COMPLEX ANALYSIS IN BANACH SPACES
NORTH-HOLLANDMATHEMATICS STUDIES Notas de Matematica (107)
Editor: Leopoldo Nachbi...
21 downloads
663 Views
19MB Size
Report
This content was uploaded by our users and we assume good faith they have the permission to share this book. If you own the copyright to this book and it is wrongfully on our website, we offer a simple DMCA procedure to remove your content from our site. Start by pressing the button below!
Report copyright / DMCA form
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
This Page Intentionally Left Blank
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 < p < 1 consider the function f
E
3C(S2;E)
Then ( 3 7 . 3 ) guarantees that f I a x ( 0 ) ) C U , whereas (37.4) guarantees that f ( a A x -6) C U. By a compactness argument we can find E > 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 < u j < u + 1 on a neighborhood of Kj. In addition 6, > 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 < I U : U S l U ) is a homeomorphism. Then
-
+
<(U) and we can find a neighborhood V of to in X and a neighborhood W of t o in E such that c ( V ) + K s W C <(U). Whence it follows that 6 ( x , t l 2 s > 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 < b for each fixed < with 151 < t. If a < c < b then
Hence
f t ( l , < )E
2 Po (Xl
for every
(A,
5) E D
and by (f)there
t
dx(fttDt?l > 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 < p . A f t e r m u l t i p h y i n g P be a c o n s t a n t w e m y P i s n o r m a l i z e d w i t h respect t o t h e v a r i a b l e s z 1 - a n , w . L e t L = K + A n f 0 ; 2 s / and c h o o s e M > 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
[11 P o l y n o m i a l s a n d m u l t i l i n e a r m a p p i n q s v e c t o r s p a c e s . S t u d i a Math. 3 9 ( 1 9 7 1 1 , 5 9 - 7 6 .
in
topoloqical
[ 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
MUJ I CA
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 .
-
[ 2 1 Holomorphic f u n c t i o n a l s a n d complex convexity Banach s p a c e s . P a c i f i c J . Math. 7 ( 1 9 5 7 1 , 8 1 1 - 8 3 1 .
in
H . CARTAN
[ l ] Idgaux e t modules d e f o n c t i o n s a n a l y t i q u e s d e v a r i a b l e s complexes. B u l l . SOC. Math. F r a n c e 7 8 ( 1 9 5 0 1 , 29 - 6 4 . [ 2 1 C a l c u l D i f f e r e n t i e l . Hermann, P a r i s , 1 9 6 7 . [3
1
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 .
.
D . CLAYTON
[ l ] A r e d u c t i o n of t h e c o n t i n u o u s homomorphism problem f o r F - a l g e b r a s . Rocky Mountain J . Math. 5 ( 1 9 7 5 1 , 337 - 3 4 4 . G . COEURE
[ l ] A n a l y t i c Functions and Manifolds i n I n f i n i t e Dimens i o n a l S p a c e s . North-Holland M a t h e m a t i c s S t u d i e s , v o l . l l . N o r t h H o l l a n d , Amsterdam, 1 9 7 4 . J . F. COLOMBEAU
[ l 1 D i f f e r e n t i a l C a l c u l u s a n d Holomorphy. North-Holland Mathematics S t u d i e s , vol. 6 4 . N o r t h - H o l l a n d , Amsterdam, 1 9 8 2 . I . CRAW
[ l ]A c o n d i t i o n e q u i v a l e n t t o m e c o n t i n u i t y o f charact e r s o n a F r G c h e t a l g e b r a . P r o c . London Math. SOC. ( 3 ) 22(197J),
452
- 464.
J . DIESTEL
[ 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
[ 1 1 F o u n d a t i o n s o f Modern A n a l y s i s . Academic P r e s s , York, 1960.
New
S. D I N E E N
[ 1 1 The C a r t a n - T h u l l e n t h e o r e m f o r Banach S c u o l a N o r m . Sup. P i s a ( 3 ) 24 ( 1 9 7 0 1 , 6 6 7 - 6 7 6 . 2 1 Bounding s u b s e t s o f a Banach s p a c e . (1971), 61 - 70.
spaces.
plath.
