Hardy Classes on Infinitely Connected Riemann Surfaces

Lecture Notes in Mathematics Edited by A. Dold and B. Eckmann

1027 Morisuke Hasumi

Hardy Classes on Infinitely Connected Riemann Surfaces

Springer-Verlag Berlin Heidelberg New York Tokyo 1983

Author

Morisuke Hasumi Department of Mathematics, Ibaraki University Mito, Ibaraki 310, Japan

A M S Subject Classifications (1980): 30 F99, 30 F 25, 46 J 15, 46 J 20, 31A20, 3 0 D 5 5 ISBN 3-540-12729-1 Springer-Verlag Berlin Heidelberg New York Tokyo ISBN 0-387-12729-1 Springer-Verlag New York Heidelberg Berlin Tokyo This work is subject to copyright.All rights are reserved,whetherthe whole or part of the material is concerned,specificallythose of translation,reprinting,re-use of illustrations,broadcasting, reproductionby photocopyingmachineor similar means,and storage in data banks. Under § 54 of the GermanCopyright Law where copies are madefor other than private use, a fee is payableto "VerwertungsgesellschaftWort", Munich. © by Springer-VerlagBerlin Heidelberg 1983 Printed in Germany Printing and binding: Beltz Offsetdruck, Hemsbach/Bergstr. 2146/3140-543210

PREFACE

The purpose on i n f i n i t e l y Already "Hardy print

connected

in this Classes

on R i e m a n n

Surfaces"

analytic

when

H a r d y himself,

disk

J. E.

Speaking

roughly,

sively

in the case of the unit

as well as simplicity. or not,

has

drawn m u c h

Opposed

to this,

our

laid

stress

made

discussed

their

amount

have

been

now be

in the

growth

of func-

The theory

of these

G.

i~ the work

Szeg~

among

of l i t e r a t u r e studied

from the cradle

in

years.

debut

the mean

[15].

appeared

should

its f o u n d a t i o n

growing

classes

which

subsequent

F. and M. Riesz,

disk

topics.

monograph

most

of

others.

in this inten-

for its i m p o r t a n c e

The case

of f i n i t e l y

oonnected

attention

and e n j o y e d

considerable

our k n o w l e d g e

connected

classes

deals m o s t l y

bearing

on our present

analytic

methods

of c o m p l e x class

for example.

theory

theory thus

of f u n c t i o n

surfaces,

planar

progress

it,

still

an i n d e p e n d e n t of general

so we should finitely

this

in

theory

from G a m e l i n ' s

the r e a c h

field

"For w h i c h

known

class

can one get a f r u i t f u l

classes?" at least,

In the present to this

[I0]

surfaces, and

and balls.

is not yet very well

surfaces

partial

connected

book

of the new t h e o r y

of p o l y d i s k s

question:

of H a r d y

try to give an answer,

in the

and the a b s t r a c t

grown to form the core of the n e w l y -

surfaces

Riemann

direct

functional-

applications

classes

in the

of Hardy

not have m u c h

the case of i n f i n i t e l y

basic

small

theory

downwards,

as e v i d e n c e d

lies b e y o n d

Riemann

open

sion of the c l a s s i c a l we will

Hardy

study as in the case

begin with

connected

has

algebras,

1950's

successful

including

created

to be r e l a t i v e l y

The c l a s s i c a l

disk and does

From

their

Nevertheless,

as I u n d e r s t a n d

structure

problem. found

seems

surfaces.

with the unit

have

function

born t h e o r y

needs

Hardy

some r e l a t e d

years.

case of i n f i n i t e l y

Hardy

classes

was

and still

area.

that

in his pa p e r

Littlewood,

classes

Heins,

seen d u r i n g

Hardy

of functions

A n d we now have a large

recent

natural

G. H. H a r d y

on the unit

classes

and

of H a r d y

we have a b e a u t i f u l by M a u r i c e

we have

recognized,

in 1915,

surfaces

series

It is t h e r e f o r e

As g e n e r a l l y

very useful

is to give an account

Riemann

Notes

on some new a d v a n c e s

literature tions

notes

open

Lecture

in 1969.

placed

of these

The and

of inextennotes

question.

IV

Our

idea for a t t a c k i n g

any nice

surface

But what

do we mean

as most

should

promising

(abbreviated in Chapter

to

V.

in the

class

perhaps

motives.

result

class

An open

if every

bounded

holomorphic

of the unit results

almost

every

converges (D) origin

by M.

reason

here

of P a r r e a u - W i d o m will

[70]

[52]

in 1971

which

in

from very

later

why we are

following

type

be given

Parreau

definitions,

in the

functions.

turned

interested

fundamental

of this

following, positive

Green

where

measurable

R

denotes function

other

R

space

has

origin

to the given

boundary

function

along

Almost

Green

every

converges

to a point,

indeed

for

R

(D) means

that

PWS's

for w h i c h

in detail

in the M a r t i n

vertex

behavior

general

at almost

to g e n e r a l i z e

consists

the direct theorem.

the

Cauchy

whose

theorem--(DCT)

Concerning

these,

line.

A

every

of

with r e s p e c t on

solution

boundary

A of the

function

line.

class

maps

prob-

for any PWS. of surfaces

solution. the

following:

point from

of

More-

in R

A

can be

can be

disk.

Cauchy-Read

of two statements,

for

Green

measure

answer

so as to have

of a n a l y t i c

R

which

the B r e l o t - C h o q u e t

has a p o s i t i v e

as in the case of the unit

It is p o s s i b l e

Green

affirmative

first

problem

that

on

boundary

the usual

measurable every

along

from any fixed

and the h a r m o n i c

almost

relevant

R.

every

b~,

(D) can be r e f i n e d with

issuing

Furthermore,

form the

this

almost

R

in p a r t i c u l a r

properties

solution,

on

any b o u n d e d

[5]) has a c o m p l e t e l y

Stolz regions

Cauchy

with

in

lines

is m e a s u r e - p r e s e r v i n g

to the data along

and the b o u n d a r y

respectively,

~ ÷ b~

on the one hand,

statement

The t h e o r e m

Z say

on the other.

statement

genus

line

nice

a limit

of Green

has a unique

0

if and

a PWS. on

function

problem

type

has n o n c o n s t a n t

We list most

boundary

It seems

analysed

R

many

from any fixed

infinite

(E)

over

surfaces.

on the

(see Brelot

defined

inherit

issuing

converges

the

bundle

problem

for the point

The

kind

connected

harmonic

and the c o r r e s p o n d e n c e

A

is of P a r r e a u - W i d o m

line

line

0

Dirichlet

R

complex

sections.

to the Green m e a s u r e

verse

surfaces

The D i r i c h l e t

any b o u n d e d

over,

says that

to put f o r w a r d

The d e f i n i t i o n

The main

surface

disk or f i n i t e l y

Every

(C)

lem

and

of h o l o m o r p h i c

by H. W i d o m

different

same.

simple

we wish

introduced

is e x p r e s s e d

flat

surfaces

in the

(B)

on

first

independently,

Riemann

unitary

Moreover,

R

of R i e m a n n

following).

They used the

family

The c a n d i d a t e

was

of surfaces

is very

of Widom:

(A) only

an ample

is the class

"PWS"

out to be e s s e n t i a l l y in this

carry

by this?

This

1958 and also, different

the p r o b l e m

theorem

refinements for

we first

to PWS's. are called,

short--and have:

the

in-

V

(F)

The

inverse

The c o n v e r s e (G)

If

of b o u n d e d

R

holomorphic

the

utmost As we rect

type.

refinement shall

direct

there

theorem

Cauchy

we have:

(H)

Every

PWS

This,

expectation a PWS can Cauchy

state a n o t h e r

H~(R)

cussed

if

R

exist

genus

planar

of

(R)

R,

surfaces

Cauchy

formula--is

H

is a PWS.

characterizes

of

theorem--an

not always

valid.

for w h i c h

the di-

PWS's

for w h i c h

some

say about

algebra

the

fact

theorem without like

as an open

answer,

in this

is r e g u l a r

in the sense

PWS.

sub-

The

contrary

connection

satisfying

look

interesting

H~(R).

an a r b i t r a r y

has a n eg a t i v e

A curious

with

homeomorphieally

Banach

two things

R

in the

space

to the

is that

the direct

independent.

Finally,

notes.

Riemann

surfaces

tively,

PWS's

Apart

our last r e s u l t

out,

surfaces

shows

relevant

problem

subsets

this

we

that plane

of

of plane

are good regions

theory,

on

R

then

in the

R.

of PWS's

to be dis-

book c o n t a i n s

a detailed

regions

all that h a p p e n

in the c a t e g o r y which

functions

properties

from PWS's,

almost

can h a p p e n

are

on compact

about

of the c l a s s i f i c a t i o n As it turns

of p o t e n t i a l

of all h o l o m o r p h i c

convergence

been t a l k i n g

in these

account

the points

the direct

can be studied

of the

PWS's

the corona

of u n i f o r m

classes.

we have:

if the set

fact:

is dense

We have

space

have.

These

If a PWS

topology

almost

of i n f i n i t e

is all one can

one m i g h t

satisfy

separates

can be e m b e d d e d

for general

theorem.

(I)

R ideal

however,

problem

and

if and only

there

for PWS's

First

set in the m a x i m a l

PWS's

Namely,

surface

integral

while

PWS.

true.

fails.

problem

results.

corona

theorem

exist

holds,

theorem

The c o r o n a

R

On the other hand,

of the Cauchy

see,

Cauchy

on

t h e o r e m holds

Cauchy

for any

is also

Riemann

functions

Cauchy

inverse

Parreau-Widom

t h e o r e m holds

statement

is a h y p e r b o l i c

then the inverse Thus

Cauchy

of this

in terms

of Hardy

in the c a t e g o r y

of plane regions.

in some

sense

of

Intui-

or other.

can be as i l l - b e h a v e d

So

as one

can imagine. In w r i t i n g faces

of P a r r e a u - W i d o m

behaved

surfaces

It is hoped this

t h e s e notes

feeling

study

that

type

been

probably

our d e s c r i p t i o n

somehow

aims

or other. adaptation

faces.

It also

general

nor too special.

at finding

~ust refleet

led by the

of Hardy

I wish

of the e x i s t i n g some new facts

The present

notes

personal

pages

knowledge

will

of well-

justify

is that

our

of R i e m a n n

are n e i t h e r

are not c o m p l e t e interest

the sur-

is concerned.

to note

which

that

family

classes

in the f o l l o w i n g One thing

the a u t h o r ' s

feeling

form the widest

as far as the t h e o r y

is not a mere

sense but

I have

too

in any

in the field.

sur-

VI

At all event I hope that our effort w o u l d help not only extend the theory of H a r d y classes but also increase our k n o w l e d g e of R i e m a n n surfaces in general. The p r e r e q u i s i t e s

f o r ' r e a d i n g these notes are the f u n d a m e n t a l s of

a d v a n c e d complex function theory and some k n o w l e d g e of f u n c t i o n a l analysis.

As for the function theory~

we assume that the r e a d e r has some

a c q u a i n t a n c e with the facts to be found in Chapters Ahlfors and Sario

I, II, Iii, V of

[AS] and also in the first four chapters of the book

[CC] by C o n s t a n t i n e s c u and Cornea. Chapter i of Hoffman

As for the f u n c t i o n a l analysis,

[34] may be useful,

if not sufficient.

We now comment on the contents of the present notes. theoretic prerequisites

are sketched

The f u n c t i o n -

in Chapter I without proof.

order to deal r e a s o n a b l y with Hardy classes on m u l t i p l y - c o n n e c t e d surfaces,

we rely on two concepts:

and Martin c o m p a c t i f i c a t i o n . III, respectively.

multiplicative analytic

These are e x p l a i n e d

Chapter IV contains

In open

functions

in Chapters

II and

preliminary observations

on

H a r d y classes, where the b o u n d a r y b e h a v i o r is our p r i n c i p a l concern. The main body of this book begins

in Chapter V.

There,

the d e f i n i t i o n

of surfaces of P a r r e a u - W i d o m type is given after Widom. by means of r e g u l a r i z a t i o n , valent to Parreau's,

that this d e f i n i t i o n

p r o b l e m for PWS's

gebra of

L~

(see

(B),

(see (D)).

(F) is e s t a b l i s h e d by using Green lines and, as an

it is shown that

H~

is a m a x i m a l w e a k - s t a r closed subal-

on the Martin boundary.

(DCT) is p r e c i s e l y stated.

Next, the direct Cauchy t h e o r e m

We prove a w e a k e r v e r s i o n of (DCT), which

is valid for any h y p e r b o l i c Riemann surface, ation.

(C)) and then solve the Two types of Cauchy theo-

and i n v e r s e - - o n PWS's form the main theme of Chapter VII.

There, the statement application,

On the other hand,

t o g e t h e r with some applic-

(DCT) itself fails sometimes.

not be seen until we know something about invariant are studied in Chapter VIII. submodules of

But this can-

subspaces,

subspaces of

LP--on the Martin

C o r r e s p o n d i n g to the known results

for the case of

the unit disk, we c o n s i d e r two p r i n c i p a l types of invariant which are called doubly invariant and simply invariant, As for doubly invariant

which

In Chapter VIII we c l a s s i f y closed H ~-

LP--(shift-)invariant

b o u n d a r y of a PWS.

any PWS.

fundamental

In Chapter VI we discuss the D i r i c h l e t

p r o b l e m on the space of Green lines rems--direct

is e s s e n t i a l l y equi-

we present a d e t a i l e d proof of Widom's

t h e o r e m m e n t i o n e d in (A) above.

Brelot-Choquet

A f t e r showing,

subspaces,

respectively.

subspaees the situation is r a t h e r simple for

But the so-called Beurling type t h e o r e m for simply invariant

subspaces is not always valid for PWS's.

It is proved in fact that the

VII

Beurling

type

nection,

examples

types

theorem

is v a l i d

in C h a p t e r

of constr~etion:

Myrberg

type)

planar

PWS's

the third

for w h i c h

gives

same c h a p t e r we first bolic

prove

the

first

(DCT) (DCT)

fails

the

statement

(DCT).

classification (E) will not

are three

appendices

if

(DCT)

PWS's

the

is.

(G), w h i c h

of plane

theorem

a couple

just

by using

and a list of r e f e r e n c e s ,

PWS's

among

Hardy

IX hyper-

on PWS's

problem

classes.

in Chapter

which

and

In the

of c o n d i t i o n s

sketched

(of

In C h a p t e r

XI we solve Heins'

regions

but

three

of

holds;

is false. (I).

characterizes

in Chapter

con-

genus

a family

theorem

(H) and

collect

be proved

yields

but the corona

the c o r o n a

In this

We give there

of i n f i n i t e

second

statements

and then

Finally

statement

defines

holds;

for w h i c h prove

surfaces,

to

concerning

PWS's

we also

Riemann

equivalent

the

for w h i c h

if and only

X m a y be interesting.

The

VI.

There

is by no means

ex-

haustive. My i n t e r e s t was v i s i t i n g this

I am i n d e b t e d

It was

like

classes

to thank

after

thesis

Professor

when

primitive

the notes

written

thanks

expand

for a series

Metropolitan

notes

are due to P r o f e s s o r s and made

who

my k n o w l e d g e

Most

Sakai

some v a l u a b l e

and

remarks

from Professors

to w h o m

I wish

is my first

teacher

couragement

have r e m a i n e d

Mito,

Ibaraki

July,

1983

this

on the r e s e a r c h

book

who

Year

chapters topics

in The have

were reat Tokyo

Particular who o r g a n i z e d

I have much

benefited

and H. W i d o m

and

my appreciation.

to P r o f e s s o r

level

right

classes.

5, 1982.

Z. K u r a m o e h i

to express

thus

New d i s c o v e r i e s

S. Yamashita,

suggestions.

by very h e l p f u l

like to d e d i c a t e

of H a r d y

of the main

of July

study

I would

its B e u r l i n g

then.

I

For

L. Carleson,

I gave on the present

Dr. M. Hayashi, I would

to a serious

surfaces.

during

while

L. Kelley.

showed me the thesis

was w r i t t e n

the week

M.

and J.

to P r o f e s s o r

Institute

subsequently.

during

Riemann

aroused

in 1962-64.

H. H e l s o n

led me in 1972

thanks

to deepen

of lectures

University

lectures

that

first

at B e r k e l e y

Bishop,

L. A. Rubel,

of these

here was

connected

I also owe

I was able

version

E.

[45],

me to the M i t t a g - L e f f l e r

1976-77,

made

treated

of C a l i f o r n i a

on i n f i n i t e l y

its completion.

invited

subject

to P r o f e s s o r s

C. N e v i l l e ' s

of Ha r d y

the

in the

the U n i v e r s i t y

and w h o s e

Zir6 Takeda, inspiration

who

and en-

with me as fresh as ever.

Morisuke

Hasumi

CONTENTS

PREFACE

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

CHAPTER

I.

§i.

§2.

§3.

Topology

Riemann

2.

The

Homology

3.

The

Fundamental

Classical

SURFACES:

Surfaces

Groups

Potential

Theory

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

4

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

4

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

S

6.

Potential

Theory

Differentials

9.

Cycles

Definition

Class

F

and

Riemann-Roch Cauchy

9

Subclasses

on

9

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

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

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

Compact

Bordered

Surfaces

. . . . . .

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

II.

MULTIPLICATIVE

The Line

3.

Existence

Cohomology

Bundles

and

of

Basic

5.

0rthogonal

Group

of

Structure

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

23 23

Analytic

Sections Functions

Functions

.

31

. . . . . . . . . .

33

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

Definition

2.

Integral

3.

The

33 36 38

COMPACTIFICATION

Compactification i.

28

. . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . . . . . . . . MARTIN

14 17 22

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

Decomposition

12

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

Multiplicative

Harmonic

ii

FUNCTIONS

Functions

Holomorphic

Structure

4.

ANALYTIC

Analytic

First

2.

III.

its

Theorem

Kernels

Multiplicative

CHAPTER

7

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

and

Differentials

ll.

Notes

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

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

I0.

Lattice

i 2

Problem

Dirichlet

i

3

The

The

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

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

5.

Basic

REVIEW

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

Functions

8.

QUICK

Group

Superharmonic

7.

A

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

4.

i.

51.

RIEMANN

Exhaustion

CHAPTER

§2.

of

OF

1.

Notes

§i.

THEORY

iii

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

Representation

Dirichlet

Problem

39 39

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

40

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

43

X

§2.

§3.

Fine

Limits Definition

5.

Analysis

Covering

IV.

§3.

Some

Harmonic

Harmonic

the

Unit

Definitions

3.

Boundary Some

5.

The

on

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

Functions Measures

Disk

Behavior on

B-Topology

RIEMANN

57

. . . . . . . . . . .

59 63

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

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

of

Riemann Hp

and

Multiplicative

Surfaces hp

. . . . . . . .

Functions

Analytic

. . . . . .

Functions

and

Basic Widom's

3.

Regularization

Definitions

Widom's

of

on

5.

Proof

Necessity

of

Widom's of

(I)

Regular

of

. . . . . . . . . . .

Parreau-Widom

Type

....

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

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

Theorem

6.

Review

Principal

7.

Modified

Green

8.

Proof

of

Sufficiency

9.

A

Direct

(II)

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

Operators

Functions

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

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

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

Consequenoes

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

. . . . . . . . . . . . . . . . . . . . . . . . . . . GREEN

The

Dirichlet

i.

Definition

2.

The

The

Space

3.

The

4.

Limit

Green

73

83 83 85 86 90 90 95 99 99 102 iii 117 118

LINES Problem of

Dirichlet of

73

TYPE

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

Subregions

66

75

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

Surfaces

64

82

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

Theorem

Analysis

VI.

Properties

Characterization

4.

Few

PARREAU-WIDOM

Fundamental

2.

of

OF

64

74

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

SURFACES

i.

of

50 57

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

V.

49

. . . . . . . . . .

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

Results Hyperbolic

Results

Definitions

CHAPTER

§2.

of

on

Classes

Notes

§i.

of

Classical

4.

Proof

Behavior

CLASSES

Classes

Basic

Proof

49

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

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

HARDY

2.

Hardy

Limits

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

i.

CHAPTER

§2.

Maps

Preservation

Hardy

Fine

Boundary

7.

Notes

§i.

of

Correspondence

CHAPTER

§2.

of

6.

Notes

§i.

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

4.

the

Star

Lines

of

Green

Lines

. . . . .

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

. . . . . . . . . . . . . . . . . on

Regions

Green

Space

Lines

Problem

Green

along

on

Green

Lines

a

Surface

of

Parreau-Widom

Type

119 119 121 124

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

124

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

129

XI

§3.

The

Green

5.

Convergence

of

6.

Green

and

7.

Boundary

Notes CHAPTER §i.

§2.

§3.

VII.

Statement Proof

The

Direct

3.

Formulation

4.

The

§2.

of

of

. . . . . . . . . .

132

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

132

Boundary

13S

Analytic

Common The

Theorem Results

Cauchy

IB

. . . . . . . . . .

Maps

. . . . . . . . . . .

140 143

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

144

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

144

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

145

Theorem of

the

Cauchy

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

Condition Theorem

of

151

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

iSl

Weak

152

Type

. . . . . . . . .

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

Weak-star

7.

Maximality

Inner

of

Factors

Ha

155

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

155

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

Orthooomplement

of

156

H~(dx)

157

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

VIII.

SHIFT-INVARIANT

Preliminary

Generalities

2.

Shift-lnvariant

Invariant

Subspaces

Subspaees Invariant

4.

Simply

Invariant

S.

Equivalence

160

of

on

the

Unit

Disk

160

. . . . . . .

162

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

167

Subspaces

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

167

Subspaees

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

169

( D C T )a

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

177

. . . . . . . . . . . . . . . . . . . . . . . . . . . CHARACTERIZATION

Inverse

Cauchy

i.

Statement

2.

A Mean

3.

Proof

Conditions

of

Value of

the

General

S.

Functions

OF

Theorem

the

Main

Theorem Main

Equivalent

4.

Notes

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

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

Doubly

IX.

159

SUBSPACES

Observations

i.

The

Martin

of

Theorem

Direct

6.

CHAPTER

Boundary

Lines

THEOREMS

Cauchy

Applications

Notes

§i.

CAUCHY

2.

3.

Martin

the

Behavior

i.

Notes

§2.

the Green

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

Inverse

CHAPTER

and

Lines

The

5.

§i.

Lines

Discussion mP(~,a)

SURFACES and

OF

Surfaces

Result

PARREAU-WIDOM of

178 TYPE

Parreau-Widom

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

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

Theorem to

the

. . . . . . . . . . . . . . . Direct

Cauehy

Theorem

.....

. . . . . . . . . . . . . . . . . . . and

(DCT)

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

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

Type

179 17g 183 187 198 198 200 207

XII

CHAPTER §i.

§2.

§3.

§4.

X.

PWS

of

§3.

Infinite

Genus

PWS's

2.

Verification

Plane

of

3.

Some

Existence

Further

of

of

Simple

Embedding

6.

Density

The

Corona

7.

(DCT)

8.

Negative

for

. . . . . . . .

208

Which

(DCT)

Fails

215 217

. . . . . . . . . . . . . . . . . Ideal

Space

. . . . . . . . .

Hardy-Orlicz

223

PWS

227

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

Theorem:

Positive

Examples

....

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

Classes

i.

Definitions

2.

Some Sets

of

3.

Preliminary Existence

N~

of

5.

Lemmas

6.

Classification

7.

Majoration

Null

233

234 234 235

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

238

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

238

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

Regions

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

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

by

229

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

Sets

Plane

227

REGIONS

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

Lemmas of

Classification

Notes

PLANE

Properties

Class

4.

OF

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

Basic

221 221

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

Corona

CLASSIFICATION

215

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

Maximal

Examples

for

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

PWS

for

Type

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

XI.

208

213

(R)

the

Holds

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

of

H

(DCT)

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

the

Problem and

Which

TYPE

Type

Lemmas

into

PARREAU-WIDOM

(DCT)

Theorem

of

OF

Parreau-Widom

Properties

5.

Null

SURFACES

Myrberg

Regions

4.

CHAPTER

§2.

OF

i.

Notes

§i.

EXAMPLES

Theorem

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

Quasibounded

Harmonic

Functions

. . . . .

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

247 253 253 256 260 261

APPENDICES A.I.

The

Classical

A.2.

Kolmogorov's

A.3.

The

F.

References Index Index

of

and

Fatou Theorem

M.

Riesz

Theorem on

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

Conjugate

Theorem

Functions

. . . . . .

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

. . . . . . . . . . . . . . . . . . . . . . . . . Notations

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

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

262 267 269 272 276 278

CHAPTER

I.

Some basic

THEORY

results

OF RIEMANH

in the theory

here

for our later reference.

them

can be found

Sario,

Riemann

Riemannsoher

§i.

together

Surfaces

Fl~chen,

OF R I E M A N N

We refer

to

[AS]

A QUICK REVIEW

of R i e m a n n

surfaces

They are

with

stated w i t h o u t

complete

proofs

or in C o n s t a n t i n e s c u

referred

TOPOLOGY

SURFACES:

to as

[AS] or

for most

basic

definitions

not be given here,

e.g.

parametric

disk,

number,

surfaces

i.

IA.

Every in

structure regular in

R.

Unless

concerning

conformal

otherwise

etc.

Riemann

structure, In what

stated,

local

follows,

all Riemann

to be connected.

Every

region

and

in

R

creasing

R

in

Any sequence

number

a (sub-)region

(or

to have the c o n f o r m a l

D

in

compact,

R

is called

the b o u n d a r y

of n o n i n t e r s e c t i n g

a gD

of r e g u l a r

surface,

then there

regions

n = i,

Existence R

admits

and hence

regions

exhaustion

of

of a r e g u l a r a locally

that

R

of

analytic

components.

of r e g u l a r

for each

is called

A region

if it is r e l a t i v e l y

{Rn}n~ 1

a regular

IB.

disks

R.

is an open R i e m a n n

Rn+ I

R

is supposed

2,...

in and

R

exists

an in-

such that

CI(R n)

R : Un= I R n.

([AS],

12D)

is called

surface

R

has no compact

sequence

is included

set in

in

of a finite

R\ D

If

open

region

from that of

consists

Theorem.

II~

connected

induced

R

curves,

Ch.

and

R~nder

Exhaustion

domain)

D

intersection

surface.

are assumed

Ideale

SURFACES

which will

a Riemann

in Ahlfors

of

[CC] below.

surfaces,

denotes

either

and Cornea,

variable, R

are c o l l e c t e d

proof but most

R

n

in

R

having

shows

that

this

property

R.

exhaustion

finite

covering

can be r e g a r d e d

consisting

every Riemann of p a r a m e t r i c

as a polyhedron,

i.e.

a tri-

angulated

surface,

Theorem.

K = K(R)

hedron

if and

IC. of

K

(ii) An

Let

K

called every

has

I,

that

{Rn}n~ I

every

K

if

does

not

Riemann

which

P

a single

P

to

and

contour.

of

is the u n i o n

surface.

poly-

surface.

subcomplex

subcomplexes

belong

46A).

is a p o l y h e d r o n

and has

K

permits

Every

2A.

open

K

of

P

is and

n

if

Pn"

Then

the

a canonical

polyhedron

exhaustion.

K ([AS],

Let

R

hedron

K

regarded

Riemann

([AS],

2B.

surface ([AS],

the

looking

into

in p a i r s

= b. × b.

= 0,

I,

23D).

R

region

D

contour.

admits

1-simplex

Ch.

34A).

I,

properties

a regular

a. x b.

bi,

= i

exhaustion

or i n f i n i t e

and

a. × b. i

number

which

R.

sequence

for

in

K

is

gives

the

K

poly-

rise

to

canonical

can be o b t a i n e d

a canonical : 0

by

of this

This

is c a l l e d

HI(R)

surface

we d e n o t e

group

1-simplex

surface

of the g r o u p

1-dimensional

By IC the

homology

oriented

is c a l l e d

the

33B).

polyhedron,

on the

i

intersection

HI(R) I,

HI(R) , w h i c h

A finite ai,

and Ch.

Every

onto

by i

the

a regular

a single

1-dimensional

HI(K)

HI(K).

3

R

be the

Ch.

of

labeled

of

as an o r i e n t a b l e

([AS],

Thus

we m e a n

subregions.

group

as a s i n g u l a r

isomorphism

R has

surface

of c a n o n i c a l

HI(K)

an i s o m o r p h i s m

in

R \ D

be a R i e m a n n

can be r e g a r d e d Let

of

Groups

homology

= K(R).

l

(sub-)region

component

The H o m o l o g y

a × b

Pn+l

a subdivision

consisting

singular R

of

bordered

A finite

(i)

of c a n o n i c a l

I,

29A)

Corollary.

2.

of

polyhedron. if

Ch.

It is a f i n i t e

or c o m p a c t

is i n f i n i t e

}"

be an o p e n

By a c a n o n i c a l such

K \ P {Pn n=l

([AS],

polyhedron.

subcomplex

exhaustion

R

K = K(R)

is a c o m p a c t

of

simplex

Let

by

be an o r i e n t a b l e

sequence

border

= K(R)

R

a canonical

a canonical

Theorem.

Ch.

if

component

increasing

we d e n o t e

is an o r i e n t a b l e

only

is c a l l e d every

which

of c y c l e s sequence

in

by K,

a m. × a . 3

if

i ~ j, d e n o t i n g

by

]

of i - c h a i n s

a

and

b

([AS],

Ch.

I,

3iA). If hedron which

R

is a c o m p a c t

and t h e r e forms

exists

a basis

for

surface,

then

a canonical Hi(K).

K

is a f i n i t e

sequence

Hence,

ai,

d i m Hi(R)

bi,

orientable

poly-

i = 1,...,

g,

= d i m Hi(K)

= 2g.

The number If Cq_l,

g

R

q ~ i, then there exists in

i = i,..., of

is called the genus of the surface

is a compact bordered surface with

Hence,

The number

2C.

a canonical

contours

CO,... ,

sequence

ai, bi,

g, which, t o g e t h e r with all but one contours,

HI(K).

31D).

K

R.

q

dim HI(R) g

= dim HI(K)

= 2g + q - i

is again the genus of

Finally let

R

([ASJ, Ch. I,

R.

be an open surface,

so that

K

is an orient-

able open polyhedron.

Let

closure

is r e g a r d e d as a compact b o r d e r e d surface,

CI(D)

of

D

D

forms a basis

defines a finite p o l y h e d r o n sion to of

K.

K

be a r e g u l a r subregion of

K(CI(D)).

K(CI(D))

So we have a natural h o m o m o r p h i s m of R \ D

it

is a subcomplex

HI(K(CI(D)))

into

HI(K).

are compact, we see that the h o m o m o r p h i s m

is in fact an i s o m o r p h i s m and therefore that fied with a subgroup of

Since the

By applying a suitable subdivi-

if n e c e s s a r y we may assume that

Since no components of

R.

HI(K).

HI(K(CI(D)))

Turning to the region

is identi-

D

itself, we

have Theorem.

Let

D

be a r e g u l a r subregion of

is regarded as a subgroup of with one in In case that

HI(D)

HI(R)

R.

Then the group

HI(D)

by identifying every 1-chain in

D

R. D

is a canonical

subregion of

R, it is seen m o r e o v e r

is a direct summand of the free abelian group

HI(R).

See

[AS], Ch. I, §§31-32 for a d e t a i l e d discussion.

3.

The Fundamental Group 3A.

Let

mental group

0

be a point in

R, which is held fixed.

F0(R) , r e f e r r e d to the "origin"

The funda-

0, is defined to be the

m u l t i p l i c a t i v e group of h o m o t o p y classes of closed curves issuing from 0

([AS~, Ch.

I, 9D).

Every closed curve

as a singular 1-simplex. onto

y

from

the c o m m u t a t o r subgroup

phism of

y

Fo(R)/[Fo(R)]

[F0(R)]

of

F0(R).

By a c h a r a c t e r of an abstract group G

into the circle group

plex numbers of modulus one).

of

HI(R)

F0(R)

([AS], Ch.

G

F0(R)

to its h o m o l o g y class.

Under the natural h o m o m o r p h i s m the h o m o l o g y group

isomorphic with the quotient group

3B.

can be considered

We thus get a natural h o m o m o r p h i s m of

HI(R) , which takes the h o m o t o p y class of

Theorem.

0

is

modulo

I, 33D)

we mean any h o m o m o r -

(= the m u l t i p l i c a t i v e group of com-

The set of all characters of

G

forms

a group group by

with

thus

G*.

The

F0(R)*.

Let

such

§2.

4.

4A.

Let

s(z)

on

a

such

D

s

4B.

exists D;

and a

on

(ii)

on

an

s3

for

D, t h e r e

z

D

and

s(a)

=> ~

120

V

S

(i)

with

s

in

S

exists

an

s'

in

V; and

s

S.

A collection

if

-S = {-s:

tions.

The

character

surface

and of

every

F0(R).

reason

surface

An

extended

if

(i)

if

with -s

S

s E S}

pair

S

such

are

is g i v e n

V

S

has s"

and

with

A

A function

subharmonic.

on

Sl,

~ s3(z)

D

is c a l l e d

s2

in

for

disk

[e i t - z~

V

S

all

there z

with

in CI(V)

dt

a subharmonie

on

D

such

subharmonic

useful

by the

disk.

that

is a P e r r o n very

unit

functions

parametric

s(elt)Re

0

and

of e l e m e n t s

every

of

disk

is s u p e r h a r m o n i c .

superharmonic

and

(iii)

-~ <

semicontinuous;

the o p e n

m ~ n { s l ( z ) , s2(z)}

function

families

R.

superharmonic

of s u p e r h a r m o n i c

for e v e r y

S

Perron

Riemann F0(R)

a parametric

is i d e n t i f i e d

in

tions.

some

of

is l o w e r

exists

is s u b h a r m o n i c

a subharmonie

in

with

s(reit)dt

if it is b o t h

if

s

~ D

exists

family

(ii)

CI(V)

every

in

of

is c a l l e d

there

' s (z) ~ ~

for all

The

is d e n o t e d

isomorphic

of an o p e n

is a s u b g r o u p

D

A collection family

is

and

2C we h a v e

in a R i e m a n n D

s ~ +~;

in

that

is h a r m o n i c

a Perron

G

THEORY

s

0 < r ~ i, w h e r e

function on

of c h a r a c t e r s .

of

HI(R)*

in

subregion

F0(D)

be a r e g i o n

D

every

center

for

D

for

that

remark

group

Functions

function

~ +~

(iii)

shows

last

multiplication

character

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

POTENTIAL

Superharmonic

real-valued

the

Then

F0(D)

CLASSICAL

the

be a c a n o n i c a l

0 E D.

of

pointwise

theorem

with

D

that

character

to the

is c a l l e d

preceding

Combined

Theorem. R

respect

obtained

mlnorant,

that

s > s"

functions

family

of

on

there all

is a P e r r o n

superharmonic

in c o n s t r u c t i n g

following

D

i.e. for

harmonic

funcfunc-

Theorem.

Let

and set D.

S

u(z)

be a Perron fdmily of s u p e r h a r m o n i c functions on

= inf{s(z):

Then

u

D

is a h a r m o n i c f u n c t i o n on

([CC], p. 14)

4C.

In particular,

s u p e r h a r m o n i c majorant, on

s E S}.

D, then the set

if a subharmonic f u n c t i o n

i.e. a s u p e r h a r m o n i c f u n c t i o n

Su

of all s u p e r h a r m o n i c m a j o r a n t s

Perron family and the m i n i m u m element ceding theorem,

u

in

on

D

s

with of

has a s ~ u

u

forms a

Su, which exists by the pre-

is the least h a r m o n i c f u n c t i o n that m a j o r i z e s

u

on

D.

This is called the least h a r m o n i c m a j o r a n t of

u

LHM(u).

Then the p o i n t w i s e max-

imum

Let

uI

max{ul,

and

u 2}

superharmonie)

on

u2

D.

So, if

superharmonic majorant

5.

uI v u 2

max{ul,

void.

(resp.

Let

D

be a r e g i o n in

the set of s u p e r h a r m o n i c

is bounded below;

(ii)

min{ul,

R, whose b o u n d a r y

u2}) has a

then it has the

-S(-f;D). H[f;D]

of functions in

H[f;D])

in

S(f;D)

S(f;D)

and

s"

so that

numbers,

then

If

and

in

~D.

A function

S(f;D)

f

fl' f2

~ifl + ~2f2 ,

both

on

~D

D

on

H[f;D]

R

which

~D, we denote by

~

supremum)

s' ~ s" S(f;D)

on

D

for every and

S(f;D)

are h a r m o n i c on

D

is called r e s o l u t i v e if H[f;D]

= H[f;D].

The

and is called the S o l u t i o n

el' ~2

and

=

are nonvoid and let

are r e s o l u t i v e and f2 )

s

(iii)

S(f;D)

f.

max~fl,

(i)

denotes the R; and

with the b o u n d a r y data ~D

is non-

such that

(resp.

H[f;D]

are nonvoid and

in

We set

Since

and

r e s o l u t i v e and satisfy the following, H[.;D]:

D

S(f;D)

S(f;D),

H[f;D]

of the D i r i c h l e t p r o b l e m for (a)

in

S(f;D)).

common f u n c t i o n is then denoted by

Theorem.

a

S(f;D)

(resp.

and every

H[f;D] ~ H[f;D].

on

on

be the p o i n t w i s e infimum

S(f;D)

are Perron families,

f

s

c o m p a o t i f i c a t i o n of

for every

Suppose that both

(resp.

function

functions

~D

lim i n f D ~ z ÷ ~ s(z) ~ 0, where

lim infD~z÷ a s(z) ~ f(a)

of

(resp.

(resp.

u I A u2).

point at infinity of the o n e - p o i n t

both

u 2}

is subharmonic

(resp. the g r e a t e s t h a r m o n i c minorant),

For every e x t e n d e d r e a l - v a l u e d

~(f;D)

and

u2])

The D i r i c h l e t P r o b l e m 5A.

s'

D.

min{ul,

(resp. a subharmonic minorant),

least h a r m o n i c m a j o r a n t is denoted by

be h a r m o n i c on

(resp. m i n i m u m

and is denoted by

min(fl,

f2 )

in which we write

are real

are also

H[.]

in place

H I , i f I + ~2f2 ] = ~ i H [ f l ] + ~2H[f2 ], H[max{fl,f2}] (b)

Let

on

8D

with

the

sequence

: H[f I] v H [ f 2 ] ,

{fn}n~l limit

H[min{fl,f2}]

be a m o n o t o n e

sequence

= H[f I] A H[f2].

of r e s o l u t i v e

functions

f = lim

f . Then f is r e s o l u t i v e if and n n c o n v e r g e s . If this is the case , t h e n

{H[fn]} n:l ~

only

if

H[f]

= lim n H[fn]. (c) included and

Let in

f D

be a r e s o l u t i v e and

= H[f;D]

on

define DA

f'

~D'.

function

on

~D'

Then,

on

by

SD.

Let

D'

f'

= f

setting

H[f';D']

= H[f;D]

on

be a r e g i o n on

~D n ~D'

D'. ([CC],

5B.

A point

real-valued

a

in

function

~D

f

is c a l l e d

on

f

is c o n t i n u o u s

at

in

SD

superharmonic

a, t h e n

in

~D

if

(i)

neighborhood Theorem. Then

the

Let

The

(b)

of D

every

bounded

inequality

regular

on

D

= 0

and

: f(a).

is c a l l e d

is c a l l e d (ii)

implies

irregular.

a barrier

A positive

for a p o i n t

i n f z e D \ V s(z)

> 0

for

a every

a.

be a r e g i o n

in

R

statements

are

equivalent:

point

a

There

exists

There

exist

is a r e g u l a r a barrier

and

let

a

boundary

be a p o i n t

point

of

a

with

for the p o i n t

in

~D.

D. respect

to the

D.

(c) harmonic

function

If the then

s

this

: lim H [ f ; D ] ( z ) z÷a

is not

l i m z ÷ a s(z) V

folZowing

(a)

region

which function

for

~ lim sup H [ f ; D ] ( z ) . Dgz÷a

lim H [ f ; D ] ( z ) z÷a A point

if,

22)

~D,

lim sup f(b) ~Dgb÷a If

regular

p.

a

5C. constant Theorem. tinuous

u

on

connected

is a r e g u l a r

A region positive If

D

function

a neighborhood V nD

such

component boundary

D

in

R

of point

V that ~D of

is c a l l e d

superharmonie

function.

is a h y p e r b o l i c

region

on

~D

of

is r e s o l u t i v e .

a u(z)

and ÷ 0

containing D.

([CC],

hyperbolic

in

R, t h e n

([CC],

a positive

p.

as

super-

z ÷ a.

a

is a c o n t i n u u m ,

p.

23)

if it c a r r i e s

every 27)

bounded

a non-

con-

5D.

Let

D

be a h y p e r b o l i c region in

R

the set of all r e a l - v a l u e d bounded continuous is a linear space over

~

p l i c a t i o n of functions. the s u p r e m u m n o r m every

f

in

function

H[f;D]

on

on

on

D.

functions on

ZD.

be This

It becomes a Banach space if we equip it with If(b)l.

By the p r e c e d i n g t h e o r e m

is r e s o l u t i v e and thus determines a h a r m o n i c D.

In view of Th.

Cb(~D,~)

5A, the map

f + H[f;D]

is a

into the space of bounded h a r m o n i c functions

D, which is o r d e r - p r e s e r v i n g , 0

Cb(~D,~)

under the usual a d d i t i o n and scalar multi-

llffl = suPbE$ D

Cb(~D,~)

linear map from

and let

i.e.

For every fixed point

a

is a bounded linear functional on

f ~ 0 in

D

Cb(~D,~)

on

~D

implies

the map with

H[f;D]

f ÷ H[f;D](a)

rH[f;D](a) i ~ JJfll .

By Riesz's r e p r e s e n t a t i o n t h e o r e m there exists a unique n o n n e g a t i v e D ~a' on ~D such that

regular Borel measure,

(i)

H[f;D](a)

= I

f(b) d ~ ( b ) ~D

for every H[I;D]

f

< i

in

Cb($D,~)

with compact support.

on

~D

solutions,

~D(~D)

The inequality

It may h a p p e n that the Dirichlet D p r o b l e m has only the trivia] solution. In that case the m e a s u r e s wa vanish i d e n t i c a l l y for all a in D. When the Dirichlet p r o b l e m has nontrivial

shows that

< I.

the measure

Da

is called the harmonic m e a s u r e

at the point

a

(with respect to the r e g i o n

A function

f

on

Theorem.

~D-summable. a

~D

is r e s o l u t i v e

D).

if and only if it is

If this is the case, then the equation

(I) holds.

([CC],

p. 28) We now slightly extend the d e f i n i t i o n of

H[f;D].

Namely,

be an extended r e a l - v a l u e d

function defined on a subset of

the b o u n d a r y

f0

We call

f

~D

and let

be the r e s t r i c t i o n of

r e s o l u t i v e with respect to

sense d e s c r i b e d above,

i.e.

f0

is

D

if

f0

H[f0;D]

the b o u n d a r y data

6.

H[f;D]

f

to the set a

~D.

in the

in

D.

to denote

of the Dirichlet p r o b l e m for the region

D

with

f0"

Potential Theory 6A.

Let

let

including

is r e s o l u t i v e

D - s u m m a b l e for some ~a

If this is the case, then we again use the symbol the solution

f

R

ga(Z)

In the rest of §2, = g(a,z)

R

denotes a h y p e r b o l i c Riemann surface.

be the Green function for

This is c h a r a c t e r i z e d by the following

R

with pole

a

in

R.

Theorem.

For e v e r y

harmonic

function

(AI) harmonic z

in

in

R

in any p a r a m e t r i c function

V, w h e r e (A2)

for

ga : H [ g a ; D ]

6B. we

a

u V

on

every on

V

exists

satisfying disk

V

subregion

center

the

of

R

positive

super-

conditions:

ga(Z)

with

D

a unique

the

with

such that

is i d e n t i f i e d

a

there

exists

= - l o g Izl + u(z)

open unit with

disk

< i};

we h a v e ([CC],

For a n y g i v e n

finite

positive

regular

f : | g(z,w) JR

d~(w)

Borel

a

f o r all

{Izl

a ~ CI(D)

D.

measure

p.

~

32)

on

R

set

U~(z)

for

z

in

R.

everywhere

We c a l l

infinite.

function

on

Moreover

we have

Theorem. only

if

does

not

A subset

U~

E

of

on

E.

A countable a condition

a subset

A

of

R

condition

holds

on

Theorem.

Let

u

A

If

U~ £ u R.

As

exists

sets

points

on t h e

closed

D

R

if a n d

potential

set

if t h e r e to

is

exists

+~

E ~ A

a

identi-

a polar

(abbreviated

of

Z.

35)

is a g a i n

set

of

in

Every

p.

it is

superharmonie

support

is e q u a l

superharmonic

p.

~, u n l e s s

set

~.

a polar which

a polar

at t h e

q.e.

([CC],

of

([CC],

of p o l a r

be a p o s i t i v e

on

R

closed

quasi-everywhere

except

u

6D.

on

union

if t h e r e

support

is c a l l e d

holds

a potential. everywhere

R

function

the

o n an o p e n

measure.

by

is a p o s i t i v e

outside

the

positive

generated

U~

is h a r m o n i c

intersect

superharmonic

say that

potential

is h a r m o n i c

by a u n i q u e

6C. positive cally

R, w h i c h

D

it t h e

So a p o t e n t i a l

A potential

generated

We

there

z + ga(Z)

set.

to q.e.)

such that

on

the

E. function

support

on

of

R

and

U~

~, t h e n

U~

we h a v e

the

37)

f o r the r e g u l a r i t y

of t h e

Dirichlet

problem

following Theorem. set.

Then

([CC],

p.

6E. u

on

R

Let

D

the

be a r e g i o n

irregular

in

boundary

R

such that

points

of

R\ D

D

is n o t

form a polar

a polar set.

42)

For a subset

E

we

function

define

the

of

R

and

a positive

R[u;E]

on

R

superharmonie to be t h e

function

pointwise

infimum

of p o s i t i v e

u(z) The

on

E.

(resp.

We R uE

function

Let

function

on

(a)

v

H[u;R\E] points

of

is not

R

of

be a c l o s e d

u

set

v = u~.

R\ E

s

on

R

such

that

s(z)

and to

relative in

Then

R,

the

u

on

E

E.

a positive

following

function

u

to

on

superharmonic

hold:

R, w h i c h

except

R.

at the

is e q u a l

irregular

to boundary

R \ E. is the

smaller If

6F.

smallest

than E

Theorem.

Every as the ([CC],

q.e.

positive on the then

is d e t e r m i n e d

Finally

uniquely

u

is c o m p a c t ,

E, w h i c h

R.

function)

and

functions

u~(a) = lim inf R[u;E](z) for all a in m z÷a R [ u ; E ] ) is c a l l e d the b a l a y a g e d f u n c t i o n

is a s u p e r h a r m o n i c

v

(e) on

E

on

(b)

§3.

set (resp.

the r e d u c e d

Theorem.

on

superharmonic

we

state

positive

set

uniquely.

the

for

some

([CC],

decomposition

superharmonic

function

on

R

that

E.

v : U~

sum of a n o n n e g a t i v e p.

superharmonic

p.

nonnegative

theorem

function harmonic

measure

43)

on

of F. Riesz: R

can be w r i t t e n

function

and

a potential

41)

DIFFERENTIALS

7.

Basic 7A.

Definition Let

R

be a R i e m a n n

differential

~

b dye

(resp.

2-forms

local

variable

are

on

= x

z

complex-valued

disks

Ve

and

R

is d e f i n e d c dx dy + lye, "

functions

V6,

surface.

on

for

that

with

each

(i)

z (V)

respectively

(resp.

as a c o l l e c t i o n

), one such

A first

of 1 - f o r m s

parametric

a~

and local

second)

and

(ii)

b

a dx

disk

V

(resp.

for any

variables

order

z~

two

+ with

c a)

parametric

and

z6,

we

have ~x a ~

aB : on

V

8x

~xB

N V 6.

(resp.

+ • ~Y~ D~Z-~6' Such

cdxdy).

a differential A differential

(or a c k - d i f f e r e n t i a l ) , are of class Basic as

follows:

~(x,y)

~ + b ~Y~ b8 = a ~ Y 8 ~YB

k : 0,

(resp.

on

i,...,

R

) = c~(xB'Y~)

is e x p r e s s e d ~

cB

symbolically

is said

if all

the

as

adx + bdy

to be of c l a s s coefficients

Ck

of

C k.

operations

on f i r s t

order

differentials

~ = adx + bdy

are

10

(a)

multiplication

(b)

complex

(c)

conjugate

differentia3:

(d)

exterior

differentiation:

is of class (e) for

b.dy 3

e x t e r i o r product: j = i, 2.

~b/ax

7B.

order

for

Let are

be a first

y

function

assumed

value

of the

each

being

arc.

borhood,

V

R.

result ~

independent arc

on

with

The defined

being

y

integration similarly.

value

again

borhood,

then

the

of line

ease

~3.

3

:

a . d x +

if

dm = 0,

(~f/~x)dx +

follows, Let

arc

variable

and

local

y

arcs

and

~ = a d x + bdy is i n c l u d e d

z = x + iy the

and

in

is

integral

fy m

subdivision

order

coordinate

of s u b a r c s and we

neigh-

yj

so that

The

integral

set

we choose.

by l i n e a r i t y . C0-differential

be a s i n g u l a r

coordinate

is d e f i n e d

number

parametrization

Yj

1-chain

A

and

in a s i n g l e

neighborhood

of the

let

(t))dt,

variables

a finite

of a s e c o n d

~ = edxdy

2-simplex

neighborhood

V

by a d i f f e r e n t i a b l e

with

map

in

local

(t,u)

is

R.

If vari-

÷ z(t,u),

we d e f i n e

ff A the

in w h a t

0 £ t £ i, t h e n

D

to any

Namely,

0 £ u £ t £ i, t h e n

parametrization.

(=

smooth.

If the

included

coordinate

in a s i n g l e

z = x + iy

R.

if

to be c l o s e d

~ : df

+ b(z(t))

into

independent

extended

it is i n c l u d e d able

when

C 2.

and

local

of b o t h

is not

we d i v i d e

is t h e n

if

Here

Y

of

is said

exact

= x(t) + iy(t),

is in a s i n g l e

the

~a/~y)dxdy

formula

If the

then

yj

f~ = (fa)dx + (fb)dy;

a2bl)dXdy

-

of c l a s s

(a(z(t))at(t)

the

~

f

cl-differential

by the

(alb 2

:

to be p i e c e w i s e

neighborhood t ÷ z(t)

is d e f i n e d

~i~2

be an arc on

order

by

d~ = ( $ b / ~ x -

It is c a l l e d

some

always

a coordinate defined

f:

9" = -bdx + ady;

cl-differential

= ~a/ay.

($f/~y)dy)

1-chains

functions

~ = a d x + [dy;

CI;

A first i.e.

by n u m e r i c a l

conjugation:

being When

we h a v e

m =

ff O £ u £ t £ 1

independent A only

integrals,

is not

.

c v z ) ~

of the

the

choice

included

to a p p l y

dtdu,

a suitable

result

of

local

in a s i n g l e

being

variables

coordinate

subdivision

independent

of

to

A

and neighas

in

subdivision.

11

The integral of

~

over any singular

arity, where each simplex definitions

any

first

8.

The Class 8A.

order

Let

+ Ibl 2) dxdy

cl-differential

on

R

and

any

singular

and its Subclasses

~

be a first order cl-differential.

II~II = II ~ * F I : FI(R)

with

function

co

is a second order cl-differential.

We denote by

2-chain

×,

~*

= (lal 2

Then

We set

= II (lal2+ IbI2)dxdy"

the totality

11~ll < ~.

(2) defines

With these

formula

F

(2)

co

is assumed to be differentiable.

we have the Stokes

for

2-chain is then defined by line-

Then

a norm.

F1

of first order cl-differentials

is a linear space,

Indeed,

in which the

this norm is induced

from the

inner product (3)

(~I'~2)

for

~i' ~2

in

F I.

For a set

the set of conjugate in

FI

(resp.

if

analytic)

We denote by

d~ = 0

differentials

FI

(resp.

8B. face

FI(R)

belongs

the bomder in

C2

R

Theorem. (resp.

the class of closed

is the interior of a compact bordered

sur-

(resp.

F I.

R

is the set of are of class

have the obvious meaning.

C I up

We say

e

and that

with

f = 0

~ on

F~0(R) belongs

~R.

(3), which defines

FI(R) *) in

FI(R)

whose coeffieients

FI(R)

to the class

F~0(R) C

It is

dif-

on

and

DR

inner product

are closed.

exact)

C

f

FI

to the class

first order differentials m

~*

A*

A differential

The set of all harmonic (resp. FI). is denoted by F hi a

Then we use the following notation:

to the border. that

and

A.

co* = -i~. in

F I)

~

we denote by

in

e

belonging

Suppose that

R.

of differentials of elements

if both

and

C

ferentials

A

differentials

is called harmonic

called analytic

= IS ~I~2"

(resp. FI(R).

if

~ • F~(R)

and

~ = 0

along

to

FI~(R) if ~ = df for some eu i The class F (R) is equipped with the the o r t h o g o n a l i t y

F~0(R)) ([AS],

relation there.

is the orthoeomplement Ch. V, 5A)

of

FI(R)*e

12

8C.

If

R

is an o p e n

Riemann

F~0(R)

(resp.

compact

support.

In this

Theorem.

FI e0 in

is o r t h o g o n a l

of

F e0 I *

norm

FIe0 : F~0(R))

FI .

ease

([AS],

8D.

We d e n o t e

(2),

so that

by F

the

F

fundamental

Theorem.

the

If

result

BE. shall

Among

use the

direct

basic

sum in the

(4)

and

formulae

following

to h a v e

hood

define

respect Fe

to the

CI(FI), e

=

Fe0 the

Fh

is the

%

c* '

following Fhl

such

([AS],

decomposition

denotes

F

the

that

Ch.

of

V,

9A)

F, we

orthogonal

= re0

re0 ~{

•

we h a v e

If a f i r s t V

order

Cl(rl).

:

([AS],

Ch.

V,

10C)

C

if a.

cl-differential,

of a p o i n t

a singularity

of

FI = cO with

and D i f f e r e n t i a l s

neighborhood

harmonic)

We

with

there exists a ~i in i Fh can be i d e n t i f i e d .

C

9A.

FI

class

in w h i c h

r

Cycles

F~)

is the o r t h o c o m p l e m e n t

of

space.

the

~ rc

(5)

9.

by

(resp.

F:

r = Fe0

As a c o n s e q u e n c e ,

FI c

for o r t h o g o n a l

later,

space

we d e n o t e FI c

: (Fe*) ± , w h e r e the c l o s u r e and in the space F. We set Fh = F c n

, then

Fh

in

Fc0

concerning

~ E Fc~Fc*

and

completion

is a H i l b e r t

~lll = 0; n a m e l y ,

lJw-

F I* e 6C)

V,

= CI(F ~0 ) , F c : (Fe0*) ± , and orthocomplementation are t a k e n The

then

~

we have

to

Ch.

surface,

set of

~

at

a

a.

is a n a l y t i c

It is c a l l e d

in The

R

say

except

singularity

(resp.

removable

R

is d e f i n e d

a, t h e n

is c a l l e d

harmonic) if

~,

at

in some

~

analytic

deleted

is c o n t i n u e d

in a is said (resp.

neighbor-

to the

point

a

as a c l - d i f f e r e n t i a l . Take

a parametric

consider

a holomorphic

fdz

some

for

local

variable

disk

function

The

set

~a(~)

order

on f

on

about

a point

V \ {a}. V \ {a},

a

in

So we h a v e i.e.

in t e r m s

R

= inf{k:

c k ~ 0}

is i n d e p e n d e n t

of the

[

ckzk,

and call choice

9 =

0 < Iz[ < 1. it the

of local

order

of

variables.

~

at

and

of the

we h a v e

f(z) : We

< i}

differential

holomorphie z

V = {Izl

a.

Moreover,

13

the c o e f f i c i e n t the

C_l , the r e s i d u e

local v a r i a b l e s

and is d e n o t e d

Resa(~) for

any

point

cycle a.

0.

The

ential

on

R

V

by

a

Res a (~). = ~

is

subset

with

In fact,

number

1

removable

of

we h a v e

with

if

and

If

~

only

if

such t h a t

the

Va(~) cl-differ -

singularities exists

to

may be con-

is a c l o s e d

R, t h e n t h e r e

singularities

respect

singularities

of a n a l y t i c

of

independent

y

fact:

with a finite number

T

a, is a l s o

with prescribed

of the f o l l o w i n g

differential

at

winding

at

off a compact

~

= C_l

having

differentials

by m e a n s

holomorphie monic

in

singularity

Analytic

structed

Ch.

y

of

and is

a unique

T - ~ E Fe0

har-

([AS],

V, 17D).

9B. p, q

We c o n s t r u c t

be two d i s t i n c t

Suppose metric

first that t h e s e disk

positive Let

v

V = {Izl

numbers

e ~ i

define

p,

rl, r 2

= d(ev)

we use the r e s u l t T - ~ E Fe0.

q

q

for

and

and

which

c

and

to c y c l e s

an arc

I~lJ,

I~2[

of the f u n c t i o n let

e

E 0

on

and c h o o s e

log{(z-~2)/(Z-~l)}

R \ {Izl

on

< r2}.

= 0

R

such

F u r t h e r we

otherwise

y

in

p

for

R.

Then

~

with

differential

not p a s s i n g

T = 2~i(c x y).

I

Let q.

< r I < r 2 < i.

be a C 2 - f u n c t i o n

Izl < I, and

1-cycle

R. to

~ = { ( z - ~2 )-i - ( z - ~ l ) - l } d z

by s e t t i n g

for any

on p

in a s i n g l e p a r a -

~2 = z(q),

in 9A to find a u n i q u e h a r m o n i c

Thus,

joining

are c o n t a i n e d

satisfy

and

rI £

related c

~i = z(p),

branch

{IzI £ r I}

a differential

R

Set

r I < Izl < i

on

Izl < rl,

in

< i}.

be a s i n g l e - v a l u e d

in the a n n u l u s that

some d i f f e r e n t i a l s points

and

q, we h a v e

([AS],

Ch. V, 19C)

Y If metric

p,

j = i,..., within Tj T and

q

disk,

and

c

n, so t h a t

a parametric

for e v e r y

are not n e c e s s a r i l y

t h e n we d i v i d e

j

is a h a r m o n i c

each

number

of s u b a r c s

cj,

ej

joins a p o i n t

Pj-I

to a p o i n t

pj

For e v e r y

P0 = p

and set

differential

T - ~ E Fe0.

in a single p a r a -

into a f i n i t e

disk with

as a b o v e

contained

c

with

1-cycle

and

Pn = q"

~ = [j ~j

and

singularities y

We d e f i n e T = [j ~j.

o n l y at

not p a s s i n g

p

~j

and

any of the

and

Then q, pj's,

we h a v e

I

T = I "f

The d i f f e r e n t i a l

y T

Q = [ I ~. = [ 2~i(c. x y ) = J Y J j is seen to d e p e n d

2~i(cx7).

o n l y on the h o m o l o g y

class

of

c.

14

We w r i t e

T(c)

~(c) Then,

in p l a c e

¢(e)

= ~ ( c ) + ~--~-~. at

p

The

We

can

We

with

e

has

iT--~*)/2.

differentials

only

-i

set

: (T--~+

analytic

¢(c)

construction

our

works

~(e)

definition

Suppose

Moreover,

to

we

~(c)

residues

by l i n e a r i t y .

C F h.

and

are

see that q

is a cycle.

T(c)

~(c)

differentials

extend

belongs

Further

on

R

two

singularities,

i

respectively,

and

and

T(c)

simple and

~(e)

holomorphic.

above

1-chains c

and

and

is e v e r y w h e r e

resulting

T.

: (T(c) + i T ( c ) * ) / 2

both

poles

of

Fe0 ,

that

of

T(c), has

~

belongs

in the

~(c)

%(e) this

since

T(e)

even

and

case

are %(c)

no

p = q

compact

to

F h N Fc0

the

and

to a r b i t r a r y

case.

has

then

holomorphic.

~(e)

singularities

is the

and

everywhere

if and

Then

support

¢(c) and

only

E Fa

since

if

and T(e)-

(= Fh0 , by d e f i n i t i o n ) .

By s e t t i n g

~(c) = !~T(e), we get the Theorem.

following If

harmonic

c

is a c y c l e

differential

on

o(c)

R, t h e n

in

Fh0

there

such

exists

a unique

real

that

(~,~(c)~) : Ic for

each

~

in

F . e

(6)

If

c

(~(c),~(y)*)

and

y

= I

are

~(c)

any

cycles

on

= c x y.

R, t h e n

([AS],

Ch.

V,

20)

Y iO.

Riemann-Roch 10A.

Here

Fix a c a n o n i c a l intersection n = i,..., harmonic

we

consider

basis

numbers g

(see

~(A n) m

a compact

of 1 - c y c l e s vanish

2B).

differentials

IB and

Theorem

except

By our

Riemann

An,

Bn,

the

cases

observation

and we h a v e

= A n × B m : @mn'

for all

IA

~(Bn) m

surface

n = i,...,

R

of g e n u s

g.

g,

for w h i c h

all

A n × B n = -B n x A n = i, in 9B m,

~(An),

~(B n)

n

= Bn × A m : -@mn'

are

15

S

A

Theorem.

(a)

a(A n)

and t h e r e f o r e (b) periods,

dim

a(An)

: I

o(B n) B

m

and

O(Bn) ,

= 0.

m

n = i,...,

g,

form a basis

for

Fh

F h = 2g.

dim

F a = g~ and

i.e.

(fA 1 ¢ .....

every fA

~

in

9).

Fa

([AS],

is d e t e r m i n e d Ch.

V,

by

its A-

24A-B)

g 10B.

A divisor

D

on

(7)

R

is a f o r m a l

finite

sum

D = n l a I + --- + nkak,

where

a. are p o i n t s J is c a l l e d a p r i n c i p a l funetion

at the

f,

i.e.

point

a,

in

R

and

divisor

D = (f)

n.

are

= ~aER

Va (f)a'

to be

inf{k:

is d e f i n e d

integers.

if it c o m e s

from

Such

a nonzero

where

Va(f),

c k ~ 0}

a divisor

D

meromorphic the

with

order

f(z)

of

f

=

~k ek(z - a)k" Let

D

divisors) tively. the

R.

The

makes

that

R. by

R

deg

function

We also

is a n o t h e r

(~)

on

for all

R

functions

f

dimension

of

divisor

meromorphic

of the

sueh

belongs

to

Hence

addi-

group

form

(7)

function

sphere,

that

we

deg

on

on see

is a

= (~'/~)

of

D2

L(D)

that

(7)

(f) by

defined

e'/e

E DO.

the

D I- D 2 space

by If

Thus,

(~)'s

belong

class.

integral

if

n.j => 0 For

is i n t e g r a l .

of all m e r o m o r p h i c

is a m u l t i p l e dim D, w h i c h

~

in 9A.

is a m e r o m o r p h i e

canonical

is e a l l e d

if the

differential

of

D.

depends

The

only

complex

on the

D.

divisor

~(D)

class

of the

Riemann

was

then

called

form

by

is d e n o t e d

D

in

differentials

a fixed

class

denote

R

L(D)

we t a k e

Z = (an).

we

~a(~)

differential, (e') - (~)

D

D

principal written

D/D 0 .

of a m e r o m o r p h i e

divisor

divisor

therefore

where

is a m u l t i p l e

of

of the

and

divisor

D

every

DO

ma (~)a'

on

class

For

group

so t h a t

DI

surface

in

group

meromorphic

the

same

divisor

nonconstant

= [aER

A divisor j.

D

meromorphie

and t h e

10C.

quotient

define

~'

covering

every

the

D, of a d i v i s o r

every

(resp.

Abelian

is c a l l e d

deg

Since

divisors

is a free

D/D 0

degree,

for

set of all

DO)

group

a complete

D = 0

formula

every

The

on t h e

funetion

(resp.

[j n..3

the

to one

D O ) be the

D

quotient

surface

is d e f i n e d R

(resp.

on

nonzero if and dim

e

D

we d e n o t e

such

meromorphic only

~(D)

that

(~)

by

~(D)

(~/~0)

= dim

( D - Z).

space

is a m u l t i p l e

differential

if

the

s0

on

is a m u l t i p l e

of

of D.

If

R, t h e n of

D- Z

with

16

Theorem

(Riemann-Roch).

divsor

Z

For

every

D

in

D

and

every

canonical

we h a v e

(8)

dim D = dim

(-D-Z)

- deg

D - g + i. ([AS],

10D. = dim

By s e t t i n g

Fa = g

D = -Z

(see Th.

10A),

69) for

in

canonical

Theorem. pole

There

at a n y

other

Proof.

any

(8)

that

This,

together

with

(8),

implies

function

point

We c a n p r e s c r i b e

in

R.

([AS],

further,

be a n y

dim

~(0)

in

Ch.

J

subset such

from

we h a v e

V,

f

which

has

the

a simple

location

of

28B)

we n e e d

infinite

a I .... , ag

funetions,

fact

27A)

Z = 2g- 2

Z.

as well.

J

the

a meromorphic

al,... , ag

Take

by

f

Let

points

constant + i

of

To p r o c e e d

Theorem. tinct

exists

prescribed

poles

10E.

divisor

using

V,

we have

deg each

(8) a n d

Ch.

J.

1 ~ dim

of

R.

that

Then

dim

Sinee

there

exist

9(al+.-.+ag) L ( - ~Ij~s-

(-~j~l- a~)j : d i m

dis-

= 0.

a.)~o c o n t a i n s ~(~j~l- a~)j + s - g

and t h e r e f o r e s

(i0)

d i m ~(

We now prove nothing first

the

to p r o v e ,

theorem

for

~(0)

then dim

by

_> g - s .

induction.

= F a = {0}.

If

g = 0, t h e n

Suppose

that

there

g > 0.

is

We n o t e

that dim

for

[ a.) j=l ]

each

a

in

g ~ dim

R.

~n fact,

(a 0 - Z)

= dim

(a 0 - Z) - deg

meromorphic the point conformally

( a - Z)

onto

~(a)

~ g- i

if t h i s

were

not

~(a 0) ~ d i m

(-a 0) - g + i = 2.

function a 0.

= dim

f

on

As n o t i e e d the

R

with

in 10B,

Riemann

Thus

sphere

F a = g. there

only

this

the

case

By

(8)

would

implies

that

g

some

dim

a0,

(-a 0) =

be a n o n c o n s t a n t

one pole--a

and thus

for

simple f

should

pole--at

would be

map

zero,

R

a

contradiction. Thus, take

for

a nonzero

s = i, w e h a v e

d i m ~ ( a I)

= g - I.

~i

Since

zeros

in

~(al).

the

of

If ~i

g > i, t h e n w e is f i n i t e

in

17

number

and

J

is i n f i n i t e ,

we can find

a2

Then

d i m ~ ( a l + a 2) ~ d i m 9(a I) - i = g - 2.

gives

us

d i m ~ ( a l + a 2) : g - 2.

f i n i s h t h e proof.

10F.

Let

..., ag aj,

in

91,... , %g R

g.

ii.

Cj : fjdz

Then

Cauchy IIA.

{Aj, Bj: R 0 : R\

and

Kernels Let

R

in w h i e h

the

is c l e a r

of

dim ~(al+...+a (fj(zk))

}l(a2) (i0),

~ 0.

this

this a r g u m e n t

to

F a.

g

from the d e f i n i t i o n s .

Take d i s t i n c t

disk

{Vj,

) = 0

z}

points

al,

for each p o i n t

if and only if

~ 0,

z k : z(ak).

on C o m p a c t

Bordered

Surfaces

be a c o m p a c t R i e m a n n

j = i,..., {%gl(Aj

which

one p a r a m e t r i c

det where

with

We h a v e o n l y to r e p e a t

be a b a s i s

and c h o o s e

j = i,...,

J

[]

We add one m o r e r e s u l t ,

Theorem.

in

Combined with

g}

be a c a n o n i c a l

UBj)}.

R0

surface basis

can be a s s u m e d

single boundary

contour

of g e n u s

g

of 1 - c y c l e s

on

and let R.

We set

to h a v e a c a n o n i c a l

form~

is of the f o r m

A I B I A ~ I B ~ I . . . A g B A - I B -I gg g' where

A[I (resp. ] g i v i n g the o p p o s i t e Let an are

tial and Fa

p, q e

in

¢(c) q.

B[ I) is the c u r v e o b t a i n e d f r o m ] o r i e n t a t i o n ([AS], Ch. I, 4OD).

be d i s t i n c t R0

on

joining

R

in 9B imply t h a t

p

q

and

with resides ¢ ( c ) - 91

is d e n o t e d

by

-i

p'

and

p'

A. and B.. Take ] ] an a n a l y t i c d i f f e r e n -

singularities

¢(c).

o n l y at

differential

The p r o p e r t i e s

of

_, r e s p e c t i v e l y - - a n d

has

p

¢i

in

¢(e)

has o n l y two p o l e s - - s i m p l e

is seen to be i n d e p e n d e n t

be a s i m p l e q'.

and

and

B.) by 3

not on any

to find a u n i q u e as

(resp.

poles

at

zero A-

of the c h o i c e

of

c

and

~p,q.

Then

is an a n a l y t i c at

R

and d e f i n e

(~(c) has

¢ ( e ) - 91

= ~p,q, y

So

10A,(b)

Next we t a k e d i s t i n c t

Let

q

the same A - p e r i o d s

noticed

periods.

in

to

as in 9B.

T h e n we u s e Th.

w h i c h has

points p

A. J

p, q, p' , q '

m = Cp,

q,

arc j o i n i n g ~

Let

p

to

is s i n g l e - v a l u e d

differential

q'.

points

Z0

on

R0 \ y

be the t o t a l

and

~ :

q

within

on

in

fz

R0 \ y

R0

and

set

9. R0

and not p a s s i n g

and t h e r e f o r e

with possible

~

singularities

sum of r e s i d u e s

(see 9A) of

only ~m

18 +

o v e r the r e g i o n (resp.

R 0 \ y.

right-hand)

Denoting

b a n k of

2~iE 0 = [ ] Since

¢(z +) - ¢(z-)

e q u a l to periods y'

%

joining

p'

¢~ + r ~R 0

We a l s o

and

~

to

E0 = R e s p , ( } m )

within

q

and

y

~y'

y'

Let

[

We d e n o t e

liB.

by

conjugate

of

R

@P,q

be a c o m p a c t R' (see

function

there

g' p o i n t s

dh

exist

has n e i t h e r

zeros

is

all the A-

if we t a k e any arc

- ~(p')

= [

%p,

y,

arcs on

R0

,q

,.

with

~y =

= I y CP',q''

bordered

= RU DRU~ [AS],

surface with nonvoid

the d o u b l e

Ch.

II,

3E).

In o r d e r to d e f i n e the C a u c h y k e r n e l meromorphic

because

side

Then

I y,

DR.

= ~(q')

be n o n i n t e r s e c t i n g

: q' - p'.

y, we h a v e

R 0 \ y, t h e n we h a v e

,(~)

Theorem.

Let

/3R0 {~ = 0

On the o t h e r hand,

the f o l l o w i n g

and

on

s e c o n d t e r m of the r i g h t - h a n d

So we h a v e p r o v e d

q- p

z

z ) the l e f t - h a n d

(¢(z+) - ¢(z-))w.

see that

+ Res

(resp.

J.y

vanish. q'

z

at the p o i n t

= 2~i, the

2%i fy ~. of

y

by

h

on

R', w h i c h

al,... , ag, nor p o l e s

in

at

of

R, w h e r e

boundary

R

is the

Let

g'

for

R, we t a k e a n o n c o n s t a n t

be the g e n u s

is h e l d fixed. R

Then,

of

R'.

by Th.

10E,

such that the d i f f e r e n t i a l

al,... , ag,

and

such that

d i m ~ R , ( a l + - . . + a g , ) = 0, where Let

~R,(.) {Aj, Bj:

such t h a t IIA.

denotes

the

space

j = l,...,,g'}

R 8' = R' \ { u j g I ( A j U B j ) }

We m a y also a s s u m e

that

dh

as w e l l as all the p o i n t s

Th.

1OA we c h o o s e

fA

9(-)

~n = 6mn

for

a basis

for the s u r f a c e

be a c a n o n i c a l

R 0'

R'

(see 10C).

basis

of 1 - c y c l e s

on

is a c a n o n i c a l

f o r m in the

sense of

contains

al,... , ag,

{~n: n = I,...,

m, n = i .... , g'

all the

zeros

and p o l e s

in its i n t e r i o r . g'}

of

R'

Fa(R')

We fix one p a r a m e t r i c

of

By use of such t h a t disk

m

{Vj, z} terms

of the

we w r i t e det

around

each of the p o i n t s

local v a r i a b l e s

z k = z(a k)

(hj(Zk))

~ 0.

for

Since

chosen

al,... , ag, as

k = i,..., d h ( a k) ~ 0

and e x p r e s s

}n = h n ( Z ) d z g', for

on e a c h

t h e n we have, k = I,...,

by Th.

%n

in

Vj.

If

10F,

g', we see that

19 @j(ak)/dh(a k) : hj(Zk)/dd-~hz(Zk) are well-defined

and that the matrix

$j (ak) ] j,k:l ..... g' is nonsingular.

So a basis

{~ n

:

n = i,...,

g'}

of

F (R')

is deter-

a

mined by the formulas ' ¢:(a~) ~ k:l dh(a k)

:

(li)

$

CJ for

j : i,...,

from

g'.

If we take three distinct

and draw disjoint

such that

2y' = p - p'

is holomorphic

R~ \ y".

p ~ q, q"

shows that

and

we set

z

F(p,q), q

and

by u s i n g

:

(12),

, q

in R~

p,

p'

IIA implies that

,, Sp, ,p'

defined by either

side of this equa-

and holomorphic

is holomorphic

Differentiation local variables

@q,,,q(p) = f(z,q",~)dz

3 Then,

is distinct

within the region

points

7"

in

jointly

of

is locally holomorphic C

which

in the form

R 0' \ y'

in F(p,q)

q ~ p'.

Cq,,q(p)

Denoting by tively,

in

It follows that

and

~y" = q - q", then Th.

p, Sq",q

we see that the function

p

y'

writing this

(12)

when

arcs

and

Sy, ~q,,,q : Sy,, ~p,,p.

tion,

q" E R N R 0'

We now fix a point

al, .. ., ag,.

R~ \ {q"}

~k

F

in

for

p

in

in p

and

with respect q

p

when and

q to

q ~ p, q".

q, respec-

and also

J we h a v e

3

J

fA +IIp •

IA

p'

.

q)d p!

J

=I

A. ~ ]

the last equality vanish.

As

p

1

q" Cp,,p d~ =

I

A. Cp''P = 0, ]

sign being valid because

is arbitrary,

we conclude

all the A-periods that,

of

for every fixed

$p,,p p

in

20

the region

R N R~,

fA ~q~q,,

(13)

'q

=

0.

J

A similar computation

shows that

Ip'

B. ]

'

B. ]

As we shall see below in IIC, an application

'P" of Th.

IIA easily

shows

that

f

(14) for

j = I,...,

g'.

@p, : -2~i @j B. 'P p' 3 Thus we have arrived at the following:

fB. ~q~q"'q :-2~i%j

(15) for

j = i,...,

g'

We finally set ~(p,q) Since

q"

and all

everywhere on

R

on

+i

Cauehy kernel, ~(p,q)

aj

at the point

we claim that,

q.

of the differential (13) follows

=

B. J

in

jointly

and

q ÷ ~q~(p,q)

(14) and

~q(~q"

'q

in

p

p, the function p

q q

has a simple pole

R 0'

in

along

A~~

easily that the period along any

by use of (ii),

~qm(p,q)

is holomorphic

To finish the construction

To see this , we fix

First,

B. J

p ÷ ~q,,,q(p)

for every fixed

the periods Secondly,

%q,,,q(aj)

is holomorphic

Moreover,

is single-valued. from

j:l R,

%q,,,q(p)

p ~ q.

~ ' ~q,,, dh q(aj)~j(p).

-

belong to

R, while

except at

with residue

= @q,,,q(p)

of the q ÷

and compute and A. J

B~. vanishes.

(15) we get

(P)) -

k=l

dh

(ak)mk(P)

~' ~j = -2~i[~j(p)

-

~-~(ak)~ k(p)]

: O.

k=l Hence,

~qW(p,q)

valued function.

has only null periods

and so

Summing up these observations,

which is the main objective

of this section.

q ÷ ~(p,q)

is a single-

we get the following,

21

Theorem. Riemann

There

(BI) tial

exists

surface

on

R

a differential

such

for e v e r y R

with

(B2)

q

only

R,

p

in

and

set

of

q ÷ f(z0, q)

is a m e r o m o r p h i c

with

residue

IIC.

within

the

verify

ourselves region

point,

distinct

b,

fB ~ p ' , p

in

To p r o e e e d

Aj,

Bj

except

B"

b"

A I.

= B \ B'

(i)

vertex

differen-

residue

variable

z

+i;

in a

z 0 = z(p);

having

a simple

let

V

then

pole

at

3B"

B,

at a s i n g l e

R~, (14)

slightly say

and

(ii)

is equal

B \ [b} to

show

~i"

be a p a r a m e t r i c

take

of

to

BI

curve, AI

of the p o l y g o n of

of g e n e r a l i t y ,

move

with

side

loss

p'

does

the

end p o i n t s

first

intersects

= -2~i 'P

CI(V)

We

closed

only

~p,

Without

a simple

left-hand

B

We t h e n

we h a v e

j = i.

B

the

containing

be the

case

we h a v e

further,

closure

B, r e s p e c t i v e l y , and

with

with

R

(14).

to o b t a i n

Then

f its

q

a local

on

equation

f r o m any

R~.

and t h e r e f o r e

such t h a t

is a m e r o m o r p h i c

pole--at

= f(z,q)dz

function

to the

R' \ y'

the c o n d i t i o n s :

entirely

simple

~(p,q)

the

satisfying

lies

bordered

+i.

We now

we r e s t r i c t

on a c o m p a c t

p ÷ ~(p,q)

R, we c h o o s e

neighborhood

p

p

in

one p o l e - - a

for e v e r y

~(p,q)

that

not m e e t small

point B'

arcs b

= b " - b'

and

A'

and

so that

disk

y'

the

B'

of

AI

and

in

= b' - b".

fact

at

all

contained

~B'

F r o m the

centered also

V.

b cycles and

Let

Thus,

B = B' + B"

b'

setting

follows

that

B

We n o w c o m p u t e First

we note

from

y'

'P

the two that

So Th.

the IIA

B'

integrals arc

B"

implies

to deal

a modified deformation the

same

with

polygon

the

V, the

end p o i n t s

such

of arc

that

' R 0. A' A"

number

over

member

polygon

R~

separately. and

is d i s j o i n t

p, ~b',b"" B',

on the

In fact is c h a n g e d

curve

+i

Setting

= (AIXA') + A" , we c o n s t r u c t

with our

other

by m e a n s

is d i s j o i n t

is a c l o s e d Ai

of w i n d i n g

'P

right-hand

in the

b' ~ P ' ' P =

integral

in p l a c e

within

on the lies

B"

that

B" ~ P ' ' P = In o r d e r

'P

to a n o t h e r from respect

hand,

we use

of a c o n t i n u o u s

B'

arc and

A"

with

A' - A "

to the p o i n t

new polygon,

' out of R I,

b".

22

fA~ ~p,,p

analytic

differential

on

residue

+i

and

AI

' A I.

R 0' just by replacing the arc y', we have

by

R'

V

= fAl ~p,,p having

-i, respectively,

A{, A2,... , Ag,.

the cycles

Since

does not intersect

: 0.

Thus,

Let

simple poles and having by Th.

Cb", ' b'

at

b'

IIA applied

be the

and

null periods

with

b"

with

along all

to the polygon

R~, we have

I

B' CP' ,p = I .¥, Cb",b'

and hence

B

'P

B'

B"

Our construction of Cb",b' larities and has null period

'P

'y'

'

shows that @b' ,b '' + ¢b",b' has no singualong A2,...Ag,. As for the period along

AI, we have

f

!

f

AI ((~b',b" + Cb",b' ) =

=f Since the differentials we eonclude

in

that

,

AI Cb" ,b'

A' -A" ~b" ,b'

F a (R') T

f

=

A I @b",b'

~b',b" + ~b",b'

:

f

A~ Cb" ,b'

-2wi.

are determined

by their A-periods,

-2~i~i , as was to be proved.

[]

NOTES Most results and Chapters information results

come from Chapters

i, 3, 4 of Constantinescu can be obtained

in 10E,

from Kusunoki

and Cornea

from sources

10F and in Subsection

[42]

I, II, V of Ahlfors

(see also Hurwitz

[CC],

indicated

and Sario

so that more

respectively.

ii (Cauchy kernels) and Courant

[AS] The

are taken

[35] for this matter).

CHAPTER II.

M U L T I P L I C A T I V E A N A L Y T I C FUNCTIONS

In our later d i s c u s s i o n of Hardy classes an important role will be played by a certain family of m u l t i p l e - v a l u e d multiplicative

analytic functions.

analytic functions having

analytic functions called

These are defined as m u l t i p l e - v a l u e d

s i n g l e - v a l u e d modulus.

The first objective

of this chapter is to give a precise d e f i n i t i o n to such functions. fact, we will define t h e m in terms of two equivalent notions: bundles and characters

of the f u n d a m e n t a l group.

that, on a compact b o r d e r e d Riemann surface, n o n v a n i s h i n g bounded h o l o m o r p h i c observe the order structure

sections.

In

line

We shall then show

every line bundle admits The second purpose

is to

in the space of h a r m o n i c functions,

leading

to the so-called i n n e r - o u t e r f a c t o r i z a t i o n of m u l t i p l i e a t i v e analytic functions.

§i.

M U L T I P L I C A T I V E A N A L Y T I C FUNCTIONS

i.

The First C o h o m o l o g y Group IA.

Let

{Vi: i E I} {~i~}

over

modulus one, with

R

be a Riemann surface.

of V

R, we denote by with values

i.e.

V. N V. ~ ~ l 3

~ij

in

~

of complex numbers of

is a s s i g n e d to every pair

i, j

in

I

6ij<jk : ~ik V inv.3 N V k I ~.

Two l-cocycles

said to be e q u i v a l e n t i

in the group ~

V =

the t o t a l i t y of l-cocycles

in such a way that

(i) if

For every open covering

ZI(v;~)

in

I

such that

{~ij}

{~ij} , in

and

if there exists an element ~j

= 8~l~ij@j

if

Vi AVj

@i ~ ~.

is then defined to be the set of e q u i v a l e n c e classes

in

ZI(v;~) T

The set in

HI(v;~)

ZI(v;~),

which forms a group with respect to the p o i n t w i s e m u l t i p l i c a t i o n , {~ij]{~ij} = {~ij~ij}. Next, take any two open coverings {V~,:

i' @ I ~ }

fining map

~.

of

R

such that

V'

V = {Vi: i E I} is a r e f i n e m e n t of

We may regard the refining map

~

are

for each

and V

i.e.

V' with re-

as a map from

I'

24

into

I

so that

ZI(v;~).

If

the c o v e r i n g to p r e s e r v e

]'

=D V'i' A V!]' ~ ~

V'.

choice

of r e f i n i n g

that

varies

of

with

lB.

Theorem.

Proof.

The p r o o f

definitions. consists in

V of

admits

coverings

V, we m e a n

a finite

Vi(v) N V i ( v + I )

is c a l l e d

obtained order;

from

e.g.

y

of

c

if t h e r e

for e v e r y When

~.

(a) Let

~ E HI(R;~) c

of

~

and

R

case,

faithful

in a s i m p l y R

if

covering

So we r e s t r i c t

elements

c

on

..a m

of

V2,

it is

ourselves

covering,

m-l.

in

V

if it is changing

{VI,V2,V3}.

the On

we m e a n a p a r t i t i o n

say

oi,... , am, of

c

with

c : al..-o m

y, each s u b p a r t i t i o n

so

has an ob-

is said to c o v e r c

say

A c h a i n of

without of

R

arcs,

V

VI,

of e l e m e n t s

~ : I,...,

of a p a r t i t i o n •

(i)

and that e v e r y o p e n

y = {Vi(1),...,Vi(m)}

c = oI

some

connected

By a c h a i n of a g i v e n

of an arc

R.

is s e p a r a b l e ,

is a s u b d i v i s i o n

to

of

(it) for each t r i p l e

Since

we say that of

F0(R)

We b e g i n w i t h

is c a l l e d

of c o n s e c u t i v e

is a d a p t e d

and

group

steps.

~ = {Vi(1) , .... V i ( m ) }

First we d e f i n e

and

with

some of the

a partition

to some s u b d i v i s i o n

limit of this

isomorphic

is f a i t h f u l .

A subpartition

In this

c : ~l..-Om

adapted

of

of a c h a i n

number

A chain

exists

when

is c a n o n i c a l l y

for e v e r y

by r e p e a t i n g

c : a l a 2 - - - a m.

s y s t e m of g r o u p s

y : {Vj(1),...,Vi(m)}

~ @

by a p a r t i t i o n

vious m e a n i n g . c

sequence

into a f i n i t e

that

V" This

group

is i n c l u d e d

{VI,VI,V2,V3,V3,V3}

the o t h e r hand,

if

cohomology

a faithful

which

a subdivision

on the

the f i r s t

sets and

in the proof.

such t h a t V

V

open

it is c o n n e c t e d . R

Moreover,

The d i r e c t

is c a l l e d

into s e v e r a l

V IUV 2 UV 3

has a r e f i n e m e n t

to f a i t h f u l

does not d e p e n d

of the f u n d a m e n t a l

connected

whenever

R.

is seen HI(v;~)

: I(V",V')o I(V',V).

a direct

HI(R;T)

An o p e n c o v e r i n g

R

I(V',V)

I(V',V):

over

~.

is d i v i d e d

not h a r d to see that covering

in

F0(R)*

the u n i o n

{~ij } + {~i'j" }

I(V",V)

of

and

is defined.

a l-eocycle

we get a m a p

forms

The g r o u p

of s i m p l y

subset

relation,

HI(R;T)

group

'

and is a g r o u p h o m o m o r p h i s m .

coefficients

the c h a r a c t e r

{{i,j,}

of

t h e n we h a v e

forms

I(V',V)}

by

I'

Then

o v e r all o p e n c o v e r i n g s

is d e n o t e d

'

in

~(i'),u(j')

V', t h e n we h a v e

{HI(v;T);

be any e l e m e n t

j'

and t h e r e f o r e

see t h a t the m a p

maps

of

system R

.

{~ij} i'

Since the c o r r e s p o n d e n c e

We also

is a r e f i n e m e n t

Let

for some

the e q u i v a l e n c e

÷ HI(v''~), .

V3

= Vu(i, ). ~ ~

t~,j ' = ~ ( i ' ) , ~ ( j ' )

We set

V

A V[

i'

V (i,) N V ~(j')

means

~(Vi,)

V!

an arc

=C Vi(v)

o

is a d a p t e d of

to

al-''o m

y. is

y.

a homomorphism

e 6 F0(R).

~, r e s p e c t i v e l y .

of

Choose

HI(R;T)

into

any r e p r e s e n t a t i v e s

Namely,

{~}

F0(R)~. {~ij}

is a l - c o c y c l e

25

over a covering

V : {Vi: i E I}

from the origin

0.

covers

Let

and

c

is a closed curve issuing

y = {Vi(1),...,Vi(m)}

be a chain of

V, which

c, and set m

(2)

F(e;Y;{6ij})

with

i(m+l)

and

: i(1).

: ~ ~i(~)i(~+l) ~:i

We have to show that the product

(b)

does not depend on the choice of

F(c;y;{~ij})

representative covering

c.

{~ij}.

To see this,

Y1

is a subdivision

If

~ii = i, we see that {Vj(1),...,Vj(m)} to both y therefore

and

F(C;Yl;{~ij})

let

only on

of

y, then,

~ : i,..., m

with

i(m+l)

Next,

: i(1)

and

of

for

V

using the property

c = oi---o m

Vi(v) A V j (v)

~

y, fixing the

be another chain of

= F(c;y;{~ij}).

and if a partition YI' then

YI

vi(~) A V i ( ~ + l ) A V j ( ~ ) n V j ( v + l ) for

depends

~.

e

if

YI =

is adapted

~ : i, .. . , m

and

~

j(m+l)

: j(1).

Thus,

in

view of (i), we get m

F(C;Yl;{$ij))

= ]-F ~j(~)j(~+l) 9:1 m

: ~ [6j(v)i(v)~i(~)i(v+l)$i(v+l)j(~+l) ] ~:i m

= I T ~i(~)i(~+l) v=l Finally,

let

YI

be any chain of

subdivisions

y'

and

tition

c = oi--.o ~

F(C;Yl;{~ij})

YI' of

a covering valent,

c

y

{~i,j,}

of

V

and

{@i":

c.

simultaneously.

V'

Since

{~ij}

Thus we get

in place of of

and

~

associated

{~,j,}

V" = {V'~,: i" E I"}, which

with refining maps

i" E I"}

to which a par-

= F(c;y;{~ij}) ,

F(c;{~ij})

be another r e p r e s e n t a t i v e i' E I'}.

Then we can find

YI' respectively,

= F(c;y';{~ij})

So we may write

V' : {V~,:

and a system

covering

is adapted

there exist a covering

refinement

V and

= F(c;y~;{~ij})

as was to be proved. F(o;Y;{~ij}). (c) Let

of

= F(c;Y;{~ij})"

with values

in

~ T

and

: 8i"$~(i")~(j")@~'~

is a common

~', respectively,

such that

!

6~'(i")z'(j")

with

are equi-

28

for each pair {V~,,

i",

V"

j" }

(i) .... ' i " ( m ) for simplicity, i(v)

in

I"

with

be a chain

V'~,,nv'~,, ~ @.

of

V"

= ~(i"(v)),

covering

i'(v)

Now let

y"

the curve

= ~'(i"(v))

c.

for

: Write,

v = i,...,

m.

Then, B(y") = {Vi(1) .... ,Vi(m)} (resp. ~'(y") : { V i , ( 1 ) , . . . , V i , ( m ) } ) is a chain of V (resp. V') c o v e r i n g the curve. Then, u s i n g the convention

i(m+l)

= i(1)

etc.

as before,

we have m

F(e;{E~,j,})

= F(c;~'(T");IE~,j,})

= ]-m v=l

E~'(v)i'(v+l)

m

]-~ v:l

:

(@i,,(~)~i(~)i(~+l)@i,,(~+l~ I)

m

: T~ ~i(~)i(v+l) v=l : F(e;B(T");{~ij}) This m e a n s {~ij}.

that

F(c;{~ij})

So we may w r i t e (d)

Let

eI

and

If the d e f o r m a t i o n covered

by the

depends

F(c;~) c2

cI

chain,

say

and

..., c k'

issuing

for e v e r y chain.

that

~2 ~

{~ij}, using

is d e n o t e d

in

any

in

F0(R).

then they

are

= F(c2;{).

ci = el,

of c l o s e d

c k' = e 2

are c o v e r e d = F(c~;~) by

~

c{,

and such that, by the

.....

only on the h o m o t o p y

paths

F(c~;~)

class

and

same

~.

~

of

= c.

This

F(~;6). seen,

Thus

HI(R;T).

Given

{qi j} the

F0(R).

~

small,

sequence

in (2) is d e t e r m i n e d

by

As is e a s i l y in

(f)

expression

depends

of

= F(e2;Y;{~ij})

such that

of

so that

c' and e' v v+l F(Cl~ ~) = F(c~;~)

F(c;~)

(e) fixed

0

class

F(e;{~ij}).

is very V

k-l,

Hence,

quantity and

from

v = i,...,

the

on the e o h o m o l o g y of

we can find a finite

It follows

F(e2;~). Namely,

case,

c2

y, of

F(el; ~) = F ( C l ; Y ; { 6 i j } ) In the g e n e r a l

only

instead

be two r e p r e s e n t a t i v e s

between

same

: F(c;{~ij}).

This

<, q

associated

same n o t a t i o n F(~;[n)

F(~I~2; ~) = F ( ~ I ; ~ ) F ( ~ 2 ; ~)

F(.;~)

is a c h a r a c t e r

character

from with

as in

is d e n o t e d

HI(R;~)

the

F0(R)

by

F t.

same c o v e r i n g

V : {Vi:

(a), we have

m ~i(v)i(v+l)ni(v)i(v+l)

= F(~;[)F(~;n).

~I

for every

we find r e p r e s e n t a t i v e s

= F(c;y;{[ij~ij})

= ~ v:l

for any

of

i e I}.

So,

27 It follows that the map F0(R)*. (g)

~ ÷ F~

is a h o m o m o r p h i s m

We now define a h o m o m o r p h i s m

of

F0(R)*

will turn out to be the inverse of the map Let i E I} from

F

be any character

of Vi

R, we choose, and a curve

of

F0(R).

for every

ei

union for

HI(R;T)

into

into

HI(R; T), which

~ ÷ F$. Given any covering

i E I

V = {Vi:

once for all, a point

joining the origin

V i NV.] ~ 9, then we take a curve,

of

0

to the point

eij , joining

z.m

to

z.]

z. i

z i.

If

within the

V. UV.. The curve c.. is unique up to homotopy equivalence, i ] 11 V. U V. is included in a simply connected subset of R. We set i ] -i $ij : F(c.e..c.i l] ] )

for every pair eocycle over included to

i, j V.

in

I

In fact,

It follows

(cicijcjl).(cjCjkC k -

-i)

that

as was to be proved. of

sequently

into

(
V i ~V.] A V k ~ ~, then subset of R and so cieikc k-i

is a l-

V i UVj UV k elk

is homotopic

is

is homotopie

to the composite

that

-I -i = F(cl.e..c. i] ] )F(ejcj kCk ) : ~ij ~jk '

Since the m u l t i p l i c a t i o n

we see that the map

F ÷ ~F

in

F0(R)*

is defined to be

is a h o m o m o r p h i s m

of

F0(R)*

HI(R;~).

~ij = F(cicijc~l).j

and let

F0(R)* Let

c

v = i,..., m.

within

Vi(v).

subset of

Since

c = ~l--.~m

Then,

R, we see that

~ c

V

To show that F = and cij, and then

covering

of

c

with

point of ~ by z@v to zi(~)

join

Vi(v) U V i ( v + I )

It follows that the curve

c i.

zi,

be a chain of

Denote the initial z~ +I : z~.

~ = ~F"

i e I},

be any closed curve issuing from

y = {Vi(1),...,Vi(m)}

i,..., m, and set

and set

V : {Vi:

so that there exists a partition for

Then

It is a routine matter to verify that distinct

(h) Take any F in F~ , we choose, as in (g), set

I ~.

