Complex Analysis in Banach Spaces: Holomorphic Functions in Domains of Holomorphy in Finite and Infinite Dimensions

COMPLEX ANALYSIS IN BANACH SPACES NORTH-HOLLANDMATHEMATICS STUDIES Notas de Matematica (107) Editor: Leopoldo Nachbi...

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COMPLEX ANALYSIS IN BANACH SPACES

NORTH-HOLLANDMATHEMATICS STUDIES Notas de Matematica (107)

Editor: Leopoldo Nachbin Centro Brasileiro de Pesquisas Fisicas, Rio de Janeiro and University of Rochester

NORTH-HOLLAND-AMSTERDAM

NEW YORK

OXFORD

120

COMPLEX ANALYSIS IN BANACH SPACES Holomorphic Functionsand Domains of Holomorphy in Finite and Infinite Dimensions

Jorge MUJICA UniversidadeEstadualde Campinas Campinas, Brazil

1986

NORTH-HOLLAND-AMSTERDAM

NEW YORK

0

OXFORD

Elsevier Science Publishers B.V., 1986 Allrights reserved. No part of this publication may be reproduced, storedin a retrievalsystem, or transmitted, in any form orbyanymeans, electronic, mechanical, photocopying, recording or otherwise, without the priorpermission of the copyright owner.

ISBN: 0 444 87886 6

Publishers: ELSEVIER SCIENCE PUBLISHERS B.V. P.0. Box 1991 1000 BZ Amsterdam The Netherlands

Sole distributors forthe U.S.A. andCanada: ELSEVIER SCIENCE PUBLISHING COMPANY,INC. 52 Vanderbilt Avenue NewYork, N.Y. 10017 U.S.A.

Library of Congress Catalogingin-PublicationData

hjica, Jorge, 1946camplex analysis in Banach spaces. (North-Holland mathematics studies ; 120) (Notas de aia&tica ; 107) Bibliography: p. Includes index. 1. Holmorphic functions. 2. DanauLs ’ of holomorphy. 3. Banach spaces. I. Title. II. Series. 111. Series: Notas de m a d t i c a (Rio de Janeiro, Brazil) ; no. 107. QA1.N86 n0.107 [QA33i] 510 s [515.91S] 85-20922 ISBN 0-444-87886-6 (U.S. )

PRINTED IN THE NETHERLANDS

T o my t e a c h e r ,

Leopoldo Nachbin

This Page Intentionally Left Blank

FOREWORD

Problems arising from thestudy of holomorphic continuation and holomorphic approximation havebeen central in the development of complex analysis in finitely many variables, and constitute one ofthemost promising lines of current research in infinite dimensional complex analysis. This book isdesigned to present a unified view of these topics in both finite and infinite dimensions. The contents of this book fall naturally into four parts. The first, comprising Chapters Ithrough 111, presents the basic properties of holomorphic mappings anddomains of holomorphy in Banach spaces. The second part, comprising Chapters IV through VII, begins with the study of differentiable mappings, differential forms and the a operator in Banach spaces. Polynomially convex compact sets are investigated in detail, and some ofthe results obtained are applied tothe study of Banach and Frzchet algebras. The third part, comprising Chapters VIII through X, is

de-

voted to the studyof plurisubharmonic functions andpseudoconvex domains in Banach spaces. The identity of pseudoconvex domains and domains of holomorphy is established in the case of separable Banach spaces with the bounded approximation property. These results a r e e:it.r.!ided to Riemann domains in the fourth part, i.n w k i . i c : h envei.c>pes of holomorphy are also studied in detail. lived from a course taught at the Uni~ d : j dr!e Campinas, Brazil, in 1982. It presupposes versidade familiarity wl t h C n e triieory of Lebesgue integration, with the iis:ornorphi.c functions of a single variable, vi i

vi ii

and

MUJ I CA

with

of

t h e basic p r i n c i p l e s

Topics suchas vector-valued approximation property,

Banach

and H i l b e r t s p a c e s .

i n t e g r a t i o n , S c h a u d e r bases a n d t h e

a r e p r e s e n t e d i n d e t a i l i n t h e book.

The p r e s e n t a t i o n h e r e h a s b e e n a f f e c t e d by c o n v e r s a t i o n s a n d c o r r e s p o n d e n c e w i t h s e v e r a l f r i e n d s a n d c o l l e a g u e s , who w e r e n o t a l w a y s aware t h a t t h e y were s p e a k i n g f o r p o s t e r i t y .

In particu-

lar I w o u l d l i k e t o m e n t i o n Richard Aron, Klau s - Dieter B i e r s t e d t , Roberto C i q n o l i , Jean-FranGois

Colombeau, S e z n D i n e e n , Dicesar I s i d r o , Msrio Matos, R e i n h o l d

F e r n s n d e z , K l a u s F l o r e t , Josg M. Meise a n d M a r t i n S c h o t t e n l o h e r .

One p e r s o n h a s h a d more i n f l u e n c e o n t h l s book t h a n

anybody

else, long before t h i s p r o j e c t w a s evenconceived. N o t o n l y f o r a c c e p t i n g t h i s t e x t i n h i s series Notas d e Matemztica f o r having led

me into research

h e r e a c k n o w l e d g e my

on

,

b u t mainly

this beautiful subject. I

g r e a t e s t d e b t t o Leopoldo Nachbin.

a m p a r t i c u l a r l y g r a t e f u l t o my

wife

a n d c h i l d r e n , Ana

M a r i a , Ximena a n d F e l i p e , f o r t h e i r s u p p o r t

and encouragement

I

while

I was w r i t t i n g t h e book, a n d f o r m a i n t a i n i n g a t home a n

atmosphere ideal f o r s t u d y and r e s e a r c h . Finally, I am very pleased

t o t h a n k Miss E l d a Mortari f o r

h e r e x c e l l e n t t y p i n g of t h e m a n u s c r i p t .

Jorge Mujica Campinas, J u n e 1985

CONTENTS

. POLYNOMIALS 1 . Multilinear mappings . . . . . . . . . . . . .

1

.................. 3 . Polynomials of one variable . . . . . . . . . 4 . Power series . . . . . . . . . . . . . . . . .

12 18 27

CHAPTER I

2 . Polynomials

CHAPTER I1

.

5 6. 7 8 9

. . .

.

HOLOMORPHIC MAPPINGS

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

Holomorphic mappings Vector-valued integration The Cauchy integral formulas G-holomorphic mappings The compact-open topology

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

33

40 45 58 69

CHAPTER I11 . DOMAINS OF HOLOMORPHY

. .

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

10 Domains of holomorphy 11 Holomorphically convex domains 12. Bounding sets

79

85 94

CHAPTER IV . DIFFERENTIABLE MAPPINGS

.

Differentiable mappings . . . . . . . . . 14 . Differentiable mappings of higher order 15 . Partitions of unity 13

....

. . . . . . . . . . . . 16 . Test functions . . . . . . . . . . . . . 17 . Distributi.ons. . . . . . . . . . . . . .

. . . .

. . . .

99

. . . .

111 118 122 127

.........

139

CHAPTER V . DIFFERENTIAL FORMS 18 . Alternating multilinear forms ix

MUJ I CA

X

19 . 20 21 . 22 . 23

. .

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

Differential forms The Poincars lemma . . . . . . . . . . . The 3 operator . . . . . . . . . . . . . Differential forms with bounded support . The 3 equation in polydiscs . . . . . . .

144

153

. . .

156 162

. . .

168

CHAPTER VI . POLYNOMIALLY CONVEX DOMAINS

. Polynomially convex compact . Polynomially convex domains

. . . . . . . . Schauder bases . . . . . . . . . . . . . . . . . The approximation property . . . . . . . . . .

24 25 26 . 27 28 .

CHAPTER

Polynomial approximation in Banach spaces

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

VII

. 30 .

29

31 . 32 33 .

.

sets in 8 in C n . .

177

185 188

194 202

. COMMUTATIVE BANACH ALGEBRAS . . . . . . . . . . . . . . . .

. . . . .

211 214 219 227 236

. . . . . .

245

Banach algebras Commutative Banach algebras . . . . . The joint spectrum . . . . . . . . . Projective limits of Banach algebras The Michael problem . . . . . . . . .

. . . . .

. . . . .

. . . . .

CHAPTER VIII . PLURISUBHARMONIC FUNCTIONS

.

34 Plurisubharmonic functions . . . . 35 . Regularization of plurisubharmonic

. . . . . . . . . . . . . . . 36 . Separately holomorphic mappings . . . . . . . . 37 . Pseudoconvex domains . . . . . . . . . . . . . functions

38

.

CHAPTER IX 39

.

Plurisubharmonic functions on pseudoconvex domains . . . . . . . . . . . . . .

.

THE

265 273

. .

279

. .

287

IN PSEUDOCONVEX DOMAINS

Densely defined operators in Hilbert spaces . . . . . . . . . . . . . . . The 5 operator f o r L 2 differential forms .

. 41 . L 2 42 . C m 40

5 EQUATION

255

solutions of the solutions of the

. . . . . . . . . .

5 equation 7 equation . . . . . . . .

291 300 307

xi

CONTENTS

.

CHAPTER X

THE LEVI PROBLEM

43 . The Levi problem in Cn . . . . 44 . Holomorphic approximation in C n

. . . . . . . . . . . . . . . . 45 . The Levi problem in Banach spaces . . . . . . . 46 . Holornorphic approximation in Banach spaces . .

311 313 320 325

CHAPTER XI . RIEMANN DOMAINS 47 . Riemann domains .

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

48 . Distributions on Riemann domains 49 . Pseudoconvex Riemann domains . .

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

50 . Plurisubharmonic functions on Riemann domains . . . . . . . . . . . . 5 1 . The equation in Riemann domains

331 339 346

. . . .

353 359

. . . .

361

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

372

a

. . . . . . .

CHAPTER XI1 . THE LEVI PROBLEM IN RIEMANN DOMAINS 52

.

53 .

The Cartan-Thullen theorem in Riemann domains . . . . . . . . . . . . The Levi problem in finite dimensional Riemann domains

54 . The Levi problem in infinite dimensional Riemann domains . . . . . . . . . 55 . Holomorphic approximation in infinite dimensional Riemann domains . . .

. . .

380

. . .

391

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

397 400

. . .

408

CHAPTER XI11 . ENVELOPES OF HOLOMORPHY 56 . Envelopes of holomorphy 57 . The spectrum 58

.

Envelopes of holomorphy and the spectrum

BIBLIOGRAPHY. INDEX

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

421

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

431

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CHAPTER I

POLYNOMIALS

1. MULTILINEAR MAPPINGS T h i s s e c t i o n i s devoted t o t h e s t u d y o f multilinearmappings i n Banach s p a c e s . B e s i d e s t-.heir i n t r i n s i c i n t e r e s t , m u l t i l i n e a r mappings w i l l s e r v e a t w o f o l d p u r p o s e . Whereas s y m m e t r i c

mul-

t i l i n e a r mappings w i l l b e h e l p f u l i n t h e s t u d y o f p o l y n o m i a l s , a l t e r n a t i n g m u l t i l i n e a r mappings w i l l b e u s e d t o i n t r o d u c e d i f f e r e n t i a l forms. To begir. w i t h , w e e s t a b l i s h some n o t a t i o n . Throughout whole book t h e l e t t e r

iK

the

w i l l s t a n d e i t h e r f o r t h e f i e l d LPor

a l l r e a l numbers o r f o r t h e f i e l d

C of

all

complex numbers.

The s e t o f a l l s t r i c t l y p o s i t i v e i n t e g e r s w i l l b e d e n o t e d

by

fl u {O} w i l l b e d e n o t e d by no. U n l e s s s t a t e d o t h e r w i s e , t h e l e t t e r s E and F w i l l a l w a y s r e p r e s e n t Banach s p a c e s o v e r t h e same f i e l d M . W,

whereas t h e set

For each

1.1. DEFINITION.

m E W w e s h a l l d e n o t e by

t h e v e c t o r s p a c e o f a l l m - l i n e a r mappings

E,("E;F)

A : E m +. F,

whereas

w e s h a l l d e n o t e by S: ( m ~ ; ~ t) h e s u b s p a c e o f a l l c o n t i n u o u s members o f E a i m E ; F ) . F o r e a c h A E EalmE;F1 w e d e f i n e

When and

rn = 1

L('E;Fl

then a s usual we s h a l l w r i t e

= S:(E;F). When

write

EaimE;IK)

m = 1

and

= LaimE)

F = M

F = M and

E flE;F) = E a a ( E ; F )

a

then f o r short

we

shall

E f m E ; X ) = E f m E ) . F i n a l l y when

t h e n a s u s u a l w e s h a l l w r i t e E a ( E I = E*

E ( E l = E'.

1

and

MUJ I CA

2

1.2.

For each

PROPOSITION.

t h e foZZowing c o n d i -

A E Xa("E;FI

tions a r e e q u i v a Z e n t :

(a)

A

i s continuous.

(b)

A

is c o n t i n u o u s at t h e o r i g i n .

(c)

I I A II <

.

T h e i m p l i c a t i o n ( a ) * (b) i s obvious.

PROOF.

(b)

-

(c):

k

of p o i n t s Ixz, k

k

I!AIxl,...,x,III

( c ) i s n o t t r u e t h e n w e can f i n d a sequence k i n E m such t h a t maccllx. II 2 2 and j J k m for e v e r y k. H e n c e

If

..., x:) 2

X

k 1

maz II+ II 5 j and

for every (c)

with

=.

.

k , c o n t r a d i c t i n g (b) (a):

.

a = (al,. . , a n )

Let

m a z IIa .1 I 5 c 3 j

and

m a x IIz .lI

j

J

E

5 c.

,..., xmCmi E ?l

Em and cc= (x1 Then

and (a) follows. 1 . 3 . PROPOSITION.

A

+

IIAII.

d:

("E;F) is a B a n a c h s p a c e

under

the

norm

3

POLYNOMIALS

PROOF. One can readily see that the mapping A -+ II All defines a norm on a: f m E ; F ) . To establish completeness let ( A .) be a 3 Cauchy sequence in d: ( m E ; F). Then for each ( x l , . , x m ) in Em we have that

..

(1.1)

IIA.(x,,. 3

..,x mI -

Ak(x,,.

-< IIAj

-

A

. . , x m I1 ..

k II I I x , I I

It xmll

.. .

and it follows that ( A . f x , , , x m ) ) is a Cauchy sequence in F. 3 Since F is complete, the limit Afxl

(1.2)

,..., x m )

. .,x m )

= l i m A.(x,, 3

exists. One can readily see that the mapping A : Em F thus defined is m-linear. Furthermore, since ( A . ) is a Cauchy se3 quence in e f m E ; F ) there is a constant c > 0 such that II A .II 3 < c for every j . Then it follows from (1.2) that II A II < c too, and A is therefore continuous, by Propositionl.2.Finallyf if follows readily from (1.1) that IlA - A l l 0 when j a. -+

+

j

1.4. PROPOSITION. between

T h e r e i s a c a n o n i c a l v e c t o r s p a c e isomorphism

L a (m+nE ; F ) and

d u c e s an i s o m e t r y between

PROOF.

-+

L a (mE;Za fnE; F ) ) d : f m f n ~ ; ~ ) and

.

This isomorphism i n d:fmE;E f n E ; F ) ) .

One can readily verify that the mapping

defined by

has the required properties. For each m E A7 we shall denote by permutations of m elements. If 0 E S m the sign of the permutation (5.

Sm the group of all then (- 1)' will denote

4

MUJ I CA

1.5.

DEFINITION.

m

For e a c h

t h e s u b s p a c e of a l l

A

E W

w e s h a l l d e n o t e by E z ( m E ; F )

L a l m E ; F 1 which are s y m m e t r i c , t h a t i s ,

E

such t h a t

for all

x l J.

.. J

xm

E

and

E

o E Sm. Likewise, w e s h a l l denote

by E Z ( " ' E ; F ) t h e s u b s p a c e o f a l l A E E a I m E ; F ) n a t i n g or a n t i s y m m e t r i c , t h a t i s , such t h a t

for all

lalmE;FI

..., xm

which are a l t e r -

E and u E Sm' The spaces E S c m E ; F i are d e f i n e d i n t h e o b v i o u s way, t h a t i s

xl,

E

and

and

F = M

When

= E:("E), 1.6.

then we s h a l l w r i t e

LZ(mE;X)

etc.

PROPOSITION.

F o r each

A

E

la(mE;F)

let

= ESSmE), E z ( m E ; M ! a .

A S and

Aa

be

d e f i n e d by

Then:

A AS i s a p r o j e c t i o n o n t o L : f m E ; F 1 w i t h IIAsII < IIAII f o r e v e r y A mapping i n d u c e s a c o n t i n u o u s p r o j e c t i o n f r o m (mE;F 1 . (a)

(b)

The mapping

The mapping

f>i(,m E , ( v E ; F )

-+

A

-+

Aa

E

Ea(mE;F).

This

1 imE;F)

onto

is a p r o j e c t i o n f r o m

E

a

("E;F)

5

POLY NOM I ALS

onto

d::frnE;F)

IIAall < IIA II

with

f o r every

A E d:a(rnE;FI.

This

d: ( m E ; F )

onto

m a p p i n g i n d u c e s a c o n t i n u o u s p r o j e c t i o n from

E Q P EF I;. The proof o f t h i s p r o p o s i t i o n i.s s t r a i g h t f o r w a r d and is l e f t a s an e x e r c i s e t o t h e r e a d e r . For convenience w e a l s o d e f i n e , f o r

m = 0,

t h e spaces

as Banach s p a c e s . For e a c h

a = (alJ.

and e a c h m u l t i - i n d e x

n E W

. ., a n )

n

ENo

we set

...

la1 = a1 +

1 . 7 . DEFINITION.

a! = a1 !

..., a n )

...

n

E Wo

with

la] = rn

a m > 1

and

Axl

1 . 8 . THEOREM. L e t

1

... ,x n )

we define

a

n

"1

if

an !

T h e n f o r each (x1,

A E d:,frnE;FI.

a = (al,

and e a c h

E En

Let

+ an,

a

... x n n A

= A

E d:zfmE;F).

if

rn = 0. xl

Then f o r a l l

,...,

xn

E

E

we h a v e t h e L e i b n i z F o r m u l a Afxl +

...

+ xnIm = 2

5

@l

Axl

.. .

a n xn

w h e r e t h e summation is taken o v e r a l l m u l t i - i n d i c e s E

such t h a t

WE

PROOF.

and

..., an'

a = (al,

/ a / = m.

By i n d u c t i o n on

rn. The r e s u l t i s o b v i o u s f o r

m = 1 . Assuming t h e formula v a l i d f o r a c e r t a i n

rn = 0

rnzl

one

6

MUJ I CA

rn + 1. I n d e e d , i f

can r e a d i l y e s t a b l i s h it f o r

A

E

Ez(rn+lE;F)

t h e n one can w r i t e

...

A(xl +

. ..

+ x n ) rn+l = A ( x l +

...

+ xn)(xl +

+ xnIm

.. + xnI ,

and apply t h e i n d u c t i o n h y p o t h e s i s t o t h e mapping Alxl + . which b e l o n g s t o

Ez(rnE;FI.

1.9. COROLLARY.

Let

A

The d e t a i l s a r e l e f t t o t h e reader. Then f o r a l l

fz(rnE;F!.

E

x, y

f

E

we

h a v e t h e Newton BinomiaZ Formula

1.10. THEOREM.

Let

A

T h e n f o r aZT

f:frnE;F).

E

xo,

... ,xrn E E

we have t h e P o l a r i z a t i o n Formula A(xl,

..., x m )

1 -

=

m!Zrn

. . . €,A(xO

I:

A(xo +

. . .~ +ErnXrnlrn .

+X

~

E.;=?l

"

E

~

+X.

..

~ + E

x )rn =

rnrn

z

rn!

a ! 0

c1

... urn !

... C

+ c1rn =

1

...

a O J . ..,a

where t h e summation i s t a k e n e v e r a l l

+

E

By t h e L e i b n i z Formula 1 . 8 w e have t h a t

PROOF.

a

+

rn

ar n uo

E~

E EVo

m . Hence

. . . €,A(xO

+

E1Xl

+

...

+

E

E $1

X

rnrn

)

m

Clearly

whenever ai = 0

f o r some i w i t h

1 <

Axo

i < rn. S i n c e

U

. . .xrnrn

such that

POLYNOMIALS

2 E

... Em2

E;

.=:1

7

=

zm

3

t h e d e s i r e d r e s u l t follows. One c a n g e n e r a t e m u l t i l i n e a r forms o u t of l i n e a r forms t h e f o l l o w i n g manner. Given A E L a ("E;

,...,pm

E

one

E*

defines

F ) by

. . , xm )

A(xl ,.

x l , ..., x

for a l l

(PI

in

(PI ( 2 , )

...

pm(xm)

T h i s i d e a can b e g e n e r a l i z e d a s

E.

E

m

=

fol-

lows. 1.11. D E F I N I T I O N .

tensor product ( A QD B l ( x l

Givem

A E Za(mE)

A Q B E La(m+nE)

,..., xm+nl

and

B

E

their

LarnE)

i s d e f i n e d by

= A(xl ,..., x m ) B ( X , + ~ ,... xm+n

)

J

for a l l

xl, ..., xm+n

E

E.

The f o l l o w i n g p r o p e r t i e s of t h e t e n s o r p r o d u c t are clear:

(a1

If

A

and B are c o n t i n u o u s t h e n

is

A 8 B

con-

t i n u o u s as w e l l . (b)

The mapping (A,BI

(c)

(A Q B) d C

=

+

A Q B

is bilinear.

A 8 ( B Q C).

C l e a r l y w e can a l s o d e f i n e t h e t e n s o r product one of t h e mappings A o r

B

A QD B

if

h a s v a l u e s i n a Banach s p a c e .

U n t i l now w e have s t u d i e d E - m u l t i l i n e a r mappings

without

1K i s X? o r C. Now w e s t u d y t h e rel a t i o n s h i p between t h e s e two n o t i o n s .

d i s t i n g u i s h i n g whether

Let

E

and F be complex Banach s p a c e s and l e t

EX? and FB

d e n o t e t h e u n d e r l y i n g r e a l Banach s p a c e s . Then w e have t h e following r e s u l t , whose s t r a i g h t f o r w a r d proof cise t o t h e reader.

i s l e f t as an exer-

a

MUJ ICA

Let

1.12. PROPOSITION.

be c o m p l e x Banach spaces. Then

F

A E E a f E E J F I R I a d m i t s a u n i q u e d e c o m p o s i t i o n of

each A

E and

= A' + A",

mappings

A'

where

i s C - l i n e a r and

A'

and A''

t h e form

is C - a n t i l i n e a r . The

A"

are g i v e n by t h e formulas

and

for a l l

x

E

If A i s c o n t i n u o u s t h e n A '

E.

and

are

A"

eon-

tinuous as w e l l .

To g e n e r a l i z e t h i s r e s u l t t o m u l t i l i n e a r mappings

we

in-

troduce t h e following d e f i n i t i o n . 1.13. DEFINITION.

let

p, q

E

lN0

Let

with

and

F

p + q

2

E

t h e s u b s p a c e of a l l

Ea(PqE;F)

A(Xxl,.

for all

h E 6

and

. A AX^+^)

b e complex Banach s p a c e s , and 1 . Then

A

=

hp

E

E

we

E.

We

, x ~ 1+ ~

shall

denote

E ( p q E ; F ) t h e s u b s p a c e of a l l c o n t i n u o u s members of As usual w e shall write

by

a f p + q E I R , F I R I such t h a t

xq A f x c l , ,. .

X ~ , . . . , X ~ + ~E

s h a l l denote

by

6 a ( p q E ;F )

.

C a f P q E ; 6 ) = E a f p q E ) and E ( P q E ; 6 ' )

= E f P q E ) . For convenience w e a l s o d e f i n e E ( I f o o E ; F ) = E l o o E ; F ) = F. 1.14. LpI

EXAMPLE.

I...,

%+q

where t h e b a r

E

Let

E*.

E

be a complex Banach

space,

and

let

Then t h e mapping

means complex c o n j u g a t e I

belongs t o

La fpqE).

and F are complex Banach s p a c e s t h e n it i s clear t h a t E f m E ; F ) C E f m o E ; F ) f o r e v e r y m E i7Vo. The o p p o s i t e i n c l u s i o n a a i s f a r from o b v i o u s , b u t it i s t r u e , a s t h e n e x t theorem shows. If

E

9

POLYNOMIALS Let

1.15. THEOREM.

(a) spaces

i s t h e a Z g e b r a i c d i r e c t sum of t h e sub-

La(mEIR;FIRI d:

(b)

spaces

and F be c o m p l e x Banach s p a c e s . T h e n :

E

with

a (“E;F)

E (mEB ;FB

1 i s t h e t o p o Z o g i c a Z direct sum of t h e subp + q = m . Moreover, E(moE;F) = E ( m E ; F I .

with

L(PqE;F)

+ q = m. M o r s o v e r , L a ( moE ; F ) = EaimE;F).

p

m . The theorem i s o b v i o u s l y t r u e for m = 0, and by P r o p o s i t i o n 1 . 1 2 t h e theorem i s a l s o t r u e for m = 1. Assuming t h e theorem t r u e f o r a c e r t a i n m 2 1 w e prove i t f o r m + 1. By t h e i n d u c t i o n h y p o t h e s i s t h e r e a r e p r o j e c t i o n s PROOF. By i n d u c t i o n on

such t h a t

vo +

...

+ v

m

= i d e n t i t y . C o n s i d e r a mapping

Under t h i s i d e n t i f i c a t i o n

Ax E E a (

m

f o r each

En ; F B

x

E

E

and t h e n w e c a n w r i t e m

z

Ax =

vk(Ax).

k=O

Thus

f o r each

= 0,

with

..., m

k = O,..

.,m.

m = I ,

f o r each

t h e r e are p r o j e c t i o n s

j = 0,l

such t h a t

vk o A = Thus

Now, by t h e case

z(

ZL:

k 0

(vk

+

u k = i d e n t i t y . Hence 1

o A)

k + ul ( v k

o

A).

k

MUJ I CA

10

Clearly

And s i n c e

,. . .,

u k. ( v k o A ) (Axo) f X x l 3

--

Axrn)

. . . , Axrn)

X1-j - j k A u . ( v k o A ) l x o ) (Axl , 3

w e see t h a t

w9 ( A )

d: a ( T " + l - q ' q E ; F l

E

for

q = 0,.

..,m

f

1 . Thus

w e have found l i n e a r mappings

such t h a t seen t h a t

wo +

...

+ w m = i d e n t i t y . S i n c e i t can

be

readily

whenever j # k , we conclude t h a t each w is a projection. 9 Thus w e have shown t h a t L a ( r n + l E n ! ; F I R ) i s t h e a l g e b r a i c direct

sum of t h e subspace

bafPqE;F)

with

p + q

= rn +

1.

Now, s i n c e

u 0 ( A ) = u i ( v o o A ) i t f o l l o w s from t h e i n d u c t i o n h y p o t h e s i s a a d from t h e case lows that

rn = 1

that

wofA) E LalmflE;F).

Whence it f o l -

POLYNOMIALS

11

Thus (a) h a s been proved. But t h e n an e x a m i n a t i o n of t h e p r o o f shows (b) as w e l l . EXERCISES A E d: fmE;FI b e an rn-linear mapping which i s

Let

l.A.

a r a t e l y continuous i n each v a r i a b l e .

Let

(A.) be a sequence i n

Show t h a t

(b)

I f each

i s symmetric ( r e s p . a l t e r n a t i n g ) ,

show

x

E

E ~ .

A E Laf"E;F).

Aj

i s symmetric ( r e s p . a l t e r n a t i n g ) as w e l l .

that A

(c)

If each

A

ous a s w e l l . Let

l.C.

of

limit

EaIrnE;F) such t h a t

s

(a)

Principle

the

3

= l i m A .(xi e x i s t s f o r e v e r y

A(X)

the

i s continuous.

Uniform Boundedness show t h a t A l.B.

USing

sepa-

E

and

i s c o n t i n u o u s , show t h a t A

i

F b e Banach s p a c e s o v e r IK,

dimensional. L e t ( e ,

J . . .,

i s continu-

with

E

e n ) b e a b a s i s f o r E and let 5,

finite

,..., 5,

d e n o t e t h e c o r r e s p o n d i n g c o o r d i n a t e f u n c t i o n a l s . Show t h a t each A E d:a(rnE;F) can b e u n i q u e l y r e p r e s e n t e d as a sum

where

c E F j,. . j r n

.

... I: PE;F) . j,

l.D. M.

and where t h e summation i s t a k e n o v e r a l l

jrn v a r y i n g from

E

Let

n . Conclude t h a t

and F be f i n i t e d i m e n s i o n a l Banach

If E h a s dimension n and F h a s dimension

E (rnE; Fi h a s dimension

l.E.

1 to

A E d::frnE;Fl

Let

that

J

Ea!rnE;F)

=

spaces

over

p , show

that

n"p. and l e t

x ~ , . . .x n~

E

E.

If

r n < n show

12

MUJ ICA

2 . POLYNOMIALS

T h i s s e c t i o n i s d e v o t e d t o t h e s t u d y o f polynomials in Mach s p a c e s . P o l y n o m i a l s w i l l be u s e d t o d e f i n e power

series,

and

t h e s e i n t u r n w i l l b e u s e d t o d e f i n e h o l o m o r p h i c mappings. DEFINITION.

2.1.

A mapping

P : E

-, F

mogeneous p o l y n o m i a l i f there exists

= Axm

Pix)

f o r every

x

E

i s s a i d t o b e a n m-hoA E La(mE;FI

E . W e s h a l l d e n o t e by

such PafmE;F)

v e c t o r s p a c e o f a l l m-homogeneous p o l y n o m i a l s from E W e s h a l l r e p r e s e n t by

members of

t h e subspace of a l l

p(mE;FI

For each

Pa("E;F).

into

and

F = 1K t h e n f o r s h o r t w e s h a l l w r i t e PaImE;MI PfmE;llf ) = P("E1.

2.2.

THEOREM.

A^(xl = Ax"

fined by

(a)

T h e mapping

between

E:ImE;F)

for every

P =

and with

and

A

+

d

x

E

E.

d

E

PalmE;F)

= Pa(mEl

be de-

Then:

i n d u c e s a v e c t o r s p a c e isomorphism

PalmE;F).

We h a v e t h e i n e q u a l i t i e s

(b)

A E E,?E;

Given

PROOF.

for every

let

F.

we s h a l l set

P E P,I"E;FI

A E Ea("'E;F)

the

continuous

When

F o r each

that

P

E

F).

Pa(%;F)

w e can f i n d

A E Ea(mE;F)

such t h a t

A . But t h e n

AS

€

ES(mE;FI.

a

xo = 0

l o w a t once.

I f w e a p p l y t h e P o l a r i z a t i o n Formula

t o t h e mapping

AS

1.10

then a l l t h e a s s e r t i o n s fol-

13

POLYNOMIALS 2.3. COROLLARY. i f and onZy if

(a) A p o Z y n o m i a l IIP I1 < m.

E

Pa(mE;F)

i s continuous

is a Banach s p a c e u n d e r t h e norm

(b)

P(mE;FI

(c)

The mapping

between

P

and

ESfmE;F)

A

-+

IIP 1 1 .

i n d u c e s a t o p o l o g i c a l isomorphism

-+

P("E;F).

For each

2.4. PROPOSITION.

P

P

E

PaimE;F)

the following condi-

t i o n s are equiuaZent:

(a)

P

i s continuous.

(b)

P

i s bounded on e v e r y baZZ w i t h f i n i t e r a d i u s .

(c)

P

i s bounded on some o p e n baZ2.

PROOF. The implication (a) =+ (b) follows from Corollary 2 . 3 . The implication (b) * (c) is obvious. And the implication ( c ) * (a) follows from the following lemma.

2.5. LEMMA. open balZ

P

Let

E

B(a;r) then

If P i s bounded by P i s bounded b y c m m / m ! on

p,fmE;F).

c

on

an

baZZ

the

B(0;r).

PROOF.

There is

A E CZfmE;FI

such that

P =

A^.

Then if suf-

fices to apply the Polarization Formula 1.10 to A with and xl -- . . . = x E B f O ; r / m ) .

xo = a

rn

Next we extend the Principle of Uniform Boundedness to homogeneous polynomials. 2.6. THEOREM. A s u b s e t of P ( m E ; F ) i s norm bounded i f and o n l y i f i t i s p o i n t w i s e bounded.

The proof of the theorem rests on the following lemma. U be an o p e n s u b s e t of E , and l e t f f i ) b e a U i n t o F . If t h e f a m i Z y i f i ) i s p o i n t w i s e bounded o n U t h e n t h e r e i s a n open s e t V C U

2.7.

LEMMA.

Let

f a m i l y of c o n t i n u o u s m a p p i n g s from

MUJ I CA

14

where t h e f a m i l y PROOF.

i s u n i f o r m l y bounded.

Cfi)

Set

=

An

{r E

5 n

U : IIfi(rIII

f o r every

m

f o r every

n

E

mJ.

U =

Then

u

n=1

A

n

and e a c h

is closed i n

An

U. S i n c e U i s a Baire s p a c e , some A n h a s nonempty i n t e r i o r . Then t h e f a m i l y ( f i ) i s u n i f o r m l y bounded o n t h e open s e t V

--

0

An.

To prove t h e n o n t r i v i a l i m p l i c a t i o n , let

PROOF O F THEOREM 2 . 6 . (Pi)

be a s u b s e t of

PlmE;FI which i s p o i n t w i s e bounded. By Iem-

m a 2 . 7 t h e family ( P i ) ball

i s u n i f o r m l y bounded, by

B l a ; r ) . Then by Lemma 2 . 5 t h e f a m i l y (Pi)

cmm/m!

bounded by

on t h e b a l l

c

say,

is

on a

uniformly

B l 0 ; r ) . The desired c o n c l u s i o n

follows.

2.8.

DEFINITION.

where P

Pa(jE;FI

E

j

P : E + F i s s a i d t o be a p o l y m i f i t c a n be r e p r e s e n t e d a s a sum

A mapping

n o m i a l of d e g r e e a t m o s t

j = 0,

for

..., m.

W e s h a l l d e n o t e by P a I E ; F )

t h e vector s p a c e of a l l p o l y n o m i a l s from E d e n o t e by

P(E;F)

into

F.

t h e s u b s p a c e of a l l c o n t i n u o u s

We s h a l l

members of

Pa(E;F).

When

F = LX

P(E;LKI

2.9.

PROPOSITION.

(b)

(a)

PaIE;F)

with

PaImE;F),

is t h e a Z g e b r a i c d i r e c t s u m of m E JVo.

P(E;FI i s t h e a l g e b r a i c d i r e c t sum of t h e

PfmE;FJ, with

PROOF.

5lE;IKI = Pa(EI

= PIE).

and

the subspaces

then f o r short we s h a l l w r i t e

(a)

m E Wo. I t s u f f i c e s t o show t h a t i f

Po

+

PI +

...

+ Pm =

0

subspaces

POLYNOMIALS

with

P

j

j = 0 , . ..,rn,

for

E Pa(jE;FI

P o = PI = For e a c h

x E E

and A

E

M,

...

X #

0,

A f t e r d i v i d i n g by Xrn and l e t t i n g Proceeding i n d u c t i v e l y w e g e t t h a t

then

= Prn =

0.

w e have t h a t

IA1

+

m

-

we Fet t h a t &...,Po

P =O. = 0. rn

Prn-l

I f s u f f i c e s t o show t h a t i f t h e polynomial

(b)

P = P

+ PI +

0

...

+

'rn

..

P E Pa('E;F) f o r j = 0,. ,rn t h e n e a c h j Pj i s c o n t i n u o u s as w e l l . We prove t h i s by i n d u c t i o n on m, t h e r e s u l t b e i n g o b v i o u s for rn = 0. I f rn 2 1 t h e n f o r all x E E and A E M w e have t h a t i s c o n t i n u o u s , where

-

Plhxl

Choose

X

E M

t h e polynomial atmst m with

-

rn- I XrnP(x) =

such t h a t

x

-+

P(Ax1

z

j=O

(Aj - ArnlP.(x). 3

A j - Am # 0

-

Amp(,)

for j

.. , m -

1,

0,. ..,rn-

1. Since

i s c o n t i n u o u s and h a s degree

1, t h e i n d u c t i o n h y p o t h e s i s implies

j = 0,.

=

t h a t each

Pj

i s c o n t i n u o u s . Then i t f o l l o w s t h a t

Prn

i s c o n t i n u o u s as w e l l , and t h e proof i s complete.

E XERC 1SES 2.A.

L e t [Pi) b e a sequence i n

P(x) = Zim P j ( x ) e x i s t s f o r e v e r y (a)

Show t h a t

(b)

Show t h a t i f each

P

E

P U ( ' I 1 E ; F ) such t h a t t h e

x

E

E.

PU("E;F).

Pj

i s c o n t i n u o u s t h e n P i s con-

t i n u o u s as w e l l and (Pi) c o n v e r g e s t o P u n i f o r m l y on s u b s e t s of

E.

limit

compact

16

MUJ ICA

2.B.

Let

P :E

+

be a mapping such t h a t PlM

F

e a c h subspace M of 2.c.

If

P E P ~ ( E ; F )s a t i s f i e s

and

x

E,

2.0.

Let

p

E

E

show t h a t

and

T

for

P,("M;F)

E

x

P ( A ~ )= h r n p ( x ) f o r a l l

E M

P E Pa(mE;F).

X. Given S E % ( E ; F ) ,

E , F , G , H b e Banach s p a c e s o v e r

Pa I m F ; G )

E

o f dimension 5 rn + 1. Shm t h a t P E Pa(mE;F).

E

show t h a t

E,(G;H),

P o S E P a ( r n E ; G ) and

T o P E Pa(mF;H).

Given a sequence f a m ) of m d e f i n e d by brn -- d a l a 2 . . . 'm

2.E.

p o s i t i v e numbers, l e t

.

(b,)

be

Show t h e f o l l o w i n g :

a , t h e n (b,)

(a)

If la,)

converges t o

a l s o converges t o

(b)

If fa,)

is increasing (resp. decreasing) , then

a.

fb,)

i s i n c r e a s i n g ( r e s p . d e c r e a s i n g ) as w e l l .

Apply 2.E. t o t h e sequence ( 1 + 1 / m j r n t o show t h a t rn i s i n c r e a s i n g and converges t o e . sequence ( m / fl) 2.F.

2.G.

Show t h a t

2.H.

Let

p E Er

rn

I1 1II f o r e v e r y

w i t h ilpll = 1

and l e t

A E J fmE; F )

P E PlrnE)

. be defined

rn P = p .

by

A

II A II 2 e

the

=

(a)

Show t h a t II PI1 = 7 .

(b)

Show t h a t

p 8

... 8 p .

(c)

2.1.

Let

E = L1.

(a)

P =

A"

where

A E GSfmE)

is

given

by

Show t h a t I I A II = 1 .

(5,) Let

d e n o t e t h e sequence of c o o r d i n a t e f u n c t i o n a l s on P = ... 5 , .

P E P f r n L 1 ) be d e f i n e d by

Show t h a t IIPII = 2 / m r n .

<,<,

17

POLYNOMIALS

(b)

Show t h a t

P =

A^

(c)

Show t h a t

IIAII

= l/rn.'

The p r e c e d i n g

where

i s g i v e n by

A E Xs(rn&l)

two e x e r c i s e s show t h a t i n t h e i n e q u a l i t i e s

i n Theorem 2 . 2 t h e two extreme s i t u a t i o n s can a c t u a l l y o c c u r . 2.J.

Let

and

E

F

X, with

b e Banach s p a c e s o v e r

finite

E

.

,...,

d i m e n s i o n a l . L e t ( e l , . . , e n ) be a b a s i s f o r E and l e t 5, 5, d e n o t e t h e c o r r e s p o n d i n g c o o r d i n a t e f u n c t i o n a l s . Show that e a c h P E P(rnE;FI

can b e u n i q u e l y r e p r e s e n t e d as a s u n

a a P = I: e a t I 1 " ' 5 , n where

t h e summation

- (al,

..., a n )

2.K.

Let

is

taken

over

multi-indices

all

v.

such t h a t la1 = m.

E iTz

P ( m E ; F ) d e n o t e t h e s u b s p a c e of

f

all

P E

P(mE;F)

which c a n b e w r i t t e n i n t h e form

where

c

i

E F

and

sum of t h e s p a c e s

E

E'.

Let

P ( m E ; F ) with

f

P ( E ; F ) denote t h e a l g e b r a i c

f

n o . Each

m E

i s s a i d t o be a c o n t i n u o u s poZynorniaZ F = LK

f

P (E;M) = P (E).

f

When and

f

(a)

= P (E;F) i f

Show t h a t

P(E;F)

Show t h a t

P o T E Pf(E;F)

f

sional. (b) E

f

f i n i t e type. P (rnB;If( ) = P ( m E ) of

then as usual w e s h a l l w r i t e

f

P E P IE;FI

P { E ; F ) and e a c h o p e r a t o r

2.L.

Show t h a t i f

is

f o r each

T E d: ( E ; E )

P E Pa(E;F)

E

f i n i t e dimen-

polynomial

P

which h a s f i n i t e r a n k .

i s a polynomial

of

degree

at

18

MUJ I CA

P l a + Abl i s a polynomial i n

most m t h e n m

for a l l

2.Y.

X of d e g r e e a t m s t

a, b E E.

j P = P f P I + ... + Pm where P E P a ? ~ ; c " i f o r i m . Using E x e r c i s e l.E show t h a t t h e homogeneous polyPm i s given by t h e formula

Let

...,

- 0, -

nomial

Pm (xl = for a l l

I -

m! 2m

xo, x

2.N. Given equivalent:

E

c

El

. . . E,P(XO

+

EIX

+

... +

€.=?I

E 2)

m

.

3

E.

P E Pa(E;F)

show t h a t t h e f o l l o w i n g c o n d i t i o n s are

(a)

P

i s continuous.

(b)

P

i s bounded on e v e r y b a l l w i t h f i n i t e r a d i u s .

(c)

P

i s bounded on some open b a l l .

Show t h a t a polynomial P ET'iE;EI

2.0.

z

i s c o n t i n u o u s i f and o n l y

JI o P E P a ( E I i s c o n t i n u o u s f o r e v e r y

i f t h e polynomial

d, E F'.

3. POLYNOMIALS O F ONE VARIABLE W e have s e e n i n E x e r c i s e 2.L

that i f

polynomial of d e g r e e a t most rn t h e n

P

E

PafE;F)

is

a

P i a + Xb) i s a p o l y n o m i a l

i n X o f d e g r e e a t most m f o r a l l a , b E E . I n t h i s s e c t i o n w e show t h a t t h e c o n v e r s e i s a l s o t r u e , t h u s p r e s e n t i n g an a l t e r n a t i v e d e s c r i p t i o n of p o l y n o m i a l s i n Banach s p a c e s . B e f o r e p r o v i n g t h e main r e s u l t , w e p r e s e n t some p r e p a r a t o r y results on i n t e r p o l a t i o n polynomials.

...,

Let Xo, A, b e rn + 1 d i s t i n c t p o i n t s i n M a n d l e t b o , . .,bm b e a r b i t r a r y p o i n t s i n F . Then there i s a unique polynomial L E P ( E ; F ) of d e g r e e a t m o s t m s u c h

3.1. PROPOSITION.

.

that

(3.1)

L(A.) 3

= b

j

for

j

= 0 ,..., rn.

19

POLYNOMIALS

The p o l y n o m i a l

L

i s c a l l e d a L a g r a n g e i n t e r p o l a t i o n polynomial.

W e are l o o k i n g f o r a polynomial

PROOF.

ek

with

m

=

L(A)

Z ek X k=O

L

of t h e form

k

F , which s a t i s f i e s ( 3 . 1 ) . Hence w e have t o s o l v e the

E

system o f e q u a t i o n s

S i n c e t h e d e t e r m i n a n t of t h e c o e f f i c i e n t s i s a

n o n z e r o Vander-

monde d e t e r m i n a n t , t h i s system of e q u a t i o n s h a s a unique s o l u tion.

3.2.

k = 0,

For e a c h

M.

Let

COROLLARY.

..., m

nomial o f degree a t most

= 0,

..., m .

... ,A,

Ao,

let

m

m + 1

distinct points i n

Lk E PlE)

b e t h e u n i q u e poZy-

be

L (A

such t h a t

Then e a c h p o l y n o m i a l

k

j

)

P E P(ZY;FI

= 6k j

for

j

of d e g r e e at m o s t

m can b e r e p r e s e n t e d i n t h e f o r m

If we define

PROOF.

then both

and

= P(X.1 for

Q(X.)

3

3

nomials 3.3.

P

P

LEMMA.

and

Q

are p o l y n o m i a l s of d e g r e e a t most

Q

j = 0 ,..., m. By P r o p o s i t i o n 3 . 1

and

poly-

are i d e n t i c a l .

If a m a p p i n g

variable separately,

PROOF.

the

rn

then

By i n d u c t i o n on

P

P

: Mn

-+

F

is a p o l y n o m i a l in each

is a p o l y n o m i a l .

n , t h e r e s u l t b e i n g obvious f o r n = l .

MUJ ICA

20

Assuming t h e r e s u l t t r u e f o r a c e r t a i n Thus l e t

P : Mn+'

+

be a mapping which i s a polynomial i n

F

e a c h v a r i a b l e s e p a r a t e l y . For e a c h

u

s e t of a l l

such t h a t

E M

n w e prove it f o r n + l .

m

i7V L e t

E

.,An,uI

P(Al,..

Am d e n o t e t h e i s a polynomial

m i n e a c h of t h e v a r i a b l e s Al,.. . , A n . Then (Am I i s i n c r e a s i n g and it f o l l o w s from t h e induc-

o f d e g r e e a t most t h e sequence

co

lK =

t i o n hypothesis t h a t

u A m . Hence some

i s an i n f i n i t e

Am

m=l

m + 1 d i s t i n c t p o i n t s S o , ..., E m in M. For ..., m l e t L k E P l X I d e n o t e t h e u n i q u e polynomial of d e g r e e a t most rn such t h a t s e t . Choose e a c h k = 0,

(3.2)

for

L ( 5 I - 6 k . i kj

Q :

Define

+1

=

where t h e summation i s t a k e n o v e r a l l 0

to

0,.

. . ,m.

by

F

+

j

m.. Then c e r t a i n l y

Q

E

kl,

P(Xn+' ; F I

proof i t s u f f i c e s t o show t h a t P(A,uI and u E M .

..., k ,

and

v a r y i n g from

t o complete

= Q(A,u)

for a l l A

the E

d"

From ( 3 . 2 ) and ( 3 . 3 ) it f o l l o w s t h a t

for all

p E M

and a l l

jl,

...,.in

v a r y i n g from

0

to

m. If

Q(A,u) are polynomials of degree X I , ...,A Then i t f o l l o w s rn from ( 3 . 4 ) and P r o p o s i t i o n 3 . 1 t h a t P ( A , p I = Q ( X , p I for a l l A E M n and E A m . Since Am i s an i n f i n i t e set we c o n c l u d e t h a t P ( A , u ) = Q(A,u) f o r a l l A E M n and u E M . T h i s com1-1 E A m

then both

a t most m

and

P(A,pI

.

i n e a c h of t h e v a r i a b l e s

p l e t e s the proof. 3 . 4 . COROLLARY.

Let

i s a polynomial i n

f

X

: E

-+

for a l l

F

be a mapping s7ich that f ( a + XbI

a, b

E.

Then

f

I

M

E

PfM;F)

POLYNOMIALS

21

f o r each f i n i t e dimensional subspace PROOF.

IM

E

E.

.

Let ( e l , . . , e n ) be a basis for M. Then by hypothesis + Anen) is a polynomial in each of the variables separately. Then it follows from Lemma 3 . 3 that P(M;FI.

f(Xlel + hl,...,Xn f

of

M

...

3 . 5 . LEMMA.

Let

f : E

+

F

f ( n + Xbl

b e a mapping s u c h t h a t

i s a p o Z y n o m i a t i n X f o r a l l a , b E E. T h e n t h e r e i s a s e q u e n c e of homogeneous p o l y n o m i a l s Pk E P a ( k E;FI s u c h t h a t W

f(x) =

Pk(x)

for every

x

E,

E

k=O where f o r e a c h x E E we h a v e t h a t f i n i t e l y many i n d i c e s . PROOF.

(Pk(x))

By hypothesis for each in F such that

Pk(xl = 0

x E E

there

for

is

a

all

but

sequence

W

where P k ( x ) = 0 for all but finitely many indices. To prove k the I.emma it suffices to show that P k E P a ( E;FI f o r each k. In view of Exercise 2.B it will be sufficient to show that PklM E P( k M;F) for each k E l N o and each finite dimensional Hence subspace M of E . NOW, by Corollary 3 . 4 , f IM E P I M ; F l . there are homogeneous polynomials &?,a Q m > with Gk k = 0, ,m, such that E Q(kM;F) for *

*

A

,

...

m

and therefore rn (3.6)

f(Xx) =

z Xk

Qk(x)

for all

k=O

From (3.5) and ( 3 . 6 ) we conclude that

x

E

M

and

h E M.

MUJ ICA

22

Thus

5

P k ( x ) = Q (x) k

for a l l

x E M

and

k

P*(x) = 0

for a l l

x

and

k > m.

k Pk(M E P ( M;F)

for every k

A mapping

3 . 6 . THEOREM.

flo

M

and the proof i s conplete.

is a p o l y n o m i a l of P ( a + Ab) is a p o l y n o m i a l i n a, b E E .

P : E

at m o s t m if and o n l y if d e g r e e at m o s t m f o r a l l

E

E

m

+

F

degree A of

The proof of t h e "only i f " p a r t was l e f t t o t h e r e a d e r a s E x e r c i s e 2.L. The " i f " p a r t f o l l o w s a t once from t h e proof of Lemma 3 . 5 . Indeed, it s u f f i c e s t o observe t h a t i n ( 3 . 5 ) w e have t h a t P k ( x ) = 0 f o r every k > m .

PROOF.

L e t f : E + F be a mapping such t h a t f ( a + hb) i s a p l y nomial i n A f o r a l l a , b € E . Then f need n o t be a polynomial i n g e n e r a l . However, w e have t h e f o l l o w i n g theorem.

A c o n t i n u o u s mapping

3 . 7 . THEOREM. if a n d o n l y if

P : E

+

F

i s a polynomial

P(a + Xb) i s a p o l y n o m i a l in X for a l l

a, b

E

E.

To prove t h i s theorem w e need t h e followinq lemma. 3 . 0 . LEMMA.

Let

f

: E

-+

F

be a mapping s u c h t h a t

f ( a + Xbl

i s a p o l y n o m i a l i n A f o r a l l a, b E E . If f i s i d e n t i c a l l y z e r o on a n o n v o i d o p e n s e t , t h e n f is i d e n t i c a l l y z e r o o n E .

Assume f i s i d e n t i c a l l y z e r o on a nonvoid open s e t F i x a E U, x E E and J, E F ' . Then J, o f [a + h ( x - a ) ] i s a M-valued polynomial i n A which i s i d e n t i c a l l y z e r o on a

PROOF.

U

C

E.

neighborhood of A = 0 . Hence Ji o f [ a + X ( x - a ) 1 = 0 for every X E IK, and i n p a r t i c u l a r J, o f(x) = 0. Since F ' separ a t e s t h e p o i n t s of F w e conclude t h a t f ( x ) = 0 . PROOF OF THEOREM 3 . 7 . T o prove t h e n o n t r i v i a l i m p l i c a t i o n asThen sume P(a + Xb) i s a polynomial i n A f o r a l l a, b € E. by Lemma 3 . 5 t h e r e i s a sequence of homogeneous polynomials k P k E P a ( E ; F I such t h a t

POLYNOMIALS

23

W

x

P(x) =

Pkfx)

f o r every

x E E,

k=O where f o r each x

E

E w e have t h a t

Pkfxl = 0

for a l l but f i -

n i t e l y many i n d i c e s . W e want t o show t h e e x i s t e n c e o f an rn El@ such t h a t P k = 0 f o r e v e r y k > m. F i r s t we show t h a t each P k i s c o n t i n u o u s . W e proceed by i n d u c t i o n on k , t h e a s s e r t i o n b e i n g obvious f o r k = 0. Let k > 0 and assume P i s c o n t i n u o u s f o r e v e r y j < k . Then f o r j a l l x E E and X E IK, X # 0, w e have t h a t

L e t ( 1 I be a sequence of nonzero scalars which converges n z e r o . For each n E nV l e t Qn : E + F be d e f i n e d by

to

By t h e i n d u c t i o n h y p o t h e s i s each Q n i s c o n t i n u o u s . By (3.7) converges p o i n t w i s e t o P k , and i s i n part h e sequence (Q n t i c u l a r p o i n t w i s e bounded on E . Then by Lemma 2 . 7 t h e sequence (Qn) i s uniformly bounded on a b a l l B ( a , r ) . Hence Pk is bounded on B f a , r ) as w e l l , and by P r o p o s i t i o n 2 . 4 w e may conclude t h a t P i s c o n t i n u o u s , as asserted. k To complete t h e proof o f t h e theorem w e c o n s i d e r t h e sets Am

= {x E E : P k ( x I =

0

for all

k > m).

a,

E =

U A and each A m i s c l o s e d i n E . S i n c e E i s a m= 1 m B a i r e s p a c e , some A m has nonempty i n t e r i o r . Then i t f o l l o w s from Lemma 3.8 t h a t P k i s i d e n t i c a l l y z e r o on E f o r e v e r y k > m . The proof i s now complete.

Then

24

MUJ ICA

Theorems 3 . 6 and 3 . 7 reduce t h e s t u d y of p o l y n o m i a l s t o the

case where t h e domain s p a c e i s one d i m e n s i o n a l . Next w e p r e s e n t

r e s u l t s t h a t r e d u c e t h e s t u d y o f p o l y n o m i a l s t o t h e case where t h e r a n g e s p a c e i s one d i m e n s i o n a l . A mapping

3.9. THEOREM.

P : E

is a p o l y n o m i a l of d e g r e e X i s a polynornial of degree

F

+

a t m o s t m if a n d o n l y if JI o P : E a t m o s t m f o r e v e r y JI E F’.

+

The n o n t r i v i a l i m p l i c a t i o n i n Theorem 3.9 f o l l o w s a t

once

from Theorem 3 . 6 and t h e f o l l o w i n g lemma.

a t m o s t m if rn

J l o P : 1K

+

for every

..,m

k = 0,.

each

m + 1

be

let

d i s t i n c t points i n

Lk E PIXI

be t h e unique

B!. F o r

polynomial

f o r j L O , . . .,m. j ki t h e n by C o r o l l a r y 3.2 t h e oolynomial Jl o P c a n b e

o f d e g r e e a t most m If

M

+

is a p o l y n o m i a l of degree is a p o l y n o m i c z l of d e g r e e at most F

--*

E F’.

lo,..., Am

Let

PROOF.

P : IK

A mapping

3.10. LEMMA.

Jl E F‘

L (h I = 6

such t h a t

k

r e p r e s e n t e d i n t h e form

m

m

f o r every

h E 1K.

Since

F ’ s e p a r a t e s t h e p o i n t s of F w e con-

clude that

P(XI

=

m Z

Plhk)LklhI

f o r every

E

M.

k=O

Thus P i s a polynomial of d e g r e e a t most

m , as a s s e r t e d .

W e complete t h i s s e c t i o n w i t h t h e f o l l o w i n g c o n t i n u o u s v e r -

s i o n of Theorem 3 . 9 . 3.11.

THEORFM.

i f and o n l y $ E

F’.

if

A mapping $ o P : E

P : E +

M

+

F

i s a continuous polynomial

i s a c o n t i n u o u s polynoiniaZ for every

POLYNOMIALS

25

T o prove t h i s theorem w e need t w o a u x i l i a r y lemmas.

3 . 1 2 . LEMMA. L e t (P,) C PflK; F) b e a s e q u e n c e of p o Z y n o m i a Z s o f d e g r e e at m o s t rn w h i c h c o n v e r g e s p o i n t w i s e t o a m a p p i n g

P :M

-+

Let

PROOF. each

F. Then P i s a l s o a p o l y n o m i a l o f d e g r e e at m o s t Ao,...,hrn

k = 0,.

. .,m

m + 1 L k E PlIKl

be

let rn s u c h t h a t

degree a t m o s t

Then by C o r o l l a r y 3.2 e a c h

L (A

k

Pn

i

d i s t i n c t points i n

rn.

X.

For

be t h e unique polynomial o f

= 6

)

ki

for

=

j

0,

.. . , m.

can b e r e p r e s e n t e d i n t h e form

rn

Pn ( A ) = n

By l e t t i n g

Hence

f o r every

m Z p(hk)Lk(h) k=O

f o r every

i s a l s o a polynomial of d e g r e e a t most

P

A

E

M.

we get t h a t

+. m

P(X) =

Z: p n l X k ) L k ( A I k=O

h E l??.

m , asasserted.

Now w e can improve Lemma 3 . 1 0 as f o l l o w s . 3 . 1 3 . LEMMA.

a

.+

a

PROOF.

A mapping

P :M

-+

F

i s a polynomia2 for every

J,

E

$ oP

P :

$ 0

F'.

m E W l e t Bm d e n o t e t h e s e t o f a l l Ji i s a polynomial of d e g r e e a t most m . By

For each

such t h a t

is a p o l y n o m i a l if

E

F'

hy-

m

pothesis

F' =

U

Brn.

m=l

i s c l o s e d i n F'. Indeed l e t (+n) be a sequence i n Bm which c o n v e r g e s t o some Ji E F'. men the sequence I $ , o P) c o n v e r g e s p o i n t w i s e t o $ o P, and it f o l l o w s from Lemma 3.12 t h a t $ E Bm too. Thus e a c h Bm i s c l o s e d i n W e claim t h a t e a c h

Bm

F t , as asserted.

Since

F'

i s a Baire s p a c e , some

But s i n c e c l e a r l y

Bm

h a s nonvoid interior.

we conclude t h a t Brn = F t . Thus $ o P i s a polynomial of d e g r e e a t most m f o r every $ E F ' and t h e n a n a p p l i c a t i o n of U r n 3.10 catpletes Bm

i s a vector s u b s p a c e of

F'

MUJ I CA

26 the proof.

PROOF OF THEOREM 3.11. To prove t h e n o n t r i v i a l i m p l i c a t i o n l e t P : E -* F b e a mapping such t h a t o P E P I E ) for every Jl E F ’ . Then $ o Pla + Abl i s a polynomial i n h f o r a l l Jl E F ’ and a , b E E. Then, by Lemma 3 . 1 3 , P(a + Ab) i s a polynomial i n A f o r a l l a , b E E . Then, by Lemma 3.5, t h e r e i s a sequence of k homogeneous polynomials Pk E P a ( E ; F ) such t h a t $J

W

P(x) = where f o r e a c h

x

2 Pk(x) k=O

E

E

f o r every

x E E,

Pk(x) = 0

w e have t h a t

f o r a l l but f i -

n i t e l y many i n d i c e s . Hence w e g e t t h a t m

(3.8)

Jlo P(x)

= E

Pk(x)

$ 0

for all

x E E

and

$ E F‘.

k=O Jl E F ’ . Then i t f o l l o w s from ( 3 . 8 ) t h a t $ o P k E P ( E l f o r e v e r y @ E F ’ . B17 E x e r c i s e k Mter 2 . 0 w e may conclude t h a t P k E P ( E ; F ) f o r e v e r y k E no. knowing t h a t e a c h P k i s c o n t i n u o u s , w e may p r o c e e d as i n t h e proof of Theorem 3 . 7 , and show t h e e x i s t e n c e o f an m E B such t h a t Pk = 0 f o r e v e r y k m . T h i s completes t h e p r o o f . By h y p o t h e s i s

J l o P E P ( E ) f o r each

k

EXERCISES

3.A. L e t bo,. ..,b,

lo,.

. .,Am

m + 1

be

d i s t i n c t p o i n t s i n M and

be a r b i t r a r y p o i n t s i n

F.

unique polynomial of d e g r e e a t most m j

=

0,.

. .,m.

such t h a t

i s g i v e n by t h e formula

Show t h a t L

n

m

L(hl =

L E PIlU;F)

Let

z

bk

k=O

ih

- A

(Ak

-

j#k

n jfk

be

L l A .I = l i . 3 3

let the for

.)

3

X.) 3

T h i s i s t h e Lagrange i n t e r p o l a t i o n formula. 3.B. Given any i n f i n i t e d i m e n s i o n a l Banach s p a c e function

f

: E

-+

M

such t h a t

f(a

6 , find

+ Ab) i s a polynomial i n

a h

POLYNOMIALS

for a l l

27

f i s n o t a polynomial. O f c o u r s e

a, b E E, but

the

f u n c t i o n f m o s t be d i s c o n t i n u o u s . L e t iP i C

P a I E ; F I b e a sequence of p o l y n o m i a l s of degree a t most m which c o n v e r g e s p o i n t w i s e t o a mapping P : E F. 3.C.

n

+

P i s a polynomial of d e g r e e a t most

(a)

Show t h a t

(b)

Show t h a t i f e a c h

Pn

i s continuous t h e n P

m.

i s con-

t i n u o u s as w e l l .

4 . POWER SERIES

I n t h i s s e c t i o n w e i n t r o d u c e power series i n Banach s p a c e s i n t e r m s of homogeneous polynomials. W e e s t a b l i s h t h e CauchyHadamard Formula f o r t h e r a d i u s o f convergence. DEFINITION.

4.1.

A p o u e r s e r i e s from

E

into

around

F

the

m

point

a E E

i s a series o f mappings of t h e form

2 Pmlx-a), m=O

where

f o r every

Pm E P a ( m E ; F I

in f co

Note t h a t t h e power series m

m=O

AVO.

Pm(x - a )

I: A m ( x - a J m ,

w r i t t e n i n t h e form

where

also

can Am

be

=aC ~ Y " " ; F I ,

E

m=O

Am = Pm. 4.2.

The r a d i u s of c o n v e r g e n c e , o r more precisely,

DEFINITION.

m

t h e r a d i u s of uniform convergence of the

i s t h e supremum o f a l l u n i f o r m l y on t h e b a l l 4 . 3 . THEOREM.

Let

R

r

-

2

0

power series

B(a,r).

b e t h e r a d i u s o f c o n v e r g e n c e o f t h e power

a)

series

P

m=O

(a)

R

m (x

- a ) . Then:

i s g i v e n b y t h e Cauchy-Hadamard F o r m u l a 1 / R

Pm(x-a),

m=O such t h a t t h e series c o n v e r g e s

= l i m s u p IIPmII 1 / m m+

.

MUJ I CA

20

-

The s e r i e s c o n v e r g e s a b s o Z u t e Z y and unifomnZy on B(a;r)

(b) whenever PROOF.

0

5 r

< R.

First we show the inequality

Indeed, suppose R 0 and let 0 < r < R . Then the series converges uniformly on B ( a ; r ) . Set

Choose

mo

E

liV

such that m

for all

m

mo

and t and therefore

m > mo

E

x E B ( a ; r ) . Hence

and

B(0,r).

Whence

IIPm(t)ll

IIPmll 5 2 r - m

5 2

for all

for all m > m

0 '

l i m s u p IIPmll' I m - 1 / r .

Letting

r

+

R

we obtain (4.1).

Next we show the opposite inequality:

r Indeed, set L = l i m s u p IIPmIll/m, assume L < m , and take with 0 5 r < l / L . Choose s with r < s < 7 / L and choose m E W such that llPmlll/m < l / s for all m 2 mo Hence 0

.

Thus the series

I: P m ( x - a ) converges absolutely and m=O

formly on the ball

-

B ( a ; r ) . Hence

R

2 r , and letting

P

uni+

l/L

POLYNOMIALS

29

we obtain ( 4 . 2 ) . Thus w e have shown ( a ) . But i n t h e course of proving (a) we have a l s o shown ( b ) . m

4.4.

Let

PROPOSITION.

Z

Pm(x

-

a)

b e a power s e r i e s

m=O

E

i n t o F . If t h e r e i s

aZ2

then

B(a;rl,

3: E

m

r > 0

such t h a t

=

for every

Pm

0

X

m

-

Pm(x

n=O E

from

a) =O

for

No.

T h i s p r o p o s i t i o n i s an immediate consequence of

the

fol-

lowing lemma.

4.5. LEMMA.

m

( c ~ ) ~be= a~ s e q u e n c e i n

Let

there

If

F.

co

then

Z cmAm = 0 f o r a22 m=O for e v e r y m E W o .

cm = 0

By i n d u c t i o n on

PROOF. Assuming

co

=

...

m

deed, s i n c e

= cm m

X

cmr

1x1

A E M with

such t h a t

r > 0

m . Letting 1 = 0 = 0 w e show t h a t

is

5

P

w e g e t t h a t co= 0.

= 0

too.

In-

C

such

converges t h e r e i s a c o n s t a n t

rn=O

that

IIcmllrmI C

f o r every

m . Since

c0

=

...

= cm = 0

it fol-

lows that

Then f o r 0 < l h

Letting

X

-+

0

I 5

r / 2

we get t h a t

we get t h a t

= 0

and t h e lemma has

been

proved.

Let

E and F be Banach spaces over X , w i t h

s i o n a l . W e have seen i n E x e r c i s e 2 . 5 t h a t each admits a unique r e p r e s e n t a t i o n of t h e f o r m

E P

n dimenE

P(mE;F)

30

MUJ I CA

...

where c a E F and where C,, , 6 , denote the coordinate functionals associated with a given basis. Thus each power m

Z Pm ( x - a ) from E into F can be written, at

series

least

m=O

formally, as a multiple power series

where the summation is taken over all multi-indices a = (a1 , , a n ) E IN:. If the multiple series converges absolutely and uniformly on a certain set X then it is clear that

...

W

the

I: P m f x

power series

m=O . -

-

a ) also converges absolutely and

uniformly on X and they coincide there. Conversely, we the following proposition. m

Let

4.6. PROPOSITION.

m

Z €‘,(+I m=O

from el,

E

i n t o F with radius

..., e n

for e a c h

=

I: Amxm be a power s e r i e s m=O

of

convergence

...

= IIe n I1 = 1 , s e t

a = (a1 ,. . ., a n ) E A$

with ( a ( = m.

E

B

with

Ilelll

have

=

R

>

0.

T h e n we

Given

have

that m

a1 Z ~ , i ~+ .~. . e+ ~5 n e n I = catl a m=O

lcll

...

a

5,

n

...

w henever + + l C n l < R / e . Both s e r i e s converge abso0 whenever ZuteZy and u n i f o r m l y f o r 15, I + . . . + 15, I - r - r < R/e. PROOF.

By the Leibniz Formula 1.8 we have that

~ , ( < , e ~+

...

+ 5n e n ) = z Ial=rn

al m! a! E l

a

. . . c n ‘ n Arnei 1 ... e an n .

On the other hand, using Exercise 2.G and applying the sical Leibniz formula, we get that

clas-

POLYNOMIALS

If

0

5 r

t h e n t h e l a s t w r i t t e n series c o n v e r g e s

< R/e

I<,[

formly f o r

31

...

+

+ 15,1

uni-

< r . The d e s i r e d r e s u l t f o l l o w s .

t o multiple

The f o l l o w i n g lemma e x t e n d s Lemma 4 . 5

power

series. Let

4.7. LEMMA.

ca

F

E

f o r each rnuZti-index

..., a n )

a = (a

... A an,

,'a1 E

if t h e r e is

Illion.

r > 0

such t h a t t h e s e r i e s

I A, I

converges a b s o l u t e Z y t o z e r o whenever then c = 0 f o r every a. a

Repeated a p p l i c a t i o n s o f Lemma 4 . 5

PROOF.

CaAl

5 r ,. . ., I A n

I

< r-,

lead t o the result.

EXERCISES m

x

4.A. L e t

- a)"' be a

A ~ I X

power series from E

into

F.

m=O

(a)

Show t h a t t h e series h a s a p o s i t i v e r a d i u s o f conver-

Z i m s u p IIArnll ' / m <

gence i f and o n l y i f

(b)

m.

Show t h a t t h e series h a s a n i n f i n i t e r a d i u s

v e r g e n c e i f and o n l y i f

a n example of a p m r series

Give

m

4.C.

z Pm ( x - a ) = z

m=O l i m s u p IIPmII # Z i m s u p I1 A m I I .

such t h a t

ca(xI

Let 0

series from

Mn

-

a,)

into

F.

con-

Z i m It Amll ? / r n = 0. m

4.B.

of

... Let

(xn

- an)

b E Mn

'n

m=O

Arn ( x - a )

be a m u l t i p l e

such t h a t

rn

power

MUJ I CA

32

Show t h a t t h e s e r i e s c o n v e r g e s a b s o l u t e l y IxcI

for

. ..,

-

al 1 5 r l , j = 1,.

4.D.

. .,n.

L e t 15,)

Isn

- anl 5 r n

and u n i f o r m l y

whenever

0 < r

-

j

for

< [bj-ajl

This i s Abel’s lemma.

d e n o t e t h e sequence o f c o o r d i n a t e f u n c t i o n a l s o n m

E = KP

where

1

5

p <

m.

Show t h a t t h e power series C fSmfx)im

m=o x E B, b u t i t s r a d i u s o f convergence e q u a l s one. Thus t h e r e i s no analogue of Abel’s lemma f o r power series i n Banach s p a c e s .

converges a b s o l u t e l y f o r e v e r y

NOTES AND COMMENTS

Most of t h e r e s u l t s i n C h a p t e r I have been known f o r a l o n g t i m e and can a l r e a d y be found i n t h e book of E . H i l l e and R. P h i l l i p s [ l ] . I n S e c t i o n s 1, 2 and 4 o u r p r e s e n t a t i o n f o l l o w s e s s e n t i a l l y t h e books of L. Nachbin [ 1] , [ 2 ] I n Section 3

.

w e have mostly f o l l o w e d an a r t i c l e of J. Bochnak and J . S i c i a k [ 1] , which a c t u a l l y d e a l s w i t h s p a c e s more g e n e r a l than M a c h s p a c e s . The l e a s t known r e s u l t i n S e c t i o n 1 i s p e r h a p s Theorem 1.15, which I l e a r n e d from R. Aron o n e Monday a f t e r n o o n over c o f f e e , a t U n i v e r s i t y C o l l e g e Dublin. For a d d i t i o n a l r e s u l t s on t h e s u b j e c t matter i n t h i s chapt e r see t h e books of T. F r a n z o n i and E . V e s e n t i n i [ 11 , s. Dineen [ 5 1 and J. F. Colombeau [ 1 1 , t h e l a s t two books b e i n g concerned more g e n e r a l l y w i t h l o c a l l y convex s p a c e s .

CHAPTER I1

HOLOMORPHIC MAPPINGS

5 . HOLOMORPHIC MAPPINGS I n this s e c t i o n w e i n t r o d u c e holomorphic mappings i n Banach s p a c e s i n terms o f power s e r i e s e x p a n s i o n s . W e d e r i v e

several

p r o p e r t i e s of t h e s e mappings i n t e r m s of thecorresponding proper-

t i e s of holomorphic f u n c t i o n s of one complex v a r i a b l e . Throughout t h i s c h a p t e r a l l

Banach s p a c e s c o n s i d e r e d w i l l

be complex. I n p a r t i c u l a r , t h e l e t t e r s E

and

will

F

always

r e p r e s e n t complex Banach s p a c e s . 5.1.

DEFINITION.

f : U

a

E

-+

F

Let

U b e a n open s u b s e t of

there exist a b a l l

U

nomials

E.

i s s a i d t o be h o l o m o r p h i c o r a n a l y t i c B(a;rl C U

mapping

A

if

f o r each

and a sequence o f poly-

Pm E PlmE;F) such t h a t m

f(x) =

Z

Pm(x

- a)

m=O

x

uniformly f o r

E

B ( a ; r l . W e s h a l l d e n o t e by

t h e vec-

X(U;FI

t o r s p a c e of a l l holomorphic mapping from U i n t o F.Vhen F t h e n we s h a l l w r i t e 5.2.

=

cf

X(U;@/ = JC(Ul.

I n view of P r o p o s i t i o n 4 . 4 t h e sequence ( P m ) w h i c h

REMARK.

a p p e a r s i n D e f i n i t i o n 5 . 1 i s u n i q u e l y determined by and w e s h a l l w r i t e

Pa = P r l f ( a ) f o r e v e r y

m E

f

no.

and

a

The series

a3

Z P p f ( a l (x m=O

-

a ) i s c a l l e d t h e T a y l o r s e r i e s of

s h a l l d e n o t e by that

A m f ( a l t h e unique member o f

( A m f ( a )l - = P m f ( a l .

33

f

at

JS(mE;Fl

a.

We

such

34

MUJ I CA

5 . 3 . EXAMPLE.

P(E;F)

C JCfE;F).

I f s u f f i c e s t o show t h a t

PROOF.

Let

E P(mE;F).

P

= i,where

for

P E W(E;F)

Given

A E ES(mE;F).

each a, x

P E

E,

by t h e Newton Binomial Formula 1 . 9 w e h a v e t h a t P(x)

Thus

P

= Axm =

m

z (7

j=o

)Aam-j(x

-

u)’.

3

i s holomorphic on E and

m

z P,(x) b e a power series from E m=O w i t h an i n f i n i t e r a d i u s o f convergence and w i t h e a c h

5.4.

F

EXAMPLE.

Let

into Pm

continuous. I f we define f(x)

=

L:

for each

Pmix)

x

f

E,

m=O

then PROOF.

f E X(E;F).

Set

Pm

=

A,

with

Am E - C S ( m E ; F ) ,

f o r every

m E

no.

We claim that

f o r each

a E E

and

r > 0. Indeed, w e have t h a t

and t h e l a s t w r i t t e n series c o n v e r g e s , s i n c e by E x e r c i s e

w e have t h a t

Z i m IIAmlll’m

=

0.

4.A

HOLOHORPHIC

MAPPINGS

35

From ( 5 . 1 ) w e g e t on one hand t h a t m

m

and hence t h e series E

f o r each

P(JE;F)

j

2. m =J E

(

)

Amam-'

n o .On

d e f i n e s a n element

t h e o t h e r hand i f f o l l o w s

'i from

(5.1) t h a t

uniformly f o r

x

E

B ( a ; r ) . Thus

f E JC(E;F).

5.5. EXAMPLE. L e t (v,) b e a sequence i n pointwise t o zero. I f we d e f i n e

which

E'

converges

m

f(x)

=

z

ipmix))m

f o r every

x

E

E,

m=O

then

f

E

K(E).

PROOF. By t h e P r i n c i p l e of Uniform Boundedness t h e s e i s a c o n 5 c f o r e v e r y m E 2No. W e c l a i m s t a n t c > 0 s u c h t h a t IIq,II that

(5.2)

f o r each have t h a t

a

E

E

and e a c h

P

with

0

5

P < 1

/c.

Indeed,

we

36

MUJ ICA

and t h e l a s t w r i t t e n series converges s i n c e

c r < 1 and

(Pm(al

+. 0. From ( 5 . 2 ) w e s e t on one hand t h a t

m

and hence t h e series Q

.E

P ( ~ E )f o r each

m

x

E

m=j j E

3 (5.2 1 t h a t

uniformly f o r

z

(

j '1 ) I p m ( a ) I m - j qm

d e f i n e s an element

J

mo.

On t h e o t h e r hand i f f o l l o w s f m

a

B f a ; r ) . Thus

f E K(E;FI.

Many p r o p e r t i e s of holomorphic mappings i n Banach

spaces

can be d e r i v e d from t h e corresponding p r o p e r t i e s of holomorphic f u n c t i o n s of one complex v a r i a b l e w i t h t h e a i d of

the follawing

simple r e s u l t , whose s t r a i g h t f o r w a r d proof i s l e f t a s an e x e r c i s e t o the reader. 5.6.

LEMMA.

Let

U be an o p e n s u b s e t of

E , arid l e t

f E KfU;FI.

HOLOMORPHIC MAPPINGS

37

Then: (a)

f

i s continuous.

(b)

f

i s l o c a l l y b o u n d e d , t h a t is,

s u i t a b l e n e i g h b o r h o o d of e a c h p o i n t o f

f

is bounded

on

a

U.

For e a c h a E U, b E E and $ E F’ the function $ o f f a + h b ) i s hoZomorphic on the open s e t { h E 6 : a + Xb E U). (c)

A

+

To b e g i n w i t h ,

w e extend t h e I d e n t i t y P r i n c i p l e .

L e t U b e a c o n n e c t e d o p e n s u b s e t of E , and f E X f U ; F I . I f f i s i d e n t i c a l l y z e r o on a nonvoid open V C U t h e n f is i d e n t i c a l l y z e r o o n a l l of U.

5.7. let

PROPOSITION.

set

PROOF.

( a ) F i r s t assume U convex. L e t

a E V,

l e t x E U and

let

A = {A Since

E @ :

a + hfx

-

al

E

U).

U i s convex t h e open s e t A i s convex as w e l l ,

p a r t i c u l a r c o n n e c t e d . For e a c h gfA) =

IJJ

$ E F’

o f [ a + Alx

and

in

the function

-

a)]

i s holomorphic on A and i s i d e n t i c a l l y z e r o on an open d i s c A(O;E). Then g i s i d e n t i c a l l y zero on A by t h e I d e n t i t y Princ i p l e f o r holomorphic f u n c t i o n s o f one complex J, o f ( 2 )

particular p o i n t s of

F

=

g(1)

= 0, and s i n c e

w e conclude t h a t

F’

variable. separates

In the

f f x ) = 0.

( b ) I n t h e g e n e r a l case, l e t A d e n o t e t h e s e t of a l l points

a E U , such t h a t f i s i d e n t i c a l l y z e r o on a n e i g h b o r h o o d o f a. Then A i s o b v i o u s l y open, and t o complete t h e proof it suffices

U. L e t ( a n ) b e a sequence i n A which converges t o a p o i n t b E U. Choose r > 0 such t h a t B f b ; r ) C U and choose n s u c h t h a t an E B f b ; r l . Then i f f o l lows from ( a ) t h a t f i s i d e n t i c a l l y z e r o on Bfb;r). Hence b E A and t h e proof i s complete. t o show t h a t A

i s closed i n

38

MUJ I CA

Next w e e x t e n d t h e Open Mapping P r i n c i p l e . PROPOSITION. L e t U b e a c o n n e c t e d o p e n s u b s e t of E, and l e t f E X ( U l . If f is n o t c o n s t a n t o n U t h e n f l v l i s an open s u b s e t o f 6 f o r e a c h o p e n s u b s e t V of U. 5.8.

PROOF. s e t of

C l e a r l y i f s u f f i c e s t o show t h a t

convex

open s u b s e t of

for e a c h convex open s u b s e t V U and l e t x E V . c i p l e 5.7 t h e f u n c t i o n f i s n o t c o n s t a n t is a p o i n t y E V such t h a t f l x l # f ( y l . t h e open s e t @,

f l V ) i s an open subo f U. L e t V be a By t h e I d e n t i t y Prinon V and hence there S i n c e V i s convex,

i s convex as w e l l . The f u n c t i o n

gfXI = f [ i s holomorphic on

A and

3c

+ Afy

-

3c)l

gfOl = f f x ) # f ( y l = g i l ) .

By

the

Open Mapping P r i n c i p l e f o r holomorphic f u n c t i o n s o f one complex v a r i a b l e t h e set

g l A ) is open i n

w e conclude t h a t

fiVl

6. S i n c e

i s a l s o open i n

6.

A s an immediate consequence w e o b t a i n t h e M a x i m Principle.

Let

U be a c o n n e c t e d o p e n s u b s e t o f E , and a E U s u c h t h u t lfirll 5 I f f d l t h e n f i s c o n s t a n t on U.

5.9.

PROPOSITION.

Zet

f E X ( U I . I f there e x i s t s

f o r every

x E U

i s n o t c o n s t a n t o n U. Then by t h e Open Mapp i n g P r i n c i p l e 5.8 t h e set f ( U ) i s open i n @, and hence cont a i n s an open d i s c A ( f ( a ) ; r l . But t h i s i s i m p o s s i b l e , s i n c e by PROOF.

Assume f

hypothesis

lf(xl

I 5

Iffa) I

f o r every

x

E

U.

To end t h i s s e c t i o n we g e n e r a l i z e L i o u v i l l e ' s Theorem.

39

HOLOMORPHIC MAPPINGS 5.10.

PROPOSITION. I f a m a p p i n g t h e n i t is c o n s t a n t o n E .

f E J C ( E ; F ) is b o u n d e d

E

on

PROOF. L e t x E E and J, E F t . Then the f u n c t i o n g ( x ) = $ o f ( X x ) i s holomorphic on C and bounded t h e r e . By t h e classical L i o u v i l l e ' s theorem, g i s c o n s t a n t , and i n p a r t i c u l a r J, o f ( x ) = J, o f ( 0 ) . S i n c e F t s e p a r a t e s t h e p o i n t s of F w e conclude t h a t f ( x ) = f ( 0 ) and t h e proof i s complete. EXERCISES 5.A.

'

be Banach s p a c e s , and l e t V S E L f E ; F I , f E JC(V;G) and J C ( S - l ( V ) ; G ) and T o f E K ( V ; H i ) .

Given

F.

f oS

E

U be a n open subset o f E , and l e t f E J f ( U ; F ) l e t f a : U - a -t F be d e f i n e d by f o r every t E U - a . 5.B.

Let

(a) Show t h a t f a E J C ( U - a ; F ) and f o r e v e r y t E U - a and m E n o . Show t h a t t h e mapping f morphism between J C f U ; F I and X f U (b)

5.C.

Let

b e a n open

E , F, G , H

Let

subset of show t h a t

U be a n open s u b s e t of show t h a t

f , g E JCfU)

-+

-

T

E

L(G;HI,

a E E . For each faft) = f f a + t)

Pmfa(t) = Pmf(a + t )

i s a v e c t o r space iso-

fa

a;F). E.

two

Given

functions

J€(u) and

fg E

rn P r n ( f g )( x )

x

=

pm-j f ( x ) P j g ( x )

j=O

for a l l

m

E

liVo

and

x

-

a)

E

U.

m

5.D.

Let

m=O

P,(x

b e a power series from

w i t h r a d i u s o f convergence R > 0 and w i t h each Let f : B(a;R) F be d e f i n e d by

E

co

x m=O

Pm(x

- a)

f o r each

x

E

F,

Pm cmtinuous.

-+

f(x) =

into

B(a;R).

MUJ I CA

40

Show t h a t f i s holomorphic on t h e b a l l B ( a ; R / e I . t h a t f i s holomorphic on t h e b a l l B ( a ; R ) ? 5.E.

Let

(a)

Can you shcw

X be a t o p o l o g i c a l s p a c e . Show t h a t each c o n t i n u o u s mapping

f : X

+

F

i s lo-

c a l l y bounded. I f X i s m e t r i z a b l e show t h a t a mapping f : X F is l o c a l l y bounded i f and o n l y i f f i s bounded on each compact s u b s e t of X. (b)

5.F.

+

U be a connected open s u b s e t of E , and l e t f Suppose t h e r e are a nonvoid open s u b s e t V of U and a c l o s e d subspace N of F such t h a t f(V) C N. Show that f(U) C N.

E

Let

JC(U;F).

5.G.

Let

E JC(U;F).

for all 5.H.

Let

U be a connected open subset o f E , and let f I f t h e r e i s a p o i n t a E U such t h a t IIffz)ll 5 IIf(a)ll 2 E U, show t h a t I l f l l i s c o n s t a n t on U. F = B2

be d e f i n e d by

w i t h t h e norm o f t h e supremum. L e t

f ( z ) = ( 1 , ~ )f o r e v e r y

(a)

Show t h a t

f E Pf5;F).

(b)

Show t h a t

II f II

(c)

Show t h a t

f

(d)

Show t h a t

II f II

6 . VECTOR-VALUED

f: 5 + F

z E 5.

i s c o n s t a n t on

A(0; 1 I .

i s n o t c o n s t a n t on

AfO; I ) .

i s n o t c o n s t a n t on

5.

INTEGRATION

W e assume t h a t t h e r e a d e r i s f a m i l i a r w i t h

theory of Lebesgue measure and i n t e g r a t i o n , and by t h i s w e mean i n t e g r a t i o n of s c a l a r - v a l u e d f u n c t i o n s . However, throughout t h i s book w e s h a l l o f t e n f i n d d e s i r a b l e t o i n t e g r a t e f u n c t i o n s with values i n a Banach space. With t h i s i n mind w e p r e s e n t a few e1errenta-y f a c t s r e g a r d i n g t h e Bochner i n t e g r a l . These few f a c t s w i l l be the

HOLOMORPHIC MAPPINGS

41

s u f f i c i e n t f o r o u r needs. 6.1.

L e t (X, 8 , 11) be a f i n i t e measure

DEFINITION.

mapping

sets

f : X

-+

Al,...,Ak

E

z

and v e c t o r s

k

Then f o r e a c h

A

is s a i d t o b e s i m p l e i f t h e r e are d i s j o i n t

F

flxl =

space.

xA

Z j=1

j

(x)b j

bl,...,bk

E

F

for all

x

E

such t h a t X.

we d e f i n e

A E Z

The v e r i f i c a t i o n of t h e f o l l o w i n g lemma i s s t r a i g h t f o r w a r d , and i s l e f t as an exercise t o t h e r e a d e r . 6.2.

LEMMA.

L e t (X,

x , ~ b) e

a f i n i t e measure s p a c e ,

f : X F be a s i m p l e mapping. Then f o r each we h a v e t h a t : +

A

E Z

and and

let + €

F'

I

6 . 3 . DEFINITION.

(a)

L e t (X,

A mapping

x,

f : X

p) b e a f i n i t e measure s p a c e .

-+

F

i s s a i d t o be measurable i f there

e x i s t s a sequence of s i m p l e mappings v e r g e s t o f almost everywhere. (b)

A measurable mapping

f : X

:

f,

+

X

-+

F

which

i s said t o be Bochner

F

i n t e g r a b l e i f t h e r e e x i s t s a sequence of s i m p l e mappings

X

-+

F

such t h a t

lim n-+m I n t h i s case w e d e f i n e

1,

/Ifn

- flldu =

con-

0.

f,

:

42

MUJ I CA

€or each

A

f

2.

Lemma 6.2 g u a r a n t e e s t h a t t h e Bochner i n t e g r a l

is

IAfdll

w e l l d e f i n e d . I n d e e d , on o n e hand Lemma 6.2 inplies that

f.f,fndlJl

i s a Cauchy s e q u e n c e , a n d on t h e o t h e r hand Lemma6.2 g u a r a n t e e s t h a t the d e f i n i t i m of sequence I f , ) .

i s i n d e p e n d e n t o f t h e c h o i c e of t h e

J,fdp

F i n a l l y , from Lemma 6 . 2 and

the d e f i n i t i o n

of

t h e Bochner i n t e g r a l w e can e a s i l y o b t a i n t h e f o l l o w i n g propos i t i o n . The d e t a i l s are l e f t t o t h e r e a d e r . 6.4. let

PROPOSITION. L e t IX, Z, P I b e a f i n i t e m e a s u r e s p a c e , and f : X + F b e a Bochner i n t e g r a b l e m a p p i n g . T h e n :

(a)

f o r each

(b)

f o r each

6.5.

The f u n c t i o n

$I E

F’

and

The f u n c t i o n

IJJ

o f : X

+

6

i s i n t e g r a b Z e und

A E Z.

II f I l

: X

+

lR i s i n t e g r a b Z e and

A E 2.

PROPOSITION.

Hausdorff space

X.

Let

u

be a f i n i t e Bore2 m e a s u r e on a compact

T h e n e a c h c o n t i n u o u s mapping

f : X

-+

F

i s

Bochner i n t e g r a b l e .

PROOF. C l e a r l y i t s u f f i c e s t o show t h a t f i s t h e uniform lim it of a s e q u e n c e o f s i m p l e € u n c t i o n s . L e t n E liV be given. S i n c e f i s c o n t i n u o u s a n d X i s compact w e c a n f i n d p o i n t s al,

..., a k E

X

such t h a t

HOLOMORPHIC MAPPINGS

For e a c h

. . .,k

j = I,

U

Then

43

set

j

= f-'

[ B ( f ( a j ) ; l/n)l

are d i s j o i n t B o r e l sets which cover

AI,...,Ak

X. I f w e

define

k

fn(x)

then

COROLLARY.

Hausdorff space from

X

Then f

fx)f(a.) 3

f o r every

x E X,

i s a s i m p l e f u n c t i o n and

fn

II f n ( x )

6.6.

z x Aj j=l

=

into

-

flx)Il < l / n

Let X.

p

f o r every

x E X.

be a f i n i t e B o r e l m e a s u r e on a c o m p a c t

L e t f f n ) b e a s e q u e n c e o f continuous mappings

F w h i c h c o n v e r g e s u n i f o r m l y on X

t o a mapping f.

i s c o n t i n u o u s and

For each

A E z.

So f a r w e have only c o n s i d e r e d Bochner i n t e g r a t i o n w i t h re-

s p e c t t o f i n i t e p o s i t i v e measures, b u t

the extension t o real

measures o r t o complex measures may p r o c e e d e x a c t l y as

i n the

s c a l a r case. EXERCISES 6.A. fX,Z,

T h i s i s E g o r o f f ' s Theorem f o r v e c t o r - v a l u e d mappings.Let u ) be a f i n i t e measure s p a c e . L e t I f , ) be a sequence o f

measurable mappings from everywhere t o a mapping

X

into

F

which

converges

almost

f . By r e p l a c i n g a b s o l u t e values by norm

a t t h e a p p r o p i a t e p l a c e s i n t h e s t a n d a r d proof of t h e scalar E g o r o f f ' s theorem, show t h a t f o r e a c h E > 0 t h e r e e x i s t s a

44

MU J ICA

set A E Z with p(X \ A ) f uniformly on A .

5

E

and such t h a t

(f,)

converges t o

6.B. L e t (X, 2 , p l b e a f i n i t e measure space. L e t (f,) be a sequence o f measurable mappings from X i n t o F which c o n v e r g e s a l m o s t everywhere t o a mapping f .

(a)

Using E g o r o f f ‘ s Theorem 6.A f i n d a sequence o f sets A , E Z and a sequence of s i m p l e mappings g n .- X F such t h a t p ( X \ A n ) 5 2-n and Ilgn(x) - f ( x l l l 5 2-n f o r every x E A n . +

m

(b)

Let

Bj

k2j Ak

=

f o r every j formly on e a c h B

< 2 -

f o r each

j

i7V.

E

and show t h a t ( g

Show t h a t p ( X \ B . ) 3

converges t o

n

f

uni-

i’

m

(c)

B =

B Show t h a t jZl j ’ ( g ( x ) ) converges t o f ( x ) f o r e v e r y n shows t h a t f i s measurable.

Let

= 0

p ( X \ BI

U

x

E

B.

and s h m t h a t

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

6.C. L e t (X, 2 , v) be a f i n i t e measure s p a c e . Using Exercise 6.B show t h a t a measurable mapping f : X + F i s Bochner i n t e g r a b l e i f and o n l y i f Jx I(f 11dl.1 < m. T h i s i s Bockncr’s nharac-

t e r i z a t i o n of Bochner i n t e g r a b l e mappings. 6.D.

T h i s i s t h e Dominated C o n v e r g e n c e Theorem f o r Bochner in-

t e g r a b l e mappings. L e t (X, Z, P I b e a f i n i t e measure s p a c e . L e t If,) be a sequence of Bochner i n t e g r a b l e mappings from X i n t o F which converges a l m o s t everywhere t o a mapping f . Suppose there e x i s t s an integrable function

X

lR such t h a t IIfn(xJII 5 g l x ) f o r e v e r y II E IN and a l m o s t e v e r y ~ € 1Using . Bochner’s c h a r a c t e r i z a t i o n 6.C and t h e s c a l a r Dominated Convergence Theorem show t h a t

-f

o

and

g :

+

f i s Bochner integrable, $* II f, - fll du

SA fndu -,SAffdu f o r e a c h

A

E

X.

(X, 2 , be a f i n i t e measure s p a c e . L e t ( f n I be a sequence of Bochner i n t e g r a b l e mappings from X i n t o F which converges u n i f o r m l y on X t o a mapping f . Show t h a t f is Bcchner 6.E.

Let

45

HOLOMORPHIC MAPPINGS

Jxllfn

integrable,

-

flldu

and

0

-C

JAf n d u

+

IA fdu

for

each

A E Z.

6.F.

1-1 b e a B o r e 1 p r o b a b i l i t y measure on a compact Haus-

Let

d orff space (a)

X, and l e t

Given

$l,.

nj = and l e t

T

E

i,

e(Fm;

f

. .,$, $0

: X E

+

b e a c o n t i n u o u s mapping.

F

let

(FBI

j = 1,

for

fdp

..., n,

B n ) b e d e f i n e d by

~y = ($zIy),...,$nIy))

f o r every

y

E F.

Using t h e Hahn-Banach s e p a r a t i o n theorem show t h a t

where (b) point

c o ( B ) d e n o t e s t h e convex h u l l o f t h e set

B.

Using a compactness argument show t h e e x i s t e n c e of a

-

y E co(f(XI)

such t h a t

I n p a r t i c u l a r t h i s shows t h a t t h e Bochner i n t e g r a l S X f d p l i e s i n t h e c l o s e d , convex h u l l

c o (f (X))

of

f (X)

.

7. THE CAUCHY INTEGRAL FORMULAS A f t e r t h e i n t e r m i s s i o n on v e c t o r - v a l u e d i n t e g r a t i o n i n t h e mapp r o c e d i n g s e c t i o n , w e c o n t i n u e o u r s t u d y of holomorphic pings. I n t h i s s e c t i o n w e e s t a b l i s h t h e Cauchy i n t e g r a l f o m l a s

we study more c l o s e l y t h e q u e s t i o n of convergence of t h e T a y l o r series. and d e r i v e some o f t h e i r consequences. I n p a r t i c u l a r

7.1. THEOREM.

Let

U b e an o p e n

subset o f

E , and l e t

f

E

JC(U;F).

46

MUJ I CA

r > 0 b e s u c h t h a t a + r;t E U f o r a l l 5 E a(0;r). T h e n f o r e a c h A E A ( O ; r l we h a v e t h e Cauchy I n t e g r a l Formu l a Let

a E U, t E E

and

i

-

f ( a + A t ) = 2 In z

-+

:('

dr;.

Act)

lLl=P PROOF.

$ E F'

If

glr;) = $ o f ( a

then t h e function

holomorphic on a neighborhood of t h e c l o s e d d i s c

-

+

is

Ct)

By

A(0;r).

t h e Cauchy i n t e g r a l formula €or holomorphic f u n c t i o n s

of

one

complex - 7 a r i a b l e w e have t h a t

f o r each

A E A(O;r).

Since

F'

s e p a r a t e s t h e p o i n t s of

F the

d e s i r e d c o n c l u s i o n follows.

7.2.

U b e a n o p e n s u b s e t of E , and Z e t fE a E U, t E E and r > 0 b e s u c h t h a t a + Ct E U Z ( 0 ; r ) . The f o r e a c h A E h ( O ; r ) we h a v e a s e r i e s

COROLLARY.

JC(U;F).

Let

for aZZ

5

E

Let

expansion o f t h e form m

f(a

+

At) =

z m=O

where c

m

=-

c

m lm

f f a + Xtl d<

2r-L

.

I
0

For

5

s < r.

1x1

< 151 = r

w e can w r i t e

I I 5

s

47

HOLOMORPHIC MAPPINGS

is bounded on the set { a +
f(a +

I I?

ct)

1

m

d?

z

=

5 - X

Xm

m=O

fla + Gt) d?

I ? l=r

(=r

and this last series converges absolutely and uniformly €or 1x1 -< s . An application of Theorem 7.1 completes the proof. 7 . 3 . COROLLARY. 3CtU;f').

Let

for a l l

5

a E

E

Let

U b e a n o p e n s u b s e t of

U, t

E E

and

and Z e t f E be s u c h t h a t a + ?t E U

r > 0

a t 0 ; r ) . Then f o r each

E,

m E W o we h a v e t h e Cauchy

I n t e g r a l Formu l a pmf(a)(t) = I

1

I 5 I=rf ( ; m : l ? t )

dr;

.

Since f is holomorphic we have a series expansion the form

PROOF.

of

for I X I 5 for a suitable E > 0. After comparing this series expansion with the series expansion given by Corollary 7.2, an application of Lemma 4.5 completes the proof. 7.4. COROLLARY. 3CIU;F).

for a l l

Let

a

E

L e t U b e a n o p e n s u b s e t o f E , and l e t f E U, t E E and r > 0 b e s u c h t h a t a + Ct E U

5 E n(0;r).

Inequality

The f o r e a c h

m

E

Do we h a v e

the

Cauchy

48

7.5.

MUJ I CA

COROLLARY.

a, t

then f o r

P E ?'(mE;F)

If

we have t he

E

i n t e g r a l formu l a P(tJ

P(a + 3 t l

-

=

dr,

2TZ

1c1=1

PROOF.

P m P ( a ) i t ) = P i t ) . Then a n a p p l i c a t i o n

By Example 5.3,

of C o r o l l a r y 7.3 c o m p l e t e s t h e p r o o f . 7.6.

Let

COROLLARY.

open b a l l

P

P(mE;F).

E

B(a;rl then

is

P

aZoo

I f

P

i s bounded b y c on a n

bounded

by

c

on t h e b a Z l

B(0;rl. Thus C o r o l l a r y 7.6 s h a r p e n s t h e c o n c l u s i o n of Lemma 2 . 5 . Before g o i n g on i t i s c o n v e n i e n t t o i n t r o d u c e some notation and t e r m i n o l o g y . A p o Z y d i s c i n

i s a p r o d u c t of d i s c s . The open p o l y d i s c w i t h c e n t e r a = ( a l , . . . , a n ) and p o l y r a d i u s P = n ( rl,...,rn) w i l l be d e n o t e d by A ( a ; r ) . The corresponding -n c l o s e d p o l y d i s c will be d e n o t e d by A f a ; r ) . I n o t h e r words.

If

,..., 0 )

and

bn(O;l)

= An

a = 0 = (0

simply w r i t e 12 E

8

: 1zj

-

gn

r = 1 = ( 1 , . ..,I) then w e and x n ( U ; l ) = in. The s e t

a.1 = r 3

Li

for

i s c o n t a i n e d i n t h e boundary a A n ( a ; r l

aoAn(a;r).

be denoted by of An ( a ; r )

7.7. Let

a +

E

n

..., n l

A (a;r),

will boundary

and

I t i s c a l l e d t h e distinguished

.

THEOREM.

a

of

j = I,

shall

U, t l +

A E An(O;rl

Let

U b e an o p e n s u b s e t of E,and l e t f E J C ( U ; F ) . E and r l ,..., r n > 0 Pe .such t h a t

,..., t n E

...

E U f o r a12 < E i n f O ; r ) . T h e n for e a c h 'ntn we have t h e Cauchy I n t e g r a Z F orm ul a +

49

HOLOMORPHIC MAPPINGS f(a + Altl

PROOF. Rl

Since

> rl

all

...

+

5

+ A ntn )

-n A (0;r)

the polydisc

,..., Rn

>

E An(O;R).

rn

a + Cltl

such t h a t $

If

E

F'

w e can f i n d E U for 'ntn

i s compact,

. ..

+

+

then t h e f u n c t i o n

i n s e p a r a t e l y holomorphic i n e a c h of t h e v a r i a b l e s C l , ..., ' n when t h e o t h e r v a r i a b l e s are h e l d f i x e d . T h e n r e p e a t e d a p p l i c a t i o n s o f t h e Cauchy i n t e g r a l formula f o r holomorphic f u n c t i o n s

of one complex v a r i a b l e l e a d t o t h e f o r m u l a

+

$ of(a

X

f o r every

...

+

Altl

+ Antn)

An(O;r).

E

Since t h e f u n c t i o n

i s c o n t i n u o u s on t h e compact s e t

a o A n (0;

P),

Fubini's

Theorem

allows u s t o r e p l a c e t h e i t e r a t e d i n t e g r a l by a m u l t i p l e i n t e g r a l . And s i n c e

F'

separates points, the desired

conclusion

follows.

7.8.

E

L e t U be a n o p e n s u b s e t of U, t l , t , E E and r l

+

...

COROLLARY.

X(U;F).

Let

a

cltl

that

a +

each

A E An(O;r)

+ cntn

E,

,...,

...,

E

U

for a l l

and l e t f E > 0 be s u c h n

5 E zn(O;r).

Then f o r

we h a v e a s e r i e s e x p a n s i o n of t h e f o r m

MUJ I CA

50

where

T h i s m u l t i p l e s e r i e s c o n v e r g e s a b s o Z u t e l y and u n i f o r m l y f o r E

iZn(o;si

o

.<_ s

j

< r

f o r every

i

I

= r

for

j

j = I,.

. .,n

h

j.

The proof i s s i m i l a r t o t h a t of C o r o l l a r y 7 . 2 .

