de Gruyter Expositions in Mathematics 34

Editors

0. H. Kegel. Albert-Ludwigs-Universitat, Freiburg V. P. Maslov, Academy of Sciences, Moscow W. D. Neumann, Columbia University, New York R. 0. Wells, Jr., Rice University, Houston

de Gruyter Expositions in Mathematics

II

The Analytical and Topological Theory of Semigroups. K H Hofmann. J. D. Lawson, (Eds.) JS Combinatorial Homotopy and 4-Dimensional Complexes, H J. Baues The Stcfan Problem, A. M. Meirmanov Finite Soluble Groups. K Doerk. T 0. Hawke.c The Riemann Zeta-Function, A. A. Karaisuba, S. M Voronin Contact Geometry and Linear Differential Equations, V E. Nnsaikinskii, FE. Shataloy, B. Yu. Sternin Infinite Dimensional Lie Superalgebras. Yu. A. Bahiurin. A. A. Mikhalev, V M Peirogradsky, M. F Zaicev Nilpotcnl Groups and their Automorphisms, E I Khukhro Invariant Distances and Metncs in Complex Analysis, M. Jarnicki, P Pflug The Link Invanants of the Chern-Simons Field Theory, E. Guadagnini Global Atline Differential Geometry of Hypersurfaces. A.-M Li, U. Simon, 6. Zhao

12

Moduli Spaces of Abelian Surfaces: Compactilication, Degenerations, and Theta Functions,

I

2 3

4 S

6 7

8 9 10

K ilulek. C. Kahn. S. H. Weingraub 13

Elliptic Problems in Domains with Piecewise Smooth Boundaries, S. A. Nazarov,

II A Plamenei'skv 14

Subgroup Lattices of Groups, K Schmidt

15

16

Orthogonal Decompositions and Integral Lattices. A. I. Ko.ctrikuz. P H Tiep The Adjunction Theory of Complex Projective Varieties, M. C. &hrametzi. A. J. Sommese

17

The Restricted 3-Body Problem: Plane Periodic Orbits, A D. Bruno

18

Unitary Representation Theory of Exponential Lie Groups, H. Leptin, J. Ludwig Blow-up in Quasihnear Parabolic Equations. A. A. Samarskii, VA GaIa.k:ionov.

19

S P Kurdyumo;; A. P Mikhailov 20

Semigroups in Algebra. Geometry and Analysis, K II. Hofmann, J. 0. Lawson, E. B.

(Eds.) 21

2.2

23 24

Compact Projective Planes, H. Salzmann. 0. Beiten, T Grundhofer, H. Haul, K Lowe,,,

M. Siroppel An Introduction to Lorenti Surfaces, T Weinstein Lectures in Real Geometry. F Broglia (Ed.) Evolution Equations and Lagrangian Coordinates, A. M. Meirinanoi: V

Pukhnachov.

S I Slunarei' 25 26 27

28

29 30 31

32 33

Character Theory of Finite Groups, B. Hupper: Positivity in Lie Theory: Open Problems. I. Hi/get:, J. 0. Lawson. K-H. Neeb, E. B. Vmnberg (Eds.) Algebra in the Stone-tech Compactification, N Hind,nan, D. Strauss Holomorphy and Convexity in Lie Theory, K-H. Neeb Monoids. Acts and Categories. M. Kilp. U Knauer. A. V Mikhalev Relative Homological Algebra, Edgar E. Enoc/is, 03'ertoun M. G. Jenda Nonlinear Wave Equations Perturbed by Viscous Terms, Vikior P Mas!av. Petr P Mosolov Conformal Geometry of Discrete Groups and Manifolds, Boris N. Apanasov Compositions of Quadratic Forms. Daniel B. Shapiro

Extension of Holomorphic Functions by

Marek Jarnicki Peter Pflug

w DE

C Walter de Gruyter

Berlin New York 2000

Authors

Marek Jarnicki Institute of Mathematics Jagiellonian University u. Reymonta 4 30-059 Krakow

Peter Pflug Department of Mathematics Carl von Ossietzky University Oldenburg 26111 Oldenburg

Poland

Germany

pflug(4jmathematik.uni-oldenburg.de Mathematics Subject 2000. 32-02; 32Axx, 32Dxx, 32Exx, 32Txx, 32Uxx. 32Wxx Key words:

Riemann domains, Holomorphic extension. Holomorph convexity, Riemann-Stein domain, Plurisubharmomc function, Pseudoconvexity, Levi problem, Envelope of holomorphy 0 Printed on acid-free paper which falls within the guidelines of the ANSI to ensure permanenee and durability. Lthrary of Congress — Cataloging-in-Publication Data

Jarnicki, Marek. Extension of holomorphic functions I by Marek Jarnicki, Peter Pliug.

p cm. — (Dc Gruyler expositions in mathematics; 34) Includes bibliographical references and index. ISBN 3-11-015363-7 (alk. paper) 1. Holomorphic functions. I. Pflug, Peter, 1943— 11. Title. III. Series. QA331 .J37 2000 515'.98—dc2l

00-060145

Die Deutsche Bibliothek — Cazaloging-in-Publication Data

Jarmcki, Marek:

Extension of holomorphic functions / by Marek Jarnicki

;

Peter

Pflug. — Berlin ; New York : de Gruyter, 2000

(Dc (lruyter expositions in mathematics : 34) ISBN 3-11-015363-7

© Copyright 2000 by Walter de Gruyter GmbH & Co. KG, 10785 Berlin, Germany. All rights reserved, including those of translation into foreign languages. No part of this book may be reproduced or transmitted in any form or by any means, electronic or mechanical, including photocopy, recording, or any information storage or retrieval system, without permission in writing from the publisher. Typesetting using the author's TEX files: I. Zimmermann. Freiburg. Printing: WB-Druck GmbH & Co.. Rieden/AJIgbu. Binding LOderitz & Bauer-C,mbH, Berlin. Cover design: Thomas Bonnie. Hamburg.

To Mariola and Rosel

Preface

Starting from the discussion of the holomorphic logarithm in the plane, it is necessary to introduce Riemann surfaces in order to be able to deal with the maximal domain of existence of that function, but without being bothered about its possible multivaluedness. On the other hand, any domain in the complex plane is the existence domain of at least one holomorphic function. i.e. there exists a holomorphic function that is not the restriction of another holomorphic function having a strictly larger domain of definition.

In the n-dimensional situation (n 2) it was already observed by Hartogs that there are pairs of domains G c G', G G', such that any holomorphic function on G extends holomorphicaily to G'. Even more, simultaneously one has to handle the problem that the extended functions may be multivalued. This phenomenon has led to the notion of Riemann domains over C's. Exactly this category of objects and the theory of holomorphic functions on them is the subject of our book. We try to give a systematic representation of domains of holomorphy and envelopes of holomorphy in that category. The authors feel that a lot of results they are presenting have never been published in book form. We are not touching the theory of holomorphic functions on complex spaces or even

on complex manifolds. Instead, we continue investigating domains of holomorphy for special classes of holomorphic functions on special types of domains. Our interest in this area of complex analysis started directly after our studies when both of us were working on the continuation of holomorphic functions. Although we had changed our fields of interest, we were attracted by such questions all the time, and we were following the development in that direction. During the years we got the impression that there is a need of a source where the main results are collected. We hope this book can serve as such a source. The choice of topics obviously reflects our personal preferences. For example, we will solve the Levi problem via the a-problem and functions of restricted growth. Our idea is to address this book to everybody who likes to extend her/his knowledge beyond the standard course in several complex variables in C". We thed to make the book as complete as possible and to keep the results used without proving them as limited as possible. Nevertheless, for some parts we will have to use certain facts (for example consequences of Theorems A and B) without giving proofs. The same will happen also with some facts on plurisubharmonic functions (related to the Monge—Ampère operator), although we present an extended section on plurisubharmonic functions. Textbooks that support the reader can be found in the first part of the bibliography.

viii

Preface

We should point out that the bibliography is far from being complete. We included only the papers that we had studied during the preparation of the book or before. So we have to apologize if readers are interested in historical developments of presented results. The theory of extension of holomorphic functions contains a lot of questions which

are still waiting for being solved. We have put many of them into the text (marking them by The reader is encouraged to work on some of them. During the process of proofreading we got the impression that this became a neverending story and there had to be a tune for us to stop. We would be pleased if the reader would inform us of any errors he/she may have detected while studying the text.

It is our deep pleasure to thank our teachers, Professors H. Grauert and J. Siciak, who taught us the beauty of complex analysis. We would also like to thank Dr. habil. W. Zwonek who helped us in corrections of the text. We thank the following institutions: Committee of Scientific Research (KBN), Warsaw (PB 2 PO3A 06008, PB 2 PO3A 017 14), Volkswagen Stiftung (Az. 1/71 062, RiP-program at Oberwolfach), and Niedersächsisches Ministerium für Wissenschaft und Kunst (Az. 15.3—50 113(55) PL). Without their financial support this work would have never been possible. We would also like to thank our universities for support during the preparation of the book. Finally we thank Walter de Gniyter Publishers, especially Dr. M. Karbe, for having encouraged us to write this book.

KrakOw — Oldenburg, May 2000

Marek Jarnicki Peter Pflug

Contents

Vii

Chapter 1 Riemann domains 1.1

1.2 1.3 1.4 1.5

1.6 1.7 1.8

Riemann domains over Holomorphic functions Examples of Riemann regions Holomorphic extension of Riemann domains The boundary of a Riemann domain Union. intersection, and direct limit of Riemann domains

Domainsofexistence

Maximal holomorphic extensions Liftings of holomorphic mappings I 1.10 Holomorphic convexity 1.11 Riemann surfaces 1.9

1

3 18 21

25 29 39 50 54

62 75 86

Chapter 2 Pseudoconvexlty 2.1

2.2 2.3 2.4

25 2.6 2.7 2.8 2.9

Plurisuhharmonic functions Pseudoconvexity The Kiselman minimum principle a-operator Solution of the Levi Problem Regular solutions Approximation The Remmert embedding theorem The Docguier—Grauert criteria

2.10 The division theorem Spectrum 2.12 Liftings of holomorphic mappings II 2.11

96 96 129 153 159 177 184 190 195 201 208 219

224

Chapter 3

Envelopes of holomorphy for special domains

235

Univalent envelopes of holomorphy k-tubular domains

235 258

3.1

3.2

Contents 3.3 3.4 3.5

3.6

Matrix Reinhardt domains The envelope of holomorphy of X \ M Separately holomorphic functions Extension of meromorphic functions

284 302

324 334

Clwpter 4

Existence domains of special families of holomorphic functIons 4.1 4.2

4.3 4.4 4.5

Special domains The Ohsawa—Takegoshi extension theorem The Skoda division theorem The Catlin—Hakim--Sibony theorem Structure of envelopes of holomorphy

341 341

388 410 422 441

List of symbols

461

Bibliography

469

Index

483

Chapter 1

Riemann domains

Roughly speaking, a Riemann domain over C" is a 'domain spread over C", i.e. a connected complex manifold X having a global projection p: X —* C" such that p is locally homeomorphic. The class of Riemann domains over C" extends the class of subdomains of C". Riemann domains appear in a very natural way while discussing problems related to holomorphic continuation. Consider the following classical example. Let G := C \ R_ and let Log denote the principal branch of the logarithm. Then the function Log extends holomorphically to a domain X which is no longer a plane domain but a so-called Riemann surface of the function Log. The same phenomenon appears if we consider on G a branch of the k-th root. This means that the maximal domain of existence of a holomorphic function defined in a plane domain may be non-univalent — it may be a Riemann domain over C. More generally, given a domain G c C", we can consider the maximal domain of existence of a family c 0(G), where 0(G) denotes the space of all functions holomorphic on G. The most interesting case is the case when = 0(G). From the point of view of the theory of holomorphic functions the following three fundamental questions are the most important: (1) Does the maximal domain of existence always exist in the category of Riernann domains over C"? (2) What is a characterization of those domains which cannot be holomorphically extended to any larger Riemann domain? (3) When is the maximal domain of existence univalent, i.e. can it be realized as a domain in C"?

It is well known that an arbitrary domain G C C is the maximal domain of existence of the whole space 0(G), i.e. each boundary point of G is a singular point for a function holomorphic in G. This is no longer true in C" with n 2. For example, consider the following domain (cf. [Sha 19761): Let E denote the unit disc. Put P := E x (2E) C C2 and let

G := P \ (Q' I)

U Q2 U s),

For the notation used in this book see 'List of symbols'.

I

Riemann domains

where

Qi:={(x+iO,w)EP:x>0,IwI<1}, Q2 :={(0+iy,w)E P: y >0, IwI? I), S:={(x+iy,w)E P:x FwI=l}. Define

Go:=P\((0+iy,w)EP:y>_0} and notice that G0 \ G

0. Let f E

(9(G). Then,

for any

(z. w) E Go,

the Cauchy

integral

-

f(z, w) is

1

1

:=

2,r:

4 —

w

dC,

wi
independent of r (with wi < r, I < r < 2), and consequently, f E (9(G0). using the Cauchy integral formula and the identity principle, we see that on the domain =

Moreover,

f

f

{(x + iy, w) E P: x <0 or y <0) U {(z, w) E P \ Q2.

1

< wi

<2).

Hence fisa holomorphic extension of f to Go. Thus the maximal domain of existence must contain G0 and G0 G. Letfo: G —+ Cbegivenbytheformula

of 0(G)

fo(z, w)

f—(z)

ifiwl

1

f+(z)

where f'_(z)

e!

Log(—z)

f+(z)

e!

:= (?i+ii?, I Go (9(G)andfo(z, U') = f+(z). Moreover, Consequently, the << 1) we have: fo(a÷) = —fo(a_) and fo(a÷) = domain of existence of (9(G) cannot be univalent. Then fo

(0 <

q

The full description of the envelope of holomorphy of the above domain G will be given in § 4.1 (see also Remark 3.1.12(d)).

Let us mention that question (1) has a positive answer (for any family 8); cf. Thullen Theorem 1.8.4.

Problem (2) is completely solved only in the case = (9(G) (Chapter 2). In the general case, some characterizations are known for special domains and special families of functions (Chapter 4). Problem (3), even in the case 0(G), is far from being solved. It will be discussed for special domains G in Chapter 3.

I. I

Riemann domains over C"

3

1.1 Riemann domains over C" Definition 1.1.1. A pair (X, p) is called a Riemann region (spread) over C" if: • X is a topological Hausdorif space, • p: X —* C" is locally homeomorphic2'. If X is connected, then we say that(X, p) is a Riemann domain overC". If(X, p) is a Riemann region over C", then for any connected component C of X the pair (C, Plc) is a Riemann domain over C". If we replace in the above definitions the space C" by a connected n-dimensional complex manifold M then we get the notions of a Riemann region over M and Riemann domain over M, respectively. The mapping p is called the pmjection. For z E p(X) the set is called the sralkoverz. AsubsetA C Xissaidtobeunivalenl(orschlicht)ifplA: A —* p(A) is homeomorphic.

Observe that if A c X is open or compact, then A is univalent 1ff PIA is injective. Notice that, in general, a set A C X such that P1.4 is injective need not be univalent (Exercise). In the sequel we will sometimes identify X with (X. p) if it is clear from the context what the projection p is.

Remark 1.1.2. (a) If G C C" is a domain, then (G, idG) is a Riemann domain over C". This will be the standard identification of domains in C" with Riemann domains. Of course, any locally homeomorphic mapping p: G —+ C" makes from G a Riemann domain over C". For example. (C, exp) is a Riemann domain over C. (1,) If (X. p) is a Riemann domain over C", then p is an open mapping. In particular,

the set p(X) is a domain in C". For any z E p(X) the stalk is discrete. (c) If (X, p) is a Riemann domain over C". then the family (U. where U 2)

That is, each point a E X has an open neighborhood U such that p(U) is open in C" and

plu: U —* p(U) is homeomorphic. Recall (cf. [Nar 19681) that a topological Hausdorif space M is called an n-dimensional (rasp. complex manifold) if there exists a family (called an atlas) such that:

isanopencoveringofM.

•

• pj is a homeomorphic mapping of onto an open subset •foranyi.j e I, if U o,thenthemappingcoj

(U,) of R" (resp. C"), is of class Ct (rcsp. holomorphic)

fl U,,); here k E U (oo}U(w}, where w := real analytic, It is clear that any complex n-dimensional manifold is a 2n-dimensional real analytic manifold. Moreover, any Ce-manifold is an &-manifold f w (with oo w). Suppose that N is another tn-dimensional (resp. complex manifold) with an atlas (V1. We say that a continuous mapping F: X —p Y is of class & (resp. holomorphic) if • for any I e 1. j e J, the mapping tfr,, o F is of class Ct (reap. holomorphic) fl F F (9(M. N)). Put

:=

C), (9(M) := (9(M. C).

Notice that any connected manifold is arcwise connected.

4

1

Riemann domains

runs over all univalent open subsets of X, introduces an atlas of an n-dimensional complex manifold on X. Evidently, not all connected complex n-dimensional manifolds are Riemann domains; e.g. a compact complex manifold cannot be a Riemann domain. In the category of non-compact complex manifolds the situation is as follows: if n I. then any non-compact connected 1-dimensional complex manifold is a Riemann domain over C (with suitable projection p); cf. Theorem 1.11.1, — if n > 2, then there exist very regular non-compact connected complex mainfolds which are not Riemann domains over C"; cf. Remark 1.11.11. (d) If (X, p) is a Riemann domain over C", then (Y, is a Riemann domain over C" for any domain Y C X. (e)

(X x

If (X, p), (Y, q) are Riemann domains over C" and Ctm, respectively, then a Riemann domain over

Y. p x q)4) is

(f) Let (X. p) be a Riemann domain over C" and let Y be an open univalent subset

such that p(Y) = p(X). Then Y = X. Indeed, suppose that there exists a point a E Let U be an open univalent neighborhoodofa. := (pIyY1 o(plu): U Y C X. Thenço(x)=x for fl U Y. In particular, by continuity, = a; contradiction. xE Definition 1.1.3. Let (X,

p) be

a Riemann domain over C", let T

be

a topological

space, and let y: T —* C" be a continuous mapping. Any continuous mapping 32:

T

—+ Xsuchthatpo32=yiscalledaliftingofyio(X,p).

If T T.

Let

any lifting s of idr to (X, p) is called a section of(X, p) over p)) denote the set of all sections of (X, p) over T.

C p(X), then

r(T. (X,

Remark 1.1.4. (a) A set A c X is univalent if and only if there exists a section S E F(p(A), (X. p)) such that s(p(A)) = A. (b)

By Remark 1.1.2(f) there are no global sections of (X, p) over p(X) unless X

is univalent.

Proposition 1.1.5 (Identity principle for liftings). Let T be a connected topological and let 32j: T —p X, j = 1. 2, be two of a continuous mapping

space, t0 E T.

y: T —÷ p(X) such that 32i(to) =

322(to). Then

32j

= HO)); T0 is closed and non-empty. It remains E T: 321(t) toshowthatToisopen. Takeau eT0. Leta := 32i(u) = ,22(u)andletUbean Proof

Let To :=

open univalent neighborhood of a. By continuity there exists an open neighborhood V of u such that 321(V) c U,] = 1,2. Then = (pIvY1 o ylv, j = 1.2. Thus

0

VCT0.

Corollary 1.1.6. (a) If a

set T C p(X) is connected and

such that s1 (zQ) = s2(zo)for a (p x q)(x. )')

point zO

(p(x). q(y)).

T, then s1

s2

f'(T,

(X, p)) are

1.1 Riemann domains over

C X be univalent sets such that A1 fl A2 0, • the set p(A fl is connected. Then the set U A2 is univalent.

•

Notice that the conclusion of(b) does not hold if

fl A2

= 0 or p(Ai) fl p(A2)

is disconnected.

Pmof (a) follows directly from the identity principle.

i=

(b) Fix an xo E A1 CiA2. Let T := fl p(A2), s1 := 1,2. Observe that SI and s2 are sections over T such that sl(p(xo)) = s2(p(xo)). Consequently, by (a), s2, which finishes the proof. C)

Let (X, p) be a Riemann domain over C't. Since p is locally homeomorphic, we can introduce on X the notion of a ball. More precisely, let q be

:

—p

an arbitrary C-norm. Put

Bq(zo, r) :=

C": q(z — Zo) < r},

{z E

Zij

E C". r E (0, +oo].

For a E X and r E (0. +oo] let

Bq(a.r) = Bx.q(a.r) denote

an open univalent neighborhood of a (if it exists) such that p(Bq(a. r)) =

Bq(p(a), r). Observe that: • Bq(a, r) exists for small r > 0: • if Bq(a, r) exists, then it is uniquely determined; • if Bq(a, r) exists, then for any a' Bq(a, r) and 0 < the ball Bq(a', r') exists and Bq(a

/

/

r)=

—I

r'
q(p(a') — p(a)).

'

(Bq(p(a , ). r ));

• if Bq(a, r') exists for any 0
=O'zr
In particular, if we define the distance to the boundary (w.r.t. the norm q)

dx.q: X —÷ (0, +ocj, dx,q(a) :=sup{r E (0,+ooj: Bq(a,r)exists}, Notice that Rq(zo.

:= C".

a

X,

6

Riemann domains

1

then for any a E X the maximal ball (w.r.t. the norm q)

Bq(a) = Bx.q(a) := Bq(a.dx,q(a)) is

well defined. Observe that if Y is a subdomain of X, then

Ofl Y.

Put

Pa.q := The definition of dx,q(a) extends in the standard way to subsets of X, namely

dx.q(A) := inf{dx.q(x): x E A),

A C X.

If dx.q(A) > 0, then for 0 < r
Finally, put

:=

(a

X: dx,q(a) = +oo).

Definition 1.1.7. Let us fix the following two important conventions related to the above notations. is the Euclidean norm, then we skip (if possible) the index q and we If q = substitute the letter B by B and d by p. Consequently, we will write IB(a, r), (a, r), etc. B(a), Bx(a), px(a), lfq(z) = Izi = max{IziI q and we substitute B by P. e.g. we will write P(a. r), Px(a, r), P(a), Px(a), dx(a), A(T), Pa, etc.

Remark 1.1.8. (a) If (X. p) = (G, IdG), where G is a domain in coincides with the standard distance function to Observe that in general we have: d(x.p),q

\

then dG,q

G in the sense of the norm q.

(d(p(x;,id).q) 0 p.

Moreover, if X is univalent, then d(x.p),q = (d(p(x).id),q) 0 p.

(b) By Remark 1.1.2(f), if X onto = X). In other words, except the trivial case where p maps C" (in particular. bijectively X onto C", the set is empty. (c) If = 0, then dx,q(x) — dx,q(a)I

q(p(x) — p(a)),

a E X, X E Bq(a).

1.1 Riemann domains over C"

7

In particular, the function dx.q IS COfltJBUOUS. Indeed, by Corollary 1.1.6(b), for any x E Bq(a) the set U := Bq(a) U Bq(X) is a univalent subdomain of X. Consequently, we have:

Idx,q(x) — dx,q(a)I =

q(p(x)

= Id(p(u).id).q(P(X)) — d(p(u),id),q(p(a))l

j

(d) If (Xi, Pj) is a Riemann domain over C's', q3: C" = 1,2, and q(zi, Z2) q2(z2)}, then d(X1

xX2.p, xp2),q(.rI, x2) = min{d(x,,p1),q,

—

p(a)).

is a C-norm,

(xi), d(x2,p2),q2(X2)},

(xi,x2)EX1 xX2. (e) Let

K C X be compact. Then:

•dx,q(K) >0. is compact for any 0

• the set


c

Since K is Xv E Bq(av, r), av E K, v E K and p(xL.) zo E C". Moreover, we may assume that E Bq(ao, s), v > 1, where e : dxq(K) — r. Then Bq(av, r) C Moreover, = =: Bq(aO) and, consequently,

Indeed, let

1

compact, we may assume that

q(p(xo)

—

—+

p(ao))
? dx.q(K) — r (cf. Corollary 1.1.10).

•

•K=

flO
is univalent for small r > 0. (f) If K is compact and univalent, then Indeed, suppose that xv E Aq(av, 1/v), Yv E Bq(bv, l/v), E K, xL, —+ v>> 1. We may assume that —+ b0 E K. Then E K, p(xv) = yp E Bq(ao), v>> 1; contradiction. p(ao) = p(bo) and therefore ao = b(j. Thus

= 0, then for any a E X there exists a vector w E C" with PropositIon 1.1.9. If q(w) = dx,q(a) such that urn [O.I)a:-+1

dx,q((pa,qY'(p(a)+tw)) =0.

dx,q(Bq(a)) = 0. Pmof Fix an a E X. To simplify notation put d Uo 6)

dx.q.

:= Bq(p(a),d(a)) = p(Uo),

11o := Bq(a), 5

=

One can prove (cf. Remark I - I .14) that X is mctrizabk, and therefore, the compactness of may be checked via sequences.

8

Riemann domains

1

Suppose that for any vector w with q(w) = d(a) there exist e(w) > 0 and a sequence [0, I) tk(W) / I such that

d(s(p(a) + tk(w)w))

e(w),

k

? I.

Now, forany w with q(w) = d(a), fix ak0 = ko(w) suchthat (I and let E Uo. x(w) z(w) := p(a) + s(z(w)) E

(note that d(x(w))

<e(w) Uo

r(w)). In particular, duo C q(w)=d(a)

Since dU0 is compact, there exist vectors

WN

with

=

=

d(a) such that

au0 Put

U3 :=

Bq (Z(W3), e(wj)),

J

= I,...,

N,

and define U

It is clear that U is an open neighborhood of a and p(U) D (Jo. We will prove that U is univalent. Then, the ball Bq(a, r) would exist for some

r > d(a). which contradicts the definition of d(a). First, observe that, by Corollary 1.1.6(b), for any j

E

(I

N} the set

Uo,3 := Uo U Uj is univalent. Moreover, for any j, k E (I

N}, if

fl Uk

0, then the set

Uj,k := U U Uk is also univalenL Thus, it remains to exclude the situation where we have two points E

E Uk\LJOWIthp(YJ) = p(yk) and UJflUk = 0. 0, 0, and U1flUk 0, there existsapointzo UOflUJflUk. Thens(zo)= [1 (zo) E Ui and, similarly, S (zo) E Uk; contradiction. ) —' (zo) = (p10

Uf\UO,yk

UOflUk

1.1 Riemann domains over C X be compact. Then

= dx,q(K)

—

0 < r
r,

(cf Remark 1.1.8(e)).

= 0. Fix K and r. We already know that Since the function dx,q is continuous, there exists an a E K such that dx,q(a) = dx,q(K) =: R. By Proposition 1.1.9 there dx.q(x(t)) = 0, where exists a w E C" with q(w) = R such that x(t) := (pa,q1'(p(a) + tw). Let 0 E (0, 1) be such that r = OR. Put b := x(O). < 1. Consequently, and x(t) E Bx.q(b, R — r) for 0 < Then b E U = R — r and hence = dx.q(K) — r. dx,q(b) Pmof We may assume that > dx.q(K) — r.

z

Remark 1.1.11. Let y: [0,1] —+ p(X) be a curve. Fix an a Jo denote the set of all t E [0, 1] such that there exists a curve = a and po = yI[o.tI. Then we have:

•

and let [0, tI —÷ X with

0E

• If t E • =

then [0, t] C Jo and therefore Jo is an interval. for 0 < t' <1", t" 1o (by the identity principle for liftings) and, = a. —+ X of yIj0 with therefore, we have a lifting (for is open in [0, 1]. Indeed, if U is an open univalent neighborhood of • sometE lofl[0,l))and0 < ö < I —iissuchthaty([i,i+8]) C p(U),thenwe can define: I E [0,1] y-(t), ( Yr+ö

(y(t)),

1

t e [1,1 +

Consequently, either = [0, 11 (and then is a lifting of y) or J0 = [0, 1*) for some I. 0< then = 0 (in particular, P(Io) is not relatively com• If = [0,

pact). Indeed, suppose that dx(P(Io)) = r > y(I) IF(y(i), r) fort [1, 1*] and define f 1

Then

is

well defined and it is a lifting of

0.

Take a point 0 <

I

<

1*

such that

I E [0,1] I E [1,

contradiction.

Analogously as we defined the distance to the boundary (w.r.t. a norm), we can introduce the distance to the boundary in a given direction.

€C",andr

(note that

r) := {z)).

1 Riemann domains

10

r) denote a univalent set containing a (if it exists)

For a point a E X let

= (a, r) exists for small r > 0. Moreover, if (a, r) exists, Observe that the disc then it is uniquely determined (note that Ao(a, r) = (a}). X —+ (0, +oo], Define the distance to the boundary in direction

:= sup{r > 0:

a E X.

r) exists},

r) and Aq(a. a) exist, then the set is univalent (use Corollary 1.1.6(b)).

Remark 1.1.12. (a) If

the disc

IIp(a') —

a)

r) and 0 < r' < r —

0), then for any a' E

(b) If A4(a. r) exists

r) Li

At(a', r') exists and

r') = (PIAt(a,r)Y'

r')).

r') exists for any 0 < r' < r, then 14(a, r) exists and

If

U

O
0 < ro < R. Put K :=

(c) Assume that

let

U

rO) and

is univalent (cf. Remark 1.1.8(f)). Let > 0 be so small that the set A := such that x 1R>o be an open neighborhood of the point (ao, c Xx Uthedisc EU. Then forany (p(a),r) C r) exists and

r)).

r) = In particular, the function

Xx

(x.

—+

E (0. +oo]

is lower semicontinuous.

(d) Let q: C" —* with

any

=

1.

be

a C-norm. Then

r) exists if At(a. r) exists for

Moreover,

Bq(a,r)

U

EEC": q(f)=I

Hence

dx,q(x) =

e C",

= l},

x e X.

Notice that Proposition 1.1.9 says that for any a E X there exists a

= I such thatdx,q(a) =6x,t(a).

E

C" with

1.1 Riemann domains over C'1

Lemma 1.1.13. where

(,

= I, and let

E C",

:=

LI

z E C" (e > 0). 8• Then ) denotes the standard Hennitian scalar product in C" (z.

— (z,

/

when s \

0.

Proof First observe that if L: C" —+ C" is a C-linear isomorphism, then d(x.p).q = d(x.Lop).qoL_t.

Thus we may assume that

Obviously,

'5(X.p).t =

= = (0,..., 0, 1), then o L1 =

If, moreover, L is unitary and

= dx.q,, when 0 < e' <e".

8x.e, and

Now fix an a E X, take 0 < r <

r). Since K is

and let K :=

compact and univalent, there exists 0 <

(K) such that the set A univalent. Itisclearthatthereexistsane > OsuchthatU := Hence and, consequently, dx q,. ,(a) > r. =

is also

C p(A). D

For any curve y: [0, 1] —+ C" let L(y) denote the length of y (w.r.t. the Eucidean norm), i.e.

Now we introduce a structure of a metric space on a given Riemann domain (X. p). First, since X is arcwise connected, we conclude that for any x', x" E X there exists a curve [0, I] —÷ X with = x', = x" such that L(p o < +00. Indeed, if (0, 1] —÷ X is an arbitrary curve with = x', = x", then we find points 1]) such that N, where xN_I E j I := x', xN := x". The curve defined as the union of the 'segments'

p(x1)]),

j

=1

N,

is well defined and L(p o — p(x1)fl < +00. = Now, we define the inner (arcwise) Euclidean distance on (X, p)

x X —p by the formula o

[0, 1] —+ X is a curve with

= x'.

= x"},

x', x" E X.

(z. w) 8)

Observe that z — (z.

Note that

is the orthogonal projection of the vector z on the complex hyperplane

isa C-norm.

12

1

Riemann domains

Remark 1.1.14. (a) C1(X.p)(X'. x") (J(p(X).id)(P(X ), p(x")), x', x" (b) If A C X is a univalent set such that p(A) is convex, then

cyx(x,xI! )=IIp(x)—p(x II)fl, I

(c)

X.

/ // x,x EA.

r) = (x E X: ax(a, x)
px(a).

Indeed, the inclusion follows from (b). Suppose that ax(a, x) < r. Then [0, 1] —* X with there exists a curve = a, = x, and L(p o
): [0.11 —p Bx(a.r). (d) ax is a continuous distance. Observe that the topology generated by ax coincides with the initial topology of K. (e) For any metric space (Y, d) and mapping f: K —* Y the following conditions are equivalent: (1)

d(f(x'), f(x")) < ax(x', x"), x', x" E

(ii) d(f(x'). f(x)) < Ilp(x') — p(x)II. In particular, by Remark 1.1.8(c), if

lpx(x') — px(x")l

x

X, x' E Bx(x).

= 0, then ax(x', x"),

x', x"

X.

(ii) follows from (b). Suppose that f satisfies (ii). Fix [0, 1] —÷ X with = x', = x". It is clear N and 0 = to < = such that E <

The implication (I)

,'

x', x" E X and a curve that there exist N

j=I

E

N. Hence

d(f(x'), f(x")) L(p o

—

which implies (i).

(f)Letao E X.0 <e < set {x

X: px(x) >

e}

with ao E 0 the set

Let

be

the inner Euclidean distance on

(Ui. plo). Then for any t >

U1, := {x E U1:

(ao, x) < t)

is relatively compact in X.

Indeed, let! := (t > 0: Us,, X). Observe that, by (c), (0, eJ C I. To prove e, then t +e/2 E I. Thus, it IlL0 it suffices to check that if: E I,:

that! =

suffices to show that tJEJ+t/2 C

1.1 Riemann domains over C"

\

Take a point a E

[0, 1] —*

and let

13

be

a curve with

= ao,

lies in lBx(a, r) \ Bx(a, e/2). Then t

+ e/2> L(p o

=

L(p o Yi[O,r]) + L(p o

L(p ° Consequently, L(p o PIiO.rj)

+ IIp(b) — p(a)II > L(p ° PIfo,rl) + e/2.

<:.

Hence b E (Je,t and a E

s).

Remark 1.1.15. (a) It is known (cf. [Die 1944], [Bou 1965]) that for any connected manifold M the following conditions are equivalent: (i)

M is countable at infinity i.e. M =

K1,

where K1 is compact, j =

1,2,... (ii) The topology of M has a countable base.

(b) II M is a connected Riemann surface (i.e. a connected one-dimensional complex manifold), then M is countable at infinity; cf. [Rad 1925], see also § 1.11. Theorem 1.1.17 will show that the same property is shared by all Riemann domains over C".

Exercise 1.1.16. Complete the following construction showing that there exists a simply connected two-dimensional complex manifold M which is not countable at infinity (cf. [Cal-Ros 1954]). Let A C C, A uncountable, M := A x C2,

topM:={UCM:V(EA: ((x,v)EC2:(t,x,y)EU}EtopC2}. Fors,r E A,s

Cx C,

Cx C,,

is biholomorphic and (S,Xs,Y.c)

=

:= For

(s,

(t.

x,, y,) E M define

— (f,x,,y,)

((s. Xs, Ys) = (t, ((s. x5,

=

(t,

v1)) or (s

x,, v1)) or (s

i.

t,

=

y1, s + XSYS

=t+

E C,. (x5. y5) =

is an equivalence relation. Put M := and let top M be the standard quotient topology, i.e. the strongest topology on M such that the canonical projection M

(t,x.y)

[(t,x,y)1

Observe that, since M is locally compact, the following conditions are equivalent: • M is countabk at infinity; •M= U1 is relatively compact and open. j = 1.2,...; •M= K1. where is compact and K1 C mt j = 1.2

14

Riemann domains

1

is continuous;

{(x,y) E C2: [O,x,y)] EU) E topC2.

U E topM

For I E A define

U, := Jr({I} x C2),

C2 —p U,, tp,(x, y) := [(t, x, y)].

Then:

•

is continuous and bijective.

• U, E topM. • is Continuous. • M = UtEA U,, topM E T2. • (U,, Q1')IEA is an atlas of a complex structure on M. • M is connected. • The set S := ir(A x {(O, 0))) is discrete and #S = #A. In particular, M cannot be countable at infinity. • M is simply connected.

Theorem 1.1.17 (Poincaré—Volterra theorem). Let (X, p) be a Riemann domain over C". Then X is countable at infinity. In particular. •for each z E p(X) the stalk is at most countable; • the topology of X admits a countable base consisting of connected univalent and relatively compact sets.

Pmof Fix a point ao E X and let

be such that 1/k0
Vk

By Remark 1.1.14(f) Vk

that X =

and Ui,, be as in Remark 1.1.14(f). Let lc,j E N

Put

X, k

:= Ul/k,k.

k

/co.

k0. Obviously Vk C Vk+1. It remains to observe

Vk 10)

Since X is countable at infinity, it is clear that each stalk (z) must be at most countable. Now we will directly 11) show that the topology of X admits a countable base consisting of connected univalent and relatively compact sets. Let X = where K1 is compact, j E N. For any j and for any rational number 0 < r < dx(Kj), consider the covering (P(x, r))xEK1 of Since K1 is compact, we can select a subcovering (IF'(x, where Sj.r C K3 is a finite set. We claim that the family 'B := (P(x, r))JEN. O
t

Takeanarbitrarypointa E LetO <e max(ko. l/e. t}. independently of Remark 1.1.15(a).

L(pc

with k

rEQ. xES1,,

—+

withr > t. Conscquently,a E Vk

1.1 Riemann domains over

is the base we are looking for. We only need to prove that

15

is a base of the topology

of X. Take an arbitrary open set U C X and a point a E U. Let 0 < R
U

The end of this section is devoted to continuity properties of families of liftings.

Theorem 1.1.18 (Monodromy theorem). Let (X, p) be a Riemann domain over C's. Let A be locally compact, connected, and locally connected topological space and let B be a locally connected topological space. Fix an ao A. Assume that

F: A x B —* p(X) is a continuous mapping. Suppose that for each b B the mapping Fb := b) has a lifting Fb: A —÷ X such that the mapping B b —÷ Fb(aO) X is continuous. Then the mapping Ax B 3 (a,b) Fb(a)E X is continuous.

Proof First observe that F is continuous at each point (ao, b), b E B. Indeed, fix a b1 B and let W be a univalent neighborhood of F(ao, b1). There exist connected neighborhoods U C A of ao and V C B of b1 such that: F({ao} x V) C W (here we use the continuity of the mapping b —÷ F(ao, b)),

F(U x V) C p(W). Deflnea := (pIwY'oFluxv.

Then forarbitraryb VthemappingsU 30 —4 —+ o(a,b)areiftingsofthemappingU 30 —* F(a,b)that

F(a.b)andU

coincide at a = ao. Consequently, by the identity principle for liftings, they coincide everywhere. This means that F is continuous at (aO. b1). Now, fix an arbitrary point (a1, b1) e A x B. We will prove that there exist neighborhoods U C A of V C B of b1, W c X of F(a1, b1) such that W is univalent, F(U x V) C p(W) and Fluxv = (pIwY' o (which obviously implies the continuity of F on U x V). For any x A let W C X be an arbitrary univalent neighborhood of Fh1 (x). Choose a relatively compact neighborhood C A of x such that Fh1 (Ui) Since F is continuous at (a0.b1), we may assume that F(UP.I) x V) C for a neighborhood V C B of b1. Since A is a connected and locally connected Hausdorif space, there exist points xN e A and domains U1 UN such that xi = ao,

N—l(cf. N. Since F(U1 x {b,}) = p(Fb1(U1)) j =1 N, there exists a connected neighborhood V C V of b1 such

[Bou 1965]). Put W3 :=

j =I that F(UJ x V) C

j=l

(pIw1Y'

o

j =I

N. Note that F(U1 x V) C Wi. Define

Uj x V —p

X.

It suffices to show that for any

N:

xj =

(*)

16

1

= fr on

Riemann domains

j = 1.

N. Obviously (*) holds for] = 1. k forsome 1 k N —1. Fixapoint E Uk fl Uk+I. Note that F(x0,b) = ak(xO,b), b E V. Fix a b E V. Observe thatthemappingsUk+I 3x —+ x —+ F(x. b). Thus, by the identity principle for liftings, it suffices to prove that ak(xo. b) = F(x0, b) = ak÷1(xO, b). For, observe that the mappings V y —÷ ak(XO, y) and V 3 y —* (7k+l(XO, y) are two liftings of y —+ F(x0, y) that coincide at y = b,. Hence, once again by the identity V Observe that

x {b,},

Supposethat(*)istrueforj =

1

principle for liftings, we get the required equality.

El

In the case where A = B := [0. 1], ao := Owe get the following

Theorem 1.1.19 (Classical monodromy theorem). Let (X, p) be a Riemann domain over C", E X, and let F: [0, 1] x [0, 1] —÷ p(X) be a continuous mapping such that F (0. u) = p(xo), u E [0, 1]. Suppose that for each u [0, 1] there exists a lifting Define F(t, u) := u) such that [0. 11 —+ X of the curt'e = t, u [0. 1]. Then F is continuous. In particular, there exists a point Zj E p(X) such that F(l, u) = zi for all u [0. 1], then the mapping [0, 1) 3 u F( I, u) E (z i) is continuous. Consequently, since the stalk (z i) is discrete, there exists a point (z i)

suchthatF(1,u) =xiforallu E [0,1). A Riemann domain (X, p) is said to be a cover if for any point zo E p(X) there exists an open neighborhood (Jo such that each connected component U of the set is univalent and p(U) = Uo. A Riemann domain (X. p) is said to be arbitrarily continuable if for any curve y: [0. 1] p(X) and for any a E (y (0)) there exists a lifting [0, 1] —p X

= a.

with

Notice that in fact (X, p) is arbitrarily continuable if (X, p) is a cover; cf. the following exercise.

ExercIse 1.1.20. Let (X, p) be a Riemann domain over C". Prove that the following conditions are equivalent: (i)

(X.p)isacover;

(ii)

(X, p) is arbitrarily continuable;

(iii) For any metric space T such that • T is arcwise connected, • T is locally arcwise connected, • T is homotopically simply connected,

for any continuous mapping y: T —+ p(X), for any to E T, and for any a

there exists a lifting

12)

Notice that in (ii) we have simply T :=

[(1), 1] —* X such that 10. 1).

=a

12);

17

1.1

(iv) For any C-norm q:

—* R+ we

=

have d(x.p).q

(d(p(x),id).q)

o p (cf.

Remark 1.1.8(a)). Hint.

(i)

(ii):

:

Let

closed. (ii)

Hence jo

Take an arbitrary curve y: (0, 11 —p p(X) and a point a be as in Remark 1.1.11. Since (X,p) is a cover, the set lois

= I.

a,(0) =

(iii): For any t E T let a,: [0, 1] —÷ T be an arbitrary curve with to,a,(1) = t. Put8, := yoa,: [0,1] —÷ p(X) and let 8,: [0,1] —+ X

denote the lifting of 8, with 8,(0) = a. that 8,(l) is independent of at. Define is continuous. (iv): FixaC-normq: (iii)

The classical monodromy theorem implies

:= 6,(l). Now it remains

T

(iii) there exists a section

—*

X of

y :=

idT.

(X, p) over T such that

=

a.

= Bq(a.r).

Then

(1): Fix a zo E

(iv)

p(X)

2r) C p(X).

and let

Put

Then U = Bx(a, r), homeomorphically U onto Uo.

U be a connected component of

In

that

E p1(zo). Let

—÷

T := Bq(p(a). r) C p(X),

By

to prove

particular, PIu maps

Notice that

the assumptions on T may

Uo := B(zo, r). Let where a E

be weakened.

Proposition 1.1.21. Let (X, p) be a cover over C's.

Then:

=#p'(z2)foranyzl,z2 E p(X). (b)

If p(X)

Proof (a)

that

p(X),

Zi, Z2 ZI, y(l)

Fix

= = p'(zi) let f(x) =

thaty(0)

x

connected, then p

is simply

f:

injective (i.e. X is univalent).

Z2, and let

y: [0, 1] —+

p(X)

be a curve such

For any = x. It is clear is the lifting ofy with (z2). By the identity principle for liftings the mapping f is

Z2. It

(z

Zi

is

suffices to prove that

where

injective.

(b)Supposethataj,a2

Xaresuchthatp(ai) = p(a2) =: zoandleta: Let a(1) =

[0, 1] —÷

X be a curve such thata(0) = al,

F: [0,1]

x [0,1] —*

p(X)

be a continuous mapping (a homotopy) such that • F(0, .)

= F(l,

.)

•F(.,0)=poor,

= zo,

• F(., 1) = zo.

Since (X.

E [0, 1] there exists a curve In particular, by the identity prinand p o F(.. u) = F(., u). By the

p) is arbitrarily continuable, for each u

F(.,u): [0,1] —* X such that F(0,u) ciple for liftings, F(.,

0)

monodromy theorem. a2

=

= a, F(., 1) = = F(l,0) = F(1, 1)

=

18

1

Riemann domains

1.2 Holomorphic functions The aim of this section is to collect some basic properties of holomorphic mappings between Riemann domains. The local theory of hotomorphic functions on Riemann

domains over C° is of course the same as on domains in C'1. Thus the stress will be put on global properties.

DefInition 1.2.1. Let (X, p) be a Riemann domain over C". A function

f: X

C

(f E (9(X)) if for each open univalent subset U C X the function f o (pIuY' is holomorphic in the standard sense on p(U) C C" (equivalently: f is holomorphic in the sense of the complex manifold structure of X; cf. is said to be holomorphic

Remark 1.1.2(c)).

A mapping F = (F1,..., Fm): X —p the functions F1..... Fm are holomorphic.

Ctm

is holomorphic (F

t9(X,Cm))if

If ( Y, q) is another Riemann domain over then a continuous mapping F: X —÷ is said to be holomorphic (F E (9(X, Y)) if q o F 0(X, Ctm) (equivalently, F is holomorphic in the sense of the complex manifold structures on X and Y) 13'i, Y

The space 0(X) will be always endowed with the topology TX of locally uniform convergence, i.e. the Fréchet space topology generated by the semmorms

(9(X)? f —+ IIfIlK,. where (K3

X=

i = 1.2

is any sequence of compact subsets of X such that K, (cf. the Poincaré-Volterra theorem).

C mt

and

Exercise 1.2.2. Prove that TX is a Fréchet space topology. Prove, moreover, that TX is independent of the particular choice of a sequence of compact sets. Observe that an analogous topology may be introduced in the case where X is a connected complex manifold, countable at infinity.

For f E 0(X) define the j-th partial complex derivative at a: Of

—(a) := 0z1

(p(a)),

j=1

OZj

where U is an arbitrary open univalent neighborhood of a E X 14)• Observe that the

definition of

is obviously independent of U.

131 Observe that if Y is non-univalent, then there exists a discontinuous mapping F: X —* Y such that q o F const. 14) In the above formula, the operator on the right hand side denotes, of course. the standard

j-th partial complex derivative in C".

1.2 Holomorphic functions

For v =

(vi .

E

partial derivative off at a:

define the

(avf)(a) =

a

:=

19

o

a

f(a).

Clearly aVI E (9(X). By the Weierstrass theorem, the operator

0(X)

f -L

E

(9(X)

is continuous in the topology TX. We have the following Cauchy inequalities:

K

X, 0


V

E

The power series

:= 1,(a)(f

o p')(z) =

—

p(a))V

called the Taylor series of f at a. Let d(Taf) be the radius of convergence of the Taylor series Ta!, is

d(Taf) := sup{r > 0: T0 1(z) is convergent for z E IP(p(a), r)}. Observe that d(T0f) >

and

f(x) = Taf(p(X)).

f E (9(X), a E X,

x

P(a).

Proposition 1.23 (Identity principle). Let (X, p), (Y, q) be Riemann domains over and cm, respectively. Let F, G 0 (X, Y) and assume that F = G on a non-empty

open subset. Then F

G on X 15)•

Pmof. Put

Xo := {x E X: F = Gin an open neighborhoodofx}. Clearly, Xo is non-empty and open. It suffices to show that Xo is closed. Let a Xo. By continuity we get F(a) = G(a) =: b. Let U and V be univalent neighborhoods

of a and b, respectively, such that F(U) c V and G(U) C V. We may assume that U is connected. Then the holomorphic mappings

f := (qlv)° F o (pIuY*

g := (qlv) o Go

on p(U), coincide on p(Xo fl U) 0. Hence, by the standard identity principle, they coincide on the whole p(U). In particular, F = G on U. Thus defined

UcX0. 15)

Exercise.

D The result remains true in the case where X and Y are connected complex manifolds —

20

1

Definition 1.2.4. Let F C (9(X)

Riemann domains be

a vector subspace. Assume that F is endowed

with a Fréchet space topology r. We say that F is a natural Fréchet space if the convergence in the sense of r implies the convergence in the sense of TX, i.e. the identity embedding (F, r) —÷ (0(X), TX) is continuous. Obviously, (0(X), Tx) is a natural Fréchet space. The following example shows that many classical spaces of holomorphic functions are natural Fréchet spaces.

Example 1.2.5 (Natural Fréchet spaces). (a) ,7€°°(X) := the space of all bounded holomorphic functions on X with the topology generated by the supremum norm lix. Notice that (R°°(X), II lix) is even a Banach algebra. (b) := 0(X) fl LP(X) = the space of all p-integrable holomorphic Junelions with the topology induced by the norm II IILP(x) (1 p oo). Here is taken w.r.t. the Lebesgue measure A = Ax on X constructed as follows: A set A C X is called measurable (A E .C(X)) if p(A fl U) is Lebesgue measurable in C" for any open univalent set U C X. It is clear that any Borel subset of X is measurable. One can easily check that a set A C X is measurable if any point a E X has an open univalent neighborhood U such that p(A fl U) is measurable. Let X = where each U1 is open and univalent (use the Volterra theorem). Put B1 := U1, := U U For A E C(X) we put j = 2,3 := Am(p(A fl B1)). One can prove (Exercise) that is a regular Borel measure on £(X) which is independent of the choice of a sequence II

fl U) =

Moreover,

A

U

and

A

Jo

measurable function, then IA fdAx = andB1,j=l,2 areasabove. By the Cauchy integral formula, we get IlfilK

12fl I

(,rr

IfidA,

a where U1

A

0
fEO(X),

)

Hence,

lifliK where

(jrr2)"

I/p + l/q =

IIJIILP(X),

fE

K

I. In particular, L(X) is closed in

(L(X), Ii 11ux) is a natural Banach space. Obviously, Note that

with the scalar product

(f.g)—+ is

a Hubert space.

Jx

X, 0 < r which shows that = .lt'°°(X).

1.3 Examples of Riemann regions

21

where 2 is an open subset of X, with the topology := (9(e) fl (c) with K being a compact subset of c. generated by the seminorms f —÷ Observe that if f2 X, then is a subalgebra of

(d) (9(k)(X 8) := {f E 0(X): 18k1 lix < +oo} = the space of all 8-tempered holomorphic functions on X of order (degree) k with the norm f —÷ l8kfiIx, 0). Note that where 8: X —÷ (0, +oo) is such that 1/8 is locally bounded (k

6) =

6) is a Banach space and that Exercise 1.2.6. (a) Let

r,

be

a countable family of natural Fréchet spaces in

(9(X). Put F := :F and let r be the weakest topology on F such that all the mappingsid: F —÷ F,i E I, arecontinuous (i.e. fk —* foinr jk —÷ fo :

in r for any I c I). Prove that (F, r) is a natural Fréchet space. (b) Let E C be afamilyof natural Fréchet spaces. 0 E E, andlet avi Define F := {f E We endow F with the weakest topology F —÷ r such that the operators v E E. are continuous (i.e. fk —f fo in dVfo in r: for any u E E). Prove that (F, r) is a natural Fréchet space.

(c) Observe that using (a) and (b) we can produce a lot of new natural Fréchet spaces. For instance, with E := {v E R'I
:= {f E


E 3€°°(X),

:= {f E Lj'(X), lvl
U {oo}.

and

= .4(Q).

1.3 Examples of Riemann regions Example 13.1 (The sheaf of germs of holomorphic functions; see Example 1.6.6 for a more developed situation). Let a C C". Define

:=

1(U,

f

f): U

(9(U)},

where denotes the family of all open neighborhoods of a. For (U, f), (V. g) put 0a we

WCUflV, fiwgiw. It is clear that

is

an equivalence relation. Put

1 Riemann domains

22

The class [(U,

is called the germ

off at a. We write

:= [(U, f)]u. Define

[(U, f)]a + [(V, g)ja := [(U fl V, f + g)]a, [(U.

f)ja

.

[(V, g)Ja

[(U fl V, f . g)]a.

check that the operations +, X '9a (9a are well defined and that ((9g. is a commutative ring with the Unit element (the ring of germ.s of holomorphicfunc:ion.s a: a). [(U, avf)]a are well defined. In Let f — f(a) and E (9a. Then f(a) particular, we can define One can easily

-a"f(a)(z

Tat(Z)

—

= T0f(z).

The mapping

J* TatE the ring of all powerseries with centerata that are convergent in a neighborhood of a

is an isomorphism. Put

(9:= U

x {a}

a cC"

a)) = a (in the sequel we wiLl denote elements of (9 either formally as pairs (f, a) (when we want to point out that or simply as germs f). For and for any (U, f) f define and let ir: (9 —+ C" be given by the formula

t11(U,f)

z E U} C (9.

It is clear that the system (U.f)Ef is a neighborhood basis 16)• We endow (9 with the topology generated by this basis. We will show that this is a Hausdorif topology. Take (f. a), (g, b) E (9,(f, a) b). If a b, then iLf(U, f)fl t19(V,g) = øprovidedthatUflV = 0. lfa =b,thenUj(U,f)flUg(U,g)= 0 if U is an open connected neighborhood of a (by virtue of the identity principle, if g and therefore I = g).

f

Let X be a act Suppose that for each x C X we are given a non-empty family 8(x) of subsets of X such that •x eU for any U E 8(x). • for any>' C U E 8(x) there exists a V E 8(y) such that V C U,

.for any U. V e 8(x) there exists a W E 8(x) such that W C Un V. induces a topology Ton X: we say that a set Q C X is open E T) Notice that 8(x) C T, x C X. Thus the if for any .r C c there exists a U C 8(x) such that U C The above system system

is a basis of open neighborhoods in T.

1.3 Examples of Riemann regions

23

is continuous and is injective on any neighborhood U = it1 (U. f). U is homeomorphic and (JrIuY'(z) = (fe. z). Z E U.

Observe that

it —+

Moreover, irlu: Thus ((9, iT) is a Riemann region over C". It is called the sheaf of germs of holontorphic functions in C". Notice that a E C". = d(Tat), f E

Lety: [0,1] —+ C" beacurve. Suppose that foreacht E [0, l]we are givena convergent power series a (t) with the center at y (t) (or, equivalently, a germ a (t) from Oy(f)) such that the mapping [0, 1] t —+ := (a(t), y(t)) (9 is continuous. Then we say that the series a (0) can be continued analytically along the curve y (to the series a(l)). In other words, the series a(0) can be continued analytically along (9of y such that = (a(0), y(O)). y if there exists a lifting [0, II

Reniarkl.3.2. (a)DefineF: (9 —+ C,F((f, z)) := t(z). Thenforanyit = tt1(U, f) we get Fo = on U. This shows that IF is holomorphic on (9 (F (9(0)). (b) For any open set C C" and for any f 0(Q) define Q —f 0. sf(z) := (fr. z), z E Q. It is clear that Sf a section of ((9, ir) over Q. Conversely, ifs: ci —÷ (9is a section, then the function f := IF o s is holomorphic on ci and s = sj. Thus, r(cl, ((9, ir)) (9(Q). (c) Let G C C" be a domain, let f (9(G). and let f denote the connected component of 0 that contains s1(G). Suppose that for some a G there exists a curve y: [0,1] —p C" such that y(O) = a and that the series a(0) := Taf can be continued along y to a series a(l). Since [0, 1] —÷ (9 is a curve and I]) c f, which shows that a(l) E f. E sj(G), we conclude that (9 —p (9,3V(J, a) = (d) Let a). Then 8V is continuous and

f

r. In particular, dV(Sf(Q))

(e) Let n =

I

is locally biholomorphic (and, consequently, open). Note that

and let A C (9be a domain. Define

:= ([(V. g)Jb: 3((u.f))6E.4: f: U —+ One can easily prove that

V

is bijective. f(a) =

b,

g=

f}.

C Ois also a domain (Exercise).

In particular, if f: G —f D is biholomorphic (D, G c C), then

=

Sj_i (D).

Example 1.3.3(Riemann surface of LetG C Cbe adomain and let!: G —+ D be a locally biholomorphic mapping, D := f(G).

Evidently, (G, f) is a Riemann domain over C. Moreover, ((sj(G)Y', ,r) is a Riemann domain over C. Let X := {(w, z) E G

x C: f(w) = z}. We endow X with the topology induced

from C2. Let p(w, z) := z, (w. z) E X. Then, by the implicit function theorem, p: X —+ C is locally homeomorphic. Thus (X, p) is a Riemann domain over C.

24

1

Riemann domains

We say that Riemann domains (Xi. pj), (X2, p2) are isomorphic if there exists a homeoinorphic mapping a: K1 —* K2 such that P2 o a = p1: cf. § 1.4.

We claim that the Riemann domains (G. f), (X, p). and ((sj(G)Y *

are iso-

morphic. Let

G —* X. := (w. f(w)). Then w: G —+ K is homeomorphic and po = w) z) = f. This shows that (G, f) and (X, p) are isomorphic. Now let X —÷ (s1(G)Y* z). where g is a local inverse to I z) := such that g(z) = w. It is clear that is a homeomorphism z) = (g(z), z)) and ir o i/i = p. Thus (X, p) and ((sj(G)Y* r) are also isomorphic. Observe that the function s := is holomorphic on X and 1 domain (X. p) is called the Rieniann surface of

o

s = p. The

We would like to mention the following two particular cases.

1°: f(w) := w1'. w

G :=

(k E

already know that the domains (X, p) and (Ce. f) are isomorphic. Nevertheless, since p coincides with the standard projection, the domain (X, p) has a more geometric nature. For any point zo the stalk p'(zo) consists of exactly k points, say Ao k — 1. Notethat{wo.o Let wo.k_1} = A3 = (wOj, zo), j = 0 be any simply connected open neighborhood of Zo and let Jo UC 1k—I E We

(9(U) be holomorphic branches of the k-th root such that f3(zo) = WOj, J = k — I the set U3 := {(fj(z),z): z E U) 0...., k — 1. Then for each) = 0 is a univalent neighborhood of and (plu1Y1(z) = (f3(z), z), z E U. In park — I, U : C \ ]R. Let licular, let zo := I, WOJ := exp(2irij/k), j = 0 = (w,wk)). [0,1] —+ X E [0, k — 1, is a curve such that )2(j/k) = A1,] = 0 = = Ao. It is clear that

j=l

U3 fort E T1, where T0 := [o.

U

I] and T, :=

k—I.

This leads to the following geometric interpretation of the domain X. We take k

copiesofthedomainC\R...,sayCo

Letz E

besuchthatRez

0.

We say that z lies on the upper part of C3' if Im z > 0. Otherwise, we say that z lies on the slower part of C1'. Now we glue the domains Co,..., Ck.. i together, crosswise along IlL. That is, first we glue C0 to C1 in such a way that we join the upper part of Co with the lower part on Ci. next we glue the upper part of C1 to the lower of C2, and so on. Finally, we glue the upper part of Ck..I to the lower part of Co. One can prove that the above construction is impossible in 1R3 without self intersections.

20: f(w) := exp(w), w E G := C. The geometric interpretation of (X. p) is following. We take a countable family we glue to C1÷1 crosswise R. Next, for each j Observe that this along IR_ ,joining the upper part of C1 with the lower part of construction can be done in R3. (Cj)JEZ of copies of C \

1.4 Holomorphic extension of Riemann domains

25

1.4 Holomorphic extension of Riemann domains we need the following notion of a

To compare different Riemann domains over morphism (generalized inclusion).

DefinitIon 1.4.1. Let (X, p), (Y, q) be Riemann domains over C's. A continuous X —* Yissaidtobeamorphisrnifqoço=p.

If p: (X,p) —+

Y —+ X

is also a morphism, then we say that ço is an isornorphism. Obviously, the above notions extend to the case of Riemann regions.

are domains andQ: (G1, idG1) —+ (G2, Id0,) isa Observe that if G1, G2 c morphism, then G1 C G2 and p = id01,0,.

Remark 1.4.2. Let (X, p), (Y, q) be Riemann domains over

and let

(X,p) —p (Y,q) be a morphism.

(a) By the identity principle for liftings, if *: (X, p) —f (Y, q) is another morphism with p(a) = for some a E X, then (Y,q) —* (b) (Z,r) is a morphism, then oço: (X, p) —÷ (Z,r) is a morphism.

(Y,q) —+ (use (a)). then p is an isomorphism and fr = (c) is locally biholomorphic. In particular, .'p is an open mapping. (d) is an isomorphism if ço is bijective (use (c)). (e) If A C X is univalent, then is univalent. In particular.

r)) =

lPy(ço(x),

r),

E

X,

x E X, 0 < r

Consequently,

•dy dx, • if p is an isomorphism, then dy

o

= dx.

is closed in Y. Let E dy(y0). Let y = = dy(y) > E IPy(y0, rO). Then therefore, yo E lPy(y, rO) = co(Px(x. ro)) C ço(X). (g) The mapping Indeed, it is sufficient to show that

0<

2ro <

çØ*: (9(Y) —* (9(X).

is an injective algebra homomorphism.

çp*(g)

:= go ço

Take ro and,

26

I

Riemann domains

Sometimes, to simplify notation, we will write

:=

fE Inparticular,

fEO(Y). xEX. Indeed, using (e) we get Pa =

and hence, for g E 0(Y) and a E X, we

o

have:

=

= aL((g o

o

(i) In view of(h) we get d(Taf)

=

o

o

dy(çp(a)) for any a e X and f E

Now we are in a position to define the fundamental notion of holomorphic extendibility in the category of Riemann domains over Definition 1.4.3. Let (X, p) be a Riemann domain over C" and let 0 C 0(X). We say that a morphism a: (X, p) —÷ (Y, q) is an if C a*(0(Y)) there exists a g E (9(Y) such that g a a = f). (i.e. for each f E If = (9(X), then we say that a: (X, p) —+ (Y, q) is a holomorphic extension.

Obviously, if a: (X, p) —+ (Y, q) is an s-extension, then it is an for any 0 C cS. Consequently, the most important is the case cS = 0(X). The above definition of the generalizes in a natural way to the case of Riemann regions over C" but in this case, to avoid situations where the region (Y, q) is 'too large', we assume that every connected component of Y intersects (which is equivalent to the injectivity of the mapping p*).

We say that a family 0 for any i'

E

(equivalently,

C (9(X) is a-swble if E 8 for any f E for any f and j = I n).

