Entire and Meromorphic Functions

Lee A. Rubelwith James E. Colliander

ENTIRE AND

MEROMORPHIC

FUNCTIONS

Springer

Universitext

Editors (North America): S. Axler, F.W. Gehring, and P.R. Halmos

Aksoy/Khamsi: Nonstandard Methods in Fixed Point TheoryAupetit: A Primer on Spectral TheoryBoossBleecker: Topology and AnalysisBorkar: Probability Theory; An Advanced CourseCarleson/Gamelin: Complex DynamicsCecil: Lie Sphere Geometry: With Applications to SubmanifoldsChae: Lebesgue Integration (2nd ed.)Charlap: Bieberbach Groups and Flat ManifoldsChern: Complex Manifolds Without Potential TheoryCohn: A Classical Invitation to Algebraic Numbers and Class FieldsCurtis: Abstract Linear AlgebraCurtis: Matrix GroupsDiBenedetto: Degenerate Parabolic EquationsDimca: Singularities and Topology of HypersurfacesEdwards: A Formal Background to Mathematics I albEdwards: A Formal Background to Mathematics II albFoulds: Graph Theory ApplicationsGardiner: A First Course in Group TheoryGiirding/Tambour: Algebra for Computer ScienceGoldblatt: Orthogonality and Spacetime GeometryHahn: Quadratic Algebras, Clifford Algebras, and Arithmetic Witt GroupsHolmgren: A First Course in Discrete Dynamical SystemsHowe/Tan: Non-Abelian Harmonic Analysis: Applications of SL(2, R)Howes: Modem Analysis and TopologyHumi/Mlller: Second Course in Ordinary Differential EquationsHurwitz/Kritikos: Lectures on Number TheoryJennings: Modem Geometry with ApplicationsJones/Morris/Pearson: Abstract Algebra and Famous ImpossibilitiesKannan/Krueger: Advanced Real AnalysisKelly/Matthews: The Non-Euclidean Hyperbolic PlaneKostrikin: Introduction to AlgebraLuecking/Rubel: Complex Analysis: A Functional Analysis ApproachMacLane/MoerdJjk: Sheaves in Geometry and LogicMarcus: Number FieldsMcCarthy: Introduction to Arithmetical FunctionsMeyer: Essential Mathematics for Applied FieldsMines/Richman/Rultenburg: A Course in Constructive AlgebraMoise: Introductory Problems Course in Analysis and TopologyMorris: Introduction to Game TheoryPorter/Woods: Extensions and Absolutes of Hausdorff SpacesRamsayfRlichtmyer: Introduction to Hyperbolic GeometryRelsel: Elementary Theory of Metric SpacesRickart: Natural Function AlgebrasRotman: Galois TheoryRubelColliander: Entire and Meromorphic Functions

(continued after index)

Lee A. Rubel

With assistance from James E. Colliander

Entire and MeromorphicFunctions

Springer

Lee A. Rubel James E. CollianderDepartment of Mathematics Department of MathematicsUniversity of Illinois, Urbana-Champaign University of Illinois, Urbana-ChampaignUrbana,IL 61801-2917 Urbana, IL 61801-2917USA USA(deceased)

Editorial Board

S. Axler F.W. Gehring P.R. HalmosDepartment of Mathematics Department of Mathematics Department of MathematicsMichigan State University University of Michigan Santa Clara UniversityEast Lansing, MI 48824 Ann Arbor, MI 48109 Santa Clara, CA 95053USA USA USA

Mathematics Subject Classification (1991): 30Dxx, 30D35

Library of Congress Cataloging-in-Publication DataRubel, Lee A.

Entire and meromorphic functions / Lee A. Rubel with assistancefrom James E. Colliander.

p. cm. - (Universitext)Includes bibliographical references and index.ISBN 0-387-94510-5 (softcover : alk. paper)1. Functions, Entire. 2. Functions, Meromorphic. 3. Nevanlinna

theory. 1. Colliander, James E. II. Title.QA353.E5R83 1995

515'.98-dc20 95-44887

Printed on acid-free paper.

With 2 illustrations.

®1996 Springer-Verlag New York, Inc.All rights reserved. This work may not be translated or copied in whole or in part without the writtenpermission of the publisher (Springer-Verlag New York, Inc., 175 Fifth Avenue, New York, NY 10010,USA), except for brief excerpts in connection with reviews or scholarly analysis. Use in connectionwith any form of information storage and retrieval, electronic adaptation, computer software, or by sim-ilar or dissimilar methodology now known or hereafter developed is forbidden.The use of general descriptive names, trade names, trademarks, etc., in this publication, even if the for-mer arc not especially identified, is not to be taken as a sign that such names, as understood by the TradeMarks and Merchandise Marks Act, may accordingly be used freely by anyone.

Production managed by Laura Carlson; manufacturing supervised by Jacqui Ashri.Camera-ready copy prepared using the authors' AMS-TeX files.Printed and bound by R.R. Donnelley & Sons. Harrisonburg, VAPrinted in the United States of America.

987654321ISBN 0-3 87-945 1 0-5 Springer-Verlag New York Berlin Heidelberg SPIN10424824

Dedicated to the Memory of Steven B. BankStudent, Colleague, Teacher, Friend

Contents

1. Introduction ................................... 1

2. The Riemann-Stieltjes Integral ..................... 3

3. Jensen's Theorem and Applications ................. 6

4. The First Fundamental Theorem of Nevanlinna Theory . 9

5. Elementary Properties of T(r, f) .................... 12

6. The Cartan Formulation of the Characteristic ......... 16

7. The Poisson-Jensen Formula ...................... 20

8. Applications of T(r) ............................. 23

9. A Lemma of Borel and Some Applications ........... 26

10. The Maximum Term of an Entire Function ........... 30

11. Relation Between the Growth of an Entire Functionand the Size of Its Taylor Coefficients ............... 40

12. Carleman's Theorem............................. 45

13. A Fourier Series Method ......................... 49

14. The Miles-Rubel-Taylor Theorem on QuotientRepresentations of Meromorphic Functions .......... 78

15. Canonical Products .............................. 87

Viii Entire and Meromorphic Functions

16. Formal Power Series............................. 93

17. Picard's Theorem and the Second FundamentalTheorem ...................................... 99

18. A Proof of the Second Fundamental Theorem ......... 113

19. "Two Constant" Theorems and the Phragm6n-LindelofTheorems ..................................... 121

20. The Pblya Representation Theorem ................. 124

21. Integer-Valued Entire Functions .................... 139

22. On Small Entire Functions of Exponential-Type withGiven Zeros ................................... 146

23. The First-Order Theory of the Ring of All EntireFunctions ..................................... 158

24. Identities of Exponential Functions ................. 175

References......................................... 182

Index ............................................. 185

1

Introduction

Mathematics is a beautiful subject, and entire functions is its most beautifulbranch. Every aspect of mathematics enters into it, from analysis, algebra,and geometry all the way to differential equations and logic.

For example, my favorite theorem in all of mathematics is a theoremof R. Nevanlinna that two functions, meromorphic in the whole complexplane, that share five values must be identical. For real functions, there isnothing that even remotely corresponds to this.

This book is an introduction to the theory of entire and meromorphicfunctions, with a heavy emphasis on Nevanlinna theory, otherwise known asvalue-distribution theory. Things included here that occur in no other book(that we are aware of) are the Fourier series method for entire and mero-morphic functions, a study of integer valued entire functions, the Malliavin-Rubel extension of Carlson's Theorem (the "sampling theorem"), and thefirst-order theory of the ring of all entire functions, and a final chapter onTarski's "High School Algebra Problem," a topic from mathematical logicthat connects with entire functions.

This book grew out of a set of classroom notes for a course given at theUniversity of Illinois in 1963, but they have been much changed, corrected,expanded, and updated, partially for a similar course at the same place in1993. My thanks to the many students who prepared notes and have givencorrections and comments.

In order to discover and prove interesting, deep, or powerful theoremsin this area, what we most need is more examples of interesting meromor-phic functions-I would guess that the number of fundamentally differentexamples known is about 20 or 30. One promising source of such examplesis the Painleve transcendents (see [14], pp. 438-444, and [29]).

However, in spite of a growing literature on these functions, the unfor-

2 Entire and Meromorphic Functions

tunate fact is that the "proofs" are incomplete and not rigorous-indeed,there still is not a satisfactory proof that the Painleve transcendents ofeven the first kind (i.e., solutions of w" = 6w2 + z) are meromorphic inthe full complex plane. Basic notions like "fixed singularity" and "movablesingularity," however intuitively appealing, have never been given rigorousdefinitions.

It is hard to see how this lamentable situation will improve since, theworld being as it is, there is little "glory" attached to proving theoremsthat have already been "proved."

The subject of entire and meromorphic functions has been growing formany decades, and will continue to grow forever. It is hoped that this bookwill give the novice reader a good introduction to the subject, or the expertsome new insights. This book could "easily" have been four or five timesit length, since the subject is so extensive, but to use my favorite saying,"enough is too much."

LEE A. RUBEL

Lee Rubel died on March 25, 1995. As my teacher, the way his per-sonality merged into his mathematics always inspired me. I sincerely hopethat readers of this book find similar inspiration.

JAMES E. COLLIANDER

October 18, 1995

2The Riemann-Stieltjes Integral

We give here a brief summary of some of the basic facts about the RiemannStieltjes integral. Those unfamiliar with the subject are urged to readChapter 9 of Mathematical Analysis by Apostol [1].

Throughout this section, a and b are real numbers, usually a < b, and fand a are real-valued functions defined on the closed interval [a, b]. Whenf and a are suitably restricted, we will define f b f da as the Riemaun-Stieltjes integral of f with respect to a. When a(x) = x for all x in [a, b],fe da is the ordinary Riemann integral ofa f f, and many of the familiarproperties of the Riemann integral extend to the Riemann-Stieltjes integral.

Definition. A partition of [a, b] is an ordered (n + 1)-tuple

P = {z0,z1,...,xn} with x0 = a,

xn = b, and x j_ 1 < x j for j = 1,... , n.

Definition. A selection or from a partition P is an ordered n-tuple a ={t1i ... , tn} such that

xj_1 <tj <xj for j=1,...,n.

Definition. A Riemann-Stieltjes sum (of f with respect to a) is a sum ofthe form

S(P, or : J, a') = > f (tk){a(xk) - a(xk_01.k=1


Definition. P' is a refinement of P; written P C P', means that each xthat occurs in P also occurs in P'.

Definition. f is integrable with respect to a; written f E R(a), means thatthere exists a number A such that, for each c > 0, there exists a partitionP, such that P. C P; and if a is a selection from P, then

JS(P,a: f,a) - Al < e.

It is easily seen that A is uniquely determined if f E R(a), so that wemay write

Jb

fda = / b f (x) da(x) = A.a a

Theorem. If f is continuous on [a, b] and a is monotone on [a, b], thenf E R(a) and a E R(f).

Theorem (Integration by parts). If f E R(a), then a E R(f) and

frb rb

/ f da = f(b)a(b) - f(a)a(a) - J a df.a a

Under suitable hypotheses, the usual linearity properties and formulasfor "change of variables" hold, e.g.,

J(Afr

+B g)da=A J fda+B r gda

J b b bfd(Aa + BO) = A. f fda + Bf fdf3a a n

6

fa '

-

fa

+

The

fJ b f (x)da(x) =/ b f (a)a'(x)dxn fda

b jd=

h=fog, (3=aog,a = g(c), b = g(d),

where (f o g)(x) = f (g(x))-

2. The Riemann-Stieltjes Integral 5

Theorem. If a is a step function that jumps ak at points xk, k = 1,2, ... , n in [a, b] and f is continuous on [a, b], then

J

b

f(x)da(x) = E f(xk)aka k=1

Remark. Given a finite sum E akbk, we may write

E akbk =

where B is a step function that jumps bk at xk and a is a continuousfunction such that a(xk) = ak. Integration by parts now becomes "partialsummation" and we have the formula

aobo + albs + ... aNbN = Ao(bo - b1) + ... + AN-1(bN-1 - bN) + ANbN,

where Ak

First Mean Value Theorem. If a is nondecreasing on [a, b], f E R(a),and

M = sup{f (x) : x E [a, b]}

m = inf{ if (x) : x E [a, b]},

then there is a number c, m < c < M, such that

Jb f (x)da(x) = c f b da(x) = c[a(b) - a(a)].

a n

If f is continuous, then c = f(xo) for some xo E [a, b]. In particular,

m J b da(x) < I b f (x)da(x) < MJ

b da(x).a a a

Second Mean Value Theorem. Suppose that a is continuous and f isnondecreasing on [a, b]. Then there is an xo E [a, b] such that

6

f

b rxo

1.0

f(x)da(x) = f(a) / da(x) + f(b) da(x)./a

Corollary (Bonnet's theorem). If g is continuous and f is nonnegativeand nondecreasing on [a, b], then

J

b

f(x)9(x)dx = f(b)1.0

9(x)dxa b

for a suitable choice of xo E [a, b].

3Jensen's Theorem and Applications

One of our most useful tools is Jensen's Theorem, which can be used torelate the distribution of zeros of an entire function to its growth. We proveJensen's Theorem using the Gauss Mean Value Theorem.

Gauss Mean Value Theorem. Suppose u is a harmonic function in D.Then the value of u at the center is equal to the average of the boundaryvalues of u. That is,

u(e1t)dtU(O)

= 2w jProof. Form the analytic function f (z) whose real part is u(x, y). ApplyCauchy's integral formula to evaluate f at zero, then take the real part andthe theorem is proved.

Jensen's Theorem. If f is meromorphic in IzI//

< R, if r < R, and if

f(z) = akzk + ak+izk+1 +--- (ak O 0)is the Laurent expansion of f around zero, then

rx(3.2)

21rI log f (rei8) I dO =1og Iakl+ log n - log

T+k log r,

r <r P <r Pn

where the zeros of f are z5 = r3ei8' and the poles off are wi = pies#J, notcounting zeros or poles at the origin.

Proof. With no loss in generality, assume f (0) = 1 and R = 1:

T7rz-wn(3.3) F(z) =f(z)

11z - znFl ( n

n 1-2nZ

3. Jensen's Theorem and Applications 7

F is an analytic function with no zeros and no poles. Since log IFS is aharmonic function in the disk, the Gauss Mean Value (3.1) implies

(3.4)Zx J log I F(reie) I dO = log If (0)1 + E log pn - E log r,,,.

But 2x f% log I F(re'B)I dO = 2- f" r log If (re'B)I since I 1 _oz

I = 1 on

IzI =1ifIzoI<1.

Let n(r, f) be the number of poles of f in the closed disk I zI < r, counted

according to multiplicity. Thus n (r, f 'a) counts the number of a-points

[an a-point is a point z satisfying f(z) = a] of f. Let n(r) = it (r, j),which counts the zeros of f. Let k+ = max(k, 0) and k- min(k, 0), sothat k+ - k- = k. Then,

(3.5) E log! + k+ log r =fr

log t d{n(t) - n(0)} + k+ log rr,. <r +

with u = log t, du = - t and v = n(t) - n(0). An integration by partsgivest

n(O) dt + k+ log r[n(t) - n(0)] log t o+ +r+ n(t)

tIr n(t) - n(0)

dt + k+ log r.+ t

We define:

N(r,1I=_k+logr+Jr0+

jrN(r,f)-k- log r +

+

n(t,3) -n(0, f)dt = log

rr + k+ log r

t rn<r rn

n (t, f)-n(0,t) rt

dt = log + k-log r.PnPn<r

Remarks. n(r, f) counts the number of poles of f in the disk Izl < r.N(r, f) is a useful average of the counting function. We usually normalizef so that f (O) = 1, and in this case k+ = k- = n(0, f) = it 0,.!)(log IakI=0.

As a typical application of Jensen's Theorem, we prove the next result.

tThe 0+ indicates that we integrate from, say, a to r, where e is smaller than thesmallest positive modulus of a zero.


Theorem. Given z1, z2.... with 0 < Iz1I = rj < 1, there exists a boundedholomorphic function f in the unit disk whose zeros are precisely the zj ifand only if

E(1 - rj) < oo.

Proof If such an f exists, we may suppose f (0) = 1 and apply Jensen'sTheorem:

x

N (r, Flog * = i! f x log If (reie) I do.rj<r

Since f is bounded, N (r, 1) is bounded. N (r, j) is nondecreasing,

so s log ( ) < oo. But (exercise) s log (-) < oo if and only iflog(1-r5)<oo.In the other direction, suppose E log (s) < oo and let

Pn(z)=x-z91-zjz

The P,, form a normal family since I P,, (z) I < 1 for all n and all z ED. Passing to a subsequence if necessary, the Pn converge uniformly oncompact subsets of D to a bounded holomorphic function f. Can f beidentically zero? Since IP,,(0)l = I H (-zi)I = rI r we have If (0),rji° r1. Hence If (0) 1 > 0 since log rj' (J) _ E 0° log < 00.

4The First Fundamental Theorem ofNevanlinna Theory

Rewriting Jensen's Theorem, we get

(4.1) 27rf log I f (re'B)I dO = log IakI + N (r,

f) - N(r, f),

where N is a kind of average number of poles of f.For positive numbers x, let us write

log+ x = max(O, log x) = log[max(1, x)]

so thatlog x =log+ x - 1og+

1

x

We now list some simple properties of log+:(a) log+(x1 x2 ... - xn) < 1og+ x1 + log+ x2 + ... + log+ xn.(b) log+(xl + x2 + + xn) < 1og+ XI + 1og+ x2 + + 1og+ xn + log n.In particular,(c) log+(x1 + x2) < 1og+ xl + 1og+ x2 + log 2.From (c), we get

log+lx - al < log+Ixl + log+Ial + log 2

log+lxl < log+Ix - al + log+Ial + log 2,

8o that(d) Ilog+lx - al - log+IxI I < log+Ial + log 2.


We may write

log If (Te`')I d9 log+If(reie)I dB"

r"(4.2) - 2 J log

I f (reie)dB.

Let m(r, f) = i" f ""log+I f (re") I dB. Then we may rewrite Jensen's The-orem as

m(r,f)-m(r'f) loglakI+N(r'f) - N(r, f)

or

(4.3) m(r,f)+N(r,f)=1ogIakI+m(r'f)+N\r'fllI.

Notice that N(r, f) counts the poles of f (with a certain kind of aver-aging) that is the averaged number of times f takes the value oo, whilem(r, f) measures the tendency of f to take the value oo. Hence, the quan-tity m(r, f) + N(r, f) measures, in some sense, the total affinity of f forthe value oo. Similarly, m (r, -1) + N (r, t) measures the total affinityof f for the value zero. So the above version of Jensen's Theorem assertsthat the total affinity of f for oc is the same as the total affinity of f forthe value zero, modulo a bounded function of r. The first fundamentaltheorem is based on the observation that, for any constant a, the affinityof f - a for oo is essentially the same as that for f, while the affinity off - a for zero is, of course, the affinity of f for a. The theorem states thatm (r, 1 Q) + N (r, 1) is independent of a, modulo a bounded function

of r. Here we use the convention that if a = oo, then f 1. means f.Fix a E C. Then N(r, f) = N(r, f - a) since z is a pole off if and only

if z is a pole of f - a. From property (d) of log+ we obtain

I m(r, f - a) - m(r, f )l < log+Ial + log 2.

We defineT(r, f) = m(r, f) + N(r, f ).

T is called the (Nevanlinna) characteristic of f.From Jensen's Theorem, we have

T (r, f) = T (rl f) +.O(r)

T(r, f - a) = 7' r, + c(r; a),f -a)where O(r) = ¢(r; 0) = log IakI and O(r; a) = log Iak(a)I. Here ak(a) isthe first nonvanishing coefficient in the Laurent expansion of f - a at theorigin. For each a E C, S(r; a) is a bounded function of r.

4. The First Fundamental Theorem of Nevanlinna Theory 11

First Fundamental Theorem of Nevanlinna Theory. If f is mero-morphic in IzI < R, where 0 < R < oo, then

(4.4) T (r, f 1 a) = T (r, f) + 4(r; a)

where 1¢(r;a)l < log+'jal + Ilog+Iak(z)lI + log2 for all r with 0 < r < R.

Proof. This is simply a rephrasing of Jensen's Theorem using the new no-tation we have introduced.

Since it is customary to work modulo bounded functions of r, we maysometimes abuse the notation and write things like T(r, f) = T(r, f - a)when we mean only T(r, f) = T(r, f - a) + 0(1). The characteristic playsa central role in the theory of meromorphic (and entire) functions.

5Elementary Properties of T (r, f)

In this chapter, we present basic properties of the characteristic function.

(5.1) T(r, hIh2) < T(r, hl) +T(r, h2)(5.2) T(r,hi+h2) <T(r,hl)+T(r,h2)+log2

(5.3) T(r, h2) T(r, hl) + T(r, h2) + 0(1)

(5.4) T(r,±f')=nT(r,f); n,=1,2,....

The proofs of (5.1), (5.2), and (5.4) are simple consequences of the prop-erties of log+ and of the fact that N(r, f) "counts poles," while (5.3) followsfrom (5.1) and the first fundamental theorem, which implies that

T (rhJh1) = nT(r, h)+

E T(r, hj) + n log 2.0 0

The proof of (5.5) is by induction based on the identity:

n n> hjhj = ho + h> hjh'-1

0 1

to which (5.1) and (5.2) are applied in the obvious way.

5. Elementary Properties of T(r, f) 13

Definition. To each function A (r) that is positive, continuous, and non-decreasing on 0 < r < oo, we associate the class A of functions f that aremeromorphic in Izi < oo and that satisfy

T(r, f) < AA(Br)

for positive constants A and B as r --+ R. If f is entire and satisfies thiscondition, we say that f is an entire function of finite A-type.

It is easy to verify that A is a field, and we call any such field a A-field.

Remarks. Some A-fields have been studied heavily. The case A(r) =max(l,rP) is especially important and we denote the corresponding A-fieldby A,,. In case f E A,,, we say that f is of order at most p and of expo-nential type. In case p = 1, we say simply that f is of exponential type.The intersection, SIP = APB, consists of all functions of order at mostp. We shall discuss order and type in much greater detail later on. Noticethat the next theorem implies that the fields of all meromorphic functions(i) of order at most p exponential type and (ii) of order at most p arealgebraically closed in the field of all meromorphic functions on C.

Theorem. Each A -field is algebraically closed in the field of all meromor-phic functions on C.

By this we mean the following. Suppose that f, fo, fl,..., fn are mero-morphic in C, that fn is not identically zero, that fi E A for j = 0, ... , n,and that

(5.6) fo + f1f + f2f2 +. _ _ + fnfn = 0.

Then the theorem asserts that f E A. Notice that we do not prove theexistence of an f that satisfies (5.6).

Proof. 1 om (5.6) we may write

T(r,fn)=T (r_1fifi)..i=0

Then by (5.3), (5.4), and (5.5) we get

n

T(r, f) < ET(r, fi)+O(1) as r--> oci-o

and thus T(r, f) < AA(Br) of appropriate constants A and B.

Definition. Let B denote the ring of all of those functions that belong toA and are holomorphic in C.


Theorem. Each P-ring is algebraically closed in the ring of all functionsholomorphic in DR.

This follows directly from the preceding theorm.We give a detailed proof of the theorem only outlined in [12, p. 54],

reversing the notation for f and g.

Clunie's Theorem. Let f (z) be a transcendental entire function, let g(z)be a nonconstant entire function, and let <p(z) = f (g(z)). Then

7'(r, 'P) 00T(r, g)

_,

asr -+ o0.

Proof. We may and do assume that f (w) has infinitely many distinct zerosat w1i w2, --+ 00- (Otherwise, we could replace f by f - A for a suitableconstant A. We are implicity using the fact that if f is an entire functionthat takes each complex number a as a value only finitely many times, thenf must be a polynomial. Take this as an exercise. [Hint: Hurwitz's The-orem, the Casorati-Weierstrass Theorem, and Liouville's Theorem.] Thisfact is also a consequence of several of the later results in this book, likePicard's theorem.) Then, for any integer P,

PN r, 1 > N r, 1

'P V-1 g(w) - W.

because the averaged counting function N is a monotone increasing func-tion of the pole set. We also want

P /(5.8) im.(r,!) >Emf r, 1 ) -0(1).

v=1 \ g(w) - W.

Fix P and let

(5.9)

Write

(5.10)

0<6< 10 min{Iwi-w1I:iAj,

f(w) = (w - WO" ... (w - wp)mP (w),

where is nonzero at each wi, and where we also choose b so small that(w)96 Ofor 0<Iw-w;I<bfor all i-1,2,...,P. Sayl4(w)I>e>0

for all w within 5 of wi, i = 1, 2, ... , P.Define

n(5.11) E = U{z : Ig(z) - wig < 5}.

5. Elementary Properties of T(r, f)

For z E E, we have

P(5.12) log+ 1 > E log+ I 1 I- M

If (9(x))I _, g(x) - w

for a suitable constant M depending on P, 6, and e. But

1 " 1

27r _x log I g(rei8) - w;d9

is asymptotic as r -i oo, to

2a f log+ eie) I dB,E 9(

whereE; = 10: I9(reie) - w;I < S}

because the integral over the remaining part is less than log+We conclude, using (5.12), that

T (r,!) > PT(r,g)+0(1)

15

for any integer P, and the result follows.

6The Cartan Formulation of theCharacteristic

We begin with some remarks on convex functions.

Lemma. If /3 is a nondecreasing function, then B(t) = f6 /3(s) ds is aconvex function of t.

Corollary. If a is a nondecreasing function, then A(r) = for a(x) is aconvex function of log r, that is, A(et) is a convex function of t.

The corollary follows directly from the lemma since

rA(et) = J e, a(x) dx = 1 t /3(s) ds, where /3(s) = a(e°).

o x o0

Of course, we assume that a is small enough near 0 so that the integralexists: In most of our applications, we will have a(x) = 0 for 0 < x < xofor some x0 > 0. Formally, a condition that B is convex is that B'(t) shallbe nondecreasing, and here B'(t) =#(t). Similarly, rA'(r) = a(r).

Proof. After a simple normalization, it is seen that we must prove that

1fox ,0(s) ds < x1 ,D(s) ds for 0 < x < 1.o

This will be the case if I fo /3(s) ds is nondecreasing. Butz

fo /3(s) ds =fo l3(xt) dt, which is obviously nondecreasing since 0 is. Suppose y > x.Then we see

ffoI[fl(Yt)

0/3(yt) dt - Jl O(xt) dt = - /3(xt)] dt > 0.

0 p

6. The Cartan Formulation of the Characteristic 17

Roughly speaking, B is a convex function if and only if B can be repre-sented as in the lemma. Similarly, A is logarithmically convex if and onlyif A can be represented as in the corollary.

We now reformulate the first fundamental theorem following a proceduredue to Cartan.

Theorem. For a certain constant C,

X

T(r, f) = C + 2 N r, fleap

dccx

= C +l r {

2-.TTh l t, f le"W

dW dt

In particular, T(r, f) is a nondecreasing convex function of log r.

Proof. Apply Jensen's Theorem to the function f - e" P for some real con-stant Sp:(6.1)

x

2a logf(e:e) - e"P dO = log Iak((P)I + N r, f 1 4, J - N(r, f).

1-"r

Thinking of the Laurent series for f - e'w and the definitions of N andak(,p), it is obvious that

(i)Ifk<0,ak(cp)=ak and k(W) = k for all WER.(ii)If k=0,ak(cc)=ak-e'and k(W)=k=0 f o r all W E(iii) If k > 0, ak(<p) = -e"v and k(cp) = 0. Therefore, since k+(V) = 0

foripE]R,

dcp = 0.

With at most one exception [namely if f (0) = e=' for some cP E R] we have

n (0,f e"°

) = 0.

Now, by Jensen's Theorem for any constant b, we get

x(6.2) I log Ib - e"°I d7r =log+ dbl.

x

(Check the cases where IbI > 1 and IbI < 1.)Let us integrate (6.1) with respect to cp:

(6.3)

2wf* log jf(reie) - e'IPjd8 } dip

_ , " f., (rf_C1)dV+ j.- N d,-N(r,f)


Applying Fubini's Theorem to the left-hand side (LHS) and using (6.2)yields

1 " 1 "2zr, logIf(re'e)-e"°I d8 dip=

1

-- f m(r,f) dO=m(r,f)

Hence we obtain the result:(6.4)

Im(r, f) + N(r, f) =27r

,I log f ak(co)I dW +27r

/ N l r, f ess)x x

The first integral is a constant and we are done.

Remark. We could define T°(r) = 2 f' N (r, T -tan

dcp as the Car-tan characteristic of f, but by a minor abuse of notation we shall writeT(r) = - f "xN (r, f,,,) d<p since we customarily work modulo bounded2r -functions of r anyway.

Interpretation. Roughly, we have

dTr = L(r),

where r \L(r)

?

= -- / n r,1

d<p.JJJ ,r f -

eI;;

What does the function L measure? The function f (rese) may be con-sidered as a mapping of the circumference OD, = {z : IzI = r} into theRiemann sphere. n (r, f counts the number of times the point e""

is covered by this map. Hence, f ,,n (r, fir) dcp measures the totalarc length of the unit circumference (counting multiplicity) covered by themapping f. In other words, the more heavily the mapping f covers theunit circumference, the faster T grows. Thus, T measures the coveringproperties of f.

We outline here another characteristic, the Ahlfors-Shimizu character-istic TA (r, f), which behaves much like the Nevanlinna characteristic butwhich has as an enlightening geometric interpretation. For full details, see,from which our presentation is abstracted.

We defineTA(r, a) = MA(r, a) + N(r, a),

where

N(r, a) =fr nadt,

t

6. The Cartan Formulation of the Characteristic 19

as before, but

where

1za 1

MA (r, a)27r fo

log [w, a] dB,

1[w, a] I - a I

and [x, a] =1+

I

a x 1+IwIz

1+Iaz

Now [w, a] is the distance on the Riemann sphere between the points onthe sphere to which w and a correspond via stereographic projection.

It is easy to see that

IT(r,oo) -TA(r,oc) -log+If(a)II < 2 log 2.

By Green's Theorem, one can show thatj2ir

X

j21rlog 1 + If(re)I2 dO+n(r, f) _

I f ;e)p dp(1+ If(Peie)Ix]z

Denote the right side by A(r), divide by r, and integrate the resultingidentity from 0 to r to get

I rA(9 dt = N(r, f) + 2-j log 1 + if (reie)I z dO -log 1 + If (0) 12.

If we make a rotation of the Riemann sphere, which corresponds to thetransformation

1 + aww = ,w-a

where w = f (z), and call the resulting function to = F(z), we can derivethe first fundamental theorem for the Ahlfors-Shimizu characteristic.

Theorem. If f is meromorphic in IzI < R, where 0 < R < oo, then forevery finite or infinite a and r with 0 < r < R we have,

TA (r, 00) =f'.

dt = N(r, a) + MA(r, a) - MA(0, a).

For the geometrical interpretation of TA, note that if S is the Riemannsphere (of diameter 1), da is the element of area in the z-plane near thepoint z, and dA is the corresponding element of area on S, then

dA =do

(1 + Iz12)2

Hence irA(r) is exactly the area (counting multiplicity) of the image on the.R.iemann sphere of { IzI < r} by w = f (z). The rotation described aboveleaves A(r) invariant and replaces MA(r, oo), N(r, oo) by MA(r, a), N(r, a).We see that TA(r, oo) can be interpreted as an average of the spherical areaof the image of disks under the mapping w = f (z).

7The Poisson-Jensen Formula

The material we present in this chapter is a specialization of some generalresults of potential theory. Our presentation is in the context of analyticfunction theory.

We consider functions f holomorphic in DR = {z E C : IzI < R}. Wedenote u = Ref, choose r < R, and write z = re'B, w = Re''P.

The Poisson Formula.- fir

( 7.1) u(re'B) Pu =1

27ru(Re"')

R2 - r2dip'

RZ - 2rRcos(B - gyp) r2

The Poisson Kernel.

(7.2)

P = P(z, w) = P(reie, Re"p) =R2 - r

R2 - 2rRcos(B - gyp) + r2

_ JwI2 - Iz12 = Rew + zw - zJ2 w - z

Proof. Without loss of generality, assume R = 1:

1 2 __ 1- Iz12 IzI2

w-z+ l-zw (w-z)(1-zw) wJw-z12

By the Cauchy integral formula,

1 f f(w) [ 1 + xw] dw = f(z)

2ai w -i w - x 1 -

7. The Poisson-Jensen Formula 21

After parametrizing the integral with respect to the angle ep and takingreal parts, we get the Poisson formula.

Remark. For z = 0, Poisson's formula reduces to the Gauss Mean ValueTheorem, fx

u(0)2a

u(e'w) dcp.

The Poisson formula is the "invariant form" of the Gauss Mean ValueTheorem in the following sense. Choose Z E D and define

T,; : D -+ D by Tzw =w + zj+-,w

Let

for w ED.

F=foTzf U=uoTZ.If = i Yw then w = 1 w and

dw 1- zw dAw IA-z12 A

so that

u(z) = U(O) =1 U(w) dw = 1 r u(A)1 - Iz12 dA

tai w 2a: J IA - z12 A '

which is the Poisson formula in different notation.This generalization of the Gauss Mean Value Theorem has a natural

application which generalizes Jensen's Theorem.

The Poisson-Jensen Formula.Suppose that f is meromorphic in the disk DR = {z E C : IzI < R}, r < R.Then,(7.3)

x R2 - r2log l f (re'B) I °2a log If (ReiB) I R2 - 2rR cos(e - gyp) + r2

d`p

+ E log I BR(z : z1,)I - log I BR(z : w1,)i - k log RIz.I <R <R

where B is the Blaschke factor defined by

BR(z a)zR2- az

and the z are the zeros of f, the w are the poles of f, and k is the orderof the zero or pole at the origin.


Corollary. If f is holomorphic, then

log If(reie)I < J Plog IfI

In other words, if f is holomorphic, then log If I is dominated by itsPoisson integral. Notice that for z = 0 the Poisson-Jensen formula reducesto Jensen's Theorem. One way to prove the Poisson-Jensen formula is toshow that it is the invariant form of Jensen's Theorem. We choose anotherproof.

Proof of the Poisson-Jensen Formula. If f is holomorphic and has no zeros,then there is a branch F of log f and log If I = ReF so that the formulafollows as a special case of the Poisson formula. Also, if A denotes the left-hand side and p the right-hand side, notice that A(fg) = \(f) +.\(g) andp(fg) = p(f) + p(g). So it is enough to prove the formula for holomorphicf. Now consider g(z) = f(z)/IIBR(z : supposing that f(0) 36 0. SinceIBR(w : z.) I = 1 for IwI = R, the formula follows on applying the Poissonformula to g. And if f (O) = 0, consideration of f (z)/zk leads to the generalcase.

8Applications of T(r)

Theorem. If f is holomorphic and M(r) = sup[{If (z)[ : IzI < r}, thenfor anyR>r

T(r) < log+ M(r) < R + TT(R).

Proof. Since f is holomorphic, T(r) = m(r) and

xm(r) = Z f log+ If (Te`8)I d8 < log+ M(r).

Also, by the corollary to the Poisson-Jensen formula,

log If (reie) <l f P log if I.2 .I=R

It is easy to verify that

so that

0<P< R+rR-r

and the theorem is proved.

We now can prove the following extension of the Liouville Theorem,Which is left as an exercise.

Exercise. Suppose that f is meromorphic in IzI < oc and that T(r) isbounded. Then f is a constant.


Definition. Suppose that A(r) is positive, continuous, and nondecreasingfor r > 1 and furthermore is slowly increasing in the sense that a rT is

bounded. Let A' be the class of all entire functions f such that

log' M(r) = O(a(r)).

It is easy to see that A' is a ring. One of our exercises is to show thatf E A* if and only if f E t, where I consists of those entire functions f forwhich T(r, f) = O(a(r)). In the case where A(r) = max(1, rp), we see thatthe notions of f being of finite order, of order at most p, and of order atmost p exponential type are the same whether defined by the characteristicor the logarithm of the maximum modulus. It also follows that each A*ring is algebraically closed in the ring of all entire functions.

We say that a meromorphic function f in the unit disc is of boundedcharacteristic to mean that T(r, f) is bounded. Next we characterize thefunctions of a bounded characteristic in the unit disk D.

Theorem. A function f meromorphic in D is of bounded characteristic ifand only if there exist bounded functions A and B, holomorphic in D, suchthat f = A/B.

Proof. It is easy to see that if f = A/B, then f is of bounded characteristic.In the other direction, suppose that f is of bounded characteristic and,

without loss of generality, suppose f (0) = 1. Since T(r, f) is bounded, itfollows that N(r, f) and N(r,) are bounded, so that

log 1 < oo and s log 1 < 00.rn pn

Here, as before, {rnes°" } and { pneiw^ } are the zeros and poles of f.By the theorem of Chapter 4, there exist bounded functions cp and

holomorphic in D, such that the zeros of cp are the zeros of f and the zerosof ib are the poles of f. Thus, g = is of bounded characteristic and hasno zeros or poles.