PROOF. 15. 3

provided

If

lhj(

t h e n w e can w r i t e

... + Antn) . . . (5, - A n )

+

f(a + Xltl

(5, - 1 , )

al

= CAI

a,.,

.., x n

...

f(a + cltI + al+l

.-.

c1

51

+

an+1

cntn)

‘n

and t h i s m u l t i p l e series c o n v e r g e s a b s o l u t e l y and uniformly for

ISj I

= r

and I h j I 5 s J’ r j ’ A f t e r i n t e g r a t i n g t h i s series j term by t e r m , t h e d e s i r e d c o n c l u s i o n f o l l o w s from Theorem 7 . 7 .

7.9.

COROLLARY.

JC(U;FI.

that

Let

a

Let E U,

a + cltI +

...

U be a n o p e n s u b s e t of E , and l e t f E tl, . , t , E E a n d r l , . . . , r n > 0 be such

..

+

‘ntn

E

U

for a l l

m E N o and e a c h m u l t i - i n d e x h a v e t h e Cauchy I n t e g r a l F o r m u l a

each

a

E

5 E EVE

zn(O;p).

with

la

Then f o r

= m we

.. dc, . PROOF. form

By P r o p o s i t i o n 4 . 6 w e have a series e x p a n s i o n

of

the

51

HOLOMORPHIC MAPPINGS

where

c

f o r each

u

ci E

“I

... t n

= +m’A

m

with

I c t I = m . T h i s m u l t i p l e series converges

f(a)tl

-

a b s o l u t e l y and u n i f o r m l y on a s u i t a b l e p o l y d i s c

An(O;c). A f t e r

comparing t h i s series e x p a n s i o n w i t h t h e series e x p a n s i o n g i m by C o r o l l a r y 7 . 8 ,

Lemma 4 . 7

an a p p l i c a t i o n of

completes

the

proof. 7.10.

Let

COROLLARY.

A E L s f r n E ; F ) and l e t

T h e n for a l l a , t l J . . . t n E E and a l l we have t h e p o l a r i z a t i o n f o r m u l a

Apply C o r o l l a r y 7 . 9 w i t h

PROOF.

f

ci E

P = iDE

A^

P(mE;F).

E

la1 = m

with

= P.

W e r e c a l l t h a t a s e t A i n E c o n t a i n i n g t h e o r i g i n is said t o be b a l a n c e d i f closed u n i t d i s c i f t h e set

-

A

<x E A f o r e a c h x A . I f a E A then A

-

a

E

and e a c h

A

<

i n the

i s s a i d t o bea-balanced

is balanced.

U be an a - b a l a n c e d o p e n s u b s e t of E , and f a t a converges t o let f u n i f o r m l y o n a s u i t a b l e n e i g h b o r h o o d of e a c h c o m p a c t s u b s e t 7.11. THEOREM. f

of

E

XfU;F).

Let

Then t h e T a y l o r s e r i e s o f

u.

PROOF.

Let

K be a compact s u b s e t of

A = { a + ~ ( -x a )

:

x

E

U. Then t h e s e t K, 5

E

X}

MUJ I CA

52

i s contained i n

U, and f i s bounded on A . Using a canpactness argument w e c a n f i n d r > 1 and a neighborhood V o f K i n U such t h a t t h e s e t

U , and f i s bounded on B as w e l l . Hence

i s a l s o contained i n w e can w r i t e

f

Ia +

C(X

- a)]

5 - 1

= m=O z

f [ a + ~ ( -x a ) ] f1+1

and t h i s series c o n v e r g e s a b s o l u t e l y and u n i f o r m l y f o r 1 5 ) = P. A f t e r i n t e g r a t i n g o v e r t h e c i r c l e

and

x

151 = r

a p p l y i n g t h e Cauchy I n t e g r a l Formulas 7 . 1 and 7 . 3

E

V

and

we c o n c l u d e

that

and t h i s series c o n v e r g e s a b s o l u t e l y and u n i f o r m l y f o r

7.12.

DEFINITION. J C ( U ; F ) and l e t a (a)

Let E

U be an open s u b s e t of

E, l e t

3:

E

f

V. E

U.

The r a d i u s of

boundedness o f

f

at

a i s t h e supremum

-

o f a l l r > 0 such t h a t B l a ; r ) C U B ( a ; r ) . The r a d i u s of boundedness of by

and f at

bounded on a w i l l be denoted f

is

rbf(a).

(b) The r a d i u s of convergence o f t h e T a y l o r series of f a t a w i l l b e d e n o t e d by r e f l a ) . For s h o r t r c f ( a ) w i l l be r e f e r r e d t o as t h e r a d i u s of c o n v e r g e n c e o f f a t a .

(c) The d i s t a n c e from ci t o t h e boundaryof U w i l l b e den o t e d by d u ( a ) . When U = E t h e n f o r convenience w e d e f i n e d , ( a ) = m. 7 . 1 3 THEOREM.

Let

U be a n o p e n s u b s e t o f

E,

let

f

E JC(U;F)

HOLOMORPHIC MAPPINGS

and l e t

a

53

Then

E U.

rbf ( a l = m i n { r c f ( a ) , d , l a ) } .

PROOF.

W e observe a t t h e o u t s e t t h a t

and hence

rbfia)

2

d u ( a ) . Thus t o show t h e i n e q u a l i t y

rbf (a)

(7.1)

5

min { r e f ( a ) , d , f a ) 1

i t s u f f i c e s t o show t h a t r b f ( a ) < refla). Let 0 2 r < rbf(a). Then B ( a ; r ) C U and f i s bounded, by c s a y , on B ( a ; r ) . I t f o l l o w s from t h e Cauchy I n e q u a l i t i e s 7 . 4 t h a t I I P m f ( a ) l l 5 e r - m f o r e v e r y rn E Ih7 and an a p p l i c a t i o n o f t h e Cauchy-Hadamard Formula 4 . 3 shows t h a t

refla)

5 r. Letting

r

+

r b f ( a ) w e get that

rbf(a) < r c f ( a ) and ( 7 . 1 ) f o l l o w s .

Next w e show t h a t o p p o s i t e i n e q u a l i t y :

Let

that

0 - r < s < min { r e f ( a ) , d,la)).

B(a;si C U

f o r every

x

Since

s < d U ( a ) it f o l l o w s

and t h e n Theorem 7 . 1 1 i m p l i e s t h a t

E B(a;s).

On t h e o t h e r hand, it f o l l o w s from

the

Cauchy-Hadamard Formula t h a t

and hence t h e r e e x i s t s

c > 1

such t h a t

(7.4)

for e v e r y

m

E

W o . Then from ( 7 . 3 ) and ( 7 . 4 )

it f o l l o w s t h a t

54

MUJ I CA

f o r every

7.14.

x

B i a ; r l . Hence

E

and ( 7 . 2 )

r

U be an open s u b s e t of

Let

REMARK.

rbf(a)

E,

follows.

let

and

f

E

i s f i n i t e dimensional then each c l o s e d b a l l w i t h f i n i t e r a d i u s i s compact, and whence i t f o l l o w s t h a t r h f ( a ) = d, (a and r e f l a ) 1. d u ( a ) f o r e v e r y a E U. I n s h a r p c o n t r a s t w i t h t h e f i n i t e d i m e n s i o n a l s i t u a t i o n w e have t h e f o l l o w i n g reJC(U;F).

If

E

sult

7.15 PROPOSITION. S u p p o s e t h e r e e x i s t s a s e q u e n c e (q,) i n E ’ l i m q ix) = 0 f o r every s u c h t h a t llqmll = I f o r e v e r y m a n d m x E E . Then t h e r e e x i s t s a f u n c t i o n f E J C ( B ) whose r a d i u s of boundedness a t t h e origin e q u a l s one. By Example 5 . 5 t h e f u n c t i o n

PROOF.

i s holomorphic on a l l of Formula t h a t

rcf(0)

E.

I t f o l l o w s from t h e Cauchqr-Hadamrd

= 1 . An a p p l i c a t i o n o f Theorem 7 . 1 3

com-

pletes the proof.

7.16. C

E‘

EXAMPLE.

E = c

0

or

Lp

(1

5

p <

m)

and l e t

d e n o t e t h e sequence of c o o r d i n a t e f u n c t i o n a l s .

1151,1

i s clear t h a t 3:

Let

E E.

= 1

Thus t h e s p a c e s

f o r every and

co

Rp

m and (1

5

p

Em(x)

-+

w)

0

15,) it

Then

for e v e r y

satisfy

the

hypothesis i n Proposition 7.15.

7.17.

THEOREM.

JCfU;F).

T.hen

for a l l

m, j

Let Pmf

E

E

no

U b e an o p e n s u b s e t of E, and J C f U ; P ( m E ; F ) ) and

and

a

E

U.

Let

f

E

55

HOLOMORPHIC MAPPINGS

PROOF. ( a ) F i r s t w e assume 0 E U. Choose B ( 0 ; 3 r ) C U and f i s bounded on B 1 0 ; 3 r l .

for

It t I1

>

0

such t h a t

By Theorem 7.13 the

f a t t h e o r i g i n c o n v e r g e s t o f u n i f o r m l y on E > 0 w e c a n f i n d k o E I N such t h a t

T a y l o r series of

B(0;Zr).

r

Then g i v e n

5

and

2r

.

k > ko

By a p p l y i n g t h e Cauchy Inqualities

7.4 we get t h a t

for

/ I t / (5 r

k

and

2 k0

. Thus m

Prnflti =

P r n [ PjfIOll ( t l j=O

and t h i s series c o n v e r g e s u n i f o r m l y f o r 1 I t II 5 r . Since Prn[$f(0)]

= 0

whenever

rn > j

w e c a n even w r i t e m

Prnflti =

2 P"[ P"+JfIOll It). j=O

(b)

I n t h e g e n e r a l c a s e t a k e an a r b i t r a r y p o i n t

If we define

t

E

U - a

fa : U - a

+

by

F

fa(t)

= fla + t l

t h e n by E x e r c i s e 5.B w e have t h a t

and P m f a ( t l = P m f ' ( a

+ tl

f o r every

-

t E U

Moreover, i f f i s bounded on B ( a ; 3 r ) B I O ; 3 r ) . Hence u s i n g ( a ) w e g e t t h a t

=

z: j=O

then

P"1P"fjfa(0)l

-

a

E

U.

f o r every - a;F)

f a E JcfU

a fa

I t )

and

rn

E

INo.

i s bounded on

56

MUJ I CA

Iltll 5 r.

and t h e l a s t w r i t t e n series c o n v e r g e s u n i f o r m l y f o r S i n c e Pm [ P m + j f l a ) ] E P ( j E ; P ( m E ; F ) )

t h e proof i s complete.

COROLLARY. L e t U b e an open s u b s e t of E, and l e t If f o r e a c h rn E I N o a n d t E E we d e f i n e PTf : U 7.18.

T o end t h i s s e c t i o n w e g e n e r a l i z e t h e

f E JCIU;FI. -+

classical

by

F

Schwarz'

Lemma as f o l l o w s .

7.19.

THEOREM.

pose t h a t exists

m

U = B(a;r)

Let

Ilf~xIII < c E

Ilf(x)ll

W

5

f o r every

such t h a t 113:

c(

-

P'fla)

a l l rn )

C

and l e t f E X ( U ; F ) . S u p x € B ( a ; r l and s u p p o s e t h e r e = 0 for every j m. T h e n E

for every

x E B(a;rl.

PROOF. F i x x E B ( a ; r ) w i t h x # a , and f i x 11 II = 1 . L e t g be t h e f u n c t i o n of one complex

+

$

F'

E

with

variable

de-

f i n e d by

By Theorem 7 . 1 1 t h e T a y l o r series of

wise t o f on t h e b a l l

f

at

a

converges p o i n t -

B ( a ; r ) . Whence t h e f u n c t i o n

g

can

be

w r i t t e n a s t h e sum of t h e power series m

g(h) =

z

o p j f i a ) (x - a )

j =m

/ Ilx - a l l ) . I n p a r t i c u l a r g i s holomorphic on t h a t d i s c . Take s w i t h llz - a II < s < r . Since IIfIl 5 c on t h e d i s c on

A(0;r

B ( a ; r l i t follows t h a t

HOLOMORPHIC MAPPINGS

for

IA I

= s / l l z - a II,

and t h e r e f o r e f o r

57

IX I

t h e c l a s s i c a l Maximum P r i n c i p l e . By a p p l y i n g with

X =

< s /Il x -

this

- a II

by

inequality

we get t h a t

1

s

After l e t t i n g

+

r, an a p p l i c a t i o n of t h e Hahn-Banach Theorem

completes t h e proof.

EXERCISES 7.A. most

E P(E;FI

b e a c o n t i n u o u s polynomial

a, t

Let

E

E,

r > 0

f E 7CIE;F).

a constant

c > 0

and

k > m.

Suppose t h e r e i s an i n t e g e r m E I N o

such t h a t

II f(xlII

f

7.C.

U be an open s u b s e t of

Let

pose t h e r e i s a c l o s e d s u b s p a c e

7.0.

x

E

5 c

112 Ilm

for all

i s a polynomial of d e g r e e a t most

Show t h a t

every

at

of d e g r e e

m. Show t h a t

for a l l 7.B.

P

Let

U. Show t h a t

Show t h a t i f

N

E , and l e t

of

F

x

and E

E.

m.

f

Sup-

E K(U;F).

such t h a t f f x )

€

N

for

f E Jc(U;N).

U i s a p r o p e r open s u b s e t of

E

then

U be an open s u b s e t of E, and l e t f E X f U ; F ) . Show t h a t e i t h e r r f l x ) = 00 € o r e v e r y x E U ( i n t h i s case U = E l , 7.E.

Let

b

5e

MUJ 1 CA

or else

rbffxl

for a l l x, y

E

m

for every

x

E 11,

and in the latter case

U.

7.F. (a) Let E be a separable Banach space. Using Cantor's diagonal process show that each bounded sequence in El has a U(E',E)-convergent subsequence. (b) Show that each infinite dimensional, separable Banach space satisfies the hypothesis in Proposition 7.15. 7.G. (a) Let E be a reflexive Banach space. By considering a suitable separable, closed subspace of E show that each bounded sequence in E has a ofE,E')-convergent subsequence. (b) Show that each infinite dimensional, reflexive Banach space satisfies the hypothesis in Proposition 7.15

8.

G-HOLOMORPHIC MAPPINGS

In this section we show that a mapping is holomorphic if it is continuous and its restriction to each complex line is holomorphic. This is a useful characterization, and in m y situations this is the easiest way to check that a given mapping is holomorphic. 8.1. DEFINITION. Let U be an open subset of E . A mapping f : U F is said to be G - h o l o m o r p h i c or G - a n a l y t i c (G for Goursat) if for all a E U and b E E the mapping X f ( a + Ab) is holomorphic on the open set {A E @ : a + Xb E U}. We shall denote by J C G f U ; F ) the vector space of all G-holomorphic mapJEG(U;@i pings from U into F. If F = @ then we shall write = JC,(U). +

+

8.2. EXAMPLE.

PROOF.

P(a

PafE;F)

C

JCG(E;F).

+ Ab) i s a polynomial in X for

all

a, 0

E

E.

59

HOLOMORPHIC MAPPINGS

REMARKS. (a) The Identity Principle, the Open Mapping Principle, the Maximum Principle and Liouville's Theorem, all of them established in Section 5 for holomorphic mappings,= actually true for G-holomorphic mappings. A glance at the corresponding proofs shows this at cnce. 8.3.

(b) An examination of the corresponding proofs shws that Theorem 7.1 and Corollary 7.2 are still valid for G-holomorphic mappings.

(c) Finally, an examination of the corresponding proofs shows that Theorem 7.7 and Corollary 7.8 are still valid for those G-holomorphic mappings whose restrictions to finite dimensional subspaces are continuous. PROPOSITION. L e t XG(U;F). F o r e a c h a

8.4. E

U b e a n o p e n s u b s e t of E , a n d l e t E U and m E W o L e t P m f ( a ) : E --t

f F

be d e f i n e d by

where

r

is c h o s e n s o t h a t

0

a + < tE U fop a l l

5 E

a(O;r).

Then:

(a)

The d e f i n i t i o n o f

(b)

f o r all

(c)

P m f ( a ) ( t ) i s i n d e p e n d e n t from t h e

r.

choice o f

The m a p p i n g

t

E

E

and

m

P f l u ) i s m-homogeneous,

i.1 E C.

If U i s a - b a l a n c e d

s e r i e s expansion

t h a t is

t h e n f o r each

3:

E

U

we have t h e

60

MUJ I CA (a)

PROOF.

Let

a + r;t

that

E

U

t

E

b e g i v e n and l e t

E

for a l l

r;

E

r

0 < s

be

such

X ( 0 ; r ) . Then by Remark 8.3(b) we

have t h a t

f o r every

X

E

A(0;sl.

f o r every

m

E

Do.

t

Let

(b)

and

p E 6

be g i v e n .

If

i s suf-

r > 0

t h e n a g a i n by Remark 8 . 3 ( b ) w e have t h a t

ficiently s m a l l

for every

E E

By Lemma 4.5 w e c o n c l u d e t h a t

Then a n o t h e r a p p l i c a t i o n

XEA(0:r).

of

Lemma

4.5

y i e l d s t h e d e s i r e d conclusion.

(c)

r;

E

a.

Given

LC E

a + c ( x - a)

w e have t h a t

U

By a compactness argument w e c a n f i n d

a + <(x - a )

E

U

for all

r;

E

r > 1

U

for all

such t h a t

Then by Remark 8 . 3 ( b ) we

E X(0;r).

have t h a t

f o r each

X

E A(0;r).

Letting

X = 1

we get the desired

con-

clusion. 8.5. that

Under t h e s e t t i n g o f P r o p o s i t i o n 8.4 it i s

REMARK.

Pmf ( a )

E

PalmE;F)

for a l l

a

E

U

and

m E Do,

proof of t h i s f a c t , w i t h o u t a d d i t i o n a l h y p o t h e s e s on have t o w a i t u n t i l S e c t i o n 3 6 , f o r i t rests on a deep of Hartogs on s e p a r a t e

analyticity.

true but

a

f, w i l l theorem

61

HOLOMORPHIC MAPPINGS

8.6. E

Let

PROPOSITION.

JCG(U;F). T h e n

f

be an o p e n s u b s e t of

U

i s c o n t i n u o u s if and o n l y i f

and l e t f i s locally

E,

f

bounded.

To b e g i n w i t h w e remark t h a t Schwarz'

PROOF.

v a l i d f o r G-holomorphic mappings.

Lemma

is

7.19

I n d e e d , the sane proof

amlies

p r o v i d e d w e use P r o p o s i t i o n 8 . 4 1 ~ )i n s t e a d o f Theorem 7 . 1 1 . Now, l e t Given - c

a

f : U

for all

mapping

F

be G-holomorphic and l o c a l l y bounded.

w e choose r > 0 and c > 0 s u c h t h a t 1 I f(x)ll x E B ( a ; r ) . By a p p l y i n g Schwarz' Lemma t o t h e

U

E

+

-

f(x)

f ( a ) we get t h a t

f o r a l l x E B f a ; r l , p r o v i n g t h a t f i s c o n t i n u o u s a t the p o i n t a . S i n c e t h e r e v e r s e i m p l i c a t i o n i s c l e a r , t h e proof i s complete.

Now w e can e s t a b l i s h t h e

characterization

of

holomorphic

mappings announced a t t h e b e g i n n i n g of t h i s s e c t i o n . 8.7.

Let

THEOREM.

mapping

f : U

--t

F

U be an o p e n s u b s e t of

E.

(a)

f

i s holomorphic

(b)

f

i s c o n t i n u o u s and G - h o t o m o r p h i c .

(c)

f

i s c o n t i n u o u s and

f

f i n i t e dimensional subspace M

of

PROOF.

The i m p l i c a t i o n ( a )

( b ) * ( c ): and l e t

Let

Then

f

:

U

-+

I

each

U n M

i s hoZomorphic for each

E.

* (b) i s clear. P

be G-holomorphic and continuous,

M be a f i n i t e d i m e n s i o n a l s u b s p a c e of

and l e t ( e l , ..., e n ) be a b a s i s f o r

M.

E. Lst

+ Alel +

...

+ A e I = n n

"1

z1 c a X l c1

aE U nM

Then by Remark 8 . 3 ( c ) *

have a series e x p a n s i o n of t h e form f(a

for

the following conditions are equivalent:

"n ... in

67

MUJ I CA

where t h i s m u l t i p l e series converges a b s o l u t e l y and on a s u i t a b l e p o l y d i s c Pm E P l m M ; F )

An(O;rl.

I f f o r each

uniformly

rn E W o we define

by

t h e n w e have a power series e x p a n s i o n W

+

fla

Alel

+

... +

Xn e n ) =

w i t h uniform convergence on

(c) * ( a ) : subspace of

E

Let

U. I f

C

... +

M i s a f i n i t e dimensional

c o n t a i n i n g a t h e n by h y p o t h e s i s

-

M

-

a) from M i n t o

U

power

M and N a r e two f i n i t e

dimen-

such t h a t

F W

x

f(x) =

I

M is series

f

holomorphic and t h e n by Theorem 7 . 1 1 t h e r e i s a .Z P ( x m=O m

Anen)

T h i s shows ( c ) .

An(O;r).

B(a;r)

+

Z Prn(Alel m=O

M

- a)

Pmlx

m=O x E M n E(a;r). If

f o r every

a t h e n i t f o l l o w s from t h e M N = P,(t) uniqueness of t h e T a y l o r series e x p a n s i o n t h a t P , ( t l for a l l t E M N and a l l rn E W o .L e t Pm : E' F b e deM = P rn ( t ) i f M i s any f i n i t e dimensional subspace f i n e d by P , ( t l of E c o n t a i n i n g a and t . Then Pm E ? ( m E ; F ) by E x e r c i s e a

s i o n a l s u b s p a c e s of

E

containing

+

2.B,

and W

f(x) =

f o r every

a ball Ilf(x)ll

-

x E B(a,r).

E(a;s)

5

L'

C

Now s i n c e f and

E(a;r)

f o r every

Z Pm(x m=O

x

E

- a)

i s continuous w e can

a constant

B ( a ; s ) . Given

t

l e t M b e any f i n i t e d i m e n s i o n a l s u b s p a c e of and

0

c E

E E

such with

find that

Iltll < 1

containing

t. Then by t h e Cauchy I n t e g r a l Formula 7 . 3 w e g e t t h a t

a

63

HOLOMORPHIC MAPPINGS and i t f o l l o w s t h a t

IIPmII

5 cs-m . Hence e a c h P

m

i s continuous

W

and t h e power series

2

Pm(x

-

a ) has a r a d i u s ofconvergence

m=O

g r e a t e r than o r equal t o

s . T h i s show ( a ) and t h e theorem.

U be a n open s u b s e t of

Let

Then e a c h

Cn.

mapping f : U F is separately f ( i , , . . ., r n ) i s holomorphic i n e a c h

G-holomorphic

hoZomorphic,

+

that

is,

when t h e o t h e r v a r i ‘j a b l e s a r e h e l d f i x e d . The f o l l o w i n g r e s u l t on s e p a r a t e l y h o l o -

morphic mappings p a r a l l e l s P r o p o s i t i o n 8 . 6 .

U be a n o p e n s u b s e t of C n , and l e t f : U - + F be s e p a r a t e l y holomorphic. T h e n f is c o n t i n u o u s if and only if f is locally b o u n d e d . 8.8.

LEMMA.

PROOF.

Let

f

Let

U F b e s e p a r a t e l y holomorphic and l o c a l l y a E U choose r > 0 and c 0 such that f o r e v e r y 5 E A n ( a ; p ) . Then f o r e a c h 5 E A n ( a ; r ) :

+

bounded. Given

llfl<)ll < c w e can w r i t e

n -

2

[ f(a,,

j=l

. . . , a j - l , c j J . . ., 5 , )

- f (al,.. . , a j J c ~ + ~ ,...,<,)I.

I f f o l l o w s from t h e h y p o t h e s i s t h a t t h e d i f f e r e n c e

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

Cj

when t h e o t h e r v a r i a b l e s are

h e l d f i x e d . Furthermore, IIg . f 5 .)I1

< 2c by a p p l y i n g Schwarz’ Lemma t o e a c h g j 3

s

for

\ci -

Q

j we get that

1 < r . Hence,

f i s c o n t i n u o u s . Since t h e o p p o s i t e i m p l i c a t i o n i s c l e a r , t h e proof of t h e lemma i s

f o r every

5

E

A n ( a ; r ) . T h i s shows t h a t

64

MUJ I CA

complete. 8.9. f

LEMMA.

: U

-+

Let

t n . !!'hen a m a p p i n g

U be a n o p e n s u b s e t of

i s h o Z o m o r p h i c i f and onZy i f

F

f

i s separately holo-

m o r p h i c and c o n t i n u o u s . T o prove t h e n o n t r i v i a l i m p l i c a t i o n l e t

f : U F be s e p a r a t e l y holomorphic and c o n t i n u o u s , and l e t a E U. Thenthe p r o o f s of Theorem 7.7 and C o r o l l a r y 7 . 8 a p p l y and hence w e obt a i n a s e r i e s e x p a n s i o n of t h e form PROOF.

"1

...

f i a + A ) = E caAl

a

+

"n An

w i t h a b s o l u t e and uniform convergence on a s u i t a b l e Hence

f

polydisc.

i s holomorphic.

Lemma 8.9 e x t e n d s a t once t o a r b i t r a r y Banach s p a c e s . 8.10.

U

E I J , . .,E

Let

PROPOSITION.

n y

b e an open s u b s e t o f

El

x

i s h o Z o m o r p h i c if and o n l y i f

...

X

F

b e Banach s p a c e s , and l e t

En

.

Thehen a mapping

f: U

+

F

i s s e p a r a t e l y h o l o m o r p h i c and

f

continuous. PROOF.

To prove t h e n o n t r i v i a l i m p l i c a t i o n , l e t

f : U

-+

F be

s e p a r a t e l y holomorphic and c o n t i n u o u s . L e t a = ( a l,...,an)E U and

b = ( b l , . . .,bnl

El

E

x

. ..

x

En.

Then t h e mapping

i s s e p a r a t e l y holomorphic and c o n t i n u o u s , and t h e r e f o r e morphic by Lemma 8.9. -+

holo-

Whence t h e mapping

f ( a l + Ably

i s holomorphic as w e l l .

Thus

..., a n

+ Abnl

f i s G-holomorphic and continuous,

and t h e r e f o r e holomorphic by Theorem 8.7. The proof i s c o m p l e t e .

I n S e c t i o n 3 6 w e s h a l l prove t h a t t h e h y p o t h e s i s of cont i n u i t y i n Lemma 8.9 and P r o p o s i t i o n 8.10 i s s u p e r f l u o u s . T h i s

HOLOMORPHIC MAPPINGS

65

is a deep theorem of Hartogs. By introducing the notion of G-holomorphic mapping we are reducing the study of holomorphic mappings to the case where the domain space is one dimensional. Now we introduce a notion that reduces the study of holomorphic mappings to the case where the range space is one dimensional.

8.11. DEFINITION. Let U be an open subset of E . A mapping : U F is said to be w e a k l y h o l o m o r p h i c or t l e a k l y a m Z y t i c if ri, o f is holomorphic for every ri, E F’. Likewise f is said to be w e a k l y G - h o l o m o r p h i c or w e a k l y G - a n a l y t i c if $ 0f is G-holomorphic for every + f F’. f

-+

8.12. THEOREM.

(a)

f

L e t U b e a n o p e n s u b s e t of

i s G-holomorphic

and Zet

E,

i f and o n l y i f

f

is

f : V +F.

weakly

G-

holornorphic.

(b)

f

is h o l o m o r p h i c i f and o n l y i f

f

is

weakly holo-

rnorphic.

Before proving this theorem we establish the followinglem ma. 8.13. LEMMA. L e t U b e an o p e n s e t i n 6. T h e n a m a p p i n g f : U F i s h o l o r n o r p h i c i f and o n l y i f f i s w e a k l y h o l o m o r +

phic.

PROOF. weakly Given ri, E F ’ tegral we can

To prove the nontrivial implication let f : U F be holomorphic. First we shall prove that f is continuous. X o E U we choose r > 0 such that h ( X o ; 2 r ) C U. Let and let X E n ( h o ; r l , X # Xo Using the Cauchy InFormula for holomorphic functions of one complex variable write -+

.

66

MUJ I CA

and it f o l l o w s t h a t

M = sup {

where

1)

of(<)

I

:

I < - X0 I =

Z r } . By t h e P r i n c i p l e of

Uniform Boundedness t h e r e i s a c o n s t a n t

- f(Xol

flh)

II

X

f o r every

f

K(A,;rf

-

h

with

A,

h #

II

Ao.

5

c > 0

such t h a t

c

This

shows t h a t

is

f

continuous a t Now w e can prove t h a t

f is

holomorphic, p r o c e e d i n g as

t h e p r o o f s of Theorem 7 . 1 and C o r o l l a r y 7 . 2 . C

U

in

h(Xo;r)

Indeed, i f

then f i r s t w e g e t t h a t

f o r each

h E A(Ao;r),

an& from t h i s w e g e t t h e s e r i e s

expan-

sion

w i t h uniform convergence on e a c h p o l y d i s c s < r . t h i s shows t h a t

PROOF O F THEOREM 8.12. :

U

+

F

with

A(Xo;s)

0

5

f i s holomorphic. P a r t ( a ) i s a n immediate consequence of

Lemma 8.13. To prove t h e n o n t r i v i a l

f

-

implication

be weakly holomorphic. Then

in

(b)

let

f i s weakly G-holomor-

p h i c , and t h e r e f o r e G-holomorphic by ( a ) . W e c l a i m t h a t

f is l o c a l l y bounded. T o show t h i s l e t K b e a compact s u b s e t of U. S i n c e f i s weakly holomorphic t h e s e t fIKl i s weakly bounded,

and t h e r e f o r e norm bounded by t h e P r i n c i p l e of Uniform Roundedn e s s . Thus

f i s bounded on e a c h compact s u b s e t of

i s t h e r e f o r e l o c a l l y bounded, by E x e r c i s e 5 . E .

Thus

U , and f f i s con-

t i n u o u s by P r o p o s i t i o n 8 . 6 , and holomorphic by Theorem 8.7. The

67

HOLOMORPHIC MAPPINGS proof i s complete.

EXERCISES Cn

8.A.

U b e an open s u b s e t of

Let

E.

L e t (fnlnz1

be a sequence

of holomorphic mappings from U i n t o F which c o n v e r g e s mapping f : U F u n i f o r m l y on e a c h compact subset of Theorems 8 . 7 and 8 . 1 2 show t h a t f is holomorphic. +

to

a

U. Using

m

8.B.

8 P (x rn m=O

Let

-

a ) b e a power series from E

r a d i u s of convergence

R > 0

and w i t h e a c h

W

f(x) =

Show t h a t t h e sum

holomorphic on t h e b a l l

8 Pm(x m=O

-

Prn

i n t o F with continuous.

a l o f t h e power series i s

B ( a ; R I . T h i s improves

t h e conclusion

i n Exercise 5.D. 8.C.

U b e an open s u b s e t o f

Let

(a)

E,

and l e t

Using Theorem 7 . 1 3 and E x e r c i s e s 7 . E

f E JciU;Fi. and

8.B

show

that

f o r each (b)

f o r each

a

E

U

and e a c h

x

E Bla;dU(a)).

Show t h a t

u E

U

and e a c h

x E B ( a ; 1‘4 d U ( a ) I .

-

(c) I f u i s c o n n e c t e d show t h a t e i t h e r r c f ( x ) = for € o r e v e r y x E U , and i n t h e every x 6 U , o r else r e f i x ) l a t t e r case

f o r each

a

E

U

and e a c h

x E B(a; % d U ( a i ) .

68

MUJ I CA

8.D.

u b e a n open s u b s e t o f

Let

E . By a d a p t i n g t h e prcmfsof

Lemma 8 . 1 3 and Theorem 8.12 show t h a t a mapping

x

holomorphic i f and o n l y i f t h e f u n c t i o n i s holomorphic f o r e a c h

y

More g e n e r a l l y , l e t

8.E.

an open s u b s e t of

U

E

f : U +

+

F' i s

f(xl(yl

CC

E

F.

E

E,F,

E , and l e t

G b e Banach s p a c e s , l e t f : U

-+

flF;G).

be

U

Show t h a t t h e

f o l l o w i n g c o n d i t i o n s are e q u i v a l e n t :

i s holomorphic.

(a)

;f

(b)

The mapping

each

y

x

U

E

--*

j~'x)(yjE G

i s holomorphic f o r

F.

f

The f u n c t i o n

(c)

y

for e a c h

and

E F

rl

x

E

f

G'.

U

-,rl

flx) ly)l

E CC

is holomorphic

Show t h a t t h e same c o n c l u s i o n i s t r u e when t h e s p a c e flF;G) i s r e p l a c e d by t h e s p a c e 8.F.

PlmF; G )

U be an open s u b s e t of

Let

sequence i n

. E,

U be an open s u b s e t o f

Let

E,

f : U

let

1

5

defined

co

-+

p <

and l e t

m,

m

m

(fn)n=I

be a sequence i n

03

d e f i n e d by Let

sup c lfn(xl j p x E K n=l U. Show that the mapping f : U

J C I U I such t h a t

f o f e a c h compact s u b s e t K of

8.H.

be a

X ( U I which c o n v e r g e s t o z e r o u n i f o r m l y on each cow

p a c t s u b s e t of U. Show t h a t t h e mapping m by f ( x ) = ( f n ( x l I n = , i s holomorphic. 8.G.

m

(fnlnZl

and l e t

f(xJ = lfnlx)/n=l

+

Rp

i s holomorphic.

U be an open s u b s e t of

E,

and

let

sequence of holomorphic f u n c t i o n s from U i n t o

00

lfnln=l

be a

C which a r e u n i that the i s holo-

formly bounded on each compact s u b s e t of U. Show mapping f : U Rm d e f i n e d by f i x ) :( f , ( x ) I ~ = , -+

morphic.

8.1.

Let

V C 11

be open s u b s e t s o f

E, with

U c o n n e c t e d . Let

69

HOLOMORPHIC MAPPINGS

g E 3CfV;FI

and suppose t h a t f o r e a c h

function

f$

(a) that

f

Using E x e r c i s e 8.D.

1

Ir

(b)

9.

3CIu) such t h a t

E

1

f

I

qI

there

J, E F '

a

exists

V = qI o g .

f i n d a mapping

such

f E JC(U;F"I

g.

Using E x e r c i s e s 5 . F and 7 . C show t h a t

f E 3C(U;F).

THE COMPACT-OPEN TOPOLOGY

Our a i m i n t h i s s e c t i o n i s t h e s t u d y

of

the

completeness

and compactness p r o p e r t i e s o f c o l l e c t i o n s o f holomorphic p i n g . With t h i s i n mind w e

map-

i n t r o d u c e t h e compact-open topology,

which i s t h e most n a t u r a l t o p o l o g y on t h e s p a c e of holomorphic mappings. 9.1.

DEFINITION.

Let

C(X;F)

c o n t i n u o u s mappings from a t o p o l o g i c a l s p a c e space

F = C

When

F.

pact-open

topology

Oi-

of

denote t h e v e c t o r space

we shall w r i t e

cfX;g/

X

all

a Banach

into

= C ( X ) . The comis t h e

t o p o l o g y of c o m p a c t c o n v e r g e n c e

l o c a l l y convex t o p o l o g y t h e seminorms o f t h e form

C ( X ; F ) which i s g e n e r a t e d

on

Tc

f

s u p II f ( d l l ,

+

where

K

by

varies

XE K

among a l l compact s u b s e t s o f 9.2.

DEFINITION.

i f a set

A

A t o p o l o g i c a l space

i s open whenever

X

C

X.

each compact s u b s e t

K of

X

i s s a i d t o be a k-space

i s open

A n K

in

K

for

X.

Every f i r s t c o u n t a b l e s p a c e i s a k-space. Every

9 . 3 . EXAMPLES.

l o c a l l y compact space i s a k-space. The v e r i f i c a t i o n of t h e s e examples, as w e l l as

the

proof

of t h e f o l l o w i n g lemma, a r e l e f t as e x e r c i s e s t o t h e r e a d e r . 9.4.

Let

LEMMA.

X

be a k - s p a c e and l e t

poZogicaZ s p a c e . T h e n a m a p p i n g

o n l y if

9.5.

f

1

K

f :X

+

Y y

b e a n a r b i t r a r y to-

is c o n t i n u o u s i f and

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

PROPOSITION.

K

of

X.

If X is a k - s p a c e t h e n f C ( X ; F ) , .re) is complete

70

MUJ I CA

f o r e a c h Banach s p a c e

F.

L e t ( f i ) b e a Cauchy n e t i n ( C ( X ; F ) , T ~ ) . Then

PROOF.

i s a Cauchy n e t i n F

x

f o r each

I f we define

X.

E

(fi(xll f : X

+

F

f l x l = Z i m f . ( x ) t h e n it i s c l e a r t h a t ( f i ) c o n v e r g e s t o f

by 2 u n i f o r m l y on e a c h compact s u b s e t o f t i n u o u s f o r e a c h compact s u b s e t k-space w e c o n c l u d e t h a t

9.6.

DEFINITION.

of

K

IK

f

Hence

X.

and s i n c e

X ,

con-

is

is

X

A topological space

X

i s J,ernicotryact o r countm

abZe a t i n f i n i t y i f t h e r e e x i s t s a sequence ( K n j n Z I

X

p a c t s u b s e t s of

such

c o n t a i n e d i n some

9.7.

that

each

compact

com-

of

of

subset

X

is

Kn.

Each open set

EXAMPLE.

a

is continuous.

f

U

Crn

C

i s hemicompact.

Indeed

i t s u f f i c e s t o c o n s i d e r t h e compact sets K n = I x E U : 11x11 5 n

and

dU(x)

2/n).

The f o l l o w i n g r e s u l t i s c l e a r .

9 . 8 . PROPOSITION. I f X i s a h e m i c o m p a c t s p a c e t h e n ( c ( X ; F I , -cc) i s m e t r i z a b Z e f o r e a c h Banach s p a c e F . Let

be a t o p o l o g i c a l s p a c e and l e t

X

Then as u s u a l from

X

into

Fx F.

F be a Banach space.

denotes t h e vector space

of

all

mappings

The t o p o Z o g y o f p o i n t w i s e c o n v e r g e n c e i s

l o c a l l y convex t o p o l o g y seminorms of t h e form

T

P

f

+

on

Fx

the

which i s g e n e r a t e d by t h e

s u p II f(x:IIl, where

A

varies

among

XE A

X. The t o p o l o g y o f p o i n t w i s e convergence

a l l f i n i t e s u b s e t s of on

9.9.

i s n o t h i n g b u t t h e Tychonoff p r o d 3 x t t o p o l o g y .

Fx

DEFINITION.

Let

X be a t o p o l o g i c a l s p a c e and l e t

F

be

a Banach s p a c e . (a) each

a

F

A family E

X

and

E

C

> 0

Fx

i s s a i d t o be c q a i c o n t i i / i m i ~ si f f o r

t h e r e i s a neighborhood

V

of

a

in X

71

HOLOMORPHIC MAPPINGS

(b) each

a

- ffaIll 5

11 f ( x I

such t h a t

E

for all

x

E

V

and

f E F.

F C F x i s s a i d t o be ZocaZZy bounded i f for X t h e r e are a neighborhood V of a i n X and a c > 0 s u c h t h a t II f(x)II 5 c for all x E V and

A family E

constant f E F.

The proof of t h e f o l l o w i n g lemma i s l e f t as a n e x e r c i s e t o t h e reader.

9.10. LEMMA. L e t X b e a topoZogicaZ s p a c e and l e t F be a Banach s p a c e . If a f a m i l y F c F' is e q u i c o n t i n u o u s (resp. ZocaZZy bounded) then the cZosure F of F for t h e topoZogy of p o i n t w i s e c o n v e r g e n c e is e q u i c o n t i n u o u s ( r e s p . ZocaZZy b o u n d e d ) as weZZ.

9.11.

X b e a topoZogicaZ s p a c e and Zet F b e a Banach s p a c e . Then t h e t o p o Z o g y of c o m p a c t convergence and the topoZogy of p o i n t w i s e c o n v e r g e n c e i n d u c e t h e same topoZogy on each e q u i c o n t i n u o u s s u b s e t of C I X ; F I . PROPOSITION.

Let

F be a n e q u i c o n t i n u o u s s u b s e t of C ( X ; F I . W e always have t h a t -rP 5 T ~ and t o show t h a t t h e s e two t o p o l o g i e s co, i n c i d e on F l e t K be a compact s u b s e t o f X and l e t E > 0 . S i n c e F i s e q u i c o n t i n u o u s e a c h p o i n t a E K h a s aneighborhood Va such t h a t IIfIz.) - f i a l l l 5 E f o r a l l x E Va and f E F . S i n c e K i s compact t h e r e i s a f i n i t e s e t A C K s u c h t h a t K C V { V a : a E A ? . Whence i t follows t h a t PROOF.

Let

s u p II f izc)II XE K

2

s u p 11 f I z ) l l x€ A

+

E

F. B y a p p l y i n g t h i s argument t o t h e s e t F - F (which i s also e q u i c o n t i n u o u s ) w e can f i n d a f i n i t e s e t B C K

f o r all

f E

such t h a t

for a l l

f, g E

F.

I t follows t h a t

MUJ I CA

for each

fo E

F and the proof is complete.

Now it is easy to prove A s c o l i ' s Theorem. 9.12. THEOREM.

Let

X

b e a t o p o Z o g i c a l s p a c e . Then e a c h e q u i -

c o n t i n u o u s , p o i n t w i s e bounded s u b s e t of compact i n

ClX)

is

re2ativeZy

C l X ) f o r t h e compact-open t o p o Z o g y .

PROOF. Let F be an equicontinuous, pointwise bounded subset of C ( X i , and let 'i denote the closure of F in ex. Then F is clearly pointwise bounded, and therefore compact in C A by Tychonoff's product theorem. Now, the set 'i is equicontinuous by Lemma 9.10, and hence the product topology and the compactopen topology coincide on F by Proposition 9.11. Thus F is a compact subset of ( C ( X ) , T ~ ? ) and the proof is complete. After establishing some topological properties of the spaces of continuous mappings, we devote our attention to the spaces of holomorphic mappings. 9.13. PROPOSITION. I f U is an o p e n s u b s e t of E t h e n K(U;FI is a c l o s e d v e c t o r s u b s p a c e of i C ( U ; F . J , T,). I n particular (JC(U;FI, -re) i s complete.

PROOF. The proposition is essentially a restatement of Exercise 8.A. Let If . I be a net in J C ( U ; F I which converges to a 2 mapping f E C i U ; F ) for the compact-open topology. Given a E U, b E E and $ E F ' set gilXl = $ o f i ( a + Ab) and g(h) = $ o f ( a i Ab) for every X E A = { A E g : a + hb E U}. Then each g i is holomorphic on A and the net ( g i ) converges to g uniformly on each compact subset of A . By the well known theorem of Weierstrass for holomorphic functions of one complex variable, the function g is holomorphic on A. Then it follows from Theorems 8.7 and 8.12 that f E K I U ; F ) . The last assertion in the proposition follows from Proposition 9.5.

73

HOLOMORPHIC MAPPINGS

COROLLARY.

9.14.

U

If

a n o p e n s u b s e t of @

7 : s

n

t h e n (JC(U;F),T~)

i s a Frechet s p a c e . 9.15.

PROPOSITION.

F

family

JC(U;FI

C

U b e a n o p e n s u b s e t of

Let

E . The? for each

the f o l l o w i n g c o n d i t i o n s a r e e q u i v a l e n t .

(a)

F

is b o u n d e d i n

(b)

F

is l o c a l l y b o u n d e d .

(c)

F

i s e q u i c o n t i n u o u s and p o i n t w i s e b o u n d e d .

(JC(U;FI,

T ~ ) .

( a ) * ( b ) : I f F i s n o t l o c a l l y bounded t h e n w e c a n a E U , a s e q u e n c e (f,) C F and a s e q u e n c e ( a n ) s u c h t h a t / l a n - all < l / n and Ifnlan) 1 > n for every

PROOF.

find a point C

U

n. I f w e set

then

K

i s a compact s u b s e t of

unbounded on (b)

K. Hence

U

and

the

sequence

(f,)

is

F i s n o t bounded i n (JCfU; F1, . c c l .

F is l o c a l l y bounded, t h a t i s uniformly U. Then F i s u n i f o r m l y bounded o n e a c h compact s u b s e t of U, F i s bounded i n I J C I U ; F I , T ~ )

4

(a):

Assume

bounded on a s u i t a b l e n e i g h b o r h o o d of e a c h p o i n t o f clearly that is (b)

3

(c):

If

F i s l o c a l l y bounded t h e n

and

f o r every

c

x

E

;0

be such t h a t

B ( a ; r l and

f

E

F.

Then

it follows

Cauchy i n e q u a l i t i e s t h a t m

~ ~ f i z- i f i a i l l

5

Z: I I p r n f ( a ) ( x m=l

c IIx < -

is obviously

F i s equicontinuous let a E U , B(a;rl C U and I1 f(x)ll < c

p o i n t w i s e bounded. T o show t h a t

r > 0

F

r-

- all

112-

all

U ~ I I

from

the

74

MUJ I CA

for every (c) is

c

x E Blu;ri

-

f E F.

and

0

c + I

-

i s equicontinuous.

a E U. S i n c e F i s p o i n t w i s e bounded there such t h a t I1 flail1 5 c f o r e v e r y f E F . S i n c e F i s

(b):

Let

e q u i c o n t i n u o u s t h e r e i s a neighborhood

IIf(x)

F

Hence

5

ffa)ll

for a l l

1

x

E

f

a in

F . Then

U such t h a t Ilf~'xiII 5

LC

E

V

and

f

E

F , completing t h e p r o o f .

V

and

V of

for all

By combining A s c o l i ' s Theorem 9 . 1 2

E

and P r o p o s i t i o n s

9.13

and 9.15 w e o b t a i n a t once t h e f o l l o w i n g r e s u l t , which e x t e n d s t h e c l a s s i c a l M o n t e 1 's T h e o r e m .

9.16. PROPOSITION. b o u n d e d s u b s e t of

Let lX(UI,

U b e a n o p e n s u b s e t of E . ~ i) s r e Z a t i v e 2 y c o m p a c t .

Then

each

T

EXERCISES

Show t h a t each c l o s e d s u b s p a c e of a k-space i s a k-space.

9.A.

Show t h a t e a c h

a Hausdorff

apen s u b s p a c e of

k-space

is

a

k-space. 9.B.

UI

Let

and

K

b e a compact s u b s e t of a Hausdorff s p a c e b e open s u b s e t s of

U2

compact sets

KI

C

and

Ul

K2

such t h a t

X C

U

2

K

such t h a t

X.

Let

u U2 ' Find K = K1 u K2 . CUI

U 2 b e two open s u b s e t s of E . Show t h a t t h e s p a c e ( X ( U l U U 2 ) , r c l can be c a n o n i c a l l y i d e n t i f i e d w i t h a c l o s e d v e c t o r subspace of t h e p r o d u c t ( M I U a ), T ~ x) (X(UzI,r e ) . 9.C.

Let

Ul

and

G e n e r a l i z e t h i s t o an a r b i t r a r y f a m i l y ( U i l i E I of

9.D.

of open subsets

E. L e t (xi)

be a n e t

in a

X tvith

t o p o l o g i c a l space

the

p r o p e r t y t h a t e v e r y s u b n e t of (xi) h a s a s u b n e t whizh c o n v e r g e s t o a fixed point 9.E.

Let

x. Show t h a t

(J:?:)

converges t o x.

U be a connected open s u b s e t of

bounded sequence i n / X ( U ) , T ~ ) and suppose ( f n l l t ' ) l converges i n

6

f o r every p o i n t

E.

that x

Let

the

(f,.' !>e a sequence

i n a nonvoid

open

HOLOMORPHIC MAPPINGS

set

V

C

(a)

U. U s i n g M o n t e l ' s Theorem 9.16 a n d E x e r c i s e f ( x ) = lim f (xi e x i s t s €or e v e r y n

that the l i m i t (b)

75

Show t h a t

compact s u b s e t of

converges to

(f,)

f

show

9.D.

x

uniformly

E

U.

on

each

U.

T h i s r e s u l t e x t e n d s t h e classical V i t a l i ' s Theorem. 9.F.

Let

U be a n o p e n s u b s e t of a s e p a r a b l e B a n a c h s p a c e

E.

U s i n g C a n t o r ' s d i a g o n a l p r o c e s s show t h a t e a c h b o u n d e d sequence i n ( X ( U ) , T ~ )h a s a c o n v e r g e n t s u b s e q u e n c e .

This sharpens

the

c o n c l u s i o n i n M o n t e l ' s Theorem 9 . 1 6 . 9.G.

Show t h a t i f

i s a n o p e n s u b s e t of

I/

i s a complemented s u b s p a c e o f t h a t for each

a

E

p r o j e c t i o n from ( K ( u ; F ) , 9.H.

Show t h a t i f

Show t h a t i f

Banach s p a c e

~

-+

.

T,)

m

and

hVo

E

t

More p r e c i s e l y , show

P"f(a)

o) n t o I P ( " E ; F / ,

t h e n (P(mK;F), T ~ ) is a

continuous

T ~ ) .

E

t h e n t h e mapping

E B.

i s an open s u b s e t o f a f i n i t e dimensional

U

i:' t h e n t h e mapping

i s c o n t i n u o u s €or e a c h

9.J.

T

f

U i s a n o p e n s u b s e t of

i s continuous f o r each 9.1.

(K(U;F),

t h e mapping

U

E

Show t h a t i f

s i o n a l Banach s p a c e

I/

DJ0.

rri F

i s an open s u b s e t of an i n f i n i t e

K , and i f

i s n o t c o n t i n u o u s for a n y

rrt

E

F # {O},

dimen-

t h e n t h e mapping

M. F u r t h e r m o r e , show t h a t i f

i:'

76

MUJ I CA

s a t i s f i e s t h e hypothesis i n Proposition rn E lN

7.15

o n e c a n even f i n d a sequence ( f y L ) i n

v e r g e t o zero i n

( J C ~ U P; ? E ; F I

d o e s n o t con-

)

), T ~ ) .

A l o c a l l y c o n v e x s p a c e i s s a i d t o be b a r r e l l e d

9.K.

each

such t h a t

JCIU;FI

c o n v e r g e s t o zero i n ( J C ( U ; F ) , T ~ )b u t ( P m f

(f,)

for

then

c l o s e d , convex, b a l a n c e d , a b s o r b i n g set i s

if

each

a neighborhood

of

zero. (a)

Using t h e

Category

Baire

Frgchet space i s b a r r e l l e d . relled i f space

Theorem

Conclude t h a t

show

that

each

is

bar-

(JE(U;FI,T~)

U is an open s u b s e t of a f i n i t e dimensional

Banach

E.

(b)

Show t h a t i f

m e n s i o n a l Banach s p a c e

U i s a n o p e n s u b s e t of a n i n f i n i t e d i E l and i f

F # 101,

then f o r each n E U

t h e set

i s a closed, convex, b a l a n c e d , a b s o r b i n g s u b s e t o f b u t i s n o t a neighborhood of zero.

( 3 C ( U ; F ) , T~:),

is

Hence I J C ( U ; F I , -ri,l

not

barrelled.

NOTES AND COMMENTS Most o f t h e r e s u l t s i n C h a p t e r 11 h a v e b e e n

known

l o n g t i m e a n d can a l r e a d y b e f o u n d i n t h e book o f E . R.

Phillips I 11

.

Among t h e r e s u l t s t h a t a p p e a r e d

for

H i l l e and

within

l a s t t w e n t y y e a r s w e m e n t i o n Theorem 7 . 1 3 , d i , e t o

a the

Nachbin

L.

I , P r o p o s i t i o n 7.15, d u e t o S. Dineen I 4 I , and P r o p o s i t i o n 9 . 1 6 , n o t i c e d by H . A l e x a n d e r 11 1 . I t w a s a l s o 13. Alcxnnder 1 1 I who showed t h a t (ZftUi, I(,) i s n e v e r b a r r e l l e d when LI i s a n 12

open s u b s e t o f a n i n f i n i t e d i m e n s i o n a l Banach s p a c e ,

a result

t h a t w a s l e f t t o t h e r e a d e r a s E x e r c i s e 9.K. P r o p o s i t i o n 7.15 h a s an i n t e r e s t i n g s e q u e l , f o r 7.F

and

7.G

that

each

raised

Indeed,

we

separLib1.e,

or

a n a t u r a l q u e s t i o n i n t h e t h e o r y of Banach s p a c e s .

know from E x e r c i s e s

it

77

HOLOMORPHIC MAPPINGS

r e f l e x i v e , i n f i n i t e d i m e n s i o n a l Banach s p a c e s a t i s f i e s t h e hypothesis i n Proposition 7.15.

I t i s t h e n n a t u r a l t o a s k whether

e v e r y i n f i n i t e d i m e n s i o n a l Banach s p a c e s a t i s f i e s t h e hypothesis i n Proposition 7.15.

T h i s q u e s t i o n w a s answered i n t h e

affir-

mative by B. J o s e f s o n [ 2 1 , and i n d e p e n d e n t l y b y A . N i s s e n z w e i g [ 1]

.

T h i s i s a deep r e s u l t and t h e i n t e r e s t e d r e a d e r i s refer-

r e d t o t h e o r i g i n a l p a p e r s of B. J o s e f s o n [ 2 I and A. Nissenzweig

[ l ] , o r t o t h e r e c e n t book of J. D i e s t e l [ l ] , f o r a proof o f t h i s theorem. Many of t h e r e s u l t s i n S e c t i o n s 5,7 and 8 can b e found t h e books of L. Nachbin [ 1 ]

,

[ 2

J

and

T.

Franzoni

and

in E.

V e s e n t i n i 111. Our b r i e f p r e s e n t a t i o n of t h e Bochner i n t e g r a l i n S e c t i o n 6 f o l l o w s e s s e n t i a l l y t h e book of J. D i e s t e l

[

Our p r e s e n t a t i o n o f t h e compact-open t o p o l o g y i n S e c t i o n 9

11. is

q u i t e s t a n d a r d and can be found f o r i n s t a n c e i n t h e book of S. Willard [ 11. For t h e p r o p e r t i e s o f holomorphic mappings between l o c a l l y convex s p a c e s t h e r e a d e r i s r e f e r r e d t o t h e books of M. [

1 1 , P . Noverraz [ 3 I ,

Colombeau

I

1]

.

G.

H e r d

Coeurs [ 1 1 , S . Dineen [ 5 ] and J. F.

CHAPTER I11

DOMAINS OF HOLOMORPHY

1 0 . DOMAINS O F HOLOMORPHY

I n t h i s s e c t i o n w e i n t r o d u c e t h e n o t i o n s o f domain lomorphy a n d domain

of e x i s t e n c e , and s t u d y t h e i r

of ho-

elementary

properties. W e b e g i n by p r e s e n t i n g some e x a m p l e s t o m o t i v a t e t h e d e f i n i t i o n s . A s i n t h e p r e c e d i n g c h a p t e r a l l Banach spces considered

w i l l b e complex. EXAMPLE.

10.1.

2

and

V

U and

If

V

a r e t w o o p e n sets i n

connected, then t h e r e i s a function

V

h a s no h o l o m o r p h i c e x t e n s i o n t o Since

PROOF.

the function

flzl = ( z

f E J C l U ) d e f i n e d by

holomorphic e x t e n s i o n t o

with

U

V.

i s connected t h e r e is a p o i n t

V

C

f E J C ( U ) vhich

-

a

V n aU. Then

E

has

a)-’

no

V.

For holomorphic f u n c t i o n s of

n

2

2

variables the

situa-

t i o n i s e n t i r e l y d i f f e r e n t , as t h e f o l l o w i n g e x a m p l e shows. 10.2.

where

EXAMPLE.

c’

-q

r . 3

R .

a unique e x t e n s i o n

figure i n PROOF.

2 D = A (0;R)

Let

5

m

-

for

f E JC(D).

and l e t

j = 1,2.

Then e a c h

f

E

X(H)

has

The p a i r (H,DI i s c a l l e d a Hartogs

c2.

Choose

pI

with

rl < p 1 < RI .Given 79

f

E

at(f1) define

80

MUJ 1 C A

f o r every

i n the polydisc

z

(cl -

A f t e r expanding

z,l

i n t e g r a t i o n shows t h a t each

g

-1

= A 2 ( D ; R ' ) where R ' = ( p l , R 2 ! .

D'

. i n powers o f

f i x e d . On t h e o t h e r h a n d ,

z2

t h e i n t e g r a l s i g n w e see t h a t

zI

z 2 f o r each

zI, a term

z1

by d i f f e r e n t i a t i o n

for

under

i s a holomorphic f u n c t i o n

g

fixed. Since g

term

by

is a holomorphic f u n c t i o n of

of

i s c l e a r l y l o c a l l y bounded, a n

a p p l i c a t i o n o f Lemmas 8.8 and 8.3 shows t h a t

g

by t h e Cauchy I n t e g r a l Formula for h o l o m o r p h i c

E

Now,

7C(D').

functions

of

o n e v a r i a b l e , w e have t h a t g l z l = f ( z l f o r e v e r y z E C 2 w i t h I z 2 I < r 2 , and t h e r e f o r e f o r e v e r y z E D' n R, I z l I < PI and since

D ' n 11

i s c o n n e c t e d . Then t h e f u n c t i o n

= f on H

f i n e d by

and

7

= g

-

f E ?C(D)

de-

on D ' i s t h e r e q u i r e d exten-

s i o n . The u n i q u e n e s s o f t h e e x t e n s i o n i s c l e a r . T h i s example m o t i v a t e s t h e f o l l o w i n g d e f i n i t i o n . 10.3.

DEFINITION.

set V o f

E

Let

containing

U be an open s u b s e t of

U i s s a i d t o be a h o l o m o r p h i c extension

o r hoZornorphic c o n t i n u a t i o n of

7

extension

E

E . An opensub-

U i f each

f

E

M ( U ) has a unique

K(v).

W e want t o s t u d y t h o s e open sets

U i n E which a r e i n saw

s e n s e t h e l a r g e s t common domains o f d e f i n i t i o n f o r a l l the func-

tions

f

E

XtUl.

These

open

sets w i l l be

holomorphy. How s h o u l d w e d e f i n e domains o f

called

domains o f

holomornhy?

We

might be i n c l i n e d t o d e f i n e a domain of holomorphy a s a n

open

s e t i n F: which h a s no p r o p e r h o l o m o r p h i c c o n t i n u a t i o n , but such

a d e f i n i t i o n would t u r n o u t t o b e i n a d e q u a t e . A c t u a l l y , s u c h a d e f i n i t i o n would be a d e q u a t e i f w e e n l a r g e d t h e c l a s s

of

ob-

j e c t s u n d e r d i s c u s s i o n by r e p l a c i n g open s e t s i n Banach s p a c e s

b y Riemann domains o v e r Banach s p a c e s . W e s h a l l i n d e e d do t h i s i n Section 52, but f o r the t i m e being

we

s h a l l restrict

s t u d y t o domains o f holomorphy i n Banach s p a c e s , and

case t h e d e f i n i t i o n i s t h e f o l l o w i n g .

in

our this

81

DOMAl NS OF HOLOMORPHY

U i n E i s s a i d t o b e adomain of h o l o m o r p h y i f t h e r e are no open sets V and W i n E w i t h t h e DEFINITION. An open set

10.4.

following p r o p e r t i e s :

i s connected and n o t c o n ta in e d i n

(a)

V

(c)

For e a c h

U.

I

f

unique) such t h a t If

f E JclUl

= f

on

there exists W.

i s a domain o f holomorphy

U

(necessarily

f € Jc(V)

then

U

clearly

has

110

proper holomorpic c o n t i n u a t i o n , b u t t h e converse i s n o t t r u e i n general.

1 0 . 5 . PROPOSITION.

Let

U

b e a n o p e n s u b s e t of E . Assume t h a t

for e a c h s e q u e n c e ( a . ) i n U w h i c h c o n v e r g e s t o a p o i n t a 3

there e x i s t s a function Then

f E J c t U l w h i c h i s u n b o u n d e d on

E

aU

la .1. 3

U i s a d o m a i n of hoZomorphy.

PROOF.

Suppose

i s n o t a domain o f holomorphy, and let V and

U

W b e t w o open s e t s s a t i s f y i n g t h e c o n d i t i o n s i n D e f i n i t i o n 10.4.

By t h e I d e n t i t y P r i n c i p l e w e may assume t h a t in

W

i s a connected

W

V . By E x e r c i s e 1 0 . F t h e r e i s a s e q u e n c e ( a .) 3 which c o n v e r g e s t o a p o i n t a E V n aU n aW. By h y p o t h e -

component o f

U

fi

sis t h e r e is a f u n c t i o n hand

f ( a .) = f l a

10.6.

COROLLARY.

3

f

E

KCfUl which i s unbounded

F ( a . ) converges t o

Then on one hand

3

.)

3

i s unbounded.

E v e r y o p e n s e t in

?(a),

and

on

fa.). 3

on t h e 2 t h e r

This is impossible. &

i s a domain of holomorphy.

L e t U be a n open s e t i n @ a n d l e t f a . ) b e a sequence 3 U which c o n v e r g e s t o a p o i n t a E aV. Then t h e f u n c t i o n f ( z ) = ( z - a )- 1 i s h o l o m o r p h i c on U and unbounded o n ( a . ) .

PROOF. in

3

10.7.

COROLLARY.

ho Zomorphy PROOF.

E'uery c o n v e x o p e n s e t i n

E

i s a domain

of

.

Let

U be a convex o p e n s e t i n E

and l e t

(a.) 3

be

a

82

MUJ I CA

a E aU.By t h e Hahnsuch t h a t R e q ( x l Req(a) f ( x ) = [ q(x - a ) ]- I i s ho-

sequence i n U which converges t o a p o i n t Banach Theorem t h e r e e x i s t s

P E E? f o r e v e r y x E U. Then t h e f u n c t i o n lomorphic on U and unbounded on ( a . 1 . 3

If

U i s a domain of holomorphy t h e n f o r e a c h p a i r of open s e t s V and W s a t i s f y i n g t h e c o n d i t i o n s ( a ) and (b) i n Eefinibe t i o n 1 0 . 4 t h e r e e x i s t s a f u n c t i o n f E s C ( U ) , which c a n n o t 3: E K ( V ) such extended t o V i n t h e s e n s e t h a t t h e r e i s no t h a t 7 = f on W. I n g e n e r a l t h e f u n c t i o n f depends on t h e f for open s e t s V and W. I f w e can t a k e t h e same f u n c t i o n a l l V and W t h e n w e s h a l l s a y t h a t U i s a domain of e x i s t e n c e . More p r e c i s e l y , we have t h e f o l l o w i n g d e f i n i t i o n . 10.8. DEFINITION. An open s e t U i n E i s s a i d t o be t h e domain of e x i s t e n c e of a f u n c t i o n f € J C ( U ) i f t h e r e are no open sets

I/

and

i n E and no f u n c t i o n

W

f

E X(V)

w i t h t h e follai-

ing properties: (a)

V

(c)

7

i s connected and n o t c o n t a i n e d i n

= f

on

U.

W.

C l e a r l y e v e r y domain o f e x i s t e n c e i s a domain of En t h e two phy. The n e x t theorem shows t h a t i n

holomorconcepts

coincide. 10.9.

THEOREM.

E v e r y d o m a i n of h o l o m o r p h y i n

En

i s a domain

of existence. W e s h a l l p r e s e n t l y g i v e an e x i s t e n c i a l proof of Theorem 10.9

we s h a l l g i v e a c o n s t r u c t i v e proof of a theorem of H . C a r t a n and P . T h u l l e n , which i m p l i e s Theorem 1 0 . 9 . The key t o t h e first proof of Theorem 1 0 . 9 i s t h e f o l l o w i n g l e m m a .

based on t h e B a i r e Category Theorem.

Let

10.10. LEMMA. Banach s p a c e

E.

In

t h e next s e c t i o n

U be a domain of h o l o m o r p h y i n a s e p a r a b l e

Let

F d e n o t e t h e s e t of

a l l f u n c t i o n f E JC(UI

83

DOMAINS OF HOLOMORPHY

f. T h e n F i s a

U i s n o t t h e domain o f e x i s t e n c e o f

such that s e t of

the f i r s t category i n (JCIUl,

PROOF.

For e a c h p a i r of open sets

T ~ ) .

V

and

satisfying

W

c o n d i t i o n s ( a ) and ( b ) i n D e f i n i t i o n 1 0 . 4 l e t t h e v e c t o r s u b s p a c e of a l l

f

E

the

X ( U , V, W) d e n o t e

7

x(Ul f o r which t h e r e e x i s t s

x(V/ ( n e c e s s a r i l y u n i q u e ) such t h a t 7 = f on W. S i n c e U i s a domain o f holomorphy, J c t U , V , W ) i s a p r o p e r v e c t o r subspace

E

of

x(Ul. For e a c h

m E mT

let

K m ( U , V, WI d e n o t e t h e set of a l l

such t h a t 171 5 m on V. W e claim t h a t JC,(U,V,Wl be a n e t i n i s a c l o s e d s u b s e t of ( X ( U l , r c l . I n d e e d , l e t If,) X m ( U , V, Wl which converges t o some f i n ( K ( U ) , Tc). S i n c e I fi I < m on V f o r e v e r y i an a p p l i c a t i o n o f M o n t e l ' s Theorem f E Jc(U,V,Wl

9 . 1 6 y i e l d s a s u b n e t of

(Fi)

which c o n v e r g e s t o a f u n c t i o n g Whence it f o l l o w s t h a t f E 3Cm (U,V, Wl and f = g , I

i n (JctVl,- r e ) .

and o u r claim h a s been proved. S i n c e v e c t o r subspace of

3C(U/

Jc(U,V,Wl

NOW,

a proper

and h a s t h e r e f o r e empty i n t e r i o r

( J C ( U I , T ~ ) ,w e c o n c l u d e t h a t t h e smaller s e t

c l o s e d , nowhere

is

d e n s e s u b s e t of

xm(U,V,W)

in

is

a

Let

V

( 3 C ( U ) , -re*).

l e t D d e n o t e a c o u n t a b l e d e n s e s u b s e t of

d e n o t e t h e c o l l e c t i o n of a l l open b a l l s

all.

V whose c e n t e r s belong

are r a t i o n a l . L e t P d e n o t e t h e c o l l e c such t h a t V E V and W i s a c o n n e c t e d component of U n V. C l e a r l y P i s c o u n t a b l e , and t o complete t h e p r o o f o f t h e lemma w e s h a l l show t h a t F i s t h e union of t h e s e t s JcmIu,v,wl w i t h I V , W l E P and m E IN. L e t f E F. Then w e can f i n d open sets V and W i n E and a f u n c t i o n f E X l V ) s a t i s f y i n g t h e c o n d i t i o n s (a), ( b ) , ( c ) i n D e f i n i t i o n 10.8. Without l o s s of g e n e r a l i t y w e may assume t h a t W i s a connecV . By E x e r c i s e 10.F t h e r e i s a p o i n t a t e d component of U E v n a u n aw. Choose V ' E V such t h a t a E V ' C V and 171 i s bounded, by rn s a y , on V ' . S i n c e a E a W t h e r e i s a p o i n t b E w n v'. L e t W ' d e n o t e t h e c o n n e c t e d component o f U n V' which c o n t a i n s b . Then I V ' , W ' l E P and f E 3Cm ( U , V ' , W ' ) , c o m t o D and whose r a d i i

t i o n of all p a i r s ( V , W )

I

pleting t h e proof. PROOF O F THEOREM 10.9.

Let

U be a domain o f holomorphy i n

61".

84

MUJ I CA

Then ( J c t U l , ~ ~ i) s a F r g c h e t s p a c e a n d , by Lemma 1 0 . 1 0 , t h e s e t

f 6 J c t U ) s u c h t h a t U i s t h e domain o f e x i s t e n c e of f , i s o f t h e s e c o n d c a t e g o r y i n ( J c ( U ) , T,) , and i s i n p a r t i c u l a r

of a l l

nonempty

.

Theorem 1 0 . 9 d o e s n o t g e n e r a l i z e t o a r b i t r a r y Banch spaces. I n d e e d , A . H i r s c h o w i t z [ 1 ] h a s g i v e n a n example of a nonsep a r a b l e Banach s p a c e whose open u n i t b a l l i s n o t a domain of e x i s t e n c e . But t h e f o l l o w i n g p r o b l e m r e m a i n s open. 10.11. PROBLEM.

Let

E be a separable Banach s p a c e . Is

domain o f holomorphy i n

E

every

a domain of e x i s t e n c e ?

In Sectjon 45we s h a l l present a p a r t i a l positive

solution

t o Problem 10.11.

EXERC ISES

10.A.

Let

where

n

2

D = An(O;R)

and

and l e t

0 < r

< R < w j j D i s a holomorphic c o n t i n u a t i o n of 2

a Hartogs f i g u r e i n

j = I,.

for H.

z" ( a ; r ) .

Showthat

The p a i r (H,DI i s called

Cn.

10.B. L e t V be a c o n n e c t e d open s e t i n -n A ( a ; r ) be a compact p o l y d i s c c o n t a i n e d i n V \

. .,n.

Show t h a t

V

8

with

n

and

V,

2

2. Let

let

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

U = U.

10.C. L e t U and V b e t w o open s u b s e t s o f E w i t h U C V . Show t h a t V i s a h o l o m o r p h i c c o n t i n u a t i o n of U i f and o n l y i f e a c h f E X ( U ) h a s a n e x t e n s i o n f E K ( V ) , and e a c h c o n n e c t e d compon e n t of V c o n t a i n s p o i n t s of U. be a h o l o m o r p h i c c o n t i n u a t i o n o f an open s e t

10.D.

Let

in

Using E x e r c i s e 8 . 1 show t h a t i f

E.

then each

V

f

E X(U;F)

F

i s any Banach

has a unique extension

-

f

E

JCIV;FI.

U

space

DOMAINS Show t h a t an open s e t

10.E.

85

OF HOLOMORPHY

U i n E i s a domainofholomorphy

( r e s p . a domain o f e x i s t e n c e ) i f and o n l y i f e a c h connected cam-

U i s a domain of holomorphy ( r e s p . a domain o f e x i s -

ponent o f tence)

.

U and V be open s u b s e t s o f E , w i t h V c o n n e c t e d U. L e t W be a c o n n e c t e d component o f U n V . Show t h e e x i s t e n c e o f a p o i n t a E V n aU n aW. 10.F.

Let

and n o t c o n t a i n e d i n

L e t (H,D) be a Hartogs f i g u r e i n

10.G.

E).

(D \

Show t h a t

U h a s no p r o p e r

8

2

,

and l e t

holomorphic

U = H u

continuation,

U i s n o t a domain o f holomorphy.

but

L e t Ui be a domain of holomorphy i n E f o r e a c h i €I. 10.H. Show t h a t t h e s e t U = i n t n Ui i s a domain of holomorphy a s -LEI

well. 10.1.

q E X i E ) and l e t

Let

U = cp-'(A)

the set

A

b e an open s e t i n

i s a domain o f holomorphy i n

8 . Show that E.

10.J. Given open sets A ] , . . . , A n i n 8 show t h a t t h e p r o d u c t U = A l x . . . x A n i s a domain of holomorphy i n tn. 10.K.

Show t h a t an open s u b s e t U o f

tence of a function f o r every 10.L.

Let

x

E

f

E

JC(UI

i s t h e domain of exis-

5 duix)

rcf(xl

i f and o n l y i f

U.

U b e an open subset o f

pose t h a t f o r e a c h b a l l connected component of

E , and l e t

B ( a ; r ) with center

f i c i e n t l y small r a d i u s , t h e function of e x i s t e n c e of

E

a

f €

E

X(U).

Sup-

and

suf-

aU

i s unbounded on e a c h U n B ( a ; r l . Show t h a t U i s t h e domain f

f.

11. HOLOMORPHICALLY CONVEX DOMAINS I n t h i s s e c t i o n w e i n t r o d u c e t h e n o t i o n of holomorphicconv e x i t y and e s t a b l i s h a c l a s s i c a l theorem o f H.

Cartan

and

P.

86

MUJ I CA

T h u l l e n , which c h a r a c t e r i z e s domains of holomorphy i n terms of holomorphic c o n v e x i t y . To m o t i v a t e t h e d e f i n i t i o n of h o l o m o r p h i c a l l y convex domains

w e b e g i n by p r e s e n t i n g some p r o p e r t i e s of convex sets. 11.1. PROPOSITION.

Let

c o n t i n u o u s a f f i n e forms o n

A^ceEr

d e n o t e t h e v e c t o r s p a c e of a l l

C @ E'

A b e a s u b s e t of E and l e t

let

E,

denote t h e s e t

Then: The s e t

(a)

i s a l w a y s c o n v e x and c l o s e d , and i n

ACeE,

-

p a r t i c u l a r contains the closed, convex h u l l

-

A

(b)

I f

A

i s bounded t h e n

AC

c o ( A ) of A .

@

E , = co(A).

(c) If A i s bounded ( r e s p . c o m p a c t ) t h e n ed ( r e s p . c o m p a c t ) a s w e l 2 .

-

W e s h a l l prove t h a t

PROOF.

C

=(A)

A l l t h e o t h e r a s s e r t i o n s are clear. L e t

Banach Theorem t h e r e e x i s t

p

E

E'

$

and

A^c3Er

is bound-

when A i s bounded.

9 a

Z I A ) . By the HahnE

1R

such

that

R e V ( y ) f o r all x E = ( A ) . S i n c e Q ( ZA () ) is bounded t h e r e i s a d i s c A ( < ; r ) such t h a t P ( Z ( A ) I C z ( < ; r )

ReV(x) < a

f E c CB E' be d e f i n e d by f ( x ) - p ( x ) - 5 s u p l f l 5 sup 5 P < f f y ) , proving

and q ( y ) 9 7 i ( < ; r ) .L e t f o r every

3:

E

If(

Then

E.

A

B A^C

that

y

11.2.

PROPOSITION.

ZG(A)

@El.

F o r an o p e n s u b s e t

U of E

the

foZ:owing

conditions are equivalent: (a)

U

i s convex.

(b)

c U

(c)

z C C B En, U

f o r e a c h compact s e t

K C U.

i s compact f o r e a c h c o m p a c t s e t

K

C

U.

a7

DOMA 1 NS OF HOLOMORPHY

( a ) * ( b ) : I f K i s a compact s u b s e t of U t h e n t h e r e i s a b a l l V = B ( 0 ; r ) such t h a t K + V C U. Hence K c a E I = e o ( K I C c o ( K ) + V = c o ( K + V ) C U. PROOF.

A

(b) * (c):

zgeEl

This i s obvious s i n c e

each compact s u b s e t K

( c ) * ( a ):

of

segment

K = Ix,y).

and set

Let x , y E U

t i o n 11.1 t h e l i n e

1

[z,y

KcaEl.

=

equals

can w r i t e [ z , y ] = A u B , where A K6 a E l are two d i s j o i n t compact sets. S i n c e [ x , y ] conclude t h a t

must be empty. Thus

B

i s compact

for

E.

[x,y]

(-I

C

By

Proposi-

we B=KtBEl\U Hence

U and i s connected w e U and U i s con-

vex. With t h i s m o t i v a t i o n i n mind w e i n t r o d u c e t h e f o l l o w i n g & finition. 11.3.

DEFINITION.

Let

U b e a n open s u b s e t o f

J C ( U I - h u l l of a s e t

(a)

The

(b)

The open s e t

A

C

U

E.

i s d e f i n e d by

U i s s a i d t o b e holomorphically c o n v e x i s compact f o r e a c h compact s e t K C U.

if & C i U l

U b e an open s u b s e t o f

Let

E . W e s h a l l set

d ( A ) = inf d u ( x f

x€A € o r each set clear that Since

-

KJC(u,

-

i s a compact s u b s e t i s c o n t a i n e d i n t h e compact

A C U. I f

K

A

is c l e a r l y closed i n

K~(,,,,

i s compact i f and o n l y i f

du(isccu,’

p h i c a l l y convex i f and o n l y i f

set 11.4.

U w e conclude t h a t > 0 . Thus

U

KJC(UI

i s holomor-

du(K^JCiui) > 0 f o r e a c h compact

K C U. THEOREM.

F o r a n open s u b s e t

U of

Zoving c o n d i t i o n s : (a)

U

i s a domain of e x i s t e n c e .

E

consider the

fol-

MUJ I CA

88 U

(b) A

such t h a t

j

For e a c h s e q u e n c e ( a

(c)

a

E

is t h e u n i o n of a n i n c r e a s i n g s e q u e n c e of open s e t s d u ( ( b j ) x ( uI) > 0 f o r e v e r y j . .)

3

there e x i s t s a function

aU

U w h i c h crnvergss t o a point x(UI w h i c h i s unbounded on

in f

E

(a. ) . d

i s a domain of h o l o m o r p h y .

(d)

U

(el

du(~Xtu)I = d U (KI

(f)

U

f o r e a c h compact s e t

K C U.

i s hoZomorphicaZZy c o n v e x .

Then t h e i m p l i c a t i o n s ( a )

always t r u e . I f

(b) * ( c ) * ( d ) * ( e ) * ( f ) are

=*

i s separable then (a)

E

(b).

( a ) * ( b ) : Suppose U i s t h e domain of e x i s t e n c e of

PROOF.

a

f E X I U ) . C o n s i d e r t h e f o l l o w i n g open sets:

function

B

j

= Ez

E

u

= {x

E

B

: If(z)l <

j3

and

A

j

.

j -

d g (xi > l / j } .

j

m

Then

U =

U

j=1

A

j

and

A

C Aj+l

j

f o r every

j. F u r t h e r m o r e , the

+ E(O;l/j), and j l P m f ( x ) (t)I 5 j f o r e v e r y .c E A j and whence i t f o l l o w s t h a t > 1 / j for every y E t E Z ( O ; l / j l . W e claim t h a t r c f ( y ) function f

i s bounded by

T o show t h i s l e t

x(U)

E

< jm+l

-

y

j on t h e s e t

(zi)K(u).

.

E

it follows t h a t

A

and

(Aj)x(u)

I P l f (yl I

t

E

% ( O ; l / j ) . Since

< s u p I P:f

A.

I

< j . Thus II Pmf (Y

3

f o r e v e r y m and i t f o l l o w s from t h e Cauchy

Formula t h a t

rcf(y) 2 I/j,

as

PTf

asserted.

Hence

the

Hadamard

series

m

Z Pm f l y ) ( t ) m=O

defines a function

f

Y'

holomorphic on t h e

b a l l B(y;Z/jl, and which c o i n c i d e s w i t h f on a neighborhood we o f t h e p o i n t y . S i n c e U i s t h e domain of e x i s t e n c e of f conclude t h a t

B ( y ; l / j ) C U. T h i s shows t h a t

and ( b ) i s s a t i s f i e d .

du((ii),iul) 2 l / j

DOMAl NS OF HOLOMORPHY

89

( b ) * ( c ) : By h y p o t h e s i s U i s t h e union of an sequence of open sets

= (Zj)x(u,

every

j. S e t

= B

L e t ( a .I b e a sequence i n 3

i'

B~

du((Aj)x(uii

such t h a t

Aj

f o r every

increasing >

for

0

and note that (fi.)x(u)

j

3

U which c o n v e r g e s t o a p o i n t

aU. A f t e r r e p l a c i n g ( a . ) and ( B , ) by s u i t a b l e subseqwnces, 3 i f n e c e s s a r y , w e may assume t h a t a 9 B j and a j E B j + c l f o r in

j

every

in

j. S i n c e

a

j

such t h a t

JcfU)

9 B~~ = ( E j ) J C t u ) we can f i n d a sequence (9 .I 3

s u p 19 . I 3

Bi

[9.(a.)l

1

3

f o r every

3

j .

By

t a k i n g s u f f i c i e n t l y h i g h powers of e a c h p j w e c a n i n d u c t i v e l y and f i n d a sequence If.) i n J C ( U ) such t h a t supIfjl 5 2 - j 3

Bj

m

j. whence it f o l l o w s t h a t t h e series

f o r every

v e r g e s u n i f o r m l y on e a c h Bi t o a f u n c t i o n f >

j

f o r every

(d) * (el: L e t

Y

XtU) and

E

fines a function f w i l l coincide with

Proposition 10.5.

U and s e t r = ( a ) =. ( b ) w e s h a l l prove t h a t t h e series P m f ( y ) ( t ) de-

K b e a compact s u b s e t o f

d l i ( K i . By modifying t h e proof of

f

Y

zJCiu)

E

f

on a neighborhood of

and

ixiu,. Given

y E

E B(0;~).

find

E

and t h e r e f o r e

U

C

Then

> 0

m=O B(y;r).

holomorphic on t h e b a l l

K + apt

t

dU(Kx(Ui) = r .

E B(0;r)

choose

B = K

+

fix

NOW,

'iZpt

+

B

= ep

-m

p t

U and w e can is con-

B(0;pc.l

U and f i s bounded, by e s a y , on B . I f t h e n i t follows from t h e Cauchy I n e q u a l i t y 7 . 4 t h a t

sup

f E JC(V)

such t h a t

p > I

tained i n

<

Y

U is

s h a l l conclude t h a t

i s a compact s u b s e t of

such t h a t t h e set

Since f

and s i n c e

y,

we

by h y p o t h e s i s a domain of holomorphy, B(y;r)

conj=; x(U) and l f ( a j l 1

j. T h i s shows ( c ) .

( c ) * ( d ): This i s t h e c o n t e n t of

f o r each

E

I: fj

.

h

€

B(O;d

90

MUJ I CA

Thus f o r each

there exists

t E B(O;r)

E

>

such

0

t h a t the

m

series

2 Pmf(y)(t m=O

+ h ) converges u n i f o r m l y f o r

h

BIO;El.

E

m

T h i s shows t h a t t h e series

2 Pm f ( y ) I t ) d e f i n e s a holomorphic m=O

function f

Y

on t h e b a l l

B ( y ; r ) and t h e proof o f

(d)

=$

(e) i s

complete. S i n c e t h e i m p l i c a t i o n (el t o show t h a t ( b )

=$

( f ) i s o b v i o u s , i t o n l y remains

D b e a count a b l e dense s u b s e t of U. For e a c h x E D l e t B(x) denote t h e l a r g e s t open b a l l c e n t e r e d a t x and c o n t a i n e d i n U , t h a t i s B ( x ) = B ( x ; d U ( x ) / . By modifying t h e p r o o f of ( b ) * (c) w e s h a l l c o n s t r u c t a f u n c t i o n f E X ( U ) which i s unbounded on B ( x ) f o r e v e r y x E D. NOW, l e t (x .I be a sequence i n D w i t h t h e 3 p r o p e r t y t h a t e a c h p o i n t of D a p p e a r s i n t h e sequence (2.) 3 i n f i n i t e l y many t i m e s . By h y p o t h e s i s U i s t h e union of a n i n c r e a s i n g sequence of open s e t s Aj s u c h t h a t d u ( ( b j ) l r i u i )> 0 f o r every

=*

j. set

( a ) when B i s s e p a r a b l e . L e t

B~ = ( 2 j ) x ( u , f o r e v e r y

j . Note t h a t

Blx)

B . f o r e a c h x E D and j E ilv. Hence, a f t e r r e p l a c i n g (B<j) by a subsequence, i f n e c e s s a r y , w e c a n f i n d a sequence ( y3- ) i n f o r every U such t h a t y j E B ( x3. ! , y j 9 B j and y j B i + l j. Then, p r o c e e d i n g as i n t h e proof o f ( b ) * ( c ), w e can cons t r u c t a f u n c t i o n f € K i l l ) such t h a t l f ( y j ) l 2 j f o r e v e r y j. W e claim t h a t f i s unbounded on t h e b a l l Biz) f o r e v e r y x E D. I n d e e d , g i v e n x E D w e can f i n d a s t r i c t l y i n c r e a s i n g sequence (j,) i n JI? such t h a t x = x f o r every k . Hence jk y E B ( x ) f o r e v e r y k and f i s unbounded o n B ( x 1 , as as-

$2

jk s e r t e d . To complete t h e proof w e s h a l l

prove

that

U is

the

domain of e x i s t e n c e of f. I n d e e d , suppose t h e r e e x i s t o p e n sets V and W and a f u n c t i o n E X ( V ) satisfying the conditions i n Definition 10.8. any

r > 0

B ( a ; r ) . Then

Take a p o i n t

E V

n aU

aW

and

consider

B ( a ; 2 r ) C V . Choose a p o i n t x E D n W n

such t h a t du(x)

a

r

and

B l x ) C B ( a ; 2 r ) C V . Since

B(x)

U n V , w e conclude t h a t B i z ) C W. Thus = f i s unbounded on B ( x i , and t h e r e f o r e on B ( a ; B r ) . S i n c e r > 0 can be t a k e n a r b i t r a r i l y s m a l l w e c o n c l u d e t h a t 3 i s n o t l o c a l l y bounded a t a , a c o n t r a d i c t i o n . Hence U i s i s connected and c o n t a i n e d i n

91

DOMAi NS OF HOLOMORPHY

t h e domain of e x i s t e n c e of

f , and t h e proof of t h e theorem i s

complete. Now i t i s e a s y t o prove t h e Cartan-ThuZZen T h e o r e m :

11.5. THEOREM.

For a n open s u b s e t

U of

t h e foIZowing oon-

6'"

d i t i o n s are equivalent:

(a)

U

i s a domain of e x i s t e n c e .

(b)

U

i s a domain of holornorphy.

(c)

U

i s holomorphically convex.

PROOF. hold.

* ( b ) * ( c ) always * ( a ) c o n s i d e r t h e compact s e t s

By Theorem 1 1 . 4 t h e i m p l i c a t i o n s ( a ) To show t h a t ( c )

Kj = { x

E

U

:

llxll 5 j

co

Then

U =

U

and

du(xl

2

l/j}.

0

and Kj C Kj+l

K

f o r every

j . Since

U i s ho-

j=l

Lomorphically convex t h e set 0

I f we set

> 0

A

i

= K

f o r every

i

(ij)xcu i s )compact

f o r every j .

m

then

U =

U

A j,

Aj

C

Aj+l and d U ( ( Aj I J C ( U I )

j=l

j. By Theorem 1 1 . 4 ,

U i s a domain of existence.

Theorem 1 1 . 5 does n o t g e n e r a l i z e t o a r b i t r a r y Banach spaces. I n d e e d , B. J o s e f s o n

[ 1]

h a s g i v e n a n example

of

a holomor-

p h i c a l l y convex open s e t i n a n o n s e p a r a b l e Banach s p a c e

i s n o t a domain of holomorphy. But t h e f o l l o w i n g problem

which

re-

mains open.

11.6.

PROBLEM.

Let

E

b e a s e p a r a b l e Banach s p a c e .

h o l o m o r p h i c a l l y convex open s e t i n

E

Is

every

a domain of e x i s t e n c e , o r

a t l e a s t a domain of holomorphy? I n Section 4 5 w e s h a l l p r e s e n t a p a r t i a l p o s i t i v e t o Problem 1 1 . 6 .

solution

To complete t h i s E e c t i o n w e g i v e two a p p l i c a -

t i o n s of Theorem 1 1 . 4 .

MUJ I CA

92

11.7.

PROPOSITION.

Let

and Z.et

U =

T E L(E;FI

V be a domain of e x i s t e n c e i n -1 T ( V ) . Then:

i s a domain of hoZomorphy.

(a)

U

(b)

I f

E

i s separabZe t h e n

U i s a domain of e x i s t e n c e .

One can r e a d i l y check t h a t

PROOF.

f o r each set

A C U. NOW,

s i n c e V i s a domain

Theorem 1 1 . 4 y i e l d s an i n c r e a s i n g sequence of and a sequence of 0-neighborhoods m

u B

and f B j ) x ( V ,

j=1

j

and

U

u =

let

F,

j

= T

-1

+

V

C

Vj

in

f o r every

V

F

of

j . Set

u

A

j

B V

j

3

j . Then u s i n g (11.1) w e g e t

and

€ o r every

n

j =

= T-1 ( B . 1

A

m

j=l

sets

open

such t h a t

j

( V . ) f o r every 3

existence,

. By

that

Theorem

1 1 . 4 w e may conclude t h a t a r b i t r a r y , and t h a t

U i s a domain of holomorphy if E i s U i s a domain of e x i s t e n c e i f E i s sepa-

rable. Next w e show t h a t i n t h e case of s e p a r a b l e Banach

spaces

t h e c o n c l u s i o n of C o r o l l a r y 1 0 . 7 c a n bc improved as f o l l o w s .

11.8. PROPOSITION.

E v e r y c o n v e x o p e n s e t i n a s e p a r a b l e Banach

s p a c e i s a domain of e x i s t e n c e .

U be a convex open s e t in a s e p a r a b l e Banach space Then U i s t h e union of t h e i n c r e a s i n g sequence of open sets

PROOF. E.

Let

A . d e f i n e d by 3

Using t h e i d e n t i t y

93

DOMAINS OF HOLOMORPHY

f o r a l l a , B 2 0 w i t h a + B = I , w e can see t h a t e a c h A j convex. Then i t f o l l o w s from P r o p o s i t i o n 11.1 t h a t (2.)

JC(UI

J

is C

j . Thus an a p p l i c a t i o n of Theorem 1 1 . 4 c o m p l e t e s t h e p r o o f .

EXERC ISES Show t h a t an open s u b s e t U of E i s h o l o m o r p h i c a l l y con-

ll.A.

vex i f and o n l y i f e a c h c o n n e c t e d component of

U

i s holomor-

p h i c a l l y convex.

ll.B. each

Ui

Let

i = I,

2.

b e a h o l o m o r p h i c a l l y convex open s e t i n E Show

t h a t t h e open s e t

U = U 1 n U,

for

i s holo-

m o r p h i c a l l y convex as w e l l . Given a h o l o m o r p h i c a l l y convex open s e t U i n E

ll.C.

function

f

E

J C ( U l show t h a t t h e open s e t

a V = {x E U: If(z)l< l l and

i s h o l o m o r p h i c a l l y convex as w e l l .

11.D. L e t V b e a n open subset of F , l e t T E 8 l E ; F I and l e t U = T - 1 ( V ) . Show t h a t i f V i s h o l o m o r p h i c a l l y convex t h e n U i s h o l o m o r p h i c a l l y convex as w e l l .

U i b e a h o l o m o r p h i c a l l y convex open s u b s e t of a Banach s p a c e Ei f o r i = 1,2. Show t h a t U l X U, i s a h o l o m o r p h i c a l l y convex open s u b s e t of E l x E 2 . ll.E.

ll.F. for

El

x

Let

L e t Ui b e a domain of e x i s t e n c e i n a Banach s p a c e Ei i = 1,2. Show t h a t U l x U 2 i s a domain of holomorphy i n E2.

If

ll.G.

U i s a convex open s e t i n E show t h a t

= d U f K i f o r e a c h compact set

dU(zceE,)

K C U.

11.H. L e t U b e a h o l o m o r p h i c a l l y convex open s e t i n C n , and l e t l a .) b e a sequence i n U such t h a t e a c h f E J C ( U I is bounded 3

MUJ I CA

94

on (a.). 3

Show t h a t t h e s e t

(a)

8 =

If

E

XfUI

: sup1

f (aj)1 5 1 ) i s

a c l o s e d , convex, b a l a n c e d , a b s o r b i n g s u b s e t o f ( J C ( U ) , T ~ ) . Using Exercise 9 . K

(b)

set K i n U s u c h t h a t (c)

aj

show t.he e x i s t e n c e of

E ,(,?i

f o r every

a

compact

j.

Using P r o p o s i t i o n 1 0 . 5 conclude t h a t

U is a

domain

g i v e an

alteron t h e

of holomorphy. T h i s exercise, t o g e t h e r w i t h Theorem 1 0 . 9 ,

n a t i v e proof of t h e C a r t a n - T h u l l e n Theorem 1 1 . 5 , based F r g c h e t s p a c e p r o p e r t i e s of ( J C ( U ) , T ~ ) .

1 2 . BOUNDING SETS

I n t h i s s e c t i o n w e i n t r o d u c e t h e n o t i o n of b o u n d i n g s e t a n d s t u d y i t s c o n n e c t i o n w i t h domains o f holomorphy and domains of existence.

DEFINITION. L e t U be an open s u b s e t o f E . A s e t i s s a i d t o be a b o u n d i n g s u b s e t of U , o r . JCIUI-bounding, each f E J C ( U ) i s bounded on B . 12.1.

B C

U if

U b e a n open s u b s e t o f E . Then e a c h rel a t i v e l y compact s u b s e t of U i s JCfU)-bounding. Moreover, for e a c h compact s u b s e t K of U t h e s e t K x ( U I i s JCIUI-bounding.

12.2. EXAMPLES.

Let

&

U b e a n o p e n s u b s e t of E , and L e t B U. T h e n , for e a c h i n c r e a s i n g s e q u e n c e ( A . ) of o p e n s u b s e t s o f U w h i c h c o v e r U, t h e r e e x i s t s j s u c h

12.3.

PROPOSITION.

Let

b e a b o u n d i n g s u b s e t of 3

that

B

PROOF.

C

(A^j)JC(u,.

Set

B~

=

(Alj)xtu,

f o r every

3’.

If

j t h e n , a f t e r r e p l a c i n g ( B . ) by a s u i t a b l e 3

n e c e s s a r y , w e c a n f i n d a sequence Iz . ) i n 3

and

xj

E

Bj+]

f o r every

B

R

Eli

f o r each

subsequence,

if

such t h a t x

3’ 9 Bj

j. Then t h e proof of t h e b r p l i c a t i o n

95

DOMAINS OF HOLOMORPHY

( b ) * ( c ) i n Theorem 1 1 . 4 y i e l d s a f u n c t i o n

f

E X t U ) which

unbounded on ( x . ) , and t h e r e f o r e unbounded on

B , a contradic-

3

tion.

12.4.

COROLLARY.

For an open s u b s e t

is

U of E c o n s i d e r t h e f o l -

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

(a)

U

(b)

dU(B) > 0

(c)

U

Then

(a)

i s a domain of e x i s t e n c e . f o r each bounding s u b s e t

B

of

U.

i s a domain of holornorphy.

* (b)

(c).

Apply Theorem 1 1 . 4 a n d P r o p o s i t i o n 1 2 . 3 .

PROOF.

1 2 . 5 . THEOREM. E v e r y b o u n d i n g s u b s e t of a s e p a r a b l e Banach space

i s r e Z a t i v e Zy c o m p a c t . PROOF. L e t B b e a bounding s u b s e t o f a s e p a r ' a b l e Banach E . L e t (a.) be a d e n s e s e q u e n c e i n E . 3

Given

E

> 0,

space

l e t ( A . ) be 3

.i

t h e i n c r e a s i n g s e q u e n c e of open s e t s d e f i n e d by m

Then

E =

u

and by

A

A . = u B(ai;El. 3 i=I

Proposition 12.3 there e x i s t s j

such

j

j=1

Then, by P r o p o s i t i o n 11.1, B C ( A ^ j ) 6 e E , that B C (A^jIJc(E). c o ( A . ) . S e t K = c o f a l , . . . , a . } . Then c o ( A . ) C K . + B ( O ; E ) and

s

j

=(A

B C

that

B

Thus

B

.) C

3

K

j

3

+ B ( 0 ; 2 ~ ) . Since

3

Kj

3

i s compact w e

c a n b e c o v e r e d b y f i n i t e l y many b a l l s

conclude

of r a d i u s

i s p r e c o m p a c t and t h e r e f o r e r e l a t i v e l y compact i n

3E.

E.

U b e a domain o f e x i s t e n c e i n a separable Banach s p a c e . T h e n e a c h b o u n d i n g s u b s e t of U i s r e l a t i v e l y com12.6.

COROLLARY.

pact i n

Let

U.

T h e o r e m 1 2 . 5 d o e s n o t g e n e r a l i z e t o a r b i t r a r y Banach spaces. I n d e e d , S . Dineen [ 2 ]

. ., O ,

(0,.

has

shown t h a t t h e u n i t v e c t o r s

I, 0,. . . 1 form a b o u n d i n g s e t i n

Em.

un =

96

MUJ I CA

E XE RC ISES 12.A.

U b e an open s u b s e t of E , and l e t B b e a bounding U. Show t h a t e a c h f f X ( U ; F ) i s bounded on B .

Let

s u b s e t of

12.B. E

U be an open s u b s e t of E , l e t A C U and l e t y 1 I ffu)II 5 s u p IIf1x)II f o r e v e r y f E JC(U;FI.

Let

A^x(uI.

Show t h a t

XE

12.C.

U b e an open s u b s e t of

Let

f JCfU;FI,

s u b s e t of

(a)

E , and l e t

F be a bounded

T ~ ) .

U i s t h e union of an i n c r e a s i n g sequence such t h a t t h e f u n c t i o n s f E F a r e u n i f o d y

Show t h a t

of open s e t s

Aj

bounded on e a c h (b) tions

A

If

B

f E F

(c)

Aj. i s a bounding s u b s e t of

U , show t h a t t h e fun-

are u n i f o r m l y bounded on

B.

B i s a bounding s u b s e t of U w i t h d U f B ) > 0, f E F are u n i f o r m l y bounded on t h e B + B f O ; € ) , for a suitable E > 0. If

show t h a t t h e f u n c t i o n s

set

U b e a b a l a n c e d open s u b s e t of E . Show t h a t t h e b a l a n c e d h u l l o f e a c h bounding s u b s e t of U i s a l s o a bounding s u b s e t of U. 12.D.

Let

12.E.

Let

subset of

U be an open s u b s e t of

s u b s e t B of x

E , and l e t

F . Given a bounding s u b s e t A V , show t h a t

A x B

V

be an

open

of

U , and a bounding i s a bounding s u b s e t o f U

v.

U and V be open s u b s e t s of E. s u b s e t A of U and a bounding s u b s e t B o f B i s a bounding s u b s e t of U + V . 12.F. L e t

12.G.

Let

U be a convex open s e t i n U.

f o r e a c h bounding s u b s e t B of

Given a boundi.ng V , show t h a t

E . Show t h a t

du(BI

A +

> 0

DOMAINS OF HOLOMORPHY

97

NOTES AND COMMENTS The proof

of

Theorem 1 0 . 9

b a s e d on t h e B a i r e

Theorem i s t a k e n from t h e book of L. Nachbin [ 3

Category

1 . The c h a r a c -

t e r i z a t i o n of domains of holomorphy i n Theorem 11.5 i s due H.

C a r t a n and P. T h u l l e n [ 1 1 .

[ 1]

to

Theorem 1 1 . 4 i s due t o S. Dineen

and A. H i r s c h o w i t z [ 3 1 , and r e p r e s e n t s an a t t e m p t t o ex-

t e n d t h e C a r t a n - T h u l l e n Theorem t o i n f i n i t e d i m e n s i o n a l Banach s p a c e s . Bounding sets were i n t r o d u c e d by H. Alexander [ l ] , who o b t a i n e d Theorem 12.5 f o r s e p a r a b l e H i l b e r t

spaces.

Theorem

12.5 i s a s p e c i a l case of more g e n e r a l r e s u l t s o b t a i n e d by

S.

Dineen [ 4 ] and A . H i r s c h o w i t z [ 2 1 . The proof of Theorem 1 2 . 5 g i v e n h e r e i s due t o M.

Schottenloher [ 2

I.

This Page Intentionally Left Blank

CHAPTER IV

DIFFERENT IABLE MAPPINGS

1 3 . DIFFERENTIABLE MAPPINGS T h i s s e c t i o n i s d e v o t e d t o t h e s t u d y of d i f f e r e n t i a b l e m a p p i n g s between Banach s p a c e s . U n l e s s s t a t e d o t h e r w i s e , t h e letters F w i l l r e p r e s e n t Banach s p a c e s over t h e same f i e l d M .

E and

1 3 . 1 . DEFINITION.

f

:

U

+

F

U b e a n open s u b s e t o f

Let

mapping

A

i s s a i d t o be d i f f e r e n t i a b l e i f f o r each p o i n t a

t h e r e e x i s t s a mapping

A E

I1 f ( x )

lim

Let

REMARKS.

-

f

fE;F)

such t h a t

-

- all1

f(a) II x

x+a 13.2.

E.

Alx

=

- a II

11 be an open s u b s e t o f

(a)

Eacn d i f f e r e n t i a b l e mapping

(b)

The mapping

-+

U

0.

Then:

E.

f : U

E

F

is

continu-

ous. A E E(E;FI

1 3 . 1 i s u n i q u e l y d e t e r m i n e d by

f

t h a t appears and

in

Definition

a . I t is c a l l e d thedif-

a t a a n d w i l l be d e n o t e d by Dffal. d i f f e r e n t i a b l e mapping f : U + F i n d u c e s a mapping f e r e n t i a l of

Thus

a

Df : U

+

f

L((E;F).