V, z. or c. only result in equivalent cocycles and coni i that the above definition gives a unique element, say 6F'

HI(R;T).

pointwise,

V i NVj

and therefore

~ik : F(cicikCkl)

in

if

in a simply connected

oijcjk.

choices

with

the curve

is homotopio

to

z~ for v = by a curve c'v

in a simply connected c~ci(~)i(~+l) c'm+l-i "

to

c{ Ci(l]l( ci(1)ci(1)i(2)ci(2) -i )--'(ei(m)Ci(m)i(1)ci(1) -l)ci( l)Cl,-I Thus we have

F~(e)

= ~i(1)i(2).--~i(m)i(1)

m

c,

~v ~ Vi(v)

is included

is homotopic

0

: ~ F(ci(v)ci(v)i(v+l)Ci(v+ v=l

i]i)

28

T -1

F(ci(1)e I

=

y

celci(1) I) = F(ei(1)c i l)F(c)F(clci(1) I)

= F(c) , as was to be proved. (i) that

Finally,

~ = ~F"

take any

Let

{~ij}

~

in

HI(R;T)

and set

be a representative

F = F$.

l-oocycle of

~

We show associ-

ated with a covering V = {Vi: i E I}. Again take zi, c i and cij as in (g). For every j in I, let us choose a fixed chain yj = {Vi(l,j),...,Vi(mj,j )} is the same for all V.i NV.] ~ ~.

which covers

j, say

i(0).

cj.

Here we assume that

Take any pair

i, j

in

I

i(l,j) with

Then

(~F)jk = F(cjcjkCkl) = ~i(l,j)i(2,j)'''~i(mj,j),j~jk~k,i(mk,k)'''~i(2,k)i(l,k). Writing @j = ~i(l,j)2(j,j)...~i(mj,j),j, we have (~F)jk = 8jSj k0k I for j, k in I. Namely, {(~F)jk } equivalent to {~ij}. Hence, ~F = ~" This completes the proof.

2.

Line Bundles and Multiplicative 2A.

Analytic

We now consider line bundles over

is []

Functions R.

We do not define them

as topological spaces with oertain structure but just say that every element in H I ( R ; T ) determines a line bundle (precisely, a unitary flat complex line bundle)

over

(AI) Every l-oocycle a line bundle ~;

R.

({~ij},

Using representatives, {Vi})i,jC I

in

we say:

zl({vi};~)

defines

(A2) Two l-coeycles ({~ij }' {V'})',jell z and ( { ~ j } , {V~})i,j~i, define the same line bundle, if there exists a common refinement {V"}e~EI" of {V.}l and {V~} with refining maps ~: I" ÷ I and + , { ~ ,'( ~ ) ~ , ( ~ )} I" I' respectively, such that {~(~)~(B)} and equivalent {@ } el, ,

l-eocyeles in

~

over

{V"} r~ ; namely,

there exists a system

satisfying ~ ', ( ~ ) ~ ' ( B )

(3) for every pair

B': are

~, 6

For a line bundle

in

I" ~

= @~

with

over

1~(a)~(B)eB

V"~AV~

~ ~.

R, we define a section

f

of

~

over

29

R

as follows:

for

when a representative

~, the corresponding

{Vi})ie I

({6ij},

representative

of complex-valued

functions

of

{Vi})i,je I

f

fi'

is chosen

is a collection

fi

being defined

({fi },

on

V i,

such that (4)

fi(z)

' } ' {V~}) i,jel' is another VoINV.] ~ ~. If ({~ij and if {~ij} and {~ij} are connected by the

on V I. NV.3 whenever representative of ~ relation

(3)

is defined ({~j},

then

({fl},

{Vi})iCI,

to be the representative

{Vi})i,jEi,

when we have f ,,( ~ )

(5) on

: ~ijfj(z)

V".

The relations

modulus any

Ifi(z) I

i

(4) and

with of

f!

f

a function

with respect

for every = 8~if(

being

on

V!

to the system

~ e I"

~)

(5) imply that,

for every

is the same for any representative

z E R, the

{fi }

of

f

and

with

z E V.. The common value is ealled the modulus of f i and is denoted by If(z)I. A section f of ~ is called holomorphic (resp. meromorphic) if every fi is holomorphie (resp. meromorphie) on V I• for a representative

({fi }, {Vi})ie I.

choice

of representatives.

set of holomorphie 2B. morphic

(resp.

sections

Theorem.

Let

F0(R)

HI(R;~) section

of

~

~

analytic

0, called

to

F0(R)*. and let

~

Let

from

issuing

0

of

over

~

fi(0)

over

the

R. or mero-

For this purpose,

we take

and have the following

R

and

be any branch of

of the

M(R,~))

of holomorphic

~.

the origin,

independent (resp.

F = F~

under the canonical

fi(0)

curve

Proof.

sections

f = ({fi }, {Vi})

continuation

c

~(R,~)

continuation

be a line bundle

Then the analytic

equal to

is clearly

by

of the given line bundle

corresponding onto

We denote meromorphic)

We next diseuss

once for all a point

of

The definition

of

be any meromorphic

of

f

with

is possible

and the resulting

the eharaeter

isomorphism

function

0 E Vi(0).

along

each closed

element

at

0

is

F~(e)fi(0). Let

y = {Vi(1),...,Vi(m)}

curve

e

with

direct

analytic

tion element is exactly F~(c)fi(0)

i(1)

= i(0).

continuation

to be obtained

be a chain of

The formula of

fi

to

by continuing

{V i}

(4) shows that Vj.

It follows

fi(0)

~i(1)i(2)...~i(m_l)i(m)~i(m)i(0)fi(0), in view of the definition (1). []

eovering ~ijfj

that the func-

analytieally which

the is the

along

is equal to

c

30

As

far as a n a l y t i c

HI(R;~)

and the

tified,

so that

2C.

a

on

can a l w a y s

log u

Let

if t h e r e

phic) bundle the

~

The

that

2D.

A

~

such

defined

saw

curve

e

equal

to

sense,

bundle

f.

~.

0

M0(R,~)

in

M0(R, ~)

whose

principal

u

for

on

R

every

a meromorphic definition,

a locally

is an l.m.m.,

function

(resp.

If(z)I

then

call

on

l.a.m.) (resp.

on

we

R

if and

holomor-

R.

The

line

the

bundle

(or

f0'

and

called

f

origin

forms

have

point

the

is

R.

of

f.

f0 + go"

we call

if the

R

a

the p r i n -

branch along element

F0(R).

Let

us d e n o t e

at

0.

any

closed

at

0

is

In this

functions

sum of two

multi-

are

of

as the m u l t i p l i c a t i v e

on

R, w h i c h

for all,

functions

function

meromorphic

space

and

principal

as a c h a r a c t e r

is

over

To e v e r y

once

analytically

resulting

f

~f.

0, w h i c h

same

is c a l l e d

is an l.m.m, bundle

meromorphic

character

a linear

is d e f i n e d

of

R

For any m u l t i -

by

we a s s o c i a t e ,

at the

~f

the

on

Here,

Ifl line

is d e n o t e d

R, the

the

f

extension.

can be c o n t i n u e d 0

regarding

branch

function

a unique

and

in

straightforward.

its m o d u l u s

set of all m u l t i p l i c a t i v e

f, g

u

real-valued

=

analytic

f

if t h e y f0

from

is also

of

say

2B,

~f(c)f0,

the

in this

is c a l l e d

u, w h i c h

Two m u l t i p l i c a t i v e

same

issuing

M0(R,~)

If,

is s i n g l e - v a l u e d .

f,

function

branch,

in Th.

~f

by

2C, d e t e r m i n e s

fix a p o i n t

to be the

As we

acter

line

of

if, and

and a m e r o m o r p h i c u(z)

Ifl

function

by Th.

meromorphic

branch

function

a

If

it is r a t h e r

its m a x i m a l

meromorphic

single-valued cipal

F0(R).

of

is an l.m.m. ~

meromorphic

to be

Let us

u

that

for

always

plicative

u

extended

uniquely

(multiple-valued)

the

V.

in

u.

if its m o d u l u s

is c a l l e d

on

$

can be iden-

of

"l.m.m.") V

"l.a.m.").

bundle

is o m i t t e d ,

and t h e r e f o r e ,

real-valued to

bundle

F0(R)*

0 < u < ~.

multiplicative

plicative

to

Then

a line

of

of

assumed

Ifl

a line in

a character

to be h o l o m o r p h i c ,

is d e t e r m i n e d

proof

u =

be a n o n n e g a t i v e

f

character)

extended

(abbreviated

vanishing.

section

to d e n o t e

a neighborhood

wherever

exist

F~

exists

(abbreviated

u

identically

only

be u s e d

such

is h a r m o n i c

concerned,

modulus

be c h o s e n

modulus

Theorem. not

V

are

character

a nonnegative

R, t h e r e

f

analytic

will

meromorphic in

function f

$

We call

a locally point

functions

corresponding

by

of char-

functions

meromorphic

function

31

2E.

So far, our sections

values.

of a line bundle

Besides these, we can define a section

are first order differentials. the numerical ~i

on

Vi

ones:

The definition

in 2A.

differential,

(resp. meromorphic)

differential

(resp. meromorphic)

denoted by

3.

F~(R,~)

Existence 3A.

and

Z

R

g

and

cycles

in

(resp.

with

R, where

Yk'

{Aj, Bj:

only with

all others.

Let

Bj

fi

~i

[.

~-i

Ck

~.

Let

g}

R

is

is homologous

to

Yk

among these cycles, Aj) and every

for

7

on

~¥(q)

: I

R

Y

with

j :

sequence

A.

(resp.

is disjoint on

in

k : i,...,

every

Ck

g

basis of l-

is a canonical

be the Cauchy differential

For any arc

R

B.)

from

defined by

we set

~(P'q)"

of

~y

for

y = Aj, B.j R \ Aj,

and

C k-

so that the

disjoint

from

A..

we have

I

d~A.

B.

j = i,...,

g.

at the intersection of

over

Aj, Bj

be a canonical

~A. is single-valued and holomorphic on B differential d~A. has no periods along any cycles

Bj!

[

surface with genus

(i)

for

by a differential

The set of all holo-

Riemann

j = i,...,

We shall look into the properties

Moreover,

similar to

is a ho]omorphic

sections of

k : i,...,

(resp.

~(p,q)

Th. liB, Ch. I.

whose values

is called a holomorphic

section of

k = i,...,

Here we may assume that,

intersects

~

~

Sections

components

Ck

of

F~{(R,~)).

the sense of Ch. I, 2B and Z-I.

then

be a compact bordered

boundary

i,...,

~

is entirely

If every

differential

of Holomorphie

Let

take only numerical

i.e. we have only to replace

in the definition

(resp. meromorphic) mcrphic

[

A.~

and

In fact,

pj

of

let

Aj

= 2~i

j

]

{V, z}

and

Bj, respectively,

be a parametric

Bj

and take

containing

pj

small

disk centered

ares

Aj'

and contained

in

and V.

Then we have

~A'(q) ]

The s e c o n d on

R

off

term

of the

= I

A!

~'(P'q) + I

right-hand

Aj \ Aj' , s o t h a t

side

A.\A!

is

~(P'q)"

single-valued

and holomorphic

32

Bj

Bj

]

A! 3

B! ]

B!-B? q A[ 3 3 3 is a continuous deformation

where

B? 3 has the same end points

A! ]

B[, within V, such that BY 3 3 as B! and is disjoint from A! and such that 3 ] is a closed curve of winding number +i with respect to the

BV 3 3 terminal point of B[-

senting

p

and

where

R(z,%)

of

A!. In terms of local variables 3 q, respectively, in the parametric w(z,~)

z2

B.\B[ ] 3

in two variables.

~

repre-

V,

Denoting by

zI

and

' we thus have Aj,

the initial and the terminal points of

~(p,~)

and

= [(z- 6)-i + R(z,~)]dz,

is holomorphic

I

z disk

= log [(z 2 - ~)/(z I - ~)] + RI(~),

A[ ] with some holomorphie

I

d~A" B. ]

function

= I

3

RI(~).

{( 1 B!-B:'

]

i z l ) d ~ + dRl}

~-z2

]

So we get

= 2~i,

-

as desired. (ii)

Similarly,

the one along

the only nonzero period of

I (iii) Yk'

Finally,

k = i,...,

not with other

Cj

let

Yk'

is seen to be

daB. = -2~i. 3

A.

on

d~B. ]

A.: 3

be an arc from a point on

h-l,

intersecting

and

yj,

j ~ k

with and

Ck

y~

to a point

at a single point and

j =< ~-i.

We set

~'Ck(q) =

,(q). Supposing that Yk is positively oriented with respect Yk we see that the only nonzero period of d ~ k occurs along Ck:

f Ck for

k = i,...,

prescribed

=

-2~i

Z-I.

These observations Theorem.

d~k

to

prove the following

There exists a function holomorphic periods

along cycles

on

in the canonical

R

and having any

basis.

R,

33

3B. tions

We are now able

with

Theorem. has

Every

holomorphic

in

Aj,

R, w h e r e

tified

with

f o r all

~j,

there of

§2.

h

set

section

for

Basic

Tc

0.

~j

OF H A R M O N I C

~

We then

h

for

i.e.

of 1-cycles ~

boundary

c a n be

iden-

Choose, the

once

origin

#

sense,

e. ] a real

define

is

By t h e p r e c e d i n g

on

R

every

it is an i n v e r t i b l e

<, as was

and

A! joining ] In an o b v i o u s

= exp(2~i{j).

to

g IB,

Fo(R).

sueh that j : I,...,

bounded

to be p r o v e d .

the 2g+Z-l.

holomorphic

[]

FUNCTIONS

C(R,~)

be t h e

on a R i e m a n n convergence

linear

surface

space R.

on c o m p a c t

o f all r e a l - v a l u e d

We give

subsets

of

the

space

= HP(R)

denote

the

set o f all n o n n e g a t i v e

HP

is a c o n e

HP

and

every v

Theorem.

in t h e

nonnegative

in

HP.

The

cone

HP

sense

real

is

that

number

locally

is

R.

subset,

R.

C(R,~)

C(R,~)

dense

on

Y

then

basis

genus

group

an a r e

surface

sections,

bounded.

By Th.

function

is e q u a l

bundle

STRUCTURE

of u n i f o r m

and

~(ej)

Aj

Let

countable HP

2g+~-l,

j, b y

Riemann

also

has R.

c. : A ! A . A ! -I ] ] ] ] curve issuing from

along

functions

Let

R on

fundamental

a holomorphic

line

are

set

every

func-

Structure

4A. uous

j = i,...,

bordered

be a c a n o n i c a l

that

bundle

of the

f = exp(2~ih),

LATTICE

4.

as a b o v e

holomorphic

holomorphie

f-i

2g+~-l,

exists

of the

multiplieative or c h a r a c t e r .

bounded

such that

be a l i n e

A. and ] as a c l o s e d

theorem

If we

~

every

number

bundle

on a compact

j = i,...,

on

considered

period

f

a character

for

to a point

bundle

we a s s u m e

Let

line

(or i n v e r t i b l e )

sections

Let

contours.

u

line

nonvanishing

Proof.

to c o n s t r u c t

any prescribed

(i) ~

(ii)

with

Since

contintopology

R

has

a

to be m e t r i z a b l e . harmonic

eu E HP

and

compact

seen

the

functions

for e v e r y u + v E HP

respect

to t h e

u

in

for any

topology

C"

Proof. pact

By u s e of t h e H a r n a c k

subset

stant

E

C > 0

of

R

inequality

and every

depending

only

on

sup{u(z): for

any

u

in

HP.

So,

point E

and

it is s e e n a

in a

that,

R, t h e r e

and

for

exists

every

com-

a con-

satisfying

z e E} S Cu(a)

for every

fixed

a

in

R

and

every

fixed

real

84

numbers

0 ~ a ~ 6, t h e

set of f u n c t i o n s

equicontinuous

on e v e r y

subset

with respect

of

follows

HP

at once.

Corollary.

a

HP(a)

4B.

compact

HP'

C(R,~).

sented

as t h e

The c o n e Namely, This HP'.

defines

HP'

in

lower

uA v

Ch.

I, 4C.

HP' above

(resp.

(resp. cone

the

has

~ ~ u(a)

~ 6]

is

therefore

is a c o m p a c t

From

the

this

theorem

itself

= i)

lower

element

bound)

of

HP

which

HP'

if

ones

pair the

space

HP'

u, v

we

in

see t h a t

upper

from

bound

In p a r t i c u l a r ,

to the o r d e r

of

greatest

is b o u n d e d

least

way.

E HP.

we d e f i n e d

which the

usual

structure

every and

R.

v-u

or otherwise

has

be r e p r e on

in the

uv v

HP'

with respect

can

space

Thus,

the

in t h e

functions

linear

HP of

HP'

in the

a lattice

in

exactly

in

R

bound

subset

of

if a n d o n l y

lattice.

are

every

~

the

upper

compactness

i.e.

by some

least

which

of l o c a l

forms

a vector

the

on

u ~ v with

span

harmonic

relation

we d e f i n e

HP',

complete,

linear

of f u n c t i o n s

is c o m p a t i b l e

in

below)

the r e a l

space

becomes

greatest

HP

u(a)

of t w o n o n n e g a t i v e

HP',

HP'

By use

is o r d e r

T c.

= {u ~ HP:

an o r d e r i n g

in

relation

of f u n c t i o n s bound

set

the

denotes

It is the

In fact,

R

and

and metrizable.

u, v

ordering

R

topology

in

difference

HP for

{u E HP:

of

to the

= HP(R;a)

: HP'(R)

space

subset

[]

For e v e r y

is c o n v e x ,

compact

defined

the by

HP.

4C.

Let

E

be a l o c a l l y

a convex

cone

C

in

that

the r a y s

H

all

not p a s s i n g

The

set

B

also B

suppose

A point

that

x

in

E

vertex

from The and

having

result

at t h e C

is c a l l e d

convex

and this

in t h i s

origin

a base

property respect

0

of

of

B

if a n d

is

E.

Suppose hyperplane by

C.

define

We c a n

a point

We

x

in

if it h a s

the

for called

following

B.

y - x e C.

only

y = ~x

sometimes

is t h e

and take

is d e n o t e d

whenever

imply

space

a closed

it is s e e n t h a t

set

y ~ x

C NH

x S y

Then

linear

intersect

intersection

by s e t t i n g

of t h e

y C C

topological

0) in

is c o m p a c t .

point

C

0.

in

B

property:

fundamental

with

convex

relation

is an e x t r e m e

following

The

through

is t h e n

an o r d e r i n g

E

(issuing

convex

some

~ ~ 0.

minimal.

theorem

due to

Choquet: Theorem.

Suppose

metrizable.

Then

that the

the

base

following

B

of a c o n v e x

hold:

cone

C

is c o m p a c t

and

3B (a) of

B

The

and

on

B

sure

E

every

on

integral

B

of e x t r e m e

x

in

supported

Z

that

set

such

is d e f i n e d

for e v e r y

B

by

points

is the

E, i.e.

that in the

continuous

= i

sense linear

B

is a n o n v o i d

barycenter

there

~(E)

of

of a p r o b a b i l i t y

exists and

G~-subset

a probability

x : fE y dz(y),

of the w e a k

topology

functional

~

on

where

of

E

measure

Borel

mea-

the

E, m e a n i n g

we h a v e

~(x)

=

fE ~(Y) d~(y). (b) fined

If

by

C

is a l a t t i c e

C, t h e n

probability

every

measure

~

Our o b s e r v a t i o n the

requirements

Corollary. points nonvoid

a

of a u n i q u e

in

R

measure

= rE(a) v(z) d~(v)

4D.

We c o n s i d e r

disjoint

discrete

union

of the

of

R.

define

uI ~ u 2

if

Ul(Z)

R.

relation

~

space

S/~

same

HP'(R\

tical

in

HP'(R\

The

sum

SP' SP'

Z)

= Ul(Z) + u2(z) u2(z)

outside

Theorem. ations

SP' defined

in e v e r y The if

u

proof

a logarithmic

So e v e r y ui

(the

for some

every

z

Z).

SP'

subset

lattice

It i n d u c e s SP'

is s t r a i g h t f o r w a r d H P ' ( R \ Z) singularity

with of

is a

= i,

i.e.

In p a r t i c u l a r ,

we

Z

u2

discrete

two

only

with

the say

S

over in

the

all

S

subset

and

we Z

of

quotient

elements

if they

is r e g a r d e d

of)

runs

Ul,

S

Let

in the

are

iden-

as a s u b s p a c e i = i,

function uI ~ u2

of

2, in

(u I + u2)(z) if

Ul(Z)

R.

with

respect

original

to the vector

pointwise lattice

oper-

structure

complete.

and

discrete u

E(a)

barycenter

~(E(a))

in

H P ' ( R \ Z i)

the

is o r d e r

some

if and

of

all

set of e x t r e m e Then

Z), w h e r e

We

satisfies following

singularities.

Clearly

class

HP

R.

relation

H P ' ( R \ Z) in

with

of e l e m e n t s

outside

in

deone

the

is the

HP(a).

with

HP'(R\

the

= I}. HP(a)

in

R \ (Z I U Z 2 ) .

discrete

above.

of

= SP'(R).

being

in

in

z

pair

equivalence

is a v e c t o r

HP'(R\

is in

identical

Z).

as

HP(a)

on

cone

E(a) u(a)

~

= u2(z)

are

by

functions

SP'

the

u

space

relation

of an at m o s t

so we h a v e

every

For e v e r y

by

u I + u2,

is d e f i n e d

denote

is an e q u i v a l e n c e

is d e n o t e d

that and

topology

for

harmonic

subsets

The

and

ordering

E.

= {u e HP:

in the m e t r i c

u(z)

be the

and

HP(a)

probability

by

theorem

HP(a)

to t h e

is the b a r y c e n t e r

4B shows

Choquet

set

of

u = rE(a) y d , ( v ) have

be

respect

B

in 4A and

convex

G6-subset

with

in

supported

in the

Let

of the

x

is o m i t t e d . Z, every

or a r e m o v a b l e

We note point

one.

in

finally Z

that,

is e i t h e r

36

5.

0rthogonal 5A.

lute v

Decomposition

For

value

u

(in

in

SP'

SP')

are o r t h o g o n a l

u • v.

If

u l±u2,

then

u

Typical

examples

if

in the

of

we u.

set

lu| A Iv|

SP'

as

= uv

u, v

= 0.

be

(-u) in

If t h i s

is e x p r e s s e d

expression are

lul

Let

as

is c a l l e d

and

SP'.

call We

is t h e

case,

u = uI + u2

in

an o r t h o g o n a l

it the

abso-

say t h a t

u

and

then we write SP'

with

decomposition

of

u.

follows: +

(a) which

For each

we call

the

u

in

positive +

We then have (b) all

u

in

is a b a n d

SP' of

SP',

And we have

uI + u2 and

the

u 2 ~ 0.

of

SP ~ This

Q(R))

are

onto

and

is an e a s y

~uasibounded

u

Q(R)

Theorem.

has

only removable

has

the

are

SP ~ ,

parts

of

For every

assume

as

u ~ 0

Y±,

v

form

u ~ 0

the

(resp. bands

of t h e

Prl(u)

then

above

in

to

ym.

= Y± + Y±±. Namely,

implies

if

both

function

u =

uI ~ 0 i

on

Functions

SP'

and

SP'

projections are b o t h

general

and

the

Y±,

theory

prQ(u)

are

is t h e

pr I

and

prQ

positive. of v e c t o r called

lat-

the

inner

u, r e s p e c t i v e l y .

u

in

SP ~

so t h a t

its q u a s i b o u n d e d the

inner

part

part

prQ(u)

Prl(u)

of

If w e

set

u. without

loss

of g e n e r a l i t y .

u = u An, n = i, 2,..., t h e n u a r e b o u n d e d a b o v e a n d so a r e h a r n n m o n i c on R. The sequence {u n} is m o n o t o n i c a l l y i n c r e a s i n g a n d is majorized exists If

and

by the

superharmonic

is h a r m o n i c

s E I(R),

then

function

everywhere

lUnl A Isl

R,

in

quasibounded). of

The

ym

in

belongs

SP'

constant

Then

and has

bounded

SP'

set o f

Y.

SP'

some

is p o s i t i v e .

respectively,

singularities,

singularities

of

in

= I(R) ± = {I} ±i.

bands.

Q(R),

element

of

then

Q(R)

consequence in

summand

inner

to be t h e

lattice

of the

of o n l y

and

called

and

I(R)

We may

each

.

yl

for

Y'

in t h e

= (-u) v 0,

± u

subspace

subset

u 2 E Y±±,

sum of these

For any

to e v e r y

u

u, r e s p e c t i v e l y . -

u

lul =< lvl

Y'

and of

we d e f i n e

sum decomposition

I(R)

direct

same

of

= {i} ±

and the

Proof.

bound

Y

tices.

5B.

and

consists

onto

SP'

and

is a l i n e a r

When

Both

orthogonal

~

0

parts +

+ u-

orthogonal

and

I(R)

(resp.

Theorem.

are

of

u I ~ Y±

then we write I(R)

Y

projection

with

= uv

negative

: u

for a n y n o n v o i d

a direct

u

+

lul

u ~ SP'

upper

set

and the

,

i.e.

if

(ii)

least

Moreover,

which

(i)

and

the

- u

subset

SP',

properties: u E Y±;

any

we

-

u : u

For

SP'

: Un

on A Isl

u.

R.

We ~

So

limn÷ ~ u n

claim that

n h Isl

:

O.

(= v,

say)

v = prQ(u). So

u

n

e

Q(R).

u

37

Since

Q(R)

have

is c o m p l e t e ,

( u - v) A i = 0.

u

v S u- u n

w+

(u A n )

for

÷

0.

This

prQ(u).

n = i,

S n + I

w S U n + l - - U n.

5C.

2,...

{u n}

that

argument u

in

prQ(u)

5D.

We

apply

l.m.m°

u

belongs

to

(resp.

Q(R)).

Let

prQ(lOg

u)

b y Th.

harmonic

on

R

the

: lim

of

u, r e s p e c t i v e l y .

uI

5B,

and

bounded

characteristic

Suppose

that

branch

this

factors Finally

bounded u S i uI

and

is c a l l e d

bounded u

in

0.

(see

2C).

characteristic outer)

harmonic

are

called

UQ

prQ(lOg

These

if

if

An log u

log u E I(R)

characteristic.

and

Prl(lOg

= exp(prQ(lOg

the

u)

inner

has

function

and

u)

Then is

u)).

the

outer

factors

no s i n g u l a r i t i e s ,

f

on

R

corresponding

fl

(resp.

function the

fQ)

the outer

one

l.m.m,

0

fl

and

uI =

denotes

and

are

is s a i d

to be of

u =

Ill.

be a m u l t i p l i c a t i v e

such that

suffix

functions

of m o d u l u s

on

an i n n e r

is b o u n d e d .

of c o m m o n R.

inner

say t h a t u2/u I

characteristic J

l.m.m.'s

of b o u n d e d

and

Let

speak

l,a.m.'s

we

to

fQ

called

IflI, the

are the

UQ

=

principal

determined inner

and

f, r e s p e c t i v e l y .

we w i l l

then

v =

following

(resp.

u))

if so is the

factor

Hence,

[]

set

meromorphic

any bounded

u2

the ease, uI

of

inner for

We

holomorphic)

point

up to a c o n s t a n t outer

results

= (fl)0(fQ)0 , where

at t h e

we h a v e

Un+ I- u n

l.a.m.

occurs.

(resp.

u - v E I(R).

the

w+ un

Thus

we have

observations.

we

w

[(-m) v ( n A u),].

everywhere

UQ

an

A multiplicative

f0

lim

l.m.m,

is.

Since

is a l w a y s

and

shows

that

= U n + I.

sequence,

so

these

inner

be a n y

log u

functions

IfQl

and

It f o l l o w s ( n + i)

hand,

Then,

we have

preceding

u

The

meromorphie

also

It is c a l l e d

wherever

UQ

R

SP'

u I = exp(Prl(lOg

factor

w £ i.

w + u n =< u A

from

On the o t h e r

w = ( u - v) A i.

is s a i d to be of b o u n d e d

SP'.

is,

set

and

on

is c l e a r

For every

v E Q(R).

we

is a c o n v e r g e n t

w : 0

theorem

above

see t h a t

therefore

Since

The

Theorem.

and

means

The

we

To see t h i s ,

uI

l.a.m,

on

R

Then,

u.

divides

is n e c e s s a r i l y factor

inner

factors.

We b e g i n w i t h

of

the

For two bounded u2

if

a bounded

u 2.

such that

Let us

trivial

Let the

A = {log UT:

J

u 2 / u I S i. inner

consider

remark inner

that l.a.m.'s

If t h i s

l.a.m.,

is

so t h a t

be a set of l . a . m . ' s

inner

factor

u E J)

forms

uI

of

of every

a subset

of

38

I(R),

which

is a b a n d in

is b o u n d e d

of

SP'(R),

Then

u0

SP'(R), which

above the

l.a.m,

uI

with

u

in

J;

UI

with

u

in

J, t h e n

the

sreatest

satisfies factor

uniquely common

the

has

on

(ii)

such

uI

divides

in

(i)

inner

u 0.

of

necessarily

J.

Set

u0

It is s e e n This

then

I(R)

say

v0,

u 0 = exp v 0.

divides uI that u0

If a b o u n d e d

(ii),

Since

bound,

l.a.m,

two p r o p e r t i e s .

factor

0.

upper I(R).

that

if a b o u n d e d

by these

inner

element

R

function

least

every

divides

all

such a

u0

is c a l l e d

inner

it is c a l l e d

l.a.m,

! common

u0 inner

J.

A parallel functions

and

(i) but n o t

of

constant

A

is a n o n p o s i t i v e

is an i n n e r

is d e t e r m i n e d

by the

set

on

description

R.

But t h e

is p o s s i b l e

details

for multiplieative

look obvious

and

are

meromorphic

thus

omitted.

NOTES Gunning dles

[13]

on R i e m a n n

tions.

The

Some Chapter

discussion

basic

facts

theorem

E47]

contains with

we r e f e r

for c o h o m o l o g y Weyl

in 3A is a d a p t e d

and

the reader

a discussion a proof

source

See a l s o

on v e c t o r

2 of Constantinescu

tion

together

is o u r m a i n

surfaces.

4D.

groups

[CC].

and

for m u l t i p l i c a t i v e Kusunoki

of harmonic

to P h e l p s '

on h a r m o n i c

of T h e o r e m

from

lattices Cornea

[68]

lecture

note

with

func-

[42]. functions

For C h o q u e t ' s

functions

line bun-

are

in

representa-

[53].

Neville

singularities,

CHAPTER

Compactifieation on n o n c o m p a c t a right

way

in the

of M a r t i n ' s . the m o s t unit

As we

case

A naive

surfaces

and

chapter

we d e f i n e

discuss

boundary

compactification.

book

of C o n s t a n t i n e s c u

some

ingredients later

relation shown

use. between

idea

with.

Among we are

of the

several going

is that

open

unit

problems

indeed

suggests

kinds

of

to m a k e

it p r o v i d e s disk:

the

use

us w i t h

closed

Martin's

under

behavior results

Cornea

of h a r m o n i c

in

~§i and

[CC],

6, 13 a n d here

eompactification

are

harmonic

covering

[CC]

standard.

related

be f o u n d

out

Riemann

and r e o r g a n i z e

§3 we d i s c u s s

covering

on M a r t i n ' s

to

in the

for c o n v e n i e n c e

In

and

measures

of o p e n

functions

2 can

but we p i c k 14 of

compactification

that

this

Martin's

maps.

of the

It is

boundaries

are

maps.

chapter

R

will

denote

an o p e n

Riemann

surface.

COMPACTIFICATION

Definition IA.

in the

Let

every

be a set of c o n t i n u o u s real

space

usual

IQ

points

to in

find

a way

f(z) R

that

the

in the

sense in

compact

line

[-~,

+~].

C

with

that,

Since

for any

of

We t h e n

t(z)

of

For

R

z E R CK

points

at these

IQ

is e q u i p p e d

of

with

distinct values

set

theorem I

in

values

define

which

a map

functions

two

distinct

the

Tychonoff

Define

coordinate

taking

with

support.

f E Q UCK,

classical

space.

f E Q u c K.

R

C K = CK(R,~) with

If

f-th

on

R

By the

Hausdorff

functions

and

on

of all

for any

a function

+~]

be the

topology.

is a c o m p a c t

in such

equal

If

product

weak

[-~,

functions

let

cartesian

the

line

continuous

f E Q UCK

as the with

Q

extended

real-valued

can

and

The r e s u l t s

Throughout

§i.

Most

of C h a p t e r s

in p a r t i c u l a r

preserved

IQ

for a t t a c k i n g

vague

surfaces,

for our c h o i c e

compactification

this

i.

dealing

tool

this

disk. In this

our

see,

to R i e m a n n

reason

COMPACTIFICATION

a powerful

shall

we are

fitted

natural

MARTIN

is o f t e n

spaces.

compactification

III.

the into is

separate in

R, we

points,

the

40

map

1

is an i n j e c t i o n

I Q.

So we

space

IQ

space

I Q.

compact

can

identify

under

and

Q-compactification

the

f-th

pology in

f E Q

implies

[-~,

identified tion

f

z

to

tinuous

I Q.

IB.

Let,

a point

Let

ga(Z)

(Ch.

I,

We use

the

with

Q = QM

noted

by

that

of

R*

tended

by the

z @ R,

is c a l l e d

0),

which

kb

is a p o s i t i v e

in

2.

sure

R

in the

I Q,

RE

is a

We call

we d e n o t e

definition

for

z e R

is a c o n t i n u o u s

is dense

for all,

in

RE,

the

by

of the on

IQ

and

RE

f

thus

Ff(y)

= yf

product

to-

with

since

extension

Green

function

on

the

R

be a h y p e r b o l i c fixed

that

values

we h a v e

of the

has

func-

a unique

= i.

con-

R

with

be the

the The

A = A(R)

surface

the o r i g i n pole

is s y m m e t r i c family

parameter

at

of

a

= R* \ R

of

functions

dense

through of

and

is c a l l e d

a countable

R.

a C R

runs

R

and

of

in b o t h

Q-compaetification

compactification

contains

the

subset,

R

is deMartin it f o l l o w s

space. member

compact

symbol

b ÷ k(b,a)

space

b ÷ k(b,a),

fixed

b E R*

function

written

harmonic

for

QM

Riemann

is c a l l e d

g(a,z)

Let

Martin

metric

and

R, w h e r e

k(b,0)

Martin

÷ k(b,z)

R

is h e l d

~ g(z,a).

For e v e r y

is s o m e t i m e s

(b,z) [0,

= f(z)

Ff

to the

same

R*.

y E IQ

is c o n t i n u o u s

in IA e v e r y

by c o n t i n u i t y

over

into

compact

R

subset.

Ff

difference

is a c o m p a c t

is d e n o t e d

every

in p a s s i n g

Since

remark

dense

function

= k(b,a)

The

of

space

The

be the

is c a l l e d

ranges

tion

and

g(a,z)

R.

By the

For

convention

R*.

boundary

compact

R

R E.

We n o t e

b ÷ g(b,a)/g(b,0)

of the

of

of the

closure

y.

Since

= g(a,z)

i.e.

I(R)

be the

point

l(z),

once

subspaee

RE

as an o p e n

Ff(~(z))

to

is a h o m e o m o r p h i s m

R.

0 E R, w h i c h

6A).

variables,

R.

the

with

extension

take

R

of

Since

the

Let

subspace

of the

that

+~].

with

be fixed.

coordinate

in fact,

I.

includes

the

Let

R

the m a p

As a c l o s e d

space

and,

as

function

with

R*. where the pole

in

The

QM

the

b

(and w i t h

If

b E ~, t h e n

the

on

R.

We

that

function

b

see

also

on

R* x R

now

z ÷ k(b,z),

k b.

is a c o n t i n u o u s

function

variable

function at

can be ex-

extended

origin

function the

with

func-

values

+~].

Integral

Representation

2A.

Theorem.

~,

supported

For any by

A,

u E HP(R)

such

that

there

exists

a nonnegative

mea-

41

k(b,z)d~(b) = I J4

u(z) for

every

Proof.

z E R.

Let

{Rn:

n = i,

2,...}

be a r e g u l a r

exhaustion

of

R

with

0 ~ RI ( T h e o r e m IA, Ch. I). T h e n by T h e o r e m 6E, Ch. I, the f u n c t i o n R uE with E n = C I ( R n) is a p o t e n t i a l on R and is equal to u on n s u p p o r t e d by ~R n such R n • So t h e r e e x i s t s a n o n n e g a t i v e m e a s u r e ~n that

u(z) : u~ (z) = I g(b'z)d~n(b) n for

every

z E R

.

By s e t t i n g

z ER

n

.

n

We fix

u(z) Clearly, set

~n

{~n }

there

= I g(b,z)d~n(b)

bounded

and in the

a subsequence

to a n o n n e g a t i v e

every

see that

4C.

z E R.

measure

the m e a s u r e

2B.

We

Let

b E A.

with

kb(0)

HP(0)

are

combine

on

two

every

there

function

belong

to

QM"

k(b',a)

points exists

in

CK

of and

We

that set

kb

A = j(A).

j

R*.

~R n

observation kb

and

since

the m a p

converging

that

the m a p

£

Since

b ~ b'

is a c o m p a c t

If,

and

into

let

that

is HP(0)

b, b' of

~ f(b'). above

4A-

function

A

÷ k(b,z) A

such

shown.

kb(0)

Ch.

topology

A, the

cannot

[]

both

of

f(b)

a @ R

4.

harmonic

(b,z)

of the

to be

n ÷ ~, we

of

In fact,

boundary

a point

HP(0).

with

that

Since

for w h i c h

on the

exists

as

boundary

j: b ÷ k b

definition K

A

with

HP(0).

The

into

to

is a p o s i t i v e

to

k b ~ kb, , as was

Then

The

so t h a t

{~n }

It f o l l o w s

tends

injection.

A.

kb,

of

by the M a r t i n g

f E QMUC

there

4

on

2,...}

is an

vanishes

and t h e r e f o r e

see also

on

belongs

that

in

of m e a s u r e s

i = i,

~

spaces

an

Namely,

is a h o m e o m o r p h i s m

= u(0).

function

Moreover,

for w h i c h

R*,

boundary

see

n

= I k(b,z)d~(b)

kb

metric

large

Sd~ n = f g(b,0)d~n(b)

preceding

the

all

, we h a v e

space

is c a r r i e d

our

Then

dis-tinct

that

~

4 x R, we

is c o n t i n u o u s .

says

the

= i, so that compact

continuous

any

Since

consider

= I k(b,z)dZn(b).

{~n(i):

u(z)

for

and

= g(b,O)d~n(b)

is n o n n e g a t i v e

is thus

exists

vaguely

z E R

d~n(b)

be R*

Since f

must k(b,a)

So the m a p

= kb,(0)

j

= i, we

be p r o p o r t i o n a l .

subset

of

HP(0)

and

it

42

follows

from Theorem

ability

measure

~

2A t h a t , on

~

for each

such

u(z)

for

every

of

HP(0).

z @ R.

set o f e x t r e m e the

totality

on

R.

Points

4C,

Ch.

II,

Theorem. (b) sure

~

This

We d e f i n e

of the

b E A in

thus

shows

(a)

The

For

every

supported

that

~

convex

kb

called

the

following

AI

AI

contain

inverse HP(0).

exists

there

The

a prob-

A

extreme

under

set

in

A.

(and a l s o

a unique

points

j

A1

harmonic

points

of

exists

all

image

is a m i n i m a l

the m i n i m a l

is a G 6 - s u b s e t

u E HP(R) by

set

are

set

should

as the

for which

A1

there

v(z)d~(v)

A 1 = £1(R)

points

of

[J

:

means

u E HP(0),

that

of the

is e x a c t l y function Corollary

of

R*).

nonnegative

mea-

such that

u : I

k b d~(b), AI

where

the

HP(R).

integral

converges

In p a r t i c u l a r ,

with

every

topology

of

= [ J

kb(Z)d~(b) A1

z E R. +

Let using

to the m e t r i c

we h a v e

u(z)

for

respect

u E HP'(R).

the a b o v e

Writing

result,

we

u = u

_ -u

+ with

u

, u- E H P ( R )

and

see t h a t

u : J

k b d~(b) A1

for

some

finite

statement mined

(b)

Borel

above

uniquely

by

u.

the canonical

measure

2C.

M ( A I)

Let

It is a l i n e a r £ ~

for

subset the

E

measure

shows

space

This for

~, ~ E M ( A I) of

A I.

correspondence

on

A I.

is d e n o t e d

~

set of f i n i t e a n d has

if a n d o n l y

if

gives

a linear

is o b v i o u s l y

order-preserving.

and only

~u

are mutually

and

~v

result

~

is a l s o

on

by

real

Zu

Borel

a natural

M(AI) , which if

uniqueness A1 and

in t h e deter-

is c a l l e d

u.

In v i e w o f the r e m a r k u ÷ ~u

The

such a measure

measure

be the over

~

that

~(E)

~ ~(E)

at the

singular.

u ± v

on

for e v e r y

of in

A I.

relation,

end of 2B, w e

isomorphism So

measures

ordering

i.e.

Borel

see t h a t

HP'(R) HP'(R)

onto if

43

Let

X = XR

be the measure

stant

function

sure

(on the b o u n d a r y

u ~ HP'(R) defined

identically

is called

in Ch.

II,

equal

~i ) of

R

singular

if

5A.

in

M(A I)

to

i.

corresponding

We call

X

at the point

0.

A harmonic

u E I(R) N H P , ( R ) ,

Our o b s e r v a t i o n

in Ch.

II,

to the con-

the h a r m o n i c

where

mea-

function

I(R)

5A shows

the

was follow-

ing Theorem.

The c o r r e s p o n d e n c e

isomorphism

of

HP'(R)

if and only

if

~u

respect

AI

every

gives

M(AI).

u

is a b s o l u t e l y

to the m e a s u r e

Thus on

onto

u ÷ ~u

is q u a s i b o u n d e d

continuous

(resp.

linear

(resp.

singular)

singular)

with

X-

u E Q(R)

such that

an o r d e r - p r e s e r v i n g

determines

d~ u = f*d X

or,

u : I

a unique

X-summable

function,

f*,

equivalently,

f*(b)kb

dx(b)"

A1 Indeed,

kb(a)dx(b) ,

at the point A I.

a.e.

We will

an extended

consider

function

functions

s

let

on

R

is bounded

s

lim infRgz÷ b s(z)

and

= -~(-f).

H[f]

(i)

S(f) and

are Perron

are n o n v o i d

~ H[f]. and

H[f]

is called

the

pactifieation

R*

3B. following

that

on

A1

when

for

R*.

it is

Let

Let

S(f)

f

be

be the class

such that

for every both

(resp.

s' E [(f)

families;

(iii)

= H[f].

The common

solution

Riesz's

and

H[f]

if both

(resp.

supremum)

(ii)

~(f)

are h a r m o n i c S(f)

is denoted

problem

function

theorem

are n o n v o i d

s" E S(f);

function

of the Diriehlet

S(f)

infimum

and

resolutive,

the b o u n d a r y

and

Then we have the f o l -

H[f]

f

b e A.

~(f)

S(f)).

We call

with

problem

be the p o i n t w i s e

for any

We r e f o r m u l a t e form.

R

X.

A = A(R). R

~ f(b)

S(f)

s' ~ s"

H[f]

a.e.

of

on the b o u n d a r y

below,

H[f])

in

on

on

Suppose

(resp.

of functions

lowing:

is satisfied

the D i r i c h l e t

(A2)

S(f)

measure

among m e a s u r e s

to the measure

(AI)

We set

and

as basic

Problem

real-valued

of s u p e r h a r m o n i e

R

X

with respect

The Dirichlet 3A.

and

a E R, is seen to be the h a r m o n i c

We regard

So we say that a c o n d i t i o n

satisfied

3.

a.

S(f)

by

H[f]

for the M a r t i n

f.

(Theorem

6F,

Ch.

on

and

I) in the

com-

44

Theorem.

Let

u

a closed

subset

positive

measure

be a p o s i t i v e of

R. M

superharmonie

Suppose

supported

that

u(0)

by the

function

< +~.

closure

on

Then

E*

of

of

U

R

and

there E

E

exists

in

R*

a

such

that

uR(z)

for

every

mined

Here,

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

~IR

to

R

is d e t e r -

uniquely.

Proof.

Let

subsets s(z)

z E R.

: I k(b,z)du(b)

{E

of

= lim

E n÷~

: n = i,

2,... }

n such that u R (z)

U ~

n=l

for

: E.

En

z E R.

n superharmonic

nonnegative

be a n o n d e e r e a s i n g We h a v e

It is e a s y to

function

and

s(z)

sequence

uR

n

< uR

n+l

see t h a t

R < UE(Z).

of c o m p a c t < u R.

s(z)

Since

Set

is a

uR

> u n

q.e.

on

En

__> u(z) s

for

q.e.

every

on e v e r y

n

by T h e o r e m

En

6E-(b),

and therefore

on

Ch. E.

I, we

see t h a t

It f o l l o w s

s(z)

that

uR <

and hence R R u E = s = lim u E n÷~ n

(i)

By T h e o r e m

6E-(c),

Ch.

I,

all

u~

are

So e v e r y

potentials.

E

n

car-

n

ries

a positive

measure

~n

such that

R (a) uE

= I g(a,z)d~

n (z).

n

Since

we h a v e

(2)

] g(0,z)d~n(Z)

: u~

(0) ~ u(0)

< +~,

n

~n

has

no m a s s

get a f i n i t e

at t h e

measure

point

on

0.

E

Setting

dMn(Z)

= g ( 0 , z ) d m n ( Z ) , we

satisfying n

uE

(a) n

From

(2)

sures

on

sary,

we m a y

on

(3)

E*

follows the

that

compact assume

{~n: set that

We t h e r e f o r e

I k(z,a)du(z)

n

E*. {~n }

=

f

k(z,a)d~n(Z).

= 1, Thus,

2,...}

forms

passing

converges

to

vaguely

a bounded a

to

have

: lim N÷~

set

subsequence

| m i n { k ( z , a ) , N} d ~ ( z ) J

a positive

of if

meaneees-

measure

45

= lim lim 5 min{k(z,a), N} dPn(Z) N+~

n÷~

< lim f k(z,a)dPn(Z) n÷~ : lim UER ( a ) : n ÷~ n On the other hand, large that

K

a E R

be held fixed and then take

= {z E R: g(a,z)

a potential which (Theorem

let

uER(a).

is harmonic

~ e}

is compact.

except at points

6B, Ch. I) a positive measure

1

on

in

Since

e > 0

~K , there

~K

so

min{ga,~}

is

exists

such that

min{g(a,z), e} = f g(z,b) dl (b). Setting

z = a, we see that

f dl

f

(4)

By Fubini's

(b) : i.

theorem we have

flf

We note that I

is continuous ting

n ÷ ~

k(z,b)dl

min{~(z,a), ~} g(z,0)

(b) :

and finite throughout in (4) and using

R*

as a function

in

z.

So let-

(i), we get

(5) = I min{~(z,a), ~} dp(z)

g(z,0)

R uE

Since

is lower semicontinuous ~

are shrinking to the point

a

as

f dX

: i

to

eonvergenee

f k(z,a) d~(z)

desired

integral

Theorem

6B, Ch.

Corollary. that

Let

b ~ E*.

as

~ + +~.

(b).

Combining Finally,

these with

unicity of

(3), we get the

B[R

follows

from

[]

b E AI(R) Set

1

theorem shows that the last member of (5) tends

expression. I.

of

~ ÷ +~, we see that

u~(a) £ lim inf f u~(b)dl The monotone

and the supports

and let

E = E* h R .

Then

E*

be a closed

(kb) R E

subset of

is a potential.

R*

such

46

Proof.

By the p r e c e d i n g

(6)

(kb)~(a)

for some p o s i t i v e v(a) kb

t h e o r e m we h a v e

measure

: S k ( z , a ) d~'(z) is m i n i m a l ,

kb(a)

for

we h a v e

~ 0, t h e n we set

that way.

Thus

3C. write

~ = 0

subset

u

e

~' : pI A

with

support

of

does not con-

point

kb

(kb) ~

is a p o t e n t i a l .

harmonic

is a p o s i t i v e b ÷ (kb)~(a)_

be w r i t t e n

function

measure

If

expression

v

cannot

and

Since

0 ~ ~ ~ i.

a contradictory

the c l o s e d

~

We set

on

[]

on

R

A.

is i n t e g r a b l e

in

and

If

E

for e v e r y

and

a

of

R \ E, t h e n

If

a E R \ E

a, then,

and h a v e

R, t h e n

uER(a) = I If

E*.

0 ~ v £ (kb) ~ ~ k b.

be a p o s i t i v e

where

of

(7)

Proof.

Then

and t h e r e f o r e

Let

by

for some

In fact,

u = fA kb d~(b), a E R

a E R.

v = ~k b

so t h a t the e x t r e m e

Theorem.

is a c l o s e d fixed

b

k(z,a)dD(z)

supported

dv = ~-id~'

= ~ k(z,a) d~(z).

t a i n the p o i n t

p

f

=

is an i n t e r i o r u~(a) and

as we h a v e

uR(a)

= u(a)

if

G

k

point and

of

E

or a r e g u l a r

(kb)~(a)

is the c o n n e c t e d

seen in T h e o r e m

: f u(z)d~G(z):

([CC],

(kb)ER(a)d~(b)"

fA(f

Ch.

of

44-45)

boundary

= kb(a) , so that

component

6E-(a),

pp.

R \ E

point

(7) holds. containing

I,

kb(Z)d~aG(Z)] d~(b)

A Finally, Theorem E.

let 5B,

a E E Ch.

be an i r r e g u l a r

I, the

singleton

{a]

boundary forms

point

of

a connected

R \ E.

Then,

component

So we can find a s e q u e n c e

by

of

{G : n = i, 2,...} of J o r d a n d o m a i n s in n N n=l ~ C R \ E, CI(Gn+I) = C G n and CI(G n ) = R such that a E Gn, ~Gn = {a]. Let ~ be the h a r m o n i c m e a s u r e of G at the p o i n t a. Then n n we h a v e

(kb)R(a)

= lim I ( k b ) R ( z ) d ~ n ( Z ) n -~

and t h e r e f o r e

the f u n c t i o n

b ÷ (kb)R(a)s

= nlim f -~

is m e a s u r a b l e .

Hence,

47

n÷ ~ ~ A

[lim I ( k b ) ~ ( z ) d ~ n ( Z ) ] dB (b)

: IA "n÷~ : I

(kb)~(a)dz(b)' A

as was to be proved.

3D.

[]

We are n o w in a p o s i t i o n

to p r o v e

the m a i n r e s u l t

of this

section. Theorem. cation

boundary of

R

The M a r t i n

in the A

Let

f

The h a r m o n i c

is g i v e n by

be a n y r e a l - v a l u e d

it to a c o n t i n u o u s f.

a

We m a y a s s u m e

function that

R*

is a r e s o l u t i v e

every real-valued

is r e s o l u t i v e .

at the p o i n t

Proof.

compactifieation

sense t h a t

on

measure,

kb(a)dx(b). continuous

R*, w h i c h

0 ~ f ~ 1

continuous

on

with

function

on the

support

([CC],

is d e n o t e d

R*.

eompactifi-

function

p.

on

in

AI,

140) A.

We e x t e n d

by the same l e t t e r

For e v e r y

n = i, 2,...

we

set

A i = {b e Al: E9 : {a e R*: 1

( i - ~) =< f(b) f(a)

<

( i + 7)},

< i - i} U {a e R*: = n

f(a)

> i + i],= n

E. : E g A R , l l and

ui = I

kb dx(b) A. 1

for

i = 0, i,...,

by C o r o l l a r y Theorem

n.

3B

For e v e r y

( k b ) ~. 1 3C we see t h a t

is

(ui)R Ei is a l s o a p o t e n t i a l . f < (i + l)/n

on

Since

fixed

i

a potential

: I

A. 1

(kb)~

we see

for

.i

(ui) ~. = u i i R \ E~, we h a v e

any

A iNE~

b E Ai .

= ~

and so

By u s e

of

dx(b) q.e.

on

Ei

and

(i - l)/n

<

48

i - l(u i q.e.

on

R

R - (Ui)E) 1

i + i =< fu i =< - - U n

R i + (Ui)E. 1

and t h e r e f o r e n

n n i - l(u i (u i R i + I n )E. ) ~ f ~ [ --n--ui + [ i=0 l i:0 i=0 q.e.

on

R.

So we f i n d a p o s i t i v e

superharmonic

R (Ui)E. 1

function

s'

on

R

such that n n ~ i+ lu. + [ --£-- i i=0 i:0

f < =

on

R

for any

to the class As

E > 0

e > 0. S(f)

n

i=0 n + ~

)E.

l

+ ~s'

shows that the r i g h t - h a n d

and t h e r e f o r e

is a r b i t r a r y

It is a s i m p l e m a t t e r

as

This

(u i R

majorizes

member

the f u n c t i o n

and

[i=0n (Ui)E.R is a p o t e n t i a l , 1 n ~[f] ~ ~ +__~lu. i " i=0 n l

belongs

H[f]

on

R.

we h a v e

to see that

i+ lu. = n l i=0

A. i

i + I k b dx(b ) ÷ I kbf(b)dx(b) n AI

and h e n c e f ]

~[f]

kbf(b)dx(b)" A1

Similarly,

we h a v e f H[f] ~ ]

kbf(b)dx(b)" A1

By c o m b i n i n g Corollary.

these

inequalities

A function

X-summable.

If this

f

on

r : J )

is r e s o l u t i v e

result.

if a n d o n l y

[] if

f

we have

f(b)kb(a)dx(b) A1

a E R.

The p r o o f omitted

AI

is the case,

H[f](a) for e v e r y

we get the d e s i r e d

(cf.

is s i m i l a r to t h a t of the c l a s s i c a l

Ch.

I, 5D).

case and is thus

is

49

§2.

4.

FINE LIMITS

D e f i n i t i o n of Fine Limits 4A,

Let

b E A I.

b a l a y a g e d function is a potential.

A subset

(kb) ~

u

of

is different

R

is called thin at

from

In fact, by Riesz's t h e o r e m

superharmonic function function

E

if the

In this case

(Theorem 6F, Ch.

(kb)RE

I), the

is the sum of a n o n n e g a t i v e h a r m o n i c

(kb) ~

and a p o t e n t i a l

k b.

b

U.

Since

u ~ k b, the m i n i m a l i t y of

kb

implies that u = ~k b with 0 ~ ~ ~ i. If ~ > 0, then (kb) ~ = ~k b + R U ~ ~(kb) E + U q.e. on E and t h e r e f o r e everywhere on R. But this leads to a c o n t r a d i e t i o n If

E

R (kb)Et~,

and

E'

(kb) ~ = k b.

are thin at

b, then so is the union E U E ' , for ~ R Let G(b) be is m a j o r i z e d by the p o t e n t i a l (k b) + (kb)E,.

the c o l l e c t i o n of sets of the form seen that

G(b)

is a filter,

and

G(b)

is closed under finite

V NR

E G(b)

R \ E

E

intersection.

for any n e i g h b o r h o o d

We see further that,

with

thin at

i.e. every m e m b e r of

V

of

and

i, 2,

G2

E ~ R

be two components b e l o n g i n g to

R \ Gj

is thin at

b

so that

3B says that

is thin at G(b).

G(b).

R (kb)RkGj

We have

in the space

has a unique c o n n e c t e d component that belongs to GI

b.

is n o n e m p t y

Corollary

b E 41

if a closed set

G(b)

R*. b,

R \ E

In fact,

Then,

let

for each

is a potential.

j :

So

R R + R k b ~ (kb) R ~ (kb)R\Gl (kb)R\G 2 and thus

4B. b E 41

kb

should be a potential,

Let

f

be a map from

G(b)

= N~CI(f(D)):

is a filter,

taken outside any compact effective if

f

be the set of b E D(f)

X

we denote by

subset of

for w h i c h f(b)

the fine limit of

For any open set G E G(b).

X

For any

implies that

R.

f

f^(b)

b.

41(G)

X G ~ R

R.

is a singleton.

the point c o n t a i n e d at

f^(b)

So the d e f i n i t i o n of

is defined off some compact subset of b ~ AI

function, then we take as

with

X.

and depends only on the values of

The f u n c t i o n

is called the fine b o u n d a r y function for

line.

space

D e G(b)~.

the compactness of

a nonvoid compact subset of

f(b)

into a compact

we set f^(b)

Since

R

a contradiction.

f.

If

in

f^(b)

Let

is

f is

D(f)

For every

f^(b)

and call

f

defined on

f

is a n u m e r i c a l

D(f)

the Riemann sphere or the extended real we denote by

is a Borel subset of

41(G) 41 •

the set of

b ~ 41

To see this, we may

B0

assume Then

that

G

is c o n n e c t e d .

G • G(b)

if and

= f kb(Z) d ~ ( z ) £ × R, we AI(G)

and

see that

only

the m a p R (kb)R\G(a)

measurable

Let

X

be a c o m p a c t

function

from

R

into

and

Proof.

therefore

For

of c e n t e r X

each x

and

in

A.

We

can

shows

It is easy

the

the

right.

k = i, set

=

desired to

b

there

a convergent

closed,

we h a v e

k

that

such

have

the

we

latter

as was

be any

exists

Analysis

by

W(f)

5A.

functions set then

supremum) are also W(f)

f

Perron that

that

f^(b)

So

f

be a c o n t i n u o u s

D(f)

+ X

is m e a s -

£ I.

([CC],

denote

the

subset

of

n = i,

p.

147)

open X.

2,...}

ball Since

which

is

W(f)) R

on

Since

A.

there

For

every

to the

k = i,

2,...}

x.

Since

exists

f-l(B(Xn;k-l))

an

e

and

the r e v e r s e

A

is

index we

E G(b),

As

b • D(f)

shows

in

belongs

by

~ Cl(B(x;e)). Hence

This

side.

{Xn(k):

Then

is a B o r e l

is i n c l u d e d

b

is d e n o t e d

c > 0.

f^(b)

(8)

that

is c o m p a c t ,

= {x}.

in

of

right-hand

such

limit

any

so

side

is a r b i f(b)

: x,

inclusion,

Behavior

the

such s) in

h[f]

families,

and

set

that R.

(resp.

real-valued

s ~ f

(resp.

If both h[f])

[(f)

as the

in

W(f)

(resp.

h[f]

and

We ~[f]

call

function

of s u p e r h a r m o n i e

both

~ [[f].

are n o n v o i d

Al(f-l(B(Xn;k-l)))

[]

of f u n c t i o n s

h[f]

X

be an e x t e n d e d

on

we d e f i n e

of

{Xn:

in the

~ B(x;~). and

contained

(resp. s

let f:

be a c l o s e d

left-hand

whose

Take

of B o u n d a r y

(depending

and

B(x~r)

n = n(k)

• G(b)

shown.

Let

on

function.

Al(f-l(B(Xn;k-l))),

point

an

B ( X n ( k ) ; k -I)

being

5.

fixed.

(kb)~\G(a)

is c o n t i n u o u s

measurable

subset

for e a c h

Since

x E A.

conclude

to be

space

let A

U n=l

the

subsequence

f-l(B(x;c))

trary,

A k=l

result,

Al(f-l(B(Xn;k-l)).

has

Let

is h e l d Since

A I.

a sequence

see t h a t

So let

2,...

r.

÷ kb(Z)

function

r > 0

find

f-l(A)

set.

the

which

< kb(a).

claim

(8)

which

of

is a Borel

and

radius we

(b,z)

metric

Then

D(f)

x • X

is s e p a r a b l e ,

dense

X.

a • G

is a B o r e l

subset

Theorem.

urable

any

(kb)R\ R G (a)

since

b ÷

is a B o r e l

Take if

f

= hill.

W(f)

pointwise

W(f)).

infimum

Since

W(f)

if b o t h

is the

We d e n o t e

a compact

are n o n v o i d ,

are h a r m o n i c

harmonizable If this

R.

subharmonic)

s ~ f) o u t s i d e and

h[f]

on

(resp.

case,

on

(resp. and R.

W(f)

W(f) We

see

and

we d e n o t e

the

51

common

function

by

h[f].

if it is h a r m o n i z a b l e denote

by

Wiener

functions

not

W(R)

assume

Theorem. there on

the

and

a compact

f

f r o m the

K s =C R.

set

Let

f

n = i,

By the

2,...}

a*

K

C R n

function

such

that

i,

2,...}

s

=

n

:

Theorem

IA,

Ch.

Then, u(a)

for

u- U ~ f ~ u+ U

with

s O be t h e

in

K

sO = U2

any

such that some

u + U' ~ f

u - U" ~ f

on

set R

p.

on

R,

54)

u = h[f]. h[f],

outside function

It is a l m o s t

there

exists

a

< ~,

.

n

sum

[n=l (Sn - u)

we h a v e

We c h o o s e

K]. C = Rn

R (s n - U ) C I ( R

[ n:l

a compact

a regular

is

set

for

exhaustion

j = i,...

2n

(see

6E-(c),

~ u(a)

~ f

Ch.

on

R

I.

U'

and,

K'.

Similarly,

and,

for any

j = I,

(s n

be a n y 2,...,

positive we have

- u R (a) ) C I ( R n)

n the Ifl

compact on

Since

set

m

s (a) ~ f(a).

outside of

Let

with

+ m -I 2m~2j n=m+2j

majorant

If we

), n

U 2 = (s O ) ~ I ( R 3 ).

CI(R2).

compact

a harmonic

([CC],

So t h e

a E Rm+2j \ Rm+ j

superharmonic

on

and

such that

R.

such that

by Theorem

u + m-iUl

by s e t t i n g

u.

For every

n=m+2j

U2

E HP'(R)

n

+ m-iUl(a)

we have

h[f]

R

W(f)

m- 1 2m~2j

Namely,

of

we do

Set

is a p o t e n t i a l

integer.

in t h a t

exist

of

outside

R

UI : which

[CC]

if t h e r e

with

=

of

I).

We

h[f] - U ~ f ~ h [ f ] + U

function

R.

R

R.

h [ f ] - sU ~ f £ h [ f ] + EU

definition

point

on

> f n

{Rn:

we h a v e

R

h[f]=

in

is a n y p r e s c r i b e d

a superharmonic

on

(s ( a * ) - u ( a * ) ) n

n:l where

on and

be a W i e n e r

on

on

Our definition

book

such that

Conversely,

u E HP'(R). {Sn:

R

U

R.

in t h e

s > 0,

function

on

function

majorant

f. f

on

and a potential

that

sequence

U

is a W i e n e r

functions one

of

for a n y

f

a superharmonic

function

a potential

(i)

clear

has

quasicontinuity

is a W i e n e r

Proof.

say t h a t

Ifl

set o f W i e n e r

such that,

u E HP'(R) then

the

differs

For any Wiener

exists

R

We

and

= U 1 + U2,

for a n y we can

s > 0,

R

CI(R2)

and

CI(Rm+I).

define

then

U'

Ifl =<

is a p o t e n t i a l

u + sU'

£ f

a potential

U"

u - cU" ~ f

Let

a potential

C = R3, w e h a v e

e > 0, find

set

outside

outside such that

a compact

52

set of

K". the

So

(ii) tive

U

first

= U' + U"

half

Next

of

suppose

superharmonic

identically. f RR\ R

n

see

= i,

a potential,

that

at

which

meets

the

requirement

theorem.

function

We

for

is

the

f

is

whose

once

2, . . . .

a potential

greatest

that

f

Then,

by

U,

i.e.

harmonic

> h[f]

> h[f]

Theorem

6E,

f

is

minorant > 0.

Ch.

I,

a posi-

vanishes

We

set

sn

is

sn super-

n

harmonic sn

= f

since h[f]

R,

on

u = lim

u+

on

such

R \ Rn. s

.

We

Then,

= 0.

(iii)

Finally

with

The

following

(a)

The

f f

: u.

The

belong

W(R)

u ~

f,

we

a Wiener

function

function

on

a potential

properties

R

U.

finishes

h[f]

that

is

a

Set hand,

0.

Hence,

: 0.

u-

U ~

implies

f

that

[]

consequences

f ÷ h[f]

and

n.

other

with

(ii)

R

every the

u ~

proof.

easy

on

have

such

Then

the

are

On

should

a

This

for

u ~ h[f].

is be

s n =< f

Rn,

=> S n + 1 => 0

E ~(f),

correspondence

map

to

f ÷ h[f]

of

linear

Theorem

map

of

5A.

W(R)

W(R),

then

(c) the

where

Let

f E W(R)

decomposition

be

of

harmonic

and

h[f]

= 0

if

and

if

Then

the

(d)

Let

Proof.

h[f"]

only

and

i.e.

min{f,f'}

We

number

of

only

state

easily.

the

Take

any

= ~(~i , + ~,,)~

f'

f'

and

f"

potentials + U"

on

are U'

R.

and

~.

if

f

belong

theorem

and to

(Theorem

L

=

1R L

is

a potential

{z E R:

proof

of

(d),

real

for

numbers

= m a x { ~~,- ~

f'

Let

Then

set

64)

U"

h[f']. f = u+

6F,

u

Ch.

U

I),

= h[f].

We

have

a potential.

p.

Wiener

= h[f]A

function.

a potential.

([CC],

two

and

Since

Riesz's is

is

function

h[min{f,f'}]

superharmonic

by U f

f E W(R)

balayaged

countable

Since

homomorphism,

max{f,f'}

h[f'], a

f

is

exist

a lattice

both

= h[f]v

u

rather

is

and h[max{f,f'}]

have

Sn

on

HP'(R). (b)

be

n and

and

5B.

f'

h[f]

harmonic

clearly

R

let

u E HP'(R)

is

s

Namely,

and

onto

have

on

f E W(R)

Theorem.

sn

since

n÷~ n u is h a r m o n i c

: h[f]

U

that

and

< ~" f"

Theorem

such

f'

= 1

on

L

,

~ E

an

follow

and

~.

at

most

from

5A

set

f - ~~ , 0}. = max{~,-~-T~_ 5A

< h[f']

(resp.

for for

others

functions, that

: ~]

except

the ~'

0)

f(z)

f"

shows + U' = 1

that and on

there f" L

< ,,), w e

53

i[,

J h[f'] + U'

iR L ,, < h [ f " ]

(resp.

+ U")

and therefore (g)

min{l[

As we have,

~,

in v i e w

, iR } ~ min{h[f'], ha,,

of

= h[f'] A h[f"] = h[0]

left-hand

side

h[l~ Since

,

with

~ E •

functions.

~,

of

(9)

] Ah[l~

are

arbitrary,

forms

a family

consequently

Let

be an e x t e n d e d of

If

we have

It f o l l o w s

that

iR }] = 0. ~, ' L ,,

shown

of m u t u a l l y

R

] =

that

the

orthogonal

set of all

h[l

nonnegative

]

harmonic

G

be an o p e n

to each

case

for c o u n t a b l y

set

in

continuous

connected

is a p o t e n t i a l , the

h[l RL ] =< i

v

except

real-valued

f

IFR

in p a r t i c u l a r p.

] = h[min{l~

h[l R ] = 0

Lemma.

restriction tion.

= 0,

So we get

5C. f

+ U".

: h[min{f',f"}]

is a p o t e n t i a l .

~,,

h[l

and

+ U'

(b),

h[min{h[f'],h[f"]}]

sn the

h[f"]}

when

then F

R

many

and

function

a.

set

on

F = R \ G.

R

such

component

of

~

a.e.

on

AI(G).

subset

of

R.

exists

is a c o m p a c t

G

[]

Let

that

is a W i e n e r

the func-

This

is

([CC],

150)

Proof.

Let

applied

to each connected

L

= {z E R:

a countable

dense

on

every

G'

for

If we set

and,

E S : 0

set

on

F'

define

F.

at

~ E S

Then

points

S

f~

= ~}

of

and

= {a E R:

by Theorems

and

except

subset

f(z)

with

component, ~

such

every l~(a)

we h a v e

R

by

the

belonging

~ E ~.

of

G

By T h e o r e m

in place G,

that

IG,NL

of

5B-(d)

R, we f i n d

is a p o t e n t i a l

G' ~ 1/2},

6D a n ~ 6 E - ( a ) , on

G',

Ch.

I,

setting

following:

to a polar

set;

then

F'

F~F'G,

is a n o n v o i d

is

f~ = I F , ~Lq ,~A (a)

f

a polar on

set. G'

is c o n t i n u o u s

(b)

0 ~ f

is u n b o u n d e d

on the

closed

£ 1; a n d

Let and on

R

(c)

GT

f

IG'

~ IG,NL Let

U

, which

is a p o t e n t i a l .

be a p o t e n t i a l

which

set of

irregular

54

boundary

points

of

G

and

s = i RL

set

Moreover

.

we

[U + s J 1 R U + s F + min{s-

S

Then

s

is not R,

is s u p e r h a r m o n i c larger

so that,

than

U+ s

on

R,

on

in the n o t a t i o n h[f

on

G'.

Since

G' IG,AL

on

G'

Since

h [f ] ~ s

for

G. of

R SF, s

G' 1G,AL

is l o w e r

Furthermore,

5A, we h a v e

] ~ U + s~

= U + s

and

are

tion

on

R

potentials with

h[f

on

R.

] = 0.

on

on

This

G',

F

Gt

majorizes and

we h a v e

and

means

Hence,

on

] < s

Rsince

and

f

on

therefore

that

by T h e o r e m

h[fa]

U

i r r e g u l a r b o u n d a r y p o i n t s of G, h[fa] ~ U+ s F on ~[f] R We thus c o n c l u d e that U + s~ ~ U + 1 F on R. R IF

F

semi-continuous

s

h[f

on

G' IG,AL

+

is a p o t e n t i a l

}

set

is u n b o u n d e d

F.

] = 0, for

is a W i e n e r

5A,

at

Consequently,

h[f

f

~ U + s~

f

is

U

func-

majorized

by a p o t e n t i a l . We t h e n

claim

that R

(1O) q.e

on

R.

This

is c l e a r

the

definition

G"

be a c o n n e c t e d

of

ILR This

shows Next

~ 2(1

) + fa

on

F' n L

and

and the

component

: H[I~

(i0), we

F'

IL

;G"]

which

fact

of

in turn

means

0 SA

subset

~ E S.

kbdX(b)"

To

Noting

0 < u = As

shown

x(A a)

A

above,

ILR

that

of

A1

is

=< 2(i~)

iR L

that

is

F'

in v i e w

a polar

we h a v e

;G"]

that

on

set.

of Let

G"

+ f~.

a potential

on

R.

x(A

) =

0 ~ u ~ 1

=

is a p o t e n t i a l

= 0.

set

= k b}

in 4B.

we c o n s i d e r

A : U{A

Then

we h a v e

the h a r m o n i c

and

using

Theorem

(kb)R

dx(b)

= u~

A

We f i n a l l y x(A)

(kb)RL

as n o t e d

see this,

kb d×(b)

=

= 0.

follows

Then

of

set

is a Borel

for a n y

FAF'

.

=< H [ 2 ( I ~ ) + f

A a = {b E AI(G): which

that

G \ L

outside

La

and

: a E S}.

As

~ 1L . a

u = 0. S

u =

3C, we have

a

therefore

function

Hence,

is c o u n t a b l e ,

it

55

To c o n c l u d e (k b)

< kb

c o m p o na e n t

of

longs to

G(b)

either

in

[a, +~]

the proof,

and so

R\ L

R k L

(a, +~]

or in

[-~,

Hence,

h[f]

Theorem.

(a)

function h[f]

on

Let

f

< kb

), w h i c h So,

a E S.

(see 4A).

be-

G

is c o n t a i n e d

is c o n t a i n e d

f^(b)

Then

connected Thus

is c o n n e c t e d ,

f^(b)

shows that

AI(G) \ D(f) ~ A

either

is a s i n g l e t o n ,

and c o n s e q u e n t l y

in

i.e.

X(AI(G) \ ~(f))

=

[]

connects

for a W i e n e r

R.

f(G

Take any

be the u n i q u e

G

R G (kb)R\

This

The next r e s u l t

function

Let

[-~, e).

a].

0, as was to be proved.

5D.

b • A I ( G ) \ A.

s u c h that

and the r a n g e

or in

b E 0(f).

let

• G(b).

the fine limit

function

f

and the h a r m o n i c

f.

be an e x t e n d e d

real-valued

T h e n the q u a s i b o u n d e d

continuous

part of the h a r m o n i c

Wiener function

is g i v e n by

I

~(b)kb

dx(b).

A1

(b) defined positive Proof. and

If

f

a.e.

on

is a b o u n d e d AI, t h e n

harmonic Let

f

X(A I \ ~(f))

continuous

fu

function

u

on

be a c o n t i n u o u s = 0.

R.

function

on

E, = {a • R: f(a) 1

< i-i} = n

R

and if

f

is

for any q u a s i b o u n d e d

p. 151) R

number

!(in - 7)i =< f(b)

on

function

([CC],

Take any n a t u r a l

A i : {b • N(f):

function

is a W i e n e r

such that n

0 ~ f ~ i

and set

< 2(ni + 7)i },

U {a e R; f(a)

> i + l},= n

and

ui for

i : 0, i,...,

is a p o t e n t i a l .

n.

If

Using

: I

As

kb~(b)dx(b)

b E Ai, t h e n

Theorem

(ui)R = Ei is also a p o t e n t i a l .

A.

R \ E i 6 G(b)

so t h a t

3C, we see t h a t

A. i

1

R ~(b)dx(b ) (kb)E. l

in 3D we h a v e

n i - l(u " _ ( u i ) R ) n 1 i:0 l .

< fu < = =

n n [ i+ 1 • + [ n Ul i:0 i:0

(kb)~.

( u i ) ER

i

56

q.e. R

on

R.

So t h e r e

exists

a positive

superharmonic

function

s'

on

such that n

holds

on

class

R

~

=

i=O

for any

W(f)

n

fu <

i+ lu. + --n---

E > 0.

i

[ i:0

(u i R )E.

I

The r i g h t - h a n d

+

ss'

member

thus b e l o n g s

to the

and so n

~[fu]

<

[ i + 1 ui' n i=C

=

for

[ n (ui) ~ i=0

is a p o t e n t i a l .

Similarly,

1 n

[ i - l u . < h[fu]. u-:~ l n i = Combining

these

inequalities,

we see that

fu

is h a r m o n i z a b l e

n n [ i+l : lim ~ i+lu. : lim ~ ] kb n ÷ ~ i:0 n l n+~ i=0 A. n

h[fu]

and

u(b)dx(b)

i

= I kbf(b)~(b)dx(b)" This

shows

(b) and also

(a) as w e l l

In o r d e r to show the that

f > 0.

We set

:

bounded

part of

h[f]

lim n÷~

f

n

statement = min{f,n}

for a b o u n d e d

for

f, we m a y a s s u m e

n = i, 2, . . . .

The q u a s i -

is t h e n g i v e n by

(h[f] A n) : lim h [ f n] : lim I k b f n ( b ) d x ( b ) n÷~ n÷~

= I kbf(b)dx(b)" 5E.

f.

(a) for a g e n e r a l

As a s p e c i a l

[]

case of the p r e c e d i n g

theorem,

we get the fol-

l o w i n g result. Theorem.

(a)

quasibounded

If part

u E HP'(R),

then

prQ(u)

u

of

prQ(u)

~

exists

a.e.

on

AI

and the

is g i v e n by

= I

kb~(b)dx(b)" AI

In p a r t i c u l a r , 0

a.e. (b)

on

if

u

is a s i n g u l a r h a r m o n i c

function

on

AI . If

u*

is a X - s u m m a b l e

function

on

A], t h e n

R, t h e n

~ =

57 r

u = H[u*]

=

kbU* (b)dx(b)

I

J £1

is a quasibounded Proof.

(a)

evidently

6.

This is immediate

That

u

in 2C.

It follows

§3.

on

Al"

theorem,

for

u

is

u = h[u].

~

again from Theorem

a.e.

from the preceding

is a quasibounded

So, by (a),

G = u*

harmonic

exists

a,e.

2C that

function was already

and

u* = G

in-

u : fA I kbG(b) dx(b). a.e.

[]

COVERING MAPS Correspondence 6A.

face

Let

R

of Harmonic

(R, 0) (R', 0')

~

(A2)

every point

onto

(R, 0)

is a h o l o m o r p h i c

every connected

of a hyperbolic

0 E R, called the origin of

(AI)

under

Functions

be a pair consisting

and a point

another pair

component

of

Riemann

A map

~

sur-

from

is called a covering map if

function

a E R

R.

from

R'

onto

R,

has an open neighborhood ~-I(v)

is conformally

V

such that

isomorphic

with

V

~, (A3)

~(0')

The triple

= O.

(R', 0', #)

triple of

(R, 0).

(R, 0)

which

triple

(R', 0' , ~)

(R, ~)

onto

(R',

is conformally

Let

monic,

~

(resp.

(resp.

a conformal

of the open unit disk

(R, 0)

with origin and

and

(R', 0')

(~,

0, ~R,))

isomorphism,

super- or sub-harmonic)

on

(R', 0')

f

if R'.

on f o ~

isomorphic with

R

Riemann

onto

~R = ~ ° ~R'"

a function

(R, ~)

The uni-

and the origin

be the universal

(R', 0')) such that

super- or sub-harmonic)

~

Ch. I, 18C).

be hyperbolic

a covering map of

of

~'

isomorphism and is called the

(R, ~)

IIG).

harmonic,

up to a conformal

theorem says that 0)

of

Such a surface

([AS],

Ch. III,

0, ~R )

$ = ~ o ~'

(R, 0)

(D,

(R, 0)

(R, ~, $)

one can find a covering map

satisfying

surface

is called a coverin ~

triple

in the sense that for each covering

(R, 0)

0')

a covering

of

the pair

6B.

of

uniquely

covering

formization

having these properties

There exists

is the strongest

is determined universal

(~,

function and

a Wiener function and

(b) dieated

harmonic

0

surfaces

(R, 0).

covering Since

@

Let

triple of is locally

is h o l o m o r p h i c

is holomorphic

([AS],

(resp.

(resp.

har-

58

Let

TR

be t h e

multiplicative onto

itself

F = f o ~R that

F

group

such that

on

transformations linear

S R o T = SR"

for any

~

of c o v e r

of f r a c t i o n a l

is a f u n c t i o n

Fo T = F

tion on

group

on

D,

T E

T R.

is e x p r e s s e d

If

f

which

~R'

form

every

f o SR

i.e.

T

is a f u n c t i o n

is T R - i n v a r i a n t

Conversely,

in t h e

for

transformations

D

on

R, t h e n

in t h e

sense

TR-invariant

with

the

of

func-

a function

f

R.

Theorem. into

(a)

The map

HP'(R').

and only

if

(b)

So,

uo $

Let

s

is a p o t e n t i a l (e)

Proof.

on

function (i)

We

covering

Let

is q u a s i b o u n d e d

R

for any

on

f

R'

v ~

on

first

consider (D,

i = i,

i : i,

R

(resp.

function, ¢].

the

(R',

when

of

if

0',

R.

Then

on

R'.

then

$)

s

f o $

is the

is

uni-

(R, 0).

2, a n d w e

Since

HP'(R)

R'. on

is a p o t e n t i a l

is a W i e n e r

ease

on

function

h[f] o ~ = h[fo

of

singular)

set

v = (u I o S R ) v (u 2 o %R ).

v ~ u i o %R

for

i = i,

2, w e h a v e

T E TR

vo T = v on

isomorphism

singular)

so $

0, SR )

(u I v u 2) o SR 0

voT for

if

R

and

triple

(resp.

superharmonic

if a n d o n l y

u i E HP'(R),

Clearly,

is a l a t t i c e

is q u a s i b o u n d e d

be a p o s i t i v e

If a f u n c t i o n

a Wiener

versal

u ÷ uo $

u E HP'(R)

2

and therefore

for any

such

~ uioSRoT

that

T E T R.

v o T ~ v. Thus

v : u o %R"

= uio $R

there

Since exists

It is e a s y

to

TR

is a g r o u p ,

a harmonic

see t h a t

we have

function

u ~ u Iv u 2

u and

so (u I v u 2) o $R ~ u o ¢R = v ~ Consequently,

we h a v e

(u I V U 2) o SR : (u I o SR ) v (u 2 o SR ), w h i c h

that

the

correspondence

onto

the

set of T R - i n v a r i a n t

we

see t h a t

singular) Next v

be t h e

same

u + u o SR

a harmonic

if a n d o n l y let

s

greatest

if

u o SR

be a p o s i t i v e harmonic

as a b o v e

we

harmonic

function

It is e a s y to

show that

theorem

is a p o t e n t i a l

is.

(Theorem

u

Thus

is t h e 6F,

Ch.

on

Since

HP'(R)

I o @R : i,

is q u a s i b o u n d e d (a)

shows

of

(resp.

is p r o v e d . and

let

~.

By u s i n g

the

is T R - i n v a r i a n t .

So t h e r e

exists

of v

isomorphism

HP'(~).

superharmonic

see t h a t

if a n d o n l y

in

u E HP'(R)

minorant

argument

u

is a l a t t i c e

elements

function

a nonnegative

By R i e s z ' s

(u I v u 2) o SR"

function

s o SR

R

greatest

with

on

the

harmonic

I) a p o s i t i v e

if its g r e a t e s t

on

R

property minorant

v = u o @R" of

superharmonie

harmonic

minorant

is

s. function

B9 identically

zero.

This

(b).

proves Finally

exists

So

let

f

s

is a p o t e n t i a l

be a W i e n e r

a potential

U

on

R

if and o n l y

function

such

that

on

R.

if

By T h e o r e m

So we h a v e

h[f] o ~ R - U o ~R ~ f o ~R ~ h[f] o ~R + U o ~R'

to

HP'(~)

above.

By T h e o r e m

= h[f] o CR'

whieh

(ii) take

5A

We

v ~

servations

f o @R

proves

consider

u i C HP'(R),

before

U o @R

is a p o t e n t i a l

is a W i e n e r

general

i : i,

(i),

5A t h e r e

where

by what

function

on

h[f] o ~R

we h a v e

on

~

R.

and

seen h[fo

%R ]

(e).

the

(u I v u2) o 4. in

is.

h[f] - U ~ f ~ h[f] + U

belongs

and

s o %R

2, and By u s i n g

case set

¢:

(R',

0')

÷ (R,

0).

Let us

v = (u I o ¢) v (u 2 o ~).

Then

fact

our

the

CR = ~ o @R'

and

as ob-

we h a v e

(u I V u 2) o ¢R : (u I v u 2) o ¢ o @R'

~ v o %R'

= {(u I o ¢) v (u 2 o ¢)] o CR'

= (u I o ¢ o ¢R,)v (u 2 o ¢ o ~R,) = (u I o CR)v (u 2 o CR) = (u I V u 2) o CR" Thus

we h a v e (u I v u 2) o ¢ = v : (u I o ¢) v (u 2 o @),

from which on

R

if and

o %R'' R'.

7.

(a) follows. only

U o ~R

So

(b)

metric uous

U o }R

follows.

space,

use

there on

on

so by T h e o r e m

R

and

function

on

AI(R').

We

of

R'.

notations

in

a countable

AI(R'), D(~). we h a v e

C

the restriction 6B-(c)

It f o l l o w s

And

since

CR :

is a p o t e n t i a l

(a) and

(b).

on

[]

the

fiR f o ~

from

6B.

Since C

points

R*

is a c o m p a c t

of r e a l - v a l u e d of

R*.

is a c o n t i n u o u s is also

Lemma

5C

Wiener

a continuous

~

contin-

Let

is d e f i n e d

f E C. function

Wiener a.e.

on

set

is c o u n t a b l e DO

is a p o t e n t i a l

D. Uo %

family

separates

D O : A { D ( f o ~): Since

U

Measures

which

5D-(b)

if and

(i), on

is a c o n s e q u e n c e

our

By T h e o r e m

in

if

exists R*

shown

only

of H a r m o n i c

We a g a i n

functions

(c)

fact

is a p o t e n t i a l

is a p o t e n t i a l

Preservation 7A.

if

By the

has

In fact,

and

negligible let

each

D(f o ¢)

eomplement

b' E D O .

f e C}.

Since

in

every

has

negligible

AI(R'). f E C

We

complement

claim

that

is c o n t i n u o u s

in DO

on R ~,

80

@^(b') Since

C

separates

o n l y one point.

~

N fEC

{b • R*:

the p o i n t s

So

b' • D(@)

(ll)

R*, the r i g h t - h a n d

side can c o n t a i n

and

f($(b')) : ~ ( b ' )

for any

f • C.

Next we c o n s i d e r h(z')

for

z' • R'

w i t h the G r e e n

the f u n c t i o n values a.e.

in

on

h

AI(R').

and

Theorem. maps Proof.

= D(~)

is a B o r e l

into

Theorem

above

that

R

such that

G(b'),

$(b')

• ~(R). 7B.

i.e.

(a)

see

If

By T h e o r e m

6B-(b)

It is a l s o c o n t i n u o u s h

exists

with

and v a n i s h e s

of

h(b')

£1(R'),

= 0}.

XR,(D0(%))

= I, and

$

and

$ ~

D(h)

is a B o r e l

Take any

~ > 0

Since

h(b')

D(%)

are B o r e l set in

b' • D0(%).

sets and t h a t

AI(R').

If

$(b')

and an o p e n n e i g h b o r h o o d for

z • V.

%-I(v)

So

h(z')

is e a s i l y

~ ~ > 0, w h i c h

V

~ ~

We saw • R, t h e n of

$(b')

for a n y

z'

seen to be a m e m b e r

is a c o n t r a d i c t i o n .

in

Hence,

[]

$

XR,($-I(A)) (b)

R.

5D-(a),

• D(h):

N0(~)

= i.

The m a i n r e s u l t

Theorem.

{b'

subset

So

gR(0,z)

%-I(v).

we thus

N

4B shows that

XR,(D0(~))

to

R'

for

&(R).

find a c o n s t a n t

belonging

on

gR

so, by T h e o r e m

is B o r e l m e a s u r a b l e .

we w o u l d

function

We d e f i n e

N0(%)

DO(~)

the f u n c t i o n

: gR(0,%(z'))

is a p o t e n t i a l

[0, +~]

DO(~)

in

in

A = f o @(b')}.

f(b)

in this

part

is the f o l l o w i n g

is a m e a s u r e - p r e s e r v i n g = XR(A)

for any

f* • L I ( d x R ), t h e n

m a p of

XR-measurable

DO(~) subset

f* o $ • L I ( d x R , ) ,

into A

A(R), of

A(R).

Llf*liI = llf* o SUl,

and (12)

H [ f * ] o ¢ = H [ f * o $],

where

il-JJ1

fining

the

Proof.

(i

and extend

denotes solution Let

f*

the L l - n o r m and of the D i r i c h l e t

H[-] problem

be any r e a l - v a l u e d

it to a c o n t i n u o u s

denotes

function

as g i v e n

continuous f

on

the o p e r a t i o n

R*.

de-

in 3A.

function

on

By a d j o i n i n g

A(R) f

to

61

the class

C

complement

in

to

u s e d a b o v e we see that AI(R')

D0(~) N D ( f o

~).

is q u a s i b o u n d e d

and

D ( f o %) N D 0 ( # )

f*($(b'))

Since

f o #

with boundary

= f~(b')

values

~

(13)

H [ f * ] o 9 = h[f] o % = h[f o ~] : H [ ~ ] (12) h o l d s

function which

next that

is b o u n d e d

belonging h [ f O %]

h[f o ~] :

6B

for any c o n t i n u o u s

Suppose

b'

function,

and t h e r e f o r e

So we h a v e by T h e o r e m

(ii)

for any

is a b o u n d e d W i e n e r

H[~].

Hence

has a n e g l i g i b l e

f*

= H[f* o $].

function

on

f*

A(R)

on

A(R).

is a l o w e r

b e l o w and X R - S u m m a b l e .

semieontinuous

We take a n o n d e c r e a s -

ing s e q u e n c e converging

{f*: n = i, 2,...} of c o n t i n u o u s f u n c t i o n s on A(R) n p o i n t w i s e to f*. Since HI-] is a l a t t i c e i s o m o r p h i s m

the space

LI(dxR;~)

XR,-)

summable

bounded

(reap.

functions

harmonic

LI(dxR,;~))

onto the

functions

on

x[f*]

space

R

(reap.

: H[sup

since

f * o $ = SUPn

(13)

and

Theorem

H[f*] o * : ( V n

(reap. of q u a s i -

M[f~]

n

(f~o,)l

: v HEf~o*l

n of

XR-

Q(R'))

(f~ o $),

H~e~o Sj : ~Esup By u s e

(reap.

R'), we h a v e

f~] : v

n and,

of r e a l - v a l u e d Q(R)

of

n

6B we g e t

H[f*]) o , = V n n

= HLsup

(H[f~] o 0) =

V n

H[f* o $] n

(f~o $)I = H~f*o SJ.

n So

(12) h o l d s

continuous (iii) A(R).

for our

function Let

of u p p e r g~

exist

is b o u n d e d

functions

h~)

for any u p p e r

above

semi-

and X R - s u m m a b l e .

defined

XR-summable

sequence

{g~:

and a n o n i n c r e a s i n g

semicontinuous is b o u n d e d

(12) h o l d s

above

a nondecreasing

of l o w e r

(reap.

Similarly,

be any e v e r y w h e r e

semicontinuous

n = i, 2,...} every

which

f*

Then there

f*.

functions (reap.

on

function

sequence A(R)

below),

on

n : i, 2,...} {h~:

such that

g~ ~ f* ~ h~

and

lim S

n÷~ It f o l l o w s

that

and

=< H [ f * ]

H[g*]

{H[gn]}

f f°

lim f

is n o n d e c r e a s i n g ,

=< H[h~'~]n for

n : i,

n dXR"

n÷~

2,...

{H[hn]} .

is n o n i n c r e a s i n g

Moreover

62

~[h~](O)

- H[g~](O)

: I (h~- g~) d×R ÷ o.

So we have V n Applying and

4, w h i c h

h~

H[g~]

= H[f*]

is a lattice

A n

HEh*]. n

isomorphism,

and usin Z (12) for

g~

we get V n

H[g~o

$] =

=

v n

(H[g~] o %) = H[f*] o % =

A

(H[h*o

Furthermore,

since

see f i n a l l y

H[f*] o % = H[f* o 2]-

function (iv)

characteristic function

on

A(R),

A ~ A(R) function

of the

f~, that

H[g~ o 2] ~ H[f* o $] ~ H [ h ~ o

f*

Let

set

provided

A.

$-I(A).

Then,

(H[h*]n o ~)

f = J f*(b)dXR(b) : H[f* o $](0')

n = i, 2 ..... we for every

is e v e r y w h e r e subset

f* o $

defined.

and let

which

follows

= H[f*](0)

XR-SUm-

f*

be the

is the c h a r a c t e r i s t i c

(12),

From this

$],

(12) holds

f*

We see by

H[f*] o ~ = H[f* o 2].

XR(A)

Hence,

be any m e a s u r a b l e of

A n

$]).

n

n

mable

=

is v a l i d

for this

that

= (H[f*] o %)(0')

= I (f* o $ ) ( b ' ) d X R , ( b ' )

= XR,($-I(A)). Hence, f* o $ holds

(a) holds. maps when

tions.

f*

This 7C.

ing map

~R:

joining

Theorem.

TR

÷ R from

1-cycles

in

0 0. R,

the c o r r e s p o n d e n c e LI(dxR,;~)

= 0.

Given

T(0)

within

Since

covering

transformations

@R(0)

different

CR(y ')

func-

maps.

First we

for a u n i v e r s a l

T E TR, we take ~.

choices

determines

null

(12)

[]

case of u n i v e r s a l

with

f* +

and that

class m o d u l o

of the theorem.

of cover

with

into

as an e q u i v a l e n c e

the p r o o f

the o r i g i n

issuing

homotopic YT' of

~

in turn that

isometrically

We now look at the the g r o u p

l-cycle

implies

is r e g a r d e d

finishes

identify y'

This

LI(dxR ;~)

Then for

a unique

y'

cover-

an arc

~R(7 ')

is a

result

element,

in

say

F0(R). The map

the f u n d a m e n t a l

T ÷ YT

group

gives

F0(R).

an i s o m o r p h i s m

([AS],

Ch.

I, 19A)

of the group

TR

onto

83

7D. that

Using the notations

D0(%R)

theorems

R' = ~

and

~ = %R' we see

Besides the properties

mentioned

Let

LP(dXR)

0 < p S ~.

isometrically

Lebesgue measure only if

Then the correspondence

onto the subspace

ing of all TR-invariant

elements,

LP(do) T

where

on the unit circle.

f~ ÷ f * o SR

do

of

denotes

Moreover

LP(do)

maps consist-

the normalized

f* E HP(dXR)

if and

f* o SR e HP(do).

Proof.

We first note that the Martin compaetifieation ~

is the closed unit disk,

cumference origin

in the

in 7A and 7B, we have the following

Theorem.

disk

in 7A with

is TR-invariant.

T

its Martin boundary

and the corresponding

is equal to the normalized

of the open unit

harmonic measure

Lebesgue measure

is the unit ciron

T

do(t)

at the

= do(e it) =

dt/2~. (i)

Suppose that

p = i.

H[f*] o ~R = H[f* o SR]. for the TR-invariant a TR-invariant invariant

Since

harmonic

function

function.

on

function on

function

on

R

Theorem 6B-(a)

rem 5E,

v = HEv]

~. ~.

(ii)

v o ~R = H[u*].

Consider the general

by (i), we have

Conversely,

let

with

f* E LP(dXR )

and

in

Since

H[u*]

v

of

If*l p e LI(dxR)

h* o SR = u*p f* o SR = u*

LP(do) T.

u*

If

and

Let and

v. a.e.

a.e.

Letting

Again by That

7B.

f* E LP(dXR). I11f*lp O SRUl :

lif*o Smllp = llf*llp.

u* ~ 0, then we find, a.e.

is quasiSo, by Theo-

u* = ~ o SR

0 < p < ~.

be a T R-

7C, we find a harmonic

function

and hence

by (i), an

f* = (h*) I/p, we have

The result for an arbitrary

as a linear combination

The rest is now obvious.

to)

is a TR-invari-

is quasibounded.

f* o SR E hP(do)

is then deduced by writing elements

v

case

u* e LP(do) T.

h* E LI(dxR )

u* E Ll(da) H[u*]

is an isometry was shown in Theorem

If* O SR Ip : If*l p O SR'

li!f*IPllI

let

function

for the fine boundary

f* ÷ f* o SR

function

H[f*] o ~R' it is (equivalent

In view of Theorem

implies that

by (12), we have

is the fine boundary

Conversely,

v O ~R : H[~] o ~R : H[~ o SR ]

the map

Since

f* o SR function

such that

bounded, (12)

f* E LI(dXR ), then,

Then the harmonic

ant harmonic v

If

u*

of four positive

[]

NOTES The results

on Martin eompactifieation

13, 14 of Constantineseu

and Cornea

[CC].

improved version of those given in Hasumi

are adapted The results

[17].

from Chapters in §3 form an

6,

CHAPTER

Some on the

basic

unit

results

disk

and

boundary

behavior

surfaces

are

deduced

maps

given

covering h~--the

§i.

i.

harmonic

HARDY

IA.

