]. Suppose that g is a primitive ofJ Then the following very simple formula holds:

103

1. Line Integrals

If y' is continuous on [a, b], we have L f d z = Iab(9O 7)' dt and hence (1.4). We assume, as we may, that the domain of y is [0, 11. It suffices to replace y by y[a (b - a)t], 0 I t I 1. Suppose that A is of the form A(c; r), so that I y ( t ) - CI 5 p < r, 0 5 t I 1 for some p, 0 = Y ( f & ) t & + l- t&+ Y ( l k + l ) ~

~

- f&'

f&+l

k = O ,..., m - 1 .

tk

+S Y W ) (z - w) + 44 (z - w),

where E is continuous on 0 and

E(W)

= 0. Hence

Z(T'";f) = ICT'"'; E(Z) ( z - w)],

since f( w ) +f '(w)( z - w) has a primitive. We conclude that II(T;f)l 14"Mnd(T'"') &T'"')= M , d(T) E(T),

where M,, is the maximum value taken by I E ~ on the convex hull of T(").The theorem follows from the fact that limn+ M , = 0.

EXERCISES Establish the stated properties of l(T("))and d ( V ) . 2.2. Given a, 6, c, d E R, u I b and c I d. We denote by [u, b ; c, d ] the set 2.1.

{[u(l - s)

+ bs] + i[c(l - t ) + 4:0 I s, t 2 1).

2. The Simplicial Cauchy-Goursat Theorem

107

Suppose that [a, b ; c, d] c 0. Show by Theorem 2.1 that iff is analytic on 0, then J(b) = 0, where 4

+ ci)(b + ci), (b + di)(a + di),

+ ci)(b+ di), y4 = (a + di)(u + ci) y z = (b

y l = (a y3 =

_____,

(Cauchy-Goursat theorem for rectangle). Establish the asserted equality when f’ is continuous on 0 by showing that J’(b) = 0.

Existence of primitive for the case of a disk. Suppose that f is analytic on A(u; r). Theorem 2.1 yields the existence of a primitive off very rapidly. It is apparent that the convexity of A ( a ; r ) is to our advantage. We define g(z), z E A(u; r ) , as the line integral o f f along & Given b E A(a; r), by Theorem 2.1 we have g ( z ) - g(b) =

Jzf(w) dw,

noting that g(z) = -

J&

f(w) dw.

It follows that g ( z ) - g(b) =f(b)

( z - b)

+ J&[f(W) -f(b)l

dw.

We conclude using the continuity off at b and the inequality of Theorem 1.1, this chapter, that g’(b) =f(b). We state the result formally.

Theorem 2.2: A function analytic on a disk possesses a primitive.

EXERCISES 2.3. Show that iffis a complex-valuedfunction whose domain is the convex hull of {a, b, c} c C, whese lim, [ f ( w )-f(z)]/(w - z ) exists at each point z of the domain of A then I ( T ; f ’ ) = 0 where T = (a, b, c) and the I notation is understood to be taken in

the present context. Given f : 0 +C where 0 is an open subset of C and f is continuous on 0. Suppose thatfhas the local primitive property: for each z E 0 there exists Y > 0 for

2.4.

V. Cuuchy Theory

108

which A(z; r) c 0 such that flA(z; r) possesses a primitive. Show that the conclusion of the simplicia1 Cauchy-Goursat theorem holds. Hint: Make sufficiently many repeated quadrisections. 2.5. Let b E 0 where 0 is an open subset of C. Suppose thatfis analytic on 0 - {b} and that, in addition, f’ is continuous on 0 - {b}. Suppose that

b € A ( a ;r ) c A(a; r ) c 0,

0 < r < +a,.

Let

For r E ]0,1] let

y(s, r ) = b(1 - r )

+ (a + re‘%

OIsI2n,

O 0. Let n be a positive integer so chosen that 0,(,/2/n) < m. We introduce Zjk

= q ( j / n ,Wn),

sjk

= [ j / n ,( j

+ l)/n]

... , n ; j , k = 0,. . ., n - 1.

j , k = 0, x [k/n,(k

+ l)/n],

t Given A , E c X a metric space with metric p, by the distance between A and B, denoted dist(A, B ) , is meant inf (p(a,b): (a, b) E A x B } .

V. Cauchy Theory

110

We have d s j k )

A(zjk;

m) O.

Now let g j k denote a primitive off I A(zjk; m). It is trivial that the sum of the quantities A j k , wherej, k = 0, .. .,n - 1, is 0,each being itself 0: Ajk

= [gjk(zj+l,k)

+ [gjk(Zj+l,k+l) - gjk(Zj+l,k)l - g j k ( z j + l . k + l ) l + [gjk(Zjk) - g j k ( z j , k + l ) l .

- gjk(zjk)l

+ [gjk(zj,k+l)

We now observe the important fact that on A(zjk; m) n A ( z j + l , k ; m),which is a region, g j k ( Z ) - g j + l , k ( Z ) is independent of z, j + 1 < n. A corresponding observation holds for gjkand g j , k + l , k 1 c n. Hence

+

- gjk(Zj+l,k)l

[gjk(Zj+l,k+l)

=o,

+ [gj+l,k(Zj+l,k)

- gj+l,k(Zj+l,k+l)l

j+l r, then (Z - b)-’

dz = 0,

r)

$,(a;

first with the aid of Theorem 3.3 (deforming r,(t)= a circles” to the constant r,(t)= a), and second using

+ rezn“ through “concentric

3.5. Given f analytic at each point of A(u; r), 0 < r
b}, u E C(0; l), b E R, are convex, they are homotopically simply connected. From the point of view of analytic functions on homotopically simply connected regions the following theorem is fundamental. Theorem 4.1: Given f analytic on a homotopically simply connected region A c C , there exists g analytic on A such that g’ = J That is, f possesses a primitive. PROOF: We fix a E A and let r denote the class of rectifiable paths y in A with domain [0, 11 and initial point a. It is easy to see that A = {?(I): E r).

(4.1)

A homeomorph of a topological space is its image with respect to a homeomorphism.

1Other equivalent formulations may be given in terms of the fundamental group of X (cf.

W).

4. Simple Connectivity,Primitives

125

Indeed, the argument that was used to show that a region in C is arcwise connected shows that, given b E A, there exists a rectifiable path y in A with initial point a and terminal point 6 . We propose

By the homotopic version of the Cauchy integral theorem and (4.1) we see that g is a function with domain A. Further, given z E A(b; r ) c A , we see that

g(z) - g(b) =

I

f ( w >d w

and we conclude that g'(b) = f ( b ) (cf. the proof of Theorem 2.2, this chapter). Theorem 4.1 follows.

EXERCISES Show that the conclusion of Theorem 4.1 persists when the following condition replaces homotopic simple connectivity: there exists a E A such that all paths y in A with domain [0, I ] satisfying y(0) = y(l) = a are homotopic to the constant path y,(t) = a, 0 I t 5 1, relative to A. Actually this condition is equivalent to homotopic simple connectivity. 4.2. Show that f : 0 + C,where 0 is an open subset of C,possesses a primitive if and only iff is analytic on 0 and 4.1.

IY

fdz=O

for every closed rectifiable curve y in 0. *4.3. Let R be a homotopically simply connected region in R2.Let g, h E C'@) and satisfy gv = h, . Show that there exists f E C ( R ) such that fx = g, f, = h. It is suggested that Theorem 4.1, this section, serve as a model, the line integrals being of the form

1s d x + h d r ,

' Y

where the y are rectifiable paths in R issuing from a given point of R. Of course, the essential definitions concerning line integrals of the indicated form are wanted. This exercise is intended for readers with the necessary background.

Holomorphic simple connectivity. A region in C is said t o be holomorphically simply connected provided that every function analytic on it possesses a primitive. Theorem 4.1 says that a region in C that is homotopically simply connected is holomorphically simply connected. The converse is true, but it is

V. Cauchy Theory

126

not so immediate. A proof of the converse may be based on the Riemann mapping theorem (cf. $1, Ch. XVII of this book), which may be stated: if A is a holomorphically simply connected region which is a proper part of C, there exists a univalent analytic function with domain A and image A(0; 1); and on the fact that the inverse of a univalent analytic function is analytic (for the latter theorem, cf. Theorem 9.1, this chapter). The following theorem is useful for establishing holomorphic simple connectivity. Theorem 4.2: Let A and B be holomorphically simply connected regions in C with A n B # 0, connected. Then A u B is holomorphically simply connected. PROOF: A u B is a region. Suppose that f is analytic on A u B. Let a E A n B; let g denote the primitive off I A satisfying g(a) = 0 ; let h denote the primitive off I B satisfying h(a) = 0. Then g ( z ) = h(z), z E A n B, since g(a) = h(a) and g’(z) =f ( z ) = h’(z). Hence g u h is a primitive off. The

assertion follows. Examples. C - { a + t : t E R’}, a E R, is holomorphically simply connected. It admits the representation

[{Im z < 0) u {Re z < a } ] u {Im z > O}.

(4.2)

Similarly, C less two slits, one { a + t: t E R’}, the other { b - t: t E R’}, b < a, is holomorphically simply connected. To see this it suffices to replace {Re z < a } by { b < Re z < a } in (4.2).

EXERCISES 4.4.

Show that C - {0}is not holomorphically simply connected by calculating

fc(o;l,z-l dz*

4.5.

Let a denote the primitive of (1 {i(r+l):tER+}

+ z2)-l on A = C less the slits and

{-i(t+l):tER+}

satisfying a(0) = 0. Let f ( z ) = tan[a(z)] at the points z E A for which a(z) is not of the form (n + +)r,n E Z. Show that

for z in the domain offand conclude that aLf(z)] that f’is the identity map of A onto itself.

= a(z) for z

small and thereupon

5. Analytic and Continuous Logarithms and Powers

127

5. ANALYTIC AND CONTINUOUS LOGARITHMS AND POWERS Let f be analytic on an open set 0 c C and such that f ( z ) # 0 for all 0. We seek a necessary and sufficient condition for f to possess an analytic logarithm. Suppose that g is an analytic logarithm off. From f = exp 0 g we obtain, using the chain rule, f ' =fg'. Hence a necessary condition for f to possess an analytic logarithm is that f 'lfpossess a primitive. This condition is suficient. For suppose now that g is a primitive of f ' l f . We see that lfexp 0 (-g)]' is the constant zero and hence f exp (-9)= c, where c is constant on each component of 0, that is, the restriction of c to a component is constant. Let k be constant on each component of 0 and satisfy c = exp 0 k. Then g k is an analytic logarithm off. To sum up we have z

E

0

+

Theorem 5.1: Givenf analytic on an open set 0 c C such that f ( z ) # 0, z E 0. Then a necessary and suficient condition that f possess an analytic logarithm is that f'lfpossess a primitive. From our earlier study of logarithms and powers we see now that iff as above possesses an analytic logarithm and h is analytic on the domain o f f , then there exists an analytic hth power off. In case 0 is a holomorphically simply connected region we are guaranteed that, for everyf analytic on 0 not taking the value 0, the logarithmic derivative f 'If admits a primitive, and hencefhas an analytic logarithm. It is to be noted that this assertion appeals to the analyticity off '. EXERCISES Let f be a continuous map of a topological space X into C - {O}. Suppose that g is a continuous logarithm off: Suppose that a E X has the property that for each neighborhood U of a there exists a connected neighborhood of a contained in U.Then there exists a neighborhood V of a and an analytic logarithm L of the identity map on A [ f ( a ) ; If(a)I] such that g(x) = L [ f ( x ) ] ,x E V. 5.2. Let c; akZk= Log( 1 + el), z small. Let p denote the radius of convergence of c,"akzk. Show that the sum of the series on A(0; p ) is an analytic logarithm of 1 + ez restricted to A(0; p ) and determine p explicitly from properties of the exponential function. 5.3. Show that I - z 2 , z E C - [ - 1, 11, possesses a unique analytic square root S satisfying lim, iz-' S(z) = 1 and show by the Cauchy integral theorem and formula for a semicircle with two indentations that 5.1.

dx

=

ni[S(a)]-1,

The usual convention of real analysis applies to 1/(1

a € C - [-1, - x2).

11.

V. Cuuchy Theory

128

Let X be a connected topological space and let a E C. Show that the continuous ath powers of the constant 1 on X are the constant functions taking the values e2n'ak, k E C.

5.4.

We now consider continuous logarithms in situations where the domain has a special simple character and we shall obtain the essential apparatus to treat the notion of the topological index (= order of a point with respect to a curoe) which enters very naturally into the Cauchy theory as we shall soon see. Suppose then that f is a continuous map of [a, b ] into C - { 0 } , -a < a 0. Now let n denote a positive integer satisfying o f [ n - ' ( b - a)] < m. Let Xk

= a f n-'k(b

- a),

k = 0, ..., n.

Since If(x)

-f(Xk)l

<m 5

lf(xk)l,

xk+ 1, k = 0,

xk 5 x

., n - 1,

we see that f(x)EA[f(xk);

lf(xk)ll,

xk

5

5 Xk+l-

The composition of a continuous logarithm of the identity map on A [ f ( x k ) ;If(xk)I] with f l [xk, & + I ] yields a continuous logarithm ik of f I [xk, X k + l ] . We notice that I k + l ( X k + l ) - ik(Xk+l) is an integer multiple of 2ni, k = 0, . . ,n - 2. Starting with lo fixed we adjust the successive ik to be concordant by adding appropriate constant integer multiples of 2 d s o that the resulting i k satisfy ik+l(xk+l)= / k ( X k + l ) , k = 0, . .. , n - 2. Thus 1, is replaced , We see that with the ik so modified UOskSn-l Ikis by ll [lo(xl)- l l ( x l ) ] etc. a continuous logarithm off. Now that the case of the segment has been treated, it is easy to treat the rectangle and certain important unbounded domains. Let f now be a continuous map of [a, b] x [c, d ] into C - { 0 } , where a, b are as above and - 00 < c < d < +a.The procedure is the same as that for the case of a segment up to the point of adjusting the individual continuous logarithms associated with the subrectangles. Thus we suppose that a positive integer n has been found such that

.

+

f I Cxj, x j + l l

b k , yk+ll

has a continuous logarithm i j k , 0 lj,k 5 n - 1. Here xi = a + j n - ' ( b - a), yk = c kn-'(d - c ) , j , k = 0, . . .,n. We note that

+

zjk(xj+l,

Y ) = lj+ l,k(Xj+

1,

Y)

+ 2ni m ( j , k),

where m(j, k ) is an integer, yk I y I yk+l. Hence iok, . .. ,

may be

5. Analytic and Continuous Logarithms and Powers

129

rendered concordant so that

u

ljk

Osjsn-1

is a continuous logarithm L k off I [a, b] x Lk(x,

y k + 1)

= Lk+

l ( X 3 y k + 1)

where p ( k ) is an integer, k rendered concordant so that

= 0,

[yk

+ 2xi d k ) ,

.. . ,n - 2.

u

, y k + l ] . This time a

x 5 b,

Hence L o , .. .,Ln-l may be

Lk

Osksn-1

is a continuous logarithm off. Two other important cases are those where the domain of the function in question is either R or R x [0, 11. Suppose that f is a continuous map of R into C - (0). Let c1 denote a given logarithm of f(0). For each positive integer n let ln denote the continuous logarithm off I [ -n, n] satisfying 1,(0) = tl. Then, as is easily seen,

is a continuous logarithm off. The same argument holds when the domain is

R x [0, 11, the normalization now being made at (0,O) and [ -n, n] x [0,1] replacing [ - n, n]. Application to closed curves. We recall that the existence of continuous logarithms implies the existence of continuous arguments. Let y be a closed curve and let a be a point of C not on y. Let 0 be a continuous argument of y - a. Because of the fact that y ( t + 1) = y ( t ) , t E R, we see that

2x is an integer independent o ft. It is also independent of admitted 8. We term the common value of (5.1) the order of a with respect to y or the topological index of y at a and denote it by O ( a ; y). Let us now see that O ( a ;y) is a homotopy invariant; that is, when yo is homotopic to y1 relative to C - { a } ,then O ( a ; yo) = O ( a ;y l ) . For this purpose suppose that f is a continuous map of R x [0, 11 into C - ( 0 ) satisfying f ( x + 1 , y ) = f ( x , y ) , ( x , y ) E R x [0, 11. Let cp denote a continuous argument off. Then

V. Cauchy Theory

130

is an integer independent of (x, y ) . If we now consider closed curves yo and y1 where yo is homotopic to y1 relative to C - { a } , we see that there exists f of the stated type such that f(t, 0) + a = yo(t), f(t, 1) a = yl(t), t E R. We have

+

O ( a ; Yo) = 4f) = o(a ;71)

and the asserted invariance follows. Suppose now that b E A(a; r ) c C - y. We define f on R x [0, 11 by f(s,

0 = Y ( S ) - [(I - ?)a + tbl,

and considering a continuous argument off we conclude that O(a;y) = O(b;y). By the standard connectedness argument we see that O(z; y ) is independent of z on a component of C - y. We shall be more precise about the unbounded component. For if la1 > maxR (71, noting that

we see that O(a; y) = 0 since Log[l - y ( t ) a-'1 is a continuous logarithm of 1 - y(r) u-' left unaltered by the translation taking r into t + 1. In $6, this chapter, we shall see the importance of the notion of the order of a point for the Cauchy theory.

EXERCISES

+

Let f b e a continuous map of [a. 61 into C - {O}, -a < a < b < co. Let be the increasing sequence whose image is a given partitioning of [a,b]. Let a denote a continuous argument off and let ak denote a continuous argument of f I [ f t , t k + ~ ] , k =,..., O n-1.ShOwthat 5.5. (t&

c

n-1

4)- 44

=

0

[ak(tr+

- ak(tS1.

This simple rule is very convenient for applications. Let y(t) = exp(2nir), t E R. Show that O(0;y) = 1 and conclude that C - (0) is not homotopically simply connected. ~ Iz(< r 2 } , 0 I r~ < r2 Ico. Show that if yo and yl are closed 5.7. Let A = { r < curves in A, then y o is homotopic to ylrelative to A if andonly if O(0; y o ) = O(0;yl). Consider the corresponding question for paths in A which have a given point a E A as both initial point and terminal point.

5.6.

+

A lemma on order. Let y be a closed curve in C - {O}. Suppose that Re y takes the value 0 only a finite number of times on [0,1[. Re y is said to change sign at t provided that

5.8.

sg[Re y(t

+

E)

Re y(t - &)I= -1

6. Certain Integrals Representing Anahtic Functions

131

for E small and positive. Show that 2 lO(0; y ) 1 does not exceed the number of points on [0, 1 [ at which Re y changes sign (R. J. Backlund). 5.9. Determine with the aid of the results of this section the continuous (respectively, analytic) maps fof C into itself satisfying f(r + w) = f ( z ) f ( w ) , I, w E C; of C - {0}into itself satisfying f ( z w ) = f ( z ) f ( w ) , z, w E C - {O}. Hint: For the first part consider a continuous logarithm off’in the nontrivial case, for the second part consider f o exp. 5.10. Let f be a continuous map of A(0; 1) into itself. Show that there exists a E A(0; 1) satisfying f ( a ) = a . [This is a special case of the Brouwer fixed point theorem.] Hint: Suppose the assertion false and consider O(0; y r ) and O(0; rr), 0 < t < 1, where yr(s) = f ( t e Z n i s ) teZnir, ~

r,(s)= tf(eZzis)- eznia ,

SER.

rhis ingenious argument is due to A. P. Morse. Let -a, < a < b < co. Let A and B be continuous real-valued functions with domain [a, b]. Let w : [a, b] + C - {0} be such that w” exists and w satisfies w” + Aw’ + Bw = 0. Let 0 denote a continuous argument of w. Show that if 0 is not constant, then either e(s) < 0(t) for all s, t satisfying a I s < t I b, or e(s) > 00) for all s, t satisfying a I s < t Ib.

5.11.

+

6. CERTAIN INTEGRALS REPRESENTING ANALYTIC FUNCTIONS The Cauchy integral formulas are special instances of integrals representing analytic functions. As a preliminary to the study of such integrals we first consider functions defined on a product set which are “analytic in the first variable and are continuous in both.” Such functions were examined systematically in a euclidean setting by B6cher [136] who termed them semianalytic. We shall see that they have interesting properties. To be specific, let 0 be an open subset of C and let X be a compact metric space. Let F be a continuous map of 0 x X into C and for each p E X let the function fp assigning the value F(z, p) to z E 0 be analytic on 0. We define Gon 0 x 0 x Xby

Theorem 6.1:

G is continuous on 0 x 0 x X.

PROOF: The Cauchy integral formula for a disk plays an essential part. We consider continuity at a point for which the first two components are equal,

V.

132

Cauchy Theory

for the continuity is evident in the other case. Let p , q E X,let A(a; 2r) c 0, 0 < r < +a,and let b, c E A(a; r), b # c. We have G(b, C , 4) - G(a, a, P)

From (6.1) we conclude that IG(b, c, q) - G(a, a,p)I does not exceed r-' times the value of the modulus of continuity of F I [A@; 2r) x XI at max{Ib - al, Ic - a l l + P(P, 9). Reference should be made to Ex. 5.10, Ch. 111. We employ the metric in 0 x Xfor which the distance between (z,p ) and (w, q ) is Iw - zI p ( p , q). It induces the topology assigned to the product space. Similarly, in 0 x 0 x X the metric is that which assigns the distance

+

- Zll + Iw2 - z21 + P(P, 4)

1%

between (zl,z2,p ) and (wl, w 2 , q). We have used the fact that if z E [b, c], then

Iz - a1 I max{lb - al, Ic - al}. We conclude that given (a, p ) E 0 x X and q positive there exists 6 positive such that when Ib - al + Ic - al

+ P(P, 4) < 6,

we have IG(b, C, 4) - G(a, a, P)I < v subject to the stipulation b # c. The case b = c now follows by referring to the definition of G and taking the appropriate limit. The theorem follows.

- -

There is an important corollary. We note that G I A(a; r ) x A(a; r ) x X is uniformly continuous. Hence, given q positive there exists 6, 0 < 6 < r, such that IG(a, b,

P) - G(a, a, P)I < V ,

P E X,

(6.2)

when Ib - a1 < 6. We see that (6.2) implies the uniformity with respect t o p of the differentiation process. Let us apply the results we have obtained to the special case where

6. Certain Integrals Representing AnaQtic Functions

133

X = [0, I ] and consider functions generated from F by integration. Specifically, let p: [0, I ] --+ C have finite variation. Let

f(4 = JolF(z, 0 440,

z E 0.

Thanks to (6.2) we see by direct examination that for each a E 0 we have the existence off'(a) and the equality

f'@)= JolG(u,

a, 0

440.

It follows th a tf is analytic on 0. EXERCISES Let Fl(z,p) = G(z, z , p ) , z E 0,p E X . Then Fl is continuous on 0 x X. 6.2. Givenfanalytic on an open set 0 c C. Suppose that Kis a nonempty compact subset of 0. Let r be a positive number less than dist(K, C - 0).We define 6.1.

g(z, w) = [ f ( z

+ w) -f(z)lw-l,

z E K, 0 < IwI Ir,

z E K.

g(z, 0) = f ' ( z ) ,

Show that g is uniformly continuous. Show more generally that if n E N,the same holds true for g. with domain K x A(0; r/n) defined by gdz, w) =

[

(;)(-

I)"-kf(z

1

+ kw) w - " ,

w # 0;

gdz, 0) =f'"'(z).

+

6.3. Letf: [a,b] C, - m < a < b < m. We suppose thatf'(x1 exists, a Ix I b, one-sided derivatives being employed at a and b. Let --f

aI x ,y

Ib,

and let g(x, x )

= f'(x),

a Ix I b.

Show that g is continuous on [a, b] x [a, 61 if and only if f' is continuous on [a, bl. 6.4. Let n E N.Let X be a compact subset of C and let 0 = C - X . We introduce F(z, w ) = ( w z)-", ( z , w) E 0 x X. Clearly Theorem 6.1 of the present section is applicable. However it is desired to show directly that, given K compact c 0, the following holds: For each positive 7,there exists 6 positive such that ~

V. Cauchy Theory

134

when (z, w ) E K x X , 0 < 15 - zI < 6. The elementary identity, n- 1

is of service. Unessential computation is to be avoided. 6.5. We consider the expansion of F ( z , p ) of Theorem 6.1, this section, about c E 0 for given p E X, namely

Let 0 < r < +a, and suppose that A(c; r ) c 0. Then the series converges uniformly and absohtely on A(c; r ) x .'A Show that if Xis a segment (c R)containing a and

exists for each z E 0, the limit process being uniform on each compact subset of 0, then &'(a) exists, k E N,and m

Ft(z, a) =

2&'(a)

(z - c)~,

z E A(c; r ) c 0.

0

[Bkher, loc. cit.] We now turn to some special cases of integrals of the type considered which are of particular importance. The first is the integral of Cauchy type. The data consist of a rectifiable path y in C and g : E -, y c E c such that g y takes values in C and is continuous on its d0main.t By the integral of Cauchy type associated with (y, g, z), z E C - y, is meant

c,

I

2ni

e,

0

P

g(W) (W - z)-' dw.

From the preceding developments of the present section we see that f is analytic on C - y and by an evident inductive argument that

S,

f("'(z) = - g(w) (w - z)-"-' dw, Z",!i

z E C - y,

n E N.

Let us replace y by a rectifiable closed curve r in C and specialize g to be the constant 1 on C. We see that the function f given by the integral

t We recall the notational license used in the case of paths.

6. Certain Integrals Representing Attabtic Functions

135

has a derivative equal to zero at each point of C - r, since for fixed z E C - r, - z)-', w E C - {z}. However, we may establish a much sharper statement:

(w - z ) - ~is the derivative of - ( w

f(z) = o(z; r),

E

c - r.

To see this, we fix a E C - r and consider A(b; r ) c C - { a } . If y is a rectifiable arc with domain [ct, j?]lying in A(b; r ) and I is an analytic logarithm of z - a, z E A(b; r ) , we have s,(w - u1-l dw = 4")1 - rCr(41.

To treat our assertion for r, we let m = min, Ir - a1 and take n a positive integer so large that wr(n-') < m. We see that

r(t)E ACr(kn-');

m],

kn-' I t I ( k

+ 1)n-l

On considering the contribution to (6.4)

from [kn-', (k + l)n-'] and summing, we obtain the value

where ctk is a continuous argument of r(t)- a, kn-' 5 t I ( k + l)n-', k = 0,.. . ,n - 1, the contributions from the real part of the summands being zero because of the periodicity of r. Now by Ex. 5.4, this chapter, we see that (6.4) is simply O(a; r).We now see how the notion of order enters the Cauchy theory significantly (cf. the terminal remarks of 53, this chapter).

EXERCISES 6.6. Letfbe analytic on A, a holomorphically simply connected region in C. Let y denote a closed rectifiable curve in A and let a E A - y. Then O ( a ; y ) f ( a ) = -1 j 2m

f(z)(z-a)-'dz.

6.7. Let y denote a closed rectifiable curve in C. Show that O(a: y ) proceeding from

pv

-a)-l dw = -u-I

j

@kdw. y

o

= 0 for

la1 large

V. Cauchy Theory

136

The analyticity of O(a; y) on C - y implies the vanishing of O(a;y) on the unbounded component of C - y. 6.8. Given

an integral of Cauchy type. Let a E C - y. We have n

=

(z 0

k!

1

g(w)

z-a

"+'

w--z(F ~ dw. )

6.9. Calculate for n a positive integer

if 2m

W-" C(0; 1) w - z

-dw

for IzI < 1, for Iz( > 1. Given g: E C,[0, 11 c E c C,g continuous on E, g analytic at each point of 10,1[, and y the identity map of [O, 11 onto itself. Show that the functionfdefined by the integral of Cauchy type associated with (y, g, z ) has the property that for each a E l O , l [ there exists tp analytic on a disk centered at a such that tp u f+ is analytic wheref, = f l {Im z > 0). A similar result holds forf- = f l {Im z < 0). 6.11. Calculate 6.10.

--f

Jz - z)- * dr, (t

z E C - [0, 11,

and study its limiting behavior at points of [0, 11. Two other classical instances of integrals involving parameters which occur in the theory of analytic functions are those of Euler and Laplace. They appear in the study of certain linear ordinary differential equations of the second order which arise in problems of mathematical physics. It is not our intention to develop the theory of these integrals in detail but rather just to illustrate the use of integrals representing analytic functions. For details concerning the Euler and Laplace integrals, cf. [34].

The Euler integral. It will be seen that the subject offers excellent experience in the use of continuous and analytic powers. Our data consist of the following: a E C, y a rectifiable path in C with domain [0, 13, a disk A@; r ) having no points in common with y, g: E --t where y c E c and g 0 y is a continuous map of [0, I ] into C.In what follows we are treating the matter locally when a 4 Z. One writes formally

e,

c,

6. Certain Integrals Representing Anahtic Functions

137

but naturally the notation must be interpreted. What is intended by an Euler integral associated with ( y , g, a, z), z E A(a; r), is meant J;g

O

fit)

K(z9 t ) & ( t )

where K is a continuous ath power of y ( t ) - z, (2, t ) E A(a; r ) x [0, 13. It is not hard to see that such K exist. In fact, K ( z , t ) = exp[a L(z, t ) ]

where L is a continuous logarithm of y ( t ) - z, (z, t ) E A(a; r ) x [0, 11. For each t E [0, I ] the function assigning L(z, t) to z E A(a; r ) is analytic. We conclude that K is analytic in z on A(a; r ) for each t E [0, 11 and further that

K,(z, t ) = - - ~ 1 K ( z , t ) [ y ( t ) - z]-’, K , denoting the derivative of K with respect to z. For the derivative of the functionfgiven by the above Euler integral we have the formula

which is useful in practice. For specific applications, cf. [34]. The Laplace integral. Here the data consist of y and g as above. By the Laplace integral associated with ( y , g, z), z E C, is meant

s,

f ( z ) = g(w) ezwdw. Here no questions of interpretation arise. The functionfis analytic on C. It is immediate that f’(z)=

S,.

g(w) ezw dw.

If g is analytic at each point of y, we have

s,

zf(z) = g[y(l)]ezY(l) - g[y(0)]ezy(o)- g’(w) erw dw. It is to be remarked that we may consider e-zw in place of erw,as is often done. The two exercises that follow show how the Laplace integral may be applied to the study of certain differential equations. It will be seen that skill in choice of y and g plays an important part in making the method effective.

V. Cunchy Theory

138

EXERCISES

.. + 1,

6.12. Let n EN,ut E C, k = 0,. ,n

y

U.+I

# 0. We seek solutions of

sky"' = 0

(6.6)

0

of the form (6.5) by a suitable choice of y and g. Let

Here al,..., a, are distinct complex numbers, p l , ..., pv are positive integers. Clearlyfis a solution of (6.6) if and only if / ? P ( w ) g(w) ezwdw = 0. Show that a nontrivial solution lies at hand when g ( w ) = (w - a J ) - kk, = 1, .., ,p J , la,[ < r, and determine the solution explicitly with the .. ,Y, and y(t) = aid of the Cauchy integral formulas for the derivatives of an analytic function. 6.13. Suppose that the coefficient Lzk of (6.6) is replaced by ak btz, where ak,b, E C. Suppose that g is analytic on { p < lz( < +a), p E R+,and that y is a rectifiable closed curve in { p < IzI < w). Show thatfof (6.5) is a solution of (6.6) so modified if and only if

j = 1,

.

+

+

[7W"w)

- QWlg(w) -

g'(w)k'" dw = 0

where Q(z) = Do+bkzk. The case of the Bessel equation of order 0 is interesting here. For it P(z) = z and Q(z) = z 2 I. Show that if S is the unique analytic square root of 1 z2 with domain C - [-i, il satisfying lim, z-' S(z) = 1, then (2riS)-' serves for g. Show that

+

+

1 2m

Jo(z) = -

$

C(O;,)

e="

-dw, S(w)

1 r r be suchthat { r k < / z - akl I S k } c D. Let Ek denote the function E and the function I of Ex. 1.3, this set, relative tofl {rk< ) z - a k / < s t } . Show that fadmits a representation of the form

where g is analytic on A(u; r). Hint: Consider n

+

. . . , n. Show that a representation of the type g(z) c7 Ek(z), where g is analytic on A(u; r ) and Ex is analytic on c - Dk ,lim, %J = 0, k = 1, . . .,n, is unique. Conclude from (1.3) that if s < r satisfies [at - a1 s k < s, k = 1, .. ,n, then z E A@,; s,), j = 1,

+

.

n

and

n -

2 1

f(w) (w - z)-l dw, z E A(u; s) 2T1 C(0k;Sk)

U

lakan

A(ah;SJ.

These two results comprise of course the Cauchy integral theorem and integral formula for a closed disk less the union of a finite number of disks whose closures lie in the interior of the given disk and are mutually disjoint. 1.13. Let f ( z ) = (ZEakzk)/Q(z)where Q(z) = z"+' c5 bkz', n E N,with domain C - { Q ( z )= 0). Apply (1.3) of the preceding exercise to show that f admits a representation of the form

+

where cl, . . . , cm are the distinct roots of Q(z) = 0 and Pk is a polynomial of degree. 5 n(c,; Q ) satisfying Pk(0)= 0. [Recall that the polynomial that is identically zero has degree - co. Generally, if p ( z ) ckzk where sup{k: Ck # o} < 0 3 , the

=c;

+

158

VI. Laurent Expansion. Meromorphk Functions

degree of P is just sup{k: ck # o}.] Show that if c”, a k Z k = 0 has no roots in common with Q(z) = 0, then the degree of Pk is exactly n(ct; Q). Show that generally if clr.. . , cmare distinct points of C, d,, . ..,d, are distinct points of C, PI, . . .,P, and Ql, .. . , Q, are polynomials of positive degree such that

Pi(0) = * * * = Pm(0) = Qi(0) = * * *

= QN(0)

=0

and

then { c l , ...,Crn} = {dl, . .. ,d,} so that p = m. The mapj(k) of < 1 , m > onto itself Satisfying dk = C j ( k ) , k = 1 , . . ,M , is Such that Qk = P j ( k , , k 1 , . . .,m.

.

1.14. Obtain the representation (1.4) for 1

f ( z ) = (z - l)(z - 2 ) y Z

- 3)3 *

Obtain the Laurent coefficients of f l A ( 0 ; rlrr z ) for the following (r,, r 2 ) : (1, 2), (2,3), (3, +a). 1.15. Suppose that f is analytic on C and satisfies: ( 1 ) f ( z 279 = f ( z ) , z E C, (2) there exist positive numbers A and B such that If(z)I I AeBIzl,z E C. [fsatisfying (2) is said to be of exponential type.] Then f = g 0 exp for a g of the form g ( z) =&, & t k , where n E N (Boas). Hint: Introduce

+

h = {(ez,f(z)):z E C}, show that h is analytic on C - {0},show that its kth Laurent coefficient is given by 1 -

2r

1

2n

f(x

+ yi)e-k(x+yi) dy,

x E R,

0

and estimate the size when Ikl > B. 1.16. Let (bt): Z +C. Suppose that c, d~ C satisfy 0 < IcI < Id\ < co and that limn*mc”-. bkzkexists and is in C when z = c and when z = d . Show that the components of the Laurent pair z?: bkzkconverge, respectively, when IzI < Id1 and when IzI > IcI. 1.17. Let bk = (lkl f l)-’, k E Z. Show that the intersection of the convergence sets of the components of the Laurent pair x22 btzk is C(0; 1 ) and that the Laurent sum is continuous.

+

2. ISOLATED SINGULARITIES OF ANALYTIC FUNCTIONS Let a E C and let f be analytic in A(a; r) - {a} for some r > 0. Under this condition we say that f has an isolated singularity a t a.7 The Laurent sequence, ( b k ) k & , off I [A@; r) - { a } ] is independent of r > 0 for which f is analytic in A(a; r ) - {a}. We term the common sequence in question the a-laurent f Of course, it is not excluded that fmay be analytic at a.

2. Isolated Singularities of Analytic Functions

159

sequence o f f . We consider LY = inf{k: b, # 0 } and introduce the following definitions. We say that f has a removable singularity at a provided that LY 2 0; that f has a (polar singularity or) pole at a provided that - 00 < a < 0; that f has an essential singularity at a provided that LY = - 00. When f has a pole at a, we term la) the order of the pole at a. The first two possibilities are easily studied. (1) Removable singularity. It is clear that f has a removable singularity at a only if there exists g analytic at a such that for some r > 0 we have f ( z ) = g(z), for all z satisfying 0 < Iz - a1 < r. The function f given by f ( 0 ) = 1, f ( z ) = z, z # 0, has a removable singularity at 0 as do z with domain C and z with domain C - (0). (2) Pole. I f f has a pole of order m at a, then when r > 0 is such that f is analytic in A(a; r ) - {a}, the series b-,+,(z - a), converges when Iz - a1 < r. With g denoting its sum, we have

1;

Hence lim, f

= 00

since g(a) = 6 - , # 0.

(3) Essential singularity. We come to a famous theorem that gives information on the behavior off near a when f has an essential singularity at a. It is the following theorem of Weierstrass.

fme

Theorem 2.1 : I f f has an essential singularity at Q, then = for every V contained in the domain o f f which is of the form U - { a } , U a neighborhood of a.

PROOF: Suppose that the assertion is false. Then there exist r positive, m positive, A E C, such that f is analytic in A(a; r ) - {a} and If ( z ) -A(2 m when 0 < Iz - a1 < r. It is to be noted that if 00 $f ( V ) , then A q! f( V ) for large A E C. By the Riemann theorem, Ex. 3.10, Ch. V (a second proof, Laurentian in character, is indicated in Ex. 2.1 below), we see that C f ( z ) - A1 - = ( z - a)” g e l ,

(2.1)

0 < ) z - a1 < r, where n E N and g is analytic on A(a; r ) and does not take the value 0. Hence taking reciprocals of both sides of (2.1) we see that (b,), the a-Laurent sequence off, has the property that b, = 0, k < -n. The hypothesis is contradicted. The theorem follows.