(c)

If

E

and

F

are comDlex Banach s p a c e s t h e n w e

t o d i s t i n g u i s h between t h e complex d i f f e r e n t i a b i l i t y of C

E

-+

F

arid t h e r e a l d i f f e r e n t i a b i l i t y of

f

:

U

I t i s c l e a r t h a t complex d i f f e r e n t i a h i l i t y i m p l i e s

C

EB

real

have f : U .+

FIR.

dif-

f e r e n t i a b i l i t y , b u t t h e converse i s n o t t r u e . Indeed, t h e function

e

E

C

+

z

E

6:

Is

07-differentiable

99

without

being

MUJ I CA

100

@-differentiable. We shall soon study the connection between lR- differentiable mappings and C-differentiable mappings. 13.3. EXAMPLE. If f : E F differentiable and Df(x) = 0 then A is differentiable and

is a constant mapping then f is for every a E E. If A E f ( E ; F 1 DA(a) = A for every a E E.

+

1 3 . 4 . EXAMPLE.

where PROOF.

Every P then

A E LS(mE;F)

P(mE;F)

E

is differentiable. If P = i m-1 for every a E E .

D P f a l = mAa

By the Newton Binomial Formula

for all a , x E E . If we s e t

then it is clear that

p ( x ) / II x

- a II

+

0

when

x

a.

--t

1 3 . 5 . EXAMPLE.

Let E and F be complex Banach spaces and let U be an open subset of E . Then every f E K ( l J ; F l is C-differentiable and D f ( a ) = P 1 f ( a ) for every a E U.

PROOF.

Let

a E U

and l e t

1 f(x) = f ( a ) + P f ( a ) ( x

for every

x E B(a;r).

0 < P <

r b f ( a ) . Then we can write m

-

a)

+

Z: p m f ( a ) ( x - a ) m=2

If we set

then using the Cauchy inequalities one can readily p P ( x ) / I I x - a l l -+ 0 when z a.

show

that

+

Additional examples of differentiable mappings are given in the exercises. Next we generalize the classical C h a i n R u l e .

D I F F E R E N T I A B L E MAPPINGS

101

13.6. THEOREM. L e t E, F , G b e Banach s p a c e s o v e r IK. L e t U C a n d V C F be two o p e n s e t s and l e t f : U + F and g : V + G be t w o d i f f e r e n t i a b l e mappings w i t h f l u ) C V . Then the composite E

g of

mapping

:

U

-P

= D g ( f ( a ) l o Dflal PROOF.

and

Let

i s d i f f e r e n t i a b l e a s w e l l and

f o r every

a E U

= Dglb)

B

G

Dlgo f l l a l

a E U.

b = f ( a ) E V, A = Df(a) E l f E ; F ) Then f o r a l l x E U and y E V w e

and s e t

E d:(F;G).

can w r i t e

and

= g(f(a)l + BoA(x - a) + p(x) where

p ( x / = B ( c p ( x l / + $ ( f ( x l i . Then

and s i n c e I1f ( x ) II x

- f(x)ll

-

- a II

IIA(x- a ) + cp(x)II llx

II 9(x)ll

- IIrlll+

- all

.

II x

- a II

II P ( X ) 1 I

it follows t h a t When

E = M

lim x + a

= 0,

completing t h e p r o o f .

llz-all

t h e n t h e n o t i o n of d i f f e r e n t i a b i l i t y takes the

MUJ I CA

102

f o l l o w i n g , more f a m i l i a r form.

13.7.

PROPOSITION.

ping

f: U

+

h E U. In t h i s c a s e

e x i s t s f o r each

X

E

u

and

U b e an o p e n s u b s e t o f M .

Let

Then a map-

F i s d i f f e r e n t i a b z e i f and on2y i f t h e derivative

D f ( X ) (p) = p f ' ( A )

f o r a22

p E llf.

The proof of t h i s p r o p o s i t i o n i s s t r a i g h t f o r w a r d l e f t as an e x e r c i s e t o t h e r e a d e r .

It w i l l

be

is

and

useful

in

the

p r o o f of t h e n e x t r e s u l t , which g e n e r a l i z e s t h e c l a s s i c a l Mean V a l u e Theorem.

THEOREM.

13.8.

Let

U b e an o p e n s u b s e t of E , and l e t

b e a d i f f e r e n t i a b l e m a p p i n g . If a l i n e s e g m e n t e n t i r e l y contained i n

U

$ E Fr

f i n e d by

g(X) = $ o f(a + A t ) .

E

(0,ll.

(U,ll

and

lK = B . Then

g r ( X ) = $1 D f f a + A t ) ( t l l

By t h e c l a s s i c a l Mean Value Theorem

sup l g r ( h l 1

is

w e c o n s i d e r t h e f u n c t i o n g : [ U , 1 ] --t B deThen g i s c o n t i n u o u s on [ O , l 1 ,

f o r each

X

F

-*

then

PROOF. I n view o f Remark 1 3 . 2 ( c ) w e may assume t h a t

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

f: U

[a,a + t ]

f o r every

] g ( l l - g10)

I 5

and t h e d e s i r e d c o n c l u s i o n f o l l o w s .

U<X
13.9.

COROLLARY.

Let

U be a n o p e n s u b s e t of E ,

be a d i f f e r e n t i a b l e mapping. I f a l i n e segment e n t i r e l y contained i n

U

and l e t f : U

[a, a +

t 1

F

-+

is

then

PROOF. I t s u f f i c e s t o a p p l y T h e o r e m 1 3 . 8 t o t h e mapping g : U

+

F

DIFFERENTIABLE MAPPINGS

d e f i n e d by

glx) = flxl

103

- Df(al(x - al.

Next w e g i v e some a p p l i c a t i o n s o f T h e o r e m 1 3 . 8 and C o r o l l a r y 13.9.

U be a n open subset of E and Zet f: U + F b e a d i f f e r e n t i a b l e mapping. I f U i s c o n n e c t e d and D f l x : ) = 0 f o r e v e r y x E U t h e n f i s a c o n s t a n t mapping: 1 3 . 1 0 . PROPOSITION.

PROOF.

Fix

a

E

Let

and l e t

U

A

denote t h e set of

all

x

E

U

such t h a t f l x l = f l a l . S i n c e f i s c o n t i n u o u s A i s c l o s e d i n U. To show t h a t A i s open, t a k e x E A and choose r > 0 s u c h t h a t B ( x ; r ) C U. Ey a p p l y i n g Theorem 13.8 t o t h e l i n e segment [x, y ] we see t h a t f i y l = f ( x / = f i a ) f o r e v e r y y E B ( y ; r l . Thus A i s open and c l o s e d i n U and t h e r e f o r e A = U. 13.11. PROPOSITION. L e t ( e l , . of

Mn

5, , . . . , 5,

and Z e t

.. , e n )

d e n o t e t h e canonicaZ b a s i s

d e n o t e t h e corresponding coordinate

U b e a n o p e n s u b s e t o f - X n and Z e t f : U

functionals. Let

+

F

b e a mapping

(a)

If f i s d i f f e r e n t i a b l e t h e n t h e p a r t i a l

e x i s t s and e q u a l s

D f ( a ) ( e .l f o r e a c h 3

a

U

E

and

derivative

j = 1,.

..,a.

Hence

for a22

a E U

and

t

E

an.

If a l l t h e p a r t i a . 2 d e r i v a t i v e s af/aEj exist are c o n t i n u o u s on U t h e n f i s d i f f e r e n t i a b l e t h e r e . (b)

PROOF.

The p r o o f o f

( a ) i s s t r a i g h t f o r w a r d and i s l e f t a s

e x e r c i s e t o t h e r e a d e r . To show (b) t a k e

a

E

U.

and

an

Then u s i n g

104

MUJ I CA

C o r o l l a r y 13.9 w e can w r i t e ,

n

-

Itj

af (a) -

j=1

ac j

+

t near t h e o r i g i n :

for

v 3. ( t ) ]

where

for

..,n.

j = I,.

w e see t h a t

Since t h e f u n c t i o n s

It v j ( t l l l / II t II

-+

0

are

af/aSj

t

when

-+

0

and

continuous the

desired

conclusion follows.

U b e a n open s u b s e t o f E . Then amapF i s s a i d t o be c o n t i n u o u s l y d i f f e r e n t i a b l e i f ping f : U f i s d i f f e r e n t i a b l e and t h e mapping D f : U E(E;F) i s con13.12. DEFINITION.

Let

-+

+

tinuous. 13.13. PROPOSITION. L e t U be a n o p e n s u b s e t o f E , and let (fi! be a n e t o f c o n t i n u o u s 2 y d i f f e r e n t i a b l e m a p p i n g s from U Assume t h a t :

into F . (a)

The n e t ( f 2.. ) c o n v e r g e s p o i n t w i s e o n

(b)

The n e t ( O f i )

U to a

!iitzpping

f. s e t of

U t o a mapping

Then

PROOF.

c o n v e r g e s u n i f o r m l y o n e a c h c~ornpcri~i subg.

f is c o n t i n u o u s l y d i f f e r e n t i a b l e o n U and Let

a

E

U

and choose

Then by C o r o l l a r y 13.9 f o r e v e r y

r > 0

t

E

such t h a t

Df = g.

B f a ; r ) C U.

B ( 0 ; r l and e v e r y

i

we

DIFFERENTIABLE MAPPINGS

105

have t h a t

Then u s i n g ( a ) and ( b ) w e g e t t h a t

converges t o g

Now, s i n c e ( O f i )

u n i f o r m l y on e a c h compact sub-

s e t of U, and s i n c e each D f i i s c o n t i n u o u s , we c o n c l u d e t h a t g IK i s c o n t i n u o u s f o r each compact s u b s e t K of U. S i n c e U i s a k-space w e c o n c l u d e t h a t g i s c o n t i n u o u s . Hence, given E > 0 w e can f i n d 6 w i t h 0 6 < P such t h a t Ilg(xl -g(alll <E whenever IIx - a II 5 6 . Using t h i s w e g e t t h a t

5

whenever I I t I I 13.14.

6 . The d e s i r e d c o n c l u s i o n f o l l o w s .

PROPOSITION.

Let

U be an o p e n s u b s e t of E , and l e t 1 ~ be a X - v a l u e d B o r e l m e a s u r e on a c o m p a c t H a u s d o r f f s p a c e T. L e t ~p E C l U x T ; F ) and l e t f : U F be d e f i n e d by -+

for e v e r y

x

E

U. T h e n :

(a)

f

(b)

If t h e mapping x E U -+ ( P ( x , t ) 6 F i s d i f f e 2 . e n t i a b l e t E T, and t h e mapping ( 2 , t ) E U X T -+ D x ( P ( x , +), E

f o r each d: ( E ; F )

i s continuous.

is c o n t i n u o u s , t h e n f

for e v e r y

x E U.

i s c o n t i n u o u s l y d i f f e r c n t i a b i e and

106

MUJ 1 CA

PROOF. ( a ) To show t h a t f i s c o n t i n u o u s l e t a E U and E > 0 be given. Since 9 i s continuous, € o r each b E T t h e r e i s a neighborhood W b of (a,bl i n U x T such t h a t Ilp(x,ti f o r e v e r y ( x , t ) E Wb . Without loss of g enerality p(a,blll 5 E w e may assume t h a t W b - U b x V b where U b i s a neighborhood of a i n U and vb i s a neighborhood of b i n T. Then, using t h e t r i a n g l e i n e q u a l i t y , w e g e t t h a t I l 9 ( x , t l - P ( a , t ) II 5 2 s f o r a l l x E U b and t E V b . S i n c e T i s compact t h e r e are blJ , bm E T such t h a t T = V U .. U V b . S e t ua = bl m n . Then U a i s a neighborhood of a i n U and 'b,

.

... ...

ubl

for all

x

E Ua

and

t

E T.

Whence i t f o l l o w s t h a t

A(xf =

D x 9 ( z , t l d p ( t ) f o r every x E U. By JT a p p l y i n g ( a ) t o t h e mapping + ( x , t l = D x $ ( x , t ) we see t h a t fb)

Set

We s h a l l prove t h a t f i s d i f f e r e n t i a b l e and D f i a l = A ( a l f o r e v e r y a E U. L e t a E U a n d E > 0 be given. By a p p l y i n g (13.1) t o t h e mapping + ( x , t l = DLC+(x,tJ w e can f i n d 6 > 0 such t h a t A E C(U;C(E;FIl.

whenever ll h I1 2 S

and

t

E T.

Then by C o r o l l a r y 1 3 . 9 w e

that

whenever

I1 /z II

5 6 and

t

E

T.

Hence

have

DIFFERENTIABLE MAPPINGS

whenever

II h II < 6.

107

The d e s i r e d c o n c l u s i o n f o l l o w s .

As w e have a l r e a d y remarked, e v e r y @ - d i f f e r e n t i a b l e mapping The n e x t p r o p o s i t i o n t e l l s u s when an

is R-differentiable.

B- d i f f e r e n t i a b l e mapping i s @ - d i f f e r e n t i a b l e .

13.15 PROPOSITION.

Let

and

E

F

be c o m p l e x Banach s p a c e s , l e t

U be a n o p e n s u b s e t of E , and l e t f : U --t F b e a n I ? - d i f f e r e n t i a b l e m a p p i n g . L e t D f ( a ) d e n o t e t h e r e a l d i f f e r e n t i a 2 of f a t a , and l e t D ’ f f a ) and D r r f ( a l be d e f i n e d b y

t

f o r every D”f(a)(t)

=

f

0

E.

i s @ - d i f f e r e n t i a b l e if and onZy if a E U and t E E . In t h i s case Dffal

Then f

f o r every

= D ’ f ( a ) i s also t h e eoinplex d i f f e r e n t i a l o f PROOF.

By P r o p o s i t i o n 1 . 1 2

f

a t a.

D f ( a ) i s C - l i n e a r i f and o n l y

if

D ” f f a l = 0 . Thus i t s u f f i c e s t o a p p l y t h e d e f i n i t i o n o f 6 - d i f f e r e n t i a b i l i t y and t h e u n i q u e n e s s of t h e d i f f e r e n t i a l a t a given point. The d i s t i n c t i o n between

B-differentiability

and

C-dif-

f e r e n t i a b i l i t y i s emphasized by t h e f o l l o w i n g theorem.

13.16. THEOmM.

U

Let

E

be a n o p e n s u b s e t of

and E.

it

f e r e n t i a b l e if and o n l y i f D f ( a ) = P f f ( a ) f o r every

PROOF.

F

a

be c o m p l e x Banach spaces, and l e t

Then a mapping

f : U

+

In

i s hoZornorphic.

F

is 6-dif-

this

case

follows

from

E (1.

Suppose f i s C - d i f f e r e n t i a b l e .

Then

it

108

MUJ I CA

t h e Chain Rule 13.6 t h a t t h e f u n c t i o n

g o t ) = $ a f ( a + Xb)

is

t - d i f f e r e n t i a b l e on t h e open set A = (1 E U : a + Xb E U ) f o r e v e r y a E U , b E E and I/J E F’. Hence g i s holomorphic, by the c o r r e s p o n d i n g r e s u l t f o r f u n c t i o n s of one complex variable. Thus

f

i s G-holomorphic, by Theorem 8.12,

and t h e r e f o r e holanorphic, s i n c e f i s c l e a r l y c o n t i n u o u s . S i n c e t h e re-

by Theorem 8.7,

v e r s e i m p l i c a t i o n was e s t a b l i s h e d i n Example 1 3 . 5 , of t h e theorem i s complete. 13.17.

Let

COROLLARY.

E

and

F

the

proof

be c o m p l e x Banach s p a c e s , and

U be a n o p e n s u b s e t o f E . T h e n a m a p p i n g f : U hoZomorphic i f and o n l y if f is i R - d i f f e r e n t i a b l e a n d let

+

F

is

D”f

is

F

and

i d e n t i c a l l y zero.

EmRCISES 13.A.

U be a n open s u b s e t of

Let

g : U

f

:

U

+

b e two d i f f e r e n t i a b l e mappings.

F

+

E , and l e t

af + Bg

i s d i f f e r e n t i a b l e for all a , D ( a f + B g l l a l = c l D f ( a ) + B D g ( a l f o r e v e r y a E U.

(a)

Show t h a t

(b)

Show t h a t i f

and

13.B.

L e t E and F

F = Fl

X

...

X

F = D(

t h e n t h e product

D f f g ) ( a ) = g(a)Dffa)

f e r e n t i a b l e and a E U.

j

(j = 1 , .

Fm. Let

.., m )

b e Banach s p a c e s ,

U b e a n open s u b s e t of

.

fg

+ f(a)Dg(a)

.

BE

dif-

is

for

iK

every

and

B , let

let

fj

:

( j= 1 , . . , m ) and l e t f = (f,,. ., fm) : U F. Shm t h a t j f i s d i f f e r e n t i a b l e i f and o n l y i f e a c h f,. i s d i f f e r e n t i a b l e .

U

-+

F

+

J

Show t h a t i n t h i s case a

E

D f ( a ) = (Dfl(a),

..., D f m ( a ) )

for e v e r y

U.

13.C.

Let

...

E

j

( j

= I,.

-+ F m‘ and is differentiable

A : El

X

. ., m )

and F b e Banach s p a c e s ,

and

let

b e m - l i n e a r and c o n t i n u o u s . Show t h a t A

DIFFERENTIABLE MAPPINGS

,..., am) and

a = lal

for a l l

t = Itl

,...,

109

tm) i n

E1 x

... x E m .

...,

13.D. L e t E , F . ( j = l , rn) and G b e Banach spaces. L e t U 3 F ( j = l ,..., m) b e d i f be an open s u b s e t of E , l e t fj : U i f e r e n t i a b l e , and l e t A : F l x . . x Fm -+ G b e m - l i n e a r and continuous. Show t h a t t h e mapping g = A o (f,, ,f m ) : U --t G -+

.

. ..

i s d i f f e r e n t i a b l e and

f o r every

a

Let E

13.E.

product (x

I

and

U

E

Let

-+

F

E

E.

be a r e a l o r complex H i l b e r t s p a c e , and

with

function

U b e a connected open s u b s e t o f

E , and l e t

b e a d i f f e r e n t i a b l e mapping whose d i f f e r e n t i a l

L IE;F)

i s a c o n s t a n t mapping. Show t h e e x i s t e n c e o f

and

E

b

F

such t h a t

inner

f(z) = IIcc1I2 is B-difDf(a) (tl = ZReltlal f o r a l l a, t E E.

Show t h a t t h e

y).

f e r e n t i a b l e on E 13.F.

t

f(x) = Ax + b

f o r every

x

f : U

Df : U

+

A E LfE;F) E

U.

n E = X m and F = X , and l e t G be any Banach space o v e r X. L e t U C E and V C F b e t w o open sets, and l e t f : U + F and g : V G be two d i f f e r e n t i a b l e mappings witlrl f(U) C V. Let E l , . . .,Ern d e n o t e t h e c o o r d i n a t e f u n c t i o n a l s of El and l e t q l , . . . , n , d e n o t e t.he c o o r d i n a t e f u n c t i o n a l s of F. Show t h a t i f we w r i t e f = i f , , . . ,fn) t h e n 13.G.

Let

-+

.

f o r every

a E U

..

and

j = 1,.

. . ,m.

Let z I , . ,z n d e n o t e t h e complex c o o r d i n a t e functionals denote t h e corresponding real C n and l e t x l,yl, . . ,xn, y, c o o r d i n a t e f u n c t . i o n a l s . L e t U be an open s u b s e t of C n l l e t F

13.H.

of

.

110

MUJ ICA

be a complex Banach s p a c e , and l e t

af/axj

t i a b l e mapping. L e t

and

f : U

af/ayj

+

F bean W - d i f f e r e n -

d e n o t e t h e real partial

d e r i v a t i v e s i n t h e s e n s e of P r o p o s i t i o n 1 3 . 1 1 a n d l e t and be t h e f u n c t i o n s d e f i n e d by

af/azj

af/azj

a

f o r every

(a)

U.

Show t h a t

a

f o r every

on

E

t

and

U

E

(b)

Show t h a t

U for

j = I,.

f

. ., n .

6

8.

i s holomorphic i f and o n l y i f

af/ai. = 3

0

These are t h e Cauchy-Riernann equations.

Note t h a t t h e o p e r a t o r

j u s t defined coincides,when j a p p l i e d t o holomorphic f u n c t i o n s , w i t h t h e complex p a r t i a l d e i n t h e s e n s e of P r o p o s i t i o n 1 3 . 1 1 . r i v a t i v e a/az 3/32

i

13.1.

f

:

U

Let +

5.

function 13.J.

Let

If

be an open s u b s e t of

E.

Show t n a t

i s B - d i f f e r e n t i a b l e i f and o n l y -

f : U

-+

5.

is

if

a

function

the c o n j u g a t e

B-differentiable.

U be a n open s u b s e t of

Cn

a n B - d i f f e r e n t i a b l e € u n c t i o n . Show t h a t

and l e t

f : U

+

6

be

DIFFERENTIABLE MAPPINGS

f o r every

j = 1,

111

..., n .

1 4 . DIFFERENTIABLE MAPPINGS OF HIGHER ORDER T h i s s e c t i o n i s d e v o t e d t o t h e s t u d y of d i f f e r e n t i a b l e m a p p i n g s of h i g h e r o r d e r . U n l e s s s t a t e d o t h e r w i s e , t h e l e t t e r s

w i l l r e p r e s e n t Banach s p a c e s o v e r t h e same f i e l d

and F

14.1.

DEFINITION.

f : U

+

Let

U b e a n open s u b s e t o f

t i a b l e and t h e d i f f e r e n t i a l aF

Df

:

U

+

d:(E;FI

A mapping

E.

i s s a i d t o be t w i c e d i f f e r e n t i a b z e i f

F

E

aC.

f

is differen-

is differentiable

well.

1 4 . 2 . REMARKS.

Let

U be a n open s u b s e t o f

E

and l e t f : U + F

be a t w i c e d i f f e r e n t i a b l e mapping. Then t h e d i f f e r e n t i a l of the mapping

t i a l of

a t a point

Df

f

at

a

a E U

i s c a l l e d t h e second d i f f e r e n -

and w i l l b e d e n o t e d by

DZf(a).

Thus

D2f(a)

2

d: ( E ; C f E ; F ) ) or L ( E; F ) , in view of t h e c a n o n i c a l isomorphism g i v e n i n P r o p o s i t i o n 1.4. I f 2 t h e i n d u c e d mapping D f : U L( 2 E ; F ) i s c o n t i n u o u s , t h e n f i s s a i d t o be t w i c e c o n t i n u o u s Zy d i f f e r e n t i a b l e .

may b e r e g a r d e d a s a member o f

-+

The n e x t r e s u l t g e n e r a l i z e s t h e c l a s s i c a l Schwarz 14.3.

THEOREM.

Let

U be an o p e n s u b s e t of

E

’

Theorem.

and k t f : U + F

be a t w i c e d i f f e r e n t i a b l e m a p p i n g . T h e n t h e b i l i n e a r mapping 2 2 D f l a ) E 5 1 E;F) i s s y m m e t r i c f o r e a c h a E U. I n o t h e r words,

D 2 f ( a l ( s , t l = D 2f ( a l ( t , s ) f o r a l l PROOF. For

Let

a E U

s , t E B(O;rl

and c h o o s e

s, t

r > 0

E

E.

such t h a t

B ( a ; 2 r ) C U.

w e d e f i n e t h e mapping

g ( s , ~ )= f ( a

+

s + tl

- f(a +

s)

- f(a +

t) + f(a).

The i d e a o f t h e p r o o f i s t o e s t i m a t e t h e d i f f e r e n c e g(s,tl 1 D ’ f ( a l ( s , t l f o r s and t n e a r z e r o . F i r s t n o t e t h a t i f w e s e t

MUJ I CA

112

for

IIIp/x)II /

then

E U

3:

113:

-

a It

-+

0

when

x

-P

a.

Hence

w e can w r i t e

Now, f i x

each

x

and d e f i n e

s E B(O;r)

Dhlx) = D f f x + s )

B ( a ; r , J . Noting t h a t

E

g f s , t ) = h(a

h(xl = f ( x + sl

+ t l - h l a l , and u s i n g C o r o l l a r y

= l l h l a + t)

-

< II t II

IIDhfa + A t )

sup

h(aj - Dhla)ftill

- Dh(a)ll.

Now,

+ A t ) - Dh(al

= [ D f ( a + At

+ sl

[ Df(a + sl 2

= [ D fla)fAt +

- D f l a l ] - [ D f l a + At) - Dflall -

- Dfla)] s ) + cp(a + A t + s ) ]

2

[ D f ( a l i s ) + iola + s)

1

= cpfa + At + s l - cp(a + A t ) Thus w e g e t t h e estimate IIg(s. t )

-

2

D f l u ) is, t l l l

for and

13.9 we g e t t h a t

O_ < A_
Dh(a

- ffxl - Df(x)

113

DIFFERENTIABLE MAPPINGS

Now, g i v e n

w e can f i n d

> 0

E

IIlp(x)II

/ I I x - all 5

s, t

BI0;61

E

for a l l BI0;rl

t, s

E

the

such t h a t for

all

for all

s, t

Then

r o l e s of

s

And s i n c e

BIO;6).

and

t we get t h a t

gfs,t)

= g(t,sl

we conclude t h a t 2

-

II D f l a ) I s , t l f o r all

0 < 6 < r

z E B(a;261.

we g e t t h a t

By i n t e r c h a n g i n g

E

6 with

f o r every

E

s,

t

2

( 3 s (11s II + II t II j

D f f a ) (t,s)ll

2

s, t E E . D f l a ) Is, t l =

but then, obviously f o r a l l

E BIO;6/,

Since E > 0 w a s a r b i t r a r y , we conclude t h a t 2 D f ( a ) f t , s l f o r a l l s, t E E , a s w e w a n t e d .

2

U be a n open s u b s e t o f E . B y i n d u c t i o n f : U .+ F t o b e k times d i f f e r e n t i a b l e i f f i s k - 1 times d i f f e r e n t i a b l e and i t s I k - 1 ) t h k-1 d i f f e r e n t i a l Dk-'jf : U I f E ; F ) i s d i f f e r e n t i a b l e . A mapp i n g f : U + F i s s a i d t o be i n f i n i t e Z y d i f f e r e n t i a b z e i f i t i s k t i m e s d i f f e r e n t i a b l e f o r e a c h k E lN. 14.4.

on

k

DEFINITION.

Let

w e d e f i n e a mapping

+

14.5.

REMARKS.

be a

k

U be a n open s u b s e t o f

Let

E , and l e t f : U - + F

times d i f f e r e n t i a b l e mapping. Then t h e d i f f e r e n t i a l o f

t h e mapping

f e r e n t i a l of

Dk-lf f

at

at a point a

a

E

U

i s called t h e

k t h difk D fIal k -C( E ; F ) ,

and w i l l b e d e n o t e d by D k f ( a l . Thus

may be r e g a r d e d a s a member o f

L(E;l(k-lE;F))

or

114

MUJ I CA

under t h e i d e n t i f i c a t i o n g i v e n , b y P r o p o s i t i o n 1 . 4 . I f t h e i n k k d u c e d mapping D f : U + C ( E ; F I i s c o n t i n u o u s t h e n f i s s a i d

t i m e s continuousZy d i f f e r e n t i a b z e . For convenience w e

k

t o be

Dof = f .

also d e f i n e

Now w e c a n e a s i l y g e n e r a l i z e Theorem 1 4 . 3 . 14.6.

THEOREM.

Let

U b e an o p e n s u b s e t o f

and l e t

E

f : U

+

F

be a k t i m e s d i f f e r e n t i a b Z e m a p p i n g . T h e n t h e k - Z i n e a r mapping D k f ( a ) E J ( k E ; F ) i s s y m m e t r i c f o r e a c h a E U. k = 2

k . For

By i n d u c t i o n o n

PROOF.

this

is

just

Theorem

k > 3 a n d assume t h e t h e o r e m t r u e f o r k - 1. If k t h e n D f l u ) i s t h e d i f f e r e n t i a l a t a o f t h e mapping g

14.3. L e t

a E U

--

D k - l f , w h i c h , by t h e i n d u c t i o n h y p o t h e s i s , t a k e s i t s

in

k

D f ( a ) ( t l ,t,,

and hence

t2,.

. .,t k .

values

Thus

ls(k-lE;F).

. . . ,t k ) i s

symmetric i n t h e

variables

k On t h e o t h e r h a n d , D f (a) i s t h e s e c o n d d i f f e r e n t i a l

h = Dk-2f,

a t a of t h e mapping

and i f follows

from

Theorem

14.3 t h a t

k

D f(a)(tl,tz,

Hence

..., t k ) i s

symmetric i n t h e v a r i a b l e s

tl

t 2 , and t h e d e s i r e d c o n c l u s i o n f o l l o w s .

and

The n e x t t h e o r e m s t r e n g t h e n s

the

conclusion

in

Theorem

13.16. 14.7.

U

THEOREM.

Let

E

b e an o p e n s u b s e t o f

and E.

F

b e c o m p l e x B a n a c h s p a c e s , and Let

Then f o r e a c h mapping

f o Z Z o w i n g conditions are equiuaZent: (a)

f

i s holomorphic.

(b)

f

is C - d i f f e r e n t i a b l e .

f : U +F

the

DIFFERENTIABLE MAPPINGS

(c)

i s infinitely 6-differentiable.

f

Dk f ( a l = k ! A k f f a l

t h e s e conditions are s a t i s f i e d t h e n

If

a E U

f o r every

k

and

E

lN.

I n view o f Theorem 1 3 . 1 6 it i s s u f f i c i e n t t o show t h a t

PROOF.

f : U

if

115

i s holomorphic, then

F

-+

t i m e s C-differen-

is k

f

Dk f ( a ) = k ! A k f ( a ) f o r e v e r y

t i a b l e and

a

E

U

and

k

E W.

W e p r o c e e d by i n d u c t i o n on

k , t h e statement being t r u e f o r

= 1

k

by Theorem 1 3 . 1 6 . L e t

Pk-'f

E

every

E

U. Hence

Dk-l f = ( k

-

I)!

Ak-lf

and by Theorem 1 3 . 1 6 w e may c o n c l u d e t h a t

D k f l u ) = P1(Dk-'f)(al

f e r e n t i a b l e and f o r every

a

E

U. Thus f

f

is

k-1

tims

By Theorem 7 . 1 7 ,

P 1 ( P k - ' f l ( a ) = Pk-l(Pkffa))

~ C I U ; P ( ~ - ' E ; F ) ) and

a

and assume

2

Dk-' f = ( k - l)! A k - l f .

8 - d i f f e r e n t i a b l e and

is

= k !

k

k

k

E

for

x(U;d:s ( k - 1 E ; F ) ) , is

Dk-'f

6-dif-

= (k - I ) ! P1(Ak-'f)(a)

t i m e s @ - d i f f e r e n t i a b l e and

k

A f(a)t

.

k k S i n c e D f i a ) and A f ( a l a r e b o t h s y m m e t r i c , w e c o n c l u d e t h a t k k D f i a ) = k! A f ( a ) , a s asserted. 14.8.

k

E

W

pings

b e an open s u b s e t o f E . For each k w e s h a l l d e n o t e by c ( U ; F I t h e v e c t o r s p a c e o f a l l mapLet

DEFINITION. f : U

+

U

which a r e

F

t i a b l e . W e shall d e n o t e by mappings f : U

+

F

k

t i m e s continuously

Cm(U;F)

which b e l o n g t o

B-differen-

t h e v e c t o r space of all k C (U;FI f o r e v e r y k E T?.

For c o n v e n i e n c e w e a l s o d e f i n e C o ( U ; F ) k of C ( U ; F ) , where k E W o U { m } , are

= C(U;FI. often

The members

c a l l e d mappings

116

MUJ I CA

of class k

ck .

F = M

When

thenweshall w r i t e for short

Ck(U)

= c IU;M). THEOmM.

14.9.

and

Let

E , F , G be Banach s p a c e s o v e r

M . Let

U cE

b e two o p e n s e t s , a n d l e t f : U F and g : V + G k be two mappings of c l a s s c , w i t h f ( U ) C V . Then t h e composite V

+

F

-+

g of

mapping

:

u

+.

G

i s a l s o of c l a s s

c ~ .

we proceed by i n d u c t i o n on k , t h e theorem b e i n g obk = 0. Let k > 1 and assume t h e theorem t r u e f o r k - 1. By t h e Chain Rule 1 3 . 6 t h e rmpping g o f : U G is B - d i f f e r e n t i a b l e and D ( g o f ) ( x ) = D g l f l x ) ) o D f ( x ) f o r e v e r y x E U. Thus t h e mapping Dlg o f ) : U +. E ( E ; G ) i s o b t a i n e d by

PROOF.

vious f o r

-+

composition o f t h e mappings

The mapping Cm,

Df : U

+.

and

IT

d e f i n e d by

T i s b i l i n e a r and c o n t i n u o u s , a n d h e n c e i s o f class

by E x e r c i s e 1 4 . B .

On t h e o t h e r hand, t h e mapping

i n a product

i t s values

S

d: I E ; F )

and

p o t h e s i s t h e mappings

and i t s components

Dg o f : U Df

and

-+

Dg o f

a g a i n by Hence

takes

are t h e mappings

By t h e i n d u c t i o n hyk- I are b o t h or c l a s s c ,

6 IF;G).

and then it f o l l o w s from E x e r c i s e 1 4 . A

of c l a s s Ck-' t o o . Then, D ( g o f ) i s of class Ck-'.

S

t h a t t h e mapping

the induction

g o f

i s o f class

S

is

hypothesis, C k and t h e

proof i s complete.

EXERCISES

E' and F (j = 1 , . . . , m ) b e Banach s p a c e s , and l e t j x F . Let. U be an open s u b s e t of E, l e t f : U F = Fl x ... m j P ( j = 1 ,..., m ) and l e t f = (f,,..., f,) : U + F . Show t h a t Jf i s k t i m e s ( c o n t i n u o u s l y ) d i f f e r e n t i a b l e i f and o n l y i f each f i s k t i m e s (continuously) differentiable. j 14.A.

Let

+

117

DIFFERENTIABLE MAPPINGS 14.B.

Let

E

...

j

(j = 1 ,

..., m )

and

F

be Banach s p a c e s l a n d l e t

be m - l i n e a r and c o n t i n u o u s . Show t h a t A i s i n f i n i t e l y d i f f e r e n t i a b l e and D k A = 0 f o r e v e r y k > m + 1.

A : El

x

x

Em

F

+

E i s a r e a l o r complex H i l b e r t s p a c e t h e n 1 1 ~ 1 1 i ~s o f class Cm on E .

Show t h a t i f

14.C.

the function

3:

+

14.D. L e t U be an open s u b s e t of E , let f : U times d i f f e r e n t i a b l e mapping, and l e t [ a , a + t ] segment which i s e n t i r e l y c o n t a i n e d i n U.

+

be a

F

a

be

(a)

Using t h e c l a s s i c a l T a y Z o r ' s f o r m u l a show t h a t

(b)

Applying ( a ) t o t h e mapping g(x)=f(x)-

k

line

1 k k D flal(x-a) k.'

show t h a t

U be an open s u b s e t of

14.E.

Let

space

C"(U; F ) ,

endowed

with

E , and c o n s i d e r t h e v e c t o r

the

locally

convex

g e n e r a t e d by a l l t h e seminorms o f t h e form where

j

E

liVo

f

and K i s a compact s u b s e t of

s u p II D 3 f ( d l l , xE K U. Using Propo+

s i t i o n 1 3 . 1 3 show t h a t t h e l o c a l l y convex s p a c e always complete. Show t h a t i f

14.F.

( e l , . . . , e n ) d e n o t e t h e c a n o n i c a l b a s i s of

Let

5 , , . . . ,en

(a)

f : U

Show t h a t t h e p a r t i a l d e r i v a t i v e

-+

e x i s t s and e q u a l s j ,

E

(1

and

;Xn

j l a

a

k

... a S j

akf ( a ) / a S

D K f ( a ) ( e ,..., e ) f o r every jl jk

,..., n l .

be

F

1.

,...,

then

d e n o t e t h e c o r r e s p o n d i n g c o o r d i n a t e functionals.

L e t U be an open s u b s e t of M n and l e t t i m e s d i f f e r e n t i a b l e mapping.

j,

is

Cm(U;F)

i s f i n i t e dimensional

i s a Frgchet space.

Cm(U;F)

let

E

topology

E

U

k and

118

MUJ ICA

(b)

akf/aCj

show t h a t

1

... 3 5

i s a symmetric f u n c t i o n j k

of t h e i n d i c e s Using t h e L e i b n i z

(c)

Formula 1 . 8

show t h a t

Dkf(u

f o r every

use t h e n o t a t i o n index

a with

14.G.

Let

t = (tl,.

and

u E U

. ., t,)

k

"1

= a fiat,

a"f/ag"

la1 = k .

U be an open s u b s e t of

: U

+

F

rivatives on

Ck

i s of class

of o r d e r

aaf/ax"

.

x l , . . , 3: n iRn. Show t h a t a mapping

iRnl

denote t h e coordinate f u n c t i o n a l s of

f

an.For s h o r t w e shall 01 . . . atnn f o r e v e r y m u l t i E

and l e t

i f and o n l y i f a l l t h e p a r t i a l dela1 2 k e x i s t and are c o n t i n u o u s

U.

1 5 . PARTITIONS O F U N I T Y I n t h i s s e c t i o n w e i n t r o d u c e p a r t i t i o n s of u n i t y , a fundamental t o o l which i n many d i f f e r e n t s i t u a t ' i o n s i s used t o cons t r u c t o b j e c t s w i t h c e r t a i n g l o b a l p r o p e r t i e s by patching together o b j e c t s which have l o c a l l y t h e same p r o p e r t i e s . Before s t a t i n g t h e main r e s u l t w e g i v e two p r e p a r a t o r y lemmas.

15.1. LEMMA. that

~ ( 3 : )

= 0

b

a

If i f

strictZy increasing i f

PROOF.

then there i s a f u n c t i o n

x 5 a,

~ ( x = ) 1 a 2 3: 5 b .

S t a r t with t h e f u n c t i o n f

i f

x > b

p e CmiEJsuch

and

p(z)

is

f ( z l = e -l/z i f C m i l R i by E x e r c i s e

d e f i n e d by

and f f z ) = 0 i f 3: 5 0 . Then f E f i s s t r i c t l y p o s i t i v e for x 0 and f i s i d e n t i c a l l y z e r o f o r x 5 0 . Then t h e f u n c t i o n gfx) = f [(x - a ) ( b - 2 1 1 i s

x > 0 15.Al

a l s o of class

Cm

on

R l i s s t r i c t l y p o s i t i v e on t h e i n t e r v a l

a < z < b , and i s i d e n t i c a l l y z e r o o u t s i d e t h a t i n t e r v a l . Then the function

b

pP(3:) = J z y ( t ) d t / J a g ( t l d t

is

identically

zero

DIFFERENTIABLE MAPPINGS

x L b.

x < a , and identically one for

for

/

g(x)

that

Iab g ( t l d t ~p

Since

we conclude that IP is of class Cm on

is strictly increasing for

15.2. LEMMA.

119

Let

cp'(x) =

lR and

a 5 x 5 b.

E b e a H i l b e r t s p a c e , a n d l e t 0 < r < R. Then p E C m ( E ) s u c h t h a t p(x) = 1 i f IIxII

zr,

there i s a function p(x) = 0

i f IIxII 2 R

r < IIxII < R.

0 < p ( x 1 < 1 if

and

PROOF. By Exercise 14.C the function f (x) = - II x I1 2 isof class Cm on E . By Lemma 15.1 there is a function g E C m ( i R ) such 2 that g ( t ) = 0 if t 5 - R , g l t ) = 1 if t 2 - r 2 and 0 < g ( t ) < 1 if - R 2 i t < - r Z . Then the function P(x) = g o f i x : ) 2 = g i- II x II l has the required properties. 15.3. DEFINITION. a Banach space.

Let X be a topological space and let F be

(a) The s u p p o r t of a mapping f : X F is the closure of the set {x E X : f ( x l # 0). The support of f will be denoted by s u p p f. +

(b)

A collection ( f i j i E I

of mappings from X into F is

said to be Z o c a l l y f i n i t e if each point of X has a neighborhood which meets only finitely many of the sets s u p p f i . (c) A p a r t i t i o n o f u n i t y on X is a locally finite collection ( p i ) i E I of continuous functions from X into [ 0, 1 ]

c

such that

pi(")

= 1

for every

x

E

X.

i E I

(d)

A partition of unity ( p i ) i E I

s u b o r d i n a t e d to an open cover i U i ) i E I

for every

i

E

on

X

is said

of X if

to be

s u p p pi C U i

I.

If X is an open subset of a Banach space then it makes sense to talk about Cm p a : q t i t C o n s of u n i t y on X : thismeans of course that each member of the partition of unity is a Cm function. Actually this is the situation we shall be primarily interested in. Indeed, we have the following theorem.

MUJ i CA

120

1 5 . 4 . THEOREM. L e t U be an o p e n s u b s e t of a s e p a r a b l e Hilbert E . Then f o r e a c h o p e n c o v e r ( U i ) i E I o f U there i s a Cm

space

on

partition of unity

which

U

to

is s u b u r d i n u t e d

.

(UiliE1

i s a sequence

of open b a l l s B ( a n ; r n ) whose u n i o n i s U and s u c h t h a t each B ( a n ; 2 r n l i s c o n t a i n e d i n some U i . By t h e axiom of c h o i c e there Since U i s a Lindelof

PROOF.

is a function n

every

0 < f,(x)

and

< 1

in

2

s u p p g,

r n < IIz

- anll

< 2rn.

by

g1 = f ,

and

CmiE/

...

Then i t i s clear t h a t

2.

=

if

- f,!

gn = ( 1

n

for

By Lemma 1 5 . 2 t h e r e i s a s e q u e n c e (f ! i n ?(El n fn(xl = 1 i f 1 1 2 - anll 5 r n , f n ( z ! = 0 i f I I z - a n II

another sequence ( g n !

if

B(an;Brn) C UTin)

I such t h a t

-+.

liV.

E

such t h a t > 2rn

: liV

T

space t h e r e

S”PP

f,

=

=

Bn(an;2rn)

more, one c a n r e a d i l y p r o v e by

f o r every

n. Since

fn = 1

on

-

(1

0

5

Define

fn-l’fn

gn 5 1

UT&

on

f o r every

and t h a t Further-

E

n.

induction t h a t

-

B ( a n ; r n ) i t follows from(l5.1)

that (15.2)

g,

...

+

-

ig

n = 1

on

B(an;rnf

and

(15.3)

g j =

on

B(an;rn)

f o r every

j > n.

@

Thus ( 1 5 . 3 ) g u a r a n t e e s t h a t t h e s e q u e n c e (9,) i s l o c a l l y f i n i t e in

U , whereas

every

x E

(15.2)

guarantees t h a t

U. F i n a l l y we d e f i n e

vi(x!

z

nexi =

g (xi

T(n)=i

=

g,(x)

1

for

f o r each

DIFFERENTIABLE

and

x

each

MAPPINGS

121

U. S i n c e t h e s e q u e n c e ( g ) i s l o c a l l y f i n i t e i n U, n pi i s w e l l defined and belongs t o Cm(U). Furthermore, E

t h e set

x

every

suppgrL is closed i n

U T

U and t h e r e f o r e

supp

(pi

( n )=i

U,

E

(viiiE1

i s t h e required p a r t i t i o n of unity.

U b e an o p e n s u b s e t of a separable HiZbert b e two d i s j o i n t c Z o s e d s u b s e t s o f U. Then t h e r e i s a f u n c t i o n p E C m ( U / s u c h t h a t 0 _.< P z 1 o n U, IP = 1 o n a n e i g h b o r h o o d of A i n U, and (p = 0 on a neighborhood of B i n U. 1 5 . 5 . COROLLARY.

space

Let

E.

Let

and

B

By Theorem 1 5 . 4 t h e r e are t w o n o n n e g a t i v e f u n c t i o n s V ,

PROOF. $ in

A

$ = 1

U. Then

on

(p

contains

A , whereas

contains

B.

s u p p J, C U \ A and cp + on t h e o p e n s e t U \ s u p p JI, which

supp 9 C U \ B,

C m l U l such t h a t

= 1

= 0

(p

on t h e o p e n s e t U \ s u p p

which

(p,

EXERCISES 15.A.

Consider t h e function

x

and

0

(a) and

k

E

f(x) = 0

(b)

P Z k ( t l i s a polynomial i n

where

Using L ' H o s p i t a l r u l e show t h a t

15.C.

k

E

x > 0

t of d e g r e e 2 k .

f ( k ) ( 0 ) exists

I?. Conclude t h a t

and

f E CmllR7R).

Cm x 5 0

Find an i n c r e a s i n g sequence o f convex, i n c r e a s i n g

functions and

for

f ( k ) ( x ) = e - 1 / x P 2 k ( l / x ) for e v e r y

equals zero f o r every 15.B.

f ( x J = e -7/x

d e f i n e d by

x 5 0.

for

Show t h a t 1Iv,

f

n

(pn

E W,

Let

L e t (U,)

: B

and

-+

B such t h a t

Zim n+

(pn(zl =

(pn(xl = 0 m

for e v e r y

€or every z

0.

U be a n open s u b s e t o f a s e p a r a b l e H i l b e r t

space.

be a n i n c r e a s i n q s e q u e n c e of open s e t s whose union i s

U , a n d l e t ( e n ) be a n i n c r e a s i n g s e q u e n c e o f r e a l numbers. Using

MUJ I CA

122

Theorem 15.4 f i n d a f u n c t i o n on

and

U3

cp

2

on

en

Un \

C”tU; iR) such t h a t

cp E

Un-l

f o r every

n

2 el

cp

2.

U be an open s u b s e t of a s e p a r a b l e H i l b e r t s p a c e . find a function g E Cm(U; R) Given a f u n c t i o n f E C ( Y ; i R ) s u c h t h a t g 2 f on U. 15.D.

Let

15.E.

Let

f : X?

+

R

be a f u n c t i o n which i s b o u n d e d a b o v e o n

e a c h i n t e r v a l o f t h e form f -

m,

b).

Find a f u n c t i o n g ECm(B;?R)

x

0

g ( x ) = constant f o r

such t h a t every

x

15.F.

Let

E

-

and

g(x)

2 f1x)

for

iR. K be a compact s u b s e t o f a H i l b e r t s p a c e

and

E,

l e t U be an open neighborhood of K . By a d a p t i n g t h e proof of Theorem 15.4 f i n d a n o n n e g a t i v e f u n c t i o n cp E C m ( E ) such t h a t s u p p cp C U, 15.G.

Let

cp

5

1

on

E

and

= 1

cp

on a neighborhood o f

K be a compact s u b s e t of

a H i l b e r t space

.

such t h a t

...

+

+ cpn =

1

s u p p pj

C

Uj,

cpl

+

on a neighborhodd of

...

p D n- 1

+

and

E,

l e t Ul,. . ’ un be open s u b s e t s of E which c o v e r K . E x e r c i s e s 9 . B and 15.F f i n d n o n n e g a t i v e f u n c t i o n s ( P I , E Cm(E)

K.

on

Using

...

3

‘n

E , and

K’.

K be a compact s u b s e t of a c o m p l e t e l y r e g u l a r Hausbe open s u b s e t s of X which d o r f f s p a c e X , and l e t U I J . . . , U n Find n o n n e g a t i v e f u n c t i o n s p I , . . .,pn E C t X i such cover K . 15.H.

Let

supp p j

that

C

U

pl +

j’ on a neighborhood o f

= 1

...

+

p,

5

7

on

X , and p I +

...

+

‘n

K.

1 6 . TEST FUNCTIONS

I n t h i s s e c t i o n w e i n t r o d u c e t h e space

of

test f u n c t i o n s

and e s t a b l i s h some p r o p e r t i e s t h a t w i l l be of f r e q u e n t u s e

in

t h i s book.

16.1.

of a l l

DEFINITION. f E Cm(lRn)

W e s h a l l d e n o t e by

D(3RM,

t h e v e c t o r space

which have compact s u p p o r t . Each f

E

D(Bnl

DIFFERENTIABLE MAPPINGS

is called a t e s t function.

123

U i s a n open s u b s e t o f

If

Rn t h e n

w e s h a l l d e n o t e by D I U l t h e v e c t o r s p a c e of a l l f E D(Wn! such t h a t s u p p f C U. L i k e w i s e , i f K i s a compact s u b s e t of Rn t h e n w e s h a l l d e n o t e by D ( K ) t h e vector s p a c e of a l l f E s u p p f c K.

such t h a t

DtlRn)

I t i s n o t a t a l l obvious t h a t t e s t f u n c t i o n s e x i s t ,

from t h e z e r o f u n c t i o n . But t h e r e s u l t s i n t h e p r e c e d i n g

apart

sec-

t i o n g u a r a n t e e t h e e x i s t e n c e of l a r g e c o l l e c t i o n s o f t e s t func-

U be a n open s u b s e t o f R n l l e t i U i ) b e an open c o v e r of U , and l e t ( 9 ; ) be a Cw p a r t i t i o n o f u n i t y on U , subordinated t o t h e cover I l l i ) . I f each Ui is relatively compact i n V t h e n t h e e n t i r e c o l l e c t i o n ( p i ) i s c o n t a i n e d i n D t V ) . Another example of a t e s t f u n c t i o n . which i s v e r y u s e f u l f o r it s e r v e s t o g e n e r a t e new t e s t f u n c t i o n s , i s t h e following. Indeed, l e t

tions.

16.2.

EXAMPLE.

II x II < 1

if

p : Rn

Let

and

plxi = 0

if

lR b e defined by P I X ) =ke-"iz-'lsrl'21 II x II 1 , where t h e c o n s t a n t k

p d h = 1 , and the l e t t e r

i s chosen so t h a t

0

-+

lRn

d i m e n s i o n a l Lebesgue measure. Then

n

1 5 . A , and

let

x

E

s u p p p = B ( 0 ; l ) . More g e n e r a l l y ,

p6 E D(Bnl

Bn.

for

p6dX = 1

Then

lRn

then we

standsfor

, by E x e r c i s e each

p g ( x ) = 6-np(x/6) and s u p p p g = B ( 0 ; 6 1 .

be d e f i n e d by

U i s an open s u b s e t of

If

p E C"(lRn)

A

3

for

s h a l l denote

6 > 0 every

by

1

I U i t h e Banach s p a c e of a l l e q u i v a l e n t classes of Lebesgue 1 U. We s h a l l d e n o t e by L ( U , l o c ) t h e v e c t o r s p a c e of a l l e q u i v a l e n t classes of Lebesgue m e a s u r a b l e f u n c t i o n s on U which are i n t e g r a b l e o v e r e a c h compact s u b s e t L

i n t e g r a b l e f u n c t i o n s on

of

u.

U b e a n open s u b s e t of B n l l e t 6 3 0 and l e t U g = { x E U : d , l x ) > 6). Given f E L ' I ? J 3 l o c i and M 9 E D ( B ( 0 ; 6 ) 1 w e d e f i n e t h e i r ( ~ ( i n o o l u t i o n f * l p : U6 by 16.3.

DEFINITION.

Let

-+

if

* p i (xi =

I

R(0;61

f(x - y i p i y i d h i y i =

1

f(yl9Px

'Glx;61

- gidh(y)

MUJ I CA

124

x

for every

E

Ug.

16.4. PROPOSITION. L e t U b e a n o p e n s u b s e t o f 1 L (U, l o c l a n d l e t 9 6 P ( B ( 0 ; 6 ) ) . T h e n :

f *p E D(U6)

In particular

p a c t s u b s e t of

PROOF.

U2&

For each

-

Ug

x E

Bn, l e t

i f t h e s u p p o r t of

f

f E

i s a corn-

we can write

By differentiation under the integral sign we 1 E C ( U s ) and

get that

for every j . Then (a) and (b) follow by induction. Since is clear, the proof of the proposition is complete.

f * l p

(c)

16.5. PROPOSITION. L e t U b e an o p e n s u b s e t o f E n and f E CtU) b e a f u n c t i o n w i t h c o m p a c t s u p p o r t . T h e n f * p 6 c o n '(2t

oergss t o

f

uniformly on

Li

when

6

0.

+

PROOF. Since f is continuous and has compact support, it is uniformly continuous on U. Hence, given E > 0 we can find > 0 such that s u p p f c U Z 6 and IJ: - y i - f ( . c i [ 5 E for

If

0

x

every

E

Ug

and

y

j '

3: E

U

6

we have that r

E B(0;60).

If

0

6

then for every

DIFFERENTIABLE MAPPINGS

125

16.6. PROPOSITION. L e t U b e an o p e n s u b s e t o f Bn f E L 1 (U) be a f u n c t i o n w i t h compact s u p p o r t . T h e n :

PROOF. ( a ) Since f * p6 vanishes o u t s i d e of t h e F u b i n i Theorem shows t h a t

U6

and

let

an a p p l i c a t i o n

( b ) S i n c e t h e c o n t i n u o u s f u n c t i o n s w i t h compact s u p p o r t a r e dense i n

L 1 (U!, g i v e n

E

w e can f i n d a f u n c t i o n g E C ( U ) ,

> 0

w i t h compact s u p p o r t , such t h a t

< dUfsupp gl and l e t

K = sup

1 6 . 5 w e can f i n d

with

< E/X(K) 0 .< 6 < 6

JIu l f

*p6

whenever 0

PROOF.

of

u

-

fldh 5

Let

E.

g + B ( 0 ; r l . Then by

0 <

0 < 6 < 60.

lg

6o < r

such t h a t

0 < 2r

Proposition ]g *P6

-

g ]

Then, u s i n g p a r t ( a ) we g e t f o r

that

- fJdX

16.7. COROLLARY. dense in

6,

1,

If U is a n o p e n s u b s e t of W

n

then

PtUl is

1 L (U!.

L e t (K.) be an i n c r e a s i n g s e q u e n c e o f compact subsets 3 w h i c h c o u e r U. If f E L’ f u ) t h e n I U ] f x , f dX 0

-

j

+

MUJ I CA

126

when

j

m,

-+

by t h e Dominated Convergence Theorem. S i n c e

fX

of t h e f u n c t i o n s

h a s compact s u p p o r t ,

each

it s u f f i c e s

to

a n . Then

the

Kj apply Proposition 16.6.

16.8.

DEFINITION.

K be a compact s u b s e t of

Let

D(K) w i l l

b e always endowed w i t h t h e l o c a l l y c o n k vex topology g e n e r a t e d by t h e seminorms f -+ s u p 1 I D f fx) 11,

v e c t o r space

xE K

where

k varies over a l l n o n n e g a t i v e i n t e g e r s .

Using E x e r c i s e 14.F w e see t h a t t h e t o p o l o g y of

f

a l s o g e n e r a t e d by t h e seminorms o f t h e form where

v a r i e s over a l l m u l t i - i n d i c e s i n

c1

nV:.

+

s;p

P(K)

I t f o l l o w s from

D f K ) i s a Frechet space. Actually

Proposition 13.13 t h a t

i s a c l o s e d vector s u b s p a c e i f

Cw(lRn),

is

1 a"f/az'( 0110

which i s also a EYGchet

s p a c e , by E x e r c i s e 1 4 . E . 16.9.

DEFINITION.

Let

u be an open s u b s e t of

lRn. '&en the vec-

t o r s p a c e D l U ) w i l l be endowed w i t h t h e f i n e s t l o c a l l y

on

t i o n t o t h e Frgchet space subset

K

of

D(K)

DCU) i s conwords, a D(Ul i s c o n t i n u o u s i f and o n l y i f i t s r e s t r i c -

t o p o l o g y such t h a t t h e i n c l u s i o n m a p p i n g , t i n u o u s f o r e a c h compact s u b s e t K of seminorm p

convex

U. I n o t h e r

D(K) i s c o n t i n u o u s f o r e a c h compact

U.

It i s clear from t h e d e s c r i p t i o n o f t h e t o p o l o g y of

t h a t a l i n e a r mapping convex s p a c e Y

T t o each

T : DlU)

+

Y

from

D(U)

P ( U ) into a locally

i s c o n t i n u o u s i f and o n l y i f t h e r e s t r i c t i o n o f

D ( K ) i s continuous.

EXERCISES 16.A.

Let

Lp(U)

and

function

U b e an open s u b s e t of B n , l e t 1 2 p < L P I U , l o c ) be d e f i n e d i n t h e obvious way. f E L p ( U ) w i t h compact s u p p o r t show t h a t :

a,

and l e t a

Given

DIFFERENTIABLE MAPPINGS

1 If

(b)

- flPdh

*p6

.+

6

when

0

127

0.

--*

U

U i s an open s u b s e t of

If

i6.B.

L' (u,L O C I

16.C.

let

f o r every

P

2

U be an open s u b s e t o f

Let

Bn. For e a c h

of Lebesgue measurable f u n c t i o n s d A <

IPe-'

f

:

U

16.1).

U i s a n open s u b s e t of

Show t h a t i f

ping 9 € V ( U I index a .

1x1

<

E

let

define

with

(pi

x

--*

that

=

CmIU;RI.

cp €

multi-

K = s u p p v + ~(U;E).

by.

' . x Ae

.I

~ ( x )

Using C o r o l l a r y 1 3 . 9 show t h a t u n i f o r m l y on

R n t h e n t h e map-

and l e t

> 0

E

E DIKI

PIX +

CphlX)

(b)

such

V l U l i s c o n t i n u o u s for e a c h

aclip/azcl €

.+

9 E DllRnI,

1 6 . ~ . Let

a9/axj

Cm(u;mi

E

Using E x e r c i s e 1 5 . C show t h a t LpfU,Zoc) i s the

m.

LP(U,pI,

(a)

9

M

-+

union of t h e s p a c e s

For 0 <

L p ( U , Zoc) C

d e n o t e t h e Banach s p a c e of a l l e q u i v a l e n t classes

LpIU,cpI

I,If

show t h a t

lRn

1.

K when

X

+

converges

(PA)

to

0.

Show t h a t ( v x l c o n v e r g e s t o

aP/axj

VIKI

in

when

0.

16.F.

Let

(a)

U be an open s u b s e t of Show t h a t

Rn.

a"f *

-

f E Cm(UI and

(b)

$ E

P in

on

ug

f o r every

D(B(0;G)l.

Using ( a ) and P r o p o s i t i o n 1 6 . 5

converges t o

$

ax

ax"

D t U I when

6

+

0

fv *

show t h a t

f o r every

9

E

PGl

V(UI.

1 7 . DISTRIBUTIONS In t h i s section we introduce distributions

and

establish

128

MUJ ICA

some properties that will be frequently used in this book. The idea is to define a new class of objects, called distributions, which should include all continuous functions on l R n , and should have, in some sense, partial derivatives of all orders. The partial derivatives of distributions should again be distributions, and in the case of cm functions, the partial derivatives in the sense of distributions should coincide with the partial derivative in the classical sense.

17.1. DEFINITION. Let U be an open subset of l R n . Thena d i s t r i b u t i o n on U is a continuous linear functional on Q t U ) . In other words, a distribution on U is a linear functional on P(Ul whose restriction to Q I K I is continuous for each compact D'IUi the vector subset K of U. Thus we shall denote by space of all distributions on U. 17.2. EXAMPLE. Let U be an open subset of lRn. Then each function f E C I U ) defines a distribution T E v r ( U l by T (pl

1,

=

f

f o r every

VfdX

p E

f

Q(u~.

Using Proposition 16.5 one can readily show that the mapping C(U) T S E D'tUl is injective. Hence we shall identify each f E C ( U ) with its image Tf E D r fUi and speak of the d i s tribution f. f

E

-+

17.3. EXAMPLE. Let U be an open subset of l R n . Then more 1 generally, each f f L ( U , Z o c i defines a distribution T f D'IUI

by

T (rpl

I

f

for every

qfdh

v

E

Q l u l . Using Propo-

f U 1 sition 16.6 one can readily show that the mapping f E L (U,Zocl Tf E D ' I U ) is injective. Hence we shall identify each j' E

+

1

L ( U , Z o c l with its image

tion

f.

T

f

E

D'(U! and speak of the distribu-

Thus we see that in particular we can identify C i m i U i with a vector subspace of D ' ( U l . We would like to extend the differential operators an/aza from C*(UI to all of ~ r i u ) . TO motivate the definition take f E C m ( U l and p E D C U l . Since ~p

D I F F E R E N T I A B L E MAPPINGS

129

has compact support it follows from the Fubini Theorem and the integration by parts formula that

for every

j. Then by induction we get that

for every multi-index tion.

a. This motivates the following defini-

17.4. DEFINITION. Let U be an open subset of Bn and let T E D ' l U ) . For each multi-index c1 we define a"T/az" E P ' ( U ) by

f o r every

v,

E

DIU).

Since the mapping

9 E DfUl

+

a"lp/aza

E

D(Ul is linear and

continuous, it is clear that a"T/asa is indeed a distribution on U. It is a l s o clear that if f E C m l U ) then the derivative a"f/az" in the sense of distributions coincides with the derivative a"f/as" in the classical sense. Next we introduce a locally convex topology on D l t U ) . The topology we choose is very simple and is sufficient for a l l o u r needs in this book. DEFINITION. Let U be an open subset of B n . Then the vector space D ' ( U ) will be always endowed with the topology of p o i n t u i s e c o n v e r g e n c e , that is I with the locally convex toplcqy generated by the seminorms of the form T supp I T ( ' ) 1, where 17.5.

-+

cpE @

varies over all finite subsets of

DtUl.

Then the following result is clear.

MUJ I CA

130

17.6. PROPOSITION. mapping T E D ' ( U ) i n d e x a.

U b e ax o p e n s u b s e t of l R n . T h e n t h e aaT/aza E D r ( U ) i s continuous for each m u l t i -

Let +

In the preceding section we proved that D ( U ) is dense in L (U), and using Exercise 17.A one can actually prove that DlU) is dense in L p ( U ) for every p 2 1 . Now we would like to obtain a similar result for distributions, namely we would like to prove that DtU) is dense in D ' C U ) . With this aim in wind we shall introduce two new operations with distributions. First we shall define the multiplication of a distribution and a cw function, and afterwards we shall define the convolution of a distribution and a test function. 1

17.7. DEFINITION. Let C m l U l and T E D'(U)

E

for every

U be an open subset 05 B n . Given f we define the product fT E D ' ( U ) by

cp E D l u ) .

Since the mapping 9 E DlUl ~f E DlUl is linear and continuous, it is clear that f T is indeed a distribution on U. 1 If g E L ( U , l o c ) then it is also clear that the product fg in the sense of distributions coincides with the pointwise product fg. +

17.8. PROPOSITION. Cm(U)

and l e t

for e v e r y PROOF.

n U b e an o p e n s u b s e t of IR , l e t

Let

T E D'lU).

j = 1,

For every

Then:

..., n. v

E

D ( U ) we have that

f

E

D I F F E R E N T I A B L E MAPPINGS

Let

V

C

U

be two open s u b s e t s o f

131

B n . Then

D(V/ C D(UI

U d e f i n e s by r e s t r i c t i o n a d i s t r i b u t i o n on V . If T E o r(U1 t h e n w e shall s a y t h a t T = 0 on V i f T ( q ) = 0 f o r every cp E 0111). I f f o l l o w s from t h e n e x t propos i t i o n t h a t f o r e v e r y T E D r t U ) t h e r e e x i s t s a l a r g e s t open s e t V C U ( p o s s i b l y empty) where T = 0 . and e v e r y d i s t r i b u t i o n on

17.9.

Let

PROPOSITION.

T E o r ( U ) . Let

(ViiiEI

T = 0

such t h a t

for e v e r y

on Vi

i

E I.

Then

T = 0

on t h e

vi .

u

V =

open s e t

U be an o p e n s u b s e t of lRn and l e t be a c o l l e c t i o n of o p e n s u b s e t s 3f U

i EI PROOF.

L e t (+i)iEI

be a

Cm

o r d i n a t e d t o t h e cover ( V i l i E I .

p a r t i t i o n of u n i t y on If

cp E P t V )

then

q

V , sub-

=

9Qi

and t h e sum h a s o n l y f i n i t e l y many n o n z e r o t e r m s , s i n c e cp compact s u p p o r t and t h e c o l l e c t i o n

i s also clear t h a t z T ( q ~ $ ~= ) 0

17.10.

T

E

cp$i

E V(Vi)

(qii is

f o r every

locally finite.

i.

Hence

has It

T(cp) =

and t h e proof i s c o m p l e t e .

DEFINITION.

Let

U

be an open s u b s e t of

IRn

and

let

V i s t h e l a r g e s t open s u b s e t of U where T = 0 U \ V i s c a l l e d t h e s u p p o r t of T and w j . 1 1 be supp T.

p r i U j . If

then t h e set d e n o t e d by

17.11.

EXAMPLE.

CIU).

Then

U b e an open s u b s e t of s u p p Tf = s u p p f . Let

lRn

and l e t

f E

n U b e a n o p e n s u b s e t o f B . Then t h e d i s t r i b u t i o n s on U w i t h compact s u p p o r t a r e d e n s e i n D ’ ( 1 I j . 1 7 . 1 2 . PROPOSITION.

PROOF.

Let

Let

( K .I be a sequence of compact s u b s e t s of U such t h a t 3

132

MUJ I CA

m

0

U =

U K and K. C j=l j 3 t h e r e i s a sequence ($i) i n

borhood o f

K

' P $ ~= 9

then

j. By

D ( U ) such t h a t C Kj+]

on a n e i g h b o r h o o d of

K

j hence t h e s e q u e n c e ( 9 $ k l c o n v e r g e s t o 9 sequence ( q k T I converges t o

T

Since

Zk+],

in

15.5

Corollary

= 1 csn a neigh-

$J

j f o r e v e r y j. I f

0

supp Jij

and

j

f o r every

9 t

D(K.1

k 2 j, D ( K . I . Hence

f o r every

in 3 D ' f U ) f o r every

3

and the

9 E v'(UI.

n

s u ~ p l $ ~ TC l s u p p JI,

C

t h e proof i s complete.

Next w e want t o d e f i n e t h e c o n v o l u t i o n of a d i s t r i b u t i o n 1 and a t e s t f u n c t i o n . W e recall t h a t i f f E L (U,locl and 9 E I ) ( B ( i ) ; 6 ) 1 t h e n t h e c o n v o l u t i o n f ' * p : U6 + IK i s g i v e n by ( ~ * P ) ( x )=

r

f(y)+'(J:

- y)dX(yl

JU

f o r every

17.13.

DEFINITION. and

E l?'(U)

+. IK

x E U g . T h i s motivates t h e f o l l o w i n g d e f i n i t i o n . Let

U be a n open s u b s e t of l R n . G i v e n T w e d e f i n e t h e i r c o n v o l u t i o n T * V : U6

9 E D(z(O;6))

by

x E U6. Here T [ p(x - y ) ] means t h a t t h e d i s t . r i b u Y T i s a p p l i e d t o t h e t e s t f u n c t i c n ' P ( X - y ) r e g a r d e d as

f o r every tion

a f u n c t i o n of

y

for

z fixed.

f E L 1 ( U , l o c ) t h e n it i s clear

If

that

the

convolution

f*'P

i n t h e s e n s e of d i s t r i b u t i o n s c o i n c i d e s w i t h t h e c o n v o l u -

tion

f

17.14. E

Vr(U)

*9 in

t h e sense o f functions.

PROPOSITION.

and l e t

Ip E

U b e a n o p e n s u b s e t of D ( B ( O ; S I ) . Then:

Let

IRn,

let

1

D I F F E R E N T I A B L E MAPPINGS

* CP)

supplT

(c)

T *p

In particular

p a c t s u b s e t of

E

D ( U 6 ) i f t h e support of

Indeed, choose

< 0

E

B(a;E

+ 6 1 . If we set

and y

€

lRn

* cp

T

such t h a t

i s continuous a t each a

-

B(a;rl

= 9lx -

$,(y)

t h e n it i s clear t h a t

fore ( T * 9 l i x ) = T 1 $ x )

j = 1,.

.

in

$,

-+

NOW,

by E x e r c i s e 1 6 . E

(pX

+

D(E(o;E

in

ap/axj

Y

*

ax

( a ) . Hence

i

x E g(a;e) D I K ) and there-

I T * 9 ) ( a ) e x i s t r ; and equd~s( T * -Ia9 3X dx

whence it f o l l o w s t h a t IT * q A )( a ) = T [ v A l a

= IT

U 6' K =

(a)

j

where

0 <

E,

y i f o r each

QX

let

and

U6

j . ,n. I n d e e d , f i r s t w e observe t h a t

for

IAl <

C

E

T ( Q a ) = ( T * V j ( a ) , as w e wanted.

-+

Next w e show t h a t f o r every

i s a com-

T

U2&.

F i r s t w e show t h a t

PROOF.

+ i(O;61.

supp T

C

133

a ( T * 9 )( a ) ax

-

+ 6))

and

y ) l + T [ -av (a-yl]

Y axj

exists and equals (T

a9 * -)(a), ax

j

Li

as w e wanted. From t h e f i r s t p a r t o f t h e p r o o f w e c o n c l u d e t h a t E

Ci(U6)

and

aza3,(T * 9 )

= T

* *ax

Then i t f o l l o w s b y i n d u c t i o n t h a t

on

U6

for

*9

E

cm(U6)

*

9

j-I, ..., n.

3

T

T

and T(.-

aa

*

9)

ax:

= T *

a"9

f o r every multi-index axa (b) we f i r s t observe t h a t

a . To c o m p l e t e t h e proof of

134 for

MUJ I CA

j = I,.

. .,n.

Whence it f o l l o w s by i n d u c t i o n t h a t

- d9* f o r every multi-index ax'

-

T

aq ' * -

ax" from

a . S i n c e ( c ) i s clear

t h e d e f i n i t i o n s , t h e proof of t h e proposition i s complete. If

$ E V(Bn)

t i o n d e f i n e d by

E DfBn) w i l l d e n o t e t h e t e s t func-

then

;(XI

= $(- XI

f o r every

x E Rn.

With t h i s

n o t a t i o n w e h a v e t h e f o l l o w i n g lemma.

17.15. LEMMA. L e t U be an o p e n s u b s e t of and l e t $ E D ( B ( 0 ; 6 ) ) . T h e n ( T * $) (9) =

IRnJ TI9

let

* G)

T E D'(U) fGP

every

9 E V(U2&

PROOF. Note t h a t 1" * $ and l e t L = K + z ( O ; 6 ) .

where

c c o ( U 6 ) and

q

*

E

D ( U 6 ) . Let K = s u p p q

Then

V i L ) d e n o t e s t h e c o n t i n u o u s mapping d e f i n e d by = q ( x ) $ ( x - y ) € o r e v e r y x E K and y E # . W e would

f : K

f(x)(y)

E

+

l i k e t o interchange

T and t h e i n t e g r a l s i g n i n t h e l a s t e x b u t s i n c e D i ~ li s n o t a Banach s p a c e

pression f o r ( T * $ ) ( V ) ,

w e c a n n o t a p p l y P r o p o s i t i o n 6 . 5 t o g u a r a n t e e t h a t f i s Bochner i n t e g r a b l e . However, s i n c e D ( L ) i s a F r g c h e t s p a c e , t h e closed convex h u l l o f e a c h compact s u b s e t of VILI i s compact as well, a n d t h e n t h e argument i n E x e r c i s e 6 . F a p p l i e s . Thus there exists a vector S E

0 E D l L ) such t h a t

SlH) = IKS(f(xI!dA(T)

D ( L ) . Then on one hand w e h a v e t h a t

and on t h e o t h e r hand w e have t h a t

f o r every

DlFFERENTiABLE MAPPINGS for e v e r y y

0 = p

Rn. Thus

*

135

* $ ) (9) =

a n d IT

* $1,

T(p

as asserted. Now i t i s e a s y t o p r o v e t h e f o l l o w i n g r e s u l t . 17.16.

T E

n

L? , and

L' b e a n o p e n s u b s e t o f

Let

PROPOSITION.

D ' l U ) be a d i s t r i b u t i o n w i t h compact s u p p o r t . Then Q ( U ) tThen 6 > 0 i s s u f f i c i e n t l y s m a l l , and T * p 6

E

D'(Ul

6

when

PROOF. and

supp 9

whenever

9 C

E

D(Ul

U2?.

choose

Then

r > 0

T ( p ) when

6

+

p6

in

supp T C U2r

such t h a t

T * p g E D(Url

and

cp * p 6

DIUrl

E

0 < 6 c r . U s i n g Lemma 1 7 . 1 5 a n d E x e r c i s e 16.F,

observing t h a t

*

0.

+

Given

+

let

9 T

p6

=

we g e t t h a t

(T

*

p 6 j (9)

= T(p

and

*

+

0.

From P r o p o s i t i o n s 1 7 . 1 2 a n d 1 7 . 1 6 we g e t a t o n c e t h e

fol-

lowing c o r o l l a r y .

If

1 7 . 1 7 . COROLLARY.

i s dense i n

U i s a n open subse't o f

IRn

then

D(Ul

D'lU).

W e end t h i s s e c t i o n w i t h t h e f o l l o w i n g r e s u l t , which

will

be needed later on.

1 7 . 1 8 . PROPOSITION. --t

Let

U b e a n o p e n s u b s e t of R

n

.

Let

f:U

b e a f u n c t i o n w h i c h is L i p s c h i t z c o n t i n u o u s , tizut is, t h e r e

1K

i s a constant

k > 0

all

x , y E U. T h e n

and

p

If l x l -

such t h a t

f(yl

I 5

a f / a z j E Lp(U;Zoc) f o r e v e r y

-

k IIz y II for j = 1 , ... , n

1.

T o prove t h i s p r o p o s i t i o n w e n e e d t h e f o l l o w i n g lemma.

17.19.

LEMMA.

Let

Cf.)3

be a sequence o f

f u n c t i o n s d e f i n e d on a measurable space

zE X PROOF.

for w h i r h t h e s e q u e n c e For each

m,n

E

X-valued

measurable

X. Thcn t h e s e t of p o i n t s

(f.(x)) c o n v e r g e s is m e u s , r a b Z e . 3

IN c o n s i d e r t h e m e a s u r a b l e s e t

136

MUJ I CA

S i n c e t h e s e t of p o i n t s m

m

n

U

m=l

n=l

f o r which ( f . ( x i l c o n v e r g e s i s

x E X

3

, t h e desired conclusion follows.

Amn

PROOF OF PROPOSITION 17.18.

n = 1 . W i t h o u t loss U i s a bounded open interval.

F i r s t suppose

of g e n e r a l i t y w e may assume t h a t

Then t h e L i p s c h i t z c o n d i t i o n i m p l i e s t h a t t i n u o u s / and i n p a r t i c u l a r o f bounded

i s a b s o l u t e l y con-

f

v a r i a t i o n . By a t h e o r e m

f ' ( a l e x i s t s f o r almost e v e r y

o f Lebesgue t h e d e r i v a t i v e

a E

If'lall 5 k w h e r e v e r i t e x i s t s . N O W / s i n c e f i s a b s o l u t e l y c o n t i n u o u s , t h e i n t e g r a t i o n by p a r t s formula

U, and c l e a r l y

cp

-

f 'dx =

(p'

fdx

is v a l i d f o r every

Dill),

(p E

and

hence

f i n t h e s e n s e of d i s t r i b u t i o n s c o i n c i d e s with

t h e d e r i v a t i v e of

t h e c l a s s i c a l d e r i v a t i v e . The d e s i r e d c o n c l u s i o n f o l l o w s .

n

Next s u p p o s e

W e s h a l l prove t h a t

2.

af/ax,

E LpIU,locl

p 1 . The p r o o f f o r af/ax i s analogous. Without j loss o f g e n e r a l i t y w e may assume t h a t U = A x B , where A i s ;? bounded open i n t e r v a l i n B and B i s a bounded open set i n f o r every

nn- 1 . L e t

C d e n o t e t h e s e t of a l l p o i n t s ( a , b l

that the partial derivative

E

A

( a , b l e x i s t s . Then

such

H

x

is

C

a

axl

measurable s u b s e t of

let

denote

Cb

the

A

x

set

B , by Lemma 1 7 . 1 9 .

of

a l l points

For e a c h

a

E

b

E B

such

A

that

aS

( a , b l e x i s t s . Then, a g a i n by Lemma 1 7 . 1 9 , e a c h C b i s a 3x1 m e a s u r a b l e s u b s e t of A , and c l e a r l y C U iCb x { b } ) . Let DEB

C' d e n o t e t h e complement of t h e complement of

u

(C6

x

Cb

in

C

A,

in

A x B , and l e t

f o r each

b

E

B.

Ci

denote (7 '

Then

I b F ) . NOW, s i n c e f i s L i p s c h i t z c o n t i n u o u s ,

it

= is

bkR

c l e a r t h a t f o r e a c h b E R t h e f u n c t i o n f i z l , b l i s absolutely c o n t i n u o u s , a n d i n p a r t i c u l a r of bounded v a r i a t i o n on A . Then i t f o l l o w s a g a i n from Lebescjue's Theorem t h a t t h e s e t C i has o n e d i m e n s i o n a l Lebesque measure z e r o €or e a c h h t B . Then a d i r e c t a p p l i c a t i o n o f t h e F u b i n i Theorem shows t h a t t h e s e t C has

71

d i m e n s i o n a l Lebesgue m e a s u r e z e r o . Thus t h e p a r t i a l de( a , b l exists for

rivative i)x1

almost e v e r y i a , b l

E

ll,

and

137

DIFFERENTIABLE MAPPINGS

1 aax f(a,b) I

clearly

I

5 k

wherever it e x i s t s .

f o l l o w s from t h e a b s o l u t e c o n t i n u i t y of f ( x l J b l

rivative

before,

As

that the

it de-

i n t h e sense of d i s t r i b u t i o n s coincides w i t h af/ax, i n t h e classical sense. This completes

af/axl

the derivative t h e proof.

EXERCISES

U b e a n open s u b s e t o f B n , l e t a E U and l e t : P t U ) + IK be d e f i n e d by G a l c p l = c p ( a ) f o r e v e r y cpE V t U ) . Show t h a t G a i s a d i s t r i b u t i o n on U , c a l l e d t h e D i m e measure 17.A.

Let

a.

at

17.B.

Let

Y : LP YIxl =

b e t h e H e a v i s i d e function,

B

+

which

is

x < 0, and Y t x l = 1 i f x > 0. S b t h a t Y d e f i n e s a d i s t r i b u t i o n on IR. Show t h a t t h e d e r i v a t i v e d e f i n e d by of

if

(I

i s t h e Dirac measure

Y

17.C.

Let

U b e a n open s u b s e t o f

T E D'lUl

tion

60.

iR

n

Ti E

Let (UiliE1

D'(1l.l

E

U(Ui

E

PIUi)

i

E

T E o'(Ul

and e v e r y

i

E

Ui n U

Let

Ti(cpl

#

j such t h a t

I. Show t h a t

U be a n open s u b s e t o f Show t h a t i f a f / a x j = g f o r some o f d i s t r i b u t i o n s , t h e n af/axj = g 17.E.

U

of

there exists a distribution

I

with t h e property t h a t

n Uj), whenever

a distribution

distribuT extends

b e an o p e n c o v e r o f a n open s u b s e t

Suppose t n a t f o r e a c h

LPn.

Show t h a t a

c"(u).

as a c o n t i n u o u s l i n e a r f u n c t i o n a l t o 17.D.

.

h a s compact s u p p o r t i f and o n l y i f

@.

= T.(cp)

for

3

every

Using Theorem 1 5 . 4 f i n d

Ttcpl = Tilcp) T

i s unique.

lRn

and l e t

j (j = 1 , .

f o r every

cp

f, g E c ( U ) .

..,n)

i n the sense

i n t h e classical sense.

NOTES AND COMMENTS

The m a t e r i a l i n t h i s c h a p t e r c a n b e f o u n d i n many s t a n d a r d

138

MUJ I CA

books. We have included only that material which is required for the rest of the book. The material in Sections 13 and 14 can be found, for instance, in the texts of J. Dieudonn6 [ 1 1 , H. Cartan [ 2 ] and L. Nachbin [ 4 1 . The remaining three sections constitute a brief introduction to the theory of distributions. The standard reference f o r the theory of distributions is of course the book of L. Schwartz 1 1 1 . The book of L. Hormander [ 1 ] containsa concise introduction to the subject. The proof of Proposition 17.18 given here is taken from the book cf S. M. Nikol'skii [ 1 1 .

CHAPTER V

DIFFERENTIAL FORMS

1 8 . ALTERNATING MULTILINEAR FORMS I n t h i s s e c t i o n w e introduce t h e exterior product

of

al-

t e r n a t i n g m u l t i l i n e a r f o r m s . T h i s i s i n d i s p e n s a b l e f o r the study

of d i f f e r e n t i a l f o r m s , t o b e g i n i n t h e n e x t s e c t i o n .

18.1. DEFINITION.

Given

A

E

XaimE)

t e r i o r p r o d u c t o r wedge p r o d u c t

and

A I\ B

B E Ea(nE)

t h e i r ex-

E a ( m + n E ) i s d e f i n e d by

€

I n o t h e r words,

The e x t e r i o r p r o d u c t t h e mappings A

or

the mapping i A , H I 18.2. (-

PROPOSITION.

?irnnA

A

A

A B

c a n also b e d e f i n e d i f o n e o f

B t a k e s v a l u e s i n a Banach space. -+

A

If

A B

Clearly

i s b i l i n e a r and c o n t i n u o u s .

A E XaIm,Y)

arid

H

LainE)

then B A A =

B.

PROOF: I f c1 d e n o t e s t h e p e r m u t a t i o n t a k i n g 11, . . . , m + n ) i n t o i n + 1 , . . . , n. + m , 1 , . . . , n i t h e n i t i s c l e a r that (- ] = l a

i-

u " ~ Hence . 139

M U J I CA

140

18.3. PROPOSITION.

If

A E ea("El,

5

E

g"(ng)

and

C t Lff('EI

then

T h i s p r o p o s i t i o n i s a n immediate c o n s e q u e n c e o f t h e f o l l o w i n g lemma.

18.4. LEMMA.

Let

A E EemEl,

B E d : ( n E ) and

(a)

( A eP B l a = 0

(b)

(Aa 8 Bia

= (A 8 B a j a = ( A 8 B i a .

(C)

[ ( A 8 B)'

8 C l a = [ A 8 (B 8

whenever

A'

= 0

C E E ( I P E ) . Then:

or

Gala

B"

= 0.

= (A 8 B Q C ) ' .

( a ) W e show t h a t ( A 8 B l a = 0 whenever An = 0. The o t h e r statement i s p r o v e d s i m i l a r l y . L e t T d e n o t e t h e s u b g r o u p

PROOF.

which l e a v e m+n i s o m o r p h i c t o Sm and

of a l l

T E

S

m + 7,

..., m + n

f i x e d . Then T i s

DI FFERENTI AL

..., rn + n .

j = 3,

Sm+n

a

( x ~ , . . . , x ~1 + ~

-

(b) Since ( A a and t h e r e f o r e

0

= (A

Q

i s t h e u n i o n of t h e cosets a T , and since

nT are e i t h e r d i s j o i n t o r i d e n t i c a l , w e get t h a t

(rn + n)! ( A 8 P )

=

141

Then

Since t h e group t h e cosets

FORMS

A)a

= 0,

(a) implies t h a t

( A a 8 B l a = (A 8 B l a .

[ (Aa

- A)

8 B1'

The e q u a l i t y (A Q Ballz

is proved s i m i l a r l y .

B)'

( c ) f o l l o w s a t o n c e from ( b ) . If

A E GaimE)

and

gains) then the tensor

R E

i s a l t e r n a t i n g i n t h e f i r s t m v a r i a b l e s and

A 8 B

i n the last

product

alternating

n varia.bles. This motivates t h e following d e f i n i -

tion.

1 8 . 5 . DEFINITION. s p a c e of a l l

rn

A

€

W e s h a l l d e n o t e by firn+nE;F)

E a m n (rn+nE; F)

v a r i a b l e s and a l t e r n a t i n g i n t h e l a s t

that

sub-

n variables.

denote t h e set of a l l permutaticns

o

Snt+n such n i ? ) < ... elm) and a i m + 1 ) ... olrn + n ) . Note S,, n a s (rn + n l ! / i n ! n ! e l e m e n t s . Then w e h a v e t h e fol-

Let

that

the

which are a l t e r n a t i n g i n t h e f i r s t

Smn

E

l o w i n g p r o p o s i t i o n , whose p r o o f i s s t r a i g h t f o r w a r d and

i s left

as a n exercise t o t h e r e a d e r .

18.6. P R O P O S I T I O N . be d e f i n e d by

For e a c h

A

E L(rn+nE;F)

let

E E(rn+nE;F!

MUJ I CA

142

2'hen

Aamn

mapping

A

= A

a

for every

A E Lamn(m+nE.J. I n p a r t i c u l a r

amn m+n continuous p r o j e c t i o n from E I (E;F)

induces

-+

the

t a("+"E; F ) .

onto As

an immediate c o n s e q u e n c e w e g e t a n o t h e r , somewhat

dif-

f e r e n t formula f o r t h e e x t e r i o r product.

1 8 . 7 . PROPOSITION.

If

and

A E La(n'E)

then

B E LafnE)

In o t h e r w o r d s , (A A B)(xl,

..., x m i n

W e end t h i s s e c t i o n w i t h t h e f o l l o w i n g r e s u l t . , which paral-

l e l s Theorem 1 . 1 5 .

18.8. THEOREM, If E a n d F a r c c o m p l e x fic?n,i(.h sp,;<.e:. t h c n a m d: ( ER; F R ) is t h e t o p o l o g i c a l , d i r e c t s u m of rhc subspaces L ~ ( ~ ~ ( E ;w Fi t Ih PROOF.

p + q = m. m

By Theorem 1 . 1 5 e a c h

written in the form

A = A

o

A E d:( E B ; F

...

+ A] +

+ Am

can be

uniquely rri-k, k where .qk€ LI L'; F )

IR )

f o r k = 0 , l , . . . , ni. To p r o v e t h e t h e o r e m i t i s s u f f i c i e n t t o show t h a t i f A i s a l t e r n a t i n g t h e n e a c h A k i s a l t e r n a t i n g a s w e l l . T o show t h i s p i c k

ni

t h e i n t e r v a l (0,IT) and s e t for I'

-

*

j = 0 ,..., m . Then, f o r E E w e have t h a t

. Gm

+

1 A. J

j

d i s t i n c t numbers

= c

i0.

0,

:

and

...,

,R,

. . . ,Ow..

in

-

5 ti

111

= X ./A. = c -s,% I,

ti

and a r b i t r a r y

ECtOiS

143

DIFFERENTIAL FORMS

,<, ..., 5,

S i n c e t h e complex numbers

are d i s t i n c t , t h e d e t e r -

m i n a n t of t h e c o e f f i c i e n t s o f t h e s y s t e m of e q u a t i o n s

...

(A,

(18.1)

t i m e s a n o n z e r o Vandermonde determinant. Then i t f o l l o w s from C r a m e r ' s Rule t h a t e a c h A k ( x l , . . , x m / can b e w r i t t e n i n t h e form equals

Am)rn

.

...,

rn

A ~ ( x ~ , x I =

m

Z

ckjA(A x l ' . . . , A

j=O

where t h e complex numbers

.J:

J r n

c k j are i n d e p e n d e n t from

xl, . . . , x V T . whence i t i s clear t h a t e a c h

tors

nating i f

J,

Ak

the is

vec-

alter-

is alternating.

A

EXERCISES

X, w i t h E f i n i t e d i m e n s i o n a l . L e t i e l , .. . , e n I be a b a s i s f o r E a n d l e t 5, ,..., 5, 18.A.

Let

E

and

F be Banach s p a c e s o v e r

be t h e c o r r e s p o n d i n g c o o r d i n a t e f u n c t i o n a l s . laimE;F) = I01

(a)

Show t h a t

(b)

Show t h a t i f

1

-

m

5

n

whenever then

rn > n.

each

A

E

ea(rnE;F)

c a n be u n i q u e l y r e p r e s e n t e d as a sum

E F and t h e summation is t a k e n o v e r a l l multij,. . . j r n i n d i c e s i j , , . . , , j m ) s u c h t h a t I 5 jl ... jm 5 p i .

where

c

(c) m

-.

Show t h a t

h a s dimension

Given

and

i

n

)

whenever

0 -

( j = i ,

..., n )

n.

n

(d)

AELajnR;F)

z.= J

xikekE E

j=1

MUJ I

144

CA

show that A ( x l ,..., x )

n

18.B. of

Let

zl,

...,

2

=

det(x

jk

I A ( e , ,..., e n ) .

be the complex coordinate

n

functionals

and let F be a complex Banach space.

Cn

(b) Show that if 0 p 5 n, 0 2 4 5 n and p + I 2 1 then each A E d : a ( p q C n ; F ) can be uniquely represented as a sum

where

and the summation is taken over a l l

multi-indices (j],. . .,jp,k , , n. and 1 5 k l i . . . i k

. . . ,k 4 )

such that

i<j,

,

..

<

jp(n

4 -

(c)

0 5 p 5 n

Show that if

has dimension I

.'

P

) (

4

and

0 -<-

q 5 n

then Ea(pqt?)

I.

19. DIFFERENTIAL FORMS

In this section we introduce differential forms in Banach spaces. We define the exterior differential d and derive its elementary properties. Throughout this section the letters 6 and F will represent real Banach spaces. 19.1. DEFINITION. If U is an open subset of E then we shall denote by RmfU;F) the vector space of all mappings f

Each degraee

f

E

Rm(U;F)

m on U.

:

U

+

EnimE;F).

is called an F - v a l u e d If

k E flo u

{m}

diffcrcntial,

then we shall

forit7i

af

denote

by

145

DIFFERENTIAL FORMS

k

C m ( U ; F ) t h e vector s u b s p a c e of a l l

class

If

F

C

k

.

I n o t h e r words,

8 ( r e g a r d e d as a r e a l Banach s p a c e )

i s LR o r

shall w r i t e

= Rm(U)

Rm(U;Fi

19.2. DEFINITION.

Let

f i x ) A g ( x ) f o r every

t h e d i f f e r e n t i a l forms +

f A g

19.3. PROPOSITION. g E R,(VI

E

g

h

U b e a n o p en s u b s e t of E R (Ul. Then:

Clearly

the

Let

E.

f

E

P

g A f = i- I l m n f A g.

(b)

If A g ) A h = f A (g A h ) .

(c)

If f a n d

g a r e of class

Ck

is

aZso

follows

from where

the?? f A g

ck. ( a ) f o l l o w s from P r o p o s i t i o n 1 8 . 2 .

PROOF.

dif-

(f A g l ( x l =

o r g i s Banach-valued.

f

Given

c a n a l s o b e d e f i n e d i f one of

(a)

of c l a s s

E.

R n (Ul , their m t c r i o r product

i s b i l i n e a r and c o n t i n u o u s .

Let and

A

f

g

i s d e f i n e d by

R,+,(UI x E U.

The e x t e r i o r p r o d u c t mapping ( f , g !

and

we

then

k

Crn(U;FI = C m ( U / .

and

f A g E

o r wedge product

k

V b e a n open s u b s e t of

f E Rrn(UI

f e r e n t i a l forms

llrn(U!,

are o f

which

f E Rrn(U;F)

P r o p o s i t i o n 18.3.

(b)

T o show ( c ) w e o b s e r v e that f A g = T o S

and

Since

f

and

g

are of c l a s s

by E x e r c i s e 1 4 . A .

The mapping

and h e n c e i s o f class

C7'

Ck

t h e mapping

T is bilinear

by E x e r c i s e 1 4 . B .

S

i s of c l a s s

and Hence

ck

continuous, f A g = ToS

146

MUJ 1 CA

cm

i s o f class

by Theorem 1 4 . 9 .

From now on w e s h a l l b e p r i m a r i l y i n t e r e s t e d i n t h o s e d i f f e r e n t i a l forms which are d i f f e r e n t i a b l e , and t o s i m p l i t y matters w e s h a l l r e s t r i c t o u r a t t e n t i o n t o Cm d i f f e r e n t i a l f o r m s . W e want

t o a s s o c i a t e w i t h e a c h d i f f e r e n t i a l form

a d i f f e r e n t i a l form df

Cm(U;lQlmE;F))

x

Note t h a t f o r e a c h

E

U

E

f

E

Ci(U;F)

=

C Z + l i U ; F ) = c'Oil ( V ; l a I m+l E ; F ) ) .

w e have t h a t

we

and t h e r e f o r e , w i t h t h e n o t a t i o n o f t h e preceding s e c t i o n , have t h a t

Thus w e are r e a d y t o g i v e t h e f o l l o w i n g d e f i n i t i o n . 19.4.

DEFINITION.

ferential form m

C,+7(U;F)

f o r every

f

Let

U be a n open s u b s e t o f

m

E

Cm(U;FI,

Given a d i f -

E.

i t s e x t e r i o r differential

d f

E

i s d e f i n e d by

x

E

U. From t h e s e c o n d e x p r e s s i o n f o r

d f ( x ) w e see

that

where t h e h a t o v e r The mapping

tj

means t h a t

t

j

is omitted.

m

i s c l e a r l y l i n e a r . To e s t a b l i s h o t h e r p r o p e r t i e s o f t h i s o p e r a t o r w e need a n a u x i l i a r y d : CZ(U;F/

-+

Cm+i(U;FI

lemma. But f i r s t w e g i v e a d e f i n i t i o n . 19.5.

DEFINITION.

W e s h a l l d e n o t e by

.L""O

I

L;Fi

the

sub-

s p a c e o f a l l A € -CC(m+nE;Fi which are a l t e r n a t i n g i n t h e f i r s t .Cue" (m+nE ; F ) the m v a r i a b l e s . L i k e w i s e , w e s h a l l d e n o t e by

147

D l F F E R E N T t A l FORMS

subspace of all n variables.

For e a c h

19.6. LEMMA.

A"'~

and

E

which are alternating in the last

A E d: ( m + n E ; F l

A E l(m+nE;F)

Let

Aamo

Earno m + n (

E;FI

be d e f i n e d by

Laon(m+nE;Fi

AamO(xl,...,x m+n )

and

A a o n (xl,...,x

m+n

I

Then : The m a p p i n g A Aamo i s a c o n t i n u o u s p r o j e c t i o n from xamo m+n I E ; F ) . L i k e w i s e , t h e mapping A Aaon

(a)

+

( m + n ~F ;) o n t o

-+

i s a continuous projection from Aa = 0

(b) <Jr

-

Aaon

(c)

for every

A

E

E ( ~ + ~ E ; oFn~t o

l(m+nE;F1

E~~~

such t h a t

I m+nE ' ; F ) . Aamo

= o

0.

(Aamo)"

=

= A*

( A " O ' ~ ) ~

for every

A

E

L("+~E;F).

(a) is clear. To prove (b) and (c) it sufficesto repeat

PROOF.

the proofs of parts (a) and (b) in Lemma 18.4. 19.7. cvary

PROPOSITION.

f

E C"'(U;F) rn

Let aBd

U b e a n o p e n s u b s e t of E . :"hen g E C Z f l J ; F . ) w e h a v e t h e formula

for

03

PROOF.

By Proposition 19.3,

Exercise 13.D we get that

f A g

E

Cm+n (U;F).

snd

using

148

Similarly we y e t t h a t

and t h e r e f o r e

MUJ ICA

as w e wanted. If U is a n o p e n s u b s e t of

19.8. PROPOSITION. dldfl = 0 PROOF.

for every

Since

E

then

2

d f =

f E CIfU;Fl.

d f ( x i = i m + 111 D f i z l ] a , an a p p l i c a t i o n

of

the

Chain Rule shows t h a t

f o r every

x

E

and

U

t

E

E , and t h e r e f o r e 2

D(dfl (xl = f m + l l [D f ( x j ]

f o r every

x

E

ail, m + l

U. Thus, u s i n g Lemma 1 9 . 6 w e g e t t h a t

Now,

s i n c e , by Theorem 1 4 . 3 ,

D 2 f ( x 1 E l S ( 2 E ; l a ( m E ; F ) ) . Thus

by Lemma 1 9 . 6 , and t h e r e f o r e

= 0,

[ D2f ( z j l a

d 2 f i x l = 0 , as asserted.

To c o m p l e t e t h i s s e c t i o n w e i n t r o d u c e a n i m p o r t a n t

opera-

t i o n €or d i f f e r e n t i a l f o r m s . 19.9.

DEFINITION.

Let

E, F , G

be r e a l Banach s p a c e s , l e t U C E

MUJ I CA

150

and

V

C

F

be o p e n s e t s , a n d l e t

Then f o r e a c h

for all

f

x E U

let

Rm(V;G)

E

cp E C m ( U ; F I

with

9”f E f i m ( U ; G )

cp(UI

V.

C

b e d e f i n e d by

t t , . - . tm E E.

and

J

Rrn (U;G) is c l e a r l y linear . Other The mapping P * : R m ( V ; G I p r o p e r t i e s o f t h i s o p e r a t o r are g i v e n n e x t . -+

19.10. let g E

For e v e r y

(f A g l ( x 1 It I , .

19.11. C

Let

Cm(U;F) w i t h R n ( V ) we h a v e t h u t

PROOF.

lp*

PROPOSITION.

p E

3: E

..

PROPOSITION.

E, V C F

and

$ E Cm(V;GI

with

f o r every

f

E

J

U

and

U C E

lp(U1

C V.

and

V C F

b e o p e n s e t s , and

Then f o r e v e r y

t l , . . . ,tm+n E E

f

E

R m f V I and

we have t h a t

tm+n I

Let W C G

lp(1J)

RmiW;H).

C V

E, F, G, H

be r e a l Aantzch

be o p e n s e t s , and and

let

$ ( V ) C W. T h e n

S ; ~ L ~ - ~ l eSt ,

p € Cw(U;FI

and

DIFFERENTIAL

PROOF.

For every

x

E

I!

FORMS

t l ,..

and

151

.,t m E

E

w e h a v e that

19.12. PROPOSITION. L e t E , F, G he r e a l Banclch s p a c e s , let U C E and V C F b e open s e t s , and l e t 9 E C * ( U ; F I with P(UI C V.

Then:

PROOF.

( a ) If

f E C:lV;Gl

then

9 * f = T o S , where

and

a r e d e f i n e d by

f o r every

x

E

U

and

f o r e v e r y (A,B, ,..., h rn ) E f E E. C l e a r l y S i s of c l a s s (m

i

ll-linear

a m

( F;GI

Cm,

and continuous.

x

[d:(t';FI]

and so

Hence

9*f

is

m

and

t l ,... ' t m

T , since

it

is &so of class

is

c",

MUJ ICA

152

as asserted.

W e have t h a t

(b) dIP*fl

( X )

m I t o J . .., t m l = 2

(-

l ) ' D ( V * f ) 1 x 1 (tj)I t o , .

. .,^t
J

t?,1.

j=O

p*f = T o S

Since

it follows from E x e r c i s e 1 3 . D t h a t

m

n

Hence

Since

f o P I T ) i s a l t e r n a t i n g and

2

I) p i x ) i s s y m m e t r i c , it

fol-

lows that d (P * f I I x l I t o J

where

a

jk therefore

=

. . - ,t m

- a

whenever

ki

j # k . Hence

2 jfk

( Z j k = u

and

DIFFERENTIAL F O R M

153

completing the proof.

EXE RC ISES

Let U be an open subset of B n , let

19.A.

xl,.

. .,xn

denote

the coordinate functionals of W n , and let 1 < rn 5 n. Show that each f E R I U ; F I can be uniquely represented as a sum m

f = Z f . J1"'jm

where

f j l - . .in

:

U

F

+

...

A

dx

jl

A dx jm

and the summation is taken

multi-indices ( j l , ...,jm ) such that

l'j,

...

over

all

< j m I n . This

representation will be often written in the abbreviated form f -

f J d x J . Show that the differential form f is of class

if and only if each of the mappings f

c. n

19.B. Let U be an open subset of L?? the coordinate functionals of W n . (a)

Show that if

f

d f =

(b)

Show that if

df

1

Z df,

E

,

k

if of class

and let x l ,..., x

n

denote

Cm(U;F) then n I:

j=l

x

d

ax j

x

j

.

f = Z: f J d x J E C mm ( U ; F I then J

A dxJ = Z

n

af,

I: - d x , A dx,.

J k=l

J

20.

-

J - 'j1...jm

k

C

axk

THE POINCARE LEMMA

This section is devoted to the proof of a single theorem, a partial converse to Proposition 19.8. As in the preceding section, the letters E and F represent real Banach spaces. 20.1. DEFINITION.

Let U be an open subset of E. A differential

154

MUJ I CA m

g E Cm(lI;F)

form

i s s a i d t o be c l o s e d i f d g = 0,

and i s s a i d m

t o be e x a c t i f t h e r e i s a d i f f e r e n t i a l form t h a t df = g .

f

E

Cm-l ( U ; F ) such

P r o p o s i t i o n 1 9 . 8 asserts t h a t e v e r y e x a c t d i f f e r e n t i a l form i s c l o s e d . W e s h a l l p r e s e n t l y prove a p a r t i a l c o n v e r s e when the open set U s a t i s f i e s a c e r t a i n c o n d i t i o n . 20.2.