Let

with

ON THE

~ by

0

be the g~(a,z)

being

can be e x t e n d e d the unit

that

ference

UNIT

DISK

open

unit

Martin disk

both T

(or

0

For any

CI(~)

with and

~). with

z = re ie 0

H~(~) let

the

For each pole

at

t)

following,

0 e p < ~

f(z)

on

denotes

the

functions or,

The h a r m o n i c

in the

For

N(~)

Hp into

results the

on

dual

of

B-topology.

are

point

e it

be the f(z)

equivalently,

on

~ space

let such

of b o u n d e d

Nevanlinna D

HP(~) that

such

Lebesgue or

denote If(z)I p analytic

class

~n

that

log+If(z)l

log Ifl • s P ' ( ~ )

(Ch.

usual

the We

unit

Martin

b ÷

CI(~),

see circum-

function

Poisson

kernel

(on

of

2

(8-t) + r

d~(e It)

topology.

the the

d x ~ ( e it)

normalized by

with

to the

by

function

is i d e n t i f i e d Euclidean

on func-

given

disk

equal

l-r

are

the

closed

e it • T

measure

plane

unit

is e x a c t l y

=

g~

so the M a r t i n

z • •

~

usual

function

and

complex

to the

of

AI(D)

is seen to be the

is d e n o t e d ,

IB.

look

on R i e m a n n

by u s i n g

to the

a I}

fixed

I- 2rcos

functions

Results

The G r e e n

of the

continuously

P(r,8-

origin

classes

we d e t e r m i n e

respect

disk.

compactification

A(~)

z ~ k~(elt,z)

with

Hardy

We m a i n l y

disk

Finally,

H~--with

the o r i g i n

= g~(b,z)/g~(b,0).

closed

§3.

unit

= log { I i - ~ z l / I z -

k~(b,z)

also

classes.

for the

Ill,

of

k~(b,z)

so that

in such

concerning

surfaces.

Definitions

is g i v e n tions

in Ch.

CLASSES

here

Riemann

from those

version

HARDY

recalled

on g e n e r a l

of f u n c t i o n s

CLASSES

Basic

are

IV.

~)

measure

~

at the

dt/2~,

which

d~(t).

the has

space

of a n a l y t i c

a harmonic

functions

~ , i.e.

II,

2

the has

5D).

on

space

majorant. ~.

a harmonic Then

Further,

of a n a l y t i c majorant

6S

N(D) for

0 < p < q < ~.

~ HP(~)

For

~ Hq(~)

f ~ HP(~),

l[fU

~ H~(~)

0 < p < ~, we

set

= ( ( L H M ( I f l P ) ( 0 ) ) I/p, P

where set

LHM

stands

IIfJl : s u p { I f ( z ) l :

subharmonic and

for any

therefore

it has

IC.

Along

that

Lemma.

If

f(z)

If(z)l p

has

Proof.

We

harmonic

Since

u

¢

f

¢o f

The least

rf(z)] p

any

0 < p <

majorant

is

provided

harmonic

if

any

on

functions h~(~)

Iz- z01 v

analogous

hP(~), f(z)

I ~ p on

denotes

the

com-

~.

¢(x,y)

is a c o n v e x

function

If(z)I p

on

is s u b h a r -

i £ p < ~.

v = Im(f),

Let

consider let

In p a r t i c u l a r ,

and

~ r0

so t h a t

be any

u

closed

and disk

v

are r e a l

in the

domain

are h a r m o n i c ,

I

=

we

Namely,

majorant,

and

and

f(z 0)

T

(U(Zo+roeit)'v(zo+roeit))d°(t)"

= ¢(u(z0),v(z0))

is seen

that lemma

harmonic

as b e f o r e sup {If(z)l

r0

e it-))da(t)

(¢o f)(z0+r0eit)do(t).

to be

subharmonic.

(x 2 + y 2 ) p / 2

is a c o n v e x

then

that,

implies

majorant

+

¢(u(z0+r0eit),v(z0

= I

fact

harmonic

we

that

and

HP(D)

functions

and

I

the

D

f E H~(D)

is convex,

¢o

Hence

on

least

classes

subharmonic.

(u(z0)'v(z0))

Since

in p a s s i n g

f

functions.

a harmonic

harmonic

u = Re(f)

functions.

f(z).

the H a r d y

harmonic

set

the

For

majorant.

has

is

majorant.

We note

function

is h a r m o n i c

¢o f

for any

z e D}.

of c o m p l e x - v a l u e d

bounded

then

monic

space

harmonic

of h a r m o n i c

If(z)I p

plex-valued

~2,

with

consisting

< ~, be the such

that

least

analytic

a superharmonic

classes

of

for the

whenever

The

in case it has

latter

function

fact

if

i ~ p < ~,

a harmonic

follows

i ~ p < ~. Ill p

has

majorant.

llfJl = ( ( L H M ( I f l P ) ) ( 0 ) ) 1/p for i ~ p < ~ and P : z E ~}. A complex-valued harmonic function

from [] the

We d e f i n e llfll = f

is c a l l e d

66

quasibounded

if b o t h

by

h~(~)

the

2.

Some C l a s s i c a l 2A.

along

Re(f)

Theorem.

(a) values

Let

f* @ LP(do).

every

For

If(z)l p

is the P o i s s o n

to

f*

ie

The m a p 11fllp

LP(do).

is a norm,

We d e n o t e

to

hl(~).

HP(D)

has r a d i a l

limits

integral

If

on

f ÷ f*

{fr:

LP(da)

as

of

= JI

Then

f

for a l m o s t 0 e r e i} r ÷ I- 0

admits

all

radial

e i0 @ T

with

and the

and

fr(e ie)

=

LHM of

If*I p, i.e.

F(r,e-

t) If*(eit) IPdd(t )

it is q u a s i b o u n d e d .

If

f ~ O, t h e n

~. is an i n j e c t i o n

11f*llp, w h e r e

of

HP(~)

llf*llp d e n o t e s

into

the u s u a l

i =< p ~ ~, t h e n the f u n c t i o n a l If.If p HP(~) into a B a n a c h space.

LP(do)

n o r m of the defined

in IB

which makes

We d e n o t e HP(~)

belonging

0 < p ~ ~.

the net

and t h e r e f o r e

is s u m m a b l e

(b)

space

E ~,

with

in

(LHM( If IP)) (z)

satisfying

in

= l i m r ÷ I 0f(re le)

0 e p e ~

converges

log If*I

function

f E HP(D)

f*(e i6)

f(re ie)

z = re

are q u a s i b o u n d e d .

elements

all radii.

boundary

for

Im(f)

Results

We show t h a t

almost

and

set of q u a s i b o u n d e d

by

HP(d~)

is i s o m e t r i c a l l y

the

set of

isomorphic

The p r o o f w i l l be s t a t e d

in 2B-2E,

0 < p < ~, for the ease

p = ~

f*

with

with

f E HP(D),

HP(do)

under

so that

the m a p

f + f*.

w h e r e we o n l y deal w i t h the case

is e a s i l y

covered

by the d i s c u s s i o n

to

follow.

2B. let

uf

First be the

r 2 < ... < i f(r e it) n

for

be a f i x e d

n = i, 2, . . . .

IT

Ifn(eit)12do(t) So

and c o m p a c t

{fn ) Hilbert

with respect

has a c o n v e r g e n t J = i, 2,...}

case

majorant

sequence

for

is a s e p a r a b l e

{fn(j):

the s p e c i a l

least harmonic

n = i, 2, . . . .

L2(d~)

{fn )

consider

p = 2.

of

tending

Let

Ifl 2 to

i

on

f E H2(~) D.

Let

and set

and

0 < rI <

fn(e it)

=

Then,

~ S

uf(rneit)da(t)

= uf(0)

is a b o u n d e d

sequence

space,

closed ball

every

to the w e a k t o p o l o g y

subsequence. and a f u n c t i o n

Namely,

in

L2(do).

and t h e r e f o r e there

F E L2(d~)

exist

Since

is m e t r i z a b l e the

sequence

a subsequence

such t h a t

@7

S for any

k E L2(do)

re i@ E D,

as

fn(j)k do ÷ I j + ~.

Setting

Fk do k(e it) = P(r,@- t)

with

z :

we have f(rn(j)z)

: I

fn(j)(eit)p(r,@ -t)do(t)

+ I

F(eit)p(r,@ -t)do(t)

and therefore f(z) = I

F(eit)p(r,@ -t)do(t).

Fatou's theorem (Theorem AI.2, Appendix) then implies that, for almost i@ every e , f(re l@) tends to F(e i@) as r tends increasingly to i. For any such 8, define f*(e i@) = F(ei@), so that f* E L2(do) and f*(e i@) = lim f(re ie) a.e. We thus have r+l-0 f(z) : for

z = re i0 6 ~.

(I)

I f~(eit)p(r,O -t)do(t)

By use of the Schwarz inequality

If(z)12 =< S

If*(eit)[2p(r,% -t)do(t). T

The right-hand side, say (2)

u, gives a harmonic majorant of

if[2

and so

If(z)r 2 ~ uf(z) ~ u(z)

on ~. Since u is seen to be quasibounded, so is uf. Thus uf is equal to the Poisson integral of its radial boundary function. Taking radial limits in (2), we have If*(eit)I2 ~ u~(e it) ~ u*(e it) = If*(eit)I2

a.e. and therefore

u = uf.

IIfI'~ = (LHM(IfI2))(0)

: S

It follows that If*'2d° = 'f*'~"

T Namely, f + f* gives an isometry of H2(D) into L2(do). from (I) that the L2-convergence of f*'s implies the almost convergence of f(z)'s on ~. Hence, H2(~) is a Hilbert which is canonically isometrically isomorphic with a subspace L2(do).

It follows uniform space, of

68

2C. order

Suppose that

m ~ 0

f ~ H2(~)

with

f ~ 0.

at the origin, then we set

If

f

has a zero of

f0(z) = f(z)/z m.

Then

f0(0)

0, f0 E H2(D) and indeed If0(z)I 2 ~ uf(z). Let al, a2,.., be the zeros of f0' repeated according to multiplicity and ordered so that 0 < fall ~ la21 ~ . . . .

By Jensen's

r

[

formula

(ef. Rudin [59], p. 330)

+ IT l°g[f0 (reit) Ido(t)

l°g;~--~I k : -loglf0(0)

lak I
+ IT If0 (reit)12d°(t)

-loglf0(0) Letting

r ÷ I - 0, we get

(3)

[ k ~: l l ° g (i/lakl)

~ k=l

lakl

< ~

which is equivalent to:

is convergent.

This in turn implies that the Blaschke product with zeros B(z) = ]~ k=l converges almost uniformly on IB(z)l ~ i

on

D

To see this,

and

~

lakl

ak

ak

{a k}

z

i - ak z

and is an inner function,

IB*(elt) I = i

a.e. on

i.e.

~.

let

B (z) denote the n-th partial product of B(z). n Then, every B n is analytic on the closed unit disk and IB~(elt) I ~ I on T. Moreover, for m > n, lIBm_ Bnll22 : IT IB~ (~it) - B*(eit)I2do(t)n

: 2 IT (i- Re( B~(eit)/B*(eit) n )do(t)

: 2(1 - Re[

(B*(emt)/B*(emt))do(t)]) m n . .

i = 2(1-

Re[B

: 2(i-

77

m

(0)/B

n

(0)])

m

JakJ) ÷ 0

k=n+l as n ÷ ~, because of (3). The sequence {Bn(Z)} is thus a Cauchy sequence in H2(D). By our observation in 2B, there exists an element F E H2(~) such that B + F in H2(~) and B* ÷ F* in L2(d~). n n

69

F, we c o n c l u d e

Since B c o n v e r g e almost u n i f o r m l y to n So, there exists a s u b s e q u e n c e of B*

which

n

Hence,

IB*I

2D.

= i

a.e.

Continuing

n = i, 2,...

for B(z)

almost

h(z)

on Let

and

h(z) on

so that

n

of 2C, we set

= f0(z)/B(z).

h (z) converge n is a n a l y t i c on D.

Then,

for any

IBn(Z) I ~ I

for

B n (z)

almost

1 - 6 ~

Izl ~ i.

uf(z)/(l-

g)2

majorized

on

that

Since

everywhere D

lh(z)l 2 ~ uf(z).

(4)

lhn(Z)l 2

on

by the

B*.

Izl ~ i

D.

same Hence,

As

E

a

~ > 0

e) 2 lhn(Z)l 2

is a r b i t r a r y ,

for every

h e H2(~)

n.

lhn(Z)l 2

From this

is

follows

with

lh*(eit) I : I f ~ ( e i t ) I / I B * ( e i t ) : If* (elt) I

exists

to

to

and so

is s u b h a r m o n i c ,

uf(z)

converge

uniformly

0 < e < i, there

I- ~ ~

lhn(Z)l 2 ~ Ifo(Z)12/(l - E) 2 =< u f ( z ) / ( l for

to

h n (z) = f 0 ( z ) / B n (z)

Since

~,

h

be fixed. i- e ~

F : B.

a.e.

~.

the d i s c u s s i o n

uniformly

D,

such that

on

that

converges

I :

[f~(eit) I

a.e.

We thus have f(z)

(5) where

h(z)

has no zeros

Finally, lh(z)I2/2

in

we show that

~ uf(z)/2

is a n o n n e g a t i v e

harmonic

directly),

exists

z = re ie E D.

the a b s o l u t e l y v*

function

=

Fatou's Since

2E.

Hence,

Turning

by

f

in

log lh(z)l

~,

~.

theorem

part

(4),

By T h e o r e m

measure

d~

on

2A, Ch. T

III

(or

such that

P(r,8 - t)dp(e It) then

says that

of the m e a s u r e

v*(e it)

- log lh*(eit) I + If~(eit)12 be summable.

Since

has no zeros

on

a nonnegative

continuous

is summable.

is integrable.

h(z)

= - log lh(z)l + uf(z)/2

v(z) for

~. log If*l

and since v(z)

there

= zmB(z)h(z),

and

: -log since

case,

d~(e It)

f* e L2(d~),

:

log lh*(eit)I

is summable.

let us take any

is

and t h e r e f o r e

lh*(eit) I + u } ( e i t ) / 2

log If*(eit) I

to the g e n e r a l

v*(eit)do(t)

f E HP(~),

must

70

f ~ 0, with

0 < p < ~.

be the h a r m o n i c

Let

u

be the LHM of

of

u

on

conjugate fl(z)

Then

fl

~

and

: exp(-~(u(z) P

is a n o n v a n i s h i n g

analytic

f2 : ffl'

f2/fl. f~

Since

exist

follows

H~(D)

a.e.

that

we have

function

where

B

where

h(z) Set,

in

~

= £(z)/fl(z)

as before,

n

al,... , a n .

for

Bn

on

T.

is the finite

n = i, 2,...

[h*(eit) I =

Since

lim r÷l-0

If(z)l p

-fT If(remt)IP d~(t) I

formed

= f(z)/z m

and thus

as

in

f~ E LP(d~) h

h 0 E H2(~) shown

in 2B,

negative

and so

has no zeros such that u0

harmonic

hn(Z)

f~

and

summable.

It

We then

= zmB(z)£(z),

~ 0 f(z)

of

f2' and

= zmB(z)h(z),

= f0(z)/Bn(Z),

product

in 2D then

B(reit)l

on

formed

shows

that

[hn(Z)I p

Hence

h • HP(~)

[f(r

ei t n

and

a.e.

we see that

By Fatou's

n =

of the first

: ]f*(eit)l

~,

= IT (lim n÷~

the i n t e g r a l s

lemma we have

)[P)d~(t)

f(rneit)IPd~(t)

u(O) < ~.

h* E LP(d~). in

~,

there

h = h02/P

is q u a s i b o u n d e d function

f =

~.

r ÷ i - 0.

<

Since

that

f2(z)

So we have

lim inf [

Hence,

are

and

summable.

of the zeros

t

" If*(emt)[Pdo(t)

in 2D shows

and get

and

is s u b h a r m o n l c

and

[]f2[] ~ i

lh(z)I p ~ u(z).

If(reit)I/rm

increase

O,

log If~[

Blaschke

Our a r g u m e n t

Let

11flll~ £ i

is also

f2

II£II ~ i.

has no zeros

f0(z)

where

zeros

and

(5) to

product

with

i, 2,..., u(z)

log If~l

formula

is the B l a s c h k e

has no zeros

f2 ~

= log [ f ~ l - log If~I

apply the f a c t o r i z a t i o n

with

our o b s e r v a t i o n

and that both log If*l

~.

] f l l -~

:

f2 @ H ~ ( ~ ) '

q H2(~),

on

+ i~(z))).

If] ~ u I/p ~ e×p(u/p) Setting

''|f|P

set

on

Let and

~.

are positive,

exlsts u0

be the LHM of

Since

SP'(~)

both

nonnegative.

Moreover,

since

-Prl(v 0) S 0

and thus,

in view of Fatou's

the p r o j e c t i o n s P r l ( v 0)

is q u a s i b o u n d e d , theorem

and

function lh012

v 0 = - log lh01 + u0/2

in the space

u0

an a n a l y t i c

As

is a nonpr I

p r Q ( v 0)

Prl(lOglh01) (Appendix

and are =

A.I.3),

prQ

71 [h0(z) I : exp(logrh0(z) I) = exp[Pri(logah0(z)I) + prQ(lOglh0(z)])] exp(prQ(lOglh0(z)I)) = exp[I~ P(r,e - t)log Ih~(eit) i dg(t)] for

Z : re i8

For the function

f

we get

If(z)l p ~ lh(z)JP = [h0(z)i 2 exp[[

for

I

lh~(eit)12p(r,g - t)do(t)

= I

I f * ( e i t ) e P P ( r , e - t)da(t)

z = re i0 E ~.

Just as in 2B, we infer that

(LHM(IflP))(z) for

z = re it E ~,

This means HP(~)

m(r,0 - t)log ]h~(eit)I2 d a ( t ) ]

= f

which is quasibounded,

finally that,

for

into a Banaeh space. 2F.

Theorem. (a)

Let

with

f*

belongs

(b)

If

i < p ~ ~, then

integral of

f*

If

quasibounded The map

with

part, f ÷ f*

lif]Ip

f

Since the case and set

Let

a.e.

on

T

and the

p = =

and is the Poisson

iif*iJp f

and

i < p ~ ~, and of

i ~ p < ~.

Then the following hold:

f*(e it)

is quasibounded

is an isometric hQI ( ~ )

Proof.

i

1 ~ p £ ~.

LP(do).

prQ(f), I of

if

ing to

to

Ilf*iJp

we have the following

values

p = i, then the Poisson

LP(da),

case

hP(]D)

has radial boundary

function

llfirp

II-li is a norm, which makes P This completes the proof of Theorem 2A. []

in

f ~ hP(~)

and therefore

i £ p < ~,

As for functions

f

(c)

[f* (emt) IPP(r, g - t)da(t) ,

integral

of

isomorphism of onto Ll(do).

is rather obvious,

0 < r I < r 2 < ... < I

f (e it) = f(r e It) n n

f*

is equal to the

i]prQ(f)iiI = IIf*iLI.

Then

hP(D)

onto

we only consider

be a fixed sequence

the tend-

72

I where

u

{fn } in

t u(r n eit)da(t) =< J~

is a harmonic majorant

quence in space;

ifn(ei t )IPda(t)

LP(da).

If

of

If P.

i < p < ~, then

so, by passing to a subsequence converges

LP(da)

weakly

in

LP(da),

{fn }

LP(da)

is a bounded

is a reflexive

if necessary,

se-

Banach

we may assume that

i.e. there exists an element,

f*,

such that fn(eZt)k(elt)da(t)

for any

So

: u(0) '

+

f*(eZt)k(elt)da(t)

k C LP'(da)

with p-i + p,-i = i. ie z = re E ~, we get

with any fixed = [ i

f(rnZ)

f (eit)p(r,9 - t ) d a ( t )

Setting

÷ r

f * ( e i t ) p ( r , e - t)da(t)

j

n

k(e it) = P(r,0 - t)

and therefore f(z) By Fatou's that

f

theorem,

= ]

limr÷l_ 0 f(re ie) = f*(e i9)

is quasibounded. If(z) Ip <

and the argument

bounded

let

topology Namely,

of

f

of finite

is compact

w(M(~),C(~)), d~

on

the sequence ~

k E C(~).

side gives the least

{fnda:

Borel measures contains

{fn(i)da:

n = 1, 2,...} on

~.

Since the

with respect

a convergent j = i, 2,...}

is

to the weak

subsequence. and a finite

such that

I k(eit)fn( (eit)da(t) y J) for any

shows

llfllp = I1f*llp.

and metrizable

there exist a subsequence

Borel measure

This also

If*(e it) IPP(r,9 - t)da(t)

Then the sequence

M(T)

M(T)

a.e.

inequality

Ifl p, which implies

p = i.

in the space

closed ball of

By H~ider's

in 2B shows that the right-hand

harmonic majorant Finally

f*(elt)p(r,9 - t)da(t).

By setting

÷ I

k(eit)d~(eit)

k(e it) : P(r,e - t),

z : re i0

]D,

we

get as above f(z) By Fatou's

theorem again,

=

f f*da

°

P(r,e - t)d~(e±t). is exactly the absolutely

continuous

73 part

of

d~,

so that

prQ(f)(z)

This

implies

§2.

HARDY

3.

CLASSES

Boundary 3A.

with the

at once

origin unit

p < ~,

0.

disk

Ifi p

analytic

•

rJprQ(f)liI = i]f*ilI.

of

Hp

at H a r d y The

and

the

on

hp

SURFACES

on a h y p e r b o l i c are

for the

the

sake

set of a n a l y t i c majorant.

R.

For

[]

Functions

definitions

repeated

a harmonic

functions

RIEMANN

classes

basic

denotes

has

f*(elt)p(r,0-t)d~(t).

f

ON H Y P E R B O L I C

but are

HP(R)

that

that

Behavior

We look

:

Riemann

same

as

in the

of c o m p l e t e n e s s . functions

H~(R)

f E HP(R)

surface

denotes with

f

on

the

R

case

of

For

0 <

R

such

set of b o u n d e d

0 < p < ~, we put

11flip : ( L H M ( i f B P ) ( 0 ) ) I/p, where

LHM

stands

f e H~(R)

we

set

0 < p ~ ~, are the

complex

for the

monic

field.

functions

denotes

We also

functionals

and

on

z e R}. and

define

hP(R), f

the

A complex-valued Re(f)

and

H~(R)

R

such

elements

R.

that

For

HP(R),

is an a l g e b r a

that

quasibounded.

in

hl(R).

has

bounded

function

are

f And

We d e f i n e

over

space

Ifi p

of c o m p l e x - v a l u e d harmonic

Im(f)

surface

i ~ p < ~, to be the

on

space

on the

It is c l e a r

that

functions

set of q u a s i b o u n d e d

By u s i n g

a universal

Theorems

6B and

a

har-

is c a l l e d hQI(R) the

7B,

covering Ch.

III,

map we

%R:

~

easily

÷ R

with

deduce

the

~R(0)

= 0

following

and

result

from

2A.

Theorem.

(a)

boundary

values

the LHM of

Let

f E HP(R)

f(b)

Ifl p

and t h e r e f o r e on

majorant

ll.it even on hP(R), 1 ~ p ~ ~, by m e a n s of the a b o v e P It is t r i v i a l to see that HP(R) ~ hP(R) for 1 ~ p ~ ~.

formulas.

Theorem

spaces

h~(R)

R.

if b o t h

the

linear

harmonic

majorant,

quasibounded

harmonic

llfll = s u p { I f ( z ) i :

complex

of c o m p l e x - v a l u e d harmonic

least

a.e.

with

on

is e x a c t l y

the

is q u a s i b o u n d e d .

AI

0 < p < ~. and

Poisson If

Then

f E LP(dx ). integral

f ~ 0, then

f For

admits

H[Ifl p] log

ire

fine

0 < p < of

Ifl p

is s u m m a b l e

AI . (b)

isfying makes

The map ilfiip

HP(R)

f + f

iLfilp into

is an i n j e c t i o n

If

a Banach

i < p < ~ space.

then

of

HP(R) it.ilp

into

LP(dx)

is a norm,

which

sat-

74

We d e f i n e

a subspace HP(dx)

which

is i s o m e t r i c a l l y

3B.

Similarly,

Theorem.

Let

function

f

= f.

f = H[f]

S

surjective

4.

Some

Let

is b o u n d e d the

f E hP(R),

into for

HI-]

to

f ÷ f.

the

LP(dx)

and

fine

LP(dx).

boundary Put

such

i < p ~ ~,

is the

p = i

injective

application

has

L H M ( u p)

(b)

u

(c)

(log u) V 0

(d)

S(f)

that

in w h i c h

inverse

case

of the m a p

is i s o m e t r i c

S;

as w e l l

as

on each

Analytic of the

on LHM's

covering

II,

3C)

majorant

for

i ~ p ~ =.

Functions map

~R:

of m u l t i p l i c a t i v e

(Cf.

a harmonic

HP(R)

on for

R

~

÷ R,

analytic

such

some

that

we

functions.

either

u

Then

i ~ p < ~.

is q u a s i b o u n d e d ,

The

is of b o u n d e d

inner

if

p ~ ~.

characteristic,

logu

i.e.

E SP'(R).

is q u a s i b o u n d e d . factor

is a b o u n d e d

u I = exp(Prl(lOgu))

l.a.m.

~Ulll~ : i.

Proof.

In v i e w in the

ments

(a),

see

tion

of T h e o r e m

ease

(c) are

we n o t e

- log u - h

quasibounded v 2 E Q(R). v 2 $ 0. -v I ~ 0.

that

harmonic

of T h e o r e m

Ch.

III,

we h a v e

We m a y

also

assume

almost

Ch.

minorant,

Since

the

h

Moreover,

II,

i.e.

clear

-log u

is n o n n e g a t i v e

5A,

parts,

Since

6B,

R = ~.

(b) and

(d),

quasibounded use

and ~

be an l.a.m.

up

the m a p

following

belongs

hP(R)

for

under

following:

ments

To

u

or

the

and

i.e.

setting

hA(R) ;

result

(a)

and

of

on M u l t i p l i c a t i v e

following

Theorem.

we h a v e

space

HP(R)

surjective

f E hP(R),

As a n o t h e r

the

map

41

by

f E HP(R)},

for e a c h

on

and

is i s o m e t r i c

Results

4A. prove

Then,

is n o r m - d e c r e a s i n g

S

with

2F i m p l i e s

a.e.

is a l i n e a r

on the

(c)

Theorem

is i s o m e t r i c

for any

(b)

LP(dx)

: {f e LP(dx):

i ~ p ~ ~.

S S

of

isomorphic

is d e f i n e d

Then (a)

HP(dx)

to p r o v e

f r o m our

p < ~.

discussion

is a s u p e r h a r m o n i c e.g. and

therefore this

- log u - h = v I + v 2

projections

is q u a s i b o u n d e d ,

are

positive,

we h a v e

i n f z E R Vl(Z)

belongs

into

the

with

to

having

and

v I E I(R)

vI

the

a

func-

SP'(D).

inner

we h a v e

statestate-

in 2A-2E.

Thus

By

the and

vI $ 0

in p a r t i c u l a r

= 0, for

the The

function

h = -p-I(LHM(uP)).

we d e c o m p o s e

we h a v e

only that

and

Prl(lOgu)

is inner.

=

75

Hence

u I : e x p ( - v I) ~ i

and

SUpzER

Ul(Z)

= i, as d e s i r e d •

[]

^

4B.

We a l s o

Theorem. on

AI,

If

compute

u

v E SP'(R),

then

= prQ(v)

where

Vq

Since

v = v

for

u C SP'(R).

~

exists

is t h e

a.e.

on

quasibounded

AI

part

and of

~ = Oq

a.e.

v.

+

Proof. v

is n o n n e g a t i v e .

bounded

parts

- v Now

vi

in v i e w let u s

and

Vq,

of Ch.

II,

decompose

i.e.

v

into

v = v i + Vq.

v. > 0 a n d v > 0. By T h e o r e m i = q = on R and the singularities of

vi,

if any,

be a s i n g u l a r i t y

vi

has

we h a v e where

vi(z) h(z)

tive,

we

h

v i.

= c log Izl + h(z)

see t h a t of

c ~ 0 a.

by R i e s z ' s

As

is i n n e r ,

Ch.

a

II,

therefore

suppose

inner

and

that quasi-

v ~ 0, we h a v e

v

is e v e r y w h e r e h a r m o n i c q are isolated. Let a E R

only

The

its

Since

logarithmic

in a p a r a m e t r i c

function. and

disk

singularities,

centered

function vi

is

v. being m superharmonic

at

a,

nonnegain a

v. is s u p e r h a r m o n i e o n R. l I) we h a v e v. = h + h , w h e r e l s p harmonic function and h is a p o t e n t i a l . Since p h . C o m b i n e d w i t h L e m m a 5C a n d T h e o r e m 5D, Ch. III,

theorem

is a n o n n e g a t i v e

s v.

5B,

Since

is a h a r m o n i c

neighborhood Thus,

of

5A, we m a y

so is

l

is a r b i t r a r y ,

(Theorem

6F,

Ch.

S

these

observations

AI •

So

a.e.

again

to

vi

Qq

exists

Corollary.

on

5C,

AI .

If

u

a.e.

on

that

Ch.

AI,

where

Proof•

Set

preceding ~q

a.e.

is a n o n z e r o AI

v : log

theorem with

Vq

The

We w i l l

purpose

convex complex

v

vanish

Moreover,

AI .

Hence,

and

exists

a.e.

a.e.

Vq

on

exists

and

is e q u a l

l.m.m,

of bounded

characteristic,

then

= i

and

u : UQ

u.

shows

Since that

and

UQ

= exp(prQ(lOgu)).

v E SP'(R)

Q

uI = i

a.e.

exists

by the

a.e.

Hence

we h a v e

a.e.

[]

on

assumption,

AI

and

~ = expQ

the

is e q u a l

: exp @ q

to

= ~Q.

It

B-Topology

5A. this

exist

p

and

= prQ(v).^

that

h

a.e • on

III.

u I : exp(Prl(lOgu))

is n o w t r i v i a l

5.

and

s

[]

uI on

h

and vanishes

by L e m m a

a.e.

exists

imply

we

linear field.

describe

first

spaces.

the

recall Let

Precisely

B-topology

some E

basic

f o r the

facts

be a l o c a l l y

speaking,

this

space

in the

convex

is d e f i n e d

h~(R).

theory

linear b y the

of

space

For locally over

following

the

78

conditions: (AI)

E

(A2)

is a l i n e a r

E

plication

is a H a u s d o r f f

are

the o r i g i n of

space

space

each continuous

0

has

a basis

over

the

complex

such that

in b o t h

field;

addition

variables

of n e i g h b o r h o o d s

and

jointly

consisting

scalar

and

multi-

such that

of c o n v e x

subsets

E. Let

E'

denote

space

of a l l

each

x' E E'

spondence

continuous

E

meaning

(resp.

of

E')

we

{x E E:

A

in

E'

E')

satisfying

is c a l l e d < ~

every

x E E).

and then of

only

the

totality

of p o l a r s

of

E

a basis

by

E'.

s(E',E)

The

dual

E

and

A°

of the

bounded)

that

a subset

The

strong

convex

A

of

over

0.

strong

dual

topological

linear

space

The of

w(E,E') topology (resp.

for

is subsets

for

all

if

E'

such

bounded

space and

E'

subwith

is d e n o t e d

is d e n o t e d

We c a l l

by

Identification

x' ÷ < x , x ' > E

on

semireflexive

the w e a k - ( E , E ' )

(resp.

w(E',E)),

for

(resp.

E

E".

is c a l l e d

(resp.

E')

if

gives

the

x E E

us a n a t u r a l

inclu-

E = E".

weak-(E',E))

is d e f i n e d

for

E'

of each

topology,

as t h e w e a k e s t

that makes

every

locally

functional

written

as

convex x'

E E'

x E E) c o n t i n u o u s .

Theorem. if e v e r y

A locally bounded

harmonic first

We n o w

linear

of

[37],

E

p.

on

R.

is the u s u a l

space

E

is s e m i r e f l e x i v e

is r e l a t i v e l y

if a n d o n l y

w(E,E')-compact.

(cf.

190)

l o o k at the

functions one

convex

subset

and Namioka

5B.

The

s(E',E)

E E'

(resp.

< ~

on b o u n d e d

topology

ranging

number E

is b o u n d e d

topology

the

functional

E =C E". Finally,

Kelley

of

sup{i<x,x,>I:

E

convergence

locally with

of

polar

bounded

S

of

a linear

sion

A

if

of

x E A}

a positive

a subset

is

= {0}

A

it t h e

is c a l l e d

S

bidual with

which

s u p { l < x , x ' > I : x' @ A}

of n e i g h b o r h o o d s

is c a l l e d

E

and

x' E E'}

call

corresponds hand,

E

corre-

E × E'

for a l l

and

of

weakly*

of uniform

it is the

sets

topology

E

(resp.

bounded.

that

the

(resp.

topology

words,

forms

in

in p a s s i n g

if it is w e a k l y as the

0

On the o t h e r

x' E E'

We n o t e

In o t h e r

of

bounded

for every

defined E.

U

A

x E the

For a s u b s e t

~ 1

x' E A})

linear

each Then

for all

= {0}.

I<x,x'>I

A subset

For

f o r m on

: 0

x E E}

for all

E).

E.

as t h e

x'(x).

a bilinear

A ° = {x' E E':

in

on of

{x E E: < x , x ' >

A ~ eU.

weakly

x ~ A}

in p l a c e

for all

I<x,x'> I ~ I (resp.

is d e f i n e d

functionals

defines

that

if to e a c h n e i g h b o r h o o d > 0

E, w h i c h

<x,x'>

= 0

set

of

linear

(x x') + < x , x ' >

{x' E E': < x , x ' >

(resp.

dual

we w r i t e

nondegenerate, and

the

space

h~(R)

Let us d e f i n e norm

topology,

of c o m p l e x - v a l u e d in it two k i n d s which

is g i v e n

bounded

of t o p o l o g y . by the

sup-norm

77

Ilhll

(6) for

h E h~(R).

: sup{ih(z)I:

z • R}

The second one is the B-topology

(or the strict topol-

ogy), which is the main objective of this section. this,

let

tions

f

C0(R) on

In order to define

be the space of all c o m p l e x - v a l u e d continuous

m

compact for any

i.e.

e > 0.

forms a Banaeh space with

Clearly,

C0(R)

{z • R:

func-

that vanish at infinity,

If(z)l ~ e}

is

respect to the usual a d d i t i o n and scalar m u l t i p l i c a t i o n of functions and the s u p - n o r m of the form (6). Now, for every setting

Nf(h)

f • C0(R)

= llfhll .

we define a seminorm

The totality

{Nf: f • C0(R)}

determines a locally convex t o p o l o g y for topology.

In other words,

f

ranging over

for this topology.

in

h~(R)

by

of seminorms

h~(R), which we call the B--

the c o l l e c t i o n of sets of the form

Vf = {h • h ~ ( R ) : with

Nf

llfhrl < i},

C0(R) , makes up a basis of n e i g h b o r h o o d s of

The space

h~(R)

0

e q u i p p e d with the B-topology is

denoted by h ~ ( R ) .

5C. dual of

We want to determine the dual of h~(R).

Let

Borel m e a s u r e s on

Mb(R)

R.

hB(R) , which we call the B-

be the space of all c o m p l e x - v a l u e d bounded

This forms a Banach space with respect to the

usual a d d i t i o n and scalar m u l t i p l i c a t i o n of m e a s u r e s and the total variation norm tional on

IJ~II : fR C0(R)

Id~I"

(7)

for

Each

~ e Mb(R)

determines a linear func-

by means of the formula

: I hd~ JR As we see easily by means of Riesz's r e p r e s e n t a t i o n h E C (R). 0 is identified via (7) with the dual C0(R)' the space Mb(R)

theorem,

of the Banach space N(R) and denote by

C0(R).

= {~ E Mb(R): M{(R)

the

On the other hand, we set = 0

(algebraic)

for all

h E h~(R)]

quotient space

Mb(R)/N(R).

Then

we have the following: Theorem.

The space

(algebraically)

h~(R)

is s e m i r e f l e x i v e and its dual

identified with

h~(R)'

is

M{(R).

Proof. The proof is divided into three parts. (a) First we show that the dual h~(R)'

is a l g e b r a i c a l l y equal

78 to

M{(R).

Let

is c o n t i n u o u s i, 2, ... } Un=l ~ Kn

hB(R).

of c o m p a c t and

Urysohn's R

~ E Mb(R) ; t h e n the l i n e a r

on

To see this,

subsets

such that

fn : 0

on

R.

to

on

a measure

R\ Int(Kn+l) ,

K0

=

[ n=0

= i

fn' on

defined

d~'

{Kn: n = R =

We t h e n use

n = i, 2,...,

Kn,

and

on

0 =< fn =< i

by

2-kf k

2 -n-I £ f £ 2 -n+l

denotes

by s e t t i n g

IB'J(R)

[ k=l

and in fact

where

~'

n : i, 2, . . . .

functions

T h e n the f u n c t i o n

C0(R)

n = D, i,...,

for

F : h ÷

a sequence

K n =C Int( Kn+l) ,

such that

continuous

f = belongs

R

I ~ I ( R \ I n t ( K n )) =< 4-n

l e m m a to d e f i n e

everywhere

of

functional

we c h o o s e

the e m p t y

= f-ldu.

i~'I(Kn+ I \ K n) ~

on

subset.

Then

[ n:0

Kn+ I \ K n

Finally,

for

we d e f i n e

~' E Mb(R) , b e c a u s e

2n+li~l(Kn+ I \ K ) n

co

< 2 1 ~ I ( K I) +

It f o l l o w s

that,

for every

[ n:l

2n+1-4 -n < ~.

h E h~(R),

iF (h) I : i I = l i < Jlfhll ilp'il = I[~' IINf(h) . Hence, have

F

is ~ - e o n t i n u o u s .

s h o w n that

M~(R)

S i n c e this

In o r d e r to see the r e v e r s e linear that

functional IF(h)I

functional setting

F

on

~ Nf(h) FI

= F(h).

This

IF(h) l ~ Nf(h)

= llfhll = 0.

continuous

fh~(R)

theorem space

on

FI

without

dual of

C0(R)

measure

~" E Mb(R)

Then there h • h a (R).

~ C Mb(R) , we

fh~(R)

an

for

the bound.

Fl(fh)

F2

Therefore,

the f u n c t i o n a l

= F2(fh)

= I

C0(R)

by

implies FI

is

By use of the H a h n - B a n a c h on the B a n a c h

As we r e m a r k e d

Mb(R).

which represents

of

fh = 0

such

a linear

]Fl(fh) I ~ llfhll ,

to a l i n e a r f u n c t i o n a l

with

f E C0(R)

= {fh: h E h ~ ( R ) }

is w e l l - d e f i n e d ,

changing

z

exists

8-continuous

We t h e n d e f i n e

Since we have

is i d e n t i f i e d

F(h)

we take any

in the n o r m t o p o l o g y .

can be e x t e n d e d

C0(R)

inclusion,

h~(R).

for e v e r y

on the s u b s p a c e

Fl(fh)

is true of any

~ h~(R)'

above,

the

we can find a F 2.

So we h a v e

fhd~"

R Setting

d~ : fd~",

we see that

~ • Mb(R)

and

F(h)

= fR hd~

for all

79

h • h (R). (b)

This means that

hB(R)'

the B-topology--if

subset of

sup{Nf(h):

h~(R).

and therefore

A

let

A

: M~(R),

sup{llhll

be any B-bounded

sup{lI:

for every fixed

~ E Mb(R).

: h • A}

<

subset of

via the formula

h~(R).

Then

A

As we have seen in (a)

Fh(~)

h • A} <

On the other hand,

be viewed as a linear functional, = .

say

every

h • h~(R)

Fh, on the Banach space

Since

Mb(R)

contains

can Mb(R)

all point mea-

R, we see that

IIFhll for every

Our property Mb(R)'

and therefore

embedded

in the dual

(8) then says that

By the principle

[8], p. 52), norm-bounded (e)

~ e Mb(R) , I1~11 S i} = Ilhll

= sup{II:

h E h~(R)

isometrically

A

that the Banach space Mb(R)'

A

in the space

in the space

h~(R),

h~(R)

in (b) as well as Theorem

that the ball

B : {h • h~(R):

let

{hl: I E A}

closed unit ball,

say

functions

BI, of

(Dunford and Schwartz

Mb(R)'

L~(dx )

is semireflexive.

B. for

Then,

the net

hl's

is contained

(Theorem

by means

(Dunford and Schwartz

there exist a subnet such that

{hl,}

{hl,:

converges

l' E A'} to

h*

BI

3B).

LI(dx ),

theorem

is

of

By the to prove

is w(h~(R),Mb(R))-compact.

the dual of the Banaeh space of Alaoglu's

and hence

5A, it is sufficient

IlhlI ~ i}

be any net in

of fine boundary

Mb(R).

as claimed.

Finally we will show that

consisting

is

bounded as a subset of

of uniform boundedness

is norm-bounded

ha(R)

of the Banach space

is weakly*

fact mentioned

namely,

f • C0(R) , we have

it is an easy matter to verify that

(8)

Thus,

A

: h • A}

in the weak topology w(hB(R),hB(R)').

h~(R)'

sures on

for any

in

Let

is 8-bounded.

Conversely, is bounded

is B-bounded--bounded

in the norm topology.

Then,

h • A} : sup{Ufhll

< Ilfll

as was to be proved.

h~(R)

and only it is bounded

be any norm-bounded

that

~ M{(R),

Next we show that a subset of

{hl:

Since

I • A} in the

L~(dx )

is

is w ( L ~ ( d x ) , L l ( d x ) ) - c o m p a c t {h l}

[8], p. 424).

and an element

So

h* • B I

in the topology w(L~(dx),LI(dx));

we have

I

(hl'- h*)f*dx ÷ 0

£I

for every

f* E LI(dx).

Since the map

h ÷ h

is a isometric

linear

80

map of the normed space h~(R) onto L~(dx) (Theorem 3B), there exists an h E h~(R) such that IIhll ~ I and h = h* a.e. Now we take any E Mb(R)

and set f*(b) = [ kb(Z)d~(z) JR

for any

b E A I. ;

Then,

f*

belongs to

'f*(b)'dX(b)~

;

gI

[; gI

LI(dx); in fact

kb(Z)'d~(z)l]dx(b) R

: IR[I

kb(Z)dX(b) ] Ida(z) I AI

JR

We thus have I

(hl,(z)-h(z))dg(z)

: SR[I

R

(hl,(b)-h(b))kb(Z)dX(b)]d~(z) A1

= S

(hl,(b)- h(b))f*(b)dX(b) A1

= I

(hk,(b) - h*(b))f*(b)dx(b)

+ 0.

A1 Namely, the subnet w(h~(R),Mb(R)). desired. [] 5D.

{hk,]

converges to

This implies that

B

h

with respect to the topology

is w(h~(R),Mb(R))-eompact,

Here is another characterization

of the dual of

h~(R).

Theorem. A linear functional F on h~(R) is B-continuous only if there exists a function f* E LI(dx ) such that (9)

F(h)

= [ J

for

h E h~(R).

dual Proof.

h~(R)' Let

F

hi(R)

h(b)f*(b)dx(b) F ÷ f~

onto the space

be any 8-continuous for

h E h~(R).

is

a bijection

from the

B-

Ll(d×).

linear functional on

by Theorem 5C, there exists a measure fR h(z)d~(z)

if and

£1

The c o r r e s p o n d e n c e

of

as

~ E Mb(R)

ha(R).

such that

Since Theorem 3B shows that

Then,

F(h) =

81

h(z)

: I

h(b)kb(Z)dX(b) £i

for

h E h (R), we have by use of the Fubini theorem F(h)

with

f*(b)

: fR kb(Z)d~(z)

Conversely,

let

it suffices

{h E h~(R): h~(R)

F(h)

h(b)f*(b)dx (b) AI

e LI(dx).

f* E LI(dx)

by means of the formula uous,

= I J

(9).

In order to prove that

to show that the kernel = 0], is closed

is equal to

M{(R)

h~(R).

Banach space

it is sufficient space of

Mb(R)'

Schwartz of

respect

Ker(F)

So it is enough to show that we know already that prove that {hl:

B

Ker(F) A B I @ A}

in

As before, w(L

(dx)).

Mb(R)'

B'

of If

Mb(R)' B

of the 5C, then sub-

(Dunford and is compact with

denotes

the closed unit

IIhll ~ i} = Ker(F) NB. is w(h~(R),Mb(R))-compact.

is w(h~(R),Mb(R))-compact,

Ker(F) N B

such that

if and

w(h~(R),Mb(R)) -

theorem

is w(h~(R),Mb(R))-closed.

I E A]

Since the dual of

if (and only if) the intersection

Ker(F) ~ B

converge

suppose that a

to some element

k E B

We then take fine boundary {hl,:

{hl,}

I' E A'}

converges

In partlcular,

to

of h*

As

so we have only to Thus,

in the closed unit ball

there exist a subnet

h* E B I (dx),L

{hl:

is B-continF, given by

then

the weak topology w(h~(R),Mb(R)). to form a net

h~(R)

is a W ( M b ( R ) ' , M b ( R ) ) - c l o s e d

to the topology W(Mb(R)',Mb(R)). h~(R),

on

is B-closed

in the dual

with the closed unit ball space

F

(b) of the proof of Theorem

Ker(F) N B' : {h E Ker(F):

ment

Ker(F)

In view of the Krein-Smulian

ball of the normed

net

5C,

h~(R)

[8], p. 429), this happens

Ker(F)

F

of

or, equivalently,

as in the part

to prove that

Ker(F)

in the B-topology.

by Theorem

If we embed

Mb(R)

(9) holds.

and define a functional

only if it is w ( h ~ ( R ) , M { ( R ) ) - c l o s e d closed in

Hence,

BI

of

{hl]

in

functions L~(dx ).

and an ele-

in the weak*

topology

we have

S

( h l ' - h*)f*dx ÷ 0

Since

hl, E Ker(F),

AI

for the function and therefore dition

h0 = h*

f*.

fA I h*f*d X = 0. a.e

If we determine

, then the subnet

topology w(L~(dx),LI(dx)).

fA I

{hl,}

,f*d X : F(hl,) h 0 E h~(R)

converges

to

So, as in the proof of Theorem

= 0

by the conh0

in the

5C, we have

82

hl, ÷ h 0

in the topology w(h~(R),Mb(R)).

It then follows

that

k = h0

and thus

F(k)

: F(h0):

I

h0f*dx

: I

AI namely,

k C Ker(F).

h*f*dx

= 0;

AI

Consequently,

Ker(F) N B

is w(h~(R)~Mb(R))-closed,

as was to be shown. Finally,

the map

(h: h • h~(R))

F ÷ f*

coincides

given by (9) is injective,

with

As an easy consequence

L~(dx )

3B.

[]

of the theorem we have the following

Corollary.

A subspace of

h~(R)

is B-closed

in

under the map

h ÷ h

is closed

L~(dx )

for the set

as shown in Theorem

if and only if its image

in the weak* topology

w(L~(dx),LI(dx)).

NOTES There

is a large literature

for §i is Chapters

3, 4, 5 of Hoffman

are taken from Hasumi concerning [SI].

on the

[34].

[17] and Neville

harmonic majorants

Results

fied form,

on classical

6-topology

Results

[47].

are adapted

from Rubel and Shields

Hardy classes.

from Heins

in Subsection [47].

Results

Our source

in Subsection in Subsection

3 4

[31] and Parreau

5 are taken,

in a modi-

CHAPTER

The m a i n first

theme

objective

to p r o v e

stated

every

arbitrary the

type

with

type,

deleting

we are

by M.

Parreau. at the

Unless face

and

point

§i.

i.

A few

IA.

For

principle

every

(A2)

0 • R

~ > 0

in Ch.

R

and

that

follows

that

for any

gous

zero

in the

case

~ ~ ~'

Hl(R(e',a)) Let

B(~,a)

number

the

Green

fact

indicates

same

as the

of W i d o m ' s

by

R

that

an

a surface theory, that

of by

the

one g i v e n theorem

a hyperbolic

function

is f i x e d

FUNDAMENTAL

for h a r m o n i c

in

to

from

fol-

down

defi-

earlier

are men-

and

for

R

is c a l l e d

Riemann

with

pole

the o r i g i n

sur-

at a of

R.

PROPERTIES

Definitions

property

region

we d e n o t e

= g(a,z)

and

R(~,a) The

shown

of

is

proof

It is a l s o

of p o t e n t i a l

and

chapter.

otherwise,

AND

the

The

to w r i t e

is o b t a i n e d

consequences

type

theorem

one.

Our

surfaces

but we try

sense

This

is e s s e n t i a l l y

A point

DEFINITIONS

type

in the

chapter.

of

Widom's

the o r i g i n a l

of W i d o m

this

definition

to Widom.

subset.

direct

end of this

ga(Z)

a • R.

Basic

discrete

with

TYPE

of P a r r e a u - W i d o m

The

modification.

is r e g u l a r

adopting

surfaces

than

as that

of P a r r e a u - W i d o m

which

stated

by

weaker

lines

begins

theorem.

according

some m i n o r

a certain

nition

tioned

same

surface

same

is g i v e n

OF P A R R E A U - W I D O M

notes

Riemann

fundamental

slightly

the

detail

SURFACES

is to d e f i n e

in a f o r m

almost

RIEMANN

of the p r e s e n t

H. W i d o m ' s

Parreau-Widom

lows

V.

~ 0

be the

of g e n e r a t o r s

show

R \ ~(e,a) ~ ~ 0

surface

a singular R

if and

a natural

among

singular Betti

of the

that

has

set

function every

1-chain only

and

the m a x i m u m

R(~,a)

no c o m p a c t

is a c o n n e c t e d

components.

in

R(~,a)

if it is so in

inclusion homology

number

group

we > el.

6A of the G r e e n

functions

first

a • R

: {z • R: g(a,z)

I,

we h a v e

~ HI(R)

every

relation groups

of the r e g i o n

HI(R(~,a)).

So,

It then is h o m o l o R(e,a).

In

HI(R(~,a))

(Theorem R(e,a), when

2C,

Ch. I).

i.e.

a point

the a @ R

84

is h e l d

fixed,

for all

sufficiently

isomorphic

B(e,a)

with

Definition.

is a n o n i n c r e a s i n g

large

an o p e n

disk

A hyperbolic

Parreau-Widom

type

~,

function

for the r e g i o n

for all

Riemann

(abbreviated

S

large

surface

e

and

vanishes

is c o n f o r m a l l y

e. R

is c a l l e d

to a P W - s u r f a c e

B(~,a)d~

~

in

R(e,a)

a surface

or PWS)

of

if

<

0

for

some

a ~ R.

IB. choice Let

We

of

V

first

be p a r a m e t r i c

the b o u n d a r y we can

hold

find

~G

of

every

(Theorem

6A,

taken

I).

any

ZG.

for we k n o w If we d e n o t e

on the

boundary

This

means

not

such

that

and

s0

the m i n i m u m

therefore

verge

tial as

{z E R:

Theorem.

If

a',

~ g(a,z)

inequalities

ga'

= H[ga';G]

of

g(a,z)

0 < ~ < e0 B(~,a)

~ A

and

we have

~ B(A-I~,

a')

B(a,a)d~,

R

R

at

of

is r e g u l a r ,

S

~ B(~,a)d~ 0

a E R.

Such

the

function

Green or,

for any

which

in the

con-

~ > 0.

This

w C R

z = w.

we r e p e a t

property

Z(a;R)

according

= ~{g(a,w):

is a PWS

w e Z(a~R)}

if and

only

if

to

zero

region

is a c r i t i c a l Let

of p o t e n -

tends

the

then

a surface

d~

sense

ga(Z)

equivalently,

A point at

f0 B ( ~ a ' )

to be proved.

regular

= 0

ga'

and

is c a l l e d

a.

= ~ga/~ of

as was

infinity

is c o m p a c t choice

f0 B(~,a) d~

time,

a E R

point

points

(i)

for e v e r y

some

~ga/~Z

critical

same

surface

~ ~}

of the if

integrals

at the

to the

g(a,z)

ga(Z)

of all

two

if for

tends

independent of

that

A Riemann

theory z

Since and

~0

or d i v e r g e

IC.

respec-

that

B(~,a')d~

therefore

a

same

and

for any

~0

and

a',

A-ig(a',z) the

= H[ga;G]

by

and

contain

of fact,

~G, t h e n

~ R ( A - I ~ , a ') ~ R ( A - 2 ~ , a )

B(A-2~,a).

ga

a

of the

a, a' E R.

G : R \ (CI(V U V ' ) ) .

does

A > 0

As a m a t t e r

points

centers

set

and

is i n d e p e n d e n t

distinct

with

and

is c o m p a c t

definition

two

disks

constant

z @

z E G,

Ch.

above

take

closures,

G

a positive

g(a',z) R(~,a)

disjoint

for any

for

the

V'

having

Ag(a',z)

that

In fact,

and

tively,

note

a ~ R.

is a g a i n point

be the

set

to m u l t i p l i c i t y .

88

[{g(a,w):

(2) for some

(and h e n c e

Proof.

Since

2,...}

of p o s i t i v e

level

h

bordered to

Z(a;R)

curves

We set

all)

is d i s c r e t e ,

numbers

= Cl(R(an'a))"

g ( a , z ) - a n.

gn(a,z).

extended

Let

~n(a,z)

gn'

denotes

points

in

Rn, then

deg(T)

=

readily

I~a

denote

2.

(9) in Ch.

= gn' = N n

gral

B(~,a)da

:

_anB(a n

,a) -

if

~ g(a,w) foB(~,a)da

e ~, t h e n < ~,

n ÷ ~, and thus Parreau

inequality on G r e e n

of

single-valued

R'n deg(T)

has

but

and can be

Nn

in v i e w of Ch. 2, . . . .

of

Rn . : 2 g ~ - 2,

critical

zeros and two poles, n = i,

is e q u a l

so that I, 2B

It f o l l o w s

+ [{ga(W):

f0 B ( ~ , a ) d a

inition

clature

type.

w e Z(a;R), e ~

ga(W)

Conversely,

This

B(a,a)da

< ~.

> an}. if the inte-

÷ 0

H e n c e we get the e q u a l i t y

regular

in sueh surfaces.

for the s u r f a c e s

Riemann

surfaces

to d i s c u s s As we shall

e n o u g h to c o v e r is p r o b a b l y

we are d e a l i n g

[]

for w h i c h the problems

see in 3B below,

in e s s e n c e

enough

R

Diriehlet

(i).

all s u r f a c e s

to l e g i t i m a t e

based

his defof

our n o m e n -

with.

Characterization

A remarkable

a hyperbolic

~

His a i m was

is i n d e e d g e n e r a l

Widom's 2A.

[ g(a,w)

[52] c o n s i d e r e d

lines

~dB(a,a) n

then

(2) holds.

Parreau-Widom

2.

I ~a

n

anB(an,a) as

a

conjugate

and h a v e

is c l e a r

Z(a;R).

is a c o m p a c t

on the d o u b l e

gn(a,z)

for

n = i,

that

= -anB(an,a) So,

If 2N n

gn' = B(an,a)

B(an,a)

that

R'.n

h

is s i n g l e - v a l u e d T

in

w i t h pole

the h a r m o n i c

I, 10D,

s h o u l d have

Since

every

is not n e c e s s a r i l y

the genus of T

any p o i n t

gn(a,z)

differential

{an:

to zero such that the

is r e g u l a r ,

d(gn(a,z) + i~n(a,z))

where

we c o n c l u d e

R

function

~n(a,z)

to a m e r o m o r p h i c

-

decreasing

do not c o n t a i n

Since

So we can use the f o r m u l a

2N n

= a n}

its G r e e n

the d i f f e r e n t i a l

we can find a s e q u e n c e

strictly

and

The f u n c t i o n

<

a C R.

{z E R: ga(Z)

surface

w • Z(a;R)}

Riemann

characteristic

surface

property

due to W i d o m

is a PWS if and o n l y

if it has

says that sufficiently

88

many

analytic

functions.

we n e e d

the n o t i o n

fact, in Ch.

to give

a precise analytic

statement

of this

functions

explained

II.

Given ~(R,~)

a line

bundle

of h o l o m o r p h i c

Theorem that

In o r d e r

of m u l t i p l i c a t i v e

2C,

Ch.

if(z)i p

II,

~ • HI(R,T)

sections

says

that

(0 < p < ~)

over

of the

R, we c o n s i d e r

bundle

if(z)l

6.

If

is an l.a.m,

is a s u b h a r m o n i c

on

function

the

space

f • ~(R,~), R

on

and R.

then

therefore

We

set

surface

R.

llfllp = llfllp,0 : { ( L H M ( B f B P ) ) ( 0 ) } I/p, where

LHM

stands

for the

Ifi p

has

no h a r m o n i c

least

harmonic

majorant,

then

majorant

on the

is d e f i n e d

Nfli

to be

+~.

If When

P p = ~, we

set z •

ilfil : s u p { I f ( z ) i We t h e n

define

for

every

R}.

0 < p ~

~P(R,~)

= {f • ~ ( R , ~

:

iifiL

~}.

<

P Obviously,

~P(R,~),

to the n o r m in Ch.

II,

i ~ p ~ ~,

iR-ilp, if a d d i t i o n

The

fundamental

Theorem.

The

following

multiplication

result

of W i d o m

statements

are

reads

~ {0}

for any

line

bundle

~ E HI(R,~);

(c)

~A(R,~)

~ {0}

for any

line

bundle

~ e HI(R,~).

The

original

as

is a PWS;

statement flat long

form

(b)

and

We

regular

PWS.

Lemma.

Let

Then

bundle

is g i v e n

show

First R

not

in

that

we p r o v e

there

finite

exist

R.

§§2 and

for

asserts

line

The

every

much

bundles

proof

more

but

and

also

of the a b o v e

in fact

for any

theorem

is

3.

of P a r r e a u - W i d o m

the

be a h y p e r b o l i c are

theorem

only

over

of S u r f a c e s

shall

B(~,a)

of W i d o m ' s

is v a l i d

vector

Regularization

R.

defined

equivalent:

R

numbers

respect

as follows:

~(R,~)

3A.

with

are

(b)

rather

in

scalar

space

(a)

unitary

3.

and

Banach

2D.

2B.

the

is a c o m p l e x

PWS

Type

is n a t u r a l l y

embedded

in some

following Riemann

for all

a regular

surface

~ > 0

and

hyperbolic

for w h i c h some

Riemann

the

Betti

(and h e n c e surface

all) Rt

and

a

87

a discrete cerning The

subset

Betti

regular

E

of

numbers

surface

R*

and R%

such that

R

R*

has

is c o n f o r m a l l y

is d e t e r m i n e d

the

same property

isomorphic

uniquely

by

R

with

con-

R % k E.

up to a e o n f o r m a l

isomorphism. Proof.

(a)

First

is f i n i t e . of

R

Then

satisfying (i)

sets

of

the

we assume

there the

(ii)

for e a c h

curve,

CI(V.) i

onto

Vi

is c o n t i n u o u s l y

the

{w e ~: r i < (iii)

Ji'

Betti

number

of o p e n

are m u t u a l l y

which

of

R

subsets

disjoint

itself VI,...,

Vm

noncompact

sub-

m.

Take

form a formal

parametric

disk of

maps

lwl

Ji

i, t h e n

< I}

onto

R

is a s i m p l e

(0 ~ r i < i)

the

bordered itself

closed

homeomorphism

of

C I ( V i)

unit

circle;

Riemann

is e a s i l y

hi

such that onto

surface. s e e n to be r e g -

that

union

R T = RUB.

Va, w h i c h

R.

Vi

r. = 0 for s o m e i. F o r the sake of m that r. = 0 for i = i,..., s and r. > 0 f o r i i a set of s e l e m e n t s , say B = {b(1),..., b(s)},

we assume

i = s+l,...,

of

a eonformal

to a h o m e o m o r p h i s m

is a c o m p a c t

for all assume

~V i

exists

{w e ~: r i <

extendable

R \ (u'm~l=± Vi)

simplicity,

boundary

and there

annulus

lwl ~ i}

r. > 0 i So w e m a y

structure

i, t h e

say

of

and

first number

following:

closures

hi

ular.

the

a finite

R;

analytic

If

that

exist

For

To e a c h

is c o m p a t i b l e

each

i : i,...,

point

with

the

s, we put

a E R

we a s s i g n

given

conformal

Vb(i)

= V i U {b(i)}

a

and

h*(z)

We regard follows

(Vb(i) , h~)

from our

fine a conformal structure the

of

to t h e

So the

Green

extended clearly that

the

Take the

disk

full

open

function

by c o n t i n u i t y

surface

the R

put

is e q u a l R

induces

known,

every

0 <

disk

lwl

< i}

with

of

together

the

harmonic

a E R

for

R %.

R ¢. We m a y

It de-

original function

a harmonic

pole

function

boundary

R

b(i).

c a n be c o n t i n u o u s l y

so as to h a v e R

disks

on

bounded

point

on ex-

function.

can always This

function

thus

conclude

be

is r e g u l a r .

~ > 0 R

the

parametric

Green

ideal

disk about

these

to t h e

any

(a),

z = b(i).

which

for

We now consider

in

for

R%

g(a,z)

(b)

and

z E V. I

that

{w e ~: unit

on t h e R%

on

is w e l l

unit

vanishes

surface

shown

structure As

for

as a p a r a m e t r i c

construction

R.

punctured

tended

= {~ i(z)

to

general = R(~,a). B(e,a)

c a n be c o m p l e t e d

case. Since and

Let us the

fix a point

first

so is f i n i t e .

to a r e g u l a r

surface

Betti

a E R.

number

of

As we h a v e RT

by a d d i n g

88

a finite

number

0 < 6 < e, region

the

of the

of p o i n t s surface

to

Re .

surface

t {z e R~: where this

g6(a,z)

denotes

identification

It is not

gB(a,z)

the G r e e n

for all

e,

Rt

the

is a r e g u l a r

of t h e

B

with

(R t,

hyperbolic

It is a r o u t i n e

Z)

Riemann

We are

Theorem.

Let

numbers

R

Then

boundary

are

finite

the r e g u l a r

A(R)

only

a, and,

set

surface

of

Z

R

from

which

those

of

satisfies

uniqueness

a.

With

Rf

then

the p r o p e r t y

and

surface

Let

of the

pair

following

surface

~ > 0

hyperbolic

such that

the

Riemann

for any

is at m o s t

for each

Z

for w h i c h

some

Rt

be t h e

a point

into

countable

b E [, t h e r e

there

Rt

relative

R*

exists

obtained

b E A(R)

the

(and h e n c e

subset

Betti all)

in L e m m a

3A

of the M a r t i n

belongs

to

~

if

of

and

independent

a neighborhood

R

VAA(R)

with

a uniquely

Rt

> 0.

exists

of

a hyperbolic

topology

determined

Riemann

of

choice

V

b

of

= {b}.

conformal

surface,

as a s u b s p a c e

of the

and

in t h e

On the u n i o n structure

is c o m p a t i b l e

R*,

of

with

satisfies

the

which the fol-

properties:

(BI) of

Rt R%

(B2)

is a r e g u l a r induces

If

intersection bounded

pole

[]

to p r o v e

as f o l l o w s .

compactification

R t = RU Z

ture

with

if

the

lowing

R t6

for

induced

l i m sup g ( a , z ) Rgz÷b

makes

sub-

6},

to s h o w t h e

be a h y p e r b o l i c

B(e,a)

R.

Martin

if

the

~ > 0}.

isomorphism.

in a p o s i t i o n

c a n be c o n s t r u c t e d

Then

with

0 < 6 < ~, w e put

structure

matter

up to a c o n f o r m a l

3B.

and

conformal

to s h o w that,

identified

lemma.

(c)

in

R%

> ~

function

R t : u{Rt: If we g i v e

difficult

R % m a y be c a n o n i c a l l y t~ R B d e f i n e d by

V

to the

point

to t h e

conformal

(B3)

The

b

hyperbolic

R

reduces

function

u

so as to get structure

Green

surface

the o r i g i n a l

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

V A&(R)

harmonic

on

of

function

of

to the on

b E ~ single

VNR

a harmonic

and the

conformal in

R*

point

c a n be

conformal

structure such

b, t h e n

extended

function

on

V

of that

strucR. the

every

by c o n t i n u i t y with

respect

R t. gt(a,z)

for

Rt

with

pole

a E R

is

89

o b t a i n e d by e x t e n d i n g the Green function in

~

function 0 E R

kt(b,z)

for

Rt

with pole

in the t o p o l o g y of (B4)

b

for

and

face

0

R

Similarly,

to the points the Martin

and with respect to the origin

to the points

k(b,z)

in

~

for

R

by c o n t i n u i t y

R*.

The M a r t i n c o m p a c t i f i c a t i o n of

Rt

can be identified with

AI(R ) = AI(Rt ) U ~.

(B5) Rt

The h a r m o n i c m e a s u r e at the origin

0

dx t, supported on

at the point (B6)

Proof.

If

Let

(BI) and

AI(R t), of the sur-

is nothing but the restriction,

AI(Rt) , of the h a r m o n i c m e a s u r e R

R*.

is o b t a i n e d by extending the Martin function

with respect to the same origin

R*

g(a,z)

by c o n t i n u i t y in the t o p o l o g y of

dx, supported on

to the set

AI(R) , of the surface

0. R

is a PWS, then so is

(R t, ~)

R t.

be a pair given by Lemma 3A.

(B2) are clearly satisfied.

Then the properties

So we prove here that

Rt

has

the r e m a i n i n g properties. Property in

(B3):

The p r o p e r t y

R, the Green function

(B2) implies that,

z' ÷ g(z',z)

for

R

extended by c o n t i n u i t y to the Green f u n c t i o n So the d e f i n i t i o n of the Martin functions z E R, the function continuity,

Property on

Rt

(B4):

say

Rt

is o b t a i n e d by extending,

with

z E R, the function

z E R

also separates points of

R #.

~ = A(R) \ A(Rt).

Since E

Rt

b E Z, then the f u n c t i o n

tiple of have

gion

gt(b,z)

~ ~ AI(R)

b ÷ kt(b,z)

b ~ k(b,z) R #.

Hence,

R#

R* = R #.

is regular,

on

Since the set

separates points of

R#

and

b ÷ kt(b,z)

with

can be i d e n t i f i e d with In particular,

we

we conclude that a point

= gt(a,b)

z ÷ k(b,z)

> 0.

is equal to a constant mul-

and so is a minimal h a r m o n i c function on and t h e r e f o r e

R.

(B5):

This is clear from the above observation.

Property

(B6):

Let

Bt(~,a)

= {z E Rt: gt(a,z)

We thus

AI(R ) = ~I(R t) U ~ .

Property Rt(~,a)

by

E.

if and only if

lim g(a,z) Rgz÷b If

function on

zt ~ Rt

R, i.e.

in

function on the Martin compacti-

R t, the subfamily of functions

the Martin c o m p a c t i f i c a t i o n of

belongs to

to the points

and t h e r e f o r e the function

R

have

for each fixed

R

since

b E ~(R)

R t.

on

b ÷ kt(b,z t)

is dense in

shows that,

z is

for

Rt

can also be extended to a continuous

of functions

z E R

on

For each fixed

R #, of

with pole

z' ÷ gt(z'~z)

b ÷ k(b,z)

can be extended to a continuous

fication, R

b ÷ kt(b,z)

the f u n c t i o n

for each fixed

be the first Betti number of the re> e}, where we assume

a E R.

Then,

90

Bt(~,a)

S B(e,a)

So

is a PWS wh~enever

RT 3C.

to

R

for all

We now c o m p a r e

and

~ > 0 R

f0~ B t ( a , a ) d ~

and thus

is.

S /0 B(e,a)de.

[]

the i n t e g r a l s

of Betti

numbers

corresponding

Rt .

Theorem.

Let

critical

points

of the set

R

be a PWS and let of the f u n c t i o n

Z(a;R t)

I~ B ( ~ , a ) d ~ 0

that

a • R.

Then the

z ÷ g(a,z)

lie in

: I v Bt(~ ,a)de 0 : [{gt(a,w):

R

set

consists

Z(a;R)

of those

of

elements

and we have

+ [ {gt(a,w) : w • E}

w e Z(a;Rt)}

+ [{g¢(a,w):

w • ~}.

for n = Proof. Let E : {Wl, w2,...} and put R n = R t \ {w I ,. ..,w n} i, 2, . . . . Denote by B (~,a) the first Betti n u m b e r of the s u b r e g i o n n we have Rn(~,a) : {z E Rn: gt (a,z) > ~}. Since R n = Rn_ I \ {w n} ~B n l(a,a) = ~Bn-l( ~ _ ,a) + i

Bn(~,a)

for

e > gt(a

) 'Wn ~ < gt(a,Wn).

for

So we have

F

0 Bn(~'a)d~

:

:

Since

Bn(a,a)

the d e s i r e d

§2.

PROOF

result.

4.

stated

Analysis 4A.

section

~ Bt(a,a )d ~ 0

n~_ gt(a,wk ) . k=l

monotonically

THEOREM

B(~,a)

as

n + ~, we get

we prove

(I) the i m p l i c a t i o n

on R e g u l a r

(a) ~ (b) in W i d o m ' s

the-

Subregions

We begin with the f o l l o w i n g

[31],

to

+

in 2B.

For an i n t e r e s t i n g Heins

+ g t ( a , w n)

[]

OF W I D O M ' S

In this orem

converges

i

0 Bn-l(~'a)d~

p.

75.

proof

of this

classical

crucial

result

theorem we r e f e r

of C a u e h y - R e a d . the r e a d e r

to

91

Theorem. IA and

Let

let

(a) at a l m o s t

G

be any

i ~ p ~ ~. Every

f E HP(G)

every

b E

arc-length

measure

differential (b)

w

Let

a.e.

Then

on

on

4B. line

Let

G

connected.

which

~

union

every

is r e p r e s e n t e d

~P(G,~IG).

Since curves,

Assume

value p'

i.

= p/(p-

(3)

a.e. that

a simple

on

i < p < =

I).

Then

and

for any

inf{ll~llp,,~ ° :

Y

p'

line

we d e n o t e

the

exponent

-1

denotes

every

line

element

is bundle

to

bundle G \ C,

The

set

by the of a f i n i t e

nontangential same

symbol

vanishing residue exponent

~ E HI(R;~)

I~(a)l

inverse

CI(G).

of the

consists has

conjugate

: sup{li~(a)l: ~ • ~P(G,~-ln-lIG), where

of

function.

with

be a

CI(G) \ C

is d e n o t e d

is n o w h e r e

the

f.

a restriction

by the

a E G \ C

bundle

of

that

G

of

set of m e r o m o r p h i c

connected,

of

= 0

is e q u a l

< E HI(R,~)

the

has

(G,nlG)

•

f*

neighborhood such

f~G um

u

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

ZG

f*(b) to the

neighborhood that

let

~ E ~P(G,~IG)

at a p o i n t

equation

holomorphic

the b o u n d a r y

ZG, w h i c h

Let

and

~P(G,{IG)

<0 E F M ( C I ( G ) , ~ ) pole

the

such

curves

in

respect

and

~. has

only

of a b s o l u t e of

p,

i.e.

we h a v e

: l}

IlfilIp,~ : 1}, in the

group

HI(R;T)

and

I lql l) ljq

r I1~11

We n o t e in some

that

for

1

q

for

q : ~.

is d e f i n e d

for any

<

= j/

[suP{If(z)I:

the n o r m

neighborhood

of

~G.

II-II q,~

z •

G}

I,

for any h o l o m o r p h i c

function

in some

~(G,~IG)

every

= 0

FM(CI(G),~)

denotes

in

value

in some

R

is s i m p l y

of s e c t i o n s

of a n a l y t i c

one pole,

in

by a s i n g l e - v a l u e d

symbol

Lemma.

by

61G

section

number

values

boundary

region

G \ C

boundary

of Ch.

CI(G).

satisfy

defined

defined

If

sense

with

f*~ of

f C HP(G)

of a n a l y t i c

Since

of such r e s t r i c t i o n s

boundary

~

a unique

We d e n o t e

is trivial.

G, t h e n

f~G

i ~ p ~ ~,

of

in the

f* E LP(3G)

and we have

differential

R.

R

a nontangential

that

be a r e g u l a r

over

G \ C

to

of

hold:

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

exists

be a f i n i t e

simply over

subregion following

to the n o n t a n g e n t i a l

bundle

C

has

~G

differential-sections Let

the

such

u e LP(~G),

there

8G

~G

defined

for any h o l o m o r p h i c CI(G).

regular Then

m

given

only

92

Proof.

Take any

G.

~£P(G,<-in-ilG) and

~ •

the e o r r e s p o n d i n g

section.

So if we denote by

Ga

Then

lh(z)l

let

~P(G,g-lq-IIG)

h •

is a subharmonic

the harmonic measure of

G

be

function on

at the point

a

(Ch. I, 5D), then

(4)

lh(C)Id~(C)

[~(a)l: [h(a)l £ I ~G

max ( w G ( < ) / l ~ ( < ) ] ) ~EBG a Since

~

is n o n - v a n i s h i n g along

ferential along

~G

~G,

and therefore

[ Ihl [~lj ~G

!~l

is a strictly positive dif-

max{E~G(~( I ~ ) )/ ] ~_( ~ )

< ~.

Thus

by a p p l y i n g the H61der i n e q u a l i t y to the integral

in the last m e m b e r of

(4) we see that the e v a l u a t i o n map

is a continuous

functional on

~P(G,~-In-IIG)

norm of the functional

aa

llmall : sup{l~(a)I:

~ a : ~ + ~(a)

with respect to the norm

in

~P(G,~-Iq-IIG)

with its boundary values on

can be viewed as a subspaee of

I/F/I p' = I[a a II

~(a)

-

F • LP'($G,Iwl/2~)

By the such

1 I ~G F~I~I

~ • ~P(G,~-In-IIG).

we can find a h o l o m o r p h i c

differential

represent a.

CI(G).

in a n e i g h b o r h o o d of

simple pole at

a ~

with residue on

G \ C.

Since

w

[~/~ •

to a r e g u l a r s u b r e g i o n c o n t a i n i n g

section

v a n i s h i n g on some n e i g h b o r h o o d of

the point

~G, so that the space LP(~G,Iwl/2~).

2~i

By a p p l y i n g Theorem 3B, Ch.-II, CI(G)

We identify

and

(5) for every

The

~ • ~P(G,~-In-IIG) , ll~llp,~ =< i } ,

H a h n - B a n a e h theorem there exists a function that

ll.llp, .

is then given by

which is exactly equal to the r i g h t - h a n d side of (3). ~¢P(G,<-Iq-IIG)

linear

CI(G) i

k • ~f~(G,n-llG), which is nonLet

~

be any m e r o m o r p h i c

with only one pole which is a

(Theorem liB, Ch. I).

We may suppose that

~

Finally let

has residue

i

at

is nowhere vanishing,

~(G,~-ln-IlG) ~ ~:'P(G,g-ln-IlG)

and so, by (5), we have

If

a

is any h o l o m o r p h i e d i f f e r e n t i a l on some n e i g h b o r h o o d of

CI(G),

93 then if

~ +~ ~

has the

is r e p l a c e d

this h o l d s

same p r o p e r t y by

for all

~ + ~.

is e q u a l to the b o u n d a r y

Moreover

we h a v e i

We set

= i.

that

(6) r e m a i n s

]~G F ~ l ~ r / ~

e, T h e o r e m

function

f~

4A-(b)

to be v a l i d = 0.

Since

implies

of some

that

f0 @ HP'(G)"

:

= ~

f : f0/k.

If(a)l

9, so that

such d i f f e r e n t i a l s

F~I~I/~

fo(a)

as

It f o l l o w s

~G

Then

~G

f • ~P'(G,~IG) ,

Ifl = IFI

on

ZG

and

Hence,

inf{li~Up,,~:

~ • ~ P ' ( G , q I G ) , If(a)i

So in (3) the l e f t - h a n d Conversely, ~P(G,~-In-IIG) single-valued

if

side

is at m o s t

~ • ~ P ' ( G , n i G)

= i} =<

liFilp, =

liEali.

equal to the r i g h t - h a n d

with

Jf(a) I = 1

with

II~II = i, t h e n ~Z~ P,~ d i f f e r e n t i a l on G and so

and

side.

~ •

can be e x t e n d e d

to a

? i

~G This

shows the r e v e r s e

4C. gion

G

in

Theorem. which

The

inequality,

statement R.

Let

=

be any r e g u l a r

is h e l d fixed.

Then,

sup[inf{Uhli

p,a

g(a,.)

group

H I ( G ;T)

denotes

IIf IIp ,a :

{

The t i l d e h e r e d e n o t e s

n o t e that

in p

: h • ~(G,~),

R with

for any r e g u l a r

and t a k e a p o i n t

re-

a • G,

i ~ p £ ~,

rh(a)l

: i]]

w • Z(a;G)]],

function

for

I f ( { ) i P d g ( { , a ) ) I/P

suP{If(z)l:

Before

[]

G,

<

ranges

o v e r the

and [-2~

identical

P '~

2B is v a l i d

region

for e v e r y

the G r e e n

P'~

we show the f o l l o w i n g

= exp[[{g(a,w): where

'~

as d e s i r e d .

(b) in T h e o r e m

More precisely, G

P

proceeding

conjugate

harmonic

to the proof,

w i t h the one g i v e n Z(a,G)

z • G}

set.

i ~ p <

for

q = ~.

function.

we n o t e that the a b o v e

in 2A in case

is a f i n i t e

for

R = G

and

n o r m is

0 = a.

We also

Proof.

Consider

has a simple to

Z(a;G),

the m e r o m o r p h i c

pole at

a

the order

multiplicity

in

elements

Z(a;G).

in

differential

with r e s i d u e

of each

Z(a;G).

1

zero being

Let

dg(.,a) + idg(-,a),

and zeros

j

which

belonging

the same as the c o r r e s p o n d i n g

{Wl, w2,... , w m}

For each

at points

be an e n u m e r a t i o n

of

we set z

Aj(z) which

: exp[g(z,wj)

is m u l t i p l e - v a l u e d

fines a line bundle,

say

Ch.

set

II,

2D.

We then

but has

+ i

f

dg(',wj)], a

single-valued

<j, over

G

m ~' = (j~=l*=A j ] [ d g ( . , a ) + Clearly,

~' • FM(CI(G),~)

vanishing

on

a.

the r e s i d u e

Since

CI(G)

j•

with

of

~'

m = exp[~-

Aj(a)

-

co : r -I ~], along

i < p < ~

:i ~j"

Moreover,

simple

has a b s o l u t e g(a,wj)]

the h y p o t h e s i s

in case llfll q,a

Let

and set

it is n o w h e r e

pole--at

the

point

value

(= r, say),

of Lemma

4B.

Since

[~'I

= -dg(.,a)

q ;~ :

i1~ilq,~'

= rl/qEi~li q,~"

p' = p / ( p - i ) .

inf{ liflip, ,a : f e ~P'(G,T]), : rl/P'inf{U~lip,,

Then

Vf(a)i

Lemma

4B shows

that

: i}

: ~ e ~fP'(G,~),

1~(a)l

: i}

= rl/P'suP{i~n(a) I: ~ • ~{P(G,~-In-I),

iJ~nilp,~ = i}

: r l / P ' s u p {I h(a) I : h • ~ P ( G , ~ - I n -I),

UhUp, a : r I/p}

= sup{lh(a) I: h • ~ P ( G , ~ - I n -I), for any line b u n d l e

in

j=l

satisfies

~G, we have

a

it de-

idg(.,a)].

one p o l e - - a

at

so that

to our o b s e r v a t i o n

-]-]-j m

~ =

and has only

modulus,

according

q

over

G.

llhilp,a : r]

A similar

computation

is p o s s i b l e

for

p = I, so that we have (7)

inf{ iif ilp, ,a : f e ~fP' (G,q) = sup{lh(a)i:

for any line b u n d l e

over

If(a) i = i}

h • ~P(G,~-IT]-I), G

and for any

p

iihli

p,a

with

= r}

i ~ p < =.

Since

95

ihl p

is s u b h a r m o n i e ,

given

by

<-in-i to

(7)

is at m o s t

: nn -i : the

r, t h e n

side

of

we h a v e

h(a)

lh(a)[

equal

to

identity

: r

of

<

llhll p,a Setting

r.

Hi(G;T).

rlhfi : r. This p,a at this n. Hence

r

sup[inf{llflr p' ,a : f • ~{P'(G,n), n Writing all

p

p

in p l a c e

with

p',

we w i l l

~ = ~ii~ -I

seen

of the

case

that

so that,

by

__>r value

r

is e a s i l y

is r e a c h e d

seen

5A.

We are form.

Theorem.

Let

with

holds

for

~-lq-i

Letting

: ql'

, llhlil,a : r } If(a)I

= i}

= 6 -1

~i

h • ~(G,D),

and

f - i.

Thus

llhlll,a = i ] = r -1, to

h • ~(G,n),

lh(a)l

: r~

= i}]

[]

of N e c e s s i t y

following

p

at

to be e q u i v a l e n t

to be proved.

Proof

theorem

-i

sup[inf{llhJJl,a: n

5.

= r.

, llhlll,a : i}

: f • ~(G,~ll<-l),

inf[sup{lh(a)l: n

as was

the

equal

the r i g h t - h a n d

(7),

h e ~(G,nl)

= r-linf{IlfJl

which

that

: i}]

p = i.

= r -i sup{lh(a) I: h • ~(G,~I)

the

is i d e n t i c a l l y

If(a)l

we h a v e

dispose

sup{]h(a)J:

Here

h

means

I < p < ~.

Finally, we have

of

t h e r e f o r e the v a l u e ~-i n = in (7), we h a v e

If

and

(7) r e a c h e s

and

going

R

to p r o v e

(a) ~

be any h y p e r b o l i c

(b) of W i d o m ' s

Riemann

theorem

surface.

Then,

in the

for e v e r y

i ~ p ~ ~, sup[inf{IJfll

p,a

= exp

: h • ~P(R,$),

(I

B(a,a)da

If(a) I = i}]

),

0 where to

the

supremum

is t a k e n

{ ( L H M ( I f J P ) ) ( a ) ] I/p

if

over

all

p < oo

[ • HI(R,~), and

to

and

sup{If(z)l:

i[fH

p,a z • R}

is e q u a l if

p = =.

96

Proof.

Let

sisting

of c a n o n i c a l

{R : n : i, 2,...}

We then

denote

be a r e g u l a r

exhaustion

of

R

(Corollary

IC,

Ch.

con-

n

by

Rn(~,a)

first

gn(a,z)

the r e g i o n

Betti

number

We first p < ~

by

subregions

of

: inf{Iifll

Mn(~,a)

p,a

M(a)

definition

Rn

with pole

and by

Namely,

If(a)I

Bn(~,a)

we fix any

: i}

f • ~P(mn,~) , If(a)I

= sup{M(~,a):

Mn(a) With this

p.

for

> ~}

I).

p

a, the

with

1

<

notations:

: f • ~P(R,{),

= inf{iifiip,a:

function

gn(a,z)

R (~,a). n any finite

consider

a • RI

the Green {z • Rn:

and use the f o l l o w i n g

M({,a)

with

= i}

for

~ • HI(R;~),

for

~ • HI(Rn;T),

~ • HI(R;Y)},

: sup{Mn(~,a):

< • HI(Rn;~)}"

we c l a i m

(8)

M(a)

= lim M

(a). n

To see this,

let

~ • HI(Rn+I;~);

Mn(~IRn,a) Since Ch.

Rn

is c a n o n i c a l ,

II, that

bundle

over

surjection above

every R.

~ Mn+l(~,a)

over

HI(R;T)

inequalities

R

3B, Ch.

H I(Rn+ I;~))

Mn(a)

lim Mn(a)

so that

I, and T h e o r e m

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

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

(resp.

imply that

~I Rn • H I ( R n ;~)

~ Mn+l(a)-

we see by T h e o r e m

line bundle

In o t h e r words,

from

then

map

to

~ Mn+l(a)

IB,

of a line

~ ÷ ~IR n

HI(Rn;~).

is a So, the

£ M(a)

and c o n s e q u e n t l y

suppose

that

~ M(a).

n+~

In order

to get the r e v e r s e

inequality,

we may

lim M (a) < ~. n

n+~

~ @ HI(R;~).

Let us take any is d e f i n e d {V~}).

by means

Since

and that all

R V'

of an open

is separable,

closure

CI(V

tion of

Mn(~IRn,a)

an element

i n Ch.

covering

{V~}

we may assume

that

II,

1A, t h e

in the form {V '}

is

bundle ({~B},

countable

as well as all n o n e m p t y

Then we take a r e f i n e m e n t )

As e x p l a i n e d

{V }

is a compact implies

fn • ~ P ( R n , ~ I R n )

of the

subset that,

of

V' ~ R are simply connected. e n covering {V'} such that the V'

for each

satisfying

for each

~.

n : i, 2,...,

Ifn(a)l

: i

and

The d e f i n i there

exists

llfnllp,a

97

Mn(~IRn,a)

+ n -I.

Let

{V~ A R n } ~)

{fn }

{ V , A R n }~

to the c o v e r i n g of

~JR n.

fn~

is d e f i n e d

Let

un

be a r e p r e s e n t a t i v e of

Since

Rn

V ' AR n

each n o n e m p t y

as a s i n g l e - v a l u e d

be the LHM of

Ifn Ip

holomorphic

on

i I H fnt1$, a : -2--~,DR

of

fn

with respect

and the r e p r e s e n t a t i v e

Rn,

({~B},

is simply

function

on

connected,

V' ~ R n.

then

=

Ifn(~)iPdgn(~'a)

Un

(a).

n Take any compact equality, SUpzEK

K

included

in some R • By use of H a r n a c k ' s inn C > O, d e p e n d i n g on K, such that we have

we find a c o n s t a n t

v(z)

Applying

set

~ Cv(a)

this

for any p o s i t i v e

fact to the f u n c t i o n s

Ifj(z)l

harmonic u. ]

with

function

v

on

R n.

j ~ n, we have

c 1 / P u j ( a ) 1/p = c1/Pllfjllp,a

~ uj(z) I/p ~

< cI/P(Mj(~BRj,a)

+ i_)

=

]

c 1 / P ( l i m M~(a) + 1) < ~ for every

z @ K.

on any compact of some

such that,

uniformly

on

{fn(k),~"

k = i, 2,

belonging

to

thing

~P(R,~)

to see is that

ifn(k)(Z)I choose

{Ifj(z)l:

of

R n.

+

any

CI(V

).

{fn(k),~:

f

{f~} if(a)i

CI(V a)

Jf(z)J

uniformly is fixed

clearly = i.

lf(~)[

DR

on every

~,a) =< - 7 4

of the

compact

{fn(k):

a section

f

First we note that

subset

Then,

of

R.

for any

Next we

~ > 0

ifn(k)(Z)IPi

< e

[fn(k)(~)[Pdgm(~,a)

DR

con-

subsequenee

the only n o n t r i v i a l

majorant.

for a moment.

m

subset

represents

Here,

exists k 0 with m < n(k 0) such that Jif(z)I p DR for every k > k 0. It follows that m 1 2~

bounded

is a c o m p a c t k = i, 2,...}

be the limit

has a h a r m o n i c

m, w h i c h

is u n i f o r m l y

to find a s u b s e q u e n c e

a,

Let

with

every

process

for each

}; then Jfi p

j = n, n + l , . . . }

Since

Rn, we use a d i a g o n a l

k = i, 2,...} verges

Thus

subset

there on

+ e

m

if

< -7~

I f n ( k ) ( ~ ) J P d g n ( k ) (~'a)

+ s : llfn(k)lipP,a + s

DRn(k) < (Mn(k)(~IRn(k),a)+ As

s

and

n(k)

n--~)P

are a r b i t r a r y ,

+ s < ( M n ( k ) ( a ) + n ~ 7 ) p + e. we have,

by letting

s ÷ 0

and

n(k)

9B

_ I__[ 2~ DR

If([)IPdgm(~,a)

~ lim M (a) p. n÷~ n

m

We n o w m o v e

m

and set

1 I 27 DR

Vm(Z ) _

If(%)IPdgm(%'z) m

on

Rm

with

see that

If(z)l p ~ Vm(Z)

for all sequence is

majorized

If(a)

m : i, 2, . . . .

l :

m.

Thus,

£ Vm+l(Z)

1

and

Theorem

establishes

subregion

process.

values

of

Cl(Rn(~a))

the e q u a t i o n

(8), the d e s i r e d

4C by a l i m i t i n g

from critical

regular

and

Vm(a)

~ lim n M n ( a ) P

the limit

v(z)

R.

Since

Rn+ I.

because

This m e a n s

ular subregion o t h e r hand,

of

gn(a,z)

since

which

cycles

let

CI,...,

are

t h e y are

in

included

independent

valid.

Since

values,

(9) h o l d s

Combining convergence

each

for all

(8),

a > 0

Rn(a,a) on

dis-

is a

CI(Rn) , we

have a natural

map:

Since

Rn

subregion is a r e g -

B (e,a) < B(~,a). On the n = i n c r e a s i n g l y to g ( a , z ) , we h a v e the e q u a t i o n

= lim B (~,a). n n+co

with

k < B(~,a) Since

Rn,(e,a) ~ > 0

be any f i n i t e

R(~,a)

=

U

Bn,(~,a)

an at m o s t

finite

with a countable

set of inde-

co

n=l Rn(~'a)

Rn(e,a) , say

and thus

admits

(9), T h e o r e m s

Then

So (9) is

number

number

these

Rn,(e,a).

> k.

of c r i t i c a l

of e x c e p t i o n s .

IC and 4C, and the L e b e s g u e m o n o t o n e

we f i n a l l y h a v e

: l i m Mn(a)

from

that

in one of the r e g i o n s in

number

each

< gn+l(a,z)

in t u r n i m p l i e s

HI(R(~,a)).

gn(a,z)

theorem,

M(a)

Ck

M(a)

can be d r a w n

is seen to be a r e g u l a r

converge

B(e,a)

1-cycles

Thus

=< B n + l ( e , a ) .

Bn(e,a)

gn(a,z)

that

÷ Hl(Rn+l(e,a)),

Rn(~,a)

that

(9)

In fact,

gn(a,.).

R, we see s i m i l a r l y

~ Rn (~,a), = U n=l

pendent

conclusion

~ R n + l ( ~ , a ) , and t h e r e f o r e

is i n j e c t i v e

R(~,a)

<

of the

(8).

Take any p o s i t i v e

of any

Hl(Rn(~,a))

of

Rm

theorem,

we e a s i l y

EIfll

N o w that we h a v e

which

on

by the H a r n a c k

is s u b h a r m o n i c ,

{v (z)} is h a r m o n i c on R v(a) < lim M (a) p and If(z)l p m ' = n n by v(z) everywhere on R. Hence, we h a v e f E ~P(R,[),

lim n Mn(a) , w h i c h

have

Ifl p

< v(a) 1/p < lim M (a). This implies p,a = = n n S l i m n Mn(a). As ~ is a r b i t r a r y , we c o n c l u d e that

M(~,a)

tinct

Since

: lim e x p [ [ { g n ( a , w ) :

w e Z(a,Rn)}]

99

: lim exp n+~ This establishes The case

,a)d~)

in the case

is entirely

PROOF OF WIDOM'S

THEOREM

Review of Principal 6A.

Since

of

boundary of part

Rj.

Two partitions

say

RI,... , R£.

6(G)

P(G)

{Rj \ CI(G"):

and

Rj e P(G)}

of

R \ CI(G").

each equivalence P

6j(P)

class,

P(G')

nonvoid open sets.

P(G)

consists

denotes

of a fi-

the relative

are called equivalent

containing

CI(G)

{R~ \ CI(G"):

This relation ~,

defines

= {P(G)}

be a partition

has the same number of parts,

in every

P(G)

and

CI(G')

R~ • P(G')}

a partition

if there such that

define

is an equivalence

that

For every fixed

j,

and If

u

P(G) • P,

6(~)

the same

relation and

of the ideal boundary,

= (61(P),...,

is a harmonic

for

6(R).

say

with

£, and,

using the

j = i,...,

£.

6~(P)~ with

6£(P)).

function

defined in some

fSj(p)

*du

R. • P(G) with 3 have the same value for any

*du

denotes

*du = -($u/~y)dx + (~u/~x)dy. u

P(G)

if we arrange parts

we may always assume,

CI(G) ~ G', where

du:

called the flux of

Then each member

R. \ CI(G") = R! \ CI(G") for any 3 3 6j(P) denotes the collection of

then the integrals

P = P(G') • • ferential

of

in some suitable order,

above notation,

P • ~

disjoint

P = P(G)

R.

Let P

we denote by

in what follows.

and G"

in

R

into mutually

exists a regular region

6(R), of

in

R. in R, which is oriented negatively with respect to the 3 The union U.3 6j(P) is the positively oriented boundary of

G, which is denoted by

partition

G

has finitely many components,

nite number of parts,

(c) ~ (a)

Operators

R \ CI(G)

R \ CI(G)

[]

is to prove the implication

Given any regular region

any partition

for the ar-

does not occur here.

(II)

The objective of this section in Widom's theorem.

6.

1 < p < ~.

similar but much easier,

gument concerning harmonic majorants

§3.

B(~,a)d~). 0

the theorem

p = ~

= exp

0

over the cycle

6~(P)

the conjugate

dif-

The common value is

and is denoted by the

J

symbol

f6j(P) *du.

Finally,

P

is called the identity partition

of

6(R)

if each

100

member

P(G)

6B.

Let

set w h o s e lytic

of

P

W

be a n o p e n

boundary

curves.

consists

uous on

a linear

function CI(W),

on

LI : i,

(B2)

f ~ 0

(B3)

Lf = f the

on

F

which

that

number of the

Lf,

is a c o m p a c t

of n o n i n t e r s e c t i n g ideal

operator

assigns

a function

R \ W

boundary

associated

R

in

W

with

with

to e a c h r e a l - v a l u e d

harmonic

ana-

of

and

W

contin-

continuous

the

Lf

over

every

B.(P) ]

vanishes.

following

L W

Lf % 0,

~W,

f l u x of

Let in

such

of a f i n i t e

L

~W

R

By a n o r m a l

implies

(B4)

Theorem.

in

part.

that

(BI)

harmonic

set

B~(P)).

operator

f

such

T h e n we h a v e

of a s i n g l e

We f i x a p a r t i t i o n

B(P) = (BI(P),..., we m e a n

consists

be a n o r m a l

operator

and continuous

on

on

CI(W),

I

*ds

W

and

let

s

be a f u n c t i o n ,

such that

: 0,

~(P0 ) where

P0

denotes

a harmonic

the

function

identity

p

on

R

partition

which

of

B(R).

Then

there

exists

satisfies

p - s = L ( ( p - s)l~W) on

CI(W).

([AS],

The

Ch.

6C.

III,

number

p

normal

In o r d e r

bordered

operators

to c o n s t r u c t

surface

W

of n o n i n t e r s e c t i n g

border

B(W)

parts

~,

Every

Bj

oriented Lemma. function

is n o t

El,...,

closed

each

is o r i e n t e d

u

up to a n a d d i t i v e

continuous

harmonic

in

such operators, border,

we d i v i d e

it i n t o

and

is a u n i o n

with to

function W

6(W), curves.

of w h i c h

respect

kind,

analytic

positively

with

of s p e c i a l

whose

connected,

B%,

negatively For a n y

is u n i q u e

constant.

3A)

We n e e d

operators. compact

function

respect

called

let us f i r s t consists

at l e a s t

W

take

that

the

two n o n v o i d curves.

whereas

e

is

W. f

on

continuous

~

there

on

W

exists

with

the

a unique following

properties: (a)

u = f

(b)

u = c.]

on on

~, 8 j'

c.]

being

a

of a f i n i t e

Supposing

of component to

principal

a constant,

for

j : i,...,

~,

101

(c) The

]6j

*du

function

6D.

u

: 0

set

in

to the R

consists

of a f i n i t e

and

P

let

form

R \ W.