160

VI. Laurent Expansion. Meromorphic Functions

Some remarks are in order. First, the conclusion of Theorem 2.1 is frequently given the following equivalent form : For each w E there exists (Zk)k& such rhar limk..+m zk = a, each zk f a and is in the domain o f f , limk-,mf(zk) = w. It is routine to verify that the assertion just stated implies the conclusion of Theorem 2.1. In the other direction, we note that for n a sufficiently large positive integer A(a; n-') - { a } is contained in the domain offand that for M ? ECandsuchnthereexist~z~satisfyingO < Iz, - a1 < n-', If(z,) - WI < n-', and that for such n there exists z, satisfying 0 < Iz, - a1 < n-', If(z,)I > n. Hence the conclusion of Theorem 2.1 implies the above assertion and equivalence follows. Second, the propositions converse to those stated in (1) and (2) and to Theorem 2.1 hold because of the disjunction of the conclusions. (i) Thus, if g of ( I ) exists, thenf has a removable singularity at a (as, of course, is immediate directly) since f cannot have a pole or essential singularity at a. The hypothesis of the Riemann theorem is a sufficient condition for a removable singularity, and we shall see in Ex. 2.1 below that a weaker hypothesis suffices. (ii) If lim,f= co, thenf has a pole at a since, by (1) and Theorem 2.1, f cannot have a removable singularity nor an essential singularity at a. (iii) Iff fulfills the conclusion of Theorem 2. I , then by (1) and (2), f cannot have a removable singularity or pole at a. Third, the following condition, readily verified, is useful: iffhas an isolated singularity at a and a is a cluster point of {f(z) = 0}, then either f has an essential singularity at a orf(z) = 0, z E A@; r ) - {a}, for some r > 0.

e,

EXERCISES 2.1. Given fanalytic in A(u; p) - { u ) , show with the aid of the Cauchy inequalities for the a-Laurent coefficients that iffdoes not have a removable singularity at a, then limo r M(r)exists and is greater than 0 (possibly co). Here

+

M ( r ) = max{If(z)l: z E C ( a ; r)}.

The Riemann theorem is an immediate consequence. For a removable singularity the condition limn+ r. M(rJ = 0 suffices where (r.) is a decreasing sequence of positive numbers satisfying ro < p, limn-, r. = 0. 2.2. Let f ( z ) = exp( -z-~),z # 0. Show that fhas an essential singularity at 0 and explain why this fact shows that the Taylor series about 0 of the function F: R + R given by F(0) = 0, F ( x ) = exp( -x-'), x # 0, which has derivatives of all orders on R, cannot have sum F I E on E where 0 < inf E < 00, inf E qt E. 2.3. Let f ( z ) = exp[(l - z ) - ' ] , z # 1. Show that f has an essential singularity at 1 and determine its 1-Laurent sequence. Determine the Laurent coefficients of f l A ( 0 ; 0 , l ) and of f l A(0; 1, fa). 2.4. Let fbe analytic in A(0; r ) - { 11, r > 1, and let fhave a pole at 1. Let E;o"atzk be the power series expansion off in A(0; 1). Show that, for some m E N, we have un# O for n 2rn and lirnh+m~ ~ + k + , / u , , , = + ~ 1.

+

2. Isolated Singularities of Analytic Functions

161

2.5. Let fand g be analytic on A(a; r ) , r > 0, and suppose that f ( u ) # 0, g(u) = 0, g ( z ) # 0 when 0 < Iz - a1 < r . Let h [email protected] h has a pole of order u = n(u;g ) at a and the term bk, k 2 - v, of the a-Laurent sequence of h is

(v

+! k ) ! ( ~ ) ‘ ‘ ” ’ ( a ) ’

where g1 is the function analytic on A(a;r ) satisfying g ( z ) = ( z - a)’g,(z),

Iz - a1

< r.

Obtain the terms b - 2 and b - I of the 2-Laurent sequence of (z - l)-I(z

-

2 ) - * ( z - 3)-3.

2.6. Let n denote an integer >1. Show with the aid of the Riemann criterion for a removable singularity that there does not exist a continuous nth root of z with domain A(0; r ) , r > 0 (or even A ( 0 ; r ) {O}). ~

Isolated singularities at co. It will be convenient to introduce the notation A(co; r ) for the set (z: r - l < lzl I +a},where r > 0.t We are led, of course, to consider functions analytic in some A(co; r ) - {a}and their behavior near co. Parallel to what was said at the beginning of this section, a functionfis said to have an isolatedsingularity at co provided that it is analytic It is natural to consider the Laurent coefficients of in some A(m; r ) - {a}. f I [A(co; r ) - {co}], r small and positive, forf having an isolated singularity at co. Each Laurent coefficient b, is independent of r for which f is analytic in A(co; r ) - {a}. We term the common sequence (bk)the co-Laurent sequence off. We note that g defined by 0 < IZI < I,

g(z) =.f(z-l),

(2.2)

r small positive, is analytic and ( c k ) , the 0-Laurent sequence of g , satisfies k E Z. On this basis we introduce /? = sup{k: bk # 0} and say thatf has, respectively, a removable singularity, pole, essential singularity at co according as /? I0,O < p < co, /? = co. In the second case we term p the order of the pole at 03. Givenf: U -, where U is a neighborhood of co, we say that f is analytic at co provided that f has a removable singularity and is continuous at co. We extend the terminology “analytic in (on) 0”to open in the first case meaning “analytic at each point” in the subsets 0 of sense now available, and, in the second, requiring in addition that 0 be the domain of the function. It now follows that the statements made for a function having an isolated singularity at a point of C hold, mutatis murandis, for a function having an isolated singularity at co. c-, = b,,

+

c,

t We recall the definition 100 I = +a.

c

+


162

EXERCISES 2.7. 2.8.

Verify the last assertion through the mediation of g of (2.2). Suppose that f has an isolated singularity at co. Let M ( r ) =max{If(z)I: IzI = r } ,

+ r large and positive. Show that lim, [log M ( r ) ] /log r exists and that its possible values are in N u { 03). Further the limit is 0 if and only iff has a removable

+

singularity at co, it is a positive integer n if and only iffhas a pole of order n at a, it is co if and only iff has an essential singularity at 00. 2.9. Letfbe analytic on A(a; r, co), let ck denote the kth Laurent coefficient off, and let ( b k ) denote the co-hurent sequence off. Show that

+

+

sUp{k: bk # o} = sup{k: Ck # o}. Show that c - l = b - l and determine c - t in terms of .. ., b - , for k > 0. What are the corresponding formulas for ck , k 2 0, when f has a pole at co ? 2.10. Using the open mapping property of a nonconstant analytic function show that a function f analytic on C and univalent is a polynomial of degree 1, f(z) = az b, a # 0. [It is immediate thereupon that f ( C ) = C.] Hint: Consider f l [A( co ; 1) - { co I]. The converse question is trivial. 2.11. Let n E N. Let

+

n

m

1

the ak being distinct and the vk being positive integers. Determine the co-Laurent sequence (bk) of P'/P in terms of the ak and in terms of the tLk and v k . Show that a. , . .,a. may be represented in terms of b - 2 , .. .,b - ( " + 2 ) and exhibit the determining equations. 2.12. Let f be analytic on C. Let a denote the infimum of the set of nonnegative real numbers A satisfying the condition that there exists r E R + such that If(z)l e"~z~, IzI > r. When a < a,we say that f i s of exponential type a. (1) Show that if g is analytic on { Iz( > r } and y is a rectifiable closed curve in A(0; r , co), then

.

+

+

f(z)

= Jyezwg(w)dw,

z E C,

is of exponential type a r. (2) With the aid of the Cauchy inequalities show that iff is of exponential type a and a < /3 < ico, then ak = f ( k ) ( 0 ) / ksatisfies !

x; k! akzksatisfies

for large k and conclude that the radius of convergence p of p 2 (ae)-' and that the sum F of this series has the property that 1

f

f ( z ) = 2?ri , , , : where l/p < r
0. Exercise 2.5, this chapter, will be found useful. ~ , a E C. (b) (1 z"+')-', n E N. (c) zm(l z2')-', m,n E N. (a) ~ - ( " + ~ 'ne E"N, (d) sin az/[z(z' l)], a E C. (e) tan TZ. (f) csc TZ =(sin m - l . (g) (cosh z ) - ' where coshz = ( e z + cz)/2.(h)(1 z2)-(n+1),,.N.(i)eu'/(l +e'),aEC.(j)f(z)(l +z)-I where f is the analytic ath power of the identity map on C - R + specified by the

+

+

+

+

166

VI. Laurent Expansion. Meromorpliic Functions

condition that f ( z ) = exp(a Log z), Im z > 0. Here a E C. (k) (Log z)(l - z ) - ~ . (I) Let P be a polynomial of positive degree. Determine the requested information for P'(z)/P(z).(m) Givenfanalytic on A(0; r) and P as in (I), determine the requested information for f ( z ) / P ( z )under the assumption that P(z) = 0 has only simple roots, that is, when P(z) = 0, n(z; P)= 1. An invariance property of the residue. Actually, the notion of a residue is best formulated in terms of differentials on Riemann surfaces. This approach goes beyond our immediate program (cf. Ch. XVIII). However, we can bring to light at this stage the action of analytic changes of variable on the residue which bears directly on the Riemann surface aspect of the question. Let a, A E C , let f be analytic on A(A; r ) - { A } ,and let rp denote the restriction to A(a; s) - { a } of $: A(a; s ) + A(A; r), $(a) = A , t j analytic on A@; s) and univalent. We have

Theorem 3.2: res(A; f)= res[a; (f0 rp)rp'].

PROOFS:(1) Using integrals. Clearly res[a; (f

0

cp)rp'] =

fdz,

0 < p < S,

where r , ( t ) = u + pe*"'. We note that rp(4 - A = (z - a ) C p l ( Z ) ,

where rpl is analytic on A(a; s) and does not take the value 0. Hence O ( A ; rp

r,) = o ( ~r,) ; =1

and we conclude by Ex. 5.7, Ch. V, and the homotopic version of the Cauchy integral theorem that res[a ;(f

0

'f

rp)rp'] = -

2ni

fdz,

O.

174


Then VglO(0)

- v g 1 0 ( 0 3 ) = vhIO(O) - vh10(03)*

(The values of the valences entering into the equality arefinite.)

PROOF: The finiteness of the valences in question is a consequence of the fact that g(z), h(z) E C - ( 0 ) when z E fr 0. If either { g ( z ) = 0) or { g ( z ) = 03) were infinite, then g would take the value 0 or 03 somewhere on fr 0. A similar remark holds for h. We note that the set of cluster values of (h I O)/(g I 0) is exactly (4.2)

D+e

Indeed, in general, if cp: is continuous on 0 and meromorphic in 0, then the set of cluster values of cp 1 0 is precisely cp(fr 0)(cf. Ex.4.19, this section). Since the set (4.2) is contained in A(1; l), we see that if a denotes the divisor of (h I O)/(g lo), then

the latter two sums being taken over the zeros and poles of g and of h. The assertion of the theorem follows when we note that i$,lo(z) is just vhlo(0)- v h l O ( m ) and that a corresponding statement holds for g.

1

EXERCISES Let n E N. Then z"+' + e' has n + 3 zeros counted by multiplicity in A(0; el. Let 0 be as in Theorem 4.2. Let F be a continuous map of d x [O, 11 into C such that Ftakes values in C - (0)at points of (fr 0)x [O, I]. For each t E [O, 11 let ft ,the function with domain 0 given byJ(z) = F(z, t ) , be meromorphic on 0. Then vlt(0) - Y,,(oo) is constant. Does one need the full force of the hypothesis on F ? Does such a theorem hold when [0, I ] is replaced by a connected compact metric space? 4.23.

4.24.

e.

They are exactly We return to the study of functions meromorphic on the continuous functions on whose restrictions to C are the quotients of polynomials P/Q where the degree of Q is nonnegative and P and Q have no zeros in common. The factorization theorem for polynomials leads at once to the conclusion that M(e) less the zero constant is the multiplicative group generated by the constants other than zero and the functions T. = z - a, z E U E C. From an additive point of view we see from the partial fraction

c

e,

4. Meromorphic Functions

175

decomposition that M(C) is the additive group generated by the functions C T ~ " ,CT,("+ '),a, c E C, if E N.

EXERCISES _Show that the members of M(Z') having valence one are the members of M(C) admitting a representation of the form (az b)/(cz d) on C where a, b, c, d E C and ad bc # 0. Such functions are-called Mobius (or linear fractional) transformations or conformal automorphisms of C. Show that the Mobius transformations with the operation of composition constitute a group. Show that if a and b are distinct points of there exists a Mobius transformationfsatisfyingf(a) = 0, f ( b ) = a. Further properties of Mobius transformations will be developed in Ch. XV. 4.26. Show that D is a derivation on M(C) which takes the constants into the zero constant if and only if there exists g E M(C) such that D ( f ) =f'g, f E M(C). 4.27. Show that the members of M(C) of bounded valence are the restrictions to C of the nonconstant members of M(c). Show that the valence of such a member of M(C) has a constant value save at one point. What is the value of the valence at the exceptional point? 4.28. Show that the members of M(Z') having constant valence two are those representable in the form a 0 T~~ 0 fi where a and are Mobius transformations. Hint: Given thatfE M(C) has valence two, consider its set of poles and proceed according as it has one or two members. The notation T~ is that of the preceding text. 4.29. s = i(z l/z), z E C, has constant valence two. Show that s(z) s(l/z) and that s-'([ - I , 11) = C(0; 1). Let P be a polynomial of nonnegative degree n and let 4.25.

+

+

~

c,

+

1

M=max{lP(x)I: -1 < x < l J .

+

Show that IP 0 s(z)l I M Iz(", 1 I Izl < 00, with the aid of the followingboundary maximum principle applied to z-"P0 s(z), IzI > !: Let 0 be an open proper nonempty subset of C. Let f:0 + C be continuous on 0 and analytic in 0. Let g =f 1 fr 0. Then maxo

If1

= maxrro

Ifl.

[Proof: If If(a)l maxo If I for some a E 0, then by the maximum principle for analytic functions (cf. the sentences following (5.1) of Ch. IV)f i s constant on the component of 0 containing a and hence If(a)l 5 maxr, If 1. The asserted boundary maximum principle follows readily.] Remark: The maximum Qrinciple for analytic functions remains in effect when the domain is a region in C. The strong inequality in 1 I 1 1 < co " is there solely because M (zl"is not defined when M = 0, z = co. To be sure the question is trifling. Show that for each positive integer n there exists a unique polynomial of degree n. say P,such that :

"

z"

+ -Z"1 = P

0

s(z),

z 75 0,a.

+


176

This is t o be done without any formal computation. Hint: Consider

f=

((s(z), z"

+);

: z E i-)

and show that f E M(C) and has a unique pole which is at m and has order n. 4.30. Given f:[ - I , 1]+C, a necessary and sufficient condition that f be the restriction to [ - 1, I ] of a function g analytic at each point of [ - 1, I ] is that, pmbeing the infimum of the set of elements s u p { l f ( x ) -P"(x)l: -1 I x l l h

where P,, is a polynomial of at most degree n, we have

+ 1I-l log p n + l< 0. (S. Bernstein). [log(+ m) is defined as + m and log(0) as lim sup(n n-tm

00 .] Prove this theorem with the aid of the results of the preceding exercise. To show the condition necessary, consider g s(z) with domain { r < IzI < r - I } ,0 < r < 1, r sufficiently near one, and estimate the Laurent coefficients. Note that if (bk) is the sequence of Laurent b k zk = P. s(z) where P . is a coefficientsin question, br = b - k and conclude that P-,, polynomial of at most degree n. Estimate I f ( x ) - P,(x) 1, 1 < x 5 1. To show the condition sufficient, note that there exists a sequence of polynomials (P.) where P. is at most of degree n and satisfies I f ( x ) - P,(x)l < r " , - 1 < x < 1, where 0 < r < 1, for n large. Apply the inequality of Ex. 4.29 to P. + - P , to conclude the uniform convergence of (Pn) in s({r < IzI < r - I}) and note that f ( x ) = limn+ P,(x), -1 I x 4 1. -

0

0

-

Limits of sequences of meromorphic functions. There remain t o be considered among the more elementary questions of the theory of meromorphic functions limit problems and a useful and reasonable definition of unijbrmity. A convenient approach is to introduce following CarathCodory the so-called chordal metric for It induces the standard topology of We first introduce the stereographic map 0 of the sphere S in R3 with center (0, 0, f)and radius 3,that is, {x12 x2' x32- x3 = 0 ) onto C.t It is specified by the stipulation that it take (0, 0, 1) into 00 and each other point ( x l , x2, x3) of S into the image with respect to the canonical map D of (yl, y 2 ) ,where (yl, y, ,0) is the point of intersection of {x3 = 0 } and the line passing through (0, 0, 1) and

c.

+

c.

+

(XI, x2 3 x3).

It is routine t o verify that for (xl, x 2 , x3) # (0, 0, 1) we have

w,,x 2 ,

XJ

=

x1

+ ix2

-.1 - x3

It is easy to see that 8 is a univalent continuous map of the compact set S onto Consequently cp, the inverse of 0, is a homeomorphism of onto S.

c

e.

t Here R3 is

{ ( x l ,x 2 , x 3 ) :x l , x 2 ,x 3 E R}.


177

Given a, b E C, by the chordal distance between a and 6, denoted x(a, b), is meant the euclidean distance between cp(a) and cp(b), that is,

where cp(a) = ( x l , x 2 , x3)and q(b) = (yl, y , ,y 3 ) .That x is a bona fide metric

FIG.6.1

c

on follows at once from the facts that x(a, b) is just the euclidean distance between cp(a) and cp(b) and that cp is a univalent map. We call x the chordal metric of C .

EXERCISE 4.31.

Given z E C , show that

and conclude that x(a, co) = ( I

+ laI’)-”*

VI. Laurent Expansion. Meromorphk

178

Functions

and ~ ( a6),

=

Ib

- al(1

+ ~ ~ [ ' ) - ' ~ '+( 1( b 1 2 ) - 1 / 2 ,

a, b E C.

Show that for a, b E C we have the inequality x(a, 6) I 16 - 01,

and when a E C and x(a, b) < x(a, a),the useful inequality

Hint: To establish the latter inequality proceed from

+

and the inequality (1 + 1 ~ 1 ) ' p , z E C, n 2 m. If, however,fis meromorphic, it is not hard to see that f is even analytic on A . For iff' is meromorphic on A and a E A, there exists r positive such that A(a; r ) c A and f ( z ) # 00, z E C(a; r ) . [C(co; r ) = { IzI = r - I } . ] From the compactness of C(a;r ) and the local uniformity property established in Theorem 4.3, we see that (f,)tends uniformly to f on C(a;r ) in the sense of the euclidean metric. Hence by the boundary maximum principle enunciated in Ex. 4.29, this section, we see that (f,) is uniformly Cauchy on A(a; r ) and conclude that f is analytic at a. We see further that ( f . ) tends uniformly to f in A in the sense of the euclidean metric.

EXERCISES 4.32. Prove that f of the preceding paragraph is analytic when it is meromorphic by appeal to Rouche's theorem. 4.33. Given ( f . ) . . ~where each f. is a continuous map of A , an open subset of

into C. Show that (fn)tends uniformly tofin the sense of the chordal metric in A if

VI. Laurent

180

Expansion. Meromorphic Functions

and only if for each a E A there exists r positive such that A(a; r ) c A and either ( f , + & ~tends uniformly to f o n A(a; r ) or (l/fm+n)ns~ tends uniformly to l / f o n A@; r ) , in the euclidean sense, for some m E N, it being understood that the terms involved are meaningful in the sense of C. Show that the stated condition may be replaced by the following: For each a E A there exists r positive and a Mobius transformation Msuch that ( M of.) tends to M ofuniformly in the euclidean sense on A(a; r ) ; the terms involved are to be meaningful in the sense of C. Can the continuity hypothesis on the f. be dispensed with? 4.34. Let A = A(0; 1). fn(z) = l/[z (n l ) - I ] , &) = -(l/z). Show that although (fn)and (9.) have X-uniform limits in A , (f.+g.) does not. Let h.(z) = z - (n l)-'. Show that (f.h.) does not have a X-uniform limit in A . These examples show that algebraic operations on meromorphic functions may cause serious upsets for X-uniformity. 4.35. For later applications the following situation is of interest. Eachf. is meromorphic on an open subset A in C. For each a E A for at most one n does f. have a pole at a. Show that f. is X-uniformly convergent in A and has a meromorphic sum if and only if for each compact C c A there exists m E N such that f,+. is analytic at each point of C, n E N, and xo"fm + " I C converges uniformly in the sense of the euclidean metric.

+ +

+

x?

A theorem of Hurwitz. An important question concerning sequences of meromorphic functions is the relation of the sets {fn(z)= b } t o {f(z) = b} where b E and (jJnEN is a sequence of functions, each meromorphic on a region A in which tends uniformly in the sense of the chordal metric in A t o f. Iff is the constant function with value b, nothing much can be said. Otherwise we can make a fairly precise statement which is the substance of the following theorem of A. Hurwitz.

c c,

Theorem 4.4: Under the stated assumptions, f not being the constant b, i f f(a) = b and r is a positive number such that A(a; r ) c A and f(z) # 6, z E A(a; r ) - {a}, then there exists m E N such that the value of the valence of S. I A(a; r ) at b is precisely n(a;f)for n 2 m.

FROOF: It is based on Roucht's theorem, as we might expect from the very statement. Let M denote a Mobius transformation satisfying M ( b ) = 0. We take s, 0 < s < r, such that IM 0 f(z)I < 1, z E A(a; s), and introduce p

= min{lM

of(z)l: z E C(a; s)}.

By the. hypothesis on r we have p > 0. Since ( M of,) tends puniformly t o M 0 f i n A , when n is large, M ofn(z) E C,z E A(a; s) and

IM ofn(z) - M of(z)I < p,

z E C ( a ; s).

Hence by Roucht's theorem

M

ofn

=M

of+

(M

Of"

-M

of)


181

has as many zeros in A(a; s), counted by multiplicity, as does M of when n is large, so that for n large the valence a t b offn I A(a; s) is n ( a ; f ) .For n large, f,(z) # b, z E A(a; r ) - A(a; s), a set which is compact a n d such that f does not take the value b on it. The theorem follows.

EXERCISES Let 0 and g satisfy the conditions stated in RouchC’s theorem with f the constant zero. Suppose that g.: 0 + is continuous on 6 and is meromorphic in 0, n E N, and that (9.) tends X-uniformly on 0 to g. Then for n sufficiently large v d 0 ) = v,,10(0) and v , l 0 ( ~ )= v , , l 0 ( ~ ) . What can be concluded about vJnwheref. E M(c) and (f.) tends X-uniformly on to f? 4.37. Let A and B be regions c Let ( f & be ~ a sequence such that f.: A -+ B and is meromorphic on A, n E N, and (f.) tends X-uniformly to f i n A. Then either f ( A ) c B or f is constant. 4.38. Let (fn)ns~ denote a sequence of functions meromorphic on a region A in tending X-uniformly to f in A. Suppose that f is not constant and that a < v/(w). Then a < v J n ( w )for n sufficiently large. Conclude that if 4.36.

c

e

c.

c

sup {uf”(w):w E

c}I p < + co

c,

for all n , then either v f ( w )2 /I,w E or f is constant. In particular, if each f. is univalent, then f is univalent or constant. 4.39. We have seen that algebraic operations need not combine well with the taking of X-uniform limits. The object of the present exercise is to remedy the situation and to account for the breakdown by “coalescence.” Specifically, we suppose that (f.) tends X-uniformly to f and (9.) X-uniformly to g in the region A c C, f., f, g., g being meromorphic on A. Then ( f . g.) tends X-uniformly to (f+ g) in

+

A

-

{ f ( z >= a l n Mz) = a},

and ( f a g . ) tends X-uniformly to f g in A

-

{[{f(z)=a,}n{s(z)=O}lu[{f(z)=O}n{g(z)=co}l}.

e

4.40. Let A be a region in and X a compact metric space. Develop a theory of F: A x X - . continuous on A x X and such that for each p E X the function f,(z) = F ( z , p), with domain A, is meromorphic on A (cf. 56, Ch. V). In particular, investigate (1) the function f,’(z), with domain A x X, and (2) the properties of vJPfor p near q E X , formulating and establishing appropriate theorems.

e

BB Chapter VII GG

FURTHER APPLICATIONS O F T H E CAUCHYTHEORY

PREFATORY REMARKS The present chapter treats two consequences of the Cauchy theory fecund in their applications to certain calculations of classical analysis. One is the residue theorem relating under appropriate conditions line integrals of analytic functions with isolated singularities to the residues at the singularities. The proof of this theorem is simple. In situations where the residues are easy to compute, the identities afforded by the theorem may be used to calculate definite integrals (proper and improper) under favorable conditions, as we shall see by a study of examples. The application of the residue theorem for such purposes parallels (indeed it may be thought of as generalizing) “the method of the Cauchy integral theorem” used to calculate the Fresnel integrals and the other integrals treated in the exercises of Ch. V. As we shall see in the examples of 42, this chapter, successful application depends upon the combination of a “manageable” function, an astute choice of a path of integration, and care in the use of basic definitions. The problems will be found to offer good practice in the use of elementary complex function theory. An authoritative text on the residue theorem and its many applications is the useful monograph of Lindelof [77]. The other consequence of the Cauchy theory to which we have referred is a corollary of the residue theorem and is known as “the method of Cauchy.” It gives an answer to the question: What do the Cauchy integral formulas yield 182

1. The Residue Theorem

183

when the hypothesis of analyticity is relaxed to the extent of allowing a finite number of isolated singularities? The resulting compensated formula will lead to representation theorems for such meromorphic functions as the cotangent.

1. THE RESIDUE THEOREM

We shall content ourselves (following the precedent of Saks and Zygmund) with a residue theorem satisfying fairly simple hypotheses insofar as a fully proved theorem is concerned. For applications the theorem given is abundantly adequate. As in the exposition of Ch. V, we shall indicate how more general theorems may be obtained. Let A be a holomorphically simply connected region in C,let a,, .. ., a, be n (E N) distinct points of A , let f be analytic on A - {a,, .. . , a,}, and let y be a rectifiable closed curve lying in A - {a,, . . . , a,}. By the residue theorem we understand

Theorem 1.1: Under the stated conditions

s,

f dz

= 2ni

z

O(ak; y) res(ak;f).

PROOF: Let Pk denote the principal part of f at ak. We see that g(z) =f(z) - c1 Pk(z)with domain A - {a,, . . . , a,} has removable singularities at a,, . .. ,a,. Noting that

J:

dz

= 0,

and using the fact that

S,.

dz = 2ni O(ak; y ) res(ak; Pk) = 2ni O(ak; y )

res(ak;f),

k

=

1,. . ., n,

we conclude the residue theorem. What may we expect when the hypothesis that A be holomorphically simply connected is dropped? When we refer to the situation considered at the end of 93, Ch. V (cp being univalent), but now allow f to have a finite number of isolated singularities, all in cp(int E ) , that is, iff is given as analytic on 0 - X where X is a finite subset of cp(int E ) , we see that the use of the

VII. Further Applications of the Cauchy Theory

184

auxiliary g is available and we conclude

J f d z -2 J n

B

n

1

2 res(a; f),

n

f dz p*

=2ni~

aeX

1;

where E is the common value of O(z; fi) O(z; bk),z E cp(int E ) . The only matter that is not settled is that, in fact, E is either 1 or -1.

EXERCISES 1.1. Develop a residue theorem in the geometric setting of Ex. 1.12, Ch. VI, isolated singularities being allowed. 1.2. Let 0 < p < r . Let f be analytic on A(0; r ) . (a) Calculate

where a l , . . .,a. E A(0; p), distinguishing between the case where the sequence is univalent and the case where it is not. (b) Suppose that a E A(0; p), that f'(a) # 0, that f ( r ) # f(a) for z # a satisfying IzI I p, and that n E N.Calculate (ak)l s L L n

1.3. Let g be meromorphic on A of Theorem 1.1 and suppose that {a,(z) # 0) is a finite subset of A - y, y being as in the theorem, Calculate with the aid of the t heorem

Ch. I, VI Note that the former integral was treated in quite a different manner in & (Argument Principle). 1.4. Let X be a topological space. Let a,, , .. .,an:X + C, n E N,each being continuous at q E X. Suppose that

+

Let 0 < r < co and suppose that w is the unique root of Q(w, q) = 0 in A(w; r ) and that Q w ( w ,q ) # 0. Show that, for p sufficiently near q,

is the unique root of Q(w, p) = 0 in A(w; r ) and that subject to the restriction in question it defines a function continuous at q.

185

I . The Residue Theorem 1.5.

Let y be a rectifiable closed curve in C,let a, b E C - y, let m,n E N.Calculate

Jy(z - a)-.l(z - b)-" dz. 1.6.

Summation formulas. The residue theorem permits us to evaluate limits of the

form

whare P is a polynomial of degree 2 2 having no zeros in Z and either ~ ( k=) 1, k E Z, or ~ ( k=) (- l)k. Develop the details of the following approach to the question: (a) With 0 < r < co and fmeromorphic on C,let

+

~ , ( r ; f ) = s u p ( ~ / ( z ) ~ :k&Uz ~W~ -; ~ ) ] , and let ~ ~ (=fsup( ) \f(z)l:z E

+

u fr[- (n + t),n +t ; -(n + t),n + 41).

ncN

Show that M , ( r ; f ) , M 2 ( f )< co, when f(z) = cot mz, z E C, or f(z) z E C. (b) Apply the residue theorem to

= csc nz,

and csc 7Tz

Bn =

I,"po

dz*

where /3. is the positively sensed boundary path of [-(n

+ t), + t ; -(n + t),n + tl,

n being large. (c) Show that limn+ A. = limn+ B. = 0 and conclude the desired summation formulas. Apply t o the case where P(z) = z 2 a, 0 < a < 00. Consider also the general question when P may have a zero in Z and interpret the consequences of limn+ A. = limn+ B. = 0. Apply to determine x t (k l)-'("+ *), n E N,obtaining the values of the sum explicitly for n = 0, 1,2.

+

+

+

In the following theorem, forming the base of the method of Cauchy, the hypotheses and notations of Theorem 1.1, this section, are in effect. Theorem 1.2:

Under the stated conditions

VII. Further Applications of the Cauchy Theory

186

PROOF: Let F bef(w)/(w

- z) with domain A - {z, a,, . . . ,a,}. We have res(z; F) = f ( z )

and, usingf(w) = g(w)

+ C; pk(w),

res(ak; F ) = res[ak; Pk(W) (w - z)-’]

=

-Pk(z),

k = 1, . . ., n.

The theorem follows on applying Theorem 1.1 to F. It is immediate that on differentiating the identity of the theorem m times we obtain

EXERCISES 1.7. Furnish the details of the proof of the equality

res[at; Pt(w) (w - z)-’I = -Pk(z). Hint: Apply Theorem 3.1, Ch. VI. 1.8. Apply Theorem 1.2 in the special case:f(w) = v ( w ) / Q ( w ) with v analyticon A, Q ( w ) = nl (w -at), to conclude

a formula due to Hermite. Show with the aid of this formula that if A = A(0; r ) , (ah): N -+A(0; r), (ak) is univalent, and limt, at = 0, then the sequence of polynomials (Pn)nrN, where P. is of degree at most n and satisfies the interpolation condition Pn(at)= P(ak), k = 0, .. .,n, tends uniformly to rp in A(0; r ) . [To be utterly strict, “(P. I should be read for “(Pn)n.~.”] Extend the first part of the exercise replacing Q by a polynomial of nonnegative degree whose zeros are all in A. Formulate and prove a correspondinginterpolation theorem in this setting.

2. APPLICATION OF THE RESIDUE THEOREM TO THE CALCULATION OF INTEGRALS We shall show by the study of several examples how the residue theorem may be employed to calculate certain integrals. It will be seen that in standard situations a suitable approach readily suggests itself. Skill in the use of the residue theorem for such calculations is a matter of practice and experience.

2. Application of the Residae Theorem to Calculation of Integrals

187

Example I. We seek the value of 1

dx.

This example was deliberately chosen because it is simple and because the result is well known from the elementary integral calculus. Indeed, the principal determination of the arctangent (=the inverse of tan - 71/2,42[) is a primitive of the integrand. The value of the integral is seen to be n. Let us approach the question from the point of view of the residue theorem. We have to do with the rational function f = 1/(z2 + l), z E It has two poles, each simple, one at i, the other at - i. Given 1 < r < + 00, we introduce

I]

e.

W )=

J-

( -r ) ( r )

f d z + s,cdz,

where yr(8) = re”, 0 I 8 I n, and conclude from the residue theorem that

I ( r ) = 22ri res(i;f) = 2ni(1/2i) = n. Note that the order of - i with respect to the closed curve associated with the indicated line integrals is 0. The absolute value of the second integral does not exceed the length of yr times maxyr If 1, that is, nr/(r2 - 1). We conclude that the value of (2.1) is n. It is to be noted that the Cauchy integral formula for a semicircle applied to (z + i)-’ also yields the value of I(r). Example 2. Let 0 < a < 1. We seek the value of + m xa-l

joI f x d x * Here, of course, f-1

- exp[(a -

- 1) log x],

0 <x
0}, and ,I, = the analytic logarithm of the identity map on C - {t: t I 0) which agrees with 1, on {Imz < O } . We note that Il(eie)= id, 0 I 8 < n, A2(eie)= it?, 0 < t? < 271, R,(eie)= i0, 71 < 0 < 371. Hence A,(z) = A,(z) + 2714 z in the domain of 1,.Let 0 < p < 1 < r < + 00. Consider

188

VII. Further Applications of the Cauchy Theory ir

-I

0

- ir

FIG.7.1

where the yk are the positively sensed boundary paths indicated in the adjoining figure. The first and third integrals in the sum are zero, while by the residue theorem (or Cauchy integral formula) the value of the second is 2ni exp[(cr - 1) A,( - I)]

=

-2nieani.

On taking into account the relations among A l , ,I2, A,, we see that (2.3) is equal to the sum of r xa-l

(1 - e2ani) -dx

,l+x

and contributions, one from arcs on C(0;p), the other from arcs on C(0; r ) . The absolute value of the former contribution does not exceed [P"-'(l - P)-lI(2nP),

and the absolute value of the latter contribution does not exceed [re- l(r

- 1) - 1](2nr).

Taking limits we conclude that the value of (2.2) is hie""' R -elmi 1 sin(ctn)' This example illustrates the care that is called for in the use of analytic logarithms. A second approach will be given in Ex. 2.5 below.

2. Application of the Residue Theorem to Calculation of Integrals

189

Example 3. Let a E C, la1 # 1. We seek the value of (2.4) We note that the integral may be represented as the line integral

P

1

q o ; 1) ( z

*-dz=

- a)(Z - G) izl

1

PC ( 0 ; 1) i(z - a)(l - az>

dz.

Using the residue theorem (or Cauchy integral formula) we see that the value of (2.4) is 2n/(l - laI2) when la1 < 1 and is 2n/(laI2 - 1) when la1 > 1. An independent determination of the value of (2.4) when la1 < 1 is given by the Poisson integral for the real part of the constant one.

Example 4. Let a E R,let 0 < c, t < where

+ 00.

We seek the value of lim+,.,, I,

(2.5)

When a 50, the Cauchy integral theorem for a semicircle serves. Indeed, I(t) =

s

'/'

exp[a(c + re")] ite" do. c + te"

If a = 0, lim, oIi I = ni. This result could also have been obtained by introducing an analytic logarithm of the identity map on {Re z > 0}, which is a primitive of l/z. If a < 0, we see from the estimat.e

that lim, I = 0.7 Suppose now that a > 0. We take t > c and note that by the residue theorem (or the Cauchy integral formula for a semicircle) +

J3ff/2

exp[a(c

n/2

+ re")]

c

+ te"

I

= 2ni.

Noting the estimate

we conclude that lim, Cf. Ex. 3.24, Ch. V.

VII. Further Appkations of the Cawhy Theory

190

Integrals of the present type occur in the theory of the Laplace transform. It is to be remarked that we could have carried out all the integrations of these examples with the aid of a Cauchy integral formula as well as with the aid of the residue theorem. This happened because we wished to keep our illustrative examples simple. In the exercises that follow the occurrence of more than one pole will not be unusual.

EXERCISES In this set the indicated integrals and limits are to be evaluated. ..+m

1

+m

X"

dx, m,n E N, 2n - m 2 2;

(e) lim+, I, whereZ(r) = J L t f ( x ) d x , f: {Imz20}-+~beingcontinuouson its domain, meromorphic in {Im z > 0}, and satisfying lim, z f ( z ) = 0, f(R) c C ; (f) I h + , I, where Z(r) = J L t einSf(x)dx, 0 < a < + co, f satisfying the above conditionssave that the limit requirement is replaced by lim, f = 0; sin ax x(x' +a,

2.2.

*(a)

+

dx, a E R .

1)2

log x dx;

(b) ~ + m ~ ' ( ( l + x ) - ' d x -, 1 < a < 3 . 0

2.3.

laRIP(e'31 do, P a polynomial having no zeros on C(0; 1); (b) j,? [~(e>]- * dB where Tee) =2 i cos ke + B~sin ke), Ak , E~ E R, (a)

-2

(Ak

T(B)#Ofor

BER;

~ ( c o 8,s sin 8) dB where P(x, y ) and Q(x, y ) are polynomials with (') 10 Q( cos 8, sin 0) real coefficients and Q(cos 8,sin 0)# 0 for 0 E R; 2n

1- & +m

(d)

dx, where P is a polynomial of positive degree with all its

zeros simple and having positive imaginary part.