A subset

DEFINITION.

with respe ct t o a point x E A and 8 E l 0 , l I .

of

A

E

if (1

a E A

i s s a i d t o be s t . u r - s h a p e d

- 0)a + ex

E A

forevery

Now w e can prove t h e f o l l o w i n g theorem, which i s known

as

t h e Poincarg L e m m a .

20.3.

Let

THEOREM.

U b e a n o p e n s u b s e t of t’ w h i c h i s rn ? 1

s h a p e d w i t h r e s p e c t t o o n e of i t s p o i n t s . I f

star-

then

thc

m

df = g

has a s o l u t i o n f E Cm-I (U;F) f o r ench g E Cm(U;F) such t h a t dg = 0. I n o t h e r words, every closed fom i n

equation

m

00

Cm(U;F)

i s exact.

PROOF.

Without loss of g e n e r a l i t y w e may assume t h a t U i s star-

shaped w i t h r e s p e c t t o t h e o r i g i n . For e a c h d e f i n e a mapping

K : Cm(U;F)

rn

-+

rn > 1

we

shall

C ~ - , ( U ; F ) such t h a t

f o r e v e r y g E C Z ( U ; F ) . The c o n c l u s i o n o f t h e theorem follows a t once from t h i s . Now, g i v e n g E C mm ( U ; F ) we cl-efine Kg E

Rm-l

fU;F)

by

f o r every

x

E

U. O r more e x p l i c i t l y

f o r every

x

E

U

and

t

,,...,

E

E. :E w s e t

in-

p(..r,01= 0

1

g(0xlz

DIFFERENTIAL FORMS

155

then p i x J 8 ) is a Cm function of x in U for each 0 in [ 0 , 1 ] , k and each of the mappings ( x J 8 1 -+ DXp(x, 8) is continuous on U

.

By repeated applications of Proposition 13.14 we conclude that Kg is a Cm differential form and x

[ O J 1]

for every

x

E

U. Whence

for every

x

E

U

and

:0

o n the other hand,

t E E . Thus

MUJ I CA

156 Thus w e g e t t h a t

and t h e p r o o f i s c o m p l e t e .

EXERCISES

20.A.

Let

E, F, G be r e a l Banach s p a c e s , l e t U C E

b e open s e t s , and l e t p(UI = V

that

9-l

E CmIU;FI

Let

U = lR

2

E Cm(V;Ei.

\ (0).

V

C

F

be a n i n j e c t i v e mapping such

I f every closed

i s e x a c t , show t h a t t h e same i s t r u e f o r

Cm(iJ;GI rn

20.B.

and

v

and

form

in

Ci(V;G).

Show t h a t t h e d i f f e r e n t i a l form

g E

CYIU) d e f i n e d by X

is closed but i s not exact.

21.

THE

-

a

OPERATOR

I n t h i s s e c t i o n w e i n t r o d u c e complex d i f f e r e n t i a l forms and 3 o p e r a t o r . Throughout t h i s s e c t i o n t h e l e t t e r s E

define the and

F w i l l r e p r e s e n t complex Banach s p a c e s .

If

U i s a n open s u b s e t of

s e n s e , s i n c e w e may r e g a r d

E

E

and

then t h e space

RT U I U ; F ) makes

F a s r e a l Banach s p a c e s .

157

DIFFERENTIAL FORMS

21.1.

DEFINITION.

d e n o t e by

R

P4

If

fix) E EalPqE;FI C

by

k (U;FI P4

P4

U. L i k e w i s e ,

z E

t h e subspace o f a l l

PROPOSITION.

then we s h a l l

E

R

P+ii

we s h a l l

f E: C p + q I ' U ; F l

denote

such t h a t

f(xl

If

U i s a n o p e n s u b s e t of E

then:

p + q = m.

with

(U;F)

k CmIU;FI (b) k C (U;F) with p P4

i s t h e a l g e b r a i c d i r e c t s u m of t h e s u b s p a c e s

= m.

iq

By Theorem 18.8 t h e r e are c o n t i n u o u s p r o j e c t i o n s

PROOF.

such t h a t

uo +

...

+ um = i d e n t i t y . T h i s i n d u c e s

such t h a t

Go

...

+

Ck

I U ; F I such t h a t

i s t h e algebraic d i r e c t s u m o f t h e subspaces

Rm(U;F)

(a)

R

for e v e r y

f E

FI.

E la(pqE;

21.2.

i s a n open s u b s e t of

11

t h e subspace of a l l

(U;FI

+

um

= identity. I f

t h e n it i s clear t h a t each

If

f E C:lU;F)

and

x

E

Q m m ( U ; F ) is of class

i s a l s o of c l a s s

u of 4

U

E

f

projections

D f i x ) can b e decomposed i n t h e form

D1f(x)

where

D'fIxI

k

.

t h e n i t f o l l o w s from P r o p o s i -

t i o n 13.15 t h a t iDr'S(x),

C

E LiloE;Ea(mE'B;FB

Df(xl

1 and

=

D"f(xl

E 1( o z ~ ; ~ a ( m ~F I ~R ;) . S i n c e i n p a r t i c u l a r

w e are r e a d y t o g i v e t h e f o l l o w i n g d e f i n i t i o n . 21.3.

f

in

DEFINITION. C:(U;FI

let

Let

2f

U be a n open s u b s e t oE and

-

af

in

m

IU;FI

E.

For each

be d e f i n e d by

MUJ ICA

158

f o r every

x

U. In other wordsl

E

and

21.4. PROPOSITION. (a)

Cmi1(U;F)

df =

(c)

a

af + 3f

maps

ernP , q + l (U; F ) . (d)

c z (U;F ) . (e)

m

:

E

then:

00

c m ( U ; F ) -,cm+l ( U ; F I

m(i

2 : C~(U;FI

are Zienar.

(b)

into

a

Both mappings

m -+

U i s a n o p e n s u b s e t of

If

cm

F4

=

If

f E C;(U)

PROOF.

(u;F)

-2 a f = o

a 2f

0,

f o r every

nnd

into

rind

g

f E Cm(U;FJ.

m

coo

? + l J4

aZf

E

(U;FI.

+ $ 2 ~= o

-

a

m a p s C" I U ; F I P4

f o r pvery

f

E

CzfUI t h e n

(a)I (h) and (c) are straighforward consequences of the definitions. In view of Proposition 21.2 it :is sufficient t o show (d) for each f E C;qfU;F!. Then by (b) and Proposition

19.8 we have that

159

DIFFERENTIAL FORMS

NOW,

by ( c ) we have t h a t

and

a2f

E

c ; + ~ (u;F:, ,~

a'f€

c ~ , ~ + ~ ( u ; F Hence ) .

a2f

=

aTf + aaf

0,

Taf

g

5-s

co cp+l,q+l (U;F)

+ a a f = o and

= 0. F i n a l l y , it i s s u f f i c i e n t t o p r o v e ( e ) f o r f m

E

T2f

W

E C

P4

(U!

and

(U). Then by P r o p o s i t i o n 1 9 . 7 we have t h a t

Then u s i n g ( b ) w e g e t t h a t

m

By (c) the f i r s t term b e l o n g s t o C p + y + l , q + s co

t e r m belongs t o 21.5.

U

and

C V.

PROOF.

P*

(a) I f

maps

f

S i n c e t h e mapping P4

Let

E, F , G

be c o m p l e x Banach s p a c e s ,

be o p e n s e t s , and l e t

V C F

cp E S C ( U ; F )

with

let

q(U)

Then:

(a)

R

( U ). The d e s i r e d c o n c l u s i o n follaws.

Cp+r,q+s+l

PROPOSITION.

C E

(UI, whereas t h e s e c o n d

R

E

R

P4

P4

IV;GI

into

(V;G)

then

R

P4

DqPizi i s 6 ' - l i n e a r ,

(U;GI.

it f o l l o w s t h a t

q*f

E

(U;G).

(b)

I f s u f f i c e s t o show ( b ) f o r

s i t i o n s 1 9 . 1 2 and 2 1 . 4

f

E

Cicl

(V;G).

B y Propo-

MUJ ICA

160

By P r o p o s i t i o n 2 1 . 4 ,

-

a(v*f) -

v*(af) E

(u;G) whereas

%v*f)

W

Ip*taJ? E

If

cp,4+z ( U ; GI.

f E Cw(U;Fl

The desired c o n c l u s i o n follows.

is a

Cm

-

af =

mapping t h e n

Drrf

and

it

f E JC(U;FI i f and o n l y i f U. T h i s g e n e r a l i z e s t o d i f f e r e n t i a l forms as f o l -

f o l l o w s from C o r o l l a r y 1 3 . 1 7 t h a t -

af

= 0

on

lows. 21.6. E

PROPOSITION.

C m (U;Fl. PO

-

U be an o p e n s u b s e t of

Let

af =

Then

G

on

if and onZy

U

f

E E

f

ai7d Let

JC(U;d:a(Pt';E,'!.

I n view of C o r o l l a r y 13.17 it s u f f i c e s t o p r o v e t h a t a f ( x l = 0 i f and o n l y D " f ( x l = 13. S i n c e D"f(x) = 0 c l e a r l y i m p l i e s t h a t a f ( x l = 0, w e h a v e t o p r o v e t h e c o n v e r s e . I f w e assume a f ( x ) = 0 t h e n f o r a r b i t r a r y v e c t o r s to,..., t E E and P X E @ w e have t h a t numbers Xo,

PROOF. -

...,

P

(21.1)

0 = -a'f ( z l ( h o t o , .

Choose

p + 1

k = 0,.

. ., p .

. . xP t P l J

d i s t i n c t numbers

Then t h e complex numbers

t i n c t , and so are t h e numbers with

A.

Bo,.

x

= 0

J * * * '

P

= cij

P

..,tiD Bo,.

i n ( 0 , ~ ) and

. . ,BP

proof.

are a l l d i s -

. By a p p l y i n g (21.1) SF w e g e t a system of equations so,...,

which h a s a n o n z e r o Vandermonde d e t e r m i n a n t . follows t h a t

set

D N f ( z l ( t o ) ( t 7... ,

,t

P

=

I),

I n p a r t i c u l a r it completing

the

DIFFERENTIAL FORHS

161

E XE RC ISES

...,

21.A. L e t U be a n open s u b s e t o f gn and let zl, z denote n t h e complex c o o r d i n a t e f u n c t i o n a l s o f Cn. Show t h a t i f 0 5 p < n, 0 5 q 5 n a.nd p + 1 L 1 t h e n e a c h f E Q p q f U ; F I can by u n i q u e l y r e p r e s e n t e d as a sum

='

dz

fj l...j kl...k

P

.

where

f j , . .j

P

q

.. .

A

:

U

A dYk

A dz

jP

31 +

A

A dTk 9

and t h e summation i s t a k e n m r

F

kl"*k 4

. . . , j p , k l , . . . ,k 4 I s u c h t h a t 1 5 j , ... c k 9 - n. T h i s r e p r e s e n t a t i o n

a l l m u l t i - i n d i c e s in',, I 5 kl < j p 5 n and

c

be o f t e n w r i t t e n i n t h e a b b r e v i a t e d form

f =

K

J,

i f e a c h of t h e mappings

f , ,

i s o f class

(a)

Show t h a t i f

f

(b)

Show t h a t i f

f =

E

Cm(U;FI

Ck

JK

<

...

will

-

dzJ A d z , .

i f and o n l y

c . k

L e t U be a n open s u b s e t o f B n and l e t z l

8.

t h e complex c o o r d i n a t e f u n c t i o n a l s of

f

2:

Show t h a t t h e d i f f e r e n t i a l form f i s o f class

21.B.

.. .

1

,..., z n

denote

then

m

f d;:

Z J,

K

JK

J

A dzK E C

P4

(E;FI then

and

21.C.

TJet U b e a n open s u b s e t of

02

(U;F).

C p0 every

J.

Show t h a t

-

af

Bn

and l e t f = 2 f J d z J E J

= 0

i f and o n l y i f

f,/ E J C ( U ; F )

for

162

MUJ I CA

DIFFERENTIAL FORMS WITH BOUNDED SUPPORT

22.

-

I n t h i s s e c t i o n w e b e g i n t h e s t u d y of t h e e q u a t i o n a f = g , which c o n s t i t u t e s one of t h e c e n t r a l p r o b l e m s treated i n t h i s book. A s i n t h e p r e c e d i n g s e c t i o n t h e l e t t e r s E and F w i l l r e p r e s e n t complex Banach s p a c e s . To b e g i n w i t h , w e g i v e a defin i t i o n which p a r a l l e l s D e f i n i t i o n 2 0 . 1 .

22.1.

DEFINITION.

t i a l form

Let

U be a n open s u b s e t of i s s a i d t o be

g E CPmqfU;F) -

and i s s a i d t o be

Crn (U;FI P J q-1

a-exact

such t h a t

E. A differena-cZosed if ag = 0,

-

-

i f t h e r e i s a d i f f e r e n t i a l form f E

af =

g.

By P r o p o s i t i o n 2 1 . 4 e v e r y 7 - e x a c t form i s 7-closed and would l i k e t o know w h e t h e r t h e c o n v e r s e i s t r u e . I n t h i s

we

sec-

t i o n w e s o l v e t h i s problem i n t h e s i m p l e s t p o s s i b l e case, t h a t

i s , when t h e g i v e n form g h a s bounded s u p p o r t . To solve problem we s h a l l need a n i n t e g r a l f o r m u l a

for

this

differentiable

f u n c t i o n s which g e n e r a l i z e s t h e Cauchy i n t e g r a l f o r m u l a . 22.2.

THEOREM.

Cm(U;F).

Let

for att

5

E

Let

U b e a n o p e n s u b s e t of

a E U, b

E

E

and

X

crnd

be such t h a t

r > 0

a i 0 ; r ) . Then f o r e a c h

E,

E

7 c t

f E

a + gb E U

A i ' 0 ; r ) we h a v e t h e s o l -

l o w i n g g e n e r a l i z e d Cauchy i n t e g r a 5 f o r m u l a

To p r o v e t h i s theorem w e need t h e f o l l o w i n g lemma.

22.3. Let

LEMMA. -

Let

U be an open s e t i n

A ( a ; r ) C U . Then f o r each

X

E

t, Z e t

p € CrnlU)

arid

A ( a ; r l we haoc t h e forrnu7.a

DIFFERENTIAL FORMS PROOF. where

and set 17P = 5 E D :

D = A(a;r), fix XE D

Set

163

0 < p < d D ( X ) . By S t o k e s ' t h e o r e m i n t h e p l a n e

that

v(51 -dc=

/

v(51 d(-

dr;).

Using Exercise 21.B w e g e t t h a t

and t h e r e f o r e

Since (g

- A)-'dg

A dr

d e f i n e s a f i n i t e measure on e a c h bound-

e d s u b s e t of t h e p l a n e , t h e Dominated v o n v e r g e n c e Theorem guaran-

tees t h a t

On t h e o t h e r hand,

and s i n c e

9

i s c o n t i n u o u s , it f o l l o w s t h a t

(22.3)

If we let

p

-, 0

i n ( 2 2 . 1 ) and u s e ( 2 2 . 2 ) a n d ( 2 2 . 3 )

,

thenthe

d e s i r e d conclusion follows.

PROOF OF THEOREM 2 2 . 2 .

Let

IJJ

E

F'

and c o n s i d e r t h e f u n c t i o n

164

VP(5)

MUJ I CA

= $ o f l u + S b l , which i s d e f i n e d and i s o f class

a neighborhood of t h e d i s c TI E ct w e have t h a t

h ( 0 ; rl

.

Cm

on

Then, by t h e Chain R u l e , f o r

each

and u s i n g E x e r c i s e 13.H w e g e t t h a t

Hence, a f t e r a p p l y i n g Lemma 2 2 . 3 t o

f o r every

X

E

A(0;z-l.

Since F'

9, w e g e t t h a t

s e p a r a t e s t h e p o i n t s of

F , the

d e s i r e d conclusion follows. 22.4.

THEOREM.

F o r e a c h a-rz.osi-.d differcwt?hl

fo-m g c

W

c

fE;FI, Pl9+1 w i t h b o u n d e d s u p p o r t , t h e r e e x i s t s a d i f f e r e n t i a 2 form f E W C ( F ; F ) such t h a t af = 9. In. o t h e r words, e v e r g 2 - c l o s e t 1 d i f P4 ferential form i n (E;F ) , u i t h b o u n d e d :::ipport, is a-exact. PJ q + l

ern

PROOF.

The p r o o f i s s i m i l a r t o t h e p r o o f o f t h e P o i n c a r e Lem-

ma. F o r e a c h d i f f e r e n t i a l form

g

E

CM

P, q + 1

s u p p o r t , w e s h a l l d e f i n e a d i f f t x e n t i a l form

1 P; F ), w i t h bounded

Kg

E

Cm ( i . ; F ) i'cf

such

that

The c o n c l u s i o n o f t h e t h e o r e m f o l l o w s a t once from t h i s .

NOW,

165

DIFFERENTIAL FORMS fix

b

Crn PJ q+l Kg E R

b # 0 . Then, g i v e n a d i f f e r e n t i a l form

with

g E (E;F), w i t h bounded s u p p o r t , w e d e f i n e a d i f f e r e n t i a l €om E

E

P(7

by

(E;FI

Kgix) =

f o r every

x

E

E.

1 2712

x,tl,.

dr;

n d7 5

O r more e x p l i c i t l y

K g ( x l ( tl J . . . , t I = 2 1n i P+4

f o r a11

g ( x + Cblh

. .,tPf9 E

i

g ( ~+ c b ) ; b J t l , ... , t

I

dz; h d 5

5

P+4

6

E. S i n c e g h a s bounded s u p p o r t ,

for e a c h

U i n E w e can f i n d r > 0 s u c h t h a t g(x+
Thus w e may a p p l y P r o p o s i t i o n 1 3 . 1 4 t o conclude t h a t

class

for a l l

t

Pf(7f-2

Cm

x E E

on

E

Kg

i s of

U and

U

and

t

E

w e have t h a t

On t h e o t h e r hand,

E.

Thus f o r a l l

x

E

U

and

tlJ...

MUJ I CA

166

. . ., tP+cz+,

K ( 5 g ) (x)I t , ,

I

Thus w e g e t t h a t

and u s i n g Theorem 2 2 . 2 w e c o n c l u d e t h a t

-- -

1 2Tri

Thus w e have shown t h a t

Kg

6

Cm ( U ; F )

and

3iKq)

+ Kt5-g)

= g

E, on

we

P4 on U . S i n c e U w a s a n a r b i t r a r y bounded open set i n c o n c l u d e t h a t Kg E Cm ( E ; F I and a(Kq) + K(%gi = q

P4

E.

T h i s c o m p l e t e s t h e p r o o f of t h e theorem.

22.5. THEOREM. If dim E > 2 t h e n f o r earh ? - c l o s e d d i f f e r e n cu ti02 forni g E C p 2 ( R ; F I , w i t h b o u n d e d s u p o i q t , therzc ' > . r I ' u i s a u n i q u e differential f o r m f E C i 0 ( A ; PI, w i t h b o u n d e d s u p p o r l t , s u c h that

-

af

= g. T h e differential f o r n i f zs holomorphii.

on

167

DIFFERENTIAL FORMS

E \ s u p p g , and i s i d e n t i c a l l y z e r o on t h e u n b o u n d e d

of

component

supp g.

E

b e a 7 - c l o s e d d i f f e r e n t i a l form w i t h

g E Cw f E ; F l Pl

Let

PROOF.

f = Kg E C

bounded s u p p o r t , and l e t

00

PO

bethedifferential

(E;FI

form d e f i n e d i n t h e p r o o f o f t h e p r e c e d i n g t h e o r e m .

of

such t h a t

E

=

E

q

E

E'.

subspace

x E E can be uniquely

M @ 6 b . Then e a c h

x = T x + ufxlb,

r e p r e s e n t e d as a sum

shall

We

M be a closed

p r o v e t h a t f h a s bounded s u p p o r t . L e t

where

and

T E e(E;Ml

f l x l c a n be r e p r e s e n t e d i n t h e f o r m

Then

o r i n t h e e q u i v a l e n t form

s i n c e t h e Lebesgue measure i s t r a n s l a t i o n - i n v a r i a n t .

is t o p o l o g i c a l l y isomorphic t o t h e product

x

E

IT, t h e r e e x i s t s

such t h a t

c > 0

for a l l

and

y E M

then f i n d

p > 0

and e v e r y

3 E M

that

M

Since

fix) = 0

5 E 6. S i n c e g h a s bounded s u p p o r t w e c a n

such t h a t with

g(y

+

= 0

Cb)

II y I1 > p . Then it f o l l o w s

x

f o r every

E

E

from

o t h e r hand,

af =

guarantees t h a t

g = 0

on

E

s u p p g,

f i s h o l o m o r p h i c on

E

CC

(22.4)

IITxII > p , t h a t i s ,

with

i s i d e n t i c a l l y z e r o on a n unbounded o p e n s u b s e t o f -

5

f o r every

E . On

f the

and P r o p o s i t i o n

21.6

.

fol-

E \ supp g

Then it

l o w s from t h e I d e n t i t y P r i n c i p l e t h a t

f i s i d e n t i c a l l y zero on

t h e unbounded component of

and i n p a r t i c u l a r f h a s

E \supp g,

bounded s u p p o r t , as a s s e r t e d . m

To e s t a b l i s h uniqueness, l e t

f, ,fa

E

cpo (BFF!

f, - f ,

i s h o l o m o r p h i c on

E l and

af

= 2 = 9. f, - f , i s i d e n t i c a l l y

f e r e n t i a l forms w i t h bounded s u p p o r t s u c h t h a t then

af,

be t w o d i f -

168

MUJ I CA

z e r o o u t s i d e a bounded s e t . By t h e I d e n t i t y P r i n c i p l e ,

fZ - f 2

E.

i s i d e n t i c a l l y z e r o on

AS an i m p o r t a n t a p p l i c a t i o n o f Theorem 2 2 . 5 w e

prove

the

f o l l o w i n g e x t e n s i o n theniqem of H u r t o g s . Let

22.6. THEOREM.

be a s e p a r a b l e Hiltxrt space w i t h dim E 2 2 .

E

U b e a n o p e n s u b s e t of E ,

Let

subset of

contained i n

E,

Then f o r each g = f

on

f E X(U \ A;FI

U \ A i s connected.

g

such that

E JC(U;FI

U\A.

A c V

C

7

an open s e t

is

U, where t h e c l o s u r e

C

i s bounded, w e may assume t h a t

since A

b e a c l o s e d , bounded

A

there exists

S i n c e E i s n o r m a l , w e can f i n d

PROOF.

that

and l e t

U, and s u c h t h a t

such

V

taken i n

And

E.

i s bounded t o o .

V

a p p l y i n g C o r o l l a r y 1 5 . 5 t o t h e c l o s e d sets A

and

E \ V

By

h~

can

such t h a t 0 5 cp 5 1 on E , Lp = 1 s u p p cp c V . L e t h E C m ( U ; F ) be on U 1 A and h = 0 on a n e i g h b o r d e f i n e d by h = ( 1 - L p p J f m such t h a t hood o f A . W e w a n t t o f i n d a f u n c t i o n u E c ( U ; F ) m h - u E J C ( U ; F ) . Thus w e s h o u l d f i n d a f u n c t i o n u E c ( U ; F ) s u c h find a function

cp E C m f E l

on a n e i g h b o r h o o d of

au = a h

that on

U. If w e d e f i n e

on

v =

U and

C suppcp C V,

A , and

on

E

so t h a t

v

0

supp

Lp,

v

m

E

then

CO,IE;F) a v = 0 on

v =

by

-

ah

and .?upp v

E

h a s bounded s u p p o r t . By Theorem

22.5

t h e r e i s a u n i q u e f u n c t i o n u E C m ( E ; F ) w i t h bounded s u p p o r t s u c h t h a t au = v on E . Hence d u = ah on U and g = ii - U E

X(U;F I

.

To c o m p l e t e t h e p r o o f w e s h a l l show t h a t

By Theorem 2 2 . 5 , E \ s u p p Lp.

W

Since

n (U \ s t i p p

Since

U \A

Lp),

u = 0 h = f

g

1

f on U \ A .

W , t h e unbounded component of U \supp Lp, w e see t h a t g = f on and t h i s i s a n o n v o i d open s u b s e t of U \ A.

on

on

is connected w e conclude t h a t

g = f

on 1' \ A and

t h e proof i s complete.

23. THE

7

EQUATION IN POLYDISCS

I n t h i s s e c t i o n we give a proof

of

the

Dolheault

which i s a complex v e r s i o n o f t h e P o i n c a r g Lemma. preceding s e c t i o n

the

letters E

and

F

As

Lemma, in

the

r e p r e s e n t complex

D I F F E R E N T I A L FORMS

169

Banach s p a c e s .

23.1. THEOREM.

0 < r

If

d i f f e r e n t i a 2 form

R 5

c;.,q+l

then f o r

__ 2

each

(AnIa;R); F ) t h e r e e x i s t s

cZosed a

dif-

-

j ' e r e n t i a l form

i A n f a ; r i ; FI s u c h t h a t

a f = g on A*la;r).

Before p r o v i n g t h i s t h e o r e m w e show t h e f o l l o w i n g lemma. 23.2.

Let

LEMMA.

t i o n of

M @ 6b @ N

b e a d i r e c t sum

decomposi-

U C M and V C N b e o p e n s e t s . L e t 0 < r < hir = U + A ( 0 ; r l b + V a n d WR = U + Ai0;R)b + V .

Let

E.

=

E

R --< n , a n d l e t m

Let

c p , q + l IWR;F) b e a d i f f e r e n t i a l f o r m s u c h t h a t

g

f o r aZl

x

t

and

E WE

E E,

and

g ( x l I s , t l , . . . , t p + q )= 0

(23.2)

x

f o r alZ

E idR,

f

f e r e n t i a l form

s

E

E

cL9 ( W r ; F I

(23.3)

a f c x c , (s, t l , .

for a l l

x E

PROOF.

j

wr, x

Every

n ( x ) b + Tx, where

s E

E

E

M

and

. ., t P+9 M @ Cb

t E E . Then t h e r e e x i s t s a d i f j such t h a t

= g i ~ I iS , t l J . .., t and

t

j

Q

P9

9

=

1

E.

S E ~ ( I E ; M ) ,T E l ( E ; N )

on

I

can b e u n i q u e l y w r i t t e n as a sum

loss of g e n e r a l i t y w e may assume t h a t such t h a t

E

P+4

Ai0;rl

and

and

R <

supp 9

m.

C

rl E E ' .

Let

X =

Without

9 C

A(0;R).

sx +

cm(6)

Define J E

( W p ; F ) by

where t h e i n t e g r a n d i s d e f i n e d t o b e z e r o

when

x

+ < b 9 WR.

x + gb B W R then [nix) + < ( R and t h e r e f o r e 9 ( q ( x l + 5) = 0. Since t h e inteqrand vanishes outside t h e d i s c Note that i f

MUJ I CA

170

5 R + r

for every x E W r we may apply Proposition 13.14 to conclude that f is of class Cm and 15)

for every 11: E Wr E we get that

and

t

E

E. Hence for all

x

E

Wr

and

t

j

E

Note that the last written sum is nothing but

and the last written term vanishes by hypothesis (23.11, since Cpfrlix)

+ 51 = 0

whenever

3:

+ 52,

9 WR.

Thus

171

DIFFERENTIAL FORMS

If t o = s lies in M then by hypothesis (23.2) all the terms vanish, in the last written expression for af ( x i I t o , . . . ,t P+9 and therefore -

(23.4)

a f i x i i s , t , ,..., t

P+4

)

=

o =

g(x)(S,tl

,...,

t

P+4

I

for all x E w F , s E M and ti E E . On the other hand, the preceding expression for a f t x l f to , . . . , t can be rewritten P+4 in the form af(xi(to' " ' i t

P+4

i

If to = b then all the terms in the first sum vanish and follows that -

af

( X I

ib, t l , . . . ,t

P+4

it

I

Hence an application of Theorem (22.2) shows that

for all

3: E

Wra

and

t

j

E

E . Since (23.3) follows at once from

172

MUJ ICA

( 2 3 . 4 ) and ( 2 3 . 5 ) , t h e p r o o f o f t h e lemma i s c o m p l e t e . W i t h o u t l o s s of g e n e r a l i t y w e may a s s m

PROOF O F THEOREM 23.1.

n (A iU;Ki; F i , foreach

m

a = 0, so t h a t

that

g E Cp,q+l

k=l,

...>

n

set Dk=tz

Cn; /zj/ < r

E

k = I,.

For each

for

. .,n

j

5 k

we i d e n t i f y

C n g e n e r a t e d by t h e f i r s t

k

f,

Cw I D 1 ; F I

E

af,(zi(s,tl

z

for all

D1 >

E

I € ~

fk

,...,

s E C1

Cm I D k ; F l P4

E

-

afk(x)(s,tl

for all

3: E

,...,

t

of

M = 10)

yields a differential

t

P+4

i = g(zi(s,tl,.

and

t

j

... t P+4

g n . By r e p e a t e d applications

E

such

f,, . . . , f n

and

P +4

Dk, s

j < k}.

for

with t h e subspace

o f Lemma 23.2 w e c a n f i n d d i f f e r e n t i a l forms that

< R

such t h a t

P4

-

Izj(

v e c t o r s of t h e c a n o n i c a l b a s i s . A n

a p p l i c a t i o n o f Lemma 23.2 w i t h form

and

E

I = (g(x)

Ck

and

t

-

j

k-1 -

z

j=1 E

a f . i . ? , ) i i ~ , t ,,..., t J

Cn. Then

f=f, +

P+4

...

i

+

fn

i s t h e r e q u i r e d d i f f e r e n t i a l form.

I f g i s a $-closed

(p,q

+

F-valued

d i f f e r e n t i a l form of

1 ) on a n e i g h b o r h o o d of a compact p o l y d i s c

bidegree

zn(a;r) then

Theorem 2 3 . 1 g u a r a n t e e s t h e e x i s t e n c e o f an F - v a l u e d d i f f e r e n -

t i a l form f o f b i d e g r e e ( p , q l

af

g

on a

neigh-

x n ( a ; r ) . Note t h a t w e may assume t h a t f i s d e f i n e d

borhood o f

f o r otherwise it s u f f i c e s t o m u l t i p l y

gn,

on a l l o f function

such t h a t

p

E

CmiCn)

such t h a t

CP

is identically

by

f

one

a

on a

-

neighborhood of

Ania;ri,

and t h e s u p p o r t o f

i n t h e domain o f d e f i n i t i o n of

cp

is contained

f. These r e m a r k s w i l l

p e a t e d l y used i n t h e s e q u e l .

Next w e improve Theorem 2 3 . 1 a s f o l l o w s .

be

re-

173

DIFFERENTIAL FORMS t h e r e e i x s t s a d i f f e r e n t i a l form

( A n f a ; R I ;F I

Ec€34( A n ( a ; R I ; F I PROOF.

f

-

such t h a t

af

= g.

m

L e t ( r j i j Z 1 b e a s t r i c t l y i n c r e a s i n g s e q u e n c e o f psi-

t i v e numbers t e n d i n g t o

R.

Consider s e p a r a t e l y t h e

following

two cases.

3

I n t h i s case w e s h a l l inductively f i n d a sequence ( f j I j = , i n Cm ( C n ; F I s u c h t h a t af. =g on a P4 3 n e i g h b o r h o o d o f A " l a ; r . ) f o r e v e r y j 2 I , and fj = fj-l on

(a)

q

F i r s t assume

1.

00

3

n

A (a;ri-l)

f o r every

f,

j > 2 . The e x i s t e n c e o f

is

t e e d by Theorem 2 3 . 1 and t h e p r e c e d i n g r e m a r k s . L e t

guaran2

j

and

f , > .- .,fj-,

assume t h a t

h a v e a l r e a d y b e e n f o u n d . By Theorem 2 3 . 1 t h e r e e x i s t s u E Cm ( C n ; F I s u c h t h a t au = g on a n e i g h P4 -n borhood of A ( a ; r . I . But t h e n a ( u - fj-l/ = 0 on a n e i g h 3

--n

borhood of

A ( a ; r j - l ) J and a n o t h e r a p p l i c a t i o n of Theorem23.1 y i e l d s a d i f f e r e n t i a l form v E Cm ( C n ; F I such t h a t av = P J q+1 - fj-1 on a n e i g h b o r h o o d o f a n ( a ; r j - 1 I . Then t h e d i f f e r e n t i a l form f = u - av E ern ( C n ; F ) h a s t h e r e q u i r e d p r o p e r t i e s . P4 Once t h e e x i s t e n c e o f t h e s e q u e n c e (f .I h a s b e e n established, 3 n ( A n ( a ; R I ; P i by f = f on A ( a ; r .I f o r e a c h we d e f i n e RP4 j 3 j E D . Then i t i s c l e a r t h a t f E C ; q ( A n ( a ; R I ; F I and a f = g

*

on

An(a;R).

(b)

Next c o n s i d e r t h e case

If 3. ) I = l

i n d u c t i v e l y f i n d a sequence

-

af

2-J

. = g

on a n e i g h b o r h o o d o f

f o r every

tence of

f,

assume t h a t

x

E

g co

-n A (a;rj-,)

=

0.

in

an ( a ; ? 3.I

and

j

2

I n t h i s case w e s h a l l

Cm ( E n ; F I and 2.

II f . ( X I 3

Let

that

- fj--?(xI/I 5

As before,

i s g u a r a n t e e d by Theorem 2 3 . 1 .

f,, . . ..J'j-l

such

PO

j

h a v e a l r e a d y been f o u n d . By

t h e exis2

and

Theorem

2 3 . 1 t h e r e e x i s t s u E Cm (U'";F) s u c h t h a t a u = g on a n e i g h PO -n borhood o f A (a;~~-~ But i . t h e n a ( u - f j - 1 ) = 0 on a n e i g h borhood

V of

a" ( c z ; r j - l I ,

and P r o p o s i t i o n 21.6

g u a r a n t e e s that

174

24

-

MUJ ICA

fj-1

s u c h t h a t IIu(x1

P(6n;La(p6n;FI)

-n

x E A

every C

m

PO

ICn;FI

Hence w e c a n f i n d a p o l y n o m i a l

JC(V;Ea(P@n;FI).

E

-

-

fj-,fxl

P E

P I ~ ) I5 I 2-j

for

-

f = u

( a ; r j - l ) . Then t h e d i f f e r e n t i a l form

P E

has t h e required properties. h a s been estab3 I n p a r t i c u l a r w e see t h a t t h e d i f f e r e n t i a l form f . - fj-l

Thus t h e e x i s t e n c e of t h e s e q u e n c e ( f . ) lished.

3

n

JC(A ( a ; r j - l ) ;

belongs t o

Ea(p@n;FI/ f o r e v e r y

j

and t h e

2 2

m

I; ( f j - f j - l 1 c o n v e r g e s u n i f o r m l y on e a c h

series

j=2

m

z

Hence t h e series

(f.- f j - ] ) b e l o n g s t o X(An(a;rk/;Ea(PQ?;FII

j=k+l f o r each

n

A (a;rk).

a.

k E

3

Define

RPq

( A n ( a ; R ) ; F ) by

m

k E W we can w r i t e

S i n c e f o r each

m

f o r each

k

-

n ( A (a;rk);FI a n d

w e see t h a t

af

= ?f, = g on An(u;rkI

liV. The d e s i r e d c o n c l u s i o n f o l l o w s .

E

To end t h i s s e c t i o n w e show h o w Theorem 23.3 c a n b e u s e d t o

s o l v e t h e so c a l l e d f i r s t C o u s i n probZem on a p o l y d i s c . 23.4.

THEOREM.

where

0 < R

ii, j

E

Let ( U i l i E I

c __

a.

be an open cover of a polydisc

Suppose t h e r e ayae functions

fij

E

KIU. n U . ; F )

I) s u c h t h a t on

(23.6)

fij

(23.7)

f i j + f j k + Jk7: = 0

+

fji

= 0

u. n

on

11.
0 . n 11. n i l k . I.

An(a;R), 2

.

3

DIFFERENTIAL FORMS

T h e n t h e r e a r e functions

on

F i r s t we s h a l l f i n d functions

PROOF. (23.9)

LL

-

Lrj

(i E I ! such t h a t

fi E JC(Ui;F!

f,: - fj = f i j

(23.8)

175

= fij

on

n U

Ui

j

ui Ui

.

Cm(Ui;Fl

E

such t h a t

n U

j’

To a c h i e v e t h i s l e t I q i ! be a Cm p a r t i t i o n of unity un An(a;Ri, subordinated t o t h e cover ( U . I . If we d e f i n e 2

where

fik

lpk

i s d e f i n e d t o b e z e r o on

w e l l d e f i n e d and b e l o n g s t o

Ui \ U,

,

then

ui

is

m

C (Ui;FI.

Furthermore, it follows

from ( 2 3 . 6 ) t h a t

ui n

on

11

j ’

as a s s e r t e d . T o c o m p l e t e t h e p r o o f o f t h e t h e o r e m

v

it s u f f i c e s t o f i n d a f u n c t i o n

= u

+

E Cm(bn(a;R);F!

s u c h t h a t fi

i. Thus w e want t o f i n d a f u n c t i o n v E C w ( A n ( ~ ; R ) ; F I s u c h t h a t av = - au i on Ui f o r e v e r y i. S i n c e f i j i s h o l o m o r p h i c on U i n u (23.9) i

v

i s h o l o m o r p h i c on

implies t h a t

w e may d e f i n e

-

-

= au

aui

w

E

on

Ui

Ui

j C;l(An(a;R);F)

f o r every

n U

j by

j’ f o r a l l i and j. Hence

w = 2u i

on Ui f o r every -

i. Then i t f o l l o w s from Theorem 2 3 . 3 t h a t t h e e q u a t i o n av = - w h a s a s o l u t i o n v E C m ( A n ( ~ ; R ! ; F !and t h e p r o o f of t h e t h e o r e m i s complete.

EXERCISES

23.A. L e t i U i l be an open c o v e r o f $. Suppose t h a t f o r i t h e r e i s a f u n c t i o n f i meromorphic on U i such t h a t

each

fi -

i s h o l o m o r p h i c on Ui n U j €or all i and j . Using Theofj r e m 2 3 . 4 f i n d a f u n c t i o n f meromorphic on E s u c h t h a t f - f i

176

MUJ I CA

i s holomorphic on 23.B.

Let

for every

Ui

i.

( a i l be a s e q u e n c e of d i s t i n c t p o i n t s i n

6 withoot

L e t ( P i f z ! I b e a c o r r e s p o n d i n g s e q u e n c e o f non-

cluster points.

zero polynomials i n

z

without constant t e r m .

Using

Exercise

23.A f i n d a f u n c t i o n

f meromorphic on 5 , whose p o l e s are -1 Pi [ ( z - a . I ] i s the p r e c i s e l y t h e p o i n t s ai , and s u c h t h a t s i n g u l a r p a r t o f f a t ai. T h i s i s t h e Mittag-Leffler Theorem.

NOTES AND COMMENTS

Our p r e s e n t a t i o n o f a l t e r n a t i n g m u l t i l i n e a r forms a n d r e a l d i f f e r e n t i a l forms i n S e c t i o n s 1 8 , 1 9 and 2 0 ,

follcws

essen-

t i a l l y t h e book o f H . C a r t a n [ 3 1 . Our p r e s e n t a t i o n o f complex d i f f e r e n t i a l forms i n S e c t i o n 2 1 f o l l o w s e s s e n t i a l l y a n article o f E. Ligocka

[ 21.

The r e s u l t s i n S e c t i o n 22 on e x i s t e n c e of -

af = g, when g h a s bounded s u p p o r t , are s t r a i g h t f o r w a r d g e n e r a l i z a t i o n s of r e s u l t s o f L. Hormander [ 3 ] , i n t h e case o f t n l and E . Ligocka [ 2 ] , i n t h e case of

s o l u t i o n s of t h e equation

Banach s p a c e s . The p r o o f of H a r t o g s ' Theorem 2 2 . 6 , Theorem 2 2 . 5 ,

i s due t o L. Hormander

of L. E h r e n p r e i s t o P. D o l b e a u l t

[ 11. Finally,

I. 1 I .

[ 3

based

1 , following an

on idea

Theorems 23.1 and 23.3 a r e due

CHAPTER V I POLYNOMIALLY CONVEX DOMAINS

24.

POLYNOMIALLY CONVEX COMPACT SETS I N

tn

T h i s s e c t i o n i s d e v o t e d t o t h e s t u d y o f p o l y n o m i a l l y convex tn. W e show t h a t t h e a e q u a t i o n h a s a s o l u -

compact s e t s i n

t i o n on a n e i g h b o r h o o d o f e a c h p o l y n o m i a l l y convex c o m p a c t s e t . W e a l s o p r o v e a theorem on p o l y n o m i a l a p p r o x i m a t i o n

t h e O k a - W e i l theorem.

The l e t t e r s E

and

known

as

F w i l l a l w a y s repre-

s e n t complex Banach s p a c e s . If

g

i s a $-closed

d i f f e r e n t i a l form o f b i d e g r e e ( p , q + 1 ) -n A ( a ; r l t h e n Theorem

on a n e i g h b o r h o o d o f a compact p o l y d i s c

2 3 . 1 g u a r a n t e e s t h e e x i s t e n c e of a d i f f e r e n t i a l fonn f of bidegree -

i p , q ) such t h a t

af

= g

on some n e i g h b o r h o o d of

-n

A ( a ; ri)

.

In

t h i s s e c t i o n w e e x t e n d t h a t r e s u l t t o a l a r g e r c l a s s o f compact

s e t s , namely t h e p o l y n o m i a l l y convex compact s e t s i n

Cn.

To

b e g i n w i t h w e d e f i n e p o l y n o m i a l l y convex compact s e t s i n B a n a c h spaces.

o f a set

(b) v e x if

A C E

convex h u l l

i s d e f i n e d by

A cornpact s e t "(E)

PIE)-hull or p o Z y n t m i a l l y

( a ) The

2 4 . 1 . DEFINITION.

K

C

E

i s said t o be poZynorniaZZy

c3il-

= K.

The f o l l o w i n g examples o f p o l y n o m i a l l y convex compact s e t s can be r e a d i l y v e r i f i e d by t h e r e a d e r . s i t i o n 11.1.

To p r o v e ( a ) u s e Propo-

178

MUJ I CA

24.2.

EXAMPLES.

( a ) Every convex compact s u b s e t o f

E

is ply-

n o m i a l l y convex. If

(b) and

Pi

K

i s a p o l y n o m i a l l y convex compact s u b s e t o f

P l E l f o r each

E

i

E

I

E

t h e n t h e compact s e t

i s p o l y n o m i a l l y convex as w e l l .

(c)

In particular, i f

i s a f i n i t e s e t , and

pi

E

D i s a compact p o l y d i s c i n C n , I P(Cn) for each i E I, then the

compact s e t

i s p o l y n o m i a l l y convex. These p o l y n o m i a l l y convex compact s e t s

are c a l l e d compact poZynomia2 poZyhcdra. Next w e i n t r o d u c e t h e n o t i o n of germ o f

a function

on

compact s e t , a n o t i o n t h a t serves t o d e s c r i b e t h e b e h a v i o r

a of

a f u n c t i o n on s u f f i c i e n t l y small n e i g h b o r h o o d s o f a compact set.

24.3. DEFINITION. I f K i s a compact s u b s e t o f E t h e n w e s h a l l d e n o t e by XIK;Fl t h e s e t o f a l l e q u i v a l e n c e classes of F-valued f u n c t i o n s which are h o l o m o r p h i c on some open n e i g h b o r h o o d Of K , under t h e e q u i v a l e n c e r e l a t i o n f % g i f f = g on some neighborhood o f F-valued If

mapping

K.

The

members o f

hoZomorphic f u n c t i o n s o n

U i s a n open n e i g h b o r h o o d o f XIU;F)

f E JC(U;FI

+

J C ( K ; F ) a r e c a l l e d germs of

K.

K then t h e r e is a n a t u r a l

JClK;F) which a s s o c i a t e s w i t h

t h e c o r r e s p o n d i n g germ. The s e t

each

function

n a t u r a l way a v e c t o r s p a c e , and t h e c a n o n i c a l mapping X(K;Fl

i s l i n e a r f o r e a c h open n e i g h b o r h o o d

U of

a

JC(U;F)+

K.

One s h o u l d a l w a y s remember t h a t t h e e l e m e n t s of e q u i v a l e n c e classes o f f u n c t i o n s . H o w e v e r , it i s

in

is

3CIK;F)

KcIK;FI a r e

oftzn

con-

v e n i e n t t o r e g a r d t h e s e e l e m e n t s a s b e i n g f u n c t i o n s by identifying

179

POLYNOMIALLY CONVEX DOMAINS e a c h e q u i v a l e n c e c l a s s w i t h a s u i t a b l e r e p r e s e n t a t i v e , and

we

s h a l l f r e q u e n t l y do so. The p r e c e d i n g c o n s t r u c t i o n c a n b e carried o u t e q u a l l y w e l l w i t h i n t h e realm o f

Cm

f u n c t i o n s i n s t e a d o f h o l o m o r p h i c func-

t i o n s . Thus we h a v e t h e f o l l o w i n g d e f i n i t i o n .

K i s a compact s u b s e t o f E then we s h a l l of Fv a l u e d f u n c t i o n s which are o f class Cw on some open n e i g h borhood o f K , u n d e r t h e e q u i v a l e n c e r e l a t i o n f % g if f = g on some n e i g h b o r h o o d o f K . The members o f C w f K ; F I are c a l l e d germs of F - u a Z u e d Cm functions on K. 24.4.

DEFINITION.

If

m

C (K;FI t h e set of a l l e q u i v a l e n c e classes

d e n o t e by

C l e a r l y a l l t h e remarks concerning

w e l l to 24.5.

C ~ ( K ; F I . In particular

C

d e n o t e by

P4

P4

24.6.

DEFINITION.

the F-Cousin -

ay

= g

-

af

= g

for

f

K. E

Cm (K;FI and g

on some n e i g h b o r h o o d o f A compact s u b s e t

p r o p e r t y i f f o r each

= (7 there7 e x i s t s

Cm dif-

f K ; F I a r e c a l l e d germs of F - v a Z u e d

Thus t o s a y t h a t

af

equally

is a vector space.

(K;FI

J e r e n t i a Z forms of b i d e g r e e ( p , q ) o n

means t h a t

apply

If

cm

The members of

-

Jf(K;FI

K i s a compact s u b s e t o f E then= shall ( K ; F ) t h e vector s p a c e

DEFINITION. m

c

m

f

E

ciqix;i.’)

K

g

of

E

PJq+l

(K;FI

K.

E

is s a i d

m

E

Cm

cp,qcl IK;FI

to

have

such

that

-

such t h a t

af =

g.

Thus Theorem 2 3 . 1 a s s e r t s t h a t e a c h compact p o l y d i s c i n 6 h a s t h e P-Cousin p r o p e r t y f o r e a c h Banach s p a c e

F.

n

We shall

soon e x t e n d t h a t r e s u l t t o a l l t h e p o l y n o m i a l l y convex compact

sets i n

En.

F i r s t w e show t h a t e a c h p o l y n o m i a l l y convex

pat

in

8

s.rit

polyhedra.

can be approximatec!

by

compact

com-

polynomial

180

M U J I CA

24.7.

Let

LEMMA.

and l e t

b e a p o Z y n o m i a 2 l y c o n v e x c o m p a c t s e t i n ?d

K

T h e n t h e r e i s a com-

U b e a n o p e n n e i g h b o r h o o d of K .

pact polynomial pozyhedron

L

K

such t h a t

C

L

C

U.

L e t D be any compact p o l y d i s c c o n t a i n i n g K . If D C U t h e n i f s u f f i c e s t o t a k e L = D. If D 9 U t h e n f o r e a c h p i n t a E D \ U t h e r e i s a polynomial P E P I C n ) such t h a t s i p I P I < PROOF.

1 < j P ( a ) I . Since t h e set D \ U i s compact w e can f i n d Pm E P f C n ) such t h a t s u p l P j i < 1 f o r nomials P I .

....

2,.

..,m

polyj =

K

and

Thus it s u f f i c e s t o t a k e L = ( X E D : [ ~ . f x ) 2l I 3

for

j=7

,.... ml.

By Lemma 2 4 . 7 w e may c o n c e n t r a t e our a t t e n t i o n on

compact

polynomial p o l y h e d r a . 24.8.

THEOREM.

1,et

L =

D

[ X E

: IP.ixi] 3

5

2

for

b e a c o m p a c t poZynomCnL p o l ’ y h e d r o n i n

i = l

,...,

rnl

Cn, a n d l e t

1-1 d e n o t e t h e

mapping

Then:

(a)

L

h a s t h e F-Cousin p r o p e r t y for eucli R a n n ~ i i spucs F.

The k e y t o t h e proof of Theorem Lemma.

24.8

is

the

following

POLYNOMIALLY CONVEX DOMAINS

24.9.

Let

LEMMA.

181

be a c o m p a c t s u b s e t of

K

Let

En,

P E

P(8)

and L e t

Let p denote t h e mapping

and a s s u m e t h a t

f,

(a)

Kp

(b)

For e a c h

Cm IK P4

E

t h e F-Cousin p r o p e r t y .

has

x

f

E

Ciq

-

A;FI such t h a t

x

such t h a t

f

af, =

af

Cn

on

and hence

n

-2

-

af

= 0

on V . If

exists

of

C IT

-1

'+?6

+

6

(n (V);FI and P" n o u i s the iden-

Moreover

IV).

71:

-I

C

E

Note t h a t

IVl.

p(Kp)

thryze

u * f Z = f.

and

0

d e n o t e s t h e c a n o n i c a l p r o j e c t i o n t h e n n*f

a(n*f) = rr*(%f) = 0

= 0

V be an open neighborhood

and

E C;q(V;F)

-

with

fKp;F) -

(b). L e t

-

t i t y on

Then:

h a s t h e F-Cousin property.

F i r s t w e prove

PROOF.

Kp

K

Kp= ~ - l i KX

n)

and

Choose

9

u(Kpl and

f

CColCn+'I such t h a t supp p

C

IT-'(V).

Define

-

f, = 9 n * f where

Q E PlEn+7i

= 1

9

i s d e f i n e d by

a

on

neighborhood

Ciq

f, E

IK

P9 9 r*f

= P(x) -

Q(x,<)

a ; F ) w i l l be d e t e r m i n e d so t h a t

x

n ; F ) by

Qu,

-

CM Iic

X

of

i s u n d e r s t o o d t o be z e r o o u t s i d e

2fS, = 0 TI

-1

<

and

u E

(the product

( V ) , a remark t h a t

w i l l n o t be r e p e a t e d a g a i n i n s i m i l i a r s i t u a t i o n s ) . The c o n d i -

tion

-

af,

= 0

reduces t o t h e equation u

1

= -

Q

7iq

n

IT*f.

-au

= v , where

MUJ 1 CA

182

-

Since

v a n i s h e s on a n e i g h b o r h o o d o f

alp

v(Kpl w e

see

that

i s a w e l l d e f i n e d member o f C m (K x a ; F ! satisfying the P J q+l h a s the F-Cousin c o n d i t i o n a v = 0. S i n c e by h y p o t h e s i s K x p r o p e r t y t h e e q u a t i o n a u = u h a s a s o l u t i o n , as d e s i r e d . It u

a

i s clear t h a t

fZ

= n*f

it follows t h a t U*fl

on a n e i g h b o r h o o d o f

= U*w*f

= f

p f K p ) a n d whence

on a n e i g h b o r h o o d of K p . This

shows ( b ) . T o show (a) l e t

g E Cm

CU

exists

g1 E Cp,q+l

w i t h ag = 0 . BY (b) there s u c h t h a t a g I = 0 and U*gl = g .

P Jq+l

IKp;F/

h;F) S i n c e by h y p o t h e s i s K x A has the F-Cousin p r o p e r t y there exists m f , E Ch7 ( K x A;F) s u c h t h a t a f , = g 1 . S e t f = p * f E C ( K ;F). P9 1 P9 p Then Jf = = p * g l = g and t h e p r o o f i s c o m p l e t e .

(K

x

p*(af,!

PROOF O F THEOREM 24.8. We p r o c e e d by i n d u c t i o n on t h e number m of p o l y n o m i a l s i n v o l v e d . I f m = 0 t h e n p a r t ( b ) i s o b v i o u s and p a r t ( a ) i s t r u e by Theorem 23.1. L e t m 2 1 and assume t h e theorem i s t r u e f o r compact p o l y n o m i a l p o l y h e d r a i n v o l v i n g

less than m polynomials. I f w e set

then

and

Then

K

s i s , and (2).

To

x

n L

has the F-Cousin p r o p e r t y by t h e i n d u c t i o n h y p o t h e -

has theF-Cousin p r o p e r t y by Lemma 2 4 . 9 .

show (b) c o n s i d e r t h e mappings

Tnis

shows

183

POLYNOMIALLY CONVEX DOMAINS

u2

so t h a t

= 1-1. L e t

f E Ca) ( L ; F I w i t h

01-1~

24.9 t h e r e e x i s t s

f, E

P4

ca)

(K x a;FI

-

af =

such t h a t

P4 = f. B y t h e i n d u c t i o n hypothesis there exists fi

p;fl

such t h a t

-

af,

= 0

and

pif,

-

fl

.

Hence

Lemma

0. By

-

af, = E

and

0

C" I D x Xm;F) P9

1-I*f2=1-12(uif2)=

pif,=f

and the p r o o f i s c o m p l e t e . Theorem 24.8 h a s many c o n s e q u e n c e s . 24.10.

THEOREM.

h a s t h e F-Cousin PROOF.

24.11.

Each p o l y n o m i a l l y c o n v e x c o m p a c t s u b s e t of dr p r o p e r t y f o r e a c h Banach s p a c e

for all PROOF.

F.

Apply Lemma 2 4 . 7 and Theorem 24.8. THEOREM.

Let

be a c o m p a c t p o l y n o m i a l p o l y h e d r o n i n JC(L;F)

n

there e x i s t s

x

f, E X ( D

X

i n some n e i g h b o r h o o d o f

T h i s i s Theorem 24.8

d?.

Then f o r each

f

E

Km;FI s u c h t h a t

L.

(b) with

p = q = 0.

The f o l l o w i n g theorem on p o l y n o m i a l

approximation isknown

as t h e Oka-WeiZ T h e o r e m .

K be a polynomially c o n v e x c o m p a c t subset of Cn. T h e n f o r e a c h f E J C ( K ; F I t h e r e is a s e q u e n c e of p o l y nomials P 6 P(Gn;FI w h i c h c o n v e r g e s t o f u n i f o r m Z y o n K . j 24.12.

THEOREM.

Let

U b e a n open n e i g h b o r h o o d o f

such t h a t

PROOF.

Let

3CIU;FI.

By Lemma 2 4 . 7 t h e r e i s a compact p o l y n o m i a l p o l y h e d r o n

K

f

E

184

MUJ I CA

K C L

such t h a t

U. By Theorem 24.11 there exists f,

C

E

X(DXEm;F)

such t h a t

f o r a l l x i n some n e i g h b o r h o o d o f polynomials

R

.

E

3

on t h e p o l y d i s c

series of E

f,

P(Cnjm;Fl D x

Km,

L . There i s a sequence

which c o n v e r g e s t o

f,

of

uniformly

namely t h e p a r t i a l suns of t h e T a y l o r

a t t h e c e n t e r of t h e polydisc.

If w e d e f i n e

P

j

P ( g n ; F ) by

t h e n i t i s c l e a r t h a t ( P . 1 c o n v e r g e s t o f u n i f o r m l y on 3 w e wanted.

L , as

EXERCISES

24.A.

show t h a t

z

Kp(Ej

f o r e a c h compact s u b s e t K o f

KX(E)

E.

24.B.

Show t h a t i f

A

A

i s a b a l a n c e d s u b s e t of

AP ( E )

i s b a l a n c e d as w e l l .

24.C.

Show t h a t i f

-

AP(MI 24.D.

'P(E)

E

then t h e set

M i s a complemented s u b s p a c e of

f o r each subset

A

E

then

M.

of

Show t h a t e a c h f i n i t e s u b s e t of

E

i s p o l y n o m i a l l y con-

w

vex. More g e n e r a l l y , show t h a t if (arn )m = l

i s a sequence i n

which c o n v e r g e s t o a p o i n t

{am : m 24.E.

E

Let

a }u

a, t h e n t h e compact s e t {a) i s p o l y n o m i a l l y convex.

K be a c o n n e c t e d compact s u b s e t of

Oka-Weil Theorem 2 4 . 1 2 show t h a t t h e set

zpiBn,

Cn. Using

E

K =

the

is c o n n e c t e d

as well.

24.F.

Let

K be a p o l y n o m i a l l y convex compact s u b s e t of

En.

185

POLYNOMIALLY CONVEX DOMAINS

Using the Oka-Weil Theorem24.12show that each open and closed subset of K is polynomially convex as well. 24.G. Using the classical theorem of Runge on polynomial approximation for holomorphic functions of one complex variable show that a compact set K in 5 is polynomially convex if and only if the open set 5 \ K is connected.

25. POLYNOMIALLY CONVEX DOMAINS IN

tn

This section is devoted to the studyofpolynomially convex domains in g n . Most to the results are straightforward consequences of the results from the prLceding section. To begin with we define polynomially convex domains in Banach spaces, keeping in mind the definitions and results from Section 11. 25.1.

DEFINITION.

Let

U

be an open subset of

E.

(a) U is said to be poZynomiaZZy c o n v e x if is compact for each compact set K C U. (b)

U

A

KPiRl

C

U

KplEl

is said to be strongly poZynomialZy c o n v e x for each compact set K C U.

n u

if

One can readily see that U is polynomially convex if and only if d U i i ' 2 P i a i 0 Ul > 0 for each compact set K C U. The following examples of polynomially convex domains canbe easily verified by the reader. 25.2. EXAMPLES. (a) Every convex open set in E polynomially convex. For each P e P ( E l the open set U = {a is strongly polynomially convex. (b)

is strongly

E E :

1Pfx)l < 11

(c) The intersection of two (strongly) polynomially convex open subsets of E is also a (strongly) polynomially convex open subset of E . (d)

If U is a (strongly) polynomially convex open subset

186

MUJ I CA

of E then U n M is a (strongly) polynomially convex open subset of M for each closed subspace M of E .

E

8

Clearly every strongly polynomially convex open subset of is polynomially convex. The next proposition shows that in the converse is true.

25.3. P R O P O S I T I O N .

U of

A n open s u b s e t

Cn

is

polynnrniuZZy

c o n v e x if a n d o n Z y if U is s t r o n g E y poZynomiaZZy c o n v e x . To prove the nontrivial implication let U be a

polynomially convex open subset of an and let K be a compact 9 U then we can write 2 = A u B, subset of U. If P l P ) Pitn) where A = ?, and B = Z \ U. Then A and B are P(P) P(Cnl two disjoint compact sets. If we define f = 0 on a neighborhood of A and f = 1 on a neighborhood of B then f € X I 2 ) P id?-) and an application of the Oka-Weil Theorem 24.12 yields a p l y nomial P E p ( C n ) such that IP % on K^ Since PROOF.

.

fl

P(d?l

K

C

A

IPfbl

I

we conclude that ( P ( x l 1 < h ' for every x E K whereas > '/i f o r every b E B . This is impossible, f o r B C

U be a p o l y n o r n i a % % yc o n v e x o p e n s u b s e t of C n . T h e n f o r e a c h f 6 J C ( U ; F I t h e r e i s a s e q u e n c e ( P . ) i n Pi6?;F) 3 w h i c h c o n v e r g e s t o f uniforrnZy o n e a c h corripac?t s u b s e t o f U. 25.4.

THEOREM.

Let

PROOF. By Proposition 25.3 the open set U is strongly polynomially convex. Hence we can find a sequence (K .) of polynomially J

03

convex compact sets such that

U =

u K

j=l

and

K. 3

C

ijiI for

each j. By the Oka-Weil Theorem 24.12 for each j we can find P E P ( C n ; F I such that IIP - f II 5 l / j on K . . Whence it folj j 3 lows that (P.) converges to f uniformly on each compact sub3 set of U. 25.5.

THEOREM.

Let

U b e a p o l y n o r n i a i l y convec o p c n s u b s e t of

187

POLYNOMIALLY CONVEX DOMAINS m

Cn. T h e n f o r e a c h

f

cpmq f U ; F I

E

Cp,q+l

IU;F!

such that

ag = 0 t h e r e e x i s t s

-

af

such t h a t

= g.

U as t h e u n i o n of a n i n c r e a s i n g s e q u e n c e o f p o l y n o m i a l l y c o n v e x compact s e t s , t h e p r o o f of Theorem 25.5 i s almost a r e p e t i t i o n o f t h e p r o o f o f Theorem 2 3 . 3 , but using A f t e r writting

Theorem 2 4 . 1 0

The d e t a i l s are l e f t t o

i n s t e a d o f Theorem 2 3 . 1 .

t h e r e a d e r as a n e x e r c i s e . W e a l s o l e a v e as a n exercise t h e a n a l o g u e of Theorem f o r p o l y n o m i a l l y convex domains i n

23.4

Cn.

EXERCISES IT E 1 f E ; F I

and l e t

U = T - ' I V l . Show t h a t i f V i s ( s t r o n g l y ) p o l y n o m i a l l y t h e n U i s ( s t r o n g l y ) p o l y n o m i a l l y c o n v e x as w e l l .

convex

25.A.

25.B.

Let

b e a n o p e n s u b s e t of

V

Ui

Show t h a t i f

F,

let

i s a ( s t r o n g l y ) polynomially convex

s u b s e t o f a Banach s p a c e

Ei

for

i = 1,2

then

a ( s t r o n g l y ) polynomially convex open s u b s e t of 25.C.

Show t h a t i f

Ul El

U i s t h e union o f a n i n c r e a s i n g

of s t r o n g l y p o l y n o m i a l l y convex open s u b s e t s o f

E

is

U2

x x

open

E

2'

sequence

U

then

is

s t r o n g l y p o l y n o m i a l l y c o n v e x as w e l l . 25.D.

Show t h a t i f

s u b s e t of 25.E.

E

then

U i s a s t r o n g l y p o l y n o m i a l l y convex open d u i k P i s , ) = dU ( K I f o r each cornpact s e t K C U.

E

i s a s t r o n i j l y p o l y n o m i a l l y convexopen

U;

Show t h a t i f

subset of

for each

i

E

I

t h e n t h e open s e t

Show t h a t a n o p e n s e t

i f and o n l y i f in 25.G.

(JCiUi,

U

U in

Cn

n Ui -7

i s s t r o n g l y p o l y n o m i a l l y c o n v e x as w e l l .

25.F.

U = int

€I

i s polynomially convex

i s h o l o m o r p h i c a l l y convex and

P ( C n l is dense

T ~ ~ ) ,

Show t h a t if U

i s a p o l y n o m i a l l y convex open s u b s e t o f

188

MUJ I CA

t h e n t h e u n i o n o f e a c h c o l l e c t i o n o f components o f

Cn

is

U

p o l y n o m i a l l y convex as w e l l .

26. SCHAUDER BASES

The most i m p o r t a n t r e s u l t s from S e c t i o n s restricted t o

and

24

were

25

and t h e c o r r e s p o n d i n g p r o o f s do n o t s e e m t o

Cn

a p p l y i n t h e case of a r b i t r a r y Banach s p a c e s . T h i s s e c t i o n

is

d e v o t e d t o t h e s t u d y of Banach s p a c e s w i t h

a Schauder b a s i s , and i n S e c t i o n 28 w e s h a l l e x t e n d some of t h e r e s u l t s from Sec-

t i o n s 2 4 and 25 t o t h e case o f Banach s p a c e s w i t h

a

Schauder

basis. 26.1.

DEFINITION.

A sequence ( e )

a,

in

n n=l

x

S c h a u d e r basis i f e a c h

is s a i d t o b e

E

a

h a s a u n i q u e series r e p r e s e n t a -

E E

03

t i o n o f t h e form

x =

<,lx)e

2

n=l

<,13:1

where

n’

E

M for every

n . A Schauder b a s i s ( e n ) i s s a i d t o b e n o r m a i i z e d i f

IIenll

= 1

f o r e v e r y n . A S c h a u d e r b a s i s ( e n ) i s s a i d t o be m o n o t o n e n I1 X < . ( x ) e . I I 5 113: II f o r a l l x E E and n E i7d

if

jIl

3

3

If tionals

has a Schauder b a s i s ( e n ) t h e n t h e c o o r d in a te

E

<,

: x E E

-+

(3:)

E

func-

are c l e a r l y l i n e a r , and so are

1K

n

t h e mappings

Pn

: x E E

s u b s p a c e g e n e r a t e d by

E

onto

+

2 E j ( x , J e E E. j=l j

elJ...,e

then

n

If

denotesthe

Tn i s a p r o j e c t i o n from

En.

I f ( e n ) i s a rnonotone S c h a u d e r b a s i s t h e n

1 E,iz)/

En

Ilcnll

5 2 II x II for

all

x

E

E

and

IITn.rll

5

II 3: I/ and

n E B . The following

theorem t e l l s u s t h a t every S c h a u d e r b a s i s

is

rnonotone

with

r e s p e c t t o some e q u i v a l e n t norm. I n p a r t i c u l a r t h e mappings and

T,

are a l w a y s c o n t i n u o u s .

E‘,L

189

POLYNOMIALLY CONVEX DOMAINS

s s q u e n c e s w h o s e u n i t v e c t o r s f o r m a m o n o t o n e S c h a u d e r basis.

of

m

Let

PROOF.

be a Schauder b a s i s f o r

E and l e t

(nnjnZl

y =

t h e v e c t o r s p a c e of a l l s c a l a r sequences

be

F

W

suchthat

_I

t h e series

converges i n

qnen

E.

W e claim t h a t

n=l

F

is

nj

e

a

n

Banach s p a c e w i t h r e s p e c t t o t h e norm

IIy II = s u p I1

n

Z

j=l

j

11.

a3

Indeed, l e t i g p ) p = I P W

m

i s a Cauchy sequence i n M

and hence ( q E ) p = I

=

rln

Zim

for e a c h

il;

and s e t

yp =

p . One can r e a d i l y see t h a t

f o r each

n)n=l

be a Cauchy sequence i n F

f o r each

n. Set

n. W e s h a l l prove t h a t t h e sequence

y

p + m

-

b e l o n g s t o F and t h a t f y p ) c o n v e r g e s t o y

Indeed, s i n c e ( y p ) can f i n d

E

> 0 we

Hence i t f o l l o w s t h a t

p,.

(26.1)

p

F, given

F.

such t h a t

p,

f o r a l l p, q

for a l l

i s a Cauchy sequence i n

in

p,

n II t: (q; j=l

-

and a l l

n

rl . i e . II 3 3

E IN.

5 Fix

E

p

2

p,

arid choose

no

such t h a t

f o r a l l m, n > n

m

n

0

.

Then i t f o l l o w s from ( 2 6 . 1 ) and ( 2 6 . 2 ) t h a t

m

190

MUJ I CA m

2

f o r a l l m, n

n o . Hence t h e series

(nn)

y =

and t h e r e f o r e

belongs t o

1 2 6 . 1 ) t h a t l y p ) converges t o

B qnen converges i n E n=l F. Then i t f o l l o w s from

y , and t h u s

i s complete,

F

as

asserted. Next w e show t h a t t h e u n i t v e c t o r s form a monotone Schauder basis i n and l e t

Indeed, l e t

F.

denote t h e n t h u n i t v e c t o r i n F ,

f,

y = ( q n ) E F . Then

and t h e l a s t w r i t t e n e x p r e s s i o n t e n d s t o z e r o when

n

+

m.%~is

m

shows t h a t

y =

I: q j f j . To show t h a t t h i s s e r i e s r e p r e s e n t a j=l 03

t i o n i s unique i t s u f f i c e s t o show t h a t i f

rj

= 0

0

there exists

f o r every

j . NOW, i f

C j f j = 0

B

then

j=l

m

Cjfj

C

= 0

t h e n f o r each

E

>

.i = 1 72

II Z

such t h a t

no

5

Lj$II

n >

f o r every

E

j=1

n 0

.

Hence i t f o l l o w s t h a t

Isrn( Ilemll

and t h e r e f o r e

m I1 Z: 5 . e . I I 5 j=7 9 3

5 2~

f o r every

E

f o r every

rn

m

AJ. S i n c e

E E

> 0

5, = 0 f o r e v e r y m E fl, i s a Schauder b a s i s f o r F, and s i n c e

as

E

was a r b i t r a r y w e c o n c l u d e t h a t

w e wanted. Thus if,)

for a l l

y = (Q.) 3

E

Td

F

and

n

E

DV,

w e conclude t h a t

is a

(fnj

monotone Schauder b a s i s , as a s s e r t e d .

If

m

(Sn)n=I

i s t h e sequence of c o o r d i n a t e f u n c t i o n a l s

as-

m

then we define

s o c i a t e d with t h e b a s i s m

Ax = ( E n ( x l ) n = I

f o r every

isomorphism from E o n t o F

x

E

and

E.

Clearly A

A : E

+

F

by

i s a v e c t o r space

191

POLYNOMIALLY CONVEX DOMAINS

x

f o r every

a constant

By t h e Banach Open Mapping Theorem t h e r e

E E.

c

1

such t h a t

5

IIAxII

c 11x11

x

f o r every

E

is E.

The proof of t h e theorem i s now complete. 26.3.

COROLLARY. L e t E b e a Banach s p a c e w i t h a Schauder b a s i s , and l e t (T i d e n o t e t h e s e q u e n c e of c a n o n i c a l p r o j e c t i o n s . Then: n (a) f o r aZ1

There i s a c o n s t a n t

x E E

(b)

and

n

The s e q u e n c e

E

2

c

such t h a t

1

l/TnxlI

5 c lIx/l

W.

converges t o t h e i d e n t i t y

(T,)

o n e a c h c o m p a c t s u b s e t of

unifomly

E.

W e remark t h a t w e d i d n o t g i v e t h e s h o r t e s t p o s s i b l e p r o o f

of Theorem 2 6 . 2 ,

b u t t h e p r o o f g i v e n shows more t h a n h a s

been

s t a t e d , and t h i s w i l l be e x p l o i t e d i n t h e n e x t s e c t i o n . Next w e g i v e a u s e f u l c r i t e r i o n

to

recognize

a Schauder

basis. 2 6 . 4 . THEOREM. A s e q u e n c e ( en of n o n z e r o v e c t o r s i n E is a S c h a u d e r b a s i s if and o n l y if t h e foZZowing t w o c o n d i t i o n s hold. The c l o s e d v e c t o r s u b s p a c e g e n e r a t e d b y i e ) i s n

(a) of

a2L

E. (b)

sequence

Theuae i s a c o n s t a n t

c

s u c h t h a t f o r e a c h scaZar

1

(1.) and p o s i t i v e i n t e g e r s 3

m < n

we h a v e t h a t

I f i e n ) i s a Schauder b a s i s t h e n ( a ) i s o b v i o u s and (b) f o l l o w s from C o r o l l a r y 26.3. Conversely assume t h a t ( a ) and ( b ) h o l d . L e t M be t h e v e c t o r subspace of a l l x E E which admit a PROOF.

m

series r e p r e s e n t a t i o n of t h e form prove t h a t

M i s closed i n

E.

x =

Z .

j=l

Xj e j .

I n d e e d , l e t (zp) b e

We

shall

a sequence m

i n M which c o n v e r g e s t o some each

p

E

a.

Given

E

> 0

we

x

E

can

x p = j = 1 Ajp e j for po E W such t h a t

E . Write

find

MUJ I CA

192

llzp

- xqll 5

for a l l p , q

E

2

p

.

W e claim t h a t

(26.3)

2

p,

no

E

f o r a l l p, q and choose

f o r every n

0

2 n

Indeed, i f we f i x p , q -

> Po

such t h a t

JV

n II B (A;

then

-

j=1

0

5 2~

X!)e.II 3 3

f o r every

n

2

and ( 2 6 . 3 ) f o l l o w s from c o n d i t i o n ( b ) . I t f o l l o w s from ( 2 6 . 3 )

that iiV.

n

and a l l m E JV.

-

IA:

IlemII 5

Xm =

Thus t h e l i m i t

for a l l p , q

4ce

Z i m A:

2

p,

and a l l

e x i s t s f o r each

m

m E lN

E

and

P+m it follows f r o m (26.3) t h a t

for all

p

2

and

p,

m E IIV.

Since

p , w e can e a s i l y g e t t h a t

f o r every

W

and hence t h e series (26.4)

that

IIxP

-

Z

j=1

h . e 3 j

c o n v e r g e s . Then i t f o l l o w s from

co

B

jZ2

X.e.11 < n 7, -

~

C

Ef

or a l l

m

t h a t (zp! converges t o

M = 6

A . e .. Hence 3 3

and t h u s e a c h

u)

t h e form

2

p,,

proving

rn

x =

j=1

X

e

j j

and

M

E, as a s s e r t e d . I n view of c o n d i t i o n ( a ) w e con-

i s closed i n clude t h a t

S

jz1

p

x =

C X.e j=] 3 j

.

x € E

can be r e p r e s e n t e d i n

193

POLYNOMIALLY CONVEX DOMAINS

To show t h a t t h i s s e r i e s r e p r e s e n t a t i o n i s unique

i t suf-

m

2 A e = 0 j=1 3 j

f i c e s t o prove t h a t i f

then

A

j

= 0

for

every

m

j. NOW, i f such t h a t

B

X . e

j=l

a

n

= 0 t h e n f o r each

j

II B A.e . I 1 jz1 3 3

5

E

n

f o r every

t h e r e e x i s t s no

> 0

E

n

0

.

Then i t

follows

m

from c o n d i t i o n ( b ) t h a t

I XIq I

and t h e r e f o r e

II 2 X .e . II 5 c j=l 3 3

llemli 5 2 c s

= 0

m.

f o r every

co

I n t h e e x e r c i s e s a t t h e end o f

and

where

1

5 p

m

m.

I t i s clear t h a t

R

c[

-W,

E

and

E

> 0

the

normalized

L p , where we

t h i s section

examples o f Schauder b a s e s f o r t h e s p a c e s c, co

m

Since

lN.

E

The u n i t v e c t o r s form a monotone

EXAMPLES.

Schauder b a s i s i n e a c h o f t h e s p a c e s p <

m

f o r every

w a s a r b i t r a r y w e conclude t h a t A, proof o f t h e theorem i s complete. 26.5.

f o r every

E

1

2

give

0 , 2 ] and L P [ O , l ] ,

c a n n o t have a Schauder

b a s i s , s i n c e e a c h Banach s p a c e w i t h a Schauder b a s i s

must

be

separable. S. Banach

[ 11 posed t h e problem whether e v e r y

separable

Banach s p a c e h a s a Schauder b a s i s . T h i s problem, known as

the

b a s i s p r o b l e m , remained open f o r a l o n g t i m e and w a s s o l v e d i n

t h e n e g a t i v e by P. E n f l o

[ 1

I

.

EXERCISES

26.A.

Let

E be a Banach s p a c e w i t h a Schauder b a s i s , and l e t

be a f i n i t e d i m e n s i o n a l subspace o f

E.

f o r M i s p a r t o f a Schauder b a s i s f o r

E.

M

( a ) Show t h a t each sequence IY,)

2 6 .B.

E

Show t h a t each basis

c

can

be

= fa,) + t e n ) , where f u n ) i s i s a sequence b e l o n g i n g t o c o .

w r i t t e n a s a sum f V , ) sequence and (6,) (b)

Use

( a ) t o f i n d a Schauder b a s i s f o r

c.

uniquely

a constant

194

MUJ I CA

The Haar s y s t e m

26.C.

=

m ()On)n=l

i s d e f i n e d by

C Lp [ 0,l ]

Y J ~ ( X )

and

1

1

I

0

if

z

(2k

E [

-

2)2

-j-1

, I 2 2 - 1 / 2- j - 1 I

otherwise

t

Show t h a t t h e v e c t o r s u b s p a c e g e n e r a t e d by t h e

(a)

Haar

s y s t e m c o n t a i n s t h e c h a r a c t e r i s t i c f u n c t i o n s of a l l t h e i n t e r v a l s o f t h e form

I (k -

1 )2-j,

I.

k 2-i

Conclude t h a t t h e Haar system

s a t i s f i e s t h e f i r s t c o n d i t i o n i n Theorem 2 6 . 4 . Show t h a t t h e Haar s y s t e m s a t i s f i e s t h e s e c o n d

(b)

con-

c = 1 . Conclude t h a t t h e Haar s y s t e m i s a S c h a u d e r b a s i s f o r L P [ 0 , l ] whenever 1 5 p < m.

d i t i o n i n Theorem 2 6 . 4 w i t h

26.D.

co

I f IVn)n=l

m (i)n)n=I

2

Jo

Vrl-l

C

C [ 0,1]

(t)dt

(a)

i s t h e Haar s y s t e m t h e n t h e S c h a u d e r s y s t e m i s d e f i n e d by

f o r every

n

2

$l(x:)

1

[ 0,l

=

2.

Show t h a t t h e v e c t o r s u b s p a c e g e n e r a t e d by the Schauder

system c o n t a i n s a l l t h e continuous p ie c e wis e l i n e a r on

$,lx)

and

1 w i t h v e r t i c e s a t t h e p o i n t s o f t h e form

functions k2-j.

Con-

c l u d e t h a t t h e Schauder system s a t i s f i e s t h e f i r s t condition i n Theorem 2 6 . 4 . (b)

Show t h a t t h e S c h a u d e r s y s t e m s a t i s f i e s

the

second

= 1 . Conclude that the Schauder

c o n d i t i o n i n Theorem 26.4 w i t h

c

system i s a Schauder b a s i s f o r

C [ 0,l

1.

27. THE APPROXIMATION PROPERTY T h i s s e c t i o n i s devoted t o t h e s t u d y of Banach s p a c e s w i t h t h e approximation property or with

the

bounded

approximation

195

POLYNOMIALLY CONVEX DOMAINS

re-

p r o p e r t y . I n t h e n e x t s e c t i o n w e s h a l l e x t e n d some o f t h e

s u l t s from S e c t i o n s 2 4 and 2 5 t o t h e case o f Banach s p a c e s w i t h t h e approximation p r o p e r t y o r with

the

bounded

approximation

property.

27.1.

E i s s a i d t o have t h e a p p r o x i m a t i o n property

DEFINITION.

i f f o r e a c h compact set rank o p e r a t o r

T

K

C

E

and

there is a

> 0

E

- x I1 5

11 T x

such t h a t

fE;EI

E d:

E

finite

for

every

x E K.

W e r e c a l l t h a t an o p e r a t o r T E d : I E ; F I i s s a i d t o have f i T f E I is f i n i t e d i m e n s i o n a l . One can

n i t e r a n k i f i t s image

r e a d i l y see t h a t an o p e r a t o r only i f t h e r e e x i s t s

plJ..

T E t e ( E ; F I h a s f i n i t e rank if and

. ,(P,

E

El

b,

and

,... , b ,

E F

such

n

=

that

Tx

27.2.

PROPOSITION.

Z:

j=l

qj(xIb

j

x E E.

f o r every

For a Banach s p a c e

the folZowing condi-

E

t i o n s are equivalent:

(a)

E

(b)

Each

has t h e a p p r o x i m a t i o n p r o p e r t y . T E LIE;El

c a n b e u n i f o r m l y approximated on com-

p a c t s e t s by o p e r a t o r s o f f i n i t e r a n k .

F each

For e a c h Banach s p a c e

(c)

T E L ( E F ) c a n be uni-

formZy a p p r o x i m a t e d o n c o m p a c t s e t s by o p e r a t o r s of f i n i t e rank.

For e a c h Banach s p a c e F

(d)

T E d : ( F E) c a n be uni-

each

f o r m Ly a p p r o x i m a t e d on compact s e t s b y o p e r a t o r s of f i n i t e rank. PROOF.

(a)

s u b s e t of operator E

K.

Hence

-

(c): Let

and l e t

E

S E d: I E ; X I II T o S x

T # 0,

T E d:(E;F), E

> 0.

II S x

such t h a t

- T x II

5

-

x I1

f o r every

E

l e t K b e a compact

(a) there

3y

is

5

E/

x

E

a finite

rank

I1 T II f o r e v e r y K.

Since

t

T o S has

f i n i t e r a n k l ( c ) h a s been proved.

(a) * (d): F , and l e t

E

Let > 0.

d : ( E ; E ) such t h a t

T E d: ( F ; E I

,

let

K

be a compact s u b s e t o f

By ( a ) t h e r e i s a f i n i t e r a n k o p e r a t o r

IISx - z I I

5

E

f o r every

x

E

TIKI.

S E

Hence

196

I1 S

o

MUJ I CA

-

Ty

Ty 1 I

5

f o r every

E

y E K.

Since

has

S o T

finite

r a n k , ( d ) h a s been proved. Since t h e i m p l i c a t i o n s ( c ) * ( b ) , ( d )

( b ) =- ( a )

( b ) and

=)

are o b v i o u s , t h e proof o f t h e p r o p o s i t i o n i s complete. Next w e i n t r o d u c e t w o s t r o n g e r v e r s i o n s of t h e

approxima-

t i o n property.

27.3. DEFINITION. ( a ) E i s s a i d t o have t h e m e t r i c a p p r o x i m a t i o n p r o p e r t y i f f o r e a c h compact set K C E and E > 0 t h e r e i s a f i n i t e rank o p e r a t o r T E e ( B ; E l such t h a t IIT II 5 I and I I T z - xII 5 E f o r e v e r y I E K . (b) E i s s a i d t o have t h e b o u n d e d a p p r o x i m a t i o n property i f there exists a constant c > 1 s u c h t h a t f o r e a c h compact

set

K

C

and

E

such t h a t

d: ( E ; E I

27.4. THEOREM.

> 0

E

t h e r e i s a f i n i t e rank o p e r a t o r

IITII 5 c

II T x - z II <

and

For a s e p a r a b l e Banach s p a c e

E

E

T

E

f o r every z E K . the

follouing

conditions are equivalent.

(a)

E

(b)

There i s a sequence o f f i n i t e rank o p e r a t o r s

d: f E ; E l

h a s t h e bounded a p p r o x i m a t i o n p r o p e r t y

l i m Tnx = x

such t h a t

(C)

f o r every

T h e r e a r e a s e q u e n c e (a,)

in

x

Tn

E

E 6‘.

E and a s e q u e n c e

(pnl

a,

/lanII = 1

i n E ’ such t h a t

f o r every

n

E

LV and

Z

~p,izla,

n=l J:

7

f o r every

(d)

E

x E E.

i s t o p o l o g i c a l l y i s o m o r p h i c t o a compZemented sub-

s p a c e of a Ranach s p a c e u i t h a m o n o t o n e S c h n u d e r basis.

Before p r o v i n g t h i s theorem w e g i v e two p r e p a r a t o r y results. 27. 5 , PROPOSITION. there are 6

( J

T’r

El,..

Let

E

b e a Ranach . s p r ~ coJ” dz‘nicn;iiin

n vectors

.,En

f o r norm o n e i n

o f norm o n e i n E’

such t h a t

n. Then

E and

n

S j ( ek

=

uce-

6

J-k

POLYNOMIALLY CONVEX DOMAINS

j , k = 1 ,..., n .

for a l l

Fix a b a s i s i n

PROOF.

197

Alxl

,..., x n )

of

xk

= detla

)

E and l e t A E E ( n E l by d e f i n e d by where a l k ,... , a n k are t h e coordinates

j k w i t h respect t o t h e f i x e d b a s i s . L e t ( e l J . . . , e n )

n - t u p l e o f v e c t o r s o f norm o n e i n I f we define

<,,. . . , E n

then t h e n-tuples

(el,.

El

E

E where

A

b e an

a t t a i n s i t s norm.

by

. ., e n )

(El,

and

. . ., 5,)

have the required

properties.

27.6.

Let

LEMMA.

are oper ator s n2 Z B.(x) j=1 J

k = 1,

=

b e a Banach s p a c e o f

E

Bl, ..., B 2 n for e v e r y

x

E

L(E;EI,

of

dimension n. Then there

rank one, such

k

x E E

II Z B . I I j=l d

and

5

for

2

that every

..., n 2 .

By P r o p o s i t i o n 27.5 t h e r e are o p e r a t o r s A l , . . . , A n E n d: (E;E), of rank o n e , such t h a t Z A .x = x for e v e r y x E E j=I 3 PROOF.

= 1

and

IlA.11

and

( E l , ...,En

3

f o r every

1

and

1

5

A.x = < . ( x ) e

B

3

3

= n-IAS j s 5 n . Then

Then w e d e f i n e

-

(el

,..., e n )

are t h e n - t u p l e s g i v e n by P r o p o s i t i o n

)

then i t s u f f i c e s t o def ine j = I, ..., n .

n

j = 1 ,..., n . Indeed, i f

if

j

f o r every

j = r n + s, where

27.5

X E E

0

and

5 r 5

198

MUJ I CA

and i f follows t h a t

Z B. x = z j=z 3

and

II

5

B .xII 3

E

j=I

2 Ilx II

,

as

w e wanted.

(a)

PROOF OF THEOFEM 2 7 . 4 .

E b e a s e p a r a b l e E5anad-1

(b): L e t

=+

s p a c e w i t h t h e bounded a p p r o x i m a t i o n p r o p e r t y . L e t (x.) be a 3 s e q u e n c e which i s d e n s e i n E . Then t h e r e are a c o n s t a n t c L 1 and a s e q u e n c e o f f i n i t e r a n k o p e r a t o r s T n E f ( E ; E l s u c h t h a t IITnII 5 c

and

E

.

- x 3. II 5 Z/n f o r j = I , . . , n . L e t n j be g i v e n . F i r s t c h o o s e j s u c h t h a t 111; -

and

> 0

IIT z

and n e x t c h o o s e

no

n 2 n0

Then f o r

m a x {j, Z / E } .

z

E

E

3:11

5

E

we

have

that

and h e n c e

llTnz

-

x II t e n d s t o z e r o .

( b ) * ( c ) : Suppose t h e r e i s a s e q u e n c e o f f i n i t e rank operat o r s T n E E ( E ; E l s u c h t h a t Z i m T nx = x f o r e v e r y x € E . I f

we define

Sl = T I

and

Sn =

Tn

-

for

*n-1

L

n

then each

2

m

i s a f i n i t e r a n k o p e r a t o r and

Sn

Z

n=l E.

Set

Mn

=

S,(E)

Lemma 2 7 . 6 f o r e a c h

and

mn = ( d i m M n l '

f o r every

for every

t h e r e are o p e r a t o r s rnn E L(Mn;Mnl, of rank one, such t h a t C B x = x jZI nj k E Mn and II Z B n j I1 5 2 f o r e v e r y k = I , . . . , m n . j=l and d e f i n e A k = Bns 0 S n if k = rn 0 + . . . + rn n- 1 n

E

iT

k

2 A.= j = J~

and

1

n

Sn x = x

E Z V

x

whence k

E

Bnl'

* *

J

n- 1

n- 1

=

2 Sr r=I

Brim 71

f o r every Set

+ s,

x

rn 0 = 0

where

S i

E

By

DJ.

5 s 5 rn n . Then

rnr S 2 z BrtoSr+ BrtoS r=z t = l t=l

n-1

n

x

i 2 BPt) o S n . t= 7

199

POLYNOMIALLY CONVEX DOMAINS

and t h e l a s t w r i t t e n e x p r e s s i o n t e n d s t o z e r o when n + m.Thus w e have found a sequence of o p e r a t o r s A n E e ( E ; E l , of rank m

o n e , such t h a t

Z An x = x

x

f o r every

Then

E.

E

f o r each

n=I

n

E

=

1

w e can f i n d

mJ

an

E

Anx = 9 ( x l a n n

and

and n f o r every x A (El

Thus

E.

E

lla,II

such t h a t

E'

E

Ipn

(c)

h a s been

proved.

( c ) * ( d ) : Suppose t h e r e are a sequence sequence ( 9 n ) i n m

C

= x

Vn(xlan

IIa II = 1 n

E' such t h a t

x

f o r every

(a,)

i n E and a

f o r every n

and

E. L e t F be t h e v e c t o r

E

space

n=I

03

y = (1,)

of a l l scalar sequences converges i n

E , and endow

F

such t h a t t h e series

Z Qnan n=7

n

w i t h t h e norm II y II =sup11 z 11 .a.II. n j=I J J

Then t h e proof of Theorem 2 6 . 2 shows t h a t

F i s a Banach space

whose u n i t v e c t o r s form a monotone Schauder b a s i s . Define A : E m

+

P

by

Ax

= ( V r L ( x ) ) .S i n c e

9,(x)an = x

f o r every x

E

E

n=I

t h e P r i n c i p l e of Uniform Boundedness g u a r a n t e e s t h e e x i s t e n c e n of a c o n s t a n t C > 0 s u c h t h a t II B 9 . ( x ) a .II 5 C II x II f o r all j11 J J x

E

E

and

n E D.

Hence

IIAzll

5 ClIx II f o r e v e r y

x

m

A E L(E;F).

Next d e f i n e n

B E L(F;E).

I t i s a l s o clear t h a t

and ( A B ) f A B ) y = A B y

B : F

f o r every

+

E

By =

by

= x

BAx

E and A(FI and

p r o j e c t i o n from F o n t o

AIEI.

AB

and

E

.

f o r every

y E F. Hence

c a l isomorphism between

Z rln a n n=l

E

Then

z

E

i s a topoloqiis a continuous A

Thus ( d ) h a s been proved.

( d ) * ( a ) : S i n c e e v e r y Banach s p a c e w i t h a Schauder

basis

h a s t h e bounded a p p r o x i m a t i o n p r o p e r t y , t h e d e s i r e d c o n c l u s i o n f o l l o w s from E x e r c i s e 2 7 . A .

The proof

of

the

theorem i s now

complete. C l e a r l y e v e r y Banach s p a c e w i t h a Schauder b a s i s

has

the

MUJ I CA

200

bounded approximation property, and every Banach space with a monotone Schauder basis has the metric approximation property. Next we give more interesting examples. 27.7. EXAMPLE. If X is a compact Hausdorff space then has the metric approximation property.

PROOF. Let K be a compact subset of C l X ) and let K is compact we can find functions f,, . . . , f, f K m

K C

B (jCi;

U

. For each

E)

a

E

X

c(X)

> 0.Since

E

such that

consider the open set

j=1

X

Since

is compact we can find points n

ak

k=l

,...,

each

and

~1

f

E

By Exercise 1 5 . G

...

€unctions c p l , 1

.

u U

X =

that

E

cpI(x)

+

C l X ) define

we

al,...,an

can

X

E

find nonnegative

C t X l such that

.. . Tf

+ E

p

n

(T)

= 1

for every

C f X ) by (Tf)I z ) =

such

n

z

x

f

for

k =

X.

For

f ( a k ) p i c ( z )for

k=1 x E X. Then clearly T is a finite

every IITfll

5

/ I f II

for every

for every 3: E X B(f .;E) for some 3

for every

f

E

and ,j

f

E

C(XI.

rank operator It is also clear that

j = I,.. . , n .

= 1,.

.

.,in

Since each f we conclude that

E

K

and

lies in

K.

If X is a completely regular Hausdorff space ther. the Banach space C b ( X ) of all bounded continuous functions 27.8.

EXAMPLE.

on X has the metric approximation property. PROOF.

If

PX

denotes the Stone-Cech compactification of

X

P O L Y N O M I A L L Y CONVEX D O M A I N S

f E C b IX) h a s a u n i q u e e x t e n s i o n

then each t h e mapping

CbIX)

f E

+

$f E C I B X )

20 1

Bf

E

C($X). C l e a r l y

i s a v e c t o r s p a c e isomor-

phism and an i s o m e t r y . Then t h e d e s i r e d c o n c l u s i o n f o l l o w s from t h e p r e c e d i n g example. 27.9.

EXAMPLE.

h a s t h e metric a p p r o x i m a t i o n p r o p e r t y .

Rw

n

R

PROOF.

c a n be

i d e n t i f i e d with

if

Cb(lN)

i s endowed

_W

with t h e d i s c r e t e topology. G r o t h e n d i e c k [ 1 ] p o s e d t h e p r o b l e m w h e t h e r e v e r y Banach

A.

T h i s p r o b l e m , known

as

approximation problem, w a s s o l v e d i n t h e n e g a t i v e by

P.

space has the Enflo

[

the

approximation property.

11.

EXE RC ISES 27.A.

Show t h a t i f E h a s t h e (bounded) a p p r o x i m a t i o n

property

t h e n e a c h complemented s u b s p a c e of E h a s t h e (bounded) a p p r o x i mation p r o p e r t y as w e l l . 27.B.

Show t h a t i f e a c h c l o s e d s e p a r a b l e s u b s p a c e

t h e approximation property then

E

of

has

E

h a s t h e approximation property

as well.

m

simple f u n c t i o n s such t h a t

j = 7,

li {x E X :

..., m.

(a) such t h a t each

f, . . . , f m be f.(X) # 0 1 < for

L e t ()., C, p ) be a measure s p a c e a n d l e t

27.c.

Ak

(b)

3

n d i s j o i n t s e t s A I , . . . JAn €or e a c h k , e a c h f is constant Li n

Show t h e e x i s t e n c e of 0 ,: w ( A k )

and e a c h For e a c h

M

fj

i s i d e n t i c a l l y z e r o on

p with

define a simple function

1

5

Tf E

p <

and e a c h

w

LP (X,

Z,

11) by

X\ j”

E

’ on

u Ak ‘

k=l

Lp(X,

8,

w)

MUJ I CA

202

Show t h a t IIT II

with

i s a f i n i t e r a n k l i n e a r operator

T

< 1. F u r t h e r m o r e , show t h a t

. .,m

j = I,.

every

Show t h a t

(C)

k = 1,.

and

property f o r each

L

P

(X,

x,

p)

1

5

with

p

= fj

7'fj

L P (X, Z, p )

on on

for

Ak

. .,n. t h e metic

has p <

approximation

m.

28. POLYNOMIAL APPROXIMATION I N BANACH SPACES I n t h i s s e c t i o n w e p r e s e n t Banach

space

versions of

the

r e s u l t s on p o l y n o m i a l a p p r o x i m a t i o n e s t a b l i s h e d i n S e c t i o n s 2 4 a n d 2 5 . The l e t t e r s

and

E

will

F

a l w a y s r e p r e s e n t complex

Banach s p a c e s .

P I E ; F ) d e n o t e s t h e vector s p a c e of a l l f c o n t i n u o u s p o l y n o m i a l s o f f i n i t e t y p e f r o m E i n t o F. This space w a s introduced i n Exercise 2.K. With t h i s n o t a t i o n w e L e t us r e c a l l t h a t

h a v e t h e f o l l o w i n g r e s u l t , w h i c h e x t e n d s Theorem 2 5 . 4 . E b e a Banach s p a c e w i t h t h e ,zppro.cir?ation U b e a p o l y n o m i a l l y conve.l: o p e n s u b s e t of' E . Then f o r e a c h f E J C ( U ; F ) and e a c h compact s e t K C U t h e r e i s a s e q u e n c e of p o l y n o m i a l s Pj E P f 6 ; F l w h i c h c o n v e r g e s t o f

28.1.

Let

THEOREM.

p r o p e r t y and l e t

f

u n i f o r m l y on

K.

Let

PROOF.

be given. Since

0

E

compact w e c a n f i n d I l f ( y ! - ~ ( z J I I<

E

f

such t h a t

fi > 0

x

for all

E

+

K

and

K

i s continuous and RlO;di

K

U

C

Since

y E B(x;d).

is and E

has t h e approximation p r o p e r t y t h e r e i s a f i n i t e rank o p e r a t o r such t h a t

T E l(E;EI

T(K) C U

follows t h a t

x

E K.

II TZ

s i o n a l , Theorem 2 5 . 4 €

P(T(EI;F),

Since

J:

It

x

f o r every

< 6

and I i f o T i x )

-

By Example 2 5 . 2 t h e i n t e r s e c t i o n

m i a l l y convex open s e t i n

P

-

T I E ) . Since

c

ffx)ll

IIPly)

-

JIg)II

j

E

F

and

~ p . E

3

(Y(i0)

I .

every

?'(El i s f i n i t e dimen-

5

r.

Whence

polynomial

f o r every y E T ( K I .

T I E ) is f i n i t e dimensional, Exercise 2 . K

L'

Whence it

for

g u a r a n t e e s t h e e x i s t e n c e of a

such t h a t

K.

U n T ( K ) i s a polyno-

P i s of f i n i t e t y p e , t h a t i s , P i s of t h e form

where

E

if

g u a r a n t e e s that m .

P

2

9j<1

j follows

that

20 3

POLYNOMIALLY CONVEX DOMAINS

P

o

T = Z c . (p

0'

3

f o r every

x

m. o Tl

E

K,

.

P fE;Fl

E

f

j

Furthermore, w e have t h a t

and t h e proof i s complete.

to the

Next w e want t o e x t e n d t h e Oka-Weil Theorem 24.12

case o f Banach spaces. The k e y t o t h e p r o o f

the

is

following

theorem. 28.2.

THEOREM.

of

ar,d let U b e a n o p e n n e i g h b o r h o o d of

E

Let

K

be a p o l y n o m i a l l y c o n v e x c o m p a c t s u b s e t

a s t r o n g l y p o l y n o m i a l l y convex open s e t

Then t h e r e

K.

V such t h a t

K

C

V

C

is U.

-

Since t h e c l o s e d , convex h u l l

PROOF.

t h e p r o o f of Lemma

PI' .

-

*

,pm

E

24.7

shows

P I E ) such t h a t

we claim t h e r e e x i s t s

the existence

IPjl < 1

6 > 0

c o f K ) of

on K for

K i s compact, of

polynomials

j = 1,.

. .,m

and

such t h a t

Otherwise t h e r e i s a sequence f a ) such t h a t n

f o r every that

I1 a n

n. Choose f o r e a c h

- bn

II

<

I/n.

Since

n an element bn . E Z i K l such c 0 i K . J i s compact t h e sequence ( b n )

h a s a s u b s e q u e n c e ( b l which c o n v e r g e s t o a p o i n t b E G f K I . nk But t h e n t h e c o r r e s p o n d i n g s u b s e q u e n c e ( a ) of ( a , ) also connk v e r g e s t o b . Whence i t f o l l o w s t h a t li E {x E

Coi~.) :

I P . ,I( x ) /

c

I

for

.i

= I ,..., rn) \ U,

c o n t r a d i c t i n g ( 2 8 . 1 ) . T h i s shows t h e e x i s t e n c e of 6 > 0 s a t i s f y i n g

MUJ I CA

204

128 2 ) .

I f we define

V =

K

then

C

V

a n d i t f o l l o w s f r o m Example 25.2

U

C

that

is

V

s t r o n g l y polynomially convex. Now i t i s e a s y t o e x t e n d t h e Oka-Weil Theorem

24.12to the

case of Banach s p a c e s . 28.3.

Let

THEOREM.

b e a Banach s p a c e w i t h

E

t i o n p r o p e r t y and l e t

of

P.

nomials

3

E

P (E;F) w h i c h c o n v e r g e s t o f

E

approrirna-

t h e r e i s a sequence o f

X(K;Fl

f

By Theorem 28.2 w e may assume t h a t

PROOF.

U

f

T h e n for e a c h

E.

the

b e a polynomially c o n v e x compact s u b s e t

K

i s a p o l y n o m i a l l y convex open s u b s e t of

poiy-

uniformly o n

K. where

f E X(U;F),

containing

E

K.

Then i t s u f f i c e s t o a p p l y Theorem 2 8 . 1 . 28.4.

PROPOSITION.

tion p r o p e r t y .

Let

F:

b e a B a n a c h s p a c e w i t h t h e approxima-

T h e n an o p e n s u b s e t

U of h' is p o l y n o m i a Z l y convcx

if and o n l y if U i s s t r o n g l y p o l y n o m i a l l y c o n v e x . PROOF.

I n v i e w o f Theorem 2 8 . 3 t h e p r o o f o f P r o p o s i t i o n

25.3

applies. I n t h e case of s e p a r a b l e Banach s p a c e s w i t h t h e boundedapp r o x i m a t i o n p r o p e r t y t h e c o n c l u s i o n o f Theorem

28.1

can

be

sharpened as follows. 28.5.

Let

THEOREM.

E

b e a s c p a r u b l c Ban
a p p r o x i m a t i o n p r o p e r t y and l e t ; subset of

Then f o r each

E.

poZynomiuls

P

i

E

i/ E

b e a p o l . y n o m , i u l l y c o n v e x opt:ri JCIU;Fl

t h e r c i s a s + ? q u c ; n ~ ?of e

P ( 6 ' ; F ) w h i c h convcr~ge': t o

f

e a c h c o m p a c t s u b s e t of PROOF.

f

.f

xnifoi-iri'y

on

1J.

( a ) L e t u s assume f i r s t t h a t

E h a s a monotone S c h a u d e r

b a s i s . L e t ( e n ) denote t h e b a s i s , l e t (!l'nl d e n o t e t h e sequence of c a n o n i c a l p r o j e c t i o n s a n d l e t

EYl

Y',iEl

for

n = I,::,..

..

