Multivalent Functions, Second Edition (Cambridge Tracts in Mathematics 110)

MULTIVALENT FUNCTIONS Second edition

W. K. HAYMAN Professor Emeritus in the University of York

CAMBRIDGE UNIVERSITY PRESS

Published by the Press Syndicate of the University of Cambridge The Pitt Building, Trumpington Street, Cambridge CB2 1RP 40 West 20th Street, New York, NY 10011-4211, USA 10 Stamford Road, Oakleigh, Melbourne 3166, Australia 0 Cambridge University Press 1958, 1994

First published 1958 Second edition 1994

Printed in Great Britain at the University Press, Cambridge

A catalogue record of this book is available from the British Library Library of Congress cataloguing in publication data available

ISBN 0 521 46026 3 hardback

TAG

Contents

Preface Preface to the second edition

page ix xi

Elementary bounds for univalent functions Introduction Basic results Elementary growth and distortion theorems Means and coefficients Convex univalent functions Typically real functions Starlike univalent functions Asymptotic behaviour of the coefficients

1 1 4 9 11 13 14 15

2.0 2.1 2.2 2.3 2.4 2.5 2.6 2.7

The growth of finitely mean valent functions Introduction A length-area principle The growth of multivalent functions Some averaging assumptions on p(R) Simultaneous growth near different boundary points Applications Functions of maximal growth Behaviour near the radius of greatest growth

28 28 29 32 37 40 42 45 48

3 3.0 3.1 3.2 3.3 3.4

Means and coefficients Introduction The Hardy—Stein—Spencer identities Estimates of the means /AO Estimates for the coefficients A counter-example

66 66 67 69 71 76

1

1.0 1.1 1.2 1.3 1.4 1.5 1.6 1.7 2

V

1

vi

Contents

3.5 3.6 3.7 3.8

Coefficients of general mean p-valent functions Growth and omitted values k-symmetric functions and Szees conjecture Power series with gaps

4 4.0 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 4.9 4.10 4.11 4.12 4.13 4.14

Symmetrization Introduction Lipschitzian functions The formulae of Gauss and Green Harmonic functions and the problem of Dirichlet The Dirichlet integral and capacity Symmetrization Symmetrization of functions Symmetrization of condensers Green's function and the inner radius The principle of symmetrization Applications of Steiner symmetrization Applications of circular symmetrization Bounds for If(z)1 and If(z) Bloch's Theorem Some other results

103 103 104 105 107 109 112 116 119 122 127 128 130 133 136 140

5 5.0 5.1 5.2 5.3 5.4 5.5 5.6 5.7 5.8 5.9

Circumferentially mean p-valent functions Introduction Functions without zeros Functions with a zero of order p at the origin Regularity Theorems: the case a = 0 The case a > 0: the minor arc The major arc Proof of Theorem 5.5 Applications: the case A. =1 Functions with k-fold symmetry Some further results

144 144 145 148 150 152 154 155 158 159 162

6 6.0 6.1 6.2 6.3 6.4 6.5

Differences of successive coefficients Introduction The basic formalism An application of Green's formula Estimates for the first term in (6.19) A 2-point estimate Statement of the basic theorem

165 165 167 169 172 176 180

78 94 95 98

Contents

vii

6.6 6.7 6.8 6.9 6.10

Proof of Theorem 6.2 Coefficient differences of k—symmetric functions Asymptotic behaviour Starlike functions The theorems of Dawei Shen

183 185 186 188

7

The Ltiwner theory

7.0 7.1 7.2 7.3 7.4 7.5 7.6 7.7 7.8 7.9 7.10 7.11 7.12 7.13

Introduction

197 197 198 200 203 204 207 209 211 215 222 224 226 228 229

8

8.0 8.1 8.2 8.3 8.4 8.5

Boundary behaviour in conformal mapping Transformations Structure of infinitesimal transformations The class S i Continuity properties The differential equation Completion of proof of Theorem 7.1 The third coefficient Coefficients of the inverse functions The argument of f (z)/ z Radii of convexity and starshapedness The argument of f '(z) Conclusion De Branges' Theorem Introduction Legendre polynomials Proof of Milin's conjecture: preliminary results The Milin—Lebedev inequalities Proof of de Branges' Theorem Some further results

Bibliography Index

191

230 230 231 236 243 247 248 255 261

Preface

Suppose that we are given a function f(z) regular in the unit circle, and that the equation f(z) = w has there (a) never more than one solution; (b) never more than p solutions; or (c) at most p solutions in some average sense, as w moves over the open plane. Then f(z) is respectively univalent, p—valent or mean p—valent in lz I < 1. It is the aim of this book to study what we can say about the growth of such functions f(z) and, in particular, to obtain bounds for the modulus and coefficients of f(z) and related quantities. Thus our aim is entirely quantitative in character. The univalent functions represent the classical case of this theory, and we shall study them in Chapters 1, 7 and 8. By and large the methods of these chapters do not generalize to p—valent or mean p—valent functions. The latter two are studied in Chapters 2, 3, 5 and 6. The theory of symmetrization is developed in Chapter 4, both for its applications to Chapter 5 and for its intrinsic interest. This chapter could reasonably be read by itself. Chapter 7 could be read immediately after Chapter 1 by the student interested mainly in univalent functions. Otherwise the chapters depend on prvceding work. The majority of the material here collected has not, to my knowledge, appeared in book form before, and some of it is quite new. I am, however, extremely indebted to G. M. Golusin's tract [1947] for the contents of each of Chapters 1 and 7. Montel [1933] should also be mentioned, though his approach is rather different from mine. In a tract of this size it is not possible to be exhaustive. Thus I have not been able to find space for Schiffer's variational method, nor Jenkins' theory of modules, both ix

x

Preface

of which have recently scored fine successes in the general field of this book, but I have tried to give references to these results as far as possible. The variational method is developed in Schaeffer and Spencer [1950] and Jenkins has covered his theory in a tract [1958] in the Ergebnisse Series. The book does demand certain previous knowledge of function theory. Most of this would be contained in the undergraduate course as given, for instance, in Cambridge. When something further is required I have tried to give references to Ahlfors [1979, C. A.] or Titchmarsh [1939] where the results in question can be found. Apart from such references it has been my aim to give detailed proofs of all the theorems. In several cases there is a rather difficult key theorem, from which a number of applications follow fairly simply. In such a case, the reader may omit the proof of the basic theorem on a first reading, until convinced of its value by the application. Finally I should like to thank all those persons who have helped me with this book, and in particular, Professor Kennedy, Dr. Smithies, Dr. Kiivari, Dr. Clunie and Mr. Axtell for much patient criticism in the proof stage and earlier, and Mr. Barry, who kindly prepared the index for me. I am also grateful to the editors for allowing me to publish this book in the Cambridge Tracts series, and to the Cambridge University Press for their patience and helpfulness during all the stages of the preparation of this work. W. K. H. LONDON

January 1958

Preface to the Second Edition

In the last 35 years the subject of multivalent and particularly univalent functions has developed rapidly so that it seemed necessary to expand this book considerably. In my choice of new material I have tried to concentrate particularly on fundamental results that are not contained in the two most important books on the subject, that by Pommerenke [1975] and Duren [1983]. Chapter 6 has been devoted to Lucas' bounds for coefficient differences of mean p—valent functions and some new results by Leung and Dawei Shen. I have included Eke's regularity theorems for the maximum, means and coefficients which make it possible to extend results to areally mean p—valent functions which had previously been obtained only for the much narrower class of circumferentially mean p—valent functions. The most important event in the area of this book has been the proof by de Branges [19851 of Bieberbach's conjecture. Chapter 8 has been added to deal with this subject. I have also included in Chapter 3 results by Clunie, Pommerenke and Baernstein on the coefficients of univalent functions which are bounded or have restricted maximum modulus. Here we have the unusual phenomenon of a type of theorem where the results for univalent functions are significantly stronger than those for mean univalent functions. Now that Bieberbach's conjecture is proved an analogous conjecture due to Goodman [1948] for the coefficients of p—valent functions constitutes perhaps the most interesting challenge in the area. It is mentioned at the end of Chapter 5. The conjecture is plausible but looks like being extremely difficult and has been proved only in a few very special cases. Thus there are two completely new chapters and the other chapters all contain new material. All the chapters now contain examples to test the reader's understanding of the material. xi

xii

Preface to the Second Edition

I would like to express my debt to Professor Baernstein for his advice on new material, to Professor Pommerenke and Professor Duren for their stimulating books and to May Ghali and Detta Dickinson for their painstaking work in providing Cambridge University Press and me with a camera-ready manuscript. Any remaining mistakes are entirely my own. David Tranah of CUP has been very helpful and supportive throughout. W. K. H. YORK 1993

1 Elementary bounds for univalent functions

1.0 Introduction A domain is an open connected set. A function f (z) regular in a domain D is said to be univalent in D, if w = f (z) assumes different values w for different z in D. In this case the equation f (z) = w has at most one root in D for any complex w. Such functions map D (1,1) conformally onto a domain in the w plane. In this chapter we shall obtain some classical results, which give limits for the growth of functions univalent in the unit disc izi < 1. Most of the rest of this tract will aim at generalizing these theorems by proving corresponding results for p-valent functions, i.e. those for which the equation f (z) = w has at most p roots in D, either for every complex w, or in some average sense as w moves over the plane. If f (z) = an zn is univalent in Izi < 1, then so are f (z) — ao and o O. In fact if a l were zero, f (z) would (f (z) — ao)/a i , since al = f(0) take all values sufficiently near w = ao at least twice. We thus study the normalized class CS of functions w = f (z) = z + a2 z 2

+ ...

univalent in 1z1 < 1. The two equivalent basic results here are due to Bieberbach [1916] and state that, if f(z) e S, f la21 < 2, and that f (z) assumes every value w such that 11/t < ,i . This latter theorem had been previously proved with a smaller absolute constant by Koebe [1910]. The results of Bieberbach are best possible. We shall first prove them and then develop some of their main consequences.

1.1 Basic results f

We have

This symbolic statement stands for l(z) belongs to the class

a.

2

Elementary bounds for univalent functions

Theorem 1.1 Suppose that f(z) e s. Then a21 2, with equality only for the Koebe functions z f o (z) = = z +2z2ei° 4_ 3z 3 e210 + . . . (1 — zei°) 2 We need the following preliminary result: -,-Foo

Lemma 1.1 Suppose that w = f(z) = a?is regular in a domain n—co containing 1z1 = r, and that the image of 1z1 = r by f(z) is a simple closed curve J(r), described once. Then the area A(r) enclosed by J(r) is +09

7E

=— co

na

' 2n r

We write w = f (rei° ) = u(0) + iv(0), where u(0) =

1 2

[an einG

+ an e—i nlrn,

v(0) =

1 2i

e0

_ ane—in0p.n .

Thus

A(r) =

=

dv u— d0 d0 1.27r ['OD

f2'

1 4 Jo

(am eini° + am e—im° ) m= —corm + GO

rn(aneine

x[

+ an e—ine ) d0

n=—co

= =

(—na, + nr2nan )+ an (nr2nan — nã_)] n

2r 2n

9

since E nana, = E nana_n = 0, as we see on replacing n by —n in the summation. Thus the lemma is proved. Suppose now that

w = f (z) = z + a2z 2 + ... e S. Then so does F(z) = [f(z2 )]1 = z + 02z 3 + ... In fact f(z 2 ) does not vanish except at z = 0, where it has a double zero, and if F(zi) = F(z2),

3

1.1 Basic results

then f(z?) = f(zi), and so zi = zi, i.e. zi = +z2. But F(z) is an odd function, so that z1 = —z2 gives F(z 1 ) = —F(z2). Hence we must have z 1 = z2. Also since f(z 2 ) has only a single zero of order two, F(z) is regular. Therefore F(z) is univalent. Next write co 1 1 1 1 g(z) = F(z) = — — —a2z + ... = z 2 z

n=1

Then g(z) is univalent in 0 < 1z1 < 1, and so the image of lz = r by g(z) is a simple closed curve for 0 < r < 1. Hence by Lemma 1.1 CC

1 4_ E onl 2 r2n . + AO _ r2 n=1

does not vanish for 0 < r < 1. The left-hand side is clearly negative for small positive r and so for 0 < r < 1. As r --+ 1 we deduce that

00 Enlbn 12 .... 1. n=1

Thus we have Ibi 1 = 1021 bn = 0 (n > 1), and in this case

1, and equality is possible only if

1 z z g(z) = — — ze 10 , F(z) = f(z) = z 1 — z 2 ei0 ' (1 — ze 19 )2. This proves Theorem 1.1. We deduce immediately

Theorem 1.2 Suppose that f(z) E S and that f(z) w in 1z1 < 1. Then Iwi 1. Equality is possible only if f(z) is given by (1.1) and w = —e -119 . Since f (z) * w wf(z) = z + (a2 + —1 ) z 2 + ... E w — f (z) w

a.

Thus Theorem 1.1 gives 1 < 2, w

1 <_ 2 + la2 1 w

4,

1w1

1 4-,

as required. Equality is possible only if a2 = 2e 10 , w- ' = —4e i9 , and then f(z) must be given by fo(z) in (1.1).

4

Elementary bounds for univalent functions

•;

d „... .....

Fig. 1.

We note finally that the function z

fo(z) = (1 —z) 2

1 ( 1 -F z ) 2 4 — z)

1

maps 1z1 < 1 (1,1) conformally onto the w plane cut from —1 to — oo in 1z1 < 1. along the negative real axis. Thus fo(z) E S and fo(z) # Hence the functions fo(z) = fo(ze ie ) of (1.1) also belong to S and fo(z) Ae -i0 . Thus Theorems 1.1 and 1.2 are best possible. We shall see that these functions fo(z) are extreme in for a variety of other problems also. In 1984 de Branges [1985] solved the outstanding problem in the theory, by proving the Bieberbach conjecture that Ian < n holds for f(z) E S and n> 1 with equality only for f(z) = f o (z). We shall give the proof when n = 3, which is due to L6wner [1923] in Chapter 7. In Chapter 8 we shall prove de Branges' Theorem in full generality.

1.2 Elementary growth and distortion theorems We can develop an interesting further group of inequalities as a direct consequence of Theorem 1.1. Theorem 1.3 havet

Suppose that f(z)

(1 + r)2

E

f(z)

a.

Then, for 1z1 = r (0 < r < 1) we r (1 — r)2 '

Bieberbach [1916], Gronwall [1916], Szegii [1928].

(1.2)

1.2 Elementary growth and distortion theorems

5

1+r 1—r / (z) -f (1— 03' (1 + r)3 1 r f'(z) 1 +r < <• r(1 +r) — f(z) — r(1 — r)

(1.3)

-

—

(1.4)

Equality holds in all cases only for the functions fe(z) of (1.1). We assume that Izo l < 1, and set Zo Z 0(Z) = f ( 1+20z ) = bo + b i z + b2z2 +

(1.5)

Then clearly 4)(z) is univalent in 1z1 < 1. Further

bo = f(zo), b1 = 00) = (1 — 1z01 2 )f (zo), b2 = 10" (0) = 1(1 — 1411 2 )2 .114) — 4(1 — I zo1 2 )f (zo). We apply Theorem 1.1 to (4)(z) — bo)/bi in S and obtain 1b21

"(zo)( 1— Izol 2 )2

—

220f'(zo)(1

Izo1 2 )1

—

4(1

—

21b11, i.e.

Izo1 2 )If t (zo)1.

Writing zo = pew we deduce flzo)

Zo

f'(zo)

2p 2 1 — p2

4p

(1.6)

Since

0

uP

log It(Pel°)1=

9leter(peit9) f'(peit9)'

we obtain at once

1 — p2

0 log It(pet9)1 Op

2p + 4

1

—

p

2.

On integrating this from 0 to r with respect to p, we deduce (1.3). We deduce immediately that

If(re16 )1

f

Iff( pe to)Id p

fo

r 1 +p)3 P dp = (1 _ 02'

and this is the right-hand inequality in (1.2). To obtain the lower bound for If(re i9 )1, we assume without loss of generality, that f(rei°) = Re', where R < since, otherwise there is nothing to prove. It then follows from Theorem 1.2 that the straight line segment 2 from 0 to Reo lies entirely in the image of zf < 1 by f(z). Hence A corresponds to a path 1 in 1z1 < 1, which joins z = 0 to re. Thus if t = izi we deduce

6

Elementary bounds for univalent functions

from (1.3)

1—t J r dw Idzi __. f1 (1 + o3 at = (1 + 02 , R . ldwl = f f dz and this completes the proof of (1.2). Finally, we apply (1.2) to [0(z)— bo]/bi, where 4)(z) is defined by (1.5). This gives f

( Zo -1 +2øZ)f (zo) l+

Ibi I

lz1 0 - iz 1)2 •

Putting z = —z o, b 1 = (1 — (z01 2 )f (zo), we deduce (1.4). It is easily seen that the functions f 0 (z) of (1.1) yield equality in the right-hand inequalities of (1.2)—(1.4) when z = re—i8 , and in the lefthand inequalities if z = —re—i° . On noting that by Theorem 1.1 equality is possible in (1.6) and hence in the subsequent inequalities only if 4)(z) reduces to one of the functions fo(z), we easily see that no other functions can give equality in Theorem 1.3.

We develop now another proof of some of the results in Theorem 1.2.1 1.3, which is based on Theorem 1.2, rather than 1.1. Once Theorem 1.2 has been extended to more general classes of functions, the present proof will also generalize. To make this evident we introduce the following: Definition Let f(z) = z + a2z2 +... be regular in 1z1 < 1. We shall say that f(z) E So if, given any complex zo, with lzol < 1, and any function co() univalent and satisfying Ico(C)1 < 1, co() # zo in ICI < 1, we have for 0g) = f [0)()1 100)1

40(0)1 + If (zo)l).

We note that S is a subclass of So. For if f(z) E a, f(zo) =-- wo say. 0g) = f [o(C)] is univalent in IC < 1, and 4)(c) We write 0(C) = bo + b1 +..., and apply Theorem 1.2 to (0() — bo)/bi which belongs to Z and never takes the value (wo — bo)/bi. Thus wo — bo b1

1

4 ; IN

= 10' (0 )1

4Iwo — bol

4[1f(zo)I+ 10(0)1

and so f(z) E So. We shall see in Chapter 5 that the class So is effectively a good deal larger than S, and so the results of Theorems 1.4 and 1.5 will apply to a significantly more general class of functions than the univalent ones.

7

1.2 Elementary growth and distortion theorems Theorem 1.4

Suppose that f(z) z a 2z 2

G

So. Then

(1.7)

2.

la21

Further, we have for 1z1 = r (0
It(z)1

If (z)1

(1

(1.8)

02'

1+ r r(1 —rr)If (z)1 <(11+ — r)3.

(1.9)

Finally, the equation f(z) = w has exactly one root in 1z1 < 1 if iwi <

In return for our greater generality we have lost only the left inequalities in (1.3) and (1.4). This is inevitable, since the derivatives of functions in So may well vanish in 1z1 < 1. To prove Theorem 1.4, put z

Z

4d

(1 — z)2 = = ( 1 — ) 2 ' where d = r/(1 +02 for some fixed r satisfying 0 < r < 1. Then zI < 1 cut from —r to — 1 along the negative real axis is mapped (1,1) conformally onto the Z plane cut from —d to — oo along the negative real axis and so onto ICI < 1. Thus if we write z = co(), 45 (C) = f [o(C)], then co(C) is univalent, co() —r in ICI < 1, and so, since f(z) E So, we have 100)1

4 {10(0)1 + If( -01} ,

i.e.

4d1t(0)1

41f(—r)1.

Since f'(0) = 1, this gives If(-01 d, and on applying the argument to f (zei°), which belongs to So if f (z) E So, we have the left inequality of (1.8). It follows immediately that f (z) 0 in 1z1 < 1 except at z = O. Next it follows from Rouche's Theoremt that if 1w1 < r(1 +0 -2, f (z) and f (z) — w have an equal number of zeros in 1z1 < r, i.e. exactly one. Making r f(z) = w has exactly one root in tend to 1, we deduce that, for 1w1 <

lzl < 1. Next choose 0 so that a2et0 = —1a2 1. Then as r

0,

If (re )1 = r + a2e 10 r2 + 0(r 3 )1 = r — 1a 2 Ir2 + 0(r3), t See e.g. Ahlfors [1979, hereafter called C.A., p. 153.]

8

Elementary bounds for univalent functions

and r

> (1 + r)2 = r — 2r2 + 0(r3 ), If (ree )1 — by the left inequality of (1.8). This gives 1a21 2. It remains to prove the inequalities (1.9) and the right inequality of (1.8). We put

Z=

1 4_ c ) 2

z

(1

—

z) 2

=k

(1

—

C

'

where k =

r

(1

—

) 2.

Here r is a fixed positive number such that 0 < r < 1. Then 1C1 < 1 corresponds (1,1) conformally to the Z plane cut along the negative real axis and to 1z1 < 1 cut along the real axis from —1 to O. We again write z = co(), 0 (C) = f [co ( )]. Then w(C) # 0 in 1C1 < 1, and since f(0) = 0, f (z) E ao, we have

100)1= (1 r )3 4k1f(r)1 1+r

410(0)1= 41f(r)1.

Since e—i° f(ze iû ) e So also, we deduce

.1 + r 1+r If(re 18 )1 = r(1 — r) k(1 — r)3

If (ret6)1

and this is the left inequality of (1.9). Hence

a„ loglf(rel° )1 <

f'(rei9 ) 1+r < f (reio) — r(1 — r)'

(1.10)

and integrating this from r i to r2, where 0 < r i
f(r2eiû ) < f (new)

[

2

( 1 + r )dr r(1 — r)

= log

[(1 — ri)2r21 ri(1 — r2)2 i '

or (1 — r2)2 r2

09 If(r2e )1

(1 —r i )2 ie ll. If frie ri

Making ri tend to 0 in this we deduce the right-hand inequalities of (1.8) and (1.9) with r = r2. This completes the proof of Theorem 1.4. From the point of view of later applications it is worth while to note the following consequences of (1.9):

Theorem 1.5

Suppose that f(z) = z + a2z2 +... e So and set

M(r, f) = max1f(z)1 (0 < r < 1). 1z1=r

1.3 Means and coefficients

9

Then, unless f(z) = fo(z) = z(1 — ze 10 )-2 , (1 — r) 2r-1 M(r,f) decreases strictly with increasing r (0 < r < 1), and so tends to a as r 1, where a < 1. Hence the upper bounds for Ifiz)1,1f(z)1 given by (1.8) and 0 (1.9) respectively are attained only by the functions fo(z).

To prove Theorem 1.5 note that equality can hold in (1.11) only if equality holds in both the inequalities of (1.10) for r i
f (rei°)

1 r

r(1 — r)'

and so

r(rei° ) ' = 0 (r i f(rei°) i.e. z

f' (z) f (z)

r < r2),

1 + ze -i° 1 — ze -i°

for z = rei° (ri
(1 —r1)2 (1 —r2)2 0 M(r2 , f) < If fr ie )1 ri r2

(1 —ri)2 ri

Hence, unless f(z) = fo(z),tp(r) = (1 — r) 2r-1 M(r, f) decreases strictly 1 by (1.9). Thus ip(r) < 1 with increasing r (0 < r < 1), and tp(r) (0 < r <1), so that the upper bounds for If(z)l in (1.8) and for Ir(z)1 in (1.9) are not attained and lim p(r) = a < 1. This proves Theorem 1.5. r-■ 1

1.3 Means and coefficients We referred above to de Branges' Theorem that lan l < n holds when f(z) E S and n > 2. We proceed to prove the simpler inequality Ian ' < en due to Littlewood [1925 ].

Theorem 1.6

E S. Then

Suppose that f (z) = z + a2z 2

I i(r, f) =

f

2n 0

and so

<

If (re16 )00 <

r

1—r

n — 1 ,4.1 < en

n

(n

(0 < r < 1),

2).

(1.12)

(1.13)

10

Elementary bounds for univalent functions

As we saw when proving Theorem 1.1,

0(z) = [f (z 2 )] 1 = z + b3z 3 + b5z 5 + . . . E S, and by Theorem 1.3, applied to f(z), we have 14)(z)1 1z1 < r. We note that

r/(1 — r 2 ) for

27r

1

27r 0/

10' fre le )1 2d0

1 2m 1

27r

f1o0 27r4i(reit9 )4Y(reie)d0

2n

co

E nbn rn—l ei(n-1)0

(n=1

mb m rm- 'e-i("1-

')°

dû

(m=1

co

= E n 2lbn 12r2n-2 .

n=1 Thus

oo

7r E

on i2 r2n

n=1

= f PdP r

r2n

0

10'(Pele )1 2d0

{area of transform of zl < r by w = 4)(z)}

r 1 — r2

<

(1.14)

For since 4)(z) is univalent, the area of the transform is at most nR 2 , where R is the greatest distance of the transform from w = O. Integrating term by term from 0 to r after division by r we obtain 00 E Ibn1 2 r 2n < r2

1— r 2.

n=1 But 1

Ii (r2,f) =

27r fo If(r2e)Ide =

27r 127r

2n fo

Equality is clearly excluded here.

27r f0

I Ore )0(rei° ) dB

E Ibn 12 r2n .n=1 t

2rt 10(re it9 )1 2 de

11

1.4 Convex univalent functions

Thus I i (r2 , f) <

r2

1 — r 2.

Writing r for r2 we have (1.12). Again writing r = 1— 1/n we obtain 1

2n

f izi=r

2ir

f (z)dz

1

7 n+1

2nrn

1 — li(r,i) rn

fo

f (re ie )e—in° dO

1 r(1 — r)

1 n < en. n— (1 + This gives (1.13) and completes the proof of Theorem 1.6. Baernstein [1975] has proved that I i (r, f) and more general means of f (z) attain their maximum value when f (z) = f 0(z). In particular l i (r,f)

1 : r2

0<

r <1.

Baernstein's Theorem is beyond the scope of this book. ( For an account see e.g. Hayman[1989, Theorem 9.8, p. 674.])

1.4 Convex univalent functionst It is quite simple to obtain the exact bounds for the coefficients of functions belonging to certain subclasses of S. In this section we consider functions mapping 1z1 < 1 onto a convex domain D. Such functions we shall call convex univalent. A domain D is said to be convex if given wi and w2 in D the straight line segment joining w 1 to w2 also lies in D. It is then easy to prove by induction that the centre of gravity In (w 1 w2 + + wn ) of n points 14, 1 to wn in D also lies in D. Theorem 1.7

E 00n=1gn zn is convex univalent and hn zn be regular in 1z1 < 1 Let w = h(z) = E n=1

Suppose that g(z) =

oo

maps 1z1 < 1 onto D. and assume there only values w which lie in D. Then Ihl particular gf,1 for n 1.

and in

Consider hl tp(z) = g—l [h(z)] = — z . . . gi Then tp(z) is regular in lz I < 1, satisfies lip(z)1 < 1 there and Ip(0) = O. Hence Schwarz's Lemmas yields _< 1, i.e. I hi I t

The results in this section are due to Lbwner [1917]. See e.g. C.A. p. 135, Theorem 1.3.

12

Elementary bounds for univalent functions

Next let rik (1 < k

m) be the mth roots of unity and consider

1 H(z) = — Eh(qkz11m). hm z + h2m z 2 + m k=1

instead of h(z). Since D is convex and h(z) assumes only values inside D so does H(z), and we deduce hml = 2, 3,...). This proves Theorem 1.7. We deduce that if g(z) = z + g2z2 + ... is convex univalent then Ign1 < 1 for n > 1. These inequalities are sharp for every n, as is shown by w =g(z)= 1 -z

= z + z2 +

which maps lz I < 1 onto the half-plane 9iw > We can also sharpen Theorems 1.2 and 1.3 for the functions g(z). In each case the function z/(1 — z) is extremal. Theorem 1.8 Suppose that w = f(z) = z + ... is convex univalent. Then f(z) assumes every value in the disc 1w1 <

Let D be the image of lz I < 1 by f (z) and let 14,0 = re be a point of smallest modulus lying outside D, By considering —e — i° f(—ze ie ) instead of f(z) if necessary, we may suppose that wo = —r. Then 9iw > —r in D. For suppose, contrary to this, that D contains a point w i , such that < —r. By the convexity of D that segment, say L, of the line through wo, w 1 , which lies on the side of wo opposite to w 1 , lies entirely outside D. But L contains points of modulus less than r, and this gives a contradiction. Now

w = g(z) =

2rz = 2rz + 1—z

maps lz I < 1 onto Do = {w : 91w > —r}. Since f(z) = z + ... assumes values in Do only, Theorem 1.7 applied with f(z) instead of h(z) gives 2r > 1. This proves Theorem 1.8. We have finally

1.5 Typically real functions

13

Theorem 1.9 Suppose that w = f(z) = z +... is convex univalent. Then we have for zl= r (0< r <1)

r 1— r , 1 It(z)l .- (1_ 02 ,

r 1 + r -- If(z)I 1

f'(z) 1 < r(1 + r) — f(z) -- r(1— r) • All these inequalities are sharp with equality holding when f(z) = z/(1 — z) and z = -Fr. We omit the proofs which are exactly 1, analogous to that of Theorem 1.3, starting with the inequality la21 which follows from Theorem 1.7, instead of la21 < 2.

1.5 Typically real functionst Following Rogosinski we shall call f(z) typically real if f(z) is regular in lzi < 1 and f (z) is real there if and only if z is real. We have Theorem 1.10 Suppose that f(z) = z + a2z2 + ... is typically real. In that case, and in particular if f(z) G S and has real coefficients, we have lanl n (n = 2, 3, ...).

We write f(z) = u + iv and suppose that f(z) is typically real. Then f(z) is real on the real axis and so has real coefficients. Thus Go

v(rei° )=

E an rn sin nu. A

1

Also v(rei°) has constant sign for 0 < 0 < n and lanrn I

so

n 2 = — f v(re i° )sin n0d0

no 2n ' < — f Iv(re"9 ) sin 040 = nr (0< r <1), no

since 1 sin n01 < n sin O. Letting r tend to 1, we deduce 1 an 1 n. If f(z) has real coefficients then f(2) = f(z). Thus if f(z) has real coefficients and is real for some complex zo, we have f(zo) = f (2o), which is impossible for a univalent f (z). Thus if f (z) E S and has real coefficients, f (z) is typically real and the above argument applies. t Rogosinski [1931]. See also Dieudonné [1931] and SzAsz [1932] for the present proof.

14

Elementary bounds for univalent functions

1.6 Starlike univalent functionst A domain D in the w plane is said to be starlike (or star-shaped) with respect to a fixed point 0 in D, if for any point P in D the straight line segment OP also lies in D. If f(z) G CZ and maps 1z1 < 1 onto a starlike domain with respect to w = 0, we shall call f(z) starlike univalent. We have

Theorem 1.11

If f(z) = z + a2z 2 +... is starlike univalent then lanl -.- n

(n = 2, 3,...).

Let G be the map of 1z1 < 1,G, the map of 1z1 < r, by f(z). We show first that G r is starlike with respect to w = 0 for 0 < r < 1. In fact if w E G, then tw E G for 0 < t < 1, and so the function

ip(z) = r i [tf (z)] is regular in 1z1 < 1 and satisfies .ti)(z)1 < 1 there and tp(0) = O. Thus we have from Schwarz's Lemma Itp(z)1 1z1 < 1. Suppose now that w1 G Gr. Then w i = f(zi), where Izil < r. Hence Iri (twi)I = Iti)(zi)I ___ zil < r. Thus twi = f(z2) with 1z21 < r and so twi E Gr for 0 < t < 1. Hence Gr is starlike. The boundary of G,. consists of the curve w = fire) (0 < 61 < 2m). Since G,. is starlike the radius vector from w = 0 to f(re i° ) lies in G,. and so arg f(rei° ) increases with O. Thus 0 — 00

f(rei61) =

[re° f'(reit9 )] 93

arg

f (reie)

> 0 (0 < r < 1, 0 < 0 < 27r).

(1.15)

Write now OC

E

f'(z) z = 1+ am zm. f(z) m=1 Then it follows from (1.15) that f'(z) w = h(z) = z f(z)

00

1 =Earn?' m=1

takes values lying entirely in the convex domain Do = {w : 9iw > —1} . t Nevanlinna [1921]. For a generalization of Theorem 1.11 to functions mapping lz I < 1 onto a more general class of domains introduced by Kaplan [1953] see Reade [1955].

15

1.7 Asymptotic behaviour of the coefficients Also

w = g (z) =

2z 1—z

= 2z + 2z2 +

maps 1z1 < 1 onto Do . Hence by Theorem 1.7 land < 2 Again we have

E anzn (

1

+ E am

zm

. E nanzn,

m= 1

n=1

= 1, 2, .

1

and equating coefficients we deduce

na n= an +

+... + Œn-1 •

Thus (n — 1)1anl =

2(1 + la21 +... +

an-11

lan-11).

If we assume that ak < k (k = 1, 2, ... , n — 1), we deduce

n(n — 1),

(n — 1)lapd

and the proof of Theorem 1.11 by induction is complete. We remark finally that the function

(1 — z)2

=z+ 2z 2 + 3z2

is both starlike univalent and typically real, so that the inequalities of Theorems 1.10 and 1.11 are sharp.

1.7 Asymptotic behaviour of the coefficients While we defer the full proof of de Branges' Theorem to Chapter 8, we can obtain an asymptotic form of the result at this stage [Hayman 1955]. Theorem 1.12

Suppose that f (z) = z +

— —)

as

a2z2 + n

oo,

a.

Then

(1.16)

where oc is the constant of Theorem 1.5. In particular lanl n for n> no(f)-

Our proof of Theorem 1.12 is elementary but rather lengthy. Extensions covering more general functions are given in Chapters 2 and 5. They use similar arguments. For another proof of Theorem 1.12 see Milin [1970]

16

Elementary bounds for univalent functions

and Duren [1983, P. 165]. An argument of Pommerenke [1975, p. 76] based on an inequality of Fitzgerald [1972] is shorter but only yields lim lanl < 1, n---+Dc n unless f (z) is a Koebe function. Our argument is a refinement of Littlewood's proof of Theorem 1.6. We first deal with the case a = 0, which is relatively simple. If a > 0 we use the Cauchy formula: an

27r

1

= 27rpn fo

f (pe i° )e—i ne dO .

(1.17)

We first prove the existence of a (unique) radius of greatest growth arg z = 00 such that

(1— 0 2 1f(re i°0 )1 --0 a, as r --0 1 while argf (rewo) varies slowly. This allows us to establish a corresponding asymptotic behaviour on the major arc y' =I arg z — Ool --.- K(1 — 0, where K is a fixed constant. This argument works for the class So of Theorem 1.5. We then use the univalence to show that the corresponding minor arc y contributes relatively little to the integral in (1.17). In this way we obtain the required asymptotic formula for an . Theorem 1.12 shows that if, in the class S, lan l has its maximal growth for large n, then Ian 1 behaves in a rather regular manner. Results of this nature occur frequently in function theory and are called regularity theorems.

1.7.1 The case a = 0 We use the terminology and notation of Section 1.3 and suppose that a = O. Thus if E > 0 there exists ro, such that ro < 1, and E2 r

M(r, f) < (1 — 02' r o2 < r < 1. i Recalling that 0(z) = {f(z 2 )} 2 , we deduce that

er M (r, 0) < 1 _ r 2 '

ro < r < 1.

Hence the inequality (1.14) can be sharpened to

E O n Ir2n < ( 1er— r2 )2l' n=1

ro
17

1.7 Asymptotic behaviour of the coefficients Continuing as in the proof of Theorem 1.6 we obtain, as r —> 1

ocs

1 i (r2 , f ) = E bn1 2 r 2n =

i

=

=

L /0

r ( 0c E2ion12p2n-idp) 1 ro

fr

+ < e2

ro

c2 r2 (1 — r2)

r 2pdp fo ( 1 — P 2 )2 + 0(1)

+ 0(1).

Choosing r = (1 — 1/n) we obtain for large n

Ian ' _-_

1 _rn i I i (r, f) < 2e2 en.

Thus

lan l .— 0, n

and Theorem 1.12 is proved in this case.

1.7.2 The radius of greatest growth

We suppose from now on that

a > O. We proceed in a series of steps. In this section we prove the existence of a radius arg z = 90 , on which If (r ei°°)1 has maximal growth and arg f (re ieci) varies rather slowly. We call arg z =190 a radius of greatest growth (r.g.g). Lemma 1.2

If a > 0 there exists a real 00 such that

f (re i°°) = R(r)ei'l(r)

,

where R(r) ,,, a(1 — r)-2 as r -- 1 and

1L

(1 — r).1/ (r)2dr <.

(1.18)

(1.19)

In particular /1(1'2) — /1(ri) —4 0,

(1.20)

uniformly while r i —÷ 1 and ri < r2 < ri +k(1—ri), where k is a constant, k < 1.

Elementary bounds for univalent functions

18

We recall that by (1.11) (1 — r)2 r— ilf(rei° )I decreases with increasing r for fixed 0 and so tends to a limit a(0) mfr0 as r —)__+ 1c.it ,To prove (1.18) we show that there exists 00 such that a(190) = Œ. We choose rn = 1 — 1/n, and On , 0 < On < 27r, such that an =

( 1

—

rn)

2

=

( 1

—

rn )2

as n -- co.

1f(rne i6 n)1

rn rn We use (1.11) again and deduce that (1 — r)2

r

, !Preion11 -?- an,

0 < r < rn .

Let 00 be a limit point of the sequence On, and suppose that as k —+

We fix r such that 0 < r < 1. Then as 0 —+ (1 — 02 r

If(re i°°)i =

11111

Thus

R(r) = If(rel°°

(1 — r)2

through the sequence

Onk

If (rei 'ik )1 __ lim an = a.

ar )1 -- (1 — r) 2 '

0
and by Theorem 1.5

R(r) M(r, f) <

a + o(1) (1 — r) 2

as r —> 1.

This yields (1.18). Next we prove (1.19). For this we use (1.9). We write

log f (re') = log R(r) + 0.0 = Or) + i).(r) say, where 0.'02

+ /11 02 < { 1 + r y r(1 — r)

by (1.9). We write a = (TV), b = .1'(r) and c = (1 + r)/ { r(1 — r)}. Then b2 ..._ (c — a)(c + a) Thus

2c(c — a).

b2

— < c — a, 2c — and so 2(1 + r)

.1'(r)2 __

1 +r r(1 — r)

1.7 Asymptotic behaviour of the coefficients

19

We integrate both sides from ri to r2 and deduce that f.1r2

J

.2 r(1— r) , (1 + ri) fr2 (1 — r);C(r)2dr < .i.'( )2 dr ri f 1+r r2 (1+ ri) f f r2 1 + r , , dr — f cqr)dr} <2 ri tin r( 1 — 1') ri r2 r 1 = [log ri (1 — r)2 R(r) j

Letting r2 tend to 1, we deduce that

{If (r i ei°0)1(1 — r i )2 } < oo.

4

2 dr _.<_ — log r),'( ri

ri a

We choose r i = 1 and note that AV) is continuous in the interval [0, fl. This yields (1.19). Finally we suppose that E > 0, that 0 < r < 1 and choose ro so close to 1 that i E2 (1-02'0 2 dr < log k .

1

Then if no < r i k(1 — r i ) we deduce that 2f f 1r

I)-(r2) — *1)1 =

{

<

E

(r)dr

2

lo g -11-,-

lo

{

1 7

1 }

f

r2

14,

=

fr2

(1 — r).1i(r)2 dr

dr 1 1 1— r f

E.

This proves (1.20).

1.7.3 Behaviour on the major arc Our next aim is to find asymptotic formulae for f(z) and f(z), valid on an arc 10 — NI < KO — p), where Z .----- pei°. For this purpose we define rn

=1

—

Zn

= rvnnc.i00,

n> 1

where 00 is an r.g.g. It will turn out that the r.g.g. is unique for univalent f, but we do not assume this for the time being. Given a fixed positive E we denote by A n = the domain given by

An(e)

An =

E 1 1 — <11— ze —i0 {z: °I< — , arg (1 — ze—it9 °)< — rc — n En 2

E} .

20

Elementary bounds for univalent functions

We also define -2

= n f(z),

atn fn(z) = (1— ze — ie0)2. Thus an, fn (z) are defined for n

1 and by (1.18)

—> cc, as n Lemma 1.3

co.

We have as z —> eie° through the regions A n (e) f(z) , f n (z)

and

f'(z)

f ni (z).

(1.21)

We shall suppose without loss of generality that 00 = 0, since otherwise we may consider f(ze iN), f n (ze ie°) instead of f(z), f n (z). Then if 1 z = ln (Z) = rn + — Z, n

1 n

(1 — z) = — (1 — Z),

(1.22)

the domain A i (e) corresponds in the Z plane to

We write g(Z) = (1— Z) 2 f Un(z)] f(zn) •

Then, for a fixed E, A n (E) lies in lz I < 1 when n is sufficiently large and so g(Z) is defined in A i (c) for all large n. We now define

Nn = An(10, A = An( -41- E). Then the corresponding domains in the Z plane are = A 1 (1E) and A" = Let d be the distance of A' from the complement of A". Then the distance of Nn from the complement of Nn' is d/n, and hence the distance of A'n from izi = 1 is at least d/n for large n. Thus by Theorem 1.4 there exist constants no and C1 such that

If (z)1 < CO 2 , if n > no and z E Hence if Z E A' and n > n1 we deduce that If {ln(z)} I < c1n2 ,

Ifizn)1 >

1.7 Asymptotic behaviour of the coefficients

21

and so there is another positive constant C2, such that

Ig(Z) I <

C2,

if n

n 1 , and Z

E

A'.

Suppose now that Z is real, —1 < Z < 0, and let z = p correspond to Z by (1.22). Then p rn _<_ (1 + p)/ 2 and so by (1.18) and (1.20) (1 — rn) 22 , .i.e. (1 — P)

f (P) f (rn)

f (p) —

an

( 1 — P )2 '

so that gn(Z ) = (1 — Z) 2 f(P) = (1 — P)2f(p) . 1, —1
<

O.

Œn

Thus gn (z) is uniformly bounded in A' for n __ no , and g(Z) —> 1 cc for —1 < Z < 0. It now follows from Vitali's convergence as n theoremt that

g(Z) —> 1, g(Z) —> 0 (n —> c/o) uniformly for Z

E A1(E).

Translating back into the z plane we deduce

1 d n2 (1 — z) 2 f (z) , 0. f n (z) —' ' n dz f(rn ) f(z)

1

The first of these limiting relations is the first relation in (1.21). The second gives

(1 — z) 2 f' (z) — 2(1 — z)f (z) = o(n) i.e.

f' (z) =

f (z) ± o(1 — z) -3 , 1z since z E A(E). Using the first relation in (1.21), we obtain the second one and Lemma 1.3 is proved. 1.7.4 Behaviour on the minor arc In order to complete the proof of Theorem 1.12 we show next that f (z) grows relatively slowly away from the r.g.g. For this we need f to be univalent, at least in an average sense.

Lemma 1.4

Suppose that f(z) E S, a > 0 and that 00 is as in Lemma 1.3. Then given a positive ii, there exist positive constants, Co, C1 and ro, where ro < 1, such that if zo = reie , where

ro

r < 1, C0(1 — r)

t (Titchmarsh [1939], p. 168)

10 — NI < m,

(1.23)

22

Elementary bounds for univalent functions

we have or else

If (z0)1 < C1,

fi(zo)

f(zo)

<

11

(1.24)

r1—

We note that (1.21) and (1.24) are incompatible for the same z, so that the r.g.g. is unique. We suppose that E is as in Lemma 1.3. It follows from that lemma that, as z —> 1 through the union of the sectors A(c),

11(z )!

(1.25)

' all — Z1 -2, If (z)I - 2OE11 - z1 -3 .

Thus (1.25) holds uniformly as z —> 1 in the Stolz angle TE

S: 1 arg (1 — z)i < — — E,

2

(1.26)

II — Z1 < sin e.

If R > Ro(e) we consider the image D(R) by w = f (z) of the sector i 1 2 a S(R) : { R(1 Œ_ ) } ' < I 1 zl < { I arg z1 < — 7r E. ER(1 + E)} ' 2 —

We suppose that Ro is so large that S(R) lies in S. It follows from the first relation in (1.25) that, if Ro is sufficiently large, D(R) lies in ER < iwi < R.

Next the area A(R) of D(R) is given by

If' (z)I 2 Idzi 2 .

A(R) = f S(R)

We set z = 1— pe i(1) , so that idz 1 2 = pdpd(/), and deduce from the second relation in (1.25) that, for large R, {a/ER(1+017

A(R) > 41:42 (1 — e)

i—E i pdp f p-64 — i -Fe f{Œ/Ro —01 7

= (7r — 2 0 - E )R 2 [(1 — E )2 — E 2 (1 ± e)2] )

. where E 1 tends to zero with e. Suppose next that i __ izoi < 1, that the disc iz — zol < 1(1— Izol) does not meet S(R) and that if (zo)i = R/e. Then by (1.9) we have, if lz I = P, f'(z)

1+p < f(z) — P( 1 — Pr

1.7 Asymptotic behaviour of the coefficients

14(1 — Izol), we have

In the disc Iz — zoI f'(z)

f(z)

<

7

1 + 1\

p

23

4 (31zoI + 1) so that

1

< 27 4 4 )1— p 11 3(1 — IzoI) < 1 —zo

Thus if lz — z o l < J4 (1 — Izoi), we deduce, integrating along the segment from zo to z1, that

4 1 — Izol =1. 1 — Izo l 4

f'(z)

log f (zi) f (zo)

f(z)

Hence Ifizi)I < elf (zo)i, and so the image d(R) of Iz — zol < ,14- (1 — IzoI) lies in Iwl < R . Since f(z) is univalent in Izi < 1, d(R) is disjoint from D(R) so that the area a(r) of d(R) satisfies

a(R) < n R2 — A(R) <

El

R2.

Thus (z —zo)
If' (z)12

(1.27)

< E1R2 .

If we expand f'(z) in a power series 00 f'(z) = bn (z — zo)" o

we deduce that 1 f2.7r 2n 0

(zo + pei° )I 2 dO =

Ibni 2 p2

11

1b01 2 = fi(z0)1 2 .

Thus

2

f

If / (z)I 2 1dZ1 2 > n 4 (1 — Izol)] If' (4)1 2

iZ —Z01<4( 1— IZOI)

[1

Now (1.27) yields 11'(z0)12 <

16e 2Ei n( 1 —16141) 2 EiR2

ri2 lf (Z0)1 2

n( 1 — 1z01) 2if (4)12 < (1 — IZOI) 2'

if ei < nri 2 /(16e2 ). We suppose that E is so small that Ei satisfies this inequality and deduce that z = zo satisfies (1.24) provided that R elf(zo)I > Ro(E) and that the disc Iz — zol < 1( 1 — Izol) is disjoint from the Stolz angle S given by (1.26). Suppose contrary to this that z = pei° lies in S and also that lz —zol < i(1 — Izol). Thus

3 5 —(1 —Izol) < 1 —p < —(1 — I zol). 4 4

24

Elementary bounds for univalent functions

Suppose further that 1 —Izo i and consequently 1 — p is so small that

101 < C(1 — p), where C is constant depending on E. In fact this will be the case if C > cot E for p close to 1. We also have, for p close to 1

10— arg zol < C(1 — P). Thus

5 larg zo I < (C + 0(1 — p) < — (C + C)(1 — Izol). 4 Hence if (1.23) holds, where ro is sufficiently close to 1 and Co is sufficiently large, the disc lz — zo l < :If (1 — zo ) is disjoint from S and (1.24) holds. This proves Lemma 1.4.

1.7.5 Completion of the proof of Theorem 1.12 We now suppose that Œ > 0 and 610 = 0 in Theorem 1.12. We write r- n = 1 — 1/n, fn(z) = an(1 — z) 2 , where an = n2 f (rn ). Then

an =

f it

1 27Ern

Ctn(n + 1) =

r

1

27-crn

f( r ee)e—in°c10, i n

ir —it

Thus an

—

an (n

+ 1) = —2n1 f7'_7, { f (rn ew ) — f n (rn e18 )} e -1n0d0.

(1.28)

We divide the right-hand integral into the ranges K .1 1 :101 < — and n

K , < iui -- 7E,

/2 : — n

where K is a suitably large positive constant. In /1 we have by Lemma 1.3,

(1.21) f (Ne i') — fn(rne 10 ) = o(n). Thus

1

{f(rne i° ) — f n (rne ie )} e —i n0d0 = o(n), as n —> oo.

(1.29)

25

1.7 Asymptotic behaviour of the coefficients

and suppose that ro, Co, Ci are the corresponding Next we fix ri = constants of Lemma 1.4. We suppose also that Co > 1, that Co(1—r) <101 < n, and that 1f(re ie )1 > Ci . Let r i be the smallest number such that If(te i° )I C i for r i t r, further Co(1 — ri) IOI and ro r1Then we deduce from Lemma 1.4 that 1 , ri < t < r. 2(1 — t)

f'(tel e ) f(tei°)

0 log If (te16 )1 = Ot

Integrating this inequality we obtain If (rei° )1

If (riew )1 ( 1 —1.1 1— r

Our conditions for r i ensure that r i = ro or 1f(rie i8 )1 = Ci or C0(1 — ri) = 101, since otherwise we could replace r i by a smaller quantity. In the first case we obtain If (re1° )1

*NJ)

(1_ ro)' 1 —r

In the second case we have 1 — r1 )7 1—r

1f(re'e )1

7

1 — ro (1 — r

The third case yields

(rei° )1

<

If (ri e'° )1 (1 — ri)—

(1 —1. 1 ) 2 1— r 7

21 (—

1 —r

(1 —r) -1 (1 C,g101 -4 (1—r ). Thus we obtain in all cases Ifire w )1 < C2 (1 — r) -1 100, if Co(1 — r)

101

7t,

ro < r < 1,

where the constants CO, C2 depend only on the function f. We also have

ifn (re10 )1

11 — rei° 1 -2

C3 101 -2 , Co(1 — r) <101 < 7r.

26

Elementary bounds for univalent functions

We now choose r = rn = 1 — 1/n and K > Go> 1. Then

fK Irl

iel <7r

If(rne ie )— f n (rn ei9 )} e —i n° d0 1

7E

2C4n

.._ C4 f ni 0 —3 dO _.._ 2C4ni ( n ) = K Id • K In

(1.30)

Given a small positive (5 we choose K so large that 2C4/K i < 6. Combining (1.28) to (1.30) we obtain for sufficiently large n 27r (1 — —1 ) n tan — Œ n (n + 1)1 < 26n. n

Thus

an -- not,

(1.31)

and (1.16) is proved. In fact when a > 0 (1.31) is stronger than (1.16), since (1.31) gives information about the argument of an as well as Ian I. We note that either a = 1, and so by Theorem 1.5 Ian ' = n for all n, or else 0 < a < 1 in which case lan l < n for n > no. This completes the proof of Theorem 1.12.

Examples

1.1

If A > 0 and CO

bA( z).

(1—

z) -2 .Edn(A)z", o

prove that d(A)= n

1.2

r(2 + n)

n11-1

F(A)F(n + 1)

F(A)

as

n —> co,

where F(x) denotes the Gamma function (cf. Titchmarsh [1939, p. 58]). Deduce that, if A > 1, 09

217r fir

Ifre'')1 bA 'n 11-c/0

=

Edn( , ) 2 r 2n 0

,

I F(/1 — 1) (1 _ 0 1-2), 2,\/7r F(A)

as

r —> 1.

(Use the duplication formula for F(x), Titchmarsh [1939, p. 57])

27

Examples

1.3

With the hypotheses of Theorem 1.12 we define I Jr, f) = — 1 f Ifirel f3 )1/1d0. 2m

Prove that, if )>

r(il -

( 1 — r)2-1 I Ar, f) 1.4

2 \prF().)

r

1.

Obtain corresponding asymptotic formulae for

fr

2n

12 (r, f') and A(r,f) = f

0

1.5

as

dû

0

If / (tet° )1 2 tdtd0.

Prove that, if 2> 1, there exists ro such that 0
12(r, fo(z)), ro < r < 1 (1.32)

where fo(z) = z(1 — ze 10 )-2 . Obtain corresponding results for 12 (r, f') and A(r, f). (It follows from Baernstein's Theorem, referred to in Section 1.3, that (1.32) holds for 0 < r < 1.)

2 The growth of finitely mean valent functions

2.0 Introduction

We saw in the previous chapter that the assumption that f(z) is univalent in lz I < 1 imposes a number of restrictions on the growth of f(z), such as were proved in Theorems 1.3, 1.5, 1.6 and 1.12. In this and the next chapter we see what results we can obtain under the more general assumption that the equation f(z) = w has at most p roots in some average sense, as w moves over the plane. In the present chapter we confine ourselves to obtaining bounds for f (z)I. We set M(r, f) = max I f (z)1

1z1=r

(0 < r < 1),

(2.1)

and show that under suitable averaging assumptions M(r, f) = 0(1 — r) 2P (r

1).

(2.2)

The first result of this type is due to Cartwright [1935], who proved (2.2) when p is a positive integer and the equation f(z) = w never has more than p roots in lz I < 1 for any w. Such functions are called p-valent . Her method, based on a distortion theorem of Ahlfors [1930], was extended by Spencer [1940b] to the more general case. If f(z) = ao + aiz + ... has q zeros in 1z1 < 1, bounds for M(r, f) can be obtained ,depending on Jug

(2.3)

= v=0

This dependence is essential. In fact any polynomial of degree p is p-valent and has at most p zeros in iz 1 < 1, but bounds for M(r) must clearly depend on all the coefficients. Our method also leads to a number of theorems, which show that f(z) 28

2.1 A length-area principle

29

cannot grow too rapidly near several points of lz1 = 1 simultaneously, and in particular that, if M(r) attains the growth (1 — r)-2P, then if (r )1 attains this magnitude for a single fixed value of 0, and is quite small for other constant 0 as r 1. Results of this type appear to need the full strength of the methods of this chapter even for univalent functions, and were first proved by Spencer [19401)]. We complete the chapter with Eke's [19671)] regularity theorem for the maximum modulus.