It is enough to show that g has a representation g = A/B. Note thateven though g is holomorphic with no zeros and T(r,g) = 0(1), it does notfollow that g is bounded; witness g(z) = 11 Z'

Proceeding with the proof, there exists a function h, holomorphic in D,such that g = eh. Writing h = u + iv, we have

1 J A lu(re`9)1 dO < M < oo for r < 1,27r

since m(r, g) and m(r, 9) are bounded, and IgI = exp u. Now for R < 1,writing

hR(z) = h(Rz) = UR(z) + ivR(Z),

8. Applications of T(r)

we have1

( 'IuR(reie)I dO < M for R < 1.J

27r ,,,.

Now we write

Let

uR = 71R - U-; 4 = maX(0,UR).

IF tip

pR(z)27r, eiv

-z_x

x"P + z(etP) ef u- d/p.aR(z) R -

27r e. zx

It is easy to verify that aR and 6R are holomorphic in D. Let

/9R = eRP ` iAR

ABR

R

where

AR = exp(-aR),

BR = exp(-/3R),

and AR is an appropriate real constant. Now

so that

ReaR>0 and ReIR>0

25

IARI<1 and IBRI<1.Since the families {AR} and {BR} are uniformly bounded, they are

normal families; and since the unit circumference is compact, we may write,for a suitable sequence of R approaching 1,

limAR = A, limBR = B, IAI < 1, IBS < 1, limAR = A.

It is easy to see thatlim gR = g.

We therefore get the required representation

BProvided only that B is not identically zero. But

IBR(O)I = exP(-I3R(O))

and

qR(0) = t dO < M.

dco

The proof is complete.

9A Lemma of Borel and SomeApplications

Definition. A set E of real numbers has length < t, written IEJ < e,means that there is a countable union of intervals [an, bn], an < bn, thatcontains E and such that

E(bn - an) < e.

Definition. tEJ = inf{t : JEJ < e}.

Lemma (trivial). If JE1J < el and JE21 < e2i then

IE1UE2l<e1+e2.

Bore! Lemma. Suppose that µ(r) is defined for all r > ro, that u isnondecreasing, and that p(ro) > 1. Then for each a > 1

µ (r + aA(r)

except in a set Ea such that AEQI < a61

Remark. The inequality of the Borel Lemma estimates p at a point greaterthan r by using the value that µ takes at r. Intuitively, if the inequalityfails in "too big" a set, the function u will become infinite "too soon" andwill not be defined for all r.

9. A Lemma of Borel and Some Applications 27

P r o o f . Let E = E,, = {r > ro : µ (r + ) > ap(r)}. Let c > 0 begiven. We proceed to define a sequence {rn} and an allied sequence {r,}by induction.

Let r1 = inf{r : r E E}. Suppose rl,...,rn_1 have been constructedtogether with numbers Ek > 0, k = 1, ... , n- I so that El +C2 +- .. En- 1 < Eand rk + Ek E E. Let r' = rk + Ek + µ(Tk+Ck

Now define rn = inf{r : r E E and r > rn_1}. Choose c,, > 0 so thatEl + + En < E and so that rn + En E E and proceed.

This procedure will terminate after n steps if and only if there does notexist an r E E satisfying r > rn. We have

(9.2) rn < rn + En < rn < ra+l

Now (r,, rn+1) fl E is empty by construction.

Claim. There exist only finitely many rn or else rn - oo. For otherwise,there would be a finite r such that rn -> r and, by (10.2), r, -+ r. However,by the construction we have for all k

1 1rk - rk = Ek +

A(rk + Ek) - p(r) >0,

which is a contradiction.

Claim. E C Un=1[rn, r ,1.Pick an arbitrary x E E. Let rno = max[rn : rn < x]. This makes sense

by the previous claim. Now rno < x. Suppose x > rno. Then rno+1 < x,an immediate contradiction. Therefore rno < x < rno, which proves theclaim.

So we have constructed a countable collection of intervals whose union

contains the set E. We estimate E(rn - rn). Notice that

IA(rn) = it rn + En + 1 > ap(rn + En) > al,(rn)A(rn+En)) - -80 that p(rn+1) > ap(rn). Therefore, p(rn+1) > anp(r1) > an. Hence

(rn - rn) En +1

E +i

- U(rn + En) /<

N'(rn + En)

But1 1 -E00

1

µ(rn + En) {u(rn) n=1 en-1 a -

It now follows that IEJ < as l + e, and the lemma is proved.


Corollary. Under the same hypotheses \on µ and a,

(9.3)

//

µ (r l l + µr)) J < aµ(r)

except for r in a set En, where EQ has logarithmic length < as-a

(By this we mean that Ea = exp Ea, where IEaI < t. We write IEalbg =

Proof. Let µ1(y) = µ(exp y). Then µ (exp (y +v

) ) < ap(exp y) fory f Ea, where IEa1 < aal by the Borel Lemma. But

1 1exp

µ(exp y)> 1 +

u(exp y)

so that

µ I 1 +u(exp y)

I exp, y) < aµ(exp y) for y V Ea,

and the result follows on writing r = exp V.

Application of the Borel Lemma to Nevanlinna TheoryWe already have proved that if f is entire, then

T(r) < log M(r) < R +TT(R) if R > r.

Choose

to get

R=r 1+ 1T(r) )

log M(r) < (2T(r) + 1)T (,r(i+ T(r)) )

Unless f is a constant, T(r) --+ oo so that 2T(r) + 1 < ZT(r) for large r.Applying the Borel Lemma we find for entire functions f,

(9.4) logM(r) < 3(T(r))2

except for a set of finite logarithmic length.

Definition. We say that A(r) -' L means that there is a set E of finiteeff

logarithmic length such that Jim A(r) = L as r oo.r¢E

We attach a similar meaning to expressions like A(r) a B(r), A(r)= B(r) , etc.eff

We have, in effect, proved the next result.

9. A Lemma of Borel and Some Applications 29

proposition. If f is a nonconstant entire function, then

log log log M(r) - log log T (r).eff

In a certain sense, this says that T(r) and log M(r) have the same sizefor most of the r. The log log takes a lot of punishment.

propostion. Suppose that a real function f has a continuous, increasingderivative on [1, oo], and that lim f (x) = oo. Thens-00

log log(f(x)) - log log(x f'(x)).

P oof. Write v(t) = tf'(t), and suppose, without loss of generality, thatf (l) = 0. Then

f(x) =f f'(t) dt j u(t).Hence f (x) < v(x) log x. But for some e > 0, we must have v(x) > ex forx large, so that

(9.5) f(x) < (v(x))2 for large x.

In the other direction, if y < x, then

f (x) ?fZ

v(t) d > v(y) log (x )y

Choose x = y + ftv) to get

v(y) < f (11 )f (y) log (1 + !(v))

Since log(1 + t) >2t

for t near 0 and positive, we have, if y is large,P(Y) <_ 2f (y)f (y + f v) . Applying the corollary of the Borel Lemma, weget

(9.6)v(y) < e(f(x))2.

eff

Equations (9.5) and (9.6) together imply the result.

10The Maximum Term of an EntireFunction

We will give in this chapter a proof that a suitable entire function can growas fast as we please.

Let f (z) = E a"z" be an entire function; ao 0 0

An=lanlIzl = r.

For each r, the sequence A0, Ajr, A2r2, ... converges to zero. Thereforewe can define

(10.1) B(r) = max(Ao, A, r, A2r2, ... ).

B(r) is called the maximum term for r. A term Akrk is a maximum termif Akrk = B(r).

Since each Akrk is a nondecreasing function of r (increasing if k 0,Ak # 0), B(r) is nondecreasing. B(r) is also continuous and unbounded.

We define the rank of the maximum term as

(10.2) u(r) = sup(n : A"r" = B(r)).

It follows immediately that if n < µ(r), then A"r" < AM(r)rµ(r), and ifn > µ(r), then A"r" < A,(r)ra(r). Therefore we also can write the rank ofthe maximum term as

p(r) = sup(n : A"r" > B(r)).

10. The Maximum Term of an Entire Function 31

Clearly, ja(r) is a nondecreasing, integer-valued function of r. Let

(10.3)

( g = - log A, if A,, 0

gn=oo if

Since f is entire, we have

(10.4) lim A ° = oo so that lim gn = oo.n_00 n-00 n

Let cn be the point (n, g(n)) on the plane. From (10.4) it follows that belowany straight line of finite slope there is only a finite number of points cn.

91

C4

GI

C3

1 2 3 4 5 6

The Newton Polygon

This property of the c, enables us to construct the Newton polygon ir(f)of the entire function f.

We construct the polygon as follows:Among the segments oc, consider those of minimal slope. From these

segments of minimal slope choose the one that is the longest; denote thisSegment by cc-k,- Repeat this selection procedure starting with the pointck, to obtain the point ck21 and so on.

The vertices of the polygon are 'yo, -yi, 7i, . , where ryi = ck; _(ki, g(ki)) for i = 1, 2.... and -yo = cka = co. The x-coordinates of thevertices -to, -rl,... are called the principal indices.

Let Gn be the y-coordinate of the point on r(f) whose x-coordinate isn. Let

(10.5) An = exp(-Gn).


An is called the logarithmic convezifcation of An. Since An = exp(-gn),it

followsimmediately that

Gn = gn if n is a principal indexGn < gn for all n.

Given r > 0, we have

log April < log B(r) = log Anr", where n = µ(r).

Since

we have

or

(10.6)

But

(10.7)

log Ap = -gp

log An = -9n,

p log r - gp < n log r - 9n

gp?9n+(p-n)logr.

y=gn+(x-n)logr

is the equation of the straight line through cn with slope log r.Equation (10.6) says that all points of 7r(f) lie above the line (10.7); this

line is "tangent" to rr(f ). Call this line D,. Thus, µ(r) is the rightmostpoint of contact of D,. with rr(f) [because µ(r) is the largest value of n atwhich we can have equality in (10.6)]. Hence the values of p(r) are theprincipal indices.

Since, for x = 0, (10.7) yields y = gn-n log r = - log Anrn = - log B(r),it follows that D, cuts the y axis at - log B(r). Two immediate conse-quences of this are:

(a) Given fl, f2 entire such that rr(fl) = 7r(f2), then

R(r : fl) = p(r : f2)1 B(r : fl) = B(r : f2)

(b) Among all entire functions, h(z) _ Anzn is the largest functionthat has the same µ and B as f.

Let

(10.8) Rn-An_1

An

We call the Rn the corrected ratios.


Geometrically, log Rn is the slope of the side of 7r(f) joining the pointswhose x-coordinates are n - 1 and n. Rn is nondecreasing and Rn -0 ocas n --* oo.

Without loss of generality, assume that Ao = ao = 1 [note that p(0) = 0].From the definition of Rn, it follows that eGn = R1 R2 Rn.

Since p(r) runs through the principal indices, gu(r) = G,u(r), it followsthat

Taking logarithms,

rn(r)B(r) = Ri.R2...Rn

p(r) 11 (r) rlog B(r) = p(r) log r FFlogRk = log Rk .

ButAr)

j1og = log t dp(t), where p(t) _ 1.

Rk <t

Integrating by parts we get:

jlogd,L(t)1or,(t)l+ t) dtt t o+

Irt

or

(10.9) log B(r) = f pt) dt,0

and by the lemma in Chapter 6 it follows that B(r) is a convex function oflog r.

Relation between B(r) and M(r)From Cauchy's inequality, lanl < r-"M(r), we get

(10.10)fanIrn < M(r) or

B(r) < M(r).

Remark. M(r) < F(r) A' rn.Note: Choose p > p(r). Then R,, > r because log R. > log r for p > p(r),as we see on interpreting log Rp as a slope. Then for q > p we can write

r9-P+1A r9 = e-Ggrq = eG,-irP-1

Since e-Go-, rP-i < B(r), we have

rq-P+1 r/

9-P+1`44r4 = B(r)1VrP+1 < B(r)

RPP


because the slopes of the edges of the Newton polygon are increasing. Wehave:

P-1 00

F(r) _ e_c"r" + : e-c°r"0 p

00 / 4-PtlpB(r) + B(r) (k)=4 P

r r.= pB(r) + B(r)Rp

Consequently,

F(r) < B(r) [p + Rpr rl provided p > µ(r).

As a heuristic guide, let us try the choice p = µ(x) for x > r, supposingfor the moment that µ(x) > µ(r). Then

I

1F(r) < B(r) [µ(x) +

Rµ(

r_ r] < B(r) I µ(x) + r1

The last inequality is justified as follows: R,,(...) is the slope between (p -1,Gp_1) and (p,Gp). This slope grows without bound, so there exists anx so that for p > x we have the inequality. Write x = r + y; y > 0. Then

F(r) < B(r) [ILr+Y)+].

Now write y = so that

F(r) < B(r) [µ (r + t) + t] .

We try to make

r\ I r It=µ(r+tr+µ µ rr+f )

by choosing t.As a first approximation we choose t = p(r). For that choice z =

r + ut . Since we want to guarantee that p > µ(r), our actual choice is

p = µ (r + n ; ) + 1. We then get:

F(r) < B(r) I p fr+ pr)J

+ $L(r) + 1J


or

F(r) < B(r) L2µ 1 \\r + A('.) f + 11 .

/ 1Using the Borel Corollary (Chapter 9) we find

F(r) < 3B(r)p(r)eff

so that

(10.11) B(r) < M(r) 3B(r)p(r).eff

For functions of finite order, say order p, we have

log M(r) = o(r") as r - oo if p' > p.

In this case,

log B(r) = o(rP) as r -i oo if p' > p.

We have

JrlogB(r)=ptt) dt=o(nl )

andwe know that

Hence,

(10.12)

p(r) log 2 < f2r M(t) dt < f2, p(t) dt= o(r°)

p(r) = o(r°).

Together with (10.11), this implies that if either log M or log B is o(r' ),0:5 p, then both are. Now, from (10.11) we have

logM(r) log3+logB(r)+logp(r)eff

=log3+o(r')+p1ogr.

From B(r) < M(r) we get

(10.13) limsuplogB(r)

Also,

< 1 .r-+oo logM(r)

log M(r) < log 3 + log B(r) + log p(r) < 1 + log 3 + log p(r)log B(r) ;ff log B(r) log B(r)


Writing F(r) as

F(r) Anrn = E AnRn(R)

n

and using the definition of B(r), we get for R > r:

F(r) < B(R)(R)n B(r)RR r

so that

(10.14) B(R) < M(r) < B(R)RR-R r

Let us choose R = r + Brf and again apply the Borel Corollary (9.3):

MrB(r) < M(r) < B r+ Br

(r)

rsWr

B(r) < M(r) < B [ r+ B(rr))(B(r) + 1).

Hence,B(r) < M(r) < 2B(r)2.

eff

Taking logarithms twice we obtain

(10.15) log log M(r) log log B(r).eff

From (10.11) we get

B(r) <M(r) 1.

If p(r) = o(rP), p < oo, then p (r + µ(,) < µ(2r) = o(rP) so that

log B(r) < log M(r) < log B(r) + p log r + constant.

ThenlogB(r) < 1 < logB(r) + plogrlog M(r) - - log M(r) log M(r)'

We shall prove that = o(1), which implies:

10. The Maximum Term of an Entire unction 37

Theorem 10.1. If f is of finite order, then

log B(r) - log M(r).

Assertion: iog1M r) = o(l) (unless f is a polynomial).

proof. Otherwise, we can find a sequence {rn } such that rn oo and suchthatlogM(rn) < clogrn or M(rn) < rn. But Cauchy's inequality says that

lakl <M(rn < rn

rnrn .

It follows that ak = 0 for k > c, and hence f is a polynomial.

In general, we have:

Theorem 10.2. If liminfr. ogr = c < oo, then f is a rational func-tion.

Proof. We again find an increasing sequence {rn}, rn -+ oo such that

T(rn) :5 clog rn

so thatN(rn, f) < clog r.-

We may suppose without loss of generality that f (O) 0, oo. Then:

Irn(t'f) dtc1ogrt

and for s > 0 : f e " EiLLt dt < c log r,,. Then

rn(s, f) logs < c log rn or

n(s' f) < clog rn

log 8

Let rn -- oo, and we see that n(s, f) < c.Therefore, f has at most c poles al, ... , ak with k < c. Now, multiplying

f by(x - al) (z - ak) and applying the previous result to the holomorphicfunction so obtained, we complete the proof of the theorem.

Suppose we are given M'(r) such that log M'(r) is logarithmically con-'0". We shall assume that M'(r) can be written in the form

M'(r) = 1 r 2(t) dt; 7(t) increasing.0


We already have proved (in Chapter 7) that fo 4 dt, where 'y(t) is in-creasing, is logarithmically convex. Suppose further that log M'(r) = o(rP)for some p < oc. Then there exists an entire function f such thatlog M(r, f) - log M'(r). Thus, for any such M'(r) there is an entire func-tion f whose maximum modulus grows essentially like M'(r).

The idea of the proof is the following: Take 1A(t) = ry(t). Define B(r) _for L(tl dt and draw a Newton polygon associated with p and B. This givesthe An. Let f be the associated function f = E An z". Then

log M(r) - log B(r) = log M'(r).

Since µ(t) must be integer-valued, for the actual proof we take

µ(t) = [ry(t)] where

[x] = greatest integer not exceeding x.

We have then7(t) - 1 < Fi(t) 5 7(t)

Since log M'(r) = o(rP), it follows that -1 = o(rP) so that µ(r) - log B(r).It remains to be shown that

log B(r) = log M'(r) + O(log r).

But

so that

p(t) - 7(t)-log- dt < 0<ro

- ro t

IlogB(r) - log M'(r) l = O(log r).

Hence log M(r) '- log M'(r), and the proof is complete.

Dropping the hypothesis of finite order, we can still get the followingresult: Given that log M'(r) is logarithmically convex, it is possible to findan entire function f such that M(r) > M'(r).

Proof

log B(r) < log M(r)

log B(r) > log M'(r) + O(log r).

The term O(log r) is easily disposed of by multiplying by a suitable poly-nomial.

Another result along this line is the following:


Theorem 10.3. Given any continuous function M'(r), there exists anentire function f such that M(r) > M'(r).

proof. It is easy to see that any such function M'(r) has an increas-ing majorant such that log M'(r) is logarithmically convex, and, indeed,log M'(r) = fo ry=f dt. Now proceed as in the previous theorem and theproof is complete. Thus, there exist entire functions that grow as rapidlyas we please.

We conclude this chapter with some more estimates on B(r). We knowthat:

r+ 11(10.16) B(r) < F(r) < B(r) I r +

A(r)

and that for R > rF(r) < B(r) RR r.

We can refine our results using the fact that

log B(r) = J r(t)

dt.t

Since f," 11 dt < for

dt, we get

rlogB(R)=1RFU(t)) dt>1 RAM dt>µ(r)logR

so that

(10.17) AN <log B(R)

log R

Choose R = r + Ioe r R Then we have

log B r+ log B r )k(r) <

log (1 + iog r )

<2logB r+ r logB(r).log B(r)

From the Borel Corollary it follows that

(10.18) µ(r) < 3log2 B(r).eff

But then (10.11) gives

F(r) < 3B(r) loge B(r).

Hence,log F(r) - log B(r).

eff

11Relation Between the Growth of anEntire Function and the Size of ItsTaylor Coefficients

Let F be an entire function and M(r) be its maximum modulus for lzi = r.Suppose A is a positive continuous increasing function for r >_ 1 such that

W r is bounded.

Definition. If log M(r) = O(a(r)), we say f is of finite A-type and writef E A

Proposition 11.1. Eanzn is of finite A-type if and only if there exists aconstant K such that lanl < `Tn(*) for each n.

Proof. Suppose Eanzn is of finite A-type. Then, since log M(r) = O(a(r)),there exists a constant K such that M(r) _< eKA(r). Now the Cauchyinequality gives Ianl < M , which gives the result.

Conversely, suppose IanI <_xA(r) xa(r)2r n = so that

IanIrn < 2-ne"J1(r).

ThusElanlr' S eKX(" ,

so that M(r) < eKa(r) and log M(r) = O(a(r)).Definition. An entire function f is of order p thereexist constants A = A(p) and K = K(p) such that

I f(z)I < AeKIa1' for all Z.

11. The Growth of an Entire Function and Its Taylor Coefficients 41

Definition. An entire function f is of order p if it is of order < p but notof order < po for any po < p.

We have the following facts immediately from the definitions:

log M(r) < log+ A + Kr"

log+ log+ M(r) < log+ log+ A + log+ K + p' 1og+ r + log+ 2.

Thus we have 1o +Iogrm r < p+ o(1) as r -+ oc.

Now let A = lim sup log+ Og+ M(r). Then by the above observation wer-00

have A < p' for all p' > p and hence A < p. In the other direction, we have+

r,K(r)< A + o(1) orlog

Therefore,M(r) < ee(X+o(1))1oB* = erxro(').

Thus, for every A' > A and r large (r > ro) we have M(r) < erg' or, moregenerally, for all r, M(r) < Aerx'. Hence f is of order < A' for all A' > Aand thus f is of order A. We therefore have proved:

Proposition 11.2. For any entire function f,

log+ log+ M(r)P = lim sup

r-oo log r

Definition. Suppose f is an entire tunction of order p and that If (z)I <AeKkI° for all z. Then we say f is of order p, type at most K. The type 7-

of f is the infimum of those numbers K such that f is of type at most K.

Definition. We say f is of finite-type if r is finite; we say f is of minimal-type if r = 0; and we say f is of mean-type if it is of finite type but not ofminimal-type.

Definition. We say f is of growth (p, r) if either it is of order < p or it isof order p and type < r. A function of growth (1, r) is called a function ofexponential-type.

Proposition 11.3. For any entire function f of order p,

r = lim suplog+ M(r)

r-oo rP

The proof is straightforward.proving the following propositions we shall use the elementary fact

thatRy RR/e R/em, y=(RJR/e-e

e


Proposition 11.4. Given an entire function f, let

n log na = limsup I .

n-.oo log T;;---

Then a = p.

Proof. Take o > p. Then Janlrn < M(r) < er' for large r. Thus IanI <r-n er' or log j an j < r°- n log r. Now choose r = (o) ° which is large forn large. Therefore we have

log jan l < n - n lognn

or log1 > n log n - n .

0' or v JanI o, a o,

Hencen log n< n log n

< a+ 0(1) as n --> oo.log -alogn-a-

TG.j o

Therefore a < o and, since o can be chosen arbitrarily close to p, a < p.Now take ,0 > a so that n I n < Q for large n and, without loss oflog e

generality, for all n. Thus n log n < Q log 1 or nn/ft < or Ian I < ;;;70

Hence IanIrn < it and therefore B(r) < sup. x , where B(r) is themaximum term of the series Elanlrn (see Chapter 10). If we let y =

vand R = L, we have B(r1iP) < supx

=/9= sup

v

v" v = sups v = eR/e =;UY

er/Pe. Therefore B(r) < exp(rP/efe), log B(r) < " and log log B(r) <

Q tog r - tog Qa Hence lim supr_ao 1-g+ log' a(r) <Q But log M(r) -log y

log B(r), and so we have p < Q and hence p < a. We therefore must haveP = a.

Proposition 11.5. If f = Eanzn is of order < p, then

r = 1 lim sup njanIoln.ep n-oo

Proof. If f is of order p and type r, take r' > r. Using the Cauchyinequality we haveIanl < m( < SL'' for large r. We now minimize the expression on theright. Its logarithm is r'rP - n log r, and setting the derivative of this equal

1/pto zero, r'prP-1- r = 0, we choose r = (?P) . Thus Ian < _

() . Hence nJanIP/n < er'p, so that we have limsupnjan1P/n < er'pn-.oo

and therefore lim sup nI an IP/" < er p.n-.oo

11. The Growth of an Entire Function and Its Taylor Coefficients 43

Now take,3 > a limsupn. nlanl1/n. Then for large n we have Ian) <n/p *+/P n/P`/ so that Ianlrn < (n rn. Hence B(r) < rn or

B(r) < max.., "Op " rs. If we let y = v and R = eflr, we have

B(rl/P) < max(e#p)vyrv

= max I eyr)v

= max RY = eR/e = epr.

Hence B(r) < efirv and thus lira supr_,,. log B r < 6. But log B(r)logM(r) so that r < Q and thus r < eP nlanIP/n, which com-pletes the proof.

Proposition 11.6. The orders and types of an entire function f and ofits derivative f are the same.

This follows easily from the formulas for order and type in terms of thepower series coefficients.Corollary to Proposition 11.5. Write f (z) = E nz". Then f is ofexponential-type if and only if Eanl;" has a finite radius of convergence.

Proof. We use Stirling's formula n! - n"e " 2rrn.

Now we have nl nIl/n N nl n a-"

Il/n N elanll/n Furthermore, the ra-

P1 a" ". Hence, since n nI1/ndius of convergence of EanSn is19uieu 11/

elanll/", f is of exponential-type by Proposition 11.5 if and only iflimuP,,-. Ianll/" < oo, as was to be proved.

Definition. To the function f given by f (z) = En z" we associate thefunction 4'(w) = Ean . Then f is of exponential-type if and only if 0za-Fris holomorphic at oo and 4) is called the Borel transform of f.

Indeed, it is easily seen that we have a one-one linear correspondencebetween entire functions of exponential-type and functions fi that are holo-morphic near oo with 4)(oo) = 0.

Proposition 11.7. The Borel transform off is an analytic continuationof the Laplace transform of f. More specifically, 4'(w) = f+°° f (t)e-'w dtin some right half-plane.

Proof. First, fo n!a-tw dt = ZT-FT if Rew > 0. Now we estimate

f+°°[f(t) - sn(t)]e'tw dt, where sa(t) is the nth partial sum of Eat'.

We have, for ak = k

00

00

I/(z) - sn(z)I = Iaklrk = E IaklRk GO k < M(R) E (-r)kn+1 n+1 n+1

(r n+l 1 (r n+l Rr, R= lRl M(R)1- R - \RJ a R-r'

44 Entire and Meromorphic Rinctions

where T' > T. Choose R = 2r so that we get I f(z) - sn(z)I (2e2r'',so that if (z) - s71(z)I < e21 IZI . Now If (t) l < er't and Ie-tw I = e-t", whereu = Rew; thus we have convergence of the integral and we can interchangethe summation and integration if we take u > 2T'. Thus we have

+00 +oo{

1(t)e-tw dt = r (E i t') a-tw dt

0 l ` J J+00 n

Ea:nJ nl a-tw dt = E an = fi(w)

0Wn+l

in some right half-plane, it > 2r', as was to be proved.

As our final result in this chapter we shall give a direct proof that theorder and type of f and f' are the same (Proposition 11.6).

Proof. Let M1(r) = suPe I f'(re'o)I. Then we have from the Cauchy in-tegral formula, f'(z) = tai if R > r = IzI, that Mi(r) <M(r) R R

. Now take R = Ar, where A > 1. Then we get Ml (r) <M(ar) r a-\

1} < M(ar)(A

a1 for r > 1. Hence, for r > 1,

log+ log+ M1 (r) < log+ log+ M(Ar) log AT + log+ log+ la 113 + log+ 2

log r - log .1r log r log r log r

and thus p1 < p.

In the other direction, supposing without loss of generality that f (0) = 0,we have f (z) = fo f'(w) dw, where we shall integrate along a ray passingthrough the origin. It follows that M(r) < rM1(r). Thus,

log+ log+ M(r) < log+ log+ r + log+ 1og+ M1 (r) + log+ 2

log r - log r

and hence p < p1. Therefore p = p1, as desired.

Similarly for type, we have

log+ ,'l11 (r) < log+ M(ar) AP+ log+ 1A -1i2rP (Ar)P rP

Hence 71 < APT for any A > 1 and thus T1 < T. Also, M(r) rMl (r), sothatto + M(r) < log+ r to + M, r

rp - r, + and hence T < T1. Therefore r = T1.

12Carleman's Theorem

Let f be holomorphic in Re z > 0 and suppose f has no zeros on z = iy.Choose p > 0 so that p < (modulus of the smallest zero off in Re z > 0).Let {zn = rneie^ } be the zeros of f in Re z > 0. Define the following:

E(R) = E(R : f) = E (r, - R2 cos Bn

(proper multiplicity of the zeros taken into account);

I (R) = I (R : f) = 1n

2 12log If (it)f (-it)I dt,

2rr (t R }r

1 "/2

J(R) J(R : f) ;R fx/2 log If (ReiB)I cos 0 d9,

where the integral is taken along the semicircle of radius R centered at 0.Then

E(R) = I(R) + J(R) + 0(1).

proof. Let r be the boundary of the "horseshoe" bounded by the semicircleof radius R, the semicircle of radius p, and the two vertical lines connectingthem:

S 27ri Jlog f (z)

LI T2 + R2

Jdz,r


where we assume that f has no zeros on IzI = R and that log f (z) denotesa branch of log f on F, i.e., log f (z) is some continuous function on Fsatisfying exp(log f (z)) = f (z). The proof proceeds by evaluating thecontour integral

r(12.1) J = 0(1).

Izj=P

Re z>0

Along the negative imaginary axis z = -iy, y > 0, dz = -idy, so(12.2)

R

27rlogf(-iy) [R2 - y2]

rRdy = 2 J logf(-iy)

[- R2] dy.

On z=iy, y>O, dz=i dy, and we have(12.3)

1 !P

tar J Rlog A W [)2+]R2

Rdy 2a<I logf(iy) [y2 - R2

P

On z = Re's (the large semicircle), dz = iReie dO, so we have

(12.4)

1a/2

1

tarlog f (Reie)

R2(e-2'8 + 1) e'BR dO

_w/2x/2

= 2

2

2 f log f (Re'B) cos 0 dO.IrR x/2

The sum of the real parts of (12.2) and (12.3) is

R

tar log I f (iy)f (-iy) j (2 - R2 dy.

ThusRe S =1(R) + J(R) + 0(1).

Now integrate S by parts:

Hence,

u = log f (x) dv = (+ I dzz2 R2

du= f(z) dx v= R2 - zAZ)

dy.

_ 1 f'(z)_,r g f()r z 1

finish1 z

dS2ari

{ 1o zLR2 2J Istart 2ari J (R2 - Z f(z)

z

1 x 1 f'(x)dz.= purely imaginary -

2a- (R2 z f(z)

12. Carleman's Theorem 47

On taking real parts and evaluating the remaining integral by the theoryof residues we find

ReS= E 1-- rncos0 .nrn T2

Remark 1. There is an obvious extension to meromorphic functions.

Remark 2. A formula of Nevanlinna (which stands in the same relation toCarleman's Theorem as the Poisson-Jensen formula to Jensen's formula)gives the value of f inside a semicircle from its values on the boundary andits zeros and poles. (See [5, p. 2].)

Next we present two applications of Carleman's Theorem.

Carlson's Theorem. Suppose f is entire and of exponential-type < it(i.e., If (z) l < AeBIz1, B < ir) and f (n) = 0 for n = 1, 2,... . Then f - 0.Remark. Carlson's Theorem also is called the "sampling" theorem. Sup-pose W(t) is a continuous signal function with support in the interval[-B, B] with B < 7r. Then the Fourier Transform of .p is

P(t)e-'zt dt.f(z) = O(z) = 1(v(t)e :zt dt = 1

fBvl'2ir R 27Then, differentiation under the integral and a simple estimate shows thatf is an entire function of exponential-type B < a. If f vanishes on thepositive integers, then f vanishes everywhere by Carlson's Theorem andtherefore W =- 0 as well. By linearity, then, if we know f on the positiveintegers, we know it everywhere.

Proof. Suppose f does not vanish identically and satisfies the hypothesesof Carlson's Theorem. Then we may assume that f has no zeros on theimaginary axis. Otherwise, translate the plane z H z - e for an appropriateE > 0. Recalling the notation in Carleman's Theorem, we observe that

E(r) (_j)logR_o(1)n

I(R) _ (R (t2 RZ Bt dt < rR B dt < B log RP \ if

BR 2BJ(R)aR

2 = = O(1).

Now applying Carleman's Theorem we see

B log R + O(1) < log R.

7rBut a < 1; eventually, this is a contradiction. f must vanish identically.In Chapter 22 we will present a theorem of Malliavin-Rubel that gives

a very sharp form of Carlson's Theorem.For the next application, we prove a result that has applications to

1Olynomial approximation theory.


Theorem. Let a be the difference of two monotone functions on [a, b] andlet f (z) = fQ eZt da(t). Suppose that f (An) = 0 for each An in a sequence

A of numbers in the right half-plane satisfying ERe \an) = oo. Then

f (z) = 0 for all z.

Remark. By use of the Hahn-Banach and Stone-Weierstrass Theorems, itcan be shown that finite sums of the form Ean exp Ant, An E A, are densein the space of all continuous functions on [a, b] in the uniform topologyif and only if there is an a (actually, we must allow complex a, but thereis no significant change) for which f exp(Ant) da(t) = 0 for each An E Aimplies that f exp(zt) da(t) = 0 for all z. Our theorem above then willimply that if Er;1 cos Bn = oo, then the sums Ea,, exp(Ant) are dense.

Proof. It is easily verified that f is an entire function. Unless f vanishesidentically, we may suppose that f has no zeros on the imaginary axis,since we could otherwise translate the y axis. Now f is bounded on theimaginary axis, say log If I < M there. Hence

I(R)<2M,IR(t2-R2)

dt = O(1).

Now, I f(z)I < Amax(Iell : a < t < b}, where A is a constant. If c >max(jal, IbI), then If (z)l < Ae" so that J(R) < 0(1) also. Hence E(R) _O(1).

We have

E(R) > n ( - R2 J cos 9n > 4 rE (rncos On

.because M1 - - >4

r if rn < R

Thus1 cos On < oo, and on letting R - oo;rn

we get a contradiction.

13A Fourier Series Method

The idea presented in this chapter is the following: If f is a meromorphicfunction in the complex plane, and if

ck(r,f)= 2x,I (logIf(Te:e)I)e-:ke

dOx

is the kth Fourier coefficient of log If (re") 1, then the behavior of f (z) isreflected in the behavior of the sequence {ck(r, f)}, and vice versa.

We prove a basic result in Theorem 13.4.5, which characterizes the rateof growth off in terms of the rate of growth of the ck(r, f) and the densityof the poles of f, generalizing Theorem 1 of [35]. We apply this theorem asin [35] to obtain estimates for some integrals involving if (z) I and to obtaininformation about the distribution of the zeros of an entire function frominformation about its rate of growth. Our presentation follows (36].

By these means, we make a study of certain general classes of mero-morphic and entire functions that include many of the classically studiedclasses as special cases. Let A(r) be a positive, continuous, increasing, andunbounded function defined for all positive r. We say that the meromorphicfunction f is of finite A-type to mean that there exist positive constantsA and B with T(r, f) < AA(Br) for r > 0, where T is the Nevanlinnacharacteristic. An entire function f will be of finite A-type if and only ifthere exist positive constants A and B such that

I f(z)I < exp(AA(BIzI)) for allcomplex z.

If we choose A(r) = rp, then the functions of finite A-type are preciselythe functions of growth not exceeding order p, finite exponential-type. We

50 Entire and Merornorphic Functions

obtain here complete answers to certain basic questions about functions offinite A-type. For example, in Theorem 13.5.2 we characterize the zero setsof entire functions of finite A-type. This generalizes the well-known theoremof Lindelof that corresponds to the classical case A(r) = r". We obtain inTheorem 13.5.3 a corresponding result for meromorphic functions. Then,in Theorem 13.5.4, we give necessary and sufficient conditions on A thateach meromorphic function of finite A-type be the quotient of two entirefunction of finite A-type. In Chapter 14, we give Miles' proof that theseconditions always hold.

The body of the chapter is divided into five sections, the last two ofwhich contain the main results. The first three sections are concerned withvarious elementary, although sometimes complicated, results on sequencesof complex numbers. The first section discusses the distribution of thesesequences. The "Fourier coefficients" associated with a sequence are definedin the second section, and several technical propositions involving thesecoefficients also are proved there. The third section is concerned with theproperty of regularity of the function A, which is closely connected withthe algebraic structure of the field of meromorphic functions of finite A-type. The fourth section contains the generalizations of the results of [35].Finally, in the fifth section, the results about the distribution of zeros areproved.

We urge that, on a first reading, the reader read §4 first and then §5,referring to §1, §2, §3 for the appropriate definitions and statements ofnecessary preliminary results. After this, the complex sequence theory ofthe first three sections will seem much more natural.

13.1. An Analysis of Sequences of Complex Numbers

W e study h e r e the distribution o f sequences Z = {zn}, n = 1, 2, 3, ... , withmultiplicity taken into account, of nonzero complex numbers z,a such thatzn --+ oo as n -f oo. Such sequences Z are studied in relation to so-calledgrowth functions A. We denote by A and B generic positive constants. Theactual constants so represented may vary from one occurrence to the next.In many of the results, there is an implicit uniformity in the dependenceof the constants in the conclusion on the constants in the hypotheses. Fora more detailed explanation of this uniformity, we refer the reader to theremark following Proposition 13.1.11.

Let Z = {Zn} be a sequence of nonzero complex numbers such thatlimzn=ocasn -- oo.

Definition 13.1.1. The counting function of Z is the function

n(r, Z) = E 1.IZ,IST

13. A Fourier Series Method 51

Definition 13.1.2. We define

N(r, Z) =Lr n(t' Z)

dt.t

proposition 13.1.3. We have

N(r,Z) log ICI

proof. Note that

E logr = J r log (r) d[n(t, Z)].Iz+.l<_r

Ixnl o t

The proposition follows from an integration by parts.