205

POLYNOMIALLY CONVEX DOMAINS a3

Finally let

denote t h e dense subspace

Em

Em

= u

En.

Since

n=l Em i s d e n s e i n

and s i n c e

E

f i n d a sequence ( a

in

,)

3

Em

U i s a L i n d e l o f s p a c e ws? can e a s i l y a n d a s e q u e n c e ( r . I of p o s i t i v e 3

numbers s u c h t h a t

j. I f we s e t

f o r every

k Ak =

Bla .r,) j’ 3

U

j=l

and

Sk

= min { r l , . . .,

k , t h e n ( A k ) i s an i n c r e a s i n g sequence

f o r every

‘k’ of

bounded

open s e t s s u c h t h a t

k , let

for every

n k ) be a n i n c r e a s i n g s e q u e n c e

i n t e g e r s such t h a t

a

i

clear t h a t

for

E E

j = I,

nk

..., k .

of

positive

Hence

it

is

a n d t h e r e f ore Tn(Ak) = A

(28.4)

k

n En

for

I t f o l l o w s from ( 2 8 . 3 ) a n d ( 2 8 . 4 ) t h a t

n > n. -

k

Tn ( A k ) k

i s arelatively

is a k p o l y n o m i a l l y c o n v e x open s u b s e t o f t h e f i n i t e d i m e n s i o n a l s u b -

compact s u b s e t o f

U

f o r every

L‘

k . Since

U n

En

nk

s p a c e E,

‘k Pir E P I F (Ak).

Tn

,

nk’

Theorem 2 5 . 4 g u a r a n t e e s t h e e x i s t e n c e of a p o l y n a n i a l

.FI s u c h t h a t

Thus

llPkiy) - fiylll

5 l/k

f o r every

y E

E P (F,’;FI a n d t o c o m p l e t e t h e p r o o f we k S i c o n v e r g e s t o f u n i f o r m l y on e a c h compact

P k o Tn

K

show t h a t ( P k o T

“k

s u b s e t of U. L e t K b e a c o m p a c t s u b s e t o f U a n d l e t E > 0. Then f i r s t c h o o s e 6 > 0 s u c h t h a t K + B ( O ; 6 ) C U and IIf(y) -fi.x)II < E

206

MUJ I CA

for all <

E,

K

x C

K

E

and and

Ak

y E B ( x ; 6 ) . Next c h o o s e \ITn

0

Then f o r e v e r y

x

E

K

x k and

- XI/ k

T h i s c o m p l e t e s t h e p r o o f when (b)

< 6

2 ko

for all

k, 5 E

such t h a t l / k o

K and k

k,.

we g e t that

E h a s a monotone S c h a u d e r basis.

C o n s i d e r n e x t t h e g e n e r a l c a s e , i n which

E i s a sepa-

r a b l e Banach s p a c e w i t h t h e bounded a p p r o x i m a t i o n p r o p e r t y . By

w e may assume t h a t 6 i s a complemented s u b s p a c e G which h a s a monotone S c h a u d e r b a s i s . Let IT E E I C ; E l b e a c o n t i n u o u s p r o j e c t i o n f r o m G o n t o E. I f U i s a p o l y n o m i a l l y c o n v e x o p e n s u b s e t of E t h e n I T - ' ( U ) i s a p l y -

Theorem 2 7 . 4

o f a Banach space

n o m i a i l y c o n v e x open s u b s e t o f then

f o -ir E J C ( n - ' ( U l ; F I

polynomials

Q . E P (G;FI

J

f

i

E

P IE;FI

f

p a c t s u b s e t of

and

by E x e r c i s e 25.A.

I f f EJC(U;FI

a n d by p a r t ( a ) t h e r e i s a sequence of

fo

which c o n v e r g e s t o

IT

uniformly

-1

IT ( U i . If w e s e t P . = 4 . 16 t h e n 3 3 ( P . ) c o n v e r g e s t o f u n i f o r m l y on e a c h com-

on e a c h compact s u b s e t o f

P

G

3

U. The p r o o f of t h e t h e o r e m i s now c o m p l e t e .

BY c o m b i n i n g Theorems 28.2

a n d 28.5 w e c a n s h a r p e n t h e con-

c l u s i o n o f Theorem 2 8 . 3 a s f o l l o w s .

To c o n c l u d e t h i s s e c t i o n w e p r e s e n t a f i r s t Problem 11.6. solution.

I n S e c t i o n 4 5 w e s h a l l presc,nt

attempt t o solve

more s a t i s f a c t o r y

POLYNOMIALLY CONVEX DOMAINS E h a s a monotone S c h a u d e r

( a ) Assume f i r s t t h a t

PROOF.

and l e t

207

U be a p o l y n o m i a l l y convex open s e t i n

basis

E . I n the proof

of Theorem 28.5 w e found a n i n c r e a s i n g s e q u e n c e of boundedopen

Ak

sets

such t h a t m

U =

Ak

U

Ak +

and

U

BIO;sk) C

k=l

k . I f we c a n p r o v e t h a t

f o r every

h

f o r every main for

of n

Tn(

NOW, by

(28.4) we have t h a t

and t h i s i m m e d i a t e l y i m p l i e s t h a t

nk

(28.6)

U i s a doTn(Akl c Ak n En

t h e n Theorem 1 1 . 4 w i l l g u a r a n t e e t h a t

existence.

(A^k)

A

(Ak

PIEj)

n

for

'n)p(En)

2 nk ' -

Fix

E

R(O;sk

for a l l since

> 0

small, Since

-

C

E)

k

sional. Since

C

U

we g e t t h a t

Ak

+

and t h e r e f o r e

U

and

U n En

Ak + B(O;sk)

n.

NOW,

U n En

i s a domain of h o l o m r p l y i n E n ,

i s p o l y n o m i a l l y convex and -

Ak n E

n

En

i s f i n i t e dimn-

i s a compact s e t i t f o l l o w s from Theorem

11.4 that

Since

P(EKj

is dense i n I X ( U

see t h a t ( A k n k ' n I j ( i ( i n i ; n j

and t h e r e f o r e

= (Ak

En),

'Tc), by Theorem

25.4,

we

A

1 1

Whence w e g e t t h a t

20 8

MUJ ICA

UE = {x

where R(O;s

k

-

E

U

L €1.

: dl,(x)

Let

x

( b k i P i E ) and

E

y

E

UE C

U,

2 ~ ) . Then it f o l l o w s f r o m ( 2 8 . 6 ) a n d ( 2 8 . 7 ) t h a t

n 2 nk. Letting

f o r every

n

-

-+

we g e t t h a t

x+ y

E

proving t h a t

Since

t

> 0

w a s a r b i t r a r i l y small ( 2 8 . 5 ) f o l l o w s . T h i s

p l e t e s t h e p r o o f when

E

h a s a monotone S c h a u d e r b a s i s .

I n t h e g e n e r a l case w e may assume t h a t

(b)

p l e m e n t e d s u b s p a c e o f a Banach s p a c e Schauder b a s i s . L e t E

G

which

E

has

is a

b e a c o n t i n u o u s p r o j e c t i o n from

i s a p o l y n o m i a l l y c o n v e x , o p e n s u b s e t of

domain o f e x i s t e n c e , by p a r t (a). B u t t h e n

a domain o f e x i s t e n c e i n

E

U

com-

a monotone

U b e a p o l y n o m i a l l y c o n v e x o p e n s u b s e t of

and l e t

n-’iU)

71

com-

G

onto

E.

Then

and h e n c e

G

= n - l i l l l n l?

by P r o p o s i t i o n 1 1 . 7 .

The p r o o f

a is

of

t h e t h e o r e m i s now c o m p l e t e .

EXERCISES

28.A.

An o p e n s u b s e t

dense i n (3C(U),

T ~ ~ )Show .

nomially convex i f 28.B.

U

of

E‘ i s s a i d t o be Runge i f

t h a t an open s u b s e t

U

of

P(E) i s i s poly-

E

U i s h o l o m o r p h i c a l l y c o n v e x a n d Runge.

An open s u b s e t

U of

E

i s s a i d t o be j’iriiteZy poLyrznmiaLly

c’oriv+?x ( r e s p . l i n i t i : % W K u n g c ) i f

U n M

i s p o l y n o m i a l l y convex

( r e s p . Runge) f o r e a c h f i n i t e d i m e n s i o n a l s u b s p a c e C o n s i d e r t h e f o l l o w i n g c o n d i t i o n s on a n o p e n s u b s e t (a)

U

i s polynomi.ally convex.

(b)

U

i s f i n i t e l y p o l y n o m i a l l y convex.

(c)

U

i s f i n i t e l y Runge.

M U

of of

L.

E.

POLYNOMIALLY CONVEX DOMAINS

i s Runge.

U

(d)

Show t h a t t h e i m p l i c a t i o n s ( a ) ways t r u e .

Show t h a t

property. and

if

E

h a s t h e approximation

F n a l l y s h o w t h a t i f E has t h e approximation p r o p e r t y

E

then t h e

L e t E be a Banach s p a c e w i t h t h e a p p r o x i m a t i o n

property

a ) , ( b ) , ( c ) and ( d ) a r e e q u i v a l e n t .

conditions

U be a ( f i n i t e l y ) p o l y n o m i a l l y convex open s u b s e t of

and l e t E.

(c) * (d)

* ( b ) and ( b ) * ( c ) a r e a l -

i s a h o l o m o r p h i c a l l y convex open s u b s e t of

U

28.C.

209

of

Show t h a t t h e union of e a c h c o l l e c t i o n

U i s ( f i n i t e l y ) p o l y n o m i a l l y convex as w e l l .

nents of 28.D.

c o n n e c t e d compo-

L e t E be a Banach s p a c e w i t h t h e a p p r o x i m a t i o n p r o p e r t y

and l e t

be a compact s u b s e t of

K

2p(El

component of

E.

Show t h a t e a c h c o n n e c t e d

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

K.

L e t E be a Banach s p a c e w i t h t h e a p p r o x i m a t i o n p r o p e r t y

28.E. and l e t

b e a p o l y n o m i a l l y convex compact s u b s e t of

K

E.

Show

t h a t t h e union of each c o l l e c t i o n of c o n n e c t e d components of K

i s a l s o p o l y n o m i a l l y convex. 28.F.

E be a s e p a r a b l e Banach s p a c e w i t h t h e bounded ap-

Let

proximation p r o p e r t y .

Show

that

an

p o l y n o m i a l l y convex i f and o n l y i f

open s u b s e t

U

of

E is

U i s f i n i t e l y polynomially

convex. 28.G. set in 28.H.

U i s a b a l a n c e d p o l y n o m i a l l y convex K p ( E l C U f o r e a c h compact s e t K C U.

Show t h a t i f E

then

L e t E be a Banach s p a c e w i t h t h e a p p r o x i m a t i o n

property,

U be a polynornially convex open s u b s e t of E and l e t ( J c f l I ) , ~ ~ ) --t t be a c o n t i n u o u s a l g e b r a homomorphisr..

let

(a) jh(f)

1 5 (b)

Show t h e e x i s t e n c e of a compact s e t

s;p

If

1

f o r every

f

K C U

open

h :

such t h a t

E SC(U).

Using t h e Mackey-Arens Theorem show t h e e x i s t e n c e of

a unique p o i n t

a E E

such t h a t

h(PI = P i a ) f o r each

P E Pf(E).

MUJ I CA

210

(c)

= flal

Using Theorem 2 8 . 1 t w i c e show t h a t f o r every

a

E

and

U

h(f)

f E X(UI.

NOTES AND COMMENTS

Theorems 24.11 a n d 24.12 are d u e t o K . 2 4 . 1 2 e x t e n d s a n e a r l i e r r e s u l t o f A. W e i l

Oka [ 1]

[

.

1 1 .

Theorem

Our presenta-

t i o n i n S e c t i o n s 2 4 and 2 5 f o l l o w s e s s e n t i a l l y t h e book

of

L.

Hormander [ 3 1 . Schauder bases were i n t r o d u c e d by J . S c h a u d e r who gave t h e examples of S c h a u d e r b a s e s i n Banach

[

I

,[

1 ,

2

and C [ O , l l

Theorem 26.2 i s due t o

which a p p e a r i n E x e r c i s e s 2 6 . C and 2 6 . D . S.

L p [U,2

[ 11

1 1 , w h e r e a s Theorem 26.4 h a s b e e n t a k e n from

book of J. L i n d e n s t r a u s s and L. T z a f r i r i

the

11 1 .

The a p p r o x i m a t i o n p r o p e r t y w a s i n t r o d u c e d by A. Grothendieck [ 11

.

. Theorem 2 7 . 4 i s d u e t o A. P e l c z y n s k i [ 1 ] Proposition due t o H. Auerbach, c a n be f o u n d i n t h e book of S. Banach

27.5,

[I]. R.

Aron and M.

Schottenloher

[ 1]

o b t a i n e d Theoren 28.1 and

t h e r e s u l t i n E x e r c i s e 28.B, t h u s e x t e n d i n g e a r l i e r r e s u l t s of S. Dineen E.

Ligocka

[ 3 ] [

and P. Noverraz [ 2 1 .

1 1 . Theorem 28.3,

Theorem 28.2 i s d u e

d u e t o M. S c h o t t e n l o h e r

[ 5

to

1,

e x t e n d s a n e a r l i e r r e s u l t o f P . Noverraz [ 2 1 . The s i m p l e proof g i v e n h e r e i s due t o J. Mujica [ 3 1 . Theorem 28.5 i s d u e t o C. Matyszczyk [ I 1 , w h e r e a s Theorem 28.7 i s e s s e n t i a l l y d u e t o S . Dinnen and A. H i r s c h o w i t z [ l due t o J. Mujica [ 1 ]

I I].

I . The r e s u l t i n E x e r c i s e

, extends

28.H,

a n e a r l i e r r e s u l t of J. M. Isidm

CHAPTER V I I

COMMUTATIVE BANACH ALGEBRAS

29. BANACH ALGEBRAS T h e r e i s a close i n t e r p l a y b e t w e e n t h e t h e o r y

of

p h i c f u n c t i o n s a n d t h e t h e o r y o f c o m m u t a t i v e Banach

holomoralgebras.

I n t h i s s e c t i o n w e p r e s e n t some b a s i c f a c t s a b o u t Banach algeb r a s , n o t n e c e s s a r i l y commutative. 29.1.

A i s s a i d t o b e a Ban
DEFINITION.

a l g e b r a ( a l w a y s assumed complex a n d w i t h u n i t e l e m e n t

e # 0)

a n d a Banach s p a c e s u c h t h a t

5

II x I/ /I y II

(a)

Ilsy II

(b)

II e II = 2 .

A Banach a l g e b r a

A

for all

x, y E A ;

i s s a i d t o be c o m m u t a t i v e

xy = y x f o r

if

a l l x, y E A. 29.2,

EXAMPLES.

(a) Let

X b e a compact H a u s d o r f f

m u l t i p l i c a t i o n i s defined pointwise then

space.

If

C ( X I i s a commutative

Banach a l g e b r a . (b) and l e t

Let

K b e a compact s u b s e t o f a complex Banach s p a c e

P ( K I d e n o t e t h e c l o s u r e of

P(E) in

ClKl

.

Then

P(
i s a c o m m u t a t i v e Banach a l g e b r a , as a c l o s e d s u b a l g e b r a o f C I K ) .

(c!

Let

E b e a complex Bariach s p a c e .

i s d e f i n e d as c o m p o s i t i o n t h e n i s n o t commutative u n l e s s

If

multiplication

d : ( E ; F : ) i s a Banach alqebrawhich

E has dimension one.

The n e x t t h e o r e m g i v e s some basicl p r o p e r t i e s o f t h e s e t of 21 1

MUJ I CA

212

i n v e r t i b l e e l e m e n t s of 29.3.

Let

THEOREM.

(a)

a Banach a l g e b r a . b e a Banach a l g e b r a . T h e n :

A

x

For e a c h

w i t h IIxIl m m

A

E

=

i n v e r t i b Z e and ( e -x)-'

X x m=O

<

e-x is

t h e eZement

1

.

U of a Z l i n v e r t i b Z e e l e m e n t s of A i s o p e n .

(b)

The s e t

(c)

The mapping

x

E

U

+

x-'

E A

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

(a) L e t

PROOF.

x

m

<

t h e series

w,

with

E A

Z

x

m

112

I1 < 1 .

m

B ~~.zmil < B IIxIlm m=O m=O

Since

I f we set

converges absolutely.

Sn

2

m=O

e

+

x

+ x

e - x nf 1

2

+

. ..

+ x

n

f o r every n then ( e - x l S n

and a f t e r l e t t i n g

n

= S ( e - xl=

-

we get t h a t (e

+

x ) z xm = m=O

m

xm ) ( e

( B

- xl = e , as w e

wanted.

m=O If

(b) z

+ t =

x(e

z i s i n v e r t i b l e t h e n it f o l l o w s from

+ x

-1

t

t l is i n v e r t i b l e f o r every

/I t 1 I < 1 / I I X - ~ I I . T h i s shows ( b )

.

E

(a)

that

such t h a t

A

But i t f o l l o w s a l s o from ( a )

that

II t II <

and t h e l a s t w r i t t e n s e r i e s c o n v e r g e s u n i f o r m l y f o r -1

whenever

0 < r < 1 / Ilx

f i n e d by

Pm(tl =

(-

ll.

S i n c e t h e mapping

Pm

x-~~)"x-' c l e a r l y belongs t o

c o n c l u d e t h a t t h e mapping

x E

U

+

x-'

E

A

: A

+

P

de-

A

P(mA;A) w e

i s h o l o m o r p h i c and

t h e proof of t h e theorem i s complete. Next w e d e f i n e t h e s p e c t r u m of a n e l e m e n t of a B a n a c h a l g e bra andestablish its basic properties. 29.4.

DEFINITION.

The s p e c t r u m of

L e t A be a Banach a l g e b r a and l e t 3: x , t o b e d e n o t e d by o i z i , is t h e s e t o f

E

A.

all

213

COMMUTATIVE BANACH ALGEBRAS

x

of

x - Xe

such t h a t

h E 6

i s n o t i n v e r t i b l e . The s p e c t r a l radius

i s t h e number

Since

- Ae =

Hence

x - Xe p i x ) 5 IIx I1

29.5.

THEOREM.

that

-

- x/Al,

i t f o l l o w s from Theorem 29.3 i s i n v e r t i b l e f o r e a c h X E S such t h a t 1x1 > 11x11.

x

Ale

x E A.

f o r each

Let

b e a Banach a l g e b r a and l e t

A

(a)

The s e t

(b)

Am E a l x m ) f o r e a c h

(c)

plxl =

x E A. Then:

ulxl is c o m p a c t and n o n e m p t y . h E o l x ) and

m E

w.

l i m I I x ~ I I ~ / ~T h. i s i s t h e s p e c t a l r a d i u s f o r m+m

mula.

PROOF.

(a) S i n c e

pfxl <

115

I1

t h e set

i s b 0 u n d e d . B ~Thee-

a(x)

r e m 29.3 t h e s e t o f n o n i n v e r t i b l e e l e m e n t s of A i s closed. x - he E A i s c o n t i n u o u s , t h e s e t S i n c e t h e mapping X E S a ( x 1 i s a l s o c l o s e d , and t h e r e f o r e compact. Suppose t h a t a i x i +

i s empty and c o n s i d e r t h e mapping

-

he)-'.

Then

- Xf(X)

= (e

- x/A)-'

=

(X

Ihl

+

e

+

and i n p a r t i c u l a r

m

f : 6

IAl

when f

+

00.

29.3. Thus

i s bounded on

i d p n t i c a l l y z e r o by L i o u v i l l e ' s Theorem 5 . 1 0 . possible since (b)

f(0l

xm

If

-

= x-I. Thus

Xme

d e f i n e d by

A

+

f E X ( S ' ; A ) by Theorem

f(A)

Furthermore, f(A)

6.

+

when

0

Hence

But t h i s i s

f

is

im-

a ( x l must be nonempty.

i s i n v e r t i b l e t h e n i f f o l l o w s from t h e

f a c t o r i z a t i o n formula

xm

m- 1

m

-

A e = (x - X e ) ( x

x - Xe

that

(c) every (29.1)

m

+ ~xm-2 +

...

+

m- 2

3:

+ Xm-'e)

i s i n v e r t i b l e too.

I t f o l l o w s from (b) t h a t

Hence p ( x l 5 lim i n f IIxm II l / m . rnjm

E W.

p(zl"

< p(xm) < IIxmII

for

MUJ I CA

214

and i s h o l o m o r p h i c f o r and i s holomorphic f o r

IX IX

< l / p ( x ) . ' Indeed, g f h l i s

defined

< I / IIx11 , by Theorem 29.3, and since

From ( 2 9 . 1 ) and ( 2 9 . 2 ) w e g e t t h e d e s i r e d c o n c l u s i o n . W e conclude t h i s s e c t i o n with t h e Gelfand-Mazur

Let

THEOREM.

29.6.

A

be a Banach a l g e b r a i n w h i c h e a c h

z e r o e l e m e n t 7:s i n u e r t i b Z e . T h e n A to

non-

is i s o m e t r i c a l Z y i s o m o r p h i c

@.

A

o f x ) i s nonempty f o r e a c h x E A by Theorem 29.5. 01x1 t h e n x - Ae and x - pe are n o t

The s e t

PROOF. If

Theorem.

and

belong t o

p

i n v e r t i b l e , hence

x

- Xe

and

- pe

x

X = LI.Thus f o r e a c h

and t h e r e f o r e

point

x

a r e both equal t o zero, E A

t h e set

a f x ) con-

x = X ( x 1 e . Then t h e mapping

X ( x ) and

i s o b v i o u s l y a n a l g e b r a i s o m o r p h i c and

f o r every

x

E

112

I1

A.

30. COMMUTATIVE B NACH ALGEBRAS T h i s s e c t i o n i s d e v o t e d t o t h e s t u d y of i d e a l s

and

homo-

morphisms of commutative Banach a l g e b r a s . 30.1. subset

DEFINITION.

I of

subspace of

# A

A A

Let

A

be a commutative Banach

i s s a i d t o be an i d e a % of A

and

xa

E I

for all

x E I

if and

algebra.

A

T i s a vector a E A . A n ideal

i s s a i d t o b e a p r o p e r i d e a l . A p r o p e r i d e a l whjch i s n o t c o n t a i n e d i n any l a r g e r p r o p e r i d e a l i s s a i d t o be a n i m i I

ma% i d e a l .

215

COMMUTATIVE BANACH ALGEBRAS

I of A I c o n t a i n s no i n v e r t i b l e e l e ments. A s t a n d a r d a p p l i c a t i o n of Zorn's lemma shows t h a t e v e r y

Let

be a commutative Banach a l g e b r a . An i d e a l

A

i s a p r o p e r i d e a l i f and o n l y i f

p r o p e r i d e a l of

A

i s c o n t a i n e d i n some maximal i d e a l o f

I i s a c l o s e d p r o p e r i d e a l of

then t h e q u o t i e n t space

A

i s a commutative Banach a l g e b r a and t h e q u o t i e n t mapping

A.If A / I A

-*

i s a c o n t i n u o u s a l g e b r a homomorphism ( w e s h a l l always req u i r e t h a t an a l g e b r a homomorphism map t h e u n i t element of the A/I

f i r s t a l g e b r a o n t o t h e u n i t element o f t h e

sscond

one).

The

p r o o f s of t h e s e a s s e r t i o n s are s t r a i g h t f o r w a r d and are l e f t t o t h e r e a d e r as exercises. 30.2.

PROPOSITION.

Let

A

b e a c o m m u t a t i v e Banach algebra. Then:

(a)

The c l o s u r e of e a c h p r o p e r i d e a l of A is a proper i d e a l .

(b)

Each m a x i m a l i d e a l o f

PROOF.

i s closed.

A

( a ) L e t I be a p r o p e r i d e a l o f

I

is easily U of a l l i n v e r t i b l e i s open and i s d i s j o i n t from I. Hence U and 7 A.

Then

s e e n t o be an i d e a l . F u r t h e r m o r e , t h e s e t

elemelits of

A

are a l s o d i s j o i n t and t h e r e f o r e (b)

A

I # A.

M be a m a x i m a l i d e a l o f

Let

A.

By ( a )

is a praper

= M.

i d e a l and t h e r e f o r e

If

-

i s a Banach a l g e b r a t h e n w e s h a l l d e n o t e by S(A) t h e

s e t of a l l c o m p l e x homomorphisms o f A , t h a t i s , t h e s e t of a l l a l g e b r a homomorphisms h : A -, iT. The s e t S ( A I i s c a l l e d t h e spectrum of

A.

multiplicat?ve

The complex homomorphisms o f

hie) = 1

convention,

f o r elrery

A

are a l s o called

R e c a l l t h a t , by a

linear functionals.

pr. vious

h E S ( A I . This i s , o f course,

j u s t a m a t t e r of convenience. Our n e x t theorem shows t h a t i n t h e case o f Banach a l g e b r a

A,

t h e r e i s a one-to-one

t h e s e t of complex homomorphisms of i d e a l s of

A

a

commutative

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

and t h e s e t of

maximal

A.

3 0 . 3 . THEOREM.

Let A be a commutative

Runach a l g e b r a .

Then:

216

MUJ ICA

(a) The kerneZ maximal idea2 of A. (b)

each c o m p l e x homomorphism

G , ~

Each maximal idea2 of

c o m p l e x homomorphism cf

A

of

A

a

is

is the kernel of a unique

A.

PROOF. (a) If h : A g is a complex homomorphism then each z E A can be uniquely decomposed in the form x = he + z , with A E 8 and z E Ker h. Indeed, X = h (x) and z = x - hlx!c. Whence the ideal K e r h has codimension one and is therefore a maximal ideal. -+

(b) If M is a maximal ideal of A then A / M is a field and is therefore isomorphic to C by the Gelfand-Mazur Theorem. 8 Let T : A + A / M be the quotient mapping, let cp : A / M be the isomorphism given by the Gelfand-Mazur Theorem and l e t h = cp n T. Then h is a complex homomorphism of A and M = Ker h. To show uniqueness let h l and h;, be two complex homomorphisms of A such that Ker h, = Ker h 2 = M . From the uniqueness of the decomposition x = Xe + z , with A E iT and z E Ker h l = Ker h, = M, we see that h , i x l = hZlzl = X for each x E A . +

30.4. COROLLARY. x E A.

(a) E

z

Let

A

b e a commutative Banach algebra and l e t

is invertible i f and on7.y

hlx) # 0

f o p each

h

StX).

each

h

E

1 5

plzi < Ilx I1 S l X l is continuous.

lhlxl

(c)

f o r each

h

E

SlXJ. In particular

PROOF. ( a ) If xz-' = e then h(x)h(x-'J = h l e ) = 1 and therefore hlxl # 0 for each h € S ( X 1 . Conversely, if hix) # 0 for each h E S(X) then 3: is not contained in any nnxi-1 ideal

and is therefore invertible. (b) If A t u ( x ) then ~ 1 ' - X P is not invertible. Tnen by (a) there exists h E S i x ) such that ],(I. - XQ) = O and therefore

COMMUTATIVE BANACH ALGEBRAS

A = h ( x ) . Conversely, since

x

-

h ( x I e E Ker h ,

not invertible by (a), and therefore (c)

217

x

- h(xIe

is

hlxl E a(x).

This follows at once from (b).

30.5. DEFINITION. Let A be a commutative Banach algebra. The formula G ( h ! = h ( x l ( h E SIA)) defines a function 2: S t X ) -+ 6 for each x E A. The function is called the Gelfand transform of x. Let A^ denote the algebra of all 2 with x E A . The spectrum S(AI of A will be always endowed with the weakest topology that makes each E A' continuous. This topology is called the GeZfand topozogy. Because of the one-to-one correspondence between complex homomorphisms of A and maximal ideals of A, the topological space SIAI is often called the maxima2 i d e a l s p a c e of A . 30.6. THEOREM.

Let

A b e a c o m m u t a t i v e Banach a l g e b r a . T h e n

(a)

S(A) i s a c o m p a c t Hausdorff s p a c e .

(b)

The GeZfand mapping

x

t i n u o u s a Z g e b r a h c v o m o r p h i s m , and

E

A

-+

3c^

E C(S(AII

is

a con-

l l ~ l l = p ( x ) for euery I

€

A.

PROOF. By Corollary 30.4(c) S(A) is a subset of the closed unit ball of the dual A' of A . Furthermore, the Gelfand topology on StAl is precisely the topology induced on S ( A ) by the weak topology a(A',A) of A'. By Alaoglu's Theorem, the Since one can closed unit ball of A ' is a(A',A)-compact. readily see that S l A ) is a(A',A)-closed in A ' , (a) follows.

5 E C(SIAI) is clearly an alThe Gelfand mapping x E A gebra homomorphism. And it follows at once fromCorollary 30.4(b) that I1 II = p { Z I 5 115 II for every z E A . This shows (b) -f

.

To conclude this section we characterize the spectra of s c m Banach algebras. 30.7. PROPOSITION. (a)

If

F

C

Let

X be a c o m p a c t Hausdorff s p a c e .

C ( X i i s a f a m i l y of f u n c t i o n s w i t h o u t eommon

218

MUJ I CA

f,, . . . , f n

zeros then there e x i s t

E

F

-al,.

and

..,pn

E C(Xl

such t h a t fZ(xlgl(xl f o r eilery

x

E

h E S(c(X)I

PROOF. ( a ) For e a c h 0

there e x i s t s a unique

h l f ) = f i a l for every

such that

#

+ f (xlg 1x1 = 1 n n

X.

For e a c h

(b)

...

+

and h e n c e

a

f # 0

E

f

f,, ...,f n

F

E

n

z

f .(xi f.ix) j=I 3 3

i s s t r i c t l y p o s i t i v e on

f o r every

E

such that

f

neighborhoods

n

x

E

f(a)

a. S i n c e X c a n we c a n f i n d

w i t h o u t common z e r o s . Hence t h e f u n c t i o n

f(x) =

foreach

f

on some n e i g h b o r h o o d o f

be c o v e r e d by f i n i t e l y many s u c h

X

E

C(X).

E

there exists

X

a

X

and

=

L: l f . ( x l \ j=I 3

2

X. If w e d e f i n e g . ( x ) f . ( s / 1 fir) 3 3 t h e n f,(x)gI(xi + . . . + fn (x)gn (z)= 1

.

j = I,. .,n

x E X.

If h E S ( C ( X ) ) t h e n Ker h i s a p r o p e r i d e a l of C ( X ) . Then by ( a ) t h e r e e x i s t s u E X s u c h t h a t f ( a ) = 0 f o r e v e r y f E K e r h . Hence f ( a ) - h ( f ) = 0 f o r e v e r y f E C I X I . Unique(b)

n e s s of a i s c l e a r s i n c e c o n t i n u o u s f u n c t i o n s separate the p i n t s

of

x.

30.8. PROPOSITION.

Let

6 be

a p p r o x i m a t i o n p r o p e r t y and l e t pact subset of (a)

such that

CL

r o m p l c x R a n a r h s p a c r with K

the

b e a p o l y n o m i a l I y ronvp3: c-om-

E.

For e a c h

h

E

S(P(K)) t h e r e e x i s t s a u n i q u e

h i f ) = f l a ) for e v ~ r y f E P:K).

a

E

K

COMMUTATIVE BANACH ALGEBRAS

z E

K.

(a) l e t

h

f o r every PROOF.

E

S(P(KIl.

lhifll

(30.1)

5 Ilfll

By C o r o l l a r y 30.4

= sup

If1

K f E P(K).

f o r every E

Ih(p)

I 2

sup jpJ f o r every

K

and by t h e Mackey-Arens Theorem t h e r e e x i s t s a

E’

a E E

In particular

hlqi = 9(al

such t h a t

f o r every

cp E E ’ .

unique it i s

Then

clear t h a t h i p ) = Plai

(30.2)

P E P ( E ) . N o w , by Theorem 2 8 . 1 e a c h

for every

f

P E PIE)

can

K by p o l y n o m i a l s b e l o n g i n g t o P IE). Then it f o l l o w s from ( 3 0 . 1 ) a n d ( 3 0 . 2 ) t h a t ( 3 0 . 2 ) is f v a l i d f o r e v e r y P E P(EI. Hence ( P l a l ( = I h ( P ) ( 5 s u p (PI f o r be u n i f o r m l y a p p r o x i m a t e d on

K

every

P

dense i n

E

P ( E i and t h e r e f o r e P ( K ) we conclude t h a t

a

= K.

E

Since

P(E)

h i f l = f l a ) f o r every

f

E

is PIK)

and ( a ) h a s been p r o v e d .

(b)

I f t h e d e s i r e d c o n c l u s i o n w e r e n o t t r u e t h e n t h e ideal

F would be a p r o p e r i d e a l . Then IlFl would be c o n t a i n e d i n t h e k e r n e l o f some h E S i P ( K l I . Then by (a) a l l t h e f u n c t i o n s f E I I F ) would v a n i s h a t a c e r t a i n p o i n t a E K, I i F i g e n e r a t e d by

a contradiction.

31. THE J O I N T SPECTRUM This s e c t i o n i s devoted t o t h e study of t h e j o i n t spectrum o f e l e m e n t s o f a commutative Banach a l g e b r a . T h e connection with

are

S e c t i o n 24 i s a p p a r e n t . The main r e s u l t s i n t h i s s e c t i o n proved with t h e a i d If

.T

spectrum

of

O k a ’ s E x t e n s i o n Theorem 2 4 . 1 1 .

i s a n e l e m e n t of a commutative Banach a l g e b r a then t h e

aix) of

x i s t h e set o f a l l

h

E

C such t h a t

L(:

- he

g e n e r a t e s a p r o p e r i d e a l . T h i s may b e g e n e r a l i z e d as f o l l o w s :

220

MUJ I CA

3 1 . 1 . DEFINITION.

Banach a l g e b r a

x2,

Let

...,in be

elements of a c o m u t a t i v e

o(xl,

A . W e s h a l l d e n o t e by

,..., A,)

..., x n l

t h e set of

,...,

6“ s u c h t h a t t h e e l e m e n t s x l - Ale xn g e n e r a t e a p r o p e r i d e a l . The s e t a ( x l , . . ,x n ) i s called

X = (Al

all

-

E

.

A e n the joint s p e e t r u m of

Let

PROPOSITION.

31.2.

A.

Banach a l g e b r a

X

If

PROOF.

E

x1>...,xn. xi,.

. .,xn

be e l e m e n t s of a c o m m u t a t i v e

Then

a(xl,.

. ..xnl

then t h e elements .r2 -Ale

,..., xn -

Ane

g e n e r a t e a p r o p e r i d e a l and a r e t h e r e f o r e contained i n t h e kernel o f some

(h(x,l

Whence i t f o l l o w s t h a t as w e wanted.

h E S(Al.

,..., h ( x n ) l ,

X # alx

Conversely, i f Y1’*

*JY,

*

E A

(AlJ

are

then there

... ,An)

=

elements

such t h a t

n

E ( x j - X.elyj = e.

j=l

3

Hence n

and t h e r e f o r e

..,An)

(Al,.

# (hix,),

S (A),

completing t h e proof.

31.3.

THEOREM.

g e n e r a t e d by

(a)

f o r every

h

E

A be a c o m m u t a t i v e Banach a Z g e b r a w i d & is

Let

n elexents

J : ~ , . .

., x n .

Then:

The mapping 9 : h E S(A)

i s a homeomorphism from

(b)

. . ., h l x n l )

A point

A

E

+

( h ( x 7 ) ,. . . , h ( x n ) )

SlAl onto Cn

E

6

n

a(xl ,..., s,?).

belongs to

0 ( x l , ...,.r n ) if and onLy if

COMMUTATIVE BANACH ALGEBRAS

.

a(xl,. .,xn2) i s a p o l y n o m i a l l y c o n v e x c o m p a c t s u b s e t

(c) of

22 1

P.

PROOF. (a) The mapping CP is continuous by the definition of the Gelfand topology. 9 maps S I A ) onto a I x I J . . . , x n ) by Propsition 31.2. To show that 9 is injective assume

S I A ) . whence it follows that

with h , h r

E

for every

P E P(Cn,J.

Since the subalgebra

is dense in A we conclude that h = h ' . Since p:SSIAl+afx,, ...,x n I is continuous and bijective, and S I A ) is compact, we conclude that q is a homeomorphism. (b) each

If

X = IhIx,,

P E PICn!

. . . ,h I x n ) )

for some

h

E

S l A ) then for

we have that

Conversely, assume If we define h : M

IP ( X ) j 5 -+

C

II p ( x l , .

. . ,xn)ll

for every P

E

P(P).

by

for every P E P ( C n ) , then h is a continuous complex homomorphism on M and hence has a unique continuous extension to all of A . Hence X = (h(x,l h ( x n ) l belongs to a ( x l ,..., X n i .

,...,

MUJ I CA

222

(c)

x

If

E

[ a ( x 7 , ...,xn)l

-

then for each P

E

PC8)

P(P) we have that

E u ( x l , . . . , x n ) by (b), and Thus convex, as asserted.

31.4.

THEOREM.

Banach a l g e b r a

Let A,

x1,...,x and L e t

n f

on a n o p e n n e i g h b o r h o o d o f ments

xn+l,..

., x u

E

A

a ( x l ,. . . ,xn) is plynomially

be e l e m e n t s

of

a

commutative

b e a f u n c t i o n w h i c h is holomorphic

~ ( x ~ , . ,2., ),.

and a f u n c t i o n

g

Then t h e r e a r e

ele-

h o l o m o r p h i c on an open

neighborhood o f t h e p o l y d i s c

such t h a t

Before proving this theorem we give t w o auxiliary lemmas. The first one reduces the proof of the theorem to the case of finitely generated algebras. are elements of a commutative Banach algebra xl, . . . , x n A , and B is any closed subalgebra of A containing x1,...,x 'n then we shall denote by u B ( x l , , . . , x n ) the joint spectrum of X 1' * - , x n relative to B . If

-

31.5. LEMMA. algebra

A,

Let and l e t

xl, ... ,xn b e e l e m e n t s of a commutatiue Banach U b e a n o p e n n e i g h b o r h o o d of ~ i z : ~ , . . . , ~ ~ ) .

223

COMMUTATIVE BANACH ALGEBRAS

Then

there

exist

u B ( x l , . . .,xnl

xn+m

J

such

A

E

A

P,

E

n such t h a t

A

-

(x

Z

j

j=I

subalgebra of

9 a(xl,..

= e.

X .e)y

3 L i

g e n e r a t e d by

A

.,x

xl,

n

) . m e n there are y l

,..., y,

be

closed

EX

Let

the

... , x n > Y I J ' " J y n *

VA

n

which i s d i s j o i n t from (b)

x

~

* j* *

If

J

Let

u

If

xn.

(x,,

. . ., x n )

..

each p o i n t

f i n d open s e t s algebras

VI,.

. . ,B k

B1,.

v3. n

(31.1)

g e n e r a t e d by

t h e n t h e r e is nothing t o prove.

C U

t h e n by a p p l y i n g t h e argument i n ( a ) t o

of t h e compact s e t

A

A

BO

( x ~ , . ,xn) @ U

IJ

of

n

b e t h e c l o s e d s u b a l g e b r a of

Bo

X 9

Then

u B A ( x l ,..., x I a n d h e n c e t h e r e i s a n open n e i g h b o r h o o d

X

that

U, w h e r e B is t h e c l o s e d subalgebra of A generated

C

(a) L e t

PROOF. E A

* . *

..., x n+m'

zl,

by

x n+l '

elements

.

(xl,. . , x n ) \ U

o

..,Vk

can

and f i n i t e l y g e n e r a t e d c l o s e d

A , each containing

of

we

BO

,..., x n )

I J ~( x 1

=

+

Bo,

j = 1,

for

sub-

such t h a t

..., k

j

and (31.2)

(XI,.

0

.. , x n ) \ u

. ..

c Vl u

u

Vk

.

BO

Let

B b e t h e c l o s e d s u b a l g e b r a of

Then

B

sions

B 3 Bi

3

c a B I x , ,... x

)

)

j

c

0

n

( X I BO

,..., x n )

j = 1 ,..., k. Then i t follows from ( 3 1 . 1 ) a n d ( 3 1 . 2 )

,.. . , x n l

uB(xI 31.6.

... U Bk.

From t h e i n c l u -

it f o l l o w s t h a t

Bo

a g I x l ,..., xnl

for

g e n e r a t e d by B I U

A

i s f i n i t e l y g e n e r a t e d and c o n t a i n s B,.

LEMMA.

g e n e r a t e d by

C

U, c o m p l e t i n g t h e p r o o f .

Let

n

that

+m

B

be a c o m m u t a t i v e Banach a l g e b r a w h i c h

elements

x,,.

. .,x n+m'

is

L e t D denote t h e p o l y d i s c

MUJ I CA

224

and Let

T

: 6n+m

-+

n 6

denote t h e projection onto the f i r s t n

.,

c o o r d i n a t e s . T h e n f o r e a c h o p e n n e i g h b o r h o o d U of ~ l x ~ , . .x n > i n en t h e r e a r e p o Z y n o m i a Z s P I , . . , P k E P(6n+m) s u c h t h a t

.

If

PROOF.

D C a-'(U)

D $ ac1(lJl I and l e t

t h e n t h e r e i s n o t h i n g t o p r o v e . Suppose

h E D \T-'(U).

by Theorem 3 1 . 3 t h e r e e x i s t s IP

0

TlA)

I

> IIPlX1

P

Then E

J . . )xn)ll =

IIP

such t h a t

0 TIfXl

we

Thus by c o m p a c t n e s s of D \ T I - I ( U ) PI, P k E P ( t n+m I such t h a t

9 ~ (I,..., x xn ) a n d

TI(?,)

P(Gn)

,..., x n+m / I I .

can

find

polynomials

...,

The d e s i r e d c o n c l u s i o n follows.

neighborhood of E

A

an

open

..x n+m

.

ag(xl,

such t h a t

a l g e b r a of A

0

Let

f

. . . ,xn)

g e n e r a t e d by

E

X(U),

where

is

(xl,. .,xn). By kmna 31.5 we can find xnil,..

PROOF OF THEOREM 3 1 . 4 .

U

U, where B i s t h e c l o s e d subxl,. . ,x n+m . L e t D d e n o t e the polyC

.

disc

@n+m

denote t h e p r o j e c t i o n onto t h e f i r s t n P I , . ,P k E P f g n i m ) c o o r d i n a t e s . By Lemma 31.6 w e c a n f i n d and l e t

TI

such t h a t L =

{c

E

-+

.

L C T-'(U),

where

D : lPi(<)l 5 I I P . ( x l ,

xn+m+j function $

Set

:

3

= P . ( x ,..., ~ x 3

0

n

...' " n i m 111

for

j =I,..

) f o r j = I ,..., k . Since n+m is h o l o m o r p h i c on a n open n e i g h b o r h o o d o f

.,kl. ?he L,

COMMUTATIVE BANACH ALGEBRAS

225

an application of Oka's Extension Theorem 24.11 shows the existence of a function g, holomorphic on an open neighborhood of the polydisc

and such that

for every a(xl

in a certain neighborhood of L. Note that C L by Theorem 31.3. Hence, by applying (31.3)

5

,..., x n+mI

with

5 = (Zl(h),

h

for every

f

..., i?n +m ( h ) ) we

S(A!.

get that

This completes the proof.

..

31.7. THEOREM. L e t x ~ , . ,x n be elements of a c o r n t a t i t r e Banach a l g e b r a A , and l e t f b e a f u n c t i o n w h i c h i s hoZomorphic on an o p e n n e i g h b o r h o o d o f a ( x l , . . .,in). T h e n t h e r e e x i s t s y E A such t h a t

PROOF. By Theorem 31.4 there are elements

X ~ + ~ , . . . , X ~E

A

and

a function g , holomorphic on an open neighborhood of the p l y disc

and such that g

If

Z ca

a

5

01

0

G I , .. . ,Z N I = f 0 G l , . . . , E n ) .

is the Taylor series

of g at the origin then

MUJ ICA

226

c1

and hence the series y

E

caxll

A . Then f o r each

h(yl =

’ converges to an

element

S ( A ) we have that

h

E

z

c,h(xll

a

a

.. . x N @l

... h l x , )

clN

completing the proof. 31.8. THEOREM. L e t A be a c o m m u t a t i v e Banach algebra and assume t h a t S l A ) is t h e u n i o n of t w o d i s j o i n t c o m p a c t s e t s H and K. Then t h e r e e x i s t s y E A s u c h t h a t G l h l = 0 f o r e v e r y h E H and y i k ) = 0 f o r e v e r y k E R .

PROOF.

Since H and K are disjoint, for each pair ( h o . k o )

E

there exists x E A such that 2 ( h o ) # i ( k o ) and hence $ ( h J # 2 ( k ) for all ( h , k l in a suitable neighborhood of Iho,ko). Since H x R is compact, it can be covered by finitely many such neighborhoods. This yields elements X I , . * . , x n of A such that for each ( h , k ) H x K there exists some j (j = I,. . , n ) . ( h ) # G . (k). Hence the sets such that H

X

K

.

3

3

and

n

are two disjoint compact subsets of C whose union is the joint spectrum a ( x l ,..., x,) of x l,..., x n . Then the function f which is equal to zero on an open neighborhood of and equalto one on an open neighborhood of i , is holomorphic on an open neighx n ) . By Theorem 3 1 . 1 there exists borhood of a ( x y E A such that

],...,

COMMUTATIVE BANACH ALGEBRAS

$(h) = f(Zl(h),

y^(k) = 1

f o r every

k

E

..., G n ( h ) )

$(h) = 0

h E S ( A ) . Hence

f o r every

227

h E

€or every

and

H

a s w e wanted.

K,

32. PROJECTIVE LIMITS OF BANACH ALGEBRAS Very o f t e n w e have t o d e a l w i t h t o p o l o g i c a l

algebras

of

f u n c t i o n s which are n o t Banach a l g e b r a s . A t y p i c a l example

in

t h i s book i s ( X ( U ) , T ~ ) ,where

U i s a n open s u b s e t of a complex Banach s p a c e . I n t h i s s e c t i o n w e i n t r o d u c e an i m p o r t a n t c l a s s of t o p o l o g i c a l a l g e b r a s whose s t u d y may be reduced t o a

large

e x t e n t t o t h e t h e o r y of Banach a l g e b r a s .

i s s a i d t o be a t o p o l o g i c a l

32.1.

DEFINITIONS. ( a ) A

if

i s a complex a l g e b r a and a t o p o l o g i c a l v e c t o r

A

algebra space

in

which m u l t i p l i c a t i o n i s j o i n t l y c o n t i n u o u s .

A

(b)

i s s a i d t o be a

locally multiplicatively

a l g e b r a , or f o r s h o r t a l o c a l l y m-convex a l g e b r a ,

if

convex

is

A

a

t o p o l o g i c a l a l g e b r a whose t o p o l o g y i s g i v e n by a family of seminorms

p

i such t h a t

and

p . f e l = 1.

(32.2) 32.2.

2

EXAMPLES.

(a)

Let

X be a t o p o l o g i c a l s p a c e . I f m u l t i -

p l i c a t i o n i s defined pointwise then (CtXl,r,l

is a commutative

l o c a l l y m-convex a l g e b r a .

(b)

Let

Then ( J C ( U ) , . r c )

U be an open s u b s e t of a complex Banach

c l o s e d s u b a l g e b r a of If

A

space.

i s a commutative l o c a l l y m-convex a l g e b r a , a s a (C(U), T ~ ) .

i s a (commutative) t o p o l o g i c a l a l g e b r a t h e n

d e f i n e t h e s p e c t r u m of a n e l e m e n t of

A , i d e a l s of

A,

we

can

and t h e

MUJ I CA

228

j o i n t spectrum of e l e m e n t s o f

A , e x a c t l y as

i n the

case

of

(commutative) Banach a l g e b r a s , f o r t h e s e a r e p u r e l y

algebraic

n o t i o n s . B u t w e have t o d i s t i n g u i s h between t h e s e t

S,(A)

a l l complex homomorphisms of of

A,

of

A , called t h e aZgebraic spectrum

S l A ) of a l l c o n t i n u o u s members of S a ( A l r

and t h e s u b s e t

c a l l e d t h e s p e c t r u m of A , o r more p r e c i s e l y ,

the

topological

s p e c t r u m of A . L e t A be a l o c a l l y m-convex a l g e b r a and l e t p t i n u o u s seminorm on A

be a

con-

s a t i s f y i n g ( 3 2 . 1 ) and ( 3 2 . 2 ) . L e t

(A,p1

l e t A p denote t h e rP : A -P A denote t h e P of A p i s a Banach q u o t i e n t mapping. Then t h e c o m p l e t i o n A

denote t h e al gebr a

A , seminormed by

P

i s a c o n t i n u o u s homomorphism from o n t o a d e n s e s u b a l g e b r a of A p . Many q u e s t i o n s a b o u t A can

a l g e b r a and t h e mapping A

p,

and l e t

normed a l g e b r a ( A , p ) / p - ’ ( O ) ,

TI

p

A

be reduced t o q u e s t i o n s a b o u t t h e Banach a l g e b r a s

. To

AP l u s t r a t e t h i s p o i n t w e prove t h e f o l l o w i n g p r o p o s i t i o n . A

I f

32.3. PROPOSITION.

il-

is a c o m m u t a t i v e l o c a l l y m-convex a l -

g e b r a t h e n e a c h c l o s e d p r o p e r i d e a l o f A i s c o n t a i n e d in k e r n e l of a c o n t i n u o u s c o m p l e x homomorphism o f A .

the

PROOF. L e t I b e a c l o s e d p r o p e r i d e a l of A . Then there exists a c o n t i n u o u s seminorm p on A s a t i s f y i n g ( 3 2 . 1 ) and (32.21, and t h e r e e x i s t s E > 0 such t h a t I i s d i s j o i n t from t h e o p e n s e t U = { x E A : p(x - e ) < E } . Hence r p ( I )i s a n i d e a l of A which

P

i s d i s p o i n t from t h e b a l l v ( U 1 .

P

Hence i t f o l l o w s t h a t t h e i s a c l o s e d p r o p e r ideal of A^

v ( 1 1 of TI ( U l i n P’ P P i s a Banach a l g e b r a , t h e i d e a l k m is contained 2 i n t h e k e r n e l of some h E S ( A 1 . Hence t h e i d e a l I i s conP P which b e l o n g s t o t a i n e d i n t h e k e r n e l of t h e mapping h O T P P’

closure Since

S(A1.

32.4. COROLLARY.

If

A

g e b r a t h e n t h e mapping

i s a c o m m u t a t i v e l o c a l l y rn-convex alh Ker h g2vc.s a one-to-one correspondence +

b e t w e e n t h e c o n t i n u o u s c o m p l e x homomorphisms of A and the cZosed muximal i d e a l s of

PROOF.

If

A.

h E S I A ) t h e n t h e proof of Theorem 3 0 . 3 shows t h a t

229

COMMUTATIVE BANACH ALGEBRAS

i s a maximal i d e a l of

Kerh

continuous. Conversely, i f

M

A , which i s c l o s e d

since

h

is

i s a c l o s e d maximal

ideal

of

A

h

such t h a t

t h e n by P r o p o s i t i o n 3 2 . 3 t h e r e e x i s t s C Ker h .

E

S(AI

i s maximal w e c o n c l u d e t h a t

Since M

t h e p r o o f o f Theorem 3 0 . 3

M

M = Kerh.

And

h i s unique, completing

shows t h a t

the proof. To s t u d y more c l o s e l y t h e c o n n e c t i o n b e t w e e n l o c a l l y m-convex a l g e b r a s

and

we

Banach a l g e b r a s

introduced t h e following

definition. 32.5.

indexed

I. Suppose t h a t f o r e a c h p a i r o f i< j t h e r e i s a c o n t i n u o u s mapping TI :X i j j

by a d i r e c t e d s e t

indices i , j +

b e a f a m i l y of t o p o l o g i c a l spaces,

DEFINITION. L e t ( X i ) i E I

with

with t h e following properties:

Xi

(a)

rii

(b)

vij o vjk =

i;

i s t h e i d e n t i t y mapping f o r e v e r y TI

i 5 j 5 k.

whenever

i k

Then t h e c o l l e c t i o n ( X i , r i j ) i s s a i d t o b e a p r o j e c t i o n s y s t e m

o f t o p o l o g i c a l s p a c e s . The s e t

x =

{(xi)

E

lTXi

:

TI

(x.) = x

i j

3

i

whenever

i 5 j},

nXi,

endowed w i t h t h e t o p o l o g y i n d u c e d by t h e p r o d u c t t h e p r o j e c t i v e Z i m i t of t h e t o p o l o g i c a l s p a c e s noted by by

projXi.

The c a n o n i c a l mapping

X

+

Xi Xi

is called and i s de-

is

denoted

ri. I f each

Xi

is a topological

l o g i c a l a l g e b r a ) and e a c h

71

i j

. ’

vector X

j

+

X

i

space ( r e s p . a topoi s l i n e a r (resp. an

a l g e b r a homomorphism) t h e n t h e c o l l e c t i o n ( X i , r i j )

is s a i d

to

be a p r o j e c t i v e system of t o p o l o g i c a l v e c t o r spaces

(resp.

projective system of topological algebras). In t h i s

case

projective l i m i t

X = proj X

i i s endowed w i t h t h e vector

s t r u c t u r e (resp. algebra s t r u c t u r e ) i n d u c e d b y t h e product

a the

space

nXi.

32.6. THEOREM. E v e r y c o m p l e t e H a u s d o r f f l o c a l l y m-convex aZgebra

230

MUJ I CA

is topologically i s o m o r p h i c t o a p r o j e c t i v e l i m i t of Banach algebras. PROOF. Let A be a complete Hausdorff locally m-convex algebra. If p I , . , p , are continuous seminorms on A satisfying (32.1) and (32.2) then the seminorm p = max { p I,...,p,] has the same properties. Thus the topology of A is given by a directed family P of seminorms satisfying (32.1) and (32.2). If for p 5 q we denote by TI : A A the canonical mapping,

..

P4

and by

i? : Aq P4

clear that

+

A^

P

4

+

P

its continuous extension,

then

it

is

TI ) is a projective system of normed algebras, P’ P 4 whereas (A^ * ) is a projective system of Banach algebras. We P’ nP4 claim that the projective limit B = p r o j A p is a dense subalgebra of the projective limit B^ = p r o j x . Indeed, let $ = P (y^ ) E 5 . For each p E P and E > 0 we first xp E A p such P that pix P - y^ P ) E, we next choose x E A such that TI (xl = P

(A

and then we define y P E = ( T I I x ) j q E P E B . Then one can P’ 4 readily prove that the net ( y P E ) converges to y^. Next consider the mapping

x

Ciearly TI is an algebra homomorphism. TI is injective since A is Hausdorff. And clearly TI is a homeomorphism between A and its image in n A We claim that - r r ( A ) = B . Indeed, given y = P ( y P ) E B we choose for each p E P an xP E A such that

.

IsP) = y

.

Then ( x p l is easily seen to be 2 Cauchy net in A , and since A is complete, the net ( x P J converges to some x E A . Then one can readily see that TI (xi = y for each p . P P Thus T I ( A ) = B and B is therefore complete. Since we already know that B is dense in 8 we conclude that niA) B = 6 and the proof of the theorem is complete. TI

P

P

32.7. PROPOSITION.

Let

A = projAi

t o p o l o g i c a l a l g e b r a s . Then an e l e m e n t in

A

be a p r o j e c t i v e l i m i t

x = (xi/

EA

of

i s invertibie

if and o n l y if xi i s i n v e r t i b l e i n A?: j‘or e v e r y

i.

231

COMMUTATIVE BANACH ALGEBRAS

The " o n l y i f " p a r t i s c l e a r . To show t h e " i f " p a r t l e t

PROOF. yi

t h e i n v e r s e of

be

in

xi

t h e uniqueness of t h e i n v e r s e t h a t

y = (yil

Hence 32.8.

belongs t o

Let

PROPOSITION.

A

A

and y

x

h

E

whenever i ' j . m i j ( y 3.I = yi i s t h e i n v e r s e o f x.

be a complete, Hausdorff, comtative

l o c a l l y m - c o n v e x a l g e b r a , and l e t (aj

i. I t f o l l o w s from

f o r each

Ai

x E A.

is i n v e r t i b l e if and o n l y i f

h(x) # 0

each

for

StX).

PROOF.

By Theorem 3 2 . 6

is a p r o j e c t i v e 1imitofcommGtative

A = proj A

Banach a l g e b r a s , s a y S(AI

A

i' Then i t i s clear t h a t

= { hz. o TI i : h Z. E S f A i l ,

i E I}.

Hence ( a ) f o l l o w s from P r o p o s i t i o n 32.7 a n d C o r o l l a r y 3 0 . 4 ( a )

.

T o show ( b ) , t h e p r o o f o f C o r o l l a r y 3 0 . 4 ( b ) a p p l i e s . Let

A = projAi

be a p r o j e c t i v e l i m i t o f c o m m u t a t i v e top-

l o g i c a l a l g e b r a s and l e t

I be a p r i n c i p a l i d e a l o f

f o l l o w s from P r o p o s i t i o n 3 2 . 7 t h a t g e n e r a t e s t h e improper i d e a l of

Ai

I = A

A . Thenit

i f a n d o n l y i f ni(Il

f o r each

i. I f i n s t e a d of

a p r i n c i p a l i d e a l we consider a f i n i t e l y generated i d e a l , then t h e p r o b l e m i s much more d i f f i c u l t , f o r i f n n tion x j y j = e h a s a s o l u t i o n Iy 2 , . . . , y n ) j=i

2

2

and the equa-

for

( x 3 , . . . ,x n ) , t h e n t h e s o l u t i o n i s n o t unique. I n t h e

a

given

case

of

f i n i t e l y g e n e r a t e d i d e a l s w e have t h e f o l l o w i n g theorem. 3 2 . 9 . THEOREM. L e t A = p r o j A i b e t h e p r o j e c t i v e l i m i t of a s e q u e n c e of c o m m u t a t i v e Ranach a l g e b r a s , a n d a s s u m e thnt the

homomorphism

ri : A

-+

Ai

has dense image f o r e v e r y

i. I f

A such that niiI) generates for e v e r y i then I = A .

i s a f i n i t e l y g e n e r a t e d i d e a l of t h e i m p r o p e r i d e a l of

A;

I

The p r o o f o f Theorem 32.9 r e s t s o n t h e f o l l o w i n g t h e o r e m .

232

MUJ I CA

X = projX

Let

32.10. THEOREM.

i b e t h e p r o j e c t i v e l i m i t of a

s e q u e n c e o f c o m p l e t e m e t r i c s p a c e s , and a s s u m e t h a t t h e mapping 'IT

i ,i+l : Xi+7 ni

piny PROOF.

: X

has d e n s e image for e v e r y

Xi

-+

W e show t h a t

b e t h e metric o f

di

?rZ3

: X3

continuous w e can f i n d

x3 E X 3

i,j-1Ixj-1 I ) 5 m

,... .

i . Since

mI2

: X2

X1

--f

m12

: X2

+

is

XI

x

j

X

E

i = 1, . . . , j -

for

E/Zj-I

...

f o r each j = 2 , 3 ,

j

I.

i s a Cauchy s e q u e n c e i n Xi f o r every

W e c l a i m t h a t ( r i j l x .) I j z i

i = l,2,3

mi

be given.

> 0

such t h a t

such t h a t 1~

E

such t h a t

x2 E X2

Proceeding i n d u c t i v e l y w e c a n f i n d

di ( m i j (xj . I ,

and

h a s d e n s e image and

X2

-+

rl E XI

f o r each

Xi

h a s d e n s e image w e c a n f i n d

Since

i.

h a s d e n s e image. T h e p r o o f t h a t

ni

h a s d e n s e image i s s i m i l a r . L e t Let

i . T h e n t h e map-

h a s d e n s e image f o r e v e r y

Xi

-+

3

k > j > i

Indeed, f o r

w e have t h a t

m

(32.3)

di('ITik(x 1, k

'IT

Ix . I ) L: d i ( r i , ? + , i j j r=j m

< -

Thus t h e s e q u e n c e for each

('IT

i = 1,2,

j > i, l e t t i n g T h i s shows t h a t (32.3) that

k

i j

(x

... . + m

L:

m

m

dIi.rrlk(xk),xl)5

=

E

.z-j+l.

c o n v e r g e s t o a n e l e m e n t yi

nij o n

we g e t t h a t

y = (yili_]

. 2-r

r= j

3

Since

E

( x ~ + ~ n)i,r ( x r ) )

jk

n i j ( y 3.I

belongs t o E.

= vik(xk) for

( xk

Letting

= yi If

X.

k

+

m

for

j

E

Xi

k -

i.

f o l l o w s from

we

get

that

COMMUTATIVE BANACH ALGEBRAS

5

dl(yl,x,l 32.11.

E

A

> 0

: A

+

and t h e proof i s complete.

Let

LEMMA.

n

let that

E,

B

233

and

A

be two c o m m u t a t i v e Banachalgebras,

B

be a homomorphism w i t h d e n s e i m a g e , and suppose

.

is g e n e r a t e d b y n e l e m e n t s al,...,an and n e Z e m e n t s y , , . . . , ~ , E B s u c h t h a t

xl, . . . , x n

we c a n f i n d (32.5)

E

A

alxI

+

...

y.Il

5

E

Then

given

such t h a t

+ anxn = e

and

PROOF.

-

II.rr(5.l 3

(32.6)

Since

3

al,...,a

j = I

for

J . . . ,n .

generate A w e can f i n d

n

s1....,snEA

such t h a t

alsl +

(32.7)

let

= t

x j

j = 1 ,..., n .

so t h a t

+ j

s .(e

n 2 aktkl

n

s a t i s f y ( 3 2 . 5 ) . On t h e o t h e r h a n d , i t fol-

lows from ( 3 2 . 4 ) t h a t n

and d e f i n e

k=l

3

n

x 2 , ..., xm

-

A

I t f o l l o w s from ( 3 2 . 7 ) t h a t

n

Hence

+ an S n = e

t l , ..., t n b e a r b i t r a r y e l e m e n t s o f

(32.8)

for

...

n

234

for

MUJ i C A

j=I,.

. .,n.

Thus w e c a n a c h i e v e ( 3 2 . 6 ) b y c h o o s i n g

yj

sufficiently close t o

f o r each

. .,n.

j = I,.

Since

.lr(tjl T

has

d e n s e image, t h i s i s a l w a y s p o s s i b l e . L e t al,...,a be a set o f g e n e r a t o r s f o r n t h e i d e a l I. It f o l l o w s f r o m t h e h y p o t h e s i s that T i ( a l ) , . . .,ni(an)

PROOF OF THEOREM 3 2 . 9 .

f o r e a c h i E W . L e t X be t h e set o f a l l (xI,...,xnl n E An such t h a t Z a x = e . L i k e w i s e , f o r e a c h i E JV l e t ;=I j j n n be t h e s e t of a l l (z,, . . ., x n l E Ai s u c h t h a t z Ti(ai)x2=e. Xi j=I generate

Ai

Then e a c h

Xi

i s n o n v o i d by h y p o t h e s i s a n d

matric s p a c e , a s a c l o s e d s u b s e t of

A:.

i s a complete

Xi

S i n c e A i s t h e projec-

t i v e l i m i t o f t h e Banach a l g e b r a s A i l t h e s e t

X i s i n a natural way t h e p r o j e c t i v e l i m i t of t h e metric s p a c e s Xi. It follows from Lemma 3 2 . 1 1 t h a t t h e n a t u r a l mapping image f o r e a c h

i

E

IIV.

E

mJ,

-f

has dense

Xi

32.10

X Xi h a s d e n s e image f o r e v e r y and i n p a r t i c u l a r X i s n o n v o i d . Hence I = A , a s as-

t h a t t h e n a t u r a l mapping

i

Xi+]

And t h e n i t f o l l o w s f r o m Theorem -f

serted. 32.12.

DEFINITION.

A c o m p l e t e m e t r i z a b l e l o c a l l y m-convex a l -

gebra is c a l l e d a Frechet aZgebra. With t h e

a i d o f Theorem 32.6 a n d 32.9 w e c a n e x t e n d P r o p

s i t i o n 31.2 t o c o m m u t a t i v e F r g c h e t a l g e b r a s . 32.13.

PROPOSITION.

t i v e Frechet a l g e b r a

PROOF.

xI, .

Let A.

By Theorem 32.6

. . , xn

b e eLemanLs o f a commuta-

Then

A

is t h e p r o j e c t i v e l i m i t of

q u e n c e o f c o m m u t a t i v e Banach a l g e b r a s , s a y E

a ( x l J . . .,x ) t h e n t h e i d e a l n

A = projAi.

I g e n e r a t e d by

xl-Ale

a

seIf

,..., x n -An"

X

235

COMMUTATIVE BANACH ALGEBRAS

i E W

i s a p r o p e r i d e a l . By Theorem 32.9 t h e r e e x i s t s t h a t the set

ni(Il

there exists

hi E S l A i )

h = hi o

IT

then

i

..., A x )

that (Al,

i s c o n t a i n e d i n a p r o p e r i d e a l of

h

E

IT~(C I ) Ker hi

such t h a t

S ( A l and

I

C

Ker h .

= ( h ( x c Z )..., , h l x n l ) , as a s s e r t e d .

t h e converse it s u f f i c e s

t o r e p e a t t h e proof

A . . Hence

.

I f w e set

it

Whence

follows

To

of

such

prove

Proposition

31.2. A s an a p p l i c a t i o n w e can p r o v e t h e f o l l o w i n g r e s u l t .

32.14. PROPOSITION. L e t U b e a poZynomialZy c o n v e x o p e n s e t gn. L e t f,, ...,f E X l U l be rn f u n c t i o n s w i t h o u t common

in

m

.

gi, . .,gm E X i V ) is i d e n t i c a Z Z y one o n U.

z e r o s . Then t h e r e a r e f u n c t i o n s

fp, +

+ fmgm

. * -

such

PROOF. By E x e r c i s e 28.H, f o r each c o n t i n u o u s complex phism

h

lJC(UI,-rcl

of

t h e r e e x i s t s a unique p o i n t

h l f ) = f l a ) f o r every

that

f

a

b e l o n g to

E

U

h E S l X i u l , T ~ ) . Then

such

it

. .,0 ) does not

f o l l o w s from P r o p o s i t i o n 32.13 t h a t t h e p o i n t (0,.

. ., f n i .

homomor-

Hence t h e set { f 2 , . . . , f m }

E X(UI.

i s n o t c o n t a i n e d i n t h e k e r n e l of any

a(fl,.

that

Hence t h e f u n c t i o n s f,,...,f

generate

n

t h e improper i d e a l .

EXERC IS E S

32.A.

Let

A

by a f a m i l y where e a c h p"

be a t o p o l o g i c a l a l g e b r a whose t o p o l o g y i s g i v e n

P of seminorms s a t i s f y i n g (32.1). L e t P " = { p : p E P } , i s d e f i n e d by

p(x)/p(e)< @(xl 5 p f x ) for

(a)

Show t h a t

(b)

Show t h a t e a c h

p"

E

P"

x

E

A.

s a t i s f i e s ( 3 2 . 1 ) and (32.2).

Conclude t h a t a t o p o l o g i c a l a l g e b r a i s l o c a l l y m-convex i f and o n l y i f i t s t o p o l o g y i s g i v e n

by

a

family

of

seminorms

MUJ I CA

236 s a t i s f y i n g (32.1)

.

A s u b s e t U o f a n a l g e b r a i s s a i d t o be i d e m p o t e n t if U 2 C U. Show t h a t a t o p o l o g i c a l a l g e b r a i s l o c a l l y m-convex i f and only i f it has a b a s e of convex, idempotent neighborhoods 32.B.

of zero. W

Let

32.C.

be a n i n c r e a s i n g s e q u e n c e o f s u b a l g e b r a s o f m

and a l g e b r a A

satisfying

A

=

U

A

n=l

n

.

Suppose t h a t e a c h A n i s

a Banach a l g e b r a a n d t h a t e a c h i n c l u s i o n mapping

A n C-t An+l

c o n t i n u o u s , w i t h norm n o t g r e a t e r t h a n o n e . Endow A w i t h

is the

f i n e s t l o c a l l y convex t o p o l o g y s u c h t h a t e a c h i n c l u s i o n mapping

i s c o n t i n u o u s . Show t h a t A

A n k A

i s a l o c a l l y m-convex a l -

gebra. 32.D.

L e t A be a c o m m u t a t i v e , complete, Hausdorff locally mcon-

v e x a l g e b r a . Show t h a t g i v e n

ti

E SIA)

32.E.

Let

such t h a t A

h

E

S,(A)

and

x

E A

one can find

&(xi = h l x ) .

b e a commutative F r g c h e t a l g e b r a . Show t h a t given E S ( A ) such t h a t

h E S , ( A ) and xIJ...,xn E A o n e c a n f i n d h"(z.1 = h(x.) f o r j = 1 ,..., n .

<

3

3

32.F.

Show t h a t i f

spectrum

S(AI

of

i s a commutative F r g c h e t a l g e b r a then the A i s a hemicompact space f o r t h e G e l f a n d A

topology.

33. THE M I C H m L PROBLEM T h i s s e c t i o n i s d e v o t e d t o t h e s t u d y o f an u n s o l v e d problem

of E. M i c h a e l c o n v e r n i n g t h e c o n t i n u i t y of t h e complex homoto morphisms o f a c o m m t a t i v e F r g c h e t a l g e b r a . W e show t h a t solve t h e M i c h a e l p r o b l e m f o r a l l commutative F r g c h e t algebras, i t i s s u f f i c i e n t t o s o l v e t h e p r o b l e m for a s u i t a b l e F r g c h e t a l g e b r a o f h o l o m o r p h i c f u n c t i o n s on a Banach s p a c e . A l l Banach s p a c e s i n t h i s s e c t i o n a r e assumed t o b e complex.

237

COMMUTATIVE BANACH ALGEBRAS

linear mapping T : X Y between two topological vector spaces issaid to be boundedifit maps bounded sets into bounded sets. Clearly each continuous linear mapping T : X Y is bounded. If X is metrizable then one can readily prove that each bounded linear mapping T : X Y is continuous. A

+

+

+

The following two problems, posed by 1952, remain open.

E. Michael

in

[l]

33.1. PROBLEM. Is every complex homomorphism of a commutative Frschet algebra continuous ? 33.2. PROBLEM. Is every complex homomorphism of a commutative, complete, Hausdorff locally m-convex algebra bounded ?

Clearly a positive solution to Problem 33.2 implies a positive solution to Problem 3 3 . 1 . We shall soon prove that the reverse implication is also true. m

To begin with we introduce some notation. Set where

for every

n

E

W.

Let Let

z

E

m

Am

E

and

ct

= ( a3 ej)= 1

E

W

n=l In,

I we define

be a Banach space with a Schauder basis

= (zn)n=I

U

If A is any commutative ring then for each

m

a = (aj)j,l

I=

E JC(E)nv

functionals, let E n and let Tn : E + E n

denote the sequence

W

of coordinate

denote the subspace generated by e l , . .. , e n’ denote the canonical projection. Let f

Since the sequence (T,) converges to the identity Uniformly on compact sets, it follows that the sequence f f o Tn) converges to f uniformly on compact sets. If for each ct E In we set E

JCIE).

238

MUJ I CA

where

. . . ,Rn

> 0,

Rl

t h a t each

>

it f o l l o w s from C o r o l l a r y 7 . 8

then

0,

i s i n d e p e n d e n t from t h e c h o i c e o f

caf

and t h e m u l t i p l e series

Z:

eaf2

c1

Rl,...,RnJ

converges a b s o l u t e l y

and

In

c1E

u n i f o r m l y on compact s u b s e t s o f E t o t h e f u n c t i o n f o T n .

Thus

w i t h u n i f o r m c o n v e r g e n c e o n compact sets. A f t e r t h e s e p r e l i m i n a r i e s w e can prove t h e following theorem. 33.3.

THEOREM. L e t

E

b e a Bunach space w i t h a normalized

and l e t ( z n l ; = ,

b a s i s (en):=],

be t h e sequence o f

Sehauder

coordinate

f u n c t i o n a l s . L e t A b e a c o m m u t a t i v e , c o m p l e t e , Hausdorff ZoealZy m

m-eonvex a l g e b r a . L e t (an):=, <

m

be a sequence i n A such t h a t

a c o n t i n u o u s a l g e b r a homomorphism T(1)

= e

Tlz,)

and

1

a

n

The t o p o l o g y of

PROOF.

T : /JCIE),T,I

for e v e r y

n

E

Jpia,)

n=l Then t h e r e e x i s t s

on A .

p

for e v e r y c o n t i n u o u s s e m i n o r m

+

A

such t h a t

W.

i s g i v e n by a f a m i l y P of semiriorms

A

s a t i s f y i n g ( 3 2 . 1 ) and ( 3 2 . 2 ) . Given

p

E

P

choose

E

> 0

such

m

that

E

< 1.

2:

Set

rn =

E

R n = E-'

Jp(an),

n=l m

r = ( r l n n=1'

m

m

R = (Rnlnz1

and p ( a ) = ( p ( a n ) ) n = l

.

W a T ,

Observe t h a t

07

rnRn = p ( a ) f o r every

n

Since

?a.

Rn <

2:

n=l

00

t h e set

m

L =

Z c n e n : 5, n=l

i s a compact s u b s e t o f t h e compact s e t

E

c, 15,l

5 Rn

for every

n)

E , s i n c e i t i s t h e c o n t i n u o u s image of

COMMUTATIVE BANACH ALGEBRAS

239

f E J C I E I t h e n i t f o l l o w s from ( 3 3 . 1 ) t h a t

If

f o r every

CI

n E W w e have t h a t

E I. Then f o r e a c h

03

Since n=l

e

rn =

w e see t h a t

< 1

r

03

z n=l

1

n

- rn

r

03

< -

e

z - = nn=I 1 - e

1-8

and h e n c e t h e i n f i n i t e p r o d u c t m

03

n

c o n v e r g e s . Hence t h e r e e x i s t s a c o n s t a n t

c > 0

such t h a t

(33.2)

f o r every

f E JC(E)

and

lows t h a t t h e m u l t i p l e s e r i e s

I: c a f a, CIE

in A X

f o r each

eafa

a

f

E

f o r each

JC(E).

Since A

n E JV.

converges a b s o l u t e l y

I

Define

f E JC(E).

i s c o m p l e t e , it fol-

T

Clearly

: (JC(E),T~)

T

--f

A

by

i s l i n e a r . And

Tf = T is

CX€I

a c t u a l l y a n a l g e b r a homomorphism, s i n c e f o r a l l w e have t h a t

f, g

E

JC(E)

MUJ I CA

240 T(fg) =

c (fg)aY =

Z

y€I

Z

(

L

in E

B

a+B=y

yE I

P t h e e x i s t a comc > 0 suchthat p(Tf) (csuplfl

I f follows from ( 3 3 . 2 ) t h a t f o r e a c h p a c t set

c g ) aY

C a f

and a c o n s t a n t

p

E

L

for

every

f

J C ( E ) . T h i s shows t h a t

E

it i s clear t h a t

T(1)

= e

and

T(zn)

T

is continuous. Since

= an

n

for every

E

m,

t h e proof of t h e t h e o r e m i s c o m p l e t e . 3 3 . 4 . DEFINITION. E

Let

JCb(E)

be t h e FrGchet a l g e b r a of a l l

X ( E ) which are bounded on bounded

sets, w i t h

the

f

topology

g i v e n by t h e seminorms

33.5.

THEOREM.

and l e t

A

Let

E

b e a Banach s p a c e w i t h a S c h a u d e r basis,

b e a c o m m u t a t i v e , c o m p l e t e , H a u s d o r f f l o c a l l y m-convex

algebra. I f e v e r y c o m p l e x homomorphism o f ( 3 C ( E ) , . r e ) i s bounded, (a) t h e n e v e r y compZex homomorphism of A i s bounded a s w e l l .

If e v e r y compZex homomorphism of X b ( E ) is continuous, A i s bounded.

(b)

t h e n e v e r y c o m p l e x homomorphism o f

W i t h o u t loss of g e n e r a l i t y w e may assume t h a t

PROOF.

m

normalized Schauder b a s i s 00

(zn)n=l

h

, i n which case t h e

of c o o r d i n a t e f u n c t i o n a l s i s e q u i c o n t i n u o u s .

a bounded s e q u e n c e (b,)

a n = 4-nbn on A

in A

such t h a t

f o r every

there exists

Suppose

A . Then there exists

i s a n unbounded complex homomorphism of

n. Set norm p

E has a

sequence

( h ( b n )I > 8" f o r e v e r y

n. Then f o r e a c h continuous semi-

c > 0

such t h a t

p ( a n ) (4-'c

for

m

every

n. Thus

on

Then by Theorem 3 3 . 3

A.

C /plan) < n=l

m

for e a c h c o n t i n u o u s seminorm there

is a

continuous

p

algebra

24 1

COMMUTATIVE BANACH ALGEBRAS

homomorphism T - a,, for every

A such that T ( 1 ) = e and T ( z n ) h O P is a complex homomorphism of

: IJC(EI, ‘Tc)

n. Then

+

I J C ( E I , T ~ which ) is unbounded, since Ih o T ( z n ) I = Ihla,) I > 2 n for every n. For the same reason, the restriction of h o T to JCb(BI is an unbounded complex homomorphism of J C b ( B l .

Theorem 33.5 are equivalent. all commutative problem for the with a Schauder

shows in particular that Problem 33.1 and 33.2 It also shows that to solve Problem 33.1 for Frechet algebras, it is sufficient to solve the algebra J C b ( E I , where E is some Banach space basis.

To conclude this section we shall prove that every complex homomorphism of the Frgchet algebra ( J C ( C n ) , T c ) is continuous. This will be shown with the aid of the following proposition. 33.6.

Let

PROPOSITION.

b e a n a l g e b r a of c o n t i n u o u s complex-

A

v a l u e d f u n c t i o n s on a t o p o Z o g i c a 1 s p a c e

X

with the

folZowing

properties:

(a)

If

f,,.. . , f m

then there are functions

E A

gl,

a r e f u n c t i o n s w i t h o u t common zeros

. . ., g m

E A

such t h a t f , g ,

+

.. . + fmgm

= 1.

(b)

{x

x

E

: q,;(x)

.. , A n )

(Al,.

There a r e f u n c t i o n s

= A

j

for

.

PI,. . , q n E A s u c h t h a t the s e t j = 1 , . . ,n 1 is c o m p a c t f o r e a c h

.

E lP.

T h e n f o r e a c h compZex homomorphism x

E

X

such t h a t

h ( f ) = f(xl for every

h of f

A

t h e r c i s a point

E A.

Let h be a complex homomorphism of A and set Z ( f ) = {x E X : f f x ) = 0 3 for every f E A . Then it follows fromconn dition (b) that the set K = n Z ( q j - h ( q . ) i is compact. On PROOF.

j=l

3

the other hand it follows from condition (a) that

m 0

,{ =1

Z ( f .) is 3

nonvoid for all f Z , . . . , f p E Ker h . Whence it follows that the is a collection of closed family F = {K n Z ( f ) : f E Ker h subsets of K with the finite intersection property. Hence n F

MUJ I CA

242

is nonvoid, and this implies that n { Z i f ) : f E Ker h 1 is nonvoid. If x belongs to this intersection then f ( x ) = h l f ) for every f E A. 33.7. PROPOSITION. L e t U be a p o l y n o m i a Z Z y c o n v e x o p e n s e t i n C n . T h e n f o r e a c h compZex homomorphism h o f K ( U ) t h e r e is a point

a

U

E

such t h a t

h ( f l = f f a ) f o r every

p a r t i c u l a r e a c h e o m p l e x homomorphism o f

f

E

In

JC(U).

K ( U / is c o n t i n u o u s .

It suffices to apply Proposition 33.6, for condition (a) is true by Proposition 3 2 . 1 4 , whereas condition ( b ) is clearly satisfied by the coordinate functionals z I , ... , z n .

PROOF.

EXERCISES

33.A. Let U be an open subset of E . Show that a complex ho) bounded if and only if for each momorphism h of l J c ( U ) , T ~ is co there increasing sequence (A,) of open sets with U = u A n exists

n

E

such that

Ih(f)I

n=l

5 suplfl

for every

fE

K(u).

An 33.B. Let U be an open subset of a separable Banach space E , and let h be a bounded complex homomorphism of ( K ( U ) , T ~ ) . (a) Using Exercise 3 3 . A show that the restriction of h to each equicontinuous subset of J c ( U ) is continuous for the topology of pointwise convergence. (b) Using the Banach-Dieudonnd Theorem conclude that the restriction of h to EL = f E ' , T c ) is continuous. (c)

a point

Using the Mackey-Arens Theorem show the existence of a E E such that h ( P I = P l a ) for every P E P (E).

f

33.C. By adapting the proof of the Banach-Dieudonn6 Theorem show that the topology of compact convergence is the finest t.opology on p f m E ) which coincides with the topology of pointwise convergence on each equicontinuous subset of P(mEl.

243

COMMUTATIVE BANACH ALGEBRAS

33.D.

E b e a s e p a r a b l e Banach space w i t h t h e

Let

t i o n p r o p e r t y , and l e t (JC(E), -re).

a point

approxima-

h b e a bounded complex homomorphism of

U s i n g E x e r c i s e s 33.B a n d 3 3 . C show t h e e x i s t e n c e o f

a E E

h ( f ) = f f a ) f o r every

such t h a t

f E JC(El. I n is

(X(E),'rcl

p a r t i c u l a r e a c h bounded complex homomorphism o f continuous. 33.E.

Let

A

b e a commutative F r g c h e t a l g e b r a ,

A'

endowed w i t h t h e G e l f a n d t o p o l o g y , a n d l e t

SIA)

be

C c ( S ( A ) ) be t h e

a l g e b r a of Gelfand transforms of elements of

(a)

let

A.

U s i n g P r o p o s i t i o n 3 2 . 1 3 show t h a t t h e a l g e b r a

A^

al-

ways s a t i s f i e s c o n d i t i o n ( a ) i n P r o p o s i t i o n 3 3 . 6 . (b) t h e set

.

xl,.. ,xn E A

Suppose t h e r e are e l e m e n t s { h E SIA) :

f o r each ( A l , .

. . ,A n )

h(xjl = X E C

n

.

j = 1,

for

..., nl

such t h a t

i s compact

j Using P r o p o s i t i o n 33.6 and E x e r c i s e

32.D show t h a t e a c h complex homomorphism o f

A

i s continuous.

NOTES AND COMMENTS S e c t i o n 29, 30 a n d 3 1 c o n s t i t u t e a b r i e f i n t r o d u c t i o n t o t h e t h e o r y o f c o m m u t a t i v e Banach algebras.

The i n t e r e s t e d reader is

r e f e r r e d t o t h e book o f I . G e l f a n d , D.

Raikov

[ 11 o r t o t h e book o f W.

Zelazko

Theorem 3 1 . 7 w a s p r o v e d b y G.

11

[

Shilov

n i t e l y g e n e r a t e d a l g e b r a s , a n d b y R.

and

G.

for further

Shilov

reading.

i n t h e case of

[ 1]

Arens

and

fi-

Calder6n

A.

1 1 i n t h e g e n e r a l case, u s i n g a n i n t e g r a l f o r m u l a o f A. Weil [ 1 1 . The i d e a t o u s e O k a ' s E x t e n s i o n Theorem 2 4 . 1 1 i s d u e t o L . W a e l b r o e c k [ 1 ] . Theorem 31.8 i s d u e t o G . S h i l o v 11 . [

L o c a l l y m-convex a l g e b r a s were e x t e n s i v e l y s t u d i e d M i c h a e l [ 1 1, who p r o v e d C o r o l l a r y 3 2 . 4 , Theorem 32.6 s i t i o n s 32.7,

3 2 . 8 and 33.6. P r o p o s i t i o n 32.3 h a s

from t h e book of A.

Guichardet

[

Bourbaki

[ 1]

Theorem. I . C r a w

been

1 1 . Theorems 32.9 a n d

and P r o p o s i t i o n 32.13 are due t o R. Arens

[

1

by

E.

and P r o p taken 32.10,

1 , though

N.

r e f e r s t o Theorem 32.10 as t h e M i t t a g - L e f f l e r [

1 1 ,

D.

Clayton

[ 11

a n d M.

Schottenloher

244

MUJ I CA

6 ] have shown that to solve the Michael problem forarbitrary commutative Frschet algebras it is sufficient to solve the problem for certain algebras of holomorphic functions.By adaptting their ideas we have obtained Theorems 3 3 . 3 and 3 3 . 5 . We point out that Theorems 3 3 . 3 and 3 3 . 5 are still true if E is any infinite dimensional Banach space. Indeed, R. Ovsepian and A . Pelczynski [ l I have shown that if E is any infinite dimensional Banach space then there are a sequence ( e n ) in E and a sequence f z n ) in E' such that z m ( e n l = 6 m n , IlemII = l and lizmII 5 2 0 for all m and n in B , and the proofs of Theorems 33.3 and 3 3 . 5 work equally well in this case. Finally we mention that the equivalence of Problems 33.1 and 3 3 . 2 was established first by P. Dixon and D. Fremlin [ l ] [

.

CHAPTER V I I I

PLURISUBHARMONIC FUNCTIONS

3 4 . PLURISUBHARMONIC FUNCTIONS I n t h i s s e c t i o n we introduce plurisubharmonic f u n c t i o n s i n Banach s p a c e s a n d e s t a b l i s h t h e i r b a s i c p r o p e r t i e s . T h r o u g h o u t t h i s c h a p t e r t h e l e t t e r s E a n d F w i l l r e p r e s e n t c o m p l e x Banach spaces. W e recall t h a t i f f :

X

--t

Tx E .r! : f i x )

function

f :

i f t h e set

X i s a topological space t h e n a f u n c t i o n

is s a i d t o b e upper semicontinuous i f

[-m,m/

i s open i n

c)

X

{x

--t

E

X

(-

i s said t o be

m,m]

:

X f o r each

f ( x ) > c}

c E X?.

lower

i s open i n

X

set

the

Likewise, a

semicontinuous

f o r each

c

E

B.

I n t h e n e x t p r o p o s i t i o n w e g i v e some p r o p e r t i e s o f u p p e r s e m i c o n t i n u o u s f u n c t i o n s . The p r o o f s a r e s t r a i g h t f o r w a r d

and

are

l e f t t o t h e r e a d e r as e x e r c i s e s .

34.1.

PROPOSITION.

(a)

Let

X

be a t o p o l o g i c a l s p a c e .

The i n f i m u m of a f a m i l y o f u p p e r s e m i c o n t i n u o u s func-

t i o n s on X

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

If f i s an u p p e r s e m i c o n t i n u o u s f u n c t i o n on X t h e n K of X t h e r e e x i s t s a E K s u c h t h a t f ( a ) f o r e v e r y x E K. The s e t {z E K : f ( x ) = f l u ) } i s

(b)

f o r e a c h compact s u b s e t fix)

5

compact. (c)

and

g :I

Let +

I be a n o p e n i n t e r v a Z i n

[--,-I

[-m,m).

If

f : X

-+

I

a r e u p p e r s e m i c o n t i n u o u s , and g is increasing,

then

g o f

34.2.

PROPOSITION. I f X i s a m a t r l - space then ccch upper semicontinuous

i s upper semicontinuous.

246

MUJ I CA

function

f

X +.

:

is t h e p o i n t u i s e l i m i t o f a d e c r e a s i n g

[--,OD)

sequence o f continuous f u n c t i o n s If

PROOF.

n

each

-

f

* X - + R .

*

f 1 - n n Then c o n s i d e r

t h e n it s u f f i c e s t o t a k e

OJ

IN. Thus w e may assume f - m. g = @-f : X + l O , m ] and d e f i n e

E

function

g

(34.1)

f o r each n

E

n

f o r each -:m

fn

n

x

iiV and

and

n

E

< g l s ) . Since

X.

x

E

I t i s clear t h a t gnfx)(gn+I(xl'g(x) Since

X.

and

i7?

x

x E X.

X.

E

f f

-

we see t h a t

m

We claim t h a t x E X

Indeed, l e t

g

and

(2)

n

>

gnlx)

i s lower s e m i c o n t i n u o u s t h e r e e x i s t s

g

glz) > c

such t h a t

E

and

W

E

[ g f z ) + ndfx,z)l

x

zE

E W

€or every

f o r every

= inf

(2)

n

for the

0

0

<

c

6

>

0

z E B ( x ; 6 ) . Then it f o l l o w s from

f o r every

(34.1) t h a t

n

f o r every

gnlx)

that

as a s s e r t e d . But i t a l s o f o l l o w s from (34.2)

E W ,

2 c

for a l l sufficiently large

n , and t h i s shows

converges pointwise t o

t h a t t h e sequence fg,)

g.

Next

we

X

we

claim t h a t (34.3)

Ignlx)

for a l l

n

x, y

and

JV

E

-

gnly)l E

X.

5 ndlx,yl

Indeed,

f o r every

2

E

have t h a t

g fx) < g(z) + nd(x,z) n Whence

g

r o l e s of f i n e d by

n x

lx)

5

gnly) + ndlx,y), y ,

and

fn =

-

--*

[-m,m)

c o n t i n u o u s and

+ ndfy,z)l

+ ndlx,yl.

and a f t e r i n t e r c h a n g i n g

the

( 3 4 . 3 ) f o l l o w s . Then t h e s e q u e n c e (f,) de-

l o g gn

We recall that i f f : U

5 [glz)

,

has t h e required p r o p e r t i e s .

U

i s an open set i n

i s s a i d t o be subharmonic i f

a function i s upper s e m i -

6 then f

PLURISUBHARMONIC FUNCTIONS

247

-

for each a E U and r > 0 such that A ( a ; r l erally, we have the following definition.

C

U. More

gen-

3 4 . 3 . DEFINITION.

Let U be an open subset of E . A function f : U [-w,w) is said to be p l u r i s u b h a r m o n i c if f is upper semicontinuous and -+

Observe for each a E U and b E E such that a + Eb C U. that f is Bore1 measurable and, by Proposition 3 4 . 1 , f is bounded above on each disc a -i Ab C U . Thus the integral is well defined, though it can be

We

- w .

shall denote by PA ( U ) the set of all plurisubharmonic functions on U. We shall denote by PAC(U) the subset of a l l functions f .- U 227 which are plurisubharmonic and continuous. -+

3 4 . 4 . EXAMPLE.

Let U be an open subset of E , and let X ( U ) . Then R e f , I m f and belong to P b C l U j .

f

E

If1

PROOF.

If

a

+ ab

C

U

then

by the Cauchy Integral Formula 7.1. Everything this. 34.5.

PROPOSITION.

Let

(a) I f f , g E PA(U) t h e n af + Bg E Pb(U).

U be a n o p e n s u b s e t of and

a, B

follows

from

E.

are nonnegative

constants

248

MUJ I CA

sup fi 2

i s u p p e r s e m i c o n k i n u o u s on

U, t h e n

sup fi

E Ph(U1.

ie I

EI

Let

(c)

f

: U +.

be u p p e r s e m i c o n t i n u o u s . T h e n

[-m,m)

f

i s p Z u r i s u b h a r m o n i c if and o n l y i f t h e r e s t r i c t i o n of f t o U i s p l u r i s u b h a r m o n i c for e a c h f i n i t e d i m e n s i o n a l s u b s p a c e M of E . n M

If f is t h e p o i n t w i s e l i m i t of a d e c r e a s i n g sequence

(dl

(fn) C P h t U ) t h e

f E Ph1UI.

PROOF. Assertions (a), (b) and (c) are clear. T o show (d) we first observe that f is upper semicontinuous by Proposition 34.1. If a + a b C U then

is bounded above on a + a b , an for every n . Since f l plication of the Monotone Convergence Theorem shows that 2.rr

ap-

27

f l u + eieb)de = l i m

fn(a + e

i0

bld8,

n + m Jo

and the desired conclusion follows from (34.4). 34.6. COROLLARY.

Let

I1 f II

U be a n o p e n s u b s e t of

belongs t o

JC(E;F/.

Then

PROOF.

By the Hahn-Banach Theorem

E,

and l e t

f e

PhclU).

x E U. Since $ o f E X ( U ) for each $ E F ' the conclusion follows from Example 34.4 and Proposition 34.5.

f o r each

34.7. PROPOSITION. 2et f E ?b(lI).

(a)

Let

U be a connected open s e t i n

If t h e r e e x i s t s

a

E

U

such t h a t

flxl

5

E,

f(a)

and

for

PLURISUBHARMONIC FUNCTIONS

x

every

E

PROOF.

(a) Since

i s closed i n W e claim t h a t

find

y

U then

C

A.

f

=

set

A = U

it s u f f i c e s t o p r o v e t h a t r > 0 such t h a t B f b ; r ) C U.

and l e t

E A

B(b;r)

such t h a t

> 0

E

U.

Otherwise w e c a n f i n d

x

Bfb;r)

E

f i s upper s e m i c o n t i n u o u s ws can B ( x ; E ) C B ( a ; r ) and f l y ) ffa) for

f f x ) < f f a ) . Since

such t h a t every

E

i s upper s e m i c o n t i n u o u s , t h e nonvoid

f

b

x

on a n o n v o i d o p e n s u b s e t of

m

U. To show t h a t

i s open. L e t

A

-

f

(b) I f on U.

-

f f x l = f f a ) f o r every

then

U

249

E

B f x c ; ~ ) .T h i s c l e a r l y i m p l i e s t h a t

a contradiction.

(b)

Let

A

be t h e s e t of a l l

a

E

such t h a t

U

f

- m

a . Then A i s open and nonvoid. To comp l e t e t h e proof w e show t h a t A i s c l o s e d i n U. L e t f a n ) b e a sequence i n A which c o n v e r g e s t o a p o i n t a E U. F i r s t choose r > 0 such t h a t B ( a ; r ) C U, n e x t choose n E lib' such t h a t an

on a neighborhood of

E

B ( a ; r ) , and t h e n choose

and

= -

f

Indeed, i f

x

E

E

> 0

such t h a t

B r a n ; € ) . W e claim t h a t

on

m

€ ! ( a ; € ) then

x

+ ( a n - a)

Bfa,;E)

f

-

C

on

E B(an;E)

Bfa;r) Bfa;E).

and

it

follows t h a t

f(x) 5

1

iZnf[x + eiB

fan - alldB =

-

m.

JO

Thus

a

E

A

and t h e proof i s complete.

P r o p o s i t i o n 3 4 . 7 ( a ) i s o f t e n c a l l e d t h e m a x i m principle for plurisubharmonic f u n c t i o n s . 34.8.

THEOREM.

Let

U be an open subset of E, and l e t f : U

-f

[-a,,)

MUJ I CA

250

be a n u p p e r s e m i c o n t i n u o u s f u n c t i o n . T h e n t h e f o l l o w i n g c o n d i t i o n s are equiualent: For e a c h

(a)

and

a

E

U

a

E

U, b

b

E

such t h a t

E

a + hb

U

C

we h a v e t h a t

For e a c h

(b) such t h a t

0 < r < 6,

a + rab

C

6 > 0

and

E

E

U

there

r

exists

and

(c) F o r e a c h P E P(6) and e a c h a E U and b E E s u c h t h a t a + a b C U t h e i n e q u a Z i t y f ( a + Ab) - R e P ( X ) for 1x1 = 1 i m p Z i e s t h e same i n e q u a l i t y f o r I h l 5 1 .

PROOF.

The implication (a) * (b) is obvious.

(b) * (c): Let P E P(6) and let a E U and b E that a + a b C U. Set u(X) = f ( a + A b ) - Re P ( X ) for Suppose that u(X1 5 0 €or every X E a A but v ( X ) > some X E A. Then M = s _u_ p v > 0 and the set A = { X E

a

E

such

A( < 0

1.

for

u(X) = M )

*i

ii

is a nonvoid compact subset of A , by Proposition 34.1. Thus d(A,aA) = 6 0 and we can find Xo E A such that d(Ao,aA) = 6 . By (b) we can find P such that 0 < r < 6 , a+Aob + r a b C U and 2n (34.5)

f f a + Xob) 5

By our choice of

Xo

1

2n

f ( a + Aob + r e

= Jo 1

bide.

it is clear that 2n

2n

(34.6)

i 0

v(Xo + r e

i 0

)dB <

I 2n

I,

From (34.5) and (34.6) it follows that 2Tr

f

M d 0 = M.

PLURISUBHARMONIC FUNCTIONS

a contradiction, since

Xo

E

A.

-

( c ) * ( a ) : L e t a E U and b E E such t h a t u + Ab C u. L e t 9 be a c o n t i n u o u s , r e a l - v a l u e d f u n c t i o n on [0,27~] s u c h ~ ( 0 )= ~ ( 2 7 ~and )

that

W e s h a l l prove t h a t

271

f(a)

(34.8)

1

1

2 2.rr

Ip(eide.

10

Let

be g i v e n . By t h e S t o n e - W e i e r s t r a s s

E > 0

f i n d a t r i g o n o m e t r i c polynomial

5 cpple) +

cple) 5 $ f e ) Pfh)

= eo

2

f

n

L: e k X

for

E

k

.

Since

0

=

Z e k ei k O k=-n 0 5 2 ~ D . efine P

$(€I)

5

ek -

/

I

k=1

and s i n c e

E

> 0

was a r b i t r a r y ,

pOn : [ 0, 2711

-+

23

such E

P(6)

27T ik0 $(0)e d0,

can that by

w e see

0

( 3 4 . 8 ) f o l l o w s . N o w , by P r o p

s i t i o n 34.2 w e can f i n d a decreasing functions

Theorem w e

n

sequence

such t h a t

of

continuous

p n f 0 ) = ~ ~ ( 2 7 ~ )and

Zim r p n ( 9 ) = f l a i eieb) f o r e v e r y 0 E [ 0,2a 1 . Thus e a c h 9, n-rm s a t i s f i e s ( 3 4 . 7 ) and t h e r e f o r e s a t i s f i e s (34.8) as w e l l . Then an a p p l i c a t i o n of t h e Monotone Convergence Theorem shows t h a t

and t h e proof of t h e theorem i s c o m p l e t e .

252

MUJ I CA

34.9. f

:

E

U

f

IV

Let

COROLLARY.

U

E.

A

function

i s p l u r i s u b h a r m o n i c i f and o n l y i f for e a c h

[-cop)

--*

be an o p e n s u b s e t o f

U

t h e r e e x i s t s an open neighborhood

of

V

x in

x

U s u c h that

i s plurisubharmonic.

34.10.

COROLLARY.

Let

U be an o p e n s u b s e t o f

E.

( a ) W e v e r i f y c o n d i t i o n ( c ) i n Theorem 38.8.

PROOF.

and b E E s u c h t h a t a + Zb Z o g l f ( a + Xb)l RePIX) f o r

= ( e P i h ) /for

C

1x1 5

lows that

i Xbll

loglf(a

P

= 1 . Then

P(C) s u c h t h a t f ( a + X b ) l eRePlX)

1 , by t h e Maximum P r i n c i p l e .

< ReP(X) for

a EU

E

( e- p ( X ) f ( a+ ~ b 1 )< 1

( A ( = 1 . Thus

and t h e r e f o r e f o r

and l e t

U

1x1

Let

/A[ 5

I,

f o r (XI=], It

fol-

and (a) h a s

been proved. (b)

Since

it s u f f i c e s t o a p p l y ( a ) and P r o p o s i t i o n 3 4 . 5 . W e recall that i f @

: I

+

34.11. -+

LEMMA.

E

I Let

0 < X < 1.

and

I

b e an o p e n i n t e r v a l i n

R, and l e t

p : I

be a c o n v e x f u n c t i o n . T h e n :

W

(a) (b)

-

then a f u n c t i o n

i s s a i d t o be c o n v e x i f

W

a, b

for a l l

I i s an i n t e r v a l i n R

p(a)

rp

i s upper s e m i c o n t i n u o u s .

For e a c h

klx

a

E

I

- a) for e v e r y

there e x i s t s

x

E I.

k E lR s u c h t h a t p(xl In p a r t i c u l a r cp i s l o w e r

PLURISUBHARMONIC FUNCTIONS

253

sernincontinuous and t h e r e f o r e c o n t i n u o u s . PROOF.

a

(a) L e t

such t h a t

cp(a + A t ) = c p [ (1

(34.9)

P l a ) < c.

and s u p p o s e

I

E

[a - t , a + t ] C I . Then f o r

-

t

Choose

0

5

A

5

Ala + A!a + t ) ]2 ( 1

-

A)cp(al + Acp(a

>

0

w e h a v e that

1

+ tl

and

Using ( 3 4 . 9 ) w e can f i n d

for

< c

0 < A < 61

0 < 1 5 ~< 1

.

(b)

Let

x, y

E

0 <

with

:1

such that p ( a + A t )

And u s i n g ( 3 4 . 1 0 ) w e c a n f i n d

P(a

such t h a t

6 = min { 6 1 , 6 2 )

set

61

- At)

then with

I

for

0 <

p(a + A t ) < c

for

< c

X < 62.

