Complex Analysis: An Invitation: A Concise Introduction to Complex Function Theory

IL

A Concise Introduction to Complex Function Theory

COMPLEX ANALYSIS AN INVITATION

Murali Rao Henrik

S

COMPLEX ANALYSIS AN INVITATION

A Concise Introduction to Complex Function Theory

COMPLEX ANALYSIS AN INVITATION

Murali Rao Department of Mathematics University of Florida USA

Henrlk Institute of Mathematics University of Aarhus Denmark

%bworId Scientific

Singapore • New Jersey • London • I-long Kong

Published by World Scientific Publishing Co. Pie. Lid.

P0 Box 128, Farrer Road, Singapore 9128 USA office: 687 Hattwefl Street, Teaneck, NJ 07666 UK office: 73 Lynton Mead, Toucridge, London N20 8DH

Library of Congress Cataloging-in-Publication data is available.

COMPLEX ANALYSIS. An Invitation CopyTig$n e

1991

by World Scientific Publishing Co. Pie. Lid.

All rights reserved. This book, or pilLs thereof. may not be in any form orbyanymeans, electronic or mechanical. includisg photocopying, recording orany storage and retrieval system now known or to be without written pirmi sswnfrcen the Publisher.

ISBN 981-02-03756 981-02-0376-4 (pbk)

Printed in Singapore by JBW Printers & Binders Pte. Ltd.

Preface This textbook is a rigorous introduction to the theory of functions of one complex variable. Much of the material has been used in both graduate and undergraduate courses at Aarhus University. ft is our impression that courses in complex function theory these days at many places axe postponed to leave space for other topics like point set topology and measure theory. Thus we have felt free to assume that the students as background have some point set topology and calculus of several variables, and that they understand

e6 arguments. From Chapter X on we even use Lebesgue's dominated and monotone convergence theorems freely. But the proofs are meant for the students and so they are fairly detailed. We have made an effort to whenever possible to give references to literature that is accessible for the students and/or puts the theory in perspective, both mathematically and otherwise. E.g. to the Chauvenet prize winning paper [Za] and to the fascinating gossip on Bloch's life in [Ca] and its sequel [CFJ. Our goal is not to compete with the existing excellent textbooks for historical notes and remarks or for wealth of material. For that we refer the interested reader to, say the monumental work [Bu] and the classic [SG].

We have tried to reach some of the deeper and more interesting results (Picard's theorems, Riemann's mapping theorem, Runge's approximation theorems) rather early, and nevertheless to give the very basic theory an adequate treatment. Standard notation is enforced throughout. A possible exception is that B[a,r] is the closed ball with center a and radius r in analogy with the notation for a closed interval. An important part of any course is the set of exercises. We have exercises after

each chapter. They are meant to be doable for the students, so we have quite often provided hints about how to proceed. A couple of times we have succumbed to the temptation of making a digression to an interesting topic, that will not be pursued, e.g. Tauber's theorem, Fladamard's gap theorem and the prime number theorem. We hope that the reader will be irresistibly tempted as well. The authors welcome correspondence with criticism and suggestions. In particular about literature on the level of students that axe about to start their graduate studies in mathematics.

V

Contents Preface

facts. ._.4 •

V

.............. 1

Chapter 1. Power Series ...

Section 1 Elementary

1

Section2ThetheoremsofAbelandTauber Section 3 Liouville's theorem Section 4 Important power series Section 5 Exercises

7 8

9

Chapter 2. Holomorphic and Analytic Functions .........

.13

Section 1 Basics of complex calculus Section 2 Line integrals Section 3 Exercises

13 17

22

Chapter 3. The Exponential Function, the Logarithm and the Winding Number 23

Section 1 The exponential function Section 2 Logarithm, argument and power Section 3 Existence of continuous logarithms Section 4 The winding number Section 5 Square roots Section 6 Exercises

24 28 31

35 37

Chapter 4. Basic Theory of Holomorphic Functions ... 43 50 53 58 60

Section 1 The Cauchy-Goursat integral theorem

Section 2 Selected consequences of the Cauchy integral formula Section 3 The open mapping theorem gap theorem Section 4 Section 5 Exercises

Chapter 5. Global Theory .......

.........

1

Section 1 The global Cauchy integral theorem Section 2 Simply connected sets Section 3 Exercises

Chapter 6. Isolated Singularities.....

..........

Section 1 Laurent series

71

75 77 ............

......................

79

79 Vu

Contents

Section 2 The classification of isolated singularities Section 3 The residue theorem Topic 1 The statement Topic 2 Example A Topic 3 Example B Topic 4 Example C Section 4 Exercises

Section 1 Liouville's and Casorati-Weiersuass' theorems

.82 84 84 85 87 89

92

99 100 106 108 112

Section 2 Picard's two theorems Section 3 Exercises Section 4 Alternative treatment

Section5Exercises

Chapter 8. Geometric Aspects and the Riemann Mapping Theorem ...................... 113 113 115 120 122 125 125

Section 1 The Riemann sphere Section 2 The Mobius transformations

Section 3 MonteL's theorem Section 4 The Ricmann mapping theorem Section 5 Primitives Section 6 Exercises

Chapter 9. Meromorphic Functions and Runge's Theorems..... Section 1 The argument principle Section 2 Rouches theorem Section 3 Runge's theorems Section 4 The inhomogeneous Cauchy-Riemann equation Section 5 Exercises

Chapter 10. Representations of Meromorphic Functions ..... Section 1 Infinite products Section 2 The Euler formula for sine factorization theorem Section 3 Section 4 The r-function Section 5 The Mittag-Leffler expansion Section 6 The g- and p-functions of Weierstrass vu'

................. 129 129 131

135 140 144 .......... 147

147 151

153 157 161

163

.165

Section 7 Exercises

Chapter 11. The Prune .169 .173

Section 1 The Riemann zeta function Section 2 Euler's product formula and zeros of ç

Section3Moreaboutthezerosofç

176

Section 4 The prime number theorem Section 5 Exercises

177 181

Chapter 12. Harmonic Functions .............

......

..........

..183

..

Section 1 Holomorphic and harmonic functions Section 2 Poisson's formula Section 3 Jensen's formula Section 4 Exercises

183 187 192

Chapter 13. Subharmonic Functions................ Section 1 Technical results on upper semicontinuous functions Section 2 Introductory properties of subharmonic functions

199 199 201

Section3Onthesetwhereu=_oo

203

Section 4 Approximation by smooth functions Section 5 Constructing subharmonic functions Section 6 Applications Topic 1 Radó's theorem Topic 2 Hardy spaces Topic 3 F. and R. Nevanlinna's theorem

205 208

195

210 210 211

215 216

Sectionl Exercises Chapter 14. Various Applications.............

........

Section 1 The Phragm6n-Lindelöf principle Section 2 The Riesz-Thorin interpolation theorem Section 3 M. Riesz' theorem Section 4 Exercises

......

........

..

219

219 221

223 229

References

231

Index

237 ix

Chapter 1 Power Series SectIon 1 Elementary facts This section contains basic results about convergence of power series. The simplest

and most important example is the geometric series

which converges absolutely for any z in the open unit disc. It emerges in so many other contexts than complex function theory that it may be considered one of the fundamental elements of mathematics (See the thought-provoking paper [Ha]).

Definition I. A power series around

E C is a formal series of the form 00 k

(1) k=O

wherethe coefficients 0k E We

Care fixed and wherez E C. =

shall very often only consider the case (2)

0, i.e.

>.akzk

because it will be obvious how to derive the general case from this more handy special case. We shall in Chapter Vl,* 1 encounter Laurent series, i.e. "power" series in which the summation ranges over Z, not just N. A Laurent series will be a a power series in z and another in the variable l/z.

Proposition 2. Considerthe power series (2), and let us assume that the set {a&(klk = 0,1,2,...) is bounded/or some C. the power series (2) converges absolutely and umformly in the Then/or any p < closed disc B[0,]] = {z E CIIzI p}.

M for some M and all k. Thus if

Proof: Because of boundedness IakII(kI Izi

p

<

1(1

then

Iakz9 < IakIICIkpkICI

k

Chapter 1.

Power Series

So the series is dominated by a convergent series (the geomethc) and the proposition follows from Weierstrass' M-test. 0

Proposition 2 says that the power series (2) either converges for all complex z or there is a number m such that sup akmk I = oo. We define the radius of convergence k

p E [0, co), of the power series (1) by (3)

p:=inf{m>olsuplakmkl=oo}

the circle of convergence as {z E CIIz — zol = p} , provided p < oo . The reason for this terminology is that (1) converges at each interior point of the circle of convergence, and diverges at each exterior point. A side remark: If z is not a complex number but, say a matrix, then the series and

may converge even if the norm of z is bigger than p; that can happen in exponentiating a matrix. Theorem 3. Let p denote the radius of convergence of the power series (I). (a) (The Cauchy-Hadamard radius of convergence formula) 1

(4)

limsuplaki k—co

(b) (The quotient formula). If

0 for all n, then an

hm p= n—co

provided that the limit exists in [0, oo] (c) (1) converges for any 11 < p uniformly on the disc B[zo, R) ,and it converges absolutely on B(zo, p). Proof. (a) We prove that 1

I

limsuplakir

—

P

k—co

by contradiction: If

urn sup

<

k—co

then

P

there exists m > p such that urn sup k—co

<

m

Section

1.

Elementary facts

So except for finitely many k we have

<

1

,

i.e.

Iakmkl is bounded,

contradicting the definition of p. Next we shall prove that

limsuplakl

S—

urn sup

S

P

k—co

or equivalently

for any b < p. In this case we note that there exists an M > 0 such that Iaklbk 5 M for all k. Hence 1

k—oo

k—co

(b) is left to the reader as Exercise 2.

0

(c) is part of Proposition 2.

As examples we note that the geometric series has radius of convergence 1, and that the series

has radius of convergence p = 00. Corollary 4.

If the power series E akzk

has

radius of convergence p. then so does the formally

differemuiaied series Eak+1(k + l)zk.

Proof Note that E ak.tl(k + l)zk converges if E a&kz' does. Then use CauchyHadainard and that lirn

k—co

0

=1.

Recalling that a uniformly convergent series of continuous functions has continuous

sum, we get Proposition 5. The sum of a power series is continuous inside its circle of convergence.

Much more is true inside, as we will discover in the next chapter, in which we discuss differentiability. The behavior of the power series at the convergence circle is more delicate, and aspects of it will be treated in §2. 3

c_ I. An Nth order polynomial which is 0 at N +1 points is identically 0. A primitive extension of this to power series, viewed as polynomials of infinite order, is the

following:

Proposition 6. Assume that the power series (1) has positive radius of convergence, and let f denote its sum.

(a)!! f(z) = coefficients

Ofor z in a set which has zo as an accwnulation point, then the are all 0.

(fi) In particular. f dezennines the coefficients uniquely.

Our final version of Proposition 6 will be the "Unique Continuation Theorem" (Theorem IV.l 1).

Proof of the proposition: (8) is an immediate consequence of (a), so we will only

do (a). We also take

0.

be a sequence from C, such that zk —,

Let

f( z,) =

=

0

0

as k

oo .

0 and

for k = 1,2,.... By the continuity of f we get no = f(0) = urn f(zk) = 0 k—oo

But then

= so

replacing 1(z) by f(z)/z and by repeating the argument we find that

Proceeding in this way we get successively that a,, =

0

for n

0, 1,2,....

=

0.

0

For more detailed information on power series we refer the reader to [KnJ.

Section 2 The theorems of Abel and Tauber In this paragraph we will in special cases study the behavior of the power series (2) on its circle of convergence. Unfonunately, in general almost anything can happen.

One can here mention a famous example, due to L Fejér, of a power series that converges uniformly, but not absolutely, on the closed unit disc. (See p. 125 if of (ZaJ or p.122 of [Hil). Of course, absolute convergence on the circle of convergence holds either everywhere or nowhere. In case of absolute convergence the power series even converges uniformly on the closed disc B[0,p] = {z E Clizi p} , where p is the radius of convergence. We are thus left with the case of nonabsolute convergence which is technically unpleasant.

Section 2. The theuems ol Abel

Thuber

Theorem 7 (Abel's theorem). Assume thai the power series

f(z) = converges for z =

Then

ular, f

E

1

and hence for

each Izi

1.

converges uniformly to 1(x)

on the

closed interval (Oil. In partic-

Ls continuous on (0,11 ,so

as x—il in

[0,11

Proof: Note that the sequence

m>k

converges to 0 as k —.

oo.

In particular it is bounded so that the series Erkzk

1. Now, for any z such that Izi

converges whenever Izi

1 and any natural number

N we have N

00

00

k=N+1

k=N+1

k=N+1

k=N+l

f(z)_>akzk= k=O

= — z)

=

from which we for z = x E 10,1] get the estimate N

00

f(x) — k=O

Thus

xk(l

sup {IrkI} k>N

—

+ IrNI sup {IrkI} + rNI k>N

k=N+l

we have for any x E [0, 1] that

2 sup {IrkI}

—

k=O

the case x =

1

being trivial.

0

Chapter 1.

Power

Examples 8. A combmadon of Abel's theorem with the identities for arctan z

=

zl<1

for Izi <

1

yields the following pretty formulas

Log2

which are called respectively Brouncker's series and Leibniz' series. Our next theorem provides a partial inverse to Abel's theorem.

Theorem 9 (Tauber's theorem).

111(z) = EaLz' converges on the unit disc, kak —' 0 as k —' oo and limf(x) exists

for x

1_ , then E 0k converges.

Proof. Introducing

and w(n) := sup {klakl}

:= k=O

we find since

1_xkk(1_x) and that

(5)

If(x)_snIElakI(1_xk)+w(n+1) kn+1

k=1

<w(0)(n+1)(1—x)+ The expression on the right hand side of (5) is of the form At + satisfies So when = B/A it equals (1—x

w(n+1)

)2

(n + 1)2w(O) 6

and when

Section 3.

Ucuvile's theorem

we have from (5) that + 1)

—

0 RemarL More difficult is Uttlewood's Tauberian theorem which requires only } is bounded. For an accessible proof (due to Karamata and Wieland) see that { p.129 if of [Za]. Even though the following identity is easy to state and prove it is often useful in manipulations with explicitly given series. We used it, without actually saying so, in the proof of Abel's theorem.

Proposldon 10 (Abel's partial swnmation formula). Let

k

ckand dk , k = O,1,•",n be complex numbers, and put 3k

k

:=

Ed1 for 1=0

= O,1,"•,n. Then

(6)

SectIon 3

>ckdk

LIouville's theorem

Liouville's theorem (that a bounded entire function is constant), is not only interesting in itself, but it is also useful in many unexpected situations. It has a generalization which says that a power series that grows ax most polynomially at oo is a polynomial. We present the generalization here:

Theorem 11 (Liouville's theorem). Let the power series

1(z) = h.we infinUe radius of convergence.

(i) 1ff is bounded then f is constant. (LI) More generally, if f for some nonnegative constants a, b and m satisfies the estimate

for oil IzI >

jf(z)I then

m. 7

b

Chapter).

Power

Of course Liouville's theorem does not hold for smooth functions of one or more real variables. For example, the function f(x) = sin x is a bounded, smooth function of one real variable, but it is certainly not constant.

Liouville's theorem can be generalized to harmonic functions (See Chapter XII. Exercise 10). Pmof. It suffices to prove (ii). The coefficient found by the formula

J

=

may for any n =

for any R >

0,

1,2,..- be

o

which crops up when we introduce the power series expansion of f on the right hand side. In particular we get the estimate

lani

<

Substituting the estimate for I from (ii) into this we find that

forall R>b Letting R .—' 00 we see that

thepolynomialf(z)=

N

n0

=

0

whenever n > m, so the power series reduces to

,whereNistheintegerpartofm.

0

For proof without calculus (!) see [Le].

SectIon 4 Important power series In this paragraph we collect the most important power series expansions. Some of them will be derived later. In particular we will discuss the exponential function in the next chapter.

Exetcises

Section 5. 00

1

IzI<1

for

n0 00

for zEC ez=>— n! n=O

e+e_ix

cosz=

2

n=o 00

—

e

S1flZ

fl

00

=

2i

— —

for z E C

(2n) (—1)

n0

n

(2n+l)!r

2n-i-1

z

C

for z

C

for

00

cosh

z=

2

n0 00

—

sinhz=

1

—

2

n0

(2n+l)!Z

2ti+1

00

z" for IzI<1 n1 00

(1

+Z)a =

for Izi <

>

n0 00

arctanz

(—1)

=>

00

arCSiflZ=>

n

for IzI <

+1

n=o

1•3•5

2.4•6

n=O

(principal branch)

1

1

(2n—1) 2n+i Z 2n

for

JzI<1

2w

J 0

00

(

=

SectIon

for

n ÷ 1)

n=0

z

C (Bessel function)

5 Exercises

1. Determine the radius of convergence and study the behavior on the circle of convergence of the following four power series 00

k=1

where

k

00

k=0

n is a positive integer.

2. Let

>0 and bn E C\{0} for n = 9

j

00

Chapter I.

Power Series

(a) Show that

hminf— n—oo


n—.oO

is Jim

(,9) Show that the radius of convergence of

,

provided

n=n

that the limit exists in [0, ooj 3. Let a 0. Show that the radius of convergence of the power series

+...

1

is

1/(eIbI).

4. Find an example of a power series 5. (a) Let

that converges only at z = 0.

be power series with radii of convergence p and

and

a respectively. Consider the product series

,

defined by

for n=0,1,•. mm (p, a) , and that

Show that its radius of convergence r satisfies r

=

(/9)

whenever

Let E a,1 and E

be convergent series of complex numbers, and let E

n=O

n=O

n—O

denote the series given by

cn=>akbfl_k for Show that

provided that E

n0

is convergent.

6. (EnestrOm-Kakeya)

(A)

< min(p,a)

If

...

ai

>0 ,

then

0 for all IzI > 1

10

Section 5.

Exercises

(Hint: Abel's partial summation formula). (B) Deduce from (A), that

> an_I >

> ai >

> 0 implies that EOL.Zk

hasall its roots in the open disc IzI <1. For extensions of the Enestrom-Kakeya result consult e.g. [DO]. The result has applications in digital signal analysis, where it is used in examinations of stability of filters. 7. In this problem we generalize the Alternating Series Theorem from the real

domain to the complex domain (point (a)). We consider a power series E

where

{ak} is a sequence of positive numbers decreasing to 0. We let 1(z) denote its sum. (a) Prove Abel's test (due to E. Picard): The series converges everywhere on the unit circle except possibly at z = 1. Hint: Abel's partial summation formula. Or use the geometrical proof in the short note [Be]. (8) Give an example in which the series diverges at z = 1 and another in which it converges at z = 1.

as z—' 8.

Let 1(z) := Eaaz' have radius of convergence p > 0. Show for each

=

that

1

and such that

1 for all n

as not a root of unity [i.e.

such

11 that

=0 uniformly in z on compact subsets of B(0, p). Deduce the Cauchy inequaluy

sup lf(z)t foreach rE]0,p(

zlr

9.

(A generalization of Abel's theorem)

Let us assume that the power series f(z) := >2akzL converges for z = hence for each Izi <

converges

.

Show

to f(z) uniformly on the set

c} u (1)

B(0,1)

(sketch it

and

1

(a) Let C > 0. Show that Eakzk

(.8) Let B E

1

that Eokzk converges to f(z) uniformly on the set

!)

{zEB(0,1)II1—zIcos9 , 10. theorem

Show by dint of an example that the assumption kak cannot be deleted. 11

—.

L—oo

0

in Tauber's

Chapter 2 Holomorphic and Analytic Functions SectIon 1 BasIcs of complex calculus

f

Definition 1. Let f be a complex valued function defined on an open set is said ta be complex differentiable at E C if the limit (1)

lim

in the complex plane.

f(zo+h)—f(zo) h

h—O

exLct.s in C.

The limit is called the complex derivative off at zo and is denoted is complex differentiable a: each point of Il. The Lv holomorphic on derivative of I or just the derivative. If C is then called the complex : —. function f there exists a holomorphic function F defined on such that F' = f. we say that F is a primitive off. III is holomorphic in all of C then f is said to be entire.

f

Like in real variable theory we find that I is continuous on an open set Q if it is holomorphic on ft By routine calculations the usual rules for differentiation also hold in the complex case:

Theorem 2. (a) 1ff andgare holomorphic on Il then so are f+gandfg, and(f + g)' =

(fg)'

f'g + fg'.

(b) Let f and g be holomorphic on ft If g is not identically 0 on c then f/g is holomorphic on the open set {z E O}, and on this set

(fV_ f'g—fg' g2 (c) The chain rule: 1ff is holomorphic on g on then the composition g o f is holomorphic on fl, and

(go f)'(z) = g'(f(z))f'(z) for all

and f(

is contained in IZ1,

E fl

As is easy to see from the theorem, the set of functions which are holomorphic on fl forms an algebra over the complex numbers. We denote this algebra

Examples. (a) Constants are holomorphic and their derivatives are 0. 13

Chapter 2.

Holomo$ic

and

Analytic Functions

is holomorphic and its derivative is 1. is holomorphic on C for n = 0,1,2,... , and on C\{0} (C) The function z for n = -.1, —2,.... In both cases with derivative (z")' = (b) The function z

z

(d) A polynomial p(z) = Eaazk is entire and p'(z) = Theorem 3.

Let 1: Q —' C

, where is an open subset of the complex plane, and write u and v are real valued functions. f= (a) 1ff it holomorphic on ci then all the first order partial derivatives of u and v exist in ci and the Cauchy-Riemann equations are satisfied there, i.e.

u

+ iv .

where

OuOv ,40u

Ov

or in a more compact notation

Of

=0, where 0 := 110

.0

+z

(b) 1ff E C'(f 1) satisfies the Cauchy-Rie,nann equations in ci then f is holomorphic

Proof (a) This is seen from (I) by letting h tend to 0 along the real and the imaginary axes respectively. (b) Left to the reader as Exercise 4.

0

Remarks on the Cauchy..Rlemann equations: The differentiability conditions in point (b) of the theorem can be relaxed considerably. One generalization is the LoomanMenchoff theorem:

Let f

C(ci). Then f

HoI(f), if Of/Ox and Of/Oy exist in all of ci and

satisfy the Cauchy-Riemann equations there (See Theorem 1.6.1 p.48 of [Na]). Another generalization: It suffices that the Cauchy-Riemann equations are satisfied in the sense of distributions (See [Tr,Theorem p.36]). For more information see [GM].

The Cauchy-Riemann equations say roughly speaking that I does not depend on

1, and so it is a function of z only. If f is holomorphic on an open set then (use the Cauchy-Riemann equations) = Igradul2 = Igradvl2 , and lgradul2 = (Ou/Ox)2 + (Ou/Oy)2. In particular, 1' = 0 throughout ci implies that f is constant on each connected component of ci. Not all functions are holomorphic: E.g. z is not. As we saw in an example above, the functions fo = 1 and fi(z) = z are entire. Hence so is each polynomial in z. And each rational function, i.e. each function of 14

Secdcn 1.

Basics at conipici calculus

form / = P/Q where P and Q are polynomials, is holomorphic off the zeros of the denominator Q. It turns out (Propositions ilLS and 11L20 of the next chapter) that continuous roots and logarithms are holomorphic, too. So are power series; we give a direct proof here. A shorter, but more sophisticated proof is given below following Lemma 12. the

ProposItion 4. The swn of a power series is holomoiphic inside its circle of convergence, and its derivative can be go: by term-by-term differentiation: If

1(z) =

—

has radius of convergence p > 0 , then I is holomorphic on B(zo, p) and

f(z) = >aRn(z

z

—

B(zo,p)

In particular all the complex derivatives f', f",... exist in B(zo, p). Proof: Observe that the formula for f makes sense by Comflaiy 1.4. Let us for convenience of writing assume that zo = 0. The technical key to the proof is the following inequality: + h)R — ZR —

+

validfornEN,h,zEC suchihat O<JhIo. f(z+h)—f(z) =

—

—

as h—eO The inequality follows from the binomial formula:

(z +

h)R

—

Zn —

— z'1 —

=

=

k=2

=

{t2 15

and Analytic Funcdons

Chapter 2.

so

I(z +

—

—

k=2

=

k=2

k=2

= (Izi +

Definition 5. Let be an open subset of C. Afuncnon f : —. C is said tobe analytic on cZ if there exists open for each point zo C disc B(zo,p) in Q in which can be written an as the swn of a power series centered at zo, i.e. I can be written

f

1(z) =

— zn)"

for z E B(:o,p)

We have already observed that the coefficients are uniquely determined by f (Proposition 1.6). By say what the coefficients are

= So,

for n

} in the power series around zn help of Proposition 4 we can even

= 0,1,2,...

restricting to the real line, we see that the power series expansion of an analytic

with

function coincides

complex numbers the always exists a C°° in fact

(Sec

its

power

function

Taylor

series

expansion. may or

series E on the

Given any

sequence (an) of

may not converge around

real line with E an

as

its

Taylor

0. There

series. Many

e.g. [Me]). But such functions will in general not be restrictions of power

series to the real line.

A consequence

of Proposition

4 is that an analytic

function

is holomorphic, a result

which is nice, but certainly not surprising. What is surprising and remarkable is that

the converse

is

true (as will be proved

assumption that I

is

in

Chapter IV). Under the (apparently) humble

complex differentiable, we shall infer that I

is

infinitely often

even that f around each point in its domain of definition can be expanded in a power series. The situation is vastly different in the case of functions on the real line. There the differentiable and

derivative of a differentiable function need not even be continuous, let alone expandable

in a power series.

complex plane the word "holomorphic" is synonymous with the word "analytic". However, until we have proved that result, we must distinguish between the two concepts. But for

functions in

the

16

Sccuon 2.

Line integrals

SectIon 2 LIne Integrals In this paragraph we introduce the line integrals which axe important tools in complex function theory. We shall until further notice not require any finer theory of integration and convergence theorems. All we need is to consider piecewise continuous functions on the real line and uniform convergence.

Definition 6. [a, bJ —. C where —oo < a < b < oo, is said A curve (i.e. a continuous map) to be a path it is piecewise thfferentiable. This means that there are points •

such that the derivative exists in the open subi,uerval Jt,, t,+i[ and extends to a for each j = 0,1,2,.... continuous function on the closed subinten'al [t1, The length 1(7) of the path is the number 1(7) :=

I

17'(t)Idt E [0, oo[

To be precise, we distinguish between the curve which is a map, and its image -(([a, bJ) which Is a compact subset of C. We will say that a complex number z belongs (or does not belong) to the curve We express the same thing by saying that 7 passes through z z E 7' (resp. z (resp. does not pass through z). The curveis saidto be closed -,'(a) = Is contained in a subset (1 of C then we will say that is a curve in (1.

If'

Examples 7. (a) The positively oriented circle

around

(t) = zo + Re" for I

C with radius R> 0, i.e. the curve (0,2x1

is a closed path with length 2,rR. We will often write Iz — zoI = R instead of . (b) The line segment [A, BJ, where A and B are complex numbers, i.e. the curve

[A,BJ:=(1—t)A+tB for IE[O,1J isapath of Length l([A,B1) = lB—Al

(c)LetA,B,CE C. and C (the order is important!) we understand the closed curve defined by

([A,BJ(t) fcrt€[0,1J [B,C](t—1) foriE[1,2] { L[C,A](t—2) 17

fortE[2,3]

chapter 2. Holomoqthic and Analytic Functions

his apath of length

= IB—AI+IC—BI+IA—Cf.

The length of a path does not depend on its parametrization. More formally:

Proposition 8. Let-y: (a,b) C bea path. monotone map of [c,dJ onto [a,b].

Ic,dj Then

[a,b] bea

1(7) = 1(70

Proof:

0

The change of variables theorem.

Definition 9. Let [a,b] —. C be apath, and letf : = -,((a,bJ) function. The line integral of f along 7 is the complex nwnber :

C be a continuous

Ji = J f(z)dz :=J To emphasize what the variable is, we sometimes write f f(w)dw or the like instead

of ff(z)dz. The next proposition collects properties of the line integral which will be useful for us in the sequel.

Proposition 10. Let (a, 6) —. C be

a

path, and let

f be

a continuous, complex-valued function

ony'. Then (a)

J f(z)dz

sup zEir

7

(b) (A primitive version of the fact that the line integral is invariant under orientation-

preserving changes of parameter). Let —oo < c < d < 00. If 4: [c, d]

(a, bJ has the form

c6(s)=ks+1,sE(c,d),wherek,1ER and if ,naps (c, d] onto (a, b]

then

Ji = sign(lc)J i 18

Section 2.

Line integrals

For two points A and B in C we have in particular

Jf=-Jf IA,Bl

(C) LetcE]a,b[

LB,A)

and define paths 7J and 72

by

:= 7(1) for t E [a,cj 72(t) := 7(1) for I E [c,bl Then

If C

Jf=Jf+Jf E [A, BJ' , where A and B are points of C, then in particular

1+

11= 1 tA,Bl

[A,CJ

fi

IC,B)

(d) If {f,j is a sequence of continuous functions on -y' that converges wufonnly on

to f, then

/ /

j

as n —+

(e) If f is defined on a neighborhood of-y and has a prisninve, say F, there, then

Ji = F(7(b)) Ifflirthennore -y is closed then ff =

—

F(7(a))

0.

7

0

Proof: Left to the reader.

We conclude this chapter by some applications of line integrals. They are investments for the future. Definition 11. A subset

oil

A of

is said to be starshaped, if there exists a point a E A such that

the segments

{tx+(1—t)aItE[0,1J}

,

xEA

are contained in A; we say that A is slarshaped with respect to a. 19

Chapter 2.

Holocnopldc and Analytic Functions

Clearly any convex set is starshaped. An interval is a simple, but important special case.

Lemma 12. Let ci be an open subset of C which Is starshaped with respect to a point a E ci.

Letg E C(ci)have:hepropersythat f g=O,wh never

in Il witha

8A

as one of its vertices. Then the fisnction

zEfZ (a,zJ

is holomorphic on Cl and G' =g. Inparticularghasaprhnitive. Proof: Let z E ci. If hE C is so small that [z,z+h) is contained in Cl then we get from our assumption that

G(z+h)—G(z)= J g—Jg= J 9=hJ9(z+th)dt (a,zJ

ts,a+hJ

0

Division through by h and use of the continuity of g at z gives us the lemma. We can now give the promised simpler proof of Proposition 4: Proof of Proposition 4: Let us, as in the earlier proof, for convenience take zo =0. The power series

g(z) converges according to Corollary 1.4 uniformly on compact subsets of B(O, p) , so term by term integration is permitted and yields

1(z)

=

J g(w)dw+ao for zEB(O,p)

I0,zJ

By Lemma 12 it now suffices to prove that the integral of g along any triangle B(O, p) vanishes. But that is easy: Indeed,

/ g(z)dz =

>aitnJ z"1dz = 0

because the integral of z"t vanishes (z"1 has a primitive, viz. zn/n). 20

in

Sec*ioz 2.

Exercises

Lemma 13. Letçó: —.

Thenthefuncuon ,

zEC\7'

is anoiytic, hence holomorphic and

f(")(z)

=

for n = 0,1,2,...

(w

If C\7' then the power series expansion off around zo converges everywhere in the open disc aivund zo with radius dist(zo, As support for our memory we note that the formula for differentiate f under the integral sign.

appears

when we

Proof:

If

—

zol
the geometric series

, then

00

converges

uniformly for w E y' to 1 W—ZO

so

writing ( W

j(w—zo)

— ZO

we find

2irf(z)= f w—t dw= J

q%(w)

0\w—zoJ

J

dw

7

W

W

ZO

dw zO

7

=

dw}(z —

(w

which exhibits the power series expansion of f around zo. 21

zo)R

0

Chapter 2. Holomoq,hic and Analylic Funcuons

SectIon 3 ExercIses 1. Let f be holomorphic on an open subset Cl of the complex plane. Define a function f on C := (a Cl) by f(z) := f(s) for z Cr.

Show

that f is holomorphic on Cr.

Suppose that f has the power series expansion f(z) = Eanzv* on Cl. Find the

power series expansion of f' on Cr. 2. Show that the function 1(z) := 1z12

has

a complex derivative at z = 0 but

nowhere else.

Is the function f(z) = entire ? 4. Let u and v be real valued functions on an open subset Cl of C. We assume that 3.

their first partial derivatives with respect to x and y exist and are continuous. Show that f := u + iv is holomorphic in Cl if the Cauchy-Riemann equations hold. 5. Show that

exp

1(z)

z

C\{0}

z=0

is a discontinuous function, that nevertheless satisfies the Cauchy-Riemann equations everywhere.

6. Show that the pair of functions

u(x,y)=22 and v(x,y)=—22 the Cauchy-Riemann equations on C\{0} Hint: You need not differentiate. 7. Let -y be a closed path in C. Show that f

satisfies

8. Assume that the power series 1(z) = f is analytic on B(O,p).

E

is purely imaginary. converges in B(O,p). Show that

Hint: 1 I

j

w—z

dwwhenlzl
lwkR 9. Let f be holomorphic on an open subset Cl of the complex plane. :J — 1,1(—+ Cl be differentiable at t = 0 Show that f o is differentiable at i = 0 and that

d(foY)(0)

22

Let

Chapter 3 The Exponential Function, the Logarithm and the Winding Number SectIon 1 The exponential function assume that the reader is familiar with the exponential function, the logarithm and the cos and sin functions on the real line. We define the exponential function exp : C —. C by We

for zEC By Theorem 1.3(c) the power series converges absolutely and uniformly on compact

subsets of C, and by Proposition 11.4 its sum 1(z) = exp (z) is entire and satisfies f' = f(O) = 1. On the real line exp reduces to the well known real exponential

f,

function.

Let us study the exponential function on the imaginary axis in C : For any r E R we have =

exp(ix) =

+

n0

m=O 2m

00

=

00

+

,n=O

(2m+1)! 2,n+t

+ 1)!

= cos S + j Sin S

:= exp (ix) maps the real line onto the

From this we see in particular that

unit circle {zECIIzI=1} most important property of the exponential function is that it couples the additive and multiplicative structures of C as follows: The

Theorem

I (The Addition Theorem). exp(a + b) = exp(a)exp(b) for all a.b E C

Proof. When we differentiate g(z) := exp(z)exp(a + b — z) we get g'(z) = so g is constant. The contents of the theorem are g(0) = g(a).

0

0

Note that exp never vanishes (by The Addition Theorem). The power series for exp has real coefficients. Therefore exp (ii) = exp (a), which by the addition theorem implies that Iexp(a)12 =

exp(a + 23

=

=

Chapter 3. The Exponential Function, the Logarithm and the Winding Numbcr

so Iexp (a)I = exp (na). From this we deduce that

the form w =

a study of the exponential function on C : Writing w E C\{O} in we can by the above results express w as

w=exp(x)exp(iB)=exp(z+i9) for some x,OER so exp(C) = C\{O}. Finally,

exp(a)=exp(b)

a—bE2iriZ

Proof: <=: An immediate consequence of The Addition Theorem.

=>: Lctexpa=expb. Thenexp(a—b)=1,andsoa—bEiR,saya—b=iO. Then cos9 + isin9 = expi8 = exp(a — b) = 1, 50 8 E 2irZ. 0 Lemma 2. Let 4,(8) := exp (16) for 8 E R. Then 4, is 1-1 on any ha If open i,uerval of the form [a, a + 2ir[ , and it is a homeomorphism of the open interval Jo, a + 2ir[ onto the arc {z E

IzI

= 1,

z

exp(ia)}

A homeomorphism is a bijection which is continuous both ways. The lemma expresses the geometrically obvious fact that the angle of a point depends continuously on the point. Proof. The only non-trivial statement is the one about the homeomorphism property. 4' is, however, clearly continuous, so left is just the continuity of the inverse of 4,, i.e. the following claim:

where 8n,8E]a,a+2ir[ then

If

—.8

90 must belong to To see this claim, let 90 be any limit point of the sequence and exp(i8) = exp (i90). Now, Oo cannot equal any of the closed interval Ea, a + the end points because exp there equals exp(ia) exp(i8), and because exp is 1-1 on we must have 8 = 8o. 0 )a,a +

Section 2 Logarithm, argument and power By § I the restriction of the exponential function to C is the familiar real exponential function. Its inverse mapping is (by definition) the logarithm, which here will be denoted

by Log. If z E C\{O} we can write

(1) z =

= exp (Log IzI) exp(iO) = exp (Log Izi + 24

iO)

Seclion 2.

Logarithm. aiBument and power

with B real. From §1 we see that 0 is unique if we restrict it to lie in any given half-open interval of length 2,r.

Definition 3. Let z

C\{O}. The sets

(1)

argz:={oeRIexp(io)=f1}

(2)

Iogz:={aeClexp(a)=z}

anti

are called the argument of z and the logarithm of z, respectively. We will, however, use this terminology in an ambiguous way: Any real nwnber in the set (2) will be called an argument of z and denoted by arg z; any complex number in (3) will be called a logarithm of z and denoted by log z. Jr will be clear from the context which interpretation we hove in mind.

The geometrical meaning of arg z is the following: Let I be the ray in C starting at the origin and passing through z. Then arg z is the set of values of the angles from the positive real semi-axis to the ray I. We see from the definitions (1) and (2) that every non-zero complex number has an infinity of logarithms and arguments. Any two arguments differ by an integer multiple of 2,r, and any two logarithms by an integer multiple of 2,ri. We can write

for zEC\{O}

(4)

Definition 4. 1ff is a never vanishing complex valued function, defined on some topological space,

then by a continuous logarithm of f (or a branch of log f) we mean any continuous complex valued function F such that exp F = f, and by a continuous argument of I (or a branch of arg f) we mean any continuous real valued function 0 such that I = Iflexp(iO). Of course, if 1: X oo[ is continuous, then Log I is an example of a branch of log 1. From the discussion above we get a uniqueness result : If f is a never-vanishing continuous complex valued function on a connected topological space, then any two continuous logarithms of f differ by a constant integer multiple of 27ri; and any two continuous arguments of f differ by a constant integer multiple of 2,r. We now turn to the question of regularity of logarithms and arguments. The following theorem tells in particular that any branch of log z automatically is holomorphic. But more is true: In Chapter IV we shall present a generalization in which we replace exp by any non-constant holomorphic function (Proposition IV. 18).

25

Chapter 3. The Exponential Function, the Logarithm and the Winding Number

Theorem S. Let f be holomorphic on an open subset Q of the complex plane, and let F be a continuous logarithm off in

Then F is also holomorphic in fl, and F' = 1,/f. Proof:

We shall show that F has a complex derivative at each point z E fl. From the defining relation expF(a) = 1(a) for any a E fl we get

as h—iO and division through by exp F(z) = 1(z) yields (5)

exp(F(z + h) — F(z))

— 1

f'(z)

as h —'

0

Expansion of the exponential function in (5) shows that

F(z+h)—F(z) h so

f'(z)

{F(z+h)—F(z)}"

1

+h

n2

it suffices to prove that

(z+

(6)

Now Ie° —

ii

for h—.0

is small. Indeed,

IaI/2 when

C° — 1

a

1

as a —. 0

Therefore we find, putting a = F(z + h) — F(z)

in (5), that IF(: + h)

—

F(:)I/IhI is

bounded for small h, i.e. there exist C > 0 and S E JO, 1[ such that IhI

By help of this we may for Ihi

+ h) — F(z)I S

<6 estimate the left hand side of (6) as follows:

26

Logaruhm. argument and power

Section 2.