8

and

Remark 1.4.4. Observe that for any family 8 C (9(X), the smallest a-stable subalof (9(X) with 8 U gebra C [.8] may be described in the following way 17):

[S]:= {P(f1

IN): N

N.

P

.IP(CN),

Ii

IN

.8'},

where p,,}. 17)

Notice that I E [clI.

18)

P(CN) denotes the space of all complex polynomials of N complex variables.

18)

27

1.4 Holomorphic extension of Riemann domains

Remark 1.4.5. Let a: (X, p) —p (Y, q) be an 8-extension. Put

= lfa: f E

:= (a)

.8}

=

0(Y):

(g

If $: (Y. q) —÷ (Z, r) is an

then

goa

o a: (X, p) —+ (Z. r) is

an 8-extension.

(b) Iffi: (Y, q) —f (Z, r) is a morphism such that oa: (X, p) —* (Z, r) is an i-extension, then (Y, q) —* (Z, r) is an p-extension.

= (9(X), then

(c)

a vector space (resp. an algebra) if = Moreover, by Remark 1.4.2(h),

is a vector space (resp. an algebra). In particular, is ax-stable if is

ay -stable.

(e)d(Taf) = d(Ta(a)fU) dy(a(a)) for any a E X and f E (f) a: (X, p) —* (Y,q) is an [81-extension (cf. Remark 1.4.4). Therefore, in the theor-y of holomorphic extension we can always assume that [-3] = a-stable algebra with P1 E

,

i.e.

is a

.

If = 0(X), then g(Y) = g(a(X)) for any g E 0(Y). Indeed, suppose that a E g(Y) \ g(a(X)). Let

f:= Then (goa —a) .1 on

1 on X, and therefore, by the identity principle, (g —a). f(Y

contradiction.

= Jgoallx,g E

(h)1f Consequently,

"=

3t'°°(Y) and a*:

—k ,Y°°(X) is an isometry of

Banach algebras.

Indeed, suppose that there exists an a E g(Y) with IaI > hg o alix. Then the function f defined as in (g) belongs to .J€°°(X) and we conclude the proof as in (g). (i) Suppose that is a natural Fréchet space (cf. Definition 1.2.4). We already is a vector subspace of 0(Y). Now we introduce a topology on know (by (d)) that

Assume that the topology r of is given by a sequence (qj )0= of seminorms such that qj qj÷i. be an arbitrary sequence of compact subsets of Y such that L3 C Let and L3 = Y. a*; is a seminorm on Put := o observe that j E N. Now consider on the following family of seminorms U (II

This family defines a metrizable locally convex topology (gv

aa

o a and

on

such that

28

1

In particular, the topology

Riemann domains

is independent of the choice of admissible sequences

Observe that Ta is a topology of a Fréchet space. Indeed, if is a Cauchy ra). then o a)' is a Cauchy sequence in (& t) and sequence in is a Cauchy sequence in ((9(Y), Tx). Hence there exist functions Jo E and go (9(Y)

Itremainstoprovethatgooa = Jo. Recalithat

—f-.

r is stronger than TX. Hence, for any x X, we get fo(x) = oa)(x) = go(a (x)). The mapping —* (& r) is obviously continuous (recall that it is an algebraic isomorphism). Since a Fréchet space, the Banach theorem

ra) implies that a*: compact L c Y there exist Jo E

r) is a topological isomorphism. i.e. for each N

and c>

0 such

that

(j) By virtue of (i), if .3 is a closed subspace of 0(X) (in the topology TX). then is closed in (9(Y) (in Ti). Moreover, for each compact L C Y there exist a compact K c X and a constant c > 0 such that gE

(k) In the special case, if.3 is a closed subalgebra of (9(X), then is a closed subalgebra of 0(Y) and for each compact L C Y there exists a compact K C X such that

(gilL < lIgoalIx,

gE

N. Hence < oallK for any g E and k Lettingk —* +oo,weseethatwecantakec= 1. (I) The estimate in (k) may be written in the form Indeed, by (j) we have

IIXIIL

—"-'

Lca(K)

a

where for a compact set H C Y and a family 7 C 0(Y) we put HT

(yE Y: VJET: lf(y)I

IIfJIH}.

is called the 7-hull of H. The set is called the holomorphic hull of H. 7-hulls will play a very important role in the characterization of 7-extendibility

The set

1.10.

Exercise 1.4.6. Prove that Remark 1.4.5(h) remains true for.3 C

f In

particular, we can take.3 :=

a

.3.

(X), cf. Exercise 1.2.6(c).

such that

1.5 The boundary of a Riemann domain

29

1.5 The boundary of a Riemann domain Let (X, p). (K q) be Riemann domains over C" and let

X —*

Y

be a morphism.

Our aim is to define an abstract boundary 8 X of X w.r.t. the morphism rp. The idea of such an abstract boundary is due to Grauert (cf. [Oka 19841. [Gra 1956]. [Gra-Rem 1957], [Gm-Rem 1956], [Doc-Gra 1960]). In the case where (X, p) = (G. id) (G is a domain in C"), (Y. q) = (C". Id),

=

=id

Id, the abstract boundary 8 G will coincide with the set of. so-called, prime ends

of G. At first let us recall some facts and notions from topology. Let X be a topological space. We say that a non-empty family is afluter if: •AE B E AC B ; • Aj. A2 E =* A1 fl A2 E

of subsets of X

•0 A non-empty family

of non-empty subsets of X is said to be a filter basis if: CA1 flA2. It is clear that for each filter basis the family := (A C X: B C A) A

is a filter.

We say that a filter is convergent to a point a E X if each neighborhood of a belongs to We shortly write a E urn We say that a filter basis is convergent to a if a E urn (equivalently, each neighborhood of a contains an element of we put lim := lim We say that a is an accumulation point of a filter (resp. filter basis if a A for any A E (resp. A E Let us recall a few elementary properties of filters (cf. [Bou 19651):

• If

C

are filters and if a is an accumulation point of

accumulation point of

a is an

j

•!fa

• If a is an accumulation point of

then

then there exists a filter

such that

aE

• a E A if there exists a filter basis

consisting of subsets of A such that

aE • Let Y be another topological space and let X Y. Then is continuous if for any filter basis in X the filter basis := (çp(A): A E satisfies the relation: q(limq3) C • X is Hausdorif if any filter in X converges to at most one point. If X is a Hausdorif space and = (a), then we write = a. For a E X and A C X let A) denote the family of all open connected neighborhoods U of a with U C A. Put := X). Let us come back to the situation when (X, p), (Y, q) are Riemann domains over X —+ V is a morphism. C" and

1 Riemann domains

30

DefinitIon 1.5.1. We say that a filter basis a of subdomains of X is a co-boundary point ofX if: (1) a has no accumulation points in X, E Y such that urn ço(a) = (2) there exists a point (3) for any V E there exists exactly one connected component U =: e(a, V) such that U E a, (4) for any U E a there exists a V E such that U = e(a, V). Let d X denote the set of all co-boundary pomts of X. We put

X :=XUaX ==Q

and we extend co to co: X

.

—÷

Y

=

by puttmg co(a) := yo if a and yo are as above.

:= q o In the special case when (Y, q) =

Moreover, we put =

and

we write ax

=

and

(C's. id) and co = p we =p x and X . respectively.

skip the superscript"

=p

X instead of a

Note that: • If (Y, q)

then the co-boundary is empty. (X. p) andco = • If G = E\[0, 1), then dG consistsofpoints from (aG)\(0, 1] with 'multiplicity' 1 and of points from (0, 1] with 'multiplicity' 2.

Remark 13.2. If ci satisfies (2), (3), and (4) and a has an accwnulation point xo X, then tim a = such that Indeed, first observe that = limco(a) = co(xo). Let U E 19)• Take is injective. Observe that U is a connected component of with V co(U). We know that xo E e(a, V) =: A. Since A is VE and E A, we conclude that A C U. Thus U E connected, p(A) c which finishes the proof.

Instead of fitters, one can describe the co-boundary using sequences of curves: C X such that: Consider the family of all sequences = has no accumulation points in X, (a) (b) there exists a yo E Y such that lu co(xk) = yo, (c) for any neighborhood V of yo there exists a ko such that for any k, £ k0 there

=xk, yk,I(l)Xt.

existsacurve yk.!: [0.11 —f

Lenuna 1.53. Let,r = C X, following conditions are equivalent: (i)

C

(0, 1), fk < tk+I' tk —* I. Ilsen the

satisfies (b) and (c);

We argue as in Remark 1.1.2(f): Let U0 be a connected component that := maps U0 into U and = id on U. Consequently, by the contains U. Then identity principle. = id on U0, i.e. U0 = U.

1.5 The boundary ofaRiemann domain (ii) there exists a curve

31

= Xk. k >

1,

and the limit

=: such that y(t) E V fort

Let V be Then

[0, 1) —÷ X such that )2(tk)

o

Proof (ii) (i): Clearly, an arbitrary neighborhood of for any £ > k > ko the curve

(xk) = Take k0>> to

I

connects

in p

with

beabasis of neighborhoods of yo with Vj+1 C

(ii):

exists a k(j) such that for any k k(j) the points xk, Xk+I can be j connected in qr'(Vj) by a curve (we may assume that k(j)
a curve connecting the points Xk(j) in j 2 1, and let be an arbitrary curve connecting xI rk(l) in X. Now, we glue the curves step by step U U U... and, after an appropriate change of parametrization, we get a curve o [0, 1) —f X with D = xk, k > 1. It is clear that = yo.

Remark 1.5.4. Let

tk /

1

and a curve

=

C X be a sequence such that there exist [0. I) [0. 1) —÷ X with the following properties:

•Xk=}'(tk),k> 1, •dX(xk) —÷0, exists,

• yO

• the liinitlim,..41

=

zo

E C" exists 2lt

6. Indeed, by Lemma 1.5.3, it suffices to check that qo = yo. Observe = pop. Henceq(yo) = zo. Let Vbeanarbitraryunivalentneighborhood

ofy0.

q(V)for:

E V fork

to

Consider the curves

:=

qoy1110 I)

Ito. I) —+

(qlvr'(p 0 )I(:o.1)) Observe that a1(t0) = a2(to) = the identity principle for liftings, a1 Remark 1.5.5. If

[tO, I) —* V.

and q o at =

=

q °

(12. In particular,

o

Hence, by

E V fort

has an accumulation point

satisfies (b) and (c) and (xk

=xo(cf.Remark 1.5.2). Let U E xk, =

E

to.

Indeed, assume that be such that 'Plo iS injective. Note that U is a connected component of Let k0 be such that for any k, £ > k0 the points xk, Xe can be joint in Now by a curve let uO be such that ? ko and Xk1, E U for v> uo. In particular, any point with k > /co can be joint with in Hence xk E U.k > k0. For 20) 21)

=

6 we write

-'-

whenever:

Equivalently: o extends to a curve y: 10. 11 —+ Y. Equivalently: the curve p o [0. 1) —÷ C'T extends continuously to [0. 1J.

32

1

Riemann domains

=:

=

•

ko the points by a curve. xk and .4 can be connected in Note that ; is equivalent to each of its subsequences. Observe that "—" is an equivalence relation. Put yo there exists a k0 such that for any k

:= '3/—

and define p : ax —÷

Y,

:=

cp(xk).

Proposilion 1.5.6. There exists a canonical bijection E: 8 X —* 8 X such that =

Proof Leta E 8X. Vk+I

beabasis of connected open neighborhoodsof yo

C Vk,k> I. Foreachktakeanarbitrarypointxk E Uk :=e(a,

Vk)

let;

Obviously, limk++C,C,ço(xk) = yo. Observe that Uk+l C and, therefore, xk can be connected with in for any k, £ k0. Thus; and

satisfies conditions (b) and (c) (from the definition of the family'3). Suppose that has an accumulation point xo in X. Take an arbitrary U = e(a, V) E a and ko. Inparticular,xo E U. let Vk0 C V. Then Uk0 c U andhencexk E U fork Thus xo is an accumulation point of a; contradiction. We have proved that; E 6. = Now let be another basis of neighborhoods of yo and let be constructed with respect to (.4 E (4 := e(a, Vi), k > I). We will show and let U that; ;'. Take an arbitrary neighborhood V E C V. Then U C e(a. V) and, consequently, for any k k0 the points xk,.4 can be connected in e(a, V) C Thus

we can define a mapping E: a x —+ a x, =

(a) :=

Obviously

=Q

.p

is injective. Suppose that = 3(a'). In particular, of connected neighborhoods of yo with = =: yo. Fix a basis k > I, and let xk E Uk := e(a, Vk), E (4 := 1, Vk), k Vk+I C We know that; := := '. Take an arbitrary U = e(a, V) and let Jco be such that for any k ko the points xk,.4 can be connected in çH(V). Let k > k0 be such that Vk C V. Then Uk U (4 C U. Hence e(a', V) = U and therefore U E a'. Thus a = a'. It remains to show that is surjective. Let; = E 6. Observe that for any connected neighborhood V of Yo := q.(xk) there exists exactly one connected component, say of such that Xk E fork >> 1. It clear that G v remains the same if we substitute; by an equivalent sequence. Obviously, if W c V,then Gw C Gy. Puta: = {Gv: V E ttisclearthataisafilter basis and that = yo. We will show that a has no accumulation points in X. We will prove that

Suppose that xO is an accumulation point of a. Then, by Remark 1.5.2, lim a = x0. In

1.5 The boundary of a Riemann domain

33

such that Gv C U.

particular, for any neighborhood U of xo there exists a V E Consequently, Xk —+ xo; contradiction.

Finally, directly from the definition, we get quence of is equivalent to

(recall that any subseD

=

Our next aim is to endow X with a Hausdorff topology which coincides with the initial topology on X and is such that the mapping is continuous. Let a E .9 X. By an open neighborhood ot the point a we will mean any set ot the form

U0:=UU{bE .9X: UE

a.

Lemma 1.5.7. (a) For a', a"

.9x we have: a'

a" jff there exist U'

a', U"

a"

such that U' fl U" =0.

=40•

(b) The lopology of X is Hausdorff

is continuous.

(c) The mapping

Pmof (a) The implication is trivial. For the proof of suppose that U' fl U" 0 for any U' E a', U" a". Let Choose V' := limço(a'), := Suppose that Y), V' fl V'1 = 0 and let U' := e(a', V'), U" := e(a", V"). Obviously, U' fl U" = 0; contradiction. Thus = Now take an arbitrary U' = e(a'. V). Let U" := e(a", V). Since U' fl U" 0, we get U' = U" and therefore U' a". Consequently, a' = a"; contradiction. (b) Take a'. a" E X

• a' :=

a'

U'

,

a". It suffices to consider the following two cases:

a'

Since a' is not an accumulation point of a". there exist X, a" E = 0. X), U" E a" such that U' fl U" = 0. Then U' fl

• a'. a" E

By (a) there exist U'

a', U"

a" with U' fl U" = 0. Suppose

0. Then there exists a b E dx such that U', U"

that U', n U",

0 = U' fl U" E

Consequently,

contradiction. —

(c) U2

Let

Let a E .9X.

yo

Take V1, V2 E !2c(yO. Y) with V2

V1 and let

= z := co(b)

:= e(a, V2). Suppose that there exists a b (U2)a n ax with vi. V3 Y) be such that V3 fl V2 = øand let U3 := C(b, V3). Then 0 =

U3 flU2

contradiction. This shows that

c

V1.

0

The following continuation problem will appear several times in the sequel. Let T be a topological space, let S be a nowhere-dense subset of T. and let

f: T \ S —÷ x

34

Riemann domains

I

be a continuous mapping such that of extends continuously to a mapping T —÷ Y. Does

the mapping f itself extend continuously to a mapping T —÷ X

?

Proposition 15.8. Let T be a locally connected topological space and let S be a nowhere-dense subset ofT such that for any domain D C T the set D \ S is connected. Let T \ S —. X be a continuous mapping such that o extends to a continuous

f:

f

mapping g: 1 —÷

Y

Then

f extends to a continuous mapping f: T —* X.

Observe that f is uniquely determined.

Proof Take t0 e T. Let yo := g(to) and let V E ¶Bc(yO). Since g is continuous, there existsaD e c V. Recall that D\Sisconnected. Hence

f(D \ S) c

is also connected. Let Gv denote the connected component of

that contains f(D \ S). Observe that Gv is independent of the choice of with g(D) C V. Define a(ID) := {Gv: V E It is clear that

DE 0(10) is a filter basis and there = yo. Moreover, for any V E exists exactly one connected component U of '(V) such that U E o(t0) (simply U := Gv). Note that if to S. = IPx (1 (to). r) for 0 < r dx(f(to)); in particular, tim a(to) = .f(to). There are the following two possibilities: -

=9'

has noaccumulationpointsmX. Thena(to) E Weputf(ro) := 0(ttj). • 0(10) has an accumulation pointxo E X. Then lim 0(b) = xO (cf. Remark 1.5.2). In this case we identify 0(10) with xO and we define f(to) = xci. -

=9'

have defined an extension f: T —f X off such that of = g. It remains to prove that f is continuous. Fix a U = Gv0 0(10) and let Do E be such that U is the connected component that contains f(Do \ S). Obviously, U E a(t) for any I E Do. We

Hence for I E D0 we get:

• if 0(t) E X, then f(t) E U; =4' =9' • if 0(10) E ax and 0(1) E ax, then 0(1) Ua(10). Now we only need to observe that if X. then 0(t) X for t Do := a(rq) provided U is small enough. In fact, let U := lPx(x0, r) = with r := o g on Do \ S. Hence dx(xo), and let f(Do \ S) C U. Observe that f = Thusa(t) = E Gv for anyl €Do and V E IEDO. —9,

The following result charactenzes the geometry of d X. =9'

=4'

Proposition 1.5.9. For any a d X and for any neighborhood U0 C X there exists a neighborhood W0 C such that dx = du on W. In urn

35

1.5 The boundary of a Riemann domain

Proof Let Yo := and suppose that U = C(a. V), where V E Fix V and let U3 := e(a, ll'y(Yo, r/j)), j > I. 0< r < such that Py(y0. r) We have U

U2.... There are the following two possibilities:

U1

p(U10): Put W := Take xo E W. Suppose that dx (xQ)> Ip(xo) — q(yo)I. Recall that p = q In particular. — q(yo)I < r/(2j0). Hence we find an s such that p(xo) — q(yo)I < s < r/(2jo) and s), we have: s)) = lPx(xo, s) exists. Since Q(IPX(XO, s)) = p(Px(xo, s)) C P(q(yo), nj0). Recall that cc(xo) E Py(0. r/(2j0)). Consequently, s). This means that s) C s) C IP'y(y0, r/jo) and yo E and contradiction. U30 that E ThUS < Ip(xo) — q(yo)I. By the same method as above we easily show thatPx(xo) C U10. Finally dx(xo) 10. There exists a Jo such that

2°.

yo = q'(xj) with x1 E U1 for any j

Put W := U4. Take xo E

1:

W and suppose that dx(xo) r/2. Then IP(q(yo), r/4) C P(q(yo), r). Hence Py(y0, r/4) C r/2) C Py(yo. U4

C Px(x0, r/2) C Ui. In particular, x1 =

X4 for

r/2)) C r) and

consequently, 4. This means that X4 is an

j

accumulation point of a; contradiction.


= [1

Remark 1.5.10. Let c C X be a domain and let ifr

:

22)

Consider the

boundary of the domain pb). We will show that there exists a canonical continuous mapping

=*

=9'

T:c —tX such that T( d Q) C

U

=9',

=

X)

and

oT=

=

Take a E a Q. For any U E a, let V C Y be a connected neighborhood of isaconnectedcomponentofi/r'(V). LetGu.v denotethe yo := connected component of qr'(V) that contains U. Define T(a) := {Guv: U E a, U is a filter basis of domains in X and urn ça(T(a)) = yo.

If T(a) has no

=9,

accumulation points in X, then T(a) E a X. Suppose that E X is an accumulation point of T(a). Then Q(xo) = yo and xO = lim T(a) (cf. Remark 1.5.2). Observe that XO

E

Moreover, we put T(a) := a for a e —+ 23)

denotes the standard boundary of

(in the topology of X).

36

1

Riemann domains

*

9'

Thus we have defined the mapping T: —+ X such that T( a C (axc2) U =9' = = —S.(d X) and o T = Since T(Ua) C (GU,V)T(6), the mapping 1 is continuous. =9'

*9'

Now we would like to determine a 'maximal' domain X C X such that (X, p I,p) is

a Riemann domain. The idea is to take *40

X

A set P

=40

=XU('thin'partsofdX).

X is called thin if for any point a E X there exist U E 0, such that P flU C denote the class of all closed thin subsets of X.

and

C

f E (9(U), f Let

Proposition 1.5.11. Let (X. p), (Y, q) be Riemann domains over C" and let p:

(X,p) —÷ (Y,q)

be a morphism. =9'

=9'

(a) Assume that a E 8 X is such that there exists a neighborhood U C X of a with the following properties:

•V

ço(U) is open in V. =9'

—

• P :=ço(Ufl 8X) =40 •rp: U\ 8X —+ V\Pisbiholomorphic. u:

Then the mapping

U —p V is homeonwrphic. =9'

(b) Let E denote the set of all points a E a X which satisfies the above conditions. Put *40

X :=XUE. Then. *40

• (X. p =

x

• EE

I

x

is a Riemann domain over C",

(Y,q)isamorphism,

(X. p I.Q) *9'

x

*

In the case (Y. q) = (C", id),

Proof (a) Put Uo := U \

=

p. we will writeX instead of X.

=40

a

x. Let us start with the following remark.

=9'

=

(*) Suppose that a U fi a x, U' E a, yo V' E Then there exists a V") such that U" C U' fi Uo and V" C V' fi V. U" = Indeed, let U' = e(a, V1) and let U2 = e(a, V2) be such that (U2)0 C U. Let V"

V'flV1flV2flV)andputU"

e(a, V"). NotethatU'flU2flU"

Since ip(U") C V1 fl V2. we get U" C U' fl U2 C U' fl

Uo.

0.

1.5 The boundary of a Riemann domain

First we prove a' a" and

that there are a'. a" E U such that

=:

=

yo.

—

By Lemma 1.5.7(a) there exist U' E

may assume that a', a" E U fl

We

37

with U' fl U" = 0. By (*) we may assume that U' = e(a', V'), U" = e(a", V') with (I' U U" c U0 and V' c V. Observe that V' \ P is connected and U"

E

a"

therefore 0

\

P) C U' fl U"; contradiction.

is continuous. Take a E U fl dX and let yo := Now we prove that 0. Hence V) and let U1 := e(a. V,). By (*) U1 fl Uo V1 Wehavetoshow \P) CU1. C

Take

=4'

that if b E Un dX is such that

= E

U2=e(b,V2)withU2 cUoand V2 C Hence 1)2

Indeed, by (*) there exists a

Vi,then U1 E

U1flU2.

V1.

C Ui, which shows that U1 E *4'

(b) First, observe that if a

E

and U is as in (a), then U C X. Consequently.

open in K. By(:). the mapping

p = q o w. Hence p

'49

is

is locally homeomorphic. Recall that

locally homeornorphic and therefore (X, p

(K, p

Riemann region over C". Obviously,

is a

—÷ (Y, q) is a morphism.

maps biholomorphically E fl U onto

Observe that if U, V. P are as in (a), then

*49

*49

*49

P. Thus E

x

.49)

In particular, X is dense in K, which implies that X must be

E

0

connected.

Examplel.5.12. Considerthefollowingelementaryexample. Let(X, p) := (Es, id), (V. q) := (C, id), ço := id. Then: coincides with the topological boundary 8E U (0) of E, E

=

=

(0}

E. *49

The

following proposition will show that X is

in some sense maximal.

Proposition 1.5.13. Suppose that W C X is an open subset such that: •ço(W) = V \ P. where V is an open subset ofYandP E

•

W —÷ V

\

Then there exists

P is bihalornorphic. *40

—

an open set U C X such that W C U and

U —*

V is

biholomorphic.

Proof Let Ei denote the set of all points x E X such that

• y :=

E P.

• there exist W' morphic,

V' E

V) such that

W' —÷ V' is biholo-

1 Riemann domains

38

•(qIwY1(V' \ P) C colw.

Fory

W.

C W, and that mE1 = 0. Put

is an open subset of X,

Observe that W U

P0 := P \Q(E1)

let

{V(V'): V' E

0(y)

V)),

(V') that contains the set

where V(V') denotes the connected component of

Notice that Po is a relatively closed subset of P. Step

1°. a(z) E

ax,

z

P0:

P0. It is clear that a(yo) is a filter basis satisfying conditions (2), (3), Fix a (4) of Definition 1.5.1 (urn = yo). Suppose that a(yo) has an accumulation point Xo E X. Obviously, co(xo) = yo. Observe that xO E W. Otherwise, we would find W1, W2 V1, V2 E V) such that V2 V1, W1 fl W = 0, W1 —+ V1 is biholomorphic, and (p(W2) = V2. Put - 1(V2 \ P) C U2. Since W2 fl U2 0 (xo is U2 : = V( V2). Observe that W3 := an accumulation point of a(yo)), we get W2 C U2. Fix an xl E W3 and connect this point with xo inside U2 by a curve y. Then o y lies inside V2 although y crosses W1 \ W2; contradiction. V) such that W' —+ V' is Now choose W' E E biholomorphic. We will show that P) C W. Fix an x' E W'fl W and \ suppose there exists an E W' \ W with (p(x") E V' \ P. Connect x' with x" by a curve y inside W' \ (cplw'r'(V' (1 P). Then q o y is a curve inside V' \ P. So o o y) is well defined. By the identity principle for liftings both curves are equal. In particular, x" = y(l) E W; contradiction. This shows that xo E Zi; contradiction.

Define

U := W U E, U {a(z) : z

Po}.

We claim that U satisfies all the required conditions. Step 2°. U —+ V is bijective: The suijectivity is clear. Take x'. x" E U, x' x", such that = =: The only case which not obvious is that when x', x" E E1. Then there are disjoint neighborhoods W' E V) such that maps biholomorphically and W" E and V' E W' onto V' and W" onto V'. Take a point y E V' \ P. Then y has two preimages in

is

W; contradiction. *4,0

Step 3°. U is an open subset of X First observe

that U

thin and the mapping

C X. Indeed, the set Po as a relatively closed subset of P is

U \ a x —a. V \

Po

is biholomorpluc.

Thus, it suffices to show that U is an open subset of X

1.6 Union, intersection, and direct limit of Riemann domains

V2

39

V), Po. Choose V1. V2 E Let a(yo) be one of our filter basis with yo V1, and let U1 := V( V1). Take as a neighborhood of a(yo) the following set

U :=

fl

will prove that U c U. First we prove that U1 C W U E1 =: W. Otherwise, we can find points x' E Uj fl W, x" E U1 \ W. Let y be a curve in U1 connecting x' and x". Let to be the first parameter such that y(to) E dW. Put Observe that o y is a curve El. It is clear that E y(to). Then e V1 \ P. in V1 starting at the point and V3 E V \ P) := P. Choose W3 Suppose that 0, we get W3 —f V3 is biholomorphic. Then, since W3 fl W such that W3 C W (use the identity principle for liftings); contradiction. We

f :=

So we know that yt E

P

fl V. Now, as in Step 1°, we show that x' E

E1:

contradiction. Hence weknow that U1 C W.

Un 3X. We want to show that a E

U. Put y' :=

=

Of course, then fl V2. V1. Observe that y' E P y' E For, otherwise y' E V1 \ P. Hence there exists exactly one point w' E W with p(w') = y'. We claim that in such a case w' must be an accumulation point of a. W) and U" = e(a, ço(W')). Observe that if For, fix a U' E a. Let W' E We already know that U1 C W. U" fl W 0, then U" = W'. Recall that U1 E 0 and hence Consequently, U" fl W 0. Since mt E1 = 0, we get U" fl W U" = W'. In particular, U' fl W' 0. Take a E V2

Assuming that y' E p(Ej) also leads toa contradiction. =: y' E So we start with the information that a

Po.

We claim now that

a(y').

Otherwise there are U' E a and U" = V(V") E a(y') with U' flU" = 0. We may assume (cf. property (*) from the proof of Proposition 1.5.11) that U' = e(a, V") and that V" c V1. In particular, U' C U1. Hence U' = U"; contradiction. 0 The proof of Proposition 1.5.13 is completed.

1.6 UnIon, intersection, and direct Limit of Riemann domains The union of Riemann domains. Let (X, p) be a Riemann domain over C" and let 0 ic C (9(X). Consider a family a,: (X, p) —* (Y1, q), i E 1, of ic-extensions. We would like to define an ic-extension a: (X, p) —÷ (Y, q) which behaves like the union of the above extensions (cf. [Gra-Fri 1976], [Ohg 19791). DefinItion 1.6.1. We say that ic weakly separates points in X if for any xl, x2 E X such that 3"f(xi) with xj and p(xi) = p(x2) there exist f E ic and v E

40

1

f

that

equivalently, for any (as

Riemann domains x2

there exists an f E

E X with

such

24))•

formal power series

Observe that the condition is empty if X is univalent. In particular, if G is a domain in C", then any family weakly separates points in G. If is 8-stable, then the weak separation means that the family separates points in stalks. Consequently, if is 8-stable and p E i", then weakly separates points in X if it separates points in X. We say that an s-extension a: (X, p) —+ (Y, q) satisfies condition (S) if the family weakly separates points in Y.

Proposition 1.6.2. There exists an -extension a: (X, p) —p (Y, q) satisfying (S) such that:

oaj =a, •foranyi El there existsamorphism çoj: (Y,.q1) —+ (Y,q) U,€1 'p,(Y1) = • • if/3: (X, p) —+ (Z,r) is an 8-extension satisfying (S)such that there exist o a = I E I, then there exists a (Y1, q') —+ (Z. r), I E 1, with inorphismcr: (Y,q) —÷ (Z,r)suchthatcroço1 = E l(inparticular,croa morphisms

Moreover, the 8-extension a: (X, p) —* (Y, q) satLsfying the above conditions is uniquely determined up to an isomorphism.

Observe that the morphisms'p,: (Y1, q) —+ (Y, q), i E 1, are uniquely determined. We write

(a: (X, p) —+ (Y, q)) =

Uaj: (X. p) —÷

(Ye, qi))

jEl and

we say that a: (X. p) —p

(Y, q) is the union of the family

i El.

a1: (X.p) —* Proof For any i

E I and f E

(9(Y1) be such that f1 o a = f (i.e.

let

:= faa). Let x{i}. 1€!

For(y1,i1),(y2,i2) E Yweput

(yi,ii) (in particular,

qj1

=

(y2,i2)

(vi) =

and V

f: Y —* 'p 24)

a

1'

Y1

:=qj(y).

f

C, Y,

'p,(y) := [(y, i)]._,

Recall that two formal power series

and a, = h, for any v e

We define

:=

q: Y —* C".

:

fE fE

a,(Z —

i E 1. b,(z

—

are

equal if

1.6 Union, intersection, and direct limit of Riemann domains

Observe that q

41

and / are well defined. Moreover,

qo.p,=qj, Obviously,

=

(*)

Observe that for any x X we have: (a,1(x), i1) E I, 1. Indeed, Ta1(x)fj = for arbitrary i1, I, and therefore we can define

co,(Y1).

i2) for any ii, Thus çoj1 o a,1 =

f

1

o

a: X —* Y,a :=Wj oa1. Clearly,q oa =

I oa = f, f E

p,

endow Y with the strongest topology such that all the mappings q,, i I, are continuous. In particular, by (*), the mappings q, f are continuous. Since Y = is connected and a(X) C for any i 1, we conclude that UE1 We

Y is connected.

We will prove that the mappings

Fix anii

iE

1,

are open.

IandanopensetUi C Y,1. Wehavetoprovethatforeachi2 is open in Y,,. Fix an i2 E U2 := {y2

1.

is open in Y12. Fix a y2° E U2 and let y? U1 be such that Let 0 < r < Recall that q,1 dy,, = let

l?y,

r) be such that q1 (Yl) = T1 f11

r) C U2. Take a E Then for any f

=

=

Ithe

(y2,i2)}

Y12: 3y1€U1 :

lPy,1 (vp, r) C Ui. We will show that Py,,

E

We want to show that the set

ia).

be such that r) and lPy,, we have:

=

(here we identify the Taylor series T)ofik with the function z

k=

1,

2). Thus we have proved that

—÷

is open.

Now we will prove that the topology of Y is Hausdorif. Consider (yr. ii) Since and çoj, are open, it suffices i2). Let 0 < 2r <min{dy11 dy1, to show that the sets Uj := çoj1 r)) and U2 := r)) are disjoint provided r > 0 is small enough. The case q1 (y?) q,, is obvious. Assume that = q,,(y20) and suppose that for some Yj E j = 1,2, we have (y2,i2). Then we get

=

(Yo)(TYIfiI) =

1,2) =

contradiction. We have proved that Y is a Hausdorif space.

=

q,,

i €1. Thisshowsthatforanyi €1 andforany (u))t =

univalent open set U C Y1 the set p (U) is univalent (and open) and that (q

1 Riemann domains

42

Recall that Y = UIE! p,(Y1). In particular, (Y, q) is aRiemann domain (Y1. q) —p (Y, q), i E 1, are morphisms. Note that f o for any univalent open set U C }'j. = 1: o Hence f E 0(Y). Thus a: (X, p) —+ (Y. q) is an 4-extension. p,

over C" and

Nowletzi =

f

4. Then

f

= Tz2fforany

eY

= =

° and, consequently, (1 ° 4. Thus zI = Z2. We have proved that the 4-extension a: (X, p) —+

(Y, q) satisfies (S).

Now let /3: (X, p) —+ (Z, r) be another 4-extension satisfying (S) such that o a, i E I. there exist morphisms *,.: (Y,, qj) —+ (Z, r) with /3 = Define a: Y For, take ii) ((/3*)

a([(y.

:= *1(Y)- Observe that a is well defined. It is enough to show that for any! E 4. Since /3* = a7 o we get Z,

ia).

1(f)) =

T

)

(f)) =

= Tr,fj2 =

1(f)) =

1(f))•

We have a o p, = i E I. Hence a is continuous, Obviously, r o a = q. Thus (Z,r)isamorphism. a: (Y,q) Suppose that

a': (X,p) —* (Y',q'),

—+ (Y',q'),

i El,

is another system with the above properties. Then there exist morphisms

a: (Y,p)

(Y'.q'),

a': (Y',q')

—÷ (Y,q)

such that a o cp = çp. a' o = çoj, I E I, which implies that a is an isomorphism (cf. Remark 1.4.2(b)). 0 and a' =

Remark 1.63. Observe that we can apply Proposition 1.6.2 to the family consisting of one element id: (X, p) —÷ (X. p). Then the proposition tells us that for any Riemann domain (X, p) and for any family 0 4 C 0(X) there exists an 4extension a: (X. p) —÷ (Y. q) satisfying Hence, using Remark 1.4.5(f), we conclude that in the theory of holomorphic extensions we can always assume that: • 4 is a a-stable subalgebra of 0(X),

• p E 4", • 4 separates points in X. The kernel of Riemann domains. Let (X,, Pi)IEI be an arbitrary family of Riemann domains over C". We are going to define the 'intersection' of this family. 25)

Cf. the definition before Proposition 1.6.2.

43

1.6 Union. intersection, and direct limit of Riemann domains

Proposition 1.6.4. There exist a Riemann region (X, p) over C" (including X = 0) and morphisms ca: (X, p) —÷ (X,, p), I I. such that: (a) the fwnily

separates points in X; (b)for any Riemann region (Y, q) over C" for which there are morphisms

i El,

*: (Y.q)—+ (X1,p,), there exists a morphism

a: (Y.q) —÷ (X.p) such that

o

a=

i

(c)dx =

I; oWl) (if X

the

0),

system ((X, p), (çoj )IEJ) is uniquely determined by (a) and

(b)

up to

an isonwrphism.

We write

= intl')(Xj, 1)

((X. p).

iEI and

we say that (X, p) is the kernel of the family (X,, Pi)iEI.

Proof

Put

X := {x = (x,),E,: X

X,

E I,

p(xj) =

r(x)

:

a, i

I,

lflf(dX, (X1)} >0). in!

0 is elementary. We only need to check (b). Suppose that (Y, q), (Y, q) —* (X,, p), 0, is a Riemann region such that there exist morphisms i El. > dy,i El. Consequently, E X for any y E Y. Hence Y = 0; contradiction. The case where X = Y

Thus, we assume that X

0. For x E X and for 0 < r
U(x, r) :=

z

IP(p(x), r)}.

Note that U(x, r) C X. The family (U(x, r))xnX, O 0 is sufficiently small. If p(X) = p(x'), I with x10 Consequently, by the identity principle, then there exists an io

for any z

!?(p(x), r) with r

min{r(x), r(x')}. This

shows that U(x, r) fl U(x', r) = 0. Moreover, PIUx.r is a homeomorphism onto P(p(x). r). Thus (X, p) is a Riemann region over C".

44

Riemann domains

r) = U(x, r), x E X, 0 < r
Notice that dx(x) = r(x) and For io I define

X 3 (x,)IE, —+ It is clear that

E

is a morphism of (X, p) into (X10,

points in X. Let (Y. q) be a Riemann region over

and the family (c°j)jEx

separates

and let

iEI. be rnorphisms. Then

(Y))jEJ

X for all y

Y. Consequently, the mapping E

is

a morphism of(Y, q) into (X. p) with qij o a =

that ((X', p'), (W)IEJ) Then there exist morphisms Suppose

is

E

I.

another system with all the above properties.

a': (X,p) —p (X'.p')

a: (X',p') —p (X,p),

p,i 1.

i

= poa' =

1.

Hence, by (a),coa' = idx. The same argument shows thata'oa =

U

Remark 1.6.5. (a) Observe that if G1 C i 1, is a family of domains, G, then ((S, is the kernel of (G1, idG, (b) Let a: (X1, p) —* (Yj, qi) be a morphism, i I, and let ((X, p), (Wi)ieI) := mt 1)(Xj, p),

:=

q).

((Y, q), (lfri)IEI) :=

tel

jEt

Then, by Proposition 1.6.4 (applied to the morphisms a1 o pj: (X, p) —+ (Yj. qj). (Y,q)suchthat i/i oa a3 i E I), thereexistsamorphisma: (X,p) i E I. Observe that if all the morphisms a,, i E I, are injective, then a is injective. In particular, if X is a subdomain of Y, and a = idx1 i I, then the kernel of (X1, qi),el may be realized as an open subset of Y.

The following Riemann region (0W,

will be very useful in the sequel (cf.

the proof of the Thullen theorem 1.8.4).

Exainplel.6.6. Let I :=((9,ir), I (cf. Example 1.3.1). Then we define the sheaf of I-germs of holomorphic i fri nctions

:= mt 1"I(Xj, p,). IEI

45

1.6 Union, intersection, and direct limit of Riemann domains The region ((9(1),

may be described also directly:

Let

:= ((U, (fi)IEI): U For (U.

(fi)iEI), (V.

is

i

aE

1}.

we put 2WE!J3(a): W C

(V.

(U,

0(U),

E

E

Un V.

i E 1;

an equivalence relation. Put

The class

= [(U. is called the 1-germ of (ft)aei

at a.

Finally, we put

U For any 1-germ f

E

(!)

x la}

a (f is as in (!)). (as in (!)), the set

U := (([(U, (fi)iEi)lz, z): Z E U)

(!!)

isa univalent open neighborhood of (f. a); the projection ir maps horneomorphically U onto U. For any i E I we define a function IF,: —+ C. F,((f, a)) := f,(a) o on U, the mapping IF1 is holomorphic on the (f is as in (!)). Since F1 = region Notice that

= inf{d(Taf,): i €1),

(9(1)

aE

C's.

Connected components of (9W are universal models of hotomorphic extensions, namely: C (9(X). We Let (X, p) be an arbitrary Riemann domain over define a morphism

a= (1 :=

(X, p) —+

Namely.

a(x) := ([(p(U), (f o

p(x)),

X

E X,

46

1

Riemann domains

where U C X is an arbitrary univalent neighborhood of x (it is clear that the definition o a = p. Moreover, a is continuous. is independent of such a U). Obviously,

Indeed, if U c X isaunivalentneighborhoodofapointa,thena(U) = p(U)(p(U) sense of(!!)). Thus a is a morphism. Moreover, Fj o a = f for any f E 4. Consequently, if X that contains the set a(X) and fl := denotes the connected component of is taken in the

then

a: (X,p) —+ (k,15)

is an 4-extension. We will see (Thullen theorem) that this extension is in some sense maximal. Recall that

= inf{d(T0f): f E 4),

a E X.

Observe that, after standard identification, the mapping a may be rewritten in the form

a(x) := (Tp(x)(f Proposition 1.6.7. Let (X. p)

be a

X E X.

0

Riemann domain over

a: (X,p)

and let

(Y,,q,)

be an 41-extension, i E I. Put

((ko,fr),('pt)IE,) := iEl

Let x: (X. r) —p (Xo. j3) be a morphism such that q, o x =a,, i E I (cf Proposilet X denote the connected component of Xo that contains x (X). tion 1.6.4). Then

x: (X,p)—+

(X,13)

is an 4-extension with 4 := U El

is bijective if the morphism x: (X, p) —+ (X. (t)for any x', x" E X with x' x" and p(x') = p(x") there exists an io E I such that Inf{dy,

Then go E 0(X) and let g E (9(Y,) be such that g oa1 = f. Consequently. x: (X, p) —p (X, j5) isan 4-extension. = inf(dy, i E I). Recall (Proposition 1.6.4) that

Proof Take an f E and

oat: 1€!). 4,

= goa, =

Assume that x is bijective. The above formula shows that (t) is true. Moreover, since

pjoX=ai, iEI,

(+)

and the family ('Pi)IEI separates points in X (Proposition 1.6.4), we get (t). Now assume that (j-) and hold. Then, by Remark 1.4.2(t), is surjective. Using once again (+), we easily conclude that x must be also injective. 0

47

1.6 Union, intersection, and direct limit of Riemann domains

The direct unit of Riemann domains. Let (1. • i i,

be

a directed set, i.e.

•(i • Yi,J€I 3kE,: i

k, j

k.

Assume that we are given a family (X,, Pi)EJ of Riemann domains over C0 and a family of morphisms aj.j: (X1, j, such that —÷ (X3, p3), i

•

=

i

E

I,

= ak,, I

• ak,J

j

k.

Proposition 1.6.8. There exist a Riemann domain (X. p) over

and morphisms

q',: (X,, p,) —÷ (X, p), i E 1, such that: j (in particular, for i j); I (a) çoj o aj,j = C (b) X = UIEJ (c) for any compact set K c X there exist Jo E I and a compact set K30 C X10 such that K is homeomorphic; (d)for any pair ((Y, q). )iEI), where • (Y, q) is a Riemann domain over C",

•

(X1, p1) —* (Y,q) isamorphism, i El, and 1/tj = there exists a morphism a: (X, p) —* (Y, q) with a o p, = i (e) Any

pair ((X, p),

i

I.

satisfying (a), (b), (d) is uniquely determined up to

an isomorphism. We

write

((X, p), (coi)iEI) = lim((X,, Pi)iEI, and

we say that ((X, p), (Y'j)jE,) is the direct limit of ((X1, Pi)iEI,

Pmof Let X := (x =

: x, E

X,,

iE

I,

:

=

10

i}.

Forx = (X,)jEJ, y = (Yi)iEI E X let

=yj, Clearly, '--" is an equivalence relation. Put

:=

X

Observe that ifx, =

p(x) :=

x = [(X()jEJJ...

IEI

for io

i,

then

p,(Xj)

=

X.

pì(cx1,,0(x10))

=

p10(Xj0) and

therefore p: X —+ C" is well defined.

Foranyi El • x3 E

is

X —* Xby the

arbitraiy, otherwise.

formula

:=

where

48

I

Riemann domains

=p,,i El. k we where for j X, we get çoj(aj,j(x)) = have xk = (Yk,J(aJ.i(x)) = crk.i(x). Consequently. Pj o = 'pa, which gives (a). This x= X, where If x = proves (b). We define the topology of X as the strongest topology for which all the mappings 1, are continuous. 'pt, i We prove that the mappings çoj, i E I, are open. Fix an i and an open set U C X,. We want to show that the set 'p(U) is open. This means that for each j E I the set 'pJ1('pi(U)) is open in X1. Fix a j. Take an arbitrary a E V := Vj and let I with i io, j b E U be such that 'pj(a) = 'p,(b). Then there exists an io E Let W be an open neighborhood of a in X1 such that such that cr0,j(a) = It is sufficient to show that W C V. Let .r E W and let y E U C k. Then cik,J(x) = Let io = be such that = V. and consequently, x E Thus = E crk.io(aio.i(y)) = ak,(y). the set çp,(X,) is open in X. In particular, for each i is an open univalent set, then the set 'p1(U) is open and univalent in X If U c X, Consequently, p is a local homeomorphism. and = p, o To prove that (X, p) is a Riemann region it remains to show that the topology of X is Hausdorif. Suppose that p, (x') = y' y" = çoj (x") 26), We are going to show that there exist open univalent connected neighborhoods U', U" of x' and x", respectively, 0. such that p(x") and, consequently, we can take U' and p(y"), then p,(x') If p(y') U" such that p(U') fl p(U) = 0. If pu") = p(y") =: zo, then let U', U" be disjoint open univalent and connected neighborhoods of x' and x", respectively, such that p(U') = p(U") =: D. Define Then s', s" are two sections of X over D 5" := o oj 0 by the identity principle for liftings, s'(D) fl s"(D) = 0. and s'(zo) s"(zo). Hence, = 0. This shows that We have proved that (X. p) is a Riemann region over C". Observe that by (a) and (b) X is connected. is as in (d). We define a(x) := lim,E, Suppose that ((Y,q), i, then io x = Rxi)iEIh E X. The limit exists because if x, =

j,

If i

=

then for x

= qoa=

j0

1.

Moreover. q(a(x)) =

p. A similar argument shows that a oq'j =

i

I.

In particular, a is continuous, and therefore it is a morphism.

We pass to the proof of (c). Let K C X be a compact set. Then, by (b), there exists an io such that K C 'p10(X10). Since çoj0 is locally homeomorphic, we can find K. Put L := a,10(L10), io i. Observe a compact set L0 C X,0 with 'p10(L,0) that 'p(L1) = 'p10(L10) K. It suffices to prove that there exists a J0 E 1, io Jo, is injective. such that

j

26)

Observe that, by (a), for any y', y" E X, there exists an i E

I such that y'. y" E

49

1.6 Union. intersection, and direct limit of Riemann domains

Suppose that for any io = x7 E L,0 with j there exist points pj(aj,,0(x7)). In particular, p10(x) = (p10(x7). Since L0 is compact there exists a —+ (I, "subsequence" t: (I'. (i.e. for any ko E I there exists a I' such i') such that = r(i') for any i' E I' with E L1,, that !co = Consequently, there exists xç(1.) = x) L0. Clearly, I with io ko such that = Let U', U" be open a lco univalent connected neighborhoods of and respectively, such that ako,jo (U') = =: V. Let k( E I' be such that k0 and U', U".

We get: 0

=

=

0

ar(k')ko 0

=

0

=

0

Recall that çok0 is injective on V. Hence

=

0

= ako ,0(x

(k

Consequently,

0

=

aT(k')kO

=

oak0

contradiction.

Suppose that (X', p') and (X,, —+ (X'. p'), I E I, is another system with the same properties as ((X, p), Then, by (d). there exist morphisms a: (X, p) —p (X', p'), a': (X', p') —p (X, p) such thata op, = = I. Hence (a' 00)0 (p = 0 = i E I. Consequently. a' 00 = id on (X1), I

I

I,andtherefore,by(b),a'oa =

Thesameproofshowsthata oa' =

This gives (e).

U

(I',

Exercise 1.6.9. Let (I,

j')

• (I'

be directed

r(j')),

(z(i')

sets and let r: I' —÷ I be such that:

•VIOEJ Let

(X1, p), I e I, be a family of Riemann domains over C" and let

as,,:

(X,, Pi)

(X3, p1),

I

be a family of morphisms such that i

Put

j

k.

(X1', Pi') := (Xt(I'). Pr(i')),

((X', p'),

:= ar(J')r(,'), I

:= iii((X,,

j'.

I'.

Let

)i'E,",

Prove that there exists an isomorphism a: (X'. p') —* (X, p) I'

idx'.

with

a 0 (p, =

50

1 Riemann domains

1.7 Domains of existence Recall that in the classical theory one considers domains G C C'1 and families.8 c (9(G) such that not all functions from .8 can be holomorphically continued (via power

series expansion) across a boundary point of G. We will discuss an analogue in the category of Riemann domains.

Definition 1.7.1. Let (X. p) be a Riemann domain over C'1 and let 0 We say that (X. p) is the .8-domain of existence if

dx(a) = iflf(d(Taf): f E

.8},

C (9(X).

a E X;

equivalently, for any r > dx(a) there exists an f E .8 such that d(T0f)
Obviously, if (X, p) is an .8-domain of existence, then (X, p) is a 7-domain of In particular, any .8-domain of existence is a domain existence for any family 7 of existence. Remark 1.7.2. (X, p) is an .8-domain of existence if for any .8-extension

a: (X,p) —" (Y,q) we have dy o a In particular, by Remark 1.4.2(f), if (X, p) is an .8-domain of existence, then for every .8-extension a: (X, p) —÷ (Y, q) the mapping a is suijective. Indeed, assume that (X, p) is an .8-domain of existence. Then, by Remark 1.4.5(e), we get

dx(a) = inf{d(Taf): f E .8} > dy(a(a))

dx(a),

a E X.

Consequently. dy o a

Conversely, let a: (X, p) —+ (X.

be the .8-extension constructed in Exam-

ple 1.6.6. Then

a€X.

inf{d(Taf):

Remark 1.7.3. (a) If (X, p) = (G, jdG), where (3 is a domain in C'1, then ((3, idG) is an .8-domain of existence if

there are no domains Go, G C C" with 0 # Go C G fl (3, G such that for each f E 27>

Observe that if of existence for any

0$

.8

there exists an f

(9(G) with

f=

G,

f on Go.

(i.e. (X. p) = (C". Id) up to an isomorphism), then (X, p) C (9(X).

(*) is

an

1.7 Domains of existence

51

r for

G and r > dG (xO) such that dG(xO)), G := P(xo, r), and fz) :=

Indeed, suppose that there exist xo

any f e

&

Put

G0 :=

IP(xo,

G.

z

Thus (*) is not fulfilled.

Conversely, suppose that (*) is not fulfilled and let Go, G be as in (*). By the identity principle we may assume that G0 is a connected component of G fl G. Take := r. Then for any! E we get anxo E G0 such that dG(xO) =

= r.

>

(b) Any domain G ç C' is an {G

of existence, where

—: a

z

G}.

C' 28) is an s-domain of existence, where

(c) Any fat domain G

z—a In particular, any fat domain G C C' is an

fl (9(G)-domain of existence,

where

O(G):=

U

O(U)IG.

u LI open irCt

of existence, I

(d) Let (X1, p,) be an

((X0, p),

1. Put

:= iEI

1€!

Let X be a connected component of Xo. Then (X, p) is an s-domain of existence. For, by Proposition 1.6.4, we get

dx(x) =

i f€ 45}.

i

Observe that if

I. then

weakly separates points in X,, I

weakly separates

points in X. In fact, if x' x", p(x') = p(x"), then there exists an i0 E I with Thus pj0(x"). Hence there exists an f E with T,1,,0(x')f o

o çoj0).

(e) Let (Xi, Pj) be an

j=

X:=XIX'•-XXN, 28

A domain G in a topological space isfat if G = mG.

I

N. Put N},

52

1

Riemann domains

x XN —÷ X3 is the standard projection. Then (X, p) is an 8where m1: X1 x domainofexistence. Inparticular, ifG1 aredomainsinC',thenGi x. is a domain of existence in C". Indeed, N)

XXN(xI

j=

fE IE

=

= inf(d(T(X)

1

N}

N, then weakly weakly separates points in X3, j = Observe that if separates points in Xi x x XN. IN E 0(X). Put (f) Let (X, p) be a domain of existence and let 11 1

Yo := (XE X: 1f1(x)l <

I,

j=

N}

I

and let Y be a connected component of Yo. Then (Y, ply) Indeed, take an x E Y. If dy(x) = dx(x), then

dy(x) = If r :=

dy(x)

thereexistsaj E (1 is holomorphic in Y and

f

is

a domain of existence.

f E 0(Y)}

dy(x).

fl dPx(x, r). Consequently, N}withlfj(b)l = I. Hencethefunctiong :=

then there exists a point b E

= r.

0. Put Y : (g) Let (X, p) be a domain of existence and let fo E 0(X), fo (recall that, by the Riemann removable singularity theorem, Y is connected). Then (Y, ply) is a domain of existence. We proceed as in (f). Take an X E Y. The case dy(x) = dX(X) is the same. If fl aPx(x, r). Thus the function r dy(X)
X\

theorem R(Y) = (h) Let G be a domain in C". Assume that for any point a E dG there exists a = 1. ThenGisanJ€°°(G)domain functionf,, E of holomorphy. Indeed, suppose that G is not an G be as in (*) (from Remark 1.7.3(a)) with

of holomorphy and let Go and We may assume that G0 is a connected component of G fl G. Take an a E G fl and let be as above. I in G (cf. Proposition 1.4.5(h)) and lfa(a)l = Ifa(Z)l = 1. limGoaz...,a Then. fat Consequently, by the maximum principle, fat I. In particular, hal = i on G0;

=

contradiction. 29) 29)

It is natural to ask whether the above result holds if ía exists only for a E A, where A is a

1.7 Domains of existence

53

(i) Any convex domain G C is an .7?°°(G)-domain of existence. and fix an a aG. Since G is convex, there Indeed, take a convex domain G exists an affine function £: —+ R such that 0 ( on
Let

ER. Define L(z) Jo := eL. Then Ifol =

Obviouslyt

= et <

1

= ReL.

on G and Ifo(a)I = I. It remains to

apply (h).

Exercise 1.7.4. Let (X, p) be a Riemann domain over and let Y := X \ F, where F is a closed subset of X such that F is locally contained in a real (2n — 1)-dimensional & surface. Show that (9(Y) fl C(X)Iy = (9(X)Iy (cf. Remark 1.7.3(g)). Hint. Complete the following sketch of the proof.

First, we reduce the proof to the case where X =

+ iy0):

F = 1(z',

E°'

x

=

where Q is the square (—1. 1) x (—1, 1), E C x (— I, 1)), and ço(Q', 0) = 0. We will show that f is holomorphic in a neighborhood of 0. the function For each z' E is holomorphic on Q. To prove that

J is holomorphic in a neighborhood of 0 it suffices (by the Hartogs Lemma; cf. Lemma 1.8.14) to show that there exist 0
Go

Lemma 1.7.5.

+ iy0:

xnl


of existence

(X, p) is an

fe 8}forx

< IYnI


exists

a dense

F.

subset A C X

E A.

Letx0 E Xandletr > > dx(xo). rforanyf E there exists an open neighborhood U of xO such that d(Trf) > ro > dx(x) for

Proof

Then

all f E 8 and x E U;

D

contradiction.

F c 0(X) be a natural Then the following conditions are equivalent: PropositIon 1.7.6. Let

Fréchet space (Definition 1.2.4).

(X, p) is an F-domain of existence;

(i)

dense subset of 8G. It is clear that the answer is positive as long as A fl G fl aGo 0 for any G and G0 as above. This holds if, for mstance. 8G is of class Ct. To see that in the general case the situation is much more complicated, consider the following example (due to W. Jarnicki):

•

:=

(rk(1) —ra,rk(I)

• k(i) is the lowest index such that rk(j)

•e>

0,e,

Q, is so small that

c

(0,

1),

= 0, and

< 1/4.

Observe that a = to, 1j x [0,4). Take G := x (2. 3). Then any connected component G0 of G nO must be of the form x (2, 3) for some I, and therefore G fl 8G0 has an empty interior in 8G.

54

Riemann domains

1

(ii) The set 91(F)

:= {f

E

F: (X, p) is the domain of existence of f}

is of the second Baire category in F;

(iii) ¶11(F)

0.

Proof Obviously. (ii) (i). : (iii) (I) (ii): For a E X and r > 0 let Fa.r := {f E F: d(T0f) r}. It is clear that Fa,r is a vector subspace of F. Let r denote the topology of F. We endow the space Far with a topology Ta, in such a way that: Fa.r fu if fo E Fa,r jfl ,•

:

fo in r and Ta Ta In in the topology of the locally uniform convergence on P( p(a), r). One can easily verify that endowed with this topology is a space. The identity embedding id: (Fa,r, Ta,,-) —÷ (F, r) is continuous. Hence, by the Banach theorem either = F or .'Ea.r is of the first Baire category in F. Since (X, p) is an F-domain of existence, Fa,r is of the fIrst category if r > dx(a). Now let A C X be countable and dense in X. Put

(J OEA ,kEN

Then .Y( is of the first Baire category in F. By Lemma 1.7.5, F \ J( = 91(F).

0

Exercise 1.7.7. Let G1 C C" be a domain of existence of a function f1 E 0(G1),

j=I

N. Assume thatG0 := G1 n...r1GN

OandletGbeaconnected

component of Go. ForA = (A1 AN) let FA := AIIIIG +XNINIG (9(G). Prove that there exists a A E such that G is the domain of existence of

1.8 Maximal holomorphic extensions O C C" (n 2) such that any function holomorphic on G extends holomorphically to G and G is a domain of existence. We are going to discuss this phenomenon in the context of Rieman.n It is well-known that there are pairs of domains G

domains.

Definition 1.8.1. An 4-extension a: (X. p) —* is called maximal if for any 4-extension (X, p) (Y. q) there exists a morphism a: (Y, q) —* (X, such that a o = a.

Recall that in this case a: (Y, q)

(X, j3) is an

(cf. Remark

1.4.5(b)). lfX, Y arcFréchctspaces. L : X

Y

islinearandcontinuous,theneitherL(X)= Yor

L(X) is of the first Baire category in Y: cf. [Gof-Ped 19651, § 5.8.

1.8 Maximal holomorphic extensions

55

Observe that the maximal s-extension is uniquely determined up to an isomorphism.

If a: (X, p)

(X. j3) is the maximal i-extension, then we say that (X, ,3) is an 45-envelope of ho!omorphv of (X, p) and we put (X, jä) =: E((X, p), 45) or shortly

X =:

C(X. 45). 31)

If (X, is an (9(X)-envelope of holomorphy of (X, p), then we shortly say that p)) or (X, j) is an envelope of holomorphy of(X, p) and we put (X, ,5) =: shortly X =: We say that (X, p) is an 45-domain of holomorphy if for every 45-extension a (X, p) —÷ (Y, q) the mapping a is an isomorphism. If (X, p) is an (9(X)-domain of holomorphy, then we simply say that (X, p) is a domain of holomorphy. Remark 1.8.2. (a)If a: (X, p) —* (X, ,3) is the maximal 45-extension, then (X, ,3) is an of holomorphy. Indeed, suppose that (X, ,3) —+ (Y, q) is an 45w-extension. Then

floa: (X.p) —÷ (Y,q) is an 45-extension (see Remark 1.4.5(a)) and therefore there exists a morphism a: (Y, q) —÷ (X, j3) such that a o o a) = a. Consequently, is an isomor-

=a). then .7('°(X) = R(X)a (cf. Remark 1.4.5(h)). Hence (X, j3)

(b) If 45 = is an

of hotomorphy. (c) Observe that if f E ( E xC), then f = 0. Therefore, any (E x C)-function extends holomorphically to C2. On the other hand, for any 0 R the function is

fo: E xC —÷ C, fo(z) := (Zi —e'0)', holomorphic on E x C and is not holomorphically extendable across

z2).

Z2 E C.

Let s1 :=

<(1 GE

and let e :=

2

be

a sequence of positive numbers

Put

:= r2(o.

i)) UU{z E C2: Izil <Sf,

1z21 — ii <ej}

1=1

extends holomorphically toE xC. 0 E IR, In particular, if we choose the e"s sufficiently small, then f0 E of G but it is not an where G := Hence E x C is a maximal (E x C)-domain of holomorphy. Then

31)

is a Reinhardt domain and any f

Recall that

p).

is

uniquely detennined up to an isomorphism.

56

I

Riemann domains

Proposition 1.8.3. Under the notation of Proposition 1.6.7, if for any i E I the Se-extension a,: (X. p) —÷ (Y1, q) is maximal, then the S-extension

x: is iE (X, p) I, (X. p) —+ (Y. q) is an S-extension and therefore there exists a morphism (Y,, q) such that yj o = a,, i E I. By definition of the kernel of y,: (Y. q) a family of Riemann domains, there is a morphism a: (Y, q) —÷ (Xo, ,) such that oa= i E 1. Hence, o (a o fi) = y, o fi = a = çoj o x' i I, and, finally, 0 by Proposition 1.6.4(a), a o = x. The following result shows that in the category of Riemann domains (and, consequently, in the category of Riemann regions) over C" the maximal holomorphic

extension (w.r.t. any family of holomorphic functions) exists. One should mention that the maximal holomorphic extension exists also in other special categories of complex analytic spaces (but not in the whole category of complex analytic spaces) cf. [Vig 1982] for detailed discussion.

Theorem 1.8.4 (Thullen theorem). Let (X, p) be a Riemann domain over C" and let 0 S C (9(X). Then (X, p) has an S-envelope of holomorphy.

Proof I" version: Let a,: (X, p) —÷ (Y1, q), i E I, denote the family of all S-extensions (this family is not empty because id: (X, p) —* (X, p) is a trivial S-extension). Let (a: (X, tion 1.6.2. Let i

-

p) —* (X, j3)) =

(X, p) —+

qj)); cf. Proposi-

(Y1, q,) —+ (X, j3) denote the morphism such that

o a, =

a,

El.

(X. p) —+ is maximal. Let We only need to show that a: (X, p) —* (X, (Z, r) be an arbitrary S-extension. Then there exists an i0 E I such that fi = a0, Then a: (Z. r) —+ (X, fr) is a morphism such (Z. r) = (Y,0. q0). Let a := that a o = a. 2nd version: By Propositions 1.6.7 and 1.8.3 it suffices to consider only the Define a1: X —÷ (9, (X, p) —+ ((9. it) is a morphism and F o aj

case where S = {f}.

aj:

32)• := It is clear that = f. Let X be the connected com-

ponent of (9 that contains a1(X). Then a1: (X, p) —+ (X, it) is an {f)-extension. Suppose that (X. p) —p (Y,q) is another {f}-extension. Let g E 0(Y) be such that g o = f. By the same method as above we construct a (g}-extension ag: (Y, q) —÷ (Y. it), where Y denotes the connected components of (9 that conObserve that ag o fi = a1. Indeed, tains

=

°

= Tp(x)(f o p;-') = a1(x),

x E X.