Ann.
Ann. 192
423
BIBLIOGRAPHY
[ 3 1 Runge domai.ns i n Banach s p a c e s . P r o c . R o y . I r i s h Acad. 7 1 (1971) , 85 89.
-
[ 4 1 Unbounded holomorphic f u n c t i o n s on a Banach s p a c e . J . London Math. SOC. ( 2 ) 4 ( 1 9 7 2 1 , 4 6 1 - 4 6 5 . [ 5 1 Complex A n a l y s i s i n L o c a l l y Convex S p a c e s . North-Holland Mathematics S t u d i e s , v o l . 5 7 . North-Holland, Amsterdam, 1 9 8 1 . S . D I N E E N and A .
HIRSCHOWITZ
[ l ] Sur l e th6orsme de Levi banachique C.R. ris 272 (19711, 1 2 4 5 P.
D I X O N and D.
- 1247.
Acad. S c i . P a -
FREMLIN
111 A remark c o n c e r n i n g m u l t i p l i c a t i v e f u n c t i o n a l s on LMC a l g e b r a s . J . London Math. SOC. ( 2 ) 5 ( 1 9 7 2 ) ,
231-232.
P . DOLBEAULT
I 1 1 Formes d i f f 6 r e n t i e l l e . s e t cohomologie s u r une v a r i e t 6 a n a l y t i q u e complexe I . Ann. of Math. ( 2 ) 6 4 ( 1 9 5 6 ) , 8 3 - 1 3 0 . L . EHRENPREIS [
Amer.
11
A new p r o o f and an e x t e n s i o n of H a r t o g s ' Math. SOC. 6 7 ( 1 9 6 1 1 , 5 0 7 - 5 0 9 .
theorem.Bul1.
P . ENFLO
[ 1 1 A counterexample t o t h e a p p r o x i m a t i o n p r o o e r t y Banach s p a c e s . A c t a Math. 1 3 0 ( 1 9 7 3 ) , 3 0 9 - 317. M.
ESTEVEZ and C .
in
HERVES
[ 1 1 A p p l i c a t i o n d e s mesures g a u s s i e n n e s complexes aux f o n c t i o n s p l u r i s o u s h a r m o n i q u e s e t a n a l y t i q u e s . B u l l . Sci.Math. (2) 105 (19811, 7 3 - 83. J.P.
FERRIER and N .
SIBONY
[ 11 Approximation pondGr6e s u r une s o u s - v a r i 6 t 6
rnent r 6 e l l e de 1 0 1 - 115.
g n . Ann. I n s t . F o u r i e r Grenoble
26,2
totale(19761,
T . FRANZONI and E . VESENTINI
1 1 Holomorphic Maps - ?d I n v a r i a n t D i s t a n c e s . North-Holland Mathematics S t u d i e s , 1 7 0 1 . 4 0 . North-Holland, Amsterdam, 1 9 8 0 . I . GELFAND, D .
R A I K O V and G .
SHILOV
[ 1 1 Commutative Normed Rings. C h e l s e a , N e w York, 1 9 6 4 . A.
GROTHENDIECK
[ 1 1 P r o d u i t s T e n s o r i e l s Topologiques e t Espaces Nucl6aires. Memoirs of t h e American Mathematical S o c i e t y , number 1 6 . " m e r -
i c a n Mathematical S o c i e t y , P r o v i d e n c e , Rhode I s l a n d , 1 9 5 5 .
MUJ I C A
424
L . GRUMAN
[ 1 1 The L e v i problem i n c e r t a i n i n f i n i t e d i m e n s i o n a l vect o r s p a c e s . I l l i n o i s J. Math. 15 ( 1 9 7 4 ) , 2 0 - 2 6 . L . CRUMAN and C . KISELMAN C.R. A.
1 1 L e problgme d e L e v i d a n s l e s e s p a c e s d e Banach Acad. S c i . P a r i s 2 7 4 ( 1 9 7 2 1 , 8 2 1 - 8 2 4 .
GUICHARDET
[ 1 1 Special Topics i n Topological Algebras. B r e a c h , New York, 1 9 6 8 . R.