We u s e

regular

For WAG

let

us d e n o t e

n

on

~,

the (n)

on

a constant

and

= B(Gn) N R j .

operators

real-valued Dirichlet

following f

denotes

preceding on

WAG

W.

functions problem

cj,

The

boundary

on

f

on

on

W

~

on

=

shows

such

i

by

L (n)

be a

the n e g a t i v e l y

on the

continuous in

set

func-

WAG n

are w e l l - d e f i n e d

to the p a r t i t i o n

that

W, we the

Z(~,W).

solution

Take

on

C I ( W n Gn)

cj

on

Rj \ CI(Gn) ,

constants that

which each

any

make u

e. such

un

n

~(G

= - I

)

n

S

n

~

for the

The

set of

f.

We

continuous

is h a r m o n i c

n is d e f i n e d on CI(W) \ 8(G ). n n for a d i f f e r e n t i a l ~ on W, we h a v e

n

continuous

H[f~W]

du

n

P.

consider

up to the b o u n d a r y

L(n)f

implies

WnG

n < n'

that

to the w h o l e e

~, are f

"Un":fS For

curves

.-.

operator

for e v e r y

is h a r m o n i c

Clearly,

II~ll~ = [S w ~[*

denotes

harmonic

function

W

containing

GI ~ G2 C

the n o r m a l

let

a representative

region

conditions:

is d e n o t e d

j = I,...,

assumption e.

@

5)

boundary

analytic

has

Let

C G I.

L (n)

set w h o s e

regular

6A.

corresponding

L (n)

Un where

which

Ill,

Namely,

closed

the

lemma

n

to e x t e n d

functions

8(R),

in

by

6B.

Ch.

and CI(WNG ) such that (L(n)f)I~ = f (L(n)f)IB~(G ) n ' n ] B j ( G n ) * d ( L ( n ) f ) = 0 for j : i,..., Z, w h e r e Bj(G n)

The

In o r d e r

such

L

of

CI(G)

each

continuous

normal

with

W.

in

is a c o m p a c t

is a f i x e d

described

of

([AS],

considered

R\ W

G

boundary

satisfying f

R

is p o s i t i v e .

of n o n i n t e r s e c t i n g

partition

the n o t a t i o n

n tion

that

where

of

f

situation

such

P = P(G),

exhaustion

oriented

if

number

be a f i x e d

of the

j : i .... , ~.

is p o s i t i v e

We r e t u r n

be an open

for

n

on

up to the

Writing

n

n

Un*dUn"

we h a v e

WAG

n n

n

G

,\G n

n' n

set

'

102

= I

:

Thus,

lldUnllW

Un'*dUn'

iEdunilW2 -

- I

i[dun ' iSW" 2

are m o n o t o n i c a l l y

decreasing

itdunil~ as

n, n' ÷ ~.

It f o l l o w s

each parametric an L 2 - C a u c h y

disk

V

sequence,

Un*dUn

iIdun,ii~ ÷

that

and t h e r e f o r e 0

This m e a n s

iidun,- dUnll W ÷ 0.

the d e r i v a t i v e s

(resp.

~u /~x n

n, n' ÷ ~.

{~Un/~X}

~u /~x and ~u /$y are h a r m o n i c , we see that n n {~Un/~y} c o n v e r g e a l m o s t u n i f o r m l y on V. As V is

we c o n c l u d e

verges

almost

n, the

sequence

which Since

u

shows that

{Gn} , we can d e f i n e

CI(W).

du E F I determined

an o p e r a t o r

The l i n e a r o p e r a t o r

of b o u n d e d (BI)-(B4)

harmonic

CI(W)

CI(W)

by

f

d(Lf)

all h a r m o n i c

the

CI(W)

@ Fh0(CI(W)) , where

differentials

II~IiW < ~

The o p e r a t o r associated

~

such that

on

Modified 7A.

L

details

§§i and

consists

con-

for all

= f.

say

u,

A simple

du C F < ( C I ( W ) ) . of e x h a u s t i o n s

Lf = u.

space

E(~,W)

and s a t i s f i e s

the class

CI(W)

for any

is c a l l e d

w i t h the p a r t i t i o n

For f u r t h e r

7.

ula

and thus

by s e t t i n g

{du n }

la : f

n to a f u n c t i o n ,

with

into the

space

the c o n d i t i o n s

Let

we r e f e r

the P

Fh0(CI(W))

(harmonic

consists

of

up to the b o u n d a r y

dh C F I ( c I ( W ) ) e

W

III,

u

independently

maps

on

since

uniformly

on

L L

functions

of d i f f e r e n t i a l s

Moreover,

as w e l l as

(DI)

~) w i t h

on

{u } c o n v e r g e s a l m o s t n seen to be h a r m o n i c on

is u n i q u e l y

Theorem.

that the s e q u e n c e

uniformly

is e a s i l y

computation

÷ 0

Since

and

arbitrary,

form

i.e.

I I v { l ~ - ~ ( U n ' - U n ) I 2 + l~-~(un, - U n ) 1 2 d x d y as

that in

~u / ~ y ) n

a (second)

principal

and is d e n o t e d

the r e a d e r

to A h l f o r s

operator

sometimes and

Sario

by

on

W

( ~ ) L I.

[AS],

Ch.

~R'

in

2.

Green R'

Functions be a r e g i o n

of a f i n i t e

number

in

R

such that the b o u n d a r y

of n o n i n t e r s e e t i n g

closed

analytic

curves

R

103

and

CI(R')

the

is n o n e o m p a c t .

complement

of

monic

measure

monic

function

boundary

or,

Let

let

be the

Then,

G

for any

Proof. with

be any

on

constant

C

vanish

oriented

Lemma. U

negatively regular

for any

G' D R , Since

such

that

~,

u E U.

Let

h(z)

the

with the

*dh

> 0.

to

K

show

Choose

(i0)

be

positive

We n o w take vanishes

u > 0

harmonic any

on

and

(ii) on

(i0)

v

6'

and

since on

*du £ c. and

u(z)

there

Hence

u(a)

= 0

set

~'

c > 0

the

and

G DR,

and

*du £ i. a positive

G'

in

R

such

that

8' = B(G') D R ' . for the

and

to

i

along

on

~',

Then, such

We

region 8'

we h a v e

by use

of

that

v(z)

G DR' at

the

in

C' > 0

G' D R ' .

is n o t h i n g

Since

= h(z)

CI(G' D R ' ) .

on

~ G

[e,

region

problem

< cmin zES'

~'

C'

on

in

u = 0

u(a)h(z)

everywhere S~,

< v(a)

If

G DR'

a

har-

by

u E U.

increases

a constant

function

u E U.

identically

everywhere

v(z)

and

0

har-

bounded

exists

a constant

h

positive

DR'

and

there

regular

Dirichlet to

that

e'

<

if

R'.

every

DR')}

of

has

harmonic

on

a region

point

find

such

for

exists

equal

fixed

we can

to

functions,

for any

conjugate

any

R

£ C

of the

data

in

R'

We d e n o t e

respect

CI(G'

such

solution

c -Imax zES'

for any

u

G'

z E

of

hyperbolic

a nonconstant

G DCI(R,),

u(z)

that,

is c a l l e d

DR'

u = 0

of

~ G, t h e r e

harmonic

inequality

that

SUPzEK

the b o u n d a r y

local

Harnack's

region

such

subset

on

with

of p o s i t i v e

~ G' ~ CI(G')

by

carries

collection

compact

R'

boundary

R'

identically

max{u(z):

denote

ideal

CI(G D R , ) ,

It is e n o u g h DR'

in the

a region

equivalently,

which

DR'

continuous

DR'

Such

some

point

in

to prove.

second

half

G DR',

then

So we a s s u m e

of

(i0)

implies

< cu(z) on

e',

It f o l l o w s

that

~ c / ( S ~ , *dh).

we

see that

(ii)

0 ~ u(a)fe,

Next

we use

holds

*dh ~ c x

the

first

half

of

get

max

By the m a x i m u m by the c o n s t a n t

u(z)

~ c u(a)

principle

u(z)

c2/(~,.

*dh),

~ c2[i[

is m a j o r i z e d as was

*dhj~ -I

everywhere

to be proved.

[]

on

CI(G' D R')

104 7B.

We use

Lemma.

(a)

that

]a' (b)

monic (cf.

the n o t a t i o n s

There

*du

> 0.

If

R'

is a p o s i t i v e

v

and

and

then

on

function

there

CI(R')

exist

such

u

on

bounded

that

dv,

CI(R')

such

positive

har-

dv'

E Fh0(CI(R'))

and

S Proof.

v'

above.

harmonic

is h y p e r b o l i c ,

functions 6D)

given

Let

*dv

a T

GI ~ G 2 ~

and

> 0

.-.

S

*dv'

a T

be a r e g u l a r

< 0.

exhaustion

of

R

with

DR' $ G I

set

8' = B(G ) A R , for each n. Let v be the s o l u t i o n of the n n n p r o b l e m for R' N G w i t h the b o u n d a r y data equal to 0 on n a' and to i on 8'. Then v is h a r m o n i c on CI(R' N G ) and the n n n S e, *dv n , is p o s i t i v e . We set un = flux dn of vn over ~', i.e.

Dirichlet

v /d . T h e n n n is e q u a l to

u i.

n : i,

2,...}

By the

diagonal

convergent

fact

vanishes

the p r e c e d i n g

process

almost

When

v' ~ i

Thus

is u n i f o r m l y

the a s s e r t i o n

nonzero,

is p o s i t i v e ,

n

lemma

bounded

we can

uniformly

on

a'

and

shows

on any

that

compact

CI(R').

Its

flux

the

of

{Un(k):

limit

u

over

a'

sequence

subset

find a s u b s e q u e n c e

on

its

{Un:

CI(R').

k = i,

clearly

2,...}

satisfies

(a).

R'

is h y p e r b o l i c ,

~ i

and

and

]a'

Sa , *dv

*dv'

the

limit

> 0.

< 0.

The

v

of

Setting final

v'

{Vn:

n = i,

: i - v, we

assertion

follows

: lim n÷~

*dv

2,...}

see that now

from

is 0

the

that

[[dv[]°K~'

= lim n÷~

[

:

]]dv n[]~<2AG *dv

<

n

~.

~[ ] e,

n

[]

J~v 7C.

Returning

Theorem.

Let

partition

associated

The

G

negatively

given

any r e a l

there

exists

(12)

for e v e r y

to the n o t a t i o n s

be a r e g u l a r with

G, w h o s e

boundary

numbers

a., ]

I

j.

region

oriented

a nonnegative

Moreover,

in in

Rj

j = i,...,

Bj(P) the

*du

R

parts

of

harmonic

6A, we get and

functions

let

~, w i t h

(mod

u

u

following P = P(G)

are d e n o t e d is d e n o t e d

function

~ a.

the

by

be any

RI,... , R~.

by

B~(P). Then, J ~j~ a. ~ 0 (mod 2~), i ] on R such that

2~)

for all p o s s i b l e

choices

of

105

{a.} J every

can be found fixed

Proof.

a C R

If ~ > i.

tains

a hyperbolic

Lemma

7B t h e r e

that

f61(p)

fine uI

Since

Then,

uj•

the

is h y p e r b o l i c ,

{u(a)}

set of v a l u e s

For on

the c o n d i t i o n .

one

We a s s u m e

CI(R.) ~

of the RI

harmonic

each

to be

for

function

such

f Bj (p)

*ds~~ = 0, we can a p p l y

on

L

*du. ~

conBy

= i.

by

such

a positive We de-

setting with

6B to the

j = 2,

s.

=

i ~ i, j.

open

corresponding

p~,

suppose

CI(R I)

exists

CI(R.)

Theorem

s. and the o p e r a t o r J exist harmonic functions

on

there

~ = 2,..., : 0

function

So t h e r e

uI

~

that

So we

R.'s at least J such a part.

j = 2,..•,

s. on R\ G for each J : u• on CI(R.) and

fB(G)

the

R \ CI(G).

R

is a n o n n e g a t i v e

CI(RI) , since

that

satisfies

component•

function

R \ CI(G),

u ~ 0

*du I = -I.

a function on

a way

is b o u n d e d .

~ = i, t h e n

that

harmonic

in such

set

to

3,...,

W = ~, on

J

R

such

that pj - sj = L((pj - sj)l~G)

on

R \ G.

Since

L((p~J - sj)I~G)

and

(B2)

in

are

seen

to be b o u n d e d

6B and

if n e c e s s a r y , tion

on

s.

since

we m a y and

the

;

sj

are

below

assume

on that

property

*dPk

k : 2,..., Finally

(mod

2~),

clearly

and

they

(B4)

Then

there

tions:

in v i e w

the

on

R.

pj

a constant

Our a s s u m p -

the

following:

-i

if

j = i

i

if

j = k

0

if

j ~ i, k

=

(BI)

functions

by a d d i n g

positive

of

Bj(P)

(12).

Moreover,

0 < b-

=

for any

bkPk(a)

~ 2~ ~ k=2

k=2

Theorem.

there,

in 6B i m p l y

*dSk

R \ G

~.

u(a)

7D.

are

b• be r e a l n u m b e r s w i t h ] u set u : [k[2 bkPk" Then

and

on

therefore,

let

satisfies

as desired.

bounded

nonnegative R

: I

6j(P)

for

are

< 2~

and

b.

~ a.

R

and

is n o n n e g a t i v e

on

a E R

the b o u n d

we h a v e

Pk(a),

[]

Here

we w i l l

construct

Let

g(-,~)

be the

exists

a function

modified

Green

Green

function

g0(.,~)

functions.

for

satisfying

R the

with

pole

following

~ E R. condi-

106

(a)

g0(-,~) - g(.,~)

is a p o s i t i v e

d(g0(.,~) - g(.,~)) e Fh0(R) ; (b) f $ j ( p ) *dg0(.,6) E 0 is a given (e)

partition g0(z,~)

Let

G

P = P(G)

are h y p e r b o l i c

hyperbolic,

we have

tive h a r m o n i c i,...,

Z',

2,...,

6'.

such that

have only viously

~.

to set

while

and

~

such that

~

CI(Rj)_

*du I : -i

% @ G

contains

fBj(P)

*duj

= - I

holds:

(i)

a.

: g(-,6)

and the c o n d i t i o n s

sj,

j = 2,...,

and

: O

on

]

just

stated.

~', by s e t t i n g CI(R i)

~j(P)

with

]

s. : u. ] ] i ~ i, j.

on

j : i,

F h 0 ( R \ G).

Moreover,

for every

adding

us h a r m o n i c

up to the b o u n d a r y is b o u n d e d

each

j = 2,...,

3j

~',

a suitable

and

is b o u n d e d

constant

In v i e w of the p r o p e r t y

of and

*dpj

that

k = 1

.:{1 if k:j

Bk(P) 0

6D

belongs

dpj E Fh0(R).

*ds]

: I

on

dsj E Fh0(R k G).

and

if

pj

R \ G.

pj

otherwise.

We

are pos-

in 6B of the o p e r a t o r

,-i

Bk(P)

on

R \ G, T h e o r e m

if n e c e s s a r y , (B4)

functions

see that

(i4)

~' > i.

CI(R.), ]

d(L((pj - s j ) l ~ G ) )

is b o u n d e d

pj

are ob-

BI(P)

pj - sj = L((pj - sj)l~G)

~'

s. is h a r m o n i c ] L((pj - s j ) I S G )

R.

j =

Then

such that

itive on

for

(a)-(c)

j : 2,...,

may assume,

j =

for

So assume

R

Thus,

: i

> 0

6B gives

to

posi-

bounded

*dg(.,~)

Bj(P)

of T h e o r e m

that

is

and set

an a p p l i c a t i o n

shows

RI, R

duj e F h 0 ( C I ( R j ) ) ,

so that

each

a parti-

that

Since

?B we take

with and

Then the f o l l o w i n g

B(G)

for

in any c o m p a c t

suppose

are not.

By use of Lemma

in v i e w of the facts

CI(R I)

Since

with

~, where

lying

We may

R6,+I,... , R 6

on

fBl(p)

g0(-,~)

satisfied

on

R

j = ~' + i,..., ~; and (iii) a3. = 0 for i a.] = 27. for a moment, we p r o c e e d further. If ~' = !, then we

Then we define = uI

fixed

region

a. = a.(~) ] 3

..., ~',• (ii) A s s u m i n g these

z

j = i,...,

on

B(R);

= {RI,... , R6}.

uj

We take any

j = i,...,

boundary

for

6' ~ i.

functions

(13) for

for

function

z.

be any r e g u l a r

tion of the form ..., R6,

(mod 27)

ideal

is b o u n d e d

set not c o n t a i n i n g Proof.

of the

harmonic

L, we

107

For

j : Z' + I,...,

~

We now define

go

(15)

we set

pj ~ 0

on

R.

by the formula

g0(.,~)

=

[ aj(~)pj j=2

+ g(-,<).

The property (a) is then clear. The property (b) also holds, simple computation using (13) and (14) shows that -2~ IBk(p) When

~

lies in

fore, in view bounded on R Take a regular denote by C n

*dg0(''~)

G, the same

= { 0

pj,

if

k = i

if

k > i.

j = 2,...,

for a

~, can be used and there-

of the properties (i)-(iii) of aj, g0(-,<)- g(-,{) is by the same constant C not larger than 2~2 llpjtl . exhaustion G I ~ G 2 ~ .-. of R with CI(G) ~ G I and the constant C corresponding to G n. Then the property

(c) is satisfied

if

to have g0(-,~) Finally, we with i £ j ~ Z 6j(P). Then fj

- g(.,~) ~ C n. have to prove the properties (i)-(iii). Take any j and let f. be the restriction of g(',~) to the set J > 0 and, by Theorem 6A, Ch. I, g(-,%) : H[fj~Rj] on

CI(Rj).

g0(.,%)

is constructed

Take again a regular

exhaustion

for

< E G n \ Gn_ I

G I ~ G 2 C ...

of

R

so as

with

CI(G) =C GI, and define Gnj = GnAR.3 and Bj(Gn ) : 6(Gn) AR.,3 so that SGnj = 8j(G n) - 8j(P). Further, let gn (resp. $n ) be the solution of the Diriehlet problem for the open set Gnj with the boundary data equal to fj (resp. i) on Bj(P) and to 0 on Bj(Gn). Then gn and its derivatives converge almost uniformly on CI(R.) to g(-,{) and its corresponding

derivatives,

respectively.

Thus

I Bg(P) ~'~dgn: I Bj(P) Cn*dgn - I Bj(G n) Cn*dgn

ffG

ffG

n]

nj

IBj P)

gn*d@n - I

=f Bj(p ) fJ *d~n Letting

n ÷ ~, we see that

Bj(G n )

gn*d~n

108

(16)

I

*dg(.,~)

= I

Bj(P) : H[I;Rj].

where Bj(P). R. 3

Moreover,

f.*d~, ]

6j(P)

Since

0 ~ ~ ~ i, we see that

*d~ < 0

on some c o m p o n e n t

Since

f. > 0, it f o l l o w s ]

is h y p e r b o l i c .

I

of

*d~ ~ 0 8j(P)

from

along

if and o n l y

if

(16) that

< 0

*dg(.,{) Bj(P)

if and o n l y if follows

at once

in T h e o r e m

7E. Lemma.

R. is h y p e r b o l i c . This p r o v e s (i) and (ii). (iii) 3 f r o m the l o c a l e x p r e s s i o n of the G r e e n f u n c t i o n g i v e n

6A, Ch.

We n e e d Let

I.

some f u r t h e r

{V, z}

and its p a r a m e t r i c V

such t h a t

to

~0'

coordinate, I~I

of

{Izl

g.

disk,

< r I}.

in w h i c h we i n d e n t i f y

~0 , %0'

and let

< r I < r 2 < I.

the d i s k

with

property

be a p a r a m e t r i c

I~ol,

within

tersecting

[]

Let

Then

Proof.

*d(g(.,~)

- g ( . , % 0 )) =

differential

We use the n o t a t i o n

arg{(z-

joining ¥

in ~0

not in-

f ]

~ 2 ~

o(y), c

is a r e a l

single-valued

points

be an are

for any 1 - c y c l e

y

o(y)

be d i s t i n c t

c

c

I where

a point

branch

in Ch.

in

Fh0

I, 9.

of the f u n c t i o n

given

Let

in T h e o r e m

v(z)

9C, Ch.

(resp.

s(z))

~)}

(resp.

log{(z - %0)/(z-

I.

be a

~0)/(z- ~)}) in the a n n u l u s r I S Izl < i a n d let e(z) be 2 C - f u n c t i o n on R such that e ~ I on {Izl < r I} and

a real-valued 0

on

R \ {Izl

{Izl £ i}

and

< r2}.

k(z)

We also

= g(z,~)

of the G r e e n

Izl < i, w h e r e

is a h a r m o n i c

*dk = d { a r g ( z for

Izl < i, w h e r e We f i r s t

(17)

h2

function

= logl(z-

Then

~0 ) - a r g ( z -

is a h a r m o n i c

on

d{(l - e)k} • Fe0(R).

follows

function.

{0)/(z- ~)I

that

k = w+ h I

for

Thus, ~)}

+ dh 2

conjugate

of

hI

in

{Iz

< i}.

see that d ( k - ew)

: d{(l - e)k} + d { e ( k - w)} :

and so

w(z)

- g(z,~0).

F r o m the d e f i n i t i o n hI

set

* d ( k - ew) @ FeO ~{.

d{(l - e)k} + d ( e h I) E Feo(R) On the o t h e r hand,

dk + i * d k - d(ev)

E F e.

109

In fact,

on

V

we h a v e dk + i * d k -

which

is closed.

Next,

d(ev) on

: d{(l-

R \ {Izl

dk + i * d k which

is a n a l y t i c Let

a real

y

and

d(ev)

contained

differential

for any c l o s e d

S r 2}

we h a v e

= dk + i*dk,

so is closed.

be a 1 - c y c l e

harmonic

e)v] + d(h I + ih2) ,

in

o = o(y)

differential

~ E F

R \ {Izl E Fh0

< rl].

such

(Theorem

9B,

Then

that Ch.

there

is

fy ~ = (m,o*)

I).

Since

dk-

C

d(ev)

is exact

f

*dk

on

R \ {Iz[

= -i f

¥

< rl},

dk+ i*dk-

we have

d(ev)

= -i(dk+ = A,

Now,

from

(17)

follows d(k-

This

i*dk-

differential

say.

that ew) + i * d ( k -

is e q u a l

d k - d(ew) + i ( * d k -

ew) @ Fe0 + F e0 * "

to

e*dw-

w *de)

: d k - d(ew) + i ( * d k = dk + i * d k -

Since

Fho" ±

(Fe0 + Fe0*)

by

(4)

in Ch.

( d ( k - ew) + i * d ( k and

d(ev),o*)

y

I,

eds - w ' d e )

d(ev) + i(sde - w ' d e ) .

8E, we h a v e

ew),o*)

= 0

thus A : -i(i(w*de

- sde, o*)

: ( w ' d e - sde, o*)

rl< I z l< r 2 sded wde°*

d(ew)(]*

+ ffrl < zl
- ff

edw(~* rl< ]z
rl
rl<]Zl
110

:_f

w~o_ff I z I:r I

To c o m p u t e

edwo~-f I~ I
rl<

this,

we add p u r e l y

s~-ff

I z I: imaginary

rI

terms

rl<

eds~. I~ I
as follows:

_if z I=r I s~ iffrl< Iz I
is this:

-flzlvoo_ ff :r I

rl<

=

i

Ii z

:

i

Ii z

I z l
e~v~ + if

v ( a + ia*)

a

+ i IIrl
vda --r 1

f~;

da

z

= -2~r

~0

where

%0

f

z

(a+icr*),

e

is an a n a l y t i c

function

I

on

V

with

*dk = -2~ I y

7F.

Concerning

Theorem.

Under

fY where %0

and

Proof.

the f u n c t i o n

We set

pj

the f o r m u l a

(13).

Thus

with

[ j=2

we have

fc

k 0 = go (. , {0' ) - g 0 ( - , ~

are h a r m o n i c

Hence,

[]

as in Lemma

dependent

k0 = k + where

go

the same c o n d i t i o n

is a c o n s t a n t

o.

da = a + io*.

c

*d(g0(.,~)-g0(.,%0))=-2~

K. 3,Y {~.

rl< I z I < r 2 edv(a + ia ~':)

--r 1

I z l=r I

= -2~r

edvo

z I =r 1

of

7E, we have

{o(y) - j:2 [ K.],yO(Bj(P))}, j

0)"

and

y

Then by

but not of points

(15) we have

(aj(~') - aj(~0))Pj 0

dpj e Flhe(R)

by Lemma

the f o l l o w i n g

7E

and

a.(%)'s]

are g i v e n

by

111

aj(~6)-aj(~0 ) = - I

*dk : 2z I 8j(P)

It follows

that,

f

for any cycle

f

*dk 0 : y

y

*dk +

=-2~

in

[ j=2

y

R \ {Izl

< rl},

(aj(~6) - aj(~0))

{(~(y)c

8.

o(Bj(P)). e

~

(

f

*dpj y

*dpj)~(gj(P))}.

j=2

[]

y

Proof of Sufficiency 8A.

We are now in a position to prove the implication

in Theorem

2B.

We do this by contradiction.

is a hyperbol~e

Riemann

for some

Namely we suppose that

surface but not a PWS and therefore I~ B(~,a)d~

(and hence all)

(c) ~ (a) R

that

:

a E R.

We set m(~,a)

= sup{If(a)I:

for any line bundle

~ m(a)

In view of Theorem show that

m($,a)

over

R

llflll,a ~ i}

and then

= inf{m(~,a):

5A we have : 0

f e ~(R,~),

m(a)

~ e HI(R;~)}.

= exp(-f~ B(~,a)de)

for some line bundle

$.

: 0.

We should

For this end we still

have to prove two more facts. 8B. R£}

Let

GO

be a canonical

the set of all connected

j ~ £, the boundary respect to ~£-i

GO.

Let

CI(V) ~ G O ..., b N

V

in

R

of

R \ C I ( G 0 ) , and

contour of

basis

cycles

be any parametric

and all

Yn'S

N' I k=l

for

YI'''''

~i'''''

V.

Then,

~N'

~dg0(-,~ k) ~ b n Yn

CI(G 0)

{RI,..., ~j,

i

in

consists

of

~I'''''

~N"

disk included in

lie off

there exist points

(Ch. I, IC),

Rj, which is oriented positively with

Then a homology

and certain nondividing

Lemma.

region

components

GO

such that

for any real numbers V

such that

(mod 2~)

b I,

112

for

n = i,...,

Proof.

Let

N, a n d

Tj

R

N.

~0

number

be the

center

= o(6j(P)),

representing

write

The

Yn

of

V

a n = UndX + vndY

6j(P), and

~n IN Cn nU + ~j_£_2 c'u'3J : 0

~

e

on

V.

This

j = 2, . . . , £, are

+

u

~

n n

means

is

constant on

are

linearly

was

to

be

the

c:u'

V,

harmonic

] J

in

~0

n I

bn.

n = i, ... ,

V.

on

P = P(G0).

We

Then,

n =

Un,

independent

on

V.

In

then

c

3-{

numbers

differentials

where

:

j:2

of the

a n = a(Yn) ,

linearly

on

~0

n i

Z

eTj

a

+

n n

~

e."~

j:2

3 j

that

ca

tically

let

T3. = uj'dx + vj'dy

fact,

n:l

and

£, be

and

o :

independent

respectively,

i,... , N, if

is

j = 2,..., and

uj, '

N'

on

V

R.

Since

and

independent,

shown.

It

Tj

with

n = i,...,

we h a v e

follows

j=2

In N 1 e n a n + ~j_£ -2 c'T. J 3

therefore an,

+

n n

c

N

= c[ = 0 3

n

and

vanishes

iden-

j = 2,...,

for

all

n

V.

So we

and

£

j, as

that

#

~-

u n : u n - j[ K. u' :2 3 'Yn 3 for

n = i,...,

induction,

N

are

linearly

independent

%1' .... ' ~N' C V

points

such .

on

.

We

set

defined IXnl Since

the

eontains

for

n = i .....

Jacobian an

open

includes

an N - c u b e

have,

Theorem

by

of set

consider

~

0

N}

with

at

0

in

19N.

with

sides

the

al'''''

of

in

19N

mapping

k 1 ~

given

by

a

)

U = {(Xl,...,XN)"

r = min{l-]~'I: n = i , . . . , N}. n ¢(U) is e q u a l to A ( 6 { , . . ., ~N') ¢ 0,

For of

a large

length

7F,

), ....

k=l~y I

and

= (k:iJ~

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

< r

0.

K. T. 3 'Yn 3

~(Xl'''''XN)

by

°

,...,~)--

a# an [ _£ n = - j-2

find,

.

•

A(~{

can

that

Nf

k i 7N

integer

27.

On the

M > 0, other

(2~M)~(U) hand,

we

113

M(

:

*dg0(-,~ ~) ..... k=l Y1

In view of the d e f i n i t i o n

of

with

sides

when

any

(bl,...,b N) E ]RN

of length

2~

M ~ k=l for

n = i,...,

there

If M

f • ~{~(R,~)

Then

in Ch.

Since

[F0(G0)] and

Hence,

for

such that

~n'

to

V

+x

n

'

n

i,...

=

and

~

e of

YI'''''

as in the p r e c e d i n g ,

there

over

II,

(mod 27) n

exists

GO

satisfying

e, of the f u n d a m e n t a l

"''' e~-i

there

2B, the bundle F0(G 0)

is i d e n t i c a l l y [0(G0), YN'

a constant exist

ILflL ~ I

group 0

and ~ to

elements

such

an e x t e n s i o n If(a)l

initial i

is d e t e r m i n e d

for these

m > 0

is given

with

equal

we have

~

to

~ m. by a character,

point

0, where

on the c o m m u t a t o r by its values

generate

at

~i'

the h o m o l o g y

HI(G0,%) = F 0 ( G 0 ) / [ F 0 ( [ 0 ) ] . Here, the paths from 0 by p e r f o r m i n g obvious d e f o r m a t i o n s .

exp(-2~iaj)

for

..., N, where the region there

a E GO.

for any line b u n d l e

subgroup

belongs

~k ) - b

same n o t a t i o n s

As n o t e d

an N-cube

U.

(mod 27)

are the points

'

the

t 0 = CI(G0).

group issue

*dg0(" Yn

8C.

Proof.

over

(Xl,...,x N) • U

- bn

~k ! s

[]

and an

say

~MN

then

N.

Suppose

a point

x k)

n = i,...,

Theorem. R

exists

side covers

ranges

r

I

that

(Xl,...,x N)

~i'''''

times,

k=l

Using

- 2~M%(Xl,...,XN).

M, the r i g h t - h a n d

*dg0(.,~+

MN

for

[ *dg0(-,~)) k:l YN

Yn

N.

N, each r e p e a t e d

Nf

j = i,...,

aj

and

bn

GO, we should

exists

~

and

0(yn)

are s u p p o s e d to We set 8(e.) = ] : e x p ( - 2 w i b n ) z for n = i,

are real numbers. have

[j=l a. - 0 ] harmonic function

a nonnegative I

*du ~

(mod

a.

Since (mod u

[j=l ~j

2w). on

bounds

By T h e o r e m

R

7C

such that

2~)

] for every

j, where

u(a)

aj.

Next we use Lemma

~MN

from a fixed

k=l

compact

Yn

~ C < ~,

C

8B to choose set

being

a finite

K

lying

*dg0(.,~ k)

~ bn -

in

f

Yn

a constant number G O \ {a}

*du

independent

of points

(mod

such that

27)

of

~i'''''

114

for

n = i,...,

where

~

N.

denotes

is h o l o m o r p h i c

We set

v = u+ [~i

the h a r m o n i c

on

R.

g0(''~k )

conjugate

of

and

v.

Since the d e f i n i t i o n

f = exp(-v-

We see first

of

go

(Theorem

i~),

that

f

7D) shows

that

I for any

*dg0(.,%)

and any

6 • GO

-

~ O

(mod 2~)

] j

= 1,...,

rI J

*dv

~,

~

we

see

that

(mod 2~).

-a.

]

We also h a v e MN -

*dv

:

-

*du

Yn Thus, ~

f

-

[

Yn

is a s e c t i o n

of

~

,~k )

-b n

-

Yn

of a line b u n d l e

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

If(a)l

*dg0(-

k=l

to

GO .

= exp(-v(a) -

~

over

R, so t h a t the b u n d l e

Moreover,

MN ~ g0(a,~k)) k=l

> m > 0,

where m : e x p ( - C - MN sup g 0 ( a , ~ ) ) %EK is i n d e p e n d e n t

8D.

of the c h o i c e

of

~.

[]

For each line b u n d l e

T

striction

of

GO .

Theorem.

Suppose

T

to the r e g i o n m(a)

= 0.

over

R

we d e n o t e

by

p0(~)

the re-

Then

Then,

for any line b u n d l e

T0

over

[0'

we h a v e inf{m(~,a): Proof.

Since

bundle

T'

the p r e c e d i n g a section where

m

m(a)

over

= 0, we see that

R

with

''-I)

is a c o n s t a n t

belongs

l(f0h)(a)l

m(T',a)

: T0}

for any

< ~.

to

~(R,T~

~ m(~',a)

such that

independent If

of

< E.

Since

e > 0

say

llf011~ ~ i ~.

h • ~(R,6)

''-I) : ~ / ( R , < ' )

= 0.

We set

t h e o r e m we f i n d an e x t e n s i o n ,

f0 e ~ ( R , ~

P0 (T) = P0 ( T )'P 0 (6'') = ~0" foh

p0(~)

and

Set with

there

exists

a line

T~ : P 0 ( T ' ) - I T 0 • T", of and

T~

to

R

By and

If0(a) I ~ m > 0,

~ = ~'T".

Then,

llhl11,a =< i, t h e n

llf0hlll,a =< I , so that

If0(a) I £ m, we have

lh(a)l ~ m-ls,

115

which means that sired result. 8E. that

~ m-it.

= 0

for some

s

is arbitrary,

open subset

V

(and hence all) of

We then choose a regular nonical regions

(Ch.

(resp.

Pjk

with

bundles

over

R

R

a @ R.

and a countable

2B, we suppose

We take a relatively

dense

subset

S

of

V.

exhaustion

j ~ k) the restriction Gk).

~n

over

We fix a line bundle

~j

mn = inf{mn(~n'b):

to

Gn

(resp.

Gj) of line

Finally we set

: sup{If(a)l:

for any line bundle

n ~ j

we get the de-

{G : n = i, 2,...} of R by can I, IC) such that CI(V) ~ G I. We denote by On

(resp.

mn(~n,a)

for

As

To finish the proof of (c) ~ (a) in Theorem

m(a)

compact

m(6,a)

[]

f • ~ ( G n , ~n ), llflll,a =< i]

Gn

and any

over

Gj

a • G n-

and take any

~n e HI(Gn;~),

b E Gj.

We set

Pjn(~n ) : ~j}

and = inf{m(~,b):

~ • HI(R;T),

pj(~)

= ~j}.

We claim (18)

lim m

n

: m.

This can be shown in the same way as in the proof of (8) (see 5A). fact,

if

belongs

~n+l E HI(Gn+I;T) to

HI(G

;~)

with

Pj,n+l(~n+l ) = ~j, then

In

Pn,n+l(~n+l )

and therefore

n

mn(Pn,n+l(~n+l)'b)

~ mn+l(~n+l'b)

~ mn+l"

Since every line bundle over see that

~ mn+ 1 ~ m

G. is the restriction ] and hence limn÷~ {n =

In order to show the reverse We take

~ E HI(R;T)

before that

<

in the form

({ ~ }

V' n G n

with

pj(~)

inequality, = 6j

and

we may assume that m(~,b)

is defined by means of a countable , {V~})

such that all

are simply connected.

that there exists an

V'

The definition

fn E ~ ( G n , P n ( ~ ) )

of one over

< ~.

R, we m < ~.

We assume as

open covering

{V~}

as well as all nonempty of

mn(Pn(~),b)

with the properties

shows llfnlIl,b

i f

and Ifn(b) I ~ m n ( P n ( ~ ) , b ) - i/n. Let {fn } be a representative of with respect to the covering {V' A G } of G and the representn a n~ n ative ({~eB}, {V~ n G n } ) of pn(~). Let {V } be a refinement of the covering

{V'}

such that

CI(V

)

is a compact

subset of

V'

for

116

every

~.

Since

IffnIll,b ~ I, t h e r e e x i s t s on

Gn

on the set

CI(V

n > i =

subsequence f o r m l y on formly

)

by a c o n s t a n t

are u n i f o r m l y {nk: V

on each

V .

{f }

Since

defines

Ill ~ h,

Finally,

~

C

independent

on

V

such that

{fnk

and such that

= lim f k÷~ nke a section

and

n. ~.

}k

{hnk} k

Thus,

f n~

So t h e r e

is a

is c o n v e r g e n t is c o n v e r g e n t

uni-

uni-

h : lim h k÷~ nk

of the b u n d l e

f • ~(R,~)

and

~, w h i c h we d e n o t e

llflil,b ~ h(b)

: k÷~lim Ifnk(b)l

by

= l i m k + ~ hnk(b)

f.

~ i.

=> limk÷~sup{mnk (pnk ~),a) - nk -I}

> lim sup m =

~

of

for each

we have

If(b)l

As

hn,

We set f

Then

bounded

k = i, 2,...}

for e a c h

function,

h n (b) =< i. Take any ~. If Gn i n e q u a l i t y shows that hn is b o u n d e d

on G n with Ifnl ~ hn includes CI(V ), then the H a r n a c k with

a harmonic

and

is a r b i t r a r y , By T h e o r e m

we have

8D we h a v e

k÷~

= lim m nk

n+~

. n

m > lim m and t h e r e f o r e (18) holds. = n+~ n b C m = 0 for any ~j e H I (Gj ;~) and any

Gj, and thus (19)

lim[inf{mn({n'b):

~n • H I ( G n ;T)'

P j n ( ~ n ) = ~}]j

= 0.

n+~

Let us take a s e q u e n c e appears

infinitely

n I < n 2 < • .. that

{bj:

often.

j = i, 2,...}

By use of

and line b u n d l e s

in w h i c h

each p o i n t

in

S

(19) we find a s e q u e n c e


over

{n.} 3 i = i, 2,...

for

Gn. l

with such

-i Pni'ni+l Then there

exists

(~ni+l)

a unique

= ~n. l

and

m

line b u n d l e

~

ni

(~n

l

,bi) < i

over

R

satisfying

pn (~) = i

~n.

for all

i, so t h a t we have

m(~,bi)

i

any

i

b E S.

T h e n our a s s u m p t i o n

for i n f i n i t e l y that e v e r y f ~ 0

~ mno(~n

on

many

i.

f E ~(R,~) R.

So

m(~,b)

vanishes

H e n c e we h a v e

on

{bj} = 0

e i -I.

Take

l

says t h a t we h a v e for all

at all p o i n t s

~I(R,~)

~bi)

b E S. in

S

b l.

= b

This m e a n s

and t h e r e f o r e

= {0}, as was to be proved•

[]

117

9.

A Few D i r e c t 9A.

Consequences

First

Theorem.

Let

we r e s t a t e R

be any

Theorem

PWS.

5A as

Then,

inf[sup{If(a)I:

follows:

for e v e r y

f e ~P(R,~),

S

= exp(-

B(~,a)d~)

p

with

IIfll p,a

= i}]

i ~ p ~ ~,

> 0,

0

where the

~

ranges

common

over

value

all

line

is e q u a l

The

last

statement

for any

Corollary. common

point

If

zeros

R

line

Theorem.

Let

(a)

For each

pair

with

(b)

For

a simple

zero

Proof.

comes

is a PWS,

9B.

f e H~(R)

{g(a,w):

f(a)

every at

from

Theorem

R

tion.

Take

because of

let

then

the

bundle

an

R

surface

R.

statement

h(z)exp[-g(a,z) H~(R).

Since

this

the in

theorem

is

following

~(R,~)

have

no

Then

points

the

following

a, b E R

hold:

there

exists

an

a E R

there

exists

an

f E H~(R)

which

has

bounded

be the

line

bundle

-I)

such

- i~(a,z)]

is e a s i l y

and

seen,

the p r o p e r t y :

over

R

extend

choose

by this

This

function,

and

f(b)

-I)

with

~ 0.

func-

is p o s s i b l e

a function

it a n a l y t i c a l l y

resulting = 0

determined

lh(b) l ~ 0.

We t h e n

the

f(a)

function

ig(a,z))

that

Corollary.

analytic

element

to the w h o l e say

f, b e l o n g s

This

proves

~ 0.

Then,

the

(a).

To p r o v e

in

then

~ f(b).

point

h E ~(R,~

As

and has

IC.

functions

be a PWS.

of d i s t i n c t

preceding

h(z)exp[-g(a,z)

H~(R)

is r e g u l a r ,

~ E HI(R;~).

the m u l t i p l i c a t i v e

~

of the

R

a.

Consider

and

If

in p a r t i c u l a r

z ~ exp[-g(a,z)on

R.

w e Z(a;R)}].

a ~ R, we h a v e

for any

over

to

exp[-[

valid

bundles

(b),

we

choose

- ig(a,z)]

Since

h(a)

h E ~(R,~

gives

~ 0,

f

rise has

h(a)

to a s i n g l e - v a l u e d a simple

zero

at the

function point

f a.

[]

118

NOTES For the fundamental brilliant

Parreau's

[70],

Most

two

in 1976, paper

although

Subsection

definitions

except

some

those

was

was

chapter

shown

out

coined

in S u b s e c t i o n s

from Hasumi

are

have

[18].

contains

taken

[52].

when

from

[18].

[18]

[19].

his

this

Essential

was

As a written.

to me the r e l e v a n c e

with my work

then

which

by H a s u m i

results

pointed

small m o d i f i c a t i o n s

3 are a d a p t e d

[70],

is in P a r r e a u

Parreau's

in c o n n e c t i o n type"

Widom

in this

definition

Z. K u r a m o e h i

[52]

results

follow

results

I did not k n o w

of P a r r e a u - W i d o m All

of PWS we

Parreau's

of t h e s e

of fact,

Afterwards,

faces

theorem.

paper.

equivalence matter

definition

The n a m e

of "sur-

by M. H a y a s h i . 3 and

6 are due

b e e n made.

to W i d o m

The r e s u l t s

in

CHAPTER

Every bounded almost means

harmonic

of the

Poisson

function

integral.

has a striking type.

discovered

Parreau

by M.

paper of Parreau time.

of Parreau the case

that

problem asking

of surfaces

positive

harmonic

issuing of Green

between

show that

harmo n i c

faces

i.

limits

V), results

Green

Green

in

sections

boundary. lines

on

are devoted

We show in 52 that

every

almost

line

every Green problem

data m e a s u r a b l e

on the

lines)

of functions

space

problem

and the fine

including

to

with re-

in §3 the B r e l o t - C h o q u e t

(along

for a class

objective type,

Definition Let

of Green R

g(a,z).

= r(a;z)

PROBLEM

and

8(z)

hyperbolic

is to i n v e s t i g a t e

3B, Ch.

limits

bounded

Green

assumption

Riemann

sur-

lines

on sur-

is e v i d e n t l y

V.

ON THE SPACE

OF GREEN

LINES

Lines

be a r e g u l a r Let

only r e g u l a r

this r e g u l a r i t y

in view of T h e o r e m

THE D I R I C H L E T

function

for any bounded We solve

for the

answer,

and the M a r t i n

and that the D i r i e h l e t

we c o n s i d e r

of P a r r e a u - W i d o m

IA.

r(z)

chapter

As our u l t i m a t e

legitimate

§i.

along

in that

appeared

functions.

In this faces.

same

type.

were

Brelot-Choquet

The other

has a limit

surfaces

direction

2B, Ch.

surfaces.

is r e s o l u t i v e

the radial

lines

by

as the Fatou

affirmative

to the

along

function

It is indeed type

is given about

of P a r r e a u - W i d o m

function

the

in this

(Theorem

type,

Green

known

discussion

spect to the Green measure.

are e s s e n t i a l l y

theorem

limits

limit

to all Riemann

in 1958.

of P a r r e a u - W i d o m

from any fixed point lines

result,

us to give a c o m p l e t e l y

Riemann

the case

famous

of P a r r e a u - W i d o m

with Widom's

relations

hyperbolic

from its radial

[52] already

In §i an i n t r o d u c t o r y general

in the unit disk has

generalization

surfaces

Combined

LINES

The first main results

then permit

of surfaces

This

direct

of P a r r e a u - W i d o m

and

GREEN

all radii and can be r e s t o r e d

theorem,

first

VI.

a E R

hyperbolic be held

= e(a;z)

Riemann

surface

fixed and define

by equations

with

Green

two functions

120

dr(z)/r(z) respectively. r(z) the

In the

= exp(-g(a,z)) point

which

a

large

the

following for the

~ 0

and

8(z)

set of G r e e n

~ > 0

conformally

the

de(z)

we d e n o t e

by

any m a x i m a l

issuing

region

with

the

: -*dg(a,z), r(z)

equation.

arc

a.

CI(R(a,a))

special

line

from

(= {(a),

for

from a

on

short)

For a s u f f i c i e n t l y

= {z e R: g(a,z)

unit

solution

issuing

issuing

$(R,a)

from

closed

the

A Green open

is c o n s t a n t .

lines

closed

isomorphic

and

first

is by d e f i n i t i o n

de(z)

denotes

= -dg(a,z)

disk

under

~ ~}

is

the m a p

z ÷ e~r(z)exp(i0(z)). Thus

each

= ~8 Using

Green

in such the

R

a way

that

and

the

set of G r e e n in

~

longs

to

Green lines

~0(a)

e

{(a)

in

So

on

only

called

regular

zero.

We d e n o t e

by

{(a).

Since

is r e g u l a r ,

R

infimum, L(R,a)

The u n i o n Green

$'(a)). the

The

region

function

E L(a)

lB.

by

point

a

~'(a)

the

S(f;~(a))

all a

set.

the

point

of be-

of the

A Green ~

unit

which

~ E {(a)

over

line

~

is e q u a l

We

s

(B2)

lim

set

is b o u n d e d

Green

the p o i n t

problem

real-valued

lines

in

is d e n o t e d

connected

as a g l o b a l

S(f;{(a)) pointwise

$(a)

is c a l l e d

z

on

~

on the

function

by

{'(R,a)

and

(resp.

in

on it.

For

for w h i c h

set f

~(a)

on the

functions

(or

and we can use every

g(a,z)

~ f(~)

let

for

Suppose G[f;{(a)]

supremum)

on

R

dma-a.e, that

both

(resp.

on

s

L(a), on

R

L(a).

[(f;{(a)) G[f~{(a)])

of f u n c t i o n s

in

=

of G r e e n set

below;

= -S(-f;~(a)).

are n o n v o i d infimum

to

lines

region

coordinate

set of s u p e r h a r m o n i c

inf ÷ 0 s ( z ( ~ ; ~ ) )

S(f;~(a))

is

Green

that (BI)

as

be the

CI(~)

set of r e g u l a r

and

is a s i m p l y

Dirichlet

the

= ~0(a)

a line

g(a,z)

the

with

= d8/2~,

closure

in a c r i t i c a l

2w)

= ~ N ~R(~,a).

{(a)

~0(R,a) the

see that

z, of

and at

denote

an e x t e n d e d

{z)

[0,

= L(a) UE0(a).

centered

z(~;~)

We d e f i n e Given

in

8 6

we h a v e

z ÷ r ( z ) e i0(z)

let

lines. denote

of the

star r e g i o n

to

= L(a)

with

identify

is a c o u n t a b l e

~(a)

the

that

if it ends

E0(a)

if the

Let

such

It is easy

2~)

with

d m a ( ~ 0) = dma(8)

{(a).

{(a)

set.

if and

g(a,z).

a measure

measure ~

is p a r a m e t r i z e d

8(z) ~ 8 (mod i0 ÷ £8' we can

introduce

is a c o m p a c t

function

in

correspondence

circumference is e a l l e d

line

and be the

S(f;@(a))

we such

121

(resp. and

S(f;{(a))).

s" E S ( f ; @ ( a ) ) , so that

lies,

G[f;@(a)] if

function

G[f;@(a)]

problem

G[f;@(a)]

5A,

Ch.

Dirichlet

f

on

R

are

nonvoid

An

only

is c a l l e d of G r e e n

with

the

following,

real-valued

which

if t h e r e

function

exist

s

on

function

a harmonic R

such

solution

lines.

and

common of the

It is c l e a r

that

is an a n a l o g u e

of T h e o r e m

for any

E > 0.

Proof.

The

If this

proof

sequence

{s~:

with

n : i,

function

u

to that

in

then

a*

superharmonic

super-

u : G[f;~(a)].

of T h e o r e m

5A,

~(f;{(a))

such

Ch. then

IIl.

In fact,

there

exists

a

that

< ~,

n

is any p r e s c r i b e d

is a p o s i t i v e

is r e s o l u t i v e

and a p o s i t i v e

u = G[f;@(a)],

(s'(a*)-u(a*))

n=l

L(a)

u - Es E S ( f ; @ ( a ) )

case,

solution

2,...}

on

and

is the

is s i m i l a r

is r e s o l u t i v e

f

that

u + ss e S ( f ; { ( a ) )

m > i

the

The

HP'(R).

extended

(i)

where

resolutive.

and

Problem

We b e g i n

harmonic

f

fami-

S(f;@(a))

is c a l l e d

{(a)

s' e S ( f ; { ( a ) ) are P e r r o n

are h a r m o n i c

and

and

space

for any g(f;~(a))

~[f;@(a)]

G[f;~(a)]

to

and

IIl.

Theorem.

if

by

s' ~ s"

S(f;{(a)) then

for the

belongs

2A.

if and

and If

= G[f;@(a)],

is d e n o t e d

Dirichlet

The

we h a v e ~(f;{(a))

~ G[f;~(a)].

G[f;{(a)]

2.

Since both

point

in

function

R. on

So the R.

Then

sum

s'

for any

= ~n=l (s~ - u) fixed

integer

we h a v e

=

u(z) + m - l s ' ( z )

~ u(z)+

m -I

(s n'- u)(z) n=l

=

for e v e r y ilarly, that

z @ R.

there

u - m - l s ''

superharmonic Suppose, positive

It f o l l o w s

exists

belongs

to

S(f;@(a))

that

function

s n'(z) that

u + m-ls ' E S(f;@(a)).

superharmonic

s = s' + s"

conversely,

superharmonic

-i ~ n=l

readily

a positive

function

m

for any satisfies

we h a v e s

function integer the

a harmonic

satisfying

s"

on

m $ I.

conditions function

(I).

Then

in u

SimR

such

So the (i). and

a

122

u : lim c÷O wherever

s

conclude

that

Using usual

(u + es) ~ [ [ f ; { ( a ) ]

is finite. f

Corollary.

s

is r e s o l u t i v e

this,

Dirichlet

Since

we can e a s i l y

~ G[f;{(a)]

~ lim s÷O

is i n f i n i t e

and

( u - ss) : u,

o n l y on a p o l a r

u = G[f;{(a)].

show the f o l l o w i n g

set, we

[]

as in the case of the

problem.

(a)

If

fl'

f2

are r e s o l u t i v e

functions

on

L(a)

and

el'

~2 are r e a l n u m b e r s , then ~ifl + a2f2, m a x { f l , f 2} and m i n { f l , f 2} are r e s o l u t i v e and s a t i s f y the f o l l o w i n g , in w h i c h we w r i t e G[.] in place

of

G[-;{(a)]: G [ ~ i f I + e2f2 ] : ~ i G [ f l ] + e 2 G [ f 2 ],

G[max{fl,f2}] (b)

Let

functions

= G[f I] V G [ f 2 ] ,

{fn:

on

with

Let

limit

sequence

G[f]

is the case,

2B.

n = i, 2,...]

~(a)

if a n d o n l y if the

G[min{fl,f2}]

= lim

f

{G[f

G[f

n÷~

n

n ].

= G[f I] A G[f2].

be a m o n o t o n e

sequence

of r e s o l u t i v e

= limn÷~

Then

is

fn"

]: n = I, 2,...}

f

: n = i, 2,...} be a s e q u e n c e of p o s i t i v e n u m b e r s n d e c r e a s i n g to zero, such that DR(en,a) N Z ( a ; R ) = ~. We set

strictly

n = I, 2, . . . .

Since the

surface

R

is r e g u l a r ,

family

{R : n = i, 2,...} forms a r e g u l a r e x h a u s t i o n n sense of Ch. I, IA. For e v e r y n the G r e e n f u n c t i o n

with pole

a

is e q u a l to

function

length measure R

n

on

DR

u(a)

=

g(a,z)which

n

and if

w i t h the b o u n d a r y

(2)

u

1 ; 2~ ~R

So if

is s u m m a b l e

is the

function

an.

solution

f

=

Lemma.

Let

s

the g r e a t e s t [(~

)(~) n

T h e n we h a v e

lr2

in the

gn(a,z)

is an e x t e n d e d

with respect

for

Rn

real-

to the a r c problem

for

1 I DR = -~-~

f(z) *dg(a,z) n

f(z(go;an))dO

be a s u p e r h a r m o n i c harmonic

R

the

f, t h e n

f(z) * d g n ( a , z )

-)--~-JO

of

of the D i r i c h l e t

n

u

If this

{a

R n = R(~n,a) ,

valued

resolutive

converges.

minorant

=

I ~(a)

function of

= lim inf s ( z ( £ ; ~ n ) ) , n÷~

s

on ~(Z)

f(z(g;an))dma(~)"

which R.

is b o u n d e d

b e l o w and

We set

= lim inf s ( z ( £ ; e ) ) . e÷O

123

u(a) => I]L(a) S(c~n )(£)dm a (£) => I]L(a) -s(£)dm a (£). Proof.

Let

un

be the greatest

u n = H[s;R n]. Then, of (2) we have

: lim Un(a) n÷~

u(a)

=>

2C.

harmonic

minorant

u n => Un+ I => --- => u

I]L( a )

on

of

Rn

s

and

on

Rn, i.e.

Un ÷ u.

By use

: lim IL s(z(£;~n))dma(~) n+ ~ (a)

S(en)(Z)dm

We apply the preceding

a

(£) => I]L( a )

result

s(£)dma(Z). --

to the Dirichlet

~

problem

defined

in IB. Theorem. that

Let

f(~)

~(f;{(a))

G[f;{(a)](a) ~

and

Proof.

If

I

f(£)dma(~)

s

f(~)dm

L(a)

such

(£) > G[f;~(a)](a), a

the lower and the upper

and if

2B implies

=

--

integrals

with respect

> I]L( a )

u

is the greatest

harmonic

minorant

that

s(Z)dm a --

(£) > T]L ( a )

f(£)dm

a

(£).

is arbitrary, G[f;¢(a)](a)

The remaining tegral

~ r

on

dma, respectively.

s 6 ~(f;{(a))

u(a)

function

Then we have

//u(a)

denote

s, then Lemma

Since

real-valued

are nonvoid.

JL(a)

to the Green measure

of

S(f;{(a))

> r =

where

be an extended

and

inequality

is always

inequalities.

7

> | f(~)dm (£). = JL (a) a

can be obtained

similarly.

larger than the lower integral,

As the upper in-

we @et the desired

[]

Corollary.

If an extended

resolutive,

then

f

real-valued

function

f(~)

is dm a -sunlmable and G[f;{(a)](a)

: [

~m (a)

f(Z)dm

a

(~).

on

]L(a)

is

124

§2.

THE S P A C E OF G R E E N

In w h a t Riemann

3.

follows

surface

The G r e e n 3A.

gion

ON A S U R F A C E

OF P A R R E A U - W I D O M

in this c h a p t e r we a s s u m e

of P a r r e a u - W i d o m

that

R

TYPE

is a r e g u l a r

type.

Star R e g i o n s

We c h o o s e

notations

LINES

given

a point

in IA.

D = D(a)

a • R, w h i c h

We map the G r e e n

in the open unit d i s k

is h e l d fixed, star r e g i o n

~

and use the

¢'(a)

onto a re-

in the c o m p l e x w - p l a n e

con-

W

formally

and u n i v a l e n t l y

by m e a n s w : ~(z)

where

r(z)

= r(a;z)

9 • [0, 27) ~(a) UE0(a

with )

and

e(z)

of the g l o b a l

= Q(a;z). Since

Se,

8 E ~0' d e n o t e s

the end p o i n t

the

slit

of the G r e e n

e @ ~0'

s h o u l d be a c r i t i c a l

Theorem

iC , Ch.

V, shows,

In v i e w of the fact of the slits

theorem

there

exists

in the c o m p l e x

be the set of all we h a v e

¢(a)

~ r ~ i},

z(e)

:

point

of

[{g(a,zj):

he

g(a,z),

e • ~0'

onto

$8,

D

~D

this

is finite. ~

of

D

Each

z(e),

z(e) E Z(a;R). < =

As

and therefore

< ~. that the t o t a l

By the R i e m a n n m a p p i n g the o p e n u n i t d i s k

and u n i v a l e n t l y .

is the u n i o n

The a b o v e

a.

implies

which maps

conformally

0 • ~0"

than

i.e.

zj • Z ( a ; R ) }

Se,

a function

r(z(8))

other

z]. E Z(a;R)}

= exp(-g(a,z)),

boundary

slits

e E ~0 },

r(z)

E-plane

the t o p o l o g i c a l and all the

~0

is r e g u l a r ,

{reie:

line

I{1 - e x p ( - g ( a , z j ) ) :

length

Let

R

and so D = D w \ U{S8:

where

function

= r ( z ) e i0(z)

%~ ~ E 0 ( a ) .

(3)

being

coordinate

of the c i r c l e

observation

~

Clearly, ~D

shows that

w ~D

is r e c t i f i a b l e .

3B0 ~.

We are g o i n g

First we prove

Theorem. ci(D

~

to m e n t i o n

some u s e f u l p r o p e r t i e s

of the f u n c t i o n

the f o l l o w i n g

can be e x t e n d e d

continuously

to the c l o s e d u n i t d i s k

).

Proof.

$

set of

e E [-7,

By T h e o r e m

is a b o u n d e d 7)

2A, Ch.

hoiomorphic

function

such that the r a d i a l IV, the set

~

on

limit

~%.

Let

~

l i m r ÷ I $(re i0)

has a n e g l i g i b l e

complement

be the exists. in the

125

interval in

I-z,

D~

and

a point orem,

in

~).

by ~D

let

c

C0 : i

for the

in

to

in

~D.

sequences • -. ÷ 0

curve

sake

in

r ÷ i.

be the

{@"(n)}

> 8"(2)

>

in

~

G n : {z @ ]D :

@'(n)

G'

: ~(G

converges

In o r d e r

to p r o v e

to a point

on the

in

contrary

in

such

that

We take < arg

~D.

n

~. ZD.

Set

that

c'

points

We then

@'(i)

any

take

< @'(2)

and

to

the the-

S0 E

to a p o i n t

set of a c c u m u l a t i o n

included

.-- ÷ 0.

0 £ r < i}

c$

converges Suppose

E $ ~D

{rei@:

~, then

converging

= ~(e)

is a c o n t i n u u m and

8"(i)

the r a d i u s under

as

~

e'

let

E

{@'(n)}

c@ c@

of s i m p l i c i t y .

and

Then

by

of

@ 6 9

show t h a t

D

and

image

every

be any

only

c'

the

for

We h a v e

oscillates

If we d e n o t e

c~

of two

<

set

z < @"(n)}

and

n

Then and

G'n the

@'(n)

is a r e g i o n

part

of the

bounded

boundary

< @ < @"(n)}

on

Gn,

the

the r e g i o n

G~,

so that

Let

u(w)

E

be the values

Diriehlet that

v

[-~,

or

u(w)

tends

tends

to

to

Hence,

3C. same

We d e n o t e

letter

we h a v e Theorem.

the

~.

should

the

So

~I~

problem

in

is a r e g u l a r

u > 0 on

w

the

to

as

maps

on

We @

point

e@'

to the

boundary

n + ~.

This

to a p o i n t

extension ~

of

with

tends

e@.

the Since

v = u O ~,

so

to e i t h e r

to

~D \ ~G'n

along

with

that the

~D, as d e s i r e d .

T

to

continuously

CI(D~) onto

~D.

and So

' c@.

It f o l l o w s

contradicts in

G'.n

for the

belongs

curve

along

D

in

of

w E ~ D \ E.

set

in

is c o n t a i n e d

for

if

D. If

~

converge

continuous

= 0

~.

goes

goes ÷ 0

e'

comes

on the b o u n d a r y

and

E

{el@:

eventually

image

lies

Dirichlet

then

as

~

E

c

c@,,(n ) '

fact []

by the Then

following

The

its r a d i a l

fined

c'

~),

zero as

of the

e@.(n),'

to the arc

curve

w E E

and h a r m o n i c

< (@"(n) - @ ' ( n ) ) / 2 ~

u > 0.

of the

and t h e r e f o r e

zero

the

part

if

curves

corresponding

continuum

point

(@"(n),

thus

v(0)

= i

every

problem

v(~)

and

solution

is b o u n d e d

@'(n))

the

).

convergent

Since

oscillating

f(w)

is a c o n t i n u u m ,

~D

~C.

the r e g i o n

boundary

by two

n

derivative limit

~'(6)

is a l m o s t

belongs everywhere

to the H a r d y equal

to

class

~T/~,

HI(~) which

by 8~. i@ ) : lim ~-~<e t÷0

(~(e l(@+t) ) - ~ ( e i @ ) ) / ( e i(8+t) _ ei@).

is de-

126

is of bounded Proof. Since ~D is rectifiable,, the function. ~I~I3% variation. We set d~(e it) = (ielt)-id~(e~t) on ~I3 . Then I and for

eitd~(eit)

= i-lf

d~(eit)

= 0

n : 2, 3,... I

enitdB(eit ) = i-i I ~D{

e(n-l)itdY(e it) 3D%

(n-i)[

~(eit)e(n-l)itdt

= 0,

J because ~({) is holomorphic in I]~ and continuous up to the boundary. By the F. and M. Riesz theorem (see Appendix A.3.2 and A.3.3) dz is absolutely continuous with respect to the arc-length measure and u(rei8 ) = i

0

P(r,O - t)d~(e it)

belongs to the Hardy class HI(ID ). Setting have by the Fatou theorem (Appendix A.I.2) f(e it )

= ~~(

eit

)

d~(e it) = f(eit)dt, we

a.e.

So the radial limit of u(~) is equal to ~ I ~ a.e. on ~]D . On the other hand, since du is orthogonal to e nit , n = i, 2,..., we have I~D

e -it + ~ e-l~_~d~(eit)

= f~]D~

d~(e it)

and therefore

u(~)

=

i I i [e it + ~ e -it + ~] ~ [e-~-- + d~(e it) ~ ~9~ ~ e_mt_ ~j

= 4~ = ~i =

f

•

[elt+ ~ ~D [e±t_ ~

+I

i d~(e it) = 2~

f

~]D

it

e__

el~_

I ~13 -e ~ I - % d~(eit) = 2-~ i ; ~I] (e~(eit) it- ~)2 d(eit )

~'(~),

which implies the desired result. [] Combined with the Fatou theorem, the preceding theorem shows:

127

Corollary.

If

~Y/Z%

verges to the value

exists at a point (~Y/~6)(~0)

any fixed Stolz region 3D.

in

~{

as

with vertex

in Appexdix A.I.2.

For

the connected

S(~0;~,p) {z ~ ~:

to which the point

~0 ~ ~ '

S(~O;~,O)

through

~0

of Stolz regions

0 < ~ < ~/2 component

is adherent.

con-

Y'({)

tends to

~0"

and

given

0 < 0 < i

we de-

of the set

-~ ~ a r g ( l - Z~o) ~ ~, 0 ~ ~0

then

%

We now have to modify the definition

note by

region

%0 ~ ~ '

uniformly

Izl < I},

In order to make the new Stolz

simply connected,

we always assume that

P ~ sin ~. Theorem.

(a)

every angle (b)

(~/~6)(~0)

exists and is different

80 E [0, 2~) \ ~0' where

Let

w 0 : exp(i80)

80 E [0, 2~) \ ~0 = ~(%0 ).

Then

satisfy the condition

SD

have the same tangent at the point

with

0 < ~ < ~/2

the region

Stolz region with vertex

p

depends

S(w0;a,p) ~0

{0

~0' lies in

~%

We showed

Poisson

integral of

Theorem

in Theorem

function 2A, Ch. IV.

point in

the

in

~

3C that

Stolz region

(< ~) in the disk sides which, at

%0

and ~0 ei(~+~),

Since

~-i

Stolz region

~/~ ~'(~)

Hence the statement

~6

any

except the vertex to the two adjacent

respectively.

exists a.e. and is not identically

can vanish only on a set of measure

c

which,

(a) holds

~'

is the

zero, its zero by

for almost every

~0"

~

satisfies

the condition

except the initial point

and has a tangent at

tial to the circumference tinct from

2~

80 E [0, 2~) \ ~0

Draw a smooth curve ~

Under the map

~.

Suppose that disk

Z~/$~. ~T/~

D) and

2e.

and such that the tangents ~0 ei(~-e)

of

for any

a Stolz region of the form ~.

is sent to a curvilinear

triangle with smooth

Proof.

Consequently,

(b) above we mean by a curvilinear

sides have directions

boundary

and

and angular measure

closed eurvilinear

w 0.

contains w0

and angular measure

In the statement with vertex

D on

in (a) and set

(see (3) for the definition

~w

S(w0;~,0) , where

from zero at almost

%0 = ~-l(exp(i80))"

t 0. .

Let

We suppose that ~i' ~2

We have

~(~i ) - ~(~2 ) =

T'(~)d~,

in (a).

~0' lies in the c

is not tangen-

be two points on

c

dis-

128

where

the r i g h t - h a n d

and let

%2

side is the i n t e g r a l

tend to the p o i n t

~0

along

along

c.

the c u r v e Then

c.

Fix

tl

~(~2 ) ÷ ~(%0 ), so

that

I

~(t I) - ~(t 0) = Since

~'(t)

tends

to

(3~/3~)(~0)

shown that the d e r i v a t i v e

of

is e q u a l to

(~/~t)({0).

arg{~(%0)}

u n d e r the map

angle

between

with respect boundary tangent Hence

~D to

two c u r v e s

to the c i r c u m f e r e n c e

~.

at

cI

{0

between

~D

and

at

~

÷ ~0

~(e).

~(~0)

to

tends

As

~D

Since

~(t ~ )

to

~0' we h a v e

the c u r v e e

issuing

under a rotation the c i r c l e

and

along

at

c

{0 ~0

(3~/~)(~0)

~ 0, the

is o b t a i n e d

e and

e 3~ w .

f r o m the

arg{~(t0)}.

= w 0 6 3~ w

{~n*: n = i, 2,...}

the

nontangentially

by the a n g l e

( = w n, say)

by

preserves

and the curve

~(~0)

e x i s t s and

is r o t a t e d

~(t)

from

w 0 = ~(~0 ), w h i c h

we can find a s e q u e n c e

~ ~0'

t E c

%@

and t h e r e f o r e

e2

~Dt

as

~ + ~(~) and

the a n g l e

tifiable,

at

~'(t)dt.

So the t a n g e n t

has a t a n g e n t

that b e t w e e n

~

{l

is e q u a l to SD

~C ~

is r e c -

such that

Then,

wn ÷ w 0

and

~ ( t ~ ) - ~(t O) w - wO % ~ - ~0 = lim n n÷~ ~tO

= lim n÷~

We thus h a v e a r g t0~r[3~~ " )} = a r g w 0 - arg ~0 This

shows

w 0.

The first h a l f of

in p a r t i c u l a r

c'

~D

and

a Stolz-region

be any one of its sides

issuing

arg(1-w~ for

w E c'

D t.

Set

We s h o u l d

with

w i t h the c o m m o n

for

~i 6 ci,

above,

c!l

equations c'

c

arg(l-

initial

i : i, 2.

converge

to

~ 0 ) = e.

a r g ( l - w i w 0) = ~l.

point

~0

c

from

for i n s t a n c e

to

t0

for

w0:

c

in

at

D

is a s m o o t h

and

cI

arg(l-

and

determined

lies in a Stolz r e g i o n w i t h v e r t e x

in

As s h o w n

w0, w h i c h

w.l ~ c!l' r e s p e c t i v e l y . w0

~I' c2

t i ~ 0) = ~i

c i' = ~(c.). ± at

in

at

we take any

segments

such that we set

curve

and has a t a n g e n t

To see this,

and have t a n g e n t s

lies in a Stolz r e g i o n w i t h v e r t e x

and c o n s e q u e n t l y

included

and d r a w two

And f u r t h e r w0

S(w0;a,p)

so t h a t

converges

0 < ~i < e < e2 < 7/2

~t

h a v e the same t a n g e n t

w

0) =

c = ~-l(c' \ {w0}),

show that

t0 , for w h i c h we h a v e ~2

$~

(b) is n o w clear.

Let us now c o n s i d e r let

that

(mod 2n)

satisfy

the

So the c u r v e c I'

by ~0

and

c~,

determined

129

by

cI

to

e, we h a v e

and

satisfying of the

c 2.

the

Since

shown

equation

statement

4.

Limit

X.

For a g i v e n

Let

can be t a k e n to

~0

~ 0 ) = ~.

arbitrarily

and has

This

proves

close

a tangent the

at

latter

~0

half

Lines

be a f u n c t i o n Green

u~(~)

e2

converges

[]

Green u

and c

arg(l-

(b).

along

4A.

el

that

line :

on

R

with

~ ~ ~(a)

n

we

Cl[u({z(~;B):

values

in a c o m p a c t

space

set o

~ < ~})].

<

~>o If

u~(~)

dial

consists

limit

Theorem.

of Let

Green

star

~(~)

exists

on

(i)

~ in

~

u

that,

Since

to

E

this

property.

~-io

¢

Hence

Then

(ii) {'(a)

Let

u

is s i m p l y

uniquely

under

the

in any So on

zero.

is a b o u n d e d

holomorphie every

function

~0

function

through

under

with

vertex

in

for a l m o s t

every

so that

its r a d i a l

limit

harmonic

the h a r m o n i c ~(a)

= 0.

: exp(-u(z)

function ~ E L(a).

on

for a l m o s t function

conjugate Sinc~

u

say)

the map in

E.

~ e L(a). vanishes

every

on ~

{0

possess

~80

exists

~ 0

~%

80

Stolz

we see

(= {0'

Let

~, ~0

fixed

zero,

~-l(exp(ie0))

line

on

of such

of m e a s u r e

n0).

and

point

any

totality

region

~(~)

limit

~'(a)

every

Stolz

,

on the

the

on

~(~) •

condition

function

of

Green

~(~).

dma-measure.

sets

2~) \ n0,

of the

u

be the

definition

the ra-

is dm a - m e a s u r a b l e vanishes identically

holomorphic

to

E

by

~ E L(a)

for a l m o s t

preserves

be any p o s i t i v e

the

for a l m o s t

tends

Hence

connected,

tends

Let

[0,

image

h(z)

exists

%

t 0.

80 E

v ~ 0

of m e a s u r e

as

÷ ~D

exists.

u ~ 0, t h e n

is a b o u n d e d

3A for the

is c o n t a i n e d

on a set

~

every

(see

~(Ze0 )

limit

harmonic

then

holomorphie

v(~)

vertex

~:

for a l m o s t

v

is c a l l e d

is d e n o t e d

~ ÷ ~(~)

holomorphic,

theorem

point

every

function

on a set of p o s i t i v e

Then

with

and

for a l m o s t

be a b o u n d e d

Fatou

~

the

be a p o s i t i v e

The

is b o u n d e d

then

line

u

Then

vanishes

~%

.

belongs

If

u

to a f i n i t e

region

let

{'(a).

point,

Green

is finite.

Let

by the

one

the

and

v = u o ~ - i o ~.

so that

in

and

~(Z)

Proof.

in

a E R

If

whenever

along

region

L(a).

set

u

of o n l y

{'(a). of

only

~ ~ L(a).

u

Since

exists

is p o s i t i v e ,

- i[(z))

~'(a) Since

and t h e r e f o r e ho ~-io ~

does

by

(i)

not

h(~)

vanish,

130

its r a d i a l

limit

that

h(~)

~ 0

~(Z)

= -log

4B. fact,

lh(Z) I

In the

Let

case

f(~)

on a set of m e a s u r e

and

~ E ~(a).

is f i n i t e

of a PWS,

Theorem

and

Consider

•

be a b o u n d e d the r a d i a l

for a l m o s t

Rn

with

every

in 2B.

limit

{R

is d e f i n e d

the o r d i n a r y

Then

fn

a.e.

We

Dirichlet

bounded.

It is e a s y

bounded

equioontinuous

subsequence monic

data

set

clear

see

In

that going

S~

to

S~:

+ S 0) if

solve

w

rectifiable

boundary

let can

slit

the

and

of

with

of the

can be v i e w e d

Then is equal

Rn

Dirichlet

=

problem

conformal

limit

So,

forms

of

a uniformly passing

converges

to a

to a h a r -

map

}

of the

we d e f i n e d

in

to b e l o n g w' E D

D.

For

of its

to

two

Se

satisfying

common

point

exp(-g(a,z(0))

sense

3A.

Pn = It is

as the u n i o n

only

i.e.

Green

z E R'n, w h e r e

for the r e g i o n

of

The

in an o b v i o u s

~R n,

is u n i f o r m l y

D n = (pn D) D D w .

e E ~0'

> 0).

2,...}

which

for

is d e f i n e d

as the

on

solution

R.

problems

$0,

is the

{u n}

on

D : D(a),

slit

R

the

Consequently,

= pn ¢(z)

w E $8

I m ( w ' w -I)

of

Se

+ i0).

as a J o r d a n

region

Then with

L, w h e r e

L = U{S +e U S~:

v

n Un+2,...}

).

uniformly

Dirichlet

is r e g a r d e d

D

exists

function

: n = I,

n a s s u m e that

Cn(Z)

each

.. }

measurable

{u

CI(R

the r e g i o n

a point

the r e g i o n

Now

on

we can

and

I m ( w ' w -I) < 0 (resp. + and S 8 is the v e r t e x

L(a).

= f(~)

{Un+l,

we c o n s i d e r

we r e g a r d

and

on

where

Then

that

almost

onto

.

u n = H[fn;Rn] , which

family

hand,

2,

we c o n s i d e r

be the image of Cn' i.e. D I =D D 2 ~ .. • =D D n =D . . . D D .=

Dn

purpose,

(resp.

to

u,

~'(a)

Let

We are

n ary

lh(z)1,

strengthened.

function

i,

=

n

fn'

problem.

R'n = {'(a) D R n

exp ~n"

edges

say

other

star r e g i o n

this

It f o l l o w s

= -log

[]

2C is m u c h

is a b o u n d e d set

if n e c e s s a r y ,

function, On the

then

zero.

u(z)

G[f;{(a)](~)

: n

n

For e a c h

the b o u n d a r y

~ E L(a).

which

We

a.e.

measurable

fn(Z(~;en)) for

As

~ E L(a).

an e x h a u s t i o n

R(~n,a) , as g i v e n for

only every

exists

f(~)

f is r e s o l u t i v e

Proof

vanish

we have:

Theorem.

to

can

for a l m o s t

0 E ~0}

U {e ie : 8 E

[0,

2w) \ ~0 }.

v be the r e s t r i c t i o n of u o ¢-i to the r e g i o n D. Then n n n c l e a r l y be e x t e n d e d c o n t i n u o u s l y to the b o u n d a r y L, the b o u n d -

values

being

denoted

by

f~.

If

Z = ~e E ~ ( a ) ,

then

we h a v e

131

+

f~(e 18)

point

= f(£).

If

Zn(W) 6 R

sequence

{zj:

~n(Zj) ÷ w tinuity.

D.

W E S;

(or

converges

w

of

f~(w)

C R'

=

f*(W)n

the

is

sequence

£ = £8 E l ( a ) ,

of

z 6 S + US;. e D supported

see that of

L.

d~ w

exists

there

3

÷ zO(w).

+ U(Zo(W)).

function

f*,

is

z. ÷ z (w)

a

a

and

n

determined

by c o n -

(w): n = I, 2,...} n to a p o i n t z0(w) 6 R.

: Un(Zn(W))

Thus,

Namely,

stays In

if {f~}

i.e.

: f(£)

and

f*(w) if

properties

uniquely

~-l(pnlw')

to a w e l l - d e f i n e d

there

and (ii)

{z

converges

f*(e ie) if

S8) , then

= Un(Zn(W))

with

w' ÷ w, t h e n

then

pointwise

and

(or

Zn(W)

n

the

R

and

S:),

(i)

2,...}

such

subset

w' E D

w e S;

Such a p o i n t

For e a c h

if

and

such that

j : i,

in

in a c o m p a c t

fact,

8 e ~0

: u(z0(w))

For any

given

w E D

on

Since

L

L.

is a b s o l u t e l y

let

d~

w is r e c t i f i a b l e ,

continuous

be the h a r m o n i c it is not

with

respect

: S

f~d~w'

hard

measure to

to the a r c - l e n g t h

As we have

(U n o }nl)(w)

: Vn(W)

L so we d e d u c e , vergence

by l e t t i n g

n ÷ =

and

using

the

f

(u o ¢-l)(w)

= I

f* dm

JL Namely, D

u o ¢-i

is e x a c t l y

corresponding As we h a v e

i}

with

theorem

dominated

con-

8 6 shows

to the seen

[0,

the

solution

function

in T h e o r e m

2w) \ ~0

of

r ÷ i

of the

problem

3D, ~w

almost

every

is o r t h o g o n a l

radius

for a l m o s t

every

8 E

every

£ E ~(a),

Every

we can

[0,

as was

to

~D.

So the

= f(£8 )

2z) \ ~0"

This

means

that

= f(£)

to be

shown.

[]

show:

dm a - s u m m a b l e

for

{re ie : 0 ~ r <

that

As a c o n s e q u e n c e Corollary.

Dirichlet

f*.

lim u ( z ( £ ; ~ ) ) for a l m o s t

• w

( u o # - l ) ( r e i 8 ) + f*(e i8) as

Lebesgue

theorem:

function

on

L(a)

is r e s o l u t i v e .

Fatou

132

§3.

5.

THE

GREEN

LINES

Convergence 5A.

bolic

the

function kb(0)

For

of G r e e n

surface

with

the M a r t i n

R

at

for a n y

III.

£ E

pole

Let

L(a)

and

The

and of

A.

sists and

CI(£)

means

Z U {a} We

say that

of a s i n g l e

to the

a detailed tive

5B.

We a s s u m e

for

z E R, w h e r e

poles

shall

and

IC,

Ch.

~g(a,z)

V,

set

then

the H a r n a c k

We d e f i n e

covering

s(z)

function {V

}

of

R. R

R*,

Martin

so that

have

lines

been

given

Z E L(a).

i.e.

Since ez

R*

is c o m p a c t

is a n o n v o i d

is c o n v e r g e n t

standing Green

that

that

problem

line

if

subset

e~

due

con-

to B r e l o t

in

L(a)

(with

See

Brelot

[5]

the p r o b l e m

has

for

an a f f i r m a -

R

is a r e g u l a r

PWS.

We t a k e

: 6g(a',z)/6g(a,z)

function

on

R.

Since e ~.

: ~{g(z,zj): shows

R

If

It is easy

to

a ~ a',

then

is a PWS, So,

if we

we h a v e

see that it has by T h e o r e m

set

zj E Z(a~R)},

that

is a p o s i t i v e

s(a)(z)

< ~

for

superharmonie

z ~ Z(a;R)

function.

Set

and = g(a)(z)exp(-g(a',z)).

= s(a)(z) + g ( a ' , z ) ,

on

R*.

= 2(Zg(a,z)/~z)dz.

u(a,a';z)

harmonic

kb

R,

of PWS's.

follows

inequality

and t h e r e f o r e that s (a) (a) g (z) = e x p ( - s ( a ) ( z ) ) (5)

show

zj e Z ( a ; R ) }

s(a)(z)

in

every

case

Z(a;R) U {a'}.

~{g(a,zj):

in

of

set

is a m e r o m o r p h i c

in the

of

see that

is a long

P(a,a';z)

P(a,a';z)

origin

Green

Z

boundary

b 6 A, be the

dm a ) is c o n v e r g e n t .

in the

in w h a t

a' E R

of

~ E L(a)

almost

We

at least

point

line

measure

£

R, we

Here

whether

Green

the

consider

end

of

in

a Green

exposition.

solution

another

closure

point.

asking

kb,

properties

and

the

0

of a h y p e r -

A I) be the M a r t i n

Let

by

R*

: CI(~) \ ( ~ U {a]),

is n o n - c o m p a c t

Choquet

respect

the

(resp. A).

basic

denote e~

where

A of

be f i x e d

e£

compactification

We d e n o t e

b E A. a E R

let

points

b.

let

BOUNDARY

Lines

set of m i n i m a l

= I

in Ch.

THE M A R T I N

Let us c o n s i d e r

Riemann

(resp.

AND

In fact,

consisting

which

is a l s o

it is a p o t e n t i a l . of s i m p l y

connected

a positive We t a k e open

sets

super-

an open V

133

such that,

whenever

V

disk which contains associate notes

a function

a harmonic

A V B ~ ~,

of the f o r m

conjugate

t a i n s no s i n g u l a r i t i e s

of

is a u s u a l h a r m o n i c critical

point,

say

UV B

is i n c l u d e d

of

s(z) + i~

siV a .

s, t h e n

conjugate

z E V , where

u

of

in a p a r a m e t r i c

s(z).

(z),

W i t h each

z E V , where

To be m o r e p r e c i s e ,

s(z) of

is h a r m o n i c

s

on

zj, of m u l t i p l i c i t y s(z)

for

V

at m o s t one s i n g u l a r i t y

V .

mj,

If

V

we

~

de-

if

V

con-

on

V

and so

V

contains

a

then

= -mj log Iz - zj I + u(z)

is h a r m o n i c ,

and so

$ (z) = -mj arg ( z - zj) + ~(z), where

~

is a h a r m o n i c

conjugate

of

u

on

V s.

Then we set

f (z) = e x p ( - ( s ( z ) + i~ (z))) for

z E V , which

that,

a constant union ({V

~B

defines

with

any n o n z e r o

a line b u n d l e ,

2B, Ch.

element,

V s.

associated

w i t h the c o v e r i n g ({V

section,

say

belongs

f to

function

on

Since

of

Let us

with norm

of this e l e m e n t

}, {he}).

a nontrivial

R, w h i c h we d e n o t e by

By the

~ {0}.

~(R,~a,a,)

({V

If I

~(R,{a,a,).

~(R,~a,a})

h (a'a'),

defines

So the p a i r

< a , a , , ( ao,vae,r ) R, for w h i c h

{V }, i.e.

}, {f h })

A V B ; for the

V

set.

T h e n one can find a r e p r e s e n t a t i v e

that the p a i r

It is easy to see -i is e q u a l to

f (z)f6(z)

connected

V) we k n o w

say

i.

F

on

~ ~,

say

f(a,a')

less t h a n

morphic

NV~

one on the i n t e r s e c t i o n

forms a h o l o m o r p h i c

(Theorem

function

V

in a s i m p l y

~, we see that

Widom t h e o r e m choose

~

is i n c l u d e d

}, { ~ B } ) for each

e,

of m o d u l u s

UV B

V

({V }, {f }) i

is a h o l o m o r p h i c

for each p a i r

It is i m m e d i a t e

single-valued

F(a,a';z).

holo-

The p r o p e r t y

of

that we n e e d is this:

(6)

IF(a,a';z)l Let

{an: n = i, 2,...}

decreasing

We m a y

Green f u n c t i o n

for

exists

for

R

n a constant

~gn(a',z)/~g(a,z) -(2~i)-16g(a,z)) the p o i n t

be a s e q u e n c e

to zero w i t h the p r o p e r t y

Rn = R(en,a). function

~ u(a,a';z)

a'

R

suppose here with pole

n w i t h pole c < o

along (resp.

on SR n a).

~Rn,

that a'.

o n l y on for

of p o s i t i v e

a' E R I. Since

Let

ga- ~

n inequality

a, a'

and

the h a r m o n i c

u(a,a';z)

~ i

on

numbers : ~

and

strictly set

gn(a',z)

be the

is the G r e e n shows that t h e r e

R, such that

-(2~i)-l~gn(a',z)

represents Since

R.

~R(~n,a ) AZ(a;R)

a, the H a r n a c k

depending

on

0 <

(resp.

measure

of

R, we h a v e

Rn

at

134

(7)

0 S u(a,a';z)(6gn(a',z)/Gg(a,z))

on

DR

. Since n the m e r o m o r p h i c

and

since

u(a,a';z), function

(6) holds,

defined

by

(5),

vanishes

~gn(a'~z)/6g(a,z)

the

S c

with

at e v e r y

the

same

pole

of

multiplicity

function

F(a,z';z)(6gn(a',z)/6g(a,z)) is a h o l o m o r p h i c It f o l l o w s

function

from

on

R

and

n

is c o n t i n u o u s

CI(Rn).

on

R, we h a v e

Since

6gn(a',z)

converge

to

R.

Since

both

F(a,a';z)

holomorphic

on

are

harmonic

bounded

and

R, the real

line

limit

~(a,a';~)

z(~;~)

in

0.

in

A.

We d e n o t e

parametrization Theorem. forms

The

Take

function almost dial nate

Along

set

A(a)

has

~ E L(a).

z = x + iy

Z

a' E R

P(a,a';~)

we t h e n

radial

thus

A

lines

of

for

exists.

in such

b~

set

when

every

radial

Green

that

dx = d g ( a , z )

and

each

dy

if topology

of c o n v e r g e n t denote

~ = ~e

lines

the

Green

limit

of

according

radial

z E ~

to the

= dx

from

a

one. above,

the

P(a,a';~)

for w h i c h

we take

= *dg(a,z).

have

seen

limit

~ E L(a)

= (~xg(a,z))dx

issuing

dm a - m e a s u r e

As we h a v e

finite at any

At

Martin

in IA.

~ a.

a way

6g(a,z)

F(a,a'~z)

almost

finite

to the

b~

of G r e e n

a'

look

the let

defined

a nonzero We

along

and

functions

to be c o n v e r g e n t

respect

= A(a)

L(a)

with

limits

bounded

4A both

a nonzero

is said with

~ E A(a) be

has

are

of t h e s e

~ E L(a).

L(a)

in

parts

So by T h e o r e m

of c o n v e r g e n t

subset

P(a,a';z)

every

limit

any

R.

every

A(R;a)

write

of G r e e n

a measurable

Proof.

by

For e v e r y

sometimes

on

uniformly

I ~ c

imaginary

nonzero

~ E

to a p o i n t

L(a). We

line

almost

F(a,a';z)P(a,a';z)

the

P(a,a';z)

for a l m o s t

Green

converge

a ÷

lines

have

~ E L(a).

A regular

as

functions

F(a,a';z)P(a,a';z)

Green

and

and

I ~ c

6g(a',z)

IF(a,a';z)P(a,a';z) on

boundary.

(7) that IF(a,a';z)(6gn(a',z)/~g(a,z))

on

up to the

a local

the

for ra-

coordi-

135

and

6g(a',z) where

g(a',z)

uniquely g(a,z)

~(a,a';£)

tends as

to

shows

(9)

we (i) and

Then

the

both

left-hand

g(a,x+

g(a',z),

We m a y

assume

Then

side

iy 0)

we h a v e

of

and

of the

of

(9)

{g(a',x+

exists such

to

determined that

(8) has

g(a',x+ surface

a limit

iy 0) R,

a'

run

there

for e v e r y

a' E A.

on

density

R, the

portional.

As

eL

consists

6.

Green 6A.

exists

iY0)/g(a,x+

a sequence that

tend

x

to

zero

a measurable

and

on

b

eL

in

£

subset

A

We

Theorem. (a) f E LI(dx )

Since of

shown

So the

of

A

in

of

L(a)

R.

such

A

and

shall and

Let

the

in

R

establish

a C R z:

eL,

b. Then that

exists

that

this

each

with

b E A

kb

and

means

that

£ 6 A.

[]

is c o n t i n u o u s k b'

are

b = b'.

pro-

Namely,

Boundary

further

connection

between

the

space

of

boundary.

~ ÷ b£

function

2B,

for

be fixed.

= G[f o z;~(a)].

kb

implies

III,

the M a r t i n

the M a r t i n

The map

function

point

final in

= kb,(a')/kb,(a)

the

in Ch.

with

is a r b i t r a r y

as a f u n c t i o n

dense

subset

zn

z n ÷ b.

Since

a countable

iY0)/dx}

iY0)).

of p o i n t s

xn ÷ 0

is c o n s t a n t

through

of a s i n g l e

Lines

lines

H[f]

as

l'Hospital's

m a (A) = i; (ii) for each a' e A and ~ e A, P(a,a';~) is finite. So, if b, b' E e£ with £ E A, t h e n

Green

x =

on

+ iY0)/dx).

iYo)/dx)/(dg(a,x +

kb(a')/kb(a).

kb(a')/kb(a)

let

see that

= lim x+0 there

is e q u a l

quotient

{dg(a',x +

kb(a')/kb(a)

and

£.

= lim x÷0

" 0 x n + ly

coordinates

Now

factor.

along

of

that

b E e£.

member

conjugate

of the r e g u l a r i t y

Re(P(a,a';£))

Let

+ i~xg(a',z))dx,

: (dg(a',x + iY0)/dx)/(dg(a,x

Since

in v i e w

constant

= constant

exists,

zero.

x ÷ 0

rule

the

Y = Y0

Re(P(a,a';z))

Since

the h a r m o n i c

up to an a d d i t i v e and

(8)

the

denotes

= (~xg(a',z)

of

Then

we h a v e

A(a)

f o ~, d e f i n e d In p a r t i c u l a r ,

into

the A

dm a -a.e. f o ~

following: is m e a s u r a b l e . on

A(a),

For any

is r e s o l u t i v e

is d m a - s u m m a b l e

and

136

I

f(b)kb(a)dx(b)

= I

A (b) ~(~)

If

u

is a p o s i t i v e

for a l m o s t

function

u

(c)

every

function

exists

a.e.

A(a)

on

R, then

u(b~)

=

The same is true of e v e r y m e r o m o r p h i c

characteristic,

P(0,a;b) The set

harmonic

~ E A(a).

of b o u n d e d

P(0,a;b)

(d)

f(b~)dma(~)" L(a)

on

i.e.

AI

= kb(a)

log lul E SP'(R).

and a.e.

can be c h o s e n

on

A I.

in such a w a y that the m a p

in (a) is i n j e c t i v e . (e)

The map

(A(a),

dm a)

Proof.

(a)

in 5B.

Then

n

~

and

is an i s o m o r p h i s m

(A, dXa)

dXa(b)

We take a s e q u e n c e ~

is the

goes to i n f i n i t y ,

any d x - s u m m a b l e finitions

with

We also have

4.

Put

problems

~ S(g;~(a))

H[f] ~ G [ g ; { ( a ) ] .

it is r e s o l u t i v e

by C o r o l l a r y

J that

G[f;~(a)]

(b) t i o n on

R, i.e.

have

G(b)

dial

limit

is d e f i n e d Suppose

the s t a t e m e n t Suppose

: 0

for e v e r y

R.

on

set

on

~ E B. so that,

to be d x - s u m m a b l e , = H[f]

By C o r o l l a r y

at once. Then,

positive

by T h e o r e m

by T h e o r e m

~(~)

= 0

4B,

~ e L(a),

a contradiction,

cB

function

we h a v e u

on

u(z)

L(a).

~ > 0

such t h a t cB

is r e s o l u t i v e

for

4A, the ra-

function which

number

dma-measure

func-

III, we

dm a -a.e.

exist a p o s i t i v e

of p o s i t i v e

harmonic

5E, Ch.

dm a - m e a s u r a b l e

We c l a i m that

by C o r o l l a r y

2C we h a v e

(~).

is a s i n g u l a r

that t h e r e

for any

and

a

T h e n the c h a r a c t e r i s t i c

is c l e a r l y

H[f]

On the o t h e r hand,

B ~ L(a)

in IB and Ch. III,

H[f] ~ G [ g ; ~ ( a ) ] .

i.e.

g(~)dm

as a n o n n e g a t i v e

L(a).

~(~) ~ ~CB(~) This

4 I.

exists

on the c o n t r a r y

dma-summable Since

a.e.

a.e.

u

the de-

is a s s u m e d

= H[f].

: [ J L(a)

(a) f o l l o w s that

as be

f(b)kb(a)dx(b).

u E HP(R) A I(R).

~(~)

a measurable

on

first

f

41

= ~[g;{(a)]

G[g;{(a)](a) From these

f III,

let

By c o m p a r i n g

given respectively

Since

given

~ ÷ z(~;~ n)

Secondly,

g : f o 7.

3D, Ch.

spaces

the p r o p e r t i e s

functions

and t h e r e f o r e

H[f](a) = [ It f o l l o w s

satisfying

it is m e a s u r a b l e .

on

of the D i r i c h l e t S(f)

{a n]

the m e a s u r e

= kb(a)dx(b).

limit of c o n t i n u o u s

so that

function

3A, we see that

between

of on

and

~(~) B

is

~(a).

~ G[BCB;~(a)](z)

is s i n g u l a r

while

the

137

function erty

G[~cB;~(a)]

(a) i m p l i e s

-I(A)

is q u a s i b o u n d e d

in p a r t i c u l a r

is dm a - n e g l i g i b l e .

for a l m o s t

every

Suppose

f(b)

f(b)

Then

= 0.

E A(a).

By

G[g;~(a)].

u

= G(b) f

for

(a) a b o v e

on

and T h e o r e m

As an a n a l o g u e

s

for any

e > 0.

on

R

for any

~ ~ A(a).

a.e.

L(a).

For any

on

A(a).

u - Es E S(f)

every

characteristic,

~ E A(a).

bounded holomorphic

result

Re(~(0,a;~)) ~ E A(a).

bounded holomorphic

Since

from (12)

every

~ E A(a).

has a d x - n e g l i g i b l e

(i0) and

and

= prob-

superu - ss E S(f)

so that

u(£)

u(~)

exists

exists,

dm a-

we have

~ f(b~). a.e.;

inequality

so we h a v e

is o b t a i n e d

~(~) f r o m the

: ~(b~) if

u

is a m e r o m o r p h i c

it can be e x p r e s s e d

follows

theorem

f r o m the a b o v e

P(0,a;z)

of

(Theorem

2B,

consideration.

: k(b~,a)

is e x p r e s s e d

the p r o p e r t y

as a q u o t i e n t

(b) i m p l i e s

of

that

: P(0,a;b~)

The p r o p e r t y complement

(a) i m p l i e s

in the space

in p a r t i c u l a r A.

(ii) that Re(P(0,a;b))

function

as a q u o t i e n t

5B we s h o w e d that

: k(b~,a)/k(b~,0)

P(0,a~)

~(A(a))

for e v e r y

u = H[f]

= f(b~)

in v i e w of W i d o m ' s

functions,

(ii) for a l m o s t

otherwise,

= f(b~)

a positive

limit

is f i n i t e

Finally then

In the p r o o f of T h e o r e m

for e v e r y

function.

exists;

~ S(g;~(a)),

~ g(~)

s(z(Z;e))

: f(b~)

functions,

so the d e s i r e d

(i0)

= 0

and h e n c e

for a l m o s t

(c)

~(f)

The r e v e r s e

of b o u n d e d V);

= ~(b~)

for the D i r i c h l e t

exists

for w h i c h

lim inf ÷ 0 s ( z ( ~ ; ~ ) )

g(~)

III, we have

4A the r a d i a l

~ e A(a)

~(~)

Ch.