Recall that t9a may be identified with the ring of the Taylor series at a; cf. Example 1.3.1.

57

1.8 Maximal holomorphic extensions

In particular, ag(9(X)) = aj(X) and therefore Y = X. Thus is the morphism we are looking for. version: It suffices to prove that the .8-extension

(Y, q)

(X, it)

a.3: (X,p)—+ (X.15), constructed in Example 1.6.6, is maximal. Let fi: (X, p) —b (Y, q) be another .8-extension. By the same method (Exam-

ple 1.6.6) we construct a morphism

(Y,q) —b it (0 , it ) C X (up to an isomorphism). ((9

,

Remark 1.8.5. Observe that a,

thatdk(a.3(x))=inf(d(TXf):

is

f

=

). Moreover,

Consequently,

0

injective if.8 weakly separates points in X. Recall .8),x E X.

E

Proposition 1.8.6. Let a: (X. p) —* (V. q) be an .8-extension such that (V. q) is an of holomorphy. Then the extension is maximal.

Proof Let &: (X, p) —k (X, j) be a maximal .8-extension (which exists because of the Thullen theorem). By definition, there exists a morphism a: (Y, q) —k (X. ;3) such that a o a = &. Then a: (I', q) —÷ (X, 3) is an Since (Y, q) is 0 an of holomorphy, a is isomorphic.

Examplel.8.7. ForO
Zn) E

Iz.iI
n

I

— I, p < Iz,,I <

Observe that = E". is biholomorphic to Tr/a,p/fi. Then i.e. Id: (Tr.p. Id) —+ (E'7, id) is the maximal holomorphic extension. Indeed, is a domain of holomorphy (cf. Remark 1.7.3(e)). If f E then f can be represented by a power series f(z) = (the series zE converges locally normally). It is well known that such a series must be convergent in E'1. Thus f extends holomorphically to E'7. Put Tr,p :=

Remark 1.8.8. Recently. E. M. Chirka proved the following essential generalization of the above result for the case n = 2 (cf. [Chi-Ros 1998], [Chi 1999]). Let p: E —+ E be an arbitrary continuous function. Then any function holomorphic in a connected neighborhood of the set (dE x E) U {(z, p(z)): z E E} extends holomorphically to a neighborhood of E x E Observe that Example 1.8.7 reduces to the case

=

0.

1 Riemann domains

58

The result is not true for continuous mappings ço: E —' E2, i.e. there exist a continuous mapping q: E E2 and a function f, holomorphic in a neighboihood of the set (8E x E2)U{(z, z E E}, such that! does not extend holomorphically to E3 — cf. [Rosa 1998].

The following proposition will be useful in § 1.9.

Proposition 1.8.9. Let fi: (Y, q) —÷ (Y. 4) bethe maximal 7-extension (0 7 C 0(Y)). let (X. p) be a Riemann domain over C", and let r: X —f I be a locally biholomorphic mapping. Then fi o r: (X. q a r) —+ (Y. 4) is the maximal r*(T). extension.

In particular

a: (X, p) —f (Y. q)

is an 4S-extension and if

$: (Y.q) —÷ (Y,4)

then fib a: (X,p) —÷ (Y,4)

is the maximal

is the maximal 8-

extension.

Proof ltisclearthatfior: (X, qor) —÷ (Y, 4) isa

Toprovethe maxitnality we may assume that fi: (I, q) —÷ (Y, 4) is the maximal 7-extension

i.e. V C (9w, fi =

realized in

(as in Example 1.6.6). Let

ar(T) (X, q o r) —+ (X, be the maximal r * (7)-extension realized in that fi o r = Indeed,

b8(r(x)) = (Tq(T(x))(f o

We only need to prove

o r) o (q o

r);'))gEr.(T) = at.(T)(x)

Proposition 1.8.10. Let (X, p) be a Riemann domain, 0 following conditions are equivalent: (1)

(ii)

E

o

0(r(fl)

u

C (9(X). Then the

(X, p) is an i-domain of holomorphy; weakl,v separates points in X and (X, p) is an s-domain of existence;

(iii) there exists a dense subset A C X with A =

•for any x', x" E A with x'

(p(A)) such that: x" and p(x') = p(x") there exists an f E

such

that

•for onyx E A we have dx(x) =

fE

Proof By Remark 1.8.5 we get (1) (ii). (iii) (ii): Lemma 1.7.5 implies that (K, p) is an i-domain of existence. To prove that weakly separates points in X suppose that for some x', x" X with :

x"andp(x') =

=

(z)) for z from an open neighborhood of p(x'); contradiction.

0

59

1.8 Maximal holomorphic extensions

PropositIon 1.8.11. Let (K. p) be a Riemann domain and let F C (9(X) be a natural Fréchet space. Then the following conditions are equivalent: (i)

(X, p) is an F-domain of holomorphy;

(ii) The set

91(F) :=

{f F:

(X, p) isan in

(iii) 91(F)

F;

0.

Proof Obviously, (ii) (iii) (i). (i) (ii): For x', x" E X let Observe = := {f F: that is a closed subspace of F (here we use the fact that r is stronger than the topology of locally uniform convergence). Thus by the Banach theorem, either is of the first Baire category in F. By Proposition 1.8.10, if = F or :

x" and p(x') = p(x"), then is of the first category. Now let A c X be a countable dense subset such that A = p1(p(A)) (first we take a dense countable subset Ao C p(X) and then we put A := Put

x'

:=

U

U Fa,dx(a)+1/k U aEA,kEN

x',x"EA. p(x')=p(x")

where Fa,r is as in the proof of Proposition 1.7.6. Then R is of the first Baire category

in F. By Proposition 1.8.10 F \

= 91(F).

D

Remark 1.8.12. Let F C (9(X) be a natural Fréchet space such that F separates points of X. Then for any at most countable set A c X the set

(f F: hA is injective) is of the second Baire category in F. Indeed,

{f E where

F:

fIA is not injective}

=

U

x'.x"EA:

:= {f E F: f(x') = f(x")}:

F is of

the following natural problem: Let (X, p), (Y. q) be Riemann domains over C (9(Y). Put

12= thE

(9(X x Y): VXEx:

f(x,.)

E

F

C

7.

f(., y) E

60

1

Ricmann domains

(Y.q) —*

Supposecr: (X.p) —p

sions, respectively. We ask whether the morphism

ax

(Xx Y, p x q) —p (Xx Y,15 x

is a V-extension. The general answer seems to be not known sition 1.8.15. we will discuss the following two particular cases:

Below, in Propo-

7 = 0(Y), fi = id, = (9(X), 7 = (9(Y). Lemma 1.8.13. Let(X, p), (Y, q)beRiemann domains and Ctm, respectively. (a) Assume that (X, p) is an s-domain of holomorphy and (Y, q) is a 7-domain Then (X x Y. p x q) is a 'U-domain of holomorphy, where U := of denote the standard projections. (b)Assumethata: (X,p) —÷ (X,j3),,8: (Y.q) —÷ 7-extensions, respectively. Then a x fi: (X x Y, p x q) —÷ (X x Y, J5 x maximal 'U-extension, where 'U is as in (a).

is the

Proof (a) Take (x', y') (x". y") with p(x') = p(x") and q(y') = q(y"). We may assume that x' x". Then there exists an f E d such that and ° By Remark 1.7.3(e). (X x Y, p x q) is a hence T(X',V)(f o ,rx) # ti-domain of existence. Thus, by Proposition 1.8.10, (X x Y, p x q) is a 11-domain of holomorphy.

-

It is clear a x fi: (X x Y. p x q) —÷ (X x Y, f x is a U-extension. By (a), (X x Y, ,ô x is a domain of holomorphy w.r.t. the family Now the (b)

result follows from Proposition 1.8.6.

U

Lemma 1.8.14 (Hartogs' type lemma). Let (X, p), (Y, q) be Riemann domains over C'7 and C'", respec:iveh; and let D be a domain in X x Y such that for each x X the fiber D

• f(x..)

is such that:

x E X,

•flu E (9(U),whereU C DisanopensetsuchthatU,

X.

Then f E (9(0).

Proof ThecaseX := E'7, Y := E'78,D := E'7+P71,U := E'7 x(8E)m (0<8<1) reduces to the standard Hartogs lemma on separate analyticity (cf. [Kra 982]). In the general case let 0 denote the maximal open subset of 0 such thatf E

-

Obviously (I C D. Observe that if Px(xo,r) x IPy(yo,s) C D and lF'x(xo,r) x C Dforsome0 <e < s.thenwe can applythe special casetothefunction

t

1.8 Maximal holomorphic extensions

61

(z,w) —÷

E" x Etm

x l?y(yo.s) C I).

Suppose that

for some

E

X. Since

0

C

and

connected,thereexistsapointy1

is

D

s). Then there exist 0 < r < r0and 0 < e s such and let Yo E that iPx(x0, r) x lPy(y0, e) C 15. Consequently, Px(x0, r) x Py(yo. s) C D. Since wegetacontradiction. 0 (xe, yi) lF>x(xo,r) x PropositIon 1.8.15. (a) Let (X. p) be a Riemann domain over C" and let C (9(X). Suppose that a: (X, p) —÷ (X. j3) is the maximal 8 -extension. Then for any Riemann domain of holomorphy (Y, q) over Ctm

axid:(XxY,pxq)—÷(X x Y,j3xq) is the maximal V-extension, where 'V

:= (f

O(X x Y):

y)

As(b) Let (X, p), (Y, q) beRiemann domains over C" and Ctm, sume that a: (X, p) —÷ (X. j3), $: (Y. q) —* (Y, are maximal holo,norphic is the maximal holoextensions. Then a x fi: (X x Y, p x q) —+ (X x Y, j3 x are envelopes of holomorpkv morphic extension. In other worcLc, ( X, j3) and (Y, is an envelope of holomorphv of (X. p) and (Y, q), then (X x Y. j5 x of(X x Y, p x q).

Proof (a) By Lemma 1.8.13 we only need to show that

ax Id: (Xx Y,p xq)—* (Xx Y,13 xq) is a V-extension. Take an f E V. Define

X x Y —+ C. Note that ft o (a x idy) = fi (9(X x Y).

f.

In particular,

y) := (f(•. y))a(x). E

O(a(X) x Y). By Lemma 1.8.14,

(XxY,pxq) —÷ (Xx is a holomorphic extension. Take an f O(X x Y). Define fj: X x Y —f C as in the proof (a). Then Ii (9(X x Y). Now we repeat the same procedure with respect Observe tothesecondvariableandletf2: Xx Y —+ C,f2(x,y) := (f1(x, that f2 o (idx x fi(Y)) and hence, by Lemma (9(X x Finally, 12 o (a x 0 1.8.14, 12 = o (a x idy) = f.

62

1 Riemann domains

1.9 Liftings of holomorphic mappings I Let us consider the following general problem. Assume that: (1) (X, p), (Y, q) are Riemann domains over

(2) 4 c (9(X), 7 C 0(Y), (3) a: (X, p) —÷ (X, j3) is the maximal 4-extension, (Y. is the maximal 7-extension, is a holomorphic mapping such that J r*(7) We ask whether there exists a holomorphic lifting of t, i.e. a holomorphic mapping

(4) fi: (Y. q)

(5) r: X

Y

1:X—* Ywithloa'=8or.

ii

ji

Y

Observe that the lifting I is uniquely determined (if

12 are two liftings of r, then a, and hence, by the identity principle, Notice that if r is constant, t(x) = x X, then obviously the mapping I(s) := fi(yo), x X, is the lifting of r.

oa=

o

In the present section we study the problem of existence of the lifting under special assumptions on r (but with general families and 7). The case where r is arbitrary (but 4 = 0(X). 7 = 0(Y)) will be presented in Theorem 2.12.1.

0(X)) the lifting I need not exist. For example, X := E5, p := idx, Y := C \ E, q = idy, r(x) := l/x, x E X (r is biholomorphic), 4 := 7 := J(°°(Y). Then idx,E: (X, id) —* (E, id) is Example 1.9.1. Notice that in general (if 4

the 4-envelope of holomorphy. Moreover, Y is an domain of holomorphy (Y is fat) and therefore idy: (Y, id) —+ (Y, id) is the 7-envelope of holomorphy. The lifting of r should be a holomorphic extension oft to E. It is clear that such an extension does not exist.

Proposition 1.9.2. Assume that (1)—(5) are satisfied and that there exists a T Aut(C'1) 36) such that q o r = To p. Then r has a holomorphic lifting I: X —* such that o I = T o X

q

Recall that r(g) := g o r,

Y &E

Y

q

0(Y).

Aut(G) denotes the group of all biholomorphic mappings T: G —÷ G. Observe that r and must be locally biholomorphic.

Y

63

1.9 Liftings of holomorphic mappings I

Moreover, (ft is biholomorphic and 45 = r*(T), then

is biholomorphic.

Notice that if r is a morphism (i.e. T = idc"), then is a morphism.

Proof First, observe that

r: (X,Top)—*(Y,q)

a: (X,Top)—+

Moreover, by Proposition 1.8.9, fi o t: (X, T o p) —+ (Y, (X, T o J3) —÷ Hence there exists a morphism is the maximal (Y, such that f oa = o r. are

If r is biholomorphic and 45 = r*(7). then we apply the above argument to the mapping a := (Y, q) (X. p) and we get a holomorphic mapping = a oa. Hence (& oI)oa = & ofi or = a oa or = a. ô: Y —+ X withô :

Thus, by the identity principle, & o

=

Similarly

=

o

0

Lemma 1.9.3. Assume that (1 )—(4) are satisfied. Let B be a locally connected topological space and let

r: Xx B —*

x B —÷

T:

Y,

be continuous mappings such that: B, (6)T(•,b)€ (7) q(r(x, b)) = T(p(x), b), (x, b) E X x B 38)

(8)45 J{gor(.,b): gET. bE

B}.

Suppose that for each b E B the mapping b) has a lifting and define I(x, b) X x B —÷ Y is continuous r(•, b)(x), (x, b) X x B. Then

CxB

pxid

XxB

oxid

'.

>XxB

pxid

rj q

/3

Y

Y

Proof First note that f(a(x), b) =

b)), (x, b) X x B. Consequently, since b) = a islocallyhomeomorphic, f iscontinuousona(X)x B. b)) = q(r(x, b)) = T(p(x), b) = T(1(a(x)), b). (x,h) c X x B. Hence. by the identity principle, o f(x, b) = T(j(x), b), (x, b) E X x B. Now, to get we use the monodromy theorem (cf. Theorem 1.1.18) the global continuity of with ((X, p), A, B, F, ao) = ((Y, ti), X, B, F, a(xO)) with F(x, b) := T(j3(x), b), (x, b) E X x B, and where xo E X is arbitrary. As a direct consequence of the above lemma and Proposition 1.9.2, we get 38)

Inpartlcular.r(•.b)Ee)(X.Y).bEB. We say that

is a lifting oft.

El

64

1

Riemann domains

Proposition 1.9.4. Assume that (1 )—(4) are satisfied. Let B be a locally connected topological space and let

i:XxB—÷Y, T:C"xB—.C". be continuous mappings with (6)—(8) such that T(, b)

E

b E B. Then the

Xx B —+1. Proposition 1.9.4 may be slightly generalized in the following way.

Proposition 1.95. Assume that (1 )—(4) are satisfied. Let B be a locally connected topological space and let

be continuous mappings with (6) and (7), such that there exists a locally connected topological subspace B0 of B for which: • b) E Aut(C"), b E Bo,

•for each b E B \ B0, the mapping r

the lifting f: X x B —+ Y.

The most typical applications of the above propositions are gathered in the following Remarks 1.9.6 and 1.9.8.

Remark 1.9.6. Let G c C" be a domain, let

C

(9(G), and let

a: (G,id) —+ (X,15) be the maximal s-extension. (a) If G is starlike (i.e. tz E G for any t E [0, 1] and z E G) and for any f E and E (0, 1] the function G z —+ f(tz) belongs to 8. then there exists a continuous mapping H: X x [0. 1] —* X such that H(a(z), t) = a(tz), (z, t) E G x [0, 1] t)) = and (x. t) E X x [0 I] 41) (b) Observe that in (a) the envelope (X, j5) must be univalent.

In fact, let (Z, r) be a Riemann domain over C" such that there exists a continuous E Z x [0,1]. Then (Z, r) is univalent.

mapping H: Z x [0.11 —+ Z with r(H(x.t)) = tr(x),(x.t)

Indeed, first observe that H(•, 0) = const =: xo. Hence since H(•. I): Z —* Z is a morphism and H(xo, 1) = thatH(•. xo

42) Now,

40)

In

41)

In

42)

Because [0, 11 31 —* H(x0, r)

particular. H(•, 1) = particular, H(•. 0) = const because X ax —+ H(x, 0) E is continuous.

is

= const = xo, we conclude

continuous.

65

1.9 Liftings of holomorphic mappings 1

Letx1,x2 E Z besuchthatr(x1) = r(x2) =: zo. Thenthemappings and H(x2,.) are two liftmgs of the mapping [0. 1] t —p tzo with the same value att = 0. Hence H(xi,•) H(x2, .). In particular. = H(x1, 1) = !i(x2, 1) = x2. (c) By virtue of (a) and (b) we get: Let G C C" be starlike and let .8 C (9(G) be such thatfor any f E .8 and t E (0, 1] the function G

f(tz) belongs to.8. Then

z

8) is a starlike domain in C".

E G)andforanyf E 4and (d)IfGiscircular(i.e.Cz the function G z —÷ f(Cz) belongs to 4, then there exists a continuous —+ X such that H(a(z), and mapping H: X x = a(Cz), (z, E G x 3(H(x, = Cft(x), (x, E X x 8E.

E dEandz

E

E

Example 3.1.20 will show that in the above situation (X, j5) need not be univalent.

(e) By virtue of (c) and (d) we get: be balanced (i.e. E (3 for any E E and z Let G C be such that for any f E .8 and E E \ {0} the function G to 4. Then 8) is a balanced domain in C".

E

G) and let4 C (9(G)

z —+ f(çz) belongs

(f)IfG isn-circled(orReinhardt, i.e. E G forany E (8EY' and z G for any f 8 and (aE)" the function G z z) belongs to 4, then there exists a continuous mapping H: X x (8EY' —÷ X such that H(a(z), C) = (x, E X x (dE)". E G x (aE)", and (z, = Corollary 1.9.18 will show that in the above case (X, j3) must be univalent. Nevertheless, the exists a non-univalent Riemann domain (Z, r) which possesses .

.

.

a continuous mapping H: Z x (dE)" —* Z such that H(., 1) = r(H(x, = . r(x), (x, E Z x (8E)" — cf. Example 1.9.7.

and

Properties (d) and (f) will be generalized in Proposition 1.9.15.

Example 1.9.7. Define

:= ((2E) \ E) x E, Di := ((2E) x (2E)) \ ((dE) x D2

:= (2E) x E.

Note that D0, D1, D2 are 2-circled domains and Do is a connected component of D1 fl D2. Put Z := (11) x D1) U ({2} x 1)2) and let denote the equivalence relation which identifies {l} x Do with (2} x D0. Define Z := r([(i,z)]-..-) := z. It is clear that (Z, r) is a non-univalent Riemann domain over C2. Define Ii: Z x —* Zby the formula

H(I(i,

C) := [(i, C z)i-..

C E (aE)2, z E

ThenHiscontinuous,H(., 1) =idz,andr(H(x,C)) z

Recallthatl=(1

1)EN".

That is, we glue D1 to D2 along D0.

i=

C.r(x),(x,C)

1,2.

Zx(dE)2.

66

1 Riemann domains

Remark 1.9.8. Assume that 1 k n — 1, D is a domain in subdomain of D x Assume that for any z E D the fiber

and

G is a

:= (wE Ck: (z, w) E G} is

non-empty. Let T c 0(0 x Ck) be such that D x Let

phy

C 0(G) be such that TIG C

is a T-domain of holomor-

Moreover, let

a: (G,id) —p extension. Observe that idGDXCk: (G, id) —* (0 x C', id) is the maximal TIG-extension. Since a: (G, id) —÷ (X. ) is a TIG-extension, there exists a morphism a: (X, ,3) —÷ (0 x Ck, id) such that a a a = idGDXck. Clearly, be the

a' = J3. Write

=

(13n-k, pk): X

D X Ck.

(a) If G is a Hariogs domain with starlike fibers (i.e. for any z E D the fiber is a starlike domain in Ck 47)) and for any f E and t E (0, 1] the function tw) G belongs to &, then there exists a continuous mapping (z. w) f(z, H: Xx [0. 1] —÷ X such that H(a(z, w), 1) =a(z, tw), ((z, w), t) E G x [0, 1] and j3(H(x, t)) = (13n_k(X). tPk(X)), (x, 1) E X x [0, 11. Indeed, we will proceed as in the proof of Lemma 1.9.3. Let r: G x [0, II —+ G, r((z. w). 1) := (z. 1w), T((z, w). 1) x Ck) x (z. tw), ((z. w), 1) E [0. 1]. By Proposition 1.9.2, for any I E (0. 11 the mapping r(•. t) has a lifting

H(..fl: X —÷ k.

EL Wehave

H(a(z, w), 0) = a(j3(a(z. w)), 0) = a(z, Ow), (z. w) E G. Thus H(•, 0) is a lifting of r(. 0), and consequently, by Lemma 1.9.3, H is continuous (and satisfies all the required identities). (b) The domain (X. j3) in (a) must be univalent.

Indeed, let xi, x2 E X be such that = 13(x2) = (zo, wo) E 0 x Ck. Then the mappings H(x1..) and H(x2,.) are two liftings of the mapping [0, 1] t (zo,two). Moreover, H(x1,0) = a(zo,0) = H(x2,0), which implies that I) = H(x2, 1) =x2. H(x2, .). Finally, xi = (c) By (a) and (b) we get:

JIG

and

are as in (a), then C(G,

is univalent.

C D x Ck, and

is a Hartogs domain with starlike fibers. (d) If G is a Hartogs domain with circular fibers (i.e. for any z E D the fiber

C(G.

is

acircularopensetinCk)andforanyf E E dEthefunctionG (z,w) —' —* X çw) belongs to then there exists a continuous mapping H: X x f(z, such that H(a(z. w), = a(z, E G x dE, and 1ô(H(x, ((z, w), = (x, 46

48)

E X x dE.

where 7 70 In particular, D x (0) C G. In particular, H(., I) = idk.

For example.

C (9(D) is such that D is a To-domain of holomorphy.

1.9 Liftings of holomorphic mappings 1

The result follows directly from Proposition 1.9.4. Notice that (X, univalent — cf. Remark 3.1.8. (e) By (c) and (d) we get:

67

need not be

D the fiber JIG is a Hartogs do,nain with balanced fibers (i.e. for anyz is a balanced domain in Ck) and for any f E and E \ (0) the function isunivalent,C(G, 0 (z, w) —* C DxC1', is a Hartogs domain with balanced fibers. and E(G, (t) Similarly as in (d), directly from Proposition 1.9.4 we get that if G is a Hartogs domain with k-circled fibers (i.e. for any z E D the fiber is a k-circled open set in Ck) and for any f E and e (dE)" the function G ?(z. w) —p f(z,C . w) belongs to 8, then there exists a continuous mapping H: X x —p X such

.w), ((z. w), C) E G x (dE)", and j3(H(x, C)) =

that H(a(z, w), C) = a(z, C

Pk(X)), (x, C) E X x (dE)'.

More developed characterizations of envelopes of holomorphy of Hartogs domains will be given in Section 3.1.

We come back to the general context (I )—(5) from the beginning of this section.

PropositIon 1.9.9. If 8 = (9(X), ping r: X —+ Y has a

7=

(9(Y), then any locally biholomorphic map-

Moreover if t is biholoinorphic, then I is biholoniorphic.

Proof By Proposition 1.8.9, weknowthatflor: (X, qot) —+(Y, isthemaximal rn): X —÷ C". Since r is r := (rI r*(O(Y))..extension. Letr3 := (q1 o locally biholomorphic, we have det I

1

L

azt

(x) I

J

X.

x

0.

Observe that

or)] 3Zk

Hence, by Remark 1.4.5(g), .r

X,

dzk

and, consequently,

r: X —+ C" is locally biholomorphic. Observe that

a: (X,qor) —÷ (X,r) is a

Hence there exists a morphism

I: (X,r)—÷ such that 1 o a = ° r. The case where r is biholomorphic is the same as in the proof of Proposition 1.9.2.

0

68

1

Riemann domains

Remark 1.9.10 Assume that = 0(X), 7 = 0(Y), and is biholomorphic. of r. Let By Proposition 19.9 there exists a biholomorphic lifting 1: X —* Then r := (po r')13 E (D(Y, Ce). Observe thatr = j3 o • (Y. r) isa Riemann domain,

•r: (X.j3) —* (Y,r)isanisomorphism, • o r: (X, p) —+ (Y, r) is the maximal holomorphic extension. X

rj Assume

of a Lie group. Let g be a connected real Lie group (cf. [Var that g acts continuously on C's. i.e. there exists a continuous mapping Action


such that

E

9 —)

X

for anyg E gand the mapping

g —p
Example 1.9.11. (a) g := Il'." (with addition); g) (with multiplication); (b) g := (C) g :=

g

Z.

Let (X. p) be a Riemann domain over C". We say that g acts continuously on g —+ X such that

(X. p) if there exists a continuous mapping

p(
Cxg C"

Proposition 1.9.12. (a) The mapping (b)
93g is


pxid

Xxg

(P

x

is uniquely determined. g and the mapping

g) E Aut(X)

a group homomorphism. 49)

x€ X, g E 9.

Here 0 denotes the neutral element of the group

(t)

69

1.9 Liftings of holomorphic mappings 1

Proof (a) follows from the identity principle for liftings. g) op. Hence (b) The mapping cDx(., g) is continuous and po g) = 4>x(. g) is holomorphic for any g E g. Let '+' denote the group operation in g. Fix an xo and observe that

ga)) =

=

+g2).

gi.g2 E 9.

Thus the mappings

E X,

9 X

are

+g2) E X

(gI.g2)

gxg liftings of the mapping

—p

g x 9

with

£(O,

0) =

r(O,

0) =

xo.

E C'1

Hence, by the identity principle for liftings, we get

gi + g2) for any

gi), g2) =

gi,

E 9.

=

=

In particular,

Aut(X)

+gz)

Consequently, Dx(,g) E

0

and the mapping (t) is homomorphic.

Remark 1.9.13. By the monodromy theorem (Theorem 1.1.18) the following conditions are equivalent: (i)

g acts continuously on (X, p);

(ii) for any xci E X there exists a lifting

Proposition

(0) = xo

g)

g

9

with

X of the mapping

g

E C'1

50)

1.9.14. Suppose that there exists a zo E C" such that

(zo. g) = zo 0. Then the following

for any g E g. Assume that (X, p) is such that conditions are equivalent:

(i)

g acts continuously on (X. p):

(ii) for any g

9 there exists a ljfting Fg: X

X of the mapping

EC"

X

withFg(x)=XforaflyX Proof Letdxbeasinthedefinition. PutF8 :=

Itisclearthat

Moreover, foranyx

Fg

(zo, g) =

zo.

Consequently, for any x E p g

g

(20). the mapping

g) E

50) In particular, if (X. p) is a cover and g is simply connected, then g acts continuously on (X, p) (cf. Exercise 1.1.20).

70

1

Riemann domains

must be constant. Recall that = x. Hence 4x(x, g) = x, i.e. anyx E (ii) (I): By the monodromy theorem, the mapping

Xx g

(x,g) —k F9(x)E

= x for

X

is continuous. It remains to prove that F0 = (I'x(., 0) = idx. Recall that Fo is the lifting of 0. Hence, by the 0) = p with Fo(x) = x for any x E identity principle for liftings. we conclude that Fo = idx. 0 By virtue of Proposition 1.9.4, we get

Proposition 1.9.15. Let a: (X, p) —+ (X,

be the maximal 45-e.xtension. Assume

that g acts continuously on (X. p) and

g

acts continuously on (X. fr) and

g) =

X9( pxid Xxg

C

g)), axid

x E X. g

Xxg

jxid

g.

C xg

't,c.,

a

Let (X, p) be a Riemann domain over C'1. We say that (X. p) is n-circled if the group (dE)" acts continuously on (X, p), i.e. there exists a continuous mapping ctx X x (aE)" —÷ X such that 1)

= idx,

C)) =

C

p(x),

x E X, C

(dE)":

ci. Remark 1.9.6(0.

Remark 1.9.16. (a) A domain G C C" is n-circled in the classical sense if (G, idG) is n-circled in the new sense. (b) If (X. p) is n-circled, then p(X) is n-circled in the classical sense. (c) By Proposition 1.9.15, if (X, p) is n-circled and a: (X, p) —÷ (X. )5) is the maximal i-extension with = {f o (dE)'1}, then (X, ,3) is ç): f ,z-circled.

(d) Recall that there exist non-univalent n-circled Riemann domains (Example 1.9.8).

1.9 Liftings of holomorphic mappings 1

71

ProposItion 1.9.17. Let (X, p) be an n-circled Riemann domain over C's. Then any

function f E (9(X) can be represented by a Laurent series of the form

f(x) = In particular; 0(X) =

x E X.

p*(0(p(X)))

Consequently, any n-circled Riemann domain such that 0(X) separates points in X is univalent.

Pmof First we will prove that: any point a

X has an univalent open neighbor-

ho(Xl Ua such that p(Ua) is an n-circled domain.

Foranyx E

=

(dEY'IetC'(x) = (C(x)

be defined by the formula

Obviously

.

I

4

I

I

if p1(x) ifp1(x)=O,

j=l

n.

p(x) = ç'(x) . p(x). Moreover, 4>x(x,

=

(**)

C'(x)).

Indeed, fix an xo E X and observe that the mapping —+

E

is continuous. Let s(C) := C). s'(C) := 4x(xo, C E (aE)'t. Then s and s' are two liftings of the mapping p(xo). Since s(1) = C C they coincide for all C.

Now, suppose that p(4x(xo. C)) = for some C. E (dE)" and Then = . p(xo) and therefore C'(xo) . p(xo) = 1?'(xo)• p(xo). This shows Consequently, C'(xo) = l7'(xo). Thus, by (**), C) = 4)x(xo, E X.

that for any xo E X the mapping P10(x0)

injective, where

O(xo) := {CDx(xo, C): C

Observe that O(xo) is compact. Fix an a E X. For 0
Ua(r) :=

C):

XE P(a,r), C E (dE)"} =

U 0(x). xEP(a,r)

72

1

Riemann domains

Note that for any E (aE)". In particular, the domain = p(Ua (r)) is n-circled in the classical sense. To prove (*) it suffices to show that there exists 0 < r < dx(a) such that PIU0r) is injective. Suppose that for any k N there exist

1/k) and

E

:= ctx(xk, —f '1k '10. r,o) =: xo. In

'1k E (aE)" such that

=

but

We may assume that (k

'1k) =: Then to)) = 4o) = 'io)). Hence particular, E P(x0) for k>> I; contradiction. The proof of (*) is completed. Fix an f E (9(X). For a E X let Ua be as in (*). Then the function := o Hence is holomorphic in the n-circled domain p(Ua). 1

Z E p(Ua).

aL,(a)z,

ga(z)

Thus

f(x) =

av(a)(p(x))",

X E

a (v E Zn).

We only need to show that Observe that

I

= =

I

=

l•

I—ri

f .

(2ir,)" J(ae)"

.

... . .

E (IR>oY'. where c is an arbitrary point from Ua such that p(c) = fri Let a, b X be such that Ua fl Ub 0. Then there exists a point c E U0 flUb (IR>0)". Consequently, V E 7L". such that p(c) = lfa, h X are arbitrary, then we may easily find points cO, . . ,CN E X such that fl 0, j = I N. Finally, av(a) = av(cO) = = a, eN b and .

D

Corollary 1.9.18. Let (X, p) be an n-circled Riemann domain over C" and let d C (9(X) be such that 8 = (f o Then the E of holornorphv of (X. p) is univalent, and, consequently. it may be realized as an n-circled domain in C".

f

Proof Let a: (X. p)

(X, j3) be the maximal 8-extension. Recall that (9(X) separates points in X. We know that (X. is n-circled (cf. Remark 1.9.16(c)). On the other hand, by Proposition 1.9.17, holomorphic functions on X do not separate points in stalks. Consequently. X must be univalent. 0 Recall the following criterion for an n-circled domain G C C" to be a domain of

holomorphy.

1.9 Liftings of holomorphic mappings 1

73

Proposition 1.9.19. Let G C C'1 be an n-circled domain. Then the following two conditions are equivalent: (i)

G is a domain of holonwrphy;

(ii) G satisfies the following two geometric conditions: (li)1 the logarithmic image logG := R'1: is convex and x {O} x C'1'1) (ii)2for ever jE (I, .. , n}, :fG fl .

G}

0, then

xEx{l}'1'andzEG.

G

.z

Note that (u)2 is automatically satisfied in each of the following two particular cases:

(1) G in an n-circled domain contained in (2) G is a complete n-circled domain, i.e. G for any e E'1 and z E G. z Observe that the proposition implies that any n-circled domain of holomorphy G with 0 e G must be complete n-circled.

Proof (i)

(ii): Let f

(9(G). Then f may be represented by the Laurent series G. Moreover, f(z) = (a) for any compact K G there exist C > 0 and 0 <0 < I such that z

(*)

<

for any z e K and a E

Z'1

(in particular, the series is locally normally convergent in

x{0}xC'1J) such that

=Oforanya =(a,

E

<0.

Takex = (XI b := er")

t,1), y G fl

= (yj

logG. Leta := (eXt

G fl (Ca)" be neighborhoods ofa and b, respectively. By (a) there exist C > 0 and 0 < 6 < 1 such that (*) holds for UU,, anda E Z'1. Define anyz and let Ua, Ub

LJ:={(eOhJziIIhIwiIt (zi

E [0,

Ui,,

l]} C

U and a Z'1. Consequently, the series is locally normally convergent in U and its sum extends f to U. Since G is a domain of holomorphy, U c G and therefore [x, yJ C log G, which gives (u)1. Put := C)' x x C'1', j = I n. By (b), if G fl 0, then the series is locally normally convergent in the domain One can easily check that (*) holds for all z

GW :=

z:

E

{l}J1 x E x

and its sum defines a holomorphic extension of f to (u)2 is satisfied.

Z

E G)

Hence we conclude that

74

1

(ii)

;

logG =

(i): Since log G is convex, there exists a family A C (R"). x R such that where

ER":

{(xi Put

G :=

Riemann domains

+xj,, < c}.

xh8I

where

mt

:= {(zi

Zn)

C"(fi):

:=

x

.....

<e'},

x

ift>O ift
IC

Suppose that we already know that each domain is a domain of holomorphy. Then, by Remark 1.7.3(d), G is a domain of holomorphy. Define

U Vj.

M:=

jE41

n}

GflV1=e

ItisclearthatG\Mo = G\Mo. ByRemark 1.7.3(g), G\Misadomainof holomorphy. Using (u)2, we conclude that G = G \ M, which finishes the proof. It remains to show that each is a domain of holomorphy. We may assume that c = 0. Moreover, we may assume that th > 0. fi,, <0 With o .c n. Consider the biholomorphic mapping

C5 x (C4)"5

Zs, l/Zs+i

(zi

(zi

E

C5 x

Let Gfl :=

Observe that is a domain of holomorphy if D(Gp) is a domain of holomorphy. Hence, using once again Remark I .7.3(g), we conclude that it suffices toshowthatforanyyj y,, > Otheset D := {(zj

Z,,) E C":

.

.

< l}

..

is a domain of holomorphy. Define

f(z) :=

z E D, t=I

...

where [xJ stands for the integer part of x. The function f is well defined and holomorphic on D. Moreover, one can easily check that f cannot be holomorphically extended to a complete n-circled D D (recall that the envelope of holomorphy of D must be univalent and complete n-circled; Corollary 1.9.18). 0

1.10 Holomorphic convexity

75

1.10 Holomorphic convexity The aim of this section is to characterize domains of holomorphy in terms of the holomorphic convexity. DefinitIon 1.10.1. Let (X, p) be a Riemann domain over C'7 and let 0 For a compact set K C X define (cf. § 1.4) K8

{x E

X:

If(x)I < II! lix, f

The set K4 is called the i-hull of K. The set

C (9(X).

E

is called the holomorphic

hull of K. if K = K.t. If

= (9(X), then we simply say thatK is holomorphically convex. We say that (X, p) is if K4 is compact for any compact K C X. If = 0(X), then we shortly say that (X, p) is holomorphically convex. We say that K is

The definition extends in a natural way to Riemann regions over C".

Remark 1.10.2. (a) K C

CT, then

Jç4

(c) K4 = KCIO(X) 4

51)

j

k8

is closed.

=

In particular, if K C C". then

(d) If K8 is compact, then K8 = K8.

Proposition 1.10.3. Lef(X, p)beaRiemanndomainoverC'7andletO

C 0(X).

Then the following conditions are equivalent: (i)

X is i-convex;

(ii) there exists a sequence

K3 C mt

of i-convex compact subsets of X such that

and X =

Pmof It is clear that (ii)

(I). To prove the opposite implication, let

he an

and X =

L3. Put

arbitrary sequence of compact sets such that L3 C

Lf. Since X =

K1

K2

:=

mt

Now we take au >

there exists a 12> 1 such that K1 C mt L12. Put such that K2 C mt etc. 0

Theorem 1.10.4 (Cartan—Thullen theorem). Let (X. p) be a Rieniann domain over C" and let C 0(X) be a d-stable closed subalgebra such that p E .8". Then the

following statements are equivalent: (i)

.8 = 0(X) and (X, p) is a Riemann domain of holomorphv;

(ii)

(X, p) is an .8-domain of holomorphy; c10( X) 4

denotes the closure of 4 in the topology of locally uniform convergence on X.

76

1

(iii)

separates points in X

52)

Riemann domains

and for any compact K C X we have

=

(K); (iv) 8 separates points in X andfor any compact K C X we have dx(K4) > 0;

(v)

8

= 0 there

separates points in X and for any set A C X with

function f E

.8

such that SUPA fl

exists a

+00;

(vi)

.8

separates points in X and X is .8-convex;

(vii)

.8

separates points in X and for any infinite set A C X with no limit points in X

there exists afunction f E .8 such that SUPA

ft

= +00.

Remark 1.10.5. In particular, in the case.8 = 0(X), the following conditions are equivalent: (a)

(X, p) is a domain of holomorphy;

(b)

0(X) separates points in X and X is holomorphically

Any Riemann domain over C" satisfying (b) is called a Riemann—Stein domain over C". A Riemann region (X, p) over C" is a Riemann—Stein region if X is countable

at infinity and (b) is satisfied We will prove in Theorem 2.5.7 that in the case.8 = 0(X) the separation of points is a consequence of the holomorphic convexity. Thus, in the case.8 = 0(X), each of the above conditions (a), (b) is equivalent to the following condition (c)

X is holomorphical!v convex.

Pmof Notice that the proof essentially simplifies if(X, p) = (G, jdG), where G is a domain in C". Moreover, the proof is simpler if .8 = 0(X) (then obviously (i) (ii)). First observe that the implications (i) (ii), (iii) (iv) are : (iv), and (vi) trivial.

The scheme of the proof is the following one ('k°' denotes the k-th step of the proof):

(i)

(ii)

I"

(iii)

)0

(iv)

(v)

3°

(ii)

6°

40

(vi)

50

(vii)

70

(vi)

(i)

Step 10 (ii) (iii): By Proposition 1.8.10 the family .8 separates points in X. The remaining part of (iii) is a consequence of the following lemma.

Lemma 1.10.6. Let (X, p) be a Riemann domain over C", let.8 C 0(X) be a d-siable family, and let K C X be a compact set. Then 52) Notice that under our assumptions: .8 separates points in X .8 weakly separates points in X (cf. Definition 1.6.1). Observe that if (X, p) = (G, idG), where G isa domain in C". then I

always separates points in X. Notice the following general definition: An n-dimensional complex manifold M is said to be a Stein manifold if: • M is countable at infinity, • (9(M) separates points in M, • for any point a M there exists a holomorphic mapping F: M —÷ C" such that F is injective

in a neighborhood of a. • M is holomorphicaiiy convex.

1.10 Holomorphic convexity

aE > f 4. (b) lf(X, p) is an 4-domain of holomorphy, then

Pmof of Lemma 1.10.6. (a) Take a point a

0
E

77

=

a function f E

4,

and a number

Recall that by the Cauchy inequalities we have

hay! IlK <

If

PE

Hence, by the a-stability of 4, we get II

VE

r. Thus d(Taf)

In particular, d(T0f)

(b) By (a), for any point a E Hence ?

K's,

we get dx (a) = inf{d(T0 f):

fE

4}

dx (K).

0

(v): Let be a sequence of compact subsets of X such L,. Since dx(A) = 0 and that L, C mt and X = > 0, for each j there exists a point A \ 14. Put K1 := Lp, ai bi. Let 12 be the smallest L,,, := bk. Let index > such that ai E L1. Put K2 such that etc. Thus, we get a E L,. Put K3 := > sequence of compact subsets of X with K1 C mt C A such that a1 E and a sequence j E N. Consequently, for \ each j, there exists a function f1 E 4 such that If, ) I > Ill, II iç• Taking cff instead Step 2° (iv)

j j

:

1

off, (4 is an algebra) we may assume that If,(aj)I I + I + I If, lix, < 21, j N (use induction). In particular, the series Moreover, uniformly convergent and therefore I := E 4

?

Ifk(aJ)h

j+l—

k=j+I

k=I

fk(a,)l and is locally

2 j.

j EN.

k=j+I

Leta: (X, p) —* (X, j3)bethemaximal4-extension. Since 4 separates points in X, the mapping a is injective. Suppose that yo E a(a(X)) and let = 0. C X be such that a(xk) —÷ yo. Take A {xk: k E NI. Then Indeed, suppose that dX(Xk) 2r > 0, k E N. We may assume that 2r r). Then Take lco such that E E r) = a(IPx(xk0. r)) C a(X) (cf. Remark 1.4.2(e)); contradiction. Let

fE

4 be such that supA Ill

= +oo and let f

f(a(xk)) —+ f(yo); contradiction. Thus (ii) Recall that

(iii) is closed.

(iv)

(v).

Then f(xk) =

78

I

Remark. The implication (v)

Riemann domains

C 0(X).

(ii) remains true for any family

Step 4° (vi) => (vii): Let

be a sequence of compact subsets of X such that C mt L3÷1 and X = L1. Since A is not relatively compact, for each j there exists a point (v). E A \ L. Now we can continue as in the proof of (iv) Step 5° (vii) (vi): Suppose that K C X is a compact set such that the set K1 contains a sequence without limit points in X. Let f E be such that sup = +oo. On the other hand, IIfDx, v I; contradiction. ,'

Thus (vi)

(vii).

Remark. Observe that in the case (X,p) = (G, idG) the implication (iv)

(vi)

is obvious. Indeed, in this case the set K1 is relatively closed, bounded (because the coordinate functions belong to and has a positive distance to the boundary of G. Consequently, K has to be compact. Thus, in the case (X, p) = (G, ida) we have (ii) (iii) (iv) (v) (vi) (vii) and, consequently, we can skip Step 6°. (vi): This is the main part of the proof. The idea is due to

Step 6° (ii) E. Bishop [Bis 1963].

C 0(X) be

Lemma 1.10.7. Let (X, p) be a Riemann domain over C", let 0

a d-stable subalgebra such that (X, p) is an 8-domain of holomorphy. Assume that

K C X is a compact set. Then

(Kl)(r) C K(")1,

0 < r < dx(K)

(observe that, by Lemma 1.10.6(b), the set

is well defined).

Proof of Lemma 1.10.7. Take 0 If IlL. Replacing f by cf with a suitable c E C, we may assume that f(a) = 1. Hence If < 1 on L1 and therefore the function g := 1/(1 — f) is defined on a neighborhood U of L1. In particular, since K C L1, the function g is holomorphic near b. Let r < r' be such that K(T') c U.

such that a

I

By the Cauchy inequalities, we get VE

Putgk := 1+f+...+fk,k EN. Observethat,foreachv E 3 K1 3 K. Since at'gk E

uniformly on

=

urn

lim

k—p+oc

......÷

we get

=

r'

Thus d( Thg) > r' > r; contradiction (evidently, the function g cannot be holomor-

phically continued to a neighborhood of a).

0

79

1.10 Holomorphic convexity

Lemma 1.10.8. Let (X, p) be a Riemann domain over C'1, let C 0(X) be a a-stable subalgebra with p such that (X, p) is an i-domain of holomorphy. Suppose C", that X is not .6-convex. Then there exist a compact set K C X, a point zo a sequence xi fork I, and a number r > 0 such that C p'(zo), Xk dX(xk) > r and IP'(xk, r) C is not compact Pmof of Lemma 1.10.8. Let L C X be a compact set such that E .6", we have limit point. Since p be a sequence without any and let C is k N, and therefore the sequence J = 1 n, < 2r (Lemma 1.10.6(b)), there exists an Xk P(ak, r) By Lemma with p(xk) = zo. Obviously, dx(xt) > r, k E N. Now, let K := 2r) c 1.10.7 we have IF'(xk, r) C c k's. It remains to prove that #(xk: k N}=oo. Suppose it does not hold. Choosing a subsequence we may assume that xk = 0 for any k,l > ko. Hence, the for k Ico. Then lP(ak, r) fl P(aj, r) = P(ak, r) is univalent (cf. Corollary 1.1.6(b)). Consequently, set U := U contradiction. (pIuY'(p(ak)) —+ Lemma 1.10.9. Let (X, p) be a Riemann domain over C'1 and let 6 c 0(X) be a closed vector subspace such that.6 separates points in (z) for every z E p( X).

that zo E p(X), r > 0, and (xk)r C

Xk

N. Then there exists afunction f that dX(Xk) > r, k for any k 1. Moreover,

IP(zo,r) :

Pmof of Lemma 1.10.9. Let .6k, := {f

k E

# fork

1, are such

6 such that f(xk) # f(xj)

N}
.6: f(xk) = f(x,)}, k

I. Then .6k.,

I. Hence .6k,, is of the first .6 for any k a closed subspace of.6 and .6k,, Baire category in .6 (cf. the proof of Proposition 1.8.11). Consequently, the set .6o := Uk$, .6k.I is of the first Baire category, and therefore there exists a function f E \ 8o• Thus f(xk) f(x,)foranyk #1. = #1. ThesetVk,isanalytic LetVk, := {z IP(zo,r): P(z0, r). In particular, = 0. Let V := in P(zo, r) and Vk,, Vk,,. Then = 0. Now we only need to observe that is

I

{z E

r)

:

(z)): k

N}

< oo} C V. U

(C") denote the space of all complex polynomials F of n-complex Let P(d1 ) variables such that (1.10.1)

80

1

Riemann domains

denote the set of all FE 1, Define

Let

L(F.r.i)

P(r): IF(z)I <

fz

of the form (1.10.1) such

F€

r"},

Lemma 1.10.10. For any 0 < r <

I then' exist constants c > .d)(C) withd ? d0 we have

ihatfor any FE

0< r,t < 1.

a)(C)v 0

and d0 E N such

0<1<1.

logt

Proof of Lemma 1.10.10. Fix an arbitrary polynomial F E and let V E fl [0, dJ)" be such that = v!. Put p = p(r) := (I — r)/3. To simplify notation put:

z w := (ziwi

z

zn). w =

= (zi

eZ :=

z =

(zi

(w1

C'1,

E

We get

>

JP(p)

f

=1

?f

[O,pj'

=

JIO.pl" pVn+2

vj+2 Consequently, there exists a point a E

F(a)I

Let a = a(r) :=

(1

..

>

/

\fl

I

\d+2

such that

2p" .

(1.10.2)

+ 2r)/3. By the mean value theorem for plurisubharmonic

1.10 Holomorphic convexity

81

functions 55) and by (1.10.2) we get

log

(d + (d + 1)". Since IP(a, a) C

we get

onl?>(a,a). Now, observe that L(F, r, t) C P(r) C IP(a, a) and therefore

I J F.rj) log L(

Finally, 2pd

2n

=:—

c(r.d) logt

and lim

c(r. d) = —(ira2 (r)f'n logp(r),

which directly implies the assertion of the lemma.

U

Lemma 1.10.11. Let (X. p) be a Rienzann domain over C", let f be a compact subset of X. Then there exist 0 < 9 < 1 and

d>s>

there e.xists a polynomial F E

IIF(pi

Proofof Lemma 1.10.11. For F Fix 0 < e < There exist N

d.s

(9(X), and let K N such that for all

E

(C" x C) with 0

S

I/n

xC)define

N and xj

xN

:= F(pi f). K such that K C

Cf. § 2.1. Let f e O(IP(a, a)). Then log lf(a)I JP(a.a) log If! the result is well known for n = 1. The general case follows by induction: log

lf(a)I

<1

J

dA2,,_2

(do)

J

Pi(ana)

Indeed.

log If(z', Zn)!

dA2,,_2(z')

=

JP(a.o) log Ill

dA2,,.

82

1

< I besuchthatK C

•UIF(XN, e). LetO

P(x1,

Riemann domains

&e)U•• •UIP(XN, 8€).

Put

c := max{ 1, flpi II

IIPn

'If

'

}.

In particular,

$ (d + 1)"(s +

FE

d,S)(C x C).

(1.10.3)

d, s) such that d > s N. Recall that dim x + 1). Let k denote the greatest positive integer such that N/c" < (d + I )" (S + I). Consider the system of equations

Fix an arbitrary (d

C) =

(d

+

8"cI)F(Xj) = 0.


v

(I

N}, F E 11'(d

(s + 1) coefficients of the polynomial F. Therefore there exists a non-zero polynomial F E P(d X C) such that 81C1F(XJ) = 0 for all v with IvI
1

(d+ l)"(s+

l)cnd+3

Consequently, since k + I

(IP(x) _P(xJ)l)k

j=I

XE

N.

N'1"(d + l)(s + 1)", we get < (d + IY'(s +

< (d + IY'(s +

(d + l)"(s +

=:

where

logO(d.s) = N"' logt5

+ds"[nlog(d+I)+log(s+I)+(nd+s)logc—logS] and, finally,

lim

d>s—.. +oo

0

Recall that we are going to prove (ii) (vi). We already know that 8 separates points in X. Suppose that (X, p) is not i-convex. Then, by Lemma 1.10.8, there exist a compact K C X, a sequence C K4,

l,a point zo E C", and a numberr > 0 such that p(xk) = zo, > r. P(xk. r) C K4, k E N. Observe that if we substitute the projection r), p by p — zo, then the situation does not change: IP(x,P_ZO)(a, r) = xk

dX(Xk)

1.10 Holomorphic convexity

i", etc. Thus we may assume that zO = o. of course, we may also assume

E

p—

thatr <

83

1.

By Lemma 1.10.9 there exists a function f E such that =0, where V := {z P(r) #f(p'(z)OK')
thatV ={ZEP(r):#4wEC: (z,w)E E}
s

so there exists a polynomial

0

<

E

IIFd.s(p1

I

and so E N such that for all

d.S)(C

x C) with

f)IIK

p,,,f) E

Hence

(1.10.4)

IIFd,sIIE

Write

Fds(z, w)

(z. w) E

X

C.

= Fix a j(d, s) E 40

logodS

log: :=

d)(C). Let i =

:= FdSJ(d$) E

s} such that t(d, s) E (0. 1) be given by the relation

=

is" logO.

Let Ldç :

By Lemma 1.10.10 there exist constants c > 0, d0 E N L(Gdc, r, (depending only on r and it) such that

2c .

s

In particular, there exists an si E N, si ? d0. such that <

x :=

d

s

Takes := max{so, si } and let

A :=

U (P(r) \ Ld.5).

j.LS d=p

> x. Hence, by virtue of the definition of A, for each z E A the set is infinite. It follows that for arbitrary z E A there exists an {d: d > s: z infinite set D(z) C fl [s. +oo) such that

Then

}

Cdc(Z) :=

j=0

s} >

d

D(z).

84

1

Riemann domains

Put Cd,s,j(Z) :=

Fd.5,J(z)

z E A, d E D(z).

Cd.c (z)

Then by (1.10.4)

Z E A, (z, w) E E, d E D(z),

(1.10.5)

and

max(Icd5J(z)I: j =0

s}

= I,

z E A, d E D(z).

—* +oo and numbers Hence, for each z E A, there exist a subsequence co(z) ,...,cc(z) such that —+ j = 0 s. Obviously

s} = I,

max{1c3(z)l: f =0

z E A.

In view of (1.10.5) we conclude that

= 0,

z E A, (z. w) E E.

In particular, A C (z E IP(r): #{w E C: (z, w) E E} <s} C V; contradiction Thus (ii) Step 70 (vi)

(iii) ;

(iv)

(v)

(vi)

(vii).

(i):

Lemma 1.10.12. Let (X, p) bea Riemann domain let C (9(X) bea vector space such that p iV' and separates points in X. Assume that K = K' C X is an i-convex compact set. (a) For any open neighborhood Uo of K there exist an open set U with K C U C Uo, a number N N, and F E such that Flu: U —÷ EN regular and infective. (b) For any open neighborhood Uo of K, for any function f E 0(Uo), and for any e > 0 there exist N N, a polynomial P of N-complex variables, and functions f,.,' E 8 such that hf — P(fi fN)IIK 1' Pmof of Lemma 1.10.12. (a) We may assume that Uo X. Since K = K5, for any point a duo there exists a function fa such that 1f0(a)l > I > If0 Ilk. Let

be an open neighborhood of a such that hal >

is compact, there exist ai That is. That is, rank

ar

dUo such that dUo C

(K) is compact for any compact K C EN. = n for any x

u.

1

Ofl c2a.

U ...

Since dUo U Put

85

1.10 Holomorphic convexity

U := {z EUo: Ifaj(Z)l < 1. J = 1. r}. It is clearthatK CU C Uo and that the mapping F1 := (fai'..., far): U —+ Er is proper. n. Put F2 := p. Then the Let c > 0 be so small that IIcp1llu < 1, j = I mapping (F1, F2): U —+

is proper and regular. P(x) x P(x)). Since 4 separates points in X. for any

Let L := U x U \

(a, b) E L there exists an E 4 such that fa.b(a) fah(b) and IIfa,bIIU < Let be an open neighborhood of (a, b) in X x X such that fa.b(X) fa.b(Y) for any (x, y) E c2a.b.Since L is compact, there exist (at, b1) Put F3 that L C c2atb, U . . U (fai h1,.., .

(F2, F3): U —÷

is

(as,

E L such

Clearly, the mapping

injective. Finally, F := (F1, F2, F3): U —÷

is

proper, regular, and injective.

(b) Let U, N, and F = (ft fN) be as in (a). Then V := F(U) is a closed n-dimensional complex submanifold of EN. Put := f o F'. Clearly,q E (9(V). Let 0
— PIIP(r)

Thus

—

P(F)llK = lI'p — PIIF(K)

e.

El

By the above lemma the set

is

dense in (9(X). Since 4 is an algebra, we get 7 = 4. Since 4 is closed, we get

4= (9(X). The proof of Theorem 1.10.4 is completed.

[1

Proposition 1.10.13. Let (X, p), (Y, q) be Riemann domains over C" and let

F: X —*

I

be proper and holomorphic. Assume that (Y, q) is holomorphically convex. Then (X, p) is holomorphically convex.

Proof Let K C X be compact. Then L := F(K) C V is compact and, consequently, is compact. Thus, to prove Lao') is compact. Since F is proper, the set that k0(x) is compact it suffices to show that C L°°'). Fix a point and a function g E (9(1). Then go F E (9(X) and hence aE

Jg(F(a))I < max which implies that F(a) E 110(Y)

Ig o

Fl

max lgI.

El

1 Riemann domains

86

1.11 Riemann surfaces The following result due to R. C. Gunning and R. Narasimban (cf. [Gun-Nar 19671) can be thought as an example of a Riemann— shows that any open Riemann surface Stein domain over C'. Theorem 1.11.1. Any open Riemann surface X admits a locally biholomorphic mapC, such that (X, p) is a Riemann—Stein domain over C.

ping p: X —p

To see the main line of proof from the very beginning we observe that it suffices to find a holomorphic 1-form w 59) without zeros on X and a holomorphic function

f: X —s C such that

j

etw =

0 for any closed piecewise C' curve y in X.

(*)

In fact, put p: X —p C, p(x) := e1w, x c X, where xo is a fixed point in X and where integration is along an arbitrary piecewise & curve in X connecting xo and x. Because of (*) it is clear that p is well defined and holomorphic on X with dp = e1w. i.e. p is locally biholomorphic. We first recall a few facts from the theory of open Riemann surfaces we assume to be known to the reader (cf. [Fors 1981]).

Lemma 1.11.2 (Runge Approximation theorem). Let Y bean open subset of an open Riemann surface X. Suppose that X \ Y has no compact connected components. Then an)' holomorphic function f E 0(Y) is the compact uniform limit of functions I

E 0(X). Recall that an open set Y C X is called to be geometrically Runge in X if X \ is without compact connected components.

V

Remark. This generalization of the Runge approximation result is due to Behnke and Stein (cf. [Beh-Ste 19481). The main tool in their proof is a general Cauchy kernel whose existence was proven by using techniques from the theory of compact Riemann surfaces (e.g. Schottky-Verdopplung). in the meantime other proofs using methods from real analysis have been found (cf. [Nar 1968] or [Fors 19811). Moreover, we have the following exhaustion result:

Lemma 1.11.3. Every open Riemann surface can be exhausted by a sequence of relatively compact .cubdomains Xk

X. Xk

Xk+ 1' that are geometrically Runge

in X and have smooth boundaries. For a proof the reader is asked to contact [Fors 1981]. As usual, an open Riemann surface is a one-dimensional non-compact connected complex manifold.

Torecaliholornorphic I-forms see [Fors 19811.

1.11 Riemann surfaces

87

As a standard consequence of Runge's theorem one proves that any -form on X can be written as 131 with f E C°°(X) (cf. [Fors 19811). From here one is led to the generalization of the classical theorems of Mittag—Leffler and of Weierstrass. As a simple consequence we obtain: Lemma 1.11.4. Let X be an open Rien:ann surface and fix an xo E X. (a) Then there exists a merotnorphicfiinction f on X with xo as its only zero such that the order of vanishing off at xO is one. (b) There exists a holomorphic I -form w on X without zeros.

Remark. In order to prove part (b) of Lemma 1.11.4 one uses the generalization of the Weierstiass product theorem. Claim (a) in Lemma 1.1 1.4 which is a simple consequence of the Mittag—Leffier theorem is also true if we substitute 'meromorphic' by the word 'holomorphic'. Nevertheless, we will apply Lemma 1.11.4(a) only in the form it has been formulated.

One of the main results the proof of Theorem 1.11.1 relies on is the Mergelyan— Bishop theorem.

Theorem 1.11.5 (Theorem of Mergelyan—Bishop). Let K be a compact subset of X. Suppose that X \ K has no relatively compact connected components. Then any continuous function f: K —÷ C that is holomorphic in mt K can be uniformly app roximated on K by holomorphic functions on X.

Remark. This fact was first established by Mergelyan (cf. [Mer 1954]) in 1954 in the case where X is the complex plane and extended by Bishop (cf. [Bis 1958]) in 1958 to open Riemann surfaces using highly measure theoretic methods. Proofs of the classical result may be found in different textbooks (e.g. [Rud 1974]). Therefore, we will present only the proof of the general form. We point out that the proof given here does not use the Cauchy kernel of Behnke—Stein. But having once Theorem 1.11.1 it becomes easy to introduce a global Cauchy kernel on open Riemann surfaces. Using that kernel one can obtain a lot of interesting results in approximation theory on open Riemann surfaces (cf. [Gau-Hen 19821).

Proof Fixapoinix E Kandchooseacoordinateneighborhoodir: U = U(x) —+ E, biholomorphic. Put V := ir1 E). In virtue of the topological property of K we concludethat U\(VflK)isconnected. sois Therefore,C\ir(V()K) is connected.

Put g := Jo Observe that the pair (,r(V fl K),g) satisfies the assumptions of the classical Mergelyan theorem. Thus there is a sequence g1 (9(C) o ir E (9(U). Obviously, with gj —* g uniformly on ,r(V fl K). Put := fJ —*funiformlyonVflK. What we have done so far is that we have proved a local version of the Mergelyan— Bishop theorem using already the Mergelyan theorem in C. In order to complete the proof of Theorem 1.11.5 the following localization theorem is needed.

88

1

Riemann domains

Theorem 1.11.6 (Localization theorem). Let X be an open Riemann surface. K C X a compact subset and f a continuous function on K. Suppose that for every point XE KthereexistaneighborhoodU = U(x) andasequence(fj)JEN. fj E such that f3 —+ uniformly on K fl U. Then f is the uniform limit of a sequence (fj K )JEN with fj

f

I

Proof of the localization theorem. The proof will be based on the following solution of the d-problem

with estimates.

Lemma 1.11.7. Let G0 X be open subsets of X. G finite nu,nber of coordinate neighborhoods E, 1 U1

j

choose a

N, such that

Uj. Then there is a universal pos itive constant Co such that the following is GC true: wheneverw E 1)-form on G that is locally given =

(here z1 means the co?nplex variable for the coordinate neighborhood Uj)

(w1

I < j < N, then there exists an f E C°°(G0) with

such that O)ft0 and

Proof Fix a point a E Go. In virtue of Lemma 1.11.4(a), we find a meromorphic function on X such that vanishes only at a and its order of vanishing at a is one. for some r, I < r < N, and Then we choose a neighborhood Va = V(a), V0 r0 > 0 such that ._.+ raE is biholomorphic, fa(G \ V0)fl (r0E) = 0. Using the compactness of Go we find points ai E Go, meromorphic functions on X and neighborhoods x
= I on Go. we put on X with compact On := (PxW. i.e. we obtain a global support inside of Ur(x). Thus we find functions E such C denotes the complex for the coordinate where variable that =f neighborhood given by (Vi, fr). Then, by the classical theory in the complex plane, we find global functions given as E I

f

gx(z)

IT Jc

17),

= + iii,

Z E C,

—Z

such that = on C. Observe that this integral formula implies that

(a) (b) 60)

is holomorphic on C \ E; II

(Ic. where the constant

is independent of w and x.

Here we simply mean thc problem to solve the equation af

=

o for

a given

E

(G).