Proposition 13.1.4. We have

n(r, Z) = rdr

N(r, Z).

Proof. Trivial.

Definition 13.1.5. We define, for k = 1, 2,3,... and r > 0,

k11S(r; k : Z) _

klz,.l<r

Definition 13.1.6. We define, for k = 1, 2,3.... and r1, r2 > 0,

S(r1i r2; k : Z) = S(r2; k : Z) - S(r1; k : Z).

When no confusion will result, we will drop the Z from the above nota-tion and write n(r), S(r; k), etc.

Definition 13.1.7. A growth junction A(r) is a function defined forO< r < oc that is positive, nondecreasing, continuous, and unbounded.

Throughout this chapter, A will always denote a growth function.Definition 13.1.8. We say that the sequence Z has finite A-density tomean that there exist constants A, B such that, for all r > 0,

N(r, Z) < AA(Br).


Proposition 13.1.9. If Z has finite A-density, then there are constantsA, B such that

n(r, Z) < AA(Br).

Proof. We have

n(r, Z) log 2 <Zr n(t Z)

dt < N(2r, Z).1, t

Definition 13.1.10. We say that the sequence Z is A-balanced to meanthat there exist constants A, B such that

(13.1.1) IS(r1,r2ik : Z) I _< k

AA(Brj)+

AA(Bkr2)

r1 r2

for all r1, r2 > 0 and k = 1, 2,3,. ... We say that Z is strongly A-balancedto mean that

(13.1.2)AA(Br1) AA(Br2)

IS(ri, r2i k : Z) I <_ krk + krk1 2

for all r1,r2>0andk=1,2,3,....

Proposition 13.1.11. If Z has finite A-density and is A-balanced, thenZ is strongly A-balanced.

Remark. Using this result for illustrative purposes, we make explicit herethe uniformity that we leave implicit in the statements of similar results.The assertion is that if Z has finite A-density with implied constants A,B, and is A-balanced with implied constants A', B', then Z is stronglyA-balanced with implied constants A", B" that depend only on A, B, A',B' and not on Z or A.

Proof of Proposition 19.1.11. We observe first that, if r > 0, and if we letr' = rk1/k, then

(13.1.3) IS(r,r';k)I < 3kr,)k

To prove this we note that

I S(r, r'; k) I k f r tk dn(t),

from which (13.1.3) follows after an integration by parts. Now, for r1,r2 > 0, let r'1 = rlkl/k and r2 = r2k1/k. Then

IS(ri, r2; k)I _< IS(r', r2; k)I + IS(ri, ri; k)I + I S(r2, rz; k)I.


On combining this inequality with (13.1.3), Proposition 13.1.9, and the factthat k11k < 2, we have

I S(rl, r2; k)I I S(r' , r2i k)I + krk AA(Br1) + krk AA(Br2).1 2

But, by hypothesis,

IS(ri, r2; k)I < kri AA(Br1) + krzAA(Br2)

for k = 1,2,3,....

Definition 13.1.12. We say that the sequence Z is A-poised to meanthat there exists a sequence a of complex numbers a = {ak }, k = 1, 2, 3....such that, for some constants A, B, we have, for k = 1, 2,3,. .. and r > 0,

(13.1.4) jai, i S(r; k : Z)AA(Br)

rk

If the following stronger inequality:

(13.1.5) Iak + S(r; k : Z)I <AA(Br)

krk

holds, we say that Z is strongly A-poised.

Proposition 13.1.13. If Z has finite A-density and is A-poised, then Zis strongly A-poised.

Proof. The proof is quite analogous to the proof of Proposition 13.1.11,based on the substitution r' = rk1/k. We omit the details.

Proposition 13.1.14. A sequence Z is A-balanced if and only if it isA -Poised, and is strongly A-balanced if and only if it is strongly A-poised.

Proof. We prove only the second assertion, since the proof of the firstsssertion is virtually the same. If it is first supposed that Z is stronglyA-Poised, where {ak} is the relevant sequence, then we have

I S(r1, r2; k)I = I S(r2; k) + ak - ak - S(r1; k)I

Iak+S(r1;k)I +Iak+S(r2;k)I,

so that Z is strongly A-balanced.Suppose now that Z is strongly A-balanced, with A, B being the relevant

Cftstants. Let

p(A) = inf{p = 1, 2, 3, : lim inf Ar) = 0}.r-oo rp


Naturally, we let p(A) = oo in the case liminfA(r)rP > 0 as r --> oo foreach positive integer p. For 1 < k < p(A), we have inf r-'A(Br) > 0 forr > 0. Thus, there exist positive numbers rk such that

A(Brk)<

2 A(Br)

k

for r > 0 and 1 < k < p(A). For k in this range, we define

(13.1.6) ak = -S(rk; k).

For those k, if there are any, for which k > p(A), we choose a sequence0 < P1 < p2 < with pj ---* oo as j --* oc such that

limA(Bp')

= 0.j moo PjPW

For values of k, then, such that k > p(A), we define

(13.1.7) ak = jAm -S(pj; k).-00

To show that the limit exists, we prove that the sequence {S(pj; k)}, j =1, 2,..., is a Cauchy sequence. Let

Aj,m = S(Pj; k) - S(pm; k) = S(pm, pj; k)

We have )AA(Bpm) + AA(Bpj

kpkm kp;

Since pk pP(a) for p > 1, it follows from the choice of the pj that 0

as j, m - oo. We now claim that

I ak + S(r; k) I <3Ak (Br)

For, if 1 < k < p(A), then

I ak + S(r; k)I = IS(rk, r; k)I <AA(Br) + AA(Brk) < 3AA(Br).

krk krk - krk '

if k > p(A), then

I ak+S(r; k)I = lim I S(r, pj; k) I <AA(Br) +limsup

AA(Bpj) AA(Bp)j-.oo kr j_.oo kpj kp'°


Definition 13.1.15. We say that the sequence Z is A-admissible to meanthat Z has finite A-density and is A-balanced.

In view of Propositions 13.1.11 and 13.1.13, the following result is im-mediate,

proposition 13.1.16. Suppose that Z has finite A-density. Then thefollowing are equivalent:

(i) Z is A-balanced;(ii) Z is strongly A-balanced;(iii) Z is A-poised;(iv) Z is strongly A-poised;(v) Z is A-admissible.

In Proposition 13.3.3, we give a simple characterization of A-admissiblesequences in the special case A(r) = r°.

13.2. The Fourier Coefficients Associated with a Sequence

We now present the sequence of so-called Fourier coefficients associatedwith a sequence Z of complex numbers, and study its properties. Wewill use it in §5 to construct an entire function f whose zero set coin-cides with Z, and to determine some properties of entire and meromor-phic functions whose growth is restricted. The reason for calling them"Fourier coefficients" will become apparent on comparing their definitionwith Lemma 13.4.2.

Definition 13.2.1. W e d e f i n e , f o r k = 1, 2, 3, ... ,

kS'(r;k:Z)=k lrl

1Z-1:5r

Proposition 13.2.2. We have

S'(r; k : Z)I < -N(er, Z).

PrOof. It is clear that IS'(r;k : Z)j < n(r)/k, and we also have

n(r) <'Cr n(t)

dt < N(er).J(r


Definition 13.2.3. Let a = {ak}, k = 1,2,3,,.., be a sequence ofcomplex numbers. The sequence {ck (r; Z : a)), k = 0, ±1, ±2,..., definedby

(13.2.1) co(r; Z : a) = co(r; Z) = N(r, Z),

(13.2.2)

rkCk(r;Z:a)=2

{ak+S(r;k:Z)}

- 2 S'(r; k : Z) for k=1,2,3...,

(13.2.3) c_k(r;Z : a) _ (ck(r;Z;a)) for k=1,2,3 ....is said to be a sequence of Fourier coefficients associated with Z.

Definition 13.2.4. A sequence {ck(r; Z : a)} of Fourier coefficientsassociated with Z is called A-admissible if there exist constants A, B suchthat

(13.2.4) Ick(r : x; a) I <AA(Br) (k = 0, 4:1, ±2.... ).IkI + 1

Proposition 13.2.5. A sequence Z is A-admissible if and only if thereexists a A-admissible sequence of Fourier coefficients associated with Z.

Proof. Suppose that Z is A-admissible. Then, by Proposition 13.1.16, Z isstrongly A-poised. Let a = (ak), k = 1, 2, 3,..., be the relevant constants,and form {ck(r; Z : a)} from them by means of (13.2.1)-(13.2.3). NowDefinition 13.2.4 holds for k = 0 and some constants A, B since Z hasfinite A-density. For k = ±1, ±2, ±3, ..., we have

rIki 1 ,ICk(r;Z:a)I <2

Iak+S(r;k)I+2IS(r;k)I.

Then an inequality of the form (13.2.4) holds by Proposition 13.2.2 sinceZ has finite A-density, and because Z is strongly A-poised with respect tothe constants {ak}.

On the other hand, suppose that (13.2.4) holds. Then

N(r) = co(r) < AA(Br),

so that Z has finite A-density. Moreover,

2 (ak+S(r;k))I =1ck(r;Z:a)+2S'(r;k)

AA(Br) + N(er) < 2AA(eBr)IkI + 1 2k k

so that Z is strongly A-poised. By Proposition 13.1.16, it follows that Z isA-admissible.


proposition 13.2.6. Suppose that Z and a = {ak } are such thatIck(r; Z : a)I < AA(Br). Then {ck(r; Z : a)} is A-admissible. In par-ticular, there exist constants A', B', depending only on A, B, such that

Ick(r; Z : a)I <A'A(B'r)IkI+1I

proof. For k = 1, 2, ... , we have

(13.2.5) Ick(r)I <2

I ak + S(r; k)I + 2IS'(r; k) I

and

(13.2.6) 2 Iak + S(r; k) I < Ick(r)I + 1 IS'(r; k)I.

Since co(r) = N(r) < AA(Br), Z has finite A-density. Then, by Proposition(13.2.2), IS'(r;k)I < (1/k)O(A(O(r))) uniformly for k = 1,2,3,..., bywhich we mean that there are constants A", B" for which IS'(r,k)I <(1/k)A"A(B"r). From our hypothesis and (13.2.6), it then follows that

r'I ak + S(r; k)I = O(A(O(r))) uniformly for k > 0.

Then, by Proposition 13.1.13, we have that

rklak + S(r; k)I S O(A(O(r))) uniformly for k = 1, 2,3....

Then, using (13.2.5), we have

Ick(r)I < O(A(O(r))) uniformly for k = 1,2,3....

Since c_k(r) = (ck(r)), and since Z has finite upper A-density, the propo-sition follows immediately.

Definition 13.2.7. The quadratic semi-norm of a sequence {ck(r; Z : a)}of Fourier coefficients associated with Z is given by

1/Z

Ez(r; Z : a) { I ck(r; Z :a) 121

(

00lk__ao


Proposition 13.2.8. The Fourier coefficients {ck(r; Z : a)) are A-admissible if and only if E2(r; Z : a) < AA(Br) for some constants A,B.

Proof. First, if

Ick(r; Z : a)I <Al (Bir)

I+1then E2 (r; Z : a) < AA(Br), where B = Bl and

00 1 2 1/2

A=A1 > (Ikl+1fk=-00

On the other hand, suppose there are constants A, B for which E2(r; Z :a) < AA(Br). Then it is clear that Ick(r; Z : a)I < AA(Br), so that byProposition 13.2.6, {ck(r; Z : a)} is A-admissible.

13.3. Sequences That Are A-Balanceable

In this section, we are concerned with the process of enlarging a sequence Zso that it becomes A-balanced. Growth functions A for which this is alwayspossible are called regular and give rise to associated fields of meromorphicfunctions with special properties; for example, see Theorem 13.5.4. Theprincipal results of this section are Propositions 13.3.5 and 13.3.6, whichgive the simple condition that A be regular. In addition, we give in Propor-tion 13.3.3 a simple characterization of A-admissible sequences of the caseA(r) = rP.

Definition 13.3.1. The sequence Z is A-balanceable if there exists aA-admissible supersequence Z' of Z.

Definition 13.3.2. The growth function A is regular if every sequence Zthat has finite A-density is A-balanceable.

Proposition 13.3.3. Suppose that A(r) = rP, where p > 0. Then(i) the sequence Z is of finite A-density if and only if lim sup r-Pn(r, Z) <

0o as r -+ oc;(ii) if p is not an integer, then every sequence of finite A-density is A-

admissible;(iii) if p is an integer, then Z is A-admissible if and only if Z is of finite

A-density and S(r: p : Z) is a bounded function of r;(iv) the function A(r) = rP is regular.

Proof. To prove (i), we have that n(r) = O(rP) whenever Z has finite A-density. On the other hand, if limsupr-Pn(r) < oc, then n(r) < Ar" forsome positive constant A, so that

t-n(t) dt < Ap lr".N(r) = fo"


To prove (ii), suppose that N(t) < At. Then so long as k 54 p, we have

(13.3.1)ill

r2

k dn(t) (A + Akl) a(k2)1

t \ Ip - k1 rl r2 1

For, on integrating by parts, we have that the integral is equal to

n(k2) - n(ri)+ k J r2 tk+i dt.

2 r 1 ri

n(r2) < A . = AA(r2)

T2

r2 r2

n(r1) AA(ri)k kr1 T1

f + ) ,Ip-

kI \and the inequality (13.3.1) follows. Hence, so long as p is not an integer,every sequence Z of finite rP-density is rP-balanced.

To prove (iii), suppose that Z has finite rP-density and that p is aninteger. Then, by (13.3.1), we see that all the conditions that Z be A-balanced are satisfied except for k = p. For this case, the condition thatS(r1,r2; p) be bounded by ri PAA(Br1) + r2 PAA(Br2) for some A, B isprecisely the condition that S(r; p) be bounded, as is quite easy to see.

To prove (iv), we observe first that if p is not an integer, then A(r) = rPis trivially regular by (ii). If p is an integer and Z has finite rP-density,let Z' be the sequence obtained by adding to Z all numbers of the formw`1Z, where wP = 1, but w 0 1. Then Z' has finite rP-density andS(r; p : Z') = 0 for all r > 0. Hence, by (iii), Z' is rP-admissible, and itfollows that A(r) = rP is regular.

The next two results give simple conditions, both satisfied in case A(r) _", that imply that A is regular.

Definition 13.3.4. We say that the growth function A is slowly increasingto mean that A(2r) < MA(r) for some constant M.

If A is slowly increasing, it is easy to show that for some positive numberA A(r) = O(rP) as r -- oo.

p&oposition 13.3.5. If A is slowly increasing, then A is regular.

imposition 13.3.6. If log A(ex) is convex, then A is regular.

The proofs of these results use the next lemma.


Lemma 13.3.7. The growth function A is slowly increasing if and onlyif there exist an integer po and constants A, B such that

(13.3.2) P(+)+) dt <AA(Br) for p > po > 0.

If A(r) = ro, then we may choose p0 = [p] + 1.

Proof. Suppose first that (13.3.2) holds. We may clearly suppose thatB > 1. Then

AA(Br) > r0 ,\(t)dt > r- Adt >

A(2Br)prP Jr tP+1 2Br tP+1 p(2B)PrP

whenever p > po. Taking p = po, we have A(2Br) > MA(Br), where

M = A(2B)PO,

so that A(r) is slowly increasing. Suppose next that A(r) is slowly increas-ing, say A(2r) < MA(r). Then

00foo A(t)dt = rkr+lr +i dt < PrP(2k)P < M E (M)k.

k-0 J k=0 k=0

Hence, if po is taken so large that 2PO > M, we have an inequality of theform (13.3.2). In case A(r) = rP, we have M = 2P, and the final assertionfollows.

Proof of Proposition 13.9.5. Let A he slowly increasing, and let Z be asequence of finite A-density. Choose po as in the last lemma so that, forp>po,

1coA(+1+i

dt < `4 ( r)PrP

DefinePO

Z' = U W-kZ,k=0

where no = po+l, w = exp(2iri/no), and w-kZ = {w-kzn}, n = 1, 2, 3, ...Then we have S(rl, r2; k : Z') = 0 fork = 1, 2, ... , po, since

,pok =0

so long as wk 0 1, and this is true for k = 1, 2,. .. , p0. Hence, to prove thatZ' is A-balanced, we need consider only k > p0. For such k, with r < r',we have

JS(r,r;k:Z)j= I l <J k dn(t,Z).

k r< r' `Zn

k

I k rr t '


On integrating by parts, we have

Jr, I

k dn(t, Z')I < n(r,j) k') + n( ,,k ') + kI r'

n(t, Z') dt.r

Since Z is of finite A-density and n(r, Z') = (po + 1)n(r, Z), we haven(r,Z') < A1A(Blr) for some constants Al, B1 by Proposition 13.1.9.

Since A is slowly increasing, we have A(BBr) < A2A(r) for some positiveconstant A2 > 0. To complete the proof of the proposition, we have onlyto prove that

f r n(t, Z') A'A(B'r)f tk+l dt < krkr

for some constants A', B'. However,

1rn(t, Z') f A(t) AA2A(Br)

tk+1 dt- A2 tk+1 <

krk

since k > po.

Proof of Proposition 13.3.6. It is no loss of generality to suppose thatr-PA(r) -> oo as r -+ oc for each p > 0, since otherwise A is slowly in-creasing by Lemma 13.3.7, and then Proposition 13.3.5 applies. Now forp = 1, 2,3,..., let R1, be the largest number such that

A(RP) = inf A(r)

19 r>0 rP

and let R0 = 0. Then we have that Ra < R1 < R2 < ... , and that R. - ooas p -+ oo. Further, by Lemma 13.3.7, r-Pa(r) decreases for r < RP andincreases for r > RP. We also have the inequality

(13.3.3) 2PA(r) < 2A(2r) if r > Rp_1i

since, by the above remark,

A(r) A(2r)

rP-1 - (2r)P-1'

Now let Z be of finite A-density. For convenience of notation, we supposethat N(r) < A(r) and n(r) < A(r), since we could otherwise replace thefunction A(r) by the function AA(Br) for suitable constants A, B. We thenclaim that

(13.3.4) / r k1 dn(t) < 4A(r)

r


if k > 2P and r < r' < Rp.To prove (13.3.4), we first integrate by parts, replacing the integral by

r')knrk) 1- k

/r tk+l dt.

(

Nown(r') < A(r) <

A(r)(1J)k - (ri)k - rk

since r < r' and r-kA(r) is decreasing for r < Rp < Rk. Also,

Jr n(t) dt rA(1 dt <A(r) 1 1

tk+1 r rP tk+l-P rP (k - p) rk-P'

since t-Pa(t) < r-Pa(r) for r < t < r' < R. Thus,

f1

dn(t)<

A(r)

+A(r)

+k A(r)

Fk- rk (k - P) rk

We have k/k(k - p) < 2 since k > 2P, and (13.3.4) follows.We now define Z' as follows. For each zn E Z with R.- j < I zn I < RP,

we introduce into Z' the numbers

1 1Zn, Zn5 ...,wm-1Zni

where m = m(p) = 2P and w = w(m) = exp(27ri/m). We make thefollowing assertions:

(13.3.5) n(r, Z')-n(Rp_1i Z') = 2P(n(r)-n(Rp_1)) if Rp_1 < r < RP,

(13.3.6) n(r, Z') < 2Pn(r) if r < Rp,

(13.3.7) N(r, Z') < 2PA(r) if r < Rp,

(13.3.8)

jldn(tz')<kjldn(t) if k > 2P and r < r' < Rp.

(13.3.9)S(r, r'; k : Z') = 0 if r, r' > Rp and k is not a multiple of 2p.

The assertions (13.3.5) and (13.3.6) follow immediately from the defini-tion of Z', while (13.3.7) follows from (13.3.6) and (13.3.8) follows easily


from (13.3.5). To prove (13.3.9), it is enough to prove that S(r, r'; k : Z') =0 if R;_1 < r < r' < R;, j > p, and k is not a multiple of 2P. But, in thiscase, we have

S(r, r'; k : Z') = 7S(r, r'; k : Z),

wherery= 1+wk+w2k+...+w(m-Uk'

where m = m(j) = 2' and w = w(m) = exp(2iri/m). Since k is not amultiple of 2P, k is therefore certainly not a multiple of 21, so that wk # 1.We then have

_ 1 - wkmti

1-wk =0,

and our assertion is proved.We now prove that Z' is A-admissible. To see that Z' has finite A-

density, let r > 0 and let p be such that 4_1 < r < RP. Then, by (13.3.7)and (13.3.3), we have that N(r, Z') < 2PA(r) < 2A(2r). To see that Z' isA-balanced, let k be a positive integer and suppose that 0 < r < r'. Writek in the form 2Pq, where q is odd. Then, by (13.3.9), S(r, r'; k : Z') = 0 ifRp < r < r'. Suppose that r < R.P. Then S(r, r'; k : Z') = S(r, r"; k : Z'),where r" = min(r', RP), by (13.3.9). However,

IS(r, r"; k: Z') I< k J r I t dn(t, Z').

By (13.3.8), this last term does not exceed

tdn(t),T

and this, in turn, does not exceed 4r-kA(r), by (13.3.4). Consequently, wealways have IS(r, r'; k : Z)j < 4r-kA(r), so that Z' is A-balanced, and theproof is complete.

13.4. The Fourier Coefficients Associated with a MeromorphicFunction

In this section, we associate a Fourier series with a meromorphic functionand use it to study properties of the function. As we mentioned at thebeguning, the results of this section are generalized versions of the resultsof the earlier paper [35], and the proofs are essentially the same. Ournotation follows the notation of [35] and the usual notation from the theory0(meromorphic and entire functions. Our presentation still follows [36].We first recall the results from the theory of meromorphic functions thatgill be needed.


For a nonconstant meromorphic function f, we denote by Z(f) [respec-tively W(f)] the sequence of zeros (respectively poles) of f, each occurringthe number of times indicated by its multiplicity. We suppose throughoutthat f (O) 0 0, oo. It requires only minor modifications to treat the casewhere f (0) = 0 or f (0) = oc. By n(r, f) we denote the number of poles off in the disc {z : IzI < r}. By N(r, f) we denote the function

N(r, f) =J

r n(t, f) dt,t

and by m(r, f) the function

m(r, f) = 2xJ x log+ I f (re`B) I d6,

where log+ x = max(log z, 0). We have, of course, that n(r, f) = n(r, W (f) )and N(r, f) = N(r, W(f)). The Nevanlinna characteristic, which measuresthe growth of f, is the function

T(r, f) = m(r, f) + N(r, f).Three fundamental facts about/T(r, f) are that

(13.4.1) T(r,f)=T(r,f/+loglf(0)1,

(13.4.2) T(r,

fg)<\\T(r,f)+T(r,g),

(13.4.3) T(r, f + g) <T(r, f)+T(r,g)+log 2.An easy consequence of (13.4.1) is that

1 -W

(13.4.4) loglf(re`B)II dO<2T(r,f)+logIf(0)I-

This follows from (13.4.1) by observing that the first term is equal tom(r, f) + m(r,1/ f ), which is dominated by T(r, f) + T (r, 11f).

For the entire functions f, we use the notation

M(r, f) = sup{I f (z)I : z = r}.

The following inequality relates these two measures of the growth of f incase f is entire:

(13.4.5) T(r,f) Slog+M(r,f): R+rT(R,f)for 0 < r < R. We will use (13.4.5) mostly in the form

(13.4.6) T(r, f) < log+ M(r, f) < 3T(2r, f),which results from setting R = 2r in (13.4.5).

The following lemma, which is fundamental in our method, was provedin [9] and [351.t We reproduce the proof of [35] here.

tI have a vague memory of seeing this formula in a paper of Frithiof Nevanlinnapublished around 1925, but was unable to find it on a recent search.


Lemma 13.4.1. If f (z) is meromorphic in I zI < R, with f (0) # 0,oo, and Z(f) = {zn}, W(f) = {wn}, and if log(f(z)) _ Ek o akzk nearz-0, then for0<r<R we have

(13.4.7) log I f (re'a)I _ E ck(r, f)eike,00

k=-oo

where the ck(r, f) are given by

co (r, f) = log If (0) I + log r - logr

Ixn

I

(13.4.8)IEn,_/r

\\l Iwnl<r Iwk

=loglf(0)I+NIr, fI -N(r,f).

For k = 1,2,3,...,

(13.4.9)

A;

k((

llk

ck(r,f) = akr + 2k [(Zn) - l rsllEnl<rk k1 (r \ - rgr

2kIwnl<r

lwn/) r

Fork=1,2,3,...,

(13.4.10) c_k(r, f) = CA: (r, f)).

There are appropriate modifications if f (0) = 0 or f (0) = oo.

Remark. Observe that in the notation of §1 and §2, formula (13.4.9) be-comes

ck(r, f) =2akrk + 2 {S(r; k : Z(f)) - S(r; k : W(f))}(13.4.11)

- 2 {S'(r; k : Z(f)) - S'(r; k : W(f ))}.

PrOof. We may suppose that f is holomorphic, since the result for mero-morphic functions then will follow by writing f as the quotient of twolbbmorphic functions. We may suppose further that f has no zeros on(z : IzI = r}, since the general case follows from the continuity of bothAides of (13.4.8) and (13.4.9) as functions of r. Formula (13.4.8) is, offturse, Jensen's Theorem, and (13.4.10) is trivial since log If I is real. ToMove (13.4.9), write

Ik(r) = ! I [log f (re'®)J cos(k8) dO


for some determination of the logarithm, and k = 1,2,3,.... Then, byintegrating by parts, we have

This may be rewritten as

4(r)- 2irki Jf (z) xk - rk } dz.

This last integral may be evaluated as a sum of residues, and on takingreal parts we get the kth cosine coefficient of log If 1. Similarly, consideringthe integral

Jk(r) =k

J [log(f (reè))J sin(kO) d8,

where the integration is now between it/2k and 2w + it/2k, we get the kthsine coefficient. On combining these, we get (13.4.9).

We now define the classes of functions that we shall study.

Definition 13.4.2. Let A be a growth function. We say that f (x) is offinite A-type and write f E A to mean that f is meromorphic and thatT(r, f) < AA(Br) for some constants A, B and all positive r.

Definition 13.4.3. We denote by AE the class of all entire functions offinite A-type.

Proposition 13.4.4. Let f be entire. Then f is of finite A-type if andonly iflog M(r, f) < AA(Br) for some constants A, B and all positive r.

Proof. This follows immediately from (13.4.6).

Note in particular that, if A(r) = rP, then f E A if and only if f is ofgrowth at most order p, exponential-type.

We also note that by inequalities (13.4.2) and (13.4.4), A is a field andAE is an integral domain under the usual operations.

The main theorem of this chapter is the following.

Theorem 13.4.5. Let f be a meromorphic function. If f is of finiteA-type, then Z(f) and W (f) have finite A-density and there exist constantsA, B such that

(13.4.12) Ick(r, f)1<AA(Br) (k = 0,±1,±2,... ).jkI+1


In order that f should be of finite A-type, it is sufficient that Z(f) [or W (f )]have finite A-density and that the weaker inequality

(13.4.13) Ick(r, f)I < AA(Br)

hold for some (possibly different) constants A, B. Thus, in order that fshould be of finite A-type, it is necessary and sufficient that Z(f) have finiteA-density and that (13.4.12) should hold. It is also necessary and sufficientthat Z(f) have finite A-density and that (13.4.13) should hold.

Proof. The order of the steps in the proof will be as follows. We first showthat if f satisfies an inequality of the form (13.4.13), and if either Z(f) orW(f) has finite A-density, then f must satisfy an inequality of the form(13.4.12). We then show that if f is of finite A-type, then Z(f) and W(f)are of finite A-density, and f satisfies an inequality of the form (13.4.13).Finally, we prove that if Z(f) [or W (f )] has finite A-density and if f satisfiesan inequality of the form (13.4.12), then f must be of finite A-type.

We shall suppose that f (0) = 1. The case f (0) = 0 or f (O) = 00causes no difficulty since we may multiply f by an appropriate power of zand the resulting function still will be of finite A-type. This is because ifliminf(A(r)/ log r) = 0 as r - oo, then by the exercise in Chapter 8 theclass A contains only the constants.

Let us suppose that either Z(f) or W (f) is of finite A-density and thatIck(r, f )I = O(A(O(r))) uniformly for k = 0, ±1, ±2,.... On consideringthe case k = 0, we see that both Z(f) and W (f) have finite A-density. Itis enough to prove that f satisfies an inequality of the form (13.4.12) fork = 1,2,3,..., since c_k is the complex conjugate of ck. We prove thisexactly as Proposition 13.2.6 was proved. From (13.4.11), we have

(13.4.14))I <-

2I«k + S(r; k : Z(f)) - S(r; k : W(f))IIck(r, f)1:5-r"

+ 1IS'(r;k: Z(f))I + 1IS'(f;k: W(f))I,2 2

and

(13.4.15) 2IC,k + S(r; k : Z(f)) - S(r; k : W(f))I < Ick(r, f)I

+ IS'(r;k : Z(f))I + 1IS'(f; k : W(f))I

Then, by Proposition 13.2.2, for Z = Z(f) or Z = W (f ), we have

IS'(r; k : Z)I = k-'O(A(O(r))) uniformly fork > 0.

BY (13.4.14), it is therefore enough to prove that

IQk + S(r; k :Z(f )) - S(r; k : W(f ))I = k-lr-kO(A(O(r)))uniformly for k = 1, 2,3,....


But, we already have from (13.4.15) that

Iak+S(r;k : Z(f)) - S(r;k: W(f))I = r-kO(A(O(r)))

uniformly for such k. Replacing r by r' = kl/kr and observing that r' < 2r,we have that

Ic k + S(r'; k : Z(f)) - S(r'; k : W(f))I = k-1r-kO(A(O(r)))

Thus, the assertion will be proved if we can show that, for Z = Z(f) andZ = W(f), we have

I S(r, r'; k : Z)I = k-'r-kO(A(O(r))).

This was proved in Proposition 13.1.11 [see (13.1.3)].Now suppose that f has finite A-type. Then

N(r, W(f)) = N(r, f) <T(r, f),

so that W (f) has finite A-density. By (13.4.1), the function 11f also hasfinite A-type. Hence, Z(f) = W(11f) also has finite A-density. To see thatan inequality of the form (13.4.13) holds, note that

Ick(r,f)I=IZxf x{loglf(Te")I}e-:kedel

:5-f f l log If(re16)Il dO < 2T(r, f) + log If(O)Ix

by (13.4.4).Finally, suppose that W (f) has finite A-density and that (13.4.12) holds.

If Z(f) has finite A-density, we apply the argument below to the func-tion 1. Then N(r, f) = O(A(O(r))). It remains to prove that m(r, f) _O(A(O(r))). However,

lioglf(reied dB,m(r, f) <_27r fir

which, by the Schwarz inequality, does not exceed

log If(re`B)112

dB27r Tr

By Parseval's Theorem, we have, for suitable constants A, B,2 00

loglf(re:e)I dB= Ick(r,f)I2k=-oo

< A2(A(Br))2k. (IkIl)2Hence, m(r, f) = O(A(O(r))), which completes the proof of the theorem.

Specializing Theorem 13.4.5 to entire functions, we have the next result.


Theorem 13.4.6. Let f be an entire function. If f is of finite A-type,then there exist constants A, B such that

(13.4.16) Ick(r, f)I <A(A(Br)) (k = 0,±1,±2,... ).

IkI + 1

It is sufficient, in order that f be of finite A-type, that there exist (possiblydifferent) A, B such that

(13.4.17) Ick(r, f)I < AA(Br) (k = 0,±1,±2,... ).

Thus, in order that f should be of finite A-type, it is necessary andsufficient that (13.4.16) should hold, and it is also necessary and sufficientthat (13.4.17) should hold.

Proof. This result is an immediate corollary of Theorem 4.6 since W (f) isempty in case f is entire.

Definition 13.4.7. For a meromorphic function f, we define

7, Y

Eq(r, f) _ {j-jlogIf(re19)Ix I

Iq

d0j

Not ice that if f is entire with f (0) = 1, and if a = {ak} is such thatck(r,f)=ck(r;Z(f):a),k=0,±1,±2,...,then E2(r,f)=E2(r;Z(f)a), where this last quantity is the one defined in Definition 13.2.7.

Theorem 13.4.8. Let f be an entire function. If f is of finite A-typeand 1 < q < oo, then

(13.4.18) Eq(r, f) < AA(Br)

for suitable constants A, B and all r > 0.Conversely, if (13.4.18) holds for some q > 1, then f is of finite A-type.

Proof. If f is of finite A-type, then by the Hausdorff-Young Theorem ([511,p. 190), the L9 norm of log If (reie)I, as a function of 0, is bounded by theP norm of the sequence {ck }, where (p) + (q) = 1. By Theorem 13.4.6, this

norm is dominated by an expression of the form AA(Br). Conversely,using Holder's inequality,

Ick(r,f)I <_ 2; JxllogIf(re's)IIdO,

r r l< 2 {

1I l o g I f(rei8)IIgd0 } < AA(Br)

for suitable constants A, B, and it follows from Theorem 13.4.6 that f hasfinite A-type.


Theorem 13.4.9. Let f be a meromorphic function of finite A-type, withf (0) 36 0, oc. Then for each positive number a there exist positive constantsa, /3 such that, for all r > 0,

x(13.4.19)

2_ j exp A

(()logIf(reb8)Ix

IdO<1+e.

Remark. We have as a consequence that, for all r > 0,

1J x

1 dO < 1+ e,27r x If (Tese)I011iBr)

which is somewhat surprising, even in case f is entire, since it is by nomeans evident that the integral is even finite.

Proof. There is a number /3 > 0 such that

Ick(r,f)I MA(Qr)

<(kI -h 1

(k = 0, ±1, f2, ... ).

Let

Then

F(O) = F(O, r) =A(Ar)

log If (Te'B)I.

F(O) = >ykeike, where yk = ck(r,f)A(Qr)

We may also suppose that the constant M satisfies

IF(0)I dO < M21r

by Theorem 4.9. By a slight modification of [49] (p. 234, Example 4), weknow that for any such F there exists a constant a > 0, where a dependsonly on M and e, such that

1

J

xexp(aIF(0)I) dO < 1 + e,

21r

from which (13.4.19) follows.

13.5. Applications to Entire Functions

We present in Theorem 13.5.2 a simple necessary sufficient condition ona sequence Z of complex numbers that it be the precise sequence of zerosof some entire function of finite A-type. The condition is that Z shouldbe A-admissible in the sense of Definition 13.1.15. This generalizes a well-known theorem of Lindelof (see the remarks following the proof of the


Theorem 13.5.1, for constructing an entire function with certain proper-ties from an appropriate sequence of Fourier coefficients associated witha sequence of complex numbers). We also prove in Theorem 13.5.4 thatA has the property that each meromorphic function of finite A-type is thequotient of two entire functions of finite A-type if and only if A is regular inthe sense of Definition 13.3.2. Accordingly, Propositions 13.3.5 and 13.3.6give a large class of growth functions A for which this is the case, includingthe classical case A(r) = re'. Even this case seems to be unknown.

We turn now to our first task, the construction of an entire functionf from a sequence Z and a sequence {ck(r; z : a)} of Fourier coefficientsassociated with Z. We recall that we have assumed that Z = {zn} is asequence of nonzero complex numbers such that zn --+ 00 as n - oc.

Theorem 13.5.1. Suppose that {ck(r)} = {ck(r; Z : o!)}, k = 0, ±1,12,..., is a sequence of Fourier coefficients associated with Z such thatfor each r > 0, E Ick(r)12 < oo. Then there exists a unique entire functionf with Z(f) = Z, f (0) = 1, and ck(r, f) = ck(r) for k = 0, ±1, ±2,....

Proof. We define

,P(pe"°) = E ck(p)eikw.

Since E Ick(p)12 < oo, this defines as an element of L2[-w, 7r]for each p > 0 by the Riesz-Fischer Theorem. For p > 0, we define thefollowing functions:

(13.5.1)

(13.5.2)

(13.5.3)

zn p(zn - z)B°(z' Z.) _ Izn1 p2 - znz

Pp(z) = H B,, (z; zn),I z. I

K(w; z) _w + zw-z

(13.5.4) Q(z) = expJ

K(w z)4i(w)dw 1

p 21ri 1w1=pw ,

(13.5.5) fp(z)PP(z)QP(z)

We make the following assertions:

(13.5.6)

The function fp is holomorphic in the disc {z : Iz) < p},and its zeros there are those zn in Z that lie in this disc.


(13.5.7) ff(0) = 1.

(13.5.8) ff r < p, then ck(r,f) = ck(r).