Example

6.

R,

Consider, for given Go

:=

the open set E

exP(iOo)}

C\{O} I

Il denote the value of arg z in the open interval }Oo, Go + 2ir[. Then (a) z —, 0(z) is a branch of argz in (8) z —+ Log Izi + iO(z) is a branch of log z in Q. In particular it is a holomorphic

Let 0(z) for z

function.

Proof:

(a) is a consequence of the last statement of Lemma 2, while (/3) follows from

0

(a).

Example 7. Let us take = —ir in Example 6, so that fI = C\{: E RI: <0). Let Arg: for z fl denote the value of arg z in the interval — it, ir[. Then z .4rg: is a branch of arg z in il. It is called the principal branch of arg z. The corresponding continuous logarithm of z in Q is called the principal branch of log: and denoted Log: . So

Log z is holomorphic according to Theorem 5. Note that the principal branch of log z extends the function Log on R from earlier so that the notation is unambiguous. If x E R+ and nothing is specified to the contrary, we will very often be sloppy and write log x instead of the more precise term Log x.

Note also that (Log z)' =

and

Now we can define complex powers : Let a we have a countable infinity of choices 1(z)

0. For any fixed choice of log a -

exp(zloga)

is called the zh power of a. f is clearly holomorphic in the entire complex plane, i.e. is entire. The Addition Theorem tells us that f satisfies the relations

f

f(z+w)=f(z)f(w),

f(0)=1,

f(1)=a

In particular

f(n)=a" for n=0,l,2,•.• so it is natural to use the notation aX for 1(z). Note also that a" is independent of the choice of log a. We see that exp z is one of the zth powers of e, viz the one with the 27

Chapter 3. The Exponential Function, the Logarithm and the Winding Nwnbcr

choice loge = 1 , and we shall from now on often write e' instead of exp z generally, if a > 0 we shall always put

.

More

az=exp(zLoga) for :EC

Section 3 Existence of continuous logarithms In this section we shall discuss whether branches of log f exist for given function

f. Before studying general functions f we will examine the important special case 1(z) =

z. To indicate the kind of difficulties we may encounter, we present an example to the effect that log z cannot be defined in a continuous way in all of C\{0) . However. to make the reader happy again, we note that there do exist branches of log z in the complex plane minus any half-ray starting at the origin (Example 6 above).

Example 8. We claim that it is not possible to find a continuous log z on := (z E C Izi = 1) I

let alone on the punctured plane C\{0}. It suffices to prove that there does not exist any branch of arg z on S' - after all arg z is the imaginary part of log z. It is intuitively

obvious that no branch of arg z exists on S1: Let us keep track of the argument as we go around the unit circle counterclockwise starting at I. From the geometrical interpretation of the argument we see that arg z increases continuously, and when we have gone all

the way round it has increased by 2,r. And so it does not take its former value at I as it should by its continuity at 1. Thus arg z cannot be defined in a continuous way all around the unit circle. Here is a short rigorous proof of the noh-existence of arg z: As most proofs of non-existence it goes by contradiction, so we assume that there exists a continuous argument 0 : S' R , so that

z =exp(iO(z)) focal! zES1 The relation shows in particular that 0 is 1-1. Thus the mapping I : S' given by

f(z)

{1, —1},

for z E S'

—

is well-defined and continuous. But since 1(z) = —f(—z) it maps the connected space S1 onto the disconnected space {1, —1). Contradiction! 0

We shall now ueat the question of whether a given function f has a continuous logarithm log f, i.e. the existence question. We hide most of the technicalities away in the proof of the following proposition. 28

Existence of continuous logarithms

Section 3.

Proposition 9 (The homowpy lifting property). Let X be a topological space and let F : X x [0, 1) —. C\{0} be a continuous Assume that the restriction of F to X x (0) has a continuous logarithm, say X x {0) —' C. Then F has a continuous logarithm that extends Proof: We

= çii(x,0)

the abbreviation

use

for

x E X.

step. We

to verify the

will prove that it suffices

version of

following localized

the

proposition

To every x E X there exists an open neighborhood N(x) of x map N(x) x [0,1) —* C ,such that Ox(Y,O)

=

all yE

for

N(x)

,

in X, and

a continuous

and

cxp(97)=F on N(x)x(0,1) Indeed,

E

assume that

N(x) fl

92(xo,i) —

The

is

left side

function

on

version

the localized

is true.

for any x, y

E X and any

= F(xn,t) = exp(9,(xo,t)) so O,(xo,i) = 2irin(l) , where n(l) E Z

of 1, so t —. n(i) is a a constant. Taking the condition

a continuous function

[0,11 ,

hence

we

see that the constant is 0.

We have thus shown that so the collection

continuous, integer valued

=

= into account

Then

N(y) we have

and

agree

on

their

common domain

of definition, and

z E X patches together to a continuous function 9: X x [0, 1] —'

which is the desired logarithm

of

C,

F.

2nd step.

We

shall

verify the

localized

neighborhood N = N(ro) of ro

version E

X such

F(Nx [tk,tk+I)) is contained in a

We

subset of

C \

of the proposition is true,

{0) on

,

so

let

that each

k=0,1,2,•••,n—1

which

there exists

a branch

of

now define inductively continuous maps

Ok:Nx[tk,tk+1J—.Cfork=0,1,2,...,n-—1

in the following way: 29

log z

E X. By

chapter 3. The Exponential Function, the Logarithm and the Winding Number

Since there exists a branch of logz on F(N x we can find a continuous function A0 : N x [0, ti]

we can form logF there, i.e. C such that

exp(Ao) = F on N x [O,t1] Noting that

exp(Ao(x,O)) = F(x,O) = exp(44x)) so that Ao(x,O)) = 1 we define

9o(x,t) := Ao(x,t) — Ao(z,i) + çt(z) Then

: N x [0,t1) —. C is a continuous function that satisfies

exp(9o)=F ,and 9o(x,O)=çS(x) for xEN Arguing in exactly the same way we find successively for k = 1,2,.. , n—i continuous C such that functions 9k : N x [tk,tk+1) .

exp(Ok)=F Obviously

the

,and Ok(x,tk)=Ok_1(x,ik)

for zEN

9k glue together to form a continuous function 0

: N x [0, 1)

C

Remark. Proposition a covering projection

Proposition

with

0

the desired properties.

is

case of a result from point set topology, See e.g. [Sp;Theorem 2.2.3 p.671.

9 is a special

a fibration.

viz that

10.

If A is a starshaped subset of R'1 then any continuous mapping I : A —' C\{0} has a continuous logarithm. Proot With notation from Definition 11.11 we consider the continuous map F: C\{0} ,defined by Ax [0,1)

F(z,t):=f(tz+(1—t)a) for (z,t)EAx[0,1) Since x

F(z, 0) = f(a) is

by Proposition 9 so does

a constant function it has a continuous logarithm, F. A fortiori so does x —. F(z, 1) = f(z).

hence

0

As we shall see later (Theorem V.9), Proposition 10 extends to from starshaped to simply connected spaces. Here we will as an application of Proposition 10 present an interesting digression: The 2-dimensional version of the famous Brouwer's fixed point theorem (Theorem 11(c) below). 30

Section 4.

The winding number

Theorem 11.

(a) There is no Continuous mapping r: B[O,1) —.S' with the pmperty thatr(z) =:

for all z

f:

Any continuous mapping f: B[o,

a fixed point. (c) Let K be a non-empty, convex, compact subset of C. Then any continuous map K —. Khasaflxed point. (b)

1] —* B[O, 1] has

Proof: (a) Let us assume that such a map r exists. BID, 1] is starshaped, so we can by Theorem 10 find a continuous function R B[O, 11 —. C such that exp(R) = r. Restriction to S1 yields that exp (R(z)) = z for all z S'. so R is a branch of log z on S'. But such one does not exist (by Example 8), so we have arrived at a contradiction. (b) If f has no fixed point then we can construct a map r as in (a) as follows: Let r(x) be the intersection point of S' with the ray from f(x) through x. B[O, —* (c) We may without loss of generality assume that K ç B[O, 1] . Let

K be the mapping that to z

E B[O, 1] associates the nearest point to z in K. By (b) the mapping f o has a fixed point. And that point necessarily belongs to K, because the 0 image of f is contained in K.

SectIon 4 The winding number In this paragraph we introduce the notion of winding number (= index) which will turn out to be of central importance.

Definition 12. Let 7 : [a, b]

C be a closed curve. If z E

then the winding number or

index of 'y relative to z is the integer Ind7(z)

where I

log(-y(b) — z) — log(7(a)

—

z)

2,rz

log (7(1) — z) is any continuous logarithm oft —* 7(1)

Pertinent

—

z.

remarks.

(1) By Proposition 10 there exists a branch of log (7(1) — z). (2) Since any two continuous logarithms differ by a constant, we see that Ind7(z) does not depend on our choice of branch of logarithm. (3) The reason why Ind7(z) is called the winding number is the following: Let us write

log (7(1) — z) = Log

— zI + iO(t)

where 1 -.. 6(1) E arg(-y(t) — z) is a continuous argument of t —* 7(1) = O(b)—6(a)

31

—

z.

Then

Chapter 3. The Exponential Function, the Logarithm and the Wmdmg Number

Now, 0(t) is geometrically the angle between the positive half of the x-axis and the ray from z through 7(t), so Ind7(z) measures the net increase in that angle as the parameter

t ranges from a to b. Thus Ind7(z) is the number of times (counted with sign) the ray from z through 7(t) performs a full rotation (in the countemlockwise direction) as t increases from a to b, i.e. how often the curve winds around the point z. (4) The concept of winding number of a closed curve has been generalized to higher dimensions, where it is replaced by the concept of degree of a map.

Example 13. We will compute be the triangle with vertices I, and Let (defined in Example 11.7(c)) relative to the the winding number of its boundary origin 0. y

x

Note that t —' Arg 8 (i) is a continuous arg z-function along the two sides (t) is a continuous arg c-function along , and t —' Arg1O O 112,3) Iio,ij and 0 the remaining side, where we let Arg1 denote that branch of arg: that is defined in . Now C\(0, oo[ and takes values in

(ArgOA(t)

for 0 t < 1 for 1
is a continuous arg z-function along

,

so

t

=

3

1.

Example 14. Consider the positively oriented circle around zn E C with radius R, i.e. the closed curve 7 : [0, 27r) —' C , defined by

7(t) =

+ Re" 32

Secuon 4.

The winding number

It is easy to compute the winding number of -,' relative to zo: Indeed, t —. Log R + it , so by definition is a continuous logarithm along 7(t) — Ind.1(zn)

(LogR+i2ir)—(LogR+iO) =

Example 15. The three curves

2iri

:

[0,27r) —'

C\{0}

,

=

1

given by

7,(t) := exp(ijt) for j = 1, —1,2 have the same image, viz the unit circle, but nevertheless different winding numbers relative to 0, viz 1, —1 and 2. The example shows in particular that the index of a but also on how the moving of the curve curve depends not just on the image

point 7(t) traces the image 7. A trivial example: A constant curve has index 0. We have in the above examples pedantically computed the index analytically, although the results are geometrically evident. From now on we will not bother to do so. If a closed curve has so simple a look that it is geometrically obvious what its index is, then we will not elaborate on the matter. Before we list properties of the winding number we call the attention to a topological

concept that will be handy now and later. Definition 16. Let be a topological space. (a) Two closed curves 70,72 [a, b] —. ii are said to be homotopic in Q, jf there exists a continuous map (a homotopy) r: [a, b] x [0, 1] —. Q such that

r(t,0)=70(t) and 1'(t,1)=71(t) for all tE [a,b] and r(a,s) = F(b,s) for all sE [0,1] (fi) A closed

curve

is said to be

null.homotopic,

4f it is homotopic to a constant curve.

The geometrical contents of (a) are that the curve 70 continuously is deformed into 71 as the parameters increases from 0 to 1. The curves t 1'(t, s) are closed curves for each fIxed 3 E (0,1] and they form intermediate steps in the deformation of 70 into 71•

Theorem 17 (Properties of the

number). (a) The winding number is independent of the parametri zation of the basic interval as long as the orientation is unchanged: More precisely, [c, dj —+ (a, b] is continuous and = a and = b, then winding

Ind7o#(z) = Ind7(z) 33

for all z E C\7

Chapter 3. The Exponential Function, the Logarithm and the Winding Nwnber

(b) The winding number is a homotopy invariant, i.e. two closed curves and r are homotopic in C\{zo) then Ind7(zo) = (c) Let be a closed curve in C. Then the mapping z — Ind7(z) of C\7' into Z is constant on each connected component of C\7', and is is 0 on the unbounded component. (d) Let 'y be a closed path that does not pass through z E C. Then we have the

following formula for the winding number:

1 dw — 2ir&J w—z 1

Ind7(z) =

7

Proof (a) This is left to the reader. (b) Let r: [a, b) x [0,1] —' C\{zo} be a homotopy between and r. The function (t,3) r(t,s) — zo has a continuous logarithm F since its domain of definition is starshaped. Now the function F(b,s)

—

F(a,s)

= Indr(.,)(zo)

is a continuous, integer-valued function on [0,1) , hence a constant. (c) Let : [a,bJ —. C . Let E C\7' aixi let r := dist(zo,-1)

> 0. The

C\{O) given by f(z,t) 7(t) — z has a continuous logarithm F since its domain of definition is starshaped. The formula

continuous mapping f: B(zo,r) x [a,bJ

Ind7(z)

F(z,b) — F(z,a) fix z E B(zo,r) = 2,ri

reveals that Ind7(z) is a continuous function of z. Being also integer-valued it is constant on the connected components of ft is left to prove that Ind7(zo) = 0 for just one point zo in the unbounded component of C\7', say the point zo = 1 + sup {17(t)I E [a, bJ}. The homotopy

r(t,3) := (1— 3)-y(t) for (t,s) E [a,bJ x [0,11 contracts

to the constant curve r =

0,

so by homotopy invariance

Ind,(zo) =

0

(d) If h is a continuous logarithm along 7(t) — z, i.e. if exp (h(t)) = y(i) — z then we find formally by differentiation that I

'y(t)—z 34

Section 5.

Square roots

which tempts us to define the function h by

h(t):=J

y(s)—z

ds+c

a

where the constant c E C for later purposes is chosen so that exp(c) = 7(a) — Note that h is continuous in [a,b] and that h is differentiable wherever is. An easy calculation reveals that the continuous function

4(t) := (7(1) — has

a vanishing derivative at each point where 7'(t) exists, so 4 is a constant. Since = 1 by our choice of c we have thus eh(t) = 7(1) — z for all I E (a,b)

so

h is a continuous logarithm along 7(1) —

z. Finally

h(b)—h(a) —

1 —

2irz

1 dw

2iri.j w—z

0

Section 5 Square roots An imponant concept, related to the concept of logarithm, is that of

root. We concentrate on the special case of square roots and leave the generalization to nih roots to the reader. Definition 18. Let f X —i C be a complex-valued function on a topological space X. By a continuous square root of f (or a branch of we mean any complex-valued COntinUOUS function r: X C satisfying r2 = f.

An example is the standard square root function [0, oo[—* R with the usual convention that is the unique positive number r(x) satisfying r(x)2 = x. Of course is another example of a continuous square root. a —i The uniqueness of square roots is the topic for the next proposition. Then comes regularity and finally existence. Proposition 19. Let and T2 be two continuous square roots of a never vanishing function, all defined on a connected topological space. 35

Chapiar 3. The Exponential Function, the Logarithm and the Winding Number

Then either ri =

or ri =

.

In particular, if rl and T2 agree at one point

they agree everywhere. Proof:

r2(x) Thus

rj/r2 is a continuous integer-valued function on a

connected space, hence a

constant, i.e. either + 1 or —1.

0

Proposition 20.

Let 1: —' C\ {O) be a holomorphic function defined on an open subset Q of the complex plane. Then any continuous square root off is holomorphic on ft Remark. It is actually superfluous to assume that f never vanishes. See the remarks prior to Theorem IV.18. Proof: Let r be any continuous square root of f in

Let B be any open disc in Q. Now

f has a holomorphic logarithm F in B (Theorems 10 and 5), and so exp(F/2) is a holomorphic square root of f in B. But rIB is by Proposition 19 either exp (F/2) or exp(—F/2). 0 Proposition 21. Any holomorphic, never vanishing function on a starshaped open subset of C has a holomorphic square root. Proof:

Such a function has according to Proposition 10 a continuous logarithm F. A continuous square root is provided by exp (F/2) by Proposition 20.

,

and

that square root is holomorphic

0

As we shall see later (Theorem V.9), Proposition 21 extends from starshaped to simply connected sets. Examples 22. (a) The function z has on C\) — cc, 0) the continuous square root

= which on the positive axis reduces to the ordinary square root. We call this square root

for the principal square root of z. It is holomorphic according to Proposition 20. 36

Section 6.

(b) We claim that there is no continuous square root of z on all of C\{0), not even It has the be one, and consider the curve on S1. Indeed, let 72 property that = r , where

r(t) = Now 2 Ind,(O) =

=

for t E [0,1J

which contradicts the fact that the index is integer

1 ,

valued. (c) As a third example we will study the function z2 —1 and its holomorphic square

roots which enter naturally in many considerations. Consider for example arcsin z. This is a multiple valued function defined by sin w = z. On any open subset of the complex plane where w = arcsin z a holomorphic function w, satisfying sinw(z) = z , can be defined, we must have w (z) So

1

1

= cosw(z) =

—

a look at the holomorphic square root of z2 — 1 will not be a waste of time. Our result is that the function z2 — 1 on C\[—1, 1) has exactly one holomorphic

square root which is positive for z E]1,oo[. ,because — 1 on C\(—1, Proof of that: It suffices to exhibit a continuous the statements about uniqueness and holomorphy are consequences of the Propositions be the principal square root of z on C\ J — oo, OJ (described in (a) 19 and 20. Let above). If z C\[—1, 1] , then by a small computation 1—

so

we can form the composite map

z2 E C\j — 00,0) —

.

r(z)

The function

z E C\[—1, 11

is the desired continuous square root of z2 —

1

on C\I—1, 1).

Section 6 Exercises
Chapter 3. The Exponential Function, the Logarithm and the Winding Nwnber

y

x

4. A topological space X is said to be conwactibie c

if there is a continuous map

: X x [0, 1] —. X and a point xo E X such that

c(x,O)=x and c(x,l)=zü forall xEX Show that the circle 5.

is not contractive.

Show that for

<1

Hint: Theorem 5. 6.

Show that

as n—'oo uniformly for z in any compact subset of C. 7. (a) (J. Bernoulli's paradox) Is anything wrong in the following pretty argument: Taking Logs in the identity (_z)2 = we get 2Logz = 2Log(—z), so Logz = Log(—z). Compute Logi and Log(—i) (b) (c) Is the formula Log(z1z2) = Log(zj) + Log(z2) correct ? S. Let

be complex numbers such that

for j=1,2,.•,n

1a11<1 and Show that

Lo9{H(1+afr)} 38

Section 6.

Exercises

Find the values of i'. 10. Does there exist a branch of log z in

9.

(a) {ZECI1
11. Let f,g C :

f and g are holomorphic. Hint: + 92 = (1 + ig)(f — ig). 12. Let y,p : [0,1) —' C\{0} be closed curves. Show that

(a) (SB)

(-y)

= Ind,(0) + Ind_7(O) = —Ind7(0) , where (—-y)(1) = Ind.,(0) + Indp(0) ,where — —

13. Let

:

that Ind7(0) =

[0,2irJ —.

y(1

—

t)

f7(t)

if tE (0,1) jp(t—1) ifte[1,2J

C\{0} be the closed curve -y(i) := 3cost + isint. Show

1.

Let A be a starshaped subset of C. Let be a closed curve in A and let z E C\A. Show that = 0. 15. Let A : C C be an R-linear isomorphism and let -y be a closed curve in C\{0}. Show that IndA07(0) = sign(det A) 16. A merry man walks with his dog on lead 10 times around a lamp post. He 14.

never allows the dog so much leash that it can get around to the other side of the lamp post of the man. Assuming that they both return to their initial positions after the man's 10 rotations, show that also the dog has been 10 times around the lamp post. Hint: The problem is a special case of the following mathematical problem: Let 7, a: [a, b] C\{0} be two closed curves satisfying —

a(t)I

I-v(t)I for all

t 6 [a, b)

=

(note for example that and a- are homotopic). 17. (Extension of Exercise 16). Let a- : [a, bl —+ C be closed curves satisfying

Show that

17(t) — o(t)I < Show first that neither

I(t)I + Ic(t)I

for all t

E

[a,

b]

nor a passes through 0, and then that Ind,(0) = 1n4(0).

Hint:

never takes values in] — oo,0) , so we may form Log(y/a-). 18. Prove the Fundamental Theorem of Algebra: Any complex polynomial p(z) = z" + has a: least one zero + +

in C, if n 1

39

Chapter 3. The Exponential Function, the Logarithm and the Winding Number

Hint: Assume not. Choose R so large that Ip(z) — z"J <

for all Izi

=R

consider the closed curve

for tE[0,1] and combine Theorem 17(b) with Exercise 17. 19. Show that the complex valued function

1(z) = z17

—

16i sin (931z122) z12 + 2

has a zero in the disc B(O,2). [For heaven's sake, don't try to compute it!I

20. Let A be an invertible 3 x 3 matrix whose entries all are 0. Show that 113) such that A has a positive eigenvalue and a corresponding eigenvector (t't,

0,

t.'2

0, V3 0.

This result is known as the Perron-Frobenius Theorem, and it is true for n x n matrices.

Hint: Consider the compact convex set {

and

E R1 I

+ X2 + X3

= 1,

0,

0, X3 o}

the mapping x

Ax

(Ax)1 + (Ax)2 + (Ax)3

21. Let C\] — 00,0] —* C be the principal square root of z. For which values of z does the equation = z hold? 22. Consider the square root v'z2 — 1 from Example 22(c). Find — 1 for z on the imaginary axis. Show that v'z2 — 1 has a limit as z —* 0 through the open upper half plane, and find it. Same question when Z --4 0 through the open lower half plane. 23. There axe many ingenious ways of showing the non-existence of a continuous on the unit circle For the benefit of the reader we will below sketch 4 different methods. They all proceed by contradiction. So we assume that there exists a branch We r of on 51, i.e. a continuous function r satisfying r(z)2 = z for all E may without loss of generality assume that r( 1) = 1. Method 1: The one from Example 22(b). Method 2: Show that the function := r(z)/r(—z) is a constant (indeed c8(z)2 = —1). Derive a contradiction. Method 3: Consider the smooth function f(i) := r(exp(it)) for t E R. Differentiate the identity f(i)2 = exp (if)). Method 4: (a) Show first that r is a continuous injective mapping. 40

Section 6.

Exercises

(8) Show that r(S') = S' [Either by a point set topology argument ( r(S') is homeomorphic to S' and hence not starshaped (Cf Exercise 4) , or by showing that r(S') is a subgroup of the multiplicative group S']. (y) Choose a point zo E S' such that r(zo) = —1 . Obtain a contradiction using the injectivity of r2. 24. Does there exist a branch of the root of z on S' when n > 0? Exhibit a specimen if your answer is yes. 25. Let h : S' C\{0} be continuous, and consider the closed curve 7(t) := h(exp(it)) for t E [0,2ir] (a) Show that h can be written in the form h(z) = z"exp(d(z)) , where n is an integer and h(e") and C(S1) . Hint: Let F be a continuous logarithm of t n = Ind7(O), so that h(c*t) Note that the exponent

a function

=

=

:= F(t) — mt satisfies that

=

and so defines

on S1.

(8) Show that Ind7(O) is even if h is even. In particular, if h is odd then Ind7(O)

(y) Show that

0.

26. We will in this exercise present a proof of the 2-dimensional version of the so-called

Erouwer's Theorem on Invariance of Domain: Let be an open subset of the complex plane, and let f : C be continuous and 1-1. Then f(Q) is open in C. (a) Verify that the theorem is a consequence of the following Lemma. If the function f : B[0, 1] C is continuous and 1-1, then 1(0) is an interior point of f(B[0,1)). We proceed with a proof of the lemma: (fl)Showthatwemayassumethatf(0) = 0,andthatthenr := dist(0,f(S')) >0. (y) Show that it suffices to lead the following assumption to a contradiction: There exists a point zo B(0,r)\f(B[0,1)). (6) Consider now the closed curve'y(t) := for t [0,2irl. Show that 0 = Ind7(zo) = Ind7(0).

(e) Show that F: S' x [0,1]

C\{0}

,

F(z,3) := f(j-f)

given by —

is a homotopy in C\{0} between I Is' and the odd function

h(z) := F(z,1) =

—

(() Use Exercise 25(7) to arrive at the desired conclusion. 41

Chapter 3. The Exponential Function, the Logarithm and the Winding Number

27. (a) Show that there at any given time are opposite ends of the earth which have the same weather, i.e. the same temperature and the same barometer pressure. This is a meteorological interpretation of the 2-dimensional version of Borsuk-Ulam's Theorem. x E R3 I Let S2 = 1) be the unit sphere. 1ff: S2 —+ C is continuous then { there exists an such that f(xo) = f(—zo). E (fi) Prove the Borsuk-Ulam theorem. Hint: Consider the map !i : B[0, 1] —. C given by

h(x + iy) := f(z,y, s/i —

—

y2)

—

x2 — y2)

the closed curve 7(t) := h(exp(it)) in C. (-y) Prove the following : If a balloon is deflated and laid on the floor, then there exist two antipodal points which end up over the same point of the floor. and

Hol(fZ) and assume that the zeros of I all are 28. (Cf Proposition 20) Let f isolated. Let r be a continuous square root of 1. Show that r E

42

Chapter 4 Basic Theory of Holomorphic Functions While Chapter III essentially teeats continuous functions and a bit of topology of the complex plane, this chapter specializes to holomorphic functions and their surprising properties, derived from the Cauchy integral theorem. The contents of this chapter are dealt with in all introductory text books on complex function theory.

SectIon 1 The Cauchy-Goursat Integral theorem The main results of this paragraph are the Cauchy-Goursat integral theorem and

some of its surprising consequences: The integral of a holomorphic function I around a closed curve, the interior of which is contained in the domain of definition of f, is zero. A holomorphic function is analytic, i.e. it can be expanded in a power series. Theorem 1 (Cauchy-Goursat). Let be holomorphic in an open subset of the complex plane. Let a, b.c E Q and with vertices a, b and c is contained in ft Then assume that the triangle

f

Jf(z)dz =0 Proof.

C

a

b

Divide the triangle as in the figure, using the vertices and the midpoints of the segments [a,bJ, [b,cJ and tc,a). We get 4 smaller triangles and as 43

Chapter 4. Basic Theory o( Holomorphic Functions

shown in the figure. Orienting all the triangles in the counter clockwise direction it is a matter of direct verification that

J f(z)dz=> J Denoting the absolute value of the left hand side for a, we have for at least one small triangle call it that

Note that where I denotes length of the path in question. = We can repeat the process to successively obtain triangles such that

=

and

There is a point, say

f is differentiable at

in the intersection niJI

fi > 0 we have from a certain n on, in the entire triangle (*) 11(z) — f(zo) — f'(zo)(z — zo)I

so given

that — 201

The constant function f(zo) and the function z — have primitives, so (by Proposition H.1O.(e)) the line integrals of these along any closed path vanish. Thus where (*) holds: for any

= where we have used that zo for any z and that Iz — zol ( a. Since this holds for any > 0 we conclude that a = 0.

Now,

0

By a bit of ingenuity we can get a stronger looking version of Theorem I Out by allowing the function f to have singularities

Corollarj 2. Let be an open subset of C, and let f be holomorphic in Then fiirther,nore thai (z — zo)f(z) —, 0 as z —,

Jf(z)dz =0 44

Q\ 4

}.

Assume

Section 1. The Cauchy-Cioursat integral theorem

for any triangle

in Il such that zo

Proof:

c

b

a

Let 6 be a small triangle inside of & containing zn in its interior. Subdividing into smaller triangles we get by the Cauchy-Goursat theorem that

Jf(z)dz =Jf(z)dZ Let us now consider a sequence Oi, of equilateral triangles, each with Zn as its center, and with diameters shrinking to 0. By elementary geometry, when we choose By assumption so that = 1/n then = where

lf(z)I

c(z)

—' 0

Let e > 0 be given. Choosing n so large that Ic(Z)I

= The

statement

[i

of the

as z

—'

e for

=

sup

corollary

follows

we get

all z E

since this is true

for

any e >

0.

D

From Corollary 2 we deduce a remarkable and important regularity property: A holomorphic function cannot possess a minor singularity at a point. If has a point singularity it is a serious one. This result is called "The removable singularity theorem" or "Riemann's extension theorem": it

45

Basic Theory of Holomorphic Functions

chapter 4.

Theorem 3 (The removable singularity theorem). Let be a point in an open subset Q of:he complex plane, and let f : Q\{zn} C be holomorphic in 1Z\ (Zn). fin particular 4ff is bounded near then f can Jf(z — zo)f(z) —. 0 as z —+ be extended to a function which is holomorphic in all of Q.

Proof Without loss of generality we take be a triangle, = 0 in the proof. Let contained in Q, with positively oriented (Cf = 0 in its interior, and its boundary Example 111.13) so that (Theorem 111.17(d))

1=

fdw

1

2irij w—z for any z E I

each of the two singularities 0 and z of the function

For any fixed z E

on

satisfies the condition of Corollary 2. Subdividing

into two triangles, each with only

one of the singularities, we get

j

f(w)

f(z)d =

which implies that

1(z)

for any z

=

The right side is analytic and hence holomorphic off extension F may unambiguously be defined by

(1(z)

(Lemma 11.13), so the desired

for z

I.

OA

0 We see that Corollary 2 actually does not strengthen Theorem 1, because the exceptional point is not a singularity after all! The corresponding result is not true in one real variable: The function t is not differentiable at 0. During the proof of Theorem 3 we noticed the remarkable fact that a holomorphic function is analytic. We shall in Theorem 8 below prove that its power series converges in the largest possible disc. 46

Section 1. The Cauchy.Oowsat integral theoitm

As a by-product of "The removable singularity theorem" we note the following example which will be used several times, both in the near and more distant future. Example 4. If f is holomorphic in the open set

g(z) :=

and a E Q then the function for z E

1

forz=a

If'(a)

is holomorphic in all of Theorem 5 (The Cauchy Integral theorem). Let f be holomorphic in an open starshaped subset f(z)dz = 0 for each closed path in Q.

of the complex plane. Then

Proof. When we combine the Cauchy-Goursat theorem with Lemma 11.12 we see so we may refer to Proposition 11.10(e). that f has a primitive in 0

Theorem 6 (The Cauchy integral formula). Let f be holomorphic in an open starshaped set have for any closed path in the formula

Ind7(z)f(z)

Proof Fix z E

of the complex plane. Then we

bra!!

=

z E Q\'y

Then the function for w E

for w=z

tf'(z)

is according to Example 4 holomorphic in fl, so by the Cauchy integral theorem its integral along 'y is 0, i.e.

=0 Recalling the formula (Theorem 111.17(d))

Ind7(z) =

1 1 dw 2irzjI w—z

7

we get the theorem.

47

Chaper 4. Basic Theocy 01 Holomorphic Functions As a special case we note the Cauchy integral formula for a disc:

Theorem 7.

1ff is holomorphic in a neighborhood of the closed disc Bizn, r], then we have for each z E B(

r) the formula

fH' J

w—z

'w—.zo'=r

The purpose of the next chapter is to extend the Cauchy integral theorem and the Cauchy integral formula to more general domains than starshaped. The following remark (which will not be used later) points in another direction.

Remark. There is an extension of the Cauchy integral formula to general functions. It is often called the Cauchy-Greenfornuda since Green's theorem is used to prove it (For a proof, see e.g. [HO;Theorem 1.2.11): Let w be a bounded open domain in C with a smooth positively oriented boundary -y.

If u is C' in a neighborhood

u(z) =

1 u(w)

1

2irz

j7

w—z

then dw

1

—

—I ir

jx—z

dm(x) for z E w

dm denotes Lebesgue measure on C = R2. Note that the last term on the right hand side drops out if u is holomorphic, so that the formula in that case reduces to the ordinary Cauchy integral formula. A version of the Cauchy integral theorem can be derived from the Cauchy-Green theorem (replace u by (w — z)u(w) etc.): where

11 u(z)dz

j

=

_1fOu

j F(x)dm(z)

There is a striking contrast between R- and C-differentiability: There are examplea of functions on R with the property that f is continuously differentiable and f' is not differentiable anywhere. But as we noticed after Theorem 3 any C-differentiable function, i.e. a holomorphic function, is in fact even analytic and hence in particular infinitely often differentiable. In the future we will use the words analytic and holomorphic interchangeably. What is new in Theorem 8 below is that the power series converges in the largest possible disc.

48

________

Section 1. The Cauchy-Goursat integral

Theorem 8. Let f be holomorphic in an open subset fl of the complex plane. Then I is analytic in any zo E fI the power series expansion off around zo converges in the largest open disc in around zo. In particular, I is infinitely often C-differentiable and all derivatives f', f",... are holomorphic.

partial derivatives of all orders with respect to the real variables x and y exist and are continuous.

Also I

i.e.

Proof. Combine Lemma 11.13 and Theorem 7 with the uniqueness of power series expansions. For the last statement we note that L =

and

= if'

from which we by induction get ç

OxmOyn

— —

0 It may be remarked that Theorem 8 does not hold for an analytic function of a real variable. To take an example, the function

f(x)

1+x2

for xER

is analytic on all of the real line, but its power series expansion

f(z)= 1 converges only for —1 < z < 1. The explanation is that if we view f as a function of a complex variable 1(z) :=

1+'z2 for zEC

then f is not holomorphic = analytic on all of the complex plane, but only on the domain C\{i, —i} , and the biggest disc around 0 in that domain is B(O, 1). We next present a converse to the Cauchy-Goursat theorem, viz. Morera's theorem, which is often expedient in determining whether a given function is holomorphic. It can be used in situations where a direct resort to the definition - estimation of difference quotients etc. - is hopeless or at least very complicated. For a more thorough discussion of Morera's theorem we refer to the beautiful article (Za]. For the latest news see [GIl.

4°

Chapter 4.

Baiic Theory ci Holomorphic Fimcucns

Theorem 9 (Morera's theorem). Let f : — C be a continuous function on an open subset Q of the complex plane. Assume that each point z E has a neighborhood ç Il with the property that f f( z )dz = 0 for all triangles contained in U1. Then f LS holomorphic in

Proof: Let z E and choose r > 0 so small that B(z,r) c U1. Ii suffices to prove that f is holomorphic in B(z,r) for each z E f has by Lemma 11.12 a primitive F in D(z, r). Being C.differentiable means that F is holomorphic. By

0

Theorem 8 so is its derivative f.

Section 2 Selected consequences of the Cauchy integral formula It is a sad fact from the theory of functions of a real variable that a uniform limit of a sequence of differentiable functions need not be differentiable, although ii is continuous. Indeed, there exists a continuous nowhere differentiable function on 10,1], and (by the classical Weierstrass approximation theorem) it is a uniform limit of a sequence of polynomials.

The situation is quite the contrary for holomorphic functions, as demonstrated by the following surprising result, in which no assumptions are made on the derivatives: Theorem 10 (Welerstrass' theorem). be a sequence of functions which are holomorphic in an open subset Let I i, of the complex plane. Assume that the sequence converges, wufornily on each compact

subset of Q, toafunctionf. Furthermore, the sequence Then f Lc aLso holomorphic in converges, again wufonnly on each compact subset of to

of derivatives

f.

We shall later (Exercise VI11.10) meet a stronger version of Weierstrass' theorem, viz. Vitali-Porter's theorem. The assumption about uniform convergence is important; poiniwise convergence does not suffice. For a discussion see [Dii] and [Za* 111.

Proof

LetzoEQandchooser>Obesosmalltha*B[zo,r]çcl. Accordingtothe Cauchy integral formula (Theorem 7) we have (*)

1

t J

dw

50

forall

of the Cauchy integral formula

Section 2. Selected

and going to the limit n

we get

00

f

1 (**) f(z)=— 2irz

i

j

f(w) w—z

dw forall ZEB(zo,r)

I

which [by Lemma 11.131 shows that f is analytic and hence holomorphic in B(zo,r).

Now f E Hol(fl), the disc being arbitrary. Concerning the last statement we may now assume that f = 0 [if necessaiy we replace f, by — f] By a compactness argument it suffices to show that {f } converges uniformly to ç Q. Differentiating (*) we get such that B[zo, 0 as n — oo on any ball D(zo, [directly or using Lemma 11.131 that

(2dW

J

forall

1w—

z

<

E B(zo,

[Cf Proposition I1.10.(a)J:

27rr

1

=

sup 7t (r/2) Iw—zol=r

sup B(zo,r/2)

The circle {w E C 11w — zol =

r

sup r Iw—zoI=r

sup

r} is a compact subset of

convergestoOasn—ioo.

{Ifn(w)I}

so the right hand side

0

An easy corollary is the fact (Proposition 11.4) that a power series in its disc of convergence defines a holomorphic function and that its derivative can be found by term by term differentiation (Sec Exercise 1). More generally one can apply Weierstrass' theorem to infinite series, the terms of which are holomorphic functions. Example: The Riemann (-function (Exercise 26 below and Chapter XI.*1). We next generalize Proposition 1.6.

II

Theorem (The unique continuation theorem). Let Q be an open connected subset oldie complex plane, and let g E Hol((Z). (a) If the set of points z E Cl where f(z) = g(z) hasaliinit point in Cl, thenf = g. (fi) In particular, the zeros off have no limit points in Il, unless f is identically 0. We remind the reader, that a point p is a limit point of a set A, if every neighborhood

of p contains at least one point of A distinct from p. We emphasize that the limit point from (a) and (fi) must belong to Cl; limit points outside of Cl do not suffice. 51

Chaplu 4.

BasIc

Thccry c( Holomorphic Fwictions

Proof:

(fi) is a consequence of (a), so it suffices to prove (a), and it is even enough to 0. Consider the set

do so with g =

forall n=0,1,2,...}

N := Its complement in

1l\N =

•

i.e. E

çllThere exists an a such that

is continuous. On the other hand, if a is open because each from the power series expansion

o}

N•

then we see

1(z)

=

f (and hence each of its derivatives, too) is identically 0 in a neighborhood of a, so N is also open. By connectedness of Cl either N = Cl or N =0. It suffices to prove that the limit point zo E Cl belongs to N, so that the first possibility takes place. We shall in other words show that in the power series expansion 1(z) =

—

of f around zo all the coefficients

are 0.