2.1 A length-area principle Let f (z) be regular in an open set A and let n(w) be the number of roots in A of the equation f (z) = w. We write 1

f2m

p(R) = p(R, A, f) = — 27r 0 n(Rei 4 )4.

(2.4)

The integral exists as a finite or infinite Lebesgue integral, since n(w) > O. The function p(R) will play a dominant role in the sequel. It is clear that p(R), like n(w), increases with expanding domain A. Further if z = t(C) maps A1 (1,1) conformally onto A, then if p(R, A1) refers to f [t( O], we have p(R, A) = p(R, A 1 ). We might make the averaging assumption p(R) p (0 < R < oo).

This is certainly satisfied if p is a positive integer and f (z) is p-valent in Izi < 1. We shall use this assumption in Chapter 5; at present weaker hypotheses will be sufficient. We shall base our results on a length—area inequality. A result of this type was first proved by Ahlfors [1930], and its use in this context is due to Cartwright [1935]. Such results play a key role in conformal and quasi-conformal mapping (see Lelong-Ferrand [1955], Ahlfors [1973, Chapter 4]).

Theorem 2.1

Suppose that f(z) is regular in an open set A and that p(R) = p(R, A) is defined by (2.3). Let l(R) be the total length of the curves in A on which if(z)1 = R, and suppose that the area A of A is finite. Then we have

l(R) 2dR Jo Rp(R)

27" '

1. It will be shown in the course of Lemma 5.2 that the sets n(w) finite plane. Thus n(Rei 0) is certainly measurable.

K are open in the

30

The growth of finitely mean valent functions

where the inte grand is taken be zero if l(R) = 0 or p(R)= +co. In particular, l(R) < +co for almost all R for which p(R) < +oo.

2.1.1 We first prove Theorem 2.1 when A is an open rectangle and f (z) is univalent and not zero in a domain containing A and its sides. We say that f (z) is univalent and not zero on A. Then any fixed branch of

s(z) = log f (z) =

a + it

is also univalent on A and maps A onto a domain Q in the s plane. The boundary of Q is the image of the boundary of A by s(z) and is a sectionally analytic Jordan curve. Let 0, be the intersection of Q with the line a = constant. Then 0, consists of a finite number of straight-line segments -C

1 < -C < 1-1 1,

T2 < 1" < "C12,

...

As s = a + iv describes 0,, z = x + iy describes the set y, in A on which If (z)1 = e' . Since f(z) = es(z ) is univalent, t Tv < 2n. On the arc of y, corresponding to the segment T v
has no roots if t To < T v + 2n and one root if Tv
1 p(e, A) = — E(T /y — 2n

Tv )

0(a)

=- r.

We now have, using Schwarz's inequality, t 2 [ f dz l(ea , A) 2 = dd [19, ds

f

dz 2 dT dT 0, fOc ds dz 2 dz 2 = 0(a) of, dt = 27rp(e, A) /64, dt. ds ds

If al, a2 are the lower and upper bounds of a on Q this gives

f +" l(ea , A) 2

L. p(ec, A) du =

f„i p(ea, A)

1

'2 l(ea , A)2

„I L --d-s-

62 da < 27r f da

dz 2

dT

= 27rA,

22 t IE ab _< E lal 2 E Ibl 2 or If fgdx1 __ f Ifl 2dx f Igl 2dx. For a proof see, for example, Titchmarsh [1939, p. 381].

2.1 A length-area principle

where A is the area of A. Writing

eci =

31

R we have

r l(R, A) 2 dR < 27rA jo p(R, A) R — '

(2.5)

as required.

2.1.2 To extend our result to the general case we need the theory of the Lebesgue integral. We may assume without loss of generality that f (z) is regular and f (z) 0, f' (z) * 0 in A. For otherwise we may consider f(z) in the set Ao, consisting of A except for zeros of f (z) and f (z), without affecting any of the quantities p(R), l(R) and A in Theorem 2.1. Next we express A as a countable union of non-overlapping rectangles Av (y = 1 to oo).t At each point zo of such a rectangle f(z o ) # 0, and so there is a neighbourhood of zo in which f (z) is univalent. Thus we may subdivide each A, into a finite number of smaller rectangles on each of which f (z) is univalent. We therefore assume without loss of generality that f (z) is univalent, f (z) * 0 on each A v . The set of curves yR, on which If (z)I = R, meets a side of one of the rectangles A. in only a finite number of points, unless If (z)I R on the side. Thus if we omit the finite or countable set of values of R for which If(z)1 R on some side of a rectangle A v , we have 00

p(R, A) =

Ep( R,Av),

(2.6)t

v=1 00

l(R, A) =

E l(R, A,),

(2.7)§

v=1 since the sides of the A, do not then contribute to p(R, A) or l(R, A). Also if A v , A are the areas of A v , A respectively, we haves 00

A =

E Av . v=i

We now set 2

l(R, Av ) 2

v

P(R, Av)

a =

t Burkill [1951], B, p. 8. * This follows from (2.3) and [B], p. 41. * [13] , 11 35 . 1 [B], p. 21.

,

2

bY

= p(R, Av)

32

The growth of finitely mean valent functions

in Schwarz's inequality (E a,,b,,)2 < E a,2, E b2, and sum over those values of y for which p(R, Av),1(R, A v ) are not zero. Using (2.6) and (2.7) we obtain l(R, A,,) 2 l(R, A) 2 p(R, A) V =1 p(R, A v ) where indeterminate terms are taken to be zero. We now integrate from 0 to co and obtain, using (2.5) applied to each A y , l(R, A r )2 ] dR A ,) R E v— 1 p(R"

l(R, A) 2 dR Jo p(R, A) R

z--e v =1

l(R, A v )2 dR < jo p(R, A v ) R

Av

27A.

v=1

The inversion of integration and summation is justified since the terms are non-negative. t This completes the proof of Theorem 2.1.

Examples

2.1

If A is the rectangle a <x < b, c 0, show that l(R) = d — c, p(R) =

2.2

a(d — c)

ife"
and l(R) = p(R) = 0 for other values of R. Deduce that in this case equality holds in Theorem 2.1. If A is the disc 1z1 < r, and f (z) = azn where a is a complex constant, a 0, and n a positive integer, show that l(R) = 27(R/lal)

and p(R) = n, for 0 < R < lalrn

and l(R) = p(R) = 0 otherwise. Deduce that again equality holds in Theorem 2.1.

2.2 The growth of multivalent functions We now prove a useful result relating the change in If(z)1 from one point of A to another to the function p(R). We suppose that

00

1(z) =

E a zn 0

t [B], p. 41.

2.2 The growth of multivalent functions

33

has zeros z 1 , z2 ,..., z q in A = { lz I < 1} and define ,u = inf {E lay laol/ lzv I } v=o v=i

(2.8)

where an empty product is taken to be 1. Then we shall obtain a bound for the maximum modulus M(r, f) in terms of /2 and p(R). The proof is adapted from a theorem of Jenkins and Oikawa [1971].

Theorem 2.2 Suppose that f(z) is regular in A with q zeros, counting multiplicity, that p is defined by (2.8) and p(R) by (2.4). Then if

= m(r,f)= sup If(z)1, 1z1=r

and M > p, we have

dR Rp(R)

< 2 log

1 + r 7r2 Ao < 2 log , 0 < r < 1, 1— r 2 1 —r

where Ao = 2e2 /4 = 23.58 ... We need

Lemma 2.1 Suppose that f(z) is regular in A and has zeros z 1 ,...,z q there counting multiplicity. Let y be a simple closed Jordan curve which surrounds the origin and lies in A. Then inf If (z)I

p

ZE'y

where p is given by (2.8). Landau [1922] has proved a sharp version of this result, when y is a circle. The present simple argument is due to Hayman and Nicholls [1973]. We consider first the second term on the right-hand side of (2.8). We define zv F(z) = f (z) / H( z 1— f v z) v=1

Then F(z) yields

0 in A and so the maximum principle applied to 1/F(z)

inf If(z)1 zey

inf IF(z)1 zey

IF(0)1 =

IaoI/ftz t, I. V =1

34

The growth of finitely mean valent functions

Suppose next that

if(z)i > E V

=0

on y. We write g(z)=

Eavzv, v=0

so that

tez)i E lad v=0 for z G A and in particular for z p. 153] f(z) and

E y.

Thus by Rouches Theorem [C. A. co

f(z)— g(z) =

E

av zv ,

v=q+1

have equally many zeros inside y, i.e. at least q -1- 1. This contradicts our hypotheses and Lemma 2.1 is proved.

Lemma 2.2

Suppose that f(z) satisfies the hypotheses of Theorem 2.2. Let zo = reie be a point on = r, such that

If(zo)1 = M. Then if It < R < M and R is not one of a finite or countable exceptional set F of values, there exists an analytic open arc y R , which meets the line segment 1 : [0, zo] and approaches the boundary lz1 = 1 of A as we move along y R in either direction. Also If (z)I = R on y R •

We note that 11(0)1 ti by (2.8). Thus there exists p, such that If(Pe 1e )1= R and 0 < p < r. Let E1 be the set of zeros of f` (z) in A and let E2 be the set of points z = pei° on 1, such that d•1

dt

If(te e )1 2 = 0, when

t = p.

(2.9)

Then, unless f is constant in A, E1 is finite or countable. Also unless the analytic function ç(p) = (pe1t9 )1 2 is constant for 0 < p r, the set E2 is finite. In all cases the image F of E1 U E2 by If(z)1 is finite or countable.

2.2 The growth of multivalent functions

35

We assume that It < R < M and that R is not in F. Let p be the least positive number such that

If (P?)I = and let yR be that component of the level set if(z)i = R, which contains z = pei° .

0 on yR so that y R is an analytic curve. In fact By hypothesis f(z) f(z) is locally univalent on y R and so yR is locally the image of w 114, 1 = R by the regular function z = f-1 (w). We obtain the whole of yR by moving along the circle w = Re 10 from wo = f (pe i° ) = Reio° say, in the direction of increasing or decreasing 4). If yR goes through no point of A more than once, y R must approach 1z1 = 1 as 4) —> +oo or 4) —> —oo. Otherwise there exist Oh 4)2 such that (Rei 02), while f — '(Re 10 ) is (1,1) for 4)1 < (Po < 4 and f (Re id'i) = < 02. In this case yR is the image of the interval [4)1, 02] by 4)i f 1 (Re) and so yR is a closed Jordan curve. It remains to show that this case cannot occur. Suppose contrary to this, that y R is a closed curve and let D be the interior of y R . The curve yR cannot touch the segment 1 at pei° since R F, so that (2.9) is false for t = p. Thus the segment 1 crosses yR at z = pei° . Also since p is the smallest number such that If (pei° )I = R, we must have t < p.

If (teie )I
(2.10)

Thus we cannot have

a —

io If (te )1 < O

at

at t = p, and so

0 • — 1f(te'9)1 > 0, et

t = p.

It follows that, for t slightly larger than p, If (tei° )1 > R. By the maximum principle If (z)I < R in D, so that points tei° lie outside D for t slightly larger than p. Since the segment 1 crosses yR at pei° , it follows that tiei° must lie in D for tl slightly less than p. By (2.10) the segment [0, hell does not meet y R and so the whole of this segment lies in D. In particular z = 0 lies in D and this contradicts Lemma 2.1, since If (z)1 = R > II on YR. This completes the proof of Lemma 2.2.

36

The growth of finitely mean valent functions We can now complete the proof of Theorem 2.2 . We write ( = log

(1 + ze — ie )

(2.11)

1 — ze — ie ) •

Then C maps A (1,1) conformally onto the strip

(2.12) so that 1 corresponds to the segment

L : 0 < < (21 -= log

1+r , 1—r

II = O.

Let So be the subdomain of S consisting of all points distant less than In from 1. Thus So is given by n —— < 2
Itil

<

n < (:1 -i- — 2,

o

The total area of So is

a = 7r0 + -7r3- . 4 Consider the function g(C) = f[z(C)] = f {el ° ( e !, +

1)}

,

in So. Let p(R, S0, g) be defined by (2.4) with respect to g. Since So corresponds by (2.11) to a subset of A, the equation g(() = w cannot have more solutions in So than the equation f(z) = w has in A. Thus p(R,S0,g) p(R, A, f) = p(R). Again it follows from Lemma 2.2 that for all R, such that p < R < M, R çt F, the level set Ig(CY = R contains an arc F going through a point P of L and approaching the boundary of S in both directions; F contains a subarc FR passing through P, lying in So and approaching the boundary of So in both directions. The length of FR must be at least 7r. Hence l(R, S 0 , g) n,

it < R < M, REF.

2.3 Some averaging assumptions on p(R)

37

Now Theorem 2.1 yields LAI

<

Rp(R)

dR fm lw

dR Rp(R, So, g) 7r 2

2a < _ . no

n

l(R, So, g)2 dR ni2 fi, m p(R, So, g) R 1+ r n 2 ± —. r 2

This proves Theorem 2.2.

2.3 Some averaging assumptions on p(R)

If

p(R) p (0 < R < oo),

(2.13)

Theorem 2.2 shows immediately that

1 M ±r) log — <2 log ( e n 2 /4 1 P 14 1—r)' 2P /4 1 + /1 2 (e' II < ' M 1— r) which is the type of result we are aiming at. Following Spencer [1940b, 1941a] we show in this section how we may weaken (2.13). We write

p(R) = p + h(R), R

W(R) = f P(P)d(P 2 ) 0

R

= pR2 + f h(p)d(p2 ) = pR2 + H(R), o

say, and have the following inequality: Lemma 2.3 1R2

With the above notation, and if R1 < R2 we have

dp

11

JR, PP(P)

P

R2

°g R1

1 1 H(R2) 2 2 2pRi

1

1R2

P hi

H(P)dP 3 .

P J

We have R2

J

R'

dp

PP(P)

R,

dp _ f R2 dp [1 JR, P[P + h(p)} iRi P LP 1R 2 1r h(p)dP . > — log — p Ri p- R. P f

.

h(p) h2 (p) 1 + P2 P2 {P + h(P)] i

38

The growth offinitely mean valent functions

Again 1R2 h(p)dp _ hi p —

f R2 dH(p) _ H(R 2 ) H(R i ) ± f R2 H(p)dp RI 2p — 2Ri

2R?

hi

p3 •

Also W(R) is necessarily positive and so H(Ri) > —p4 Now Lemma 2.1 follows. Following Spencer we shall in this and the next chapter call a function f(z) mean p-valent in a domain A, if f(z) is regular in A, p is a positive number, and W(R) pR 2 (0 < R < co).

Using the definition (2.4) we see that W(R) =

1 7

[27r fR

JO JO

n(pei(P)pdpd4),

and so mean p-valency expresses the condition that the average number of roots in A of the equation f (z) = w is not greater than p, as w ranges over any disc I w 1 < R. We can now prove the theorem of Cartwright [1935] and Spencer [1940b].

Theorem 2.3

Suppose that f (z) = ao ± al z + ...

is mean p-valent and has q zeros in Izi < 1. Then 0 _..ç_q p and M(r, f) < A o2Pe p(1 — r) -2P,

0 < r < 1,

where Ao = 2 exp {7r2/4} <47.2, and ti is defined by (2.8).

The result was first proved by Cartwright [1935] for p-valent and by Spencer [1940b] for mean p-valent functions but with e2402P replaced by an unspecified constant depending on p. Jenkins and Oikawa [1971] first obtained the above form of the theorem, with a somewhat larger value of the constant Ao. To prove it we note that if f(z) has q zeros in A then f(z) assumes all sufficiently small values at least q times in A. This contradicts mean p-valency if q > p. Thus q p. By Lemma 2.3 m

M 1 dR 1{ > — log — II — — } 2 ' Rp(R) p

39

Examples

since f is mean p-valent. Now Theorem 2.2 yields

< fm dR Ao _-.. 2log 1 - r . -1 {lo g --M 1} p y 2 - f tiRp(R) Thus

M ii <

1 exp { - -I- 2p log A° } 2 1-r

= eiA02P(1 - r) -2P. This proves Theorem 2.3. The constant Ao cannot be replaced by any number less than 2 in Theorem 2.3, This is shown by the function

w = f(z) = ao

(1+Z Z

2P

) '1-

which maps Izl < 1 onto the sector (possibly self-overlapping) in the w plane given by

w arg — < pm ao

(0 < lwl < oo)},

and for which

p(R)= p, W(R)= pR 2 (0 < R
m(r, f)

to ± 02p o 0-2" ,...., 22p li(1 r)-2p, r -- 1.

Examples

2.3

With the notation of the previbus section, suppose that f(z) is regular in A, f(z) 0 and

f' H+(R)dR < co, where H(R) = max(H(R), 0). fi R3 Prove that in this case

lim sup(1 - r)2PM(r, f)< GO, r--,1

where M(r, f)= max If(z)I. 1z1=r

(2.14)

40 2.4

The growth of finitely mean valent functions

If f (z) is regular in A, f (z) # 0 and

hm sup

R --+x

H(R) = 0, R2

(2.15)

prove that, for every positive e, lim(1 — r)2P'M(r, f) = 0.

2.5

Suppose that f i (z) = f(z) C where C is a constant and that W(R), Wi(R) refer to f(z), f 1 (z) respectively. Prove that

W(R —1c1) W1(R) W(R +10). Deduce that if f(z) satisfies (2.14) or (2.15) then so does f i (z). Show also that f i (z) may be mean p-valent or satisfy (2.13) without f (z) doing so.

2.4 Simultaneous growth near different boundary points We have seen that a function f (z) mean p-valent in lz I < 1 satisfies (z)1 = 0(1 — r) -2P (1z1 = r). However a function can be as large as this only on a single rather small arc of Izi = r. We shall close this chapter by investigating in what way f(z) can become large near several distant points of 1z1 = r. Our basic result is

Theorem 2.4 Suppose that f(z) is mean p-valent in a domain A containing k non-overlapping circles lz — z n 1 < rn (1 n k). Suppose further that 2Plf (zn)i Ri, If (zn' )1 R2 > eRi, where „ ir5n rn — iZ n > 0,

rn and that f(z) # 0 for lz — z n i < -IN (1

f

L

—1

n k). Then

2p log(R2/R1) — 1 .

lOg 611

Let An be the circle lz — zn 1 < r and let p,1 (R) =

A , f)

be defined as in (2.4). Consider

4)(0 =

f(zn

+ N().

41

2.4 Simultaneous growth near different boundary points Then p,1 (R) corresponds to

Og) and

ICI

< 1.

ZI — Z n n

± rnCn —

Zn

We choose ç so that rn

Then 2P10(0)I _- Ri, 10(Cn)1 R21 4) (C) * most p zeros in ICI < 1. Thus p _2P1(1)(0)1

IC! < -1. and 4)(C) has at Ri and Theorem 2.2 gives

0 for

f R2 dR Ao A0 <21 og = 2 log — on . 1 — 1CnI iR I RPn(R) Now we have from Schwarz's inequality D

(log

„ 2

2

f R‘ aR ) < f R2 Pn(R)dR f R2 dR — j R1 RI R h. r n R

)—(

1‘2

Ri

and so 1 log

<

2

L

R2

(t )

< dR _

RPn(R)

2

f

pR2 n tR)dR .

R

2 JR i

D

•

‘1 ) (log lR2

Adding we deduce

E [log -d AO k

n=1

—1

f R2 [ k 2 dR Ep(R) ] [log(R2/ROP ./,R n=1 R.

u

(2.16)

Now k

k

Using the notation of

dR p(R) R fR i R2

§2.3

R2 p log

.

R2 p log ki. + I I ( R2)

I I (R1

2Ri

[log

R2) (--R7

2/??

(2.16) and (2.17),

_1

2p

< —

+

f R2 H(R)dR R3 I

1

O. Using

E [log (t)]

)

[log(R2/Rifl

we deduce

+

(2.17)

p-valent in A,

dR

.

k

n=i

R2

IT

< p since —pR 2 H(R)

we have, since f(Z) is mean

+ L i h(R) R R2 1f R 2 dH(R) p log --1-? + -- RI R2

.

= p(R).

10,A)

Epn(R)=E p(R, An) n=i n=1

p

Pog(R2/RA2

42

The growth of finitely mean valent functions 2p log(R2 /R1 ) — l'

< and this proves Theorem 2.4.

2.5 Applications We shall make two applications of Theorem 2.4. We first apply it to the concept of order of a mean p-valent function at a boundary point. Suppose then that f (z) is mean p-valent in jz1 < 1 and that for C = e 0 there is a path y(0), lying except for its end-point 4 in Izi < 1, and also a positive 6 such that

lim(1 — lz1) 6 If (z)I > 0, as z -4 ( along y(0). Then we define the order cx(C) of f(z) at C as the upper bound of all such 6. If no path y(0) and positive 6 exist, we put oc(C) = O. We can then prove the following result of Spencer [1940b]. Theorem 2.5 If f(z) is mean p-valent in 1z1 < 1, then the set E of distinct C on ICI = 1, such that a(C) > 0, is countable and satisfies EE a(C) -_ 2p.

It suffices to show that if (1, C2, ..., Ck are distinct points of ICI = 1, then k

E a(Cn ) _<_ 2p. n=1 For then the set EN of C, such that ICI = 1 and oc(C) > N -1 , is finite for N = 1,2,3,..., and so the set E consisting of all the EN is countable. Letting k tend to oo in the above inequality we have E (x(c) 2p as required. Suppose then that Theorem 2.5 is false. Then by the above remark we can find a finite set of points C15 C2, ...5 Ck and a positive e such that

k

E Œ ) . 2(p ± ke). (

n

n=1 For each C„ there exists a path Yi, approaching

C„ from IzI < 1, on which

( 1 — 1zrlf(z)1 > 1, where ni, = *CO — e, and so

k n=1

> 2p.

(2.18)

2.5 Applications

43

Hence there exists a positive Ro, such that for zn' ---- rn en on yn , such that

If (4)1 = R2>

R2 >

Ro we can find

n k).

( 1 —1 rnr (1

(2.19)

Choose now 6 so small that

(i) f(z) has no zeros for 1 — 26 < lzi < 1, min, ICm — CnI, (ii) 46 < 1<m
lzm' —zn'l > 46 (1

m < n < k),

since z n —0 Cn as R2 —0 00, and if we put zn = circles lz — z n I < 6 are non-overlapping. Also

roe

it follows that the

R i = 2P M(ro, .f) and if (4)1 = R2,

2Plf (zn )I

where we may suppose that with 6n =

R2 >

eRi. Thus we may apply Theorem 2.4

6 — (r n — ro) = 1 — rn 6 ' 6

and obtain k

r

,

( A06 \ 1 — '

E rg 1 — r n Li n=1

2p

• log(R2/eR1) • --•

In view of (2.19) this gives k

E[log(A06R21/1")]-1 <

k

n=1 lin

E tin log[A0 6] ± log R2

<

n=1

As

R2 —0

00

2p log(R2 /eR i )'

2p log R2 — log(eR i ) •

the two sides of this inequality are asymptotically equal to

E nn log R2

and

2p log R2

respectively, and so we obtain a contradiction from (2.18). This proves Theorem 2.5.

44

The growth of finitely mean valent functions

D

1r.

frœ 1

,I

Fig. 2.

2.5.1 An example Theorem 2.7 is sharp. Suppose in fact that a n is any sequence of positive numbers for 1 < n < k, where k may be finite or

infinite, such that

E ar, = 2. E

N

Set So = 0, SN = an (1 < N < k), and let D be the domain n=1 consisting of the w plane cut from lw 1 = 1 to cx) along the lines

arg w = nSN (0

N < k).

By Riemann's mapping theorem t we can find a function z = (/)(w) which maps D (1,1) conformally onto lz 1 < 1. As w moves along the boundary of that part DN of D which lies in the angle 7cSN_1 < arg w < rcSN , z moves along an arc of the unit circle. Simultaneously W = w -1 /7n moves along a straight line segment through the origin and so by Schwarz's reflection principle can be continued as a regular function of W near W = 0, and conversely if z = z n corresponds t See, for example, C.A. p. 230 t See, for example, C. A., p. 172.

2.6 Functions of maximal growth

45

to W = 0, W becomes a regular function of z near zn . Thus

W = c i (z — zn )+ c2(z — z n )2 + ..., near z = zn, where ci

O. This gives

w — [c i (z — Zn)]" (z ---* Z n ). Hence w = f(z) = 0 -1 (z), which maps lz I < 1 onto D, is a univalent function having the orders an at the points z n for 1 < n < k, and the an are arbitrary subject to E an = 2. Again for any positive p the function [f(Z) — ly is mean p-valent in lz 1 < 1 and has orders pan at the points zn, where pan is now an arbitrary sequence satisfying > p an = 2p.

2.6 Functions of maximal growth p-valent in lz I < 1 then M(r, f) = 0(1

We have seen that if f(z) is mean

—

r)2P (r --- 1).

We complete the chapter by investigating more closely those functions for which this order of growth is effectively attained so that a = limr, i (1 — r)2P M (r, f) > 0.

(2.20)

We show that f (z) attains this growth along a certain radius and that If(z)1 is quite small except near this radius. Theorem 2.6 Suppose that f(z) is mean p-valent in Izi < 1 and that (2.20) holds. Then there exists 00 (0 < 00 < 27r) such that

ao = lim( 1

—

r-1

02/9 1f(reie° )I > e(

Π)p

Since f(z) can have only a finite number of zeros in lzi < 1 we may suppose that f(Z) * 0 in 1-26 <1z1 < 1 where 6 > 0. We set ro = 1 —0. Then there exists a sequence Cn = rn ei°n, where

ro < rn < 1, 0 < On < 27r, such that rn —* 1 (n --- oo) and If() 1 > -,2(1 — rn)_2P . We now apply Theorem 2.4 with k = 1, z

(2.21) Cn, z i = reien, where

46

The growth of finitely mean valent functions

ro
eRi

eRi or

( Ao ) 2P 61 )

(2.22)

so that (2.22) holds in any case. Using (2.21) we obtain

R 1 = 2P 1f(re 10

)1 >

R (5 2,P , > ( — r) -2P, ro
If 90 is a limit point of the sequence On , we deduce by continuity, taking r fixed, that

2P1 f (re* )1

a(1— 2eA02P

, ro < r < 1.

(2.23)

This proves Theorem 2.6.

2.6.1 It follows from Theorem 2.5 that f(z) must have zero order at all points of lz I = 1 other than eiN, so that eiN is necessarily unique in Theorem 2.6. However, we can prove more than this. Theorem 2.7 With the hypothesis of Theorem 2.6 and given e such that 0 < e < 2p, we can find a positive constant C and ro, 0 < ro < 1, such that 1 (2.24) If (re it9 )1 < (1 — 0E10 — 0 01 2P-6 for ro < r < 1, C(1 — r) 10 — 0 0 1 < n. Further we have uniformly as r --- 1, while e <10 — 0 0 1 < n, i (2.25) log If (rei0 )1
2.6 Functions of maximal growth

47

Suppose then that f (z) is mean p-valent, f (z) * 0 in

1— 26 < jz I < 1, and that

If (res(9°)I .- —21 ŒO(1 — r)2P (1 — .5
(2.26)

These assumptions are satisfied for all sufficiently small (5. We also write

2PR1 = M(1 — (5,f) = lzmax 6 If(z)I,

(2.27)

z i = (1 — 6)e`e°, z 2 = (1 — (5)e ° , and assume that 27(.5 < IO — 0131 _< m. Then the circles lz — z i I < (5, tz — z2I < (5 are disjoint. We now suppose further that

(2.28)

If(r2e i° )1 = R2 > eRi,

where 1 — (5 < r2 < 1, and put z2 = r2e10 , z'i = ri ei°°, where r i is chosen to be the smallest number such that

If(z)1= Iffrie91 = R2. Such a number exists by Theorem 2.6 and (2.27), (2.28). Then r i > by definition of RI. We now apply Theorem 2.4 with

61 =

(5 — [ri — (1 — 6)] (5

1 — ri (5 '

(52 =

1— 15

1 — r2 .5 '

and obtain A 06

[log ( 1 _ r 2)]

—1

2p log(R2/Ri) — 1

—1 A06 [log ( 1 _ ri )1 .

(2.29)

Now by (2.26) R2 > -1 ao(1 — r1 ) -2P,

and by Theorem 2.3 I c-2p Ri _._ A o2pe2po .

Thus

Ri

.-_ Ci

,

Where Ci, C2, .... will denote constants depending on f(z) and p only.

48

The growth of finitely mean valent functions We deduce from this and (2.29) that

[

-1 2p log A06 < 1 — r2 ] — log(R2/R1) — 1

2p log(R2/Ri) + C2 '

and this gives 2

1? ) — I} log (1 406 ) C3 {log (-2RI 1 — r2 Thus

i

R2 log — ._ç_ 1 + C4 [ log (, C5' )] 2 • R1

(2.30)

1 - r2

Taking 6 and R1 fixed, we deduce (2.25), provided that (2.28) holds, and the inequality is trivial otherwise. It remains to prove (2.24). We may again without loss of generality suppose that (2.28) holds. Then (2.30) gives, if 27(6 =10 — 001 1 — r2, i 2 R2 < R1 exp 1 + C4 [log (1 056r2) ] } < C6 10 - 00 I-2P

<

exp {C4 [log 1 056r2 ] 1}

1 10— Ool2P-E (1 - r2 )'

provided that

1

(1 0 — 001 y > C6 exp {C4 (log [Co- 00 11 7 [2741 r2) j) } ' 1 — r2 —

C(1 — r2 ), where C depends on p,e and f(z) which is true for ID— NI only. This proves (2.24) and completes the proof of Theorem 2.7. 2.7 Behaviour near the radius of greatest growth It follows from Theorem 2.6 that if f(z) is mean p-valent in A : 1z1 < 1 and has maximal growth there, so that (2.20) holds, then f(z) attains this growth along the radius arg z = 00. Further, by Theorem 2.7, f(z) is quite small except in the immediate neighbourhood of the radius, which we shall call henceforth the radius of greatest growth (r.g.g.). There is a general principle in the theory of functions, that if under of certain hypotheses there is a bound on the growth of a class functions, then those functions in which have the extremal growth

a

a

2.7 Behaviour near the radius of greatest growth

49

display a regular behaviour. Such theorems were first found by Heins [1948] for the class of entire functions with bounded minimum modulus and are called regularity theorems. We had an example for univalent functions in Theorems 1.5 and 1.12. We now proceed to prove some regularity theorems due to Eke [1967a,b] for mean p-valent functions. It is convenient to use the transformation

a

= + iri = -1- log {

ze — leo

(2.31) 2

(1 — ze — A ))2 } '

where arg z = 00 is the r.g.g. Then the domain Doo, consisting of the unit disc cut along the radius arg z = 00 + TC is mapped onto the strip S, given by (2.12). If g(C) = f [z ()] then g( ) is mean p-valent in S and by (2.23) we have log Ig(C)1 = 2p, + 0(1), as C --- +co

(2.32)

along the real axis. We shall, following Eke [196713], prove Theorem 2.8 Suppose that g(C) is mean p-valent in S and satisfies ( 2.32), where C = ç + ill. Then we have as --- +oo, uniformly for 1 111 < 21 — 6, where 6 > 0, log Ig(C)I = 2/;1 + fl + o(1),

(2.33)

where fl is a real constant and

g'(C)/g(C) —) 2p.

(2.34)

We note that g(C) is mean p-valent in S and so in any subdomain Si of S. We first consider g(C) in a rectangle n S i : (:, — n < < i + n, Irli < — 2' We choose 1:) so that k > it and so that all the zeros of g(C) lie in Then it follows from (2.32) that there is also a positive constant Co, such that log Ig(01 < 2/;, 0 + Co,

o — 2n _-

(2.35)

_-

and I log Ig()1 — 21) 1 < Co,

(2.36)

> c).

In fact (2.36) is an immediate consequence of (2.32) and

g ) * 0, (

and

50

The growth of finitely mean valent functions

(2.35) follows from the fact that lg()1 is continuous and so bounded on any compact subset of S. We note that if (2.35), (2.36) hold for a given 6, with a constant Co, then they also hold with the same Co when o is replaced by a large number. We shall need a sequence of lemmas for our proof of Theorem 2.8.

2.7.1 Construction of some level curves Lemma 2.4 We define Ci = Co + Ci, that 1)(1 — 6)

Co

+ 1 + 165p n, suppose that

2k0 + CI < log R < 2g i — Cl

(2.37)

and that R does not belong to a certain exceptional set F, which is finite or countable. Then there exists a level curve TR on which igg)i = R, and such that TR joins the sides ri = -Pr f of S i in S i and so crosses the segment [6,U of the real axis an odd number of times. Let F be the (finite or countable) set of values of R such that R = ig(01 for some value of ( for which e( ( ) = 0, or R = Ig( ) 1 where is real and Pg(t) I = 0 at t = ç. We suppose first that R is not in F, that 210(6 — CI) C2 = Co + C1 and that

2k0 + Co < log R < 2p6 — CO.

(2.38)

Then by (2.36) there is an odd number of values of ç such that o < c < 6 and lg(01 = R. The level curves Ig(01 = R through these points meet the real axis only on the segment [6, M by (2.35), (2.36) and (2.38). Hence at least one of these level curves TR crosses the real axis an odd number of times and so either joins the boundary segments ri = +3- of Si in Si or else meets one of the vertical boundary segments = 6 7E, or = i + it of S i . In the latter case the length l(R) of TR is at least 3n/2, since TR contains one arc going from a point ç on [6, M to one of the above segments, whose length is at least Tr, and another arc going to the boundary of Si whose length is at least In. Suppose that this latter situation occurs for each R not in F and satisfying (2.38). We shall show that this leads to a contradiction. We apply Theorem 2.1 and Lemma 2.3 to the function g( ( ) which is mean p-valent in Si, so that H(R) < 0 in Lemma 2.3. We write R1 = exp{2g0 + Co}, R2 = exp {2gi — Co} and obtain —

2.7 Behaviour near the radius of greatest growth T R2(37/2)2dR 97E2 { 1 R2 — 4p log IT — 27E2 {6 — o + 2n} > =

51

( Rp R) j Ri 972 9 72 1 2Co + , 4p — (i — ()) — — 2

i.e. 7r2

57E2

2

(i — o)

9 9C0 + 4

16p 9 18C0 5 + -1-6+ --- < C29

+ 8p , 2W i — ())

and this contradicts 2p(i — ID) C2. Thus there exists R satisfying (2.38) such that TR joins n --= +3 in Si. C2. We define 6, 4 by Suppose now that p(i sD) —

4,

P6 =

1

1

,

No + —2 C2, Phl = NI — —2 C2-

By what we have just proved, there exist RI, R2 such that 2g0 + Co < log Ri < 21)6 — Co,

2k4 + Co < log R2 < 2gi — Co,

and such that yR„ 7R 2 joins ri . + 3 in SI. Hence if R1 < R < R2 and R does not belong to F then a level curve yR exists, which crosses the segment [121, d an odd number of times. This level curve separates 7Ri from yR2 and so also joins ri = + 1 in SI. The condition is certainly satisfied if 2g3 — Co < log R < 2p1 4 + Co, i.e 2/34$0 + C2 — Co < log R < 2gi + Co — C29 i.e. if (2.37) holds. This proves Lemma 2.4.

Our next aim is to prove that when R is large the levels curves yR, whose existence is asserted in Lemma 2.4, are close to vertical segments. We define

2.7.2 Basic estimates

i (R)= inf 91(, Eyi?

2,(R) = sup 9i(,

(2.39)

GyR

co(R) = 2(R) —

(2.40)

and note that the length l(R) of yR satisfies

1 (R) 2

72 + CO(R) 2 .

We leave this result as an exercise for the reader.

(2.41)

52

The growth of finitely mean valent functions

Examples 2.6

Show that if y is a Jordan arc which meets all four sides of a rectangle Q, then the length of y is not less than that of the diagonal of Q.

It is a consequence of (2.34) that Ig(01 is finally increasing so that Y R is unique, when R is large. For the time being we do not assume this, but choose for T R the first such arc which we meet on moving along the real axis from 6 in the direction of increasing Given a sufficiently large R, we may define 12) , 6 by

2Po = log R — Ci — Co, 2 pi = log R + +Co, since (2.35) and (2.36) continue to hold if 6 is replaced by a larger quantity. We may then apply Lemma 2.4 and deduce that

co(R) :5_

— 6 + 2n =

2C 1 2C0 +27r = C3, 2p

(2.42)

provided that R > Ro say. Our next result is Lemma 2.5 R2

27E 2

f

If R2 > R > Ro we have

( n2 +00)2)dR Rp(R)

-

g 2( R2)

1(R1)}

-

{(R2) — (R1) + w(R2) + (0(R1)} 1 R2 <— 2p log — (2.43) RI + C4, where (R) denotes the least value of and C4 = Co/ p 2C3.

at which T(R) meets the real axis,

We apply Theorem 2.1 to that subdomain S' of the strip S which is bounded by TR I , TR 2 and the lines ri = +7r. In S' we have 1(R1)

2(R2)

so that the area A of S' satisfies A :5_ n(2(R2) — Also if l(R) denotes the length of TR then (2.41) holds, and if R1 < R < R2, (2.41) remains true if we replace l(R) by the total length of all levels curves Ig()1 = R in S'. Now the first inequality in (2.43) follows from Theorem 2.1.

2.7 Behaviour near the radius of greatest growth

53

Next if = (R) denotes the least real value of on the level curve y R, we have

R

1(R) w(R),

2(R)

(R)

1(R)

R2

by (2.40). Thus 2(R2) —c1(R1)

(R2) — (R1) + w(R2)+ w(Ri).

This proves the second inequality in (2.43). Finally we deduce from (2.36) that if = (R)

I log R — 2/g I < Co,

R1 < R < R2.

Applying this with R = R1, R2, we obtain + 2C0 1 {log (R2) — (Ri) — RI 2p

}

.

Using also (2.42) we obtain 1

—

2p

log

R2

Co ± — + 2C3. p

as

R

This completes the proof of Lemma 2.5. Lemma 2.6

We have R

dp 1 pp(p) = -; log R + fl + o(1),

oo

(2.44)

where fl is a real constant. Also f ' (02(p)dp

(2.45)

<00.

JR° pp(p)

To prove (2.44) we use the argument of Lemma 2.3 and of its proof. With the terminology of Section 2.3 we have R2

f

dp

Ro p p (p)

1 p

log

R2 1 { H(Ro) H(R2) + Ro p2 2R(?) R2 h2(p) d p

R2 II(P)dP JR, P3 (2.46)

fR o P(P + h(P)) .

Here h2 (p),—H(p) are nonnegative for p Ro. It follows from Lemma 2.5 that

R2dp I

PP(P)

-

P

log R2 + 0(1),

as

R2 —4 CO-

54

The growth of finitely mean valent functions

Thus h 2 (p )dp(p)). s3,

H(p)dp = s2,

I

ho 132 P (P +h P3 where 132, 133 are finite nonnegative constants. Next we show that

H(R2)/Ri —> 0, as R2 —4 00.

(2.47)

(2.48)

Suppose that (2.48) is false. Then there exist arbitrarily large values p = pn such that 11 (P1)< —Op,

where 6 is a positive number. Also W(R)= pR 2 +H(R) is a nondecreasing function of R. Thus pR 2 H(R) < ppn2 +11(Pn) < (P (5 )p2 ,

R < p,

i.e. H(R) < (p — (5)pn2

O p 2n,

pR2 <

if R < pn and

p(pn2 _ R2) < _1 (5pn2 , i.e. (pn _ R) < 2

6 Pn2 2P(Pn R)'

and so certainly if 6Pn pn— — < R < pn. 4p

This yields Pn

H(p)dp

f

p,(,-614p)

2

fPn

< — TON j

P3

dp

P 1(l -6 14P) P3

—2 94n2 {[ P n ( 1 —

44113- )]

6{ i (

6 \ -21

4

473 ) f

=

say. This inequality holds for some arbitrialy large pn contrary to the conclusion that the first integral in (2.47) converges. Thus (2.48) holds. Substituting (2.47) and (2.48) in (2.46) we deduce (2.44). Next we prove (2.45). We deduce from (2.43) and (2.44) that fR2

CO2 (R)dR

27E 2 JR° Rp(R)

< 2p

log R2

f R2 dR 2 f Ro Rp(R)

ow.

2.7 Behaviour near the radius of greatest growth

55

This proves (2.45) and completes the proof of Lemma 2.6. Lemma 2.7

We have as R

-4

co w(R)

-4

(2.49)

0

and 1

fi

j = 1, 2

(2.50)

where fl is a real constant. We shall prove (2.50) and note that (2.49) is an immediate consequence of (2.50) and (2.40). We define 1

fl4 = R—>OED lim sup {2(R) —

log R},

and 1 = lim inf { 1 (R) — — log R—co 2p R} It follows from Lemma 2.4 that X I and )15 are finite. Also since 1 (R) -. 2(R) we have )15

1 lirn sup { 1 (R) — — log R} R—> cc 2p

134

1 2(R) — — log RI 2p

)65.

and lim inf

R—co

{

Thus )1 4 )15. Hence to prove (2.50) with —/3/2p = )14 = )15 it is enough to show that /35 = )64. Suppose then contrary to this, that )14 > )15. We define e by $4 —

)65 = zle

1 (1 + - ) . P

We chose R,'2, so large that

L

oo w(R) 2 dR < e3 , Rp(R) 27,'

(2.51)

and also that for R > R' > R,'0 we have

T R' dp

1l

R (2.52)

J12

PP(P) P og R'

We can satisfy (2.51) by using (2.45) and (2.52) by using (2.44).

The growth of finitely mean valent functions

56

Next we choose R i > R;), so that 2 (R1 )

1 > — log Ri + )14 — 2p

6,

and then R2, so that R2 > R1 and 1 1(R2) < — 2p log R2 + #5 +

Subtracting we deduce that 1 R2 + fl5 — )64 + 26 2(R1) < — log 2p 1 R2 — log -IT 2p

1(R2)

(2.53)

We next note that by (2.51) f e' (0 2 ( p )d p

JR,

63

pp(p)

while by (2.52)

f RI eL

dp

g

E

JR, PP(P)

>

----

P

2P

=

2p •

Thus there exists p = R , such that w(R) < E, and Ri < R < RieE.

Similarly there exists R'2 , such that w(R) < e, and R2e -' < R2/ R2.

If Ri < R2, ), R I separates —oo from TR2 in the strip S and so 1(R), 2(R) are strictly increasing functions of R. We deduce that 2(fe2) — 1(R'1) = <

1(R) — 2(Ri)+ w(R) +w(R) 1 (R/2 ) — 2 (/?) 26 < 1(R2) — 2(R1) + 2E.

On the other hand Lemma 2.5 and (2.52) show that dp 1 R' > log 2 2pp(p) 2p R1

2(R) —1(R)

1

2p

Thus 1 R2 26 — log — — — p 2p

R2

E

E

p 4p

p

2.7 Behaviour near the radius of greatest growth

57

This contradicts (2.53). Thus fl4 = )6'5 and Lemma 2.7 is proved with )64 ------

$5 = — I3/(2p).

2.7.3 Proof of Theorem 2.8 We deduce from (2.50) that on the level curve yR we have — fl 1 = — log R — + o(1). 2p 2p Since R = Ig(()I this gives (2.33) on the level curves YR. To complete the proof of (2.33) we need a final lemma.

Lemma 2.8

If g( ( ) is mean p-valent and 00 * 0 in the disc IC —Col < 6,

then

g'Go) < A i (p)

WO

6

(2.54)

'

where Ai(p)= 2e1(2A0) 2P depends only on p.

We may suppose without loss of generality that Co = 0, 6 = 1, since otherwise we consider g((0 + (5() instead of g(C). It then follows from Theorem 2.3 with q = 0, that

Ig(01 < (2A 0 )2Pei g(0), for ICI

1.

Now Cauchy's inequality yields

2ei (2A0) 2P 1g(0)1,

Ig'(0)1

and this proves Lemma 2.8. Suppose now that Ci = çt i + iii, where i is large and lqi 1 < 11 — O. We suppose that E > 0 and choose R, so that

1

5fl

<E

where fi is the quantity in Lemma 2.7, and so that the level curve yR exists. This is possible by Lemma 2.4. It follows that yR contains a point (2 -= 2 ± bi t , since yR joins the lines ri = +nI2, and by Lemma 2.7 we have, if c i is sufficiently large,

1 2p

< E,

58

The growth of finitely mean valent functions

so that I2 deduce that

U < 2E. Integrating (2.54) along the segment [Ci, C2] we

lloglg(C2)1 — log 1g(C i )11 = 1 log g( 1 ) — log RI <

2A1(p) E 6 ,

so that, if ci is sufficiently large, depending on e, we have I )6' i - - loglg(C1)1+ — < 2p 2p E

1+ AO)) (

136

Thus

log 1g(C)1 — 2k —* fl, as

—* +oo, while 1ril < ; — 6.

This proves (2.33). To prove (2.34) we again suppose that g > 0 and choose o so large that

1 log 1g(C)1 — 2k — )61 < L 26 when lq1 < -; - 6,

> ().

(2.55)

Suppose now that CI — i + iqi, where i > o + 6 , irld < — 26. Consider F(z)= log g(Ci + z) — 2p(Ci + z) — fl = u + iv, Then

1z1 -. 6.

00 F(z)= Ean zn,

0

We write an = an + On and note that cc u(6e i° ) = E {an cos(0) —

z

fin sin(n0)} (5n .

0

In particular al = PICO = — 7r16

r u( 6e it9 )(cos 0 — i sin 0)d0,

so that

l ai l

1 fir — ir lu(6el° )1(10 < e n6 _

by (2.55). Hence given positive numbers 6 and e, there exists 6) = 12) (6, E) such that

gig i

)

n 2p < e, if i > 63 + (5, and 1 1111 < — — 26.

2 gi g) This proves (2.34) and completes the proof of Theorem 2.8.

59

2.7 Behaviour near the radius of greatest growth

We shall obtain an analogue of Theorem 1.5 for mean p-valent functions. In order to do so we proceed to prove

2.7.4 A bound for )(I

Suppose that g(C) is mean p-valent in S, and satisfies (2.33)

Theorem 2.9 and also that

—oc, through real values.

loglg(01 = 2pC + o(1), as C

(2.56)

Then fl 0 in (2.33). Equality holds if and only if

g( ( ) = exp(2pC + i2),

(2.57)

where 2 is a real constant.

Since (2.56) holds we deduce that for all positive R we can find such that

ig()1 > R, Thus if R does not belong to a countable exceptional set F the level curve y R joining ti = +.1 in S exists and meets the real axis in an odd number of points in the interval [4 We can thus apply the inequalities of Lemma 2.5 for positive R1, R2. We apply Lemma 2.3 with a fixed R2 and allow R1 to tend to zero, and deduce that dR 1 R2 _ - log — + 0(1). 1 Rp(R) p R1

R1 2

Thus (2.43) becomes

1 R2 Jo

w2

pp<

PP(P)

so that I (02(P)dP < co. PP(P) Jo

Also Lemma 2.3 shows that f R° H(p)dp P3

Jo


H ( p) p 2 —> 0, as p —> 0 and as p

oc.

We now apply the proof of Lemma 2.3 with R1, R2 where R 1 --> 0, R2 00 and deduce that

1 R2 dp JR1

PP(P)

1 R2 I I (R1) log + p R1 2p2 R?

I I (R2) 2p2 Ri.

1 R2 - log — + o(1), as R1 -413, R2 -4 GO. p R1

60

The growth of finitely mean valent functions

We allow R1 to tend to zero and R2 to tend to oo through sequences of values, such that w(Ri ) -> 0, w(R 2 ) (X). Then Lemma 2.5 yields 1 1 R2 dp +

(1?2)

(1?1)

2

L

i

op(0)

1 f R, (0 2 (0)dn

PP(P)

272

r + o(1).

2 + f R2 0-)2(P)dP + o(1). 1 log 2p RI Ri

Using (2.55) and (2.33) with that

R2 =

P(P)

R1 =Ig((R I ))1, we deduce

1 (log R2 - log =— 2p

) - — + o(1) 2p

Thus ,6

1

f oc

2p +22 jo

w 2 (p)dp < 0 PP(P) - •

In particular fl 0, with strict inequality unless w(p) = 0 for almost all p. In the latter case almost all the y R are vertical segments so that 2 - log + irOl= 0, on all these y R and so identically. OP! If 4)(0 = log g() = u+ iv, we deduce that 41g) is purely real, so that (//() is a real constant q. Hence

log g() = q( + C. Now (2.56) shows that q = 2p and C = iA. This completes the proof of Theorem 2.9.

2.7.5 Return to the unit disc We can now write down our regularity theorems.

Theorem 2.10

Suppose that f(z) is mean p-valent in IzI <1. Then a = lim(1 - r)2P M(r, f)

(2.58)

r--+1

exists and 0 < a < co. Further if a > 0 there exists a (unique) radius of greatest growth arg z = 00 such that (1 - r)2Pjf(rei°°)1

a, as r -÷ 1.

(2.59)

Also if C is a positive number, and z = re° f'(z)

2p

f(z)

(ei°0 — z)

, as z -÷ ei0°, whilelarg(ze -i°91 < C(1 - r).

(2.60)

2.7 Behaviour near the radius of greatest growth

61

Theorem 2.11 Suppose that f(z)/zP is regular t and equal to 1 at z = 0, and that f(z) is mean p-valent in 1z1 < 1, cut along a radius. Then the conclusions of Theorem 2.10 still hold and a <1, unless f(z) = zP(1— ze —i N) -2P. Also we have

0< r<1

M(r,f) < Co rP(1— r)-2P,

(2.61)

where Co = exp {(pn 2 + 1)/2} With the hypotheses of Theorem 2.11 the proofs of Theorems 2.6 and 2.7 still go through. Thus either

(1 — r)2P M(r, f)

0, as r

1

in which case a = 0 and there is nothing to prove, or (2.21) holds. Thus we may suppose that (2.21) holds. In this case there is a radius of greatest growth arg z = 00 so that Theorems 2.6 and 2.7 hold. We now set

1

z

(2.62)

= — log 2 { (1 — ze—i00)2}

Then the domain D(00 ), consisting of the disc 1z1 < 1 cut along the radius arg z = 00 + 7E, is mapped onto the strip

=

S

where 101 < C(1 — r) for some constant C

Suppose that ze — w° = re and that r 1. Then

= log

(2.63)

1 711 <

r 1 /2

— rei(15 1'

= tan-1

{

1+r 1— r

tan !(/) 2}

so that if r is close to 1,

1 711 < 2 tan-1 C =

— 6, say.

Thus in this region we can apply (2.33) and obtain

1oglf(z)1

cf. Example 2.8.

= log 1,g(01 = 2/3 + 16 1 = 2p log 1— rei0

o(1)

62

The growth of finitely mean valent functions

i.e. efl as r -- 1, while larg z — NI 11 — ze1 — i00 2P '

C(1 — r). (2.64)

We write a = exp iq, and deduce (2.59). Next we choose g = p in Theorem 2.7 and choose C so large that C —P < la. Then by Theorem 2.7 we have, when z = rei° , C(1 — r) < 10 — 001 < 7E, and r is sufficiently close to 1, 1 la 1 . If (re`0 )1 < CP(1 — r) 2P

(1 —

while if 10 — 001 _-_ C(1 — r), (2.64) yields

(2.65)

If (re d) )1 < 1 + O(1) (1 —

Thus (2.65) holds for all 0 when r is close to 1, so that a + o(1) (1 — r) 2P •

M(r, f) <

On the other hand by (2.59) a + o(1)

M(r, f) >

(1 —

This proves (2.58). Finally if larg z — 001 C(1 — r), we have n — — 6 for some positive 6, and as z —+ e10°, —÷ +co. Further 1 171 2 4 1 -cl—z = E

1

+

-z

1 ,

ei 00 — z '

and by (2.34)

g(c) + — 2p g'(C) Hence dz

4 d dz 4

log f (z) = — — log g(ç) --

2p ei00 — z

This proves (2.60) and completes the proof of Theorem 2.10. Next if f (z) satisfies the hypotheses of Theorem 2.11 then g() satisfies (2.56), since as z —3 0 1 1 = — loglz1+ log I — z1 = I log 1z1 + o(1), 2 2 I

so that log I()1 = log If(z)1= p log Izl + o(1) = 2g + o(1).

2.7 Behaviour near the radius of greatest growth

63

Thus, by Theorem 2.9, /I = log a O. Equality holds if and only if

f (z) = g( ) = exp(2K + i.1.) = eii- {

z CA) P ' (1 - ze -4 )2

Also we must have ii, ---- p0o, since f (z)/ zP -> 1 as z -+ O. It remains to prove (2.61). We use the transformation (2.62), but without assuming that 00 is a radius of greatest growth. Then g ( ) = f [z ( )] is regular in the strip S given by (2.12) and is p-valent and non zero there. We suppose that g(0)1 = R 1 , Ig(o) 1 = R2, where R1 < R2, and proceed as in the proof of Theorem 2.2. Then, since g(()* 0, there are no closed level curves in S and so for R1 < R < R2 there is a level curve Igl = R meeting the segment

L 0< :

and going to the boundary of the domain So consisting of all points distant at most n/2 from L. This level curve has length at least n. The area a of So is 7-(0 + n 3 /4 and so Theorem 2.1 and Lemma 2.1 yield

2.7r 2

f R2

n 4 > n2 dR > 7r2 { 1 2 R2 ± _.= co o- R i - 2, } ' RI Rp(R) - p 2 -

since g is mean p-valent. Thus nn 2 + 1 .g(o) < 2i) ci ± r log g(0) 2 By a simple translation in S we obtain for -co < i < 2
pn 2 + 1 log 1.g(2)1 < log ,g(1)1 + 2p(2 — 1) ± We let

2

.

i tend to -co in this. Then since f (z)/ zP -÷ 1 as z -> 0, log Ig(i)l

-

2ki --3 0, as i -- -co.