-6

62 w i t h I f we

< A < 6.

x < a < y. Then w e c a n w r i t e

y-a a=-x Y-X

+

a-x

Y. Y-X

i s convex w e have t h a t

Since

and i t f o l l o w s t h a t

Hence w e c a n f i n d

k

E Z?

such t h a t

x>a

x < a and (b) f o l l o w s .

34.12.

PROPOSITION.

P b ( U ) and

let

cp :

IR

Let -+

IR

U be an o p e n s u b s e t o f

E,

let

b e a co?:vex i n c r e a s i n g f u n c t i o n .

f E I f

254

MUJ I CA

PROOF. By Lemma 34.11 and Proposition 34.1 the function cp o f is upper semicontinuous. Let a E U and b E E such that a + Ab C U. By Lemma 34.11 for each t o E lR there exists k E R

2

cpltol + k ( t

such that ticular

pit)

for every

0 E [O, 27l

1 271

/Bn lo

I

I

-

t o ) for every

t

E

IR. In par-

and therefore 271

c p [ [ f a ieieb)ldO

If we choose

to

LcpP(to)+ 271

1

= 2Tr

fla

1 271

k

f(a

+ e ieb l d t 2

ie

ie

b)dt

-

to].

f l u ) then we obtain

0

27l

cp[fla

ie i e b l l d 8

L

cp(tol

2

cp(flull,

and the proof is complete.

EXERC ISE S

34.A. Let X and Y be topological spaces. Show that if f : X + Y is continuous and g : Y + [ - m J m ) is upper semicontinuous then g o f is upper semicontinuous. 34.B. Let U be an open subset of E , and let f : U +. [ - m , m l be an upper semicontinuous function. Show that given a compact set K C U and E > 0 we can find 6 > 0 such that K + B ( O ; S i C

U

and

fly) < flxl +

E

for each

x

E

K

and

y

E

B(x;&).

34.C. Let U be an open subset of E l and let f, g : U [O,m) be two functions such that Zog f and Zogg belong to P b t U ) . Show that l o g ( f + g ) E P 6 l U ) . +

34.D.

Let U be an open subset of E and let

f

E JC(U;FI.

255

PLURISUBHARMONIC FUNCTIONS

(a)

Show that each of the functions u m ( x l =

is real-valued and continuous on

U.

(b) Show that the function harmonic on U.

- l o g rc f(xl

k

s u p II P (z)11 k l m

is

plurisub-

3 5 . REGULARIZATION OF PLURISUBHARMONIC FUNCTIONS

This section is devoted to the study of plurisubharmonic functions of class C 2 . We show that each plurisubhmnic f u n c tion on an open subset of gn is the pointwise limit of a decreasing sequence of plurisubharmonic functions of class Cm. 35.1. LEMMA. L e t t i o n f E C 2 (U;Z?)

U b e an o p e n s e t i n S. T h e n f o r e a c h f u n c the following conditions are equivalent:

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

(a)

f

(c)

For e a c h

a E U

the integraZ

i s an i n c r e a s i n g f u n c t i o n of PROOF.

(a) * (b):

Let

-

U.

r.

A l a ; r ) C U. Since f

NOW, by Taylor's formula,

is subharmonic,

256

MUJ I CA

where

ere E [ a , a + r e

and after dividing by

ie 1.

Thus

r2/2,

an application of

Convergence Theorem shows that

TI[

a2f

(a)

the

+

Dominated -

ax

aY

we wanted. (b) =. (c):

One can readily verify that

Then condition (b) implies that

0. Thus [ r M ' ( r ) ]' = r M " f r ) + M ' ( r ) that is, M"(r) + r-'M'(r) > 0 and r M ' ( r ) is an increasing function of P. Since it is clear that r M ' ( r ) 0 when r 0, we conlude that rM'(r) > 0 for every r > 0. Hence M ' ( r ) > 0 for every r > 0 and M ( r ) is an increasing function of r . -+

+

Since clearly (c) * (a), the proof of the plete.

lemma

is

com-

If U is an open subset of E and f E C 2 ( U ; l R ) then it follows from Exercise 3 5 . A that D ' D ' ' f ( a ) is an Hermitian form for each a E U. In other words, D ' D ' ' f ( a ) ( s , t ) is a-linear in

6-antilinear in t, and D ' D ' ' f ( a ) ( s , t ) E = 8 then one can readily prove that s,

1

D'D''f(a)(t,s).

If

n D'D''f(a)(s,t)

=

Z

j,k=I

a2faz .azk

(ais

.tk

3

3

for each a E U and s, t E En. The Hermitian form D'D''f(a) serves to characterize plurisubharmonic functions as follows.

257

PLURISUBHARMONIC FUNCTIONS

U be an o p e n s u b s e t o f E , and l e t f E C2(U;R). T h e n f is p l u r i s u b h a r m o n i c o n U if and only if the H e r m i t i n n form D ' D ' ' f ( a ) i s p o s i t i v e f o r e a c h a E U, t h a t i s , D'D"f(a)(t,t) > 0 for e a c h a E U and t E E . Let

35.2. PROPOSITION.

Given a E U and t E E consider the functions = f ( a + S t ) , which is defined on a suitable disc A ( 0 ; r ) . it follows from Exercises 35.B and 35.D that

PROOF.

u(5)

Then

Hence the desired conclusion follows from Lemma 35.1. 35.3. DEFINITION. Let U be an open subset of E. A function f E C 2 ( U ; I R ) is said to be s t r i c t l y p t u r i s u b h a r m o n i c on U if the Hermitian form D ' D ' ' f f a ) is strictly positive for each a E U, that is, D ' D " f ( a ) ( t , t l > 0 for each a in U and each t # 0 in E . The following lemma will be useful later on. U be an o p e n s u b s e t of E , and l e t f b e U. T h e n t h e r e e x i s t s s t r i c t l y p o s i t i v e f u n c t i o n cp E C m ( U ; R I such t h a t

35.4. LEMMA.

Let

s t r i c t l y plurisubharmonic f u n c t i o n on

for e a c h

and

a E U

t

E

a a

Cn.

PROOF. increasing sequence of relatively compact open

subsets of

U

m

such that

U =

I>

i=l

U it follows that

I/;.

Since f is strictly plurisubharmonic on

258

MUJ I CA

+

for every i E AJ. If we set U, = then by Exercise 15.C there exists a function J, E C m ( U ; l R ) such that $ 2 l / c i on ui U i - 1 for every i E JV. If we define cp = l / J , then cp E

‘

,

Cm(U;IR)

for every follows

.

is strictly positive and

cp

a

E

ui

\ Ui-l

and

t

E S.

The

desired

conclusion

PROPOSITION. L e t U b e a c o n n e c t e d o p e n s u b s e t and l e t f E P h ( U ) . I f f $ - m t h e n f E L ’ ( U , l o c ) . 35.5.

PROOF.

of

En

We first show that

A ( a ; r ) C U. Indeed, if ( e l , . for each compact plydisc -n denotes the canonical basis of g n then for 0 .: p 5 have that

. . ,e n ) P

we

2lT

flu) I1

jo f ( a + p e i e e l ) d B ,

and it follows that

Since f is bounded above on each compact subset of U,F’ubini’s Theorem allows us to replace the iterated integral by a double integral. Thus

By iteration we get that

259

PLURISUBHARMONIC FUNCTIONS

and another application of Fubini's Theorem yields ( 3 5 . 1 ) . Now we show that f is Lebesgue integrable on a neighborhood of every point. Indeed, let a E U and choose r > 0 so that -n A ( a ; Z r ) C U. Then we can find a point b E A n ( a ; r ) such that f ( b l > - 03, for otherwise f would be identically - m on U , by Proposition 34.7. Since a n ( b ; r l C a n ( a ; 2 r ) C U, the function f is Lebesgue integrable on a n ( b ; r ) , by ( 3 5 . 1 ) . Since a E A n ( b ; r ) , the proof is complete. 35.6.

let

f E Pb(U),

(a)

f

* pg

(b)

f

*

f o r each

z E

U be a c o n n e c t e d o p e n s u b s e t of f f - m. T h e n : Let

THEOREM.

m

E

C (U61 n PbfU6!

p 6 i z ) decreases t o

for e a c h

f f z ) when

6 > 6

Cn,

and

0.

d e c r e a s e s t o zero,

U.

PROOF. Since f E L 1 ( U , l o c ) , by Proposition 3 5 . 5 , the convolution f * p 6 is well defined and belongs to C m ! U 6 ) , By Proposition 1 6 . 4 . To show that f * p 6 E p b ( U s ) let a E U6 and b E C n such that a + x b C U6. Then an application of Fubini's Theorem shows that

proving that f * p 6 is plurisubharmonic. Next we show that f * p 6 converges pointwise to f when 6 0. Indeed, given a -n E U and c > fia) we choose r > 0 such that A fa;r) C U and -+

260

MUJ I CA

-n A ( a ; p ) . W e claim t h a t

on

f ' C

for

6

0

5

Indeed, s i n c e t h e f u n c t i o n

P.

p 6 i C l depends only

..

of I I,. , 15, I , t h e p r o o f o f ( 3 5 . 2 ) i s almost a r e p e t i t i o n o f t h e p r o o f of ( 3 5 . 1 ) . We l e a v e t h e d e t a i l s t o t h e r e a d e r . W e s t i l l h a v e t o show t h a t (f * p g l ( a ) i s a n i n c r e a s i n g func-

6 f o r each

t i o n of

a

E

U. W e f i r s t show t h a t f o r e a c h

E.

> 0

the integral (35.3)

i s a n i n c r e a s i n g f u n c t i o n of

6 . I n d e e d , t h i s f o l l o w s from re-

p e a t e d a p p l i c a t i o n s of t h e c o n d i t i o n ( c ) i n Lemma 3 5 . 1 t o f u n c t i o n f * p, , which i s p l u r i s u b h a r m o n i c a n d of c l a s s on

U E . Next w e c l a i m t h a t f o r e a c h

6 > 0

the

integral

the Coo

in

(35.3) converges t o t h e i n t e g r a l r

(35.4)

when

f

E

0.

+

when

E

-+

I n d e e d , o n o n e hand

f

*

p,

converges pointwise t o

0. On t h e o t h e r h a n d , t h e estimates i n ( 3 5 . 2 ) a l -

l o w u s t o a p p l y t h e Dominated Convergence Theorem t o

get

the

d e s i r e d c o n c l u s i o n . S i n c e t h e i n t e g r a l i n ( 3 5 . 3 ) is an increasing function of

6

f o r each

E

> 0,

t h e i n t e g r a l i n (35.4) i s a l s o

6 . This completes t h e proof of

an increasing function of

the

theorem. 35.7. JC(U;F)

PROOF.

THEOREM.

and

Let

U

g E PdtVl,

C

E

with

( a ) F i r s t assume t h a t

c i s e 35.D w e h a v e t h a t

and

V C F

be o p e n s e t s ,

$111) C V . T h e n

g

i s of class

g of

E

let

f

E

PbtUl.

C2. Then by Exer-

PLURISUBHARMONIC FUNCTIONS

26 1

and the desired conclusion follows from Proposition 3 5 . 2 . (b)

Next assume that F

is finite dimensional.%

Theorem Vj with

3 5 . 6 we can find an increasing sequence of open sets m

V

=

v

j=l

n Cm(v

and a decreasing sequence of functions

V

j'

which converges pointwise to g in

.)

3

= f-l(v.1

then

3

g. o f

Pn(U

E

3

V.

gi

E

Ph(V.) 3

If we set

U

j

for every j by part (a). Since

.)

3

the sequence ( g of) is decreasing and converges pointwise to j g of in U , we conclude that g o f E P h ( U ) by Proposition 34.5. (c) We finally deal with the general case. By Exercise the function g o f is upper semicontinuous. We still have to show that 34.A

2Tl

(35.5)

g of(al

jo

1

5

g o f(a + e

i 0

bld0

for every a E U and b E E such that a + E b C U. Since f ( a + Xb) is a holomorphic function of h on a neighborhood of the disc m m A, we have a series expansion f ( a + A b ) = z cm A , where the m=O

coefficients e /,,m lie in F and the series converges uniformly m on A . Consider the partial sums S m ( X ) = I: c .'1 Since j=o 3 f ( a + a b ) C V we can find m E lU such that S m ( a ) C V for 0 every m 1. m 0 . Let F m be the vector subspace of F generated is subby e o , c l ,..., e m . By part (b), the function g o S m -

harmonic on an open neighborhood of

g o f l u ) = g oSm(OI 5

(35.6)

1

A , €or each

\

m

m 0 . Hence

2n

i0 g oSm(e )d0

'0

for every c?.n find < g(x) + m 1 -> m 0

and

m

E

L

mo. Let

E

> 0

be qiven.By Exercise

we

34.B

6 > 0 such that f ( a + a b ) + B(O;6) C V and g ( y ) for every x E f ( a + a b ) and y E B ( x ; S ) . Choose

such that

m 2 m 1 ' Then

IISm(Ai

- f(a +

Ab)ll

< 6

for

all

X

E

a

MUJ I CA

262

g o S m ( X ) < g o f ( a + Xb)

(35.7)

X

for all

a

E

m > ml. From ( 3 5 . 6 ) a n d ( 3 5 . 7 ) it f o l l o w s

and

that 2T

gof(a)

I

5

g of(a + e

i eb ) d e

+

E,

'0

and s i n c e

E

w a s a r b i t r a r y , ( 3 5 . 5 ) f o l l o w s . The p r o o f

> 0

of

t h e t h e o r e m is now c o m p l e t e .

35.8. PROPOSITION.

U be u n o p e n s u b s e t of

Let

be a s e q u e n c e of p l u r i s ~ b h a r m o n i c f u n c t i o n s o x

(a)

If 3- ) i s

The s e q u e n c e

s u b s e t of

(b)

There e x i s t s

such t h a t

each

3n

.31imp,;.ct

5

Z i m sup f . ( z ) j + W

3

c

z E U.

such t h a t Let

such t h a t

lR

c E

Then f o r e a c h compact s e t

PROOF.

U slnch t h a t :

U.

f o r every

jo

bounded above

Ifj)

Let

Cn.

fjh' < c +

K and -n A fa;3r)

generality that

E

K

C

for e v e r y

E

be given. L e t

C

U.

fj 5 0

and

U

a

E

j E

K

there e x i s t s

> 0

jo

.

and choose

r

By ( a ) w e may assume w i t h o u t loss -n on A i a ; 3 r ) f o r e v e r y j. Then

>

0 of

an

a p p l i c a t i o n of F a t o u ' s Lemma shows t h a t

Hence w e c a n f i n d

€or every

j

jo

jo. Let

such t h a t

0 < 6 < Y'

and l e t

z

E

x n ( a ; 6 ) . Then

PLURISUBHARMONIC F U N C T I O N S

263

Nowl by ( 3 5 . 1 ) w e have t h a t fj(z)nnir + 6

1< ~

~

-n

A (z;r+6)

f o r every

j. S i n c e

f o r every

z E an(a;6)

fj -

f i c i e n t l y small. Since

on

0

and

-n A ( a ; 3 r ) it f o l l o w s t h a t

j - jo

,provided

6

is

0

suf-

K can be c o v e r e d by f i n i t e l y m a n y p l y -

d i s c s I t h e proof i s complete.

EXERC ISES 35.A.

Let

U b e a n open s u b s e t of

E l and l e t f

2

E

C (U;F).

Show

that

2 2 4D'Dr'f(a)(s,t) = D f ( a )(s,t) + D f ( a (is, it) 2 + D f(a1(s,it)

for a l l

35.B.

a E

Let

U

and s , t

E

E.

U be an open s u b s e t of

Show t h a t

for a l l

a E U

and s , t

E

- i D 2 f a ) (is,t )

P

n

.

Cnl

and l e t

f E

2 C (U;FI.

264

MUJ I CA

35.C. V

C

F

flu)

C

Let E, F, G be complex Banach spaces, let U C E and be open sets, and let. f E C 1 ( U ; F ) and g E C l ( V ; F ) with V . Show that

and

x

for every 35.D. V

C

f(U)

F C

E

U.

Let E, F , G be complex Banach spaces, let U C E and be open sets, and let f 6 X ( U ; F I and g E C 2 ( V ; F ) with V . Show that

for all

x

E

U

and

s, t E

35.E. Let U C E and V C 2 ( U ; B ) and g E C 2 ( V ; l R ) (a)

for all

(b)

E.

C B

with

be open sets, f(Ui

C

and let

f

E

V.

Show that

x E U

and

s, t E

Show that if f

and increasing then

gof

E. is plurisubharmonic and g

is convex

is plurisubharmonic.

(c) Show that if f is strictly plurisubharmonic and g is convex and strictly increasing then g o f is strictly plurisubharmonic. 35.F.

Let U be a connected open subset of

En.

265

PLURISUBHARMONIC FUNCTIONS

(a)

Show t h a t i f

{ z E 11 : f i x ) =

t h e set

Show t h a t i f

(b)

t h e set

U

{z E

: flxl

is not identically

E PhlU)

f

-

m)

h a s Lebesgue measure z e r o .

f E 3C(U;F)

= 0)

then

--m

i s n o t i d e n t i c a l l y z e r o then

h a s Lebesgue measure zero.

36. SEPARATELY HOLOMORPHIC MAPPINGS I n t h i s s e c t i o n we prove t h a t each s e p a r a t e l y mapping i s h o l o m o r p h i c , a r e s u l t w h i c h h a d

holomorphic

been

promised

in

S e c t i o n 8.

L e t U b e a n o p e n s u b s e t of C n . T h e n a m a p p i n g i s h o Z o m o r p h i c i f and o n l y i f f i s s e p a r a t e Z y h o -

36.1.

THEOREM.

f

+

:

U

F

Zomorphic. T o p r o v e t h i s t h e o r e m w e n e e d t w o a u x i l i a r y lemmas.

36.2. C

F

Let

LEMMA.

E, F,

G

b e Banach s p a c e s ,

be o p e n s e t s , and Z e t

f

: U x

V

-f

G

tinuous m a p p i n g . T h e n f o r e a c h c o m p a c t s e t L n o n v o i d o p e n s e t s A and B , w i t h A C U and that

is b o u n d e d o n

f

U

Zet

C

E and

V

be a separatelyconC

there

V

are

L C B C V , such

A x B.

Consider t h e sets

PROOF.

Since

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

f (x,y)

fixed, each

An

i s closed i n

U.

And s i n c e

x

when f(x,y)

y

is

held

is a

con-

m

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

y

when

t h e B a i r e C a t e g o r y Theorem some

B E ( a ; r ) . Thus

IIfll

5 n

on

v An. n=l an open

x is held fixed, U = contains

An

BE ( a ; r )

x

[ L + BF(U;l/n)],

as

By ball

as-

serted.

n 2 and l e t f : A n ( a ; R ) F be a mapp i n g u h i c h i s h o l o m o r p h i c i n z , = i z l , . . . , z ~ - ~ f )o r z n f i x e d , 36.3.

LEMMA.

Let

--).

266

MUJ I CA

and h o l o m o r p h i c in z n f o r z ' f i x e d . I f f i s bounded *n-l ( a ' ; r I x Afan;RIJ w i t h 0 < r < R, t h e n f po Zydisc lornorphic o n

hn ( a ; R I

E

a

i s ho-

.

Since f is holomorphic in z ' for z , A (a;R) we have a series expansion

PROOF.

z

on

fixed,

for

each

n

where the summation is taken over all multi-indices a =(a,, E Dn-l, and

.. .,an-1 j

0

Since f

is holomorphic in z n

for z '

fixed, we see that each Thus, to show that f is holomorphic on An(a;R) it is sufficient to show that the n series in (36.1) converges uniformly for z E A (aJR1) whenever 0 < R l < R.

c a ( z n ) is a holomorphic function of

Let 1

0

such that

2,.

Rl < R 2 < Rg < R. By hypothesis there exists Ilfll

5 c on A n - l ( a ' ; r )

x

A(an;Rj.

Hence

Since for each z n fixed the series in (36.1) converges formly on (a';R3j it follows that

Consider the subharmonic functions

u

a

From (36.2) and (36.3) it follows that

(Z

n

I =

la1

c 2

uni-

l o g I1 caizYl/lI .

PLURISUBHARMONIC FUNCTIONS

267

Then an application of Proposition 35.8 shows that ua(zn) 5 - Z o g R 2 for l z n l 5 Rg and la\ sufficiently large. This

I

IIeiylznJIJRZal < 2

means that

for

lzn/ 5 R3

and

suf-

ficiently large. Whence it follows that the series in (36.1) converges uniformly for z E A n ( a ; R l ) . This completes the proof. We proceed by induction on n , the theorem being obviously true for n = 1 . Let n 2 2 and assume the theorem is true for n - 1 . Thus we have a mapping f : U + F which is holomorphic in z ' = ( Z ~ , . . . , Z ~ - ~ )for z n fixed, and holomorphic in z n for z ' fixed. Let a E U and choose R > 0 such that A n ( a ; 2 R i C U . By Lemma 3 6 . 2 there is an open on polydisc A n - ' ( b ' ; r ) C A n - ' ( a ' ; R ) such that f is bounded An- 2 ( b ' ; r ) x a ( a n ; R ) . By Lemma 36.3 f is holomorphic on An(b;RI, where b = ( b ' , a n ) . Since clearly a E A n ( b ; R I , the proof of the theorem is complete. PROOF OF THEOREM 36.1.

Now we can prove a result promised in Section 8. 36.4. THEOREM. JCG(U;F).

U b e a n o p e n s u b s e t of E , and l e t f E a E U t h e r e is a u n i q u e p o w e r s e r i e s

Let

Then f o r each

co

B p" f ( a ) ( x m=O

- a ) from

f l x ) f o r every

E

,

into

F which converges pointwise t o

where

Ua

d e n o t e s the l a r g e s t

open s e t which i s c o n t a i n e d i n

U.

The p o l y n o m i a l s

x

Ua

E

a-balanced Pmf(a) are

g i v e n by t h e f o r m u l a s

where f o r

each

f o r every

5

E

t

E

is c h o s e n s o t h a t

E, r > 0

a

+

E

U

Lto;r).

PROOF. In view of Proposition 8.4 it is sufficient to prove that the mapping P m f l a ) : B F defined by (36.4) belongs to +

268

Pa(

ut

MUJ I CA

m E

U

s, t E E

Fix

E;FI.

1x1

for

< r

and c h o o s e and

r > 0

a + As +

'such t h a t

IpI < P. Then t h e mapping

(p(X,p)

f ( a + A s + p t ) i s s e p a r a t e l y h o l o m o r p h i c , and t h e r e f o r e

morphic, for

1x1

and

< P

I p ( < P , by Theorem 3 6 . 1 .

=

holo-

Thus

we

h a v e a series e x p a n s i o n

w i t h a b s o l u t e and u n i f o r m c o n v e r g e n c e o n e a c h p o l y d i s c 1x1 5 p, ( P I 5 p , w i t h 0 c p < r. Given rl E 6 c h o o o s e p > 0 s u c h that

0 c p < P

plnl <

and

p.

Then a t e r m by t e r m i n t e g r a t i o n

shows t h a t

T h i s shows t h a t

a t most m

m P f ( a ) f s + rit) i s a p o l y n o m i a l i n

q

o f degree

s, t E E . By Theorem 3 . 6 w e c o n c l u d e t h a t m P - f ( n ) i s a p o l y n o m i a l of d e g r e e a t most m. NOW, by P r o p o s i f o r each

t i o n 8 . 4 w e have t h a t and

p E 6. Thus

P m f ( a l ( u t ) = pmPmf(a)ltl f o r a l l

Pmf ( a )

E

palm^;^)

t

E E

and t h e p r o o f i s c o m p l e t e .

W e o b s e r v e t h a t i n v i e w of Theorem 3 6 . 1

the

conclusions

o f Theorem 7 . 7 and C o r o l l a r i e s 7 . 8 a n d 7 . 9 a r e v a l i d f o r G-hol o m o r p h i c mappings. d e n o t e by [ Amf(al]

If

f

E JCG(U;F)

A m f ( a ) t h e u n i q u e member

- = Pmf ( a ) .

W e n e x t want t o e x t e n d

and of

a

E

U

EzlrnE;FI

then we s h a l l such

that

Theorem 3 6 . 1 t o t h e case o f Banach

s p a c e s . A key t o o l i n t h e p r o o f i s t h e f o l l o w i n g t h e o r e m .

U be a c o n n e c t e d o p e n s u b s e t o f E , and l e t f E J C G I U ; F ) . If f i s L o c a Z l y b o u n d e d at some p o i n t of U t h e n j is locaZly b o u n d e d at o v e r y p o i n t of U, a n d ?'s t h e r l e f o r e ho l o m o r p h i c . 3 6 . 5 . THEOREM.

Let

269

PLURISUBHARMONIC FUNCTIONS

Before p r o v i n g t h i s t h e o r e m w e g i v e t w o p r e p a r a t o r y r e s u l t s .

Let

LEMMA.

36.6.

Then f o r e a c h

f : B(a;R)

b E B(a;RI, k

t

and

I: ( E ) A m f ! a ) ( b

-

we h a v e t h a t

E

E

m

k

P f(bl(t)=

(36.5)

b e a G-holornorphic m a p p i n g .

F

-+

E Z V

a)

rn-k

t

.

k

m=k

-

0 < r < R

If

PROOF.

-

IIb

all t h e n

f o r each

t E B l O ; r l we have t h a t

(36.6)

f(b + t) =

B ( b ; r ) C B ( a ; R ) and hence

m

- a + t)

Z Pmflal(b m=O

m

m

I: ( ~ ) A r n f ( a l (-b a ) m-k t k .

= I:

m=O k=O

Choose

+ vt

E

B(0;R) for

5

/A1

-

plllb

such t h a t

p > 1

and

p

Atb

- a)

Since Corollary

7.9

all

+ r ) < R . Then

11-11

5

p.

a p p l i e s t o G-holomorphic m a p p i n g s w e h a v e t h a t

m-k Amf(a)(b-a)

By Theorem

IIf

[a

+ A(b

k

t

36.1

-

f [ a + A(b - a ) + p t l Xm-k+l k + l 1-1

( m - k l ! k!

= m! ( 2 n i )2

there

a) + 1-1tlII

is a

5 c

constant

1x1 5

for

>

c

that

such

0

and

P

dAdp.

11-11

5

P.

Whence

i t follows t h a t

T h i s shows t h a t w e c a n i n t e r c h a n g e t h e o r d e r

of

summation

in

( 3 6 . 6 ) . Thus m

(36.7)

f(b + t) =

Since for every m

C m=k

r) E

( il l l A m f ( a i ( b - a )

m

-

2 2 (:)Amf(al(b k=G m=k

t

m-k

we have t h a t k

i n t ) II 5

a

m=k

clnl p

m-k

a)

m-k

t

k

.

k

= p

(- 101

P

)k

C P

p - I ’

MUJ I CA

270 m

t h e series

Z

l Am f ( a ) ( b

(

-

a)m-k

c o n v e r g e s p o i n t w i s e t o an

m=k

e l e m e n t of

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

Pa(kE;F).

( 3 6 . 7 ) w i t h t h e series

e x p a n s i o n i n Theorem 3 6 . 4 y i e l d s ( 3 6 . 5 ) .

PROPOSITION. L e t f : B ( a ; R ) F b e a G-holomorphic mapm P f l u ) is c o n t i n u o u s f o r e v e r y m E i7? t h e n f is holomorphic. 36.7.

-+

ping. I f

b

Let

PROOF.

B ( a ; R l and l e t

E

ma 36.6, f o r e a c h

k E liV

and

-

0 < r < R

t E E

- all.

IIb

By Lem-

w e h a v e t h e series

ex-

pansion m

k

(36.8)

P f(bl(tl

= z f:)

Amf(alIb

- a)

rn-k

t

m=k S i n c e t h e mappings

m

A f ( a ) are continuous, an

Lemma 2 . 7 y i e l d s a n o p e n s e t

V

in

k

.

application

of

E w h e r e t h e p a r t i a l sums of

t h e series i n ( 3 6 . 8 ) a r e u n i f o r m l y b o u n d e d . Hence t h e p o l y n o k P f l b l i s bounded o n V a n d i s t h e r e f o r e c o n t i n u o u s . N m ,

mial

f o r every

t

E

B ( 0 ; r l w e h a v e t h e series e x p a n s i o n m

f(b + tl =

(36.9)

2 Pkf(bl(tl. k=O

Then a n o t h e r a p p l i c a t i o n of Lemma 2 . 7 shows t h e e x i s t e n c e of a ball

B(c;pl C BfO;rl

u n i f o r m l y bounded.

where a l l t h e polynomials

Pk f f h ) are k By C o r o l l a r y 7 . 6 t h e p o l y n o m i a l s P f ( b ) a r e

u n i f o r m l y bounded o n t h e b a l l Cauchy-Hadamard

Formula

B ( 0 ; p i . Then it f o l l o w s from the

t h a t t h e series i n ( 3 6 . 9 ) h a s

o f c o n v e r g e n c e g r e a t e r t h a n or e q u a l t o

p.

radius

T h i s completes t h e

proof.

PROOF OF THEOREM 3 6 . 5 . T o prove t h e t h e o r e m i t i s c l e a r l y s u f f i c i e n t t o show t h a t t h e s e t o f p o i n t s w h e r e f is locally b o u n d e d i s closed i n U. L e t a E U be t h e l i m i t of a s e q u e n c e o f p o i n t s w h e r e f i s l o c a l l y b o u n d e d . Choose r > 0 s u c h t h a t B ( a ; 2 r ) C U , and c h o o s e b E %(a;r) s u c h t h a t f is l o c a l l y b o u n d e d a t b . T h i s i m p l i e s t h a t P m f ( h l i s c o n t i n u o u s €or every

PLURISUBHARMONIC FUNCTIONS m E IN. Since

271

U, an application of Proposition 3 6 . 7 shows that f is holomorphic on the ball B(b;r,J. Since a E B(b;Ya) we conclude that f is locally bounded at a , and the proof is complete. B(b;rf

B(a;2rl

C

C

Now we can generalize Theorem 36.1 as follows. 36.8. THEOREM.

El

Let

b e a n o p e n s u b s e t of

,..., E n , ...

El

X

b e Banach s p a c e s , and l e t

F X

En.

Then a mapping

f : U

+

U F

i s hoZornorphic if and o n l y if f is s e p a r a t e l y h o l o m o r p h i c . PROOF. To prove the nontrivial implication assume f : U is separately holomorphic. It clearly suffices to prove theorem for n = 2. And without l o s s of generality we may

sume that

U = UI

x

where

1J2,

Uj

is an open ball in

F the

+

as-

E . for 3

j = 1,2.

We claim that f is G-holomorphic on U. Indeed, given l a , , a 2 ) E U, x U 2 and ( b l , b 2 1 E E l x E 2 , the mapping ( A I , A 2 )

+ A 2 b 2 ) is separately holomorphic, and hence holomorphic by Theorem 36.1. Hence the mapping A f ( a , + Abl,a2 + hb 2 ) is holomorphic, and f is therefore G-holomorphic,as asserted. Since f i s separately continuous, an application of Lemma 36.2 shows that f is bounded on some nonvoid open subset of U, x U 2 . By Theorem 36.5 we conclude that f is holomorphic on U, x U 2 , completing the proof.

+

f ( a l + Albl,a2

+

We conclude this section with a characterization of holomorphic mappings in Banach spaces with a Schauder basis. 36.9. THEOREM. ( e n ) , Zet I!

En

Let

E

be a Ranach s p a c e w i t h a S c h a u d e r b a s i s

be t h e subspace g e n e r a t e d by

m o r p h i c if and o n l y if f morphic f o r each

PROOF.

n

E

elJ...,en,

and L e t

f :U F is h o b is continuous and f I U n E n is holo-

be a n o p e n subset of E.

Then a mapping

+

D.

To prove the nontrivial implication, suppose that f is

continuous and f I U 0 E n is holomorphic for every n E n . T o prove that f is holomorphic it suffices to shew that f is

MUJ I CA

272

G-holomorphic.

a

Let

t h a t t h e compact s e t T,

: E

Tnx f

liv

+

U and b E E and choose r > 0 K = a + 2rzb i s contained i n U.

E

be t h e c a n o n i c a l p r o j e c t i o n f o r each

En

n

E

such Let

U.Since

converges t o such t h a t

f o r each n - n

x u n i f o r m l y on compact s e t s w e c a n f i n d no T n { K l C U f o r e v e r y n 1. n By C o r o l l a r y 7.2, 0 and I h l 5 2 r w e have a series expansion

.

m

(36.10) where

f o r every

1x1 5

P

n

E

To c o m p l e t e t h e p r o o f w e s h a l l show

Do.

that f o r

w e h a v e a series e x p a n s i o n

+ Abl =

f(a

(36.11)

w k Z A ckj k=O

where

f o r every

n E Do.

In view

of

( 3 6 . 1 0 ) , t o p r o v e ( 3 6 . 1 1 ) it is

s u f f i c i e n t t o show t h a t (36.12)

lim f o T n ( a + Abl = f ( a + Ab) n-+w

and m

m

(36.13)

/XI < r. L e t

i s continuous and K i s compact w e c a n f i n d 6 > O such t h a t K + B(O;6) C U a n d 11 f ( y l - ffxlll 5 E f o r x E K a n d y E B ( x ; S l . I f w e c h o o s e n I 2 no s u c h t h a t IITnz - x II < S f o r every LT E K then I1 f o T n ( x ) - f(z!ll < E f o r e v e r y c. E K a n d n 2 n I . This for

E

> 0

be given. Since

J”

PLURISUBHARMONIC FUNCTIONS

273

already shows (36.12). Furthermore

for every n 2 n l of the theorem.

, and (36.13) follows. This completes the proof

3 7 . PSEUDOCONVEX DOMAINS

In this section we introduce the notion of pscudoconvex domain in a Banach space and give several necessary and sufficient conditions for a domain to be pseudoconvex. We recall that if U is an open subset of a Banach space and x E U then d u ( x l denotes the distance from x to the boundary of U. We now introduce another distance function. DEFINITION. Let U be an open subset of (0, - 1 is defined by function 6U : U x E

E.

37.1.

Then

the

+

Thus, for each

t

6,,,(x,t)

E E,

tance from x to the boundary of Observe that if U = E then d U 37.2.

PROPOSITION. (a)

Let

2 E

-

U b e a p r o p e r o p e n s u b s e t of

IdUfx) - dU(yll 5 d,

ticuZar, the f u n c t i o n

may be regarded as the disU along the direction t. and A U = m.

IIX

- ~ I for I a22 x,

i s c o n t i n u o u s on

E . Then:

y E U. ~n p a r -

U.

i s Zower s e m i c o n t i n u o u s on

(b)

The f u n c t i o n

(c)

d (xi = inf { S U l x , t l : t U

6u

E

E, IItll

= 21

for

U

X

E.

each

u.

PROOF. (b)

(a) This is easy: it was left as Exercise 7.D. Let ( x o , t o l

E

U

x

E

and let

0 < r

s < 6U (xo J t o ) '

274

MUJ I CA

Then xo + A t o E U for 1x1 2 s and a standard compactness argument yields a neighborhood V of xo and a neighborhood W of t o such that x + At E U for every x E V , t E W and 1x1 < s . Hence 6 ( x , t l 5 s > r €or every ( x , t l E V x W. U -

(c) Since B ( x ; ~ = ) u { x + rzt : t desired conclusion follows at once.

E

= 11,

E , II tII

the

An open subset U of E is said to be p s e u - l o g 6u is plurisubharmonic on U x E.

37.3. DEFINITION.

d o c o n v e x if the function

in order to give other characterizations of domains we introduce the following definition. 37.4. DEFINITION. Let U be an open subset of set A C U we define

pseudoconvex

For

E.

each

and A

A Pn c (U)

= {x

E

u

: f ( x )

5 sup f

for all

f

E

Pncfu)}.

A A

Note that the set A P n c ( U ) is always closed in U , but it is not clear whether the same is true f o r A P b ( U J . Note also that since I E P d c f U ) whenever f E J C ( U ) , we always have the inclusions A c 2 Pb (UJ APnctul %c(uJ-

If

=

37.5.

For a n o p e n s u b s e t

THEOREM.

U

of E

t h e folkozoing eon-

d i t i o n s are equiuaZent:

(a)

is p s e u d o c o n v e x .

U

(b) f o r each

The f u n c t i o n

(c)

The f u n c t i o n

(dl

dU(A^PdCfUl

t

E

- log

6u(*,

t i i s p l u r i s u b h a r m o n i c on U

E.

- log

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

I = dU ( A )

f o r each s e t

A C U.

U.

PLURISUBHARMONIC

(e)

The s e t

(f)

The s e t

compact s e t

K

C

-

KPhC(uI

i s compact f o r each compact s e t

KCU.

is r e Z a t i v e Z y c o m p a c t i n U f o r each U.

If IH,D) is a H a r t o g s f i g u r e i n t 2 t h e n e a c h J C ( H ; E ) w i t h f ( H ) C U has a n e x t e n s i o n 3 E J C ( D ; E ) w i t h (g)

c

275

FUNCTIONS

f

E

Y(D)

u.

-

-

PROOF. The implications (a) (b), (c) (d), (d) * (e) and (el * (f) are clear. The implication (b) * (c) follows from Proposition 37.2(c). Thus it only remains to prove the iqlica(a). tions ( f ) * ( g ) and ( g )

-

(f) * ( g ) : We have that

and

R and 0 < s < R . Let f E JC(H;E) with f(H) C U where 0 < r be given. By Example 10.2 and Exercise 8.1 there exists 3 E J C ( D ; E ) such that 7 = f on H. We still have to show that Y~I!?)C U. Fix b with r < b < R . For each t E ( G , R ] consider the set

and let

We shall prove that T = (0. '91. This means that T(DR) C U. And since f ( D \ DR) C ?(HI = f ( H I C U, ( g ) will follow. The set T is nonvoid, since (0, s ] C T. Thus to show that T = ( ( I j R ] if suffices to prove that T is open and c l o s e d in (0, R ] . Note that t E T implies (0, t ] C T . The set T is closed in

276

MUJ I C A

,

(0, R ]

for

t,

E

T

for e a c h

n E liV

To show t h a t l' i s o p e n i n (0, R ] , f i x u

implies t h a t

s u p t nE T .

t

t < r, f i x

E

T

with

r < u < b , and consider the set

with

H I a n d K = f(J) i s a s u b s e t of U. I f u E P b ( U ) then u o f E P b ( D t ) by is 3 5 . 7 . T h u s , i f w e f i x 5 w i t h 1 6 1 < t, u o ?(A,<) h a r m o n i c f u n c t i o n of X f o r lX( < b . Then for e a c h a < c b it f o l l o w s f r o m t h e maximum p r i n c i p l e f o r

Then

i s a c o m p a c t s u b s e t of

J

compact Theorem

a subc with

subhar-

monic f u n c t i o n s t h a t

Whence i t f o l l o w s t h a t Then by ( f )

?(A,<)

there exists

E

E

2p b ( u )

for every ( A , < ) such t h a t

> 0

Dt.

E

S i n c e f i s u n i f o r m l y c o n t i n u o u s o n e a c h c o m p a c t s u b s e t of we can f i n d 6 0 with t + 6 < R such t h a t

D,

s u c h t h a t IX - A ' I < 6 .md l for a l l (A,<), (A',<') E D t+6 < 6 . From ( 3 7 . 1 ) a n d ( 3 7 . 2 ) i t f o l l o w s t h a t f ( D t + & ) C U. T i s o p e n i n (0, R ] and (9) f o l l o w s .

5'1

(9)

-

Log 6

(s,t)

Let

1x1

* (a):

i s p l u r i s u b h a r m o n i c on

U E E

P

Assuming ( 9 ) w e h a v e t o show t h a t t h e

E

x E

P(C)

U

x

E. Fix ia,b) E U

i-

Thus

function x

E

and

such t h a t

such t h a t

= 2 . S i n c e e Re

- Log 6 C , ( a + As, b + At) = I e P i A ) I,

< Re P I X ) €or

t h i s means t h a t

PLURISUBHARMONIC FUNCTIONS

(37.4)

S U ( a + As, e - ' ( ' ) ( b

+ At))

For a fixed defined by

with

p

for

1 1

We have to show that ( 3 7 . 4 ) holds for sufficient to show that

277

1x1

[ A / = 1.

< I . And it is clearly

0 0 such that if we set

and

H = H

1

u HZ,

then

f ( H 1 C U. If we set

then ( H , D ) is a Hartogs figure in C 2 , and by ( g ) there is a function E J C ( D ; E ) such that = f on H and f ( D ) C U. Then f = f on D by the Identity Principle, and therefore f ( D ) C U. Whence ( 3 7 . 5 ) follows and the proof of the theorem is complete.

7

7

i s pseudoconvex if and o n l y if U n M i s a p s e u d o c o n v e x o p e n s u b s e t of M f o r e a c h f i n i t e d i m e n s i o n a l s u b s p a e e M of E. 37.6.

PROOF.

COROLLARY.

A n open s ~ b s e t U

of

T h i s follows from Proposition

E

34.5 ( c )

and

Theorem

.

3 7 . 5 ( b ) or (c) 37. 7 .

COROLLARY.

E v e r y holomorphically e o n v e x o p e n s e t

in

a

278

MUJ I CA

Banach s p a c e is p s e u d o c o n v e x .

I n p a r t i c u Z a r e v e r y domain o f h o -

Zomorphy i n a Banach s p a c e i s p s e u d o c o n v e x .

PROOF. Since K P h c l U ) c 17,,,, €or every set clusion follows from Theorem 37.5(e). 37.8. PROPOSITION. !l'E E (E;FI and l e t

Let

K C U , the con-

V be a p s e u d o c o n v e x o p e n s e t i n F , l e t

U = T - I ( V ) . T h e n U is a p s e u d o c o n v e x o p e n

s e t i n E.

PROOF. Let K be a compact subset of verify that

U. Then one can

readily

(37.6)

By Theorem 37.5 (f) there is a 0-neighborhood B in F such that ( T ( K ) ) p A ( V ) + B C V . Then A = T - l l B ) is a 0-neighborhood in E and it follows from (37.6) that 37.5(f) again,

zPAtul+

A C

U.

By

Theorem

U is pseudoconvex.

37.9. PROPOSITION. L e t U C E and V C F be o p e n s e t s . T h e n U x V i s p s e u d o c o n v e x i f and o n l y if U and V are pseudoconvex. PROOF. If l x , y ) E U readily verify that

X

V

and

(s,tl

E

E

x

F

then

one

can

Whence it f o l l o w s that - l o g 6,, is plurisubharmonic if and only if - l o g 6 u and - l o g 6 v are plurisubharmonic. By Corollary 37.7 every domain of holomorphy in a Banach space is pseudoconvex. The converse is not true in genera1,for as we have mentioned already, B. Josefson [ 1 1 has given an example of a holomorphically convex open set in a nonseparable Banach space which is not a domain of holomorphy. But the f o l lowing problem remains open.

279

PLURISUBHARMONIC FUNCTIONS

37.10.

PROBLEM.

Is e v e r y

E be a s e p a r a b l e Banach s p a c e .

Let

pseudoconvex open s e t i n

E

a domain o f e x i s t e n c e , o r a t l e a s t

a domain of holomorphy?

€or it w a s posed by E . L e v i 1 ] i n 1 9 1 1 f o r E = tn. I n S e c t i o n 4 3 w e s h a l l n s o l v e t h e Levi problem i n t h e case where E = g , and i n Sect i o n 4 5 w e s h a l l s o l v e t h e Levi problem i n t h e case of separable Problem 37.10 i s c a l l e d t h e L e v i probZern,

Banach s p a c e s w i t h t h e bounded a p p r o x i m a t i o n p r o p e r t y .

EXERCISES 37.A.

Show t h a t a n open s u b s e t

U of E i s pseudoconvex i f armd U i s pseudoconvex.

o n l y i f e a c h connected component o f 37.B.

Show t h a t i f

s u b s e t s of

E

I i s a f a m i l y of pseudoconvex

(UiliE

t h e n t h e open s e t

U = int

n

Ui

open

is pseudoconvex

i E I

as w e l l . 8

37.C.

Show t h a t i f

(Uj)j=l

i s an i n c r e a s i n g sequence of pseum

doconvex open s u b s e t s of

E

t h e n t h e open s e t

U =

U

is

U

j=l

j

pseudoconvex as w e l l . 37.D. Pbc(U)

U be a pseudoconvex open s u b s e t of E, let f E and l e t V = ( x E U : f ( x ) 0 ) . Show t h a t V i s pscu-

Let

doconvex.

38. PLURISUBHARMONIC FUNCTIONS CN PSEUDOCONVEX DOMAINS

I n t h i s s e c t i o n w e o b t a i n s e v e r a l r e s u l t s of e x i s t e n c e p l u r i s u b h a r m o n i c f u n c t i o n s o n pseudoconvex domains.

We

of

begin

w i t h t h e f o l l o w i n g lemma.

U b e a p s e u d o c o n v e x o p e n s e t i n d?, 2et K be a c o m p a c t s u b s e t o f U, and l e t V b e a n o p e n n e i g h b o r h o o d i n U. T h e n t h e r e e x i s t s a f u n c t i o n f E PhciUl such O f K P m I 38.1.

LEMMA.

Let

280

MUJ I CA

that:

(a)

PROOF.

The s e t

Let

{z E U : f l z )

5 c)

is compact for each c

E

R.

u E P b C l U ) be defined by

s u p u < 0. Clearly the K c E R. If set f l c = { z E U : u ( z ) 5 c) is compact for each K O C V then the function f = u has the required properties. V. Since K 2 is compact we can find Suppose then that K O 6 > 0 such that K 2 C U g , where U g = { z E U : d , ( z l > 6). For each a E K O \ V we can find 9 E P b I U ) such that s u p V <

where the constant k is chosen so that

K

0 < c p l a ) . By Theorem 35.6 there is a decreasing sequence

in

P b c l U 6 ) which converges pointwise to

in

(Vj)

U g . Hence

m

K

( z E U6 : c p . ( z l 3

U

C

j=l

< 0)

JI E P b c ( U g l such that s u p $ < X is compact we can thus find

and thus we can find a function 0 < $ ( a ) . Since E

KO \ V

P o c ( U 6 ) such that

s u p $J K j

0

for

j

=

I,.

. .,m

and

m

z E IY

and

v(z) > 0

f o r every

z E KO \ V .

Set

M = sup v > 0 K O

and define

f : U

-+

iR

by

d

281

PLURISUBHARMONJC FUNCTIONS

1 < ulzl

If

then

< 2

z

E

K

and

f(z) > 0

%cltUi 'Pb ( U l

PROOF.

(z E U : f ( z )

5 c)

is

compact

for

If

The i n c l u s i o n

d,, (ul

sup f K

such t h a t complete.

E

i s a l w a y s t r u e . To show

K^P,c(U) U with

a

9 d p A i u l . By

applying

w e can f i n d a f u n c t i o n f E PbC(U) a ~ P n c ( u i a n d t h e proof is

< 0 < f ( a l . Thus

U be a p s e u d o c o n v e x o p e n s e t ~ T E, L let V = { x E U : f l x ) < 0). T h e n V is pseudoconvex Let

PROPOSITION.

and l e t

Pd(U)

a

V = U \ {a}

Lemma 3 8 . 1 w i t h

E

every

z E U \ V . And s i n c e

f o r every

the opposite inclusion let

f

for

< 0

U i s a p s e u d o c o n v e x o p e n s e t i n 61" then f o r e a c h compact s e t K C U. I n p a r t i c u l a r - %nc(UI i s c o m p a c t f o r e a c h compact s e t K C U.

PROPOSITION.

38.3.

f(z)

is w e l l

f

M - u ( z ) , so t h a t

Pbc(U). Clearly

w e conclude t h a t t h e set e v e r y c E X?. 38.2.

5 M <

V ( Z )

d e f i n e d and b e l o n g s t o

as w e l l . I f s u f f i c e s t o show t h a t

PROOF.

V n M

each f i n i t e dimensional subspace V n M.

subset of f

5

on

c

'P,

that

K.

f

5 c

'PbfUnM)

< 0

c

on

v n

V n M.

Thus

<

K^p,(unM)

M.

such

WPniUnM) d,, (v

that

V n M

0

and it

Since

p a c t , by P r o p o s i t i o n 38.2, w e conclude r e l a t i v e l y compact i n

c

for

K be a compact

E. Let

Then t h e r e i s a c o n s t a n t

Hence

(v n M I

M of

i s pseudoconvex

that

follows

i s comM)

is

i s pseudoconvex and

t h e p r o o f i s complete.

U be a p s e u d o c o n v e x o p e n s u b s e t of E. Then

38.4.

THEOREM.

Let

each

f

i s t h e p o i n t w i s e l i m i t o f a d e c r e a s i n g sequence

ifn)

E Pb(U)

C PbC(U).

PROOF.

If

f

= -

t h e n w e may t a k e

fn

E

-

n

f o r every

n ER.

282

MUJ I CA

Thus we may assume f 9 g n f s ) = inf [ g f z ) + n l l s Z E

u

-. -

Then set g = e -f and define fnfxl = - Zoggn(x) for

z I I ] and

every n E 2$ and x E U . Then the proof of Proposition 34.2 shows that each f n : U lR is continuous, and that the sequence f f n ) is decreasing and converges pointwise to f. Thus it only remains to prove that each f n is plurisubharmonic on V . Then set U x 6 is pseudoconvex, by Proposition 3 1 . 9 , and the set +

is pseudoconvex as well, by Proposition 3 8 . 3 . Observe that fx,O) E V €or every x E U. To show that the function f n is plurisubharmonic on U we shall prove that n

g n lx) = dv(ix,O)i

for every

x

E

u,

where d; denotes the distance to the boundary of respect to the norm

V

with

Observe that all these norms are equivalent and induce the product topology on E x S. Since

it is clear that

PLURISUBHARMONIC FUNCTIONS

5

d;( (x,Oll

Since c l e a r l y

ci

= B = i n f Cnllx -

d;f(x,Oll

283

we conclude t h a t

+ g(z) : z

z II

E

U) = g fxl. n

This completes t h e proof.

Now w e c a n e x t e n d P r o p o s i t i o n 3 8 . 2 t o t h e case

of

Banach

spaces.

KPn (U)

PROPOSITION. If U is a p s e u d o c o n v e x o p e n s e t i n E then f o r e a c h compact s e t K c U. I n p a r t i c u l a r - KPnc(u)

38.5,

i s c o m p a c t f o r e a c h compact s e t

2P.j IU)

K C U.

a E U w i t h a 9 ~ p n f u j then t h e r e is f E Pd I U I s u c h t h a t s u p f < 0 < f ( a ) . By Theorem 3 8 . 4 t h e r e is a decreasing K C Pbc(U) which c o n v e r g e s p o i n t w i s e t o f . Thus s e q u e n c e If,)

PROOF.

If

m

K c

u

ir

E

u

:

f n f x ) < 03

n=l and w e c a n f i n d

~PnciW

n

E W

such t h a t

s u p f n < 0 < f n ( a ) . Thus

K

a

and t h e p r o o f i s c o m p l e t e .

T o c o n c l u d e t h i s s e c t i o n we improve Lemma 3 8 . 1 as f o l l o w s .

U be a p s e u d o c o n v e x o p e n s e t i n en, l e t K b e a compact s u b s e t of U, and l e t V be a n o p e n n e i g h b o r h o o d i n U. Then t h e r e e x i s t s a s t r i c t l y plurisubhamonic 3 8 . 6 , THEOREM.

Of

KPb(U)

function

f

E

Cm(Ul

such t h a t :

is compuct f o r each c

The s e t

(b)

flz) < 0

f o r every

z E K.

(c)

flzl > 0

f o r every

z E .!/ \ 8.

E

U : f(zl

c}

By Lemma 3 8 . 1 t h e r e i s a f u n c t i o n

the conditions

u

Iz

5

(a)

PROOF. {z E

Let

( a ) , (b) a n d ( c )

: ~ ( z 5 ) c}

f o r every

c

of E

u E

EB.

Pdc(U) satisfying

t h e theorem. Set Kc = lR, and consider the functions

284

u

j

MUJ ICA

E

... )

Cm(U) (j = 0 , 1 , 2 ,

defined by

Observe that

If > 0 is sufficiently small then by Theorem 35.6 the function u i is strictly plurisubharmonic and u 0 and A 1 > 0 can 0 onaneighborhood be chosen so small that u o < 0 and u I t

of

K.

xtt) =

If we define

for

0

t < 0

and

X(t)

=!,e-'"ds m

for t > 0 then it follows from Exercise 1 5 . A that x c c (R;iR) and it is clear that x ' ( t ) > 0 and x " ( t ) > 0 for t > 0. Then it follows from Exercise 35.E that the function x o ( u3 . - j + 1 ) is plurisubharmonic on a neighborhood of K j and strictly plurisubharmonic on a neighborhood of j

E

mo

let

fj

E

Cm(U;lRI

Kj \

i j - ].

For

each

be defined by

where the coefficients c r > 0 will be chosen later. Observe that X o ( u r - r + 1 ) = 0 on Ki for r > i + 2, and hence m

fj = f j

Ki

on

j, k ? i

for

+

1.

Since

U =

U

Ki

we

con-

i=@

clude that

f

= Zim f

j

E

Cm(U;lR).

We shall inductively choose the coefficients c y S O that each fj is > u and is strictly plurisubharmonic on a neighborhood of K j . This is clear for f, = uo. Let j > 0 and suppose that fj-l is > u and is strictly plurisubharmonic on n

a neighborhood of

Set

S

= it

E

:

Itk['

=

1).

k=1

Since

X o (u

j

- j +

I ) is

> 0

and is strictly plurisubhamnic

285

PLURISUBHARMONIC FUNCTIONS 0

on a neighborhood of

K . \Kj-l , we can find J

c

j

0

such that

0

c

i

i n f { X o (u

I f j - l ( a ) I :a E K

E

> sup

j

- j + l/(a) j

\

:

a

gJ.- 1

E

Kj\ K j - l 1 0

+ sup

E

l u ( a ) I : a E K~ \ K

~ 3 -

and

Since and

fj

fj 0

K . \ Kj-l 3

.

x

- j + 1 1 , it follows that f . is j 3 is strictly plurisubharmonic on a neighborhood = fj-l

+

o (u

Using the induction hypothesis we conclude that

> u

of fj

is > u hood of

and f j is strictly plurisubharmonic on a neighborK as we wanted. j' Since f = f i + l on Ki for every i we conclude that f

is > u and f is strictly plurisubharmonic on all Hence f > u > 0 on U \ V . We also have that f = u 0 K. And since f > u it follows that

and hence the set { Z E U : f l z ) 5 c) B . The proof is now complete.

of 0

is compact for each

U. on

c E

NOTES AND COMMENTS

The results in Section 34 are straightforward generalizations of the corresponding results for subharmonic functi.ons. Theorem 35.7 is due to P. Lelong, who obtained the result for finite dimensional spaces in P . Lelong [ 11 and for infinite dimensional spaces in P . Lelong [ 2 1 . Theorem 36.1 on separate analyticity is due to F. Hartogs [ 1 1 , and its generalization to Banach spaces, Theorem 36.8, is due to M. Zorn [ 1 ].Theorem

~

286

MUJ I CA

1 . The p r o o f o f Theorem 3 6 . 5 [ 1 1 . Theorem 3 6 . 4 can be found i n t h e book o f E. H i l l e a n d R . P h i l l i p s [ 1 1 . T h e o r e m 3 6 . 9 , due t o M . Matos [ 1 1 , e x t e n d s a n o l d r e s u l t o f A. Taylor 3 6 . 5 i s a l s o d u e t o M.

Zorn

[ 2

g i v e n h e r e i s due t o P . Noverraz

[ 1]

. Our

p r e s e n t a t i o n i n S e c t i o n 3 7 f o l l o w s e s s e n t i a l l y a paper

o f H . Bremermann

[ 2 ]

,

who was t h e f i r s t t o c o n s i d e r plurisub-

harmonic f u n c t i o n s and pseudoconvex domains i n i n f i n i t e dimens i o n a l s p a c e s . Another r e f e r e n c e f o r S e c t i o n 3 7 i s t h e book of P . Noverraz

[ 3

F e r r i e r and N .

1 . Theorem

Sibony

3 8 . 4 i s e s s e n t i a l l y due t o

J.

[ 1 1 , who o n l y s t a t e d t h e r e s u l t f o r

= Cn. M. E s t g v e z and C. Hervgs

[

1]

Theorem 3 8 . 6

[ 31

and w i l l p l a y a f u n d a m e n t a l r o l e i n t h e s o l u t i o n of

2

from t h e

e q u a t i o n i n pseudoconvex domains.

book

of

Proposition

38.5.

-

E

n o t i c e d t h a t t h e r e s u l t is

v a l i d f o r Banach s p a c e s , and u s e d i t t o e s t a b l i s h

is taken

P.

L.

Hormander the

CHAPTER

5

THE

IX

EQUATION IN PSEUDOCONVEX DOMAINS

39. DENSELY DEFINED OPERATORS IN HILBERT SPACES T h i s s e c t i o n i s d e v o t e d t o t h e s t u d y o f densely defined operat o r s i n H i l b e r t s p a c e s . T h i s i s a n i n d i s p e n s a b l e tool t o s o l v e -

a

the

e q u a t i o n i n pseudoconvex domains.

39.1. DEFINITION. L e t E and F be H i l b e r t s p a c e s , and let be a l i n e a r o p e r a t o r whose domain i s a s u b s p a c e D + F A

A : DA

of

range of

A,

The o p e r a t o r in

NA

E . W e s h a l l d e n o t e by

a n d by

t h e k e r n e l of

by

A,

E . The o p e r a t o r

the

t h e g r a p h o f A,that is

GA

i s s a i d t o be d e n s e l y d e f i n e d i f

A

HA

i s s a i d t o be c l o s e d i f

A

DA i s

dense

GA i s c l o s e d i n

E x F.

39.2. DEFINITION.

and

E

Let

F

be H i l b e r t s p a c e s , and l e t A

b e a d e n s e l y d e f i n e d o p e r a t o r from t h e subspaces of a l l

y

E

E

into

DA* d e n o t e

F. L e t

f o r which t h e r e e x i s t s

F

y*

E

E

such t h a t (As

Since

DA

1

I y*i

y l = is

is dense i n

c a l l e d t h e a d j o i n t of

the vector

E,

i t e x i s t s . The o p e r a t o r A.

f o r every

A*

:

DA*

-+

E

y*

DA

.

i s unique

d e f i n e d by

Thus w e h a v e t h a t

287

x E

whenever

A*y = y * i s

2aa

MUJ I CA

39.3.

Let

PROPOSITION.

and

E

be H i l b e r t s p a c e s , and l e t A

F

be a d e n s e l y d e f i n e d o p e r a t o r f r o m operator PROOF.

to

yo

= (x

I

be a s e q u e n c e i n

L e t (y,)

such t h a t ( y

DA*

I yo)

= x0

= (x

.

I

Let

PROPOSITION.

and ( N A * I

1

-

= RA

x

x o ) for every

6 and

F

be a d e n s e l y d e f i n e d o p e r a t o r f r o m

DA.

E

Hence

Yo

DA*

. t h e n ( A x I y ) = (x I A*y)

y

= (x

and t h e r e f o r e

10)

= 0

y E (RA)

y E DA*

Hence

x

f o r every

versely, i f

39.5.

converges

be H i l b e r t s p a c e s , and l e t A 1 E i n t o F . Then NA* = ( R A )

i s a n immediate consequence. I f

.

)

I t s u f f i c e s t o p r o v e t h e f i r s t i d e n t i t y . The s e c o n d o n e

PROOF.

DA

n

F a n d (A*y ) c o n v e r g e s t o x o i n E . Then ( A x I y n ) n A * y n ) f o r e v e r y x E DA and n E a . L e t t i n g n + m w e

in

A*yo

39.4.

Then t h e a d j o i n t

F.

i s closed.

A*

get that (Ax and

into

E

Let

THEOREM.

1

t h e n (Ax

1

and

NA*

y

E

1

(RA)

.

t h a t is,

y

E NA*.

be H i l b e r t s p a c e s , and l e t

F

Con-

y t = 0 = (x,Q) for every x E

A*y = 0,

and E

E DA,

E

A

be

a closed, densely defined operator from E i n t o F . Then t h e i s a closed, densely defined operator from F i n t o a d j o i n t A* E , a n d A** = A . PROOF.

W e a l r e a d y know t h a t

A*

i s d e n s e l y d e f i n e d and t h a t Observe t h a t

E x F

and

A**

F

x

E

i s c l o s e d . T o show t h a t

A*

= A

w e p r o c e e d as f o l l o w s . are isometric H i l b e r t s p a c e s

under t h e canonical i n n e r products, t h a t i s ,

and

for

tors

xl, x

2

S :

E

E

E x F

and -+

yl, y 2

F x E

E

and

F.

Consider t h e isometric opera-

T : F

x

Z

-+

B

x

F

d e f i n e d by

5

THE

= I- y,x)

S(x,y)

Iy)

t h e e a u a t i o n (Ax

1

Ax,x)

and

- i dE

T OS =

Observe t h a t

((-

EQUATION I N PSEUDOCONVEX DOMAINS

and

x F

= I- x , y l .

T(y,xJ

= - id

S o T

= (x I y * ) i s e q u i v a l e n t t o

(y,y*l) = 0

Since

E .

the

equation

w e see t h a t

(39.1)

= (SGAl

GA*

1

.

S i n c e GA i s c l o s e d and S i s a n i s o m e t r y w e g e t t h a t = SGA = SGA . S i n c e T i s a n i s o m e t r y i t f o l l o w s t h a t 1 (TGA*) = TSGA = GA

(39.2)

yo

Let every

E

therefore sequently A

and

that

(DA*ll.

T GA**

E

ITGA*I

= A ( O ) = 0 . T h u s DA*

YO

1

1

.

is orthogonal t o

Then 1 0 , y o ) Thus ( O , y o l

y E DA*.

IGA*l

= GA,

for

A*y,y)

(-

by (39.2),

i s dense i n

F

and

and

con-

e x i s t s . A p p l y i n g ( 3 9 . 1 ) w i t h A* i n place of i n p l a c e of S w e o b t a i n , w i t h t h e h e l p o f (39.2) , 1 = ( T G A * ) = G A . Hence A** = A a n d t h e proof is

A**

complete. From Theorem 3 9 . 5 a n d P r o p o s i t i o n 39.4 w e o b t a i n : 39.6.

Let

COROLLARY.

E

and

F

be a c Z o s e d , d e n s e l y d e f i n e d o p e r a t o r f r o m E

39.7.

THEOREM.

Let

E

and

a c l o s e d , denseZy d e f i n e d o p e r a t o r from s u r j e c t i v e i f and onZy i f i t s a d j o i n t v e r s e . In this c a s e PROOF. {O}

Assume f i r s t t h a t

and

l e t (y,)

RA*

A*

I (Ax I y n ) I

=

E

A*

into

A*yn)

I5

A

be

Then A

is

i s cZosed i n E . A

i s s u r j e c t i v e . Then

DA*

NA*

= IRA)

1

=

is continuous

such t h a t t h e sequence

E . Then t h e r e i s a c o n s t a n t

1 (x 1

F.

3

Then

h a s a continuous i n -

i s i n j e c t i v e . To show t h a t (A*)-'

be a s e q u e n c e i n

i s bounded i n

into F.

be H i l b e r t s p a c e s , and l e t

F

A

be H i l b e r t s p a c e s , and l e t

c > 0

such

IIx I1 I I A * ~ , I1 5 c IIxI; f o r e v e r y

(A*yn)

that LL' E

DA

MUJ I CA

290

n E W.

and

Since

RA

= F we conclude t h a t t h e sequence ( y n )

i s weakly bounded, and t h e r e f o r e norm bounded, by t h e p r i n c i p l e o f Uniform Boundedness. T h i s shows t h a t ( A * ) - ' C o n v e r s e l y , assume t h a t A* there is a constant

yo

Let

2

I1 y II

(39.3) E

c > 0

h a s a c o n t i n u o u s i n v e r s e . Then

such t h a t

c IIA* y I1

1 ly I

F. Then

and h e n c e t h e mapping

yo)

A*y

f o r every

I 5

I

I xo) =

-

E DA**

DA

(y

I yo)

W e s t i l l h a v e t o show t h a t

be a sequence i n in F

E.

.

E

D

A* con-

By t h e R i e s z Representation

xo E FA, f o r every

y o = A**xo = A x o J

and

for every y

is a w e l l defined

yo)

RA*

Theorem t h e r e i s a u n i q u e vector and (A*y

y E DA*.

c IIA*y 11 IIyoII

ly

+

t i n u o u s l i n e a r f u n c t i o n a l on

c IIy, II

i s continuous.

RA*

Ilzoll

such t h a t

y E DA*

.

Hence

5 xo

proving t h a t

F = R

i s closed i n

E . L e t (y,)

A'

such t h a t IA*yn)

converges t o a p o i n t x

I t f o l l o w s from ( 3 9 . 3 ) t h a t f y , )

i s a Cauchy sequence i n

DA*

and hence converges t o a p o i n t

y

follows t h a t

E DA*

and

y . Since

A*y = x . Hence

is closed

A*

and

x E RA*

it the

proof i s complete. T h i s theorem w i l l b e f r e q u e n t l y u s e d i n t h e f o l l o w i n g form. 39.8.

COROLLARY.

Let

E

and

F

be H i l b e r t s p a c e s ,

c l o s e d , denseZy d e f i n e d o p e r a t o r from E i n t o be a c l o s e d s u b s p a c e of F s u c h t h a t R A C Fo i f and o n l y i f t h e r e i s a c o n s t a n t

IIy II

5

c IIA*y II

I n t h i s c a s e f o r each

IIzII < c IIy II

and

for. e v e r y

c > 0

.

F,

Zet

A

and l e t Then RA

be a O - FFo

such t h a t

y E D A x n F~

.

x E D s u c h that A 1 Ax = y . F u r t h e r m o r e , f o r e a c h z E ( N A l there Fo s u c h t h a t II y II 5 c IIz II and A*y = x . y E Fo

there exists

exists

y E DA*

PROOF.

The f i r s t a s s e r t i o n f o l l o w s from Theorem 39.7

t o t h e H i l b e r t spaces

E

and

Fo. The s e c o n d a s s e r t i o n

applied follows

THE

7

EQUATI ON I N PSEUDOCONVEX DOMAINS

29 1

from t h e p r o o f o f Theorem 3 9 . 7 . F i n a l l y , by C o r o l l a r y 3 9 . 6 and

, and

the

last

Bn. Show t h a t

for

each

= RA* IDA*

Theorem 3 9 . 7 w e h a v e t h a t ( N A l l

fl F o

a s s e r t i o n i n t h e c o r o l l a r y follows.

EXERCISES 39.A.

U be a n open s u b s e t o f

Let

multi-index

a t h e operator

a"/aaa

i n t h e sense of d i s t r i b u -

t i o n s d e f i n e s a closed, densely defined

operator

in

L2fU).

Determine i t s a d j o i n t .

4 0 . THE

7

OPERATOR FOR L 2 DIFFERENTIAL FORMS

I n t h i s s e c t i o n we s h a l l extend t h e

ern

s p a c e s of

3

o p e r a t o r from

the

d i f f e r e n t i a l forms t o c e r t a i n s p a c e s o f L2 d i f -

f e r e n t i a l forms. Our aim i s t o a p p l y t h e H i l b e r t

space

niques s t u d i e d i n Section 39 t o t h e s o l u t i o n of t h e

tech-

7 equation

i n pseudoconvex domains. T o b e g i n w i t h w e i n t r o d u c e some n o t a t i o n . L e t U b e a n open

8.Given

s u b s e t of and

g =

Z g, J,

dz,

-

A dz,

in

K

R

P4

-

f =

Z fJKdzJ h dzK J, K (.!I w) e shall set

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

and

Next w e i n t r o d u c e s e v e r a l s p a c e s o f d i f f e r e n t i a l f o r m s . I f

i s a n open s u b s e t of t h e n w e s h a l l d e n o t e by L2 ( U l pq A dz, ( r e s p . L P9 2 (U, Z o c l ) t h e v e c t o r s p a c e o f a l l f = Z f d z JK J in Rpq(U) s u c h t h a t e a c h f,, belongs t o L 2 ( U l (reap.

U

I

2 L ( U , l o c i ) . If

(J3

q E

C IU;lRj

then t h e space

i n a s i m i l a r manner. Observe t h a t

L 2 IU,q) i s P4

f U , p l is defined P4 a H i l b e r t space

L2

MUJ ICA

292

f o r t h e inner product

The f u n c t i o n

i s called a w e i g h t f u n c t i o n . W e s h a l l set

9

2

2

By E x e r c i s e 16.C, L l U , Z o c )

with

a n d whence i f f o l l o w s t h a t

cp E C m ( U ; l R ) ,

t h e u n i o n of t h e s p a c e s

40.1. DEFINITION. (a)

Let

D,

-

a

:

(b)

Do

+

Let

L

2

P4

& a zj

ag. 3

a

Crl(Lr) C D2

-

a

E

D ~ .T h e n

Let

-

af

E

Dl

L 2 ( U , Z o c l such

f in

L lU, Z n c ) €or j = 2 ,

: Dl

-f

g

..., n.

- ag

asj

-

2 = sg,. d z . i n Lol(U,loc)

j

u'

belongs to

L2 (U,Zoc)

L o2 2 f U , Zoc) be d e f i n e d b y

m

c (U)

and when r e s t r i c t e d t o

coincides with the usual definition.

a n d when r e s t r i c t e d t o

and t h e o l d d e f i n i t i o n o f 40.2. PROPOSITION.

ask

C m ( U ) C Do

t h e new d e f i n i t i o n o f

En.

2

belongs to

D l be t h e s u b s p a c e o f a l l

I t is c l e a r t h a t

L~qiU,Zoc) is

9 E CmIU,iR).

be d e f i n e d b y

L i I I V , Zoc)

for j , k = l , . . .,n. L e t

f

with

be t h e s u b s p a c e of a l l

such t h a t t h e d i s t r i b u t i o n

Likewise,

fU,cp),

U be a n open s u b s e t of

Let

that the distribution Let

i s t h e u n i o n o f t h e s p a c e s L IU,P),

W

co,(U),

the new

-

a

coincide.

U b e an o p e n s u b s e t o f a n d -2 a f = 0.

gn,

and

let

THE

PROOF.

If

f o r every

3

then

f E Do

for a l l

-

af

= Z g

-

a;,

a;

j and

k.

j

a 'f aPk a 2

Hence

j

- a 2 faz

j

=-.kfa;

E L2(U, Zocl

i

= o

az -2

a

and

E Dl

g

j

-

-

af

-

. dz ,where 3

i

j. Hence

-a -g j- -

293

EQUATION I N PSEUDOCONVEX DOMAINS

f = 0.

I n t h e n e x t s e c t i o n w e s h a l l p r o v e t h a t whenever pseudoconvex open s e t i n solution

f

Cn

f o r each

E Do

then t h e equation g

-

af =

U

is

g

has a

a

ag = 0 . I n order

such t h a t

Dl

-

S e c t i o n 39 w e

t o apply t h e H i l b e r t space technique s t u d i e d i n

s h a l l r e d u c e t h i s problem t o a s i m i l a r problem f o r t h e H i l b e r t

~ p (u,2 P)~ .

spaces

40.3. DEFINITION. ct,

B,

co

y E C (U;R)

(a)

Let

Let

DA

be t h e subspace of a l l

that the distribution Let

A

: DA

(b) 2 LO1(U,B) 2

Lo2(U,y)

-+

Let

U b e a n open s u b s e t o f

L2 (U,8) 01

DB

az j

in

f

such

L2(U,ctl

L 2( U , 8 ) f o r

belongs t o

j = l , ..., n.

b e d e f i n e d by

be t h e subspace of a l l

j, k = 1,..., n . L e t

g

=

P

L:

3

g . d: 3

ag j

such t h t t h e d i s t r i b u t i o n

€or

let

and

Cn,

be t h r e e w e i g h t f u n c t i o n s .

-- -

belongs t o

L:2(U,y)

bedefined

az

aGk

B

in

j

: DB

-+

by

If

U i s an open s u b s e t of

UP 4 ( U ) t h e v e c t o r s p a c e o f a l l such t h a t each

f,,

belongs t o

gn f

then we s h a l l

= C fJKdZJ P(U).

denote

A d,TK i n

R

P4

by

IU)

I t follows fromExercise

294

MUJ I CA

P ( U ) i s dense i n

16.A t h a t cp,

2 L (lJ,cp)

D

and whence it f o l l o w s t h a t

f o r each weight f u n c t i o n

P4

f o r e a c h weight

function

i s dense i n

Liq(U,cp)

(U)

q.

40.4. PROPOSITION. L e t U b e an o p e n s u b s e t of E Cm(U;R?) be t h r e e w e i g h t f u n c t i o n s , and l e t A Then:

operators introduced i n Definition 40.3. A i s a closed,

(a)

Cn, l e t a, f3,y and B be t h e

densely defined operator from

L'(u,~I

(u,61.

into

B is aclosed,

(b) 2 Lo2

into

fu,

2

d e n s e l y d e f i n e d o p e r a t o r f r o m L o I ( U ,@I

y).

PROOF. ( a ) I t i s clear t h a t

in

L2(U,a).

in

DA

D I U ) C DA

To show t h a t A

i s dense

DA

i s c l o s e d l e t (f")be a sequence f in L2(U,a) and ( A ? )

such t h a t ($") c o n v e r g e s t o

converges t o g

in

inequality t h a t

L ~ I ( U , O I . I t f o l l o w s from t h e Cauchy-Schwarz

If")

converges t o

f o l l o w s from P r o p o s i t i o n 1 7 . 6 t h a t

af/azj

and hence

P'tU)

in

for

j = I,

in

f

Z)'(U),

(afm/ai

..., n .

J

and

then

it

converges

.)

to

since

On t h e o t h e r hand,

J

converges t o

,..., n .

j = 1 Thus f

in

gj

Hence

E DA

and

L2fU,BI,

-

and t h e r e f o r e i n

af/azj = E L2(U,p) gj Af = g .

for

V'iVI, j = I,

for

. . . ,n.

The proof o f ( b ) i s s i m i l a r t o ( a ) and i s l e f t t o the reader as an e x e r c i s e . And t o prove ( c ) it s u f f i c e s t o r e p e a t the proof of P r o p o s i t i o n 4 0 . 2 . The p r e c e d i n g p r o p o s i t i o n t e l l s u s t h a t would l i k e t o prove t h a t tions

a,p,y.

RA = N B

RA C NB

for suitable

weight

func-

To achieve t h e d e s i r e d conclusion w e i n t e n d t o

a p p l y C o r o l l a r y 3 9 . 8 . Thus w e want t o f i n d a c o n s t a n t such t h a t

we

and

c

>

0

THE

(40.1)

Ilgll

a -<

3

EQUATION I N PSEUDOCONVEX DOMAINS

f o r every

cIIA*gIIa

295

g E DA*

NB

C l e a r l y i f w i l l be s u f f i c i e n t t o f i n d a c o n s t a n t

c

>

. such

0

that 2 lIgllB 5 e2 (IIA*gll:

(40.2)

i

2 llBglly)

g E DA*

f o r every

.

DB

we s h a l l prove t h a t f o r s u i t a b l e w e i g h t f u n c t i o n s a , 6 , y, the space P o , ( U ) i s a d e n s e subspace o f DA* n DB w i t h r e s p e c t t o t h e

T h i s w i l l be a c h i e v e d i n t h e n e x t s e c t i o n . I n t h i s s e c t i o n

norm 111 g 111 = IIg I1

a

+ IIA*gIIa

i IIBg

1 I

Y’

and t h e n it w i l l b e s u f f i c i e n t t o p r o v e ( 4 0 . 2 ) f o r e v e r y

g

Pol(UI.

pre-

Before proving t h i s d e n s i t y r e s u l t w e g i v e t h r e e

p a r a t o r y lemmas f u r n i s h i n g i n f o r m a t i o n a b o u t t h e o p e r a t o r s B

and

E

A,

A*.

U b e a n o p e n s u b s e t of Cn, l e t a, 3! E Cm(U;W) b e two w e i g h t f u n c t i o n s , a n d l e t A b e t h e o p e r a t o r i n t r o d u c e d 40.5.

LEMMA.

Let

i n D e f i n i t i o n 4 0 . 3. Then then

A*g

Z g . dz

j

J

j

C DA*

.

-

g = B g . dz 3 j Li

If

E

DA*

g

=

i s g i v e n by t h e f o r m u l a

To show t h a t

PROOF.

Do,(U)

E Q,,(U).

Dol(U)

DA*

let

e ’)dl =

-

C

Then

a

-

(f, e

g E DA*

a

L: j

3

T h i s shows t h a t

and l e t

f E DA

and t h a t

A*g =

- ea Z j

-f3 2 ( g . e )la. az J

j

a a2

j

296

MUJ I CA

To prove that this formula is valid for every

,

we proceed similarly. Indeed, for every we have that DA*

and this implies that A*g wanted. 40.6. LEMMA.

Let

a - e -a = - jB i az

U be a n o p e n s u b s e t of

three weight functions, and l e t A

and

B

in g = B g.ds j~ j f E D l U ) C DA

( g j e-’),

let

Cn,

we

as

a, B,

y

be

be t h e o p e r a t o r s in-

troduced i n D e f i n i t i o n 40.3.

A(lpfl

(b)

PROOF.

for

If

cp E

V t U ) and

= cpAf + f%p. g

E

DB

then

lpg

E

DB

and

(a) By Proposition 17.8

j = I,

..., n ,

and the desired conclusion follows.

The proof of (b) is similar to (a) and is left as an exer3 = Z g.dz E DA*. cise. To prove (c) let f E D A and let 3 j Then, using (a) we get that

THE

5

EQUATION I N PSEUDOCONVEX DOMAINS

297

and the desired conclusion follows.

f =

If

I: f J , d z J A dzK K

E

J,

I: (fJ, * 9 1 d z J A d z , ,

P4

which belongs to

J, K

=

f * 9

f * p the differential form

then we shall denote by

-

9 E VCfj(0;S)l

L 2 tU,Zoc) and

Cm (U61, by P9

Proposi-

If the support of f is a compact subset of U and m 6 > 0 is sufficiently small then f * 9 E vp,lU). As a direct consequence of Proposition 17.14 we obtain the following retion 16.4.

sult. 40.7.

Let

LEMMA.

U be an o p e n s u b s e t of

be t h r e e weight f u n c t i o n s ,

C"(U;iR)

Cn,

let

a , f3, y

A

and

B

and Z e t

be

E

the

operators introduced i n D e f i n i t i o n 40.3.

(a)

6 > 0

with

= IAf)

s u f f i c i e n t l y small, then

f*p

E

D t U l and A(f

* 9)

*v.

(b) with

9 E D(B(O;6)),

h a s compact s u p p o r t , and

f E DA

I f

g E DE

If

6 > 0

h a s compact s u p p o r t , and

s u f f i c i e n t l y small,

then

9

E

D(E(O;6)),

g * 9 E Do,(U)

and B ( g * p )

= (Bgj * p . n

40.8. PROPOSITION. be a sequence i n

i , and large

qi

i. Let

1

U be an o p e n s u b s e t o f 47

PfU) s u c h t h a t

0

5 ni 5

1

on

.

00

Let

(n;)i.l

U for

every

o n e a c h compact s u b s e t of U for a l l s u f f i c i e n t l y

a , 6, y

2 laqil 5 e D o l ( U i i s dense in that

Let

m

E C

and DA*

(U;B)

b e t h r e e w e i g h t f u n c t i o n s such

f o r every i. w i t h r e s p e c t t o t h e norm

< eY-' IaqiI2 -

n DB

Then

298

MUJ I CA

111 g 111

(40.3)

= IIg Ilg + IIA*gllcl + IlBgII Y

We shall proceed in two stages. We shall first prove that each g E DA* r\ DB can be approximated by members of

PROOF.

with compact support. Next we shall prove that each g DA* Dg with compact support can be approximated by members of D o I I U ) . DB

DA*

(a)

Let

g

= I: g ~j . dz

E

DA*

n DB

j and we shall prove that

by Lemma 40.6,

.

Then

nig

E

DA *

1-1

DB

(40.4)

It follows from the Dominated Convergence Theorem that ZimII nig -gll B i+ m

= 0,

Z i m IloiA*g

j - r m

to prove ( 4 0 . 4 )

- A*g IIcL =

0

and

IIni

Zim

j + m

Bg

-

Bg 1 I

Y

= 0. Thus

it is clearly sufficient to prove that Zim 1 I A* f n i g )

(40.5)

- niA*gll

cl

= 0

i + m

and (40.6)

By Lemma 40.6,

-

A*(nig)

niA*g

= - e

a-8

g

a

ani .

Hence IA*(vig)

-

qiA*gI2 e

<

2 e-a t!

2

5 lgl e

-6

and ( 4 0 . 5 ) follows from the Dominated Covergence Theorem-Onthe other hand, again by Lemma 40.6, m i g )

-

niBg

= ani A g =

k,j

a qi gjdzk a:,

A di

Li

THE

7

E Q U A T I O N I N PSEUDOCONVEX DOMAINS

299

Hence

Thus

and another application of the Dominated Convergence

Theorem

shows (40.6). Thus (40.4) has been proved.

-

g = Z g.dz be a member of DA* n D B with 3 j j compact support. By Lemma 40.7, g * p 6 E P o , ( U ) if 6 > 0 is

(b)

Now let

sufficiently small, and we shall prove that

Zim Ill g

(40.7)

*

p6

-

g 111 = 0.

6-tO

lim I I g * p 6 - g l l

If follows from Exercise 16.R and Lemma 40.7 that

6-+0

= 0

and

lim I I B ( g

6+0

*

p6)

suffices to prove that

-

BgII

Y

= 0 . Thus to show

(40.7)

6

it

MUJICA

300

NOW, by Lemma 40.5, A * can be decomposed as a sum where

A*

= S + T,

Thus to show (40.8) it suffices to prove that

and

Since clearly S ( g x p 6 ) = (Sg) cise 1 6 . A . On the other hand,

x p6,

(40.9)

follows from

Exer-

Hence

and (40.10) follows from Exercise 1 6 . A . This shows ( 4 0 . 7 ) the proof of the proposition is complete.

41. L~ SOLUTIONS OF THE

and

-

a

EQUATION -

In this section we solve the a equation in the sense of distributions for L 2 differential forms. The key result is the following. m

41.1. PROPOSITION. L e t U be an o p e n s u b s e t o f d'. Let (nilizl be a s e q u e n c e in D(U) s u e h that 0 5 ni 5 1 on U f o r e v e r y i, and ni E 1 on e a c h compact s u b s e t o f U for aZZ s u f f i c i e n t l y l a r g e i. S u p p o s e t h e r e a r e functions c p , $ E C m ( U ; l R I sueh t h a t

THE

\%nil2 5 e'

f o r aZi!

7

EQUATION I N PSEUDOCONVEX DOMAINS

on U f o r every i and

a E U

d e f i n e d by

a

and

=

Ip

-

b

E

En.

a , P, y be t h e we
Let

6 = 9

2$,

-

$,

y = p,

and l e t

be t h e o p e r a t o r s i n t r o d u c e d i n D e f i n i t i o n 4 0 . 3 .

For e a c h

g

E

there exists

NB

A f = g . For e a c h

and

ilgll,

that

30 1

5 ilf (la

f E DA

1 f E (NAl

and

A*g

A

and

II f lla

5

B

Then

such t h a t

t h e r e e x i s t s g E DA* n DB

IIg II

,

such

= f.

To prove t h i s p r o p o s i t i o n w e s h a l l need t h e f o llo win g l e m -

ma.

LEMMA. L e t U be a n o p e n s u b s e t of En, and l e t 9 E Cm(U;iR). For e a c h j = 1 n , Z e t 6 : D(U) -+ V t U i b e d e j f i n e d by 41.2.

,...,

Then f o r e v e r y

f, g

E

V ( U ) we h a v e

that.

and

An i n t e g r a t i o n by p a r t s y i e l d s t h e f i r s t i d e n t i t y . And t h e second one follows r e a d i l y from t h e d e f i n i t i o n o f

".

The

d e t a i l s a r e l e f t t o t h e reader a s an e x e r c i s e .

PROOF O F PROPOSITION 4 1 . 1 .

I t i s s u f f i c i e n t t o p r o v e (41.2)

for

MUJ I CA

302

every

g

E

DOIIUI,

v a l i d i t y of (41.2)

f o r Proposition 40.8 w i l l

then

g E DA* n D B . L e t

f o r every

imply t h e g = B g. di 3

j E

D,,IU).

j

Then

hence

and t h e r e f o r e

(41.3)

On t h e o t h e r h a n d , by L e m m a s 40.5 a n d 4 1 . 2 w e h a v e t h a t A*g =

ag

(2 a @ ) = -

- e a-@ j

az

j

g j az

j

e-*

z: 16.9. + i

3 3

*I,

g j

az

i

and hence

I t f o l l o w s from t h e i n e q u a l i t y

II Lc + y 1 1 5 ~ 2 II x 1t2

2

+ 2 II y II

that

THE

7

E Q U A T I O N I N PSEUDOCONVEX DOMAINS

303

and t h e r e f o r e

(41.4)

ju

C 6 . g . ?-. i k g k e-Ip dX 5 2 I I A * g l l a2 j,k

''

I t f o l l o w s from ( 4 1 . 3 ) and ( 4 1 . 4 )

+ 2

I

I

lgl

l a + /2

2 -

e-'dX.

U

that

Now, u s i n g Lemma 4 1 . 2 w e h a v e t h a t

Hence s u b s t i t u t i n g i n t o ( 4 1 . 5 ) y i e l d s t h e i n e q u a l i t y

and u s i n g ( 4 1 . 1 ) w e g e t t h a t

2llglI;

=

2

iU T h i s shows ( 4 1 . 2 ) g

E

PA*

for every

g E Do2(U)

a n d t h e r e f o r e f o r every

D B . The r e m a i n i n g a s s e r t i o n s i n t h e p r o p o s i t i o n f o l -

low from C o r o l l a r y 3 9 . 8 .

41.3. THEOREM. f o r each

g

E

L e t U be a pseudcconuex open s e t i n C n . Then 2 2 L o I (U, 20c) s u c h t h a t %g=O t h r e e z l s k s f E L (U,locJ

MUJ I CA

304 -

af =

such t h a t

g.

B e f o r e p r o v i n g t h i s t h e o r e m w e g i v e t h r e e a u x i l i a r y lemmas. 41.4.

for a l l sufficiently

U

ni

V l U l such t h a t

sequence in

m

Q?.Let (nilizl

U be an open s u b s e t o f

Let

LEMMA.

large

5 e'

lanil'

Cm(U;R)

such t h a t

PROOF.

Let lUjlj=,

be

a

o n e a c h c o m p a c t s u b s e t of

1

i . Then t h e r e i s Q f u n c t i o n on U f o r e v e r y i .

$ E

m

be a n i n c r e a s i n g s e q u e n c e

2

= sup sup laniir! 1 j i E B x g ~ j f o r all s u f f i c i e n t l y

compact o p e n s u b s e t s of U which cover U. Let f o r each

j

E

-

ani

Since

W.

on

0

relatively

of

Uj

l a r g e i w e see t h a t c j < m f o r e v e r y j . I f w e s e t t h e n by E x e r c i s e 1 5 . C t h e r e i s a f u n c t i o n T E c m ( U ; l R l that

> c

T

U . \ Uj-I

on

i

3

uo

= 4 such

j E iX. Then t h e

f o r every

func-

= 2og-i h a s t h e r e q u i r e d p r o p e r t i e s .

11)

41.5.

Let

LEMMA.

f

:

lR

-+

2R

b e a f u n c t i o n w h i c h is

above on e a c h i n . t e r v a l of t h e form increasing function for

t < 0

and

(-m,b 1 . Then t h e r e

g E Cm(R;1R) s u c h t h a t

gltl

bounded

t

f l t l for every

g(tl

;'s

an

i s constant

IR.

E

L e t c = s u p { f ( t l : t 5 j} f o r e a c h j E B . O b s e r v e j t h a t (c.) i s a n i n c r e a s i n g s e q u e n c e . By Lemma 1 5 . 1 f o r e a c h j

PROOF.

3

LV t h e r e i s a n i n c r e a s i n g f u n c t i o n

E

9 . h ) = 0 3

t 5 j

for

-

1

and

9

j ~ . ( t l= 1 3

E

such t h a t

CmlB;lRl

for

t

2 j.

Let

IR b e d e f i n e d b y

g : lR

+

Since

9.lt) = 0

for

3

t 5 n

and

j 1. n + 1

w e l l d e f i n e d , i n c r e a s i n g and o f class

= e l . And i f

n n-I

-

1

5 t 5 n, w i t h

Ca. I f VI

E

B/,

w e see t h a t g i s

t 5 0 then

then

g(t)

THE

Thus g

3

EQUATION I N PSEUDOCONVEX DOMAINS

305

has t h e required p r o p e r t i e s .

R X? be a f u n c t i o n w h i c h is bounded Then t h e r e i s a above o n e a c h i n t e r v a l o f t h e f o r m ( - m , b 1 . 41.6. LEMMA.

Let

f :

+

convex, i n c r e a s i n g f u n c t i o n

g

E Cm(lR;lR!

g ' ( t l 1 f ( t l f o r every

t

E 2,

and

g(tl 2 f ( t )

such t h a t

By Lemma 4 1 . 5 t h e r e i s a n i n c r e a s i n g f u n c t i o n

PROOF.

f o r every

( - m,0

]

t

Since

B.

E

the derivative

I

form ( - m , b ] .

t

Then f o r e a c h

t - 0 and q(t) > i s c o n s t a n t on t h e i n t e r v a l

Ip

i s bounded on e a c h i n t e r v a l of t h e

9

Then anot.her a p p l i c a t i o n of Lemma 4 1 . 5 y i e l d s a n $ E Cm(R;R! s u c h t h a t

increasir.g function

f o r every

E

p l t l is constant f o r

C m ( B ; R )s u c h t h a t f ( t )

9

E W.

t

g

Let

w e have t h a t

W

6

be d e f i n e d by

E Cm(B;lR)

t

and

f ( t ) . Moreover, g ' f t ) = $(t)

g'lt) = $(t)

= $'(t) 2

0,

proving t h a t

g

0 and g " ( t l

as

we

ago = 0

be

i s convex and i n c r e a s i n g ,

wanted.

PROOF O F THEOREM 4 1 . 3 .

go

Let

E

g i v e n . With t h e h e l p o f P r o p o s i t i o n 4 1 . 1 weight functions

a, B , y

Cm(U;lR)

E

-

L i i (17, l o c ) with

shall

we

such t h a t

go

construct E

2 L o l (lJ,t31

and (41.6)

l l g l l ~5 IIA*gllf + IIBg 1 1 2

Y

m

Let (niliz7

be a s e q u e n c e i n

U for every

i t and

qi

1

f o r every

g

D(U) s u c h t h a t

E

D A x n DB 0

.

5 qi 5 1

on e a c h compact s u b s e t o f

U

on for

306

MUJ I CA

f. By Lemma 41.4 t h e r e

a l l sufficiently large E

)I

ge'

is

a

function

laniI2

Cm(U;Z?l such t h a t 5 e' on U f o r e v e r y i. S i n c e 2 E L o I ( U , l o c ) t h e r e i s a weight function u E CW(U;nZI such

that

ge'

L 2O I ( U , u l , t h a t i s , g

E

E

2 LO I(U,a

-

$1.

I n order

a p p l y P r o p o s i t i o n 41.1 w e s h o u l d f i n d a f u n c t i o n

9 E

to

Cm(U;B)

such t h a t

9 2 0

(41.7)

on

U

and

for all

a

E

and

U

b E gn. To f i n d s u c h

a

function

we

9

p r o c e e d a s f o l l o w s . By Theorem 38.6 t h e r e i s a s t r i c t l y p l u r i subharmonic f u n c t i o n (z E

5 t)

U : u(zl

such t h a t t h e set

u E Cm(U;Z?)

i s compact f o r e a c h

t h e r e is a s t r i c t l y p o s i t i v e function

f o r every

a

E

U

and

b

x

Cn. I f

E

t E 37. T

E

By Lemma 35.4

E Cm(U;lR)

Cm(Z?;lRl

=

Kt

such t h a t

is a

i n c r e a s i n g f u n c t i o n t h e n i t f o l l o w s from E x e r c i s e s

convex,

35.B

and

35. E t h a t

f o r every

a E

U

and

b

On. Thus i t i s s u f f i c i e n t t o f i n d a

E

convex, i n c r e a s i n g f u n c t i o n

(41.9) and

x

x o u

E

ern 2 u

W ; m ) such t h a t

THE

on

7 EQUATION IN

U, f o r then t h e function

PSEUDOCONVEX

x

9 =

(41.8). TO f i n d s u c h a f u n c t i o n w : 27 lR d e f i n e d by

x

DOMAINS

307

w i l l s a t i s f y ( 4 1 . 7 ) and

o u

v,

consider t h e functions

--f

v ! t ) = supa

w ( t l = sup

and

zf151J12 +

% t > infu,

for

v(tl = w(tl

and

=

T

0

t 5 i n f u.

for

t h e f u n c t i o n s v and w

are i n c r e a s i n g f o r

p l i c a t i o n o f Lemma 4 1 . 6 y i e l d s a convex, Ca(lR;lR)

f

such t h a t

lR. Then f o r e a c h

t E

x

Thus

Since

U

U

x

2)

t > infu, U increasing

x l t l 2 v ( t l and x ' f t l U w e have t h a t

an

ap-

function

w l t l f o r every

z E

s a t i s f i e s (41.9)

and ( 4 1 . 1 0 ) and

9

=x

o u

satisfies

-

( 4 1 . 7 ) and ( 4 1 . 8 ) . I f w e s e t a = 9 2+, 6 = cp - $ and y = q , 2 t h e n go E L O I ( U , B / , and ( 4 1 . 6 ) f o l l o w s from P r o p o s i t i o n 4 1 . 1 . 2 2 Hence t h e r e e x i s t s fo E L (11,a) C L f U , Z o c l s u c h t h a t af = 0

and t h e p r o o f o f t h e t h e o r e m i s c o m p l e t e .

SO

4 2 . Cm SOLUTIONS OF THE

5

EQUATION

In t h i s section we solve the

a

00

equation f o r C

differen-

t i a l forms.

42.1.

DEFINITION.

Let

U b e a n open s u b s e t o f en. F o r e a c h Wk ( U i t h e v e c t o r s p a c e o f a l l . f f

w e s h a l l d e n o t e by

k f JVo 2 L (U) whose d e r i v a t i v e s ( i n t h e s e n s e o f d i s t r i b u t i o n s ) of or-

2 f U l . L i k e w i s e , w e s h a l l denote by wk(U,Zocl 2 t h e v e c t o r s p a c e o f a l l f E L lU,Zocl whose d e r i v a t i v e s of 2 o r d e r 5 k b e l o n g t o L (U, Zoc). The s p a c e s o f d i f f e r e n t i a l der 5 k

belong t o

forms W k ( U ) and P4

L

W k ( U , Z o c l a r e d e f i n . e d i n t h e o b v i o u s way. P9

308

MUJ I CA

PROPOSITION.

42.2.

and l e t

U be a p s e u d o c o n v e x o p e n s e t i n

Let

g E W ~ l ( U , l o c )w i t h

2

L (U,locl

ag = 0 . T h e n e a c h s o l u t i o n

-

af = g

of t h e e q u a t i o n

Cn,

-

belongs t o

f E

Wk+l(~,Zocl.

T o p r o v e P r o p o s i t i o n 4 2 . 2 w e need t h e f o l l o w i n g l e m m a .

42.3.

LEMMA.

port. I f

Let

af/.az

E

j

f E L 2 ( Cn ) be a f u n c t i o n w i t h compact s u p 2 n L (C ) f o r j = 1 n then f E W1(blnl.

,...,

Two i n t e g r a t i o n s b y p a r t s immediately show t h a t

PROOF. (42.1)

f o r every

D(Cnl.

rp

Applying ( 4 2 . 1 )

with

= f

rp

*

p6

-

f

*

p,

gives that

2 af/aij E L tCnl,

Since when

6,

E

-+

0

there exists in

L

( 32: af

2

E

*

s u c h t h a t ($

L2(Cn)

*

p6)

*

"6)

, . .. ,n

converges t o g

p6)

i

I t f o l l o w s from t h e Cauchy-Schwarz

(Cnl.

j= 1

Hence f o r e a c h

gj

(6

Converges t o

gj

converqes to

af/azj

J

Thus

t h e l a s t w r i t t e n i n t e g r a l t e n d s t o zero

by E x e r c i s e 1 6 . A .

af/8z j = g j E L

2

n ( C I

in

for

On

D'(Cn).

in

i

i n e q u a l i t y that the

other

hand

D' (En),

k y P r o p s i t i o n 17.16.

. . ,n

a n d t h e lemma has

j = I,.

been p r o v e d . PROOF OF PROPOSITION 4 2 . 2 . ag = 0,

equation t h a t if

and l e t -

af

Let

9 =

j~

g.dg

f E L 2 ( U , Z o c ) = Wo(U,locl

= g . T o show t h a t

f E Wr(U,LocI

and

0

k

j

E Wol(U,loc) with

be a s o l u t i o n o f t h e

f E Wk+'(U,locl

5 r < k

then

f

we s h a l l prove E

Wr+l(U, l o c l .

THE

rl E P ( U l .

Let

5 EQUATION IN PSEUDOCONVEX DOMAINS

Then

7 a (qf) az j of nf. Then

and therefore

5 r

order

< r of 7 a (nf). az au/ai

j

E

2

Since

i

Let u be a derivative

Wr(gn).

E

au/a;

is a derivative

i a (nf) a; j

E

w ~ ~ P ) it ,

of

E Wr+l(Q?),

was arbitrary we conclude that proof. PROPOSITION.

W2n+k

(u,loci

c

CJC(UI

Let

and

f E Wril(UI,

U be an open

f o r every

E

order

u

E

since q E D(U! completing the

subset

of

Then

the

Fundamental

no.

If p E D ( c r n ) then it follows from Theorem of Calculus that PROOF.

of

follows that

L ( g n ) , and by Lemma 4 2 . 3 we may conclude that

v 1 ( t n ) . This shows that nf

42.4.

309

(42.2)

(a) Let f E W Z n ( g n ) be a function with compact support. Applying ( 4 2 . 2 ) with p = f * p 6 - f * P E gives that

Since f E when 6, E formly on

wZn ( 5n I , +

en

the last written integral converges to zero 0, by Exercise 16.A. Hence If * p 6 ) converges unito a function

g

E

C ( t n ) with compact

support.

310

MUJ I CA

Hence it follows that (f * P A ) converges to g in L 2 (6n ) . On the other hand, we know that (f * p 6 ) converges to f in L2(d?), and thus we conclude that f = g almost everywhere, that is, f E CISn). (b) Let f E W 2 n ( U , l o c ) . If E DIU) then it follows from (a) that 'If E C ( t n ) . Since TI E D ( U ) was arbitrary we may conclude that f E C f U ) . Thus we have shown that 8 " ( U , Zocl c C(U). (c) Finally let f E W 2n+k (U, Zocl. If u is a derivative of order 5 r of f then u E W 2 n ( U , l o c ) C C ( U ) by (b). Hence k "f E C (U) and the proof of the proposition is complete. From Propositions 42.2 and 42.4 we obtain at once the following theorem. 42.5. let

THEOREM.

g

E

Let

Cil(U) w i t h

o f t h e equation

-

af

U b e a pseudoco n vex open s e t i n g n , and 2 ag = 0 . Then ea ch s o l u t i o n f E L ( U , l o c )

= g

beZongs t o

Cm(U).

NOTES AND COMMENTS

The idea to use the weighted spaces L~

ill,^)

to solve

P4 a equation in pseudoconvex domains is due to L. Hormander

the 12

I.

Our presentation in Section 39 follows the book of F. Riesz and B. Sz. Nagy I 1 I . Our presentation in Sections 40, 41 and 42 follows the book of L. Harmander [ 3 1 .

CHAPTER X

THE LEVI PROBLEM

4 3 . THE LEVI PROBLEM IN

Q?

In this section we solve Problem 37.10 in the n E = @ . 43.1. THEOREM.

Each p s e u d o c o n v e x o p e n s e t i n

case

where

t n is a domain

of h o l o m o r p h y .

The proof of this theorem rests on the following lemma. 4 3 . 2 . LEMMA.

U b e a p s e u d o c o n v e x o p e n s e t i n en, w h e r e = U n tn-'. T h e n t h e r e s t r f c t i o n m a p p i n g is s u r j e c t i v e .

n > 2, and l e t

SCtui

+

PROOF.

Since that U. By 9

1

of

U

XtU,)

Let

Ul

Let 71 : en -+ tn-l denote the canonical projection. the set U 1 is closed in U , but is open in k n - ' , we see UI and U ', n - ' ( U 1 ) are two disjoint closed subsets of Corollary 15.5 there is a function 9 E C m ( U ) such that on a neighborhood of U l and 9 5 0 on a neighborhood \ n - i ( U l ) . Given f l E x(U,) we define f E C m ( U ) by

f =

9(fi

0

71)

-

ZnUJ

-

where -

af

u E

af

C w ( U ) will be chosen so that

= 0. The

equation

-

= 0

Since v

is equivalent to the equation

is a well defined member of

au = u , where

Cw0 1 (Ul satisfying %u = 0,

31 2

MUJ ICA

Theorem 42.5 guarantees the existence of u f E X t U ) and it is clear that

au = v . Thus

E CmfU)

f = fl

such that on U,.