0

But that is part of Proposition 1.6.

The unique continuation theorem is amazing: It tells us that a holoinorphic function

is determined by its values on e.g. any curve segment, be it ever so small; and that we cannot change the values of the function in one part of the domain without it being felt everywhere. As an illustrative example we mention that Log thus can be characterized uniquely as the holomorphic function on C\ I — oo, OJ that extends the ordinary real logarithm 10,001.

The unique continuation theorem implies that holomorphic functions obey the so-called "principle of permanence of functional relationships", i.e. if hotomorphic functions in a part of a region satisfy a certain relationship, then they do so everywhere. As an example we infer from the well known relation sin2x + oos2x = 1, which is true for all real x, that the same relation holds everywhere, i.e. sin2 z + z= 1 for all complex z (Exervise 16).

52

Section 3. The opai mapping theorem

DefinItion 12. Let f be holomorphic in open subset () of the complex plane and let E Q. Then f can in exactly one way be expanded In a power series (in the largest open ball B in Q) around zo: Co

k f(z)=2__ak(z—zo) for zEB

k=O

Clearly fhasazero at

if and only if an =0. Wedefinetheorderofthezeroas

sup{n+lIao=aj=...=a=0)ENU{oo) An equivalent characterization that does no: involve the power series expansion, is that the order is the biggest n such thai

f(zo) = f(zo) = ... =

=

0

We leave it to the reader to prove, that if f has a zero of order n <00 can be factorized as

f(z) = (z —

where g E

and g(zo)

,

then

f

0

1ff has a zero of infinite order then f vanishes identically near the zero (and hence on all of the connected component of fl that contains zo).

Section 3 The open mapping theorem The following continuous version of Rouché's theorem will be used in the proofs of the Open Mapping Theorem and of Bloch-Landau's theorem (Theorem VIL6). Rouché's theorem will be discussed in more details in Chapter IX.

Theorem 13 (Rouché's theorem).

Let F E C(B[0,r]) and let H be holomorphic on a neighborhood of B = B[0,r]. Assume that

(*) JF(C)

—

IH(C)I for all CE OB

and that H has a zero somewhere in B[0, r]. Then F, too, has a zero in B[0,rJ. If the inequality (C) is strict then F has a zero

in the open bail B(0,r). Proof (The proof is adapted from [TsJ). We shall derive a contradiction from the asswnption that F never vanishes in B[0, r]. Let

:= re'1 for 0 t 53

Chapter 4.

Since F

never

Basic

Fwicilons

Theory

vanishes we may divide (') by F

get that

to

for (EOB which shows that

H(C)/F(C) is closer to I than to 0. Hence

we may form

for 1(1

Fo-y we get that

so

= r. Letting be a continuous logarithm of is a continuous logarithm of Ho7 ,and

so lndHo7(O) = 1nd107(O).

The map r, defined by F(.qrea)

for

(t,s)

E

[0,2ir] x [0,1]

is a homotopy in C\{0) between the constant map F(0)

and

F

1nd1107(0) = Indpo,(0) = IndF(o)(0)

o -y,

so

0

We arrive at the desired contradiction by proving that Indj,0.7(0)

0:

We may write H in the form

H(z) = (z

— Z2)

—

(z —

zR)g(z)

z2,• .. , z, are the zeros of H in B(O, r) and where g is holomorphic in a neighborhood of B[0, r] and never vanishes in B[0, rJ. In particular

where

1

H(z)

4-'z—z,

g(z)

j=I

Now,

J

1107

7

dz

2irz j z —

2,rz

j

g(z)

7

1

Theorem 14 (The open mapping theorem).

Let f be a non-constant holomorphic function on an open connected subset complex plane. Then

f(fZ),

f(Q)

of the

is open.

It suffices to prove that there for each zo E fl exists a disc, contained in f(zo).

around

54

Sectkm 3. The open

theorem

Consider the function H(z) := f(z)—f(zo) for z Q. By the Unique Continuation Theorem l1(ft) there is a ball B[zo,rj in which the only zero of H is z = ZO. In

particular

R :=

inf IC—zi1r

{IH(OI} >0

Let now w be any point such that Jw — f(zo)j < R , and let us apply — w and H : For IC — zoI = r we get

to

the two functions F(z) = f(z) IF(C) —

= If(zo) — wI
by Rouché F has a zero in D[zo,r] ,i.e. w belongs to the range off. Since w was arbitrary in B(f(zo), R), we have proved that any element of B(f(zo), R) belongs to the range of f, or in other words that B(f(zo), I?) ç 1(Q). 0 so

A topological proof of Theorem 14 can be found in ICS]. An immediate consequence of The Open Mapping Theorem is 15 (The Maximum Modulus Principle). Let f be holomorphic in open connected subset Q of C and assume that Ill has a local maximum in CL Then f is constant in Cl.

0

Proof Make a sketch! In applications the following consequence is useful.

Corollary 16 (The Weak Maximum Principle). Let Cl be a bounded, open subset of C and let f be holomorphic on Cl and continuous on the closure of Cl. Then 111 takes its supremum value at a boundaty point of Cl.

Proof: Since f and is compact, Ill takes its supremum value M in = CluOCl. Let us assume that it happens at an interior point zo E Cl and then deduce that Ill takes the same value at some boundary point, too. By the Maximum Modulus Principle f is constant in the biggest open ball B(zo, R) in Cl with center Since the ball is the biggest, its boundary contains at least one point zi of &CZ. By continuity If(z')I = M. 0 We shall in the chapters XII and XIII extend these maximum principles to harmonic and subharmonic functions.

As an example of how to use the maximum principles we shall prove Schwarz' lemma.

55

Chaplcr 4.

f

Basic Theory of

Functions

Theorem 17 (Schwai'z' lemma). Let 1: B(O, 1) (J[O, 1] be holomorphic with f(O) = 0. Then (ür) If(z)I IzI for all z E B(0, 1), and If'(O)I 1. (fi)If there existsazo E B(0,1)\{0) such that If(zo)I = Izol ,oriflf'(O)I = 1,then has the form 1(z) = cz for all z E B(0, 1) , where c is a constant of absolwe value I.

Proof. The function

()

._ •

for zEB(0,1)\{0}

lf'(O) for z=0

is holomorphic in 13(0, 1) (Cf. Example 4 above). For fri Weak Maximum Principle that


if(w)i r


ig(z)I

max {Ig(w)I} = max f Iu1r IwI=r

1

we infer from the

r

letting r —. 1_ we get lg(:)I 1, i.e. (a). assumes its supremum in B(0, I), In each of the cases of (19) the function so by The Maximum Modulus Theorem g is a constant function, g(z) = c. And 0 = 1. This proves (j9). id =

and

The conclusion If(z)i IzI of Scbwaiz' lemma expresses that f does not increase modulus. It implies in particular that if z ranges over the subdisc 13(0, r) then the

values f(z) also lie in that disc. Such considerations form the basis for the so-called Principle of Subordination. For more information on that topic we refer the reader to the paper [Mac). Another important application of Schwarz' lemma is to the uniqueness question of conformal mappings, so we will return to Schwarz' lemma during our treatment of the Riemann Mapping Theorem. We have already earlier (Theorem 111.5) observed that any continuous function g satisfying (exp og)(z) = z, i.e. any continuous logarithm, automatically is holomorphic. And we have claimed that an assumption in Proposition 111.20 (that f never vanishes) is superfluous. Both these results axe special cases of our next theorem. Theorem 18. Let g : ci .-. C be a continuous function on an open subset ci of the complex plane, and let f be a non-constaju holomorphic function, defined on an open connected subset of the complex plane containing g(ci). If the composite map f o g is holomorphic on ci, then so is g. Proof: Since holomorphy is a Local property we may in the proof assume that fl is connected. Let a ci be arbitrary. We shall show that g is complex differentiable at a. 56

Section 3. The open mapping theorem

Since f is holomorphic around g(a) we can write

f(w) — f(g(a)) = (w — g(a))k H(w) 0. So there is an open disc D around g(a) where H is holomorphic and H(g(a)) in which H never vanishes. By Proposition 111.21 H has a holomorphic kth root fo on D. So we write f(w) — f(g(a)) = [(w — g(a))fo(w)Jk for all w near g(a)

Here k is an integer 1, fo E HoI(B(g(a), 6)) for some 6 > 0 and fo(g(a)) 0. Replacing w by g(z) and using the abbreviation h for the holomorphic function h = o g — f(g(a)), we get for z in a neighborhood of a, say for z in B(a, €) ç ci,

f

that

h(z) = [(g(z) — so we see that the bracket on the right hand side is a continuous kth root r of the holomorphic function h on the left hand side. There are now two cases: 1) h is identically 0 in B(a, e). In this case g(z) = g(a) near z = a, so g is clearly complex differentiable at a. 2) h is not identically 0 in B(a, e). In this case we know by the Unique Continuation Theorem (Theorem II) that a is an isolated zero for h. According to Proposition 111.20 the root r is holomorphic in B(a, e)\h'(O) , so a is an isolated singularity of r. But r is continuous in B(a, e), so by the Removable Singularity Theorem (Theorem 3) r is actually holomorphic in a neighborhood of a. Dividing the identity r(z) = (g(z) — g(a))fo(g(z)) by z — a and noting that r(a) = 0 we get

r(z)—r(a) =

9(Z)_9(a)f(())

The left hand side has a limit as z .-. a, and, since fo(g(a)) difference quotient of g, i.e. g is differentiable at a. Terminology 19. A function is said to be univalent,

0, so does the

0

it is holomorphic and infective.

Theorem 20.

1ff is univale,u in an open set Il, then f(1l) is open and

is holomorphic on f(1l).

Being open, Cl is a union of open balls. f is injective so it is not constant on any such ball. Hence f(cl) is open by the Open Mapping Theorem (Theorem 14). Since f is an open map, f_1 is continuous, so we can apply Theorem 18 with g = 0

f*

The reader might note that a result corresponding to Theorem 20 is false for functions of one real variable: The inverse of a differentiable function need not be differentiable (Example f(x) = x3). 57

Chapter 4. Bask Theory o( Holomorphic Functions

Theorem 21. Let f be holomorphic in an open set of the complex plane and let E there exists a neighborhood of zo on which f Is univoJetu and only if f'(zo)

Then

0.

Proot Then Let us first assume that f is univalent on the open neighborhood Q of g = ' is holomorphic on f(1Z) by Theorem 20, so we may differentiate the identity g(f(z)) = z . We find via the chain rule that g'(f(zo))f'(zo) = 1, so f'(zo) 0. Let us conversely assume that f'(zo) 0. Then the Jacobian of the mapping

I

f=u+iv : R2—iR2 0 (use the Cauchy-Riemann equations), so by the inverse function theorem f is 0 injective on a neighborhood of zo. is

Remark: Theorem 21 extends to the case of holomorphic functions of several complex variables. An induction proof on the number of variables can be found in the note [Ro].

SectIon 4 Hadamard's gap theorem As application of Theorem 8 we make an unnecessary, but nice, digression back to power series.

Theorem 22 (Hadamard's gap theorem). Let fl > 0 be the radius of convergence of the power series

1(z) = where {nk 1k = 1,2,•• .} is a sequence of natural nwnbers satisfying nk+1 (1 + O)nk

fork = 1,2,••. for some 8 > 0. Then IzI = R is a natural boundary for f in the following sense: If there is a holomorphic extension off to a connected open set Q 2(0, R). then Q = B(0, fl). Proof.

We assume for convenience that R = 1 , and denote the extension of f by F. It suffices to derive a contradiction from the assumption 11(0, 1). Under that assumption Q fl S' 0, say 1 E Q. Choose an integer p 1/8 and consider for w E C the convex combination WI' + WP+l

2

58

Section 4.

Hadamard's gap theorem

We have

zEB(o,1)ccz if IwI1

,and

and

z = 1 if w = 1 so z E fl for all w in a neighborhood of B[O, 1]. The composite function 2

is therefore defined and holomorphic in that neighborhood of B[O, 1J, and so the radius power series expansion around 0 is strictly bigger than 1. Let us of convergence of find the power series expansion of 9S: For w E B(O, 1) we have

=

(*)

+

+

+

The exponent of the last term in the kth bracket, viz. + , is strictly less than the exponent of the first term in the next bracket: Indeed, since p 1/0, we see that Pflk+1 — (pnk

+ nk) p(l + O)nk — (pnk + nk) (p0 — 1)n* > 0

which means that no power of w occurs more than once. We infer that the power series expansion of is gotten by simply omitting the brackets in (9: Indeed that power series converges, at least for <1, because

+... +

+ 00


)

2

since

+

<1

2

And by (9 the power series converges to As we mentioned above, the radius of convergence of

is strictly bigger than 1.

So its power series converges for some real number r> 1 , implying that 00

2

k=O

converges.

)

But then the power series of f converges at rk + rk4i

59

>1

chapter 4.

Basic

and so the radius of convergence R

Theoiy i of

Functions

f is strictly bigger than 1.

0

ThatcontradictsR=1. As an example we mention that the function

for zEB(O,1)

1(z) :=

to a continuous function on all of the complex plane, but that it cannot be prolonged to an analytic function on any domain bigger than B(0, 1). extends

SectIon 5 Exercises 1. In the proof of Weierstrass' theorem [Theorem 10] we used the fact that I analytic implies I holomorphic, i.e. Proposition 1L4, so the remarks about power series following Weierstrass' theorem are really part of a circular argument. Repair the argument by showing directly from (**) in the proof of Weierstrass' theorem that I is holomorphic. 2. For which n E Z will the function z —, possess a primitive 1) on all of C\{O} 2) locally on C\{O}? 3. Let Q be an open convex set, and let f be holomorphic on Il with > 0 on Q. Show that I is univalent on Q.

Hint: f(b) — 1(a) =

(6—

a)ff'(a+i(b— a))dt. 0

4. Let ')' formula

:

[a, 6] —, C\{O) be a path, and let f be continuous on

Prove the

Jf(z)dz = in

which the path : [a,bJ Show in particular that

—,

C is defined by 7_1(i) = 1/7(i).

J f(z)dz= J

:;')dZ

Compute

2,n

f(z—a)'dzfornEZ

j 1:1=,

6. Let a E C. Show that the principal branch of (1 + (1 + 0)° = 1 has the following power series expansion

(1 +

for IzI < 1

= 60

,

ie. the branch with

Section 5.

7. Let a

Excrcises

C. Is the following formula correct? =

8.

Let I and g be continuous in the closed unit disc and analytic in the open disc.

Write them as

f(z) = >anz" and g(z) =

for IzI <

1

(a) Show for any Izi < 1 that

f

Iwt=I

(fi) Show Parseval's identity 2r

00

Hint : Show first for any r

10, 1[

that

>IanI2r2n = and F. Riesz) Let f be continuous on the closed unit disc and analytic in the interior. Show that 9. (L. Fej6r

2JIf(x)I2dx Hint: Apply the Cauchy integral theorem for the semi-circular disc to the function z —, 10.

Let t E [—1,1]

(a) Show that 1 — 2tz + z2 does not vanish in Izi < 1. (fi) Show that there exists a continuous square root of z —. (1

—

2t: +

z2)' in

IzI
(i — 2tz + +

can be expanded in a power series converging for

=

61

<1

Chapter 4.

Basic Thexy of

Rolomorphic Functions

are polynomials in i of degree n (They are the so-called

(6) Show that the Legendre polynomials).

(e) Show that

is a solution of the equation — 2t1/

(1 —

+ n(n + l)y =

0

Hint: Term by term integration yields the following explicit formula for

J IzI=r

11. (a) Show that z —' exp (—z2 — 2zw) for any w E C can be expanded in a power series

exp (—z2 — 2wz)

whose

=

radius of convergence is 00.

is a polynomial in w of degree n (The so-called (8) Show that polynomial of degree n). Show that the Hermite

polynomials have

-

Herinue

the following properties:

= (—1)"H,1(—w) = —2(n + 1)H8 =0 +

=

J Hn(x)Hm(x)exp (—x2)d.r =

12.

Consider

the closed path

0

if n

in

composed of the four paths sketched below:

62

Section 5.

Exercises

y

x

(a) Show that =

I (8) Show that

—---+0 as R —' 00, and that

J

e

J

as r—.O

z

Cl

('y) Show finally that R

jI

—dx —' x

— as R —. 2

0

13. LetO
63

=

LetCandc

Chapter 4.

Basic Theory o( Holoinorphic Puncuons

y

C

x

(a) Show that

/

—. 0+

z—1

(8) Show that

T LogIl +e"Idi

z—1

—T+o

0 as v

0+

f

Show that

I

Log(sin O)dO —'

—

2 as r —i

14. Does there exist a holomorphic function on B(0, 1) that in the points 1/2, 1/3,

1/4, 1/5, ..

takes

the following values

(a) 1, 0, 1, 0, (8) 112, 0, 1/4, 0, 1/6, 0, (y) 112, 1/4, 1/4, 1/6, 1/6, 1/8, 1/8, (6) 2/3, 3/4, 4/5, 5/6, ... 15. Let f be holomorphic and non-constant in an open connected subset Q of the complex plane. Show that I is non-constant in any open subset of Q. 64

Section 5.

Exercises

16. We recall that

=

(2n+l)!Z

Show that sin2 z + cos2 z = 17. Show that

1

(—1)2,I

and cosz =

for all z E C.

J_=Log{z_Lfl

for zEC\[—1,11

and ci2 be two open subsets of the complex plane such that Q1 fl 18. Let is connected. Assume that every holomorphic function on Q, has a primitive on

for j =

1,2.

U Q2, has a primitive on

Show that every function which is holomorphic on ci1 U

19. Let f be holomorphic on an open subset Q of the complex plane. Let g be a continuous kth root of f on Show that g, too, is holomorphic on B(g(a), R). Prove the following inequality which generalizes 20. Let g : B(a, r) Schwarz' lemma: Ig(z + a) —

IzI

Derive Liouville's theorem from the inequality. 21. Show that the maximum of sin on the square [0, and is attained at z = + 2iri.

< r

x [0, 27r] equals cosh (2ir)

22. Let f be an entire, non-constant function and consider the corresponding "analytic landscape" {

(x,y,If(x+iy)12)ER3IX,yER} in R3

Show that there are no highland lakes in such a landscape: When it rains the water collects in puddles around the zeros of f, not higher up, or runs off to infinity.

65

Chapter 4. Basic

o( Hotomorphic Functions

23. Prove the Strong Maximum Principle: Let be a bounded, connected, open subset of the complex plane, and let I be a holomorphic, non-constant function on Let

M := sup Then If(z)I < Mforallz E 24. Let be a bounded open subset of the complex plane. Let {f,1) be a sequence of functions which are continuous on and holomorphic on Assume that the sequence

converges uniformly on the boundary of Show that {f,j converges uniformly on towards a function which is continuous on and holomorphic on Q. 25. Show that there exists a holomorphic function A: B(O. 1) C with A(O) = 0 satisfying the identity sin (A(z)) = z for all Izi < 1. Hint : Use Theorem 21 to get hold of A near 0. Differentiate the identity to get

A'(z) 26.

=

Show that the Riemann (-function

is holomorphic on the open half-plane {z E C I > 1). 27. Show Schwaiz' Reflection Principle: Let be a connected open set which is symmetric with respect to the real axis. Put

= {z E = {z

Let f :

U flo —. C be continuous on

> 0}

<0} U

,

holomorphic on

and such

that f(Qo) C R Then there exists exactly one function F: It satisfies

—'

C which is holomorphic on Q and

equal to f on

F(z) = F(s) for all z E 28. Assume that both f E C(R) and the function

, x ER

x

66

Section 5.

Exercises

are compactly supported. Show that

J e"f(i)dt = 0

for all x E R

29. What is the maximum of the product of the four distances between a variable point in a square and the vertices of the square? 30. Let f : B(0, 1) B(0, 1) be an analytic map which is bijective and maps 0 into 0. Show that I has the form 1(z) = cz for some constant c E S' constitute a Why doesn't the function z —. 31. Show that k/TI = counter example to the Removable Singularity Theorem? 32. Let f be a non-constant entire function. Show that the range of I is dense in the complex plane. Hint: Assume not. Use Liouville. C, is of exponential 33. We will say that an analytic function f, defined in ci type if there is a T E R and a constant C > 0 such that

If(z)I

S

CeTN for all z E ci

Prove

The Phragmdn-Lindelbf Theorem: 1ff is analytic and of exponetufal type in a sector ci of angular opening strictly less than x, iff is also condnuous in the closed sector ?L and if If I < M for some constant M on the boundaiy of 1.1, then (f I < M throughout ci.

34. Let f be holomorphic on an open subset ci of the complex plane. [a,bj —' ci be paths such that = P,Q E ci and let = P 70(b) = 71(b) = Q. :

Let and

Assume furthermore that are fixed end point homotopic in ci, i.e. there and exists a continuous map H : [a, b] x [0,11 ci such that

H(a, s)

P,

H(b,

s) = Q for all

f(z)dz

=Jf(z)dz

s E [0, 11

Show that

35. Let ci be an open connected subset of C. Let

be a sequence of continuous

functions on ci, converging locally uniformly to f E Hol(Q). Assume that none of the f vanishes at any point. Show that I vanishes everywhere or nowhere. 36. Derive Brouwer's fixed point theorem (Theorem 111.11) from Roucbe's theorem. 67

Chapwr 4.

Basic Theory of Holomorphic Functions

37. In this exeivise we will derive a formula which as special cases includes the Poisson inugra!

Je_z2dx =

and

the Fresnel iiuegrais

iirnf con (x2)dx =

and

iirnf sin (z2)dx = To that purpose we consider the meromorphic function

f(z) :=

,

C

E

—

(a) Show, to ease your computations below, that

1(z) + f(—z) =

+

Consider

and

+

—

for each R > 0 and fixed a

A, B, C, D , in which

A=-B=-Re'°, E=

F=

—G

=

68

jO,

=

[

the

following parallelogram

Section 5.

Exercises

y C

D x

A

Note that f is holomorphic on the interior of the parallelogram and on its boundary only has a singularity at z = 0. (b) Show that

I

f(z)dz—.0 and

IB,Cl

J f(z)dz—.0

as

R—'oo

IA,DJ

(c) Show that

J f(z)dz = —ie'° Jexp {it2e2'° }dt IC,D1

and that

J f(z)dz + J f(z)dz =

—&°

r

IF,Bl

IA,G)

fexp {ii2e2to

Hint: Use (a). (d) Show that the integral of f along the small half circle on the figure from G to F as r —' 0+ converges to i (e) Show that R

urn fexp {ii2e2ba}dt =

R—ooj

0

69

1

+

Chapter 4.

Basic

Theory of Holomorphic Funcuons

(f) Derive the Poisson and Fresnel integrals. (gJ Show the following more compact form of the result of (e):

I

=

for a

is taken on the right. where the principal value of The exercise is adapted from the paper [Ra].

70

0

Chapter 5 Global Theory Section 1 The global Cauchy integral theorem In this section we will state and prove a general version of the Cauchy integral formula and the Cauchy integral theorem. Theorem IV.6 states the Cauchy integral formula for a closed path in a starshaped domain. We would of course like to remove the condition of starshapedness, but we must be careful, for if we do so then some kind of restriction on the paths in question is necessary: Take as domain the annulus A := {z C I 1:1 < i} and consider functions which are holomorphic in an open set containing the closure of A. It is simply not true that

f

for

zEA

Kl=1 as

the function f(z) =

demonstrates. On the positive side there exists a version of the Cauchy integral formula in A, viz.

J

J IcI=1 (Exercise:

for zEA

Derive it from the Cauchy integral formula for starshaped domains by

inserting suitable auxiliary segments). In the case of a starshaped domain Q the Cauchy theorem states that

(s) for any closed path 7 in and any function f which is holomorphic in Q. Let now be any open subset of the complex plane, and let be a closed path in Q. If (*) holds

for all f E Hol(cl), then in particular

forall zEC\Q Our general version of the Cauchy integral theorem is that the converse holds. The

crucial necessary condition on the path 7, i.e. that Ind,(z) = means intuitively that does not circle any "hole" in

71

0

for all z

C\Q,

Chapter 5.

Global Thcoy

Theorem I (The global Cauchy theorem and integral formula). Let 0

be a closed path in an open subset Il of the complex plane such that Ind7(z)

broil

z

=

C\ft.

Then we have for any f E Hol(ft) that

(a)

Jf(z)dz=o ,and forall

(fi)

Example 2. Here is a drawing of a path that satisfies the condition of l'heorem 1, but which nevertheless is not null-homotopic [Q = C\{O, 1)]

Proof of Theorem I: (due to [Dii). (a) follows from (8) when we replace I by so from now on we may concentrate on (8). ((— By the formula for the index (Theorem 111.17(d)) (8) is equivalent to

—.

lJf(z)_f(w)dwo z—w

2in

forall

zEft\7'

7

Introducing the function h

h(z,w) :=

x Q —, C given by

for z

1

Lf'(w)

w

(orz=w

it thus suffices to prove that

(oral! zEQ

72

Section 1. The global

integral Ihecwem

We start by noting some of the properties of h and g. h is continuous in all of This is obvious off the diagonal in fl x fl. To prove that h is continuous on the diagonal as well we compute x

1(z) - 1(w)

+ i(z - w)))di = (z

=J

- w)

I

f'(w + i(z - w))dt

which implies that h near the diagonal can be expressed as

f'(w + t(z — w))dt

h(z, w)

= a formula that also holds on the diagonal. This formula proves the continuity of h on the diagonal. So g makes sense and is continuous in fl. By help of Morera's theorem (Theorem IV.9) we will prove that g is even holomorphic in fl

If A is a triangle in fl then

=

=

Now, the function z —, h(z, w) is holomorphic in fl for fixed w [Example !V.4J, so according to the Cauchy-Goursat theorem

,andhence

Jg=O

so according to Morera g is holomorphic in fl. By our assumption on the path the set

:= {z E C\y' Ind7(z) = 0) is an open subset of C satisfying fl U f1O = C For

z E fl fl flo we get:

lJ

g(z)=-J_fh(z,w)dw=_LJ f(Z)d 2ir:

z—w

27r:

= -1(z) Ind,(z) + 27r1

2irz

f(w)

f w—z dw =

w—z

27r1

7

7

The function

go(z) :=

1

(7(w)

I

f(W)dW z—w

w—z

7

73

dw for z

fl0

dw

chapter s.

Theoy

is holomorphic on (b (Lemma IL 13); the computation just made shows that g and go patch together over (1 fl to an unambiguously defined function G on (1 U (bo = C, i.e.

C given by

fcvzEQ

' ' is

for

entire.

The unbounded component of

is by Theorem 111.17(c) contained in Qo, so for

z out there we can estimate C as follows:

..L. sup{If(w)IIwE

IG(z)I =

= which shows that G(:) —+ 0 as z —' 00. But then C vanishes identically by Uouville's

theorem (Theorem 1.11). In particular g is identically 0 on (1, and that fact is exactly the contents of 0

An inspection of the proof reveals that it applies to the case of several paths, not just one. This comes in handy for, say, an annulus as mentioned in the introduction of this section. The precise statement is Theorem 3 (The global Cauchy theorem and integral formula). Let

72,•

be closed paths in an open subset (1 of the complex plane such thai

,

bra!! zE C\Q Then we have for any I E Hol(Q) that

(a)

=

0

,and

)=17,

(/3) 1(z)

Ind7,(z) =

forall Proof: Replace

g

in the proof of Theorem 1 above by

Qo by

go

by

J=I 74

Simply connected sets

Section 2.

The details of the easy modification of the proof of Theorem 1 are left to the

0

Corollary 4. Let f be holomorphic in an open subset of:he complex plane. Let 71,72, ,7N and let n i, closed paths in , and be and mi, m2,••• , nN , OM be integers such that M

N

forall z E

>2njIndi,(z) = 1=1

En,Jf = >2 miff j=1

1=1

Corollary 5. Let f be holomorphic in an open subset which are homotopic in closed paths in

of the conzple.x plane. Let y and p be two Then

/1=/fl

In particular,

is null-homo:opic in Q. then f I =

0.

7

Section 2 Simply connected sets The topological concept of simple connectivity is intimately relaxed to homotopy

(Definition 111.16).

Definition 6. A topological space is said to be simply connected, if each closed cu,ve in it Lx null-homotopic.

It is obvious that any starshaped space is simply connected. R3\{0} is an example of a topological space which is simply connected, but not starshaped. It is intuitively

obvious that R2\{0) is connected, but not simply connected (For a rigorous proof see the remark following Theorem 9). B(0, 1) U B(3, 1) is simply connected, but not connected.

75

chapter 5.

Global Theoiy

Theorem 7. Let fl be a simply connected open subset of C,

be a closed path in fl and let

fEHol(ffl. Thenff=Oand 7

f(z) Ind7(z) =

for all

J

z

E fl\7

0

Proof: Obvious from Corollary 5. Theorem 8.

Let fl be a simply connected open subset of C. Then any f E Hol(Q) has a (holomorphic) primitive on fl. Proof: It suffices to consmict a primitive on each of the connected components of fl, so we may as well from the start assume that fl is connected. Choose zo E fl and

Then

F(z) := Jf(w)dw is according to Theorem 7 independent of the choice of path from zo to z, so F is well 0 defined on fl. Proceeding as in the proof of Lemma 11.12 we see that P = 1. We can now as promised extend Theorem 111.21 and 111.10 to simply connected sets.

Theorem 9. Let fl be a simply connected open subset of C and let f be a never vanishing holomorphic function on Il. Then f has a holomorphic logarithm and a holomorphic square root on Il. Proof: It suffices to consuuct the logarithm and the square root in each of the so we may as well assume that fl is connected. Let F is be a primitive of f'/f in fl; such one exists by Theorem 8. The function I

connected components of fl, constant because

(fe")' = we see that F — w is a holomorphic c= f= logarithm of f. A holomorphic square root is exp [(F — w/2)]. 0 76

Ewcises

Seczion 3.

In passing we note that this gives us a rigorous proof that R2\{O} is not simply connected: Combine Example 11122(b) and Theorem 9. For other results on simply connected sets see the discussion around the Riemann mapping theorem (Theorem Vul.20), Chapter VflI.*4, Remark 1)1.11 and Exercise IX.20.

SectIon 3 ExercIses 1.

Consider the path -y in Example 2. What is

Jz 7

f

dz

J z—f

J

f>Ois small?

when

e _ed

2. Let f be holomorphic in C\{O}

J

Show that

i=Ji

lzl=R

3. Show that the function 1

.

1z11

has a holomorphic square root in any simply connected open subset of the complex plane which does not contain { —1, 1) . Find the possible values of —

z2

dz

f

J y'1—z2 7

is a closed path in such a domain. 4. Let be an open, connected and simply connected subset of the complex plane, and let zo,al,a2,•••,aN E fi be distinct points. (loose €>Oso small that the closed balls B[ai,e),. . . are contained in Cl and so that a' B[afr,E) when j k. For h Hol(Cl\{ai,a2,.. ,aN}) we introduce the constants when

J h(w)dwEC lw—a, l=(

Let

=

be any path in Cl\{aj,a2,. . . ,aN} from

following function of z

fish(w)—

J

I.

w—ai

—•••—

7

77

CN

1

w—aN)

to z. Show that the

Chapter 5.

Global Theory

is independent of the choice of path from zo to z. S. Theorem 9 deals with holomorphic functions, but the result is really a topological

one. The purpose of this exercise is to establish the following topological version of Theorem 9: Theorem 10. be a simply connected open subset of the complex plane and let 1: C\{O} be continuous. Then I has a continuous logarithm on Let

—,

It suffices to prove the theorem for each of the connected components of so we may as well assume that Q is connected. (a) Prove that Ind;07(0) = 0 for all closed curves in be continuous curves such that 7i(0) = 72(0) and : [0,1) —.. (fi) Let and 1072 10,1) —+ C be continuous logarithms of 10 71(1) = 72(1). Let respectively, starting at the same point = Show that = and wo E C such that exp wo = f(zn). Find for any z E a Fix zo curve such that 7(0) = zo and 7(1) = z and take a continuous logarithm of 107 such that = wo. does not depend on the choice of the curve so that we unamShow that biguously may define F(z) := 7(1)

(6) Show that F is a logarithm of f. (e) Show that F o is continuous for any curve (C) Show that F is a continuous logarithm of 1.

78

from (y).

0

Chapter 6 Isolated Singularities We begin the present chapter with a study of functions which are holomorphic in an annulus; we show that they can be expanded in Laurent series in much the same spirit as functions which are holomorphic in a disc can be expanded in power series. The special case of a holomorphic function, defined in a punctured disc is the topic of the remainder of the chapter. The center of the disc is in that case said to be an isolated singularity of the function. We classify isolated singularities into removable singularities, poles and essential singularities. We finally prove the Residue Theorem

and use it to evaluate definite integrals of various types; this is certainly one of the high points of any introductory course on complex analysis. A deeper study of essential singularities can be found in the next chapter in which the two Picard theorems are derived.

SectIon 1 Laurent series Definition 1. A Laurent series is a series of the form (1)

for n E Z

where the coefficients

are conzple.x numbers, and where z is a complex

number different from 0.

If the two series (2)

=

and

(3)

both are convergent with sums S_ and

respectively, then we will say that the Laurent

series (1) converges and that its sum is S_ + If at least one of the series (2) and (3) diverges, then we will say that the Laurent series (1) diverges. The concepts of absolute convergence and uniform convergence are defined simBy definition the convergence of the Laurent series (1) means convergence of the two series (2) and (3) which are power series in 1/: and z respectively. So it is to be expected that results, analogous to those for power series, hold for Laurent series.

79

Isobted Singularities

Chaixer 6.

Theorem 2.

Let

and R:= rn—co

Assume

that 0 < r


urn sup , and let A denote the annulus

A

:= {z

E

C Ir< IzI< R}

Laurent series (1) converges absolutely at each point of A, and uniformly on any compact subset of A. When Izi R the Laurent series (I) diverges. The function f, given by The

for

1(z)

z E

A

is holomorphic in A, and its derivative can be found by term

f'(z)

i.e.

by term

for z E A =

We know that a function which is holomorphic in an open disc may be expanded in a power series there. We will now show the corresponding result for a function which is

holomorphic in an annulus. Theorem 3. LetfbeholomorphicinanannzdusA

Roo.

:= {zeCIr
(r<

Then there exists exactly one Laurent series (1) such that it converges at each point

of A and such that

for z E A

1(z)

= The coefficients of this Lawvzt series can be found from f as follows

J

1w1p where

p is arbitrary in the open interval )r, R[.

Proof: Assume that f is the sum of a Laurent series (I) in A. By the previous theorem the series converges uniformly on the circle IwI = p 1

1 1(w)

j

IwI=p

=

1

1

J

IwI=p

80

dw

=

, so

=

Section

Laurent series

1.

This proves the uniqueness of a possible Laurent expansion and that the coefficients will be the ones indicated in the statement of the theorem. It is left to show that f is the sum of a Laurent series in A. Choose r' and R' such that r
jf

f(w)d

f j Iwl=r

w—z

I

w—z

The first integral, i.e. the function

f

G(z)=—1-

w—z

J

IwI=R'

holomorphic in {z E C I Izi R'} it has a power series expansion is

,

hence

in particular in the disc B(O, R'), and so

for IzI
g(z) =

may for IzI >

1 —-———

2irz

[ f(w) (itt'

j lwI=r'

w—z

r' be treated as follows: g(z) =

2irs

f

1(w) dw = ! 27rzz J w—z

jf

fwlr' I

I

1(w)

dw

Iu'l=r'

fl

F

I

J

n=o

lwl=r

11

00 1

,

f(w)wdw j lwIr

which exhibits a Laurent expansion of g. We conclude that f = G + g has a Laurenc

expansion in the annulus {z E C r' < Izi
0

Aandhasthesumfthere.

Laurent series constitute an important tool in digital signal processing in form of the Z-uansform: A signal is a two-sided sequence {an},,€z of complex numbers; its 81

Chapter 6.

Isolated Singularities

Z-transform is by definition the sum of the Laurent series (1). This is a generalization of the corresponding Fourier series

but the Laurent series may converge even if the Fourier series does not. The Z-transform has many pleasant properties: E.g. the Z-transform of a convolution is the product of the Z-transforms. For more information on the Z-transform and its applications the reader may look in any book on digital signal processing, say the classic monograph [OS].

Section 2 The classification of isolated singularities Definition 4. If a function f is holomorphic on an open set of the form B(a, r)\{O} , we say thai

a is an isolated singularity of f. As examples of isolated singularities the reader can have the following three functions in mind

sinz

—,

cosz

— and

11

Each of them has an isolated singularity at the point z = 0. The classification of an isolated singularity of f is based on the behavior of f at the singularity a. There are three mutually excluding cases: In the first case f remains bounded near a. This case is taken care of by the Removable Singularity Theorem (Theorem IV.3): f extends across a to an analytic function on all of B(a, r). We say that a is a removable singularity of 1. An example: We will always let f denote z = 0 is a removable singularity of the function also the extended function, even if we don't say so explicitly. So for example we give the function the value 1 at z = 0. In the second case f has a pole at z = a meaning by definition that If(z)I —.oo as z —. a. Here the function g(z) := 1/1(z) has a removable singularity at z = a where it takes the value 0, so we can write it in the form

remains bounded near z =

a. The smallest integer n >

0

such that (z —

a is called the order of the pole. It equals the order of the zero of

1/f. If a pole has order I we call it a simple pole. An example of a simple pole is A more general example of a simple pole is a quotient of the form 1(z) = g(z)/h(z) where g and h are holomorphic near z = a, h(a) = 0, h'(a) 0 and g(a) 0. 82

Section 2.

The classificauon of isolated singularities

The third and final case is the one in which urn If(z)I as z in R U {oo}. We say that f has an esseiuial singularity at a.

a

does not exist An example is

f(z) = exp(z_'). The beautiful relation between the coefficients of the Laurent expansion of f and the behavior of f at the isolated singularity is taken up in Exercise 1 below. Definition S. Let a be an isolated singularity of the holoniorphic function f : B(a, r)\{a} By the residue of f at a we mean the complex number

C.

J f(z)dz I

e is any number in the interval 10, r(. By Corollary VS the residue off at a is independent of the choice of e. where

The next sections will show that residues are important for the evaluation of definite integrals. So we must be able to compute residues. The simplest instance occurs when I has a removable singularity at a. Here Re.s(f; a) = 0 by the Cauchy integral theorem (Theorem IV.5). Another particularly simple case - which even occurs quite often - is the case of a simple pole. This is singled out as a special instance in the following recipe for computing the residue at a pole.

Proposition 6.

(a) 1ff has a pole of order n < oo at z = removable singularity w z = a. and

Res(f;a)=

1

a

a then the function (z —

{(z—a)

(b) In particular, 4ff hasasimple pole atz =

a then

Res(f; a) = {(z — (c) 1ff and g are holomorphic near z = a,

Res(L;a) = g

g(a) =

0

and g'(a)

0

•

then

g(o)

Proof: We prove only (b) and (c) and leave the easy generalization (a) to the reader. (b) The definition of a simple pole implies that g(z) := (z — a)f(z) remains bounded near z = a, so that g has a removable singularity at z = a. Now, let's write

83

ChapW 6.