Hence we obtain pn 2 ±

Ig(1)1

2132 +

2

log 1•

Writing z = rew° and using (2.62) we obtain

loglf(z)1 and this proves (2.61).

p log

r (1

pn 2 ± 1

— 02 ± 2

64

The growth of finitely mean valent functions

Examples

2.7

If a = 0 and p > in Theorem 2.10 or Theorem 2.11 prove, using the argument for Theorem 1.6, that / i (r, f) = o(1 —0 1-2", as r

2.8

2.9

1,

and so that an /n2P-1 --+ 0 as n cc. If f(z) is regular in 0 < 1z1 < 1, and f(z)/z P remains regular at z = 0, show that in any sector T : 0 1 < arg z < 02, where 0 1 < 0 < 02, the value of p(R, T) is independent of the branch of zi` chosen. Deduce that, if 02 = 01 2n,p(R, T) is also independent of 0 1 . We say that f (z) is mean p-valent in 1z1 < 1, if f (z) is mean p-valent in 0 1 < arg z <01 -I- 27r for some, and so for every real 01 . Verify that Theorems 2.6 to 2.11 and Example 2.7 remain valid. If 1(z) =

En" n .

n

,

0

is mean p-valent in z < 1, if a > 0, and arg z = 00 is the r.g.g. show that

f(re'°)

2.10

a(r) as r (1 — rei( 0-00 ) ) 2P '

1, for 10 — 001

C(1 — r),

where C is a fixed positive number and a(r) = (1 — r)21 f(rei°0 ) (use (2.60)). Using Cauchy's formula

1 4)(z)dz 1 , where r =1 — an — — 2ni 1z 1=r z n+ 1 applied to 0(z) = f(z)/zR, and comparing this with the corresponding formula for 00 F(k -I- 2p) zke-ikoo Oo(z) = a(r)(1 — ze-1°°) -2P = a(r)V F(2p) k=0

show that if p> 1, a > 0, and arg z = 00 is the r.g.g, then a (1 —

11 2P-1 e -1(n+11)°°

F(2p)

,

as n

co.

(Use Example 2.9 and Theorem 2.7. This theorem eliminates the need for the analogue of Lemma 1.4.)

2.7 Behaviour near the radius of greatest growth

2.11

If f (z) is mean p-valent in IzI < 1, where p> show that lim Ian I n--■ oc

2.12

2.13

65

=

n2P-1

a F(2p)'

where a is given by (2.58). (Hayman [1955], Eke [1967b]). If f (z) EZ.^,- , show that lanl n, n > no(f). (Distinguish the cases a = 1, a < 1.) See Hayman [1955] for the above results for circumferentially mean p-valent functions and Eke [1967b] for the general case. We shall show in Chapter 5 how to extend the conclusions to the case p> . They fail when p < Prove, with the hypotheses of Theorem 2.8 that, if q> 1,

Grzi

d q

as --+ -Fcx) uniformly in 1 ri I (2.60) can be extended to

logg(C) --t 0

<

3 -5. Hence prove that,

if q > 1,

f (q ) (z) 2p(2p ± 1)...(2p ± q — 1) as z --+ ei0°, (eioo — z)q f (z) '--'

while I arg(ze —'°°)I < C(1 — 1z1). Deduce that for q > 0

If (q) ( 7)1 , 2p(2p + 1)...(2p + q— 1)c as z —> 1 2p+q lee, —

2.14

z

while I arg ze -0°1 < C(1— If q > 1 prove that, with the hypotheses of Theorem 2.10, lim(1 — r)2P+q M(r, f(q ) ) = 2p(2p + 1)...(2p ± q — 1)a. r--, 1

(Consider separately the cases a = 0, a > 0.)

3 Means and coefficients

3.0 Introduction function

In the last chapter we investigated the growth of a 00

f(z) =

E an z n , 0

mean p-valent in lz I < 1. We showed in Theorem 2.3 that the maximum modulus M(r, f) satisfies M(r, f) < A(p),u p (1 — r) 2" (0 < r < 1).

(3.1)

In this chapter we estimate the order of magnitude of the coefficients an and show that, if f(z) is mean p-valent in 1z1 < 1 and C, fi are positive constants such that C > 0 and fi > , then the inequality M(r,f) < C(1 — r) — fl (0 < r < 1),

(3.2)

lan I < A(P, fi)C(1 + n)" (n = 0, I, 2, ...).

(3.3)

implies

It will follow at once that, if f(z) is mean p-valent in 1z1 < 1, so that (3.1) holds, then

Ian ! < A(p)pp n2P-1 (n > 1),

(3.4)

provided that p> 1. The functions 00

f(z) = ( 1 — Z) -213 =

E bn,Pzn ,

(3.5)

0

which are mean p-valent in 1z1 < 1 and for which

f(n

bn ' P

+ 2p)

n213-1 — oo), r(2P)r(n + 1 ) -%' F(2p) (n >

66

(3.6)

3.1 The Hardy-Stein-Spencer identities

67

show that the order of magnitude of the bounds in (3.4) is correct. The method used by Littlewood [1925] for proving Theorem 1.6 is sufficient to show that (3.2) implies (3.3) if f(z) is univalent and fl > 1. The idea for extending this to the case fl > I occurs first in a joint paper of Littlewood and Paley [1932]. The argument was extended to p-valent functions by Biernacki [1936] and to mean p-valent functions by Spencer [1941a]. We shall show further, by means of an example of Spencer [1940b], that (3.2) does not imply (3.3) for a general mean p-valent function f (z) if fl < and that (3.4) is false in general if p < However, we shall prove in Section 3.5 a theorem of Baernstein [1986], which yields an extension for univalent functions. In the final sections of the chapter we shall give some further applications of our main results by estimating the coefficients of certain classes of mean p-valent functions for which more restrictive bounds than (3.1) can be obtained.

3.1 The Hardy-Stein--Spencer identities We suppose now that f(z) is regular in lz I < 1 and further that A > 0 and 0 < r < 1. Let n(r, w) be the number of roots of the equation f (z) = w in Izi < r and write 1

f2n

Pfr, R) = -ii 0 n(r, R e itP)dip. Thus p(R) = p(r, R) is defined as in (2.4) when A is the domain Izi < r. We also write 1,1(r, f)= l t ?)1/140. fo 2nIffr

We then have the following remarkable triple identity:t Theorem 3.1

With the above notation f 2n 22 f r d r-d--1:1,1(r) = - -7- r- - 0 pdp 0 If(pe i° )1 A-2 1f(pe w )1 2d0

= 22 f p(r, R)R i.-1 dR. o Suppose first that f(z) has no zero on lz = r and write f (rei° ) = Re14) . 1. The first equality is due to Hardy [1915] and Stein [1933] and the second to Spencer

[1940a].

68

Means and coefficients

Then near a fixed point of 1z1 = r, we have 1

11b

log f =

R

log R i(1) ' by the Cauchy-Riemann equations. Thus d

f+n

r— If(re l° )1' de =f R dr j, 1z1=r

r dO =f ledeo. or I:1=r

We make a transformation (not conformal) by writing, when w =

(3.7) ,

P = R, tp = (1), W = Pe'.

Then the right-hand side of (3.7) reduces to p2d4:1).

Now 1P 2d0 is a sectorial element of area in the W plane, and so the right-hand side of (3.7) represents 2.1 times the area in the W plane corresponding to 1z1 < r, multiple points being counted multiply. This is quite evident if f(z) is univalent in 1z1 < r, so that area is the interior of the simple closed Jordan curve which is the image of 1z1 = r. In the general case we can prove our result by splitting the disc into a finite number of regions in each of which f(z) is univalent and noting that

f p2dq)

taken over the boundary of the region is additive and so is the area in the W plane. Now the area in the W plane is equal to

r°'

f

JO

2n

. ,

v(Pe"' )P dP dtp,

0

where v(Pew) is the number of points in 1z1 < r, corresponding to W = Pe'. Thus v(Peiv') = n(r, P2Re iv'), and we obtain

2À] R 'I d(1:1 ,1z1=r

)1,

= /1,

f d(P 2 ) o

f

1

2n v(Peindtp

0

2it d(R) f n(r, Reidy)

R=0

x

0

= 22t/12 i p(r, R)R Â-i dR, .n

3.2 Estimates of the means I)(r)

69

and on combining this with (3.7) we see that the first term in the identity of Theorem 3.1 is equal to the third. Again If(pe w )1 2 pdpd0 is the area of the image of a small element of area, p < Izi

for

2n

oo

2n

le -2 n(r, Re iv)dip. pdp f _flpe i° )1 2 1f(pe i° ) -2d0 = f RdR 0 o 0 In fact both sides represent the total mass in the w plane of a mass density 1w1 11-2 is spread over the image of lz I < r by w = f (z). The right-hand side becomes 00 p(r, R)le -1 dR, o and so we have the identity of the second and third terms in Theorem 3.1, and that theorem is proved on the assumption that f(z) has no zeros on Izi = r. The result follows in the general case from considerations of continuity. In fact the continuous function /;,(r) has a continuous derivative

2n f

2nr 22 S(r) r = 2nr 0 PdP 0

If( pei0)12-2Lo pei0)12do ,

except possibly at certain isolated values of r. At these latter values r0, 11 (r) clearly remains continuous and so we see that the equation d = S,1(r) dri'jr) r— continues to hold, by using the strong form of the mean-value theorem

/),(r i ) — / ;,(ro ) = (log r 1 — log r0 )S2(p), and making r 1 tend to ro from below or above.

3.2 Estimates of the means h(r)t f (z)

Suppose again that oc

=

E

anzn

0

is regular in izi < 1. Then we have for 0 < r < 1 'Il an ! =

1 f f'(z)dz 2ni j1 , 1=,. zn

<

I i (r, f') rn-1 •

t The results from here to Section 3.5 inclusive are mainly due to Spencer [19414

70

Means and coefficients

We choose r = (n — 1)/n for n > 1, so that

=

< e,

and deduce

Ia
(n

J.')

(3.8)

1).

It is thus important to be able to estimate I 1 (r, f'). For this purpose we use Theorem 3.1. It follows from this theorem that 12 (r, f) is an increasing convex function of log r, when 2 > 0. We have further

Theorem 3.2 Suppose that f(z) is mean p-valent in izi < 1 and set A = max(2, 22 ), when 2> 0. Then d

SAr, f) = r— lAr, f) pAM(r, f) ;- (0 < r < 1), dr

(3.9)

and f)

Mfro, f) À + pA

r

M(t, f) ldt

(0
(3.10)

fro

We have by Theorem 3.1 oo

M(r,f)

S,i(r) = 22 f p(r,R)R ;t-l dR --

2.2

f

p(r,R)R 1-1 dR.

Also since f(z) is mean p-valent in 1z1 < 1 and so à fortiori in izi < r, we have, using the notation of §2.3,

W(R) = f p(r, p)d(p 2 ) pR 2 (0 < R < co). Hence, writing M = M(r, f), we obtain

1 fm p(r,R)1-d io

le-2dW(R)

im,i-2w(m) 2 — 2 [41 2 2

—3

W(R)dR.

There are now two cases. If 2> 2, we deduce, since W(R) > 0,

p(r, R)R 11-1 dR

M/1-2 W(M)

1-;-M

P M(r, f) A .

2

3.3 Estimates for the coefficients If 0 <

2 we deduce

—

1 mA-2pm2 ± 2 2 2

p(r, R)12).-1dR

m A.

71

1

/1 )

RpR 2 dR P

i`v

/1 1" '" This gives (3.9). Also

IA(r,f) =

dt

f) T M(ro, f + pA f M(t, f

+f

and this yields (3.10).

3.3 Estimates for the coefficients

Theorem 3.3 and that

We now prove our basic result. 00

Suppose that f(z) =

M(r, f)

C(1 — r) — fl

anz" is mean p-valent in 1z 1 <

(0 < r < 1),

(3.11)

where C > 0 and )6 > 1. Then we have

(3.12)

A i (p, fl)Cn" , (n > 1) where A i (p,#) depends on p, )3 only.

We shall need the following preliminary result: Suppose that f(z) is mean p-valent in 1z1 < 1 and that r < 1, 0 < A. < 2. Then there exists p such that 2r — 1 < p < r

Lemma 3.1 1

and 1

2n

r2n

frl If(Pel° )1 2 1f(Pele ) 2-2 01 < 4PM(r .1(1 — r) '

(3.13)

We deduce from Theorems 3.1 and 3.2 that 27E f r pdp f It(pel° )1 2 1f (pe )1 2-2 de S(r) < M(r f)a - 2 A2 A Ljr 2 r — 1 0

Hence we can choose p so that 2r — 1 < p
If (Peie )1 2 1f (Pei9 )1 .-2d8 ;

2n

This proves Lemma 3.1.

< 4pM(r, 14/-2 — (2r — 1) 2] — A(1 — r) •

72

Means and coefficients

We now suppose that r > and that (3.11) holds with C = 1 and fi > We set). = (2/1— 1)/(2/3), so that /1(2 = # + > 1 and choose p so that (3.13) holds. Then f

1

(pei°)Ide

27r jo

(p,

\

2

(27 'n 127r ift(Pel9 )1 2 1f(Pelt9 )1 /1-2de) 1

X ( 217r fo 2n I fipe o )1 2—).do ) 2 by Schwarz's inequality. We now take ro = r r p, we deduce

/2_1(p,f)

in (3.10) and, noting that

/2—Ar,f) < (1 _ ro)4(2—,1)

p(2 — A) r0 P(2fl + 1) fr

=

f

—

13(2- dt

dt

J0 (1-

# < A2(1_ro,

(3.14)

,

where

A2 = 2/3/1

2P(2fi + 1)

#(2fl — 1) •

We write r 1 = 2r — 1, so that r 1 < p < r and deduce from (3.13), (3.14) and the above that

Ii(ri,f')

1 1( 9f ) =

4.,34A2} (1 - 0-1(10A-0-1) {1

A3 (1 —

= 2$243(1 —

where

A3 =

4p

)1

=

(

2

2fl — 1

Here r1 may be any number such that 0 < r i <1. Now (3.8) yields //lard < 2fieA 3 nfi, n > 1.

73

Examples

This proves Theorem 3.3 if C=1, with

Ai(p, i6) = 2fl e { 813)6 ( 2p - 1

2fl+ 1, ± 2p(2fi + 1) 1 1 fi(2,8 - 1) ) f '

If 0< C
Examples 3.1

3.2

3.3.1

Show that if f(z) is mean p-valent in 1z1 < 1 and satisfies (3.11) and 0 r < 1, then li.(r,f) < A(p, fl, 2)01 - 0 142 if P > 1; Or, f) < A(p,2,)(1 + log TIT.) if 131 = 1; and h(r,f) < A(p, 13, :1) if P < 1. By considering f (z) = (1 - z) -2", show that orders of magnitude in Example 3.1 cannot be improved.

However we have

Theorem 3.3 fails if fi <

Theorem 3.4

Suppose that f(z) =

Eo anzn is mean p-valent in 1z1 < 1 OD

and satisfies (3.11). Then if /3 = -, we have

lan l < A i (p)C CI log(n + 1),

n > 1,

while if 13 < -, lard < Ai(P,13)C n-1 , n> 1 and

i

a, = o(n 2

),

as n -* cc.

The inequalities (3.16) and (3.17) are due to Pommerenke [1961/2]. We shall shown by an example in the next section that they are sharp even when f (z) is bounded, i.e. 16 = 0. The case fi = 1 remains open. However for univalent functions we shall prove, following Baernstein [1986] that the implication from (3.11) to (3.12) remains true for 16 > .491. To prove Theorem 3.4 we need Lemma 12

Suppose that f(z) is mean p-valent in 1z1 < 1. Then

1

fo

pdp

27r

fo

If (Pel° )1 2 pdpd0 < pit 5 x 2' - 4 1 ± If(Pei° )1'

2'1 - 4 ' /1 > 2

(3.18)

74

Means and coefficients

and if A = 2 and ( 3.11) holds with C =- 1,fl = 1,

r f 27 I f (pew ) 1 2 o pdp 0 1+ If(peo)1 2 pdpc/61 < pn (5 + 3 log

f

1 ), 0 < r <1. 1- r

(3.19)

En be

the subset of 1z1 < 1, where If(z) I __ 1 if n = 0, and where 2n— ' < If (z)1 __ 2, if n 1. Then, since f(z) is mean p-valent Let

In = f If' (pel6 )1 2 pdpd0 pn4n = 4pn 22(n-1) . En

Thus if

fn=

If (pes° )1 2pdpd0 In 1 + If(Peie W ' '

we have

Jo

_ 41)71.2(2—an—I) , n > 1. < 2,1(1—n) 1 n < lo < pit, and in _

Thus, if A > 2,

00

00

E in < ion 1+ 4 E 2(2-)(n-1) n=0

pn 1 +

n=1

4

5 x 2'1 — 4 22 — 4

1 — 2 2- 11 )

This proves (3.18). If 2 = 2, we have .10 < pm, I n < 4pg, n > 1. Also since M(r, f) 15_ (1 — WI , En does not meet 1z1 < r if 2' > (1 — r) - 1, i.e. if

n> 1 --1- [

1 1 log ] = N(r), say. 2log 2 1—r

Hence

j 27 PaP of 1 +Ippeio)12 pm {1+

r

f

o

If'(pe`')12pdpc/0 N(r)

4

1

= mr[1+4N(r)],

and this yields (3.19). Suppose now first that fl = I-. We suppose that r > ri = 2r —1. Then using Theorem 3.2 and (3.19) we obtain

3 -4:'

and write

27-c

21ir fr: pdp fo It(pe` () )40} 2 {

<

17r f f If (Pe l(9 )1 2 pdpd0) (1 yi f f (1 + If(Pew)12)Pdpa) (2 i i 1 +If(pe i0)12 (5 + 3 log

1 1— r

) fr (1 +12(P,f))dp ri

3.3 Estimates of the coefficients

75

1-3-i (5 + 3log 1 1 r) ( r ri ) (2 + 4p log 1 1 r ) <._ 5p(1 + p)(1 —r i ) (1 +log 1 1 1.1 ) 2

Thus

-1 (r2 —r?)11(ri,f) < {5p(1 +p)} 1 (1 —r i ) (1 +log . 1 1 — ri ) • 2 Setting r1 = % I if n > 2, r1 = 1 if n = 1, and using (3.8), we deduce

(3.15). Next suppose that )8 < 1. We choose A = (2fl + 1) 1 (2/3). Then 1

2n

r

-s-' r 1 pdp fo Ir(pez ° WO i < f 1 f I1 If 12 pdpd0 — 217r f (1 + I frl )pdpc/01 2 1 27r f f 1+ IfV) f {A(P,i3 )( 1 — 01 / . In fact if p __ 1 f 2n

2n 0

we have by (3.10) and (3.11) with C = 1, ro = 1

.

P

I f(pe` e )l'Icle :5_ 2fiii. +2132 f i (1 — t) udt 7

< 20. ± 2 P/1

.2 fi + 1 ± 2p(2fi + 1)

1 - 16A

/3(1 — 2/3) •

Using also (3.18) we find there exists p, such that r1 = (2r — 1) 5_ p < r, and

ii(r i ,f) <11 (p,f) < A(p, /3)(1 —p) - 1 :5_ 21,4(p, fl)(1 — r i ).

(3.20)

Choosing r1 = (n —1)1n, and applying (3.8) we obtain (3.16). Next we note that by (3.18) 1

fr

i

pdp

fo

27

1r(Pez9)1 dû —> 0 as ri 1 + If (Pei6)12

This enables us to sharpen (3.20) to

i i (ri ,f) = o(1 — Again choosing ri = (n — 1)/n and applying (3.8) we obtain (3.17). This completes the proof of Theorem 3.4.

76

Means and coefficients

3.4 A counter-example

The results (3.16), (3.17) of Theorem 3.4 are best possible. Even if f (z) is mean p-valent with p as small as we please and continuous in lz < 1 nothing stronger than

= o(n)

(3.21)

is true in general. To see this let nk be a rapidly increasing sequence of integers and put f(z) = 1 + Ean zn, n=1

where an = E2—k n- 1, if n= nk (k =1, 2, ...), a,1 = 0 otherwise. If is any pre-assigned sequence of positive numbers, tending to zero as n oo, however slowly, we suppose nk oo so rapidly with k, that (k = 1, 2, ...). Thus for the infinite sequence of values of n given by n = nk we have An

(3.22)

an > --T.

On the other hand, we have for lz < 1, 00

11(z)-11

1

= E,

n=1

k=1

so that the series for f (z) converges uniformly and so f(z) is continuous in lz < 1. Also the area, with due count of multiplicity, of the image of lz < 1 by f(z) is

r 2ir rdr

Jo

co

co

If(re )1 dO = rt E ni a n I 2 '0 2

n=1

6'2 2 -2k = ne k= 1

2

3

(see §1.3). If 71W(R) denotes the amount of this area which lies over I w < R, then W(R) = 0 if R < 1— e W(R) < ire2 /3 otherwise. If we choose e < 1, it follows that ,

W (R) IT R2

7te 2 3702

4E2 3

(0 < R < co),

and we can make the right-hand side as small as we please by choosing 6' small enough. Thus f(z), which is continuous in f(z) < 1, can be made mean p-valent there with p as small as we please. Clearly, for any function f (z) bounded and mean p-valent in lz < 1, nlan 1 2 converges and so

Examples

77

(3.21) holds. Nevertheless (3.22) shows that nothing stronger than this need be true.

Examples

33

If f(

cc

an zn is mean p-valent in 1z1 < 1 and M(r, f) = o(1 —

, as r

1

where fl > 0, prove that 1,(r, f) = o(1 — r) 1-fi'l as r --+ 1, provided that f3> 1. Deduce that, if /3>

I i(r, f') = o(1 — r)-fi as n

co,

and

= o(n 3.4

1 ) as n --+ co.

(Use Theorem 3.2 and Lemma 3.1.) Suppose that g(R) is positive increasing for R > 0 and that

" RdR Jo g(R)


Prove that °C)

n=0

22n

g(2n)


Deduce that if f(z) is mean p-valent in jz1 < 1, then 1

fo

3.5

, f 2 I (pe19 )1 2d0 < co o elf (Pei° )I)

PaP

Suppose G(R) to be twice continuously differentiable for R 0, G' (0) = 0, and that f(z) is regular for 1z1 < 1. If u = G(If (z)I), prove that

a2u

a2u V2 /4 = a x 2 -— ay 2 =

(z)1 2g(If (z)I), 1z1 < r

where g(R) = Gu(R) + R -1 G'(R), if R > 0, and g(0) = 2G"(0). If IG(r, f)

= 27cf0 27r ufrew )de,

78

Means and coefficients

deduce from Green's Theorem that if d SG(r, f) = G(r, f) then

sdr, f =

pr 2n Jo PdP 0f If f (Pei° )1 2

(Pe1° )140

Rg(R)p(r, R)dR. By writing G(R) = (e 2 ± R2 ,) 14 , and letting e tend to zero, deduce another proof of Theorem 3.1 (Flett [1954]).

3.5 Coefficients of general mean p-valent functions and 3.4 give immediately

Theorems 2.5, 3.3

an z n is mean p-valent in lz1 < 1.

Suppose that f(z) — Then we have for 1 < n < co

Theorem 3.5

< A(p),up n2P- ',

(3.23)


Ian { < A(P)laoln -A ,

(3.24) (3.25)

and a, = o(n I), as n

oo (0 < p < i),

where A(p) depends only on p and pp =

(3.26)

a.

For by Theorem 2.3 we may write C = A(p)p p , = 2p in Theorems 3.3 and 3.4. The order of magnitude in (3.23), which is due to Biernacki [1936] for p-valent functions, and to Spencer [1941a] for mean p-valent functions, is best possible. The example of the last section shows that the right-hand sides in (3.25), (3.26) cannot at any rate be replaced by en n- i, where en is a fixed sequence which tend to zero as n cc.

3.5.1 The case of univalent functions The example of Section 3.4 shows that (3.12) fails if 16 < 12- for bounded mean p-valent functions. However, Clunie and Pommerenke [1967] have shown that for bounded univalent functions the result can be strengthened.

3.5 Coefficients of general mean p-valent functions

79

We first give the Clunie-Pommerenke Theorem in a sharpened form recently obtained by Pommerenke [1985a] and then obtain a corresponding extension of Theorem 3.3 due to Baernstein [1986]. Theorem 3.6 11>

Suppose that f(z) c

r10(/1 ) = -- + A + (i — A + 4.12 )

S. Then if 0 < .3. < 1 and

we have

1,1(r, f') < 24(q, 2)(1 — r) -'

0 ._ r < 1.

(3.27)

— /1), we have

Hence if )6 > Po = {lo(A) + (1 —

I i (r,LI ) < AW)(1 — 0 -13 , f

< r < 1. 21 —

(3.28)

Choosing .1= .07 we obtain (3.28) with fib < fl < .4905525 <

Corollary

If f(z) is univalent and If (z)1 < 1 for 1z1 < 1, we have

I i (r,f')< 2A(fi)(1 — r)-13 , 1 __r < 1,

(3.29)

laid < 2eA(Mn I3-1 , n > 2.

(3.30)

We write 1 f 2it 1 (r) = 1,4r, t) =

0 Inre`e )1/140.

(3.31)

We apply Theorem 3.1 to f' and deduce that

fr

f2n rr(r) = -sr- 0 pdp 0 It(peie )1 11 flpe /3.2

2 )

fi (p e1°)

de,

or _ 2,1.72T fo lt If velo )1;1, d r ir- (rIf (r)) = r f"(Pell9. ) ft(pee)

de.

Since 1'0> 0, we deduce, writing z = re'', 2 yrt

/12

f2n

r i (r)__ — i If(z)1 4 2n 0

f "(z) 2r 2 ) (

2r2 2 de. (3.32) f'(z) 2 1 r ) ± 1— — r2

We define z

f'"(z) f'(z)

2r2

2r2

1— r2 =a+ib, 1 — r2 =

c.

80

Means and coefficients

Then using (1.6) we have I a+ib+c12 = a2 +b2 +c2 +2ac

16r 2 + 4r4

4r2

I f"(z) (1 — r 2 )2 + 1 r2 "1 z (z)

16r 2 — 4r 4

4r2

f"(z))

f(z) ) •

(1 — r 2 )2 + (1 — r2 )"

To deal with the last term we differentiate (3.31) under the integral sign. We write

0(z) = log f' (z) = u + iv, so that, with z = rei° ,

eu f"( z) = 9{z • 7/7- = "tz0z) .f (z) Thus

r r(r) =

f2ir

27(

d 1

r—— f dr 27r 0

'd0 ') = —

e

r=e1u(reie)de

Or

27r 0

(3.33)

ff" ((zz)) ) dû.

2 / 2n If (z)1 )-

277

Substituting in (3.32) we obtain finally

4r 3 ;, 16r2 4r4 r2 (r) < it 1(r) + (1 — r 1(r). — (1 — r 2 )22 2) Given a positive g we can therefore find ro = ro(e) such that ro < 1 and I"(r) <

2 .1 + E ( 1 — r)

+

3;t2 + E (1 — r) 2

1(4 ro < r < 1.

Let )6' = fl(e) be the positive solution of fl(fl + 1) = (2). + E)fl + 3/1 2 +E. Then the comparison function v(r) = C(1 — r)— fl satisfies the equation v"(r) =

3 ,12

2 /1 + 1 —

:

v'(r) +

e v(r ).

(1 — r)2

It follows for instance from (1.3), (1.6) and (3.33) that ' (ro)

Co,

1 1 (ro)

(3.34)

fiCo,

where Co depends on )„ f3, ro and e only. Choosing C = Co and writing h(r) = I (r) — v(r) = I(r)— C0 (1 —

ro r <1

3.5 Coefficients of general mean p-valent functions

81

we have h(ro ) < 0, h'(ro) 0, and h"(r) 0, ro r < 1. We deduce that 10r) 0, ro
I(r) < C0(1 — r)— fl, ro
(3.35)

Using (3.32) we see that (3.35) remains valid for 0 r < 1 possibly with a larger C0. Here ie = fi(E) and C0 depends on /3. Also ME) fl(0) = rio(A) as E —÷ 0. Thus if j > rio().) we can choose E so small that fl(e) < , and now (3.35) implies (3.27). We note that OA) < A for small A. In fact 3A2 as A

1/0(/1 )

O.

Pommerenke [1985b] has also shown that the best rio(A) satisfies /3(A)> CA2 for small positive or negative A, where C is a constant. We next prove (3.28). We suppose that 0 < A < 1, 13 170. Further we define 6, p, q by

6=

2 P = 2—A' 1+

2

,

11

We also write a = a(0) = It(re`° )1 6 , b = b(0) =

rif(re`6) )1 1-6 If (rei6 )1 •

We now apply Holder's inequality (Titchmarsh, 1939, p. 382) 1 f 2n -27 r0 a(0) b(0) cle

2n (1 (277c 0 a(0)Pd0) -T7r- f0 b(0)qd0) . 1 f 2n

P

We note that g(z) = f(z) 2 (z/f(z))q is regular in Izi < 1 so that, by Theorem 3.1, 1 1 (r, g) increases. Also since f E S we deduce from Theorem 1.4 that

I f(z)1

>

Thus, if r >

.1 1/ 2 b(0)q dO =--- Ii(r,g) (1 _ r2) I 1(1), g)pd p 27t Jo jr

11.1 7r(1 — r2 ) 1 f — r2 )

f2R

g(pel ° )Ipdpde 0

izigif'(z)1 2 1dzi 2 If(z)V1

82

Means and coefficients 1

A1

(z)1 2 1dzI 2

+ 5q

Tc(1 — r 2 ) J J17

, 1 , 1

1 + If(z) lq1

-

r

by Lemma 3.2, where A i , A2, . . denote constants depending on A and /3. Again 2n 2n

a(0)Pd0 = f Ifi(ren9 )rd0
fo by (3.27). Thus fo 2n

a(0) b(0) de

=

12' nreie) r dO
f (reie) A 3 (1 — 0 -13 .

This yields (3.28). Choosing 2 = .07 we have fio
3.5.2 The results of Carleson and Jones At this stage we ought to mention a recent paper of Carleson and Jones [1992]. Let S i be the class of functions f( z ) .

Ea

7n

considered in the corollary to Theorem 3.6, i.e. such that f (z) is univalent and If(z)1 < 1 in 1z1 < 1. Let S2 be the class of functions CO

g(z) =

z

considered in § 1.1, which are univalent in z1 < 1. Write

A n = sup fes- i

Bn = sup

The authors prove that there is a positive constant Co such that A,,

n —

Co sup /1 fEai

( n

1

, f) , B n Co sup gEa- 2

(n — 1

, ,g

(The inequality in the opposite direction follows from (3.8)). They further prove the existence and equality of the limits — log lAn j — logq3,1 = lim n---,co log n n

y = lim

) •

3.5 Coefficients of general mean p-valent functions

83

and characterize y in terms of a conformal dimension of the boundary on of the image S/ of Iz < 1 by f(z) or g(z). Further they show the existence of f, g in Si, S2 respectively such that -,--- log Ian I n,oc log n

log I bnl = -,— urn, log 11 =

.76, sharpening the previous Finally they obtain the upper bound y bound y < .83 of Pommerenke [1975, p. 133]. Carleson and Jones also conjecture that y = .75. Their method however only gives y > .50245 compared with Pommerenke's y > .59944 of Theorem 3.6, corollary. The methods of Carleson and Jones are based on fractals and iterations and are unfortunately outside the scope of this book. Instead we shall in the next section develop a result of Baernstein [1986].

3.5.3 Baernstein's extension of Theorem 3.3

We proved in Section 3.5.1 < 1 and that (3.27) for the class S with constants ri, /1., where 0 < < P., yields (3.29), i.e. i(r,

— r) -1 +K

<

for functions f(z) univalent and satisfying If(z) I < 1 in Iz I < 1. Here

i.e. /I—

2,1

(3.36) 4— From this Baernstein [1986] has deduced a corresponding extension of Theorem 3.3. We proceed to prove Baernstein's result. K=

Theorem 3.7 Suppose that (3.27) holds for all f(z) in S with constants .1 such that 0 < < 1 and 0 < < P.. Then if 09 f (z) =

Ean zn 0

is univalent in 1z1 < 1 and satisfies M(r,f)

—r) -fl,

0 < r < 1,

(3.37)

where fi > — lc and ic is defined by (3.36), and in particular if fi > .4905525, we have A 1 C(1-0 -13 ,

0 < r < 1,

(3.38)

84

Means and coefficients

and hence

lanl < A2Cre6-1 .

(3.39)

From now on A i , A2 = eA. I , A3, . . . will denote constants depending on q , )„ fl only. In particular the conclusion holds with fl > .4905525 and absolute constants A1, A2. We note that if 13 > 1, the conclusion of Theorem 3.7 follows from Theorem 3.3. Thus we assume from now on that /3 < 1. The proof of Theorem 3.7 is rather complicated and makes essential use of Theorem 2.4 as well as a localized version of Theorem 3.5. We proceed by a number of steps.

3.5.4 Some auxiliary results We assume now that f (z) is univalent and f (z) * 0 in 1z1 < 1. Let E be a set of points z = re's lying on 1z1 = r, and suppose that M = suplf (z)1. zEE

We proceed to estimate

1(E) = I i(r, f , E) = f 1 flref ° )40. E

Our first result

Lemma 3.3

We have for 1 < r < 1

LI

f (reio )1 2 de < A3m 2 / (1 _ I. ).

We note that, since f (z) * 0 for 1z1 < 1,

0(z) =

f (z) — f (0) E s

(3.40)

and 0(z) * — f( 0)/ f '(0). Using Theorem 1.2, we deduce 1 f (0)I / If'(0) I 1. We apply this conclusion to f {zo + (1— r)z} , where zo = re° and deduce that 41 f (zo)I

1— r ' and so

—Or0 log lf (rei ° )1

<

4 1— r

(3.41)

85

3.5 Coefficients of general mean p-valent functions

Integrating this inequality from rei° to pei°, we deduce that if (1 — r) __ (1— p) 2(1 — r), we have log If (p? )1

log If (re1° )i + 4 log

1—p 1— r

log if (rei° )1 + 4 log 2,

so that f (Pe)1

161f(re ie )1 ._ 16M.

Let A be the set defined by

(1— p) _-__ 2(1 — r)}.

A : {z1 z = pe` ° , rei° E E and 1 — r

Since f (z) is univalent and If(z)i < 16M in A, the area of the image of A by f (z) is at most n(16M) 2 = 28 7rM2 , i.e.

.1.

r

2r-1PdP fE

(3.42)

lf(pe` ° )1 2 d0 -_ 28 7-cM2 .

Next we recall the inequality (1.6) which yields

4 2p + 4 .0 a < )1 lf'(Pel 1 — p2 — 1 — p . 0p Integrating this we obtain log if(re`° )1 __. log If(pe i° )1 +4 log 2, 2r — 1 : f) -_ r. i.e.

Inrel(4 )1

1 6If (Pe 18 )1, 2r — 1 < p < r.

Substituting this into (3.42) we obtain

1 [r2 — (2r — 1) 2 ] i If' (re`e )1 2 de ,2 E <

r

pdp f Inreit4 )1 2 d0 12r-1

E r

28 f pdp f inpe i° )1 2 d0

2r-1 G 7r216m2.

E

Thus

f

71 218 m2 7r2171142 r) _< 1 _ r , since r Inrei° )1 2 d0 < (3r _ 1)(1 _

This proves Lemma 3.3. We deduce

1 i.

86

Means and coefficients

Lemma 3.4

fE

With the hypotheses of Lemma 3.3 we have

It(reie)40

A4f(c ) li./(2-i.)A4(2-2),)/(2-i.)(1 —

We choose a positive constant B and divide E into the subsets E 1 : Ir(rete )1._ B, and E2 : inre 1e )1_ B. We recall (3.40) and that If(0)1 yields

41f(0)1 and apply (3.27) to 4)(z). This

2n

— 0 -11

27r1.11 0WA(r1,

(reVde

A5lf(0)1(1 — r)". Thus

Li

f (re i6 )610 < B" f Inre ie WdO A5B"If(0)1 11 (1 — rr,

while Lemma 3.3 yields

IE2

1 —mQ f Ir(re` ° )1 2 d0 < A3M2 - B(1 — r) . E2

(rele )Ide

Choosing B so that M2

B"If ( 0)1 2 ( 1

= B(1 — r )

{f 0*(1

(")/(2-A) A/12 ( Az-A) -1111 0

r

i.e. ={

M2 1f(0)1 )-(1 — 0 1— '7 }

9

we obtain Lemma 3.4 with A4 = A3 + A5. We need a localised version of Lemma 3.4. For this purpose we now assume that E lies in an arc T : 0 1 < 0 < 02 of 1z1 = r. We write 1/1 = 0 2 — 0 1 , assume that

1(1 — r) <

< e-9,

(3.43)

e8 111, z(I)= pei4) .

(3.44)

and define

1 = —2 ( 0 1 + 02), p =1

—

Thus

1 > (1 — p) = e8 1.11> e7 (1 — r).

(3.45)

3.5 Coefficients of general mean p-valent functions

Lemma 3.5

L

87

With the above hypotheses we have 51 - 1C

Ifl(reie)Ide

._

A6 ( 1 - P

)

'

We suppose without loss of generality that 4) = 0, so that / is symmetrical about the real axis, since this may be achieved by a rotation. Thus z(/) = p. We now consider the bilinear map p 1 — pz Z -

C=

(3.46)

followed by the radial projection Z=R

C IF'

where R =

r—p 1— pr .

(3.47)

Writing r = 1 — E, p = 1 — ô, we have e < e-7 6 < -1- e-7 by (3.45). Thus R=

6—e 1— e 7 > > .998. (5 + E(1 — (5) 1 + e-7

Let El be the image of E in the ( plane and let e be the radial projection are related by (3.46). of E on 1z1 = R. We write f(z) = g((), where z, Then, since r> 1 by (3.45), we have

1

Inz)Ide -- 2 f Iff(z)IldzI = 2 E

f 1001141-

(3.48)

I

We note that for lz = r, we have z—p 1— pz

2

(1 — r 2 )(1 — p 2 ) 11 — pz1 2 =

Thus E l lies outside ICI = R. If

(1 — r2 )(1 — p 2 ) . (1 — pr) 2

(3.49)

= te' then, as z = rei° varies on /, we

have

4

1 dt P — + idip = dz { + t z —p 1 — pz J — 1 + p = izdO { z — p 1 — pz 1 1 1. = idO { P + z — p 1 — pz f

(3.50)

We take imaginary parts in (3.50) and write z = x+iy,1/1 = 2y. By (3.43), y < le-9 on /, and r>1— 4.y. Thus

x = rcost9 > (1 —4y)cosy > (1 —4y)(1 — ly)> 1— Sy > p

Means

88

and coefficients

since p =1— 2e8 y by (3.45). Thus — ' ox de. 1 1— PzI 2

dip = de I P(x — P) + 1— Px 1 > 1 1 lz — PI 2

1 1— PzI 2

We recall that p = 1— 6, r =1 —

E

so that y = e -8 6 by (3.44). Thus

11 —pz11—pd-plz-111—pd-(1—r)-1-101

ô1-+y < 1.016 (3.51)

and 1 — px _._ 1 — p = 6. Hence

dip

(5 1 dO > (1.012 > 1.036 .

(3.52)

On the other hand lz — PI

r—p=6—E> .996, 11 — pz I

1 — px 6.

Thus (3.50) yields 4

de

1 I 11 1+ 61

2.02 .9J<

(3.53)

6•

On combining (3.52) and (3.53) we obtain lc/CI < 3 dip

(3.54)

in (3.48), where Z = Re' in (3.47). Next we compare Ig'(C)I and 10Z)1, where C, Z are related by (3.47). We have C = tel.', Z = Re', where by (3.49), (3.51)

2 ( 1 - r2 )( 1 — p 2 ) (1 + r)(1 +p)6e 3(1 — r) 1t — = > > 11 — pz1 2 — (1.01) 26 2 since r > .99, p> .63. Also (1 — r)(1 +p) 2(1 —r) 1— R = < and 1 1 — pr 1—p

R>

1— r 1—p •

(3.55)

Thus 3 (1 — t) > — (1 — R). 4

(3.56)

The function h(w) = g{Re iw + (1— R)w} is univalent in iwi < 1. Hence (1.3) in Theorem 1.3 yields for 114,1 < 1 1 + lwl lh / (0)1. Ik(w)1 ( 1 — Iw1) 3

3.5 Coefficients of general mean p-valent functions

89

We define w by

Re"" -1- (1 — R)w = = teiw, so that

t—R 1 1w1 = 1 — R < -4 by (3.56). Thus

5/4 ,

1001

Ig (Z)1 < 3 10Z)1.

On combining this with (3.48) and (3.54) we obtain

JE If' (Ode < A7 f (Re nP )Idtp, with A7 = 18. We apply Lemma 3.4 with g instead of f, R instead of r, g(0) = f(p) and M as before. Thus

If' (z)1(10 < A4A71f(P)1 27(2-)) A 1(2-2A)/(2-2 `) ( 1 — R)K— /. Using also (3.55) and K <

we deduce Lemma 3.5.

3.5.5 Proof of Theorem 3.7 We can now complete the proof of Theorem 3.7. We assume for the time being that f(z) 0, and that f(z) satisfies (3.37) with C = 1. We suppose that -21- < r < 1 and write M = M(r, f). We choose a so that -1—K < a < and a < fl. We define fl —a = y > 0. We recall that /1, ri are constants such that (3.27) holds and K is defined by (3.36). Let k be the smallest nonnegative integer such that

2-k M < e8 (1 — r)-v.

(3.57)

We define

< 27r and 1f(re i° )1 2-k M}

Bk =

and, if k > 0 and 0 < j < k,

Bj = (010

2n, and 2-1-1 M <

We apply Lemma 3.4 with E = Bk. Since D011 C = 1, we obtain, since a > K —

If (rei6 )1d0 < A4(2- k m)(2-2i.)/(2-).) (1

1 by (3.37) with

Means

90

and coefficients

If 2-k M < 1, we deduce that

(re16 )40

A4(1 —

A4(1 —

If 2-k M > 1 we obtain

If(rele )Ide

< A4e8 (1 — r)-'"

A42 -k M(1 —

= A4e8 (1 — r)4 .

(3.58)

Thus (3.58) holds in all cases. If k = 0, Bk = [0, 2m] and (3.38) is proved. Thus we now suppose that k > 0, and consider Bi for 0 j < k. Hence 2-1 M > e8 (1 — r)'. We fix such a value of j and define

p = 4(1 — r)-'6 21 M -1 ,

(3.59)

Using also (3.37) with C = 1, we deduce that 4 X 2i < p < 4e-8 (1 — 0'13 = 4e-8 (1 — r) -".

(3.60)

Let 1 be the largest number of the form 2n /n, where n is a natural number, such that

1 (3.61) < —(4e-8 ) 1 /' < e 16 16 by (3.60) and since a < Thus (3.43) holds if I is an arc of length 1. By (3.61) we have n> 1. Thus 1 2m > — (1 — r)p1/2, (n — 1) 16 so that using (3.60) we have n — 1 27E (3.62) > 1 (1 r)p l /' > (1 — r). /= nn— 1 32 Thus (3.43) holds, when I/I = /. We divide the interval [0,2m] into n arcs I of length 1 = 2Tc/n and apply Lemma 3.5 with Ej = I n Bj. We recall that by (3.37), (3.44) and (3.62) 1

If (Pell

1 —(1 — r) 16

(1 — p )

( e / 8 ) fl < {32e 8 (1 — r) i r i/œ lfl •

Also by (3.61)

1—p 1— r

e8 1 1< 1— r

e8 1/, 16

3.5 Coefficients of general mean p-valent functions and

—

K<

91

a. Next we have on Ei

M = 4(1 — r)— fir l

(r ei6 )1

by (3.59). Hence Lemma 3.5 yields If(re9 )1d0 Ç A6 ( - p) 16

< A8(1—

{32C 8 p'—

1

4

—1 1 1302-A) r

p 1—r a(2—.1))—(2-2(2—.1)

(2-22)/(2-A)

p(1 —

= A8(1 — r) —fl p < A82-7-i(1 — r) —'6

(3.63)

by (3.60), where y =

( A

v

2- 2) a 2 — In order to complete the proof of Theorem 3.7 we need a final Lemma.

Lemma 3.6

For each fixed j at most Ag of the sets Ei are nonempty.

Assuming Lemma 3.6 we deduce from (3.63) that

f

Inrei° )Ide A 8 A9(1 — r)2'1, 0

j < k.

BJ

On combining this with (3.58) we deduce (3.38) on the assumption that f (z) O. It remains to prove Lemma 3.6. For this purpose we use Theorem 2.4. We label as Iv the arcs I which meet B i and assume the I„ with increasing v to be arranged in anticlockwise order along lz I = r. Thus on each arc 1, there is a point in 131, i.e. 4 = rei°v ,

If (4)1 > M2'

R2

(3.64)

say, while by (3.37) with C = 1

z„ = pe'8v satisfies If (zv)I < ( 1 — P)

= R1

(3.65)

say. We define Q = 19 000 so that Q — 1 > 27re8 . Also, by (3.61), (Q — 1)/ < 1. Then (1 — p) = e8 1 < (Q — 1)11(2n) < 1-, so that p > Suppose that, for some integer rio, there are at least noQ sets E. We consider the discs lz — zvl <1— p, y =1,Q +1,...,(no — 1)Q + 1,

(3.66)

Means and coefficients

92

and note that they are all disjoint. In fact if p, p' are 2 distinct values of y in (3.66) we have = peiû' where 0 + (Q —1)1 < 0' 0 +27c — (Q— 1)1.

= pei° , This yields

— zp, I

2p

since (Q —1)1 <

Tr,

(Q-1)1 sin (19 — 0') 1 > 2p sin > it 2 ) 2

and p> On the other hand

2(1 — p) = 2e8 1 < (Q 1)1 . 7t

Thus the discs (3.66) are all disjoint. Next we have from (3.45), (3.62), (3.64) and (3.65) R2

= 2—( J+1)M(e 8 1)

= 2—(1 +1) M(1 —

R1

)fl

32

(1 —

pfl/'

(3.67)

2 ( T;) fl by (3.59). Since p> 1, a

> .4,

and fl > a, we obtain 8)A

e 2 ( _i_i

21.3.2 > e 2 . =e

Now Theorem 2.4 yields no/log ( A°(1 P) ) < 1— r log

4 log(R2/Ri)'

2

(t)

—1

i.e. 4 log {

no < log (-/k) By (3.67) we have R2 10g (- ) > 2 + fi —œ logp > -1± logp,

a

while by (3.61) log p > a log

16e-8 (1 — 0) 16/ " > = a log 1—r 1—r

{log 1

P 6} .

3.5 Coefficients of general mean p-valent functions

93

On the other hand by (3.45) log

(1 1 Pr )

a, 1 -p t , > 7, so that log p> og 7

1-r

Thus V 1-p R2 v , log — > - log p > - log R1 a 7 1-r and

4 log = 11-re+4 log A0 < 28 ± 2 log Ao. V log /3-2log lk Ri

no <

This proves Lemma 3.6 with A9 = 19 000{2 log Ao + 28v -1 ± 1 } , since otherwise we obtain a contradiction from our assumption that at least noQ of the Ei are non-empty. To complete the proof of (3.38) with C = 1, we need to eliminate the assumption f(z) * O. It follows from (3.37) with C = 1, )6 < 1, that If(z)1 < 2 for 1z1 < -1-, so that by Cauchy's inequality ail = If(0)1 4. Suppose that the image of lzl< 1 by w = f (z) contains the disc 1w1 < R. Then z = f-1 (w) gives a univalent map of lw 1 < R onto a subdomain of lzl< 1 and now Cauchy's inequality yields

1

1 4 - al

=

1

dz dw

i.e. R < 4.

4, and f(z) # w o in Izi < 1.

Thus there exists wo, such that lwol Hence if

F(z) = f(z) — wo 5

then F(z) is univalent, F(z) # 0 in lzl< 1 and

IF(z)I

'

If(z)1

5

4 4 C + -5 <- 5 - ( 1 - IzIrfl + -5 < - CO lz1) 3 .

Thus we can apply what we have just proved to F(z) and obtain (3.38) with

1

/i(r,r) = - /1(rd) 5 instead of / i (r,f). This proves (3.38) with 5,4 1 instead of A 1 , when C = 1. We obtain the result for general C by considering f (z)/ C instead of f (z). Finally (3.39) follows from (3.8) and (3.38). This proves Theorem 3.7.

94

Means and coefficients

3.6 Growth and omitted values We give next an application of Theorem 3.4 to a problem raised by Dvoretzky [1950]. We show that if a univalent function f (z) omits a set of values which is fairly dense in the plane, then the effect on the coefficients is nearly as strong as if f (z) is bounded.f Theorem 3.8 Suppose that f(z) = E opp anz n is univalent in 1z1 < 1 and maps 1z1 < 1 (1,1) conformally onto a domain D in the w plane. Let d(R) be the radius of the largest disc, lw — wol < p, whose centre lies on Iwol = R and which lies entirely in D. Then if for some constant a < .245 we have

d(R)

al? (0 < R
(3.68)

we deduce

lard < A(Œ)laoln —

' 509

(n = 1, 2, ...).

(3.69)

We define zo = re° and 0(z) = f

zo

+z

1+20 z

= bo +biz +...

Then 0(z) is univalent in 1z1 < 1 and maps lzl < 1 onto D. Hence there exists w 1 outside D, such that lvvi — bol Thus

4)(z) — bo =z b1 is univalent in fzI < 1 and omits the value (w1 — bo)/bi. Now Theorem 1.2 gives wi — bo > — b1 and so Ib i

4albo l, i.e. (1 —r2 )1f(ret ° )1

4alf(reie )1.

We deduce 0 4« — loglf (retc )1 (0 < r < 1), Or 1 — r2 and integrating with respect to r, we obtain

e

f(r ) < ( 1 + r f(0) 1 r —

f For generalizations to mean p-valent and other functions see Hayman [1952].

3.7 k-symmetric functions and Szegei's conjecture Since a <

95

4 this gives M(r, f) < V21a01(1 — r)-2Œ (0 < r < 1),

and now Theorem 3.8 follows from Theorem 3.7, with .491 instead of /3. If 0 < A < 1, then the function w = f(z) =

(1 + zV

maps IzI < 1 (1,1) conformally onto the angle I arg WI < Pril. and satisfies d(R) = Rsin(Pril.). It does not satisfy (3.69) if A> .491. Thus in (3.69) .245 cannot at any rate be replaced by .7> sin(.246n).

If fk (z) is regular in

3.7 k-symmetric functions and Szegii's conjecture IzI <1 and of the form fk(Z) = Z ± ak+iZk+1 ± a2k+1Z 2k+1

•••9

(3.70)

we say that fk(z) is k-symmetric. If further fk(z) E S where k is a positive integer then also

f (z) = {f k (Z k )} k G S. In fact since fk (z) is univalent fk(z) z

1

,

,

= 1 -I- ak+1,,

k _L

m a2k+iz

2k

is regular and non zero in IzI < 1, and so is

4)(z) = 1 + ak-nz + a2k+1z 2 +... =

fk(zi/k) z l/k •

Hence so is

1(z) = z4)(z) k = z + kak + 1z 2 +... It remains to show that f(z) is univalent. Suppose then that zl, z2 are distinct numbers in Izl < 1, such that f(zi) = f(z2) = w. Since f(z) = 0 only for z = 0, we may assume that w 0. Let C i , C2 be k th roots of z 1 , z 2, so that VI' = z1, ( 21( = z2 . Then W = fk(Ci)} k

= f(Cli) = f(CI = 1fk(C2)} k . )

Thus fk(C1) = Wl, fk(C2) = W2

96

Means and coefficients

where wl2 = w = w, so that w2 = w 1 w, where co = exp Now by (3.70) fk(c)(:1) = fk(Ci

Since fk is univalent we deduce that C2 = WC'. Thus z2 = C 2/' = so that f (z) is also univalent, i.e. f (z) G Z. We can now prove

= Zi,

Theorem 3.9 (Szeg6's conjecture). If fk(z) G e, where k = 1,2,3,4 and f(z) has the expression (3.70) then

jan l < An211-1 ,

n > 1.

(3.71)

The result was obtained by Littlewood [1925] for k = 1, by Littlewood and Paley [1932] for k = 2, by V. I. Levin [1934] for k = 3 and by Baernstein [1986] for k = 4. Littlewood [1938] showed that (3.69) is false for large k and an example of Pommerenke [1967, and 1975, p. 133] proves that it is false for k > 12. The conjecture (3.71) is attributed by Levin [1934] to an oral communication of Gabor Szeg6. We note that since fk(z)= f(z k ) 1/k

we have M(r,f k ) = M(rk ,f ) 1 /1` < r/(1 — rk )2Ik

0< r<1

by Theorem 1.3. Now (3.71) follows from Theorem 3.3 for k = 1, 2, 3 and from Theorem 3.7 for k = 4. We see that the cases 5 < k < 11 remain open. To prove (3.71) for k = 5 the range of Theorem 3.7 would need to be extended to )6 > .4 instead of fi > .491. This seems well beyond techniques available at present. If fk(z) is k-symmetric and p-valent a similar argument, based upon Theorem 3.3, shows that < A(p,k) ppn (2P 10-1

(3.72)

provided that 1 < k <4p (Robertson [1938]) and the proof was modified so as to apply to mean p-valent functions by Spencer [1940b, 1941a]. For this latter case (3.72) breaks down when k > 4p, as is shown by a suitable form of the examples in §3.4. For this type of proof is essential that all the coefficients vanish, except those whose suffixes form an arithmetic progression of common difference

3.7 k-symmetric functions and Szeg6's conjecture

97

k. We proceed to prove (3.72) under somewhat weaker assumptions, 'basing ourselves on the fact that the functions fk(z) satisfy Ifk [re i(°+27" /k) ] = Ifk(re i8 )1 ( 0 < 3.7. 1

k —1 ).