89

1.11 Riemann surfaces

In particular, could be thought as a function on C that is holomorphic at inunity. (It was already mentioned that this argument could be substituted by starting directly with the Weierstrass product theorem on open Riemann surfaces.) Since o are holomorphic mappings into C we obtain global functions on := X. Observe that is holomorphic on Go \ Therefore, we get on Go that

=

o

f

f E C°°(X) with duG0

=W1G0.

Moreover,

Ill IlGo the last inequality we used the coordinate transformations between the ir1 's and the f, 's and the assumption we made on the estimate of w. Hence, the lemma has been verified. 0 In

Now we turn to the proof of the localization theorem: First, let us choose a finite number of coordinate neighborhoods —+ E, biholomorphic, I
such that, putting U1 := we have that K C can U1 and be uniformly approximated by functions belonging to fl K). Fix an i > 0. Then we choose open sets U, and holomorphic functions U1 fl K C with < N. Let us fix some open set G0 — fIIo,nK < e. < satisfying K C G0 U1 and a partition of unity for Go with respect to the covering i.e. we have functions I. such 0 C

j

1

that

=

I

on Go.

Forapairofindicesj,k E {l

N}wedefine

p1(x)(f1(x) — f*(x))

ifx

:=

E

if x E czk \

clk n STZ1

UJ

Uk. It

is easy to check that hjk

:= fl7, Then we obtain the
Put

E

function hk :=

h1k

with —

we reach the following estimate for the form 8hk Ic'

sup

8(hk azk

=

a(h&o2Tt)

—

dzk):

Ct2e.

90

domains

1

is independent of k. where the positive constant we see that K C Therefore, by continuity, there Since Uk fl K C such that is an open set G, K C G G0

a(hk °-

sup

The fact that hk —

h3

jr') (nk(x))
azk

= fj

—

ft

fl Go implies the existence of a

fl

on

satisfying a = —dht on Go fl a on Go function u on G such that According to the above claim there exists a

au =

—8ht on G fl

and IIUIIG

:= ht + u on

where C* and Co are independent of e. Thus, function satisfying

<2€ + 3C*Coe Observe that — g1 = hk — h3 = 1) of a holomorphic function F on G,

If(x) — F(x)I <

+ Ifk(x)

—

1k

<3C*Coe,

=

(2

fl

on

=

flG is a holomorphic

+ 3C*Co)e. fl G which implies the existence

+ ft such that on K (1

f(x)I < (2 + 3C*Co)e + e = 3(1 + C*Co)s.

Thus we conclude that for any e > 0 there exists an F E (9(G), K C G, such that <3(1 + C*Co)e. i.e. f can be uniformly approximated on K by functions If — belonging to (9(K). Applying the Runge theorem fur the approximating functions the 0 localization theorem is verified. Hence the proof of the Mergelyan—Bishop theorem has been finished.

D

Remark. According to further applications we reformulate the geometric condition in the Mergelyan—Bishop theorem:

X \ K is connected if and only if

= K (cf. Definition 1.10.1).

For a proof of that statement use the Runge theorem.

Lemma 1.11.8. Let D, D' be relatively compact subdomains of X with smooth boundaries. Suppose that D c D' and that D and D' are geometrically Runge in X. Then there are simple closed piecewise C1 curves y, Yp in D' such that their homology classes a ) a (Yp) E H1 (D', D) form a basis of the relative homology group H1 (D'. D) 61)• Here ci denotes the second morphism in the following exact For the homology theory, the reader is asked to contact any book on algebraic topology.

1.11 Riemannsurfaces

91

homology sequence: 0 —* Hi(D) —÷ Hj(D') —* Hi(D', D) —÷

0.

Moreover,

Proof Recalling the classification theorem for compact orientable surfaces (cf. [Spr 1957], chapter V) with smooth boundaries we know that every connected component of of D' \ D is topologically equivalent to a 2-sphere with a finite number deleted. of open discs i = I handles attached and a finite number a canonical For each handle we select a pair pair of cuts on the handle. For different handles we may assume that the cuts are their boundaries disjoint. Suppose that for the first di,) of the discs a slightly 1 < <cu, are disjoint from OD C D'. Now let For the larger disc containing Hence is a closed curve in D', 1 j < we choose simple arcs such that joins remaining discs
j

j

Hi(D', D).

-

's. Since their endpoints lie in D they can be extended through D to form simple closed curves. y,, to represent a basis. Within all of these simple closed curves we select yi PutK := DUy1 U.•.Uy,,. InordertoshowthatK = K itsuffices,since D' is Runge, to prove that D' \ K has no relatively compact connected components. If there The only curves not closed are the

were such a component U, then its boundary 8U =:y is a cycle in K, i.e. y can be written as y = n1y1 + A, where A is a cycle in D. Passing to Hi(D', D) we get ni = = = 0 and therefore y C D. Hence U is a relatively compact connected component of X \ D which contradicts the hypothesis that D is geometrically Runge

inX.

D

Lemma 1.11.9. Let f: [a, b] —+ C be a continuousfunction and fixa c E C. Then there exists a continuous function g: [a. b] —* C, supp g C (a, b), such thot b

b

= c,

L

L

0.

Proof First we will prove the existence of a step function g with the desired properties and then we will use some approximation. In the case c 0 fix 10 E (a, b) and e > 0 such that — e, + e) c

Then we choose g to be a

non-zero constant on (to — e, to + e) and zero elsewhere to have both equations. In the case c = Owe fix points a
92

1

Riemann domains

such that

fit 0

riZ±€2

J

0.

J

Fix g tobeconstant0on(11 —

on (t2 —€2, t2 +62), andzeroelsewhere

t1

such that p11

+

e° I

Jil —€1

j

= 0.

/

Ji2—f2

b

= seX f

0.

Hence Lemma 1.11.9 is proven for a step function g. We approximate the step function g from above by continuous functions

gv: [a.bJ —÷ C,

C (a,b),

uniformly on compact subsets of the complement of a finite subset of[a, b]. Then we obtain the following entire functions coy and b

b

f

co(A) := f Obviously, the sequence

'p(l) =

c and

'p'(l) =

there exists a point A0 E C,

f

A0

converges compact uniformly on C to 'p. Observe that 0. Therefore, for a sufficiently large v 0, such that = c and 0, i.e.

b

Thus the function

b

=

c,

f

0.

has all the desired properties.

The purpose of Lemma 1.11.10 is to provide a way to modify, step by step, a zero free holomorphic 1-form w such that the modified form (9(X), has zero integral along any closed piecewise C' curve in X.

f

Lemma 1.11.10. Let D, D' be relatively compact subdomains of X with smooth boundaries; suppose that D C D' and that both are geometrically Runge in X. Let w be a nowhere vanishing holomorphic 1 -fonn on X with w = 0 for any closed piecewise C1 curve y in D. Then, for any e > 0, there exists a holomorphic function f E 0(X). If(x)I <e for x E D, such that e1w = Ofor all closed piecewise & curves y in D'.

Proof Using Lemma 1.11.8 for the pairs (D, D') and (D, 0) we find simple closed piecewise C' curves in D' and Yp+I yg in D such that

93

1.11 Riemann surfaces

y,., lead toabasis in Hi(D'. D),

•

• Yp+l' .. ., )/q give a basis in H1(D),

•K

compact subsets of X. w

=O,p+l

ApplyingLemmal.ll.9

j

we find continuous functions u-, I <j
. .

. U Yq such that

1
•Uj iszeroonb,

<j q. The (1

wi s are entire functions which satisfy the following properties for a 1.0

f

f

•

Wj(a)

•

wj(a)=fwehJ+hiP=fw=0, p<j
=

• • •

=

=

= L(a)

=

= 0.

f

u1wet4'

0.

1

f

j

1

=

0,

p.

k

<j
—(a)= f dço1

is non-singular. = 0 and the Jacobian matrix For j = I p we extend the function uj to be identically zero on D. Then

Thus

continuous on K and holomorphic on mt K. Therefore the Mergelyan—Bishop theorem implies the existence of a sequence of functions t' E N, holomorphic converge on X and uniformly converging to uj on all of K. In particular, the uniformly to zero on D. The remaining functions u, p < j q, are uniformly approximated on L by functions v1, holomorphic on X. Observe that L is a holomorphically convex compact subset of X and that L' 1-) U Yp) is a finite set of points outside of L. Then, fore > 0, it is easy to construct a function he, holomorphic on X, helL' = 0, and — LI < e on L. Therefore we may assume that the functions v3 vanish at the points of L' (multiply vj by he's). u1 is

94

1

Riemann domains

We define a new function v on K by setting v(x) := v1(x), wheneverx D, and := 0, x E K \ D. These functions are continuous on K and holomorphic on mt K. Again applying the Mergelyan—Bishop theorem the v can be approximated uniformly on K by functions holomorphic on X. Thus there exist sequences of functions holomorphic on X and uniformly converging on U .. U Yq to the limit function p q. If /2 E then we put

<j

C" —+ C",

:

+Zqtt'q.p

J

is a holomorphic mapping and. if/2 has been chosen sufficiently large, we conclude from above that j=1

=

det det

q

(J

<j
"+Up

p <j
J

0•

,...,q

Now we introduce the following entire functions

*p.j.v(Z)

j' converges compact uniformly in

Observe that the sequence

to the

function Therefore, for any positive t5 and any sufficiently large v there is a point zo E = 0, 1 q. llzo — all
j

f(x) :=

+

... +

+ zo.P+IwP+i.,L(x) +

... +

satisfies the following two properties

[we' =Oforalli, if v>> I and8 <<1, then lf(x)l
1.11 Riernann surfaces

95

Proof of Theorem 1.11.1. Take a non-vanishing holomorphic 1-form w on X. Using Lemma 1.11.3 and Lemma 1.11.10 we find an exhausting sequence of subdoX, and holomorphic functions 1k: X —÷ C, mains (Dk)k€N of X, Dk Dk÷l wefI+"+fA such that = 0 for all closed piecewise & curves y on Dk, IfkI in Dk+l. Putf := 1k. 1 is holomorphic on X and, if y is a closed piecewise C1 curve in X. then we1 0. Finally, we recall that any open Riemann surface is holomorphically convex and that (9(X) separates points in X (ci [Fors 1981]). Hence Theorem 1.11.1 has been verified.

Remark 1.11.11. In general, there is the following result due toO. Forster (cf. [Fors — 1970]): Any n-dimensional Stein man can be immersed 62) into The proof of this theorem relies on the above Theorem 1.11.1. For details the reader is asked to contact [Fors 1970]. While any open Riemann surface can be immersed into C, there is a concrete 20} which z2) E P(C2 ) : zo" + z12 + z22 does not admit an immersion into C2, i.e. X is not a Riemann domain over C2 (cf. [Fors 1970]). Thus the class of Riemann—Stein domains over C's, n > I, is a strict subclass of all Stein manifolds of dimension n. dimensional Stein manifold X := {(zo,

62) A holomorphic map 1: X —* Ck, X a complex manifold of dimension n, n k, is called an immersion if its Jacobian is of maximal rank. If n = k, then f is locally biholomorphic.

Chapter 2

Pseudoconvexity

In Chapter I domains of holomorphy were characterized with the help of properties of holomorphic functions (e.g. the holomorphic convexity). Now we want to give a characterization via more geometric objects (e.g. properties of the distance to the boundary).

2.1

Plurisubharmonic functions

Holomorphic convexity may be regarded as a generalization of standard geometric convexity. In the same spirit, subharmonic and plurisubharmonic functions may be considered as a generalization of convex functions. Plurisubharmonic functions are one of the most important tools of modem complex analysis. Therefore and for the reader's convenience, we decided to present an extended list of their properties (although not all of them will be explicitly used in the sequel).

If u: X —+ then for any Let (X, p) be a Riemann domain over a E X and C'1 we define a function Ua,f of one complex variable by the formula

(A €C: DefinItion 2.1.1. A function u: X —p is called plurisubharmonic (shortly if: ps/i) in X (U E • u is upper semicontinuous on X (u Ct(X)) 2), • for every a E X and is subharmonic in a neighborhood C'1 the function of zero: cf. [Rad 1937]. [Hay-Ken 1970], [Ku 1991]. We say that a function u: X —+ IR.). is logarithmically plurisubharmonic (log-

psh)iflogu 2)

Recall that for a function U: X —p

the following conditions are eqwvalent:

•i' E slim u(y) = u(x) for any x E X; • there exists a sequence C C(X. R) such that

u pomtwise on X.

97

2.1 Plurisubharmonic functions

The definitions extend in a natural way to Riemann regions 1)

u(X) C !},

E

As usual we put

IC

Remark 2.1.2. Directly from the theory of subharmonic functions (cf. [Via 19661) one gets the following properties of psh functions (Exercise). (a) For an upper semicontinuous function U: x —+ the following conditions are equivalent: (i) (ii)

U

E

IU=l

u(a) = (iii)

<

2ir —

0

dO,

J


< R;

30
f

<

0
< R;

IfJ=l 20
(iv)

f, jf

A —p (u o

is subharmonic; (vii) for any univalent open set U C X the function u o

=

+

(b)

(d) if C In particular, if

(e) If (f) If Ui tion 2.1.16).

and u, C

C

psh on p(U)

= ?SR(X).

R>0

f

(c)

is

(9(X).

\

u pointwise on X, then u E PS3€(X). [—oo, 0]), then u1, ,'PS.Y(X).

u locally uniformly in X, then u

and

UN E .'PSR(X), then max(ui

(g)(Liouvilletypetheorem)IfU E Even to complex manifolds: Let M

uN}

E

u

E

(cf. Proposi-

<+oo,thenU

as const.

an n-dimensional complex manifold with an atlas is said to be plurisubhannonk on M if for any I E I the function u o co7'is psh on Wi in the classical sense. 4) in particular, the local theory of psb functions on Ricmann domains reduces to domains in C". (Ui. coi)iEI (cf. § 1.1). A function u:

M —.

be

98

2 Pseudoconvexity

I —* IR be convex and increasing. (h) Let I C R be an open interval and let Then o u E .'PSR(X) for every u I). Consequently: If u E .'PSR(X), then e" E .'PSR(X) (in particular, any log-psh function is psh). If u E then 5'SR(X) for every p I. (1) If are log-psh, then + U2 is log-psh. U2 ProposItion 2.1.3 (Maximum principle). Let (X, p) be a Riemann domain and let U .'PSR(X). Ifu
< sup(limsupu(w):

E aD),

Z E D.

Pmof. Let X0 := {x E X: u(x) = u(a)}. Observe that X\

Xo

=

X: u(x)
{x

and therefore Xo is closed. Let xo E Xo. Applying the maximum principle (for subharmonic functions) to each of the functions with = 1, we conclude that lPx(xo) C X0. Thus X0 is open and therefore X = X0. 0

Letu: U —+

Ctm

bedefinedinaneighborhoodUofapointa E X. Wedefine

formal derivatives of u at a au

(p(a)),

-

aZ3

(d)a Observe that if f

If u: U —÷

(p(a)),

j=

n

1

has all real partial derivatives at p(a)).

(which exist if u o

az1

—(a) :=

o

•o

o

o

az1

••o

a, fi

E

= Dv.Of.

(9(X), then R is such that u 0

is twice IR-differentiable at p(a), then we

define the Levi form of u at a

j.k=I

Notice that £(11p112)(a;

=

Zj Zk

for any a E X and

E

C". Observe that

Consequently, we have the following Observe that

=

n. Moreover, a function! E

It clear that

J.k =

= 8j.k (the Kronecker

&(X, C) is holomorphic 1ff

Oforj =

1

n.

99

2.1 Plurisubharmonic functions

Proposition 2.1.4. Let u

C2(X, R). Then

£u(a: fl > 0.

u E ?SR(X) Remark 2.1.5. Assume that I C2(I, IR). Then

c

R is an open interval, u E C2(X, I), and

E

n

=

£(ço o u)(a;

+ j=J

a

X,

E C".

'J

Notice that the above formula and Proposition 2.1.4 give a direct proof of Re-

are of class C2.

mark 2.1.2(h) for the case where u and

Remark 2.1.6. Let (Y, q) be another Riemann domain over Ctm and let F E 0 ( Y. X). Then for u C2(X, R) we get o

= .Cu(F(b); F'(b)(ij)).

F)(b:

bE

E Ctm,

Y,

where

17(lfl

urn) ECtm

j=I denote the formal derivatives in Y).

F

Consequently, if u D'S.X(X) fl O(Y, X) —cf. Proposition 2.1.26.

&(X, IR), then u o

F

?SR(Y) for any

DefinItion 2.1.7. A function u C(X, R) is called strictly plurisubharmonic if for every domain Y X there exists an e > 0 such that the function u — eIIpfl2 is psh on Y.

Proposition 2.1.8. A function u .Cu(a;

is strictly psh iff

C2 (X,

> 0,

a E X,

E

= £u(a;

Pmof Fore > 0 put VE := u — Note that Fix a domain Y X and put

(observe that e

tion2.l.4, VE

= l}

a E Y,

e

> 0). Then £vf(a;

> 0,

aE

—

Y,

E

C". Hence, by Proposi-

E

Fix ana be such that

Lu(a;

XandletYbearelativelycompactneighborhoodofa. Then

=

which finishes the proof.

+

>

>0

Lete >0

0,

U

The following proposition gives a very useful tool for constructing new psh functions.

2 Pseudoconvexity

I 00

Proposition 2.1.9. Let that

C X be open. let

Y

v E

U E .'PSJt°(X).

limsupv(x)

4

Assume

E aY.

Yax—. .

Put

max(v(x), u(x)},

-

u(x) :=

xE Y

xEX\Y.

Then u E

Proof It clear that ü a vector E C'7 with

Obviously ü is psh on X \ I = 1. and 0
Take a point a

e1(X).

ü(a) = u(a) <

i

i

— 2,rj0I

dO < — I

aY,

üat(re'9) dO

2irj0

and we can apply Remark 2.1.2(a).

U

We will use the following two notations: eZ z

:=

z=

w := (ZIWI

Leta = (al If

z

= (ZI

(Zi

E

W=

Zn).

E C's.

E C", r = (ri

d&P1(a. r) —*

6)

is

bounded from above and measurable 7),

then

we

define

P(u; a,

r: z).

2

(ft

Irje'oj_(zj_aj)12)

+r.

z = (zi I

J(u;a,r):=P(u;a;r;a)= (2r)" If u: IF,, (a, r) —f

A(u; a. r) :=

lfr=(r

(IrrI).. .

E

J

is bounded from above and measurable, then we define

J

P(a,r)

u

=

-i--

f u(a + r E"

r)wesimplywrite: P(u:a,r:z).J(u;a.r),A(u:a,r).

6 7)

e'9)

That is, the function 10.

x ... x 9 —+ u(a -F- r . e'°) is Lebesguc measurable.

2.1 Plurisubharmonic functions

101

be upper semiconRemark 2.1.10. Let D be a domain in C" and let 14: D tinuous. Put := {(z. r) D x (IR>0Y': P(z, r) C D}. (a) The functions

—+

c2

A(u;z.r)

are upper semicontinuous (in particular, measurable). Indeed, by Fatou's lemma we have:

J(u; z. r) =

tim sup

urn sup

f

u(z + r

tim sup

u(z + r

(2it)

(z,r)—.(zo,ro)

J

[O.2,rl" (z.r)—.(zo.rn)

I

f

(2ir)" = J(u; ro),

A(u; Z, r) =

lim sup

( u(z + r

lim sup (:.r)—+(Zo,ro)

(z.r)—+(zo.ro)

J

ir

I

Jr" JE"

urn sup

u(z + r-

u(zo +

= A(u; zo, ro). (b)

a, r) =

2

. .-

r1

I

JO

2

fri I

--

I

J(u; a, (TI

Tn))ti .

. .

...

t,,

Jo

I J(u; a, (Tin,...,

.

-.

dr1

...

JO

PropositIon 2.1.11. Let D be a domain in C" and let u E rh'), 0 < r. r7
a E D, r' = n. Then

j(u;a,r')_<J(u;a,r")forr'_
2 Pseudoconvexity

102

Proof By Remark 2.1.10(b) it is enough to consider only the function J(u: a, •). First, we will prove that

,J(u; a, r') <J(u; The case n

=

I

a,

is well known. Hence

J(u(z', •, z"): a3, r) < j(u(z',

z"): aj. r,"), (z', a3, z") E P(a.

dD(a)). j

= 1, ..

.

,

n.

Consequently, using a finite induction, one can easily get the required inequality. By Fatou's lemma we have

u(a)<

IimJ(u;ar)<

I

f

r—.O

whichprovesthatJ(u;a.r)

Su(a), r-+O

\ u(a)whenr \0.

0

2.1.12. Let u2 E .'PSJt9(X). Ifui = U2 Ax -almost everywhere in X (cf Section I.2forthe definition of the measure Ax), then U2. ProposItion

Proof WemayassumethatX = DisadomaininC'1. Fixana ED. Sinceui =u2 everywhere, we get A(ui: a. r) = A(u2: a, r) for any 0 < r < d0(a). Hence, by Proposition 2.1.11, 0 = u2(a).

If D C is a domain and L: Aff(C", C'1)), then we define DL := we put UL

a o L: DL

an afline C-isomorphism (L —+ L1(D). Moreover, for any function u: D —+ is

+

Note that U E S'S.Y(D) if uj, E PropositIon

2.1.13. Let D C C" be a domain and let u: 1) —÷ Then the following conditions are equivalent:

semicontinuous.

(i) (ii)

u VrE(OR)n:

VOEDI

UL(Z) < P(UL;

a, r; z),

Z E P(a, r);

V,.E(OR,n:

(iii)

uL(a) <J(uL; a, (iv)

r)

VaED, 30
uL(a)

A(uL; a, r).

be upper

2.1 Plurisubharmonic functions

103

then the inequalities in (ii), (iii), and (iv) hold for such that P(a, r) C DL.

Moreover, every r = (rj

E

Proof

FixL e Aff(C", C??),a E DL,r = (r1,... P(a,r). Theresultiswellknownforn = Zn)

DL,afldZ = (Zi UL(W', zj, w")

P(uL(w', ., w");

r) 1.

C

In particular.

rj; Z1),

j=

n.

I

Hence, after finite induction, we get the required formula. The implication (ii) (iii) is evident. (iii) (iv) follows from Remark 2.1.10(b). (iv) (i): Fix a E D, E C's, = I. It is sufficient to show that

<

I u(a +

irr

JTE

forr > Osufficiently small. LetL E XE C. Now, by Proposition 2.1.11, forO <

u(a) = uL(a)

=

A(uL;a, (rj

urn

I

tim TI

= —i

(

=

I.


(ai an_I,

0

+

Exercise 2.1.14. Let (X. p) be a Riemann domain over C" and u E tPS,lt'(X). Prove

that 1

14(a)

<

in Sketch

f

Ax(Bx(a,r))

u dAx,

a

X,

0
px(a).

Bx(a.r)

u is subharmonic on X in the real sense.

of the proof We may assume that X = 18(a. r) C C", (p = id). Then

r))

=

I

JB(a.r) I

I /

u(a +

I

Jo JP(C"')JO >

I

,

jf o

P(C"')

.

=

where j,t denotes the standard measure in the complex projective space P(C" I)•

2 Pseudoconvexity

104

andu

PropositIon 2.1.15. If u E = 0. lar.

L'(X, bc); inparticu-

—00, then u

Proof We may assume that (X, p) = (D. idD), where D is a domain in Suppose that there exists a point a E D such that u = —00 for any neighborhood = —oo for any z P(a, r). U of a. Let 2r := do(a). Observe that J'P(z.r) u Consequently, A(u: z, r) = —00, z E IP(a, r). Hence

u=

—oo

in P(a, r). Let

Do := {z ED: u = —ooinaneighborhoodofz}. We have proved that D0

0. The same method of proof shows that D0 is closed in

0

D. Thus D0 = D — contradiction.

Proposition 2.1.16. Ifafa,nilv

C PSR(X) is locally bounded from above,

then the function

a := (sup IE I

ispsh in X. In particular. max (u

UN }

E IPSJt'(X) for any

UN

denotes the upper regularization of v. v*(x) := urn

Here

.?S.X(X).

v(y). x E

x Proof The problem is local, so assume that (X, p) = (D, idD), where D is a domain

We use Proposition 2.1.13(u). Fix L, a C DL, and r = (ri We have n. Note that UL = dD, (a), j = I in

supP((u1)L; a, r: z) < P(uL; a. r; z), LE1

rn), 0 <Ti <

z e P(a, r).

(El

Since the function z —÷ P(uL; a, r; z) is continuous, we get UL(Z)


Proposition 2.1.17. if a sequence above, then the function

z E P(a, r).

0

is locally bounded front

C

U := (limsupuv)* v-. ôc

ispsh on X. Notice that in general u*: X —* [--oc. +ocl is upper semicontinuous on X. If v is locally Ct(X, R), v = bounded from above, then v: X —p and v = ç' E C(X. R), v :

2. I

Plunsubharmonic functions

105

Proof We may assume that (X, p) = (D, jdD), where D is a domain in C'1. We use Proposition 2.1.13(u). Fix L Aff(C'1, C'1), a DL, and r = (ri, .

0
Note

that UL =

.

. ,

Then

V++OO


z E

P(a,r).

%'—•

Hence, using the continuity of the function P(UL; a, r;

we get

uL(z)
[1

Psh functions may be considered as a generalization of holomorphic functions in the sense that log If is psh for any holomorphic function f. Consequently, the class where u is psh, extends the class of thin sets. It will of sets of the form turn out that these sets will behave similarly w.r.t. psh functions as thin sets do w.r.t. holomorphic ones. I

DefinItion 2.1.18. A set M C X is called pluripolar connected neighborhood ha and a function Va

if

any point a C M has a

:PSR( Ua) with Va

—00,

Mfl Ua

C

We say that a pluripolar set M C X is complete if any point a E M has a connected neighborhood Ua and a function Va E [PS3C(Ua) with Va (—00).

—00, MflUa =

By Proposition 2.1.15, if M is plunpolar, then Ax(M) = 0. It is clear that any thin set M C X is pluripolar. Remark 2.1.19. (a) We would like to mention the following two results characterizing complete pluripolar sets. A. Zeriahi [Zer 1989]: Let M be a pluripolar subset of a pseudoconvex 10) domain

E D\Mthereexists

DC C'1.

au such that u = —00 Ofl M and u(a) > —oo. Then M is complete pluripolar in D, i.e. there is a v C S'SJ((D) such that M = {z E D: v(z) = —oo}. J. Wiegerinck [Wie 19991: Let D C C be a domain and let S C D be a sequence

in D without accumulation points in D. Let f E \ S). Then M := {(z. w) C (D \ S) x C: w = f(z)} is complete pluripolar in D x C. (b) (Pluri-thin sets) Let

A :=

[0. 1) —÷

be real analytic with w(O) = 0. Put

xe (0, 1)} CC2

and let u be psh in a neighborhood U of A. The function p extends to a holomorphic function in a complex neighborhood V C C of [0, 1). We may assume that A e V} C U. Put v(A) := A E V. Then 10)

See

§ 2.2 for the notion of pseudoconvexity.

2 Pseudoconvexity

106

SR( V) (cf. Corollary 2.1.26). Hence, by the Oka theorem (cf. [Via 1966])

v

u(0) = v(0) =

=

urn SUP(()

Consider the following two subsets of C2:

A1 :=

XE (0,1)1,

XE (0,1)),

A2 :=

where a E R>0 \ Q. In [Sad 1981] A. Sadullaev asked whether the set Aj is pluri -thin at 0 e C2, i.e. whether there exists a function u1 psh in a neighborhood of A1 such that U1(O) > urn supA a'—.O U1(Z), j = 1,2. The positive answer was given by N.

Levenberg and E. A. Poletsky in [Lev-Pol 1999] for A1 and by J. Wiegerinck in [Wie 20001 for A2.

Proposition 2.1.20 (Removable singularities of psh functions). Let M be a closed pluripolar subset of X. (a) Let u E IJ->S.If(X \ M) be locally bounded from above in X

Define

ü(x):= limsup u(y), XEX (notice that

is well-defined). Then ü E (b) For every function u E .'PSR(X) we have

u(x) := limsup u(y),

x

X.

X\May-.x

(c) The set X \ M is connected.

Proof (a) Since the problem is local, we may assume that (X, p) = (D, jdD), where D is a domain in Cz. Observe that ü = (uo)*, where uo

In particular, ü

:=

Iu

onD\M

(—oc

onM.

Ct(D). Moreover, a = u on D \ M. It remains to prove that ü is

psh on D. We assume additionally that M is complete (see [Ku 199!] for the general case). We may assume that M = v'(—oO), where V PSX(D), v —00, and v 0 in

D.

Fore >

0 let

u5(z)

:=

I u(z)+ev(z),

(—00, It is easy to see that and that from above in D. Observe that = a =

ZED\M ZEM.

the family Hence,

is locally bounded by Proposition 2.1.16,

E

That is every point a E X has a neighborhood Va such that u is bounded from above in Va \ M.

2.1 Plurisubharmonic functions

107

(b)Let ü(x) By (a), ü S'83((X). Moreover, use Proposition 2.1.12.

tim

ü=

sup u(y), u

x

X.

on X \ M. Now, since Ax(M) =

0,

we can

(c) Suppose that X \ M = U, U U2, where U, and U2 are disjoint and non-empty open sets. Then, in view of (a), the function u(z) := j for z E U3 would extend to a 0 psh function on X, which contradicts the maximum principle.

Corollary 2.1.21. Let M be a closed pluripolar subset of X. Let f E (9(X \ M) be locally bounded in X. Then f extends holomorphically to X.

Pmof We may assume that X D is a domain in Cit. Write f = u + iv. Then U E 5'SR(D\M) and u is locally bounded from above in D. Consequently, by PropoD), ü+ (a) = tim sition 2.1.20, u extends to a function u (z), a E M. We can repeat the same for the function —u. Thus —u extends to a func-

tionü_ E

=

=

—liThiflfD\Maz_.aU(Z),

PSR(D) and ü÷ + ü. = u — u = 0 on D \ M. Hence, a E M. Then ü÷ + u... 0, which implies that u extends to a function by Proposition 2.1.12, + ü_ ü E C(D). By Proposition 2.1.13, for any a E M and 0 < r < dD(a), we get ñ(z) = P(ü; a, r; z), z E P(a, r). In particular, ii is of class (3°° in D. The same argument may be applied to v. We have shown that f extends to a function f E C°°(D). Since the interior of M is empty, we see that f must satisfy 0 the Cauchy—Riemann equations in D and therefore f E (9(D). Remark 2.1.22. There are other, more general results on removable singularities of psh functions that we will only quote. Let M be a relatively closed subset of a domain G C C". The following two results are due to B. Shiffman (cf. [Shif 1968] and [Shif 1972]).

C G,theneachu E M), locally bounded from above on G, extends to a psh function ü moreover,

= ü(x),

x E G.

\ M) is locally bounded from above (b) If 3?2"2(M) = 0, then any U E on G, and so it extends to a psh function on G. > 0, then (b) becomes, in general, false; e.g. take Obviously, if G := C, M = (0}, and u(z) := 1/Izi, z E G \ M. So the problem remains which sets M of locally finite (2n — 2)-Hausdorif measure are removable singularities for

\ M). If M is a connected (real) &-submanifold of codimension 2, then the question was solved by N. G. Karpova (cf. [Kar 1991], see also [Pfl 1980] and 12)

denotes the k-dimensional Hausdorff measure (cf. § 2.12).

2 Pseudoconvexity

108

[Siu 1974]). The precise result is the following: M is a complex submanifold or if not, then any u E 9SJ((G \ M) extends to a psh function on G. For a sufficient criterion to extend not necessarily bounded psh function through a complex surface see [Kar 1998].

C 5'S3?(X) beasequence locally

Proposition 2.1.23 (Hartogs lemma). Let bounded from above. Assume that for some m C mit. v-_p

Then for every compact subset K C X and for every e > 0, there exists a

such that

Proof WemayassumethatX = DisadomaininC". Takean€ >0. itissufficient to show that for every a D there exist 6(a) > 0 and v(a) such that u m + e in v(a). Fix a and 0 < R
<Jim sup sup A(uL,; z, R + 5)

urn sup sup

:EP(a..)

a, R)


Ri"

R)

—

RTh

Ri"

<m+e,

0

provided that 8 is sufficiently small.

Recall that smooth psh functions may be easily described by properties of their Levi forms (Proposition 2.1.4). Thus it is important to be able to approximate (at least locally) a given psh function by smooth psh functions. The required approximation may be given by the following procedure.

Let 4(Zi

Zn)

:=

.

. ..

'1'(z,),

z=

E C", Where 4' E

(zi

is such that:

.supp4' = • 4'(z) = 'l'(IzI), Z E

.1 4' dA2 =

C,

1.

E

e>

0.

109

2.1 Plurisubhai-monic functions

Notice that: ITh • E '-o • supp = P(e), •

=

• •

=

Z

=

(zI

Zn) E

1.

Let

:= (x

X: dx(x) > e},

>0.

For every function u E Lt(X, bc), define

I = j(u

ue(x) : =

The function X

is

—

p(y)) dAx(y)

JPx(x)

+ ew)c1(w)

o

x E X5.

called the e-regularization of u. Observe that for a E

E lPx(a, dx(a) — e) we get

I

JPx(a)

=

j(u

+ ew)4(w)

o

Remark 2.1.24 (cf. [Sch 1966]). (a) Ut (b) If E e"(X), then Da,flu€ =

E

(t)

e > 0. for any al +

< k and e > 0

(k E

(c)If K := suppu

In for 0 < e < X, then suppue C C forO < e C < —÷ uniformly onXwhene —÷ Oforany (d)Ifu E then Daflue uniformly on X k. In particular, if u E al + when e —÷ 0 (for any a, fi). Lt(X) (1 < t < +oo) and K := suppu X, then ut E Lt(X) and (e) If u

particular,

E

forO <e < (f)

Moreover,

u when e —÷

0.

)e2 = (Ut,)tj on

fl

e u

pointwise

Pmof It is clear that

in X when e \ E

0.

(Remark 2.1.24(a)). By the second part of (t) we

get 1'I

= (2jr)" JO

p1

... Jo

J(u; a, e(r1

tn)ti .

. .

dt1 ... dry.

110

2 Pseudoconvexity

Consequently, by Proposition 2.1.11,

willapplyRemark2.1.2(a). Fixa E 2ir J —

= 2

+

o

j

f(u

Proposition 2.1.26. Let Then u o F

u. It remains to prove that is psh. We E C", = 1,and0
dO

+

o

f(u o

\

+ ew)

+ ew)D(w)

=

0

ue(a).

(Y, q) be a Riemann domain over Ctm

and let F

(9(Y, X).

any u E

Proof We may assume that u —cc. We already know that the result holds if U E C2(X) (Remark 2.1.6). Let denote the s-regularization of u. Put F1(XE). Then o F and u o F E 0

Corollary 2.1.27. Let u: X —+ be upper semicontinuous. Then u ispsh on X ifffor any anah'tic disc ço: E —÷ X the function u o is subharmonic in E.

Lemma 2.1.28. Let D c C" be a domain and let u E log-psh:ffforanya E Candj E (1 n}lhefunction D

z

Iltf). Then U is

Ieozulu(z)

ispsh.

Proof We only need to prove that if v,,,, is psh (for any a and j), then log u is psh. By definition, we have to check that for any zo E D and e the function A log u(zo + is subharmonic (in the region where it is defined), equivalently (cf. [Via 1966]), we have to prove that for any wo C, the function

+

A

0. Put a := Let j be such that 1v0J(zo + A'). Thus ço is subharmonic provided Vaj is psh.

is subharmonic.

Proposition 2.1.29. (a) Any C-norm q: C" —p (b) Let/i: C" —+ R÷ be such that

h(Az)=IAIh(z), Then

h ispsh

is log-psh.

is log-psh.

AEC, ZEC".

Then

=

0

Ill

2.1 Plurisubhannonic functions

Proof (a) Given arbitrary a,

E

C", the function

C?

A

—÷ q(a

+

is convex, and consequently subharmonic. Thus q E .'PSR (C"). = Now we can apply Lemma 2.1.28 because hand side is psh by Proposition 2.1.26. (b) follows from the proof of (a).

If u

Proposftion 2.1.30. (a)

U

—00, then Lu > 0 in X in the

u

E

and the right

distribution sense, i.e.

dAx(x)

u(x)Cço(x;

0,

E

R÷),

E

(b) If u E '(X, bc) is such that Lu > 0 in X in the distribution sense, then there exists afunction v E ?SJt°(X) such that u = v Ax-almost everywhere in X. Observe that if u

E

0 for any x E X and

C2(X. IR), then Lu 0 in the distribution sense ill Lu(x; E C" (use the Stokes theorem).

Proof (a) Let

denote the s-regularization of u. By Proposition 2.1.4, Lue in the classical sense and, consequently, also in the distribution sense, i.e.

0 in

EC". u (Proposition 2.1.25), we get the required condition. Now, since (b) The problem has a local character and therefore we may assume that (X, p) = denote the e-regularization of u. First (D. idD) where D is a domain in C". Let IR+) we have For every function E we prove that E

f

=

f —

= ID, (ID

w:

?:0,

C",

IDE

which shows that

E

Now we show that u6 \ when e \ 0. Let 0 < ue,(z) =

0

urn

I

(

= urn

I

(I

=

< e2. Then, by Proposi-

we have

tion 2.1.25, for z E

= urn

I u(z + eu' + e2fl)1(11) JE"

(z) > urn

6—'O

(z) =

urn (uE1 )E(z) = Ut, (z).

2 Pseudoconvexity

112

Let v :=

Then V

E

?SR(D). It is well known that

u

in D. Hence u

u

=v

0

As-almost everywhere on D.

Exploiting the above properties, one can simplify the original description of psh functions, namely:

PropositIon 2.1.31. Afunction u: X —÷ is psh if C X of a and a function (a) for any a E X there exist an open neighborhood V

E

such that u < v in (b) for every a E X and E

the function

is subharmonic in a neighborhood

of zero.

Remark 2.1.32. (1) There is a conjecture that the above proposition remains true without condition (a), i.e. u is psh if for every a E X and E C" the function is

subharmonic in a neighborhood of zero ft]

(2) Proposition 2.1.31 under the assumption that u is locally bounded from above was first proved by Lelong in [Lel 1945].

(3) The following general theorem is true; cf. [Ru 1989]. Let U C Rk, V C Rt be open, k, £ > 2, and let u: U x V —p

be such that

(a)for any (a. b) E U x V there exist an open neighborhood c C U x V of(a, b), such that u
= +oo is due to Avanissian LAva 1961].

The case r = I is due to Arsove EArs 1966]. (4) Notice that the separate subharmonicity (without any additional assumptions) does not imply the global subharmonicity; cf. [Wie-Zei 1991]. Proof of Proposition 2.1.31. We use induction on n. The case n = 1 is trivial. Assume that the result is true for any Riemann domain over C"'. Let (X, p) be a Riemann domain over satisfy (a) and (b). The problem has a local and let u: X —+ character and therefore we may assume that (X, p) = (D, idD) where D = U x V is

a bounded domain in C" =

x C. We may also assume that u

v, where

v E L '(D). Observe that for any k E N the function Uk := max{u, —k) also satisfies u as k / +oo. Consequently, it suffices to prove that (a) and (b). Moreover, for any k E N. Thus we may assume that u —k in D, and finally, uk E taking u + k instead of u, we may assume that u 0. We have to prove that u is upper semicontinuous. We will need the following lemma.

Lemma 2.133. Let A C B U: A x B —* R+ be such that

C R'l be bounded Lebesgue measurable sets and let

2.1 Plurisubharmonic functions

(i)u(x, .)

I 13

is upper semiconrinuous on Bforanyx E A,

(ii)u(.,y) is measurable on Aforanyy E B. Define

ü(y) := I u(x,

y E B.

JA

Then

ü is Lebesgue measurable on B.

Proof Let Uk := min(u, k}, k E N. Observe that Uk satisfies (1) and (ii). Put

yEB,

kEN.

k/

+00. Thus, it suffices to prove that Uk is upper semicontinuous on B for any k. Take a yo B. Since Uk is upper bounded, the Fatou lemma gives:

By the monotone convergence theorem Uk / ü as

limsupük(y)

J

=

A

f

=

0

A

First we prove that u is locally bounded from above. Take (a. b) D and let r > 0 be such that b), 2r) D. By the Fubini theorem, L'(U) for A2-a.a. w E V, say for w E B. w) Consequently, by the inductive assumption, u (., w) e PSR(U) for w e B. Notice that u(z, .) is subharmonic in V for any z U. By Lemma 2.1.33, the function B

is

I

irr- Jp(,7r)

<

irr

u(z,

E ?(a, r). Consequently, for any

measurable for any

irr

I

w —÷

j

f

(jrr2)"

w)dA2(w) =

( (irr

f

)"

( (,rr—)" v(z,

f

Jrr

I

E

w)dA2(w)

u(z, w)dA2ii_2(z))dA2(w) v(z, w)dA2n_2(z))dA2(w)

P(E.r)

w).

JP((a,b),2r)

Now we prove that u is upper semicontinuous in D. Take (a, b) D and let ro > 0 be such that IPUa, b), ro)

D. Suppose that u(a. b) < M. Since u is locally bounded from above, the inductive assumption yields

2 Pseudoconvexity

114

that u(.. h) is psh. In particular, u(., b) is upper semicontinuous. Hence there exists 0 < ri
A(u(•,b);a,r1) =

1

2

u(z, •) is subharmonic, we get

A(u(z, .); b, r) =

I

irr

u(z, w)dA2(w) \ u(z, b) as r

Jp(b.r)

\ 0,

z E U.

that for each 0 < r < ro the function z —÷ A(u(z, .); b, r) is measurable. Moreover, since u is bounded on IP((a, b), ro), we easily conclude that the integral Recall

I

A(u(z, .); b, r)dA2,,_2(z)

JP(a.r)

is finite for any 0 < r
JP(a,ri)

(lrrI

Finally, for 0
irr2

and

f

P(b,r2)

E IP((a,

b), r), using the mean value theorem (and

the fact that u > 0), we get ,j)

(ir(ri — r) 2 —r

\JT(r2

—

u(z,

r) 2 1

JP(b,r2)u(z,

r2 — r

r)2 M, which shows that u <M in P((a, b), r) provided that r is sufficiently small.

Remark 2.1.34. Let c

be open and let u E

C

R). Then

41u((x + iy); (a + ib)) = .7(u((x, y); (a, b)) + Ru((x, y); (b,

—a)),

where R denotes the real Hessian IfU N

82

x

Q) k

U, Q = (Qi

QN)

RN.

0

2.1 Plurisubharmonic functions

ProposItion 2.1.35.

C C's,

13(x

(if v is convex on U

Then 1)

and let v: U —+

L.e: U be a domain in

U := U +

115

+ iy) := v(x),

Define

x + jy e U.

14)

Proof First consider the case where v is of class e2. By Remark 2.1.34 we get

4.Ci)(x + iy; a + ib) = .lev(x; a) + Jev(x; b), which, of course, implies the required result.

In the general case, assume that i3is psh and let of 13. Observe that + iR" = Hence UE C R". Moreover,

+ it) =

I

13(z

=

denote the s-regularization for an open set +

+ it +

J E"

=

1(z

+

=

zE

tE

J E"

=

Hence,

UE —÷

E

IR.

\

v.

By the first part of the proof, u6 is convex in UE for any e > 0. Consequently, v is convex.

v (in R"):

Conversely, assume that v is convex and let v(x +

vE(x) := f

{x E U: B(x, e) C U},

x

JBU)

where 0 is a

function in R" Put Ue : UE + iR" C C", By the first part of the Note that 13. 0 for any e > 0. Consequently, 13 is psh in U.

\

13(x+iy) := v,.(x), x +iy E proof,

is psh in

PropositIon 2.1.36. Let D c C" be an n-circled do,nain and let u: D —f such that Zn) E D. U(ZI, . .. ,Zn) = u(IztI (ZI Put IDI

:= {azil

:=u(rI logD:={xERtt:eXED}, ü(rI

14)

Zn) ED) C

(zi

(ri

IDI.

ü(x):=u(eX),

x€logD.

rn),

Ifv E C2(U, R). then v isconvex 1ff Rv(x; Q) Oforanyx EU and Q ER". I.e. 9 E SUpp9 = B(1). fa(I) 9 = 1, and 9(x) = for any x = (xI e R". It is known that if v is convex in U, then is convex in and e \ v.

be

116

Foranyl=(ii

.

2 Pseudoconvexity

ik)withl

k). the intersection of D with the (n — k)-dimensionalsubspace (Zj1 = 0: j Let uj denote the We will identify Di with an (n — k)-circled open set in restriction of u to D,. is convex. liz the case where D C we gel: u E zffü (a) (b) In the general case, u E .'PSJt"(D) (i) ü is upper semicontinuous on IDI, ix — land! = (i1 (ii)foranvk =0 i1 < ... <
(iii)for any j =

1

n,

I

if x

{(r? then

x

IDI,

the function

is increasing.

Proof (a) First z E

and

assume that u

D and 9 E R'1.

Consequently, for

E

E C2(D, IR). Recall that u(z) = = £u(e10 z; e'0 . for any z we get: any z E D and 0

E

£u((JziI

? 0.

Hence .Cu(z:

£u(z;

0

One can easily check that 4Lu(z; E)

+

d.jrk

=

Z = (zi

Zn)

E D. r = (rj, ..

.

:= (IziI

,

In particular.

4LuV;

=

arjark

+

j=I

=

(e%)e_x1(al j,k=I

j=I

x€logD, On the other hand,

Rü(x:

(2= j,k=I

j=1

x€logD,

z) for any D, 9 E

2.1 Plurisubharmonic functions

117

Finally,

4Lu(ex;a+ib)=.u(x,e_%.a)+J?n(x,e_X.b),

xElogD,

which implies the required relation.

\

Now, let u be arbitrary and assume that u is psh. Let denote the s-regularization is n-circled and for any z E and is psh and u). Observe that of u o W' we get ue(et8 z)

=

z

+ ew)cD(w)

JE"

u(e'0 z + ee'0 .

=

= I u(z +

=

J

Zfl)=UE(IZII of the proof, ü is convex in log(DE). Since

Conversely, assume that v :=

ü

\

UE(z).

Bythefirstpart ü, we conclude that ñ is convex.

is convex in G := log D. Let

denote the

standard e-regularization of v:

XE

v(x JB(I)

is convex in

and

\

G€

:=

B(x,e) C G}

{x E G:

v). Define

:= {(Zi := VE(logIzlI

C't: (loglzil log

Ge).

(zi

Zn) E

D and uS \ u. By the first part of the proof, ut E Hence U

E

Then it is clear that (i) is satisfied. Observe

(b) First assume that u

that log D = log(D fl (Ce)"). Hence, by (a), condition (ii) is satisfied for k = 0. n — I, the function uj is psh on D,. Consequently, we can repeat Fork = 1

the above argument and so (ii) is satisfied for any k. To prove (iii) observe that the function

rI_i,A,rI+i is

well-defined, radial, and subharmonic. In particular, the function [0,

is increasing.

rj —÷ J(v; 0,

=

rI_i, rj. rI+i

2 Pseudoconvexity

118

Now, assume that (i), (ii), and (iii) are satisfied. Then obviously u is upper continuous on D. Moreover, by (a), u is psh on D fl and, more generally, each function is psh in D, 11 = (13 1k).

=

with

1.

We want to prove that

u(a) — 2,r j0 1

u(a +

dO

forsuflicientlysmallr >0. Wemayassumethatai•...•at =a,, =0 0 with 0 < £ < n — I. Let 0 < r < dD(a) be so small that ... e a Take! I Recall ispshin for(zt n). that Ui D, fl

I). Hence, using (iii), we get

u(a)< — I 27r Jo 2,r

/ j0

0)dO

+

+

dO

f2iT

0

=— 2ir /

The following results on psh functions will be needed in Chapters 3 and 4 (the proofs of a part of them are beyond the scope of our book).

Theorem 2.1.37 (Choquet's lemma). Let X be a metric space whose topology has I. Assume that the family (uj)JEJ is a countable basis. Let u: X —. locally bounded from above. Then there exists an at most countable subset of! such that

= (supuj)*. lEt

Moreover,

each u is lower semicontinuous, then U,

lElo

Proof We may assume that #1 2: bounded from above, put

S(v) := {(x, t) E X x IR: v(x) > t},

=

iEi

can be found so that

U.

For any function

T(v) := ((x, 1)

v:

X —+

X x R:

locally

v(x)

> t}.

Observe that S(v) is open if v is lower semicontinuous and T(v) is closed if v is upper semicontinuous. Moreover, T(v*) = S(v) 16)• Since v

observe that

is uppersemicontinuous, T(v) is closed. Obviously S(v) C T(v). It remains to C S(v)

2.1 Plurisubharmonic functions

119

For A c I define uA := SUPIEA u,. We want to find a countable set lo C I such that u0 = u. It is equivalent to the equality = T(u). Since S(uA) = we only need to prove that UEA S(u,),

U S(u,) = Usui EEl

IEIO

for a countable set Jo C I. Let } be a countable dense subset of S(uj). For any U E N take an i(i') E I such that Now, it suffices to put E Jo := {i(1), i(2). ... Assume that each u is lower semicontinuous. Consider the open covering (S(u,))jEJ of S(u,). By the Lindelöf theorem, it has a countable subcovering Then S(uj0) S(u,) and consequently, uj. U . . .

Proposition 2.1.38. Let (X, p) be a Riemann domain over C's. (a) Let (uj)jE, C ?S.X(X) be locally bowuled from above. Put u :=

u.

Then the set {x E X: u(x)
=

u1(xo).

In fact the set {x E X: u(x)

is

pluripolar—cf. Theorem 2.1.41.

Proof (a) We may assume that (X. p) = (D, id), where D C is a domain. By Choquet's lemma there exists a countable set Jo C I such that where = u0 Observe that (x E X: u(x)
u(z)

P(u; a, r; z).

z E P(a, r), 0 < r < r.

Hence

zEIP(ar), 0
P(P(u;a, r; .);a, r';z) = P(u:a, r: z).

ZE

r'), 0< r' < r < r.

Thus

P(u*; a, r; z)

In particular, J(u*: a. r) A(u*; a, r).

P(u; a, r, z),

z E IP'(a. r), 0 < r
J(u: a. r), 0 < t < r. Consequently, A(u; a. r) =

2 Pseudoconvexity

120

u and

b) Let Vk := SUP1,>k

(x

E

X: u(x)
E X: vR(x)

< v(x)}

0

and we apply (a).

The problem whether any pluripolar set can be described by one global psh (cf. [Let 1957]) function was open during many years and was finally solved by Josefson in 1978.

Theorem 2.1.39 (Josefson theorem; cf. [Jos 1978]; see also Proposition 4.1.11). If M C C" is pluripolar, then there exists a v v —00, such that MC Notice that a similar result is true for plunpolar subsets of an arbitrary Riemann domain over C", cf. [El 1980]. Proof. We will follow a simplification of the original proof given by J. Siciak in [Sic 19831.

Step 10: Suppose that for each a M there exist r(a) > 0 and Va Va 0 —oo,suchthaiMfllP(a,r(a)) C tç'(—oo). Thenthereexisisav E —00. such that M C v_1(—oo). v Proof of Step 10. By the Lindelöf theorem, there exists a sequence C M such that M C r(a1)). Put V3 := v01, = 1,2 Since for each j the set

j

vJ'(_oo) is of zero measure, there exists a point b IP(I) such that v1(b) > —oo for all j. We may assume that v3 0 on P(j) and —2', j E N 17) 1)efine v := v1. Then v v(b)> —1, and M C 0 5'SR(P(a, r)), Thus fix an a M. By definition there exist r > 0 and u —00, such that M fl IF(a, r) C u'(—oo). u We may assume that a = 0, r = 4, and u <0. Step 2°: (Cf. the theorem of Bremermann, Proposition 4.4.13.) There exists a sequence C (9(P(4)) such that IIft'IIP(2) 1, ii = 1, 2 and u = P(4), where (2.1.1) V

Proof of Step 2°. Define a Hartogs domain

G := ((z, w) E P(4) x C: wi < ft suffices to substitute v1 by a function of the form r,(vj — c3) with Cj := an appropnatc > 0.

Vj

and

2.1 Plurisubharmonic functions

121

It is well known that G is the domain of existence of a function f E (9(G) (cf. [Via 1966]; see also Corollary 2.2.15 and Theorem 2.5.7). Observe that

K := P(2) x E

G.

may assume that ft

I on K. The function f may be represented by a Hartogs (z. w) G,where fv (9(IP>(4)), V =01,2 and the series is locally normally convergent in G. In particular, by the Cauchy I on IP(2), v = 0, 1,2.... Obviously inequalities, u. Finally, since G is 18)• the domain of existence off, we easily conclude that u = 0 Since u —oo, we have 5UPP(l/4) > —oo. Hence there exists a point zo E with > —oo. Consequently, we find an infinite set C N and q N, 2, such that Ifp(zo)I > VE q We

series f(z,w) =

Define

gv(z)

+ zo),

:

z E IP(3),

E

V

Observe that

Step 3°: Foranyq,k,

V

>

1.

Jg1, IIP(

E Nwithi' > >

4q

VE

>8,

(2n+2)!,

(2.1.2)

1,

(2.1.3)

and for any g E f9(P(3)) wit/i IIgIIP(I)

Ig(O)I >

d :=

there exists a polynomial P

E P(l/2): Ig(z)I <

Proof of Step 3°. Put K := k'1,s

g(z) =

such that

<

I< {z

(2.1.4)

c (z

P(1/2): IP(z)I < Write Z

E P(3).

Let ü := O := ((z. w) P(4) x C: wi l'hen G is a Hartogs domain with G C G. Take 0 <(1 < 1 and put G0 := {(z, w) E P(4) x C: <9). Then, by the Hartogs lemma (Proposition 2.1.23). for any compact L n O9 there exists a such that "wi < 9 for (z. w) E L and v Thus the series is locally nonnally convergent in G to a function I E (9(G). Hence G = G and, therefore. = u.

2 Pseudoconvexity

122

By (2.1.3) we get

ae

IaaI < 1,

(2.1.5)

Let A := (Z4. fl [0. vK))", B := (Z÷ fl [0, s))". For any x =

Q(x,z) :=

C C, put

H(x.z) := Q(x,z)g(z) = fl€A

where

and au_p =0 ifa —

:=

The main part of the proof is to show that there exists an x = (XP)fiEA such that

La(X)=0, aEA\B, maxlxpl=1.

(2.1.6) (2.1.7)

>

L(x) := max

(2.1.8)

aEA

Assume for a moment that such an x is already constructed. Define

W(z) := crEA

P(z) := (l/L(x))W(z) = >2(La(X)/L(x))Za,

Z

E C".

aEA

By (2.1.6) P Moreover, by (2.1.6),

By the Cauchy inequalities, we get that IIPIIP(l)

lIP IIPU)

#B

=

.s"

Observe that by (2.1.5) and (2.1.7) < (KvY'. Now, let z such that Ig(z)I Then, by (2.1.6), (2.1.7), and (2.1.8), we get

IP(z)l <

<

z)I

+ IH(x, z) — W(z)l) + \A

=

+ 2" — 2"(I

(Kp)ne

/2(e_KI)

<

(2.1.2)

<

where (*) follows from the Bernoulli inequality. Indeed,

+

/6(e_Av13 +

<

t9)

E A.

I

—

E

P(1/2) be

2.1 Plurisubharmonic functions

123

Thus it remains to construct x (with (2.1.6), (2.1.7), (2.1.8)).

A =: Bo, A,1 := A \ {aI

as"}. A0

Write B = {a'

=

?. The system (of linear equations in the variables x)

La(x)=0,

CXEAI,

has less equations than variables. Consequently, there exists a solution x' = (x),)PEA = I for some B0. such that maxficA = 1. We may assume that Put B1

Now consider the system

Bo \

La(X)O, a€A2. =0. The

system has a solution x2 =

some

such that max$EA

I

for

E Bj.

Put B2 := B0 \

:=

Using this procedure, we get

B€_i :=Bo\{13'

s'1,suchthat

I

La(Xt)0,

x,— £

maxlxfl —1. i

flEA

We only need to show that

max{ILE(xt)I: a E A. Suppose (2.1.9) does not hold,

D(U. where U

=

a

(2.1.9)

A.

£

=

1

s".

(2.1.10)

c Awith#U=#V=:m> l,let =:

V) :=

and V = (SI

{y1

> e_I
1

i.e.

ILa(X')I

ForanysetsU,V

=

£

in the lexicographical order.

By (2.1.5)

I

(Ky)"

v2" <

+ I)!

e".

s"

20)

(2.1.11)

2 Pseudoconvexity

124

We claim that

D(A,, Bt)I >

£ =

0. Sn.

(2.1.12)

Observe that (2.1,12) finishes the proof of Step 3°. Indeed, using (2.1.11) with

£ = s", we get

+s"Ki'/2 < v(Ku)°, which implies that < 2(q + 1)v(KvY'; contradiction 21) We move to the proof of (2.1.12). For £ = 0 the matrix Faa_fl la.flEA is triangular. Hence, by (2.1.4). > ID(Ao, Bo)I =

<s". Put

Suppose that (2.1.12) is true for some £ with 0

if(a.$)

b

if(a, fi) =

—

'I

Then

= 0,

22)

r

flE

Consequently, detlbufilaEAf ,ficHt = 0. On the other hand

det[baflj =

Bt) ±

Be+i).

Now, using (2.1.10). we have

JD(At÷i,

Be+i)I

>

2: ID(At,

which finishes the proof of Step 3°.

Step 4°: Main pmof First we apply Step 3° to and k E N with v E such that with d(v, k) := (2n + 2)!. We get a polynomial E

K

=

k'1

>

such

that <

I

(z E P(1/2):

C {z

(2.1.13)

P(1/2): (2.1.14)

21

Becauseq > 1+! Since

1.

= 1, we have

=

=0.

2.1 Plurisubharmonic functions

Put

— Zo). Then

and by (2.1.13) we have231

E

>

2

IIQu.kIIpi) 2

125

(2.1.15) < .jdR.k)

IIPV,kIIP(5/4) <

(2.1.16)

Moreover, by (2.1.14), we get

<

(z E P(1/4):

C (Z E IP(1/4). gv(z — zo)I

C {z E P(1/4).

—

< e_Kt}

zo)I < <

C {z E P(1/4):

(2.1.17)

4r, r > Observe that by (2.1.16) a locally bounded family of log-psh functions. Put

1.

Thus

is

vk

4r on P(r), r >

Then v is log-psh and Uk We claim that

1.

(2.1.18)

VZEMnP(I/4): v(z)

Vr
Indeed, suppose that there exist T < 0 such that (2.1.18) does not hold. Then there exist k and E M fl P(1/4) such that k'2 > (2n + 2)!, T > and —÷ zo such that 2 eT+I. By Proposition 2.1.38 there exists a sequence > eT/2 for IL 2 Consequently, = —÷ v(zo). Hence for any ji 2 there exists an infinite set C E such that A

> 23)

>

nqv

=

t'j12

vE

We will apply the following elementary inequality:

Ip(:)I

pE

Indeed, fix p, r, and z as above. Let C

:= IIPIIP(r). Consider the function

0< Al c

v(A) := log Observe that v is subharmonic in

z C C". Izi ? r > 0.

K(lzl/r) \ {0) and v(A) 0 for Xl = lzi/r. Thus. v extends to a function subharmonic in K(IzI/r). zl/r. In particular, v(l) <0, which gives OforalIlM

Moreover, < Hence, by the maximum principle, v(A) the required inequality.

126

Pseudoconvexity

2

?

which implies

YE Hence, by (2.1.17), 24) contradiction. and finally u(zo) > e_ Now, we are going to prove that

? 3/10.

(2.1.19)

k —*

< 3/10 on 2(1). Then, by the Hartogs lemma Suppose that urn (cf. Proposition 2.1.23) there exists a ko such that

forzElF(7/8), Thus, urn

<3/9 forz E 2(7/8),

sup

k > k0.

li—.

Using once again the Hartogs lemma, we conclude that

< (3/8)d(v.k) forz

E

2(3/4),

V

V > Lk)(k), k > ko.

E

Consequently, by (2.1.15),

[(3/8)(4/3)I"

<

.k)

for v E

V>

k > /co;

contradiction. Let be as in (2.1.19). By (2.1.18) (with T := > (2n+2)!suchthat <

on

M

— I) we get

fllF(1/4)

E N

with

(2.1.20)

and lim

i1—*+cx)

> 3/10.

v

(2.1.21)

Define v := I

Obviously, by (2.1.20), v

=

—oo

on Mn 2(1/4). Recall that on 2(r), r>

log v which

implies that v

v

—

log 4r

=

is

<4r.

a global psh function

log

and

1, we get

E N,

By (2.1.21),

the series consists of negative psh functions on P(r).

2.1 Plunsubharmonic functions

127

C X be

Proposition 2.1.40. Let (X, p) be a Riemann domain over M1 is j E N. Then M

pluripolar,

Proof We may assume that (X. p) = (D, jdD), where D is a domain in C". By —co, such Josefson's theorem, for each j E N there exists a v1 E

c

that

(—00). Now we argue as in Step 1° of the proof of Josefson's theorem. D

As already mentioned, now we are going to list a few results on psh functions without proofs.

Theorem 2.1.41 (Bedford—Taylor theorem; cf. [Kli 1991], Th. 4.7.6). (a)Assume that u,. Then afamily(u,),E, C 5'SR(X) is Iocallyboundedfivm above. Putu := X: u(x) < u*(x)} the set {x C PSJ€(X) is locally bounded from above.

(b) Assume that a sequence (u


E X:

Putu

Proof (a) is beyond the scope of the book. We show that (a) (b): Let vj :=

u

< v(x)}.

E X:

E

j=l

0

It remains to use (a) and Proposition 2.1.40. Theorem 2.1.42 (cf. [Bed-Tay 1976]). There exists a (non-linear) operator ?SJ?(X) fl

bc)

—÷

u E

is a non-negative Borel measure on X}

26)

such that: U E

(b)

bc)

if ,9'SJf(X) fl

u,,

\

fl L°°(X. bc),

u E

then

DefinitIon 2.1.43. The operator the Mon ge—Ampere operator on X.

Definition 2.1.44. A function u E Y

Xandanyfunctionv

As usual Notice that a function u

in the weak sense.

fl

bc)

a —*

is called

called niaxirnal if for any open set uoni)Y,thenv u in Y.

bc) denotes the space of all locally bounded (measurable) functions on X. belongs to L°°(X. bc) if it is locally bounded from below.