2 base.
GUNNING and H.
Cordon
and
ROSS1
[ 1 1 A n a l y t i c F u n c t i o n s of S e v e r a l . Complex V a r i a b l e s . P r e n t i c e - H a l l , Englewood C l i f f s , N e w J e r s e y , 1 9 6 5 .
F. HARTOGS
[.&I Zur T h e o r i e d e r a n a l y t i s c h e n Funktionen Peherer unabhangiger Veranderlichen, insbesondere u b e r d i e D a r s t e l l u n g d e r s e l b e n d u r c h R e i n e n , w e l c h e n a c h P o t e n z e n einer Ver2nderlichen f o r t s c h r e i t e n . Math. Ann. 6 2 ( 1 9 0 6 ) , 1 - 8 8 . M.
HERVE
[ 1 1 A n a l y t i c a n d 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 i n F i n i t e and I n f i n i t e Dimensional S p a c e s . L e c t u r e N o t e s i n M a t h e m a t i c s , vol. 198. Springer, B e r l i n , 1971. Y . HERVIER [ 1 1 S u r l e problgme d e L e v i p o v r l e s espaces 6 t a l e s banac h i q u e s . C.R. A c a d . S c i . P a r i s 2 7 5 ( 1 9 7 2 1 , 8 2 1 - 8 2 4 . E.
HILLE and R . PHILLIPS
[ 1 1 F u n c t i o n a l A n a l y s i s and Semi-groups, r e v i s e d e d i t i o n . American M a t h e m a t i c a l S o c i e t y Colloquium P u b l i c a t i o n s , ~ r o 1 . 3 1 . American M a t h e m a t i c a l S o c i e t y , P r o v i d e n c e , Rhode I s l a n d , 1 9 5 7 . A.
HIRSCHOWITZ
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.
infinie.
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.
BIBLIOGRAPHY
425
Acta Math. 113 (1965), 89 - 152. [ 3 1 An Introduction to Complex Analysis in Several Variables, second edition. North-Holland Mathematical Library, vol. 7. North-Holland, Amsterdam, 1973. J. M. ISIDRO [ I 1 Characterization of the spectrum of some topological algebras of holomorphic functions. Advances in Holomorphy, pp. 407 -416. North-Holland Mathematics Studies, vol. 34. NorthHolland, Amsterdam, 1979.
B. JOSEFSON
1 1 1 A counterexample in the Levi problem. Proceedinqs on Infinite Dimensional Holomorphy, pp. 168 -177. Lecture Notes in Mathematics, ~701. 364. Springer, Berlin, 1974. [ 2 1 Weak sequential convergence in the dual of a Banach space does not imply norm convergence. Ark. Mat. 1.3 (1975), 79
- 89.
P. LELONG
[11 Les fonctions plurisousharmoniques. Ann. Sci. Ecole Norm Sup. (3) 62 (19451, 3 0 1 - 3 3 8 . [ 2 1 Fonctions plurisousharmoniques dans les espaces vectoriels topologiques. SGrninaire Pierre Leloncj 1967/68, pp. 167-190. Lecture Notes in Mathematics, vol. 71. Springer, Berlin, 1968. E. LEV1
1 1 1 Sulle ipersunerfici dello spazio a 4 dimensioni che possono essere frontlera del campo di esistenza di una funzioni analitica di due variabili complesse. Ann. Mat. Pura Appl. (3) 18 (1911), 69
- 79.
E. LIGOCKA [ l ] A local factorization of analytic functions and its applications. Studia Math. 47 (1973), 239 -252. [ 2 1 Levi forms, differential forms of type ( 0 , l ) andpseudoconvexity in Banach spaces. Ann. Polon. Math. 33 (1976) , 63
- 69.
J. LINDENSTMUSS and L. TZAFRIRI
1 1 1 Classical Banach Spaces I. Ergebnisse der Mathematil. and i:,rer Grenzgebiete, Band 92. Springer, Berlin, 1977. M. MATOS [11 A characterization of holomorphic mappings in Banach spaces with a Schauder basis. Atas do 190 Semingrio Brasileiro de Anzlise. Soci.edade Brasileira de Matemstica, Sao Paulo,1984.