3. Application of the Method of Cauchy c+it

2.4. 2.5.

lim+mZ,Z(t ) = l ear J'_t" 1+e'

191

em' ~.,,dz,wherea~R,n~N,O O}. We proceed by induction. When n = 0, the assertion is trivial, thefbeing the nonzero constants of M(%). If the sum in question is n + 1, n EN, let a(a) > 0, a(b) 0, and let m denote a Mobius transformation satisfying: m(a) = 0, m(b) = m. Then 8 - 8, satisfies the stated conditions with sum in question now being n rather than n 1. Hence withfsatisfying a, = a - a,, = a. The remainder of the assertion is immediate. As far as we have principal parts are concerned, it is clear that if a,, ...,a, are n distinct points of %, and PkE M(%)has a sole pole, in fact at ak, then there exist f~ M(%), analytic in % - {a1,.. .,a,} and having the same principal part at ak as

-=

+

194

Prefatory Remarks

Pk,for example,

c;

195

Pk,and the difference of such functions is a constant. Thus the totality consists of the functions of the form c + Pk, c constant. the situation changes When we turn to a region A , a proper part of radically. Here we have to cope with the possibility of infinitely many zeros and poles. I f f € M(A) is not the constant zero, then the zeros off do not cluster at a point of A . We conclude that 8, takes values in Z and {a,(z) # 0} clusters at no point of A . The general existence theorem of Weierstrass ($1, this chapter) implies that a sufficient condition tnat 8: A + Z be the divisor of a member of M(A), not the constant zero, is that {d(z) # 0} cluster at no point of A . The Mittag-Leffler theorem (51, this chapter) implies that if B c A clusters at no point of A , then there existsf€ M(A) having an assigned principal part at each point of B and analytic in A - B. I f f is such a function, then the totality of such functions consists of the functions of the formf + g, g analytic on A . In these theorems A is replaced by an arbitrary proper nonempty open subset of The two cited theorems are the coarse qualitative results of the theory. They lead to important conclusions. We shall meet the following applications of the Weierstrass theorem: (1) When A # M(A) is the field generated by A(A) = the ring of functions analytic on A . (2) If 6 is an isomorphism of A(A) onto A(B), A and B being regions which are proper parts of and 8 “preserves” the constant taking the value i, then there exists a unique univalent map cp of B onto A , meromorphic on B such that O ( f ) =f cp, f E A(A). A corresponding assertion holds when 8 satisfies the alternative condition of taking the constant with value i into the constant with value - i. This striking theorem (2) was given by L. Bers [135]. We shall give a proof due to Nakai [144]. The cited paper of Nakai extends the theorem of Bers to the setting of Riemann surface theory. Recently, great advances were made by H. Iss’sa [138] who obtained the corresponding theorem for field isomorphisms. The resources of his study go beyond the program of this book. We have emphasized the qualitative character of the theorems of Weierstrass and Mittag-Leffler. When conditions of a quantitative nature (growth conditions) are imposed and we restrict our attention to special domains, more refined results emerge. It is intentional that we introduce infinite products of complex numbers and complex-valued functions just before turning to the more refined problems. No special advantage is gained by treating infinite products earlier, and one might even say that the special restrictions in the theory of infinite products become significant when viewed in terms of limit questions of the kind encountered in the proof of the Weierstrass theorem. We shall consider in the latter part of the present chapter the distribution of the zeros of a function analytic on a disk or a half-plane and bounded, as well as the distribution of the zeros of an entire function f satisfying

1;

e,

e.

c,

c,

0

196

VIM. The Zeros and Poles of Meromrplrc Funciions

I f(z)I IA exp(lzlB), Z E C, A and B being positive numbers. These are questions where growth conditions are present. 1. THE THEOREMS OF MITTAG-LEFFLER AND WEIERSTRASS We shall treat the Mittag-Leffler theorem first. It states

e,

Theorem 1.1: Let 0 be a nonempty proper open subset of let B c 0 cluster at no point of 0,let there be assigned to each b E B a function gb E M(2') having a pole at b and no others. Then there exists f E M(O), analytic in 0 - B and having the same principal part at b as 96. Iff is such a function, the totality of such functions consists of the functions f + g, g analytic on 0.

PROOF: We note first that we may assume that 00 E 0 - B. To see that this is so, we let a E 0 - B and let m denote a Mobius transformation satisfying m(a) = 00. If the restricted theorem is valid, then there exists f E M[m(O)], analytic in m ( 0 - B) and satisfying the condition that for each b E B the principal part offat m(b) is equal to that of& m-'. Hencef ( m 10)E M(O), its poles are the points of Band since for given b E B, f - (gb m-') I m ( 0 )is analytic at m(b), we conclude that f 0 (m 10)- gb 10 is analytic at b and f o (m1 0)has the desired property of having the same principal part at b as gb . We proceed, as we may, assuming that co E 0 - B. We also assume that B is injinite, for the case where B is finite is immediate, xbEB(gbIO) yielding a function of the desired type. Given b E B, there exists a point c E C - 0 such that 0

0

0

Ib - CI = dist({b}, C - 0) since C - 0 is nonempty and compact. Now gb is analytic in 2'- A(c; r), r = 16 - cI. Considering the Laurent expansion of gb in {r < Iz - CI < +a}, we see that, given E positive, there exists h E M(e) having a pole at most at c and such that Ig&z) - h(z)l < E ,

Iz

- cI

> 2r.

Now let (bJnoNdenote a univalent enumeration of B. It is to be noted that given 6 positive the s'et of b E B satisfying dist({b}, C - 0) 2 6 iscompact, and, since it clusters at no point, it is finite. Let c, and r, have the same context relative to b, as do c and r relative to b in the preceding paragraph, and let h, take over the role of h in the present context where b = b, , c = c, , r = r, , and E is taken as 2-". We propose to establish the desired existence assertion

1. The Theorem of Mittag-Leger and Weierstrass

197

by verifying that

0

has the required properties. From the observation of the second sentence of this paragraph we see that r,, = 0. Now suppose that A(a; p ) c 0, a E C, and note that for n large we have 2r,, < dist(A(a; p), C - 0). Hence (1.1) converges X-uniformly on A(a; p ) since Igb,(Z)

- hn(z)l

2-",

(1.2)

z E A(u; p), for n sufficiently large. It is to be noted that (1.1) converges uniformly (in the sense of C) on A(o0; p ) when l/p > Ic,I + 2 r n ,n EN.We conclude that (1.1) converges X-uniformly in 0 and has a meromorphic sum f (cf. Ex. 4.35, Ch. VI). It is clear that f has the same principal part as gb at b. The last sentence of the theorem is immediate.

EXERCISE 1.1. Suppose that g, satisfies instead of the stated condition the requirement that gh be analytic on C - {b}.Prove the existence off analytic on 0 - B such that f and gb have the same principal part at each point of B.

The Weierstrass theorem will be formulated as follows.

e.

Theorem 1.2: Given 0 a proper nonempty open subset of Let a: 0 -+ Z be such that {a(z)# 0 ) clusters at no point of 0. Then there exists f E M(0) such that a, = a. Further the totality of such functions consists of the functions f g where g is analytic on 0 and has no zeros. PROOF: We suppose that co E 0 and d(co) = 0, for we may employ the reduction used in the Mittag-Leffler theorem, considering a E 0 such that a(u) = 0, and F E M[m(O)] satisfying aF 0 (m I 0) = a. We suppose that B = {a(z)# 0 } is infinite, for when B is finite, it suffices to let w E - 0 and to introduce a member of M(e) whose divisor takes the value 0 in the complement of B u {w} with respect to and takes the value a ( ~ at ) z E B and - d at w, where d is the sum of a over B. We continue to use the notation of the proof of the Mittag-Leffler theorem. It will be convenient to introduce E,, , the Weierstrass primary factor with

e

e

198

Vlll. The Zeros and Poles of Meromorphic Functions

index P(E N), given by E,(z) = (1 - z) exp($

);.

Clearly, E, has a simple zero at 1 and no others. A special feature of Ep is the fact that 1, = z k / k with domain A(0; 1) is an analytic logarithm of E, I A(0; 1). For each n E N let m(n) E N be so chosen that

-x;+

We consider the sequence ( F J of functions meromorphic on 0 where

Here

Taking A(u; p ) as in the proof of the Mittag-Leffler theorem and taking q E N such that /lk[lA(a;p)]

A(0; 3)s

k24

+ 1,

we see that

converges uniformly in the sense of the Euclidean metric of C and has a bounded sum. Since q+n+ 1

2

&(k)

b k

1 A(u ;p)1

4+ 1

is an analytic logarithm of (F,+,,+,/F,) I A(a; p), we conclude that the sequence whose nth term is (F,+,,+,/F,) I A(u; p) tends uniformly to an analytic function with no zero. The proof of the theorem is now readily completed. We see that (F,,) tends X-uniformly in 0 to a meromorphic function with divisor a. It is to be noted that since UnEN A(c,,; 2rn)is bounded, (F, I A(c0; p)) tends uniformly to alimit not taking the value zero for some positive p. The last assertion of the theorem is immediate.

1. The Theorems of Mittag-LeSper and Weierstrass

199

EXERCISES 1.2. It is easy to construct a sequence (fn) of functions taking values in C which tends uniformly to its limit in the sense of the metric of C but is such that (exp of.) does not tend uniformly to exp of: Example: fn = x ( n 1) - I , 0 < x 5 1, n E N. Validate the assertion of the text concerning the uniform limit property of ((Fq+ n + 1 /Fq) I A(a ; PI). 1.3. Show that when p > 0, there exists a positive number A such that

+ +

lE,(z)l 5 exp(A lzIp+l),

< exp(A 1~13,

Iz1 I1 ; IZI > 1.

Hint: Consider [E,(z) - l ] ~ - ( ~on+ A~( )0 ; l), the appropriate definition being made at 0. Make a direct _estimateof IE,(z)I when (zl > 1 . 1.4. Given g E M(C) having a sole pole at b E C - {O}, g(co) = 0, and n E N, we define

Suppose that B is an infinite subset of C clustering at no point of C and that for each b E B there is given gb having a sole pole at b. Let ( b " ) "denote .~ a univalent enumeration of B. Show that there exists v : N + N such that %' Tv(k)(gbt)converges xuniformly in C to a meromorphic sum with poles in B, the principal part at b being given by gb . Carry out the corresponding program for the Weierstrass theorem using products

c.

Let 0 be an open nonempty proper subset of Show that there exists N + 0 univalent such that fr 0 is the set of cluster points of {c.: n E N}. Show that there existsfanalytic on 0 such that a,(c,) = 1, n E N ,and conclude thatfis not the restriction to 0 of a function meromorphic (let alone analytic) at each point of 0 and at some point of fr 0. 1.6. Let A be a region that is a proper subset of Show that M(A) is the field generated by A(A). Hint: Given f~ M(A) introduce g E A(A) satisfying ag = -a,-. *1.7. Let A be a region in C. Let ( f . ) " . be ~ a sequence of functions meromorphic on A , none the constant zero, satisfying: (1) there exists a E A such that (f.(a)) possesses a limit in C - {0},(2) (f.'/f.) tends X-uniformly in A to a function g meromorphic on A and satisfying a,(z) 2 - 1, z E A , a&) 2 0. Show that (f.)tends X-uniformly in A to a function F meromorphic on A which satisfies f'/f = g (*) and that the totality of functions meromorphic on A and satisfying (*) consists of the nonzero constant multiples of F. Hint: Apply a connectedness argument introducing E, the set of z E A for which there exists 0 open, z E 0 c A , such that (f,10) tends X-uniformly to a function meromorphic on 0, not the constant zero. What happens when the hypothesis on a, is dropped? Show by example that when (fn) satisfies instead of (2) the condition that (f.)tend X-uniformly in A to f meromorphic on A , it is not necessarily the case that (fn'/fn) 1.5.

(c"):

c.

VIII. The Zeros and Poles of Meromrphic Functions

200

tends X-uniformly in A tof’lf. What is a necessary and sufficient condition on such (f.) and f for (filf.) to tend X-uniformly in A to f’/f? 1.8. Apply Ex. 1.7 to obtain from the expansion of R cot RZ given in 53, Ch. VII, sin Rz = lim nz i-m

fi

(1

- k-222).

1

2. THE ISOMORPHISM THEOREM OF BERS

Thanks to the Weierstrass theorem, Theorem 1.2, this chapter, we are in a position to give an account of Nakai’s proof of the theorem of Bers. In what follows A and B are regions in 8 is an isomorphism of A(A) onto A ( B ) ; that is, 8 is a univalent map of (AA) onto A(B) satisfying

c,

Kh) = Nf) m>,

fxf+ 9) = S(f) + Q),

f, g E A ( 4

(2.1)

The theorem of Bers states

Theorem 2.1:

If A #

e, then either there exists a unique cp: B + A such that e(f) = f o

or else there exists a unique

‘p : B

=

--*

f

0

f E NA),

cp,

A such that

f E 44.

cp,

Further ‘p is a univalent map of B onto A , meromorphic on B in the former case, and such that the conjugate of cp is meromorphic on B in the latter case. PROOF: It is understood that by the conjugate f of a function f, taking values in is meant a of, where a(z) = 5, z E C,a(o0) = 00. The conjugation map a is clearly a homeomorphism of onto itself. We start the proof by studying the behavior of constant functions under 8. To prevent confusion of meaning we shall introduce special notation for constant functions. Thus given c E C, by cA will be meant the function with domain A taking at each point the value c. We see at once from (2.1) that 8(OA)= 0,, O(lA)= 1 . It follows by induction from (2.1) that 8(n,) = n B , n E N, and thereupon by standard arguments that 8(qA)= q B , q E Q. Since [8(iA)I2= (- I),, , 8(iA),which is analytic, is either i , or ( -i),. Thus either 8 maps each constant of A taking a rational complex value (that is, of the form a + bi, a, b E Q ) onto the constant of A(B) taking the same value or 8 maps each such constant cA onto (I?),. Now a constant is the nth power of a constant, n a positive integer. Hence given c E C and n a positive integer, 8(cA)is the nth power of a member of A(B). Hence if c # 0, 8(cA)does not have any zeros. Hence if c is complex but not rational complex and q is

e

e

,

2. The Isomorphism Theorem of Bers

201

rational complex, 8[(c - q)A]does not have a zero. Consequently 8(c,) omits all rational complex values. By the open mapping property of nonconstant analytic functions we see that 8(c,) is a constant. That 8 maps the set of constants of A(A) onto the set of constants of A@) follows on noting that 8-' maps a constant of A(B) onto a constant of A(A).t To establish the theorem of Bers we employ the notion of a maximal principal ideal, which is a concept of fundamental utility in questions concerning rings of functions. In the case of a commutative ring with identity, %, a principal ideal turns out to be a set of the form { a x : x E %}, a E %.I By a maximal principal ideal of % is meant a principal ideal Zproperly contained in % but not properly contained in a principal ideal J itself properly contained in %. We proceed by determining the maximal principal ideals of A(A) and the action of 8 on them. Given a E A, Z(a) = {f E A(A),,f(a)= 0)is a principal ideal of A@). For let g E A(A) have a simple zero at a and no other zeros. From f = (f/g)g we see that Z(a) is a principal ideal. Suppose that J is a principal ideal, say {fh: f E A(A)}, h E A(A), containing ](a) properly. Now h(a) # 0. I f f € A(A),

and hence f E J, so that A(A) = J. We conclude that Z(a) is a maximal principal ideal. Let us reverse the question and show that every maximal principal ideal of A(A) is Z(a) for a unique a E A. Unicity is easy. Z(a) has a member having a as its sole zero. Hence Z(a) = Z(b) implies that a = b. Let g E A(A) and suppose that Z = { fg:f E A(A)}. If g has no zero, we have Z = A(A), since h = (h/g)gE Z, h E A(A). Hence if I # A(A), g(a) = 0 for some a E A. Hence Z c Z(a) and we conclude that the maximal principal ideals are precisely the Z(a). The action of 8 on Z(a). Since 8 is an isomorphism mapping A(A) onto A@), we see that O[Z(a)]is a principal ideal of A(B) and is a proper part of A ( B ) as well. Hence by the argument of the preceding paragraph 8[Z(a)] c J(b) for some b E B, J(b) denoting {g E A(B),g(b)= 0). From Z(a) c O-'[J(b)], the maximality of Z(a), and the fact that 8-'[J(b)] is a principal ideal of A(A) properly contained in A(A), we conclude that Z(a) = O-'[J(b)Jor, equivalently, O[Z(a)] = J(b).We introduce the set of ordered pairs (a, 6) E A x B satisfying

+,

e,

t If we abandon for the moment the hypothesis A # and suppose on the contrary that A = C, we see that A ( B ) consists of constants. Since A(E) has nonconstant members if B # t?, for example, rn I B where m is a Mobius transformation taking some point of - B into 03, we see that B = In this case, 6 may be extremely complicated.

e.

f: The unlikely reader unfamiliar with the notion of an ideal is to take the preceding

sentence as the basis for a stopgap definition.

202

VIII. The Zeros and Poles of MeromorpMc Functions

fl[Z(a)] = J(b)and show that $ is a function and maps A univalently onto B.All further developments depend upon the properties of $. Clearly,

{a: (a, b) E $} = A ,

and, since J(b) = J(c) implies b = c, JI is a function with domain A. If $(al) = $(a2), then I @ , ) = Z(a2) and hence a, = a 2 . Applying the above reasoning to 8-' we conclude that given b E B we have e-'[J(b)] = Z(a) for some a E A, so that b = $(a). Hence $(A) = B. We note that f- f(a)A E I@). From elf- f(a)J E J[$(a)],we conclude

If O(f) ($(a)) is rational complex, we have e(f) ($(a)) = f ( a ) in case O(i,) = i, and e(f)( H a) ) in case O(iJ = - i B . To continue our study of J/ we consider (a,,): N -+ A univalent and such that {$(a,,):n EN} clusters at no point of B. Let g E A(B) and satisfy a,[$(a,,)] = 1, n E N,the existence of such g being assured by the Weierstrass theorem of the preceding section. Let f = e-'(g). Clearly f # 0, and f(a,,) = 0, n E N. Hence {a,,:n E N} clusters at no point of A . Obviously, $-' enjoys the same property, mutatis mutandis. We next show: if a E A, then lim, J/ exists and is in B. Indeed, there exist r positive and K compact c B such that A(a; r ) c A , $[A@; r ) ] c K. Otherwise there would exist (a,,): N -+ A univalent and such that limn+- a,, = a but {$(a,,):n E N} clusters at no point of B.7 Hence if lim, $ does not exist, there are two univalent sequences (a,,), (a,,) with elements in A and limit a such that (b,,) and (b,,), where b,, = $(a,) and b,, = $(a,,), possess limits in B, say b and b, respectively, b # b. We suppose, as we may, that a, # a,,, m, n E N, and that a # a,,, a # a,,,b,, # b, 6, # P, b,, # b, 8, # fi for each n EN. We shall see in the next paragraph that the conditions on b, b,, , b, b,, imply that there exists g E A(B) satisfying: g(b) # g(p); g(b,) and g(&) are rational complex, n E N. With f = K1(g), by the remark following (2.2) either f(4 = dbn) and f(a3 = d b n ) , n E N, or else f(4= dbn) and !(an) = g(Bn), n E N. In either case the continuity off at a is violated. Hence lim, $ exists, and using the compactness of K we see that lim, $ E B.

=fo

Existence of g . For each n E N we introduce auxiliary functions u,, and w,,in A(B) satisfying: u,,(b)= 0, u,,(bk)= 0 for k = 0, ,n - 1, u,,(b,,)= 1,

...

t To obtain (a") we introduce K = Q{x(zlc)2r} and take a sequence, of positive r with limit 0. The Riemann surface situation is treated differently, but the point is not a serious one.

2. The Isomorphism Theorem of Bers

203

. ..

.. . .. .

, n ; W,(bk) = W,(b) = 0 for k = 0, ,n, = U , ( b ) = 0 for k = 0, w,,(B) = 0, w,(Bk) = O for k = 0, , n - 1, w,,(B,,) = 1. Let h E A(B) and satisfy h(b) = 0, h(B) = 1. We seek to construct g as a function of the form u,(Bk)

where y,, 6, E C are to be so chosen that uniform convergence in B to an analytic function is assured and that the function g satisfies: g(b,) and g(B,) are rational complex for n E N. To that end, we introduce for each n E N the set K,, as the set K of the second paragraph back with r = (n + 1)-’.f’ Clearly K, is a compact subset of B. We choose y o , 6, so that IYO

uo(4

+

60

wo(z)l 5 2 O ,

z E KO ,

and h + ( y o uo + 6, wo) takes rational complex values at bo and Po. We proceed inductively, and having chosen Y k , for k = 0, ...,n, we take Y n + l , 6 n + t SO that IYn+ 1 un+ 1 ( z )

+ 6 n + I wn+ l(z)I

5 2-(”+

‘’3

z E Kn+1,

and

With y, and 6, so chosen the takes rational complex values at b,+, and desired convergence properties as well as the stated requirement for g are assured. Action of O on constants. We proceed under the assumption that O(i,) = iB. The case where O(i,) = ( - i ) B admits parallel treatment. Let b = lim, II/. Letf be a nonconstant member of A(A). Let (a,,): N 3 A be such that limn+ma, = a and f(a,) is rational complex, n E N. The existence of (a,) is assured by the open mapping property. By (2.2) we conclude

e(f>( b ) = f ( a ) .

(2.3)

Takingfnonconstant,f(a) = 1, and taking c E C, we see, on considering cfin place off; that CO(c,)

so that O(c,)

= c B . The

(b)ICW) @)I= c f ( a ) ,

complex constants are “preserved.” On appeal to

t Here, too, the Riemann surface situation would be treated differently, but the point is not serious.

VIM The-Zeros and Poles of Meromorphic Functions

204

(2.2) we reach the basic conclusion

w-1($(a>) = f(a),

(2.4)

u E A , ~ A(A), E This relation permits us to complete the proof rapidly as far

as the present setting is concerned. It suffices to take O ( f ) as univalent and analytic on B; $ is then meromorphic on A for $ = O(f)-' of. We take q = $ - I . The proof is completed by noting that if rp,: B - , A satisfies f q1 =f rp, f E A(A), then rpl = rp, for if al and a2 are distinct points of A, there exists f~ A(A) such that f ( a J # f ( a 2 ) . However to have a proof available also in a Riemann surface setting where such univalent functions may be lacking, we indicate an alternative attack. We take g E A ( B ) having a simple zero at b = lim, $ and no other zeros and conclude from (2.3) and (2.4) that $(a) = b. Consequently $ is continuous on A . To show that $ is meromorphic, we continue with g and take positive r so that g I A(b; r) is univalent and thereupon positive p so that $[A(a;p ) ] c A(b; r). Hence $ I A(a; p ) is the composition of the inverse of g 1 A(b; r) and f l A(a; p ) and consequently is meromorphic. The remaining details are readily supplied. 0

0

EXERCISES 2.1. Treat the case: e(iJ = ( - i ) B . It is convenient to consider the compositionof functions that are either meromorphic or the conjugates of meromorphic functions. 2.2. Show that if v is a univalent meromorphic map of a region B c onto a region A c then f 9 with domain A(A) defines an isomorphism of A(A) onto

e,

e

0

A@).

3. INFINITE PRODUCTS

It is well to keep in mind in the study of infinite products that when we deal with nonzero terms we want convergence to correlate with convergence of some infinite series whose terms are logarithms of the factors involved. The same goes for uniform convergence with factors not taking the value 0 or 00. Thanks to this introductory observation it will be seen that large sections of the initial parts of the subject make completely straightforward exercises at this point. Dejinitions and elementary facts. Let there be given (a,): N + C.By the infinite product associated with (un), denoted a,, is meant the sequence .t We say that a, is p-convergent provided that for some m E N

(n:

n:

fl:

t The notion may be introduced for more general algebraicstructures where a multiplication is present.

3. Inflnite Products

n;

the sequence used to denote

205

am+,,possesses a limit in C - (O).? In this case,

nz

a, is

which exists in C and is termed the product of the a,, n ranging from 0 to 00. We say that a, is p-divergent if it is not p-convergent. We say that a, is p-Cauchy provided that for each positive E there exists m E N such that

n;

n$

nz n;

a, is p-convergent if and The following theorem is easy to establish: a,, is p-Cauchy. Consequently, if a, is p-convergent, then only if a, = 1. First suppose that n g a, is p-convergent and that m E N is such that n g am+,= b E C - (0). By the continuity of x/y with domain C x (C - (0)) at (6, b), we see that given positive E there exists p E N such that

n;

for q > p 2 p

+ m, that is,

I1 n; n;

for admitted p and q. Hence a, is p-Cauchy. The fact that limn-rma, = 1 is a consequence. Suppose now that a, is p-Cauchy. We fix m such that

Hence q+(m+l)

p+(m+l)

m+ 1

m+ 1

We conclude that (nL;(;I+l) a&N

t“p

”

is for “ product.”

p+m+

m+ 1

is a Cauchy sequence and from the choice

VIII. The Zeros and Poles of Meromorphic Functions

206

of m that n g u ( , , , + ~ )possesses +~ a limit in A(1; 4). Consequently n g a, is p-convergent . We say that anis absolutelyp-conuergeentprovided that flg (1 la, - 1I) is p-convergent and that n g an is absolutely p-Cauchy provided that n g (1 + la, - 1I) is p-Cauchy. These conditions are, of course, equivalent to the conditions: the sequence

+

\

possesses a limit in R (resp. is Cauchy). We are now prepared to consider the first set of exercises on infinite products.

EXERCISES 3.1.

Given ak E C, k

= 1,

n

I

. .., n. Show that n

I

n

5 ex,(

2

IM) - 1.

3.2. fl; a. is absolutely p-convergent if and only if X:o"la,, - 1I is convergent. Each of these conditions is equivalent to: for some m E N, urn+,,# 0 for all n E N and (LogU,,,+~I is convergent. If &' a. is absolutely p-convergent, it is p-convergent. 3.3. r k an isp-convergentif and only if there exists m E N such that an# 0, n 2 m, and X:o"Log a, is convergent. 3.4. If azn= (n 1)-ll2, = -(n l)-l/z, n E N,then n2 (1 an)p-diverges but Xi? a. converges (however not absolutely). If

.

+

~

2

= n -(n

+

+

(a)

0

(b)

ff [

[I

0

+

(n+ l)(n+ l 3) l '

1 - (n

+

+ +

n E N, (n 1)-1'2 (n l)-', a.) is pconvergent (however it is not absolutely

~ z n + = l

then Xi? a. diverges but n:o"(1 + p-convergent). 3.5. Evaluate

+

3 1

+ 2)(n + 4)

'

3. Injnite Products n

(d) lim n-m

0

k2-k+A k2+k+A

where k 2

207

+ k + A # 0,

k EN.

+

Using sin TZ = TZ n2 [l - (n 1)-2z2],represent I/T, l / ( ~ 4 2 and ) , 43/(2rr) as infinite products (cf. Ex. 1.8, this chapter). The case z = 4 yields the Wallis formula for 4 2 . 3.7. Determine the sets of p-convergence (absolutep-convergence) of (1 z" + I ) , fl; (1 z2"), n2 (1 cnz)given 1c.I convergent. Show that the products define functions that are analytic on the respective convergence sets. 3.8. Given (a"):N C.Let p denote the supremum of 3.6.

+

+

n;

xo"

-

--f

x,"

Show that p is the radius of convergence of a,z". Show that if p > 0, then the function given by fl; (1 a.z") is analytic in A(0; p). 3.9. Show that when IzI < 1 we have

+

m

and

3.10. Let (pn)nEN denote the sequence of prime positive integers in increasing order: p o = 2, p1 = 3, Then the domain of absolute p-convergence of

. ...

m

is {Re z > 1) and for z in this set we have

=x?

where [(z) exp( -z log n). Hint: Use is to be made of the geometric series and of the multiplicative representation of a positive integer pi("), Y : N + N, { n : v(n) # 0) finite. 3.11. Let la1 < 1. Let f(z) (1 -u"+'z). The function f is analytic on C. Determine with the aid of the identity f(z) = (1 az)f(az) the coefficients of the

n;

=n;

-

power series expansion of 1: 3.12. Let 0 < (an\< I , n E N, and suppose that 2;; (1 that

-

la.]) is convergent. Show

Vlll. The Zeros and Poles of Meromorphic Functions

208

pconverges for Iz( < 1 and show that the product represents on A(0; 1) a function f analytic on A(0; 1) and having modulus less than 1 and such that a,(z) is the number of elements of the set { n : a. = z}. What happens when IzI > 1? Hint: If la1 < 1, 121 = 1, we have

Let (urn): N + C - {O}, let p E N, and suppose thatz; la.( - ( p + l ) is convergent. Show that n2 Ep(z/a.) represents a function fanalytic on C such that a,(z) is the number of elements in the set { n : a. = z). Show that the product is absolutely p-convergent using the hint of Ex. 1.3, this chapter. Show that If(z)l <exp(A I z I p + * ) , A a positive number, using the result of the exercise just cited. 3.14. The gamma function. We consider the product of Ex. 3.13 with an= -(n I), p = 1 (for p = 0 the condition is not fulfilled). Letfdenote the function represented by the product. Show that 3.13.

+

(z

+ l)f(z + 1) = f ( z ) c Y

where y = lim n-rm

[(It) 1 - log(n

+ 1) .

[ y is the Euler constant. It satisfies 0 < y < 1. The existence of the limit and the inequality just stated may be established with the aid of the representation

l o g x = ~ l x1- d t ,

O<X p 2 p.

When X is a Hausdorff space, we shall use the earlier “ i n ” convention in the present context. It is easy to check that (1) f, is uniformly p-convergent on X if and only if is uniformly p-Cauchy on X. It is also easy to see that (2) nFf, is uniformly p-convergent on X if and only if, for some m E N, the functions fndo not take the value 0, n 2 m,and Log f,+, is uniformly convergent on X. The sufficiency is concluded by using the continuity of the exponential function at 0. The necessity may be concluded as follows. For some p E N, we have

nr

nzf,

0

whence f,(x)# 0, x

E

X,n 2 p

+ 1. In addition, we have

This may be seen by induction, since the assertion is trivially true for q = p and

n4,’+

+1

{ f, whose imaginary part takes values in ] - n, n[ since is a logarithm of the imaginary part of each of the two terms of the sum takes values in

VIII. The Zeros and Poles of Meromorplric Functions

210

] - n / 2 , 4 2 [ . We thereupon conclude that, given positive E , 6 E 10, 1[ being such that lLog zl < E for Iz - 1I < 6, if 11 E N is now chosen so that

then

I$

Log

Of,+,(X>

I

E,

x€

x, p 2 p + 1.

The necessity follows. f, is uniformly p-convergent on X , then for m It is to be noted that if sufficiently large the sequence fm+k)nsN tends uniformly on X (in the ordinary sense) to a function not taking the value 0. The example where X = 10, l],f;(x) = x ( " + ~ ) - n* ,EN,shows that the converse is false.

n: (n:

EXERCISES 3-15. Establish the last assertion. 3.16. If ngf. is uniformly p-convergent on X , there exists P E N such that (n?,f,,,)..~converges uniformly on X (in the ordinary sense) to a function F, bounded from 0 and m, m 2 p. [That is, there exist u, /3, 0 < a < /3 < co, such that u < IFm(x)II@, x E X . ] In the opposite direction, if there exists m E N such that fm+&N tends uniformly in the ordinary sense to a function bounded from 0, then fn is uniformly p-convergent on X. I f m + . - 1 I converges uniformly on X, then 3.17. If for some m E N the series l: fn is uniformly p-convergent on X. 3.18. Test the products of Ex. 3.12 and 3.13, this section, for uniform and absolute p-convergence.

+

(na n$

x:

4. SUMS AND PRODUCTS

It frequently occurs in representations of functions by sums or products that the given enumeration of the terms is irrelevant. The best way to approach the matter is via a general notion of a sum or product for a function. It is not our object to develop the subject at length here, but rather to have ready a few convenient modes of expression and formulation.

Sums. Given a set A and f:A --t C,for X finite c A, we define s ( X ;f)as 0 when X = 0 and as the common value of c : j ( x k ) where (xk): is a univalent enumeration of X in case X # 0.There is at most one a E C such that for each

4. Sums and Products

211

positive E there exists B finite c A such that i s ( X ; f ) - a1 < E ,

X finite, A 3 X 3 B. When such a exists, we say that f is summable and term a the sum off, denoting it S ( f ) .When A is finite, the notion of a sum agrees with that given earlier [Ex. 3.1, Ch. V]. Suppose thatf(A) c R+.If in this case the set of s ( X ; f ) is bounded,fis summable and S ( f )is just the supremum of the set of s ( X ; f ) . If in the same casefis summable, then the set of s ( X ;f ) is bounded. Returning to the unrestricted case of complex-valued functions, we note that i f f , g: A -,C are summable and a E C, thenf’+ g and af are summable and S(f + 9 ) = S(S>+ S(g),

W f =)a S(f 1.

If 0 r f ( x ) Ig(x), x E A , and g is summable, then f is summable and S ( f ) s S(g). Further, if If1 is summable, then so isf’and IS(f)I IS(lf1).This may be seen as follows. Let g = Ref, h = Imf. Then g + < I f ] , -9- IIfl, h+ Ifl, - h - IIfl.7 Since g = g+ +g-, g is summable. The same observation holds for h ; hencef= g ih is summable. The inequality follows on noting that I s ( X ; f ) l 5 S(lf1). lff’is real-valued and summable, thenf’ and f - are summable and hence If1 is. It suffices to take E = 1 and to consider s ( X ; f + ) ,X 3 B. Iff is summable, then Ref and Imf are summable. Hence i f f is summable, If1 is summable. We conclude that iffis summable and B is now an arbitrary subset of A , then f I B is summable. We denote S ( f I B ) by f ( x ) . Here on occasion we shall write ‘ ‘ P ( x ) ” for “ X E B” where P ( x ) is a sentential function such that B = { P ( x ) } . The same remark holds for products. We recall that a set is said to be countable provided that it is finite or may be mapped univalently onto N and note that iffis summable, then {f ( x ) # 0 } is countable. Iff is summable and { f ( x ) # 0) is infinite, for every univalent (x,) mapping N onto { f ( x ) # 0 } , given positive E there exists m E N such that

+

xxeB

for n 2 m. What has been said for numerical-valued functions may be paraphrased for a map f from A into the family of complex-valued functions with a given domain B. The image of a E A will be denoted indicially f, and the proximity is to be unijorm in the sense of the Euclidean metric on C. When a sum exists, we say that f is uniformly summable on B. The sum is denoted

t Here g + is the composition of x + , X E fi, with 9 ;g -

is similarly defined.

VIII. The Zeros and Poles of Meromorphic Funciions

212

xaEA

fa. As in earlier considerations of convergence, in case B is a Hausdorff space, we also consider f uniformly summable in B, the definition being phrased in terms of the functionsf", fa" =f, I K , K being compact c B. Iff is uniformly summable on a Hausdorff space B and each fa is continuous at b E B, then fa is continuous at b. If B is an open subset of each fa is analytic on B, and f is uniformly summable in B, then fa is analytic on B. We may consider sums for meromorphic functions. Suppose that B is a region c and that fa is meromorphic on B for each U E A . Uniform summability off in B is referred to the chordal metric. That is, f is termed X-uniformly summable in B provided that there exists g: B + such that for each positive E and compact K c B there exists X finite c A such that for Z finite, X c 2 c A , we have

xaEA

xaEA c,

c

e

xC@;S) (4, &)I < &,

z

E k'.

Clearly such g is unique-it is termed the X-uniform sum off in B. Further, it follows from arguments drawn from our earlier studies concerning sequences of meromorphic functions that either g is the constant 0 3 ~or else is meromorphic on B. If for each b E B there exists positive r such that A(b; r ) c B,f, is analytic in A@;r ) save for a in a finite subset X of A and g is uniformly f, is summable in A(b; r), where go =f, I A(b; r), a E A - X , then meromorphic on B.

xaeA

EXERCISE 4.1.

Validate the unproved assertions made thus far in this section.

Products. Parallel definitions are to be made for products. Starting with

n;

f:A + C we introduce, as above, p ( @ ; f )= 1 and p ( X ; f ) = f(xk). We say that f is productible provided that there exist a E C - (0) and X finite c A such that for each positive E there exists Y finite c A that

- X with the property

Ip(Z;S) - 4 < E

+

for 2 finite, Y c Z c A - X.It is easy to see that 1 If - 11 is productible if and only if If - 11 is summable and that f is productible if 1 + If1I is.? Iff is productible, there exists unique b E C such that for each positive E there exists X finite c A such that Ip(Y;f) - 4 < E

t Here

"

1 " for

"

1":'

5.

The Jensen Formula

213

for Y finite, X c Y c A. We term b the product o f f and denote it by n o d

f(4-

EXERCISE Validate the above assertions concerning products. To pass from “absolute 1 I productible) to productibility, establish a “Cauchy ” productibility ” (1 condition for p ( Z ; f ) and introduce (Z&N satisfying 2, finite c A, Z. c Z n + , , ( p ( Z - Z n ; f )- 1I < 2 - ( ” + l ) Z , finite, Z. c Z. Does productibility imply absolute productibility ? 4.2.

+ If-

Products with function factors. Again the definitions paraphrase those for sums. We allow (looking to our function-theoretic situation)f, to take values in We say that f is uniformly productible on B provided that (1) there exists X finite c A such thatS,(b), a E A - X,is productible for each b E By (2) for each positive E there exists X finite c A such that

c.

b E By Y finite, X c Y c A. Again, “uniform in” is formulated viafaK. In the meromorphic case uniform productibility in B (open c implies that there exists a unique function g meromorphic on B such that given positive E and compact K c B there exists Y finite c A such that

e)

xCp(Z;.f) (4,&)I < E

for z E K , Z finite, Y c Z c A. We denote g by zero, then f, is not.

naEA

naeA f,. If no

fa

is theconstant

EXERCISES 4.3. 4.4.

a,

=

Develop the above remarks. In the meromorphic situation, if g

LA a,, .

=n.,,

/ is not the constant zero, then

In the meromorphic situation, if for each b E B there exists positive r such that A(b; r ) c B, f. is analytic in A@; r) save for a in a finite subset X c A and g., a E A - X , is uniformly summable on A@; r) where g. = - 1 /)I A(b; r), then /is uniformly productible in B. 4.5.

(If.