Isolaled Singularities

and observe that the first term on the right hand side is holomorphic across the singularity z = a by the Removable Singularity Theorem. So

f(z)dz=

J

J Iz—aI=

I

-±_=O+g(a)2,ri

J Iz—aI=

proving (b).

(c) Since f/g has a simple pole (if 1(a) 5&

f(a) =

0)

or a removable singularity (if

0) at z =a we get from (b) that

\g

= /

g(z)

z—a

f(z)

=lim (g(z)

f(a)

— g(a))/(z —a)

g'(a)

0 As a final example of how to compute residues we consider the function exp which has an essential singularity at z = 0. Since the series

fl\

00 1

—n

converges uniformly on compact subsets of C\{O) we can integrate term by term and find

Res(exp

=

=1

= Of course, by the same method we see around z = a has a Laurent expansion

1(z) has

Res(f;a)

Section

more generally

that any function f

which

-

=

a_i.

3 The residue theorem

The statement. The residue theorem generalizes both the Cauchy integral theorem and the Cauchy integral formula. Our version of it runs as follows:

84

Section 3.

The residue theorem

Theorem 7 (The Cauchy residue theorem). Let be an open subset of the complex plane and let

that Ind.,(z) =0 for all z E

be a closed path in fl such Let I be holomorphic in Q except for finitely many

isolated singularities at,

E fl\-y. Then =

The hypothesis on -y is clearly satisfied if -y is homotopic in to a constant curve, say if is simply connected. In most examples 'y will be a simple closed path traversed counterclockwise, and Ind,(z) will be 1 or 0 according to whether z is inside or outside

Proof. Choose closed paths -yr, in circles around such that each the singularity once and does not contain any of the other singularities of f. More precisely choose them such that

Ind,,(z) =0 forall

and

j=1,2,.,N

and

Ind.7,(aj)=6,k for j,k=1,2,.",N Then N

for all z E C\(Q\{al,a2,...,aN}) j=1 so

by the Global Cauchy Theorem V.3 (or better its Corollary V.4)

fi = Dividing

through by 2,ri we get the Residue Theorem.

0

The Cauchy Residue theorem can be used to evaluate definite integrals. How successful one is depends largely on correct choice of contour and representative function. The techniques will be illustrated below by various examples that will provide us with recipes for evaluation of definite integrals of different types.

Example A. An integral of the form

JF(cos 0, sin 0)dO 85

Chapter 6.

Isolated Singularities

is by the substitution z = exp (19) convened into a complex line integral as follows:

J

0

1:1=1

As a non-trivial example we evaluate

7

sin2O

a+

for

dO

con 9

0

where the restriction on a is chosen so that the denominator in the integrand does not

vanish anywhere. Using the above substitution we find

( sinO

J0

1

1

2i J

a+cos9

(z2 —1)

2

z2(z2+2az-i-1)

1:1=1

where the integrand, i.e. the function —

1(z)— has

a pole of order 2 at z =

0

(z2_1)2 z2(z2+2az+1)

and simple poles at

z=—a+ Va2_i and z= —a— with corresponding residues —2a, 2v'a2 — 1 and — 2v'a2 — 1 , where we to avoid ambiguity agree to take 'Va2 — 1 as that branch of the square root of a2 — 1 which is positive for a > 1. (Cf Example 111.22(c)). To apply the Residue Theorem none

of the poles

h(a) :=

—a

—

—1

Ih(a)I = 1

,

0 , we

then g(a)

—a

+ Va2 —

1

h(a) and g(a) are mots of the equation —2a. Now if, say, is the complex conjugate of h(a) , so

may belong to the unit circle Izt

+ 2az + 1 =

and g(a) :=

1

.

Since

have h(a)g(a) =

1

and h(a) + g(a) =

= h(a) +h(a) = h(a) +g(a) But this means that a is real and in the excluded cases a E [—1, 1)

— 1 .

=

—a

is purely imaginary, and that happens only

Since Ih(a)I never 86

takes the value 1 and is > 1

The residue theorem

Section 3.

for a > 1 we see that Ih(a)I > 1 for

Ia

all a considered, and so

< 1 for all

By the Residue Theorem

a E C\[—1, 1]

sin6 9d9 =

+

i} =27r{a_ Va2_ i}

=

The method of this example can be applied to any integral of the form 2r

f

P

and Q(cost,sint)

Q

0 for all t E

[0, 27!J.

Example B. In this example we want to evaluate integrals of the form f R(x)dx where 1? is a rational function. We state our result as a proposition: Proposition & Let P and Q be polynomials in one variable with Q has no zeros on the real line. (I) If4tz > 0 then

P <deg Q and assume that

R

IP(x) I —e" dx R-.ooJ Q(x) urn

—R

exists, and

iirnf wherethe

=

Q(w)=O}.

ranges over

(u) If deg(P) < deg(Q)

— 1 .

then 00

[

J Q(x) -00 converges

absoiwely, and

J

.!.dx = 87

Chapter 6.

Isolated Singularities

> 0, Q(w) =

where the sum ranges over {w E C

0}.

Proof (1) Note that there exist two positive constants A and B such that

Q(z) Choose

of Q

R>

Izi

when Izi B

so large that the triangle in the picture below contains all the zeros

in the upper half plane.

y

The distance from the origin to each of the two skew sides of the triangle is

for z E [R iRJe u

hR1 —R)

With the usual parametrization of the segment y := [R. ii?]

,

i.e.

= [R,iR](t) = R(it + 1 —1) for 0 t < 1 we find

J

=

R,iRJ

/

j

—

0

=

(i -

-+0 as R

In exactly the same way we see that the integral along the other skew side [iR, —RJ of the triangle tends to zero as R 00. The result (i) is then an immediate consequence of the Residue Theorem. 88

Section 3.

The residue theorem

(ii) Proceed in the same way except for noting the stronger estimate

P(z)

A

large.

0 Example C. As an example of an integral involving a multivalued function we consider

f

J0

P(x) dx x°Q(x)

assume that 0 < a < 1 and that P and Q are polynomials in one variable such that = exp(a Log x) deg P < deg Q and such that Q has no zeros on [0, oo[. Of course, in the integral. We

It turns up to be a good idea to introduce that branch of the logarithm which is defined in := C\[0, oo[ and has its imaginary pan in the open interval 1O,2ir[. So for the remainder of this example, if z E fl we let

logz = + iO(z) , where 0 < 9(z) <2ir, and = exp(—alogz) = exp(—aLoglzf — ia9(z)) We let f denote the function

for zEQ

f(z):_—z It is holomorphic in

except at the zeros of Q.

Consider the following contour in which R> 0 is chosen so large, and r and €, r > e > 0, are chosen so small that the area inside contains all the zeros of Q. denote the two horizontal paths

where

for r—
Isolated Singularities

Chapter 6.

1

4

Then

R-t

R-e

J 1.

fdz

=

J (i + r—e

+ IE)di i€)° P(i Q(t + ie) =

P(i +

J

Q(t + ie)

di

F—c

and similarly

J

fdz = e2h10

J

e_oLog(t_u)

P(t —

Q(i—i€)

dt

T—c

'—1

so that (*) Let

,

urn

{Jfdz

_J fdz} = (1_

denote the paths indicated in the figure. Using the abbreviation

E := we get from the Residue Theorem that I

(**)

Jfdz_ Jfdz -El

si 3=1

7—.

90

I

The residue theorem

Section 3.

so we shall estimate the eight integrals

Let us first consider the paths c2, C3, C4 on the boundary of the large square. Here we infer from the assumption deg P < deg Q that there exist constants A > 0, B > 0 such that Q(z)

Izi

IzIB

forall

Noting that the distance from the origin to the boundary of the large square is B] for any outer path that we get [choosing

We shall next consider the boundary of the small square. Here we estimate P/Q by a constant C> 0 and find for any inner path that

Substituting these estimates into (**) we get that

RA + C

fdz _J =

we read from (*) that

C1 and C2. Letting e —'

(1 — e_2Tba)

Finally, letting r —, 0 and R

—

oo

we see that the right hand side converges to zero, so

(1 —

—

The final result is now 00

C1R—a +

p =

C)

91

E

=

0

Chapcr 6.

Isolated Singularities

Section 4 Exercises I. Let z = 0 be an isolated singularity of a function

and

f(z) = its Laurent expansion. Show that (or) 0

= 0 for all n <0.

is a removable singularity of I 1ff

and a,, =0 forall n<—m. 0 is an essential singularity of f 1ff a,, 2.

0 for infinitely many negative n.

Evaluate

JeV')do 3. Find the value of 1 1 — ji

.

fl\

szni—jdz

\zJ

1z11

4.

Let N be the year you were born. Find the value of

J tan(,rz)dz IzIN S.

Show that

for n=0,1,... 6. LctPbeapolynomialofdegree2. Showihat

f

dz

J

I:I=R when

= R surrounds all the zeros of P.

R is chosen so large that the circle

7. LetP bcapolynomialofdegree n 2withdistinctzems Z1,Z2,",Zn. Show that 0

8. Let f be holomorphic in a neighborhood of z = a, and assume that 1(a) = 0

and f'(a)

0. Compute for e > 0 small the integral

f

J

Ix-alt 92

Section 4.

9. Let m and

Exercises

be integers 0. Find

n

1 —

2iriJ(

m

1

z—a )

z—a)

z—a+————

dz when

IzI=1

Deduce Walks' formula 2w

11

(2rn)!

2ir

(rn!)2 1)

10. Show that 2w

t

1

2ir

2rcose+ r2 =

1—

Ii

when r E R\{1,—1}

—

1)

11. Show that 2w

27r

dO

J 1 + acosO =

—

for a2

0
0

12. Show that 2w

J

27r

dO

for p ER

+ /4 2

=

1

1)

and derive the formula 2w

dO

J a2cos2O + b2sin2O =

2,r

for a >0, b >0

0

13. Show that w

dO

J

when a>0

o

14. Show that 00

I

for

-00 15.

Show that w

for bER\{0) 0

93

Chapter 6.

Isolated Singularities

16. Pinpoint the error in the following reasoning : Let be the half circle in the upper half plane with center 0 and radius R> 0. Since the function (called Sinus Cardinalis)

:= S1flZ —

Ssnc(z) is

in all of C), it follows from the Cauchy integral theorem

entire (i.e.

with 1(z) = (Sincz) that

J f(x)dx =

iimf f(x)dx = iimJ —

Estimating sinus by 1 we get

= so that

R-.oo

(Stncz)dx=0 jI. 00

2

-00 Show

that the correct value of the integral isn't 0, but

17. Show for n =

that

J

=

—

—

1)!

18. Show that 00

f

z3sinx

00

19. Show that 00

fxsinx dx=—e

I Jx4+1

.

fl

2

0

20.

Show that

7 Jcoshx

cosh(E

Hint: Consider the contour of height 94

Section 4.

Exercises

0

-R

R

21. Show for n = 2,3,••• that

[dx J by help of the following contour in which P = Rexp(2iri/n)

p

Compute the more general expression

I

X

1+

dx where

—1
Answer: yr/n

sin((i 22. Show that Logx f (x2+i)2 Hint : Consider the contour 95

4

Chaper 6.

-R

Isothted Singularities

r

•r

R

An elementary deduction of the result can

be

found in [AGRI.

23. Show that

tLogz Hint: Consider the contour

24. In this problem we will deduce the following Theorem.

If f is a rational function such that 11(z) = 1 whenever z = 1, then f has off its poles the form I

I

1(z) =

(0,1,2,...), and where are complex constants such that IakI ]forallk. Proceed as follows for the proof: a. Show that we in the proof without loss of generality may assume that 0 is neither a zero nor a pole of I [divide by an appropriate power of z]. where c is a consta?u of modulus 1, m is an integer, n

96

Seclion 4.

b. Let a

Exercises

E C . Show that forall zES1\{a}

denote the zeros off inside the unit disc, and c. Let Z1,Z2, ,Ps its poles there, zeros as well as poles repeated according to their multiplicity. Show that the function

F(z) := f(z)fl 1ZkZ H is holomorphic and zero-free in a neighborhood of B[O, 1] and that IF(z)I = 1 whenever

IzI=l. Show that F is a constant in B(O, 1) and hence everywhere. e. Finish the proof. d.

97

0

Chapter 7 The Picard Theorems In this chapter we shall prove Picard's "big" theorem on the strange behavior of holomorphic functions near essernial singularities, and as a consequence derive his "little" theorem on entire functions. These two theorems have been the spur for much development in complex analysis, in particular for the so-called Nevanlinna theory which is concerned with the distribution of the values of meromorphic functions. Let us cite the two theorems:

Theorem I (Picard's little theorem). The range of a nonconstant entire function omits as most one complex number.

Theorem 2 (Picard's big theorem).

Leif be holomorphic in

is an open subset of the complex plane

{ zo }, where

and Zn E ft 1ff has an essential singularity as zo then f( Q\ Zn)) is either C or C with one point removed.

SectIon 1 Uouvilie's and Casoratl-Weierstrass' theorems Liouville's theorem (Theorem 1.11) gives a first weak hint to the truth of Picard's little theorem1 so we repeat it here:

Theorem 3 (Llouville's theorem). Let f be entire. (i) 1ff is bounded then I is constant. (II) More generally, if f for some real constants a, b and m satisfies the estimate If(z)I

broil IZI > b

then f is a polynomial of degree < m. As an important application of Liouville's theorem we derive the fundamental theorem of algebra which asserts that every algebraic equation over the field of complex

numbers has a root. More generally that it takes any complex value and thus shows the validity of Picard's little theorem for polynomials. The fundamental theorem of algebra is important because it implies that any polynomial can be written as a product of linear factors. Corollary 4 (The fundamental theorem of algebra). Let an, ai,••• , a,, C where n 1 and a,, 0 , and let p denote the polynomial

p(z) :=

an

+ aiz f... + a,,z" for z E C 99

Chapter 7.

The Picaid Thcorcms

C such thai p(zo) =

Then there exists a

0.

Although the fundamental theorem of algebra looks like an algebraic result, being

concerned with polynomials, it also involves the completeness of the reals: It is obviously false if C = R + iR is replaced by Q + iQ, where Q denotes the rational numbers. For a discussion and many references see p.117-118 of the monography [Bu].

Proof: The proof goes by contradiction, so we assume that p never vanishes. Then the function f(z) := 1/p(z) is entire and converges to 0 as Izi —' 00. In particular f is bounded and so a constant by Liouville's theorem. But then p is constant, contradicting

0

thatthedegreeofpis1.

The Casorati-Weierstrass' theorem is an immediate consequence of Picard's big theorem. We give a direct proof though, since it is quite simple. In the Russian literature the theorem is attributed to Y.V. Sokhotskii. For a detailed history of it see [NeuJ. Theorem 5 (Casorati-Weierst,uss' theorem). Let f have an essential singularity at E C. Then for any w and —. w as n cc. sequence such that z,, —.

C there exists a

Proof: The proof goes by contradiction: If the assertion were false then there E C and an > 0 and a neighborhood Q of zo such that

would exist a

If(z)—wnle forall The function

g(z) := is then analytic in cZ\{ Zn) and

g.

1

1(Z) —

,z E

bounded (by lIE),

so zo is

a removable singularity for

Thus

is either analytic at z =

0), or has a pole at z = zo (if g(zo) = 0). In zo (if g(zo) any case it contradicts our assumption about being an essential singularity. 0

SectIon 2 Picard's two theorems Our key to the proof of Picard's big theorem is the following remarkable result on the range of functions which are analytic in the unit disc.

100

two theorems

Section 2.

Theorem 6 (Bloch.Lassdau's theorem). There exists a positive number A with following property: Let f be any function. 1. Then the range off contains holomorphic in the unit disc and such thai If(0)I

a disc of radius

A.

The surprising feature of the theorem is of course the existence of the universal constant A in spite of the vast class of functions involved. For our purposes any

positive constant wiil suffice. Remark. Let

us for

I E 4' := {gE Hol(B(O,1))IIg'(o)I 1) put

L(f) :=

sup

{r >

f(B(0, 1)) contains a disc of radius r}

The theorem above then tells us that Landau's constant L :=

inf{L(f))

The exact value of Landau's positive. Our proof of Theorem 6 reveals that L constant is not known, but it has been ascertained that 0.5 L 0.56. is

6 is an immediate corollary of an even more surprising quantitative discovery on the range of an analytic function: Theorem

Theorem 7 (Block's theorem). 1. Let f be any function which is holomorphic in B(O, 1) and has lf'(o)I Then there exists a subdisc D1 of B(0, 1) such that 1(D1) contains a disc of radius 0.43 and f is

univalent on

D,.

Remark. Let D(f) be the supremum of all r > 0 for which there exists a region D on which f is univalent and such that f(D) contains a disc of radius r. Then Bloch's theorem tells us that Bloch's constant

B := inf{B(f)I 1€ Hol(B(0,1)) and If'(O)I 1) is bigger than 0.43. The exact value of Bloch's constant is not known, although it is

known that 0.43 B 0.47. For interesting gossip on Andre Bloch's life see the essay [Cal and the follow up [CFJ. 101

Chapter 7.

The Picard Theorems

Proof of Theorem 6: We may assume that If'(O)I = 1 . We will first treat the special case where I is holomorphic on a neighborhood of the closed disc B(O, 1). The (unction a : (0, 1) [0, oo[ given by

a(r)

:= (1 — r)

for 0

sup

r :S 1

continuous. Since a(0) = 1 and a(1) = 0, there exists a largest number s E (0, 11 such that a(3) = 1. Choose E B(0, 1) such that 1(1 = 3 and If'(OI = sup {If'(z)I} IzIs Consider for R := (1 — s)/2 the function is

F(z)

2(f(Rz +

F is holomorphic on B(0, 1), F(0) = IF'(O)I

for

() —

1

0, and

= 2RIf'()l

= 2Ra(:)

=

1

Furthermore, since a(r) < 1 when 3 < r, we get I

Rsup

{

I

R

R

=

l_s_R0(3+R)<

2. so IF"(z)I In the following lemma we shall show that the range of F contains a disc of radius 1/6. Prom the definition of F we see that the range of f then contains a disc of radius 1/12. This takes care of the special case. In the general case we consider for any 0 < p < 1 the function —' f(pz)/p. It satisfies the conditions of the special case, so its range contains a disc of radius 1/12. But that means that the range of f contains a disc of radius p/i2. Taking p = 12/13 we see that the range of f contains a disc of radius 1/13. 0

Lemma &

Let F E HoI(B(O, 1)) satisfy that F(0) = M is a constant. Then

F(B(0, 1)) 2

0 ,

2(M+

Proot The function Z —4

F'(O) =

M+1 102

)

1

and IF'I

/t'!, where

Section 2.

Plcard's two theorems

satisfies the conditions of Schwarz' lemma, so

IF'(z)—lI (M+1)IzI

all IzI <1

Integrating FV(z) -1 along the segment (0, zJ we get the inequality

IF(z) — zI It says in particular that we for

If

w

M+ 1 1z12

on the circle

= 1/(M + 1) have IF(C) —

B(0, 2(M+1)) then

I(F(() - w) - (( - w)I = IF(C) - CI 2(M+1)
soRouché'stheozem(TheoremlV.13)tellsusthatthefunctionz-iF(z)—whasa zero in the ball B(0, 1/(M + 1)). Since w was arbitrary we have shown that

2B(O,2(M1+l))

0

which is more than required.

We need one result more prior to Picard's big theorem, viz. Schottky's theorem. Theorem 9 (Schotthy's theorem).

Le:M > Oandr €J0,1[begiven. ThenthereexistsaconstantC > Osuchthat the following implicadon holds: 1fF is holomorphic in B(0, 1) , omits 0 and I from Us range, and (fIF(0)I M, then IF(z)I Cfor all z

B(0,r).

Proot Since F does not assume the value 0 in the starshaped set B(0, 1), F has a holomorphic logarithm log F [Proposition 111.101 , which we choose such that 19(logF(0))I < x. The function A := not assume integer values, and be holomorphic square roots because F never takes the value 1. Lct

ofAandA—1 (They exist by Proposition 111.21). Then B := /A—y'A—l is holomorphic in B(o, 1), vanishes nowhere and cannot assume the values farn = 1,2,... [Indeed, if

forsomezandn then taking reciprocals we get

A(z) = n, which is excluded.]. 103

Chapter 7.

The Picard Theorems

Since B never vanishes there is a holomorphic branch H of log B, and H cannot assume the values

for n = 1,2,... and mE Z

1)

Log

an,m

We leave it to the reader to verify that eveiy disc of radius 10 contains at least one of the points an,m , so that the range of H cannot cover any disc of radius 10. If z B(0, 1) and H'(z) 0 , then the values of the function

H(C)—H(z) for (EB(z 1—1:1) H' must fill a disc of radius (1 — IzD/13 (by our proof of the Bloch-Landau theorem), so the values of H fill a disc of radius f H'(z)I(l — IzD/13. This quantity cannot exceed 10, so

(1) IH'(z)I(l — IzI)

130

Although (1) was derived under the assumption H'(z) 0 , it is clearly also valid when H'(z) = 0 . Since H(z) — H(0) is the integral of H' along the segment (0, z), the estimate (1) leads to

(2) IH(z)I IH(0)I + l30Log(1

—


for

IzIY'

IH(0)I — l3Olog(1

—

r)

Now recall the definition of H:

-

exp H =

Taking reciprocals and adding the result we get 3)

/IogF(z) V

2iri

—

+

—

2

so that

F=

+

and IFI

In view of (2) the theorem follows once we establish that IH(0)I
is a constant depending only on the bound M on F(0). We will first treat those functions that satisfy the extra requirement F we get from (3) that there is a constant C2, depending only on M such that

C2

+

—

2

2

104

Picard's two theorems

Sectico 2.

Similarly for a lower which gives us an upper bound of the desired type on The imaginazy part poses no problems since we always may choose bound on H = logB such that 19(H(O))I ir . We have now proved the theorem under the

extra requirement that IF(O)l If IF(O)I

then we apply the just derived result to the function 1 — F instead of

0

F.

We finally arrive at Picard's big theorem, which is a remarkable generalization of the Casorati-Weicistrass theorem.

Theorem 10 (Picard's big theorem). If F has an essential singularity then the range of F omits at most one comp!e.x number.

The example F(z) = exp(1/z) shows that it is too much to hope that the range

of F is all of C. Proof: By translation in C we may assume that the singularity is situated at z = 0,

and by dilation that F is holomorphic in B(O, e2z)\{O). We will show that if F omits two complex numbers, say a and b, then 0 is either a pole or a removable singularity; that will prove the theorem. We may even assume that F omits the special values 0 and 1 - if not we replace F by

F(z)-a

Z —4

b—a

Casel: IF(z)l—'oo as z—.0. In this case 0 is a pole of F. Case 2: There exists a sequence

0

such that {IF(:n)I} remains bounded,

M < oo for all n. say IF(zn)I Passing to a subsequence we may assume that

1>IZ'I>.">IZnI>IZn+'f>"PO Consider for fixed n the function (

which is holomorphic in B(O, 1),

omits the values 0 and 1 and has M . By Schouky's theorem there exists a constant C, depending only on the bound Iv!, such that

C for all (

13[0.

I

In particular

2r't


for all I E

105

1

1

Chapter 7.

so that IFI that IFI

The Picard Thcoccms

bounded by Con the circle {z E C is independent of n we get by the weak maximum principle C on all of B(O, IziI)\{O} . But then 0 is a removable singularity of F. 0 is

The next result, Picard's little theorem, is an extension of the fundamental theorem of algebra and of Liouville's theorem. Theorem 11

(Pkard's Wile theorem).

1fF is a non-constant entire function then the range of F omits at most one complex number.

The function F(z) = exp z demonstrates that it does happen that a complex number is absent from the range of F.

Proof: Consider the function G(z) :=

F(!)

for z

C \ {0}

If 0 is an essential singularity for G then we are done by Picard's big theorem. If 0 is a pole of order m or a removable singularity of C (a pole of order in 0) then C can be written in the form G(z) = zmH(z) where H is holomorphic in all of C, also at z = 0. Now F(z) = zmHc3) so that (IH(O)I +

IF(z)I

for

large

By Liouville's theorem F is a polynomial, and so its range includes all of C by the

0

fundamental theorem of algebra.

Section 3 Exercises 1.

Let F be an entire function with the property that

F(z)

0 as

z

Show that F is a constant. 2. Let F be an entire function with the properly that IF(z)I

1

for all

z

C

Show that F is a constant. 3. Let F be an entire function with the property that Show that F is a constant (Hint: Consider exp F). 4. (Injectiveness of the Fourier transform. Cf. (New)). 106

is bounded from above.

Section 3.

Exercises

Let f be continuous and absolutely integrable over R and assume that

for all

(a) Show for each fixed a

R that =—

Je

and let denote this common value. (b) Show that the left hand side for fixed a E R can be extended to a function which is continuous and bounded in the upper half plane 0 and is holomorphic in the open half plane >0 Show similar statements for the right hand side. (c) Show successively that F0 for fixed a R is holomorphic in all of C, constant, even 0.

(d) Show that f =

0.

(e) Show by induction on n that the same result holds for functions f which are integrable over 5. Let F be an entire function that never takes values in JO, oo(. Show that F is a constant. 6. Let F be a rational function, i.e. a quotient between two polynomials. Assume that its poles occur at ai, (22, ... , with orders , Show that there exist a polynomial p and constants C such that C,&

(z

—

a,) k

for

7. Let F be holomorphic in C with the exception of finitely many poles. Assume that F(1/z) has an inessential singularity at z = 0. Show that F is a rational function, i.e. a quotient between two polynomials. 8. Show the following version of Picard's little theorem: Let F be a non-polynomial, entire function. Then there exists a C such that F takes every value in C\{zo) infinitely often. Hint: Assume the conclusion fails. Show by Picard's big theorem that the function z —' F(1) only has an inesscntial singularity at z = 0. Apply Liouville's theorem. 9. Let F be a non-constant, entire function. Assume that F does not take the value a E C. Let b C\{a). Show that F takes the value b infinitely often. 10. Let F be entire and injective. Show that F is a polynomial of degree I. 11. Define for fixed e >0 the function I B(0,l) —. C by :

1(z) :=

for

107

chapter 7.

The Picard Thecrems

(a) Show that f has f(O) = 1 and hence satisfies the conditions in Bloch-Landau's theorem.

(fl)Showthaz -e/2doesnobelongtotherangeoff.

(y) Thking e = 1/13 we see that f(B(0, 1)) does not contain a ball around 0 with theorem? radius 1/13. Doesn't that contradict

I2L6sdenotethesetoffunctionsfEHal(B(O,1))whichsatisfyf(O)=O, 1, and f(z) = 0 only if z = 0. We want to show that there exists a universal constant a > 0 such that / E So

f'(O) =

implies that B(0,a) ç f(B(O,1)). We prove it by contradiction. So assume that the implication Is false.

(a) Show that there exist sequences {a,.} ç C\{O} f,(B(O,1))). {f.} c So such that

such

that a.

—,

0

and

(b) Show that

1_i!.: a. B(0,1)—.C\{O) such that L.(0) =0. has a holounorphic logarithm (c) Show that the sequence {L,.} is uniformly bounded on B(O1f2). is a bounded sequence. (d) Show that (e) Derive a contradiction

SectIon 4 AlternatIve treatment The following alternative treatment ci Schouky's and Picaid's theorems is borrowed

from the paper [MSJ.

0282 denotes the Laplace operator = the Lq,lacian on R2. In polar coordinates (r, 0) it takes the form

82ioi82

We shall a couple of times use that = 0 when F is a holomorphic function. That is a consequence of the Cauchy-Riemann equations (See Theorem Xlii for more information on this point). Definition 12. Let Q be an open subset of the complex plane. A continuous function p:

—,

[0, oo[

Ameuiclsa function in

)0, oo[), so a metric Is also a pseudo-metric.

Section 4.

Alternative treauncnt

The (Gaussian) curvature of a pseudo-metric p is the function ,c(z,p) •—

inthe subset of

—

p( z)

defined by

2

where p(z) >0.

Examples 13. (1)

)tR(z) := is

2R R2

a metric on B(0, R) with constant curvature (ii)

o(z) is

— 1z12

—1.

9

1+IzI 2

a metric on C with constant curvature +1.

The formula in the following lemma expresses that the curvature is a conformal invariant, i.e. is preserved under holomorphic mappings, and so it is a natural object

that ought to be studied. Lemma 14. Let f : be a holomorphic map between two open subsets ci and ci' of the complex plane, and let p be a metric on ci'. Then fS(p) := pof is a pseudo-metric

onfland

K(z,f(p))=n(f(z),p)forali zEIlforwhich f'(z)

0

Proot It is obvious that f'(p) is a pseudo-metric, so left is the formula for the curvature, i.e. that —

poflf'12

—

p(f(z))2

or in other words that

of) + Log If'I }(z) =

Log p)(f(z)) 1f1(z)12

Here the term If'I) on the left hand side vanishes, because Log 11,1 (locally) is the real part of a holomorphic function (branches of log f'), so it is left to show the formula

f)}(z) = (t,.F)(f(z)) If'(z)12 a

formula that may be left to the reader. 109

The Picaid Theorems

Chaptcr 7.

Theorem 15 (Ahlfors' lemma). A1 is the biggest pseudometric on B(0, 1) among the ones with curvature at most —1.

Proof: Let us for arbitrary but fixed r EjO, 11 consider the function v := p/A, Ofl 1 , i.e. that p A, because we then obtain the B[0,r]. It suffices to show that v desired conclusion by letting r —. 1 Since v is non-negative and v(z) —. 0 as Izi —' r , we see that the continuous function v attains its supremum in at some point zo in the open ball B(0, r). We shall show that in < 1. If m = 0 we are done, so we may as well assume that p(zo) > 0. is a maximum for v and hence for Logy, we get that 0 2 (A Log Since Now,

0? (ALogv)(zo) = (A Log p)(:a) — (ALo9A,)(:o) —ic(zo,p)p(zo)2 + K(zfl,A,)A,( Zn)2

=

—

A,(zo)2

By the assumption on the curvature of p. we find 0 2 p(Zo)2

p(zo)/Ar(zo) S

1

—

A,(z0)2

,

50

in =

0

.

PropositIon 16. There exists a metric p on C \ { 0, 1) with the following two properties: (i) p has curvature at most —1.

(ii) P 2 cc for some constant c > 0. Proof: Using the Laplacian in polar coordinates we find by brute force computations

for any a E Rand z

C\{0} that

A(Log(l + Izl°)) = a2

2

(1 +

)

Furthermore A(Log IzI) = 0 because Log Izi locally is the real part of a holomorphic function (branches of log z). Combining these two facts we get for any a, E R and Z

C\{0) that 1

1

t Since the Laplacian A =

+

1z10)

82/0x2+82/8y2 has constant coefficients it satisfies chat

(AF)(z — so

(1

J

= A(w

F(w

—

—

in particular forz E C\{1} 1

A
1 + Iz—111 Iz—

>=a

J

110

2

IzhI (1 + I:— 110) 2

Alternative ircaimcnt

Seclion 4.

We

claim that for suitable o > 0

> 0 a positive multiple of the metric

,

1+IzIl+Iz—it can

be used as the desired metric p. By a straightforward calculation 2$

Z

ic(z, r) = — 0 2

(1

+

Iz

Iz — lIa+2$_2

(1+ Iz so

C\{0,1}

for

lz—1I

Izi

-

Iz12'9

(1+ Izr)2

the curvature is everywhere negative. Thking a = 1/6 and fi = 3/4 we find that (1)

Inn

r(z)

and (2) limlt(z,r)=—oo when z—.O,loroo

IzI—oo

Combining (2) with the fact that the curvature is negative we find that there exists a has curvature constant k > 0 such that ic(z, r) —k on C\{0, 1}. Then p := at most —1, which is the first property. From (1) we see that urn

xI—.oo c(z)

0

which implies the second property.

Theorem 17 (Scholiky's theorem). To each M > 0 and r E JO, 1[ there exLus a constant C > 0 such that the following implication holds: If I E Hol(B(0, 1)) , 11(0)1 M and the range of f omits 0 and 1 , then sup {If(z)I) C. IsIr Proof.

Let p and c be as in Proposition 16. Then f(p) is a pseudo-metric on

B(O,1) with curvature at most —1 (Lemma 14), by Ahlfors' lemma f(p)

f'(u) c'f(p) <

,

i.e.

If'(z)I 1

<

C

for all ZEB(O,1)

+ If(z)12 — 1 —

and so

lf(z)I

1+lf(z)I 2

for all

111

E B(0,r)

and so

The Picard Theorems

Chapter 7.

denotes the constant c1 = [c(1

where Since

f never takes the value 0 ,

—

the

function t

If(tz)l is continuously

—..

differentiable for any fixed z E B(O, r) and cj

< cj

J

larctan If(z)l — arctan 11(0)11 from which we get

arctan If(z)I
S

+ arctan M

0

+arctanM).

From here we proceed exactly as before to prove Picard's big theorem etc.

SectIon 5 ExercIses 1.

Show that the metric p on B(a, r), given by 2r

p(z) —

has

lz — al

for

2

E B(a,r)

curvature —1.

2. Derive Schwarz' lemma from Ahifors' lemma. Hint: Consider the metric f(A1) and note that d

= 3.

1 —

In this exercise we shall give a proof of Liouville's theorem based on Ahifors'

lemma.

Let f E Hol(C) be bounded, say 1(C) ç B(0, r) for some r E JO, for any R> 0. (a) Show that f'(A?)

(fl) Show that f(A,) =

0.

(y) Derive Liouville's theorem. 4. In this exercise we give a proof of Picard's little theorem, based on Ahifors'

lemma and the metric p from Proposition 16(i).

So let f : C

holomorphic.

on any ball B(0,R). (a) Show that f'(p) (fi) Show that f(p) = 0. ('y) Derive Picard's little theorem. 112

C\{0, 1) be

Chapter 8 Geometric Aspects and the Riemann Mapping Theorem We start this chapter by studying geometric aspects of analytic maps, in particular

of the so-called MObius transformations. We finish by proving the Riemann mapping theorem which states that simply connected regions contained in C but different from C, are conformally equivalent.

SectIon

1

The Rlemann sphere

We have earlier studied analytic functions near isolated singularities. To handle such singularities it is aesthetically pleasing and often even convenient to introduce the Riemann sphere = the extended complex plane. As an example consider the map which has an isolated singularity ax z = 0 and whose image does not contain z 0.

map into a homeomorphism of the extended complex plane. Definition I. The extended complex plane consists of the complex plane plus an point 00 (called infinity). We give the topology which is defined by keeping the topology on C and by letting the sets (C\K) U (oo} , where K runs thmugh the compact subsets of C, be the neighborhoods of Co. The sets

,O
constitute a neighborhood basis for the point oo.

It may be remarked that in any book on point set topology.

C

as

found

We adopt the following natural rules for computing with the symbol oo and a complex number z:

—=0, 00+z=z+00=00 ftranyzEC ,

We interpret

zoo=ooz=oo

geometrically by exhibiting an explicit homeomorphism between

the unit sphere

= {(x,y,z) E R31x2 + V2 + Z = i} in ft3 and the extended complex plane As usual we identify the complex plane with the xy-plane in ft3 such that the real axis is identified with the x-axis and the 113

Chapter 8. Geometric Aspects and the Riemann Mapping Thcorcm

imaginary axis with the y.axis in R3. Stereographic projection from the north pole N = (0,0,1) establishes a bijection of S2\{N) onto C. Putting ir(N) := cc we see 52 onto that is a bijection which in coordinates is given by

ir(x y z) =

I

for (x,y,z)

L.oo

Its

inverse ,r' :

—, S2

ir1(z)

=

E

S2\{N)

for (z,y,z)=N

has in coordinates the expression ( (2Rez,2Imz,1z12—1)

forz

1(0,0,1)

forz=oc

E C

z

C It is geometrically obvious (and also easily verified from the explicit expressions above) that both and are continuous so that indeed is a homeomorphism of onto We call for the Riemann sphere when we have in mind the Connection between which is established above by stereographic projection from the north pole. and

Definition 2. A circle in point cc added.

C

A pretty, geometric relation between S2 and

a straight line in C with the

is expressed by

Proposition 3. Under the stereographic projection ir the circles of the sphere the circles of 114

are mapped onto

SectIon 2. The Motxus transfonnadom

Proof: The verification consists really just of elementaxy computations that are better left to the reader. We shall here only note that a circle on is the intersection

0

betweenS2andaplaneinR3.

which is a bijection of onto itselL The Let us return to the map z -. corresponding bijection of the Riemann sphere onto itself is rotation by the angle around the x-axis [seen by elementary computations from the formulas for and

inparticularitmapsneighborhoodsofOontoneighborhoodsofooandviceversa. In the light of these remarks we introduce the following extensions of previous definitions:

w

Definition 4. (i) Afwzcnon f; B(oo; r) C is said to be holomorphic around oo if the function -. f(w') is holomorphic around 0. (ii)A function 1: B(oo;r)\{oo} -. is said to have an isolated singularity at oo

if f is holomoiphic in B(oo;r)\{oo). Such a singularity at oo is said to be removable I a pole of order k I an essential has a removable singularity I a pole 0/order klan essential singularity if w — singularity at 0. (iii) Let be an open subset of . A/unction f which is holomorphic on for isolated singularities, all of which are poles, is said to be memmorphic in

except More

precisely, to each z E there must exist an open disc B(z, r) ç such that either I is holomorphic on B(z,r) or f is holomorphic on the punctured disc B(z, r)\{z} and

z is a pole. A bolomorphic function is in particular meromorphic. In short, a meromorphic function is holomorphic except for poles. Essential singularities are not allowed. Meromorphic functions occur naturally as quotients between holomorphic functions: If f and g are holomorphic in an open subset of and if furthermore g does not vanish identically on any of the connected components of (1, then f/g is meromorphic We shall later see that the converse is true as well, i.e. any meromorphic function can be written as such a quotient (Exercise X.9). A rational function, i.e. a function of the form P/Q where P and Q are polynomials in z, is meromorphic in all of If f is meromorphic in an open subset of C, then we shall from now on ascribe in

thevalueootofatanypointofQatwhichfhasapole. Thenfisdefinedonall of

and is a continuous mapping of Q into

In particular, any rational function

is a continuous mapping of C, into

SectIon 2 The Möbius transformations In this paragraph we introduce and study the Möbius transformations of the extended

complex plane. They are important examples of rational maps with surprising and beautiful properties. 115

Chapter 8. Geometric Aspecis and the Riemann Mapping Thcorcm

Definition 5. A mapping f :

C(, of the form

forzEC forz=oo where the four complex numbers a, b, c, d E C satisfy

detsia

d)

is called a MObius transformation (other authors use terms like fractional linear trans formation, bilinear transformation, homographic transformation or homography).

Theorem 6. The Möbius transformation (1) is a homeomorphism and a biholomorphic map of Coo OntO C00.