2 Pseudoconvexity

128

bc). Then u

Theorem 2.145 (cf. [Ku 199fl, Th. 4.4.2). Let u E = 0.

is maximal

The following class of open sets with 'good' psh exhaustion functions will be useful in the sequel.

Definition 2.1.46. We say that an open set f function u E 1L) such that

X is hyperconve.x if there exists a

t<0. c

Let

be open and let N c

Define

:=sup{u: u The

that 0 < that

(*)

u
u<

is called the plurisubha rmonic measure of N relative to c. Observe I and (cf. Proposition 2.1.16). It is clear hN.c2 = 0 on mt N.

Proposition 2.1.47. Let D C be a bounded domain. Then D is hyperconvex if there exists a continuous function u E S'SR(D, IlL) with (*). Proof (Cf. [Zah 1974]). Let U: D —* [—oo.0) be a psh function with (*). We will construct a continuous psh function vo: D —+ (—oc, 0) with (*). Fix a ball and v 0 in K := B(a. r) C D and let u := hiD. Recall that v

3(a, r). By the Oka theorem (cf. [VIa 19661), for any point b E dB(a. r) we get v(a + t(b — a)) = 0. Thus v = 0 on K. v(b) = The maximum principle implies that v(z) < for any z E D. Fix a t0 > 0 such that u < —zo on K and put uo (l/to)u + 1. Then uo E v. Consequently, the uo < 1, and uo 0 on K. Hence uo hK,D function := v — I satisfies (*). We will show that v is continuous (then satisfies 1

all the required conditions).

Fora E (0,1) let Da := {z E D: v(z)
Da /

D

when a /

= 0 on K. Observe that

I. Moreover,

V

On

Indeed, define h

v}

J

on Da

onD\Da.

1

Then urn sup

D(Z)

a

E

Hence, by Proposition 2.1.9, h e 5'SR(D). Obviously h < I on D and h = 0 on K. s h < v on Da. Thus h hK,D < v. In particular,

2.2 Pseudoconvexity

129

Fix a point zo E D. We want to prove that v is continuous at zo. Let fl(a) := maxb v. Observe thata <$(a) < I. In particular, fl(cr) —+ I when a —÷ 1. Fix

> 0 and a E (0. 1) such that

E Da and fl/a — I < where fi psh functions defined in a neighborhood of Da, C D, such that VE \ v on when e \ 0 (Proposition 2.1.25). One can find e > 0 such that for w := yE E :PS,7€() fl we have w> v on on w K, and w fi + on Consequently, - Let (VF)O
(w —

q)/fl

<

h<.0 on Da.

Hence,

Now, by the continuity of w, there exists a neighborhood U of Zo, U C Da, such that 0 forz E U. Iw(z) — w(zo)I s r?forz E U. Finally, Iv(z) — v(zo)I <

Theorem 2.1.48 (cf. [Kli 1991], Prop. 4.5.12). If is compact, then

c C" is hyperconvex and K C

\ K) = The measure

is

o.

called the equilibrium measure for K.

Proposition 2.1.49 ([Ku 1991], Lemma 5.6.2). Let 2 C C" be hyperconvex, let K

=

(P) = 0. Then

be compact, and let P C K be such that

2.2 Pseudoconvexity Let (X, p) be a Riemann domain over C". Let

For a compact set

C

K C X define

:= {x

E

X:

u(x) 5 supu}. K

Definition 2.2.1. We say that (X, p) is pseudoconvex if for any compact set K C X the set is relatively compact. The definition extends in a natural way to Riemann regions over C".

Remark 2.2.2. (a) Let (X, p) be a Riemann region over C". Then (X, p) is pseudoconvex if for each connected component C of X the domain (C. Plc) Indeed, if (X, p) is pseudoconvex and K C C is compact, then

c

c.

IS

pseudoconvex.

2 Pseudoconvexity

130

Conversely, if any connected component is pseudoconvex and K C X is compact, then K is contained in a finite union of connected components, say K c C1 U. LJCN.

N.Then

c

x.

u. . . u

c

ko(X). Consequently, if (X, p) is holomorphically convex, then (X. p) is pseudoconvex. (b)

Recall (Theorem 1.10.4) that (K, p) isa domain of holomorphy if (9(X) separates points in X and (X, p) is holomorphically convex; cf. Theorem 2.5.7. (c) If X is hyperconvex (Definition 2.1.46), then is pseudoconvex. be such that u(x) < t) for {x E Indeed, let u any z < 0. Take a compact K c c and let o < 0 be such that K C Then

c

a

In view of (b) it is natural to ask whether any pseudoconvex Riemann domain is a domain of holomorphy. This is the famous Levi Problem. The problem, formulated by E. E. Levi in 1910, was solved by Oka only in 1942 for n = 2 and in 1954 by Oka, Norguet, and Bremermann for n > 2. A solution of the Levi problem. via d-techniques, using the theory of holomorphic functions with tempered growth, will be presented in § 2.5.

First we like to discuss pseudoconvex open sets X with smooth boundaries. If u is a function defined in a neighborhood of a point a X, then we put

az1

where

——(a)

denote the formal derivatives; cf. § 2.1.

Let c X be open. Recall that ac is smooth of class Ck (or C(C.smoolh) aS a poini a E ac if there exist an open neighborhood U of a and a function u E Ck(U, IR) such that

c2flU={x€U:u(x)<0}, U\c2={xEU:u(x)>.o},

(')

()

for u is called a local definingfunctionfor means that u is real analytic.

at a; here k E N U {oo) U {co}. where U E

Observethatifu &c(U,R)satisfies(*)andgradu(x) (), i.e. u is a local defining function. We say that is ek.sniooth if is &-smooth at any point a E

u satisfies

U,then

2.2 Pseudoconvexity

131

Put

=oJ.

eCu:

XE

Zj

at x; notice that the

is called the complex tangent space to

The space condition

=0 The means that I grad u(x) in the sense of the Hermitian scalar product in is independent of u (Cf. the proposition below). Observe that if definition of n = 1, then (0}.

Proposition 2.2.3. Let c

X be open, a

ac2, and let U be an open neighborhood

of a.

(a) Let uj, u2 E Ck(U, R) be two local defining functions (k E N U {c'o}). Then U2 = vui with v E C"'(U,IR>o). 27) C' (U, (b) The space T5c(ac) is independent of the local defining function u E

un (c)Letui. U2 E C"(U, R), k> 2, be two local defining functions with u2 = R>0) i.c as in (a). Then

where V E

£u2(x;

= v(x)Cu1(x;

x

U fl ac,

E

where £ denotes the Levi form (cf § 2.1).

(d) (Cf. [Kra-Par 1981]) Letk

E

NU{oo}, k >2, and let cz be Cr-smooth. Define —pcz(x),

xE

0,

PX\c(X),

p) with respect to the to the boundary of and a Euclidean norm: cf § 1.1). Then there exist an open neighborhood U of

(recall that pg denotes the

jr: U —* dQ such that:

•S

•

E

II grad

• ,r(x)

S(x) II = 1, x E df2 (consequently, is a local defining function). and is the only point from ac that realizes the Bx(x,

distance to ac, i.e. Bx(x, Iö(x)I) ñ x U.

= {,r(x)) and

= IIp(x) —

(e) The above result does not hold fork = 1. (f) Let k E N U (oo). Then the following conditions are equivalent: 27) The proof will show that also the following result is true: If u2 e Ck(U, R) are such 1(U. R÷). that ti1 is a local defining function andu2 satisfies (*) and (). then u2 = vu withy E

2 Pseudoconvexity

132

is eks,,woth.

(i)

(ii) (here exists afunction U E CJC(X, R) such that

=

{x E

X:

u(x)
={x E X: u(x) >

for u

is called a global defining function for

Proof (a) Using a local C"-diffeomorphism we reduce the problem to the case where U = (—1, 1)" with N = 2n, u1(x) = ul(x',xN) = xN, and > Oforany x E U. Then U2(X) = v(x)xN with

(b) Let

(c) Take x E U fl

XNO. be two local defining functions. Then, by (a),

u2 E C8 (U.

where v e C"1(U

XN # 0

XN),

v(x', xN)

=

R>0). In view of the proof of (a), we get gradu2(x) =

and

Then

a

dzi1

—

j.k=I dv

a

=y

duj

d2ui

+ v(x)Luj(x;

=

= v(x)Lui(x:E) (the latter equality follows from the fact that if =0).29) = =0, then 28>

29)

E C' and

E C are such that

2 the fonnula may be obtained by direct differentiation. Notice that the case k 2 3 can be so'ved by direct differentiation: Notice that in the case k

It

j.k=I

8

tiu

-

It

=

dui

dv

+

I 8v

-

= 0.

2.2 Pseudoconvexity

(d) In fact, the problem has situation to the case where:

= M := {(t, f(t)):

•

(—r, r)N c RN

•

f:

t

a

133

local real character. Thus we can easily reduce the

N + I = 2n, i\N(r) :=

C

E

:= (—r', r') C R,

—*

with f(O) = 0,

is a

0, j =

N, We

:= (sgn(y—f(x))dist((x,v),M),(x,y) E AN÷,(s) C want to prove that for small 0 < s
such that: •öE

=

'pN):

—p

IR),

• 1 grad8(x, v)ll = I for any (x, v) E Mfl • 82(x y) = (f(ço(x, y)) — y)2 +

—

x3)2

and co(x, y) is the only

pOint in AN(r) with the above property (for every (x, y) E Recall that

82(x y) = minj(f(t)

Fk(X,y,t) := (f(t) We

—

—

v)2

+

—

y)±() +fk

X)2:t

Xk,

axk

(s)).

k

E

=

N.

I

want to findi = ço(x, y) such that Fk(x, y, t) = 0.

Observe that the functions F1 and I

k=

N.

I

A(r') x

x

FN are of class Jk,t=I

(t)

N

Consequently, by the implicit mapping theorem, there exist small r1 > 0 and a Ck_ I mapping'p: AN+l(5I) —f suchthatt = 'p(x, y) istheonlysolutionof(t)in (TI). Now lets > 0 (s <SI) be so small that for any (x, y) E I (SI) X I (s)

the distance of the point (x, y) to M is realized by a point from x A(r')) fl M (consequently, by the point ('p(x, y), f(ço(x, y)))). In particular, since

(f('p(x, y)) — dXk

y) —

= 0,

k=

I

we get

8(x, y) = (f('p(x, v)) — y)(l +

y))))'.

(x, y) E

N,

134

2 Pseudoconvexity

Hence, using (i), we obtain (Exercise)

—(x, y) =

(I —(x, y) = dy

which shows that 8

1/2'

2

dXk

k=1

N,

) —1 2

(i +

y))) )

12' /

R) and that 1 grad 8fl = I on M

E

fl

(e) It suffices to consider M := {(t, It t E (—1, 1)) (0 < e << 1) and 8(0, y) for small y > 0; Exercise. (f) Assume that (i) is satisfied. Let U and 8 be as in (d). Put := 8 (recall that uo E Ck(U, IR) is a local defining function). Take an e > 0 so small that K := (x EU: )uo(x)I
t<—e,x(0)=0,x(t)=efort>e,andx'(t)>OforItI<e.Define —C.

xE

uE

0

lR) is a global defining function.

DefinItion 2.2.4. Let

X be an open set. We say that

is strongly pseudoconvex

at a point a E acz if there exist an open neighborhood U of a and a local defining R) such that function U E

Lu(x:

> 0,

x EU

n

E

Observe that the definition is independent of u (cf. Proposition 2.2.3(c)). We say that is strongly pseudoconvex if is strongly pseudoconvex at any

pointa E dQ. The notion of the strong pseudoconvexity in invariant under local biholomorphic mappings (cf. Remark 2.1.6). We will see (Proposition 2.2.25(a)) that any strongly pseudoconvex open set is hyperconvex (and, consequently, pseudoconvex). Obviously, if n = 1, then any C2-smooth open set X is strongly pseudoconvex.

ProposItion 2.2.5. Let X be open. (a) Assume that ac is C2-smooth at a E Let U be an open neighborhood of a and let u E e2(U, R) be strictly psh with () and (). Then u satisfies (v'). In particular. u is a local defining function.

2.2 Pseudoconvexity

135

and let u E C"(U, IR), k 2 2, be a (b) Let U be an open neighborhood of local defining function with (u). Then there exists a c > 0 such that for the function — 1) we have: u,, := > 0,

x

E

E

In particular, u,, is strictly psh in a neighborhood of (notice that u,, is a local function). (c) For k 2, the following conditions are equivalent:

(i)

and strongly pseudoconvex:

is

(ii) there exist an open neighborhood U of ac and a strictly psh function u Ck(U, R) with (*) and (). is strongly pseudoconvex for any boundary point a (d) (Narasimhan)

ac

there exist an open neighborhood and a biholomorphic mapping F: U —÷ F(U) is strongly convex at each point fivm C" such that the open set G := F(U fl In

C" is strongly pseudoconvex.

any strongly convex open set

Recall that an open set G C" is said to be strongly convex at a point 1' E 8G if there exist an open neighborhood V of b and a local defining function v C2(V, R) for G such that

3(v(z; Q) >

E V fl

0,

Q

E

(recall that ,7( denotes the real Hessian; cf. § 2.1), where TZR(aG) denotes the real (2n — 1)-dimensional tangent space to at z. It is clear that the definition is independent of v (cf. the proof of Proposition 2.2.3).

Proof (a) If uj C2(U'. R), U' C U, is a local defining function, then u = v E C'(U'. the proof of Proposition 2.2.3(a)). Obviously, grad u(x) = v(x) grad u 1(x), x E U' fl In particular, grad u(x) 0 provided v(x) 0. Suppose that v(b) = 0 for some b U' fl ac2. Consequently, grad v(b) = 0. Then we get 82u -

a

/

azJ

'

8v

8uj

8Zj

dZk

(b)=—(ui-—--)(b)+—(b)—--(b)+v(b) 8Zk'

a

- (b) aZJaZk

= aZJ

(if(p E C',

E C

and

=

(b) It is clear that for any c >

=

0, then

0

the function

=

0); contradiction.

is a local defining function.

Moreover, one can easily check that n

azk

2 ,

xEd,

2 Pseudoconvexity

136

Let K :=

E

= I.

x C'1:

<0}.

Observe that K is compact and

Kfl Now we only need to take a c > 0 such that

K) K) (c) follows from (a) and (b). (d) First observe that if G is strongly convex at each point from M := F(U fl

then analogously as in (b) one can show that for any b E M there exist an open neighborhood V C F(U) and a defining function v E C2(V, R) such that

Rv(z;Q)>0, Then

v is

ZEVflM,

strictly psh in a neighborhood of b. Consequently,

is strongly pseudo-

convex at

and let Conversely, assume that c is strongly pseudoconvex. Fix an a E be a local strictly psh defining function (use (c) and (a)). The U E e2(U. 1k), a E U, problem has a local character and therefore we may assume that C's, a = 0, and 0). Write gradu(0) = (1,0

u(z) = 2 Re(zi + B(z)) + Lu(0; z) + where B(z) :=

I

"

j.k=1

Zj

and define

F: C'1 —* C".

F(z) :=(zI +B(z),z2

Zn).

Observe that

• F(0) =

0,

• F is biholomorphic in a neighborhood Uo of 0, •uo = 2Re + Lu(0; w) + o(flwII2). is strongly convex at each point of F(U 0 In particular, G := F(U 0 any small neighborhood U C Uo of 0.

for D

We say that u: X —÷ is an exhaustion function for X if for any I E R the X: u(x) < t} is relatively compact.

set {x

_______

2.2 Pseudoconvexity

137

Proposition 2.2.6. Assume that (X, p) is a holomorphically convex Riemann domain over C's.

(a) Let K C X be compact and let U be an open neighborhood of

Then

there exists a strict!)' psh real analytic exhaustion function u: X —+ IR such that

u Oon X \ U. (b) if U: X —+ IR is a strictly psh real analytic exhaustion function (as in (a)), then there exists a sequence tk / +00 such that tt u(lx E X: grad u(x) = k = 1,2 In particular; jfwe put

Xk:={xEX:u(x)
Proof (a) There exists a sequence of holomorphically convex compact sets (K1 such that K1 = *0(X) K1 C mt K1; cf. the proof j 1, and X = ofRemarkl.1O.3. l,suchthatU1 C U,K1 CU1 C 1. Fix a j > I. Any point x E j \ Uj does not belong to K1. Thus IlK,. We may assume E (9(X) such that there exists a function > I for be a neighborhood ofx such that that > I > IlfIlK1. Let is compact, there exists a finite number of points xI Since Kj÷2 \ y tk(j) U I such that KJ+2 \ U Vxk(J). Define fjt := where £(j) is so big that k(j)

k(j)

<

E

If1

Define

> j,

\

xE

oo k(j)

:=

—1

j=1 Observe that the series converges locally uniformly in X. It is clear that v E 9SJt"(X), v is an exhaustion function. To prove that v is real analytic we proceed as follows. Consider the Riemann domain (X, j5) with X := X and

n. Observe

that f

function X x X

E

(9(X) (x, y) —÷

f €(9(X). In particular, for any f

E

(9(X) the

f(x)f(y) is holomorphic. Define s'p: X x X —÷ 00 k(j)

p(x, y) := —1 + J=I v=I

C.

2 Pseudoconvexity

138

The series converges locally uniformly in X x X and therefore q E (9(X x X). Since v(x) = x E X, we conclude that v is real analytic. where e > 0 is so small that u <0 on K. It is clear Now we put u := v + that u satisfies all required conditions. (b) We only need the following aaxiliary lemma (which is a direct consequence of the Sard theorem; cf. [Fed 1969], Th. 3.4.3).

Lemma. Assume that (X, p) is Riemann domain over C" and let u E &"(X, Then the set u((x X: grad u(x) = O}) C R is of Lebesgue measure zero.

IR).

Recall that pseudoconvex domains were defined via properties of psh functions. Now we are going to describe the pseudoconvexity in terms of properties of the boundary

ProposItion 2.2.7. Let (X, p) be a Riensann domain over C. —

log

Then the function

is subharrnonic on X.

Proof First suppose that (X. p) = (D, idD), where D is a domain in C. We may assume that D ç C. Then

_logdo(z)=supf log

1

z€D.

Recall that — logdD is continuous. Thus — logd0 is subharmonic. In the general case fix an a E X and observe that it suffices to prove that there exist a connected neighborhood U C X of a and a domain D C C such that p(U) C D

anddx = dDopOn U(then —logdx = (—logdD)opon U andtherefore, —logdx is subharmonic on U). Define

U := P(a, dx(a)/3),

U P(x),

D

p(Y).

XEU

Observe that

(we have

=

dy on U. Moreover, by Corollary 1.1.6(b), the set Y is univalent 0 E IP(x2) for any xI, x2 E U). Hence dy = dD 0

In the following lemma we collect a few conditions which assure that — log dx,q

is psh on X (where q is a norm on C"). The lemma will be useful in the proofs of Theorems 2.2.9 and 2.9.2.

(X. p) —+ (Y, q) be a morphism of Riemann domains such Lemma 2.2.8. Let that (Y. q) is pseudoconvex. Let q: C" R÷ be a complex norm. Then each of the conditions below implies that — log dx.q E —+ X, where (a) For any sequence of holomorphic mappings CC 1, the following implication is true: is a neighborhood of E, X;

2.2 Pseudoconvexity

(b) (X, p) ispseudoconvex;

139

—

(c) There exists a neighborhood U of dx 30) such that —log

q E .'PS3t(U flX);

o f)(t,.) E

(d) For any continuous mappings f: [0, 1] x E —* X such that 0(E) for any t E [0. 1], the following implication is true:

f(([o, 1) x E) U ({1} x aE)) C X —

:

f((0, 1] x E) C X;

(e) For any continuous mapping f: [0. 1] x E —÷ X for which the mapping o f extends to a holomorphic mapping g: D x E C", where D C C is a

neighborhood of[0, 1J, the following implication is true:

f([0, 1) xE) cx

f({l} xE) c aXorfUl} xE) C

;

X.

(f)Foranydomain T = Tr,p 31), every biholomorphicinapping 1: T —÷ f(T) C X extends to a holomorphic mapping f: E" —÷ X: (g) (Y, q) is a domain of holomorphy (cf Remark 2.2.2(b)) and for any domain T = Tr,p and a biholomorphic mapping f: T —* 1(T) C X such that of extends to a biholomorphic mapping 1: E" —p E") C Y, there exists a holomorphic

mappingf: E" —p Xwithf= fonT. Observe that

in conditions (d)

and (e) the function

.) must be holomorphic on

f(t.

Eforanyt E [0,1). Proof To simplify notation. let (*) denote the condition '— log dx., E :PSJe(X)'. Put moreover u := q• (*): Suppose that (*) does not hold. (a) Then (cf. Remark 2.1.2(a)) there exist: a point a E X, a vector v E C" with q(v)= u(a),anumber0 < < I,andafunctionh1 harmonic inr1Eandcontinuous —Iogu

0,

o

r,E

where xi(A) :=

+

E. Let

A

r1E be such that a =

A1


—loguox,(At) =h2(X1).

Cf. § 1.5. 31)

Recall that(cf. Example 1.8.7) 1',,,

((zi

Zn) E

U {(Zi

forO < r 32)


:=

C":

1z11

Zn) E C":

:= TL), where < r,

j

=

n — I,

I I

n

znl <

— 1, p < Iz11 <

140

2

Pseudoconvexity

LetO
R

X(A) := XL(rA) = Let

+ Arv). A E

E (9(RE) be such that h3 = Re

uox U

Let b

h2(rA), A E RE,

X(AO) and let w E

> IPI

(l/r)E,

Define p

Ao := Aj/r.

exp(—p3). Then

onaE, onE,

(2.2.1)

(2.2.2)

X(AO) = p(Ao).

(2.2.3)

with q(w) = u(b) be such that

+ tw))

0:

(2.2.4)

cf. Proposition 1.1.9. Define

4'(r,A):= poX(A)+f

p(A) p(Ao)

w= p(a)+Arv+t

p(A) p(Ao)

w,

x RE.

=

x and 'P(:, Ao) = p(b) + rw. By (2.2.2) and (2.2.3) we have X(A)) rlp(A)l
q(4'O. A)

S := ([0, 1) x E) U ([0, 1] x

=: Si US2

such that

(t,X)€G.

q('lI(t,A)—pox(A)) < In particular, we define

4(:, A) :=

A)),

(f, A) E U.

where U(A) Bx.q(X(A)). Observe that (by the identity principle) any point A°) E G has a neighborhood W C G such that

(I)(t, A) = (pIu(Ao))'('P(A, t)),

(t, A) E W.

Consequently, 4) is continuous. Moreover, 4)(r,-) E 0(E) for any t e [0, 1). Ob+ 1w), t E [0, 1). Since viously (by the identity principle) 4)(:, A0) = X, condition (2.2.4) implies that there exists a sequence [0, 1) /1 = such that u(4)(tk, A0)) —÷ 0. u o 4'(tk,) const on E 33), and infE u o Becawic

u(4(fk.

—4 0 and

U

1flh4(S2) U > 0.

2.2 Pseudoconvexity

141

1. We may assume that Ak —p u o 4)(fk, Ak) for some Ak E E, k := 1. Notice that u(xk) —f 0. Ak), k Now take 1. Then we get a contradiction with (a). .), i'

E. Let

(*): Suppose that (*) does not hold. We keep all the previous notations. we have Observe that, by the maximum principle, for any function a E (b)

sup a = supa o 4) = sup supa o cD(t,.) = sup supa o aE

E

S1

supa. L

where L := 4)(S2). Hence 4)(S) C X (because (X, p) is pseudoconvex). On the other hand u < u(xk) —÷ 0; contradiction. (*): Suppose that (*) does not hold. We keep all the above notations. (c) ; First, similarly as above, the maximum principle implies that q(4)(S)) C Y. with L := *p(cD(S2)). Since (Y, q) is pseudoconvex, we conclude that 4)0 y: [0, 1) —+ X, In particular, we may assume that co(xk) —p Y. Let where y: [0, 1) —÷ S1 denotes the curve obtained by joining the segments

k = 1.2

[(tk, Ak). (tk+i. Ak÷I)i,

k > 1. It is clear that y extends 1. Thus xk = so that y(tk) = (1k, Ak), k continuously to [0. 1] (we put y(l) (1, to)). Hence, by Remark 1.5.4, E

ax.

Recall that, by Proposition 1.5.8 =q'

4):[0,1]x E —+ X

.

1)xE extends continuously to a mapping =", -

Obviously,bycontinuity,cD = 4)onSand p ocD = '1'.

Notice that (D((l, so)) = Let U be a neighborhood of

such that

—

log u E fPSR(U fl X). By continuity

(of 4)), thereexistsane > Osuchthat4)((l —e,1)x

CU. Letk0

Then, applying the be such that (tk, Ak) E (1 — e, 1) x + eE) for all k > maximum principle to the subharmonic function E A —p — logu 0 4)(lk, + eX) with k 2 ko, we conclude that u(4)(tk, + eA)) = u(xk) for any A E and k ko. E such that there exist e > 0 and Ico such that Let Eo denote the set of all E and k > ko (observe that 4)(1, + eE) C + eA)) u(xk) for any A =40

a X). We have proved that

Eo. Obviously, E0 is open. The above argument

shows that Eo is relatively closed in

contradiction (because 4)(l, dE)

q) =

f :=

4)

E and hence 4)(l, E) C

dx;

X).

(*): Suppose that (*) does not hold. We apply the above argument to

(d) (Y.

E. Thus E0 =

= 4)(l, dE)

p up to the moment when 4) is defined. Then, taking (C's, id) and we get a contradiction.

With 10, 11 x E as the topological space and 111 x E as the singular set.

2 Pseudoconvexity

142

(e)

(*): Suppose that (*) does not hold. We argue as above. Observe that the

function of = 'P extends holomorphically to C x RE. Thus we get a contradiction with (e). (f) (*): Suppose that (*) does not hold. We argue as in the first part of the proof. Observe that the vectors v and w must be linearly independent. Suppose they are not. LetXo := Xo —÷ Cbegivenbytherelationp(x) =

p(a)+po(x)v,x E Xo. Then(Xo, po)isaRiemannregionoverC. Clearly,

'x >

OIl d(x.p).q = u on X0 36)• Moreover, x: (l/r)E —÷ Xo and (cf. (2.2.1)). By Proposition 2.2.7 we conclude that — log d(X0,,,0) E S.X(X0). Hence o x > on E. In particular, d(X0p0)(b) > Ip(Ao)I = u(b). On the other hand, property (2.2.4) implies that = d(x.p).q(b) = u(b); contradiction. Fix V1 Vn_2 E C" such that the vectors VI Vn_2, w, v are linearly independent and define

+

%(A) := p(a) + A1v1 + ... + An_2vn_2 + An_i A

=

0. t, A,,) =

Observe that 'Po is injective and

A,,). Moreover,

+ IA,,_ilu(x(A,,)).

IAiIq(vi) + . .. +

q('Po(A) — p0 X(An))

x RE.

E

Let

G0 := {A E C": q(%(A) — P0 x

x (E x aE)) c G0.

x Put

(plu(A)Y'('Po(A)),

A E G0.

By the same method as for 4), one can easily check that

is continuous. Moreover,

0, :, A,,) = CD(t, A,,). Since p ° 4>o = 'Po, we see that 4)o is injective. Consequently, Go —÷ 4)o(Go) C X is biholomorphic. Fix an e > 0 such that

(eE)"2 Take

x

((1 +e)E) x {l —e<

an arbitrary I — <0 < I and let

<1 +e} C Go.

E (O.e] be such that

x (9E) x (1 + B(X.p).q(xo.r)flXo = P(xo.po)(xo,r),xo E X0.

C Go.

9/(1 + e) and

2.2 Pseudoconvexity

143

A(A) (1 + (eAt,..., EAn_2, (1 + T —÷ X. Noticethatf: T —+ f(T)isbiholomorphic. Then,by (f), f extends to a biholomorphic mapping f: —+ f(E") C X. This means that

Define T

f

there exists a holomorphic mapping

4>0: (eE)"2 x (1 +e)E2 —÷ X such that = 4>o on G1 U G2. In particular, the mapping 4>: [0, 1] x E —÷ X defined by 4>(r, := 4>o(O 0. t. A,) gives a continuous extension of 4>; contradiction (because u(4>o(0 0. tj, Ak)) —* 0). (*): Suppose that (*) does not hold. We argue as above. Observe that (g) the mapping r := p o f: T —+ Y is locally biholomorphic. By Proposition 1.9.9, r extends to a locally biholomorphic mapping f: E" —÷ (recall that E" is the envelope of holomorphy of T). Note that

I

q of =qoçoof = pof = 'I'ooA on T. Hence q o = 'Po o A on E" and, therefore. f is injective. Consequently, by (g), f extends to a biholomorphic mapping f: E" —÷ f(E") C X and we conclude the proof as in the previous step.

U

Theorem 2.2.9. Let (X. p) be a Riemann domain over C". Then the following conditions are equivalent: (i)

for any compact K C X the set

(ii)

(X, p) is pseudoconvex;

is compact 37);

for any C-norm q: C" —+ the function — log dx.q is psh on (iv) for any E C" the function — log is psh on X (cf § 1.1); (v) there exists a C-norm q: C" —÷ R+ such that the function — log dx.q on X (cf § 1.1); (iii)

is

there exists an exhaustion function u E .'PS3f(X) fl C(X);

(vi)

(vii) there exists an exhaustion function u E (viii) there exists a strictly psh exhaustion function u E

(X).

Proof (vii)

(ii)

I (I)

(ii)

(iii)

(iv)

(v)

(vi)

I (viii) To simplify notation we put

:=

fl C°°(X).

(i)

psh

144

2

The implications (i)

Pseudoconvexity

(vii)

(ii), (vi)

;

:

(i) are elemen-

(ii), and (viii)

The implication (ii) (iii) follows from Lemma 2.2.8(b). The implication (iv) follows from Lemma 1.1.13. The implication (iv) (v) is a direct consequence of Lemma 1.1.12 and Proposition 2.1.16. ',

(iii)

:

(v)

:

.

(vi):

Step 10: Fix a point ao X. For 0 < e < px(ao) let Lie be the connected componentoftheset{x E X: dx(x) > e)thatcontainsao. By Remark 1.1.14(f), for any r E IR the set

:= {x E

<

is the inner Eucidean distance on pIus). We will prove that for any 0 < e << I there exists a strictly psh function E such that for any t R the set

is relatively compact in X, where


(x

is relatively compact in X.

Fix 0 <e << I and let ,j := e/3. Then C {x

P(x, z7) C

Put w(x) := 7u,,2(aO, x) and define

:= f w(p'(p(x) + =

f

w(y)c1(PW —

where ct' is a regularization function in C" (as in § 2.1) and {x E

fl

< t} C {x E U2,,: P(x,

xE It is clear that

01 C

E

X,

t ER.

:= + where C > 0. has the same properties. It remains to prove that there exists a C > 0 such that is strictly psh on It

Observe that the function suffices to show that

2

<

x E (Jr,

E

C",

'1

Zn)

fc 'L'di\2 = 1.

=

...

'1'

C°°(C,R÷), W(z) = 'V(lzI), supp'l' =

2.2 Pseudoconvexity

145

where Co is an absolute constant (then we put C := 2con2/zi). We have

_!

= dZk

+

J

L4, k = I,..., n.

x dzk

Recall that

x") =

lw(x') — w(x")j

x'. x" E lP(x, ii). x E

— p(x")II.

(cf. Remark 1.1.14). Hence a2

(x)

—

aZJaZk

=

! lim!

j

(w(p;' (p(x) +

E" dZk

(p(x) +

+ tei)) —

XEUe, j,k1

dz

E

which finishes the proof of the required result. It is clear that, taking cq (with c > 0) instead of q, we may assume that dx, q (aO) > 1. For v I let denotes the connected component of the set {x X: dx, q (x) > l/v} that contains ao. Observe that for any v 2: 1 there exists an 0 such that C IJ(. Consequently, for any v 2: 1 there exists a strictly psh function vv such that (x 0, v N, such that if

:= {x

:= {x E Y1,+ii,_: Vv÷2(X) <

Vv+2(X) < c1,},

then

+ l},

UKvx yEN

Now we construct inductively functions > on \ L,1, = I,..., v — 1, and

Weputui :=

v2.

Supposethatu1

\ Let

such that

on

arealreadyconstructed. Itsufficesto

> v—I onLL\Lt,_I,uy(x)> v—2

E

on

fl

=

and Uy = Uy_I on be defined on by the formula c°L

:=max{—Iogdx,q—log(v—l),vt÷i

= {x

Then >v—

I

on

\

and I

> 0 such that 0}. In particular, there exists a on Now it suffices to put > X

x €

\

2 Pseudoconvexity

146

Finally we put u := exhaustion function. (viii): Let (vi)

ut, and we easily conclude that u is the required

,'

v€Z,

0 for u —1. Let 0 on X. Since u is be a strictly psh function such that u as continuous, u locally uniformly on X. For each u E Z let e(u) > 0 be such that and u < UE(V) < u + I on For simplicity let := Ue(v), cc t' Z. Let x: IR —* [0, I] be an increasing function such that Taking u + const instead of u, we may assume that

e\

uE E

• x(t) = Ofort <—3/2, • x(t) = I fort > 2, • x'(:) >0, x"(t) > Ofor frI <3/2. Define I

vEZ.

LI Hence

•

ifx then + 2—i' > u(x) + 2—u = 2. ± 2— u) = forx in_an open neighborhood of Indeed,ifx E
Hence + 2 — u) = 0. is psh in an open neighborhood of Indeed, if x 2— u< u(x) + I + 2—u 1. Hence, by Remark 2.1.2(h), near —2.

•

• V1 is strictly psh in an open neighborhood of

\

then

I>

U1,(x)

+ 2— v > u(x) + 2— v

\ —1. Thus

then

+

+ 2—i') is psh Indeed, if x E C's,

0,

then

>0 U = x"(U1,(x)+2—u)i grad (recall that x' > > Oin(—3/2,3/2)). We pass to the main part of the proof. We show that there exists a sequence C [I. +oo) such that for each v N the function := cjV1 + ... + We proceed by induction over v. is psh in an open neighborhood of The case v = I is obvious (we may take Cl := 1). Suppose that are already constructed. It is clear that for any 1 is psh in an open neighborhood Thus the function W1,÷1 = + we have to chose such that is psh in an open neighborhood of i\ This is possible because is strictly psh on \ = W Observe that on X. Moreover, {x X: W(x) < v} C X for any v u we get = N. Indeed, if x 2V, then for ci V1(x) + ... + c1, = CI+ V. 112. To get a strictly psh exhaustion function it suffices to take W + D

________ 2.2 Pseudoconvexity

147

Remark 2.2.10. Let (Y, q) be a Riemann domain over and let X be an open subset of Y. Observe that Theorem 2.2.9 remains true for the Riemann region (X, qlx). Corollary 2.2.11. Let (X, p) be a Riemann domain over C". Asswne that X = UVEN

where XL. is a pseudoconvex open subset of X with X is pseudoconvex.

0

Proof dx1, /

Corollary 2.2.12. Let (X, p) be a Riemann domain over C". Assume that Y = mt flLEN X'., where pseudoconver.

is a pseudoconvex open subset of X, V E N.

Then Y is

Proof dy = Corollary 2.2.13. Let (X1, Pj) be a pseudoconvex Riemann domain over C", N. Then X1 x

Proof dx,x...XXN(xI

j

= 1,

x XN 1S pseudoconvex.

xw)=min{dx,(xj):j=l

0

NJ.

Corollary 2.2.14. By Proposition 2.2.7, any Riemann domain over C ispseudoconvex. Corollary 22.15. Let (X, p) be a pseudoconvex Riemann domain over C" and let u be psh on X. Put Y := (x E X: u(x)
Proof First assume that u is additionally continuous. Let K C Y be a compact set and let e > 0 be such that u —e on K. Then

C {x E X: u(x) <—e} fl Hence £J'SJt!(Y) C {x

E X: u(x) < —e} fl

Now, let u be arbitrary. Let denote the e-regularization of u. By the first part of the proof (applied to the function — log dx + loge) the set

—logdx(x)+loge<0}

/

is pseudoconvex. Moreover, for each e. the set YE := {x E uc (x) <0} is pseudoY when s \ 0. Now, it remains to use Corollary 2.2.11. convex. Observe that

0

Corollary 2.2.16. Let (X. p) be a pseudoconvex Riernann domain over C" and let Y C X be an open set such thatfor any point a E d Y there exists an open neighborhood Ua such that Y fl Ua ispseudoconvex. Then Y ispseudoconvex; cf Theorem

Proof Let u

be

a continuous psh exhaustion function on X.

Put X,, := lx

E

X: u(s) < v},v €N. Then Xv ispseudoconvex (Corollary2.2.15)and Xv / X. By Corollary 2.2.11 it suffices to show that Y fl X,, is pseudoconvex for each v c N. Fix

avE NandletZ :=

Takeana E

Ifa E Y,thenbyCorollary2.2.12,if

2 Pseudoconvexity

148

:= iPx(a, r) C Y, then Zfl Ua = fl (Ia is pseudoconvex. If a E dY, then by our assumptions, there exists an open neighborhood (Ia such that Y n Ua is pseudoconvex. Hence, by Corollary 2.2.12, Zfl Ua = Y n X, n U0 is pseudoconvex. Observe that for any a E dZ there exists an open neighborhood Va C (Ia of a zn V0. Hence there exists a compact set K C Z such that x = such that — Iogdz E PSR(Z \ K). Take an N > 0 such that N > —logdz on K and put v := max{— logdz, N}. Then v is a continuous psh exhaustion function on Z (recall that Z is relatively compact in X). 0 we take

Corollary 2.2.17. Let (X. p) be a pseudoconvex Riemann domain over C0 and let M be a pure (n — 1)-dimensional analytic subset of X. Then X \ M is pseudoconvex Riemann domain. Proof For any a E Mthere exist 0 < r dx(a) and a function f0 E O(18(a, r)), 0, such that M (cf. [Chi 19891, § 2.8). By Remark 1.7.3(g) r) = f0

13(a, r) \ M is a domain of existence and, therefore, pseudoconvex. Now, we apply 0 Corollary 2.2.16.

Corollary 2.2.18. Let (X. p) be a pseudoconvex Riemann domain over C C" be a complex k-dimensional affine subspace and let T: H —* C' be an affine isomorphism. Put V := p1(H), q := To ply. Then (Y,q) ispseudoconvex. Proof If u is a continuous psh exhaustion function for X, then uiy is a continuous

0

psh exhaustion function for V. Using the same method of proof, one can easily show the following property.

Corollary 2.2.19. Let (X, p), (V. q) be Riemann domains over C" and Ctm, respectively, and let Z C X x V be a pseudoconve.x domain. Then for any yo E V the fiber Zv() := {x E X: (x, yo) E Z} is apseudoconvex open subset of X. Corollary 2.2.20. Let (X, p), (V. q) be Riemann domains over C" and Ctm, respectivelv and let f: X —* V be holomorphic. Assume that X is pseudoconvex. Then for any pseudoconvex domain Z C Y the region f'(Z) is also pseudoconvex.

Proof Put

:=

Takeacompact K C f2. Then

c It suffices to show that any V E

E f(K)

C

we have: ?SJt(Z

cc Z,

x.

cn

—+ xo. Then for

Take v

v>

v> 1, and, therefore, f(xo) E Z.

Proposition 2.2.21. A domain D C C" is pseudoconvex if the function

D x C" is psh on Dx C".

—÷

1.

Hence

0

2.2 Pseudoconvexity

149

is trivial. Assume that D is pseudoconvex. By Theorem 2.2.9(iv), it suffices to show that

Proof The implication

D

xC",

(2.2.5)

:= (0,0, 1) E C" x C" xC.

where G C Dx C" xC is a pseudoconvex open set and Define

xC

ED x

D x

E

C".

:= f'(D);

note that g(D x C") C G. Since D x C" x C is pseudoconvex, Proposition 2.2.20 implies that G is pseudoconvex. It remains to check (2.2.5): Put G

=

sup(r >

c

0: z + K(r) =sup{r

D}

= sup{r >0: {z) x

x

K(r) c

G}

0 Let D C C" be a balanced domain (i.e. ED C D; cf. Remark 1.9.6(e)). Then there exists a uniquely determined function h : C" —p R÷, with

h(Az)=IAIh(z),

XE C .z E C",

that D = {z E C" : h(z) < l}; observe that h is upper semicontinuous. The function h = hD is called the Minkowski function of D. such

Proposition 2.2.22. (a) Let D C C" be a balanced domain and let h = be its Minkowskifunction. Then D is pseudoconvex Wh E log h E (C") (cf Proposition 2.1.29(b)). (b) Let G C D x Ck be a Hartogs domain with k-dimensional balanced fibers. i.e. D C C"_" is a domain (1 < k < n — 1) and for each z E D the fiber

if

{w E Ck: (z. w) E G} is a non-empty balanced domain (cf Remark 1.9.8(e)). Let H(z, w) = HG(Z. w) := h0.(w), (z. w) E Dx Ck, wherehG. is the Minkowski

function ofthefiber

(notice that H is upper semicontinuous). Then G is pseudoD x C"). In particular, in the special case, let

convex

D is pseudoconvex and log H E

G = ((z. w)

E

D x C": h(w)

where D C C"_' is a domain, u: D —÷ is upper semicontinuous, and h: Ck 0, is an upper setnicontinuous function with h(Aw) = AIh(w), R÷, h A 40) Then G is pseudoconvex if D C, w E C" h PSR(C"), andu E 5'SR(D). Proof (a) If log h E that

then D is pseudoconvex by Corollary 2.2.15. Observe

= l/h(U,

E

C".

11G(Z. w) =

Observe that in the case k = I. h(u')

Ju'I. we get a classical complete Hartogs domain.

150

2 Pseudoconvexity

Consequently, if D is pseudoconvex, then logh is psh by Proposition 2.2.21. (b) If D is pseudoconvex and log H is psh, then G is pseudoconvex by Corollary 2.2.15. Assume that G is pseudoconvex. Then D = (z (z. 0) E G} is pseudoconvex by Corollary 2.2.18. Moreover, 0)

(z, w) E D x Ck.

= 1/H(z, w),

0

Hence, by Proposition 2.2.21, log H is psh. So far, pseudoconvex domains were characterized by the the function — log dx. In the case of smooth open subsets c more, namely:

plurisubharmonicity of X we can say even

Proposition 2.2.23. Let (X, p) be a Riemann domain over C", let X be a C2smooth open set. Then (c2, pIe) is pseudoconvex (iffor any local defining function u E C2(V, R) we have:

£u(x;

> 0,

x E V fl

(Levi condition)

E

Notice that the Levi condition is independent of u (Proposition 2.2.3(c)). Proof Supposethat Since therefore

E

is pseudoconvex and let S and U be as in Proposition 2.2.3(d). we conclude that —log(—8) .'PSJt'(U fl and

£(— log(—8))(x;

—

> 0,

j.k=I (+) Fix a point xo E a vector a vector E C" such that

E

For any x E Q near

=

one can find

0

and E(x) —p when x —p xo. By (+) we conclude that .tS(x; by continuity (recall that S E (32(U)), we get £S(xo; 0.

0. Hence,

Conversely, assume that the Levi condition is satisfied. It suffices to prove that is psh near (cf. Corollary 2.2.16). Let 8, U, and be as in Proposition 2.2.3(e) (note that the Levi condition is satisfied for 8). Suppose that — log

<0 for some x

U fl

and E E C".

2.2 Pseudoconvexity

fl Bx(x, px(x)/2),

ir(x) E

Put

:=

151

:= p(xo) — p(x) (note that

=

Write

IogpQ(s(p(x) + Ce)) =

+ Re(AC + BC2)

+ 211C12 + o(1C12)

ICI<< 1,

(++)

and define

s(p(x) +

+

Observethat = Indeed, by (++), we have:

id << 1.

+ BC2)).

= 0. Moreover,S(11(C))
—8(*(C))

+ BC2)Il

—

+

>

ICI << I.

BC2)I

= lIliiIIexp(Ad + BC2)I(exp(e1C12)

— 1)

>0,

0 < IdI << 1:

contradiction. Hence

0=

= C

:= *'(O) E

(i.e.

0

Moreover, in view of the Taylor formula for 5 o

=Re(a

we get

0>

+

at 0

+°(k1).

')

?'

a recent paper (cf. [Zai-Zam 2000]) D. Zaitsev and G. Zampieri present a characterization of pseudoconvexity for domains whose boundaries satisfy a weaker condition than to belong to C2. Their boundaries are allowed to have edges. Such a domain Q C" (or X, (X, p) a Riemann domain over C'1) is pseudoconvex

if: (a) the Levi form is positive semidefinite outside of a set M C Hausdorif measure zero and

with (2n — 2)-

(b) the tangent cone (in the sense of Whitney) at any edge point outside M is convex.

2 Pseudoconvexity

152

Proposition 2.2.25. Let X be strongly pseudoconvex. (a) If is -smooth (k > 2), then there exist an open neighborhood U of and a strictly ps/i defining Jluwrion u E R). In particular, any strongly pseudoconvex open set is hype rconvex. (b) For any open neighborhood U of c there exists a strongly pseudoconvex C U. C

Proof (a) Let Uo be a neighborhood of

and let uo E ho be a strictly psh defining function (cf. Proposition 2.2.3(c)). Without loss of generality, we may assume that for a sufficiently small e > 0 we have

K :={x EU0:

U0.

Il') be such that x(t) = —r for t Let x E —e, x(O) = 0, x'(t) > 0 for t > —e, and x"(t) > 0 for any (cf. the proof of Proposition 2.2.3(f)). Define U := U U0, —C,

xE E

V

a a'

:= {x

e"(U, IR) is a local defining function, ui is psh, and UI (Jo: uo(x) > —e).

Let u2 be a hR such that

strictly psh exhaustion function for

{x EU: Let1, E

< —€/2}

CL :={x E

I]) be such that irfr = U3(X) :=

I

1

is

strictly psh on

(cf. Proposition 2.2.9). Take

u2(x)

r}.

inaneighborhoodofL. Define xE

10,

Then for any c > 0 the function u = UI + CU3 is strictly psh in a neighborhood of L (because u = Ut +cu2 andu2 is strictly psh) and in (because u = and Uo \ supp c V). Moreover, it is clear that for small c > 0 the function u is strictly psh in a neighborhood of Land < Oon Thus, forO 0 such that U. Let E be such that 0 — us m on (cf. Proposition 2.1.25). Put L42 := u1 + '7211p112 > 0); U2 is a strictly psh function on Taking > 0 sufficiently small, we may assume that Th.

e/2
2.3 The Kiselman minimum principle

Remark 2.2.26 (Cf. [For 1976]). For any there exist: G

153

strongly pseudoconvex domain

•N> •a

strongly convex domain D c C"', • an open neighborhood U of G, • a holomorphic mapping F: U —÷ C" such that: • F maps biholomorphically U onto a complex submanifold of CN, • F(G) C

•F(U\G)cC"\D, • F(U) intersects

transversally.

2.3 The Kiselman minimum principle We say that a domain G c is a tube domain if G + = G. Then we have the following necessary shape of pseudoconvex tube domains. Theorem 2.3.1. Any pseudoconve.x lube domain G C C" is of the form G = G1+i R", where G1 is a convex subdomain of R". Proof We may assume that G = to prove that G1 is convex.

G1

+ iR" for a certain domain

c R".

We have

Fix two points aij,ai E G1. Since G1 is connected, there is a continuous curve [0, 1] —÷ G1, y(s) = a•1 for s = 0. I. By S1 we denote the segment Lao, aT] y(r). Put S := {r E [0, II: ST C G1}. It is clear that 0 E S and that with a1 S is an open subset of [0, 1]. Take a ro E [0, 1] such that there exists a sequence c S such that = to. Observe that the function —logdist(., 8G) is plurisubharmonic. Since it is independent of Im z, it is a convex function (cf.

y:

Proposition 2.1.35). Therefore, if x E

then

> c> 0,

dist(x,dGi) > where c is independent off. Hence,

C Gi.

Since [0. 1] is connected, it follows

thatS=[0,l];inpasticular,Sj=[ao.u1JCG1.

0

The main goal of this section is to prove a minimum principle for plurisubharmonic function which has been found by C. Kiselman [Kis 1978]. The proof we will present here is taken from [Kis 1993J. The precise formulation of the result is as follows.

Theorem 2.3.2. Let (X, p) be a Riemann dotnain over C" -k and let G be a pseudoconvex subdomain of X x Ck such that for each x E X, the fiber

:={z€Ck: (x,z)€G}

2 Pseudoconvexity

154

is a non-empty (connected) tube domain (i.e. G is a Hartogs domain over X with be such that non-empty co,mected tubular fibers). Let U E

u(x, z) = u(x, z + it),

(x, z) e G, t E

Then the function

g: X—*

g(x):=inf{u(x,z):

x

is plurisubharmonic.

are tubular but not necessarily connected (they must consist of convex components), then the function g is not defined on X but on a Remark. (a) If the fibers

Riemann domain over X. For more details see [Kis 19781. x Ct. Then {0} + iRk may be thought as a closed subgroup of (b) Let G = +). Therefore, one is led to study psh functions on certain Lie groups invariant under a closed subgroup. For a minimum principle in that context see [Boud 1983] and [Loe 1985]. As an immediate consequence of Theorem 2.3.2 we derive the following -k x Corollary 2.3.3. Let G C w x Ct C be a pseudoconve.x domain such that each fiber

z' E w, is a non-empty tube domain. Then w is pseudoconvex.

Pmof Observe that the following function U: G —÷ u(z)

= u(z'. z") :=

max{— log dist(z, aG), O} + IIz'112.

z

=

(z'. z") E G,

is plurisubharmonic and independent of Im z". Therefore, the function g: co —÷ isaplurisubharmonicexhaustionfuncfiononw4D, g(z') := inf{u(z', z"): z" i.e. w is pseudoconvex. 0

Remark 2.3.4. Observe that the projection of a pseudoconvex domain is, in general, not pseudoconvex. For example (cf. [Kis 1978]) put

G := {(zI,z2,z3) EC3:

Z2Z31 < l}.

Obviously. G is a pseudoconvex balanced domain in C3. Then

In: C2 x C —+ C2 is the projection onto the first factor. From Theorem 2.2.9 it is clear that n1(G) is not pseudoconvex. Other examples may be found in [Kas where

1980]. [Pfl 1978]. In fact, the situation is much more delicate, namely, it turns out that any domain in , n > 2, is the projection of a pseudoconvex domain in xC (cf. [Shc 1983]).

The proof of Theorem 2.3.2 will be given along a series of lemmas. Some of them will be also of independent interest. Let us start with a result from linear algebra (without giving its proof). IIz'112.

2.3 The Kisehnan minimum principle

155

Lemma 23.5. Let L be a positive semidefinite Hermitian (it x ,:)-matrix. Then there exists a Hermitian (n x n)-matrLx M with LML = L. We say that such a matrix M is a Hermitian quasi-inverse of L.

Lemma 23.6. Let F: C"

R.

F(z) := j=I

i.j=l

bounded from below, where L := <,,<,, is a positive semidefinile Hermizian matrix and b := (b1 E C". If M is a Hermitian quasi-inverse of L, then be

LMb' = b'

and

F(z)> —bMb' = F(_(Mb*)r), z E C".

42)

Observe that LML = L implies that LMIj = 1,, j n, where denotes the j-th column of the matrix L. We claim that LMb' = b'. Otherwise, Proof.

1

1,}cC". ThuswefindavectorcEC"suchthatl1.c=Oforall

j, but b . c

0. Using these properties we obtain

F(Ac)=

A EC:

j=I

i.j=I

contradiction (because F is bounded from below). Using now the identity LMb' = we write F in the following form:

F(z) = Hence

(Zr

+ Mb*)IL(z( + Mb*) — bMLMW,

E c".

F(z)> —bMLMb = F(_bM*) = —bMb'.

Lemma 2.3.7. Let G be a domain in

9

and let u E

x

Moreover, let M(z, w) denote a quasi-inverse of

L(z, w) := (

- (z. w))

1
.

(z, w) E G.

Then

> bMb' 42)

b'denotesthetransposed(n x

on G,

:=b'.

fl C2(G).

2 Pseudoconvexity

156

where b

(b1

Pmof Fix a point

. :=

( 8zdwi

.

E G. Since u E

d2u

): G

C'1.

it follows that

+ for

all a =

E

C'1.

Thus, in virtue of Lemma 2.3.6, putting a

we obtain

= Now,letU C Cbeanopenset,andlety: U y(z)) = 0,

(z, v(z)) E G,

C" bea&-functionsuchthat

j

1

n, z E U.

where u and G are as in Lemma 2.3.7.

Define g: U —b R, g(z) := u(z, y(z)). Differentiation of g with respect to z leads to

= y(z)) +

=

Y(Z))Yjz(Z)

y(z)).

Here we used the assumption on the function y. The equality implies that the function Xz

is still of class (5>1 on U.

Differentiation with respect to

=

yields

+

y(z)) +

With the abbreviations

a :=

b

i,..., (Ylz

Ynz)

the above equation has the following form

=

y(z)) + a(z, y(z)) • fl(z) + b(z, y(z)) • a(z),

z E U.

2.3 The Kiselman minimum principle

157

(z. y(z)) =

To detennine the functions a and b, we differentiate the equation

k=

0,

Forallz cU weobtain o = ak(z, y(z))

k<

1

and

1
H(z) := (Hk3(Z)) := (

82u a Wk a

(z.

y(z))).

a' U__(z. := ( aWkaw1 y(z))).

L(z) :=

Summarizing, the following identities hold for z

z

u.

U:

a(z, y(z)) = —a(z)H(z) — b(z, y(z)) = —$(z)H(z) — &(z)Lt(z) (recall that H =

11'). Keeping the notations we have used so far. we formulate the following criterion for g to be a subharmonic function on U.

Proposition 23.8. Let M be a matrix-valued function on U sue/i that for all z E U the M(z) is a Hermitian quasi-inverse of L(z). Then >

—

Z

E U.

In part icula, the function g is subharmonic on U, if the right-hand side of this inequality is never negative on U.

Proof In virtue of Lemma 2.3.7 we know that (LMb')(z. y(z)) for y(z)) z E U. Recall from the calculations which were done before Proposition 2.3.8 that a = —aH — and b = — &L' and from Lemma 2.3.6 that LMb' b ='. Then, for z C U, we get

= UZE(Z. y(z)) + a(z, y(z)) • fi(z) + b(z, y(z)) • a(z) > bMb' + + ba' = —1311Mb' — aL'Mb + a$' + ha' = — ab' + a13' + ha' + = fiHMH13' + + aH'13' + aL&' — aHf3' — flLfl' = 13(HMH

= $(HM'H

— —

—

—

L')13t + /3H(ML — L)13 + (—b —

aL')(ML

—

= 13(HM'H

where the last equality uses again the fact that LMW = b'.

—

L)fi' + 0, D

2 Pseudoconvexity

158

Corollary 2.3.9. Let the situation be as in Pmposition 2.3.8. Moreover, assume that the following properties are fulfilled: if z E U and I E IR", then (z. w + it) E G and u(z, w) = u(z, w + it). Then g: U —* R is subharmonic on U. Pivof. Observe that here 11(z) = L(z), z E U.

U

The remaining part of the proof of Theorem 2.3.2 consists of eliminating the smoothness assumptions made so far. Proof of Theorem 2.3.2. We start mentioning that the function g 43) is obviously upper semicontinuous on X. Moreover, for the future we may assume that, without toss of

generality. X = E. i.e. k = n — 1. It suffices to prove that u is subbarmonic on =: E'. In the first step we even assume that u satisfies the following additional condition: (*)

where G(r) := {(z. w) E G: u(z, w) < r}. R÷) as in Section 2.1 and put for

We choose a radial cut-off function '4' E (0, 1]:

e

/(Z,W)\

\ el

I

n—I

(z,w)EC

j,

e

e e

1, yields a plurisubharmomc

ue(z. w) := (u *

w) +

Re wit2

on

:= {(z. u') E G:

Observethalif(z. w)

E

w), e) C G}.

w+it)

andt

E

and u6(z,

w+it) =

ue(z, w). Next, we choose positive numbers a
G(a) +

and fix an

Let e

E (0.

such that G(a) C G(c) C

(0. e21 and Zo

E'. Then we see that

inf{u€(zo, w): (Zo, w) E G(a))
ge(zo) := On

the other hand we mention the following inequalities on w) > u(ZO, w) > c. Thus we conclude that

= and

\

G(c), namely:

w): (zo, w) E

that the infimum is attained at a point (Zo. w) E G(c). Now g is the function from the minimum principle. i.e. from Theorem 2.3.2.

2.4 a-operator

159

From the definition it is clear that the functions

(z, .), z E E', are strongly convex

on

(z, w) E GE} =: By our assumption is connected, z E E'. Hence, in virtue of Theorem 2.3.1, they are convex. Therefore, also the sets

:= (WE

C G}.

z

are convex for sufficiently small e. Observe now that

fl

:'Ez+eE In particular, this implies that the fibers

z

E', are convex domains for all

sufficiently small e. Because of these two remarks we conclude that for any z E E' there is exactly one point Moreover, applying the implicit E such that gt(z) = Ut(Z, function theorem yields that E' —+ is a &-function. Thus we are in the situation of Corollary 2.3.9, which gives us that the functions are subharmonic on E'. Observe that g; hence, also g is subharmonic on E'. What remains is to treat the general case. Put Uj(Z, w)

u(z, w)} + (1/j)(max{0, — logdist((z, w), dG)} + 1 Re w112),

G. Then the functions satisfy the assumption (*), so they lead to subharmonic functions gj on E'. Since \ u, it follows that gj \ g. Hence g is a (z, w)

subharmonic function on E', which finishes the proof of Theorem 2.3.2.

0

2.4 3-operator The a-operator has been introduced by L. Hörmander ([Hör 1965], [Hör 1990]). We

will mainly follow his approach. Let (K, p) be a Riemann domain over Let A = denote p= X —+ IR be a continuous function. Define the Lebesgue measure on X and let L2(X, q) to be the space of all A-measurable functions f: X —+ C such that

I

Jx instance, if

i

<+00.

Observe that the space L2(X, may be also defined for more general weights ç, for X —f is upper semicontinuous.

2 Pseudoconvexity

160

with the scalar product

Observe that L2(X,

(f. g)

:= j

(f,

is a complex Hubert space. Put

f E L2(X,

:= (f, It is well known that the space 2)(X) :=

C) is dense in L2(X,

Moreover,

one can easily check that

L2(X,loc)= U We endow the space v

Z+, then

qo

X

U

with the standard Fréchet topology: if 'h E in Coc(X) locally uniformly on it' —+

E

£)(X). v E Z+. We say that Let —÷ Jo in compact set K c X such that supp C K for any v Z.4. and uniformly on X for any a, X) —*

Remark 2.4.1. If

if there exists a

then

for arbitrary a,

(a)

(b) The operator x is

continuous, i.e. if ij,,

(ii, f) —+

E

Jo. then

and

iiofo.

A linear operator C: £)(X) —* C is called a distribution on X (C E .V'(X)) if —k C(fo) whenever Let CR

£)'(X), v e

We say that

Co(f) for any f 46

Shortly.

'70.

1>(X)

Shortly.

—p

Shortly,

—÷ Co.

i)(A)

In.

b

—÷ Co in 2Y(X) 47) if C,,(f) —*

2.4 a-operator

161

Remark 2.4.2. (a) For u E Lt (X, bc) put

[u](f) :=

ufdA,

IE

Moreover, [uJ = 0 in £Y(X) iffu = 0 in L'(X,Ioc). Observe

Then Eu] E that if

I

Jx

L1(X,ioc)

£Y(X)

then

a subspace of (b) For C E .TY(X) and a, fi e

.

tuoj. Thus we can identify Li (X, bc) with

define

f e £)(X).

:= Then

E

.

co, then

Moreover, if

Note that if u E &(X) then

=

lal +

k.

(c) For ,i E C°°(X), C E 2)'(X) define

(,1C)(f) :=

f E £)(X).

E .Z)'(X) (Remark 2.4.1(b)). Moreover, the operator

Then

x £i'(X) is continuous, i.e. if

qo and

(ij. C) —+

E

Co, then

Let C1,C2 E £)'(X)and let U c Xbe open. We say that C1 =

C2

on U if

C1 (1) = C2(f) for any f E

Recall the following localization principle for distributions (cf. [Sch 1966]): C1 = C2 if there exists an open covering (U1)1€1 of X

suchthatC1 =C2 mU, foranyi El. bet supp C denote the supporl of C, i.e. the set of all points a E X f2 E .0(X) 0 on U for any neighborhood U of a. Observe that if are such that f' = f2 in a neighborhood of supp C, then C(f1) = C(f2). denote the space of all differential forms C on X of Fix r, s E Z÷. Let type (r, s) with coefficients in .0'(X) (cf [Nar 1968]), For C E such that C

C

=

>

Cj,jdp'

(2.4.1)

where

{l = C11 E .0'(X),

ik) E Nk: 1
< Ik

:=djij, A...AdpJ,.

2 Pseudoconvexity

162

The representation (2.4.1) is called the canonical representation of C. To simplify notation we will frequently wiite

C= IIi=r. li—s

=

Obviously,

(O}

if r >

n

ors >

n.

Let 'V(X) be a vector subspace of £)'(X). We define V(,5)(X) as the space of all such that C,,j V(X) for all I E! and J In such a forms C way we define

bc) j

C C(r.s)(A') C

Moreover, if

v E Z±, then we say that

E

so).

—* Co in V(,,3)(X)

inthetopologyofV(X)foranyl

—÷

E

we define the support of C:

E

suppC Similarly as in § 2.1, L1(X, bc); UE is defined in

U

suppC,,j.

will denote the r-regularization of a function U E := {x X: > e}; cf. Remark 2.1.24. A bc). then we define the eE

If u =

regularization of u as the form

:=

(u,,J)Edpt

E

C > 0.

111=,. IJI=s

Observe that Remark 2.1.24 generalizes in a natural way to the case of differential

forms. From now on we assume that a 'regubarization function' 4' is fixed (cf. § 2.1).

Lemma 2.4.3. If f E L' (X, bc) and f is holomorphic in f(a)foranyO < e <.r.

(a, r), then ft (a) =

Pmoj By the mean value theorem for hobomorphic functions we get

fE(a)

= =

+ ez)4'(z) dA?M(z)

J

f f

10.11"

(Jo

(J

+ rr e'°)dAn(O))4'(r)ri . . ... .

= ro.ir

0
Shortly,

. .

,,(X)

Co.

