Analytic Capacity and Rational Approximation

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Lecture Notes in Mathematics A collection of informal reports and seminars Edited by A. Dold, Heidelberg and B. Eckmann, Zi.irich

50

•

!

Lawrence Zalcman Massachusetts Institute of Technology, Cambridge

Analytic Capacity and Rational Approximation 1968

Springer-Verlag- Berlin. Heidelberg-New York

All rights reserved. No part of this book may be translated or reproduced in any form without written permission from Springer Verlag. © by Springer-Verlag Berlin - Heidelberg 1968 Library of Congress Catalog Card Number 68-19414. Printed in Germany. Title No. 7370

PREFACE

The purpose of these notes is to make available in English a reasonably complete discussion of some recent results in rational approximation theory obtained by Sovie~ mathematicians.

More specifically,

we shall be concerned

with recent theorems of Vitushkin and Melnikov concerning (qualitative) approximation by rational functions on compact sets in the plane.

Accordingly,

we shall have nothing to

say about problems of best approximation,

or of approximation

on "large" planar sets, or of approximation in the space of n

complex variables

(n > 1).

Each of these subjects is an

active discipline in its own right and deserves its own (separate) treatment.

On the other hand,

since problems of

rational approximation have a "local" character,

our theorems

are related to questions of approximation on regions on Riemann surfaces~ however, we shall not pursue that line of thought to any extent. Since our principal desire is exposition, to keep the prerequisites a minimum.

we have tried

for understanding the material at

A knowledge of basic function theory and functional

analysis plus a willingness to pursue a few references given in the text are all that is required.

On occasion,

it has

been convenient to suppress the details of a proof in the interests of exposition~

in each such case the reader will

supply the missing steps easily.

In general, however, when

a point has seemed to me obscure,

I have chosen to say more

rather than less by way of explication. Although this is not primarily a research paper, contain some new material:

it does

a few of the examples and the

unacknowledged contents of sections 7 and 8 have not appeared in print previously.

There are also, as might be expected

in a work of this sort, a (small) number of simplifications of proofs,

etc.

No attempt, however,

has been made to take

specific notice of such minor improvements.

At times I have

followed the papers of Melnikov and Vitushkin quite closely~ in other instances, considerably.

the original material has been reorganized

The reader who consults the original papers

will easily identify the sections in question. These notes are based in part on a lecture given at the Brandeis-Brown-M. I.T. joint function algebra seminar at Brown University and on a series of lectures given at Professor Kenneth Hoffman's

function algebra seminar at M.I.T.

I would

like to thank Professor T. W. Gamelln, who first interested me in giving the series of talks mentioned above~ his help, encouragement,

and enthusiasm at every step of the way have

been invaluable.

In particular,

he read and criticized a

preliminary version of the manuscript and made available some very useful notes of his on material covered in sections ll and 12.

I am also very grateful to Dr. John Garnett for many

helpful conversations and for permission to incorporate several

unpublished results of his into the text~ I am especially indebted to him for his careful reading of the final draft. And thanks are due Jane Jordan, who typed the manuscript. Finally,

I owe a very real debt of gratitude to the National

Science Foundation, which provided me with fellowship support during the preparation of these notes.

CONTENTS

ii

Introduction ...........................................

2.

Peak points . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

3.

Analytic

4.

Some useful

5.

Estimates

6.

Melnlkov's

7.

Further

8.

A p p l i c a t i o n s . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57

9.

The p r o b l e m of rational a p p r o x i m a t i o n ................. 65

10.

AC c a p a c i t y . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 77

ll.

A scheme

12.

Vitushkin's

13.

Applications

14.

Geometric

15.

F u n c t i o n a l g e b r a m e t h o d s . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ll7

16.

Some open q u e s t i o n s . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 130

Appendix

I.

Appendix

II.

1

capacity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ll facts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23

for integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27 t h e o r e m . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45

results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52

for a p p r o x i m a t i o n . . . . . . . . . . . . . . . . . . . . . . . . . . . . 86 t h e o r e m . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 100 of V i t u s h k i a ' s

t h e o r e m ................... 108

conditions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ll2

Logarithmic

capacity ........................ 132

A n a l y t i c capacity and the r e m o v a b i l i t y of singularities ............... 137

Bibliographical

notes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 139

Bibliography ..............................................

147

i.

INTRODUCTION Let

X

be a compact set in the plane and let

the set of all functions on

X

X .

set of all functions continuous on

X

X .

R(X)

is quite old:

and

A(X)

denote

uniformly approximable by

rational functions with poles off

the interior of

R(X)

Let

A(X)

be the

and analytic on

X° ,

The study of questions relating to one can trace interest in

rational approximation back to the days of Appell [5], [6], Weierstrass, and Ruoge [78]. p. l#l).

(For further references see

In recent years, particularly the last two decades,

the development of the subject has been especially active. Russian mathematicians have studied rational approximation from the point of view of the constructive theory of functions of one complex variable.

The techniques involved in this

approach have been, quite naturally, classical; they consist mainly of the construction of suitable kernels and the ingenious application of the complex form of Green's formula. The resulting proofs, while delicate and sometimes quite involved, provide explicit constructions for the approximating functions.

These methods have been used with notable success

to settle many interesting and important questions [67], [89],

[91]. In the United States, recent interest in qualitative rational approximation sprang up from a different corner.

-2-

Functional analysts functions to examine

studying certain Banach algebras of

(uniform algebras, R(X)

and

A(X)

of a more general theory.

sup norm algebras) were led as interesting

special cases

Their approach has been to try

to find abstract or semi-abstract rational approximation theory.

proofs for theorems of

Such efforts have resulted

in some beautiful proofs of known theorems

[38], [16], [34].

They have met with less success in proving new results (see, however, Recently,

[107] and [37]). Melnikov obtained a necessary and sufficient

condition for a point

[63], [64].

x 6 X

Since questions

from considerations function theory,

to be a peak point of

R(X)

concerning peak points arise

of functional analysis rather then

one can regard Melnikov's work as a synthesis

of the two approaches mentioned above; here the abstract theory asks the questions, answers.

and the constructive methods provide

Slightly earlier,

necessary and sufficient coincide with

A(X)

.

Vitushkin

condition on

[94] X

had obtained a for

R(X)

This remarkable achievement

to generalizes

earlier well-known work of Mergelyan and Vitushkin. With these two results, Vitushkin,

the theorems of Melnikov and

the theory of rational approximation on compact

planar sets approaches

a certain completeness.

to be sure, interesting and challenging problems subject

There remain, in the

(see section 16); but the feeling persists that the

"'3-

battle lines for future attacks on these problems have been drawn and that there is every hope that final victory will be attained in the relatively near future.

Unfortunately,

however, because of the language barrier and the inaccessib~1~ty of complete proofs, the results and techniques of [64] and [9#] have attracted less attention than they deserve.

It is

with the hope of rectifying this situation that these notes have been prepared. A word or two is in order concerning the organization of the material.

Sections 2 through 8 deal with material

relating to the work of Melnikov, while sections 9 through l# contain an exposition of Vitushkin's theorem and its amplifications.

The reader interested principally in the

approximation problem can read section 3 and then skip to section 9.

Section 15 contains some applications of "function

algebra" methods to rational approximation, of the fact that

R(X)

including a proof

is local; it depends only on section 2.

Some open questions are mentioned in section 16, and in t~o appendices we have collected,

for the reader's convenience,

relevant facts concerning logarithmic capacity and the removability of singularities for analytic functions.

Finally,

there is a substantial bibliography with notes~ this section is essentially independent of the text.

-4-

2.

PEAK POINTS Let

X

be a compact set in the plane and let

We shall say that exists and

f E R(X)

(3) If(Y)l

x

is a peak point for

such that

< i

for

R(X)

y E X \ {x] .

sense for any sup norm algebra

The concept of peak

Now fix

x E X

is a continuous

to

C(X)

any collection domain of definition).

will not concern us here.

and let

Lx(f ) = f(x) , f E R(X)

linear functional on

which,

R(X) ; it makes

however,

By the Hahn Banach theorem, L~

(or, indeed,

functions with a con~non

This generalization,

if there

(1) II fll~ = 1 ; (2) f(x) = 1

point need not be restricted to the algebra

of complex-valued

x E X .

Lx

R(X)

.

having norm i .

has a norm-preserving

extension

by the Riesz representation theorem,

given by a finite complex Baire measure

Lx

~x

is

supported on

X .

We then have

Lx(f) =

f d~x

f 6 C(X)

.

X

Since

= ~x(~) o ~x~ ~ x -~ ~x Id~l =, ~tl = " ~xll : ~, it is clear that

~x

is a positive measure of total mass I .

We shall call such a measure a representing measure for

x .

More precisely,

is

a representing measure

a Baire probability measure

for

x

~ , supported on

on

R(X)

X , such that

-5-

f(x) = ~

f dw X

for every

f E R(X)

Clearly, to-one of

the r e p r e s e n t i n g

correspondence

Lx to

m~y

.

points

x

can have of which

is i n t i m a t e l y

Indeed,

Let

[13]).

extensions

The q u e s t i o n

measures

considerations.

(Bishop

are in one-

a g i v e n point

measures.

representing

up w i t h peak point

T h e o r e m 2.1.

in general,

representing

have unique

for

with the norm p r e s e r v i n g

C(X) ~ thus,

different

measures

X ~ ~

bound

we have

be compact,

x E X .

These are equivalent-

(i)

x

is a p e a k point

(2)

x

has a unique

(3)

Given any exists

x

on

(i) =~

(relative)

f E R(X)

R(X)

.

(2) .

w i t h this property. Let

c = ~((x~)

is a representing Let

f

measure

such that

if

supported

U

of

for

X .

II fll~_~ 1 , f(x) > 3/4 ,

y ~ X \ U the point mass

~ # 6x

0 _~ c < 1 .

measure

on

x , there

8x

represents

show it is the only p r o b a b i l i t y

Suppose

; then

.

neighborhood

Obviously,

We must

R(X)

representing

and If(Y) l < I/~ Proof.

for

x

also represents The measure

which

be a function that peaks at

x .

x .

~ = (1-c)-l(w-CSx)

satisfies Then

measure

~(~x3)

fn_~ 0

= 0 .

point-

-6-

wise b o u n d e d l y dominated

almost

everywhere

convergence

dv

as

.

n--*~

By the

theorem

I = fn(x) = ~ fn d~ ~ 0 , X

a contradiction (2) =>

.

(3) •

Let

U

be given

such that

v _ < 0 , v < log(i/4)

It follows

from

v(x)

{Re g(x)l

= sup

assume

that

Im g(x)

written R(X)

g E R(X)

of

(3).

[uj}

of

Let

VI = UI .

for

U = VI

choose

for some

Moreover,

as a u n i f o r m l y

(3) => ( 1 ) .

v E CR(X)

X \ U , and v(x)

and

Re g_< v}

g E R(X)

Then the function

is uniformly

hoods

on

Choose

f = eg

f E R(X)

convergent

.

We may thus

such that satisfies

, since

f

series

in

power

Choose a decreasing

x

such that

Take

and set

V n c U n n Vn_ I

Igj(Y)l

the can be g , and

closed.

fl

U1 = X

satisfying

sequence and

fn

~ Uj = {x} j=l

the c o n d i t i o n

gl = [ f l (x)]-l fl " and

of neighbor-

For

inductively

 Iog(3/4).

(and the proof of the Hahn Banach theorem)

v = Re g

= 0 .

conditions

(2)

.

y 6 Vn , j < n

.

that

-7and

II

Let

gn = [fn(x)]-If n .

Let

f n l ~ _< 1 , f n ( X ) > 3/4 ,

(b)

gn(X) = 1

(c)

II gnll= _< 4/3

(d)

Ign(y)l < 1/3

2 jgj .

f =

By

If n(y) i

< 1/4

for

y ~ vn .

Then

for

y ~ Vn

(b) , f(x) = 1 .

We must show that

j=l if

y ~ x

let

n

If(y)l < I .

Suppose,

then, that

be the largest index for which

y ~ x , and

y E V

.

Then,

n

using

(a),

(c), and

(d), we have

n-1

If(y)l _( i ~2-Jgj(Y)l + 12-ngn(Y)l + I ~ 2-Jgj(Y)i J =i

j =n+l

< (1-2-n+1)(1+2 -3n) + 2-n(4/3) + 2-"(1/3) < 1

as required. As a consequence of 2.1 we obtain

Corollary 2.2. %B x

Let

(0 < a < 6 < l)

x 6 X .

Suppose there exist numbers

such that for any neighborhood

there exists a function

f 6 R(X)

satisfying

U

of

(1) llfll ~ < 1 ;

-8-

(2) f(x) > ~ ; and x

(3) If(Y)l

is a peak point for

Suppose

Proof.

not.

Then,

exists a representing U([x])

= 0 .

R(X)

of the corollary.

IIU f dul +

~x

Theorem 2.4 of points positive

Let

is a

of

P

G~ . for

x

(Bishop X

G6

x

such that

II XkU f dul < (6-~) + ~ =

provide R(X)

If

.

some motivation

for our

.

be the set of peak points

for

R(X)

x 6 X , there is a representing

which satisfies

If

([13]).

ux(X \ P) = 0 .

R(X)

C(X)

which are not peak points

(Lebesgue planar)

Theorem 2.3 deserves is a

of

proof of 2.2 is given in [39]

study of the peak points of

Theorem 2.3.

such that

satisfy the conditions

f E R(X)

The next two theorems

measure

U

x

.

A constructive

P

for

there

Then

< f(x) = I X f d._<

a contradiction

Then

as in the proof of 2.1,

measure

and let

y £ X \ U .

.

Pick a neighborhood

u(U) < ~ - ~ ,

Then

if

< a

for

then the set R(X)

has

measure. some comment.

was noted by Bishop

The fact that

P

[13] and follows easily from

.

-92.1 ~ indeed,

if

U

exists an

in

R(X)

and

f

If(y)I < 1/4

n

x 6 X

is the set of satisfying

for

y

for which there

II fII~ <_ 1 , f(x) > 3/4 , Ix-yl _> 1/n , then

such that

GO

Un

is open and

nN_l U n = P .

The importance of this obserP

ration is that it insures that

is a Baire set, thus

allowing us to avoid unpleasant problems of measurability. The second part of 2.3 is also due to Bishop, who proved the result for an arbitrary sup norm algebra on a compact metric space.

A discussion of this theorem and its generalizations

is given in [73].

What we should like to point out is this.

By the maximum modulus principle, attains its maximum on

8X , the boundary of

that the restriction map in a~y

C(3X) x 6 X

.

each function in

f-*fI8 X

embeds

X . R(X)

R(X) It follows isometrically

The reasoning preceding 2.1 now shows that for there exists a measure

is supported on

3X .

Since

~x

representing

x

which

P c 8X , 2.3 provides a (strong)

generalization of this fact. We shall postpone the discussion of 2.4 until section 15, where we shall give a proof. Since

R(X)

is a Ba~ach algebra it is natural to ask

for the identification of its maximal ideal space and Shilov boundary.

These questions can be answered without difficulty.

It is easy to see that if a point from the complement of

X

x 6 X

is linearly accessible

then it is a peak point for

Since such points are clearly dense on

R(X).

3X , it follows that

-10-

3X

is the Shilov boundary of

ideal space of

R(X)

is

X .

R(X)

.

Moreover,

This follows from the obvious

fact that any (nonzero) complex homomorphism of determined by its action on the function Finally,

the maximal

R(X)

is

z .

let us note explicitly that all the definitions,

remarks,

and theorems of this section remain valid if we

replace

R(X)

by

A(X) ~ the proofs of 2.1, 2.2, and 2.3

remain unchanged while 2.4 becomes trivial. 8X

is the Shilov boundary of

that the maximal ideal space of is no longer elemeatary.

A(X)

.

A(X)

Since

R(X) ~ A(X),

It is also true is

IT]

X , but this fact

-ll-

3.

ANALYTIC

CAPACITY

We shall an open

denote

subset

is a n a l y t i c

of

on

z = ~ .

f'(~)

U .

~nen

is c o m p u t e d

let

G(K)

analytiq

K

using

capacity

of

K

the sup is t a k e n (1)

f

(2)

II flL _< 1

We c a n also

(3)

a ~ ~

the

Let

U

function

be f

-- o ,

note that

the d e r i v a t i v e

the c o o r d i n a t e set in the component

1/z

.

(finite) of

= supjf' (~)I ,

over all on

G(K)

satisfying

f ;

f' (a)

z-a , w h i l e

is g i v e n by

is a n a l y t i c

require

Alternatively,

to the c o o r d i n a t e

the u n b o u n d e d

y(K)

where

One should

for

be a compact

denote

S2 .

has a n e x p a n s i o n

f' (~) = a I .

- f(~)).

with respect

N o w let

by

= a o + al/z + a2/z 2 + ...

% f' (~) , since

is c o m p u t e d f' (~)

f

By definition,

= l i m z(f(z)

l i m f'(z)

sphere

S 2 , ~ E U , and s u p p o s e

f(z)

at

the R i e m a n n

plane,

S2 \ K .

and The

-12-

since if (i),

g = If - f(~)]/[l

(2), and

satisfying

function for admissible

K

g •

satisfies A function

(i) - (3) will be called an admissible

or, when the context

function.

the existence

Ig'(=)l >_ If'(-)l

(3); and

properties

- f---~Tf] then

is clear,

A normal families

of an admissible

function

simply an

argument ~

such that

~' (~) = ¥(K) ; this function is, in fact, unique call set

~ S

the Ahlfors

function for

K .

establishes

[49].

We

For an arbitrary

define

¥(S)

= sup

¥(K)

K

compact

.

Kc_S

Analytic in some detail

capacity has been studied by various authors ([4], [49], [74], [88]).

selves with a discussion or-less

relevant

Proposition

of those properties

to the discussion

3.1.

If

S I c S2

then

Proposition 3.2.

If

KI

~

G(KI)

= ~(K2)

depends

then

and

3.3.

Let

Y(Sl) _< y(S2).

are compact

Y(Kl) = y(K2)

a 6 C .

which are more-

following.

.

only on the "outer boundary"

Proposition

We shall content our-

Then

(Monotonicity).

sets such that

In other words, of

?(K)

K .

~(aS) = laIy(S),

(Homogeneity).

-13-

Proposition 3.4.

Let

.

aE~

Then

y(S+a) = y(S) . (Translation

invariance). Propositions 3.1 and 3.2 are trivial~ we leave the easy proofs of 3.3 and 3.4 to the reader.

Proposition 3.5.

For any

S,

cap(S) ~ y(S) .

denotes the logarithmic capacity of

Proof.

Here

cap(S)

S .

It is enough to prove the inequality for compact sets.

For such sets we have (see Appendix I)

cap(K)

= suplf' (~)I ,

where the sup is taken over all functions analytic but not necessaril[ slngle-valued on valued and which satisfy single valuedness of is well defined).

Proposition 3.6. where

Proof.

d

Ifl

IfCz)l

_~

whose moduli are single1

,

f(,~)

insures that

It is now clear that

If

C

is a continuum,

is the diameter of

Since

O(K)

G(C)

= o

.

(The

If' (oo)1 = llmizilfCz)l

cap(K) > ~(K) cap(C) = y(C) > d/4 ,

C .

is simply connected , cap(C) = y(C) .

For continua, it is well-known

([84]) that

cap(C) > d/4 .

-14-

Proposition 3.7.

Let

K

be compact and let

~

be a

rectifiable contour which has winding number I around each point of

K .

Then

¥(K) = suPl2~ ~ f(z)dzl , where the

sup is taken over all admissible functions.

1

Proof.

~ f(z)dz = f' (-) . Q

Proposition 3.8. Thus

Let

y(K) < inf ~

a

be as above.