5. THE JENSEN FORMULA Let f be meromorphic on a region A in but not the constant zero. Suppose that r is positive and such that A(0; r ) c A . The Jensen formula, which we

214

VIIL The Zeros and Poles of Meromrphic Functions

shall develop below, owes its importance to the fact that it relates in a way that turns out to be useful the following: 8, I A(0; r), the term of index d,(O) of the 0-Laurent sequence off, (log Ifl) I C(0;r). As preliminaries for developing the Jensen formula we shall want: (1) the Gauss mean-value formula for the real part of an analytic function, which follows from the Cauchy formula for the disk, on taking real parts, as well as from the theory of harmonic functions, (2) the formula (which we shall treat presently) (2n)-1 Jo2'10g Ire'' - a1 de = max{log r, log lal>,

(5.1)

(3) the Mobius transformation L, given by a-z LJZ) = 1 -Liz'

(5.2)

where la1 < 1. We recall that IL.(z)l = 1, IzI = 1 (cf. Ex. 3.12, this chapter).

Theformula (5.1). This can be verified very simply with the aid of the Gauss mean-value formula when la1 # r. Thus when la1 > r, we see that the left side of (5.1) is log la1 and we conclude (5.1) in this case. When la1 < r, the left side of (5.1) is

and applying the Gauss mean-value formula, we conclude (5.1). When la1 = r, we see that the left side of (5.1) reduces t o log r

+;n

jo2=log 11

eiel do.

The second term is seen to be 0 by applying the Cauchy integral theorem to the sum of the integrals of [Log(l - z ) ] / zalong the paths

-

(e-la)(eia),

va(e)= ei',

a I8 5 2n

- tl, o
a ( f ) we have for sufficiently large r. In the opposite direction, it is easy to see that the

t An elegant proof of the Jensen formula in the version (5.5) may be given with the aid of the theory of subharmonic functions, based on convexity properties of means of subharmonic functions. The version (5.7) brings us to the threshold of R.Nevanlinna's First Fundamental Theorem [5 1, 88, 891. $ If Xis a topological space, a E X adheres to X - {a}, and Fmaps X or X - {a} into fi, by the limit superior of F at a,denoted lim sup. F, is meant inf{sup F ( V - {a}):V a neighborhood of a}. In (6.1) we consider r > 1.

6. Entire Functions of Finite Order

217

infimum of the set of positive fl such that (6.2) is satisfied for large r is the order of j : To say that f does not grow roo rapidly means simply that a ( f >< +co. It is not hard to determine the order of some of the standard entire functions we have encountered. Thus for the exponential function M(r) = e' and the order is 1. For the sine from lsin zI IIsin(i 1z1)1 we have M(r) = (e' - e-')/2 and the order is again 1. For a polynomial the order is 0. Forf(z) = exp(ez), a ( f ) = +a. For

For

where ,u has finite variation, we have If(z)l _< AeBlzl,A and B being positive, and consequently at all events ~ ( f I ) l.t

EXERCISES 6.1. Show that for J o , the Bessel function of order 0, M ( r ) = Jo(ir)and conclude

from the power series expansion that sinh r

-< M ( r ) < e', r

r positive, and that not only is a(Jo)= 1 but also lim+ r d l log M ( r ) = I . 6.2. Given f ( z ) a k z kentire. Let p ( r ) laklrk.Show with the aid of the

=x;o"

=c,"

Cauchy inequalities that

M ( r ) l p ( r ) < M ( q r ) ( l -q-')-',

1 0, we have 1

+ Iz12 < + z12 + ( 1 - z'(

< 3(1

+1 ~ 1 ~ ) .

Hence

Consequently, d is the divisor of a function analytic on {Re z > 0}, bounded, not the constant 0, if and only if d(z) (Rez)(l + lzI2)-' with domain {Re z > 0} is summable.

228

VIIL The Zeros and Poles of Meromorphic Functions

EXERCISES

+

Given Re a positive. Show that h.(z) = (a - z)/(cS z) with domain {Re z > O } is a univalent map of {Re z > 0 } onto A(0; 1). Show that f i s a univalent function analytic on {Re z > 0 } with image A(0; 1) satisfyingf(a) = 0 if and only i f f = & , 7 E C(0; 1). State and prove an analog of the lemma of Schwarz for g analytic on {Re z > 0)whose modulus takes values less than 1, g ( a ) being 0. 7.6. Given Re b positive. Suppose that CP is the set offanalytic on A(0; 1) satisfying Ref(z) > 0, 1zI < 1, and f ( 0 ) = b. Show that { If'(0)l: f~ CP} has a maximum; determine its value and the members of @ which yield the maximum. Hint: Consider hb f and apply the lemma of Schwarz. 7.7. Suppose that 8 mapping {Re z > 0 } into N satisfies the condition that 8(z) = 0 for z small and the condition of the text: a(z)(Re z)(l )zI2)-' summable. Show that [(cS/a)h.]'(")with domain {Re a > 0 } is uniformly productible in {Re z > O } . Prove a corresponding result when a(z) = 0 for z large, 1 replacing ci/a. 7.8. Show directly that iffis analytic on {Re z > 0 } ,takes values of modulus less than 1, and is not the constant 0, then when E is a finite subset of {Re z > O } , f/n.., (h.)'f(") is analytic on {Re z > 0 } and takes values of modulus at most 1. Conclude that the divisor condition of the text cited in the preceding exercise characterizes the divisors of the functions analytic on {Re z > 0)which are bounded and different from the constant 0. 7.5.

0

+ ~

B> Chapter IX @@

T H E GAMMA AND ZETA FUNCTIONS. P R I M E NUMBER T H E O R E M

PREFATORY REMARKS This chapter-the final one of the first part of the book-extends the examination of the gamma function and Riemann zeta function initiated in the exercises of $3, Ch. VIII. In part, our object is to prepare the way for a proof of the Prime Number Theorem of Hadamard and de la Vallte Poussin, which states: With n(x)denoting the number of positive primes not exceeding the positive number x , we have

log x lim n(x) -= 1. X

+m

This important theorem of number theory is of great interest at the present stage of the student’s experience in complex function theory because the preparation for its proof brings into play a number of basic concepts and results developed thus far in the book. We cite the following: the Cauchy integral theorem, the residue theorem, the identity theorem for meromorphic functions, analytic logarithms, infinite products, the Hadamard representation theorem. 229

230

I X . The Gamma and Zeta Functions. Prime Number Theorem

The gamma function, introduced in the eighteenth century by Euler, has been the subject of numerous investigations. The Riemann zeta function is of basic importance in analytic number theory. The celebrated conjecture of Riemann concerning the distribution of its zeros has for a century challenged the efforts of some of the most eminent mathematicians. We shall see that properties of the gamma function enter the study of the Riemann zeta function in an essential way. Prominence will be given in this chapter to the elegant characterization of r I P (we recall that P is the set of positive reals) due to Harald Bohr and the beautiful treatment of the gamma function based on Bohr’s characterization given by E. Artin in a short monograph now available in English translation [7]. It would be an overwhelming task to survey the literature on the gamma and Riemann zeta functions and one not consonant with our objectives. For the gamma function cf. [31]; [77], [130]; for the Riemann zeta function cf. [118]. Accounts of the Prime Number Theorem are given by Dinghas [38] and Valiron [120] in their treatises. Of course, treatments are to be found in works on analytic number theory.

1. THE GAMMA FUNCTION

We recall that the gamma function r is the reciprocal of the entire function g given by

where y is the Euler constant. At this moment we see that g(z) = eYz P(z), where P is the Weierstrass canonical product associated with the divisor a taking the value 1 at the nonpositive integers and the value 0 elsewhere in C and that the genus of a, ~(a),and a(g) .- . are all equal to 1.t We recall also that r satisfies the functional equation r(z

+ 1) = W).

(1.2)

Let f = log 0 (rI P).It follows from (1.1) that f “(x) > 0, x positive. In fact,

2 (n + a,

f”(x)=

x)-Z.

0

t Only the assertion a(g) = 1 calls for attention. That a(g)I 1 follows from the facts that u ( P )= 1 and that the order of the product of entire functionsfi,f2 does not exceed the maximum of the orders offi and f 2 . In the other direction we know that u(g)2 ~ ( a ) by Theorem 6.1, Ch. VIII. Hence U(g) = 1.

1. The Gamma Function

231

Further g(1) = 1 as we see from (1.1) on considering the partial products, account being taken of the definition of the Euler constant. Trivially, it follows that r(1) = 1. H. Bohr showed that r 1 P is characterized by the conditions: (1) it takes positive values, 1 at 1, (2) it satisfies the functional equation (1.2) for positive z, (3) its logarithm (in the sense of real analysis) has a second derivative taking positive values everywhere on P;and he showed that for sufficiency the condition (3) may be replaced by the requirement that the logarithm of the function in question be convex. We shall establish the sufficiency part of the Bohr theorem after we treat the notion of the convexity of a real-valued function of a real variable. We have already treated convexity on C. For R2 we merely paraphrase (for that matter the notion is available for vector spaces over R or C) and say that E c R2 is convex provided that (a(1-t)+ct,b(l

-t)+dt)~E,

0 I 1 I 1, whenever (a, b), (c, d ) E E. Given f:C + R,where C is a connected subset of R,we say thatfis a convex function provided that the set

E(f)

= {(x,

Y ) : x E c, Y

2f(x)l

is convex.? The following facts are easily verified: (1)fis a convex function if and only if

a I x I b, where a, b E C, a

< b. (2)fis a convex function if and only if

f ( b ) - f(a) < f ( c ) - f ( b ) b-a C-b ’ a l}. We speak provisionally because we shall obtain relations permitting us to infer that the function so specified is the restriction to {Re z > I } of a function meromorphic on C. The latter function will be termed the Riemunn zeta function and will be denoted by 6. Let us first consider the function + m ezlogl

F=Jo

e'-ldt

with domain {Re z > 0). The integral is absolutely convergent for such z. Using the identity

we obtain

IX. The Gamma and Zeta Functions. Prime Number

238

Theorem

Making the change of variable carrying t into tlk, we find

so+ I-

,(zlogr-kr)dt

=

qZ+ l ) e - ( z + l ) l o g k

(2.2)

O0

Given E positive, with x = Re z, we see that the absolute value of the last integral on the right-hand side of (2.1) is less than e

exlog'

I - e-'

dt

+ 1) + (1 - ede)-' ( nr (+x 1)X+"

The estimates involved are straightforward, the second factor of the second term arising from an integral of the form occurring in (2.2), k being replaced by (n + 1). We conclude that

and hence by the arbitrariness of E that (2.3)

We now introduce a function defined by an integral, as we did the functions Hl and H , , but this time a parameter that will be significant enters. Given positive p, not an integer multiple of 271, we introduce H P ( z )=

s

-p

@log

-oo

111 - x i )

ipe" dB

I - e-'

dt. On combining the first and third integrals, we obtain &log p

+ i0) ipe" do. (2.4)

W ( z ) = 2i sin nz

Now the first integral in (2.4) is readily seen to define an entire function. We introduce G(4

Clearly IG(z)l =

iJl

+m

+a

=Jl

eI logt dt.

er l o g t - 1 D d t l I ( 1 - e-l)-lr(Izl

+ 1).

(2.5)

2. The Rienrann Zeta Function and its Functional Equation

239

We proceed to obtain an upper estimate for the order of the entire function G and thereupon of H P . We can use the Stirling formula to control M(r; G). Actually a much cruder estimate of r I P is adequate. From the convexity of log o (rI P), the nonconstancy of r, and the fact that r(1)= r(2), we know that r is increasing on [2, + a[.We obtain the crude inequality

r(x) < r([x]

[XI = max{k: k E Z, k I

+ 1) < [ x ] [ ~ ]I eX'Ogx,

x}, for x 2 2. Using this estimate and (2.5) we obtain

log M ( r ; G) I -log(l - e - ' )

+ ( r + 1) log(r + l),

r 2 1, and hence a(G) 5 1. It is easy to see that a ( H P- 2i sin nzG) I 1. Hence we conclude ct(HP) I 1. Interpreting the integrals involved in the definition of H P as line integrals, as we did in the case of H I , we see with the aid of the Cauchy integral theorem that H P ( z ) is independent of p, 0 0, and obtain, using (2.3), (2.4), and the obvious H(z) = limo H P ( z ) ,the important relation

+ 1) 2 m

~ ( z=)2i sin nz r ( z

e-(Z+l)logk.

(2.6)

1

1;

From (2.6) we reach the basic conclusion that e-'logk with domain {Re z > I } is the restriction of a function meromorphic on C, which function, [, satisfies the relation

the usual conventions holding.

EXERCISE 2.1. Show with the aid of (2.7) that & ( l ) = -1, &(z) 2 0 , z # 1. Show from (2.4) that H ( 0 ) = 27ri and conclude that res( 1 ; 6) = 1.

We consider now H ( * " + l ) ' , nN~ . From (2.4) we see that for fixed z, Re z < - 1, we have limn+mH(2nf1)'(~) = 0. It is to be noted that

(

0 < inf 11 - ezI : z E

I

U C[O; (2n + 1)71] .

noN

Further, by virtue of the line integral interpretations and the residue theorem,

240

IX. The Gamma and Zeta Functions. Prime Number Theorem

H(2"+1)n(~) - H(z) is simply 2ni times the sum of the residues of the meromorphic function ez Log w

1 - e-w

with domain C - { - t : t E R+}at the points 2nik, 1 I Ikl 2 n, k E Z. That is, n

Hence we see that for Re z < - 1 we have XZ

H ( z ) = -4ni cos - ezlog "[( - z ) ,

2

and hence for Re z < 0 we have nZ

H(Z

- 1) = -2i sin - eZlogZn [(I - 2 ) . 2

By appeal to (2.7) and the identity theorem for meromorphic functions we obtain the functional equation for the Riemann zeta function

valid throughout C with the usual conventions prevailing. This important relation yields information on the zeros of the Riemann zeta function. Thanks to the Euler product formula for [ ( z ) ,Re z > 1, of Ex.3.10, Ch. VIII, we see that [ has no zeros in {Re z > l}. Using (2.8), the fact that r has no zeros, and our knowledge of the zeros of the cosine, we see that a,(z)= 1, z ~ T = { - ( 2 + 2 k ) : k ~ Nand } a,(z)=O, R e z < O , z # T . The members of Tare termed the " trivial" zeros of [. We conclude: the nontrivial zeros of 5 lie in the strip (0 I Re z I I}.? We introduce two traditional auxiliary entire functions. The first, 5, is to satisfy < ( z ) = + z ( z - l)e-(z'ogn)'2 T(2/2) c(z),

z # -2k,

k E N.

(2.9)

From (2.7), the fact that a ( H ) I 1, and Ex. 6.4, Ch. VIII, we see that > 2,

a(() I 1. We shall now show that a ( t ) is exactly 1. Indeed for r

&r) 2 e-("ogn)'2r ( r / 2 ) .

t Note that the term "strip" is used with a certain amount of license.

2. The Riemann Zeta Function and its Functional Equatwn

Suppose that y > 2. Using the convexity of log

0

241

(rI P) we have

log r(x)- log 1-(2), log r(y) - log r(2) x-2 Y-2 for x 2 y. Hence for r 2 2y we have

It follows from the fact that lim+m[log I-(y)]/y= + co that a( 0, 1 < x < a, for some a satisfying a > 1. By (3.2)

and hence @' Q

2 A(n) e-x'ogn[cos(tlog 00

Re - (x) = - 2

n)

+ 112 5 0

2

when x > 1. The contradiction is manifest. Consequently {Re z = l} and the theorem is established.

has no zero on

4. THE PRIME NUMBER THEOREM OF HADAMARD AND DE LA VALLRE POUSSIN The exposition of this section follows substantially N. Wiener's proof of the Prime Number Theorem in the formgiven by Landau [141]. That there is an intimate connection between the Riemann zeta function and the set of positive primes is, of course, clear from the Euler product formula of Ex.3.10, Ch. VIII. The point is to take advantage of the relation. An important link is furnished by the function $(x) = A(n) with domain P, A retaining the meaning given to it in the preceding section of this chapter. Let 3 5 x < + 03. Let p be a positive prime. Let

xnsx

m(p) = max{k: k E N , pk 5 x}. Clearly m(p) I(log x)/log p . Hence =

2

m(p>log P Ixw log x,

(4.1)

the sum being taken over the set of positive primes. When 1 c y < x,

the sum being taken over the set of primes in ]y,XI. Now setting

244


y = x(log x ) - ~ ,which satisfies the above condition, and noting that ~ ( yc) y , we obtain

logx

x

n(x)

log x x log x - 2 log(l0g x ) '

1

$(XI

l + If(b>l + If'(x)l dx. jab

The asserted inequality is now evident. The second result that we shall want is a very special instance of a Fubini theorem. It states: Let - co c a < b < + 00, - co < c < d c + co. Let f: [a, b]-,C satisfy Y [f ] < + co. Let g be a continuous map of [a,b] x [c, d ] into C. Then

t That is, letfE

C'[u, b].

4. The Prime Number Theorem of Hadamard and de la Vallge Ponssin

245

the integrals involved existing.t The proof will follow from Ex. 4.1 below.

EXERCISE 4.1. (a) With f as in the above-stated result and h a continuous map of [a,b] into C, f h possesses an integral in the sense of Riemann and

laf h

b

b

f dH,

dx = a

where

H ( x ) = r h ( t )dt. a

(b) The function given by

is continuous on [c, d ] .(c) Let n be a positive integer and approximate the integral on the left side of (4.4) by a sum of the form

where yk = c

+ n-'k(d - c), k = 0, . .. , n. Note that

where M = sup[o.b] I f

I and conclude (4.4).

We now turn to the proof of (4.3) to which, as we have seen, the Prime Number Theorem reduces. We make a change of variable for convenience introducing K(u) = J/(eU) e-", u E R,and note that (4.3) is equivalent to lim K = 1.

(4.5)

+m

t We are not presupposing a knowledge of such results, though to be sure some readers are familiar with fairly general forms of the Fubini theorem. It is worth emphasizing the important role played in analysis by theorems that permit interchange of order of integration or summation.

246


We introduce the function G meromorphic on C,satisfying

K

G(z) = - -(z)

I'

+z-1 z

when z # 0, 1. For Re z > 1 we have by (3.2) of the preceding section

and hence, by the change of variable t = e", 1 e-'" d$(e") - z - 1'

Integrating by parts and noting that $(x) 5 x log x , x 2 1, we obtain 1 $(e") e-'" du - z-1

-

= jO+OD[K(u) l]e-('-')"

du.

We note that G is analytic at each point of {Re z 2 l } since [ has no zeros on {Re z 2 1) (cf. Theorem 3.1, this chapter) and has a pole only at 1, which is harmless as far as G is concerned. By the above form of the RiemannLebesgue lemma applied on [-2,0] and [0,2] we see that for each positive 1 there exists a positive number C(L) such that

for y positive, 0 5 E 5 1. Let G,(z) =

1':

[ K ( u ) - l]e-('- 'Iu du,

n E N.

Applying the ancillary theorem of Fubini type we obtain

It is to be observed that no restriction need be imposed on E , A, y in (4.7).We

247

4. The Prime Number Theorem of Hadamard and de la Vallie Poussin

continue assuming 1,y positive and 0 < E 5 1. It is routine to see that

when u # y (and is = A when u = y). Introducing the change of variable A - l u , we see that each side of (4.7) is equal to

u =y

+

We note that G, tends uniformly to G in {Re z > I}. Hence by (4.6), (4.7), and (4.8) we are led to

and, since K(u) = 0 for u negative,

(4.9)

We next show that (4.9) remains valid when E is replaced by 0. Indeed, when 0 < E I 1, from (4.9) we obtain

- co

< s -
- CO. Show that with definitions analogous to those for A of addition and multiplication B is a field. 1.3. Let (an)E A and ~ ( ( a . ) = ) (b.) E B where b. = 0, n < 0,b. = a n ,n E N. Show that is an isomorphism of A into B and that the subfield of B generated by v ( A ) is precisely B. What happens to the above results when F = C and A and B are a.z" taken as the sets of sequences ( U & ~ N and (b,),.z, respectively, such that and b,z' have radii of convergence greater than zero and inf { n : b. # 0} > 00 ? 1.4. If (a")is the 0-Laurent sequence off E M(C), there exists (b"):Z C,6 . = 0 for n < 0, bo = 1, b. = 0 for n 2 m for some positive integer m such that the components of (u,,)(bJ with indices 2 m are zero. Show that if (a"): Z + C satisfies inf { n : a. # O} > - co and there exists (6,) of the above kind such that (un)(bn)has the stated property, then (an)is the 0-Laurent sequence of some f E M(e).

ct

c,"

~

--f

2. SUMS, CAUCHY MULTIPLICATION

EXERCISES 2.1. Let f: X + C be summable. Suppose that '$ is a collection of subsets of X satisfying U Y . g Y = X ; if Y , Z E'$ and Y # Z , then Y n Z = 0 . WedefineF(Y)as s(fl Y). Show that F is summable and that

2

F(Y) =

YOV

2 m.

xsx

A special case was considered in Ex. 3.1, Ch. IX. 2.2. Let fi X - t C and g: Y + C be summable. Let h(x, y ) = f ( x ) g(y) with domain X x Y. Show that h is summable and that

2

( x , Y )E x x

2.3.

h(x, v) = r

[ 2 f W ][ 2 xsx

g(y)].

YSY

Apply Exs. 2.1 and 2.2 to the situation of absolutely convergent serieszt an,

c,"b. to conclude that c,"c. is absolutely convergent and

Throughout this exercise (c,) is the product (u,,)(bn)in the sense of the preceding section. a. is absolutely convergent and Xt b. is Show that under the assumption that convergent, c. is convergent and has sum a.)(C,"b").Show that the theorem is false if the hypothesis of absolute convergence is dropped for X : u.. It suffices to take a. = b. = (-l)"dn where (d.)..~ is a nowhere increasing sequence of positive d.(Q, dk)= 03. Give an explicit numbers satisfying: limn-- d. = 0, instance of such a sequence (d,,). Show, however, that if both a, and b. are convergent, then

ct

c,"

(c,"

+

x,"

fbeingx; c.x" with domain [0, 1[.

c,"

3. Holomorphic Functions of Two Complex Variables

269

3. HOLOMORPHIC FUNCTIONS OF TWO COMPLEX VARIABLES

With the experience of one-variable complex function theory at our disposal, it is not difficult to treat the elementary aspects of the local theory of complex-valued functions of several complex variables. The difficulties of the several variable theory, which are serious, come at a later stage. The present section develops some of the simpler aspects of the two-variable theory starting from the power series point of view. A modern treatment of analytic functions of several complex variables is to be found in [48]. We start with (a,,,,,): N 2 + C and study the set of (2,w)eC2 for which amnzmwn, domain N2, is summable; that is, we are concerned with the summability set of the family of functions a,,z"w" with domain C2.t Suppose that a,,,, ampis summable, q? # 0. Then la,,,, ~ ' " p is )summable and consequently ~amnz"wnis uniformly summable on A(0; I N ~ ) x A(0; IBI).$ It follows that R, the interior of the summability set of amnzmwn, is the union of a family of bicylinders A(0; r ) x A(0; s). Further, at each point ( Z , W ) of the exterior of the summability set, sup{~a,,Z'"W"~:(m,n) E N2} = +co. Indeed, if Z W # 0 and lamnZm W"I IM
l2 - - M r Log( 1 2s

): + (do

-

g),

(4.6)

IzI c p, as we see by integration. We are led to construct an allowed V as follows. We observe that Z2

Z--

2s

with domain A(0; s) is univalent and we introduce cp, its inverse.t We take p, 0 c p c r, so small that the right-hand side of (4.6)takes values in the domain of cp, IzI c p, do being given on [0, s[. We propose for V the composition of cp and the right-hand side of (4.6)with domain A(0; p). Clearly, V is analytic on A(0; p). So defined Vsatisfies (4.6)and V(0)= d o .By its definition V[A(O; p ) ] is contained in A(0; s), the image of cp. Differentiating (4.6)we see that Vis a solution of the differential equation in question. The verification of (c) is complete.

EXERCISE Suppose that in the situation just studied the Taylor coefficients umnof U&) are such that u m n z M is summable ~ to a sum of modulus less than s for (zI < h < r, < p. Let &(z) denote the value of the sum at (2, ,9). Show that 4.1.

SBTZ) =4 2 ,

for each ,9 satisfying

SB(Z)l.

121

< A,

< p.

The Cauchy majorant calculus, which we have explicated in the case of the above existence problem, may be profitably used in numerous situations. An important example is furnished by the implicit function theorem. We continue with A as above, but further restricted to satisfy: A(0,O) = 0, AJO, 0) # 0. We wish to obtain information on the solution set

2l = { ( z , w): A(z, w) = O } , in particular, on the nature of 2l n [A(O; t ) x A(0; u)] for t and u small. The implicit function theorem asserts that for A so specified there exist t , u satisfying 0 c t I r, 0 < u I s, such that

2l n [A(O;

t) x

A(0; u)]

(4.7) is a function analytic on A(0; t). Naturally, we are referring to the complex version of the theorem. The real variable form of the theorem varies some-

t We could be very explicit about y and represent it with analytic square roots. It is not necessary to do so.

4. Cauchy’s Majorant Calculus

277

what from this as one would expect on comparing differential analysis in the real domain with that in the complex domain. We shall show that (4.7) enjoys the property of being a function for suitable small t and u and shall leave the study of the analyticity by means of the Cauchy majorant calculus as an exercise. We have, when 0 < t < r, 0 c u I s, the equality A(z, ~

2

-) A ( z , w i )

=

JWIW* +

Adz,

W)

dw,

JzI< t, Jw,I, Iw21 < u, and hence, when t and u are so small that I A W k

w) - a011 < la0,1/2

for Izl < t , IwI < u, IA(z,

w2)

-

Wl>l

2

la011 I w 2

-W11P

The function property follows, but it is to be observed that we have not even shown that the domain of the function is a neighborhood of 0.

EXERCISES Apply the Cauchy majorant calculus to establish the existence of a function U analytic at 0 and satisfying U(0) = 0 as well as I U(z)l < s and 4.2.

for It1 < p for some p, 0 < p < r. Show that this result taken with the above remarks of the text yields the implicit function theorem. 4.3. Show that the objective of Ex. 4.2 may be attained by applying the existence theorem of the text for differential equations to obtain U analytic at 0, and satisfying U(0) = 0 and (with due regard for admissibility)

4.4.

Show that the implicit function theorem may also be proved by considering

for u suitably chosen and z small. Hint: Take 0 < ( w (I u, and thereupon z so small that

u, 0 < u I s,

so that A(0, w ) # 0,

I W I = u. < IA(O,w)l, 4.5. Treat the existence and uniqueness theorem for a system of the form

IA(z, w ) - 4 0 , w)l

W:

= A ~ ( zw , ,,

. .., w J ,

k

=

1,.

. . ,n,

XII. Topics in the Theory of Power Series

278

where each Ar is holomorphic on [A(O; r)Y+’. It is to be understood that “ holomorphic” is defined in C”+’ by a direct paraphrase of the C2 definitions. 5. VECTOR-VALUED FUNCTIONS

Numerous aspects of the classical theory of analytic functions may be extended to the theory of analytic functions taking values in a complex Banach space-a notion that we shall presently define. Our object in considering briefly a theory of vector-valued analytic functions is to show how natural and useful the power series methods are. We shall see that the methods to be developed lead to a treatment of the existence theorems of systems of linear differential equations which is formally elegant and avoids a plethora of notation. For a thorough account of vector-valued analytic functions cf. [56]. For future reference we agree to state explicitly that we are dealing with vector-valued functions when this is the case. We start with the notion of a right vector space over C. Given a set V and maps + : V z + V, * : V x C + V, we say that ( V , +, * ) is a right vector space over C provided that ( V , +) is an abelian group and that * ,right multiplication by scalars, satisfies : (x

+ y)a = xa + yay

x(a

+ p) = xa + xp,

xl = x, x(a8) = (xa)p, where x, y E V, a, p E C. Of course, we can introduce left vector spaces as well. Solely on notational grounds we shall consider the “right ” theory. The usual practice of omitting * will be followed. Further, the earlier conventions made for algebraic and topological structures are to prevail, so that we term V itself a vector space. The elements of V are termed vectors. We shall denote by 0 the neutral element of V considered as an abelian group. A map of V into R-the image of x E V with respect to it will be denoted Ilxll-is termed a norm provided that the following conditions are fulfilled : (a) If llxll = 0, then x = 0. (b) IIxall = llxll I4 ( 4 IIX + Yll 5 llxll + Ilull. Here x , y E V and a E C. We term llxll the norm of x. It is immediate from 00 = 0 and (b) that 11 0 I( = 0, and from this fact and (c) applied to x + ( -x) and (b), that llxll 2 0. V is endowed with a metric space structure by p(x, y) = Ily - xII. The composite ( V , f, , 11 11) is called a normed (right) vector space over C. It is termed a (right) complex Banach space provided that ( V , p) is a complete metric space. The designation “ right ” will be understood in what follows. For the remainder of this section V will be a complex Banach

-

5. Vector-Valued Functions

279

space. We shall be especially concerned with maps from a subset of C into V. By a bounded endomorphism of V is meant a map T : V + V which is linear, that is, satisfies T(x

where x , y

E

+ y ) = T ( x )+ T(y),

T(xu) = T(x)u,

V , c1 E C, and which also satisfies the condition

N ( T ) =sup{JIT(x)(l:1 1 ~ 1 1= l } c

+CO.

N ( T ) is termed the norm of T. We have IIT(x)ll I N(T) Ilxll, bounded endomorphism T as we see from considering

XE

V, for a

T ( x ) = TCx(llxIl)-'l llxll when x # 0. We introduce B(V), the set of bounded endomorphisms of V, and observe that B(V) is itself a (right) complex Banach space with the following agreements. Given S, T E B( V ) , u E C, S T is the map of V into itself which takes x E V into S(x) + T(x)and Sa the map taking x into S(x) a. In addition, N is proposed as a norm. It is routine to verify that with these agreements B( V ) is a complex Banach space and that further B( V ) is closed with respect to composition so that when S, T E B( V ) we have S 0 T E B( V ) . Indeed, we have N(S T ) I N ( S ) N(T).

+

0

EXERCISE 5.1.

Furnish the details omitted in the above discussion.

We note that the usual notions of elementary real and complex differential and integral calculus may be introduced for functions taking values in V when the norm replaces absolute value. Thus, in particular, the notion of the derivative at a of a map f from a neighborhood (in the sense of the R- or C-topology) of a into V lies at hand.? The following lemma is very useful in the elementary theory of vector-valued functions. Lemma 5.1: Let - c r , < a < b < + o o . a 5 t Ib, then f is constant.

Let f : [ a , b ] + V .

If

f'(t)=O,

PROOF: We introduce g ( t ) = 11 f ( t ) -f(u)II and note that

Ids)

- s(0l I

IIf(4 -f(t)ll.

It follows that g is the constant zero and consequently f is the constant f ( a ) .

t We introduce [ f ( x )-f(u)](x - u)-I

and take the limit at a.


280

E XERCISE Extend the developments of Ch. V to construct a Cauchy theory for functions taking values in a complex Banach space, the starting point of the discussion being the study of Riemann-Stieltjes integrals j : f d g whereftakes values in Vand g in C. Lemma 5.1 will be seen to play the same important role as its cognate in the theory of complex-valued functions. 5.2.

In what follows we shall concentrate on the power series aspects of vectorvalued analytic functions. We suppose that (u,,): N + V. We are led to a paraphrase of the theory of power series for akzk. Here we consider complex z. Among the extensions that are automatic are the property of radius of convergence encountered at the outset in Ch. 11, the CauchyHadamard theorem, the identity theorem, and term-by-term differentiation. It is to be noted that if (b,,):N + B( V) and ak zk and bkzk converge on A(0; r), r > 0, having, respectively, sums A and B on A(0; r), then

1;

c;

1;

m

where

(Cauchy " multiplication "). The asserted equality may be established by appeal to the fact that Ilukll 1zlk and N(bk) lzlk are convergent for IzI < r and by considering the limiting behavior of

1;

1;

Account is to be taken of the fact that when IzI < r, limn+mIlc,,z"ll

= 0.

EXERCISE 5.3.

Let A be an open subset of C,let a E A and let8 A + V be of power series type.

Let rn

for small z. Show that the radius of convergence of x : ckzk is at least d = dist({a},C - A) and that its sum on A(0; d) is f(a + z). The present exercise


281

extends Ex. 6.7, Ch. IV. Show that the method indicated in that exercise is available in the present setting with appropriate changes, principally in the question of treating differentiation under the integral sign.

Geometric series. Let Zdenote the identity map of Vonto itself. It is trivial that Z E B( V ) . Let b E B( V ) . We shall show the identity ( I - bz) o

($

bkzk)= I

=

($

IzI < [ N ( b ) l - ' .

bkzk)0 ( I - bz),

(5.1)

Here bk denotes the kth iterate of b (bo = 1, bk+' = b o bk). Indeed, it is easy t o see that

The identity (5.1) follows on taking the limit. The first half of the identity (5.1) shows that I - bz maps V onto itself a n d the second half shows that Z - bz is univalent where in each instance IzI < [N(b)]-'. Further under this condition ( I - bz)-', the map of V onto itself inverse to Z - bz, is equal to bkzk. We continue with b E B( V ) and introduce Eb: C --t B( V ) given by

C:

2 bk W

Eb(z)=

0

Zk

-

k!

Clearly Eb(0)= I and Eb'(Z) = b o E(z) = E(z) 0 b. With Cp(Z) = Eb(Z) 0

Eb(-Z)

and

$(Z) = Eb(-Z)

0

Eb(Z)

we have the equality

Cp'(z) = 0, = $'(z),

z

E

c,

where 0, is the neutral element of B( V ) . Hence we conclude that Eb( z )

O

Eb( - z ) = I

= Eb( -z )

O

Eb( z )

(5.2)

and see that for each z E C the endomorphism E&) is invertible, that is, it is a member of B( V ) which is a univalent map of V onto itself whose inverse is also a member of B(V).t Suppose that a~ V and that we seek y : C + V satisfying y(0) = a and y'(z) = b[y(z)],z E C.If y does satisfy these conditions, we see that Eb( - z ) [y(z)]with domain C has a derivative taking the value 0

t We are not presupposing a knowledge of the theorem of Banach space theory which states that a bounded univalent linear transformation of a Banach space onto a Banach space has a bounded inverse.


282

at each point of C. Hence Eb(- z ) [y(z)] = a and we conclude with the aid of (5.2) that y(z) = E b ( Z ) (a). In the opposite direction, it is immediate that &(Z) (a) fulfills both of the imposed conditions. It is the unique solution of the proposed problem.

EXERCISE 5.4. Develop the preceding discussion with particular attention to the formulas for the derivative of f(z) [g(r)Jwhere g takes values in V and f i n B( V ) and also for f(z) g(z) where now bothfand g take values in B( V). Suitable hypotheses insuring the validity of the formulas in question are to be given. 0

Let us now see how power series methods for vector-valued functions are applicable to the study of the theory of linear systems of differential equations. We shall start with the problem in its original setting and thereupon shall subsume it under a problem stated in vector terms. Let n be a positive integer. Let ujkand cj be analytic on A(0; r ) , where r O,j, k = 1, . .., n. We wish to study the existence and uniqueness of functions wl,. . . , w, with domain A(0; r ) satisfying wj'(0) = uj , j = 1, . . . , n, where the aj are assigned complex numbers, and also

=-

Of course, the problem just stated, which is not settled by our earlier local study of $4, this chapter, without further considerations, can be treated in the same setting in which it is formulated; but the question may be reformulated in the language of a suitable Banach space. We shall see that a twofold advantage will be gained. Not only will the formalism be very simple but also the class of problems rendered accessible to the methods to be used will include certain problems involving infinite systems of differential equations. Indeed, let V = C". Let + be given by componentwise addition so that if u = ( u , , . . ., u,,) and u = ( u l , . . . , u,) are members of C", then u + u is defined as that member of C" whosejth component is uj + uj , j = 1, ... , n. Let scalar multiplication be defined also by componentwise multiplication so that, with u as above and a E C, we define uu as that member of C" whosejth component is uj u, j = 1, . .., n. We introduce a norm by

With these agreements we see that V is a complex Banach space. Let A(z)


2a3

denote the map taking U(E V ) into the element of V whose j t h component is

. ..,n. A ( z ) is a bounded endomorphism of V . Let C(z) denote the member of V whose jth component is cj(z),j = 1, . .., n. The problem stated above is equivalent to investigating the existence and uniqueness of W : A(0; r ) + V satisfying W(0)= (a,, . .., cr,) and the differential equation

j = 1,

W ’ ( z )= A(z) [W(z)]

+ C(z),

IzI

< r.

(5.4)

It is to be observed that A and C are vector-valued functions given as sums of power series on A(0; r ) . Here A takes values in B( V ) and C in V. Let us now put aside the restrictions of finiteness and the special nature of the Banach space under consideration. We suppose then that we have at hand an arbitrary complex Banach space V and A and C, sums of power series on A(0; r ) , the coefficients, a,, of the first being in B( V ) and the coefficients, c, , of the second being in V, and that wo E V. We seek to study W : A(0; r ) + Y satisfying (5.4) and the initial condition W(0)= wo . We see from the availability of the Cauchy theory that the problem reduces to the study of solutions which are sums of power series on A(0; r ) . If a solution bkzk is available, then by the standard arguments (term-by-term differentiation, Cauchy “ multiplication ” given earlier in this section, the identity theorem) we obtain

1;

It follows from (5.5) that the problem has at most one solution. We note that if

1; bkzk has radius of convergence at least r, where (bk):N + V is the unique

sequence satisfying ( 5 3 , then W,its sum on A(0; r ) , is a solution of our problem, for the initial condition is assured by the first equation and the remaining equations imply that W ‘ ( z )and A(z) [ W(z)]+ C(z) are the sum of the same series, JzI < r. To control the radius of convergence of bkzk we proceed as follows. Let 0 < s < t < r, and let

1:

p,

= max{

11 bk(lsk: k

= 0,

. .., n},

1;

n E N. It suffices to show that (p,) is bounded for then llbkll pk converges when 0 Ip < s and hence, given the arbitrariness of s, when 0 I p < r. The convergence of b, zk for IzI < r follows. From (5.5) we obtain

1;

284


N(ak)t k and c g Ilckll t k are convergent, we conclude that there Since both exists CI positive such that

We put aside the trivial case where all p,, are zero and notice that by (5.6) for n sufficiently large llb,,+l11 s"" p,, and consequently p,,+' = p,, . The boundedness of (p,,)follows.