We need the following preliminary result:

Theorem 3.10 Suppose that f(z) = and that there exist k points k > 2, such that (i)

E ooc an z" is mean p-valent in 1z1 < 1

on Izi = r, where 0 < r < 1 and

(1

i < j < k)

and

If(4)1

(ii)

R (1 < i

k).

Then we have R < A(p)0 2p(i/k—i) (1 _ r)-2p/k ,

where p is given by (2.8). We suppose that 6 > 4P+2(1 — r).

(3.73)

For if this is false, we have by Theorem 2.3

M(r, f) < A(p)p(1 — r) -2P 471_2

2p-2plk

A(p)p(1 — r)-2P/k (

6 ) < A(p)p(1 — r) — 2plk 6 2(p/k) -2p so that Theorem 3.10 holds. At least one of the annuli 1 — 4'6 < IzI <1 — 4—(v+ 1) (5 (1

y

[p] +1)

is free from zeros of f(z), since f(z) has q p zeros in IzI < 1. For such a value of y we choose ri = 1— 1-4'6 = 1 — On, say, and note that f(z) has no zeros in 1 — 200 < lz < 1 — 1 613. Using (3.73) we deduce further that

6

60 = 4>4P+2—''(1 — r) > 2(1 — r),

(3.74)

98

Means and coeffi cients

since 1 < y < p + 1, and hence that r 1 < r. Suppose now that

=

/T

O

(1

/

j

k).

We write z i = ri ei6J (1

j < k),

and apply Theorem 2.4 with R1 = 2PM(r1 ,f), using the number R of Theorem 3.10 (ii) instead of R2, and 60 instead of rn . This gives

— — zi I

=

=

bo

1—r

(1

bo

j k).

By construction the circles lz —zi l < 1-60 contain no zeros of f(z). Also the circle lz — z1 1 < 60 contains the point zi and has diameter 260 < by (3.74). In view of hypothesis (i) of Theorem 3.10 it now follows that the circles lz — z1 1 < 60 (1 j k) are non-overlapping and Theorem 2.4 is applicable. We obtain -1

-1 k [log A° ' °

2p log

1—r

R2

{ A0(50

eR i

1—r

— R2 eR i

}2p 1 k

\2p/k

< A(p)

(1— r

by (3.74).t Since r1 = 1 — (50 Theorem 2.3 gives further

R1 = 2P M(ri, f) <

—2p A(P)/ 1 6 0

. A(p)ii ( 2 x 4% ) 2P < A(p) 6

and we deduce Theorem 3.10 on writing R instead of

2P,

R2.

3.8 Power series with gaps It is easy to see that the theorems quoted in §3.7 follow at once from Theorems 2.3, 3.3 and 3.10. We can, however, prove more. Theorem 3.11

Suppose that N —1

f(z) =

Eanzn + aNz N + aN±kz" + aN+2kzN +2k ± o

t We may assume that otherwise.

R2 > eRi. Since

( 5 Al —

r) >

16 by (3.74), our result is trivial

3.8 Power series with gaps

99

is mean p-valent in 1z1 < 1. Then M (r, f) < A(p,k, N)p(1 — r) -2P/k (0 < r < 1),

(3.75)

and if 1 < k <4p we have further

ianl < A(p,k, N)pn (2P/k)-1

(n = 1, 2, ...).

(3.76)

Write N-1

g( z

00

an zn,

)

h(

aNd-knZ

)

Nd-kn

,

n=0

and for r > I choose 00 so that If (rei4 = M (r , f). Then if 2nv 0„ = 00 + k

(0 v

k — 1),

we have 27rivN h(reiev) = h(re16°)exp ( k)

.

Again for lz I = r < 1, we have from Theorem 3.5 N-1

E Ian' ._

Ig(z)1 __

n=0

if p >

N-1

EA(p)1in2P — '

._ A(P, N)P

n=0

1 and N-1 Ig(z) 1

..__

E A lao 1 = ANIaol n=0

if p <

1. Thus 1.f(rev)I >

I h(relev )1 - Ig(rev )1

> Ih(rell - A(P, N)ti > M(r, f) — 2A(p, N)p (0 _._ v _._ k — 1).

We may therefore apply Theorem 3.10 with 6 . Ire2gi/k _ 11 > 1 le2giik _ 11= A(k),

and

R = M(r, f) — A(p, N)p, and obtain M(r, f) < A(p, N)p + A(p)pA(p,k)(1 — r) -2Pik ,

100

Means and coefficients

and this gives (3.75). The inequality, proved on the assumption r > clearly remains valid if r < 1, since M(r,f ) increases with r. Also (3.76) follows from (3.75) and Theorem 3.3. This proves Theorem 3.11.

3.8.1

If k = 2, we can sharpen Theorem 3.11.

Theorem 3.12 Suppose that f(z) — an zn is mean p-valent in zI < 1 and that an = 0 whenever n = bm+c, where b, c are fixed positive integers and m goes from 1 to oo. Then M(r, f) < A(p, b, c),u(1 — r) — P (0 < r < 1),

(3.77)

and so, if p>

< A(P, b, c)imP -1 (n

1).

(3.78)

To prove Theorem 3.12, consider f ( z ) E abm+rzbm+v m=0

b — 1),

and write co = exp[2ni/b]. Then we have 1 f t (z) = —

b- 1

f(coPz). p=0

In fact the coefficient of zn on the right-hand side is b- 1

b- 1

w —pv copn = _Eko n—v i p

an E p=0

11 =0

and this is a„ or 0, according as n — y is or is not a multiple of b. Hence if y is so choosen that c — y is a multiple of b, we have from the hypothesis of Theorem 3.12 for z < 1 b—i

E

b+

co—Pvf ( ai`z ) =

p=0

v

b

Ea

A(p, b, c)1.1= K,

n=0

say. Choose now, for < r < 1, zo so that f ZOf = r,

If(zo)1 = M(r, f).

Examples

101

If M(r, f) < 2K, then (3.77) follows. Otherwise we obtain b-1

b-1

Ew —Pvf(WPZO)

--

I f (z o ) 1 — E w — Plf(WPZO) p=0

P= 1

__

If (zo)1

—K

..?- 1.t. (zo)I = -1M(r, f).

Hence we can find p. (1 _. it __ b — 1) such that

If (co" zol __

M(r, f) 2(b —1)'

and

i Ir ) _. -. ko Pzo — zol = ill — wP > I —-11 21 — wl = sin (--2b b We can thus apply Theorem 3.10, with k = 2, z'i = zo, z; = coPzo, 6 = b-1 and R = -1M(r,f)/(b — 1). We obtain

M(r, f) < 2(b —1)A(p)ttbP(1 — r) -P, and this proves (3.77) if r > Since M(r, f) increases with r, the result for 0 < r < -I also follows, and (3.78) follows from (3.77) and Theorem 3.3. Thus Theorem 3.12 is proved. It is worth noting that the above argument does not require the full strength of our hypotheses. It would be sufficient to assume that on

1z1 = r cc

= E abm+,,L,bm+v = 0(1 — r)-P

as r -- 1,

m=0

in order to obtain M(r, f) = 0(1 — r) -P. A similar remark applies to Theorem 3.11. The conclusion of Theorem 3.11 continues to hold if an = 0, except for a sequence nv such that n +1 — n, > k finally (instead of n„ +1 — n,, = k finally) [Hayman 1967]. Similarly that of Theorem 3.12 holds if an = 0 for a sequence n = nv , such that n„ +1 — n, < b [Hayman 1969]. These results involve some Fourier series arguments to show that the hypotheses of Theorem 3.10 are satisfied and are beyond the scope of this book.

Examples

3.6

Show that if f(z) is univalent and satisfies the hypotheses of Theorem 3.11 for k = 4, then Ia n ' < A(N)tn. (Use Theorem 3.7.) Note that this conclusion extends Szegb's conjecture for

k = 4.

102 3.7 3.8

Means and coefficients

If f(z) is univalent as well as mean p-valent in Theorem 3.12, show that (3.78) continues to hold for p > .491. If k > 4p in Theorem 3.11, or p < -1 in Theorem 3.12, show that i a, = o(n) and that this conclusion is sharp. (Use Theorem 3.4 and the example of Section 3.4.)

4 Symmetrization

4.0 Introduction

In this chapter we develop the theory of symmetrization in the form due to 1361ya and Szegei [1951] as far as it is necessary for our function-theoretic applications. Given a domain D, we can, by certain types of lateral displacement called symmetrization, transform D into a new domain D* having some aspects of symmetry. The precise definition will be given in §4.5. 1361ya and Szeg6 showed that while area for instance remains invariant under symmetrization, various domain constants such as capacity, inner radius, principal frequency, torsional rigidity, etc., behave in a monotonic manner. We shall here prove this result for the first two of these concepts in order to deduce Theorem 4.9, the principle of symmetrization. If f(z)= ao + aiz +... is regular in 1z1 < 1, and something is known about the domain Di- of values assumed by f(z), this principle allows us to assert that in certain circumstances lad will be maximal when f(z) is univalent and Di- symmetrical. Applications of this result will be given in Sections 4.10-4.12. Some of these will in turn form the basis of further studies of p-valent functions in Chapter 5. Some of these results can also be proved in another manner by a consideration of the transfinite diameter (Hayman [1951]). The chapter ends with a recent proof of Bloch's Theorem by Bonk [1990]. Since the early part of the chapter is used only to prove Theorem 4.9, the reader who is prepared to take this result for granted on a first approach may start with §4.5 and then go straight to §4.9. We shall need to refer to Ahlfors [1979] (which we continue to denote by C. A.) for a number of results which space does not permit us to consider in more detail here. A good set-theoretic background is provided by C. A., Chapter 3, §1, and we use generally the notation there given.

103

104

Symmetrization

We shall, however, in accordance with common English usage, call an open connected set a domain and not a region. The closure of a domain D will be denoted by b. A domain D has connectivity n if its complement in the extended plane has exactly n components; if n = 1, D is simply connected, if n = 2 doubly connected, etc. (C. A. pp. 139 and 146). If the boundary of D consists of a finite number n of analytic simple closed curves (C. A. pp. 68 and 234), no two of which have common points, we shall call D an analytic domain. We shall assume a right-handed system OX,OY of rectangular Cartesian axes in the plane. Points in the plane will be denoted in terms of their coordinates x, y either by (x, y) or by z = x iy, whichever is more convenient. Accordingly, functions u will be written as either u(z) or u(x, y).

4.1 Lipschitzian functions Let E be a plane set and let P(z) be a function defined on E. We shall say that P(z) is Lipschitzian or Lip on E if there is a constant C such that

P(z2)

— z21

(4.1)

whenever z2 lie in E. It is clear that a Lip function is continuous, and further that if P,Q are Lip and bounded on E, then PQ is Lip on E. Suppose that E is compact and that P(z) is Lip in some neighbourhood of every point zo of E. Then P is Lip on E. For if not, we could find sequences of distinct pairs of points z n , z n' (n > 1) on E such that

IP(zn )— P(4)1 1zn — 41

00

(n —> co).

(4.2)

zo , By taking subsequences if necessary we may assume that z n z z'o , where zo, 4 lie in E. If zo, 4 are distinct, we at once obtain a contradiction from our local hypothesis and (4.2), since P is bounded near any point of E. If zo = 4, then zn, z'n finally lie in that neighbourhood of zo where P is Lip and this again contradicts (4.2). If P(z) = P(x, y) is defined in a disc y (C. A. p. 52) and has bounded partial derivatives there, then P(x, y) is Lip in y. Suppose first that P(x,y) is real. If (xo,yo), (xo +h,yo +k) both lie in y, then either (xo, yo + k ) or (xo + h, yo) also lies in y. Suppose, for example, the former. Then if M is a bound for the absolute values of the partial derivatives in y we have from the mean-value theorem

4.2 The formulae of Gauss and Green

105

IP(x0 + h, yo + k)

P(xo, Yo)I IP(xo + h, yo + k) — P(xo, yo + k)I + IP(xo, yo + k) — P(xo, LP)

h

(

Ox (x o +Oh, yo+k)

M(1111+1k1)

k " \ °Y

(x 0 ,y0+6q)

2M../(h2 + k2 ).

Thus P is Lip in 7. For complex P we can prove the corresponding result by considering real and imaginary parts. If follows that if P has continuous partial derivatives in a domain containing a compact set E, then P is Lip on E. For in this case P has continuous partial derivatives in some neighbourhood and so bounded partial derivatives in some smaller neighbourhood of every point of E. Conversely, we note that if P(x, y) is Lip on a segment a < y < b of the line x = constant, then P is an absolutely continuous function of y on this segment and so OP/Oy exists almost everywhere on the segment and is uniformly bounded. t Thus

P(x, b)— P(x, a) =

OP(x, y) dy. a OY

(4.3)

4.2 The formulae of Gauss and Green We proceed to prove these formulae in the form in which we shall require them in the sequel.

Lemma 4.1 (Gauss formula) Suppose that D is a bounded analytic domain in the plane and that its boundary y is described so as to leave D on the left. Then if lc, y), Q(x, y) are Lip in b, we have

OQ OP (Pdx + Qdy) = f f (— — —) dxdy. D °X OY Let z = a(t) (u < t < b) give an arc of y. Then a(t) is a regular function of t and i(t) 0. The tangent is parallel to OY at those points where ce(t) is pure imaginary, and this can be true only at a finite number of points, since otherwise a'(t) would be identically pure imaginary and so this arc of y would reduce to a straight line. t This is, however, impossible, since y consists of a finite number of analytic closed curves. Thus there are only a finite number of tangents to y which are parallel to 0 Y, and Burkill [1951], Chapter IV. The real part of a' (t) is a regular function of t for a < t < b and so has only isolated zeros or vanishes identically.

106

Symmetrization

we assume that these are

x = xm (1 m < M), where x1 <X2 < < xm . For xm_i < < xm, the line x = ç meets y in 2n points Y = Yi(),

-•

Y = Y2n(),

where n depends only on m and the y v ( ) (1 < y < 2n) are differentiable functions of ç for xm_i < < xm . Also the part Dm of D lying in xm_ i <X < xm consists of n domains Dm,v

< x < xm (1 < V < n),

y2 v- 1 (x ) < Y < Y2v(x),

since at the intersections of x = and leaves D. Consider now

with y, the line x =

alternately enters

I = f P(x, y)dx taken along y so as to keep D on the left. Then dx > 0 on the curves y = y2v-i(x) and dx < 0 on the curves y = Y2v(x). Thus if /m is the integral taken over those points of y which lie in xm_ i < X < xm, then xm n

Im =

El/3[x, y 2v_,(x)] _ P[x, y2 v (x)]}dx

f

Xm-1 v =

1

Xm

=

—f

dx

E= 1 v

ry2v

OP(x,y)dy ay

_OP

L-1 v=1

.1

dxdy

CY

by (4.3), since P(x, y) is Lip. Adding over the separate ranges xm_i <x < xm and all the domains Dmo, we obtain OP

Pdx = f f — dxdy. D

CY

Similarly we prove

aQ 1Qdy = f f dxdy, D OX

and Lemma 4.1 follows. We deduce Lemma 4.2 (Green's formula) Suppose that D is a bounded analytic domain with boundary y, that u is Lip in b, and that y possesses continuous

4.3 Harmonic functions and the problem of Dirichlet

107

second partial derivatives near every point of b. Then

,,-))1

a2v ) 4_ (P_IL1 -!-) 4- 'ill- -°2 u °v ds = — f f [--u °21j ( ax24_, oy 2 ' o x Ox Oy cy j j dxdy, D fy where a/an denotes differentiation along the normal into D, and ds denotes arc length along y. In fact u, avgx, av/ay are Lip in b in this case and hence so are u(av lax), u(av/ay). Let (xo, Yo) be a point of y and let 0 be the angle the tangent to y makes with OX, where y is described so as to keep D on the left. Then the normal into D makes an angle 0 + -In with OX and

Ov -an

lim h-+0

v [xo + h cos(6 + i'r), yo + h sin(0 + -1 Tc)] — v [xo, Yoi h

av . av = cos 0- — sin 0—. 0y Ox If ds is an element of length of y at (xo, RI), its projections on OX, 0 Y are dx = ds cos 0 and dy = ds sin O. Thus

0v

av

av

— ds = —dx — — dy. an ay ax We now apply Lemma 4.1 with P = u(av/ay), Q = —u(av/ax) and Lemma 4.2 follows.

4.3 Harmonic functions and the problem of Dirichlet A function u(x, y) is harmonic in a domain D if u has continuous second partial derivatives in D which satisfy Laplace's equation 02 u

OX2

02u 2

ay2

It follows from this that (1)(z) = Ou/ax — i(au/ay) has continuous partial derivatives which satisfy the Cauchy-Riemann equations, and so 4)(z) is regular in D. Also in any disc with centre zo that lies in D, u is the real part of the regular function

z f (z) =

f70

(/)(C)dC + u(z0).

Thus u possesses continuous partial derivatives of all orders in D. Also, by the maximum modulus theorem applied to exp{+f (z)}, u can have no local maximum or minimum in D unless u is constant in D. Suppose now that D is a domain in the open plane and that u(C) is

108

Symmetrization

a continuous function of , defined on the boundary y of D considered in the extended plane. Thus, if D is unbounded, y includes the point at infinity. The problem of Dirichlet consists in finding a function u(z) harmonic in D, continuous in b and coinciding with u(c;) on y. If ul, u2 are two functions satisfying these conditions, then u 1 — u2 is harmonic in D, continuous in b and zero on y, and so by the maximum principle /4 1 — /42 vanishes identically in D. Thus a solution to the problem of Dirichlet is unique, if it exists. A solution does not always exist. If D consists of the annulus 0 < 1z1 < 1, then any functions bounded and harmonic in D can be extended to be harmonic also at z = 0 and so in 1z1 < 1. Thus boundary values assigned on 1z1 = 1 determine u(0). The question of existence can often be settled by means of the following result, for whose proof we must refer the reader to C. A. p. 250. Theorem 4.1 Suppose that for every boundary point z o of D there exists w(z), harmonic in D, continuous in b, and positive in I), except at z o , where co(z o ) = 0. Then the problem of Dirichlet possesses a solution for any assigned continuous boundary values on the boundary of D. From this we deduce the following criterion:

Theorem 4.2 The problem of Dirichlet always possesses a solution for the domain D if, given any boundary point of b, there exists an arc of 0 straight line or circle containing zo and lying outside D. We shall deduce Theorem 4.2 from Theorem 4.1. Suppose first that zo is finite. Let the arc c have end-points z o and z i . The function = ei

a Z — Zo z1 — z

maps the exterior of this arc c onto the plane cut along a ray through the origin, so that z = zo , z 1 correspond to = 0, co. By suitably choosing oc, we may arrange that the ray is the negative real axis. Then

0 —1

w= 1 + i

V+1

maps this cut plane onto the circle lw — 11 < 1. Also z = zo corresponds to w = 0, and D maps onto a domain whose closure lies in lw — 11 ._-. 1, and such that only z = zo correspond to w = 0. Thus co = 9iw is the

4.4 The Dirichlet integral and capacity

109

harmonic function whose existence is required in Theorem 4.1.t If zo is infinite and some ray from a finite point to infinity lies outside D, the argument is similar. This completes the deduction. A plane domain satisfying the criterion of Theorem 4.2 will be called admissible (for the problem of Dirichlet).

4.4 The Dirichlet integral and capacity

Historically Dirichlet attempted to base a proof of the existence of a solution to his problem on the problem of finding the minimum for the Dirichlet integral

= f fp [(4 1x4 ) 2 + (4) 1 2

ID(U)

dxdy,

for all functions having the assigned boundary values and satisfying certain smoothness conditions. In favourable circumstances this minimum is attained by the required harmonic function. However, even for continuous boundary values on the unit circle, the harmonic function inside the circle with these boundary values may have an infinite Dirichlet integral, so that the minimum problem has no solution. We shall need the Dirichlet minimum principle only in a special case, where its validity can be proved without difficulty.

Theorem 4.3 Let D be an admissible domain in the open plane whose complement consists of a compact set E l and a closed unbounded set E0, not meeting E l . Let v(z), co(z) both be continuous in the extended plane, 0,1 on E0, E i respectively and Lip on every compact subset of D. Suppose, further, that co(z) is harmonic in D. Then D [V(Z)1

D [W(Z)] Ya

On

ds,

where ya is the set {z : co(z) = a}, and a is any number such that 0 < a < 1 and Oco/ Ox—i(Oco/ y) # 0 on ya . In this case ya consists of a finite number of analytic Jordan curves. The system consisting of the domain D and the sets E0, E 1 will be called a condenser, and 'D [W(Z)] will be called the capacity of the condenser for evident physical reasons. In the special case when E0, E1 are continua, so that D is doubly connected, a simple function-theoretic interpretation t A point of b on the arc through ZØ,

zi corresponds to a pair of complex conjugate values of w. Thus co is uniquely defined and continuous even on this arc.

110

Symmetrization

is possible. We may then map D (1,1) conformally onto an annulus Atz : 1 < 1z1 < RI. (For a construction of the mapping function in terms of w(z) see C. A. p. 255.) The Dirichlet integral is clearly invariant under this transformation so that A has the same capacity as D. The harmonic function satisfying our boundary-value problem in A is

fl(z) =

log1R / z1 in A, log R

and hence the capacity of A is 2ir 1 pd0 1, i=i, log R p = log R . The quantity log R is frequently called the modulus of the doubly connected domain D. It is thus a multiple of the reciprocal of the capacity of D considered as a condenser. For our purposes it is, however, essential that E0, E1 need not be connected.

4.4.1 Proof of Theorem 4.3 Let w(z) be the function of Theorem 4.3. Then (/)(z) = aw/Ox — i(Oco I ay) is regular and not identically zero in D, and so (/)(z) vanishes at most on a countable set of points in D, which we shall call the branch points of co(z). Suppose that 0 < a < 1 and that the set Va, where co(z) = a, does not pass through any branch point of co(z). Clearly Va is compact and lies in D by our hypothesis. Also, near any point z0 of y a, CO is the real part of a regular function

w = f (z) = co + iwi = a + ib + ai(z — zo) + . . . . Here a l # 0, since 4)(z) * 0 on Va. Thus the inverse function z = f 1 (w) is also regular and the part of Va near zo is the image by a regular function of a straight line segment, i.e. an analytic Jordan arc. If we continue along this arc we must finally return to our starting point, since otherwise Va would posseses a limit point not of the kind described above. Thus Va consists of analytic Jordan curves, no two of which have common points. There can only be a finite number of these curves, since otherwise there would be a point of Va any neighbourhood of which meets an infinite number of curves. This again contradicts the local behaviour of Va established above. Suppose again that Va, Yb contain no branch points, that 0 < a < b < 1, and that Do = {Z : a < w(z) < b } . Then Do consists of a finite number of bounded analytic domains, of which Va, Yb form the complete boundary. Also if 010n denotes differentiation along the normal into D a, b

4.4 The Dirichlet integral and capacity

111

then Oco/On> 0 on ya and Ow/en < O on yb. Now, on applying Green's formula to the regions of Dad, separately and adding, we obtain aw n ds =0, f T aw n ds+ f T

I

a

ib

and since Ow/en has constant sign on Ow iYa

y a ,yb,

Ow

ds =

an

an

Yb

we deduce ds = I,

say. Also a second use of Green's formula gives aw ds— f w— aw ds = IDb [co(z)]. (b — a)I = — f w—

an

Ya

an

Yb

Since every point of D belongs to some Do, we deduce, on making a tend to 0 and b tend to 1 that

lim

D[W(Z)1 =

D a,b [W(Z)1

a—).0,b—■ 1

=1.

Suppose now that v(z) is the function of Theorem 4.3 and set h = v—w. Then h(z) is continuous in the plane and vanishes on E0, E 1 and so at infinity. Hence, given e > 0, the set

el

E = {z :1h(z)1

is a compact subset of D. Let ao, bo be the lower and upper bounds of w(z) on E, so that 0< ao
f

La b raxh a(1)x ± 6 'awddxdy a ,

a CO

n—as

T

fYa+y b

an

Ow ds = 2e1. an Thus IDa, b (h, w) =

ha

[Oh Ow j,

Lax ax +

Oh Ow

Tv

]

dxdy 0

(a —> 0, b —> 1).

Also 1Da, b ( V

)=

b

(

a)

)

b

(h)

21 Da b ( h,

Making a tend to 0, b tend to 1 we deduce /D(v) = /D(w) + /D (h). This proves the inequality of Theorem 4.3 and completes the proof of that

112

Symmetrization

theorem. We note also that equality is possible only if /D (h) = O. In this case it is not difficult to see that h(z) vanishes identically so that v(z) must coincide with w(z). We shall not, however, make any use of this.

4.4.2 Capacity decreases with expanding domain

We make an immediate deduction from Theorem 4.3, which has independent interest:

4,

E be pairs of closed sets characterizing Theorem 4.4 Let E0, El; two condensers as in Theorem 4.3. Let I, I' be their capacities and suppose that Eo c E0, Ei c El. Then I' __ I. Let w(z) be the potential function corresponding to the first condenser as in Theorem 4.3. If 0 < a < b < 1, we define a function v(z) as follows: v(z) = 0 where co(z) ._ a, v(z) = 1 where co(z) > b, and v(z) =

co(z) — a . , in Va,bb— a

4,

E respectively. Thus if Then v(z) is Lip in the plane and y = 0, 1 on D' is the complement of Ec,' and 4 Theorem 4.3 gives I I' < 'DO) --= — f - an s = 1 I D[w(z)] = d b — a. b— a ya, Yb

Making a tend to 0, b tend to 1, we obtain I' < I as required.

4.5 Symmetrization We shall now consider the process of symmetrization introduced in the middle of last century by Steiner and developed by 13■51ya and Szegii [1951]. Let 0 be an arbitrary open set in the open plane. We shall define a symmetrized set 0* in two ways as follows.

4.5.1 Steiner symmetrization

In this case we symmetrize with respect to a straight line, which we can take to be the x axis of Cartesian coordinates. For any real ç, the line x = ç meets 0 in a set of Mutually disjoint open intervals of total length l( ) say, where 0 _-_ l( ) __ co. The symmetrized set is to be the set

o. = {(x, y)

:

II < -/(x)}.

4.5 Symmetrization

113

Fig. 3. Steiner symmetrization In this case we symmetrize with respect to a half-line or ray, which we take to be the line 0 = 0 of polar coordinates r, O. We define the symmetrized set 0* as follows. Consider the intersection of 0 with the circle r = p, 0 < p < co. If this intersection includes the whole circle or is null, then the intersection of 0* with r = p is also to include the whole circle, or to be null respectively. Otherwise let 0 meet the circle r = p in a set of open arcs of total length pl(p), where 0 < l(p) < 2n. Then 0* is to meet the circle r = p in the single arc 4.5.2 Circular (Nlya) symmetrization t

1 0 1 < ]1 (P). We proceed to discuss some properties of symmetrization. Unless the contrary is stated results are true for either kind of symmetrization.

4.5.3 The symmetrized set 0* is open

We prove this only for circular symmetrization. The proof for Steiner symmetrization is similar. Suppose that (po, it) lies in 0*. Then 0 contains the whole circle r = po . Since 0 is open, 0 therefore contains some annulus po — 6 < r < Po + 6 and this annulus also lies in 0*. Thus (po, 7r) is an interior point of 0*. A slightly modified argument applies if the origin lies in 0*. ,

t

Nlya [1950]

114

Symmetrization

d

Fig.

4.

Circular symmetrization

Suppose next that (po, 00) lies in 0*, with 0 < po 21001. Choose x so that 21001 <x < l(p0 ). Then we can find a finite number of open arcs of total length greater than pox on the circle r = po and in 0. By diminishing these arcs slightly, we may take them to be closed and still lying in 0 and of total length greater than pox. If av < 0 < fi, is such an arc and 6, is its distance from the complement of 0, let 6 be the smallest of the 6,. Then the arc cx, < 0 < fl, of every circle r = p for Po — (5 < P < Po ± 6

lies in 0. Thus

l(p) _- x > 2100 1 for po — 6 < p < po+ 6, and so (r, 0) lies in 0* for

Po

—

6 < r < po d- 6 and 01< -1- x,

4.5 Symmetrization

115

where Ix > 100 1. Thus (p0,00 ) is an interior point of 0* and so 0* is open.

4.5.4 The Steiner symmetrized

set D* of a domain D is either the whole plane, or a simply connected admissible domain Suppose that the lines .x = Ç1, 2, where i < 2, meet D* and so D. Then x = meets D for i < < 2, since otherwise D could be expressed as the union of the disjoint non-null open subsets lying in the half-planes x < and x> respectively. Thus the segment < x < 2 of the real axis lies in D* in this case. Two points ( 1 , n i ), (2, 172) in D* can be joined in D* along the polygonal arc (1, qi), (2, 0 ), (2, 0), (2, 112). Thus D* is a domain. Suppose next that D* is not the whole plane, so that V) < +co for at least one . Then the complement of D* meets x = ç in the pair of rays ly1 > 1-1( ) . Thus the condition for admissibility in Theorem 4.2 is satisfied for every point (, n) in the complement of D* and also at co. Further, every point in the complement of D* can be joined to infinity by a ray, and so this complement is connected in the extended plane. Thus D* is simply connected and is admissible. The results for circular symmetrization are not so simple. If, for instance, D consists of an annulus pi
4.5.5 The circularly symmetrized set of a domain D is a domain D* • If D

is simply connected, so is D* . If D is admissible or D* is simply connected, D* is admissible or D* is the whole plane The proof that D* is connected and is a domain is similar to that for Steiner symmetrization. If D is simply connected and contains the circle r = p, then D must contain the interior of this circle, since otherwise the complement of D would contain points inside and outside the circle (e.g. infinity) but no points on the circle. Hence if D contains r = p, D and D* also contain the disc r < p in this case. Let po be the upper bound of all p for which this holds. If po = d-oo, D and D* contain the whole open plane. Otherwise the complement of D contains some point of every circle r = p > po, and so the complement of D* contains the ray r > po, 0 = it. Any boundary point of D* on this ray and the point at infinity clearly satisfy the criterion of admissibility. If (p,00) is any other point in the complement of D*, then p > po, 1 0 01 < it, and the arc IN! < 0 _-_ it of

116

Symmetrization

the circle r = p lies in the complement of D* and joins (p, 00) to the ray 0 = 7r, r _-_ po in this complement. Thus the complement of D* is connected and D* is simply connected. The criterion of admissibility is satisfied in all cases for every boundary point of D* not on the line 0 = 7r. The above argument shows that if D* is simply connected the criterion is also satisfied for boundary points (p, 7r). It remains to show that if D is an admissible domain, so is D*. By the above argument it suffices to establish the criterion of Theorem 4.2 for the boundary points (p0, 7r) of D*, and for co. Since D is admissible the complement of D contains a ray, which must meet every sufficiently large circle r = p. Thus (p, 7r) lies in the complement of D* for large p, and so this complement contains a ray r > p, 0 = 7r. Thus our criterion is satisfied at cc. Suppose next that (po, 7r) is a boundary point of D*. If an arc a < 0 < )6 of r = po lies in the complement of D, then an arc of r = po bisected by (po, 7r) lies in the complement of D* and the criterion is satisfied. If not, let (po, 0) be a frontier point of D. Since D is admissible there is an arc of a straight line or circle containing (po, 0) but lying outside D and so by our hypothesis not on r = po. This arc must meet r = p for all p either in some interval po — 6 ._ p _._ po or Po •-- P po +6. Thus the interval [Po — (5, Po] or [Po, Po + 6] of 0 = it lies in the complement of D* and contains (po, it), and so D* is admissible.

Let u(z) be a function, real, continuous and bounded in the plane. We symmetrize u to obtain a new function u*(z) by simultaneously symmetrizing all the sets 4.6 Symmetrization of functions

Da = { z : u(z) > a}(—cc < a < +co).

These sets are open since u is continuous. More precisely, let Da* be the symmetrized set of Da with respect to some straight line or ray. For any point z in the plane we define u*(z) as the least upper bound of all a for which z lies in D. In practice we shall be concerned only with the case where u(z) is nonnegative and vanishes continuously at infinity. In this case the sets Da are bounded for a > 0 and their closures Da are compact. Now a function u(z) continuous on a compact set E is uniformly continuous on E (C. A. p. 65). In other words if f2(6) is the upper bound of lu(z i ) — u(z2)1 for z 1 , z2 on E and 1z1 — z21 < (5, then Q( (5 ) is finite and Q( (5 ) —> 0 as (5 --- O. The quantity Q( (5 ) is called the modulus of continuity of u(z). Clearly u(z) is Lip on E if and only if i/((5) < C(5 for some positive C and 0 < (5
4.6 Symmetrization of functions

117

u(x, y) = a

I I

Fig. 5. Symmetrization of functions

4.6.1 Symmetrization decreases the modulus of continuity We shall be able to show that u*(z) is 'at least as continuous as u(z)' by proving Theorem 4.5 Suppose that, with the above definitions of u(z), u* (z), the set E = ti : a < u(z) < b} is bounded and that u(z) has modulus of continuity f2(6) on E. If E* = { z : a < u*(z) < b}, then u*(z) is continuous on E* with modulus of continuity fl*(6) < Q(6). In particular if u is Lip on E, u* is Lip on E*. We shall prove this result for Steiner symmetrization. The proof of circular symmetrization is similar, but the details are a little more complicated. We need

Lemma 4.3

With the above hypotheses let 1(x, t) denote the total length of the y intervals on the line x = constant where u(x, y) > t. Suppose that

118

Symmetrization

b — fl(6) and jx2 — xi i _._ 6. Then

a _-_ t2 < ti —S1(6)

1(x2, t2) .1(xi, 0+106 2 —

(X2 - X1) 21

Let F1 be the set of all y such that u(xi, Y) > ti and let F2 be the set of y for which u(x2, Y) > t2. Then Y2 E F2 if yi E F1 and (x2

—

xi)2 + (V2

For otherwise we could find, on the line joining (x i , Yi) and (X2, Y2), two points at distance at most 6 from each other at which the continuous function u takes the values ti, t2 respectively, and this would contradict the definition of Q(6). Taking y2 = Yi we deduce at once that F2 contains F1. Further, let y' be the upper bound of F1 . Then u(x i , y') = ti . Hence it follows that u(x2, y) > t2 if (Y

—

02 + (x2

—

xi)2

and in particular y E F2 if y' __ y __ y' + V[6 2 — (x2 — xi) 21.

Similarly, if y" is the lower bound of F1 , y E F2 if y” — .06 2 - (X2 - X1) 2 1 --- y

y".

Thus F2 contains the whole of F1 together with two intervals, each of length V[62—(x2—x1)2], which do not belong to Fi. Since /(xi, ti), i(x2, 1.2) are the lengths of F1 , F2 respectively, the lemma follows. Suppose now that Theorem 4.5 is false. Then we can find (X19 Y1), (.X2, y2 ) on E*, such that for some positive (5 (x2

—

x1)2 + (y2 - Y1) 2 < 62,

and

a

u* (x2, y2) < u * (xi, Yi) — n(6) _<_ b — Q(6).

Choose ti, t2 so that u*(x2, y2)
Then it follows from Lemma 4.3 that 1()C2,

t2)

i(Xl, 0+ 2V[6 2 - (X2 - X1) 21-

Also by the definition of u*(x, y) we have /(xi, ti) > 2 1Yil,

i(x2, t2) __ 21Y21,

119

Examples

since (x i , Yi) lies in Dt*i , but (x2, y2) does not lie in D72 . Thus I Y2 — Yi

IY21 — IYI >

and we have the contradiction which proves our theorem.

Examples

4.1

Prove Theorem 4.5 for circular symmetrization.

4.7 Symmetrization of condensers Consider a condenser satisfying the conditions of Theorem 4.3. Let w(z) be the potential function of Theorem 4.3. We symmetrize the function co(z), and write = 1z :

(z) =

(i = 0, 1).

By Theorem 4.5 of (z) is continuous and so Eo*, E i* are closed. It also follows that the complement of Eo* is the symmetrized set of the complement of E0, and that E i* is obtained by symmetrizing just as in the definitions 4.5.1 and 4.5.2 except that open set and open interval must be replaced by closed set and closed interval. The complementary set to Eo*, E i* consists of the open set D* = {z : 0 < co*(z) < 1 };

and this is again a domain. Suppose, for example that we are symmetrizing with respect to a half-line 0 = 0. Let ro, ri be the lower and upper bounds of r in D. Then every circle r = p (ro < p < ri) meets D* in a single symmetrical arc y(p) of one of the forms IBI < l(p), 101 l(p) <101 TE or in a pair of arcs y +(p), y_(p) given by /1(p) <101 < 1 2(P). One of the former cases certainly holds unless r = p meets both E0 and Ei, and so holds if p is sufficiently near ro or ri. It follows from the openness of D*, that if p is sufficiently near to po, y +(p) and Y+(p0) (or y(p) and y(p0)) can be joined by a straight line segment in D*, and the connectedness of D* follows. We can further show just as in 4.5.4 and 4.5.5 that D* is admissible, since D is admissible. We have thus obtained a new condenser, which is called the symmetrized condenser of the original condenser. We now prove the following result of Pellya and Szeg6 [1951] : Theorem 4.6 Suppose that a condenser C and its symmetrized condenser C* have capacities I, I* respectively. Then I* < I.

120

Symmetrization

By Theorem 4.5 co*(z) is continuous in the plane, and since co(z) is Lip on the set ba,b = {z : a < co(z) < b} if 0 < a
1 = I D [co(z)].

I D. [CO s (Z)],

To complete the proof of Theorem 4.6 it thus remains to prove that I D• [CO * (Z)1

(4.4)

D [COW].

This result represents the basic tool of the 1361ya—Szeg6 theory.

4.7.1 Symmetrization decreases the Dirichlet integral We shall prove the inequality (4.4) for circular symmetrization. The proof for Steiner symmetrization is similar and even a little simpler. We express the Dirichlet integral in polar coordinates: /D(CO)

2 pdp L p [(-F°(;) + 2 (-°4011c10. 0 2

=

Jo

P

Here Œp = 10 < co(peie) < 1} Consider the expression J(p) =

OW

2

[(Tr)) +

1 (Ow ) — p 2-

2

O.

If w is constant on every circle with centre the origin, then D must be an annulus r 1 < p < r2 and D* coincides with D and w(z) with w*(z). We ignore this case. Otherwise, since w(z) is harmonic in D, co(z) can be constant and different from 0,1 on the circle r = p only for isolated values of p,t which we may omit from the range of integration. If w(z) is not constant on r = p, co(z) cannot be constant in any interval of ep . For the end-points of such an interval would have to lie outside ep, and so w(z) would be 0 or 1 in the interval, contrary to the definition of ep . We thus assume that co(z) is not constant in any interval of ep , and so 0a)/00 = 0 only at isolated points in Œ. The values of co(z) at these points will be called stationary values of w(z). They can have at most 0, 1 as limit points. We can thus arrange the stationary values in order of magnitude as a sequence tm , where the lower bound of m is finite or —co Otherwise 0o,)/00 would be identically zero.

121

4.7 Symmetrization of condensers

and the upper bound of m is finite or d-oo. If 0 is the lower bound of w(z) on ep and 0 is not a limit point of the tm , we include 0 as the smallest tm ; similarly for the value 1. Let t, n , tm+i be successive stationary values. Then there will be n open intervals in ep in which w(p, 0) increases from tm to tm±i and n intervals where w decreases from tm± i to tm , where n is a positive integer, and these intervals occur alternately on the circle r = p, since , co is continuous on the circle. Let us denote these intervals in the order in which they occur on the circle r = p by Tm, v (1 < y < 2n). We suppose that w(p, 0) increases in Tmo, for odd y and decreases in T,n,v for even v. Also the totality of intervals Tm, ,for varying m, v make up ep , except for isolated points.

Thus 2n(m)

[ 0 co 2

f

J(p)=E m v= Tm, ,

(-) + ( L)) p p

00

2

] O.

We now change the variable of integration from 0 to t = co(p,O) in each interval Tm, v . Since Ow/00 0 in Tm ,v the relation between 0 in Tmo, and t is (1,1). We regard 0 as a function 0 = 0(t, p) of t, p. The partials of this function are given in terms of the original Ow/00, 0w/Op by the formulae

_ 00 at

0(t) 1/_

00,

=

sow aw 7 and dO Op / 00

01 9 dt Ot

Thus if Ov(t, p) denotes the value of 0 in Tm,,, such that w(p, 0) = t, we have tm+t 2n(m) f o v ia

(p) m

jtm

0O/ 0t

1

100v gt1

}

dt.

(4.5)

If tm < t < t m+ i, the set of intervals on r = p where w(p, 0) [021 (t) — 0 2v-1(01v=1

Clearly l(p, t) decreases strictly with increasing t in tm < t < tm+ 1. Thus (p, 0) satisfies w* (p, = t, where 4 = +Op, t) if tm < t < tm+1.

122

Symmetrization

Thus if J*(p) corresponds to of (z) as J(p) corresponds to co(z) we have

J*(p) =

E2 In

)2 + 1 t.,if(4/

tfm

1

} (4.6)

dt,

p2

where

0 = RO 2 — ) + + (02n —

0 2n-1)].

Thus

ao

t

1 2n

aot,

2 v=1 E( 1)v

at

2n

— 2 Z--d1

2n

ao ap

aev at

ae,

2 L-•

Thus the arithmetic—harmonic mean theorem gives

2

4

ao = 2n at 2.-w at 1

4 < — (2n)2

2n

1

aev at

Also Schwarz's inequality gives (E (2)2 < (E a 2 /b)(> lb) and so

2(4/4) 2 la0/atl

(

O/cp

lae v /atl

f la0v/aP1 2 1

z-a 1 lat9v/atI

Substituting these inequalities in the expressions (4.5) and (4.6) for J(p) and J*(p) we deduce r(P) J(P),

and hence on integrating with respect to p we deduce (4.4). This completes the proof of Theorem 4.6.

4.8 Green's function and the inner radius Suppose that D is a domain in the complex z plane, zo a point of D, and that there exists a function g[z, z o , D], continous in the closed plane and harmonic in D except at 41, vanishing outside D and such that

g [z, zo, D] + loglz — z o l remains harmonic at z = zo. Then g[z, z o, D] is called the (classical) Green's function of D. Clearly g[z, zo, D] is unique if it exists, for if g i (z) is another function with the same properties, then g — g i is harmonic in the whole of D, In estimating the integral / D . [of(z)] we may ignore the sets where of = tnt . For such a set either has zero area or else Ow* / Op,a(o• 100 vanish almost everywhere on it.

4.8 Green's function and the inner radius

123

continuous in D, and zero outside D, and so g — g i vanishes identically by the maximum principle. If D is admissible and bounded g[z, z0 , D] exists. For let h(z, zo) be harmonic in D and have boundary values loglz — 41 on the boundary of D. Then

g(z, zo) = h(z, zo) — loglz — zo I is the required Green's function in D. Suppose now that g(z,zo) exists. Then g > 0 in D, since otherwise g would have a minimum in D at a point other than z0 , and this is impossible. Since

g(z, z o )± log lz — zol remains harmonic at zo, the limit

-y = lim g(z, z o )± log lz — zol Z->Z0

exists. We write y . log ro

and call ro the inner radius of D at zo. To explain this terminology, suppose that D is simply connected and that

w = 1P(z) = z — ao ± b2(z — a0) 2 ± — maps D (1,1) conformally onto a disc lw 1 < ro, so that tp(a0) = 0, ipi(a0 ) = 1. Then ro g(z, ao ) = log IIP(z)1 is the Green's function of D at ao, and the inner radius of D at ao is ro. For g(z, ao ) is harmonic in D except at ao and as z approaches the boundary of D in any manner lip(z)I O. Also near ro and so g(z) z = ao we have

g(z, ao ) = log

ro

lz — aoll 1 + 0(01

= log

1 ± log ro + o(1), lz —a I

as required. We also note that if

z = f(w)= ao + ai w +...

124

Symmetrization

maps 1%4)1 < 1 onto D (1,1) conformally, then w= f

_1

(z) =

z — ao +... ai

near z = ao, so that a 1 f —1 (z) has properties of tp(z) in the above analysis. In particular, la1 1 is the inner radius of D at ao in this case. We remark specifically, however, that our definition of the inner radius applies also to multiply connected domains D, which cannot be mapped onto a disc, and even in the simply connected case we do not need to assume the existence of the mapping. We note that if the domains D,Di possess Green's functions g(z, zo), gi(z, zo), and if D c D1, then the difference

gi(z, zo) — g(z, zo) is harmonic in D and non-negative on the boundary of D. Thus the difference is non-negative in D, and so if r, r 1 are the inner radii of D, D1 at 41 , then log r i > log r and so r < r1 . Thus the inner radius increases with expanding domain. We accordingly define the inner radius ro at a point ao of an arbitrary domain D, containing ao, as the least upper bound of the inner radius at ao of all domains containing ao, contained in D, and possessing (classical) Green's functions. We shall have 0
4.8.1 The inner radius and conformal mapping

Our application of the

inner radius to function theory derives from

Theorem 4.7

Suppose that w = f(z) = ao + ai z +... is regular in 1z1 < 1 and takes there values w, which lie in a domain D having inner radius ro at ao. Then lall ro . Equality holds if f(z) maps 1z1 < 1 (1,1) conformally onto D. 1"

We discussed in the previous section the case when f(z) maps 1z1 < 1 (1,1) conformally onto D. In the general case take 0 < p < 1 and choose p so that f(z) * 0 on 1z1 = p. This can be achieved by increasing p slightly if necessary. Let C(p) be the curve which is the image of 1z1 = p by w = f (z). Thus C (p) is a closed analytic curve which may cross itself. The set of values w assumed by w = f (z) inside 1z1 < p forms a domain D(p) whose boundary consists of certain arcs of C(p). If two such arcs f This is, in fact, the only case of equality. See, for example, Hayman [1951] for the limiting case ao = 00.

4.8 Green's fUnction and the inner radius

125

touch at a point P, we add the interior of a small circle of centre P to D(p). In this way we construct a domain Do which contains D(p), is contained in D, and whose boundary consists of a finite number of analytic arcs no two of which touch each other. Thus Do satisfies the condition of admissibility of Theorem 4.2. Further, Do is bounded and so possesses a Green's function

gp (w, ao) = log

1 + log r(p)+ o(1) w — ao

near w = ao, where r(p) is the inner radius of Do at ao. Since Do lies in D, we have r(p) ro, Consider now the function h(z) = gp (f(z), a o ) + log P

Then h(z) is non-negative on 1z1 = p and harmonic in lz 1 p except possibly at points where f(z) = ao . At all such points, except possibly the origin, h(z) becomes positively infinite. At the origin we have h(z) =

=

log

1 f(z)— ao

1z1 + log r(p)+ log — + P

ow

r(p)

+ o(1). lailP Thus h(z) remains harmonic at the origin, unless a l = 0, in which case Theorem 4.7 is trivial, and log

r(p)

. lailP Thus h(z) > 0 in 1z1 < p, since otherwise h(z) would have a negative minimum somewhere in 1z1 < p, and at such a point h(z) would be harmonic, which is impossible. Thus h(0) = log

h(0) = log

r(p) >

lailPI r(P) P

When p —+

0

—

'

ro P

we have the inequality of Theorem 4.7.

4.8.2 The inner radius and symmetrization Theorem 4.7 becomes a powerful tool in the theory of functions, when combined with the following result of Nlya and Szegii [1951] :

126

Symmetrization

Theorem 4.8 Suppose that ao is a point of a domain D in the w plane and that D* is obtained from D by symmetrizing with respect to a line or half-line containing ao . Let ro , ro* be the inner radii of D, D* at ao . Then ro
1 W — C10

+ log ro + o(1)

near w = ao . We have -° g(ao + rei8 ) = ----1 + 0(1) (r -- 0). r Or Thus if K is sufficiently large, there is exactly one value of r for each 0 such that g(ao d-re'° ) = K, and the set of these points ao + relû forms an analytic Jordan curve yK surrounding ao . We define r = 1,i2 — ao l. On YK we have r = e—K [1 + o(1)]ro for large K. In other words, given e> 0, we have for sufficiently large K ro e—(K+ e) < r < roe—(K— e) on

(4.7)

7K .

Consider now the condenser formed by keeping yK and its interior at unit potential and the boundary of D at zero potential. The corresponding potential function is g(w)/K, and hence by Theorem 4.3 the capacity of the condenser is 1 f Og , -Ftias, -

17 71( f

where 0 < K'
ds = —

1 Og , ds = 2nr[— + OW ] — r fiw _ao l =r Cr

2n

(r -- oo).

Thus the capacity of our condenser is 27r/K. Consider next the capacity c(r) of the condenser formed on replacing YK by the circle 1w — aol = r. By Theorem 4.4 we have c(r) > 2ni< —1 or c(r) __ 27rK -1 according as 1w — aol < r includes or is included in the interior of t Jenkins [1955] has shown that when D is simply connected ro < ros unless D* coincides

with D or is obtained from D by rigid rotation.

4.9 The principle of symmetrization

127

yK • Thus we have from (4.7) for large K

271 c(roe—K— e) < — < c(roe—K+e ), K or ro 2n ro log — — e < < log — r r c(r)

e

for all sufficiently small r. Thus 1

1 ro = — log — c(r) 27r r

o(1) (r

0).

If we now symmetrize our condenser with respect to a line or half-line through ao, then the disc lw — a0 < r, being already symmetrical, is unaltered. The domain D is replaced by its symmetrized domain D*. If c*(r) denotes the capacity of the new condenser, we have by Theorem 4.6 1 c(r)

1 c*(r) .

On applying the above asymptotic expression for 1/c(r) we deduce ro log — r

o(1)

log — r°

o(1),

and this gives ro ro*. We have thus proved Theorem 4.8 when D is bounded and admissible. For a general domain D in the open plane we argue as follows. Let ro be the inner radius of D at c/o . Let D 1 be a domain in D, having a Green's function g(w, ao) and inner radius r 1 at ao, where r 1 < ro — e. Suppose that e is so chosen that Og .g u Ou Ov on the curves ye = {w : g(w) = e}, where w = u + iv. Let D2 be that domain in D i , bounded by some of the analytic Jordan curves y e, which contains ao. Then D2 is analytic and so admissible, the Green's function of D2 is g — e, and the inner radius, r2, of D2 at ao is r1 e. Thus if D*, D; are the symmetrized domains of D, D2 and ro*, r; their inner radii at 0/0 , we have ro* > r; > r2 > e'(ro — e). Since e is arbitrary, the inequality ro < ro* of Theorem 4.8 is proved.

4.9 The principle of symmetrization t We may combine Theorems 4.7 and 4.8 in the following result, which is in a form suitable for applications: For the remaining results in Sections 4.9 to 4.12 see Hayman [1950].

128

Symmetrization

Theorem 4.9

Suppose that w = f(z) = ao+ai z +... is regular in IzI < 1, and that D = D f is the domain of all values w assumed by w = f(z) at least once in IzI < 1. Suppose further that the symmetrized domain D* of D with respect to a line or ray through ao lies in a simply connected domain Do and that 0(Z) = ao + a'l z + a2r z 2w= + ... maps IzI < 1 (1,1) conformally onto Do. Then Iall

Idi I.

Let r, r* and ro be the inner radii of D, D* and Do at ao. Then we have

Icill < r _. r*

ro = Idil.

The first inequality follows from Theorem 4.7, as does the equation lai l = ro. Again we have r _.. r* by Theorem 4.8 and r* < ro, since, as was shown in Section 4.8, the inner radius increases with expanding domain. This proves Theorem 4.9. Using the work of Jenkins [1955] and Hayman [1951] quoted above, one can show that strict inequality holds in Theorem 4.9, unless D coincides with Do and D*, and f(z) with 0(ze ii- ). However, we shall not need to use this result. In Theorem 4.9 we may allow f (z) = ao + al z +... to vary over a class R of functions for which ao is fixed and D; remains inside Do. Then if (/)(z) G R, Theorem 4.9 shows that la'l l gives the exact upper bound for We shall give some examples of l ai I, when f(z) varies over the class this type of result, which may have independent interest, in the next two sections. Frequently the extremal functions 4)(z) give also the exact upper bounds for M(r, f) and M(r, f') when f(z) G R, and in the last section of the chapter we shall give a proof of this in a typical case, which is important for Chapter 5.

a.

4.10 Applications of Steiner symmetrization We suppose that the hypotheses of Theorem 4.9 are satisfied and that ao is real, and we symmetrize with respect to the real axis o = 0 in the w = u + iv plane. Let 0(u) be the total measure of the intervals in which Di- meets the line u = constant. Then D* is the domain {w:v<

0(u), —co < u < +co}.

As we saw in §4.5.4, D* is necessarily simply connected in this case, and so we might choose Do = D* in Theorem 4.9, unless D* is the whole

4.10 Applications of Steiner symmetrization

129

plane. For by the Riemann mapping theorem (C. A. p. 230) any other

simply connected domain may be mapped (1,1) conformally onto 1z1 < 1 by a function 45(z) = ao + di z +..., where ao is an assigned point of the domain.

We shall not appeal to this general result, but simply give two examples of the method where the extremal function 4)(z) can easily be calculated explicity so that numerical inequalities result. Evidently many other examples could be given. Theorem 4.10t Suppose that f(z) = ao + ai z + ... is regular in 1z1 < 1 and D1 intersects each line u = constant in the w = u + iv plane in a set of intervals of total length at most I. Then

lad

21 — 7r •

Equality holds for f(z) = ao + log{(1 + z)/(1 (1,1) conformally onto the strip Iv 3a0 1 < —

P.