The p r o p -

harmonic

u + Es E ~(f)

(u + e s ) ( z ( ~ ; ~ ) )

By T h e o r e m

~(b)

2A is t r u e there

k n o w that

~(~) + c l i m i n f ~÷0

fact

~(£)

positive

Set

5E, Ch.

satisfying

We a l r e a d y lim inf

2B,

positive.

is d x - n e g l i g i b l e ,

shown that

A.

of T h e o r e m

function

dma-a.e,

A

b E AI, w h e n e v e r

is d x - s u m m a b l e

harmonic

f(b~)

~

is a q u a s i b o u n d e d

compactification,

By Lemma

A

H e n c e we h a v e

lem for the M a r t i n

on

and s t r i c t l y

if

~ E A(a).

next that

We d e f i n e

that,

= k(b,a)

a.e.

on

A7 .

that

So it f o l l o w s

138

Changing

the role

(13)

0

and

Re(i/P(0,a;b))

It is easy b E 41 on

of

to

= Re(P(a,0;b))

see that

satisfies

both

a, we get

P(0,a;b) (12)

: i/k(b,a)

is r e a l

and

(13).

and

Hence

a.e.

on

A I.

is e q u a l

to

P(0,a;b)

= k(b,a)

k(b,a),

when

a.e.

41 . (d)

Let

f

is r e s o l u t i v e

be a b o u n d e d

by T h e o r e m

measurable

(14)

G[f;~(a)](Z)

for a l m o s t

every

E A(a).

~ E L(a).

In v i e w

of

(c),

to the

we look

parametrization

Of c o u r s e we h a v e

at the

gible on

the

This

A(a)

and

the

Let

u

5D).

every

if

~ = he

according

we h a v e

= 8

the

Let set

Theorem

eonsequenee

A'(a) A'(a)

of

z:

(a) and

6A to m u l t i p l i c a t i v e

be an l.m.m,

Then

the

of b o u n d e d

following

be the

set of

has

a dm a - n e g l i Z ÷ b£ is o n e - t o -

(d).

analytic

[]

functions,

we

~(b)

exists

a.e.

on

A I.

exists

a.e.

on

A(a).

(c)

~(b~)

: ~(~)

We

set as and

by C o r o l l a r y now

a.e.

in Ch. UQ 4B, that

IV,

for

~ @ A(a).

4B

v = log u,

= exp(Vq). Ch.

on

R says

on

R

(see

Then

Ul

v i = Pri( v), = i

and

and

v

Vq

u = UQ

: prQ(v) , a.e.

on

IV.

v > 0.

So

=

ing t h e o r e m

characteristic

hold:

~(~)

Suppose

(16)

for a l m o s t ~ ~ A(a)

: 8

correspondence

(a)

harmonic

every

Then

£ = £8" Then

is an i m m e d i a t e

Applying

u I = e x p ( v i) &l

with

holds.

is true

f0(~)

in IA.

(b)

Proof.

it

following

Theorem. II,

(15)

in

Then

A'(a).

6B.

Ch.

Z E A(a)

complement

(e)

get

every

for w h i c h

~(a).

: f(~).

G[f0;~(a)](bz)

E A(a)

one

(14)

for a l m o s t

function

mentioned

(15) for a l m o s t

on

= f(£)

G[f;{(a)](b~) In p a r t i c u l a r ,

function

4B and we h a v e

by T h e o r e m

v. i

5B,

Ch.

II,

> 0 =

the

that ~q(~)

> 0. q

: Oq(b~)

Since

=

statement

v

is q

(b) of the

preced-

139

for almost every

£ E A(a).

On the other hand, we set

# £) = lira sup v i (z(~;a)) vi(b for

£ e A(a).

Since

v. > 0, we have

v'#(b£ ) l

> 0 = 0i(b ~)

last equality sign being true by T h e o r e m 4B, Ch. IV. bounded l.a.m, fined by

h

h = exp(-vi . - v q ) and denote by q the line bundle deThen we take any nonzero l.a.m, k < i,

(Ch. II, 2C).

whose line bundle is

q-i

(Theorem 2B, Ch. V).

We set

0

and

Vq'

This is possible in view of Widom's theorem - l o g k = v i' + v'q

~q(£)

for almost every Since

with

v'i : P r i ( - l o g k) =>

pr Q (-log k) => 0, and see as above that

=

(17)

h

and

7(£) = f(b£)

almost every

: Q'(b~)q

Z 6 A(a). k

c o r r e s p o n d to m u t u a l l y inverse line bundles,

there exists a nonzero have

a.e., the

We look at the

f 6 H=(R)

such that

for almost every

hk : IfI

£ E A(a).

on

R.

We thus

This means that for

£ 6 A(a) lim h ( z ( ~ ; ~ ) ) k ( z ( ~ ; e ) )

exists and is equal to

(18)

If(m)l

: If(bm)l =

exp(-Qq(b~))exp(-~'(b~))q

by means of Corollary 4B, Ch. IV. almost every

: ~(bm)~(bm)

From (16) and (17) follows that,

for

£ E A(a), lim e x p [ - V q ( Z ( ~ ; ~ ) ) - v'(z(~;~))] ~+0 q

exists and is equal to

exp[-vq(b£) - 0~(b£)]

~ 0.

Therefore

lim exp[-v.(z(~;e)) - v!(z(£;e))] ~+0 1 1 exists

and

E A(a). value

is

equal

On t h e

to other

If(b~)[expEOq(b~) hand,

since

+ O~(b~)]~

v.

1

and

v!

1

for are

almost

(18) is m a j o r i z e d by lim inf e x p [ - v i ( z ( ~ ; ~ ) ) ] ~+0

Combined with

every

nonnegative,

: exp[-v~(b£)] $ i.

(18), this gives

i ~ If(b~)lexp[0q(b£) + 0~(b£)] ~ exp[-v~(b~)] ~ i.

the

140

Hence, to

v~(b~)

0

: 0

for a l m o s t

a.e., every

which means ~ E A(a).

that

on

A(a).

This e s t a b l i s h e s

The g e n e r a l to the p o s i t i v e

7.

Boundary 7A.

R

Behavior

Let

R

be a r e g u l a r

with vertices

in the M a r t i n

We d e n o t e

by

our a s s e r t i o n s

parts

of A n a l y t i c

~(a,a')

and is e q u a l

= 0

case can be s h o w n by a p p l y i n g and the n e g a t i v e

exists

Thus

~'(£)z = Oi(b~) a.e.

~i(~)

of

when

log u $ 0.

the a b o v e

log u.

consideration

[]

Maps

PWS.

We first d e f i n e

boundary

the set of

41 .

e E [0,

Let

2~)

Stolz r e g i o n s

a, a' E R

satisfying

in

be fixed.

the f o l l o w i n g

conditions: (i) (ii)

~8 E A(a); (~I~¢)(~)

exists

and

is d i f f e r e n t

from

0

for

6 =

~-l(exp(ie)); (iii)

both

P ( a , a ' ; Z 8)

and

P ( a , a ' ; b 0)

exist,

are f i n i t e

and

satisfy P(a,a';~8) (iv)

both

l i m i t s at

e ie

respectively, Moreover,

: P(a,a';be)

F0 o ~-i

where

(F 0 o ¢ - l ) ( e i e )

F0(z)

and

and

have nonvanishing

(F 0 o % - i ) ( e i 8 )

= F(a,a';z)

F 0 o ~-l(w)

~

A(a)

and

and

and

Fl(Z)

with

~

were

were g i v e n

and

finite radial

(F I o % - I ) ( e i 8 )

= F(a,a';z)P(a,a';z).

F I o %-l(w)

respectively,

S(eie;~,p)

statements,

F(a,a';z)

by

(F I o ~ - l ) ( e i S ) ,

through any Stolz region

z),

FI o ¢-i

w h i c h we d e n o t e

the f u n c t i o n s

In t h e s e

and

: k(be,a')/k(bs,a) ;

tend uniformly

as

w

tends

to

to e ie

0 < ~ < 7/2. defined

in 5B.

in 3A, w h i l e

Our p r e v i o u s

P(a,a';

observations

s h o w the f o l l o w i n g Lemma.

The

7B. included

set

Let

set

D(a)

This m a y be c a l l e d be .

We c o n s i d e r (Theorem

complement

in

[0, 27).

any Stolz r e g i o n

3D), w h e r e

0 < ~ < 7/2

S(eiQ;e,p) and

0 <

set S(bs;~,Pia)

point

has n e g l i g i b l e

e E ~(a,a').

in the

p < i, and

~(a,a')

: S(be;~,p)

a Stolz r e g i o n

To see this,

we d e n o t e

in by

= %-l(s(eie;~,p)). R

if it c l u s t e r s e(be,e)

the

o n l y at the

set of c l u s t e r

141

points

in

A

of the

is i n d e p e n d e n t in the

proof

Lemma.

set

of the

of T h e o r e m

For any

S(be;e,p).

choice

We take

a fixed

Then

Lemma

negligible

7A shows

we have

that

the

equation

with

distinct

= b 8.

any

in

A1

of

A1

(19)

Namely,

a • R

with

number

one

8 • ~l(a)

(20)

r(a~z) a unique

e(be,~) to t h a t

solution

z

(Ch.

consists

set

subset

of

[0,

27)

For any p o i n t is d e n s e

in

Martin

used

III,

2B),

of a s i n g l e We thus

we

Al(a)

be the

Al(a)

is a m e a s u r a b l e

0 < p < 1 and

such

e(a;z)

~'(R,a)

that E d

mod

for any

b8

have for

b8

subset

For any

exists

the

see

set of p o i n t s

following: there

should

point

set

the

R, we

have:

let

0 < ~ < 7/2

b' E

functions

the

satisfies

in

and

and

any

: c

R

Since

0 < ~ < 7/2.

and

of

A

with

and

with

Since a' • R.

proportional

Then

A

8 • ~l(a).

a' • A.

be f i x e d

p = p(b@,~)

similar

a' • A}.

e(be,e) ~

8 • Ql(a).

of d x - m e a s u r e

Al(a)

has

set

any

let

for all

never

subset

is a m e a s u r a b l e

Now

holds

are

the

and

Let with

~l(a)

for any

(19)

poles

e • ~l(a)

Theorem.

that

dense

: n{Q(a,a'):

complement.

e(bs,~)

see that

= k(bg,a')/k(be,a).

countable

~l(a)

with

to

By an a r g u m e n t

we h a v e

k(b',a')/k(b',a)

7C.

It is easy

p.

5B we h a v e

b' • e(bs,e)

(19)

b'

of

b8 •

a positive

system 2~

pair

of n u m b e r s

(C, d)

satisfying (21)

p £ c < i

where (20)

r(a;z) and

and has

cluster

We call

the

set

in

R

region

lows.

a). Let

there

S(ei8

point

Then,

a positive

;~l,~2,p)

set of

i(d-8)

in

(or, be this

= {z • ~:

z

for the

we d e f i n e d

system in 7B,

A. more

exactly,

and angular definition

for any number

£ ~,

solutions

S(be;~,p) , which

be

vertex

generalize

8 E ~l(a). exists

set

S(bs{~,p) with

We can

The

to the

one

Stolz

-~ £ a r g ( l - c e

: exp(-g(a,z)). is e q u a l

only

origin

7/2,

(21)

and

~i' p,

~2

measure

of S t o l z with

0 < p < i,

~I £ a r g ( l -

S(be;~,pla))

ze

2~

-i8

(with

regions

-~/2 such

a

as fol-

< ~i < ~2 < that

) £ ~2 }

the

set

142

is i n c l u d e d

in the r e g i o n

S(be;~l,~2,pia) Then,

S(bs;~l,~2,p)

D(a).

We set

: S(bs;~l,~2,p)

has a u n i q u e

cluster

v i e w e d as a k i n d of S t o l z r e g i o n s . choices

of

7D.

a

We can e x t e n d

describe

some r e s u l t s

in C h a p t e r

space

X.

Let

Let

~i' ~2

-~/2 < ~i < ~2 < 7/2,

p

the

with

intersection 0 < p < i.

Theorem.

surface

subset ~ A',

7E. points

Theorem. plane

of

R'*. A

If

A

Let

fV(b)

R, t h e n

side is t a k e n

f

f

fV(b)

for a l m o s t analytic

the

every

m a p of

for all p o s s i b l e

b E Al(a). R

of

set of

fv, we d e f i n e

such that

e ie

Then one.

exists

into a R i e m a n n R'

A'

and

b E Al(a)

A2(a)

a

for w h i c h

m a p of

is a m e r o m o r p h i c = {f(b)}

for i n s t a n c e

R

such that

according b

in

function

for a l m o s t

the

into the e x t e n d e d

w0 E [

every

to be the is t h i n at

set of b0

is seen to be a s u b s e t

is r e p e a t e d

for a l m o s t

A2(a)

R \ ¢'(R;a)

We t h e n h a v e

a point

summation

= {~(b)} If

for any

b E Al(a).

compactifieation

be an a n a l y t i c

If t h e r e the

into a

< '~i < ~2 < ~/2},

every

denotes

e E ~l(a)

of d x - m e a s u r e

¢.

We w i l l

is d x - n e g l i g i b l e .

is thin at

Corollary. on

is n o n e m p t y

a metrizable

then

w 0} < ~, w h e r e then

for a l m o s t

be a n o n c o n s t a n t

with

Al(a)

set,

-~/2 < ~i < ~2 < ~/2}.

R'*

R

put

In o r d e r to speak of be

D w \ D(a) set

on the r i g h t - h a n d

f~(b)

[CC].

S(b;el,e2,p)

and

: U{f(b;~l,~2):

R',

f~(b)

so that

b C Al(a)

fV(b)

f

of a n a l y t i c

m a p of

-~/2

(a)

polar

behavior

= N{f(b;~l,~2):

Let

(b)

distinct

= n Cl(f(S(b;~l,e2,p))) ,

We f u r t h e r

~ f~(b)

and m a y be

Stolz r e g i o n s .

fA(b)

f^(b)

Corollary.

A

and C o r n e a

be fixed,

Take any

f(b;~l,e2)

in

be a c o n t i n u o u s

a C R

S(b;~l,e2,pia).

where

on b o u n d a r y

f

always means with

be

equivalent

19 of C o n s t a n t i n e s c u

a few of t h e m here.

compact metric

point

It is not h a r d to see that

give r i s e to e s s e n t i a l l y

maps a p p e a r i n g

= ¢-l(s(eie;~l,~2,p)).

and

of the

following complex

[{g(a,z):

f(z)

to m u l t i p l i c i t y ,

~(f) n A 2 ( a ) . of b o u n d e d

every point

characteristic

b C A2(a).

=

143

NOTES The general d i s c u s s i o n in §i on Green lines and the c o r r e s p o n d i n g Dirichlet p r o b l e m is adapted from Brelot and Choquet Sario and Nakai

[6].

See also

[62; pp. 199-209].

The results

in §2 are due to Parreau

were r e d i s c o v e r e d by Hasumi

[52], a l t h o u g h some of them

[17, 18].

The B r e l o t - C h o q u e t p r o b l e m - - t h e main topic of §3--can be found in Brelot

[5;

§14].

The result in this section were obtained by Hasumi

[17, 18], except T h e o r e m 6A, due to Hayashi in Subsection

[27].

(c).

The latter is an important c o m p l e m e n t

Stolz regions on PWS's and related things sketehed

7 are d i s c u s s e d at length in Hasumi

contains a cluster value theorem on PWS's.

[19].

Niimura

[48]

CHAPTER

The this

Cauehy

chapter.

ity w o u l d them

be a m a t t e r

in a f o r m m o r e

far f r o m verse

being

Cauchy

teristic direct

of

surfaces

Cauehy

theorem--is In direct

Among

other

weak-star

closed

Unless

(= PWS),

§i.

THE

Statement IA.

a E R of

INVERSE

be

Let

the

utmost

and

prove

shall

of

of the

We

algebra

IX.

§2 the

which

are

the

integral

in Ch. in

of

charac-

hand,

state

variation,

the

surfaces

Cauchy

again

is

in-

is true

given

H~(dx )

in

for

§3.

is a m a x i m a l

L~(dx ).

stated,

R

is r e g u l a r

CAUCHY

that

The

is i n d e e d

On the o t h e r

A few a p p l i c a t i o n s

show

for any

this

theorem.

its w e a k e r

surface.

we

IX,

of

problem

analysis.

form holds

be d i s c u s s e d

Cauchy

So the

detailed

generalization

and w i l l

inverse

denotes

in the

a surface

sense

of P a r r e a u - W i d o m

of p o t e n t i a l

theory.

THEOREM

R

be a r e g u l a r and

let

{Zl,

the

critical

to m u l t i p l i c i t y .

We use

PWS w i t h z2,...}

points the

of

j:l

(a)( z)

: exp

S (a)

Green

function

be a f i x e d

g(a,-),

following

s(a)(z) : [ g(z,zj) g

situation.

rather

valid-

of R e s u l t s

fixed

Z(a;R),

thinking

type.

subalgebra

which

we are

of P a r r e a u - W i d o m

otherwise

type

in fact,

in Ch.

valid

in

their

see

shall

be d i s c u s s e d

sense,

But,

deserves

will

classical

general

Riemann

things,

inverse

in its m o s t

theorem

any h y p e r b o l i c

its

in the

As we

theorem--an

§i we p r o v e

and

taken

THEOREMS

to o u r p r e s e n t

and

not a l w a y s

Cauchy

theorem were

fitted

type.

CAUCHY

of t r i v i a l i t y .

evident

theorem

Parreau-Widom

i.

integral If t h e s e

VII.

which

(- s(a) (z)),

= exp(-s(a)(z)

we r e p e a t

notations:

s(a)(z) : '

g(a,z).

enumeration

n

[

g(z,zj),

j:n+l gn(a)( z)

= exp ( - s(a) (z))

- i[(a)(z))

Let

of m e m b e r s

according

145

and s(a)(z) n where

~

denotes

tion

[(a)

= 0.

lytic

function

which

S (a)

IB.

: exp(_s(a)(z) _ i[(a)(z)), n n

the h a r m o n i c S (a)

on

S (a))

S (a)) d e f i n e s n

Our o b j e c t i v e

in this

of

u

normalized

{ (a) over

section

_(a)) %n

(resp.

is to p r o v e

and w i l l be p r o v e d

Theorem.

be fixed.

u E LI(dx)

S

If

line b u n d l e

the f o l l o w i n g ,

Cauehy theorem

a E R

the

ana-

R.

is c a l l e d the i n v e r s e Let

by the c o n d i -

is a b o u n d e d m u l t i p l i e a t i v e

We d e n o t e n by

R.

(resp.

conjugate

(resp.

h(b)u(b)k(b,a)dx(b)

which

in 2D below.

satisfies

0

AI for any and on

h, m e r o m o r p h i c

h(a)

R, such that exists

an

lhig (a)

f E HI(R)

is b o u n d e d

such that

on

R

f = u

a.e.

AI . IC.

theorem

By the way we w i s h to state w i t h o u t for an a r b i t r a r y

Theorem. Rt

on

= 0, t h e n t h e r e

Let

R

be a (not n e c e s s a r i l y

the r e g u l a r i z a t i o n

gt(a,z)

be the G r e e n

ical p o i n t s according

in

Rt

of

R

function

for

morphic

on

then t h e r e

If

R, such that

This f o l l o w s

an

regular) and

PWS w i t h o r i g i n

Z ( 0 ; R t)

z ÷ gt(0,z),

{gt(z,w):

satisfies

h(0)

f E HI(R)

easily

Cauchy

3B, Ch.

V.

0

the set of c r i t -

w h i c h we r e p e a t

= 0

w e Z(0;Rt)})

fA I hud X = 0

and

Ihlg t(0)

such that

from Theorem

u = f

3B, Ch.

for a n y

is b o u n d e d a.e.

on

h, m e r o on

R,

A I.

V and the p r e c e d i n g

the-

orem.

2.

P r o o f of T h e o r e m 2A.

To p r o v e

w h i c h we state

IB

Theorem

in t h r e e

and

Let

Set

u C LI(dx )

exists

Rt

of the f u n c t i o n

to m u l t i p l i c i t y .

z E R.

the i n v e r s e

in the s e n s e of T h e o r e m

g t(O)( z) : e x p ( - [ for

proof

PWS.

IB we n e e d

lemmas.

some p r e l i m i n a r y

The f i r s t

observation,

is the f o l l o w i n g

146

Lemma.

^(a)( g b)

Proof.

Since

= 1 s(z)

R,

F. R i e s z ' s

u

is a n o n n e g a t i v e

orem Ch.

5A,

Ch.

III,

is a p o s i t i v e

(Theorem

harmonic

III,

s

A1

and

by

fA I s ( b ) d b d X ( b ) .

from

s

part

n : I,

s - s n.

Since 2, . . . .

finite

except

at

zj,

u = 0

on

and

consequently

R

2B.

Next

we

u = his]

in Ch.

Lemma.

Let

V = {I~I

fixed.

Then

there

VI,

< i}

exists

on

we

~',

Proof.

We c o n s i d e r

in Ch. {an:

VI,

~'

= ~.

and

~", h

on

a constant

on

R

a.e.

on

A I.

C

disk

such

h(% , ) - h(~")

/el < 1/4}

is g i v e n and

his]

S'n :

= h[Sn ]

Since

except

on

at

for

s(z)

is

z]..

Hence

[]

further

gn(~",z)

in

{Rn:

any

let

a

be

z • R \ CI(V).

n = i,

tending

to

0

that

CI(V)

~ R I.

Green

functions

for any r e a l

2,...),

decreasing

be the Then,

and

< CI~' - ~"I

for a s t r i c t l y

numbers

R

that

and

exhaustion

of p o s i t i v e

Rn,

so

2, . . . .

zero

be a p a r a m e t r i c

respectively.

h[s]

a.e.

function

Rn = R(en,a)

and

exists

5B.

a regular

We a s s u m e

s

5B,

real-valued

Sn : s(a)n and

By The-

by T h e o r e m

function

set

on

= ~g(a',z)/~g(a,z),

= {~ • V:

i.e.

2,...)

gn(~',z)

function

(1)

5B,

n = i,

Z(a;R) let

~" e V'

and

that

= 0

IP(a,~'; z) - P ( a , ~ " ; z ) I g (a)(z) for a n y

R

n = i,

to

function

s = u + U, w h e r e

as an e x t e n d e d

III,

hisS]

£ Sn,

the

that

is a p o t e n t i a l .

harmonic

s = 0

investigate

already

U

Ch.

2,. ..

converge

P(a,a';z) appearing

5D,

is a p o t e n t i a l ,

sn

and

function

of the

n = i,

Thus,

superharmonic

I) shows

is c o n t i n u o u s

Theorem

For

Ch.

function

Since

quasibounded

s n'

6F,

is a W i e n e r

= u.

it f o l l o w s

the

A I.

= s(a)(z)

theorem

h[s]

function,

a e. on

with

Let

~, for

as d e f i n e d sequence 3R(~n,a ) N

~" E V' Rn

quasibounded

with

and poles

harmonic

we have

i ~a - -~-~-~ 1

h(z)[

i~gn(~' ,z) ~gn(~",z)] ~g(a,z) ~g(a,z) ] 6 g ( a , z ) .

n

Put

h+ = h v 0

quasibounded

and

h-

harmonic

= (-h) v 0.

functions

on

Then, R

n

.

h+

and

By the H a r n a c k

have c(r)-i

h-

~ h + ( ~ ' ) / h + ( ~ '') S c(r),

are

positive

inequality,

we

147

where

the

r =

I%' - ~"I

function

h

.

and

Ih(¢')-h(¢")l

(2)

c(r)

= (3 + 4r)/(3 - 4r).

The

same

is true

of

So

~ Ih+(C ' ) - h + ( ¢ ' ' ) ]

+ Ib-(C')-h-(C")I

< 8 r ( h + ( % '') + h - ( ~ " ) ) ~gn({",z) )h(z) I ~ - ~ [ ] ~

: 8r(- 1----) I ~R 2~i

6g(a,z)

n

8rl(

-i---) 2~i S~R

Ib(z)i~g(a,z), n

where

is

1

Combining

a constant

(i) and

(2),

depending

~gn(~',z) ~g(a,z)

~R

n

.

We

v(z) v < i.

It f o l l o w s

(a)

~R

.

and

R, and

not

on

n.

function

on

R

~gn(~",z) I 6g(a,z) ~ 81r

(z)exp(-g(6',z)-

Since

the

g(~",z)).

of a m u l t i p l i c a t i v e

analytic

that

6gn(~',z) 6g(a,z)

(3)

on

= g

is the m o d u l u s

Then,

V

a,

set v(z)

with

on

we h a v e

] on

only

-

left-hand

6gn(~",z) 6g(a,z) member

.v(z)

of

(3)

)

so that

~ 81r

is the m o d u l u s

of a m u l t i -

n

plieative

analytic

function

remains

to be v a l i d

when

on z

CI(R

ranges

n over

R n.

IP(a,{';z) - P ( a , ~ " ; z ) I v ( z ) on

R.

Since

CI(V')

is a c o m p a c t

the

subset

inequality

Letting

sign

n ÷ ~, we h a v e

~ 81r

of

V, the

set of f u n c t i o n s

exp(g(~' ,z) + g(6 " ,z)) with

~',

~" E V'

R \ CI(V).

Hence,

2C.

Lemma.

R \ Z(a;R) I~)

< 1/4},

and

forms there

Let J

where

any

a uniformly exists

V

a desired

be a p a r a m e t r i c

closed

a e R

bounded

rectifiable

is fixed.

Pj(z)

constant

on the C.

disk whose curve

Set

= I P(a,~;z)d~ J

family

region

[]

closure

contained

in

lies

in

{6 E V:

148

for

z E R \ (Z(a;R) U CI(V)). (a)

Pj

a n a l y t i c a l l y to (b)

Pj(a)

(c)

Pj

R \ (Z(a;R) U C I ( V ) )

and can be extended

R \ Z(a;R). = 0.

is m e r o m o r p h i c with poles in

ity such that (d)

Then we have the following:

is h o l o m o r p h i c on

IPjig (a)

Pj(b)

Pj(b)

counting multiplic-

is bounded.

exists a.e. on

(4)

Z(a;R)

= I

A1

and

(kb(~)/kb(a))d%

a.e. on

A I.

J Proof.

Since the poles of

Z(a;R),

the function

1/4}).

if

Pj

P(a,{;z)

~ ~ ~', where

in

~'

h(~,~')

~c,h(~,~')

with

is symmetric

in

singularity at

~

and

~', is h a r m o n i c

~' = ~.

So we have

= -(~ - ~')-id~' + 5~,h(~,~'),

is an analytic d i f f e r e n t i a l

in

~' E V.

For

V.

can be c o n t i n u e d a n a l y t i c a l l y to the whole

regarded a n a l y t i c on Since

5g(a,z)

are c o n t a i n e d in

R \ Z(a;R).

the function

V, so that it can be

This proves the statement

has a pole at

Z(a;R)

Hence,

a,

Pj(a)

= 0.

counting multiplicity.

(a).

The poles of Since

J

pending only on

a,

J

and

(S) R.

In fact,

let

a

R, where

c(a,~)

is fixed and

the fact that say, when

~

~

~ E J. (a)

depends only on

a

I~I < i/4},

range over

J

5B,

~ c(a,~)e g(~'z) and

~

and is b o u n d e d when

ranges over any compact set, e.g.

z

de-

~ e

Then, as we saw in Ch. VI,

(z)IP(a,~;z)I

J ~ {~ E V: and

c

with

IPj(z)Ig(a)(z)

g on

R

Pj

is compact,

the d i s c u s s i o n in Ch. VI, 5B shows that there exists a constant

on

~' E V

1/4 < I~'I < i, we have

the r i g h t - h a n d m e m b e r being analytic t h r o u g h o u t Pj

{~} U

I~I

: - l o g I~- ~'I + h(~,~')

and has a r e m o v a b l e 6~,g(c,~')

where

R \ (Z(a;R) U {~ E V:

~, ~' @ v, then g(~,~')

for

are c o n t a i n e d in the union

is analytic on

and

g(~,z)

J.

In view of

is bounded above, by

R \ {z' e V:

Iz'l < 1/2},

c'

149

respectively.

Therefore

g(a)(z)IPj(z) I ~ {max~ej where

z

ranges

o v e r the o u t s i d e

c(a,~)}{length(J)}.e

of

{z' • V:

Iz'I

o~

,

< 1/2}.

On the

o t h e r hand, Pj is a n a l y t i c on CI(V) and so is b o u n d e d there. (a) g IPjI is b o u n d e d e v e r y w h e r e on R, w h i c h shows the i n e q u a l i t y Hence

(5).

(c) is proved. The i n e q u a l i t y

of b o u n d e d a.e.

Thus

on

(S) also

characteristic,

A1 .

such

Pj

logiPjI

is a m e r o m o r p h i c

• SP'(R),

In o r d e r to show the e q u a t i o n

a fixed parametrization continuous

shows that

i.e.

on

J

of the c u r v e

for any f i x e d

J.

(4),

Since

and thus

let

y:

function Pj

exists

[0, I) ÷ J

a' ÷ P ( a , a ' ; z )

z • R \ (Z(a;R) U C I ( V ) ) ,

be

is

we h a v e

for

z,

Pj(z) where

n = lim ~ P ( a , ~ n , j ; z ) ( ~ n , j - ~ n , j _ l ), n -~ j=l

~n,J~ = y(j/n),

be a m e a s u r a b l e

subset

b • 4', we h a v e every

n

and

and Lemma

j = 0, i,...,

n-l,

of

X(A I \ A')

-g(a)(b) j.

41

with

= i

and

S u c h a set

and

~n,n = ~n,0"

P(a,~n,j;b)

&'

exists

= 0

Let

such that,

A'

for each

= kb(~n,~)/kb(a)j

in v i e w of T h e o r e m

for

6A, Ch. VI,

2A.

Take any exists

an o p e n

eluded

in

b • &' set

R \ CI(V)

D

Then,

for any

• G(b)

0 < c < i

(see Ch.

n and

III,

4A)

and any

n, t h e r e

such that

D

is in-

n

Ig (a) ( z ) P ( a , ~ n , j ; z ) - k b ( ~ n , j ) / k b ( a ) I < for any

z • Dn

and any

j = i,...,

n.

Thus,

n Ij=l ~ g(a)(z)P(a'~n' j

;z)(~ n

'J - ~n'j-l)

for

z • Dn,

n kb( ~ ~n,~ ) j=l kb(a) (~n,j - ~n,j-l)I

e length(J). We take an i n t e g e r

nO > 0

in such a w a y t h a t

contained

in a disk of d i a m e t e r

Put

Jn,j

= y([(j-1)/n,

no.

Then,

of L e m m a f I

since

j/n]),

I~ - ~n,jI

for e a c h

j = i,..., for each

n.

y([(j-l)/n, n ~ nO Let

~ • Jn,j'

and

j/n])

-

z • Dn

with

we have,

n [ g(a)(z)P(a'~n,j;z)(~n,j j:l

is

j = i,..., n =>

in v i e w

2B,

g(a)(z)P(a'{;z)d% J

< g

g

- 6n j-i )I '

n.

150

n If

{g

_-< ~

j=l

(a)

(z)P(a,%;z) - g ( a ) ( z ) P ( a , 6 n , j ; z ) } d ~

I

Jn,j

Cs.length(J). Since for

a' + kb(a')

is c o n t i n u o u s

on

R, there

exists

an

n kb(~n, j ) l kb(~) I ~ kb(a ) ( ~ n , j - ~n,j_l ) - j k b ~ d ~ I j=l J Hence,

for

n => max{no, - f

Ig(a)(z)Pj(z)

n I}

a.e. ^(a) g

and is equal

AI

= i

a.e.

on

2D.

Proof

such that,

AI,

z E R.

f

be any p a r a m e t r i c

their

coordinate

fiable

curve

and Lemma

e-length(J) + Ca-length(J)

to

IB).

To prove

f = I J

disk

Theorem

contained

in

+ e.

g(a)pj

By Lemma

exists

2A we have

equation

(4).

[]

R.

Let

IB, we set

u(h)k(b,z)dx(b) AI

included are

for

the d e s i r e d

is a q u a s i b o u n d e d

variables

function

fj ( k b ( ~ ) / k b ( a ) ) d ~ .

so that we obtain

(Theorem

Then,

<

< E.

z E Dn, we have

the fine b o u n d a r y

f(z) for

for

kb(~ )

shown that

on

and

j kb--~7 d~I

Thus we have

V

nI

n > nl,

in

harmonic

R \ Z(a;R),

identified.

{z E V:

Let

IzI < 1/4}.

function

on

in w h i c h

points

J

be any c l o s e d

Then the Fubini

and recti-

theorem

2C show that

SJ f z dz fA I ifJ k--~--~ dz ]u ( b ) k ( b , a ) d x ( b ) = S

Pj(b)u(b)k(b,a)dx(b)

= 0.

AI By the M o r e r a tinuous hence monic f = U

on f

theorem

R, every

f point

is e v e r y w h e r e

majorant, a.e.

on

so that At.

is a n a l y t i c

[]

in

on

Z(a;R)

holomorphic f E HI(R).

R \ Z(a;R).

is a r e m o v a b l e

on

R.

Since

f

is con-

singularity

Clearly,

In view of T h e o r e m

Ifl 5E,

and

has a harCh.

III,

151

§2.

THE D I R E C T

3.

CAUCHY THEOREM

Formulation 3A.

of the C o n d i t i o n

First we e x p l a i n w h a t we m e a n

To do this,

we b e g i n w i t h two

Lemma.

Let

istic.

Then

f

Proof.

We m a y

be a m e r o m o r p h i c f

is a q u o t i e n t suppose

by d e f i n i t i o n , positive

simple

that

f u n c t i o n on

of b o u n d e d

f

uI

and

R

u2

SP'(R)

character-

functions

identically.

As we saw in Ch.

in

C a u c h y theorem.

of b o u n d e d

holomorphic

does not v a n i s h

log IfI E SP'(R).

elements

by the d i r e c t

lemmas.

II,

such that

on

R.

We have,

5A, t h e r e

exist

log If[ : u 2 - u I.

Thus we h a v e f : c . e x p ( - u I - i ~ l ) / e x p ( - u 2 - i~2) , where

c

is a c o n s t a n t

of m o d u l u s

functions

exp(-uj - i~j),

j = i,

say.

h

element

Let

of W i d o m ' s and

be a n o n z e r o

theorem

(Theorem

f2 = h ' e x p ( - u 2 3B.

Lemma.

Ifig (a)

i~2)'

If

f

has a h a r m o n i c

belongs

to

Proof.

The h y p o t h e s i s

by L e m m a fl/f2

with

fj

Since

~(a)

is a h a r m o n i c 5E, Ch.

3C.

(DCT a)

If

V).

on

f

is s i n g l e - v a l u e d ,

the

same line b u n d l e ,

A~(R,~01),

which

Setting

is a m e r o m o r p h i e

shows t h a t

exists

function f

[0

i~l)

f = fl/f2.

on

R

exists

the

because

fl : e h ' e x p ( - u l -

expression

R, t h e n

values

= i

a.e.

majorant III,

so is

f

is of b o u n d e d

of b o u n d e d

By T h e o r e m

analytic

5E,

Ch.

a.e.

on

AI

on

AI

by L e m m a

of

[]

such that a.e.

on

AI

and

characteristic.

functions,

III,

both

and t h e r e f o r e

Iflg (a)

f

2A, we h a v e

Since

u

e.g.

fl

and

exists

So, f = f2

a.e.

on

IfI ~ u, w h e r e

is s u m m a b l e

a g a i n by

i, as was to be proved.

We m e a n by the d i r e c t

a harmonic

in

Ch.

majorant

bounded.

A I.

following

2, g e n e r a t e

we get a d e s i r e d

3A, it is a q u o t i e n t

fine b o u n d a r y

Theorem

Since

LI(dx).

have

u

2B,

one.

Cauchy

theorem

for a point

a E R

such that

Iflg (a)

the

statement: f

is a m e r o m o r p h i c

majorant

on

function

on

R

R, t h e n

f(a)

: I

f(b)kb(a)dx(b)"

has

152

As

indicated

theorem that,

for a PWS,

a n d so w e m a y ious

4.

The

Cauehy

Although

statement

(DCT)

the remaining

part which

a E R {an:

n : i,

function

n = i,

2, . . . .

Un: I R n = R. Lemma. u

does

Let

on

R

g(a,.). Rn

We a l s o

set

hold

in Ch.

n

}

a shown

for a l l

a E R

a.

Var-

X.

for P W S ' s ,

we d e n o t e sense

by

tending

do n o t

there

Riemann R

theory.

decreasing

to

zero

contain

is a

surface.

In

any hyperbolic

of p o t e n t i a l

a strictly

numbers

Rie-

Let

sequence

such

that

any critical

the

points

set

R : R(~ ,a) : {z E R: g ( a , z ) > ~ }, n n n are Jordan regions, C I ( R n) ~ Rn+ I and 6g(a,z)

be a c o n t i n u o u s some

be g i v e n

always

choose

= e

We

or false

be

Type

in the

and

it w i l l

to a n y p a r t i c u l a r

for a n y h y p e r b o l i c

section,

g(a,z)

Then,

F

for

not

of p o s i t i v e

{z E R:

of the

of W e a k

is r e g u l a r

2,...}

curves

will

(DCT a ) is n o t

chapter

true

referring

(DCT)

is v a l i d

be f i x e d a r b i t r a r i l y

level

statement

simultaneously

to

of this

the

In the n e x t

without

Theorem

which

mann

surface,

(DCT)

equivalent

Direct

4A.

introduction, on PWS's.

(DCT a) a r e

write

conditions

weaker

in t h e

but a c o n d i t i o n

= 2(~g(a,z)/~z)dz.

Wiener

quasibounded

function

harmonic

on

R

function

such that

u

on

R.

IFI

T h e n we

have

(6)

- lim ~ n+°°

Proof.

We m a y

4B in Ch. have be

if

F(z)6g(a,z)

suppose

llI,

F

F

that

exists

In o r d e r is b o u n d e d . by T h e o r e m

5D,

Ch.

h[F]

is n o n n e g a t i v e . AI

a.e.

convergence is b o u n d e d

III,

and

By L e m m a

5C a n d

is m e a s u r a b l e . on

£i"

Since

Since

~

Theorem we

is s e e n

to

summable.

to s h o w t h e Then

F

a.e..on 0 ~ F ~ G

is a l s o

F(b)kb(a)dx(b).

&i

O ~ F ~ u, we h a v e

sununable,

f

=

~ Rn

of

(6), w e f i r s t

and afortiori

assume

that

F

is q u a s i b o u n d e d .

So

we h a v e

h[F]

= I

F(b)kbdX(b)" A1

By T h e o r e m for any

5A,

Ch.

III,

(7)

h[F]

outside the

there

exists

a potential

U

on

R

such

that,

~ > 0,

a compact

proof

subset

of T h e o r e m

5D,

- eU < F < h [ F ]

K

of

Ch.

Ill,

R.

+ sU

Since

shows

that

F

is f i n i t e

we m a y

assume

everywhere, U(a)

< ~.

153

Given

any

~ > 0, we take

inequality

n

(7) w i t h r e s p e c t

differential

so large to

-(2~i)-16g(a,z)

hEnCa) : I

d~n, to

Ks =C R n

that

which

SR . n

and i n t e g r a t e

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

the

of the

Since we have

hEFl
ZR n

AI

and

U(z)d~n(Z)

~ U(a),

I~R n

we c o n c l u d e

that

~Rn Since ting

U(a)

Al

is a s s u m e d

Next we c o n s i d e r F m = min{F, h[F] A m have

to be finite,

m]

the g e n e r a l

is a W i e n e r

and t h e r e f o r e

shown above,

a positive

integer

exists,

Fm(Z)d~n(Z)-

for

n ~ nO.

s > 0, a n u m b e r

Sinee

9

m ~ m 0. < u- u

F(b)kb(a)dx(b)

=

m

5B-(b),

m = i, 2,... a.e.

on

A I.

R, w h e r e

0 < I

llI,

h[F m] =

By the fact we

m = I, 2,...

and any

¢ > 0,

such that

f

Fm(b)kb(a)dx(b)l

< s

and

Fm + 9

a.e.,

there

exists,

for

such that

< I

Fm(b)kb(a)dx(b)

I

u

m

we have

= min{u,

0 ~ F ~ u

m}.

F(z)d~n(Z) - I DR

+ ~ and t h e r e f o r e that

n U m ( Z ) d Z n (z)

r

DR

It follows

Fm(Z)d~n(Z) ~R

n

u(z)dDn(Z) -]DR n

n

< u(a) - (u A m)(a). If we take

Ch.

with

AI

On the other hand, on

m}

is s u m m a b l e m 0 = m0(s)

AI

m

(6) by let-

A1

any

for

the i d e n t i t y

By T h e o r e m

for

for any

n o = n0(m,¢)

~R n

I

case.

function

Fm = min{F,

there

f

F

we have

s ÷ 0.

m ~ m0(s)

and

n ~ n0(m,s) , then we o b t a i n

F-

154

II

F(b)kb(a)dx(b) - I AI

Since

u

implies

F ( z ) d ~ n ( Z ) [ < 2e + u(a) - ( u A m ) ( a ) . ~R n

is q u a s i b o u n d e d , the d e s i r e d

4B.

we h a v e

result.

By the d i r e c t

(u A m)(a)

+ u(a)

as

m ÷ ~, w h i c h

[]

Cauchy

theorem

of w e a k type we m e a n

the f o l l o w -

ing result: Theorem.

Let

..., z I

R

be a r e g u l a r

be any f i n i t e

subset

g0(z) If

f

is a m e r o m o r p h i c

majorant

on

R, t h e n

= exp(on

R

a.e.

on

= I

surface

and let

Zl,

We set

I ~ g(zj,z)). j=l

exists

f(a)

Riemann

Z(a;R).

function

f

(8)

hyperbolic of

s u c h that AI,

has a h a r m o n i c

Iflg o

is s u m m a b l e

and

f(b)kb(a)dx(b)" A1

Proof.

Since

Theorem

4A, Ch.

surface

R

compact

set

z~

and

that

c

R \ K.

IV,

is an l.a.m, shows that

being regular, K

in

go ~ e -i

harmonic bounded

]flg 0

u

R

on

R\ K

on

Re(f)

We then m o d i f y

functions

of T h e o r e m on

R.

(resp.

fl

on

R

Since

with

III,

-

by

X(A I \ N ( f j / u ) )

f2/u = 0

j = i,

the p r e c e d i n g

if

lemma

(fl + i f 2 ) ( z ) 4 g ( a ' z ) 8R n

Im(f))

for

and

is

it is q u a s i function K

on

so as to

By a p p l y i n g

A I \ ~(Re(f)) and

set

We see

f2 ) on the s u r f a c e

n

lim ~ n+~

(resp.

and a

Zl,... ,

R \ K.

c-lu,

on the

f(z)6g(a,z) 8R

on

Re(f)

If21 S c-lu).

fl/u

R, Our

c > 0

is thus a W i e n e r

Im(f))

on

the p o i n t s

]fl ~ c - l u

(resp.

fj = (fj/u)u,

Ifjl S e-lu,

-lim ~ n÷~

that

number

contains

Im(f))

(resp.

function

It f o l l o w s

5D, Ch.

a positive

Since

majorant

is q u a s i b o u n d e d .

in m o d u l u s

G = R \ K, we see that

are n e g l i g i b l e . uous

R \ K.

(resp.

Ifll S c-lu

III, w i t h

on

Re(f)

a harmonic

Int(K)

and is m a j o r i z e d

R \ K.

the p r o p e r t y

exists

So we have

is q u a s i b o u n d e d

on

have a real c o n t i n u o u s

Ch.

there

such that R \ K.

having

u = L H M ( I f l g 0)

R

Lemma

with 5C,

A I \ N(Im(f))

are b o u n d e d j = i,

2, are W i e n e r shows that

2.

continIn v i e w

functions

155

= I

(31+ if2)(b)kb(a)dx(b)

= S

AI

f(b)kb(a)dx(b)' AI

where

{R : n = i, 2,...} is an e x h a u s t i o n of R m e n t i o n e d in 4A. If n n is so l a r g e that R includes K, t h e n f(z)6g(a,z) is a m e r o m o r n phic d i f f e r e n t i a l in z on the c l o s e d r e g i o n C I ( R ) w i t h o n l y one n p o l e at the p o i n t a, w h o s e r e s i d u e is equal to -2~if(a). H e n c e , we get the f o r m u l a

§3.

5.

(8).

[]

APPLICATIONS

Weak-star 5A.

which

(see H e l s o n Theorem.

that

extends

[33],

p.

H~(dx ) Let

algebra

of

s* = 0.

L~(dx )

that

bounded

among weak*

generated

S (a)

is b o u n d e d

on

there

and analytic

closed

and let

by

C

H=(dx ) to

we p r o v e the

on

R.

on

R.

a

function

Since

C

closed f*.

Suppose

is any f i x e d p o i n t

in

B E H~(R)

B

is c l o s e d

in

H ~ ( d x ), we h a v e

I

B(b)u(b)(f*(b))ns*(b)dx(b)

under

such R. such

is o b t a i n e d

~(R,(~(a))-l).

sub-

to s h o w that R

function

u

L~(dx).

on

Such a function in

of

be the w e a k l y *

It is s u f f i c i e n t

a nonzero

element

subalgebras

and the f u n c t i o n

C.

R, w h e r e

exists

and a n o n z e r o

t i o n by f u n c t i o n s

theorem,

in the ease of the unit d i s k

we take any m e r o m o r p h i c

IB(z) I ~ g(a)(z)

tiplying

a fact w e l l - k n o w n

is o r t h o g o n a l

is a PWS,

Cauchy

27).

To see this,

R

of the i n v e r s e

f* ~ L=(dx ) \ H~(dx)

Iulg (a)

Since

H~

is m a x i m a l

s* E LI(dx)

that

of

As an a p p l i c a t i o n

following,

Proof.

Maximality

by m u l -

Thus

Bu

is

the m u l t i p l i c a -

: 0

AI for

n = 0, i, . . . .

each

n = 0, i,...,

hn : B s * ( f * ) n m a p of

R

by T h e o r e m

with

a.e.

By the i n v e r s e a function on

Al"

%R(0)

= a.

7B, Ch.

h

Let Then

n

Cauchy

~R:

D

on

~D. of

there

exists,

for

such that

÷ R

h (a) = 0 and n be a u n i v e r s a l c o v e r i n g

h n o %R E H I ( D ) ,

h n o %R ( O) : 0

and,

III,

(h n o ~R )^ = ( ( B s * ) o S R ) ( f * o

C'

theorem

E HI(R)

Thus

(Bs*) o SR

L~(do)

generated

is o r t h o g o n a l by

H~(do)

and

SR )n

a.e

to the w e a k l y * f* o SR"

closed

Since

subalgebra

f* ~ H ~ ( d x ),

156

it is e a s y to see, H (do)

C' = L=(do) 0.

that

B ~ 0

Ill,

closed

Common

(Bs*) o SR = 0 a.e.,

Lemma.

a.e.

we c o n c l u d e

~ H~(do).

Since

L (do), we see that This

shows that

Bs* =

s* = 0, as was to be

We b e g i n w i t h the f o l l o w i n g Let

~

Proof.

Let

~C

be a line b u n d l e

common

no n o n c o n s t a n t

inner

common

Q0

over

factors

inner

factor,

common

--~{-(R,~-I~ I)

i n n e r f a c t o r of a n y e l e m e n t ~(R,~

0)

of i n d u c t i o n

2, . . . .

T h i s means,

k : i, 2 . . . . .

constant

function,

6B.

has

Q0"

Then

of

and

~ ( R , q -I) -i ) :

~{~(R,~

common

inner

0)

should

be d i v i d e d

by

Q0"

and t h e r e f o r e ~ ( R , n 0) = QoH~(R). k = Q0 H (R) for each k = i,

that

a point

)

~ ( R , ~ k) 9A,

Ch. V, that

IQ0(a) I = i

as was to be shown.

Let us c h o o s e

has no

~ (R,~

has no n o n c o n s t a n t

in v i e w of T h e o r e m

It f o l l o w s

Then

inner factor to

~(R,~

Q0

we see that

~{~(R,~)

5D).

-i) =c S{=(R,q -I),

in

contains

By m e a n s

such t h a t II,

too.

associated

~(R,n0)~(R,n-ln0

Obviously,

R

(see Ch.

be the g r e a t e s t

be the line b u n d l e

Since Q0 ~ (R,n -i ~i)f a c t o r s and since

for

f* O S R of

Inner Factors

nonconstant

the

7D, that

subalgebra

[]

6A.

let

weakly*

and t h e r e f o r e

As we k n o w

proved.

6.

in v i e w of Ch.

is a m a x i m a l

and so

Q0

is a

[]

a E R, w h i c h

is h e l d

fixed,

and set

Jn (a) = Jn = s ( a ) ~ ( R ' ( n ~n(a))-l)' S

n

= S

(a) n

and

~n = ~n~(a) for

n = i, 2, .

. (see . . IA)

T h e n we h a v e

oo

Lemma.

Un=l Jn

Proof.

First we look at e a c h

common

has no n o n c o n s t a n t

i n n e r f a c t o r of

c o m m o n i n n e r factors.

~=(R,~n) ,

~{~(R,~n) , then

were nonconstant,

t h e n it s h o u l d

n.

w o u l d have a c o m m o n

So

~(R,~

)

n = i, 2, . . . .

Q

should divide

v a n i s h at some zero,

If S n.

Q

is a If

Q

z. E Z(a;R) with j > ] c o n t r a d i c t i n g to C o r o l l a r y

157

9A, Ch. V.

This means that

~(R,~

)

has no nonconstant

common

inner

-n

factors.

By the preceding

lemma

inner factors.

Thus the function

factor of

Since the eommon

Jn"

n = i, 2,...}

are only constant

~ ( R , [ [ I)

has no noneonstant

S (a) ~is the greatest n

common inner ~(a)

inner factors of the sequence functions,

we are done.

common i~ n

:

[]

oo

7.

The Orthocomplement 7A.

space

of

H (dX)

We want to determine

Ll(dXa ).

the orthocomplement

For this purpose we fix a point

following notations

besides

those given in IA.

S n'(z) = s(a)(z)s(a)(z)-lexp(-g(a'Z)n for

n : I, 2,...

of

H~(dXa )

a E R

in the

and use the

We set - ig(a,z))

and

S'(z) whose line bundles

: S (a)(z)-lexp(-g(a~z)

are denoted by

~n

and

- i~(a,z)), ~', respectively.

We also

set

for

Let in

n : i, 2,...

Jn Jn

(resp. (resp.

III (resp. in

n

K'(a)

= K' = S'~I(R,~'-I).

and

Kn )

Lemma

lowing result

= Kn

be the set of fine boundary

functions

for elements

Kn) , whose existence has been shown in Theorem

the weak* closure Un=l = Kn

Kn(a)

3B). of

Ll(dXa)),

We further denote by U n=l ~ Jn where

concerning

in

L~(dXa )

J(a)

dXa = kb(a)dx(b).

orthocomplements,

= J

(resp.

5E, Ch.

(resp.

K(a)

the closure

= K)

of

Then we have the fol-

which will be used later and

has its own interest. Theorem.

For any PWS

R

(a)

K : H~(dXa )-L,

(b)

~ : (~')±,

we have the following:

where the o r t h o c o m p l e m e n t a t i o n (Ll(dXa), L~(dXa)). Proof.

is taken with respect to the dual pair

(a) If f E Kn, then f is meromorphie on n If(z)iexp(- [ j=l g(zj,z)) has a harmonic majorant.

R,

f(a)

= 0

By Theorem

and

4B we

158

have

I

f(b)h(b)dXa(b)

: f(a)h(a)

: 0

A1

for any

h E H~(R) •

So ,

Kn C = H ~ ( d X a )±

(9)

for e v e r y

and t h e r e f o r e

n

K ~ H~(dXa)1. N e x t we take any

i, 2~...,

f~ E K±

= L ~ ( d X a )) . (C

K'Jn C = Kn

Since

for

n =

we have

f

^^

f~hk dXa

: 0

AI for any

h E K'

theorem

(Theorem

f~k.

The

bounded u/k

and

IB) t h e r e

function

and thus

a.e.

analytic

k 6 Jn"

f*k

u 6 H~(R).

If

on

u E HI(R)

k ~ 0, t h e n

k ~ 0

f*

can be v i e w e d

as the b o u n d a r y

f, of b o u n d e d

characteristic.

R

7B, Ch.

on the disk,

are c o m p l e t e l y

is d e t e r m i n e d

f2 E H~(R)

determined

uniquely

by use of L e m m a

III, w i t h analytic

3A.

by

R' = •

f*.

a n d so

function

f* =

for an

As seen for in-

of b o u n d e d

boundary

We w r i t e

So we h a v e

u =

is also

and p r o p e r t i e s

functions

by t h e i r

u

a.e.

say

Cauchy

such that

AI, we see that

Namely,

functions

f

a function

By the i n v e r s e

function,

istic on that

exists

f*k ± K'.

being bounded

stance by u s i n g T h e o r e m analytic

Thus,

functions,

f = fl/f2

seen that

of

character-

with

so fl'

k f l / f 2 E H~(R)

~ ~ J n has, by L e m m a 6B, no n o n c o n s t a n t for any k C U n=l Jn . Since Un=l c o m m o n i n n e r f a c t o r s , the i n n e r f a c t o r of f2 m u s t be an i n n e r f a c t o r

of

fl"

It f o l l o w s

belongs closed

to

H~(dXa )

in the

space

that

f = fl/f2

belongs

and c o n s e q u e n t l y

to

H~(R).

K± ~ H ~ ( d X a ).

L l ( d X a ), we c o n c l u d e

Hence,

Since

K

f* = is

that

: ~±± ~ H~(dXa )Combining (b) and

this w i t h

(9), we get the d e s i r e d

T a k e any

f E K',

Iflg (a)

meromorphic a harmonic

has a h a r m o n i c on

R,

majorant.

(fu)(a)

so that majorant = 0

By T h e o r e m

f

and

f on

R.

If

on

u E Jn'

l(fu)(z)lexp(-[j~l

R,

f(a)

then

fu

g(zj,z))

= 0 is has

4B we h a v e

^^ AI fu dXa = (fu)(a)

So

identity•

is m e r o m o r p h i e

= 0.

f 6 J± and as n is a r b i t r a r y f 6 ~ Thus n ' ' " N e x t we take any f~ e 9 ± (~ Ll(dXa)). Since

~v C J

n

J±.

is an ideal of

159

H~(R),

h E Jn

and

u E H=(R)

imply

I

hu E Jn

and so

^ ^

A1 f*hu dXa = 0.

f*h E H ~ ( d X a )± = K

Namely,

n : I, 2 .... So there

and since

exists

the a r g u m e n t belongs

is closed

a function

used

(a) .

in view of

K'

in

u E ~/(R,~ '-I)

in (a), we deduce

such that

from this

7B.

Theorem.

Let

R

be a PWS

for every

to

K _C K,.

f*h : S'u.

observation

to ~ ( R , ~ '-I) and thus f* b e l o n g s ^± ^ ± J = K' or, e q u i v a l e n t l y , J : K'

Hence,

K n =C K'

Since

Ll(dXa ), we have

that

By

u/h

K'; namely,

5± C K,.

[]

for w h i c h

( D C T )a

holds.

Then

~'(a) : ~(a) : H~(d×a )±. Proof.

We have

is i n c l u d e d any

in

h E H~(R)

which

is equal

shown

in the p r o o f

K'(a).

and thus, to

K.

of the p r e c e d i n g

Now we take any by

(DCTa) ,

Hence,

f E K'

theorem

Then,

f£1 fh ^^ dXa

: 0"

K' ~ K, as desired.

that

K(a)

fh E K' So

for

f E H ~ ( d X a )_L '

[]

NOTES The c l a s s i c a l IV of Heins Neville tion.

[31].

theorem Cauchy

inverse

[47] has a v e r s i o n It is p o s s i b l e

pactification, false

as was

of this

to f o r m u l a t e indicated

in its full g e n e r a l i t y .

strong

restriction.

The w e a k - s t a r lished

[27].

Cauchy-Read The

note

[20].

A weaker maximality The r e s u l t s

is d i s c u s s e d theorem

depending

in [18].

version

in

in C h a p t e r

in H a s u m i

6 and

com-

theorem

[17] under

4B) is in H a s u m i L~(dx)

[18].

compactificaof W i e n e r

Cauchy

in H a s u m i

(Theorem

H~(dx)

in terms

The direct

in S u b s e c t i o n s

in detail

shown

on a n o t h e r

our r e s u l t s

It was p r o v e d of

was

is

a

[17].

is in an unpub7 are due to H a y a s h i

CHAPTER

The themes

ence of A.

theory

Beurling's

began

the

such modules and simply is strong simply

role.

tion that

every

plicative)

As a m a t t e r

function.

covering

as a preliminary. doubly

Cauchy

theorem,

Beurling a E R

subspaces

chapter,

will

and their

is not

all

simply

be given

(multi-

in Ch.

here

in §2.

the direct

subspaces

that the statements

X.

to the

is r e c a l l e d

Assuming

invariant

plays

to the condi-

over a PWS are d i s c u s s e d

show that

so

to sur-

(DCT)

by some

liftings disk

theorem

As for

extends

is equivalent

of the unit

subspaees

Cauchy

theorem

is g e n e r a t e d

on the

As usual,

subspaees.

theorem

are determined.

it is proved

invariant

the p r o b l e m

subspaces

in Ch.

PRELIMINARY

let

measure III.

sense of p o t e n t i a l

i.

Invariant

Finally,

the h a r m o n i c

as defined

51.

The case

type.

inverse

Cauchy

discussion

subspaces

surface.

we next

interesting.

are of

(DCT a ) with

are all equivalent. In this

dX

(DCT)

H~(R)

Further

invariant

invariant

type.

of

sur-

defined

invariant hand,

Gamelin Riemann

LP(dx)

doubly

the direct

of fact,

ideal

over

influ-

it very

out that the

on the other

Here

of Helson,

of P a r r e a u - W i d o m

all doubly

type.

6-closed

inner

First,

It turns

of

to see when and how Beurling's

In §i we define universal

R

the decisive

subspaces

found

into two classes:

ones.

of P a r r e a u - W i d o m

a crucial

in books

already

surface

subspaces,

We have

under

H~(dx)-modules

to d e t e r m i n e

invariant

simple. faces

are divided

enough

we have

closed

of a given

invariant

[2], as seen

SUBSPACES

has been one of the central

classes

study of invariant

only recently,

boundary

subspaces

of Hardy

paper

Although

Our aim is to c l a s s i f y Martin

SHIFT-INVARIANT

study of s h i f t - i n v a r i a n t

in the m o d e r n

and others. faces

VIII.

R on

denote AI

a hyperbolic

for the point

We assume

as before

Riemann

0--the

that

R

surface

origin

of

is r e g u l a r

and R--

in the

theory.

OBSERVATIONS

Generalities IA.

By T h e o r e m

We are going 3A, Ch.

to c o n s i d e r

IV, the map

LP-spaces

f ÷ f

gives

LP(dx)

in what

an isometric

follows.

linear

injec-

161

tion of space

HP(R) HP(R)

into

LP(dx)

for every

p

is equipped with the norm

with

((LHM(IflP))(0)) I/p ]IflIp : {sup{if(z)l: The space

HP(R)

is denoted by

A subspace is closed

M

Moreover, we set

of

if

for

LP(dx)

(weakly* closed,

invariant subspace variant)

z • R}

if

0 < p <

if

p = ~.

if

is thus identified with a subspace of

HP(dx).

H~(d X) = {f: f • H~(R)}

and

0 < p ~ ~, where the

(the quasinorm,

M

H~(dx).M

H~(R)

0 < p < i)

LP(dx), which

: {f • HP(R):

f(0) = O}

0 < p ~ =.

is called an invariant subspace if it

if

p = ~) and if

H ~ ( d x ) - M ~ M.

Such an

is said to be doubly invariant

(resp.

is dense

M.

(resp. not dense)

in

simply in-

When

p = ~,

the density should also be taken in the sense of the weak* topology w(L=(dx),LI(dx )) Let

f

of

L~(dx ) .

be any m e a s u r a b l e function on

able subset

E

of

AI

supports

there exists a m e a s u r a b l e supports

f.

the support of

M

f.

smallest measure such that

IB.

Let

% = %R

T : TR

in

in

A I.

AI

LP(da).

L~(dx)

or in M

EM

[.]

up to equivalence

set

functions,

EM ~ AI

f • M.

of the

The set

EM

is

up to equivalence.

a(~-l(A)) [-]p

be a subspace of

D

÷ R

with for

= x(A)

~(0) ~.

T

for any m e a s u r a b l e

LP(da)

M

LP(dx).

Let

{M}

M o $.

Then

with set

A LP(dx)

LP(da) T

is the set of T - i n v a r i a n t

P {M]p

is the closure

of the linear envelope of

in

M o $ : {f* o ~: f* • M}

be the invariant subspace of

H~(dx ) o $

LP(da)

(weak* closure,

if

H ~ ( d a ) ( M o $).

is equal to the set

H~(da)T

H~(da).

be a doubly invariant subspace of

is a doubly invariant subspace of

LP(da).

LP(dx); then

=

By

denotes the closure o p e r a t i o n in

LP(da)T, where

LP(da).

of T - i n v a r i a n t elements Let

f

of m e a s u r a b l e

is a m e a s u r a b l e map defined a.e. on

We note in passing that

Lemma.

Then

denotes the o p e r a t i o n of taking weak* closure in

which is generated by p = ~) in

M

supports every M

A I \ E.

L~(da).

is a subspace of elements of

$

0 < p < ~,

or in

For a set

be a universal covering map:

such that

For

Let

a.e. on

be the group of cover t r a n s f o r m a t i o n s

Theorem 7B, Ch. III, values

f = 0

Ef, of the smallest measure that

is defined to be a m e a s u r a b l e

again d e t e r m i n e d u n i q u e l y by

and let

if

say

We say that a measur-

Such a set is d e t e r m i n e d u n i q u e l y by

and is called the support of

0

f

subset,

£I"

{M}p

162

Proof.

Since the map

LP(da)

and since the linear

linear Mo $

envelope

of

f* ÷ f* o $

H0(da)T(MO

~)

is dense

dense

in

M.

[M] 2

p = =, so that

The linear

with respect

note that

H~(dx).M

envelope

in

in

L2(dx),

M ~ [H~(dx)-M]2w Since the operations L2(dX)

[lin(H~(dx).M)] 2. Theorem

[']2

(Dunford

and

Passing

of

it thus

H0(dx).M

To see this, in

M.

If

that

the lemma

is w(L~(dx),LI(dx))

topology.

-

is dense we first

[']2w

de-

then

~ [lin(H~(dx).M)]2w. coincide

for any convex

[8], p. 422), we infer that

to the universal

7B, Ch. III, we see that

M, the

implies

is invariant,

H0(dx).M

[']2w

and Schwartz

into in

This

This proves

IIn(Ho(dX).M)

to the L2-norm

LP(dx) is dense

M o $.

is w(L2(dX),L2(dx))-dense

notes the weak closure

of

H0(dx).M

in ei@{M} • Since ei@{M} P i8 P and therefore e {M} P = {M} P .

Next we suppose

of

of

is included

includes {M}p for p < ~.

in

is an isometry

envelope

covering

surface

Mo $ ~ [lin(H~(da)T.(Mo

subset M

and using

$))]2

and a

fortiori M o $ E [lin(HV(d@).(M o $))]2" Clearly,

the right-hand

side is an invariant

subspace

of

L2(d~).

So

we have

{M} This shows that

~ {M} 2 ~ [ l i n ( H ~ ( d a ) . { M } 2 ) ] 2 = ei@{M}2 . {M} 2

to the L2-elosure implies

then that

invariant. IC.

of

is doubly {M}

{M}

Here

= {M}2 nL~(da).

M

is another

Let

(resp.

S) be the support CE

invariant

and also that

In view of Lemma

2B below,

Hence,

{M}

{M} 2

is equal

the latter

fact

must be doubly

[]

Lemma. where

.

simple

fact:

be any set of measurable

(resp.

of

C S) denotes

M

(resp.

functions M o 2).

the characteristic

on

Then

AI

and let

CS = C[O $

function

of

E

E a.e., (resp.

S).

2.

Shift-lnvariant

Subspaces

on the Unit Disk

2A. A subspace M of LP(da) is invariant if and only if it is closed (weakly* closed, if p = ~) and ei@M ~ M.

163

Lemma.

If

Proof.

Let

f E M

and

quence

{hn:

n = i,

2, ... }

tend

to

M ~ LP(do)

h

2, . . . .

a.e.

If

on

is i n v a r i a n t , h E H~(do).

T.

for

By A p p e n d i x

P(T) M

hM ~ M

such

for any A.I.4

h E H=(da).

there

is a se-

IIhnII~ =< IIhlI~

that

is i n v a r i a n t ,

h f E M n

and

for

hn

n = i,

0 < p < ~, t h e n

n : i,

belongs

in

Since

l h f - hnfiP. =

vergence

then

2,...

theorem to

=< 2PIIhII~IflP E Ll(dg)

lh- hn[Plf[P

and we h a v e

h f - h f + 0 a.e. By L e b e s g u e ' s n ~ I h f - h n f i P d o ÷ 0. Since M

M, as desired.

If

p = ~,

then,

for any

dominated

con-

is closed,

hf

k E Ll(do),

l ( h f - h n f ) k [ ~ 211hll IIfll Ik I e Ll(da) for as

n : i, n ÷ ~.

Hence

2,... This

hf E M.

2B.

and

(a)

0 < p < ~, then,

for

of

M

(b) then,

and If

M

for any

Proof.

(a)

Lq(d~).

p'

sure.

of

we

show

and to

M

M

n = I, $ n},

set

{IfI

topology

÷ 0

of

L (do).

that

The a b o v e

latter let

half

we h a v e

M

the we

in

with

subspace

with

0 < p ~ ~,

Lq(do)

M ALq(d~)

is w e a k l y *

is an in-

LP-closure see that

of the

u n = max{log+Ill

always

(a),

- logn,

£ p-l]flP/nP.

0 £ u n =< p - I i f i P / n P

on

the

In e i t h e r

take 0}.

T

that clo-

Suppose

to the

contains

Since

When

LP-norm

f E M.

belongs

is w e a k l y *

statement

shows

to its

that

f

f E M.

M AL~(do)

any net

w(LP(d~),LP'(do))

spaces

we c o n c l u d e

take

in the

f ~ L~(d~).

is e q u a l

that

q < ~, t h e n

closed,

to an

of B a n a c h

shows

If

is c l o s e d

topology

LP(d~)

LP(do),

thus

LP(do)

LP(do).

is closed.

that

theory

in

0 ~ u n = log(Ifl/n) < n},

LP(do)

of

convergent

argument

is L P - c l o s e d ,

in

of

in the w e a k

in

of

is an i n v a r i a n t

M NLq(d~)

f

Since

and

N

MNL~(do)

to

subspace

M = NNLP(do).

is w e a k l y *

is c l o s e d

2,...,

subspace

and

MNL~(da)

M A L~(do)

see the

{[fl

f (hf- hnf)kdo

MNLq(do)

closure

The d u a l i t y

of

of

IlfIiq i m p l i e s

MAL~(do).

belongs

each

Lq(do)

converges

p ~ i.

closure

To

of

= p/(p - i).

Since

closure

invariant

which

closure

now t h a t

So

M NLq(do)

closure

To see that

C MNL~(do)

the w e a k

a.e.

in the w e a k *

is an i n v a r i a n t

p < q ~ ~,

IIfI!ip ~

p > i, the net with

M

is the

First

inequality

{fl}

If

is an

subspaee

space

÷ 0

h f + hf n

0 < q < p, the

variant

the

that

D

Lemma.

Lq(do)

(hf- hnf)k

means

L 2-

L 2-

case,

f

closed. any

f E M.

Then,

on the

un = 0 and

thus

For set

on the

184 I u do < p-ln-Pi[fl[P ÷ 0 T n = p as

n ÷ ~.

Let

~n

T h e n by A p p e n d i x

be the h a r m o n i c

A.2.2

in

belongs

conjugate to

of

LS(d~),

un

with

~n(0)

= 0.

0 < s < i, and

iJ[nlls ~ CsllUnill,

(i) where

c s = 2 ( s + l ) / S ( c o s ~s/2) -I/s.

Fixing

an

s

with

0 < s < i, we

see that

ill II ÷ 0 as n ÷ ~. This m e a n s that u + i[ ÷ 0 in m e a ns n n So, t h e r e e x i s t s a s u b s e q u e n e e {Un(j) + i~n(j): j = i, 2,...}

sure.

converging

a.e.

to

llhjii

and

hj ÷ i

~ i

since

h .] f E M

0.

We set a.e.

by L e m m a

2A

'

hj

= e x p ( - U n ( j ) - i[n(j)).

Since

ihjfl

Then we h a v e

= Iflexp(-Un(j))

hjf E M A L ~ ( d o )

C MALq(do).

~ n(j)

and

Moreover

=

,

if - hjfl p = i ( l - hj)fl p £ 2Plf] p e Ll(do) for e v e r y vergenee

j

f - h . f ÷ 0 a.e. 3 f I f - h j f I P d o + 0 as

theorem,

the c l o s u r e (b)

and a l s o

of

M N Lq(do)

Clearly,

l a t t e r half,

N

fn ÷ f and

in

we m a y a s s u m e , a.e.

min{Ifi,n} + I

tfl < i

for

or

n : i, 2,...

Then

I I ~ [Ifn - fiiq We n o w c o n s i d e r

for

M

con-

is equal

LP(do).

to

To see the

exists

Since

a sequence

fn ÷ f

in m e a -

if n e c e s s a r y ,

- log ]fl , 0}

Ifl > n.

Then,

if

that

i =< Ifl =< n

[fnleXp(-Un)

and

I undo =< I [l°g+Ifni - l ° g + i f I l d °

= I I + I2,

of

to a s u b s e q u e n c e

Un = m a x { l ° g + I f n l

if

Hence

so that t h e r e

iif- fnliq ÷ 0.

with

by p a s s i n g

We set

= log+Ifn I

subspaee

f E N A LP(do), M

j ÷ ~.

dominated

LP(do).

is an i n v a r i a n t

we take any

{fn : n : i, 2,...} sure,

in

By L e b e s g u e ' s

+ I

ifi>n

l°g+ifld°

say. q ~ i

the case

and

~ q-lilf n - fit2 ~

p < ~.

for

0 < q < i.

T h e n we see that

JflPdo. Ifl>n Thus,

I I + 12 + 0

be the h a r m o n i c

as

n + ~

conjugate

of

and c o n s e q u e n t l y u

for

{Un( j) + i [ n ( j ) } , w h i c h

0 < s < i. converges

T h e n by

Let

with

~ (0) = 0.

As in

(a), we f i n d a s u b s e q u e n c e

n

llUn~is ~ Cs]lUnllI

f U n d o ÷ 0.

(i) we h a v e

n

to

0

a.e.

on

~.

We set

~n

h 3• :

exp(-Un(j) - i~n(j)) a.e. and

and see that

hjfn(j)

[hjfn(j) - fl p ~ 21fl p + i e Ll(da).

c o n v e r g e n c e theorem, LP(da), we have

hjfn(j) + f

f E M

in

and t h e r e f o r e

We finally suppose enough, then

12 = 0

p = ~.

So

and thus

we see that

w(L~(da),Ll(da)).

f E M

and so

NNL~(da)

and we are done.

2C.

is bounded.

f Unda + 0

M

C M.

as

If

n + ~.

= M,

n

M.

is large

P r o c e e d i n g as

hjfn(j) + f

~ f

in

L~(da)

a.e. and

in the weak* topol-

is weakly* closed in

L~(da), we have

The converse inclusion is true as before

[]

As is easily seen, an invariant

doubly invariant

is closed in

A g a i n by the d o m i n a t e d c o n v e r g e n c e

hjfn(j)

Since

M

is included in

we have got the desired result.

hjfn(j) e M A L ~ ( d a )

theorem, we conclude that ogy

Since

N N LP(da)

f

[hjfn(j) - fl £ 21fl + i ~ 211fll + i.

hjfn(j ) ÷ f

By Lebesgue's d o m i n a t e d

LP(da).

The converse inclusion being trivial,

above,

e M AL~(da),

(resp.

simply invariant)

subspace if

M

of

ei0M : M

LP(da)

(resp.

is

eiSM

M).

Theorem. LP(da),

Let

M

be a closed

(weakly* closed,

(a)

M

S

of

T

such that

a c t e r i s t i c function of

S.

(b)

M

q E L~(da)

M = CsLP(da),

The set

up to a set of Lebesgue m e a s u r e

S

where

CS

is the char-

is d e t e r m i n e d u n i q u e l y by

with

(i)

subspace of

zero.

lql = 1

a.e.

such that

M : qHP(da).

We begin with the case L2(da).

p = 2.

Suppose first that

The function

factor of modulus one. Let

M

be an invariant

ei0M = M.

Then let

orthogonal p r o j e c t i o n of the constant function i E L2(da) in0 space M. So i- q w e q for all integers n, i.e. I T for all integers Hilbert space

n.

Since

L2(da),

(i

I

P(T) + P(~). hq E M we have

e-i0M = M Since

for any

M

{e ine}

be the

forms an o r t h o n o r m a l basis of the a.e.,

subset,

is closed in

q

to the sub-

0

say

so that S, of

and therefore that

h ~ L~(da)

qL2(da) ~ M.

q)qeineda

( I - q)q = 0

istic function of a m e a s u r a b l e implies that

M

is simply invariant if and only if there exists a function

is d e t e r m i n e d u n i q u e l y up to a constant

Proof.

p = ~) subspaee of

is doubly invariant if and only if there exists a measur-

able subset

q

if

0 < p ~ ~.

q T.

hq E M

is the c h a r a c t e r Our a s s u m p t i o n for any

L2(da), A p p e n d i x A.I.4

and c o n s e q u e n t l y

for any

To see the reverse inclusion,

h

in

shows that

h ~ L2(da). take any

f E M

So

166

which

is o r t h o g o n a l

to

qL2(da).

i- q ± ein0f

for all i n t e g e r s

Hence

a.e.,

f = 0

a null

set,

M, there

ei@M ~ M.

a function

Then

q ± qe in@

is equal to a c o n s t a n t Jql = i

a.e.

qH2(da) space for n

Since

~ M.

Then

a.e.

that

qlq2

to a c o n s t a n t Finally,

fq

a.e.

qlq2

a doubly

L2(da).

This p r o v e s

the case

Suppose

tersection of

CsL2(da) suppose (i),

N = CsL2(do) It f o l l o w s

Finally doubly (resp.

for

It f o l l o w s

that q ± fe in0

w i t h all f = 0

integral

a.e.

= [q2J

that

to the

and also

JqlJ

Hence

= i

of

a.e.

qlq2

is equal

(resp.

qH2(do)

invariant

subspace

By

(i)

subspace

or to

by L e m m a

and

M

with of

qH2(da)

2B that

of

in the s e n s e d e s c r i b e d qHP(da)

of

M.

: qH2(da)

M : N nLP(da)

CsLP(da)

subspace

or

to

lqJ : i

= qHP(da).

be the L 2 - c l o s u r e S ~ T

either

with or

subspace

2B, the in-

is the c l o s u r e

is e q u a l

M : CsLP(da) N

As for

be an i n v a r i a n t Then,

MnL2(da)

S ~ T

invariant

M

0 < p < 2.

and let

CsLP(da).

2B and the case

Jql 2

p = 2.

p ~ 2, and let

from Lemma

q) is u n i q u e

is o b v i o u s

simply)

with a measurable

simply)

that

then implies

and h e n c e

CsL2(da)

(resp.

we h a v e to show that

(resp.

with

of

as desired.

we find that

2 < p S ~

subspace

To see the u n i q u e n e s s

H2(da)

is an i n v a r i a n t

LP(da).

LP-closure,

that

i a.e. qHP(da).

Lemma

in

with a measurable

By t a k i n g

to

one,

first that

M nL2(da)

MNL2(da)

a.e.

S, up to

is o r t h o g o n a l

a.e.,

M = qH2(da). = q2H2(da)

gives

LP(da).

lql : i

belong

a.e.)

of

~ M.

which

for

Since

of m o d u l u s

0 < p S ~,

qP(~)

f E M

it is easy to see that

Let

we have

is o r t h o g o n a l

This m e a n s

lJqil2 = i

n : 0, i,... in0 is o r t h o g o n a l to e

Jq] : i

(ii)

n = i, 2, . . . .

take any

qlH2(da)

and

of

is a c l o s e d

JfqJJ2 = i, w h i c h

Our a s s u m p t i o n

and c o n s e q u e n t l y

q, we s u p p o s e Then b o t h

for

f ± qe in0 So

vanishes

M ~ qH2(da)

ei@M

Since

eieM ~ M, we have

n = i, 2, . . . . and h e n c e

Moreover, (i - q)f = 0

The u n i q u e n e s s

q @ M, w i t h

a.e.

Conversely,

qH2(da).

a.e.

obvious.

next that

exists

ei@M.

to

fq = 0

and c o n s e q u e n t l y

w h i c h we had to show.

is a l m o s t

Suppose

Then n

with

(resp.

qHP(da))

LP(da)

and a l s o that

the r e s u l t

p = 2, w h i c h has a l r e a d y

Next

Then,

= CsLP(da )

in the t h e o r e m .

a.e. by

JqJ = or

:

is a CS

The r e s u l t

follows

from

been established.

[]

167

§2.

3.

INVARIANT

Doubly 3A.

SUBSPACES

Invariant

We now c o n s i d e r

tions m e n t i o n e d Theorem.

Let

in IA. E

variant

subspace

Proof.

Suppose

H~(dx).M

the case of R i e m a n n

function

of

of

LP(dx )

that

subset

Z.

for

of

Then

AI

0 < p < ~.

Since

then

its b o u n d a r y

on a n e g l i g i b l e

subset

£ I.

So t h e r e

h(b) -I

if

b e E

for any g i v e n

such that

with

f • M

f

n

fn • H ~ ( d a ) . M .

I f - fnl ~

and

If[

t h e o r e m we h a v e M.

The case

3B.

Here

Theorem.

that

i ~ p ~ ~;

(b)

0 < p < i

Since

L e m m a IC.

3C.

let

X(Z)

M ~ CzLP(dx )

S

is r e g a r d e d

Suppose

B • S(0)~(R,(~(0))

IB

which

on

=

A I.

And

is b o u n d e d ,

n

un

implies

dominated H0(dx).M

that convergence

is d e n s e

is omitted.

in

[]

section.

Let

M

be a d o u b l y

support

of

the p r o b l e m

is also d o u b l y

M.

invar-

Then

M =

is to p r o v e invariant

for some m e a s u r a b l e

as the s u p p o r t

f i r s t that -I)

true,

{M]p

is n o w d i v i d e d

be any e l e m e n t

with

Un(b)

< i.

{M}p : C s L P ( d a )

The p r o o f

of

in this PWS.

M. only

= M

Hence

be the

in

holds:

is a l w a y s

By L e m m a

2C-(a),

Z

E ~ AI Set

and the p r o o f

is a r e g u l a r and

and

inclusion.

by T h e o r e m

p / ( p - i)

R

LP(dx)

u

By L e b e s g u e ' s n ÷ ~.

similar

if one of the f o l l o w i n g

Since

T.

of

(a)

reverse

a.e. as

is a l m o s t

in-

is c l o s e d

vanishes

otherwise

Since

~ CELP(do)

is our first m a i n r e s u l t

Suppose

iant s u b s p a c e cELP(dx )

f - fn ÷ 0

E.

: 0

The d e f i n i t i o n

llf- fnlIp ÷ 0

p = =

and

h

a set

on

M

is d e n s e

function

exists

: hu f. n

Unf • UnCELP(do) and t h e r e f o r e

denote

is a d o u b l y

H0(dx)'M

0 < lh[ < ~

[h(b) I ~ i/n

we set

CE

it is c l e a r that

h • H~(R); of

and let

M = c z L P ( d X)

~ M, we h a v e only to show that

complement

and use the n o t a -

0 < p ~ ~.

Take a n o n z e r o

negligible

surfaces

First we show the f o l l o w i n g

be a m e a s u r a b l e

the c h a r a c t e r i s t i c

and

Subspaces

of

M ° $,

i ~ p ~ ~.

Let

is o r t h o g o n a l

in the n o t a t i o n

subset

CE o $ = CS

into two cases

(a) and

described

M.

S

of

a.e.

with

p'

Take any n o n z e r o

in Ch.

by

(b).

s* E LP'(d×) to

the

and so,

VII,

IA.

If

=

168

u

is any m e r o m o r p h i c

then

Bu E H~(R)

function

and so

on

R

Buf* @ M

(2)

[ i

such that for any

lulg (0)

f* E M.

is b o u n d e d ,

Thus we h a v e

B u f * s * d X = 0. A1

By the i n v e r s e tion

Cauchy

k E HI(R)

theorem

such that

(Theorem

IB,

k = Bf*s*

a.e.

Ch.

VII)

on

A I.

there

exists

Setting

a func-

u = 1

in

(2), we h a v e

(3)

t : ~ J

k(0

On the o t h e r hand,

the f o r m u l a

~ d X : 0. AI

(12)

in Ch.

III shows that

H [ ( B f * s * ) o $] = H [ ~ f * s * ] o ~ = H[~] o ~ = k O % 6 H I ( ~ ) . For any

v E H~(D)

we thus have,

by

(3) and T h e o r e m

I~ v ( e i S ) ( ( B f * s * ) o $ ) ( e i S ) d o ( 8 )

:

By t a k i n g

LP-limits

in

I

for any B

is n o n z e r o ,

Hence

s* = 0

have

S

{M}p = C s L P ( d g ) , we c o n c l u d e

on

Z.

Lemma.

This

implies

ment Proof.

f* 6 LP(dx ),

~ e H~(dx) (Lemma)

0 < p < ~.

such that Let

a.e.

~f*

that

that on

(~s ~) o

E.

zero

s* I C z L P ( d x ) .

Since in

A I.

We thus

(a) is v e r i f i e d .

In o r d e r to deal w i t h the case Let

Bs* = 0

o n l y on a set of x - m e a s u r e

C E L P ( d X) ~ M, so that the case 3D.

0.

:

and c o n s e q u e n t l y

can v a n i s h

a.e.

~)(o)

= v(0)(kO

v - ( f * o $), we see

Since

on B

III,

((as*) o ~ ) ( e i S ) f l ( e i S ) d g ( @ ) = 0

fl E {M}p.

m u s t v a n i s h a.e.

v(0)k(0)

7B, Ch.

(b), we p r o v e

Then there

exists

a nonzero

ele-

is b o u n d e d .

u* : l o g + I f * [ .

Then

u*

is p o s i t i v e

and

summable,

for

0 ~ u~ : log If~l : p-flog If~l p ~ p-ilf~IP on the

set

R

u*

of

{If*I

> i}.

and let

v

Let

u : H[u*]

be the h a r m o n i c

be the h a r m o n i c conjugate

of

u.

extension Then

u

to is a

169

positive m. on

harmonic

R.

If

function

6

denotes

on the

R

and t h e r e f o r e

line b u n d l e

e -u

is a b o u n d e d

associated

to

l.a.

e -u, t h e n t h e r e

exists

an e l e m e n t h I ~ ~ (R,6 ) with llhl~l~ = i, for R is a PWS. -u-iv Then h l e d e t e r m i n e s a s i n g l e - v a l u e d f u n c t i o n , say h, b e l o n g i n g

to

H (R)

and we h a v e

Let whose

M

be a d o u b l y

support

an i n v a r i a n t vanishes and

N

E

subspaee

of

2C, that

The a r g u m e n t

{N} 2

zero,

namely

S

given

Since

o(S)

m u s t be d o u b l y

in 3C t h e n

shows t h a t

Since

LP(dx)

3E. LP(dx)

= [L2(dX)]p,

(b) is proved. Problem. with

Can one show that

4.

Simply 4A.

point

In w h a t

consider lowing

then

element

lemma

= X(Z)

is H~(dX)

implies

that

M

N ~ C z L 2 ( d x ).

in 3B is the s u p p o r t < I, we see,

invariant.

N

of

Hence

of

{N}2,

so

in v i e w of T h e {N} 2 = C s L 2 ( d ~ ) .

and t h e r e f o r e

) ~ M ~ czLP(dx). result

M = CzLP(dx).

So

that

M

is a d o u b l y

and that the s u p p o r t

invariant

of

M

subspace

coincides

of

with

A I.

?

Subspaces

is h e l d

R, c o n s i s t i n g a function

0 < p < i,

So we h a v e

f o l l o w s we w o r k on a r e g u l a r

a @ R, w h i c h

group of

the a b o v e Z.

we get the d e s i r e d

M = LP(dx)

Invariant

LP(dx),

[]

Suppose

0 < p < i

of

[]

N = MNL2(dx);

N = CEL2(dx)

C E L 2 ( d X) : M n L 2 ( d x

the case

Set

S i n c e any n o n z e r o

o n l y on a set of m e a s u r e

the set

~ i, as desired.

subspace

< i.

L2(dx).

have the same support,

{N} 2 ~ C s L 2 ( d o ) .

orem

invariant

has x - m e a s u r e

On the o t h e r hand, that

lhf* I ~ e-U*If*l

Q

fixed.

of c l o s e d

We d e n o t e curves

on the p r o d u c t

PWS by

Let us c h o o s e a

Fa(R)

issuing

space

R.

the f u n d a m e n t a l

f r o m the p o i n t

A I × Fa(R)

with

a, and

the fol-

properties:

(AI)

for each

(A2)

there

y E Fa(R) ,

exists

b ÷ Q(b;y)

a line b u n d l e

~

is m e a s u r a b l e

over

Q(.;yl ) = ~(yl)~(y2)-iQ(.;y2 ) for any

YI' Y2 ~ Fa(R)"

corresponding

character

Such a f u n c t i o n tive")

on

A I x Fa(R)

scribe

the b o u n d a r y

Q

Here at the

~

is c a l l e d

of c h a r a c t e r values

say that two m - f u n c t i o n s

represents

same time

(Ch.

~

and

Q2

on

AI

a line b u n d l e II,

AI~

such that

a.e.

an m - f u n c t i o n

and the

2B). (m for

and w i l l be u s e d

of a m u l t i p l i c a t i v e QI

R

on

analytic

are e q u i v a l e n t

"multiplicain o r d e r

to de-

function.

We

if t h e y h a v e the

170 same bundle

(or character),

say

~, and there exists a

Y0 E Fa(R)

such that Ql(-;y) for every

y E Fa(R).

= $(y0)Q2(-;y)

We write

a.e. on

QI e Q2

if

QI

AI and

Q2

are equiva-

lent. To each as follows: $'(a)

f e ~P(R,~)

we associate

take a branch,

= {'(R,a)

say

and denote

it by

Fa (R).

For each

tity of the group

f0(z;y) for

z E $'(a).

Then,

most every Green line

for

b = b£,

determined

£ E A(a),

a.e.

on

AI

A I.

tions

fl(z;y)

f0

and

fl

function

where

y E F a (R)

on

A I × Fa(R)

l

denotes

the iden-

we set

= $(¥)f0(z;l) 4A, Ch. VI,

% E A(a).

We set

if the right-hand

f0(%;y)

exists

side exists.

So

for al-

f0(b;y)

is

and

: ~(y)fo(b;1)

fl(b;y)

say

as above,

= ~(y0)fl(z;y)

are equivalent.

for

4B.

and

f0(z;y)

f

on the Green star region

f0(z;l),

If we choose another branch,

such that

f

by Theorem

fo(b~y) on

an m-function

f0' of

and

fl' of

f

and define

then there exists f0(b;y)

a

Y0 E Fa(R)

= ~(yO)fl(b;y);

We call any one of

~

func-

namely,

the boundary m-

f.

In the following we consider

shall see, results

are independent

the case

a = 0.

of the choice of

In fact,

a.

as we

We first show

the following Theorem.

Let

IQ(.;y)I

= i

for some Proof.

R

with

uj(z;y)

procedure

be an m-function for each

is a PWS, there exist

and

~j(b;y),

and

~2(b;y)Q(b;y)

are independent

functions

and

in

Then

of character

~

M = {f* e LP(dx):

uI E ~(R,~)

lu2(0) I ~ 0

a = 0. of

Then

7 e F0(R)

and

(Corollary

j = i, 2, be constructed

in 4A with v~

A I × F0(R)

with f*/Q

I ~ p ~ ~, is simply invariant.

lUl(0) I ~ 0

described v~

on

y e F0(R).

h E ~P(R,~-I)},

Since

~ ( R , ~ -I) Let

Q a.e.

u2 E 9A, Ch. V).

from

Ul(b;y)/Q(b;y) and define

L~(dx ), respectively.

uj

by the and

single-valued

Ul(Z;Y)U2(z;y)

is

171

also independent

of

y

and defines

We now take any If

h(b;y)

denotes

the boundary

f~(b)/Q(b;y)

and if

corresponding

to

and defines HI(R)

h(z;y)

h(b;y),

an analytic

say

u3, in

f*/Q

~ h

for an

m-function

for

h

Ul(Z;y)h(z;y)

function,

say

hl,

H~(R).

h • ~P(R,[-I).

satisfying

is the multiplicative

then

analytic

h(b;y)

is independent

in

HP(R).

:

function of

For any

y k •

we have k(b)f*(b)v~(b)

= ~(b)(f*(b)/Q(b;y))Q(b;y)v~(b)

~(b)~(b;Y)~l(b;~) = ~(b)fil(b)

:

a.e.

a function,

f* • M, so that

on

AI

(cf. Theorem

6A, Ch. VI) and therefore

;

~(b)f*(b)v~(b)dx(b)

= I

41

k(b)hl(b)dx(b)

: k(O)hl(O)

: 0-

41

Namely,

v[ ± Ho(dx).M.

I

But, on the other hand,

v~(b)v~(b)dx(b)

= I

hl

we have

v~ E M

and

u2(b;y)Ul(b;y)dx(b) AI

= I

u3(b)d×(b)

= u3(0)

~ 0,

AI for

]u3(0) I : lUl(0)llu2(0) I f 0.

M, as was to be proved. 4C. (DCT a)

In order to proceed

(Ch. VII,

Theorem.

Suppose

space of

LP(dx),

4 l x F0(R )

Thus

H~(d X).M

cannot

be dense

in

[] further,

we need the direct

Cauchy theorem

3C), which we use here as a hypothesis. that

(DCT 0) holds.

If

M

I ~ p £ ~, then there

of some character

~

with

is a simply

exists

invariant

an m-function

IQ(-;y)I

= i

a.e.

Q

on

subon

AI

such

that M : {f* e LP(dx): The proof is divided an m-function

Q

on

f*/Q

of modulus

M ~ {f* e LP(dx):

f*/Q

character

And in the second

4D. p ~ ~.

of

Q.

Let Then

were doubly

M

for some

into two parts.

A I × F0(R) ~ h

~ h

for some

be a simply

h e ~P(R,~-I)}.

In the first part we define one a.e.

h • ~P(R,~-I)}, part we verify

invariant

subspace

of

in such a way that where

~

is the

the reverse

inclusion.

LP(dx)

i

with

{M} is a simply invariant subspace of LP(do). If it P invariant, then M would turn out to be doubly invariant

172

by our a r g u m e n t L~(do)

with

in 3C.

So, by T h e o r e m

lql = i

a.e.

on

(4)

~

{M)p

c(T)q

a.e.

complex

is T - i n v a r i a n t , on

~

number

the u n i q u e n e s s

for e v e r y

of m o d u l u s

T ÷ y~

numbers

of

T

a character T E T. to Ch.

F0(R)

~ E F0(R)*

This II,

of m o d u l u s

onto ~

it by

N(z;~),

on

~

w E ~

F0(R)

7C, Ch.

by s e t t i n g

~(yT)

as a line b u n d l e

group

isomorphism

= c(T)

over

which

star r e g i o n

where

l

is the i d e n t i t y

as in 4A.

on

and

Moreover,

= IN I o ~

= N(~(w);y

R

is h e l d

for e v e r y according

fixed. 6'(0)

of the g r o u p

we d e f i n e

for some

N1

We take and de-

F0(R).

an a n a l y t i c

Then

function

then,

= N(0;I). for any

everywhere

on

F • HP(~).

in NIF

By the

= c(T)NI(W)

f* • M.

Multiplying

same t r a n s f o r m a t i o n

function

T • T,

~.

Q, we take any n o n z e r o

o b e y the

k E HI(R).

NI(0)

) = ~(yT)N(~(w);~)

N1 o T = c ( T ) N 1

T

and

%(w) E 6'(0);

(5)

HI(~),

N1

Then on b o t h

= k o

formula

(12)

in Ch.

follows

that

~ o $ : ~i ~

a.e.

(f*/i) o $ : ( f * o $ ) / ~ ] ~

on

T

= q/Nl

f* o $ = sides,

rule w i t h r e s p e c t

so that

III we h a v e

H[k o $] = H[k] o ~ : k o ~ : NIF : H [ ~ I g ] . From this

defined

is e a s i l y

III, we can d e f i n e

N • ~(R,~),

is a T - i n v a r i a n t

for some

q o T =

into the m u l t i p l i o a t i v e

By use of the c a n o n i c a l

on the G r e e n

In o r d e r to d e f i n e

q

T ÷ c(T)

N

satisfy

and t h e r e f o r e

(6)

q G

by the c o n d i t i o n s

N 1 o T(w)

Since

T

that

is a u n i q u e l y

of

branch

INII

NIF

of

is a l s o r e g a r d e d

are d e f i n e d

qF a.e. we get

shows

c(~)

g i v e n by T h e o r e m

of

q

The c o r r e s p o n d e n c e

one.

Take any n o n z e r o m e m b e r

Let

a function

2B.

a single-valued

N(z;y)

of

~ E T, w h e r e

one.

seen to be a g r o u p h o m o m o r p h i s m of c o m p l e x

Nl(W)

exists

{M}p = qHP(da).

Since

note

2C, t h e r e

such t h a t

and t h e r e f o r e a.e.

to

T,

173

This m e a n s

that

the n o t a t i o n

f*/k

is i n d e p e n d e n t

introduced

Q(b;y) Then,

for e a c h

dependent

of the c h o i c e

of

f* E M.

on

AI

Using

in 4A, we set

y ~ F0(R) ,

of the c h o i c e

: f*(b)N(b;y)/k(b). Q(b;y)

of

is d e f i n e d

f* E M.

Moreover

a.e.

we h a v e

and is in-

Q(b;y)

: ~(y) ×

Q(b;~). Let implies

u = INI . that

E A(0).

Since

u

exists

a n d is e q u a l to

We thus have,

for a l m o s t recall

](£)

every

is a b o u n d e d

for e v e r y

£ E A(0)

Theorem

u(b£)

6B,

for a l m o s t

Ch. VI, every

y E F0(R) ,

and so

IQ(b;Y)I

= 1

a.e.

on

We n o w

A I.

the e q u a t i o n

(7)

f~(b)/Q(b;y)

where

k

is d e t e r m i n e d

[NIFI/INI[ longing

uniquely

~P(R,~-I).

Lemma.

Let

Q

and

by (5).

on

Since

k(z)/N(z;y)

fe/Q

To p r o v e the r e v e r s e

for some

a.e.

£i'

Ik(z)/N(z;y)I o % =

is a s e c t i o n ,

say

h, be-

Hence

M ~ {f* 6 LP(dx):

4E.

= k(b)/N(b;y)

= IFI, we see that

to

(8)

f*/Q

l.a.m.,

~

~ [

for some

inclusion,

we f i r s t n o t e

be as in 4D and let

f* E M}.

Then

J

h 6 A'P(R,<-I)}.

the f o l l o w i n g

J = {h e ~ P ( R , ~ - I ) :

h

has no n o n c o n s t a n t

common

inner

for e v e r y

Since

I(R)

factors. Proof. band Ch.