Then

y(K) < ~

length(Q).

length (~), where the inf is taken over

the class of all such curves.

y ( K ) = s u p I ~ I f(z)dz, < ~

Proof. i 2rr

length (~)

~ ds =

"

Proposition 3.9.

Let L

y(L) = re(L)/4 , where of

If(z)IIdzl < ~

be a compact set on a line. m(L) = the Lebesgue

Then

(linear) measure

L .

A proof of this result is in [74] .

Proof.

will require only the weaker estimate m(L)/N . axis.

By 3.3 we may assume that

Actually, we

([4]) m(L)/4 ~ ¥(L) L lles on the real

It is clear from the preceding proposition that

¥(L) ~ m(L)/~ . function

For the other inequality,

consider the

-15

ef(z)-I =m~) i+ g(z) : el(Z)+ 1

Z

"'"

where

f(z) = ~

Lz-'x = 2Z

"'"

Now dx

i

~

dx

@@

<

=

= arc tan ~ o

--co

Thus

C = f(z)

maps

S2\L

Since

Proposition K, K n

below, ~n

S2\L

IIm C l S ~n

into

into the right half plane,

3.10.

are compact

Proof.

Let

maps

g(~) = 0 , g

Since

2

0

is admissible

Let

~

o ~

sets

.

Then

[Y(Kn)]

o...o K

II gll~ < I .

and

(~K n = K, where

Y ( K n ) - + y(K)

for

.

sequence bounded

exists; moreover3

function

ef(Z)

.

is a decreasing

its limit obviously be the Ahlfors

and

so that

Kn .

lim y(Kn) ~ y(K) Then

{~n]

forms

-16-

a normal family some subsequence function {~nj]. K .

admissible T

Let

for

of which converges

K .

to a

Denote this sequence by

be a large circle about the origin surrounding

By 3.7 we have

~(z)dzl

= lim

I~

(z)dzl : lim

Y(Knj)

=

lim ¥(Kn)

Thus

¥C K) = lim yCKn)

function is unique, ~n(Z) -- ~(z)

we can prove even more:

uniformly

It follows y(K) = inf ¥(U)

If we use the fact that the Ahlfors

-

on compact

sets of

easily from 3.10 that if

namely, n(K)

K

that

.

is compact,

where the inf is taken over all open sets

UDK

containing sets,

¥

K .

This expresses

is "continuous

for open sets at least,

the fact that,

from above." ~(U) = y(U)

for compact

One might think that, •

Example 3.11 shows

that this is not the case.

Example 3. ii. discs

An

Let

of radius

L = [0, I] . rn

Take a sequence of open

such that

-17-

(1)

xj

n~k=~

if

j ~k, oo

(2)

L

is the set of limit

points

of

[J

An

,

n=O (3)

~rn<

1 ~ • ao

Let

U =

~J A n ; then n=O

YS)

> ?(L) = ~ 1 .

U = L U nUO An " =

By 3.1 and 3.9,

On the other hand, from 3.8 and the

definition of analytic capacity for open sets we have

v(U)

_~ ~

~ 2~r n = ~ rn < ~ • n=O n=O

Proposition S.12. set, V(K)

Proof.

(Pommerenke

the area of

K .

Following Pommerenke,

First of all suppose

K

F(z)

[?4]). Then

K

be a compact

V(K) < w[y(K)] 2 .

we merely sketch the proof.

has finitely many components.

= z + _a + Z

GCz)

Let

b = z + -z +

Let

. •

"

. • .

be the (unique) functions functions which map

Q(K)

onto a

horizontal slit region and a vertical slit region respectively• Let

@(K) = - a-b ~- .

In [80],Schiffer showed that

V(K) < ~ ( K )

On the other hand, Ahlfors and Beurling have shown [4] that

.

-18-

~ ¥(K) ~ thus

V(K) ~ W[y(K)] 2

finitely many components.

for compact sets with

Applying 3.10, we obtain our

result. There is one more elementary property of analytic capacity we should like to mention, bear directly on what follows. of a compact set

K

though it will not

Define the P a i n l e v ~ ~

by

$(K) =

inf lim length

(~On) ,

where the inf is taken over all possible of finitely connected domains which exhaust

~(K)

Proposition 3.13.

[~n}

(with rectifiable boundaries)

.

Suppose

analytic function on

G(K)

$(K) < ~ .

is a finite

Let

such that

f(z) = [

where

sequences

C- z

f

be a bounded

f(~) = 0 .

Then

•

(complex) Baire measure supported on

8K .

Proof.

(See [#9]).

such that we have

Let

lim length(8~n)

(giving

8~m

[~n]

be an exhaustion of

= ~(K) + E < ~ .

the appropriate

For

orientation)

G(K)

z E fin

-19f(C)(C-z) -I d c .

f(z) : (2~i) -I ~

5~n Let

D

be a closed disc whose interior contains

large

n, 3Gn c D .

For

g 6 C(D)

~n(g) = (2~i) -I ~

K .

For

define

f(C)g(c)dC

•

3Gn Then

II ~Anll = ,[Dld~n I = 13~nldUnl = (2~) -I ~3~nlf(C)l ds _<

It fll= length (~On) 2~ Here we have used the fact that the linear functional can be identified with the measure to the of see

3On

(2~i)-lf(~)d~

[Un ]

in the weak

that

over, if so that

restricted Since

and that their respective norms coincide.

II ~nll are bounded, there exists a cluster

U

is supported on

z E ~(K) IK

* topology. K

~n

point

It is easy to ,

in fact on

3K

.

More-

is fixed, the function h(~) = (~-z) -I 6 C(K) :

is a cluster

-20-

point of the

~D(C-z)-id~n(~)

•

Clearly, then,

(C-z) -I d~(~) = f(z) . K

Finally,

IK Idol

= I1~11 _< ~

II ~nll _~ (2r') -1

Ilfll®

(4(K) + ~) .

Using 3.13 we can prove

Proposition 3.14.

Suppose

4(K) < ~ .

Then

7(K) = sup l S dul u K where the sup is taken over all (finite complex Baire) measures

~

supported on

K

which satisfy

sup rlj (C-z)-id~(=)l < 1 n(K) Proof.

Obvious.

An interesting, apparently open, question is whether 3.14 can fail if if

4(K) = ~

4(K) = ~ .

Vitushkin has shown [92] that

then 3.13 is no longer valid.

Finally, we will need the following result due to Mergelyan.

-21-

([81]).

Proposition 3.15.

Let

K ~

[Iz-zol <

~

.

Then

for any function

fCz) =

admissible for

K

_.,<

ianl Fix

zI

~(z)

if

_.< ev(K)Rn-ln

such that

x-R - z~Zi

...

.

we have

If(z)l

Proof.

al + a2 z-z o (Z-Zo)2

"

lZ-Zo~ I x > R

2, 3, ...

n =

~Zl-Z 01 >_ x > R .

f(z)-f(Zl)

x-R

l-rCqTf(z) -

~Xo f(zz) +

Let

N

If(z)l < i if z ~ o(K) n [Iz-zol <

Clearly.

by the maximum modulus principle, for

K

so that

f

~] .

Therefore,

is an admissible function

{~' (-){= lim {z~(z)l = (x-R) If(zl)l <_ y(K) .

This proves the first assertion.

To get the coefficient

inequality, write

~n - ~

Iiz_~oi=~R~(z)¢Z-Zo)n'Idz

n

= 2,S. ....

-22-

Then

lanl _~ y(K) x n (x-R) -I .

The right hand side of this

inequality is maximized by setting

x = nR/(n-l) .

Since

(l+i/n) n < e , we are done. It is not hard to improve the coefficient estimate of 3.15~ indeed, Vitushkln [88] has shown that where

A

is a universal constant.

refinement in the sequel.

lanl _K Ay(K)R n-1 ,

We shall not need this

-23-

4.

SOME USEFUL FACTS This section will be devoted to establishing two rather

well-known facts about the Cauchy integral;

since references

are somewhat scattered, we include them for completeness. It will be convenient to set some notation. let

Accordingly,

r = [{z{ = i] , A = [Iz{ < i] , A = [Iz{ ~ I ] .

This notation will persist throughout the rest of the paper.

Proposition 4.1.

Let

F(C) 6 C(r) , F

Let

C = eie •

Then

2~

~r ~

Now where

Re ~

where

dc = ~

e i8 1 reie+z eie_ z = ~ L e l ~ z + l] PZ

u

;0

eie F ( e l e ) ee ~ - z

so that

~-z dC

z .

½ Cu(z)÷

is the harmonic extension of

lu(z)l < max IF(C)I , r

d~

eie Re[~]

is the Poisson kernel for

;F

z £A

r ~'(c) ¢-z d¢l _< ~ xr I~(¢)1

IRe~ Proof.

Then for

real .

: I[P~(e)

Thus,

F(e ie) de] ,

F

to

A .

Since

the assertion of the proposition

+ i]

-2#-

is now obvious. Clearly• if

F 6 C(F)

is a purely imaginary function,

the argument of #.i shows that the imaginary part of its Cauchy integral is bounded in modulus by

Propositlon #.2.

Let

F(C) E ~ ( F )

IZ' I < 1 , Iz"l > 1

Then for

lira f~, . Z, z,_~oert ~z ) -

Proof.

1 [

Then

Let

,. ~(z )]= F(CO) .

F

is real.

Let

As before,

rei0+z dc = 1 ' e - ~ - z + 1] de =

~ . ~

Let

.

we have

By linearity, we can assume

z = pe iX .

.

maxlF(c)~ F

z t = re it

l-p, 2 - i 2psln(e-x) + 1] 1-2pcos (e-x)+p 2 1-2pcos(e-x)+p ~

•

z"

=

R

e iT

(r i) .

•

-25-

Re {~(z,) 2n

~(z)] = F(eie)[

I { ~ I0

2

l-r l-2rcos(e-t)+r ~

I-R2

I-2Rcos(e-T)+R2 ]de]

-

i = ~

{~ ~0~ F(eie)[

l'r 2

]de}

l-2rcos (e,-t)+r2'

i~ cos(e-T)+~

+

1 [u(reit) + u(~ eiV)],

where

u is the harmonic extension of

r, R - ~ I

and

t, T-~arg Co we have

F

to

A . Letting

lim Re{F(z' )-F(z")] = F((o )" Zt--~O Z"-~o

Similarly,

~{9(z' ) 9(z')]

=_ [ ~

= -[~

F

_ R sln(e-T) ]de] )[i 2Eos(e T)+r2 l-2Rcos(e-T~R 2 r sin(e-t)

_ ~i sin(e-T) ~0~F(eie~[ r sin(e-t) • l_2rcos (e_t)+r2 i- ~cos (e-T)+R%]de]

1 {v(reit) --~

where

~0~ (eie

v

v(~ eiT)s ,

is the harmonic conjugate of

u

such that

v(O) = O.

-26-

Since

F

is continuously differentiable,

exists uniformly for

~ 6 F .

which completes the proof.

lim v(pe ik) ~ei~-~

Hence z'li~ z "~o ~{~(z' )-F(z")] = o,

-27-

5.

ESTIMATES FOR INTEGRALS In this section,

proof of Melnlkov's and theorems,

we shall lay the groundwork

t h e o r e m by proving a sequence of lemmas

some of w h i c h

(5.7, 5.8, 5.11) are of con-

siderable interest in themselves. Melnikov's

development

Lentm2 5.1.

Let

K

be analytic off > 0 .

Proof.

Then

Let

for the

Our treatment follows

[64] quite closely.

be a compact set in the plane.

K , f(~) = 0 , and If'(~)l _< 2~¥(K)

f(z) = ~

A n z -n

Let

f

Re f(z) < a , where

.

at

Then the admissible

~ .

n=l functior~

fl(z) = f(z)[f(z)

fl(z) = (-Al/2~)z-1 + ... 2mlf l'(~)l < 2my(K) ,

Lemma 5.2.

Let

- 2a] -I at

a

I! f ll,,.

1

If' (a)ip(p-~,(K))

Proof.

By 3.10, we may assume that

Then

I =

to

K .

a 6 G(K) , and let

the distance from .

If'

Thus

as required.

be compact,

K

- .

has the expansion

Let

f

~

be analytic off

_< 2y(K)

K

be K ,

.

has a boundary con-

sistiug of finitely many smooth closed curves.

(For if

the

lemma is proved for this case the general result follows by taking limits).

Also, by 3.2, we may assume that each component

-28-

of

K

is simply connected.

For large

R

we have, by the

Cauchy integral formula,

f, ¢a) = ~

'[I.~I=R ¢¢-a) ~

~K

The first integral on the right is R-*~

(<~I 2

0(R -I) , so letting

we obtain

~K (l:-a)2 dC Thus

If' (a)i = I ~

I~ K ~ d C l

=

(~-~32 i ~ / ' ~ K ~-f-(~ >-~(al

~-a

Now the function

F(z)

-- r ( z ) - f ( a )

f'Ca)

(z_a)2 is obviously analytic off at

z

=a

there also.

K

z-a

except perhaps at

F(z) = (1/2)f"(a) + ...

Clearly. F(®)

=

o ; and

by the maximum modulus theorem.

so that

F

z = a ; but is analytic

I1~11._< ~2 +

Therefore , by 3.7,

P

°

-29-

F(¢)d¢l < [2

i,'

:

This is just a restatement Note that if of 5.2 to read

Lemma 5.3. K*

Let

unit circle

and

K

we can strengthen .

be compact.

r2y(K *) < 2y(K)

two cases.

_< S1 (since

K* ~ I:lzl

Denote by

If

.

0 # G(K) ,

1 < ¥:t)

•

Hence

K

must

r < y(K)

r 2~(K*) <

as required. 0 E G(K)

function for

K* , ~(z) -- f(!/z).

~' (0) = y(K*)

.

~ (with

the conclusion

It is then easy to see that

Suppose then that

to

< y(K)

Izl < l]

(z-+ l/z). Then

2y(K)

"

under inversion with respect to the

the origin.

¥(K*)

r < 2r <

K c {r <

o

of the assertion of the lemma.

f(a) = 0

We distinguish

encircle

+

If' (a)IP(P-Y(K))

the image of

Proof.

lf'(a)l] y(K)

-

Applying 5.2

a = O)

f(z)

Let

Then

be the Ahlfors

~(o) = o

and

(and the remark following

we obtain

r2y(K *) < y(K) + ry(K*)y(K)

.

l~'(0)Ir(r-~(K))

< 2y(K)

,

since

it)

< y(K) , so

1 y(K*) < ~ .

We can now prove

Lemma 5.4.

Let

K c ~ , 0 ~ K .

Let

K*

be as above.

Then

-30-

¥(K U K*) < c[¥(K) + 7(K*)]

where

c

is a u n i v e r s a l

Proof.

If

K

separate Let

0

[Kn}

whose

Again, that

0

is t h e n

from

~ .

from ¥(K*)

~ .

the i n e q u a l i t y So suppose

K

Then we may assume that

be a d e c r e a s i n g

interiors

consists

constant.

separates

since the 1.h.s.

,

contains

sequence of compact

sets each of

of f i n i t e l y m a n y smooth closed curves,

KI* D ~ *

o ... o K*

a(K n) and

= Se \

Kn •

NKn* = K* .

does not ~(K) = S 2 \ K.

K , such that for each

we m a y a s s u m e that

is obvious,

n and

3K n nK n = K .

It is clear We w i l l show

that

y(K U K*) < c[y(Kn) + Y(Kn*)]

for all

n , where

the l e m m a w i l l

c

depends

neither on

t h e n f o l l o w by 3.10.

n

nor on

K

It is enough to prove that

[f'(~)[ < c[y(Kn) + ¥(Kn*)] for

f

admissible

N o w let modify S2

f

without

n

for

K U K* .

be fixed and

so that it becomes disturbing

f

as ~bove be given.

We can

a smooth f u n c t i o n on all of

its values on

$2' \

(K n U Kn*)

.

-31-

Let us assume this modification accomplished.

Let

f(z) + f(I/~) , y_(z) = f(z) - f(i/~)

~+

.

Then

and

are smooth functions analytic off

K n U Kn*

Also,

i s imagirmz-E there.

~+

is real on

F

and

~_

r

and

~+(=)

11%11.

=

~_ _< 2

Let

Izl 1

~-z

r

~-z

dC

Izl 1

are continuous functions on

is analytic off

Kn

and

~2

By 4.1 we have

IRe ~l(Z)l _<

211 ~+tI~_<

IRe ~2(z)l _< 211 ~+!1~_< 4

S2 .

is analytic off

.

-32Also,

I~2(~)I _< e

~l (~) = 0 ,

~l

,

+

o2

=

'P+

•

Applying

5 •i, we get

I~+' (®)l _< I~1' (=)1 + 1~2' (®)1 __<8[~(K~) +

Y(Kn* )]

.

Proceding in a similar manner, we obtain

IV_' (a')l _.< 8[¥(K n) Since

f =½

Y(Kn*)] •

÷

[~+ + ~_] , we have

I~' (®)I _< 8[~(~) + ~(~*)] , as required.

Lemma 5.5.

Let

be continuous

on

K~

~ r<

Izl _<

be compact.

I}

Z , analytic on

A \ K , with

Let

f

II f1~ < 1 .

Then

I 'F f(¢) d¢l -<7c ~(K), where

Proof.

c

is a universal

constant.

As in 5.4, we may assume that

f

is smooth on all

-33-

of

S2

Let

Izl < 1

f(z) ~+(z)

=

Izl>

i

I~! < 1

f(z) ~_(z) =

Let

~1(z)

~+(z)-

i

~(~~ : ~-("~ ~

'r

C-Z

4 ;(~ -T~ °~

It is easy to check, using 4.2, that continuous on all of are analytic off

S 2 ; moreover,

K U K* .

Ixm ~iCz)l < 3 Therefore,

by 5. i,

~i

and

~2

are

it is clear that they

Applying 4.1, we see that

~Re ~2(z)l < 3 •

-34-

1%01' (=)I _< 6v(K U K*)

!%o2 ' (oo) I __<6v(K U K*) ,

so by 5.4

1%01'(')1 _~ c [ v ( K ) + v ( K * ) ]

By 5.3,

v(K*) < % . v ( K )

1%02' (~)1 ~ e [ y ( K ) + v ( K * ) ]

.

, so that

r

1%0~'(=)1 _<~ ~(K), 1%02'(=)1 _<~ ~(K) r

r

Now

%0Z' (=) + %02'(=) = llm z(%01(z) + %02(z))

z I =+~

f(c) dC • F

Thus

as was to be proved.

Exercise.

Justify the first sentence in the proof of 5.5.

-B5Lemma 5.6. F .

Let

Let f

and such that

K c ~

lie at a positive distance

be continuous on llf~ _< 1 .

~ ,

analytic on

r

from

A \ K ,

Then

lJ'r f(~)dCl _ l - r

.

is analytic on

Izl > l-r . ByLahrent. s theorem, we have

f(z) = fl(Z) + f2(z)

Thus, we may continue

f2

1-r < lzl < i

to a function analytic on

It is also clear from the above equality that extended continuously to z E £

we have, easily,

.

~ (and

f2

to

If2(z)l _< 1/r .

fl

S 2 \ K) . Thus

S2 \ K .

can be Now for

-36-

I! f21~

=

IIf-f 11~ _< II fl~ + II flllr

Ilfl~ + IIf-f211r

=

__< llf lk + Ill lit + llf2 lit _< 2 +~

1

.

Then

l;rf(~)d{I = llr[fl({)+f2(C)ldCl = llrf2(_()d{1_< (2~ir)Y(K) • Lemma 5.6 can be improved considerably: remains true w h e n even if we a l l o w

T h e o r e m 5.7. f E C~)

Let

~ K

be analytic on

c

Proof.

r .