—

z)}, which maps Izi < 1

We may, without loss of generality, suppose ao real by subtracting an imaginary constant if necessary. We then symmetrize with respect to the real axis and note that Df* lies in the strip 1,1 < l. Since

1 21 1+z 4)(z) = ao + — log = ao + — z + . . . 27r it 1— z maps 1z1 < 1 (1,1) conformally onto this strip, the inequality lai I _._ 21/7r follows from Theorem 4.9. This inequality is sharp, since 4)(z) satisfies the hypotheses for f (z). As another example we prove Theorem 4.11 Suppose that f(z) = ao + ai z + ... is regular in 1z1 < 1 and that Df meets the imaginary axis in the w plane in a set of intervals of total length at most 1. Then

lall ...ç.. [4001 2 + /2 ] . Equality holds if ao is real and f(z) maps 1z1 < 1 (1,1) conformally onto the domain Do consisting of the w plane cut from —1il to —ioo and from l2 il to +ioo along the imaginary axis. t For univalent f(z) this an unpublished result of Rogosinski. For the general case see Hayman [1950, 1951].

130

Symmetrization

Suppose first that ao is real. We may then symmetrize with respect to the real axis and then Et; lies in Do. Thus if

4)(z) = ao ± di z + ... maps 1z1 < 1 onto Do we shall have la l 1
1 maps 1z1 < 1 onto the closed C plane cut along the segment (-2, 2) of the real axis. Thus il W =

-,

maps 1z1 < 1 onto Do. We now choose r so that —1
zi + r 1 + rzi . Thus we obtain a more general map of 1z1 1 <1 onto Do, given by iz =

W =

/(z i + r)(1 + rz i ) (1 — 4)(1 — r 2 )

1(1 + r2 ) (1 — r 2 ) + (1 _ r 2) zl +....

Ir

We may choose r so that 101 — r 2 ) = ao. Then , io + r2) ai = = 044 + 12 ),

and Theorem 4.11 follows for real ao. If, finally, ao = a + ifi, we consider f(z)—i13 instead of f(z) and deduce

la1 I

04Œ2 + 12 )

04 1c/o1 2 + 12 ).

This completes the proof.

4.11 Applications of circular symmetrization Circular symmetrization is more powerful than Steiner symmetrization, and any result obtainable by the latter method can also be obtained by the former on taking exponentials, though this may be less direct. When we use circular symmetrization the domain D; is not necessarily simply connected. On the other hand, D; cannot reduce to the whole plane unless Df does, whereas in Steiner symmetrization D; consists of the whole plane as soon as Df meets every line perpendicular to the line of symmetrization in a set of intervals of infinite total length. We proceed to give some examples

4.11 Applications of circular symmetrization

Theorem 4.12

131

Suppose that 0 < a < 2, that f (z) = ao + al z +

is regular in 1z1 < 1 and that Df meets each circle 1w1 = p (0

We may suppose that ao is real and positive, since this may be achieved by considering ei1f(z) instead of f(z). With this assumption we symmetrize Df with respect to the positive real axis. Then I); lies in the domain Do = lw : argw < out , 0 <114,1
w = 4)(z) = ao ( 1 + z ) 1 = ao(1 2az ...) maps 1z1 < 1 onto Do and now Theorem 4.12 follows from Theorem 4.9. Our key result for further applications is

Theorem 4.13 Let f(z) = ao + ai z + ... be regular in 1z1 < 1, and let R = Rf be the least upper bound, supposed finite, of all p for which Df contains the whole circle 1w1 = p. If there are no such p, we put R = 0. Then R). Equality holds when ao > 0 and f (z) = ao +

4(ao R)z (1 — z) 2

which maps Izi < 1 onto Do consisting of the w plane cut from —co to along the negative real axis.

—

R

We again suppose that ao > 0 and symmetrize with respect to the positive real axis. Then, since D1 does not contain the whole circle --= p for p> R, if; does not contain the point w = —p for p> R and so D* lies in Do. Since ç1(z) = ao +

4(ao R)z = ao 4(ao (1 — z) 2

R)z

evidently maps 1z1 < 1 onto Do (see §1.1), Theorem 4.13 follows. Various special cases of Theorem 4.13 are of interest. Thus if f (z) = z + a2 z2

(1z1 < 1),

we may take ao = 0, al = 1 and obtain R > 14 . This is a sharp form of

132

Symmetrization

a theorem of Landau [1922]. The result also includes Theorem 1.2 of Koebe—Bieberbach. In fact if f (z) is, in addition, univalent, Df is simply connected and so its complement is connected. Thus if this complement contains a point wo, it must meet every circle lw I = P for P > lwol, and so Iwo l > R > Thus Df contains the disc lw 1 < Again if R = 0, we obtain la l 1 < 41ao l, a limiting case of Theorem

4.12. It is easy to obtain inequalities when in addition to the other hypotheses f (z) is bounded above. As an example we state Theorem 4.14 Suppose that f(z) satisfies the hypotheses of Theorem 4.13 and that in addition If(z)1 <M in 1z1 < 1. Then 4(M — laol)(M 2 + lao1R)(1a01+ R) (M +R)2 (M + laol) Equality holds when ao > 0, and f(z) maps 1z1 < 1 (1,1) conformally onto Do, consisting of 1w1 < M, cut from —M to —R along the negative real axis.

The proof is left to the reader (see Example 4.2 at the end of the chapter). Clearly Df* lies in Do, and the determination of the map of 1z1 < 1 onto Do, which gives the extremal value for lall, is an elementary exercise in conformal mapping. Finally, we obtain the following corollary of Theorem 4.13:

Theorem 4.15 Suppose that w = f(z) = ao +ai z+... is regular inIzI< 1, that 0 < 1
e2na0ll 1 ± 2na 1

/

z

+...).

Then g(z) is regular in 1z1 < 1, and since 2 log g(z) * — r (u+ iv + in!) (—cc < n < +co), /7 27c g(z) * exp [--1—(u + iv)] .

4.12 Bounds for if (z)1 and if (z)1

133

Since u is arbitrary subject to —co < u < +co, we can for every positive p find a number w, such that lw 1 = p and g(z) # w in 1z1 < 1. Thus we can apply Theorem 4.13 to g(z) with R = Rg = 0, and obtain

2nai <4. 1 Equality holds for real ao, when

f (z) = ao + —1 log ( 1 + z ) which maps lz I < 1 onto the strip iv I < 1/. This proves Theorem 4.15. A comparison of Theorems 4.10 and 4.15 illustrates the relative power of Steiner and circular symmetrization. It may be seen that Theorem 4.15, while containing Theorem 4.10 as a special case, is significantly stronger. For, under the hypotheses of Theorem 4.15, D1 may well meet each line u = constant in a set of intervals of infinite length, provided that a single sequence of points in arithmetic progression of common difference / is on the line and outside Df . The results also suggest that the hypotheses of Theorem 4.15 might be further weakened to the assumption that D1 meets no line u = constant in any interval of length greater than /. Dubinin [1993] has proved that the inequality la i l < 21/it still holds under this assumption.

4.12 Bounds for If(z)1 and If(z)1

The functions which give the maximum for !al l under the hypotheses of Theorems 4.10-4.15 also give the maximum values for M(r,f) and M(r,f') for 0 < r < 1, and we can even prove in some cases that they are essentially the only functions that do so. The proofs are similar and we confine ourselves to the case of Theorem 4.13, which has most applications.

Theorem 4.16

Suppose that f(z) satisfies the hypotheses of Theorem 4.13. Then we have, for 1z1= r (0< r <1),

if (z)i < laol + it (z)1-

4(laol + R)r (1 — 02 '

(4.8)

4(R + I f(z)l) < 4(001 + R)(1 + r) 1 — r2 (1 — 03 '

unless

4(Re i2 + ao )ze —i° (1 — zeif 9 ) 2 where 0, A are real and arg ao = 2 or ao = 0.

f (z) = ao +

,

(4.9)

134

Symmetrization

We suppose that lz I < 1 and consider d)(z) = f

( zo + z ) = f (zo) + (1 — Izo1 2 )f ' (zo)z + . . . , 1 + 2oz

instead of f(z). Then Do = Df, since 0 is obtained from f by a map of the unit disc onto itself, and so we can apply Theorem 4.13 to 0(z) with the same value of R. This gives

( 1 — Izo1 2 )1f / (zo)1

4(R + If (zo)1),

and, dropping the suffix, we obtain the left inequality of (4.9). We next write zo = rel9 , f (z o ) = T(r)e"i-(r) where 0 is fixed. Then el° f(rei° ) = ei'l(r) [T'(r) + i).'(r)T(r)]. Thus we have

4(R + T(r)) (4.10) 1 — r2 ' and equality is possible only if AV) --= 0 or T(r) = 0. We may write (4.10) T'(r) -_

as

dlog[R + T(r)] _. d [2 log (

r )] .

Thus

ip(r) =

(1 + r ) 2 [R + T(r)]

(4.11)

cannot increase with increasing r. If Ip(ri) = tp(r2), where 0 ._ ri
1,(r) = 1 = constant. This gives e

'(z) =

4[Rei'l + f(z)] 1 — (ze - w) 2

for z = reie (ri < r < r2), and hence by analytic continuation throughout 1z1 < 1. Thus

4e-i° d — log[Re + f(z)] = 1 — (ze - i9 )2 ' dz Re i1 + f(z) Rei2 + ao

(1 + ze-w ) 2 1 — ze - i°

'

4.12 Bounds for If(z) I and Inz)1

135

and since arg f (rei9 ) 2, we must have arg a0 = 2 unless a0 = O. Thus in this case f (z) reduces to one of the extremals given in Theorem 4.16. They clearly satisfy the hypotheses of that theorem and give equality in the inequalities (4.8) and (4.9), when z = rei° . In all other cases the function ip(r) in (4.11) decreases strictly with increasing r for 0 < r < 1, and in particular

WO < (PM = R + laol, i.e. If (re"9 )1 < (R + la01)

(1+ r 1— r

) 2

_R

'

This proves (4.8), and by substitution we obtain the second inequality of (4.9). This completes the proof of Theorem 4.16. Two special cases of Theorem 4.16 are worth noting. If a0 = 0, then (4.8) gives M(r, f) <

4Rr (1 — 0 2.

We thus obtain the following sharp form of a theorem of Bohr [1923]: If f(z) is regular in 1z1 < 1, f(0) = 0, and M(r,f) = 1 for some r such that 0 < r <1, then Df includes a circle 1w1 = p, where (1 — 0 2 P > 4r '

except when f (z) =

(1-0 2 ze i'l r (1 — ze — i9 )2.

For, except in this case, (4.8) yields

R>

(1 —r2 4r

and we may choose p so that

M(r, f) =

(1 — 0 2 4r '

,--- '(1 — r) 2 < p < R, and lw I = p lies in

Df . We also collect together the results when R = 0: Theorem 4.17 Suppose that f(z) = ao+aiz +... is regular in 1z1 < 1 and that for each p (0 < p < co) we can find w = pe149(P) such that f(z) * w in 1z1 < 1. Then unless f(z)= ao[(1±ze 1e)/(1— zei°)] 2 for some real 0 we

136

Symmetrization

have for Izl= r (0 < r < 1) (1 +r y ( 1 — rV laol 1 + r ) < If(z)1 < laol 1 — r ) '

(4.12)

It(z)1 __ 1 —4 r2, Iff(z) < 41a& (11 —+ rr)3'

(4.13)

and {(1—r)/(1+r)} 2 M(r, f) decreases strictly with increasing r (0 < r < 1).

In this case we take R = 0 in Theorem 4.16, and (4.8) and (4.9) give the right-hand inequality of (4.12) and (4.13). Also [f(z)] -1 satisfies the same hypotheses as f(z), and we obtain the left-hand inequality in (4.12) by applying (4.8) to [f (z)] -1 instead of f(z). To prove the last statement of Theorem 4.17, suppose that 0 __ r 1
1 + r2

M(r2,f)

= <

(1—r2 ) 2

1+ r2

io If(r2e )1

(1 —riy (1 —r i )2 M(r, f). If(rie io )1 __ 1 + ri 1 + ri

This completes the proof of Theorem 4.17.

4.13 Bloch's Theorem

Although rather on the boundary of the area of multivalent functions the following proof of Bloch's Theorem recently obtained by Bonk [1990] is too elegant to be missed out.

Theorem 4.18

Let

f (z) = z + a2 z2 + ...

(4.14)

be regular in 1z1 < 1. Then f(z) maps some subdomain Ao of I z I < (1,1) conformally onto a disc of radius B, where B > = .433...

1

The result with some absolute constant B is due to Bloch [1926]. The bound B > 4 .,./3 is due to Ahlfors [1938]. This was sharpened by Heins [1962] to B > 14- \/3. Bonk [1990] proved that B > . .\13 ± 10-14 . In the opposite direction Ahlfors and Grunsky [1937] showed that B __ .472 and their upper bound is widely conjectured to give the correct value for t-

137

4.13 Bloch's Theorem

\13. The proof that B. We follow Bonk's argument to show that B B > 1.\/3 +10-14 lies rather deeper and we omit it. Lemma 4.4 Let T;0 be the class of functions f(z) regular in 1z1 < 1, having an expansion (4.14), and satisfying

(1— lz1 2 )1.0z)1_ 1,

(4.15)

1z1< 1.

Then if f(z) E T40 1

1—z

for iz1

1 —1z1/0)3('

93{f'(z)1 and w = f(z) maps 1z1 <

1 -7-i-,

(4.16)

onto a domain D, containing the disc

lwl < 14.j

We define 11—w F(w) = 0 1 — wo and

9 1 G(w) = - w(1 — - w)2. 4 3

Then 1G(w)1( 1 — 1F(w)1 2 )= 1, when 1w1= 1.

(4.17)

In fact if w = e 0 , we have

3 ( 1 — w) (1 — iv) 2 1F(w)1 = (3 — w) (3 — 17v)

6(1 — cos 0) 10 — 6 cos 0

3 — 3 cos 0 5 — 3 cos 0'

so that 2 1 — 1F(w)1 2 = 5— 3 cos 0 ' and

9 — 11-) ) 1G(w)1= -4 (1 — -1 ;-) (1 \ 3

=

1

- ( 5 — 3 cos 0). 2

This proves (4.17).

We now consider h(z) = I f'(F(z)) 1 G(z)

Since f(z)

E Zo,

we have

1}

z

(z —1) 2.

(4.18)

138

Symmetrization

If f"(0) = pe

and let t tend to zero. This yields

we set z = te

1

pt 0(t 2 )

1 + t2 + 0(t4 ),

and we deduce that p = 0, i.e f"(0) = O. 1 and, by (4.17), The functions F(w), G(w) are regular for 1, IF(w)I < 1 there. Hence h(z), given by (4.18), is regular for 1z1 except for possible poles at z = 0,1. The zero of G(z) at z = 0 is cancelled by the factor z. To study the behaviour of h(z) at z = 1, we put z = 1 -I- i and see that, for small F(z) =

—1/ N/3

2—

, f' {F(z)} = 1 + 0 {F(z)} 2 = 1

since f"(0) = 0, and

G(z) = (1 + ri)(2 — ri)2 = 1 + 0( 1 2 ). Thus h(z) remains bounded and so regular in 1z1 (4.17) we have for z = , 0 < 0 < 2n

f'(F(z)) G(z)

-

1. Also by (4.15) and

1 =1 ( 1 - I F(z ) 1 2 )IG( z) I

while

1 = -- cosec 20 <0. 2 (z — 1)2 4 -

Thus

91h(z)

0.

By continuity this remains true for z = 1, and so for 1z1 < 1, by the maximum principle. In particular we obtain for 0 < w < 1

(F(w)) >_ 1; 9if' {F(w)} G(w)

G(w), since G(w) > O.

Setting z = F(w), we see that z decreases from -?/-3- to 0, as w increases from 0 to 1, while w=

1 — z .\/3 1 — z/ \/3'

G(w)

1 — z \/3 (1 — z/ \/3) 3.

This yields (4.16) when z > 0. If z = teiŒ, we apply the conclusion to f (z el') instead of f (z). This completes the proof of (4.16).

4.13 Bloch's Theorem

139

We deduce that, with the hypotheses of Lemma 4.4, f(z) is univalent 1/0. For if zi, z2 are two distinct points in 1z1 1/0, write in 1z1 = z1 + pel'. The segment

0< t
T : z = + te,

lies in Izl< 1/V3, and so by (4.16) 9ir(z) > 0 in T. Thus

Z2 (z2)

-

f (zi)} =

fz i

f'(z)dz

f t(Z1+ tei')dt > 0.

=

(4.19)

Thus f(z) maps 1z1 < 1/V3 (1,1) conformally onto a domain D. The boundary OD of D is the image by f(z) of Izi = 1/V3. w = f ((1/ V3)el is a point of this image, (4.16) and (4.19) yield

1w1

w = f

.1/3

f(te `Œ)dt >

f \/3

(1 — tV3)dt

0 (1 -

0

t/.0)3

If

\/3 4

Thus D contains the disc lwl < and Lemma 4.4 is proved and so is Theorem 4.18 if f(z) E 00Suppose next that f(z) is not in 00. Then there exists zo, such that

(1-141 2 )1.04)i > 1 somewhere in 1z1 < 1. We choose p such that 0 < p < 1, and

(1 — 1zo1 2 )Plf(Pzo)1 > 1. This is possible by continuity. We now define tp(z) = f(pz) and

1 . (4.20)

p(z)= ( 1 — 1z1 2 )1V(z)1= ( 1 — 1z1 2 )P 1.r(Pz)1, 1z1

Then it(z) = 0 for 1z1 = 1, since p < 1, and so p(z) attains its maximum value po in 1z1_-_ 1 at z i where Izil < 1. Also po p(zo) > 1. We suppose that arg ip/(z i ) = a and consider

+

0(z) =

(-±

z

1)} •

We note that (1)(z) is regular in 1z1< 1 and 4)(0) = 0,

0'(0) =

e-12 ( 1 — 1z11 2 )17P'(zi) PO

1.

140

Symmetrization

Also we have for 1z1 < 1 (

1

-

1z 1 2 )10z)1 =

(1— 1z1 2 ) PO

+ z ) 1 — iz 1 1 2

1

Y)

2 =

1

1 + f1Z

ip

1

11 + f IZ1 2

-

+ Z1Z

(1+

,

( Z1 +

1 + fiZ

<1.

by (4.20). Thus 4)(z) G 23.0, and so, by Lemma 4.4, 0(z) maps 1z1 < 1[0 onto a domain containing the disc 11471 < .\/3/4. Hence tp(z) maps some subdomain of 1z1 < 1 onto a disc of radius po ,./3/4, and so f(z) maps some subdomain of iz1 < p onto a disc of radius po V3/4 > \/3/4. This completes the proof of Theorem 4.18.

4.14 Some other results We close the chapter by mentioning some further developments. Baernstein [1975] has proved that, with the hypotheses of Theorem 4.9, we have further I Ar, f)

I ( r, 0),

0

r < 1,

0 < 2. < oo

(4.21)

where

1/). 1 f 2ir , if I;.(r, f) = { 7, 0 If(rel 0 )r} and 1(r, f) = M(r, f). If f(z) E S he proved that I(r, f) -_ 1,1 (r, k), where k(z) is the Koebe function (1.1). Next Weitsman [1986] proved that Theorem 4.9 extends to arbitrary domains D, whose complement in the extended plane contains at least 3 points. If D* = Do and D* is multiply connected, 4)(z) maps 1z1 < 1 onto the universal covering surface over Do . It is not known whether the analogue of (4.21) holds in this more general case. The case when the complement of D* consists of the points (0, —1, cc) is of particular interest, since it leads to sharp forms of the theorems of Landau [1904] and Schottky [1904]. This special case had previously been dealt with by Hempel [1979] and Lai [1979]. For proofs of the above results the interested reader is referred to the original papers. A connected account is also given in Hayman [1989, Chapter 9].

141

Examples

Examples If 4.2

(1+W) 2 1-1,1, )

( 1 — r) 2 1 + r)

K 7 1+Z 2 1— z ) '

show that the C plane cut along the negative real axis corresponds (1,1) conformally to lz I < 1, and to lwl < 1 cut from —1 to —r along the real axis. Hence complete the proof of Theorem 4.14, first when M = 1, and then generally by considering f(z)/M, RIM, 1 instead of f(z), R, M respectively. The following examples are generalisations of Koebe's Theorem 1.2.

Examples

4.3

4.4

We denote by Wi the class of functions meromorphic in 1z1 < 1 and having s an expression (4.14) near z = O. Show that if f G 9J/ and a, b are positive numbers such that 1/a — 1 /b > 4, then D1 contains a circle 114,1 = r, such that a < r < b, (Hayman [1951]). Show by an example that this result fails whenever 1/a — 1/b = 4. (Assume the result is false and apply Theorem 4.9.) If w = f(z) E 9j1 and 0 < r < 1, show that D1 meets lwl = r in a set of linear measure at least 2rŒ, where r = cos (a/2), and for each r give an example for which this lower bound is attained. (Show that the map

z=

4.5

11— weli — (1 — w e —i')1 (e--11 — w)1 — (eil — w)

maps the complement in the closed w plane of the arc w = e 0 , a < 101 7E, onto 1z1 < 1.) If W = f(z) G Z where S is the class of Chapter 1, prove that, for 1 < r < 1, D = Df meets 1w1 = r in a set of linear measure at least 2rŒ where r = cos4(a/4). (If the conclusion is false, the complement of D* contains the ray —oo, —r of the real axis as well as the arc a < i arg z1 < it of lw 1 = r. To find the extremal map, set z Z =4k (1 +Z)2 ' (1 + z) 2 where k=

cos 3 (1 + cos i)2'

142

4.6

Symmetrization and W = rw,w , = f(z), is the map of Example 4.4. (Netanyahu [1969] has shown that a function of this type yields the extremal value 1a2 1 d = 2/3 where d is the radius of the largest disc with centre 0 contained in the image of lz1 < 1). If w = f(z) E Ti and f(z) is regular in lz1 < 1, show that for 0 < r < 1, Di meets 1w1 = r in a set of linear measure at least 2ra, where a is the (unique) solution, in 0 < a < 7r, of

r=

sin2 -1 2 log (sec -'21 ) '

This example needs the full strength of Weitsman's Theorem. The symmetrized domain D* lies in the domain Do, whose complement in the closed plane consists of the arc w = re'9 , a < WI < it and the point at oc. Thus, by Weitsman's Theorem the extremal map is w = ip(z) = r A i z+..., where tp(z) maps lz1 < 1 onto the universal covering surface over this doubly connected domain D o . In the map w = f(z) of Example 4.4, 1z1 < 1, punctured at z = c, corresponds to Do (1,1) conformally, and z = c = cos(a/2) is mapped onto w = cc. We now set

Z =

1 log z i + K c—z , K = log — . where z 1 = c log z i — K ' 1— cz

Then 1Z1 < 1 is mapped onto the universal covering surface of the unit disc 1z1 < 1 punctured at z = c. Thus the combined map w = ip(Z) = f(z(Z)) gives the required extremal map with tp(0) = 0,

Ilp'(0)1

r 2 log 1/c . 2r log 1/c = c (1/c— c) 1— c2 .

If Ip(z) E 9N, then, as required, 1 ____ e2

r=

2 log 1/c •

The function 1 — C2 2 log 1/c

increases monotonically from 0 to 1 as c increases from 0 to 1 and so as Πdecreases from it to O. Thus the equation

r=

sin2 (a/2) 2 log sec(a/2)

has exactly one solution in the range 0 < cc < it.

143

Examples

We denote by 93 the class of functions regular in 1z1 < 1 and such that

/20 = sup(1 —1z1 2 )1f(z)1 < oo.

(4.22)

Izi <1

The functions in 93 are called Bloch functions [Hayman 1952] and have been widely studied. (See e.g. Pommerenke 1975, p. 269 et seq.)

Examples

4.7

4.8

If f(z) E 93 show that f(z) maps some subdomain of 1z1 < 1 (1, 1) conformally onto a disc of radius R, provided that R < Po B, where B is Bloch's constant, i.e. the quantity of Theorem 4.18. Suppose that f(z) is regular in 1z1 < 1 and that Ro is the upper bound of the radii R of discs lw — w0 1 < R onto which some subdomain of 1z1 < 1 is mapped by w = f(z). Then Ro is finite if and only if tro is finite and

po B

4.9

Ro

(To obtain the right-hand inequality consider the map z = f-1 (w) from a disc lw — w01 < R into 1z1 < 1.) 00 1. If f(z) =V a n zn E 93, show that Ian I __ letto, n (Prove that the exact bound for lan l is

4.10

(4.23)

tha.

,u0A n , where Ai =

2, log A n = 3 log ( 11+ /nD + log (1 — ,n 2).) (Landau's Theorem). If f (z) = ao+aiz +... is regular in 1z1 < 1, and f (z) * 0,1 there, we define 1 g(z) = -27ri log f(z), h(z) = g(z)+ .\/(g(z) 2 — 1), b(z) = log h(z).

Show that, with a suitable choice of the square roots and logarithms at z = 0, the functions g(z), h(z) and b(z) are regular in 1z1 < 1, and b(z) ±logIn + .\/(n2 — 1)1 + 2mni

for any positive integer n, or integer m. Deduce that b(z) E 93 and that R0 < 2.0. Hence show that lai I < 8001{110g 1 ao l 1 + 270. (For the best result lail < 21 ao I { I log lao 1 1 + 4.37...} see Hempel [1979] and Lai [1979] or Hayman [1989, p. 702].)

5 Circumferentially mean p—valent functions

5.0 Introduction We consider again functions f (z) regular and not constant in a domain A, define n(w) = n(w, A, f) as the number of roots of the equation f (z) = w in A, and write as in (2.4) 1 f2n

p(R) = p(R, A, f) = y7-c 0 n(Re)d4. In Chapters 2 and 3 we considered functions satisfying the condition

L

R

p(p)d(p2 ) __ pR 2 (R > 0),

where p is a positive number. In what follows we shall call such functions areally mean p-valent (a.m.p-valent). We shall consider in the first part of this chapter some consequences of the more restrictive hypothesis

p(R) p (R > 0). (5.1) Functions satisfying this condition were introduced by Biernacki [1946] and are generally called circumferentially mean p-valent (c.m.p-valent). We first prove some sharp inequalities restricting the growth of functions c.m.p-valent in 1z1 < 1 and such that either f (z) 0, or f (z) has a zero of order p at z = 0, where p is a positive integer, basing ourselves on Theorems 4.13 and 4.17 of the last chapter. In the second part of the chapter we prove some regularity theorems for a.m.p-valent functions in 1z1 < 1. We proved in Theorem 2.10 that if 00 f (z) = V an ? is such a function, then a = lim (1 — r)2P M(r, f) r — ■ 1—

144

5.1 Functions without zeros

145

exists. We shall prove that, if further p> j4 , then also

a

iim lanl n2P -1

A generalization to positive powers of mean p—valent functions and funcak,+,,z k n' will also be proved. As an application tions of the type z

dn=o

—

of these results, we shall prove that if 1(z) = z + a2 z 2 + • •

is a.m. 1—valent in 1z1 < 1, then la rd

5.1 Functions without zeros

n for n > no(f) (cf. Example 2.12).

In this section we prove

Theorem 5.1 Suppose that f(z) = ao + alz + • • is c.m.p—valent and f(z) 0 in 1z1 < 1. Then a1 ^4pao.

Further, unless f(z) = ao [(1+ ze ie )/(1— ze i9 )] 2P for a real 0, we have for p (0 < p <1) ( 1 P ) 2P < 1+ p

<

( 1 + P ) 2P 1—p

and

It(z)1

4p 4 1aolP( 1 P) 2-1 /3 p21f(z)1 < (1 _ p )2p+i 1

Also [(1 — p)/ (1 + p)] 2P M(p, f) decreases strictly with increasing p (0 < p < 1), and so approaches a limit ao as p 1 where cxo < laol.

We need two lemmas: Lemma 5.1 Suppose that ri > 0, that f(z) is regular in a domain A and that tp(z) = f(z) is single valued there. (a) If f(z) is c.m.p—valent in A, then tp(z) is c.m.(rip)—valent in A. (b) If f(z) is a.m.p—valent in A, then tp(z) is a.m. (lop)—valent in A, where rio = max(ri, ri 2 ). Hayman [1950, 1951, 1955]

146

Circumferentially mean p—valent functions

The result is sharp (see Example 5.1 at the end of the chapter). We write ip(z) = W = Re i4), f(z) = w = ITO, so that W = wq. Thus we may write R = rq, (1) = If on passing along an arc of a level curve If(z)1 = r = constant, 0 = arg f(z) increases strictly by an amount 60, then this level curve contributes A-60 to p(r, A, f). Also the corresponding contribution to p(R, A, tp) where R = rq, is fl-,(50. We call 60 the variation of arg f (z) on the arc. We may express the set of level curves If(z)1 = r in A as the union of a finite or ennumerable set of Jordan arcs, no two of which have more than end-points in common and in each of which arg f(z) varies monotonically. Then 27rp(r, A, f) is the sum of the variations of arg f(z) on these Jordan arcs, and so we obtain by addition

(5.2)

f).

10,A, TP) =

Now (a) follows from (5.1). To prove (b) we write

p(r) = p(r, A, f), P (R) = p(R, A, tp), w(r)= f r p(p)d(p2 ), W(R)= f P(s)d(s 2 ). Thus (5.2) yields P(s)= rip(sl iq) and

W(R) =

Ro R fo r ip(s1 lq)2sds = 2172 foi A top2r1-1 dp

2 R?7 p 2q-2 dw (p) f = 112 [p2i/ W(p)] oRi/n

2(217

Since f is a.m.p—valent we have 0 1 we obtain

2)

w(p)

fo

R 111

p2q-3 w(p)dp.

(5.3)

pp, 0 < p
R I in

W(R) pri 2 R 2 ri2 (2 —2q) f pp 2'-1 dp = pqR2 , so that (b) holds in this case. If ij > 1 we obtain at any rate

W(R) pri2R2, since the integral on the right-hand side of (5.3) is nonnegative. This completes the proof of Lemma 5.1.

5.1 Functions without zeros

147

Lemma 5.2 If f(z) is c.m. 1—valent in 1z1 < 1, then there exists a number 1 = If > 0, with the following property. If 1w1 < I then the equation f(z) = w has exactly one root in 1z1 < 1, while if R 1, we can find wR in 1z1 < 1. wR = Re 14 ), such that f(z) Let n(w) be the number of roots of the equation f(z) = w in 1z1 < 1. We note that the sets {w : n(w) K } in the w plane are open for every finite positive K. In fact if n(wo) > K, then there exist a finite number of distinct points z1,z2,...,z q in 1z1 < 1 at which f (z v ) = wo with multiplicity K v > K. We can then, given a small positive 6, choose e K, and v=i so small that for 0 < w — w0 < e the equation f (z) = w has exactly Kv roots in lz — < 6.f If 6 is sufficiently small all these roots are distinct and so

E

n(w)

Kv K

if lw — w0 <

E.

v=1

Thus the set where n(w) > K is open. It follows that if n(Re*) > 1, then n(Re 10 )> 2 in some range 4) - 4)o < e. Since also 1

2n

f 21 n(Re ki)d0

1,

by hypothesis, we deduce that n(Re 10 ) < 1, i.e. n(Re) = 0, for some other value of 4). Thus if n(Re) > 1 (0 < < 2n), then n(Re 10 ) = 1 in this range. In this case the circle 1w1 = R corresponds (1,1) continuously to a set y in 1z1 < 1, which is therefore a simple closed curve. Take jwo l < R. Then since If (z)1 = R on y, it follows from Rouché's Theoremt that the functions f (z) and f(z) — wo have equally many zeros inside y, No say. Since f (z) is not constant by the introductory hypotheses of this chapter, f(z) assumes some values w such that lw < R inside y, and so No 1. Thus n(w) No (wl< R), and so n(w) = 1 for 114, 1 < R. Let now 1 = lf be the least upper bound of all R, if any, which satisfy the above hypotheses. If there are no such R, we put 1 = O. Then n(w) = 1 for 1w1 < 1. This shows that 1 < +co, since otherwise the inverse function z = f -1 (w) would provide a map of the whole w plane onto 1z1 < 1, which contradicts Liouville's Theorem. Also for R > 1, we can find 4) such that n(Re 10 ) = O. This remains true for R = 1. In fact the set where n(w) > 1 is open, as we saw above, and so if it includes the circle t C. A. Theorem 11, p. 131. C. A., p.153

148

Circumferentially mean p—valent functions

1w1 = 1 it will also include 1w1 = 1+E for sufficiently small the definition of L This proves Lemma 5.2. We can now prove Theorem 5.1. We write

E,

contradicting

al z + • .1. ty(z) = Lf(z)1 14 = a,VP [1+ — Pao

0 in 1z1 < 1, we can choose a singleSince f(z) is regular and f(z) valued branch of tp(z), which is therefore c.m. 1—valent by Lemma 5.1. Also tp(z) 0 in z1 < 1, and so I = = 0 in Lemma 5.2. Thus for R > 0 we can find w = wR such that I wR I = R and ip(z) wR in Izl< 1. We may therefore apply Theorem 4.17 to p(z). (4.13) gives 4 tp'(z) tp(z) — 1 — lz1 2 ' and, putting z = 0, we deduce

4pla0l. Also unless

up (1 + ze ie ) 2 tp(z) = ao 1 — zee we deduce from (4.12) 1 p )2 11 a0 14 (

1

p

<1111 (z)i <

(1+ ) 22 P

= 14,

p)/(1 + p)]2 M(pop) decreases strictly with and for 0 < p < 1, [(1 increasing p. Writing f(z) = [1p(z)]P we deduce Theorem 5.1. The extremal functions —

ao [(1 + ze'° )/(l — zel ° )] 2P are c.m.p—valent and not zero by Lemma 5.1, since their pth roots are univalent and map 1z1 < 1 onto the plane cut along a ray from the origin to infinity. We remarked in §2.3 that if f(z) is a.m.p—valent and hence a fortiori if f(z) is c.m.p—valent then f(z) has q < p zeros. Thus the condition f (z) 0 of Theorem 5.1 is a consequence of mean p—valency if p < 1.

5.2 Functions with a zero of order p at the origin consider a function

f(z) = zP + ap+izP+1 + • • ,

In this section we

5.2 Functions with a zero of order p at the origin

149

c.m.p—valent in z < 1, where p is a positive integer. In this case f (z) has no zeros in 0 < z <1 and so [f(z)] 1 /P = z (1 + aP+1

-)

is single—valued and hence c.m. 1—valent in IzI < 1 and vanishes only at z = O. Thus we can reduce our problem to the case p = 1. We show that the methods of §1.2.1 are applicable.

Theorem 5.2 Let f(z) = z a2z 2 + • • • be c.m. 1—valent in 1z1 < 1. Then f(z) belongs to the class So defined in §1.2.1. We suppose that 141 < 1 and that z = w(C) gives a univalent map of ICI < 1 into IzI < 1 in such a way that co() # zo. Consider

(C) = f [o(C)]. Then the equation v(C) = w never has more roots in ICI < 1 than the equation f(z) = w has in IzI < 1, since w(C) is univalent. Thus tp(C) is c.m. 1—valent in 1 < 1. We write f(z o ) = Re*. Then it follows from Lemma 5.2 that if the equation f(z) = Re* has a root in z < 1 other than zo, then we can find 4:1)1 such that f (z) Re kli in IzI < 1. In this case ip(C) Re' in ICI < 1. If, on the other hand, f(z) Re* for z then tp(C) Re* in ICI < 1 Re' l) in Thus in either case we can find (i) = OR) such that ip() < 1. Now Lemma 5.2 shows that if p R there exists q = 4)(p) such that Ip(C) peiO(P) in ICI < 1. Hence Theorem 4.13 gives lip' (3)1

4[4(0)1 +

= 4[11'(0)1 + If (z0)1].

Hence f (z) E So and Theorem 5.2 is proved. We deduce immediately t

Theorem 5.3 Suppose that f(z) = zP ap+izP+1 < 1, where p is a positive integer. Then lap+il for Izl= r (0 < r < 1) rP -1.f(z)1 (1+ r)2P t

is c.m.p—valent in 2p. Further we have

rP (1 — r) 2P'

In this form the result appears in Hayman [1950]. The inequality lap±i I 2p has been proved for a.m.p--valent functions by Spencer [1941b], and the last statement of the theorem was extended to a.m.p—valent functions by Garabedian and Royden [1954].

150

Circumferentially mean p—valent functions P1 1

r ) 1f(z)I

prP-1 (1 + r) (1 — r) 2P+ 1

Finally the equation f(z) = w has exactly p roots in 1z1 < 1 if iwi <

Theorem 5.4

4-P .

With the hypotheses of Theorem 5.3

r"(1 — r)2P M (r, f) decreases strictly with increasing r (0 < r < 1), and so tends to a <1 r 1, unless f(z) = zP(1 — ze i° )-2P. Hence the upper bounds for If(z)1,1f(z)1, given in Theorem 5.3 are only attained by these functions.

We write

n+1 2 tp(z) = [f(z)] 1 h) = z + ar Z -I--

•••.

By Theorem 5.2 ip(z) G So, and so Theorems 1.4 and 1.5 are applicable with t,v(z) instead of f(z). We deduce the results of Theorems 5.3 and 5.4 by taking pth powers.

5.3 Regularity Theorems: the case a = 0 For the rest of this chapter we concentrate on the more general hypothesis that f (z) is a.m.p—valent and shall say that f(z) is mean p—valent in this case. We start with the regularity Theorems 2.10 and 2.11 for the maximum modulus and radius of greatest growth of f (z) to obtain corresponding results for the coefficients of f(z) 2 if 13,) > The results fail for p2 < as the example of Section 3.4 shows. The extension to a.m.p—valent functions is due to Eke [1967b] and was made possible by his regularity Theorems 2.10 and 2.11. For c.m.p—valent functions the results are contained in Hayman [1955]. 0 in an annulus Suppose now that f(z) is mean p—valent and f(z) 1 — 26 < z < 1. Put 4)(z) = [f(z)] )- . Then 4)(z) may be analytically continued throughout the annulus 1 — 26 < zI < 1. Also if 4)2 (z) is obtained from a branch 4) 1 (z) of 4)(z) by analytic continuation once around the annulus in the positive direction, then 14)2(z)/01(z)1= and so by the maximum modulus principle 4)2 (z)/4) 1 (z) =a real constant. t For this and subsequent results of this chapter see Hayman [1955].

where ji is

5.3 Regularity Theorems: the case a = 0

151

Thus 4(z)/z' 2 remains one-valued in the annulus and possesses a Laurent expansion, so that +00 0 ( z ) . zii E bn zn.

(5.4)

n=—oo

We shall prove Theorem 5.5 Suppose that f(z) is mean p-valent in 0 < Izi < 1, that p),1> 1, and that 0(z) = [f(z)]' possesses a power series expansion (5.4) in an annulus 1-26 < izi <1. Thent

fim

Ibn l

a'l

=

n—H-oo 112P"

F(2p 2 )'

where a is the constant of Theorem 2.10. We shall use the formula (n+ p)b, =

41(z)dz 1 f 2m:j1z1=p zn+P p 1 — n —y

fir

27r

o t (peio )e-i(n+y- o° de,

(5.5)

—n

where 1 - 26 < p <1. In this section we prove Theorem 5.5 when a = O. We choose a positive constant t such that

2 1 0 < t < 2, t > 2 - -,it-, and t > 20 .

(5.6)

This is possible since p/1, > 1. Then Schwarz's inequality gives 7.1-

zir

f ir 141(pe)Ids9 —7r

t

t

ir

< ( 217r f l ope/6 )1 2 10(peiordo) 2

(

—n

1 r

2n

L

I 4)(pe w )1 140) 7

,

_i

1( 1

/ 21r f

/0 2

ie (2—tr-2

1

2

A de) 2 X (

1 f 2n If(Pe i° )1 14 d0) . zno Since oc = 0 we can, given E > 0, find ro (E) < 1 such that A — 27r 0

If (pe )1 If(pe )1

M(r, f) .- E(1 — 0 -219 (ro () < r < 1).

(5.7)

Since f(z) is mean p-valent and (5.6) holds, we may apply Lemma 3.1 f

Here and subsequently F(x) denotes the Gamma function. Relevant properties will be found for instance in Titchmarsh [1939] pp. 55-8. We need assume only that f(z)/z' remains one—valued for some real y (cf. Example 2.8).

Circumferentially mean p—valent functions

152

with (2 — 02. instead of 2. Hence if r0 (E) < r < 1 there exists p such that 2r — 1 < p < r and 1 1 2m 4 ,,E ;.(2—t) (pei0)121fi pei0)1(2-0).-2de <

2n 0

—

(1

2(2

—

—

r)-2p).(2-0-1

t)

Next we have from inequality (3.10) of Theorem 3.2 applied with At instead of 2, and (5.7), for 0 < p < r, ro < r <1, f)

1 121' 2,7E 0

(pei °

dO

m(ro, f )" + ';[( ,;-

e(1 — ro)- 2t

(1 _ 0 1-2p;.t , if ro >

02

frro M(X'f)4t x

d

x

P n 2 + 1] fr e t (1 — x)-2pAt dx — [t) 2 X ro

e is chosen so small that pe t [(2t)2 +1] < 1 , (2p4 t —1)

and then r is chosen sufficiently near 1. With this choice of ro, E, r, p we deduce that f2m

27r 0

1(1)'(pei° )Id0

A(p, >it ) e ).(2—t) (1

r ) ;11-213.1t-2p1(2-0-1]

= A( p, ))( 2_t) (1 _ 0-2p, where p is some number such that 2r —1 < p < r. We choose r = 1-1/n and apply (5.5). Then if n is sufficiently large

Rn + ti)bn

-(1 —

2 —" 1 / 2' 0 141(pei° )1(10 n

A (A /1, )E ).(2t)n2P).. Since E may be chosen as small as we please, we deduce

bn = o(n2P;' -1 ) (n —+ +0r4, and this proves Theorem 5.5 when a = 0.

5.4 The case a > 0: the minor arc When a > 0 it follows from Theorems 2.10 and 2.7 that f (z) has a unique radius of greatest growth arg z = Oo and that If (re )1 is relatively small except when 0 is near Oo.

5.4 The case a > 0: the minor arc

153

We shall deduce Theorem 5.5 from formula (5.5) in this case by obtaining an asymptotic expansion for O'(peie ) on a major arc {0 :10 — 00 1 < K(1 — p)}, where K is a large positive constant, and showing that the complementary minor arc y = {O : K(1 — p) <10 — 0 0 1 < 7r1 contributes relatively little to the integral in (5.5). It is this latter result which we prove first. Lemma 5.3 Suppose that 4)(z) is defined as in Theorem 5.5, that a > 0 and that 00 satisfies (2.59). Then given ri > 0, we can choose K > 0 and a positive integer no with the following property. If n > no, there exists p in the range 1-1/n < p <1— 1/(2n) such that

1 1 14i(ped0 < 27r

where y is the minor arc {0 : K(1 — p) <10 — 0 0 1<_7r}. We again define t to satisfy the inequalities (5.6) and have

1 2m

10' (pe 1° )IdO _< A

f if v el0w lf (peie )1 (2-0A-2) 2 1 2 X (1 f

2m

If(pe i° )1 À`d0) .

(5.8)

Since f (z) is mean p—valent, f (z) satisfies (3.11) with a suitable constant C and 18 = 2p. Thus we may apply Lemma 3.1 with r = 1 — 1/(2n), fi=-- 2p and 2(2 — t) instead of A. This gives, for some p such that 1 — 1/n p 1 — 1 1(2n), 1

1

If ,, (pc )1 L I ppeie )1 (2-0;.-2 de

27r

4pC)-(2—t) (2n) 2P1(2-0 +1 . ) (2 — t)

(5.9)

To estimate the second factor on the right—hand side of (5.8) we note that the number 00 of (2.59) certainly has the properties of 00 in Theorem 2.6. We may thus apply Theorem 2.7. Hence if K is chosen sufficiently large, depending on E, we have for ro < p < 1 If(Pe1651<

1 (K(1 — p) <10 — 001<m). (1 — p)'I0 — Ool 2P— E

154

Circumferentially mean p—valent functions

Writing 10 — 0 0 1 = x we deduce

1 f P) Ut r X —)1(2P—E) dx. 27r j 1.f(Peg))dO < 2(1 27r iK(1—p) We choose

E

so small that i,t(2p —

7r 21 f7

E) >

1, which is possible by (5.6). Then

f

(1

.6,

If (pel )1 -, dO

TE

x —Ât(2)- E) ax

JK(1—p)

K 1—m2p—E) ( 1 _ 01 -)1(2p -e) -E).t n[2t(2p — v) — 1] K 1—)1(2p—e) n[Àt(2p — 8) — 1]

We note that when 2, t, C, ri and chosen so large that

224pc )-(2-0 2(2

—

E

(2n)2t-1

(5.10)

have been fixed, K may then be

K 1--i.t(2p—E)

t) n[242p — E) — 1]

24P).

With this choice of K Lemma 5.3 follows from (5.8), (5.9) and (5.10).

5.5 The major arc Our next aim is to find an asymptotic formula for f (z) , f'(z) and 0(z) on the major arc

F

Fn : 16/

001

K ( 1 — 14

(5.11)

of 1z1 = p. Thus F is complementary to 7. For this purpose we use the asymptotic formulae (2.59) and (2.60) of Theorem 2.10. We define rn = 1 — 1/n. Let p = p„ be the number of Lemma 5.3 and write

= pn ei6), n > 1 = (1 — Pn) 2P f (zn),

(5.12)

and fn(z) =

(1 — ze — i°0) 2P •

(5.13)

We deduce from (2.59), (2.60) that

ccn

n-2P f (rn e'°°)

(5.14)

and kLnI --+ a as n

Using (2.60) we prove

co.

(5.15)

5.6 Proof of Theorem 5.5

Lemma 5.4

155

We have

f (z)/ f n (z) —*

1,

(5.16)

t(z)/f(z)

1

(5.17)

--+

uniformly as n —> cc, while z G F. We write z = pn ei° and recall that by (2.60) of Theorem 2.10 we have

2p

f'(z) f (z)

o(1)

ei°0 — z

uniformly for z on Fn as n —> co. Integrating along F, from z = zn to pn ew , we obtain

log f (p

n eo)

= 2p log (

f (zn)

ewo — z n ei°0 — pnew +

2p log (

lp nePon 00) —

+ o(1 ),

-

1 — pn 1/(2n) and the length of Fn is at most since lei00 — zl 2K(1 — p n ) 2K /n, by (5.11). Thus pne 0-00) 1 2P

f (pneio)1

f (zn)(1 — Pn) 2P

Writing z = Ne i° and recalling (5.12) and (5.13) we obtain (5.16). Next we deduce from (5.13) and (2.60) that f'(z) f n (z) f (z) f(z)

=

f'(z) e l°0 — z 1 f (z) 2p —*

as z ---> ew° on the union of the arcs Fn . Multiplying this by (5.16) we obtain (5.17). This proves Lemma 5.4. An alternative approach generalising the method of Lemmas 1.2 and 1.3 is provided for c.m.pvalent functions by Theorem 5.1 (see Example 5.5).

5.6 Proof of Theorem 5.5 We now conclude the proof of Theorem 5.5 when a > 0, by proving the stronger

Theorem 5.6

Suppose that 4)(z) = z

E +cc

bn zn satisfies the hypothe-

ses of Theorem 5.5, that a > 0 and that 00 satisfies (2.59). Then

nbn

(I) [(1

-'1F 1 )(e2i °°p) ) e—itn+11)°° (n —> +co).

Circumferentially mean p-valent functions

156

We have by (5.14), (5.15)

Ø

f [ (1 -

[( 1 —1 ) el l

oc;-n2Pj- (n —> )

Thus Theorem 5.5 follows at once from Theorem 5.6. The latter result is significantly stronger, since it gives information about arg bn as well as Ibnl. We suppose, as we may without loss of generality, that 00 = 0, since otherwise we may consider f (zewci), 4)(ze 10°) instead of f (z), 0(z). Write

(1 —

E

=

CnZ .

n=0

Then 2p2(2p2 + 1) ...(2p/1+ n — 1) Cn

1.2 ... n F(n 2p/i) (n F(n 1)F(2.1p) F(2/3),.)

-Foo).

(5.18)

We also set O n (z) = f n (z) A , so that On(Z)

=

E cni zm, m=0

and hence

p 1—n 2.7r

1 44(z)dz ncn an = 27riZn 1 1=P

7E

otn (pe i0) e—i(n-1)0 do (0 < < 1) p

.

Finally we recall (5.5): f-Fiz

(n + P)bn =

2n ,

4)'(pei6 )e-i( n+P-1)e d0 (1-26

< p <1).

Thus 27rpn-1 [(n

p.)pPbn — ncna )ri] — illn (peie )]

—

-1)° d0.

(5.19)

We now suppose that ri > 0 and choose K so large that for n> no we such that can find p = pn in the range 1 — < p <1—

A

1

10/ (pel61 )1d0

5.6 Proof of Theorem 5.5

157

where y denotes the arc K(1 — p) <10 n. This is possible by Lemma 5.3. Next we note that =12p/ioc;,t111 — z1 -2PÂ-1 .

Also for z = pei° , where

< p < 1, K(1 — p) <101

7t, we have

—z1 2 = 1 — 2pcose + p 2

202 0 = (1 — p) 2 + 4p sin2 > 2 — n2

while 12p2a 1 l < C1. Here C1, C2, . . . denote constants independent of K and n. Thus 14) ln (Pei° )4 0

C2 f 01 -1-213/1 (10 = 2 C2

[ — 0 -2PA

[ 2p2 .1 K(1— p)

Thus if p = pn we deduce that

144(pe` ° )Ide < C2

1 [K (

—P)]-21';'

<

C2(1K)-2131 n2p/i. 2

P/1,

since (1 — p)> 1/(2n). Hence if K is sufficiently large we obtain .17 0(pe`e )Id0 < qn2P1. Thus y

f

( i0 )1e dO [0.1 (pe i0 )e —410 —4) fnpe

fy

[l ov eio ) 1 + kyn(pe io ii dû < 2nn2P4, )

(5.20)

if K is sufficiently large, depending on j. Having fixed K we now apply Lemma 5.4 on the complementary arc = Fn : 0 101 K(1— p). Since 0(z) = f(z) 1 , 4l(z) = and similarly 41n (z) = kf n'(z)f n (z) )--1 , we deduce from Lemma 5.4 that, uniformly on y',

o v eie )e—ikto

ov eie )

ofn(peie ),

where p = pn . Thus o v eio )e—itio — 4(pee) = o { n(pe io ) } =

— pr i-219,1 _ o(n2P'+1)

158 as n

Circumferentially mean p—valent functions

oo. Hence w (pe io )e

o

o ln(pe io e —i(n—i )0 do

fyi

f 14Y

(pei° )Icl0

(Pei°

= o(n2P'+')2K(1 — p)= o(n 2P').

(5.21)

Since y, y' together make up the whole range —7r < 0 < 7r, we deduce from (5.19), (5.20) and (5.21) that + ft)pP bn — nc n aijI < [211 + o(1)]n2P'

127r p" for large n. Also

(1

1

) n

1

ri is arbitrarily small, and (5.18) we have ncn a'n'

bn

(n

(n

4

n

2),

—+ 1 as n --+ +co. Finally, using (5.14) and

+

11 )Pt`

F(2p2)

[f(rn )? nF(2p).)

(P(rn )

nr(2/3/1)

(n --+ +oo),

where rn = 1 — 1/n. This proves Theorem 5.6.

5.7 Applications: the case .1. =1 The case = 1 of Theorem 5.5 is of most interest, since it refers directly to the coefficients of mean p—valent functions. If 00

f(z) =

an zn

is mean p—valent in lz I < 1 and p> 1, then lanl --+ 11 2 P-1

(n —+ co),

F(2p)

where a is the constant of Theorem 2.10. Whenever we can obtain bounds for a, we obtain correspondingly sharp bounds for the asymptotic growth of the coefficients. Thus we have for instance

5.8 Functions with k—fold symmetry

Theorem 5.7

159

Suppose that p is a positive integer and that cc

E

f(z) = zP +

an zn

n=p+1

is mean p—valent in 1z1 < 1. Then

urn

n— ■ ce

n2P-1

=

a 2p— 1)!•

where a <1, except when f(z)= zP(1— ze i° )-2P.

In fact the limiting relation holds by Theorem 5.5 and a = lim (1 — r)2P M(r, f).

Also, by Theorem 2.11, a < 1 except for the functions zP(1 — z ei°) 2P. The proof was suggested in Example 2.11. For univalent functions the result is contained in Theorem 1.12. However the conclusion of Theorem 5.7 now extends to real p, such that p> We have similarly Theorem 5.8

f(z)

Suppose that f(z) =

0 in 1z1 < 1, where p>

E cc

n=0

anzn

is c.m.p—valent

and

Then

hm Ian I n —>G0 n2 P-1

a F(2p)'

where a < laol4P with equality only for the functions

ao

(1+ zei° ) 2P 1 — zed° •

The limiting relation is again a consequence of Theorem 5.5. The inequality for a follows from the last statement of Theorem 5.1.