If c =

then f is a biholomorphic map of C onto C. If c

0

0 then f is a

biholomorphic map of C\ { — } onto C\ { — As is easy to check, the MObius transformation

9(z) :=

gz+h

IC

hj

(c dj

is the inverse of f. Being rational both f and g are continuous from

into C00 .0

The next two theorems are included simply because they are pretty. We won't need

them later on. Theorem 7. If A

a

= {

we let IA denote the Möbius transfonnation fA(Z) :=

az + b for z E C,, cz + a

The map A -, IA is a group-homomorphism of GL(2,C) into the group of homeomorphisms of So the Möbius transformations form a subgroup of the group of homeomorphisms of C,

0

Proof Left to the reader. Theorem 8.

The set of Möbius transftrmations equals the set of bijective meromorphic mappings

of C00 onto C00. 116

Section 2. The MObus transfaimations

Proof: The only statement which is not yet verified, is that any bijective meromoris a MObius transformation. We verify it: —. Since oo is either a pole or a removable singularity, I can near oo be written as

phic mapping f:

f(z)=z N

forsomeNE{O,1,2,...}

where h is holomorphic near 0. In particular there exist constants C and R such that

If(z)I CIzl'1

f has only finitely many singularities in B[O, R] and since these are poles, there exists a polynomial q 3& 0 such that qf is entire. Combining this with the estimate above we get by Liouville's theorem that qf is a polynomial, say p. So f = p/q is rational. We may of course assume that p and q have no 1" order factor in common. If the degree of the polynomial p is bigger than 1, then p takes the value 0 at two different points or has a double root in any case f is not injective (Cf. Theorem IV.21). Thus p is a polynomial of degree 1 or 0. Similar arguments reveal that q is of degree I or 0. So f = p/q has the form where

If dei{ a

a,b,c,dEC

} = 0 then (a, b) is proportional to (c, d), so f reduces to a constant. But

0

f was a bijection, so the determinant is not 0. Theorem 9. Any Mobius transfonnation maps circles in

onto circles in

Proof: Let us note that any MObius transformation (1) is composed of rotations z —'

e'z where e

E

R

dilations z —, az

where a> 0

translations z z

.

z

+ zo where

This is trivial if c = 0, and for c

0 ii is apparent from

the easily verified formula

a

f(z)=——

ad—bc

1

c

Obviously rotations, dilations and translations map circles in C onto circles in so it is just left to show that the inversion does too. But that is a consequence of Proposition 3 and the remark following it. 117

Chapler 8. Geomeuic Aspects and the Ricmann Mapping Thcorcm

Here is an analytic proof for the reader who dislikes geometry: Lines and circles

in C

are

sets of the form

{z E cla(z12+$z

= o}

where a and 7 range over R and 5 over C subject to the condition Replacement of z by l/z transforms the equation

aIzI2 +f3z +

>

-y = 0

into

a+

fiz + 71z12

0

0

which is an equation of the same type.

Another important and useful property of MObius transformations is that they preserve angles. To formulate that properly and to clarify the concepts we introduce a definition:

Definition 10. Leta : I —. C and 5: J —i C be wo paths and let (to,so) 0 and 5'(.so) 0 a(to) = 5(m)) and that The angle from a to 5 at (to, SO) is the discrete set

argf3(so)—arga(to)=arg

I x J. Assume that

(so

a(to)

of real nwnbers. Theorem 11. Let fl be an open subset of C and let 1: —' C be complex differentiable at Z() E with f'(zo) 0. Then is angle preserving at in the following sense: = 5(so) and Let a : I C and 5 : J C be two paths with =

a'(t0) Oandfl'(sn) 0. Then the angle from foa to fofi a(to, so)

equals the angle from o to 5 at (ta, so).

Pivof. The computation (1 ofl)'(so) (Jo a)'(to)

—

—

f'(5(so))8'(so) — f'(zn)5'(so) — f'(a(to)) a'(to) — f'(zn)o'(tn) — a'(t0)

0

reveals that the proof is simpler than the statement

Coroilaiy 12. The Möbius transformation (1) is angle preserving everywhere in C\ I

{—

}

The Møbãus transformations

Section 2.

Proof. Differentiation of (1) yields

f'(z)=

(cz+d)

A mote sophisticated argument runs as follows : f is injective (Theorem 6), hence f'(z) 0 (Theorem IV.21). 0

The above results enable us to examine any given Möbius transformation by a minimal amount of computations, using in particular that Möbius transformations preserve circles in CQ and angles. Example 13. The Möbius transformation

for zEC,, occurs as the so-called Cayley in the theory of operators on a Hilben space. Since f maps circles in onto circles in we get from 1(0) = —1, f(1) = —i

and f(oo) =

that f(Ru {oo)) is the unit circle. Combining the injectivity of f with f(i) = 0 and f(—i) = 00 we conclude that f 1

maps the upper half plane {z E C I cz > O} onto B(0, 1) and the lower half plane onto B(oo, 1). The mapping of the upper half plane onto the disc is also interesting from another point of view: It establishes a connection between two models of non-Euclidean geometry. In fact f is an isometry with respect to the non-Euclidean distances.

We proceed with another interesting example of a MObius transformation that we shall need below in our proof of the Riemann mapping theorem.

Example 14. Let A(a;.) for a E B(O, 1) denote the MObius transformation

z—o — for 1—az claim that A(o;.) is a univalent map of B(0,l) onto itself with A(a;a) = 0. And that any other such map is a constant (of modulus 1) multiple of A(a;.). Note for later reference ti at A(0;.) is the identity map, that A(—a;.) = A(a; We

that A'(a; a) =

(i

—

lol2)

Proof of the claim : When z

IA(a;

=

A'(a; 0) =

and

E

=

1 —

.9B(0, 1) then

z—a

—=

—

1—az

1—az

119

=

1—a1 —

1—az

=1

Chapter 8. Geomeu*c Aspeota and the Riemann Mapping Theorem

so by the weak maximum principle A(a; B(O, 1)) c B[O, 11 , and further by the open mapping theorem A(a; B(O, 1)) c B(O, 1) Since the inverse mapping A(a; .)' = A(—a;.) of A(a;.) is of the same form as A(a;) we see that A(a;•) maps B(O, 1) univalently onto B(O, 1). Let g be a univalent map of B(O, 1) onto itself with g(a) = 0. Then both := go {A(o;.y'} and its inverse satisfy the conditions of Schwarz' lemma, so If(z)I IzI and 1f'(z)I Izi . But then

lf(z)I so

IzI = 1f'(f(z))I If(z)I

by the equality part of Schwarz' lemma there exists a constant cEC with IcI=1,

suchthatf(z)=czforallz E B(O,1). Andtheng(z)=cA(a;z)forallz E B(0,1).D

SectIon 3 Montel's theorem An important ingredient in our proof of the Riemann mapping theorem is the following compactness result which in the literature is often called the condensation principle. In functional analytic terms it slates that the holomorphic functions on an open set constitute a Montel space.

Theorem 15 (Montel's theorem (1907)). Let F be a sequence of holomorphic functions on an open subset Q of C. and assume that F Lc locally wuformly bounded in Then

F has a subsequence which converges locally uiuformly in

to a holomorphic

function.

The corresponding theorem is not uue on R as the example {sianx

= 1,2,...)

shows.

Proot Let A be a countable dense subset of fl. By the standard diagonal sequence from F which converges at each point a E A. argument we extract a subsequence { converges uniformly on each compact subset K of The program is to show that This will prove Montel's theorem because the limit function automatically will be analytic [by Weierstrass' theorem IV.lOI. Since C(K) is complete it suffices to show that is a Cauchy sequence in C(K). So let e > 0 and a compact subset K of be given. Since K is compact, dist(K, > 0, so we can fix R E JO, dist(K, C\(Z)[, and

let M be a bound for F on the compact subset {z E C I dist(z,K) R). For a E C, z E K, Iz — aI < R and f E F we get by applying Schwarz' lemma to the function w —' [f(z + wR) — f(z)J/2M that 11(z) — f(a)I

<2MIZ

120

—aI

Section 3.

Monteli theorem

e/4

By the compactness of K we can cover it by finitely many discs B(zi,r),B(z21r),... ,B(z,,r) with centers z1,z2,• •. ,z, in K and with (the same) any z

K there is an a2 such that

.1

Iz—o,I

Re

so by the estimate above — fm(z)I

+

Ifn(z) —

—

fm(a,)I + Ifm(a,) — fm(z)l

S

Ifn(aj) — f,n(a,)I

+ which shows that

+ — fm(aj)I

IIfn — fmlloo,K <

sequence {f,(a,)} converges for each fixed j by the vezy choice of {f,j, so when n and m both are sufficiently large. We have thus proved — fmfloo,K 0 is a Cauchy sequence in C(K) as desired. that The

The following useful result is a corollary of Mantel's theorem. It says that analyticity is preserved when the parameter in a family of holomorphic functions is integrated away. Proposilion 16.

wherelis an interval on the real line and Il is an open subset of the complex plane. If

f(t,•)EHol(1Z) for eachfzxed tel, and sup / zEK

If(t,

dt <00 for each compact subset K of 11

then

is holomorphic on 121

Chapter 8. Geometric Aspecta and the Riemann Mapping Theorem

Proof. If I is compact then Fubini and Morera secure the conclusion, so left is the non-compact case. Here we take an increasing sequence {Ia) of compact subintervals

I, such that their union is I. Then as just stated

is holomorphic on Q. By the assumptions {FR } is a locally uniformly bounded sequence.

It converges pointwise to

F(z) := by the dominated convergence theorem. Montel's theorem implies that F is holomorphic

0

onQ.

Section 4 The Riemann mapping theorem Definition 17. Two open subsets V and W of the complex plane are said to be conformally V W, i.e. a univalent map equivalent 4f there exists a conformal equivalence

of V onto W. Conformal equivalence is an equivalence relation [by Theorem W.20].

Examples 18: 1. The Cayley transformation Z —4

z—i

z+t

is a conformal equivalence of the open upper half plane onto the unit disc. < onto the upper half 2. z -. e5 is a conformal equivalence of the strip 0 < plane and takes iir/2 into i. Combining with the Cayley transformation we see that z —.

1 + ic5 CS

+

is a conformal equivalence of the strip onto the unit disc. <2tr conformally onto the plane cut along maps the strip 0 < 3. z the negative real axis. Definition 19. A region is an open, connected and non-empty subset of C.

In

Section 4.

The

Rieniann mapping theoern

V and W are of course homeomorphic if they are conformally equivalent. So topological properties may prohibit domains to be conformally equivalent. For example,

an annulus and a disc are not homeomorphic, and so a fortiori not conformally C and the unit disc

equivalent. But topology alone is not the complete story

B(O, 1) are homeomorphic, but C is not conformally equivalent to any bounded domain

[Liouville's theorem], in particular not to the unit disc. So much the more chocking is the fantastic statement that B. Riemann enunciated in his Gottingen dissertation of 1851:

Theorem 20 (The Riemann mapping theorem). Evesy simply connected region other than C is conformally equivalent to the open unit disc. The Riemann mapping theorem does not provide us with a formula for the conformal equivalence between the given simply connected domain V and the unit disc. For an

explicit solution of a problem it will be necessary to have more information about ô than its mere existence. Explicit formulas for when V is bounded by polygons can be found in Section 17.6 of [Hill. For a discussion of an possible extension of to a homeomorphism of V Onto B[O,l] see [No). An easy, but surprising consequence of the Riemann mapping theorem is that any two simply connected regions of C are homeomorphic. If : V B(O,1) is a conformal equivalence then f —' fo is an algebra isomorphism between Hol(B(O, 1)) and Hol(V). So any problem about the algebra Hol(V) can be carried over to a problem on the unit disc, and the (possible) solution then carried back from there to Hol(V). We will prove the following version of the Riemann mapping theorem. The earlier one is of course a corollary. Theorem 21 (The Riemann mapping theorem). Any connected non-empty open set ç C which Is not the entire complex plane, and on which never vanishing holomorphic function has a holomorphic square root, is confonnally equivalent to the open tutU disc. In fact, given a E there exists exactly

one conformal equivalence H :

—'

B(O,

1) such thai H(a) = 0 and H'(a) >

0.

Proof: For the sake of clarity we divide the proof into 5 steps. We let .7 be the set of all univalent maps of into B(0, 1). Step 1. We show that.7 is not empty. Choose any zo (recall that is not all of C). The function z z — zo does not vanish anywhere in fl, so it has a holomorphic square root g:

(g(z)12=z—zo forall aearly g must be univalent. It is also immediately verified that —g(Q) and empty intersection. is open by the Open Mapping Theorem. Let 123

have E

Chapter 8. Geometric Aspects and the Riemann Mapping Thcorcm

is univalent and maps the and B(zi,r) ç —g(Q) . The map i,b(z) := r/(z — complement of B(zi, r) (in particular g(Q)) into the unit disc B(0, 1). The composite map o g then belongs to ,7. Thus ,7 is not empty.

Step 2. We find H. Every g E .7 is univalent on Il and therefore g' does not vanish on Q. Thus

:= sup Ig'(a)I > 0 9EJ

EJ for n = 0,1,2,... such that lim = q. The f,, are uniformly bounded on so Montel's theorem applies. By choosing a subsequence if necessary we may assume that {f,, } converges locally uniformly on to a holomorphic function H. By Weierstrass' theorem (Theorem IV.lO) converges locally uniformly to H', so that IH'(a)i = . In particular H is not constant. Choose

Step3. HE,7. Suppose not, so that there exist x,y E Q, x y such that H(.r) = H(y). Choose a disc B(x,r) whose closure is in Il, that does not contain y and satisfies that

inf{IH(z)—H(x)IIlz—xl=r} >0 This is possible because zeros of nonconstant holomorphic functions are isolated. If the is as in step 2 we get for large enough n that sequence —


liz — xl = r)

—

If

does not vanish in B(x, r) := — to the weak maximum principle. So

h,,(w) = f,,(w) — f,,(y) =

0

then

1/h,, is a counterexample

for some w E B(.r, r)

But that contradicts that f,, is univalent.

B(0, 1) and {f,,} converges to H, we see that Since f,,(1) is open. Thus H(1l) B(0, 1) and H EJ. H is not a constant function Step 4. H(a) = 0 and H has range B(0, 1). Put now a = H(a) and letg be the composite map g := A(a,H). Then g EJ. By a small computation (a)I = Thus

H(a) =

IH'(a)I > 2= 2 1—lal 1—lol

unless a =

0

0

Suppose H is not onto B(0, 1) and let b E B(0, 1)\H(fl) . The function A(b, H) does not vanish in Q, so it has a holomorphic square root 4, in Q: (4,(z)J2 = .4(b. Clearly 4' E,7. Recalling that H(a) = 0 and IH'(a)I = , we compute that

k/(a)I = 124

Section 6.

Now,

Exercises

also belongs to J, and we find that

:=

l+IbI

>'l

which is a contradiction. Hence H is onto.

Step 5. The uniqueness. The uniqueness of H is an easy consequence of the uniqueness statement in Example 0 14 above.

SectIon 5 PrImitives We have seen in Theorem V.8 that every holomorphic function on a simply connected domain has a primitive there. There are open sets for which this fails. For example the plane punctured at the origin has not got this property : The function l/z defined on C\{O} does not possess a primitive [Because if it did, the line integral of l/z along the unit circle would be zero, but it is 2,rij. It is a most remarkable fact that a connected open set has this property if and only if it is homeomorphic to the open unit disc.

Theorem 22. For any non-empty connected open set ci the following are equivalent: a) Every holomorphic function on ci has a primitive. b) ci is homeomorphic to the unit disc.

Proof: a) => b) If ci = C, the map z

(i —

L.

a homeomorphism of C onto the open unit disc. If ci C then the hypotheses of Theorem 21 are satisfied: Let f be a non-vanishing holomorphic function on ci. By assumption 1,/f has a primitive, say qS. It follows that f = const Adjusting 0 by adding a suitable constant we see that f has a holomorphic logarithm, and in particular is

a holomorphic square root. By Theorem 21 ci is conformally equivalent to the unit disc.

b) =>

a)

If ci is homeomorphic to the unit disc then it is simply connected, and so we may appeal to Theorem V.8. 0

SectIon 6 Exercises 1. When

is a Möbius transformation T a projection, i.e. when is T2 = T? 125

Chapcr 8. Gcomcuic Aspects and the Ricmann Mapping

Theorem

I to 00, the unit circle (except I) to the imaginary axis. > 0) 3. Find a MObius transfonnation that maps the semi-disc {z E 11(0,1)1 onto the positive quadrant. 4. Find a MObius transformation that maps the open right half plane onto 11(0, 1) in such a way that = 0 and > 0. 5. Let and g be holomorphic mappings of 11(0,1) onto a subset of the complex plane. Assume that f is a bijection of 11(0, 1) onto Q and that 1(0) = g(0). Show that 2. Show that the MObius transformation which takes —ito 0, 0 to I and

takes

f

g(D(0,r)) ç 1(11(0,1)) for any r E10,1[ 6. Let f: 11(0,1) B(0, 1) be holomorphic and fix more than one point. Show forall z E 11(0,1). that 1(z) = z 7. (a) Why can't you display an analytic function f: 11(0, 1) 11(0,1) such that = 3/4 and f'(i/2) = 2/3? (b) Can you find an analytic function f : 11(0, 1) —. 11(0, 1) such that f(0) = 1/2 and f'(O) = 3/4 ? How many are there? —' be analytic. Assume be the open right half plane, and let 1: 8. Let . Show that If'(a)I < 1 that f(a) = a for some a E 9. Let f,g : 11(0,1) —' C be holomorphic functions, g univalent. Assume Show that If'(O)I g(B(0, 1)) Ig'(O)I and that 1(0) = g(0) and 1(11(0,1)) equality sign implies 1(11(0, 1)) = g(B(0, 1)). 10. Prove Vitali-Porter's Theorem: Theorem (Vitali 1903, Porter 1904). Let fi be a connected open subset of C and let {f,, } be a sequence of holomorphic } is locally uniformly bounded in Il. and Assume that the sequence functions on that has a limit as n —. oofora set of z E f with an accumulation point in fl. exists locally uniformly in Then lim

f(i/2)

.

Hint : Montel's theorem. 11. Let {f,,) be a locally uniformly bounded sequence of holomorphic functions in a connected open subset of C, and assume that I,, is zero-free in Q for each n. Show that urn I,, = 0 locally uniformly in Q if there exists a zo Q such that

=

0

Hint: Apply the maximum principle to 1/f,, and refer to the Vitali-Porter theorem. 12. In this problem we generalize a pan of Schwarz' lemma.

bea bounded, connected, open subset of C, leta E a holomorphic function with f(a) = a. Let

be

and letf :

—'

We put for

that

(a) Show If'(a)I

that there exists a constant

K, depending on

K. 126

and a but not on 1'

such

Section 6.

(48) Show that If'(a)I 1 Assuming that f'(a) =

Exercises

Him: Compute 1 , prove that f(z) = z for all z f(z)—z=crn(z—a) m +(z—a) rn-fl h(z) .

E

Hint: Write

compute the coefficient cm. (8) Assuming that lf'(a)I = 1 , prove that f is a bijection of Q onto Q. Hint: Find a sequence of integers oo as k oo and a function g such that and

/ j(a)

—41 and

is a conformal equivalence of the strip 13. Show that z the plane cut along the negative real axis. 14. Find a conformal equivalence of

<

<

onto

{zECIIzI < landlz—(1+i)l <1) onto the unit disc. 15. Find a conformal equivalence between the semi-disc >O} and the unit disc. 16. Find an explicit formula for that conformal equivalence H between {z E C (

—1

<

<

1}

and B(O, 1) which satisfies H(O) = 0 and II'(O) > 0. 17. Show that {z E 0 < Izi < i} is not conformally equivalent to any annulus {z E r
Show that f is either 1-1 or a constant (Hurwitz 1889), and show that both possibilities can occur. 19. Let f be holomorphic and bounded in {z

that f(x)

—i 0

as x

E

—1

< 1} and assume

oo.

Prove that

limf(x+iy)=O foreach yEJ—1,1[ and that the passage to the limit is uniform when y is confined to any interval of the form [—a,a] where 0 < a < 1. Hint: Consider f(z + n) for = .r + iy in

the square {z+iyIIxI <1, IvI< 1). 20. Show the following generalization of Schwaz-z' lemma:

A! for 1

and let 1(0) = a. Then for

127

IzI<1

IzI

=

Chapter 9 Meromorphic Functions and Runge's Theorems SectIon 1 The argument principle In this section we derive an important and beautiful result called the argument principle, which expresses the number of zeros of an analytic function inside a closed curve in tenns of a winding number. More generally, it is a formula for the difference between the number of zeros and poles of a meromorphic function. To reassure ourselves that we will not encounter meaningless expressions like 00—0° we make the following observation.

Observation 1. Let f be meromorphic in an open set Cl ç C , and asswne that there is a compact subset of Cl that contains all the zeros (resp. poles) off in 11. Then the nwnber of zeros (resp. poles) off in Cl is finite.

0

Proof: Left to the reader.

We remind the reader that index = argument increment divided by 2,r (Remark (3) following Definition m.12).

Theorem 2 (The argument principle). Let Cl be a simply connected open subset of C, a closed curve in Cl and f meromorphic in Cl. Let be oil zeros and poles of f Z2," , z, and pi with rnultiplici:ies n m, respectively. Suppose thai none and mz, , of the zeros and poles lie on Then

nj

1nd101(O)

—

mfr

= If7 is even a closed path then

= >njlnd,(z,)—>2m&Ind,(p&) See also Exercise 1.

Proof: Write

f() =

(z

(z —

—

129

Chapier 9.

Functions and Runge's Theorems

where g is holomorphic and vanishes nowhere in Q. Choose any continuous logarithms for — , y — PL and g (recall that g does not vanish in ()). Then

— z,)—

is a continuous logarithm for f o

The

proof follows from the definition of md

(Definition ilL 12) and the formula for the index of a path (Theorem 111.17(d)).

L]

To understand what the above means, let be a circle. For w inside the winding number = 1 and for w outside it is zero. The argument principle thus says that the change in argf(z) as z traces equals Z — P where Z is the number of zeros and P the number of poles of f inside multiplicities counted.

Example 3.

The argument principle can give information about the location of the zeros of a polynomial. We present here a typical example of the kind of problems that can be solved:

0, b 0 and

The polynomial f(z) + a? + bz + c, where a exactly 2 zeros in the first quadrant.

If t 0 then 1(t) >

>

0 and

0,

> 0 has

so f has no zeros on the positive

semi-axes.

Let R> 0

and

consider the quarter circle

z = Re',

0 9

On this quarter circle f is

f(Re*9) = R sufficiently large the parenthesis has positive real pan, so a continuous argument function for f along the quarter circle is 9 —, 89

+ Arg

}

where Arg denotes the principal determination of the argument.

Consider now the curve indicated on the following figure 130

Seclion 2.

Rouche's theorem

y

iR

x

We have observed that > 0 when t 0 . So as z varies from iR to o along the imaginary axis the argument of f increases by Arg 1(0) — Arg.f(iR) = —Arg f(iR) , and so the argument increment is in absolute value less than ir.

The argument of f is constant (viz, equal to 0) from 0 to R. Finally from R to iR along the circle it increases by where < ir. So along the curve + = + on the figure the argument off increases by 4,r + something , where But the argument increment is an integer multiple of 2,r, so here it is exactly 4,r. By the argument principle there are exactly 2 zeros of f (counted with multiplicity) inside the contour, and hence (since R> 0 can be arbitrarily large) in the first quadrant.

SectIon 2 Rouché's theorem The following clever lemma will be useful in the proofs of Rouch6's and Runge's theorems. It is an integral representation theorem like the Cauchy integral formula. In contrast to the earlier results it asserts the existence of a curve with certain properties, not that every curve has those properties. The power of integral representations has already been demonstrated in connection with the Cauchy integral formula; it led to local power series expansions.

Lemma 4. Let Il be an open subset of C and let F set of piecewise linear closed paths 7i, 'Y2,' 131

Cl be compact. Then there exists a finite , y,

in Cl\F such that for any function

Chapter 9. Meromophic Functions and Runge's Theorems

IE =

0

and 1=17

aEF j—1

Proof: Consider

the following grid (i.e. set of squares) on R2

j range over the integers, and where the positive integer N is chosen so large that any square of the grid having a point in common with F is contained in We orient each of the squares the counter clockwise direction. Denote by Qi, Q2, Qm the set of those squares in the grid that have a point in common with F. The union of these Q1 contains F that much is clear. Note also that if for example a vertex of a square belongs to F, then all the four squares having this point as a vertex are included in the system ,Qm. A similar remark applies to a point on an edge. Let L1, L2,. .. , L, denote the boundary segments of this system of squares, i.e. L1, L2,. .. , L, are those edges of the system which are edges where i and

,

-

of exactly one square. Then L, must for each j be contained in Q\F

.

Since each

square is oriented, each segment is oriented, too. Now let f be holomorphic in ci. Since ci we get by Cauchy's theorem that (1)

ff(z)dz=0

for

1
aQk

And if a belongs to the interior of the square Qk (it can belong to only one)

(2)

and

J aQk

f(Z)df()

for the other squares

f(Z)df(

(3)

k

k

ÔQ

132

,

,

Section 2.

theorem

If an edge is common to two squares, then the orientations induced on it are opposite

and so the net contribution is zero. We get

>Jf(z)dz = 0 Using the same reasoning we get from (2) and (3) that if a belongs to the interior of one of the squares then (4)

to the interior of one of the squares. (4) is therefore by continuity valid

forall aEF.

We shall finally show that the oriented line segments

= [a,, b,) where j = 1,2,••• , r can be grouped together to a collection of closed paths 71,72,• For any polynomial p we get from what we have already shown that

0=

=

so

= for any polynomial p

(5)

E C let

For

:= := Choose

i.e.

the

number of j for which a =

the

number of j for which

a polynomial p such that p(zo) =

1

and such that p =

0

at the other points,

at

..,b,}\{zo}

{al,a2,. We

=

conclude that the a's nit

the

same as the b's; more precisely that there exists

a permutation of (1,2,.. .,r) such that b1 = for j = 1,2,••• ,r , and so = [a,, for j = 1, 2, ... , r Any permutation is a product of disjoint cycles, and each of the cycles of gives

0

rise to a closed path of L's.

133

Chapter 9. Meromoiphic Functions and Runge's Theorems

Lemma 5. Suppose a/Let be open, let F ç Il be compact and let f be meromorphic in the zeros and poles off are in F. Then ef the are as in Lemma 4 we have

1,

where

Z and P are the number of zeros and poles off, counted with multiplicities.

Proof: The proof is similar to that of Theorem 2. Represent f as there with the as the zeros and poles of f with the corresponding multiplicities ii, and z, and Differentiating we find

I Lemma 4 applied to the holomorphic function f =

7 3

Since

I

1

gives

= 1 for all a E F

7,

g'/g is holomorphic in Q we get the desired result from Lemma 4.

0

theorem) has many applications as is apparent from The following result the exercises to this chapter. Another version of it was proved in Chapter IV (Theorem IV.l3). theorem). Theorem 6 Let K be a compact subset of C with interior Int(K). Let the two mappings f , g K —. be meromorphic in Inz(K), continuous on K, finite-valued on OK and there stuisfying

(*) 11(z) — g(z)I < If(z)I + Ig(z)I

forall z

OK

Let Z(f) and P(f) (resp. Z(g) and P(g)) denote the number of zeros and poles of f (resp. g) in Int(K), counted with multiplicity. Then Z(f), P(f), Z(g) and P(g) are finite, and Z(f) — P(f) = Z(g) — P(g). expresses the natural condition that f and g should be The crucial condition "close" to one another, if only on OK. Proof: By (*) neither f nor g has zeros on OK. Since they are finite-valued there, we see that there is a neighborhood U ç C of OK such that f and g are finite-valued on U fl K and the inequality () is still true for all z E U ri K. In particular all the 134

Section 3.

Runge's theorems

poles and zeros of f and g belong to the compact subset F := K\U of Int(K). Hence their numbers are finite (Observation I above). Let us for )h E 10,

1]

consider the meromorphic function hA :=

Al +(1 —

A)g, which

has all its poles and zeros off U, i.e. in F. By Lemma 5 we see that the quantity

is integer for each A E [0, 1]. Being continuous as a function of A it takes the same

0

valueatA=OandatA=1. A special case of Rouch6's theorem is the following: Theorem 7 (Rouché's theorem). Let f and g be holomorphic in an open set containing a compact set K. If

< If(z)I +

11(z)

forall z E 8K

then f and g have the same (finite) nwnber of zeros in K.

SectIon 3 Runge's theorems We know that a function which is holomorphic in a disc, can be expanded in a power series, and so be approximated uniformly by polynomials on any compact subdisc. If a function is holomorphic in an annulus, say in B(0, 1)\{0) , then it cannot necessarily be approximated by polynomials. The obvious counter example is z —. But we can in an annulus resort to a Laurent expansion and hence approximate it by rational functions. Runge's theorems deal with approximation of holomorphic functions by polynomials or rational functions on more general sets than discs and annuli. We remind the reader that we have introduced the concept of a pole at 00 (Definition Vffl.4). To set the stage for Runge's theorem on approximation by rational functions we note that a polynomial (non-constant to be pedantic) is a rational function with a pole at oo. And that a rational function whose only pole is at 00, is a polynomial.

Theorem 8 (Runge's theorem on approximation by rational functions). Let F be a compact subset of C. Let S ç be a set which meets each connected component of COO\F. Let be holomorphic in an open set containing F. Then to each e > 0 there exLus a rational function P/Q with poles only in S such

f

P(z)

<eforallzEF

Proof: In the proof it will be convenient to interpret expressions of the form where P is a polynomial, as polynomials in z. 135

Functions and Runge's Theorems

Chapter 9.

Letting set

_V1,_V2,••

denote the open set in question there is according to Lemma 4 a finite of oriented segments in Q\F such that

forall

(5)

zEF

7

so the theorem will be proved once we show that each summand on the right hand side of (5) can be approximated uniformly on F by finite sums of the type where the P's are polynomials and the c's belong to S. Let us just treat the first summand;

the other terms may be handled in exactly the same way. The line integral

J can be uniformly approximated by Riemann sums

where b,E-y'.

b_z

Again let us just treat the first term (bi —

, the rest being handled similarly.

does not meet F so b F. Let E S as b, and assume first that c 00. Join b and c by a curve contained entirely in this component. Since this is disjoint from F, there is a positive distance between this curve and F. We can find points We drop the subscript and write 6 for b1.

be in the same connected component of CQO\F

b

=

co,

,c,,, =

c

on the curve such that

< dist(curve,F)

(6) 2Icj

Then writing z—b = z—ci +ci —bwe get = (z

—

civ' >(b — ci)k(z

the series converging uniformly for (7)

(z —

E F by (6). So we may choose N so that

z

b)'

—

N —

>(b— ci)k(z — k=O

is uniformly small for z E F. We can use the same argument on each term of the sum

in (7) to replace

by C2. Namely (z —

= (z

—

C2+

—

cj)_k

which can be expanded in a series that - because of (6) - converges uniformly for E F. In other words we can find a polynomial in (z — c2)' which approximates (: — uniformly on F. This procedure can be repeated until we reach c = Cm. 136

Section 3.

Runges theorems

The argument is thus complete except when the point c chosen in S is oo. To treat where 6 belongs to the unbounded component of C\F. this case consider (z — Choose din the same component such that ffl < for all z E F. From what we approximates (z — have done above there is a polynomial P such that occurring in the polynomial can be each (z — uniformly on F. Since f expanded in powers of z/d, i.e. there is a polynomial which approximates (z — uniformly on F. The argument is thus complete.

0

For the next result (Runge's polynomial approximation theorem) and for the discussion of the inhomogeneous Cauchy-Riemann equation we need a fact from point set topology:

Proposition 9. Let be an open subset of the complex plane. Then there exists an increasing of compact subsets of Q such that sequence K1,

(a)K1UK2U••• (b)

ç mt

for all n =

1,2,

(c) Every compact subset of is contained in for some n (d) Every connected component of COO\K,I contains a connected component of

= The last statement (d) says that K, has no holes except the ones coming from the

holes of

(make a drawing!).

Proof: The cases fl = 0 and = C arc trivial, so we shall concentrate on the remaining case in which 0. We will show that we for n = 1,2,... may take

K, := This K, is a closed and bounded subset of C, so it is compact. It is now almost

obvious that (a), (b) and (c) arc satisfied, so left is only point (d). Here we begin by noting that if a connected component of , say U, intersects a connected component of say V, then U contains V: Indeed, since CQO\KII the connected components of COO\KII cover in particular they cover V. They arc open and disjoint, so by the connectedness of V we get U V. Hence (d) boils down to proving that each connected component U of intersects To do so we notice that

=

U B(z, !) 137

UB(oo, n)

Chapter 9.

Meromorphic

Functions and Runge's Theorems

Let w E U. Then w lies in one of the balls, say w E B(zo, The ball is a connected subset of COO\Kn, so B(z0, , so U intersects component. Now, zo E U fl

ç

U

,

where

U being a connected

0

Theorem 10 (Runge's polynomial approximation theorem). Let bean open subset of C. Every holomorphic function on can be approximated uniformly on compacts by polynomials

is connected.

Proof: We will first show that the condition is sufficient. Proposition 9 tells (c) that it suffices to approximate the given holomorphic function f on any given (d) and so it that COQ\KTh has only one connected component. That one contains in Runge's theorem on approximation by contains oo. Take now S = {oo} and F = rational functions (Theorem 8) to get polynomials to approximate f uniformly on is not connected. Then there exist two Now for the necessity: Suppose that We may suppose that closed, disjoint, non-empty sets F and F1 with union 00 E F1 so that F is a compact subset of C. Clearly

is open. Put 0 =11 U F. F is a compact subset of the open set 0. Let be paths in O\F = as in Lemma4. Let a E F. The function z —,

(z

holomorphic on Q. Assume now that all holomorphic functions on are uniform limits on compacts of polynomials. Then we can in particular approximate the function z —. (z — 0)_i . These by a sequence {Pk) of polynomials on the union of the paths , polynomials are of course holomorphic in 0 and hence by the first pan of the conclusion of Lemma 4 we have =

0

'dz =

0

, which in the limit as k

oo

becomes

.1

0

But that contradicts the second pan of Lemma 4.

Remark 11. If Q be an open subset of C, then

is connected dJf is simply

connected.

is connected. Any function f which is holomorphic on may by Runge's polynomial approximation theorem be approximated uniformly by polynomials on compact subsets of So if is a closed path in Q we conclude Proof: Suppose first that

138

Section 3.

that ffdz =

0

Runge's theoscins

which implies that f has a primitive (use the proof of Lemma 11.12)

and hence that ci is simply connected (Theorem V11L22). Assume conversely that is not connected. We then want to show that ci is not simply connected. We assume it is and arrive at a contradiction as follows: Let F and F1 be closed, disjoint, non-empty sets with union . Without loss of generality we may suppose that oo E F1 so that F is a compact subset of C. Clearly

=I1UF isopen.

PutO:=f1UF. Fisacompact subset of the open set 0.

a)'

be the closed paths in 0 \ F =

Cl from Lemma 4. Let a E F. Since z (z — is holomorphic in Cl we get according to The Global Cauchy Theorem (Theorem V.7) that

•1

0

which contradicts the second part of Lemma 4.

Remark 12. Let K be a compact set. Then the unbounded component of C\K is precisely those a E C for which the function z (z — a)' is a uniform limit on K of polynomials (This follows from the proof of Theorem 8). Runge's theorem on polynomial approximation is not the last word in these matters.

Clearly, if a function f on a compact set K is the uniform limit of a sequence of polynomials then f E C(K) fl Hol(Int(K)) , so this is a necessary condition for approximation by polynomials. Mergelyan's theorem states that it is also sufficient:

Theorem 13 (Mergelyan's theorem (1952)). Let K be a compact subset of C with C\K connected. Let I C(K)flHol(Int(K)) Then there exists a sequence of polynomials converging to I wuformly on K. Note that when K has empty interior then the only condition remaining on f is that C(K). So the classical approximation theorem of Weierstrass for an interval is a particular case of Mergelyan's theorem. For a comprehensible proof of Mergelyan's theorem the reader can consult Chapter 20 of [Ru]. Another extension of Weierstrass' approximation theorem is provided by

I

Theorem 14 (Carle,nan's theorem (1927)). C(R) then there exists to every continuous function e: R function F with the property that

1ff

11(x) — F(x)I <€(x) 139

for all XE R.

JO, oo(

an entire

Chapter 9.

and Runge's Theorems

Mercmorphic

A proof can be found in IBu;Chap.VIII,*5]. For further generalizations (Arakelian's theorem, Keldysh-Lavxent'ev's theorem) see the references in the notes to Chapter VIII of [Bu], the note [Ga], ERR] and the survey article [Vi].

Section

4

The inhomogeneous Cauchy-Rlemann equation

In this section we shall use Runge's theorem to show that the inhomogeneous Cauchy-Riemann equation has a solution on any open subset ci of C. We recall from Chapter II that the Cauchy-Riemann operator is the first order differential operator

8

.8\

118

on

C=R2

and that a function u c C'(Q) is holomorphic on ci 1ff it satisfies the homogeneous Cauchy-Riemann equation Ou

on ci. In the case of ci = C an explicit formula for a solution can be displayed: Proposition 15.

If I

E

C8°(R9 and u(z)

if

w)d() for z

f(z—

C

denotes the Lebesgue measure on R2, then u E C°°(R2) and

where

= f.

Although the function w has a pole at w = 0 it is nevertheless integrable over any ball, so that the integral in the proposition makes sense: Indeed, using polar coordinates we get

I J

B(O,1)

Iwl

=

2r

jjr 1

dp(w)

1 11—rd4'dr=27r
0

0

the equation 0 on has a solution u E That is known from the thecry of partial differential equations. We may for example choose u = * E , where E is a fundamental solution of D. A fundamental solution of the Cauchy-Riemann operator is E(z) = ; that produces the formula of the proposition. We will, however, not appeal to the theory of partial differential equations, but prove directly that the formula of the proposition works. For any constant coefficient differential operator D

Du =

f

E

f

140

Section 4. The Inhomogeneous Cauchy-Riemann equation

Proof of Proposidon 15: is locally integrable we sec that u is continuous and Since the function to that we may differentiate u by doing so under the integral sign. So for any differential operator D with constant coefficients we get

(Du)(z)

=

if

(Df)(z —

w)d()

which shows that u E C00(R2). To prove the other statement we fix z E C. For simplicity in writing we delete the subscript R2 on the integrals and put F(w) := f(zo — to). Now,

i (Of

.9u

d14w)

—(zo)=— I irj8i — —

to

i f

i

H

/

—

Using that is of first order (so that the usual formula for differentiating a is holomorphic, we get product holds) and that to

8 f F(w) 1

- OF —

+ F (to)

O(w')

— —

OF

i

so

Ou

i fO(F(w))

1.