PROOF OF THEOREM 43.1. By induction on n. By Corollary 10.6 each open set in 5 is a domain of holomorphy, and hence the > 2 and assume t-hat the theorem is true for n = 1. Let n -

theorem is true for n - 1. Let U be a pseudoconvex open set in k n and assume that U is not a domain of holomorphy. Then there are open sets V and W in k n such that:

# U.

(a)

V

is connected and

(b)

W

is a connected component on

(c) For each W.

f

E

V

X t u ) there is j;

U n V. E

Xtv) such that j; =

f on

Choose a E V r~ a l l n aW, choose r > 0 such that B ( a ; r l C 8, and choose b E W n B ( a ; r ) . Fix a coordinate system in 6 such that a and b both lie in gn-'. Let U I = U n 5'-', let

V l be the connected component of V n 5n-1 which contains b , and let W 1 be any connected open subset of W n 5n-1 containing

b. We claim that:

(bl) b E W 1

C

(c1) For each j;, = f, on W, .

U1 n Vl fl

E

.

X(U,)

there is

f,

E

XlV,) such that

It follows from the definition of V l that B ( a ; r l n Cn-' V l , proving (all. (bl) follows at once from the definition of Wl. To prove (cl) let f, E X t U , ) . By Lemma 43.2 there is f E K ( U ) such that f = f, on U l . By (c) there is f E X(V) such that f = f on W. Then y l = 7 1 V l E 3C(V1) and f , f l on W l , proving (cl). Thus U, is a pseudoconvex open set in

C

p- 1

which is not a domain of holomorphy, contradicting the induction hypothesis. Theorems 11.5 and 43.1, and Corollary 37.7, can be summarized

313

THE LEV1 PROBLEM

as follows. 43.3. THEOREM.

For an open s u b s e t

t h e fozloiding con-

U of

d i t i o n s are equivalent:

(a)

U

i s a domain of e x i s t e n c e .

(b)

U

i s a domain of h o l o m o r p h y .

(c)

U

i s hoZornorphically con vex.

(d)

U

i s pseudoconvex.

44. HOLOMORPHIC APPROXIMATION IN

Cn

In this section we shall prove an approximation theorem that generalizes the Oka-Weil Theorem 24.12. The key is the following lemma. 44.1. LEMMA. L e t U be a p s e u d o c o n v e x o p e n s e t i n Cn. L e t u Cm(U) be a s t r i c t l y p l u r i s u b h a r m o n i c f u n c t i o n s u c h t h a t t h e s e t K c = { z E U : u f z ) 5 c} i s compact f o r e a c h c E B . T h e n f o r e a c h f E K ( K o ) t h e r e is a s e q u e n c e ( f .I i n 3 C ( U ) such that

E

3

PROOF. We have to show that J C ( K o ) is contained in the closure 2 of X ( U ) in L (KO). Thus it suffices to show that JCfU)' C 1 J C I K o l . In other words, it is sufficient to show that each f o E

2

L ( K O ) satisfying

also satisfies

[

(44.2)

Fix

f,

E

L

2

foFdh = 0

for every

f

E

X(Ko).

( K O ) satisfying (44.1), and define

fo

=

0

on

314 U \KO

MUJ I CA

, so that

f, E L 2 ( U ) . Let a, B,

y E Cm(U;lR)

be

three

weight functions, and let A,B be the corresponding operators f E NA introduced in Definition 40.3. By Theorem 42.5 each belongs to Cm(UI and hence NA C J C ( U ) . Thus it follows from (44.1) that

foea

E

fNA)'.

If it happens that

then by Corollary 39.8 there exists g A*g = f e

(44.4)

E

DA*

such that

a

0

and II g II

(44.5)

If we write

then

h

E

g =

L 2 (U, 01

2

-

2 II foe a II c1

.

g . dz then it follows from Lemma 40.5 3 j

that

f3) and (44.4) and (44.5) can be rewritten in

the form

and

Let ( q i l be a sequence in D ( U ) such that 0 5 q i 5 1 on U for every i , and qi I 1 on each compact subset of U for all Q E sufficiently large i By Lemma 41.4 there is a function C m ( U ; I R ) such that lTqiI2 < e Q on U for every i. Then the proof of Theorem 41.3 shows the existence of a convex, increasing function x e C m ( L R ; l R i such that

.

THE LEV1 PROBLEM

315

for every a E U and b E Cn. By Exercise 15.B there is an m increasing sequence of convex, increasing functions e p E C IR;B) such that 6 (tl = 0 for every t 5 0 and lim 6 ( t i = m €or

P

every

t

P

xP

0. If we define

it is clear that

x

P

P’” -

x

i

OP

for every

p E

a!

then

x p is convex and increasing, and Zim x I t ) = €or every P

E cmi2?R;IRi,

xP

It) = X ( t ) for every

t - 0

P””

t > 0. Moreover, it follows from Exercise 35.E and (44.6) that

If we define

(Pp

= X

P

ou,

ct

P

= cp

-

P

2$,

Bp =

cp

P

-

$

and

-

yP

the for every p E W , then Proposition 41.1 guarantees validity of the inequality (44.3) for the weight functions ct P’ . Then the preceding argument shows the existence of a

%

%’

sequence of differential forms

hP = B h? d z j Li J

E

2 LOlIU,

- BPI

such that (44.7) d

and (44.8)

I

z

Ju j

x

ou-J)

dh.

lh$I2e

U

for every p E PI. Since f0 = 0 on U \ K O and since Xp(t) = x ( t ) for every t < 0 and every p E IN, it follows that

0

f0

I 2 e x 0 u - 2 $ d h . Since the sequence

316

MUJ I CA

(X 1 is increasing we conclude that

P

This shows that the sequence (hP)"

2

P=Q

is b u n d d in Lol(U,rj,-x

9

ou)

for each q E ZV. Since each bounded sequence ina Hilbert space has a weakly convergent subsequence, an inductive argument m

in n L c2l ( V , $ - x q h = I: h . dz j J 3' y=l which is a weak cluster point of ( h P / " in L o2l l u , $ - xq P=Y for each q E W . Then it follows from ( 4 4 . 9 ) that yields a differential form

o u)

(7

to

-

Since the sequence (X (t)) increases to for each t > 0, 4 an application of the Monotone Convergence Theorem shows that h = 0 almost everywhere in U \ K . On the other hand, it fol0 lows from ( 4 4 . 7 ) that

for every

f

E

V ( U l and

p

?N.

Whence it follows that

( 4 4 . 10)

U

Since

KO

is a compact subset of U, and

(44.10) implies that

In particular

f

0

= h = 0 on U \ K 0'

THE LEV1 PROBLEM

317

T h i s shows ( 4 4 . 2 ) and t h e l e m m a . c o n v e r g e n c e and u n i f o r m c o n v e r g e n c e

Lp

a r e connected

by

t h e following lemma.

pact s e t constant

PROOF.

U b e a n o p e n s e t i n 5n.

Let

4 4 . 2 . LEMMA.

K and each open s e t c > 0

V C U, t h e r e

C

is

a

such t h a t

U b e a n open s e t i n

(a) L e t

K

V with

T h e r e f o r eachcom-

5

and l e t

D l and DI C D C 2

two r e l a t i v e l y compact open d i s c s s u c h t h a t W e claim t h a t t h e r e i s a c o n s t a n t

c > 0

be

D2

o2

C

U.

such t h a t

V ( D ) s u c h t h a t cp E 1 o n a n e i g h b o r h o o d 2 o f D1. Then by a p p l y i n g Lemma 2 2 . 3 t o t h e p r o d u c t qf w e obt a i n t h e formula Indeed, choose

cp E

-

f o r every

f(al

1

f E K(U)

and

neighborhood of f o r every

and

a

a E Dl, (b)

DI

Dl

E

fo a

E

2

E

f z l d z A dz

aq -

Dl. S i n c e

there exists and

* a;

a;

-

on

=

2

such t h a t ( z - a 1

6 > 0

.

supp 7 a' a z

we

If

set

c

a 6 =

as a s s e r t e d .

U b e a n open s e t i n

C n , and l e t

DI and D2 be two r e l a t i v e l y compact open p o l y d i s c s s u c h t h a t D I C D 2 C

-

D2

C

Let

U. Then by r e p e a t e d a p p l i c a t i o n s o f

constant

c > 0

such t h a t

(a)

w e can f i n d

a

318

MUJ I CA

(c) Since each compact set in 5n can be covered by finitely many polydiscs, the lemma follows at once form (b). 4 4 . 3 . THEOREM.

U be a pseudocon vex open s e t in E n . b e a compact s u b s e t of U s u c h t h a t = K . Then

K

each

f

Let

X ( K ) t h e r e i s a sequence

E

(f .I 3

in

Let for

X ( U I whichconverges

t o f uniformly o n a s u i t a b l e n e i g h b o r h o o d o f

K.

PROOF.

Choose an open set V such that K C V C U and f E strictly plurisubharmonic function u E C m ( U I such that:

J c i V ) . By Theorem 38.6 there is a

c

E

0 0

K, =

The set

(a) IF?.

{ z E

U : ulz) 5 c)

(b)

u(z) < 0

for every

z E

K.

(c)

u(z)

o

for every

z E

u\

>

Since u ( z ) < 0 for every z such that u ( z ) < c for every C

K

C V.

0

E

is compact for each

V.

K, there is a constant c < K. Thus K C 2, C K c C

z E

By Lemma 44.1 there is a sequence (f

Ifj

such that

-

2 f I dX

-+

.)

.7

x(U)

in

0. But then Lemma 44.2 implies that

0

sup 1 f j

-

f12

+

0, and the theorem has been proved.

Kc

n L e t U b e a p s e u d o c o n v e x o p e n s e t i n 6' Let b e a n o p e n s u b s e t of U s u c h t h a t E p J l u l C V f o r each com-

.

44.4. COROLLARY. V

pact s e t 44.5.

K C V . Then

THEOREM.

Let

X 1 U I is d e n s e i n ( X ( V l , T ~ ) .

U be a hoZomorphicaZZy c o n u e z o p e n s e t i n

En. T h e n f o r e a c h o p e n s e t

V C U

t h e f o l l o w i n g c o n d i t i o n s are

equivalent: *

(a)

KJC(U)

(b) - & u ,

(c)

V

8 3 V

i s compact f o r e a c h compact s e t

C V

f o r e a c h compact s e t

is h o l o m o r p h i c a l l y c o n v e x and

K

C

K

C

V.

V.

X(V)

?:s

2ensc: i n

THE LEV1 PROBLEM

319

PROOF. (a) * (b): Let K be a compact subset of V and suppose V . Then = A U B , where A = K X ( U I n V and that KJCfu)

-

zJCcUi

L

\V.

= %U)

Then A and B are two disjoint compact sets.Let

to zero on a neighborhood of A and equal to one on a neighborhood of B . By Theorem 4 4 . 3 there is a function g E X ( U I such that ( g - f ( < 7 / 2 on Gf(UI * Hence lgl < 2 / 2 on K and l g ( > 2 / 2 on B. This is impossible, for B C ? x ( u ) . f E JC(iJC(,,,)) be equal

(b) * ( c ) : Clearly V is holomorphically convex. And it follows from Corollary 4 4 . 4 that J C f U ) is dense in ( J C ( V ) , T ~ ) . (c) (a): Let K be a compact subset of is dense in I J C I V ) , ~~l one can readily see that =)

V . Since

V . Since V

n V

is holomorphically convex we conclude that is compact, as asserted. THEOREM.

44.6. KPbfU)

PROOF.

=

U is a p s e u d o c o n v e x o p e n s e t in f o r e s c h compact s e t K C U. If

L

1

KxfU)

The inclusion ‘PA(u!

%fU)

JCfU)

&UVI - %U)

is clear.

Sn

n

zxiul then

To show the

reverse inclusion let V be an open neighborhood of

-

KPA

( u ) in

U. By Lemma 38.1 there is a function u E P d c f U ) such that u < 0 on K and u > 0 on U \ V . Set W = { z E U : u ( z ) < 01. Then W is pseudoconvex by Proposition 3 8 . 3 . Clearly K C W C V, C W for each compact set L C W. Thus X(U) is and L p A ( u ) dense in ( K I W I , -re) by Corollary 4 4 . 4 . Since U and W are both pseudoconvex, it follows from Theorem 4 3 . 3 that U and W are both holomorphically convex. Then an application o€ Theorem 4 4 . 5 shows that L ~ C (W for ~ each ~ compact set L C W, and a

a

a

in particular

KJCfU)

neighborhood of L

K P b (U)

C

ZPa f U)

W

C

Since V was an arbitrary

V.

in

U

we

conclude

that

and the proof of the theorem is complete.

open C

MUJ I CA

320

EXERCISE 44.A. Let U be a ho-Dmorphica ly convex open set in t n l and let K be a compact subset of U such that K x t u , = K. Show that if L is an open and closed subset of K then

-

Ljclul

=

L.

44.B. Let U be a holomorphically convex open set in C n l and let K be a compact subset of U . Show that each connected component of Zxiui contains points of K. 44.C. Let U be a holomorphically convex open set in @?, and let V be an open subset of U such that ,?xiul C V for each compact set K set of V then

Show that if W is an open and closed subzx(u, C W for each compact set K C W.

C V.

44.D. Let E be a Banach space with the approximation property, and let U be a pseudoconvex open subset of E which is finitely - ZPtEI for each compact subset K of Runge. Show that -

u . In particular U is polynomially convex.

45. THE LEV1 PROBLEM IN BANACH SPACES In this section we solve Problem 37.10 in the c a s e of separable Banach spaces with the bounded approximation property. In this and the next section all Banach spaces considered will be complex. Corollary 44.4 motivates the following definition. 45.1. DEFINITION. Let U be a pseudoconvex open subset of E . An open subset V of U is said to be U - p s e u d o c o n v e x if i?pn(v, C V for each compact set K C V .

The following examples can be readily verifiedbythe reader.

45.2. EXAMPLES.

Let U be a pseudoconvex open subset of

E.

THE LEV1 PROBLEM

(a)

Each convex open set

(b) For each is U-pseudoconvex. (c)

f

E

If N i t i E I

then the open set

321

is U - p s e u d o c o n u e x

V C U

V = {x

PAfUl the open set

E

U : f ( x ) < 0)

is a family of U-pseudoconvex open sets,

V = in+

Vi

is U-pseudoconvex as well.

i E I

4 5 . 3 . LEMMA. let

Let

U be a pseudoconvex open s u b s e t o f

V b e a U-pseudoconvex open s e t . Then:

(a)

V

(b)

K(U r~ M) i s d e n s e i n ( X ( V n

i s pseudoconvex.

dimensional subspace

PROOF. (a) If KPA

and

E,

(U)

K

M

of

M),Tc)

finite

E.

is a compact subset of

U

then

ipAiu,

c V. BY Proposition 3 8 . 2 the set A

f o r euch

KPA

is

(V)

C

compact.

is relatively compact in V and V is therefore

Hence KPA(V) pseudoconvex.

K

is a compact subset of V n M then one can n M c V n M. Then an ap'PAfU) readily see that 'PA ( U n plication of Corollary 44.4 yields the desired conclusion. (b)

If

Let E be a Banach space with a monotone Schauder basis En denote the subspace generated by e I J . . . , en' and

( e n ) . Let

let

Tn E E(E;En)

denote the canonical projection.

II T n z - x I1 5 2 I1 z II for every see that the function

w.(xl = 3

is continuous on E for each ing result. 4 5 . 4 . LEMMA.

Let

E

x

E

s u p IIT n> j

j

E

and

E

2 -

W.

n E IN

Since

one can readily

XI1

Then we have the follow-

b e a Banach s p a c e w i t h a m o n o t o n e Schauder

MUJ I CA

322

b a s i s f e n ) , and Z e t A

3’

B . and 3

A

j

i

=

U b e a p s e u d o c o n v e x o p e n s u b s e t of

E. Let

be t h e f o Z Z o w i n g o p e n s e t s :

C

U : s u p IIT

{J: E

n’j

J:

n

-

< dufx)l,

511

Then:

j, n

U

(b)

T n f A .) C U

(c)

B j n En

E

=

m

m

m

(a)

U

A

i

j=l

3

=

U

j=l En

B

i

=

and

U

j=l

C

j ’

T n ( C .) C B n E whenever n )j. 3 i n

A j n En

i s r e l a t i v e l y compact i n

f o r each

W.

Each

(d)

A

i s dense in ( X ( A

j

j

i s U-pseudoconvex. (3 E , ) , T ~ ) f o r e a c h

In p a r t i c u Z a r j, n

E

X ( V n En)

l7i.

Assertions (a), ( b ) and (c) are clear. To show (d) it suffices to observe that the function Zogw.(x) = sup log 117’ nz-xll n’>j is plurisubharmonic on E and apply Example 45.2(b) and Lemma

PROOF.

45.3(b).

Let

4 5 . 5 . LEMMA.

E

U be a p s e u d o c o n v e x o p e n s u b s e t of E . Then

b a s i s ( e n ) , and l e t f o r each

f,

such t h a t PROOF.

E XfU

If

- f,

Since

particular

f,

be a Banach s p a c e w i t h a monotone Schauder

n E n ) and o Tn(

Tn(An)

C

5 U

o T n E JC(An

E

E

on

> 0

Cn

there e x i s t s

f

E

XfU)

. f, o Tn E J C f An 1 and in Since 3 c ( U n E n i 1 1 is dense

we have that

n Enil).

in ( X ( A n n E n + I ) , T ~ ) , we can find

fn+l

E

X f U 17 E n i l )

such that

THE LEV1 PROBLEM

323

and hence

Proceeding inductively we can find a sequence (f.), with 3 J c ( U n E . ) , and such that

fj E

3

j > n . Whence it follows that

for every

m

-

I f k O T k

5

f j O T j j

z

E

2-r+n-l

=

E . 2

-j + n

on

C

r=j

for every k 2 j formly on each C

n . Thus the sequence ( f . 0 T . j

convergesunito a function f E X ( U ) . Moreover, 3

j

for every

j

3

j 2 n , completing the proof.

E be a Banach s p a c e w i t h a monotone Schauder and l e t U b e a p s e u d o c o n v e x open s u b s e t of E. Then B^ j' PA c ( u ) and i n p a r t i c u l a r dU( ?j)X(U) - l / j f o r Let

every

j E iN.

PROOF.

Let

2 E

(Fj)jctui

and choose

k

j

such that x E Ck.

We claim that A

(45.1)

'nX

(Bj

E n ) x ( U nE ~ ) €or every

n

2

k.

To prove (45.1) let n 2 k and let f n E X ( U n E n ) . Given E > 0 , by Lemma 45.5 we can find f E K i l l ) such that - f, o T n I < E on C n . Since c U (21 c ck c Cn we get that j

If

'

fn

oTn(x)I

5 If(e)

+ E

5 ~uplfl+ cj

E

MUJ I CA

324

Since sup B

3

.

~

E

was arbitrary we conclude that I f n o T n ( x ) l -< and ( 4 5 . 1 ) follows. Thus by Theorem 4 4 . 6 we have

> 0

Ifn/ E ~

that

for every n 2 k . Letting

n

+

we get the desired conclusion.

The following theorem improves Theorem 2 8 . 7 and partial solutions to Problems 10.11, 1 1 . 6 and 3 7 . 1 0 . 45.7.

THEOREM.

Let

E

b e a s e p a r a b Z e Banach

space

provides

with

the

bounded a p p r o x i m a t i o n p r o p e r t y . T h e n e a c h p s e u d o c o n v e x open s e t in

E

i s a domain of e x i s t e n c e .

(a) Assume first that E has a monotone Schauder basis, and let U be a pseudoconvex open subset of E . By Lemma 45.6 there is an increasing sequence of open sets C.;U such that U

PROOF.

m

=

u j=l

C j

and

du((e.)

3 JC(Ul

I

> 0

for every

j. By Theorem 1 1 . 4

we may conclude that U is a domain of existence. (b) If E is a separable Banach space with the bounded approximation property then by Pelczynski's Theorem 2 7 . 4 we may assume that E is a complemented subspace of a Banach space G which has a monotone Schauder basis. Let 71 be a continuous projection from G onto E l and let U be a pseudoconvex open subset of E. Then T - 1 ( U ) is a pseudoconvex open subset of G by Proposition 3 7 . 8 , and thus n - ' ( U ) is a domain of existence -2 in G , by part (a). But then U = 71 ill)n E is a domain of existence in E by Proposition 1 1 . 7 , and the proof of the t h e rem is complete. Theorems 1 1 . 4 and 45.7, marized as follows: 45.8.

THEOREM.

Let

E

and Corollary 3 7 . 7 ,

can

be a s e p a r a b Z e Bunach s p a c e

by

sum-

with

the

325

THE LEV1 PROBLEM

bounded a p p r o x i m a t i o n p r o p e r t y . Then f o r e a c h o p e n s u b s e t U o f E

the following conditions are equivalent:

(a)

U

i s a domain of e x i s t e n c e .

(b)

U

i s a domain of holornorphy.

(c)

U

i s holomorphically convex.

(d)

U

i s pseudoconvex.

EXE RC ISES

45.A. Let E be a Banach space with a monotone Schauder basis, and let U be a pseudoconvex open set in E . (a) By adapting the proofs of Lemmas 43.2 and 45.5 show is surjective that the restriction mapping X ( U ) +. X ( U n E n ) for each n E Dl. (b)

Using Exercise 26.A show that the restriction mapping n M ) is surjective for each finite dimensional subspace M of E .

X(U)

+.

X(U

45.B. Let E be a separable Banach space with the bounded approximation property, let U be a pseudoconvex open subsetof E , and let M be a finite dimensional subspace of E. (a)

Show that the restriction mapping

XIU)

+.

X ( U n M 1 is

surjective.

46. HOLOMORPHIC APPROXIMATION IN BANACH SPACES In this section we obtain infinite dimensional versions of the approximation theorems from Section 44. We &gin by extending Corollary 44.4.

326

MUJICA

46.1. THEOREM.

Let

b e a s e p a r a b l e Banach space with the bounded

E

a p p r o x i m a t i o n p r o p e r t y . L e t U b e a p s e u d o c o n v e x o p e n s e t i n E, and l e t V be a U-pseudoconvex o p e n s e t . T h e n X(U) i s s e q u e n t i a l l y dense i n (Xtv), .re).

(a) First assume that E has a monotone Schauder basis. B and Let U be a pseudoconvex open set in E , and let A j’ j be the open sets constructed in the preceding section. Let Cj V be a U-pseudoconvex open set, and let B ! and C! be the J 3 following open sets:

PROOF.

m

Then it is clear that

V

=

latively compact in

n

En

C B!

3

whenever

V

n 2 j . Let

3 . 0

B! 3

U j=j

=

U

j=I

for every j,n f

E

B ! n En 3

C’ j’

a,

E

re-

is

and

X ( V ) be given. For each

Tn(Cg)

n E JV

Jc(U n E n )

is dense in ( J c ( V n E n ) , ~ c ) . Hence we can find g n E 3C(U n E n ) such that I g n - f I 5 l / n on B A n E n and therefore

By Lemma 4 5 . 5 for each that

n E N

we can find

f,

E

JctUl

such

We claim that If,) converges to f in ( J c t V l , T ~ ) . Indeed, given a compact set K C V and E > 0 we first choose 6 > 0 such that K + B ( 0 ; 6 ) C V and I f ( y ) - f ( x ) I 5 E for all x E K and y E B ( x ; 6 ) . Next choose no E W such that I/no 5 c,K C CA and IIlfytx - X I / < 6 f o r every x E K and every x E K and n 2 no we have that

n

2

y!

.

0

Then

for

327

THE LEV1 PROBLEM

T h i s c o m p l e t e s t h e p r o o f when E h a s a monotone S c h a u d e r b a s i s .

i s a s e p a r a b l e Banach s p a c e w i t h t h e bounded a p p r o x i m a t i o n p r o p e r t y t h e n by Theorem 2 7 . 4 w e may assume t h a t E i s a complemented s u b s p a c e o f a Banach s p a c e G which h a s a monotone S c h a u d e r basis. L e t U be a pseudoconvex o p e n s e t i n E and l e t V b e a U-pseudoconvex open s e t . L e t IT be a con-1 t i n u o u s p r o j e c t i o n from G o n t o E a n d l e t U ' = I T (U), V ' = n - ' ( V ) . Then U r i s a pseudoconvex open s e t i n G , a n d V r is Ur-pseu(b)

If

E

doconvex, so t h a t

is s e q u e n t i a l l y dense i n

J€(Ur)

by p a r t ( a ) . NOW, l e t

f E X I V ) be given.

Then

(~C(V'),T~), f o F E 3C(Vr)

n i n % I l l r ) which c o n v e r g e s t o U = U r n E and V = V r n E wemay X ( U I for e v e r y n , and t h e n f f n ) converges

and h e n c e t h e r e i s a s e q u e n c e ( g i n (3C(Vr),-rc).

f O F

define

f

n

= gn

I

U

Since

E

t o f i n (3C(V), . r e ) . 46.2.

let

THEOREM. L e t U b e a p s e u d o c o n v e x o p e n s u b s e t o f E , a n d K b e a c o m p a c t s u b s e t of U s u c h t h a t K p n ( u ) = K . Then for

each open s e t set

V with

W such t h a t

r < dUfK/

PROOF.

Let

Then

i s compact and

L

0

f o r each p o i n t that

K C V C U

t h e r e i s a U-pseudoconvex open

K C W C V.

a

s u p fa < 0

K

L\ V

E

and s e t

c L c U. S i n c e

K =

t h e r e is a function

f a ( a ) . Since

-

2?b (U) -- K ? b c ( U ) ' fa

E

P h c ( U ) such

i s compact w e c a n

L \ V

find

K functions

j = I,

..., m

I,,.. . , f m

E

Pbc(U)

such

that

and rn

L \ V C

u

j=z

C X E

u : f.(x) 3

sup f j < K

> 01.

0

for

328

MUJ I CA

f = sup

If we define f

PbC(UI,

K

on

0

{ - log dU

+ log

r , fl,.. ., f,}

6 > 0

we

E

can

show

the

such that

W = ( Z ( K ) + B ( O ; 6 ) ) n {x

Then the open set

f

and

Proceeding as in the proof of Theorem 28.2 existence of

then

E

U

: f ( x ) < 0)

has the required properties.

By combining Theorems 46.1 and 46.2 we obtain the following theorem, which can be regarded as a generalization of Theorem 44.3. 46.3. THEOREM.

Let

E

b e a s e p a r a b l e Banach

spcrce

with

the

U be a pseudoconvex open s u b s e t of E , and l e t K b e a c o m p a c t s u b s e t of U such t h a t * = K . T h e n f o r e a c h f E X ( K ) t h e r e a r e an o p e n s e t V K P n (Ul w i t h K C V C U, and a s e q u e n c e (f .) i n X ( U I such t h a t f E X ( V l J and ( f .) c o n v e r g e s t o f i n ( J c ( V l , . r c l .

bounded a p p r o x i m a t i o n p r o p e r t y . L e t

3

46.4. THEOREM.

Let

E

be

a s e p a r a b l e Banach

space

with the

U b e a hoZornorphicaZly T h e n f o r e a c h o p e n s e t V C U t h e fol-

bounded a p p r o x i m a t i o n p r o p e r t y , a n d l e t convex open s e t i n

E.

lowing c o n d i t i o n s are e q u i v a l e n t : (a)

KX(Ul

9

V

i s c o m p a c t f o r e a c h compact s e t

[b)

KWUl

C

V

for e a c h c o m p a c t s e t

(c)

V

i s h o l o m o r p h i c a l l y convex and

K

C

K C V

V.

K ( U ) i s sequentially

dense i n ( K ( V ) , . r c ) . (d)

(X(VI,

Tc).

V

is h o l o m o r p h i c a l l y e o n v e x and

JC(UI

is d e n s e i n

3 29

THE LEV1 PROBLEM

PROOF. In view of Theorems 4 6 . 1 and 4 6 . 3 , 4 4 . 5 applies. 46.5.

THEOREM.

Let

E

b e a s e p a r a b l e Banach space w i t h t h e bounded

a p p r o x i m a t i o n p r o p e r t y , and l e t in

E.

Then

PROOF.

Let

KPh(ul

the proof of Theorem

be a pseudoconvex open s e t

U

f o r each compact s u b s e t

= K ^ W

V be any open set such that

-

c

Kphiu,

K

v

c

of

U.

u.

By

Theorem 4 6 . 2 there is a U-pseudoconvex open set W such that C W C V . By Theorem 4 6 . 1 JC(UI is dense in ( J c ( W I , ' T o ) . KPh (U) Since by Theorem 4 5 . 7 both U and W are holomorphically con* vex, we may apply Theorem 4 6 . 4 to conclude that K X t U i c w c v . Since V was arbitrary we conclude that

KJC(U)

c

t p h l uSince ).

the opposite inclusion is clear, the proof is complete.

EXERCISES 46.A.

Let E be a separable Banach space with the bounded ap-

proximation property, let of

E, and let M be a complemented subspace of (a)

+

be a pseudoconvex open subset

U

JCiU

(b)

E.

restriction mapping K i U l M i for the compact-open topology.

Show that the image of the M ) is dense in Show that

-

JclU

KxIUi

0 A

-

-

KjCIU 3

MI

for each compact set

K

C U f i M .

NOTES AND COMMENTS Theorem 4 3 . 1 was obtained by K. Oka [ 3 ] for n = 2 in 1 9 4 2 , and by K . Oka [ 4 ] , H . Bremermann [ 1 ] and F. Norguet [ 1 ] for arbitrary n in 1 9 5 3 - 1 9 5 4 , thus solving a problem posed by E. Levi [ l ] in 1 9 1 1 . The proof of Theorem 4 3 . 1 given here is due to L. Hormander [ 3 ] . Theorem 4 4 . 3 , due to L. Hormander [ 2 ] , improves an approximation theorem in domains of holomorphy obtained by K. Oka [ 2 1 . Theorem 4 5 . 7 is due to L. Gruman [ 1 1

330

MUJ I CA

in the case of separable Hilbert spaces, to L. Gruman and

C. Kiselman [ 11 in the case of Banach spaces with a Schauder basis, and to P. Noverraz [ 3 ] in the case of separable Banach spaces with the bounded approximation property. Theorems 46.1 and 46.2 are due to J. Mujica [ 4 1 . Theorems 46.3, 46.4 and 46.5 sharpen results of P. Noverraz [ 4 1 . The result in Exercise 46.A is also due to P. Noverraz [ 4 ] .

CHAPTER X I

RIEMANN DOMAINS

4 7 . RIEMANN DOMAINS I n t h i s s e c t i o n w e e s t a b l i s h t h e b a s i c p r o p e r t i e s o f Riemann d o m a i n s o v e r Banach s p a c e s . T h r o u g h o u t t h i s c h a p t e r a l l Banach s p a c e s c o n s i d e r e d w i l l b e complex.

X and Y are t w o t o p o l o g i c a l s p a c e s t h e n

W e recall t h a t i f

a mapping each

J: E

that

f f U )

f : X

+

is said t o b e a l o c a l homeomorphism i f f o r

Y

t h e r e i s an open set

X

i s open i n

Y

f

and

I

U i n X containing x such U : U ftU) i s a homeomor+

p h i s m . C l e a r l y e v e r y l o c a l homeomorphism i s c o n t i n u o u s a n d o p e n .

47.1.

DEFINITION.

such t h a t

X

A Riemann domain over

is

E

a pair

i s a Hausdorff t o p o l o g i c a l s p a c e and

5

:

(X,C)

X

i s a l o c a l homeomorphism. A c h a r t i n X i s a c o n n e c t e d o p e n U

on

C

X s u c h t h a t
:

+

E

set

U S i U l i s a homeomorphism. An a t Z a s f u i i i E I o f c h a r t s w h i c h c o v e r X. +

(X,<) i s a Kiemann domain o v e r E t h e n X i n h e r i t s many

If

of t h e t o p o l o g i c a l p r o p e r t i e s of

I n p a r t i c u l a r it is clear

E.

i s a f i r s t c o u n t a b l e s p a c e , and t h e r e f o r e a k-space.

thatX

i s also clear t h a t

X

e a c h c o n n e c t e d o p e n s u b s e t of.

X i s p a t h w i s e c o n n e c t e d . Furtlier-

X i s l o c a l l y compact i f a n d o n l y i f

more,

It

i s l o c a l l y p a t h w i s e c o n n e c t e d , and hence E

i s f i n i t e dimen-

sional. 47.2.

DEFINITIONS. ( a ) L e t (X, 5 ) a n d (Y, r)) b e two Riemann d o m a i n s

o v e r E . A c o n t i n u o u s mapping T : X + Y if

if

rl o T 7

= 5 . A mapping

T :

i s a b i j e c t i o n and both

X

-+

T

Y

and

i s s a i d t o be a morphism

i s s a i d t o b e a n isomorphism T

-1

are morphisms.

332

MUJ ICA

More g e n e r a l l y , l e t ( X , c )

(b)

b e aRiemann domain

over E ,

l e t ( Y , r l i be a Riemann domain o v e r F , and l e t T E E(E;F). T : X Y i s s a i d t o be a T-rnorphisrn

Then a c o n t i n u o u s mapping ~ I O T=

if

-+

7'05.

The n e x t two lemmas w i l l be v e r y u s e f u l . 4 1 . 3 . LEMMA. b e s u b s e t s of

Let ( X , c ) X

b e a Riernann domain o v e r

such t h a t

A

A, B

i s open, A n B

5 IA : 5 1A u

n
E. L e t

Then

A

+

B

is n o n v o i d , < ( A ) < ( A ) a n d 5 I B : B -+ is i n j e c t i v e .

5 ( A u B i s i n j e c t i v e it c l e a r l y s u f f i c e s < ( A n B ) = C ( A l n S ( B ) , and t o prove t h i s i t

PROOF. To show t h a t t o prove t h a t

< ( A n B i i s open and c l o s e d i n

s u f f i c e s t o show t h a t

<(A) n

<(B). S i n c e A i s open i n X, t h e i n t e r s e c t i o n A n B i s open i n B . Hence E ( A B I i s open i n S ( B ) , and t h e r e f o r e open i n < ( A ) n 51B). To show t h a t be a sequence i n y

in

Hence

-+

p o n e n t of

PROOF.

3

E.

I

Bi-'(y).

Let

A

be

S ( A l i s a e o n v e x o p e n s u b s e t of E a n d

< ( A ) is a homeomorphism.

Then

A

i s a c o n n e c t e d com-

5-l ( < ( A / ) .

I t i s c l e a r l y s u f f i c i e n t t o show t h a t

closed i n

= (5

L e t ( X , < ) b e a Riernann domain o v e r

a s u b s e t of X s u c h t h a t : A

,I

such t h a t (<(a: . ) I c o n v e r g e s t o a p o i n t

<(A) n E ( B ) . Then ( 5 \ A l - l ( y ) = l i m x j y E < ( A n B) and t h e proof i s complete.

4 7 . 4 . LEMMA.

5 IA

< ( A ) r~ <(B), l e t ( x . 1

< ( A n B) i s c l o s e d i n A n B

<-l(<(Al).

Given

a

E

A

A

i s open

w e can f i n d an open s e t

and U

X c o n t a i n i n g a such t h a t 5 ( U ) i s an open b a l l i n ?, and < 1 U : U + < ( U l i s a homeomorphism. S i n c e < ( U ) n < ( A ) i s convex, and t h e r e f o r e c o n n e c t e d , L e m m a 4 7 . 3 i m p l i e s t h a t < ( U u A i s i n j e c t i v e . Whence i t follows t h a t U n A = 11 n 5 - l (<(A)), p r o v i n g t h a t A i s open i n 5 - l ( < ( A ) I . in

To show t h a t

A

is closed i n

< - ' ( < ( A ) 1 , l e t ( x - ) b e a se-

quence i n A which converges t o a p o i n t

<-' J

x in

( < ( A ) ) . Hence

333

R I EMANN DOMA I NS

L e t fX,<) b e a Riemann domain o v e r

47.5. DEFINITION.

let

and

E,

X. W e s h a l l d e n o t e by dXfx) t h e supremum of a l l r > O U i n X c o n t a i n i n g x such t h a t 5 IU : U B E (S(x);r) i s a homeomorphism. For e a c h r w i t h O < r 5 d X f x ) w e s h a l l d e n o t e by B X ( x ; r ) t h e c o n n e c t e d component

x

E

f o r which t h e r e e x i s t s a s e t -+

of

5 - l ( B E ( 5 (xl;r) I which c o n t a i n s

47.6.

PROPOSITION.

(a) for e a c h (b)

then

5

I BX(x;r)

x

E

If

X

L e t fX,<) b e a Ricmann domain o v e r

and

: 3

Xfz;r)

0 < r

f o r every

00

E.

B E ( < ( t ) ; r )is a homeomorphism

-+

5 dxfx). f o r some a and t h e f u n c t i o n d, : X

dx(a) <

X is c o n n e c t e d and

dx(xl <

x.

x

E

X

m

E +

X R

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

(c) dx(x) =

m

dx(al = for some a E X then 5 : X -+ E is a homeomor-

If X i s c o n n e c t e d a n d f o r e v e r y x E X and

phism.

X. I f 0 < r < d , ( x ) t h e n i t f o l l o w s from t h e d e f i n i t i o n of dx(x) t h a t t h e r e i s a s e t U r in X cont a i n i n g x such t h a t F, I U r : U r BEf
PROOF.

(a) L e t

x

E

+

phism. I t f o l l o w s t h e n from Lemma 4 7 . 4

that

Ur

i s the connected

<-'(B E ( 5 f x ) ; r ) ) which c o n t a i n s x , t h a t i s U r = BX(x;rl. I f w e s e t U = U { B X ( x ; r ) : 0 r < dxfxll t h e n it component of

f o l l o w s from Lemma 47.3 t h a t open w e conclude t h a t

5

1

U :

F,

U

1 +

U

is i n j e c t i v e . Since

BE(cfx);dx(x))

is

a

U

homeo-

morphism. Then a n o t h e r a p p l i c a t i o n of Lemma 47.4 shows t h a t

= B (x;dX(x)). Thus ( a ) h a s been proved. To e s t a b l i s h ( b ) X ( c ) w e s h a l l f i r s t prove t h a t i f a E X t h e n w e have t h e equalities

is

U and in-

334

MUJ ICA

x

I n d e e d , if

B X ( a ; d X ( a ) ) then

E

B E ( < ( x ) ; d X ( a )-ll<(x: -<(a)ll)

c BE(E,(a);dX(all and ( 4 7 . 1 ) f o l l o w s . O n t h e o t h e r hand, i f 1

B ( a ; T d x ( a ) ) t h e n i t f o l l o w s from ( 4 7 . 1 )

t h a t d x ( x ) > $-dxfa)

X

> Il<(x) - <(a)ll. Hence

-

BE(E(a); d X ( a l

Il<(x)

- <(a)ll)

C

(47.1) implies t h a t t h e set

and (47.2) f o l l o w s .

xE

BE(E,(d;dX(xI)

{ x E X : d X ( x )= w }

i s open, whereas ( 4 7 . 2 ) i m p l i e s t h a t t h e s e t I x E X .-dx(x) i s open. All t h e a s s e r t i o n s i n ( b ) and ( c ) a r e t h e n c l e a r .

47.7.

Each c o n n e c t e d Riemann domain overs

PROPOSITION.

sepa-

r a b l e Banach s p a c e is s e c o n d c o u n t a b l e , and in pa27ticular separable

and L i n d e l 6 f . PROOF. F i x x E X and 0 < r < d , ( x ) . F o r e a c h n E IV let = ( y E X : d,(y) > r / m ) and l e t Bm b e t h e c o n n e c t e d comAm ponent of Am which c o n t a i n s x . S i n c e X i s pathwiseconnected m

i t i s clear t h a t

X

=

u

B

m=l

t h e s e t of a l l Bm I,

E

Bm

W e claim t h a t

u

Bmn

m, n E IN

let

and

xj

E

E m --

xo,...,x n

E

BX(xj-l;r/2m) €or j=

U

n=Z

f o r each

Bmn

m

IN. I n d e e d ,

E

i s c l e a r l y open, and t o e s t a b l i s h o u r

n=l

be

Bmn

m

m

t h e set

For e a c h

f o r which t h e r e are p o i n t s

xo = x , xn = y

such t h a t

..., n .

y

.

m

claim

W

i t s u f f i c e s t o prove t h a t

i s closed i n

u n=I Bmn

Bm.

Let

y

m

belong t o t h e c l o s u r e of

U

n=l

tion

BX(y;r/2m)

it f o l l o w s t h a t

n Bmn

Bm n

in

Bm.

i s nonempty f o r some

y E Bm,n+l

.

X

m

E

IN

and hence

ti

E

intersec-

IN and whence m

Thus w e have showntha': m

f o r each

Then t h e

X =

u Bmn m, n=l

.

Bm= u Bmn n=l

Thus t o show t h a t

i s second c o u n t a b l e it i s s u f f i c i e n t t o p r o v e t h a t e a c h Bmn

335

RIEMANN DOMAINS

i s second c o u n t a b l e f o r t h e i n d u c e d t o p o l o g y . To p r o v e t h a t each Bmn i s second c o u n t a b l e f i x m E IN. Then Bml i s second c o u n t a b l e , f o r Bml C B X ( x ; r / 2 m ) and E i s second c o u n t a b l e . I f Bmn i s second c o u n t a b l e f o r some n t h e n Bmn i s i n p a r t i c u l a r L i n d e l o f and hence w e can f i n d a sequence f a k ) i n Bmn s u c h m

that

Bmn C

B X f a k ; r / 2 r n ) . Hence i t f o l l o w s t h a t

U

CT

k=l

U

BXfy;r/2m)

B m , n+1

a,

B X f a k ; r / m ) , and s i n c e e a c h

U

C

C

BXfak;r/m)is

k=l

YEBmn

so i s

second c o u n t a b l e ,

.

Bm,n+l

Thus

each

is

Bmn

second

c o u n t a b l e , and t h e proof of t h e p r o p o s i t i o n i s complete. 47.8.

COROLLARY.

Each c o n n e c t e d Riemann domain

over

en

is

hemicompact. Each l o c a l l y compact L i n d e l o f s p a c e i s hemicompact.

PROOF. 47.9.

DEFINITION. (a)

Let

fX,O

be a Riemann domain o v e r

[a,bl i n X w e mean a s e t i n a and b and homeomorphic under 5

By a l i n e s e g m e n t

containing the points

t h e l i n e segment [ < f a ) , 5 f b ) ] (b)

By a p o l y g o n a l l i n e

in

[xo,xl,

... ,xnl

in

PROPOSITION. E.

For

x, y

E

Let f X , < ) X

let

X w e mean a z.] w i t h j 3

...,

47.10.

X to

6.

f i n i t e union of l i n e segments of t h e form - 1, n.

over

E.

be a connected

Ricmann

domain

p x f x , y ) be d e f i n e d b y

where t h e i n f i m u m is t a k e n o v e r all polygonal line:: [ xo,xl, ...,x n1

in X

X

x

such that

0

= x

and

xn = y .

which generates t h e topology o f

g e o d e s i c d i s t a n c e between PROOF.

Given

x

E

X

let

X.

x

and

y.

Am

, Bm

and

i s a m e t r i c on p x f x , y ) is c a l l e d the

Then

Fmn

px

be

+h sets i n t r d u c e d

336

MUJ I CA

i n t h e proof o f P r o p o s i t i o n 47.7. some B

rnn

. Hence

= x, xn = y

t h e r e are p o i n t s

Then e a c h x03

- *

j X

n

y E

E

X

X

belongs t o

such t h a t

xo

,...,

E BX(xj-l,dX(xj-l)) for j = i n. Hence j [ xo, ,xn] i s a p o l y g o n a l l i n e i n X j o i n i n g x and y . T h i s shows t h a t px(x,y) < m f o r e v e r y x, y E X. W e n e x t show t h a t

and

x

. ..

pX(x,y) =

if

x = y. L e t

then

0

i s a polygonal l i n e

[

xo,

. . .,xnl

0 < in

E

< dx(x).

Then

there

X s u c h t h a t xo = x, xn = y

n

k = lJ...,nJ

<

E

C

B E (<(x);E) f o r e a c h

f o r each

and w e conclude t h a t [ ~ ( X ~ - ~ ) , ~ ( X ~ ) I

k = I,

..., n.

Then by r e p e a t e d

applica-

t i o n s of Lemma 47.3 w e o b t a i n t h a t

, x k ] C BX(x;€) for e a c h k = 2 , . , n , and i n p a r t i c u l a r y E BE(";€). S i n c e E > 0 can be t a k e n a r b i t r a r i l y s m a l l w e conclude t h a t y = x. S i n c e t h e p r o p e r t i e s px(x,y) = px(y,xI and px(x,z) 5 px(.x,-y) +px(y,z) can be r e a d i l y v e r i f i e d , i t f o l l o w s t h a t p x i s a m e t r i c on X. And s i n c e i t i s c l e a r t h a t px(x,y) = 115(x) - <(y)ll if x and y b o t h l i e i n some c h a r t U i n X t h e n w e c o n c l u d e that px g e n e r a t e s t o t o p o l o g y of X. T h i s c o m p l e t e s t h e p r o o f .

.,

47.11. DEFINITION. L e t (X,<) b e a Riemann domain o v e r E . A mapping f : X + F i s s a i d t o b e h o Z o m o r p h i c ( r e s p . o f c l a s s C k ) i f t h e r e i s a n a t l a s ( U i ) i E I on X s u c h t h a t f ~ ( < l U ~ i :- ~ <(Ui) --* F i s holomorphic ( r e s p . of class c k ) f o r e a c h i €1. k W e s h a l l d e n o t e by Jc(X;F) ( r e s p . c (X;F) t h e v e c t o r s p a c e of a l l mappings f : X F which a r e holomorphic ( r e s p . of c l a s s +

k

c 1. The s e t Pb(X) of a l l p l u r i s u b h a r m o n i c f u n c t i o n s f : X can be d e f i n e d i n a s i m i l a r manner. W e can l i k e w i s e de-

+

[-m,m)

IX;F) o f a l l F-valued d i f f e r e n t i a l X , and t h e v e c t o r s u b s p a c e c k IX;FI k PC? of a l l members o f 0, (X;F) which a r e of c l a s s c A s usual P4 = Jc(X), C k fX;C) = C k (X), e t c . w e s h a l l w r i t e Jc(x;d') f i n e t h e v e c t o r space

Q

P4 forms of b i d e g r e e ( p , q ) on

47.12. DEFINITION.

.

L e t fXJE) be a Riemann domain o v e r

l e t ( Y , n I be a Riemann domain o v e r

F. A mapping

f : X

E , and --*

Y is

337

RIEMANN DOMAINS

s a i d t o be h o l o m o r p h i c (resp. of class

k

i f f i s continuous and i- o f : X F i s holomorphic (resp. of c l a s s C k ) . w e s h a l l k d e n o t e by Jc(X;P) ( r e s p . C ( X ; Y ) ) t h e s e t of a l l mappings f : X k Y which are holomorphic ( r e s p . of c l a s s C 1. C

+

+

EXERCISES L e t (X,<) a n d ( Y , n )

47.A.

respectively. L e t

T

E

be Riemann domains o v e r

E(E;F)

and l e t

be two 9-morphisms. Show t h a t i f

X i s connected, t h e n

and

a(x)

and

F,

CI : X Y and T : X + Y a ( a ) = ? ( a ) for some a E X, = T ( X ) € o r e v e r y x E X. -+

L e t (X,
47.B.

E

E. L e t

b e a sub-

A

X s u c h t h a t < ( A ) i s a convex s u b s e t of E and < I A : A < ( A ) i s a homeomorphism. Show t h e e x i s t e n c e of a n open set U i n X c o n t a i n i n g A s u c h t h a t 6 ] U : U -+ <(U) i s a homeomors e t of

+

phism. 47.C.

Banach s p a c e

(a) X

be a c o n n e c t e d Riemann domain o v e r a s e p a r a b l e

L e t (X,
Show t h a t

f o r each (b)

E.

a

i s a countable d i s c r e t e subset

<-'(a)

of

E E.

Show t h a t i f

D i s a c o u n t a b l e d e n s e s u b s e t of E t h e n

<-'(D)

i s a c o u n t a b l e dense s u b s e t of

47.D.

L e t ( X , < ) and ( Y , i - )

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

f, g

X.

be Riemann domains o v e r E

Jc(X;YI.

Show t h a t i f

E and

f

g

some nonvoid open s u b s e t of X I and X i s c o n n e c t e d , t h e n g on a13 of X. T h i s i s t h e I d e n t i t y P r i n c i p l e .

F,

on f

3

47.E. let

L e t (X,<) be a c o n n e c t e d Riemann domain f E Jc(X).

that is, f

Show t h a t i f

f

ever

E,

and

i s n o t c o n s t a n t t h e n f i s open,

maps open sets o n t o open s e t s . T h i s i s the @en

Map-

ping Principle, 47.F.

L e t (X, < ) be a c o n n e c t e d Riemann domain o v e r

E l and l e t

338

MUJ I CA

a

Show t h a t i f t h e r e i s

f E JCfX).

x

for e v e r y

then

X

E

f

I f f x ) I '1

such t h a t

X

E

i s c o n s t a n t on

ffa) I

T h i s i s t h e Maxi-

X.

mum PrincipZe.

47.G. L e t f X , C ) , f Y , r l ) and (2,s) b e Riemann domains Banach s p a c e s E , F and G , r e s p e c t i v e l y . Show t h a t : (a)

If

f E 3 C f X ; Y ) and

g

J ( f Y ; Z ) then

g of

E

Jc(X;Z).

(b)

If

f E C k ( x ; Y ) and

g E C k f Y ; Z ) then

g o f

E

Ck fX;Z).

(c)

If

f

E

PA(X).

47.H.

m E

JCfX;Yl

E

fX,O

Let

and l e t

P!f

E JCfX;F)

Let fX,E)

47.1.

morphism T : X g O T f o r every

each

x

PA(YI

E

let

X

E

g

47 .J.

L e t (ii,

E

x

x . Show t h a t f o r each t

then

E

Pmf E

E

E , let

Pmffxl

E

f

let

Y

g

E

: SCfY)

T*

+

X f Y ) . Show t h a t

and

t

E

and

j

E

let

INo

E

JCfII;F)

P f m E ; F ) bede-

U i s any

JCfX;PfmE;F)).

chart

Conclude

PTffx) = Pmf(x)ft).

E , where

E . Given a

X f X ) b e d e f i n e d by .r*fPTg)

T*g=

= P;(-r*g)

for

E.

5) be a Riemann domain o v e r

X

g o f

and f Y , n ) b e Riemann domains o v e r +

J C ( Y ) , m E INo

every

CmfX;FI,

g

be a Riemann domain o v e r

m0. For

containing

X

that

and

E

P m f ( x l = Pm[ f o f C ( U ) - - ' ] ( < ( x ) l , where

f i n e d by in

over t h e

E . For e a c h

Djffx)

E

Es(jEm;

f

E

FIR) bede-

D j f f z ) = ['D f o ( 5 1 U ) - * ] f s ( x ) ) , where U i s any chart U c o n t a i n i n g x . Endow c " f X ; F ) w i t h t h e l o c a l l y convex s u p II D3 f ( x ) l l , with t o p o l o g y g e n e r a t e d by t h e seminorms f

f i n e d by in

+

k E K

j

and

E Wo

Show t h a t i f

K

compact. Show t h a t

C X

E = 6r

n

t h e n t h e t o p o l o g y of

g e n e r a t e d by t h e seminorms

f

+

sup ?iE K

and way.

K

C

X

Cm(X;F)

compact, where

a"fc1 ax

1

5

i s complete.

Cm(X;F) (z)((, w i t h

is

also

c1 E

JV:

1 x 1 i s d e f i n e d i n t h e obvious

3 39

R I EMANN DOMAl NS

48. DISTRIBUTIONS ON R I E M A " DOMAINS I n t h i s s e c t i o n w e s h a l l e x t e n d t o t h e case o f Riemann domains s e v e r a l r e s u l t s from S e c t i o n s 1 6 and 1 7 . A key tool i s the f o l l o w i n g theorem on e x i s t e n c e of p a r t i t i o n s o f u n i t y . 48.1.

is a

L e t (X,<) b e a Riernann domain o v e r a s e p a r a b l e

THEOREM.

H i l b e r t space

E.

Then f o r each open c o v e r ( U i l i E I

p a r t i t i o n of u n i t y

C~

I

on

X

of X there

which

is s u b o r -

By r e s t r i c t i n g o u r a t t e n t i o n t o e a c h c o n n e c t e d compon e n t of X w e may assume t h a t X i t s e l f i s c o n n e c t e d . B u t t h e n

PROOF. X

i s a L i n d e l o f s p a c e and t h e proof o f Theorem adapted.

15.4

can

be

directly

X i s a t o p o l o g i c a l s p a c e t h e n w e s h a l l d e n o t e by K i X ) t h e s u b s p a c e of a l l f E CfX) which have compact s u p p o r t . If

48.2. THEOREM.

Let (X,<)

b e a Riemann domain over 6?.

e x i s t s a u n i q u e r e g u l a r BoreZ m e a s u r e

px

on

X

Then there

such t h a t

(48.1)

f o r each chart PROOF.

Let

f

U in E

KlX),

X

and e a c h

l e t (UiIiE

f E K(Ul. I

be a n a t l a s o n

X , and l e t

( p i 4€1 b e a p a r t i t i o n o f u n i t y on X s u b o r d i n a t e d t o ( U i j i E I . Then p i s = 0 f o r a l l b u t f i n i t e l y many i n d i c e s . I f w e d e f i n e

W i l t h e n one can r e a d i l y v e r i f y t h a t t h e d e f i n i t i o n of

Tf

is in-

dependent from t h e a t l a s and p a r t i t i o n of u n i t y chosen. And c l e a r l y T i s a p o s i t i v e l i n e a r f u n c t i o n a l on K ( X l , that is

340

MUJ ICA

2

Tf

0

whenever

f

2

T h u s , b y t h e R i e s z R e p r e s e n t a t i o n Thee

0.

r e m t h e r e i s a r e g u l a r Borel measure

ux

on

i,f

ux

v e r i f i e s ( 4 8 . 1 ) . And

dux

f o r every

vx

uniqueness of

Clearly

f E K(X).

such t h a t

X

f o l l o w s a t o n c e from t h e u n i q u e n e s s

Tf=

in the

R i e s z R e p r e s e n t a t i o n Theorem. i s a Riemann domain o v e r

I f IX,
we s h a l l d e n o t e by

IflPdpx <

-.

and

L p ( X ) t h e Banach s p a c e o f

classes of B o r e l m e a s u r a b l e f u n c t i o n s

I,

Cn

The v e c t o r s p a c e

f

all

X

on

Lp(X,locl

p <

1

is

then

equivalent such

that

simi-

defined

i s r e g u l a r o n e c a n r e a d i l y obtain px t h e f o l l o w i n g r e s u l t , whose s t r a i g h t f o r w a r d p r o o f i s l e f t t o l a r l y . S i n c e t h e measure

t h e r e a d e r as a n exercise. 48.3.

PROPOSITION.

Let

U be a c h a r t i n X .

let

Let tX,El

be a Riemann d o m a i n o v e r C

(X,c)

b e a Riemann domain o v e r f E C"tX)

X t h e n w e s h a l l d e n o t e by

D l X ) a r e c a l l e d t e s t f u n c t i o n s on 48.4.

DEFINITION.

6 > 0 , and l e t and bY

cp f

Let (X,C)

X6 -= (x

and

C n . W e s h a l l d e n o t e by

w i t h compact s u p p o r t .

f E R t X ) such t h a t

subspace of a l l

,

Then:

D l X l t h e subspace of a l l

i s a compact s u b s e t of

n

E

X

suppf

C

K.

DlK)

dU(xl > 63. Given

PiaiO;6I) w e define t h e i r

the

The members o f

X.

b e a Riemann domain o v e r :

If K

Cn, l e t

f E L

c ~ ~ ? i ~ o Z u i fi o* cnp

I

:

(X,lorl

X6

-

C

34 1

R I EMANN DOMA I NS

Ux = EX("; dx(x) I.

where 48.5.

Let (X,<) be a Riemann domain over En, and

PROPOSITION.

f E L 1 (X,locl and

let

aa

(b)

ax"

D(g(O;E;II. Then:

* -

= f

-(f*Cp)

small then Fix

a

E

6 >

0 is sufficientl;!

x E B (a;dX(a) - 6) we have X = <(UaI, a n d t h e r e f o r e

Then f o r e a c h

U6.

B(<(x);6) C B(<(a);dX(alI

Then ( a ) a n d ( b ) f o l l o w b y d i f f e r e n t i a t i o n u n d e r t h e s i g n . To show ( c ) l e t ( U , )

L.

Cw

p a r t i t i o n o f u n i t y on and

C $if

qif

E

D(Ui)

f * p = Z:

48.6. 1 <

$,f

*p E DlXl if

Let (lJi)

f p6

=

0

6 > 0

E

- flpd.sx

-+

0

that

is s u f f i c i e n t l y small.

Lp(X) be a function with compact

be a n a t l a s o n

t i t i o n of u n i t y o n X $if

Since

Let (X,<) be a Riemann domain over En,

-, a n d let

PROOF.

be a n a t l a s o n X a n d l e t (qi) be a X s u b o r d i n a t e d t o ( U i I . Then f =

i i t follows f r o m P r o p o s i t i o n 1 6 . 4

f o r each

If*

integral

f o r a l l b u t f i n i t e l y many i n d i c e s .

0

PROPOSITION.

p

c1

VlXl.

f *p E

-

that

f o r each multi-index

ax'

If f has compact support and

(c)

PROOF.

p E

6

when

+

sup-

0.

X a n d l e t ( + i ) be a

s u b o r d i n a t e d t o ( U . ) . Then 2

f o r a l l b u t f i n i t e l y many i n d i c e s .

let

Tt

cw

f = Zqif

follows

parand

from

342

MUJ I CA

P r o p o s i t i o n s ' 4 8 . 3 a n d 1 6 . 6 , a n d E x e r c i s e 16.A,

that

f o r e a c h i. Then a n a p p l i c a t i o n o f M i n k o w s k i ' s i n e q u a l i t y y i e l d s t h e desired conclusion. COROLLARY. If (X,S) i s a Riernanrz d o m a i n o v e r

48.7.

D(xi is

then

p

dense in

and 1 -

Cn

Lp(xI.

f = 0

almost e v e r y w h e r e on Xi € o r a l l b u t c o u n t a b l y many c o m p o n e n t s Xi o f X. Then the proof

PROOF.

Let

f E

Lp(X).

Then

of C o r o l l a r y 1 6 . 7 a p p l i e s . 48.8.

L e t (X,5 ) b e a Riemann domain o v e r

DEFINITION.

(a)

F o r e a c h compact s e t

K

C

t h e v e c t o r space

X

w i l l b e endowed w i t h t h e l o c a l l y c o n v e x t o p o l o g y

Cm(X). The v e c t o r s p a c e

Cn.

induced

DtX) w i l l be endowed w i t h t h e

(b) tions

The v e c t o r space

X.

Z)(KIC>

K C X.

Each c o n t i n u o u s l i n e a r f u n c t i o n a l o n

a d i s t r i b u t i o n on

by

finest

l o c a l l y convex t o p o l o g y s u c h t h a t t h e i n c l u s i o n mapping

U ( X ) i s c o n t i n u o u s f o r e a c h compact s e t

D(K)

DlXl i s c a l l e d

D'lXl o f a l l d i s t r i b u -

X w i l l b e a l w a y s endowed w i t h t h e t o p o l o g y o f p o i n t -

on

w i s e convergence. A s i n Section 17 t h e vector space

L 1 ( X , Z o c ) c a n be canoni-

c a l l y i d e n t i f i e d with a v e c t o r subspace of 48.9.

let

U be an open s u b s e t of

t i o n of

= TP

48.10.

If T

(b) Scp

L e t (X,{) b e a Riemann domain o v e r

DEFINITION.

(a)

T E pr(X)

to Given

D'(x).

S, T E

f o r every

PROPOSITION.

and

X.

and

D(UI. Thus

Cn

T

T

I

1

U

w i l l denote t h e

restric-

U E V'(U).

D ' (Xi w e s h a l l s a y t h a t

S

= T

on U i f

cp E U ( V ) .

L e t (X,5) be a R i e m a n n d o m a i n o v e r

8.L e t

343

R I EMANN DOMA I NS

( u i ) i e I be an open cover o f X . there e x i s t s ever

n U

Ui

such t h a t

PROOF.

T

Let

Ti

j

1

E

D'(Ui)

Ti = T

such t h a t

j

V ( X - 1 . If

($i)iEl

on X subordinated to ( U i l i E I

is a

then

qilp

ern E

E

I

Ui n U whenj

on

is n o n v o i d . Then t h e r e is a u n i q u e Ui = T i f o r e v e r y i E I .

9 E

i

Suppose t h a t f o r each

T

E

D'(Xl

partition of unity VfUi)

for every i E

and $ 2. p E 0 for all but finitely many indices. If we define T I P = Z T.(Qip) then one can readily verify that the i EI 2 definition of TV is independent from the partition of unity chosen. T is clearly linear. To show that T is continuous let K be a compact subset of X and let J = { i E I : K n Ui # $ 1 . I

Since

Tp =

Z

Ti(14i90) for every

9 E P(KIJ and since each of

i EJ

$i9 E D ( U i ) is continuous we conclude the mappings 9 E D ( K ) that the restriction of T to D ( K l is continuous. This shows that T E D ' t X ) . If V E D(Ui) then qjV E D(U n Ui) for +

j

every j and therefore

proving that T I U i = T i for every i. Finally, to show uniqueness let T be a distribution on X such that T 1 Ui = T i for every i . If 9 E D l X ) then

and the proposition has been proved. Let (X, < ) be a Riemann domain over C n and let T E P ' i X ) . Proceeding as in Section 17 we can show the existenceata largest open set V C X (possibly empty) such that T = 0 on V. The set X \ V is called the ;uppor*t of T and is denoted by s u p p T. Before defining derivatives of distributions we establish

344

MUJ I CA

the f o r m u Z a of i n t e g r a t i o n by p a r t s . 48.11. PROPOSITION.

Let

(X,t)

be a Riernann d o m a i n over

8.T h e n

PROOF. The formula is clearly true if the support of P lies in a chart U in X. To establish the formula for an arbitrary m p f Pix), let (Ui) be an atlas on X and let ( $ i ) be a c partition of unity on X subordinated to (Ui). Then

Let ( X , c )

48.12. COROLLARY.

48.13. DEFINITION. T

=

E

p r ( X ) and

(- 1 )

c1 E l N z n

be a Riemann domain over

w e define

7 aaT E D’tXl ax

i4 T ( c a %1I

f o r every

ax

The mapping

Let ( X , < I

b e a Riernann domain o v e r

T

E

D(X).

by

c. T h e n

6;”.

Given

aaT (Pi

axa

345

RIEMANN DOMAINS

continuous. If f E C"(Xl then it follows from Corollary 48.12 that the definitjon nf the derivative a"f/az" in the sense o f distributions coincides with the classical definition. 48.14. DEFINITION. Let (X, 5) be a Riemann domain over 8.Given f E c"(X) and T E V'IXI, their product fT E D ' f X l is defined

by f f T ) ( p ) = T ( f p )

for every

f

T

If

E

Cm(X) and

p E D(U).

V ' f X ) then the formula

E

can be proved exactly as in Section 17.

EXERCISES Let (X,<) be a Riemann domain over t n fand let 1 ' p < m . Show that a Borel measurable function f on X belonrjs to Lp(X, Zocl if and only if f o (5 UI-' belongs to Lp(5(UI, Zoc) for each chart U in X. 48.A.

n

48.B. Let (X,C) be a connected Riemann domain over 6 and let m 1 5 p m. For each 9 E c (X;lRl l e t Lp(X,p) denote the Banach space of all equivalent classes of Borel measurable func-

I

tions f on X such that

<

IflPe-piipX

m.

Show that LpfX,Zoci

X

is the union of the spaces

48.C.

Lp(X,p),

with

Let ( X , < ) be a Riemann domain over

an atlas on

X, and let

(

I

)

J

~

on X subordinated to the atlas

be~ a ~

)

~

c"

iUi)iEI.

distribution with compact support, and let

X

(a) $i T .

iEJ to P ' f V i ) .

9 E

Cm(X;lR).

Cnl

let ( U i l i E I

partition of Let

be unity

T E V r ( X l be a

p E P(B(0;S)).

Show that T can be written as a fiziite sum T = Show that each $ . T has compact support and belongs 2

MUJ ICA

346

(b) Show that if 6 > 0 is sufficiently small then the convolution ( J l i T ) * p can be defined in a natural way and belongs to D ( U i ) €or each i € J . Show that if we set T *Lp =

(c)

Show that

T

*

p6

-+

T

in

D ' l X ) when

6

+.

0.

49. PSEUDOCONVEX RIEMANN DOMAINS

This section is devoted to the study of pseudoconvexRiemann domains over Banach spaces. The results in t h i s section are natural generalizations of the results from Section 3 7 . 49.1. DEFINITION. Let I X , 5 ) be a Riemann domain over E, and let (x, t ) E X x E . We shall denote by 6,(xC, t l the supremum of

all r > 0 for which there is a set D in X containing z such that 5 1 D : D <(x) + Art is a homeomorphism. For each r with 0 < r < 6 X ( x , t ) we shall denote by A X ( x ; t ; r ) the connected component of E;-115(x) + A r t ) which contains x. -+

49.2. PROPOSITION.

(a)

5

I

Ax(x;t;r) : Ax(x;t;rl

m o r p h i s m f o r every ( x , t )

(b)

be a R i e m a n n d o m a i n o v e r

Let ( X , c )

The f u n c t i o n

x

E

6x : X

x

E

X

3

and E

-+

51x1 + Art

7:s

E

a

homeo-

0 < r < GX(x,t). (0,

w ]

is l o u e r

semicon-

continuous.

PROOF. ( a ) Given ( x , t ) E X X E and 0 < r < f i X ( x J t )be can find a set D in X containing x such t h a t 5 1 D : D +. c1.z) + Art is a homeomorphism. By Lemma 47.4 D = A X ( x ; t ; r ) and (a) has been proved.

347

RIEMANN DOMAINS -

5 I x X ( x O ; t o ; s ): A X ~ x o ; t o ; s ~ Efxo) + +

-

Asto

is a homeomor-

phism. By Exercise 47.B there is an open set U in X containing A X ( x o ; t 0 ; s ) such that r for every ( x , t ) E V x W and ( b ) has been proved. c(xo) + nsto

(c)

C

Let

x

E

X. Since

for every r > 0, the inequality d X ( x ) 5 i n f { 6 x ( x , t ) : t E E, I I t /I = I ) is clear. To show the reverse inequality let 0 P < inf 6 X ( x , t ) .- t E E , IIt 11 = 2 ) . For each t E E with I1 t I1 = 1

5

-

1

-+ 51x1 + F r t is a homeomorphism, and X then by Exercise 47.B there is an open set U t in X contain5 ( U , ) is a homeomorphism. ing A x ( x ; t ; r l such that 5 I U t : Ut

xX(x;t;rl : A (x;t;r/

+

U ( U t .- t E E , I1 t II = 1 ) . Since we may assume that is convex for every t, an application of Lemma 47.3 shows that
Set

U =

StU,)

-+

49.3. DEFINITION.

A Riemann domain (X, 5, over E is said to be - l o g 6 x is plurisubharmonic on

p s e u d o c o n v e x if the function

X

X

E.

49.4. DEFINITION. Let (X,<) be a Riemann domain over each set A C X we define

X

E.

For

If t X , 5 ) is a Riemann domain over E then for each set A we shall s e t d X ( A ) = i n f d , ( a i . a€ A

C

348

MUJ I CA

49.5.

THEOREM.

E the following

For a Riemann domain ( X , < ) over

conditions are equivalent:

x

(a) (b) for each

f

E

is pscudoconuex.

The function t E E.

- log Ax( - ,t) is plurisubharmonic on X

(c)

The function - l o g d x is plurisubharmonic on

(dl

d x ( G n c ( x ) ) = dX(Al for each set

(e)

dx(iPnctx) I > 0

(f)

dX(I?Pb(XII > 0

C

X.

f o r each compact set

K

A

f o r each compact set

(9) If (H,D) is a Hartogs figure i n k 2 Jf(H;X) there exists 7 E Jf(D;Xl such that

PROOF. The implications (a) * (b)I (e) and (e) =. (f) should be clear. (f)

=)

(4) :

X.

X.

C

K C X. then

for

7I H

= f.

each

(b) * ( c ) , (c) * ( d ) , ( a ) +

Consider the sets

where 0 < P < R and 0 < s < R. Given f E JC(H;X) we want to find E Jc(D;Xl such that I H = f. Fix b with r b < R. For each t E ( O , R ] let

7

7

and set

T = {t

E

( O , R ] : there is

We shall prove that

f,

E

JCIDt;X) such that f t = f

T = (O,R]. This will imply (9)I

function f defined by

7=

f

on H and

7=

fR

on

011

Dt n H}. for DR

the will

349

R I E M A N N DOMAINS

then be the required extension. Since ( U , s ] C T the set T is nonvoid. Since t n E T f o r each n E IIV implies that s u p t n E T the set T is closed in ( O , R ] . To show that T is open in fix t E T consider the set iU,Rl

Then J

with

t < R,

fix a with

is a compact subset of H

and

<

r

K = f(J)

a

b

and

is a compact

subset of X. If u E PA(X) then u o f , E P d ( D t ) by Exercise 47.G and hence u o f,(h,<) is a subharmonic function of A for 1x1 E . Set g = 5 0 f E 3 € ( H ; E ) . By Example 10.2 and Exercise 8.1 there exists E 3€(D;E) such that 3 = g on H . Since = g = < o f = 5 o f t on H n D t we = 5 o f t on Dt. Since is uniformly continuous see that on each compact subset of D we can find 6 > 0 with t + S R such that exists

E

> U

such that

5

for all (X,C), < 26.

Given

(X',C')

(A,<)

E Dt

D t+6

such that

IX - A' I

< 26,

15

-

5'

1

we claim that

for all ( X r , C r . J E D t such that Indeed, consider the convex set

11' - X I

< 26,

Then (49.1) implies that 5 o f t ( A ? C BE(< o f , ( X , < ) ; f t ( A l is connected and contains the point f,(X,
15'

-

< 26.

and since we conclude €1,

350

MUJ I CA

that

ft(A) C BX(ft(A,5);E)

we define

f t + 6 : Dt+6

+

and (49.2) has been

proved.

Next

by

if ( A r , < ’ ) E Dt satisfies I A r - X I < 6, I I ; ’ - < I .c 6. Observe that (49.1) guarantees that ;(A, 5 ) E B E ( < of, ( A ’, 5 ,); E ) . We have to show that f,+, is well defined. Indeed, let ( A , < ) E and let (A‘,<’), (Arr,<”) E D such that I X ’ - XI < 6 , Dt+6 t l i ; ’ - 5 ] < 6, ( A ” - A 1 < 6, 1 L r r - < 1 < 6. Then I X ’ - h f r l < 26,

-<“I

< 26 and (49.2) implies that f t ( X r , C r )

li’

E

B (f ( X r r , < r r ) ; ~ ) . X

t

Then Lemma 47.3 implies that 5 l B X ( f t ( A r , C r ) ; ~ )V B X I f t l A r r , 5 r r ) ; ~ l is injective and hence ft+& is well defined. It is clear that ft+6 E J C ( D t + 6 ; X ) and f,+& = f, on D t . This shows that T is open in ( O , R ]

-

and (9) follows.

(9) * (a): Assuming (9) we have to prove that the function is plurisubharmonic on X x E . It suffices to show

log6x

is plurisubharmonic that the function - l o g S x ( E I U 0 ) - ’ t x ) , y ) on [(Uo) x E for each chart U, in X such that [tUol is convex. Fix ( a , b ) E <(Uol X E and ( s , t ) E E x E such that

Let

P

E

PlC) such that

for every

A

E

an, or equivalently

for every A an. We have to show that ( 4 9 . 4 ) holds for every X E It suffices to prove that

a.

for every

A

E

n

and

0 < p < 1 . Fix

p

with

0 < p

i

and

RIEMANN DOMAINS consider the function

g

E

2

X ( 6 ! ; E ) defined by

(49.3) implies that g(a x I o I ) c ((Uo), and a compactness argument yields an open disc Do centered at zero and containing A, and an open disc Dd centered at zero, such that g(Do x D d ) C ((U,). By (49.4) and Exercise 4 7 . B for each A E a A we can -1 (a + A s ) such that find a chart U,, in X containing ( 5 I U,) Then for each A E a A ((U,,) is convex and contains g ( o \ } x we can find an open disc D,, centered at A , and an open disc I D,, centered at zero and containing A , such that g ( D X x Ul;) C CtU,,). Set

a).

We want to find a function f E J C ( G ; X ) such that

= 9.Eand f =

< o f

fine f : G + X by f = ( 5 I U0)-' 0 9 on Do x 1"; ( 5 1 U,,)-' o g on D,, x for each E aA. We must show that f is well defined. If X E an then ( 5 i V o ) - 2 ( a + As) E Uon U,, and <(Uo) n <(U,,) is convex. TheiiLemma 47.3 implies that 51Uo -1 o g = ( 5 I U,,)-' o g on u U,, is injective and therefore ( 5 I U,) ( D o x D d ) n ( D h x Di). Next consider 11 E aA such that

Di

I

( P A x !I{)n ( D x D ' ) is nonvoid. Then D,, n D P P u is nonvoid too, and this clearly implies that D,, n Du n Do is nonvoid as well. If p E D~ n D~ n D~ then

Since

on I D X X Di) n ( D x 0 ; ) n (Do x D L ) , u implies that ( 5 I U h ) - 7 o g = ( 5 I U P ) - 7 Thus f is well defined. Clearly f E serve that G contains an open set H

the Identity Principle 47.D ')I3 (D XD'!. on (D,, xDX u i-I J C ( G ; X ) and 5 o f = g . Obof the form H = H , U t i 2 ,

og

1

MUJ I CA

352

where

Thus f where

n

If

b

# 0

and by ( g ) f has an extension

= g on H we see that 5 0 7 = g set A X = f t { X } x A ) . Note that

5

Since X E

E J€(H;X)

+

on

of

1

D.

E

K(D;X),

For

each

then ( 4 9 . 5 ) is obviously true. And if b + A t then we immediately see that g 1 { X I x A : { A } x A g({X} x A! A t = 0

+

is a homeomorphism. Whence it follows that 5 I AX : AX + 5 ( A X I is a homeomorphism and ( 4 9 . 5 ) holds. This shows (a) and completes the proof of the theorem.

E XERC I S E S

Let X and Y be topological spaces and let f : X Y be a local homeomorphism. Let N be a topological subspace of Y and let M = f-'(N). Show that f I M : M N is a local homeomorphism. 49.A.

+

+

Let (1,s) be a Riemann domain over E. Let Bo closed vector subspace of E , let Xo = < - ' / E o ) and let 5 I X o . Show that ( X ,so) is a Riemann domain over Eo. 49.B.

Let ( X , < ) be a Riemann domain over

49.C.

be

5,

a =

E.

(a) Show that an upper semicontinuous function f : X + is plurisubharmonic if and only if its restriction

[-m,ml

353

R I EMANN DOMA I NS

f

1

&-l(Eo)

space

Eo

is plurisubharmonic for each finite dimensional subof E.

c-'(E0I is pseudoconvex for each finite dimensional subspace E o of E. (b)

Show that X is pseudoconvex if and only if

49.D. Let ( X , < ) and (Y,qI be Riemann domains over respectively. (a)

Show that IX

x

E

and

F

Y, ( < , n I ) is a Riemann domain over

(c) Show that X x Y and Y are pseudoconvex.

E

X

is pseudoconvex if and only if

50. PLURISUBHARMONIC FUNCTIONS ON RIEMANN DOMAINS In this section we extend Theorem 38.6 to the case of ~Xiemanr, domains. This is a key step in the solution of the 3 equation in Riemann domains. We shall give the main result after a series of preparatory lemmas. 50 1 . LEMMA.

u PU

L e t tX,<) b e a Riemann d o m a i n o v e r

b e a c o n n e c t e d c o m p o n e n t of

5n, and

let

X 6 = {x E X : d,(x)

> 6). I f U t h e n t h e s e t {x E U : i s r e Z a t i v e Z y c o m p a c t i n X for e a c h xo E U and

denotes the geodesic distance i n

pu x 0'

x)

5 c)

PROOF. Choose E with 0 c E 6/3. Fix xo E il and set K = { x E U : p;;(xo,x) 5 c ) for each c E B. Since p u is a metric the set Kc is obviously compact when c 5 0. To prove thatKc it clearly suffices to is relatively compact for every c E 2R prove that if

K,

is relatively compact for a certain

then K C + E is relatively compact as well. Before proving we show that

c

2

0

this

354

MUJ I CA

(50.1)

Indeed, given y E K c i E there is a polygonal line in U , with xrn = y , such that

[xO,...,xm]

m (50.2)

if

m E

j=l

ll<(x.) 3

- <(X~-~)II

5 c

then

pU(xo,y) 5 c and y

m

Thus we may assume that Then we can find k with that

c < 1

C

j=l

5

I/ C ( x . 1

k 5 rn

3

-

and

< ( X ~ - ~ ) I I<

x E

c

E

Kc.

+

?E.

[ X ~ - ~ , X ~ such ]

and therefore

By (50.3) p u f x 0 , x ) 5 c and hence x E K c . By(50.4) l l < ~ x . ~ - < ~ d I l 3 < 2~ for j = k , k + I , ..., m , and it follows from repeated applications of Lemma 4 7 . 3 that the points X , , , X ~ + ~ , . . . , X ~ all lie in B X ( x c ; ~ E ) .In particular y = x, E B X f X ' ; 2 & ) and ( 5 0 . 1 ) has been proved. If Kc is relatively compact for a certain c 2 0

then there are points

al,...,a

m

E

Kc

s u c h that

m U

Xr

C

m B X f a j ; & ) . Then it follows from (50.1)that ii.1

f

c_ C

j=l

.u B X ( a j ; 3 € ) 3 =1

and

KC+E

is relatively compact set as well.

rhis

completes

the proof of the lemma. 50.2.

LEMMA.

and Z e t

L e t (X,<) b e a p s e u d o c o n v e x R i e m n n n d o m u i n

,~i!er

X 6 . Tizen t h e r e is U : f i x ) ~< c} i s c o r n p a c t , f o r e a c h c E B . In p a r t i c u ~ a rt h e s e t is 'Pb e (11) c o m p a c t f o r e a c h c o m p a c t s e t X (I U. Cn,

a function

U b e a c o n n e c t e d component of

f E Pbc(U) s u c h t h a t t h e s e t

{x E

355

R I E M A N N DOMAINS

PROOF. Fix E with 0 2~ < 6 and let V be the connected component of X E which contains U. Observe that U C V E . Fix x E V and set a t x ) = pytxo,xl for every x E V . It is easy to see that

if x and y both lie in a same chart. Then it follows from 1 Proposition 17.18 that au/ax E L tv, L o c ) and I aa / dx . I 5 I j 3 almost everywhere in V for every j. Then it follows from P r o p sition 48.5 that u R p, E C m ( U ) and for each x E U and j , k = I,. . ,n we can write

.

where

Ux = Bx(x;dx(x)). Hence the functions

a 2 i a * p,)

/ ;izjagk

are uniformly bounded on U by a constant which depends only 011. E . Hence we can find a constant c E > 0 such that the function

is plurisubharmonic on

aixl - a *

P,(d

U . It follows from (50.5) that

= ij(O;d

for every z 6 U. Then by Lemma 50.1 we may conclude that the set {x E U : u E ( x / 5 e} is relatively compact in X for each 0

6 B.

We next consider the function

356

MUJ I CA

Since the function 9 ( t l = - l / t

is convex and increasing for

we see that v E PhCtXs). Set f = max { u , , v } . Then f E Pbc(U), and since v ( x .) 03 when ( x .) approaches the boundary 3 3 of U we conclude that the set {x E U : f ( x ) 5 c) is compact for each c E B. The proof of the lemma is complete. t

0

+

5 0 . 3 . LEMMA. L e t ( X , c ) b e a p s e u d o c o n v e x R i e m a n n domain o v e r En, and Z e t U b e a c o n n e c t e d c o m p o n e n t o f X 6 . L e t K be a comp a c t s u b s e t of U ‘ a n d L e t V b e an o p e n n e i g h b o r h o o d o f bPhiu,

in

U. T h e n t h e r e i s a f u n c t i o n The s e t

(a)

{x

E

U

: f ( x )

In view of Lemma 50.2,

PROOF. 50.4.

LEMMA.

over

Cn,

5 c}

i s rompact f o r earh c

E

R.

the proof of Lemma 38.1 applies.

L e t (X,6) b e a c o n n e c t e d p s i v r h * e y i z ’ ~ . I :Riemann domain

and l e t

Xr

s e t of

f E PhclU) s u c h t h a t :

r

2

s > 0 . Then

I

K

f o r each compact s e t

?

i s a~ compact~

sub-

C X r .

Let K be a compact subset of X r . Since X is connected we can find 6 with s > 6 > 0 such that K is contained in a connected component U of X s . By Lemma 50.2 K p n c l U , is

PROOF.

A

a compact subset of U. Let

Xs(K)

nected components of

which intersect I f . Then X,IKI C U ‘PhC(U) . This shows that Kphc(xs)

Xs

denote the union of those conA

A

and ‘PhC(Xs) = KPbC(Xs(KI) is a compact subset of U. But on the other hand, since K C X p there is t- > 0 such that d > r i f on K , and therefore

x-

dX

2

+

On

‘phc(x)*

Hence

z

KPhC(X,,

“^bCiX,

c

xr

and

A

is a compact subset of

KphctXs)

LEMMA.

50.5.

over {J. E

En.

X

L e t (X,S ) be a c o n n e c t e d u s , zrdoconvex Ritwunn domain

T h e n t h e r e is a f u n c t i o n

: f(xl

as asserted.

Xr,

5

c)

f

E

is c o m p a c t f o r e a c h

PhciX) s u c h t h a t t h e s e t L‘ E iR. : n p a r t i p i l l a r

~

~

357

RIEMANN DOMAINS

the s e t

KPnc(x)

is c o m p a c t f o r e a c h c o m p a c t s e t

K C X.

PROOF. By Corollary 47.8 X is hemicompact. Let ( A .) be an in3 creasing sequence of compact subsets of X such that each compact subset of X is contained in some A j . Let fr.) be a sequence 3 of strictly positive numbers decreasing to zero such that Aj is contained in a connected component U j of Xr Set K =

.

*

n Xr

'j+l

j . Then

for each

(Aj)pbC(Ui+il

j

i

is a compact

Kj

by Lemmas 50.2 and 50.4. For each j

subset

let 8 . denote 3

j

the union of those connected components of X,,: K Observe that U c v c U j , l has and V

2

j .

which intersect only finitely

j

j

of

many connected components. By Lemma 50.2 there is a v l E P b C ( V l ) such that the set {x E V l : V , ( X ) 5 c }

function is com-

pact for each c E iR. Choose e l > 1 such that u l < c 1 On Kl and set L l = {x E V I : v l (x) 5 e l 1 . Since L 1 is Contained in some A

we may assume without loss of generality that L1

j

Since ( B l ) p h C ( V 2 ) = ( ' l ) p h C ( U 2 )

A2.

of Lemma 50.3 yields a function

{x

E

V2 : v2(xl

5 el

v2

on

Al

and

on

K2

> 2

and set

v

= Kl C V 1 , 2

C

an application

P b C ( V 2 ) such that the set

E

c E IR,

is compact for each V 2 \ V l . Choose

< 0

on

such that v2 < c2

c2 > 2

L 2 = (x E V 2 : v 2 ( x )

v2

5 e2}. Proceeding induc-

tively we can find a sequence of functions ( u . I

3 j=l

m

P h c ( V . ) , and a sequence of constants ( c . )

3 j=l

3

m

with

with c

j

I

v

j

E

j , such

that if we set

then and

Aj

v

C

> j

Lj

C Aj+l

on

5

1,

for every

j

for every

V . \ Vj-l 3

j

and

2

be a convex increasing function such that and x i t l 2 t f o r t 2 1 . If we define $. = J

j

c

i=I

x o v

i

v

j < O On AJ2. Let x E c ~ ( R ; w )

Xtt) = 0

€or

t< 0

358

MUJ I CA

f,. = f k on Ai whenever j, k i and exists and belongs to P b c ( X ) . To hence the limit f = Z i m f j complete the proof of the lemma we shall show that for each

j

x

5 j)

( x E X : f(x)

(50.6)

Let

then

E lNl

E

Lk+l

X \L

Then there is

i -

for each

C Lj

k

2

and therefore

X o v k ( x ) > k. If

3:

#

Vk

x

then

THEOREM.

domain o v e r

Let ( X , c )

cfn.

Let

K

vkil(x)

an o p e n n e i g h b o r h o o d o f

Kpdixi.mThen

pZurisubharrnonic f u n c t i o n

(a)

The s e t

(b)

f1x) < 0

{x

E

X

Riemann

and Z e t V be

there exists

a strictZy

f E C 1 X l such t h a t :

X :f ( x l < c ) i s c o m p a c t f o r each

f o r every

x

E

c

E

lR.

K.

In view of Lemma 5 0 . 5 , the proofs of Lemma Theorem 38.6 can be directly adapted. PROOF.

> k + 1

f(x) > k > j ,

be a c o n n e c t e d p s e u d o c o n v e x

b e a compact s u b s e t o f

and

then v k ( x ) > k

E Vk

and therefore X o vkil 1x1 > k + 1 . In each case proving ( 5 0 . 6 ) and the lemma. 50.6.

x @ Lk

j such that

We distinguish two cases. If

*

j E lN.

38.1

and

EXERC ISES 50.A.

that

Let (X,<) be a pseudoconvex Riemann domain over Cn . Show h X ) = K^Pne(X) for each compact set K C X.

Let f X , < ) be a pseudoconvex Riemann domain over E l let f E P n t x l and let U = {x E X : f ( x l < 0 ) . Show that U isalso pseudoconvex. 50.B.

50.C.

over

Let (X,€,) be a connected pseudoconvex Rjemann domain Show that for each 6 > 0 there exists a strictly

en.

359

RIEMANN DOMAINS plurisubharmonic function f 6 E Cm(Xsl such that ( x E X 6 : f s ( x ) 5 c} is compact for each c E l??.

51. THE

7

the

set

forms

and

EQUATION IN RIEMANN DOMAINS

The definitions and properties of differential -

the a operator admit straightforward extensions to the case of Riemann domains. The necessary ingredients from differentiation and integration in Riemann domains were introduced in Section 48 so as to make these extensions clear. In particularthe have the expected spaces L 2 ( X I , L 2 ( X , 2 o c ) and L E q ( X , i p ) P9 P9 meanings, and the operator a induces operators

and

similar to those introduced in Section 40. Likewise, if a , m y E C ( X ; I R ) are three weight functions then the operator induces operators

-

a

and

similar to those introduced in Section 40. Then the resalts from Sections 40, 41 and 4 2 admit straightforward extensions to the case of Riemann domains. An examination of the corresonding proofs shows this. We restrict ourselves to state the main results in the form they will be most convenient to us. 51.1. THEOREM. domain o v e r

(a)

Let ( X , c )

En, and l e t

be a connected pseudoconvez Riernann 2 go E L o l I X , l o c ) w i t h ago = 0 . T h e n :

There e x i s t weight f u n c t i o n s

a, B,

y

E

Cm(X;lR)

such

360 that

MUJ I CA

g o E L O2 l ( X , B )

g E L:l(X,13) such that ag = 0 t h e r e e x i s t s 2 E I, (x,aI s u c h t h a t af = g a n d I1 f k l c1 5 IIg I1

(b)

f

and

For e a c h

-

L e t (X, 5 ) b e a p s e u d o c o n v e x R i e m a n n d o m a i n over

51.2. THEOREM.

m

cCn, 2

and Z e t

L (X,loci

g

E

C,,(X)

with

of t h e e q u a t i o n

-

af

-

ag = 0 . T h e n e a c h soZution

= g

b e Z o n g s to

f

E

Cm(XI.

NOTES AND COMMENTS

The results in Section 47 can be found in M. Schottenloher

1 ] or G. Coeurg 1 1 1 . Our presentation in Section 49 follows M. Schottenloher [ 3 1 , whereas in Section SO we have essentially followed the book of L. Hormander [ 3 1 . The results in Section 51 are special cases of results obtained by L. [

Hormander [ 2 ]

[ 3 ]

.

CHAPTER XI1

THE LEV1 PROBLEM IN RIEMANN DOMAINS

52.

THE CARTAN-THULLEN THEOREM IN RIEMANN DOMAINS

In this section we extend to the case of Riemann domains the notions of domain of holomorphy, domain of existence and holomorphically convex domain, and study the relationships between these notions. 52.1. DEFINITION. let F c X i x i . (a) If i Y , q ) X Y is said there is a unique T : X Y is said an X(X)-extension ‘I

:

+

+

Let I X , < )

be a Riemann domain

over

and

is a Riemann domain over E then a morphism to be an F - e x t e n s i o n of X if f o r each f E F f E 3 C f Y ) such that f o T = f. A morphism to be a h o l o m o r p h i c a r c t e n s i o n of X if T is of X.

(b) X is said to be an F-domain o f hoZornorphy F - e x t e n s i o n of X is an isomorphism. X is said to be of h o l o m o r p h y if X is an X(Xl-domain o f h o l o m o r p h y . to be the d o m a i n o f e x i s t e n c e of a function f E X ( X ) an { $ } - d o m a i n o f h o Z o m o r p h y .

x

E

if each a domain X is said if X is

is an F-extension of X then it isclear that T i X ) . We observe each connected component of Y intersects that a morphism T : X + Y is a holomorphic extension of X if g o T E X ( X ) is an aland only if the mapping T * : g E X i Y ) gebra isomorphism. If

T :

+

Y

+

DEFINITION.

52.2.

let

F

C

Let ( X , < ) be a Riemann domain

XIXI.

36 1

over

E

and

MUJ I CA

362

(a)

For each set

A

C

X

we define

(b) X is said to be F-convex if the set f F is compact for each compact set K C X. X is said to be holomorphically convex if X is X(X)-conuex. (c) X is said to be metrically F-convex if dX(zF1 > 0 for each compact set K C X. X is said to be metricalZy holomorphicaZZy convex if ‘X is metrically X(XI-convex.

x, y E X such that x # y and < ( x ) = < ( y ) there exists an f E F such that fix) # f l y ) . X is said to be holomorphically separated if X is JCIX)-separated. (d)

X

is said to be F-separated if for each

52.3. LEMMA. Let (X,5) be a Riemann donain over E. T h e n X i s hoZomorphically separated if a n onZy if each holomorphic extension of X i s injective.

PROOF. We first assume that X is holomorphically separated. Let ( Y , n l be a Riemann domain over E , let T : X Y be a holomorphic extension of X, and let x,y E X with x # y. If c(x) = s l y ) then by hypothesis there exists an f E X ( X ) such that f(xl # f(y). Hence 7 0 T(X) # 7 0~ l y land therefore T(X) # ~ ( y ) . If <(x) # < l y l then n o -r(x)# n o - r I y ) and T(X) # ~ ( y l too. Thus T is injective. +

Conversely assume that X is not holomorphically separated. Define an equivalence relation on X as follows. Let x y if Q ,

tixi = ( ( y l and f ( x ) = f(y) for every f E J C ( X ) . Let Y = X / Q be the quotient space and let IT : X + Y be the quotient mappping. Clearly 5 induces a unique mapping n : Y + E such n o r = and each f E X(Xi induces a unique function that f : Y + 6 such that ? o r = f. Since Y has the quotient toLet pology, the mapping q is continuous, and so is each f. x, y E X with r(xJ # r(yl. Then either q o r(x) = c ( . r ) # < ( g ) - n o r ( x ) ,

<

363

THE LEV1 PROBLEM I N RIEMANN DOMAINS

or else there is an f E X t X l such that f o T ( x ) = f ( x ) # f ( y l = ~ o n ( y ) . Whence it follows that Y is a Hausdorff space. We next show that the mapping 1~ is open. Let A be an open subset of X . To show that n ( A / is open in Y it suffices to prove that TI - 1 ( n ( A ) I is open in X , by the definition of the quotient topology. Let x E n - ' ( n ( A ) ) . Choose a E A such that x 2, a , and choose r > 0 such that r < dX(xI, r d x l a l and BXla;r) C A . We claim that B ( x ; r l C n-'(n(A)l. Indeed, since x % a X then for each f E X i X l and t E B E ( O ; r ) we have that

where that

cz

I

= 5

B x ( z ; d X ( z ) l for each

B X (x;r)

C

n-'(n[All

%

Whence it

z E X.

F;a'(F,!al + t l for each

and

T

t

E

follows

B E ( O ; r l . Thus

is an open mapping, as asserted. Now,

let x E X and choose an open set U in X containing x such that < 1 U : U <(UI is a homeomorphism. Since n is continuous and open, it follows that both mappings 71 I U : U v(UI and n n(U) : nlU) S ( U ) are homeomorphisms. Thus (Y,qJ is a Riemann domain over E and TI : X --t Y is a morphism. Since 7 o 71 = f, it is clear that f E X ( Y l for each f E X ( X l . Thus TI is a holomorphic extension of X which is not injective, and -+

-+

+

the proof of the lemma is complete. 52.4.

DEFINITION. Let ( X , < ) be a Riemann domain over E , let A be a subset of X such that d X ( A ) = p > 0. Then each set B C B E ( O ; p ) we shall set

where 52.5.

5, = LEMMA.

F;

1

BX(x;dX(x)

Let (X,
for each

x

E X.

be a Riemann domain o v e r

= d X ( K l . Then

a compact s u b s e t o f

X,

c o m p a c t s u b s e t of

f o r e a c h compact s e t

PROOF.

Choose

X

and l e t

0 < r c s c p

p

and for

such that

L

L

C

C

B, l e t K b e K + L is a

BE(O;pl. BE(O;r).

Since K

364

MUJ I CA

such is compact there are points xl, ..., x m E K m U BX(xj; s - r ) . Then one can easily see that

that

i(

C

j=l

and the desired conclusion follows. The next results parallels Theorem 11.4.

52.6. THEOREM.

Let

( X , 5 ) b e a Riernann d o m a i n o v e r

E. Consider

the following conditions:

i s a domain o f e x i s t e n c e .

X

(a)

i s h o l o m o r p h i c a l l y s e p a r a t e d and X i s t h e union of an i n c r e a s i n g sequence o f open s e t s A j s u c h t h a t dx((zj)x(xj) > 0 for e v e r y j . (b)

X

(c)

X

i s h o Z o m o r p h i c a l l y s e p a r a t e d a n d f o r e a c h sequence

12.) i n X such that 3

d 1x ) x j

suplf(x.ll =

such that

3

+

0

there i s a function

X

i s a domain o f holomorpky.

(el

X

i s h o l o m o r p h i c a l l y s e p a r a t e d and

(f)

E

#(Xl

m.

(d)

f o r each compact s e t

f

dx(&(xI)

dX(K)

K c X.

i s h o l o m o r p h i c a l l y s e p a r a t e d a n d m e t r i c a l Z y hoZo-

X

morphicaZly convex. (9)

PROOF.

function

i s h o l o m o r p h i c a l l y s e p a r a t e d and p s e u d o c o n v e x .

X

(a) * (b): Suppose X

f

E

3 C t X l . Then X

is the domain of existence of

is holomorphically

Lemma 52.3. Now, consider the open sets

separated

a

by

365

THE L E V 1 PROBLEM I N R I E M A N N DOMAINS m

X =

Clearly

prove that

(A^j x( x)

*

U

j=I

A

and

j

dx( (A^ .I j

A

j

C

Aj+l

l/j

X(Xi

€or

for every

every

shall

j. Indeed, let

rn E lN

t E BE(O;l/j) and

Then for each

We

j.

we

z E

have

that IPrnf(zl(tI( 5

Hence

IIPrnfizIII 5

J

sup x€ A

.rn+l

IPrnf(XI(tll

5

suplfl B

j

€or every

rn

E

IN

5 j.

j

and it follows from cu

the Cauchy-Hadamard Formula that the series

Z

Prnf (

2 )

( t ) de-

m=O

fines a function f,

which is holomorphic on the ball BE(<(zI;l/j)

and satisfies f, o 5 = f on a neighborhood of z . Let X, be the disjoint union of X and FE(<(zl;l/j), with its natural E be defined 5, = 5 on X and topology, and let 5, : Xo +

5, = i d e n t i t y on

BE(C(zl;l/jl. Clearly (Xo,<,)

is a Riemann domain over B and the inclusion mapping X c, Xo is a morphism. Let f, E K(X ) be defined by f, = f on X and f o = f , 0 on B E ( E , ( z I ; l / j , J . Define an equivalence relation on X, as y if t 0 ( x ) = S o ( y i and Pmf ( x i = P : f , i y i follows. Let x t o for every rn E lN and t E E . Let Y = Xo / % be the quotient Y be the quotient mapping. There is space and let TI : Xo E such that r) o TI = 5 0 and there is a unique mapping rl : Y 6 such that ~ O T = I f 0 .Proceeding a unique function 7 : Y as in the proof of Lemma 52.3 we can show ( Y , r l ) is a Riemann E K ( Y / and the mapping T = domain over E. If follows that --f

+

-+

7

1

X:X Y is an {fl-extension of X. Since X is the domain of existence of f, T is an isomorphism. Since clearly dY ( T ( z 1 1 > l/j, we conclude that d,(zl l/j too, and (b) has been proved. TI

-+

(b) * (cf: The proof of the implication (b) * ( c ) in Theo-

rem 11.4 applies. (c) * (d):

Suppose X satisfies (c). Let ( Y , r l ) be a R i m

domain over E , and let T : X Y be a holomorphic extension of X. By Lemma 52.3 the mapping T is injective. Since every +

MUJ I CA

366

morphism is a local homeomorphism, we conclude that T : X T(X) is a homeomorphism. If T : X +. Y is not an isomorphism then T ( X ) # Y. Since each connected component of Y intersects T I X I , there is a point Y in the boundary of T I X I in Y. Since T : X +. T ( X ) is a homeomorphism there is a sequence Ix .) in X 3 such that T I x . ) +. y . It follows that d ( x ) +. 0 and by (c) J x i there is a function f E X I X I such that supIflx.! I = m. Let E X(Y) such that 7 O T = f. Then fix.) = f o T(x.) fiyl, a J 3 contradiction. Thus T is an isomorphism and (d) follows. +

7

-+

(d) * (e): Suppose X is a domain of holomorphy. Then X is holomorphically separated by Lemma 5 2 . 3 . Now, let K be a compact subset of X , let r = dX(KI and let z E We shall first prove that for each

f E X(X) there is function f z

which is holomorphic on the ball PE ( t ( z l ; r l and satisfies f, o 5 = f on a neighborhood of z. Indeed, let f fz X t X ) . Given t 6 BE(O;r) choose p > 1 such that p t E B E ( O ; r l . Then K + a p t is a compact subset of X by Lemma 52.5. Choose F. > 0 such that d X ( K + a p t ) > P E and f is bounded, by c sayton the set K + n p t + BE(O;p€). Then

m

for every

h

E

BE(O;E)

and

n7 E D?. Hence the series

z ?f(z)(t) m=O

defines a funct.ion f, which is holomorphic on the ball BE(c(z);r) and satisfies fz o 5 = f on a neighborhood of z, as we wanted. Proceeding as in the proof of the implication (a) * (b) we can find a Riemann domain i Y , r i ) over E l and a holomorphic extension T : X Y of X such that d,i.riz)l 2 r . Since X is a domain of holomorphic, T must be an isomorphism. Hence dx(xl r and (el has been proved. +

Since the implications (e) * ( f ) and (f) * the proof of the theorem is complete.

(cj)

are obvious,

is separable and X is an open subset of E then Theorem 11.4 asserts that (b) * ( a ) . But it is unknown whether this implication holds in the case of Riemann domains. If E

367

THE LEV1 PROBLEM I N RIEMANN DOMAINS

52.7. DEFINITION. Let ( X , E ) be a Riemann domain over E . An a, increasing sequence U = t U . i of open subsets of X is said j=l

J

for every

a,

X =

to be a r e g u l a r c o v e r of X if

uj+l

iU.l 3

> 0

j ~ ,

for every

E X ~ X I: suplfl <

U

d

and

Xo0llllthe algebra

j. We shall denote by

K"
U U j=I j

j

endowed with the topology generated by the seminorms f +suplfI. U 3Cm(Ul is clearly a Frgchet algebra. If f E Jcm(UI then it is easy to see that P T f E JC"(U) for every rn E W and t E E. These two properties of Xm(Ul play a role in the next two lemmas ~

52.8.