-=

EXERCISE 5.5. With A as above apply the method of undetermined coefficients to study the differentialequations(a)S'(z) = A ( z ) S(z) and (b) T'(z) = T(z) A ( z ) with solutions taking values in B( V).* [Show that if S(0) = I , then S(z)isinvertible.Show that with S normalized by S(0) = I the totality of solutions of (a) is given by S(z) C, C E B( V), and that the solution of W'(z)= A ( z ) [W(z)](taking values in V) satisfying W(0)= w o is S ( z ) (wo). Discuss the totality of the solutions of (b).] 0

0

0

An important application of the method of undetermined coefficients is to the following problem. Suppose that A is as above and that we are interested in the differential equation W'(2) 2 = A(2) [ W ( z ) ] .

(5.7)

Here W is to take values in V. We have deliberately left open the question of what the domain of the solution is to be. The reason will become clear below. While problems of this kind have been studied exhaustively in the classical case where V = C" (cf. [13]), much remains to be done in the setting of general complex Banach spaces. Initial work in this direction is due to E. Hille (Proc. Internat. Symp. Linear Spaces, Jerusalem, 1960, 263-273). We continue under the supposition that a, # 0,. If bkzk defines a solution of (5.7) on A(0; r), then we obtain

ao(b0) = 0,

If In - a, were univalent for each n E N, the only solution analytic on A(0; r) would be the constant 0. We put this approach aside. Instead, we ask whether by a suitable choice of a complex number 1 we may find a solution of


285

(5.7) with domain a region R c A(0; r ) - {O}, where the solution is of the form U ( z )s(z) where U : SZ V is analytic but not the constant 0 and s is a n analytic I t h power of the identity map on R. Clearly it follows from (5.7) that the following condition is imposed: I , SZ, U must be such that --f

V ’ ( z )2

+ V ( z )I = A ( z ) [ U ( Z ) ] ,

z En.

(5.8)

Of course s of the stated type must exist. A converse proposition is also seen to hold. These observations lead t o the following construction of a nontrivial solution under favorable circumstances (present in the C”theory). Suppose that I E C is such that 112 - a, is not univalent and that

I[I

+ (n + l)] - a,

maps V onto itself, n E N . Let b E V - (0) satisfy ( I n - a,)(b) = 0 and let (b,) be a sequence of vectors in V satisfying b, = b,

Then bk zk has radius of convergence 2 r. Further i f R is any subregion of A(0; r ) - (0) on which there exists an analytic Ith power of the identity map of SZ onto itself; say s, and U is the sum of cF bk zk on R, then U(z)s(z) satisfies (5.7) at each point of R.

EXERCISES Prove the above result. It is convenient to treat first the radius of convergence. The argument used for the system ( 5 . 5 ) may be paralleled here. The fact that (5.9) is satisfied together with the condition on the radius of convergence yields the result that the differential equation of (5.8) is satisfied on A(0; r ) . 5.7. Show that Bessel’s equation 5.6.

+ z u’(z) + (z’

z 2 u”(z)

- X2)

u(z) = 0,

X E C, is equivalent to the system z u’(z)

= v(z),

z o‘(z)

=

(A* - z’)

u(z),

(5.10)

in the sense that the solutions of Bessel’s equation are first components of the solutions of (5.10) and vice versa, the domains being open subsets of C. Show that the method of the present section leads to solutions of z w”(z)

+ (2X + 1)

w’(2)

+ z w(2) = 0

286


which turn out to be restrictions of entire functions when Re h 2 0 (cf. Ex. 2.13, Ch. VI). 5.8. Examine the hypergeometric equation of Ex. 3.12, Ch. IV, from the present point of view at 0 and 1.

6. ANALYTIC CONTINUATION The notion of analytic continuation, which will be introduced in this section, gives complex function theory a special sense of coherence. We shall approach the subject twice, first in the present chapter, where the concept of a function element is used, and later in Ch. XVIII, where the problem is viewed as a part of the modern theory of Riemann surfaces and the advantageous alternative formulation of H. Weyl is central. The less sophisticated approach of the present section has the merit of being very accessible as well as useful in applications. We shall take care in Ch. XVIII to study the relation between the two approaches. In the exposition that follows we consider complex-valued functions. We consider ordered pairs (a, P ) where a E C and P is of the form S 0 7, where S is the sum of a power series with radius of convergence r > 0 on the interior of its convergence set and z is z - a with domain A(a; r).t By an analytic element (or, more simply, an element) is meant such an ordered pair (a,P ) . An element (b, Q ) is termed a direct continuation of (a, P ) provided that b is in the domain of P and Q(z) = P(z) for z near b (and hence for z in the intersection of the domains of P and Q). Given an element (a,P ) and b in the domain of P there is a unique element (b, Q)which is a direct continuation of (a, P). The uniqueness is obvious by the identity theorem for power series. Further (b, Q) is a direct continuation of (a,P) where Q is simply the sum of

on the interior of its convergence set. It is to be noted that the existence of a direct continuation may be considered in purely Weierstrassian terms, that is, exclusively in terms of power series concepts without reference to the Cauchy theory. It suffices to use such ideas as sums of functions and the identity theorem for power series.

t It must be emphasized that if P is entire and a # b, then (a, P) and (b, P) are distinct since they are ordered pairs with distinct first components. At this point it may seem that we are laboring the obvious, but the fact is essential when we turn to incorporating these notions into the theory of Riemann surfaces. It is only the entire P which necessitate the " ordered pair" approach.

6. Anahtic Continuation

281

EXERCISE 6.1.

Develop these remarks.

We denote the component Q of the direct continuation (byQ ) of (a, P ) by Pb and the radius of convergence of the power series whose sum is Pb(b + z), z small, by r(b). Clearly if r(a) = 00, then r(b) = 00. If r(a) c + 00, we have the following inequality:

+

+

+ Ib - al. r(b) > r(a) + Ib - al, then

r(a) - Ib - a1 5 r(b) I r(a)

(6.1)

The left inequality is immediate. If we would conclude that A[a; r(a)]c A[b; r(b)] and P ( z ) = Pb(z)for z in A[a; r(a)],and the direct continuation of (b, Pb) with first component a, would have an associated radius (in the sense of the first sentence of this paragraph) exceeding r(a). Contradiction. The right inequality follows. Under suitable conditions we may use direct continuations of an element (a, P ) to obtain a function f which is analytic on a region Q containing the domain of P properly and which has the property that P is the restriction off to the domain of P.? In any case we have

u

Theorem 6.1: Pb is a function, analytic on its domain, the union being taken over b E A[a;r(a)]. PROOF: We put aside the obvious case where P is entire. It suffices to show that Pb(Z) = P,(z) when z is in the intersection of the domains of Pb and P, . We take advantage of the convexity property of circular disks. We note that A = A [ b ; r(b)] n A [ c ; r(c)]

is open and convex. Let d E A . Since Ib - cl I Ib - dl

+ Id - cI < r(b) + r(c),

we see that U =

+

r(c) b r(b) c E A n A[a; r(a)]. r(b) + r(c)

Now P and Pb agree in some neighborhood of u. The same is true of P and P, and consequently the same is true OfPband P , .Consequently, Pb(z)= P&), z E A . The theorem follows.

7 For example, a = 0,P(z) = (1 + z)-',

XZZ. Topics in the Theory of Power Series

288

Theorem 6.1 leads to the following important conclusion: there exists w E C[a;r(a)]such that r(z) I Iz - wI,

z E A [ a ; r(a)].

If r(a) < + co, (6.2)

Continuing with the case where r(a) < +a, we say that w E C[a;r(a)] is a singularity of P provided that there exists no f analytic at each point of A [ a ; r(a)]u {w} whose restriction to A[a; r(a)]is P . It is immediate that if w is a singularity of P,then (6.2) holds relative to w. The converse is true. For if w is not a singularity of P,and (w,Q) is the element such that Q ( z ) = P(z) for z in the domain of P near w and A(w; p ) is in the domain of Q, then r(b) 2 p - Ib -

WI

> Ib - W I

for b near w. The inequality (6.2) is thereby violated.

EXERCISE

xz

6.2. Suppose that (at): N + R + and that a t z k has radius of convergence p, 0 r ( l z l ) = p - IzI. Show that this situation may occur even when the power series converges uniformly on A(0; p) by considering (k l ) - z z k .

+

x: +

Continuation along apath. Let y be a path in C with domain [a, /?I. A map n from [a, P] into the set of elements is termed a continuation along y provided that for each t E [a, /?I the first component of n(t) is y ( t ) and that for each such t for s E [a, sufficiently near t the element n(s) is a direct continuation of n(t). We say that n is a continuation from ( a , P ) to (b, Q) provided that .(a) = (a, P) and n(P) = (b, Q). If n1 and n2 are continuations along y , then either n, = n2 or else nl(t)# nz(t)for all t E [a, /?I. This assertion is readily established with the aid of a connectedness argument. Indeed, the t for which nl(r) = n,(t) clearly constitute an open set in the relative topologyof [a, PI. However if nl(t)# n,(t), the second components of nl(t)and n,(t) are different (though, to be sure, they may very well take the same value at y(t)) so that for s near t either y(s) = y ( t ) and hence q ( s ) # n,(s), or else y(s) # y ( t ) and hence nl(s)# n2(s)since the values of their second components at y(s) are different. Since-the set of t for which nl (t )# nZ(t)is also open in the sense of the relative topology of [a, /?I and [a, /?I is connected, the asserted property of continuations along y follows. It is obvious now that given y with initial point a and an element (a, P) there is at most one continuation n along y satisfying n(a) = (a, P).

6. Analyrie Continuation

289

EXERCISES 6.3. Let n be a continuation along a path y. Let A[y(t);p(t)] be the domain of the second component of n(t).Show that p is either the constant co or else p(t) < to, a It I /3, and that in the second case for each such t we have

+

+

IAd -&)I lY(4 r(t)l for s E [a, /3] sufficiently near t . 6.4. Given an element (a, P) and a path y in the domain of P.Show that there is a unique continuation n along y satisfying 44 = Ma), Pvca,). Let n be a continuation along a path y having domain [a,/3] and let v denote a continuous map of [ a l ,PI] into [a, 81, - ai < aI< 8, < co. Show that n y is a continuation along y v. 6.5.

+

0

0

It is easily verified that continuation along paths induces an equivalence relation in the set of elements. We say that (6, Q ) is “equivalent” to (a, P ) provided that there exists a continuation n from (a,P ) to (b, Q). It is routine to verify that the relation so specified among the elements is reflexive (n( t )= (a, P),0 If < l), symmetric (consider n(1 - t ) ) , and transitive (n(r)= n,(2t), 0 It s +, n(r) = n2(2t - I), 4 < t 5 I , where n, is a continuation from (a, P ) to (b, Q ) and n2 is a continuation from (b, Q ) to (c, R)). We are using here the standard domain [0, I ] for continuations. The resulting equivalence classes will be termed Weierstrass continuation classes or more simply Weierstrass classes. We shall see in Ch. XVIII that Weierstrass classes may be endowed with a Riemann surface structure. It is easily comprehended that Weierstrass classes lack a desirable maximality-for example, poles and the point at co are not taken into regard. Just how far short of maximality the Weierstrass classes fall will be considered in $2, Ch. XVIII. It will be seen that the “ oversights have an isolated character. Given a Weierstrass class W we say that a path y in C with domain, say [a, PI, is a path of singularity for W provided that there exists n: [a, B[ 4 W such that n 1 [a, 61 is a continuation along y 1 [a, 61 for each 6 E 1 4 B[, but there does not exist a continuation n, along y satisfying n,(a) = .(a). ”

EXERCISES 6.6. Using the notation p of Ex. 6.3, show that y is a path of singularity for W if and only if there exists T: [a, fl[ +W such that n I [a, 61 is a continuation along y ,.[I 61 for each 8 E ] a , /3[ and lim, p - 0. 6.7. Show that if (0, P)is an element and the domain of Pis A(0; r ) , r < co,then & is a path of singularity for the Weierstrass class containing (0, P)for at least one w E C(0; r ) .

+

290


6.8. Let f be analytic on a region A c C. Show that the set of elements (a, P), a E A, satisfyingP"(z)= f ( z ) , z near a, belong to a Weierstrass class and that, in fact, if y is a path in A, then (y(t), P y ( r )is ) a continuation along y.

Monodromy theorem. This theorem is an important instrument for constructing analytic functions associated with a given Weierstrass class. The topological background will have a familiar ring from the developments of Ch. V. We consider a homotopically simply connected region A (r C and an element (a, P ) where a E A. It is assumed that for all paths y in A with initial point a and domain [0, 11 there exists a continuation ny along y satisfying n,(O) = (a,P ) . The monodromy theorem may be stated as follows.

Theorem 6.2: Under the above hypotheses there exists a unique function f analytic on A such that f(z) = P(z)for z near a. REMARK: It is to be emphasized that it would be a flagrant error to suppose that the hypothesis on the paths y might be replaced by the weaker one that for each b E A there exists a path y in A with initial point a and terminal point b along which there exists a continuation n satisfying n(0j = (a,P) (cf. Ex. 6.12 below of this section). PROOFOF THE MONODROMY THEOREM. We show that if y and r are paths in A having initial point a and a common terminal point, then ny(l) = n,-(I).? Granting that this is so, we propose for f the set o f ordered pairs

(Y(

P Y M1)I),

(6.3)

where P yis the second component of ny(l), all admitted y being taken into account. Clearly f is a function with domain A. Thatfis analytic on A may be seen by noting that for given y we have for z in some disk centered at y(1): f ( z ) = Py(z).This may be verified by using the path r satisfying

r(t)= y ( W , O I t I f - , r(t)= ( 2 t - l ) +~ (2 - 2 t ) ~ ( l ) , f- -= t I 1 . The uniqueness o f f is immediate. Let us turn to the proof of the equality n,(1)

= Xr(1).

We note, as was remarked by Koebe, that the domain of P yis of the form A[y( 1); r ] where r 2 dist({y(l)}, C - A) since Py admits continuation along all y(1) w where Iw - y(l)l is less than dist({y(l)}, C - A). For this reason we put

t This explains the terminology

"

monodromy " = "one course."

6. Anubtic Contimution

291

aside the case where A = C since P is then entire. To continue, we introduce a continuous map cp of [o, 112 into A satisfying: q ( s , 0) = y(s), q ( s , 1) = r(s), 0 I s 5 1; cp(0, t ) = a, cp(1, t ) = b, 0 I t I 1. Let m denote the distance from the image of cp to C - A . We take n, a positive integer, so that o(n-',/2) < m where w is the modulus of continuity of cp. For each k E (0, n) we define ejk recursively by: eok = (a, P),e j + ' , k is the direct continuation of e j k which has first component cp[n-'(j + l), n - ' k ] . Because of the observation made at the beginning of this paragraph the ejkare well defined. We next note that for each j € (0, n) it is the case that ej,k+lis a direct continuation of e j k , k = 0, . . . , n - 1. This is trivial f o r j = 0. Suppose that the assertion holds for a givenj E (0, n - 1). Because of the choice of n the points cp[n-'j, n-'(k l)] and cp[n-'(j + l), n-'(k + l)] are in the disk of radius m having center cp(n-'j, n - l k ) . Hence ej+l,k+lis a direct continuation of e j k . Since ej+',k is also a direct continuation of ejk and the segment

+

lies in the disk with radius m and center cp(n-'j,n-'k), we conclude that e j + l , k +isl a direct continuation of e j + ' , k , k = 0, . .. ,n - 1 (cf. Ex. 6.8, this section). Our assertion follows. On specializing j to be n, we conclude en0

= en".

To complete the proof we note that thanks to the condition on m and the property remarked at the beginning of the preceding paragraph we have n,(n-'k) = eko, nr(n-'k) = e k , , k = 0, . .. , n, since z y ( t )is a direct continuation of n,(n-'k) and nr(t)is a direct continuation of zr(n-'k), k I nt 5 k l , for k = 0, ... ,n - 1. It follows that z,( 1) = zr( I), and the proof is completed. In the exercises below we explore some aspects of the significance of analytic continuation for complex function theory.

+

EXERCISES 6.9. Let 7rI and 7r2 denote continuations along a path y (domain [0, 11) with initial point a. Let ~ ~ (= 0(a, ) P) and ~ ~ (=0(a, ) Q). We shall use the PY notation of the monodromy theorem. Let S denote a primitive of P. Show that there exists a con) (y(l), (PY)').Prove the tinuation 7r along y satisfying n(0) (a, P') and that ~ ( 1 = corresponding result for (a, S). Treat the analogous question for (a, T ) where T(z) P ( z ) Q(z),z near a, as well as the corresponding question for products and 1

+

quotients. 6.10. Let F be holomorphic on an open set 0 c C"+ where n E N. Suppose that 7r is a continuation along a path y with domain [0, 11. Let Pt denote the second component of 7r(t). We assume that

XII. Topics in the Theory ofpower Series

292

0I t I 1, and that F[z, Po(z), Po'(z), for z near y(0). Show that F [ z ,Pl(Z), Pl'(Z),

. .. ,P6""z)l = 0, .. . ,Py"(z)] = 0

for z near y(1) (permanence principle for differential equations). 6.11. Let f be entire and not constant. For each a E C such that f'(a) # 0 let ( f ( a ) ,P,,) denote the unique element with first componentf(a) satisfyingP.[f(z)] = z, z near a. Show that the set of these elements is a Weierstrass class. 6.12. Let f = z(z - l)z, z E C.Let (0,P) denote the unique element with first component 0 satisfying f[P(z)] = z, z small. Show that if A(0; r) is the domain of P, then r < co, and that for each z E C there exists a continuation m along a path with initial point 0 and terminal point z which satisfies m(0) = (0, P). Hint: The roots off'(z) = 0 are Q and 1. Given a E C there exists a path y with initial point 0 not passing through Q or 1 such that f[y(l)] = a. Show that there exists a continuation m along f 0 y satisfying m(0) = (0, P). In fact n(t) is the unique element (fo y(t), P,)for which P , [ f o y(t)l = y ( t ) and f [ P t ( z ) ]= z, z near f o y ( t ) . This shows that the misstated hypothesis of the monodromy theorem does not assure the conclusion. 6.13. Show that the set of all elements (a,A) where X is an analytic logarithm of the identity map on A(a; lal), a E C - {0},is a Weierstrass class. Show that a corresponding statement holds for the set of elements (a, P) where P is an analytic ath power of the identity map on A(a; lal), a E C - (0).Here a is a fixed member of

+

C-N. 6.14. Let O l r < s l +co. Let ~ E CLet . A betheannulus { r < Iz-al <s}. Let b E A and suppose that (b, P) is an element with the property that for each path y (with domain [0, 11) in A with initial point b there exists a continuation my along y satisfying m,(O) = (b, P). Let c be a logarithm of b - a. Let (c, Q) be an element

+

satisfying Q(z) = P ( a e') for z near c. Show that for each path r (with domain [0, 11) in {log r < Re z O}; lim, I m f = 0, - 1 < a < 1. The Schwarz reflexion principle states: Let u be continuous on A = A(0; 1) n {Im z 20}, be harmonic in the interior of A , and satisfy u(x) = 0, - 1 < x < 1. Then there exists a unique function h harmonic on A(0; 1) satisfying u = h I A . The uniqueness is immediate (cf. Ex. 3.3, Ch. XI). To establish existence, let 0 < r < 1 and let

's'

u(z) = 271

U(reie)k(0, z / r ) do,

IzI < r,

-R

-=

where U(reio)= u(reie), 0 5 0 5 71, and U(reie)= - u ( r e - 3 , --II I 0 0. Now u(5) = -v(z) as we see on examining the Poisson integral and hence u(z) = u(z), IzI 5 r, Im z 2 0. We see that the union of the u is the desired h. In the function-theoretic situation for which it was indicated that the Schwarz reflexion principle was applicable, we see that there exists h harmonic on A(0; I ) such that h(z) = Imf(z) on the domain off. Let g be analytic on A(0; 1) and satisfy h = Im g. Then for some real constant c the function g + c has restrictionf on the domain off. Carleman's extension principle states : Let f be analytic on A = A(0; 1) n {Im z > 0) and let g be analytic on B = A(0; 1) n {Im z < 0). Suppose that lim[f(z) - g(5)] = 0,

(3.1)

a

-1 < a

< 1. Then there exists a unique function h analytic on A(0; 1) such that = h I B.

f = h I A and g

PROOF: i l f ( z ) - s(z)]and f ( z ) +g(z)are analytic on A and the imaginary part of each has limit 0 at each point of 1- 1, l[.We conclude that

f ( z ) - g(z)= F(z),

f(z)

+ z)= G(z),

where F and G are analytic on A(0; 1). Setting h(z) =

l(z) =

F(z)

+ G(z) 2

'

- G(5) - F(5) 2

'

z E A,

XIII. Harmonic and Subharmonic Functions

304

Izl c 1, we havef= h 1 A, g = I I B. By (3.1) we have h(a) = l(a), - 1 < a < 1. Hence h = 1. The uniqueness assertion is obvious. Carleman’s principle follows. It should be remarked that Carleman also gave more refined versions of his principle (cf. [25]).

EXERCISES 3.1. Let fi R C. Show that if (9. h) satisfies g analytic on {Im z > 0}, h analytic on {Im z < 0}, f ( u ) = lim,[g(z) h(i)] for all u E R, then g and h are determined additively up to the same entire function; that is, if (gl, hl) also satisfies the stated conditions with g, replacing g and hl replacing h, there exists 9) entire such that gl(z) =g(z) ~ ( z ) ,Im z > 0, and hl(z) = h(z) - ~ ( z ) Im , z < 0. If g and h are --f

+

+

considered which satisfy lim z-’ g(z) = lim z-I h(z) = 0, m

m

then they are determined additively up to a constant. 3.2. Show that iff: R C is continuous on R and bounded, then f admits a representation of the type given in the second sentence of the previous exercise. --f

4. SUBHARMONIC FUNCTIONS The theory of subharmonic functions may be approached at different levels of generality. We shall be content to proceed under a hypothesis of continuity quite adequate for the applications to be considered. Some preliminaries. Given X # 8 suppose that fk: X + R, k = 1, . . . ,n (where n is a positive integer). We define max{fi,. . . ,f,},the maximum of fi, .. . ,f,, as the function with domain X assigning to x the value max{fi(x), . . .,f,(x)}. In spite of the similarity of notation the sense will be clear from context. Givenf: X + fi and g: X + R we definef + g and f - g, as would be expected, by taking f g as the functionf(x) g(x) with domain X and f - g as f (-g).t We shall also want a simple definition for integrals of For functions which are continuous maps of a segment [a, b] into R u { - a}. such a function f by

+

+

+

Ja>dx

we shall understand the limit of the sequence the nth term of which is the It is to be observed that our earlier conventions did not cover the situation at hand.

4. Subharmonic Functions

305

integral on [a, b] of the maximum off and the constant -n, n E N.When f is finite-valued the integral as just given coincides with the definition given earlier. We consider 0 open c C. We say that a continuous map u of 0 into R u { - co} is subharmonic on 0 (more briefly, subharmonic) provided that for each a E 0 we have u(a) I -

i n Jo2n

u(a

+ re") ne

(4.1)

for sufficiently small positive r (mean-value property). Our prior experience with the Gauss mean-value property of harmonic functions should make it apparent that the apparatus of the theory of harmonic functions can be brought to bear on the theory of subharmonic functions. Since the modulus of a function analytic on 0 satisfies (4.1) as does the norm of a vector-valued function analytic on 0, we may anticipate the utility of the theory of subharmonic functions for complex function theory. We start our study of subharmonic functions with the important principle of the harmonic majorant: Let u be subharmonic on 0 and let h be harmonic on 0 and satisfy:

lim sup(u - h) I 0 W

for each

MIE fr

0 where fr is taken in the sense of u(z)

I h(z),

e. Then

z E 0.

Here " lim sup" is construed in the sense of the relative topology of 0 u {w]. We note that this result subsumes the boundary maximum principles for harmonic functions and analytic functions. Thus iffis a continuous map of 0 (in the sense of 2') into C which is analytic in 0, then max5 If I is the maximum of ( I f 1) 1 fr 0. To see that this assertion follows from the principle of the harmonic majorant it suffices to take u =(lfl) 10 and h the constant whose value is the maximum of ( I f l ) 1 fr 0. The same observation holds for harmonic functions. Let us now prove the principle of the harmonic majorant. Let u = u - h. If the assertion were false, there would exist a E 0 such that u(a) = maxO u > 0. Let R denote the component of 0 containing a. Since u is subharmonic, as we readily verify (considering separately the cases ~ ( z=) - co and u(z) > - co), we see that as a consequence of the continuity of u and the mean-value property both { z E R, u(z) < o(a)} and {z E R, u(z) = u(a)} are open. Hence u(z) = zl(a), z E R. But fr R is not empty and is contained in fr 0. Applying (4.2) at MIE fr R we obtain a contradiction. The principle follows.


306

EXERCISE

+

Show that if u and v are subharmonic on 0, then so are u v and max{u, u}. [Here addition is pointwise; u + u is the function u(z) + v(z) with domain 0.1

4.1.

Poisson modification. We continue with u subharmonic on 0 and consider A(a; r ) c 0 , O < r c + co. We define u with domain 0 by z

u(z) = u(z),

E0

- A(u ; r),

u,(u

+ re") k

(8

-

Iz-al < r,

do,

where u, is the maximum of u and the constant -n. Using the principle of the harmonic majorant and the Harnack convergence principle (Ex. 1.5, this chapter) we see that u(z) 2 u(z), Iz - a1 < r, and that u is harmonic in A(a; r ) when u takes a value > - co at some point of A(u; r), and otherwise v takes the value - 00 at each point o f A(a; r). We now show that v is subharmonic. It suffices to test the continuity of u at points of C(a; r), for the continuity and the mean-value property at other points are assured and the mean-value property of v at points of C(a;r ) will follow from the fact that v(z) 2 u(z), z E 0, with equality holding on C(a;r ) . Now

u(z) Ilim inf u I lim sup u I u,(z), I

z E C(u; r),

z

and the asserted continuity follows by the arbitrariness of n. We term v the Poisson modijcation of u associated with A(u; r).

EXERCISE 4.2. u

Show that if int{u(z) = -a}# 0, where u is subharmonic on a region A , then co. Hint: Standard connectedness argument.

is the constant

~

Perron family. Let A be a region in C.A family CD of functions subharmonic on A is said to be a Perron family provided that the following two conditions are fulfilled :(1) I f u, v E (D, then max{u, v} E CD. (2)If a E A , then for each r , 0 < r < + co,such that A(a; r ) c A , for each u E CD the Poisson modification of u associated with A(a; r ) belongs to (D. Let M ( z ) = sup{u(z): u E (D},

We prove the following theorem concerning 0.

z

E A.


Theorem 4.1: One of the following holds: M is ;he constant constant + 00, M is harmonic on A .

307

- co, M

is the

PROOF: Let a E A. If M ( a ) = - 00, then either 0 = 0 or Q, contains only the constant - co. For if u E 0 and u is not the constant - co,some Poisson modification of u takes a finite value at a so that M(a) > - co. We see that if M(a) = -a, then M is the constant - co. Suppose that M(a) = + 00. Let A@; r) c A , 0 < r < +a.Let (u,,): N +0 satisfy limn-rmu,,(a) = +a,all u,,(a) being > - co. Let u,, = max{u, , . . . , u,,} and let w,be the Poisson modification of u, associated with A(a; r). Then w,, E 0,w,,is harmonic in A, w,,(z) < w,,+,(z) for Z E A , n E N , and w,,(a) = +co. Hence by the Harnack convergence principle (Ex.1.5, this chapter) M ( z ) = + co,z E A(a; r). Consequently { M ( z )= + co} is open. If M(a) E R, given b E A(a; r), we take (u,,): N 40 satisfying limn+mu,,(b)= M(b) and construct u,, and w,, as before. w,,(b) = M(b), and consequently, since w,,(a) I M(a), by the Then Harnack convergence principle M(b) < + co. Hence { M ( z )E R} is open. We conclude that at least one of the following is true : M is the constant - 00, M is the constant + co, M takes values in R only. Of course, the connectedness of A is essential to the argument. Suppose now that M takes values in R only. We continue with A(a; r ) as u,,(a) = M(a), above, b E A(a; r). Let (u,,), (U,,):N +0 be such that U,,(b)= M(b). Let u, = max{u,, . . . , u,,}, let V,, = max{u,, ..., u,,, U,,. .., U,,},

let w,,denote the Poisson modification of u,, associated with A(a; r), and let W,, denote the Poisson modification of V,, associated with A(a; r). Now (w,,) tends pointwise on A to a function w and (W,,) tends pointwise on A to a function W . The functions w and W are harmonic in A(a; r ) and satisfy w(z) I W(z), z E A(a; r ) and w(a) = W(a) = M(a). Hence w(z) = W(z), z E A(a; r), by the maximum principle for harmonic functions. Since M(b) = W(b),M(b) = w(b), and we see that M is harmonic in A(a; r). By the arbitrariness of a. M is harmonic on A. Theorem 4.1 follows.

EXERCISE Show that Theorem 4.1 remains valid when “for each r” is replaced by “for some r in condition (2) for a Perron family. The more restrictive definition is however useful.

4.3.

”

An application. We shall apply Theorem 4.1 later appropriately recast for Riemann surfaces to solve a simple Dirichlet problem incidental to the study

308


of the problem of Riemann and Klein. For the present we show how the theorem just proved and the solution of the Dirichlet problem for an annulus permit us to treat the Dirichlet problem for a bounded region A in C for which fr A = U C(&; rk) 1 sksn

where C(Uj;

rj) n C(ak; rk) = 8,

j # k,

and all rk are positive. Suppose that f k is a continuous map of C(ak; rk) into R, k = 1, .. ., n. We seek h continuous on A, harmonic in A and satisfying h I c(ak; r k ) = fk , k = 1, . . . ,n (Dirichlet problem for A). To that end we introduce for each k an annulus A, c A where fr A, consists of C(ak; r k ) and C(ak;sk) for some positive s, and J j n A, = 8,J # k . We let m denote a real number less than all the values taken by thefk and the ( -fk).Let u be defined on A by the requirements that u I be the solution of the Dirichlet problem for A, reducing tofk on C(ak; T k ) and to the constant m on c(ak;sk), k = 1, . . .,n, and that u take the value m at each point of A - Uls;k$n A k . Let w be similarly defined with (-fk) replacing fk . Both u I A and w 1 A are subharmonic. Applying the principle of the harmonic majorant to (u w) I A with comparison function the constant 0 we see that v(z) 5 - w(z), z E A . Now let @ denote the family of functions u subharmonic on A satisfying:

+

v(z) 5 u(2) I- w(z),

(4.3) Clearly u I A E 0. Now @ is a Perron family. Indeed, it is immediate that if u l , u2 E 0,then max{u,, u 2 } E 0,as we see by reference to (4.3). If u E 0 and U is a Poisson modification of u, then (4.3) with U replacing u holds. The inequality on the left is obvious and the principle of the harmonic majorant applied to U + w I A with comparison function the constant 0 shows that the inequality on the right holds. By Theorem 4.1 the function M is harmonic on A . Since M replacing u satisfies (4.3), we see that M u Ulsksnfk is the desired function h. z E A.

EXERCISES These two exercises give further experience with some elementary aspects of the notion of a subharmonic function. 4.4. Suppose that u E C"[A(O; r ) ] and is real-valued. Let


309

0 Is < r. Show that p‘ is continuous on [0, r [ . Show by integrating Au(seie)with respect to 0 on [0,27r] and employing standard theorems of the differential calculus that if Au(z) = 4u,;(z) 2 0, (zI < r , then p’(s) 2 0,O I s < r, and if Au(z) > 0, IzI < r , then p’(s) > 0, 0 < s < r . Conclude that if now u E C ” ( 0 )and is real-valued, then u is subharmonic if and only if Au(z) 2 0, z E 0. Here 0 is an open subset of C. The C” notion in the present setting is to be appropriately formulated. 4.5. Let -- 00 Ia < b I co and let f b e convex on ]a, b[ and satisfy f ( x ) f(x)l, namely, if

(xk, Yk)

E

E, pk 2 0, k

= 1,

. . .,n, and P1 pk = 1, then

BB Chapter XZV @@

COMPLEMENTS T O THE CAUCHYTHEORY

PREFATORY REMARKS This chapter is a little Cauchy anthology, the items of which have been chosen to show the broad scope of the theory. We start with Mordell's very ingenious treatment of the celebrated theorem on Gaussian sums (which contains as a by-product the evaluation of the Gauss integral 2 exp( -xz) dx). The second to fourth topics are: the problem of characterizing the divisor and the residues of a doubly periodic meromorphic function ; the Weierstrass preparation theorem, which sheds light on the local behavior of { ( z , w):f(z,w) = 0) wherefis holomorphic at a point (a, b) E C2; the refinement of the Runge theorem permitting us to characterize open subsets 0 of C having the property that functions analytic on 0 may be approximated uniformly in 0 by polynomials. In the proof to be given of the Weierstrass preparation theorem the notion of the logarithmic derivative will be central. We consider also two topics important for applications. One is the boundary behavior of integrals of Cauchy type; however we consider the question under severely restricted hypotheses. Our object is not to develop a refined study of the question, but only to indicate its function-theoretic aspect in a simple setting. The other topic is the adaptation of the Cauchy integral theorem to Weierstrass classes. We shall apply the results obtained to an example arising in the theory of ordinary linear differential equations of the second order. The treatment of the Cauchy theorem in this framework is pragmatic in intent and execution.

s'

310

1. Gaussian Sums

311

1. GAUSSIAN SUMS

In this and the following section we shall have occasion to calculate integrals along the sensedperiphery of a parallelogram. Let us be precise. We recall that given a, b, c E C,Im(hc) # 0, the parallelogram determined by (a, b, c) is the image of [0, 1]* with respect to q(s, t ) = a + bs + ct. By the integral off (suitably given) along the (a, 6, c)-sensedt periphery of this parallelogram is meant the sum of the integrals off along A

(a)(a

+ b),

(a

+ b)(a + b + c),

>

(a

+ b + c)(a + c),

-

(a

+ c)(a).

It will be important to know the order of q(s, t), 0 < s, t < 1, with respect to the closed curve formed by taking the above four paths in succession and making a periodic extension. It is clear that the order is independent of a. We take a = 0. Since the order is unaltered when b-'q replaces cp and (1, b-'c) replaces (b, c), we are led to the case of the parallelogram determined by (0, 1, b-'c). If im(6c) > 0, the segment [b-'c, i] lies in {Im z > 0} and we see, making a continuous deformation, that the order in question is that associated with the datum (0, 1 , i ) , which is 1. If Im(6c) < 0, in the same way we arrive at the order associated with datum (0, 1, - i), that is, - 1. Mordell's treatment of Gaussian sums. Let n be a positive integer. By the Gaussian sum Snis meant n- 1

Gauss gave a simple formula for Sn, many proofs of which are known. Among these the following one of L. J. Mordell [143] is remarkable for its elegance. It contains incidentally-gratis-the evaluation of the Gauss integral. Given a E R,by L(a;f ) we shall understand lim +m

f f(a + teni/4>e

dt,

-r

assuming, of course, f admissible. The interpretation in terms of line integrals is obvious. Mordell's proof starts with the introduction of the function f meromorphic on C which is simply n- 1

exp[2ni

t We shall use

"

+]

sensed " in place of " (a, b, c)-sensed."

XZV. Complementsto the Cauchy Theory

312

divided by eZnir- 1. An immediate calculation shows that j ( z + 1) - j ( z ) = eZniz2/n(eZniz

+ 11,

(1.1)

the standard conventions prevailing. We introduce Z(r), the integral off along the sensed periphery of the parallelogram determined by (-+ - reXiI4,I, 2reni/4),

r positive, and verify with the aid of the residue theorem and the observation concerning order made in the first paragraph of this section that Z(r) = S, . On noting the existence of L(+;f)and L( -+; f), and observing that lim +m

1”’

+ reni/4)dt = 0,

f ( t - renil4)d t = lim

-1/2

+m

we conclude that Sn = L(3;f ) - L(-+; f>.

By (1.1) the right side of (1.2) is equal to

i

L [ - ;e x p E z 2 ) ]

+ (- i)” L( -

;e x p E ( z

+

:)’I}.

(1.3)

We introduce J(r), the integral of exp(2niz2/n) along the sensed periphery of the parallelogram determined by ( a - reni/4,b, 2reLiI4),a E R,band r positive. By the Cauchy integral theorem J ( r ) = 0. From

+m

we conclude that L[a;exp(2niz2/n)] is independent of a. From this observation applied to (1.3) using (1.2) we obtain S , = [l

+ (-i)”]

L

=[1+(-i)”]By (1.4) and the fact that S, = 1 we have S,,

s, =-s1 = and

1+(-i)” 1-i

J.,

2. Doublj Periodic Functions

313

2. DOUBLY PERIODIC FUNCTIONS Let y, 6 E C and Im(jj6) # 0. Let R denote the additive subgroup of C generated by {y, 6). The object of this section is to penetrate further into the study of the functions f meromorphic on C which satisfy f(z w) =f(z), o E R. Such functions are termed the meromorphic doubly periodic functions associated wifh R. Their totality D(R) constitutes a field. Let a E C. It is immediate from Ex. 3.6, Ch. X, that z E C admits a unique representation of the form

+

a

+ ys + 62 + w

where 0 Is, t < 1, o E Q. We term the parallelogram determined by (a, y, 6) a fundamental parallelogram of R. It follows from the observation just made and the theorem of Liouville that iff E D(R) is entire, thenfis constant. In the exercises that follow we shall be concerned with the properties of d,(z) and res(z;f) forfE D(R) not the constant 0.