Now (13.5.6) is clear from the definition of fP. Also,

fP(0) = PP(0)QP(O) = QP(0) RIznl

IE^1<PP

However,

QP(0) = exp21 ,

O(w)dw

= exp{co(p)} = H nl1ri IW1=P IznISP

and it follows that ff(0) = 1.To prove (13.5.8), we see by 13.4.1 that it is enough to show that, near

z = 0,

log ff(z) _ E akzk,k=1

where the ak are such that a = {ak) and ck(r) = ck(r; Z : a). That is,near z = 0,

(13.5.9)r (z)

= E kakzk-1.k=1

We now make this computation. First we have that

BP' (z; zn) Izn12 - p2 xn - 1

BP(z;za) (zn - z)(p2 - znz) p2 - xnz zn - z

00 00 k(2nzk-1 - zk-1

-p2 znk=1 k=1

Thus,

where

PD(z) - Uk,PZk-1near z = 0,

1'(z) k=1

Uk,P =

(.)k>

(i)C

I ..I_P I .I_P

F o r k = 0,1, 2, ... , we write w = pe"° and ck(p)e'k`° = I kwk. Then by thedefinition of ck(p) we see that

Po = N(p, Z)

13, A Fourier Series Method

and

f2k 2nk +2k E zn)k -

(2n)klIz, I{SZkwk + S1kwk}k=1

00

N(p, Z) + E {kwk + 1ZkP2k(i)k}

k=1

so that

where

But

and

Hence,

where

27ri KIWI=P-'(w)KK(w, z) w 27ri (w

twz)2 dw,2

I,,.I=pwk dw = kzk-1 for k = 0,1,2,...,

W-i

1

21 (w - z)2

1 r r l 1 dw = 0 for k=0,1,2,....2arz IwI=P w (w - z)2

00

Q' (z) = c Vk zk_1,x L °

QP( ) k=1

Vk,P = 20k = (xk + k n) k - \ p2) kIzn I SP

Hence, near z = 0 we have

00f°(z) _ P(z) k 1

fP(z) PP(z) + QP zkakz

() k=1

and (13.5.9) is proved

It next follows from (13.5.6)-(13.5.8) that

(13.5.10) if p' > p, then f,,' is an analytic continuation of fP


For if we define for Izi < p

F(z) = f,(z)fa(z)'

then

Ck(r, F) = ck (r, fP') - Ck(r, fP) = Ck(r) - ck(r) = 0

for 0 < r < p, and therefore IF(z)l = 1. On the other hand, F(0) = 1, andit follows that F is the constant function 1.

We now define the function f of Theorem 13.5.1 by setting f (z) = f f(z)if p > Izj. It is clear that f is entire and, by (13.5.6), that Z(f) = Z. Also,f(0) = 1, and ck(r, f) = ck(r, fp) for p > r, so that ck(r, f) = ck(r). Anargument analogous to the one used in proving (13.5.10) proves that f isunique, and the proof of the theorem is complete.

We now characterize the zero sets of entire functions of finite A-type.

Theorem 13.5.2. A necessary and sufficient condition that the sequenceZ be the precise sequence of zeros of an entire function f of finite A-typeis that Z be A-admissible in the sense of Definition 13.1.15, that is, that Zhave finite A-density and be A-balanced.

Proof. If Z = Z(f) for some f E AE, then by Theorem 13.4.6 the se-quence {ck(r, f)} is a A-admissible sequence of Fourier coefficients associ-ated with Z and thus Z is A-admissible by Proposition 13.2.5. Conversely,suppose that Z is A-admissible. Then by Proposition 13.2.5 there exists aA-admissible sequence {ck(r)} associated with Z. Then by Theorem 13.5.1there exists an entire function f with Z = Z(f) and {ck(r, f)} = {ck(r)}.Then by Theorem 14.4.7 and the fact that {ck(r)} is A-admissible, it followsthat f E AE, and the proof is complete.

Remark. This theorem generalizes a well-known result of Lindelof [201,which may be stated as follows.

Theorem 13.5.3. Let Z be a sequence of compex numbers, and let p > 0be given. If p is not an integer, then in order that there exist an entirefunction of growth at most order p, finite-type, it is necessary and sufficientthat there exist a constant A such that n(r, Z) _< Are. If p is an integer.it is necessary and sufficient that both this and the following condition besatisfied for some constant B:

< B.znIz,.I<r

This result follows immediately from Theorem 13.5.2 and the character-ization of rP-admissible sequences given in Proposition 13.3.3. Our resultshows that, in general, the angular distribution of the sequence of zeros


of a function, and not only its density, is involved in an essential way indetermining the rate of growth of the function.

We turn now to the second problem of this section, that of determiningwhen A is the field of quotients of the ring AE. We first prove the followingresult.

Theorem 13.5.4. In order that a sequence Z of complex numbers be theprecise sequence of zeros of a meromorphic function of finite A-type, it isnecessary and sufficient that Z have finite A-density.

Proof. The necessity follows immediately from the fact that if f is a mero-morphic function, then N(r, f) < T(r, f). For the sufficiency, we remarkfirst that the method used in proving Theorem 13.5.1 can be used to con-struct suitable meromorphic functions. Indeed, suppose that we are giventwo disjoint sequences Z, W of nonzero complex numbers with no finitelimit point and constants ryk, k = 1,2,3,..., such that the coefficientsdefined by

co(r) =N(r, Z) - N(r,W),

ck(r)= 2 {yk+S(r;k:Z)-S(r;k:W)})- 2 {S'(r; k : Z) - S'(r; k : W) (k = 1, 2,3....

c-k(r) =(ck(r)) (k=1,2,3.... )

satisfy E Ick(r) 12 < oc for every r > 0. Then, by defining

Dp(z; w..) = J B,(z; w..)Iwnl Jzi has zero sequence Z, pole sequence W, andFourier coefficients {ck(r)}. It is therefore enough to prove that, given asequence Z of finite A-density, there exist a disjoint sequence W of finiteA-density and constants ryk, k = 1,2,3,..., such that the ck(r) satisfyIck(r)l < AA(Br) for some constants A, B and all r > 0. For then, bythe first part of the proof of Theorem 13.4.5, the ck(r) must satisfy thestronger inequality

Ick(r)I :AA(B'r)

(r > 0)Jkl + 1

for some constants A', B', so that the function f synthesized from the ck(r)Must be of finite A-type by Theorem 13.4.5.

76 Entire and Meromorphic Minctions

Supposing now that Z = {zn } has finite A-density, we define W = {wn }by wn = zn + En, n = 1, 2, 3, ... , where the En are small complex numbersso chosen that Iwn I = Izn I, n = 1,2,3,..., all of the numbers wn and zkare different, and such that

IEnI < A(O).

Iznl

Then, N(r, W) = N(r, Z) so that W has finite A-density. Hence,

I S'(r; k : Z)I = k-'O(A(O(r)))

and

We define

IS'(r; k : W)I = k-'O(A(O(r))) k=1,2,3,-...

rk k>/n`k- (1)k}

It remains to prove that

2I'Yk+S(r;k:Z- Sr;k:W

uniformly for k = 1, 2,3,.... Now

2 Ilk+S(r;k:Z)-S(r;k:W)Irk

2 k n/ k -()k}

nl>r

rk

2 [IznI>r

(wnzn)k < 2 1[-` I(wn)k - (zn)_I

Izn I2kInI>r

1 [ (wn)k - (zo)kIr

However, I(wn)k - (zn)kl C klEnllznik-1, so that we have

2 Iyk+S(r;k:Z)-S(r;k:W)I

rk IE j< 1 IEnl < ' A(0) < A(r).

21znl>>r

Iznl 2Jz,. >r Iznl

2


Theorem 13.5.5. The field A of all meromorphic functions of finite A-type is the field of quotients of the rings AE of all entire functions of finiteA-type if and only if A is regular in the sense of Definition 18.3.2, that is,if and only if every sequence of finite A-density is A-balanceable.

proof. First, suppose that A is regular and that f E A. Then Z(f) has finiteA-density by Theorem 13.5.3. There then exists a sequence Z' D Z(f) suchthat Z' is A-admissible. [We may suppose, by the remarks preceding theproof of Theorem 13.4.5, that f (0) 54 0, oo]. Then, by Theorem 13.5.2,there exists a function g E AE such that Z(g) = Z. Since we have thenthat Z(g) C Z(f), the function h = g1 f is entire. However,

T(r,h) <T(r,g)+T (r, f )=T(rg)+T(r,f)-logjf(O)j

by (13.4.1) and (13.4.2), so that h E AE, and f = g/h is the desiredrepresentation.

Conversely, suppose that A = AE/AE. Let Z have finite A-density.Then, by Theorem 13.5.3, there exists a function f E A with Z(f) = Z.We write f = g/h with g, h E AE. Then Z(g) is A-admissible, and Z(g) DZ(f) = Z, and we have proved that A is regular.

14The Miles-Rubel-Taylor Theoremon Quotient Representations ofMeromorphic Functions

Let f be a meromorphic function. In this chapter we describe the workof Joseph Miles, which completes the work in the last chapter concerningrepresentations of f as the quotient of entire functions with small Nevan-linna characteristic. Miles showed that every set Z of finite A-density isA-balanceable. As a consequence of this and the work of Rubel and Taylorin the last chapter, there exist absolute constants A and B such that iff is any meromorphic function in the plane, then f can be expressed asfl/f2, where f, and f2 are entire functions such that T(r, f;) < AT(Br, f)for i = 1, 2 and r > 0. It is implicit in the method of proof that for anyB > 1 there is a corresponding A for which the desired representation holdsfor all f. Miles' proof is ingenious, intricate, and deep. Miles also showedthat, in general, B cannot be chosen to be 1 by giving an example of ameromorphic f such that if f = fl/f2, where fl and f2 are entire, thenT(r, f2) i4 O(T(r, f)). We do not give this example here.

In the previous chapter, namely in Propositions 13.3.5 and 13.3.6 usingTheorem 13.5.2, we have obtained the above theorem for special classes ofentire functions. Results in this direction for functions of several complexvariables appear in [16], [17], and [10]. Quotient representations of func-tions meromorphic in the unit disk are discussed in [2]. The presentationbelow follows Mile [24].

We state the theorem.Theorem. There exists absolute constants A and B such that if f is anymeromorphic function in the plane, then there exist entire functions fl and

14. The Miles-Rubel-Taylor Theorem on Quotient Representations 79

f2 such that f = f, /f2 and such that T(r, f;) < AT(Br, f) for i = 1, 2 and>0.Suppose Z = {zn } is a sequence of nonzero complex numbers with jzn I -->

co. We include the possibility that zn = z,n for some n # m. As in theprevious chapter, let

(14.1) n(r, Z)IZn Kr

and

(14.2) N(r, Z) = f r n(tt Z)dt.

0

It was shown in the previous chapter that the following lemma is suffi fdent to establish the theorem.

Lemma. Suppose Z = {zn} is a sequence of nonzero complex numberswith IzzI -- oo. If A(r) = max(1, N(r, Z)), then there exist absolute con-stants A' and B' and a sequence Z = {-;n} containing Z (with due regardto multiplicities) such that

(1) N(r, Z) < 5A(4r) r > 0and,(ii) for j = 1, 2, 3.... ands > r > 0,

(;)'A'A(B'r) A'A(B's)

0 8

The argument of the last chapter which shows that this lemma is suffi-cient to prove the theorem may be summarized as follows. Without loss ofgenerality we may assume f has infinitely many poles and that f (0) 54 oo.Let Z be the sequence of poles of f . Recall from the last chapter thatcondition (i) of the lemma says that Z has finite density with respect tothe growth function A(r) and condition (ii) says that Z is balanced with re-spect to the growth function A(r). Let A1(r) denote an arbitrary increasingunbounded function defined on (0, oo).

In Theorem 13.5.2 we characterized the zero sets of entire functionssuch that T(r, ¢) < aa1(lr) for some constants a and Q and all r > 0as those sets Z* which both have finite density and are balanced withrespect to Al (r). This characterization combined with the above lemmaguarantees the existence of constants Al and B and of an entire functionf2 having zeros precisely on the set Z (counting multiplicities) such thatT(r, f2):5 A1A1(Br) for all r > 0. Hence,

(14.3) T(r, f2):5 A1N(Br, Z) < A1T(Br, f)


for r > ro(f). Letting fl = f2 f, we see that fl is entire and that T(r, fl) <(A1 + 1)T(Br, f) for r > ro (f ). For an appropriate complex constant c,0 < Icl < 1, we have for i = 1, 2 that

(14.4)

and

(14.5)

T(r,cf=)=0 r<ro(f)

T(r, cf:) < T(r, f;) r > 0.

Letting A = Al+1, we see that f = cfl/cf2 is the desired representation.It is implicit in the methods of the last chapter and in the proof of the abovelemma that A and B are absolute constants and that to each B > 1 therecorresponds an A for which the representation holds for all f .

We now prove the lemma. For each integer N we let

ZN=Zl{z:2N <IzI<2N+1}.

If ZN # 0, we relabel the elements of ZN as simply z1, z2,.. -, zk, witheach number being listed in this sequence as often as it appears in Z. For1 < n < k we define pn E (0,1] and 0,. E (0,27r], so that zn = 2N+PReieWe do not indicate in the notation the obvious dependence of k, zn, pn,and On on N. We let

k oo

(14.6) YN(8) _ -2 E E 2j(1+Pn)n=1 1i=1 I

and

(14.7)

Clearly,

(14.8)

hN(0) = RgN(B).

k o0

IhN(0)I :5 2 {2_ci+}Pn)n=1 j=1

< (n(2N+1 , Z) - n(2N, Z))

Letting

(14.9) fN(O) = hN(O) + 2(n(2N+1' Z) - n(2N, Z)),

we have

(14.10) 0 < fN(O) < 4(n(2 N+1' Z) - n(2N, Z)).


We now define sets ZN for all integers N. If ZN # ¢, we have from(14.10) that

(14.11) 0 < 12"

fN(O) dO < 4(n(2N+1, Z) - n(2 N, Z)).fo

If [ ] denotes the greatest integer function and LN = [Zx fo"fN(0) do],

we define for 0 < n < LN a monotone increasing sequence 9'n in [0, 27r] bychoosing 9'n to satisfy

rAn

(14.12) 2- j fN(O) dO = n.o

In addition, we let BLN+1 = 2ir. For 0 < n < LN, we let z'n =and Z' = {2N-lesen : 0 < n < LN}. As before, we do not indicate thedependence of 61. and en on N. If ZN = 0, we let ZN = 0 and for thatvalue of N do not define LN or numbers 9n and z;,.

We let Z' = UNZ'N and Z = Z U Z'. From (14.11) and the definition ofZ it follows that n(r, Z') < 4n(4r, Z) and hence that n(r, Z) < 5n(4r, Z)for all r > 0. From this fact it is immediate that Z satisfies condition (i)of the lemma.

We now consider a positive integer j and a value of N for which ZN ¢.Let ZN = {z1, z2,. .. , zk}. From this point until inequality (14.27), weregard j and N as fixed. Although many of the quantities to be defined(S, T, P, U0, UE7 V0, and Ve) depend on both j and N, for simplicity wesuppress this dependence from the notation.

A key step in showing Z satisfies condition (ii) of the lemma is to estab-lish

(14.13)

k

1 I< 48j2-i(N-1).

We first observe that(14.14)

k 1 1

+ -12,,

dOnZn 2if

k 1 21r k eii(8-e,.)e-sje2-i(N-1)2irJo 2i(l+v..)- (L)jn=1 n=1

k 1 i k2-i(N+p.)e-il(O.) = 0.z

d9


Thus (14.13) is equivalent to(14.15) ()I

1

f2lr-ij9 )(N-1) j(N-1)e 2 fN(B) d8 < 48j2E xn 21r

I Z" In

That the quantity on the left side of (14.15) is small can be seen intu-itively from the observation that the sum is essentially an approximatingR.iemannn sum for the integral. We first estimate

(14.16)

\ j - 2r r2 e-1792-7(N-1) fN(O) dB

=2t'_, ( I

= 2-j(N-1)112;,1=2N-,

The function cos jB decreases from 1 to -1 on [ix , mi 1 "] for m even,

0 < m < 2j - 1, and increases from -1 to 1 on [ *'!x , m' 1 x 1 for m odd,

O<m<2j-1. Letl,n= (Jx, "`+jlx) forO<m<2j-1. Definesto be the set of all n, 0 < n` < LN, such that [88, B'+1] is not containedentirely in some IT1f 0 < m < 2j - 1. Since each point mir/j belongsto [B'n, B,,+1] for at most two values of n, we see that S has at most 4jelements. Let T be the set of all n, 0 < n < LN, such that either n or n -1belongs to S. Thus T has at most 8j elements.

Let P = {0,1, ... , LN} - T. If n E P, it follows that [Bn+l, On] and[91L1 On+1] belong to the same interval for some m, 0 < m < 2j - 1. LetU,, be the set of all n E P such that [Bn+l, On] and [On, 0n+1] are containedin In for some odd m, and let U. be the set of all n -1 such that n E P andthe intervals [B'n_1, On'] and [On, Bn+1] are contained in Im for some even m.Certainly U,, fl Ue = 0, for if n E Uo fl UE7 then [B'n, Bn+1] is contained in I,,,for both an odd value of m and an even value of m. Letting U = U,, UUei weconclude that U and P have the same number of elements. Since 0 V P, wesee that -1 Ue and hence U C {0,1, ... , LN}. Thus, {0,1, ... , LN} - Uhas at most 8j elements.

If n E P is such that u E Ua, then cos jB is increasing on [Bn, Bn+1] andhence, from (14.12),

1

I fN(0) cos jB dB.(14.17) coS jAn <27r Bn

If n E P is such that n- 1 E Ue1 then cos jB is decreasing on [Bn+l, On'] andthus

(14.18) cos jOn < 1 r fN(0) cos jB d8.2w Bn-,

2x-

2afN(B) cos jB dBcos jOn I0


The definition of U together with (14.17) and (14.18) implies

,

(14.19)1

cos j0n < - j6.+ifN (0) cos j0 d8.

;,nEP nEU

It follows that(14.20)

IYn1=2N-1

_ ECosje,+E Cosje",nEP nET

60+2 8;.+- - ( fN(6) cos jO d9 - 2- fN(0) cos j9 dO

nEU21r e; O<n<LN Bn

niU

fBn+l

COSOn - E 1 J fN(9) cos jO dOnET O<n<LN

27r at.

nU<8j+8j = 16j.

We now obtain a lower bound forj21r

(14.21) - fN(0) cos j8 d8.2aIV .

Let Ve be the set of all n E P such that [9n_1, e'n] and [B'., n+1] arecontained in I,,, for some even m, and let Vo be the set of all n - 1 suchthat n E P and the intervals [On_ 1, On'] and [en, n+1] are contained in In forsome odd m. Using the same reasoning as before, we see that V. fl Vo = 0and hence, if V = Ve UV0, then {0, 1, ... , LN} -V has at most 8j elements.

If n E P is such that n E Vef then cos j0 is decreasing on [8n, Bn+1] and

(14.22) cos j8'n > 1fal.

gn+l fN(9) cos jB dB.2a

If n E P is such that n -1 E V0, then cos j8 is increasing on [8n_1, On] and

an(14.23) cosjen > a J

fN(9) cos j9 d8.Bn-1

NM (14.22), (14.23), and the definition of V,

+1fN(9) cos j8 d8.(14.24) cos j8'n >-

1

10.10'

2nEV

2AnEP

zx1

cos jO'n -2-

f fN(6) cos j9 d80


Thus(14.25)

1 2x

f fN(0)Cos jO dOCos j0, -21r

I z:,1=2 N -'o

_ cosjOn + cosjOn

nEP nET

-21r

ffN(6) cosjO dO - E 2 J A*+ fN(8) cos jO d8nEY0<n<Lv Bn

n V

1 Bnt:

fN(0) cos j8 d8> cos jAn - -fon'nET 0<n<LN

2nqY

> -8j - 8j = -16j.

Combining (14.16), (14.20), and (14.25), we conclude that(14.26)

J

f2-7+je -j(N-1) j(N-1)e- 2 fN(8) dB < 16j2-

1=2Nzn

27r -

The same discussion applies to the imaginary part of the above quantity.The only minor modification is that we must divide [0,2w] into 2j + 1subintervals on which sinjO is alternately increasing and decreasing. Since2j + 1 < 4j, this causes no difficulty. We obtain

27r

fzxe-'j''2-j(N-1)fN(0) dO < 32j2-j(N-1).

Combining (14.26) and (14.27), we obtain (14.15), which in turn establishes(14.13).

We are now in position to show that 2 satisfies condition (ii) of thelemma. Suppose that s < 8r. We then have trivially for all positiveintegers j

(14.28)

1 (i)1<n(sZ),

z' - rj7 r<Iznl<e n 7

< n(8r, Z) < N(8er, Z)<

5A(32er)jrj - jr.? jr'

Suppose that 8r < s. In this case there exist integersq2 - 3, such that

(14.29) 2q' < r < 2q'+1 < < 2 q' < s < 2q'+i

q1 and q2, q1 5


Thus, for(14.30)

9

YT<l<8 'Zn <IZn9, +4 ( 'Znq2-ql-1J1 1 1

+ +-v=2 2vl+°<Iznl<2q,+..+1 Zn Izn' l=2°l}D-'

zn

11 11 )i + 1 1

2v2 < zn1<s xn 3 lz;.i=2q2-.

xn

(;:)}+I 1.11=2q2

1

(n')jlWe remark that if Zq,+ = 0 (and hence Z,,,+,, = 0) for some integer v,then the terms corresponding to that value of v in (14.30) are of courseomitted.

For any positive integer j we have

(14.31)

1 1 3 < n(4r, Z)77-

< +2 z++ jr<Iz, <2r1

N(4er, Z) <) (4er)jr' - jri

From (14.13) we have

q2-ql -1

(14.32)

Ill

1 1-z

+z7

241+V V

<Iz,.1<2Q1+ +l n z, 2qi+.-1 "1z'

q2-qt-1< 48 E 2-.i(ql+Y-1).

=2

Also from (14.13) we have(14.33)

I1 1 i 1 1 J+ - E !

2v2<Iznl<a ZnI41=2°2-1

Zn

3

7

Un)1 1

<Iz..l<<2 2 IZ,.l=2 7 s<Iz,.l<<2v2+1 Gn)n 2q2+1 Z

< (48)2-i(q2-1) + (j8i'

) < (48)2-i(q2-1) + ads).


Finally,(14.34)

1 (1\j

IZr 1=2v2xn

iz z

1 1

2az+'<Iznl<2az+z nIz;.1=2°z Y2az+i<I=,.I<29z+z n

(48)2-i(92) +n(2vz+2, Z)

< (48)2-j(9z) +A(4es) .

jsi jsi

Combining (14.30) through (14.34), we have

(14.35)

1 (i)'!<Ian1<d

< L(4+(48)

9z-9t+12-j(92+Y-1)+2A(4es)jrj E jsi

< A(4er) + 96 + 2A(4es) < A(4er) + 96A(r) + 2A(4es)

jrj jsi jr' ri jsi

From (14.28) and (14.35), we see that

(14.36)1 1

x

< A'A(B'r) A'A(B's)rj + Si

for all positive integers j and all 0 < r < s if A' = 100 and B' = 32e. Thiscompletes the proof of the lemma.

15Canonical Products

We shall suppose that a countable set Z = {zn} of complex numbers isgiven. For convenience, we shall suppose that 0 0 Z and that Z has nofinite limit points. More generally, we consider "sets with multiplicity."This means that some of the zn may be counted multiple times. It ispossible to make this notion rigorous, but at the price of clumsier notation.For convenience in this chapter, we shall exclude the null function F(z) = 0from consideration.

Definition. If F is an entire function, we write Z(f) for the set of zeros(other than the origin) of F.

Definition. If p is a nonnegative integer, we define E(u,p), the Weier-8tnass primary factor of order p, by

// \up/E(u,p)=(1-u)explu+ u2

2 p/E(u,0)=1-u.

Lemma 15.1. Given any Z, there exist nonnegative integers An suchthat the product

lf(z) = TIE ( - 'An 1

c1n

converges uniformly on compact sets to an entire function f such thatZ(f) = Z.

We omit the easy proof which follows directly from Estimate A later inthis chapter. Any sequence An such that An --i oo will work. The nexttheorem is a corollary of Lemma 15.1.


Weierstrass Factorization Theorem. Given an entire function F, thereexist nonnegative integers An, a nonnegative integer in, and an entire func-tion g such that

00F(z) = zm exp(g(z)) II E ( n , An/n=1

Definition. Given any Z, the convergence exponent p1(Z) is defined by

P1(Z) = inf{a : EIznl-" < 00}-

Lemma 15.2. If E z , < oo, then n(t) = o(t"), where n(t) _ 1.Izn'St

Proof. Choose a large number a, and write

E 1 > n(t) - n(a)" - t"IZnI

Lemma 15.3. E . 1 a < oo if and only if f 0m i, t dt < oo.

Proof. Write1 =

Jt-" dn(t)

Iz= Iznl" 0

and integrate by parts to prove the lemma.

Corollary. pi(Z) = inf{a : n(t) = O(t")}.

Lemma 15.4. p, (f) <_ p(f)

Proof. We must show that if p = p(f), then for each e > 0, n(t) = 0(t°+`)But by the Nevanlinna first fundamental theorem, we know that

N(t) = J t n(s) ds = 0(t°+').0 s

But N(t) > n (2) log 2, by the usual argument, and the result follows.

Definition. The genus of Z, p = p(Z), is defined by p(Z) = inf{q : q isan integer, E zj ; < oo}.

Definition. The canonical product of genus p over Z is defined by

P(z) H Ez

+P

z EZ n

15. Canonical Products 89

Definition. The canonical product PZ over Z is defined by

zPZ(z) = [J E - P

ZnZ

where p = p(Z).

znE

Theorem 15.5. The order of Pz is pl(Z).

The next result is a corollary of Theorem 15.5.

The Hadamard Factorization Theorem. Given an entire function f,

f (z) = PZ(z)zm expQ(z),

where m is a nonnegative integer and Q is a polynomial of degree < p(f).Here, Z=Z(f).Deflnition. The genus of the entire function f is defined by

P(f) = max(n,p),

where n = deg Q.

Examples. Suppose zn = n2, n = 1, 2,3,... . If f (z) = rj (1 - n ), thenf is of genus 0. If f (z) = ez rj (1 n ), then f is of genus 1.

We first shall show how the Hadamard Factorization Theorem followsfrom Theorem 15.5. Without loss of generality, suppose f (O) 96 0. In fact,assume f (O) = 1. Let

9(z) = f(z)PZ(z)

Then g is an entire function without zeros, so that there exists an entirefunction Q such that

z = exp Q.

We must prove that Q is a polynomial of degree < p(f).

Lemma 15.6. If F and G are meromorphic, then p (c) < max{ p(F), p(G)}

Proof. We have

P(f) = lim sup log T(r, f )r_.oo log r

T `r, 1) T(r,f)+O(1)T(r, fg) = T(r, f)+T(r,g)+D(1),


from which the lemma is easily proved.

It then follows that exp Q is an entire function of order at most p. Writ-ing Q = u + iv, we have

Iu(Te'B)I dO = 0(r°)

for each p' > p. Since, with IzI = r and R = 2r, we have

w + zuw

Q(z) - Im Q(O) = 2 i ,u('w)

it follows that

Hence

IQ(z) - Im Q(0)I = 0(r°)

IQ(z)I = 0(r,"),

and it follows that Q is a polynomial of degree < p.

Proof of Theorem 15.5. Our proof uses the Fourier series method of Chap-ter 13. It is shorter and less tedious than the standard proofs.

Estimate A. If Iznl > 21z1, then

log E (.!n , p) 1:5 21

n

IP+1 < IIP.

Let u = z- . Now log E(u, p) E +1 k . Here Jul < a. Hence

00 00

I log E(u,p)I <_ E Iulk < Iulp+1 E 2-k = 2Iut"1.P+1 0

It follows from Estimate A that the product

converges uniformly on compact sets to an entire function f, so thatlog I f (z) I = E log I E ( p )1We write

00

log If(z)I = E c,, e'-O,

m=-00

15. Canonical Products 91

where Cm = c,,,(r) is the mth Fourier coefficient of log If I. We have seen inLemma 13.4.1 how to calculate c,n:

/ \/1klogE l

xP

1 r.zn ) = k

k_=p,+l

(xz

n

so that

Hence

00 00 1z

log f(z) _ > - > k (zn

)k)n=0 k=p+1

1 0 fork<pan =j

1 -21On E* fork>p+1.

Recall the notation from Chapter 13; near z = 0

00

log 1(z) = Eakzkk=0

Using the formulas in (13.4.18), we therefore getrnW co = >2 log r

ISn I <r

1 (r)ml l(n)=2m u (rI ifm<pISnI<r I.. 1<-r

-Tm 1 n l m

(m) cm = 2m ` ()mL>2 \ r J if m > p + 1.ISnI>r ISnICr

Choose p' > p. We must prove that for some M = M(p')

Ic,n(r)I <- Mrp

ImI + 1form=0,1,2,....

But this estimate is easily derived, as in Chapter 13, from the fact thatn(r) = O(rP ), once we know that the cm are given by formulas (i), (ii),and (iii).

The next theorem follows easily from the Miles-Rubel-Taylor Theoremof Chapter 14 but is included here due to its simple deduction from Theo-rem 15.5.

Theorem. If f is a meromorphic function in the plane with p = p(f) <00, then there exist entire functions g and h, with p(g) < p and p(h) < p,such that f = g/h.

Proof. Let W = {Wn} be the poles of f, let pi = pi (W) be the exponentof convergence of W, and let p = p(W) be the genus of W. Since N(r, f) <


T(r, f) = O(rP+E) for each e > 0, we have pl < p. By Theorem 15.5,p(Pw) = pi. We now let g = f Pw and observe that g is entire. Sincep(g) < max{p(Pw),p(f)} = p, it follows that f = g/Pw is the desiredrepresentation.

It was proved by similar means by Rubel and Taylor that if A(2r)/A(r)0(1), then every meromorphic function of finite A-type is the quotient oftwo entire functions of finite A-type. However, the proof is long and issubsumed in Miles' proof of the last chapter, so we omit it.

Laguerre's Theorem on Separation Zeros. If f is a nonconstant en-tire function with only real zeros, has genus 0 or 1, and is real on the realaxis, then the zeros of f' are real and are separated by the zeros of f, andthe zeros off are separated by the zeros of f'.

Proof. We have either

f (z) = CzKeaZII 1- x

xn

or

f(z) = CzKe4zII 1 -xz ez/Z",xn

where the xn are real, and c and a are real (possibly a = 0).It follows that either

f'(z) =k+ a + 1

f(x) z Z - Zn

or

f'(z) = k -F a + Ez

f(z) z zn(z - zn)'

On writing z = x + iy and recalling that a is real, we get, in both cases,

I.Rz)_-y{k + 1 1f(z) x2 + y2 (x - zn)2 + y2

which does not vanish except for y = 0, so that the zeros of f' are real. Sincef is real for real z, the theorems of calculus apply. By Rolle's Theorem,there is a zero of f' between two consecutive zeros of f, so that the zerosof f are certainly separated by the zeros of f'. To see that the zeros of f'are separated by the zeros of f, note that

k(x) =

x2 - (x - xn)2< 0,

so that /LLI is decreasing in any interval free of zeros of f, so that f'cannot have two zeros in any such interval. The case of repeated roots ishandled by a suitable convention.

16Formal Power Series

We consider the formal power series

f(z) = ao + a1z +a2z2 + ...,

which we usually normalize by ao = 0.Let fn(z : a) = a + a1z + a2z2 + + anzn; fn is a polynomial of

"degree" n (possibly an = 0). We adopt the convention that fn has n zeroszi, z2, ... , zn, where, if am # 0 but am+1 = am+2 = ... = an = 0, then2nn+1 = Zm+2 = - - - = Zn = oo. For later use, we shall make the conventionoo/oo = 1.

Let rn(a) = max{jzI : fn(z : a) = 0}, that is, rn(a) is the modulus ofthe largest root of fn(z : a) = 0. Note that if an = 0, rn = oo. In a certainsense, rn(a) measures the disaffinity of f for the value -a.

Theorem 16.1.

(16.1)

if b = 0, we have

where

Given any formal power series f , then

lim suprn(a)

< 2, (b 0 0);n-oo rn(b)

p rn (a) p ( 1 1 "-"lim su < 1 + lim su

ILn-.oo rn(0) n-»oo rn(a)

f (z) = amzm + am+l zm+l +..., a. # 0.

Note that if the roles of a and b are interchanged, then (16.1) yields< lim inf rn (a)n-.

94 Entire and Meromorphic Function

Proof Let al, a2, ... , an be the zeros of fn(z : a) arranged so that (a1 I <Ia2I < Ianl. Let be the zeros of fn(z : b) arrangedso that IP'I < I02I < < IP.I. Thus fn(z : a) = anH(z - ak) andfn(z : b) = anH(z - Pk), where we shall take an 0.

Now,

(16.2) fn(z : b) - fn(z : a) = b - a = an[II(z - Pk) - Il(z - ak)]

Suppose there exists a sequence of n for which {an}, {fin} is such thatIan I = An IPn with An > A > 1. Otherwise, the conclusion of the theoremis clearly true. Therefore, rn(a) = Anrn(b). Now,

Ian - Pkl -Ianl-IPk1=Anl/nl-IPkl - (An-1)IPkl.

Letting z = an in (16.2) gives b - a = anH(an -,Ok). Hence,

Ib - al >- Ianllrr(an - Pk)l >- IanIfI(An - 1)IPkI-

Dividing through by b and taking nth roots gives

an [i'n(A_1)IflkJ] n1->[

Now suppose we had An - 1 > 1. Then

a n Ianl nL

I1-bil>[IbI(1+E)IIIPkI,".

But rrlPkI = a so that supposing An - 1 > 1 implies I1 - e ° > (1 + E);a

a contradiction. Hence, lim sup An < 2, so that we have lim sup r. b) < 2.rn(n-ooSuppose now that b = 0. Write f (z) = amzm +an+lzm+1 +..., am # 0.

n-mWe now have fn(z : b) = zman jj (z -,6k). Thus, fn(z : b) - fn(z : a)

k=1n-m n

-a = anlzm fT (z - Pk) - f1 (z - ak))k=1 k=1

n-mWe again have Ian-PkI = (An-1)IPkj and I-al = Ianan f1 (an-Pk)I

k=1n-m

lanl[rn(a)]m fl (An - 1)IQkl:k=1

_ rr n-m ^-*"(16.3) Ial Llanlfrn(a)ln` H (An -1)IPA; I

L k=1

16. Formal Power Series

Now suppose for contradiction thatn-m m

\l ml (Thenthere exist {En} such that En > 0 and

An - 1 = lal(^(laml)n-m

(rn(a)mss-

(1+En).

n-mUsing this expression in (16.3) and that n 113kl =

k=1

lal ^ m >lal n'tn (1 + En)n-mi

a contradiction.Hence,

S-I gives

n m2 1 - 1An-1<lal()

laml rn(a)Consequently,

95

lim suprn (a)

< 1 + lira sup 1 ,rn(0) n-00 rn(a)

which completes the proof. Theorem 16.1 has the following corollary.

Corollary. If f is holomorphic in a neighborhood of 0 (that is, f has apositive radius of convergence), then

limsuprn(a) < 2n-.oo rn(b)

for all b.

Definition. Tn (f) _ - log rn (1) is called the discrete characteristic of f.

Theorem (First Fundamental Theorem) 16.2. Tn(f) =Tn(f -a)+O(1)for all a with at most one exception.

The proof is immediate from the definition of Tn (f) and Theorem 16.1.Now let us examine the case of our exception more closely,

mlim sup rn(1) < 1 + lira sup 1n-.oo rn(0) n-+oo rn(1)

Suppose fn(z : 1) = 1 + a1z + + anzn has zeros l3n arranged inorder of increasing moduli. Thus, anlI(-Qk) = 1 and hence IIlakl =Accordingly, [rn (1)]" > r , and therefore

1 ^ n-m 1 n-M --[rn(1)]n-M > (-)lan1or :5(-) (lanl^) ^ m

which gives us the next result.


Corollary. If f is holomorphic, then Tn(f) = Tn(f - a) + 0(1) for all a.

Theorem (Kakeya) 16.3. Let f be a formal power series not identicallyzero and R(f) its radius of convergence. Then R(f) < limn rn < 2R(f

n-oo

Proof. Choose R' < R(f ). The fn have only a finite number of zeros in thedisk DR, and { fn(z)} - { f (z)} uniformly in DR.. Hence, past a certainno we have, by Hurwitz's Theorem, that the functions fn have the samenumber of zeros in DR, as f does. Therefore, the largest zero of fn mustleave DR.. Thus R(f) < lim inf rn, as desired.n-oo

Alternatively, we have seen that rn > Ian! w and hence liminf rnn-oo

R(f).To prove the second inequality in the theorem, we use the following

result.

Lemma. Let P(z) = bo + b1z + - - - + bnzn = 0. Then

IzI < 2max Ibb ll, Ibzn21 ,...,16(

Proof of Lemma. Clearly, IbnllzIn < Ibol+Ibol+IbiIIzl+---+Ibn-1IIzIn-1.

Writing IzI = r, w e h a v e rn 2 m a x (Ib^ I ' I

bÎ ' . . . ' I I

Then r < 2 r +2-(n-1)rn + + 2-1rn = rn(a +1 +... + 2;,) < rn, which is impossibleand the lemma is proved.