5.8 Functions with k—fold symmetry integer and that

Suppose now that k is a positive

fk(z)= akz k + a2kz2k +... + ankzn k + • • •

is circumferentially or areally mean p—valent in Izi < 1. Then 0(z) = fk(z i/k )= akz a2kz 2 + • • •

160

Circumferentially mean p—valent functions

is mean (p/k)—valent there and conversely. For to each root zo of the equation 4)(z) = w there correspond exactly k roots of the equation fk(z) = w, given by z k = zo. Thus for every w n[w, fk(z)] = kn[w, 4)(z)], and so for every positive R

More generally if

y

1 P[R, (P] = — P[R, fk]. k is real and

fk(z) = f' + ak+vz k+ v +

2k + * * * Cl2k+vZ+v

(5.22)

(5.23)

in 0 < 1z1 < 1 so that If(z)1 is one-valued, then if 4)(z) = fk(z 1 /k )

the equation (5.22) still holds. To see this we note that each arc F of a level curve 14)(z)1 = R, which lies in 00 < arg z < 00 + 2m, 0 < lzi < 1, and on which arg 4)(z) increases by 6, corresponds by z = wk to k level curves yv , one of which lies in each sector (00 + 2vn)/k < arg w (00+ 2(v + 1)n)/k and on each of which argfk(w) also increases by 6. The 7,, contribute a total of k6/(2n) to p[R, fd, while F contributes 6/(27r) to p[R, 4)] and this yields (5.22). Thus, if fk(z) is mean p—valent in 0 < 1z1 < 1, 4)(z) is mean p/k—valent and we may apply Theorem 5.5 with p/k instead of p, ;i, = 1 and 4)(z) instead of f. Also a

= lim(1 — r)2P/k M(r, q5) r--*1

= 11111(1 - rk )2p/k mfrk , 0) r-0 = liM(1 - rk ) 21h1k M (r, fk) r--41 = k2P/k 11111(1 - r)2P/k M(r, fk) r--+1

(5.24)

and bn = ank +v . We deduce

Theorem 5.9

If fk(z), given by (5.23), is mean p—valent in Izi < 1, where 1 < k < 4p, then a ilM l ank-1-1, 1 fl_-

F(2p/k)'

where a is given by (5.24). We note the special case p = v. In this case 4)(z) satisfies the hypotheses of Theorem 2.11 with p/k instead of p. In particular a < 1. We deduce

5.8 Functions with k—fold symmetry

Theorem 5.10

161

Suppose that

fk(z) = zP a p+kzP +k7P+2k , ap4-20,

•••

is mean p—valent in 1z1 < 1 and that 1 < k < 4p. Then a F(2p/k)

lap-Enkl

lim n2p/k-1 n , ce

and a < 1, with equality only for the mean p—valent functions

fk ( z ) = z po _ z k eio )-2p/k . In fact by Theorem 2.11 we have a < 1 with equality only when

4 (z)

=

Z PR O—

ze i°

so that f k ( z ) = zP(i

z k ei0)-2p/k .

We note the special case of an odd univalent function

f 2 (z) = z

a3 z 3

a5 z 5 + • • • .

Taking p =1, k = 2, we deduce from Theorem 5.10 that

lim n—*oo la2n+1 < 1 in this case, except when f2(z) = z la2n 4-1

z3e10

z5e20 i +.... In particular,

1 (n > no(f)).

Nevertheless, for any fixed n > 2 an odd univalent function f 2 (z) can be found t for which I a2n+1 > 1. We remark finally that if fk(z) is c.m.p—valent or univalent and given by (5.23) with y = 1, then f(Z)= {fk(Z i/k )} k = 0(Z)k = Z (1+

E

akn fiZ

n)

-

n=1

is also c.m.p—valent or univalent respectively and conversely. For c.m.pvalent functions the result follows from Lemma 5.1 (a). For univalent functions we recall the argument for the proof of Theorem 1.1 for the case k = 2. For a.m.p—valent functions f (z) we can still, using Lemma 5.1 (b), assert that fk (z) = f (zk ) 1 /k is a.m.p—valent but the argument for the converse fails. That is why, in order to prove Theorems 5.9 and 5.10, we found it convenient to work with 4)(z) directly instead of 0(z)k = f (z). Schaeffer and Spencer [1943] (see also Duren [1983, p. 107]). Earlier Fekete and Szeg6 [1933] showed that the exact upper bound for ja51 in this case is 1/2 + e-213 = 1.01... We shall prove this result in Chapter 7.

162

Circumferentially mean p—valent functions

5.9 Some further results

Some inequalities for the class

a of functions

f(z) = z + a2z 2 ---

circumferentially mean 1—valent in 1z1 < 1, which go beyond Theorem 5.3, have been proved by Jenkins [1958]. He obtained the exact upper bound for 1f(r)I, when 0 < r < 1 and either la21 or 1f(—p)1 is given, where p is fixed and 0 < p < 1. In either case the extremals are univalent functions with real coefficients. It follows from Theorem 1.10 that for these extremals Ian 1 _< n and so If(p)1 +

If(-01

= f(P)— f(—)9) = 2(p + a3p 3 + a5p 5 • • •)

<

(2n + 1) p2n+1,

2 n=0

If(P)1 +

<

PP

( 1 + 02 ( 1 — 0 2.

This latter inequality remains valid for all f(z) G and hence la31 < 3 for f (z) E R. An unpublished example of Spencer shows this to be false in general for a.m. 1—valent functions, though la21 _< 2 remains true. We shall prove la31 < 3 for univalent f(z) by Lôwner's original method in the next chapter. Jenkins' results also imply that if 1a21 is given and f (z) G then

a

a = lim (1 — r)2 M(r,f) < tp(a) = 4b -2 exp(2 — 4b-1 ), r-+1—

where b = 2— (2 — la21)1, and this inequality is sharp for 0 la21 2. If f(z) = zP-f-a p_fi zP+1 +• • is c.m.p—valent in 1z1 < 1, then [f(z)1 1 /P G and from this Jenkins deduced the sharp inequality la+21 p(2p 1) in this case. It also follows that the number a in Theorem 5.10 satisfies the inequality cx^

{tp

klap+kl}

p/k

P which is again sharp for given lap±k 1. These results are described in Jenkins [1958]. Regularity theorems for the means I f) and their derivatives will be found in Hayman [1955] at least for the case of c.m.p—valent functions. (See Examples 5.2 and 5.4 below.) Finally mention should be made of the following conjecture by A. W. Goodman [1948].

163

Examples Suppose that p is a positive integer and that X

f (z) —

bn z"

is p—valent in lzl < 1, i.e. that the equation f(z) = w never has more than p roots in 1z1 <1. Then for n > p

k=1

2k(n+ p)! 0 kl. (p + k)!(p — k)!(n — p — 1)!(n 2 —k2 )

Goodman showed that equality is possible for arbitrary values of 1b 1 1 to Ibp 1, with f (z) a polynomial of degree p in z/(1 — z) 2 . The case p = 1 is de Branges' Theorem. For p> 1 only a few general results are known. If

ai = a2 = • • • = ap_ i = 0 the conjecture holds for n = p+1 [Spencer 1941b] and n = p+ 2 [Jenkins 1958]. The simplest general open case is I b31 -_ 51b11 +41b21 for 2—valent functions. Goodman and Robertson [1951] proved the conjecture for functions typically real of order pt. Livingston proved it for functions close to convex of order p for n = p + 1 in [1965] and for general n if al = a2 = • • • = ap_2 = 0 [1969]. These results are all sharp. Now that de Branges' Theorem is known, Goodman's conjecture provides a fascinating challenge.

Examples 5.1

If 0 < R < 1 let A be the cut annulus A : R < 1w1 < 1,

larg w1 < n

and let f(z) map Izi <1 uniformly onto A. Show that, if q > 0, the function

fq (z) = f(z)q is c.m.p—valent in 1z1 < 1 and it is a.m.p—valent for p = q(1 _R2) , but for no smaller value of p. By comparing f q (z), f q (zr show that Pip cannot be replaced by any smaller number in Lemma 5.1 (a), and that /lop cannot be replaced by any smaller number in Lemma 5.1 (b), (choose R = 0 or R close to 1). t For definitions of these classes of functions we refer to the quoted papers or the interesting historical account by Goodman [1979]

164 5.2 5.3

Circumferentially mean p—valent functions If k is a positive integer prove that f(z) G S if and only if fk(z) = f(z k ) 1 /k E S, and fk(z) is of the form (5.23) with y = 1. If f (z) = (1 — z ) 2 P, where p > 0, prove that, if p), >

IAr,f) ,,, 21 \/7rF(4) (1. "( 4

as r 5.4

1. (Use Example 1.2 at the end of Chapter 1). Prove that, with the hypotheses of Theorem 5.5, and if —

(1 — r)2P"I;(r,f)

2.\/7rF ( i,p)

a

>

;

as r —) 1. (Harder). Show also that, if /3) =

log

5.5

—> — r 1.

(Apply Theorem 3.1 with f(z);12 instead of f(z) and A. = 2, considering separately the cases a = 0, a > O.) Corresponding results for the derivatives can also be obtained in suitable cases. We confine ourselves to the case of I2(r), where complete results are possible. Prove that with the hypotheses of Theorem 5.5, we have, if and q = 1, 2, ...

(1— r) 4P)-+2q-1 /2 (r, ow) )

5.6

oc2,; F(20 + q)F(2/3.1

q—

247rF(2p2) 2

Extend this result to p > 0, for q = 1 and hence for q > 1 by using Lemma 1.1 and Theorem 3.1. (If q = 1, a > 0 and kp.1 < mimic the argument for Lemma 1.4.) 0 in If f(z) is regular and c.m.p—valent and f(z) 1 — 2(5 < z < 1, prove that

f'(z) < 4p6 f(z) — (1 —1z1)(1z1+ and deduce that ((1 — r)/(r

— 1)

, 1 — 6 < izi < 1,

26 — 1)) 2P lf(pe'°)1 for fixed 0 and ((1 — r)/(r + 26 — 1)) 2PM(r,f) are decreasing functions of 0, for 1 — 6 < r < 1. (Apply Theorem 5.1 to f(zo + g), where zo = (1 — (5 )e'0 ). Hence show how to extend the proof of Theorem 1.12 to c.m.p—valent functions.

6 Differences of successive coefficients

6.0 Introduction

In this chapter we consider a function

f(z) =Ean zn.

(6.1)

We suppose that p> 14 , since for p < 1 we cannot improve on Theorem 3.5. We also suppose that either

(i) f(z) is regular and areally mean p—valent in A : Izi < 1, in which case n goes through the integers from 0 to oo in (6.1), or (ii) if p is not an integer we also allow the possibility, as in Theorem 2.11, that f (z)/zP is regular in A and that f(z) is mean p—valent in A, cut along a radius z = pe, 0 < p < 1. In this case n = m + p in (6.1) where m goes through the integers from 0 to co. With these hypotheses we shall obtain estimates for

I lan+i I — lanll. We note that if f(z) is a pth power of a Koebe function

f(z) = (1 where p

zP _) 2P'

1 and n = p + m, we have an+1 — an

=

F(2p + m) F(2p — 1)F(m + 2) m 2p-2

ri2 P-2

F(2p — 1)

F(2p — 1)'

while if

f(z) =

zP (1 — z 2 )P' 165

as

n

166

Differences of successive coefficients

and n = p +m, where m is even, then an+i = 0, and F(p + ,112) an = lan+i — anl =

nP-1

2P-1 F(P) .

F(P)F( 1 +

This suggests the conjecture that

I lan+i

0(n 2P-2 ), p

=

—

p <1

= 0(n 1 ),

Ilan+11 —

(6.2)

1

(6.3)

and, if p < (6.4)

—

In fact the examples of Section 3.4 show that we cannot in any case expect more than (6.4). We shall in this chapter prove (6.2). It is not known whether (6.3) is true. We prove the best known result, namely = 0(n 2P-2,/P),

lan-Fi — while (6.4) holds for p <

For p =

I lan-Fi —


(6.5)

Theorem 3.5 yields = 0(n -3: log n)

and we are unable to sharpen this. The above results are due to Hayman [1963] when p = 1 and to Lucas [1969] for other values of p. Suppose that f2(z) = z + c3 z 3 + • • •

is univalent and odd in A. Then (cf. Section 3.7) f (z) = f2(z 1 ) =

(1 + c3z c 5z 2

-)

is mean -valent and so we obtain 11C2n+11

1C2n-1 = 0(T1 1- N/ 2 )

in this case. This estimate due to Lucas [1969] is the best that is known but El Hosh [1984] has shown that if f is close to convex, and in particular starlike, we have 1c2n+i — 1c2n-111 =

(6.6)

As we saw above this order of magnitude is attained for the 4-symmetric function f2(z) =

(6.7) (1 — z 4 ) .

6.1 The basic formalism

167

which is starlike and univalent. Shen [1992 ] has shown that (6.6) also holds if f G S and

M(r, f2) = 0(1 — r). He obtains a slightly weaker conclusion if, for some 00 lim inf(1 — r) 1 1f (re`e°)1 > 0. r--41

Both these conditions are satisfied by the function (6.7). If

lim(1 — r)M(r, f2) = a > 0 r--*1

Shen proves, using Baernstein's technique (cf. Section 3.5.3), that c2n+1

1C2n-1

=

These results show that functions f2 (z) failing to satisfy (6.6) are likely to have M(r,f2) oscillating around the order (1 — r)- 1 . The proof of (6.2) and (6.5) will occupy Sections 6.1-6.6. In Section 6.7 we consider k—symmetric univalent functions, in 6.8 asymptotic behaviour, in 6.9 starlike univalent functions, and in 6.10 the results of Shen.

6.1 The basic formalism We suppose from now on that f is mean p—valent in A and of the form (6.1). We define

=

E

Next we choose a fixed n such that n > 6 and write

rn = 1 —

and M1 = M(rn, f) = maxlz1=r, If(z)1.

(6.8)

We also choose z 1 = rneie° so that I f (z 1)1 = MI . Then we define Aiv = 21 - V MI,

1 < y < ce.

(6.9)

Next we write

Ev = E(n) = { z 1 — < 1z1 < 1 — , and Mv+i < If(z)1 mv}. We note that, by (6.1), we have

(n — 1)a n_1 — nzia n = (n — 1)(a 1 -1 — e l%n) 1 1 (z — zi)f' (z)dz 27ri jizi=p Zn

1—

n

< p< — 1 — 2n'

(6.10)

168

Differences of successive coefficients

Thus r 2rt

(n — 1)la n _1 — e ° anl

I !Pei° 27rpn o

(6.11)

z1llf'(pe` ° )Ipd0.

Also, since n> 6,

n

p—n < (1 — —3 ) n < (1 — 1 ) 6 = 64 and n— 1 2 n

. 6 5'

We integrate (6.11) from p= 1 — 3/n to p= 1 — 2/n. This yields 27r f 1-2/n 32n I pe i° — ZIllf (pe 1e )lpdpd0 dO 1-3/n (n — l)n /0

liki an_i — e n l -.

40 <— 7"

L 1z

Ef

_z o t(z) l pdpd0

,

v=i

< Ê

f

( f

' 'z ' 2 P dP dO) ' (f E', f(z)

v=1

lz — z11 2 1f(z)1 2 pdpd 0)

by Schwarz's inequality. We note that, by (6.10) and since f is mean p—valent in Ev , 2 f'(z)

IL

f(z)

v+Y f LIf'(z)12pdpd0 < (Mi

pdpd0

A42

< irp ; = 4np. 114 +1 Thus we obtain

I an_

— eiO0a n

cc

<80

(

11

TE

)2

{E v=1

f E' f

(6.12)

pdp f — z11 2 1f(Pe it1 )1 2 d0 }

In Ev we have If(z) I < M v and so

< 2M1 If(z)1 2 + Ifiz)1 2. Thus (6.12) yields finally

lanl — lan-il

I an_

—

el °`' n —

2 MW(Z) 2

+ I f(Z)2 P0}

.

(6.13)

6.2 An application of Green's formula

169

We need to transform the right-

6.2 An application of Green's formula hand side of (6.13) further.

Lemma 6.1 Suppose that v(z) = v(x,y) is twice continuously di fferentiable in 0 < 1z1 < p, and that y remains continuous at z = 0 and t ev(te i° )/et log(l/t) 0 as t O. If ‘72

=(

\2 + ( \ 2 Oy

we have p j2it

v(pe 10 )d0 = 2nv(0)

f dt

V 2 v(te i° )t log )2 d0

o

(6.14)

and so, for 1- p < 1,

2n

fo 2n

v(pe w )d0 fo -

v(le i° )d0 t(log 2p)dt +

Jo

7

t log —P dt t }

fo

V 2 v(te 1° )d0. (6.15)

We write

I(t) = f v(te ie )d0, and apply Green's formula, Lemma 4.2, to v in the annulus

E < t < S.

This yields s277 SOS)

—

er(E) = f tdt f V 2 v(te )d0, 0 <

E

<S

< p.

We divide by s and integrate both sides from s = E to s = p. We obtain

I(S)

—

I(e)— el / (e) log

=

ds s f — f tdt f V 2 v(te i° )610 s 0

LE

tdt

P ds f —

V 2v(te w )d0

s=, 5 0 2.7r

t log — f V 2 v(te i° )d0. t

170

Differences of successive coefficients

We let g tend to zero, and note that by hypothesis EI

V) log f2e —* 0, /(E) -- v(0)

in this case and now (6.14) follows. We deduce (6.15) by subtraction. We shall apply Lemma 6.1 to the right-hand side of (6.13). To do so we need If f(z) is regular in 0 <1z1 < 1, If(z)1= R and

Lemma 6.2

m2 R2

G(R) = Av + R2'

(6.16)

4m4(m2 — R2) (M2 + R2

(6.17)

then

V 2 G(R) =

) 3

If v(z) = lz —

z 1 1 2 G(R),

(6.18)

then

V 2 v(z) 4G(R)+ lz — zil2V2G(R) +

8 1z — zilM 4 R 1 fq z)1. k

(6.19)

Also if f(z) is given by (6.1) subject to the conditions (i) and (ii) of the introduction, v(z) and G(R) satisfy the hypotheses of Lemma 6.1.

We write R = If (z)1 = eu , so that u is harmonic in 0 < 1z1 < 1, except where R = O. Also

a ex

a

)

a eu G(R)= — G(eu) = e —C(eu), ex

al

2

(6.20)

ex

)a2u

x G(R) = [e2uG"(eu)+ euC(eu)] (— + euC(eu . (-F Ox 2 ex

Differentiating similarly with respect to y and adding we have

au) 2 V 2 G(R)= [R2 G"(R)+ RC(R)] {(— ± aX except at points where R = O. Also

eu

ex

f'(z) eu d — i— = — log f(z) = f(z) , ey dz

(6.21)

6.2 An application of Green's formula so that

171

If'R(z2)1 2 . 2 = Gay / (eauxY 4-

Thus

V2 G(R) = [G"(R) + ° Ili' (z)1 2 , R except perhaps where R = 0. Using (6.16) we obtain (6.17) when R # 0 and the result remains true by continuity at points where R = 0. Again (6.20), (6.21) yield for 1z1 = p in case (ii)

(0 .0 --_,-‘, - t .1)-- )G(R)

= If'(z)IC(R) = 0(pP-1 )pP = 0(p2P-1 ), as p ---÷ 0

and G(R) = 0(p2P). Thus G(R) satisfies the hypotheses of Lemma 6.1. Also

a

—V(Z)

ex

= iz — Z11 2

0 G(R) + 2(x - x i )G(R) = 0(p2'') ex

and

—0 v(z) = O(p21 ) ay similarly. Thus v(pet ° ) ---÷ 0 and

plogp—a v(pe 10 )= 0(p2P) loo 1- -+0 Op / i° P as p -* 0 in case (ii), and in case (i) all the partials of v are continuous at z = 0. Thus v(z) satisfies the hypotheses of Lemma 6.1. It remains to verify (6.19). We set X = lz - z 1 1 2 , Y = G(R) and note that + 2 (OX a Y + 0X 0Y V 2 (X y) = Y V 2 X + XV 2 Y ex ex ay ay )

= 4G(R)±1z - z 1 1 2 V 2 G(R)+ 2 By Schwarz's inequality

( OX 0Y OX eY V ex ex + ay ay )

<

(

ax aY±ax Ox Ox

ay . ey ay )

)1 [(

[(aaxx ) 2 + (oaxy 2

= 41z —z1 2 G'(R) 2 1.0z) 2

a yx ) 2 +

(6.22)

Differences of successive coefficients

172

by (6.20) and (6.21). Substituting in (6.22) and noting that

2M4 R OR) = (m2 ± R2)2' we deduce (6.19).

6.3 Estimates for the first term in (6.19)

In order to prove (6.2) and (6.5) we apply (6.13) and estimate the terms on the right-hand side. For each p, such that 2 3 1, 1 -- < p < 1 — — we apply Lemma 6.1 with v(z) given by (6.18) and M = M. We obtain three integrals corresponding to the three terms on the right-hand side of (6.19). We denote by C a positive constant depending on p only, not necessarily the same each time, and by CI, C2, ... particular constants C. Lemma 6.3 Suppose that f(z) is as in (6.1), mean p—valent in 0 < iz I < 1 and normalized so that

=

E a = 1;

(6.23)

z = re'''. Then we

further that G(R) is given by (6.16) with R = have for 0< r < 1 27r

G(If(re i° ))dr9 < C 1 min mop-1)/(2p), (1

— 0 1-4P

}

,

p>

1 -

(6.24)

4

and 2n

10

(1 —

r)rdrf G(if(res(9 )1)d0

3 (6.25) p> _ . 4 We apply Lemmas 6.1, 6.2 with v(z) = G(R) and use (6.17). Suppose first that M < 1. Then < c2 min

{M (4P -3)/(2 P) , (1 _ p)3-4p}

G(R) <

Azi2 < m2-1/(2p) < (1 — 0 1-4".

Thus (6.24) holds with C 1 = 2n. Suppose next that M> 1 and r < Then If (rew )
1 < M < (1 —

6.3 Estimates for the first term in (6.19)

173

Let Fr be the set of points in < 1z1 < r, where M2' < if(z) I < M2 1-v. Then Theorems 2.3 and 2.11 yield if (z)1 < C3(1 —10 -2P.

(6.26)

Thus in Fv we have M2' < C3(1 — z1) -2P i.e.

Izl> 1 _ c4m-i/(2p)2v/(2p), where

C4 =

Since also tlog(r/t) < r — t <1— t, we have for z = tei° in F, r— 1 t log - < <2(1 — t) <2C4(2' /M) t

t

2 .

Also, if R =If(z)1, we have

m 4 (m2

R2)

1

<1.(m2

+ R2)3

Thus

f

f

4A44(m2 —

j F., (m2

± R2)3R2If') (z)1 2 log :1 tdtd0 t

<8G4 < 8C4

, ( 2v () 1 / 2P)

IL

If t (z)1 2 -A-1

(2''1, \

tdtdO

p7r(m2i-v)2 2

= C5 (M2- v ) 2-1 /(2P) ,

(6.27)

since f (z) is mean p-valent in Fv . Summing from y = 1 to oo we obtain, writing z = tei°, R =If(Z)1 fr 1.27t 4m4(m2 R1 2.

f 11 2

( M2 + R2 )3 1 I f' (

)I 2 t 102 '' -rt dtd0
p>

'

To estimate the integral over lzi < 1, we note that by (6.26) 3 if (z)I < C7, for 1z1 < 4 so that Cauchy's inequality yields in case (i) 1 it(z )1 < 4G7 , Izl < •-•• i

i

1 (6 28) 4. .

174

Differences of successive coefficients

In case (ii) we have by Schwarz's Lemma f(z) zP

3 < G4) P C7, Izl< 4.

This yields

f(z)

f(z)

4P-1-1

P zP+1 < 3P C7,

zP

2

i.e.

It(z)1 < CsIzI P-1 ,

1 1z1 < i .

Hence we have in all cases in 1z1 < 1

It(z)1 < C91z1Y, where y = inf(0,p - 1). Thus [27r

f1

1

1 7

1 If/(z)1 2 t log :1 dtc/0 < 27cCr f t2;'+1 log -t dt < c10.

t 0 J0 On combining this with (6.28) we obtain

f

m 4 (m 2

R2 ) ,

R 2 )2 I L l
If

0

r

1 ..2-02p) , (z)1 2 t log -t dtcle < Cii m P > -4.

On applying Lemma 6.1, with v(z) = G(1.f(z)l) and using (6.14) and (1 - r)-2P. If v(0) < If(0)1 2 1 we obtain (6.24) if p > and M M > (1 - r)-2P, (6.26) yields for IzI r R __ M(r,f) < C3(1 - 0 -21) = C3M0

say, and m2R2 , MriR 2 (1 + CD = (1 + Ci)Go (R) G(R) = (m2 + R2) < R` < - Mo2 + R2

say. Applying (6.24) with Mo instead of M we obtain

L

2n

2n

G(R)d0 _<_ (1 d - CD f G0(R)d0 o < C12MO2-1/(2P) = C12(1 __

0 1-4p .

This concludes the proof of (6.24). To obtain (6.25) we integrate (6.24). Suppose first that M < 1. Then G(R) < m2 <

175

Examples

so that (6.25) holds with C2 = n/3. Next if M __ (1 — p) 2P, we deduce from (6.24) that 2n

P

io

(1—

r)dr f G(If(rei ° ))d0 Jo


fo P (1 — 02-4 dr < 4pC1— 3 (1 — p) 3-4 (6.29)

as required. Finally if 1 < M < (1 — p) 2P, we define po by

M =(1 — po) 2P, so that 0< Po < P. Then (6.24) and (6.29) applied with p instead of po yield 2n

P

io

(1 —

Po

r)dr f G(If (re))dt9 = f + 0

0

f

p

Po

P

cl

< 4p — 3 ( 1 — Po) 3-4P + C1 f M(4P-1)/(2P) (1 — r)dr Po
f (1 _ p 1 4p1— 3'

) 3-4P + _1 (1

po2m(42p)} = C 2 M ( 4P- 3 2p ) .

2

This proves (6.25) and concludes the proof of Lemma 6.3. We leave the case < p __ -43 as an exercise.

Examples

6.1

Prove that, if M > 1, P2n

jo (1— r)rdrJof G(If(re i° )1)d0 < 6.2

{ C13

minllog(eM), log

Ci3,

Prove that if M < 1, p > 0

L

2n

P (1—

r)rdr .1 G(Iffre i° )Dd 0 < Lr A4 2 . 3 o

176

Differences of successive coefficients

6.4 A 2—point estimate In order to deal with the second and third terms on the right-hand side of (6.19) we need a two point estimate for mean p—valent functions, which has independent interest. Theorem 6.1 Suppose that f(z) satisfies the hypotheses (i) or (ii) of Section 6.0 and is normalized so that (6.23) holds. If z1 = z2 = p2e2 are two points in lz1 < 1, such that (6.30)

I f (z 2 )

If (zi)I

and a,b are positive numbers, then we have

if(zi)1 02 +2abv2p

if(Z2)1 b2/2P

< Co(p, a, b)

1 (l

p' 2

1 ( 1

1

2ah — P2)b2 lz1 — z2

(6.31)

where the constant Co(p,a,b) depends only on a, b and p.

The result is due to Hayman [1963] when a = b, and in the general case to Lucas [1968]. To prove Theorem 6.1 we shall use the case k = 2 of Theorem 2.4. In the proof C14, C15, ... will denote constants depending on p, a and b. Suppose first that

If (z2)1 = If (zi)I = R, and that

pi

P2 •

(6.32)

We define (6.33)

= lei°2 — ei0'

and first prove (6.31) with i instead of 1z2 — z 1 1. We may also assume that ii

> 3P+ 2 (1

10 1 ) .

(6.34)

For if this is false we have by (6.26)

If(z2)1 (a2 )/2p lf(zo pab+b2 )/(2p) _ R(c1) 2 /2p < m(pi,f) (a+b)2 /2p f4a+b) 2 /2P —(a+b)2 u3 ( 1 — P1)

C14

(1

pi ) 2 (1

p2 )112 11 2ab'

(6.35)

which is the desired conclusion. We suppose therefore from now on that (6.34) holds, so that pi > 4, since ri < 2. We choose ro to be the smallest number P2 such that < Pi, 1 - < ro 1 — (6.36)

177

6.4 A 2—point estimate and

f (z) # 0 for ro — 1(1 — ro) <1z1 < ro +(l — ro).

(6.37)

We start by showing that (6.36) and (6.37) are compatible. We recall (see Section 2.3) that, since f (z) is mean p—valent, f (z) has q zeros in 0 <1z1 < 1, where q p. Thus at least one of the q + 1 annuli

r/

<(1 —1z1) < 111 3 1—v , y = 0 to q 8 is free from zeros of f(z). We choose the smallest associated value of y and define 1 ro = 1— Thus (6.37) holds. Also

— > 3—t 1 — ro > — 3—q 4

>1

‘ P+

—

pi

This yields (6.36). 1— by (6.34), while ro We now apply Theorem 2.4 to the discs

An : { — roewnl < 1 — ro}, n = 1, 2. These discs are disjoint since, by (6.36), role if92 — e

2(1 — ro).

=

Also by (6.37) f(z) # 0 in lz — r0e 0 < p—valent in A 1 U A2. We take z i , z2, z 1 , z2, rn in Theorem 2.4. Then (5 n —

1 p 1 — ro

— ro) and f(z) is mean roe'92 , (1 — ro), instead of

(6.38)

,

and R2 =f (z2) I = If (z 1)1 = R

by (6.32) while

2P lf (roew.)1

2P M(ro,f)= R1

(6.39)

say. Now Theorem 2.4 yields, if R2 >

--1 ( llog

Ao + (log — (5 2

--1

2p

< log(R2/Ri) — 1.

We write log

Ao = X 61 a

Ao Y log .75;- =

(6.40)

178

Differences of successive coefficients

and note that, by Schwarz's inequality,

a2 b2 Y X (X±Y)(-x—+-37 ) = a2— + b2— + a2 b2 X Y 2ab

a2 ±b2 = (a + b)2 ,•

y X ) 2

with equality when Y /X = bla, i.e. (,40 /62 )b = (A0/61)a. Now (6.40) yields

(a + b)2

2p X ± Y < log(R2leR1) •

Taking exponentials and using (6.26), (6.36), (6.38) and (6.39) we obtain R(a+b)2 12p <

<

(eRi)+b)212p (A0)a2 61 )

(1_ roi

15 0 ro)—(a+b)2

I

ro) 2abo

C 161,1 -2ab

ra2 0

p i ra2

b2

P2

Pl C15 (1

— ro

p2 rb2

p2 )—b2 .

(6.41)

We have assumed that R > eR i . If R < eRi we have R(a+))2 12p

since ro

pi

C17 ( 1 17 ( 1

r0)'2 ro )2ab ( 1 C

p ) -a2

p 2 rb2

p2 by (6.32) and (6.36), so that (6.41) always holds.

6.4.1 The case If(zi)i < If (z2)I To complete the proof of Theorem 6.1, we now suppose that If (zi)I < If (z2)I. We recall (6.26), so that (pei° ) < C3 (1 — p) 2P, 0

p < 1, 0 < 0 < 2m.

Since f (z) = zP (1 — z) -2P satisfies our hypotheses, C3 that

lf(z i )1

C3(2 x 3P+ 1 )2P.

Then (6.31) is a consequence of (z 2)1
So we suppose from now on that

2P lf (zi)I = Rt,f (z2) I = R2

(6.42)

1. Suppose now

179

6.4 A 2—point estimate where R2 > 2—P R1 > C3(2 X 3P+1 ) 2P > 1.

(6.43)

Then if(0)1 < R1 by (6.23) and so we can find z =

, such that

< < p2

2f (z) = R1 and 1 —

by (6.42) and (6.43). We choose the smallest such p'1 , and define

6 = 1 — p'1 , so that 3P±1 6 < 1

(6.44)

We note that, since f has q zeros in 0 <1z1 < 1 where q < p, at least one of the annuli

1 3"6 < (1 —1z1) < -13'+ 1 6, v = 0,1,...,q is free from zeros of f(z). We choose the smallest such y, set z o = (1 — 3v 6)e i°2 , and apply Theorem 2.4 to the single disc lz — zol < 36, with zo, instead of z 1 , z . Then 211f (Z0)1

R1, if(z2)

Z2

1 — P2

I = R2 ,

=

and f (z) # 0 for z — z o l < -13v 6 . Thus the hypotheses of Theorem 2.4 are satisfied and we obtain, using (6.44), either R2 < eR i or

2p )2p AO R2 < eRi ( ) —< Ci 8R 1 (1 Pi 61

(6.45)

so that (6.45) holds in any case. We next apply (6.41) with z'1 instead of z2. This is legitimate since If(zOl = 2PR 1 = lf(z 1 )1 and we have proved (6.41) in this case. We obtain kia+b)2/2P < c16 (1

1 pi )a2 (1

p/i )b2 ri 2ab •

Combining this with (6.45) we obtain R (a2 +2ab)1(2p) D b2 1(2p) 1

kia+b)2 /(2P)(R2 /R i )b2 /(2p)

Ci9( 1 /( 1 — Pi ) a2 ( 1

P2)b2 q 2ab ).

(6.46)

To complete the proof of Theorem 6.1 we need to replace ri by 1z2 — in (6.46). Suppose first that ip2 — p l Then

1z2 — zi

IPi ei81 — P2e i°1 + I P2e 162 — Pze lel

P2 — p1I + < 2j

180

Differences of successive coefficients

so that (6.31) follows. Suppose next that 11

If p 2 yields

<1P2 -

1 < max(1 - p2, 1 - P 1).

pi, we have ri <(1 - p2), and 1z2 -1-2ab)/(2P) R (a2 1

zi 1 < 2(1 - p2). Then (6.42)

D (a2 +2ab)/(2p) , r(a 2 +2ab)/(2P)0 _ p2 )—(a 2 -F2ab) -"2 --- t-- 3 c(a2 +2ab)/(2P) 22ab( 1 _ p i ra2 1z2 _ 1-2ab Z1 1 3

< —

<

If pi < p2, we have 1z2 R (a2 +2ab)/ (2p)

-

-

z1 1 < 2(1 - pi). Hence

<

c (a2 -1-2ab)/ (2p) 2 —lab ( 1 - Pi)-a 3 (a 2 +2ab)/(2P)22ab( 1 — 1 —2ab C3 1Z2 — Z11 p i, '

1

-,

<

(6.47)

So (6.47) holds in both cases. Also, again by (6.42), R2b2 /(2p) < c3b2 /(2p) ( 1 _

On combining this with (6.47) we obtain (6.31) and the proof of Theorem 6.1 is complete.

6.5 Statement of the basic theorem We now state our result on coefficient differences. Theorem 6.2 Suppose that f(z), given by (6.1), is mean p-valent in the sense (i) or (ii) of the introduction. Then for n > 1 flan-Ell -a lnlf where p =

{ C20

it n2P-2 ,

C20 ,U112 P-2 ■/P,

E lay', and the constant V

C20

if p > 1 if < p <1

depends only on p.

p

We continue to assume that p = 1, since otherwise we may consider f (z)/ p instead of f (z). In this section we deal with the second and third term on the righthand side of (6.19) by means of Theorem 6.1. We shall put our results together to prove Theorem 6.2 in the next section. We define s(p) =

p>1 {3, (2.0) - )2 1, 1 < p <1

(6.48)

and Y(P)= s(p)/(4p).

(6.49)

181

6.5 Statement of the basic theorem Lemma 6.4

We have

f f z — z 1 1 2 1t(z)1 2 (1— r)rdrd0 <

(6.50)

C2ins(P) 2 -2".

Here the integral is taken over all points, z = rew for whichlf(z)j< M v , and 0 < r
If(z)1 < M.

We recall the definitions (6.8) and (6.9), and have If(z i )1 = A41 and If(z)I > Mk+1 = M1 2—k in Fk. Thus Theorem 6.1 applied with z, z i instead of z1 , z2 gives Conb2

uf (a2 +2ab)/(2p) A,,b2/(2p)

Ivik+1

(6.51)

(1 — r) a2 Z1 1 — z12ab

Suppose first that p > 1. In this case we choose a = b = 1 and obtain in Fk

(1— r)lz i — z1 2 < C0nMk+311(2P) M i

2p)

Also since f is mean p—valent and III < Mk in

conw 2h323k1(20 . Fk

we have

(6.52)

(z)1 2 rdrdt9 < TcpMi = 47.cpM2-2k .

f fFk Thus 2/p 231020 7.cpmi2 -2k 1t(z) 1 2 1 z — z11 2 ( 1— r)rdrd0 <4ConA1 1 /1 2-2yk . i = 4ConnpM21 -2/P 2(3/(2P)-2 )k = 4Co nrcpM 2-2

Summing from k = y to co we obtain 1-2yv

ff

i z i
It(z)1 2 1z — zi1 2 (1 —r)rdrd0 < 4ConnpM12-2IP 1 — 2 -2Y .

lf(z)I<m,

Also, by (6.26), M 1 < C3 n2P. Now (6.50) follows. Suppose next that 14 < p <1. We apply (6.51) with a = 1, b = This yields in Fk mi2.2—k(14,1p-1)/(20)

ur (4.0)-1)(2p),,,(2Vp-1) 2 /(2p) < k+1

Ivi 1

GOP) (1 — r)lz — z 1 1 4.1P-2

so that (1 — r)lz — z i l 2 < 4(1 — r)lz — z i l4 VP -2 <4Cons(P)2k(4 Vp-1)/(2p)w2 ,

— 1.

182

Differences of successive coefficients

since 0 < 4 — 4.03 < 2, and lz — z i l < 2. Using this and (6.52) we obtain

IL

It(z)1 2 1z —zii 2 0 — Ordrd0 < .

167cpCons ( P) 2—k(2.0)-1)2 /(2p )

167I pCo ns(P) 2-27k .

Summing from k = v to co we obtain (6.50) also in this case. This completes the proof of Lemma 6.4. It remains to deal with the third term on the right-hand side of (6.19).

If z = rete , If (z)1 = R and M = Mv we have

Lemma 6.5

Iv =

f

f 1z1
M4 R lz — zdIf (z)I (m2 4.. R2)2 (1 — r)rdrde (6.53)

< C22ns(P) 2"-I

' v .

We have by Schwarz's inequality Iv<

(f

21 flz,
2 2 lz — zi1 1f(z)1 (1 — r)rdrd0)

11 x (I fizl
(6.54)

For the first integral we apply Lemma 6.4 with y = 1 and obtain

f

lz — z i 1 2 1f (z)1 2 (1 — r)rdrd0 < C21 ns(P) .

(6.55)

izi
To estimate the second integral we note that m2R2 M8R2 (m2 + R2)4 (m2 + R2) = G(R), and apply Lemma 6.3 and Examples 6.1 and 6.2. Suppose first that M > 1. If p> i, we note that s(p) > 4.p — 3. Then by (6.26) < ms(p)/ (2p) .: (A 41 21—v)s(p)I(2p)

< c23n s(p)2-2yv .

(6.56)

If p ._ i we use Example 6.1. Then

max{1, log(eM)} < C24 M s(P)/(2P) < C24 C23ns(P) 2-2" . If M < 1 we use Example 6.2. Again M2 < Ms (P)/(2P) < C23ns(P) 2-2Yv. Thus in all cases

f

ii m8R2 k lz1
1

-

r)rdrd0 < C25ns(P)2-2yv .

6.6 Proof of Theorem 6.2

183

On combining this with (6.54) and (6.55) we obtain (6.53).

We write z = reie , R = If (z)I and

6.6 Proof of Theorem 6.2

v v (z) = lz — z i 1 2 Gv (R), where mv2R2

Gv(R)=

m2 + R2

Suppose that 1 — 3/n

=

f

2n

2

tdt f (log max (t,

0

We note that if

v v (-1e10 ) dû

V 2 v v (te`19 )d0. o


p t

p— t <2(1 — t) t

1 = log log <

and since log(2p) < log 2 < 1, this inequality remains valid for t < We deduce from (6.17) that 0 < V2 Gv (R) < 41f(z)1 2 , if If (z)1 < M, and V2 Gv (R) < 0 otherwise. We define x v (z) = 1, where If (z)1 < Mi,, and x(z) = 0 elsewhere. Now (6.19) yields p2n .

10 V v (pei° )d0 — f v t, ( 12 ei° ) dO 0

2 f (1 — r)rdr f {4 41z — z11 2Xv(z)if'(z)1 2 G,(R)d81z — ziIM R + R2)2 If/(z)1} dO (6.57)

= 4,1 + 42 +

say. By Lemma 6.3, (6.25) and Example 6.1 we have, if Mv > 1,

c m (4p-3)/ (2p) (1 — r)rdr

G(R)de < 0

2 r Ci3 10g(eM v

3

p> 4 P -

184

Differences of successive coefficients

Using (6.56) and noting that s(p) > 0, we deduce that / v ,i <

C26 ATI! (P)1(2P) <

C26C23n"2.

If M,, < 1 we have G% (R) < M so that m lv ,i < 16111,2,7r f (1 - t)tdt 8 7r ms(p)/(2p) < 3 3

8. z ir ' 3

c23n s(p)2-yv

Thus in all cases < C 27 ns(P) 2-7 v.

Next Lemma 6.4 shows that =

8 f f x v (z)lz - z11 2 If '(z)1 21.(1 - Odra+ < 8C2i ns(P ) 2-".

Finally Lemma 6.5 shows that 1v ,3 < 16C22ns(P) 2-".

Now (6.57) yields 27r

fo Again for Izi =

v y (peie )d0 <

(e10 ) dO C28ns(P) 27"

o

we have by (6.26) and (6.56) if M,, > 1

VV(z) < 4R 2 < 4Ci(16)P <4016)PM 1+1 C1 C23 ns(P) 2-2'''',

2P) <(16)

and if Mv <1 v v (z) < 4M

4M,s, (P)/(2P) < 4C23ns(P) 2-2".

Thus we obtain finally 27r

fo

3 2 v y (pele )d0 < C29 ns(P) 2 -", 1 - - < p < 1 n

On substituting this in (6.13) we have

CI c29 E (n x

ilanl

80

This proves Theorem 6.2.

7r

)

v=1

s(p)_1 2_,v ) 71

< c20 0 sky)

2 .

6.7 Coefficient differences of k-symmetric functions

185

6.7 Coefficient differences of k-symmetric functions We apply Theorem 6.2 to k-symmetric functions. Suppose e.g. that f (z) is mean p-valent in A :IzI < 1 and has a power series development p+2k +'•• 1(z) = Z P + ap-Fk ZP-Fk + ap-F2kZ

(6.58)

Here p > 0, k is a positive integer and we assume as in (ii) of the introduction that, if p is not an integer, f(z)/zP is regular in A and that f(z) is mean p-valent in A cut along a radius z = rel° , 0 < r < 1. We make a substitution C = Z k and note that F(Ç) = f(C 1 /k )= CP/k (1+ ap+kC + ap-E2k 2 + • • .) is mean (p/k)-valent in ICI < 1 in the above sense. (cf. Example 2.8, or Section 5.8.) In fact a sector 2m S : 0 1 < arg z < 0 1 + — 0 < Izi < 1 k ' corresponds (1,1) conformally to S': kO i < argC < kO i + 2m.

Since F(Ceii) = 0F(ç) we deduce that p(R, S', F) = p(R,S, f) is independent of 0 1 . On applying this conclusion in turn with 0 1 + 27ry/k instead of 0 1 , where y = 0, 1, 2, ...,k — 1, we deduce that p(R, S', F) = p(R, A, f)/ k.

Thus, since f is mean p-valent,

L

R

p(t, S' , F)tdt = k1 f0 RP(t„ntdt S _. kP R 2, 0 < R < oo

so that F is mean (p/k)-valent. We may now apply Theorem 6.2 to F( ) and deduce Lucas' [1969] Theorem 6.3

If f(z) is mean p-valent in IzI < 1 with the expansion (6.58)

then lap+nk —

ap-- -F(n-1)kl <

C30n2P/i(-2 ,

i c30 n2p/k-2(p/k)7 ,

if p _- k

if < p < k .

186

Differences of successive coefficients

If p = k = 1 we obtain the right order of magnitude [Hayman 1963] Ilan+i I - larill <

(6.59)

C30

for the coefficient differences of mean univalent and in particular univalent functions. If p = 1, k = 2 we obtain the bound 1 la2,11-1 I

—

I a27—i I 1 < C 30 n l-

. c30 n -.4142...

/2

(6.60)

for the coefficient differences of odd univalent functions. If p = 1, k = 3,

we obtain 1 la3n+1 I — I a3f1-21 1 <

C3on213-21 `i 3 = C30n-.4

880..

(6.61) •

If k = 4 we have from Theorem 3.7

l ia4n+ii - la4n-311


if f is univalent, and from Theorem 3.5 Ila4n+i I - I a4-3I 1

< And

log(n + 1),

if f is mean univalent. The example of Section 3.4 shows that in the latter case we cannot at least replace n- 1 log(n + 1) by anything smaller than i o(n). The results (6.60) and (6.61) are the best that are known even for univalent functions, and no other proof than that of Lucas exists as far as I know. However in the univalent case Grinspan [1976] has obtained the best known bounds -2.97 < lan-Ft I - Ian I <3.61.

6.8 Asymptotic behaviour It is natural to ask what functions attain the maximal growth in Theorem 6.2, at least when p > 1, so that the conclusion of Theorem 6.2 is sharp. If a > 0, p = 1 in Theorem 2.10 or 2.11 and f is circumferentially mean p-valent I proved [1963] that

lan-Fil - lanl

cx

(6.62)

and in fact lan+1 —

ei°°anl-* a

(6.63)

as n -> co, where arg z = 00 is the radius of greatest growth and p > 1. The argument extends to a.m.p-valent functions, now that Eke's Theorems 2.10 and 2.11 are known. Also if p > 1 and a = 0, (6.62) and (6.63) still hold. It is merely

187

Examples

necessary to modify the estimates of Lemmas 6.3, 6.4 and 6.5 to the case when (6.26) is replaced by M(r, f) = o(1 — r) -2P as r --- 1. This conclusion breaks down when p = 1 as the examples fe(z) =

1_

00 z „sin nO = Ez sin 0 2z cos 0 + z 2 o

(6.64)

show. However Eke [1967b] showed that the only mean univalent functions for which lan+i I — Ian ' does not tend to zero are the functions with positive a, for which (6.62) holds, and functions with maximal growth on two rays for which lan+ il— lan l oscillates. Finally Hamilton proved [1982] if f E S, and [1984] if f is mean univalent and f'(0) = 1, that lim supf lan+i l— lan if < 1 n--.00

unless there is a real 0 such that z or eiç b f (ze-4) = (1 — z) 2 where fo(z) is given by (6.64).

Examples 6.3

If fo(z) is given by (6.64) prove that

I lan+i I — Ian' I

1.

Show also that, if 0 is a rational multiple of 7t, equality holds infinitely often while, if 0 / n is irrational, strict inequality holds for all n, but lim sup(a n+ i — an ) = 1 n—oce

and lim inf(a n+i — an ) = —1. n-0oo

Eke's Theorems are refinements of the regularity theorems of Chapter 2 and their proofs would take us too far. However some other results can be proved fairly simply.

188

Differences of successive coefficients

Examples 6.4

Suppose that f(z), given by (6.1), is mean p—valent in the sense of the introduction, where p > 1, and that a > 0 in Theorem 2.10 or 2.11 respectively and arg z = 00 is a radius of greatest growth. Show that lan+ 1 — an e i6° I = {

a

F(2p — 1)

4- o(1)}

11 2P-2 ,

as n .— co.

(Apply the Cauchy integral formula in the form

an+ i — ?clan =

(z — ei°°)f (z)dz

Zn+I

/1: 1=rn

to f(z) and f n (z) = an (1 — ze —i(9°) -2P, where rn = 1 — 1/n, and a, = n-2Pf(rn ei°0 ), and subtract. Use Theorem 2.10 on the major arc 10 — NI < Kin, where K is a large constant, and Theorem 2.7 on the minor arc.) Extend the conclusion of the previous example to the case p> by using the formalism of Section 6.1, and in particular (6.12), on the minor arc. (Deduce from Theorem 2.7 that

6.5

JE,

lz — z 1 } 2 1f(pe' e )} 2 pdpd0 = 0 {n4 P-4 243/(2P— ')-21} .)

If p > 1 and a =0, show that

6.6

Ian+ i I — Ian I = o(n 2P-2 ) as n -- co. (In this case I, =--- o {n4P-3 2-7 ' } in (6.57) for j = 1, 2, 3. Where does the argument break down for p = 1?)

6.9 Starlike functions Leung [1978] has shown that for starlike functions the behaviour of the functions (6.64) is extremal. We prove his elegant Theorem 6.4

If CC

f(z) = z and

f

an zn

ES

is starlike then for n = 1, 2, ...,

1 lan+1 I — Ianll

1.

(6.65)

6.9 Starlike functions

189

We shall see in the next chapter that this conclusion fails for general f 1n. We need a lemma of MacGregor [1969]: Lemma 6.6

Suppose that oc

oz) = 1 + E cnzn ,

and

00 tp(z) = E A n cn zn 1

are regular in A : Izi < I, that A n 0 and that for z E A

910(z)

0,

911p(z) _..

M.

Then oc

E Anicni2

2 M.

(6.66)

n=1

We write cn = xn + iyn and for 0 < r < 1, 0 < 0 < 2n, 09 u(r,0)= 910(rei° ) = 1+E(xn cos ne — y n sin nO)rn n=1

and

00

v(r,0)=9itp(rei° ) = E A n (x n cos nO — y n sin nO)rn. t Then since u(r, 0) > 0 and v(r, 0) < M we have 2n

2TcM =

2n

Mu(r,O)d0 f u(r,O)v(r,O)d0

io

Jo

GC)

.

n E 1n(xn2 ± yn2 ),.2n . n=1

Letting r tend to 1 we obtain (6.66). We deduce Lemma 6.7 If 4)(z) is as in Lemma 6.6 and n > 0, there exists C, such that ICI = 1 and

190

Differences

We choose

Ak =

of successive coefficients

lik for k < n,

Ak =

0 for k > n in Lemma 6.6 so that n

1

E 1k-ckz k ,

tp(z) =

k=1

sup 9itp(z). Izi---1

M= Then, by Lemma 6.6,

n

n

n

l ei<

— k 1 2C

1 — Ickl 2 k

=

k=1

— 291tp(C) ±

E

1 }

k=1

n

1

< 2M — 2911,v(C) k=1

Choosing C so that 9100 = M, we obtain the desired result. We can now prove Theorem 6.4. If f(z) is the function of that theorem we write GC zf'(z) k Ck Z =1+ OW = f(z)

E k=1

and recall from (1.15) that 910(z) > 0. Let C be the point whose existence is asserted in Lemma 6.7. We have log

f(z)

=

z

fo z

OW — 1 t

co

Ck k dt = E — z . k k=1

Thus GC

log {(1 —Cz) f (z) Z

1 = Eakzk,

where

1

Œk

k=1

= —k (ck

—

rk \ )

4D

*

On the other hand

f(z) (1 — Cz) z

00

= Ep0kz , where fik = ak-1-1 — tak. k

1....

k=0

Thus Do

.

E fikz k = exp (E akz k ) . k=1

k=0

It now follows from a theorem of Milin [1971] that

k 2 lockI 2 — 1 } IfinI 2 .. exp { Ê k k=1

(6.67) •

6.10 The theorems of Dawei Shen

191

We defer the proof to Chapter 8. The inequality (6.67) is a special case of Theorem 8.4, with Ak = Œk, Dk = fik, A = 1. Using Lemma 6.7 we obtain ! fl

= I an

—

Can I

1,

and this yields Theorem 6.4. We note that the functions (6.64) are starlike and yield equality in Theorem 6.4 if sin nO = 0, i.e. 0 = kit/n, where k is a positive integer. Equality also holds for all n, when f (z) = z / (1 — z) 2 .

6.10 The theorems of Dawei Shen We conclude the chapter by proving some of the results of Shen [1992] referred to in the introduction. We start with Theorem 6.5

If

f(z)= z c3z 3 c5z 5 - • • E

a

(6.68)

and a=

— r)M(r, f) > 0

then I C2n-F1 I

IC2n-11 =

0(n") as n

co,

if [3 > flo, and flo = .490... is the constant of Theorem 3.6. We write F(z) = f (z 1 ) = z(1 ± c3 z c5z 2 ± • • •),

(6.69)

so that F(z) is mean -1—valent in lz I < 1 in the sense of the introduction. Also F(z) has maximal growth and so a unique radius of greatest growth arg z = Oo by Theorem 2.11. Again (6.11), with n = m z 1 = (1— 2/(2m + 1)) eitlo and p = (2m — 1)/(2m + 1) yields (Pr/

1 — —) 2

We choose a positive number IF(peiÛ )I

1 2npm -3/2 fo

c2m_1 — e ie °c2m+i l E,

— e 16° II

(peie )40 .

(6.70) and note that, by Theorems 2.7, 2.11

Co (1

— p)Ele — 001 1- E

nCo (1—— Oor

if p o

192

Differences of successive coefficients

where Co is a constant. Also lei' — e i0°1 <10 — 001. Thus foo+re

n co

l ei() _ 100 e 11r (pem9 )1d0

reo +II

/

IF (Pe )1

(1

00-7T

p)8 J00 _, IF(Pe1°)I d0

rap+

_ rc C0 2( 1 — p )P

f(Plel i° ) d0

—n

)

by Theorem 3.6. Here Co, C1 are constants depending on fl, e and we can choose fi + E as close as we please to fio. Now (6.70) yields Theorem 6.5.

6.10.1 Some weaker growth conditions enable Shen [1992] to obtain (6.6) or something close to it.

Theorem 6.6

Suppose that f(z) is as in (6.68). If

M(r, f) = 0(1 —r),

(6.71)

icn i = 0(n-1 )

(6.72)

we have

and hence

I le2n4-11 —

= 0(n-1 ).

On the other hand if there is a radius arg z = 00 such that

lim inf(1 — r) 1 If (re:6°H > 0, r—ol

(6.73)

we have

I

IC2n+1 — 1C2n-1 =

log

as n —0 co.