= ——j

fOG

j

IWI>(

where we, again for the sake of simplicity, use the notation G(w)

F(w)/w. Intro-

ducing the definition of the Cauchy-Riemann operator we find by a small computation that OG

so Green's theorem for the annulus

=dzv(G,zG)

<

Iwl

vanishes out there, implies that

J 141


where

R is chosen so large that G

Chapter 9. Memmorpluc Fwicdons and Runge's Theorems

where

denotes the usual arc length measure on the circle Iwl = e. Now,

I I

=

IwI=c

IwI=e

j =

J IF(w) - F(O)]dA(u') + F(0)}

{

=0+F(0)=F(O)=f(zo)

0 Theorem 16.

Let be an open subset of the complex plane. The inhomogeneous Cauchy-Riemann equation Ou/8i = I has a soluüon u E any I E C°°(Q). Proof: be an increasing sequence of compact subsets of Let K0 = 0 ç K1 c K2 ç Q as in Proposition 9. Note: Any function which is holomorphic on an open set containing K,,, can be approximated uniformly on K,, by elements from Hol(Q). (Combine Runge's theorem (Theorem 8) with the property (d) of Proposition 9). Choose E C$°(1) such that q5,, = 1 on K,,. (even in C00(R2)) such that By Proposition 15 there exists uj E = Since and U2 E C°°(Q) such that 0U2/01 = (= 1) on K1 we get =

O(u2—uj)

=0 on K1

so u2 — u 1 is holomorphic on Int(K1). Trivially 1

because K0 =

0.

142

on K0

Section 4. The Inhomogeneous Cauchy-Ricmann equation

Assume now that we have produced uI,u2,

Hol(IniK,) and

u,+i

vE

such that

E

•

We will extend the system {uI,u2,... such that

by adding one function more: Choose

Ov

uz = Since qS,, =

4

on

we get

=

and so v —

0

E Hol(Int

By the note we may find a w E such that Iv — u,, := v — w to the system. In this way we produce a sequence from

—

wI <'2" on Ks_i. We

add

such that

N

WnJ

z

and

E —

Any compact subset of

<

1

Ofl Ks_i

is contained in some KN (by property (c) of Proposition

9), so the sum —

converges uniformly on any compact subset of Q. Thus

u :=

—

+

is a well defined function on (1. For any fixed N we have whenever n N , so the right hand side of

—

E Hol(Ini KN)

(*) is

on mt KN the limit of a uniformly convergent sequence of holomorphic functions,

so it is itself holomorphic on ml KN (Weicrstrass' theorem). , and since N is arbitrazy, u E 143

In particular u E

Chapter 9. Meromotphic Functions

Runge's Thootems

to (*) on mt KN the right hand side vanishes (it is holo-

When we apply morphic), so

8(u —UN)

=

0

on mt KN

so

Since

N is

c3u

OUN

liZ

liZ

on IfliltN = I throughout Il.

we read that

0

SectIon 5 Exercises 1. Generalize The Argument Principle as follows: Replace the condition about simple connectivity by the condition that 'y is a closed path in Q with the properly that

Ind7(z)=0 forall zEC\Q 2. Show that the polynomial z —+ first quadrant. 3.

Let f be

z4

+ z3 + 5z2 + 2z + 4 has no zeros in the

meromorphic in a neighborhood of B[o, 1], and assume that there are . Let a E C satisfy

no poles on the unit circle Izi = 1

al> sup lf(z)l lzI=1

Show that f has the same number of poles and a-values (both counted with multiplicity) in B(0, 1). 4. Consider the holomorphic function 1(z) := sinh(irz) in the half strip {x + iy >0, 0) exactly once. S. Find the number of zeros of the polynomial z —. zT — 5z4 + in the annulus

{z E C 1 < IzI <2). 6. Show that the polynomial z —.

3z15+4z8+6z5 + 19z4 +3: + 1

has 4 zeros in lzl < 1 and 11 zeros in 1 < Izi < 2. 7. Find a ball B(0, R) such that the polynomial z3 — roots in it.

8. Let a E C , lal > e and let n E {1,2,3,• 1(z) :=

—

az's

4z2 + z — 4

has exactly two

Show that the function .} has exactly n zeros in B(0, 1), and that they are different from .

one another.

= 1 has exactly one root: in 9. Let A> 1 . Show that the equation 1], and that the root belongs to the open interval 10,11 = 0 has exactly one root :o in 10. Let.\ > 1 . Show that the equation A — z — the half plane {z E C I 0) , and that > 0. What happens as A —+ 1 ? 144

Section 5.

Exercises

11. Show that all four roots of the polynomial z4 — z + 5 lie in the annulus 1.35 < fri < 1.65} , and that there is exactly one root in each open quadrant. where 0 < ao
{z E

13. Use Rouché's theorem to prove that an nth order polynomial has exactly n roots. 14. Use Rouché's theorem to prove that the positions of the roots of an 0th order polynomial depend continuously on the coefficients of the polynomial. 15. Prove the following lemma: Lemma. are holomorphic functions on the disc B(0, R), and suppose Suppose 02,• ,

that wo E C is a simple root of the polynomial

Then there exist an r EJO, R[ ,and a

+ ai(z)(c?(z))"1

Hol(B(0,r)) such that

+

=

0

for all z

=

u'o

and

B(O,r)

16. Let f be analytic on a neighborhood of B[O, 1] and assume that f(B[0, 11) c B(O, 1). Show that f has exactly one fixed point. 17. Prove the following theorem: Theorem. be a sequence of functions which are holomorphic on an open set Let Assume that —, fo locally uniformly on Q as n —. 00. Assume furthermore that

fo has a rem of order no at zo r > 0 there exists an N = N(r) such that n > N => the function Then to zeros in the set ii B(zo,r). has at least Hol(Q) are univalent 18. Let fl be a connected open subset of C. Assume that uniformly on compacts as n —' 00. —'

for n = 1,2,... and that

I

Show that f is either univalent or constant. 19. Show that the open set C\S is simply connected, where S is the infinite spiral (Archimedes' spiral)

S:={z=retnIOr
(y) C,\Cl is connected. (6) ff = 0 for each f E Hol(Q) and each closed path 1

(e)

Every I

HoI(Q) has a primitive. 145

in Cl.

Chaper 9. Meromarphic Functions and Runge's Theorems

f

(C) Every never-vanishing E Hol(Q) has a continuous logarithm. Every never-vanishing f E has a continuous square mot. is homeomorphic to B(0, 1). (0) 21. Show the following theorem: Theorem.

Let K be a compact subset of C with the property that C\K is connected. Let f be holomorphic in an open neighborhood of K. Then f is on K a uniform limit of polynomials.

22. Let Q

and

let I E Hol(Q). (a) Does there exist a sequence of polynomials converging uniformly to f on

?

(fi) Given a compact subset K of Q, does there always exist a sequence of polynomials which converges uniformly to f on K ? Does there always exist a sequence of polynomials which converges uniformly to f on Q, if f is even holomorphic on an open set containing the closure of ? 23. Does there exist a sequence { P, } of polynomials such that for all E C we have as n —, 00

(1

>0 0

—'

146

Chapter 10 RepresentatIons of Meromorphic Functions Infinite products and construction of meromorphic functions with prescribed zeros

and poles are the main themes of this chapter. The Euler product for sine, the partial fraction expansion of cotangent and the r-funcuon will be the showpieces.

Section 1 InfInite products It is easy to construct a holomorphic function which has zeros exactly at finitely many given points z2,• , ZN E C. Simply take the function N

k=

With an infinity of points

1

is tempted to take

(z — zk). Although this kFl product does not converge, a suitable modification (where we insert suitable convergence one

factors) does, as we shall see in Weierstrass' factorization theorem. So we are led to study infinite products. An infinite sum behaves in most respects like a finite one. Analogously an infinite product should behave like a finite product; in particular

= 0 should mean that

at least one of the factors ck vanishes. However, examples like 11 1 lu

——...—(nfactors) =0

n—.oo22

2

demonstrate that we must be cautious, at least if the limit is zero. That is the background for the following rather involved definition of an infinite product.

Definition 1. Let C2,.•• be a sequence of complex numbers. The infinite product (1)

is said to converge .7 there to each >0 exists an N N such that for all vu N

andp 1:

n+P

(2)

H ck—l

If the infinite product (1) converges, then the numbers C, :=

form a bounded sequence. Using (2) we find that {C,, } is a 147

cg

,

vu =

1,2,..

sequence and

Chapter 10.

therefore has a limit in C.

We

Reptesentalions of Meromorphic Functions

use the notation fl

also for this limit and say that

it is the value of the infinite product (1). If a finite number of factors is inserted in or removed from a convergent infinite product then the result will still be a convergent infinite product (and the value of the result will be the obvious one). Taking p = 1 in (2) we see that 1 as n .—. 00 if the product (1) converges. That is the reason why it is often convenient to write an infinite product in the form

11(1 +ak). Also note that if H

is

= 0 if and only if at

a convergent product, then H

least one of the factors is zero. We leave it to the reader to verify the following two quite elementary properties of infinite products: dk both convergent

and

k=I

H

fi

converges

and H ckdk

convergent and H

=

(ñ dk)

0

k=1

converges and

H! =

1

fl

k= 1

The following result turns out to be very useful, because it connects infinite products with sums.

PropositIon 2. The infinite product fl (1 + Ok) converges 41 E IakI <00 , and in that case (3)

Proof: We first show the inequality (4)

fl (1 + Ok) —1 is a polynomial in

with positive coefficients, say

k=i

p(a1,a2,...,aii)=fl(1+ak)— 1 148

Section I.

Infinite products

p has positive coefficients so

which is the inequality (4).

Noting that 1 + x 9 fx any real x we put x =

Iak

I

in (4) and deduce the finite

version (3') of (3): (3')

to

—1

Coming now to the pmof of the proposition we get from the inequality (3') applied that

/n+p

\

IakI)_1

H (1+a&)—1 k=n-fl If

E 101,1 <00 then the right hand side is small, independent of p 1 if only n is

large.

We next turn to infinite products where the factors aie analytic functions. First a natural definition. Definition 3. be complex valuedfunctions defined in an open subset (f 1 of C. We say Let ai, that the infinite product fl (1 + a1,) converges locally uniformly in (1 + ( z))

converges a: each z E and if furthermore to each compact subset K of ci and each > 0 there exists an N such that for all z E K and n N

+ak(z))—[J(1

Lemma 4. Let

be holomorphic on an open subset Ii of C.

(a) If the infinite product fl

limit function F(z) :=

(1 + 01,) converges

locally uniformly in ci, then the

[1(1 + a1,(z)) is holomorphic on ci.

F vanishes precisely at those points where a: least one of the factors 1+

vanishes.

In particular F doesn't vanish identically on a connected open set unless at least one of the factors does so. 149

Chaptcr 10.

(fi) If as N infinite product

00 the

Represenladolis of Meromorplüc Functions

sum E an(z)I converges locally wuformly to 0, then the

(1 + ak) converges locally wuformly in 11.

Proof. The last statement (8) the inequality (3).

of

the lemma is a consequence of Proposition 2 and

0

It has already been observed that a convergent infinite product vanishes at a point 1ff at least one of its factors does so. If F vanishes identically on a connected open subset 11o of 11 then at least one of the factors vanishes at uncountably many points of In that case the zeros of that factor has a limit point in flo. By the Unique Continuation Theorem (Theorem IV.ll) the factor is identically 0 on Qo.

Definition 5.

1fF is holomorphic then P/F is called the logarithmic derivative ofF. Placing ourselves in the situation of Lemma 4

= [1(1 + ak(:)) we

where

k=I

compute formally to get log F(z) =

+ (lk(Z)) which by formal differentiation

yields

''

(5\ F'(z)

00

g

ak(z)

—

F(z)

or zE

f

though it would be tricky to justify this way of deriving the formula (5) for the logarithmic derivative of an infinite product, the formula is true, and it is possible to prove it directly. That is done in the next proposition. Even

Proposition 6. Let at, be holomorphic in an open subset 11 of C and suppose that the sum E Ia,s(z)I as N —+0° converges locally uniformly in 11 to 0. Let F denote the value

of the inflniteproduct F(z)=fl(1+ak(z))andpw

1

+ak(Z)

converges absoiwely for any z E 11o, and it converges locally uniformly on

the meromorphic function P/F. so

"

'

F'(z) F(z)

00

=

k—I

Ck(Z)

1+ 150

ak(z)

jor z E

11 0

towards

Section 2.

The Eu1ex formula for sine N

Proof: The finite product FN(z)

= k=1 fl (1+ ak(z)) converges as N —'oo locally

uniformly on to F (Lemma 4). By Weierstrass' Theorem (Theorem JV.lO) the sequence of derivatives } converges locally uniformly on fl to F'.

Let K be a compact subset of flo. Then PN/FN converges uniformly on K to

F'/F.

Since

rM(z)

N

FN(Z)

kl

aL(z)

+ ak(z)

forzEflo

we have only left to show the claim on absolute convergence, i.e. that 00

a&(zo)

Ii + for each zo fin. Given E fin there exists an N such that Iak(zn)I <

converges

for k N, and hence

for k N. Thus it suffices to verify that that the denominator Ii + ak(zn)I such
=

1

and

= 14(zo)I

.

By assumption the series E cLak converges

locally uniformly on fi, hence (Weierstrass' Theorem) so does its derivative

In particular it converges at z = zn, which is the desired statement.

£

0

Section 2 The Euler formula for sine The infinite product

F(z) := converges (by Lemma 4) locally uniformly in the entire z-plane, so that F is an entire function. The zero-set of F is Z and each zero has multiplicity one. The function z sin irz has the same properties. Hence z —. sin (rrz)/F(z) is holomorphic and zero free in the entire z-plane. We can write

(6) sin irz = where A is entire. The main difficulty is the determination of A. To find A we take logarithmic derivatives in (6) 00

(7)

1

151

/

1

1

Chapter 10. Rcpresentatiom of

Functions

The series on the right hand of (7) converges locally uniformly on the open set C\Z. Differentiating (7) we get = A"(z) —

(8)

and since

£

+

(Zfl)2)

converges, we may rewrite (8) in the form

(9)

1

A"(z)=

(z

+ n) 2

sin

The right hand side in (9) is unaltered if we replace z by z+1. In other words, the entire function A" has period 1. To show that A" is a constant it is thus by Liouville's theorem l,IyI 1). For

such z we have 00

00

-1

00

+E((1 Since 0

x 1 we estimate as follows

(x+n)2n2 and and get

=

If y > N (N positive integer) then

2>2'N2 For the other term of the right hand side of (9) we estimate as follows for y > 0 (The case y <0 can be treated similarly):

— e"

2

= =

2

eTtxe_TP —

=

2 —

2

e2T*xe_nI

2 2

2

feTy — &TY\2 2

152

)

2

Section 3.

Weiersuass'

theorem

so for any y we have sin2

—

sinh2

These inequalities say that A"(z) tends to 0 as z tends to infinity in the pciiod strip

0

1, so we conclude that A" = 0, Ic. that A' is a constant. By (7) A' is odd, so A' = 0, i.e. A is a constant. Finally, dividing both sides of (6) by z and letting z tend to 0 we evaluate the constant to be and arrive at the beautiful Euler formula for sine: (10)

=

rzfl (i 4) —

Let us note that we on the way also found (formula (7)) the partial fraction expansion

of cotangent:

1÷

(11) ircotarz=

for zE C\Z

As an application of (10) take z = 1/2 to get

00,

00

2 4n—i

4n2

n1 from which we obtain Walls' product: —

2

00

4n 2

14n2—1

—

00

(2n) 2

1(2n—a)(2n+1)

22 4•4 1•3 3.5 5.7

As another illustration divide both sides of (10) by 1 —

z

and let z —.

1

to get

2fl2(n) For another treatment of the topics of this section sec [Wal].

SectIon 3 Welerstrass' factorizatlon theorem In contrast to the example of the last section consider the problem of constructing an

enme function with simple zeros exactly at the points 0,—i, —2, —3,... The product z

jj

(1

+

not do, simply because it does not converge. To ntas such cases

Weierstrass introduced certain explicitly given factors - the so-called primary factors - into the product so force convergence. Bcforc we sake up the general case let us illustrate the idea by the following lemma: 153

Chapter 10. Representations of Meromophic Functions

Lemma 7. Let al,

be different complex nwnbers with the property that

001

<00

n=1

Then the infinite product

(12)

converges locally uniformly on C and its value is an enüre function with simple zeros precisely azak , k = 1,2,3,...

Proof: We start with an estimate, valid for all IaI <

1

(13) I(1_a)ea_1I1a12 To

prove (13) note that we for lal <

1

have

J(1_a)eO — 11=

(n— 1)!

—

=

1a12

1)! —

Now,

1an12 < 00 SO

oo

as

n —' 00. For z in a fixed

< 1 uniformly in z for all large n. Using (13) we

compact set K we have realize that the series

as N

must tend to infinity

converges to 0 locally uniformly on C. Lemma 4 now takes over.

In the general case we will modify the infinite product fl(i — each factor (1 —

0

by replacing

by a factor of the form (14)

(1

—

We choose the functions in such a way that the new factors are so close to 1 that the product converges. That (1 — is close to 1 means that )i(z) approximates . The technical details arc as follows: log ((1 — z)—') 154

Soction 3.

Wcicrzuass

factorization theorem

Definition 8. The primary factors of Weierstrass are the entire functions

E(z,O) =

1—z

for Note that the primary factors vanish only at z = We will need the estimate

1.

for n =

(15) 1E(z,n) — 11<

and fri

1

which shows that E(z,n) is close to I for large n, even though it is zero at z = To prove (15) we find by a simple calculation that

so the power series expansion of —E'(z,n) around z =

0

1.

starts from z" and all its

coefficients are nonnegative. From

1—E(z,n)= J (—E'(w,n))dw lO,zl

we then see that the power series expansion of 1 — E(Z,n) starts from

nonnegative coefficients as well, i.e. 1—

So for izi

1

=>

E(z,n)

(ZLZk where all

0

we get

>

—

akizlk

=

=

>

—

—0) =

which is (15).

1:

Andthenforanyrn0,n (16) IE(z,n)m —

ii

2mIE(z,n) —

where we have used the identity

and

the fact (evident from (16)) that 1E(z,n)i 2 for fri 155

1

E(1,n))

and has

Chapter 10. Representations of Mcromorphic Functions

9—k

(17) IE(z,m +k)—

With these preliminaries to a side we can prove Weiers:rass' factorization theorem:

Theorem 9 (Weierstrass' factorizadon theorem). be a sequence ofthffereiupoiiusfrom Q Let Q be an open subset of C • let be a sequence of positive integers. with no cluster point in Q. and let finally m for at There exists a function g E Hol(fl) such that g has a zero of order suchthatgvanishes nowhere else in Il. k = 1,2,..., and

Proof: not cluster in

Since the sequence {zk} does

there

exist a E

and r > 0 such

B(a,r)

fl and such that B(a,r) does not intersect the set {:k}. Let us for convenience assume that a = 0 and r = 1. Under the transformation : —. 1 the open then goes into an open set V. If := 1. set E V and {ck} that

ç'.

does not cluster in V (Recall that 0 does not belong to V). Choose for each k a point a& E C\V closest to ck. Since {ck} does not cluster in V we get (18)

tim

— CkI = 0

k—oo

[Indeed, if for some 5 > 0 and subsequence {k,) we have that ak. — ci,I 5 , where tends to c (recall that {ck} is bounded), then c must belong to C\V. But by choice we have of lc&,

— ci Ica. —

a contradictionj.

Now put

(19) f(z) :=

for

zEV

If F is any compact subset of V. then by (18) for all large k, 2Ich — aal < z

uniform convergence of the infinite product in (19) now follows from

(17) and Lemma 4. By assumption B(0, 1) we have

so {zIlzl> 1) —

z—ak Using (17) and (19) we see that


(20) If(z)I

is

whenever

V. Since Iakl land lckI 1, Iz 3

bounded for lzl 3 156

The r-function

Section 4.

at 2k and f(1/z) for a 0. Then g has a zero of order Now define g(z) vanishes nowhere else in Q\{O} . (20) says that g is bounded in B(0, 4)\{O). 0 is therefore a removable singularity of g and g can hence be defined to be analytic at 0 as well. If g(0) 0, g is the required function. If 0 is a zero of g of order n, then 0 fulfills all the requirements.

Section 4 The r-function Taking

=

—n

in (12) we find that

(21) g(z) := etzz[1

(i +

is entire with simple zeros precisely at a = 0, —1, —2,.... Here is a constant chosen so that g(1) = 1. is called the Euler constant. The reciprocal of g is meromorphic in C with simple poles at 0, —1, —2,... This is the famous F-function (gamma-function): (22)

and it is of utmost importance. The F-function was introduced by Euler in 1729 (See [Ds] for a readable account of its history). Let us derive some of the most important properties of the F-function. Since q is entire, F has no zeros. Writing

(23) g(z) = and

+

1)e'H { (i +

exp

}

replacing a by a + 1 in (21), the following computation is valid for any a g(z) _e zg(z+1) —

(

z

1

z

+!)} =

n(n1+

1)} fi

}

Using that

=1 we find further that (24)

157

Chapter 10.

Representations

of Meromoiphic Functions

Thus g(z) = czg(z) for z O,—1,—2,•••, where c is a constant. By continuity this holds for all z E C. To find the constant c we note from (21) that lim

z

z—O

=

1

Since also g(1) =

1 we see that the constant is 1. Translating from g to 1' we find the functional equation for r:

(25) F(z+1)=zI'(z) for Since

=

1,

(25) leads to

(26) r(n-l-1)=n! for which explains why the r-functjon is also called the factorial function. We proceed by deriving another fundamental relation for the r-fiinction, called the formula of complementary arguments. Usc (21) with z and —z and multiply to get

g(z)g(—z) =

_z2[J

Appealing to the Euler formula (10) for sine and to (25) the above relation reads in

terms of 1'

(27) r(z)r(1 — z)

=

for z

Z

(27) is the formula of complementary argwnents.

In particular since r'(x) > 0 for z > 0, we get taking z = 1/2 in (27) that (28)

The reader might in other connections have encountered the Mellin transform (= which is the function F9 Euler integral) of a complex valued function g on defined by the

:= Jtz_19(t)dt wherever the integral converges. is of special interest for us here because this The particular case of g(t) = equals the r-function. This result is called The Euler integral formula

(29) F(z) = Jt1_1e_tdt for

158

>

()

Section 4.

The F-function

To derive it we use Prym's decomposition (30)

Jtt_1e_tdt =

+ Je_ftt_Idt = P(z) + Q(z)

/

The first integral on the tight, i.e. Q(z) is entire. Expanding e_t and integrating term by term we find that the second integral equals (31)

The right hand side of (31) converges uniformly in any compact set not containing the points z = 0,—i,--2,•.., and we see that

(32) F(z) is a meromorphic function in the entire complex plane with simple poles at = 0, —1, —2,.... Ii also equals the integral in (29) for > 0. Using this, routine integration shows that

(33) zF(z) = F(z + 1) when 3tz >

0

But then (33) must hold in the entire z-plane minus the points 0,—i, —2,.•• (The Unique Continuation Theorem). We know that r satisfies the identity (33) in form of (25) and that F never vanishes. Thus from (22) we see that

(34) 0 :=

= Fg

is entire and has period 1, i.e. G(z+1) = G(z). Since F(1) = F(1) =

1,

G— 1

vanishes at the (real) integers. Now, sin wz has period 2, has simple poles at the integers and vanishes nowhere else. Comparing 0 with sin irz we see that

0(z)—i

(35) H(z)

srnirz is entire and has period 2. We will prove that H is zero by showing that H is bounded

in the period strip 1 3tz <3, and then applying Liouvile's theorem. F is bounded in this strip, as is evident from its integral representation (30). Let us estimate g in the strip. If z = a + iy is in the strip under consideration then

I(i +

X)2

=

+

(1 +

Using this estimate in (21) and separating products we get:

(i

Ig(z)I lzIe7'{[J (1 + =

+

=

I

a

159

y

Chapter 10. Representations of Meromorphic Functions

where the last equality comes from the Euler sine formula (10). Using this estimate it is elementary to show that (remember that g(x) is bounded for 1 x 3 and that

= isinh(iry)) (36)

g(z)

—'0

sin irz

as z—'oo in

From (34) and (35) g(z)

IH(z)IIF(z)I

+

Using (36) we conclude that H is bounded in the period strip and hence everywhere.

By Liouville's theorem H is a constant. Since H(z) tends to 0 as z tends to oo in the period strip, the constant is 0, so H is identically 0, i.e. F = r. We have thus derived

0

(29).

From (28) we then get by a change of variables the important formula

=

(37) J

J

=

As an application of the Euler integral formula (29) we shall derive Stirling's asymptotic approximation of the factorial function:

Theorem 10 (Stirling's approximation formula). —,

V2,r as n —' 00

The proof is adapted from the one in [Pa). Introducing x = as new variable in (29) we find —

=

2J (i +

Since 1 + y expy we get for x> (except for the factor exp (—x2))

the following estimate of the integrand

2,*— 1

(1

+

< exp

—

1)} exp

e so the integrand above is bounded uniformly in n by the integrable function e exp (— z2). 160

Section 5.

The Minag-Leffier expanston

We next show that the integrand converges poiniwise, so x is fixed for a moment. Using the power series expansion of the logarithm we find for n 00 that

x\

/

x

/1

x2

and so log

{(i +

= (2n — 1)log (i +

+

= (2n —

=

—

—

+

—

Finally, by the dominated convergence theorem 00

2J

e_Z2e_X2ds

=

as n —.

00

00

0 As the end of this paragraph we mention the famous dupilcanon formula of Legendre

(see the exercises for a proof):

=

(38)

1r(z)r(z +

Section 5 The Mittag-Leffier expansion The theorem of this paragraph is analogous to the Weierstrass theorem on zero sets in §4. Instead of zeros we deal with poles. Before we state and prove the theorem we remind the reader of the following fact from the theoiy of Laurent expansions:

Theorem 11. Let f be holomorphic on the punctured disc B(O, R)\{O}. Then f may in exactly one way be written in the form

1(z) = g(z) + h

for z

B(O, R)\{O}

where g E Hol(B(O, R)) and where h is an entire function such thai h(O) = 0.

Proof This was seen during the proof of Theorem VU.

0

More generally, if a E C is an isolated singularity of a holomorphic function f then f can be written in the form (39)

f(z)=h(_!_) +g(z) 161

Chapcr 10.

Representations of Meromorphic Functions

where h is an entire function with h(O) = 0, and where g is a holomorphic function on the domain of definition of f and a is a removable singularity for g. The decomposition (39) is unique. The first term, i.e. the function z —' h((z — aY') is called the

principal part offal a. Example. The function (Cf formula (9) above) 2

sin2 has

for each n

E

1

=

— n)2

N the principal part (z — n)2

at

n.

=

z

However, if we are given a sequence of principal parts, then their sum will normally not converge. Mitiag-Leffler's theorem says that a meromorphic function with the prescribed principal pans nevertheless exists:

Theorem 12 (Miltag-Leffler's theorem (1884)). be an open subset of C. Let pi, be a sequence of different points from without cluster point in Il. and let P11P2,... be a sequence of polynomials without constant term.

Let

Then there exists a function which is meromorphic on fl, has

of poles in

is Pk((z

and whose principal part at

} as its set

PLY')for all k = 1,2,...

—

Remark. If we only specify that the desired function should have poles at P2 of given orders mi, then it can be found by help of Weierstrass' theorem on zero sets from §4: Indeed, if f is a holomorphic function with zeros at p1,P2,••• of orders then 1/f satisfies the requirements.

Proof: LCtB(pk,rk),k=1,2,.. be

and choose for each k

disjoint balls in

a function

4'k E

r&))

such that

is identically I on the smaller ball Bk :=

so

it satisfies

f

£ the requirements of the theorem except that it is not holomorphic. To

function

is smooth on

.

.} and reduces on B& to fk.

remedy that we consider the function F

g is clearly

on

f

for z E cZ\{pi,p2,. . for z E And on the punctured disc Bk\ {p,, } the function

f reduces to a holomorphic function (viz 1k). so Of Ofk 9 — — — oz uz 162

Secdon 6. The

on Bk\{pk} . Thus

and p-fimctions o( Weiers*rass

g

The inhomogeneous Cauchy-Ricmann equation = g has a solution u E C°°((l) (Theorem IX.16). by the veiy definition of g. Now, h := — u is holomorphic in On Bk\{pk} we have h = 1k — u. As observed u E Hol(Bk) , so h has the desired 0 principal pan

f

Wcierstrass' factorization theorem (Theorem 9) told us how to construct a holomorphic function whose zero set is a prescribed sequence of points. The next result is that what can be done with zeros can be done with any sequence of values:

Coroilar, 13 (Gerasay'z interpolation theorem). be an open subset of the complex plane. Let {zl,z2,...) be asubset of Cl w2,••• be a sequence of complex numbers. without limit points in Cl, and let Then there exists a function f E Hol(Cl) such that f(zk) = Wk for k = 1,2,...

Proof: By Weierstrass' Factorization theorem (Theorem 9) there is a holomorphic function 0 fork = 1,2,... By the 10 with simple poles at the zk, i.e. fo(za) = 0 and Mittag-Lefficr theorem there is a function h E Hol(Cl\{zi,z2,. .}) such that .

Wk

fo(zk)

1

z —

holomorphic in a ball around zk for each k. Now, f := foh is holomorphic on Cl, we find that because the zeros cancel the singularities, and for z is

f(z) = fo(z)h(z) = fo(z){h(z) —' 0

+

— ft,(zk)

z — Zk

}+

z — Zk

= Wk

sof(zk)= Wk fork= For a generalization of Corollary 13 see Exercise 17.

Section 6 The (- and p-functions of WeIerstrass We have examples of periodic holomorphic functions. E.g. exp, cos and sin. Liouville's theorem rules out that a nonconstant holomorphic function can have two independent periods. We shall now give an example of a doubly periodic meromorphic function, viz Weierstrass' p-function. Let w and w' be non-zero complex numbers with non-real quotient, i.e. they are linearly independent over the reals. Let G be the group generated by w and w', i.e. G consists of the complex numbers of the form g = mw + nw' where m and n are integers. If w and w' are periods of a given function then so is every g E G. 163

Chapter 10. Representatiom of Metamorphic Functions

By Minag-Leffler's theorem there is a meromorphic function with principal part g)1 at each element g E G. However we do not have to appeal to that theorem, since an example is given by the so-called Weierstrass (-function (Weierstrass zeta(z —

function):

(42)

To prove that (is such an example we prepare a lemma. Lemma 14. 1

(43)

>

where ImI+JnI1. If

Proof: = k either

ImI + ml

k/2; if, say lml

or frzl is

+ nw'l =

k/2 then

ka

=

+

where a = Now there axe at most 4k pairs (m,n) such that lml + ml = k. Thus

>2

191

k=1 ImI+1n1b

0

which proves (43).

Let us now return to the right hand side of (42). Consider the disc B(O, 1?). There arc only finitely many g G satisfying R, as is clear from f.ex.. (43). Since 1

1

z—g

g

z

(z—g)g

g

if

=

,

,and and Izl

IgI

R

the series

>2 IgI2R

(1 z

1

9

g

z g

converges uniformly in B(O, R) and represents a holomorphic function in this disc. It follows that ((z), given by (42) is meromcrphic in C with simple poles precisely at

the points of G. 164

SectIon 7.

Exercises

Differentiation of (42) termwise (which is permitted) leads to the

p-

function (Weicrstrass pc-function):

(44) p(z) = .-('(z) =

+

{ (z—g)

2 —g1

}

The p-function is meromorphic in C with double poles precisely at g E G. Let us now show that the p-function is doubly periodic with periods w and Differentiating

(44) we get

(45) p'(z) =

2

2

—

— —

g)

— g)

For any h E G the series for p' is unaltered by a change from z to z + h because C is a group. In other words, p'(z + h) = p'(z), implying that

(46) p(z+h)—p(z) =C(h) where C(h) is a constant, perhaps depending on h. But the series for p shows that is an even function : p(z) = p(—z). Now it is immediate from (46) that

C(h+g)=C(h)+C(g) forall h,9eG Using these facts we find that

But C(h) + C(—h) C(O) = 0, so we must have C(h) = 0 and so p(z + h) = p(z). We record this in the final statement of this section: p and p' are doubly periodic with periods andw', and = —p.

Section 7 Exercises 1. Show that 00

2

3

2. For which z E C will the infinite product :[•J

(i

+

converge? Show that the value of the product is (1 — Hint:

(1— z)J1[ (i +

=

165

1—

Chapter 10. Representations of Metamorphic Functions

N

3. Assume that the limit urn

N—co

0. Show that the product

exists and is

fl

n1 4. Assume that the infinite product fi (1 +

converges. Discuss convergence

of 5. Let

the sequence

} of complex numbers satisfy that

(i) Show that the infinite product ( a so-called Blaschke product)

B(z) := a function which is holomorphic in B(O, 1). Find its zeros. as above with the property that every point on the unit (ii) Find a sequence is a cluster point of circle Iz I = I 6. We shall in this exercise present another way of dealing with the entire function A from §2 (Herglotz' trick):

A(z)=ircot(irz)—

/1

00

2z

(a) Show that A' satisfies the functional equation

+A'(!j_!)}

A'(z) =

(8) Use the functional equation to show that A' is a constant. (Hint: The maximum modulus principle).

7. Show that 2

by help of the partial fraction expansion of the cotangent.

8. Show that

i-ri COslrz=Ilcl— ( 9. Let f

a quotient I

be

4z 2

(2n—1) 2

meromorphic on an open subset Cl of C.

Show

that f

= g/h, where g and h both are holomorphic in Cl. 166

can be written as

SectIon 7.

10. Let t

Exercises

R\{O} . Show that ,T.2

Ir(it)I = 11.

(i) Show that Euler's constant equals

f

1

1

(ii) Derive Gauss' formula N!NZ

r(z)= lam

for

N—co

and from it deduce Wallis' product

2•2 4.4

21.3 3.5 5.7 (iii) Prove the Legendre duplication formula

= 22z_1r(z)r(z ÷ 12. Find the residue of rat z = 0. More generally at z = 13. Let C be the following contour

—n

for n =

y

C

-ì

Prove Hankel's formula

r(z) = For which z E C is Hankel's formula valid?

14. Let fl and B = {b1, b2, . . .} be as in the Mittag-Leffler theorem. Let fk for k = 1,2,••• be entut functions with fk(O) = 0. 167

Chapter 10. Representations of

Functions

Show that there exists an f Hol(Q\B) such that the principal part of f at bk fk((z—bk)) for all k = 15. With the notation of §6 show that

a(z) :=

is

(l(z)2)}

—

is entire with simple poles exactly at the g e G. The logarithmic derivative of is Weierstrass' (-function. a is called the sigmafiuzctlon of Weierstrass.

Show the following generalization of Proposition 2: Let be a sequence of reals with 11k > 0 for all k. Let cl,c2,••• be a sequence of complex numbers from the open unit disc, and assume that 16.

<00 Show that the product

[1(1 converges (principal determinations). 17. Generalize Corollary 13 to the following theorem: Theorem. Let Cl be an open subset of the conzple.x plane and let { z 1,22,

.}

be a subset of CI

withow limit points in Cl. Let there for each Ic = 1,2,..• be given finitely many complex Then there exists afunction I Holffl) such that nwnbers wk,1,wk,2,••

for On

and

Ic

=

Hint: Modify the proof of Corollary 13.

18. Derive the formula (37) directly without resort to the r—function as follows: Let

i :=J e_t'dx show that

j2

=J

and introduce polar coordinates. Another elegant way of proving (37) can be found in [An). 168

Chapter 11 The Prime Number Theorem Section 1 The Riemann zeta function The famous prime number theorem states that ir(r), the number of primes less than

or equal to x, is asymptotically the same as z/logz. This was conjectured by Gauss and about a hundred years later proved by Hadamard and de la ValIde-Poussin. It is this proof that we present below. An "elementary" proof was given by A. Sclberg [Sc). The proof uses the celebrated Rieniann zeta function defined for z in the open half

1) by:

plane {w E C

(1) ((z) := The (-function was studied already by Euler, long before the time of Riemann (See [Ayl). Euler found e.g. formulas for ((2n), n = 1,2,•.•. These formulas are derived in an elementary fashion in [Ber]. The special case ((2) = ir2/6 has a short elegant proof [Ap2]. Very little is known about the values of the (-function at other natural numbers, except that it has been established thai ((3) is irrational. See the informal report [Po]. We shall presently show that the function given by (1) for 1 is the restriction to this open set of a meromorphic function on C with a simple pole at z = 1. It is the extended function that is usually called the (-function. It is remarkable that the properties of this function should relate so decisively to the prime number theorem. For more information on the (-function we refer to the monograph [Ed).

Theorem 1. The series (1) converges wufornily on every half plane of the form {z E C

I

a)

where a> 1. There isameromorphic function(onCsuch that ((z)—(z—1)1 is entire and such that ((z) is given by (l)for > 1. Proof: We concentrate on the second statement and leave the first one to the reader. For > 1 we have

r(z) so

=

J t'1edt = nt

7t*_1eltdt

therefore 00

(2)

((z)r(z) =

00

=

11=10

169

/ 0

—

Chapter II.

The Prime Number Theorem

where interchange of sum and integral is justified by the dominated convergence theorem. Write

((z)r(z)=

(3)

Noting that the second integral is entire in z, we concentrate on the first one. The

function z/(ex — 1) is holomorphic in the ball Izi 00

Z

—1

>:" for IzI<2ir

The series converges in particular for z = 2, so a,,2" —, 0.

such that Jani

we may expand it

so

Thus there is a constant C

C2" for n = 0,1,2,.... Since an = 1, we have 00

1

1

e:_1 =—+Lanz z n= I

—

Using this expansion and term by term integration (justified by uniform convergence) the first integral in (3) can be written

Ii 1

(5)

I

J

00

dt=—+) z—1 1

a,,

0

Thus we may for

> 1 write 00

(6) As said before the third term on the right in (6) is entire; the second term is by our estimate on a,, meromorphic in the complex plane with simple poles at —(n — 1) if a,, 0 , n = 1,2,.... And the first term has a simple pole at z = 1. Writing g(z) = 1/r(z) we may define ( on the complex plane by

(7)

Since

g is entire with zeros at 0, —1,

.

the right hand side of (7) is meromorphic

on all of C with at most one pole, namely one at z = 1. Since g(1) = 1 ,the function ((z) — 1/(z — 1) is entire. 0 Riemann discovered a remarkable functional relation for the (-function from which many of its properties can be deduced. 170

Section 1.