Now let bk = akRk, P(z) = fn(Rz) = ao + a1Rz + - + anRnzn. Thenthe roots of fn(Rz) = 0 are times the roots of fn(z) = 0. By the lemma,

10-1n < 2 max I an-I.R - ( Rn

anRn anRn ,.la"_2Rn-zl,...,l as Il

Choose R> R(f); it follows that {IanIRn} is unbounded. Therefore, for asuitable subsequence, IanIRn > IakIRk for all k < n. In that subsequence,

< 2. Hence, lim inf rn < 2R. Therefore, lim inf rn < 2R(f), as was ton-.oo n-oo

be proved.

Corollary (Okada (28)). A function f is entire if and only if lim Tn(f) =n-oc-00.

16. Formal Power Series 97

eorem (Tsuji [45]) 16.4.

I- en a(f) = P(f)

Let a = o(f) = lim suplog n =1im sup

log n

n.oo Tn n-.oo log rn

proof. We know that R(f) = lim supn log n

110g Ian

I

from Proposition 11.4. Take

pl > p; then, for large n, p' >n log1

Hence, IanI < 1-fl and thuslog

lanlnp,

to > R > n'' , or rn > n o1r. Consequently, o < p' for large n;hence, a no ), rn > nor . Choose

2K-

M > 1 so that for K = 1,2,..., no, laid < m (K1) We prove by

2ninduction that for n > no, IanI < m

(n!) °First,

Ianlrn < Ian-l lrn 1 + ... + Iallrn + 1

since &(z : 1) = 1 + aoz + + anzn = 0. Hence,

IanI < Ian-1I

n

+Ian-2Ir 1 + rn n

< Ian-1l + Ian-2l + ... + tall 1+ s}n nT n°

Therefore,

20Ia.l<_M

2n-1 + 2n-2 +...+ 2' + si-n° [(n- 1)!]° n° [(n- 2)!] n ° (1!)° n° (0!)

< M2n-1 2n-2 + 2'

+20

(n!)°7 + (n!)°+

(n!) I' (n!

< M <M 2n ,

(n!)° - (n!)

as desired. This now leads to a contradiction of the fact that f is of order

p, for we have Ta7 - M 2^° Consequently,

109 I1I ? constant + log n! - n log 2an


so that

n log n c n log n _,log constant + p, log n! - n log 2 - p + 0(1)

since log n! ti n log n. This implies that p(f) < p' < p = p(f ); a contradic-tion.

17Picard's Theorem and the SecondFundamental Theorem

In this section we shall prove Picard's Theorem, state the second funda-mental theorem of Nevanlinna (leaving the proof for the next section), andderive some of its consequences.

We shall use two conventions that will greatly simplify our notation.First, we shall always work "modulo 0(1)". For example, A = B andA < B shall mean that A - B is bounded and that A - B is boundedabove, respectively. Second, we shall use a notation of Weyl, writing 11 infront of a statement to mean that the statement holds with the possibleexception of a set of finite length.

Picard's Theorem. If f is a nonconstant meromorphic function on thecomplex plane, then f does not omit three values on the sphere.

It may happen that f omits two values. For example, ez omits 0 and 00.Our Proof of Picard's Theorem is patterned after our proof of the secondfundamental theorem of Nevanlinna, and illustrates the main features ofthe method in a simpler context.

Proof. Without loss of generality, we may assume that f omits 0, 1, andoo, so that f, 1, and 1 are entire. The next lemma is referred to as thelemma of the logarithmic derivative.

Lemma. 11 T (r, f) <- K' log r + K" log+ T(r, f) for f omitting 0, 1, andoo.


From Poisson's formula (7.1),

I 7r vlogf(z)=- I logIf(pei )I ;±zdco;

21r .,

f() 27rf xlo If(Pe"°)I(pe;x

If (z)f(z) (p? r)2 2w f xlogIf(pe"°)IdW

log+ I f'(z) I< log+ p + 2log+ p1 r

+log+ 2 J loglf(Pe"°)IdW (modO(1)).

Recalling the definition of m(r, f), we find

f 1 .log+f (z) I <_ log+ p

+ 2 log+p

1

r + log+ m(r, f) + log+ m (r,

Since f 0 0 or oo, we have m(r,f) = T (r, f) and m (r, f) = T (r, f) .Moreover, by the first fundamental theorem of Nevanlinna, we have

T(r,f)=T (71) .

Hence,

log+ I f()I< log+ p+ 2 log+ p r+ 21og+ T (p, f).

Since the right-hand side is independent of the argument of z, it followsthat the same estimate will hold for the average,

m I r, f I< log+ p+ 2log+ p 1 r+ 2log+ T (p, f ).

Now choose p = r+ logT f)fi and use the Borel lemma on 2 log+ T(p, f)to find,

T (r, fi ) < log+ r + 2 log+ log+ T (r, f) + 3log+ T (r, f)

f) .< log+ r + 4log+ T (r,

17. Picard's Theorem and the Second Fundamental Theorem 101

This proves the lemma.Now let

F(z) -1 f(z) f'(z) + f'(z)

f(z) f'(z) f(z) f(z) - 1

IF(z) I is large whenever f is near 0 or f is near 1. Of course, the set of values

of z where these occur is disjoint: hence, m(r, F) > m(r, f) + m (r,ff1 1) .

Therefore,

(17.1) T(r, F) > T I r, 1) +T (r, 1 I since f 54 0, 1.f f-1/Notice that

Thus,

(17.2) T(r,F)<T(r,f)+T(r,f')+T(r, )+T(r, ff'j).From the first fundamental theorem we have

TIr, f)=T(r,f)=T(r,f-1)=TIr, f1 I

and

T (r, T'/=//lr,Therefore, on combining (17.1) and (17.2\\) we obtain

2T(r,f)<T(r,f)+2T(rL) +TI r,fConsequently,

(17.3) T(r,f)<2TIr, f)+T(r, ff ).The lemma above shows that

TIr,f I < k'logr+k"logT(r, f).

Applying this lemma to f /and f - 1 in (17.3) givesT(r, f) <k'logr+k' log T(r, f).

Of course f is not a rational function since it omits three values. Conse-quently, by Theorem 10.2,

T(r'f) -iooas

But we have

1F(z)

+f(z) A z) - 1

log rr -ir

II T(r,f) <k'+

k" log+T(r,f)log r log r '

which is impossible, and the proof by contradiction is complete.


Theorem. (Second Fundamental Theorem of Nevanlinna). Suppose f (z)is a nonconstant meromorphic function in the plane (or in Izi < R, R <oo). Let a1i a2, ... , a4 be q distinct finite complex numbers. Then

(17.4) m(r,f)+>2rral r, f 1 ) <2T(r, f)-N(r, f)+s(r),v=1

where

and

N1 (r, f)=N(r, ' ) +2N(r,f)-N(r,1')

S(r)=mlr,f +m r,> f\\ f! i_1 f - av

Interpretation. It will be shown that

11 S(r) = O(log T(r, f))+O(logr).

Accordingly, we may think of "S" as standing for "small." To interpretthe function N, (r, f), suppose f (z) = (z - za)Kg(z), K E Z, g(z) # 0in some neighborhood of zo. We analyze the contribution to N1 due tothe behavior of f at zo by examining the counting function n1(r, f) =n (r, t, ) + 2n(r, f) - n(r, f'). If K > 0, then f vanishes at zo to order K.Consequently, n(r, f) and n(r, f') are unaffected by the behavior of f nearzo. The contribution to n1 near z equals the contribution to n (r, f). Of

course, j has a pole of order K - 1 at zo. If K < 0, then f has a pole oforder -K at zo and a similar analysis reveals that n1 "counts" this poleK - 1 times. The case K = 0 is trivial. In general we find that poles oforder K and zeros of order m are "counted" by n1, respectively, K-1 andm - 1 times.

The gist of the second fundamental theorem is that for most values of a,the contribution of m (r, ffl a) to T (r, f 'a) is much smaller than the con-

tribution of N (r, f 1) to it. Thus, for most values of a, N (r, f11) makes

the preponderant contribution to T (r, f Q- . Recall that T (r, f 1 a) -T(r, f) = 0(1) by the first fundamental theorem.

Remarks. The following weaker form of the theorem also gives us the PicardTheorem:

(17.5) m(r,f)+Emlr, 1 } <2T(r,f)+S(r).-1 \\\ f -ate/


For suppose f omits three values, say 0, 1, and oo. Then, since f is entire,m(r, f) = T(r, f) and, since f omits 0 and 1, we have

m(r,f T (r, f ) = T(r, f)

and

in (r, 1 J = T (r, 1 J = T(r, f - 1) = T(r, f )f-1/Thus (17.5) implies that 3T(r, f) < 2T(r, f) + S(r) and hence T(r, f) <S(r)-

By a lemma to be proved in the next section we assert that

m (r, f) < 61og+ T (r, f) + 41og+ r.

Hence,

and thus

11 S(r) < 6(q + 1) log+ T(r, f) + 4(q + 1) log+ r

T (r, f) < S(r) implies that limT(r' f) < 00,

r-oo 109 rwhich is only true, by Theorem 10.2, for rational functions. Since f 96 00,f is a polynomial and therefore does not omit three values; a contradiction.

Consequences

Definition. Let n(t, f) be the number of poles of f in the closed diskDt = {z E C : IxI < t} counted once, no matter what the multiplicity. LetN(r f) = fr n(t,f)-n(e,f)dt.

o+ t

1 1m r, N (r,Definition. b(a) = limr

T(r,f) a

- 1 - limr_ 00 T(r f)a

6(a) is called the deficiency or defect of the function for the value a.

Definition. 9(a) = licnr N(r,

f 1a) - N (r, f 1 a)T(r, f)

Definition. The branching index is defined to be 0*(a) where

1 1N r, lr, JN@'(a) = 1 - limn..oo T(r, f)

a = 4M,-. 1 - \T(r f) a


What contributes to 9* (a) is a-points that are taken on by f with multi-plicity greater than or equal to 2. So a value a that is taken on often withhigh multiplicity will have a large branching index.

Remark. 0*(a) > 9(a) + b(a). Then,

9*(a)-r m I1- NJ = urn f N-1VT 1-NJN]00 L T r-.°° l

m> lir-oo

m INTNJ+li [1 TN1

= 9(a) + 5(a).

The next result follows from the second fundamental theorem.

Theorem. Let be any finite collection of distinct complex numberspossibly including oo. Then,

(17.6) 2.

Proof. First suppose f is not a rational function. From the second funda-

mental theorem we have m(r, f) + q m (r, iaV) < 2T (r, f) - Nl (r, f) +V=1

q

S(r). Adding N(r, f) + > N (r, 7.7t o both sides, we get:

(q + 1)T(r, f) < 2T(r, f) + N(r, f) + EN \(r, f 1 ) - N, (r, f) + S(r).V=1 - aV

Hence,

(q-1)T(r,f) <EN (r, 1 -N (r, f,)+N(r,f')-N(r,f)+S(r).

f-atev=1

Wherever f takes the value aV with multiplicity k, n (r, i ) -n (r, f,)counts 1. Also, at a pole of f, n(r, f') - n(r, f) counts 1. Therefore, wemay write

(q - 1)T(r, f) < N (r, i-/ + N(r, f) + S(r).v_1 f - a

Hence,

q N (r, ) R (r, f)(17.7) (q - 1) 5 E T(r, f)y + T(;7) +,.-i 7'(rrf)


Now f is not a rational function, and therefore lim,- , T r = 0 sinceIIS(r) = O(log T(r, f)) + O(log r) by Lemma 18.3 in the next section.

Now, using the fact that lim (A + B) < Em A + urn B, we have

qN(r, f lam(4-1) lim T(r,f) + LI o

N(r,

from which

v=1

1

- rQ lim N(r,T.

-lim N(r, f) < 1 - q.L T(rf) T(rf),,Adding q + 1 to both sides gives

=1

q

11- lim NT(rf)J

+ 1 -p

< 2.L

,

Hence, E 2 holds if f is not a rational function.V=1q

Now if f is a rational function, the same claims hold up to (17.7). Wemay suppose If (z) I - IzI', where p is a nonzero integer. Then T(r, f) _[PI log+ r. Also,

/ , 9 , 4 zyS(r)=m[r, f+m(r,Ef f =m(r,p)+m(r,t p V-a,, z i=1 z - a\\ /// \` ff `Now m (r,

P)= 1 f,, log+ I P I dO and, for IzI sufficiently large, m (r, P) _

27r ZQ p-1

0. Similarly, for IzI large, we have m (kr,E = 0. Therefore, itv=1

follows that lim Tarf = 0, and the rest of the proof follows as before.

Corollary. Let be any finite collection of complex numbers possiblyincluding oo. Then,

v

2.

V=1

Definition. a is called a deficient value of f if b(a) > 0.

If f never takes on the value a, then b(a) = 1. The same is true if f takeson the value a very infrequently. In general, b(a) measures the tendency off to omit the value a and 1 - b(a) measures the tendency of f to take onthe value a.


Corollary. There are at most countably many deficient values.

Corollary. There are at most two values for which 6 > 3 .

Definition. a is said to be a fully branched value of f if f takes the valuea with muAiplicity either zero or > 2.

Remark 17.1. If f is entire, f can have at most two fully branched valuesand this is the best possible, since, for example, ±1 are fully branchedvalues of sin z. Also, if f is entire, then there are at most two values of afor which 6(a) > 2.The reader will find it instructive to work out b(a), 9(a), and 9*(a) forsome specific functions, such as eZ, tan z, and the Weierstrass p function.

Remark. If a is fully branched, then

1 1

9* (a) =1im 1 - NTr(r,f)

> lim 1-N

Nf -a since T > N.

Also, if a is fully branched, then

9*(a) > 2. In general, there are

a meromorphic function.

N(r, f ia) <2, and

1

N(r,fa)therefore

at most four fully branched values for

The next result shows that Nevanlinna theory may be used to study cer-tain types of exponential identities. As an example, consider the functionalequation

of + e9 = eh,

where f, g, and h are entire functions. Do there exist f, g, and h so thatthe identity holds? What is the relationship between f, g, and h? We maywrite

of-h + eg-h = 1.

Then the entire function of -h omits 0. But eg-h omits 0 as well, thereforeof -h must also omit 1 for the identity to hold. By Picard's Theorem, ef-hmust be a constant so f = h+ const. Similarly, g = h+ const.

Lemma (Hiromi-Ozawa [15]). Let ao(z), a, (z),... , an(z) be meromorphicfunctions and let g1(z), ... , gn(z) be entire functions. Further, suppose that

n `T(r,ai) = o Em(r,e9°)J j =0,1,...,^

v=1

holds outside a set of finite logarithmic length. If an identityn

2av(z)e9I(=)

= ao(z)

V=1


holds, then we have an identity

n

E c.a.(z)esvW = 0V=1

where the constants c,,, v = 1, ... , n are not all zero.

Proof. On writing f' = f f, we may use the lemma on the logarithmicderivative to estimate T(r, f) when f is entire. For notational simplicity,let Then we have

n

(17.8) E G (z) = ao(z).V=1

By differentiating both sides we obtain

n

(17.9) G(") (z) = a( 0")(z),

which may be rewritten as

ng)

(z) = a(µ) z - 1, ... , m - 1.(17.10) >2 G., (z)(z)

o (), p -V=1

We regard this as a system of simultaneous linear equations in the G.Now we have

G(IJ) (z) = P,(a,,,a;,,...Ia(" ), (µ))eg.,(z)

with a suitable polynomial P, in the indicated functions. Thus we have

(m(re9))(17.11) TY-1

outside a set of finite logarithmic length.Suppose, for the simultaneous equations (17.8) and (17.10), that the

determinant A 34 0. By solving (17.10) with respect to G j = 1, ... , n wehave by Cramers' rule

where

0=

1 ... 1

GI/G1 ... G,1,/Gn

Gin-1)IGn . _ _ G(n-1)IG.

108

and

Entire and Meromorphic Functions

1 ... 1 a0 1 ... 1

Gj-1 lG .1G1 ... Gj-1

a0 Ci+1

....G

Aj=

Gj_-1_ Gin l1, (n-1) ___ ____GI

... GI_' a0 Gj+t ... Go

Since

(17.12) T (r, C I = o m(r, e9) f ,

we have

n

T(r,A)=0 m(r,e9°) ,T(r,Aj)=o( m(r,e9°))=1

for j = 1, ... , n outside a set of finite logarithmic length. Thus we have

m(r,e9") =T(r,e9°) <T(r,a.)+T(r,G.)

o (m(r1e9)v=1

and hence r (Em(re9))!2m(r,e) =oV=1 v=1

outside a set of finite Lebesgue measure. But this is a contradiction. Con-sequently, the Wronskian 0 =_ 0 and the result follows.

We say that two meromorphic functions f and g share the value a (a = 00is allowed) if f (z) = a whenever g(z) = a and also g(z) = a wheneverf (z) = a, counting multiplicities in both cases. A famous theorem ofR. Nevanlinna, which will be proven shortly, implies that if two nonconstantmeromorphic functions f and g on the complex plane share five distinctfinite values (ignoring multiplicity), then it follows that f = g, and thenumber 5 cannot be reduced. We consider here the special case g = f', thederivative of f, and prove the following result.

Theorem. If f is a nonconstant entire function in the finite complexplane, and if f and f' share two distinct finite values (counting multi-plicity), then f' = f.

In other words, a derivative is worth two values. We show at the endof the chapter that the number 2 of the theorem cannot be reduced. We


do not now know whether there is a result corresponding to our theorem ifone ignores multiplicities, or if one considers meromorphic instead of entirefunctions.

Proof of Theorem. To fix the ideas, we suppose that f and f' share thevalues a and b, where a = 1 and b = 2. Other choices of a and b make noreal difference, except if a or b is zero, in which case the analysis becomeseasier, and is left to the reader. We may write then

f'-1 k(17.13) 'f-1

(17.14)f' - 1 _ eke '

f-2where k1 and k2 are entire functions. We solve (17.8) and (17.9) for f toget

(17.15)1 + ekl - 2ek2

f = ek1 - eke

We now differentiate both sides of (17.10) and substitute (17.8) to get(17.16)

2k' 2k2 + kl+k2 - 2k1+k2 k1+2k2 k' - . k22e +e +(k2-ki-3)e a +e +kle k2e = 0.

We shall make repeated use of the lemma of Hiromi and Ozawa.Now, in (17.16), divide by eke to get

(17.17) 2e2k1-k2 + eke + (Jc2 - k' - 3)ek - 22k' + ek'+k2 + k2.

We now apply the lemma to get(17.18)

ele2kl-k2 + c2ek2 +c3(k' - k' - 3)ekl +c4e2k' +c5ekI k2 +csklek'-k2 = 0,

where c1, ... , cs are constants that are not all zero.The hypotheses of the Hiromi-Ozawa Lemma are satisfied because, for

example, ki is the logarithmic derivative of ekl and we may use the lemmaof the logarithmic derivative. It follows that T(r, k') = 0(T(r,ek)), outsideof a suitably small exceptional set. At any rate, we divide in (17.17) by ek1

to get

(17.19) clek'-k2 + c2ek2-k' + c4ek' + c5ek2 + csk1e-k2 = c3(k2 ' - k' -3),

and we may use the lemma once more to get

(17.20)diekl-k2 + d2ek2-kl

+ daekl + d4ek2 + dskie-k2 = 0


for suitable dl , ..., d5. Multiply by eke to get

(17.21) dleki + d2e2k2-ki + d3ek'+k2 + d4e2k2 = -dsk'

and apply the lemma yet again to get

(17.22) ulek' + u2e2ki-k2 + u3ekt+k2 + u4e2k2 = 0,

where u1i u2, u3, u4 are constants that are not all zero. Now, by successiveapplications of the lemma, we reach a contradiction, unless possibly one ofthe five following conditions holds for some constant C:

k1=k2+C,12=C,kl=2k2+C,k2=2k1+C,k1=C.

We now rule out these possibilities unless f' = f. First, it is easy to seethat k1 = C (and similarly k2 = C) is consistent with (17.13) and (17.14)unless d = ec = 1, in which case f' = f. For if (f' - 1)/(f - 1) = d,then f = (d - 1)/d + bedz for some constant b, and hence f' = bdedz. Thisclearly contradicts (17.14) unless d = 1, for we would have

(17.23)f'-2_bdedz-2- k2f-2 bedz-2

-e '

which is impossible unless d = 1 (remember that f is not constant, so thatb36 0).

Next, we rule out k1 = k2 + C. We go back to (17.16) to get

(17.24) m1e2k1 + rn2e3k' = kim4ek'.

Apply the lemma again after dividing by ekl to get

(17.25) n1 ek' + n2e2k1 = 01

which implies that k1 = 2k2 + C (and similarly, k2 = 2k1 + C) is impossibleunless f = f. From (17.24) we would get

(17.26) d2e-c + (dlec + d4)e2k2 + d3eceSk2 = -2d5k2

and apply the lemma for the last time to get

(17.27) B1 + 12e2k2 + 13e3k2 = 0,

where P1, £2i t3 are constants that are not all zero. In other words, P(ek2)0, where P is a cubic polynomial, so eke is a constant, which we alreadyhave ruled out unless f = f'. This completes the proof of the theorem.

IT Picard's Theorem and the Second Fundamental Theorem 111

Finally, it is easy to see that there exists a nontrivial entire function thatdoes share one value with its derivative. For example,

z) =efZ

=

f( - e)dt

satisfies (f' - 1)/(f - 1) = ez so that f and f' share the value 1. Thisshows that the number two of our theorem is the best possible.

Now we come, as an application of the second fundamental theorem,to a truly beautiful and surprising theorem of Nevanlinna. Recall that,given two meromorphic functions fl(z) and f2(z), and a complex number(finite or infinite) w, we say that fl and f2 share the value w (ignoringmultiplicity) if every z for which f1(z) = w also satisfies f2(z) = w, andvice versa. We use the same terminology if both functions omit the valuew.

Theorem. If two functions, meromorphic in the whole complex plane C,share five distinct values, then the two functions must be the same.

Note that es and e_2 share 0, oo, 1, -1, so the number five is sharp.

Proof of Theorem. Given a number w, finite or not, let no(r, w) be thenumber of common roots of f1 (z) = to and f2 (z) = to contained in the disc{jz) < r}, each counted only one time. Then put

No(r,w) =rr no(t'w) t no(0,w) dt+no(0,w)logr0

and

w)-2No(r,w).N12(r,'w)=N(r'fi1 w)+IV (r,12

1

Now, taking for wl,... , wq distinct finite complex numbers and applyingthe second fundamental theorem to the functions fl and f2i we have, off apossible exceptional set of finite length,

q l(q-2)(T(r,fi)+T(r,f2))<E(N(r'f1 1 )+N(r'f2iwY))

+O[logrT(r, fl)T(r, f2))]q q

_ 21: No(r,w,,)V=1 Z-1

+ O[log rT(r, fi)T(r, f2)].

Now, if the functions f, and f2 are not the same, every common root of theequations fi = w,,, f2 = w,,, is a pole of the function i

z. We deduce

from this that/ \

J <T(r,fi-f2)+O(1).Y-1 \ h-f2/


On the other hand,

T(r,fi - f2) <T(r,fl) +T(r,f2) +O(1).

We conclude

q

(q - 4) (T(r, fl) + T (r, f2)) < E N12(r, w.)V_1

+ O[log(rT(r, fl)T(r, f2))]

off a set of exceptional segments of finite total length.(If one of the w is infinite, just apply a linear fractional transformation

to fl andSuppose to begin with that one of the two functions, say fl, is transcen-

dental. Then among the five given values w = w,,, v = 1, 2,. .. , 5, thereare at least three that are taken by fl in an infinite number of points z.By hypothesis, the same is true for f2. Hence, f2 is also transcendental.Suppose for a moment that fl and f2 are not the same. Apply the lastinequality above, which shows that for q = 5, the expression N12 vanishingfor the five given values,

T(r, fl) + T(r, f2) < O(log[rT(r,fi)T(r,f2)1)

This implies that

Urn Tl) < oo and lim Tlo' 2) < 00.r- oo log r r-co g

This is impossible since fl and f2 are both not rational functions. Thetheorem thus is proved by contradiction in the case that one of the givenfunctions is transcendental.

But the conclusion is obvious if both of the functions are rational, sincea rational function f is uniquely determined by the roots of f (z) = w forany three distinct values of w.

18A Proof of the SecondFundamental Theorem

We now begin the proof of the second fundamental theorem of Nevanlinna.We continue to use the convention that all equalities are to be read "mod-ulo 0(1)." There are a number of other proofs of the second fundamentaltheorem, some of them leading to generalizations of it.

Lemma 18.1. Let f be a meromorphic function in the plane, and letal,..., an be distinct complex numbers. Let

F(z) =1(z) - a,,

Thenn

m(r, F) > E m (r

Proof We first remark that the lemma is intuitively obvious, for when f (z)is "close to" a it contributes to m (r, lam) , but not to any m (r, )with j # Y.

To prove the lemma in detail, we introduce the following notation. Letd > 0 be given with 2b < min{ la - a,1 : 1 < v < j < n}. We require6<1. Let

E. = f{w: Iw - a,,I < b} = {z: If(z) <b}E,', (r) = E, fl {z : Izl = r}

E,,(r) _ {B : re's E[0, 2x] - Ea(r).

n


In other words, E (r) is the set of all 9 for which If (reie) - a,, < 6, i.e.,those 9's for which f (reie) contributes significantly to m (r, f la-) . It isclear that E, (r) and Ej(r) are disjoint provided that v 54 j.

Moreover,

1 " 1

21rr log+ IF(re`B)IdO >_ 2 r log+ IF(reie)Id8,

x J

where f ` is over the set Now

f* log+ IF(reio)Id927r J - 27r ,l E.. (rl

But

n 1

>2a JE (r)

log+If (reie) - a , I d9

v-1

_ 1 log+27r E.. (r) f(reè) - aj

j=1hAm

1

=

1 "1o

+ 1m r, f - a 2a ,j g f (reie) - ad9

do.

1og+ I ie I d9 + 1J

log+ I ie I d9.27r

E.(r)f(re ) - a 27r

ca(r) f(re ) - a

27r JET r lOg+ I f(re`) - a,, I °IB + O(1).

Also,

1 log+27r

Hence we see that

n 1

L f(reie) - ajj=1j6v

m(r, F) > 2a f log+

1 log+27r E.,(r)

<nal=0(1).

1

f(Teie) - ad9

f(reie) - ajj=1

I j #vn \Em r, 1 I+0(1),

v_1 f-°'vwhich proves Lemma 18.1.

d9

18. A Proof of the Second Fundamental Theorem 115

Definition. N, (r, f) = N (r, f,) + 2N(r, f) - N(r, f'). Nl (r, f) is inter-preted in the previous chapter after the statement of the second fundamen-tal theorem.

Lemma 18.2.1

> m(r,f-av i <2T(r,f)-Nl(r)-m(r,f).1 //

( lln

+m r, f'!

- +m r,Ef V=i f a

Proof. Let F(z) be as in Lemma 18.1 and write

zn

F(z) 7_(z) . f ) f(z)(x)a"

Then

1 n '(18.1) m(r, F) < m (r, -l +M r, .t + m r, f

f / ` f' I Y_1 f

Butm(r, )=T(r,f)-N (r,1)andm(r, f)=m(r, f)+N(r, f)-N (r, .). Hence,(18.2) / // )+N(r,L)-N \

m(r,F)< T(r,f)-N(r,f)+mir, f )lr,

f - aY=1

Now observe that if we define

co(r,9)=N(r,9)-N \r, 9/then

co(r, 9h) _'P(r, 9) + co(r, h)

Using this observation in (18.2) we see that(18 3).

m(r,F)<T(r,f)-Nl r, f l +m(r, fll +N(r,f')+NI r, fl

y=1

" /1).-N r, ;1 l -N(r,f)+m r,/


If we now add and subtract T(r, f) = m(r, f) + N(r, f) on the right-handside, we obtain

(18.4)

or

(n

m(r, F) < 2T(r, f) - m(r, f) + m r, :c +M (r, /')f) `/ L=1 lav

-(2N(r,f)-N(r,f')+N(r,

n ,m(r,F)<2T(r,f)-m(r,f)+mr, f)+m r,E ff av -Ni(r).

\\ v=1

By Lemma 1, m(r, F) > EV_1 m (r, f av ), and Lemma 18.2 follows.

We are now ready to prove the second fundamental theorem of Nevan-linna.

Theorem. Let f be a meromorphic function in the plane and let

n

f v=lf -avC f,) f,

Then

n

m(r, f) + E m (r7 f 1 av) 2T(r, f) - N, (r) + S(r).

Proof. This is just Lemma 18.2.

We saw previously that in order to deduce the Picard Theorem from thesecond fundamental theorem, we needed an estimate on the size of S(r).Such an estimate follows from:

Lemma (Lemma of the Logarithmic Derivative) 18.3. If f is meromor-phic in the plane and 0 < r < p, then

m (r, f I) 4 log+ p + 3 log+ 1 + 4log+ T (p, f ).f p - r

Proof. Without loss of generality, we suppose that f (0) 96 0, oo since, ifg(z) = Z' f (z),

m (r, 9) - m \r f 0(1).


Let z = reie. Then for a suitably defined branch of the logarithm we haveby the Poisson-Jensen formula,

(18.5)

1 r" peswlogf(x)=2 J_ logIf(pe"°)I t!T,+ E log BP(z, z,) - E log BP(z, iA,

Izv1<P 1W-1<P

where BP(z, a) is the Blaschke factor mapping the disk of radius p onto theunit disk and a onto 0, A is a real constant, and, as usual, z and w arethe zeros and poles of f, respectively. [Equation (18.1) holds without any0(1) terms.] Thus, by differentiating (18.5), we obtain(18.6)

fi(x) x2pe"°

p2 - IzzI2

f (x) 2w ..A1°g if

(Pe.P)I

(peiV - x)2 d`p + (z - zz)

E 2 - Iw. I2Iw

I<P(z_'rW,)(p2-ww).

Taking simple estimates, we have(18.7)

rIRY (p - r)2 21r

I log If (Pe"°)IIdco +

p2 - IAvI2

IAIx - AYIIp2 -

YI<P

where A now runs over both the zeros and the poles of f. Observe that

p2-Avx p(p2-IAYI2)I p(z-A,)I

p(p2 - IAYI2) < 1 p- I BP(x,A,.)I BP(z,AY)I (p-r)2

The last step follows from Ip2 - p2 - IAv IIzI p2 - pr =P(P - r).If we use this estimate in (18.7), take log+ of both sides of (18.7),

and apply the additive inequality satisfied by log+, (log+(a, + . + a,) <10g+ a, + + log+ a + log n), we then find that(18.8)

log+ I f(x) I

<_ log+ p + 2log+p

1 r + log+ 2J x Ilog+ If (rei ')I I

E log+ IBP(zAL)I

+ log+ (n(p, f) + n (p, f)).Iavi_ 1, we see that log+ IB(=,a. = log IBp(z,a-


Therefore, by Jensen's Theorem,(18.9)

1 J log+ I 1 I dO =1 1°g log la f if r

21r ,, BP(re=B, logla l

if r.

To simplify the notation, let n(r) = n(r, f)+n(r, j) and likewise N(r) _N(r, f) + N(r, f). If we let z = re`B and integrate (18.8) over the circle ofradius r with center at 0, we obtain(18.10) f :_ _ 1 )r, f - ( log+ ff( (re

) ' d821r d

7r

re'er

< log+ p + 2 log+ p 1 r+ log+ 2 J x 11og+ If (T e' ') I f dip

r+ 21r ,1-

x

xlog+ I Bv(ree, ,\v) dO + log+ n(p).

l a..l <P

We recognize that i f' Ilog+ If(PQse)IIdcp = m(p, f) + m (p,.I) =O(2T(P,f)) and

r` +L 21r log B.(r;`°, A.,) dla.I<P

_ log (FAP-'/ L log ( a,.N(P) - N(r).

Ia.,ISP I IA.I<r I I

Hence,(18.11)

m(r, f,/ <log+p+2log+p1 r+log+T(P,f)+log+n(P)+N(P)-N(r).

We now will estimate the term log+ n(p). Choose a number p' withp' > r. Then

n(p) = n(P) P dt < 1 P n(t) dtlog(P) fv t log(o) JP t

N(p) 2T(p', f) + C-log(p)

-log(p)

where C is an appropriate constant. Now

P(P - P) :5 Jtt = log P - p (P - P)-p,


Hence,

so that

Hence,

n(P) <2T(#',', f) )C

1og+ n(p) < log+ p' + 1og+ 1 + log+ T(P', f )p,_P

M (r,fc/I < log+ p + log+ p' + 2 log+ p 1 r + log+ p, p + 1og+ T (p, f)

+ log+T (P , f) + N(p) - N(r).

Therefore, with r < p < p' we have(18.12) \

M 1 r, f I < 2log+ p' + 2log+ p 1 r + log+ p, 1 p + 21og+ T (p!, f)///

+ N(p) - N(r).

We now want to make the term N(p) - N(r) small. To do this we usethe logarithmic convexity of N. [N is logarithmically convex, since n(t) isincreasing and N(r) = for ntt dt]. Since r 1.P-rNow, considering 2 (P ) [2T(p', f) + C] as a function of p, we see

that it vanishes for p = r and is greater than 1 for p = p'. Since it is aContinuous function of p, we may choose p so that

P'(P - r) [2T(P', f) + C] = 1.r(p' r)

For this choice of p, we thus have

N(p) - N(r) < 1 and(18.13)

(p - r) =r(p' - r)

P'[2T(P',f)+C]


and

Hence,

Therefore,

(p'-p)=(P -r) 1- rp' ]2T (p', f) + C]

1og+ 1 < log+ p' + log+ T (p', f) + log+1

p-r - p' - r

(18.14) log+1

< log+ p' + log+T(p', f) + log+1

p-r p' - p.Also, we see that

log+1

< log+1

+ log+ 1 rp' - p - p' - r

1 + log+ 1 11 -2Tp',f+C

Hence,

(18.15) log+ 1 < log+1

Using (18.13), (18.14), and (18.15) in (18.12), we have

(18.16) m (r, r) < 4 log+ p' + 4 log+ T (p', f) + 3 log+p' -

which proves thelemma.

We will use the notation 11f (x) < g(x) to mean f (x) < g(x) with Borelexceptions (i.e., a set of finite length).

Proposition.

I]m(r, 4log+ r + 8log+ T(r, f ).

Proof. Taking p\= r + fl in the lemma, we have

m (r, f,!

< 4log+ 1 r + log+7,(r f) + 31og+ 1og+ T (r, f)

+ 41og+ T r+i 'f

log+ T (r, f)

Hence, by the Borel Lemma, we get

11m (r, f i < 4log+ r + 8log+ T(r, f ).

Corrollary. IIS(r) = O(1og+ T(r, f )) + O(log+ r).Proof. This follows immediately from the proposition since S(r) is a finitesum of terms of the form m(r, 4-).

r'

19"Two Constant" Theorems and thePhragmen-Lindelof Theorems

The Two Constant Theorem. Suppose that f is holomorphic in D =1z: IzI < 1} and continuous in D\{1}. Suppose further that if I N in Dand if 1< M in oD\{1}. Then If I < M in D\{1}.

First Proof. Choose a > 0 and let

g(z) _ (Li)u

f (z)

f o r a suitable branch of (1 2 1)". Now g is analytic in D and continuous inD if we define g(1) = 0. By the maximum modulus theorem,

suupplg(z)I = m xlg(z)IzaDBut Ig(z)I<Mfor zEBID. Hence, Ig(z)I<Mfor zEID,and so

0If(z)I<MII zI for zED.

Now let a - 0 to get the result.

Second Proof. This proof uses the Cauchy integral formula. Since there ex-ists a linear fractional transformation U(z) = e'9 taking an arbitraryYPoint z0 E D into 0, and mapping I)/{1} conformally onto ID/{1}, we needonly prove that if (0)I < M. But choosing 00 with 0 < 00 < 7r,

if(o)i = I 1 I f(w)dwl1 1 If(rei8)Ide+

l I If(Te`B)Id0,27r w=R w 21r 27r 2)


where R < 1 and

L)fo<1e1<- J(2) - 1101!50o'

But

and as R-+1-

Oo N'(2)

C

1 f < (M+0(1)) 21r2200

(1)

Hence, letting R -+ 1-,

If(0)I! oN+Mf 1- 0)

Now letting 00 - 0 we again obtain the desired result.

Third Proof. By Jensen's Theorem,

log If (0)1 < 2,-f log If(reze)IdO.

7r

The argument of the second proof works here, too.

Fourth Proof. By a conformal mapping, we may replace the unit disk bythe upper half-plane. We have If I < M on the real axis and if 1:5 N abovethe real axis, and must conclude that If I < M above the real axis. ChooseA > 0 and let

Note that

and that

g(z) = (ZiA)f(z).

A I A

z + iA IzI

Ig(x)I < M for all real x.

Hence, if R is large enough, we see by the maximum modulus theorem thatIg(z)I < M for z in the semicircular region bounded by the real axis andthe are I z I = R, 0 < arg z < 7r. Hence, Ig(z)I < M in the upper half-plane.Letting A -+ oo, we complete the fourth proof.

19. "Two Constant" Theorems and the Phragmen-Lindelof Theorems 123

A Phragmen-Lindelof Theorem. Let f (z) be holomorphic inside anangular region of opening xr/a, where a > 1, and continuous on the closureof the region. Suppose that If (z)I < M on the boundary and that

If(z)I <-KeiZI-3

in the region, for some constant /3 with Q < a. Then If (z) I < M in theregion.