0

2.4 a-operator

163

Now, for C E £)'(X), we define the e-regularizarion of C by the formula Ce(X) :=

.

x

CVE(p(x) — p)),

0,

C 2)(X) is such that Xx = 1 Ofl lFx(x, e). Observe that the definition is independent of a particular choice of Xx• where Xx

Remark2.4.4(cf. [Sch 1966]). (a)Ifu E L'(X,loc),then [uk = [uE],r >0. = for arbitrary a, and e > 0. and (b) E

(c)lf K :=suppC (d)IfK :=suppC

C

X,4d:=dx(K),then = C((xf)€), f

2)(X) is such that x =

where x

foranyo <e <

C

I

0


on

C when e —÷ 0.

K

The definition of the e-regularization extends in the natural way to forms from

:=

E

IIl=r, 111=-s

X, a E X, and let Lemma 2.4.5. Let C E £)'(X). Assume that K := suppC n. = 0 on Px(a. 3r), j = I 0 < t 4dx(a)}. Suppose that Then

C=

[CC]

on Px(a, t)forO <e
Proof Since

= 0 on Px(a. 2t).

=

j=

1

n, the function

is

holomorphic on Px(a, 2r), 0 < < t. Hence, by Lemma 2.4.3, (C1),, = (C,1)1 =
distribution, i.e.

Fixana X,letO < 7r 2r. Hence, by Lemma 2.4.5, C is holomorphic Proof

inlFx(a,r). For u, v

LI

bc) put

:= (u, u) =

(u. v) III=r, IJI=s

Note that (ii, v) C E 0(U)

L1(X, bc), hull E L2(X, bc). 3UE0(U): C = tul on U.

lu,.j12. II=r. IJI=c

164

2 Pseudoconvexity

bc), v E bc); recall that

Remark 2.4.7. Observe that for u E

belongs to

product U A v

(

v,',j'dp" A di")

A dr') A (

>J'

II=r, IJI=s

bc), their wedge

II'I=r'. J'I=s' IKI=r+r', Lj=s+s'

where WKL

J, I

:=

,

J

)u;,Jvf',j'

IlI=r.

J'I=s' IUJ'=K, JUJ'=L

and

J. I', J') E (—I, +l} is the number such that

dp'

A

=

A dp" A

J, 1'. J')dp" A

The notion of the wedge product extends in the natural way to the case where

(cf. Remark 2.4.2(c)).

and v E

UE

Remark 2.4.8 ([Fed 1969], § 1.7). For any r, s, r', s' E such that lu A vii < MlIuIl

lvii,

bc),

uE

there exists a constant M

v E

bc).

Let Mrsrs' denote the minimal constant with the above property. Then

(r ± < I.

r + s < I or r' + s' < 1, then Exercise 2.4.9. Prove that in general Mr.5,r',s' > I. For U, v E

IR)) put

E

(u.

=

f x(u. :=

Note that the space

space and that the space

=

(uj,j, III=r, IJI=s

Jx

= lII=r.

(with the above scalar product) is a complex Hubert is dense in so).

2.4 8-operator

165

(of form (2.4.1)) put

For any C E

>'

3C:=

1l=r, IJ!=s jt

:=

>1

III=r.

JI=sj=I

Zj

Zj

Adp'

Note that the expression on the right hand side of the formula for 8C (resp. 8C) is not canonical form of is the the canonical form of (resp. SC). For instance, the following one: 0C

=

(2.4.3)

WJ.Kdp' !I=r. IKI=s+I

where

8C1.j

a(1, K, J

wj,x :=

dz1

j5J=(k1

anda(1,K,J,j) E {—l,+1}issuchthat d131 A dp' A = a(I, K, J. j)dp1 A dji". Remark 2.4.10. (a) d:

—f C(T±IS)(X).

d:

-..—-+

CfrS÷I)(X).

8:

+

8: C(rS)(X)

(b)aoa

=0,aoa=0,aoa =

(c) By Proposition 2.4.6, for C E £Y(X) we have: C E (9(X)

8C

= 0.

Remark 2.4.11. 807 A C)

=

(8z1) 17

E

AC +

A (dC). C

C E

is The equation 8u = v, where v is given and u E unknown, is called the non-homogeneous Cauchy—Riemann equation or 8-problem. Observe that, by Remark 2.4.10(b), a necessary condition for the existence of a solution u is that at' = 0. In the case where v has some additional regularity (e.g. v bc) or V bc)) one can ask whether the 8-problem has a solution with the same or similar regularity. Our aim is to prove that the a-problem has a solution if (X, p) is pseudoconvex. The proof will be based on the following result from the Hilbert space theory.

2 Pseudoconvexity

166

Lemma 2.4.12. Assume that

R2 are complex Hubert spaces. Let T

3(I 3 Dom T —+

densely defined operator and let F be a closed subspace 013(2 a linear such that Range T C F. Then the following conditions are equivalent: be

There exists a constant C > 0 such that

(i)

IIfIIx2

CIIT*fII.,(1.

f

E F fl Dom

T*.

(2.4.4)

(ii) For any v E F there e.xistsa u E DomTflRangeT* such thai Tu = vand IIuIIJt
(iii) Range T = F. 3 Dom T be a linear closed densely defined operaRemark 2.4.13. Let tor. Then: (a) Let f E R2. By the Riesz representation theorem we get

f

Dom T*

3M>O VUEDOmT

(b) (cf. [Wei 1980]) The operator

(c) KerT =

(f.

<

is densely defined and T** = T.

(Range

Pmof of Lemma 2.4.12. (i)

(ii): First observe that by Remark 2.4.13(b), (c) we

have:

Ker T* = (Range T**)l = (Range T)1 3 F1.

Consequently,iff =

Ii +f2 E DomT*,wherefi

Fandf2 .1. F,thenf2

E Dom T* and T*f = T*fi. This implies that Range T* = Ker T*. In particular. T*(F fl Dom T*). We will show that Range is a closed subspace of Ri. —* uo 3(1 Then, by Indeed. let C — > 1. Consequently. there exists (2.4.4). 1111'— such that an Jo E —÷ fo. Obviously Jo E F. Let U E Dom T. Then (us.

= lim (T*fv.

=

= (b.

urn (ft,. v-*+oo

Thus Jo E Dom T* and T* Jo = uo.

Fixav E FandletL: RangeT*

Cbegivenbytheformula

L(T*f) := (v. f)ae2. 50)

fo =

Tue.

That is. if

DomT x

(us..

f E Dom

—* (uo.fo)

x

then

C DomT and

2.4 a-operator

167

By (2.4.4) L is well defined and

IL(T*f)I
I E Dom

which shows that L is continuous. By the Riesz representation theorem (recall that CIIvII.,e, such that Range T is closed) there exists a u E Range T* with L(T*f) = (u, T*f)jf1, f E Dom Tt. Thus ii Dom T = Dom T and T**U =

=

Tu

v.

(iii) is trivial.

(ii) (iii)

(1):

We only need to prove that the set

A := (f E FflDomT*: IIT*fII.


is bounded. Since F is closed, it suffices to show that A is weakly bounded in F. Take a v E F. Since F = Range T, there exists a u Dom T such that v = Tu. Hence

for any f E A we get:

=

I(v,f)x21 =

U

IuII.x1.

EC(X,R),j = l,2,3,define

j = 1,2,3,

4S+J_I)(X. D0mT:={uE,7t,: auE3t'2},

Tu:=au, u€DomT,

DomS:={f€R2:df€3€3},

Sf:=df, feDomS.

Remark 2.4.14. (a) D(r.s)(X) c Dom T and T and S are densely defined. (b) The operators T and S are (c) Range T C Ker S (cf. Remark 2.4.10(b)).

C Dom S. Consequently,

Now Lemma 2.4.12 (with F = Ker S) gives the following Corollary 2.4.15. Assume that

I EDOmT*flDomS. Then for

any

with av = 0 there exists a u E Dom T fl Range T*

v E

such that du = v and Itu 114?1

Wefixasequenceih

E C2(X,R)such

1]). V E

E

that

112

a (ui,.

Let Dom T X

Consequently.

—+ duo in

Thus fo = auo and finally.

E

eq',

v

— (uo, fo)

(2.4.5)

1,

1lvIK

(2.4.6)

N. X

Then Uv

On the other hand.

Dom T and fo = Tu0.

Uo in

—÷ fo in

2 Pseudoconvexity

168

Exercise 2.4.16. (a) Find a sequence tion (2.4.5).

C

(b) Prove that for any sequence IR) with (2.4.6). a function E

LO,

I]) satisfying condi-

[0, 1]) with (2.4.5) there exists

C

=4,.

ForçaE e2(X,R)putq,1

Theorem 2.4.17 (Hormander s L--estimates; [HOr 19651, [Hor 19901). If £4,(x:

>

c E C(X. R>0), then

f(c

I

+ 11Sf

—

(2.4.7)

If + ed'),

c then

If

+

f

11Sf

E

Dom

Tt fl Dom S.

(2.4.8)

Assume for a moment that the above result is true. We will show how it applies to solve a-equations on pseudoconvex Riemann domains.

Lemma 2.4.18. Assume that (X, p) is pseudoconvex. Let w be a strict! psh e.xhaustionfuncuon 52) fr E & (X, R), V E bc). increasing function. Then there exists a convex be an arbitrary Let Xo: that increasing R —+ R such function x: • X >: Xo. • u x ow — vi). 112, x C(X, bR>0) such that C", where c X, • £(x o w)(x: fl >

+

c>

Proof Let co: X —+ R>o be a continuous function such that

> For any convex increasing

112,

function x:

IR

x

52)

Cf. Theorem 2.2.9(viii).

C".

—+ R we have

L(x o w)(x: U = x"(w(x))I grad w(x) •

?

X.

+ x'(w(x))_Cw(x;

2.4 a-operator

169

Put

L,:={xEX:w(x)
f

11v112

dA,

u E N.

i

Consequently, we only need to choose x in such a way that

•X

Xo,

•

< +00,

• x'(t) >

+

t

D

R.

Now Corollary 2.4.15, Theorem 2.4.17, and Lemma 2.4.18 give the following

fundamental result.

Theorem 2.4.19. If (X. p) is pseudoconvex, then for any r, s > 0 and bc) with av = 0 there exists a u 6

v

bc) with au = v.

The proof of Theorem 2.4.17 will be divided into a few steps. First, observe that is dense in Dom T* fl Dom S in the sense of the graph norm if the space

DomT*flDomS3f_+ then (2.4.8) is a consequence of (2.4.7). The above mentioned density will be proved

in Propertiesl—7. Next, we will prove (2.4.7) using Properties 8,9.

Property 1. Assume that f E Dom S. Then S(ih,f) —+ Sf in X3

Proof Since d(,7f) = qf E Dom S. For

dq A

17f

E Dom S for any 17

f + qdf, we easily conclude that 8(77f) E R3.

.0(X). Thus

we get (by Remark 2.4.8 and (2.4.6))

<

—

Hence, by the Lebesgue convergence theorem, —* Sf in Finally —+ Sf in

Property 2. Let where

E

f 6 Dom S be such that supp f

—

—4 0 in 3€3. Clearly

0

X. Then

Sf

in

denotes the s-regularizarion off.

Proof By Remark 2.1.24, Sf5 = (Sf)5, 0 <s << 1. Hence Sf5 —+ Sf in By virtue of Properties I and 2 we get The sequence

is as in (2.4.5) and (2.4.6).

0

2 Pseudoconvexity

170

Property 3. (a) The space !)(r,s)(X)

is dense

+

U

(b) The space .D(r,s+I)(X)

in Dom T in the norm

in Dom S in the norm

iS

I —+

+ 11Sf

Property 4. Assume that f e Dom T*. Then Moreover.

T*(,hf)

E

1)(X).

in .lti.

Proof By Property I we know that u Dom T we get

= (f.

(ref,

Dom T* for any

Dom T for any u

= (f,

—

(f,

Dom T. Moreover, for

A

=

— (f,

A

Hence

+ If

071.

U E Dom T. x

E Dom Tt. Moreover,

Consequently, by Remark 2.4.13(a), (T*(lif).U)ç1 —

=

f

If lie

for all u E Dom T and, consequently, for all u e

This implies that

Ill

IIT*Oif) — (Exercise). In particular. IIT*OJL,f) —

s

If

and, therefore, by the Lebesgue theorem, T*Olvf) —

—÷ 0 in .7(i. Hence, U

in .Jt'i

Let! E define 11,1K

For! E

(k,

k3)

E (1

n}

:=Oifj E {k1

:= sgn(r)f,.L if j

r(j,ki

=

E .D'(X) by the formulas

k5)=L E

k5}, where r is a permutation such that

2.4 a-operator

Property S. (a)

C Dom T*.

=

fE

II1=r. )KI=s

>-

DomT. In particu!a1

e*T*f = Af + Bf, where n

Af :=

(>2-[fi.Jx])dp'

>2'

A d15".

j=I

KI=s n

Bf:=(—1)T

fEDomT*.

>2'

III=r. IKI=s

Notice

j=t

that A. B: Dom T —+

Proof Fix an f E

bc).

and let

g :=

AdIK

>2'

E

III=r, KI=sj=I For U

>2'

UJKdp' Ad1öK E

IIJ=r. KI=s

we get

(g,

=

>2(_ly_'

>2'

II=r. IKI=sj=1

Jx

aZJ

n

=

!.jK

>2'

aUIK

dA = (1.

X

By Property 3, the above equality holds for all u E Dom T. This shows that f E

DomTtandg = T*f.

Now, take an arbitrary f E Dom T*. Let

g := T*f =

>2'

III=r, KI=s

gJ,Kdp'

E

2 Pseudoconvexity

172

Take a u E

Then

'°'dA = (T*f,

L I =r

= (f,

= X

III=r.IKI=sj=1

Zj

= I!I=r.

=

(

IIrr,

j=1

Consequently,

(_1)t_1

=

which gives the required formula.

Property 6. Let

D

f E Dom T* be such that supp I

X. Then T*fE

in

Proof Using Property 5 and Remark 2.4.4 we easily conclude that it suffices to prove that —÷ bc). We have in L2

Af + Bf =

=

0

Properties 3, 4, and 6 imply

is dense in Dom T* fl Dom S in the sense of the

Property 7. The space graph norm

DomTtflDomS9f—+ Remark 2.4.20. Let ag

aço

j=1 =——g—. azj Observe that

=

aZk

Property 5 gives

8ZkdZj

g,

gE

n.

2.4 a-operator

173

Property S. n

= (_l)r_1

A

I1I=r. IKI=si=1

+(_l)T_I

>'

fE

dzJ

I!I=r, IKI=sj=I

in

Ix III=r E' j.k1 +

2f

f

f

we have

E

azk

111r.

1I=r,

Pmof Recall that

st= I1I=r.IQI=s+2

71.Q.J.J)—--—)dP'AdiQ.

(

dZJ

{i.ii

wherea(/, Q, J, j)

E

(—1, +l} is such that A dp1 A

Moreover,weputa(i, Q,

dji' = a(i, Q, J, j)dp1 A

J,j) :=Oif{j,j1 a(i,

Oslo2 =

Q, J

Hence J)

af,.j

—a(i. Q, L,

III=r. tQI=.c+2

j,IE(I (j)uJ=(t)uL=Q 2

=

x

X 'J(l, (j.

£,

aZJ

III=r.

II=r, IJI=s+1 K>).

j)a(1. (j, £, K>), (j. K>), £) sgn(t(e.K)) sgn(r(j.K)) afl.JK

2

= fl=r, Jl=.r+I

Zf

af,j

= fl=r, JI=.c+1

I!I=r, —

U

jfl=r. IKI=sJ.k=1

(JhI,jK afl.kK ôZk

3Zj

174

2 Pseudoconvexity

recallthatQ =(j..e, {j, k1,..., k5J, {ji

=(j,K>)E = (i. k1. .. , Ac3), .

is the permutation mapping (j, Ac1

Property 10

(The

Ac3)

onto

respectively;

.

moreover, r(j,

J.

K)

0

main part of the proof of Theorem 2.4.17). Inequality (2.4.7) is

true.

Proof By

Properties

8 and 9

we get —

I

af,.JK

1I=r. KI=.cj.k=1

Zj

x III=r, IJI=s+lj=1

+2j II!

+ Hence, by Remark 2.4.20, we have

j

IJ=r

>.L' l!I=r j=1

j.k1

IKI=s

+2f

+

(2.4.9)

0

which directly implies (2.4.7).

Proposition 2.4.21. Let (X, p) be a pseudoconvex Riemann domain R>0) are such that that y E C2(X), c E

£y(x; Let v E

y)

be

such that

II

Then there exists a u E

x E X.

Assume

E

dv = 0 and

f I 11v112—dA <+00. C Jx y) such thai du =

v

and (2.4.10)

2!.

L

Proof We may assume that I = 1/2. Let 6 be a CCO strictly psh exhaustion function d for X such that £6(x, > E

X: 8(x)
t ER;

2.4 b-operator

175

X, is a pseudoconvex region (cf. Corollary 2.2.15). We will show that: bc) such that 8u1 = (*) For each t R there exists a u, E

I

J xl Suppose

v

and

< 1.

lu1

(2.4.11)

for a moment that (*) is already proven. Fix a sequence

/ +00.

is bounded in By (2.4.11), for any compact K c X the sequence u weakly Consequently, by the Banach—Alaoglu theorem, we may assume that bc). In particular, au = v and with u E

I

< 1,

Jx, which finishes the proof.

a:

lit Suppose that the functions E satisfy (2.4.5), (2.4.6), and, moreover, (notice that such functions do exist). = 1 on v E N, and * = 0 on convex increasing function such Put := y + o (5. where x: IR —÷ IR+ is a Then — 2* > y and that x = 0 on (—oo, z), x oS > 2*, and X'o15 ? ))2 := — := 0 ö)(., E C". Put q'i := q) — 2*, 2: and consider the a-operators We move to the proof of (*). Fix [0, 1]), v E N, and * E

L

D Dom S

wa).

By virtue of Theorem 2.4.17 we get

fE D0mT*flDOmS.

+ 11Sf

Observe that v )(v,

Consequently,

f f I

f Recall that (KerS)1 C

0 = (v.

and v

Dom

fl KerS.

KerS. Thus

f I KerS.

(2.4.12)

2 Pseudoconvexity

176

Finally, by (2.4.12),

< IIT*fIIwi,

f

E

Define L: Range T* C, L(Ttf) := (v, the Riesz theorem, there exists a u, E Range T* C

L(T*f) = (u,,

DomT. Then, L is well defined and, by WI) such that

f E Dom Tt,

and (2.4.13)

< 1. Hence

= Tu, =

v.

0

Condition (2.4.10) follows directly from (2.4.13).

The following auxiliary function will simplify our notation:

+ 11z112

Notice that — log

E

Theorem 2.4.22. Let (X, p) be a pseudoconvex Riemann domain over C", let E and let v E W). ôv = 0. Then there exists a U E bc) such that au = v and

j

o p)4dA

f

(2.4.14)

with av =

In particular, if p(X) is bounded, then for any v E exists a u E W) such that Ôu = v and

0 there

(I whered :=

XE X}.

Pmof Step 1°. Assume that J2y(x:

E

L(—4logSoop)(x:

e2(X). Put y :=

—

4logSo o P. Then we get

=:

x E X,

y) with du =

Consequently, by Proposition 2.4.2 1, there exists a u E

o p)4dA

=

j

2f I$vII2f_dA =

L

E C". v

and

2.5 Solution of the Levi Problem

Step 2°. The general case. Let

:= {x E X:

177

\ ço as e \ 0.

fl

E

is a pseudoconvex region (cf. bc) such that Corollary 2.2.15). Hence, by Step 1° there exists a ur E where

=

v in

> e}. Note that

and

f

o p)4dA

f

(2.4.15)

L

\ 0. Observe that the sequence is locally uniformly bounded from above in X. In particular, by (2.4.15), for each compact K C X the sequence is bounded in Consequently, we may assume that there —+ uweakly. Inparticular,du = vinXand existsau E Fix a sequence

f

o p)4dA =

K

urn fK


o

X, e >0.

K


X

K

which directly implies (2.4.14).

2.5 Solution of the Levi Problem Recall (cf. § 2.2) that the Levi problem is to decide whether any pseudoconvex Riemann domain over is a domain of hobomorphy.

Theorem 2.5.1 ([Jar 1979], [Pfl 1982]). There exists a constant C > • for any pseudoconve.x Riemann domain (X, p) over

•foranyço

E

with

E

0 such that:

L'(X, bc), and

X and 0 < r
• for any a

X

<(I

P(a,r)

1,

f(x) =

Ofor x E 2C

\ (a}, )fl

r S0(p(a))

C'onsequent!v if (X. p) is a pseudoconvex (in particular a holomorphically convex) Riemann domain overC', then 0(X) separates points in X (cf Remark 1.10.5) Proof Fix a function x E C Izi > 1/2. Letc :=

[0, 1]) with x(z) = 1 for IzI S 1/3, x(z) = 0 for Define Xr(Z) := X(z/r), r > 0.

18c.

In fact, we know more: if (X. p) is pseudoconvex, then for any E PSJ?(X) with o L1(X. bc) the space (f E (9(X): < +ac} separates points in If stalks.

2 Pseudoconvexity

178

Fix (X, p),

a,

and r. Let

Xk:={xEXJpJ(x)—pj(a)I
k=0

n},

n.

Note that P(a, r) is the connected component of Xo that contains a (cf. Remark C X,, = X. It is clear that (Xk, p) is a 1.1.2(f)). Obviously C Xo C n (cf. Corollary 2.2.15). n } there exists a function fk E O(Xk) such p'(p(a)) \ (a) and

pseudoconvex region over C", k = 0 We will show that for each k E (0

that fk(a) =

=

1,

0 forx

f

E

2C

<(f

r

P(a.r)

Xk

The case k = n gives the required result. We proceed by induction over k. Put

fo(x) :=

I,

0,

xEP(a,r) xEXO\P(a.r).

It is clear that fo satisfies all the conditions.

Suppose that 1k-I is already constructed (for some 1

k

n). Define

pk(a))

j

10 where

XkI: Jpk(x) — pk(a)I < r/3}, := {x Xk: pk(X) — pk(a)I > r/2}.

ilk := (x

E

It is clear that dv =

Then v c

0.

Moreover,

p)6(k_I)dA

<

I

—

dZ

pk(a))

1k—I

p)6(kI)dI\

pk—pk(a)

r

By Theorem 2.4.22 there exists a function u E L2(Xk, bc) such that du = v and

f u E O(Uk). Define 1k

J Xr(Pk

—

pk(a))fk_I

(Pk

—

pk(a))u

on X*i

— (pk

—

pk(a))u

on

—

2.5 Solution of the Levi Problem

dfk = 0 in

Then Uk

/

2/

IXr(Pk

(thus we may assume that Itremains to estimate the integral.

179

and

E

1k

= 1k—I in

o

—

o 2 2

80(p(a))

o Xk

0

)=Mk. Applying the same methods one can get the following result.

Theorem 2.5.2. For any d > 0 there exists a constant C(d) > 0 such that:

•for any pseudoconvex Riemann domain (X, p) over

with

XE X}

•foranyço E S'SR(X) with •foranya E X, 0
E

L'(X, bc),

and

anf E 0(X)suchthatf(a) = 1, f(x) =Oforx E

/ In particular

C(d)

(fP(ar)

X

= 0), the space

__) r

(X) (cf Example 1.2.5(b)) separates points in

stalks.

Let

8x:=min{px,öoop}=min{px,

1

+

11p112

By Theorem 2.2.9, if (X, p) is pseudoconvex, then — bogSx E Recall that for any function 6: X —÷ we put

0(k)(X 8) := {f E 0(X): the function

6'f is bounded on X},

k>

0.

We know (cf. Example 1.2.5(d)) that if 1/8 is locally bounded, then 8) with the norm f —* 116"! lix is a natural Banach space. Observe that if 6 is bounded, then

8) C

8) fork

2 Pseudoconvexity

180

Lemma 2.5.3. Let (X. p) be a Riemann domain over C" and let 8: X —÷ be

such that 8

Px and

x E X, x'

IIp(x') — p(x)II,

18(x') — 8(x)I

Bx(x);

(2.5.1)

cf Remark 1.1.14(e)(.fore.wmple8 = ox). Then

q >0, j E 0(X),

k where (k

c(n.k,q) := In

particular (k = 0,q =

2k+2n

I/q

+ C

2),

Proof Fix k, q, f, and a E X. Since 1j1q inf ml

we get (cf. Exercise 2.1.14)

f

f

—

inf

I

I

f

r)k Já(a.r)

—

—

0<0<1

Ifr'dA}

B(a,r)

—

<

IJq

0)k

I

ix

f

= (c(n. k.

D

Proposition 2.5.4. Let (X, p) be a pseudoconvex Riemann domain over C" and let 8: X —+ ILo be such that •8 • condition (2.5.1) is

•

— logO E .'PSJ((X).

Then:

For any k > 0, a E X there exists a function f {a}, and f(a) = 1, f(x) = Oforx where c(n.k) > Odependsonlyonnandk. (a)

In particular. (b) Ifsup{llp(x)II :x

functionf

0(k+4n)(X 5) such that

8) separates points in X. X)

<

d <+oo, then for any k

0, a E X there exists a

0(X.8)suchlharf(a)= l,f(x)=Oforx p1(p(a))\{a},and


where c(n, k. d) > 0 depends only on n. k, andd. 8) separates points in X.

Proof (a) By Theorem 2.5.1 with çü := —2k logO, r := &8(a), there exists a (a), and function f 0(X) such that f(a) = 1, f(x) = 0 forx

J

X

In2o2k800p6fldA

(j'P(a,r

2c

r 80(p(a))

5

2.5 Solution of the Levi Problem

181

where d(n, k) > 0 depends only on n and k. Hence, by Lemma 2.5.3 (with q = 2) we get [c(n, 2k + 6n, 2)]2 j

If < ?(n, 2k + 6n, 2)d(n,

X2. The case To prove that 8) separates points take xi, E X, xI p(x2) follows from the proposition (with k = 0). If then there exists a) E {l n} with Pj(XI) and p1 E (9m(X. 8x) C

p(xi) =

0(4ht)(X. 8).

0

(b) Use Theorem 2.5.2.

ProposItion 2.5.5. Let (X, p) be a pseudoconvex Riemann domain over C'1 and let 8 be as in Proposition 2.5.4. Then:

8)-domain ofholomorphyforanyk > 6n. (a)(X, p) isan (X. p) isan 0(k)(X 8x)-domain ofholomorphyforanyk > 6n. (b) If p(X) is bounded, then (X, p) is an

of holomorphy for

anyk> 2n. (X, p) is an

In

8x)-do?nain of holomorphy for any k > 2n.

Remark 23.6. In the case (X, p) = (G, id), where G is a domain in C'1,the result may be improved. Namely, any domain of holomorphy G C C'1 is an := min{1, pG}So (cf. § 2.12). Observe of holomorphy for any k > n, where

that if G is bounded, then

SG)

C (9(k)(G 6G) C

=

SG); if G is unbounded, then

8G).

Proof of Proposition 2.5.5. (a) Fix a k > 6n. Let a: (X. p) —p (X, j3) be the 8) separates points in X (Proposition 8)-extension. Since maximal 2.5.4(a)), the morphism a is injective and therefore we may assume that X C X, p= a = idx.

PutT := {f

0(X): fIx E

8)). Suppose that there exists an xo E

Let U bearelativelycompactunivalentneighborhoodofxo. Thenthereexistsac0 >

0

such that

If IIu < coIIökfIix,

fE7

(cf. Remark 1.4.5(i)). By Proposition 2.5.4(a) for any a

with f0(a) = I and il8kfa lix
Thus Px

coc(n,

X there exists an fa

E

7

Consequently, a E U fl X.

> 0 on U fl X for some &j > 0; contradiction.

(b) Use Proposition 2.5.4(b).

[]

2 Pseudoconvexity

182

Theorem 2.5.7 (Solution of the Levi Problem). Let (X, p) be a Riemann domain over C'7. Then the following conditions are equivalent: (i)

(X, p) is a domain of holonwrphy;

(ii) (9(X) separates points in X and (X, p) is holomorphically convex;

(iii) (X, p) is holomorphically convex;

(iv) (X, p) is pseudoconvex. Proof By Theorem 1.10.4 we know that (i) (iv) are obvious. The implication (iv)

(ii). The implications (ii) ; (iii) (i) follows from Proposition 2.5.5. 0

From now on the following notions will be used synonymously: • (X. p) is a domain of holomorphy. • (X. p) is holomorphically convex, • (X, p) is a Riemann—Stein domain, • (X. p) is pseudoconvex. Notice that the above result is not true in the category of ramified Riemann domains — cf. [Gra-Rem 1956), [Gra-Rem 19571.

Proposftlon 2.5.8. Let (X. p) be a Riemann domain over C'7. Let k > 6n, and let Ox)-extension such that (Y. q) is a domain of a: (X, p) —+ (Y, q) be an holomorphy. Then (V. q) is the envelope of holomorphy of (X, p). In other words, if all functions from Ox) can be holomorphically extended to a domain of holomorphy (Y, q), then (V. q) must be the envelope of holomorphy

of(X,p). The above result has interesting applications in quantum field theory; cf. [Bor 1996]. Moreover, it permits to give an elegant proof of the classical Bochner theorem (the envelope of holomorphy of a tube domain G = R'7 + iB C C'1 coincides with the convex hull of G); cf. [Pfl 1982bJ.

Proof Weonlyneedtoshowthata*(O(Y)) = (9(X). Recallthatpyoa ?: px. Hence Oyoa 8x and, consequently. Sy)) C Ox). By Proposition 2.5.5 451)-domain of holomorphy. Thus a: (X, p) —+ (V. q) is the (V. q) is an maximal Ox)-extension. Let $: (X. p) —p (X. be the maximal holomorphic extension. By the same (X, p) —p (K, J5) is the maximal argument as above Ox)-extension. Consequently, there exists an isomorphism a: (V. q) —÷ (X, p5). 0

As an application of Proposition 2.5.8, we will prove the following important theorem.

Theorem 2.5.9 ([Gra-Rem 19561). Let (X, p) be a Riemann domain over C'1 and let a: (X, p) —÷ (X, be the maximal holomorphic extension. Let M be a pure (n — 1)-dimensional analytic subset of X. Put M := Then alx\M is the maxima! holomorphic extension. (X \ M. —k (X \ M,

2.5 Solution of the Levi Problem

183

Proof (Cf. [P11 1 982b]). By Corollary 2.2.16 (X \ M. j5) is a Riemann—Stein domain. Thus, by Proposition 2.5.8, it suffices to prove that a: (X \ M, p) —÷ (X \ M, j3) is an

\ M, Sx\M)-extension. := f, E (9(X), i = flEl Mj, VllI.A.1855>). Putf, := oa E 0(X),i El. Fix an arbitrary function f e Let

(cf. [Gun-Ros 19651, Th. 0

be such that

C.

First we show that for each I E I the function

f is locally bounded on X.

Indeed, fixi E landa EM. Observethatforanyo < r 3x(a, r/2) \ M

px(a)andx E

we have

PX\M(X)

= dist(p(x). p(13x(a, r) fl M)).

=

Let 0
xE

r) \ M.

It is clear that there exists a c, > 0 such that —

Hence,

forx

.f,(x")I

c, IIp(x') —

x" E Bx(a, r).

Bx(a, r/2) \ M, we have

(x)f(x)I

(ct

If(x)I

inffllp(x) — p(x')II: x' E 8x(a. r) fl

= Now, by the Riemann removable singularity theorem, the function toafunctiong1 E 0(X). E oa

1f extends

By the identity principle, for arbitrary Ii, i2 E I the functions and g,2/J coincide on k \ (M,, U Thus the function f given by formula onX\M1,i E Evidently,foa = f. D The following extension theorem will be useful is Sections 3.1 and 4.5.

ProposItion 2.5.10. Let (X. p) be a Riemann—Stein domain over C" and let H C C" beacomplexaffinesubspacewithl PutM sider (M, PIM) as a Riemann region over Ck with k := dim H (cf corollary 2.2.18). Then 0(M) = 0(X)IM, i.e. any function f E (9(M) extends to afitnction f 0(X). Proof Using induction we reduce the proof to the case H = (Zn = 0} (k = n In fact, one can find I with #1
—

1).

2 Pseudoconvexity

184

Fix a function f E 0(M).

Ua€M Px(a). Write

Put U

xC. Take an a M. Observe that p(M fl Px(a)) = P((p'(a),O),dx(a)) fl = 0). Consequently, we define the local extension fa(x) := 0)), x Px(a). One can easily prove that Ja = Jb on lPx(a) fl Px(b). Thus we get a function

fEO(U)suchthatflM =1.

=

Leti'p E

U. Define

-

onU

g.1Jcof

Observe that g is well defined, g

C

I

C°°(X), and g =

f on V. Put

onX\M

lo

onV.

Then v is well defined, v and av = 0. By Theorem 2.4.22 there exists a function u L2(X, bc) such that du = v. In particular, du = 0 in V and, consequently, we may assume that u 0(V) (cf. Proposition 2.4.6). Define — Thenf E boc)flO(V), 1= Jon M, anddf = OmX. Using f := g

0(X).

once again Proposition 2.4.6, we conclude that f

D

2.6 Regular solutions Let (X. p) be a Riemann domain over C's. Our aim is to prove that if (X, p) is pseudoconvex, then for any a-closed form V E i.e. av = 0, there exists a U with du = v (cf. Theorem 2.6.4). For k E U {oo} let bc) be the Sobolev space of all u L2(X, bc) such that for any a, k the derivative belongs to with al + flu L2(X, bc), i.e. there exists a Ua8 L2(X, bc) such that in the = sense of distribution (cf. [Nar 1968]). Let further bc) denote the space of all forms of type (r, s) with coefficients in .Wk(X bc).

1968]). In particular,

bc)) C

(b)

k>

bc) C

Remark 2.6.1. (a) (Sobolev lemma)

bc) = bc), k >

0.

Defined:

c',jdp'

aCI.JK

:=

j1

II=r

JI=s+I

KI=s

)d I

0 (cf. [Nar

2.6 Regular solutions

recall that C,,1 K :=

IfJEK

10 I,

o(l, J, K, j)Cj,j

and a(I, J, K, j)d

where J := (j,

if j K, A dp' A df? = dp' A

=0 Remark 2.6.2. (a) Ifs > 1, then Indeed, for C E we have C,k11, =

j,k= I

185

(Ill = r, ILl =

—CJ,JkL

s — 1,

n)andhence

I

k

bc) =

0.

bc)).

bc): Ou E

(u

r

0, k 0.

bc)

(b)

=

bc): au E 'W(rs+2)(X, bc), Oti

E

'Wfr$)(X, boc)},

r,s>0, k>0. bc) is such that

Pmof. (a) The inclusion C' is trivial. Suppose that u E

au = >au, Adp1 E Wfr1)(X,lOC). III=r Then

au,

bc) for any I and, therefore, we may assume that r = 0.

We use induction on k. in

[0, 1]) be such that x = I k = 0: Fix a compact set K c X and let x E X, and neighborhood U of K. Put v := x u. Then v E L2(X), suppv

a

EL2(X),j1 For any w E

0w

—

2

n.

we have:

=1( ———dA=—II' ix ix OwOw

f

x OZ1 dZJ

0Z3

f thdA=—I ix

02w

L'-X)

,

j=1

n.

-

wdA (2.6.1)

2 Pseudoconvexity

186

Let VE denote the e-regularization of v, 0 < e << marks 2.1.24, 2.4.4) that

I

(cf. § 2.1). Recall (cf. Re-

8v

I aV \ —=1—I

j=l

— whene—+0,

n.

By (2.6.1) we get dUE"

L-(X)

dZj

Hence there exists

—÷ 0,j =

a

I

=

dv('

dUE"

,

dZj

dZj

function uj e L2(X) such that n. This implies that

=

U

n.

,

L2(X)

urn E—.O dZJ

=

urn

=

[v1].

—+ Vj in L2(X) when

j=I

n

n. Hence v E W'(X) E L2(X), j = I and, therefore,u E W1(U). Consequently.u C W1(X,loc). - Assume that the result is true fork—I (k > 1). Let u L2(X,loc) be such that be). We already know that u Wk(X, bc). For any al + du E
we get

j=I Applying the case k = Oto the function

n.

we conclude that DCE$u E W1 (X. toe).

Thusu is trivial. Suppose that u e (b) The inclusion bc) and t)u E bc). 814 E We use induction on k. k = 0: Fix a compact set K c X and let x

bc) is such that

[0, 1]) be such that x = 1 in a neighborhood U of K. Put v := x u. Then v E suppv Au = Bu + where B is W(r.5+2)(X) and -Ix8u = (8x) a

linear operator with coefficients depending on x 56)• Observe that the main inequality (2.4.9) in the proof of Property lOin § 2.4 is true

for*

=0(noticethatthen

=

Hence,foranyw

we get n

H

1I=r If I=s+ I 56

flu := I

OL2(X)

(X)

+ 118W112(X).

(2.6.2)

2.6 Regular solutions

187

denote the e-regularization of v, 0 < r << I. Recall that Moreover, dye —+ dv in and —+ when e —÷ 0. By (2.6.2) (put w = and next e —* 0) we get

=

Let

=

L(X)

IIL2(X)

ii

IlI=r

and in

+

3

JI=s+I Hence, by (a), v E

bc).

Consequently. u E

Assume that the result is true for k — I (k such that du E bc) and Ou E UE bc). For any at + flu k let

1).

bc) be Let u E bc). We already know that

:= I

We have:

=

= Applying the case k = 0 to the form Dcr

we get

E

'W(',.S÷I)(X, bc).

0

bc).

Consequently, u E

Theorem 2.6.4. Assume that (X, p) is pseudoconvex. Then

bc)) =

bc): dv = O}.

{v E

r,

s>

0,

k>

0.

in

= Pmof Fix a v E The case s =

0

{v

dv = 0},

E

r, s >

0.

bc) with dv = 0. is simpler: by Theorem 2.4.19 there exists a u e

bc)

such that u E WJ)(X, bc). Assume that s 1. By Corollary 2.4.15 and the proof of Theorem 2.4.19, there exist functions IR), j = 1, 2, such that if E

T:

thenthereexistsau we have:

Tt(f) =

jDomT v.

ByProperty5fromj 2.4

f E DomT*.

2 Pseudoconvexity

188

= 0, we get

Since u E Range(T*) and

= 0. Hence

= Au,

(2.6.3)

coefficients. is a linear operator with In particular, E bc). Now, by Lemma 2.6.3(b) (with k = 0) we conclude that u E bc). Suppose that we already know that u E bc) and bc) with 0 < f < k — I. Then, by (2.6.3),

where A

hence, again by Lemma 2.6.3(b), u

bc). Induction over I finishes the

E

0

proof.

n > 2, Theorem 2.6.5 ([Ser 19531). Let (X, p) bea Riemann—Siein domain =0. Then there existsau E £)(X)suchthat and Ietv E V

= du.

Remark. Notice that the above result is not true for n =

Indeed.take(X, p) := (C, Suppose that v = for a function u E

1.

v dA2 0. Then, by the Stokes theorem, for

E

sufficiently large R > 0 we have 0

=

f

udz

=

J

A

dz =

B(R)

2i fB(R) v dA2

0;

contradiction.

—* Recall Pmof Consider the Linear operator T = ft: (Theorem 2.6.4) that T is swjective. Observe that Ker T is a closed subspace of

cf. § 2.4). Hence, by

(in the standard Fréchet topology on the Banach theorem, the induced operator

T:

—*

is homeomorphic. Consider A:

Ker T

C,

A([f]) :=

f

VA

f.

Observe that A is well defined.

Indeed, first of all, the integral exists because supp v is compact. Moreover, if (recall that f KerT. then by Theorem 2.6.4 f = 8/i for some h E n > 2 (!)) and, consequently, by the Stokes formula, we get:

JvAf=JvAah=_fa(VAh)=o.

2.6 Regular solutions

189

Put

j=l . n. and define

C: C°°(X) —* C,

C(g) :=

The operator C is linear and continuous and, therefore, it is a distribution with compact support (cf. [Sch 1966]) Let us calculate 8C in the sense of distributions:

=

=

=

= By Theorem 2.6.4 we get v = aw for a w E v

We have proved that

=

= —J —v

in the sense of distributions.

Thus a(w + C) =

0

in the sense of distribution. Consequently, w + C is a holomorphic function (cf. Proposition 2.4.6) and hence C E Finally, we put u := —C. D As an application, we present the following extension theorem.

Theorem 2.6.6 (Hartogs theorem). Let (X, p) be a Riemann domain over C" such that (9(X) separates points in X and let K C X be a compact such that X \ K is connected. Then any function f E (9 (X \ K) extends holomorphically to X. We do not know whether the above result is true without the assumption that

(9(X) separates points

Proof Let

a:

(X.13)

be the maximal holomorphic extension. Since (9(X) separates points, the mapping a is injective (cf. Remark 1.8.5). Consequently, we may assume that X is an open subset of X, p = and a = id. Fixanf E (9(X\K). Let92 E [0, = inaneighborhood U of K. Then the function 1

w :=

onX\K onU

10

is well defined and belongs to C°°(X). Note that w =

Since C is

IC(g)l

f in X \

I

onX

10

onX\suppq.

Put

continuous, there exist a compact K C X. k E and d > for anyg C K.

0

such that

190

2

Pseudoconvexity

Then the form v is well defined, belongs to (X), and = 0. By Theorem 2.6.5, thereexistsau E £)(X)withau = v. Inparticular,u Let Ydenote the union of those connected components of X \ supp that intersect X \ supp u. By the identity principle, we get u = 0 on Y. -

fonXflY. ObservethatXflY 0. \ supp 1' Y and let y: [0, 1] —* X be a curve with y(l) = b. if y([O, 1]) C X \ then y([O, 1]) C Y and hence w—u. Then 7€ (9(X)andf =

PutJ Indeed, y(O) = a,

take a

X

a x flY. Otherwise, let to max{t [0. 1]: y(t) Obviously, to < 1 and y((to, 1]) C Y. Lete > Obesosmallthatc := y(:o+e) e X. ThencE XflY. Since X \ K is connected, the identity principle gives f = in X \ K. 0

f

Let us notice that L. Ehrenpreis [Ehr 1961] was the first who observed that the proof of the above Hartogs theorem may be based on solutions of the d-problem with compact support. See also [For 1998] for an analysis of the proof of the Hartogs theorem.

2.7 Approximation The a-theory provides the following possibility of approximation of holomorphic functions on pseudoconvex domains.

PropositIon 2.7.1. Let (X, p) be a pseudoconve.x Riernann do,nain over and let u be a strictly psh C2 -exhaustion function on X. Then for any function Jo holomorphic

in a neighborhood of the compact set Ko := (x X: u(x) C 0(X) such that —p — fojIL2(K0)

0) there exists a

sequence

Proof Let F1

:= 0(X)1x0,

F2

0(K0)

U 0(U)IK0; UJK0 U open

and F2 are subspaces of L2(Ko). We have to prove that F1 is dense in F2 in the topology of L2(Ko). By the Hahn—Banach theorem, it suffices to show that if E (L2(Ko))' is such that = 0 on F1, then 0 on F2. in other words, since (L2(Ko))' = L2(Ko), we have to show that for any function v L2(Ko) if F1

=Oforanyf

J'K0JvdA =Oforanyf

Fix a function v E L2(K0) with fK0 fi3dA = 0 for any f

F2.

F1. We extend v to

X by putting v =OonX \ Ko. Suppose that the functions 'h satisfy (2.4.5) and (2.4.6). Let co: X

£u(x:

>

112,

x

X,

r(t) := max {

E N, and be a continuous function such that

[0, 1]), i' IR>o

C". Define 112 + e*)J,

t E IR

2.7 Approximation

191

where K, := (x E X: u(x) t}. Recall (cf. the proof of Lemma 2.4.18) that if x: IR —+ R÷ is a convex increasing C°° function such that x' r, then o

Thus, for

if we put

+

u)(x:

x

X.

:= x o u, the condition from Theorem 2.4.17 is satisfied. Consequently, := := := then we can use Hormander's

L2-estimates. Take a sequence Xv: IR —* R÷. v? I. of convex increasing that +00 for = xi(t) fort < 0 and >

/

corresponding functions L2(X,

E C's.

1, j =

v

1,

r>

functions such 0. Define the

2, 3, and a-operators

3

3 Dom

42)(X,

Dom S1,

By Theorem 2.4.17 we have:

f

+

> I.

E

C (9(X). Thus

Recall (cf. Proposition 2.4.6) that Ker

I

u> I.

Now, by the proof of Lemma 2.4.12 (with F = Ker E Dom

fl Ker

for each u there exists a

such that

= Let

=

Then by Property 5 from § 2.4 we have:

= in the distribution sense. Define

:= Then V

=—

Moreover, we get (recall that Xv

—i-, aZJ

I)

> 1. —

Xv+1)

j = L0

f = =: C,

v> 1.

(2.7.1)

2 Pseudoconvexity

192

Consequently, after selecting a subsequence, we may assume that

—* ho weakly

Obviously,

in

in the distribution sense. Moreover, ho = 0 on X \ Ko (use (2.7.1)). We have:

fi)dA

=

fi3dA

f

=

f

is holomorphic in a neighborhood U of Ko, then we apply the above formula to the function defined as where f on U and 0 on X \ [0, 1]). = near K0. Consequently, fK0 fi3dA = 0. E 0 1

Proposition 2.7.2 (cf. Proposition 2.2.6(a)). Let (X, p) be a pseudoconvex Riemann domain over C", let K c X be a compact set, and let U be an open neighborhood of Then there exists a strictly psh exhaustion function u E such that K C {x E X: u(x) <0) C {x E X: u(x) 0) C U.

Proof Let uo be a continuous psh exhaustion function on X (cf. Theorem 2.2.9). We may assume that uo < 0 on K. Define L := (x E X: uO(x) 2), M := {x X \ U: uQ(x) 0}. The sets L and M are compact. Suppose that M 0. Since M fl = 0, for any point a E M there exists a function Ua such that Ua < 0 on K and ua(a) > 0. Let := {x X: dx(x) > e) and let (Xe) denote the regularization of (cf. § 2.1). It is clear that (U0 there exists an 6(a) > 0 such that L C (ua)E(a) <0 on K and (uQ)E(O)(a) > 0. Put Va := (Ua)E(a). Since v0 is continuous, there exists a neighborhood V0 C X6(0) of a such that Va > 0 in Va. Since M is compact. there exist points ai aN M such that M c Vaj U ... U VaN. Define è := maxfr(a1) e(aN)}, w fl ,..., } E and C on M. Let c := max{ I, maxj. and put f

max{w(x),cuo(x)}

ifuo(x) <2

v is a well-defined continuous psh function on X. Observe that (v 0, then v(x) > cuo(x) > 0; ifx E X \ U and uo(x) <0, then x M and therefore v(x) > w(x) > 0. Thus v > 0 on X \ U. If M = 0 we put v := uo. It remains to smooth v. Put

2)

QL.:={xEX:u(x)
2.7 Approximation

Let

193

be the regularization of v and let

E

Then V5 is strictly psh and v5 \ v as e \ 0 on X. Since V is Continuous, V5

V

let r(v) > 0 be such that locally uniformly on X. For each v CC 10, 1]) be such that and V5(v) <0 on K. Let V < < V + I on on Define := = 1 in a neighborhood of and suppq)1 C is a well-defined C°° function on Observe that = 0 on X \ XE(v) and v in a neighborhood of X and vt be a C°' increasing convex function such that X(t) = 0 for Let x: R —÷ R÷). Observe t <0 and x'(') >0 fort >0. Define := x("v + I — v) that:

-

(a) V,, is psh in a neighborhood of

(cf. Remark 2.1.2(h)).

Indeed,ifx v(x)+l+1—v Indeed, if x E - (c) V,, is strictly psh and> 0 in a neighborhood of c2L \ 0. Hence V,,(x) > 0 and for + I — v > v(x) + I — p > then \ (C" )* we get

LVL(x;

> 0

x'(v,,(x) + 1 —

C JR>o such that for each Now we are going to construct a sequence v and strictly psh in a := N the function We proceed by induction over v. Put W0 := vo and suppose neighborhood of 0 (this condition is empty for are already constructed for some v that Cl is strictly W,, + v = 0). By (a), for any ct÷l > 0, the function is We have_to find such that psh and > v in a neighborhood of V

Fix an A > 0 such strictly psh and> v in a neighborhood of H := \ x E H. C". In virtue of (c) there exists a constant that

B>

C". Hence = E C", which shows > 0 is strictly psh on Recall that >> 0 the function that with >> 0, then W,,÷1 > v on H. on H (cf. (c)). Hence, if > js(use(b)). Thusu = then we get strictly psh function on X. If x KC is a well-defined u (x) = W1 (x) = vO(x) + q x (vj (x)) = vo(x) < 0. Moreover, u > v and therefore E u is an exhaustion function.

>

E

E

(—A

Corollary 2.7.3. Let (X, p) be a pseudoconvex Riemann domain over C" and let 5'SJ? flC In particular. the set K C X be a compact set. Then K is closed.

Take an a Proof Obviously, c X \ {a}. Then by Proposition 2.7.2 there exists a function u such that u <0 on K and u(a) > 0. Consequently, a

and let U := fl C°°(X)

0

Proposition 2.7.4. Let (X, p) be a pseudoconvex Rie,nann domain over C" and let K C X be a compact set. Then for any function fo holomorphic in a neighborhood of there exists a sequence C 0(X) such that lIfe — fouR —+ 0.

2 Pseudoconvexity

194

Proof Take an fo E (9(U), where U is a neighborhood of sition 2.7.2 there exists a strictly psh exhaustion function u

By Propo-

C°°(X) such that

Kc{XEX:u(x)
Iv

C

0(X) such that 0.

fo II L2(Ko)

Let 0
0

— fOllL2(K0) —±

— 10 IlK

Proposition 2.7.5. Let (X, p) be a pseudoconvex Riemann domain over C" and let

Y C X be an open set such that (Y, p) is also pseudoconvex. Then the following conditions are equivalent: (i)

the space 0(X)lt',(y) is dense in 0(Y) in the topology of locally uniform convergence, i.e. (Y, X) is a Runge pair;

(ii) for any compact K C Y we have:

=

(iii) for any compact K C V we have: ka(X y = y

y

(iv) for any compact K C Y we have:

Proof Obviously (1) (iii) (ii) and (iii) (iv) by Theorem 2.5.7. Suppose that (iv) is satisfied. Fix a compact K C V. Let Lo := L1 = \ Y. Observe that Lo, L1 are compact and disjoint. In particular, they have disjoint open neighborhoods U0 and U1. Take a function f E 0(Y) and let .'

,'

fo := f on (Jo and fo := I on Ui. Then fo is holomorphic on a neighborhood of L := Ka(x). Obviously, L = Hence, by Proposition 2.7.4, there exists

C 0(X) such that b —f

a sequence

Jo

uniformly on L. In particular,

if we take f 0, f on Lo j K, which shows that (iv) (i). <Moreover, we get a holomorphic function g E 0(X) such that 1/2 on K and > 1/2 on ;

Recall that L1 C

Hence

L1 = 0, which shows that (iv)

(ii).

0

Proposition 2.7.6. Let (X, p) be a pseudoconvex Riemann domain over C" and let = kO(X). K C X be a compact set. Then

c

Proof Obviously, hood of K

=: L. Let U be an arbitrary open neighbor-

By Proposition 2.7.2 there exists a strictly psh function u E C°° (X)

X: u(x) < t} X for any t Rand K C Xo C U. By Corollary 2.2.15 Xo is pseudoconvex. Moreover, by Proposition 2.7.4 (applied to compact sets {x E X: u(x) t}, t < 0), any function holomorphic on Xo can be approximated locally uniformly in Xo by functions from 0(X). Thus (Xo, X) is a Runge pair. Hence, by Proposition 2.7.5 we get C Xo C U, which gives the opposite inclusion. 0 such that: X, := {x

Proposition 2.7.7. Let (X, p) be a Riemann domain over C" and let X be strongly pseudoconvex (cf Definition 2.2.3). Then the space is dense in (in the topology of locally uniform convergence).

2.8 The Remmert embedding theorem

195

Proof In virtue of Proposition 2.7.4 it suffices to find a pseudoconvex open set

CX

for any compact K C with C such that C C U (cf. Let a E C2(U, R) be a strictly psh defining function for f2 with Proposition 2.2.25(a) with k = 2). We put := (x E U: u(x) <e} with sufficiently D smallO<e<<1.

2.8 The Remmert embedding theorem FN): X —* and let F = (F1 Let (X, p) be a Riemann region over N n, be holomorphic. Recall (Cf. the proof of Lemma 1.10.12) that F is regular on A

C X if

raFi __Lj(x) = n for any x E A. dZIC

If F

is

regular (on X), injective, and proper, then we say that F

Notice that if F: X —+ dimensional submanifold of CN.

is an embedding, then F(X)

is an embedding. is a complex n-

Theorem 2.8.1 (Remmert embedding theorem). Let (X, p) be a Rie,nann—Stein rea holomorphic embedding F: X —* gion over C? 58)• Then there a compact Lemma 2.8.2. Let (X, p) be a Riemann region over C set such that 0(X) separates points in K. Then there exist N E N, N > n, and a holonwrphic mapping F: X —* C's' such that F is infective and regular on K.

Proof Since p is locally biholomorphic, there exists a neighborhood U C X x X of the diagonal {(x,x): x E K} such that if (x', x") E U and p(x') = p(x"), then

= x". 0(X) separates points in K, for any pair (x', x") E K x K \ U there exists C X x X be Let an f(x',x") E 0(X) such that f(X',X")(X) any (y', y") E for a neighborhood of (x', x") such that f(x'.x')(Y') f(x',x")(Y) (x,x') such that U(x'.x"). Since K x K \ U is compact, there exist K x K \ U c U(X'X") U U(X'X",. Li Now we only need to take N := n + £, F := (p, Since

Lemma 2.8.3. Let (X, p) be a Riemann region over C' and let

F = (Ft

FN÷1): X —+

be a holomorphic mapping. For a = (at

Fa := (F1 —aIFN+1 Cf. Remark 110.5.

aN) e

let

FN —aNFN+t): X —* C".

196 FL':

2 Pseudoconvexity

a compact set K C K. Then: (a) If N 2n and F is regular on K, then A2N({a E

Fa

is not regular on K)) =0.

(b) If N > 2n + I and F is injeclive on K, then A2N({a E CN: Fa isnot infective on K)) = 0. Pmof (a) Observe that

CN: Fa is not regular on K}

(a

a(F1

= (a E CN: = (a E CN:

_ai FN+,)

jiaf, j = I

3XEK

= (a E CN: (a, I)

=0, j = I,..., N)

E

x

N + 1),

=: T,

where 1,

X

xC"

j=l

N+l.

k=I

Since N + I > 2n, we get A2(N+I)(4(K x Ca)) = 0. Observe that

(a ECN: (a,t)E 4(K xC")}=:T, Hence the Fubini theorem implies that A2N(T)

t EC1.

= 0.

(b) We have (a E

=

(a

Fa is not injective on K) CN: Fj(x')—aJFN+J(x')= Fj(x")—aJFN÷l(x"),

j=1 = (a E C

N

= (a E CN: (a. I) c

F1(x ) —

xK x C)}=: 7',

)=

j=I

N) N + 1)

2.8 The Remmert embedding theorem

197

where 1,

4):

4)(x', x", t) := t(F(x') — F(x")).

XxXxC

Since N + I >

2,:

+ I, we get A2(N+1)(4)(K x K x C)) =

(a,t)e 4)(K xC")}=tT. Hence the Fubim theorem implies that A2N(T) = 0.

0.

Observe that

tE D

Lemma 2.8.4. Let (X, p) be a Riemann region over C". countable at infinity, such that 0(X) separates points in X. Then: (a) If N 2n, then the set (F

F is no: regular) is of the first Baire

category in 0(X)".

(b)IfN > 2n+1, the,: the set (F€ of the first Baire category in

Fis not infective or not regular) is

Proof (a) Since X is countable at infinity, it suffices to prove that for any compact FisnotregularonK} is nowhere dense in K c X the set E := (F

0(X)N. Fix a K. Observe that E is closed in 0(X)"'. It remains to prove that the set 0(X)" \ E is dense in 0(X)N. Fix A E L X, and e. We are looking for

a mapping F (9(X)N such that F is regular on K and max!. IF — Alt e. By Lemma 2.8.2 there exist M and B E (9(X)M such that B is injective and regular on K. Consider the mapping (A, B): X —÷ CN+M. Then, by Lemma 2.8.3(a) (applied M times), there exists an (M x N)-dimensional matrix C with IC Ii such that the mapping F := A + BC: X —p Ct" is regular on K. s/maxL liBli e. Moreover, maxK hF — All = maxL llBCii

:= (F (b) The proof is as in (a). We only need to observe that the set and next proceed 0(X)N: F is not injective or not regular on K) is closed in [J as in (a) (we use M times Lemma 2.8.3(a,b)). Definition 2.8.5. Let (X, p) be a Riemann region over C" and let F: X —* CN be holomorphic. Any relatively compact union of connected components of the set is called an analytic polyhedron of order N. Lemma 2.8.6. Let (X, p) be a Riemann region over C" and let K C X be a holornorphically convex compact set. Then for any open neighborhood U of K there exists an

analytic polyhedron P with K C P

U.

Proof Cf. the proof of Lemma 1.10.12(a). DCII denotes the operator norm of C. i.e. UBCV < IIBIIIICD.

D

2 Pseudoconvexity

198

Lemma 2.8.7. Let (X, p) be a Riemann region over Ce!, let K C X be a compact set, and let P be an analytic polyhedron of order N + 1 such that K C P. If N ? 2", then there exists an analytic polyhedron P of order N such that K C P C P. Proof Assume that P is a relatively compact union of connected components of Q := such

thatK C

FN+,) E O(X)N4* FixO

F'


Consider the Riemann region X := X \ Ff41 (0) and the compact set K := {x E

P: IFN+II 2 c}. Since N 2n, the proof of Lemma 2.8.4(a) (applied to X) implies that there exists an (M x N)-dimensional matrix C with arbitrarily small entries such that the mapping F := (FI/FN+I

FN/FN+I) + BC

is regular on K, where B E O(X)M is injective and regular on K. Put

(G,

GN) := FN+IF = (F1

FN)+ FN+IBC E t9(X)N,

GN,FN+I), It is clear that if IICII is small then K C

and U

Qk := (x E X:

I

P. Put N}, k EN.

Let Pk denote the union of those connected components of Qk that intersect K. LetL := {x E 8U: IFN+I(x)I c}. RecallthatLisacompactsetandKflL =


LetO <

0.

Since L c K, rank F'(x) = n for any x E L. Consequently, there exists a constant x > Osuch that for anyx EL > < dk, 2bk < ct, and a k >> I (at least so big that: 2a1' < bk, bk + I/k2
Our aim is to prove that Pk C U for k >> I (then P := Pk will be the required

polyhedron of order N). Suppose that Pk U. Then Pk fl following two possibilities. Xk

L. Then

<

0. Fix an xk E Pk fl

bk + ck < d", j = I

There are the

N. Hence Xk

8U;

contradiction.

E L. Let yk: [0,1] —* be a curve such that yk(O) = xk, E K. fl K = 0, there exists a point Yk 1]) such that (xk, r0) fl Yk ([0, Since IPx (xk, E IPx p(xk)I = I/k2. Notice that Yk Pk and, therefore, E Ip(yk) — xk

N.

I

Moreover, Gk(xk)

IF (xk) — II = I

1/2 < I — (b/c)k < 1F1(xk)Ik

—

(xk ) I

k

< (b/c)k,

I + (b/c)k <2,

j=I

N.

2.8 The Remmert embedding theorem

199

N}besuchthat

.

> We have:

= IFN+l(yk)IkIF(yk)

—

FN÷I(yk) k( Vk

= k

1

+

kf

-

/l\ /1 1+O(—) (—k \k/ \2

ckfl

x

+

I

k

Fj(xk)

I(' +

F() i

FJ(xk)

fl

i) +

—

FN÷l(Xk) c

—

—

k

+

—

—

— ii)

I

+01—) —I— 'k21

'c

bk

- (-) xck

k >————i l=———>b 2\8k \cJ / 16k 2

k>>l

0

contradiction.

Proof of Theorem 2.8.1. Recall that (9(X) separates points in X (cf. Theorem 2.5.1). Hence by Lemma 2.8.4(b), there exists a holomorphic, injective, and regular mapping

G: X —+ Assume

that we have already constructed a holomorphic mapping F: X —+

such that {x E

X: IF(x)I

X,

kEN.

Then, by Lemma 2.8.3, there exists an (2n + 1) x (2n + 1)-dimensional matrix C with ICGI s IGI such that the mapping H := F + CG is injective and regular on X 60)• 60) Indeed, let (K1 be a sequence of compact sets with = X. Consider the injective and regular mapping If := (F, G) and N = k := (2n + I) + 2n. By Lemma 2.8.3 for each j there exists a zero measure set C Ce" such that the mapping

(H1+aIHk÷I

Hk+akHk÷1)

(*)

is injective and regular on for anya 1). Taking T := 1) we get azero measure set such that the mapping is injective and regular on X for any a T. Fix an a T with 101 e. Now we repeat the same procedure with respect to the mapping (*) and N := k — 1. We find b.c E Ck_l with e such that the mapping Ibi, Id

(H1+blHk+cIHk÷1 is regular and injective on X. Finally, finite induction finishes the proof.

200

2 Pseudoconvexity

Moreover,

(XE X: IH(x)I
{x E X: IF(x)I
kEN,

consequently. H is the required mapping.

We move to the construction of F. Recall that X is holomorphically convex and therefore there exists a sequence of holomorphically convex compact subsets of X such that K3 C mt and X = U'°=, K3. By Lemmas 2.8.6, 2.8.7, there exists a sequence of analytic polyhedra of order < 2n such that K3 C P3 C j 1. Assume that we have found a holomorphic mapping F: X —+ such that

IF(x)I > k + Mk+I,

XE Pk+1

\

k EN,

(t)

where Mk := SUPPk GI. k N. Then for any k E N we get {x

X: IF(x)I
X.

Thus we turn to the construction of a mapping F with (t). Assume that we have constructed a holomorphic mapping

F: X —+

such

that

IF(x)I>k+M&+i,

XEdPk, k€N.

(j)

Then put

Ak := {X Pk+I \ Bk := {x Pk: IF(x)I

F(x)I
Condition (1) shows that for any k the sets Ak, Bk are compact (and disjoint). Directly from the definition we conclude that fl = Ak U Bk. By the approximation theorem (Proposition 2.7.4) there exists a function 1k E 0(X) such that

flXIfkI

1/2".