The Ricmann zeta function

Theorem 2 (The functional equation for the zeta-function). The following relation is valid: (8) ((1 — z) = (2ir)

or written in a more synvne:ric form ((1 —

(8')

Proof: Integrate by parts in (2) for

1

to obtain 00

00

((z)F(z)

=J

di =

/

so that

(9)

-

z((z)I'(z) =

Now from the formula (8) of Chapter X we get (10)

The function 1 —

1

1

=

4sin2

001

1

+

t(it +

(it — 2,rn)2

sin2 z Z2

/ \

sin z \

I

Z/\

sin Z

like z2/3 near z = 0. Hence the first integral in (9) is holomorphic for

—1.

for

+

behaves like z2/3! near z = 0, so 1—

behaves

j2

The last integral is in fact entire. Thus (9) holds for

<1 =

—

(11) zC(z)1'(z) =

+

171

> —1. However,

Chapter ii. The Prime Number Theorem And now we can use the expansion in (10). We may interchange the sum and the integral because of dominated convergence: Indeed, if —1 < 0, then

=

E Il

(2irn)

(

s

=

+

—

I

4T2n2

—

n—i

3 s < 00 1

+

Thus using (10) in (11), changing the summation and integration around and using the

substitution i =

we get

(12) z((z)I'(z) =

}d

+

+iu)2

(1

— z). The integral - which is The sum on the tight hand side is just independent of n - can be evaluated as follows: The integrand is

2

so

uZ

(1 + u2)2

integral equals

the

2

1—

00 1

J

00

1 — U2

(1+u2)

2

I

I

1+u

j

00

00

I

J

du =

(1+u2)

I J

u

1+u2

du

where we have just integrated by parts in the second integral. The integral on the right hand side is for —1 < < 1 evaluated below to be 00

(13)21 j

1+u2

du=

cos(!2!)

0

<0. Since Using (13) in (12) we conclude that (8) is valid in the strip —1 < both sides are meromorphic in all of the complex plane the relation is an identity. 0 Proof of

(13): The substitution v =

converts

the above integral to 00

00

f uT I—dv=I

and

the

j1+v

J

0

0

1+v

dv

required result is a particular case of the formula 00

f J I

dx =

"

r(p+q) 172

when

>0,

>0

SectIon 2.

Euler's product formula and zezus of

combined with the formula of complementary arguments X.(28). To derive the formula we write

r(p + q)(1

=

Jte(1

=

J

9—;

:';rqdt

e_t(l+z)gp+q_lzp_ldg

Now we integrate both sides with respect to x from 0 to ooand change the order of integration on the right hand side (which is justified by absolute convergence):

r(p + q)J(1

(le_tz#_;dx)

=

(le_'vu'_;dv) di

r(p)r(q)

SectIon 2 Euler's product formula and zeros of ( Euler's product formula relates the (-function directly to the prime numbers. Simple consequences are some results on the zeros of the (-function.

Theorem 3 (Euler's product fonnula).

(14) ((z)=[J(1

_x)

Wizen

where p in the infinite product ranges over all primes p> 1. Proof: Let us first prove that the infinite product converges: If we for each prime p write 1

then

=

it suffices by Proposition X.2 to show that

Now 00

p

p11=1 173

pn=1

Chapter 11.

The Prime Number Theorem

The right hand side is a subset of the terms from the series

>1P_* +p_21 To

=

+

prove that the value of the infinite product actually is E

we consider the

1

finite product

P(N)

:= fi

for NEN

pN We can multiply this product out and rearrange the terms as we please without altering the result

it is just a product of (finitely many) absolutely convergent series. A typical term is of the form because

—a*z

—aix —a2z

Pj

P2

. ,p,, are different primes and ai, By the fundamental theorem of arithmetic P(N) =

where

,

E

are non-negative integers. where A = A(N) consists

nEA

N. Thus

of those n E N whose prime factors all are

n=1

nED

where B consists of those n having at least one prime factor > N. In particular

n1 The

n>N

right hand side converges to 0 as N

E

n)N

oo, so

asdesirecL

As an amusing commentary on (14) we may note that there are infinitely many primes (that fact was already known and proved by Euclid): Indeed, if there were but finitely many then the (-function would not diverge at z = 1. Not only are there infinitely many primes, but as observed by Euler (1737) there art so many that the series formed by the reciprocals of the prime numbers diverges, i.e.

174

Section 2.

Euler's Foduct formula

zeros of C

(Sec Exercise 1). For elementary derivations of this fact consult [AG] or [ApI ;Theorem 1.13].

Coroila,y 4.

1. For

The C-function doe.s no: vanish for points —2m , m =

<0 is vanishes only at the

Proof: The first assertion is inuncdiatc from (14). Since the r-function has no 0 zeros, the second assertion follows from the first and the functional equation (8). Taking logarithmic derivatives in (14) we obtain for

Logp

(15) —

1p_Z

=

where

> 1 that

= EA(rn)m_z

=

A(m)= fLogp ifmisapower(>O)ofp otherwise

0

It is permitted to rearrange tenns because of the absolute convergence (Cf Proposition X.6).

The fundamental link between the C-function and the distribution of primes is provided by the following function (called Chcbyshev's function)

A(m) for x E [1,oo[

(16) &(x) in

The main result, known as the Hadamard-de Ia Vallée Poussin theorem asserts that

(17)

We will first show that the prime number theorem is a consequence of (17):

Given a prime p the number of k such that pk z is less than or equal to (Logz)/(Logp), so recalling that A(m) = Logp if m is a power of p we get from the definition (16) that

Logp = ir(z)Logz

(18)

pz x(x) = the number of primes less than or equal to x. On the other hand, if 1 < y < x then

where

(19)

1 175

Chaper 11. The Prime Numbu Theorem

Taking y = x/(Logx)2 and noting that

x>e


for

any

that

logz

(20)

x

log x — 2 log

the prime number theorem is equivalent

to (17) is now

z That

log x

log z

evident

from

(18) and

0

Beforeweembaikonthepmofof(17)weneedonemorefact,namelythat( has =1. That result is due to Hadarnard and de la Vailde Poussin.

no zeros on the line

Theorem 5. The (-function has no zeros on the line

=

1.

(is meromorphic, hence so is ('/(. Thus, for any complex number w, urn (z — w)ii(z) exists and is an integer. It is positive when w is a zero of (, negative when w is a pole of (, and zero otherwise. Proof:

Ifz=1+e+itwheree>OandtERwehavefrnm(15)thal

++

cos(i logm)

= —

m2

Using

0(1 +cos6)2 =

1

+2cos6+coa26=

we then get + e)) +

(21)

=

+ e

+ it)) +

—EA(rn)m'll +cos(tlogm)12

+e

+ 2it))

0

Now multiply the above bye and letelendtoO. Ifl-i-itwereazeroof(then

wegetthatthelimitofthelefthandsideof(21)is8—3+4+0=1.

0

This contradiction proves Theorem 5.

Section 3 More about the zeros of ( In this section, which is

of (.

a digression

we mention couple of results on the zeros

have seen that ( has no zeros in the closed half plane rtz

The functional equation (8) then tells us that the only zeros in the half plane Rz 0 arc -2m, m = 1,2,.... These latter zeros are called the trivial zeros. All the non-trivial < 1. This is called the critical strip. zeros - if any - are thus in the open strip 0 < We

176

1.

Sec*ion 4.

The prime number theorem

Westartwiththeformula (2) or rather

the consequence 1

C(z)F(z)

which we for

>

1

=Jet_idt+Jet

1dt

rewrite as

((z)I'(z) = ]tx_i(, The

co

jz—1

i

above formula is derived for > 0 it is valid in

meromorphic in

+

—

> >

1

0.

1, but since the right hand side is For 0 < t we have

1

et —1

so the first integral above is negative when z > so the third term is at most

0.

For t 1 we have et — 1 t2

Finally,forO< z<1 wehave 1/(z—1)< —1,soC(z)r(z)
0

One of the most famous still unsolved problems in mathematics is the

Riemann hypothesis: All non-trivial zeros of

lie on the line

=

Nwnerical computations support the validity of the Ricmann hypothesis. In [LR] it is shown that the C-function has exactly 300 000 001 zeros whose imaginary parts lie between 0 and 119 590 809.282 and all of them have real part 1/2 See also [Wag). On the theoretical side we will here just mention that t has infinitely many zeros on the line = a result due to Hardy (1914).

SectIon 4 The prime number theorem We now set out to prove the prime number theorem. It is not easy to explain why or how the method caine about. Let us start with the right hand side of (15). Using Abel's partial summation formula (Proposition 1.10) we get

177

Chapter 11.

The Prime Number Theorem

<x the last term above tends to 0 as N

is defined in (16). Since by (18) when > 1. Thus where

Since

for m t < m + 1 the last sum can be written

equals

>z J j_Z_ldjb(flz) = zJ

=

where we have just introduced y = log t as new variable - it is much more convenient to deal with the interval [O,oo[. Thus from (15) we get

(22) —

z((z)

=

Je

x—l)*H(j)dt when

>1

0

where H(t) := Note that (17) is equivalent to lim H(t) = We can further rewrite (22) as

(23) A(z) :=

—

=

1.

J

—

1)dt

The reason for doing so is: We know that ((z) — (z — iy' is entire (that is Theorem 1). A simple computation then shows that the function A is analytic wherever z((z) 0; in particular (by Theorem 5) in a neighborhood of the line = 1. The right hand side of (23) is the Laplace transform of the function H—i. Referring back to the fact that (17) is equivalent to H(t) — 1 —. 0 as t —4 cc, we now realize that we need to prove a type of Tauberian result (A Tauberian result derives the asymptotic behavior of a function from the behavior of its averages). We define p : R —' R by when ku when

p(y) := I

S

2

and let u > 0, A > 0 be real parameters that later on in the proof will converge to 00. With z = 1 + E + jAy, where e > 0, we multiply both sides of (23) by and integrate with respect to y to get

(24) J A(1 + +

=J 178

J(H(t) {

— 1

)e

dy

}

Section 4.

The

prime number theerem

Interchange of order of integration is permitted on the right hand side of (24) because

1 + H(t) = 1 +

IH(i) —

< 1+t

by (18), so the double integral is absolutely convergent. Effecting this interchange and evaluating the elementary inner integral we find

fA(1 + +

(25)

=

J(H(t) —

The left hand side of (25) is continuous in e at = 0, because A is holomorphic on the line = 1, and p is bounded. The limit as e \ 0 of the left hand side therefore exists. Also

7sin2(u_—_At)df

J (because

> 0 and u

(u—At)


real), so by the monotone convergence theorem we get from

is

(25) that

'

00

IH(t) sin (u—At) dt
J 0

Going to the limit

0

in (25) we get

JA(1 +

=

J(H(t) —

The above holds for A > 0 and any real u, so we may replace u by Au and change variable to get

=7 (w(u

AJA(1 +

—

—

Writing the integral on the left hand side above as a sum of the integral over [—2,01

and that over [0,2) and integrating by parts we see that the integral on the left hand side is bounded by C/lu I for a suitable constant C = C(A). In particular the left hand side converges to 0 as u — oo (The reader may possibly recognize the statement as a special case of the Riemann-Lebesgue lemma). Thus for A > 0, (26)

tim

JH(u

—

=I

179

Chapter 11.

The Prime Number Theorem

It is not necessary to know the value of the last integral, so we just call it B (Actually B = by Exercise VI.16). Observe that e'H(t) = is increasing in t, so that we for b > 0 have

+ b), i.e. H(t)

(27)

Now, for 11(u)

H(i + b)e'

<

H(u+

—

H(u +

S

—

—

which implies that exp

1

2\

I sin2 t

1

1

J

+

J

t \ sin2 t

1 —

A(II+*) i

1

J Using (26) and letting u / 00

above

sin2t

we get

B

exp (—i) limsupH(u)J This relation holds for all ..\ > 0, so letting .A

oo we deduce that

(28) limsupH(u)

1

In particular H is bounded, say H M. Let e > 0 be arbitrary, but fixed. For sufficiently large

I Replacing

u by u

—

* in (26) we get for all .A > A(e) that

B=lim

J

e+liminf J 180

say.\ > A(e) we have

Exercises

5.

From (27) we get for

(31)

that

<

H(u —

S

S

H(u)e*

—

so we may continue the estimates as follows

B e + liminf J

2

<e+ e?TliminfH(u)

J -is;

t

2

<e +

e + IiminfH(u)B ,

Letting A —, oo we find B 1

I

and

so, e >

0 being

arbitrary,

liminfH(u). Comparing with (28) we have finally arrived at Theorem 6 (The prime number theorem). denotes the number of primes less than or equal to n, then

ir(n)Iogn n

1 as n —4 00

5 ExercIses

SectIon

1. Show that

where

the summation ranges over all primes p. Hint: If not, then the infinite product

converges, 2.

and you may take z = 1 in the procedure of the proof of Theorem 3. R be the Mobiusfwwtion, i.e.

Let,i : N

p(l) = -

1

. - p7k) =

if

pr,---

,Pk are distinct primes and

p(n)=O otherwise Show that the series (a so-called Dirichiet series)

n1 181

= ... =

=1

Chaper 11. The Prime Number Theorem

> 1 and that

converges absolutely when

((z) when

=

1

> 1.

Hint: Note first that

(1 ifn=1 if n>1

din

3. Let

: N —, C have the pmpeiiies (i)

whenever (n,m) =

=

1

(ii) > k(n)I Show that

(n) =

fi {i+

+

+

where p in the product ranges over all primes. , where z E C is fixed with Apply this result to := is the Möbius function from the previous exercise, to get

As another application take 4. The result

=

> 0 and ji

1.

21 j 1+u2 du=

for

from the proof of Theorem 2 can also be verified by calculus of residues. Do

182

Chapter 12 Harmonic- Functions Harmonic functions are solutions to the Laplace equation

=

0 , where

0202 Ox

Oy

The Laplace operator crops up in so many practical and theoretical situations, that it is hard to overestimate its importance. In the context of holomorphic functions every

harmonic function is the real part of a holomorphic function in a simply connected domain.

Subharmonic functions, which are the topic for the next chapter, are very useful in the study of holomorphic functions. It may be said that this largely is due to the fact that when f is holomorphic then log If(z)I is subharmonic. We will approach this via Jensen's formula. Harmonic and subharmonic functions are very closely related in much the same way as linear and convex functions.

Section 1 Holomorphic and harmonic functions Definition 1. Let be an open subset of harmonic = 0, where

A real valued function u E C2(Q) is said to be

8202

02

is the so-called Laplace operator.

The Laplace equation = 0 occurs not just in mathematics, but also in many different connections in physics and mechanics, in particular in descriptions of stationary situations and equilibrium states.

Theorem 2. Let be an open subset of R". (a) 1ff E Hol(1Z) then and

are harmonic in ii. ($) Assume thai fl is simply connected. If u is harmonic in

then there exists an

I E Hol(f 1) such that u = The theorem yields a lot of interesting harmonic functions, e.g.

u(x,y) = Log(x2 + y2) defined for (x,y) E R2\{0,0}

183

chapter 12.

Harmonic Functions

Proof of Theorem 2: (n) The Cauchy-Riemann equations tell us that Of/81 = 0 , so the result is a consequence of the formula

8(Of

M (fi) The function

h := is

C' in

Since

Ou.Ou -

:

—+ C

and satisfies the Cauchy-Riemann equations, so it is holomorphic in is simply connected ii has a primitive f (Theorem V.8), and (again by the

Cauchy-Ricmann equations)

h—

Ox

Oy

Comparing with the definition of h we get Ou

0x so u =

—

Ox

and

—

Oy

(hj

+ some constant. When we absorb the constant in f the proof is complete.D

Corollary 3.

A harmonic function Lc

The function u(z, y) = Log(x2 + y2) is not the real part of an analytic function in all of C\{0} : The analytic function in question would be 2 log z, which, as we

know, cannot be defined in all of C\{O) . However that is the only regrettable thing that may happen: Theorem 4 (The Logarithmic Conjugation Theorem).

Let K be a connected compact subset of B(0, R). where R E 10,001. Let zo E K. and

flu is harmonic on := B(0, R)\K then there exist an analytic function f on a real constant c such that

forall zEQ The applications of the Logarithmic Conjugation Theorem are typically to ring shaped domains, i.e. the K of the theorem is a closed ball, possibly degenerated to a point. Proof. As in the proof of Theorem 2(o) we note that Ou

.Ou

Ox

Oy 184

harmonic functions

Secdon 1. Rolomoiphic

is holomorphic in (1, because h satisfies the Cauchy-Riemann equations.

chooser EIo,R(suchthatKcB(o,r),andlCtCdcnOtethecirclc Iz—zol= r. Let us from now on take zo = 0 for simplicity. be any closed path in (. If we put m := Ind,(zo) Let

, then

Ind,(z) =

forallzElC,Kbeingconnccted. Now,

m

Ind7(z) = mlndc(z) for all z E so by the global Cauchy Theorem (Corollaiy V.4) we get for any complex number c that

J (h(z)_

=mJ (h(z)_

=m(Jh(z)dz_2iric)

choosing

fh

jC we get

/

(h(z)_

= 0 for all closed paths7 in C\{0)

It follows that z h(z) — c/z has a primitive in each connected component of fl, hence in all of fl. Letting f denote such a primitive we have f'(z) = h(z) — c/z. We shall soon need that c is real, so we prove thai now:

c=

=

J h(z)dz =

-.

/

{

(e*)

—

}

so

c is real.

Now,

+

=

+cLoglzI} = 185

+

=

=

Chapter 12.

Harmonic Functions

and similarly

8

=

so that

+ cLoglzl = u +

g

where g is constant on each connected component of fI. Absorbing the constant in the proof is finished.

f 0

In the above version of the Logarithmic Conjugation Theorem the complement of only one bounded connected component, viz. K. In general each bounded connected component contributes with a logarithmic term. has

Let us use the Logarithmic Conjugation Theorem to prove the following classical result. A simple proof, that uses the maximum principle, but not the relationship between harmonic and analytic functions, can be found in LPe;Satz 111.30.6).

Theorem 5. If a harmonic function Lc bounded in

a neighborhood

of an isolated singularity then

that singularity is a removable singularity.

Proof: We may assume that the harmonic function u is bounded in a punctured disc JY around 0. By the Logarithmic Conjugation Theorem u has the form

u(z) =

where I

+

We will first show that c =

0.

Hol(D') and c E R

If not we may divide through by it and so assume that

—' 00 as z 00. —oo as z —+ 0, so to balance it c = 1. Now, fhasapoleatz=0,andso as z—.0,i.e. lnpanicularweseethatff(z)I—.oo f may be written in the form

where N 1 and h(0) is not 0. Choosing

as

an Nth root of —h(0)/n we find

—

=

— h(0)

— h(O)

00

as z —0,soc=0 , 186

andthusu =

Scction 2.

formula

If we assume that f has a pole at 0 we may derive a contradiction in the same way as above. If we assume that 0 is an essential singularity then we get a contradiction by Picard's big theorem or just by Casoraii-Weicrstrass ( Remember = u is bounded). So left is the desired case of 0 being a iemovable singularity for 1. 0

For information on the Logarithmic Conjugation Theorem in finitely connected regions and its applications we recommend the paper [AxJ from where the above treatment is taken.

SectIon 2 PoIsson's formula The

important Poisson formula expresses harmonic functions in terms of their

boundary values:

Theorem 6. (PoIsson's Formula). flu is continuous in B[O,RJ and harmonic in B(O,R) then

u(z) =

for

u(Re'd')

Izi


In particular u has the mean value property, i.e. u(O) =

Proof: The proof is from [Aa]. We will assume that u is harmonic in an open ball containing B[O, RJ and leave the derivation of the general case to the reader.


Let Izi

Z —4

1(z) — za

in the disc B[O,R] is

f(z)

—

1

f

1(w)

1

J

dW

IwI=R —

f(Re'#)

1

2irJ

Res*

d—

0

Taking

1

f

f(Re'#)

1

d

0

z = a here we get Poisson's formula. 187

0

Chapter 12.

Harmonic FUnctions

If we as center of the circle take z instead of 0 then the resulting formula is

Ju(z + Res)d4,

u(z) =

i.e. the value of u at the center of a ball is the mean value of the values of u at the boundaiy. We say that u has the mean value property.

The function 1

R2__1z12

for

and IzI
called the Poisson kernel for B(O,R) and is of great value in the study of harmonic functions. We list a couple of its properties that will be useful in the sequel: is

Properties of the Poisson kernel: (i)

PR > 0 (ii)

when IzI
+

=

=

In particular PR(ç6, z) is harmonic in B(O, R) for fixed (iv) 1 1 R—IzI

—

(v) If 0 < 6 < ir/2 then

1-R2

•2 (5) am Proof: The proofs of (i), (iu), (iv) and (v) are elementary estimates and computations left to the reader. You get (ii) by putting u = 1 in Poisson's formula. 0

The following theorem is a consequence of the mean value property. Maximum principles are among the most useful tools employed in differential equations. See (PW]. 188

Section 2.

Poisson's fcrmula

Theorem 7 (The strong maximum psinciple). Let u be harmonic in an open connected subset Cl of R2, and asswne thoi u takes its supremwn, i.e. there exists a zo E Cl such thai u(zfl) = sup {u(z) Iz E Cl). Then u is constant. Proof: It suffices to show that the set Co := {z E Cl I u(z) = u(zo)} is both closed and open in Cl. It is closed because u is continuous. Let z E Co ; for the simplicity

of writing, assume that z =

0.

If B[O, R] ç

Cl,

then we get from the mean value

property for any 0 < r R that 0=

u(O)

=

—

{u(O)

—

and - since t:(0) u(rel#) - we get u(O) = u(re'#) for all w E B(O, 1?). This shows that the set Co is open.

So u(0) = u(w) for all

0

We shall next formulate an inverse to Poisson's formula:

Theorem 8. Let f be continuous and real valued on the circle IzI = R. Then the function u, defined by

for IzI
u(z) :=

for IzI=R

1.1(z)

is continuous in the closed ball B[0, R] and harmonic in the open ball 13(0, R).

Proot The procedure of proof is borrowed from EMil. We treat only the case R = 1 and leave the derivation of the general case to the reader. For Izi < 1 we have 2t

2r

u(z) = Jf(et*)Pi(c5,z)d4

=

::

dçb

The last integral can be expanded in a power series in z, so u is the real part of an analytic function and is hence harmonic. Left is continuity at the boundary Izi = 1. To motivate the procedure we recall the MObius transformation A(z; w) from Example VIII. 14. If our theorem is correct then the mean value property for the 189

chapter 22.

HarmonIc Functions

harmonic function u o A(—z,.) implies that

z)dt = u(z) =

=

[u

o A(-z, )J(O)

u(A(_z,e'9))dO

= The ciucial point of the proof is to establish the identity

(*)

for

=

<1

To do so we note that the map w —, is analytic and nowhere vanishing on a sufficiently small neighborhood of [0,

has a continuous and hence analytic logarithm there, say ie. We may choose we have the identity = 0. Letting 0 :=

=

so it so that

— 1 —

ic't

1— z

A differentiation of this identity with respect to t shows that

0'(t) =

= Ic

—zi

From property (ii) of the Poisson kernel we infer that 0 is an increasing bijection of [0, 2ir) onto [0, 2ir). And when we introduce t as new variable on the right hand side

of () we get (*) by the change of variables formula. Since the restriction of u to the boundary is continuous (being equal to f there), it suffices to finish the proof of the theorem to show that —. f(e'°°) for any } from the open ball B(O, 1) converging to the arbitrary point e'9° of the boundary. But is continuous, hence bounded, so this is an immediate consequence of (*) and Lebesgue's dominated convergence theorem. D

sequence {

f

At this point we must mention the Dirichiet problem. Given an open subset fl of and a continuous function I on the boundary O1. Find afunction u such that u is continuous on the closure of It, u = 0 in It and UI8()

= 1. 190

Section 2.

Poisson's fonnula

The Diiichlet problem in It3 may be interpreted as finding the electrostatic potential u inside a hollow cavity (1, given the potential at the boundary 0(1.

ProposItion 9. Let (1 be a bounded open subset of R2. Then the Dirichlet problem has at most one solution, given the bowidasy values.

Proot We shall show that u E C(fl),

that u = Since

u

is harmonic in (land ulan = 0 implies

0.

say at zn E (1. If O(=u(zo)) everywhere. If zo E (1then u is by the maximum

is compact, u assumes its supremum somewhere in

E 0(1 then u

principle constant throughout the connected component of zn, taken at some boundary point. Thus u 0 once again.

so

the value u(zn) is also

0

Replacing u by —u we get u =0.

Theorem 8 shows that the Dirichlct problem for a disc has a solution. Even better, it provides us with an explicit formula for the (unique) solution.

Theorem 10 (Hansack's monotone convergence theorem). Let Il be an open connected subset of R2. Letu1

bean

increasing sequence of functions that are harmonic in (1. Assume that there exists a point Zn E (1suchthwsup{u,1(zo)In=1,2,...} < Then converges locally wuformly in (1 to a harmonic function. Proof.

Let B[zo, RJ c (1. We will for n,p E N estimate the positive function := — in B[zn,R] . Choose e >0 such that B[zo,R+e] ç (1, and let us for the sake of simplicity assume that zn =0. From the Poisson formula we get for any z E B(0,R-4-e) that

V,,.p(Z)

and

= / VRP((R +

so, from Property (iv) of the Poisson kernel, the estimates

/

+

1

1

:

/

,#\R+e+IzI

191

Chapter 12.

Harmonic Functions

which by the mean value property reduce to

R+e+IzI Vn,p(O)

R+e—IzI R+e+

R+e—

These inequalities show that converges uniformly in B[zo,R) The argument actually shows that if converges in a point z, then it converges uniformly in any closed ball B[z,R) in fl. Thus the set {z E converges) is both open and closed in Q, and so by connectedness it equals or the empty set. But zo belongs to the set. We have now shown that converges uniformly on any closed ball in Since any compact set in Q can be covered by finitely many such balls the convergence is uniform on compact subsets of Since is continuous, so is the limit function u. Let E Cl and B[zi,R] ç Cl. Letting It —. 00 in the formula

= / un(zi + we

—

get for z in the ball B(zi,R) that u(z) =

Ju(zi +

—

It now follows from Theorem 8 that u is harmonic in B(zj,R). But

was arbitraiy

0

inQ.

SectIon 3 Jensen's formula If f is holomorphic and without zeros in an open set containing the ball B10, II] then log Ill is harmonic inside the ball, so by the mean value property log 11(0)1 =

In the case where f has zeros Jensen's inequalny

log 11(0)1

flog fr(Re1')

holds as a consequence of Jensen's formula (Theorem 11 below). Jensen's inequality also follows from the theory of subharmonic functions (next chapter). 192

Secüon 3.

Jensen's (otmula

Jensen's formula tells what the difference is between the right hand side and the left hand side of Jensen's inequality. It is useful tool for providing information on the zeros of analytic functions on a disc, when restrictions on their sizes axe imposed.

Theorem ii (Jensen's formula). Let Pt,P2,

f be ineroinorphic in an open set containing B[0, R]. Let Z2," , and ",Pm be the zeros and poies respectively off in 11(0, R) (counted with multi-

Assume none of them Lc zero. log If(Rei#)I is integrable over Then 4'

=

log 11(0)1 + log

Proot Let

and

Jiog f (Ret') k4' be

• ,Zf,1 and

the zeros and poles

respectively of f in the closed ball BED, RJ, counted with multiplicity. The function

is holomorphic and zero free in a neighborhood of BED, RJ, so

flog

log IF(0)l = Putting

z =0, we find from the expression for F that IF(0)l =

so

log IF(0)I is the left hand side of the desired formula.

we find that

I

so

11

=

Ii

taking logarithms we get log

= log If(&'9) I + j=m+1 193

—

II

Il

—

Chapler 12.

Harmoinc Functions

To finish the proof it now suffices to show that — ir,ir[ and that

—.

log

—

is incegrable over

(Now, this is a book on complex function theory, so we shall here evaluate the integral by help of the Cauchy theorem. But the integral can actually be computed by elementary means [AGR),[Yo]). is continuous is locally integrable, and the function (1 — Since x log and never 0 on (.—lr, irj , the function +1og141

is integrable over ) — ir, 7r[. lb show that the value of the integral is 0 we note that z •—*

Log z z—

1

is analytic in the right half plane, so chat its integral along any circle I r E JO, 1[ , is zero by the Cauchy theorem, i.e.

0=!

J

—

= r, where

—f

Taking real parts we get

We will apply the dominated convergence theorem to this for a sequence The desired result is then an immediate consequence, once we dominate the incegrands by an integrable function. By differentiation we see that the function 1—

ret*

1 —

> 2

0,so 1_ret

194

1+r

1

Section 4.

Exercises

and so

log2 > log

log Ii —

—

6i431

— Iog2

0 Section 4 Exercises Show that u(Q) is open, unless u is constant. Hint: 1. Let u be harmonic in Connected subsets of the real line are intervals. 2. Derive the maximum principle for holomorphic functions from the strong maximum principle for harmonic functions. 3. Let f be a continuous, bounded and real valued function on the real line. Define

forx E Randy>Oafunctionuby

lIt (

u(z + iy) :=

00

(z—t)7+y2

dt

f

V

0

fory=0

(1(x)

(a) Show that u is bounded and harmonic in the upper half plane. Show that u is continuous in the closed upper half plane. 1).

4. Let u be harmonic in fl and identically 0 in an open, non-empty subset of (This property of harmonic functions is called the Show that u = 0 in all of

weak unique continuanon property). Hint: The unique continuation for holomorphic functions.

5. Define the map u : {x +iy E dy >0)

by

u(z) := the angle under which the interval [0,1] is seen from z. Show that u is harmonic. Hint: Consider the function Log((z — 1)/z). 6. Show that the function

u(x,y) :=

x2 + y2

is hannonic on R2 except at the origin. 7. Is harmonic for some 8. Show that the polynomial u(x, y) :=

an I

x is harmonic on ft2. Find

Hol(C) such that u =

Same problems with u(x, y) sin x cosh y 9. Show the following formula for r E [0,1[ and 0 1 rcos0=—

E

ft 2d4

27rj

Hint: Solve the Dirichlet problem when the boundary function is f(x + iy) = 195

x.

Chapler 12.

Harmonic Functions

10. (Liouville's theorem for harmonic functions) Let u be harmonic on ft2 and assume that u is bounded from below. Show that u is a constant. Hint: Property (iv) of the Poisson kernel. See also [Bo] and [Chi. 11. Show that the Dirichlet problem on {z E CO < Izi < 1) with the boundary function 1(z) = 0 for IzI = 1 , 1(0) = 1 has no solution. 12. Let u be harmonic in the annulus (z E dR1 < Izi
is a linear function of logr, i.e. there are real constants c and d such that for all

r

EJR1,R2[

= clogr + d Hint: The Logarithmic Conjugation Theorem. 13. Let R ejO, oo[. Inversion in the circle IzI = R is the map PR : R2\{0} —. R2\{0}

defined by

PR(Z) :=

for z E R2\{0}

Note that pR(z) = z when IzI = R, that points inside the circle are mapped outside and vice versa, and that

PR0PR —identity onallof R2\{0} (ar)Lctu be harmonic in R2\{0}. Show that so iSuOpR. (fi) Define and prove similar statements for the circle Iz - zol = R centered at

zo E ft2. 14. Let u be harmonic in all of R2, positive on the open upper half space and 0 on the real axis. (a) Let zo be a point in the open lower half plane. Let p denote inversion in the circle Iz — zol = R (See the previous exercise). Let z be a point in the open upper half plane such that Iz — zol < R. Show that u(z) u(p(z)). Hint: Apply the weak maximum principle to u 0 p — u in the domain D indicated on the figure 196

Section 4.

Exetcises

y

x

zo

The curve Cl is the image by p of the interval [A,BI.

Let 0 < vi <112. Show that

forall x1Ix2ER (y) Show that u(x, y) is independent of x for y > 0 (6) Show that u(x, y) = cy where c is a constant. (e) Let f be an entire function that maps the upper half plane into itself and is real on the real line. Show that I is an affine function, i.e. has the form f(:) = az + b where a and b are constants. The exercise is taken from [ii]. See also [Bol. 15. Prove the following theorem, due to Kellogg. It says that in special situations to get harmonicity we do not need the mean value property for all sufficiently small circles around the points of the domain, but only for one circle around each point. Theorem (Kellogg's Theorem). Letu E C(B[0,1))andasswnetharthereforeachz E B(0,1)existsanr = r(z) E 10,1 — IzIE such that

u(z)= —w

Then u &c harmonic on B(0, 1).

Hints to the proof: Show that we may assume u = 0 on the unit circle by subtracting a suitable harmonic function. Assume u(z) > 0 for some E B(0, 1); 197

Chapter 12.

Harmonic Functions

obtain a contradiction by applying the assumption of the theorem to a point in {z

B[O, lfl u(z) = IIuIL0} closest to the boundary. 16. As a curiosity derive the fundamental theorem of algebra from Jensen's formula.

198

Chapter 13 Subharmonic Functions Section 1 TechnIcal results on upper semlcontlnuous functions Technical results on upper semicontinuous funcdon& Definition I. Let X be a topological space. A fluzc:ion u X [—oo, oo( is said to be upper semi-continuous (usc) if the following holds: To every xo X and M > u(xo) there exists a neighborhood U of zo such that

M > u(x)forallx

E

U.

A continuous real valued function is clearly usc. Note for later usc that an USC function is Borel measurable - indeed, the set {z E ClIu(z)
Lemma 2.

If and u2 are usc then so are u 1 + u2 and max { u1,u2). IfuisuscandA E IfuG isuscforeachainanon-enzptyindexsetAthen inf

oEA

is also usc.

Theorem 3. An upper semi -conu nuous function on a compact topological space attains its supre-

mum. In particular it is bounded from above.

0

Proof: As the standard proof for continuous functions. Theorem 4. Let u :

is an open subset of the complex plane. Then u is usc if and only jf there exists a decreasing sequence of u2 continuous real valued functions on fl such that u.(z) u(z) as n — oofor each — [—oo, 001 where

zE Proof: The theorem is trivial in case u(z) = —oo for all z, so we will from now on assume that there exists a such that u(zo) E R. E The if direction of the theorem is pan of the lemma above. So assume from now

on that u is usc. We wifl first treat the special case of u being bounded from above, say u(z) < MforallzEftl nE N givcn by sup {u(y)

—

niy —

199

z

fl

Chaplcr 13.

u,,

Subhxmonic

is finite valued, since M uR(z) u(zo) — nizo — zi. We next prove that uR is continuous: Given z E (1 and

0 we choose y E Q

such that

u,1(z)
{u(y)—nly—z'I}

ufl(z)—uR(z') <

= n{Iz' — vi — Iv — zi) + e and since

> 0 is arbitrary we

—

+

that

see —

—

z'i

lnteivhanging z and zi we get Ju,,(z)

tt,,(z')I

—

—

z'i

proving the continuity. Next we note that

u(y)

which implies that uN(z)

—

u(y) — (n +

— zI ,

i.e.

— zI

that {uN} is a decreasing sequence.

Since

u,,(z) = sup {u(y) — nly — zI) u(z) — nfz I

—

zI

= u(z)

it is left to prove that limuR(z) u(z) , i.e. for any prescribed K > u(z) that K for sufficiently large n. For that we use that u is usc : There is a ball B(z,6) such that u(y) < K for all y E B(z,6). Now, = suplu(y)— nly — zI)

I

sup lu(y)—nly—zI), sup {u(y)—nly—zI)

We shall finally deal with the general case in which u is no longer necessarily bounded from above. Choose an increasing homeomorphism (-oo,OJ 1-00,001 (E.g. = —exp(—t)). The function u is usc and bounded from above, so the construction in the special case treated above yields a sequence { v,, } of continuous real valued functions with the properly that as n —' 00 for all Z E 51. It .1 suffices to verify that —oo < vR(z) < 0 for all z E 51, because we may then choose

:=

0

200

Section 2. Irnroductoy propenies of suthannonic functions

The left inequality is trivial because

is finite valued. To settle the other inequality

we use the definition of v,,, i.e.

= sup

—

I

nIy

—

zI}

< 0, and by upper semi-continuity of

By hypothesis u(z) < oo , so a

ou

there exists 6 >0 such that

forall yEB(z,6) Thus

va(z)=max

sup

sup

I,—zI6

<0

Section 2 Introductory properties of subharmonic functions introductory properties of subharmonic functions. Throughout the remainder of this chapter subset of the complex plane.

denotes an open, connected, non-empty

DefinItion 5. A map u : [—oo, oo[ Lt said to be subharmonic in if (a) u is upper semi-continuous (usc) and no: identically -oo. (b) For each z E there is a bail ç fl such that

u(z)

for all r EJ0,R1[

S

Remark: The integral in (b) makes sense: Fix z and r >0. Since u is usc, then the function 9 u(z + re'9) is Borel measurable and bounded from above. Thus u is essentially a negative function. The value of the integral may a priori be —oo; as we shall see in Proposition 9, it is actually finite. There we shall also see that the inequality (b) holds for all r such that B(z, r) ç not just for sufficiently small r.

A harmonic function is subharmonic by the mean value property. 201

Chapser 13.

Subbxmonlc Functions

asaximu principle). Theorem 6 (The If there is a zo E Let u be subhannonic in

such that u(zO) = supu(z), then sEQ

u is a constant.

u

In particular, is compact, u subharmonic on 0 thmughow Il.

in

and

0 on Oft, then

The maximum principle is one of the crucial propenies of subharmonic functions.

= u(zo)} is closed in il since u is usc. We will

Proof: The set M := {z E

showthatMisopenaswell,solctzEMbearbitrary. Claim: + rei) = u(z) for all 0 E ft and r E JO, R5( We prove the claim by contradiction. If it is false then there are

E [0, 2ir[,

E JO, R3[

and e > 0 so that



+

=

'I'

=u(z)_Er which contradicts (b) and hence proves the claim.

The claim shows that u equals u(zfl) in a ball around z so M is open. By connectedness Iv! =

0

Cl.

202

Section 3.

On the set where u = -00

Theorem 7.

Let u be subharmonic in

u(z)

and let B[a, R) be a closed boll contained in Q. Then

Ju(a+

PR(O,Z —a)dO for all z E B(a,R)

where PR denotes the Poisson kernel.

Proof: Since ii is usc it is the pointwise limit of a decreasing sequence continuous functions. The function v,1 defined by

(2ir n(

).

of

forz E B(a,R)

J

forlz—aI=R iS USC is by Theorem Xll.8 continuous in B[a, R) and harmonic in its interior. u — in B[a, R), subharmonic in its interior and 0 on its boundary. So by Theorem 6u— 0 throughout B(a, R). The result follows now from the monotone convergence theorem. 0

Theorem 7 explains the terminology subharmonic: The graph of u is below the harmonic function which has the same boundary values on the circle Iz — = R as it. So subharmonic functions are related to harmonic functions in the same way as convex functions are related to linear functions in one dimension.

Section

3

On the set where u = —00

In this § we deduce that the set of points in which a subharmonic function takes the value —00 is so small that all our integrals are finite.

ProposItion 8. A function u which is subharmonic on Q, is locally iiuegrable in Q. In particular, the set {z E I u(z) = —oo) isa null set, and u cannot be identically —00 on any non-empty open subset of Q.