Proof. Without loss of generality, we may suppose that the region is givenby I arg z I < Za . Choose e > 0, and choose ry with p < y < a. Let

F(z) = e-Ezhf(z),

so thatIF(reie)I = exp{-er7'cos7O}If(z)I

For z = rese in the closed angle, we have exp{-er'1 cosry6} < 1 so thatIF(z) I < If (z) I there. Notice also that by using the estimate on f insidethe region we have

IF(re`B)I < exp{-eR'' cosryO}Kexp{RR},

so that for R large enough we have

IF(Re'o)I < M.

It follows thatIf(z)I 5 Mexp{eR1},

and the result follows on letting a - 0.Taking for simplicity a = 1, the Phragmen-Lindelof Theorem says that

if a function is holomorphic and of order less than one in a half-plane andbounded on the boundary, then it is bounded inside by the same bound.The example f (z) = ez in the right half-plane shows that the order 1 iscritical. However, the next result shows that the analogous result holds ifwe replace the condition "order less than 1" by "growth at most order 1,zero exponential type."

Theorem. If the hypothesis is changed to read that for each 6 > 0

If(z)I 5 K6exp{6IzI°}in the region, then the conclusion follows as before.

Proof. Now letF(z) = exp{-eza} f (z),

and the same proof works.

Remark. Analogous results hold for functions holomorphic in other regionsand satisfying appropriate growth restrictions. One useful case is the paral-lel strip. Calderon has used this case in developing a theory of interpolationof Banach spaces that can be applied in the fields of harmonic analysis andPartial differential equations.

20The Polya Representation Theorem

The Polya Representation Theorem plays a central role in the theory of en-tire functions of exponential-type. We give a somewhat augmented versionof this theorem.

Before proceeding with the theorem, it will be necessary to discuss con-vex sets. We say that a set E (in the complex plane) is convex if E con-tains the line segment joining any two points. That is, if z1, z2 E E, thentzl + (1 - t)z2 E E for all t E [0,1]. The intersection of convex sets is againconvex.

Definition. Given a set A, the intersection of all half-planes that containA is called the closed convex hull of A and is denoted by K(A).

Definition. A point is an extreme point of a set if it is not the midpointof any line segment contained in the set.

Theorem 20.1. A compact convex set is the convex hull of the set of itsextreme points.

We omit the proof.

Definition. For any set E, k(8) = sup{Re(ze-'B) : z E E} is called thesupport function of E.

We note that k(O) measures the directed distance from the origin to themost remote point of the projection of E on the ray arg z = 0. Note alsothat if E is empty, then k = -oo. It is easy to show that, for a given setE,

K(E) = {z : Re(ze-'B) < k(0) for all 0}.

Remark 20.2. If zo = xo + iyo = roe'Bo and E = {zo}, then k(0) =TO cos(0 - Bo) = zo cos 0 + yo sin 0.

20. The Polya Representation Theorem 125

Remark 20.3. Let E be a circle with center at 0 and radius R. Thenk(8) = R for all 0.

Remark 20..¢. Let E be the line segment [xo, x1], xo < x1. Then k(8) _x1 cos 0 if - a < 0 < 2 and k(0) = xo cos 8 if 2 < 0 < s2

. In particular,if xo = -x1 and x1 = R, then k(0) = RI cos 01. If E is the vertical linesegment [-iR, iR], then k(8) = RI sin 01.

Remark 20.5. If E1 has support function k1 and E2 has support functionk2, then E1 + E2 = {z1 + z2 : z1 E E1, Z2 E E2} has support functionk1 + k2. From Remark 20.3, it follows then that the rectangle with vertices(±R1i ±iR2) has support function k(0) = R1I cos e] + r21 sin 01.

Remark 20.6. The convex hull of E1 U E2 has support function k =max{k1, k2 }.

Remark 20.7. If E1 with support function k1 is translated, so that thepoint originally at 0 is moved to zp = xo + iyo, then the support functionof the translated set is k(O) + xo cos 0 + yo sin 0.

Definition. Let Ho(oo)be the class of all functions that are holomorphicnear oo and that vanish at oo.

Let f be an entire function of exponential-type and write f (z) _() zn. The BoreI transform -6 of f, defined by 4'(w) = > anw mar,

belongs to Ho. Each function in Ho is the Borel transform of a uniquef. We have seen (Corollary to Proposition 11.5) that the type of f is theradius of convergence of the series >2 an war.

In Proposition 11.7, we saw that D(w) = f ow f (t)e- t7°dt, in the sense ofanalytic continuation. We also saw that

f (z) = 27ri ,14'(w)ez'°dw,

r

where r is a rectifiable curve that winds once around the singularities ofI. We call this the P61ya integral representation formula.

Definition. Let S(4') be the set of singular points of 4', and let k be itsanpporting function. We call S(4') the conjugate indicator diagram of f.Sometimes this name is used for S*(4'), the closed convex hull of S(4'). Wewill write D(f) =

Definition. The indicator function of f is

h(8) = hf(0) = limsup 1 log ]f(reie)1.r-.oo r

We now state an important part of the Polya Representation Theorem.


Theorem 20.8. h(0) = k(-0) for all 0.

The proof of Theorem 20.17 is contained in an appendix at the end ofthis chapter. We now make some remarks to illustrate this theorem.

Remark 20.9. If f is of zero-type, then h = 0 so that k = 0, and hence 0is the only possible singularity of 1.

Remark 20.10. Denoting by r(f) the type off, we have r(f) = max h(0).

Remark 20.11. If f (z) = eaz, where a is real, then

h(0) = lim sup 1 log lea` l = lim sup 1ar cos 0 = a cos 0.r.-.oo r r-.oo r

On the other hand, O(w) a" w- r = (w - a)-1 and so S(O) = {a}.Thus, k(0) = acos0 and we do have h(9) = k(-0).

Remark 20.12. If f (z) = eiz, we have h(0) = - sin O and 4)(w) = (w-i)-1.Hence, k(0) = sin 0 since S(4,) _ {i}. Note again that h(O) = k(-O).

Remark 20.13. Suppose f (z) _ > a,,ea^z, a finite sum with distinct A,,and nonzero an. Then O(w) = so that S((D) = {A,.}. Notethat only the extreme points of affect h(0). If the A,, lie on a straightline, only the endpoints affect h(0).

Remark 20.14. It is easy to see that h f.9 < h f+h9, so that D(fg) C D(f)+D(g). An interesting problem that has applications in harmonic analysisis that of finding suitable conditions under which D(fg) = D(f) + D(g).

Let Mo be the class of all Borel measures of compact support.

Definition. If dp E Mo, its Laplace transform dµ^ is defined by

dµ^(z) =J

e-zwdA(w).

It is easy to see that dµ^(z) is an entire function of exponential-type for

dzdp^(z) = f(_w)e'd/L(w)

(as can be verified on differentiating "by hand") so that d,i' is entire. And

Idi^(x)I < f Iez `Jjdµ(w)I 5 eRjzj f Idu(w)I,

where R is chosen so that the support of dµ lies in the disk of radius R

centered at 0.

20. The Pblya Representation Theorem 127

Definition. We write dµ - dv to mean that dµ^ = dv^.

It is not hard to show that dp - dv if and only if f f dµ = f fdv foreach entire function f, or for each entire function f of exponential-type. Itis clear that - is an equivalence relation.

If we take dµ = dzjr, where r is a circle, then dp^(z) = fr e-a'dw = 0by the Cauchy Theorem. Hence, dµ - 0 even though dµ 0 0. We shallsee that for any dµ E mo, dµ - dv, where dv = 4P(-w)(-dw)Ir, whereis the Borel transform of dp^ and IF is any curve that winds once aroundS(C.

Definition. [dp] is the class of measures equivalent to dp.

Definition. Mo is the class of all [dp] for dµ E Mo.

Definition. If f is continuous in the plane and dp in Mo, we define theconvolution f * dµ by

(f * dµ) (z) = J f (w - z)dp(w).

Definition. If dp and dv are in Mo, we define the convolution dµ * dv by

f*(dµ*dv)=(f*dµ)*dv.

By means of the Riesz Representation Theorem, it is easy to see thatthe above definition defines dµ * dv as a unique measure in Mo. Indeed,Mo is an algebra over the complex numbers.

Proposition. If dp1 - dp2, then (dpi * dv) - (dµ2 * dv). It thus makessense to define [dp] * [dv] = [dµ * dv]. Ma is an algebra over the complexnumbers.

Definition. Let Eo be the algebra of all entire functions of exponential-type.

Definition. Let E be the space of all entire functions in the topology ofUniform convergence on compact sets.

Definition. E', the dual of E, is the space of all continuous linear func-tionals on E.

Now E is a locally convex topological linear vector space. It will appearthat each of the spaces Mo', Eo, Ho(oo) "is" the dual space E.

Definition. For f E E and [dµ] E Mo', define the inner product (Fl, [dµ])hY

(F1, [dp]) = (F * dp)(0) = f F(-z)dp(z).


Definition. For F E E and f E Eo, define the inner product (F, f) by

(F,f) = (f(D)F)(0),

where D = dZ. This means that if f (z) = E z", then (F, f)n F(")(0)It is not hard to show that, for each f E Eo and F E E, the series defining

(F, f) converges. Indeed, the linear functional A defined by A(F) = (F, f)is a continuous linear functional on E. The same is true for A(F) = (F, dµ).

Definition. For F E E and fi E H0(oo), define the inner product (F, 4b)by

(F, _1I <P(w)F(w)dw,2ri rwhere r is any curve that winds once around S(4')

Again, A(F) = (F, -6) defines a continuous linear functional on E.

Theorem 20.15. Let [dp] E MO', f E Eo, -0 E H0(oo) be related byf = dµ^ and P be the Borel transform of f. Then dµ, f, and give riseto the same functional in E'. And given any functional in E', there is aunique [du] (or f or t) that gives rise to it.

To say that f gives rise to A is to say that A(F) = (F, f) for each F E E,with a similar definition for dp or 1b. We omit the proof of the theoremsince much of it is straightforward.

Remark. To illustrate the theorem, take f = ez so that d IA is the unit pointmass at 1 and *1(w) = (w - 1)-1. Then f (D) = eD, so that by Taylor'stheorem

(f (D)F)(0) = F(O) + F'(0) + 1 F"(0) + = F(1).

Note that f F(-z)dp(z) = F(1) also. And, by Cauchy's Theorem,

f27ri

F(w)41(w)dw = - f w(-w) dw, = F(1).

Summarizing the results of this section so far, we have the followingomnibus theorem. We call it the Pd1ya Representation Theorem, althoughthe name is not entirely appropriate.

The P61ya Representation Theorem. Let Eo be the algebra of all en-tire functions of exponential-type and Ma the convolution algebra of equiv-alence classes of Borel measures of compact support, where du dv meansthat f f dµ = f f dv for each entire function f. The Laplace transform isdefined by dµ^(z) = f e-Zwdp(w); dp - dv if and only if dp^ = dv^, so

20. The P61ya Representation Theorem 129

it makes sense to talk of [dp]^ = du^, where [dµ] is the equivalence classthat contains dµ. The Laplace transform is an isomorphism of Mo ontoFo. To invert the Laplace transform, i.e., given f E E0, to find dp E Mosuch that dµ^ = f, take dp(z) = 4)(-z)(-dz)Ir, where 4) is the Boreltransform off and t winds once around the set S(4)) of singularities of 4).The indicator diagram of f, D(f), is defined as the convex hull of S(4)). If

h(O) = lim sup '° f t e is the indicator function of f, then h(O) = k(-6),r-.oo

where k is the support function of D(f ). If E is the space of all entire func-tions in the topology of uniform convergence on compact sets, then both Eoand MO' are the dual space of E, where, if f = dp^, then for F E E

(F, f) = (f (D)F) (0) = (F, dµ) = JF(_z)d(z).

We now turn to some applications of this theorem.[Note. We repeatedly use conventional, but strictly incorrect, phrases

like "r winds once around the set S(4)) of singularities of C" A morecorrect wording is "r has winding number -1 around oo, and lies in aconnected and simply connected open set containing co in which 4 is ana-lytic."]

Definition. For an entire function f and a complex number A, let fa bethe entire function defined by

f.\ (z) = f (z + A) for all z.

Lemma 20.16. For any entire function f, -r(fa) = r(f ).

Proof. With M(r : f) = max{ I f (z) J : IzI = r}, we have

M(r : f'0:5 M(r + IAI : f)

so that

I logM(r : fa) <- 1M(r+ IAI : f) = (1+0(1))r+ICI logM(r+ IAI : f)-

Hence,

r(A) <- r(f).

Similarly,

-r(f) < T(fa) since


Lemma 20.17. If f and g are entire functions of exponential-type, thenD(f) = D(g) if and only if

T(f(z)eaz) = T(g(z)eaz)

for each complex number a.

Proof. Clearly, if D(f) = D(g), then r(f (z)eaz) = r(g(z)e6z). To provethe converse, it is enough to compute, say, h(O) from r(f (z)eaz). We showthat

h(O) =a

lien ar(f (z)e6z) - a.

Let Da be the indicator diagram of f (z)eaz. Then Da = a + D(f), thatis, D. is D(f) translated by a. Choose B < max (uhf (z) 1, Jhf (-3) 1)

Now when a is large, D. lies to the right of the origin, and Da lies inthe strip {z = x + iy : jyj < B; x < a + h1(0)}. D. also contains the point(a + h(0), 0). Hence, if T. = T (f (z)eaz), then

Ta > a + h(0)

T < {(a + h(0))2 + B2} j,

Ta - (a + h(0)) = o(1).

Lemma 20.18. If f is an entire function of exponential-type, then foreach complex number A,

D(f) = D(fa)-

Proof. We use Lemma 20.17, with g = fa. Note that

e6zg(z) = eazf(z + A) =e-aa(ea(z+a)f(z

+ A)).

So if we let F(z) = e6z f (z), then e6Zg(z) = CFA (z), where C is a nonzeroconstant. Since r(F,) = T(f ), we have r(e6z f (z)) = r(e6Z fa (z)), and theresult follows.

The P61ya Theorem has the following corollary.

Proposition. If h (2) < 0 and h (- z) < 0, then f = 0.

Proof. h (2) < 0 implies that D(f) lies above the real axis ) < 0implies that D(f) lies below the real axis. Hence, D(f) is ...consequently, 4 has no singularities.

Since -t(cc) = 0, it follows from Liouville's Theorem that 0 = 0, anohence f = 0.

The Pd1ya Representation Theorem provides an alternate proof of Carl-son's Theorem presented in Chapter 12.

20. The Polya Representation Theorem 131

Theorem (F. Carlson, 1914) 20.19. Suppose f is an entire func-tion of exponential-type with r(f) < w, and suppose that f (n) = 0,n = 0, ±1, ±2,.... Then f = 0.

Proof. Consider the function g(z) = f (z)/ sin wz. It is clear that g is entire.Since

T(r,F/G) <T(r, F) + T(r, G),

g is of exponential type. Now hg (-E) < 0, and hg (a) < 0, so that by theabove proposition, g = 0. Hence f = 0.

Remark 20.20. It is clear from the proof that the hypothesis that r(f) <x can be relaxed to h f(f7r/2) < w. The condition r(f) < r is sharp,since sin wz is of type w. The Carlson Theorem says, roughly, that anentire function of exponential-type must grow fast in a direction at rightangles to its zeros. Later in this section, we shall get a stronger version ofCarlson's Theorem. Chapter 22 is devoted to proving an extremely stronggeneralization of it.

Definition. Let k denote the set of all entire functions f of exponential-type such that h f(±w/2) < w.

Definition. A sequence {An}, n = 0, 1, 2.... is said to be k-admissiblep r o v i d e d that there exists an f E k such that f (n) = An, n = 0,1, 2, ... .

We now give a characterization, due to Buck, of k-admissible sequences.

Theorem (Horseshoe Theorem) 20.21. The sequence {An} is called k-admissible if and only if F(z) = E Anz" is holomorphic at 0 and oo andin some neighborhood of the negative real axis as well.

Proof. Suppose first that {An } is k-admissible, so that An = f (n) for somef E k. Then, if z is small,

F(z) _ f(n)zn =(2wi

fb(w)enwdw I zn2wi 1 (z) dw.

We choose 1 to be a rectangular path that winds around S(4') such thatr lies in the strip

S. = {z = x + iy : lyl < w}.

It follows that F(z) can be analytically continued in the complement of theimage of S(4') under the mapping a-", and that F(oo) = 0.

To prove the other half of the theorem, suppose that F satisfies therequirements of the theorem. We may write

_ 1 r F(z) dzAn 2wi A zn z '


where A is a circle around 0. But A is homotopic, in the complement ofS(F), the set of singularities of F, to the path AR illustrated in the figurebelow, where A' is a circle of radius R.

T U Horseshoe Contour

Hence,

Now

Hence,

IR

1 r F(z) dzAn _27riJAR zn z

-+ 0 as R --+ oo, since F(oo) = 0.

_ 1 /' F(z) dzAn

27ri A. z" z '

where A* is any curve that winds around S(F) with multiplicity -1.thus are led to define

f (W) = 21 JF(z) dz

7ri z- z

We

for an appropriate branch of zw on A*. It is easy to check that f E k, andthe proof is complete.

We now may prove the following extension of Carlson's Theorem.

20. The PGlya Representation Theorem

Theorem 20.22. Suppose that f E k and that f(n) = 0 for n

0,1,2,.... Then f = 0.

Proof. We have

F(z) _ E f(n)zn

= 2ri J(1 - zew)-1-t(w)dw.

By hypothesis, F = 0. Hence,

zF(z) = 1 J z i(w)dw = 0.2iri r 1- ze'

On letting z --, oo, we have

133

e

' J ew-b(w)dw = 0.ri J

Since f (z) = -I- Jr we have f(- 1) = 0.Now we use the fact that f E k; thus f_1 E k, where f_1(z) = f(z- 1).

We may apply the same argument to f -I to conclude that f_1(-1) = 0,that is, f(-2) = 0. Repeating the argument, we get f (n) = 0 for n =0, so that f = 0 by Carlson's Theorem.

We may prove several theorems about k-admissible sequences by meansof the above characterization using classical theorems of function theorythat relate the properties of a function to its power series coefficients.

Theorem 20.23. If {An} is k-admissible and JAnI1"" -* 0, then An= 0for each n.

The proof is trivial.

Theorem (Pringsheim). Let f (z) _ E anz" have nonnegative coefficientsand radius of convergence R. Then z = R is a singular point of f.

Theorem 20.24. If {An} is admissible and (-1)"An > 0 for n =0,1,2,..., then An=0 fore=0,1,2,....Proof. F(-z) = E(-1)"Anzn has nonnegative coefficients, but no positivereal number is a singularity of F.

Hadamard Gap Theorem. If AZ) = E anz" with an = 0 except forn = nk, where lim 1, then every point of the circle of convergenceoff is a singular point of f.

Theorem 20.25. If {An} is k-admissible and An = 0 except for n = nkwith lim nk+1 > 1, then An = 0 for n = 0, 1, 2, ... .

nk

This is a simple consequence of the Hadamard Gap Theorem.


Fabry Gap Theorem. If f (z) = > anzn with an = 0 except for n = nkand k-'nk -+ oc, then every point on the circle of convergence of f is asingular point of f.

Theorem (Szego). Suppose that f (z) _ anzn, where the an lie in somefinite set. Either IzJ = 1 is a natural boundary of f or f is a rationalfunction, and the an are eventually periodic.

As a corollary we have

Theorem 20.26. If {An } is k-admissible and the An lie in some finiteset, then the An are eventually periodic.

AppendixThe proof that h(0) = k(-0) is presented in this section. The actualproof of this assertion is fairly simple, but we prefer to give some of thebackground concerning supporting functions of convex sets. First, we givea simple necessary and sufficient condition that a function h(0) should bethe supporting function of a nonempty compact convex set. The conditionis that the function should be "subsinusoidal." Next, we prove that if h(0)is the indicator function of an entire function of exponential type, thenh(0) is subsinusoidal. Finally, we show that h(0) is the supporting functionof the conjugate of the conjugate indicator diagram. From now on, whenwe speak of a "function of 0," we mean a function that is 27r-periodic; andwhen we speak of a "supporting function," we mean a supporting functionof a nonempty compact convex set.

Our treatment is a combination of the treatments in Pd1ya [31] and Boas[5].

Definition. A function H(0) is a sinusoid if it has the form H(O) _acos0+bsinO.

Remark. Given 01 $ 02 and real numbers h1 and h2, there is a unique sinu-soid H such that H(01) = hi and H(02) = h2. We call H the interpolatingsinusoid: It is given by (0 < 02 - 01 < 7r)

sin(02 - 0) sin(0 - 01)(20.1) H(0) =

hlsin(02 - 01) + h2sin(02 - 01)

Definition. Given a function h(0) and 01 # 02, we call H the interpolatingsinusoid of h if H is given by (20.1) with h1 = h(01) and h2 = h(02).

Definition. A function h(0) is subsinusoidal if it is majorized by each ofits interpolating sinusoids, that is,

(20.2) h(02) <h(o1)sin(03 - 02) +

h(03)sin(02 - 01)

sin(03 - 01) sin(03 - 01)

whenever 01 < 02 < 03 with 0 < 03 - 01 < ir.

20. The P61ya Representation Theorem 135

Remark. The theory of subsinusoidal functions has some similarity to thetheory of convex functions.

Remark. If h is subsinusoidal, and if H is sinusoidal and H(81) > h(81),H(82)>h(02),then H(8)>h(8) if81 <0<82with 0<82-81 <7r.Remark. The sum of two subsinusoidal functions is subsinusoidal.

Remark. That h is subsinusoidal is equivalent to the assertion that thepoint h(0)eie does not lie outside the circle that passes through the points0, h(01)e",, and h(02)e`°2, where 81, 02, and 0 are in the specified range.This geometric interpretation can be used to supply geometric proofs ofsome of the subsequent results.Problem. Suppose that h1 is subsinusoidal, h2 is supersinusoidal, andh1(0) > h2(8) for all 8. Does there exist a sinusoidal function H such thath1(0) > H(O) > h2(9) for all 0?

Lemma 20.27. Suppose that h is subsinusoidal, that 01 < 02 < 83, that0 < 03 - 0 < 7r, and that H(8) is a sinusoid such that h(01) < H(81) andh(02) > H(02). Then h(03) > H(03).

Proof. Suppose b > 0 and h(83) < H(03) - b. Let H6 be the sinusoid suchthat

H6(01) = H(01), H6(83) = H(03) - 6.

Then H5(02) < H(02), since

H5(8) = H(8) - 6sin(0 - 01)sin(83 - 0&

Since h is subsinusoidal and we have

it follows that

h(01) < H5(01), h(02) 5 H6(02),

h(02) < H6(02) < H(02),which is impossible.

Lemma 20.28. A function h(8) is subsinusoidal if and only if

(20.3) h(91)sin(03 - 02) + h(92)sin(01 - 03) + h(03) sin(02 - 01) > 0

whenever 01 < 02 < 83, 02 - 01 < 7r; 03 - 02 < 7r.

Proof. Clearly, (20.3) is equivalent to (20.2) if 83 - 01 < 7r. To prove(20.3) in general, choose 04 so that 02 < 04 < 81 + 7r and let H(8) be theinterpolating sinusoidal for h at 01, 82. By Lemma 20.27, h(84) > H(04).Repeating this argument with 02, 04, 03 we get h(03) > H(03). Now

h(81)sin(03 - 02) + h(02)sin(81 - 03) + H(03) sin(02 - 01) = 0,

but sin(02 - 81) > 0 and h(03) > H(03), so the result follows.


Lemma 20.29. If h is subsinusoidal, then it is continuous and even hasleft and right derivatives at each point. The left derivative is never greaterthan the right derivative.

Proof. Choose 0 and suppose without loss of generality that h(8) < 0.Otherwise, consider h - H where H is a sinusoid such that H(8) > h(8).Choose c > 0, 6 > 0 with c + b < ir. Applying (20.2) in turn to the followingtriples cp1, V2, 03:

(i) 0 -E-6, 0-E, 0(ii) 0-E, 0, 0+E,(iii) 0, 0 + E, 0 + E + 6,we eventually obtain

h(8) - h(8 - E - 5)<

h(0) - h(0 - E)

sin(E + 6) sin f(2U 4)

< h(8 + E) - h(8) < h(0 + E + 6) - h(8)

sin f sin (E + 6)

For example, considering the triple (i), we get

h(8 - E - 6) sin E - h(8 - E)sin(E + 6) > -h(8) sin 6

so that

Hence,

[h(8 - E - 6) - h(0)] sin c - [h(0 - E) - h(8)] sin(E + 6)

> h(0)[sin(E + 6) - sin c - sin 6]1

= -h(8)2 sin E

26 [COS 2 b - cos ---j > 0.

[h(0 - E - 6) - h(0)] sin c > [h(0 - E) - h(9)] sin(c + 6),

which proves the first inequality of (20.4). The others are proved in asimilar way. Now (20.4) is precisely the assertion that h(8+ex)-h(8) is anincreasing function of x when x is small, from which all of the assertionsof the lemma follow easily.

Lemma 20.30. If h'(8) denotes the nght derivative of h(0), then

h(0) cos(W - 0) + h'(0) sin(V - 0) - h(V) < 0

for each pair 0, cp with 0 - ir < ' < 0 + ir.

Proof. Choose c so that 0 < E < it and apply (20.2) to the following triplesWP1, cp2, V3: (i) gyp, 8, 0 + E (ii) 0, 0 + E, W. This use of (20.2) is admissible ifeither 8-ir<8<.Xor0+E<V<E+7r. We get

(20.5) h(8) sin( 0,

20. The Pdlya Representation Theorem 137

which may be rewritten as(20.6)

0) - sin(cp - 0 - e) h(0 + e) - h(O)h(9)

+ sinesin(cp - 0) - h(V) < 0.

sin eNow let e - 0 to get the assertion of the lemma.

Theorem 20.31. The function h(9) is the supporting function of somenonempty compact convex set K if and only if h(0) is subsinusoidal.

Proof. Suppose first that h supports K, so that for each z = x + iy E Kand any 01, 03 we have

(20.7) xcos01 +ysin01 < h(91)

(20.8) xcos93+ysin93 <h(83).

If we now choose 01, 02, 03 with 01 < 02 < 03, 02 - 01 < ir, and 03 - 02 < 9r,we may multiply (20.7) by the positive quantity sin(03 - 02) and (20.8) bythe positive quantity sin(02 - 91) and then add to get

h(91)sin(03 -02)+h(03)sin(02 -01) > (xcos02+ysin02)sin(03-01);

but x cos 0 + y sin 0 = h(0) for some z = x + iy in K, and it follows that his subsinusoidal.

To prove that if h is subsinusoidal, then it is supporting, let

Ke={z=x+iy:xcos0+ysin9<h(9)}and then let

K=nK9.0

Each Ke is a closed half-plane in the direction 0. It follows that K is closedand convex. Further, K is bounded since it is contained in the rectangleKO n K z n K. n Kg f . We now prove that each K9 contains a point of Kon its boundary, thus completing the proof of the theorem. Given 0, then,we must prove that there exists a point ze = xe + iye E K for which

(20.9) xe cos 9 + ye sin 9 = h(0).

By the theory of envelopes, we also want

(20.10) xe sin 9 - ye cos 9 = -h'(0),

where we interpret h'(9) as the right derivative.Following this heuristic idea, we let xei ye be the simultaneous solution

of (20.9) and (20.10). But on applying Lemma 20.30, we get for all cp

xe cos cp + yo sin cp < h(cp).

It follows that zB E K, and since

h(9) = max{x cos 0 + y sin O : x + iy E K},

the result is proved.


Theorem 20.32. If f is an entire function of exponential type and

h(0)=hf(0)=r rlogIf(rei8)I,

then h(0) is subsinusoidal, and h(0) is consequently a supporting function.

Proof. The proof is a simple application of the Phragmen-Lindelof princi-ple. Suppose that 0 < 02 - 01 < x, and let h1 = h(01), h2 = h(02), andfor 6 > 0 let H6 be the interpolating sinusoid for h1 + 6, h2 + 6 at 01i 02.Let A = A6 be the complex number for which H6(0) = R{Ae-'B}, and let

F(z) = f (z)e Az so that if z = reie, then

IF(z)I = If(z)I exp{-rH6(0)}.

Now F(z) is bounded on the rays z = reie', z = re'e2, and of order 1 inthe angle between. By the Phragmen-Lindelof Theorem, F(z) is boundedin the angle 01 < 0 < 02, and it follows that h(0) < H(0) for each 0 in thisrange. Since llm H6(0) = Ho(0), the result is proved.

6-0Theorem 20.33. If f is an entire function of exponential type and h(0) _h(0) = Um I log If (re") I, then h(0) = k(-0) where k is the supporting

r-00 I'function for the conjugate indicator diagram.

Proof. Let I' = Of be the Borel transform off and let D = D f, the conju-gate indicator diagram of f , be the convex hull of the set of singularities of0. Let C be a rectifiable curve that winds once around D and that stayswithin an E-neighborhood of D, where e > 0.

Now iff (x) = 2 i I -b(w) exp(zw) dw,c

so that if z = refe, then

If (reie)I < A mEax I exp(zw) I,

where A is a constant. Hence,

h(0) < max R(weie).WEC

If we now let c - 0, we see that h(0) < k(-0).In the other direction, it is enough to prove that h(0) > k(O) since, if

we replace f (z) by g(z) = f (ze'`°), the general case follows from hg(0) >_kg(0), since hg(0) = h f(0 + gyp), 4Ds(w) = e"P4bf(we-"°) Dg = e"'Df andkg(0) = k(0 - cp). Now, asrwe have seen,

-O(w) = J co f (t)e-*w dt for w > h(0),0

so that fi has no singularity to the right of the line x = h(0) and the

inequality h(0) > k(O) follows.

21Integer-Valued Entire Functions

An integer-valued entire function f is one such that f (n) is an integer forn = 0, 1, 2, . Some examples are

(i) sin irz(ii) 2Z(iii) any polynomial with integer coefficients.In this section, we shall mainly follow a paper of Buck [7]. In outline,

a certain construction generates a special class Ri of integer-valued entirefunctions. We will be concerned with finding growth conditions on aninteger-valued entire function f that imply f E R1. The three examplesabove belong to R1.

Definition. We say that an algebraic number a is an algebraic integer ifit satisfies a polynomial equation:

(21.1)

where aJ E Z, j = 0, I,_ , n - 1. Notice that the coefficient of z" is 1.

Examples. Any n E Z satisfies z - n = 0. And ±i satisfies z2 + 1 = 0.

It is not hard to prove that the algebraic integers form a ring. If theinteger n in (21.1) is minimal, the other roots are called the conjugates ofQ. The collection of all of the roots of a minimal polynomial is called acomplete set of algebraic conjugates. It is not hard to show that if a is aroot of P (where P is not necessarily minimal), then each conjugate of ais also a root.

Consider now the polynomial

Q(x) = 1±q1x+g2z2 +... +qnz",


where qj E Z f o r j = 1, 2, ... , n. We can write

n

Q(x) _ fl (1 -$jz),j=1

where the f3j run over one or more complete sets of algebraic integers. Thismay be seen from the fact that the f3j are the roots of the polynomial

R(x) = znQ ( 2) = xn + Q1xn-1 + ... + qn

Now let P be any polynomial with integer coefficients and Q as above.Then

P(x) = E bnxn, where bn E Z.Q(z)

This follows since we can write

1 1= 1 +Q`(z) + [Q*(z)]2 + .. .

Q(x) 1 + qlx + + gnxn 1 -4;(,)

= 1+B1x+B2x2+ .

The BI are clearly integers. To express the bj in terms of the ,6j, let usfirst suppose for simplicity that the f3j are distinct, and that P = 1. Usingpartial fractions and writing

00

= r fjnxn'1 - #jx L1

n=1

we have

1 = L.rE.

M 00

= L L11(1 - Qjx) j=1 1 - f3jx j=1 n=O

1

where the Ej are the coefficients in the partial fraction expansion. Hence,

nbn =L[ Ejf31",

j=1

so that it is natural to takem

f(z) = Ej/,,j=1

1

where Q. is suitably defined.

21. Integer-Valued Entire Functions 141

Example

Q(z) = 1 + z2 = (1 + iz)(1 - iz)

Q(z) 21 iiz+21+iz 2{1+iz+i2z2 }

+

2[in + (-i)n]

1,0,-1,0,1,0,-1,0,...

f (z) = 2 [iZ + (-i)Z] = cos 2 Z.

}

In case Q has repeated roots, or if P is not constant, some minor modifi-cations must be made, and in general we have

m

b = Ef'.i(n),

so that we take

(21.2) f(z) _ Pj(z)Q,,

i=t

where the P3 are suitable polynomials (not necessarily integer-valued).

Definition. Let Rl be the class of functions f constructed above.

Definition. Let R be the class of integer-valued entire functions f ofexponential-type for which h f(±ir/2) < 7r.

Our problem is to find additional growth conditions on f that implyf E R1 if f E R. The conditions will be phrased in terms of the "mappingradius" of certain sets associated with the indicator diagram of f.

Definition. Let S be a simply connected open set containing 0 such thatthe complement of S contains at least two points. Let cp be the function(whose existence and uniqueness is guaranteed by the Riemann MappingTheorem) that maps S conformally one-one onto the unit disk D = {z :IzI < 1} and such that w (O) = 0 and cp'(0) > 0. Then cp is the normalizedmapping function of S and p(S) = ) is called the mapping radius of S.

We shall need the following deep theorem of P61ya, which we state with-out proof. The proof may be found in [8].


Theorem. If g(z) = >J 0 b,, E Z, and if g is analytic in a regionS containing 0, with p(S) > 1, then there exist polynomials P and Q, withinteger coefficients, Q(0) = 1, such that

P9=Q

We also require a simple lemma on polynomials, whose proof we leave tothe reader.

Lemma. If P and Q are polynomials with integer coefficients and Q(0) =1, then there exist polynomials P1 and Q1 with integer coefficients Q1(0) =I such that P1 and Q1 have no common factors and PI /Q1 = P/Q.

Now given f E R, let

9(z) _f(n)z'.

00n=0

As we have seen earlier,

f- zew

-b(w)dw9(z) =1

2ri r 1

1

for any curve t that winds once around D(f) = S`(4). Now let S bethe complement of the image of D(f) under the mapping e''', i.e., S =C\ exp(-D(f)).

Now g is analytic in S, so that if p(S) > 1, then g = , where P andQ are polynomials with integer coefficients and Q(0) = 1. By the lemmaabove, we may suppose that P and Q have no common factors. By theconstruction that characterizes R1, we can find a function fl E R1 suchthat f, (n) = f (n) for n = 0, 1, 2,.... By Carlson's Theorem, if we knowthat h f, (f 2) < ir, then we have f = fl, so that f E R. To prove thath f, (±M) < r, we write

M

fl (z) = E P, (z)f; .=1

By construction, the fji 1 are the roots of Q, so that the ,QJ 1 are thesingularities of g, and hence the 131 1 are in the complement of S. Hence,we may write f3j = exp(-,uj), where µj E D(f), so that

m

fi (z) _ Pi (z) exp(-FUiz),j=1

and it follows that hf, (f 2) < a since D(f) is interior to the strip Iyj < 7rWe therefore have proved the next theorem.


Theorem. If f E R, and if the complement of the image of D(f) underthe map a-' has mapping radius exceeding 1, then f E Rl.

For applications, a variant of the foregoing procedure gives a more usefulresult. We let A be the difference operator defined by

(Af)(z) = f(z + 1) - f(z),

defining

and

A°f = f

An+lf = A(Anf)Now, using Taylor's Theorem, we may write

A=eD-1An = (eD - 1)n

where

Ddz'

We define the functionals Tn and Tn by

Tnf = f (n)T* f = (Anf)(0)

To illustrate,T = f (0)

Tif =f(1)-f(0)T2*f=f(2)-2f(1)+f(0)T3 *f = f(3)-3f(2)+3f(1)- f(0).

It is easy to show that

T,a = (-1)n >(-1)k (n)Tk.0

Tn=E(k)Tk.

For example, the first identity is proved on writing

n()e)(_1y1_1c.An = (eD - 1)n =

k0


Definition. To say that a sequence {bn }, n = 0, 1,2,..., is K *-admissibleis to say that there is an f E R such that

T; ,f =b,,, n=0,1,2,....

For f E R, write

It is easily seen that

00

g(z) = E(Tn*f)zn-0

27ri 1 - z((w)-D(d)dw'g(z) = Jr

where ((w) = e' - 1 and I' is a curve that winds once around D(f). Wesee that g is analytic outside the image of D(f) under the map (e'° -1)-1.The argument may be reversed to prove the next result.

Theorem. A sequence {bn} is K`-admissible if and only ifEbnzn isanalytic on the segment [-1,01.

We may now prove the main result of this section.

Theorem. If f E R1, let E be the complement of the image of D(f)under the mapping (e' - 1)-1 and let E" be the image of D(f) under themapping eL - 1. If p(>) > 1, then f E R1 and f (z) = F, Pk(z)(1 + ak)z,where the Pk are polynomials and the Qk run through the complete sets ofconjugate algebraic integers lying in `.