EXERCISES

+

+

For such f we have: a,(z w ) = a,(z), res(z w ; f ) = res(z;f). Furthermore neither {a,(z) # O } nor {res(z;f) # 0} clusters at a point of C. 2.2. By choosing a so that f(z) # 0, 03 for all z belonging to the frontier of the parallelogram P determined by (a,y, 6 ) and integrating f(z), f’(z)lf(z), zf’(z)/f(z) along the sensed periphery of P conclude that given any a E C and {a, p } generating R, with Q denoting 2.1.

{a + as

we have : (a) LEQ res(z;n (b)

+ pt: 0

IS,

t < I},

= 0,

~ A z= ) 0, (c) L E Q z E R. Apply (b) to f - c where f is not constant a_nd c is a complex constant to conclude n(z;f) is independent of k E C. that C,IQ(r)=k 2.3. Show that condition (a) and the results of Ex. 2.1 permit us to characterize the distribution of the residues o f f € D(R). More precisely, let p : C C satisfy (1) {p(z)# O } clusters at no point of C, (2) p(z w ) = p(z), w E R, (3) CteQp(z) = 0. Then there exists f r z D(R) such that res(z;f) = p(z), z E C. Hint: Let 5 denote the Weierstrass zeta function associated with R (cf. Ex. 6.10, Ch. VIII) and consider Czca

+

--f

Show with the aid of the Weierstrass 5-function associated with R and its successive derivatives that there existsfe D(R) with assigned principal parts Pb[(z- b ) - ’ ]

XIV. Complementsto the Cauchy Theory

314

where P b is a polynomial, b E Q,subject only to the conditions that Pb is not the zero constant for only a finite set of b and that

Show that if a: C + Z satisfies (1) {a(z) #0} clusters at no point of C, (2) = a(z), w E n,(3)Cz.Q a(z) = 0, (4)ZzsQ z a(z) E R, then there existsfc D(R) satisfying 8, = a. Further such f are determined up to a nonzero multiplicative constant. Hint: With u the Weierstrass sigma function associated with a, C E Q, d =C I E Q z a(z), consider

2.4. a(z

+w)

f ( z ) = [u(z - c)]'(')-' u(z - c

+d)

[u(z - b)lacb'

b=Q-(C)

and verify that fso specified has the stated properties. 2.5. Let p denote the Weierstrass pe-function associated with R. Suppose that a is taken as 0 in the definition of Q. Show that the zeros of p' in Q are precisely u/2, p/2, (u p)/2 and that each has multiplicity 1. Hint: Use the fact that p is even. 2.6. Suppose that A and B are positive and that ct = A , 9, = iB. Show that p takes real values on fr[O, A / 2 ; 0, B/2] - {O} and that p maps int[O, A / 2 ; 0, B/2] univalently onto {Im z < O}.

+

3. THE WEIERSTRASS PREPARATION THEOREM This theorem (also known by its German name " Vorbereitungssatz ") gives

a useful insight into the local behavior of the sets of constancy of a holomorphic function of several complex variables. We treat here the case of two complex variables.? We remark that the corresponding situation in the real C" theory is vastly more complicated. In the exposition that follows we assume that f is holomorphic on [A(O; r)I2 and not constant and thatf(0,O) = 0. It is clear that the local study of a function F holomorphic at (a, b) may be reduced to the normalized case by considering the translated function F ( a z, b w) - F(a, b). The Weierstrass preparation theorem asserts

+

+

Theorem 3.1: There exists p, 0 0, (2) ak is analytic in A(0; p ) and a,(O) = 0 for k = 0, . . . , n - 1, and ( 3 ) g is holomorphic in [A(O; p)]' and does not take the value 0. ( z , w) E

t For the general situation, cf. [18].

3. The WeierstrassPreparation Theorem

315

We see that the preparation theorem reduces the study of the set { f ( z , w) = 0} to the study of the zero sets of the first two factors and for these we have a fair amount of information.?

PROOF: We start with the representation m

bk analytic on A(0; s). Let m denote the minimum of the k for which bk is not the constant 0. The proof reduces to the case m = 0 and we continue under this assumption. If b o ( 0 ) # 0, the theorem follows at once. Suppose that bo(0)= 0 and let n denote the multiplicity of bo at 0. Let s, 0 -cs < r, be such that bo(w)# 0, 0 < I w I 5 s. Now take t, 0 t < r, such thatf(z, w) # 0 for )zI s t , IwI = s. For each allowed z the equationf(z, w) = 0 has n roots in A(0; s) counted by multiplicity. For each such z let

-=

{@k(z): 1 5 k 5 n } = { W : f ( Z ,

W)

= 0, IWl

< S}

where the number of k such that ak(z) is a given member o of the set on the right is precisely the multiplicity at o of the analytic function assigning to w the valuef(z, w). We proceed by defining P(z, w) as n-

1

is analytic at each point of A(0; s) - {cll(z), . . . ,a,(z)} and has removable singularities at the clk(z). When w E A(0; s) - {al(z), . . . ,a,(z)}, the value given in (3.2) is equal to

The function defined by (3.3) on A(0; t) x A(0; s) is holomorphic on this set. Now there exists q, 0 < q < s, such that f(z, w) # 0 for IzI < t, q < IwI < s.

t For the zero set of the second factor, cf. 53, Ch. XVIII. 1Cf. Theorem 1.2, Ch. VII.

316

XIV. Complements to the Cauchy Theory

Hence by(3.2) and the integral representation (3.3) we see that Pw(z,w)/P(z,w) is holomorphic in A(0; t) x {q < IwI < s}. We have for JzI< t the representation

where each c k is analytic on A(0; t), as we see using the observation of the preceding sentence and the formula

q 2 complex variables which is holomorphic at (0, .,0) E C" and satisfiesf ( 0 , . .. ,0) = 0 and

..

z.

small, m a positive integer, g analytic at 0, g(0) # 0. Construct an example of a C" real-valued function of two real variables

3.3.

4. Runge's Theorem on Polynomial Approximation

317

which has the property that some point of the zero set belongs to the frontier of the interior of the zero set.

4. RUNGE'S THEOREM ON POLYNOMIAL APPROXIMATION We return to the Runge theorem concerning approximation by rational functions (Theorem 8.1, Ch. V) and apply it to obtain the following theorem of Runge on the possibility of uniform polynomial approximation.

e

Theorem 4.1: Let 0 be an open subset of C.Then the connectedness of - 0 is a necessary and sujicient condition for 0 to have the property that given f analytic on 0, K compact c 0, and E a positive number, there exists a polynomial p such that

z

IfG) - P(Z)l < E ,

E

K.

PROOF: Let us first consider the necessity of the condition. Suppose contrary to the assertion that - 0 is not connected. Then there exist closed subsets A and B of which are nonempty, disjoint, and have union - 0. We suppose that 00 E B. If B = { co}, and a positive number r is so chosen that A c A(0; r), then ( z - a)-', where a is a given point of A , may be approximated uniformly on C(0;r ) by a polynomial thanks to the hypothesis of the necessity part of the theorem. Introducing a sequence of polynomials tending uniformly to ( z - a)-' on C(0; r ) , we see that it is uniformly Cauchy on C(0; r ) and hence by the boundary maximum principle for analytic functions it is uniformly Cauchy on A(0; r). The sequence tends uniformly on A(0; r ) to a function f continuous on A ( 0 ; r ) and analytic in A(0;r). Further f ( z ) = ( z - a ) - ' , lzl = r. Hence f( z ) ( z - a) - 1 = 0, IzI = r, and consequently f ( z ) ( z - a) = 1, IzI Ir, and this is obviously false as we see on setting z = a. The argument just given holds with slight modification when B contains points of C.In this case we introduce m = dist(A, B - {co}), which is a positive number, and take n a positive integer so that @ < mn. We let S denote the union of those squares [ n - ' j , n - ' ( j 1); n-'k, n-'(k l)], j , k E 2, which have points in common with A . Now S is a compact subset of C, S n B = 8, and fr S c 0. We now introduce (z - a)-' where a is a given point of A , and we repeat the above argument with fr S replacing C(0; r). The asserted necessity of the connectedness of - 0 follows. Suficiency. We base the proof on the Runge theorem concerning approximation by rational functions (Theorem 8.1, Ch. V) and the following lemma: Let K be a given compact subset of C. Let E denote the set of a E C - K such that ( z - a)- may be approximated uniformly on K by a polynomial. Then E is the unbounded component of C - K .

e,

e

c

+

+

e

318

XIV. Complementsto the Cuuchy Theory

To be exact, when we say that (z - a)-' may be approximated uniformly on K by a polynomial, we mean that for each positive E there exists a polynomial p such that l(z

- a)-' - p(z)l < E ,

z

EK

.

This language will be used generally without further gloss.

PROOFOF THE LEMMA: We put aside the trivial case where K is empty. If a E C - K and Ib - a1 < dist({a}, K), then using the Laurent expansion of (z - b)-' in the annulus { 16 - a1 < Iz - a1 < + co} we see that ( z - b)-' may be approximated uniformly on K by a polynomial in (z - a)-'. If a E E and P is a polynomial, then P[(z - a ) - ' ] may be approximated uniformly on K by a polynomial as we see on noting that (z - a)-k,k E N, shares with (z - a)-' the property that it may be approximated uniformly on K by a polynomial. Hence if a E E and Ib - a1 < dist({a}, K), then b E E. Consequently, E is open. Let us now show that E is closed in the sense of the relative topology of C - K. Indeed, if b E (C - K) n E, there exists a E E such that 2 Ib - a1 < dist({b}, K). We have Ib - a1 < dist({b}, K) - Ib - a1 I dist({a}, K).

Hence ( z - b)-' may be approximated uniformly on K by a polynomial. The asserted property of E follows. Clearly all points of C with large modulus belong to E. Hence E contains the unbounded component of C - K. Suppose that a E R, a bounded component of C - K, and that (z - a)-' admitted uniform approximation on K by a polynomial. Repeating the argument used in treating the first half of the present theorem we infer the existence off continuous on and analytic in R satisfying ( z - a ) f ( z ) = 1, z E fr R. The same contradiction is reached. Hence t2 n E = 0.We conclude that E is the unbounded component of C - K. The sufficiency proof is now readily completed on appeal to the earlier Rungetheorem(The0rem 8.1,Ch.V).It sufficestonotethat e-Oc Eu{co}, where now E is taken relative to a given compact subset K of 0, and that the finite poles of an approximant furnished by Theorem 8.1, Ch. V, lie in E. The remainder of the argument is routine.

EXERCISE Suppose that f ( z ) = 0, Theorem 4.1 is applicable.

4.1.

z

E

A( - 1 ; l), and f ( z ) = 1, z E A(1; 1). Show that

5. Integrals of Cauchy Type, Cauchy Principal Value

5. INTEGRALS OF CAUCHY TYPE, CAUCHY PRINCIPAL

319

VALUE

+

Let - 03 < a < b < 03 and suppose that g is a complex-valued member of C'[a,b].We are concerned with the limiting behavior of the integral of Cauchy tYPe

f (4=

jab -

dt

t-z

at points of ]a,b[. Suppose that c E ]a,b[. We consider

+

0 < E < min{c - a, b - c]. Setting g(t) = [g(t)- g(c)] g(c), we see that the limit in question is

b-c c-a

ja6h(r)dt + g(c) log -,

where h(t) = [g(r) - g(c)]/(t- c) when t # c and h(c) = g'(c). We denote the limit by P(c; g). We are concerned here with the notion of a Cauchy principal value o f an improper integral. Generally, if F is a continuous map of [a,b] - { c } into C-the continuity condition may be replaced by a weaker one-then by the Cauchy principal value of F ( x ) dx is meant

lim[ 0

dx

+I:+!

dx],

assumed existent. Suppose that g = G I [a, b] where G is analytic at each point of [a, b]. We shall show that limf I {Im z > 0} = P ( c ; g )

+ ni g(c).

(5.3)

C

Indeed, for E small and positive we have by the Cauchy integral theorem

f(z)

=

dt t-z

+

--x

+

G(c Ee") c Eeie - z

+

and we conclude that the desired limit is

XI V. Complements to the Cauchy Theory

320

and hence that (5.3) holds. Formula (5.3) is also valid when g satisfies the original C' hypothesis. For by the Weierstrass (polynomial) approximation theorem we may approximate g' uniformly on [a, b] by a polynomial. If p is a polynomial satisfying p(a) = g(a), we have

IP(c; g ) - P(c; p)I IM ( b - a) 1 + log - "),

(

C-a

(5.4)

where M = max{ Ig'(x) - p'(x)I :a I x I b}, as we see by reference to (5.2) and a routine estimate of P(c; g) in the case where g(a) = 0. Further setting q(t) = g(t) - p(t), we obtain from

the inequality

when Im z # 0, a I x = Re z I b. Using (5.4) and (5.5) we conclude that lim sup If(z) - P(c; g) - nz' g(c)l I2M(b - a) 1 C

+ n + log -3 , c-a

and hence that (5.3) holds under the C' hypothesis since p may be so chosen that M is arbitrarily small. It is to be observed that a polynomial approximant yielded an analytic approximant-a fact that put the Cauchy apparatus at our disposal. A corresponding formula for the case of the lower half-plane holds. It is limf 1 {Im z < 0 } = P ( c ; g ) - x i g(c). C

The above developments extend to vector-valued analytic functions.

EXERCISES 5.1. Given a, 6 as above and h positive. Show with the aid of the Cauchy integral formula for the closure of

[a?,bz; C z , dz1- [ah61;Cl, 41,

6. The Cauchy Integral Theoremfor Weierstrass Classes

321

where a2 < a , < b , < b, and c2 < c1 < d, < d 2 , that iff is analytic on A(h) = int[a

-

h, b f h ; -h, h] - [a, b],

then there exists a unique (fr , f e ) , wherefr is analytic on int[a - h, b is analytic on - [a, b] and satisfiesf(co) = 0, such that

+ h ; -h,

h] and

fe

+&>,

f(4 5.2.

z E A(h).

Suppose that g = G I [a, b] where G is analytic at each point of [a, 61. Let

and z, w E A(h), h small.

H(z,z) =G ( z ) ,

Then J: H ( z , t ) dt is analytic on A(h). Further b

jaH ( z , t ) dt

1

=-

27ri

p(h)

c(w) dw, w -z

z E A@),

where h(w) = Log[(w - a)/(w - b)] and P(h) is the positively sensed periphery of [a - h, b h ; -h, h]. Conclude that if J: H ( x , t ) dt = 0, a < x < b, then g is constant. Show that if q is analytic on A@), then with G = (q/h),we have

+

jab H ( z , t ) df

=q(z),

z

E

A@).

[Carleman, cf. [24, p. 1421.1 Relate this problem with the subject of Cauchy principal value.

6. THE CAUCHY INTEGRAL THEOREM FOR WEIERSTRASS CLASSES Given a Weierstrass class W we introduce two complex-valued functions having domain W: One is the center function c assigning a t o e = (a, P ) E W; the other is the valuefunction v assigning P(a) to e. The terminology is that of Saks and Zygmund (cf. [lOS]). Let us now consider cp: [O, 11’ -+ W having the following properties: (1) c 0 cp is continuous on [0, I]’, (2) cp(s, t ) is a direct continuation of q ( s o , t o ) for (s, t ) sufficiently near (so, to), (so, to) arbitrary in [0, 11’. It is easily concluded that there exists 6 positive such that cp(sl, t l ) and cp(s,, t z ) are direct continuations of one another whenever the distance between (sl,t l ) and (s2, t 2 ) is less than 6 . Suppose now that c o cp taken along the edges of [0, 112 defines a rectifiable

XIV. Complements to the Carchy Theory

322

path. More explicitly, we define C on [0,4] by C ( t ) = c O cp(4 01,

O I t I l ,

1 I t I 2 ,

=cocp(l,t-l),

l),

2 It I3,

cp(0,4 - t),

3 5 t I4,

= c 0 443 - t , =c

0

and suppose that Cis rectifiable. We define V having domain [0,4] similarly, u replacing c in the definition. We have a direct extension of the Cauchy theorem for a singular cell, namely, that under the stated assumptions j : V dC = 0. The proof is a direct paraphrase of that of Theorem 3.1, Ch. V. We take n a positive integer such that J2 < n6, reference being made to a positive number 6 allowed in the sense of the first paragraph of this section, and subdivide [0, 112 as earlier into n2 squares. Here we introduce in place of gjka primitive of the second component of cp(j/n, k/n),j , k = 0 , . . . , n - 1. The details then go as in the earlier proof. A similar result holds when cp satisfies ~ ( 0t), = cp(1, t), 0 5 t I1 , and . 0 It I1 , are rectifiable. Setting Cl(t) = c q(t, 0), C2(t)= c 0 ~ ( tl), Vl(t) = u 0 cp(t, 0) and V2(t)= u 0 cp(t, 1) we have 0

ri

Jo

ri

Vl d C , = J Vz d C 2 . 0

It is well to consider an example. The one below arises in the theory of linear differential equations of the second order. Essentially our task is one of interpretation. EXAMPLE:The integral

ezw(w2- 1)” dw,

FIG.14.1

where c1 E C - N and r is a rectifiable closed curve in C - { - 1, l } satisfying the order condition O(- 1; r) = - 1, O(1; r) = 1, as indicated in the figure. Since we are concerned with continuous determinations of the integrand, or

6. The Cauchy Integral Theorem for Weierstrass Classes

323

more strictly of exp[z r(t)] times an ath power of [r(t)I2 - 1, as is intended, there is no difficulty in definition up to a constant factor of the form e2'lak, k E N. For the present consideration the important point is the use of the Cauchy theory to establish independence of path results. Specifically,suppose that r is deformed into a rectifiable closed curve rl by $, a continuous map of R x [0, 11 into C - { - 1 , l}, satisfying $(s

+ 1, t ) = $(s, 0,

w, 0) = us),

(s,

0E R

x 10, 11;

w, 1) =

s E R.

+

There exists a continuous logarithm I of $' - 1. We have I(s 1, t ) = I(s, t ) , (s, t ) E R x [0, 11. Each continuous map f of R x [0, 11 x C into C satisfying f(s, t , z) = expCz $(s,

where g(s, t) is an ath power of [$(s,

t)l g(s, t )

r)I2 - 1 is of the form

elniakexp[z $(s, t ) + a l(s, t ) ] ,

(6.2)

where k is an integer independent of admitted (s, t , z). We observe that for fixed z there exists q (unique) mapping [0, 112 into a Weierstrass class which meets the conditions (1) and (2) of the first paragraph of this section as well as (3) c 0 q(s. t ) = $(s, t), (s, t ) E [0, l]', and (4) the second component of q(s, t) takes the value exp[zw + a &w)] at each point w of some neighborhood of $(s, t). Here 1 is an analytic logarithm of w2 - 1 with domain a small disk centered at $(s, t) which takes the value I(s, t) at $(s, t). The verification is routine. We see that u

0

q ( s , t ) = exp[z $(s, t )

+ a l(s, t ) ] .

By (6.1) we conclude that the integral we are considering is unaltered when r is replaced by rl,where it is understood that use is made of a fixed continuous determination (6.2) to specify the integrals involved. In our present example the integral yields an entire function as we see by fixing t = 0.

EXERCISES

+

+

6.1. Suppose that r(t)= 1 exrc4'- I ) , 0 5 t 2 4, = - 1 e"fcZ-4r) , i c t l f , and that I' is defined elsewhere on R so as to be periodic with period 1. Show that if Re u > - 1 the functions defined by the integral of the example using a "continuous determination of the integrand along J?" (that is,f(s, 0, z) of the text) admit a representation of the form

exp[zt

+ a log(1

- z')]

dt,

k E Z.

324

XIV. Complements to the Cauchy Theory

+

6.2. Let a E C. Let {w,(z), wz(z)} be the pair of roots of w z - 2zw 1 = 0 where z lies in an open disk D not containing either of the points 1, -1. We suppose that w1 and w Z are analytic. Let I? denote a rectifiable closed curve lying in c - 10,w l ( ~ ) W, ’ ( D ) } satisfying

o(o;r) = o[wl(z); r]= 0, O[wz(z),

r1= - 2 ,

as, for example, in the figure. Interpret the formally given

$,w=(w’

- 2zw

+ 1)-

I/’

dw.

Treat the questions of continuous determination of the integrand, independence of path, analyticity of the function defined by the integral. Investigate the nonvanishing of the integral when z is small and I’ is specified by taking the paths yl, y z , y 3 , y4 in succession as indicated in the figure and extending periodically. 6.3. Treat Jr [(z’ - l)(z - 3)11/’ dz where (a) I’ is the closed curve of Ex. 6.1, (b) r(t)= 2esnrr. In each case reduce the calculation of the integral to that of the integral of a real-valued continuous function with domain [- 1,1]. FIG.14.2

There is a wealth of material on integrals of the type considered in this section. For their further study the following references are useful: [13], [34], P71, [13Ol.

WI,

&>

Chapter X V @@

MOBIUS T R A N S F O R M A T I O N S

PREFATORY REMARKS

c,

The study of the univalent meromorphic functions with domain the Mobius transformations, was begun in Ex. 4.25, Ch. VI. These functions have long played an important part in the geometric theory of functions, Their study is a preliminary to the theory of automorphic groups and functions, a subject initiated by Klein and PoincarC and still provoking lively mathematical interest. Hence for the student wishing to pursue modern complex function theory further, especially in the two areas mentioned, the basic facts about Mobius transformations are an essential part of his equipment. We shall encounter these transformations in the study of the elliptic modular function (Ch. XVI). Our principal concern in the present chapter will be with the classification of Mobius transformations.

1. CLASSIFICATION OF MOBIUS TRANSFORMATIONS We have already noted (Ex. 4.25, Ch. VI) that if a and b are distinct points of 2', there exists a Mobius transformation f satisfying f (a) = 0,f ( b ) = co. Thus ( z - a)/(z- b) serves when a, b E C; ( z - b)-' when a = co;and z - a when b = CQ. To what extent f is determined is easy to see. If g is a Mobius transformation also satisfying the same conditions as f, g f - ' ( z ) = cz, c E C - (0}, and hence g = cf. The converse is obvious. It will be convenient to let Idenote the identity map of onto itself. 0

c

325

XV. Mobius Tramformations

326

An equivalence relation. Given Mobius transformations L and M , we say that M is equivalent to L provided that there exists a Mobius transformation S such that

M

=S

O L0 s-'.

(1 1) 9

It is immediate that a bona fide equivalence relation is defined thereby. To check this, we note (1) when L = M , I serves for S, (2) from (1.1) we have L = S - ' o M o ( S - ' ) - ' , (3) if N and Tare Mobius transformations satisfying N = T OM OT-', then N = ( T OS ) 0 L O( T OS ) - ' . Clearly, I is equivalent only to itself. Let M be a Mobius transformation, # I . We shall see that the number of jxedpoints of M (that is, the z E satisfying M ( z ) = z ) is either one or two. Suppose that M(c0) = 03, so that M(z) = az + b, a # 0. If a = 1, then b # 0 and 00 is the only fixed point. If a # 1, then the fixed points of M are co and (1 - a)-'b. Suppose that M ( w ) # 00. Then M - I has constant valence two and has 00 as a fixed point. The fixed points of M are the zeros of M - I and are at most two in number. We continue and consider the case where M has exactly one fixed point, say a. Let S denote a Mobius transformation satisfying S(a) = co.t Now S OM o S-' has just one fixed point, 00. Hence S M OS - ' ( z ) = z + 6, b # 0. Using b- [(bz) b] = z 1, we conclude: all Mobius transformations with just one fixed point are mutually equivalent. In particular, they are all equivalent to the translation through 1 (extended to in the obvious way). Mobius transformations with just one fixed point are termed parabolic. We conclude using the definition of equivalence that the set of parabolic transformations constitutes an equivalence class. There remains to consider the case where M has two distinct fixed points, say a and b. Let S denote a Mobius transformation satisfying S(a) = 0, S(b) = 03. Then S M 0 S-' = cl, c # 0, 1. When are two Mobius transformations of this special form equivalent? If d # 0, 1 and T is a Mobius transformation satisfying d l = To ( c l ) o T-', we see that T(0) and T(w) are the fixed points of dl. Hence either T = ul or T = u ( l / l )for some u # 0. If the first possibility occurs, d = c; if the second occurs, d = I/c. It is immediate that if d = c, or d = l/c, such a T exists. We conclude that d l is equivalent to c l if and only if

c

'

+

0

+

c

0

c

+ c-'

=d+d-'.

Given M with exactly two distinct fixed points, we introduce K ( M )as the common value of c + c- such that M is equivalent to cl. If N is also a Mobius transformation with exactly two distinct fixed points, then N is equivalent

'

t If a =a,we may take I ; if a f a , l/(z-

a).

1. Classification of Mobius Transformations

327

to M if and only if K ( N )= K ( M ) . If co is a fixed point of M , so that M ( z ) = cz d, c # 0, 1, then M i s equivalent to e l a n d hence K ( M )= c c-’. If 03 is not a fixed point of M , we see that S o M = cS and T M = c-’T, c # 0, 1, where S is as in the preceding paragraph with a and b the fixed points of M and T = 1/S. From the first equality we conclude M’(a) = c and from the second M’(b) = c-’. Hence K ( M )= M‘(a) M’(b). The Mobius transformations with exactly two distinct fixed points are classified by the invariant K ( M ) .The set of values taken by K ( M )consists precisely of C - (2). We term M elliptic provided that K ( M )E [ - 2,2[, hyperbolic provided that K ( M )E 12, 03 [, and loxodromic otherwise. We may sum up as follows:

+

+

0

+

+

Theorem 1.1: The equivalence classes of the set of Mobius transformations relative to the stated equivalence are: { I } ; the set of parabolic transformations; and the sets C,,, a E C - {2}, where C,, is the set of Mobius transformations M having exactly two distinct fixed points and satisfying K ( M )= a.

EXERCISES 1.1. Given a, b, c E distinct, show that there exists a unique Mobius transformation M satisfying M ( a ) = 0, M(b) = 03, M ( c ) = 1. Given (a, b, c, d) and (a,p, y, 6) where the components of each ordered quadruple are distinct members of (?, show that a necessary and sufficient condition that there exist a Mobius transformation T satisfying ?“(a) = a , T ( b )=/?,T(c) = y, T ( d )= 6, is that M(d) =L(6), where L is the unique Mobius transformation satisfying L(a) = 0, L(/?)= 03, L(y) = 1. Formulate a notion of equivalence for ordered triples of distinct points of relative to the group of Mobius transformations (as in Ex. 3.5, Ch. X). Show that all such ordered triples are equivalent to one another. Carry out the same program for ordered quadruples of distinct points of C,determining the equivalence classes and a complete invariant. 1.2. For what c is c l elliptic, hyperbolic? 1.3. Suppose that the Mobius transformation M # I does not have co as a fixed point. Show that if M is parabolic and a is its fixed point, then M‘(a) = 1. Show that if M has two distinct fixed points a and b, then M ’ ( a ) M’(b) = 1. This exercise is to be done without appeal to the auxiliary S by considering the poles of 1/(1- M). 1.4. We have seen in Ex. 4.25, Ch. VI, that a Mobius transformation M admits a representation on C of the form M(z) =

+

+

az+ b cz + d’

a,b, c, d e C, ad

-

bc # O .

Show that (az B)/(yz 6) is also such a representation if and only if there exists X E C - {0}such that a = ha, p = Xb, y = Xc, 6 = Ad. Consider the determination of the fixed points of M in terms of a, b, c, d and in the case of exactly two distinct fixed points the determination of K ( M ) .Consider the relation of a representation of L o M to representations of L and M , where L is also a Mobius transformation.

XV. Mobius Tramformatiom

328

A Mobius transformation M maps R u {a} onto itself if and only if it admits a representation of the form

1.5.

az+b M ( z ) =cz+d'

a, b, c, dE R, lad -bcl

= 1.

Show that M maps {Im z > 0 } onto itself if and only if M admits such a representation with a, b, c, d E R , ad - bc = 1. 1.6. The set of Mobius transformations mapping {Im z > 0 } onto itself for which a, 6 , c, d of a representation of the form given in Ex. 1.5 are integers constitutes a group with respect to composition. [This is the modular group to be considered in Ch. XVI.] 1.7. Given f E M(e) not constant. Show that ifg E M(C) satisfiesf 0 g = A then g is a Mobius transformation. Show also that the set of such g constitutes a finite group G relative to the operation of composition. Show that if a is a constant whose value is not a fixed point of any g E G other than I and

then f = F h for a unique F E M@). Show for a finite group C of Mobius transformations that h as _above is automorphic with respect to G, that is, h 0 g = h, g E G, and that iff E M(C) is automorphic with respect to G, then as above f = F 0 h for a unique F E M(e). Show furthermore that if hl E M(e) is automorphic with respect to G and also shares with h the property stated in the preceding sentence, then hl = M h where M is a Mobius transformation. 0

0

BB Chapter X V I @@

T H E MODULAR F U N C T I O N h . T H E PICARD THEOREMS

PREFATORY REMARKS

c,

The little theorem of Picard says that if vf(w) = 0 for three distinct w E where f is meromorphic on C, thenfis constant. The big theorem of Picard says that if the same condition holds forf meromorphic on A(a; r ) - { a } ,then f is the restriction of a function meromorphic on A(a; r ) . In this chapter we follow the old classical route of Picard to these famous theorems, proceeding via the modular function 1 to be introduced below. To be sure, the prospective specialist in complex function theory will want to go far beyond Picard’s original program and learn the beautiful Nevanlinna theory of meromorphic functions, which has roots in the Picard theorems, and its modern developments. The account of W. K. Hayman [51] renders the Nevanlinna theory accessible to the English-reading student. Apart from putting the subsequent study of the Nevanlinna theory in perspective and playing a motivating role, the approach we use recommends itself both because it brings into play notions from the earlier chapters of our discipline-such as that of the pe-function-which serve as a starting point for the study of modular functions (thereby revealing a wealth of mathematical ideas implicit in the classical developments of complex function theory) and also because it introduces important examples from the theory of automorphic groups and functions-the modular group MG and a modular function. The lively renewed interest in this theory justifies encounter with such examples, embryonic though they may be. 329

330

X VI. The Modular Function A. The Picard Theorems

Another advantage of the old road is that it furthers the student’s experience with covering maps and liftings, the general mathematical importance of which needs no apology. He has already met such notions in dealing with the exponential function.

1. THE MODULAR FUNCTION h We shall single out for study an analytic function, to be denoted A, with domain {Im z > 0}, which emerges in the course of examining the dependence of the pe-function on the entering parameters. It will be seen to enjoy interesting transformation properties relative to the modular group of Ex. 1.6, Ch. XV, which we shall denote MG.We shall be principally concerned in our study of L with showing: The set of functions f analytic on A(0; r ) and mapping A(0; r ) into C - (0, l } are the functions of the form L O g where g is analytic on A(0; r ) and maps A(0; r ) into {Im z > 0). This property of 1 is the exact counterpart of the property enjoyed by exp relative to the functions analytic on A(0; r ) which map A(0; r ) into C - {O}.? We may already anticipate the small Picard theorem, setting r = +co and applying the theorem of Liouville to l/(g + i). The general situation is reduced to the case where 0, 1, co are omitted with the aid of a Mobius transformation mapping a set of three distinct omitted values onto (0, 1, a}. We define the modular function L as the function with domain {Im z > 0) given by

where p is the Weierstrass pe-function associated with the additive subgroup of C generated by { I , u } . It is to be recalled (Exs. 2.2 and 2.5, Ch. XIV) that p attains each point of twice on {s tu: 0 I s, t < l}, the count being by multiplicity, and that p‘ takes the value 0 at each of the points: 1/2, a/2, (1 + a)/2. Hence p takes distinct finite values at these three points. We see from (1.1) that L(a) # 0, 1 and by the definition of p that L is analytic on {Im z > O}.$ To continue we shall want the first result of the following exercise.

c

+

t The factorization in terms of composition, f = h 0 g, yields g as a “ lifting” of J h is a universal covering map of {Im z > 0) onto C - (0,1). 1It is to be remarked that a certain amount of harmless freedom prevails in the normalization of h and that some variation will be detected in the literature. We refer to the possibility of forming ratios (C- A)/@ - A) with {A, B, C} = {p(1/2), p(a/2),p[(l cc)/2]} in different ways.

+

1. The Modular Function h

331

EXERCISE 1.1. Stow by studying the principal part of ( P ’ ) ~at 0 that ( P ’ ) ~= Q 0 p where Q E M(C) has a pole of order 3 at 03 and elsewhere is analytic (that is, Q I C is a polynomial of degree 3 and Q(m) = 03). Determine thecoefficientsof Qintermsof a.

Suppose that u is meromorphic on some disk A(0; r ) , has a pole at 0, and satisfies (u’)‘ = c 2 Q 0 u where Q is the function of Ex. 1.1 and c E C - (0).By examining the 0-Laurent coefficients of u, we see that u has a pole of order 2 at 0 and that the coefficient of index -2 is c-’. If similarly u is meromorphic on A(0; p ) and has a pole at 0 and satisfies (u‘)’ = d 2 Q u, dE C - {0}, we clearly have u(z) = u[cp(z)] for z small where cp is analytic a t 0 and satisfies cp(0) = 0, cp’(0) # 0. On referring to the differential equations satisfied by u and u we conclude that [(~’(z)]’= (d/c)2for z small and hence that either cp(z) = (d/c)z,z small, or cp(z) = -(d/c)z, z small. We shall use the relation we have just established to show that 1 is not constant. In order to bring to the fore the parameter involved we shall denote by pcu3the pe-function associated with the group generated by ( 1 , a}.? Suppose that A(B) = A(a). Then there exist complex constants A( # 0), B such that A2pCP1+ B attains the same values with multiplicity 2 as does pcal.Since the zeros of Q are the finite values attained bypculwith multiplicity 2, we conclude that A2pcP1+ B satisfies (u’)’ = A-’Q 0 u. Hence A’ pIB1(z) B = pcal(z/A) and we see that 1/A and P/A are in the group generated by ( 1 , a } and that A and a A are in the group generated by { 1 , B}. It follows that there exists TEMG such that /I = T(a). The fact that A is not constant follows from the countability of MG, but it may be established also as follows. If 0

+

+

az b T ( z )= cz + d ’

a,b,c,dEZ, ad-bc=1,

then Im[T(z)] =

~

Im z Icz + d1’’

cz + d # 0.

Hence Im[T(z)] Imax{Im z , 1/Im z } ,

Im z > 0.

(1.2)

Thus A@) # A(a) when Im /? > max{Im a, 1/Im a}.

t In the course of the exposition we shall use the notation P( z;wI,wz} for the value at z of the pe-function associated with the group generated by {wl,wz}.

XVI. The Modular Function A. The Picard Theorems

332

Let us now consider the transformation properties of 1. From (1.1) we conclude A(a + 1) = [A(a)]-'

(1.3)

A( - l/a) = 1 - A(a),

(1-4)

and

Ima>O. We consider the values taken by A when Re a = 0. Using the notation introduced two paragraphs back we see that

P(iz;i, ia)= -P(z;1, a) and hence that p["](z) is 00 or real when z is pure imaginary. Since p["](l- z) = pcU1(z),we see that p["](2-' it) is even, that is, it is equal to p["](2-' - it), and when t is real,pcu1(2-' - it)is theconjugateofp["](2-' it). it) is real when t is real. We conclude: Im A(@) = 0, when Hence p["](2-' Re a = 0. Referring to (1.3) we see: Im A(a) = 0, when IRe a1 = 1. Applying (1.4) with Re a = - 1 we conclude: Im A[(l eie)/2] = 0, 0 < 8 < x . Suppose that S E MG. We are interested in the action of S on 1. Let fi = S(a). The additive subgroup of C generated by { 1, fi} is the subgroup generated by { p , pa} for some p E C - (0). We see, using the differential equation of Ex. 1.1, this section, for the pe-function, that the sets

+

+

+

+

{P(1/2; 1,

fi), P(8/2; LP), P[(1 +PIP; 1, PI}

and

+

{ p - , P(1/2; 1, a), p - , P(a/2; 1, a), p - , ~ [ ( 1 a)/2; 1, a]}

are equal. We conclude on examining the ratios formed from these quantities that A@) E {Tk[A(a)]: k = 1, ..., 6 }

where T,(z)= z, T,(z)= l/z, T3(z) = 1 - z, T4= T, o T 3 , T, = T, o T3o T, , T6= T3 T,. It is readily seen that T,,...,T6 constitute the group Go of Mobius transformations which map (0,1, 0 0 ) onto itself and that Ti# Tk, j # k. Since 1 is not constant, we may take a o , Im a. > 0, such that Tj[A(ao)]# Tk[A(ao)], j # k. It follows from the continuity of A o S and Tko A that L[S(a)]= Tk[A(u)] for a unique k when tl is near a. . We conclude that 1 S = T ( S ) 1where T ( S ) is a unique member of Go. It is clear that the map assigning T ( S )to S is a homomorphism of MG onto Go as we see with the aid of (1.3) and (1.4). 0

0

0

1. The Modular Function A

333

EXERCISE 1.2. Show that there exists a nonconstant function f meromqrphic on {Im z > 0} and satisfying f S =f, S E MG. Hint: Introduce g E M(C) nonconstant and satisfying g 0 T = g, T E G o . Show that f meromorphic (resp. analytic) on {Im z > 0} satisfying f 0 S =f, S E MG, are of the form h 0 h where h is meromorphic (resp. analytic) on the image of A. [we shall see presently that the image of h is C - {0,1 }.] Obtain a necessary and sufficient condition on such h for h 0 h to satisfy h h 0 S = h A, S E MG. 0

0

0

We turn next to the study of the mapping properties of A o n the set E defined as the union of

E, = { x + y i : x ~ ] - 1 , 0 ] , y ~ P ,I x + y i + 2 - ' I

>$I,

and E2 = {x

+ yi: x E 10, I], y E P, (x + yi - 2-'I

2 3}.t

E

r

\

/

d0

-1

FIG.16.1

We introduce v E , the E-valence of 1, which differs from valence as defined earlier in that only points of E are regarded in calculating its value at a point of That is

c.

v&)

=

2

n(z; 1).