This is the content of

be a compact s u b s e t of A \ K

l~r f(~)d¢l where

is a universal

and satisfy

.

Let

IIfll~_<

1

K

is bounded by a finite number G(K) = S2 \ K .

as before,

is smooth on all of

we can assume that E CR~ )

Then

constant.

We may assume that

O(z)

.

_< cy(K),

of smooth closed curves and that

Let

it

is replaced by an absolute constant, to extend to

K

actually,

be such that

f

G(z) = 1

for

Moreover, A .

z E {~_< Izl < 1},

-3Y8(z) = 0

I~I __< 4 Then

for

z E {0_< Izl _< ½] , 0_< ~(z) _< 1 ,

. (Here, as usual, ~~

6f + (l-6)f = f.

(6f)(z) =

1

~FU~K

and

= ½ [ ~~ + i ~~] , w h e r e

z = x+iy)

By the complex form of Green~ formula

(Bf)(c) dC + C-z

~

1

(~6)f

I A\K ~

(~)~-Iz d~Ad~

-- fl(z) + ~lCZ) •

Similarly,

(l-6)f = f2 + °2 "

on all of

The function

~I

is continuous

S2 , being the convolution of a locally integrable

function and a bounded function of compact support ; the same remark holds true for to be continuous Let

K1 =Kfl

on

{½<

easy to see that

02 . by

We can therefore continue

fl = Bf - 01 , f2 = (l-6)f - 0 2 .

Izl _< l] , fj

~

moreover,

llf2I~ _< ll(1-6)flI= + II o2 II=_< c l 1 r = S

= K 0 {Izl _< ~ ] .

is analytic on

~i + 02 = 0 , fl + f2 = f ;

A \ Kj .

It is

Since

llfIII~ _< IIGfll~ + llcP III~ < c I ,

Applying 5 •5 and 5.6 with

we obtain

F

fj

F < 9ClY(K ) + 9ClY(K ) = cy(K) .

-38This completes

the proof.

At this point, comments

it is perhaps

on the t h e o r e m we have

that

for continuous careful

of Cauchy's

in c o n n e c t i o n with Morera's

(closed)

sets of zero analytic analytic

functions

reader will have

noticed

need the full h y p o t h e s i s valid if we require F ~ elsewhere,

For one thing, theorem.

theorem,

capacity

are removable

f

.

In fact,

The

5.7 remains

be continuous

we need only the boundedness

on and near

of

f .

in the proof of

theorem.

Corollary 5.8. f

II).

that we did not actually

Our next result will be used explicitly Melnikov's

On the

5.7 shows

(cf. A p p e n d i x

f 6 C~)

only that

to offer a few

just proved.

5.7 is a kind of g e n e r a l i z a t i o n other hand,

desirable

Let

be continuous

K c [ r < [z]

on the annulus

Suppose f u r t h e r that

< R]

be compact

and analytic

I] flL = m .

and let

there off

K .

Then

f(~)d~- I i=r f(~)d~l < c(~)y(K) Here

c(r/R)

is a constant

bounded a w a y from i , c(t)

Proof.

By a change

depending is b o u n d e d

of variable

only on

r/R ~ for

away from

we m a y take

t

~ .

R = I, r < i .

-39Also, we may once again assume that

K

has a nice boundary.

Then

~r f(~)d{ - ~i~ I=rf(:)d= = 13Kf(=)d=

by Cauchy' s theorem.

Let

the annulus such that 0 < iB(z) _< 1 ,

and

be a real

B(z)

6(z) = i

I 11._<

el(r)

is analytic on

~(z) = 0

where

l~K g( ~)dcl _< c~(K)

and

[Izl>

is analytic on

Izl = r ,

!1 flll.-<

c2(r) •

Then 5.8 follows from 5.7

and from the fact that g

on

is analytic on

fl

{Izl > r} \ K ,

II f211~ _< c2(r) • ~i + ~2 = 0 .

function in

As in 5.7 we can write

•

~f = fl + ~i ' (l-~)f = f2 + ~2 • A \ K , f2

r

on

C~

r] \ K .

if

11 gl5 _< 1

(The proof of this

last fact proceeds exactly like that of 5.7).

Lemma 5.9. sets lying

Let

~

be a circle and let

S1

and

(respectively) interior and exterior to

S2

be a .

Then

~(s I u s2) _< c[~(s l) + ~(s2)] , where

Proof.

c

is a universal constant

We may assume that

.

S = S 1 U S2

is a compact set with

-40-

boundary consisting of finitely many analytic closed curves. Let on

~

be the Ahlfors function for

n(S) [33].

Suppose

in such a way that

S ; then

G = [IZ-Zol= r] .

S2 c [IZ-Zol < R] ,

~

Choose

and let

is continuous R >_ 2r ~ = [IZ-Zol

= R].

We have

J" ~(z)dz I _<

,(s> = ,. (.> -- i ~

- ]" ~(z)~z]l + l~-~ J" ~(z)dzl

__< 12~1

_< c(r/~)y(s 2) + % ~(Sl) < o[¥(Sl)+Y(s2)] by 5.7 and 5.8 and the fact that

1 r/R_( ~ .

We should remark that we could have avoided using the nonelementary fact about the boundary behavior of the Ahlfors function for a set with analytic boundary; we leave it to the reader to supply an entirely elementary argument. It is not hard to generalize 5.7 to a theorem for domains with analytic boundary.

Let

G

be a domain bounded by a

simple closed rectifiable Jordan curve compact.

a

and let

K c ~

be

Define

I(a,K) = supl~ f(~)dzl , G

where the sup is taken over all functions continuous on

G ,

analytic on

5.10. ~

c G

G \ K

Let

such that

GI

be closed.

conformally onto =

w(%)

and

<_ 1 .

and

We have

be domains as above and let

Suppose the function G2

.

G2

II fll=,

w(z )

maps

0 < m < w'(z) < M < ~

for

% z 6 G I.

Then

1 1 1~ I ( G 2 " ~ ) --< I(GI'KI) -< ~ I ( G 2 " ~ )

Proof.

Let

oj = ~Gj .

Then

f(w)dw = ~

az Choose

fl

f(w(z)) w' (z)dz

~l

such that

I

fl(Z)dz _> I ( G I , ~ )

- ~ , c > 0 . Let

aI f2(Z) = mflCZ)/Wl'(Z )

and

is the function inverse to

f2(w) = f2(z(w)), w(z) .

where

Then

mT(GI'KI) < la I mf l(z) W' (z)dz + m8 : ~

-

Z(W)

~'(z)

_< I(G2, K2) Jr mE -~ I(G2, K2)

f2(w)dw + me

~2

as

e-+ 0 .

The other inequality is proved in a similar manner.

-42Let

Theorem 5.11. analytic

G

and let

curve

be analytic

be a domain bounded by a simple closed

on

G \ K

K c ~

be compact.

and satisfy

Let

II flI~ _< 1 .

f £ C~)

Then

I~ fCz)dzl _< oCG)~CK) , where

c(G)

Proof.

is a constant

Let

G

and

K

depending

is an analytic

curve

map of a neighborhood containing that

A

.

G .

be as in the statement

There exists a conformal map a

only on

w(z)

w(z) U(G)

Clearly,

of

G

of 5.11.

into

A ; since

may be extended to a conformal of

~

onto an open disc

there exist constants

0 < m < lw' (z)l < M < ~

for

z 6 U(G)

.

U(A)

m,M Let

such K 1 = w(K).

By 5.10,

1 I(G,K) _< ~ I(A,K 1)

I(U(A),K1) < M I(U(G),K)

Invoking 5.7, we obtain at a fixed distance to

U(G)

instead of

from

1

I(G,K) < ~ cy(Kl)

.

Since

K

~U(G) , we have from 5.6

(applied

A)

~(KI) _< I(U(A),KI) _<MII(U(O),K) _< M c!(O)~(K)

lies

.

-43Hence

I(G,K) i c(G)y(K)

Corollary 5.12.

.

If the sets

an analytic curve

SI

and

S2

are separated by

a • then

~(sI u s2) _< c(a)[~(s I) + ~,(s2)], where

c(q)

Proof. the

is a constant depending only on

a •

This result is the analogue of 5.9 for analytic curves;

proof

follows

that

of

5.9 , using

5.11 .

It is rather remarkable that results such as 5.7 and 5.11 are valid, i.e., that the constants involved do not depend on the compact set in question. that the constant constant

c

c(G)

One might be led to conjecture

in 5.11 can be replaced by a universal

(independent of

G) .

Such is not the case.

This is the content of

Example 5.13.

(Vitushkin [92]).

the unit square.

Let

Q : [%1]

Using horizontal lines divide

x [0, I]

be

Q

#n

into n-1

congruent rectangles n-1 S = ~) k=-0

R4k+3

.

Let

RI,~, G

...,R4n.

Let

T:

U

k:O

~k~l

be a simply connected domain with

analytic boundary such that

T c G

and

~ n S = ~ .

Let

m > 0

and set

k(C) = i

f(z) = ~sum An interchange

on

C-z

T,

k(C) = -i

|

f(z)dz

= ~/2

It is not hard to see that as

n---

shows that

c(G)

S .

Define

(~+±~ = C)

d~du

of order of i n t e g r a t i o n

@

on

gives

.

~G

must a c t u a l l y

!1 f l l . - o

depend on

G .

(K~]).

~s

-456.

MELNIKOV' S THEOPd~4 We are now p r e p a r e d

proceeding

with the proof,

set some notation. and

£n = [Izl

For

z 6 C

from

z

= 2 -n]

of

•

If

p(z,S)

S .

the diameter

however,

Accordingly,

we set

to

T h e o r e m 6.1

to prove Melnikov's

Finally,

= inf

Before

it will be convenient

let

S c ¢

theorem.

A n = [2 -n-1 ~ we shall write

to

Izl ~ 2 -n] CS = 2 \

[Iz-x I :x E S] , the distance

d(S) = sup

[Iz-gl

:z,C

6 S]

is

S .

(Melnikov).

Let

y(CX

fl {2 -n-1 _< Iz-xl _< 2 - n ] ) .

for

R(X)

x E X

and let

Then

x

Yn =

is a peak point

if and only if @O

2n

Yn = ~ n=0 Proof.

For convenience

First, positive

assume

integer

poles lie off

we may suppose

x = 0 .

that the series above N

X .

and let We modify

f

converges.

be a rational f

S .

off

function

a neighborhood

in such a way that

(1)

IIflls2_<211 fllx

(2)

IIfllrN_< 211 fllrNnX

Fix

of

a whose X

-#6-

Choose a compact set X c int K and boundary of

(3)

f

K

with smooth boundary so that

is analytic on

CK 0 [Izl _< 2 -N] •

f(0) = ~ ~

K .

Let

~

be the

Then

IIFN f(() dc - ~ ~ i

~

d~

by the Cauchy integral formula (all integrals being taken in the counterclockwise direction).

Also, by Cauchy's theorem,

I

(4) d~ - ~

~ jrn+ 1

only finitely many su~nands on the r.h.s, being nonzero. follows from 5.8 in conjunction with (i) and (4) that CO

(5) 12-~f~ f(~) dCl < cl' f"x I 2n Yn n=N where

c

is a universal constant.

Consequently, we have

from (2), (3) and (5)

(6)

If(0)l --<2~flIFNNX + cllfllx ~

2n Yn

n--N oo

Choosing

N

2n

so that

7n < i/2c , we obtain

n=N

(7)

llrNnx+

if (o) I _< 2 llf

(1/2)

~fllx•

f({) d~]l

It

~

,

-47This inequality re~eins valid for all

f E R(X)

suppose

fn E R(X) peaks at

0

f E R(X)

peaks at

0 ; then

for every positive integer

n .

Choosing

enough, we obtain a contradiction to a peak point for

R(X)

(7).

.

n

Now

large

Thus

0

is not

.

For the sufficiency,

suppose

Z 2n Yn = ~ "

Let

CO

(8)

CI = i

Z

- e

k

8-k-Ti > 0 .

K--2 We shall verify that the = C1/lO , ~ = C1/9 . CX N A n

for w h i c h

Let

of

(1/8)2 -n-1 . constant

c

Jn

condition of 2.2 holds with be a compact subset of

y(Jn) > yn/2 .

of circles we can chop independent

~-~

Jn

Using a finite number

into pieces

n ) each of which has

Jnl,...,Jn m (m

diameter

less

than

It follows from 5.9 that there is a universal such that m

(9) Y(Jn)-< c ~ y(jk)

n = 0, i,...

.

k=l For each

n

pick a set

J k n

and call it

(I0)

K

n

.

Then

K n = CX N A n

,

for which y(Jnk) > y(Jn)/mc

(ii)

d(Kn)

(12) ~

Let

~n

2n

_<

(1/8) 2 -n-I ,

and

Y(Kn)

be the Ahlfors function for

zn £ Kn .

and let

Kn

Then O0

an k

(13) ~ (Z) = k=l ~ (ZIZIn)~

where

anl = ¥(Kn) .

centered at

(14) for

zn

Since the circle of radius

covers

lank I _< ey(Kn)

k = 2,3,....

K n , it follows from 3.15 that

k

[d(Kn)]k-i

Moreover,

closed disc containing

Kn

if

z

lies exterior to a

it follows, again from 3.15,

that

(15)

ln(z)l _< Y(Kn)/P(Z, Kn)

Now let

(16)

6 > 0

d(Kn)

be fixed.

2 -N < (9/20). ~-(6-2 -N)

. Choose

N

so that

-49and then pick

M

to satisfy

M

(17)

i < ~

2 ~ y(K n) < 3 .

n=N

(This is possible by (Ii) and (12)). Let M

(18)

f(z) = (1/9)

~

~n 'n (z)

'

n=N

where the

~n

are unimodular constants such that

~n 'Pn(°) > o . We claim (*) (*')

Ir(z)l < ci/io ir

**@)

Ir(o)l > ci/9

To obtain n ,~

II r llx_ 6

z~xnAj

. Then for

J, J+_1 we have

(19)

Iq~n(Z)! _< Y(Kn)/P(z,K n) < 2 n+l Y(Kn)

by (15) and (i0) .

Combining this inequality with the

-50-

trivial estimate

Iq,k(z)i

__< 1

k = j,J_.+l

and

invoking

(17),

we have

M

(2o)

M

l%(z)l __< ( 1 / 9 ) [ 3 + 2

I f ( z ) l __< (1/9) n=N

~ 2 n y(~)] n=N

_< 1 ,

as desired. Next, suppose

Izl >_ 6 .

Since

2 -N < 6

by (16) , we

have, from (lO) ,

(21)

6-2 -e _< P(z,A N) _< P(z,K N) •

Therefore,

by (15), (21), and (16) M

(22)

M

if(z)l _< (.,/9) ~ I~,n(z)l _< (1/9) ~ ",.(Kn)/p(z.K n) n=N

n=N

M

< (1/9) ~ 2-n/(6-2 -~) < 2 2 -N -

n=N

-

9

~_2-~T

C1
which is (**) . It remains to establish

(***) .

Since

8(1/8) 2-n-1 l 8d(Kn) , we have from (13),

IZnl >_

(14), and (8)

-51-

Y(Kn) (23) I%(o)I _>T ~ -

ey(Kn) ~ k=2

Y(Kn )

~

k.[d (Kn)]k-i IZn lk

k[d(Kn) ]k-i

k--2 Y(Kn )

oo

->T~ [l-e~ 8-g~ ]k k--2

Y(Kn) 2n = Cl T ~ >- Cl Y(Kn) "

Hence, by (23)

and

(17), M

(2~)

M

f(0) = (I/9) ~ ~n mn(O) >- (C1/9) ~ 2n Y(Kn) > OI/9 ' n=N

which is

(***) .

n=N

This completes the proof.

It is worth noticing that Melnikov's theorem can be given a formulation slightly more general than that presented above.

Indeed, let

0 < k < 1

y(CX n {~n+l ~ iz_x I ~ kn)).

and set

yn(k) =

Then the proof of Melnikov's

theorem shows that

X n=O

x -n

yn(x)

= ~,

is also necessary and sufficient for point for

R(X) .

x 6 X

to be a peak

-52-

7.

FURTHER

RESULTS

In general, 6.1 directly~

it is not easy to verify the condition

therefore,

truly geometric for

R(X)

.

it is useful

condition

For

x E X

of the diameters

for a Rolnt and

r > 0

of the components

[39] , Gonchar proved

that

x

is a peak point

Let

X

is a peak point

Proof.

be the supremum

CX N {Iz-xl

< r}

for

R(X)

.

for

where

R(X)

result d m e t o

, x E X .

In

P. C. Curtis

Then if

take

.

x = 0 .

< rn])/r n > e > O c

.

Using Melnikov's

r

y(CX N {!zl

0 < & < e/c ,

d(r)

(r-* O)

be compact

For convenience,

such that

let

(cx n {Iz-x I < r} > 0 ,

lim r~O

x

to be a peak point

of

t h e o r e m we can prove a m o r e general

T h e o r e m 7.1.

a

if

lira d(r)/r > 0

then

to have available

of

is the constant

Let

rn @

0

and take of 5.9.

Then

be

[iI$].

-53-

for large k

n

kk+l _< rn < ~k , where , of course,

we have

depends upon

n .

Applying 5.9 twice, we obtain

;kk+l e _< r n g _< y(CX CI [ I z l

< rn} )

< c y(ox n {Izl < ~k+21)+ oy(cx n {~k+2 < Izl < ~k])

< c}k+2 + c2 Yk ( )") + 02Yk+l ( ~")

where

"

yn(k)" = y(CX CI [kn+l_< Izl _< } n ] ) .

Hence ,

~k+l

o

• (~-

~) _< Yk(~) + yk+l(~)

•

It follows that

@o

a-n yn(X ) = n=O

By the remark following the proof of Melnikov's theorem we are done. It is easy to see that the condition of 7.1 is not necessary for Exercise.

x

to be a peak point.

Prove this assertion.

-54We now turn to some corollaries

Theorem 7.2.

Let

x E X

and let

cx n {2 -n-I _< Iz-xl _< 2 -hI . a peak point for

Proof.

R(X)

~ =~4n

Theorem 7 •3. c x n {Iz-xl

x

Proof.

= =

4n mn = =

then

(2 n ~/'Wnn)2 _<~ (~)(2n~

r]

x E X

(We h a v e

.

and let

is

x

used

yn ) = n ~ 2 n ¥n "

the

fact

that

=(r)

be the area of

Suppose

~l0 Wr~~ Then

be the area of

3.12).

Let

<

theorem.

.

mn = ~

and

~n

If

Now apply Melnikov' s theorem. w n _< n4-n

of Melnikov's

is a peak point for

dr = = .

R(X)

.

We have

~o

Clearly, value by

=

o ~(r~~ ( - ½

7)=-~.

the first term of the r.h.s, ~/2 o

On the other hand,

71o

+

;o

r

is bounded in absolute

-55-

1 dell ~ r

Thus,

7.3

follows

Although sufficient necessary.

n-l~

2

n~O

__< ~

4n

d,i,(r) = 4

n=O

4 n ~)n "

n=O

2 -n-I

f r o m 7.2.

the

(equivalent)

conditions

for 0 to be a p e a k point, Indeed,

let

7.2 and 7 . 3 are

t h e y are far f r o m

A n = [Iz-3 -2-n-21

< n-12-n]

and

oO

set

X = A \ ~

that, but

for

An

One s h o u l d

continuous

.

problem every

discussed

Let

D

u 6

a regular

D

x 6 8D

if

CR(~X ) . point

the r e l a t i o n

A

u

u

point

as

z -+ x

[105]

showed

for the D i r i c h l e t

u

problem

of h a r m o n i c be a real

we can associate•

(continuous

is a r e g u l a r

Wiener

of the

theory

To e a c h s u c h

~(z) -~ u(x)

.

d o m a i n and let

a function

so

= ~ n -I = - ;

< ~

to the c l a s s i c a l

8D .

manner•

A point on

on

n = 4,5, ....