Proof. Let

g(z) = E(Tk f)zk,

and letF(z) = E(Tkf)zk

As we have seen,

Now,

1 1

9(z) = Jr(w)21ridw

1 - z(e"' - 1)

F(z)27ri J

ID(w)1- zew

dw.r

1 1 / 1

ewdw9(z) = 27ri 1 + z r

(w)1 - 1+z

so that

9(z)= 1+zF(1+z}


Similarly,

F(w)=I

Iwg+1 ww )-

Since p(E) > 1, we see by the P61ya Theorem that g = 11, where Pand Q are polynomials with integer coefficients and Q(O) = 1. Thus,

1 P(1 w)w) (1-w)'+1Q(1-WW) Q*(w)'

where we choose N > max(deg P, deg Q). Now, P* and Q* are polynomialswith integer coefficients and Q* (0) = 1. Thus, f E R1. As before, we seethat

f (z) = E A(z)'Y; "

where the yi are the reciprocals of the roots of Q* and the Pi are polyno-mials. If we write yi = 1 +- pi, we see that Qs 1 is a root of Q, and since theroots of Q are the singularities of g, the theorem is proved.

Using this theorem and some facts about algebraic numbers, the next tworesults can be proved easily. We state them without proof, as illustrativeapplications. For details, see the paper of Buck [7].

Theorem. If f is an integer-valued function of exponential type such thathf(7r/2) = hf(-ir/2) = 0 (that is, the indicator diagram off is a horizontalline segment), and if L = exp h f(0) - exp h f(7r) < 4, then f E R1.

Theorem. If, in addition, L < V5-, then for some polynomialsPo, Pi,... , Pn, we have

f (z) = Po(z) + Pi (z)2z + + P,n (z)nz.

22On Small Entire Functions ofExponential-Type with Given Zeros

This chapter is extracted from a paper of the same name by P. Malliavinand L. A. Rubel [22]. We obtain here a result that considerably generalizesCarlson's Theorem presented in Chapter 20.

For a sequence A of positive real numbers, we denote by F(A) the ideal,in the ring of all entire functions, of those entire functions that vanish atleast on A. (We exclude once and for all the null function f = 0 and theideal containing only the null function.) We introduce an order relationin this system of ideals, F(A) < F(A'), meaning that for each g E F(A'),there is an f E F(A) such that If (iy)I < Ig(iy) I for every real y. Crudelystated, F(A) < F(A') if it is easier to construct small entire functions thatvanish on A than those that vanish on A'.

The major problem is to decide, by elementary computations on A andA', whether F(A) < F(A'); we solve this problem here. By specialization,then, we prove as a corollary the following result.

Theorem 22.1. There exists a function f E F(A) such thatIf (iy)I < exp Trblyl if and only if

)(y)-)(x) <blog(x)+0(1), x<y,

where )(t) is the sum of the reciprocals of the elements of A that do notexceed t.

Remark. Carlson's Theorem deals with the case b = 1 and A = {1, 2,3,...The main innovation of our method is to give our entire functions suitablezeros on the imaginary axis, in addition to the required real zeros.

22. On Small Entire Functions of Exponential-Type with Given Zeros 147

We proceed now to the body of the exposition. We study sequencesA = {an} of positive real numbers,

A:0<Ao<A, <...and define

A(t) = >i an i

tA(t) >2 1=J sd.1(s).

An <t 0

Definition. A(t) is called the characteristic logarithm of A, and A(t) iscalled the counting function of A.

For simplicity, we suppose that A is an infinite sequence and that

D(A) = lim supA(t)

< oo,t-.oo t

since the problem is trivial if A is finite or if D(A) = oo. The functionW(z) = W(z : A) belonging to F(A) is called the Weierstrass product(over A) and is defined by

!W(z:A)=H 1-z2

z).\\ n

We may write

log IW(z A)l = log I 1- ezi8 l dA(t),jwhere z = reie. For 0 # 0, 7r we have, on integrating by parts,

log IW(reie)I = r J P(t,0)A(rt)I dt,00

where

P(t, 6) =2

1-t2 cos201 - 2t2 cos 20 + t4'

We define, for 0 1)

W(z:Ab)_ sin 7rbz

irbzhwb (0) = irbl sin 61.

We write A C A' to indicate that A is a subsequence of A', and remarkthat A C A' if and only if A(y) -A(x) < A'(y) - A'(x) for x < y.


Definition. A is equivalent to A', written A - A', shall mean that A'(x) -A(x) = 0(1).

Definition. A' > A shall mean that there exists a sequence A", A" D A,such that A" - A.

Definition. A < A' shall mean that there exists a sequence A"', A' C A',such that A"' - A.

Although A < A' and A' > A mean two different things, the first corol-lary of the next lemma resolves this notational difficulty.

Lemma 22.2. A > A' if and only if

(22.1) a(y) - J,(x) < A'(y) - A'(x) + D(1); 0 < x < y < oo.

Likewise, A < A' holds if and only if (22.1) is satisfied.

Corollary 22.3. A < A' if and only if A'> A.

Corollary 22.4. If A - A1i A' - Ai, and A < A', then Al < A'1.

Corollary 22.5. If Al < A2 and A2 < A3i then Al < A3.

Corollary 22.6. If A < A' and A' < A, then A - A'.

Thus, < is a well-defined partial ordering of equivalence classes under

Proof of Lemma 22.2. That A' > A and A < A' each imply (22.1) is trivial.To show that (22.1) implies that A' > A, we define

W(x) = inf{A'(s) - A(s) : s > x}.

It follows from (22.1) that V(x) > -K for some constant K.Now V(x) is constant except for possible jumps at the jumps of A(x).

Let xo be a point of discontinuity of W. Then,

W(xo - 0) = A'(xo - 0) - A(xo - 0)

and

p(xo + 0) < A'(xo + 0) - A(xo + 0).

We denote by i (xo) the jump of at xo. Then

(22.2)

We let

L 4 (xo) < AA'(xo) - AA(xo) < AA'(xo)

A*(t) = [W(t)),


where [a] denotes the integral part of a andt

-D(t) = f s dcp(s),U

and let A* (t) be the characteristic logarithm of that sequence A* whosecounting function is A*(t). The function A*(t) is constant except possiblyat the jumps of ap(t), and we have

AA* (X0) < o + i (xo)

Using (22.2), we get

(22.3) t1A*(xo) < 1 + AA, (X0).xo

Furthermore, xo0)*(xo) and xo0)'(xo) must be integers, so that (22.3)implies

(22.4) AA*(xo) < AA'(xo),

and this means that A* is a subsequence of A'.We now define

A"(x) =A(X) + A*(x),

so that A"(x) is the characteristic logarithm of some sequence A" D A. Toprove that A" - A', we must prove that 6(x) = 0(1), where

6(x) = A(x) + A* (X) - A, (X).

Now,

and

td4i(s)W(t) - (P(0) = fo

s

A*(t) = tf 1 d('F(s)]o

An integration by parts shows that

A* (t) - W(t) = 0 (i),so that it is enough to prove that 9(x) = 0(1), where

0(x) = )(x) + W(x) - )'(x) = )(x) - A'(x) + inf{A'(s) - A($) : s > x}.But it is clear that 9(x) < 0, and (22.1) is simply another way of sayingthat 9(x) > 0(1).

To prove that (22.1) implies that A < A', we put(22.5) AY"(x) = )'(x) - A*(x).

Since by (22.4) A* is a subsequence of A', there is a subsequence A" definedby (22.5) and A" is a subsequence of A'. Since we already have shown thatA"' A, i.e., 6(x) = 0(1), the proof is complete.

We now state the main result.


Main Theorem. Given A and A', the following three statement are equiv-alent

(i) F(A) < F(A').(ii) A < A'.(iii) There exists a single pair, fo, go with fo E F(A), go E F(A'),

Ifo(iy)I < Igo(iy) I for all real y and such that the only zeros of go in theopen right half-plane belong to A.

Theorem 22.1 is a direct corollary of this result. Given A and b, chooseA' = Ab and go(z) = Since Igo(iy)I - e'"bl't, the equivalence of (ii)and (iii) proves Theorem 22.1.

Proof of the Main Theorem. We leave the proof that (ii) implies (i) forlater. It is clear that (i) implies (iii); a suitable choice for go(z) in (iii) isthe Weierstrass product W(z : A'). We now prove that (iii) implies (ii).We write f and g instead of fo and go. Now we choose p with 0 p, and write oneform of Carleman's Theorem (Chapter 12), taking y > x > p as

(22.6) E(y) - E(x) = I(y) - I(x) + J(y) - J(x) + 0(1),

where

E(R)=E(R: f)= n - Rz )Cos0,,,

I(R) = I(R:f) = 1

R

(1 1 ) log If (it)f (-it) I dt,

21r t2 R2

J(R) = J(R : f) _ 7rRJ

I log If(Rese) I cos a d8.

Now

(22.7)

since

R2 cos0 = O(1)

rncos0, < rn = 1 L n- 1 n(R) = 0(1),

R2 R2 R - R

where n(r) counts the number of zeros of f whose modulus does not exceedr.

Also,

(22.8) J(R) = 0(1)


since

1 j log If (Reie)I cos Bd8R

a2

< R J log If (Reie)II dOx

5RO(R) = 0(1),

since

1 Iloglf(Re'B)II d0=rn(R,f)+m(R, 1) <2T(R,f)27r j x

and f is of exponential-type.From (22.7), we obtain

(22.9) E(y) - E(x) > A(y) - A(x) + 0(1),

and using (22.8) and (22.9) in (22.6) we get

(22.10) A(y) - A(x) < 1(y) - I(x) + 0(1).

Now, since If (iy)I < Ig(iy)I, we see that

(22.11) I(y:f)-I(x:f) :5 I(y:9)-I(x:g).On the other hand, applying Carleman's Theorem now to g, whose onlyzeros in the right half-plane are the A' , we see that

(22.12) I(y : g) - I(x : 9) = A'(y) - A'(x) + 0(1).

Combining (22.12) with (22.11) and (22.10), we get

A(y) - A(x) < A'(y) - A' (X) + 0(1),

and the proof is complete.

To prove now that (ii) implies (i), we suppose that A(y)-A(x) < A'(y)-A'(x) + 0(1), and we are given g E F(A'); we must construct a functionf E F(A) with I f (iy)I 5 I9(iy)I for all y. By Lemma 23.1, we may supposethat A(t) = Y(t) + 0(1), since A is a subsequence of a sequence A" forwhich this is true, and F(A") C F(A).

By the Hadamard Factorization Theorem we may write

9(z) = 91(z)92(z),

where

9i(z)=H(1- 3 Iexp (z),x z l

92(x) = CzIèa2rf 1 - Sn expCan


where the (n 54 0 are the zeros of g that are not counted in A'.Writing log 1gl(iy)I as a sum of logarithms, and that sum as a Stieltjes

integral, we get

(22.13) log Igl(iy)I = 2 J log (1 + t2 J t dA'(t).o \ /

The next lemma provides the main tool of our construction; it will enableus to "move the zeros" from the real axis to the imaginary axis.

Lemma 22.7. Let dL be a measure with compact support contained inan interval f f, e-1) for some small e > 0. Then there exists a function V(t)defined on (0, oo) such that

(22.14) flog (1 + t2 d0(t) = r log I1 - e I w(t) dA(t)

and

IVI < 2 sup ( 1 y 1 dL(s) I.z 0 sProof. By a contour integration, it is easy to see that

roo 2log (1+x21_ J0

log I1- tt

By Fubini's theorem,(22.15)

fiog(i+E) dd(r) = fiogii_i{Jz1t21do(t)} dw

We therefore are led to define

(22.16)2 f t2 d0(t)

W(w) _ i w2 + t2 t

and (22.14) asserts (22.13) in another form. The bound on 'p follows fromintegrating by parts in (22.15):

(22.17) 'P(w) _ 2 2

2

J f J x dx +w )0 0 \xHence,

1W(W)1 1I.1+ J 1 dx ( + w2

)I STp I J x da(t)0


since-zw is increasing, and the lemma is proved.xr+

We now choose

(22.18) 6(t) = 2 {A'(t) - A(t)}, d0(t) = t d6(t)

but cannot apply Lemma 22.7 to d0 since its support may not be compact.We truncate the support by defining

6k(t)- 6(t) ift<k{ 6(k) if t > k

Ok(t) = t d6k(t)

with the same convention for A(t) and A'(t).We now apply Lemma 22.7 to dOk and conclude that there exist func-

tions Wk(t) such that

/ z1 r(22.19) 1 1 + tJ dAk(t) log I1 - jpk(t) dt.flog

Now,

(22.20) Iwk(t)f < B,

where B is a constant that is independent of k, namely, from the bound onIV(t)l in the lemma and the equivalence of A and A',

B=2supIA(t)-A'(t)1.

On putting

/ ya \(22.21) Lk(y) = fiog

1

1 1 + Via) t dAk(t) + floe1 1 -y2

t2 I dlbk(t),

where

(22.22) d4ik(t) = cak(t) dt,

we have

/(22.23) Lk(y) = 2 flog f 1 + dA(t).

\\

Hence, by (22.13),

(22.24) kl oo Lk(y) = log I91(iy)I


At this point, the idea is to find an entire function F for which thehypothetical formula

R!log I F(iy)l = k f log 11 - z I dfik(t)

holds in some appropriate sense. First, however, the limit need not ex-ist, but a simple argument with normal families will handle this difficulty.Also, the measures d4k(t) = cpk(t)dt are unsuitable since they need not bepositive and cannot be discrete. [It is easy to see that all the d4k(t) arepositive only in case A C A', a trivial case.] But first we show that addinga constant to Yak, in order to make d<bk(t) positive, does not change L.Then we show that the resulting measure may be made discrete with littleloss of precision.

Resuming the construction, we define

(22.25) PA; (t) = B + cpk(t)

and by (22.20) conclude that

(22.26) 1/ok(t) > 0 for all t.

A contour integration shows that

(22.27) fioghi_hhl dt = 0,E!

so that

a(22.28) Lk(y) = f log (i + ta)

tt dak(t) + flog I1 -

z

t I

where d'Yk(t) = 1k(t) dt. Now let Wk(t) = [Wk(t)], the integral part ofWYk(t), and define

M!)z

(22.29) Lk(y) = f log (1 + t2)t2 d)tk(t) + flog I1- tz I dlk(t).

Lemma 22.8. There is a constant fl, independent of k, such that for ally>1

(22.30) f log 11 - to I d`l`k(t) < flog 11 - t I dWk(t) +,Qlog IyI

Proof. We apply the next lemma with %Pk(t) = v(t) and '1 (t) = n(t). ,0is independent of k because dand ITkt) -'pk()i are boundedindependently of k.

22, On Small Entire Functions of Exponential-Type with Given Zeros 155

Lemma 22.9. Suppose that v(r) is a continuously differentiable functionfor 0 < r < oo, that 0 < V(r) n(r) > v(r) - C.

Then

J log 11 - t2zI dn(t) < r log I1- e l dv(t) + 0(logy)

asy-->oc. J

Proof. For fixed r, we write L(t) = log 1 1 - 2 I and point out that L isLebesgue integrable on (0, oo):

L(0+) = +oo, L(r-) = L(r+) = -oo, L(oo) = 0

and that L(t) is decreasing and continuous for t E (0, r) and increasingand continuous for t E (r, oo). We must compare Y = fo L(t)dn(t) andZ = f °O L(t)dv(t). We will prove that Y < Z + O(log r). We assume thatv'(t) > p > 0. This involves no loss of generality since, if we replace v(t)by v(t) + t and n(t) by n(t) + t, we change Y and Z not at all, becausef0 L(t)dt = 0. We may suppose, without loss of generality, that v(0) = 0,since suitably redefining v on the interval [0,11 changes Z only by O (1),which is small compared to the allowed discrepancy O(log r).

With each large r we associate the numbers r1 and r2 such that

v(rj)=n(r)=v(r2)-C.

Since v'(t) > p, we will have r - rl < r2 - r1 < v . It is easy to see thatthe following inequalities hold:

Ifr L(t) dn(t) <

Jr1 L(t) dv(t),

0 0

J L(t) dn(t) < J L(t) dv(t).r rz

It follows that Y = X + Z, where

rs 2

X = - J log I1 - t2 I dv(t).r,

We shall prove that X < O(log r). Clearly,

rrsX<-! log I t t

_r I dv(t).r,


Since r2 - rl < p and v'(t) < B, we have

jr2 t-r rrsX<-B log-I t I dt<B(rz-rl)logrz-B J log-It-rldt

i ddd r1

so that

X< pClog(r+C)+2B.

We now consider polynomials Pk

`\defined by

flog!log IPk(z)I = 1 +tzz

I d''k(t).

Lemma 22.10. There is a constant B', independent of k, such that forall z

(22.31) log IPk(z)I < B'Iz .

Proof. Putting z = x + iy, we see that

fiogii+..i dWk(t) <flog 11

+tz

2 I dW(t)

But since 'It(t) = [ fo {B + cpk(s)}ds], and since Icpk(s)I < B by (22.20),the proof is immediate on integrating by parts.

Since the family {Pk(z)} is therefore uniformly bounded in each diskDR = {z : IzI < R}, it is consequently a normal family, and we mayextract a subsequence {Pk,(z)} that converges uniformly on compact setsto an entire function F:

(22.32) F(z) = kJim Pk.(z).1-00

Because Pk(0) = 1 for each k, it follows that F(0) = 1. By Lemma 22.10,we conclude that F is of exponential-type. Since Pk has only imaginaryzeros, so has F. Furthermore, F is an even function since each Pk is.

Let i1' = {iryn} be the zeros of F on the positive imaginary axis. SinceF is of exponential-type, 1' has finite upper density. Thus,

z(22.33) F(z) = II (1 + z2 }

'Yn

At last we can define f (z). Let

(22.34) f (z) = fl(z)F(z)gz(z),


where

(22.35) f1(z)=11(1- Zj exp z 1As

a consequence of the estimates (22.24) and (22.30) we see that

(22.36) log If (iy) I < log I g(iy) I + O(log IYD.

Now (22.36) is as good for our purposes as If (iy)I 5 Ig(iy)I, since we couldotherwise consider f *(z) = f (z)a{(iz)-1 sin(iz)}b for a suitable choice of aand b.

It is not obvious, though, that f is of exponential-type, since fl and g2need not be of exponential-type, although they are certainly of order 1.To prove that f is of exponential-type we appeal to Lindelof's Theoremproven in Chapter 13.

Let us denote by a and b the zeros, other than the origin, of f and g,respectively. Then we see that

Ibnl<Rbn ¢A'

lbnI<Rbn4A'

Ibnl<Rbn fA'

bn' + A(R) + (.\'(R) - A(R))

bn' + A(R) + 0(1)

since A'(R) - A(R) = 0(1) by hypothesis. Now the zeros of f, other thanthe origin, fall into three categories: (i) those bn not counted in A', (ii) theelements of A, and (iii) the zeros of F. If we consider

S(R) = E b.-1,

the zeros of F contribute nothing to S(R) since F is even. Hence,

S(R) = bn1,l bn I <Rbn jW

and it follows thatS(R) = O(1).

This, together with the obvious fact that the zeros of f have finite upperdensity, implies (via Lindelof's Theorem) that f is of exponential-type.The proof is complete.

23The First-Order Theory of theRing of All Entire Functions

The material of this chapter is drawn from the paper [3], "First-OrderConformal Invariants." Let E denote the ring of all entire functions as anabstract ring. Much information about the theory of entire functions ispresent in the theory of C. For example, an entire function f omits thevalue 7 if there exists an entire function g such that (f - 7)g = 1.

We show that even using a restricted logic (first-order logic), a greatdeal can be expressed in the theory of E. We shall show, indeed, that allof classical function theory can be so expressed. By the ring language wemean the first-order formal language appropriate to the structure E. Thislanguage has basic symbols for addition and multiplication of entire func-tions, as well as the usual logical symbols: A ("and"), V ("or), -, ("not")and = ("implies") as well as quantifier symbols d ("for all") and 3 ("thereexists") together with variables that range over E. For convenience we alsoinclude in the ring language a constant symbol which is a name for theconstant function i = vr---I. (Otherwise, there would be no way to distin-guish between the two solutions of f2 + 1 = 0.) Formulas and sentencesin this language are finite combinations of these basic symbols, arrangedaccording to the obvious formal rules of grammar. A key restriction is thatthe language is first-order, which means that we can only use quantifiersover elements of E and not over subsets, ideals, relations, etc. Also, theexpressions in a first-order language are always finite in length. (See [39]for a general treatment.)

The algebra language is appropriate to E as an algebra.. This is formed byadding to the ring language a 1-place predicate Const. In E (as an algebra)we interpret Const(f) to mean that f is a constant function. (Thus we are

23. The First-Order Theory of the Ring of All Entire Functions 159

identifying C with the field of constant functions in E and are using theordinary addition and multiplication of functions in E to play the role ofaddition of constants and the scalar multiplication in the algebra.)

But in E it is easy to say that f is a constant (either f = 0, f = 1, orf omits 0 and 1) so that the ring language and the algebra language areequivalent.

In dealing with rings and algebras, the expressive power of first-ordersentences is reasonably well understood. For example, to say that a ringis commutative is first-order (VxVy(xy = yx)) but, at least superficially,to say that a ring is simple is not first-order, since it seems to requirequantification over subsets (there does not exist a proper two-sided ideal),and in fact does so require. It is first-order to say that a ring has at leasttwo elements (3x3y(x # y)), but it is not first-order to say that a ringis infinite, since this requires a sentence of infinite length, as one mightsuspect.

Nearly all of the results in this chapter depend on the fundamentaldefinability results for the algebra E. We show that there are formulas inthe algebra language which define in t the set N of nonnegative integers,the set Z of all the integers, the field of rational numbers Q, the field of realnumbers IR, the ordering relation < on 1[t, and the absolute value functionon C. We also show how to interpret in the first-order language of 6 thequantifiers that range over countable sets and sequences of constants. (It isstriking that second-order concepts can be represented within the restrictedfirst-order language.)

An immediate consequence of this is that second-order number theory isinterpretable in the first-order theory of E. Other recursive undecidabilityresults are treated later. For example, we show that the first-order theoryof the ring of entire functions is recursively isomorphic to second-ordernumber theory. (This improves on a result of Robinson [33], who showedhow to interpret first-order number theory in the first-order theory of entirefunctions.)

Later we extend these definability results even further. We give enoughexamples to suggest that all of classical analytic function theory on C canbe interpreted in the first-order theory of E. This is quite surprising, giventhe apparent limitations of the first-order algebra language. We also showhow to interpret in the first-order language of E the quantifiers that rangeover countable subsets and sequences in E itself. That is, the first-orderlanguage is already as expressive in this context as the restricted second-order language.

The Ring LanguageIn this section we will begin to explore the expressive power of the first-order theory of the ring E. To begin, note that the constant functions 0 and1 are definable using their first-order properties in E. Also, the propertythat f is a unit in E is first-order expressible: 3g(fg = 1). Thus we can


express, for any definable constant c, the condition that f omits the valuec on C by saying that f - c is a unit in £. Since 1 and i are definable in £,so is each Gaussian rational number. This means that for each Gaussianrational q there is a formula Fg(x) in the algebra language such that, forany function f in £, Fq (f) holds in £ if and only if f equals the constantfunction q.

In this section we present a detailed study of certain basic definabilityquestions for the algebra e. These matters are fundamental to the generalcontent of this chapter and are of interest in their own right.

We call a function f E £ a point function if it has a unique zero on C,of multiplicity one. These will be used to represent the points of C withinC. It is easy to construct a formula P(x) in the algebra language such thatP(f) holds in £ if and only if f is a point function. That is, P(f) shouldexpress

f is notaunit,anddgVh[f = gh = g is a unit or h is a unit].

Using this we can, for example, construct an algebra language formulaA(x) such that A(f) holds in .6 if and only if f is a 1- i conformal mappingon C. That is, A(f) should express the condition:

For any constant a and any point functions g and h,if both g and h divide f - a, then g divides h.

[To say that a point function g divides f - a is equivalent to saying thatf has value a at the unique point of C where g is zero. To say that onepoint function divides another is equivalent to asserting that they are zeroat the same point of C. Of course the conformal maps of C are exactly thefunctions f (z) = az + b, a A 0.]

Next we discuss how to code arbitrary countable or finite sets of con-stants in C using first-order formulas. It is very striking that we can rep-resent second-order mathematical concepts in a first-order language.

Given f, g E C, define V(a; f, g) to mean that a is a constant, g 34 0,and there exists zo E C such that g(zo) = 0 and f (zo) = a. This can berepresented by an algebra language formula, since V(a; f,g) is equivalentto the existence of a point function h such that h divides g and h dividesf - a (together with the other conditions, that a is constant and g # 0).We think of the pair (f, g) as coding the set of constants

E = {aECI V(a; g) holds in 61.

If g = 0, then this set is empty. If g 0, then it is a countable or finitesubset of C. Moreover, by taking g to have an infinite sequence of zerostending to oc and by letting f vary over £, then we obtain for E all possiblecountable or finite subsets of C [38].


We may use this idea, for example, to find a formula IZ(x) in the algebralanguage such that IZ(g) holds in £ if and only if the zero set of g in C isinfinite while g # 0. Namely, let IZ(g) be the formula:

g # 0 A 3f [V (0; f, 9) A da(V (a; f, 9) = V (a + 1; f, 9))]

This condition asserts the existence of a function f such that the set EN. By the well-known interpolation theorem for entire functions this canhappen exactly when g has infinitely many zeros.

Theorem 23.1. The following sets and relations are all definable in £by formulas in the algebra language:

The set Z of positive and negative) integers,That set N of nonnegative integers,The set Q of rational numbers,The set Q(i) of Gaussian nationals,The ordering relations < on Z, N, Q, andThe absolute value function on Q.

Proof. This is immediate from the discussion above. To get N we use thePeano axioms: N is the smallest countable subset of C that contains 0 andcontains a+1 whenever it contains a. Precisely, the formula N(a) definingNis

a is constant A

bfd9[{V(0;f,9) AVQ(V(/3;f,9)) V(/3+ 1,f,9)}V(a; f, 9)]

Once having defined N, Z and Q are trivial:

aEZ4--* aENV -aEN,a EZAyENAy j4 QAay=p].

To define the Gaussian rationals Q(i) we note that

aEQ(i),# 3Q3ry[f EQA7EQAa= fi+yi].

To define <, on Q say, note that

a E Q,

lal=QU0<_/3A(fi=aV

Note that since N is definable in E, there is an effective procedure forinterpreting the first-order theory of (N, +, ) into the first-order theory of


E. (The operations + and are just the natural operations on £.) Actually,this interpretation extends to the second-order theory of (N, +, ), since wehave a way to discuss countable sets of constants in the first-order languageof C. It follows that the first-order theory of £ is very undecidable in thesense of recursive function theory. This will be discussed fully below.

We caution the reader that at this stage we are only able to define thefield of rational constants in E. It is possible to define the real field R, aswe show below, but there does not seem to be any easy way to do it.

Once we have a first-order definition of R in £, no matter what it is, weget immediately as a bonus a first-order defuiition of < on R and of theabsolute value on C:

IaI =&3 373S[7ERA6ERAa=7+SiA/iERAO<QA/32=72+62].

We now turn to the construction of an algebra language formula R(x)that defines the field of real constants in E. The construction of R(x) iscomplicated, but it is based on an elementary fact: A constant a is a realnumber if and only if there is a Cauchy sequence S of rationals such thatevery analytic function that carries S to an equivalent Cauchy sequenceof rationals also preserves a. Our problem has two parts: to prove thatthis equivalence is correct and to construct an algebra language formulathat expresses the right side of it. To prove the equivalence we need thefollowing fact:

Lemma A. If / E R and a E C\R, then there is an entire function fsuch that f (/3) = /3, f (a) 54 a, and f maps Q onto Q.

Proof. This argument is a simple modification of the proof given in [27].Fix Q E R and a E C\R. If ,6 E Q, then f is trivial to find, so we mayassume /3 V Q also. We take p(z) = e1, so that p(z) > elzl for all z E C.Take S = T = Q U {,0} and choose fo E EM so that

fo(3) = 2/3,I3`(fo(a) -a)I -I! 2p(IaI),

and let 6 = 2. (Here, EM is the set of all entire functions whose restrictionto R is a real, monotonically nondecreasing function.) Let us take anenumeration of S (and T) with xl = (3 (in particular, xl 54 0), and let6 > 0 be such that

fo(xi) + 6x1 E T and ISzI < 2p(Izl) for all z E C,

where fo is any function chosen from EM. We define

f1(z) = fo(z) + Sz, S1 = {xt} and T1 ={fl(at)}.


Note that Sl = {i3} and fl(p) _ /3, so that Tl = {9} also. We nowconstruct the sequences {fn}, {Sn}, and {T.} so that

fn(x) (2-1 + 2-")b and fn(Sn) = Tn

for all n E N, X E R. Suppose that fn, Sn, and Tn have been constructedand choose a polynomial g with real coefficients such that

(1) g(z) = 04= z E S. (z E C)(2) Ig(z)I 5 2 n-1P(IzI) (z E C)(3) g'(x) > -2-n-16 (x E R).

[Any polynomial of odd degree with positive leading coefficient for which(1) is valid will also obey (2) and (3) after it has been multiplied by a smallenough positive constant. The degree can be chosen odd by adjusting themultiplicity of one of the zeros.]

For each M E [0, 1] we have

(fn + Mg)'(x) = fn(x) + Mg'(x) > (2-1 - 2-n-1)b (x E ]R),

so that In + Mg is strictly increasing on R.Moreover, if we let M vary in [0, 1], then for x V S,,, (fn + Mg)(x)

varies in an interval of R that contains points of T\Tn; and for y V Tn,(fn + Mg)-1(y) varies in an interval of JR that contains points of S\Sn.

Now for n odd, let x be the point of S\Sn with smallest index and letM E [0,1] be such that (fn + Mg)(x) E T. We define

fn+1 = fn + Mg, Sn+1 = S. U {x}

and

Tn+1 =Tn U {fn+1(x)}.

For n even, let y be the point of T\Tn with smallest index and letM E [0, 1] be such that { fn + Mg}-1(y) E S. We define

fn+1 = In + Mg, Tn+1 = Tn U {y}

and

Sn+1 = Sn U {fn+l(y)}.

The following properties of the constructed sequences are easily verified:(a) Ifn(z) - fn-1(z)I <- 2-np(IzI) (n E N,z E C)

00 00

(b) U Sn = S, U Tn = T, andn=1 n=1

fm (Sn) = Tn, (m, n E N, m > n).

From (a) it follows that fn converges pointwise to a function f for which

1 f (z) - fo(z)I <- P(Iz) (z E C).

164 Entire and Merornorphic Functions

Now p(IzI) is a function of z that is bounded on compact subsets of C, sothe convergence of {f, (z)} is uniform on such sets. From this we concludethat f is an entire function.

For each n E N we have fn(x) > 16, (x c ]l2), so the same is true for f.Hence, f is strictly increasing on R. From (b), it follows that f (Sn) = Tnfor each n, and so f(S) = T. Moreover, we have insured that fn (p) = /3for all n E N and therefore f (/3) = /3. This implies f (Q) = Q also.

Finally we have

I(f(a) - a)I I(fo(a) - a)I - I(fo(a) - f(a))I> 2p(I al) - p(IaI) > 0,

which means f (a) 0 a. This completes the proof.

Now it is easy, using Lemma A, to show that our characterization ofthe real numbers is correct. In one direction, the equivalence is trivial: Ifa E P., we need only take S to be a Cauchy sequence of rationale thatconverges to a. For the converse direction, suppose that a E C\R and S isany Cauchy sequence from Q. Let # be the limit of S in R Use Lemma Ato get an entire function f such that f (/j) = ,Q, f (a) # a, and f (Q) C Q.Then f maps S to a Cauchy sequence of rationals that is equivalent to S(since both converge to Q) yet f does not preserve a. That is, a E C\Rimplies that the right side of our equivalence is false. This completes theproof that our characterization of R is correct. Now we must express itformally within the algebra language.

Theorem 23.2. There are formulas R(x), L(x, y), and M(x, y) in thealgebra language such that for any a, ,0 E E:

(i) a E HIS R(a) holds in e;(ii) For a, Q E P, a < fl = L(a, /3) holds in E;(iii) For a, /3 E C, j al = /3-#=:>=:a, 0) holds in E.

Proof. We begin by building some machinery for discussing sequences ofconstants within the first-order language of E. (Earlier we did the samefor countable sets of constants.) This is done using a triple of functions(f, g, h); g has infinitely many zeros on C, and on that zero set h takes onthe values 0, 1, 2,... . In effect, h lists the zeros of g. Then the sequencecoded by (f, g, h) is (a,,), where a,, is the value f (z,,) at the zero of gwhere h take the value n. First let Basis (g, h) be a formula in the algebralanguage which expresses the fact that h "lists" the zeros of g in the mannerdiscussed above:

Basis(g, h)t=da[V(a; h, g)t=a E NJA daVpVq[{P(p) A P(q) A p divides q

A p divides h - a A q divides g A q divides h - a}

= pdivides q].


Now we construct an algebra language formula Seq(a, n; f, g, h) whichexpresses that a is the nth term of the sequence of constants coded by thetriple (f, g, h):

Seq(a,n; E NABasis(f,g,h)A a E C A 3p[P(p) A p divides g A p divides h - n

A p divides f - a].

Using the defining formulas described in Theorem 24.1, we now cansay, using a formula in the algebra language, when the sequence coded by(f,g, h) is a Cauchy sequence of rational numbers and when this sequenceof rationals converges to 0. (Note that since we have the absolute valuefunction only on Q at this point, there is no hope of discussing convergingor Cauchy sequences outside Q. This is precisely our difficulty in this entirediscussion.) For the first of these:

Cauchy Rat Seq(f, g, h)4-- Basis(g, h)

A VaVn(Seq(a, n, f, g, h) a E Q)

AV6 EQ+3mENdi,j ENVa,Q EQ[{m <iA m < j A Seq(a, i, f, g, h) A Seq(,Q, j, f, g, h) }

* Ia-pI<_6].

Here we simply have written down in formal terms the usual version of theconcept to be defined. (Q+ is the set {q E Q 10 < q}, so it is defined byan algebra language formula.)

For rational sequences converging to 0, we have an entirely similar for-mula:

Zero Rat Seq (f, g, h)t Basis(g, h)A dadn(Seq(a, n, f, g, h) a E Q)

Ad6 E Q+ 3m E NVi E lWa E Q[{m <_ i A Seq(a,i, f,gh)}

= Jal < 6].

Consider a pair (g, h) for which Basis (g, h) holds and suppose that fl,f2 are functions so that (fl, g, h) and (f2i g, h) code Cauchy sequences ofrational numbers, say (an) and (,3n), respectively. Then it is clear that(h - f2, g, h) codes the sequence (an - Qn). [Here it is essential that thesame (g, h) be used.] Therefore, (an) and (On) are equivalent if Zero RatSeq(fl - f2i g, h) holds.

Next we face the second major difficulty in formalizing our charactriza-tion of R in the algebra language: We have no direct means of expressingthat "the value of f at a equals fl," where f E E and a, p E C. Weapproach this indirectly by introducing, as a parameter, a 1 - 1 conformal


mapping a on C. Of course a is an element of E and, as noted above, theproperty of being a 1 - 1 conformal mapping is expressible by an algebralanguage formula A(u). Now we define a formula in the algebra language,which we will abbreviate by writing f (a) = /3, by

Cr

f (a) = /3=a is a 1 - 1 conformal mapor

Aa E range(a) A f (a-1(a)) = /3

e=A(a) A a - a is not a unitAa, /3 E C A a- a divides f -/3.

This formula, which will be quite important in later sections as well, hereenables us to express the condition that the composition f o a-1 carriesone sequence coded by (f1, g, h) to a second sequence coded by (f2, g, h).The formula that expresses this is the following:

'dra E NVa, /3 E C[Seq(a, n, fl, g, h) A Seq(/3, n, f2, g, h) f (a)o 91.

Note that for a fixed 1-1 conformal mapping a, as f ranges over E thecomposition mapping f o a-1 ranges over E.

We now are ready to give the formula in the algebra language whichdefines R in E. (It is convenient to define C\R instead.) We see thata E C\R,# a E C and there is a 1- 1 conformal mapping a on C, andfor any Cauchy sequence (an) of rationale that is f E E and a CauchySequence (/3n) of rationals with the properties

(1) f (an) = /3n for all n E N);

(ii) (an) and (/3n) are equivalent Cauchy sequences of rationals;(iii) f (a) = a is false.

aIt remains only to show that this equivalence is correct. Earlier we

proved the equivalence a E C\R=a E C, and for any Cauchy sequence(an) of rationals there is an entire function g and a Cauchy sequence (/3n)of rationals with the properties

(i') g(an) = Qn for all n;(ii') (an) and (/3n) are equivalent Cauchy sequences of rationals;(iii') g(a) 0 a.Clearly, if a exists for a as above, and if (an), (/3n) and f satisfy (i), (ii),

and (iii), then we need only set g equal to f o a-1. Conversely, the rangeof every 1 - 1 conformal mapping or on C includes a and Q. Given sucha a and g satisfying (i'), (ii'), and (iii'), just take f to be the compositiong0a.