XEAk,kEN. Define FN+1

2.9 The Docquier—Qrauert criteria

Observe that

mt Bk

= X.

201

Hence the above series is locally uniformly convergent

on X and defines a holomorphic function. It is easy to check that F4v+i(x)I > k + Mk+l for x Ak, which directly implies (t). Thus we move to the construction of a mapping F with (i). Suppose that Pk is a union of connected components of Qk H1 where Since Pk...1 CC Pk, we get maxp5_1 IHkI < I and JHk(x)I = I for Hk E x aPk. Consequently, there exist ak > I and m,, N such that if we put

then

maxlllkl <

xEapk. kEN. Define

F It is clear that the series is locally uniformly convergent in X and that F satisfies (i).

0

2.9

The Docquier-Grauert criteria

The aim of this section is to localize the description of the pseudoconvexity. 61) Theorem 2.9.1. Let (X. p) be a Riemann domain over C's. Then the following conditions are equivalent:

(i)

(X, p) is a Riemann—Stein domain;

(ii) for any continuous mappmgs f: [0. 1] x E —+ X (cf Section 1.5) with o f)(t,•) e (9(E)for any t [0, 1), the following implication is true: 1) x E) U ({1} x SE)) C X, then f([0, 1] x E) C X;

(iii) for any continuous mapping 1: [0, 1] x E

Xfor which the mapping

oI

extends to a holomorphic mapping g: D x E —÷ C", where D C C is neighborhood of [0, II, the following implication is true: 1) x E)c X,thenf({1} xE) C dXorf({l) xE) C X. 61)

The reader is advised to recall the main results from § 1.5.

a

2 Pseudoconvexity

202

(iv)for any domain T =

T

a holomorphic mapping f: a

Riemann—S rein domain

(Y, q) over C" and a morphism

and a biholomorphic (X, p) —* (Y, q) such that for any domain T = mapping f: T —* f(T) C X such that o f extends to a biholornorphic mapping E" —÷ 1(E") C Y, there exists a holomorphic mapping f: E" —÷ X such that f = I on T:

(vi) there exist a Riemann—Stejn domain (Y, q) over C" and a morphism

w: (X,p) —+ (Y,q) such that there exists a neighborhood U of d X with — log

E

?S3t'(U fl X).

Remark. There is a domain of holomorphy G C C3 and a continuous curve a [0, II —÷ C2suchthatUo<,<1((a(t)}xE) c Gand(a(1),0) E G,but(a(1)}xE G (cf. LFav 1979]).

Proof of Theorem 2.9.1. By Lemma 2.2.8 and Theorem 2.5.7, each of the conditions (ii), (iii), (iv), (v), (vi) implies (i).

(ii): Put K := f([0, 1] x dE). Then K is compact subset of X. Let

(i)

For each t E [0, 1) the function v(f(t, .)) therefore, by the maximum principle, supg v(f(t, .)) vE

Hence

(i)

f([0, I) x E) C :

is

subharmonic in E and

v(f(t, .))

V.

X.

(iii): Suppose that the theorem does not hold, i.e. that there are points X,,

j = l.2,withf(l,Aj) E X,butf(1,A2) E ax. Observe that dx(A) = 0, where A := {f(l — 1/k, A2): k E N). Because of Theorem 1.10.4 we find a holomorphic function h E c9(X) such that

lim Ih o f(l — 1/k, A2)I = +00.

=10,1] x Eand B' := ([0, 1)x E)U((1,.A1)}. Forevery point (r, A) B' we choose a neighborhood U(t, A) := U(f(t, A)) C X such that U(r, A) —÷ p(U(r, A)) is biholomorphic. Since o = there exists ForabbreviationweputB

a bidisc V(r. A) c D x E, centered at (r, A), such that

g(V(r,A)) C p(U(r,A)) 62)

Recall that

:=

and

f

f(V(r,A)flB) C U(r,A).

where

n—l.IznI
Zn) E

C":


n—I, p <

< flu.

203

2.9 The Docquier—Grauert criteria

Put

f(r.A) :=

V(r. A)

o

X.

f on

Obviously, the function

V(r, A) fl B. Fix two such parameters (t', A'), (r". A") E B' such that V0 := 0; in particular, this intersection is connected. Observe V(t". A") fl V(v', A') that f(t',A')IVo = 81v0 =

f(r'.A')(To, A0) = f(ro. where (To, A0)

:=

+ r", A' + A") E

lifting it follows that

=

Put V := U(r.A)EB'

A0)

= f(r".x")(to, A0),

fl B'. In virtue of the uniqueness of the

V0

on V0.

A). Then V is an open subset of C2 with B' c V. Define the function 7: V —÷ x, := f(r.A). It is clear that fIB' = f a'. Chooser with max{JA1I, A21} < r < I. Then there is an open subset W' of C V

such that [0, 1) C W' and W' x K(r) C V. E —* K(r) with t(0) = Ai and choose Take a biholomorphic mapping neighborhoods W1 = W1(l) C D and U1 := U1(0) C C such that W1 x £(Ui) C V. We introduce the following function

F: (W' x E)U(W1 x U1)—÷ C,

F(r,A) :=h(f(r,e(A))).

Put W := W' U Wi. In what follows we study the Taylor expansion of F(r, .), r A = 0:

E W,

at the point

F(r, A)

= 1 for Let R: W —+ (0, +oo] denote the Hartogs radius of this series, then R(r) any r E W'. Since —log R is a psh function on W, Oka's theorem (cf. [VIa 1966])

leads to the following additional information:

—logR(l) = lim(—logR(r))

0:

r/1

1. in particular, we havethat R(l) We choose r' E (0, 1) in such a way that

a neighborhood

W

and uniformly on

< r'

x K(r')

1.

Then there exists

x K (r'). Thus we obtain a holomorphic

function (W' U W)

<

that

W

C,

F(r, A) :=

204

2 Pseudoconvexity

The fact that

= tim

IF(l.

IF(l — 1/k,

r' (A2))I = urn

k-*+oo

leads then to a contradiction. The implications (i) (iv) and (I) The implication (i)

:

Ihof(1 — 1/k.

= +oo

(v) follow from Proposition 1.9.1.

(vi) is obvious ((Y, q)

(X. p). p := id. U := X).

0

Let

x 8E.

x E,

The main local criteria for the pseudoconvexity are contained in the following theorem.

Theorem 2.9.2 ([Doc-Gra 1960]). Let (X, p) be a Riemann domain over C". Then the following conditions are equivalent: (I)

(X. p) is a Riemann—Stein domain (cf Theorem 2.5.7);

(p1) there exists afunction u e .'PSJt°(X) such that (x E X: u(x)
Xfor any

Xfor any compact K C X;

(ps) X =

where

is a is St rongly pseudoconvex domain with real analytic

boundary, and

v

1;

(pa) there exist a Riemann—Stein domain (Y, q) over C" and a morphism

(X. p) —* (Y,q) =q'

such that any point a d x has a neighborhood U C X such that the region (X fl U. p) is holomorphically convex;

—* X, (ps) (Kontinuitätssatz) for any sequence of holomorphic mappings where D,, C C is a neighbo rhood of E, v 1, we have thefollowing implication:

'f (p6)

X, then

for any biholomorphic mapping f: W —÷ f(W) C X. where W C C" is a neighborhood of & if

X, then

X;

(p7) there exist a Riemann—Siein domain (Y, q) over C" and a morphism

p: (X.p) —+ (Y,q) such that there is no continuous mapping f: E" —÷ X with the following properties: The notations (pi) —

(p) are taken from IDoc-Gra 1960].

205

2.9 The Docquier—Grauert criteria

(ti) (t2)

(t4)

-

C X,

nax

0,

o f extends to a biholornorphic mapping in a neighborhood of

there exist a Riemann—Stein domain (Y. q) over C" and a morphism

q: (X,p) —÷ (Y.q) =9'

=9'

such that any point a E d X has an open neighborhood U C X such that there is no continuous mapping f: E" —÷ U with the above properties (ti —t4).

Proposition 2.9.3. if f is as in

then there exists a bi/wiomnorphic mapping

f: W —* f(W)c X, where Wisan open neighborhood of&suchthatf = I on

Proof. We know that the mapping

o f extends to a biholomorphic mapping In particular, f is injective on E" and is injective on the compact f(E"). Since is locally biholomorphic on f(A), for any 0
g: V —+ g(V) C Y, where V is an open neighborhood of

Proof of Theorem 2.9.2.

(1)

(3) (i) (4) (i)

(ps)

(i)

(ps)

(p4)

(7)

*

(12)

)

(1)

(pi) follows from Theorem 2.2.9.

(1) (i) (2) (i)

(4

(i)

.'

(p7) follows from Theorem 2.2.9.

(p3) follows from Proposition 2.2.6(b) and Corollary 2.2.11.

(pd) is trivial: we take (Y, q) = (X, p) and U := X. —÷

X,v > l,beasin(p5)withA :=

X. Put K := A. It suffices to show that

c

Take a

206

2

Pseudoconvexity

and fix a v. Then v o vE the maximum principle we get:

is subharmonic in

maxv<maxv.

aE

E

and, consequently, by

K

(ps): Let f: W —+ f(W) c X be as in (p6) with cc X. (6)(p2) Take K := Let D := X. It suffices to We know that D show that f(E") c D. Since f(E'1) is open (f is btholomorphic), we only need to :

show that Take an arbitrary function v E c the maximum principle, we have:

sup v = f(E")

sup v o

E'

f

sup v o

f = sup v o f =

sup

Then, by

v <sup v. K

Thus f(E'1) c =w

Let a E ax and let U C X be such that (X fl U,p) (7) (p4) is holomorphically convex. We already know that (i) (p6). Thus the region (X fl U, p) satisfies (ps). Suppose that f: E" —--+ U is a continuous mapping satisfying (ti—ti) of (p7). By Proposition 2.9.3 there exists a biholomorphic mapping f: W —p f(W) C X, where W is an open neighborhood of such that I = I on

By

(ti) we get

X. Now condition (p6) implies that

f(E") = f(s) f(s) X which contradicts (t3). (8) (ps) : (p6): Let f: W —* f(W) be a biholomorphic mapping of an open neighborhood of such that A := X. Take an arbitrary sequence = f(z,) t'> 1. Then

f(s),

=

Zv,n) E

v>

C A and, therefore, we can select a convergent subsequence. (9) (p6)

1,

and define cov(A) C

A),

X. Now

(p7): Use (7) with U := X =c',

(10) (p7)

is tnvial: we take U := X (i): It suffices to check condition (v) from Theorem 2.9.1. Let

(11) (p7) ço: (X. p) —+ (Y, q), T = Tr,p, f: T X, and be as in(v) from Theorem 2.9.1. < I such that f extends holomorphically to We may assume that there is no r p'E. Since extends ço o fIg. Fix p < p' < and let Q := 1

Proposition 1.5.8 implies that f I extends continuously to a mapping f: Q —+ X x p'E). to the mapping f substituted by Now we can apply Note that extends 4 c f to a neighborhood of Q. Consequently, K := f(Q) C X. Since c f = is injective on K. Hence there exists a on Q, the mapping neighborhood U of K such that c'Iu is also injective. Now, the mapping (coIuY' 01 gives a holomorphic extension off to W := Finally, f must extend to some

with r'> r; contradiction.

207

2.9 The Docquier—Grauert criteria

(l2)(p)

=c (i): Take an a E aX and U as in (p). Let yo := e(a, V). where V is a connected neighborhood of

and suppose (and e(a, V) - '(V) that belongs to a). Fix 0 < r
•,

that U fl X = denotes the connected component of such that Py(yo, r) V and let := e(a, lPy(yo, r)). =* Let and let T: —÷ X be as in Remark 1.5.10. satisfies (p7) with respect to the morphism At first we show that

Letf: -

beacontinuousmappingsuchthatf(Si\)

E't —+

=

=IIfr

f extends to a biholomorphic mapping in an open

neighborhood of E".

-

:= Tof. Then ettherg(E")fl a X 0. 3!Py(y0, r) Putg

øorg(E'1)

C X and

On the other hand, since lFy(y0.

r) is pseudoconvex,

Py(yo, r);

contradiction.

Thus by (ii) — neighborhood

E

such that dx =

C

Finally, by Proposition 1.5.9, there exists a on Wand therefore — tog dx =c"

In particular, there exists a neighborhood Z C X of a x such that — log dx

fl X). Now, it remains to use Theorem 2.9.1 (vi).

E

0

The Docquier—Grauert criteria remain even true in the following much more general

context (which, unfortunately, is beyond the scope of our book).

Theorem 2.9.4 ([Doc-Gra 1960]). Let M be an n-dimensional connected Stein manifold (cf Section 1.10) and let (X, p) be a Riemann domain over M (cf Section 1.1). Then the following conditions are equivalent: (1)

X is a Stein manifold;

(X) such that {x E X: u (x) < t I

(ps) there exists a function u E

X for any

tE

Xfor any compact K C X;

(p2)

(ps) X =

where

boundary, and

is a is strongly pseudoconvex domain with real analytic v

I;

(p4) any point a e ax (the boundary ax is taken with respect to the mapping p: X —* M analogously as in Section 1.5) has a neighborhood U C X such that the X fl U is Stein;

—* (ps) (Kontinuitatssatz) for any sequence of holomorphic mappings where C C is a neighborhood of E, v> 1, we have thefollowing implication:

if

X, then

2 Pseudoconvexity

208

(p6) for any biholomorphic mapping f: W neighborhood of X, then

f(W) C X, where W C

f(s)

is a

X;

there is no continuous mapping f: E" —÷ X with the following properties: X,

(t2)

f extends to a biholomorphic mapping in a neighborhood of E?z; (p) any point a E ax has an open neighborhood U C X such that there is no continuous mapping f: E'1 —÷ U with the above properties (t I—t4). As an application of Theorem 2.9.2, we get:

Theorem 2.9.5 (Kajiwara theorem, IKai 1965]). Let

:= liin((X1. p),

((X, p),

(cf Proposition 1.6.8). Suppose that all the domains (X,, pi), i E I, are pseudoconvex. Then (X, p) is pseudoconvex.

Pmof To prove that X is pseudoconvex we use the Docquier-Grauert criterion (cf. Theorem 2.9-2(p6)). Assume that f: W —p f(W) C X is a biholomorphic mapping,

X. Wehavetoprovethat

where W

f(s)

X. Let V be a relatively compact neighborhood of

where

U

x E. By Proposition 1.6.8 there exist Jo and a relatively compact V0 —÷ V is biholomorphic. Observe that for open set V0 c X10 such that any 1/2 < t < 1 there exists an e = e(t) > 0 such that f(G,) C V. where G, G, —* V0.

Let*,

holomorphic mapping '1',: G, —* X30, where G, := (I E)" — x < I + e(t) } (O, o is the envelope of holomorphy of G,). By the identity principle we get =f on G,' fl G,'. Thus, there on G,. In particular. '1', is injective on C, and = exists a biholomorphic mapping 4': Uo —f C Xj0, where Uo C U is an neighborhood of such that Since o '1' = f on Uo. Moreover, X10 is pseudoconvex, we get 'P(s) = X. 0 X10. Finally, {

I

2.10 The division theorem F1 and let 0(X) be given. FN Roughly speaking, the division problem is to characterize those cases where there exist functionsuj (9(X) such that UN

Let (X, p) be a Riemann domain over

U°

=

... + UNFN.

209

2.10 The division theorem

In the case where the functions u0, F1,..., FN satisfy some additional regularity conditions (e.g. they are bounded), one may ask whether the functions ui. be found in the same class of regularity; Corollary 2.10.5, Lemma 4.4.6.

.

.

.

, UN

can

It is clear that to solve the division problem one must at least assume that {x E F1(x) .... = FN(x) = 0) C E X: u0(x) = 0}. Below we will present a general division method based on the so-called Koszul complex introduced in [Hör 1967] and used, for instance, in [Kel-Tay 1971], [Cno X:

1972], [Sko 1972]).

We will need the following extension theorem for forms.

ProposItion 2.10.1. Let (X, p) be a Riemann domain over C" and let S be an analytic subset of X with dim S n — 1. Suppose thatu E bc), v E bc)
•,fau=OonX\S,thenau=oonX,

• if!

(9(X

\

5) fl

bc), then f exterwLc holomorphically to X.

Pmof First consider the special case where (K, p) = id), S = Ek x {0}fl_k (0 k n — 1), u E v E We have to prove that for any çø E

f Fix such a ço. Let Obviously

çø.

= I on

[0. 1]) be such that

E

= (z', z") E Ck x

z

17E(z)

VA

e > 0, and let

Define

:= (I —

\ 5). Hence

E

(_1)r+s+1

A

JE"

j

VA

Observe that

/

/

JE"

/

JE"

VAp

e—.O JE"

by the Lebesgue theorem. On the other hand Ud(p€

JE"

=

j(l

By the Lebesgue theorem

—

11ç)U

A

U —

U

—÷ 0 as e —* 0. Put

j(I)

—

when e —+ 0. It remains to show that Note that

A

:= Ek x

1(2)1 < CIIUIJL2(A

A drk A

JE"

) .

210

2 Pseudoconvexity

a constant independent of e). Since IIu IIL2(AE) —+ 0 as e —÷ 0, we only need to show that is bounded when e —÷ 0. Indeed, we have: (C denotes

<

<

Now we move to the main proof. Let d := dim S.

If d = 0, then we apply the special case locally (after a biholomorphic change of coordinates) with k := 0 and we get the assertion.

Suppose that I sd Sfl — I. Then inaneighborhood of any regularpomt ofS we apply the special case with k := d. This shows that there exists an analytic set S' C X with dim S' cd — I such that du = v on X \ S'. Finite induction finishes the proof. Let (X. p) be a Riemann—Stein domain over C"

and let

F=(F1 be

a holomorphic mapping. F

=

0. For r, s E Z+, define

bc): au

(r.s)(X) := {u

loc)}.

E

Fort ENputM(N,t):=(Nfl[i,N]Yanddefine := {u = (UI)IEM(N,)

=

= sgn(J)uI},

V!EM(N,1): Ui E A?rs), where

E, denotes the group of all permutations of t elements 65)• Observe that

=

(0)

ifr > n ors —p

> n on

> N. Define := du1,

I E M(N, t).

Let S

:= (x E X: Fj(x) = ...= F,v(x) = 0}.

Notethat S isananalytic subsetofX withdim S sn—i. lathe sequelthefollowing

direct consequence of Proposition 2.10.1 will be frequently used. Notice that the first part of our construction of the Koszul complex may be performed also for an arbitrajy Ricmann domain. 65) The definition of says that the system (u()JEM(pj,) IS SkeW-symmetriC W5.t. 1.

Obviously.

is

the Cartesian product of N copies of

2.10 The division theorem

211

(E) Letu =

(U;); EM(N,:) andy = (VI)IEM(N beskew-svmmetricsvs:emsofforms from bc) and bc), respectively. Suppose that au, = vj in X \ Sfor any 1 E M(N, t). Then U E and au = V.

Let P:

be

given by the formulae

ift=0:

Pu:=EFjui,

if t

(Pu), :=

1:

Fju(If),

I E M(N,

:=

Note that P is well defined and linear. Additionally, we put for all u E For arbitrary E N we have:

(2.10.1)

1).

{0}

aoP=Poa.

and Pu :=

0

(2.10.2)

a

(is)

(r.s-l-I)

c+ I)

Moreover, P o P = 0 (as an operator t = is trivial. Fort > 2, u E and 1

—÷

t E N). Indeed, the case

K E M(N. t — 2) we get

N

N

(P2u)K =

= 0,

=

j=I

j,k=I

where the last equality follows from the fact that the system (u,); is skew-symmetric

and therefore U(K,j,k) = Observe also that the operator P: —+ is injective. Indeed, we may assume that FN 0. Let U E be such that Pu = 0. Then

= 0 = (Pu)(j Hence u(1

N)

N

Fju(I

N—i)

N—I.j) = FNU(I

= 0 and, consequently. u = 0.

Foru = (u,),

E

let

(

juj IEM(N.I)

112)1/2

2 Pseudoconvexity

212

(where jIu,II has been defined in § 2.4). Note that flull E L2(X, bc). Directly from (2.10.1) we get (2.10.3) UE 11P1411 IIF1I For u

define

E

if: =

j=1

(Qu)1 := FjIIFII2u.

0:

N,

i+1

if:

J E M(N, t + 1);

(Qu)j :=

> 1:

notice that the (r. s)-form (Qu)j is formally defined only on X \ S. Moreover, the system ((Qu)J)JEM(N,:+1) is skew-symmetric with respect to J (in particular, Qu E L\(rs)(X \ S) and Q = 0 fort> N). Indeed, let t > I and consider the transposition a of the form

x—l,x+l,x,x+2

t+1)—*(l

(I

t+1).

Then we obtain 1+I

=

)t+Iv1

I

II

+

F

,

F II

—2

—(Qu)j. Observe that the following estimates hold on X \ S: II

Qufl < C lull

Fjl

Ild(Qu)lI

l,

(2.10.4)

—

IlaFII . 1F112 + lauD . 11F1I1

.

uE

(2.10.5)

Here and in the sequel C denotes a positive constant depending only on n, N. r, s, t. Let

:= {u E A(rS) : It is clear that property (E). if u operator

huh

.

IlFIr2 E L2(X, bc),

.

IIb'II' E L2(X, loc)}.

Moreover, by (2.10.4), (2.10.5), and The then Qu may be considered as an element of

is a subspace of E

uD

(r,s)

(r.s)

213

2.10 The division theorem

is linear. Observe that, by (2.10.2) and (2.10.3),

'

(r,s)'\ C

(rs)

The following relation between P and Q will play the fundamental role in the

sequel. P 0 Q+ QcP where Q := 0 on

= id on

t

=

(and therefore P o Q = id fort Id if: = N). 0, then

=

(2.10.6)

N,

0

0). Notice that

PoQ =0

if t = N (and therefore Q c P = Indeed,

if: =

=

(Po Q)u =

If t =

1,

= u.

then

((P0 Q +

=

Q

=

Fk(Qu)(jk) +

+ FkIIFH

+ N.

If:

> 2, then

((P 0 Q +

Q

P)u)j =

Fk(Qu)(,.k) +

IIFIl2u!.k\{131 + FkJIFII2UI)

Fk(

=

IIFr2

+

=

1

M(N, fl.

Now we a going to formulate a genera! division pmb!em. Fix 0 < r, s 0

r

N—

1.

Put

:=min(n—s,N—t—1). Letp

Define

E

XF

log IIFII + max{log

(IIaFIIl + 11p112)), log nFl!), (u —i)xF +jlogllaFIl.

n and

214

2 Pseudoconvexity

j =0.

Note that t/Ij E

is such that

Assume that U0 E

(b) Pu0 = 0, (c) II := Ilu°lI

< +00.

Consider the following division problem: Find u E

(i) Pu = (ii) au = (iii)

II

such that:

uçj,

0, lu lle

In the case r

L2(X) < Cl. s = t = (I the above division II

=

problem reduces to a division

problem for holomorphic functions. Observe that by (c) we get

E L2(X. bc).

(d) lIu°ll

ifj.t>

By (a) and (d) we see that u0 E

I. Put

:= Q(u0) E Here and in the sequel the reader should observe that the case it = 0 needs a special interpretation of the operators P. Q, and a. For instance, if = 0. then h° is defined as a skew-symmetric system of forms from (X, bc) and dh° should be considered in the sense of distributions. Observe that by (2.10.6) and (b)

Ph° = u0.

(2.10.7)

Moreover, by (2.10.4), using the inequality XF H lihOlle

II

L2(X)

C

II

2 log II Fli, we get

u0 (II Fll

H

L2(X)

The main idea of the construction is to find an element g1

u :=

—

such that

Pg'

solves our division problem. Notice that we have Pu = u0 independently of g1. The problem is to get (ii) and (iii).

=0. Theneithers =norr = N—I. We = 0, i.e. h0 itself solves our problem. Ifs = n, then obviously dh° E = (0k.

will show that in these cases

If r = N — 1. then AN

U ence

i

= 8Ph° =

= 0. Recall that P is injective on

—

Thus we may assume that

s
2.10 The division theorem

215

On X \ S we define formally the following sequence

j = 1,2

h' Using (2.10.4), (2.10.5), and (a) we get < IIah'II

CIIu°JJ

I

j = 0. 1,2

U8FIV4'

CIIu°II

In particular, by (d),

j=

(2.10.8)

.

for j =

E

0

— I

and

for

E

Moreover,onX\Swehave

E

1.

(2.10.9)

= ah'—',

j=

N— t

1

(2.10.10)

— 1.

Indeed, by (2.10.6), (2.10.2), (2.10.7), and (a), we get inductively:

Ph' =

PQdh°

=

=

dh°

—

QPah° = ah°

—

= aW—' — QPah'—' Qa

aW—2 = dh'',

QdPh° = ah°

= ah'—' j=2

—

Qau° = ah°,

—

N—i—i.

Now we will show that

=0. Consider the following two cases: = n — s: Then ahL E

=

(2.10.11)

{0}.

= N —: — 1: Then and therefore it suffices to show that E Pahu = 0. Indeed, by (2.10.10) we get Pah,L = = = 0. By (2.10.8) we get II

II

L2(X)

FII_(2i

< C IIu°Il

0L2(X) < CL

By virtue of Theorem 2.4.22, using (2.10.11), we conclude that there exists a g'2 E such that agIL = hIL and

° Suppose

that for some j E (1

such

CR. —

1}

we have already constructed gIL,

that g

k

E

At+1+k

iir.s+k—1'

= hk —

op)2 11L2(X) < Cu,

(2.10.12)

k=

j +I

/1,

(2.10.13)

216

2 Pseudoconvexity

where := 0. We will construct g1. Observe that by (2.10.10) and (2.10.12) we have: —

=

=

—

pgJ+2) =

—

such that

Hence, by Theorem 2.4.22, there exists a and H JIg3

II

ag' =

0. h

(2.l0.l4) —

Jib' — PgJ+I II exp(—fp —

o p)211

—

and

E

—

It remains to estimate the last term. By (2.10.8) and (2.10.13), using the inequalities

obtain

h' — pg f+'

L2(X)

Cl llu°jI . ilaFil'

.

Ii F

1—(2j+

+ Cl II

U FII H

Cu + cli

lie

—2(p—-j)

. I!dFIi

L2(X)

L2(X)

P)2 HL2(x)

CL (2.10.15)

Consequently, after finite induction, we construct g1. Put u := h° — Pg1 E

Then, by (2.10.7), (2.10.14), and (2.10.15)

Pu = Ph° =

u0,

= 0,

HIIuIICHL2(X)
(2.10.16)

Thus we have proved the following

Theorem 2.10.2 (Division theorem). Let (X, p) be a Riemann—S:ein domain over FN): X —+ Cw be a holomorphic mapping, F 0. Let Let F = (F1 r.s nand0 0(depending only 0 on n, N, r, s, 1) such that

for any

E ,9'SJ?(X) and for any element u0

Pu0 = 0. du0 = 0. and 11

ltllu°it

< +00.

.

such that Pu = u0, au = 0, and

there erists a u E

< CL where ii := min{n

—

s, N

—

In the case z = 0 we get:

i — l}.

With

2.10 The division theorem

217

Corollary 2.103. Let (X, p) and F = (F1,..., FN) be as in Theorem 2.10.2. Let < r, s ii. Then there exists a constant C > 0 (depending only on n, N, r, s) such thatforanyço E and for any a-closed form u0 E with

0

I := there exists a

u = (uj,

.

< +00,

. IIFII .. ,

such that au1 = ... = dUN = 0

E

and

5CR.

where ji := min{n

s, N — I).

—

In particular if r =

s

=

then we get the following division theorem for holo-

0.

morphic functions. .p E

There exists a constant C > 0 (depending only on n and N) such that for any and for any u0 E 0(X) with II

:= IIIu°I

there exists a u = (Ui

UN)

< +00,

.

E

0(X)" such that

u0 =FIUI+•••+FNUN. Cli,

II

whereji :=min{n,N —1). The following lemma will be useful in the proof of Corollary 2.10.5.

Lemma 2.10.4. Let 8: X —+ R>0 be such that 18(x') — 8(x)I S IIp(x')

Then (a) where

lix

X

—

d(n, k, v)ll8kfllx,

d(n, k, v) :=

In particuku the operator d": anyk Oandv E L'(X), then

k >

E X, x'

0, v e

f

E

O(k)(X 8),

(k +IvI)k+IvI k

8)

8)

is continuous for

k >0, q >0, f E

218

2 Pseudoconvexity

Proof (a) Fix k. v, f, and a E X. By the Cauchy inequalities, and by (c2) we get <

8

I

<

I 15k111

e)k }

I


0

(b)is obvious.

Corollary 2.10.5 (Division theorem for functions with tempered growth). Let (X, p) be a Riemann—Siein domain over C'T and let

8: X —÷ (0.11 the following conditions: := min{px. So o p}. • 1S(x') — 8(x)I lp(x') — p(x)II, x E X. x' E — logS (cf Proposition 2.5.4).

.8

•5ae Let x > 0, N E

N,

forsonzeb>0,

and F = 0, u0

5)N 67) F

FN) E 0(X) we have (F1

0. Assume that

onX. Then there existuj v

:=

S)withu0 = u1Fj +

UN E

+ fi + js(2x + 3) + n, 1tL := min{n.

Proof Put

:=

k

I

:=

+

N .—

+UNFN on X, where

II.

Then 21i+l)5k11

<

Hence, by the division theorem (with ço := —k log 8), there exist functions UI

Notice that if (X. p) = (G. Id). where G is a domain in C". then > 0. E L2(C"), Recall that 8) < +oo}. (1 E 9(X):

and

.5

67)

UN€0(X) E

L2(X) (because

219

2.11 Spectrum

suchthatu°=uiFi

and —ILXF) ii L2(X)

CL

where C > 0 depends only on n and N. By Lemma 2.10.4 XF = log UFII + max{log (IIaFIRI + 11p112)), log IIFII}

where A depends only on x and H

IF liii

A — (2x

+ 3) logS,

Consequently.

HIIUIISIIL2(x =I

Finally, by Lemma 2.5.3, U3 E

N.

D

Corollary 2.10.6 (Division theorem). Let (X, p) be a Riemann—Stein domain over UN E (9(X) = 0. Then there exist and let F E (9(X)N be such that such that 1 = F1 + ... + UN FN. Proof By Corollary 2.10.3 we only need to find a function ço E ?S,1e(X) such that

< +00. Since

E L2(X, bc), wecanproceedas intheproofofLemma2.4.18. U

Exercise 2.10.7. Let (X, p) and F be as in Theorem 2.10.2. Assume additionally thatd:=sup(IIp(x)II: XE X} <+00. Prove that in this case the division theorems 2.10.2 and 2.10.3 remain true with XF

:= log IIFII + max{log 118F11, log IIFII}

and with a constant C depending only on n, N, r, 5, t, and d.

As a corollary prove that the division theorem 2.10.5 remains true with v :=

l)+n.

2.11 Spectrum and let A be a closed a-stable subalgebra Notice that I A. of (9(X) with p E (e.g. A = clO(X) Let 6 = (5(A) denote the spectrum of the algebra A, that is the set of all continuous A —÷ C (in characters on A, i.e. non-zero continuous algebra homomorphisms = 1). Let (X. p) be a Riemann domain over

220

2 Pseudoconvexity

Remark. In fact, any homomorphism .4 —÷ C is continuous; cf. Theorem 2.11.5. This is a particular case of the so-called Michael-conjecture.

Forf E Adefinef:6 For F = (f, 6—. C't.

IN) E

{f:f

Let,4 := E A). IN). In particular, we define

—÷

.4N

put F :=

.4 —. C be the evaluation at a, i.e. := f(a). = Observethatfoa = f, f Ed.. 6. Puta: X —+ 6,a(x)

For a E X let E

In particular, j%oa = p.

Proposition 2.11.1 (ERos 1963]). There exists a topology on 6 such that: • (6, is a Riemann region over C", • a: (X, p) (6, j5) is a morphism,

• A C 0(6), • a: (X, p) —* (X, j3) is the maximal .4-extension, where X denotes the connected component of 6 thai contains a(X). Proof First, we endow (5 with a topology. Take a

set K =

C X such that

IIfIIK,

inequalities, we get

f EA. V :=

Let

f

E (5. Then there exists a compact Hence, by the Cauchy

0
(2.11.1)

Put

f€A.

>

Note that by (2.11.1) the series is convergent. Observe that: • f, -) is holomorphic in P(zo, dx(K));

•çp(4. l,z)= 1; • •

1, z)I f + g. z)

r < dx(K); =

I, z) +

Thus for each z E iP(zo, r), 0 < r belongs to 6 and

f,

g, z); the mapping .4

f

.4. Hence

we may assume that c


E .4, k

= I; ef. Remark 1.4.5(k).

N. Consequently,

1, z)

A.

ByconunuitythcrcexistacompactK C Xandaconstantc >

f

f —÷


cUfUp, E A. Thus

221

2.11 Spectrum

r):=

z): z E 1?(zO, r)). We endow €5 with the topology O 0. — r, then we get Indeed, if z E ?(w, p) C IP(zo. r) andO < p

Let U =

•,

in which the system

(z—w) I)

=

—

=

Zo)L)(z — w)t

—

=

zoYL(z

—

(z—zoY (2.11.2)

Recall that z)) = z. Hence the mapping j)Iu is a homeomorphism onto P(zo, r) and , z), z E IP(zo, r). = The topology of €5 is Hausdorff. r) fl r) 0 Indeed, let If then 6, for small r > 0. Assume that = zO and let r > 0 be such that = r) = 0. Suppose r) fl 2r) is defined, j = 1,2. We will show that •, w) = w) for some w E IP(zo, r). Then, by (2.11.2), for f A that we have: w), 1, zo)

1, zo) =

(1)

=

u'). f, zo)

=

f, zo) =

contradiction.

We have proved that (6,

is a Riemann region over C's. Moreover, for f

we get:

z€P(zo.r). which shows that A C

0(6).

Now, we will show that a is continuous. Indeed, if x E IPx(a), then

f, p(x)) = Hencea(x)

—

.,

p(x)).

f(x) =

= a(x)(f).

A

222

2 Pseudoconvexity

Thusa:(X.p)—+(15,j5)isamorphism. is

an A-extension. Let (X. p) —* (Y, q) be any other holomorphic extension. Define a: Y —÷

a(y) :=

Then, by Remark 1.4.5(i), a is well defined. It is clear

o

anda = a. Moreover,a is continuous. Indeed, forb E y E Py(b, r) (r > 0 small) we have:

that ,5oa =

q

f q(y)) =

Y

and

(q(y) —

—q(b))"

=

=

I

= a(y)(f). Hence a(Y) C X and so a is a morphism of (Y, q) into (X. j3).

0

Let

Hom(A, C) :=

A —+ C:

is an algebra homomorphism),

C) := Hom(.4. C) \

10).

Proposition 2.11.2(Igusa theorem, cf. [Igu 1952]). Let (X. p) be a Riemann—Stein domain over C's. Then the mapping

X3 x

C)

E

any algebra homomorphism C).

is bijective. In

0(X) —+ C is continuous

and, consequently, (5(0(X)) =

Proof Since 0(X) separates points, r is injective. Fix a

E

C) and

let

:=

{f E 0(X):

We want to show that there exists an First observe that

E

=0).

X such that f(xo) = 0 for any f E

0. Otherwise, by Corollary 2.10.6, there exist UI

Hence 1

contradiction.

UN E 0(X) with

('ii)

223

2.11 Spectrum

Let zo :=

By virtue of (*) with F := p — zo we conclude that

)

p(X). Let Jo (9(X) be such that foIP-'(ZO) is injective (cf. Remark 1.8.12). Take an f E 0(X). Then by (*) with F := (f, fo. p — Zo) we see that there exists a such that fo(xi) = f(x1) = 0. Since lois injective on pomtxf E the point xj is independent of f. zo

Proposition 2.11.3. Let (X, p) be a Riemann—Stein domain over C",

E X, and let

—+ f(xo)for

E X. t E T, be a Moore—Smith sequence. Then x, —+

anyf E0(X). Proof Suppose that f(x,) —p f(xo) for any f

E

0(X). Let F: X

be a Remmert embedding (cf. Theorem 2.8.1). Put M := F(X), zo := F(xo), g(zo)foranyg E (9(M). Inparticular,takingas z, := F(x,),t E T. Theng(z,) we easily conclude that z, —p zo. Finally, since g the coordinate functions in D F: X —* M is homeomorphic, we conclude that x, —÷ xo.

Remark 2.11.4. The above proof may be essentially simplified for ordinary sequences: Let

f

such that E (9(X). Let P E N}. Then A is relatively compact. Otherwise, by Theorem 1.10.4, A := v E N} = +00. On the other hand there exists an fo E 0(X) with fo(xv) —+ fo(xo): contradiction. be an accumulation point of A. Suppose that Let xO. = Then there exists an fo E 0(X) with = # lo(xo). On the other hand, = fo(xv) = fo(xo); contradiction. E

X, v

be

Theorem 2.11.5. Let (X, p) be a Riemann domain over C" and let .4 be a closed d-stable subalgebra of 0(X) with p E .4". Then: (a) 6(A) = C). with the weak topology fmm A' 69) then (b) If we endow (6(A), fr) is a Riemann domain over C",

a: (X,p) —p (6(A).j)isanwrphism.

A = 0(6(A)), a: (X, p)

(6(A), j3) is the maximal A-extension.

Proof. (a) Let X be as in Proposition 2.11.1. First, observe that A is a closed i3is E .4 (Remark 1.4.5). Moreover, (X, stable subalgebra of 0(6(A)) with separates points in X domain of holomorphy (Remark 1.8.2(a)) and an = 0(X). (Proposition 1.8.10). Hence, by Theorem 1.10.4,

2 Pseudoconvexity

224

Since ax: (9(X) —*

is an algebra isomorphism, the mapping —÷

is well defined. Moreover, since ax is a topological isomorphism, A maps bijectively tS(.4) onto G(9(Xfl. Thus, assertion (a) follows from Proposition 2.11.2. (b) We already know that = (9(X). Moreover, by Proposition 2.11.2 we know that the mapping X is bijective. Hence, for E y —÷ r(y) = = any E 6(A) there exists a yo E X with = This simply means that = yo. i.e. X = 6(A).

Thus, it remains to use Proposition 2.11.3 to prove that the topology of 6M) (introduced in Proposition 2.11.1) coincides with the weak topology from A'.

Theorem 2.11.6. Let (X, p), (Y, q) be Riemann Stein domains over

and

0

Ctm.

respectively, and let

(9(Y) —+0(X) be an algebra homomorphism with mapping

(1)

F: such that

=

X

1.

Then there thsts a holomorphic

Y

=

Proof Take an x

X and let

0(Y)

g —L

E C.

= 1. By Proposition 2.11.2 there = g(y) for any g E (9(Y). Thus we

Then is an algebra homomorphism with exists (exactly one) point y Y such that

have defined a mapping F: X —+ Y such that E = F F = m, we only need to verify j =I that F is continuous. Let X —+ xo E X and g E 0(Y). Then, from the detinition of F. we get = = g(F(xo)). Hence, by Proposition 2.11.3,

0

F(xo).

2.12 Liftings of holomorphic mappings H Theorem 2.11.6 allows us to lift arbitrary holomorphic mappings to envelopes of holomorphy. The case of proper holomorphic mappings will be studied in Theorem 2.12.5.

Theorem 2.12.1. Let (X, p). (V. q) be Riemann domains over lively, and let

a: (X,p) be the maximal holomorphic

(X.13),

fi: (Y.q) —+

Then:

and Ctm. respec-

2.12 Liftings of holomorphic mappings II

225

(a) For any holomorphic mapping t: X Y there exists a (exactly one) holomorphic mapping 1: X Y such that i o a = o t. if(Y, q) is a Rienzann—Stein domain over Cm, then any holomorphic In

mapping r: X —* Y extends to a holomorphic mapping i: X —k Y such that 1

70)

oa=

(b) For any topological space B and for any continuous t: X x B —+ Y such that

for each b E B the mapping Tb := r(., b) : X —* V is holomorphic, the mapping

1: X x B

b) :=

V given by the formula

is continuous.

Proof (a) By the identity principle i is uniquely determined. Let

:=(a*)_l

(9(Y) —÷ (9(X),

is an algebra homomorphism and = 1. By Theorem 2.11.6 there exists a holomorphic mapping 1: X —* Y such that = f Consequently, = a* 01*, which implies that fi or = I oa (because (9(Y) separates points). (b)LetX x B (x,,l,,) —* (xo,bo) E Xx BbeaMoore—Smith sequence. By Proposition 2.11.3 we only need to show that f(I(x,. b,)) —+ f(I(xo. b0)) for any function f (9(Y). Fix an f. Recall that Obviously,

f(I(x,b)) = (a*)_I(fo.florh)(x), r is continuous, we get f

Since

0

o Tb,

—+ f 0

(x,b) E Xx B. 0

locally uniformly in X.

Hence, by Remark 1.4.5(i) weget uniformly in k, which completes the proof.

(f

locally

Now we turn to discuss proper mappings. Let (X, p), (Y, q) be Riemann domains over C" and let F: X —+ phic. Define ,

Y

be holomor-

xEX.

Theorem 2.12.2. Let (X. p), (Y, q) be Rienzann domains over C" and let

F:

X

Y

be a proper holomorphic mapping7U. Put S := (iF) — (0). Then: • S X,

•

the set

:= F(S) is analytic in V and E

70)

In other words, ifs: (X, p) —p (X. is the maximal holomorphic extension, then for any Riemann—Stein domain Y over Cm (with arbitrary n,), the morphism a gives also a holornorphic extension w.r.t. all holomorphic mappings X —+ V. 71)

Recall that F is proper if F1(K) is compact for every compact K

V.

226

2 Pseudoconvexity

• the mapping

Y\E

FIx\p—I(Z):

an analytic covering. i.e. there exists a number k E N (called the multiplicity of F) such that for any b E Y \ E there exists an open neighborhood V C Y \ E such that F- 1(V) = U1 U• - - U where U1 Uk are pairwise disjoint open sets, and —+ V is biholomorphic, j = I k, F'u,: is


will

follow [Rud 19801. We need a

Hausdorif measures. The fined by the

p-th

few

auxiliary

results.

Hausdorff measure on RN is the outer measure de-

formula

:=

sup

= lim

3€,

where

A=(JA,.

diamA1

ACRN

with respect to the Eucidean norm in RN; diam 0 := 0). It is that = #A. Notice that our definition differs from the standard one by a constant factor (ci. [Fed 1969]). (the

diameter is taken

clear

Property

(HI). If A C R" and F: A —÷ RN' satisfies the Lipschitz conditions with < p > 0.

constant L, then

Property(H2). if A cW"

and

< +00, then

> p.

Proof

3t(A)

Hence .1?(A) Property

—p 0 when 8 —+

0

0.

(H3). if A C RN then x [0, 1])

Proof LetA = number such that k1 diam

p > 0.

<8< 1,)

EN. Letk1betheminimalnatural A < 2). We find intervals

227

2.12 Liftings of holomorphic mappings II

1.

= 1,.

Observe that diam(A1 x

x [0, 1])

Then

x j=1 v=1

x [0, 1])

Hence

0

2P+23ft(A), which implies the required estimate.

Property (114). Let Q C RN' be an N-dimensional cube 72)• Then

forp
—

N,

forp>

N.

Proof By (H2) it suffices to prove only that J€N(Q) E (0, +oo). By (HI) and (H3) we may assume that N' = N and Q = [0, 1]". Fork €N divide Q intokN cubes with edge 1/k. Then

= Hence

RN(Q)

j

A with diam A3 N. Let B1 be a closed Euclidean ball Let Q = with radius diam such that A C B3 (B3 0 if A3 = 0), j E N. Let B denote the unit Euclidean ball in R" and let a = aN AN(B). Then

!AN(U A3)>

=

AN(Q)

0 Property (HS). Let (X, p) be a Riemann domain over fl

subset of X such that connected.

and let A be a closed

= Ofor any a E X. Then the set X \

A is

Proof It suffices to prove the following claim. be a convex domain and let A be a relatively closed subset of D (*) Let D C such that RN_I (A) = 0. Then D \ A is connected. 72)

That is. Q = L([O.

11N

x (o}N'_N) where L: RN'

is an affinc isomorphism.

228

2 Pseudoconvexity

Indeed, suppose that (*) is true, and let 00, b X \ A. Since X is connected, there exist £ N and points aj at X such that such that ao lP(ai), P(a1) fl

j

0, = I £ — I, andb E P(at). Observe that intA = 0 (by(H4)). in particular, there exist points £ — I. By (*) E (115(a3) 1) \ A, j = 1 the set \ A is connected for j = I £. Thus we can join ao with b1 by a curve in IP(a1) \ A. next b, with in IP(a2) \ A, and finally with bin \ A. We move to the proof of (*). Suppose that D \ A is disconnected and let a and b lie in two different connected components of D \ A. Let Q C D \ A be an Observe that (N — 1)-dimensional compact cube containing a such that [a, b] I Q A fl[b, x I 0 for any x Q (because a and bare in different connected components). Let ir denote the conic projection with center at b onto the (N — 1)-dimensional affine A fir plane containing Q. Observe that ir is Lipschitz on Ao let L be the Lipschitz constant ofJrIA0. Then, by (HI) and(H4), we get

0<

= RN_l (,r(Ao))

=0; D

contradiction.

Property (H6). Let D be a domain in C",

R2''(Z)= 0.

f

0(D), 1

0, Z

:=

Then

Proof By (H2) it suffices to prove that for any a D there exists a neighborhood K1, where is compact and U c D such that Zn U = <+oo,

j

N.

We apply induction on n. The case n = 1 is trivial. Assume the result is true for

n — 1.

Then, by the Weierstrass Preparation Theorem, we may assume that D

=

E" and

that f is an irreducible Weierstrass polynomial f(z', Zn)

=4

Let

z1

:= Z

:={z€Z:

\ Z1 =

{z E

1(z) =

1(z) = 0).

We will prove the result separately for Z1 and Zo. If a = (a',a,,) Z1, then in a neighborhood of a the set Z1 is the graph of a holomorphic function Zn = Consequently, by (Hi) and(H4), Zi = KJ'),

where K)" is compact and

< +oo, j

N.

That is. Q is an (N — 1)-dimensional cube containing a and contained in the (N — 1)dimensional auline hyperplane perpendicular to the segment [a. fri.

229

2.12 Liftings of holomorphic mappings II

On the other hand, 4 C Z' x E, where Z' := R1 (0) and

—f C

R: is

the resultant of the polynomials f K1. where K is compact and

we can write Zo =

and

a—.

By the inductive assumption, Z' =

<

+00, j

E N.

Now, by (H3),

< +00,

is compact and

KJ°>, where

JEN. C] Proper mappings. Let (X, p), (Y, q) be Riemann domains over C" and let F: X —÷

I

be a proper holomorphic mapping. First observe that F is closed.

Property (P1).

is finite for any b E Y.

Proof The set M := is a compact analytic subset of X. Let M0 be a connected component of M. By the maximum principle the mapping p is constant on Mo. Hence M0 must consist of one point. Consequently. M Consists of isolated points. Since M is compact, M is finite. U

Property (P2). LetS := (JF)1(0). Then S

X.

Proof. Suppose that S = X. Let 1

Xo:=(xEX:rankl L

3Zk

1 (x)I=r}:

J

Xo is open. Then, by the rank theorem, the set (FIx0Y' (b) is infinite for b which contradicts (P1).

E

F(X0).

0

Property (P3). Let E := F(S). Then E is nowhere dense, Y \ E is connected, and F is surjective.

Proof Since F is closed, is closed. By (P2) and (H6) 3&''(p(S fl IP(a))) = 0 for any a X. Hence fl IP'(b))) = 0 for any b E Y. Consequently, is nowhere dense and I \ E is connected (by (H5)). By the implicit mapping theorem the mapping is open. The set F(X) is closed in V. Suppose that F(X) Y. Since F is proper and FIx\s is open, we easily conclude that aF(X) C E. Therefore, by (H5). Y \ is connected. On the other hand, we have the decomposition Y

\ 8F(X) =

contradiction. Consequently, we get

(mt F(X)) U (Y

\ F(X)); C]

2 Pseudoconvexity

230

Property (P4). FIx\F-l(E): X \

—*

Y

\E

is an analytic covering. More-

(b) = kfor any b E

over, there exists a k E N such that

I \ E.

Property (PS). F is open.

Proof Fix an a E X and let Uo

X be a neighborhood ofa such that

fl On = {a}.

Let V be an open neighborhood of F(a) such that V fl F(aU0) = 0. Put U := V is proper (for any compact K C V we have Uo fl F'(V). Then Flu: U = fl Uo) and hence, by (P3), F(U) = V. 0

= £, F'(b) = {ai and letU u; Property(P6). Leib E ag, respectively. Then there exists an be arbitrary open neighborhoods of ai open neighborhood V of b such that F' (V) has exactly £ connected components t. In particular, by (P4), Ui U, wi:ha1 E U1 C UandF(U1) = V. j = 1 £
= k, then b

Moreover,

Proof Let Qi (2 C U,j =

E.

Qe be neighborhoods of t,andQ,LflQV = Oforv

at, respectively, such that

PutA := A is closed. Since F is closed and b F(A), there exists an open neighborhood V of I. b such that V fl F(A) = 0. Let := (2 fl F'(V), j = I and let T be a connected component of U := Uj. Recall Fix a j (I that V fl F(8U) = 0. Now we can easily prove (like in (P5)) that FIT: T —÷ V is 1

0.

proper. In particular, F(T) = V and soT fl

Consequently, the number of all connected components of all sets t, are connected and must be equal C. Hence j =I

j

=1

=

V,

j =

C.

Assume now that C = k. Then

U1 —÷ V must be bijective and hence

0 Proof of Theorem 2.12.2. By (P3), (P4), and (P6) we only need to prove that E is analytic.

ak}andletUl

UkandVbeasin

(P6). Define g

notice that g is holomorphic on V. Thus we have defined a holomorphic function

g: Y \

We will prove that limy\E?Y...+b g(y) = 0 for any b E E (then, by the Radó theorem (cf. [Nar 1971]), g E 0(Y) and E = Fix a b E E. Let V0 be a relatively compact neighborhood of b. Put E —÷

C := max(IJF(x)I: XE

F'(Vo)}.

.

2.12 Liftings of holomorphic mappings!!

Vc JgJ

231

.

be as in(P6). We may assume that ai Sand IJFI 5 on Ui. Then, 0 e on V \ E, and consequently, by continuity, <e on V \ E.

V0

Proposition 2.12.3 ([Ker 1960]). Let (X, p), (Y, q) be Riemann domains over

F: X —+

let

Y

be proper and holomorphic, and let k denote the multiplicity of F. Then for any function f E (9(X) there exists a manic polynomial

(y,z)€

xC,

Ck E (9(Y) such that

with coefficients Cl {z

C: P(y, z) = 0} =

(t)

YE Y.

In particular,

fk +

o

F)fk_i

o.

Proof LetEbeasinTheorem2.12.2.Takeapointb€ Y\EandletV,Uj Uk be as in Theorem 2.12.2. Put çoj := (FIu,Y': V k, and define U3, j = 1 P(y,z) :=

(z

...

—

(z

—

f(Wk(y))

=

zk

(y, z)

V x C:

notice that c1(y) is the value of a symmetric polynomial on and, therefore, cj(y) depends only on the set V. j = I k. y The functions Cl ck can be defined on Y \ E. Obviously, they are holomorphic. Moreover, (t) holds evidently on Y \ E. Let K C Y be compact. Then is compact, and consequently. the set C C is compact. In particular, the functions cl ck are bounded on K \ E. Since E is analytic, the Riemann removable singularity theorem implies that Ck extend holomorphically to Y (we denote the extensions by the same letters). Cl Observe that, by continuity, . .

k1

holds on Y.

j'

isclear. = Itremainstocheck(t)forb e E. Theinclusion and let Y \ E —f b. Then there exists numbers = {al, E Nsuchthatkj +kt =kandforv>> I weget .

j=l

232

2 Pseudoconvexity

where #Pv.j

(j =

aLjk,}, and P1..1 —* Pv.k = (cf. (P6)). Now, by continuity, we have

as

v —f +00

e

z) =

fl fl(z —

—*

j=Iu=l

fl(z

—

f(aj))k) = P(b, z),

j=1

0

which implies the inclusion c' in (t).

Proposition 2.12.4. Let (X, p), (Y. q) be Riemann domains over

F: X —.

and let

V

be a holomorphic mapping such that for any function f E (9(X) there exist k and a monic

P(y, z) =

+

(y, z)

E

N

V x C,

k E (9(Y) such that

wit/i

k

+

o

(1)

Q

I

Assume

that X is Stein. Then F is pmper.

Pmof Take a compact K C V. To prove that the set L := F' (K) is compact it suffices to show that every function f E (9(X) is bounded on L (cf. Theorem I.1O.4(vii)). Take an f E 0(X) and let k and P be as in the assumptions. Let k}.Then f(L) C {z E C: aYEK. P(y, z) = O} C (Izi

C + 1).

C]

Theorem 2.12.5 ([Ker 1960]). Let (X. p), (V. q) be Riemann domains over C'1 and let

F: X —÷

V

be pmper and holomorphic. Let

a: (X,p)—÷ (X,13),

fi: (Y,q)—÷

the maximal holomorphic extension. Let F: X —* Theorem 2.12.1). Then: be

(a)

F: X —÷

V

Y

be the lifting of F (cf

233

2.12 Liftings of holomorphic mappings II

=

(b) F(a(X)) = 13(Y), (c) FIk\a(x): X \ a(X) —÷ Y \ (d) FIa(x): a(X) —* 13(Y)

13(Y) is proper and surjecnve,

Recently, the above result have been regained in a special case in [Ber-Pin 1995]. (a) We will apply Proposition 2.12.4. Take an / tion 2.12.3 there exist 0(Y) such that Ck

Proof

(f oa)k +

0(X). By Proposi-

0 F)(f oa)k_J =0. =C3,J = 1

k. Then,bytheidentityprinciple.

we get

+

F)fk_J =0.

(b) Since F is the lifting of F, we get F(a(X)) = = 13(Y). Y) and let I 0(X) be a holomorphic function Now take $'o = 13(Yo) (vo E that separates points of the set T := F(90) (cf. Remark 1.8.12). Let k, and

j=

I

k,beasin(a)andsuchthat

=0) =

(z €C: zk

yE

Y

(cf. Proposition 2.12.3). Then

/(T) C

{z E

C:

zk

+

= 0) =

{z

e C: zk +

(f

=

= 0} C

f(Tfla(X)).

Consequently, C a(X). (c) For any compact K C Y \ 13(Y) we get

(X which, by (a), implies that

\ a(X)),

is compact. Moreover, by (a) and (b),

weget fr(k\a(X))= Y\13(Y). (d) Follows from (a) and (b).

D

Using the above results, we prove that the property to be a Stein domain is stable under proper mappings.

234

2 Pseudoconvexity

Proposition 2.12.6 ([Ker 1960]). Let (X, p), (Y, q) be Riemann domains over C" and let

F: X —÷

Y

be proper and holoinorphic. Then X is Stein iffY is Stein.

Proof The implication follows from Proposition 1.10.13 and Theorem 2.5.7. Now, assume that X is Stein. By Theorem I.10.4(vii), we only need to show that (9(Y) separates points in Y and that for any infinite set B C Y with no limit points in Y there exists a function g E 0(Y) such that sup8 Igi = +oo. Take b1,!'2 Y with b1 b2. Then there exists a function f (9(X) that separates points in the set U F'(b2). By Proposition 2.12.3 there exists a monic polynomial

(y,z)E YxC. with coefficients cj

ck

{z

(9(Y) such that

E C: P(y, z) = 0} =

fk + In particular, by (*)

o

(*)

y E Y,

F)fk_)

(**)

0.

P(b2, .). Hence there exists a Jo such that c10(bi) #

cj0(b2).

Let B C Y be an infinite set with no limit points in Y. Then A := F'(B) is an infinite subset of X with no limit points. Let I E 0(X) be such that SUPA Ill = +oo. Let P be as above. Suppose that all the function k are bounded on B. Then 0 F o F are bounded on A. Hence, by (**), the functions would be bounded on A. Consequently, there exists a Jo such that the function c10 is unbounded

f

onB.

0

Chapter 3

Envelopes of holomorphy for special domains

3.1 Univalent enveLopes of holomorphy LetG cC't

c (9(G). Themainaimofthissectionistodiscuss

conditions which assure that the

of holomorphy of (G, Id) is univalent, i.e. there exists a domain G C C G, such that idGO: (G, Id) —÷ (G, Id) is the maximal 5-extension. In such a situation we will shortly write C C't. The problem has been studied, for instance, in [Cas-Tra 1991], [Cas-Tra 1994], [Cas-Tra I 994b].

It is useful to consider also the following more general question. Given a Riemann—

Stein domain (X0, P0) over C", we like to characterize those domains X C X0 and families C (9(X) for which the i-envelope of holomorphy of (X, may be realized as a subdomain of Xo, i.e. there exists a domain X c Xo, X c X, such that idXk: (X, polx) —* (X, Polk) is the maximal i-extension; we will shortly write C(X, C Xo. The case (X0. P0) = (C". id) reduces to the previous problem of univalent envelopes of holomorphy.

Remark 3.1.1. (a) Let G C C". D C Ctm be domains,

V :=

{f E (9(G x D):

VU,ED:

C (9(G). Put

f(., w) E

By virtue of Proposition 1.8. 15 we get the following two results: If C C" and D is a domain of holomorphy, then and x D, V) = e(G, x D. If C(G) C C" and e(D) C Ctm then x D) C x

x D, V) c and

C(G x D) =

(b)IfE(X) C Xo, then (by Corollary 2.2.12)

fl

C(X)=int

z.

XcZcX0 Z is a domain of holomosphy

(c)

Let

F: X0 —÷

X0 be biholomorphic and let X. I c Xo be domains such that Then F(C(X)) c In particular, if

C X0, C(Y) C X0, and F(X) Recall that

:= £(G, 0(G)).

c I.

236

3 Envelopes of holomorphy for special domains

F(X)= X,then

and

fl

Z.

x czcx0 Z isadomain of holomorphy

Remark 3.1.2. (a) (Cf. Remark 1.9.6) Let G C C" be starlike and let 4 C 0(G) be such that for any f 4 and t E (0, 1] the function G z —p f(tz) belongs to 4. Then 4) is a starlike domain in C". Notice that if (0, l]G = G but 0 G, then £(G, 8) need not be univalent.

For example. G := C \ R_, 8

{L,: t E (0, lj}, where L,(z) := Log(rz),

z E C \ IR_, and Log denotes the principal branch of the logarithm.

(b) (Cf. Remark 1.9.6) Let G C C" be balanced and let 4 C 0(G) be such that 4 and E E \ 1O} the function G z —÷ belongs to 4. Then 4) is a balanced domain in C". In particular, if G is balanced, then

for any f

fl

DJG D is a balanced

domain of holomorphy

(C) (Cf. Corollary 1.9.18) Let G C C" be n-circled and let 4 C 0(G) be such that (8E)" the function G z —+ 4 and z) belongs to 4. Then is an u-circled domain in C".

for any f E

In particular, if G C C" is n-circled, then

fl

D.

DDG D is an n-circled domain of holomorphy

(d) Let G C C" be a complete n-circled domain (cf. Proposition 1.9.19) and let be such that for any f E 4 and E (E \ {0})" the function G z —+ G, belongs to 4. Then E(G, 4) is a complete n-circled domain in C". z), z Notice that in (a)—(d) one can take 4 :=

4 C (9(G) -

(e) Recall that if G C C" be an n-circled domain such that G fl x {0} x $ 0 for some E (1 n}, then any function holomorphic on G extends holomorphically to the domain C"—1)

j

z:

E

x E x {1}"', z e

cf. the proof of Proposition 1.9.19. Consequently, if G C C" is a n-circled domain with 0 E G, then any function holomorphic on G extends to the domain

G :=

E E", z€G};

237

3.1 Univalent envelopes of holomorphy

G is complete n-circled. In particular, if 0 E G, then

E(G)=int

fl

D.

DJG

D isa complete n-circled domain of bolomorphy

(f) (Cf. Remark 1.9.8) Let G C D x C" be a Hartogs domain with non-empty starlike fibers. Let 7 C (9(D x Ck) be such that D x is a 7-domain of holomorphy, E (0, 1]thefunction andleU c TlGandforanyf E G Then is univalent, (z. w) —+ f(z. 1w) belongs to C D x Ct, and E(G, is a Hartogs domain with non-empty starlike fibers. (g) (Cf. Remark 1.9.8) Let G C D x C" be a Hartogs domain with non-empty balanced fibers. Let 7 C (9(D x Ck) be such that D x is a 7-domain of bobmorphy, and let C (9(G) be such that J and for any f E and E E \ (0) the function G Then C(G, S) is univalent, (z. w) —f f(z, çw) belongs to C D x Ck, and C(G, is a Hartogs domain with non-empty balanced E(G,

j

fibers.

(h) Let G C D x be a Hartogs domain with non-empty connected k-circled x fibressuchthatGfl(D x C"—') øforsomej (1 k). Then any function holomorphic on G extends hobomorphically to the Hartogs domain

:=

.

xEx

w): ç

Consequently, if G c D x

(z, w)

G}.

Ck

is a Hartogs domain with non-empty connected k-circled fibers such that G fl (D x 101) 0, then any function holomorphic on G extends to the following Hartogs domain with complete k-circled fibers

a

:= {(z.

. w):

E

Ek

(z, w)

Indeed, since the fibers are connected, any function f

G}.

(9(G) may be represented

where byaHartogs—Laurentseriesf(z, w) = Z") x and the series is locally normally convergent. Now, the condition Gfl (D x (0) x Ck_J) x 0 implies that Consequently, 0 for fi Z'—1 x the series is locally normally convergent on GW. (i) Let G c D x C" be a Hartogs domain with non-empty k-circled fibers (we do not assume that the fibers are connected) such that D x (0} C G. Then any function

C''

holomorphic on G extends hobomorphically to the Hartogs domain G (defined in (h)). Indeed, any function f E (9(G) maybe represented by the Harlogs—Laurent series (9(G),foranyz E Dthefunctionfp(z..) f(z. w) = E$EZk ffl(z, w)wfi, where and the series is is constant on any connected component of the fiber E locally normally convergent. On the other side, let

Go:={(z.w)EG:VcEEA:

•W)EG)

3 Envelopes of holomorphy for special domains

238

x (0) C G, the set GOC D x Ck is a Hartogs domain with non-empty and fp(z. w) 0 on G0 for fi complete k-circled fibers. Thus fp (z, w) E G, and for (z. w) G0 and E Hence f(z, w) = Since D

consequently, the series is convergent in G. Now we will present two criteria for the univalence of the envelope of holomorphy.