~n2 n_ ¥

~#n= Wn= ~ n - 2

be a p l a n a r

function

in a c a n o n i c a l D

so that

Now

not fail to m e n t i o n

we h a v e

functions.

on

CX n A n = A n

x = 0 , Yn = n-12 -n .

w n = n-24-n~

results

Then

and) h a r m o n i c

for the D i r i c h l e t

from w i t h i n that on

x 6 8D D

D

for

is

if a n d only if

n

n=O where

c n = cap(cD

n {~n+l _< Iz-xl _< ~n} , o < ~ < 1 .

This

-56condition is independent of the choice of

k •

Wiener's condition was phrased differently,

(Actually,

~ nce the definition

of logarithmic capacity which he used is not the one in common currency today).

In particular,

conditions

implies that

(see

~ k-ncn =

[84] ).

Moreover,

any of the (inequlvalent) x

is a regular point

the conditions of 7.2 and 7.3 are

sufficient to insure that

x

A decade after Wiener,

be a regular point for

([8hi).

D

Keldysh [57] observed that one

could characterize the regular points for a domain in terms of the peaking properties of an appropriate space of harmonic functions.

More specifically,

functions continuous on showed that problem on

x 6 ~D D

D

let

~Z(D)

be the set of all

and harmonic on

D .

Keldysh

is a regular point for the Dirichlet

if and only if

x

is a peak point for

a simple proof of this fact is in [32].

Since

~(~)

it is clear that any condition which implies that is a peak point for ~) D .

R(D)

implies

x

~(D~

;

o ReR(D),

x 6 D

is a peak point for

and hence a regular point for the Dirichlet problem on Thus

theory.

7.2

and 7.3 strengthen known results from potential

-578.

APPLICATIONS Melnikov's

t h e o r e m gives us useful

in the simplest a compact

set

a sequence

(nontrivlal) X

of

cases.

constructed

open discs

information

Consider,

as follows.

even

for example,

Delete

from

A

A n = ~..[IZ-Xnl < rn] , where

(i)

I > Xl > x2 > ... - ~ 0

(2)

x I + rI < i ;

(3)

Xn+ I + rn+ I < x n - r n

;

for all

n .

~O

Then

X = A \

~

An

We shall call such an

.

X

a set of

n=l

type

(L).

Thus,

of a domain

if

(X °)

X

is of type

obtained

(L),

by deleting

it is the closure from the open unit

disc the o r i g i n and a sequence

of pairwise

discs

real axis and clustering

0 .

centered

on the positive

It is clear that

~

rn < ~ .

disjoint

closed

It is also obvious

at

that

n=l each point of

8X \ {0]

natural q u e s t i o n R(X)?

t® ask is:

Melnikov's

a complete

is a p e a k point

answer.

w h e n is

t h e o r e m allows

0

for

R(X)

.

a peak point

us to give this

The for

question

-58Theorem 8.1.

Let

X

be of type

(L) .

0

Then

is a peak

OO

point for

R(X)

if and only if

(rn/Xn) = . n=l

Proof.

Let

I = CX n [0, I] , I n = I n A n , Jn = I n A n .

First we show that

]I x-1 dx

diverges with

~

(rn/Xn) .

Indeed, we have O0

CO

~ rn ~ 2rn n=l q -< X~rn n=l

Hence,

n~l -<

=i

Jn

dx ~ dx x : I x

"

for the sufficiency it is enough to show that

is a peak point if

[

x -I dx = ~ .

Suppose,

0

then, that

I the integral diverges.

Then O0

x -I dx < _

n=O

In

~ 2 n+l ~i n d x n=O

O0

2 n length

=2

(In) = 8

n=O

2 n Y(In) n=O

OO

<8

~2nyn

,

n=O

where we have used 3.1 and 3.9.

Thus

~ 2n ¥n = ~ ' and

0

-59is a peak point.

For the other direction,

to invoke Melnikov's theorem. Then

rn/Xn-~O

rn/(Xn-rn) measure

~ rn/X n < ~ .

, so that for some integer

< K(rn/Xn)

is carried out, as usual, 8X \ F ; U

n °

for all

dU(C) = (2Ni) -I ( -i d~

wise on

Suppose

on

we do not need

K

(0 < K < ~)

Consider the (complex) 8X , where the integration

counterclockwise

on

F

and clock-

has finite total variation since O0

ds

Clearly,

2Wrnxn_rn_< 2~(1 + K ~ ~rn ) < n=l n=l n

IUI({0]) = 0 .

-.

Since

~3X r(~) d u(C) = r(O)

for every rational function with poles off

!

f(C)

d~(C)

X , we have

= f(0)

3x for every ~en

f(o)

fn(c) -~ 0

f 6 R(X) . = i

and

Suppose

f 6 R(X)

If(C)I < i

,

peaks at

¢ J o .

~us

pointwlse boundedly almost everywhere

But then, by the dominated convergence theorem,

0 .

d I~I

~60-

i = fn(o) = 5 fn(¢) du(¢)-* o

a contradiction. at

Hence

instance,

8.1 admits

if

consists

X

of

is b o u n d e d

CX n A n

can peak

depends

of type

(L)

0

exists

to conclude

of

X

fails

0 ; one might

fact,

by

Ixnl).

near the point

R(X)

More precisely,

suppose

rectifiable

3X .

However,

we have the following

boundary

circles

that if

X

0 for

near

0 .

for a set for

3X .

situation persists

then Cauchy measure on

on

(i.e.,

in question)

condition

as a finite measure

conjecture

Since

is local

to be a peak point

the o r i g i n and a u n i o n of disjoint

finite measure

xn

n , t h e n 8.1

to be a peak point

that this

of a domain whose

for

circles

if and only if the Cauchy measure

is rectifiable.

point

of

the above b o u n d e d n e s s

to 8.1,

~-l PC •

tempting

closure

replace

only on the structure

According

For

of a domain whose b o u n d a r y

independently

for a point

we need only require

(2ui)-l

generalization.

0 , and if the number of components

(of course we must

the c o n d i t i o n

at

R(X)

of the o r i g i n and a union of disjoint only at

holds.

immediate

is the closure

accumulating

3X

in

,

0 . Theorem

X

no function

as

0 , It is

whenever is the

consists

accumulating

of only

fails to be a peak 0

exists as a

this is not the case.

In

-61-

Example 8.2.

~n = { I z - 4 - n - # - 2 n [

Let

< #-2n]

•

n = 1,2,

....

00

Then

~ ~-2~/(~-~-4-2~ R(Xo) and each

i...et

X0 = A - U An n=l

.

Hence

0

let

then D

0

Y

is not a peak point for (j = 1,2,...,kn)

n,j

open discs with radii

rn, j

..d

~ (rJx.)

is not a peak point for

It is clear, therefore, that if

y o X0 n

< ~ .

of typ~ (~.)

is compact R(Y) .

For

be a collection of

such that

(i)

]~n,j c A n

for all n,j

(2)

]~n,j N 1~n,k = ~

(3)

n -2 <_~ rn, j < 2n -2

unless

j = k

for each

n

j=l

Let

Y = ~ \

~J

D

n, j

Then

y = yO • BY

consists of

n, j "

the origin and a union of disjoint circles accumulating only at

0 , and

BY

is rectifiable since k

~ rn, j < 2 ~ -n=~l n=l j=l n=l

Since

Y o X0 , 0

On the other hand,

3-~-

is not a peak point for

R(Y) .

=

-62-

kn

~

kn

rn, j

~ lxn;j, _> ~ 4 n=l j=l

n

n=l

so that the measure

4n

Z rn, j > I j=l

(24)-1C-I

n"~ = ~'

n=l

dC

does not have finite

total variation. At this point, it is perhaps worthwhile to note that the notion of analytic capacity is quite crucial for Melnikov's theorem:

even in the simplest cases it is not

possible to replace analytic capacity by the more familiar logarithmic capacity.

Example 8.3.

from

A

Let

This is shown by

Y

be the compact set obtained by deleting

the open discs

~k+l

= [ Iz

An

i

where

t2k2+8

I

2k-~ ,k +-~-3 ) I < 2k+2 (k2+31 } _

A2 (k+l) = ~Iz k = 0,1,2, . . . .

Then

1

4k2+i0

~

A2k+l U A2(k+l)

of ~ = [2 k-l_< Izl _< 2 -k] We claim

~ 2n Yn < ~

but

the logarithmic capacity of that

Yn --< 2-(n+l)

i

)] .

( k2÷3 11 < 2~+2 (k2+3

, and

Y

lies in the interior is a set of type

Z 2 n c n = ~, where CX ~ A

n

(n2+3) -I, so that

.

cn

(L).

denotes

From 3.8 it is clear 2n Yn < ~ "

On the

-63other ha0.d,

, ,- 1

t2 0.2+7

1

20"+2(~+3) 1 2 °.+2 (0.2+3)

1

t2 0.2+9 ~ 1

cap([20-2+7,20-2+9] u [4~+P,#n2+ll]) cap([-n2-~-,', 2] u [~2,0-2+2])

by the monotor~city, homogeneity, and tra0.slation invariance of logarithmic capacity.

If we use the fact that the

logarithmic capacity of a compact set coincides with its transfi0.ite diameter (see Appendix I), it is not hard to show that

cap([-~2-2,-~ 2] u [0-2, ~2+2]) = ~+i

> n

so that

~=~

I Z n=l

as required. that

0

n~+3 <

Z2°

cn"

n=l

By a remark in the preceding section it follows

is a regular point for the Dirichlet problem for

-64-

yO

(in fact, a point of stability,

to be a peak point for R(Y) CX n A n

.

see [5T]); yet

Note, however,

¥n

by

cn

fails

that if

consists of a single component for each

can replace

0

n

we

in Melnikov's theorem ; this follows

from 3.6. Finally,

let us remark that we have said nothing about

the characterization

of the peak points of

difficult problem remains open (see 16.4). Melnikov's

theorem gives a sufficient

condition is necessary as well when

Of course,

condition, and the R(X) = A(X)

to the question of when these two algebras we now turn.

A(X) ; this more

.

It is

coincide that

-659.

THE P R O B L ~ Let

OF RATIONAL A P P R O X I M A T I O N

X c C

functions in over,

be compact.

R(X)?

What can we say about the

Clearly,

they are analytic on

they belong to

X° ,

C(X) ~ more-

the interior of

X .

(This follows from the fact that each function in is a u n i f o r m limit on Denote by

A(X)

is empty.

of functions analytic on

the algebra of functions in

are analytic on X°

X

X ° ~ then

A(X) = C(X)

We have observed that

this inclusion be proper? R(X) = A(X)?

What if

succeeding

X°).

C(X)

which

if and only if

R(X) c A(X) X ° = ~?

. Can

When is

These natural questions are basic to the

qualitative theory of rational a p p r o x i m a t i o n planar sets.

R(X)

on compact

We shall treat them in some detail in the

sections.

The best understanding

we have of the problems which

arise in rational a p p r o x i m a t i o n

comes from examples

counterexamples).

most of this section will

Accordingly,

(and

be devoted to a d i s c u s s i o n of p r e v i o u s l y k n o w n results, a special emphasis on examples. essentially trivial but

with

Our first theorem is an

enormously useful o b s e r v a t i o n due

to Runge.

T h e o r e m 9.1. X . of

Then f

to

Let

f

be analytic in a neighborhood

flx 6 R(X) , where X .

fix

U

of

denotes the r e s t r i c t i o n

-66-

Proof.

Let

s c U

be a rectifiable

winding

number 1 about

each point of

1 f(z) = 2~i

Since the right hand

contour that has

~

X .

f(c) ~-z dc

Then

z~X

side is a Riem~nn

integral

it is

(for fixed z) a limit of sums of the form n

(2~i)-l

~

cj

(~j-z)-l

, ~j 6 ~ .

Since

X

is contained

j=l in the interior bounded

by

converge Let on of

uniformly ~(X)

~(X)

on all of

denote

(a n e i g h b o r h o o d

so that

of)

connected)

set

.

R(X)

the algebra

to ~ e

The content =

.

sums

X . of all functions

X , and let

with respect

~(X)

uniform

This

be the closure

norm on

of T h e o r e m

X .

Of course,

9.1 is that

situation

analytic

does

~(X) c R(X)

not persist

~n (n > i) . A second useful

component (z-~) -1 in

(not necessarily

~ , it is easy to see that the R i e m a n n

R(X) c ~ ( X )

in

of the

of

S2 \ X

observation and let

can be a p p r o x i m a t e d

(z-6) -I

(in z, if ~ = ~)

and follows argument. fix a point

in general

Thus, Pn

is this.

%~ on .

T

6 T , ~ % ~ . X

uniformly

be a Then

by polynomials

This is clear for

from a standard

in studying

Let

rational

in each component

Tn

near

connectedness

approximation of

m

S2 \ X

we may and

-67then consider only rational functions whose poles lie in the set

{phi.

Po = ~ "

In particular,

Suppose now that

function in

R(X)

if

G(X)

S 2 \ X = ~(X)

can be approximated

functions whose only pole lies at Thus,

TO =

.

we may choose Then

every

uniformly by rational

~ , i.e., by polynomials.

for sets which do not divide the plane,

a p p r o x i m a t i o n and rational a p p r o x i m a t i o n are

polynomial (qualitatively)

the same. We b e g i n with a b r i e f discussion of the situation in which

X

is "simply connected,"

is connected.

i.e.,

for w h i c h

S2 \ X

We shall delay the proofs of the results

stated below,

since these will appear as easy consequences

of a later t h e o r e m

(see section 13).

Example 9.2.

X = [a,b], an interval on the real axis.

Let

The Weierstrass in this case. line

a p p r o x i m a t i o n t h e o r e m implies that Similarly,

if

R(X) = C(X) , since if

X

{Izl < l~ Then

Recall that

and

~

A(X) = R(X)

A

C([a,b])

denotes the open unit disc Let

X = A .

One's first guess might be that the

partial sums of the Taylor e x p a n s i o n of approximate

C(X)

.

is its closure , {Izl < 1}. .

set on the

X c [a,b] any function in

can be extended to a function in

Example 9.3.

is any compact

R(X) = C(X)

f ~ however,

f 6 A(X)

at

0

it is w e l l - k n o w n that these may

-68-

fail to converge means

on

F = 8A .

On the other hand,

of the Taylor

series for

f

converge

the Cesaro

uniformly

to

f ,

and we m a y take these as the a p p r o x i m a t i n g

functions.

ExamRle

set whose b o u n d a r y

9.4.

If

is h o m e o m o r p h i c This result, by W a l s h

to the unit

clearly,

may be nowhere

closure

dense

(i.e.,

Lavrentiev

in

can be a p p r o x i m a t e d

then

is uniformly arbitrary until

Theorem function nomials

connected

A necessary

in

A(X)

is that

X = X° , i.e.

compact

concerns

A(X)

= R(X)

.

on

and

can take:

m a y be the

X X

is simply in

A(X)

The case of an

set remained

unsettled

the beautiful

and sufficient

condition

approximable

on

X

that every by poly-

be connected.

us is that if

X

by polynomials.

every function

[66] proved

one half of this t h e o r e m

What

X

X

dense then every function

uniformly

be uniformly S2 \ X

forms

[60] showed that if

by polynomials.

when Mergelyan

9.5.

Of course,

= R(X)

approximable

simply

1951,

that if

A(X)

.

is no longer elementary:

X ° = ~) or

and nowhere

[58] proved

= A(X)

of 9.3, was proved

there are two extreme

connected

connected

however,

R(X)

from conformal mapping.

is simply

Keldysh

F , then

generalization

of its interior.

C(X)

connected

circle

The argument,

on theorems

Quite X

is a simply

the natural

[96].

it depends

X

S2 \ X

is virtually is connected

trivial. then

-69If

S2 \ X

contains more than one component the situation

is, quite naturally, that if

S2 \ X

A(X)

= R(X)

for

R(X)

.

more complicated.

Mergelyan

has only finitely many components then He also obtained some sufficient

to coincide with

connectivity.

A(X)

conditions

in the case of infinite

As evidence of the complexity of the infinitely

connected case we submit the following example, Mergelyan

[67] showed

[67] and k n o w n a f f e c t i o n a t e l y

due to

as Mergelyan's

Swiss

Cheese.

Example 9.6. from

A

Let

X

be the compact

a sequence of open discs

(1)

An c A

for all

(2)

A n O ~ m : ~ " unless

(3)

X = ~ \

set obtained by removing

An

satisfying

n

n : m

OO

(4)

U An n=l

~ r n < ~ , where n=l

One way to construct of

A

has no interior ;

rn

is the radius of

•

such a set is to enumerate the points

with rational coordinates

obvious manner,

A n

always

choosing

and then to proceed in the r

< 2 -n .

We claim

n

C(X) ~ R(X)

.

Indeed,

let

C

be the measure on

F U

%; n=l

w h i c h coincides ~ith

dz

on

r

and with

-dz

on

U ~An " n=l

-70-

By (4),

is a finite measure

on

Cauchy's

X .

t h e o r e m gives

~x r(z) d~(z) = 0

for any rational

function

r

with poles

off

X .

Thus

~x f(z) d~(z) = 0 if

f 6 R(X)

there

Since

exists

U

g 6 C(X)

is clearly a nonzero measure

,

such that

~x g(z) d~(z) / o Hence

R(X) /

C(X) .

Thus, t h e r e which

exist

R(X) ~ C(X)

Rosenthal measure

[44]

.

compact s e t s w i t h o u t i n t e r i o r In the other direction,

showed that if

zero then

R(X)

The set constructed natural

to ask if

is the closure

R(X)

X

= C(X)

quite

and

(Lebesgue)

. dense.

coincide w i t h

of its interior.

if we make other,

Hartogs

has planar

in 9.6 was nowhere must

for

This

A(X)

It is if

is not the case,

strong assumptions

on

X .

X even

Before

we give an example we need

Lemma 9.7. usual

(Urysohn

Cantor ternary

[85]; Denjoy

[22]).

Let

set on the unit interval.

K

be the Then there

-71-

exists a nonzero function off

K x K

Proof.

such that

Let

continuous on

Q

and analytic

is a planar Cantor set and can

(without reference to

familiar manner.

S2

g' (~) ~ 0 .

Q = K x K .

be constructed

g

K) in the following 0 _< x,y _< i

Divide the unit square

into

nine equal squares and delete the union of the open squares containing points 1/3 < y < 2/3

.

(x,y)

such that

This process

1/3 < x <

Then

Q =

planar measure of

Qn

by

4n

closed squares of

(4/9) n ; call the union of these squares ~ Qn n=l

is a totally disconnected perfect

zero.

Zn, k

or

is iterated in the usual way.

After the nth step, we are left with total area

2/3

Qn " set of

Denote the centers of the components

(k = 1,2,...,4n),

and let

~n gn(Z)

= 4 -n~

(Z-Zn, k )-I

k=l

It can be shown that that

g

lim gn(Z) ~ g(z)

is continuous

on

off any neighborhood of

S2 .

:

Izl

by the residue theorem.

The convergence

Q , so that

g(z)dz g'

exists for all z

=

g

is uniform

is analytic off Q

0 ,

and

and

-72-

9.7 can also be proved using the more general techniques

of Arens IT].

Example 9.8. follows.