We set

(Ch.

II,

5A) and

IV, the f a m i l y

I(R),

which

that

lq'l

k(b)/N(b;y) on

u h = P r l ( l O g lhl)

R.

Ill,

is n o n p o s i t i v e . : e x p ( u o ~). with

uh S 0

If ~

Let

f* E M, then,

preserves

(7),

This m e a n s

f* o $ = q ( ~ o $)/Nl

h E J

by

be an i n n e r (7),

and h e n c e

q'

function

by T h e o r e m

divides

Since

such

: S u

6B, Ch.

the i n n e r

(k o ~)/N I E q ' H P ( ~ ) .

E qq'HP(d~).

4A, u, in

P r l ( l O g Ik/Nl)

the i n n e r p a r t s

that

say

f*(b)/Q(b;y)

Thus we h a v e

is a

by T h e o r e m

least u p p e r bound,

q' ~ H ~ ( ~ )

P r l ( l O g I(k o ¢)/NII ) ~ u o ¢ (k o %)/N I.

for e v e r y

has the

k/N e ~ P ( R , ~ - I ) .

S i n c e the m a p

f a c t o r of use of

since

{Uh: h E J}

h E J.

f* e M

So, by is a r b i -

174

t r a r y and

since

In v i e w of therefore

is i n v a r i a n t ,

(4) we c o n c l u d e that

as desired. 4F.

qq'HP(do) u

that

vanishes

q'

p'

= p/(p - i), o r t h o g o n a l

which

independent

of

side of

f~(b)/Q(b;y)

on

R.

This proves

the l e m m a

a.e.

f* E M

on

AI

s'f*

M

and

show t h a t

We take a n o n z e r o

fixed.

Clearly,

s'f*

and then c h o o s e

for e v e r y

is o r t h o g o n a l

~ K'(0).

(9)

4C, let us take any

s~ •

it is o r t h o g -

element

h0

h0(b;y)Q(b;y)

in

is

y • F0(R).

H~(dx)-module, s'f* • K(0

to

(8).

is a r b i t r a r i l y

Take any nonzero

such that

~ qq'HP(do). P of m o d u l u s one and

[]

onal to the r i g h t - h a n d ~(R,~-I),

{M}

is a c o n s t a n t

identically

To f i n i s h the p r o o f of T h e o r e m

LP'(d~),

we h a v e

= k0

exists

a.e.

AI

H ~ ( d x ).

so that

Since

y • F0(R). to

So t h e r e on

h • J

M

=

is an

By T h e o r e m

a meromorphic

h(b;y)

7A, Ch.

function

VII

k 0 • K'(0

and thus

s * ( b ) q ( b ; y ) h 0 ( b ; Y) : s * ( b ) f * ( b ) h ( b ; y ) - l h 0 ( b ; y ) = ko(b)h(b;y)-lho(b;y).

Since

h0

dent of

and y

h

and d e f i n e s

correspondence bounded h • I

have the

same c h a r a c t e r ,

a single-valued

meromorphic

z + k0(z)h0(z~y)/h(z;y)

characteristic, in v i e w of

is the c h a r a c t e r

(9). of

k I • ~I(R,6'-I).

say Since

u, w h i c h

It f o l l o w s

is i n d e p e n -

function.

is a m e r o m o r p h i c is i n d e p e n d e n t

k 0 • K'(0)

S '(0), we h a v e

h0(z;y)/h(z~y)

So the

function

of

of the c h o i c e

of

: S'(0)~(R,~'-I),

k 0 = S'(0)(z;Y)kl(Z;y)

where

~'

for some

that

~og (JuJls,(°)J -1) : log Ih01 + log l k l l -

log

Ihl

and

Pri[log

(lulls'(°)l-1)]

H e r e we f i r s t n o t e that over

J

and since

4E, we c o n c l u d e

J

: pri[log

(IhoIIkll)]

Prl[lOg (lh011kll)] has no n o n c o n s t a n t

Consequently,

Since

h

can v a r y

inner factors

by L e m m a

that Pri[log (lulls'(O)l-l)]

namely,

~ 0.

common

Ihl].

- pri[log

lul Is'(O)I -I

has a h a r m o n i c

~ O. majorant,

we have lu(z)Ig(0)(z)

~ v ( z ) e -g(0'z)

say

v, on

R;

175

In o t h e r words, lulg (0) that

u

is a m e r o m o r p h i c

has a h a r m o n i c

[ • LI(dx)

I

majorant.

function

s*(b)Q(b;~)h0(b;~)d×(b)

hand side of (8).

4G.

: I

closed

Suppose

of

if

for

for

(Ii)

(i)

%: ~

if and o n l y

÷ R

(B-closed,

on

~

i £ p < ~.

into

qHP(~) (Ifl/l) p

dX

HP(~).

is c l o s e d

on So

in

The ease (ii)

in

p = ~

This

finishes

the p r e c e d i n g

the-

i ~ p ~ ~

and let

Then,

exists

M

a bounded

M

be a

is an H=(R) inner

l.a.m.

a harmonic

majorant}

AI,

covering

map w i t h of

(i0) or

HP(R).

Since

: 0

on

D.

f ÷ fo %

function let of

is a c l o s e d

by T h e o r e m in

T h e n the

be a s e q u e n c e

in

M

the m e a s u r e

da

on

is an i s o m e t r y

to

similarly,

statement,

M0

q

fo %

in

which

from

HP(~) implies

As that

as was to be proved.

inner

function

maps

f o ~ • qHP(D),

majorant,

HP(dx )

$

converges

so

and let

qHP(~).

Since

p = ~) H = ( R ) - s u b m o d u l e

of

of

~(0)

I o ~ = lql

the map

for any

Every nonzero

of the f o r m

{fn: n = i, 2,...}

{fn o ~}

M 0 = {f: f • M}, then

M

p = ~) H ~ ( R ) - s u b m o d u l e

To show the c o n v e r s e

=) H ~ ( d x ) - s u b m o d u l e

subspace

we have o n l y to show its c l o s e d n e s s .

can be p r o v e d

if

is b o u n d e d } .

such that

HP(~),

H~(~)

(B-closed,

Ifl/l

f • HP(R).

has a h a r m o n i c

is B - c l o s e d

IV).

(8).

HP(R)

HP(R).

admits

is a s u b s p a c e Let

to some

to the m e a s u r e

closed

Let of

every

if

be a u n i v e r s a l

converges

we set

of

if t h e r e

(Ifl/l) p

M o ~ = {f o ~: f E M}

HP(R)

Ch.

(DCT 0) holds.

an H ~ ( R ) - m o d u l e ,

Suppose

T

in

to the r i g h t -

i ~ p < ~,

be an i n n e r f u n c t i o n

which

sign h o l d s

p : ~) s u b s p a c e

We show first t h a t

is a c l o s e d

set

= 0.

p = ~,

is c l e a r l y

Let

is o r t h o g o n a l

subspaces

M = {f • H~(R):

Proof.

: u(0)

[]

M = {f • HP(R):

(ii)

M

that

HP(R)

such that,

and,

and

the f o l l o w i n g

(B-closed,

(i0)

{(b)d×(b)

s*

the e q u a l i t y

4C.

As for the i n v a r i a n t

submodule I

is a r b i t r a r y ,

Hence

of T h e o r e m

orem implies Theorem.

= 0

we see

AI

h 0 • ~ P ( R , ~ -I)

the p r o o f

u(0)

(DCT0),

and

AI Since

such that

As we are a s s u m i n g

Mn

if we note that on

M

be a n o n t r i v i a l

~.

HP(R), (weakly*

3B, Ch.

cannot

qH~(~)

q

IV

vanish

i ~ p ~ ~. closed,

(Corollary

If

if

p =

5D,

identically

on

176

any n o n n e g l i g i b l e v i e w of T h e o r e m exists

subset 3B.

an m - f u n c t i o n

IQ(';Y)I

= i

a.e.

of

M0 Q

on

on AI

such that So

f • M

f*/Q

boundary

therefore

f(z)/hf(z;7)

(12)

istic

such that, every

easily

majorant

and,

shall

see in the f o l l o w i n g

that

I

is a b o u n d e d

Lemma.

on

inner

Q0

are

Proof.

We set

characteristic

u0(z)

function

y • F0(R) ,

every

for

5E, Ch.

v 2 = ( l o g u 0) v 0, and

u2

and

u 0 = U l / U 2.

functions

III.

Q0

if,

M

on

R.

Ql(Z;y)

Q2(z;y)

Q0(z;y) We n o w show t h a t

Q9

Setting

and

characterfor al-

to

M

i ~ p < ~, is bounded.

has

(Ifl/I) p As we

inner function

the d e s i r e d

inner

if

we see

so

form.

[]

function. u0

is an l.m.m,

of

= IQ(b£;Y)l = 1 l o g u 0 E SP'(R)

and

its

for = 0

We set

v I = ( - l o g u 0) v 0,

u 2 = exp(-v2).

without

there

belongs

for

Then

l o g u 0 E Z(R).

i n n e r l.a.m,

and

f.

6B, Ch. VI,

We k n o w that

Consequently,

It f o l l o w s

= IQo(Z;y)l,

IfI/l

Hence

identically,

u I = e x p ( - v I)

are b o u n d e d

f.

= Q(b~;y)

is a b o u n d e d

by T h e o r e m

So

of b o u n d e d

of

of b o u n d e d

l(z)

p : ~,

l.a.m.

Q0

p r Q [ l O g u 0] = H [ l o g ~0 ] = H[0] by T h e o r e m

of

~ P ( R , < -I)

y • F0(R).

function

f e HP(R)

if and o n l y

lemma,

~ E A(0).

part v a n i s h e s

for e v e r y

Q0(~;y)

By s e t t i n g

= [Q0(z;¥)I and,

in

= f(z)/hf(z;y),

G0(b £) = ~0(£) = LQ0(£;~)I for a l m o s t

such that

hf,

independent

is a ( m u l t i p l i c a t i v e )

bounded

quasibounded

AI

say

meromorphic

Thus a f u n c t i o n

that the l a t t e r o c c u r s

4H.

a.e.

meromorphic

f/Qo • ~ P ( R ' ~ - I ) "

has a h a r m o n i c

~

in

4C t h e r e

h • ~P(R,~-I)}.

is also i n d e p e n d e n t

for any f i x e d

~ • A(0).

and o n l y if

invariant

By T h e o r e m

of c h a r a c t e r

an e l e m e n t ,

values

Q0(z;y)

we get a m u l t i p l i c a t i v e

is not d o u b l y

invariant.

for some

exists

: Q(b;y)

whose

most

~ h

is a m u l t i p l i c a t i v e

characteristic, that

M0

simply

A I × F0(R)

there

~(b)/hf(b;y)

f(z)/hf(z;y)

and so

and

M 0 = {f* e LP(dx): For any n o n z e r o

AI

is thus

nontrivial

Then both common

exist multiplicative

such that

u I = IQII ~

inner

uI factor

analytic

u 2 = IQ2I

and

= Ql(Z~y)/Q2(z~y).

is a c o n s t a n t

function.

As we h a v e

seen,

an

177

element

f E HP(R)

(Ifl/u0)P f/u 0

has

is b o u n d e d

be c h o s e n

belongs

a harmonic on

to

M

R.

Suppose

as q u a s i b o u n d e d

by a p p l y i n g the p r o j e c t i o n -i p uf. Since P r i ( l o g IfI)

Namely, means

the

Q2

factor

if

h E I

independently

and

f

of

h(z;y)

and y

f(z)

and

But,

Q2 as

should shown

and

hence

the

same w a y

5.

Equivalence

Q2

5A. more

must

the The

p = ~,

Since

uf

can

log Ill + v I - v 2

v I A v 2 = 0, we get IfI ) + v I) ~ v I A v 2 : 0. by

QI"

then

to

M.

By w h a t

factor

has

uf

in turn

some f

f c M}. is w e l l - d e f i n e d

we h a v e

seen

above,

I f ( z ) / Q l ( Z ; y ) I p S uf.

inner

changing

This

of

= Q0(z;y)h(z;y),

7

of

h, as was

no n o n c o n s t a n t The

ease

to be proved.

common

p = ~

for a s u i t a b l e

can

inner

factor

be t r e a t e d

constant.

in

[]

(DCT)

direct

Cauehy

following

are

of i n v a r i a n t theorem.

equivalent

( D C T )a

holds

for

(b)

( D C T )a

holds

for all

(e)

an ideal

form

for

IV, we o b t a i n

inequality

factor

be a c o n s t a n t .

by just

of

the

Ch.

for

belongs

4E,

i ~ p < ~,

and,

= f(z)/Q0(z;y)

Our c h a r a c t e r i z a t i o n

about

Theorem. (a)

of

some

H~(R)

subspaees

Namely

enables

us to

say

we h a v e

for any r e g u l a r

PWS

R:

a E R; a E R;

is

B-closed

been

shown

if and

only

if it is of the

(ii).

Proof.

(a) ~

to r e p l a c e (c) ~ included

in

(c):

0

by

(b):

This

has

in T h e o r e m

4G,

for we h a v e

only

a. Take

K'(a)

to the d u a l i t y the

divide

in L e m m a

for

R

I ~ p < ~.

can be d i v i d e d

inner

lh(z;y)/Q2(z;y)I p = Thus

4A,

if,

on

that

to the

£ 0

of

only uf,

Ifl) + v I - v 2 ~ 0,

pr I

is a c o m m o n

I = {h E ~ P ( R , ~ - I ) : In fact,

first

Ifl) + v I = v I A ( P r i ( l o g

inner

that

say

by T h e o r e m

Pri(log

Pri(log

if and

majorant,

any

and

a ~ R, w h i c h

since

the

(L~(dXa),Ll(dXa))

orthocomplement

of

K'(a)

is h e l d

fixed.

orthocomplement is

H~(dXa )

is i n c l u d e d

in

of

Since K(a)

by T h e o r e m H ~ ( d X a ).

K(a)

with 7A,

We n o w

is

respect

Ch. set

VII,

178

f

M : {h • H~(R):

| J

h(b)f(b)dXa(b)

: 0

for any

f • K'(a)}.

AI Since

K'(a) ±

plies that ideal of in I

is weakly* closed in

M

is a B-closed subspace of

H~(R),

H~(R).

L~(dXa ), Corollary

for

K'(a)

h/l

Jn(a) ~ M

was defined in Ch. VII,

6B.

Ch. VII, IA) is bounded.

M

is an

(c), there exists a bounded inner l.a.m.

M = {h ~ H~(R):

We claim that

Moreover

is closed under m u l t i p l i c a t i o n by functions

By our hypothesis

such that

H~(R).

5D, Ch. IV, im-

is bounded on

for every In fact,

So for any

R}.

n = i, 2,..., where

if

]hl/g n(a)

h • Jn(a),

f • K'(a)

Jn(a) (ef

we have

n

Ifh[exp(- [ g(',zj)) j:l

:<

Iflg (a) • lhl'Ig n(a)

where the r i g h t - h a n d side has a h a r m o n i c m a j o r a n t on

f

fh

is X a - S U m m a b l e and

A^

£i fh dXa : f(a)h(a) Therefore

By the weak

R.

^^

Cauehy t h e o r e m (Theorem 4B, Ch. VII),

,

: 0.

h E M, as desired. U co

Since

n=l Jn (a)

6B, Ch. VII, we have H~(dXa ).

In

has no n o n c o n s t a n t common inner factors by Lemma I = i

and thus

M = H~(R).

Namely,

K' (a) ± --

partieular,

I

~

d× a

= 0

At for

any

f C K'(a),

(b)

~

(a):

which

This

is

is

clearly

trivial.

equivalent

to

( D C Ta) .

[]

NOTES There have a p p e a r e d so many papers on invariant case of Hardy classes~ with different

flavor are found in Gamelin

[34] and others. Forelli

the origin is Beurling

subspaces.

In the

Detailed accounts

[i0], Helson

[33], H o f f m a n

The case of compact b o r d e r e d surfaces was studied by

[9], Sarason

[60], Voichick

[66, 67] and Hasumi

As for doubly invariant subspaces, in Hasumi and Srinivasan [17].

[2].

[23].

As for simply invariant

was proved in Hasumi

an abstract general t h e o r e m is

The case of Riemann surface is in Hasumi subspaces,

the main t h e o r e m

[17] under a s t r o n g e r assumption.

sion is due to Hayashi

[16].

[27], t o g e t h e r with T h e o r e m 5A.

(Theorem 4C)

The present ver-

CHAPTER

IX.

In the invariant

preceding

subspaces

investigation these

CHARACTERIZATION

chapters

on PWS's.

further

we h a v e

We are n o w

a n d to e x h i b i t

OF P A R R E A U - W I D O M

studied going

intimate

Cauchy

to p u s h

TYPE

theorems

and

our p r e v i o u s

relations

existing

among

things. Our o b j e c t i v e

inverse

Cauchy

speaking, those the

two

OF S U R F A C E S

in this

theorem.

chapter

Here

we find

PWS's

are

characterized

for w h i c h

the

inverse

direct

Cauchy

theorem

exist

PWS's

for w h i c h

which

PWS~s

satisfy

Chapter

VIII,

5A,

(DCT).

(DCT).

theorem As we

holds.

shall

Thus

We h a v e

already

emphasis

we l o o k

fact

hyperbolic

fails.

the

First

a remarkable

among

Cauchy

(DCT)

so that

is t w o f o l d .

see

that,

should

the

roughly

Riemann

surfaces

Next

investigate

we

in C h a p t e r

it is n e c e s s a r y had

into

X,

as

there

for us to tell

a criterion

for this

n o w be p l a c e d

on o t h e r

in

conditions. In

§i we

terizes 2D. R

show that

PWS's,

And

in

satisfy

(DCT).

i.

to a v o i d

that

nonconstant

THE

R

INVERSE

IA.

point

fix

map

Martin

R

point

boundary da(9) 0

formations

A

: d8/2~

(Theorem for

~.

of on 7B,

role

theorem

which

will

F0(R)*

hyperbolic

stated

make

be p l a y e d

corresponding

complication

AND

SURFACES

notations.

covering

is a m e a s u r e - p r e s e r v i n g

sure

value

Riemann

in

a given

PWS

by c e r t a i n to

we a s s u m e

charac-

R.

through

surface

this

carrying

OF P A R R E A U - W I D O M

TYPE

Result

some

a universal

a mean

essentially

functions.

THEOREM

of the M a i n

We first

~ = ~R

principal group

a connected

analytic

theorem

of c o n d i t i o n s

unnecessary

CAUCHY

Cauchy

being

a couple the

denotes

bounded

Statement

and

There

inverse

on the c h a r a c t e r

In o r d e r

§i.

crucial

~2 we p r o v i d e

functionals

chapter

the

the

of the with

T Ch.

unit

respect

and

Let D

0

~ R

7C,

Let Ch.

be a f i x e d with

~(0)

circumference

T

TR III,

measure

be the TR

dX

group

point = 0.

in

the

Lebesgue on

R

Then

into

to the n o r m a l i z e d

the h a r m o n i c

III).

By T h e o r e m

map

AI

of c o v e r

is c a n o n i c a l l y

meaat the

transidenti-

180

f i e d w i t h the f u n d a m e n t a l A measurable acter

~ • T{

a c t e r of

Q

if

Q

F0(R). on

Q O T = ~(T)Q

is d e n o t e d

is an m - f u n c t i o n 0 < p ~ ~

group

function

Q

by

T a.e.

6Q.

on

T

with character

= {f • H P ( ~ ) :

~

~

an m - f u n c t i o n

for any

An i - f u n c t i o n

and for any c h a r a c t e r HP(~,~)

is c a l l e d

TR

T • T R.

The c h a r -

with character

such that

of

with char-

IQI = i

~

on

a.e.

For

we set

f o ~ = ~(T)f

for any

T • T R}

and HP(do,~)

= {f E HP(do):

fo T = ~(~)f

For the sake of s i m p l i c i t y function

on

stance,

F • HP(D,~)

notes

~

is c a l l e d p - o u t e r

means

if

nonconstant

common

for all

with

acter

p

HP(~,~)

acter common

HP(dx,Q)

inner factor

IB. R

on

T E TR}.

for a h o l o m o r p h i c on

while ~.

~.

For in-

F • HP(do,~)

A character

equivalently,

HP(do,~))

outer

G i v e n an i - f u n c t i o n

Q

~

de-

of

TR

has no

if it is p - o u t e r on

~

with char-

is c a l l e d

R

Q'

of

We now b r i e f l y

for

and

recall

g z E R t.

theorem

t(O)

If

meromorphie then there

q

IC,

u* • LI(dx) on

VII)

an

on

~

Cauchy

t h e o r e m on PWS's.

PWS and

Rt

V.

gt(a,z)

l.a.m,

satisfies

Let

the r e g u l a r i z a t i o n

f • HI(R)

be the G r e e n

points

according

in

Rf

of

to m u l t i p l i c i t y .

w e Z(0;Rt)}) on

R

and the i n v e r s e

Cauchy

the f o l l o w i n g :

fA I h ( b ) u * ( b ) d x ( b )

h(0)

so that

HP(dx,Q).

the set of c r i t i c a l

states

char-

the g r e a t e s t

is a g a i n an i - f u n c t i o n ,

w h i c h we r e p e a t

R, such that

exists

3B, Ch.

is an i n n e r Ch.

Q

for

(z) = exp(- [ {gt(z,w):

Then

(Theorem

if the c o r r e s p o n d i n g

the i n v e r s e

regular)

Z ( 0 ; R t)

the f u n c t i o n z ÷ gt(0,z), Set q = gt(0)IR , where

I)

e x p r e s s i o~n

sense of T h e o r e m Rt

standard

HP(do,~

is the s t a n d a r d

in the

fo $ • QHP(do,<~I)}.

For any i - f u n c t i o n

be a (not n e c e s s a r i l y

function

(ICT)

~,

It is c a l l e d

: {f • LP(dx):

is p-outer.

H P ( d x , Q Q ')

for

(or,

for any

~Q, we set

An e x p r e s s i o n

of

~

function

on

function

factors.

i ~ p ~ ~.

HP(dx,Q)

Let

value

a function

boundary

inner

on

we use the same l e t t e r

and for its b o u n d a r y

the c o r r e s p o n d i n g

a.e.

= 0

and

such that

lhlq u* = f

= 0

for any

is b o u n d e d a.e.

on

on AT"

h, R,

181

Let

Q

IQI = q O ~. i.e.

be the inner function on

•

Then the character

is identified

~Q : ~q h

says that

h O @(0)

{T(0): tion

T • T R}

D,

{T(0):

ulus one such that and

to

(hO ~)(T(0))

of

T~

onto

= { F e H (D):

= 0

for any

so that we can find a Blaschke

with the same zeros as

belongs

isomorphism

Q(0) ~ 0

and

with that of F(0)

Since

BO m

product

Thus,

namely,

have no common (~,n

~Q);

B

zeros,

for

We also

So

ho

h • H (dx,B/Q).

follows that the inverse

Cauchy theorem is expressed as

(i)

H~(dx,B/Q) ± ~ HI(dx),

where the orthocomplement

of mod-

~.

0 ~ Z(0;Rt).

in other words,

whose

product

n(T)

has character

func-

B

is a Blaschke

is formed with respect

The

= 0}.

• • T R.

B, there exists a complex number

B o T = n(T)B; B

(B/Q)H

T • TR}.

q,

F0(R).

is a subset of the zeros of the bounded analytic on

zeros are exactly

Q

such that

(hO ~)Q E H ~ ( ~ )

: 0, we have

(ho ~)Q

see that

Q

under the canonical

condition on Since

of

It

to the dual pair

(LI(dx),L~(dx)). The inverse LI(dx)

Cauchy theorem shows that the invariant

given by

H~(dx,B/Q)

H~(dx )±

~ HI(dx,B/Q),

has a special we infer from

form.

In fact,

subspace

in

since we have

(i) that

HI(dx,B/Q) ± ~ H~(dx,B/Q) ± ~ HI(dx) and therefore

that

HI(dx,B/Q) ± ~ HI(dx ) N L ~ ( d x ) : H~(dx ).

that

H~(dx) ± ~ HI(dx,B/Q).

have

{H~(dx)±} I ~ (B/Q)HI(dc~); namely,

simply invariant IC.

subspace

Using the notations

of

This means

in Ch. VIII, we thus

{H~(dx)±} I

is seen to be a

Ll(do).

Using the notations

in IA, we now state our main result of

this section. Theorem.

Consider the following

mann surface

R

three conditions

carrying nonconstant

(a)

R

(b)

There exists an inner l.a.m,

for a hyperbolic

bounded analytic

Rie-

functions:

is a PWS.

Cauchy theorem

(ICT) is valid:

I

namely,

q if

h(b)u*(b)dx(b)

on

R

for which the inverse

u* E Ll(d×)

satisfies

0

41 for any on

h, meromorphic

R, then there exists (c)

on an

H~(dx )± ~ HI(dx,Q)

R, such that f E Hl(R)

h(0)

= 0

such that

for some i-function

and u* : Q

on

lhlq a.e.

is bounded on

T, where

AI .

182

H ~ ( d x )i : {f* E LI(dx):

[ J

Then,

we h a v e If

are

= 0

for any

h C H~(R)}.

(b) <=> (c).

separates

the

points

of

R,

then

all

these

conditions

equivalent.

ID. show

We h a v e

explain

our

given

(c) ~

for r e a d e r s '

(and h e n c e

all)

ml(a)

(b) ~

(c).

proof

is not

convenience.

The p r o o f

the

of

R

(c) ~

a E R, w h e r e

instead

of

result

(b)

if m(a)

m(a).

was

More

defined

a mean

in in

if)

generally,

we

3A.

is devalue

The

3B-3J.

m(a)

> 0

in Ch.

V,

to

first

subsection

being

is g i v e n

(and o n l y

only

so short,

is s t a t e d

assumption

is a PWS

So we have

The next

the m a i n

additional

V, a s u r f a c e

now w r i t e

(a) ~ As the

observations,

in 2D.

(a) u n d e r

in Ch.

some

plans

in IB that

implication.

to p r e l i m i n a r y

theorem

saw

seen

the r e m a i n i n g

voted

of

(a) ~

H~(R)

~f*dx AI

proof

As we for 8A.

we use the

We

follow-

ing n o t a t i o n s : mP($,a) for

a E R

and

= sup{if(a)I:

shall

finally

separates steps,

on

that

R

(i)

measure

and

of the

dx).

IIflIp,a =<

i)

R.

This

~ K

will

for any

the

(iii)

every

outer

if

Blaschke

(c) h o l d s

and

by showing,

character

of e v e r y

character

to the proof,

of

~, w h e r e

finite

F0(R)

if

H~(R)

in t h r e e K > 0

Blaschke

product

can be a p p r o x i -

products.

we n o t e

of the o r i g i n

To see this,

> 0

be d o n e

character

of f i n i t e

choice

6 E F0(R)*}.

m ~( 0)

3J that

(ii)

proceeding

independent

in

m~(6,0)

by c h a r a c t e r s Before

= inf{mP(~,a):

of

constant;

is outer;

mated

prove

the p o i n t s

is a f i x e d

, ~),

~ E F0(R)* , and also mP(a)

We

f E ~P(R

(or,

that

the c o n d i t i o n

equivalently,

(c)

is

the h a r m o n i c

we d e f i n e

N a = {f* e L l ( d X a ): J r

hf*dx a = 0

for any

h E H~(R)}

AI for

a E R, w h e r e

i.e.

dXa(b)

HI(dx,Q) simply

for

b E A I.

such

that

is the h a r m o n i c

= kb(a)dx(b) some

invariant.

for

dXa

(Ch

i-function N o w take

By use

III

•

Q

any

of H a r n a c k ' s

c -I ~ s ~ c

a.e.

on

~

measure

2C)

on

T

Then

.

R

and

inequality This

we

at the p o i n t

it is c l e a r

if and o n l y

a, a' E R

A I.

of

set

{Na} I

s(b)

find

implies

if

that

N a =C is

= kb(a')/kb(a)

a constant

that

sNa,

of

dXa , = sdxa , and

therefore

that

c > O

= Na,

A

because

a,

{Na} I : (s o @ ) { N a , } l .

183

Since the multiplication of

by

so $

Ll(do), we conclude that

only if

eiS{Na'}l

defines a bicontinuous

ei@{Na} I

is Ll-dense

in

is Ll-dense in

a E R,

H=(dXa )± ~ HI(dX,Qa )

I

(c), IC, holds.

to the assumption that, for every

for some i-function

H~(dXa )± = {f* E LI(dx):

if and

{Na,} I, as was to be proved.

In what follows in §i, we assume that the condition As shown above, this is equivalent

isomorphism

{Na} I

hf*dXa = 0

Qa

for any

on

T, where

h C H~(R)}.

A1 IE.

Remark.

We just look into the excluded case in which

contains only eonstant functions. tions

f* E LI(dx )

function in

We also know that every

So if the condition AI

(e), IC, holds, then

cannot be divided into three disjoint

subsets of positive measure.

most two linearly independent harmonic this matter further.

2.

H~(R)

consists of all func-

cannot vanish on a set of positive measure with-

identically.

the Martin minimal boundary measurable

H~(dx) ±

~A I f*dx = 0.

such that

HI(dx,Q)

out vanishing

Then

This means that functions.

R

carries at

We will not go into

A Mean Value Theorem 2A.

We now begin our preliminary observation

(b) and (a) in Theorem IC. surable function,

say

QA' on

By the condition T

{H~(dx)±}I Since

{Hm(dx)±}l

with

for the proof of (e)

(c) there exists a mea-

IQAI = I

a.e. such that

: QAHI(d~).

is invariant under

TR,

QA

is seen to be an i-func-

tion and H~(dx) ± ~ HI(dx,QA ), the right-hand acter of

Q~I.

inner factor of

side being a standard expression. Then

6A

is 1-outer.

HI(d~,~A).

Let

To see this, let

~A

be the charQ'

be a eommon

Then

H~(dx )±o $ ~ QAHI(do,{A ) ~ QAQ'HI(d~) and therefore

QAHI(d~)

= {H~(dx)±}I ~ QAQ'HI(d~).

Hence,

Q'

should

be a constant function, as desired. We set I 0 = HI(dX,QA)±~ which is seen to be a weakly ~ closed ideal of

H~(dx ) .

This ideal plays an important role in the following inves-

184

tigation. In the following, [.] denotes the closed (weakly* closed, P if p = =) linear envelope in the space LP(dx). First we show: Lemma.

H~(dx )± = [10HI(dx,QA)] I.

Proof. If f* @ I0, h* E HI(dx,QA ) and longs to HI(dx,QA ) and thus f* ± h'u*.

u* E H~(dx), then h'u* beIt follows that f'h* is

orthogonal to H~(dx ). Namely, [IoHI(dx,QA)]I ~ H~(dx) ±. Next we take any f* E L~(dx) which is orthogonal to the set IoHI(dx,QA ). Then, for any h* E HI(dx,QA ), f'h* C I0 ± : HI(dx,QA )±± = HI(dx,QA ), which implies (f* o $)(HI(dx,QA ) o $) ~ HI(dx,QA ) o $. Since

~A

is 1-outer, we have

f* E H~(dx ). So obtained. [] 2B.

Lemma.

(f* o $)QAHI(do)

[IoHI(dx,QA)]I ± ~ H~(dx)

I0 o $

$ QAHI(do)

and thus

and the desired result is

has no nonconstant common inner factors.

Proof. Let Q' be the greatest common inner factor of I 0 o $. I 0 o $ ~ Q,H~(do) and therefore I 0 ~ H~(dx,Q'). So we have

Then,

10HI(dx,Q A) ~ H~(dx,Q')HI(dx,QA ) ~ HI(dx,QAQ'). Using the preceding lemma, we get QAHI(do) 2C.

= {H~(dx)±}I ~ QAQ'HI(do). Here we determine

Lemma. Let Q p ~ ~ and let HP(dx,QQ'), (2)

H~(dx )± ~ HI(dx,QA Q') Hence,

Q'

and so

is a constant.

[]

HP(dx,Q) ±.

be an i-function such that H=(dx,Q) ~ {0}. Let i Q' be an inner function on ~ such that HP(dx,Q)

the right-hand

side being a standard expression.

=

Then,

HP(dx,Q) ± = [IoHP'(dX,QA/QQ')]p,

and (3)

[IoHP(dx,Q)] p i = HP'(d X 'QA/QQ, ) ,

where p' = p/(p- l) and the orthocomplement the dual pair (LP(dx), LP'(dx)). Proof.

is formed with respect to

In view of Lemma 2A we have IoHP(dx,Q)HP'(dX,QA/QQ')

~ IoHI(dx,Q A) ~ H~(dx )±,

185

which in turn implies (4)

10HP'(dX,QA/QQ')

~ HP(dx,Q) ±

and (5)

HP'(dX,QA/QQ ,) =c [10HP(dx,Q) ]p ±.

Next, let

s* • [10HP(dx,Q)]p ±

If

f* • HP(dx,Q)

and

h* • I0, then

f*h*s ~ • H~(dx) ± ~ HI(dx,QA ). We employ the fact f* o $ = QQ'F with ,~QQ,-I) -I F • HP(do and get (s* o ~)(h* o ~)QQ'F e QAHI(d~). Since ~QQ, is p-outer, we have Lemma 2B,

I0 o $

(s* o $)(h* o $) • (QA/QQ,)HI(do).

has no nonconstant

s* o $ • ((QA/QQ,)HI(do)) NLP'(do) belongs to

HP'(dX,QA/QQ').

Moreover,

by

common inner factors and therefore

= (QA/QQ')H p (do).

Consequently,

s*

Combining this with (5), we have the T

equality

(3).

On the other hand,

suppose

s* • [10HP (dX,QA/QQ')]p, ±

If h* • I0i then h's* • HP'(dX,QA/QQ') ± = [10HP(dx,Q)] p ~ HP(dx,Q). Since I 0 o ~ has no nonconstant common inner factors, we get s * • HP(dx,Q). Namely, [10HP'(dX,QA/QQ')]p, m ~ Hp (dx,Q) , which, together with (4), implies the equality 2D.

~

such that

Qg(0) ~ 0

and

Let

Qg

be

IQgl = exp(_g 0 o ~),

g0(z) : g(0,z) is the Green function for R with pole 0. Let denote the ideal of H~(R) consisting of functions vanishing at

the origin 0. Since R lytic functions, we have

is assumed to carry noneonstant

H~(R) We see that

(exp g0)H~(R)

therefore that function

Q~

~

! H~(dX,QgQg)

Let now

QC

dard expression

consists of bounded functions on

R

and

By Lemma 2C there exists an inner

such that ±

= H~(d×,Qg

)±

=

[IoHI(dx,QA/QgQg)]I, ,

is the standard expression for be an i-function

for

bounded ana-

~ {0}.

H0(d X) = H~(dX,Qg). on

H0(d X) where

[]

We are now able to prove a mean value theorem.

the inner function on where H~(R)

(2).

HI(dx,QA/QgQ~)

such that and let

H~(dX,Qg)

HI(dX,Qc -I) ~C

is the stan-

be the charaeter of

QC" As we have i • HI(dX,Qc-I), so we see that QC represents inner function on D. We may suppose, in what follows, that Qc(O) > o.

an

186

Thus every element f* in H0(d X) is regarded as the boundary function of a meromorphic function f, of bounded characteristic, on R, so that its value at 0 is determined. Theorem.

For every

f* E H0(dx )±

we have

f(0) = I

f*(b)dx(b)" AI

Proof. Since H~(dx )m = [10HI(dX,Qc-I)]I ~ HI(dX,Qc-I), each f* in Ho(dx )± determines an F E HI(~) such that f o ~ = Qc-IF, where f is the meromorphic extension of f* into R. We define a linear functional L on H0(dx )± by setting L(f*) = F(0). Since we have IL(f*)I = IF(°)I ~ S

IF(eiS)Id°(8)

t = J

If*(b)Idx(b)

: I T

IQc(eiS)-iF(eie)id°(e)

: IIf*IJl,

AI the Hahn-Banach theorem says that L has a norm-preserving extension to LI(dx). So there exists a function u* @ L~(dx ) such that Iiu*II = IILII < i and =

F(D) : I

f*(b)u*(b)dx(b) AI

for every f* E H0(d X) ± . belongs to H~(dx )± and that I

If f* E H ~ (dx) ± and h E Ho(R) , then ~f* (fif*) o $ = (~o $)(f* o $) = (~o $)Qc-IF, so

hf*u*d X : L(hf*) = (hO %)(O)F(0)

: 0.

AI As h E H~(R) is arbitrary, f'u* E H~(dx )± ~ HI(dX,Qc-I). Therefore, Qc-IF.(u * o $) E Qc-IHI(do), where F ranges over (I 0 o $)Hl(do,~C ). Since (I 0 o $)Hl(do,~C ) has no nonconstant common inner factors, we conclude that u* belongs to HI(dx ) and consequently to H~(dx ). It thus follows that, for any f* E H~(dx )±, (6)

F(0) = S

f*u*dx = I Al

f*(u*-u(0))dx

+ u(0) I

£i

: u(0) S

f*d X AI

f*dx. A1

Since

HI(dX,$c )

has no common inner factors,

it has no common

187

zeros. over, an

Namely, since

there

I0 o $

h e I0

with

exists

an

F 0 E Hl(d~,[C )

has no noneonstant h(0)

# 0.

Let

common

f~ e LI(dx )

(~ o ~) Qc-IF0 " Then, f~ E H0(d ~ X) ± and fl is the meromorphic extension of f[ corresponding FI(0)

~ 0

we have any

to this

f[

and hence, F = QC

u(0)

~ 0.

and so, again by (6),

f* C H0(d X)

FD(0)

be defined

by

# 0.

More-

there

is

f{o $ =

fl o ~ = (hO ~)Qc-IF0 , where into R. The function FI

is thus equal to

by (6),

with

inner factors,

(hO %)F 0.

Consequently,

In particular, QC(0)

= u(0)

for

~ 0.

f* ~ i

Hence,

for

, S

f*dx

= u(0)-iF(0)

= QC(0)-IF(0)

AI =

as was to be proved.

3.

= f(0),

[]

In the proof we have Corollary.

(f o ~)(0)

shown the following

IQc(O) I # o.

Proof of the Main Theorem 3A.

We are now in a position

Theorem. Let

Let

phic on

Let

can find an function h

satisfy

R, such that

exists

Proof.

Let

be the inner l.a.m,

u* 6 LI(dx )

there

Then hand,

qc

h*

an

Ihiq C

is bounded such that Since

F E Hl(da,~C )

such that

being bounded,

F

function

h(b)dx(b) A1

the last equality hypothesis

is arbitrary, proved.

[]

we conclude

= 0

on

R

u* = f

q c ° % : IQCI.

for any

and a.e.

h(0) on

h* o $ = Qc-IF

on

h, meromor= 0~ then

A I.

R

with

= I

a.e.

on

T.

we The

i.e.

F E H~(da,~C )D. h O # = Qc-IF on

h* : h a.e. on 2D implies h*(b)dx(b)

On the other

A I.

= O'

A1

sign being true because

in the theorem

such that

is also bounded,

is bounded on R and h* E H~(dx) l, Theorem = I

R

(c) ~ (b).

HI(dx) ± ~ H0(d X) ± ~ H I (dX,Qc-I),

h* 6 HI(dx) m.

h(O)

on

/£I h(b)u*(b)dx(b)

f E HI(R)

be a meromorphic lhlqc since

to prove the implication

h* A I.

and thus

/A I u*h*d X = 0.

that

belongs

u*

to

So As

HI(dx),

h

satisfies h* C HI(dx) ±

as was to be

the

188

3B.

We now begin the proof of the implication

assumption that H=(R) separates the points of we have to look into HP(do,<) more closely.

R.

(e) ~ (a) under the For this purpose

Lemma. Let ~ • T[ with H~(do,[) ~ {O} and let Qp be the greatest common inner factor of HP(do,[), i ~ p ~ ~. Then Qp are the same up to constant

factors of modulus

one.

Proof. Take any nonzero F • H~(do,[) and set Q : [F[/F. Then Q is an i-function with character [-i. So QHP(da,[) consists of T Rinvariant functions and HP(dx,Q) o $ = QHP(do,6). Our definition of Qp shows that HP(dx,QQp) is the standard expression for HP(dx,Q), i p ~ =. Since Qp divides Qq for p < q, we have only to show that Q~

divides

QI' i.e.

QI/Q~

is an inner function.

By Lemma 2C

HI(dx,Q) ± = [ 1 0 H ~ ( d X , Q A / Q Q I ) ] (7) H~(d×,Q) ± : [10HI(d×,QA/QQ Since in

H~(dx,Q)

H=(dx,Q) m.

)]i .

= HI(dx,Q) n L~(dx ), we see that Combined

this with

HI(dx,Q) ±

is Ll-dense

(7), we have

H I (dX,QA/QQ I) ~ [10H (dX,QA/QQI)] I = [HI(dx,Q)±]I = H=(dx,Q) ± = [10HI(dx,QA/QQ~)]I . Since I0 o $ has no nonconstant common inner factors, we infer that H I (dX,QA/QQ I) ~ H I ( d X , Q A / Q Q ) . Since QI divides Q~, we thus have HI(dx,QA/QQI ) = HI(dx,QA/QQ ). Consequently, H=(dX,QA/QQI ) = HI(dx,QA/QQI ) A L ~ ( d x ) = HI(dx,QA/QQ

) NL~(dx)

= H=(dX,QA/QQ

)

and therefore (8) Since

[IoH~(dX,QA/QQI)]~ H~(dx,QQ

)

is standard,

= [10H~(dX,QA/QQ=)] so is

HI(dx,QQ~) ± = [IoH~(dX,QA/QQ

HI(dx,QQ

that

HI(dx,QQ

By (8) and Lemma 2C

)]~

= [IoH~(dX,QA/QQI)]~ This implies QI" []

).

.

) = HI(dx,QQI )

= HI(dx,QQI )±. and hence that

Q~

divides

189

3C.

Consider any

~ • T~

i ~ p ~ ~

and let

HP(d~,~).

Take any nonzero

so that Q and define

Q~

with

HP(do,~)

~ {0}

for some

p

with

be the greatest common inner factor of the space F0

from

HP(do,$)

and set

Q = IF01/F0,

is an i-function with character ~-i. We set M = HP(dx,Q) M A as the orthoeomplement, in L p' (dx), of the space

H0(dx)M , where

p' = p/(p - i).

Lemma. in 2D.

M ~ $ HP'(dx,I/QQ~Qc) , where

Proof.

Let

h* • M ~.

Then,

(h* o $)QQ6HP(do)

QC

is the inner function defined

h*M ~ H~(dx) ± ~ HI(dx,I/Qc )

and so

= (h* o $){M}p ~ {h'M} 1 ~ (I/Qc)HI(do).

This means that h* o $ • [(I/QQ~QcHI(do)] N L P ' ( d o ) Hence,

M A ~ HP'(dx,I/QQ~Qc).

3D.

Lemma.

outer, then

Let

[]

$ • T~

and

HP'(do,$-I~c ) ~ {0}

i ~ p < =

Let

F0

be any nonzero element in

= IF(0)] ~ I

If

~

is p-

~ IQc(0)i. HP(do,~)

IF0r/F 0. We then define a linear functional L setting L(f*) = F(0), where f* o $ = Q0 F with F • HP(do) ~ Hl(do), we have

IL(f*)i

be given.

and

mP(~,O)mP'(~-I$c,O) Proof.

~ (I/QQsQc)HP'(do).

IFldo = I

and set

Q0 =

on M = HP(dx,Q0 ) by F • HP(do,~). Since

IQoFldo = I

~

If*Idx AI

= ilf*liI ~ Elf*Up. By the Hahn-Banaeh that

theorem there exists a function

llv*lip, = IILII and

If f* • M so that

and

I

F(0) = L(f*)

h • H~(R),

then

hf*v*d X = L(hf*)

= fA I f*v*d X

hf* • M

and

v* • LP'(dx) for every

such

f* E M.

(~f*) o $ = (~o $)QoF,

: ((ho %)(0))F(0)

: 0.

Al As h C H (R) is arbitrary, f'v* • H0(dx) ±. This implies that v* belongs to M A. Since M m ~ HP'(dx,I/QoQc) by Lemma 3C, there exists

190

a

V e HP'(do)

with

v* o $ : (QoQc)-Iv

and therefore

Qc-IFv with FV E Hl(do,6C ). Namely, Qc-IFv meromorphie extension of (f'v*) o $ into D. (9)

QC(0)-IF(0)V(0)

: I

(f'v*) o $ =

is the TR-invariant By Theorem 2D we get

f*v*dx = L(f*) = F(0). Al

Since

$

has been assumed to be p-outer,

and thus one can find an

f* E M

HP(~,~)

such that

has no common zeros

F(0) ~ 0

with

f* o $ =

Q0 F. It follows from (9) that V(0) = QC(0) ~ 0, the last inequality sign being true in view of Corollary 2D. Finally, Since

our definition of

V E H p' (do,~-16 C)

Iv(0)I/llV*Up,.

and

mp

3E.

We know that

= IILIJ = llv*HpT. m p' (~-i~ C ,0) =>

H~(d~,~C ) ~ {0}.

Suppose now that

~ IV(0)1 : IQc(0) I > 0.

HI(dx,Q~ -I)

On the other hand,

so that outer.

mP(~,0)

Hence,

mP($,0)mP'(~-I~c,0)

1-outer.

shows that

llVllp, = IIv*11p, , we have

[]

is standard and therefore

H (do,~ C)

It then follows from Lemma 3B that

~ E T~

is 1-outer.

is

~C

QC

is

Then, because of the inclu-

sion relation Hl(d~'~)H~(d°'~C ) ~ H l ( d ° ' ~ C )' So, by Lemma 3D, m ~(~-i ,0) ~ m l ( ~ c , O )m~((<
~C

contains the inner function

~C

is also 1-outer.

~ ]Qc(O)I

H~(d~,~ -I) ~ {0}.

> O.

We further see the

following Theorem. Proof.

If

~

is 1-outer,

We have shown that

then

~-I

is outer.

H=(do,6 -I) ~ {0}.

common inner factor of H~(do,$-l). P, then 6-1~p-I is outer. Since

If

~p

Let

P

be the greatest

denotes the character of

H=(dO,~p)H=(do,~-l~p -I) ~ H=(do,$-l), H~(dO,~p)

should be divided by

PH=(do,Id), where Id of induction we have

oo ]1 H (dO,~p)

(lO) for

n = i, 2, . . . .

P.

We thus conclude that

denotes the identity character of

= pnH~(do,ld)

On the other hand,

H~(dO,$p) T R.

By use

=

191

H~(dO,~c)Hl(do,~)H~(do,~-l~p-l) In view of our h y p o t h e s i s , inner

factors

is 1-outer. i, 2, . . . .

By Lemma

n

3D and

~

3F. that Let

then

and

Let

that ~-i

Q0

Namely,

~0

above

~

~0

(a)

IlTII < i

~p-l~ C

~

for

n =

IQc(O) I > 0.

is a constant,

as was

T

the f o l l o w i n g

for some

Using

common

of

p

with

i ~ p < ~.

operator

T

is a n o n z e r o

of

H~(R) o $

in 2D, we have

So there

and

exists

liP011~ ~ 1/2.

Q0 =

function

a function

For any

in

H~(D),

h o ~ = (p0/Q0)(f o ~ - f(0)).

in

so

e x p r e s s i o n for H~(dX,Qg). -i ~0 is o u t e r and, by the

Then

too.

~ 0

factor

standard

Q0"

with

inner

the n o t a t i o n s

is the

is a T R - i n v a r i a n t H~(R)

bounded

by setting

linear

P0 ~

f E H~(R), so that we We then

Tf = h.

operator

in

H~(R)

with

and

(Ii)

T(fg)

for any

= (Tf)g

+ f(0)Tg

f, g E H~(R).

(b) Then,

P

immediately

is p - o u t e r

P0(0)

h E H~(R)

a linear

almost

is outer,

such that

(p0/Q0)(f o ~ - f(0))

Lemma.

and h e n c e

be the g r e a t e s t

be the c h a r a c t e r

find a unique define

i.e.

is 1 - o u t e r

are outer.

H~(dx,Q0 )

corollary,

H~(do,~0 )

= i

implies

H 0 ( d X) = H ~ ( d x , Q 0 ).

QgQg.

side,

~p-n~c

(i0) we get

IP(0)I

Suppose

Then both

the r i g h t - h a n d

we see that

common

[]

The t h e o r e m Corollary.

side has no n o n e o n s t a n t

~ ml(~p-n~c,0)m~((~p-n~c)-l~c,0)

is a r b i t r a r y ,

to be proved.

so does

by i n d u c t i o n

IP(0)I n = m ~ ( ~ p n , 0 ) As

the l e f t - h a n d

and t h e r e f o r e Again

~ Hl(do,~p-l$c).

Define

T~f = (Tnf)(0)

IIT~II ~ i

for

for

n = 0, i,...

f E H~(R)

and

n = 0, i, . . . .

and

(i2) j=0 for any Proof. f(0)

f, g E H~(R). (a)

= Q0F

T

is o b v i o u s l y

with IITfll

linear.

F E H~(do,~0-1). = II(Po/Qo)QoFII

For

f E H~(R)

we w r i t e

Then ~ = IIPoFII ~ ~

IIPoII

IIFII ~

fo $ _

192

= lip011 IIf-f(0)II ~ % 211P011 IIfll~ =< IIfII~. This means there

that

exists

and t h e r e f o r e

of (ii)

For

IIToII < 1.

hand,

such that

: P0(0)F0(0)

since

~0 -I

f0 o % = QoF0

~ 0.

Thus

T

is outer,

with

F0(0)

~ 0

is n o n t r i v i a l .

The

is easy and is omitted.

f E H=(R)

IT0fl so

On the o t h e r

f0 E H0(R) (Tf0)(0)

verification (b)

IITU ~ i.

an

we have

= l(Tnf)(0)l

= II

(Tnf)dxI __< llTnf'l < "fI' ; 41 (12) follows i m m e d i a t e l y from the r e l a t i o n

The i d e n t i t y

n

Tn(fg)

which

can be shown 3G.

ivation set

s

Lemma. in

by an easy

We now expand T.

Since

and

For every

< i}

For every The map For every

(14)

(b) (c) 3F.

If

with

(a)

by means

the f u n c t i o n

%

This

of the point

for

I[I < I.

% ÷ ~ f

is h o l o m o r p h i c

I~l < i, the f u n c t i o n a l on

f ÷ ~ f

H~(R).

l~I < I, is injeetive.

h E H0(R)

and

f E H (R)

we have

f) : ( T h ) ( f - ~[f)

I[[ < i. is almost

follows

Let

f0 E H0(R) with

clear,

easily

satisfy l[I,

for

from Lemma I~'I

S

= ~n=0 3F and

~nTn"

(13).

T0f 0 f 0, as in the proof

of Lemma

< i, then

¢~((I- ~T)(I- {'T)f 0) = ¢~,((I- ~T)(I- {'T)f 0) and thus (d)

derWe

~n(T~f).

with

~ ÷ ~%,

This

~{ = }~,

[ n:0

homomorphism

(h - ~Th)(TS

Proof.

H~(R)

is i n v e r t i b l e

= (S f)(0).

f E H~(R)

fixed

complex

(c)

%

in

[]

and

(d)

for any

(Tn-]f)(0)TJg,

induction.

~ f =

(b)

[ j:l

I - IT

~f

(13)

a nontrivial

+

functions

IITH ~ i,

= (I - IT) -I (a)

{I%I

: (Tnf)g

~'(T0f 0) = ~(T0f0). Since T0f 0 ~ 0, we have In view of the i d e n t i t y (ii), we have

~ = ~'

is

193

T(hS{f)

: (Th)(S%f) : (TS{f)h

so that

(TS~f)h

= (Th)(S

f- ~f).

( h - ~Th)(TS~f) = (Th)(S for

( I - ~T)S~ 3H.

of

R.

= I.

an

= h(TS{f)

Since

f0 E H~(R)

by setting then

shows

for any ation

and

and @(z)

that

= ( T h ) ( f - ~%f),

[]

Tf 0 E H~(R) 0

follows

- ~(Th)(TS~f)

f - ~ { f - ~(TS%f))

H~(R)

separates

T0f 0 = (Tf0)(0)

(Tf0)(z)

(Tf0)(0)

R \ U0

that

with

U 0 : {z @ R:

the point

+ (S~f)(0)Th, From this

We now use the a s s u m p t i o n Choose

+ h(0)(TS~f)

# 0,

= f0(z)/(Tf0)(z).

and set

# 0}.

U0

is discrete.

~ 0

the points

is an open

We define

set c o n t a i n i n g

a function

An a p p l i c a t i o n

of

@: U 0 +

(14) to

h = f0

that

f E H~(R)

functional

and ~

f(z)

: %~(z)f

z E U O.

Namely,

at the point

z.

~(z)

Choose

is e x a c t l y any compact

the evalu-

neighborhood

z

U~

of the origin

morphic

function

a disk

{I61

Theorem.

0 on

U0

< r 0}

Suppose

Proof.~

We have

that

tinct

evaluation

means

that

spect

~

not vanish

R.

Let

0.

(c),

~g

IC, holds

and that

be the c h a r a c t e r

Q~

is a c o n s t a n t

R, d i s t i n c t

It follows

coordinate

at

T0f 0 ~ 0.

is an inner

Moreover

holo-

includes

that

0.

Q~

So factor

in

(d~/dz)(0)

Consequently,

f0 of

has no zeros

R

Since give dis-

remark,

this

So, by Lemma ~ 0

with re-

~ 0 zero at the

f0' we see that R.

Then

we have

has a simple in

H~(R)

Q~.

function. z

: (d~%f0/d6)(0).(d@/dz)(0)

with

QgQN'

z

points

of

E . C o m b i n e d with the above z for any d i s t i n c t z, z' E U0.

~ ~(z')

is an injection.

at

~(U~)

functionals

~(z)

Since

0, the image

of

f0 E H~(R) 0.

is a n o n c o n s t a n t

the points

to any local

for any

at

~

r 0 > 0.

the c o n d i t i o n

of

(df0/dz)(0)

point

Since

and v a n i s h e s

only to show that

separates

3G-(c),

U~ ~ U 0.

for some

separates the points -i ~9 is outer.

H (R)

with

Q~

If this were

cannot

194

the case, any

then

Q~(a)

f E H~(R),

contrary

= 0

so that

for some H~(R)

be a s i n g u l a r

inner f u n c t i o n .

T, we c o u l d c h o o s e would

IP01

Thus

f(0)

separate

: f(a)

the p o i n t s

= 0 of

for R,

to our b a s i c h y p o t h e s i s .

To get the final c o n c l u s i o n ,

P0/Q!~

a ~ 0.

c o u l d not

and

P0 E H ~ ( d ~ , ~ 0 )

be u n b o u n d e d .

k 0 o % = IQ~I.

bounded.

Since

there would

we suppose,

Then,

k0

exist

Let Thus,

h0(0)

P0(0)

be the l . a . m . ' s ~ 0

and

h0/k 0

~ 0

with

such that

(h0/k0)(Zn)

Q~

and

h0 o % =

w o u l d be un-

a w a y f r o m zero on the c o m p a c t

z n E U 0 \ U 0'

that

of the o p e r a t o r

in such a way that

h0, k 0

is b o u n d e d

on the c o n t r a r y ,

in the d e f i n i t i o n

÷ ~.

set

U~,

T h e n we

w o u l d have I¢(Zn) [ : I f o ( Z n ) [ / I ( T f o ) ( Z n ) I ~ k o ( Z n ) / h o ( z n) ÷ O. So, for a s u f f i c i e n t l y i n i t i o n of ~(Zn).

r0

This

shows

is, h o w e v e r ,

As we n o t i c e d of the o r i g i n Corollary.

q(z)

set

large

impossible,

n

= exp(-~j=l

Lemma.

be any c h a r a c t e r of

R, t h e r e

such that

~q(y)

denotes

~q

taken

R~,...,

let

yj,

R~'

{Uk:

that

YI'''''

Y~

CI(V). Fh0(R)

o k = O(yk ),

representing

i ~ k ~ N} in

Yk

V, then,

are linearly V

such that

of p o i n t s

in

[]

q(z)

y e F0(R)

q

R

of

and

q.

of

R.

subregion

= exp(-~j~l contained

and the p o i n t s

Then,

is outer.

the p o i n t s

for any c a n o n i c a l

Let G

g(~j,z))

in

G,

can be

same lines as that of Lemma

8B, Ch.

G.

of the c o m p o n e n t s contour

of

So a h o m o l o g y

nondividing

d i s k s u c h that

Let

=

of the c h o i c e

{¢j}

be an e n u m e r a t i o n

i ~ k ~ N, in

al,... , a N

of

s u b s e t of

G.

number

of the f o r m

for any curve

and c e r t a i n

@(z*)

is an i n j e c t i o n .

(c) is i n d e p e n d e n t

separates

Then, q

the c h a r a c t e r

to

The def-

with

~q

H~(R)

F0(R).

with respect

be a p a r a m e t r i c

in

R

an l.a.m,

= ~(y)

~

z* E U 0'

be t h e c h a r a c t e r as in the t h e o r e m , ~q - i

0 ~ j ~ ~, be the b o u n d a r y

positively

VkdY ,

Let

The p r o o f goes a l o n g the

Let

off

on

f r o m any f i x e d o p e n

Proof.

of

of

exists

where

for

be a f i n i t e

g(aj,z)).

Suppose

19(Zn) I < r O.

a point

we get the f o l l o w i n g

al,... , a n

u n d e r the same h y p o t h e s i s

31.

exists

in IC, the c o n d i t i o n

and t h e r e f o r e

Let

n, we w o u l d h a v e

that t h e r e

CI(V)

R \ CI(G)

Rj, ' which

basis

cycles

$ G

of for

Y~+I'''''

YN"

I, 9B).

If we w r i t e on

V.

Let

V

lie

differentials ok = U k d X +

as we saw in the p r o o f of L e m m a independent

consists

Yk'S

i ~ k ~ N, be the h a r m o n i c (Ch.

is o r i e n t e d

CI(G)

a n d the c u r v e s

V.

and

So t h e r e

8B, Ch.

exist

points

V,

195

A(al,...,a N) = det (uj(ak))l<j,k< N ~ 0. By identifying

V

with the unit disk,

~(Xl'''''XN) which

is defined

k = i,..., ~(U)

N}.

in

= ( ~

U = {(x I ..... XN):

an open set in

cube with sides of length By use of Lemma

M( ~ k:l

~N )'

%

with

at the origin

and thus

~ k=l

*dg(.,ak),... , of

M

implies 2~

(bl,...,b N) E IRN M ~ k=l

contains

large integer

an NM > 0.

~ k=l

(Xl,...,x N) exists

*dg(. yj

*dg(-,ak))

- 2~M~(Xl,...,XN).

YN

that the right-hand

when there

*dg(.,a k + x k)) YN

an

side covers

ranges

+ x k) = b 'ak

Thus,

given

an N-cube

U.

So, for

such that

(mod 27) J

~ f yj*dg(-,~k ) --- b 3. k=l N.

over

(Xl,...,x N) E U

for j = i,..., N. If ~i''''' ~n are the points each repeated M times, then ~k'S belong to V

j = I,...,

does not vanish,

(2~M)~(U)

for a sufficiently

YI

with sides of length

for

r = min{l-[akl:

7E, Ch. V, we have

*'dg(.,a k + x k),...,

The definition

~

k=l

[Xk[ < r}

of

I~N

2z

YI = M( ~ k=l

any

~i .... '

k=l

Since the Jacobian

contains

we set

a k + Xk, and

i < k < N,

(mod 2~)

~ @ F0(R)* , we find

~i'''''

~n E V

such that

[ k=l for

y = yj,

We set

q(z)

l.a.m,

q(z).

3J. Theorem. points

of

the l.a.m,

(mod 27)

*dg(',~ k) - - a r g ( ~ ( y ) ) y

i ~ j ~ N, and therefore = exp(-[k~l Then

g(z,~k))

~q(y)

= ~(y)

for any

y 6 F0(R)

and let

~q

for any

y e F0(R)

lying in

be the character lying in

We are now able to prove our final objective

easily.

Suppose

separates

R.

that

Then

defined

R

(c), IC, holds is a PWS and

in IB.

and that

H (R)

qc = g~(0)[R'

where

g~(0)

G.

of the G.

the is

[]

196

Proof. Ch.

Let

{G : n = i, 2,...} be a c a n o n i c a l e x h a u s t i o n of R (see n 0 E G I. Take any c h a r a c t e r [ of F0(R). By Lemma 31

I, IC) with

and C o r o l l a r y F0(R)

3H there

such that

In -I

exists,

~n(y)

is also outer,

for each

= ~(y)

[n-l[c

for

n, an outer

y e F0(R)

is outer

character

lying

in

and t h e r e f o r e ,

In

G n.

of

Since

by Lemma

3D,

m "( ~n ) => IQc(0) I > 0. So we can choose, and of

Ifn(0) I ~ TR

for each

n, an

and denote

by

Fn

in some n e i g h b o r h o o d

principal

branch

we may assume, that

F

passing

Moreover,

so that

YT

such that

YT ~ Gn

that

) : [(yT)

[n(y

of

(G n} for

Letting

n ÷ ~, we have that

phic

it is clear follows

on

R

that

that

nally have

m

f e H~(R,~),

In o r d e r

qalR .t

with Let

defined admits

by

a # 0

q](z)

R

is a PWS

(resp.

qa ). f(a)

and t h e r e f o r e equal

bounded

extension : 0}, then

of

H ~% ()Ra to

= {f E H~(R):

f/qa

implies

we

holomor-

of

0, then

IQc(0) I .

holomorphic

let

fi-

Rt

V, 3B.

R t.

(resp. function So,

be

For each qa = F0(R)) on

R

if we set

= {f • H~(Rt):

f/q~

R, of f u n c t i o n s

in

is bounded}.

It

we

z • R t, and

to H~(R)

TR

°

F0(Rt)

to the whole

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

~

in Ch.

= exp(-gt(a,z)),

q~

holomorphic

In

is a r b i t r a r y ,

of the theorem,

~a ) be the c h a r a c t e r Every

of

of

N > 0

is a r b i t r a r y ,

If(0)l

and use the n o t a t i o n s

we set

an

the m u l t i p l i e a t i v e

and

assertion

and

0 £ t ~ i,

isomorphism

exists

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

(resp.

a unique

is e x a c t l y

Hence

~ • TR

: [(X)Fn.

~ e F0(R)* ,

con-

that

T E TR

~at

H~(R ¢) : {f e H~(Rt): bounded}

R

As

°

to show the second of

As

denotes

llfll ~ i

IQc(0) I .

=

the r e g u l a r i z a t i o n a • Rt

~

there

family,

IF(0)[

x(0)t,

the d e f i n i t i o n

f0 o ~ = F

0) > IQc(0) I > 0

~(

Then

f

Fn :

is the

and

let

segment

= i

it is clear

IIFILo~ ~ 1

In fact,

It follows

If

(fn)0

Then,

the c a n o n i c a l

F O T = ~(T)F.

such that

m~(<,0)

~.

: [n(YT)F n : [(yT)Fn

F E H (~,~).

function

under

n ~ N.

F n o T : [n(T)Fn

conclude

on

satisfying

n ~ N.

such that

{F } is a b o u n d e d n if n e c e s s a r y , that {F n}

is an e x h a u s t i o n ,

for all

llfn II

as c h a r a c t e r s

0, w h e r e

of the r a d i a l T

In

Since

uniformly

~

such that

and

H~(~,[n )

F • H~(~,~).

under

Since

2D).

function

is the image

F0(R).

II,

almost

we have

be the image

in

to a s u b s e q u e n c e F

[

of the origin

(Ch.

n

is a h o l o m o r p h i c

YT

onto

f

to a f u n c t i o n

IQc(G) I . let

of

We r e g a r d

the f u n c t i o n

(fn) 0 o @

verges

fn • H ~ ( R , [ n )

[Qc(0)I/(I + n-l).

is H~(R t) a

197

Let

St(z) a

tion of [a ).

: exp(-gt(a,z) - igt(a,z))

S at

to

R, so that

S at

for

(resp.

Then we have in an obvious sense

H~(R)a = S a ~ ( R , [ a - l ) .

Therefore,

z E Rt

and

Sa

the restric[at (resp.

S a ) has character H~(R t) = s at ~ ( R t a

~(R,[a-l)

[at-l) and

is regarded as the re-

striction of ~ ( R t '[at-i ) to R. Since [~-i is outer by Theorem 3H, ~ ( R t '6a¢-1 ) has no nonconstant common inner factors and so does the space

~(R,~a-l).

Let now n = i, 2,...

Hence

~a -I

{Zl, z2,...} let

is outer for any

be an enumeration of

a E R %. Z(0;Rt).

For every

n

qnt : exp(-j:l [ gt(zJ'z)) for z E R%' qn : q~[R, and let ~n be the character of F0(R) determined by qn" Then the above observation shows that each ~n is an outer character. Let Qn be the inner function on ~ such that Qn(0) ~ 0 and IQnl = q n ° %. Let f be meromorphic on R such that Iflq n has a harmonic majorant on R. As is seen easily, f has a unique meromorphic extension to R t, which we denote by the same letter. Take any Then

h e H~(R), which is also regarded as an element in

Ifhlq~

has a harmonic majorant on

R t.

H~(Rt).

It follows from Theorem

4B, Ch. VII, and Theorem 3B, Ch. V, that I

ffidx : Y AI

This means that

ffidxt = f(O)h(O)

= 0.

&I(R t )

HI(dx, Qn-l) ~ H0(dx )m ~ HI(dX,Qc-I).

outer, we see that Qc/Qn on R~ or, equivalently,

is inner.

Namely,

qc/qn

Since

~n

is

is an inner l.a.m.

n

qC(z) < exp(- ~ gt(zj,z)) j=l on R for any n = i, 2, . . . . Letting is divided by q = gt(0)IR , where g t(0) (z)

=

n ÷ co, we conclude that

qc

exp(- ~~ gt(z j ,z)). j:l

In other words, if Qt(0) is the inner function on ~ such that Qt(0)(0) ~ 0 and IQt(0) I = q O ~, then Qt(0) is an inner factor of QC"

I

On the other hand, we claim that H~(dx,11Qt(0) )± ~ H~(dx). To see this, let f* E H~(dx,I/Qt(0)) ±. This means that f* E LI(dx ) and

lg8

f£1 f*hdx = 0

for any

h, meromorphic

bounded.

By the inverse

a unique

f e HI(R)

contains

constant

Cauehy theorem

with

f* = f

functions,

f(0)

R, such that

lhlq

is

(Theorem IC, Ch. VII) there is

a.e.

on

A I.

Since

H~(dx,I/Q t(0))

we have

= I

~dx : I AI

f*dx = 0. AI

f C H I(R) 0 , as claimed

Hence,

on

We now repeat our argument

in IB and get

H0(dx )i ~ HI(dx,I/Q %(0) So, in view of the definition of QC (see 2D , [10HI(dX,Qc-I)]I HI(dx,I/Q t(0) ). Since I 0 has no nonconstant common inner factors ~C

is an outer character,

function.

this implies

Summing up, we have shown that

by a constant

factor of modulus

assumed that

QC(0)

§2.

4.

that

CONDITIONS

> 0

and

EQUIVALENT

Qt(0)/Qc

QC

one.

(Indeed,

Qt(0)(0)

> 0.)

TO THE DIRECT

and

Qt(0)

differ only

QC = Qt(0) Hence,

and

is also an inner

because we

qc = gt(0)IR"

[]

CAUCHY THEOREM

General Discussion 4A.

The direct

chapters.

count of the results "direct

Cauchy theorem was studied in §i.

Cauchy theorem"

possible

in the preceding

We are now able to deepen our discussion

choice of

bolic Riemann Definition.

Our present

which contains

{zj}.

objective

by taking

two

into ac-

is to seek for a

Theorem 4B, Ch. VII, for every

We need a definition

valid for general hyper-

surfaces. Let

Cauchy theorem

q

be an inner l.a.m,

(abbreviated

on

R.

We say that the direct

to (DCT)) holds for

f(0)

: I

q, if we have

f(b)dx(b)

J

£i for any meromorphic

function

f

on

R

such that

Iflq

has a harmonic

majorant. There are two types of (DCT), which are valid for any (i) (DCT),

It is trivial

for we have

that the constant

f E HI(R)

function

in this case.

q ~ i

R. satisfies

199

(ii) 4B, Ch.

If VII,

{Zl,

z2,...]

is an e n u m e r a t i o n

shows t h a t any i n n e r q(z)

satisfies

(DCT)

Suppose

for

Then

q

= exp(-

q

R

carries

includes

can be d i v i d e d

by

the set of i n n e r l . a . m . ' s .

for some

z.

an i n n e r

the c a s e

the

q

for all

such that

q

for

s h o u l d be d i v i d e d

that

q

sat-

n : i, 2, . . . .

It f o l l o w s

and t h e r e f o r e

We thus a s s u m e , PWS.

in Ch.

n = i, 2,...

in w h a t

VII,

by (DCT).

Lemma.

R

regular

in by

that

follows

R

should converge

in the

sense of po-

is a PWS.

in this

section,

that

C a u c h y t h e o r e m we w i s h to h a v e

3C.

be a r e g u l a r

~j=l g(zj,z)

implies

equivalent,

collectively

that

is a s s u m e d

T h e n the d i r e c t

are m u t u a l l y

Let

R

t h e n our r e s u l t

(DCT a) g i v e n

a • R

l.a.m,

(ii)

e x p ( - [ j=l n g(zj ,z))

If, m o r e o v e r ,

theory,

4B.

then Theorem

co

g(zj,z))

a regular

Z(0;R),

n [ g(zj,z)) j=l

co

exp(-~j=l

tential

of

of the f o r m

n = i, 2, . . . .

now that

isfies a (DCT) w h i c h

l.a.m,

By T h e o r e m

5A, Ch. VIII,

so that t h e s e

PWS for w h i c h

conditions

(DCT)

holds.

R

is

is just

(DCTa)'S

with

are d e n o t e d

Q(O) be Q(O)(o) >

Let

the i n n e r f u n c t i o n on ~ such that g(0) o % = IQ(0)I and (0) 0, w h e r e g was d e f i n e d in Ch. VII, IA. Then we h a v e oo

H 0 ( d x )± : H I ( d x , I / Q ( 0 ) ) . Proof. that

Let

f* E H I ( d x , I / Q ( 0 ) ) .

f* o $ = (I/Q(0))F.

is T R - i n v a r i a n t on

R

on

such that

Take any

D.

Then t h e r e

Since Thus

(I/Q(0))F

there

f o ~ = (I/Q(0))F

h • H~(R).

Then

exists on

Ifhlg (0)

•

exists

an

F E Hl(d~)

is T R - i n v a r i a n t a meromorphie and

f~ = f

has a h a r m o n i c

on

T,

function

such it f

a.e.

on

A I.

majorant

on

R, so

that f

^^

fhd x : f ( O ) h ( O )

: 0

AI by (DCT). cluded

in

Namely,

H 0 ( d ×)

T h e o r e m 3J t h a t 4C. Theorem. conditions

f* E H 0 ( d x )±

.

On t h e

other

hand,

we h a v e

H~(d×) ± ~ HI(d×,I/Q(0)).

It is n o w e a s y to p r o v e Let

and t h e r e f o r e

R

be a r e g u l a r

are e q u i v a l e n t :

our first

H I ( d x , I / Q (0))

shown in the

H e n c e we a r e

PWS w i t h o r i g i n

$.

proof

done.

characterization T h e n the

is in-

of

[] of

(DCT).

following

200

(a) (b) (c) the form Proof.

(DCT) holds. H0(d X) • = H I (dx,I/Q (0) ). Every simply invariant subspace of

LP(dx),

HP(dx,Q)

T.

for an i-function

By Lemma 4B,

satisfied.

Let

f

(a) implies

has a harmonic majorant jorant on

~

Theorem

2D,

Theorem

4C, Ch. VIII,

that

(c) holds.

subspace of

on

R.

and therefore f(0)

(b).

be a meromorphic

Conversely, function on

Then, Hence,

states that H0(d X)

I < p < ~, is of

suppose that R

such that

I( f o ~)Q(0)] (DCT) holds.

(a) implies

(b) is Iflg (0)

has a harmonic ma.

So, by

On the other hand,

(c).

Finally,

suppose

is easily seen to be a simply invariant

LI(dx ), there exists

H[(dx )& = HI(dx,Q).

on

f @ HI(dx,I/Q (0)) = H0(d X)

= fA I fd X.

Since

Q

an i-function

Q

on

T

such that

Then [10HI(dx,I/Q(0))] I = HI(dx,Q),

by use of Theorems

2D and 3J.

Lifting to

~, we see that

(I 0 o $)(I/Q(0))HI(do,~C ) ~ QHI(do,~Q-I). Since

io o $

has no nonconstant

outer,

we see that

(QQ(O))-I

common inner factors and since

is an inner function,

say

P.

~C

is

Thus

HI(dx,1/Q (0)) ~ HI(dx,I/PQ (0)) = [IoHI(dx,I/Q(0))] 1 $ HI(dx,I/Q(0)), which shows the statement

5.

Functions SA.

mP(~,a)

(b).

and

This completes

the proof.

(DCT)

We will give our second characterization

the functions

mP(6,a)

(see ID) play a principal

notations

given in Ch. VII,

Theorem.

Let

R

[]

of (DCT), role.

in whieh

By use of the

IA, we first get the following

be a regular

PWS.

Then it satisfies

(DCT) if and

only if ml(~(a),a)

(is)

for some Proof.

(and hence all) Suppose that

~(-~,~(a)) morphic

with

: g(a)(a )

a E R.

(DCT) holds and choose any

a E R.

Let

h

be in

ilhll < i and set f = h/S (a). Then f is a merol,a = function on R such that rflg (a) has a harmonic majorant on

201

R.

Setting

H[~]

and

u(z)

= lh(z)I,

i~ I = ~/~(a)

we get an l.a.m,

= @

a.e.

f(a)

t : |

on

41 .

u

such that

LHM(u)

As we are assuming

=

(DCT),

we

have

with

f(b)dXa(b),

J A1

dXa(b) = k b ( a ) d x ( b ) , and t h e r e f o r e If(a)[

~ [

If(b)IdXa(b)

: [

41 =

So

lh(a)l

= If(a)l

u(b)dXa(b) £1

llhlil, a ~ i.

g(a)(a)

~ g(a)(a).

As

h

is arbitrary,

we have

(a) g

(a) ~ sup{lh(a)]:

h • ~(R~,6(a)),

llh111,a ~ i}

= ml([(a),a). On the other hand, properties: ml(~(a),a)

Suppose We denote

such that extends

(i)

by

belongs

to

Is(a)(a)I

Hence the equation

(15) holds

B

that the equation

the set of functions

has a harmonic

to a meromorphic

function

majorant on

R.

with the

~ ( R , [ (a))

and

conversely

|h i

S (a)

=< irs(a)ll~ =< 1

JIs(a)Ell,a

~ g(a)(a).

(b) a E R.

the function

= g(a)(a). for all

Thus

a e R.

(15) holds

for some

h

on

on

defined R

and

(ii)

R \ Z(a;R) h/g (a)

In other words,

B = (exp(i[(a)))~(R,~(a)). Take any by Lemma have and,

h E B.

Then

3B, Ch. VII,

~(a)

= i

in fact,

a.e.

h/g (a) admits

on

belongs

41

is a meromorphic

fine boundary (Lemma

to the space

IBhiJB = I

values

2A, Ch. VII), LI(dx).

function a.e. so

on

on

R, which,

41 •

As we

h

exists

a.e.

B

is identified

We set

Ifi(b)ldXa(b)' A1

which defines

a norm in the linear

with a subspace setting

L(h)

of

Ll(dXa ).

= h(a)/g(a)(a)

h.exp(-i~ (a)) e ~A(R,~(a))

space

Define for

B, so that

a linear

h E B.

If

functional

on

B

h e B, then we have

and so

lh(a) l : lh(a)exp(-i~(a)(a)[ ~hllB'ml(<(a),a)

L

by

202

: lih11B.g by use of

(15).

This m e a n s

we see via the H a h n - B a n a c h sion to

Ll(dXa )

ilwll : IILII ~ i

IL(h) l ~ ilhllB

theorem

that

and that t h e r e e x i s t s

L

h • B.

: i J

has a n o r m - p r e s e r v i n g

a function

w • L~(dXa )

So exten-

with

h(b)w(b)dXa(b)

=

for

h = g(a) f

] w b) Xa

J

AI w -- i

h • B.

AI

In p a r t i c u l a r ,

i = L(g(a))

Thus

for e v e r y

and

L(h) for e v e r y

that

(a)(a)

a.e.

on

AI

we h a v e

lw b)Id a b)

i.

AI and t h e r e f o r e

L(h)

= I

h(b)dXa(b) AI

for e v e r y

h • B.

To f i n i s h R and

such that ~ = ~

f(a)

the proof, Iflg (a)

a.e.

on

let us take any m e r o m o r p h i c

has a h a r m o n i c A I.

majorant.

F r o m what we h a v e

= h(a)/g(a)(a)

: L(h)

= I

Then

h(b)dXa(b)

(DCT a) holds,

5B.

Let

positive

{an:

: I

numbers

is d i s j o i n t

tending

from

to

Z(0;R) •

0

• B

that

f(b)dXa(b)"

[]

be a s t r i c t l y

such that each

We set

follows

on

AI

as was to be proved.

n = i, 2,...}

f

h = fg(a)

seen a b o v e

AI Hence,

function

decreasing

sequence

{z E R: g(0,z)

R n = R(en,0)

and denote

by

of

= a n} gn(a,z)

the G r e e n f u n c t i o n for R . Let ~n be the c h a r a c t e r of the l.a.m. (0) n un (z) = e x p ( - [ {gn(W,Z): w • Z ( 0 ; R n ) } ) . Since gn(0,z) = g(0,z) - an, we h a v e Z ( 0 ; R ) : Z(0;R) A R for n = i, 2, . . . . We d e f i n e a m u l t i n n plicative holomorphic function B • ~{~(R,~ (0)) (resp. Bn • ~(Rn,nn)) by the c o n d i t i o n s 0

and

branch

iBn(z)i at

0.

most uniformly may assume

that

for e a c h f i x e d

B0(0)

and

IB(z)I

= g(0)(z)

: u(0)(z)) w h e r e the s u f f i x n For any f i x e d a • R, gn(a,z) on

R \ {a}. Bn(Z)

Passing

converge

l e m m a gives

(15) to an a r b i t r a r y

B(z)

converge

almost

R.

which

(resp.

denotes

(Bn)0(0)

to

g(a,z)

if n e c e s s a r y ,

as

>

the p r i n c i p a l

uniformly

~n(y ) + ~(0)(y)

a result PWS

0

to a s u b s e q u e n c e

to

y • Fo(R) , we h a v e

The f o l l o w i n g equation

> 0

on

R.

alwe So,

n ~ ~.

is an a d a p t a t i o n

of the

203

Lemma.

Let

i ~ p < ~ g (0)

for any Proof.

(0) ~

mp'

p' = p/(p - i).

Then

(~,O)mP(~-l~(O),o)

~

mP'(~,0) 61 n

denotes

= lim mP'(~In,0), n÷~

the restriction

of

~

to

R n.

Next we recall that

(7) in Ch. V, 4C, asserts

inf{llfl[p,: f • ~ ( R n , ~ [ n ) = sup{lh(0)[:

, If(0)l

: i}

h • ~ P ( R n , ~ I n - l ~ n ) ' llhllp,n

where

llhfl p,n = [-(i/2~)f$ R lhlPdgn(.,0)] I/p n then implies (17)

mP'(~In,0)mP(~In-lnn,0) We now estimate

aim we choose

the limit of

by Theorem

F0(R n)

as

n ÷ ~.

n, such that

Since

R

is the restriction

For this llhllp,n = i

is a PWS and since of some character

~ inf{mP(~',0):

~' • F0(R)*}

lhn(0) I ~ (n- l)mP(0)/n

for

>

= mP(0)

n = i, 2,...

5A, Ch. V,

on any compact

and, by passing to a subsequence

subset of

R

{lhn(Z) I }

.

seen in the proof of Theorem

sary, we may assume that the sequence

{hn(Z)}

formly to some m u l t i p l i c a t i v e

function,

immediate

that

(18)

ilh[Jp =< i

and

mP(~-l~(0),0)

Combining (19)

(16),

(17) and

analytic

of

As we have

say

bounded

if neces-

almost uni-

h, on

R.

It is

So

~ lim sup mP(~In-l~n,0).

(18), we get

In order to get the desired

~ lim sup r n. n÷~

inequalities,

side of (19) is not smaller than

g ( 0 , z ) - an, we have

0

is uniformly

converges

h • ~P(R,~-I~(0)).

mP'(~,0)mP(~-l~(0),0)

right-hand

This

3B, Ch. I and Theorem IB, Ch. If, we have

mP(~[n-lnn,0) and therefore

n

= r n.

h n • ~ P ( R n , ~ I n - l ~ n ), for each

of

-i]

=r

r n : u(0)(0). n

and

mP(~In-l~n,0)

Jhn(0) i ~ ml(~In i n n , 0 ) ( n - l)/n.

every character F0(R)

,0)

By the normal family argument we easily see that

the formula

and

m I (~( 0)

~ • F0(R)*.

(16) where

and

we first show that the

g (0)(0).

Since

gn (0,z)

=

204

log r n = ~{g(0,w) - ~n: w e Z(0;Rn)}

-

~{g(0,w) - an+l:

w e Z(0;Rn)} =

So,

lim l o g r n

and,

for any fixed

~{g(O,w) - ~n: w E Z(0;Rm)} for all to

n $ m.

g(0)(0),

Letting

which

Finally ml(~(0),0).

shows

n ÷ ~

m > 0,

~ - lim log r k ~ - log g (0)(0) k÷~ and then

m ÷ ~, we see that

the l e f t - h a n d

easily

follows

side of

from the

(19)

finishes 5C.

the proof.

We will

group

F0(R)

now study

F0(R)*

characters for each

with

with

the c o n t i n u i t y

fixed metric

space.

PWS

(DCT)

implies

(20)

Since

~ F0(R)

~ E F0(R)*

an i m m e d i a t e

and t h e r e f o r e

namely,

and any

if

~n(y)

is c o u n t a b l e , that,

{~n }

of

÷ ~(y)

F0(R)*

is a

for any r e g u l a r

= ml(~(0),0)

i S p S ~, w h e r e

of T h e o r e m

we equip

its c h a r a c t e r

a sequence

if and only

We b e g i n with r e m a r k i n g

consequence

~ ÷ mP($,0),

For this p u r p o s e

topology

mP'(~,0)mP(~-I~(0),0)

for any

).

of f u n c t i o n a l s

F0(R)*.

topology;

to a c h a r a c t e r

y E F0(R).

compact R,

group

the d i s c r e t e

the c o m p a c t

converges

than

relation

[]

i S p ~ ~, on the c h a r a c t e r the group

tend

is not larger

inclusion

~p' (R , ~)~P(R, ~-l{(O)) =C ~(R,~(O) This

rk

our claim.

we show that But this

w • Z(0;Rn+ I) \ R n}

(0)(0).

l ~ - log g

-logrn+

exists

+ ~{g(0,w) - ~n+l:

p' = p / ( p - i).

5A and Lemma

5B.

We next

This prove

is the

following Lemma. mP($,0) Proof.

If (20) holds is c o n t i n u o u s

for any on

Take any s e q u e n c e

~ E F0(R)* , then the f u n c t i o n a l

F0(R)* {~n:

for every

n = i, 2,...}

p

C F0(R)*

to the i d e n t i t y c h a r a c t e r Id. Since m~(~n,0) implies that ml(~n-l~(0),0) ~ ml(~(0),0) for lim inf m l ( ~ n - l ~ ( 0 ) , 0 ) n~

We now c l a i m

the r e v e r s e

inequality:

with

~ ÷

i ~ p ~ =. which

converges

~ i, the e q u a t i o n (20) n = I, 2,... and thus

~ ml($(0),0).

205

lim sup ml(~n-16(0),0)

£ ml(~(0),0).

n+~

If this were not the case,

then we could assume,

quence

if necessary,

ml(~n-l<(0),0)

2,...

and for some fixed

with

llfnllI = i

By passing

that

and

~ > O.

[fn(0)[

to a further

> ml(~ (0),0) + s

Then,

choose

shows our claim.

this with

(20), we have

(21) Now take any

p

for any

with

=

and using

the role of

6

These two inequalities

5D.

Let

R to a multiplicative If(0)l ~ m l ( ~ ( 0 ) , 0 ) + s/2.

i = m ~( Id,0).

I £ p £ ~.

Since

~(R,q)~(R,n-l~)

and

{z6:

let

e

inequality:

£ mP(~,0).

show that

F0(R)*.

mP(~,C)

is continuous

in

[]

j = i, 2,...}

n = i, 2,...

~ mP(n,0).

~, we get the reverse

together

group

~ mP(~,0).

(21), we have

lim sup mP(~,0)

every

n = i, 2,...

we could also assume

~, q e F0(R)* , we have

~ ÷ q

on the character

(0))

shown

lim inf mP(~,0) Changing

n = i,

: ml(~(0),0).

mP(D,0)m~(q-l~,0) By letting

for

Hence we have

lim m~(~,0) ~÷Id

~P(R,~)

to a subse-

fn E ~ ( R , < n - l {

for each

if necessary~

lim ml({-l$(0),0) ~÷Id Combining

an

> ml(~(0),0) + e/2

subsequenee

that {fn } would converge almost unformly on function f C ~ ( R , ~ (0)) with llfU1 ~ 1 and This contradiction

passing

be an enumeration

be the character

of

Z(O;R).

For

of the l.a.m.

n n

exp(-

and l e t and

Cn(Z) E ~ ( R , 8 n)

(Cn)0(0 ) > 0.

Since

~ g(zj,z))

j=l

be d e f i n e d by

IOn(Z) I = exp(-~ j =n l g ( z j ' z))

llCnllp =< llCnll~ = i, we see first that

n

exp(for

~ g(z~,0)) j=l

= ICn(0) I £ mP(en,0)

i £ p £ ~. Let

p,

i ~ p ~ ~, be fixed and take any

f E ~(R,Sn).

The

.

206

quotient f0/(Cn) 0 of the principal branches then extends uniquely to a meromorphic function, say hf, on R. Since lhflexp(-[j~l g(zj,.)), being equal to Ifl, has a harmonic majorant, the weak Cauchy theorem (Theorem 4B, Ch. VII) shows that Since

hf • LI(dx)

lhf(0) I = If(0)I/ICn(0) I

and

If(o)l ! ICn(O)I I

and

hf(0) = fA I hfdx.

ILhfllI = NfilI S IIfNp, we have

I~fldx

=

ICn(O)lFIhfll 1

A1

= ICn(O)IllfllI ~ ;Cn(O)lllfH p. As f • ~P(R,8 n) and thus

is arbitrary,

mP(~n,O) ! ICn(O)l

= exp(-[j~ I g(zj,O))

n

mP(Sn,O) for

= exp(- ~ g(zj,O)) j=l

i S p S ~• oo

Since

~j=l

of

R\ Z(0;R),

the

function

verges to

g(zj,z)

we s e e B, w h i c h

~(0)

ml(6(O),o)

If

is that

uniformly {C n}

was d e f i n e d

ml($;0)

convergent

converges in

5B,

on a n y c o m p a c t

almost

uniformly

and therefore

is continuous

in

that

on

subset R

{0 } n

to con-

~, then

= lim ml(@ O) : exp(- ~ g(zj,O)) n+~ n' j:l

: g(O)(o),

which implies, by Theorem 5A, that R satisfies (DCT). Summing up our considerations in §2, we finally get the following characterization of (DCT): Theorem. conditions (a) (b) (c) the form (d)

Let R be a regular PWS with origin are equivalent: (DCT) holds. H~(dx) ± = HI(dx,I/Q(0)). Every simply invariant subspace of

0.

Then the following

LP(dx),

i ~ p ~ ~, is of

HP(dx,Q) for some i-function Q on T. ml(~(0),0) = g(0)(0), where ~(0) is the character of the

l.a.m, g(0)(z) = exp(-[ {g(w,z): w e Z(0;R)}). (e) m~(6,0)ml(~-l~(0),0) = ml(~ (0) ,0) for any ~ • F0(R)*. (f) 6 + ml(~,0) is continuous on F0(R)*. (g) ~ ÷ mP(~,0) is continuous on F0(R)* for every p with i~pS~.

207

NOTES Most of the o b s e r v a t i o n s Hayashi

stated above are based on the idea of

[28] with a couple of m o d i f i c a t i o n by the author.

involving the function

m~(~;0)

is also in Pranger

[55].

A discussion

CHAPTER

The faces sion

EXAMPLES

objective

of this

of P a r r e a u - W i d o m is by no m e a n s

Riemann The

X.

surface

emphasis

we are

going

should

(DCT)

but the

examples

§i we g i v e

surface

of M y r b e r g (DCT).

type

three

type

for which

problem

in the m a x i m a l

shown here

functions positive corona

as w e l l

problem.

the o t h e r ,

i.

OF I N F I N I T E

PWS's

of M y r b e r g

of M y r b e r g

ing

type

surfaee

counting

over

R ÷ ~ of

in

§3 t h a t

of

Ha(R)

is d e n s e

of a l m o s t

~,

are

condition

corona

exist

PWS's

FOR WHICH

(DCT)

for which out more

s u c h a PWS a l w a y s

every

for

spaee

of P a r r e a u -

discussing

PWS

R

as an o p e n

can

that

the

our

the

be em-

subset.

It is

o f all h o l o m o r p h i c

convergence.

theorem

for which

PWS's

for a R i e m a n n

regions

constructed

one h a n d ,

the

Here

for which

PWS's

show that

in t h e

uniform

examples

on t h e

GENUS

Finally

concerning

examples

but not corona

in

(DCT)

in

§4,

the §2 c a n

and,

on

theorem

fails.

PWS's

of i n f i -

HOLDS

Type

for w h i c h

~:

show

of PWS's.

(i)

PWS's

is to f i n d

As a p r e l i m i n a r y

space

satisfy

Our o b j e c t i v e

genus,

function

we

ideal

show,

there

PWS

IA. nite

that

fails.

as n e g a t i v e We

sufficient

in

bordered

of PWS's.

§2 p l a n a r

H~(R)

so as to

problem

sur-

discus-

connectivity.

(iii)

then

PWS's,

that

and

classification and

compact

be t y p i c a l

planar

to be a PWS a n d

in the t o p o l o g y

be m o d i f i e d

§i.

for

is true;

of

foregoing

of c o n s t r u c t i o n :

(ii)

we c o n s t r u c t

(DCT)

corona

types

A remaining

a necessary

every

never

TYPE

examples

our

of i n f i n i t e

holds;

theorem

is false.

Next,

Although such can

on P W S ' s

(DCT)

some

to s h o w t h a t

different

corona

bedded also

a PWS,

at a r e a s o n a b l e

In

Widom

in vain.

for which

theorem

aiming

satisfies

a labor

be p l a c e d

OF P A R R E A U - W I D O M

is to p r o v i d e

in o r d e r

to g i v e genus

the c o r o n a

chapter

type

is e v i d e n t l y

of i n f i n i t e fails

OF S U R F A C E S

in t h i s

(DCT)

the unit which i.e.

multiplicities.

section

holds. disk

makes

• R

each point In

§i

R

is to c o n s t r u c t

A Riemann

surface

if t h e r e

exists

an n - s h e e t e d , of

~

denotes

has

R

is s a i d to be an a n a l y t i c

branched,

exactly

a Riemann

n

full

cover-

pre-images,

surface

of t h i s

209

type and

{~j} = {6j:

over which Theorem.

~

j = i, 2,...}

is branched.

Let

R

fined above.

is the sequence

surface of Myrberg type over

Then the following conditions

R

(b)

The set

(e)

The sequence

in

~D

Then the following holds.

be a Riemann

(a)

of points

D

as de-

are equivalent:

is a PWS. H~(R)

separates {%j]

the points

of

R.

satisfies

the Blaschke

(i-l{j[)

< ~.

condition,

i.e.

j=l The proof will be given in IG below,

after some preliminary

obser-

vations. IB.

Lemma.

Suppose that a bounded analytic

separates

the points

satisfies

the Blaschke

Proof.

We define,

in

~-i(~,)

where

~-I(~)

function

for every

function whole

in

= {al,..., of

n

~* • ~ \ {~j}.

on

R

Then

{~j}

< • D \ {%j},

= (-l)kak(f(al),...,f(an)), a n}

and

~k

variables.

~ \ {~j)

~.

f

condition.

Ak(%)

metric

for some

function

denotes

Clearly,

the k-th elementary

Ak

and thus can be continued

The definition

of

Ak'S

sym-

is a bounded analytic analytically

to the

implies that

f n + (A I o ~)fn-I + ... + A n O ~ = 0 holds on

~ - i ( ~ \ {~j})

discriminant

and thus everywhere

of the equation

see that the function

D(~)

and only if the equation arates the points the points

{j

in

iC. verges,

condition.

(z E R: g(a,z) choiee of distinct

to

X

n

has

D({j)

D(~) = 0.

function

H~(~)

B(~,a) > e}.

D(~) = 0

roots.

be the

in

X.

D(~)

~ 0

Since

f

We if sep-

does not vanish identically. Namely,

D(~)

on

~j's ~.

So

At

are among the zeros {~j}

satisfies

[] integral

f0 B(~,a)d~

is the first Betti number of the region Since the convergence

a, we may assume that the set points.

Let

and that

distinct

Next we want to see when Widom's where

R.

X n + AI({)xn-I + --- + An(~) belongs

~-i({,),

we have

of the bounded analytic the Blaschke

in

on

We may also assume,

conR(e,a)

does not depend on the

~-l(@(a))

consists

by applying a conformal

of

n

transfor-

=

210

mation

of the

~-i(0)

disk

contains

if n e c e s s a r y , n

distinct

h(z) Then

we h a v e

h(z)

= -log

function

(see

harmonic

everywhere

h(z)

closed

u(z) and

region

follows

that

u(z)

For e a c h first such we

Betti that

find

on

e > 0

number

let

of

{z E R:

S

S .

I~(z)I

a constant

l~(z)I find

~ r.

R.

Lemma.

Let

Proof.

< r}

is c o n n e c t e d .

such

Moreover, Aa,

Let

> 1

any

such

I~(z)I

S e = {h(z) {I~(z)I

in

S

e -e.

by

(i),

R ( ~ / A , a I) all

of

ID.

S

cannot .

By use

of

$ log p

on

see that

log I~(z)l

is a p o t e n t i a l ,

.

and

number

B ' ( ~ , a I) r,

< 0 it

the

0 < r < i,

By H a r n a c k ' s

inequality

S A g ( z , a I) a' E R

with

l%(a')l

< r < I,

that < Aa,g(z,a, )

Then

S

> ~}

< r}

and

e < h(z)

and

=C R ( ~ / A , a l )

= {[~(z) I < e-S},

which

includes

the

so is c o n n e c t e d .

I~(z)I

Conversely,

is c o n n e c t e d

< e -~,

take

any

=< A g ( z , a I)

have

any

compact

Let

0 < e < - log r.

i.e.

z E se.

z E R and

with

thus

complementary

r <

Then

B ( ~ A , a I) ~ B ' ( ~ , a I) ~ B ( ~ / A , a l ) .

So

R ( e A , a I)

l,(z)l

z E R ( ~ / A , a I).

component,

[]

Lemma.