F

C

Let

LEMMA.

be a Rieniann d o m a i n o v e r

(X,c!

E

Zet

and

Jc(X).

If X i s F - s e p a r a t e d t h e n e a c h F - e x t e n s i o n o f

(a)

X

i s

injective.

(b)

PTf E F

If

for every

f

E

F, rn

E

no

and

t E E , and

if e a c h F - e x t e n s i o n o f X is i n j e c t i u e , t h e n X i s F-sep'mzted. PROOF.

Just examine the proof of Lemma 52.3. Let iX,c!

52.9. LEMMA.

b e a Riemann d o m a i n o v e r

U of X s u c h t h a t X i s

there i s a reguZar cover rated.

T h e n for e a c h c o u n t a b Z e s e t g

that

x # y

PROOF.

E

g l x l # g(yl

XmfUl s u c h t h a t

tion

and

P C X

Suppose

E.

Jc"iUl

for all

x, y

E

S f x i = Sty).

Consider the countable set Q = ifx,yI

Since X is

Jc

m

s

P

x

P : x # y

and

<(xl # S l y ) ) .

(Ul-separated, the set

XY

=

{ g E Jc

m

fUl

: glx)

sepa-

t h e r e e x i s t s a func-

# glyl}

P

such

368

MUJ ICA

is nonvoid f o r each ( x , y ) JCc"(UI.

f

E

We claim that

X m ( U ) with

E

Q. The set S

is dense in

sxY

9 S"Y ' choose g

f

gn

is clearly open in given

1 and set gn = f + n g.

+

in

f

an in particular nonvoid.

JC"(U),

Indeed,

Jcco(U/.

E SXY

for every n and Clearly g E S xCY 3Cm(U) is a Baire space the set

is dense in

XY

Since

3CmlUJ.

This

completes

b e a Ricniann d o m a i n o v e r

E . Consider

the p r o o f . 52.10.

Let ( X , E )

THEOREM.

the following conditions:

(a)

X

(b)

T h e r e is a r e g u l a r c o v e r

is a d o m a i n of e x i s t e n c e . U of

X

such that

X i s an

Jc"( U i - d o m a i n of h o l o m o r p h y .

(c) There i s a regular cover JC ( 0 ) - s e p a r a t e d a n d dX(5 ) > 0 m

U

or

X

such t h a t

u

f o r evpry

E

X

is

X

is

U.

JC"(!f)

(d)

There i s a reguZar

T h e n (a)

3

~'uver 1

d x ( f i J C ( X I) > 0

JCwiU l - s e p a r a t e d a n d

( b i * (c)

of

X

for every

such t h a t

U

E

U.

id). if E is s e p u r n l i 7 c t h r n

these

conditions are equivalent.

PROOF. (a) * ( b ) : Suppose X is the domain of existence of function J' E J C ( X / . Consider the open sets

iU.) > 2 !I Jfl m sequence U = i U .)

Since

du.

.I

JC"'(UI,

-j- 1

j=I

( b ) follows.

for every

r:,

it is clear that

is a regular cover of

X.

sir.cc

a

the f

E

369

THE LEV1 PROBLEM I N RIEMANN DOMAINS

(b) * (c): Let

U = (U

J

m

,)

j=l

be a regular cover of X such

03

03

that X is an Jc (U/-domain of holomorphy. Then X is Jc (UIseparated by Lemma 52.8. If we set r = d (U . / then the j Uj+l 3 proof of the implication (a) * (b) in Theorem 52.6 shows that ) 2 r for every j . This shows (c). dX((tjl Jc" (UI j Since the implication (c) * (d)is obvious, it only remains m

to show that (d) +. (a)when E

is separable. Thus let U=(Uj)j=l

be a regular cover of X such that X is JCm(U/-separated and dX( tu^ .I I > 0 for every j. Let D be a countable dense 3 Jc(X)

subset of

E l let

P = 5

-1

(Dl and let

By restricting our attention to each connected component of X we may assume without loss of generality that X itself is connected. Then P is a countable dense subset of X by Exercise 47.C, and by Lemma 52.9 there exists a function g E X m f U / such that gis) # g(y) for every ( x , y ) E Q. NOW, let ix.) be a 3

sequence in P with the property that each point of P appears in the sequence ( x3.I infinitely many times. Set BX (2)=BX(x;dx(x)I for each x E X and set V = for each j E ZV. Then j

BX(xl

for each

Vj

x

E

X

(cj)x(x)

and

j

E W ,

and after replacing

the sequence ( V . ) by a subsequence, if necessary, we can find 3 a sequence ( y .I in X such that y j E BX(xjl, yj $2 V and y j 3 j for every j E D . Hence we can inductively find a Vj+1 J C ( X / such that sequence if.) in 3

M

for every

j

E D.

fj converges uniformly

Hencethe series j=l

on each V3. to a function f E X ( X ) such that If(gj1 I > jg(yj) 1 + j for every j E m . Since the set of quotients (fix)- f(y//(g(xl -g(y)), with ( x , y )

E

Q, is countable, there exists

f(xl - f ( y i f 0 1 g i x l - g i y ) ) for every

0 E ( 0 , l ) such that

(x,y)

E

Q.

If we

set

MUJ ICA

370

-

h = f

8g and 1 h f y . I 3

h E K ( X ) , h l x ) # h f y l f o r every

then

I

> j

f o r every

j

r

Q is

X

We s h a l l p r o v e t h a t

h . Indeed, l e t (Y,rl) be a

t h e domain o f e x i s t e n c e o f E , let

E ZV.

fx,~) E

Riemann

X Y be an {hl-extension of X I and let E JC(Y) w i t h $ o r = h . To p r o v e t h a t T i s i n j e c t i v e let a , b E X w i t h T ( a ) = r f b ) . Then f o r e v e r y m E JV and t E E O m w e h a v e t h a t P : h f . r ( a l ) = P:?i(.r(b)), and t h e r e f o r e Pthtal = P T h t b i by E x e r c i s e 4 7 . 1 . I f a # b t h e n w e c a n f i n d P > 0 s u c h t h a t P < dX(al, r < d X t b l a n d B X ( a ; r l n P X f b ; r ' = 8 . Then f o r e a c h t E B B f O ; r ) w e h a v e t h a t domain o v e r

E

+

S(al = c ( b l , and s i n c e D i s d e n s e i n E , t h e r e e x i s t s s u c h t h a t S l a l + t = c ( b l i t E D . I f w e set x =

Since

t

:

BE(O;r)

+ t ) and

c,'(<(a)

y = Ebl(c(b) + t l

then

(x,y)

E Q

= h(yl, a c o n t r a d i c t i o n . Thus T i s i n j e c t i v e , a n d f o l l o w s t h a t T : X -+ T ( X ) i s a homeomorphism. I f then there is a point s h a l l prove t h a t

6

d y ( b ) . Indeed, i f that

r(al

E

b

i n t h e boundary of

i s unbounded o n

dx(a)

q u e n c e (j,) i n W

such t h a t

J:

it

T ~ X I #

Y

Y.

W e

T ~ X )i n

r

= a

a

E P

such

and i t follows t h a t

By t h e d e f i n i t i o n o f ( x . 1 t h e r e i s a s t r i c t l y 3

whence

B y ( b ; 2 r ) whenever 0 < 2r <

dylbl t h e n w e c a n f i n d

0 < 2r

By(b;x3I. Hence

b u t h(xl

f o r every

increasing

se-

k . Hence

yjk

j, E B X ( a ) and t h e r e f o r e I%o.r(y

jk

)I

= lhiy

i s unbounded o n

jk

?(yj

I1 > j,

k

) E By(b;2r)

f o r every

f o r every

k.

Since

k , w e conclude t h a t

$

By(b;2rI, a s a s s e r t e d . T h i s c o n t r a d i c t i o n c o m -

p l e t e s t h e proof of t h e theorem.

52.11. THEOREM.

L e t (X,<) b e a Rienrann d o m a i n o v e r

t h e foZlowing conditions a r e e q u i v a l e n t : (a)

X

is a domain of e x i s t e n c e

6".

Then

371

THE LEV1 PROBLEM I N RIEMANN DOMAINS

(b)

X

is a domain of hoZomorphy.

(c) X i s hoZornorphicaZZy s e p a r a t e d and rnetricaZ2.y hoZomorphically convex. Without loss of generality we may assume that X is connected. Then there is an increasing sequence (If-)3 of rela-

PROOF.

m

tiveiy compact open sets such that

for each

j E IIV, m

then

and J c f X ) = Jc ( U ) . Theorem 52.10

u = ( Uj

j=l

X =

u

j=l

V

j'

If we define

is a regular cover

of

X

Then the theorem is a direct consequence of

EXERCISES 52.A. Let U be an open subset of E . Show that U is a domain of holomorphy (resp. a domain of existence) in the sense cf Definition 52.1 if and only if U is a domain of holomorphy (resp. a domain of existence) in the sense of Definition 10.4 (resp. Definition 10.8). 52 . B . Let IX, S ) be a Riemann domain over E . Show that X is a domain of holomorphy (resp. a domain of existence) if mdcnly

if each connected component of (resp. a domain of existence).

X

is a domain

of

holomorphy

52.C. Let ( X , < ) be a Riemann domain over E . Show that X is holomorphically convex (resp. metrically holomorphically convex) if and only if each connected component of X is holomorphically convex (resp. metrically holomorphically convex). 52.D. Let I X , < i be a Riemann domain over E . Show that X is holomorphically separated if and only if each connected component of X is holomorphically separated. 52.E.

Let i X , C )

be a Riemann domain over

E.

Show that if X is

372

MUJ I CA

holomorphically convex (resp. metrically holomorphically convex) then (Eel is holomorphically convex (resp. metrically holomorphically convex) for each closed vector subspace Eo of

<-'

E.

52.F. Let ( X , c ) be a Riemann domain over E . Show that if X is holomorphically separated then 5 - l I E , 1 is holomorphically separated for each closed vector subspace E o of E .

53. THE LEV1 PROBLEM IN FINITE DIMENSIONAL RIEMANN DOMAINS

In this section we show that each pseudoconvex Riemann domain over E n is holomorphically convex and holomorphically separated, and therefore a domain of existence. T o begin with we observe that the results

on holomorphic approximation from Section 44 admit straightforward extensions to the case of Riemann domains.

t X , E ) b e a c o n n e c t e d p s < ? u d o c o n v e zRie m ar !: do-

Let

53.1. LEMMA. main o v e r

8.L e t

u

e Cm(X) be

function such that the s e t f o r each

Let ( X , O

53.2. LEMMA.

each compact s e t i s

Kc

c E §?. T h e n f o r e a c h

a constant

K

c > 0

53.3. THEOREM.

7

sf-.tQinr;7y

p ~ , . i ~ ~ . ~ i ~ ~ , ~ j , ~ ~ ; i ~

= {x E X : u ( x ) 5 c} is compact f E TCIK ) t h e r e i s a s e q u e n c e

b e a Riemann d o m a i n o v e r

and e a c h o p e n s e t

V wiiii

Let tX,<)

there

3

i p , , (=X K . lThen X I X ) u h i c l : eon-

u n i f o r m l y o n a s u i t a b l e n e i g h b o r h o o d of

53.4. COROLLARY.

X

b e a p s e u d o c o n v e x R i e m a n n doma.1'~over

Cn. L e t K b e a c o m p a c t s u b s e t o f X s u c h t h a t e a c h f E X ( K I t h e r e is a s e q u e n c e C f .I i n f

C

such that

for.

v e r g e s Lo

En. Then f o r

K C V

L e t (X,

<)

K.

b e a p s e u d o c o n v e x R-i'rirrcum domain o w r

373

T H E LEV1 PROBLEM I N RIEMANN DOMAINS

En.

I: b e a n o p e n s u b s e t of

Let

each compact s e t

K

C

U. T h e n

X such that KpnlXI C U JC(XI i s dense i n (JC/U),Tc).

for

53.5. THEOREM. L e t ( X , O b e a h o Z o r n o r p h i c a l Z y c o n v e x R i e m a n n d o m a i n o v e r Cn, a n d l e t U b e a n o p e n s u b s e t of X. T h e n t h e f o l lowing c o n d i t i o n s a r e equivaZent:

(a)

KJCiX!

(b)

KJC(XI

(c)

U

C

U

is c o m p a c t f o r e a c h c o m p a c t s e t

C

U

f o r each compact s e t

is h o l o m o r p h i c a Z l y c o n v e x a n d

K

K C X.

U.

C

Jc(X)

dense i n

i s

(JCIVI, r e ) .

We cannot yet extend Theorem 44.6 to the case of Riernann domains, for we first have to prove that every pseudoconvex Riemann domain over Cn is holomorphically convex. To prove this result we need the following lemma. Let t X , c )

53.6. LEMMA. CmIXI fi

be a s t r i c t l y plurisubharmonic f u n c t i o n T h e n t h e r e a r e an o p e n n e i g h b o r h o o d

E X.

tion

b e a Riernann d o m a i n o v e r

v

E

for every

K ( v I such that

z

E

v

vial = 0

and

V

ovl

let

Cn,

X

and

u E

let

o f a a n d a func-

Rev(z)

uiz)

- uia)

\ {a}.

PROOF. Let U be a chart in X containing a . If zIJ. n denote the complex coordinates of 5 1 2 ) €or each z E U then the Taylor expansion of u can be written in the form

374

MUJ I CA

2

Since the Hermitian form

a

Z j,k

azj

a:,

(a)t

i

t

k

is strictly psi-

tive, there is a neighborhood V of a such that u ( z l z E V \ {u}.

u(z) >

iic v f z ) for every

53.7. THEOREM.

E a c h pseudoconvex Riemann domain over

Cn

is

holomorphically convex and holomorphically separated.

PROOF. Let ( X , < ) be a pseudoconvex Riemann domain over En. Without loss of generality we may assume that X is connected. By Theorem 50.6 there is a strictly plurisubharmonic function < r} is comu E C-lX) such that the set K r = ( x E X : u l x ) pact for each r E IR. To show that X is holomorphically conp for each r € W . vex it suffices to prove that ( i r ) x (= xK i

Now, fix r E 1R and let L: E X K p . Sy Lemma 53.6 there are such an open neighborhood V of a and a function u E K ( V I that v ( a i = 0 and Rev(,-) <

(53.1)

Choose

U ( X )

- u(al

such that T s u p p 3-r is a compact subset of such that 7

E D(V/

for every

^P

E

V \ {a)

on a neighborhood of a . Since V \ { a } there exists E > 0 1

We shall set U t = {x E X : 1 4 ( ^ P ) < t l for each t E B ,and if t > u l a ) then we shall denote by Ut the union of those COPnected components of U t which intersect K r U { a ) . Note that I

375

THE LEV1 PROBLEM I N RIEMANN DOMAINS

-

if t > u f a ) then Ut is a pseudoconvex Riemann finitely many connected components. Set s = u(al (53.2) For each (53.3)

lie v ( x ) < k E lN

-

E

for every

x

E

us

domain with + E . Then

supp

2-r.

set

wk = e

kv

T

- fk,

where f k E C m f f i s l will be chosen so that wk E JCffis). The equation a w k = 0 is equivalent to the equation afk=gk,where (53.4)

gk = e

kv

-

27.

By Theorem 51.1 there are weight functions a, a , y

E

cm'Us;

lRl

2 2 such that llg I1 < llA*glia + ilBgl12 for every g E DA* n DB. Since B Y each g k belongs to D ( V \{a]) it follows from Theorems 51.1

and 51.2 that we can find a sequence (fk) in C m f f i s l such that 5 Ilgklla for every k E lN. Using (53.4) afk = gk and llfkIla and (53.2) we can then find a constant c > 0 such that

-

(53.5)

llfkIIa

5

llgkIIa 5 c e

-kE

for every k E vv'. Since each fk is holomorphic on a neighborhood of a , an application of Lemma 53.2 shows that lim fklal k+m

= 0, and then a glance at (53.3) shows that

(53.6)

lirn w f a ) = 1 . k k-+

On the other hand, it follows from (53.31, (53.1) and that

(53.5)

for each t < u ( a ) , and then another application of Lemma 53.2 shows that

376

MUJ I CA

(53.7)

By (53.6) and (53.7) we can find

k

sup 'r

such that

E

/wk(all. Since w k E J C ( S~I an application of Corollary yields a function ~p E J C l X ) such that s u p l P 1 < I I p ( a ) I . <

IWkl

53.4 Thus

Kr

('r)x(x) =

Kr

as asserted. 1

To prove that X is holomorphicaliy separated, let a , b be two distinct points of X. Without l o s s of generality we may assume that u f b ) 5 u ( a ) . If V is the neighborhood of a from V.We the first part of the proof, then we may assume that b still set s = u l a ) + E and Us = {x E X : u ( x l < s}, but we now denote by Us the union of those connected components of Us which contain a or 6 . Then the first part of the proof still applies, and in particular it follows from (53.31, (53.5) and Lemma 53.3 that

Zirn w la) = 1

Since

k+m

k

there exists

k

E D

such that

wk:(:) #

w k ( b ) . Then another application of Corollary 53.4 yields a func-

tion

)I

E JCfX)

such that

$ ( a ) # $ ( h ) . The proof @ € the theorem

is now complete. Theorems 52.6, 52.11 and 53.7 can be summarizedasfollows. 53.8. THEOREM. L e t ( X , c ) b c u Kicrnanvi d o m u i n fo7lowing conditions a r e ~ q u i v n 7 e n t (a)

;i

is a domain of e x i s t e n c . c J

(b)

X

is a d o m u i n of h n 7 n m o r p h g .

PUP

8'. T h ~ nt h e

377

THE LEV1 PROBLEM I N RIEMANN DOMAINS

Now wecan repeat the proof of Theorem following theorem. 53.9.

THEOREM.

to prove

the

If I X , 5 1 is a p s e u d o c o n v e x Riemann d o m a i n

over

44.6

f o r e a c h compact s e t

K

C

X.

For later use we give the following separation theorem. 53.10.

THEOREM.

L e t (X, 5, be a c o n n e c t e d , h o l o m o r p h i c u l l y sepa-

r a t e d Riemann d o m a i n o v e r lrn. T h e n t h e r e is a n i n j e c t i v e mapN f E JC(X;lr ) f o r a s u i t a b l e N E W .

ping

To prove this theorem we need three auxiliary lemmas. 53.11.

Let t X , S I

LEMMA.

domain o v e r

En,

i s a mapping

and l e t

f E JcIX;lr

N

b e a h o l o m o r p h i c a l l y separated

Riemann

K b e a c o m p a c t s u b s e t of X . Then there I w h i c h i s i n j e c t i v e on K .

PROOF.

5 I

x

Each a E K has an open neighborhood Ua such that Ua is injective. Thus we can find open sets U l , U p in such that K C U I n . . . n U and 5 I Uj is .injective for

...,

j = I, . . . , p .

P

If we set

then the diagonal of K x K is contained in U and < ( x ) # < ( y l for every ( x , y ) E U with x # y. On the other hand, for each i a , b ) E IK x K ) \ U there exists p E X ( X ) such that V(a) # p ( b ) , and hence @(x) # p l y ) for all f x , y ) in a suitable neighin borhood of (a,b). Thus we can find open sets V l , ’ X x X and functions p l , . . . , q q E J C C X ) such that

...

p .(xi 3

and K

x

#

p..ly) d

K C U u Vl u

(s,p1,

.. .,p

4

) E

j = I , . ..,q.

Since

it is clear that the mapping

f =

for a l l I x , y )

...

u V

E

Vi

and

4 J C I X ; E ~ + ~ is ) injective on K.

378

MUJ ICA

L e t (X, 6) b e a Riemann domain o v e r

53.12. LEMMA.

b e a compact s u b s e t of the

compact s e t

f(K1

and l e t

X,

f

E

J€(X;CN).

let

Cn, If

has Lebesgue measure z e r o i n

N z n

K

then

Cn.

Without loss of generality we may assume that K is contained in a chart U in X. Let 0 < 6 < dU(X). I f we set f = (f,, ...,f,) then it follows from the Mean Value Theorem that there is a constant c > 0 such that

PROOF.

(53.8)

1f.(x 3

+ tl - f.(xl 3

< c II t II -

,...,

f o r all x E K, t B ( 0 ; 6 ) and j = 1 N. Let 0 < E < 6, and let C be a subset of X which ntersects K and which is homeomorphic under 6 to a cube in Cn with side E . Then C C U and it follows from (53.8) and Proposition 48.3 that

There are finitely many such "cubes" Cl, . . . ,CP K

C

CI

U

...

U

C P

P and

j=1

u

x

(C

i

)

in U such

that

< p X ( K ) + 1 . Then it follows

from ( 5 3 . 9 ) that k(f(K)I 5 c ( p X ( K 1 + 2 ) 2 N~ - 2 n . Since E can be taken arbitrarily small we conclude that A ( f ( K ) ) = 0, as we wanted. 53.13. LEMMA. K

L e t ( X , c j be a Co n w r tPd Riemann domain o v e r Cn , l e t

be a c o m p a c t s u b s e t of

X,

and L e t ( f , ,

b e a mapping w h i c h i s i n j e c t i v e o n K . e a c h E > 0 we c a n f i n d ( C 1 , . . . , C I v )

i

= 1,.

. ., N ,

and s u c h t h a t t h e mapping

. . .,j"+?

) E

sctx,8+1)

I f N > 2n + 1 tl:,'iz f o r EN , w i t h ( C j l < E for

379

THE LEV1 PROBLEM I N RIEMANN DOMAINS

Since X 2 x 5 is hemicompact, it follows from Lemma 53.12 that the set g l X 2 x 8 ) has Lebesgue measure zero in EN. Thus to complete the proof of the lemma it suffices to prove that the mapping

is injective on K

for each

5 = ( 5 1,...,5N)

E EN \

g(X2

C).

NOW, let x,y be two distinct points of K. If fN+l(xl = fN+l(y) then there exists j with 1 5 j 5 N such that f j ( x ) # f j ( y ) , and therefore

f .(x) 3

-

# fN+l(y), then since < j

5 N

- SjfN+l(y). If fN+l(x!

5jfN+l(x) # fj(yl

5

g(XL

*

5 1 , there exists j

with

1

such that

and therefore fj(x) - ijfN+l ( X I # f j ( y ) pletes the proof.

-

5jfN+1 (y). This

com-

PROOF OF THEOREM 53.10. Let (K,):=~ be an increasing sequence of compact subsets of X such that each compact subset of X is contained in some

If,.

Let F be the set of all f

E

JC(X;C

2n+2

)

such that the mapping (<,f) E J C ( X ; C 3 n + 2 ) is not injectiye. be the set of all f E Likewise, for each rn E IT? let F, such that the mapping (<,f) E J C ( X ; 5

JC(X;C2n+2)

3n+2

I is not in-

m

jective on

K,.

Certainly

F =

u F ~ Since .

J C ( X ; C ~ ~ + ' ) is a

m=l

Frkhet space forthecompact-open topology, to prove the theorem it is sufficient to show that each F, is nowhere dense. is closed in J C ( X ; C 2 n + 2 ) . Indeed, let We first show that F, (f.) be a sequence in F, which converges to a mapping f E 3

JC(X;C2n+2)

can find

for the compact-open topology. For each j E D we xj # y j in K, such that <(x.) = 6 1 y . J and f j ( x . ) 3

3

J

= f . ( y . ) . Without loss of generality we may assume that (x .) 3 3 3 converges to a point x , whereas ( y . ) converges to a point y. 3

380

MUJ I CA

Then clearly

C(x1 = c ( y ) ,

+ /f.(y 3

and it follows from the inequality

.)

3

- f(Yj1

that f(xl = f l y ) . We claim that x # y. Otherwise let U be a chart in X containing x = y. Then x . E U and yj E U for 3 sufficiently large j . But this is impossible, for x j # yj and t(x.1 = < f y . ) . Thus x # y and f E F,. Thus Fm is closed 3

3

in JC(X;g2n+2), and to complete the proof we show that F, has By Lemempty interior. Indeed, let f = (fl,...,f2n+Z1 E F , . ma 53.11 there exists a mapping

g =(g,,

. . ., g P 1

E

X i X; tp which 2n+2+p

) is injective on K,. The clearly the mapping (f,g) E X ( X ; C is also injective on K,. Then by repeated applications of I m ma 53.13 we can for each E > 0 find complex numbers i with jk I S j k I < E such that the mapping

Obviously ( < , 3 i E ) E X ( X ; t 3 n + 2 ) is also inis injective on K,. f jective on K,. Thus h E 9 Fm and since ( h E ) converges to whe-n E + 0 we conclude that F, has empty interior. This completes the proof of the theorem.

54. THE LEV1 PROBLEM IN INFINITE DIMENSIONAL RIEMANN DOMAINS In this section we prove that each pseudoconvex Riemanndomain over a separable Banach space with the bounded approximation property is a domain of existence. We also prove that such domains are holomorphically convex.

54.1. DEFINITION. Let ( X , c ) be a pseudoconvex Riemann domain over E . An open set U C X is said to be X - p ; L ? 13.7’t 1‘ if * KPsf X) C U f o r each compact set K C U. I’

THE LEV1 PROBLEM

Let (X, 5 ) be a pseudoconvex Riemann domain over

EXAMPLES.

54.2.

381

IN R I E M A N N DOMAINS

E. If V is a convex open set in E then the U = < - ' f V i is X-pseudoconvex. (a)

(b) If f E Pn(X) then the open set is X-pseudoconvex. (c)

If lUiliEI

X

E

U

over

i s pseudoconvex.

is d e n s e in ( X I U n t-'(EO)), f i n i t e d i m e n s i o n a l s u b s p a c e Eo of E . (b)

sets

n Ui is X-pseudoconvex as well. i€I

5 4 . 3 . LEMMA. L e t (X,tl be a p s e u d o c o n v e x Riernann domain E, and Zet U b e a n X - p s e u d o c o n v e x o p e n s e t . T h e n :

(a)

set

: flxl < U )

is a family of X-pseudoconvex open U = int

then the open set

U = {x

open

X(S-'fEo))

T

~

)

f o r each

The proofs of these statements are left as exercises the reader.

to

If E is a Banach space with a Schauder basis ( e n ) : = 2 then 00

(2 )

n n=2

will denote the sequence of coordinate functionals,En

will denote the subspace generated by e l , . . . , e n and T, will denote the canonical projection from E onto En. If (X,<) is a Riemann domain over E then we shall set is a Riemann domain over E n .

Xn = < - ' ( E n ) .

Then

Xn

54.4.

E.

Let

LEMMA.

b a s i s (en):=2 For each

. j

E b e a Banach s p a c e w i t h a m o n o t o n e Schauder

L e t ( X , 5 ) be a p s e u d o c o n v e x Riemann domain over E

W

let

MUJ I CA

382 Then: m

(AiljZl

(a) cover

Each

X.

i s a n i n c r e a s i n g s e q u e n c e of o p e n s e t s which

xj.

i s X - p s e u d o c o n v e x and c o n t a i n s

Aj

WX,)

i s d e n s e i n ( K ( A n X 1, T ~ for ) e a c h j, n E LV. i n

5

(b)

‘j

O

O

‘j+1 = ‘j+J

o T

2’. 0 5

on

3

j

=

T

on A

j

j E JV s u c h t h a t n L j.

exists

A

T

j’

j

K

C

Aj

C

j

X

xi

= i d e n t i t y on

f o r every

F o r e a c h compact s e t

(c) x E K

=~

T

j

and

and

T

E

0 <

and

JV. < d X ( K ) there

E

n (zl E BX

( x ; E ) for

ever2

and

tinuous on E for each

j

- t /I is conn n 2 j LV, all the assertions in t h e lem-

w . f t )= sup 1 IT t

Since t h e function

PROOF.

3

E

ma can be readily verified.

54.5. LEMMA. L e t E b e a Banach s p a c e w i t h a rionotone Schauder b a s i s ( e n ) ; = l . L e t (X, 5 ) b e a c ~ w n ~ 3 r t ppds e u d o c o n v e x Riemann domain o v e r

T h e n t h e r e a r e two i n c r e a s i n g s e q u e n c e s of

E. m

a,

s e t s (BjljZl

and f C j ) j = l

such t h a t : m

C C ii

(b)

dA f B . 1 > 2-j

Aj n X n

PROOF.

3

C

j

A

j

j, n

dC (C.1 2 2 j+l 3

and e v e r y

and

-

f o r every

(c) j E

j

B

f o r every

(a)

n

2

E

open

B

j

j

E W.

n Xn

X =

U

j=1

m

B

j

=

U C

j z 1 j’

i s r e Z a t i v e Z y compact

in

2N.

-j-1

and

j .

Consider the open sets

~ ~ ( c C. B1 3

j

n Xn

for

every

383

THE LEV1 PROBLEM I N RIEMANN DOMAINS m

W

Then each of the sequences

and ( Q j l j = , T ~ ( Q.I 3

X. We a l s o have that n 2 j and in particular

and covers

(54.1)

C

T.(&.)

3

3

P

j

n X

j

=

Q . n

3

X j

C

is increasing n Xn

P

for every

j

whenever

E

IiV.

Without loss ofgenerality we may assume that QI n X l is nonvoid. Fix a point x E Q 1 n X I ,let Q' be the connectedm0 j ponent of Q j which contains the point xo, and define

where

Q;.

denotes the geodesic distance in

"&J

Since

the

geodesic distance is a continuous function, each B j is open. And since X is pathwise connected, it follows that x = m

00

U

j=1

Q! =

U

B

j

j=1

3

to a point

x

E

.

If L

QJ n X

j

is a polygonal line in

QJ n X joining xo to j is connected. We claim that

Indeed, the inequality

joining x0

then it follows from (54.1) that

is a polygonal line in

lar

Qj

P ,(xOJx) 5 PQ;

x. In particu-

f x ,x)

Qj

TjfL)

is obvious.

j

To show t h e opposite inequality let E > 0 be given and let L = [ x o ,..., x ] be a polygonal line in QJ joining xo to x m

and such that m

Then and

T

. ( L ) is a polygonal line in 3

Q;

n

xj

joining

xo

to x

384

MUJ I CA

Since E. > 0 was arbitrary, (54.2) follows. From Lemma 5 0 . 1 it follows that the set

(54.2)

and

is relatively compact in A j n Xj for each j E IN. Whence it follows that B j n X, is relatively compact in A j n En for all j , n E D .

If we define

then the remaining assertions in the lemma follow at once. m

C = (CjJjZl is a regular cover of X.

Observe that the sequence ma 54.5

LEMMA.

54.6.

basis

domain over exists

on

Cn

PROOF.

Let

m

f

E

E

Let E.

constructed in Lem-

b e a Banach s p a c e w i t h a m o n o t o n e Schauder

(X,t)

b e a c o n n e c t e d pseudoconvex

Then f o r e a c h

JCmlCi s u c h t h a t

f,

E

f = f,

and

SCIX,)

>

E

Riemann there

0

\f-fn o

o n Xn a n d

Tn((E

. Consider the sets

If follows from Lemma 54.4 and 54.5 that ( B n n Xn+llpn(xn+l J, is

a compact subset of Q

A n n Xn+l

.

Whence it follows that P and

are two disjoint closed subsets of Xn+l. Choose 9

such that cp 1 neighborhood of

where

m

u E

on a neighborhood of &. Set

C (Xn+l)

P

and

will be chosen so that

0

cp

g

m

E

E

c (X,+*) on

Jc(Xn+l).

a

385

THE LEV1 PROBLEM I N R I E M A N N DOMAINS

ag = 0

The equation where

v

is equivalent to the equation

m

au

=

V,

is defined by

E ColIXn+lI

-

-

then equation au = v has a solution 51.2. Thus g E X(XniI) and it is dear

Since av = 0 on Xni1 co u E C fXni1I by Theorem

-

that g = f n o n X n . Since a u = v = 0 on aneighborhood ofthe compact set ( B n n xnil ) p- n l X n i l ) , an application of Theorem h

53.3 yields a function

then

=

fnil

Rn n X n + 2

.

g

= fn

Since

on T

nil

E

X(Xni1) such that

and

Xn

IC,)

C

Ifnil

B n n Xn+l

- fn

5

o

E

/ 2

on

we obtain that

03

such that

Proceeding inductively we obtain a sequence (fj.)j=n fj E XIX.),

for every

fj+] = f j

xj

and

j 2 n . If follows that

for every k j uniformly on each on X . and .I

on

n, and hence the sequence ( f Ci

to a function

f E XIXI.

. o T . I converges

3

3

Clearly f = f

i

386

MUJ I CA

for every j > n . From this estimate we also conclude that f is bounded on each Cj, for T . ( C . I C B n X and this set is 3 3 i j relatively compact in X

i'

Now we are ready to extend Theorem 4 5 . 7 Riemann domains. THEOREM.

54.7.

Let

to

the

of

b e a s e p a r a b l e Pana/+Jispa(?e dit?:14e bounded

E

a p p r o x i m a t i o n p r o p e r t y . T h e n e a c h p s e u d o c o n v e x Riemann over

case

domain

i s a domain of e x i s t e n c e .

E

PROOF. (a) We first assume that E has a monotone Schauderbasis. Let f X , < I be a pseudoconvex Riemann domain over E. Without l o s s of generality we may assume that X is connected. We first show that X is JCm(C)-separated. Indeed, let x , y be two distinct points in X. First choose j E IIV such that x, y E C j 1 T ~ ( x and ) and next choose n 2 j such that T , ( x ) # ~ ~ ( 2 .and -c,(y) both lie in C j . By Theorem 5 3 . 7 X, is bolomorphically f, E J C i X n ) such that separated and hence there exists there exists I f n o T ( X I - f O T i y ) l = 3~ > 0 . By Lemma 5 4 . 6 n

f E

n n such that lfix) - f ( y l 1

J€-(c)

that ted.

1f E

I

5 E on C n . Hence it follows and X is JCmICI-separated, as asserfn o

T~

Next we shall prove that

and therefore

dX( ( 2 .I

I

3 JC"ICI

this l e t

x

E

(2.1 J

for every

> 2-j -

and choose

k

L

j

j E IIV. 'ro prove

s u c h that

x E Ck.

XrnfCI

We claim that

Indeed, given f, E J C f X , ) and E 3Cm(C) such that - fn o lows that

f

If

E

> 0,

5

E

by Lemma 54.6 we canfind on Cw. Whence it fol-

387

T H E LEV1 PROBLEM I N RIEMANN DOMAINS

and since E > 0 was arbitrary, (54.4) follows. Then, Theorem 53.9 we obtain that

k . Letting n for every n and (54.3) has been proved.

We have shown that X

+

m

we get that

is 3Cm(C)-separated

x E

using

iB^j)p,c(x)

dx((?.)

and

J

for every Theorem 52.10. 1

j

0

E W .

Thus X

is a domain

of

JC"(0

existence

I

by

If E is a separable Banach space with the bounded (b) approximation property then by Theorem 27.4 we may assume that E is a complemented subspace of a Banach space G which has a G = E x F monotone Schauder basis. Thus we may assume that €or a suitable Banach space F. Then ( X x F, (5, id)) is apseudoconvex Riemann domain over G by Exercise 49.D. Hence X x F is a domain of existence by part (a). Hence X is a domain of existence by Exercise 54.B. 54.8. LEMMA. L e t ( X , E ) be a Riernann domain o v e r E . a s u b a l g e b r a of K i x ) s u c h t h a t P:f E F for e v e r y E W and t E E . L e t A b e a s u b s e t of X s u c h t h a t r

0.

Then

f o r each baZanced s e t PROOF. f E

F be f E F, m dX(zFI = Let

Let

y

E

i F ,t

R C RE(O;rl.

and

E B

F we have that m

m

0

<

0 < 1.

Then

for

elch

388

MUJ I CA

f n , taking nth root

After applying the preceding inequality to and letting n -+ m we get that lf(y

+ etil 5

SUP

A + B

Whence it follows that y + 0 t get the desired conclusion. 54.9.

Let

LEMMA. m

basis

domain o v e r

F.

The s e t ( ? . I J

and e a c h compact s e t

PROOF.

Let

and

1

+

we

Riemanrz

i s c o m p a c t f o r aZZ

'n

JCW(C)

n S-liXI

JCW(CI

cfl.

j,n

i s compact f o r each

jE

N

K C E,

To prove (a) it certainly suffices to prove that

x

(t.)

E E

fn

and since

and

there exists

> 0

If -

'n

Xrn(C)

J

each

F . Letting 8

Then:

J

(b)

B

IX,<) b e a c o n n e c t e d p s e u d o c o n v e x

The s e t fF.1

(a)

+

(A

b e a B a n a c h s p a c e w i t h a m o n o t o n e Schauder

E

Let

E

fl.

5

0 7 ~ 1

c > 0

E

on

f

ma 54.8 that

E

3C(XnI.

E JCm(C)

By Lemma

such that

54.6

f = f,

on

for X,

Hence

Cn.

was arbitrary, ( 5 4 . 5 ) follows.

To prove (b) fix j E W dxI(?.) I and 3~ < d JC"(C,

fn

and choose F: > 0 such that 3~ (C.). Then it follows from Lem-

Cj+l

J

389

THE LEV1 PROBLEM I N R I E M A N N DOMAINS

and after replacing (xi) by a subsequence, if necessary, w e m y assume that (<(xi)) converges to a point a E K. Hence we can i 0 On find io E W such that II < ( x i ) - a II < E for every the other nand, since T a a we can find n E W and b E E n n such that IIb - ciII < E. Thus IIb - t ( x i l l l < 2~ for every i i 0' and hence f o r each i i 0 there exists a unique point y i E B l x i ; 2 & I such that <(yi) = b . Then a glance at (54.6) X shows that

.

-+

z

i 2 i0

for every

n

xn

.

Since

Slyi

= b

we conclude that

E En

yi

is compact and therefore, after replacing ( y i ) by a sub-

sequence, if necessary, we may assume that (yi) converges to a point y. Observe that d X ( y l 2 E . Since Slyi) = b for every i > i we see that s l y ) = ii too. And since IIa - < ( y ) l l = 0 IIa - bII <

there is a unique point x E BXIy;€) such that 6 l ; c l = a . Choose i I -> i0 such that y i E B X (y;~) for every i > i . Since < ( x i ) +. a = <(x), we conclude that xi x. This E,

-+

shows (b) and the lemma. 54.10. DEFINITION. Let f X , c ) be a Riemann domain over E l and let F E X ( X ) . A subset B of X is said to be F - b c u q d i n g if each f € F is bounded on B. We shall say that R is a boundi n g subset of X

if

B

is

X(XI-bounding.

E b e a s e p a r a b Z e Ranach s p a c e w i t h t h e iX, ( I b c , a p s e u d o c o n v e x Riemann domain o v e r E . T h e n e a c h b o u n d i n g s u b s e t o f X is r e is c o m p a c t f o r l a t L v t - l p c o m p a c t . I n p a r t i c u Z a r th,. s e t 54.11. THEOREM.

Let

bounded approximation p r o p e r t y . L e t

L

%X)

each compact s e t

PROOF.

K C X.

(a) We first assume that E has a monotone Schauder ba-

sis. Let ( X , S 1 be a pseudoconvex Riemann domain over E l and let B be a bounding subset of X. Clearly B is contained in the union of finitely many connected components of X,and hence

390

MUJ I CA

we may assume from the outset that X is connected. Then proof of Proposition 12.3 shows the existence of j E 2? that B C ( ? . ) X I X I . On the other hand it is clear that 3 is a bounding subset of E, and is therefore relatively pact by Theorem 12.5. Thus B C ( ? j ) x ( x l n S - ' m and

the such C(B)

comis

B

relatively compact by Lemma 54.9. (b) The general case can be reduced to case (a) as in the proof of Theorem 54.7. The details are left to the reader a s a n exercise. Theorems 52.6, 54.12. THEOREM.

54.7 and 54.11 can be summarizedasfollows.

Let

b e a s e p a r a b t e Ranach s p a c e

E

bounded a p p r o x i m a t i o n p r o p e r t y . L e t ( X , < 1

over

E.

with

be a Riemann

the

domain

Then t h e f o l t o w i n g c o n d i t i o n s a r e e q u i v a l e n t :

(a)

X

i s a domain o f e x i s t e n c e .

(b)

X

i s a domain of h o t o m o r p h y .

(c)

X

is a h o t o r n o r p h i c a t l y c o n v e x .

(d)

x

i s m e t r i c a t l y hoZomorphicaZly c o n v e x .

(el

X

i s pseudoconvex.

EXERCISES 54.A. Let ( X , c ) be a Riemann domain over E , and let U be an X-pseudoconvex open set. Let X' be a connected component of X which intersects U , and l e t U' = U n 1'. Show that U'is XIpseudoconvex.

54.B. Let t X , c 1 be a Riemann domain over E . Let E, closed vector subspace of E , let X, = < - ' ( Z 0 ) and let 5 I Xo . Show the following:

a I;, =

be

a3

(a)

If

U = (Uj'j=I

is a regular cover

of

X

then

39 1

THE LEV1 PROBLEM IN RIEMANN DOMAINS

uo =

m

xo)jzl

IU. n 3

X,

separated then for

every

j

is a r e g u l a r cover of

E

IN

is

3Cm(U ) - s e p a r a t e d . 0

(b)

X

If

If

dx((oj) I

n Xol-

x

is

JC~(U/-

i'

0

>

for

0

every

JCWIUol

3

w.

j E

If

Jc"(U,

d,(tU.

then

x,.

i s a domain o f e x i s t e n c e t h e n X, i s a domain X i s a domain o f e x i s t e n c e a n d E o i s s e p a i s a l s o a domain o f e x i s t e n c e .

o f holomorphy. I f r a b l e then

(c)

Xo

U

If

i s X-pseudoconvex t h e n

U

Xo

i s Xo-pseudo-

convex.

55. HOLOMORPHIC APPROXIMATION I N INFINITE DIMENSIONAL RIEMANN

DOMAINS In

this

s e c t i o n w e extend

t o t h e case of Riemann domains

t h e a p p r o x i m a t i o n t h e o r e m s from S e c t i o n 4 6 . 55.1.

THEOREM.

Let

E

b e a s e p n r u b Z e Banach space w i t h the bounded

a p p r o x i m a t i o n p r o p e r t y . L e t (X,<) b e a p s e u d o c o n v e x Riemanndomain o v e r

U

be an X-pseudoconvex

open

s e t . Then

-eel.

i s sequentiaZly dense i n IJCIUI,

XIXI

I n v i e w o f E x e r c i s e 54.A, w e may r e s t r i c t o u r a t t e n t i o n

PROOF.

to

and Z e t

E,

those

c o n n e c t e d components o f

h e n c e w e may assume t h a t

X itself

X

U,

which i n t e r s e c t

is connected. I f

E

and has a

monotone Schauder b a s i s t h e n t h e p r o o f c a n p r o c e e d p a r a l l e l t o the

proof

case where

of t h e corresponding

case

i n Theorem 4 6 . 1 .

And t h e

i s a s e p a r a b l e Banach s p a c e w i t h t h e bounded ap-

E

p r o x i m a t i o n p r o p e r t y c a n b e r e d u c e d t o t h e p r c c e d i n g case i n t h e u s u a l manner u s i n g E x e r c i s e 54.B.

The d e t a i l s a r e

l e f t t o the

r e a d e r as an e x e r c i s e . 55.2.

THEOREM.

Let

E

be a separabZe

Banach

space

with

the

b o u n d e d a p p r o x i m a t i o n p r o p e r t y . L e t (X, t i be a pseudoconve.?: Riemann domain o v e r

KPdcIxl

E.

Let

K

be

a

compact

subset

= K. T h e n f o r each o p e n n e i g h b o r h o o d

of X s u c h t h a t U of K t h e r e i s

MUJ I CA

392 a n X-pseudoconvex open s e t

such that

V

K C V

U.

C

PROOF. ( a ) W e f i r s t a s s u m e t h a t E h a s a monotone Scha.uderbas i s . W i t h o u t loss o f g e n e r a l i t y w e may a s s u m e t h a t t h e Riemann i s c o n n e c t e d . Choose

domain X

j

such t h a t

1IV

E

K C C;.Then

t h e set

i s c o m p a c t by Lemma 5 4 . 9 , a n d c o n t a i n s K . F o r e a c h p o i n t x E L \ U t h e r e is a function f x E P b C l X ) such t h a t f , 0 on K and f , f x l > 0. S i n c e L \ U i s compact w e c a n f i n d f u n c< 0 on K f o r j = t i o n s f l , . . ., f, E P b C ( X ) s u c h t h a t I,.

f j

. .,m

and m L

\u c

u

{z 6

x

:

Q‘x’

01.

j=l I f w e set

Choose

= s u p { f l , ...,f,

f

0

E

such t h a t

) then

dx((?.i

E

J

on

< 0

f

and

K

J c t X ) I and

E

d

‘j+l

(C ,). 3

Then i t f o l l o w s f r o m l e m m a 5 4 . 8 t h a t

W e claim t h e r e e x i s t s

6 with

0 < 6 <

Otherwise w e can f i n d a sequence ( A k ) , Ak

+

0,

E

such t h a t

with

0 < Ak

C

F

and

and a sequence ( z k ) such t h a t

For each

f o r every

k

such t h a t

IIbk - <(zkill

E W .

6k

iiV t h e r e i s a p o i n t b E?OiE,(h‘.Jl k

k

E

.

Hence €or e a c h

k t LV

there i s

393

THE L E V 1 PROBLEM I N RIEMANN DOMAINS

a unique y k E B X ( xk ; 6 k I such that lows f r o m (55.2) that

<(yk) = bk '

Then it fol-

for every k E JV. By Lemma 54.9 that set is compact. Hence, after replacing ( y k ) by a subsequence, if necessary, we may assume that ( y k I converges to a point y. But then ( x k ) a l s o converges to y , and whence it follows that

contradicting (55.1). This shows the existence of 6 > 0 isfying (55.3 . Set

v = i n t iiiJ.I

sat-

X(X

Then K C V C U and to show that V is X-pseudoconvex it only remains to show that the open set i n t ( 5 . 1 is X-pseudoJ

JCIXI

convex. To prove this let A be a compact subset of f n t Choose p with

0 < p .< 6

such that

A

+ BE(O;p)

C

(i?

j X(X)'

( ?j

H(XI'

Then it follows from Lemma 54.8 that

X-pseudoconvex, as we wanted. (b) The case in which E is a separable Banach spacewith the bounded approximation property can be reduced to the preceding case in the usual manner. We leave the details to the reader as an exercise. From Theorems 55.1 and 55.2 we obtain at once the following result:

55.3.

THEOREM.

Let

E

b e a s e p a r a b l e E'anach spaze w i t h t h e bounded

394

MUJ I CA

L e t (X, 5 ) b e a p s e u d o c o n v e x Riernann do-

approximation property. main o v e r

Let

E.

K

b e a c o m p a c t s u b s e t of X

K^P h C I X )

such t h a t

= K . T h e n f o r e a c h f E JC(I0 t h e r e a r e a n o p e n n e i g h b o r h o o d U of K and s e q u e n c e ( f j ) i n X ( X ) s u c h t h a t f E X(U) and tfj) converges t o 55.4.

f

THEOREM.

i n (JC(UI, Let

E

-eel.

b e a s e p a r a b l e Banach space w i t h the bounded

approximation property. Let ( X , 5) be a holomorphically Riernann domain o v e r

and l e t

E,

convex

U b e a n o p e n s u b s e t of X.Then

the following conditions are equivalent:

(a)

"(XI

-

KJCIX)

(b)

n U C

U

i s c o m p a c t f o r e a c h cornpact s e t f o r e a c h compact s e t

U

(d)

U

U.

K C U.

is h o l o m o r p h i c a l l y c o n v e x and dense i n ( K I U ) , T ~ J . (c)

I( C

J C I X ) i s sequentially

is h o l o m o r p h i c a l l y c o n v e x and

JC(X)

7:s

dense

in

(X(U),T , ) . The proof of Theorem 44.5 applies.

PROOF. 55.5.

THEOmM.

Let

E

b e n s e p a r a b l e Banach space w i t h t h e bounded

approximation property. Let I X , S ) Riernann d o m a i n o v e r set

convex

f o r e a c h compact

= KJC(X)

"h(X)

We already know that

i P h t X ) + iXtX)

function > 0

u

and let

E Pb(X/

such that

tinuous and such that

.

KXtX)

a

6

such that

C

x^X(x,

\

u < V

4 F < d,Ij?JC(X)).

iXtX,.

x^Pb(XI. on

Since

Suppose

Then K

u

and

that

there is u(a)

is upper

a

0 . Fix

semicon-

K is compact we can easily find 6 with O < 6 < E u < 0 on K + BE(O; 26). Let T E 6 ( E ; E I be a finite

rank operator such that A

Then

holomorphically

K C X.

PROOF.

E

E.

be a

Define

a E

IIT o 51x1

~i2,,~,;~)

by

- <(x)Il

< 6

f o r every

x

E

395

THE LEV1 PROBLEM I N RIEMANN DOMAINS

a(a) = a

Then

u

whereas

and

a(x) < 0

o

ulK) C K + B,(O;26). f o r every

3: E K . L e t E, E g e n e r a t e d by t h e set ) C = < - 1 (E,). Then a(&(x)

dimensional subspace of

+ ((a), a n d l e t

X,

u

Hence

u(a) > 0

o

be t h e f i n i t e

<(a) a n d i f fol-

P o <(El - T o

x,

l o w s from Theorem 5 3 . 9 t h a t

Hence t h e r e i s a f u n c t i o n

f E sC(Xol s u c h t h a t

fl Since

Ifoo(a

I.

a n a p p l i c a t i o n of Theorem 5 5 . 3

f o a E

a function

<

g E X ( X l such t h a t

suplgl K

<

(glall, a

yields

contradic-

t i o n . T h i s completes t h e proof of t h e theorem.

NOTES AND COMMENTS Theorem 5 2 . 1 1 i s a c l a s s i c a l r e s u l t o f H . Thullen

[ 1

i s d u e t o M.

S c h o t t e n l o h e r [ 1]

* ( e ) i n Theorem 5 2 . 6 i s d u e t o

A.

. The

1 .

P.

hplication

Hirschowitz 1 3 ]

The main r e s u l t i n S e c t i o n 5 3 , Theorem 5 3 . 7 , Oka [ 4

and

1 . I t s g e n e r a l i z a t i o n t o t h e case of Banach spaces,

Theorem 5 2 . 1 0 , (d)

Cartan

is due t o

The p r o o f of Theorem 5 3 . 7 g i v e n h e r e i s d u e

Hormander [ 3

1.

Theorem 53.10

t a i n e d by R . N a r a s i m h a n I 1 ]

1

.

M.

to

K. L.

i s a weak v e r s i o n o f r e s u l t s o b -

and E. Bishop [ 1 I

.

The m a i n r e s u l t s i n S e c t i o n 5 4 , Theorems 5 4 . 7 a n d a r e d u e t o Y . Hervier [ 1

.

54.11,

Schottenloher [ 4 1 has obtained

F r g c h e t s p a c e v e r s i o n s of t h e s e t h e o r e m s . Theorems 5 5 . 1 a n d 55.2 a r e d u e t o J . M u j i c a [ 5 1 . Theorems 55.3,

55.4 and 5 5 . 5 s h a r p e n r e s u l t s

cf

M. S c h o t t e n l o h e r

4

The i d e a of t h e p r o o f of Theorem 5 5 . 5 g i v e n h e r e i s d u e t o Schottenloher

5 1 .

1

.

M.

This Page Intentionally Left Blank

CHAPTER XI11

ENVELOPES OF HOLOMORPHY

5h.

ENVELOPES OF HOLOMORPHY

In this section we introduce the notion of envelope of holomorphy. We show that the envelope of holomorphy of a given Riemann domain always exists. DEFINITION. Let ( X , C ! be a Riemann domain over E and let F C J C ( X ) A morphism T : X Y is said to be an F-envelope e f ho lornorphy of X if: 56.1.

+

(a)

T

s an F-extension of

(b) If p a morphism v :

A morphism rnorphy of X if

:

z T

T

x

x.

Z is an F-extension of X Y such that v o i-( = T .

+

-+

then there is

X Y is said to be an envelope o f h o l o is an JC(XI-envelope of holomorphy of X.

:

+

If T : X Y and T ’ : X Y ’ are two F-envelopesof holomorphy of X then the Riemann domains Y and Y ’ are easily seen to be isomorphic. In other words, the F-envelope of holomorphy of X, it it exists, is unique up to isomorphism. +

56.2. DEFINITION.

+

Let

1

be an index set. For each point

a

E

consider the collection of all pairs ( U , q ) such that U is Two such an open neighborhood of a and q = i q P i I i E I C J c ( U ) .

E

pairs ( U , q p l

and i V , + i

are said to be equivalent if there is an

open neighborhood W of n with W c U n V on W for every i E .T. We shall denote by of all equivalent classes. The members of

397

-

such that P i - +i the collection Jcd

are called gems

398

MUJ 1 CA

o f holomorphic I - f a m i l i e s at the point a . The germ of (U,9) at a wil be denoted by p a . Clearly F?a is an algebra. Next consider the collection

where the algebras Va

E

4 let

are regarded as disjoint sets. For each

N ( p a ) denote the collection of all sets

of

the

form

where ( U , V J The set

$

varies over all representatives of the germ

will be endowed with the unique topology such that

N(pa) is a neighborhood base at

xi

pa

for each

pa E

b e d e f i n e d by

56.3. PROPOSITION. L e t

71

for each

( 6 , ~i s )a Riemann d o m a i n

cp,

(9,.

E

<.

Then

:

-+

E

d

E' T(cpa)

over

= a E.

1 We first show that X E is Hausdorff. Let 9, and qb be two distinct points of x;. If a # b then we can find a representative ( U J p ) of 9 , and a representative ( V , $ ) of JI, such that il and V are disjoint. Then the sets N(U,cpI and N I V , J , I are also disjoint. Next suppose a = b . Let ( U , c p ) be a representative of cp,, let ( V , J i ) be a representative of $a and let W be a connected open neighborhood of a such that W C U V . Then the sets IWJlpl and (W,$I are necessarily disjoint, for otherwise the Identity Principle would imply that p i -- qi on W for every i E I, and therefore pa -- JI, , a contradic-

PROOF.

tion. Thus x i is a Hausdorff 'space. Since the mapping TI is clearly a local homeomorphism, the proof of the proposition is complete. 56.4. THEOREM. L e t ( X , c ) b e a R i e m a n n domain o v e r E a n d l e t F C J C l X ) . T h e n t h e F - e n v e l o p e of h o Z o m o r p h y of X cildays e x i s t s .

399

ENVELOPES OF HOLOMORPHY

PROOF.

F =

Write

(filiEI

E JC(X).

Given

x

E

X

let

U

be

a

-1

for each i E chart in X containing x , let pi = fi o ( 5 1 U) € X I be the I, let 9 = ( q i I i E I C J c ( < t U ) ) and let p5 ( a ? ) 5(21 germ of ( < ( U l , v )at <(XI. Then the mapping

is clearly well defined and a morphism. Given ( V , q ) be a representative of

z.

9a

y (q ) = qi(a)

E

let

and define

for each

a

pa

i

E

I.

Clearly each g i is well defined. Since g i = p Z.o T on a neighXi. borhood of pa we see that each gi is holomorphic on For each

i

E

I

and

3: E

X

we have that

If Y is the union of those connected components of 3$ which intersect ~ i X lthen it is clear that T : K Y is a F-extension of X. Let (Z,C) be another Riemann domain over E and suppose that II : X Z is an F-extension of X too. Then for each i E I there is a unique function h i E K I Z ) such that h 2. o u = f i ' Given z E Z let W be a chart in Z containing -I z , let $i = hi o (C I W l for each i E I, let J, = ( J , i ) i E I C +

+

K ( C ( W ) ) and let

$C(zl

E

XIs ( z l

be the germ of I c ( W l , + I

at C.(zl.

Then the mapping

is clearly well defined and a morphism. Given x E X a chart in X containing 3: and let W be a chart in taining p ( x 1 such that W = p t U l . Then

let U be Z con-

MUJ I CA

400

for every

i E I. Hence

and in particular v ( p ( ' X ) ) = ~ t x )C Y. Since each connected component of Z intersects u ( X ) we see that v l Z l C Y. This completes the proof. The envelope of holomorphy of a Riemann domain denoted by E(Xl.

X

will be

EXERCISES 56.A. Let ( X , 5 ) be a Riemann domain over E , and let T : X Y be a holomorphic extension of X. Show that Y = E ( X j if and only if Y is a domain of holomorphy. -+

56.B. Let ( X , < ) be a Riemann domain over E . Show that X = E ( X ) if and only if X is a domain of holomorphy.

57. THE SPECTRUM In this section we show that if X is a Riemann domain then the spectrum of ( 3 C ( X ) , - r c ) can be endowed with the structure of a Riemann domain. We also show that in certain cases the envelope of holomorphy of X can be identified with a subset of the spectrum. 57.1. DEFINITION. Let ( X , S I be a Riemann domain over So (XI denote the spectrum of the topological algebra Let 71 : S , ( X / E be defined by the condition

E.

Let

IxfX),-re).

-+

for every h E Sc(X) and cp E E'. And let E : x be defined by = f(xi for every f E 3 C l X j .

E X - + i E

S,(,Yl

s(f)

It follows from the Mac: ey-Arens Theorem that the

mapping

40 1

ENVELOPES OF HOLOMORPHY

is well defined, for each h by linear functional h" on

71

E

'.

p E

Furthermore, since p o ri;) E', we see that n o E = 5.

57.2.

defines a continuous h"(pI = h(lp o 5 ) €or every cp E

E Sc(X)

= p o <(XI

for every x

E

X

and

L e t C X , O b e a R i e m a n n d o m a i n o v e r E . Then t h e r e

THEOREM.

i s a topoZogy on

S c f X l such t h a t :

i s a Riemann domain o v e r

(a)

(ScIXI,

(b)

The e v a Z u a t i o n mapping

7)

E

: X

+

B.

Sc(Xl

is a morphism.

PROOF. Let h E S c i X ) . Then there are a compact set K C X and a constant c > 0 such that I h ( f l 1 5 c s u p l f l for every f E K Jc(XI.

f",

By applying the preceding inequality to

root, and letting

m

+

taking

mth

we obtain that

~0

In this case we shall write h K. Let 0 < r < d,(K). Given t E B ( 0 ; r ) choose p > 1 such that p t E E R ( 0 ; r ) . It follows from the Cauchy inequalities that E for every

f

for every

f E X(X)

E

Jc(X).

and

rn

E

IIvo.

If we define

(57.3)

for every f E J C l X l , then clearly functional on (sCixI,-rc) and

h,

is a continuous

linear

(57.4)

for every

f E JclX).

Using the formula

m Py(fg1 = ZPT-jf j=O

we see that h , 1 f g ) = h t I f ) i z t f g ? for a l l f, g is a continuous complex homomorphism of ( J c ( X I ,

- Pig,

Thus h , )and it follows

E

Jc(X,J.

T

~

MUJ ICA

402

from (57.4) t h a t (57.5)

for every

f

For e a c h

JctX).

E

h

Thus

Se(XI

E

ht

let

K + at.

-<

Nlh) denote t h e c o l l e c t i o n of a l l

sets o f t h e form

0 < r < d X ( K ) f o r some c o m p a c t s e t

where

such t h a t

W e claim t h a t t h e r e i s a u n i q u e t o p o l o g y o n

-< K .

N t h ) i s a n e i g h b o r h o o d base a t

which and

ha

E

N,(h),

Nr(h). Let

0 < r

Np(h,)

p > 0

N (ha) P

E

K be a c o m p a c t s u b s e t o f

d x ( K I . Choose a c o m p a c t s e t

and choose Then

we can f i n d

h

E

such t h a t

L

E

Sc(X), N , ( h )

E

such t h a t

Nlh,)

such t h a t

X C

h

X

Sc(Xi.

N(h)

E

Np(ha) K and

h

ha

such t h a t

p < d x ( L ) and

h

for

Se(X/

h f o r each

It is s u f f i c i e n t to prove t h a t given

C

K C X

-4

L

B E ( a ; p ) C BE(O;r).

N f h a l , a n d if w e a s s u m e f o r a moment t h a t

(57.6) f o r every

t E BE‘IO;p),

then it is clear t h a t

To p r o v e ( 5 7 . 6 ) observe t h a t f o r e a c h

w i t h uniform convergence f o r

x

E

K.

f

E

Since

N p ( h a ) C Nr(h).

JCCX) w e have t h a t

h

-<

K

it

follows

that

and (57.6) h a s been proved. W e n e x t show t h a t

h2

Sc(XI i s a H a u s d o r f f s p a c e . L e t h l a n d be t w o d i s t i n c t p o i n t s o f S c ( X I . Then t h e r e i s a f u n c t i o n

f E KiX) s u c h t h a t

Ihl(fi - h2(f)l = 3~

> 0.

Choose a c o m p a c t

403

ENVELOPES OF HOLOMORPHY

K C X

set

such that

hi

4

K

j = 2,2,

for

0 < r < dX(KI such that f is bounded, by

t

Then for each have that

If

p

E BE(C;rl,

and choose r with

c say, on K + B E f O . ; r i .

0 < p < 1

m E DV,

and

we

j =1,2

is sufficiently small then m

((77

i

1

Pt

If,

and the neighborhoods N

Fr

(h,)

and

N

Pr

( h Z i are disjoint.

We next show that T : S c ( X 1 E is a localhomeomorphism. More precisely, we show that TI maps N r ( h ) homeomorr-!~izallyonto B E ( ~ l h ) ; r )for each h e S c ( X l and each N r ( h l € h l h l . Indeed, if t € BE(O;r) and 9 € E l then using (57.3) we get that -+

Thus (57.7)

for every

v(htl h

t

E N

r

= vihi + t

( h l , and the desired conclusion follows.

We next show that E : X S c l X l is a morphism. Indeed, if X, 0 < r d X ( x l , t E BEIO;rl and f E X t X l then +

x

E

m

lz + t l - i f ) = f ( x + t l =

z

m

Ptfixl

m=O

Hence E maps B ( x ; r l into N r ( G ) , and X tinuous. Since we already know that TI o E is a morphisrn. 57.3. THEOREM.

Let

E

E

=

qf,.

is therefore con= 5 , we conclude that

(X,
E.For

each

404

MUJ I CA

(c)

The e x t e n s i o n mapping

i s a t o p o l o g i c a l i s o m o r p h i s m b e t w e e n lX(X), T c I and a cornplernented subalgebra of

(X(Sc(X/),~cl.

(a) Let f E X ( X ) and let h E S c l X ) . Choose a compact 0 < r dXiK) set K C X such that h 6 k , and choose r with such that f is bounded, by c say, on K + B E ( O ; r ) . Then

PROOF.

for each

t

E

BE(O;r), and therefore rn

for each

t E BEIO;r)

and

m

Ihl

2.

Hence the function

A

E

A

is holomorphic for each t B g ( O ; r ) , t l i u s proving that 7 is G-holomorphic. On the other hand for each t E B E ( O ; r ) and 0 c p < 1 we can write

+

f(hAtl E 5

m

proving that SciX),

(b) hl,

7

is continuous at

and clearly

E

h . Thus

-

L, compact sets h

j

K . and 3

7

is holomorphic on

= f.

Let L be a compact subset of

. . . ,hm E

such that

o

<

KZJ

. . . ,Krn

% I K 3. 1

for

SriX).

Then we can find

in X and r l , . j = l , . . .,rn,

..,I'

m > 0

and

405

ENVELOPES OF HOLOMORPHY

(57.8) For each

j = 1,

..., m

consider the compact set

We claim that (57.9) < j <m Indeed, if h t L then by (57.8) there exists j with 1 and t E B E ( O ; r . ) such that h = ( h . I Then it follows from j 3 3 t ' (57.7) that n l h ) = n ( h . ) i t , and hence t C Next we conJ j' sider the set

m

K =

U

j=l

IKj + aC.1, 3

which is compact by Lemma 52.5. If

2

5 j 5 m, t

E C

j

and f E

J C I X ) then using (57.5) w e get that

(c)

The restriction mapping

is always continuous. The extension mapping

is continuous by (b). And This completes the proof.

E*

oG(f)

= f

€or every

f

L e t ( X , c ) b e a Riemann domain o v e r 57.4. PROPOSITION. l e t ( Y , ~ I )be a Riemann domain o v e r F . L e t T : X Y +

E

W(XI.

E,

be

and

a

406

MUJ I CA

morphism, and Z e t

b e t h e r e s t r i c t i o n mapping

T*

T h e n t h e mapping

5 s a morphism and t h e f o l Z o w i n g d i a g r a m io c o m m u t a t i v e

Ex

T

X

+

Y

T

PROOF. L e t h E S,(X) a n d l e t s u c h t h a t h + K . Then

f o r every 0

p

g

E

K b e a compact

J C ( Y ) , and t h e r e f o r e

d,lTtKI),

choose

r > 0

h

OT*

with

r

subset

+ T(K). p

and

of

X

Given p w i t h

r < dx(KI. W e

claim t h a t

Indeed, f o r every

t

E BE(O;r)

and

g

E

JC(YI w e have t h a t

m

T h i s shows ( 5 7 . 1 0 ) a n d w e c o n c l u d e t h a t Then i t follows e a s i l y t h a t -

Ex

OT.

T**

T**

is

continuous.

i s a morphism and that

T * * O E ~

407

ENVELOPES OF HOLOMORPHY

57.5.

.?

L e t ( X , E , J be a Riemann d o m a i n o v e r

PROPOSITION.

d e n o t e t h e u n i o n o f t h o s e c o n n e c t e d c o m p o n e n t s of

intersect

Let

E.

S c l X ) which

E ~ X ) . Then:

(a)

E

(b)

E ;

x *-

X

+

x

i s a holomorphic e x t e n s i o n o f

: ( X ( ~ ) , T ~-+) I J C ( X I , T ~ )

X.

is a t o p o 1 o g i c a E

isomor-

phism.

(c) that

I f

T : X

T* : ( X ( Y / , T c )

+

i s a holomorphic e x t e n s i o n o f X such i s a topologicaz isomorphism,

Y

.+

/X/X),-rc)

t h e n t h e r e i s a morphism

ci : Y

+

x^

such t h a t

0 O T

= EX '

PROOF. (a) and (b) are direct consequences of Theorem 57.3. To prove (c) we observe that since the mapping T*: I X ( Y I , T ~ ) -+ / J C ( X ) , T ~ ) is a topological isomorphism, then the morphism T * * : SclY) Sc(X) given by Proposition 5 7 . 4 is a bijection, and therefore an isomorphism. Let u : Y S c ( X ) be defined by CJ = (T**)-' o E Then certainly 0 o T = Ex . Furthermore, since Y ' each connected component of Y intersects T ~ X ) , it follows that u ( Y l C 2, thus completing the proof. +

+

57.6. COROLLARY. = Sc(x) and

Let ( X , 5 1

57.7.

L e t (X,51 be a Riemann domain o v e r

2

COROLLARY.

be a Riemann domain

E.

Then

Se(f)

En.

Then

= f.

= E(X).

Without l o s s of generality we may assume that X is conIf T : X Y is a holomorphic extension of X then Y connected and hence fXtX), T ~ and I ( J C f Y I , T ~ are I Frkhet Hence T * : (x(Y),-cc/ I J C ( X I , re) is a topological isomorphism by the Open Mapping Theorem. Then the desired conclusion follows from Proposition 57.5. PROOF. nected. is a l s o spaces.

57.8. set

+

+

PROPOSITION.

A =

: f

c a l l y A-convex.

E

Let

JClXll

/ X , < ) be a Riemann domain o v e r E . I f w e then

S c f X l i s A-separated

and m e t r i -

408

MUJ I CA

PROOF. Given h , # h , in S , ( X ) we can find f E X ( X ) such that h l ( f ; # h g f f l . Thus f l h , ) # F ( h , ) and S c f X ) is A-separated. L C Sc(X), Theorem 57.3 yields a com-

Given a compact set

Choose r such that

r

d,(K).

h E

Z A and

0

> r . Indeed, for each

..

We claim that f E XfX)

dS I X I ( z A I c

we have that

-

Hence h K and therefore N,ih) E N ( h l . Since ~ ( N ~ ( h =i ) BE("ihl;rl we conclude that dSclxii h l 2 r , as we wanted. 57.9. COROLLARY.

Let

E

b e a s e p a r a b l e Banach s p a c e w i t h

bounded a p p r o x i m a t i o n p r o p e r t y . L e t (X, 5 ) b e a Riemann

over

E.

Then

2

the

domain

= E(X).

PROOF. By Proposition 57.5, E : X x' is a holomorphic extension of X. By Proposition 57.8 and Theorem 54.12, 2 is a domain of holomorphy. Then the conclusion follows fromExercise 56.A. +

58. ENVELOPES OF HOLOMORPHY AND THE SPECTRUM In the preceding section we proved that if X is a Riemann domain over a separable Banach space with the bounded approximation property then the envelope of holomorphy E(XI of X can be identified with a subset 2 of the spectrum YJ of (X(X),Tc). In this section we shall prove that actually E(XI=

sc ( X i . 58.1. DEFINITION. A polynomial m a l i z e d if

P f z l = C caza

and

a 58.2. LEMMA.

For e a c h p o l y d i s c

P

E

p ( 8 ) is said to be

nor>-

maxlcCil = 1 . a An(O;rl

with

0 < r < 1 there

409

ENVELOPES OF HOLOMORPHY

i s a constant c = c ( n , r ) > 0 w i t h t h e foZZowJing p r o p e r t y . If n E P i 6 I is a normaZized p o Z y n o m i a Z of d e g r e e a t most j i n each

P

v a r i a b Ze, and

where

then

0 < 0 < 1,

Let

PROOF.

5 -

AlS)

s = (1 - r)/2

c /Zog 0 .

Ms =

and let

sup

IPI.

If

P(z)

An(O;s)

=

cclz

c1

then it f o l l o w s from the Cauchy integral formulasthat

c1

I MsS

leal

a

with

-

ai < j

for every for

i =

a. Since P is normalized there isan 1,.

. .,n

such that

0 c s < 1 it follows that M s 2 s l c y l > sn', find a point a E A " ( 0 ; s ) such that Pis)

t = ( I + r)/2

Let

leal

= 1.

Since

and hence we

> snj2-n.

and observe that

n

Since P is the sum of at most (j + 1 ) with leal 5 I, we see that /PI < ( j + l o g [ (j + l / - n l ~ l ]

0

on

An(O;l),

and it follows that

S

On the other hand, us ng the inequality (35.1) we get that

An ( a ; t i

l o g [ (j

c1

terms of the form eclz , on A n ( O ; l / . Thus

J

\

can

MUJ I CA

410

Since l o g e <

0

it follows that

n

TI

-

log 0

I ,n (

l + r 2n log

log 0

8 I-r '

completing the proof. 58.3. LEMMA. L e t (X,SI b e a R i e m a n n d o m a i n o v e r C n . Let K be a c o m p a c t s u b s e t of X a n d l e t f E J c ( X ) . T h e n t h e r e e x i s t 0 with

0 < 0

j,k

each

nomial

f o r all

IN w i t h

E

zl,

x

0

w i t h t h e f o l l o w i n g p r o p e r t y . For

E

ZiY

k

2 ko

j

of d e g r e e

..., z n J a n d

there e x i s t s a nonzero

of d e g r e e a t m o s t

K. I f a i s a n y g i v e n p o i n t i n

E

i n each

j

a t most

- aI,

zI

Let

. . ., z n

C

the

i n w such t h a t

Cn

then the poly-

to

the

- an,w.

0 < 2~
K

poZy-

of

k

can b e t a k e n t o be normalized w i t h r e s p e c t

P

variables PROOF.

k

and

P ( z l , . . .,zn,wl

variables

nomial

1

and choose b

m u

I'

..,,ba

E

K

such that

AX(bi;ci.

i=l

Fix

j

2

k

in

l7V

and let P be the vector s p a c e of all poly-

nomials P ( z I,...,znJwI of degree at most j in each of the .,z and of degree at most k in w. Then P variables z l , n

..

has dimension ( j + I l n ( k + I ) . . Let smaller than m - ' ( j + 1 ) ( k + I functionals on P of then form

.

c1

Tia(PI

= a

be the largest integer Consider all the linear

+. . .+a

n a 1 'n az, . . azn 1

p

.

(<(bi),f(bi)),

41 1

ENVELOPES OF HOLOMORPHY

with

. .,m

i = 1,.

and

p.

la1

l i n e a r f u n c t i o n a l s , and since can f i n d a nonzero 1,

..., m

and

assume t h a t

-

a l J . . .,an

P

E

P

T h e r e are, a t m o s t m p n o f s u c h m p n < l j + I l n ( k + I) =dim P, w e

such t h a t

= 0

Tia(P.J

i =

for all

/ a 1 I such

If(z)I 5 M a n d I Si(zc) - ui I 5 M f o r e v e r y x E L a n d i = I,. . ..n. L e t Q = P o ( 5 , f ) . Then Q(x1 i s t h e sum of a t mst ( j + I I n ( k + I / terms o f t h e f o r m c ar ( S l z l - a)"flxIr, w h e r e lcarl 5 1 . Whence it a = ( a l , . . .,a,), w i t h ai 5 j , r 5 k a n d that

follows t h a t

f o r every

x

E

L.

B y o u r choice of

n o m i a l s of o r d e r l e s s t h a n Q

a t the point

bi

P, a l l thehomogeneous p l y -

i n t h e T a y l o r series expan:-;ion of

p

v a n i s h . Then i t f o l l o w s f r o m Theorem 7 . 1 9

that

for a l l

x

for all

3: E

x

E

i = I,. . . , m .

Thus

K. S i n c e ( j + 1 ) n+IM- In+ 1 ) j

can f i n d a constant

for a l l

and

E AX(bi;"3)

c = cfnJM) > 0

-f

j

-+

m

we

such t h a t

(j + l ) ( k +

.i k l / n m

>

t71

it f o l l o w s t h a t 4 ' I

when

K. S i n c e

p + l >-

0

0

cj / m ( p + l )

such t h a t

82

--f

1 /'m

1

when

j, k

+

-.

Choose

I , and next choose

ko E

8

with such

412

MUJ I CA

for every

x

and

K

E

j

2

k > k o . This completes the proof.

In the next lemma Cn is endowed withthe norm so that the balls are the polydiscs.

IIzII = maxlz.I, 3

n 58.4. LEMMA. Let (X,<) b e a Riemann domain over C . Let a closed s u b a l g e b r a of ( K ( X ) , T ~ ) such t h a t -

A

(a)

P!f

(b)

E l , . ...Sn

Assume that

€

f o r every €

f

€

A, m E lN and

E.

€

A. X

is A-separated and metrically A-convex, b u t

X

K in X, a p o l y -

is not A-convex. Then there a r e a compact set

disc A n ( a ; r l in C n , a function distinct points in X s u c h that (c)

xi E S - ' ( a )

(d)

Anix . ; r )

(e)

The

x

t

A be

3

C

and

9

iA

f E

A, and a sequence i x

d (x

X

i

f o r every

j

> r

) E

f o r every

.)

3

of

j E N.

17v.

set

has Lebesgue measure zero

Since X is not A-convex there is a compact set L C X such that LA is not compact. Since X is metrically A-convex

PROOF.

we can find quence in

r > 0

such that

(y

2r .: dX(iAi. Let

with no cluster point. Since 1<,(y

.)I J

each i = I,.

. . ,n

and

j

E

TN,

.)

n

be a se-

i ~ u p / E ; ~ for I L

we see that the sequence ( S i y j )

)

is bounded and has therefore a cluster point a in C n . After replacing i y .I by a subsequence, if necessary, we may assume 3 that t i y j ) + a and 1 1 ~ i y : j ) - a l l r for every j E Lli. Hence

ENVELOPES

OF HOLOMORPHY

413

for each j E B there is a unique point x E A y ( y j ; r ! such j that S(x .I = a . If there were a point x E X such that x . = x 3 3 for infinitely many indices, then x would be a cluster point of the sequence f y .I. Hence, after replacing ( y .I and (x .) by 3 3 3 suitable subsequences, if necessary, we may assume that = x k whenever j # k . Clearly d ( x ) > r and using Lemma 54.8 x i we get that

for every j E W . If we set K = L + a n ( 0 ; 2 r ! then (c) and (d) are verified. Since X is A-separated, the set {f E A : Is.! # 3 f(zck)} is open and dense in A whenever j # k . Since A is a n If E A : f ( x ,) # f ( x k ) ) is also dense Baire space, the set 3 jfk in A, and in particular contains a function f. Since thesets

Ax(xj;r) are pairwise disjoint, the function

is well defined and is holomorphic on the polydisc An(a;rl,and does not vanish at the point a whenever j # k . Hence the set ZBk

= {z

E

has Lebesgue measure zero whenever Since one can readily see that { z E An(a;r)

C { z E

: f(<-'(z)

A n i a ; r l :f(<-'(z!

n

j # k

by Exercise

35.F.

is f i n i t e }

n

(e) follows. The proof of the lemma is now complete. 58.5. THEOREM. L e t (X, < ) b e a R i e m a n n d o m a i n o v e r g n . L e t b e a c l o s e d s u b a l g e b r a of ( J c t X ) , -re! ,sz,ch t h a t (a)

Pmf E t

A

for every

f E

A, m

E

and

t E E.

A

MUJ I CA

414

Suppose t h a t is A-convex.

x

i s A - s e p a r a t e d and m e t r i c a l l y A-convex. T h e n X

Suppose that X is not A-convex. Then there are a compact set K in X, a polydisc A n ( a ; r ) in C n , a function f E A, and a sequence ( 3 : .I of distinct points in X with the properJ ties (c), (d) and (e) stated in Lemma 58.4. By Lemma 58.3 there exist 8 with 0 < 8 1 and ko E IN such that for each j, k E W with j 2 k __ > ko there is a nonzero polynomial P j k ( z I , ..., z 'n w l of degree at most j in each of the variables

PROOF.

zI,

. . ., z n ,

and of degree at most k

in

w, such that

(58.1) for all

w

E 5

x

E

K , and hence for all

x

For

E

z E gn

and

we can write

(58.2) ( z ) is a polynomial of degree at most j in each of 'jks the variables z l , . ,z For j 2 k 2 k , set n

where

..

.

By Lemma 58.3 each of the polynomials can be taken to be Pjk normalized with respect to the variables z 2 - a], ..., z n - a n' w . Hence for each j > k 2 ko at least oneof thepolynomials P j k s is normalized with respect to the variables z l - a l , ... , B n - a n . Since we may assume that 0 < r 6 I, an application of Lemma 58.2 yields a constant c = c ( n , r ) > 0 such that X(Sjkl 5 - c

For

/+ kiln

log 0 .

j 2 k 2 ko

In particular set

XISjk)

-+

0

when

j, k

-+

00.

41 5

ENVELOPES OF HOLOMORPHY

Then we can find every j 2 k 2 k l

and k > k I 0 If we set

.

hlT)

L

p > 0.

Furthermore, if

infinitely many values of

z E

n

i A ) .For each

T

j

2

j k

€or

p

then

z E

for

T

jkl

l/n

> 8 2

max I p j k l S ( z ) I s5 k,

for infinitely many values of f(<-’(zl

I 2

j, and therefore 1 -

(58.3)

jk

m

00

then

such that X(T

p > 0

j. Fix kl

J

z

E

and each

T

and let

s 5 j

w

E

set

and observe that (58.5)

for every j (58.4) that

kl.

It follows from (58.1), (58.2) (58.3)

and

for infinitely many values of j. By choosing a sequence of values of j for which (58.6) is true, we can find for each s 5 kl m a cluster point c s ( z i of the sequence (cjs)j=k such that 1

7,

nl

(58.7)

2

s =o

cs(z)ws

= 0.

Furthermore, (58.5) guarantees the existence of at least sl 5 k l such that Icsl(z)l = 1 . Thus, for each z E T set

f(<-l(z) r~ 17,)

has at most

kl

one the

points, the roots of the

416

MUJ ICA

X ( T ) > 0,

equation (58.7). Since

t h i s c o n t r a d i c t s t h e conclu-

X most b e A-convex

s i o n o f Lemma 5 8 . 4 . Thus

and t h e proof

of

t h e theorem i s complete. 58.6.

L e t ( X , < ) b e a Riemann domain o v e r @

COROLLARY.

n

.

Then

the s e t

i s compact f o r e a c h compact s e t PROOF.

{F

A =

Let

A-convex,

K"

that

x(X)1.

: f E

m e t r i c a l l y A-convex,

K

C

X. S c ( X ) i s A - s e p a r a t e d and

Then

by P r o p o s i t i o n 5 7 . 8 , a n d h e n c e

by Theorem 5 8 . 5 .

j? = ( c ( K ) ) i

Since

S,IX)

we

i.s

conclude

i s compact. L e t ( X , < ) b e a Riemann domain o v e r

5 8 . 7 . THEOREM.

Cn.

Then

s c i x ) = x^ = E I X ) . PROOF.

h

Since each

set

K C X ,

set

K

C

-

belongs t o

E ScfX/

it s u f f i c e s t o show t h a t

K"

C

K

for

x^

f o r each

K be a f i x e d compact s u b s e t o f

X. L e t

b e t h e c l o s u r e of J C ( X ) i n

8

Then

C(K).

some compact compact

X and l e t

8

i s a commutative Banach

correspondence w i t h K". B i s endowed w i t h t h e G e l f a n d topology

a l g e b r a whose s p e c t r u m i s i n o n e - t o - o n e

If t h e s p e c t r u m

S ( B ) of

I

then t h e canonical b i j e c t i o n and i s t h e r e f o r e

K

+

S ( 8 ) is c l e a r l y

a homeomorphism, s i n c e

S I B ) can be c a n o n i c a l l y i d e n t i f i e d w i t h

and t h e sets

17

n

x^

I? 1 x^

and

Theorem 3 1 . 8 t h e r e e x i s t s

h

E

K"

x^

and

t i c a l l y zero on

h(fl = 1

is complete.

E

\x^

K" i s c o m p a c t . i . Now, w e can

Thus write

are compact and d i s j o i n t . By €3 s u c h t h a t h ( f ) = 0 f o r e v e r y

for e v e r y h(f) = 1

K but

impossible, unless

f

continuous,

h E

K" \ ? . h

f o r every

i s empty.

Thus

I?

C

Hence j ' is i d e n E

K"

2.

This is

a n d the p r o o f

417

ENVELOPES OF HOLOMORPHY 58.8. En.

Let (X,
COROLLARY.

b e a p s e u d o c o n v e x Riemann domain over

T h e n e a c h c o n t i n u o u s compZex homomorphism of ( J C ( X I , - c c )

is

a point evaluation.

X is a domain of holomorphy by Theorem 5 3 . 8 . Hence X = E(XI by Exercise 56.B. And hence X = S c ( X l by Theorem 58.7. PROOF.

5 8 . 9 . THEOREM.

Let

E

b e a s e p a r a b l e Banach s p a c e

bounded a p p r o x i m a t i o n p r o p e r t y . L e t f X , < ) Riemnnn domain o v e r p h i s m of

(JftxI,

-eel

be

a

with

pseudoconvex

Then each c o n t i n u o u s complex

E.

the

homomor-

i s a point evazutation.

(a) We first assume that E has a monotone Schauder basis. Given h E S , ( X l we can find a compact set K C X such that I h ( f l I 5 suplfl for every f E J C f X ) . Then the compact K set K is contained in one of the open sets given by Lemma Aj 54.4. Then A i s X-pseudoconvex, and hence J C I X ) is dense in j ( J C ( A . I , T ~ ) by Theorem 51.1. Hence h can be extended to a con3 1 . T ~ ( A ~ ) tinuous complex homomorphism h j of ( J C ( A . 1 , ~ ~ Since 3 C Xn for every n > j, the mapping PROOF.

defines a continuous complex homomorphism of ( X ( X n ) , ~ ) c each n 1. j. Then, by Corollary 5 8 . 8 , for each y1 2 j exists a point a E Xn such that n

for there

€or every f E J C ( X , I , and in particular for each f E J C ( X I . If co K ( X ) then we can readily see that the sequence ( f o T n l n Z j

f E

converges to f in ( X ( A . l , - c I , and whence it follows that 3

c

In particular the sequence ff( a , ) )

is bounded for each f E JC(X),

418

MUJ I CA

and it follows from Theorem 54.11 that the sequence (a,) has a cluster point a . Then it follows from ( 5 8 . 8 ) that h ( f ) = f ( a ) for every f E 3 C t X ) . (b) If E is a separable Banach space with the bounded approximation property then by Theorem 27.4 we may assume that E is a complemented subspace of a Banach space G which has a monotone Schauder basis. Thus there is a closed subspace F of G such that G = E x F. Let h E S e ( X ) . If we identify X with X x ( 0 ) then the mapping

defines a continuous complex homomorphism of ( X ( X ( a ) there is a point ( a , b ) E X x F such that

x

F),-rc).

By

X denote the canonifor every g E J C ( X x F ) . Let 71 : X X F cal projection. If f E X ( X ) t.hen f o TI E J C f X x F ) and f o n I X = f. Whence it follows that h ( f l = f ( a ) for every f E JC(X), and the proof of the theorem is complete. +

58.10. COROLLARY.

E

Let

b e a s e p a r a b l e Banach s p a c e w i t h t h e

bounded a p p r o x i m a t i o n p r o p e r t y . L e t ( X , E ) over

6. T h e n

Se(X)

b e a Riemann

domain

= f = EIX).

PROOF. By Proposition 57.8 i is pseudoconvex. Then hy Corollary 57.6, Theorem 5 8 . 9 and Corollary 5 7 . 9 we get that ScfX) = Sc(i) = f = E(X). Theorems 54.12 and 58.9 can be summarized as follows. 58.11. THEOREM.

Let

E

b e a s e p a r a b l e Banach space w i t h

the

bounded a p p r o x i m a t i o n p r o p e r t y . L e t t X , < )

hi: a

Riernann domain o v e r

conditiorzs w e equiva-

E.

Then t h e f o l l o w i n g

lent:

(a1

X

i s a domain of e x i s t e n c e .

pseudoconvex

419

ENVELOPES OF HOLOMORPHY

X

is a domain o f ho1ornorph.y.

X

is h o l o m o r p h i c a l l y c o n v e x .

X

i s m e t r i c a l l y holornorphicaZZy c o n v e x .

X

is p s e u d o c o n v e x .

Each c o n t i n u o u s c o m p l e x homomorphism o f

( J c t X l , ~ is ~)

a point evalutation.

To conclude this section we prove that if X is a Riemann domain over Cn then each complex homomorphism of K f X ) is continuous for the compact-open topology. THEOREM.

58.12. Cn.

fl,.

Let

Let ( X ,

. ., f ,

5) be a p s e u d o c o n v e x Riernann domain over

E Jc(X)

+

...

+ f m ( x l g m ( x )= 1

be

m f u n c t i o n s w i t h o u t common zeros.

g l , ...,gm

Then t h e r e a r e f u n c t i o n s

f o r every

E

K ( X ) such t h a t

fl(x)g1(x)

x E X.

Without l o s s of generality we may assume that X is connected. Then X is hemicompact and ( J C ( X I , ' r c ) is a Frgchet algebra. By Corollary 58.8 each continuous complex holomorphisn of ( J c ( X ) ,-re) is a point evaluation. Hence the set {fl,. ,f m ) is not contained in the kernel of any h E S c ( X ) . Then it foldoes not lows from Proposition 32.13 that the point ( 0 , . . . , a ) belong to the joint spectrum of f l , . . ,f , . Hence the functions f,,. . . > f m generate the improper ideal. PROOF.

..

.

5 8 . 1 3 . THEOREM.

8 . Then

Let (X,<1

b e a p s e u d o c o n v e x Riernann d m i i n Over

e a c h c o m p l e x homomorphism o f

J c ( X ) is a p o i n t e v a l u a -

tion.

To prove this theorem we apply Proposition 3 3 . 6 . Observe that condition (a) follows from Theorem 5 8 . 1 2 , whereas condition (b) follows from Theorem 13.10. PROOF.

58.14.

COROLLARY.

L e t (X,

be a Riemann domain o v e r Cn. T h e n

e a c h comp l e x homomorphism of KiX) is c o i z i i r ~ n u r for t l z compact-cpen

MUJ 1 CA

420

topology.

PROOF.

If suffices to apply Theorem 58.13 to S c f X ) .

The preceding results can be summarized as follows. 58.15. THEOREM.

L e t (X, 6 ) b e a Riemann domain o v e r E n . Then the

following conditions are equivalent:

X

i s a domain of e x i s t e n c e .

X

i s a domain of h o l o m o r p h y .

X

i s holomorphically convex.

X

i s m e t r i c a l l y holomorphically convex.

X

i s pseudoconvex.

Each c o n t i n u o u s c o m p l e x homomorphism of ( J C f X ) ,

Tc)

is

a point evaluation.

(9)

Each c o m p l e x homomorphisn of

K ( X ) i s a point evalua-

tion. (9)

X

i s h o Z o m o r p h i c a l l y s e p a r a t e d , and

generated, proper ideal o f

euch

finitely

X f X ! has a common z e r o .

NOTES AND COMMENTS

Theorem 56.4 is due to A. Hirschowitz [ 3 1

.

Theorem 57.2 and 57.3, and Proposition 57.5, are due t o H. Alexander [ 1 ] , who adapted to the case of Banach spaces a construction of H. Rossi [ l ] in the finite dimensional case. has Theorem 58.5 is due to E. Bishop [ 2 ] . Theorem 58.7 been taken from the book of R. Gunning and H. Rossi [ l ] . Theowas rem 58.9 is due to M. Schottenloher [ 7 1 . Theorem 58.12 proved first by H. Cartan [ l ] for domains of holomorphy in Q? using sheaf theory.

B I R L IOG,RAPHY

H . ALEXANDER [ 1 ] A n a l y t i c f u n c t i o n s o n Banach s p a c e s . U n i v e r s i t y of C a l i f o r n i a , Berkeley, 1968.

Ph. D .

thesis.

R . ARENS

[11 Dense i n v e r s e l i m i t r i n g s . M i c h i g a n Math. J . 5 ( 1 9 5 8 ) ,

169

- 18%.

R . ARENS a n d A . CALDERON

11 A n a l y t i c f u n c t i o n s o f s e v e r a l . Banach a l g e b r a e l . e m e n t s . Ann. o f Math. ( 2 ) 6 2 ( 1 9 5 5 ) , 204 - 2 1 6 . R . ARON a n d M .

SCHOTTENLOHER

[ 1 1 Compact h o l o m o r p h i c m a p p i n g s o n Banach spaces a n d t h e a p p r o x i m a t i o n p r o p e r t y . J. F u n c t . A n a l . 2 1 ( 1 9 7 6 1 , S

7 -30.

. BANACH

[ l ] Thgorie d e s Opgrations L i n g a i r e s . Warsaw, 1932. p u b l i s h e d by C h e l s e a , N e w Y o r k . E.

BISHOP

83

(19611,

Re-

[ 1 1 Mappings o f p a r t i a l l y a n a l y t i c spaces. A m e r . J. Math. 209 - 2 4 2 .

[ 2 1 Holomorphic c o m p l e t i o n , a n a l y t i c c o n t i n u a t i o n a n d t h e hlath. (2) 78 (1963) , i n t e r p o l a t i o n o f seminorms. Ann. of 468 - 500.

J . BOCHNAK a n d J. SICIAK

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[ 2 1 A n a l y t i c f u n c t i o n s i n t o p o l o q i c a l v e c t o r spaces.Studi a Math. 39 ( 1 9 7 1 1 , 7 7 - 1 1 2 . N.

BOURBAKI

[11 ElGments d e M a t h g m a t i q u e . T o p o l o q i e GGnSrale, t r e s 1 2 4, n o u v e l l e G d i t i o n . Hermann, P a r i s , 1 9 7 1 .

42 1

chapi-

422

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H . BREMERMANN

[ l ] Uber d i e i q u i v a l e n z d e r pseudokn-vexen G e b i e t e und der H o l o m o r p h i e g e b i e t e i m Raum von n komplexen Veranderlichen Math. Ann. 1 2 8 ( 1 9 5 4 ) , 6 3 9 1 .

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Formes D i f f g r e n t i e l l e s . Hermann, P a r i s , 1 9 6 7 .

H . CARTAN and P . THULLEN [ 11 Zur T h e o r i e d e r S i n g u l a r i t a t e n d e r Funkkionen mehrerer k.omplexen V e r a n d e r 1i c h e n R e g u l a r i t a t s - und Kcnvergenzbereiche. Math. Ann. 1 0 6 ( 1 9 3 2 ) , 6 1 7 - 6 4 7 .

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[ I ] Sequences a n d S e r i e s i n Banach S p a c e s . G r a d u a t e T e x t s i n M a t h e m a t i c s , v o l . 9 2 . S p r i n g e r , N e w York, 1 9 8 4 . J . DIEUDONNE

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S. D I N E E N

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1 1 1 S u r l e non-plongernent d e s v a r i e t s s a n a l y t i q u e s banac h i q u e s rselles. C . R . Acad. S c i . P a r i s 269 ( 1 9 6 9 ) , 8 4 4 - 8 4 6 . [ 2 1 Bornologie des espaces de fonctions analytiques en d i m e n s i o n i n f i n i e . S s m i n a i r e P i e r r e L e l o n g 1 9 6 9 / 7 0 , pn. 21-33. L e c t u r e Notes i n M a t h e m a t i c s , v o l . 2 0 5 . % r i n g e r , B e r l i n , 1 9 7 1 . L 3 1 Prolongernent a n a l y t i q u e e n d i m e n s i o n I n s t . F o r i e r Grenoble 22, 2 (19721, 255 - 2 9 2 . L.

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Ann.

HORMANDER

[ 1 1 L i n e a r P a r t i a l D i f f e r e n t i a l O n e r a t o r s . D i e Grundlehren d e r m a t h e m a t i s c h e n w i s s e n s c h a f t e n , Eand 1 1 6 . S p r i n g e r , B e r l i n , 1963. [ 2 1 L 2 estimates a n d e x i s t e n c e t h e o r e m s f o r the 3 operator.

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abstract

429

BIBLIOGRAPHY

L. WAELBROECK

1 1 1 L e c a l c u l symbolique d a n s l e s a l g g b r e s J . Math. P u r e s Appl. 3 3 ( 1 9 5 4 1 , 1 4 7 - 1 8 6 . A.

commutatives.

WEIL

[ 11 L ' i - n t e g r a l e de Cauchy e t l e s f o n c t i o n s v a r i a b l e s . Math. Ann. 111 ( 1 9 3 5 ) , 178 - 1 8 2 . S.

de

plusiers

WILLARD [ 1] General Topology. Addison-Wesley,

Reading,

Massachu-

setts, 1970. W.

ZELAZKO

[ 11 Banach A l g e b r a s . E l s e v i e r P u b l i s h i n g Company,Amsterdam, and P o l i s h S c i e n t i f i c P u b l i s h e r s , Warsaw, 1 9 7 3 .

M.

ZORN

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in

Banach

boundedness.

This Page Intentionally Left Blank

INDEX

Abel's lemma, 31 adjoint operator , 287 alternating mapping, 4 analytic function, 33 approximation problem, 201 approximation property, 195 Ascoli's theorem, 912 atlas, ?31 i F86, , 87, 177, 274, 347, 362 balanced set, 51 Banach algebra, 2 1 1 barrelled space, 76 basis problem, 193 Bochner integrable, 41 boundary distance, 52, 273, 333, 346 bounded approximation property, 19 6 bounding set, 94 B x i ' x ; r ) , 333

closed form, 154 closed operator , 287 compact-open topology, 69 complex homomorphism, 215 continuously differentiable,104 convex function, 252 convolution, 123, 132, 340 countable at infinity, 70 Cousin problem, 174 Cousin property, 1 7 9 6, 1 CmfU;FI,

CIIU;F), C F q (U; F )

115 145

,

157 m

Cm(K;F),

Cpq(K;F)

C"(X;F),

C;q(X;F)

, 179 , 336

densely defined operator, 287 differentiable function, 99 differential, 99 differential form, 144, 336 distinguished boundary, 48 Cartan-Thullen theorem, 91, 370 Dolbeault's lemma, 169, 172 Cauchy-Hadamard formula, 27 domain of existence, 82, 361 Cauchy inequality, 47 domain of holomorphy, 81, 361 n Cauchy integral formula, 45, 47, an (a;~), T n ( a ; r ) , ? o ~i a ; r ) , 48 49, 5 0 , 2 6 7 d~ , 52, 333 chain rule, 101 6u 273, 346 -hart, 331 Dfla), 99

MUJ I CA

432

G-holomorphic, 58 generalized Cauchy integral formula, 162 geodesic distance, 335 germ, 178, 179 G A ~287

Haar system, 194 Hartogs figure, 79, 84 Hartogs’ theorem, 168, 265 hemicompact space, 70 envelope of holomorpy, 397

holomorphic extension, 80, 361 holomorphic function, 33, 336 holomorphically convex, 87, 362

equicontinuous, 70 exact form, 154

holomorphically separated, 362

exterior differential, 146

homogeneous polynomial, 12

exterior product, 139, 145

1CiU;FI

Et

1

,

33

JCG(U;F),

58 178

Em, 7 EIX), 400

JC(K;FJ,

finitely polynomially convex, 208

3Cm(UI,

finitely Runge, 208

ideal, 214

finite rank operator, 195 finite type polynoinial 17

idempotent set, 236 identity principle, 37, 337

Frgchet algebra, 234

infinitely differentiable, 113

F, 1 F-Cousin property, 179

joint spectrum, 220

x b ( E ) , 240

JC(X;F);JCfX;Y!,

336

367

F-convex , 3 62 F-domain of holomorphy, 361

k-space, 69

F-envelope of holomorphy, 397

lK, 1

F-extension, 361 F-separeted, 362

Laqranqe interpolation formula,

Gelfand-Mazur theorem, 214

26 Laqrange interpolation polyno-

Gelfand topology, 217 Gelfand transform, 217

mial, 18 Leibniz’ formula, 5

433

INDEX

Levi problem, 279, 311, 324,

Newton's binomial formula, 6

374, 386 line segment, 335

normalized basis, 188 normalized polynomial, 408

Liouville's theorem, 39

m, r n o I

local homeomorphism, 331 locally bounded, 37, 71

N

locally finite, 119

Oka's extension theorem, 183 Oka-Weil theorem, 183, 204, 318,

locally m-convex algebra, 227 lower semicontinuous, 245 .C(m~;~), 1 Z ~ ( ~ E ; F IJ ', ? E ; F )

,

123, 126, 127

P?

(u), L'

P4

291 L'(XI,

L'(x,

i ~ , ~ o c L)

~ 287 ,

393 open mapping principle, 38, 337

4

L P I U ) , L P ~ u ,l o c i , I > J ' ( u , c ~,I L~

1

2 P4

i~,Cp),

partition of unity, 119 polarization formula, 6, 51 plurisubharmonic, 247, 336 polygonal line, 335

loel, L ' I X , ~ ~ ) ,

polydisc, 48

340, 345 polynomial, 14 ( x i , L~ (x,l o c i , L 2~ ~ ( X , , ~ polynomial ) polyhedron, 178 P4 P4 359 polynomially convex, 178, 185 Poincarg lemma, 154 maximal ideal, 214 power series, 27 maximal ideal space, 217 projective limit, 229 maximum principle, 38, 337 proper ideal, 214 measurable mapping, 41 P P E ; F I12, mean value theorem, 102 P ( E ; F I , 14 metric approximation property, P r n f ( a l , 33, 338 196 P p i a i , 56, 338 metrically F-convex, 362 metrically holomorphically radius of boundedness, 52 convex, 362 radius of convergence, 27, 52 Michael problem, 237 regular cover, 367 Mittag-Leffler theorem, 176 Riemann domain, 331 monotone basis, 188 Runqe domain, 208 morphism, 331 B ,1 Montel's theorem, 76 r b j f l a l , r c f ( u ) , 52 multilinear mapping, 1 R A , 287 multiplicative linear functional, 215 Schauder basis, 188 L~

434

MUJ I CA

Schauder system, 194

test function, 122, 340

Schwarz' lemma, 56 Schwarz' theorem, 111 separately holomorphic, 63

topological algebra, 227 topology of compact conver-

simple mapping, 41 spectral radius, 213 spectrum of an algebra, 215, 228 spectrum of an element, 212 star-shaped, 154 strictly plurisubharmonic, 257 strongly poiynomially convex, 185

subharmonic, 246 subordinated to a cover, 119 support of a distribution, 131, 343 support of a function, 119 symmetric mapping, 4

sc(X)

I

gence, 69 topology of pointwise convergence, 70 7

c'

69

upper semicontinuous, 245 U-pseudoconvex, 320 Vitali's theorem, 74 weakly holomorphic, 65 wedge product, 139, 145 weight function, 292 k k W (Ul, W ( U , Z o c ) , 307 W k (U), W k ( U , l o c i , 307 P4

P4

X-pseudoconvex, 380

400 G,,,(U;FI,

Taylor series, 33

Rpq ( U ; F ) ,

tensor product, 7

R

P4

144 157

(X;F), 336