Let

X

be the compact set constructed as

Delete from the closure of the disc

a sequence of open discs

An

(i)

An c D \ Q

for all

(2)

An n Am = ~ • unless

(3)

I rn < " ;

with radii

rn

D = [Izl< 2} such that

n ;

m=n;

n=l

(4)

{An]

accumulates at every point of

Q

and at

no other points.

Then (a)

x=x

° .

(b)

~(~X)

(c)

~X \ (peak points of

(d)

R(X) ~ A(X) .

=

c(~x)

.

R(X))

has measure

It is clear from the construction that (a) holds.

0 (dx dy) .

Since

-Z3@0

Bx = BD U ( U

BAn) U Q haS planar measure

0 , (b) follows

n--1

from the theorem of Hartogs and Rosenthal quoted above 13.6); also,

it is clear that every point of

is a peak point for

R(X)

so that

R(X)) c Q , a set of measure R(X) ~ A(X) dz

on

.

Let

BD , -dz

W on

0 .

is a finite measure on

X .

f(z)

d

t) ~A n n=l

(BX \ peak points of It remains to show that

be the measure on U BA n , n=l

BD U

(see

and

0

BX on

defined as Q .

By (3),

As in 9.6 ,

(z)

- 0

X if

f E R(X) , since by Cauchy's theorem

functions with poles off 9.7.

X .

Let

g

W

kills rational

be the function of

Then 00

IX f(z) du(z) =

Izl~

g(z)dz - I ~BA n n=l

g(z)dz

each term in the infinite sum being equal to theorem.

Thus

g ~ R(X)

.

Since

Note that 9.8 shows that

R(X)

0

--

,

by Cauchy's

g 6 A(X) , we have R(X) ~ A(X). and

A(X)

can have the

same peak points except for a set of measure zero and yet fail to coincide. 9.8 to insure that

It is easy to modify the construction of R(X)

and

A(X)

fail to have the same

-74-

peak points.

Indeed, choose a point

point for the algebra on

S2

A(S2;Q)

and analytic off

Q .

q 6 Q

of all functions continuous Such a point exists by 2.3

and the maximum modulus principle.

(Actually, in order for

2.3 to be applicable it is necessary that points.

that is a peak

A(S2; Q)

separate

To verify this condition we use an idea of Wermer

and choose

Zl, z 2 6 ¢ \ Q

the functions

g(z),

separate points on

such that

g(zl) / g(z2)

[g(z)-g(zl)]/(Z-Zl) $2).

.

Then

, and [g(z)-g(z2)]/(z-z 2)

Suppose we remove the

An

in such

a w~y that

co

Z n=l

where

p(q, An)

r n

p(q, an) < ~ ,

is the distance from

cannot be a peak point for

(£ni) -I (z-q) -I dz is a finite

on

A(X)

and

to

An .

Then

q

since the measure given by

-(2ni) -I (z-q) -I dz

(complex) measure that represents

(cf. the proof of 8.1). for

r

R(X)

q

But clearly

q

q

on

on

U An n=l

R(X)

is a peak point

.

In view of 9.8, it is interesting that the following result is valid.

Theorem 9.9.

Proof.

If

If

R(X) -- A(X)

then

R(SX) = C(3X) .

R(SX) ~ C(SX) , then by 2.4 there is a set

E ~ 3X

-75of positive measure for

R(SX) ; without

closed.

such that no point of

E

is a peak point

loss of generality we may assume

E

is

Let

g(z) =

I

dxdy ~-z

& = x + iy .

E

g

is quite clearly analytic on

continuous on integrable

S2 \ E .

A = A(S2;E)

function of

of all functions

S 2 \ E ; since for

A.

points

A(X)

.

at any point of

Since

Finally,

E).

S2

As above,

A c A(X) ,

E

because

and analytic E

contains

contains peak

But since no function in

E , no function in

g ~ 0

Consider the algebra

continuous on

g E A , A ~ C .

peak points

Thus

is

function and a bounded function of compact support

g' (~) = lim zg(z) = -~ dxdy ~ 0 . z-~ E

for

g

S 2 , since it is the convolution of a locally

(the characteristic

on

Moreover,

R(X)

R(~X)

peaks

can peak on

E .

R(X) % A(X) , as required. Our final

(counter-)

is due to Dolzhenko

example is especially curious;

it

[28], whose construction was slightly

more involved.

Example measure. J ~ 8D

9.10. Let

Let D

J

be a Jordan arc of positive planar

be an open disc such that

consists of a single point

sequence of open discs

An

p .

with radii

J c D

and

Delete from rn

such that

D

a

-76(i)

~n c D

for all

n ;

(2)

An N ~m = ~

(3)

~ n fl J = [Pn ] " a singleton ;

unless

n = m ;

CO

~rn<~; n=l

(5)

[A n ]

accumulates at each point of

J

and at no

other points.

Clearly,

X° = X

and, as before,

R(X)

A(X)

.

(Here we

use the fact that the function

g (z ) =

is in

A(X)

l j C dxdy -z

but not in

in this example is that

C=x+iy

R(X)). X°

What is of special interest

is simply connected.

the connectivity of the interior of

X

Thus,

has little to do

with the possibility of approximation. The examples of this section have indicated the difficulty and complexity of studying rational approximation on arbitrary plane compacta.

We now turn our attention to some ideas

necessary to Vitushkin's ingenious solution of this problem.

-7710.

AC

CAPACITY

In studying

the peak points

make use of the analytic of its thinness. duce another

Definition Then

C(S,m)

To study

Let

S c C

f ~ c(s 2) ;

(2)

f(-) = 0 ;

(3)

f

(~)

Ilfnl._<

is analytic

10.2.

m

AC

consideration

and f

to intro-

m .

m > 0 . for w h i c h

off some compact subset of

differs

S ;

Jf'(~)I

from that of the analytic

not only boundedness

but also continuity;

analyticity

on all of

Proposition

10.3.

This

capacity

be arbitrary

a(s) = sup

in that we require

Proof.

of a set as a measure

.

This d e f i n i t i o n

under

we were led to

it is necessary

f~c(s,1)

y

y

A(X) the

R(X)

is the set of all functions

(i)

Definition

capacity

such measure,

i0.i.

of

S 2 \ S (not just

~(S) _< y(S) .

is clear

.

capacity

of the functions

further, Q(S))

.

we demand

-78Even in the simplest cases the analytic set and its if

AC

K = [O, 1]

capacity may fail to coincide. then

y(K) = 1/#

by 3.9.

theorem)

S2

is analytic

and hence is constant.

important

on all of

For instance,

On the other hand, S 2 \ [0, i] and

~(K) = 0 , since any function analytic on continuous on

capacity of a

S2 (Morera' s

There is, however,

case in which the two capacities

one

do give the same

result.

Proposition

Proof.

10.4.

Let

U

be an open set.

Then

capacity was introduced by Dolzhenko

[28] and is a

more natural tool for studying continuous analytic

of

(ordinary) analytic

9.8, 9.9,

~(Q),

.

Exercise.

AC

than

e(U) = ¥(U)

analytic indeed,

AC

For instance,

the examples

and 9.10 were based on the fact that the quantities

~(E), and ~(J)

however,

capacity.

functions

did not vanish.

Generally speaking,

capacity is considerably less tractable than

capacity.

Thus,

this would imply

it is not true that ~(K) = inf ~(U); UoK ~(K) = y(K)

for all compact

K

via 3.10 and 10.4. Although we cannot enter into a detailed discussion of the similarities

and differences between

~

and

y- - a

-79careful

rereading

of section 3 will orient

this d i r e c t i o n - -we analogues

boundary

Proposition

10.5.

K~

with

behavior.

Let

K c ~

f ~ C(K,1).

be compact,

z ~ K

I f(z) I < c~(K)/~ (z, K)

where

in

out that the following

of 3.15 and 5.7 remain valid for functions

continuous

Then for

should point

the reader

P (z, K)

[IZ-Zol

is the distance < R]

f(z)

from

z

,

to

K .

If

and

aI = - - +

a2

Z-Zo (Z_Zo)2

...

then

i anl <

Proposition

10.6.

be analytic

on

ec~(K)R n-I n

Let

A \ K .

I~

K c A

n = 2,3, . . . .

be compact

and let

Then

f(.~)d~ I < c~(K)llfll ~ , F

f ~ c(~)

-80-

where

c

Remark.

is a universal constant.

The proof of 10.5 is a simple exercise,

while the

argument of section 5 (with appropriate modifications)

can

be used to establish 10.6. We now turn to a slightly technical result which will be useful to us in the sequel.

(10.8)

First, we need

the following auxiliary

Lemma lO. 7. 0 _~ cn, k _~ 1

Let

p

for

be a positive k = 1,2,...,rip

rl,k

Proof.

a = 1,2, . . . .

Then

n, k

Exercise.

Proposition 10.8.

Let

S c

be given and suppose

is a collection of subsets of radius

integer and suppose

~(S)

meets at most

S p

a(sj) < 4oop j=l

Moreover,

if

fj E C(Sj,I)

then

~Sj]

such that each disc of of the sets

CSj] .

Then

-81GO

max ~ Ifj(z)I < 200p.

j=l

Proof.

Obviously, we may assume

renumber the sets

Sj

a double sequence

{Sn, k]

(n+l) m(S) ; to

Sn, k .

moreover,

here

~(S) > 0 .

using double indices such that

P(Zo, Sn, k)

By hypothesis,

lfn,k(Zo)I _< ~(Sn,k) for

n > I . Estimating

to obtain

p

sets

for a fixed value of Then by 10.5

Ifo, k(Zo )I

Z lfjCZo)i --< p + Z

by

I

and applying

(Sn,k)

n, k ~0

m

Since

=

zo

was arbitrary,

j=l @D

max~

j=l

n .

/ p(Zo, Sn, k) __< ~(Sn, k) / na(S)

oo

where

So, k ;

~(S) , there are at

10.7 , we have

j =i

zo

{n~(S) < JZ-Zol _( (n+l)~(S)}

Sn, k

fn, k E C(Sn, k, 1 ) .

and

is the distance from

can be covered by 20n discs of radius

Now let

n,k

zo

n~(S) _< P(Zo, Sn, k) <

there are at most

since the annulus

most 20np distinct sets

Fix

Ifj(z)I _< p + 9 p ~

•

-82-

~j £ C(Sj,I)

Now take let

~ :~

~j

.

such that

Then

: (I/2)~(Sj)

~(~)

II~II. -< P + 9 PJ~

and

so that

@@

j=l whence

m ~ 2p + 18/pm

first inequality.

.

Then

m < 400p , which is the

It follows that

O0

= x ~ l~j (z)W - 0 .

Then if

f(z) =

S c C

is a bounded set for which

f £ C(S,m)

aI

a2

+

Z-Zo

and

z° E C

we have

+

(Z_Zo)2

"

"

Set

~(S, zo, f ) = ~

a2

•

Clearly,

~(S, Zo, f ) =

i

1

~

f(C)(¢_Zo)d¢ '

-83where

o

about

is a rectifiable

each point of

S

contour having w i n d i n g

(for instance,

a large

number i

circle).

Set

S(s,z) -- sup l~(s,z,f)l •

where the sup is taken over all

f E C(S,I)

.

Finally,

let

B(S) = inf B(S,z) Z

It is easy to see that z

on

¢ ; moreover,

B(S,z)-~ 0(S)

as

a simple

z~=

such that

6(S,z)

.

B(S,O(S))

the interested

of thought

on his own.

computation

Hence,

We shall not develop fashion;

is a continuous

some point

.

the properties

reader is invited However,

shows that

there exists

= B(S)

of

G

Proof.

Let

10.9.

with

aI

,.-o (s)

this line

there are two basic observa-

~(s) _< ~(s)

~ E C(S,I)

in a systematic

to pursue

tions that should be made.

Proposition

function of

a2

(,.-o (s~) 2

....,

-84-

where

fall --> ~(S) - a .

Then

~(s) - ~(s,0(s)) _> IB(s,o(s),~2)l

Let

e -- 0

to obtain the desired result.

Proposition

i0.i0.

I~l _< ~(S)

and

such that

g' (-) = ~

Proof.

> (~(s)-~) 2 =(s)

-

Let

~,B be complex numbers

IBI <_ B(S)

It follows

.

and

such that

Then there exists

6(S,O(S),g)

from the integral

g £ C(S,20)

= B •

formula for

B(S,z,f)

that

B(S,z,f)

Take

~ £ C(S,2)

= B(S,t,f)

such that

z ° -- o(s) + B(s,o(s),~)

.

Finally,

fl £ C(S,2)

take

f2 = fl + ~

~ "

[t-z]

~' (~) = e(S)

.

and let

Then

~(S, Zo, ~) = ~(s,o(s),~)

Next choose

+ ~

+ ~

such that

so that Then since

[o(s)-z o] = o .

B(S, zo, f I) = B(S)

f~(~) + ~ ~' (~) = 0 f~(~) = 0

we have

and set

.

-85G(S~O(S),f2) = B(S, zo, f2) = B(S, zo, fl) + s B(S, Zo, ~)

= ~(S, Zo, f I) = ~(s)

Also, since

fl E C(S,2) , I~I i 2 , so that

Recalling that

B(S,O(S),~)

lO(S)-Zol _< 2~(S)

f

.

-- ~

.

= Zo-O(S )

f2 £ C(S,6)

.

, we have

Then

• + ~(s~(s)

f2 ~ c(s,2o)

is the function we seek . It is easy to see that the constant 20 of i0.i0 can be improved considerably; of the norms of where

6 > 0

~

and

indeed, by a more careful choice fl

is arbitrary.

we can insure that

f £ C(S,5+6),

It would be interesting to know

if and by how much the constant 5 can be improved.

-~6ii.

A SCHEME FOR APPROXIMATION

This section deals with constructions and estimates which will enable us to prove Vitushkin's theorem. contains

It also

(11.8) a constructive proof, due to Garnett,

of

Bishop's celebrated localization theorem. Let

K(Zo, 6 ) = {IZ-Zol ~ 6] , K°(z,6) = {IZ-Zol < 6] .

If

f 6 C($2),

by

sup wf(6) = ix_yi< 6 If(x)-f(y)I

write ,,,,

CV

its modulus of continuity

for

S2 \ V .

is given by

Let

n

C~

there exists a

if

Finally,

~--~ = ~1 (u~^

Proposition ii.I.

.

6

@ n

wf(6)

is given

v c @ ,

we shall

recall that the operator

+

&~) i u

0

.

,

where

For each positive integer

partition of unity

satisfies

(i)

o _< ,k,n(Z) _< 1 , ~k,n ~ C~(s2) OO

(2)

(3)

~ ~k,n - i . k=l

9k, n(Z) = 0

off

z = x + iy

K(Zk, n, 6n) .

{ek, n ]

which

-87(~)

No point discs

~.

Take

~(z) = 0

-

~ E CR(S2 )

for

II ~" If®-< l0 8~ ~ o

K(Zk, n,6n ) •

II ~'Pk'nll® < 20/6n

(5)

Proof.

is contained in more than 25 of the

z

such that

0 < ~(z) i , ~I ~ ( x + i y ) d x d y = I , and Divide the plane into mnit squares

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

having m u t u a l l y disjoint interiors

and set

~(¢-z)

dxdy

(¢ = x+iy)

.

% The functions

~k

clearly satisfy

(1) , and

(C-z)dxdy = i , k=l

k=l

so that

(2) holds.

at least 1 from

Qk

Qk

supported on a disc of

Qk "

25 of

the

Moreover, then

if

lies at a distance of

~k(Z) = 0 ~ thus each

K(Zk,2 ) , where

It follows that no point K(Zk,2)

z

.

zk z

~k

is

is the centroid

belongs to more than

-88-

Now fix

n

and set

~k,n(Z) = 9k(2Z/6n)

is supported on a disc

K(Zk, n, 6n)

to more than 25 of the

K(Zk, n, 6n) .

Ff~lly,

the

~k,n

satisfy

.

Then z

and no point

~k,n belongs

II

Clearly ,

li® -< lo.

(I) and (2) because the

~k

function on

is a

do .

This completes the proof. If

f

is a continuous

continuously differentiable

S2

and

g

function of compact support we

shall write

fg(z) = 1 I~ ~

~g

~-z

dxdy

aT f(C) aT ~-z dxdy

= f(z)g(z) + W

(We have used Green' s formula for second equality). f

is analytic and

ng

=

S2 \Sg .

Let

af

Sg

g

(~ = x+iy)

in passing to the

be the set of points at which

the closed support of

g .

Let

Then we have

Proposition ll.2.

Let

that the diameter of

f Sg

(1)

f g 6 C(S 2) , and

(2)

fg

and

g

be as above and suppose

does not exceed

is analytic on

fg(~)

= 0 .

af U n g

6 .

Then

2

rj.

-89-

II fgll. -< 28~f(6)I1

(3) Proof.

(I)

fg 6 C(S 2)

~g ~ 11oo

since fg - gf

is the convolution of a

bounded function of compact support and a locally integrable function.

Moreover, it is clear that

fg(=) = 0 .

(2) can

be established by a straightforward calculation, which we leave to the reader.

Ifg(Zo)l

=

11 fg 11.'

For (3), take

zo

such that

Then

f(¢)-f(zo) If rIo :

O

-< ~ ~f(8)rf ~

3g dxdy I 3Z Sg r~-Zor

Since

~ dxdy

~

= 2~6 ,

we are done. Our next order of business is a coefficient estimate (ll.6) .

The proof, which is quite involved, will be

presented in a series of lemmas.

Lemma 11.3.

Let

f E A(X)

and let

differentiable function supported on

be a continuously K(Zo, 6) .

Then

-90-

Ii ~

f(~)3~0 dxdy, < 46 • w f ( 2 6 ) " l l ~ I I ~ "

3~

Proof.

We may assume that

IIf~ II®

-

on

f E C(S 2) .

< 46 • ef(26) • II ~3'3~ I I ~

X ° U CK(Zo, 6 )

c~(CxO N K(Zo, 6))-

By (3) of n . 2

Moreover,

•

f

,

is analytic

so that, by the definition of

a , we have

I f~' (~)! _< $~ • ~0f(26) • 11 "~ II~ • ~( cx° n K(Zo,6))

.

But clearly

I

-~ dxdy .

We are done .

Lemma 11.4.

Let

f

and

~p be as in 11.3.

<

~ Proof. so that

Let

g(C) = (~-Zo)~(~) •

IIs-~_II~< 6 I~_II~ •

Then

Then

~g = (C-Zo) ~.~

Applying i1.3 with

~ replaced

oC

by

g , we obtain the desired result. Now let the compact set

X

be fixed.

Suppose that there

-91-

exist constants all

(z(CXO N K(z,6)) < m(z(CX N K(z, r6))

[~k,n ]

be the sequence of partitions

structed in Ii.i and set

Lemma 11.5. all

such that for all

z

and

6 > 0

(*)

Let

m > 1 , r > 1

Suppose

6 > 0 , and let

, 8~

where

mI

Proof.

X

J

of unity con-

Xk, n = CX O K(Zk, n, r6n)

satisfies

f £ A(X)

(*)

.

for all

.

We have

z

and

Then

(C-O(Xk, n)ldxdy I _< ml" ~f(26n)- ~(Xk, n)" 8(Xk, n) •

is a constant depending

For convenience

6 = 6n

.

only on

m

and

of notation we shall write

z o = Zk, n , ~ = O(X k,n)

.

Choose

~

r .

~ = ~k,n "

so that

_< s(xk, n) _< 2s

if

8(Xk, n) < a

= 6

if

~(x~, n) _> 6

(i)

and cover

K(Zo, 8)

by (a finite number of) discs

in such a way that each disc

K(z,B)

of radius

K(ti, ~) B

inter-

-92-

sects at most 25r 2 of the discs K(Zo, r6) a(Xk, n)

for all _<

i .