426
C
MUJICA
-
MATYSZCZYK [ l ] Approximation of analytic operators by polynomials in complex Bo-spaces with bounded approximation property.Bd.1. Acad. Polon. Sci. 20 (19721, 833-836. [ 2 1 Approximation of analytic and continuous m a r - r \ i n r r s by polynomials in Frgchet spaces. Studia Math. 6 0 (1977),223 -238. E .MICHAEL [ 1 1 Locally vultiplicatively - Convex Topological algebras. Memoirs of the American Mathematical Society, number 11. American Mathematical Society, Providence, Rhode Island, 1952. J. MUJICA [11 Ideals of holomorphic functions on FrGchet spaces. Advances in Holomorphy, pp. 563 - 576. North-Holland Mathematics Studies, vol. 34. North-Holland, Amsterdam, 1979. I21 Complex homomorphisms of the algebras of holomorphic functions on Frschet spaces. Math. Ann. 241 (19791, 73 -82. [ 3 1 The Oka-Weil theorem in locally convex spaces withth? approximation property. S6minaire Paul Kr6e 1977/78, expose nQ 3. Institut Henri Poincarg, Parls, 1979. [ 4 1 Holomorphic apnroximation in Frschet spaces w i t h basis. J. London Math. SOC. (2) 29 (1984), 113 -126. [ 5 1 Holomorphic approximation in infinite dimensional Riemann domains. Studia Math. (to appear). L. NACHBIN [ 1 1 Lectures on the Theory of Distributions. Textos de Mg temdtica, NQ 15. Universidade do Recife, Recife, 1964. I21 Topology on Spaces of Holomorphic Mappings.Ergebnisse der Mathematik und ifrer Grenzgebiete, Band 47. SDringer,New York, 1969. [ 3 1 Holomorphir Wnctions, Domains of Holomorphy and Local Properties. North-Holland Mathematics Studies, vol. 1. NorthHolland, Amsterdam, 1970. e [ 4 1 Introdugao 5 Andlise Funcional: Espaqos de Banach Cdlculo Diferencial. S6rie de Matemstica, monografia nQ 17. Organization of American States, Washington, D.C., 1976. R. NARASIMHAN 1 1 1 Holomorphically complete complex spaces. Amer.J.Math. 82 (1961), 917 - 934. [ 2 1 Several Complex Variables. The University of Chicago Press, Chicaqo, 1971. S. M. NIKOL'SKII
[11 Approximation of Functions of Several Variables and Imbedding Theorems. Die Grundlehren der mathematischen
B I BL IOGRAPHY
427
Wissenschaften, Band 205. Springer, Berlin, 1975. A. NISSENZWEIG [l 1 w* seuuential convergence. Israel J. Math. 22 (19751, 266 - 272. F. NORGUET (11 Sur les domaines d'holomorphie des fonctions uniformes de plusieurs variables complexes. Bull. SOC. Math-F'rance 82 (1954), 137 - 159. P. NOVERRAZ [11 Fonctions plurisousharmoniques et analytiques dans les espaces vectoriels topologigues complexes. Ann. Inst. FOLIrier Grenoble 19,2 (1969), 419 - 493. polynomiale [ 2 1 Sur la pseudo-convexit6 et la convexit6 en dimension infinie. Ann. Inst. Fourier Grenoble 23,l (19731, 113 - 134. [ 3 1 Pseudo-Convexit6, Convexit6 Polynomiale et Domaines d'Holomorphie en Dimension Infinie. North-Holland Mathematics Studies, vol. 3. North-Holland, Amsterdam, 1973. 14 1 Approximation of holomorghic or plurisubharmonic functions in certain Banach spaces. Proceedings on Infinite Dimensional Holomorphy, pp. 178 -185. Lecture Notes in Mathematics, vol. 364. Springer, Berlin, 1974. K. OKA [ 11 Domaines convexes par rapport aux fonctions rationelles. J. Sci. Hiroshima Univ. 6 i19361, 245 -255. [ 2 1 Domaines d ' holomorphie J. Sci Hiroshima Univ. 7 (1937), 115 - 130. [ 3 1 Domaines meudoconvexos. Tohoku Math. J. 49 (19421 , 15 - 52. [ 4 1 Domaines finis sans point critique int6rieur. Japan. J. Math. 27 (1953), 97 - 155.