(6.74)

The conclusion (6.72) is an immediate consequence of Baernstein's Theorem 3.7. Shen's deduction of (6.74) from (6.73) lies deeper. In Theorem 6.6 the hypothesis f(z) G S was only used to obtain (6.72). For the rest of the proof we assume only that f(z) is mean univalent. Following Shen we base our argument on

Lemma 6.8

Suppose that F(z) given by (6.1) is mean p-valent in 1z1 < 1 in the sense of Section 6.0 and that

lim inf(1 — p)P1F(pe10°)1 = a > 0,

(6.75)

193

6.10 The theorems of Dawei Shen

where 0 < 00 < 2n. Then there is a constant Ko, independent of z = pe such that 1F(z)1 <

Ko

0 < p <1,

(1 — p)P10 — Oo2P l

— 001 n.

(6.76)

We denote by K0,K1,..., positive constants independent of z. By (6.75) there exists p o such that 0 < po < 1, and

Ki

IF(pieiN)1>

( 1 — Pi)P

(6.77)

, if Po < Pi < 1-

Hence if R> K2 = IF(POe ie° )1, there exists pi, such that po

pi < 1 and

If(Pie i°°)1 = R.

(6.78)

Suppose that 1F(z)1= R. If R < K2, (6.76) holds. If R > K2 we choose p i to satisfy (6.78) and apply Theorem 6.1 with z i = z, instead of z1,z2,1F(zi)1=1F(z)1= R, and a = b = 1. This yields 2

R < We have, if 10

—

(6.79)

(1 —p i )(1—p)lz _z,1 2

Ool

—

= Pi + p2 — 2p pi cos(0 — 00) (P1 cos(9 — 00) — )0 )2 + pi sin2 (0 — 0) 4 p i(0 — 00) 2 ' n2

while if

<10 — 0o _< n, we have iz — z 1 1 2 > pi. Thus in all cases lz zi 1 2

_2 (0 P1

2 —0

2 (0 —

Po

2

00 2

772

'

Also by (6.77) we have 1 —

R

)

1IP

Thus (6.79) yields

R i1 P < (

1— p)(0 — 002 '

which is (6.76). We can now prove Theorem 6.6. We start with (6.11), applied to F(z) = f(z1), with n+1 instead of n, so that z1 = {(2n — 1)/(2n + 1)}ei°°,

194

Differences of successive coefficients

1 — 3/ (n

p K /

dt9

— 2n+6

This yields

1Pel" — zillr(z)lpdpde

F'(z) 2 F(z) pdpd0) (f lz — zi1 2 1F(z)1 2 )

f


4 1— 2n+1

2n

00 ei C2n+11

1c2n-1

1 — 21 (n +

7

(6.80)

By Lemma 3.2 we have

If

F'(z) 2 pdpdt9 < K log n. F(z)

(6.81)

To the second integral we apply Lemmas 6.1, 6.2 with G(R) = R 2 . Letting M tend to co in (6.19) or directly we obtain for v(z) = lz — zi 1 2 R2, V2V(Z) < 4R 2 + 41z

— Z11 2

Also Lemma 6.1 gives for

L

2m

Iff (Z)I 2 + 81z

Z1

IRT(Z)1.

(6.82)

p <1

2m

2m

v(pe w )d0 f

fp

dt9+K f (14)

V 2v(te 14) )(1—t)tdt. (6.83)

We estimate the integral on the right-hand side by means of (6.76). We have from Theorems 3.2 and 2.3 that 2m

K 1F(te le )1 2 dt
2n f Jo

C1(1)

fp IF (t el 411 2 (1 — t)tdt < K.

(6.84)

Next we have

f fl(1

where Iv (Z) = f f (1 IZDIZ E,,

Here if° < y < N, fy is the subset of lz1 < p where M1, + 1 < 1F(z)1 < Mv and Mv is defined in (6.8). We choose N to be the first integer such that MN < 1, and define EN to be the set where 1F(z)1 < MN. We recall that by Lemma 6.8, with p =

1F(te l° )1 <

Ko

(1 - W10 - 001

6.10 The theorems of Dawei Shen

195

where 00 is as in (6.73). Hence if 10 - 001 > (1—t) we have, with zi = piewo,

a = tei°, 0 < t < pl ,

lz — zil

it — Pt 1 + 'Pie° — Peitl° 1

I pi — tl +1 0 — ool < (1— t)+ le— NI <216— NI. Thus in this case we have in Ev , if 0 < y < N,

lz — z i 1 2 (1 — lzi) < 4(0 — 00)2 (1 — t) <

4K 2 16K 2 ° = Ai12° .

kr2 -

1)+1

On the other hand if 10 — 001 __-_ (1— t) we have

lz — zt i < 10 — NI + ( 1 — t) < 2 (1 — t). Thus in this case

lz — z1 2 (1 — t)Mv2 < K(1 — t) 3 M? < K. Thus for 0 < y < N we have in Fy

lz — z11 2 (1— t) <

K M'

so that

f fF

lz — z1( 1— 0111z)1 2 1dz1 2 <

= K.

Also in FN we have 1F(z)I <1 so that

f IN l z

-

zti 2 ( 1— t)Ir (z)1 2 1dz1 2

4 f IF' (z)1 2 1dz1 2
Thus, for n > 6,

f Lp (1 — lzpiz — z i i 2 ir(z)lidzi 2 < K(N +1)
MN-1

(6.85)

A ,f 1 = 22--N NI > 1, so that

2N+ 1 < 8M1 < Kn. Finally by (6.84) and (6.85) we have

f fizi

196

Differences of successive coefficients

On combining this with (6.82) to (6.85) and noting that v (e 10 ) < K we obtain 2n

3

v(pe l0 )d0 K log n, 1 — -

2

p <1 — - , n > 6. n

Now (6.80) and (6.81) yield

— eal()c2, 4_ 1 1

Kn -7 log n.

This is (6.74) and completes the proof of Theorem 6.6.

Examples

6.7

If F(z) satisfies the hypotheses of Lemma 6.8 prove that,

(If p <

0(nP-1 ), as n

co

if < p < 1;

as n

co,

if p <

—

=

—

= o(n),

choose .1 so that 2 <2 < 1/p, and use instead of (6.12)

— land

< A(p)E f

+ (z)Ifl2 zd

ffE,

— Zi1 2 (If(Z)! Â +1) IdZ1 2 } .)

v=1

6.8

If f(z) is mean p-valent in lz! < 1 and satisfies lim inf(1 — r)" if (re* )! > 0, r--■ 1

where 0 < < 2p, prove that If(pel ° ) 1

0 < 119 — 00 1 < n;

< (1 — p) 2P-119 — Ool 2Œ (apply Theorem 6.1 with a = a, b = 2p — a).

7 The Liiwner theory

7.0 Introduction We shall in this chapter give a deeper theory due to Ltiwner [1923], which enables us to obtain sharp bounds for the class S of functions f (z) = z ± a2z 2 + • • • univalent in lz I < 1. These results do not seem to be accessible by the methods of Chapter 1. We shall say that the class S' is dense in S, if S' is a subclass of S and if every function f (z) in S can be approximated by a sequence of functions f n (z) in S' so that f n (z) —) f(z) uniformly on every compact subset of jz 1 < 1 as n —) co. It will then follow that pth derivatives at an arbitrary point in I z I < 1, and in particular the coefficients of fn (z), approach those of f(z) as n —) co. Thus bounds obtained for the class S' will remain true for the wider class S. Liiwner's basic result can now be stated as follows:

Theorem 7.1 Let K(t) be measurable and complex valued for 0 < t < co and satisfy IK(t)I < 1. Then if 1z1 < 1, there exists a unique function w = f(z,t) absolutely continuous in t for 0 < t < cc, and satisfying for almost all t Leiwner's differential equation Ow = w 1 + K(t)w _ 1 — K(t)w' Ot

(7.1)

with the initial condition f(z, 0) = z. Also

g(z , t) = et f (z , t)

E e, o < r
Finally there is a dense subclass Si of S such that if f(z) E Si, there exists K(t) continuous and with IK(t)I = 1, such that the associated function

197

198

The Lôwner theory

g(z,t) satisfies g(z , t) ----o f (z)

as t --0 co,

(7.3)

uniformly on compact subsets of 1z1 < 1. We shall see that the functions g(z) in S i map the unit disc Izi < 1 onto the complement of a sectionally analytic slit. We recall that if f (z) is the limit of a sequence f n (z) of univalent functions then f (z) is either constant or univalent (C. A. p. 231). If f n (z) E a i , f'(0) = lim f,' (0) = 1 so that f (z) E S. Thus if the limit f (z) in (7.3) exists f (z) E S. In the first part of this chapter we shall prove Theorem 7.1 by (a) constructing the class S i of mappings f (z), which is dense in S; showing (b) that if f(z) E S i then there exists g(z,t) satisfying (7.1) in (7.3) for a suitable continuous K(t); and (c) given K(t) as in Theorem 7.1, the functions g(z , t) given by (7.2) belong to S. The proof of Theorem 7.1 is rather long and not easy. It is however fully justified by its many beautiful applications. In the last part of this chapter we shall use Si to obtain the exact bounds for I a31 and all the coefficients of the inverse function z = f'(w) as well as the arguments of f(z)/z and f'(z) and the radii of convexity and starlikeness. In the next chapter we use Si to prove de Branges' Theorem la,' < n for all n. Liiwner Theory is a very powerful tool for finding exact bounds for functionals, but the method makes it difficult to discuss the form that the extremals take. Sometimes other techniques, which are outside the scope of this book, are more effective for this. (See e.g. Duren [1983] and Jenkins [1958].)

7.1 Boundary behaviour in conformal mapping In this section we prove two preliminary results: Lemma 7.1 Suppose that w = 4)(z) maps a domain D 1 (1,1) conformally onto a domain D2 lying in 1w1 < 1. For any point zo let l(R) be the total length of the image in D2 of that part of the circumference lz — zol = R which lies in D 1 . Then if R 1 > 0, k > 1, there exists R such that R1 < R < kR i and l(R) < n(2/ logk)1. In particular, there exists a sequence Rn , decreasing to zero as n ----o co, such that l(Rn ) ---0 0 as n —0 co. We consider the mapping z = 4)- '(w) of D, onto D i and put 1P(w)

= 0-1 (w) — zo.

7.1 Boundary behaviour in conformal mapping

199

Then the level curves YR in D2 corresponding to lz — zo! = R are the level curves kp(w)! = R. Let l(R) be their total length. Then since ip(w) is it. This gives univalent, we may apply Theorem 2.1 with p(R) ._-. 1, A f: Ri i(R)2dR f ' l(R) 2 dR < < 2 TC A _ < 2 Tr 2 . R — 0 Rp(R) — If 1 is the lower bound of l(R) in R1 < R < kRi , we deduce /2 log k

2n2,

/ <

TC

(1o2g k ) 1 '

as required. If we now define Ri, inductively by R,'9 = 1, Rn' +1 = e'R n' . Then it follows that there exists Rn such that Rn' +1 < R < R and l(Rn ) ._-_ This completes the proof of Lemma 7.1. We shall also need

Lemma 7.2 Let y be a simple arc which lies in lz! < 1 but approaches lz! = 1 at both ends, and suppose that the length 1 of y is less than 1. Let D be the set of all those points P in lz! < 1 such that any curve joining the origin to P in lz! < 1 meets y. Then the diameter of D is at most 1.1. We may parametrize the curve y by z = a(t) (a < t < b). Let tn (n = 0, 1, 2, ...) be an increasing sequence of numbers such that a < tn
(n

1),

(n --) oo).

t n --+ b

Then

E

loc(tr) — a(tr—i)! __ 1,

r=i

and so E

00

,,,,(tr ) —o,(tr_i)1 r=1

1, E

00 r= I

[a(tr ) — a(tr—i)1 converges.

Thus a(tr ) approaches a definite limit a(b) as r --) oo and a(b) is clearly independent of the sequence tr. Thus a(t) —) a(b) as t ----) b— and similarly oc(t) approaches a finite limit a(a) as t —> a+. t

The diameter of a point set E is the upper bound of distances 1z1 —z21 of pairs of points Z1, Z2 in E.

200

The Liiwner theory

We write z1 = c((a), z2 = 0 (b ), z3 = -(z1 + z2). Then 1z11 = 1z21 = 1 by hypothesis. If z is any point on 7, then we have

lz — zi I + lz — z21

/,

since y has length 1 and so z lies inside an ellipse of centre z3 and major axis 1. Hence z lies inside the circle C of centre z3 and radius 1. This circle has diameter /, and since C contains z 1 on 1z1 = 1, C cannot contain the origin. Any point P in 1z1 < 1 but outside C can clearly by joined to the origin by a curve lying in lz < 1 and outside C, and so P lies outside D. Thus D lies inside C and so has diameter at most J.

7.2 Transformations

Consider f (z) = fl (z ± a 2 z 2 ± • • •),

(7.4)

where /3 > 0, f (z) is univalent in 1z1 < 1 and If (z)1 < 1. Such a function f (z) will be called a transformation. The transformation w = f (z) maps 1z1 < 1 onto a domain D in 114,1 < 1. We denote by S = Si. the set of all points of 114;1 < 1 not in D, and by d = df the diameter of Sf. We ignore the trivial case when Sf is null and f (z) = z. We shall say that two points z, w on 1z1 = 1 and the frontier of D respectively correspond by the transformation w = f (z), if there exists a sequence z n , such that Izlil < 1 (n

1)

and z n -- z, f (z n ) -- w (n -- ce). Let B = B1 be the set of all points oflz 1 = 1, which correspond to points of S. We note that points of lz 1 = 1 not in B can correspond only to points on 1w1 = 1. We write 6 = 6f for the diameter of B. Our aim is to study the limiting behaviour of transformations when 6 or d is small. In this case D approximates to lw 1 < 1 and f(z) approximates to z. Our first aim is to show that, if either of 6 or d is small, then so is the other, and in this case fl is nearly equal to 1.

If f(z) is a transformation given by (7.4), then we have with the above notation 1 — d < )6' < 1. Lemma 7.3

The inequality fl < 1 follows from Schwarz's Lemma. To prove 1 — d, we may suppose that d < 1. It follows from Lemma 5.2 that $

201

7.2 Transformations

if S meets 114, 1 = r for some r < 1, then S meets lwl = p for r < p < 1. Hence S has at least one limit point e'0 on jw1 = 1. Thus S lies entirely in lw — e 0 1 < d, and so in 114/1 > 1 — d. Thus D contains the disc 1w1 < 1 — d. Hence the inverse function z = f -1 (w) maps 1142 < 1 — d onto a subdomain of 1z1 <1 and

r

,

to

_ cowl =______

(1 — d)w

fi

+ ...

satisfies the conditions of Schwarz's Lemma. Thus 1 — d < 13 as required and Lemma 7.3 is proved.

Lemma 7.4

We have with the above notation

6

4ir [log(2/d)] - 1, d

47r [log(2/6 )] .

To prove the first inequality, we may assume that d < 2e-4'2 . For since 6 < 2, the inequality is trivial otherwise. Let wo be a limit point of Sf on lwl = 1. Then Sf lies entirely in I w —w0 I < d, and so for R > d that arc, cR say, of I w — wol = R which lies in 1w1 < 1, lies also in D. Let yR be the image of cR by z = f-1 (w), and let 1(R) be the length of YR. Then by Lemma 7.1 with R1 = d, k = we can choose Ro satisfying

1 _1 n (-1 log —1 ) 2 n (-1 log —2 ) 2 <1. 2 d 4 d

d < Ro < 1, l(Ro)

Next since cRo separates w =-- 0 from Sf, it follows that yRo separates z = 0 from Bf. Hence by Lemma 7.2

of -.

/(R0)

2 —1 .. 27r (log - ) , d

as required. The proof of the second inequality of Lemma 7.4 is similar, and so the lemma is proved.

7.2.1

We shall also need the following:

Lemma 7.5 Suppose that ip(z) = u + iv is regular in 1z1 < 1, that u(z) has constant sign there and v(0) = O. Suppose further that u(re'4') -- 0 (r —, 1),

202

The Liiwner theory

uniformly for (5 - 14) —

7E. Then we have

ei0° tp(z) = tp(0)[ eioo

z z

E(z)],

where 1E(z)l< 561e 10° — z1 -2 for le* — z > 2(5, z < 1. We may without loss of generality assume that 00 = 0. Then we have by Poisson's formula for 1z1< r <1,

1f +' rei0 + z tp(z) = — u(re io ) cl(/)+ iC, 2n _7, reid) — z where C is a constant in 1z1 < r, and since w(0) is real, C = O. Thus we have for a fixed z, 6 1 rei z (z)= — u(rei(l) o(1) (r 1). (7.5) 27t red' — z Writing ip(0) = a we obtain 1 27( ib

1).

a (r

u(re)(14)

(7.6)

Also

and if lz

re' 4) + z ei0 + z re'—z ei0 — z' 6 we have lei(/' — 1 < 6 and so

2(5 ,14)1

el0 + z ei4) — z

1+z

2z(e 1(1) —1) (e4 — z)(1 — z)

1—z

46 — 1 1 — z1[1 1 — z1 — (5] — 1 1 — z1 2. Thus if lz — 11> 26, 101 < 6, and r is sufficiently near to 1, then 26

re4 z !TN' — z

1—z

Hence we deduce from (7.5), as r

ip(z)1+ z 1 1 6

56

1+z

z1 2.

1,

u(re) `4) (14) <

56

1 16

lu(re l.çb )41) o(1). — z1 2 27r —6 Using (7.6) and the fact that u has constant sign in 1z1 < 1 we deduce

1— z 2ir _6

1 +z tP(Z)

if 11 — z1

œ l—z

26, and this is Lemma 7.5.

56 1Œ1 11 — z1 2 '

7.3 Structure of infinitesimal transformations

203

7.3 Structure of infinitesimal transformations We are now able to describe infinitesimal transformations, i.e. those for which 6(f) and d(f) are small. Lemma 7.6 Let fn (z) = fin (z + • • .) be a sequence of transformations of z1 <1, let d(fn ), 6(f n ) be defined as in §7.2 and suppose that either d(fn ) or 6(f n ) —> 0 as n —> co. Then

f n (z) --> z (n --> co),

(7.7)

uniformly in izi < 1 and in particular fin —> 1. Further, if zn is a point on 1z1 = 1 which corresponds to a point wn in lwl < 1 by w = f n (z), and if 0 < r < 1, then f n(z) — z , (log /3n)z

uniformly in IzI

Zn -4- z Zn — Z

(n .. co), —

(7.8)

r.

It follows from Lemma 7.4 that if 6(fn ) —> 0 then d(fn ) —> 0 and conversely. It then follows from Lemma 7.3 that fi n —> 1. Write

ipn (z) = log fn(z) = un (z) + iv n (z). z

Since f n (z) vanishes only at z = 0, pn (z) is regular in IzI < 1, and since fn(z) satisfies the hypotheses of Schwarz's Lemma, un (z) < 0 in 1z1 < 1. Let Z n = ekbn correspond to wn , where I wnl < 1, by w = fn (z). Then if I ei(1) ei(1) "1 > WO, the point z = ei0 corresponds only to points w on jwi = 1, and given 6 > 0, this will be true for 6 ._ 10 — 4)111 < it if n > no(6), since 6(f n ) —> 0. In this case —

un (re) -- 0 (r —>

1),

uniformly for 6 <10 — Old ._... IT and a fixed n> no(6). We now apply Lemma 7.5 and obtain tpn (z) = log fin

rzn [Z n

+z —

+(z)]

,

Z

(7.9)

where lEn(z)l < 50Iz — z -2 for lz — Zn InI> 26. Thus if q is fixed and positive we may choose n 1 so large that

IWn(z)I

11

if n > n1 and Iz — z,d > ti, and then

I f n (z) — zl = jz 1 le`" (z ) — 1 1 __ en — 1 < 2q

(7.10)

204

The LAwner theory

if ri <

We also suppose n so large that the end—points of the arc — zn 1 = ri on 1z1 = 1 correspond to points on 1w1 = 1 only. By (7.10) we have

Ifn(z)—

2r, and so If(z) — z n 1

317

on this arc. The values which w = f n (z) assumes for lz — zn 1 < ri form a domain A in lw < 1, which is separated from w = 0 by the image of the arc lz — zn 1 = ri by w = f n (z). Hence A also lies in 1w — zn < 3. Thus we have finally If(z) — z,1 1

317 in lz — zn 1 <

and so

f(z) —z < 4ri for z n — z

rl •

This, together with (7.10), gives (7.7) since ti is arbitrary. Next (7.9) shows that if r is fixed, 0 < r < 1, then 4'p(z) —> 0 (n co), uniformly for 1z1 < r, and so we have as n oci uniformly for 1z1 < r, f n (z)— z = z[e`t'n (z ) —1]

zy n (z) , z log fin zn z zn — z

by (7.9), which is (7.8). This proves Lemma 7.6.

7.4 The class Si

Let BP be an analytic Jordan arc lying in with one end-point P = rel° . Let Poo be the ray

z = pew , r < p < co.

izi

r and

(7.11)

We shall call the Jordan arc y : BPoo a slit. The set of points G consisting of all points w not on y will be a simply connected domain. In fact if Q i , Q2 lie in G near different points of y we can pass from Qi to Q2 in G by a curve near y, which if necessary will go round the tip of y and along the other side. Thus any two points of G near y can be joined in G, and any point in G can be joined to some point near y, for instance, by a straight—line segment to the tip of y. Thus G is connected. Further, the complement of G consists of y and so is closed and connected. Thus t G is a simply connected domain containing w = 0, and so by Riemann's mapping theorem t there exists a unique function

w = f (z) = t C. A. p. 139 t C. A. p. 230

a2z 2

• -) ,

(7.12)

7.4 The class S I

205 z

w

MO = w = 1P(z) C = g n (z) Fig. 6.

mapping IzI < 1 (1,1) conformally onto G so that f(0) = 0 and f'(0) = fi > O. Clearly f (z)/ 16 G S. We have further Lemma 7.7 The functions f(z)/fl, where f(z) is constructed as in (7.12), form a dense subclass S i of S. Suppose that f(z) E S and 0 < p < 1. Then f(pz) maps IzI < 1 onto the interior of an analytic curve, namely, the image of Izi = p by f(z). Also Pif (pz) E S 5 if 1, uniformly for Iz I r, when p (pz) -- f (z) as p

0 < r < 1. It is thus sufficient to show that the functions if (pz) can be approximated by functions in S i . Next if M is large 1 1 w = îp(z) = — f(pz)= — z + • • • M PM

maps IzI < 1 onto the interior D of a closed analytic curve F lying entirely in I w I < 1. Let P = rei° be a point of largest modulus on F. We obtain the arc PP,, of F by going along the curve F in an anticlockwise sense from P until the whole of F apart from an arc of diameter 1/n has been described. Then ooPPn is a slit F. Let it be the complement of Fn and let

f n(z) =f3 (z + a2z 2 + •

)

map Iz < 1 onto An . To prove Lemma 7.7 it remains to show that, if t < 1, f()

.1P(

)

,

as n

CO

,

uniformly for ICI

t.

(7.13)

206

The ',Owner theory

For then fn (z)/ fi n approximates p-1 f(pz) for a fixed p, which in turn approximates f(z). We consider gn(z) = fn-1 {tr(z)}.

We verify that gn (0) = 0, gn'(0) > 0, so that g(z) is a transformation. Let Sg, be defined as in §7.2. Then Sg„ consists of all those points in 1(1 < 1 which correspond to points outside D by w = f n(C). Choose now 6 so small that the circle of centre P and radius R meets D in a single arc TR for 0 < R < 6, lies in 1w1 < 1, and that the origin w = 0 lies outside this circle. Then if 1/n < R, y R corresponds to a single arc cR in 1(1 < 1 by = f n-1 (w) and all points of Sgn are separated from C = 0 by CR. By Lemma 7.1 we may choose R so that the length l(R) of cR satisfies 1 l(R) n log(n6)] 2 , and so by Lemma 7.2 the diameter d(g) of Sg„ satisfies 1

d(g) .. n [i log(n6)] 2

1

It now follows from Lemma 7.6 that

gn (z) —> z (n —> co), uniformly in 1z1 < 1. We note that w = fn (C) maps ICI< 1 — d(g) into D and so into 1w1 < 1. Given 0 < t < 1, we now choose p = 4( 1 ± t) and suppose that n > no so that d(g) < (1 — p) = -41 ( 1 - t). Then for 1C1 p we have by Cauchy's inequality 1 2 If/n(01 < 1 — d(gn ) — il < 1 — p . Hence if 1C11

P, 1C21

p, n > no we obtain

1

C2

If n(i) — fn(2)1 = Suppose now that 11

f)(1C <

—1

2 —

p

1C1 — C21.

t. Then if n > n1 say, we have gn(01 .. p. Now

IfnG) — 001= Ifn(C) — fn{gn( C)}l

2 1 — p IC — gn(C)1 —> 0 , as n

This is (7.13) and the proof of Lemma 7.7 is complete.

207

7.5 Continuity properties

7.5 Continuity properties Following L6wner we now investigate the class el and show that the functions f (z) in (7.12) can be obtained by a series of successive infinitesimal transformations from w = z. Let y be a slit given by w = a(t) (0 < t < co). We denote by ye t, the arc t' t t" of y and by y, the arc y. Let G(t) consist of the complement of y t . As t increases from 0 to co, G(t) expands from G = G(0) to the whole plane. We denote by

w = gt(z) = fi(t) (z a2(t)z 2

• -)

(Mt) > 0),

the function which maps 1z1 < 1 onto G(t) and proceed to show that gt (z) varies continuously with t, as t increases, from go(z) = f (z). We have first Lemma 7.8 If w = g(z) is defined as above, then the inverse function z = g 1 (w) remains continuous at w = a(t). Thus as w = g(z) a(t) in any manner from G(t), z approaches a point A(t) such that 12(01 = 1. Choose 6 so small that the circle 1w —a(t)1 = R meets yt in exactly one point for 0 < R < 6. The choice is possible since yt has a continuous tangent at a(t). Then for 0 < R < 6 the circle 1w — a(t)1 = R lies in G(t) except for a single point. The image of this circle is an arc cR lying, except for end—points, in 1z1 < 1. By Lemma 7.1 we can then choose a sequence Rn such that l(Rn ) 0 and Rri 0, where l(R n ) is the length of cRn . If Rn < WO, then the disc 1w — a(t)1 < Rn cut along the arc yt corresponds to one of the domains into which cR, divides 1z1 < 1, namely, that one, An , which does not contain z = 0. By Lemma 7.2 the diameter of A, is not greater than l(Rn ) and so tends to zero as n ao. Since An c An__i when R,
7.5.1

We next write h (z, t', t") = g 1 [g (z)]

(0

t'

< t" < co) .

We shall study the behaviour of h(z , t', t"), as t" — t' 0. We note that w = ge (z) maps Izi < 1 onto G(t'), that is the plane cut along y e . Also = gi; i (w) maps G(t") onto ICI < 1, and so GO corresponds to 1C1 < 1 except for the image of yer by = g; 1 (w). For Ye t, lies, except for one end—point Œ(t"), in G(t") but not in G(t'). By

208

The Liiwner theory

C

z

al

A(t1

ge, () = w =

gt, (z)

C = h(z, t', t") Fig. 7.

Lemma 7.8, = g 1 (w) is continuous also at a(t") and so on the whole of T ee, and maps ye t, onto a Jordan arc Se t, in 14- 1 1. Thus = h(z , t' , t") maps 1z1 < 1 onto 1r,1 < 1 cut along Ser. We write Be t,' for the set of boundary points on Izi = 1 which correspond to Set, by this transformation. Other points of 1z1 = 1 correspond only to 11 = 1. We note also that z = 2(0 belongs to B et, and corresponds to the tip of the cut Se t, and that the other end—point of Set, on 1"1 = 1 is = 2(0. We now have Lemma 7.9 As t" — t' 0, while either t = t' or t = t" remains fixed, both See , and Bee , approach the point 2(0 in their respective planes. Also 2(t) is continuous.

Suppose first that t" t' while t' remains fixed. Then the arc yet, shrinks to the fixed point a(e) and so finally lies in a disc of centre Œ(t') and radius 6. If E is sufficiently small it follows from Lemma 7.8 that we can choose 6 depending on E, such that if w = ge (z) and lw a(01 < O, then lz /401 < E. Thus if y e e, has diameter less than 6, Bee , has diameter less than 2E. Thus as t" t', the diameter of Bee , tends to zero and hence so does that of Set, by Lemma 7.4. Similarly, if t' t", while t" remains fixed, y'» shrinks to the point Œ(t"). It then follows from the continuity of g l (w) at w = a(t"), that Se », which corresponds to ye t- by = gT, i (w) shrinks to the point gT, 1 [a(t")] = 2(0, and so the diameter of Set, tends to zero. Thus the diameter of Be t, tends to zero also by Lemma 7.4. It now follows from Lemma 7.6 that in either case —

—

h(z , t' , t")

z

(7.14)

209

7.6 The differential equation

uniformly in Iz I < 1 as t" — t' -- 0 through any sequence of values. Thus if z lies on Bee, and C is a corresponding point on Sec , we have Iz—I < E finally. Also See, always contains 2(t") and Ber contains AO. Taking z = AO fixed, choose t" so near t' that the diameter of See, is less than e. Then IC — 2( t")I < e and I — z I < e, and so

I4t") — A(t')I < 2E. Similarly, if t" is fixed, choose ( = .1(t") and t' so near t" that the diameter of Ber is less than E. Then K —zI <8 and Iz — 2(01 < e, and so I2(t") — .1.(01 < 2e. Thus in either case .1.(t") — ;t.'(e) -- 0, and if t = t' or t" and t is fixed 2(0 and .1.(t") approach 2(t). Thus 2(t) is continuous. Since the diameters of Sec , Bee, tend to zero and these sets contain 2(t") and 2(0 respectively, both these sets approach 2(0 as t" — t' -- 0, while one of t', t" remains fixed. This proves Lemma 7.9.

7.6 The differential equation

We note that for 0 < t' < t"
h(z , t', t") =

z+•••

satisfies the conditions of Schwarz's Lemma in Izi < 1. Thus fl(t) is a strictly increasing function of t in 0 < t < oc. Again by (7.14)

fl(e) fl(t")

-- 1 (t'' — t' -- 0),

while either t' or t" remains fixed. Thus 13(t) is continuous. Hence .E. = log

fl(t) fi(0)

is a continuous strictly increasing function of t for 0 < t < cc, and we may take t for our parameter t, which has been left undetermined so far. We shall do so in what follows. With this normalization gt (z) = flet (z + a2 (t)z 2 + • • .) (0 where fi = fi(0) = f'(0). Also h(z , t' , t") = et' -t" z + » • We now prove

t < cc),

(7.15)

210

The Liiwner theory

With the normalization (7.15) suppose that t"—t' —> 0 while t = t' or t" remains fixed in the range 0 < t' < t" < co. Then we have, uniformly for z r, when 0 < r < 1,

Lemma 7.10

h(z, t', t") — z 1+ K(t)z —> z t" — t' 1— K(t)z'

where K(t) = In fact Lemmas 7.6 and 7.9 show that, for 1z1 h(z, , t") — z

r <1,

ei`k + z (t' — t")z . el) — z'

as t" —t' —> 0, where eig5 is a point of B ee, so that ekl) —> 2(t). Now Lemma 7.10 follows. We can now prove part (b) of our fundamental theorem.

Theorem 7.2 If f(z) G S i so that f(z) is given by (7.12) with /3 = 1, if t < oo, then further g(z) is defined by (7.15) and f(z,t) = [f(z)J, 0 w = f(z,t) satisfies the differential equation (7.1) and f(z,0) = z. Also g(z,t) = et f (z, t) satisfies (7.2) and (7.3). The function w = f(z) maps lz < 1 onto the w—plane cut along the slit y. Also = e(w) maps 1z1 < 1, the plane cut along the smaller slit Yt onto 11 < 1. Thus = f (z, t) maps 1z1 < 1 (1,1) conformally onto a subset of (1 < 1. Using also (7.15) we see that et f (z , t) E e so that (7.2) holds. Also, when t is large, y t is a ray given by (7.11) and w = gt (z) =

4rz

where 4r = e t. Suppose that 1z1 p < 1, so that If (z)! < K, where K is a constant. As t —> oo we see that z —> 0, w 4rz uniformly for 114, 1 < K. Thus f (z,t) = gï l {f (z)}

f (z)

4r

= e—t f (z)

as t —> co, uniformly for 1z1 p. This proves (7.3). Next write f (z, t') instead of z in Lemma 7.10. This is permissible since < 1 for 1z1 <1. Also If(z, h[f (z, t'), t', t" ] = g 1 fg e (g l [f (z)])} =

[f (z)] = f(z,t").

Thus Lemma 7.10 gives f(z,t") — f (z , t') t" — t'

, 1 + K(t)f (z, t') f (z, t ) 1 — K(t) f (z, t')

(7.16)

7.7 Completion of proof of Theorem 7.1

211

if t" - t' -> 0, while t' or t" remains fixed. If t = t' is fixed this gives the required result for the right derivative. If t" remains fixed the required result for the left derivative also follows. In fact by Lemma 7.9 and since K(t) = 1/2(t) K(t')

> K(t") (t'

-

> t"),

-

while t" is fixed, and by (7.14) h(z, t', t") - z -> 0 (t' -> t"),

uniformly in 1z1 < 1. Writing f (z, t') instead of z we obtain f (z , t") - f (z, t') --÷ 0 (t' -+ t"),

uniformly in 1z1 < 1, so that (7.16) gives, also if t = t" is fixed, 1 + K(t)f (z , t) f (z, t) - f (z , t') —+ f (z, t) 1 - K(t)f (z ,t) t - t' as t' -> t from below. This proves (7.1) and completes the proof of Theorem 7.2.

7.7 Completion of proof of Theorem 7.1 In order to complete the proof of Theorem 7.1 it remains to show that given K(t), as in Theorem 7.1, there exists a unique solution w = f (z, t) of the differential equation (7.1), such that f (z, 0) = z, and further that et f (z, t) E It is convenient to put

a.

1 W = U + iV = log (- ) . w Then (7.1) becomes

1 + K(t)e -w

3W at

1 - K(t)e - w

(7.17)

= 0(t, W),

say. We note that 90(t, W) > 0 for 91W > O. Also if U > 6 > 0,

12K(t)e-w I

11

-

140e -1'1'1 2

<

2 (1

-

e-6 ) 2

=K,

say, and

(t, W)

2 < K. 1 e -6 -

(7.18)

212

The Lôwner theory

If we integrate along the straight line segment from W1 to W2 we deduce, if 91W1 > 6, 91W2 > 6,

a oaw IdW

10 (t ,11/2)— (t, W 1 ) 1 fww:

1 KIW2 — W11.

(7.19)

Thus 0(t, W) is Lip in W in the half—plane U > 6 uniformly with respect to t. We now define a sequence of functions Fn (t, co) (n > 0) as follows. We suppose that 91co > 6 > 0 and set

Fo(t, co) co, } Fn (t, co) = co + fc; 4[T, Fn_i(T, (0)1d -c (0 < t < oo, n _. 1).

(7.20)

Since K(t) is measurable so is cf)(t,w) for fixed co. Also 0(t, co) is bounded, so that the integral exists. We note first that 9iFn (t, co) increases with t for fixed co and so remains greater than 6. For if this is true for n, then 910[T,Fn (r, co)]

>0

and the result remains true for n + 1. Next we have K n tn in

IFn(t, CO) — Fry-1(t5 (01

n!

tU

< t

—

< 00,

n > 1).

In fact if n = 1 we have from (7.18) t

t

Fi (t,co) — Fo(t, co) =

fo Kch = Kt.

fo

Also if our result is true for n, then (7.19) gives t I Fn-Fi(t, W) — Fn(t, (0 )1 = f {0[T, Fn (T, 01 — 0[T, Fn_ier, wnIch

o

t

t

Kt j K n+1 tn+1

1Fn(T,(0) — Fn—i(T, (0)IdT K f o at = n! (n + 1)! . o Thus the result is true for n + 1. It follows that the sequence Fn (t, co) converges uniformly for 9ico > 6 and 0 < t < to to a limit function F (t, co). Taking the limit in (7.20) we deduce t (7.21) F (t, co) = co + f 0[-c, F(T, co)]ch .

Kf

JO

Thus 9iF (t, co) _. 6, and so, by (7.18), the integrand in (7.21) is bounded by K. Hence F (t, co) is absolutely continuous in any finite interval

7.7 Completion of proof of Theorem 7.1

213

0 < t < to, and satisfies (7.17) almost everywhere, and conversely an absolutely continuous function W(t) = F(t,co), which satisfies (7.17) almost everywhere and W(0) = co, also satisfies (7.21). Next we prove that F(t,co) is analytic in co. Since Fn (t,co) converges to F (t, co) uniformly for fixed t and 91co > 6, it is enough to show that Fn (t,co) is analytic in co for n = 0,1,2,... We prove by induction on n that Fn (t,co) is analytic and further that, for 91w ' > 6, 9ico2 > 6, t > 0, we have

IFn (t,w2) — Fn(t,(01)I

eK `10)2 — 0)11.

These results are clearly true for n = 0 by (7.20). Suppose that they hold for n. Then (7.19) and (7.20) yield

f' H_1(t,c02) — Fn+1(t,(0 1)1

1 (1)2 — a) 11

1+

f

KeKtcl-c = eKro2 — (D i .

Also {(¢[T,Fn(T,(-0 2)] (P[T,Fn(t,(0 1)]}/( 0)2 — coi) is bounded for 0 < t < t and tends to

1 1- 0[T, Fn (T,04]1 = L ""

[ aFn n( : t C°

1 [ a w, W) L

Fn(T,(01)

as co2 —> coi through any sequence. Thus by Lebesgue's dominated convergence theorem (Titchmarsh [1939, p. 345]) Fn+i(t, W2) — Fn+1(t, (1) 1) = 1 o t [t Fn(' + I (0 2 (0 1 1

[T,

— [t F r 0)2 — (0 1

Fn (T, (0)]

0

ut

ch (0,

as co2 —> col through any sequence and so generally. Thus Fn+i(t,co) is analytic and the inductive step is proved. Suppose next that W(t) is any absolutely continuous solution of (7.17) which satisfies 91147 (0) > 6 and so 9l14/(t) > 6 (0 < t < to ) and further that

W(t i ) = F(ti ,co) for some pair (t i , co), such that 0 < t 1 < to and Who > 6. W(t) F(t,co) (0 < t < to ). In fact write M = sup 1W(t) — F(t, COI 0
Then

214

The Lôwner theory

Then (7.17) gives, for 0

t < to,

W (t) — W (t 1) = F (t, co) — F (t 1, co)

f: (19[-c, W(x)]dT,

= ft: (/)[-c, F(t, co)]d-c,

and so

It IF (t, (0) — W(t) I = {0[-c, W (t)] — 4)[T, F(t, co)lidT

(7.22)

We deduce by induction that m K nI t n!

IF(t, (0) — W WI

—

t i I"

for every positive integer n. For this holds for n = 0 by hypothesis. If it is true for n, then (7.19) and (7.22) give

IF (t, (0) — W WI

K fti

' MK n (-c — ti )n ch

MK n+1 It — ti i n+1

n!

(n + 1)!

as required. Since n is arbitrary W(t) = F(t, co). Taking t 1 = 0, we see that F(t, co) is the unique solution of (7.17) such that F(0, co) = co. We also see that F (t, co) is a univalent function of co for 91co > 0 and a fixed positive t. For if

F(t, col) = F (t, (02), the above uniqueness theorem shows that F(0,(D I ) = F(0,co2), i.e. (01 = (02. Again, if (D2 — col = 2ni, we easily see that

F(t, co2) — F(t, col) = 2ni. Thus f (z, t), given by

— log f (z, t) = F(t, — log z) (0

t

to),

is for fixed t an analytic one—valued univalent function of z for 0 < 1z1 < 1, and is the unique solution of the differential equation (7.1), which satisfies f (z , 0) = z. Also If (z, t)1 < 1 in 1z1 < 1, and so f (z, t) remains regular at z = 0 and is clearly zero there. If we apply (7.1) to g(t) = f (z, t)/ z, we obtain

ag(t) et

1 + zK(t)g = g 1 — zK(t)g'

7.8 The third coefficient

215

and putting z = 0, we deduce

ag --a-t- = —g, g(t) = e-t g(0) = e t . Thus f(z, t) = ze -t + • • • near z = 0 and so er f (z, t) E S. This completes the proof of Theorem 7 .1.

7.8 The third coefficient We proceed to give a number of applications of Theorem 7.1. Let S denote as usual the class of functions f (z) = z + a2 z 2 + • • • which are univalent in I z I < 1. In this section we prove

Theorem 7.3

If f(z) E S, then la31 < 3.t

By Theorem 7.1, we may confine ourselves to the functions et° f (z, to), 0 < to < oo, of that theorem, since they form a dense subclass of S. We write fl = e-t°, Do an zn) f (z, to ) = fl (z +

,

E n=2

and proceed to develop Li5wner's formulae for a, in terms of K(t). It is convenient to work with the function MC) defined by gt[f(z,t)] = f(z,to) (0

t

to).

We write = f(z,t) and differentiate the above relation with respect to t. This gives ag a( ag n a( et + at Substituting in (7.1) we obtain

ag ag r l + KC 0('' 1 — 1C . at

(7.23)

We have f(z,t) = e -t(z + • • .) near z = 0 and so

x

gg) =

Pet +

E Cn(tgl n=2

t A rather more detailed discussion shows that la31 < 3 except when f(z) = z(1 A similar remark applies to Theorem 7.4. See Ldwner [1923].

—

zele r2 .

The ',Owner theory

216

near C = O. We substitute in (7.23), equate coefficients of C n , and obtain for n > 2 n-1 c(t) ± c(t) = fiCn (t)

+2

(n — r)c n,(t)Pc(t)Y , r=1

i.e. n-1 c(t) =

rcr(t)Pc(t)r-r (0

(n — 1)c(t) ± 2

t

(7.24)

to).

r=1

The term-by-term differentiation with respect to t may be justified by expressing the coefficients as contour integrals in terms of g,(0. Also when t = 0, f(z,0) --- z, go(z) = f (z 'to), c(0) = an. When t = to, g(z) _= z and so c(to) = 0 (n > 2). Finally, c i (t) 1. Thus we have the boundary conditions ci(t) —= 1; c(0) = an , cn (to) = 0 (n

2). We

From this system we can successively determine the coefficients. obtain in fact for n > 2

Ito { n-1 c(t) =

—

2? 1)`

r=lrCr(t)[K(Tnn—r

-

e-(n-l)r d-c (0 < t < to).

This gives to

c2(t) = —2et

(7.25)

K(r)e-tdt. I

Also c3( t) = —2 e2t

If

to

to

i \2, —2T A ickc) e uT ± 2 f K(z)c2(t)e 2' d-c} t

{ ftt° r I ‘12 -2.r A = — 2e 2t i Kk til e tit

to — 2 (f t

2

(7.26)

K(T)e rd-c) } ,

since

d ( ft° „ 7. , dt

t

'<we - at

2

to

= _2K (t) e t ft K(r)e - rdt = K(t)c2(t)e

2t .

We deduce at once that la21 = c2(0)1 _< 2. To find the upper bound for la3 1 = 1c3 (0)1, we may without loss of generality suppose a3 real and

217

7.8 The third coefficient

positive, since this may be achieved by considering e-i49 f (zei 0 ) instead of f (z). We write KW = ei°( t ) and consider t

to

2

o

9ia3 = —2 1 [2 cos2 0(t) — 1] e -2 tdt + 4 (f cos 0(t)e -t dt)

o

2

to

fo sin 0(t)e -t dt) . —4 ( To obtain an upper bound for 91a3 we omit the third term on the right. The first term is to e -2t COS2 0(t)dt + 1 — e -2t0 , o

—4 f

and by Schwarz's inequality the second term is at most

to to e-t dt f e -r cos2 0(t)dt < 4 f e -t cos2 0(t)dt. Jo o o

4f

to

Thus we obtain to

la31 = 91a 3 < 1 + 4 f cos2 0(t)(e' — e -2t )dt o

ac < 1 + 4f (e t — e -2t )dt = 3.

o

This proves Theorem 7.3.

7.8.1 The Fekete-Szegb Theorem While Theorem 7.3 is a special case of de Branges' Theorem, which will be proved in the next chapter, Liiwner's technique can be refined to yield the following result which has interesting applications to coefficients of powers of univalent functions and in particular of odd univalent functions. (Examples 7.3 and 7.4)

If f

Theorem 7.4 (Fekete-Szegii [1933].) ja3 — oca31

G S and

0 < a < 1, then

1+

Equality holds for a function f(z) in CE and real for real z.

We follow the argument given by Duren [1983, p. 104] and start with the following lemma essentially due to Valiron and Landau [see Landau 1929].

218

The Riwner theory

Lemma 7.11 Let 0(t) be real—valued and continuous for t > 0 and satisfy et and 10(t)1

r

o(t)2dt.(,+

1 2

(7.27)

where 0 < 2 < co. Then 0(t)dt

(7.28)

1)e-4 .

Equality holds if and only if 0(t) = -f-p(t), where tp(t) = e tp(t) =

,

0
(7.29)

We note that oc 1 0(02 dt < f e-21 dt = — 2. Since ( + e-2)' decreases from to 0 as  increases from 0 to cc, 2 is uniquely determined by (7.27). We also note that if 40 = ±p(t) equality holds in (7.27) and (7.28). Finally we observe that F(t) = [tp(t) — 10(0] [2e—' — ip(t) —14)(t)1]

0, 0

t < oo.

For if t < À the right-hand side is (e—i- —10(01) 2 and if t > À e t = tp(t) and 2e—)- — tp(t) — 10(0

10(01

e t — 0(t)

O.

Hence (cc 0 <

o F(t)dt

pc oc 00 = 2e-4 {f tp(t)dt — f 10(t)idt} — f ip(t) 2 dt + f 4)(0 2 dt

2e—j- {(2 + 1)e—)- — f 10(t)Idt} 0, i.e. by (7.27) and (7.29). Also equality holds if and only if F(t) 4)(t) -Fv(t). This proves Lemma 7.11. We can now prove Theorem 7.4. We suppose without loss of generality that A = a3 —ca i is real and positive. For if we replace f (z) by eiçb f (ze -4 ), A is replaced by e-20 A, and we may choose 4 so that Ae-20 > O. We write KW = cos 0(t) + i sin 0(t). Using (7.25), (7.26) with t = 0 we obtain

7.8 The third coefficient A

219

=

or

=

ro

2

2f ( 1 — 2 cos2 0(t)) e -2t dt + 4(1 — a) U e-t cos 0(t)dt} 0

2

to

—4(1 — a) (f e t sin 0(t)dt) o < 4(1 — a) (i x 4)(t)dt) 2 — 4 f 02 (t)dt + 1, o o where 4)(t) = e- t cos 0(t), 0 7.11 and deduce that

(7.30)

t < to, OW = 0, t> to. We apply Lemma

9i(a3 — ca) ._ 4(1 — a)(2 + 1)2 e 2' — (42 + 2)e-22 + 1

(7.31)

where to

fo

1 4)(0 2 dt = (A + - ) e-2 11 . 2

The right-hand side of (7.31) attains its maximum 1 + 2 exp{-2a/(1 — a )} = a/(1 — a). Thus we obtain whenA A = 91 (a3 — ca,) < 2e-2Œl(1- ') + 1

as required. To find an extremal we need to choose K(t) = e'9( t ) so that equality holds in (7.31) with A = c/(1 —4 Setting K(t) = e( t), and using Lemma 7.11, we see that we must have

OW OW

= e- t cos 0(t)

= e-''', 0 < t <2 = e- t, t> 2.

Thus 0(t) is given by

cos 0(t) = e t_, 0 < t < A and

0( t) = 0, t > L We may suppose that -3 < 0(t) < 3 for 0 < t <2. To obtain equality in (7.30) we also choose the sign of 0(t) so that

/.

x.

10

e-t sin 0(t)dt =

o

e- t

sin 0(t)dt = 0.

220

The Liiwner theory

Clearly this can be done in many different ways. For instance we may choose 0(t) > 0 for 0< t < T, 0(t) < 0 for T < t < /1. Then 1

1(T) = f e-t \I 11 — e 2( `-;-) } dt —

e-t 11 — e 2(t--;-) } dt

is an increasing function of T and I(0) < 0 < 1(i). Thus there exists a unique value of T, such that 0 < T < ), and 1(T) = 0. (See Duren [1983, p. 1071) To find a function with real coefficients we proceed as follows. Let N be a large positive integer and define 0(t) = ON(t) by ON (t) ON ( t)

> 0,

2). 2N

< 0,

(2m+1)).

t

<

t

<

2N

< <

(2m+1)).

2N' (2m+2)).

0 < m < N — 1.

2N '

Then evidently e

• ON (t)dt sin

oc.

0 as N

(7.32)

If K(t) = ei°N ( t) , let w = f N (z, t) be the corresponding solution of (7.1) for 0 < t < A. We write = AIN and Wm = f N (z,mn), Wm+ 1 — Wm

bm =

Then (7.1) shows that to a first approximation

+ k(11111)wm wm { 1„ , i — Kktnrowm z

1/

bm

1 + ic(mq)w m „ — Kviunw m

1 w —1 1 wm 1 + w — 2w m9bc(mq) —

1

—11vvm 1 + — 2w 11 emqw

Hence, as N

oo and uniformly for 1z1 < r, 0 < t < A, where r < 1, f N (z, t)

f(z,t),

where w = f(z, t) is the solution of 1 — w2

Ow

at = 14,1 + w2 — 2wet-Â, f(z,0) = z.

—

Also et f N(z, t) E S, so that for z = 0

a f(z,t)

— az

.

= lirn

a f N (z, t)

=

(7.33)

221

Examples

Hence g(z, t) = et f (z, t) is not constant and so belongs to S. Clearly g(z, t) is real for real z and so has real coefficients. Once f (z)) has been obtained we apply (7.1) with K(t) = 1 for t> i.e. f 4.1 + f at j1—f so that for t >

et f (z, t) [1 f (z, t)] 2

[1

f( z , f (z, .1)] 2.

We recall that, as t —> co for fixed z, et f (z, t) E S and so remains bounded, so that f (z, t) —> O. Thus

g(z, t) = f( z , t) —> g(z) =

f(z, /1)

(7.34) [ 1 f (z, 2 )1 2 and g(z) E g(z) has real coefficients by (7.33) and g(z) yields equality in Theorem 7.4, since (7.32) holds. Also g(z) is given by (7.33) and (7.34).

a,

Examples

7.1

Use Lemma 1.1 to show that if f (z) E S then a3 —1231 < 1, with equality if and only if f (z) = fo(z) = z/ (1 — zel 2 (The function 1 1 g(z) = =— z f (z)

7.2

n=0

is univalent in 0 < 1z1 < 1 and b 1 = a — a3.) Show that, if a < O or a > 1 and f (z) E S, then

— aa31

7.3

co

with equality only for f (z) = f o(z). (If a > 1, note that acti — a3 1 If f (z) = z E°2c an zn E S and (

f (z))

z

=1+

14a — 31

I

— a3 1+ (a — 1 a3 )

an Ozn -1 ,

1)ai. Deduce the show that a2(2) = 2a2 and a3 (2) = 2a 3 sharp bounds la201 < 22, —oo <À < -{-oo and 1a3 (2)1 < 121 (1 + 2e21)1)), —1
222

7.4

The Liiwner theory Find also the sharp bound for a 3 ().) when I.11 > 1. (Hayman and Hummel [1986].) If k is a positive integer and pc

g(z) = z + Ebnk+iznk+i e S we recall from the end of Section 5.8 that g(z) = f(z k )1- , where f(z) E S and conversely. Deduce the sharp bounds

2 , 2(1-0 / 0 +0 } Ibk-Fi i __ - and 1b2k+11' < 1 {1 + 2e k 2

.

([Fekete and Szegii 1933]. If k = 2 we obtain 1b31 < 1, + e-2/3 = 1.013...) lb51 __

7.9 Coefficients of the inverse functions w = f(z) E Ca and let the inverse function z

Suppose

again that

= (1)(w) = f -1 (w)

be given by

x

wm

0(w) = w m=2

near w = 0. By means of Theorem 7.1 Lbwner obtained the exact bounds of all the coefficients bm . His result is

Theorem 7.5

If w = f(z) E IN, 1 _<

a

1.3 5

and z = f -1 (w) = w + Errix=2 bm wm then

(2m — 1)2m (m (m + I)!

2).

Equality holds when

f(z) = (1+z z)2 , f -1 (w) = [1— 2w —(1 — 4w)1] /(2w). It is again sufficient to consider instead of f(z) the functions et°f (z, to) of Theorem 7.1. For these functions can be used to approximate a general f(z) in S, so that their coefficients approximate those of f(z), and the coefficients of the inverse functions f -1 (w), which are polynomials in the coefficients of f (z), can be similarly approximated.

223

7.9 Coefficients of the inverse functions

Let then f(z,t) be the function of Theorem 7.1 and let z = ç5(w) be the inverse function so that

O t [f( z, t)] = z.

(7.35)

We write

Ot(W) = et [W

rri (t)W rri] (0

t '_

(7.36)

to),

m=2b

and fl = e—t°. Thus cc

ow) = oto(sw)= w +E bpn W ni m=2

is inverse to /3'f (z, t o ) in S, and so we need only prove our inequalities for the coefficients bm of this function 0(w), where

bm = fim—l bm (to). We note that the equations (7.1) and (7.35) lead again to the analogue of (7.23), namely,

a Ow) a Ot(w) w 1 + Kww at aw 1 — Kww Substituting from (7.36) we obtain just as in (7.24) m-1 b'n (t)+ bm (t) = mbm (t)+ 2 Erbrwpoim

r (0 '_ t

to, m > 2),

r=1

with the boundary conditions

NW

1 (0 < t < to),

bm (0) = 0, bm (to) = Vm+l bm (m

2).

These yield the inductive relation r {m-1 bm (t) = 2e(m-1)t i o

Erbreoo [K(T-r

e—(m-1)Tdt (m > 1).

r=1

It is now clear that, if to > 0, in > 1, bm (t0)1 attains its maximum possible value if K(t) 1. In this case all the b 1 (t) are real and positive. It is also evident that the corresponding value of bm = e— on— l )t°bm (to ) increases with increasing t o and so we obtain the upper bound for variable to in the limit as to —> oc.