It can be proved that {z E 5.32]).

in

I u(z) = —00) is a set of capacity 0 (See IHK;Theorem

Proof: We first observe that u is integrable over B(a, R) if B[a, R) is contained and u(a) > —oo: Indeed, multiplying the inequality

u(a) 203

Subbarmonic Functions

Chapter 13.

by irr and integrating from 0 to R = R —oo

2w

JJu(a + re1)d9rdr = J J u(z, y)dxdy

<

0

0

B(a,R)

As a consequence, if u is not locally integrable at a then u must be identically —00 throughout some neighborhood of a. Thus the complement in of the set {z E

lu

is locally integrable at z}

is the set {z E QIu

is

identically — oo in some neighborhood ofz}

Both these sets are clearly open, so by the connectedness of

one of them is empty. But u is not identically —oo by assumption, so the latter set is empty, and so the first

o

oneisa

The integral in the defining inequality (b) of Definition 5 may a priori have the value —00. It is, however, finite: Proposition 9. Let u be subharmonic on

Let a E

and R> 0 be such that B(a, R] c Cl. and

let I be the fluiction given by

for r E

1(r) := Then I is increasing,flnite and continuous for

r

[0,R)

JO, R]. and the map t

1(e1) is

convex in the imuer.'al ) — oo, Log RJ.

In particular

for rE[O,R] a result that extends (b) front Definition 5. Proof: To establish that I is increasing we use the notation of the proof of Theorem 7. We saw that u For r
Jun (a + Re'9) dO

Jvn (a + re'°) dO = Vn(a:1=

204

Section 4.

and

Approximation by smooth functions

the monotone convergence theorem takes over.

We next show by contradiction that I is finite valued on 10, R]. Assume that —oo for some ro €)0,RJ . Then 1(r) = —oo for all r [0,ro], because I is increasing, i.e.

I(ro) =

= —oo

for all r

[0,ro)

This implies (integrate with respect to r from 0 to ro) that

/

B(a,ro)

contradicting the local integrability (Proposition 8).

A convex function on an open interval is continuous, so left is only the convexity statement. We postpone a proof of that to the next section where we will possess more

0

machinery.

Section 4 ApproximatIon by smooth functions The following characterization of smooth subharmonic functions is very useful and connects once more the concepts of subharmonic and harmonic functions.

Theorem 10. Let u

<0 on ci.

Then u Lcsubharmonicon ci

Proot

SupposeflrstthalAu0. WeshallforzEflverifythat u(z) <

Ju(z +

for simplicity take z = 0. The expression for the Laplacian

r

in polar coordinates (r,O) is

(1) au = U1y +

1

+

1

7jU99

When we integrate (1) wth respect to B from 0 to (noting that the last term drops out because —. u,(rel#) is periodic in we get with the notation

1(r) :=

205

chapter 13. Subhannomc Fincuocu

=

(2)

i" +

=

0 we see that r —. rr(r) is an increasing function. Since I'(r) clearly is bounded (for small r), rI'(r) 0 as r 0. But then rI'(r) 0 so that r(r) 0 and thus I is an increasing function of r. Now, as desired Combining (2) with

u(0) = 1(0) <1(r) = _Ju(resO)de Let next u be subharmonic. We assume that > 0 for some z E Q and shall derive a contradiction. Take again z = 0 for the sake of simplicity. Applying the first part of the proof to —u we get that I is decreasing for small r. But u is subharmonic so I is actually increasing (Proposition 9); hence I is constant and the right hand side of (2) is zero. But that contradicts the assumption — > 0. 0 A subharmonic function need not be C2. But it can be approximated by smooth subharmonic functions:

Theorem II. Let u be subharmonic in Q. Let w be an open subset of Q with the property that > 0. Then there is a decreasing sequence u ... of subharmonic functions in such thai u(z) for all z E Q. Proof: The standard mollifying method (convolution) works. It goes as follows: Then the function Let 0 < R <

U(z)

f u(z) 10

when
is locally miegrable in R2, since u is locally integrable in Q. Let 4, E Cr(R2) be a positive radial function such that supp 4, ç B[0, Rj and f 4' = 1 and define for

n = 1,2,••.

the functions

n24,(nz) for z E R2 Then E

c 206

=

1

Section 4.

Approximation by

Let U, be the convolution of U (the integration is over R2):

and

smooth functions

4,,, i.e. the function given by the expression

UJ.(z):=JU(Y)4R(z_Y)dV for zER2 C°°(R2), because we may differentiate under the integral sign. and let 0 < us next prove that U, is subharmonic on w. Let z E Then (using Proposition 9 and that is radial) Clearly

U,, E

Let

U(z +

=

/

u(z+y)4R(y)dy

/ B(0,*)

B(0,*)

<

y)41R(y)dy

/

0

0

B(0,f)

= ...LJU,i(z+rei)de showing that (JR is subharmonic on w. We continue by proving that UR(z)

is

decreasing in n when z E

/

/ B(0,*)

R2r

= The inner integral is increasing with p (Proposition 9), so we get

R2r UR(z) The

UR+1(z)

computation just made also reveals that

= u(z)

207

r

R.

Fwiciions

Chapie 13.

so now it is only left to show that

forall zEw

lim

i.e. given K >

u(z)

K.

there exists an n so that

Bytheuscu
J

J

We shall now keep a promise: Proof of Proposition 9 (continued): For simplicity of exposition we take a = We shall show that the function

h(t)

0

as usual. Assume first that u E C°°.

:=

is convex for —oo < t < Log R. During the proof of the first part of Theorem 10 we observed that r —' an increasing function. Hence so is t

implying that h is convex. If u is no longer assumed functions u,, u. Then

is

=

rr(r)

is

= h'(t)

we approximate by a sequence of

subharmonic

a decreasing sequence of convex functions, so its limit, which is h, is convex as

0

well.

SectIon 5 ConstructIng subharmonlc functIons ProposItion 12.

(a)Ifui ($)Ifu is subharmonic intl then so is Au for any A E (O,oo[. 208

Section 5.

Cocstnicdng

functions

is a decreasing sequence of subharinonic functions in fl then lim u,, is (y) if subharmonic In (1, unless U is identically —oo.

Proof: Left to the reader. For the next result we need (a particular case of) Jensen's inequality for integrals.

Theorem 13 (Jensen's Inequalijy for lntegnzLc). 1ff is a real valued function In L1(O, such that a < on )a,b[Ja,bf, then

The cases a =

—oo

f(t) < b and

4, is convex

and b = 00 are allowed.

For a proof consult [Ru;Theorem 3.3].

Theorem 14. (a) flu is harmonic in

and 4, is convex on the range of u, then 4,ou is subharmonic

(fi) If u is subharmonic in (1 and 4, is a convex increasing function on the range of u, then 4,o u is subharmonlc in fl. (We put 4,(—oo) = Urn 4,(t)).

Proof: (a) A convex function on the real line is automatically continuous, so 4, 0 u is continuous. From the mean value property 2r

t u(a)=ju(a+re we deduce via Jensen's inequality for integrals that

(4,ou)(a) so 4,o u is subharmonic. (fi) Similar arguments to the ones from (a) work here.

0

Theoretical Examples 15.

(a) Let f E

Hol(1Z).

1ff Is not identically 0 then

{Loglfl, 0) are subharmonlc In fl. 209

and

:=

Chapter 13.

(fi)

Subhxmonic FwEdons

1ff andg are holoniorphic in then IfI'IgI' is subharmonic for all p,q 0. If u is harmonic in fl then lur is subharmonic in fl for all p 1.

Proof: (a)

is clearly usc and not identically —00. If 1(z) = 0 at a point z E fl then Loglf(z)I = —oo, so the inequality (b) from Definition 5 is obviously satisfied at is harmonic around z, hence subharmonic. z. If 1(z) 0 then I=

is the composition of

with the convex increasing function t

(fi) Assume first that neither f nor g is identically 0. Then and are subharmonic, hence so is pLogif I + qLoglgi. Left is just to compose with the exponential map. The modifications in the cases where f and/or g are identically 0 are left to the reader. Theorem 14(a).

Practical examples ate now numerous, e.g.

u(z, y) = lxI or more generally

= lxi' for p 1 SectIon 6 ApplicatIons We present three applications: One is a proof of Radô's theorem, the other is a treatment of H'—spaces, and the third is a proof of the F. and R. Nevanhnna theorem.

Rad6's theorem Our first application of the theory of subharmonic functions is a proof of Radó's theorem.

Theorem 16 (Radó's Theorem).

1ff E C(1l) is holomorphic in {z E fl 11(z)

0) then! is holomorphic in all of fl.

Proof: We may assume that f is not identically 0. We write I = u +iv ,where u and v arc real valued.

The function a :=

is subharmonic in ci (same proof as in (a) of the

Theoretical Examples 15 above). So is u + es for any e> 0 (same proof).

We will show that f is holomorphic in any ball in ci; for simplicity of notation we assume that the ball is B(0, 1) and that B[0, 11 fl. We assume funhermore that

11(2)1< 1 for z E D[0,1]. Note that then, 0 on B(0,1). 210

Applications

Section 6.

Now, for any z E B(O, 1] we have by Theorem 7 that

+

u(z)

0 we get by letting e decrease to 0 that

At those points z E B(O, 1) where f(z)

u(z) < However, u is continuous and the points where f(z)

0 are dense in B(0, 1)

(Proposition 10), so the inequality holds for all points z E B(0, 1). Our arguments work in the same way for —1. i.e. with —u instead of u. Thus

u(z) = ..i_Ju(ei)Pi(eiz)do which shows (Theorem Xll.8) that u is harmonic. In particular u

1)).

Replacing f by if we get v E C°°(B(0, 1)), so f E C°°(B(O, 1)). f satisfies the B(0, 1)11(2) 0). and hence on all 0 of B(0, 1). So f is holomorphic. Cauchy-Riemann equations on the dense set {z

Hardy Spaces Definition 17. Let f be holomorphic in the unit disc.

p E)0,00( ,

f is said to belong to the Hardy space

2r

sup O(r<1

We

t fy'e I ,9\P dO
1

I

)

J 0

leave it to the reader to verify that

is a linear subspace of Hol(13(0, 1)).

Theorem 18.

If I

E

where p

1

then there exists exactly one

f(z) =

f

for all

211

I/(S') such thai <1

Subharmontc Funcuons

Chapter 13.

Proof: We prove the theorem for p =

1;

for p >

1

the argument is similar, bui

simpler.

<1 be an increasing sequence such that


Let 0 <

—' 1 as

n —' 00,

and define

:=

for t

E

R

is a bounded subset of L2(S'). Since the unit ball of That f E H' means that L2 is weakly sequentially compact (See [DS; Corollary IV.8.4J) , we may (possibly by taking a subsequence) assume that there exists a function g E L2(S') such that g weakly in L2. Recalling that if is subharmonic (Theoretical Examples 15(,9)) we get by Proposition 9 for m < n that

where z = rme"

J and

letting n —00 it follows that

Jg(e19)p,(o,z)do By the Cauchy-Schwarz inequality we get

J{g (e'9) (P,(o,

}2d9 JP1(9. z)dO

= Jg(eb)2P,(o,z)do which we integrate with respect to t to get This inequality combined with the fact that 9m converges weakly to g implies that 9m — g

in L'. Indeed,

in L2 And then

=Jigm_giigm_gidt

J

il9m — gii2llgm + gil2

In particular the sequence

sequence {t —.

}.

=

Iffr

is

211gm — gii2lig 112

uniformly inregrable. Hence so is the

By the Dunford-Pettis compactness criterion (See EDS]) we

may (possibly by taking a subsequence) assume that there exists an f E L' (S') such —, f(ed) weakly in L'. that Letting n —. oo in the Poisson formula 2v

f(z)=_1-Jf(rnett)

2..

IzI

2

— zi

0

212

2dt

Sectioc 6.

Applications

we get the existence part of the theorem. The uniqueness will follow once we check for any L1-function f' that

J implies f' =

=

0

for all fzf < 1

It suffices to prove the statement for f real valued. In that case we get by property (iii) of the Poisson kernel that 0.

0

=

J f(e")dt +

J

}

parenthesis is an analytic function with real part 0, so it is a purely imaginary constant c, i.e. The

2w

2w

c= z

Ji'

= 0 we see that c =

/

+2 0,

e""f' (est)dt

because f' is real. Now,

J

=0 for all n

0

Since f is real, complex conjugation shows that this holds for all n E N, i.e. that all

0

the Fourier coefficients of f are 0. But then f' =0.

The following corollary which is equivalent to the theorem just proved, has been generalized extensively and is useful in prediction theory.

Theorem 19 (F. and M. Riesz's theorem). Let p be a complex Borel measure on S' for which

0 for all n = 1,2,... Then p is absoiwely continuous with respect to the linear measure on S'.

Proof: Expanding P1(t, z) in powers of z and we see that

f(z) := JPL(t,z)dp(eul)

213

and using the condition on p

for IZI <

1

Chapter 13.

is a power series in z

alone,

Subharmomc Functions

so that f is holomorphic in B(O, 1). Using Fubini we

find that f E H1. When we apply Theorem 18 and once again expand P1 in z and

J euhltdp(i) =

J

we find that

for all n E Z

Since the trigonometric polynomials are dense in the periodic continuous functions we

get for all g E C(S') that

J g(t)djt(f) =

Jg(t)f(ett)dt

0

which finishes the proof.

The f of Theorem 18 above cannot be an arbitrary function in L'(S') as the following result shows: Theorem 20. Let f and f be as in Theorem 18. Assume that I is no: identically 0. Then

and f cannot vanish on any set of positive linear measure on S1.

Proof: In the course of proving Theorem 18 we showed that the sequence converges to If (e'8)I in L'. Choosing a { may assume that

subsequence

Ii' (e'°) for almost altO

I

Since —

0

Fatou's lemma implies that

(e'9) By Jensen's inequality,

( 214

if necessary we

Section 6.

Applications

finishing the proof of the first statement.

IffhasazerooforderNatz=O we apply the above result to F(z) :=

noting

that F'(e") =

f E Hol(D) fl 11(D) for all p E Li, oo[ (Use polar coordinates). The converse is not true: Example 21. The function 1(z) := (1 —

z)'

belongs to Hol(D) fl L'(D) . but not to H1: we leave it to the reader to check that f E 11(D) for

Qearly I E Hol(D) , and all p E [1,21 We assume now that f E H1 and derive a contradiction from that assumption: weakly in L'. It During the proof of Theorem 18 we saw that f(rned) —. follows that

I

(e' ) = 1

1 —

&t

1

, 30

—'

i—

EL

1

But —

= iti

=

<

for small t, so (again for t small)

>1

1

and the right hand side is not integrable over any open interval around 0.

0

F. and R. Nevanlinna's Theorem Theorem 22 (F. and R. Nevanlinna's Theorem). Let f be analytic in the unit disc. Then f is the quotieiu between swo bounded analytic functions, 4ff

sup

o<,
I'Ll '.

<00

0

Proof: Let us first assume that (*)


sup

w is subharmonic (Example 15). 215

Chapter 13.

The harmonic function

Subbarmonic Functions

on B(O, r) that on the boundary coincides with w,

increases with r (Remark immediately after Theorem 7). The values at 0 of the functions u, are by the mean value property (Theorem Xll.6) bounded by (*), so according to Harnack's theorem (Theorem Xll.10) U,. —. u as r 1_ , where u is harmonic on B(O, 1) and nonnegative, because w u.

Choose a holomorphic function g such that u = (Theorem Xll.2(8)). Then 1 and I JhI, so both 1/h and f/h are analytic and bounded by 1. We IhI = are through, since their quotient is f. Suppose conversely that f = a/b where a and b are bounded analytic functions. We may assume that IaI 0. Then 1, 1 and that b(O) —Log Ibi and

by the subharmonicity of Log Ibi (Example 15)

fLog

—oo
b(re'8) ide

Thus

—Log Ib(0)I

and

0

the right hand side is independent of r.

Section 7 Exercises 1. For which values of a, b, c E R is the polynomial U(x,y) := ax2 +2bxy+cy2 a harmonic function? a subhaimonic function? 2. Let : be a holomorphic map between two domains in the complex —, plane. Let u be subharmonic on o is subharmonic on fl unless it is identically —oo. Hint: Show that = when is holomorphic. 3. Let K be a compact subset of Il. Let u be subharmonic on Let h be a real valued continuous function on K such that h is harmonic on ir&t(K) and u h on OK.

Show that u < h on all of K. 4. Let f E Hol(B(0, 1)) have the power series expansion for Izi <

1(z) =

216

1

Exercises

SectIon 7.

Show that

IEH2 *

>1an12
5. Let f E Hol(B(0,1)) and let p E [1,ooj. u. (a) Let f E H'. Show that there exists a harmonic function u such that Hint: Theorem XIIL2O. (fi) Assume that there exists a harmonic function u such that u. Show that

I

E

H'.

[This exercise gives us a hint as to how to define the Hardy space HP(Q) when fl is a domain in C : I E Hol(fl) is in HP(IZ) if there exists a harmonic function u on fl such that I' S u. 6. Let u be a convex function on an open convex subset of R2. Show that u is subharmonic. Hint: A convex function on an open set is automatically continuous, so

+ re'8)dO makes sense as a Riemann integral. 7. Let u be usc, and let f be continuous and increasing on the range of u. Show

that f o u is usc. 8. Show that a continuous function with the mean value property is harmonic. Hint: Theorem 7. 9. Prove

Sckwwz's reflection principle. Let Il be an open subset of R2 such that (1 is symmetric around the x-axis. Let u be harmonic on {(x,y) E (fly >0). continuous on {(x,y) E (fly? 0) and 0 on =0). {(x,y) E Then there exists a harmonic function U on fl such that U = u on {(x,y) E >0). Hint: Show that

(fix '

fory?0 1 —u(x,—y)

for y <0

has the mean value property. 10. Let u be subharmonic on an open subset fl of R2. Let BI0, r] ç Q r > 0 . Define v on (1 by for Z

v(z) := I.

E

B(0,r)

for z E cl\B(0,r)

u(z)

Show that v is subharmonic on fl. 217

,

where

Chapter 13.

11. We

have seen that hi

general statement: If a 1 and h1, h2,•

,

is

Subharinonic Functions

subharmonic if h is harmonic. Show the following are harmonic, then

(Ihii° + lh2Ia + ... + is subharmonic. 12. Use the procedure from the proof of Radô's theorem to show the theorem on an isolated singularity for a harmonic function (Theorem XII.5).

218

Chapter 14 Various Applications SectIon 1 The Phragmén-Llndelöf principle The weak maximum principle (Corollary IV.16), which says that an analytic function never exceeds its maximal boundary value, does not hold in general for unbounded domains. For example, if fl is the lower half plane, so that Of 1 is the real axis, then 1(z) := exp(iz) is bounded on Ôfl but not on Il. So we need extra

restrictions on the function and the domain to get maximum principles for unbounded domains. The so-called Phraginén-Lindelof principle is a way of finding such maximum principles. Our version is the following.

Theorem I (Phragmén-lindelUJ's principle). be afunction in

Let fl be a connected open subset of C and let which Is bounded and no: Identically 0.

1ff E

fl Hol(1l) andthe constant M

(1)If(z)I Mforailz E

(1 Hol(1Z)

that

0

,and

—+ 0 as z —+ 00 In Il for each jixedq > 0,

(2)

then If(z)I Mforallz E Proof: The case M = 0 is a consequence of the case M > 0 , so we may assume Consider for > 0 the continuous function that M >0. Let B >0 be a bound of

u,,(z) :=

for z E

which is subharmonic on (1 (Example Xll.17(8)).

choose R >

0

By the assumption (2) we may

so large that

fcrall zEIlsuchthal IzIR Since Offl fl B(0,R)) c E fl IzI = R} we get by the maximum principle for subharmonic functions (Theorem XIL6) that

uq(z)_<max{M,M}=M for zEflflB(O,R) This inequality is also true outside of the ball B(0, R) by our choice of R, so

u,,(z) M for z E fl, i.e. :S M for z E fl

0. However, the zeros of are isolated in Il (Theorem IV.1l), so the desired inequality holds by the continuity of f everywhere in 11. 0 219

chapter 14.

Applications

The connectivity assumption was only used at the very end of the proof. We may delete that assumption if we in retwn require that 4' vanishes nowhere. Corollary 2. Let C be an open subset of the complex plane. Let I e C(?i) fl

be

bounded. 11111

MonOQ,then Ill

Proof: Wemayofcourseassumethatfl Lete>0. By continuity of f there exists a point zo e such that If(zo)I < M + e and then there is even a ball B(zo,rJ ç such that Ill M + e on that ball. Applying Theorem I to the set and the function 4'(z) := (z — :o)_1 we get that

If(z)I M + e for all z E Q\B(:ø, r) This inequality also holds for z E D(zo, r] by the very construction of that ball, so

0

f(z)I<_M+eforallzEIl. Bute>Owasarbitrary.

Corollary 3 (The three lines theorem). Let := { z E C 0 < < 1). Let be bounded and holomorphic on 1 and continuous on the closure

(1)

of Q and pw

M(r) :=

f

sup —oo<,
+ iy)I}

Then log M as a convex function on (0,1]. In particular

(2)

M(z) M(0)l_ZM(1)z ford! xE

Proof: Let g(z) := g we see for any x E]O,1[ that

where

[0,11

a is a real constant. Applying Corollary 2 to

Ig(z+iy)I max{M(0),M(1)ea) and so that

max

(3)

If M(0) = 0 then (3) says that M(z) Since this holds for all real numbers a, we see that M(x) = 0 , and so (2) is trivially satisfied. The same kind of arguments works if M(1) = 0. If both M(0) and M(1) are different from 0 we choose a so that M(0) = M(1)ea (this minimizes the right hand side of the inequality (3)); then (3) simplifies as desired

M(x)

= 220

Section 2. The Riesz-Thorin liucipoladon theorem

If we apply the same arguments to the strip

where O<s
1} we find that

10

M(z)' 0

M is convex.

SectIon 2 The Rlesz-Thorln Interpolation theorem Using The Three Lines Theorem we shall now prove the M.Riesz-Thorin interpolation theorem. The interpolation theorem holds for general measure spaces, but for simplicity we will only consider Rd with Lebesgue measure. A proof for the general case can be found in [DS;VLIOJ. We need some preliminaries: A simple function is a (finite) linear combination of indicator functions for measurable sets of finite measure. We let S denote the complex vector space of simple

functions.

Indeed, if qi
, then

fIT

+ Ifl"lB

where

A :=

If

=

then

E

Rd 11(x)! I

i} and B :=

E

Rd 11(x)! > i} I

00, then we may scale f down so that we may assume that Ill

f

so

1; and

we are through also in this case.

Theorem 4 (Rlesz-Thorin's interpoLation theorem). Le:1
haveftnitenormcM0andM1 ,whenSis equipped with the norms from and LPt respectively. Fort E [0,1J we define pg and qg by the formulas

1—i t —=—+— pg po P1 1

Then T maps S into norm from L".

Proof:

and

1 1—i t —=—+— qo qi qg

when S is given the

and its norm Lc at most

<1), and forz

:= {zE 1 — = 1—z — + —z p(z)

P0

and

put

1—z z — = + — q(z) qo 1

qi

P1

221

Chapler

Furthermore we

let

the dual exponent to

rg denote

11

i.e.

—+—=1 rg Qearly

rt

which we extend *0 z E

ri

ro

by

1—z —=—+— r(z) ri z

1

ro

Note that p(i) =

q(i) = qg

and

r(t) =

andr=re.

FixtE]O,1[.

Let

f be in S. We must show that Tf E

L'

has norm less than or equal to

It suffices to prove that for all g

For

each

zE

S

we define

f and

that

g;

and

If

I= then f2

so by the linearity of T

we

=

get

= Since

on

a similar formula holds for 9z we clearly have that the function

defined

by

:= is holomorphic on

applied to

and the

(1) k(t)I

,

,

and continuous and bounded on point

ri.

By

the Three

z = t we get where No

= sup

I

222

and N1 = sup

Lines Theorem

Secdon 3.

+ iy)I

Let us estimate I4(1

+ iv)I

:

M. Riesz' theorem

By HOlder

IIT(fi +iv)IIq(1)1191+iPIly(1) A'I IIfi+iIIIp( l)Ilgl+IVII,( 1) IIp(1)

11111,

1)

IIgII'

N1

estimate N0 similarly. Substituting these estimates into (1) and recalling how p and r are defined we get We

14
0

proving the theorem.

A bounded linear operator mapping of L' into V is said to be of type (p, q). In this language Theorem 4 reads: If T is simultaneously of type (po,qo) and of type (pi,qi), then it is also of type (ps,qg) for any I E [0,1) where 1—1 —=—+-po pg P1 1

1

and

1—I I —=—-f— 1

qg

qi

qo

and we have an estimate for the the operator norm of T. A very simple consequence of Theorem 4 is Theorem 5 (The Hausdorff.Young inequaWy). lip E [1,2] andf E L'(R") then its Fourier transform Ff belongs to 11Ff II,. 11111, where

=

and

1

Proof: The Fourier transformation f —, Ff is simultaneously of type (1, oo) and 0 of type (2,2). Apply the Riesz-Thorin interpolation theorem.

SectIon 3 M. Rlesz' theorem Any harmonic function u on the unit disc is the real part of an analytic function I (Theorem XIL2). The imaginary part v off = u + iv is not unique, but any two choices of v differ by a real constant. The conjugoiefiuzcdon (or harmonic conjugate) of u is the v which is fixed by the requirement that v(0) = 0, or in other words by the requirement that 1(0) should be real. As simple examples we mention that the conjugate function of is v(z) = and that the one of u(z) is v(z) = u(z) := more generally, if u is a trigonometric polynomial with real coefficients, n=

=

cos(nO) + b,, sin (nO))

ao +

223

14.

then its conjugate function is v(re19) =

+ aRsin(nO)}

The holomorphic functions that have u as real part are characterized by Schwarz'

formula: Theorem 6 (Schwarz' formula).

f

Let = u + iv be holomorphic on the open boil D(O, R) and let u be continuous on the closed bail B[O, RJ. Then

f(z) = iv(O) +

Ju

+

for oil IzI
Proof: By Poisson's formula (Theorem XIL6)

R2_1Z12& = 2w

=

+ z)dO =

+

so the real part of the analytic function 2w

z —,

f(z)-.

is 0. Hence the function is a constant. Taking z = For any complex valued function

0

we see that this constant is iv(O).D

on the unit disc we let I

for pE [1,00[ With this notation a holomorphic function f on the unit disc is in <0o. 1ff 111 1ff is holomorphic or harmonic then If I' is subharmonic (Example XIII. 15) and so the supremum is the limit as r —. 1_ (Proposition XIII.9). If I furthermore is continuous on the closed unit disc then clearly

11111,!

for pE [1,001

= 224

3.

M. Riesz'

i.e. the nomi equals the 1/-norm of the restriction of f to the unit cixcle. Let f = u + iv be holomorphic on the unit disc and such that f(O) is real, i.e. v is the conjugate function of u. The problem that we shall now consider is whether lull,
We consider an f

1(z) = such

that 1(0) =

R. Then u(z) =

=

From the theory for power series we know that the series E

r E [0, 1[ converges uniformly with respect to 9 to course converge uniformly to u (re'°). Since

for each fixed

its real part will then of

2r

— jf e 1

e

0

we get

111

p2d0

J

/

=

,2

=

=

Ian 12

=

In a similar way we can compute (we leave the details to the reader) that 00 2

— a0 -r

i

L1 10*1

ii=1

In particular

Ill so

I

=

—

4

H2 if huh2 < oo , answering our question above affirmatively. 225

Chaper 14.

Marcel Riesz showed in 1927 that this phenomenon persists for p in the range ji, oo[:

Theorem 7 (Al. Riesz' Theorem). For each ji, oo( there exisss a positive constant C, such that

f E Hol(B(O, 1)) such that 1(0) E R

11111,

Remark: We have not specified any bound on the constant C,, except in the case works EHK]. p = 2. It can be shown that C, = (p/(p — Proof: We let for convenience R denote the set of those I E Hol(13(0, 1)) for

which f(0) is real. We divide the proof into 3 steps. Step 1: Here we prove the following claim. Claim: Let p be fixed and assume that there is a constant C > 0 such that $1111,

for all I

E

111112,

R. Then 112, fir all I E 1?.

(4C +

Proof of the claim: Consider first the case of 1(0) = 0. Possibly replacing I by z —e f(rz) for r jO, 1[ we may assume that f is holomorphic in a neighborhood of the closed unit disc, so that 2r

2r

=

J If(e")

=

=

/

+ 2iuvl'dO =

—

=2' uv+

2?

+

2*

which by our assumption this can be estimated by

JIurIvl'do

= and

further by the Cauchy-Schwarz inequality 2r

=

(2CYIIuII,11f11, 226

U2

M. Riesz' theorem

Section 3.

we get that 111112, 2CllulI2,. For a general f E R Dividing through by Ill thus have established the inequality we

Ill — 1(0)16, 2CIIu

Now, 1(0) = u(0),

— u(0)112,

so

111112, 2CIIuII2, + (1

To deal with the last term we note that so that

+ 2C)lu(0)l

1u12' is subharmonic

(Example X1ll.15(7)),

Iu(0)12P

and finally 111112,

(1

+ 4C)IIu 112,,' proving the claim.

Step 2: We know that the inequality of M. Riesz' theorem holds for p = 2 (with the constant

112). From Step I we see that it then also holds forp = 4,8,16,.... In the present step we use the Riesz-Thorin interpolation theorem to show that the inequality holds for any p in the interval [2,oo[. = u + iv , where v is the conjugate function The map that interests us, i.e. u of u, has according to Schwarz' formula (Theorem 5) the explicit expression

f

for Izl<1

(*)

if u is continuous on the closed unit disc. However, u and f are in general only := defined on the open disc, so we must consider the functions on the circle More precisely, we define for each r} and afterwards let r = 1_. {z E Izi r EJ0,1[ a linear map : L'(S') —. C(S') by 2w

:= and

/

:

shall for each p E [2, oo[ study its restriction

T,,,

:

TrILP

: Il(S1) —. 11(51)

That restriction is a continuous linear operator from LP(S1) into 11(51).

Let now 4' E C(S') be real valued and let f denote the analytic function from (1 with u replaced by 4'. Then 2w

IITr,p4'IIL,

=

11111,

m

Chapter 14.

Various Applications

If, furthermore, p is of the form p = 2N for some N = 1 estimate as follows:

Ill lip

1,

2,... we may by Step

=

where A,, is a constant. If is not real valued we split it into its real and imaginary parts to get

so

that

IITT,PIIL,Lp <2A,, for p= 2N

,

N =

By the Riesz-Thorin interpolation theorem there exists to each p E [2,001 a constant

C,,, not depending on r, such that for all

IIT,,SlI,,

E

1ff is holomorphic on a neighborhood of the closed unit disc and f( 0) is real then 11111,

= urn I1IISJILP =

If f is holomorphic on B(0, 1), but not necessarily on a neighborhood of the closed unit disc, then we apply the result just derived to the analytic function fA( z) := f( Az) and let A — 1_. With Step 2 we have proved M. Riesz' theorem for p in the range [2. oo[. Step 3: Here we derive M. Riesz' theorem in range Ji, 2[ by the standard technique of duality, knowing it to be nue in the range (2, oo[. For exponents q < 2 we consider the conjugate exponent p = q/(q — 1) > 2. Let where v(O) = 0. We may assume that f is continuous on the closed < 1 (If not consider f(rz) for r < 1), so that the norms reduce to the relevant disc Izi L'-norms of the functions on the unit circle. Since trigonometric polynomials are dense in LP(S1) we have

=+

lit'liq

=

supJ

where f denotes integration over S' with respect to and where the supremum ranges over all real uigometnc polynomials g with I/-norm 1. Let N

g(O)

=

+

cos(nO) + b,, sin(nO)}

228

Section 4.

Exctcises

be such a polynomial and consider the harmonic function given by

Its

conjugate function h is

h(ret9) =

sin(nO)}

cos(nO) +

Since g + ih is analytic so is the product f(g + ih). Its imaginary part vg + ub is harmonic and vanishes at 0, so by the mean value property for harmonic functions (Theorem XII.6)

fvg= _Juh Combining Hälder's inequality If uhi we have already proved

IIUIIqIIhIip

with the following result, which

+ we get

J which implies that flVIJq

CpIIuIIqIIgIIp

CpIIUIIq

.

Hence

+ 1)IIUIIq

Ill IIq IIhhIIq + which is the desired statement.

Section 4 Exercises 1.Let

where a

is a constant. Let f E If(z)I

n Hol(Q) satisfy that

Cexp (AIzIb) for all z

Q

where A and C axe positive constants and b E [0, a[. Show that Ill M throughout Q, if Ill M on Oil. Hint: Consider the function 4(z) := exp(_zc) for c E]b,a[. 229

Chapter 14.

2. Let

Applications

be the strip

:=

where L EJO,oo[. Assume that f If(z)I

satisfies that

C(Ci) fl Cexp

for z E ci

A and C are positive constants and a E (0, [. Show that If M throughout if Ill M on Oci. Hint: Consider for 6 E ]a,

where

I

the function

:= exp

—

230

eItIz)

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Rcfcrcnccs

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HfI E. Hille

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235

Index A

Cauchy-Hadamard radius of convergence formula 2 Cauchy-Riemann equation 14 Cay Icy transform 119 Cayley transformation 122 chain rule 13 Chebyshev's function 175 circle 114, 117 circle of convergence 2

Abel's partial summation formula 7 Abel's test 11 Abel's theorem 5, 11 addition theorem 23 Ahifors' lemma 110, 112 alternating series theorem 11 analytic 16

angle 118 annulu! 71, 80 argument 25

closed curve 17 complementary arguments 158 complex derivative 13 complex differentiable 13 condensation principle 120 conformally equivalent 122 conjugate function 223 continuous argument 25 continuous logarithm 25 contractible 38 converge 147 converge locally uniformly 149 critical strip 176 curvature 109

argument principle 129, 144

B Bernoulli's paradox 38 Blaschke product 166 Bloch-Landau's theorem 101 Bloch's constant 101 Bloch's theorem 101 Borsuk-Ulain's theorem 42 branch 25, 35 Brouncker's series 6 Brouwer's fixed point theorem 30, 67 Brouwer's theorem on invariance of domain 41

curve 17

D derivative 13

C

Dirichlet problem 190 duplication formula of Legendre 161, 167

Carleman's theorem 139 Casorati-Weierstrass' theorem 100 Cauchy inequality 11 Cauchy integral formula 47, 71 Cauchy integral formula for a disc 48 Cauchy integral theorem 47, 71, 72, 74 Cauchy-Goursal 43 Cauchy-Green formula 48

E EnestrOm-Kakeya 10 entire 13 essential singularity 83, 92, 115 Euler's constant 157, 167 237

Index

Euler's formula for sine 153 Euler's integral formula 158 Euler's product formula 173 exponential function 23 extended complex plane 113

Hausdorff-Young inequality 223 Herglotz' trick 166 Hermite polynomial 62 holomorphic 13, 115 homotopic 33 homotopy invariant 34 homotopy lifting property 29

F F. and M. Riesz' theorem 213 factorial function 158

I

Fejdr 4

index 31 infinite product 147 inhomogeneous Cauchy-Riemann equation 140, 142

formula of complementary arguments 158 Fourier transform 106 Fresnel integral 68 functional equation 158 functional equation for the zeta-function 171 fundamental theorem of algebra 39, 99

inversion 96 isolated singularity 82, 115, 186

J Jensen's formula 193 Jensen's inequality 192 Jensen's inequality for integrals 209

G gamma-function 157 Gauss' formula 167 geometric series 1 Germay's interpolation theorem 163 global Cauchy integral theorem 71 global Cauchy theorem and integral formula 72, 74

K Kellogg's theorem 197

L Landau's constant 101 Laplace operator 108, 183 Laplacian 108 Laurent expansion 92, 135, 161 Laurent series 79 Legendre duplication formula 167 Legendre polynomial 62 Leibniz' series 6

H Hadamard-de la theorem 175

Poussin's

Hadamard's gap theorem 58 Hankel's formula 167 Hardy space 211, 217 harmonic 183 harmonic conjugate 223 harmonic function 202 Harnack's monotone convergence theorem 191

length 17

limit point 51 line integral 18

line segment 17 Liouville's theorem 7, 65, 99, 112 238

mdcx

Liouvile's theorem for harmonic functions 196 Littlewood's Tauberian theorem 7 logarithm 25, 76, 77, 146

Picard's big theorem 105 Picard's little theorem 106, 107, 112 Poisson integral 68

logarithmic conjugation theorem 184 logarithmic derivative 150 Looman-Menchoff theorem 14

Poisson's formula 187

Poisson kernel 188, 203

pole 82, 83, 92 pole of order k 115 polynomial 130 positively oriented circle 17 power 27 power series 1, 15 primary factors of Weierstrass 155 prime number theorem 181 primitive 13, 76, 125, 145 principal branch 27

M M. Riesz' theorem 226 Maximum Modulus Principle 55 maximum principle 219 mean value property 188, 217 Mellin transform 158 Mergelyan's theorem 139 meromorphic 115

principal part of f at a 162 principal square root 36

metric 108 Mittag-Leffler's theorem 162

principle of permanence 52 principle of subordination 56 Prym's decomposition 159 pseudo-metric 108

MObius function 181 Möbius transformation 116

Montel's theorem 120 Morera's theorem 50

Q

N

quotient formula 2

Nevanlinna's theorem 215 null-homotopic 33, 75

R radius of convergence 2 Rado's theorem 210 rational function 87, 96

0 open mapping theorem 54 order 53

region 122

order of a pole 82

removable singularity 82, 92, 115, 186 removable singularity theorem 46, 82 residue 83 residue theorem 85 Riemann's extension theorem 45 Riemann hypothesis 177 Riemann mapping theorem 123 Riemann sphere 114 Riemann zeta function 169

P partial fraction expansion of cotangent 153 path 17

Perron-Frobenius theorem 40 principle 219 Phragm6n-LindelOf theorem 67 239

Index

three lines theorem 220 type 223

Riesz-Thorin's interpolation theorem 221 theorem 53, 134, 135, 145 Runge's polynomial approximation theorem 138 Runge's theorem on approximation

U unique Continuation theorem 51 univalent 57 upper semicontinuous 199

by rational functions 135

USC 199

Schonky's theorem 103, 111 Schwarz' formula 224 Schwarz' lemma 55, 112, 126 Schwarz' reflection principle 66, 217 sigma-function of Weierstrass 168 signal 81 simple pole 82, 83 simply connected 75-77, 138, 145, 183 Sinus Cardinalis 94 square root 35, 76, 146 starshaped 19, 30, 75 Stirling's approximation formula 160 strong maximum principle 66, 189, 202 subharmonic 201, 203

V value 148 Vitali-Porter's theorem 126

w Wallis' formula 93

Wallis' product 153, 167 weak maximum principle 55 weak unique continuation properly 195 Weierstrass' factorization theorem 156 Wcicrstrass' p-function 165 Weierstrass' theorem 50 Wcierstrass' zeta-function 164 winding number 31

z Tauber's theorem 6

Z-iransform 82

Taylor series 16

zero 130

240

1212 hc

ISBN 981—02—0375—6