= p.

is

since

~ r.

z E R(~A,al) ; then

is i n c l u d e d Then,

Green

that

given

0 < e < - log r.

set

Then,

u(z)

we h a v e h

~(z) n ~ ( a ' ~

with

We h a v e

of the

p ÷ i, we

> e}

fix a p o s i t i v e

R ( e A , a I) ~ S

connected

hand,

Since

: {z E R: h(z)

AaTlg(z

z E R

I~(z)l

we h a v e

We

A > i

a constant

for any

that

We set

: h(z) + log I~(z)I

Letting

other

on

form

0 < p < i.

curve

~ p}.

On the

local

u(z)

with

functions

A - I g ( z , a I) S h(z)

for

we a s s u m e

a = a I.

~ 0, as c l a i m e d .

(i)

we

R.

~ h(z)

the

that p

on the

I~(z)I

and

an

g(z,ak).

any

$ log p

Namely,

= 0.

In fact,

for h a r m o n i c

0 ~ u(z)

~(a)

al,...,

I) shows

Take

{z e R:

is n o n n e g a t i v e therefore

Ch.

R.

u(z)

principle

n [ k=l

=

19(z)l.

6A,

on

is p o s i t i v e ,

the m i n i m u m the

Theorem

that

points

< As

it i n c l u d e s

211

Proof.

We claim that

relatively have

compact

S

in

3U ~ 3R(~A,al).

therefore cycle

g(z,a I)

y

gion

in

If

has no components were

It would follow that bounds

in

U, which are

such a component,

on

then we would

g(z,a I) : ~A

on

3U

U, a contradiction.

and

So, if a

S , then it should bound in the re-

This shows the first inequality.

be shown similarly.

The other half can

[]

We thus have for any •-log r I B(~A,al)d~ which

U

would be constant

R(eA,a I)

R(~A,al).

\ R(~A,a I)

S .

0 < s < -log r i-log r

~

B'(~,

al)d~

r-logr B(~/A,al)d~,

~

implies A_iI-l°g r ~

B'(~,al)d~

~As

Hence,

converges

We now estimate

suppose that the circle the double

S

of

a meromorphic

S

say

We see that

~.

It is then clear that

makes

order to obtain a further Riemann-Hurwitz

relation:

branched,

= e}

Let

contains

B(~'al)d~"

~, of S

B'(e,a I)

does.

0 < ~ < -logr

and

~j's.

~(z)

We then form

can be extended to

onto the extended

an n-sheeted,

branched,

complex plane

full covering of

is exactly the genus of

information

about

if a compact

full covering

(2)

S

f0 B~(~'al)d~

no

so that the function

function, ~

if and only if

f0 B'(~'al)d~" {l~I

¢.

sheeted,

~ A;c/A

"s

f0 B(~'al)da

IE.

r-log r

I-l°g r B(~,al)d~

In

B'(~,al) , we need the

Riemann

of another

S •

surface

compact

F

surface

is an nF0, then

X = nx0 + V,

where and

X V

(resp.

X0) is the Euler characteristic

of

F

(resp.

is the sum of the orders of all the branch points of

Nevanlinna

F

Fo) (see

[44], p. 324).

Lena.

(3)

B'(~,a I) ~ b(e -e) + I - n,

where

I~1

<

Proof.

is the total order of branching

b(e -~ )

over the disk given by

e -~

Let

X

spectively;

so

and

X

X = -2.

¢, the Riemann-Hurwitz

be the Euler characteristic Since relation

S

of

is an n-sheeted,

(2) gives us that

S

and

~, re-

full covering of

~ = - 2 n + 2b(e-e).

212

Since

S

is a compact surface,

X + 2

is equal to the first Betti

number of ~ , i.e. 2B'(~,a I) (cf. Ahlfors and Sario The formula (3) then follows at once. [] The meaning of still holds when IF. with

B'(~,a I)

~S

contains

Suppose that

I~(z)I = e.

and

b(e -~)

[AS], p. 55).

shows that the formula

(3)

some branching points.

0 < g < - log r

and there is no branch point

z

Then we have

I~g B ' ( a , a l ) d a

= ~ B ' (~,a I ) ~ - [~ a d B ' ( a , a g ~

1)

~

f

= -eB'(g,al) Let

r'

be fixed with

r < r' < i.

Then

~dB'(~,al). g

dB'(~,a I) = db(e -~)

for

0 < ~ < - log r'. So, if n. denotes the total order of branching over 3 the point ~j, then we have, for 0 < e < - log r', I-log r' - .g ~dB,(e,al)

=

=

I-log r ' -~C

l

edb(e -~)

exp(-E) log r db(r)

Jr ~

= -

When

[j njlog l%jl

[ n log I~jl. r'
is convergent,

_f~logr' ~dB,(~,al)

therefore ~0 B'(e'al)de converges. integral converges. Since B'(e,a I)

Conversely, suppose that the last is decreasing in ~, we see that

eB'(g ,a I) ~ /0g B'(~,al)d~ ÷ 0 as E ÷ 0 verges. It follows that the series

-

~

converges and

nj log

and so

[0 ~dB'(~'al)

con-

l~jl

r'< I ].I
The integral

~{njlog l~j I: Since only if

Hence we have the following f0 B'(~'al)de

0 < l~j I < I}

i ~ nj ~ n-i

~j log I~j I

converges

if and only if the series

converges.

for all

j,

~j njlog [~jl

converges or, equivalently,

converges

if and

~j (I - l~j I) < ~.

213

IG. shown IB.

Proof

of T h e o r e m

in T h e o r e m Namely,

\ {%~},

9B,

Ch.

H~(R)

separates

exists

a function

points

of

the

Blasehke

IF and

the

the r e m a r k

Verification 2A.

quence

We b e g i n type.

[x2j_l, and

R2

be the tifies with

each

covering

2, . . . . along

IA,

these of

R

condition

of i n f i n i t e Theorem.

Proof.

Let

exist

positive

that

{z e R:

end

(c) ~

of ID.

(a)

for a n y

takes

n

that

follows

~* E

distinct {~j}

from

Sat-

Lemma

[]

a class

of e x a m p l e s

of PWS's

R the

numbers

r

< r}

(4)

Then

unit

closed

usual disk

each

of the

if and

only

< ~.

We n o t e

by

which

of M y r b e r g xj,

{xj} that

be

joining

fashion.

points if

R

~,

is a s u r f a c e

R2

segments

surface

in the

the

R

over

the

a Riemann

onto

(i - xj)

be a PWS direct

be f i x e d

l~(z)l

along

segments

R

is a PWS =i

Then

a E R

slit

of se-

Let identype

j = i,

satisfies

such

a surface

genus.

Let

4.

IB shows

Lemma

We d e f i n e

R. with ~. 3 p o i n t of o r d e r two

map

Thus

each

leaf

Blaschke

2B.

disk

projection

By T h e o r e m

is a l w a y s

R, then,

which

constructing

unit

crosswise

a branch

2, . . . . the

j = i,

natural

in

already in L e m m a

Ix.: j = i, 2,...} be a s t r i c t l y i n c r e a s i n g ] real n u m b e r s c o n v e r g i n g to i. Let R 1 and

of the o p e n

x2j] ,

points

(b) was

is c o n t a i n e d

f • H~(R)

Finally,

at the

(a) ~

(c)

(DCT) With

Let

of p o s i t i v e

two c o p i e s

RI

of

given

(b) ~

the

9-i(%,).

condition.

implication

fact

if

isfies

Myrberg

The

The

there

v a l u e s J at the

2.

IA.

V.

- log

once

of M y r b e r g

Cauchy and

type

theorem

(DCT)

for all.

and

Aa with is c o n n e c t e d ,

~ ( z ~ ~ @((a_)

over

As we

0 < r < I

holds saw

and

I@(a) l < r

~

with for

R.

in IC,

there

Aa > i

such

and

~ Aag(Z,a )

1-~-I~7~(z) for any

z e R

with

meration

of all

critical

I@(z)l

~ r.

so large

take

an i n t e g e r

N > 0

j ~ N.

From

(4)

that

g(z,a),

9(zj) - ~ ( a ) . . . . . . . . log i - ~ - - ~ (zj) for

j ~ N.

Since

R

{z.:

of

We t h e n

follows

Let

points

is a PWS,

j : i, counting

that

2,...}

be an enu-

multiplicities.

19(zj)l

~ r

for e v e r y

< Aag(Zj,a) =

~j g(zj,a) _

< ~

and therefore

214

@(zj) -¢(a) : j : i,

] 2,...~

i - @(a)~(z.) ] We further infer that

satisfies the Blaschke condition. I, 2,...}

l,(zj)l Bn( 6 ) : for

n = i, 2, . . . .

j:n+l ~

{¢(zj):

j :

Put

also satisfies the Blaschke condition• ¢(zj) - ~ 1-~-~3

~

Then ~(zj) -~(z)

[Bn(,(z)) I : j:n+l

j:n+l

g(z,z.)) 3

(a) : gn (z). Since we have I

I1-

B (eit)12d°(t)

: I

n

= 2- 2Re[I

: 2(1-

~

f

77 j=n+l

l~(zj)[)

be a meromorphic

has a harmonic majorant, meromorphic on R and

say

n

B (eit)da(t)] n

as n + ~, a suitable subsequence l a.e. on T. We also note that Now let

(2 - 2Re(B (eit)))do(t)

= 2 - 2Re(B (0)) n

÷ 0

{Bn(k): k = i, 2,...} converges to B ÷ 1 almost uniformly in ~. n (a) function on R such that If g

u, on

R•

If

h n = B n f, then

n

lhn(z)Iexp(-j=l [

hn

n

g(z,zj))

= IBn(z)f(z)lexp(-j=l [

If(z)Ig(a)(z) By the weak Cauchy theorem f(a)Bn(~(a))

g(z,zj))

~ u(z).

(Theorem 4B, Ch. VII) we have = I

f(b)Bn(@(b))dXa(b)" A1

Taking the limit along the subsequenoe

{Bn(k)} , we get

f(a) : lim f(a)Bn(k)(@(a))

is

215

f

:

^

f(b)[lim k÷~

AI

: I

(@(b))]dXa(b)

Bn(k)

f(b)dXa(b)" A1

This

shows

for all

§2.

each

exist

PLANE

have

3.

i.e.

(DCT)

holds

given

in

Some

plane

fails.

It w i l l

seen

3A.

denote

In w h a t axis.

£(a,b)

( b - a)/2,

so that

begin

with

Lemma.

follows

For

A(c,~)

Let

we

every the

later

the

[] we h a v e

(DCT)

is valid.

the

that

the

with

be the

subset

of

closed

disk

with

-K = A ( - b , - a ) .

respect

to the

solution

of the D i r i c h l e t

problem

for the r e g i o n

boundary

data

equal

to

Proof.

If ~\

u0(-z)

is the h a r m o n i c

to

denotes

( K U (-K)),

and

1

region

includes

on

i

on

( K U (-K)),

~\

u(0)

ZK

u

measure

-~K

Thus u(z)

= 1/2.

0

on

a e b

we

and

radius

£(a,b). ~ c}.

MoreWe

0 < a < b < ~ that

of

with

~F. ~K

u0(0)

to

= 1/2.

F a

of the

i.e.

the

(K U F )

with

the

Then

u(0)

~ 1/2.

respect

to the

with

we

to the

function i,

and has

measure

In fact,

respect

harmonic

equal

F

(KUF),

[\

= 1/2.

~ u0(z) []

[\

on

measure u0(0)

is a b o u n d e d

we have

~ Un(0)

of

region

to

identically

( K U (-K)).

of

Re(z)

be the h a r m o n i c

and

we h a v e

values

~\

let

the h a r m o n i c

then

u0(z) + u0(-z) boundary

In p a r t i c u l a r ,

may

centers

with

Suppose

in

equal

for

facts.

with

Since

problem

(a + b)/2

{z E ~:

~K

K

with

is a d i a m e t e r

"disk"

complement of

~

at

boundary

with

type

kind.

a, b

center

[a, b]

infinite

disks

numbers

connected

(-K))

that

FAILS

corona

of this

use v a r i o u s

disk

simple

K : A(a,b)

gion.

(DCT)

shown

of P a r r e a u - W i d o m

regions

of real

interval

some

closed

u0

(DCT a) h o l d s

genus,

FOR W H I C H

regions

some

shall

pair

closed

denotes

proving

an a r b i t r a r y

for

VIII,

Lemmas

by

over,

be

R.

infinite

TYPE

to c o n s t r u c t

solution

Ch.

for w h i c h

going

Simple

the real

genus

OF P A R R E A U - W I D O M

an a f f i r m a t i v e

5A,

for this

2A has

of i n f i n i t e

REGIONS

(DCT)

By T h e o r e m

surface PWS's

We are which

(DCT a) holds.

a E R,

As there

that

see that same re-

on

~\

(KU

it is i d e n t i c a l l y Since

in the

~\

domain

(KUF) of

u 0.

216

3B. A(c,~)

Lemma. are

Let

conformal

structure

the

we have

case,

Proof. which

We h a v e

symmetric

the

= 0.

disks

= L(~)

3C. with

that

Then

the

origin

disks

in t h e

A(a,b)

sense

a - 1 - b -I = c -I

to f i n d a f r a c t i o n a l and

carries

the o r i g i n • L

see t h a t

maps

and

of the

If t h i s

and

and

L(A(c,~))

are

intervals

and

-L(b)

and

A(a,b)

is

axis

symmetric

The

onto

L([c,~])

L(z)

and

b < d < c

L(A(c,~)),

L([a,b])

= L(c).

transformation

disks

the r e a l

L([a,b])

L(~(a,b))

the

linear

the

So let

disks

L(A(a,b))

-L(a)

to t h e

provided

Trivially,

if so a r e

computation.

[,

about

We a l s o

transformed

only

of

respect

the o r i g i n

(z - x) -I + d -I. L(0)

with

b- a = a2/(c -a).

only

preserves

to d i s k s

0 < a < b < c < ~.

symmetric

and

set

itself are

A(c,~) L(z)

and

diameters

respectively.

about

L([c,~]).

desired

result

This

of

So the

the origin

and

if a n d

happens

follows

=

if

by a simple

[]

Lemma.

Let

i : 0, i,...,

K 1• = ~ ( a i , b i),

0 < a n < bn < an-i

< a0 < b0 < ~

< bn_ I < -..

n,

be c l o s e d

and

set

the r e g i o n

Dn

disks

n

(5)

D

= [\(

u i=0

n We d e n o t e

by

gn(~,z)

at the p o i n t

at

points

.

x I,..

Proof. axis, on

Since

the

infinity

, x n,

~.

the r e g i o n

Dn

implies the

that,

interval

critical

the

(bi,

points

contours, theorem

for e a c h

shows

lemma.

has

(-~,

a

2,...,

with

n

of

n.

n, t h e r e

The

critical i

'

2

'''''

= 0.

critical

Ch.

V).

Dn

This

x. l xi's

has

as t h e

= 0

theorem

So t h e s e

points,

gn(~,x)

we get

the

is a s t r i c t l y

thus in are

n + i

Riemann-Roch

establishes

following decreasing

function

on t h e

interval

) . n

3D. tions,

As a t r i v i a l

we g e t

the

consequence

following

of t h e

definition

"

so t h a t

a point

since

n

to t h e r e a l

Rolle

exists

hand,

IC,

=

is r e g u l a r ,

Dn

other

n i

pole

[~gn(~,x + iy)/~y]y=0

2,...,

of Theorem

exactly ai-l'

<

with respect So

Sgn(~,xi)/~x On t h e

exactly

(see the p r o o f

point

i = I,

has x .l

<

with

[]

As a c o n s e q u e n c e , Corollary.

for

gn(~,z).

gn(~,z)

gn(~,z).

i = i,

a i _ I)

of

b I.

is s y m m e t r i c

boundary

gn(~,bi ) = gn(~,ai_l ) = 0

for

gn(~,z)

we h a v e

function

Every

function

Then

for which

so is the G r e e n

n Ki)" ~ \ (U i=0

Green

K.). 1

of the G r e e n

func-

217

Lemma.

Using

region

D

4.

the n o t a t i o n s

4A.

Theorem

We are in a p o s i t i o n

There

such that the f o r m

exists

H~(D)

to p r o v e

(ii)

a regular

possesses

in Ch.

VIII,

In v i e w of T h e o r e m

the s e c o n d m a i n r e s u l t

Let

of the

region

Dn

w i t h pole

in

[

of P a r r e a u - W i d o m

ideal w h i c h

type

is not of

VIII,

the d i r e c t

be c l o s e d

d<sks with

Cauchy Theorem

(DCT)

D.

0 < • -. < an defined

D

4G.

K i = A ( a i , b i)

the r e g i o n

region

a B-closed maximal

5A, Ch.

for such a r e g i o n

Proof.

Dn

in the

~ gn+l(~,z)

chapter.

Theorem.

fails

gn(~,z)

n+l"

Existence

present

in 3C, we have

<

b

n

<

an-i

by

(5), and

~.

Suppose

<

--.

<

a 0

gn(~,z) that

<

b 0

<

the G r e e n

an ÷ 0

and

~, function

for the

set

oo

s

= ~-\ ( u

K. u { o } ) .

i:l Finally,

let

g(~,z)

We c o n s t r u c t arbitrary

our disks

fixed closed

By C o r o l l a r y interval

be the G r e e n Ki

disk with

3C the f u n c t i o n

we c h o o s e

chosen

an+l,

disks

bn+ I

function

for

D

with pole

in the f o l l o w i n g

way.

0 < a0 < b0 < ~

and

g0(~,x)

(-~, a0) , we h a v e

that we h a v e

l

is s t r i c t l y

0 < g0(~,z) K0,... , Kn,

g s0

for

0 < an+ I < bn+ 1 < a n

with

0 < c n < 1/2,

K0

be an

s 0 = gB(~,0).

decreasing

on the

0 ~ x < a 0.

n ~ 0, w i t h

with

Let set

~.

Suppose

~i = gi (~'0)" satisfying

Then

the fol-

lowing conditions: (a)

an+ I = Cna n

(b)

bn+ I - an+ I = a n + 1 2 / ( a n - a n +

(c) where ~

n = 0, i,...}

should

This c o n s t r u c t i o n Let

u

satisfy

is a l w a y s

(resp.

the r e g i o n

tion

Kn+ I = A ( a n + l , b n + l ) , and the s e q u e n c e {Cn:

condition

by t a k i n g

on

n=0 On

sufficiently

of the D i r i c h l e t

small.

problem

for

n

Dn+ I

~Kn+ I (e) that

implies

an a d d i t i o n a l

possible

v ) be the s o l u t i o n

n

i) on

1 ) = {Cn/(l- Cn)}an+l,

i n f { g n ( ~ , z ) : z E ~Kn+ I} ~ se n w i t h is a f i x e d c o n s t a n t w i t h 0 < ~ < i

w i t h the b o u n d a r y

and to Un(Z)

that the d i s k

0

on

$ ~CnVn(Z) Kn+ I

data

equal to

~K 0 U ... U ~K n. on

Dn+ I.

gn(~,z)

It f o l l o w s

(resp.

On the o t h e r hand,

and the i n f i n i t e

disk

to

f r o m the c o n d i -

A(an,~)

LemJna 3B

are

sym-

218

metric

with respect

A(an,~) , Lemma Un(Z)

in

to the origin.

3A shows

that

Dn+l, we have gn+l(~,O)

Since

Vn(0)

K 0 U ... U K n

~ 1/2.

Since

is i n c l u d e d

gn+l(~,z)

in

= gn(=,z) -

in p a r t i c u l a r

= gn(~,O) - Un(O) < ( i - ~/2)~ =

S gn(~, O) - e a n V n ( O )

. n

Thus we h a v e by i n d u c t i o n ~n =< ( i - ~ / 2 ) n s O.

(6) By L e m m a all

n.

So

creasing Harnaek monic

theorem

function,

D.

gn(~,z)

sequence

boundary on

3D,

value

of b o u n d e d sequence

say

u, on

0

Namely,

gn(~,z) then

say

in e a c h of the

converge shows

can be m a d e a r b i t r a r i l y

small

large

that

enough.

g(~,z)

a regular

Since

tends

region.

to

~ gn(~,z) z

tends

s

is a r e g u l a r

Next t a k e any boundary total

values

l e n g t h of

(7)

for

at a l m o s t

see t h a t

in

D, w h e r e I ~K

n : i, 2,...,

n

for e v e r y

The point,

and t h e r e gn(~,z) by l e t t i n g

n, we c o n c l u d e Namely,

D

is

(i- e/2)neo < ~.

n =0

We see that

every point

f

has the n o n - t a n g e n t i a l

on the b o u n d a r y

~D.

S i n c e the

is finite,

f(z)

z

D.

that

of the o r i g i n

to the origin.

[

on

PWS.

f E H~(D).

~D

~ n

har-

identically

n = 0, i,..., (6) i m p l i e s

~ gn(~,Zn ) ~ ~ gn(~, 0) n:O n=O

[

By the

seen to h a v e

g(~,z)

g ( ~ , z n) ~

n=O

D.

vanishes

we have

= D

to

for de-

nonnegative

Moreover,

n:O

Hence,

an) ,

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

as

D

has a s i n g l e c r i t i c a l

The r e l a t i o n

g(~,z) 0

(bn+l,

on

is e a s i l y

and t h e r e f o r e

g(~,z)

intervals points.

u

decreasingly

that

e x i s t no o t h e r c r i t i c a l

n

~D

functions

to a b o u n d e d

The f u n c t i o n on

on the r e g i o n

is a m o n o t o n i c a l l y

harmonie

converges

D.

everywhere

principle

~ g(~,z)

n = 0, i,...>

positive

this

argument Zn,

~ gn+l(~,z)

{gn(~,z) - g(~,z):

= f(~)

+

each path

I = the l e n g t h of we have,

for e v e r y

[ 1 ~ n=0 ~K ~K

n n z

f

~K

f({) d~ ----2-7 c n

is d i r e c t e d

clockwise.

= ~e n l a n / ( l - C n _ in the n e g a t i v e

S i n c e we

I) < 2~c ,a = n-~ n r e a l axis,

for

219

I

f(~) d% ~- z

~K

< IIfll ]~Knl/a n < 2Wen_iIifil . . . .

"

n So the r i g h t - h a n d In:0 Cn Then

side of (7) c o n v e r g e s

is e o n v e r g e n t .

f(z)

tends

to

We d e f i n e

f(0)

as

even

f(0)

z

for

z : 0, for the series

by the f o r m u l a

tends

to

0

(7) w i t h

z : 0.

a l o n g the n e g a t i v e

real

axis.

4B. Lemma.

We n o w i n t e r r u p t If

D

is as a b o v e a n d

is a B - c o n t i n u o u s Proof.

complex

(i)

¢(f)

above,

¢(f)

= f(0)

We t h e n s u r r o u n d Yn'

~

n.

and

:

: f(0) of

lim x÷-0

is an a l g e b r a i c

each disk

n = 0, i,...,

ented clockwise; each

if

homomorphism

As we h a v e r e m a r k e d

It is t h e n c l e a r that ¢.

the p r o o f and s h o w the f o l l o w i n g

Kn

T h e n it f o l l o w s

iYnl

for any

z

in

is t r u e

each

~

CI(D)

for

on

lying

z = 0.

homomorphism

exclusive;

(ii)

(= the l e n g t h of

f r o m the u s u a l

Cauchy

¢

of

in

H (D) D

each

into

such that Yn

is o r i -

yn ) ~ 3 1 ~ K n l / 2

integral

formula

for that

f(~) d -~z Yn

in the e x t e r i o r

The p r o p e r t y

then

f(x).

Yn

1 I f(6) 1 - z 6d~ : ~ i 2-~ ~K n

this

f E H~(D),

by a c i r c l e

are m u t u a l l y

(iii)

for

H=(D).

(iii)

of

Yn"

of

Yn

In p a r t i c u l a r , implies

that

for

Yn On_ 1 °a : a n - 2 ( i _ Cn_l )

I~] => a n - ( b n - an)/2

n

3en_l/2

-

It f o l l o w s

f(%) ~

d~n(C)

n

>

a

n

/2.

iYnl l~Knl < i)f)) .a---77 =< 3Jifil~ - - -a n n

d~

=

Yn For e a c h

a

n-i

that I[ i]

(8)

1l-c

n = 0, i,...

= (2wi~)-id61y n

we d e f i n e •

3On_ I.

Since

In=0 ~ Cn

finite

Borel measure =.

fD f ( z ) d ~ ( z )

for

~n

T h e n the i n e q u a l i t y

defined

at the p o i n t

a measure

is c o n v e r g e n t , on

S i n c e the f o r m u l a f E H=(D),

~ =

D, w h e r e

~

(7) w i t h

it f o l l o w s

< 6wttf)i~ C n - I

=

on

D

"

by the f o r m u l a

(8) i m p l i e s

that

n:0 ~n + ~

is a w e l l -

denotes z : 0

from Theorem

11~nll < =

the D i r a c m e a s u r e

shows that 5C, Ch.

¢(f)

IV, that

:

220

¢

is B - c o n t i n u o u s

4C. and

¢

In o r d e r

ideal

factors.

of

To

respect

(see

Ch.

single

I,

~(~Kn)

Un=0 ~

~Kn

ease,

(ii)

Take

that

of

that

7

then

in Ch.

a polar

Vlll,

÷ ~

I

7

to

D

Since

is a B - c l o s e d common

see that

~

7

be the h a r m o n i c

~D = Un= 0 ~K n u

as

= 0}.

no n o n c o n s t a n t

of h a r m o n i c set,

we c o n s i d e r

~(f)

maxinner

cannot

4G.

let

~({0})

continuous

z ÷ 0

measure,

at the p o i n t Since

= 0

and

function

u~

U n=0 ~

through

<

u~({)dw(6)

{0}

measures).

we have

any n o n n e g a t i v e

u~(z)

4A,

H~(D), has

it is easy

7,

definition

forms

= i.

such

homomorphism

D, of the b o u n d a r y

{0}

of T h e o r e m

J = {f E H~(D):

concerning

5D for the

point

~n=0

form

claim

proof

set

We c l a i m

is the

to

the

and

complex

in the

see our

with

finish

above

H~(D).

If this

be w r i t t e n

H~(D).

to

constructed

is a B - c o n t i n u o u s imal

on

~K n

the therefore on

and

~.

~D Let

u0

be the

with

the

boundary

see that to

0

u8

as

I,

z

u0(z)

5B

tends

in

sense

tion,

say

that

lh01

an o u t e r has

D. D.

of Ch.

h0,

(cf.

no n o n c o n s t a n t

that

~(D,~

-I)

We n o w factor

tains

a function,

We d e f i n e

has

h.

k E 7.

But

divisible

by

+ 0

So t h e r e ~

inner

say

as

~(D,~ hl,

at the

l.a.m,

D

= -u 0

has

through that

inner J

factor

see that z ÷

to

by this

0 l.a.m.

holomorphic

secsuch

that

6A,

Ch.

h0

is

~(D,~) VII,

we

see

factors.

has

no c o m m o n

k = h 0 h I.

of

h 0 E ~(D,~))

means

By L e m m a

that

0

sense

is q u a s i b o u n d e d ,

clearly

common

origin

tends

determined

to

tends

function

and

(i.e.

inner

shows

We thus

a bounded

lh0(z)ilhl(Z) I ~

our c o n s t r u c t i o n h, w h i c h

whose

z ÷ 0

Look

the

in the

exists

factors.

by s e t t i n g

that

problem

D

it is easy

g(=,z)

on

This

-I)

Then

in 4A,

means

bundle

log lh01 5D).

for the r e g i o n

111.36).

D.

line

on the c o n t r a r y ,

Since

This

Dirichlet

through

bundle

II,

shown

Theorem

no n o n e o n s t a n t

k E H~(D)

k(z)

As

it is a b o u n d e d

Since

Ik(z) I = so that

2C.

problem Un= 0 ~K n.

D.

82,

0

line

Ch.

D.

be the

common

suppose,

inner

p.

~

= exp(-u0).

l.a.m.

on

for the

to

II,

of the

u~

on

Clearly, Let

Dirichlet

to

through

[65],

tends

on

z ÷ 0

in the

z

0 point

Tsuji

as

exp(-u0(z)) as

of the

equal

to

boundary

(cf.

+ ~

data

is q u a s i b o u n d e d

is a r e g u l a r Ch.

solution

a noneonstant inner

is not

common

factors,

divisible

it con-

by

h.

Then,

IlhlII~'exp(-u0(z)), D. the

is a c o n t r a d i c t i o n .

This

means

inner This

in p a r t i c u l a r

factor proves

of

k

Theorem

that

is not 4A.

[]

221

4D. the

As we e a s i l y

fact

that

disks

by any

axis;

e.g.

§3.

5.

FURTHER

Banaeh and

to H o f f m a n algebras

under

[34]

and

Let

set

the

H~(R);

namely,

M(H~(R))

space

of the

Banach

know

by the

pact

Hausdorff

function, for

every

uous

and

of

{

the

fine

{~i} if

H~(R).

theory

space.

the

For e v e r y

space

+ ~(f)

defined

each

from

H~(R)

element

a

on f u n c t i o n s the

reader

of

i.e.

H~(R),

into

the

The

identity

~ • topolspace

i, we

we a s s o c i a t e

seen

f(~)

a

= ~(f)

to be c o n t i n -

(Namely,

preceding,

= I.

is a c o m -

by s e t t i n g

is e a s i l y

in the

~(i) to

ideal

M(H~(R))

H~(R)

f.

set

to an e l e m e n t

that in

of

with

f • H~(R).

is d e f i n e d f

¢

the

relative

the m a x i m a l

has

f

transform that

fixed.

forms

of B a n a c h

for e a c h

function

Gelfand

of b o u n d e d

exposition

converges

algebras

Which

R

We r e f e r

topology

is c a l l e d

Since

The

weak

M(H~(R))

M(H~(R))

is d i f f e r e n t

boundary

to the r e a l

is h e l d

on

operations

Ha(R)

the

of B a n a c h

With

~ • M(H~(R)). is c a l l e d

on

H~(D).

ideal

with

~l(f)

algebra

f, on

depend

algebra

which

functions

for a d e t a i l e d algebra

in

thus

Banaeh

z • R}.

of the a l g e b r a

M(H~(R))

say

here

[12]

of the

~

only

not

if we r e p l a c e

respect

be a PWS,

algebraic

be the m a x i m a l

general

the

R

holomorphie

is e q u i p p e d

if and

ogical

Let

pointwise

or G a r n e t t

a net

with

goes

Space

IIfll : s u p { I f ( z ) l :

especially

M(H~(R))

4A does

proof

axis.

is to i n v e s t i g a t e

norm

M(H~(R))

same

symmetric

Ideal

on a PWS.

of all h o m o m o r p h i s m s The

are

of T h e o r e m

The

on the r e a l

of all b o u n d e d

supremum

proof

OF PWS

objective

algebra

the

which

the M a x i m a l

functions

H~(R)

our

is a disk.

intervals

into

Our

set

n sets

PROPERTIES

Embedding 5A.

see,

K

closed

closed

holomorphie The

each

the m e a n i n g

where

f

meant

function.)

point

z • R

the

evaluation

functional

e : f ÷ f(z) Z is a n o n z e r o mines on into

homomorphism

an e l e m e n t

R, the m a p

the

This

means

evaluation

set

z At this

H~(R)

in t u r n

of the

M(H~(R)).

z + s

M(H~(R)).

Then

in

is seen point

separates that

functionals

algebra Since

f

into in

to be a c o n t i n u o u s we use

our hypothesis

the p o i n t s

distinct of

H~(R)

every

H~(R).

points So,

of z

by

R in

~

and

H~(R) function that

from R

by T h e o r e m R

determine

identification

so d e t e r -

is c o n t i n u o u s R

is a PWS. 9B,

Ch.

V.

distinct

of each

point

222

z

w i t h the c o r r e s p o n d i n g

M(H~(R)).

functional

In v i e w of the e x i s t i n g

to the f o l l o w i n g (i)

ez, we r e g a r d

theory

of

R

H~(~),

as a s u b s e t of we n a t u r a l l y

come

questions:

Is the e m b e d d i n g

z ÷ E

of

R

into

M(H~(R))

a homeomor-

Z

p h i s m onto an o p e n (ii)

Is

R

The first explained corona next

subset? dense

question

right

problem,

below.

in the

space

M(H=(R))?

has b e e n a n s w e r e d As for the

the a n s w e r

which

is not d e f i n i t e .

The r e s u l t

is u s u a l l y

We w i l l

called

discuss

is

the

it in the

section.

5B.

We n o w s e t t l e the q u e s t i o n

Theorem.

If

R

maximal

ideal

Proof.

Let

n o t e by

space R

0.

is a P W S ,

(i).

then e v e r y open

M(H~(R))

be a PWS.

of

Take an a r b i t r a r y

of

0

in the

TR

~

is o p e n

in

in the

R, w h i c h we de-

each n e i g h b o r h o o d M(H~(R)).

of

0

in

R

To show this,

in §i of Ch.

IX can be u s e d h e r e

map

with

•

+ R

be the g r o u p of c o v e r t r a n s f o r m a t i o n s

for

arguments

be the u n i v e r s a l

in 3F-3H of Ch.

be the g r e a t e s t

character P0(0)

R

change.

Let let

point

space

we first n o t e that all the d i s c u s s i o n s without

s u b s e t of

H~(R).

What we n e e d to see is that

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

Q0

affirmatively.

second,

of

~ 0

element

QO"

and

covering

IX.

common

Set

H0(R)

i n n e r f a c t o r of

We t h e n t a k e a f u n c t i o n

JIP011~ ~ 1/2.

h E H~(R)

For e a c h

%.

P0

in

f E H=(R)

= 0

We r e p e a t

= {f E H~(R): H~(R)~ o ~

%(0)

f(0) and

the

= 0}. ~0

Let

the

H=(~,~0 ) there

and

such that

exists

a unique

such that h o ~ = (p0/Q0)( f o % - f(0)).

We d e f i n e

a linear

fixed

in

~

(S%f)(0). operator

(resp.

and

operator

with

As shown

Choose fixed,

{

T

I~I < i

in Ch.

IX,

in

3G,

Since

~(~)

S~

a complex homomorphism)

an e l e m e n t

f0 E H~(R)

with

by s e t t i n g

Tf : h.

s f = (I - ~ T ) - i f (resp. in

and

For e a c h }~f =

~ ) is a b o u n d e d

linear

H~(R). (Tf0)(0)

~ 0, w h i c h

is h e l d

set

v : {~ e M(H'(R)):

of

H~(R)

we set

Tf 0 E H=(R),

M(H~(R))

f0(0)

containing

= ~(f0)/~(Tf0).

= 0

and

the p o i n t

Clearly,

I~(f0) I < I~(Tf0)I}.

~

(Tf0)(0) 0.

~ D,

Define

is c o n t i n u o u s

V

is an o p e n

~: V ÷ C

subset

by the f o r m u l a

and has m o d u l u s

less

223

than

i

on

equation

V.

(14)

Take

in Ch.

any IX,

w

in

V

3G a f t e r

and

apply

setting

it to b o t h

h = f0

and

sides

of the

~ = Y(m).

We

get (9)

~(f)

for

every

the

equation

set

f E H~(R). (9)

U : VNR.

an o p e n

Since

implies Then,

subset

of

Since

~(z)

= f0(z)/(Tf0)(z)

on

i.

U, ~oo.

function

on

U

r 0 > O.

For

0 < r < r0

{m • V:

I~(w)I

< r}.

subset

U

R, e a c h

of

in

M(H~(R)).

Ur

(reap.

r0.

@

The

M(H~(R)). borhood

6.

spect

set

in the

set

is d e n s e

Theorem. (a)

Let

a subregion

(b) of

is compact.

onto

the

on

R.

M(H=(R)). R

that

0

in

Vr

for

is o p e n ~) m a p s

each

0 < r <

Ur = Vr

as a subset space

< r0}, Vr :

on the o p e n

(reap.

< r}

with

be in-

{I~I

R

for

0 <

of the

space

is a n e i g h -

is a r b i t r a r y

is o p e n

U

and

each

~

it is

holomorphic

< r}

see that

in the As

We

of

should

function

open

0

{.

= ~(Sz ).

a disk

I~(z)l

{161

T, we

is thus

of

~

Similarly,

disk

of

V,

includes

{z • U:

implies

of

the

every

in

R, we

in

M(H~(R)).

[]

R

is c o n v e x

with

functions

and

also

functions

on

R.

that More

following

if each R',

re-

holomorphic

PWS.

is a c o n n e c t e d

every

PWS

of all h o l o m o r p h i c

be a r e g u l a r

into

regular

of b o u n d e d space

bounded then

line

R'

the

following

surface

holomorphic

which

function

on

hold: includes R

R

as

extends

= R.

hull

If(z) I ~ s u p { I f ( ~ ) I :

Then

Riemann

bundle

R, the ~ ( R , ~ ) - c o n v e x {z E R:

r0

~(z)

is a u n i v a l e n t

in

of

subset

that

H~(R)

R

and

For

open

in the

R'

analytically

is open

0 < r < r0,

space

we w r i t e

is a c o n t i n u o u s

neighborhood

here

we h a v e

If

Ur

M(H~(R)),

H~(R)

We p r o v e

precisely,

K

every

of

to the

Ha(R)

Ur,

0

that

6A.

set

of into

it is h o l o m o r p h i c

~(U) =

V

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

Sz'

9

points of

I ( T f 0 ) ( z ) I } , so t h a t

~

z E U,

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

every

Density

with

So,

@

Ur

Let

words,

definition

Thus

of

conclude

we

Since

the

is i n j e c t i v e

= 0.

V r) u n i v a l e n t l y

Since

r < r 0.

The

~

In o t h e r 4(0)

z for

Since

with

If0(z) I < 0.

By i d e n t i f y i n g

than

separates

is an i n j e c t i o n

containing

U.

jective

~

U = {z e R:

R

set

less

H~(R)

that

to the

modulus

: ~(~)f

~

over of

R

and

every

compact

subset

K, n a m e l y

~ E K}

for each

f E ~(R,~)},

224

(c) R

Ha(R)

in the

is d e n s e

topology

The

of u n i f o r m

statement

(a)

(b) a n d

(c),

proof

(b) and

(c) w i l l

6B. R'

Proof

of

is c o n n e c t e d ,

nonempty.

Let

sequence [n=l

(a). the

b

Suppose

2,...}

for

some

bundle, 2B,

is seen

say

~.

Ch.

V),

such

Then

Thus,

f

R'

f

function

cannot

by

have

In o r d e r

results

functions

may

a eluster

to p r o v e

~ R.

R'

and

choose

R'

< ~

for

VII,

2A, the

l.a.m

.,

in

R'

a

and the

z e R \ {an:

function

which

is

denotes

every

z +

determines

by W i d o m ' s -I)

Since

space

g(a,z)

we find,

[ n=l

and

a

theorem

therefore

an

vanishing

at

g(an,Z))-

in

function (a)

f

be h o l o m o r p h i c

the

zeros

point

on

R

extends at the

in its domain.

holomorphi-

point

of any n o n z e r o

the

the p r o p e r t i e s

To m a k e

notations: on

X.

the

let

b.

This,

holomorphic

The

Riemann

the

statement

(a)

functions

on

R

the

points

such

following

(BI)

and

B-convex

that of

(B2).

we use

somewhat

subset

B F

two

gen-

simpler,

we

a set of c o m p l e x of

X

we d e n o t e

set:

surface

separates

(c),

be a set and

If(×) I ~ s u p { I f ( { ) I :

be c a l l e d

(b) and

description

X

For an a r b i t r a r y

following

{x • X:

hyperbolic

R'

an ÷ b

h E ~(R,~

holomorphic

should

because

of Bishop.

following

~(X,B)

which

The

[]

6C.

valued

is a PWS,

lh(z)lexp(-

bounded

Thus

is i m p o s s i b l e

the

subsection.

in the

within

that

in Ch.

section

R.

to E. Bishop.

that

R

R, w h e r e

g(an'Z)

By the h y p o t h e s i s

however,

is proved.

R

of

that

2, . . . .

into

R

to be an i n n e r

Since

is a n o n z e r o

n = i,

~n:l

a nonzero

If(z) I =

eral

in

RI

f • H~(R)

use

such

z

by the a r g u m e n t

(Theorem

tally

R

for

of

of

in

and,

due

contrary

point

subsets

on

6D.

boundary point

functions

in the n e x t

results

in

on the

holomorphic

on c o m p a c t

general

2,...}

exp(-~ n=l ~ g(an'Z))

an,

some

be d e s c r i b e d

be a b o u n d a r y

< ~

of all

is p r o v e d

function

n : i,

line

and

topological

{a n : n = I,

g(an'Z)

Green

we use

space

convergence

is easy

To p r o v e of

in the

R.

and A

hull

let

{ • F}

for all

of

Now,

A

contains

F.

let

be an a l g e b r a the

The t h e o r e m s

constant

of B i s h o p

f e B}, R

be a c o n n e c t e d

of h o l o m o r p h i c function we n e e d

are

i

and the

225

Theorem. every

(BI)

If

function,

F

can be a p p r o x i m a t e d (B2)

Suppose

all h o l o m o r p h i c gence

R'

that the

in

by elements is c l o s e d

and an a l g e b r a

A'

of h o l o m o r p h i c

element A'

(v)

e R' × R':

A

functions

conver-

Riemann

on

R'

such

as a subregion. is e x t e n d e d

precisely

of

z ~ w,

f(z)

R' × R'

holomorphically

of these

to an ele-

extensions.

for all

f e A'}

F'

of

R'

there

points

in

exists

a compact

R' × R'

F' ~ ~(R';A').

set

subset L

(z,w)

= f(w)

w h i c h has no c l u s t e r

subset

such that

with

F'

of

R'

and all points

E T.

The

and all r e l a t i v e l y

See Bishop of

R'

in

For each compact

F'

of u n i f o r m

of

set

R

w ~ L of

in

consists

subset

of

exist

F,

hold:

is embedded

For each compact

F

H(R)

Tc

the union of a compact

union

R

then

of

A.

in the space

a connected

is a c o u n t a b l e

exists

on

in

interior

in the t o p o l o g y

T = {(z,w)

subset

F A

F : ~(R;A),

on the

Then there

The

(iv)

on

that

with

R.

and

(iii)

uniformly

subsets

Every

A'

R

and h o l o m o r p h i c

of

R

(ii) ment

of

R

moreover

following

(i)

subset

on

functions

on compact

surface

is a compact

continuous

set

L

compact

[3] for the proof

of

the

set

z E R'

~'(R';A')

for which

can be taken

components

of

(BI) and Bishop

is

there

to be the

R' \ F'

[4] for the proof

(B2).

6D.

We are going

comments.

Let

H(R)

tions

on

R, w h i c h

gence

on compact

topological H(R)

subsets

algebra;

in

H(R). H(R)

Since

separates

that

the points

by Bishop's

theorem

and an algebra properties

function

(a) then rates

(i)

A'

of

implies

the points

R

Then,

exist

that of

(B2).

extends R'

of

set

conver-

its closure,

9B, Ch.

so does

(iii)

on

V,

the a l g e b r a Riemann

R'

into A'

say

A.

surface

which

R' = A.

(B2)

So, R'

satisfy

The

A,

T c-

H~(R)

~ A, every bounded

of

in

is c o n v e r g e n t

in the t o p o l o g y

a connected

H~(R)

in

H(R)

by T h e o r e m

and t h e r e f o r e T

in

H(R),

holomorphically

= R

R, the

func-

of u n i f o r m

is then a complete

algebra

functions

Since

Tc

H(R)

and a f o r t i o r i there

some general

and the m u l t i p l i c a t i o n

sequence

topological

of h o l o m o r p h $ c

R

space

Cauchy

is a PWS.

- (v) of on

The

is a s u b a l g e b r a

(B2),

We begin with

space of all h o l o m o r p h i c

the a d d i t i o n

and every

R

(c).

the

with the t o p o l o g y

R.

is also a c o m p l e t e

We now assume

phic

of

namely,

H~(R)

(b) and

as above,

is equipped

are c o n t i n u o u s

in

to prove be,

the

holomor-

statement

Since is void.

A

sepa-

226

Proof of

(b).

(v) of (B2) the

Let

set

(iO)

F

be an a r b i t r a r y

~(R;A)

is c o m p a c t ,

~(R;~(R,~))

for e v e r y

line b u n d l e

~

set

~(R;A).

(i0).

Choose

H~(R)

is dense

in

A

see that

H~(R)

F U {z}.

Suppose

exists

a point

Since

This b e i n g

where

e

hnf • ~ ( R , ~ )

for all

is a c l o s e d inclusion

and an e l e m e n t Tc, t h e r e

s u b s e t of a

relation

f E A.

in

Since

exists a sequence

{f } tends to f uniformly n b e l o n g s to ~ ( R ; H ~ ( R ) ) , we have

z

true

f

constant.

on

l h ( z 0 ) I,

Take any

f • ~(R,~).

and

in

So we

such that

< c e I :

2,...

we

Then there

~(R;H~(R)). R

in

A, we

inclusion

is not the case.

h

in

z • 9(R;H~(R)),

lying outside

~ e r)

the l i m i t s

is a r b i t r a r y

for e v e r y

function

n : i,

By t a k i n g

As

that this

holomorphic

is a p o s i t i v e

the s t a t e m e n t

Next, we show the first

z 0 • ~(R;~(R,~))

sup{lh(~)I:

is shown,

second

~ • F}.

~ ~(R;A).

on the c o n t r a r y

can find a b o u n d e d

By

~ ~(R;A)

n = i, 2 . . . . .

~ sup{if(~)I:

~(R;H~(R))

R.

We c l a i m

such that

~ e F},

z • ~(R;A).

(i0).

the

s u b s e t of

is void.

~(R;~(R,~))

in the t o p o l o g y

in

If(z)l

find that

because

T

Once this

z • }(R;H~(R))

Ifn(Z) I ~ s u P { i f n ( ~ ) l : n, we h a v e

R.

First we prove

a point

{f : n = I, 2,...} n on the c o m p a c t set

~ ~(R;H~(R))

over

(b) will be e s t a b l i s h e d , compact

compact

for

As we h a v e

z 0 • ~(R;~(R,~)),

so we get

If(z0) I = Ih(z0)nf(z0) I ~ s u p { l h ( ~ ) n f ( ~ ) I : ~ • F} e n . s u p { i f ( % ) I : ~ • F}. As

n

is a r b i t r a r y

point

z0

dicting

should

property

(c).

subset

clearly

(BI) w i t h

F

F

f of

f

zero of e l e m e n t s

Hence,

the first

be any h o l o m o r p h i c R

a n d set

F'

in

~(R,~),

inclusion

is a r b i t r a r y ,

[]

we h a v e

s u b s e t of

Moreover

F' : ~ ' ( R ; H ~ ( R ) ) .

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

uniformly

on

R

function

F' = ~ ( R ; H ~ ( R ) ) .

is compact. that

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

on each c o m p a c t as desired.

V.

implies

A = H~(R),

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

As

Let

(b) just proved,

convex hulls

F ~ F',

9A, Ch.

f(z 0) = 0.

The contra-

in (i0)

is

[]

P r o o f of any c o m p a c t

0 e c < i, so we s h o u l d h a v e

thus be a c o m m o n

to C o r o l l a r y

established.

orem

and

F'

on

By B i s h o p ' s

flF ,

of

f

in

H~(R).

shown that by b o u n d e d

on f

F

Take by the

the d e f i n i t i o n

by e l e m e n t s

uniformly

R.

Then,

to the

by e l e m e n t s

set

F'

Since in

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

holomorphie

of

the-

functions

H~(R). uniformly on

R,

227

§4.

THE

7.

CORONA

(DCT) 7A.

an o p e n zebra R if

Corona

Let

be a PWS.

R

subset

R

in

(DCT)

now

M(H=(R))?

We

be p r o v e d

7B.

Let

V

sarily

into

[7]

M(H~(V))

says

Behrens

for w h i c h Consider eluded by

that

V

that

the

the

corona

in

conjecture

conjecture

of the of

H~(V),

holds

alIs

for

R

following type

for w h i c h

This

V

complex

is the

for the o p e n

This

if,

M(H~(V))

is not n e c e s we a g a i n

L. C a r l e s o n ' s

of c o n s t r u c t i n g

plane

case

If

V

conjecture

M(H~(V)).

in 4A.

is h o m e o m o r p h i c a l l y

although

corona

holds.

extended

V.

then

is true

construction

a continuum.

subset,

By the

a method

use of o u r

includes

as an open type.

conjecture is the

Banach

problem:

holds.

points

of

of the

corona

with

unit

a variety

is e x p r e s s e d

mean

famous disk.

reLater

of r e g i o n s as follows.

a sequence

in

V

r a d ( K n)

such

space

is d e n s e

[i] d e v i s e d

{K : n = i, 2,...} of d i s j o i n t c l o s e d disks inn centers ~ w h i c h c l u s t e r o n l y on ~V. We d e n o t e n the r a d i u s of the d i s k K n. We say that the s e q u e n c e {K n} with

is h y p e r b o l i c a l l y disks

ideal

of P a r r e a u - W i d o m

statement

sult

the

is i d e n t i f i e d

of P a r r e a u - W i d o m

subregion

[\ V

the

result

in 7C by m a k i n g

separates

the m a x i m a l

embedded

conjecture

R

M(H~(R))

corona

first

region

the c o m p l e m e n t

5B,

studying the

Our

be a c o n n e c t e d

H~(V)

space

say that

corona

will

denotes

ideal

a plane

This

such that

M.

exists

Examples

by T h e o r e m

set a b o u t

M(H~(R)).

but the

for i n s t a n c e ,

the

in

Positive

Then,

of the m a x i m a l

There

fails

Theorem:

We w i l l

is d e n s e

Theorem.

FOR PWS

and the

H~(R).

dense

PROBLEM

K' n that

with

rare

in the r e g i o n

centers

~

such

n

V

that

if t h e r e K

exist

C K' C V n = n =

disjoint

for e a c h

closed

n

and

co

r a d ( K n ) / r a d ( K n)

< ~.

n=l

Theorem. U

Suppose

is o b t a i n e d

closed

disks.

from

V

Then

the

See

Behrens

7C.

Proof

in 4A does

that

the

corona

by d e l e t i n g corona

[i] for the

of T h e o r e m

the r e q u i r e d

conjecture

holds

for

a hyperbolically

conjecture

holds

V

rare

and

a region

sequence

for

U, too.

show

that

of

proof.

7A.

We are

job a f t e r

going

a slight

to

modification

the r e g i o n

in the

D

construe-

228

tion.

We r e c a l l

requirements: 0 < ~ < i

that

D = T\

first,

and

let

(Un~ 0 A ( a n , b n) U {0})

~,

a0

0 < a 0 < b 0 < ~;

n = 0, i , . . . } ,

{a

n

: n = i,

and then

2,...}

b0

the

be a r b i t r a r i l y

determine

and

with

{b

n

three

: n = I,

following

fixed

with

sequences 2,...}

{Cn:

recurrently

by the c o n d i t i o n s : 0 < c n < 1/2

(CI)

and

[ n=0 ~

(C2)

an+ I = Cna n

(C3)

bn+ I - an+ I = a n + 1 2 / ( a n -

(C4)

inf{gn(~,z):

where

gn(~,z)

for

c n < ~,

n = 0, i,..., an+ 1 ) = { C n / ( l -

z • ~A(an+l,bn+l)}

is the

Green

function

with

\ (Ui7 0 A ( a i , b i ) ) . Take a sequence {X : n = i, 2,...} -i n such that [ n=l ~ Xn < ~ We t h e n set

pole

for the

n = i, disk

a i -Cn_ 1 2, . . . .

with

By a s i m p l e

A(a~,b~)

X n r a d ( A ( a n , b n ))" n = i,

2,...

Since

the

small, Then,

are m u t u a l l y

ture

last

by the

holds

to the

holds

punctured

disk

also

the

that

ically

D.

punctured

jecture

rare

A ( a n , b n)

Since

sequence

converge

n = i,

2,...;

X

> i

for e v e r y

n,

rad(A(a~,b~))

the

disks

included

the

{A(an,bn):

domain

[\

to

A(a~,b~

0;

n = i,

=

A(a',b')n n

in the r e g i o n

D

is a r e g i o n to show

~\

theorem

same

of P a r r e a u - W i d o m

that

rad(A(a

implies

the H~

as that

n = i,

for the

2,...}

are m u t u a l l y C [\ =

nl

n

n

:

X

n:l

Ii~ n <

the

disjoint

(A(a0,b 0) U {0})

~.

corona

con-

for the

disk.

of disks

because

type,

conjec-

is e o n f o r m a l

that

,b n)

~ad(a(a~,b ,)

corona

for the a l g e b r a

and

X :

the

(A(a0,b0) U {O])

(A(a0,b0) U {0})

' b') ~ A(an' n

c 's sufficiently n all the c o n d i t i o n s .

satisfying

only

2, . . . .

by t a k i n g

D

(A(a0,b0) U {0}),

is e s s e n t i a l l y

in the

and

and are

for

the r e g i o n

Carleson's

~\

A(an'bn) for

4A,

So we h a v e

disk,

for

region

of T h e o r e m

fails.

for

see that,

see that

disjoint

can be f u l f i l l e d

get a p l a n e

proof

numbers

Dn =

if we have

condition

(DCT)

we

A ( a n , b n) to

2en(l + A n + i O n ) < 1 - Xnen_ 1

we thus

for w h i c h

with

for the r e g i o n

i + (X - l ) C n _ i / 2 n .a i -Cn_ I n

b' = n

it is easy

n = 0, i,...,

n

computation

is c o n c e n t r i c Moreover,

\ (A(a0,b0) U {0}) (C5)

and n

~

for

of p o s i t i v e

i- (Xn+l)Cn_i/2 a' = n

Cn)}an+l,

~ ~gn(~,0)

We

see

is h y p e r b o l centers with

of

229

Thus T h e o r e m

7B i m p l i e s

that the c o r o n a

D, as was to be proved.

8.

Negative 8A.

Our o b j e c t i v e

t i o n of f a m o u s

genus.

Riemann

Sn'S

of d i s j o i n t

We d e n o t e

very

for w h i c h [43],

the c o r o n a the c o n s t r u c -

in a s i m p l e w a y to

simple trick

surfaces

by

~'n

joint u n i o n

be a s e q u e n c e

%

for d e f i n i n g

considered

copies

Let

0 ~ Re(z)

w i t h an open arc in

~Sn+ I.

B n') the left surface

~n

in

h e r e are

Since

R'

can be e m b e d d e d

which

R'

is not dense,

and d e f i n e

Bn_ I

O n : {z E Xn:

We s u p p o s e be a se-

vertical

for each B~

side of

n

the side

w i t h an o p e n are

are a s s u m e d

Riemann

surface,

Riemann

to be d i s j o i n t .

say

R",

surface.

in

Although

it is p o s s i b l e

to m a k e

simple m o d i f i c a t i o n .

{~n: n = i, 2,...}

a vertical

right)

is a PWS or not,

into a PWS a f t e r a s u i t a b l e Take a s e q u e n c e

(resp.

and

in a l a r g e r

R'

~S n. 2,...}

R' : R ( { S n } , { X n } ) , out of the dis-

it is a h y p e r b o l i c

we do not yet k n o w w h e t h e r

of c o m p a c t

< i}.

and the side

an

Sn

strip

by i d e n t i f y i n g

~Sn

H e r e the arcs

n = i,

~ 2, 0 < Im(z)

(Un~ I Sn ) U (Un~ I Xn),

Bn

borders

{Xn:

of the r e c t a n g u l a r

(resp.

an'

of i n t e r i o r s

with analytic

disjoint.

We t h e n f o r m a R i e m a n n

nn < i

PWS's

shown by N a k a i

All the R i e m a n n

surfaces

are m u t u a l l y

{z:

R'

As

of B. Cole can be m o d i f i e d

{Sn: n : i, 2,...}

bordered

X n.

for the r e g i o n

to be c o n n e c t e d .

Let

quence

is to p r e s e n t

We b e g i n w i t h N a k a i ' s

PWS's of i n f i n i t e

that

here

solution.

examples

such PWS's.

supposed

holds

Examples

p r o b l e m has a n e g a t i v e

yield

conjecture

[]

slit,

Re(z)

say

of p o s i t i v e On,

in each

= i, n n ~ Im(z)

numbers Xn

with

0 <

by s e t t i n g

< i}.

We set

R({Sn};{Xn};{nn}) so

R'

m a y be d e n o t e d

Theorem. then the Proof. of

W

by

If the s e q u e n c e surface We w r i t e

= R' \ ( U o ); n:l n

R({Sn}{{Xn}~{I}). {n n}

converges

R({Sn};{Xn};{nn])

W = R({Sn};{Xn};{~n})

corresponding

to the f i n i t e

to zero

is a r e g u l a r

union

sufficiently

rapidly,

PWS.

and d e n o t e

by

Wn

that part

230

I

Si

i:l

Each

fix a sequence

creasing a E SI the

U

U Xi\ai i:l

1

U {z e X

: 0 < Re(z) =

n

W is e a s i l y s e e n to be the i n t e r i o r n a n d t h e r e f o r e has a f i n i t e f i r s t B e t t i

face We

II

to

zero

and

let

surface

R'

W

g(a,z)

g(a,z) (resp.

pole

for e v e r y

of a c o m p a c t number,

: n = i,

n

a.

bordered

sur-

we call

B n-

which 2,...}

en+l ) < ~.

g'(a,z))

with

strictly

Take

be the

Green

Since

W

a fixed

function

depoint

for

is a s u b r e g i o n

of

z E W.

~n'S,

z = x + iy

{~

Bn+l(Cn-

(resp.

to d e t e r m i n e

coordinate

numbers

[n=l

R')

< g'(a,z)

In o r d e r local

of positive such that

< i}.

in

take X

an arbitrary

which

we u s e d

n

and

consider

the

at the b e g i n n i n g ,

so

n

that

X n = {0 =< x =< 2, 0 < y < i}.

{0 < x < 2, y = 0} in it is r e g a r d e d

may

Since

be r e g a r d e d

as a r e g u l a r

the

lower

as a b o u n d a r y

boundary

point

horizontal

arc

of

of

R'

R'

side each point

In p a r t i c u l a r ,

we h a v e g ' ( a , l + iy) as

y > 0

tends

such that

to

zero.

g ' ( a , l + iy) the

So t h e r e

e en

sequence

that

First

we

{~n )

with

0 < Y < ~n"

By t h e m a x i m u m

have

g(a,z)

for a l l

g ( a , l + iy)

e ~n

exists

for all

claim

see t h a t

÷ 0 a positive

I + iy E X n

thus

obtained

~ g ' ( a , l + iy)

.

the

< cn

principle

z E W \ W

has

choose

desired

Green any

nn > 0

0 < y e ~n"

for a l l

for the

Now

number

with

property.

i + iy E X n function

~

with

in t h e that of

Then

region

the

first

Wn+l,

the

region

Wn+ I

as a r e g u l a r

Betti

number

i.e.

B(e,a)

f Our assumption

W(e,a)

en

subregion of

on t h e

W

is a PWS.

bordered

ular.

Since

every

boundary

> en" This

B(~,a)d~

sequence

As

en

R(e,a)

is i n c l u d e d

I, IA). is n o t

It f o l l o w s larger

than

that

~ B n + l ( e n - Sn+l).

{E

n

On the o t h e r

Riemann

g(a,z)

completes

(Ch.

> e}

Cn+ I }

then

~ B(~,a)d~ 0

compact

g(a,z)

Thus

I namely,

> n

= {z E W:

B(e,a)

~ Bn+ I.

we

~

n

> en+ I.

We

surface,

is c o n t i n u o u s

on

Wn+ I

boundary Wn+ I

is the

interior

of a

points

of w h i c h

is reg-

and

is less

point,

we

see t h a t

{g(a,z)

~ ~}

tend

to

zero,

surface

is t h u s

the proof.

[]

the

that

< ~;

hand,

every

implies

than

is c o m p a c t seen

an

at

whenever

to be r e g u l a r .

231

8B.

A very u s e f u l

reformulation

of the c o r o n a

problem

is g i v e n

by

the f o l l o w i n g Theorem.

A PWS

R

for e v e r y

finite

sequence

such that

is d e n s e

in the space

of e l e m e n t s

M(H~(R))

{fi:

Ifl(z) I + . . - + Ifn(Z) I ~ ~ > 0

a sequence fng n = i

{gi: on

i = i, 2,...,

in

for e v e r y

H~(R)

n}

if,

in

z e R, t h e r e

such that

H~(R) exists

flg I + --. +

R.

Hoffman

[34; p. 163] p r o v e s

The p r o o f c o n t i n u e s

8C.

n}

if and o n l y

i = i, 2,...,

to be v a l i d

this

In v i e w of the p r e c e d i n g

the f o l l o w i n g

in the case of the unit disk.

in the p r e s e n t

in o r d e r to o b t a i n

theorem

ease as well.

it is s u f f i c i e n t

a negative

example

to p r o v e

for the c o r o n a

problem. Theorem. there

There

exists

a regular

e x i s t two e l e m e n t s

PWS

R

w i t h the f o l l o w i n g

f, g E H~(R)

property:

such that

IfI + IgI $ 6 > o everywhere

on

R

but the e q u a t i o n fh + gk : I

has no s o l u t i o n s Proof.

Let

of a c o m p a c t

the f o l l o w i n g

holomorphic

H~(R).

follows

0 < 6 < i

Cole's malformed

interior having

in

The c o n s t r u c t i o n

examples. notes

h, k

in

Wm

finite

there

]gmI ~ i,

fm h + gm k = i

tained Wm"

from

We t h e n

the s e q u e n c e say

let

by a t t a c h i n g

W, is a r e g u l a r

m

de-

namely,

W is the m with analytic border

W

m two f u n c t i o n s

fm' gm'

such t h a t

h, k E H = ( W m )

on

Wm;

implies

Ikl $ m.

S

be a f i n i t e R i e m a n n m an a n n u l u s to each b o u n d a r y

surface

is f i x e d PWS.

%,

W

m

form a Riemann {qm}

exists

Ihl + SUPw

m = i, 2,... Wm

m = i, 2,...

IfmI + IgmI ~ ~

m For e a c h

surface

on

with

SUPw

same lines as that of C o l e ' s

surface:

Riemann

and c o n t i n u o u s

the

For e a c h

Riemann

bordered property:

IfmI ~ i, and such t h a t

almost

be fixed.

R({Sm};{Xm};{~m})

surface

component

let

Ym

of

as in 8A, w h e r e

in such a w a y that the r e s u l t i n g

Then

ob-

be a s i m p l e a n a l y t i c

surface, arc that

232

starts

at

joining

~Wm, passes Sm

to

are m u t u a l l y

directly

Sm+l,

disjoint

and that

F: has no r e l a t i v e l y ways p o s s i b l e and

go

compact F

Lf01 __
using

(B2)

Carleman

go = gm

lg01

and

< 1

Igl ~ 2

ous way,

R

if n e c e s s a r y , So, using

denoting

by

RAW

and

H~(R).

based

we may assume the Green

and

have

f, g

We suppose

g01

theorems W

< @/4m for the C a r l e m a n of

F

W

on

in an obvi-

has no compact

compo-

as in the p r o o f of T h e o r e m

8A and

function

for

first

R

with

Betti

z E R \ W n.

on the c o n t r a r y

sup W

in

Restricting

W \ R

(BI)

such that

pole

number

a, we see

B

and that n

fh + gk : i.

lm = fm h + gm k"

fc

=

that

Define

It follows,

as we did in 8A,

f h + gk : 1

has no s o l u t i o n

there

elements

exist

the f u n c t i o n

1

on m

ting

is al-

F.

on

neighborhood

we show that the e q u a t i o n

such that

This

~unctions

on

for instance,

that

the same

W.

on B i s h o p ' s

Ifi + Igi ~ 6/2.

gR(a,z) ~ g(a,z) < gn for all that R is a r e g u l a r PWS.

H~(R)

Ig01 => 6

functions

[36],

in

continuous

n

Finally

Ym'S

%;

If01+

the same n o t a t i o n s W

that

conditions: on

be a c o n n e c t e d

gR(a,z)

n

next

+ S U P W--m U Y m l g -

f01

(See Kaplan

Let

If[ ~ 2,

components

Define

fo : fm'

approximation.)

X m \ Om

We assume

m:l

by the f o l l o w i n g

which

that

%1

type a p p r o x i m a t i o n

m = i, 2, . . . .

nents.

m:l

seen.

S U P W--m U Y m l f -

for

~Wm+ I.

the union

in 6C, we find h o l o m o r p h i c

(ii)

the slit r e c t a n g l e

at

complementary

as is e a s i l y

on the set

through

and t e r m i n a t e s

It follows Ii - lm[

from

: SUPw

m

h, k E W

by setm

(ii) that

i h ( f - fm ) + k(g - gm) V m

= SUPw

i h ( f - f0 ) + k ( g - g0)I m

~sup w

lhi + sup w m

<

If

m ~ JJhlJ~+ JJkrJ~, then

belong

to

H~(Wm )

(ithii

oo

llml ~ 3/4

and satisfy

Iki~'4~ m

+

llkiL

on

oo

)/4m.

W m.

Thus

fm(h/l m) + gm(k/Im)

h/l m = i

on

in

and Wm.

k/l m It

233

then follows

from the p r o p e r t y

of

Ih/Xml

SUpw

fm'

gm

Ik/Xml

+ SUpw

m

that ~

m

m

and t h e r e f o r e

sup w

Ihl

+ sup w

m

As

m

of

h

can be made and

space

k.

H~(R),

arbitrarily

Hence

Ikl

3m/4.

~

m

large,

the e q u a t i o n

this

contradicts

fh + gk = i

as was to be proved.

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

has no solutions

in the

[]

NOTES Theorem such a PWS

IA t o g e t h e r

satisfies

Theorem

(DCT)

to that of Rudin

Koosis tive

example

Cole's

paper

Neville

[64].

That

[27].

but a simple

argument

In a sense our example

[7].

A simple

account

(Theorem

genus

6A)

8C)

finite

5B)

is taken

is a d a p t e d

described is similar

example

[47] contains

either

(Theorem

is due to Nakai

Riemann

solved

further

Hara

is new,

in his

furnished

by

[12] or in

while

See Gamelin

the nega[ii]

[14] c o n s t r u c t e d

conjecture

examples

[66].

in Garnett

[43].

surfaces.

the corona

7A)

[64] and

by C a r l e s o n

new proof was r e c e n t l y

may be found

for w h i c h

from Stanton

from Pranger

for the unit disk was

The p o s i t i v e

malformed

of infinite

by Hayashi

[30],

(Theorem

(Theorem

problem

A detailed [40].

found

by the author.

theorem

theorem

The corona

Wolff.

was

is due to Stanton

[68].

The e m b e d d i n g

very famous

its proof

4A is due to H a y a s h i

here has been devised

the density

with

for PWS's

holds.

of PWS's

satisfying

(DCT).

CHAPTER Xl.

CLASSIFICATION

OF PLANE REGIONS

This chapter deals with the c l a s s i f i c a t i o n p r o b l e m of plane regions in terms of Hardy classes or, more precisely, classes.

The result extends Heins'

surfaces given in bis book [31].

in terms of H a r d y - O r l i c z

c l a s s i f i c a t i o n table for Riemann

Using notations to be explained below,

our result can be expressed roughly as follows: 0 G = 0AB , : 0AS < N{0q: < U{0q:

0 < q < ~} < 0AB.

The first equality is long known p. 332).

0 < q < ~} < 0-P < 0p < 0 P+

(cf. Sario and Nakai

[62], p. 280 and

The second one concerns the q u a s i b o u n d e d m a j o r a t i o n and is

due to Segawa

[63].

Except these equalities,

all that happen in the

c a t e g o r y of Riemann surfaces can already be r e a l i z e d in the c a t e g o r y of plane regions.

The crux of our d i s c u s s i o n lies in c o n s t r u c t i n g a cer-

tain kind of exceptional

sets in the plane.

After having done this, we

shall see little d i f f i c u l t y in solving the problem. In §I we describe some basic p r o p e r t i e s of H a r d y - O r l i c z classes of analytic functions.

§2 is devoted to the c o n s t r u c t i o n of e x c e p t i o n a l

sets, b e l o n g i n g to a p r e a s s i g n e d class, of positive logarithmic capacity. A merit of our c o n s t r u c t i o n is that the size and p o s i t i o n of such sets can be at our disposal.

By making use of exceptional

sets we prove in

§3 our c l a s s i f i c a t i o n result, w h i c h implies the inclusion r e l a t i o n mentioned above as a special case.

§i.

I.

HARDY-ORLICZ

Definitions IA.

chapter, [0, ~) plane D

CLASSES

Let

be a convex function,

any nonconstant, with

~

~

~(0) = 0.

we denote by

such that

~(log+Ifl)

nondecreasing, For any region

H~(D)

by which we mean,

through this

convex function defined on D

in the e x t e n d e d complex

the set of h o l o m o r p h i c functions

has a h a r m o n i c m a j o r a n t on

D.

f

on

If this is the

235

case,

then

~(log+Ifl)

has the LHM,

We say that a set totally

disconnected

for any r e g i o n

V

the c o l l e c t i o n only c o n s t a n t If convex Hardy 0p

E ~ ~

compact

in

T

is a null s u b s e t of

including

of n o n v o i d

regions

E. D

in

of

defined

N~

and

0AB~)

NAB,,

We also d e n o t e on w h i c h

by

in Ch.

0G

IV.

follows

Every

sense that

0AB)

we o f t e n

is a

by

0~

contains

H~(D)

Q ~ t < ~, is a

coincides

w i t h the

We d e n o t e

(resp.

the c l a s s

with

of s u b r e g i o n s

(resp.

and

P AB*(D)

by

0¢)

N¢,

N

¢(t) D

no n o n c o n s t a n t

= t.

of bounded

use the c o n d i t i o n :

f E H~(D)

In fact,

~(t) ~ Cot

there for

i.e.

~(log+Ifl)

on

if

characteristic.

function

on

D

on

has a h a r m o n i c

numbers

f E H~(D)

D, then we h a v e

is of b o u n d e d

bounded holomorphic

So,

characteristic

log+Ifl

exist a p o s i t i v e

t ~ tO.

= ~.

is of b o u n d e d

log Ifl e SP'(D),

f

H~(D)

In this case we w r i t e

functions

lim ~(t) t t÷~

Hence,

E

= H~(V)

let us d e n o t e

= e p t - i,

class

He(D)

(resp.

(i)

function

if

functions).

In what

IB.

N~

D.

H ~ ( V \ E)

for w h i c h

0~, r e s p e c t i v e l y .

the class

t h e r e e x i s t no G r e e n

holomorphie

~

~(t)

and the c o r r e s p o n d i n g

HP(D)

in p l a c e

and if Moreover,

0 < p < ~, then the f u n c t i o n

class

set of class ~

on

functions.

function

(resp.

D.

for it is s u b h a r m o n i c

cO

and

and

u

log+Ifl

tO

on

LHM of the

~ c01u + t O

to any

in the

such that

is the

On the o t h e r hand,

belongs

D

majorant

H~(D).

on

D.

every Summing

up,

AB(D) ~ He(D) C AB*(D), where

AB(D)

is the

same as

H~(D).

Concerning

the c l a s s e s

0~, we

have

0G ~ 0AB, ~ 0¢ ~ 0AB. A precise 2.

Some 2A.

r e s u l t w i l l be g i v e n

later.

Basic P r o p e r t i e s We are g o i n g to state

classes

following

stated

for g e n e r a l

after

Parreau

Riemann

some b a s i c [51].

surfaces,

properties

Although

of H a r d y - O r l i c z

the p r o p e r t i e s

we r e s t r i c t

our a t t e n t i o n

can be to p l a n e

regions. Let

S

be a p r o p e r

such that the r e l a t i v e

subregion

boundary

of

S A ~G

~

and let (= Cn,

say)

G

be a r e g i o n is n o n v o i d

in

S

and con-

236

sists

of a n a l y t i c

side of

We

suppose,

moreover,

be a h a r m o n i c

function

on

that

G

lies

on one

such that

u = 0

Then

~(u +)

CO.

Lemma.

Let

on

and

CO

and

curves.

u

%(lul)

~(u-)

Proof.

have

A s was

has

a harmonic

harmonic

shown

majorant

majorants

in IB,

u

S ACI(G)

on

on

G.

both

G.

is o f b o u n d e d

characteristic,

so t h a t

+

and

u

u

Let such

exist {Sn:

that

(Ch.

n = i,

SI NC O

G AS I

such that

G n Sn

which

II,

5A).

2,...}

be a r e g u l a r

exhaustion

is n o n v o i d .

We c h o o s e

a component,

S I n ~G I ~ ~.

Moreover,

let

includes

G I.

We

set

COn

Gn

S n n ~G n

:

boundary

data

equal

to

0

on

COn

and

to

S

(Ch.

say

be t h e and

It is t h e n c l e a r t h a t COn is a p a r t of c O• + N o w let un be t h e s o l u t i o n o f t h e D i r i c h l e t the

of

GI,

of

component

problem

of

: ~G n \ COn"

Cln

max{u,0}

I, IA)

for on

Gn

with

Cln.

As

max{u,0}

on

+

max{u,0)

is s u b h a r m o n i c ,

we

see t h a t

u

is the

LHM of

n

G n.

If we d e n o t e

by

U

the

%(u +) ~G n.

Since

~(u~)

%(luI)

: ~(max{u,0})

n

on

L H M of

The

definition

< ~(Iul)

is s u b h a r m o n i c

of

on

G n.

Thus

Harnack

theorem

function, Since

{u~(z):

say

~(v)

the

on

Gn,

Theorem.

satisfying that G

G.

Let

We

If

G

CO

and

function

{u~} s~e

sense

vanishing

that

Proof.

In v i e w of the p r e c e d i n g

Let

U

be the

CO •

We d e f i n e

where any

~n = 0 a E GI

on and

~(u). n = i,

COn denote

and by

2,...} = i ~a(n)

so

z E G.

By the

harmonic + = u .

¢(u +) ~ U.

[]

let

~

be a c o n v e x

harmonic

function

majorant

exists

we m a y

~(u)

= 0

as b e f o r e on

and

on

function u

such

G, t h e n

a nonconstant

bounded

CO .

result,

Since

on

v = LHM(max{u,0})

a harmonic

there on

...

at e a c h

that

~ U.

2A a n d

has

~ U

to a n o n n e g a t i v e

that

a nonconstant

~(lul)

in t h e

{Gn:

converges

@(u-)

on

L H M of

is b o u n d e d

in

~(u~)

+

.-. ~ U

in f a c t

be as

carries

G

we h a v e

uI ~ u 2 ~

~ U, we c o n c l u d e

G

on

is h y p e r b o l i c ,

2,...}

show that

(i).

u = 0

harmonic

n = i,

we c a n

that

) ~ ~ ( U n + I) £

: l i m n ÷ ~ ~(u~)

Similarly,

2B.

implies

sequence

v, on

< U

+

un }(u

G, t h e n

=

+

G n.

on

Cln.

suppose on

and

Then

the h a r m o n i e

CO, set

that U

on

vn = H[~n;G n ]

vI ~ v2 ~

measure

u ~ 0.

vanishes

.--

for t h e

Take region

237

G

at the p o i n t

n

u(a)

f Cln

:

Since

%

p.

implies

63)

a.

Then

ud~ (n) a '

is c o n v e x

and

U(a)

I Cln

=

increasing,

(n) Ud~ a ,

Jensen's

I Cln

Vn(a ) :

inequality

d~ a (n)

(Rudin

[59],

that

If

/

U~a

1 f

a

C1 n

-<-

Cl n

< [

=

~ ( u ) d ~ a(n)/[ C1 n

d~(n)

;Cln Ud~(n)/f

Cln

Cln

a

d~(n) " a

Namely, (2)

Vn(a)%(u(a)/Vn(a))

for

n = i,

for,

otherwise,

in v i e w

of

harmonic

S

(i).

{Vn(a)}

of

say

which

Let

a harmonic

side

limit,

G

that

(2) w o u l d

v,

of

vanishes

on

be a r e g i o n

in

If

v

is b o u n d e d

on

be u n b o u n d e d

from

as

0;

n ÷

{v } is a n o n c o n s t a n t b o u n d e d n CO, as was to be shown. []

[

and

is a h a r m o n i c

majorant

away

S, t h e n

let

%

function

both

v

be a c o n v e x

on

and

S

such

that

LHM(~(Ivl))

quasibounded.

Proof.

By L e m m a

it is t r i v i a l l y there void

exists proper

then G.

on

satisfying has

means

left-hand

So the

Theorem.

function ~(Ivl)

This

the

(i).

function

2C.

are

2, . . . .

~ U(a)

G

namely, that

say

open

u

there

u0 = 0 S Ul,

is thus part v > 0 Pri(v)

subset the

exists on

CO

quasibounded.

If

and

on

S

is s t r i c t l y

[]

has

one

We

set G

on in

G.

We

S \ G.

v.

This

by

means

majorant

u = v-c

on

on

G

u0

to the

resulting The

that

this G,

the

the

such

function,

LHM of

for a n y

inner v

u1

quasibounded

is t r u e

function

on

is h y p e r b o l i c ;

u0

v.

Thus

is a n o n -

components,

extend The

Since

a harmonic

c}

of its

function

then

unbounded.

2B,

positive.

such

v

in 2A.

is m a j o r i z e d

by

is b o u n d e d ,

of T h e o r e m

harmonic

zero

v

assume

denotes

a harmonic

Hence,

If

{z C S: v ( z ) - >

mentioned

and

is m a j o r i z e d

~(v)

G

0 ~ u0 ~ I

identically

be zero.

we m a y

the h y p o t h e s i s

v

for w h i c h should

S.

and

of

v ~ 0.

such that

a nonconstant

is s u b h a r m o n i c bounded

of

that

so that

c > 0

condition

satisfies

by s e t t i n g

prQ(v)

assume

quasibounded,

a constant

satisfies

Since

whole

2A we m a y

part

should

be

238

2D. polar

Theorem.

set

(Ch.

tion

on

G

such

some

convex

a harmonic Let

be the

bounded monic

that

SO

data on

on

equal

SO.

Theorem

impossible, u

§2.

NULL

3.

for

( U + ~(livU

then

Since

to

F

OF CLASS

u

u

F

be a c o m p a c t

is a h a r m o n i c U

on

G

is e x t e n d e d

to

funcfor F

as

Let

as

%(t)

for

))/2

on

on

S O \ F.

~(iu01) G.

SO \ F

set.

F

and

SO

with

~S0,

v

is also

u0

is h a r -

Then

the

~ ¢((lul + 1vi)/2)

If

u0

should

Hence

be a c o n v e x

function

~

were

nonconstant,

be h y p e r b o l i c .

u0 ~ 0 function.

to

there

on

SO

This

and

is

there-

[]

such

= 0

and

~

as

large

t;

the r i g h t

t ÷ =

exists

We w r i t e t => to,

~(t)

tO > 0

E(t)

both

Let

E(t I) $ 4 for

number

with

property:

-~(t2) > 32

and

in v i e w that

E(t)

of

#(t)/t

t}

and

and

E l(t)

and

tI

be a r e a l

El(t I) ~ 2.

num-

We

set

t I ~ t ~ t2,

_~l(t2) > 3.

t 2 > t I + i, When

the p r o p e r t y :

( n + l ) l o g 2},

}'(t + 0)

way.

~ t I + 2 ( t - t I)

n => 3, w i t h

such

= min{#(t)/t,

so that

by i n d u c t i o n .

the

derivative

nondecreasingly,

a number

t I $ m a x { t 0 , log 2},

~ ( t n _ I) > m a x { ( n - 2 ) t n _ l ,

Then

that

in a n o n d e c r e a s i n g

tn_ 2 =< t =< tn_l,

(i).

l i m t + ~ ~ ( t ) / t 2 = 0.

it has

for

our

satisfying

(i);

t ~ tO. 2t}

function

for all

is c o n v e x ,

t ÷ ~

is a r e a l

m a x { t 2 , 41og 2},

on

contains

for

N~

¢

tends

Thus

construct

that

which

is b o u n d e d

as a h a r m o n i c

satisfies

= m i n { % ' ( t + 0), ~

S

problem

Lemmas

t, w h i c h (i).

of

and

that

is a p o l a r

~(t)

for

imply

limt+ ~ ~(t)/~(t)

t2

(i),

u

~S 0

(A3)

such

where

on

~(t)

to

let

majorant

u 0 = ( u - v)/2

(A2)

We n o w ber

~

a convex

is n o n d e c r e a s i n g

tend

and

a harmonic

Dirichlet

Since

~(t + log 2) ~ 27(t)

for all

= (t) -i

set

F

Lemma.

exists

u.

u0 : 0

2B w o u l d

SETS

condition

[

If

subregion

of the

(AI)

Proof.

has

in

G = S \ F.

satisfying

to

We

Preliminary

there

~

is e x t e n d e d

3A.

Set

be a r e g u l a r

S O \ F,

fore

be a r e g i o n S.

~(iui )

solution

(~(iui) + ~ ( i v l ) ) / 2 then

S

in

function

let

boundary

6C)

function.

Proof. v

Let

I,

E(tn_ I) > n

~(t)

2tn-i and

~(t 2) > is d e t e r m i n e d

=> tn_ 2 + i, _~l(tn_l)

> n,

239

we d e f i n e $(t) where

= $ ( t n _ I) + n(t - tn_ I)

for

tn_ I ~ t g t n,

tn

is a r e a l n u m b e r w i t h the p r o p e r t y : 2 m a x { ( n - l ) t , ( n + 2 ) l o g 2}, E(t n) ~ ( n + i) and of the m o n o t o n i o i t y

property

seen to be p o s s i b l e , ously,

~(t)

for all [0, ~)

so that

is c o n v e x

0 g t g tI with

~(0)

of

for

Z

and

~(t) t ~ tI

t => t I .

and

~(t n)

Zl(t n) ~ n + I.

Zl, t h e s e

is d e f i n e d

operations

for all

In v i e w are e a s i l y

t ~ t I.

Obvi-

it is easy to c o n t i n u e

so as to h a v e a n o n d e c r e a s i n g

convex

~(t)

function

on

: 0.

We n o w h a v e o n l y to s h o w that ties for

t n ~ tn_ I + i,

~(t)

tn_ 1 =< t ~ t n.

Let

possesses

the d e s i r e d

t + log 2 =< t n + i

Then

propertn+ 1

and t h e r e f o r e ~(t + log 2) g ~(tn_ I) + (n+l)(t + log 2 - tn_ I) = ~ ( t n _ I) + ( n + l ) ( t

- t n _ I) + ( n + l ) l o g

2

2~(tn_ I) + 2n(t - t n _ I) =

The c o n d i t i o n

(AI)

2~(t).

is thus

satisfied.

n ( t - t I) + t I ~ ~(t)

For the

~(t)/#(t)

÷ 0

shows t h a t

3B.

E(t) ~ n 2, we h a v e and

t/~(t)

The

~(t)/t 2 + O ÷ 0

as

following

as

t ÷ ~.

result

t

we h a v e

= ~(tn_ I) + n(t - tn_ I)

(n-2)tn_ I + n(t-tn_ As we k n o w that

same

~(t)

I) > (n-2)t.

=> n2t

t + ~.

and

t 2 => n2t.

The a b o v e

Thus

inequality

also

[]

shows the

importance

of the c o n d i t i o n

(AI) or, m o r e g e n e r a l l y , ~(t + log 2) = 0(i), ~Ct)

(3) which (of.

is e q u i v a l e n t Krasnosel'skii

Lemma.

Let

let

be a t o t a l l y

E

and o n l y if Proof.

~

to the A 2 - c o n d i t i o n and Rutickii

be a c o n v e x

in the t h e o r y

of 0 r l i c z

spaces

[41]).

function

diseonneeted

t ÷ ~,

which

compact

satisfies set in

¢.

(i) and Then

(3) and E E N~

~ \ E e 0~.

S i n c e the n e c e s s i t y

is t r i v i a l ,

we o n l y

show that

[ \ E 6 0~

if

240

implies V 2 E not

E E N}. and any

vanish

and

u

~(u)

~ U

LHM of

on

now divided

both

u = lim

into

and n~

Let

and

parts. that

are

uA n

and

U

E

U : lim

are b o u n d e d ,

means

of T h e o r e m

2D o r o t h e r w i s e .

monic

on

V

and

that

quently,

f

is b o u n d e d

even

(ii) In t h i s We

part,

set

Since

we

Hence,

suppose

assumption

We may

assume

case

analytic V \ En,

curves

and

now

loss

that

where

2C, Ch. II,

uA n

and

functions

and

by

U are harConse-

therefore

is h o l o -

to be p r o v e d . E

is not

F = [\ V

including (i)

shows

that

E

is n o t

of generality

a polar

is a p o l a r

F

and

that

f

set. set.

f E H ~ ( S \ F).

f ~ H~(S).

is a c o n s t a n t

of

a polar

that

region

is c o n t i n u o u s

n $ i, be a n e x h a u s t i o n

boundaries,

and

that

i.e.

in

both

by c o n t i n u i t y .

E

as w a s

implies

is a J o r d a n f

seen

5C,

is

Our

function.

f E H~(V).

suppose

V

V E 0G,

by T h e o r e m

Since

u

V \ E

proof

by Theorem

that

does

see t h a t

The

as h a r m o n i c

of

follows,

is a r e g i o n

S E 0~

without

in w h i c h

V

V, as

our proof

that

Finally,

on

Then,

Thus, V \ E.

to

f C H~(V),

S

set,

set.

It f o l l o w s

in w h a t that

Then

it is t r i v i a l (iii)

~ U

}(u).

f on

We e a s i l y

L H M of

on

extended

any region

that

}(log+Ifl)

IB).

V \ E.

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

suppose,

is a p o l a r

original

the

E.

S = [ \ E. F

Hence

on

So we

~(u)

on UA n

n~

UA n

morphic

are

is the

We take

of c o u r s e

L H M of (see

is a p o l a r

quasibounded

these

suppose

v \ E

in f a c t

three

[ \ E E 0~.

be t h e

on

suppose

U

that We m a y

U

log+ifI

V \ E

First

u

suppose

identically.

the

(i)

So w e

f E H ~ ( V \ E).

~ ~ V.

bounded

up to t h e

V k E

E =C En+ I =C En =C V,

set a n d

We b e g i n

by a f i n i t e boundary

by J o r d a n

n => i.

V ~ 0 G.

of

V.

regions

For a n y

with

number

of

Let

with

analytic

z @ V \ En

we

have

f(z)

1 SaV f(~) ~ = ~ d~

i IDE ~f({) r ~ d~.

= 2-~

- 2-~

n

The

first

and

spectively

by

respectively. that

g(z)

the

second

g(z)

members

and

F r o m the

hn(Z) , which fact

is c o n t i n u o u s

f(z)

by a c o n s t a n t

on

V \ E1

and therefore

define

on

[ \ E.

We t h u s

f(z)

lh(z)I

~ M+

positive C2

on

If(z) I

constants

M > 0.

there. CI

and

right-hand are

C2

(z) of

see a l s o

a single

o f the

on

that

on

property

denoted

and on

re-

[ \ E n,

V \ E and

all

follows n so is b o u n d e d

(z) n function,

V \ E

in s u c h a w a y t h a t

are V

on V

holomorphie

= g(z) - h(z)

By u s e

n

boundary

We

side

defined

: g(z) - h

up to the

in m o d u l u s

have

of the

h

and

(3) of

coincide h(z),

consequently ~

we f i n d

~(log+lhl)

£ CIU+

V \ E.

Let us

fix a point

z0 C V \ E1

and

denote

by

Bn

(resp.

v n)

the

241

harmonic

measure

at the

point

z0

with

(resp.

V \ E ) for n = i, 2, . . . . n a constant C > I s u c h that

find

see this,

take

a Jordan

E 1 C ~ C CI(~) a region

region

_C V \ {z0}.

D, we

for

n = i,

have

Green

and

V \ E = Un~ I V \ E n.

most

~.

in

E

So,

there

each

exists

fixed

g(a,z:D)

a polar

by a s s u m p t i o n

): n = i

boundary the

set,

for all

DE n

set,

such

Green

2,.

that

}

we can n.

To

that

function

the

g(w,z0;¢\

sequence

for

{m n}

g ( w , z 0 ; ¢ \ E) ;V\E) g(w'z0

principle

and

¢\ E

V \ E 1 C V \ E 2 =C ...

to

we

g(w

see that

z0;V\E)

al-

E n) ÷ g ( w , z 0 ; V k E )

E n) ÷ g ( w , z 0 ; ¢ \ converges

E)

uniformly

to

< oo.

i < C < ~ such = < C g ( w , z 0 ; V \ E n) for all

n, the m a x i m u m

V \ E

theorem,

tends

a constant

E n)

both

n ' "" In p a r t i c u l a r , g(w,z0;V\

max wED~

g(w,z0;¢\

analytic

by

by use of H a r n a c k ' s

Similarly,

Consequently,

i.e.

with

is not

We k n o w

V\E.

D~.

Hence

on

¢\ E n

a polar

g ( w , z 0 ; ¢ \ E n) g(w,z0; V \ E n ) <

= wmax ED~

{g(w,z0;VkE

on

~

to the r e g i o n

is not

dun =< Cd~ n

Denoting

Since

functions.

uniformly

uniformly on

2, . . . .

sequence

E

set

mn

the

respect

Since

applied

that n

m

and

to the

< C for all n, n = all w E D~. For

region

~ \ En

shows

that (4)

g(w,z0;[\

for all

w E ~ \ E n.

E n) ~ C g ( w , z 0 ; V \ E n)

On the o t h e r

hand,

we have

dUn(W)

:

i 2--~ ~ ( w ' z 0 ; [ w

d~n(W)

:

~--~

along

DE n

\ Enlldwl

and (w,z0;V \ En)idw I, w

where

D/Dn

the o u t w a r d

arc-length desired

normal

element

the d e r i v a t i v e

relative

along

to

at

[\ E n

DE . n

w C or

Combining

DE in the d i r e c t i o n of n V \ E n, and IdwI d e n o t e s the

these

with

(4),

we h a v e

the

result.

Since of

denotes

w

Ih(z) I

~(log+ihI )

on

is

bounded

¢\ E n

Vn(Z0)

on

and

¢\

En,

we c a n

define

in fact

= I

~(l°g+ih(z)l)dUn(Z) DE

n

the

LHM,

Vn,

242

< C I

~(log+ih(z)I)d~

(z)

DE

n

n < C f =

(CIU(Z) + C2)dv

(z)

DE

n

n < C I = D(VXE

(CIU(Z) + C 2 ) d v

)

n

(z)

n

= C ( C i U ( z 0) + C2). Since

{Vn:

is b o u n d e d

n = I, 2,...} at the p o i n t

to a h a r m o n i c function on

E

function,

~(log+lhl)

implies

say

on

that

h

we c o n c l u d e

is a b o u n d e d

analytic

previously, vn

n = i, 2, . . . . I, 2,...

of

of

Vn \ E

V

V

~(log+IfI )

Fix a p o i n t

Thus

n )

\ E

As

f, w h i c h

h

the

vanishes

is equal to

such that

exhaustion

E E V

of

V

to

g,

H~(V). but

with

the a s s u m p t i o n

= E $ V I.

made

n = i, 2, . . . .

Vn \ E

z0 C VI \ E

measures

majorizes

function.

satisfies

for

n on

which

it c o n v e r g e s

and our a s s u m p t i o n

and h e n c e b e l o n g s V ~ 0G

sequence

that

Let u n Vn, r e s p e c t i v e l y ,

and

and let

~n

at the p o i n t

z0

and

vn

and for

with

with respect

n =

to

V , respectively. By an a r g u m e n t s i m i l a r to n (4), we find a p o s i t i v e c o n s t a n t C i n d e p e n d e n t of n

that u s e d for

and

dVn =< Cd~n

such that

to

f E H~(V

be the h a r m o n i c

the r e g i o n s

h ~ 0. on

be a r e g u l a r

f

we see that

be the LHM's

h e H ~ ( ~ \ E)

any r e g i o n

{Vn: n = i, 2,...}

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

implies

~ \ E, w h i c h c l e a r l y

So

that

function

Now let us c o n s i d e r Let

v, on

~ \ E.

increasing

theorem

s h o u l d be a c o n s t a n t

at i n f i n i t y ,

V.

is a m o n o t o n i c a l l y z0, H a r n a c k ' s

U(z0)

on

DV n

~ Un(Z0)

for all

~ I

So

~(l°g+If(z)i)d~n(Z) DV

C-I I

n.

n

~(log+If(z)I)dVn(Z) DV n

= C-ivn(Z0) for all

n.

function, longs

to

3C. cumference

This v, on

H~(V),

shows that the

{ I z - z01

F(z0;r0) ,

= r0].

we a l w a y s

When

{v n}

tends

~(log+IfI )

as was to be proved.

We d e n o t e by

the f o l l o w i n g ,

sequence

V, w h i c h m a j o r i z e s

Izol

on

to a h a r m o n i c v.

Hence

f

be-

[]

< ~

and

0 < r 0 < ~, the cir-

z 0 = 0, we w r i t e

deal w i t h the e x t e n d e d

simply

complex

F(r0).

plane

~.

In So,

243

w h e n we write, in the Lemma. in

Let

{Iz[

for e x a m p l e ,

{Izl

0 < a < b < ~

~ b}

the h a r m o n i c

such that measure

and let

ds

denotes

the a r c - l e n g t h

positive,

Proof.

D I : {a < Izl < b},

Harnaek's

inequality

o n l y on the r a t i o A ' U ( z 2) U

on

D I.

i

on

there

For any arc harmonic e

e

and to

DI.

0

(5)

denotes

F(a)

U2(e;w) of

A(t),

t > 0,

c : (ab) I/2

= A'(a/b), such that

by

By

depending U(z I)

harmonic Uj(e;z), values

function 0 ~ j ~ 2,

are equal

~ U2(e;z)

inequality

the arc l e n g t h

max{U2(e;w):

and

A'

way,

we d e n o t e

~ U0(e;z)

By use of H a r n a c k ' s

lel

> a}

a constant

by

Then

t.

on

lel 4~a ~ A ' U l ( e ; w ) ' where

D 2 = {Izl

in

Denote

DO.

and

D. whose boundary ] elsewhere. Clearly,

Ul(e;z) on

F(a)

set i n c l u d e d

,

and any n o n n e g a t i v e

on

function

on

function

in a n o n d e c r e a s i n g

Zl, z 2 • F(c)

to

~ A ( a / b ) . - - ~2~a

element

exists

closed

is a region.

with respect

nondecreasing

a/b

for any

the b o u n d e d to

the p o i n t at i n f i n i t y

be a b o u n d e d

> a} \ F ~

z • F(a)}

is a finite, Let

F

D O = {Izl

at the p o i n t

m a x {d ~ (z): where

> a}, we i n c l u d e

set.

we see for

lwl = c

lel i + (a/b) I/2 ~ 2~a i - (a/b) I/2 e.

'

So

lwl = c} ~ A . m i n { U l ( e ; w ) :

lwl = c},

where A = A(a/b)

From

(5) f o l l o w s

el,

B(e I) ~ A ~ ( e 2)

if

and the f u n c t i o n problem

e2

max{

on

F(a)

lel]

w i t h the

= le21 , b e c a u s e

U0(ej;z) ,

F(e)

(z):

lwl : e} ~ A - m i n { U 0 ( e 2 ; w ) :

on

for the r e g i o n

U0(ej;z)

"

at once that

max{U0(el;w): for any arcs

= 2A'(a/b) .I + (a/b)i/2 i - (a/b) I/2

and to

z e F(a)}

same length. ~(ej)

j = i, 2, is the

{Izl

> c} \ F 0

(z):

This

shows

= U0(ej;~),

solution

w i t h the b o u n d a r y

elsewhere.