This finally completes the proof that R is definable in E. As was dis-cussed earlier, from this we get formulas defining < on R and the absolutevalue on C. Thus the proof of Theorem 23.2 is complete.

Theorem 23.2 is fundamental to nearly all of our other results.


More on the Algebra LanguageIn this section we present a variety of results which, when taken together,show that the expressive power of the algebra language is strong enoughto include all of the classical mathematical theory of entire functions. Itis quite striking that this should be possible, since the first-order algebralanguage seems to be so limited. These results are not used in any otherpart of this chapter, and we have not made a great effort to be completenor to give more than sketchy proofs. Our goal is to give examples thatshow what is possible.

We discussed how to deal with sequences of constants within the first-order theory of E. Now we will improve this to code sequences of functions.The idea is to fix a point zo in C and a sequence {zn } in C that converges tozo. By the uniqueness of analytic continuation, a function f E E is uniquelydetermined by the sequence of constants if (zn)}. Hence, a sequence if. Iof functions can be coded by an infinite matrix of constants If. (z.) m, n EN}. Now this is not quite enough, since we cannot refer to the points ofC in a direct way. We overcame this earlier by introducing a parametera that is a 1 - 1 conformal mapping on C. We replace the points zn bythe constants an = a(zn) and refer to the values Ym,n = fn,(zn) by usingthe equivalent, definable relationship fm(an) =#,",n. Finally, we code thesequence (an) and the matrix (f3,,,n) into a single matrix that has (a - n)as the top row.

To code an infinite matrix of constants we proceed as earlier, but use abasis triple (g, hl, h2) instead of a pair (g, h). Here we require that g(# 0)have an infinite zero set in C, as before, and that (hl, h2) map this zero setbijectively onto N x N. It is routine to construct a formula M Basis(x, y, z)in the algebra language such that M Basis(g, hl, h2) holds in E if and onlyif this basis condition is satisfied.

When M Basis(g, hl, h2) is true, any function f in E determines a matrixof constants (a,nn) by taking a,nn to be the value of f at the unique zeroz,,,n of g for which hl(z,,,n) = m and h2(z,,,n) = n. Earlier we expressedthis relation using point functions. Also, by the interpolation theorem forentire functions, every matrix (a,,,n) is coded in this way by some f, nomatter which basis triple (g, h1, h2) is used.

The matrices (a nn) that arise in coding sequences of functions are notarbitrary, of course. First, the sequence (a,nn) from the top row must be aCauchy sequence in C. Finally, for each m > 0 there must exist a functionf,n in E such that

fm(aon) = amn for all n E N.

Evidently, f,,, is uniquely determined by this condition. It is the nthfunction in the sequence coded by (a,nn) and a. As was discussed pre-viously, we take the matrix (amn) to be coded by some (f, g, h1, h2), whereM Basis(g, hl, h2), holds.

168 Entire and Meroniorphic Functions

We can construct an algebra formula Code(f, g, hi, h2, v) that holds in£ if and only if a is a 1 - 1 conformal mapping on G and M Basis(g, h1, h2)holds and the matrix (amn) coded by f using the basis (g, h,, h2) sat-isfies the conditions above. [That is, the sequence aon and its limit hein the range of a and, for each m > 0, there is an fn in £ so thatf,n(ao,n) = amn for all n E N.] Then we formulate an algebra formulaF Seq(k, m, f, g, h1 i h2, a) that holds in ,6 if and only if Code (f, g, h,, h2, v)holds, m E N, and k is the (unique) function that satisfies k(aon) = amnfor all n E N. That is, F Seq(k, m, f, g, hl, h2, a) holds if and only if k isthe mth function in the sequence coded by (f, g, h1, h2, o).

Using the absolute value on complex constants and the "evaluation" offunctions f via a (that is, in the form f (a) _ 6), we now may obtain algebra

formulas that express various types of convergence of the coded sequence offunctions. This can be done for pointwise convergence, uniform convergenceon C, or even uniform convergence on compact subsets of C. For this lasttype of convergence it is not necessary to quantify over arbitrary compactsets, but rather to use a particular exhaustion of C by compact sets. Forexample, suppose Code (f, g, h1i h2, v) holds in £ and we wish to discussconvergence of the coded sequence of functions (f,,,). Consider the setsG' C range(a) defined by

Gn={aECIlaI<n}.

These sets are definable using an algebra formula from the parameters nand a. Also, (f,n) converges uniformly on compact subsets of C if and onlyif (f,,, o or-') converges uniformly on each of the compact sets G' . Thiscan be expressed by an algebra formula using F Seq at the end to replacemention of (fn).

This method of representing sequences of functions within t enables usto define many specific sequences-for example, the sequence of powers(f'n) of a particular function. FYom this we can find an algebra formulawhich expresses that g is equal to a polynomial in f.

Next we discuss a method for interpreting in £ the lattice of open subsetsof C. (This procedure is also used later, where it is discussed in greaterdetail.) This is done by associating each open set O with the set R(O) ofall pairs (q, r), where q is a Gaussian rational, r is a rational > 0, and thedisc {a E C I ja - qj < r} is contained in O. Since 0 is the union of thisfamily of discs, we see that 0 54 £) implies R(O) # 1Z(D) for open sets 0,D. The set R(O) can be coded by a quadruple (f1, f2, g, h) by first writingR(O) as a sequence (qn, rn) for n E N and then taking (qn, rn) to be thevalue of (f, (z), f2(z)) at the unique z E C for which g(z) = 0 and h(z) = n.Here, (g, h) satisfies the Basis formula described above.

We first obtain an algebra formula 0 Basis(fi, f2, g, h) that is true in £if and only if there is an open set 0 such that R(O) = {an, /3n) I n E N},


where (an, Jan) is the sequence of pairs coded by (fl, f2) using the basis pair(g, h). This must express that every a,, is in Q(i), every fl, is in Q+; also,it must express that if q E Q(i), r E Q+ and if the disc {ry E C I I'y - qi < r}is contained in the union of the discs {ry E C I I' - anI < 8,, 1, then (q, r)is in the set {(an, Jan) n E N}. Of course, 0 is just the union of thediscs {-y E C I Iry - anI < On } for n E N. Thus we can obtain an algebraformula Open(a, f1, f2, q, h) which expresses that 0 Basis(f1, f2, q, h) holdsand that a is in the open set 0 for which R.(0) = {(an, fn) I n E N}.

Evidently we also can represent sequences of open sets by using matrixpairs (a,nn, Ann) of constants, each row of which codes an open subset ofC. This gives us another way of referring to an exhaustion (Gn) of C bycompact sets, referring instead to the sequence CGn) of open sets. Also,we can develop a way of representing all analytic subsets of C (includingall Borel sets) using the Souslin operation applied to sequences of sets. Inaddition, we can construct an algebra formula that expresses the value ofLebesgue measure applied to these analytic sets using approximation byopen and closed sets. These matters are more complicated, and we omitthe details.

Finally, we discuss how to express integration of functions and the"Nevanlinna characteristic" applied to entire functions using algebra formu-las. First we consider integration. Fix f E e and let o be a 1-1 conformalmapping on C We then can express by using algebra formulas how tointegrate f o or-' along a circular curve -y = {z I Iz - al = r} contained inthe range of a. That is, we have an algebra formula Int(f, Q, a, r, J3) whichholds in E if and only if or, a, r are as above and fr f dz = P. This canbe done by considering a sequence of successively finer subdivisions of ryand by evaluation of f o a-1 on points of ry using our expressions f (17) = 6

as above. We then get upper and lower sums and evaluate the integral bytaking limits. (Lebesgue integration can be handled also, but it is morecomplicated. In any case, our functions are highly continuous.) By a simi-lar procedure we can also treat integration over the inside of the curve 'y,with respect to area measure.

Now we consider the "Nevanlinna characteristic" on entire functions.For an entire function f this is defined by

2,K

T(r,f) = r log, If(re'BIdO.J0

We also consider the order a of f defined by

log T(r, f )a = lim sup

r-00 log r

Note that when we apply the above approach to integration, the I - 1conformal mapping o that appears as a parameter must be affine. Also,


the order of f is equal to the order of f o a-1, no matter which affine orwe choose. This shows that there is an algebra formula Ord(f, a) which istrue in the algebra of entire functions if and only if the order of f equalsa.

Obviously there is no completely natural point at which to stop thisgeneral discussion of how to express the mathematics of entire functions inthe first-order theory of E. As far as we can see, essentially all of the classi-cal theory of entire functions, including topological and measure-theoreticaspects, can be so represented.

Derivatives and Definable ConstantsWe will call a constant a E C definable if there is a formula D(x) in thealgebra language such that, for every function f E E,

D(f) holds in E4=#,f equals the constant a.

Evidently, i is a definable constant, since we have included a name for itin the algebra language. Also, it is clear that a is definable if and only ifboth the real part of a and the imaginary part of a are definable.

One way to obtain a large number of definable constants is via infiniteseries and the device for coding sequences of constants that was discussedearlier. Recall that such a sequence is coded by a triple of functions (f, g, h):a is the nth term of the sequence if f (z) = a, where z is the (unique) zeroof g for which h(z) = n. It is now a routine matter, given the defin-ability results obtained earlier, to construct an algebra language formulaSeries(x, y, z, w) such that, for any a, f, g, h E 6,

Series(a, f, g, h) holds in E if and only if (f, g, h) codes

a sequence (an) and E an converges to a.

To make this clearer, we take the first step toward the formula:

Series(a, f, g, h)= Basis(g, h) A a E CA (3k)[Vn E NV/3, -y, 6 E C(Seq(/3, n, k, g, h)

ASeq(ry,n+1, f,g,h) =Seq(/3+ry,n+1,k,g,h))A d/3 E C(Seq (Q, 0, f, g, h) = Seq(/3, 0, k, g, h) )AYbER+3nEN`dmENd/3EC(n <mASeq(l,m,k,g,h)

= I/j - al 6)l.

The middle two clauses of this formula assert that the sequence (/3n) codedby the triple (k, g, h) is the sequence of partial sums of the sequence (an)coded by (f, g, h). The third clause asserts that (fan) converges to a.

Now we will show, as an example, that e is a definable constant. This fol-lows because we can say in a first-order way that (f, g, h) codes a sequence


(an) which satisfies the recurrence relation ao = 1, an+1 = an/(n + 1),forcing an = 1/n! and so e = an. Thus the defining formula D(x) for eis obtained from:

D(a) 4= f 3g3h[Series(a, f, g, h) A Seq(1, 0, f, g, h)

A bn E N V,3 E gSeq(,0, n, f, g, h)) Seq(/3/n + 1., n + 1, f, g, h)].

[Strictly speaking, we cannot use division, but this is easily eliminated fromD(x).]

Clearly this same device can be used to obtain most of the familiartranscendental real numbers, as well as many others, for example, Liouvillenumbers such as F, 10'x'. All that is required is that the number be thelimit of a series whose terms are generated using some recurrence formula.

Theorem 23.3. Let F be the set of all definable constants from C. ThenF is a countable algebraically closed field that contains e and ir.

Proof. It is clear that F is a subfield of C. For example, if D(x) definesa = 0, then the formula

3y(D(y) A xy = 1)

defines 1/a. Also, F is countable, since there are only countably manyformulas in the algebra language. It is easy to show that F is algebraicallyclosed. For example, suppose aj is definable by D3 (x) for j = 0, 1, 2, 3.We will show how to define a root of the polynomial p(z) = aoz3 + a1z2 +a2z + a3. Consider the linear ordering on C defined by taking

a < Qt=*X(a) <_ R(f3) or

(R(a) = R(/3) and 3`(a) < 3`(Q)).

This ordering is definable by an algebra formula B(x, y) in the sense thatfor any a, 0 E C

B(a, /3) holds in Era < /3.

Using this we can define the "smallest" root of p(z) by a formula D(x).Namely, D(a) is equivalent to

3ao3a13a23a3[Do(ao) A Dj(aj) A D2(a2)

AD3(a3)AaECAaoa3+a1a2+a2a+a3=0AV/3(/3 E RAa0Q3+a112+a2f3+a3) = 0 B(a,Q)].

Other interesting definability results for the constants come from anindirect treatment of derivatives, which we now discuss. Of course, thereis no hope of obtaining a first-order definition of the relation between a


function and its derivative, since this relation is not conformally invariant.Our indirect approach comes via the use of a parameter a, which is a1 - 1 conformal mapping, as was used earlier to obtain the definition ofR. Given such a a and given f, g E E, we will show that the relation(goo-1) = (f oa'1)' is a first-order property of the triple (f, g, a). Namely,this relation is equivalent to the condition:

Va E C 3h3k3/3 E C 3y E Cq [if a - a is a nonunit, then

g=/3+(a-a)h and f =y+f3(a-a)+(a-a)2k].

Proof. Let a = a(z) for some z E C and suppose the equationsg = +(a-rr)h and f = y+/3(a-a)+(a-a)2k holdover C. Substitute01-1 in the second equation and get

foa-1(w) = y + /3(w - a) + (w - a)2k(a 1(w)) for all to E C.

Differentiating with respect to w and setting w = a yield (f oa-1)'(a) = /3.Since (g o or-') (a) = J3, we have the desired equation.

(==>) Suppose that (goo r-') = (foa-1)' and consider a E C. Expandingg o a-1 and f o a-1 about the point a yields

(g o or-')(w) = /3 + (w - a)h(w),

(f o a-1)(w) =,y +/3(w - a) + (w - a)2k(w).

Substituting to = a(z) yields the equations needed.

The exponential function is uniquely determined on any neighborhoodof 0 by its functional equation f = f and f (0) = 1. This leads to a certaindefinability result for the exponential function:

Theorem 23.4. There is a formula E(x, y) in the algebra language suchthat for any a, /3 E E

E(a, f3) holds in E if and only if a, /3 E C and e° = /3.

Proof. Given a, /3 E C, consider the following condition:

there exists a 1 - 1 conformal mapping a and afunction f, both in E, such that

(i) f(0) =1,°

(n) (foa-' )' _ (foa-'),() f(a) =°

From our previous discussion it is clear that this can be expressed by analgebra language formula. Parts (i) and (ii) imply that f o a-1 is equal tothe exponential function, and part (iii) therefore says that e" Thus,E(a, /3) implies e" = /3.

Conversely, given that e" = /3, to satisfy E(a, /3) we need only set orequal to any 1 - 1 conformal mapping on C and then set f (z) = e°(z) forz E C.


Corollary 23.5. Let Fo be the field of definable constants. Then Fc isclosed under the exponential function.

Note that Corollary 23.5 gives an alternate proof that F0 contains e.Also, it yields that F0 has infinite transcendence degree. Indeed, by theHernlite-Lindemann Theorem, if al, ... , an are algebraic numbers (hencein FO) that are linearly independent over Q, then en',...,ea^ are alge-braically independent elements of F0. (See [21].)

We close this section by noting that the other elementary functions,such as sin(z) and cos(z), can be treated in a way that is similar to ourdiscussion of el.

Recursive UndecidahilityEarlier there was presented a way of defining N in F and a method forcoding countable sets of constants. This yields an effective interpretationof second-order number theory in the first-order theory of F.

This interpretation is an example of a 1 - 1 reduction of one "problem"or set to another. If 61, 62 are sets of sentences in formal languages L1,L2, respectively, we say 61 is 1-1 reducible to 62, and write 61 <1 62, ifthere is an effectively computable 1-1 function W (i.e., a recursive function)such that for any sentence S of L1,

S E 614=(S)E62.

We say that C51 and 62 are recursively isomorphic if there is an effectivelycomputable function cp that maps the sentences of L1 bijectively onto thesentences of L2 and which satisfies '(61) = 652. Evidently this meansthat the problems of deciding membership in 61 and in 62 are effectivelyequivalent in a strong way. It is a well-known fact [34] due to MybM thatif 61 _<1 62 and 652 <1 61 , then 61, 62 are recursively isomorphic. (Theconverse is obvious.)

As an example of what type of undecidability theorem can be provedusing the results above, we consider the ring of entire functions F. Robinson(33] showed that the first-order theory of this ring is undecidable by showingthat first-order number theory can be interpreted in it. The following resultis a substantial improvement of this.

Theorem 23.6. The first-order theory of the ring of entire functions isrecursively isomorphic to second-order number theory.

Proof. Let 61 denote the first-order theory of 6 and let 62 denote thesecond-order theory of (N, +, ) We will show 61 <1 62 and 62 <1 E51 -

Let 6, denote the first-order theory of the algebra of entire functions F.Earlier it was shown that the field of constants is definable in F, whichyields a direct interpretation of 6; into 61. In particular, 6 <j 61. Aswas discussed earlier in this section, we also have a direct interpretationof second-order number theory in 6. This implies 62 <1 61, so that62 <1 61 has been proved.


To show 61 <1 62i we sketch how to give an interpretation of the first-order theory of E in second-order number theory. Each entire function fhas a power series representation centered at 0,

f (z) = E anz",

with an infinite radius of convergence. Each coefficient a" = x" + y"ican be identified with two sequences of integers, say {rn,j I j E N} and{8",j I j E N}, where r",o is the integer part of x" and r, ,j is the jth decimaldigit of xn, and similarly for sn,j and y". Therefore, f can ultimately beidentified with an infinite matrix of integers M = (mij I i, j E N) obtainedby setting mi,2j = r j and rn1,2j+1 = sfj for all t, j E N. Finally, M can beconverted to a set of integers M# _ {2i 3j 5"'' 1 i, j E N}. Evidently wecan recover the function f from the set M#. To show that this providesthe desired interpretation, one should prove that the collection of all sets ofthe form M# is definable in the second-order theory of (N, +, ) and provethe same for the relations on these sets which correspond to addition andmultiplication of entire functions. The details are tedious and routine, andwe choose to omit them.

Corollary 23.7. Second-order number theory is 1 - 1 reducible to To.

24Identities of Exponential Functions

In this chapter, which is based on [13], we take up some questions promptedby mathematical logic, notably Tarski's "High School Algebra Problem."We study identities between certain functions of many variables that areconstructed by using the elementary functions of addition z + y, multi-plication x . y, and one-place exponentiation ex, starting out with all thecomplex constants and the independent variables z1, ... , z,,. We show thatevery true identity in this class follows from the natural set of 11 axiomsof High School Algebra. The major tool in our proofs is the Nevanlinnatheory of entire functions of n complex variables, of which we give a briefsketch. It is entirely parallel to the one-variable theory presented in detailearlier in this book. The timid reader can take n = 1, at least for a firstreading.

Tarski's conjecture for a more extended class of terms than those weconsider here has been shown to be false by Wilkie (see [50]). The largestclass that we are aware of for which Tarski's axioms have been shown tobe complete was studied in [11].

We briefly recapitulate the basic Nevanlinna theory.Consider first the case of one variable, n = 1. For meromorphic functions

f of one variable defined on the complex plane C, the characteristic functionis defined for 0<r< cc by

T(r,f) = rn(r,f) + N(r, f):

this is a sum of two terms: the proximity function

n

m(r,f)27r

log+If(re'B)IdO,


which measures how close, on the average, f is to oo, and the averagecounting function

I-N(r,f)= f rn( f) dt,U

where n(r) is the number of poles of f in the disc IzI < r. Here, r rangesover the interval 0 < r < oc, and appropriate modifications must be madein the definitions of N(r, f) if f (0) = 0 or if f (0) = oc. The functionlog+(t) is defined by setting log+(t) = log(t) for t > 1 and log+(t) = 0 for0 < t < 1. The growth of the characteristic T(r, f) as r --, oo gives a veryuseful measure of the growth of f. The basic properties that we shall useare listed below.

(C2.0) T(r, f) is a nondecreasing function of r and a convex function oflog r.

(C2.1) T(r, f + g) < T(r, f) + T(r, g) + 0(1).(C2.2) T(r, fg) <T(r, f) +T(r,g).(C2.3) T(r,1/(f - a)) = T(r, f) + 0(1) for any complex constant a.(C2.4) T (r, f 1g) 5 T (r, f) + T(r, g) + 0(1).(C2.5) T(r, e9)/T(r,g) -i oo as r - oo if T(r, g) is unbounded.The other basic fact we need about the characteristic T(r, f) is the

Lemma of the Logarithmic Derivative (LLD):(C2.6)

m(f, f'/f) 0 (log(T(r, f))+logr),

except possibly for r lying in a set E of finite length.When it comes to several variables, the theory is substantially the same

and the basic properties we need are still expressed in the same form (C2.0)-(C2.6). (See [42] for the details of proofs.) Alternatively, one could usea characteristic based on the exhaustion of C" by balls, rather than bypolydisks DPti . (See [48] and [101.) For a meromorphic function f (z1 i ... , z,.,)defined on C", we define

fn W

log+ I f (re`0', ... re`s,. )jdqi ... din ,

where 0 < r < oo. Also, N(r, f) is defined much as before, as an averagedcounting function of the poles of f. Note that if f is a holomorphic functionon C", then T(r, f)=m(r,f).

(In [42], a characteristic T(T, f) is developed for a vector variable f =(r1,. .. , rn), but we use only the diagonal case r1 = . = rn = r.)

The basic properties above, including LLD, are shown to hold in [42] orfollow exactly as in the case of one variable (e.g., (C2.5)). One thing whichneeds explanation is the derivative f' that occurs in the LLD. When n > 2,we shall take f to stand for the Euler operator

of off'=Df=zlaz1 +.+znaxn.

24. Identities of Exponential Functions 177

This has the useful property that D f = 0 if and only if f is identicallyconstant. (This is because f may be expanded as a nicely converging sumof homogeneous polynomials and because DP = mP for any homogeneouspolynomial of degree m.)

Our basic tool is a lemma proved in one dimension by Hiromi andOzawa-see Chapter 17 of this book. The proof in n dimensions is similarto the proof in 1 dimension and depends only on the properties (C2.0)-(C2.6) we have listed of the characteristic function T(r, f ). (Carlos Beren-stein has recently pointed out that the proof in [13] is incomplete. Theauthors of [13] are preparing a complete proof. More varied Wronskiansare needed to establish linear dependence in the N-dimensional version ofthe Hiromi-Ozawa lemmal. See [4] for a correct version of the proof for theball-characteristic.)

Lemma 24.1. (Hiromi-Ozawa). Let ao(z),... , a,(z) be meromorphicfunctions and let gl (z),... , gn(z) be holomorphic functions defined on thedomain CN. Suppose that these functions satisfy

(a) T (r, aj) = o (m(r,e1))s-1

for each j = 0,1, ... , n; and

(b) T(r, ell) 0 O(log r)

for at least one i = 1, 2, ... , n. Undf. these hypotheses, if the identity

1: aj(z)e9i(=) = ao(z)

j=1

holds for z E CN, then there exists constants c1,... , cn (not all 0) so that

nCj . aj (z)e91(Z) = 0

jj=11

for all z E CN.

Our use of the Nevanlinna theory tools previously discussed comes en-tirely through this Hiromi-Ozawa Lemma. Indeed, we use it only in caseswhere 91,.. . , g. are holomorphic functions, so that T (r, e91) = m(r, e9')(and the prohibition that the function not take the value 0 at the originis satisfied), and in cases where ao, a1,.. . , an are slowly growing func-tions. Here we consider expressions that are built up from variables andcomplex constants using addition, multiplication, and the 1-variable expo-nential function es (where e is the usual base of the natural logarithm).


We prove a version of Tarski's High School Algebra Conjecture for theseexpressions. (This result was proved independently by van den Dries [47]and, for terms containing just one variable, by Wilkie [49]. Their methodsare quite different from ours.) We also settle positively a conjecture, dueto Schanuel, which asserts that if f is a function on C' which is definedby an expression of this type, and if f is nowhere equal to 0, then f = e4for a function g on C' which is also defined by an expression of the kindconsidered here.

Definition 24.2. E is the smallest class of terms which contains thevariables x1, x2, ... and a constant for each complex number, and whichcontains the terms s + t, s t and exp(t) for each s, t E F_

Here we interpret exp(t) to stand for et. We note that if t E E and thevariables of t are among x1,.. . , x,,, then t defines a holomorphic functionon all of C". Ifs E E also has its variables among x1i ... , xn, we write t - sto mean that t and s define the same function on C". (Various equivalentformulations of this definition are possible in special cases because of theuniqueness of holomorphic functions. For example, if t and s contain onlyreal constants, we may be interested only in the functions they define onR". But t - s will hold as long as t and s define the same function on IR",or even on S" where S C C is any set with a limit point in C.)

One has the additional useful fact that a holomorphic function f onC" has the small characteristic T(r, f) = O(log(r)) if and only if f is apolynomial. Hence, if f is a polynomial and g is any nonconstant holo-morphic function on C", then T(r, f) = o(T(r,eg)) (which is necessary aspart of the hypotheses of the Hiromi-Ozawa Lemma as we apply it). See[17, Proposition 4.4ff]. For holomorphic functions this can be proved byestimating the Poisson integral for log if I to show that f is of polynomialgrowth as a function of x3 (when xi, i 54 j, are held fixed), for each j = 1,2, ... , n. By the Liouville Theorem in one variable, then, f is a polynomialseparately in each x3. That f is globally a polynomial now follows from[30]. (There must be many other proofs of our assertion in the literature.)

Theorem 24.3. (Tarski's Conjecture for E). If t, s are any two termsin E and t =_ s, then the identity t = s is probable from the axioms

x + (y + z) = (x + y) + z, x(yz) = (xy)z,

x+y=y+x, xy=yx,x+0=x,

x(y+z)=xy+xz,

exp(x + y) = exp(x) - exp(y),

together with all axioms giving the facts of addition, multiplication, andexponentiation for constants from C.


Proof. Because we have included here a constant for -1, the operationof subtraction is available and we need only consider the case where s is0. That is, if t E E, then we must show that t = 0 is formally derivablewhenever t - 0.

Moreover, it is easy to show that for any term t E E there are termss1, ... , sk E E and polynomials Pi,. . , pk in n variables, with coefficientsin C (also realized as terms in E) so that the identity

t = p1 exp(si) + ... + Pk - exp(ak)

is provable from the permitted axioms. We will prove the theorem by induc-tion on the total cumber of symbols in the sequence s1i ... , Sk, showing thatPl,... pk are polynomials, sl,... , sk E E and pi exp(s1)+ . +pk exp(sk) -0, then p, exp(sl) + +pk exp(ak) = 0 is formally derivable. (Note thatwe allow sj to be 0.)

First suppose that k = 1: If p1 exp(si) - 0, then p1 - 0. It is wellknown that p1 = 0 is provable from the admitted axioms, since p1 is apolynomial. Hence, p1 exp(s1) = 0 also is provable.

From now on assume k > 1. Assume p1, ... , pk are polynomials,81, ... , sk E E and pl exp(s 1) + +pk exp(sk) - 0. For 1 < j < k, let arjbe the function on C' defined by exp(s,). (Choose n so that all variables ineach pi and sj are included among x1,... , x,,.) Note that we may assumeeach w1 is nowhere equal to 0 on C.

After dividing by wk we have

Pl(lr1/lrk) +"' +pk-1(lrk-1/7rk) - -Pk

Suppose first that we can apply the Hiromi-Ozawa Lemma. In this setting,this means T(r, 7r;/lrk) 54 O(log(r)) for each 1 < i < k -1. If so, then thereexist constants c1,. .. , ck_1 (not all 0) so that

c1P1(lr1/xk) + ... + ck-lPk-1(rk-1/irk) = 0,

which gives us an identity with k-1 exponentials after multiplying throughby Irk. By the induction hypothesis, the formal identity

C1P1 exp(sl) + ... + ck_1Pk-1 exp(sk-1) = 0

is derivable in the allowed system. Now we can use this identity to solvefor one of the expressions p. exp(s,,) (1 < j < k - 1) and eliminate itfrom the original expression P1 exp(s1) + + pk exp(sk). The resultingidentity (setting this expression = 0) has at most k - 1 exponentials, so itis derivable. From this one deduces the desired identity

Pl exp(sl) + ... + Pk exp(sk) = 0.


On the other hand, it may happen that for some i(1 < i < k - 1),T(r,1ri/lrk) = O(log r). Since it , irk are nowhere 0, it follows that iri - clrkfor some constant c. [By (C2.3) the same kind of "big-0" estimate holdsfor Irk/7ri, and hence both 7ri/Irk and irk/iri are polynomials.] That is,exp(si) - c - exp(sk), so that for some constant d E C, c = ed andSi - sk - d. Using the induction hypothesis, we therefore get a formalderivation of si - sk - d = 0 and, hence, also of exp(si) = c - exp(sk). Thisallows us to reduce the original identity to one involving only exp(sj) fori < j < k - 1, which will be derivable by the induction hypothesis. Againthis yields a derivation of the identity pl exp(sl) + +pk exp(sk) = 0 andcompletes the proof.

Theorem 24.3 has an interesting corollary for trigonometric functions,which we present next. Consider terms in a language with constants fora l l the complex numbers, variables x1, x2, ... , and function symbols foraddition, multiplication, and for sin and cos. Let E* be the set of all theseterms.

Corollary 24.4. If t, s are any two terms in E* and t = s, then theidentity t = s is provable from the axioms

x + (y + z) _ (x + y) + z, x(yz) _ (xy)z,x+y=y+x, xy=yx,x+0=x, 1-x=x,

x(y+z)=xy+xz, 0-x=0,

sin(x + y) = sin(x) cos(y) + coa(x) sin(y),

sin(-1 - x) = -1 - sin(x)

together with all axioms giving the facts of addition, multiplication, sin,and cos for constants from C.

Proof. We use the fact that in the context of the complex plane, ex isinterdefinable with sin and cos. Note that since the allowed axioms includethe identities sin(a/2) = 1 and cos(ir/2) = 0, we can prove cos(x) =sin(x + it/2). This in turn allows us to derive the other addition identity,

cos(x + y) = cos(x) cos(y) - sin(x) sin(y).

In E*, let EXP(x) be an abbreviation for the termIt is easy to verify that from the allowed identities in E* one can prove theexponential identity

EXP(x + y) = EXP(x) . EXP(y)

as well as all the numerical facts involving EXP.


Given any term t in E*, we define a term t# in E by replacing (induc-tively) each term of the form sin(s) by

-.5i (exp(i s) - exp(-i s)),

and cos(s) by

.5. (exp(i s) - exp(-i s)).If t is any term in E we define t* in E* by replacing (inductively) eachterm of the form exp(s) by EXP(s). Note that if t E E*, then the identityt = (t#)* is provable from the axioms allowed in Corollary 24.4.

Now suppose t, s E E* and t - s. Then t# = s#, so the identityt# = s# is provable from the axioms allowed in Theorem 24.3. Hence,(t#)* = (s#)* is provable in the system of Corollary 24.4. It follows thatt = s is also provable in that system, completing the proof.

Remark. Suppose t, s E E* and t, s only contain real constants. We do notknow if there is a proof of the identity t = s in the system of Corollary 24.4in which only real constants appear.

Next we settle positively a conjecture of Schanuel.

Theorem 24.5. Let t E E and suppose the function represented by t isnowhere equal to 0. Then log(t) is in E, in the sense that t - e$ for some3EE.Proof. Let 7r be the function (on C' say) defined by t. There is someholomorphic function G on Cn so that 7r - eG. We may suppose t is aterm of the form pl exp(s1) + + Pk exp(sk), and we argue by inductionon the number of symbols in sl,... , sk as in the proof of Theorem 24.3.Clearly we are done if k = 1. Assume k > 1, and for 1 < j < k let 7r5be the function on C' defined by exp(s,). Then we have puns + +Pkxk ° eG so that pl(lrle-G) + . + pk(irke-G) - 1. We may assumethe functions plinl a-G, ... , pkake-G are linearly independent (otherwise,we could replace t by a simpler term to which the induction hypothesiswould apply). Hence, the Hiromi-Ozawa Lemma cannot apply. It followsas argued in the proof of Theorem 24.3 that there must exist 1 < i < j < kso that 7ri/irk is identically constant. Again this permits us to reduce thecomplexity of t and to apply the induction hypothesis. This completes theproof.

We conclude with a related problem.

Problem. Suppose that f is an entire function for which there exists aterm t E E such that the function represented by t is equal to f2. Thenmust there exist a term s E E such that the function represented by s is f ?

Put more simply (but not as correctly), if f is entire and f2 E E, mustf E E? Even if one assumes that f 2 and f 3 belong to E (and hence fn e Efor n = 2, 3, 4, 5, ... ), does it follow that f (assumed to be entire) lies inE?

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Index

algebraic integer, 139

Boas, 47Borel Lemma, 26, 28, 29Borel transform, 43, 125B(r), 30branching index, 103Buck, 131, 139

Calderon, 123canonical products, i, 87-89, 91Carleman's Theorem, i, 45, 47Carlson's Theorem, 47, 130characteristic logarithm, 147characteristic, Ahlfors-Shimizu,

18, 19

characteristic, Cartan, i, 16-19characteristic, Nevanlinna, 10, 18Clunie's Theorem, 14conjugate indicator diagram, 125convex, 16convex hull, 124convolution, 127corrected ratios, 32counting function, 7, 50, 147

defect, 103deficiency, 103deficient value, 105d(a), 103

eff

in words, A(r) approaches Leffectively, 28

effectively, 185

eff'28

, 28eff

exponential-type, 41extreme point, 124

Fabry Gap Theorem, 134finite A-density, 51finite A-type, 49finite (A)-type, 13, 40first fundamental theorem, i, 9-

11, 95formal power series, 93Fourier coefficient, 49Fourier coefficients associated

with Z, 56Fourier series method, 49, 90fully branched value, 106


Gauss Mean Value Theorem, 6,21

genus, 88growth function, 51

Hadamard factorization theorem,89, 151

Hadamard gap theorem, 133Hahn-Banach Theorem, 48Hausdorff-Young Theorem, 69

Hayman, 1412, §1.5-1.6, 18

Hiromi-Ozawa Lemma, 106, 109

Ho(oo), 125Horseshoe Theorem, 131h(8), 125

Malliavin, 146Malliavin-Rubel Theorem, 47mapping radius, 141maximum term, 30Miles, 50Miles-Rubel-Taylor Theorem, i,

78, 91m(r, f), 10

Nevanlinna, 108, 111Newton polygon, 31N(r, f), 1Un(r, f ), 7N(r,f),7n(t, f), 103

Okada, 96order, 40, 41

indicator function, 125integer valued entire function, 139interpolating sinusoid, 134

Jensen's Theorem, 6

K`-admissible, 144k-admissible, 131Kakeya, 96

A-admissible, 55Laguerre's Theorem, 92Laplace transform, 43, 126A-balanced, 52, 58

Lindelof, 50, 74Lindelof's Theorem, 157Liouville's Theorem, 23, 130logarithmic convexification, 32Logarithmic Derivative, 116logarithmic length, 28logarithmically convex, 17A-poised, 53

Phragmen-Lindelof Theorems, i,121, 123

Picard's Theorem, i, 99, 101, 103,105, 107, 109, 111

mr(f), 31Poisson Kernel, 20Poisson-Jensen Formula, i, 20, 21P61ya Representation Theorem,

124, 128principal indices, 31Pringsheim's Theorem, 133

quotient representations, i, 78

rank of maximum term, 30regular, 58

sampling theorem, 47Second Fundamental Theorem, i,

99, 101, 103, 105, 107, 109, 111Second Fundamental Theorem of

Nevanlinna, 102share the value a, 108sinusoid, 134S(r), 102Stone-Weierstrass Theorem, 48

Index 187

strongly A-balanced, 52 Two Constant Theorem, i, 121,strongly A-poised, 53 123

subsinusoidal, 134 type, 41

supporting function, 124

Szeg 134 , 99o

Weierstrass FactorizationTsuji, 97 Theorem, 88

Universitext (continued)

Sagan: Space-Filling CurvesSamelson: Notes on Lie AlgebrasSchiff: Normal FamiliesShapiro: Composition Operators and Classical Function TheorySmith: Power Series From a Computational Point of ViewSmorynski: Self-Reference and Modal LogicStillwell: Geometry of SurfacesStroock: An Introduction to the Theory of Large DeviationsSunder: An Invitation to von Neumann AlgebrasTondeur: Fuhations on Ricm.Lnudn I'Aanifolds

UniversitextThe book is an introduction to the theory of entire and mero-morphic functions intended for advanced graduate students inmathematics and for professional mathematicians. The bookprovides a clear treatment of the Nevanlinna theory of valuedistribution of meromorphic functions, starting from scratch. Itcontains the first book-form presentation of the Rubel-TaylorFourier series method for meromorphic functions and the Milestheorem on efficient quotient representation. It has a concisebut complete treatment of the Polya theory of the Borel trans-form and the conjugate indicator diagram. It contains some ofBuck's results on integer-valued entire functions, and theMalliavin-Rubel uniqueness theorem. The book closes withapplications to mathematical logic. In particular, the first-ordertheory of the ring of entire functions is developed and ques-tions concerning identities of exponential functions are studiedas in Tarski's "High School Algebra Problems." The approachof the book gets to the heart of the matter without excessivescholarly detours. It prepares the reader for further study of thevast literature on the subject. which is one of the cornerstonesof complex analysis.

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9ISBN 0-387-94510-5

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Entire and Meromorphic Functions