(2+r)~ .

By

and

K(tir6)

(1)

and

10.8

K(ti, r6)

we have

It follows easily (cf. the proof

of 10.8) that each disc of radius

a(Xk, n) meets at most

of the discs

K(t i rG) , where

is a constant depending

only on

r .

Choose

K(ti B)

such that

0 _< gi < i , Z gi (z) = I i

hood of

K(z o,8) ,

and

i

Z ~1 ~ i

Since

C~

~

p

functions

gi

P

supported on

II ~g---~I[~ < 20/6 .

on a neighborThen

(~-~) dxdy =

f(~) ~8

) (ti-(~)dxdy (~gl)(~-ti)dxdy + ~ 1 ~ f ( ~ )~(~gi ~ i

G ~ 6 , we have by ll.3

(3) < 16o. wf(2 ).

Similarly, by ll.# we have

(cx°n K(ti, B)) .

-93-

(4) Iwllfl(C)._~_(~gi) (C-ti)dxdYIs~

-< 1606. wf(26)- ~(CX° N K(ti, G)).

It follows from (2), (3), and (4) that

1606- ~f(28) • Z m(CX° A K(ti'6)) i

+

160wf(28) ~ Iti-~I a(CX°AK(ti,8)) . l Since each disc of radius of the discs and

m(Xk,n) meets at most

K(tl, rS) c K(zo, r6 ) ,

it follows from

p (*)

10.8 that

Z ~(CX° N K(ti,G)) _< m Z m(CX N K(ti,r6)) i

i < m " #00p • ~(Xk

--

,n

)

°

Accordingly, by (i) ,

(6) 160~" wf(2~) ~ a(CX°nKCtl, 6)) _< CI. mp-6- wf(26)~(Xk, n) i _< C1 . mp. w f ( 2 B ) c z ( ~ , n ) ~ ( ~ , n ) ,

-9#which is the bound required for the first term on the r.h.s. of (5).

(Here

C1

is an absolute constant)

.

Estimating the second sum in (5) is more complicated. For

ti ~ ~,

(7)

choose

~i 6 C(CX 0 K(ti,rB),l)

such that

1 Iti -~I ~' i (~) : ~ ti-~ • m(CX N K(ti, r8))

and let

~ : ~ Wi "

By 10.8

II ¥ If.

q = 8K(zo'r6) ' qi = 3K(ti'rS)

"

Let

_< 2oop .

Then since

~(~,n) -

S(Xk,~ ) >_ IBCXk,n,~,~/2OOP)l , we have (8) ~(xk,n) _> [2~. 2oop. ~(xk,n)]-ll~ ~(z)(z-~)~zl > [4oo~ • p -

~(Xk, n)] -I

I ~ (ti-~)I ~i (z)dz i

~i

+ ~ ~ ~i(z)(z-ti)dzl i ~i

By (7) ,

(9) ~ (ti-(~)~ ~i(z)dz = [½ Iti-~l(~(CX N i while 10.5 gives

~i

i

K(ti, r6)) ,

I

I^

M

q

v

v

r~

01:1

0

el-

i-~.

:D

I

e

rO

O~ 0

IA

Q

v

v

PO

i,

L~O

IA

ol ct-

c~

o 0

0

cf

~

t.~

I.~°

0

('1

otto !

ro

i-~ O~ 0 v

~0

~u

ct-

0 r~

~e

I

Q

I^

q

i-u !

v

0

~o

0

0

Co

0

v

ct-

('1

+

0 0

I^

v

c+

23

Q

!

I-1

g

c+

0

~0

Co

0

0

0

H

Q

0

Q

P

13)

(1)

ro

I^

0~

I c~ i.J.

bl

v

! ~0 Ln !

-96< CI. mp. wf(26)~(~,n)6(~,n ) + C3.mpr.wf(2G)a(~,n)B(~, n) .

Taking

m I = mp(C I + rC3) and noting that (i) implies

wf(26) ~ wf(26) , we obtain the desired result.

Proposition 11.6.

Let

f , X , and

Xk, n

be as in 11.5

and let

g(z) = f(z) ~k, nCz) + W Co

s

s=l

Proof. etc.

8~k, n i ~-z dxdy fCc)~

~

a

=E Then

i

(z - O(Xk, n))s

la21 _< m I wf(26n)e(~,n)8(~,n ) As before, we shall write Then for

~k, n

"

Zo

c = ~KCZo,26)..

3~

= W ~I f(~) 3~ i =~ ~

=

i

dxdy)(z-~)dz

~ ~-z dz) dxdy

~q~ (Q-G)dxdy .

f(C)~

= Zk, n "

-97By 11.5, we are done.

Finally, we need the following result.

Lemma 11.7. open set.

Let

X

be a compact set and

Suppose that

(a neighborhood of)

f 6 R(X)

CU .

Them

and

U

f

a bounded

is analytic on

f E R(X O CU) .

We shall obtain ll.7 as a corollary to a significantly more general result that is of considerable interest in itself, R(X)

namely, the fact that a function "locally in"

actually belongs to

Theorem ll. 8.

(Bi shop ).

Suppose that for each hood

K z = K(z, 6z)

R(X) .

More precisely, we have

X c C

Let

be compact,

f E C(S 2) .

there exists a closed neighbor-

z E X

such that

fIxnK

E R(X n Kz) .

Then

Z

f

R(x) .

Proof (Garnett).

By compactness, we can take

Kz I

, .-.3

K

zn

n

such that U Kz o X .

Let

Kj = Kzj

c'(s e)

Choose

@

j=l j (j = 1,2,...,n) Kj

and

0 < ~j(z) < 1 , ~j : 0

such that

~(z) = ~

~j(z) = 1

for

z E V

, where

off V

n

neighborhood of

X

contained in

fj(z) = f(z)~j(z) + ~

l ~

U Kj . j=l

~j

f(~)~

l

Then if

c-Z dxdy

is a c l o s e d

-98n

we have

~ fj -- f + ~

~(z) = S

where

~

~

dxdy .

J=l

Now let

C =

e

> 0

be given, and set

max sup W l~j
I__-LJ&[SCT ~ E T

Choose rational functions X

such that

dxdy .

hj , i ~ j ! n , with poles off

II f-hj[[XOK.

< s/8nC

•

Then

o n a suitable

,3

closed neighborhood (and

hj

Uj

is analytic o n

such a way that

of

X 0 Kj

Uj) .

[[f-hj[Is 2 < e/4nC

Modify

is analytic on

neighborhood of

X .

off

8~j

1

ch~

C-z

Moreover ,

g = ~ gj , we have

Uj

dxdy

Uj U CKj , which is a

-< II "j-flYs2 + I~j-f&j Setting

hj

and set

gjCz) = hjCz)~jCz) +~1 ~ hi(C) By ll.2 , gj

IIf-halluj <

we have

• e < ,/2~.

in

~/4nc

-99n

llg- fjllv_<

llgj-fjllv <

.

j=l

N o w since

B~ = 0

on

V , ~(z)

is analytic on

Therefore,

by 9.1, we can find a rational function

such that

II~-rll X < s/4 .

Moreover,

analytic in a neighborhood of find a n o t h e r rational function llg-qll x

< s/4 .

The function by m o d i f y i n g if

V

Then ~

see 15.9.

r E R(X) gj

is

is also , so we can

q E R(X)

satisfying

llf-(r+q)ll < E , as required.

can be eliminated in the argument above

f (off X) to have compact support.

contains the support of

= 0 .

X , g

since each

V .

f

For then,

it is easy to see that

For another p r o o f of ll.8 and a comment on its origin

-i00-

12.

VITUSHKIN' S THEOREM

We n o w h a v e

at our disposal the tools necessary to prove

Vitushkin's theorem.

Theorem 32.1.

(Vitushkin [94]).

Let

X c @

be compact.

Then the following are equivalent

(i)

R(X) = A(X) .

(2)

=(cx n G) = =(cx ° n a)

for every bounded open set

G .

(3)

~(CX A D) = ~(CX ° A D)

(4)

For each

z 6 CX °

for every open disc

there exists

alCX° A K(z#6) I ~ CX O K(z,r6 )

(5)

There exists z

and

all

r > l, m > I

r ~ 1

D .

such that

< ~

such that for every

6

u(CX ° O K(z,6)) < mu(CX O K(z, r6)).

Proof.

We shall prove the following chain of implications:

(5) => ( 1 ) = >

(2)=>

(5) => (I) .

(3)=>

(~)=>

Let f E A ( X ) a n d

(5) • extend

f

to

~

as a function of

-i01-

compact support.

Fix

n

and let

unity constructed

in ii.I.

be the partition of

For convenience of notation we

shall suppress the subscript so that

[~k,n ]

n

in the argument that follows,

6 = 6 n , ~k = ~k,n ' Zk = Zk, n " etc.

fk(z) = f(z)~k(z)+-~ ~ f(c) ~ is analytic on

X ° U CK(Zk, 6)

X k = CX 0 K(Zk, r6 ) .

Put

and

~-z

Z fk - f

Ok = 0(Xk)

Then

dxdy Let

"

and write

O0

fk(z) =s~=I (z-0k)s ass "

By (3) of 11.2 and (5) of ii.i we have 80 wf(26)

lalkl

~,,f(28). 20/~s =

so that

= If~:(®)l

by (5).

l l~'klL-< 2.28.

_< 80wf(26)or,(CX ° n K(Zk, 6))_< 80mwf(26)~(X.k),

Also, by ii.6,

la2kl _~ ml. wf(26)~(Xk)~(X k) Let gk

m 2 = 20 max(80m, ml) . E C(Xk, m2~f(28))

According to 10.10 there exists

such that

~(~)

= f~(~)

and

-102-

~(~, o~, gk)= ~(~, ok, fk)"

of oou~se, I1%:11®<_.½,.,,f(26)

so that

II~-~ll®

<_ 2½~f(26) ;

henc e

Iz-zkfS (r6)3 Ifk(z)-gk(z)f <_ 2½~f(28) , Therefore,

IZ-Zkl = r6 .

by the maximum modulus principle,

Ifk(z)-gk(z)I _< 2m2r363wf(26)/IZ-Zkl 3

IZ-Zkl >_

r6

.

It follows that

83 Ifk(z)-gk(z)l < m3wf(26 ) min

where

m 3 = 2mer3 Now fix

Each disc

K(Zk, 6)

one and at most two of the circles

C S

•

Let

Ns

CS

intersects at least

: [Iz-z*r -- s6]

be the number of discs which touch

Then by (4) of ll.1, N s- ~62 < 25. area of

{(s-l)6 < Iz-z*I <_ (s+l)6}, s > 2

Iz-zkJ 3 }

•

z* .

(s = 1,2, .... ).

{ l,

and

so that

K(z k,6) N C s ~ ~, we have

N s < lOOs .

Also , if

fz*-zkl > (s-l)6 ,

-Io3whence

Ifk(z.)_gk(Z.)l _< m3~f(26) " 1 (s-l) 3 Therefore, @O

OO

l fk(z*)-gk(z*)l

-< m3~'f(26)N 1 + m3~f(26 ) Z

k=l

Ns (s-l)3

s=2

_<

ms.f(26) O0

w~e,e

m4

,

i~i~>

~00~ 3 (~ + Z

~s

S=2

IIf - ~ gklIs2 k=l

= lI~ fk - ~] gklIs2 = IIZ k=l

k=l

(fk-gk)IIS2

k=l

O0

_< sup z k=l

rfk(z)-gk(z)l _< m4~f(26) .

!

Let

~

denote the (finite) suan over those indices

k

for

!

which II

fk

is nonzero.

Then

(fk-gk)IIX < m4wf(26)

Z fk = f

Also, each

and we have gk

is analytic !

on a neighborhood is analytic

of

CX k = X U CK(z k,rS) ; thus

in a neighborhood

of

X .

g = Z gk

By 9.1 , g 6 R(X)

.

-i0~--

Since as

ms

n-*

~epends

only on

ml, which is fixed,

and

m , we are done.

(i) => be given.

(2) .

Suppose

R(X) = A(X)

Choose a closed set

f £ C(K,I)

such that

f E A(X) , so f 6 R(X) a neighborhood

of

, and let

K c CX ° A G

f' (~) > ~(CX ° A G) - E . by (i) .

CG ; hence

Also,

f

Clearly ,

is analytic

f 6 R(X lJ CG)

modifications

on

fn E C(CX A G, I) and

uniformly

X U CG ; then

on

If v (-)J < ~(CX N G) Letting

(after suitable

f' (co)-* f' (co) n

we have

~

inequality (2) =>

we get

~

~

~(CX ° A G) _< ~(CX A G) .

(3)

The opposite

is trivial. Fix

r > 1 .

Then

=i

K°(z, 6-*0

we obtain

lim 6-,O

f o r ~

Since

is obvious.

(3) => ($) •

Letting

fn -* f

m(CX ° A G)-e < f' (-) < a(CX N G)

~

e-* 0

r > i .

on

by ii.?.

fn

CX A G)

such that

e > 0

and a function

Choose rational functions

n

6 = 6n-* 0

~ICX° N K ( z , 6 CX N K(z,r611

< 1

.

"

-105(4) => (5) •

Suppose (5) does not hold.

Choose

z1

and

61

Then

e(CX ° O K°(Zl,261) ) ~ e(CX n K°(Zl,261)) , so that

in such a way that m(CX ° A K(Zl, 61))>a(CX0 K(Zl,361)) .

A(X n K(Zl,26 I)) J R(X n K(Zl,2~ I)) Set

X 1 = X O K(Zl,261)

the existence of

greater than

z2,62

such that

so that

~(CXl° O K(z 2, 62)) >

K(z2,562) c K(Zl,281 + 82) ,

CX 1 O K(z2,562) 62

(1) => (2).

The implication (5) => (i) guarantees

2~(cx I n K(z2,582)) . Now for otherwise

by the fact that

contains a disc of diameter

2~(CX 1 n K(z2,562) ) > 62 >_

m ( C ~1 O K(z2,62)), a contradiction. Also, K(z2,262) c K°(Zl,261 ). For otherwise,

there exists

x + 2eie62 6 K(z2,562) choice of

K(z2,262) \ K°(Zl,261)

for any real

and

8 ; but for a proper

8 , x + 2ei862 ~ K(Zl,261 + 62) , contradicting

K(z2,562) c K(Zl,261 + 62) . so that

x £

Thus, Xl° O K(z2,62) = X ° O K(z 2,62),

CXl° N K(z2,62) = CX ° N K(z2,62) .

We have, there-

fore,

e(CX ° O K(z2,62)) > 2~(CX O K(z2,562) ) .

since

562 _< 261 + 62 , 62 _< 61/2 .

Finally,

Proceeding in this manner we obtain sequences {rn}, and

[6n]

6n+ 1 _< 6n/2 , and

such that

[Zn} ,

r n -+- , K(Zn+]f26n+l) c K°(Zn,26n) ,

a(CX O N K(Zn, 6n) ) > n~(CX O K(zn, rn6n)) .

-106-

Clearly, IZn-Zol and

6 n -~ 0 . < 26 n

Let

we have

zo = N K ( Z n , 2 6 n )

.

Since

a(CX ° O K(Zo, 36n)) _> a(CX ° O K(Zn, 6n))

e(CX O K(z O, (rn-2)6n)) _< ~(CX O K(zn, rn6n))

.

a(CX O O K(Zn, 6n))

a(CX°NK(Zo, 36 n) ) > a(CX O K(Z O,(rn-2)6n))

Now

let

r

as

n -- ~

>

i

m(CX A K(zo, ran))

> --

> n . --

~(CX O K(zn, rn6n ))

a n = 36 n .

for large enough

C:(CX0 O K(Zo, an))

so that

--

be fixed and set

we have,

Therefore,

Since

rn ~

n ,

c~(CX0 O K(Zo, Sn)) .... rn_2 ~(CX O K(z O, (--3---)en))

> n ,

(#) fails to hold.

This completes

the proof of 12.1.

The construction

employed in the proof of (5) => (i)

above can be used to prove other approximation For instance,

Theorem 12.2.

theorems.

we have

Let

X c ~

be compact,

f 6 A(X)

.

Then

-i07-

f

can be a p p r o x i m a t e d u n i f o r m l y on

C(S 2) of

w h i c h are analytic

a

(x)

Proof.

on

X°

X

by functions in

and on a neighborhood

.

Exercise.

(See 13.2).

T h e o r e m 12.1 gives necessary and sufficient conditions on

X

that

(the r e s t r i c t i o n to

C(S 2) , analytic

on

X

X ° , belong to

of) every function in R(X)

.

One can pose

a slightly different question and ask what conditions (£ C(S 2)) X

insure that

f

by rational functions.

T h e o r e m 12.3. f E R(X)

Let

X c ~

z

The answer is contained in

be compact and

f 6 C(S 2) .

and all

Proof.

~(6) -* 0

[93] and

as

[95]

6-* 0 .

•

r >_ i

8 > 0

l aK(z, )f( )dCl i (CX n K(z, r6))G(6) where

f

can be a p p r o x i m a t e d u n i f o r m l y on

if and only if there exists a constant

such that for all

on

,

Then

-io8-

13.

APPLICATIONS

OF VITUSHKIN' S THEOREM

As we remarked contains

earlier,

V i t u s h k i ~ s theorem

as a special case virtually

theorem on the possibility compact planar sets.

(12.1)

every other major

of rational approximation

on

It is now time to justify this

remark and to make good

our promise to prove the results

stated in section 9. The following

Lemma 13. i.

Let

lemma is basic.

X c

be compact.

Suppose

for every

z E 8X .

lim a(CX A K(z, 6))/6 > 0

R(x)

= A(X)

Proof.

Then

.

For any

r > i

0 < lira ~ C X A K(z, 8)[ < lira %(CX n K(z~6r)) ~ --~ ~ ( c x ° n K(z,5)) Hence, of

condition

8X .

point of

Proof.

for each point

We are done.

Theorem 13.2.

Then

(4) of 12.1 is satisfied

8X

R(x)

Let

X c C

be compact and suppose that each

lies on the boundary

= A(x)

of a component

of

CX .

.

Suppose the condition of the theorem is satisfied,

-lO9and let

z 6 3X .

CX n K°(z, 6)

10.4

Then,

contains

for sufficiently

small

an arc of diameter

6

By 3.6 and

6/2 .

a(cx n K°(z, 6)) = ¥(cx n K°(z, 8)) h 8/8 .

$

Thus the

condition of 13.1 is satisfied. Theorem 13.2 has as its corollaries results

of (qualitative)

For instance,

Then

CX

(Mergelyan) consists

R(X) = A(X)

Corollar~ suppose

13.4.

Let

Let

Q(X) = S 2 \ X , i.e., A(X)

X

X c ~

be compact and

does not divide the plane.

is uniformly

approximable

on

from 13.3 and a remark in section 9.

When 12.1 is specialized obtain the following

Let

be compact and

(in z).

This follows

Theorem 13.5.

X c C

.

(Mergelyan)

by polynomials

Proof.

theory.

of a finite number of components.

Then every function in X

rational approximation

we have

Theorem 13.3. suppose

the "standard"

(earlier)

X c C

to sets without

interior we

result of Vitushkin

be compact.

~91].

The following

are

equivalent : (I)

R(X) = C(X)

.

(2)

y(G \ X) = ¥(G) , for any bounded open set

G .

-llO-

(3)

Proof. that

y(K°(z, 6) \ x) = ~(K°(z, 6)) = 6

This follows immediately R(X) = C(X)

y(U) = ~(U)

Actually,

sufficient

(3) above.

condition than

condition for

R(X)

XO =

z

and

We need only note

and recall that

R(X) = C(X)

under conditions

For instance,

(2) or (3).

to coincide with

13.1 gives a better

An even w e a k e r sufficient C(X)

is that

62

for almost all

(dxdy) z E X .