AIE(r)=w

t We recall that P denotes the set of positive real numbers.

XVI. The Modular Function 1.The Picard Theorems

334

We shall now show that

VE(W)= 1,

w E c - (0, 1).

To establish this assertion, we note that thanks to (1.2) the function rp, analytic on A(0; 1) - ( 0 ) satisfying (1.5)

rpm(eaia) = A(a)

has a pole or removable singularity at 0. Indeed, by (1.2) we see that I(a) 4 A[A(i/2;1/4)] when Im a 2 4, whence the asserted property of rp, . We let L, denote limo rp, .We introduce similarly rpo satisfying rpo(enia)= A( -a- ') and let Lo = limo cpo, and finally cp, satisfying rpl(ezia)= 1(1 - a-') and let L , =limo rp,. We now show that v E is finite-valued and continuous on - { L o , L , , L,}. The finiteness of vE(w),w # L o , L,, L, ,follows from the behavior of I in E near - 1, 0, 1, co, in the neighborhood of which points A I E takes values near one of L o , L,, L, . If vE were not continuous at w # L o , L , , L, , there would exist (w,,), limn-rmw, = w, v(w,) > v(w) for all n E N. We may suppose that w, = I(z,), z,, E E, and that limn+, z , = z E E , Im z > 0, and A(z) # w. This follows on considering the set d of points z lying in B,Im z > 0, at which I ( z ) = w and noting that the z, may be chosen to lie outside of an open set 0 1 d such that on 0 n E the function A attains no value more than vE(w)times counted by multiplicity. The contradiction is manifest. We conclude that vE takes the same positive integer value at each point of - ( L o ,L,, L,}. Hence since vE takes thevalueOat0, 1, and co, (0, 1, co} c { L o ,L , , L , } . It follows that (0, 1, co} = { L o ,L , , L , } . There remains to be shown that vE takes the value 1 on C - (0, l}. To that end we note that when a E E has a large imaginary part and /3 E E satisfies A@) = A(@, then Im /3 > 1 since Lo, L,, L, are distinct and by (1.2) we have

e

c

Im a I max(1m /3, 1/Im /3}.

+

Hence for such a, if /3 # a, we have /3 = a - 1 or /3 = c( 1; for otherwise Im /3 I1/Im a. Using (1.3) we find A(a) = - 1. This is not possible when Im a is large. Our assertion follows, since we are assured that I' is not the constant 0. We shall see that r/'(a) # 0, Im a > 0, and even more.

EXERCISES 1.3. Show with the aid of (1.3) and (1.4) that L , = 1, Lo = 0, L1 = 03. 1.4. Show that h(int E l ) = {Im z > 0) and h(int E2)= {Im z < O } . [Reference is made to the sets introduced in the definition of E.]

Our principal goal is to show that I has the following important property: Given a E C - (0, l} there exists r positive such that I I C is a homeomorphism


335

of C onto A(a; r ) for each component C of A-'[A(a; r ) ] . t To establish this assertion, suppose that Q = A@). There exists a E E such that a = A(a). Hence B = S(a), S E MG. We take p, 0 < p < Im a, so small that 1 I A(a; p ) is univalent and I[A(a;p ) ] n T A[A(a;p ) ] = 0 for T E Go satisfying T(a) # a. Using the relation 1 S = T ( S ) 0 1, T(S)E G o , established above, we see that A'(/?) # 0 and that A l S [ b ( a ; p ) ] is univalent and has as image T(S) {I[A(a;p ) ] } , a region containing a. If we take r so small that A(a; r ) c T{I[A(a;p ) ] } for each T E Go such that a = T(a), then each component C of I-'[A@; r)] is mapped univalently by 1 onto A(u; r). For C n S[A(a;p ) ] # 8 for some S E MG satisfying T(S)(a)= a and hence C is the image of A(a; r ) with respect to the inverse of I I S[A(a;p)] since C contains the image in question but cannot contain it properly (for otherwise it would contain a frontier point of the image which would be taken by 1 into a point of C(a; r). The covering property of I is established. The fact that I is a covering map leads to its lifting property. We first show: Let f be a continuous map of [ a , b ] into C - (0,l}, --OO < a 0} satisfying f = 1 0 g , g(c) = 8. 0

0

PROOF:Uniqueness. This is a consequence of the observation that if g and h are continuous maps of a connected topological space Xinto {Im z > 0} satisfying I o g = R h, then either g = h or g(p) # h(p), p E X. Indeed, by the continuity of the maps and the Hausdorff property of C the set {g(p) # h(p)} is open. If q is a point of { g ( p )= h(p)}, by the univalence of I restricted to a sufficiently small neighborhood of g(q), A[g(p)]= I[h(p)]implies g(p) = h(p) for p near q. Hence { g(p) = h(p)} is open. The asserted relation between g and h follows from the connectedness of X . It is to be noticed that the full force of the fact that f is a covering map was not invoked here, only the local univalence of I.$ 0

Existence. We consider a finite covering of the image of f by disks 6, = A ( q ; rk), k = 1, .. ., m, such that 1 maps each component of 1-'(6,) univalently onto 6,. If n E N is sufficiently large, then for each j E (0,2" - 1) there exists k ( j ) E (1, m ) such that f([tj, tj+ll)

6k(j)

where ti = a + 2-"j(b - a). The proof of this assertion by the method of successive bisections is routine. Let c E [t, , t,+ I ] and let p, denote the inverse of 1

t This says that h is a covering map of {Irn z > 0) onto C - (0,l} in the sense of topology. $ A reference to the uniqueness of the continuation of an element along a path is appropriate here.

336


restricted to the component of A-’[6k(,)] containing p. We propose g(t) as p,Lf(t)], t , I t I t v + l . Thereupon (when n > 0) we proceed to adjoining segments. Thus if v > 0, we introduce p V v 1as the inverse of 1 restricted to containing p,[f(t,_J] and propose g(t) as the component of 1-’[6k(,-l)] p,-lu(t)], I t I t , . We proceed similarly for [ f v + l , t , + z ] when v < 2”. The construction is continued in this manner and the function g so obtained fulfills the imposed conditions. For the case of a continuous map f of a rectangle [a, b ; c, d] into C - (0, l} we proceed along similar lines starting with subrectangles obtained by successive quadrisections which are so fine that the image of each with respect to f lies wholly in a 6, as above. We take the subrectangle containing the point of normalization s t i : f ( s t i ) = A@), and construct g(z) as above. A divergence from the I-dimensional case arises when we proceed to adjacent subrectangles, for two adjacent subrectangles have a segment rather than a point in common. Since a segment is connected, we see that the g for two adjacent subrectangles adjusted to take the same value at a point of the segment common to the two subrectangles take the same value at each point of the segment. We proceed first along the horizontal “ row ” which has the preferred subrectangle containing (s, t ) and thereupon proceed in the vertical direction along each “column” starting from the subrectangle of the preferred “row.” The connectedness of the intersections of adjacent vertical strips assures coherence and we obtain g continuous on [a, 6 ; c, d] having the property that f = Log andg(s + t i ) = p. We see by the reasoning employed in the study of continuous logarithms that continuous maps f of {Im z > 0) or C into C - (0, I} admit normalized liftings g as above. For the case where the domain is {Im z > 0 } we proceed, when the point wheref is normalized is c, by taking the union of the normalized liftings of

+

+

f I [-(n + 1) + Re c, (n + 1) + Re c; (Im c)/(n + l), (n + 1)(Im c)], n EN. For a continuous map f of a disk A(a; r ) into C - (0, I} we may reduce the problem to one where the domain is an upper half plane by the intervention of

a

+ r-zz +- ii ’

Imz>O,

when a and r are finite and an obvious variant when a = co and r is finite. In the case where the domain is C our assertion is now immediate. For the stated domains it is easy to see that iffis analytic on its domain, then so is g. Indeed, if g is a continuous map of a region A c into {Im z > 0 } such that 1 g is analytic on A , then g is analytic on A . For given a E A , let 6 denote a disk in {Im z > 0 } centered at g(a) so small that A 16 is univalent. We take

c

0


337

A(u; r ) c A so that g[A(a;r ) ] c 6 and let p denote the inverse of A 16. Then [(A g) I A@; r ) ] is analytic on A@; r ) and is equal to g I A@; r ) .

p

0

0

The small Picard theorem is now under control. It is to be emphasized that our object in introducing 1 is precisely to have a function analytic on {Im z > 0) which is a covering map in the indicated sense of {Im z > 0) onto C - (0, I}. The wealth of interesting properties of A which emerge is a bonus.

EXERCISES If h , is analytic on {Im z > 0) and is a covering map of {Im z > 0) onto C - ( 0 , l ) in the sense specified in the text, then A, = h 0 S where S is a conformal automorphism of {Im z > 0). 1.6. If S is a continuous map of {Im z > 0) into itself such that h = h S, then S E MG. 1.7. Givenfanalytic on A(0; 1) with image in C - {O, l}. Let 1.5.

0

m

where p > 0, and let f(z) = h[g(z)]for z small. Using the fact that h is a covering map of {Im z > 0) onto C - {O, 1 ) show that p 2 1, that Img(z) > 0, and that f(z) = h[g(z)], IzI < 1. Appeal is not to be made to the existence of a normalized analytic lifting with respect to h. Hint: Suppose that p < 1 and show using the covering map property of h that Im g(z) > 0, /zI < p, and that, for each u E C(0; l), limpg(tu) exists and is in {Im z > 0). Here g(tu) has domain [0, p[. For each such u let h. denote a function analytic on a disk with center pu such that h[h.(z)] =f(z), z on the domain of h. (assumed in A(0; 1)) and satisfying h.(pu) = limpg(tu). Thereupon consider g u

u h.,

IuI = 1

and complete the proof noting that Im g(z) > 0, (21 < 1. It is to be remarked that we could use the result of this exercise rather than the more general lifting result which was obtained for continuous f. Preference was given to the reasoning of the text because of its more gene@ character. 1.8. Let A be a homotopically simply connected region c C. Letfbe a continuous map of A into C {O, l}. Show that iff(a) = h(a), where a is a given point of A, there exists a continuous map g of A into {Im z > 0) such that f = h 0 g, g(a) = a. [Of course, g is analytic on A when f i s , as follows from the remarks of the text.] Hint: Let Q, denote the family of paths y in A with domain [0, I ] satisfying y(0) = a. Let G, denote the unique continuous map of [0, 11 into {Irn z > 0} satisfying: G,(O) = a, h 0 G, = f o y. Show that ~

{ ( ~ ( l )Gd1)): , YE

serves for g.

@I


338

2. THE BIG PICARD THEOREM

As a preliminary to the proof of the big Picard theorem we establish the following theorem of a type due to Schottky.? Theorem 2.1: Let 0 < 6 < 1. Let @ denote the class of functions analytic on A(0; 1) with image in C - (0, l } and such that xLf(O), a] 2 6 , a = 0, 1, 00. Let m(r) = inf{xlf(z), a ] :a = 0, 1, co,f

E@,

IzI 5 r } .

Then m(r) > 0, 0 I r < 1. PROOF: We put aside the trivial cases where the 0 are empty. Given f E 0, let a(f) satisfy: I [ a ( f ) ]=f(O), a(f) E E, where E is the set so designated in $1, this chapter. Let c denote the supremum of the set of positive numbers max{Im a ( f ) , [Im a ( f ) I - ' } , E @ being taken into account. It follows from the behavior of I I E that < +a. Given y , 1 Iy < +a, let p(y) denote the chordal distance from

all f c

n({y-' IIm z Iy } )

to (0, 1, co}. Now let g be analytic on A(0; 1) and satisfy: I o g =f, g(0) = a(f ) . Using the Harnack inequality (1.7), Ch. XIII, we obtain

so that

IzI < 1, a = 0, 1, co. The theorem follows. The big Picard theorem. It is immediate that the big Picard theorem stated at the outset of this chapter may be reduced to the following theorem.

Theorem 2.2: Let f be analytic on (1 < IzI < +a} and have image in C - (0, l}. Then f does not have an essential singularity at 00. Indeed, iff satisfies the hypotheses of the big Picard theorem and has image in - { A , B, C} and r is finite (as we may suppose), then it suffices

e

t The notation x below refers to the chordal distance (cf. $4, Ch. VI).

2. The Big Picard Theorem

339

to consider M [ f ( a + z-lr)] when a E C and M[f(r-'z)] when a = 00, 1 < IzI < +co, where M is the Mobius transformation satisfying M(A) = 0, M ( B ) = 1, M ( C ) = co.

p

PROOFOF THEOREM 2.2: We distinguish two cases. (1) For some 11, +a[, we have

E

f [ C ( O ;r ) ] n C(0; 2) = 0,

< +a})is connected and does not intersect C(0; 2), it is contained in A(0; 2) or in A(co; 4). When

p < r < +a.Sincef({p < IzI

f({P
l = sup(lf’(a>l:.fE

n+ m

@I,

(2) (9”) tends to a function g uniformly in A . It is to be noted that g so obtained

is analytic on A and satisfies: g(a) = 0 ;

Ig’(a)l 2 If’(a>l> 0 ,

f~

@;

1dz)I < 1, z E A .

To see that the third condition is fulfilled, we observe that at all events Ig(z)l I 1, z E A , and g(a) = 0 so that by the maximum principle for analytic functions the strong inequality holds. To see that g E@, we note that if a sequence of functions analytic on a region A and univalent tends uniformly in A to a nonconstant function, then this function is univalent (cf. E x . 4.38, Ch. VI). The image of g is A(0; 1). We shall introduce an auxiliary function to establish this. Given b E A(0; 1) we use L b (cf. 993, 5, and 7, Ch. VIII), to denote the Mobius transformation given by Lb(Z) = ( b - z)/(l - hz), z E C , Lb(oo)= 116. We note that the inverse of L b is Lb itself. Suppose that b # 0 and that cz = b. We introduce B = Lb 0 (L,)’. Since B has constant valence 2, B(0) = 0, and lB(z)l < 1 for IzI < 1, we see by the Schwarz lemma that IB‘(0)l < I . Suppose now that b E A(0; I ) is not in the image of g. Of course, b # 0. We let s denote the analytic square root of Lb 0 g satisfying s(a) = c and

342

X V l l . Some Questions Concerning Univalent Analytic Functions

thereupon define h as L, 0 s. The steps leading to h insure that h E @. Indeed, Lb g is univalent and analytic on A and its image is in A(0 ; 1) - (0) ;s enjoys these properties also and hence since s(a) = c, h E 0. Now g = B 0 h. Hence g’(a) = B’(0)h‘(a) and we conclude Ig’(a)l c Ih’(a)l.Contradiction. Hence the image of g is A(0; 1). The theorem is thereby established. 0

EXERCISES 1.1. Show that the totality offanalytic on A, univalent, and having image A(0; 1) consists of the functions of the form Lb 0 ( 7 g ) where 7 E C(0; l), b E A(0; 1). Show that there is a unique such f satisfying f ( a ) = b, sg[f’(a)J = 7,where b and 7 are given and fulfill the above conditions. 1.2. Let Y denote the set of functions analytic on A and mapping A into A(0; 1). Show that Ig’(a)l 2 If’(a)l,f~Y ,and that equality occurs precisely forfof the form 7 9 , E~ C(0; 1). Let Y odenote the subset of Y consisting off€ Y satisfyingf(a) = 0. Let b(# a) E A. Show that ( f ( b ) l , fY ~ o, and examine the caseof equality. Here reference is made to g of Theorem 1.1. 1.3. Show with the aid of the Riemann mapping theorem that a holomorphically simply connecjed region A in C is homotopically simply connected and has the property that C - A is connected. Hint: For the latter property show that

c - g[W; r)l, 0 < r < 1, is connected, when g is a univalent function analytic on A(0; 1) and use the standard theorem valid for a Hausdorff space which states that K . is cone~ is a sequence of compact connected subsets of the space satisnected when (Kn)” fying Kn+lc K . , n E N. 1.4. Show that the zeros of B (of the text) are 0 and d =L,( -c) and that B = 7IL, for some 7 E C(0; 1) where I is the identity map of C. Conclude that

u,

Show with the aid of this formula for IB’(0)I that the sequence (v.) of univalent functions analytic on A defined below tends uniformly in A to a function with image A(0; 1). Let v E @ have image not equal to A(0; 1). The sequence ( p n ) :N + CD is defined recursively as follows: vo = v, pn+ = #n 0 vn,n E N. Here #“ is a univalent function analytic on y.(A) satisfying #.(O) = 0, i,hn’(0)> 0, B.{sg[=J #.(z)} = z, z E v.(A), where B. is just B of the text with b a point of C - p.(A) nearest 0 (cf. the construction of h). The possibility of constructing such a sequence is to be verified. 2 Iv.(z)I, z E A, and show that the sequence (rn) is inHint: Show that lvn+l(t)l creasing and has limit 1 , r. being min{ Izl: z E C - vn(A)}. 1.5. Let f b e analytic on A(0; r), r > 1, and suppose further that f I A(0; 1) is univalent and thatf’(z) # 0, IzI = 1 . Show thatf I A(0; s) is univalent for somes, 1 < s < r , and that for such s a function analytic on f[A(O; s)] may be approximated uniformly on f[A(O; l)] by a polynomial. Conclude that if h is continuous on f[A(O; 1)J and is analytic in f[A(O; l)], then h may be approximated uniformly on f[A(O; l)] by a

2. Carathebdory’s Theorem on Variable Regions

343

polynomial. Hint: Consider h 0 [/ I A(0; l)] and approximate it first by a polynomial P and then consider P composed with the inverse of/ I A(0; 1). Far-reaching generalizations of approximation theorems of this kind are known (cf. [83], [125]).

2. CARATHEODORY’S THEOREM ON VARIABLE REGIONS Let ( A n ) n p Nbe a sequence of simply connected regions of C satisfying 0 E A,, n E N.? Carathkodory defined the kernel of (A,), which we shall denote k[(A,)],as follows. If no A(0; r ) lies in all A , , then by k[(A,)] is meant simply ( 0 ) . Otherwise k[(A,)] is defined as U n a N B nwhere B, is the component of int( >, A k ) containing 0. His object in introducing this concept was to treat the following problem [22, pp. 362-4051: Let (f,),EN be a sequence of univalent functions analytic on A(0; 1) and normalized by the conditions f,(O) = 0,f,‘(O) > 0, n E N . W e let A, denote the image o f f , . We seek a necessary and sufficient condition in terms of (A,,) that (f,) tend to some function f uniformly in A(0; 1). Carathtodory solved this problem (loc. cit.) by the following theorem.

nk

Theorem 2.1: r f (f,) tends to f uniformly in A(0; l), then the image o f f is k[(A,)l. A necessary and suficient condition that (f,) tend to some function f uniformly in A(0; 1) is that every subsequence of (A,,) have the same kernel and it be a proper subset of C. PROOF: Suppose that (f,)tends uniformly to f in A(0; 1). Then either f is the constant zero or f is univalent. In either case let A denote the image off. We show that A = k[(A,)],thereby establishing the first assertion of the theorem and rendering the necessity part of the second assertion immediate. We first verify that A c k[(A,)]. Suppose that a E A. The case, a = 0, is trivial and we put it aside. Let b satisfy f ( b ) = a( # 0 ) and let Ibl - ~

’ +1 [fn(z>- f ( ~ 1 1 5

t We do not need to qualify “simply connected” since we now know that the various kinds of simple connectivity used in this book for regions of C are equivalent, that is, holomorphic, homotopic, and simple connectivity defined by the stipulation that the complement with respect to i? is connected.

344

XVII. Some Questions Concerning Univalent Analytic Functions

with domain A(0; r ) , w ~ f [ A ( 0p)], ; n large. Hence a ~ k [ ( A , ) ] and we conclude that A c k[(A,)]. We next show that k[(A,)] c A . We put aside the trivial case k[(A,)] = (0). Let g, denote the inverse of f,. Let A(0; r ) c A , for all n E N, where 0 < r < + oc). By the minimum principle for the modulus of an analytic function applied to f,(z)/z and the univalence off, we have If,(z)l 2 r IzI for IzI < 1.tHencef,[A(O; 1/2)] 3 A(0; r/2). Using the integral formula representing g,, I A(0; r/2) in terms of f, I C(0; 1/2) and f,' I C(0; 1/2), we see that (9, I A(0; r/2)) is uniformly Cauchy in A(0; r/2). Since the domain of g, contains B, when n 2 m, by the theorem of Stieltjes-Vitali (Ex. 1.14, Ch. XIII), (g,+,,, I B,) tends uniformly in B,,, to a function gCm1, say. Let g = UmEN gcml.We have: g is analytic on k[(A,)], g(0) = 0, the image of g is contained in A(0; 1). Fromf,[g,(z)] = z, z E A , , we conclude thatf[g(z)] = z, z E k[(A,)]. Hence k[(A,)] c A . Putting the results of the two preceding paragraphs together we see that A = k[(A,)]. Now A # C, otherwise Liouville's theorem applied to the inverse off would furnish a contradiction. To complete the proof of the theorem, there remains to show that the sufficiency part of the second assertion holds. We introduce p, , the distance from (0) to C - A , . The sequence (p,,) is bounded, otherwise for some subsequence (A,,,,) of (A,) we would have k[(A,,,,)] = C. We introduce 0 the set of univalent functionsf, analytic on A(0; 1) and satisfying: f ( 0 ) = 0,

min(lw1: w #f(z), IzI < 1) = 1,

and note that by (1.8) of Ch. XI11 there exists M : 3 0, l [ + P such that

I4 II m l IIZI W r ) ,

(2.1)

IzI Ir c l , f ~ ( D .Applying (2.1) to p,-*f, we see that {f,:n E N} is a normal family. The proof will follow once we show that (f,)tends to a limit pointwise on A(0; 1). If this were not the case, there would exist two subsequences of (f,)tending uniformly in A(0; 1) to distinct functions, say g and h. Of these one is not the constant 0. By the first part of the theorem and the sufficiency hypothesis, the images of g and h are both equal to k[(A,)]. Hence neither is the constant 0. Further, h-' 0 g is a conformal automorphism of A(0; 1) taking 0 into itself and having a positive derivative at 0. Hence h- 0 g is the identity map on A(0; 1) and we conclude that h = g. Contradiction. The proof is complete.

t We may use Theorem 4.1, Ch. VI in place of the minimum principle.

3. Application of the Schwarz Reflexion Principle

345

EXERCISE Continuing with the setting of Theorem 1 . I let g be analytic on A(0; 1) and in addition to satisfying g(0) = 0 either be the constant 0 or be univalent and satisfy g'(0) > 0. Then (f.) tends uniformly to g in A(0; 1) if the image of g is the kernel of every subsequence of (A"). 2.1.

3. APPLICATION OF THE SCHWARZ REFLEXION PRINCIPLE The notion of a regular, Jordan, analytic arc is introduced as follows. A path

+

y with domain [a, b], - 03 < a < b < 00, is termed an analyfic arc provided that y = cp I [a, b] where cp is analytic on an open set of C containing [a, b]. The term Jordan is adjoined when y is univalent, the term regular when

# 0, a I t I 6, one-sided derivatives being understood at the endpoints. Suppose now that f is analytic on {lm z > 0 ) and that for each c E 0 where 0 is an open set of R there exist rc > 0 and a regular Jordan analytic arc y c with the property that, for each real x , Ix - c J < r c , lim, d(z) = 0, where d(z) is the distance from {f(z)} to y c . We shall show y'(t)

Theorem 3.1: Under the stated assumptions f is the restriction to {Im z > 0) of a function analytic at each point of 0 v {Im z > 0).

PROOF: Given c, we suppose, as we may, that yc is the restriction to its domain [a, b] of a univalent function cp analytic on a region containing [a, b], and that the domain of cp is symmetric with respect to the real axis. We take p , 0 0} is mapped by f into the image of cp. The Schwarz reflexion principle applies to cp-l (f I S ) and shows that cp-' o (fl S ) is the restriction to S of a function $ analytic on A(c; p ) whose image lies in the domain of cp. Hencef 1 S = (cp $) I S a n d the assertion of the theorem follows. 0

0

Another application of the Schwarz reyexion principle.

Suppose that

I-: C(0, 1 ) + C is univalent and is the restriction to C(0; 1) of a function analytic at each point of C(0; 1) and that I-'(z) # 0, IzI = 1 . By a closed,

regular, analytic, Jordan curve we understand the image of such a r and we shall, as in similar questions, denote image and map by the same symbol. We suppose, as we may, that r is the restriction to C(0; 1) of a univalent function cp analytic on an annulus {r < IzI < l / r } , 0 < r < 1 . Suppose that f is a univalent function analytic on A(0; l ) , the frontier of whose image is I-. We shall prove the following theorem concerning the boundary behavior off.

346

XVII. Some Questions Concerning Univalent Analytic Functions

Theorem 3.2: Under the stated assumptions on r and f there exists a univalent function F analytic on some A(0; s), s > 1, such that f = F I A(0; 1). PROOF: By Theorem 4.1, Ch. V1, we see that r is the set of cluster values off. Hence sincefis analytic, the component of - r containing m has no points in common with the image o f f , which is consequently bounded. Further let p, 0 < p < 1, be so near 1 that f ( { p < IzI < 1)) is contained in the image of cp. Since it is connected, f ( { p < Iz1 < l})lieseitherincp({r < IzI < l}) or in cp((1 < IzI < l/r}). Now let i,h denote the inverse of cp and consider the harmonic function

c

with domain (0 < Im z < log(l/p)}. By the Schwarz reflexion principle and the periodicity of the exhibited function we see that log Ii,h[f(z)]I = u(z), p < IzI < 1, where u is harmonic on { p < IzI < l/p} and u(z) = 0, IzI = 1. Since the logarithmic derivative of the function @ Lf(z)] with domain { p < IzI < l} is the restriction to its domain of 2u,, we conclude that 2u, is the logarithmic derivative of a function analytic on { p < IzI < l/p} (cf. Ex. 1.11, Ch. VI). Hence there exists a function g analytic on { p < IzI < l/p} satisfying g(z) = $Lf(z)], p < (21 < 1. Further llog Ig(z)ll < log(l/r),p < IzI < l/p. Consequently f u (cp g) is analytic on A(0; l/p). The assertion of the theorem will follow once we show that g I C(0; 1) is univalent and g'(z) # 0, IzI = 1. Suppose that g has multiplicity n > 1 at a E C(0; 1). From 0

lim t-"[lg(a + teie)12- Ig(a)12] = Re(Ae""), 0

where A E C - {0}, we conclude that 9 takes values of modulus greater than one and of modulus less than one in { p < IzI < l} at points arbitrarily close to a.t This is not possible by the property of f ( { p < IzI < 1)) noted in the preceding paragraph. Hence g'(z) # 0, IzI = 1. Suppose now that g(a) = g(b), where la1 = Ibl = 1 . Suppose that limn-tma, = a where p < lanl < 1. We suppose, as we may, that (a,) is so chosen that g(a,) = g(b,), n E N,where limn+a b, = b. Since either

< I Z I < 11) = { r < Izl < 11, in which case g({l I

IZI

< l/P})

= (1 I

IZI

< WI,

or else g ( I P < IZI < 11) = (1 < IZI
0) have a representation of the form

where the Pk are positive and are subject to the condition n

Let us now turn to the converse question. We shall show that a nonconstant function f analytic on {Im z > 0} which satisJ7es the condition (4.6) is the restriction to {Im z > 0} of an admitted F. T o establish this it will be convenient to introduce the simply connected region A containing {Im z 2 0} - {ak:1 I k I n } defined as

c - su {ak-iit:tER+}. ksn 1

We introduce g, a primitive of

X VII. Some Questions Concerning Univalent Analytic Functions

350

with domain A, and let h denote a primitive of exp 0 g. Since f ' ( z ) = tee('), c E C - {0},it follows thatf(z) = c h(z) + d, d E C. We note that Img I l a k ,

ak+l[

is constant, k = 1, . . . , n - 1, and that the same holds for the restriction of I m g to {x < a,} and the restriction of I m g to {a,, < x } . It follows that the image of each of these intervals with respect to h and hence with respect to ch + d lies on a line. Our assertion will follow once we show that lim, h and limakh exist and are finite. Let us consider the behavior of h near uk. Let I denote an analytic logarithm of z - ak with domain A. Then g(z) = (n-'Bk - 1) Xz) + 444 where cp is analytic at ak . On introducing the Taylor expansion of exp cp about ak, we see that

1; A ~ ( -Z a,)'

0

B E C, when z is near ak . Hence the stated limit property of h at ak holds. The behavior of h at 00 is similarly treated. It is to be noted that if I now denotes an analytic logarithm of z - a, with domain A, then g(z) = a A(z) + cp(z),for z large and satisfying Im z > 0, where cp is analytic at co and a < - 1. This follows on considering the right-hand side of (4.6) near 00 and taking (4.7) into account. The remainder of the argument parallels that just given for the behavior at a,. We introduce the Laurent expansion of exp cp in { Iz - a, I > r } , r large, and verify that a counterpart of (4.8) now holds. 0

EXERCISES 4.1. Study with the aid of the method of Schwarz-Christoffel treated above the continuous univalent maps of {Im z 2 0)u { a)} onto a rectangle which are analytic in {Im z > O}. 4.2. Let Im a > 0. Let T = {A + pa: A, p 2 0, + p I 1). Show that there is a unique Schwarz-Christoffel function Fmapping {Im z > 0)u { 00) univalently onto T and satisfying F(0) = 0, F( 1) = 1, F( a)) = a. Determine F. 4.3. Extend the study of the Schwarz-Christoffel functions to the class of continuous maps F of {Im z 2 O} u {a)}into C which satisfy the condition that there exists (&)I < ) < n , an increasing sequence of real numbers, such that (1) Fis the restriction to its domain of a function meromorphic at each point of

{ImZ>O} - {Uk: 1 I k S n } which is analytic at each real point not an ar and has a nonzero derivative at each real point (not an ak),and (2) F ( ] a k &, + I [ ) is contained in a line, k = 1, . . . ,n - 1, and the same is true for F ( { x < al}) and F ( { x > a"}).Distinguish the special cases:

5. Regions with Simple Polygonal Frontiers

351

F analytic in {Im z > 01, F locally univalent in {Im z > 01, F both analytic and locally univalent in {Im z > O}.

5. REGIONS WITH SIMPLE POLYGONAL FRONTIERS Let (tk)OSk_ 0 } having A as its image is the restriction of a univalent Schwarz-Christoffel function F. IfF(o0) is a vertex of r, the anda, associated with the preimages with respect to F of the vertices satisfr

O < a k < 2 ~ , k = 1 , . . . , n - 1, as well as -2n,U e J f (C(e2), Ue,)), then on invoking the continuity at 0 of the component functions f l , g l , f 2 , g 2 , we see that for small positive r the sets

k

=

I , 2, are disjoint. Suppose that ( C ( e , ) , V(e1))= (C(e,), Ve2)).

If f l is constant, thenf, is not, and we see at once that the sets (2.6)are disjoint for small positive r. We proceed in the same way when any of the other three functions is constant. To continue, we suppose that f l and f 2 are not constant and that they have the canonical form of f l of Ex. 2.3. We then see that if the sets (2.6) are not disjoint for any small positive r, then by Lemma 2.1 we have e, = e, . We conclude that the Hausdorff property holds. We next examine the fl maps. Each is univalent because of the univalence

X Vlll. Riemann Surfaces

364

condition imposed on allowed (f,g), and each is an open mapping by the very definition of T. The continuity of these maps may be established as follows. Suppose that (fi,gl) and (f,,g2) are allowed and that cp analytic at 0 satisfies q(0) = 0 and f2W

=f1Ccp(z>l,

92(d = glccp(z>l~

z small. On introducing ra as above relative to (fl, gl) and letting b b serve in the same capacity relative to (fi, g2), we see that for each small b we have for z small the equality bb(z) =fl

f 2

7q7(b)[$(Z)1

where $(z> = z~(~){(P[ab(z>l}*

A corresponding equality holds with g k replacing f k . Writing 0, for 8 ( f k , e k ) , 8 for O ( f , g ) , and 8” for 8 when ( f , g ) is replaced by ( f o z,, g za), we see that 0

02(z) = 01Cdz>l,

z small, and oa=eoza.

These observations show that if &(a) = 8,(b), there exists $ analytic at a, $(a) = b, such that 0,(z) = 8,[$(z)], z near a. Further $ is locally determined near a by this condition. Suppose now that 0 is an open subset of E. We wish to show that 8 - ’ ( 0 ) is open in order to infer the continuity of 0. Since T is the family of unions of finite intersections of images with respect to the 0 maps of open sets in their domains, when 0(a) E 0, then e(a) E

o1(nl)n

-

a

n e,(n,) c 0,

where each Rkis an open subset of the domain of 8 k . But from our observation concerning $ above, 0(z>E el(n1)

* * *

(7

on(%>,

z near a. Hence O-’(O) is open. The asserted continuity of 8 follows. Analytic entity. Let A be a component of E. Let @(A) be the set of 8 maps with image in A. We show that (A, @(A)) is a Riemann surface. Indeed, if el, 8, E @(A) and their images have a nonempty intersection 0 (necessarily open) and $ is

((0;

l ( P ) > 0;

%>):P E 01,

then we have U Z )

= ~,C*(Z)l,

2. Weierstrass Classes, Analytic Entities

365

z E O;'(O), and from the developments of the preceding paragraph we see that I) is analytic on its domain. Thus @(A) is a conformal structure on A. By an analytic entity is meant a Riemann surface (A, @(A)). It is immediate that center and value yield functions meromorphic o n A (which we shall denote by C and V, respectively). For if 0 = U(f,s) E @(A), we have w")l

=f(z>,

V"Z)l

= g(z),

z in the domain of 0.

EXERCISES Suppose that for an allowed (Lg) the second component g is constant. Show that the value function whose domain is the analytic entity A containing e(A g) is constant and that there exists a univalent analytic map of C onto A. Hint: Let a denote the value taken by g and consider the map taking z into the element containing the ordered pair whose first component is the function z w , w E C, and whose second component is the constant a, z E C,and taking a,into the element containing the ordered pair with first component l / w , domain C, and second component the constant a. 2.5. Show that the map 2.4.

+

v = M L g), 4 g ,f)):(L9) allowed) is a homeomorphism of E onto itself and that for each analytic entity A the map q~I A is a univalent analytic map of A onto an analytic entity. y(A) is termed the inverse of A. Justify the terminology. Continuing the preparations for the proof of Theorem 2.1 we shall establish the following lemma relating ordered triples (S,f, g) where S is a Riemann surface a n d f a n d g are functions meromorphic on S , not both constant, with analytic entities.

Lemma 2.2 : Gicen (S,f, g), there exists a unique (A, cp) where A is an analytic entity and cp is an analytic map of S into A such that f = C rp, g = V cp. 0

0

PROOF: Unicity. Let a E S be such that the multiplicity of one of the functionsf, g is 1 at a and let u be a uniformizer satisfying u(0) = a which is such that either f u o r g u is univalent. Clearly cp 0 u is a uniformizer for A. Further 0

0

e ( f 0 u, g a) = e(C cp u, 0

0

0

1/

0

cp

0

u ) = cp(a),

(2.7)

as we see introducing a 0 map, which we denote by 8, with image in A taking 0 into Via), and noting that

c

cp

U(Z)= (c 0 e) 0 e-'[cp

0

a(z)],

(2.8)

X Vllt. Riemann Surfaces

366

and 'c.' 0 cp 0 a(z) = (V 0 6 ) 0 e-l[cp

0

a(z)],

(2.9)

z small. From (2.7) we infer the unicity of A which contains e(f 0 a, g 0 a) and of cp given the arbitrary character of a. It is to be noted that the set where the multiplicity of a nonconstant meromorphic function on a Riemann surface is greater than one clusters at no point of the surface. Existence. The case where f is constant is cared for by the analytic entity A for which the center function takes the constant value off. There is exactly one A with this property. In this case the value function maps A univalently onto and V - ' g serves for cp. Suppose then thatfis not constant. Given a E S, we assert that there exists a unique meromorphic element e such that, for every (F, C) E e and every uniformizer a for S, a(0) = a, there exists h analytic at 0, h(0) = 0, such that

e

0

fo

a(z) = F"z)I,

9

=GCWI,

O

(2.10)

z small. There is at most one such e. Let el be a meromorphic element sharing the stated property with e, (F,, C,) E el, and h, be analytic at 0, hl(0) = 0, and such that the counterpart of (2.10) holds for z small when F, G, h are replaced by F,, GI, h,, respectively. By Ex. 2.3 and Lemma 2.1 we see that e = el. There is such an e. We take a uniformizer a, a(0) = a, with domain a disk A(0; r ) so thatfo a(z) =f(u) + z " ( ' ; ~ )whenf(a) , # 03, and f 0 a(z) = 1 / ~ " ( " ; ~ ) , whenf(a) = 03, IzI < r, and g 0 a has a pole at most at 0. Let rn denote the largest positive integer dividing n(a;f ) and each integer k for which ak # 0 where (ak)is the O-Laurent sequence of g 0 a. We take F and G meromorphic on A(0; p ) , where (i) p Ir'" when r < + 03, (ii) F(z) =f(u) + z"(';~)''" when f ( a ) # 03, and F(z) = 1 / ~ " ( ' ; ~ whenf(u) )/'" = 03, IzI < p, (iii) the kth0-Laurent coefficient of G is ukm, k E Z, and (iv) (F, C) is allowed. We have (2.11) f a(z) = F(z"), 9 M ( Z ) = G(z"), z small. It is now easy to verify that if (F,, C,) E e(F, C) and /3 is a uniformizer for S, p(0) = a, then there exists h analytic at 0, h(0) = 0, satisfying 0

0

f B ( 4 =F I C W I , O

9

O

B(z>= G,Ch(z)l,

z small. Hence e(F, G) is such an e. We shall produce (A, cp) satisfying the imposed conditions. Let cp: S -+ E assign to a E S the unique meromorphic element satisfying the condition (2.10) relative to a. From (2.1 1) we see that cpCa(z)l = 6 ( F , G ) ( z m ) > z small. Hence not only is cp continuous on S-a

(2.12)

fact which implies that cp(S)

2. Weierstrass Classes, Analytic Entities

367

is contained in an analytic entity A-but also cp is an analytic map of S into A. Using (2.12) and thereupon (2.1 1) we conclude that C 0 cp =f, V cp = g. The proof of Lemma 2.2 is complete. 0

EXERCISE Let R denote the additive subgroup of C generated by {wl, w 2 } where Im(Z,w,) # 0 and let p denote the associated Weierstrass pe-function. Apply Lemma 2.2 to (C, p , p') and show that ~ ( w =) v(z)if an? only if w - z E R. Show that A is compact but is not conformally equivalent to C. Obtain a necessary and sufficient condition for two such groups R to yield the same A. 2.6.