.
.
R. OVSEPIAN and A. PELCZYNSKI [ I ] The existence in every separable Banach space of a fundamental total and '-ounded biorthogonal sequence and related constructions of uniformly bound-ed orthonormal systems in L 2 . Studia Math. 54 (19751, 149 - 159.
A. PELCZYN!.:KI [ 1 1 Any separable Banach space with the bounded approxi-. mation property is a complemented subspace of a Banach space with a basis. Studia Math. 40 (19711, 239 -243. B. RIESZ and B. SZ.-NAGY [ 1 1 Leyons d' Analyse Fonctionnelle,
sixisme
6dition.
MUJ I CA
428
Gauthier-Villars, Paris, 1 9 7 2 . H. ROSS1 [ 1 1 On envelops of holomorphy. Comm. 16 (1963)
,
9
- 17.
Fure
Anpl.
Math.
J. SCHAUDER [ 1 1 Zur Theorie stetiqer Abbildungen in Funktionalraumen. Math. Z. 2 6 ( 1 9 2 7 1 , 47 - 6 5 . [ 2 1 Eine Eigenschaft der Haarschen Orthogonalsystems. Math. Z . 2 8 ( 1 9 2 8 ) , 317 - 3 2 0 . M. SCHOTTENLOHER [ 11 Uber analytische Fortsetzung in Banachraumen. Math. Ann. 1 9 9 ( 1 9 7 2 ) , 313 - 3 3 6 . [ 2 1 Bounding sets in Banach spaces and regular classes of analytic functions. Functional Analysis and Applications, pp. 1 0 9 - 1 2 2 . Lecture Notes in Mathematics, vol. 3 8 4 . Springer , Berlin, 1 9 7 4 . [ 3 1 Das Leviproblem in unendlichdimensionalen..Raumen mit Schauderzerlegung. Habilitationsschrift, Universitat Munchen, 1974. 4 1 The Levi problem for domains spread over locally convex spaces with a finite dimensional Schauder decomposition. Ann. Inst. Fourier Grenoble 26,4 ( 1 9 7 6 1 , 2 0 7 - 2 3 7 . 1 5 1 Polynomial approximation on compact sets. Infinite Mmensional Holomorphy and Applications, pp. 3 7 9 - 3 9 1 . NorthHolland Mathematics Studies, vol. 1 2 . North-Holland, Amsterdam, 1977. 1 6 1 Michael problem and algebras of holomorDhic functions. Arch. Math. (Basel) 37 ( 1 9 8 1 ) , 2 4 1 - 2 4 7 . [ 7 1 Spectrum and envelope of holomorphy for infinite dimensional Riemann domains. Math. Ann. 263 ( 1 9 8 3 ) , 213 - 2 1 9 .
L. SCHWARTZ [11 Thgorie des Distributions, nouvelle Gdition. Hermann, Paris, 1 9 6 6 . G. SHILOV [11 On the decomposition of a commutative norn?ed ring into a direct sum of ideals. Mat. Sb. 3 2 ( 1 9 5 3 ) , 3 5 3 - 3 6 4 (Russian). Amer. Math. SOC. Transl. ( 2 ) 1 ( 1 9 5 5 ) , 37 - 4 8 (English translation). A. TAYLOR (11 On the properties of analytic €unctions in spaces. Math. Ann. 115 ( 1 9 3 8 ) , 4 6 6 - 4 8 4 .
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
[ l ] C h a r a c t e r i z a t i o n of a n a l y t i c f u n c t i o n s s p a c e s . Ann. of Math. ( 2 ) 4 6 ( 1 9 4 5 1 , 5 8 5 - 5 9 3 . 1 2 1 Gateaux d i f f e r e n t i a b i l i t y and e s s e n t i a l Duke Math. J . 1 2 ( 1 9 4 5 ) , 5 7 9 - 5 8 3 .
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