~ A.min{

lwl : c}

j = I, 2,

of the D i r i c h l e t d a t a e q u a l to

Hence z • F(a)}

that

~ A-#--(-~

.

244

3D. class

The next

l e m m a is the key to our c o n s t r u c t i o n

of null

Lemma.

Let

number

0 < a2/b

of b o u n d e d

< a0 < a < b < ~

closed

1 < i < k, is a r e g i o n i, 2,...} ~(n)/n

subsets

including

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

: o(i),

n ÷ ~, and

FI,... , F k

{Izl

> b}

such that

all

F.

with

j ~ i.

sequence ~(n)

and let

of

of p o s i t i v e

> n I/2

a0exp(2~ji/n) We set polar

and

Kn, j = { I z - Wn,jl

numbers

for all large

K n : u.n]_l Kn,j.

Suppose

set and that we c o n s i d e r

in the d i s k

{Izl

Let

Pn

~

touthe

and

regions Then,

such that

for

that e a c h

only large

F.,] n's

> a 0} \ F

for any

and

s > 0, there

exists

B = B(a/b)

(6)

< E,

is a c o n s t a n t

We first

and t h e n c h o o s e

(l+s')-ip(F(a0))

where

~'

(resp.

s' > 0

< ~'(F(a'))

~") d e n o t e s

{Izl

w. ~K n

Un(Z)

with

N = N(s)

o n l y on the r a t i o

F : > 0

a/b

in a

support

the G r e e n

on

potential

Un(Z)

b = i.

Let

c >

< 2-1~(F(a0))-l~ so c l o s e

> a'} \ F

to

a0

as to have

< (l+s')~(F(a0)) ,

measure (resp.

at the p o i n t {Izl

be the G r e e n

We d e n o t e

> a"} \ F).

function

for the

by ~ the p o s i t i v e m e a s u r e n and w i t h u n i f o r m d e n s i t y on ~Kn, and

of

= I

that

with

< ~"(F(a"))

= l o g ( l l - z ~ I / I z - w I)

of m a s s by

is i n c l u d e d

i ~ j ~ n,

the h a r m o n i c

o p e n unit d i s k w i t h pole at with

s'}

< a 0 < a" < a

to the r e g i o n

g(z,w)

i

Kn

1 ~ j ~ k,

loss of g e n e r a l i t y

choose

0 < a'

(7)

Let

i =< j =< k, is not a so that

< g,

depending

m a x { ( l + s ' ) - (i+~') -3,

with respect

for

:

fashion.

We may a s s u m e w i t h o u t

be given.

w

n,j j = i,..., n.

an i n t e r g e r

~n(~Kn, j) ~ ( B n ) - l ~ ( F ( a 0 ) ) ,

Proof.

Let

n > N

I ~ ( F j ) - ~n(Fj)l

nondecreasing

n =

such that n.

(K n U F) , r e s p e c t i v e l y

[\

'j~(F(ao) ) - p n ( ~ K n ) l

where

~ \ Fi,

{~(n):

< a} and K .'s, i ~ j ~ n, are m u t u a l l y d i s j o i n t . n,] be the h a r m o n i c m e a s u r e s at the p o i n t ~ with respect

{Izl

Uj~ I Fj.

~ a0e-~(n)}

be a f i n i t e

each

Let

=

0

sets of

N~.

Vn'

i.e.

g(z,w)d~

~K n

(w) n

245

on the disk

{Izl

~ i}.

harmonic

{Izl

< i} \ Kn, and v a n i s h e s

tation

on

shows

Clearly,

Un(Z)

is e o n t i n u o u s on

F(1).

on

{Izl

~ 1},

An e x p l i c i t

compu-

that i

n

~ g(z ) j:l 'Wn'j

if

z ~ K

if

z E K

n'

Un(Z): { i

where of

Rn,i(z)

{£(n)},

(ii)

F(a')

there

seen that

K

n Let

for

i

|2~ J"

--2~

0

exists

n > N'. = (resp.

u'

(i)

U (z) n

n,

i'

By use of the p r o p e r t y converges

uniformly

on

to the value ,a0eit)d t g(a'

log

a0

N' = N'(e')

i__ a0 <

u )

i :

an i n t e g e r

(l+s')-ll°g on

+ !R .(z) n n,l

= log(ll - Z W n , i l / a 0) ~ log(2/a0).

it is easily

the circle

and

[ g(z .) + ~(n) j~i 'Wn'3 n

,

> 0

such that

i U n (z) < (l+E')log a--~

be the

solution

of the D i r i c h l e t

problem

for

n

the region equal

to

{Izl i

on

> a'} \ F F(a')

(resp.

(resp.

[ \ (K n m F ) )

aK ) and to

with the b o u n d a r y 0

elsewhere.

data

Take

N" =

n

N"(c')

so large

that,

for

n > N"

K '

Izl

a"}

<

possible

and

Un(Z)

in view of

lies

Suppose

{a' <

n

~ (i + s ' ) - l l o g ( i / a 0 )

(i).

in the annulus

on

n ~ N(s)

F(a'),

= max{N',

the latter N"}.

Then,

being by

use of (ii), Un(Z)

= i > (i + e ' ) - l ( l o g =

on

~K

Un(Z) {Izl

(z) n

and t h e r e f o r e e v e r y w h e r e on {[z I ~ i}. n ~ (I + E') -2 = (i + e')-2u'(z). Since Un(Z)

> a'} \ F

see that together (8)

i )-Iu a 0

and has the same b o u n d a r y

Un(Z) with

~n($Kn)

~ (i + E')-2u'(z)

(7), means = Un(~)

On the o t h e r hand, (9) Combining equality

everywhere

in p a r t i c u l a r

since

K

n

lies

inequalities

in the lemma.

(8),

as

on

F(a')

we have

is s u p e r h a r m o n i e u'(z)

{Izl

on

on

~F, we

> a'} \ F.

This,

that

~ (i + s ' ) - 2 ~ ' ( F ( a ' ) )

~ n ( a K n) ~ ~"(F(a")) the

values

So on

inside

~ (i + s ' ) - 3 ~ ( F ( a 0 ) ) . F(a"),

we have

~ (i + e ' ) ~ ( F ( a 0 ) ) .

(9) and using

To show the second,

(6), we get the

we take any

j

first

with

i

in-

246

Then,

j < k.

by

(6),

(7),

k __< [ i=l

I ~ ( F j ) - ~n(Fj)I

<

k [

=

i=l

(8) and

(9)

lu(Fi) - Un(Fi) I

(~(F i) - ~"(F.))

k [

+

i

(~n(Fi) - ~"(Fi))

i=l

= U " ( F ( a " ) ) - U ( F ( a 0 ) ) + ~ " ( F ( a " ) ) - Un($Kn) g ' B ( F ( a 0 ) ) + ((l+s') - (l+s')

-3 < s.

)O(r(a0))

N o w we h a v e o n l y to p r o v e the last b u n c h of i n e q u a l i t i e s . previous F(a)

observation

shows,

U (z) n

converges

uniformly

As our

on the c i r c l e

to $2~ g(a,a0elt)dt 2~

in v i e w of the fact N"

such that,

a 0 < a.

for

So t h e r e

We take any

such

n, set

Ulj , u2j) ,

1 ~ j ~ n, to be the

the r e g i o n

[ \ (K n U F )

data equal

to

1

on

m i = min{uij(z): the c o n s t a n t lemma.

of

A'

~Kn, j j

and t h e r e f o r e

monic

in

{Izl

solution

and to and

0

elsewhere•

~K n

on

Because Izl = e},

[ \ K n , we see that

On the o t h e r hand,

and is e q u a l to n (ii), we h a v e > ((i+s')log =

and thus e v e r y w h e r e

n

on

F(a),

because

)-iu

on

I

u2j(z) a

2

of s y m m e t r y , i = i, 2,

1

[Izl

n

(z)

on

i

~K . n

> ~(log =

< i} \ K n •

1 =< nM 1 =< A ' n m 1

~j:in u2j

i ]-I . U

>-- ~ ( l ° g ~ 0

< a ^u < a .

i__) -i u ( z ) a0

n

In v i e w of

I log a

> i

n (z) --> 4 log a 0 = g It

follows

that

is har-

So, by l e t t i n g

thus get

j=l

u. (resp. ] p r o b l e m for

M. < A ' ( a ) m . for i = i, 2, w h e r e i = i a p p e a r e d in the p r o o f of the p r e c e d i n g

n

on

N1

< I} \ K n) w i t h the b o u n d a r y

M i = max{uij(z):

M 1 =< A'/n.

~ u2j(z) j:l

and define

of the D i r i e h l e t

{Izl

< i} \ K

and using

integer

and we h a v e

has a l r e a d y

=< A'

c = a I/2

I X Kn,

n Ulj (z) ~ 1 ~j:l

Since

s' < 1

(resp.

Izl = c}

are i n d e p e n d e n t

a positive

1 ~ 2-11og-~-

Un(Z) F(a).

exists

n $ NI, we h a v e

(i0) on

: log l---a'

0

(I0)

we

247

n ~

A,nm 2 >nM 2 >

> i__-

u2j(z)

] ± on

F(c).

Ulj(Z) , shows

So we have

A ' m 2 ~ 1/16n ~ MI/16A'.

i ~ j ~ n, on the annulus

F(c)

{Izl

u2j(z)

S uj(z)

S

inequality

that u.(z) i

on

Since

{a < Izl < i), the above

for any

i

and

j.

< 16A'2u.(z) = ] The

same i n e q u a l i t i e s

clearly

hold on

> c} \ F, so that U n ( 3 K n , i ) : ui(~)

for any

i, j : i,...,

n.

S 16A'2uj (~) = 1 6 A ' 2 U n ( $ K n , j )

This

implies

that

Pn(~Kn,j ) ~ (16A'2n)-Ipn(~)K n) for

j = i,...,

n.

Since

for all s u f f i c i e n t l y

D(F(ao))

large

n.

~ 0, we have

Hence

pn(~K n) ~ U ( F ( a 0 ) ) / 2

we have

Pn(~Kn,j ) ~ ( B n ) - l ~ ( F ( a 0 ) ) for

j = i,...,

large.

4.

This

finishes

Existence 4A.

n, w i t h

of Null

B = B(a)

the proof.

: 16A '2, p r o v i d e d

n

is s u f f i c i e n t l y

[]

Sets

We are now in a p o s i t i o n

to prove

the m a i n

result

of this

seetion. Theorem.

Let

~

be a convex

0 < a < b < ~

and let

closed

of

subsets

taining and

all other

0 < 6 < i

Fj,

function

{Izl

> b}

Fj,

j ~ i.

there

exists

which

satisfies

i £ j £ k, be a finite

(i) and

number

(3).

Let

of b o u n d e d

such that Then

each ~ \ F. is a r e g i o n coni for any p o s i t i v e numbers e > 0

a non-polar

set

E E N$

such that

E

at the point

~

{~a ~ LzL ! a}, (il)

l ~ ( r j ) - ~E(Fj)I

< E,

i ~ j ~ k,

and

l~(r(a))-

(12) where respect

p

(resp.

~E ) d e n o t e s

to the r e g i o n

#E(E)[ < e,

the h a r m o n i c

{]z I > a} \ F

(resp.

measure

~\ (EUF))

with

F =

with

248

uj~1 ZjProof.

By a p p l y i n g

vex

function

t2

for all

find

increasing

for

functions

h(t)

The

there

t; and

number

and

determined

~(t)

exists

So the

Lemma

In o r d e r

3D.

we m a y

that

during

only

to add

that

(b/a) I/4

and

closed

disks

a finite

for all

t(s)

the

which

as to h a v e

0-th

Suppose

disks

D~,

We t h e n

Kn+ I r' n (13)

If

t

F

the

=< a} .

Each

of

we h a v e K n'

and

Up)

conditions

set,

p

{jz I £ a},

says

finished in such

This

set

we d e f i n e

that

the

We

the

n-th

a way

that

of c e n t e r

3D.

w

and

of

same

{JzJ

(resp. point

t

~0 = ~

n ~ 0.

of c o m m o n

condition

Kn+ I

is dis-

Namely,

of c l o s e d

radius

r'

Da

= {Jz-w

of t h e s e

disks.

Pn+l(Fj)-

p~(Fj)

K'n+l a

as

disjoint

fulfilled

follows.

J =< r n + l } , The v a l u e

so

n

are m u t u a l l y

is c l e a r l y

and

and

and

=< p4r~}

eonsists

of

radius

~

I £ p4r~}

p =

(K' U F ) ) . n K 0 to be e m p t y

step w i t h K'n

also

set

K n') c o n s i s t s

of the

so that disk

then

as a n e w m e m b e r

(resp.

[\

in

property,

We m a y a s s u m e

disks

(resp.

n induction,

=

{Jz-w

define

consists

(K

for

n = 0, i,...,

Kn

closed

= o(I), ~ > 0

£ ~(t)/2

in Lemma

K'n'

t ~ t I.

~(t)/t

is a p o l a r

appearing Kn,

inverse are

for

desired

F

strictly

for any

0 < ~ < (i - ~ ( F ) ) / ~ ( F ) .

one m e m b e r

1 < ~ < N(n),

F.

[\

of our

of o n l y

that

disks

from

rn+ I < r'n

the

F_I , to

disjoint,

region

step

=

joint 4 p •

satisfies with

no harm.

families

constructed

the

h(t/e)

say

constant

{Izl

that

are the

(i')

(iii')

cause

the

to the

to c o n s i s t

that

t; and

E

we c o n s t r u c t in

following:

set.

set, will

in

£

We t h e n

A(t) Then

A(t)

respectively,

functions

n ~ t I]

a set

a non-polar

=

we h a v e

{~(n):

is a n o n - p o l a r

a • The d e f i n i t i o n r0 joint f r o m the set F. '

the

large

(~ t I) such

of m u t u a l l y

t ÷ ~.

and

A(t),

get a con-

(ii)

r') in {JzJ < a} w h o s e u n i o n is d e n o t e d by K n = ' n Pn (resp. ~) we m e a n the h a r m o n i c m e a s u r e at the

By K n'). with respect As

imply

to c o n s t r u c t

included

number

(resp.

K~

now

B = B(a/b),

By i n d u c t i o n

rn

- (iii)

is so small

~(t)

and

we

t ÷ ~;

= o(i),

both

increasing

~ t I/2

our c o n s t r u c t i o n , a

that

~(t)

strictly

F

t ÷ ~(t/2),

= o(i),

t I = ma×{~(t0),A(t0)}. of

sequence

assume

we h a v e

Let

function t/A(t)

A(t)/~(t/2)

such

~(t)

a number

t ~ t(e).

(i)

(iii)

tO

as

(i)

(ii')

3A to the that

t ~ tO.

properties

t ÷ ~;

such

large

a positive

uniquely

Lemma

A(t)

by

Let

and

also

dis-

K~,

for

b/a

max{i/p,6}r~

i =< a =< N(n). rn+ I

is f i x e d

as to h a v e ~ E ~ ( F j ) / 2 n+l,

i ~ j ~ k,

The

<

family

so c l o s e

=

to

249

and ~ n + l ( ~ D ~ ) ~ ~n'($D~)/2,

(14)

In o r d e r to d e f i n e

K~+I, we take an i n t e g e r

w h i c h w i l l be f i x e d later. w

,j = w

N(n+l)

i =< e =< N(n) " N'(n+l)

~ max{n+l,tl},

We set

+ rn+lexp[2~ji/N'(n+l)] ,

= N(n)N'(n+l),

i ~ j ~ N'(n+l),

r'n+l = r n + l e x p [ - ~ ( N ( n + l ) ) ]

and n'~,j

: {Iz-w

~,j I =< r n+l },

i < j < N'(n+l)

The d i s k s

{D' .: i < ~ < N(n), i £ j < N ' ( n + l ) } t h e n f o r m the f a m i l y ~,3 = = K' We fix N'(n+l) so large as to h a v e the f o l l o w i n g : n+l" 4 , (a) the disks {Jz-w ,j J _< p rn+ I} are m u t u a l l y d i s j o i n t and

also d i s j o i n t (b)

from

F;

l~n+l(Fj)-~n+l(Fj)j

< e ~ ( F j ) / 2 n+2

for

(c)

and

This

' (~D' ) > (BN'(n+I)) -I (~D) ~n+l e,j = ~n+l e ~ = i,..., N(n); (d)

r'n+l =< 3 - 1 m i n { r n' - r n + l ' r n + l

(e)

h((n+l)(2B)n+iN(n+l))

is t r i v i a l

Lemma

3D and

for

(a) and

j : i ..

for

j = i

k;

'''''

N'(n+l)

Brn } ;

< ~(N(n+l))/2.

(d).

Statements

(e) f r o m the p r o p e r t y

(iii').

(b) and Hence,

(c) f o l l o w

by i n d u c t i o n ,

from the

families

K and K', n > 0, are c o n s t r u c t e d . n n = C o m b i n i n g (13) and (b), we get for n > i

(15)

I~n(F j) - ~n+l(Fj)l J~n(Fj ) - ~ ( F j ) < s~(F.)/2 =

n+l + s~(F.)/2

]

: e ~ ( F . ) / 2 n, 3 It f o l l o w s

from

(14) and

K' such that ~n(~D~) n-1 see that for D' E K'

I + I~(Fj)

- ~n+l(Fj)J

n+l

3

i £ j £ k.

t h e r e is a D~ E D' @ K' n R e p e a t i n g this, we > ( 2 B N ' ( n ) ) - l ~ n I(~D~) = "

(c) that

for any

n

(16)

~(~D~)

~ (2B)-n(N'(n)N'(n-I).''N(1))-IB~(F(a)) : (2B)-IN(n)-I~(F(a)).

We set

250

E = and d e n o t e

by

DE

to the r e g i o n

¢\

the

Izl

harmonic

[Cl(

U Ks)] s:n

measure

at the

point

~

with

respect

(EUF).

Our c o n s t r u c t i o n K I' =C {6a <

A n=l

< a}

and

the

and

property

Kn+ I =C K'n

(d)

for

show

that

n : i,

2,...

K' C Int(K') n+l = n . It f o l l o w s

C =

that

E =C Cl[ for we

n : i, see

2,...

for

and

U s:n+2

Ks]

C= K'n+l =C Int(K')n

afortiori

E ~

{6a

<

Izl

By use

< a}.

of

(15)

n > i n

Bn(F)

: ~(F)

+

(~s(F) - Us_l(F))

s:l : B(r)

n [ s=l

+

k (Bs(F j) - ~ s _ l ( F j ) ) j=l k

<__sB(r) :

and

+

e~(F.)/2

[ [ s=l j=l

s

3

(i + e ) B ( F )

therefore Dn (~Kn)

The

last m e m b e r

the

Let

u

: ~(F(a))-

E~(F).

data

[\

equal

positive,

(resp.

n

for the r e g i o n

boundary

=> i - ~(F) - s ~ ( F )

is s t r i c t l y

(i - ~ ( F ) ) / ~ ( F ) . problem

= i - ~n(F)

v ) be the n

(F U C I ( U

to

i

because

on

of our solution

~ K )) s=n s

(resp.

~[CI(U

K )] s:n

0

on

~F.

Clearly,

the

sequence

{u

assumption of the [\

Dirichlet

(FUK)) n

(resp.

with

~K ) and

S

: n = i,

~ <

to

n

2,...}

is m o n o t o n i c a l l y

n

decreasing

and

bounded

to a n o n n e g a t i v e we have

below,

harmonic

0 < u(z)

< 1

so that

it c o n v e r g e s ,

function,

u(z),

u(z)

on

and

= 0

on

~\

~F.

by H a r n a e k ' s (E U F ) .

We k n o w

Obviously,

that

~

Vn(Z)

on the

0.

Hence

the

Diriehlet

0

on

set.

~F.

domain

u(~)

u

(z)

>

n

of

> 0.

problem

Thus

Moreover

theorem,

with

BE(E)

we h a v e

un

The

and

that

function

u

the b o u n d a r y

= u(~)

> 0

and

v n (~)

= ~ n ( Z K n) => 1 - (I+s)B(F)

is seen data

to be the

equal

consequently

to

1 E

solution on

E

is not

>

of and

to

a polar

251

(17)

~E(E)

= u(~)

: lim Un(~)

~ lim sup Vn(~)

n~

n~

= lim sup ~n(~Kn) On the other hand, and therefore with

(17),

since

~E(F)

proves

shows

is included

$ B(F).

(12).

~E(Fj) -B(Fj) which

E

Thus

~(F(a))

~ ~E(F) -~(F)

show finally

that

E E N@.

nonzero

H ~ ( [ \ E)

element

in

To show that

is included

n $ 0

f

and

v

of

K' n

D' = {Iz- w' I £ r'}

(a) says that the annulus

to the annulus

Let

~w

{r' < Iz - w' I < p2r'}

is bounded

A'/(4~r')

on

where

3C.

Iz -w' I < p2r'},

f

we have for any

the convex

function

and by

and analytic w

f

be a

}(log+Ifl)

we first note that Let us fix ' K n.

in

measure w

inequality

is the constant

is bounded

log+If(w)I

equality,

F(w';r')

A' = A'(p -2)

Since

of

n = 0, i, . . . .

at a point

then Harnack's

F(w';p2r'),

(3), we

Let

{r' ~ Iz- w' I £ p4r'}

r(w';pr'),

by

majorant

be the harmonic

on the circle

Applying

together

satisfies

be one of the members

and

K'n \ D'.

%

function,

for

the property

of Lemma

~ ~(Fj)

~ su(F),

[ \ E E 0}.

a harmonic

meet

F

Since

3B, that

is a constant

in the interior

and let E,

ZE(Fj)

~ BE(E) , which,

: D(F(a)) - ~ E ( E )

in view of Lemma

[ \ E.

{Izl ~ a},

(ii).

We will

E

in

We also have

have only to prove, on

~ ~(F(a)) - e~(F).

does not

with respect

in it. implies

A'/(4~p2r ') appearing

Then

If

w

that

is

d~w/dS

on in the proof

on the annulus

{r' <

in this annulus

£ [ l°g+If(~)Id~w (~)" ~

to both sides and using Jensen's

in-

we get ~(l°g+[f(w) l) ~ %(I l°g+If(~)Id<w(~))

I ~(l°g+If(%)I)d~w(~))

=f

F(w';r')

+f

F(w';p2r ')

= I I + 12 . We now assume sures at

~

that

w E I'(w';pr').

with respect

Let

to the regions

~, ~'

denote

G' = {Iz - w'

the harmonic > r'} \ E

mea-

and

252

G" : {Iz - w' I > p2r'} \ E, respectively.

Since

E C K' =

~D'

F(w';r')

:

we see that q'(F(w';p2r'))

By use of (16) we also have Lemma 3C to the annuli p4r'}, we see that circle on

and

n

F(w';r')

F(w';p2r'),

where

~(~D')

~ ~(~D').

~ (2B)-nN(n)-l~(F(a)). By applying and {p 2r ' < [z- w' I <

{r' < Iz- w' I < p2r'}

ds/d~ and

~ n(F(w';r'))

is bounded by

ds/d~'

2~Ar'/~(F(w';r'))

is bounded by

A = A(p -2)

on the

2~Ap2r'/n'(F(w';p2r'))

is the constant

given in Lemma

3C.

So we have d~w(~) ds(~) ~(log+If({)l) d s - - 7 ~ ~

II :

dq(6)

F(w';r') A' 2~Ar' 4w-~r ~(F(w';r'))

AA'

I

S~G'

< 2q(F(w';r'))

#(l°g+If(~)l)dq(~) F(w';r')

%(log+If(6)l)dn(~)

C(2B)nN(n)/2, where

C = AA'v(=)/z(F(a)).

using

N',

G"

Similarly,

and the corresponding ~(log+If(w)l)

for

w @ F(w';pr'). tI

can apply the inverse

h(t)

S

large

v(~)

# 0, the right-

n ~ no, say.

and get the following

So we

inequality:

~ exp[h(C(2B)nN(n))].

this along the circle

rnexp[-~(N(n))] , we have,

by

It follows that

~ C(2B)nN(n)

for all sufficiently function If(w)I

(18)

12 ~ C(2B)nN(n)/2

As we may always assume that

hand side exceeds

We integrate

we have

estimates.

F(w';pr').

Since

by use of the property

r' = r n

(e),

if(w) ids(w ) ~ 2~@rneXp[-~(N(n)) + h(C(2B)nN(n)) ]

F(w';pr') 2~Prnexp[-~(N(n))/2],

if

n ~ max{no,

corresponding included

C}.

We now look at the union

D' E K'. n in the inside of Fn,

morphic at

to all

Then, n ~ i.

, it has an expansion

Fn

of paths

F(w';pr')

as we have seen, the set

f(z)

Since the function : ~ ~ c.z -] j:0 ]

f

E

is

is holo-

Since we have

253

#(t + log 2 ) / # ( t ) belongs vanish of

to

E);

identically,

f.

(18)

: 0(i),

H#([\

If

and

n

t ÷ ~,

so w e m a y

let

Cp,

it is e a s i l y

seen that

assume

= 0.

f(~)

p ~ i, be the

is s u f f i c i e n t l y

large,

f(z) - c 0

If

first

f

does

nonzero

t h e n we c a n u s e

also

not

coefficient

the

inequality

get

if(w) llwlP_lds(w ) n PrnbP-iN(n)exp[-N(n)i/2/2].

This

is a c o n t r a d i c t i o n ,

Hence

f

must

4B. N~

In the p r e c e d i n g

in a n a n n u l u s

struction

can

the t h e o r e m dition,

for

with

Theorem.

theorem

be p e r f o r m e d

by H e i n s Every

6 = I

closed

subset

to the

dition

et/~(t)

= 0(i),

Proof.

Let

be a n y c l o s e d

F(1),

implies

F and

that

V

N~

a region

throughout

f

H~(V).

For a general is p o s s i b l e that

there

§3.

CLASSIFICATION

5.

This form.

of a c i r c l e

having

zero

will

function

~

In

linear

con-

was

measure

satisfies

the

con-

set of

in

[

V.

zero

linear

including As H e i n s

Then

F.

measure

shows,

it is e a s y

on a circle,

The hypothesis every

to v e r i f y

on

f ~ HI(v\ that

F)

such an

[]

convex

or not.

center.

following

convex

con-

a stronger

is l i n e a r .

if t h e

E

same

in the

H ~ ( V \ F) ~ H I ( v \ F).

to

set o f c l a s s the

arbitrary satisfies

n + ~. []

t + ~.

is h o l o m o r p h i c belongs

}

i.e.

a null

Trivially,

with

When

zero as

stated

is h e r e

belongs

say

class

origin.

~ < i.

to

to be p r o v e d .

we c o n s t r u c t e d

at t h e

is p o s s i b l e ,

and

tends

as w a s

for an a n n u l u s

that

[31]

last m e m b e r

function,

center

we a s s u m e d

the value

observed

the

be a c o n s t a n t

function

But we c a n

be n o p r a c t i c a l

OF P L A N E

~

we do n o t

take

~

know whether

arbitrarily

close

to

~ = i i

so

difference.

REGIONS

Lemmas 5A.

closed and

Lemma.

subset v

the r e g i o n s

of

Let

0 < a < b < c < ~

{Izl

~ c}

be the h a r m o n i c {Izl

> b} \ F

such

that

measures and

{Iz[

and {Izl

at t h e

F

be a b o u n d e d

> b} \ F

let

is a r e g i o n .

point

~

with

respect

> a} k F, r e s p e c t i v e l y .

Then

Let to

254

log(c/b) log(c/a) Proof. {Izl

Let

u

> a] \ F

elsewhere.

be the s o l u t i o n

F

lies

{Izl

in

u

on

: u(-)

~ c},

Fj,

Lemma.

sets

Let

such that

Suppose

with

~

z • F},

with respect

let

solution

a

Pa(Fj) > 0

on

D a.

strictly

quence

{U(an;Z)}

~\

i

v, so that

e a s y to see that

v

is the

w i t h the b o u n d a r y

d a t a equal

for

{an:

functions.

say

i

and,

measure

> a} \ F.

for e a c h

at the

Then there

is,

such that

is h e l d fixed. D a = {Iz - z01

n = l, 2,...}

Let

u(a;z)

> a} \ F

above, v(z),

sequence

it c o n v e r g e s

uniformly

the p o i n t on

~F.q

z0

[ \ F. and to

se-

is mo-

by H a r n a c k ' s

on any c o m p a c t

problem 0

set

is a r e m o v a b l e In fact,

of the D i r i e h l e t on

of p o s i t i v e

the c o r r e s p o n d i n g

Since this

is h a r m o n i c

solution to

[\ F

3F. a n d to 0 e l s e w h e r e , so ] is a s s u m e d to be n o n - p o l a r , we h a v e

is b o u n d e d , v

F.

point.

on

and is b o u n d e d

v

all

be the h a r m o n i c

property

to zero and c o n s i d e r

function,

Since

~

including

0 £ j £ k.

Take any s e q u e n c e

increasing

of

{Iz - z01

< E,

Pj

of h a r m o n i c

( F U {z0}).

singularity

Since

decreasing

to a h a r m o n i c

Let

of b o u n d e d

be any f i n i t e

be the h a r m o n i c

problem

to

number

is a r e g i o n

0 ~ j £ k, w h i c h

data equal

= u(a;~).

numbers

notonically

j,

i

w i t h the a b o v e

of the D i r i e h l e t

w i t h the b o u n d a r y

in

we have

~ log(]zl/e)/log(a/c)

to the r e g i o n

to the r e g i o n

s > 0, a n u m b e r

Take any i n d e x

theorem

elsewhere,

#(F(b)).

set.

Pa

p ( F j ) - #a(Fj)

u(a;z)

u(z)

F = Uj~ 0 Fj

F.± is a n o n - p o l a r at the p o i n t ~ with respect

0 < a < i n f { I z - z01:

that

0

> b} \ F

for this r e g i o n w i t h 0

k, be a f i n i t e

for each

z 0 • [\ F

that each

measure

be the

and to

{Izl

[]

j = i,...,

[ \ F.

j ~ i, and let

Proof.

for the r e g i o n

F(a)

u(z)dp(z).

log(c/b) ~ log(c/a)

5B.

for any

on

problem and to

we s e e t h a t

inequality.

point

problem

i

to the r e g i o n

F(b)

shows the d e s i r e d

closed

to

a < Izl < c, f r o m w h i c h we get

£(b) u ( z ) d p ( z )

with

u

: [ Jr(b)

f

This

of

of the D i r i c h l e t

d a t a e q u a l to

v(r(a))

Since

data e q u a l

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

is e q u a l to the s o l u t i o n

~ v(r(a)).

of the D i r i c h l e t

w i t h the b o u n d a r y

the b o u n d a r y

for any

~(F(b))

it is for

elsewhere.

~\ F

255

Thus

v(~)

since

= ~(F.). Since U(an;~) tend increasingly 3 =< u(a;~) for a n y 0 < a < an, we have

for all

sufficiently

chosen,

the l e m m a

5C. {bn:

to

U(an;~)

Lemma.

Let

n = 0, i,...}

0 < d < i.

small

a > 0.

is proved.

~

j

and < E

was a r b i t r a r i l y

[]

be a c o n v e x

a sequence

Then t h e r e

As the i n d e x

v(~)

~(Fj) - ~a(Fj)

function

of p o s i t i v e

exist a sequence

satisfying

numbers,

(i) in IA,

0 < p < ~ < i, and

{a : n = 0, i,...}

of p o s i t i v e

n

numbers

and a s e q u e n c e

an+ 1/a n ~ p,

{E n : n = 0, i,...}

a n =< bn,

N~

En =C {6an =< Izl =< a n }

(19)

such that

for

na n < i,

n : 0, i~...,

and

d £ ~ ( - l o g ( n a n ) ) m ( E n) £ i

for

n = I, 2,...,

where

m

is the h a r m o n i c

with respect

to the r e g i o n

Proof.

{dn: n = 0, I,...},

Let

sequence measure {Jz[ an

in

with

limit

d.

at the p o i n t

i/¢(-log(na

na

n

with

E = Un= 0 E n U {0}.

by

with respect

Pn

(resp.

and

~(-log(na

n

decreasing

mn ) the h a r m o n i c

to the r e g i o n

In the f o l l o w i n g

< i

at the p o i n t

0 < d n < i, be a s t r i c t l y

We d e n o t e

~

> an+ 1 } \ (Uk~ 0 Ek)). so small that

~\ E

measure

~\

(Uk~ 0 E k)

construction )) > 0.

(resp.

we shall t a k e

We w r i t e

~

n

=

)). n

In o r d e r a0,

to c o n s t r u c t

satisfying

polar This

E 0 e N~

which

is i n c l u d e d

is p o s s i b l e

by L e m m a

chosen non-polar

E k e N}

following

E

by i n d u c t i o n ,

and

in the a n n u l u s

3A and T h e o r e m in

(C n)

0 < ak < min{bk,i/k} for

4A.

{6a 0 ~

Suppose

{6a k ~

JzJ ~ ak} ,

and

~(-log(nan))

choose

any

a n d also a n o n jzj ~ a0}.

that we h a v e

0 ~ k ~ n, w i t h the

d n ~ k ~ ~ n ( E k ) ~ ek'

T h e n we c h o o s e (21)

an+ I

(22)

so small

O ~ k £ n;

that

d n + l ~ k ~ ~ n ( E k ) ~ ~k' Vn(F(an+l)) is no p r o b l e m a b o u t

and the fact

for

I ~ k ~ n.

0 < an+ I < m i n { b n + l , ( n + l ) - l ] ,

(23)

> 0

0 ~ k ~ n - i; and

(20)

an+ 1 ÷ 0

we first

~ ( - l o g a 3) > 0,

property:

ak+I/a k ~ p

There

our set

0 < a 0 < min{b0,1}

t/~(t)

i ~ k ~ n;

> i/¢(-l°g(n+l)an+l))

= ~n+l"

(21).

is c l e a r f r o m L e m m a

= o(i),

no f a s t e r t h a n

0 < a n + l / a n £ p;

Statement

t ÷ ~, for

(-logan+l)-I

(23)

~ n ( F ( a n + l )) while

decreases

as

i/~(-log((n+l)an+l))

5A

256

decreases

much

Lemma

5B,

Fj

an+l,

and

then

in

{6an+ I ~

faster

than

= E.]

for

use

(20).

this.

For

j = 0,..., By use

IzI ~ an+ 1 }

~K

denotes

a way

to

[\

(ukn 0 E k u K ) .

of

K

such

Then

Vn(Ek)

~ ~n+l(Ek)

(Cn+ I) is s a t i s f i e d . which

4A we

to

~a

find

set,

= ~n

in

with

a set

a =

K E N~

measure

> ~n+l' at the

it is not h a r d

to

point

find

=

with

a closed

respect

subset

En+ I

that

dn+l~n+l Since

only

and

that

~ ~K (K)

the h a r m o n i c

we have

~ = ~n

of T h e o r e m

in such

Vn(F(an+l)) where

(22)

n,

satisfies

easy

to

(22)

implies

~ ~n(Ek)

for all

for each

(19),

for

By i n d u c t i o n

(C n)

see that,

~ ~ n + l ( E n + l ) ~ ~n+l"

for

can

n ~ i.

fixed

d

we

n,

construct

Set

k,

÷ d.

k = 1,...,

the

property

{En:

n = 0, i,...}

E = Un= 0 E n U {0}.

~n(Ek)

÷ m ( E k)

as

It is t h e n

n + ~.

So,

[]

n

6.

Classification 6A.

and

Let

let

For each

polar

set

that

E

The

{an:

n

included is a l s o

we call

obvious.

By

point

~

in

En

{6an

bounded,

and

with

Izl

=<

E

=< an}. and

center

to the r e g i o n s

an+i/a n ~ p

for

center

so

In the

fol-

at the

at an a r b i t r a r y the h a r m o n i c and

non-

E n U {0},

disconnected.

[\ E

numbers

disconnected, E = U n=0 ~

set w i t h

n ~ i, we d e n o t e

respect

that

set

totally

with

of p o s i t i v e

totally We

a circular

sets

mn,

Suppose

be a closed,

closed

a set

of c i r c u l a r m

be a s e q u e n c e

be c o n s t a n t s .

let

such

definition

the

n = 0, i,...}

0 < p < 6 < i

n ~ 0.

lowing

Theorem

origin.

point

is

measures

at

{Izl

> a

} \ E, ren

speetively.

We use

sification Theorem.

Then

result. Let

~

various Namely

and

~

(AI)

~(t)/~(t -s)

there

exists

E n E N~,

we h a v e

= o(i),

that

t + ~,

set

E

the

function

only

constant

Proof.

Let

0 < p < 6 < I

be fixed.

quence

{bn:

n = 0, i,...}

of p o s i t i v e

and

to o b t a i n

our

clas-

following satisfying

for any

with

H~(¢\

< i/n

in o r d e r

functions

while

b

contains

sets

the

be c o n v e x

a circular

n ~ 0, such E)

circular

fixed

center z -i

s > D.

at the o r i g i n

belongs

to

and with

H ~ ( ~ \ E),

functions. By use

of

numbers

(AI) such

we can that

find

b 0 = i,

a se-

257

0 < ~(log(6-1t)) for

n : i, 2, . . . .

of p o s i t i v e that

By L e m m a

numbers

a n =< bn

and

d ~ ~ ( - l o g ( n a n ) ) m ( E n) ~ i

same m e a n i n g

as in L e m m a

We f i r s t

En

show that

is i n c l u d e d

{an: n = 0, I,...}

{En: n : 0, i,...}

in

En C . {6a n. < Jz . I < a. n } for

n ~ i, w h e r e

m

N}

for

and

such n > 0,

d

h a v e the

5C. z -I E H ~ ( ~ \ E).

~(-log(6an)) As

t ~ i/bn,

5C we get a s e q u e n c e

and a sequence

a n + l / a n =< p,

and

£ 2-n~(log(n-lt)),

Since

a n =< b n, we h a v e

~ 2-n~(-log(nan)) ,

n ~ i.

{6a n =< Izl =< a n] , so we have

in

~(logJz-lJ)dm(z)

=

[ n=0

E

~(log[z-ll)dm(z) E n

£

[

~(-log({a ))m(Z )

n=0

n

n co

< ~ ( - l o g ( 6 a 0 ) ) m ( E 0) + =

[ ~(-log(6a n=l

£ ~ ( - l o g ( 6 a 0 ) ) m ( E 0) +

[ n:l

this shows that

z -I

belongs

to

with

n ~ i.

is i n c l u d e d measure

Since in

F(oa

(24)

).

Harnack's

d~w/dS

where

A'

= A ' ( ~ -2)

We take any [ \ E.

assume

that

f e H ~ ( ~ \ E) f

and

origin.

Namely,

£ A ' / ( 4 ~ a n) ~ A'/(4~2an

u

0 < Jz[ < ~.

)

do not v a n i s h

Since

be the h a r m o n i c at the p o i n t

W

that

F ( a 2 a n ), in the p r o o f of L e m m a

majorant

u

of

is c o n s t a n t .

expansion:

}(log+JfJ)

Since each

f(z)

En

be-

at the

= [j=0 cj z-j

is a b o u n d e d

{a n < [zJ < ~2an} , we h a v e

3C.

So we m a y

can o n l y be s i n g u l a r

f(z)

an

F(an) ,

identically. f

following

n ~ i.

~w

then i m p l i e s

appearing

f

similar

We look at any

{a n < JZJ < @p-la n } Let

on

and a h a r m o n i c

N~, we see that

Let

on

is to s h o w that

we h a v e the

t i o n on any a n n u l u s

4A.

{a n < JzJ < O2an }

inequality

is the c o n s t a n t

Our o b j e c t i v e

longs to the class

a = (~p-l)i/4.

to the a n n u l u s

d~w/dS

(25)

for

n

))

can be shown by an a r g u m e n t

in the p r o o f of T h e o r e m We set

n

2 -n < ~.

0 < p < 6 < i, the a n n u l u s

~ \ E.

with respect

on the c i r c l e

on

used

))/}(-log(na

H ~ ( [ \ E).

The l a t t e r h a l f of the t h e o r e m to the one a l r e a d y

n

analytic

func-

258

l°g+If(w)l a I l°g+If(~)la~w(~)

(26) for any

w E F(Oan) , where

Applying

~(t)

to both

}(l°g+If(w)l)

~w

sides

is the h a r m o n i c and u s i n g

< [I = F(a

+ I

)

Assume

now that

w E F(oa

).

Jensen's

F(o2a

n

measure

)

defined

inequality,

above.

we get

] ~(l°g+If(~)l)dSw(~)-

n

Using

(24), (25), (26), Lemma 3C and the -1 ~ m(E n) ~ d } ( - l o g ( n a n ) ) , and c o m p u t i n g as in the p r o o f n

fact

mn(F(an))

of T h e o r e m

4A, we see that the r i g h t - h a n d

is m a j o r i z e d

by

C%(-logan)

, where

side of the above

inequality

C : A'(o-2)A(o-l)u(~)/d.

Thus we

have (27)

%(log+If(w) I ) / ~ ( - l o g a n ) ~ C

for any

w E F(oan).

is p o s i t i v e ~ ( t 2 ) / % ( t I) N => i

Take

a positive

and n o n d e c r e a s i n g for any

satisfying

tI > tO

for

and any

-loga N > t O .

Then

If(w)l for any

w E F(aa n)

and any

number

t _> t O .

< an

n ~ N.

tO

t 2 > 0. (27)

so large

Then we have

that

~(t)/t

t2/t I < I +

We c h o o s e

an i n t e g e r

implies

-C-I Letting

and w r i t i n g

n ~ N

r =

Oan, we have i ICkl Letting

=

f(z)zk-ldz

in

z -I

small

ek : 0

Let

p ~ i, then we w o u l d

ficiently

z, say

have

¢(log+If(z)l)dm(z)

This

contradiction

constant 6B. expressed

[ n:N'

In terms

k > C + i

k-C-i an

and t h e r e f o r e

be its h i g h e s t n o n z e r o P If(z)l => 2 -1 ICpl Izl -P ~ i

~

with

[ n=N'

N' ~ N.

f(z)

coefficient. for all

suf-

Consequently,

~(log(2 - l l c p l a n - P ) ) m ( E n )

~(log(2-11Cplan-l))/%(-log(na

shows

functions.

for

ok =

c

Izl ~ aN,

E d

< rka -C-I = n

F(r)

n ÷ ~, we see that

is a p o l y n o m i a l If

I

~

)) n

that

p = 0.

Hence

H ~ ( [ \ E)

contains

only

[] of null

as follows:

classes

0~

the p r e c e d i n g

theorem

can be

259

Theorem.

Let

O~

Then

}

and

strictly

We set

~

be c o n v e x

includes

functions

satisfying

(AI) above.

0~.

0p = U{0q : 0 < q < p}

we have the f o l l o w i n g ,

where

the

and 0 + : n{0q: p < q < ~}. Then P i n e q u a l i t y sign < m e a n s the s t r i c t

inclusion. Corollary. (b) Proof. one:

(a)

0- < 0 < 0 + for P P P 0AB , < N{0q: 0 < q < ~},

$(t)/~(t)

= o(i),

(a)

(AI)

t ÷ ~, if e i t h e r

Let ~(t)

0 < p < ~. = e pt - 1

is e q u i v a l e n t

~

or

$

Indeed,

In o r d e r

satisfies

eqt/~(t) Hence

~

satisfy

0~~ < 0p, we h a v e o n l y

0 ~ t ~ 2/p,

ePt/t,

t ~ 2/p. ~

is a c o n v e x

So, by the t h e o r e m ,

= o(i),

function

~(t)

t ÷ ~, and t h e r e f o r e

= te pt

and

(AI) and t h e r e f o r e t + ~, for any

~(t)

0q < 0~

(b)

0~ < 0 .

To s h o w t h e

~(t)/~(t)

Hence

(AI),

so that

In fact,

@

and

0~ < 05.

As

U{0q:

We r e m a r k that, their

following without

proof:

the c o n d i t i o n

and

(AI),

~

= t 2

and

0AB , < 0~

¢

by the t h e o r e m .

t ÷ ~, for any

%(t)

q > 0, so that

= exp(e 2t) - e

are seen to s a t i s f y = o(i),

if two n u l l c l a s s e s

exists

H~(D)

~(t)

=

0q <

[]

of r e g i o n s ,

e.g. 0 and P H e r e we state the

functions

a region

while

and

the c o n d i t i o n

t ÷ ~, so we h a v e

is v e r y wide.

If the c o n v e x

functions,

= t. in IA.

0 < q < ~}.

eqt/~(t)

difference

$(t)

as m e n t i o n e d

0 < q < ~} ~ 0~ < 0@ ~ 0AB.

then there

only constant

¢(t)

05 = 0AB,,

is seen by t a k i n g

0q, are d i s t i n c t ,

contains

we s e t

t ÷ ~, we have

0AB , < 0~ ~ n{0q:

Consequently, 6C.

half,

functions

~ ( t ) / e qt = o(i),

latter half

exp(e t) - e.

0~.

first

= o(i),

On the o t h e r hand,

The

we h a v e

Hence,

q

are then c o n v e x

05 < 0q.

a g a i n by

On the o t h e r hand,

q > p, so that

0 <

Then these

Op < 0¢ C 0+ . = p $

If

In o r d e r to show

= e p t - i.

0p < 0~.

and the con-

0~ < 0~ = 0p.

0p = U{0 q : 0 < q < p} C = 0~ < 0 p .

0 P < 0 p, + we set

~ ( t ) / e qt = o(i),

Since

the con-

:

is s a t i s f i e d .

q < p, t h e n

to show that

p2e2t/4,

it is easy to c h e c k that (AI)

the theorem.

and

to a s i m p l e r

and

~(t)

and

0 < q < =) < 0AB.

(3) in 3B.

to set

that

U{0q:

We first n o t e that the c o n d i t i o n

dition

dition

0 < p < ~.

~

D ~ ~

and

~

satisfy

such t h a t

is i n f i n i t e

H~(D)

dimensional

260

and

contains

lined

7.

functions

in H a s u m i

Majoration 7A.

2C.

Lemma.

Let

a convex

f

Proof. in

such

its

on

that

OAS

LHM

D

connected

D

in

let

contains

only

to

possesses

show

that

a Green

which

a non-polar

set.

Thus

it is p o s s i b l e

F

each

being

and

D.

define ux(z)

Since

contains

set

Let

analytic

is not

a point gr(z0,z)

functions

Ux,

function

the

from

let

~

be

a harmonic

the

set of a n a l y t i c by a q u a s i -

class

of r e g i o n s

D

functions.

~ 0 G.

with

thus

D £ 0AS.

clearly

Since K

set

a

two

Ch.

D'

2, . . . . D'

a

We n o w a s s u m e K

is

closed and

sets

so

I, i m p l i e s

boundary

on the r e g i o n

has

= ~ \ F'

6D,

for

n = i,

[\ D carries

D ~ 0G,

into

D'

Theorem

function

for

be a r e g i o n If

D

We

Green

D

then

is a r e g u l a r

=< n]

x E D',

let

D ~ 0 G.

to d i v i d e

set,

which

be the

g'(z0,z)

and

set.

a polar

z0

So,

i.e.

disconnected.

a non-polar

F

F n = {z @ F:

point with

of the

pole

Finally by the

z0

we

formula

: g'(x,z). D' v n = (Uz0)Fn

Let Fn

(Ch.

and

are m a j o r i z e d

n,

{Vn}

monic

I,

6E).

and

as above, {Wn}

w ~ w 0.

be the

balayaged

Clearly,

vn

there

by

g'(z0,z).

and

therefore

is i n o r e a s i n g

function,

fixed,

So

and

has

is m a j o r i z e d 0AS

is a c o n t i n u u m ,

is t o t a l l y

region

denote

constant

0AS

say)

F

AS(D)

function,

(= K,

that

region

%(log+IfI )

easily

= 0 G.

bounded

F',

be d e r i v e d

If

by

only

that

and

can

on a p l a n e

log+if(z)I

We d e n o t e

noneonstant

D = D' \ F.

which

function

[

such t h a t

component

[\ D

is out-

is q u a s i b o u n d e d .

AS(D)

We have

[, w h i c h

A proof

Functions

(i) holds.

function.

for w h i c h

Theorem.

singularities.

Harmonic

following,

be an a n a l y t i c

harmonic

[

the

For a r e g i o n

functions

in

f

then

7B.

bounded

we h a v e

function

majorant,

essential

by Q u a s i b o u n d e d

First

Theorem

having

[21].

are

function

bounded

of

harmonic

Since

{F n}

converges

Uz0

relative

functions

on

to D

is i n c r e a s i n g

with

to a q u a s i b o u n d e d

har-

set

say v, on D. Next we take any y ~ D, w h i c h is h e l d D' ~' w n = (Uy) F n for n = i, 2,... and w 0 = (Uy) Then,

{Wn:

n = i,

tends In v i e w

2,...}

is i n c r e a s i n g

to a s u p e r h a r m o n i c of T h e o r e m

6E,

Ch.

function,

and

wn S w 0

say

w, on

I, we h a v e

wn

Uy

for all

n.

D',

satisfying

on

Fn

except

261

for a polar set.

Since the union of countably many polar sets is again

a polar set, we see that set. Let

by Theorem

= g'(y,.)

Y of balayaged

The definition g~(z0,z)

w : u

on

functions

be the Green function

for

F

except for a polar

shows that

D' \ F n

w = w0

with pole

on

z 0.

D. Again

6E, Ch. I, we have Vn(Y)

= H[Uz0;D' \ Fn](Y) = g'(z 0) - g~(z0,y) = H[Uy;D' \ Fn](Z 0) = Wn(Z O) ÷ W(Zo).

D, w(z O) = Wo(Z U) : Uy(Z O) = Since z 0 is a regular boundary point of Since each v is a bounded g'(y,ZO), Hence Vn(Y) ÷ g'(y,z 0) on D. n harmonic function on D, we conclude that g'(Zo,-) is a quasibounded harmonic

function on

Finally,

D.

consider the function log+if(z)i

for a sufficiently D

is majorized

Corollary.

= (z- zo )-I

on

= log+

1

~ g, (Zo,Z) + C

large constant

C.

This means that

by a quasibounded

stant, we conclude

f(z)

that

harmonic

function.

~.

Then,

log+If(z)l f

As

D ~ 0AS, as was to be proved.

on

is not con-

[]

0 G = 0AB , = 0AS.

NOTES Heins' ili].

classification

theorem

is in his lecture note

For the case of plane regions he only showed

proposed

further investigation.

were made by Hejhal sults mentioned

Afterwards,

[32], Kobayashi

[38,

above are due to Hasumi

interesting

39] and 0brock [21].

[31; Chapter

0AB , < 01

and

contributions [49].

See Hasumi

The re-

[22] for some

related results. Lemma 3B is in Hejhal Hasumi

[21].

Theorem

[32].

The fact stated in 6C is proved

7B is adapted

from Segawa

[63].

in

APPENDICES

A.I.

The

Classical

A.I.I. Theorem. set

We b e g i n

Let

p

be a f i n i t e

= p(l A (-=,x))

(a)

F'(x)

(b) Uc

exists

if we and

set Us

For a p r o o f , [59],

Ch.

A.I.2. region 2e

we mean

E0

Let

the ~

is h a r m o n i c

S(eiS;e) Proof.

(to, that

on

~.

I

and

Then

Lebesgue

textbook

on the

and ~s

measure

on r e a l

P - Pc'

singular dx

parts

on

analysis,

I.

e.g.

unit

vertex

circle

~.

E0

angular

and

By a S t o l z measure

z~ 0) < ~},

set

Then and, to

tends

= I

F'(8)

e

0 ~ r < i. the

following

Borel measure

on

T

and

set

P(r,e -t)d~(eit).

any real

~ ( { e i8 : t O ~

=

z

a n d we h a v e

complex

Take

t O + 27)

with First

interval

summable;

-~ < a r g ( l -

We

= u ( r e i6)

t O ~ t ~ t O + 27.

interval

to the

with

kernel

be a f i n i t e

F(t)

provided

~

= {z • ~ :

Poisson

(2) for

is

on an

= ( i - r 2 ) / ( l - 2r cos e + r2),

u(z)

u

x • I. and

point

disk

0 < ~ < ~/2.

(i)

Then

measure

set

we a s s u m e

Theorem.

I

standard

be a n y

open unit

the

is c a l l e d

on

respect

see a n y

P(r,8) This

for every a.e.

with

S(~0;~) where

complex

of m e a s u r e s .

8.

Let

in the

on d i f f e r e n t i a t i o n

for A C = I and Pc(A) = fA F ' ( t ) d t a r e the a b s o l u t e l y c o n t i n u o u s a n d the

U, r e s p e c t i v e l y ,

Rudin

Theorem

with a theorem

F(x)

then of

Fatou

exists

e

tO <

and

set

t})

for almost

every

8

in t h e

for any such 8, u(z) t e n d s to 2~F'(8) ie through any fixed Stolz region

0 < e < ~/2. we

see e a s i l y

that

u(z)

is h a r m o n i c

in

D.

By t h e

263

preceding only

theorem,

to

show

If

~

then

u(z)

trivially p

if

the

is

the on

is

•

proven

notation

function

loss

any

e.

for

the

of

definition

~-9)

is

differentiable

through

F(t) So

:

the

measure

theorem

modified

it

assumed

also

F

implies

P(r,e

- t)dF(t)

=

Poisson

r COS e

integral

f

:

-

P(r,e

formula

for

the

harmonic

we

any

may to

and

= 0

T, is

given

therefore simplify e = 0.

and

so

- t)F(t)dt.

function

= -

r cos e

P(r,8 - t)sint

-'If

Thus

We

In o r d e r

= F(~)

on

theorem

for

t O = -~

F(-~)

have

-'g

P(r,e - t)cos t d~(t)

|

the

proven

= 0.

that

-g

The

be

So w e

= dt/2~

that

~ - ~(T)o.

B(T)

that

a.e. regions.

do(t)

can

measure that

is

Stolz

( t - t 0 ) / 2 ~ , so

generality

of

=

F limits

Lebesgue

and

slightly,

u(re

of

normalized

for

without

The

the

existence

~ i true

it

assume

the

states:

d~(t).

-][

have

u ( r e zS) - 2 z F ' ( 0 ) r c o s

8 = -

P(r,8 -t)(F(t)

- F'(0)sint)

dt

-7

=

(-sin t

=

+

P(r,e

+

+

= I 0 + I I + 12 + I{

with we

0 < 6 < ~•

have

only

Since

to F

Since estimate

is

(3) to

zero

increasing (3)

for

and

and

this

angular

min{~/2,~-

t ÷

function 0 < t

After i

as

I0,

~F(t) 0.

So

e(t),

say,

I I'

II

(resp.

and at

_ F'(0)

there

12

12') a r e

and

similar,

12 . 0,

'

-F (s- t~)

exists

0 £ t ! ~/2,

_ F'(0)

}

a continuous with

c(0)

monotonically

= 0, w h i c h

majorizes

~/2. =< preparation, measure

2~}

and

-

+

+I~,

differentiable max{

tends

II

- t)) is-~

belongs

2~, to

take

any

Stolz

0 < ~ < 7/2. S(I;~),

then

region If

rlsin

S(I~)

z = r e ie 91/(1

with with

-rcos

8)

vertex 101

<

~ tan

therefore

(4) where

lel ~ A(l-r), A

is

a constant

depending

only

on

e.

Since

e(~)

is

monoto-

284

nically increasing, (5)

so is

0 < 60 < i

Suppose now that

6~(6) I/4.

and

Let

60

be chosen so that

c(60) < min{(2A)-4,(260)-4,1}.

z = re i6 E ~

satisfies

(4) and

i - r £ 60~(60 )I/4

Then there exists a positive number 6 with 0 < 6 < 60 i - r = 6~(6) I/4. It follows from (4) and (5) that

such that

181 ~ A ( I - r) : A6s(6) I/4 ~ 6/2. If

Jt I £ 6, then J-sin t ~ t P(r,e - t) I =

2r(l - r 2 ) J s i n t sin(e -t) 1 2 ((i- r) + 4rsin2((e - t)/2) 2

4 ( i - r)-36(le] + 6) £

4(6a(6)i/4)-3.6.36/2

= 66-ie(6) -3/4 and thus II01 £ i6 [-sin t ~ P ( r , 8 -6 < 66-1s(6)-3/4.26~(6) If

6 ~ t

~ ~,

then

3~/2

there exists a constant

- t) I s--i-~-F(t)F'(0) dt

= 12E(6) 1/4.

~ I t - el m I t l B > 0 with

(6)

[el m I t l / 2

and therefore

sin 2 8 ~____t_t~ Bt 2.

So, for

6 £ t ~ 7/2,

l-sin

t~

P(r,8 - t)J £

4(1 - r ) t < Const. t-26a(6) I/4 sin3((t - 8)/2)

and thus IIlJ £

i712 ~6

[-sint

~

F(t) P(r,e -t) I s--i-~- F'(0) dt

£ Const.6¢(6)l/4IT/2 ;6 Finally, let 7/2 £ t £ ~. in view of (6),

]~-t

By (5),

t-2dt : Const.a(6) I/4 i - r £ 60a(60 )I/4 £ 1/2

and so,

P ( r , 8 - t)[ < 4 ( 1 - r ) [ s i n ( ( t - e)/2) I = i- r = 16sin4((t - e)/2) Isin3((t - e)/2) I < C o n s t . ( l - r) < Const.a(6) 1/4

265

This i m p l i e s

that

1~21 ~

I ~/2

+tP(r,0-t)llF(t)-F'(0)sintl

< Const.s(@)i/4(maxt

Combining

these

estimates,

IF(t)I + IF'(0)I)

z

so

u(re i8) ÷ 2wF'(0),

tends

A.I.3.

to

i

through

i14

e I £ C o n s t . s ( 6 ) I/4

S(I;~),

as desired.

Corollary.

= Const.s(~)

we get

lu(re ie) - 2 ~ F ' ( 0 ) r c o s If

dt

U n d e r the

then

6 + 0

and thus

E(6) ÷ 0;

[] same a s s u m p t i o n

as in T h e o r e m A.I.2,

let u*(e i0) which

exists

a.e.

on

T

lim r÷l-0

u(rei0),

as a s u m m a b l e

f

(7)

=

P(r,e

function.

Then

.

t)u*(elt)da(t),

-

T

for

z = re ie E ~ ,

Proof.

Since

is the q u a s i b o u n d e d

u * ( e ie)

fines a h a r m o n i c

= 2~F'(e)

function

seen to be q u a s i b o u n d e d .

on

implies

that

Pc

~,

of

we m a y a s s u m e u*(eit)da(t)

u- v that

where ~s

~s = ~ - ~c

positive monic

function,

n.

Set

F'(e)

is d e n o t e d v

is s u m m a b l e , by

v

(7) de-

is e a s i l y part of

function.

For

Since T h e o r e m A . I . I

is the a b s o l u t e l y

continuous

part

P(r,0- t)dPs(eit),

part of u- v

to show that

p.

Since

B

is n o n n e g a t i v e ,

is a n o n n e g a t i v e ( u - v) A n

v

= 0

harmonic

for any

= ( u - v) A n . Since v is a b o u n d e d h a r n n the first p a r a g r a p h of the p r o o f of T h e o r e m 2F, Ch. IV

shows that

(8)

v.

is the q u a s i b o u n d e d

is n o n n e g a t i v e .

and t h e r e f o r e

So it is s u f f i c i e n t integer

= I

is the s i n g u l a r

is a l s o n o n n e g a t i v e

function.

u(z).

is an inner h a r m o n i c p

= F'(t)dt

( u - v ) ( r e ie)

and

which

To show that

u, it is e n o u g h to see that this p u r p o s e

a.e.

~,

part of

Vn(rei0)

= I

P(r,e - t ) v ~ ( e i t ) d a ( t )

266

for

re

i8

v* denotes a bounded measurable f u n c t i o n on ~. n to t h e e x p r e s s i o n (8), we f i n d t h a t v* is the n radial boundary f u n c t i o n for vn • Since 0 =< v n =< u - v on D, 0 = < v * ( e it) < u * ( e it) - v * ( e it) = 0 a.e. a n d t h u s v E 0 by (8) as w a s n = n to be p r o v e d . []

Applying

E ~,

where

Theorem

A.I.2

A.I.4.

Let

ekeik8

on

[k~0

nomials

be

the

set

is t h u s following

Then

we

Corollary.

For

any

f E L (d~)

P(T) + P(~)

such

the

Proof.

that

We m a y

for

see

suppose

(9)

have

Ch.

re i8 E ~ . to

ascending IIhnll~ ~

By t h e

f(e i8)

a.e.

sequence IIfiJ

and

there IlfIl

IV,

that

u(rei8)

tends

the

IJfnll~ ~

definition,

of a n a l y t i c

P(T) + P - - ~

~.

(for

on

P(T) T.

Fatou as

2A),

trigonometric set

exists

and

of

f

{fn }

a.e. can

n

If

be

poly-

in

f e H~(d~)

taken

from

P(T).

Set

P(r,8 - t)f(eit)do(t)

theorem

A.2.1

and

Theorem

r + i - 0.

Let

0 < rI < r2 <

tending

to

let

h

hn + f

a.e.

P(r,e-t)

polynomials

trigonometric

a sequence

fn ÷ f

then

JIfrl > 0.

= r JT

the

i

and Using

=

A.I.I~ -.-

u ( r e ie) < i

(e ie) = u ( r eie), n n expansion

the

[

reJee l](e-t),

[ j_~

c . r i J IeiJ 8 jn

be an

so t h a t

we h a v e

h n ( e i8 ) =.

(10)

with

• = c] ST for

j = 0,

formly

on

-+i, . . . . T,

one

Since

can

find

f

sequence have

c.

the

(eiS)

{fn } : 0

conditions thus for

ilhn -

right-hand

and

N[ j:-N

fnil the

)

•.

integer

_ n-• n

fulfills

j < 0

.

the an

n

satisfies

f(elt)e-l]tdo(t

side

> 0

(i0)

converges

such

uni-

that

.. c . r l J i e l] ~ n

< 2Hfl[ /n

requirement.

therefore

of

N : N(n)

f

n

and When

E P(T).

]IfnH~ <

[Ifii .

f E H~(dg), []

The we

267

A.I.5.

For c o n t i n u o u s

Theorem.

Let

sequence

{f } n

Proof. The

f

be a r e a l in

We m a y

suppose

characteristic

on

the

Using

find that we

large lh n

-

that Unl

f < h =

uniformlY.

moreover

< i/n.

+ i/n

A.2.1.

Theorem

To a n y

u. 0 < s < i.

for

=

on C o n j u g a t e

u C Re(R(T))

u + i~ e P(T)

Let

and

f ~do

with

(a)

on

First

•

with

A.I.4,

rapidly

enough,

N = N(n) the

we

so

inequality

fn E Re(P(T)), < f + 4/n.

Hence,

=

~

the

a unique

is c a l l e d

~ E Re(R(T))

the h a r m o n i c

con-

following:

u ~ 0, -I/s IIUlll;

2 ( s + l ) / S ( c o s - ~ ) -1/s UuU1 .

< I1~11s =

Proof.

u

u e Re(P(T))

(12)

branch

(9).

Functions

= 0.

II[IIs =< ( c o s ~ ) for any

+ 3/n

by

everywhere

take

satisfies

n

we associate

T h e n we h a v e

u E Re(P(T))

(ii) (b)

1

see t h a t

+ 2/n < h n

uniformly.

[]

such that

(a)

= u n

jugate

of

f

to

is a

show that

of C o r o l l a r y

We t h e n and

we

then

= f(e i8)

tends

is r e a l

÷ f

n

there

be d e f i n e d

kernel

< I/n.

Then f

u ( r e i8)

proof

{r n}

T.

and

u ( e i8)

in t h e

= Un + 2/n,

fn

and also

Kolmogorov's

and

I f - hnl

n

uniformly.

Lemma.

as

Let

Poisson

T

When

that

Setting

< f =

A.2.

of t h e

boundary

on

f < f = n

llfll > 0.

u n = [ j = -N N c j r ] n J l e ije

n

fn ÷ f

that

same notation

hn ÷ f

can assume

function

such that

properties

up to t h e

we have

continuous

Re(P(~))

is c o n t i n u o u s T.

functions

suppose

values

u > 0.

in the (u 2

R e { ( u + i~) s } $

Then

(u + i~) s

sector

has

a single-valued

larg z I < ~s/2.

+~2)s/2cOST

$

cos

So

2

and therefore

eos~which

shows

I~lSdo < the

u $ 0, w e h a v e tend

to (b)

Re{(u

inequality only

+ illS}do

(ii)

to c o n s i d e r

for

:

Re

u > 0.

u + s

with

(u + i[)Sdo

In c a s e s > 0

= (

udo) s

u E Re(P(T)) and then

with

let

0. For e a c h

n = i,

2,...

there

exists,

by T h e o r e m

A.I.5,

a

268

function

u

E Re(P(]F))

with

u+

n

max{u(eit),o}.

Then

u

-u

< u =

< u + + l/n,

n

and

u

n

have

IIEnll s

where

+ l/n)

s

shows

the

A.2.2.

inequality

Theorem.

+i/n)S+

z : re

ie

E ~.

realized by

W(0)

a.e.

and

on

~

(12).

Let

¢(z)

also

lie n - Ell s

=< C s l l U n - ull 1

Therefore,

(IIu-lll+i/n) s)

=< 2S+icSllulll, s

[]

u E Ll(do)

=

Let

= 0.

be real

and

Proof.

Suppose

first

continuous

functions

~(z)

Then

such that

un ÷ u

in

0 < s < i, to

some

if n e c e s s a r y ,

we may

fn : e x p ( - U n fn E H~(da) and

so

fn

and

be the h a r m o n i c

l i m r + l _ 0 Y(re ie)

therefore morphie

conjugate (= v ( e i e ) ,

we have

are real,

~,

we

f(re ie)

almost

all

Since

~(rei6), that

6.

max{-u,O}.

a.e.

in

each

f(0). ~.

in

follows

: l i m r + l _ 0 ~(re ie) by a p p l y i n g

~(0)

= 0

i~)

in

LS(da),

We

Ifl ~ i, weakly*

into

= u

and

D,

on

~ + i~ As

as

set

of

in

~.

T

and

is h o l o -

f(z)

E H~(~),

r ÷ i- 0

for

÷ exp(-iu#(eiS)).

a.e.

the a b o v e

in

uniformly [fl

of

a subsequence

fn ÷ f

D.

continuously exists

that

-log

exp(-iW(rei6))

8, v a r i e s

density u n ~ 0,

converge

almost

We h a v e

Since

e,

the

extensions,

fn + f

f = exp(-¢-

such

and

By t a k i n g

: e x p ( - u ( e ie) - i u # ( e i e ) )

fixed

u E Ll(do)

A.2.1

nor-

exists

.

A.I.5

the a n a l y t i c then

¢(z)

say)

÷ u and E ÷ u # a.e. # n iu ). Then, IfnI ~ i,

F r o m this

symbols,

= ~

For

n

f= e x p ( - u -

÷ f(e ie)

v(e ie)

for any r e a l

u

of

u n E Re(P(T)),

u # C LS(da).

that

so is

find

By L e m m a

say

see t h a t

with

By T h e o r e m

we can

If we d e n o t e same

- log Ifl

in

assume and

by the

fn(0)

Ll(da).

fn ÷ f

f E H~(do).

Since

u ~ 0.

Ll(da),

element,

i[n) and

f

that in

set

P(r,8 - t ) u ( e Z t ) d o ( t )

f

llvlls =< 2 ( s + l ) / S ( c o s ~ ) - i / s l l u l l l

(13)

and

we

is a r b i t r a r y ,

which

clude

=

(ii)

1

IIEII ss 5 2Sc s (llu+lll s + ,lu-ii1)

for

So by

s + ilunll s) s

< 2ScS((llu+ll =

n

and

(e zt) "

U +

> O. =

c s = ( e o s ( w s / 2 ) ) -I/s.

II u II s =< 2S(llE_Enlls

As

-u n

__< C s l l U n l l 1 =< C s ( l l u + l l l

< Cs(llu-lll+ i/n),

where

=

E Re(P(T))

as

r ÷ i - 0, we c o n -

The

argument

same to

can

be seen

max{u,0}

269

To s h o w the i n e q u a l i t y look at the a n a l y t i c Re(P(~)) ~n ÷ $ monic and

be c h o s e n almost

extension

so that

uniformly of

un

-

u

n ~,

in

(resp.

~ n ( p j e ie) ÷ ~(pje ie)

II~(Pl ei8)

(13), we take any

functions

0 < Pl < P2 < i

¢ (pj z) + i~( pjz) ,

j = i, 2

+ u

Then

in

Ll(do).

where

%n

(resp.

~n ) to

~.

Thus

uniformly

on

T

for

~n)

%

.

Let + ¢

n denotes

and un e and the h a r -

C n ( p j e i8) ÷ ~(pje i6) j = i,

2.

So

~(P2eie)ll~

=< 3S(ll~(Pl eiS) _ ~n(Pleie)lls s + ii~n(Pleie ) - ~n(P2eie)ll s + By a p p l y i n g

(12) to the c e n t r a l

II~n(P2 eie)

-

~(P2 e±e)lls)'s

t e r m of the r i g h t - h a n d

side and l e t t i n g

n ÷ ~, we get

I1~( pl eie) We set

-

~(

~p(e ie)

p2 eiS)

II s

=< 3.2

= ~(pe ie)

(s+l)/S(cos~)-i/sll~(Pl

eie) _ } ( P 2 e i S ) U l .

and

~ (e io) = ~(pe ie) P L (da) t e n d i n g to u

0 < p < i.

Since

i

as p ÷ i, we have {~ } forms a C a u c h y net in P seen that {~ } forms a C a u c h y net in LS(dd) as p ÷ I. Since P (e i@) + v(e i@) a.e. as p + i, v E LS(dd) and ~ ÷ v in LS(dd). P P The i n e q u a l i t y (12) t h e n i m p l i e s the d e s i r e d i n e q u a l i t y (13), for ¢ ÷ P u in Ll(d~) as p ÷ i. []

A.2.3. u

n to

As a c o n s e q u e n c e

E Re(P(T)) v, w h i c h

monic

t e n d to

in

exp(-u-

of

we see that,

if

~ t e n d in LS(do), 0 < s < I, n in r e l a t i o n to u. v is c a l l e d the h a r -

u @ Ll(dd)

iF) E H ~ ( d o )

theorem,

Ll(d~),

is g i v e n by A . 2 . 2

conjusate

shows,

u

of the p r e c e d i n g

and

is d e n o t e d

provided

that

by

[.

As our d i s c u s s i o n

u ~ Ll(do)

is b o u n d e d

below.

A.3.

The F. and M. R i e s z T h e o r e m A.3.1.

We will give ~ k s e n d a l ' s

We first n o t e the f o l l o w i n g , n e rfu~la.

Let

w k E ¢,

which

k = i,...,

p r o o f of the F. and M. R i e s z can e a s i l y

n, s a t i s f y

be v e r i f i e d lWk[

< i

Then n

n

ll- It (1+%)I ~ IT (l+ lwkl)-i k=l

k=l

and

theorem.

by i n d u c t i o n . R e ( w k) < 0.

270

A.3.2. circle

~

Theorem.

Let

in the c o m p l e x

~

be a c o m p l e x

plane

such that

f

(14)

e Z.n t d ~ ( e Z t )

Borel m e a s u r e

on the unit

= 0

T for

n : i, 2, . . . .

the L e b e s g u e Proof. zero,

for

i.e.

IFI = O.

number

and

~

p - p(T)o

is a n a l y t i c

hood of the closed sequence

unit

(a)

the

{gn }

istic

function

Iz-zjl shows

of

(a).

disk

~.

j = i,...,

z E T

and

÷ i

~ 6

z E F, then

on

F.

z E Ak

Iz - ( l +

< 0

a

unit

that

n : i, 2,...;

we c l e a r l y So

and

have

Ign(Z) I < 2, w h i c h

for some

< rk

1

= Pk

n "

consider

Then,

for

k

and thus

any

]~ j=l

N.

N [ ll+

p.m.] i+

$ J z - ( i + pj)zj

pjzj z

z E ~

with

Ipjzj/(z - (I + pj)zj) I < i

j = i,...,

N[ _<- 7 7 j:l

exists

~j)zj 1

- Zkl

6 > 0.

i-

F the set

to the c h a r a c t e r -

we have

Ign(Z) I :

there

it follows

~

z C T

On the o t h e r hand,

R e { p j z j / ( z - (i + pj)zj)}

A.3.1,

cover

z-z.

j:l ~

for any fixed

n, a

in

on the c l o s e d

and

on

j = 1 .... , N.

< I z - (i + pk)Zkl gn(Z)

zN

N}, a n e i g h b o r -

theorem

polynomial,

for each

Iz

dist(z,F)

length

integer

these

uniformly

for

N

Thus

that

with

and define

pointwise

for all

I:

~

Zl,...,

such that

gn

to every

In fact, If

suppose

set in

centers

By the Runge to

converges

'gn(Z)-l

~, we may

Izl < l + m i n { p j :

< Iz- ( i + pj)zjl the c l a i m

to

N z-z. ~ $ j=l z - (i + pj)zj

Ign(Z) I <__ 2

F.

with respect

for each p o s i t i v e

with

converging

We c l a i m that

of

pj : nrj

: 1-

in

of p o l y n o m i a l s

sequence

AN

We set

disk. Since ~ is o r t h o g o n a l fT g n ( e i t ) d ~ ( e it) : 0.

(b)

continuous

be a c o m p a c t

rN, r e s p e c t i v e l y ,

[jN I rj < i/n 2.

gn

F

exists,

AI,...,

rl,...,

instead

Let

So there

of disks

gn(Z) Then,

is a b s o l u t e l y

T.

n = O, too.

and with r a d i i F

on

By c o n s i d e r i n g

(14) holds finite

Then

measure

- ( i + pj)zj

] -1

By a p p l y i n g

Lemma

271

N[

= ~

j=l

< exp

=

]

i+

-i

Iz- ( i + pj)zj [

I

~ log(l + -~--) - i j:l

exp

-1

=

exp

j :i

-1

<j =i

< eXP(n~) - i, for

[ j=l N rj < i/n 2.

So,

gn(Z)

÷ 0.

Hence,

the

statement

(b) is ver-

ified. Consequently,

the d o m i n a t e d

fF d~ = limn÷ ~ f gnd~ of length respect

zero,

: 0.

we c o n c l u d e

to the L e b e s g u e

A.3.3.

Let

p

that

f E Ll(do).

~

measure

be as above.

Thus,

on

T.

Then,

for

n = i, 2, . . . .

This means

that set

continuous

A.3.2,

: f(eit)da(t)

f(elt)elntdo(t)

"

shows

[]

by T h e o r e m

the c o n d i t i o n

IT

theorem

for any c o m p a c t

is a b s o l u t e l y

dt

d~(e it) for some

convergence

As this holds

(14)

states

that

= 0

' that

f

belongs

to

Hl(d~).

p(F)

=

in

T

F with

REFERENCES

[AS]

Ahlfors, L. V. and Sario, L., "Riemann Press, Princeton, N.J., 1960.

[CC]

Constantinescu, C. and Cornea, A., Fl[chen," Springer-Verlag, Berlin,

Surfaces,"

Princeton Univ.

"Ideale R~nder Riemannscher 1963.

[i]

Behrens, M. F., The maximal ideal space of algebras analytic functions on infinitely connected domains, Math. Soc. 161 (1971), 359-379.

[2]

Beurling, A., On two problems concerning linear transformations in Hilbert space, Acta Math. 81 (1949), 239-255.

[3]

Bishop, E., Subalgebras of functions Pacific J. Math. 8 (1958), 29-50.

[4]

Bishop, E., A n a l y t i c i t y in certain Banach algebras, Math. Soc. 102 (1962), 507-544.

[5]

Brelot, M., Topology of R. S. Martin and Green lines, in "Lectures on Functions of a Complex Variable" (W. Kaplan, pp. 105-121, Univ. of Michigan Press, Ann Arbor, 1955.

on a Riemann

of bounded Trans. Amer.

surface, Trans.

Amer.

ed.),

[6]

Brelot, M. and Choquet, G., Espaces et lignes de Green, Ann. Inst. Fourier (Grenoble) 3 (1952), 199-264.

[7]

Carleson, L., Interpolation by bounded analytic functions corona problem, Ann. of Math. 76 (1962), 542-559.

[8]

Dunford, N. and Schwartz, J. T., "Linear Operators," Wiley (Intersoience), New York, 1958.

[9]

Forelli, lllinois

F., Bounded holomorphic functions J. Math. i0 (1966), 367-380.

and the

Part I,

and projections,

[i0]

Gamelin, T. W., "Uniform Algebras," Cliffs, N.J., 1969.

[ii]

Gamelin, T. W., "Uniform A~gebras Univ. Press, London and New York,

[12]

Garnett, J., York, 1981.

[13]

Gunning, R. C., "Lectures on Riemann Press, Princeton, N.J., 1966.

[14]

Hara,

[15]

Hardy, G. H., The mean value of the modulus of an analytic tion, Proc. London Math. Soc. ser. 2, 14 (1915), 269-277.

[16]

Hasumi, M., Invariant subspace theorems for finite Riemann surfaces, Cand. J. Math. 18 (1966), 240-255.

[17]

Hasumi, M., Invariant subspaces on open Riemann Inst. Fourier (Grenoble) 24, 4 (1974), 241-286.

[18]

Hasumi, M., Invariant subspaces on open Riemann surfaces, Ann. Inst. Fourier (Grenoble) 26, 2 (1976), 273-299.

[19]

Hasumi, M., The Fatou theorem for open Riemann Math. J. 43 (1976), 731-746.

"Bounded Analytic

M. On Gamelin constants,

Prentice-Hall,

Englewood

and Jensen Measures," 1978.

Functions,"

Academic

Surfaces,"

Cambridge

Press, New

Princeton

Univ.

preprint.

surfaces,

surfaces,

func-

Ann. II, Duke

273

[20]

Hasumi, M., Weak-star maximality of H ~ on surfaces of ParreauWidom type, unpublished note, 1977.

[21]

Hasumi, M., Hardy classes on plane domains, (1978), 213-227.

[22]

Hasumi, M., Remarks on Hardy classes on plane domains, Sci. Ibaraki Univ. Ser. A. Math. ii (1979), 65-71.

[23]

Hasumi, M. and Srinivasan, T. P., Doubly invariant Pacific J. Math. 14 (1964), 525-535.

[24]

Hasumi, M. and Srinivasan, T. P., Invariant subspaces uous functions, Canad. J. Math. 17 (1965), 643-651.

[25]

Hasumi, M. and Rubel, L. A., Multiplication isometries of Hardy spaces and of double Hardy spaces, Hokkaido Math. J. i0 (1981) Sp., 221-241.

[26]

Hayashi, (1979).

[27]

Hayashi, M., Invariant subspaees Widom type, preprint, 1980.

[28]

Hayashi, M., Characterization of Riemann surfaces invariant subspace theorem, preprint, 1981.

possessing

[29]

Hayashi, M., Szeg~'s theorem on Riemann J. i0 (1981) Sp., 242-254.

Hokkaido Math.

[30]

Hayashi,

[31]

Heins, M. H., "Hardy Classes on Riemann Surfaces," in Math. Vol. 98, Springer-Verlag, Berlin, 1969.

[32]

Hejhal, D. A., Classification theory for Hardy classes of analytic functions, Ann. Aead. Sci. Fenn. Ser. A. I. no. 566 (1973), 1-28.

[33]

Helson, H., "Lectures New York, 1964.

[34]

Hoffman, K., "Banach Spaces of Analytic Englewood Cliffs, N.J., 1962.

[35]

Hurwitz, A. and Courant, R., "Vorlesungen ~ber allgemeine Funktionentheorie and elliptische Funktionen. Geometrische Funktionentheorie," (4th ed.) Springer-Verlag, Berlin, 1964.

[36]

Kaplan, W., Approximation 3 (1955), 43-52.

[37]

Kelley, J. L. and Namioka, Nostrand, Princeton, N.J.,

[38]

Kobayashi, Sem. Rep.

M., Hardy classes

on Riemann

surfaces,

on Riemann

M., Seminar talk (February,

on Invariant

Arkiv f6r Mat.

Bull.

Fae.

subspaces,

of contin-

Thesis,

surfaces,

UCLA

1982). Lecture Notes

Academic

Functions,"

by entire functions,

Press,

Prentice-Hall,

Michigan Math. J.

I., "Linear Topological 1963.

Spaces,"

of plane domains,

D. Van

Kodai Math.

[39]

Kobayashi, S., On a classification of plane domains for Hardy classes, Proc. Amer. Math. Soe. 68 (1978), 79-82.

[40]

Koosis, P., "Lectures Cambridge, 1980.

on H

Spaces,"

If,

surface of Parreau-

Subspaces,"

S., On H classification P 27 (1976), 458-463.

16

Cambridge

Univ.

Press,

P

[41]

Krasnosel'skii, M. A. and Rutickii, Ya. B., "Convex Functions Orliez Spaces," Noordhoff, Gronigen, 1961.

and

[42]

Kusunoki, Y., "Theory of Funetions--Riemann Surfaces and Conformal Mappings," Asakura Shoten, Tokyo, 1973. (Japanese)

[43]

Nakai, M., Corona problem for Riemann surfaces of Parreau-Widom type, Pacific J. Math. 103 (1982), 103-109.

274

[44]

Nevanlinna, 1970.

R.,

"Analytic

Functions,"

Springer-Verlag,

Berlin,

[45]

Neville, C. W., Ideals and submodules of analytic infinitely connected plane domains, Thesis, Univ. Urbana-Champaign, 1972.

[46]

Neville, C. W., Invariant subspaces of Hardy classes on infinitely connected plane domains, Bull. Amer. Math. Soc. 78 (1972),

[47]

Neville, C.W., Invariant connected open surfaces, 151 pp.

[48]

Niimura, M., Asymptotic value sets of bounded holomorphic functions along Green lines, Arkiv fSr Mat. 20 (1982), 125-128.

[49]

Obrock, A., Null Orlicz classes of Riemann surfaces, Sci. Fenn. Ser. A. I. no. 498 (1972), 22 pp.

[50]

~ksendal, B., A short proof of the F. and M. Riesz theorem, Amer. Math. Soc. 30 (1971), 204.

[51]

Parreau, M., Sur les moyennes des fonctions harmoniques et analytiques et la classification des surfaces de Riemann, Ann. Inst. Fourier (Grenoble) 3 (1951), 103-197.

[52]

Parreau, M., Th~or~me de Fatou et probl~me de Dirichlet pour les lignes de Green de certaines surfaces de Riemann, Ann. Aead. Sci.

functions on of lllinois at

857-860.

Fenn.

Ser.

A.

I.

no.

subspaces of Hardy classes on infinitely Mem. Amer. Math. Soc. no. 160 (1975),

250/25

(1958),

Ann. Acad. Proc.

8 pp.

[53]

Phelps, R., "Lectures on Choquet's Princeton, N.J., 1966.

[54]

Pommerenke, Ch., On the Green's function of Fuchsian Acad. Sci. Fenn. Ser. A. I. 2 (1976), 409-427.

[55]

Pranger, W., Bounded sections Math. Soc. 69 (1978), 77-80.

[56]

Pranger, W., Riemann surfaces and bounded holomorphic Trans. Amer. Math. Soc. 259 (1980), 393-400.

[57]

Rubel, L. A. and Shields, A. L., The space of bounded analytic functions, Ann. Inst. Fourier (Grenoble) 16 (1966), 235-277.

[58]

Rudin, W., Essential (1964), 321-324.

[59]

Rudin, 1966.

[60]

Sarason, D., The H p spaces of an annulus, no. 56 (1965), 78 pp.

[61]

Sarason, D., "Function Theory on the Unit Circle," Inst. and State Univ., Blacksburg, Virginia, 1978.

[62]

Sario, L. and Nakai, M., "Classification faces," Springer-Verlag, Berlin, 1970.

[63]

Segawa, S., Harmonic majoration Math. 84 (1979), 199-200.

[64]

Stanton, C. M., Bounded analytic functions on a class of open Riemann surfaces, Pacific J. Math. 59 (1975), 567-565.

[65]

Tsuji, Tokyo,

W.,

Theorem,"

on a Riemann

boundary points,

Theory

surface,

Bull. Amer.

"Real and Complex Analysls,"

M., "Potential 1959.

D. Van Nostrand, Ann.

Proc. Amer. functions,

Math.

McGraw-Hill,

Soe.

70

New York,

Mem. Amer. Math. Virginia

Theory of Riemann

of quasi-bounded

in Modern

groups,

type,

Soc. Poly. Sur-

Pacific J.

Function Theory,"

Maruzen,

275

[66]

Voichick, M., Ideals and invariant subspaces Trans. Amer. Math. Soc. iii (1964), 493-612.

of analytic

[67]

Voiohick, M., Invariant subspaces Math. 18 (1966), 399-403.

surfaces,

[68]

Weyl, H., "Die Idee der Riemannschen Stuttgart, 1958.

[69]

Widom, H., The maximum principle for multiple-valued functions, Acta Math. 126 (1971), 63-82.

[70]

Widom, H., ~

sections of vector bundles over Riemann P Ann. of Math. 94 (1971), 304-324.

[71]

Zygmund, A., "Trigonometric Univ. Press, London, 1959.

on Riemann

Series"

Flache,"

(2nd ed.),

functions,

Canad.

J.

(3rd ed.) Teubner, analytic surfaces,

2 vols.,

Cambridge

INDEX OF NOTATIONS

A(a/b), A'(a/b): XI.3C B(e,a): V.IA CK(R,~): III.IA C0(R): IV.5B • : IV.IA D(a): VI.3A N(f): III.4B dx: III.2C do: IV.IA A(R) (= A): III.IB

IV.3A H~(R), H~(dx): VIII.IA h~(R): IV.5C

AI(G): I I ! . 4 B AI(R) (: AI): III.2B E0(R,a)

(= ~0(a)):

VI.IA

F0(R): 1.3A (conformal map): VI.3A (convex function): XI.IA g(a,z): 1.6A (a)( g z): VII.IA G[f;G(a)]: VI.IB ~(R,a) (= {(a)): VI.IA {'(R,a) (= {'(a)): VI.IA G(b): III.4A F, Fc, Fc0 , Fe, Fe0 , Fh: 1.8D Fh0: 1.9B; V.6D F I, Fa' I F~, F e' I Fh: i 1.8A FM(CI(G),~): V.4B F(r), F(z;r): XI.3C

h[f]: III.5A H[f]: III.3A H[f;D]: 1.5A hP(~), h~(~), hQi(~): IV.IC HP(~), H~(~): iV.IB HP(do): IV.2A H~(D): XI.IA HP(R), H~(R), hP(R), h~(R), HP(dx):

HP(~,~), HPd~,~), HP(d~,Q): ~P(R,~):V.2A ~(G,~): V.4B HP(a), HP(R): II.4A HP'(R): II.4B HI(R): 1.2A HI(R,T): II.IA I0: IX.2A I(R): II.5A k(b,z) (= kb(Z)): III.IB K'(a), K n (a): VII.7A XR (= dx): III.2C ~A: IX.2A ~C: IX.2D ~(R,a) (= L(a)): VI.IA A(R,a) (= A(a)): VI.5B m(a), m(~,a): V.8A mP(a), mP(~,a): IX.ID

IX.IA

277

M(AI):

III.2C

M(H~(R)):X.5A Mm: IX.3C {M] : VIII.IB P Np, N¢: XI.IA 0AB , 0AB,, 0~, 0G, 0p: XI.IA 0AS: XI.7B G a:

1.5D

S(~;~): A.I.2 S(%:~,p): VI.3D S(b~,p): VI.7B SP'(R): II.4D ~(c): I.gB e(a;z): VI.IA TR: III.6B w(E,E'), w(E',E): W(R): III.5A

IV.5A

~0: VI.3A P(a.a';z): VI.5B P(r,e): IV.IA; A.I.2

z(~;~): VI.IA Z(a,R): V.IC

Prl, prQ: II.5A (conformal map): VI.3A qc: QA: QC: Qg:

IX.3A IX.2A IX.2D IX.2D

QM: III.IB Q(R): II.SA R*: III.IB R%: V.3A r(a;z): VI.IA R(~,a): V.IA R 1.6E u~:

V, A (lattice operations): ^ (fine limit):

III.4B

v (radial limit):

VI.4A

(harmonic conjugation): II.II (norms): P

1.4C

A.2.1

IV.IB~ IV.IC; V.2A

INDEX

B-dual:

Behrens

IV.5B

theorem:

X.7B

II.5A

Betti

number:

V.IA

theorem:

X.6C

Blaschke

product:

Boundary

m-function:

Bounded

--

II.SD;

problem:

(: u n i t a r y bundle):

VI.SA

flat

morphic

II.IA

function: l.m.m.:

Canonical

basis: measure:

Cauchy

mero-

group:

function:

Corona

problem:

of - -

----:

direct

--

inverse

--

Covering

triple:

Critical

point:

(DCT)

( D C T )a ) :

(=

Differential:

Cauchy-Read Character:

VII.3C;

IX.4A

type):

VII.IB;

VII.4B

--

group:

-- of

-- of an --

--:

IX.4A

1.7A I.SA

--

(o(c)

singularity: problem:

):

I.gB

1.9A

I.SA;

III.3A;

VI.IB Divisor:

1.10B

VI.5A I.IA

canonical

V.4A

regular

1.3B

--:

--:

l.lC

I.IA

1.3B

function: l.m.m.:

of an m-function:

theorem:

mero-

II.2D

II.2C

IX.IA Choquet

Vll.3C~

1.8A

--:

Exhaustion:

a multiplicative

morphic

III.6A V.IC

VII.IC;

I.IIB

theorem:

III.6A

--:

-- with

End:

kernel:

III.6B

III.6B

1.7A

reproducing

III.2B

IX.IC Cauchy

II.IA

--:

map:

Dirichlet

- - --:

II.4C

VI.7E

X.5A

Covering

conjugate

1.2B

(weak

VI:7D;

XI.IA

transformation:

analytic

II.2D

II.2C

theorem

direct

theorem:

Convex

harmonic

Canonical

XI.6A

complex

II.2A

of a m u l t i p l i c a t i v e

- - of a n

Cocyle:

group

VIII.4A

XI.IB

line

value

Cover

IV.2C

characteristic:

Brelot-Choquet Bundle

set:

Cluster

Cohomology

Band:

Bishop

Circular

IV.5C

B-topology:

VIII.4A;

Fatou

theorem:

Fine

limit:

Fine

boundary

A.I.2

III.4B function:

Function balayaged

--:

boundary

m---:

1.6E VIII.4A

III.4B

279

Function convex

(continued)

Integral

--:

Invariant

XI.IA

i---:

IX.IA

m---:

VIII.4A;

IX.IA

Martin

--:

III.IB

Wiener

--:

III.5A

Fundamental

Gelfand Genus: Green

VIII.IA~

VIII.2C

simply

- - --:

VIII.IA~

VIII.2C

-- --

theorem:

transform:

X.5A

Kolmogorov

regular

1.6A

- - --:

LHM

V.TD

VIII.3B;

boundary

point:

theorem:

1.5B

A.2.2

VI.IA

Line . VI.5A;

- - --:

VI.5B

VI.IA

measure: star

VI.IA

Hardy

class:

IV.IB~

IV.3A

Hardy-Orlicz

class:

XI.IA

region:

(=

1.5D~

--

(= l o c a l l y

lus):

II.2C

--

bounded

of

--

Harmonizable:

inner

--:

II.5D --:

1.4C

Martin

compactification:

Martin

function:

--

III.2B

function:

pole

II.SA

III.2C

III.5A IX°IC~

IX.5D;

Maximal Mean

theorem: group:

III.IB

ideal

value

III.IB

space:

theorem:

Minimal

point:

Modulus

of a s e c t i o n :

X.5A IX.2D

II.4C;

III.2B II.2A

meromorphic

tion:

II.2D

1.2A

inner

factor

of

II.SD

outer

factor

of

II.5D

region:

1.5C;

rare:

V.7A

X.7B

of

bounded

characteris-

II.SD

II.5A --

factor

factor: of a n

greatest

common

--

II.5A

part:

func-

XI.3B

tic:

common

III.IB

III.IB

of a - - --:

Multiplicative

Hyperbolically

II°5D

III.2C

X.4A

Hyperbolic

modu-

characteristic:

boundary:

function:

theorem:

Homology

meromorphic

Martin

function:

quasibounded singular

modulus):

II.5D

I.SA

-- majorant:

minimal

II.2A

analytic

A.2.1

-- differential:

--measure:

majorant):

(= b u n d l e ) :

locally

quasibounded

conjugate:

X.2B;

bundle

L.a.m.

L.m.m.

VI.IA

Harmonic

Hayashi

harmonic

II.2C

Green

least

(= l e a s t

1.4C

Green

--

VIII.2C;

VIII.4C

1.2B

line:

Inner:

VIII.2A

--:

Irregular

function:

Hejhal

VIII°IA;

--

1.3A

convergent

--

subspace:

III.2B

doubly

group:

modified Green

representation:

II.5D

l.a.m.: --

Nakai II.5D

factor:

II.5D

theorem:

Normal Null

operator:

set:

XI.IA

X.8A; V.6B

X.8C

280

Origin:

II.2B

Orthogonal Outer

Riemann-Roch

decomposition:

character:

p - - - --:

II.5A

IX.IA

Riemann

theorem:

- - of M y r b e r g

IX.IA

type:

theorem:

XI.2B;

VI.4A;

XI.2C;

X.IA

- - of P a r r e a u - W i d o m regular

Parreau

1.10C

surface

VI.4B;

Riesz

theorem:

1.6F;

F. and M.

XI.2D

type:

V.IA

. V.IC III.3B

. A.3.2.

Partition identity

--:

- - of the Poisson Polar

V.6A

ideal

kernel:

set:

1.6C

Potential:

1.6B

Pranger

IV.IA;

X.6A

branch:

Principal

operator:

type):

A.I.2

II.2A

holomorphie

--:

II.2A;

meromorphie

--:

II.2A

II.2E;

theorem:

(= s u r f a c e

V.6A

Segawa

II.2D V.6D

theorem:

Standard:

of P a r r e a u - W i d o m

Stanton Stolz

II.5A;

theorem:

region:

(= q.e.):

Radial

limit:

Region

(= s u b r e g i o n ) :

canonical

--:

star

hyperbolic

Stolz Regular

I.IC --:

VI.IA

--:

1.5C;

Regularization:

= region function:

1.4A

VIII.IA

III.4A

Universal

covering

V.7A

Weak-star

maximality:

Weak

VI.7B{

point:

surface:

topology:

IV.SA

theorem:

V.2B;

Widom

V.3B

A.I.2

1.4A

III.6A

VI.3D

I.IA

VI.3D{

X.5B

VI.7B;

I.IA

--:

boundary

Support:

Thin:

Stolz

--:

--:

1.6C

VI.4A

curvilinear

regular

Subregion

X.IA;

VI.3D;

function:

Superharmonic

Quasi-everywhere

Green

IV.IC

II.5A

XI.7B

1.9A

IX.IA

Subharmonic

- - part:

--:

V.4B

Singularity:

V.IA

Quasibounded:

V.2A

meromorphic-differential

Principal

PWS

Section:

boundary:

A.I.2

1.5B

V.gA; Wiener

VII.5A

V.4C;

V.9B function:

III.5A

V.5C;