We shall not prove this

statement here; a reference is [41]. The theorem of Hartogs and Rosenthal is an easy consequence of 13.5.

Theorem 13.6. and suppose RCx) = CCx)

Proof.

Let

show that K¢

(Hartogs-Rosenthal). X

Let

X c C

has zero Lebesgue planar measure.

be compact Then

.

D6

be an open disc of radius

~(D 8 \ X) = 6

be a closed subset of

6 .

(10.3).

one can prove

even weaker than

from 12.1

implies that

for open sets

for all

Clearly, D6 \ X

6 .

We must

y(D 8 \ X) ~ 6 .

such that

Let

V(Ke) _> n62-n¢

-lll-

(0 < ~ < 02 ) ; here measure)

of

Ke .

V(Ke)

is the area (=Lebesgue planar

By 3.12,

V(Ke) < 1~[y(Ke)]2

_< ~(K~) < y(D 6 \ X) . 6 < y(D 6 \ X) ,

Exercise.

Letting

Thus

.

~ -~ 0 , we obtain

as required.

Prove that

lim zkK°~z'~)\~-) 6

for almost all z E X

implies that

> 0

R(X) = C(X)

.

- 112

14.

-

GEOMETRIC CONDITIONS

Because the situation with which it deals admits great topological complications,

the statement of Vitushkin's

theorem was framed necessarily in terms of the nonituitive, quasi-geometrical

notion of

AC

capacity.

It is natural

to expect that,

in particularly nice cases, the actual

geometry of

will play a significant role in determining

whether

X

R(X) = A(X)

.

This is indicated by the examples

at the end of section 9 as well as by 13.2.

In this section,

we shall discuss what is known in this direction. oO

Let

X

be compact and let

composition of

CX

into

CX =

U Ui i=O

(open) components.

be the deWe call

@o

rI = 3 x \

the inner boundary of if

FI

X .

U 3u i i=O

In [94] , Vitushkin noted that

consists of a finite number of points then

This shows,

in particular,

(section 8) then For instance,

that if

R(X) = A(X)

.

X

R(X) = A(X)

is a set of type (L)

Actually, much more is true.

using the results of sections 5 and 12 it is

not hard to prove

Theorem 14.i.

(Melnikov [64]).

Suppose

FI

is a subset

-i13of an analytic

Proof.

curve.

question.

For

is satisfied,

(I)

appropriate on a circle

6

sufficiently

K = K(z,6)

(1) is local,

conformal map, s .

a

Fj = 8(Tj

N o w choose (cx ° n K)

small.

, where

is small.

we may assume if necessary)

Since

(employing an that

and

C2 .

Let

£I

lies

Tj = Uj

, j = 1,2

O K) . such that

~'(~) =

Clearly,

.

By 12.2 , ~ can be a p p r o x i m a t e d u n i f o r m l y on analytic on

fix

divides the plane into two components

~ 6 C(CX ° N K,I) .

6

Accordingly,

+

TI\E ] , where

TI° = C 1 .

from 10.6 ,

E1

Moreover,

can be taken to satisfy we have,

curve in

it is easy to see that this c o n d i t i o n

(the b o u n d e d component)

lying in

r = 1

so it is enough to prove

and let

and set

(4) of 12.1 with

depending on the analytic

z # £i

and

the condition

(1/2)

m

.

m(CX ° O K(z,6)) _< m ~ ( C X N K(z,6))

z £ £I

z 6 FI

CI

R(X) = A(X)

We shall verify condition

and a u n i f o r m bound

for

Then

T I by functions

is a closed subset of the a p p r o x i m a t i n g

IIfIITl_< 1

CX O K

functions

For such functions

f

-i14-

I~ f(~)d~I = II f(~)d(I

-< ca(E1)

j c(~(CX ~ K)

,

F1

where

c

is a universal constant.

(3)

Hence

l~Fl
Similarly,

(4)

~(~)dcl < c~(CX n K) .

It follows from (2),

(3), and (4) that

~(CX ° N K) = N-II~3K ,(c)d~l

< ca(CX N K) ,

as required.

Exercise.

The preceding proof contains a suppressed compact-

ness argument.

Vitushkin

Supply it .

[95] has improved l#.l.

To state his theorem

we need the following

Definition 14.2.

Let

and suppose that

a

of Jordan curves

al'''"~n

a

be a curve without self-intersections

can be decomposed into a finite number such that for each

i

ci

has

-115-

a well-defined z 6 qi "

tangent

Suppose

arg T(Z, ai)

vector

further that for each

Then we call

- arg x(z',ai) I _< Llz-z'I ~

a

a Liapunov

Vitushkin's

Theorem 14.3.

L

complicated.

Let

FI

E ai .

as

lie on the u n i o n of c o u n t a b l y m a n y

Then

R(X) = A(X)

.

since the proof is long and

What one must establish

5.11 for I~apunov curves. 14.3 follows

z,z'

curve.

We shall not prove 14.3,

is the analogue

of

Once this has been accomplished,

from 12.1 of 9.6 shows that if

The method

R(X) = A(X)

true if the length of

I#.4.

Let Let

then

3X\F I

J c A

3XXF I

~(FI) = 0 . is infinite.

has finite This is no longer Indeed,

Ui, i = 1,2, ....,be a sequence

~i c A \ J

we have

be an arc of positive measure

such that

(i)

the function

extension of 14.1 can now be formulated

Liapunov curves.

(cf. 9.10).

i

~ > 0 :

larg v(z,~i)

Example

at every point

obeys a H61der condition with constant

and exponent

length and

~(z,~i)

for all

i ;

of open sets

-ll6-

(2)

~i 0 ~j : ~

(3)

diameter of

(~)

J

unless

i = j ;

(Ui) > ~ > 0

for some

~

independent

i ;

is the set of"Sdmlt points" of the set CO

U

~u i

i=l

•

GO

Let

X =Z

\

UU i .

By (i) and (4) , P I = J

so that

i--1

a(r z) > o . R(x) = A(X)

However, it follows from (3) and 15.10 that .

(It is obvious from (3) that

not have finite length. each

~U i

~X\F I

does

On the other hand, of course,

may be rectifiable.)

The existence of the phenomenon exhibited by 14.4 was first pointed out to me by Alfred Hallstrom .

-11715.

FUNCTION

ALGEBRA

METHODS

In the preceding sections we met with considerable success in applying constructive methods to problems of rational approximation. however, the

No discussion of this subject,

would be complete without an indication of how

(nonconstructive) methods of functional analysis can

be employed to study rational approximation.

Accordingly,

this section is devoted to the development and explaitation of this approach.

At the same time, we can prove results

of some interest in themselves.

In particular,

a proof of Bishop's localization theorem give a proof,

due to Garnett,

we offer

(15.9)~ we also

of an interesting special

case of 14.1. X

Let

be a compact set in the plane.

and

f ~ C(X)

U

is a complex Baire measure on

f , i.e.,

kills

Suppose

r

if

|

f du = 0 ,

we write

X .

If

~If

.

X If

~I

for every

f

R(X)

(resp. A(X))

for

R(X)

~

an annihilatin 6 measure

In this case we write

Now let

f E C(X)

be fixed.

is a closed subspace of the Banach space

follows

±

R(X)

Since C(X),

it

(cf. section 2) from standard theorems of functional

analysis that measure

annib~ Iates

(resp. A(X)), we say

and call

(resp. A(X)).

(resp. ~ I A(X)). R(X)

R(X)

W

f E R(X)

if and only if

which annihilates

R(X)

.

W I f Thus

for every

R(X) = A(X)

-118-

if and only if they have the same annihilating measures. particular,

R(X) = C(X)

which annihilates

measures Let K .

for

if and only if the only measure

R(X)

Following Bishop R(X)

In

is

0 .

[13] , we shall study annihilating

by examining their "Cauchy transforms."

be a finite complex Baire measure with compact support Set

^~(z) = ~ (~-z )-i d~(C) . A

Then

U

is clearly analytic off

K ; moreover,

change of order of i n t e g r a t i o n shows that with respect to Lebesgue planar measure.

Lemma 15.1.

Let

W

~

an inter-

Is l o c a ~ i n t e g r a b l e

This gives

be a positive measure on

X

and let

N(z) = [ d~(¢) l~-zl Then

N(z) < ~

almost everywhere

(dxdy)

.

A

Lemma 15.2.

~ = 0

Proof [16].

One direction is trivial.

differentiable

almost everywhere if and only if

Let

g

~ = 0 .

be a continuously

f u n c t i o n which vanishes outside a compact

containing the support

K

of

W .

By Green's theorem

set

-i19-

- ~

where

C = x + iy

whole

complex plane

orders

3~ (-z dxdy ,

and the i n t e g r a t i o n

is extended

(or any large compact

of integration,

set).

over the

Interchanging

we obtain

dw(z )] dxdy ~-Z

I ~

=

~g G(~)

dxdy

0

A

since

~ = 0

on

can be a p p r o x i m a t e d

K

g, ~ kills theorem,

almost

everywhere.

every function

Since any function

uniformly in

C(K)

on

.

K

continuous

by such functions

By the Riesz

representation

~ = 0 .

Lemma 15.3. supported

Let

on

X .

X c C

be compact

A necessary

and let

~

and sufficient

be a measure condition

that

A

W I R(X)

Proof. on

= 0

is that

Necessity

~ \ X .

It follows

Then

on

is clear. W

kills

from a p a r t i a l

~ \ X .

For sufficiency, (C-z) -I

fraction

suppose

for each

decomposition

~ = 0

z E ~ \ X . that

W

kills

-120-

any rational function with simple poles, off

X .

Since any element of

by such functions,

R(X)

all of which lie

can be approximated

the lemma follows.

Lemma 15.3 can be used to give a neat proof of the following extension of Runge's theorem

Theorem 15.4.

Suppose

be continuously f

~ R(X)

Proof.

(9.1).

f E A(X)

differentiable

and

f

can be extended to

in a neighborhood of

X .

Then

.

We may suppose

f

to be a continuously differentiable

function of compact support in the plane.

1

=

~f

&

Then

(~ = x+iy) ,

where the integral is taken over the whole plane.

Suppose

A

~a .1_ R ( X )

I

analytic

on

X°

~, = 0 ,

---- 0

on

X .

.~f.. =

Thus

f(z)d~(z) X

\ X

on

by 15.3.

Since

f

is

there~ hence, by continuity,

0 &

~f ~

then

3f •

=

0

everywhere.

= ~

Therefore,

dxdy = 0 ~

The above proof is due to Andrew Browder

(unpublished).

-121-

Using 15.1 and 15.2 we can prove 2.4.

Theorem 15.5. of

X

is a peak point for

Proof [i00]. P

(Bishop [13]).

Let

R(X)

~ I R(X)

the set of peak points of

N(z) Then

N(z) < ~

X

X

off

R(X) .

~ ~ 0

belongs to and

l({zo}) -- 0 X

so is

~ / 0 .

Denote by

Let

]" ,dlul(:3 l -zl dxdy.

Since

A~ = 0

it follows that

~ / 0

on a subset

~(Zo) ~ 0 . r

on

Now almost every point of

P ; hence we can choose

Now if

point

R(X) = C(X) .

and suppose

having positive measure.

N(Z O) < ~

l

=

then

almost everywhere

\ X (15.3) and of

If almost every (dxdy)

From

z° 6 P

N(Zo) < ~

such that it follows that

is a rational function with poles

[r(z)-r(Zo)]/(Z-Zo) ; hence rCz)-rCz O )

~x

z-z o

d~(z) = 0

or

r(z) d~(z) : r(Zo):(Zo) ~X Z-Zo

,

where both sides of the above equation converge absolutely

-122-

by the c h o i c e

of

zo .

f(z)

It f o l l o w s

that

dw(z)

= f(Zo)~(Zo)

g

be a function

Z-Z O

for a l l f £ R(X) peaks

at

zo .

.

Let

R(X)

which

Then

(*)

I ~Z - Z

for e v e r y p o s i t i v e

o n the l.h.s,

as

.

Also

~(z) = ~(zo)

O

integer

integrand k-+~

in

k

.

tends

Since 0

to

~([Zo] ) = 0 ,

almost

the

everywhere

dw

,

I Io IZZo, where

the r.h.s,

Letting

k - * ~ in

theorem,

we get

belongs

to

Ll(dlul)

(*) and a p p l y i n g

b y the c h o i c e

the d o m i n a t e d

a contradiction. An immediate

Hence

~ = 0

corollary

is

,

and

R(X)

= C(X)

zo .

convergence

A

o = ~(Zo)

of

.

-123-

Theorem 15.6.

R(X) = C(X)

is a peak point for Recently, this result observing yields

R(X)

Wilken

if and only if every point of

X

.

[107] proved a nice generalization

(see the bibliographical

notes).

that 15.5 combined with Melnikov's

of

It is worth theorem

(6.1)

the following

Theorem

15.7.

R(X) = C(X)

if and only if for almost

every

x6X QO

2n

n(X)

= .

n=0

where

Yn(X) = y(CX fl [2 -n-I _< Iz-x I _< 2-n])

This improves if

a result of Vitushkin

~ 4 n ¥n(X) = ~

Finally,

for every

.

[89], who proved that

x 6 X

then

R(X) = C(X)

let us note that the theorem of Hartogs and Rosenthal

(13.6) is an immediate

corollary

of 15.5.

In ll.8 we proved the remarkable of rational approximation

result that questions

are "local ."

Using the ideas of

this section we can offer another proof of this fact. however,

First,

we need

Lemma 15.8. K

.

and let

Let ~

~

be a

be a measure

C~

function with compact

supported

on

X .

Then

support

-124-

^

~

^

where

_ 1.~

Proof.

dxdy .

G(~)

Interchanging order of integration and applying

Green's theorem, we have

- -

rr

du t ~

X

-

1 = IX {- ~ ~I K ~

ag ~

i

= ~X ~

=-

hood

clz dxdy] d~(t)

~_~ [%____1+

~ z ] dxdy] d~(t)

[-~(t) + ~(z)] d~(t)

+

Theorem 15.9. (Bishop) f 6 C(X) .

~Z~ dxdy

Let

X c ¢

Suppose each point

be compact and let

x 6 X

has a closed neighbor-

U x (in X) such that the restricted function

belongs to

R(Ux) .

Then

f 6 R(X) .

flu x

-125-

Proof. X

Since

X

is compact we can find

such that the sets

Suppose unity

~ I R(X) {~j]nj=l

~j ~ 0

on

.

Since

~

U. = U form a covering of j xj Choose a (finite

such that

X \ Uj .

~j

in X .

C~) partition of

has compact support and

Let

I

~j = vj~

Xl,...,x n

dxdy

is supported on

(~ = x+iy)

X , it follows from 15.3 that ^

vj

is supported on

Us .

Moreover,

by 15.8 , ~j = ~j

,

J A

so that

~j = 0

off

By 15.3 , ~j I R(Uj)

Uj .

.

Therefore,

n

vj I f

j = 1,2,...,n.

Since

= ~ ~j , we obtain

~ I f •

j=l as required. Bishop never published his proof of 15.9 , the heart of which is lemma 6 of [10].

The proof

given above

is an

elaboration of the one in [34], which is there attributed to Hoffman.

Recently,

Garnett

[34] observed that 15.9 can

be used to prove the following corollary of 12.1,

first ~ a t e d

in [67 ]. T h e o r e m 15. I0. Let of the components R(X) = A(X)

.

X c of

CX

be compact.

Suppose the diameters

are bounded away from

0 .

Then

-126-

Proof.

Let

e > 0

the components neighborhood Clearly Thus,

of

CX .

f

an immediate

we have

R(x)

Mergelyan's

whose

is connected;

f 6 A(X)

15.9 that

For each

U x (in X)

S2 \ U x

if

be a lower bound

for the diameters

x 6 X

diameter hence,

choose a closed is less than

by 13.4,

A(Ux)

theorem

on rational

consequence

(13.4),

was obtained the spirit

of 15.10.

section

while

approximation Note,

in the development

as a special

of the present

Quite

recently,

case of 13.3.

14.

Garnett

His result

generalizes

in 13.2 and 14.1.

Theorem

Let

of all points infinitely set

A(X)

PTOOf.

be a

in

many

First, C~

X

X c ~

however,

is

that in approximation

of section

13 13.#

A proof of 13.# in

of

CX .

of this

of the kind discussed a theorem

be compact.

each neighborhood

components

= R(X)

from

(13.3)

showed that the methods

and is included

15.11.

.

section is in [16].

can be used to prove a t h e o r e m

in section

= R(Ux)

.

the proof of 15.10 we needed to use the polynomial theorem

E .

It follows

RCUx) •

flu x

of

Let

of V i t u s h k i n

Fo

be the set

of w h i c h intersects

Then if

F

o

is a countable

.

suppose

function

Fo

is a singleton,

such that

(a) ~(z)

say 0 .

= 1 , if

Let

Izl _<1/2 ;

-127-

(b)

~(z)

Cpn(Z)

(b')

I zl

= 0 , if

= %o(2 n z)

~n(Z)

.

> 1 ; and

(c)

(a') ~n(z)

Then

!1 ~

= X , if

izl > 2 -n ; and ( c ' )

= 0 , if

Iio= < K

Let

izl < e -n-1 ;

~''-' ti® < K2n

II

-

~z

-

By Green's theorem

i ~ ~n ~n(Z) = - W An ~

where

C = x+iy

and

= {2 -n-I

A

1 ~-z dxdy ,

__< I~i __<2-n]

•

Now

n

suppose

U ± R(X)

.

Set

~n = ~n ~

Then

~n

is supported on

1 ~n TrS[

An

~(C) dxdy .

and by 15.8 we have

=

Let

Xn

~

X i] [Izl _> e-n-l}.

_---

BE 15.3, ~-~n

A

(l-~n)U = 0

CX n .

Applying the other half of 15.3, we obtain

Since

CX n

Suppose

~-Vn I A(X n) , and so

. f 6 A(X) ; we wish to show

U = (W-v n) + ~n

(1)

Hence

and

U ± f •

U-~ n i A(X) , we have

]'X f

du = ~

Xf

d~ n

on

U-~ n 3_ R(X n) •

consists of only finitely many components

R(Xn) = A(Xn) , by 15.10. ~-~n I A(X)

~n = ~n ~ "

Since

-128-

for all

n .

Since

(2)

~X [f-f(O)] d~ n = IX f d~n

for any

f E C(X)

sequence [11 nll]

.

An easy c o m p u t a t i o n shows that the

is bounded.

SX [f-f(O)]

(3) as

~n i R(X) , ~n ~ i ; therefore,

n - * ~ , since

(2), and

f

Therefore,

ll ntl-o

d~n _( I1 f-f(O)liA n is continuous.

It follows

from (i),

(3) that

fdu=0, X as required. Now suppose given.

Let

S

Fo

is countable and let

be the set of points of

is not locally in

R(X)

.

is countable and compact,

Suppose

S ~ ~ .

E

y , a contradiction.

f

be

at w h i c h

f

S

Then since y .

must belong to

Therefore,

15.9, we are done.

Exercise.

A(X)

it has an isolated point

But the argument above shows that locally at

Fo

f

Give a constructive proof of 15.11.

S = ~ .

R(X) By

-129-

There

is an alternate

worth mentioning. if there for w h i c h of

F0 .

exists R(K)

Say

Clearly,

is a closed

studying

N(X)

is

X ,

= A(K)

.

of

of

K

x (in

of 15.11 then shows that

X) subset

N(X)

must

set.

considerable

exploited

, the nucleus

neighborhood

In conclusion, ideas

x 6 N(X)

of 15.11 that

no closed

The argument

be a perfect

formulation

in this rational

should

consult

A more

extended

let us note that more

elaborate

function algebra

than the simple

section have also proved approximation.