We are now in a position to treat Theorem 2.1. Given W and (S,f, g, cp) standing in the relation described by (2.1) to W, we apply Lemma 2.2 to (W, c, u) obtaining (A,, I),) and to ( S , J g) obtaining (A,, $,). We conclude that (A,, I),) = (A,, I), cp). Hence A, = A,. Further $, and cp are univalent. Since S - cp(W) clusters at no point of S, we conclude that $, is univalent. Thus (A,, C , V , $,) is an ordered quadruple fulfilling the conditions stated in the first sentence of Theorem 2.1. If (S,f,g, cp) fulfills the maximality condition, then Lemma 2.2 applied to (S,f, g) yields (A,, I),) and at the same time there exists an analytic map of A, into S , say u, such that C =f u, V = g 0.It follows that C = C 0 (I), 0 u), V = V 0 (I), 0 a). Applying Lemma 2.2 we see that I), 0 0 is the identity map of A, onto itself. Hence I),has image A,. It is also easy to see that a is a univalent map with image S on noting that 0 = I);' 0 (I), 0 a). The second assertion of Theorem 2.1 is now readily established. The following remark is to be made. Suppose that Weierstrass classes W, and W, determine the same analytic entity. Then W, = W, . Indeed their images in A have a point in common. The image of an element (a, P), where P has domain A(a; r ) , is the meromorphic element containing ( J g) where f is a + z with domain A(0; r ) and g is P(a z ) with domain A(0; r ) . Elements of W, and W, mapping into a given element of A are clearly equal. Hence W, and W, have a point in common. Our assertion follows. 0

0

0

+

EXERCISES 2.7. Let S be a Riemann surface andfand g functions meromorphic on S. We say that (S,A g) is extendible provided that there exist a Riemann surface S1,functionsf, and g1meromorphic on S1 and an analytic map q~ of S into S, which either is not univalent or else satisfies v(S)# S, and at the same time satisfies f = f i o v, g =gl v. Show that ( S , Ag) is nonextendible if and only if of Lemma 2.2 maps S univalently onto A. 0

X VIII. Riemann Surfaces

368

e,

Let W be a Weierstrass class. Let U E r >0, eo = ( b , P ) E W where b E A(a; r ) - { a } . It is supposed that (1) eo can be continued along every path in A(a; r ) - { a } with initial point b ; ( 2 ) only a finite number of elements with center b are obtained in this manner, say m ; (3) there exists a E such that for each positive p we have v(e) E A ( a ; p) when c(e) is sufficiently near a, e being an element obtained by continuation from eo along a path in A(a; r ) - { a } ;(4)a = 00, a = 00, or m > 1. Show that there is a unique point q E A v(W) such that for p positive and small v maps the set of e so obtained with centers in A(a; p ) - { a }onto a “disk” punctured at q, that is, the image of a uniformizer 0 with domain a disk centered at 0 less {0(0)} and that n(q; C ) = m. Show further that every point q E A - v(W) is so associated with some (a, r , eo) satisfying the above conditions. Of course, (A, v) is understood in the sense of Lemma 2.2 relative to W. Hint: When a # m , let s satisfy a sm = b and continue the element (s, P I ) ,where P l ( t ) =P(a t m ) ,t near s, along paths in A(0; r l / ” ) - (0)issuing from s. One may proceed analogously when u = 00, using in place of a t”. 2.9. Here e = exp(1). Let the element (e, P) be such that 2.8.

e

~

+

+

+

P(e”)

= el,

near 1, and let W be the Weierstrass class containing ( e , P ) . Study the set A - v(W) where (A, v) is associated with (W, c, v ) and show that there exists a univalent conformal map of W onto C - {O}. z

3. COMPACT ANALYTIC ENTITIES AND RIEMANN SURFACES On the basis of general mathematical experience we expect situations where compactness enters to be especially tractable. This is certainly true in the study of analytic entities as we shall now see. Valence. The notion of valence for an arbitrary analytic map f of a Riemann surface S into a Riemann surface T is defined as earlier for meromorphic functions: the valence v,(q) at a point q of T is simply

Suppose now that f is a nonconstant analytic map of a compact Riemann surface S into a Riemann surface T. We wish to study the valence off. We first note that v,(q) < + co,q E T. Indeed, for each p E S, there exists an open set O ( p ) of S containingp such that f attains the point q in O ( p ) at most a t p . By the compactness of S we see that S = S k S ” O(pk). Since

u,

f-’({q))

= h:1 Ik 5 n > ,

the finiteness of v,(q) follows. From the fact that.f(S) is open and compact, we conclude that f(S) = T. Given q E T, let pi,. . . ,p m denote the points of f - ’ ( { q } ) , pi # Pk for j # k . Let Okbe a region containing pk such that vfIo,

3. Compact Analytic Entities and Riemann Surfaces

369

takes the value n(pk;f)at each point of f(O,), O j n 0, = 8 for j f k.? Since S - U, < k < n 0,is compact and f omits q on it, its image with respect to f omits a neighborhood of q. Hence vf takes the value n(pk;f) at each point of some neighborhood of q. We conclude by the connectedness of T that vf is a constant taking a positive integer as its value. Suppose now that A is a compact analytic entity for which the center function is not constant. We let m denote the value taken by vc. Since the multiplicity of C exceeds 1 only on a finite subset E of A (as we see by the argument used to prove that the valence of a nonconstant analytic map with domain a compact Riemann surface is finite-valued-here for each e E A we introduce an open subset O(e) of A containing e and such that the multiplicity of C is 1 at a point of O(e) - { e } ) ,each point of - C ( E ) has exactly m preimages with respect to C. We now show that F, the set of points w E such that

1;

e

c

{ V(e):C(e) = w }

has fewer than m members, isfinite. It suffices to show that F does not cluster at any point of 2.. Otherwise F would cluster at wo E and there would exist e l , e, E A such that C(e,) = C(e,) = wo , and sequences (p,), (4.):N + A such that P n z 4" P n f el, 4" z e, C(P")= C(qn), V P , ) = @In), n E N, and limn+mpn = e,, limn+mqn = e, . N o assertion is made up to this point concerning the equality of e, and e, . By Lemma 2.1 of this chapter, we see that el = e, . However this is not possible since a map assigning (C(e), V(e))to e whose domain is a suitably small neighborhood of el is univalent. Our assertion follows. The finiteness of Fis conformally characteristic for compact analytic entities taken with their associated C and V , C not constant, in the class of ( S ,f, g ) where S is a compact Riemann surface, f and g are meromorphic on S, and f is not constant. We shall prove the following theorem.

e

9

9

Theorem 3.1: Given a compact Riemann surface S and functions f and g meromorphic on S, f not constant, then cp of the pair (A, cp) associated with (S,J g ) by Lemma 2.2 of this chapter is a univalent map of S onto A ifand only i f the set of w E for which

c

{g(p):f(p) = wl

has fewer members than the common calue taken by vf is finite. In fact, a

t It is to be observed that a region R of a Riemann surface is itself a Riemann surface when the conformal structure is just the set of inverses o f the restrictions of the members of @ (cf. 51) with domain wv intersecting R tow, n R. It is to be understood that regions of a Riemann surface are to be endowed in this manner with a conformal structure. f l Ox is to be interpreted in these terms.

370


sufficient condition for cp to have the stated property is that the set of w in question be a proper subset of

e.

PROOF: Necessity follows on noting that

M P ) : f(P) = wl = { VCcp(p)l:CCdP)I = w > .

e

Sufficiency. There exists w E such that f - ' ( { w } ) has exactly m (the common value taken by vf) members p l , . . .,p , and g ( p j ) # g(pk) when j # k . Clearly n ( p , ; f ) = 1, k = 1 , .. . ,m . Fromf= C cp we see that n(pk;cp) = 1 k = 1, . . . ,m . Further if ~ ( p=) cp(pl), then f ( p ) = f ( p i ) ; consequently, p = P k for some k . From g(pk)= g ( p l ) , we conclude p = p l . Thus v,(e) = 1, e E A, and the proof of the sufficiency follows. 0

The problem of Riemann and Klein may be formulated as follows: Given a compact Riemann surface Sand fmeromorphic on S and not constant, does there exist g meromorphic on S such that the component cp of (A, cp) associated with ( S , f , g ) by Lemma 2.2 of this chapter maps S univalently onto A? We have not established thus far that there exists a function meromorphic on S and not constant. We shall see that this is the case and that the problem of Riemann and Klein has a positive solution (cf. §$4-6). By virtue of Theorem 3.1 we may solve the problem of Riemann and Klein by taking w E such that the number of members in f -'({w}) is precisely the common value taken by vf and constructing g meromorphic on S taking distinct values at distinct I). important consequence of the existence of a points o f f - ' ( { w } ) (cf. &An nonconstant function meromorphic on S and the solution in a positive sense of the problem of Riemann and Klein is the existence of an analytic entity conformally equivalent to S. Thus if we denote the class of Riemann surfaces conformally equivalent to S by k ( S ) , we see that the set of k(A), A compact analytic entity, and the set of k ( S ) , S compact Riemann surface, are the same.

e

Formal polynomials. In order to continue our study of compact analytic entities we shall want some elementary facts concerning formal polynomials. It is to be expected that the reader is familiar with these matters (as is doubtless the case for most notions of an algebraic character employed in this book). However, for the sake of completeness we shall sketch the basic material we want. For this as well as other algebraic notions used by the worker in such areas of complex function theory we refer to [133]. In Ex. 1.1, Ch. X11, we introduced a ring A whose elements consisted of sequences (a"):N + F where F is a given field. The subset consisting of (a,) satisfying sup{n: a,, # O.} < + 03 constitutes a subring termed the ring of formalpolynomials over F. The fact that 13 is a subring of A is readily checked.

v,

3. Compact Anaijtic Entities and Riemann Surfaces

371

Given P = (a,) E 'p, by the degree o f P , denoted d(P), is meant sup{n: a, # 0). In particular, the degree of the zero element of 'p is - 00. We have already encountered the notion of an ideal in Ch. VIII in studying the theorem of Bers; it was not important at that moment to insist on precise nomenclature. Here it will be appropriate to state some simple well-known facts in the standard language. Given a commutative ring '8, I c 3 is termed an ideal of 9? provided that I is a subgroup of 9? considered as an additive group and . : 93 x I + I. An ideal generated by a single element (that is, the intersection of the ideals containing the given element) is termed principal. For a ring 93 with identity given a E %the ideal generated by { a }is {ax: x E %}. We shall denote it by [a]. We shall want to have available the following two well-known results: (1) Given P, Q E 'p, d ( P ) 2 0 , there exist unique R, S E 'p such that Q = SP R, d ( R ) < d(P). (Division algorithm.) (2) Every ideal in '4 is principal. (1) Uniqueness follows on noting that d((PIPz)= d(P,) d(P2). Since SP + R = 0 implies S = 0 and consequently R = 0, we readily conclude the asserted uniqueness. Existence. If 0 Id(P) I d(Q), then Q = CP Q , where d ( Q l ) < d(Q). In fact, with P = (ak), Q = (bk), and 1 = d(Q) - d(P), C may be taken as (ck) where ck = 0 for k # 1 and c1 = b,,,,/ad,,, . We now proceed by induction. The case d(Q) < d(P) + 0 follows from the equality Q = OP Q. Suppose that the asserted representation holds when d(Q) < d(P) n. When we consider P and Q such that d(Q) = d(P) n, we have Q = CP Q , , d ( Q J < d(Q). Hence d ( Q l ) < d ( P ) + n. Applying the induction hypothesis we obtain the desired representation. (2) is now easy to conclude. Let I be an ideal in $' 3. If I = {0}, the assertion is immediate. Otherwise let Po be a member of I having degree equal to the smallest of the degrees of the nonzero members of I. Given P E I, we have P = QP, + R , d ( R ) < d(P,). Since R E I, R = 0. We say that P E 'p is reducible provided that there exist P I , P2 E 'p, d(pk)< d(P), k = I , 2, such that P = P,P2 . Otherwise P is termed irreducible. We term a polynomial P = (ak)of nonnegative degree dnormalized provided that ad = 1. We are now ready to consider some algebraic aspects of analytic entities. Henceforth will denote the ring of formal polynomials over thefield M(e).t Given P = (ak)E 'p and (S,kg), where S is a Riemann surface, f and g are g ) annihilates P meromorphic on S, and f is not constant, we say that (S,f, provided that

+

+

+

+

2

+

+

+

m

(ak 0 f)gk = 0.

0

t We recall that M(C) is the set of functions meromorphic on c.

XVIII. Riemann Surfaces

312

Our immediate objective is the following fundamental theorem concerning the relation between 'p as now specialized and the set of compact analytic entities for which the center function is not constant.

Theorem 3.2: There exists a unique map R from the set K of compact analytic entities for which the center function is not constant into the subset 8 of '$ consisting of the normalized irreduciblepolynomials of positive degree such that (A, C, V ) annihilates R(A).The map R is univalent and has image 8. Further, if an analytic entity A (not assumed compact) with nonconstant center function is such that (A, C, V ) annihilates a member of 8 , then A is compact. PROOF: Uniqueness. Let A be a given analytic entity such that (A, C, V ) annihilates some member of Q. Let P be a member of 'p annihilated by (A, C, V ) of least positive degree and normalized. Let Q be a member of 8 annihilated by (A, C, V ) . By the division algorithm Q = SP + R, where S, R E 'p, and d(R) < d(P). Since (A, C, V ) annihilates R, we see that R = 0. Hence there is just one member of 8 annihilated by (A, C, V ) , namely P.The uniqueness of R follows. It is to be noted that thus far the compactness of A has not been invoked. Existence. Given A E K , we introduce n 1, the common value taken by v c , and for each z of in the complement of the image with respect to C of

c

+

{e: n(e; C ) > l } u { e : V(e)= a} we define a,(z),

. .. ,a&)

by the stipulation

It is to be observed that the ak define functions analytic on the set of admitted z. To see this, it suffices to consider an admitted z , say b, to introduce for each e E C - ' ( { b } ) a neighborhood U(e) of e on which C is univalent, thereupon to introduce for each e the restriction qe of the inverse of the C I U ( e ) to nc(,)=bC[U(e)],and to observe that for z in this last set sufficiently near b the right-hand side of (3.1) is equal to (3.2)

The asserted analyticity follows. We now show that each function ak has at worst poles at the points z that were not admitted and thus is the restriction of a member of M(C), which we shall denote also by a k . Let b be a nonadmitted point. Let m denote a Mobius transformation satisfying m(b) = 0 and take p a positive integer so


373

that for each e E C - ' ( { b ] )the multiplicity of mp C a t e is at least as large as -a,(e). For z near b but different from it, we obtain from the equality 0

and the fact that (m" C ) V is analytic at each point of C - ' ( { b } ) the conclusion that m(z)"'"+- & ) ak(Z) has a removable singularity at b. Hence a, has at worst a pole at b. We propose P with a,, . . . ,a, as given, a,+l = 1, u , , + ~ +=~0, k E N. We assert that ( A , C, V ) annihilates P and that P is irreducible. Indeed, taking e, such that C(e,) is an allowed z we see from (3.1) that

'

0

[V(e)I"+'

+

5

ak[C(e)l[V(e)lk= 0,

0

e near e o , and we conclude that (A, C, V ) annihilates P. Suppose now that Q = (bk) is a normalized member of v of least positive degree which is annihilated by (A, C, V ) . We see that Vm+' +$(bkoC)Vk

=o,

0

where m + 1 = d( Q). Applying the fact established earlier that { V(e):C(e) = z } has n + 1 members save for a finite set of z, we see that we may take z such that { V ( e ) : C(e) = z } has n + I members, all different from a.Since m

for e E C-'({z}), we conclude that m 2 n and hence m = n. Using the minimal property of d ( Q ) and the division algorithm we conclude that P = Q . If P were reducible, then P would be representable as the product of two members of each having a smaller degree than P ; but (A, C, V ) would annihilate one of these factors and the minimal property of d ( P ) would be violated. Hence P is irreducible. The existence of a map 71 is thereby established. n(K) = Q. Given P = (ak) E Q, d ( P ) = n + I , we introduce

v

p' = ((k f l)ok+1)?

v

the formal derivative of P. The ideal in generated by { P , P ' } is principal, say [Q], and clearly d ( Q ) 2 0, since there are nonzero members. Since P is irreducible, d(Q) = 0, and hence [ Q ]= ql. Since the ideal generated by { P , P ' } consists o f the polynomials of the form AP + BP', A and B in 13, we

374


conclude that, for some A and 8, AP

+ BP' = l . t

(3.3)

Let R denote the set of z E C such that the components of A , B, and P are all analytic at z. The complement - R is finite. The set of roots of

e

n

has exactly n + 1 members for each z E R as a consequence of (3.3). We may control the size of the roots of (3.4) for z near a point b E - R. Indeed, introducing as above a Mobius transformation m , m(b) = 0, we see that mP(n + 1 - k ) ak takes the value 0 at b, k = 0 , . . ., n, when p is a suitably chosen positive integer. By the theorem of Roucht the roots of

e

n

are all arbitrarily small when Z(E R)is sufficiently near b. But for z E 0 near b the roots of this equation are simply the roots of (3.4) multiplied by [m(z)Iw. We shall use this observation in the next paragraph. We fix zo E R and conclude, using the local existence theorem for algebroid equations, Ex. 9.9, Ch. V, the existence of an element (2, , g ) E (5 satisfying n

[g(z)ln+l +

2 ak(Z)[g(z)lk

= 0,

(3.6)

0

z near zo . The first part of the following argument will proceed in the milieu of a Weierstrass class and its conclusion will be translated into the terms of the associated analytic entity. We consider an arbitrary path y in R with initial point zo and show that (zo, g ) admits continuation along y. The argument is a standard one and proceeds by considering the supremum of an appropriate set of real numbers. We take [0, 13 as the domain of y and let 7 be the supremum of the numbers t in 10, 11 such that ( z o ,g ) admits continuation along y I [0, t ] . Clearly, 0 < T I 1. Let K denote the union of such continuations. The domain of K is [0, 71 or [0, 7[.Now lim, u K exists and is one of the roots of (3.4) with z = y(7) as we see by appeal to the connectedness of a semiinterval and the continuity of v . The element (y(7), h), where h[y(t)]= lim, v o K and h satisfies (3.6) replacing g for z near y(z), has the property that K ( t ) is a direct continuation of it for t near 7. Hence 7 = 1. For otherwise, by defining q ( s ) as K(s), 0 s s < z, and as the direct continuation of (y(r), h) for s near 7, we would conclude the existence of an allowed t greater than z. Further the 0

t Here 1 = the identity of the ring!$.


375

domain of K is [0, I ] as we see on reference to K ~ Thus . (zo ,g) admits continuation along y. We are led to the following conclusion: m, the number of elements with center z E R which are continuations of (zo,g) along paths in R, is independent of z and does not exceed n + 1. By a paraphrase of our earlier arguments concerning the existence of a polynomial annihilated by a compact analytic entity with nonconstant center function we see that the coefficients b, specified by

2

mW"

+

1

bk(Z) W k =

[W

- U(e)],

0

w E C, where z E R and the product is taken over the elements e with center z E R which are continuations of (zo,g ) along paths in R, are analytic on R. By the observation concerning the roots of (3.5) we see that the coefficients b, have removable singularities or poles at the points of - R. We denote the member of M(e) whose restriction to SZ is 6, also by bk and introduce the polynomial Q whose kth component is bk , k = 0, .. ., m - 1, whose mth component is 1, and whose others are 0. The ideal generated by {P, Q } is principal. Let it be generated by R . Now d(R) = m since, if R = (ck), we have C; ck(z) wk = O when

c

2 bk(Z)

m- 1

W"

4-

Wk

=0

0

save possibly for a finite set of z in R, and also 0 I d(R) I m.Hence the ideal generated by {P,Q} is generated by Q and we conclude that P = Q. Now let W denote the Weierstrass class containing ( z o ,9). We note that, for each z E R, C - ' ( { Z } ) contains exactly n + 1 elements (c having domain W). Indeed, this follows since c-'(C - R) clusters at no point of W and hence c-'(R) is arcwise connected. Let A be the analytic entity and cp the univalent analytic map of W into A such that c = C o cp, u = V cp. Since A - q ( W ) clusters at no point of A, the valence of the center function C for A does not exceed n 1 anywhere. It takes the value n + 1 at each point of R. We wish to show that vc takes the value n + 1 at each point 5 E - R. We introduce a disk A((; r ) , r < + 00, such that A(C; r ) - { c } c R and note that if e is a given element of W with c(e) E A(C; r ) - (0, then e may be continued along all paths in A([; r ) - (5) issuing from c(e). Let W(e) denote the set of elements which are obtained from e by continuation along such paths, that is, the union of the images of such continuations. If e l , e2 E W, and c(el), c(e2)E A(5; r ) - {c}, then either W(e,)= W(e,) or else W(el) n W(e,) = 0. Let W , , . . . , W , denote the W(e), W j # W, when j # k. The number, &, of elements contained in w k and 0

e

+

376

X Vlll. Riemann Surfaces

having a given center in A(C; r ) - (5) is independent of the center. Further C ' ; p k = n + 1. By the argument of Ex. 2.8, this chapter, there exists a point q k E A such that cp maps wk onto 8[A(O; l)] - { q k } where 8 is some uniformher for A with domain A(0; 1) satisfying O(0) = q k . It is to be noted that the condition (3) of the cited exercise is indeed fulfilled. This may be verified with the aid of the observation that if limn-tmv(e,) = a E C where enE W j,limn+mc(e,) = C, then a is a root of bk w k = 0, bk being the value at ( of the kth component of LP where d(L) = 0 and the divisor at 5 of the zeroth component of L is max{ -dJ(): k E N}. Here P = (ak). If (3) is not fulfilled, the set of such limit values a is infinite.? We have n(qk;C ) = C(k but in contrast to the situation of that exercise we must reckon with the possibility: pk = 1, C(qk) # CO, V(qk) # 0 . Further qj # q k when j # k, otherwise W j n wk # 0.We conclude that vc(() = n 1. Thus vc is constant. We observe that the following general result holds: If there exists an analytic map f of a Riemann surface S into a compact Riemann surface T such that vf is ajnite-valued constant, then S is compact. We may see this as follows. Given q E T, there exists an open neighborhood O(q) of q such that f - [O(q)] is relatively compact, for a relatively compact open subset of S containing f -'({q}) contains all the preimages with respect to f of points sufficiently near q. To see this it suffices to consider the behavior off on small disjoint neighborhoods of the points off -'({q}). By the compactness of T there exists a finite set of such O(q) with union T. The union of their preimages is S and the compactness of S follows. Applying this result to A, we see that A is compact. Using the fact that (3.6) holds we see that (A, C, V ) annihilates P. In other words n(K) = Q. Suppose now that B is an analytic entity with nonconstant C such that (B, C, V ) annihilates P. We let qo E B be such that C(qo)E R and V(qo)# co. If (C(qo),g) E (E is such that g[C(qo)]= V(qo)and g satisfies (3.6) for z near C(qo), we have g[C(q)] = V(q) for q near qo . Now (C(qo),g) E W. Hence qo E A and consequently A = B. We conclude that 'II is univalent. It is to be noted that B is not assumed to be compact. Theorem 3.2 follows. We now turn to an examination of the algebraic structure of M(A) where A is a compact analytic entity with nonconstant center function. Our object is to show that M(A) is isomorphic to a field constructed in a purely algebraic way from 1, and Q = n(A). We shall want to review some simple algebraic facts before proceeding. We introduce an equivalence relation in by the stipulation that P , is to be equivalent to P, provided that P, - P I is in

+

t This part of Theorem 3.2 can be treated elegantly with the aid of information concerning the covering maps associated with a punctured disk, but the exposition would take us outside of the program of this book.


377

[Q].We have a bona fide equivalence relation between members of 'p. Let k ( P ) denote the equivalence class containing P. We introduce ' p / [ Q ]the , set of k(P). We define an addition on by

'$/[el

W,) +W

2 )

= k(P1

+ P21,

and a multiplication by

W , )W,)= WIP,). The operations so defined are functions on the Cartesian product of ' p / [ Q ] with itself and '$/[Q] taken with these operations constitutes a field. These assertions are easily verified, and, of course, are well known. The main point to be noted is that if P does not belong to [ Q ]there exist A , B E 'p such that A P + BQ = 1. We have k(A) k ( P ) = k ( l ) , which is the identity element of 'P/[Ql. We now define a map cp: ' p / [ Q ]+M(A) by the stipulation cp =

{(W, h,): p E 'ul

where

v,

P = (bk) E and assert that cp is an isomorphism o f ' p / [ Q ]onto M(A). Here C and V are the center and value functions on A. We note that cp is afunction, for if P , and P, are equivalent members of 'u, h,, = h,,. If h,, = h,,, then h P 2 - P = , 0. Now if h, = 0, we see on applying the division algorithm that P is in [ Q ] , for otherwise (A, C, V ) would annihilate an irreducible polynomial of positive degree less than d( Q). We conclude that cp is univalent. Since h,, h,, = h,, + p2 and h,,h,, = hPIPZ,we see that cp preserves addition and multiplication. There remains to be verified that M(A) = (p('p/[Q]).This assertion will follow on showing that iff. M(A), there exist bo , . . . , b, E M(e) such that

+

n

Here n + 1 = d(Q). We take R as the set of z in C such that ( p : C(p) = z } has n + 1 members at each of which f and V take values in C and such that { V ( p ) :C(p) = z } has n + 1 members. We have

Here (ak)= Q and d o , . . . ,dn are analytic on R. Indeed, given q such that


378

C(q) = z we see thatf(q) is the residue at V(q)of the rational function defined by each side of (3.7), and hence

Here Q' is the formal derivative of Q. Now let ql, . . . ,qn+l denote the members of C-'({z0}) where zo E R and let (Pk denote the inverse of C restricted to a neighborhood of q k on which C is univalent. For z near zo using the oo-Laurent coefficients of

we represent the dk(z) in terms of these coefficients and the ak(z), and conclude that do, . . . , d,, are analytic at zo and consequently on R. The proof will be complete once we show that the dkhave removable singularities or poles at the points of - R, for k(Q') has an inverse so that there exists S E Cp such that hsh,. = 1 and by (3.8) we have

c

for some T E Cp.t To treat the behavior of the dk at a point [ E - R, we use again the now familiar device of introducing an auxiliary factor with the aid of a Mobius transformation m satisfying m([)= 0. We take p a positive integer such that (mP C ) f and (mP0 C ) V are analytic at each point of C - ' ( { [ } ) . For z in A([; r ) - {[}, r small, we have

c

0

where

and Now the dk are analytic in A([; r) - {C}. They have removable singularities at [,as we see using co-Laurent coefficients once more. The asserted behavior of the dk follows.

t Here the dk are construed in the obvious way as members of M(e).

4. The Problem of Riemann and Klein

379

EXERCISES 3.1. Suppose that a o , . .. , a n ,n E N, are analytic on an annulus { r < IzI < $1, m, and that for each z in the annulus w" X: ak(z)wk = 0 has n 1 distinct roots. Show that there exists a positive integer p such that for w E C, 0 IY < s 5 r l / l r < IzI

+

+

+

< sl/lr,we have

n

n+ 1

where b,, . . . , are analytic on {rile < IzI < sl/lr}.Show that (n + l)! is an allowable choice of p. 3.2. Develop the theory of formal polynomials over M(A), the field of functions meromorphic on a region A c Show for such a polynomial of positive degree 9)wheref(S) = A which is irreducible the existence of an annihilating triple (S,f, and Y,(z) = n + 1, z E A, n + 1 being the degree of the polynomial.

e.

4. THE PROBLEM OF RIEMANN AND KLEIN In §§5 and 6 we shall establish the existence of a function meromorphic on a Riemann surface S and having the property that it takes distinct values at points u, b E S where a # b. Such a function is said to distinguish a and b. In the present section we shall explore the consequences of the existence for given distinct a, b E S of functions meromorphic on S which distinguish a and b. Trivially, we are assured of the existence of functions meromorphic on S which are not constant. But we have more: given a compact Riemann surface S and f merornorphic on S and not constant, we may construct g meromorphic on S (and nonconstant) such that ( S , J g) satisfies the criterion of Theorem 3.1 of this chapter. The problem of Riernann and Klein is thereby solved. Construction o f g . Let p denote the common value of the valence off. If p = 1, we may take g = J Suppose that p > 1 and let w E be such that f-'({w}) contains exactly p members. Given a, b distinct and satisfying f ( a ) = f ( b ) = w, by our assumption concerning merornorphic distinguishing functions there exists h meromorphic on S satisfying h(a) # h(b). Let m denote a Mobius transformation satisfying m[h(a)]= 1, rn[h(b)]= 0, and m(a) = GO where c( is some point of - { h ( p ) :f ( p ) = w}. The function =m h takes finite values at the points off-'({w}), the value 1 at a and the value 0 at b. On taking the product of functions ha,*,a fixed and b ranging over the remaining points of f - ' ( { w } ) , we obtain a function ha meromorphic on S taking the value I at a and the value 0 at the remaining points off-'({w}). If

c

e

0


380

2 is a univalent map off -'({w}) into C, then

is an admitted g.

5. DISTINGUISHING MEROMORPHIC FUNCTIONS We propose to reduce the problem of the existence of distinguishing meromorphic functions to a problem concerning the existence of a harmonic function with a singularity (of a nature to be specified) at a given point of a Riemann surface. The latter problem will be solved with the aid of Perron methods in $6. Our first step will be the introduction of the notion of an abelian (or meromorphic) differential on a Riemann surface. The concept is of basic importance throughout the theory of Riemann surfaces. Let 0 be the conformal structure on a Riemann surface S. By an abelian differential on S is meant a map w with domain 0 such that (1) w(8), the image with respect to w of 8 E 0, is a function meromorphic on the domain of 8, (2) if the images of members 8' and 8, of 0 have a nonempty intersection 0 and if a = {(K'(P), GYP)): P E 01,

and A = 8;'(0), then

@(el)I A = [ ~ ( e , ) ala'. 0

(5.1)

An example of special interest is the differential, df, of a function f meromorphic on S given by the stipulation df(6) = ( f 0 d)', 8 E 0.The relation (5.1) is seen to hold for df by virtue of the equality 0-0

el)

IA =

(3-0

82)

0

a

and the chain rule. It is simple to establish the following properties of abelian differentials. (1) Given an abelian differential w on S, either every o(8) is the constant zero or none is. To see this we note that if o(8,) is the constant zero and 4 8 , ) is not, then the images of 8, and O2 are disjoint. From the connectedness of S only one of the asserted possibilities can occur, for otherwise S would admit a splitting. When every w(8) is the constant zero, we term o the zero differential. (2) Given an abelian differential o on S and el, O2 E 0 whose images have a point q in common, then the divisor of o(8,) at 8;'(q) is equal to the divisor of 4 8 2 ) at 8,'(4).

5. Distinguishing Meromorphic Functions

38 1

We define d ( q ; w ) , the divisor of w at q E S, as the common value of the divisor of w(0) at O - l ( q ) where 0 E 0 has an image containing q. (3) Given w1 and o2, abelian differentials on S, o2being not the zero different ial. Then

is a function meromorphic on S satisfying df(4)=

01)

- &;Iw2),

4 E s.

The verification is immediate on referring to (5.l).? We denote f by w l / w 2 and term it the quotient, w , divided by 0 2 . (4) Given a, b distinct points of S and w1 and w 2 , abelian differentials on S satisfying d(a;wI)< 0 5 d(b; wl) and d(b;w 2 ) < 0 d(a; w 2 ) , then w l / w 2 distinguishes a and b. ( 5 ) Suppose that a E S, that u is harmonic on S - {a},and that in terms of a uniformizer 0 , a(0) = a, u[a(z)] admits for z small and not 0 a representation of the form c log IzI

+ Ref(z)

where c E R and f has a pole of order n at 0. Then w defined by w(0) = ( u O), O

9

the value being by convention co at &'(a) for 0 with image containing a, is an abelian differential on S satisfying d(a;w ) = -(n + l), d ( p ; w ) 2 0, p # a. A more detailed examination of abelian differentials is, of course, essential for a full-scale study of Riemann surface theory. We have just isolated some properties that we shall want. It is now clear that if for arbitrary a E S there exists a harmonic function u with the stated behavior, then for given distinct a, b E S we are led to differentials wl, w2 such that o l / w z distinguishes a and b. Here (3), (4), and ( 5 ) are relevant. The following exercises, which are not needed in the remaining sections of the book, pursue the subject of abelian differentials further and are offered for the interested reader.

EXERCISES Given W , an abelian differential on S, a E S, lying in the images of el, O2 E 0. Show that the residue of ~ ( 0 , at ) Oil(u) is equal to the residue of w(0,) at &'(a). The common value of the residue of ~ ( 0at) O - l ( a )for 8 E 0 whose images contain

5.1.

t Here by convention (+to)

-a =

+ to when a t R .


382

a is termed the residue of w at a and is denoted resca; w). Relate the notion of the residue of a meromorphic function on a region A c C at a point of A with the notion of the residue of an abelian differential at the point by introducing (1) 0 as the set consisting of 8,, the identity map on A n C,and 8,, Z - I with domain theimage of A - {O} with respect to z - ~ (2) , w ( & ) =f I (A n C)and w(&) = ( f 0 8,)e2’ (cf. 43, Ch. VI). 5.2. Given w as in Ex. 5.1 andfmeromorphic on S, the map assigning to 6 E Q the function ( f o 8) w(8) is an abelian differential on S. It is denoted fw. If w is not the zero differential and w1 is an abelian differential on S. then w , = (wl/w)w. 5.3. Given a compact Riemann surface S, the sum

is the same for all abelian differentials w on S which are not the zero differential. The value in question will be denoted by x(S). [x(S) has topological significance. It is equal to the Euler-Poincare characteristic of S, if we follow one of the two definitions given for the Euler-PoincarB characteristic, and otherwise it is (- 1) times the Euler-Poincare characteristic of S.]Show that = -2 by calculating the sum in question for the differential of the identity map. Determine x(A) for the analytic entity A associated with (C, p, p’) in the sense of Lemma 2.2 of this chapter, p being a pe-function. 5.4. The sum of abelian differentials on S, say w1 and w 2 , is defined as the map assigning w , ( Q w2(8) to 8 E 0. It is evidently an abelian differential on S. We denote it by w1 w 2 . Show that with addition so defined and cw, c E C, defined as the map assigning c w(8) to 8, the set of abelian differentials on S is a vector space over C. *5.5. Let S be a compact Riemann surface and let a: S + Z be such that {p:.a(p) # O} is finite. Show that %,(a), the set of functions f meromorphic on S which satisfy

x(e)

+ +

+ .(P)

2 0,

P E S,

together with addition and multiplication by scalars as defined for meromorphic functions, is a finite-dimensional vector space over C. Show that a corresponding statement holds for %*(a), the set of abelian differentials on S satisfying a(p;w)+a(p)>O,

PES,

the definitions of Ex. 5.4 prevailing. Hint: For meromorphic functions consider %,(a+) and examine the Laurent expansions of f E %l(a+)composed with a uniformizer u, a(0) = a, for a satisfying .(a) > 0. With ( c k ( f ) )the 0-Laurent sequence of f o u consider the null space and rank of the linear transformation mapping f into ( C - ~ ( / ) ) L E N and proceed by induction on the number of points in {a(p) > O}. Reduce the problem for differentials to that for meromorphic functions by considering, in case %,(a) has a member w not the zero differential, the space off satisfying

%(PI

+

w)

+ a(p)l> 0,

P E S,

and noting that the dimension of the space of such f is equal to the dimension of %2(a). We are presupposing the usual standard elementary facts concerning linear vector spaces and linear transformations (cf. [SS]).

6. Existence Theorem for Auxiliary Harmonic Functions

383

Our object in introducing this exercise and Exs. 5.3 and 5.4 is to state (without proof) the Riemann-Roch theorem, which is one of the most important theorems concerning compact Riemann surfaces: x(S) 2 ‘

(5.3)

Here “dim” means “dimension.” For proofs, cf. [2], [113]. It is interesting to see what information is yielded by the Riemann-Roch theorem. On taking a as the constant zero, we see that the left side of (5.3) is 1 and thus the dimension of the space of abelian differentials on S with divisors taking nonnegative values (the analytic abelian differentials) is [2 x(S)]/2. If a(a) < - [l x(S)], a@) = 0 for p # a, then dim B2(a)= 0 and

+

+

dim In particular, when .(a)

-a)

= - [2

=

[l

-

.(a)

+ x(S)1/2 2 [2 + x(S)1/2.

+ x(S)], a(p) = 0 for p # a, then

dim % ( - a) = [2

+ x(S)1/2.

By consideringthe linear transformation introduced above taking finto which is (1, 1) on %1(-a) for this special choice of a, show that

{-a&): consists of precisely [2

fE84-4

(~-~(f))k

E~

,

- {Oil

+ x(S)]/2 members of the section

Complex Function Theory

Complex Function Theory

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Notes on Complex Function Theory

Advances in Complex Function Theory

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Geometric Function Theory: Explorations in Complex Analysis

Handbook of Complex Analysis: Geometric Function Theory

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An Introduction to Complex Function Theory

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Handbook of Complex Analysis: Geometric Function Theory

Geometric Function Theory, Explorations in Complex Analysis

Function theory in several complex variables

Classical topics in complex function theory

Lectures on Complex Function Algebras

Function Theory of One Complex Variable: Third Edition

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Complex Analysis A Modern First Course in Function Theory

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