224

The L5wner theory

We now take K(t) a: 1 in the differential equation (7.1) and obtain on integration f(z,to) _ flz (1 + z )2' 1+ [f(z,t0 )] 2 i.e.

w flOto(w (1 + 02 — [ 1 + Oto(w) ] 2. )

Writing Ow) = 4 t0 (f3w) we deduce

w (1 + f3w) 2.

Ow)

[ 1 + 46(w )] 2 Thus fl —p 0 (to —> 00) and 0(w) —> p(w), where

_

440 [ 1 + ti)(0[ 2

w'

i.e. ip(w) =

1 — 2w — .,/(1 — 4w) 2w

.E hoe, m=i

and bm =

1.3

(2m — 1)2m (m + 1)!

1

as required. The inverse function w = 1p -1 (z) = z(1+ z)-2 E S, and so tp(w) has the largest coefficients in our class and Theorem 7.5 is proved.

7.10 The argument t of f (z)/ z While the elementary methods of Chapter 1 are adequate to obtain the bounds for If (z)1, if(z)i, etc., when f (z) E S, the bounds for arg(f(z)/z), arg f'(z), etc., lie deeper. Here the function z(1 — z) -2 is no longer extremal. Following Grunsky [1932], we prove Theorem 7.6

Suppose that f(z) E e. Then we have

f (z) — log 11 + _ IzI <_ arg z

1 +1z!

log 1

izi .

(7.37)

Both these inequalities are sharp for any fixed z in Izi < 1. It is again sufficient to consider the functions etc' f (z , to ) of Theorem 7.1, and hence f (z , to ), since arg et° = O. We write f = f (z , t), K = KW for short. Then (7.1) gives 0 — logf = Ot

t If 4)(z) log 4(0)

1 + Kf 1 — Kf =

—(1

+ Kf)(1 —1(f) 11 — lcf 1 2

.

Izi < 1 and 0(0) > 0 we define arg0(z) as the imaginary part of ± foz OW 0(04, where the integral is taken along a straight line segment. 0m

7.10 The argument of f(z)/z

225

Taking real and imaginary parts we deduce

a

log If 1

a

1 — If1 2 11et — Kf1 2 '

=

(7.38)

23(Kf )

arg f =

(7.39)

11 — Kf1 2.

The equation (7.38) shows that If l decreases strictly with increasing t. Also (7.38) and (7.39) together give

dtargf

21f1 1 dr log — . If I 1 — If1 2

(7.40)

We integrate this from t = 0 to to and note that f(z, 0) = z. Thus

log ( 1 +1z1)( 1 — If(z, to)I) (1 — 1z1)( 1 + If(z, 4)1)

arg f (z, to) z

log 1 +1z1 . 1 — 1z1

This yields the upper bound in (7.37). The lower bound follows similarly. It is worth noting that the methods of Chapter 1 only lead to arg

f(z) 1 + IzI < 2 log . z 1 — IzI

We next show that the right—hand inequality of (7.37) is sharp for a fixed value zo of z in Iz1 < 1. To do this we have to find K(t) in an assigned range 0 < t < to so that the solution f = f(zo,t) of the differential equation

df 1 + icf _ , f (0 t to), 1 — KJ dt ,

with the initial condition f (zo, 0) = zo satisfies

3 (Kf) =

—III.

For in this case we shall have equality in (7.40). Also (7.38) gives in this case a log I fl — 1— If 1 2

1 + 111 2 '

so that f = f (zo, t) satisfies

If I 1 — 111 2 We define If I

= If (zo, 01

_,

Izol

e 1 — IZ01 2*

l n argf (zo, t) by by this equatio2n,dttlfhe

dt argf .

l — 111 2 '

226

The ',Owner theory

so that

log 1 + If I ,

arg ,f(zo,t) = log ( 1 +1zol )

1— I f I

—

zo

and finally K(t) by

(zo, t)1 f (zo, t)

K(t) =

With these definitions of K(t), f(zo, t) (7.38) and (7.39) are satisfied and arg[f (zo , t)/zo] can be chosen as close as we please to 10 0 1

+1zol)/( 1 —

since f (zo, t) 0 (t co). Thus the solution et° f (zo, to) of the equation (7.1) corresponding to this value of K(t) belongs to (B" and approaches the upper bound in (7.37) arbitrarily closely, so that this bound is sharp. By considering f(z o , t) instead of f (zo, t) we can show that the lower bound is sharp. This completes the proof of Theorem 7.6.

7.11 Radii of convexity and starshapedness

The function f(z) maps

1z1 = r onto a convex curve y(r), if the tangent to y(r) at the point f (reie ) turns continuously in an anticlockwise direction as 0 increases. The condition for this is that arg[irei9 (rei9 )] increases with increasing 0, so that

a

0 < 2g),

arg Prei) + 1 0 (0

i.e.

3

a —

{

00

log f' (reie )} = {reie

f" (rel f' (reie)

> -1 (0

This gives

(1z1= r).

(z)}

The inequality (1.6) of Chapter 1 gives for f (z)

r(2r — 4)

(1z1= r),

so that our condition is satisfied if 2r2 — 4r > r 2 — 1, i.e.

r2 — 4r + 1 0, 0

r

2—

0

2g).

227

Examples

Thus f (z) in S maps Izi = r onto a convex curve for 0 < r < 2 — .0. On the other hand if f (z) = z(1 — z) 2 , then

Z

f"(z) 2z2 + 4z = f'(z) 1—z 2 '

and this is real and less than —1 for —1 < r < V3-2. Thus this function does not map Izi = r onto a convex curve for r > 2 — V3. The quantity r, = 2 — V3 is called the radius of convexity.t We may ask similarly for the radius of the largest circle 1z1 = r such that the image y(r) of Izi = r by f (z) always bounds a starshaped domain with respect to w = O. The condition for this was seen in Chapter 1, (1.15) to be that

} 0, f(z) 9q z f(z)

i.e.

TE

f'(z)}

arg {z f(z)

(1Z1 = r).

If we write

f 0(z) =

(41 + z 1 + fo z

)

— f (zo)

(1— I zo1 2 ).r(zo )

'

(7.41)

then 0(z) E S if f(z) E S. On applying the inequality (7.37) to 4)(z) at z = —zo, we obtain

arg

(

f( zo) zo f' (zo))

)

...

log

1 + lzol 1 — Izol'

(7.42)

and this is sharp. Thus the radius of starshapedness, rs, being the radius of the largest circle whose interior is always mapped onto a starshaped domain with respect to w = 0 by f (z) E S, is given by

n 1 + rs n — = log I-, = tanh — = 0.65... 2 4 Examples

7.5

If f(z)

E

S and (/)(z) is defined by (7.41) show that

zo f(zo)(1 — 1z01) 2. Deduce that (7.42) is sharp and verify (7.43). —

t Gronwall [1916] * Grunsky [1933]

z 0 0'( —zo) Sb( — zo)

(7.43)

228

The L5wner theory

7.12 The argument of f'(z) result of Golusin [1936].

Theorem 7.7

As a final application we prove the following

Suppose that f(z) E S. Then we have the sharp inequalities

{ I arg f(z)I < 4 sin-1 Izl I arg f(z)I < n + log

(1z1 --v -) t i 1) .

1z1 2 1 —Iz1 2

<1z1

<

We may again confine ourselves to the functions f(z,t) of Theorem 7.1. We start with equation (7.1)

1 + Kf

0

t) = f ,

2

2

, =f + K KJ

KO - Kf) '

and differentiate both sides with respect to z. This leads to

2

0 —f(z t) = f/(z,t)[1

1

— Kf) 2 _1'

i.e.

2 ' log flz, t) = 1 Ot

2 (1 — 14) 2.

Taking imaginary parts we deduce (1 arg f (z, t) = 23 — Kf )2 11 — 41 4 . We eliminate t between this and (7.38) and deduce -:i

clt argf =

23(1 — Kf ) 2

— dt If I

11 —4J 2

If 1(1 - I f1 2 )'

as t increases. Now since IK(t)I = 1

1 3 —4)2 J = I sin[2 arg(1 — ( 1

I 1 — KfI 2

21f 101 - I f1 2 ),

<

(If I

(7.44)

(If I ___ ,) • 1,

{

Recalling tha dtlf I
t

—4cltlfl

dt ar g f ' <{

0 1 — If 1 2 )

7 1 )

—2d, if I If 1(1 - I f 1 2 )

,/1 2)

,

229

7.13 Conclusion Integrating from t = 0 to to , we obtain for lz I <

< f lz1

arg

4dx

*2

< 4 sin-1

\I( 1 — x2) —

Jf1

and for I z I >

4dx + — x2)

ar lg

1z1

2dx x(1 — x 2 )

7E + log ( lz12

1 — Iz1 2

This proves the inequalities of Theorem 7.7. To show that they are precise we must find K(t), such that if f (zo, t) is the solution of (7.1) with assigned initial value f (zo, 0) = 41 , then equality holds in the inequalities (7.44). The resulting equation enables us to calculate Kf(zo, t) in terms of If(zo, 01, hence If(zo, t)1 in terms of t by means of (7.38) and then arg f(zo, t) in terms of t by means of (7.39). We can then choose K(t), so that equality holds in (7.44). When this is done for 0 < t < to, then (7.38) and (7.39) and equality in (7.44) will hold simultaneously and we shall have [1201 4dx arg f i (zo, to) (Izol < i If(z0,01 0 1 — x2 )

1

arg f'(zo, to)

=

v2

4dx

ji

_2, + log 1 _zoI2 (1Z01 > ,/2) IZCII2

ilf(zo,to)1 0 1 — "(' )

if If(zo, to)I < and so for all large to . Thus the upper bounds of Theorem 7.7 may be approached as closely as we please and so are sharp.

7.13 Conclusion The foregoing theorems represent some of the principal successes achieved by L6wner and his successors by means of Theorem 7.1. In the next chapter we shall use the technique to prove de Branges' Theorem and some of its consequences. At this point Schiffer's variational method and Jenkins' theory of modules t should be mentioned. These methods can be used to prove the results of this chapter and some others and in particular to give more information about the extremal functions. However Liiwner Theory has so far proved to be an essential ingredient of the proofs of de Branges' Theorem. Schiffer [1943], see also Duren [1983, p. 318 et seq.] Jenkins [1958].

8 De Branges' Theorem

8.0 Introduction In this chapter we prove de Branges' Theorem [1985], conjectured by Bieberbach [1916] that, if x, anz n E a.-,, (8.1) f(z)=z+ we have Ian I < n, n= 2, 3, ... with equality only for the Koebe functions f(z) = z(1— zeie) -2 . This result had previously been proved for n = 4 by Garabedian and Schiffer [1955], for n = 5 by Pederson and Schiffer [1972] and for n= 6 by Pederson [1968] and Ozawa [1969]. (For a more detailed history see Duren [1983, p. 69].) De Branges proved his theorem by first establishing a conjecture of Milin [1971], which Milin had shown to imply Bieberbach's conjecture. Suppose that

log

f(z)

k CkZ ,

z

(8.2)

then Milin conjectured that n

E ( _4k _ 1 ( I

c k 1 2 ) ( n—k

+1) 0, n= 1, 2, ...

(8.3)

k=1

The proof of de Branges has been simplified successively by Milin [1984] and Emelyanov, by Fitzgerald and Pommerenke [1985] and Weinstein [1991]. Inevitably these simpler proofs however miss the operator theory basis of de Branges' subtle ideas. All the proofs rely on Li5wner Theory, and a positivity result for the coefficients of certain special functions. The earlier proofs used an inequality of Askey and Gasper [1976] concerning Jacobi polynomials. Weinstein's proof, which we shall follow here, uses instead the addition formula for Legendre polynomials which

230

8.1 Legendre polynomials

231

goes back to Legendre himself and seems simpler to establish. However Wilf [1993] has now shown rather surprisingly that the two results are equivalent. We start off by proving Legendre's formula, then prove Milin's conjecture, Milin's inequalities and de Branges' Theorem, which in fact generalises to give sharp bounds for the coefficients of (f(z)/z);-, when ), > 1. The analogous result fails for 0 < < 1. A number of further generalisations and consequences, including proofs of conjectures, of Robertson [1936] and Rogosinski [1943], will be given at the end of the chapter. De Branges' Theorem, both the result itself and the subtlety of its proof, represents a milestone of twentieth century analysis.

8.1 Legendre polynomials

The Legendre polynomials P(z) are defined

by the expansion 0,

1

=1

+ E hnPn (z),

(8.4)

(1 — 2zh h 2 ) 2 valid when 1111 is sufficiently small depending on z. We follow the account given in Whittaker and Watson [1946, Chapter 15] and start with the following simple lemma, that is easily proved by means of the calculus of residues.

Lemma 8.1 Suppose that A, B, C are complex numbers, such that A 0, 0, that F is a circle containing in its interior A = B 2 — 4AC 0)/(2A) but not t2 ( —B —/A)/(2A), and that F is described t1 = in the anticlockwise sense. Then dt fr At 2 Bt C

27ri \./A •

(8.5)

We shall also need the associated Legendre functions pnk ( z ) = (z 2

1 ) ,;k

d k

Pn (z), 0

We suppose for the time being that 1 < z < c and positive. We choose for F the circle t = Z

(Z 2 — 1) 2 e iu ,

— TE <

,

k

n.

so that (z 2 — 1)1 is real

O<

(8.6)

232

De Branges' Theorem

which has centre z and encloses z = 1 but not z = —1. Then if h is sufficiently small Lemma 8.1 yields 1

dt

1

(1 2zh + h2)=1 =

fr 2(t — z) + h (1 t 2 )

1 dt rci fr t2 — 1 — 2h(t — z) •

We expand in powers of h and equate coefficients in the result and in (8.4). This yields

f

1

P(z)

jr

(t2 — dt (t — z)fl+ 1

2" rri

I'

(t — z)ndt (t2 — 1 )

+1 •

Differentiating under the integral signs we obtain (z 2 — 1)

2n (Z 2 —

1) n dt ( t _ z )fl+k+i

(n + 1) • • • (n + k) 2n -orri

(1-2 —

1k

1) 2 n(n — 1) • • (n — k + 1)(-1) k rci

(t — z)n —k dt

r (t2 — 1) 1+1 •

Finally we substitute for t from (8.6) and note that Pnk (z) is real, when z> 1. Thus } pnk( z ) = (n + 1). • • (n k) {z + (z 2 — 1) COS 0 n COS k0c10 (8.7) 2m

and k

Pn (z) =

X

f

n(n — 1) • • • (n — k + 1)(-1)k 2n —n-1

+ (z2 — 1) 2 COS 0)

COS

Ode.

(8.8)

We can now prove Legendre's addition theorem. Lemma 8.2 Suppose that x, y, w are complex numbers and that Z = xy — (x2 — 1)1 (y 2 — 1)1 cos co. Then P1 (Z) = P(x)P(y) + 2E n (-1)k (n k)! P k (x)P k (y)cos kw. (n + k)! n n k=1

8.1 Legendre polynomials

233

For the proof we shall assume that x, y, co are real, x > 1, y > 1, and that the positive square roots are taken in the definitions of Z and Pnk (x), e;(Y). The general result, with a suitable definition of the signs of square roots, is then obtained by analytic continuation. We assume that h is small, and that 4) is real. Then I

{X ± (x2 — 1) 2 COS(0) - o

x

E h"

{y

n=0

i

± (y 2 — 1) 7 cos 0}

r

n+1

1

=

.

(8.9)

y + (y 2 — 1)-1 cos s 4) — h (x + (x2 — 1) 1 cos(co — We integrate both sides of this equation w.r.t. 4) from —7r to n. The integral of the right-hand side takes the form i

=

r J _ir

dck a + bcos 4) +csinck'

where

a = y — hx, b= (y 2 — 1) -1 — h (x 2 — 1) cos co, c = —h (x 2 — 1) I sin co. To evaluate / we write t = eiO , dck = dt I (it), cos ck = 1 (t + t'), sin 4) = (t — t-1 ) /(2i). This yields

I =

dt li _ i At 2 + Bt + C

2ni \IA'

by Lemma 8.1, where A = -1-(c + ib), B = ai, C = 1(—c + ib).

(Since IA I = ICI, the Lemma is applicable.) Also A

= B 2 — 4AC = b 2 ± C 2 — a2 = (y2 _ 1 ) ± h2 (x 2 1 ) 2h (x 2 — 1) / (y 2 — 1) cos co — (y — hx) 2 1

= — {1 ± h 2 — 2hxy + 2h (x 2 — 1) / (y 2 — 1) 2 cos co} = — { 1 + h 2 - 2hZ } . Thus

1 =

+27r .\/ (1 + h2 — 2hZ) •

234

De Branges' Theorem

To check the sign, we put h = 0 and obtain 77r

=

= 27r,

y + (y 2 — 1) cos (1) by (8.8) with k = n = 0. Thus for small h we have 2

=(1+

— 2hZ) —

h"P,(Z) n=0

by (8.4). On comparing coefficients with the integral of the left-hand side of (8.9) we obtain 1n {X + (X 2 —

P(Z)

= 27C fir

2 COS(W — 0)} d4)

{y2 ± (y2 ____ 1) cos

(8.10)

}n-4-1

•

The right-hand side is a polynomial of degree n in cos w and sin co and is an even function of co, as we see on making the substitution 4) = Thus 1 P,(Z)= —A0 ±

2

Ak cos kw. k=1

It remains to evaluate the coefficients Ak. We have Ak =.-

it

—77

1

Pn (Z)cos(kco)dco

i

n

_ircos(kw)dco

27,2

r

x + (x2 —

1) 2 COS(CO

o) y do ,

n+1

•

fy + (y 2 — 1) 1 cos (/)

We integrate first w.r.t. co, setting co = Ø + 0 and using (8.7). We integrate from w = — it to 4) + 7r, i.e. from 0 = —n to n. This is legitimate since the integrand is periodic in co. We also write cos kw = cos Ice cos k4) — sin Ice sin k4) and note that the integral containing sin(k) vanishes since the integrand is odd. Using (8.7), and for the remaining integral in 4) (8.8), we obtain Ak

2/3,` (x)P,;((y)(-1 )k (n+k)(n + k — 1)• • • n(n — 1)• • -(n— k +1) 2(n — k)!(-1)k P(x)Pl(y). (n+k)!

This proves Lemma 8.2.

8.1 Legendre polynomials

235

8.1.1 We complete the section by proving the positivity result which is needed for Weinstein's proof. (In a recent preprint by Ekhad and Zeilberger [1993] the authors produce a short direct argument for this.) Lemma 8.3

Suppose that t > 0, and that w = w(z) is defined by

et w (1 — w) 2

w,(0) = O.

(1 — z) 2 '

(8.11)

Then we have for z1 <1 et wk

A(t)z'

1—w where A(t)

(8.12)

1,

n=k

0, for t > 0 and 0 < k < n < X.

We write x = y = (1 — e — t) 2 in Lemma 8.2 and note that

Z=1— _ (y2

We choose (x2 — 1)

+ e—t cos co.

1) 7 =

P(x) = 13,`())) = (ie—t ) k

(I CITC)

Pn(x)

where p is real.

= ik19'

Thus ptilc( x )pk)( y ) =

02k p2 >

O.

Now Lemma 8.2 yields

P(Z)

Ak,„(t)coskw, k=0

where Ak, n (t) > O. Next 2

Pn (Z)zn}

1 —2zZ z 2 {r.,

=

Z

m

In=0 k=0Ak, ni COS kw} {Êzn

E n Ai, n COS /W}

n=0

We recall that cos kw cos lw = {cos(k — obtain 1

1 — 2zZ z 2

n=0

1=0

+ cos(k + Ow). Thus we

Bo i COS k=0

.

kco,

(8.13)

236

De Branges' Theorem

where Bk, n (t)

O. Again by (8.11)

1 z + — 2Z

1 —2zZ + z 2

1 et w e— t ( 1 w — 2) — 2Z 1+ w 2 — 2w cos w

2

1 w2 1 w 2 — 2w cos w x etw { ,k cos(kw)} 1 _ w2 + 1 2 k=1

=

=

Ag(t)z" +1 ± 2 k=0

iVi (t)zn +1 cos kw. k=1 n=k

then of zn+ 1 cos kw in this and

Equating coefficients first of z' (8.13) we obtain that

AP(1 = B,, and Afkl =

k>1

and this proves Lemma 8.3. We remark that wt (z) maps izi < 1 onto 114, < 1 cut from —1 to --- "C along the real axis, where

= 2et —1 _ 20e2t — et). Thus wt (z) is regular and Iwt (z)i are valid for lz! < 1.

for 1z1 < 1 and so the series (8.12)

8.2 Proof of Milin's conjecture: preliminary results

In this section we

prove

Theorem 8.1

If f(z)

Ee

and

log

f(z)

GC

(8.2) k=1

then we have for n = 1, 2, ...

4

— kIck1 2 ) (n — k + 1)

O.

(8.3)

k=1

Theorem 8.1 is the key step in de Branges' proof. As we shall see the result also leads to a number of conclusions which go beyond Bieberbach's original conjecture. Weinstein's proof actually shows that equality

8.2 Proof of Milin's conjecture: preliminary results

237

holds in (8.3) only for the Koebe functions fo(z) = z/ (1 — zei 6 ) 2 . He uses a refinement of L6wner Theory which applies to every function in S and not just to a dense subclass. For this we refer the reader to Weinstein [1991] and Uiwner [1923]. To prove Theorem 8.1 it is sufficient by Theorem 7.1 to consider the functions f(z) in the subclass S i of that theorem. Thus

f(z) = lim et f(z,t), where f(z,t) satisfies Uiwner's equation (7.1), K(t) is continuous and k(t)! = 1. We define g ( ) as in Theorem 7.2 by

= f(z).

gt If(z,

(8.14)

Thus go(z) = f(z). Also for large t et z

gt (z) =

(1 — zei0) 2.

We now write OC

h(z, t) = log

gt(z)

ck(t)z k .

zet

(8.15)

k=1

Thus ck (0) = ck,

2 k

and ck(t) = —e 1-4", t > to.

(8.16)

We deduce from (7.1), (8.14), just as in Section 7.8, that (7.23) holds, i.e., if g = gi (0, = f(z,t) we obtain

eg eg r 1 + et

404. 1—

K

g = g1 (z) we have

or writing z instead of

eg äg 1 + KZ —= Z et äz 1 — KZ

(8.17)

Using (8.15) we deduce

0

— h(z,t) = et

0 — h(z, t) ez

=

eat

eg/ez g

1 1 z•

Thus (8.17) yields Oh (1 0h) 1+ Kz 1+=—+ z z ez — KZ et

238

De Branges' Theorem

i.e. Oh = ( 1 + z a h) ( 1 + Kz) — 1. at az ) 1— . KZ We substitute (8.15) in this and equate coefficients in the resulting series. This yields c(t) = kc k 2k k +2

rCr K

k—r

.

(8.18)

We need some crude bounds for ck(t),c(t).

Lemma 8.4

We have for 0 < t < oo and 1 < k < oo,

1001 < 9, 14(01 < 11k 2 . We deduce from (8.15) that z

ag/Oz

=1+

k ck z k

.

Also e'gr (z) E S. Thus (1.4) and Cauchy's inequality yield 1 —k- sup r I z i =r

1(10(1

Oglez

r

<

1+r

rk (1 — r) .

Choosing r = k 1(k + 1), we obtain

< (2k + 1) (1 +

< 3ke <9k.

Now (8.18) gives k-1 14(01

2 + 9k -I- 2E9r = 9k + 2 < 11k 2 . r=1

This proves Lemma 8.4.

8.2.1 Completion of proof

Following Weinstein we prove (8.3) by show-

ing that n

(n — k + 1)} zn +1 n=1

k=1 11+1

gn(t)dt,

(8.19)

8.2 Proof of Mum 's conjecture: preliminary results

239

where

g(t) > 0 for t > 0 and n = 1, 2,... To prove (8.19) we fix z, such that 1z1 < 1, and define w = wt (z) by (8.11), i.e. et w

(8.20)

(1 — w) 2.

(1 — z)2

We recall that by (8.20)1z1 < 1 corresponds to iwi < 1 cut along a segment of the negative real axis. So Schwarz's Lemma yields Iw(z)1 < izi for 0 < t < oc. Also /I

DO

klCk( 0 )1 2 ) (n — k + 1)zn +'

=

n=1 k=1

I

to

z d j (1 — z) 2 dt f=1

/4

— kick(t)1 2 ) W k

dt.

(8.21)

For if

tp(t) =

kick(t)12) wt(z)k, k=1

we have by (8.16) 1P(t)

= 0, t

to, and i,v(0) =

( 1–c- — klCk(0)1 2 ) Z k . k=1

We differentiate the series in (8.21) term by term and integrate the result term by term. To justify this we need to show that the differentiated series converges uniformly in [0, to]. We write x' = ax/at. Then (8.20) yields ,

w=

—(1 — w)w (1+ w)

Also by Lemma 8.4

a ICk(t)I 2

c k (t)c k/ (t)

+ ck(t )4 (t ) < 198k 2 .

Thus

d f (4 Tt —klck(t)12 ) wk } =

w (4— k 2 1ck1 2 )

W k — kW k

at (Ck(t)Ck(t)) .

240

Theorem

De Branges'

1z1 < 1, the right-hand side is bounded by Ak3 1Zi k , Since by (8.20) where A is a constant depending only on z. This implies the required uniform convergence. Using also (8.20) and (8.21) we obtain

IX e t w

{ 1 W

1(z) = fo 1_ w2

1—W

k 2 ICk1 2 ) w k } dt.

k (ck(t)ck(t)) / vv k k=1

k=1

(8.22)

Next we write z 1 = ri ei° , where z < r 1 < 1 and define

G(0) =

1

ag (z i ,t)

g(zi,t)

at

=1+

ah(z i t) at

E c'„,(t)rretme

'

(8.23)

m=1

by (8.15). Thus by Lemma 8.4, and since r 1 < 1 and 114, 1 < 1, we have oc

1

E kck wwk ..= 27r /0

x

27r

G(0 ) do

k C ,"2k = E kC kt WkW 1 w k=1

k=1

Hence

kc k' (t)ck(t)(tiw) k = 1 +2 etw`n

k=1

1.2 IT

X

X

Ekck(t)wkre'kfld0

G(0) 1+ 2 E wrn m=i

27r j0

k=1

f 2rt

/

27r

l=1

k=1

m=1

k=1

1

kc' (t)ck(t)(qw) k

mc m zr — lcizi } do

G(0){2 0

m=1

{ 27r Jo

1=1

mc m zr

G09){ 2 (1 +

do — 2

—

.

(8.24)

m=

Similarly

1+w 1

—

W

OC

k ck(t)c'k (t)(r?w )k k=1

{1 127c —27r. 0 G(0) {2 (1 +

mc m zT) — lciz

d0 — 2

m=1

1=1

Next we deduce from (8.15) that

1+

ah(z i , t)

=

ag( z t)/a z g(zi,t)

mcm (t)zT.

=1+ =

.

(8.25)

8.2 Proof of Mum 's conjecture: preliminary results

241

In particular 1

I {z i ag(z h t)lazi

=1+

g(zi,t)

m=1 where Pi±i (z i ) is a power series having a zero of order at least (1 + 1) at z i = O. Thus we may multiply the integrand in the coefficient of wi in (8.24) by 1 + Pi+i (z i ) without affecting the value of the integral. Substituting for G(0) from (8.23) and using (8.17) we obtain from (8.24) 00 1+w fkck (t)c;,(t)(qw) k + 2 Ewk

1—w

k=1

k=1

=E w / 1 [ 27z 1 + K(t)zi

1 — ic(t)zi

2m Jo

1=1

mc m zr } 2 1 + n=1

00

M= 1

1 + /al

/ J 1

2mf2 o 1 — /al

1=1

mc m zr ) — /z dO, 2

2 M=

z' mcn,zr) — lici 2 1

dû

+ 1 /2 101 2 ril l . 2 We rearrange (8.25) similarly and let r1 tend to L The series all converge uniformly and absolutely for fixed z, variable t and 1z1
1 + KZ

= 1,1(Zi,t)

— KZ1

where u > O. Thus (8.22) yields finally I(z) =

Go

et wdt

fo 2

1 f2n X liM

inc,„z7i — ici z

14(Z 1, t) Lit

0

m=

" etwdt o 1 — w2

=f

dû

)

cc

Ai(t)wl,

/=1

where Ai(t) > O. On combining this with Lemma 8.3 we obtain Do

AO) EA7 (o zn+i

I(z) = f dt 0

1=1

n=1

n+1

n=1

gn (t)dt,

242

De Branges' Theorem

where

g 1 (t) =

O.

Ai(t)A(t)

This proves (8.19) and so Theorem 8.1.

8.2.2 An extension Following de Branges [1985] we deduce the following slight generalisation of Theorem 8.1 which will be useful in the sequel.

Theorem 8.2

Suppose that o- 1 ,62 ,...,o- N+1 is a sequence such that

aN+i = 0

•••

>

(8.26)

and 01 — 0 1+1

0-k+1

0-1c+29

1

k < N —1.

(8.27)

0•

(8.28)

Then E (kIcki2 — —4 ) o-k

We choose nonnegative numbers (xi to aN, multiply (8.3) by —a n , and add for n = 1 to N. We deduce that 2 4 E (klcd — — k

ak

0,

k=1

where ocn (n — k +1).

=

(8.29)

n=k

It remains to show that if the o-k satisfy (8.26) and (8.27) we can solve this system of equations for nonnegative oc n . In fact (8.29) yields ŒN = 0- N

and if k
ak+1 to aN have been chosen we obtain from Crk

Thus an

20-k+1

(8.29)i-

Cik+2 = OCk•

0 by (8.26) and (8.27). Hence (8.28), i.e. Theorem 8.2 is proved.

8.3 The Milin—Lebedev inequalities

243

8.3 The Milin—Lebedev inequalities Milin and Lebedev (see Milin [1971]) proved some subtle inequalities involving coefficients of power series. Milin showed that by means of these inequalities Bieberbach's conjecture could be deduced from Theorem 8.1. We proceed to develop these results. We suppose that

(8.30)

Akz k k=1

is a power series with a positive radius of convergence, though in fact all the results remain valid for formal power series. We write CO

4)(z) = exp{w(z)} —

(8.31)

Dkz k . k=0

We also define the binomial coefficients dk() by Do 7k

(1—zr1 = exp

{,1E

—

-

d (2)z k ; dk(;t) =

k

F(.1. + k) (8.32) F(2)F(k + 1).

Here F(x) is the gamma function. Then we have (Milin [1971, Lemma 2.2, p. 44])

Theorem 8.3

With the above notation we have for n = 1, 2, ... and 2>

n

E ID /(12 k=0

dk(2)

1

ci„(4,+ ' 1)exp { dn(Â, + 1

n

0

( k 2 lAki 2 — 22) d n_ k (2 + 1)} •

k2

k=1

(8.33) Equality holds if and only if Ak =

k

for k = 1, 2, ... , n,

where 1 ,71 = 1

.

(8.34)

Theorem 8.2 says roughly that if Ak < 2/k in a suitable average sense, then 1Dkl < dk(),) also in some average sense. We follow Milin's argument and note that z(/) / (z)

= 0 (z )zoi(z ).

Equating coefficients of Z k we obtain kDk =

vAvDk—v• r

244

De Branges' Theorem

We write a,, = vlA r l and deduce from Schwarz's inequality Ipv

1Dk1 2 < /712

2dk-v(;1-) v= i

I.=

(8.35)

dv (2) •

We apply this with k = n and deduce n

1Dk1 2

\--.

n

1'9 1(1 2 ,

-t-

;t) k=0 dk(

k=0

1 n2dn(2)

[

U + 2 n 1 1Dk1 2

=

n

k=o

E dn_k(2)at k=1

n

n+

dk

n—

1Dkl

12

k=0

EL I dn_k wa,2, n(n + A)dn(),)

We denote the term in square brackets by x, and use (8.32) and the fact that x < exp(x — 1), —oo <x < co, with equality if and only if x = 1. We obtain Dki2

n

I

k=0

I

dk(A)

dni + 1) n 1 < dn-1(2+

1Dk12

1) k =0 dk(2)

Enk=1 dn—k().)aii n(n + 1t)d1(;1-)

exp

We now write tn —

n2

1 dn(il + 1)

/12

(8.37)

d1—k(;1-‘ ± k=1

and deduce, using (8.32) with the convention d_1(;t) = 0, that tn — tn-1 = k=1

— 2 2 ) dn-k(ii) n(n + ).)dn ()) •

(8.38)

In fact by (8.32)

dn_k(2) d,7 (A)

F(n — k 2) F(n +1) F(n — k +1) F(n + 2)•

Thus

dn_k(A + 1) d(2 +1)

d 1_i_k(2 + 1) d 1_1(2 + 1) F(n —k + 2)F(n) f (n k /1-)n F(n — k)F(n + ;t) 1(n — k)(n + A) Ak dn-k(/1 ) n(n + 2) cl1 (2)

1}

245

8.3 The Milin Lebedev inequalities —

Now (8.37) yields (8.38). Further

22

n(n + A) k =i

n-1

dn—k0 d 1 (A)

n(n + 2)d 1 (2) 1=0 dP(2) A n+

A2 dn_i(A + 1) n(n + .1.)dn (.1,)

(8.39)

n—i d(A) = dn_i(A + 1) follows from

The identity p=0

(1 — z) —À (1 — z)-1 = (1 — Using (8.38) and (8.39) we can write (8.36) in the form 1

n Pic12

dn (/1. ± 1) k=0 dk(A)

exp(

<•<

1

tn)

2

n- 1

± 1) k=0 dk(A)

1

exp{—tn— 1.

Pol 2

do(A + 1) do(A) exP(—to) = 1.

(8.40)

This yields (8.33). If equality holds in (8.33), then equality must hold in (8.40) and this is only possible if equality holds in (8.36) when n is replaced by n, (n — 1),..., 1. Since x < ex—1, for x * 1, this implies

tn

= tn_ i =

= to = 0.

Since t t = 0, we have 1A 1 1 2 — A2 = 0, and now induction on k yields

k2 lAk 1 2 = A2 , i.e. kA k = A e Ol , k = 1 to n. This yields (8.34) for n = 1. Suppose next that n > 2. Then equality must hold in (8.35) for k = 2, ..., n. Suppose first that Oi = 0, i.e. A l > 0. Then, since Do = 1, D I = A 1 by (8.30) and (8.31), we deduce, when k = 2, that

0 = arg(A D ) = arg(2A2D0), so that A2 > 0. By induction on n we deduce that Ak > 0 and Dk > 0, for k = 1 to n. If O i * 0, we apply the above argument with w 1 (z) = w (ze —i°1 ) and 4)(z) = (ze —i81 ) instead of w(z), 4)(z). Then Ak, 1Dk are unchanged so that equality still holds in (8.33) but now ai (0) > 0 so that the first n coefficients of w i (z) are positive. Thus

Ak = Ake1

=

e k01 ,

k = 1,

n•

This yields (8.34) with ri = e 0' and the proof of Theorem 8.3 is complete.

246

De Branges' Theorem

8.3.1 We can also obtain a corresponding bound for Dk. This is Milin's second inequality [1971, Theorem 2.4, p. 50].

Theorem 8.4

With the hypotheses of Theorem 8.3 we have

113 ,1 ._ dn(A)exp

n

1

dn kw ( k 2 01(1 2 — 22

{ 24:1)

(8.41)

kA

)

Again equality holds if and only if (8.34) holds.

To prove Theorem 8.4 we deduce from (8.35) and (8.33) for n = 1,2,... n

1 IDnI 2 < n2

n

-

vdn_v(il) v—

v=i

1

Ip v 1 2 d,(.1.)

< dn_1(i1-1- 1) n2

n

1 dn_1(A +1)

n _ v (1) exp { v=i

n—v-1(A + 1 ) } .

(8.42) We have, since x < ex -1 n

E l a d 2dn

_ r( 2

,

)

v=1

= /12 4- 1( 2

n

1

+ 1)

{

42,dn—v().) ,1 2 dn_1(2 + 1 ) v=

< 22 dn-1(il + 1 ) ex P { ,1 2 d

}

n

1

1

(A + 1

}

1

. A 2 c/ n- 1 (A, + 1 ) ex P { /1 2 d 1 _1( 2

+ 1)

(

V

22 )

V=1

by (8.39). On combining this with (8.42) and using (8.32) we have IDnI __

n

22) (d„—v-1(/1. + 1 ) Vil

fi

= d1(2)exP 1

n

2dn(2) V=1

(a,, _ 22 ) Av

dn—v( 2 )} •

+

dn—v(1 /12

8.4 Proof of de Branges' Theorem

247

This proves (8.41). Equality holds only if equality holds in (8.42). This yields equality in (8.33) with n — 1 instead of n, and so (8.34) holds with n-1 instead of n. Also equality holds in (8.35) with k = n and this shows that (8.34) holds also for n. This completes the proof of Theorem 8.4.

8.4 Proof of de Branges' Theorem conjecture and a little more.

Theorem 8.5

We can now prove Bieberbach's

a and that = Ê anwz n—i .

Suppose that f(z) E

( f (z))'

(8.43)

z)

n=1

Then, if A > 1, and n> 2, we have

lan(4 __ d n _1(2.1) =

F(n — 1 + 22) F(n)F(22) •

(8.44)

Equality holds if and only if f(z) is a Koebe function

k8(z) =

z (1 — zei8 ) 2

, where 0 is real.

(8.45)

If 2 = 1, ci„(2) = an , we obtain Bieberbach's conjecture. We also recall from Example 7.3 that (8.44) fails for —1 <2 < 1 and n = 3. In fact the sharp bound for 1a3 (2)1 in this case is {1 + 2 e 21 )/(i1 + 1 )} >

F(2 + 22) =14 22 + 1 1 - -1d2( 24. 2F(2)

To prove (8.44) we deduce from Theorem 8.1 that (8.3) holds. Thus

= exp

ACkZ k

{:=1

,

}

where n

, 4 E (kIck l- _ —) (n — k + 1) 0, n = 1, 2, ... k k=1 We apply Theorems 8.2, 8.4 with p = 2 2 instead of A, and Ak = ACk, Dk = ak +i(A). In order to apply Theorem 8.2, we need to check that, if

248

De Branges' Theorem

ak = dN—k(P), k = 1,2,..., N +1, p our convention

2, (8.26) and (8.27) are satisfied. By

0-N+1 = d-1(p) = 0. Also o-k > 0 for 1 < k < N, and by (8.32) Crk+1

dN—k-1( 1 )

ak

dN—k(J1 1 )

N—k <1, since p> 1. p+ N—k—1

Thus (8.26) holds. Next N k ak -Ei — ak = ( p+N—k-1

= (1

1 ) uk, ak

p N — k) ak• 1N—k

Thus { Crk+1

20-Ic

ak-1 = ak

1—p

1—p p+N—k-1 1+N—k}

— 1 )(P — 2)ak >0 (N —k p-1)(N —k +1) —

if 1 < k < N. Hence (8.26) and (8.27) are satisfied and we deduce that

E(kck12—

(k2 lAk 1 2

2

4

4A2 ) dN_k(2)1.)

dN—k(P)= -AT;

k=1

O.

k=1

Now (8.41), with

instead of)., yields

laN+I(A)1= This is (8.44). If N particular we have

dN (2).), if N > 1.

1, equality is only possible if (8.34) holds. In =

=la2(1)1= 2.

Now it follows from Theorem 1.1 that f(z) is given by (8.45).

8.5 Some further results We note some consequences of Theorem 8.3, which go beyond Bieberbach's conjecture. Theorem 8.6

If f2(Z)

a21-1Z 2" G S,

then

.a2N-1, 2 < N lail 2 +1a3i 2 +... + II

1.

(8.46)

Equality holds for N > 2 if and only if f(z) = z/(1 — z 2 eie ), for a real O.

8.5 Some further results

249

The result was conjectured by Robertson [1936] and proved by de Branges. We recall from the end of Chapter 5 that, since f 2 (z) is odd and univalent, f (z) = 2 (z )1 2 is univalent. Thus f2 (z)

1 + a3 z + a5 z 2 ± • • • =

(f(z)) 4 Z

z

Hence with the notation of (8.43) we can write (8.46) in the form

E l an( ) 12

(8.47)

N.

k=1

To prove (8.47) we apply Theorem 8.3 with follows from Theorem 8.1 that

k= 1

k 2 lAkI 2 k

)'2 ) )

Ak = ..Ck

041( 1 2

n

4

k=1

1)

k

and A = 1. Then it

(n—k+1)_ O.

Now (8.33) yields

ElDk12

n + 1.

k=0

Equality holds for n > 1 only if la3 1 = 1. In this case

f (z) = z + 2a3 z 2 + • • and since f (z) E S, f (z) = z (1 — zei°) f2(z)

=f

2 by Theorem 1.1. Thus

1

( z 2) 2 =

(1

z 2 ei0) •

This proves Theorem 8.6. Theorem 8.6 implies a conjecture by Rogosinski [1943], as was shown by Robertson [1970]. Theorem 8.7 Suppose that w i (z), w 2 (z) are regular in 1z1 < 1 and satisfy < 1 there, and further w2 (0) = 0. Suppose that f(z) E and that F(z) = wi(z)f { 02(z)} =

Z

n .

n=1

Then lA n 1 < n, n = 1,2,... Equality holds if and only if F(z) = e'Â ko(z), where k 0 (z) is the Koebe function (8.45).

250

De Branges' Theorem

If co i (z) EE 1, then F(z) is said to be subordinate to f(z). This is the case for instance if f (z) maps Izi < 1 onto a simply connected domain D and if F(z) is any function regular in 1z1 < 1 and with values in D. For these functions F(z) the result was conjectured by Rogosinski [1943]. To prove Theorem 8.7 we follow the exposition of Duren [1983, p. 196]. We write

i Z.) 2 (z n

H(z) =1

=

z ± c3z 3

and proceed to show that n l Ai

E 1C2k-112

(8.48)

n

<

k=1

by Theorem 8.6. We define

(/)(z) =

H ( \iz )

= 1 + c3z +C5Z 2

\iz

±••'

Then

2 =

f(z)

z

(Pz)

and

F(z) = wi(z)f {co2(z)} = w 1 (z)co2(z) {1 ± c3w2(z) + c5co(z) 2 + • • -12 . We denote by s(z) = Ekn =1 C2k-1Z " the nth partial sum of 4)(z). Since 0)2(0) = 0 we can modify the Cauchy representation of the coefficients A n of F(z) to

An

1 = 27ri

II,

wi(z)(02(z) {sn[o2(z)]} 2 dz z n+1

.

(8.49)

It now follows from Littlewood's subordination theorem (see e.g. Hayman and Kennedy [1976, p. 76]) that 2

1:1=r

1Sn [(02(Z )] 1

1dzi

Thus (8.49) yields, since ico1(z)co2(z)i

l An 1

<

1 21Ern

1

= 2nrn

I=:1 r

isn (z)1 2 jdzi.

(8.50)

1:1=r 1,

Isn [co(z)] 2 lidz1

1 2nr2

Isn(z)1 2 1dzI

n

E

1C2k-1 i

2r2k-1.

(8.51)

k=1

Letting r tend to 1 we obtain (8.48) and Theorem 8.7 is proved, apart from the cases of equality.

8.5 Some further results

251

To deal with these we note that if n = 1

A 1 = w1 (0)o.12 (0) ze, by Schwarz's eit' and w2(0) so that 12411 = 1 only if w 1 (z) 2 Lemma. If n> 1 and lAn 1 = n, then (8.48) shows that = n, k=1 1C2k-11 and so H(z) =

z z f(z) = H(z) 2 = 1 — z2eie' (1— zei 8 ) 2

by Theorem 8.6. We absorb the constant e 0 into (.02 and so assume that

H(z) = S n (Z)

=

(1 — z 2)

E

Z" .

k=1

We next write 00

sn [0)2(z)] = > Bkzk k=0

and prove that k > n.

Bk = 0,

(8.52)

In fact if we replace sn[02(z)] in (8.49) by Un(Z)

=

E

B k zk

k=0

then the integral is unaffected. We deduce that

1=1 lan(z)1 2 1dzI =

lAn1

E 113n12. k=0

If (8.52) is false we deduce, letting r tend to 1 in (8.51) that IBnI 2

<

f

Isn (z)1 2 dz = n,

k=0

and this contradicts our assumption. It follows from (8.52) that w2(z) is algebraic and so remains continuous in 1. Also letting r tend to 1 in (8.49), (8.50) we deduce that n=

1 Isn (z)1 2 1c1z1 = n koz(z)Ilsnko(z)1121dzI 4,7 fL.,1=1
De Branges' Theorem

252 unless

Ico2(z)1= 1

on

Izl= 1,

(8.53)

so that (8.53) must hold. Thus w2(z) can be continued into the closed plane by reflection [C. A. pp. 172-3]. Either (02(z) = n2zP, where p is a positive integer and n2 is a constant, 1 112I = 1; or else w2(z) has in O < z < 1 at least one zero at z = zo of order q say. In the latter case 0) 2(z) has a pole of order of q at 1/ 2T, and

sn{(02(z)} = 1 + (02 + W +

+ (03-1

has a pole of order qn at 1/-4 and this contradicts (8.52). Thus 0) 2(z) = n2zP and we again obtain a contradiction from (8.52) unless p = 1. So w2(z) = n2z. Thus

F(z) = r120)1(z) (1

ri2z)2 .

We replace F(z) by F(z /n2) and so assume that ri2 = 1. We also assume now A„ = n, since otherwise we may consider e-14 F(z) instead of F(z). Then, since 0) 2(z) z we obtain from (8.49)

1 f w i (z)s„(z) 2 dz n — 2ni lzi=r Z n+1 Since I(01(z)I < 1, (01(z) has radial limits wi(e'° ) for almost all 0 on IzI = 1. (see e.g. Hayman [1989, Theorem 7.44, p. 515]). Thus, letting r tend to 1, we obtain n

1 2n

col

(ele) sn (ele ,) 2 e- '41d0.

Again by (8.49) and (8.50) with A n = n, we have

1 n<— 27r

10 2m

1Sn (e)

2

d0 = n

and 1 n=— 27E

f,

sn (e) 2 C ined0.

Since iw i (ei° ) < 1, we deduce that for almost all 0 wi

so that

(de) sn (ele‘) 2 e- in° = s,, (ei° ) 2 e -in0 >

253

Examples Using Cauchy's integral formula

1 f

(0 1 (z) =

w i (C)dC

2 "i ilcl=r

z

and letting r tend to 1, we deduce that w i (z) -_s-- 1 in Iz1 < 1. This completes the case of equality in Theorem 8.7.

8.5.1 Examples

We suppose that f(z)

E

and define k0(z) by (8.45).

Examples

8.1 8.2

Why doesn't Theorem 8.5 follow from Theorem 8.2, when À = ? Prove that if Iz1 < 1 and p is a positive integer we have

If (P) (z)1 8.3

r) (131 1(P - r)P+2

lzlr.

If A is an even integer prove that

I ;,1 (r, f) < I ;. (r,k o( P) ) (use the fact that if g(z) =

;

zn then 00

12(r, f) = 8.4

( 2

Prove that if p = 1, 2, 3, ...

If ( P) re) 2 rr dd0 <

k(oP) re)

(

8.5 •

(

2

rdrd4).

If À > 2, prove that

I

8.6

r2 n).

f)

I Ar,k0).

Baernstein [1975] has shown that this result remains true for all positive À and even more general means (see e.g. Hayman [1989, Theorem 9.8, p. 674]). With the hypotheses of Theorem 8.7 prove that 12 (r, F)

I2(r,1(0),

0 < r < 1.

254 8.7

De Branges' Theorem

If S).(r, f) = r "41 jr, f) prove that, for 2 > 2 SAr, f)

SA(r,ke).

(Consider f2rt

27r 0 8.8

1C/1 (rell 2 dO,

where 4)(z) = f(z)1 ;- and apply Theorems 3.1 and 8.5.) and dk (2), ak(2) are defined by (8.32), (8.43) respectively, If 2 prove that jak+1(11 )1 2

dk(22)

k=0

dN(2),

1).

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Index

This index contains page references to topics of major importance and to terminology: usually when the topic or symbol is first used, or is used in a new sense, or features in a major theorem. Bieberbach's conjecture, xi, 4, 230 Bloch's constant, 136 Bloch's Theorem, 136 Bloch functions, 143 bounded univalent functions, 78

A(p), A(p, 13) 66 i(p, fl) 71 A(p, 2), A(p, 13, 2), 73 A(p,k, N), 99 A(p,b,c), 100 A0, 38 A1, A2, 84 Ak, 243 A(r, f), 27 a2, 1, 162 a3, 215 a5, 161, 230 a4, a6, 230 an (2), 221, 247 Œ, 9, 45, 50, 150 ao, 45 an , 20 oc(0), 18 o(r), 64 a(0, 42 admissible domain, 109 analytic domain, 104 areally mean (a. m.) p--valent, 144 argument of f(z), 224 argument of flz), 226 asymptotic behaviour, 19, 60, 154, 155 asymptotic behaviour of coefficients, 15, 64, 151, 156 averaging assumptions, 19, 37, 144 A

C,C1,C2,... 172

Co, 61 cn (t), 216 CA, ck(0), ck(t), 237 C. A. x, 7, 103 circumferentially mean (c. m.) p-valent, 144, coefficients of univalent functions, 9, 15, 247 of mean p-valent functions, 65, 131. condenser, capacity of a condenser, 109 connectivity, 104 convex function, 70 convex univalent functions 11, 12 convex domain, 11 correspondence of points under a transformation, 200 D, D*, 115 Df, 128 Dk, 243 75, 104, D(00), 61 4(4 26, 243 cI1 , 200 A, 20, 33, 163 A', 20 An(e), 19 V2 u, 77, 107, 169 o 1 , 200 de Branges' Theorem, xi, 230 dense subclass, 197 diameter, 199 Dirichlet's minimum principle, 109 problem of Dirichlet, 108

B, 136 [B] for Burkill [1951], 31

B1, 200 208 b;(z), 26 )6, 55, 66, 200, 209

Po, 79

$(0), fi(t), 209 Ei, 143 3cl, 137

261

262 distortion theorems, 4, 28 domain, 104

Index length-area principle, 29 Lipschitzian, Lip, 104 LOwner's differential equation, 197

E2, 86 fk(z), 95, 161 f n (z), 20 f (z, t), 210 functions of maximal growth, 16, 17, 45 with k-fold symmetry, 95, 159, 185 without zeros, 145, 159 zero at the origin, 61, 64, 148, 158, 165 G(R), 77, 170 G(t), 207 G(0), 240 g(R), 77 gn (t), 239, 242 gt (), 215, 237 gt (z), 207 g(z, t), 210

7, 82 207 F(x), the gamma function, 26, 151 Gauss' formula, 105 Goodman's conjecture, xi, 163 Green's formula, 106, 169 Green's function, 122 H(R), 37 H+ (R), 39 h(R), 37 h(z,t',t"), 208 h(z,t), 237

I i (r, f), 9 I (r,f), 27, 67 IG(r, f ), 77

1(E), 84 /D(u), 109 1(z), 239 inner radius, 124 inverse function, 222 K(t), 210

Koebe function, 2 k-symmetric, 95, 185 l(R), 29, 199 lf, 147 A(t), 235

i(t) 207 Landau's Theorem, 143 Lebesgue integral, 31 Legendre's addition theorem, 232 Legendre polynomials, 231 associated Legendre functions, 231

M(r, f), 8, 33 p, 33, 172, 180 pq , 28

911, 141 major arc, 16, 19, 154 maximal growth, 45 mean p-valent, xi, 38, 64, 165 Milin's conjecture, 230, 236 minor arc, 16, 21, 153 modules, theory of, xi, 229 modulus of continuity, 116 modulus of doubly connected domain, 110

n(w), 29, 144 n(w, A, f), 144 n(r,w), 67 0,0*, 113 0(6), 117 w(R), 51 co(z), 109, 243 w*(z), 119 odd univalent functions, 161, 186, 248 omitted values, 3, 94, 150 order (of f (z) at 0, 42 P ( z), P4(z), 231 p(R), 29, 67, 144 p(r, R), 67 p(R, A, f), 29

p-valent, xi, 1, 28, 163 power series, 243 power series with gaps, 98, 99, 101 principal frequency, 103 Rf, 131 rn , 19 ro, ro*, 126

radius of convexity, 226 radius of greatest growth (r. g. g.), 17, 48 radius of starshapedness, 227 real coefficients, 13, 162, 220 regularity theorems, xi, 16, 49, 150 Riemann's mapping theorem 44, 204 Robertson's conjecture, 249 Rogosinski's conjecture, 250 Rouches Theorem, 147

S, 22, 36 So, 36 S(R), 22 Sf, 200 SG(r, f),

78

263

Index St, tn, 208 Sift), 69 ck, 242 a, 1 So, 6, 149 82, 197, 204 29 82 Schwarz's Lemma, 11 Schwarz's reflection principle, 44 Schwarz's inequality, 30 slit, 204 sectionally analytic slit, 198 starlike domain, 14 starlike univalent function, 14, 188 Stolz angle, 22 subordination, 250 symmetrization, 112 symmetrization, circular or Pôlya's, 113 symmetrization of condensers, 119 symmetrization of functions, 116 symmetrization, principle of, 127

Steiner symmetrization, 112 Szegb's conjecture, 95 z, 209 19(c), 30 2—point estimate, 176 torsional rigidity, 103 transfinite diameter, 103 transformation, 200 infinitesimal transformation, 203 typically real, 13 u*(z), 117

univalent, xi, 1

variational method, xi, 229 Vitali's convergence theorem, 21 W(R), 37, 76

1(R), 2(R), 51