[103]

treatment

book Uniform Algebras,

successful

The interested

for an e x p o s i t i o n will appear

techniques in

reader

of these methods. in the forthcoming

by T. W. Gamelin.

-13o-

16.

SOME

OPEN Q U E S T I O N S

To the best of my knowledge,

the following problems

remain open.

16.1.

Let

X

be compact,

set of numbers

t E [0, i]

Conjecture

J r

If

x £ X .

for which

t -I dt = ~ , x

Let

I

denote the

[Iz-x I = t] fl CX ~ ~ ,

is a peak point for

R(X).

I This is true for sets of type is valid in general it sharpens a sufficient

(L).

If the conjecture

(via the theorem of Keldysh)

condition for regularity for the Dirichlet problem

due to Beurling. It is easy to see that the condition in the conjecture is not necessary for

16.2. c

Conjecture.

such that

S1 , S2 •

x

to be a peak point.

There exists an absolute constant

¥(S 1 U S 2) < c[¥(S l) + ¥($2)]

for all sets

(Cf. 5 . 9 and 5.12).

This may well be false, but it would be good to have a proof or a counterexample.

16.3. C

such that

Conjecture.

There exists an absolute

constant

~(S 1 'U S 2) _< c[~(S1) + G(S 2)] for all sets S1, S 2 .

-131-

16.4.

Let

X

x

Conjecture.

be compact , x £ X .

is a peak point for

A(X)

if and only if

CO

2n an

, where

c~n = c~(CX0 rl [2 -n-I _< Iz-xl <_ 2-n])

.

n=O

P. C. Curtis has recently proved the sufficiency of the above c o n d i t i o n 16.5.

[ll4].

Conjecture.

A(X) = R(X)

if and only if they

have the same peak points. We have noted requirement

(9.8) that this cannot be weakened to the

that the peak points coincide except for a set

of measure zero.

16.6. components

Let of

X

be compact and let

CX .

{Ui]

be the set of

Recall that the inner boundary

FI

of

o0

x

is

r z = ax \ ( U au i) . i=O

Conjecture.

If

~(FI) = 0 ,

We have discussed

then

A(X) = R(X)

.

some special cases of this conjecture

in section 14, where we also noted that the converse is false. If the conjecture of 16.3 is correct, this p a r a g r a p h is also true

([94]).

then the conjecture of

-132-

APPENDIX

I.

LOGARITHMIC

CAPACITY

In this appendix we bring together for the reader's convenience

several

(equivalent)

definitions

of logarithmic

capacity;

we also justify a statement

treatment

of the material of this section is in [84].

Let

K

be a compact

in 3.5.

set in the plane.

(positive Balre) probability

measure

A detailed

Let

U

supported on

be a K

and

set

Let

V = inf ~ ) ,

measures. K

Then

where the inf is taken over all such -~ < V < ~ .

The logarithmic

capacity of

is given by

cap

where

(K) = e -V ,

e -V

is understood

to be

definition,

now standard,

is somewhat

used by early writers

if

in the subject.

reader should exercise literature.)

0

As usual,

V = +~ . different

(This from that

Accordingly,

the

some caution in reading the older one extends

cap to arbitrary

by

cap(S)

= sup cap(K) KcS

A second way of defining

cap(K)

is to set

sets

-133-

U (z) = I l°gl~-l~ IdW(~) '

where

is a Balre p r o b a b i l i t y

measure

on

K .

Then

v = inf sup %(z) , u

where

the iaf is taken over all such measures.

we have If

cap(K) K

= e -V .

is a compact

set and

V n = max V(Zl,...,Zn)

over all n - t u p l e s dn

We call theorem identity

T(K)

decreasing

in

of Szeg8

K .

capacity

of a compact

definition These

g(z,~)

of

K .

definitions

need not concern us here. a proof of w h i c h

function

for

A beautiful .

This of sets.

For instance,

on the use of T c h e b y c h e f f

be the Green's

.

by which the logarithmic

set can be defined.

fact,

v(K)

the capacities

hinges

need the following

dn = Vn

v(K) = cap(K)

in estimating

There are still other methods

is taken

with limit

diameter

[84] tells us that

is often useful

Then if

sequence

the t r a n s f i n i t e

let

Izj_zk I .

, where the m a x i m u m

of points

is a (weakly)

Zl, Z2,...,z n E K

= ]-K j
V(Zl'''"Zn)

Set

As before,

G(K)

one

polynomials. We will,

is in [84]

however, .

Let

w i t h a pole at

~ ;

-134-

then

lim ( l o g l z l - g ( z , ~ ) )

Proposition I.i.

Let

K

= log cap(K)

be compact.

cap(K) = suplf'

Then

(-)I

,

where the sup is taken over all functions analytic not necessarily

slngle-valued)

(i)

Ifl

(2)

If(z)l

(3)

f(®) = o

Proof.

on

(but

fl(K) w h i c h satisfy

is single-valued; < 1 , z 6 ~(K); .

If the sup is zero,

harmonic function.

2(K)

Therefore

supports no bounded

cap(K)

wise there is a positive measure

u(z}

p

= 0 .

For other-

such that

:

is a b o u n d e d harmonic function,

contrary to hypothesis.

Suppose then that

(1) through

If'(~)l > 0 .

f

satisfies

We can assume that

be the set of finite zeros of at

~

the

zn

f .

(3) above and

o ~ a(K) . Since

can accumulate only at

f

Let

[z n}

is analytic

~n(K) .

Let

{D n]

-135-

be a sequence of domains and

8D n

such that

consists of a finite number of (disjoint) simple

closed analytic function for

curves•

Dn

g(z,~)-loglz I

Let

•

fl(K) .

u(z)

8D n

fore,

since

u

D

n -+ ~

Then

as

n-~

Since

u(z)-loglzl on

fl(K) .

is positive and

n

•

Thus

gn(Z,~)-logizl

fl(K)

superharmonic is harmonic Now

gn = 0 on

function while the

u(z)-loglzl >_ gn(Z,

compact in

Now let

.

u(z)-loglzl _> gn(Z,~)-loglzl

is a superharmonic on

~ .

is a positive

it also is superharmonic

on

be the Green's

uniformly on subsets of

u(z) : -loglf(z)l function on

gn(Z,~)

with pole at

the topology of the sphere,

O0

D I c D 2 c ... ~fl(K)

u(z) >_ gn(Z,~) 8D n .

on

~D n . r. h. s.

)-loglzl

The

_> g(z,~)-loglzl

is harmonic

on

D

n

z 6 fl(K)

Thus If' (~)I

= limlzf(z)l

: limlze-U(Z)l Z--~o

= lim

exp[loglzl-u(z)]

< lira exp[loglzl-g(z,~)]

Therei. h. s.

we obtain

u(z)-loglzl

near

-- cap(K)

. Letting

-136-

by the remark preceding the proposition. ~(z) = exp[-g(z,~)-ih(z)] valued) harmonic that

, where

conjugate of

I~' (~)I = cap(K)

.

h(z)

If we set is a (multiple-

g(z,~) , it is immediate

This completes the proof.

The importance of logarithmic

capacity in function

theory stems from the fact that sets of logarithmic

capacity

zero are "negligible

sets" for harmonic

In

particular,

U

and on

suppose

cap(K) = 0 . U \ K

functions.

is an open set , K c U

is compact,

Then any function bounded and harmonic

can be extended h a r m o n i c a l l y to all of

U .

-137

APPENDIX II.

-

ANALYTIC CAPACITY AND THE REMOVABILITY OF SINGULARITIES

This brief section is devoted to pointing out the significance of analytic capacitF in problems of function theory. A compact set if

G(K)

Thus, if A

K

is said to be a Painlev~ null set

supports no nonconstant bounded analytic function.

Painlev~ null sets are totally disconnected,

~ ~ K

is a continuum the Riemann map of

since

S2 \ ~

onto

provides a nonconstant bounded analytic function on

~(K)

.

Proposition II.i.

The Painlev~ null sets coincide with the

compact sets of analytic capacity zero.

f

Proof.

Clearly,

if

On the other hand, G(K) .

K let

is a Painleve null set f

be a bounded analytic function on

We can choose a constant

m

such that

K

and

g(z) = czmf(z)

gV(~) ~ 0 .

Actually,

Thus

something

¥(K) = 0 .

C

and a nonnegative integer

is an admissible function for

¥(K) ~ 0 . stronger is true:

sets of analytic

capacity zero are removable sets for bounded analytic functions. Although this follows from the remarks after 5.7, it is easy to give a simple independent proof.

-138-

proposition

11.2.

set such that analytic U

if

on

Let

K c U . U \ K

If

y(K) ~ 0

bounded

and analytic

on

there on

analytic

S2 \ K .

on

fl

Let

In conclusion, capacity

in function

f

If this

on all of

an open and

to all of

function U

it

be bounded and analytic formula we can write

on all of

U

and

f2

is

y(K) = 0 , f2 - 0 .

successfully

The interested

for details.

function

In the other

let us note that H a v i n s o n

quite

theory.

is analytic Since

.

theorem.

Using the Cauchy integral where

his paper

S2 \ K = ~(K)

¥(K) = 0 .

f = fl + f2

analytic

analytically

exists a nonconstant

by Liouville's

suppose

U\K.

U

Then every function bounded

to a function analytic

would be constant

set and

y(K) = 0 .

Proof.

were extendable

be a compact

can be extended

and only if

direction,

K

[~9] has used

to study other problems reader

should consult

-139BIBLIOGRAPHICAL

NOTES

A bibliography of his ignorance~

is a product

the one that follows

That the list of references is probably

of an author's

is no exception.

is complete

in any strong sense

too much to hope for~ however,

my knowledge,

all the basic references

whimsy and

to the best of

have been included,

as well as a large number of papers that are, perhaps, more tangential generous

interest.

in citations

I have tried especially

of the Soviet literature.

of

to be

Also,

with

the hope of stimulating

some interest

history of the subject,

I have included a number of references

to the early literature.

in the fascinating

These references,

chosen rather subjectively

however,

have been

and are not meant to be complete.

A word or two is in order at to the actual organization of the bibliography. by author.

The entries are grouped alphabetically

The transliteration

on the transliterations notice in Mathematical

in Doklady. Reviews,

end of the corresponding announcements

of Russian names has been based

entry.

If an article has received

this reference In general,

I have not listed

of results when complete proofs have become avail-

able in subsequent

papers~

in which the announcements

exceptions

to this rule are cases

are significantly

When a Russian paper has been translated referenced

is given at the

the translated

into English I have

paper.

Included here are also some brief notes, several headings,

more accessible.

concerning

the content,

grouped under

historical

importance,

-14o-

or interrelationships by no means meant not satisfy,

of the papers

to be complete~

the reader's

appetite

referenced.

their purpose for more.

These are is to whet,

-141-

i.

Earlier work on rational approximation.

Ruoge's

original paper [78] marks the beginning of the systematic study of rational approximation.

That same year, 1885,

Weierstrass proved his famous theorem on polynomial approximation.

Earlier, Appell [5],[6] had studied some special

cases of approximability by rational functions.

Basic progress

in studying polynomial and rational approximation was made by Walsh [96], [97], Hartogs and Rosenthal [44] (cf. Tonyan [83]), Lavrentiev [60] (cf. Mergelyan [65]), and Keldysh [58].

We

have commented on the contributions of these mathematicians in Section 9~ for another discussion of this work, see the treatise of Mergelyan [67].

One should also consult the monograph of

Walsh [99], which contains other references to the early literature. 2.

More recent

(Russian) work on rational approximation.

The definitive results on polynomial and rational approximation are due to Mergelyan [67] and Vitushkin [93], [94], [95].

Mergelyan

first proved his beautiful theorem on polynomial approximation in [66]~ [67] contains a very full treatment and generalizations to rational approximation.

Vitushkin's papers

[88], [89], [91],

[93], [94], [95] constitute a triumphant march toward the solution of the rational approximation problem; there are many misprints and incomplete proofs in these papers, however. [68] and [41] are useful summaries of the state of the subject in 1961 and 1965 respectively.

The reader can also consult the book of

Smirnov and Lebedev [81], which contains a nice exposition of

-.142-

Mergelyan's

results and the earlier work of Vitushkin.

similar treatment occurs in the difficult-to-obtain of Gamelin

[31].

Analytic

[27],[28] and Gonchar

capacity.

[40].

The study of sets of analytic

capacity zero goes back all the way to Paimlev~ Appendix II). Besicovitch

[71]

Refer also to the papers of Denjoy

[8].

notes

Other papers that here deserve mention

are those of Dolzhemko

3.

A

The fascinating

(see

[22] and

question of priority in

these matters is dealt with in [20]. The actual definition of analytic Ahlfors[2],

capacity is due to

who was interested in function theoretic

problems on finitely connected planar domains. refined by Garabedian work is in Neharfs

[33].

His work was

A summary of this and related

survey article

[70].

Ahlfors generalized

his results to regions on Riemann surfaces paper

extremal

[3]; see Royden's

[77] for another treatment as well as further references

to the literature. Ahlfors and Beurlimg

[4] were the first to study analytic

capacity from a systematic viewpoint.

Pommereoke

their results and proved some interesting

[74] extended

new theorems;

part

of his work was duplicated by Ivanov [54], who also extended [55] some of the results in [4]. contains

some properties

rational approximation.

Vitushkin's

of analytic

paper

[88]

capacity that relate to

In another paper

[90], Vitushkin

-143-

exhibits a curious anomaly.

The behavior of analytic capacity

under various transformations of the domain was studied by Havin and Havinson [48].

Havinson [49] has used analytic

capacity to considerable advantage in studying problems of function theory; his work is partly an extension of the work of Ahlfors and Garabedian to infinitely connected domains. A concise treatment of the elementary properties of analytic capacity can be found in [81].

#.

AC capacity.

Literature on AC capacity is limited.

The notion was first defined by Dolzhenko in [28].

The

strongest known sufficient condition for a set to have positive AC capacity is due to Arens [7], whose result generalizes early work of Pompeiu [75], Zoretti [108], Denjoy [21], [22], [23], [2#], and Urysohn [85].

It is particularly instructive to read

Denjoy's papers referenced above in connection with the announcements of Pompeiu (not cited here) in the Comptes Rendus of that period;

[26] contains a survey of some of these results

pp. 631-636, lO13-1016,

1066-1067) as well as the complete text

of [25] (vol. I, pp. 289-367). also of interest here.

(vol. II,

Besicovitch's paper [8] is

See the book of Collingwood and Lohwater

[20] for more complete references and a discussion.

5.

Function algebra methods.

At present, the best general

references on function algebras seem to be Wermer's monograph [lO0], Royden's survey article [76], and Hoffman's lecture

-144-

notes [53].

[lO0] and [53] contain applications to rational

approximation.

All three references have extensive bibliographies

to which the interested reader can refer.

In the comments

below we shall assume the language of function algebras. Bishop [9],[1#] was the first to apply the methods of functional analysis to problems of rational approximation~ he gave a proof of Mergelyan's polynomial approximation theorem based on linear functionals.

Glicksberg and Wermer [38]

removed the remaining function theory from Bishop's argument to obtain an honest "abstract" proof: the only fact from complex variables that is needed is a result of Lebesgue [61] and Walsh [98].

A self-contained exposition of the work of

Glicksberg and Wermer (plus much more) is in [103].

Carleson's

synthesis [16] is probably the best available proof of Mergelyan's theorem~ his treatment, based on the papers mentioned above, is abstract in spirit yet avoids the machinery of Dirichlet algebras.

Glicksberg [36], using the techniques of [38], obtained

an abstract proof of Mergelyan's theorem concerning rational approximation on sets of finite connectivity~ using different methods Ahern and Sarason [1] obtained another proof of this result.

Actually, as Garnett observed [34], Mergelyan's rational

approximation theorem is a simple consequence of the theorem on polynomial approximation. The basic paper of Bishop on peak points and the minimal boundary is [13].

Gonchar's "~-~"

criterion is in [39].

-i#5-

Wilken generalized

9.7

in

trivial Gleason parts of planar) measure. if every point of

[i07]~

R(X)

have positive

It follows that X

he proved that the non(Lebesgue

R(X) = C(X)

is a part of

R(X) .

if and only

See also [112].

In [62], McKissick constructed a compact set which

R(X)

is normal and yet

disproved the conjecture that R(X) ~ C(X) R(X)

R(X) ~ C(X) . R(X)

for

Steen [82]

is antisymmetric if

by constructing a "swiss cheese"

X

contains nonconstant real functions and yet Glicksberg has proved

X

[37] that

R(X) = A(X)

for which R(X) ~ C(X). if gnd only

if the real annihilating measures of these algebras coincide~ further,

in [35] it is shown that R(X) = A(X) if they have the

same representing measures. noteworthy~

The work of Valskii

[87] is also

it marks the first adoption by a Russian mathe-

matician of function algebra methods for studying

R(X) .

Browder [15] and Wermer [104] have studied point derivations

on

R(X). Other papers worthy of note include Fisher's work [29],

[30], the seminal paper of Gamelin and Rossi paper [106].

[32], and Wilken's

Also, we should not fail to mention the two

elegant notes of Wermer [I01], [102].

6.

Riem~nn surfaces.

the first to consider

Sakakihara

[79] seems to have been

(nontrivial) approximation on a Riemann

surface~ he obtained a generalization of Walsh's theorem [96].

-146-

Bishop also studied approximation on Riemann surfaces Gusman

[42],[43]

generalized Mergelyan's

nomial and rational approximation. independently

theorems on poly-

Kodama

[59] proceeded

of the work of Gusman and obtained many of

the same results,

including the Mergelyan theorems;

proved a "localization"

theorem for surfaces,

result of Bishop's more-or-less 7.

[i0].

Miscellao~.

she also

generalizing a

implicit in [10].

See also [lll].

There are m~my papers in the bibliography

that do not fit easily into any of the above classifications. We shall mention some of these below. Bishop

[ll], [12] studied the problem of approximating

simultaneously

a finite number of continuous

polynomial and its derivatives. the corresponding

Chatskaia

functions by a

[17] considered

problem for rational functions.

The problem of representing a bounded analytic as the Cauchy transform of a measure studied by several authors. [51], and Valskii

[86].

function

(cf. 3.13) has been

We mention Havin [45], Havinson

Havin's papers

[46], [47] also touch

on this question. The problem of approximating a compact set of analytic

a continuous

function on

capacity zero by rational functions

having a special form has also attracted attention. work has been done by Havin [46], Havinson [18], [19].

See Havinson

[52] for a survey.

Such

[50], and Chatskaia

-147-

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108. L. Zoretti, "Sur les fonctions analytiques uniformes qui poss~dent un ensemble parfait discontinu de points singuliers", J. Math. Pures Appl. (6) 1 (1905), 1-51.

-155Bibliographical Addenda

109. L. Carleson, "On null-sets for continuous analytic functions", Arkiv f6r Mat. i (1950), 311-318. MR 13-23. llO. E. P. Dolzhenko, "The removability of singularities of analytic functions", Uspehi Mat. Nauk 18 (1963) (ll2),

135-142. (Russian). ~

27 ~5898.

lll. T. Gamelin and G. Lumer, "The universal Hardy class", to appear. ll2. M. S. Melnikov, "Structure of the Gleason part of the algebra R(E)", Fuact. Anal. Appl. ! (1967), 84-86. for

113. D. Wilken, "The support of representing measures R(X) ~', to appear.

114. P. C. Curtis, "Peak points for algebras of analytic functions", to appear.

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Bitte wenden / Continued

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