PropositIon 3.1.3. Let (X, p), (Y, q) be Riemann domains over

and let

r: (X,p) —* (Y,q) be a morphism. Let 7 be a a-swble vector subspace of 0(Y) and put := C1O(X)

Assume that

a: (X,p) are maximal

fi: (Y,q) —÷

and 7-extensions, respectively. Let

f:

(X,j3) —÷

and I is infective. be a lifting oft (Proposition 1.9.2). Then = clo(k) X C Y C Xo are if (Xo. P0) is a Riemann—Stein domain and in domains such that C Xo and(X, Y) isa Rungepair(cf Proposition 2.7.5), then and L(Y)) is a Runge pair. C C(X) c Xo.

Pmof We know that is a a-stable closed subspace of 0(k); cf. Remark 1.4.5(j). Since as: is a topological isomorphism (in the topologies of locally uniform is a is a dense subset of convergence), we conclude that dense subset of It remains to observe that separates points in stalks (ProposiTo prove that I is injective, recall that separates points tion 1.8.10). By the first part of the proof we conclude that in stalks. k be such that 1(n) = f(x2). In particular, Let = points xl, x2 are obviously not separated by functions from 15(0(Y)). Thus

The

=

X2.

cJ

Proposition 3.1.4. Let D C

be a domain of holomorphy and let M be a pure (n — 1)-dimensional analytic subset of D. Then for any domain C C D we have: is univalent

if C(G) is univalent, then

Proof If quently,

\ M) is univalent.

\ M) =

\ M.

is univalent, then by Theorem 2.5.9, C(G \ M) = \ M) is univalent.

\ M.

Conse-

239

3.1 Univalent envelopes of holomorphy

Now assume that E(G \ M) is univalent. Let a: (G, Id) —÷ (G, j3) be the maximal holomorphic extension. By Proposition 1.9.2 there exists a morphism

t:

that r o a = idG.D (obviously z

such

rem

(G,13) —÷

(D,id)

fr). Put M :=

Hence, by Theo-

2.5.9,

aIG\M:

(G \ M, id) —+ (G \ M,

the maximal holomorphic extension. Hence \ M). Id) is isomorphic to Consequently, G \ M must be univalent. Suppose that EM (G \ M, j3). are such that j3(ai) = p(a2) =: zo and Letr := min{dO(aI),dO(a2)}. We easily conclude that r) U r)) \ M) 115(zo, r). Hence, there exist points b1,!,2 G \ M, b1 contradiction. !2, such that 13(b1) = Thus G is univalent. U is

Using Proposition 3.1.3, one can characterize several classes of domains with univalent envelopes of holomorphy, for instance:

Example 3.1.5 (Envelopes of holomorphy of circular domains). Let G C C" be a circular domain E G for any ( E dE and z G). Put

zeG}: C(G) is a Ct-cone. Assume that for any z E G the section := G} is connected 2) and E(C(G)) is univalent Then is univalent and circled (and (E(G), E(C(G))) is a Runge pair). Indeed, by Proposition 3.1.3, it suffices to prove that (G, C(G)) is a Runge pair. Since the sections of G are connected, any f E (9(G) may be represented by a series

f(z) = where I

2nz

z E G,

I'

I

ZEG.

"

Clearly. (9(G). Moreover, the series converges locally uniformly in G. One can easily show that

= Consequently, 2)

That is,

e G,

extends to a holomorphic function on C(G). is an annulus.

Observe that 0 a G connected.

z

C(G) = C". Moreover, if G is balanced, then each section

is

240

3 Envelopes of holomorphy for special domains

Remark 3.1.6. Let G C

be an n-circled domain. Recall that the logarithmic image of G is the set

logG := {(xI

ER'1: (eX1

E G}.

c")

For a domain U C W1 put

expU := ((zi

(log Izil

Zn) E

U}.

log

Notice that exp log G = G and exp U is an n-circled domain with log exp U = U. By Proposition 1.9.19, G is a domain of holomorphy if log G is convex. As a simple corollary we get the following formula:

E(G) = exp(convlogG). where, as usually, cony A denotes the convex hull of A.

Let U C R" be a convex domain and let U1 denote the projection of U onto the

first) axes(l < j 2,thedomain

can

be written in the form U3

= ((xi

x R: (1(xi

<

where the functions £,. —L1: U1_1 —*

are upper semicontinuous and

forany(xi

<

(which will be

This simple observation leads to the following description of used in the next section).

= ((zi

Zn)

z1,)

r3(zi

(zi where

0< ri <

R1

<

< Zj—i)

R1(zi

G1,

j

< +oo and if) >2, then:

rj:

G —* [0. +00) is upper semicontinuous. R3: G1 —÷ (0. +00] is lower semicontinuous, z1_i) < R3(zi

z1—t)

forany

(zi

Zj—l) E G1.

is a (j — 1)-circled domain of holomorphy, C r1, R3 depend only on (Izil In fact, if U := cony log G and t-.

are as above, then

:=

Zj_i) :=exp(L1(loglziI

log

=1

3.1 Univalent envelopes of holomorphy

241

Proposilion 3.1.7 (Envelopes of holomorphy of Hartogs domains). (a) Let G C D x Ck be a Hartogs domain with non-empty balanced fibers. Let that for any f E and E \ {O} the function G

IC

C

0(G) be such

f(z.

(z. w) —÷

belongs to Then there exist: • a Riemann domain (D, over •amorphisma: (D.id) (D,co).and with non-empty balanced fibers • a Hartogs domain G C D x such that

•a x idck(G) C G, • for any f E there exists an f

-

(9(G) such that f(a(z). w) = f(z. w),

(z, w)€ G.

:= {f: f

•G isthe domain of existence of thefamily

In the case where D is univalent, the envelope E (G.

= 0(G) and a: (D, id) —÷

is also univalent.

(D, ço) is the maxi,nal holomorphic

extension of D, then: • (D, satisfies all the above conditions, • a x idCk: (G. Id) —+ (G, ço X idck) is the maximal holomorphic extension. (b) Let G

fibers. Let

CDx

Ck be a Hartogs domain with non-empty connected k-circled

C 0(G) be such that for any f G

andç E (aE)r the function

E

IC

(z, w) —+ f(z.

z)

belongs to Then there exist: • a Riemann domain (D. 60) over

•amorphisma: (D,id) —+ (D,ço),and • a Harrogs domain G C D x Ck with non-empty connected k-circled fibers such that

•a x idck(G) •for any f

C G,

there exists an f E 0(G) such that f(a(z). w) = f(z. w),

(z.w)€G, • G is the domain of existence of the family

-

:= {f: f

In the case where b is univalent, the envelope e(G,

is also univalent.

= 0(G) and a: (D. id) —i' (D, 60) is the maximal holomorphic extension of D, then: • (D, 60) satisfies all the above conditions, • a x idck: (G. Id) —+ (G. ço x idc&) is the maximal holomorphic extension. Remark 3.1.8. Notice that in general, if G C D x C is an arbitrary Hartogs domain with non-empty circular fibers such that is univalent, then €(G) need not be univalent.

242

3 Envelopes of holomorphy for special domains

For example, let G be the domain from the introduction to Chapter I. Then x C is a Hartogs domain with non-empty circular fibers such that E(G) is GC not univalent (see also Remark 3.1.12(d)).

Proof (a) Any function f z, w) =

E

may be expanded into a series of the form f(z.

where fp E 0(D)

j/3=v

w) =

E

the series is locally normally convergent in G. Put F := E fE Let cr: (D, id) —÷ (D. çp) denote the maximal F-extension. For f E and v E where Z+, define E 0(D) denotes the 1131=v extension of

(ffl ° a =

f .)

let D1 denote the region of normal convergence of the series It is clear that bj c 15 x Ck is a Hartogs domain with balanced

fibers (they may be empty: cf. Remark 3.1.9) and a x idc*(G) C D1.

Let D,

be the connected component of D1 that contains a x idck(G). Obviously, the sum of the series .) defines a holomorphic extension / of f to D1. i.e. /(a(z), w) = f(z, w). Notice that = fc for any E E \ (0) 6), Let G denote (G). Then G C /5 x C' the connected component of mt Dj that contains ax is a Hartogs domain with balanced fibers such that a x idc*: (G, id) —÷ (G, x id) is an i-extension. Let /5 be the connected component of the set {x E iS: 01 that contains cr(D). We want to show that 0 is the s-domain of existence. Suppose that (xo. wo) E G Define a new Hartogs is such that d(T(r0 u'O)f) > R > u'o) for any f E domain

R) x

E E \ (0), w E IP'k(wO, R)}.

We will prove that for any f E

exists a function f

w) = f(x. w) for (x, U') E be such that

u's). Take (z. w) E

E

and let

such that E E \ (01

E Pk(WO, R). Define

f(z. u') := T(x()u,o)(fc)(z, observe that f is well defined. holomorphic on c, and (x. w) E wo)

u') = f(x. w) for

That is. the set of those (x. u') E D x Ck for which there exists an open neighborhood U C D x Ck such that the series is convergent. That is, w) = f(x. for any (x. u') Dj.

i(z.

R) x

(z. w) e

j = 1.2.

=

j = 1.2. We have to show that gi = X2 on follows directly from the identity principle. If

) fl

fl

Pk(wO. R)):

Xj

is holomorphic.

fl 0. then the equality = 0. then we may easily find points N—l,and

fl

0. Consequently, we may successively apply the previous case.

243

3.1 Univalent envelopes of holornorphy

One can easily check that f(z. w) = R)),

E

and the series is convergent locally normally in

E

= fflnearxo(f of existence, we get R

where

Since(D,q)istheF-domain

db(XO). Thus

v=O

x, w) for (x, w) E U := lPb(xO, R) x Pk(wO. R). Consequently, U C D1, and finally U C G; contradiction. coincides with

In the case 5 = (9(G) we only need to prove that I = L) = E(D). Observe that D = {x E D: (x,O) E O}. By Corollary 2.2.19, D a Riemann—Stein region containing a(D). Thus D = D. may be (b) The proof uses the same methods as in (a). Any function f E where expanded into a Hartogs—Laurent series f(z. w) = Zk) and the series is locally normally convergent in G. Put 0(D) E denote the maximal Let a: (D,id) —÷ 7 := f E J, E f let D1 F-extension. Let ff3 E 0(D) denote the extension of

9)• It is clear that denote the region of normal convergence of the series is a Hartogs domain with k-circled fibers. Let D1 denote the connected Df C D x component of D1 that contains a x idc* (G). The sum of the series

f

for any to Dj. Notice that = Dj that contains the connected component of mt denote Let G E is a Hartogs domain with k-circled fibers such a x idck(G). Then G C D x that a x (G, id) —+ (G, x id) is an s-extension. Let D be the connected ø} that contains a(D). component of the set (x ED: We have to show that G is the i-domain of existence. Suppose that (xO, wo) G Define a new Hartogs wo) for any f is such that d(T(x0w0)f) > R > domain

R) x

Fixanf E

.

w:

(aE)k, w

Pk(wO. R)}.

Take(z,w)

Pk(WO. R). Define

!(z, w) := Use the fact that

f(x. u') for (x. w) E 9)

w) =

1) x Ck for which there exists an open neighborhood l'hat is. the set of those (x, w) <0 for some) E (I n}. is well defined in U (if is convergent. cCi' x x

U C Dx Ck such that

thenU

is a Hartogs domain with connected circled fibers and wo).

244

3

Envelopes of holomorphy for special domains

observe that f is well defined, holomorphic on and w) = J(x, w) for where fp is holo(x, w) E wo). Moreover, f(z, w) = EPEZk morphic in IP(Q(xo), R), E Zk, and the series is convergent locally normally in Hence, R and R) x Pk(wO, R) C G; contradiction. It remains to prove that the fibers are connected for any x E D. Consider the set c of all (xo, wO) E D x Ck for which there exists an open such that for any f E there exists a constant M(f) > 0 neighborhood U C D x such that the function U (x. w) —p is well-defined and

M(f),

U, for any

W := {(e'°' lw'1 I

7L'. Observe that

c Dx

has k-circled fibers. Directly from the definition of G we conclude that G C c. First observe that and let U be as above. We may assume Indeed, take (xo, wo) QC that U = V x W, where V is a neighborhood of xO and W is a k-circled neighborhood of wo. Now, let A = A1 x x Ak c W be a compact neighborhood of wo such that each is a closed disc or a closed annulus. Then there exists a 9 E (0, 1) such < for (x, w) E V x A and E Zk, which implies that is normally convergent on V x A. that the series Thus we only need to prove that the fibers of are connected. For, take (xO, We are going to find a domain W C with W. We may E (xo, assume that w', w" (R>o)lc. By definition, there exist a neighborhood V of xo and r > 0 with r) U Pk(wg, r) C (C5)k, such that for any f 4S there exists an M(f) > 0 such that r) U r)) < M(f) for (x, w) V x and,fl E Zk. Define (x, w)

etOk

Pk(w0,r), (wi' R"

Then W C (Ce)" is a domain containing w', w", and

t

10, 1]).

<M(f) for (x, w)

E

Vx In the case = 0(G), similarly as in (a), we have to show that D is Stein. Define := {(x, w) D x Ck: (x, e'°) G} By Corollary 2.2.20, is t ERk}andthatcisaHartogs Stein Noticethatcz={(x,w+ir): (x,w)E

domain over D with connected fibers

Hence, by Corollary 2.3.3, D is Stein.

E

Remark 3.1.9. Notice that in general Dj need not be connected and D may be smaller than D. For example, let Recall that

e" =

We apply Corollary 2.2.20 to the mapping D x Ck (z, w) —' (z, E D x Ck. isconnectedk-circled, isconnectedk-circled. Consequently, the c (R>o)* isconnected. Thisimpliesthat((Ieu't I e'°* I): wkI): w E set ((Iwj I u' E cZX) is connected, which finally shows that the whole fiber is connected. l

245

3.1 Univalent envelopes of holomorphy

D:={x+iy€C: < G := {(z. w) E D x C: {f E 0(G): 3OER: f(z, w) =

(z. w) E G}.

lyl> lxi. Iwl be a Hartogs domain with non-empty balanced Corollary 3.1.10. (a) Let G C D x fibers. Then (G) is univalent iff D) is univalent. Moreover, if E(D) is univalent, then C(G) C C( D) x is a Hartogs domain with non-empty balanced fibers. (b) Let G C D x Ck be a Hartogs domain with non-empty connected k-circled fibers such that G fl (D x {O}) 0. Then E(G) is univalent :ffE(D) is univalent. Moreover, if C(D) is univalent, then E(G) C C(D) x Ck is a Hartogs domain with complete k-circled fibers.

(c) Let G c D x Ck be a Hartogs domain with non-empty connected k-circled fibers. Then E(G) is a univalent Hartogs domain with connected fibers if E(D) is univalent. Moreover, E(D) is univalent, then x C' is a Hartogs C domain with non-empty k-circled fibers (cf Example 3.1.16). In particular, (fG c D x Ck is a Hartogs domain of holomorphy with non-empty balanced (resp. connected k-circled) fibers, then D is a domain of holomorphy

Proof The implications follow from Proposition 3.1.7. To prove the remaining implications, let G C D x Ck be a Hartogs domain with non-empty and balanced (resp. connected k-circled) fibers. Let

ax idCA: (G, id) —+ be

(G,

X

idck)

the maximal holomorphic extension of G from Proposition 3.1.7. Recall that G

is a subdomain of D x Ck, where a:(D, id) —f (D, ço) is the maximal holomorphic extension. Moreover, any fiber (x E D) is non-empty and balanced (resp. connected k-circled). Assume that we know that C(G) C D x Ck is a univalent Hartogs domain with non-empty fibers (D is a domain in C's"). Then (G. x idck) and (E(G), id) G —* are isomorphic, which implies that x is homeomorphic. In particular. ço(b) = D. Observe that

=

U &.

(*)

XEQ 1(z)

x idck is injective on O, we conclude that fl = 0 for x' x". Thus, if the fiber E(G)2 is connected, then must consist of exactly one

Moreover, since

x', x" E point.

The above reasoning completes the proof of (c). To finish (a) and (b), we only need to show that if is univalent, then it has connected fibers.

3 Envelopes of holomorphy for special domains

246 -

The case (b) reduces to (a) because in the situation of (b) C(G) = E(G), where x Ct is a Hartogs domain with non-empty complete k-circled fibers — cf.

G CD

Remark 3.1.2(h).

Thus assume that we are in (a). By Proposition 3.1.7(a), we know that is balanced for any x E D. Consequently, by (*), is balanced (in particular, connected) for any z E D'.

Remark 3.1.11. (a) Proposition 3.1.7 may be easily generalized to the case where D is a Riemann domain over

(b) Proposition 3.1.7(b) may be generalized to the case where G C D x is a Hartogs domain with non-empty k-circled fibers (which may be disconnected). In fact, by (a), it suffices to prove the following intermediate result. Let G C D x be a Hartogs domain with non-empty k-circled fibers. Let C (9(G) be such that for any E and the function

f

G

(z,w) —*

belongs to i5. Then there exist.

• a Riemann domain (D, over (D, with • a continuous mapping a: (G, idG) w)) = z, (z, w) E G,and • a Hartogs domain G C D x Ck with non-empty connected k-circled fibers such that 13) • C G, where fi(z, w) := (a(z, w), w). (z, w) E G. •for any f there exists an I E 0(G) such that /0 =

To prove the above result we proceed as in the proof of Proposition 3.1.7(b). Any

function f E

may be expanded into a Hartogs—Laurent series

f(z, w) =

1

k(/9 E Zk),andtheseriesislocallynormally

convergent in G. Put F := fi E Zk}. f For any (zO, wO) E G, for any neighborhood

x C G of (zn, wO) such is connected, and for any g F there exists a such that x = g(z, U') for all (z, w) E Let denote the sheaf of F-germs of holomorphic functions in in Example 1.6.6. We keep the notation from Example 1.6.6. Consider the mapping that

Observethatfi: (G.id0) —* (G. idG) Let := (f: f (G, çQ X idck) IS all Now we apply the generalized version of Proposition 3.1.7(b) to the Hartogs domain G C I) x C& and the family In the case = (9(D) the above procedure gives the envelope of holomorphy of D.

247

3.1 Univalent envelopes of holomorphy

a: G —÷

given by the formula

a(z. w) := ([(ui,

w)) = z for any (z. w) E G. Moreover,

Observe that a is continuous and

Fg o a =

g

for any g E F. Let D

contains a(G). Put .p :=

(z, w) E G.

z),

denote the connected component of

Notice that the construction of (D,

q,)

that is similar to

that in [Lig 1998].

E (9(D) and denote the region of normal convergence of the let D1 C D x E D1 that Let G denote the connected component of mt series >Jp€Zk contains Then G C D x Ck is a Hartogs domain with k-circled fibers such that /3: (G, idG) —* (G, x idc*) is an i-extension. Let D be the connected component Then

f

oftheset{x ED: The proof that the fibers

x E D, are connected is exactly the same as in

Proposition 3.1.7(b).

The univalence of the envelope of holomorphy is not biholomorphically invariant (like the Runge property), namely: is univalent. Assume that Remark 3.1.12. Let D c C" be a domain such that F: D —÷ G is biholomorphic. By Proposition 1.9.9, F has a locally biholomorphic extension F: E(D) —+ C's. Moreover, we know (Remark 1.9.10) that

F1: (G, id) —p

F)

is the maximal holomorphic extension. In particular,

C(G) is univalent if F is infective. Consider the following four examples:

-

The case where K := C(D) \ D is compact: Put G := F(C(D)) 15) Observe that G \ G C F(K) is compact. Consequently, by the Hartogs Theorem 2.6.6, the mapping F1: G —* D extends holomorphically to G; let F' denote the extension. (a)

Now, one can easily check that F: Thus C(G) is univalent.

(b) The case where

—* G is biholomorphic and

\ D) = 0:

Suppose that a, b

= (F)1. a

are such that F(a) = F(b) =: c. Since F is locally biholomorphic. there exist of a, b, c, respectively, such that U0 fl Ub = and open neighborhoods U0. U,,, are biholomorphic. Let V : F(Ua 11 D) fl and Flub: Ub —± U0 —+ F(Ub fl D). Observe that V 0. Now, it is clear that there exist points a' E U0 fl D, b' E tjh fl D such that F(a') = F(b') e V; contradiction. Note that G is open.

248

3 Envelopes of holomorphy for special domains

Thus in this case

is also univalent.

(c) Let U C R2 be a domain such that:

—x'I < for some (XI. xi), (xi. x') cony U. Put D := U + iIR2. Recall that E(D) = (cony U) + (Bochner theorem). Let z2). Then F is injective on D but is not injective on F(zi, z2) := In particular. E.(G) is not univalent. (d) Let

U'): 0< r < I,

D :=

E

cPE(0,lr/21 E L2jr,

5jr/2)

I

< ti'I < 2

The mapping F(z, w) := (exp(z). w) is injective on D and G := F(D) is the domain from the example in the introduction to Chapter 1. Observe that D is a Hartogs domain over R := {x + iy E C: x < 0. 0 < v < 5ir/2} with connected circular fibers. Hence D) is univalent (cf. Corollary 3.1.10(b), see also Remark 3.2.19) and C(D) J R x E. Thus F is not injective on e(D) and therefore E(G) is not univalent.

Forr,pE(0,.1)put x K(r))

H = Hr.p := Observe that C(H) =

F: H —p

U

p<

E

< l} x E).

(cf. Example 1.8.7).

c

be a bihotomorphic mapping. As above, F denotes the holomorphic extension of F to P,,. Then there is the following criterion for F to be injective (cf. [Kas 1974]).

Let

G

PropositIon 3.1.13. F is infective Prnof

fl

\ H) = 0.

'==*' is obvious. So it remains to prove

S :=

sup{s

Put

e (0, II:

injective}.

It suffices to show that S = I. Suppose S < I. In virtue of the assumption then r S < I. In the first step we shall see that FI{ZEpn: is injective. To prove this we put

F1 (F(z0)) fl (z E Obviously,

the assumption

I

= 5) is exactly one pointj.

F(H) Ii FOP,, \ H) = 0 implies that 0

T

p.

3.! Univalent envelopes of holomorphy

Suppose now that T > 0. Then there are sequences

S=

w1,j

p, and

I.

1VI / T,

249 C

with

E N, such that

fr(z') = F(w1), j

N.

may assume that both sequences converge. Put

We

urn

=:

z0,

=:

urn

w0.

i—+oo

Since F is locally biholomorphic, it follows that z0 w°. We choose disjoint neighborhoods U = U(w0) IF,, and V = V(z°) := IF,, with F(U) = F(V) such that Flu and Fly are injective. DeBn(Z°, r) fine CD := (FJuY'. In virtue of the definition ofT we know that, if z E V, lznl = S. > Moreover, using that F is locally biholomorphic and the property of S we conclude

that

lcD,,(F(z))l > S lcD,,(F(z))l

>S

if z E V.

=

5, and 1±1 > T,

(*)

if z E V. IZ,,l <S.

With this information in mind we look at the irreducible analytic subset A :=

in V. Then B := CD(fr(A)) is an irreducible analytic subset {z E V: Zn = of U, w0 E B. Moreover, the function g: C" —÷ C, g(z) := z,,, satisfies >S and Ig(w°)i = S. Then the maximum principle gives that IZ,,l = S for any z B. which contradicts (*). Hence we get T = 0. Following the recent argument we even have that F is injective on (z E IF,,

IZnIS}. \ H) = 0 and the local biholornorphicity such that F is also injective on (z IF,,: lz,,l < S +e}:

Then the assumption that F(H) fl F(IF',,

of F leads to an

0

0

contradiction to the definition of S.

So far we don't know any injective holornorphic mapping F: H —b C", such that F is not injective on IF,,

A similar statement (as in Proposition 3.1.13) remains true in the unbounded case. Here we only give the precise formulation. (The reader is asked for the proof as an exercise.)

Fixr,p E

(0, 1) and put

H:

Let F: H

x K(r)) U

p<

< l} xC).

G be a biholornorphic mapping and denote by F its holornorphic extension to C(H) = P,,_i x C =: H. Then the following is true:

250

3 Envelopes of holomorphy for special domams

if F(H)

and

IF(z')I —+ oo for any sequence (z1)jeN C H

with

I —÷ 00. then fr is infective on Whether this statement remains true if we only assume that F is injective on

H seems to be unknown Example 3.1.14 ([Cas-Tra 1991]). Let

:= G1

((re', w) E C2: I <

r

<e. 0
<

<

is a Hartogs domain over (I < Izi <e) with circular fibers which are in general is not univalent; cf. Remark 3.1.8 for another example

disconnected 16j Then of that type.

Indeed, we can argue as in Remark 3.1.12(d). Let R := (x + iy e C: 0 < x < 0 < y <4ir},

:= 1(x + iy, w) E R x C: y — D1

2

1,

< IwI
is a Hartogs domain over R with connected circular fibers. Then

E(Di)=1(x +iy,w)E

R

xC:

IwI


Observe that the mapping F: C2 —* C2, F(z, w) (eZ, w), maps biholomorphically D, onto G1. It is clear that F is not injective on e(D1). Consequently, by Remark 3.1.12(a), C(G1) is not univalent.

Remark 3.1.15. Envelopes of holomorphy of arbitrary Hartogs domains 0 C C2 are characterized by the following result by E. Ligocka [Lig 1998]. G C D x C be an arbitrary Jiartogs domain with non-empty circular fibers. Then there exist: x C with non-enlptv connected circular • a Hartogs domain of holomorphy G C fibers, where either = E or = C, and • a surf ective holomorphic mapping CD: G —* such that CD: 0 —+ E(G) is a covering.

Example 3.1.16 ([Cas-Tra 19911). Let

02 := {(re't,

Z2,

w) E C3: I < r <e.

0 <

r

<4jr, 1z21


< IwI <e'}.

Then 02 is a Hartogs domain of holomorphy over the domain

G2:={(re",z2)EC2: I
17)

(I. e). we havc = K(r/2) U {5x/2 —2 < wi < 5ir/2). an f e (9(D1). Since D1 fl (R x (0)) 0. the Hartogs—Laurent series of I is

For cxamplc. fors e Take

convergent in the domain D1 defined by the right hand side. Since the function (z. w) —* Iwi — is psh. D1 is a domain of holomorphy (Corollary 2.2.15). Thus D1.

z

3.1 Univalent envelopes of holomorphy

251

G2 has circular fibers which are in general disconnected 18) The projection G2 is not a domain of holomorphy — cf. Remark 2.3.4.

Indeed,letR:={x+iyEC:0<x
D2 := {(x + iy, Z2. w) E R x C x C5: 1z2t < y.

Observe that the mapping F(zj, Z2. w) := (en, Z2, w) transforms biholomorphically

onto G2. Since the functions

RxCxC53(zI,z2,w)—*1z21—ImzI, R xC x (z,,z2. w) —÷ Imzj —1 R xC x C5 (zi, Z2, w) —* log wi —!mzi are psh, D2 is a domain of holomorphy. Consequently, G2 is a domain of holomorphy. G2 is a Hartogs domain over (1 < zl <e} with complete circular fibers. Thus, by Proposition 2.2.22(b). to prove that G2 is not a domain of holomorphy, it suffices to observe that the function u defined on (1 < Izi < e} by the formula u(re") := I < r < e, 0< t < 2,r, cannot be subharmonic (because it has a —Iog(t global maximum). I

Example 3.1.17 ([Cas-Tra 1991]). Let

:= {(re", Z2, a;)

C3:

I

< IwI

<e').

Then G3 is a Hartogs domain over G1 (cf. Example 3.1.14) with connected circular fibers. Moreover, C(G3) = G2 (Example 3.1.16) and therefore C(G3) has disconnected fibers. In particular, by Corollary 3.1.10, cannot be univalent (which was already directly shown in Example 3.1.14). Indeed, recall that D2 is biholomorphic toG2 via F(zi, Z2, w) := (eZi, Z2, w). In particular, F1 maps btholomorphically G3 onto

D3 :={(x +iy,z2,w) ER xC x C4: y —2<1Z21< Y,

<

<eV},

where R := {x + iy C: 0 <x < 1, 0 < y <4jr}. Hence, we only need to prove that any function I (9(D3) extends holomorphically to D2. By the Cauchy integral formula, the extension f E (9(D2) may be given by the formula -

f(zi, 18>

Z2, w) :=

For example,

I

2,r:

j(kl=r

ifs E (I, e),

then

< r,

—

—2 < r < v.

Z2

= {e"12' < wi < c"12) U

< wi <

252

3 Envelopes of holomorphy for special domains

Remark 3.1.18. Let V,, := C'7' x {O} and consider the biholomorphism C'7 \

_!*

(z'. Zn)

E

C C't \ Vn be a circular domain. Then G := 4(G) is a Hartogs domain over (Z'/Zn: (z'. E G} with non-empty circular fibers. Observe that for any gives an isomorphism 3 W —+ WZn E (z'. Zn) E G. the mapping between the section G x G has connected fibers. It is clear that P(C(G)) = D

By virtue of Proposition 3.1.4, for any circular domain G C C'7, we have: is univalent

E.(G \

\

is univalent

is univalent.

Remark 3.1.19. (a) Let G C C'7 be a circular domain with connected sections such that E(C(G)) = C'7. Then is univalent and 0 E We already know that is univalent (Remark 3.1.5) and that (E(G), C'7) is a Runge pair. Suppose that 0 Fix a point a E G and consider the holomorphic

function E(G) fl (Ca) 3 (fo is well defined because 0 Proposition2.5.l0,thereexistsanf E =

E

By E(G)0.

Consequently, since C'7) is a Runge pair, the function E may be approximated, locally uniformly in by complex polynomials. Since is a circular open subset of C, we get a contradiction. Notice that if the sections are not connected, then E(G) need not be univalent (cf. Example 3.1.20).

(b) (Cf. [Cas-Tra 1991]) Let G C C'7 be a circular domain of holomorphy with connected sections. Then C(G) is a domain of holomorphy. Indeed, by Remark 3.1.18 and Corollary 3.1.10, we get the following sequence of implications: G is a circular domain of holomorphy with connected sections G

\

is a circular domain of holomorphy with connected sections

G := (D(G \ V1,) C D x

isaHartogsdomainofholomorphy

with non-empty connected circular fibers D is a domain of holomorphy is a domain of holomorphy C(G \ Vn) = cj'(D' x C(G) \ V1, is a domain of holomorphy. In particular, by Proposition 3.1.4, Ce-cone.

is

univalent. Note that E(C(G))

is a

The same argument shows that C(G) \ E is a domain of holomorphy for any complex (n — 1)-dimensional subspace of C". Thus

\

(0}

= C(G) \

{0J.

253

Univalent envelopes of holomorphy

3.1

IfO then = C's, and This completes the proof if 0 therefore, by (a), 0 E = G. Thus C(G) = C". which completes the proof. Notice that if the sections are not connected, then C(G) need not be a domain of holomorphy — cf. [Cas-Tra 1991], where it is shown that there exists a circular domain of holomorphy G C C2 (whose sections are in general disconnected) such that C(G) = (C) Let G C C" be a circular domain with 0 E G (we do not assume that sections are connected). Then C(G) is univalent. Indeed, by Remarks 3.1.18, 3.1.2(i) and Corollary 3.1.10, we get the following sequence of implications:

G is a circular domain with 0 E G

4(G \

U

C

C is a Hartogs domain

with non-empty circular fibers such that

C G'

C(G) is univalent C(G

C(G \

=>

\

V,,) is univalent Vi.,) is

univalent

is univalent.

(d) Let G C 0

be a circular domain with connected sections such that G. Then E(G) is univalent and has connected sections if C(C(G)) is univa-

lent. Moreover, if C(C(G)) is univalent, then = C(C(G)). Indeed, we already know that if C(C(G)) is univalent, then (Remark 3.1.5). Assume that C(G) is univalent and has connected sections. Then:

E(G \

is

is univalent

univalent and has connected sections

G := 4(G \ V,,) C D x

is a Hartogs domain with non-empty

connected fibers such that e(G) is univalent and has connected fibers C(D) is univalent (Corollary 3.1.10) t(C(G) \ V,,) is univalent C(C(G)) is univalent. Moreover, \ V,,) = C(t(G \ V,,)) = C = C(C(G))\ V,,. Consequently, C(C(G))\(0} = = t(C(G)\ and C(C(G)) \ (0}. ff0 C(C(G)), then C(C(G)) = C". Hence, by (a), 0 E hence 0 E C(t(G)).

Obviously, (C(D) x

Example 3.1.20 ([Cas-Tra 19911). There exists a circular domain G C C2 (with disconnected sections) such that C(G) = and is not univalent (Cf. Remark 3.1.19(a)).

254

3 Envelopes of holomorphy for special domains

Indeed, put

1/2
Let X := Xo U X1 U X2. Endow X with the topology induced from C x JR. Define X

A E C.

(A, 1)

Then (X, p) is a simply connected Riemann domain over C 19), Therefore, in virtue of the uniformization theorem (cf. LFors 1981], [Spr 1957]), there exists a injective holomorphic mapping F0: X —÷ C. Let 6,r

Z2

Finally define

G := {z E

q(z) = < 1/25).

< 1/25 or IVzJI — I +

Ilizil — I —

One can prove that G is a circular domain such that: = (A E • sections Az E G}, z E G, may be disconnected (they may have at most five disjoint components),

• C(G) = • for any z E G the element by z. say = • the mapping

E

G is

X from the definition of G is uniquely determined

(.3

z

EX

holomorphic (and p o C-) = q). is not univalent. Define It remains to show that

D := {z E

IIq(z)I

I

—

(z

ED: (0.6ir/25)flargq(z)

<

0).

Observe that D is a circular domain with connected sections, 8D C G, and Do C G.

For f E (9(G) define

f(z :=2n':1(f

ICI=c+

—f

—

Z E D, cj :=

l±t . 0 E (0, 3ir/2) fl argq(z). IIzJI

19)

The details of the construction are left for the reader.

255

3.1 Univalent envelopes of holomorphy

=f

on D0. Then f 0(D) and f Letfo 0(G). Fixaz

= F0 F0 o Pa.) o q(z) = const. Hence

we obtain fo(crz) =

Fo 0

8fo

(1—c, l+e)with0 <e << 1,

G. Thenforct o

o

p

=

o

8fo az2

zI—(z)+z2—-——(z) =0,

Fo 0

0 q(az)

=

z E G,

which, by the identity principle, implies that

zi—(z)

dfo

+ z2—(z) = 0,

8Z1

z E D.

dz2

* ( )

1/4/i), b := (13/15)a. Note thatq(a) = q(b) = —1, a. bE G, and the segment [a. b] is contained in 1). Thus, by (*), f is constant along [a, b]. In particular, fo(a) = fo(b). (—1,,r) = e(b). Hence fo(a) f0(b), On the other hand,®(a) = (—1,0) which implies that E(G) cannot be univalent. Take a :=

The following result generalizes Remark 1.9.6(a).

Proposition 3.1.21. Let (X0, be a 7-domain of holomorphy (7 C 0(Xo)). Let X be a univalent subdomain of Xo. Assume that 8 c 0(X) is such that TIx C 8. Let H: Xo x [0, 1] —* Xo be a continuous mapping such that:

H(.,t) E 0(Xo,Xo) for anyt E [0,1], 0) = idx0,

H(Xo, 1) C X,

H(X,[0, Ij) cx, a locally connected B C [0, 1] and a continuous Pnapping

there

T: C" x B —f such that

f€

T(.,t)

8,

t E B},

Aut(C")foranyt B, po(H(x, r)) = T(po(x), t)forany(t,x) E B x X, H(Xo,:) C Xfor anyt e [0, 1] \ B. Then

8) C Xo.

Proof (Cf. [Tra

phy. Note

that

1988]) Let a: (X, poix) —÷ (X, /) be the a is injecuve. Since (X0, po) is a 7-domain of holomorphy, there

exists a morphism

t: (X,Th —* (X0.p0) (r 0 a = idx,x0). We want to show that r is injective. It is clear that r is injective on

a(X).

256

3 Envelopes of holomorphy for special domains

Let H: X x B — E

[0.

X denote the lifting of HIXXB (cf. Proposition 1.9.4). For 1] \ B, put H(x, to) := a(H(r(x), to)), X E X. Observe that

H(cr(x'), 1) = a(H(x', 1)), (x', 1) e X x [0, p(H(x, 1)) = po(H(r(x). 1)). (x, 1) E X x [0, 1). '0) In particular, H is continuous on a(X) x [0, lJ, and consequently, by the monodromy theorem, H is globally continuous on X x [0, lJ. Notice that H(•. 0) = Recall that (9(X) separates points in X. Thus to prove that r is injective, it suffices to show that if r(bi) = r(b2) for some b1,h2 E X, then f(bj) = f(b2) for any

f

E (9(X).

X such that a point a = a(a') E a(X). Assume moreover that b1, b2 Eare Let f In particular, 13(b1) = j5(b2) =: (9(X). Fix = r(h2) curves = 1.2, with X, [0,11 —+ j = 1,2. j

Fix

two Let

Yf := jo

J=

1,2. Consequently. we have two curves af: [0,11 —÷ (9, yj(1)), E 10, II, j = 1,2 (cf. Example 1.3.1). We like to prove

c1(t) :=

i

=(12(I). Define

that

O
H(a,3t).

1/3<1 <2/3

I), 1).

—

2/3

H(b3, 3(1 — 1)).

=

Note that

a,

=

h3,

j=

j=

:=

1,2.

1,

1

1,2. Thus the curves

j=

1,2. induce two

newcurvesp3: [0,11 —p E [0, lJ,j = 1,2. := Note that po(X) for 0 < t < 2/3. Indeed, if 0 < t < 1/3, then t5j(') = t 2/3, 15(H(a, 31)) = po(H(r(a(a')). 31)) = po(H(a'. 3r)) E po(X); if 1/3 po(X) (because then = — I), I)) = — I)), 1))

H(Xo. I) c X). In particular, c5i(2/3) = 62(2/3) = po(H(co. I)). Recall that a(X) is univalent and jS(a(X)) = po(X). Consequently, by the identity principle for liftings. we get = (151a(x)r'(6j(t)) forO < : < 2/3, j = 1,2. Thus 6i(2/3) = 62(2/3) and

= F2(2/3).

therefore

Moreover,61(t) =82(1) =

1.

Let

0
a,

— 2u)/(3

—

4u)),

0

u

1/2, 2u/3

0
l)t), I),2u — I),

H(a, 3(2u — —

—

20

If x = a(x').

po(H(r(x).

1)):

l)(l

—

1)),

I

I — 2u/3

1—2u/3
1/2
then by the previous equality it remains to use the identity principle.

we have j(H(x, f)) =

1)))

=

3.1 Univalent envelopes of holomorphy

257

1,2. Hence po(a'), —

p(4(u t))=

2u)/(3

—

414))).

0
0 <

z

<2u/3 I — 214/3

214/3
O
Zo,

po(H(a'.

3(2u — 1)1)),

1/2

1, 0

u

I)), 2u — I)), 1/2 u — 1)(1 — t))), 1/2< u <

I

< 1/3

po(H(r(yj(3t —

1,

1/3

t

po(H(co, 3(2u

1,

2/3 <

t

is a homotopy joining yj with 1,2.

2/3 <

I

By virtue of the monodromy theorem, we get

0

Proposition 3.1.22 ([Tra 19881). Let G C D C C" be domains and let

H: D x [0,11 —+ C" be a continuous mapping

such

that:

H(',O)=idD, H(D x {1}) C G, H(.,:) C"), t E [0, 1], H(G x [0, 1]) C G. Assume that D is a domain of holomorphy Then: - (a) The envelope of holomorphy G of G is univalent, G C D, and H (G x [0, 11) C

G.

(b) if moreover

H(D x [0, 1]) C D, H(., t) is mi ectivefor each t

[0, 1),

(H(D x (t}), D) isaRunge pairfor each t

[0, 1),

then (G, D) is a Proof. (a) We argue as in the proof of Proposition 3.1.21 of the lifting follows from Theorem 2.12.1(b)). (b)

— Exercise (the existence

Fix a compact K C G. We want to show that

C G. We will prove

more, namely that x [0, 1]) c G (then

=

x {0}) C G). Let Jo := {t

[0. 1]:

x {t}) C G).

Obviously 1 Moreover, the continuity of H implies that is open. To prove that Jo is closed (and consequently, that Jo = [0, I]) it suffices to show that

\ (1) C {t e [0, 1): H(KO(Th x {t}) C L}.

258

3 Envelopes of holomorphy for special domains

where

-

L

:= (H(K x [0.

(since G is a domain of holornorphy, L is a compact subset of G). SE [0, 1) and suppose that x {t}) c G. Since H(•, t): D —÷

Thus, fix a H(D x {:})

is biholomorphic, we get x {t})

= (H(K x

O(H(Dx(tl))

Since (H(D x {t}), 1)) is a Runge pair, we get

x (t}) = (H(K x {1}))0(D) Finally, since H(KO(Th x (t}) C G, we have x {t})

= (H(K

x

C L.

Corollary 3.1.23 (cf. [AIm 19221). The envelope of holomorphy of any starlike domain G C C" is univalent and Runge. Proof

TakeD := C" and H(z,r) :=(l—t)z,t E[0, l],z

C".

0

3.2 k-tubular domains A domain G c C" is called to be a tube domain (cf. Section 2.3) if G is of the form G = B + iR". where B is a domain in R". We say that B is the basis of G. Observe that G is invariant under all mappings z —÷ z + it, t R". It is well known that the envelope of holomorphy of G can be realized as a domain in C" and, even more, it is explicitly given as = conv(B) + iR". This result was proven by K. Stein (cf. [Stei 1937]) in the case n = 2, and later by S. Bochner (cf. [Boc 1938]) in the general case (see also S. Hitotumatu [Hit 1950]). In this section we intent to generalize the definition of a tube domain and to discuss the envelopes of holomorphy of such general tube domains (cf. [Abe 1985], [Coe-Loe 1986], and [Sac 1989]). We shall see, that the results we get during this discussion also contain the classical ones.

DefinItion 3.2.1. Let k, n E N, I k
(a)ax(•,O) = (b) p(ax(x, 1)) = (pi(x), p2(x) + it), x E X, t 21)

Compare Example 1.9.11 with g = Rk.

3.2 k-tubular domains

259

We note that if X = G is a domain in C", then G is a tube domain (in the classical sense) if and only if it is an n-tubular domain over C" w.r.t. to aG: G x R" —* G, aQ(z, t) := z + it. in the sense of this new definition. Often, instead of using the term n-tubular domain, we will say that (X, p) is a lube domain over C". Moreover, we point out that the map ax is uniquely determined (cf. Proposition 1.9.12).

Remark. Let (X, p) be a k-tubular domain over C". Assume that it is isomorphic to a Riemann domain (X', p'). Then (X', p') is, in a natural way, again a k-tubular domain over C" (Exercise).

Let (X. p) be a k-tubular domain over C" and denote by ax the corresponding map. Wefixapointxo E Xandputzo := pI(xo) E =: (pi. p2). Then Ck. := isspreadoverC" viap2: denotes the connected component of the fiber p2(x) := p2(x). Moreover, if that contains the point then X(x0) := is a k-tubular domain over Ck w.r.t. the map x

—*

ô(x,t) := ax(x. t).

Therefore, a k-tubular domain may be thought as a family of fibers consisting of unions

of tube domains over Ck parameterized by points in an open set in Cn_k.

Example. Let (X, p) be a Riemann domain over Cn_k, and let c C X x Ck be then a subdomain satisfying the following property: if (x, z) E c and if t E Put aç2: c x —+ z), I) := (x, z + it). Then (x. z + it) E (p x idck)k2) is a k-tubular domain over C" w.r.t. ac2. For further use let us recall Proposition 1.9.12: Let (X, p) and ax be as above. Then ax gives a group homomorphism IR" —÷ Aut(X), ã(t) := ax(•, I); in particular. if t', t" E Rk then ax (ax (x, I'), t") = cX(x, t' + t"), x E X.

The aim of our discussion on n-tubular domains is to see the univalence of their envelopes of holomorphy. In analogy to a Riemann domain over C" we introduce the following notation: a where B is a connected Hausdorif space and where w: B —+ R" is a pair (B, local homeomorphism, is called a domain over R". Having such a domain (B, (p) over III." there is an associated n-tubular domain over C".

Lemma 3.2.2. Let (B, be a domain overlR". Put X := B x and p: X —* C", p(x. y) : = + iy. Then (X. p) is an n-tubular domain over C". Proof

Take

andtElR".

ax: X x R" —+

X as ax((x, y), t) := (x, y + 1), where (x. Y) E X U

We say that such (X, p) is a tube domain over C" with basis (B. Inversely, any n-tubular domain can be thought as a tube domain over C" with some basis (B. cQ); namely, we have the following lemma.

260

3 Envelopes of holomorphy for special domains

Lemma 3.2.3. Let (X, p) be an n-tubular domain over C" via ax: X x —÷ X. Put B := p'(p(X) fl (IR" + i(O})) and p: B —* IR", p := ir o where ir: C" =R" +iR" —+ R", ir(x+iy) :=x. Then(B,ço)isadomainoverR",and (X. p) is isomorphic to the tube domain over C" with basis (B,

Proof Set B' := p(X) fl (R" + i{O}); then B =

We

claim that p(X) =

B' + iR".

if p(x) = E + iii. then p(ax(x, = p(x) — = R" + i{0), i.e. B' or p(x) E B' + ilR". Conversely, E B' + ilR", then = p(x) for + some x E X. Hence p(ax(x, = p(x) + iij, i.e. + E p(X). In fact,

Next, we introduce the following two mappings:

F: B x R" —p X, G: X —p B xlR",

:=

'i).

G(x):=(ax(x,—Lmp(x)),Imp(x)).

It is clear that F and G are continuous mappings and that F o G = jdx and G o F = id8 Thus B x R" is homeomorphic to X. In particular, B is a connected Hausdorif space.

Let x E

B

C X. Then there is an open neighborhood U = U(x), such that

pIu U —* p(U) is a topological map. In particular, the set p(U) fl (IR" + i{O}) = p(B fl U) is open in R"

R" + i{O}. Thus (PIBn(J:

BflU

—+ ço(B flU)

is homeomorphic; i.e. (B, ço) is a domain over R'1. Therefore, B x R" —÷ C", j3(x, v) := ço(x) + iy, gives a tube domain over C" with basis (B, Observe that p o F = i.e. (B x IR", ,3) is isomorphic to (X, p) via F: B x

0 In virtue of the classical result of S. Bochner, (conv(B) + iR", is the envelope of holomorphy of (B + iR", idB+jRn), i.e. the classical tube domains have schlicht envelopes of holomorphy. We shall prove the same result for n-tubular domains over C". Here we will follow a paper of M. Abe (cf. [Abe 1985]). We mention that there is another approach to this result due to G. Coeuré & J. J. Loeb (cf. [Coe-Loe 19861).

Theorem 3.2.4. Let (X, p) be an n -tubular Riemann—Stein domain over C". Then p is infective and p(X) fl (R" + i{O}) is a convex domain in IR" R" + i{O}; i.e. (X, p) is isomorphic to the classical tube domain (p(X) fl (R" + i{0})) + ilR". 22) Proof Let B' := p(X)fl (IR" + i{0}) and B :=

Then, in virtue of Lemma 3.2.3, (X, p) is isomorphic to the tube domain (Y, q) := (B x IR", ç' + i . with basis (B, q), where := p18: B —÷ R". 22)

Here, and later on, p( X) fl (R" + 1(01) is identified with a subset of P".

3.2 k-tubular domains

261

We have to prove that q is injective. Let Then there exist a continuous curve y: [0, 1] —÷ B xI E B with xo y(l) = such that: and N e N with y(O) =

• •

is injective in a neighborhood of o is the line segment connecting

N:

j

I

:=

with

o

1 <j
o

[0, 1] —* B. = x0, We will show that there are a continuous curve ii) with the following properties: = and an open neighborhood U of

• •

with

is the line segment connecting

o

in particular, q(xo) C Obviously, we may assume that N ? 2. Fix a r E (0, 1] and put

: = {Yi + s(yo — yi) + A : = {r E (0, 1]: C B,

—

yl): 0 s.f and s + t < r}. compact, y(l/N) E K1,

such that 'PIKr injective and

= A1).

< I such that (0, To] C A. is a local homeomorphism, there exists 0 < Moreover, we point out that if is injective on a compact subset L C B, then remains to be injective on an open neighborhood of L. In particular, this implies that for some A is an open subset of (0, 1]. Therefore, either A = (0. I] or A = (0, rj E (0,1]. Suppose now that A = (0,ri). < < Then, using the uniqueness of the lifting, we obtain Let 0 < Put D K1. Then RIo: D —÷ A1, is a topological map. C U1<1, Now, we consider the following holomorphic map g: C2 —+ C", Since

:= y' + Observe that

(i

—

—

A) =

Yi) +

A) + (i/2)77A(y2 —

—

yo) for

VI).

= + irj

E

C.

C, A E C, and

E [—1,1] x

€A11

Set x [0, ri)

—+ Y,

r)),

+ Ui. t) :

—

yo)).

Then f is continuous and the equation q o f = g holds on E x [0, ti). Because of Proposition 1.5.8 there exists a continuous extension

f: E x [0.rjj

—÷ Y := YU.dY,

of Y w.r.t. q. Since where (Y, is the leads to the same equation on E x [0. TI].

of = g on E

x 10, TI), continuity

262

3 Envelopes of holomorphy for special domains

Observe that g(—l, r) = y' + r(yo y([O. 1/Nj) x {0} C Y forO < r <

f(—l, therefore, f(E, r1)

r

/

— yi). 0

r

1, and therefore f(—l, r) E

Hence

r)

y((O, 1/NJ) x

{0} C Y;

dY. In virtue of Theorem 2.9.1(11) we get f(E, r1) C X.

Sincef(1,ri)EYandf(—l,r1)EY,itfollowsthatf([—l,IJx[0,r11)CY.Put K :=jriof([—l, l]x[O,r1]),whereir1: Y = BxW1 —÷ Bdenotestheprojection onto B. Then ço(K) =

Hence we have seen that TI

E K, and çc'Ix is injective. A. Therefore, A = (0. 1]. In particular, there exists

acontinuouscurvey2: [0,1] —k Bwithy2(0) =xoandy2(l) = y(2/N),suchthat is injective on a neighborhood of y2([O. I]) and Yo

is the line segment joining

o

and V2.

Repeating the above argument we finally see that there is a continuous curve yN: [0, 1] —* B joining xo with xi, such that is injective in a neighborhood of to w(xi). In particular, YNUO. I]) and so that ço o YN is the line segment from p(xo) W(xI) and, moreover, Lw(xo),co(xi)] c w(B) = B'. So is globally El injective, and B' = w(B) = p(X) fl (R't + i{0}) is convex.

Corollary 3.2.5. Let (X, p) be a k-tubular Riemann—Stein domain over C't, and let Xo E X. Then X(xo) is a k-tubular domain of holomorphy over Ck, B(x0) := fl (Rk + i{0}) C Rk is convex, and X(xo) is isomorphic to (B(xo) + iRk id).

Remark. Already in IBre l954], H. J. Bremermann had mentioned the following extended domain' over the shell example: Let (B, p) be a

0
:= B x R2 over C2 with basis

(B,çp)isgivenas(z E C2: < example of a Riemann domain over C2, whose holomorphic functions do not separate points. In the last step of our discussion on n-tubular domains we shall determine their envelopes of holomorphy. For this reason we recall the following result (cf. Proposiw.r.t. ax. Then its envelope of tion 1.9.15): Let (X, p) be a k-tubular domain over X x Rk X, where holomorphy (X, is also a k-tubular domain over C't via Rk. t): X —÷ X,t X —+ Xis the holomorphic extension

Now we are able to prove the main result on tube domains over C".

Theorem 3.2.6. Let (X, p) be a tube domain over C" with B

:= p(X) fl (IR" + i(0}) C R".

Recall our convention that we identify JR" + i 10) with R".

23)

3.2 k-tubular domains

263

:=convBandX := B+iR'7. Thenp: (X,p) —+ (X.id*) is an envelope of holomorphy of(X, p).

Proof Let a: (X, p) —+ (X, j5) be an envelope of holomorphy of X. In view of the above remark we know that (X,

is a tube domain over C'1. Put X' :=

and

B' := X' fl and p(X) C X'. + i(O}) c R'1. Then X' = B' + is a morphism between Riemann domains over Thus p :(X. p) —+ (X', C'1. Since (X, j3) is a Riemann—Stein domain, we know that is injective and that xI is convex, i.e. X' is a domain of holomorphy. - Let (9(X). Thenthereexistsan f E (9(X) with ba = f. Put F X' C. Fisholomorphicon X'with Fop = on X. Hence (X', idx') is an envelope of holomorphy of (X, p) via p: X —÷ X'.

f€

f

Since X' is convex, it follows that p(X) C X C X'. (9(X). So there exists an h (9(p(X)); then g o p Let g O(X') with h o p = g o p, i.e. h = g on p(X). Hence every function g (9(p(X)) is the restriction of a function h (9(X'), which implies, since X is convex, that X = X'.

0 So far we have seen that every tube domain (X, p) over C" has a schlicht envelope of holomorphy, namely, conv(B) + ilR'1. where B := p(X) fl (IR" + i{O}) C R". In

particular, if X = B + iIR" is a classical tube domain in C", then conv(B) + iR" is its envelope of holomorphy; i.e. the classical result of K. Stein and S. Bochner is a special case of what has been proven here. One might ask how to describe the envelope of holomorphy for a generalized tube domain B1 + iB2 c C", where B1, B2 are domains in IR" (cf. [Kaj 1963]).

Theorem 3.2.7. A domain B1 + i B2 is a domain of holomorphy ([and only ([the B3 are convex. Proof We only discuss the non-trivial implication, i.e. we start with the assumption that B1 + i is pseudoconvex. The proof proceeds via several steps. Claim 1. Let

X (af, bj)

C

C" be a non-pseudoconvex domain and

Then X2 := B +i((ai.di) x X(aj,bj))isnotpseudoconvex.

let d1 E Otherwise,

K2 andthenalso K3 := B+i((ai +bi —dp,bi) xX(aj. bj))would

be pseudoconvex. Since (X2 \ X3) fl (X3 \ X2) = 0, it follows that K1 = X2 U X3 is pseudoconvex 24); contradiction.

Claim 2. Let X1 (as above) be a non-pseudoconvex domain and let e E (0, 24)

Exercise.

Fora proof see lPfl I975b1.

3 Envelopes of holomorphy for special domains

264

+ e) x X(aj. bj)) is not pseudoconvex.

Then B + i((O. Use

Claim I and induction over k.

Claim 3. Let X1 (as above) be not pseudoconvex and let 0 <

—

<

— ai.

x X(aj. bj)) is not pseudoconvex.

Then B +

Choosek EN such that If k > 1, observe that

Ifk = l,use directly Claim I.

<

(bI._aI fore < min{b'

—(h1 _ai)/2k. (b1

Then Claim2together with Claim I

gives the desired result.

Applying Claim 3 step by step we arrive at

Claim 4. Let B+i j

=I

bj)benotpseudoconvex andletO <

bj —ai,

n. Then B + i X(a. b) is not pseudoconvex.

Claim 5. For a non-convex domain B C R'1 there exists a positive number r0 such

that for all r >

the domain B + i X (—r, r) is not pseudoconvex. 1=' < < Otherwise we would find real numbers r1 < —÷ oo, such n

that the domains Xk := B + i X (—rk, rk) are pseudoconvex. Observe that Xk C 1='

Xk+I. Then, in virtue of Corollary 2.2.11, it follows that B + iR'1 =

is

pseudoconvex. Hence, by Theorem 2.3.1. B is convex; contradiction.

Claim 6. If the domain B + i X (aj.

is pseudoconvex, then B is convex.

Otherwise, using Claim 5 there is a positive number r with the following properties: n

• B + i X (—r, r) is not pseudoconvex,

j=l

it.

Then Claim 4 gives the contradiction. Now we are in the situation to conclude our proof. Since —i(B1 + iB2) = B2 + i(—B1) is pseudoconvex, it suffices to show that B1 is convex. Choose numbers a3.

3.2 k-tubular domains b3 so

265

that X (aj. b3) c B2. Then the domain

B1

+i X(aj,bj)=(Bi +iB2)fl(R"+iX(aj.bj))

is pseudoconvex; hence, by Claim 6, we conclude that Bi is convex.

0

Remark. It would be interesting to know whether the above result remains true in the following general situation: Let (Gd, 1, 2, be domains over R". Define X := G1 x G2 and p: X —÷ p(x) = p(xI.x2) := 9i(xj) + içp2(x2). Then

i

(X, p) is a Riemann domain over C". Assume that X is pseudoconvex. What does this imply for (G1, In [Kaj 1963), J. Kajiwara claimed that the envelope of holomorphy of a domain Bi +i B2 is given by cony Bi +1 cony B2. In [Pfl 1974] it was mentioned that, in general,

this statement is false. Instead of presenting the rather complicated counterexample of [Pfl 1974] we give here a much simpler one. we learnt from B. V. Shabat (cf. [Sha 1976]). Put B1

: = (x ER2: 1/2< lxii < I),

82: = {y = (yi. Y2) ER2:

< 1/4.

y3i

j=

1,2).

that the function f: G := + iB2 —÷ C, 1(z) := (zf + is holomorphic on G, but there is no holomorphic function F: cony Bi + lB2 —÷ C

Observe

with FIG = 1. We point out that

it is not clear to us whether G has a schlicht envelope of

holomorphy There is another counterexample due to J. Siciak (communication) which is. in addition, starlike. i.e. its envelope of holomorphy is univalent (cf. Proposition 1.9.6(c)). Namely:

LetD:=(z=x+iyeC: lxi X

<

< l}andput

:= (D x (—1, l))L1((—l, I) x D) C

We denote by h the solution of the Dirichlet problem for D \ (—1, 1) with boundary

valuesOon (—i, I)and I on 8D\(—1, 1). Observethath issubharmonicon D. Then X := (z E

Dx D:

±h(z2) < 1}

is a domain of holomorphy containing X. The function z = x + iy —+ F'(z) — y is harmonic on

and h(z) > y, but h(z) $

y

:= (z E D: y > 0) on dD+. Therefore, h(iy) > y if 0 < y < 1. As a

consequence we get that the point (1/2, 1/2)

X.

266

3 Envelopes of hotomorphy for special domains

Wechoosee E (0. l)andp > Ososmallthat(i(l—e)/2.i(1—e)/2) with B1 := {x ER2: 1x11 < I — e) and B2 := {y E 1R2: we

Xandthat, < 1—rh

have that G := B1 + lB2 C X. Then the envelope of holomorphy ç cony G.

of 6

exists as a subset in C2 and it is contained in X; i.e.

Before we proceed with the general discussion on k-tubular domains, we recall the following classical result (cf. [VIa 1966].

PropositIon 3.2.8. Let B c

n

be a domain of holoniorphy. and let

—V: B —+

v, be

2.

upper semicontinuous functions with v < V on B. Set

G :=

{z

=

+iYn) E B x C:

Then the envelope of holornorphy

E(G) = {z =

<

<

of G is univalent, and it is given as

B xC:

+i'vn) E

<Xn <

where C' (respectively V) is the largest plurLsubhannonic minorant (respectively the

smallest plurisuperharmonic 25) majorant) of v (respectively of V) on B.

Remark. Proposition 3.2.8 shows that a 1-tubular domain G need not to be a domain of holomorphy, although each of its fibers := {z,, E C: < x,, < B, is a convex domain. In order that G is pseudoconvex, it is necessary that the boundaries of G2 (as a function on B) fulfil an additional 'convexity'—condition. Proof of Proposition 3.2.8. The proof will be given in several steps. Step I: Assume that G of Proposition 3.2.8 is a domain of holomorphy. We shall show that then v and — V are necessarily plurisubharmonic functions on B: By our assumption we know that G is pseudoconvex, i.e. ço := — log dist(., is plurisubharmonic on G. Observe that is independent of the coordinate y1,. i.e.

+

=

=

for all

z

Zn) E

Bx

+

G.

Define

D :=

(z

=

<

<

C:

F: G D, F(z) := (E. exp(zn)), is a locally biholomorphic surjection. Set 1k: D —+ := Observe that 1k is well defined and that it is a plurisubharmonic exhaustion function on D.

Then

Therefore, D is a pseudoconvex domain. E B and choose a ball U = Fix a such that v(f) < < < that a function U: G —p (—oc,

r)

B and a real numbers rj

z E U.

oc] isplurisuperharmonic if —u € ?S3€(G).

3.2 k-tubular domains

267

Di := U x K(r2) and 1)2 := (U x C) fl 1) are pseudoconvex domains with 0 and (Di \ D2)fl (D2 \ D1) = 0. Therefore,

Then

Di fl 1)2

13

:=

EU xC: IznI

UD2 =

Applying Theorem 2.2.9, we conclude that for

is pseudoconvex

= —logexp(V(.)) = is psh on U, i.e. — V is locally psh on B. Hence — V Z

:= (0,... ,O, 1)

—V(S)

E

To see that also v is a plurisubharmonic on B we only mention that the mapping provides a biholomorphism from G onto the pseudoconvex = (Z. Zn)

domain {z e B xC: —

V(Tz)

<

Step 2: Conversely, assume that v and — V are plurisubharmonic functions on B.

Since B x C is pseudoconvex it immediately follows that
Step 3: Now we discuss the general case described in the hypothesis of Proposition 3.2.8: Let f be a holomorphic function on G. For E B we choose the largest strip

:= {z E C:

<x <

such that f(E..) is the restriction of an Let v1 and —V1 be the upper semicontinuous regularization of v[ and of _v(, respectively. Then v[ < vj t' < V < V1 < V(, and f7- is holomorphic on the strip :=

lz E C: vf(z) <x < V1(z)}.

Hartogs' theorem on separate analyticity (cf. Lemma 1.8.14) it is easily seen that the function Using

f: G1 :=

(z

BxC:

< Xn <

—+ C,

1(z) :

is holomorphic on Gj. So far, we have seen that any f E (9(G) is the restriction of a function f

GcG1. 27)

28)

Compare the footnote to the proof of Theorem 3.2.7. Observe that z —' — and z — — belong Observe that such a strip always exists.

to PSR(B)

(9(Gj),

268

3 Envelopes of holomorphy for special domains

Assuming that v1 and — where

V1

are piurisubharmonic on B it is clear that G C B x C:

G :={z On

<

in virtue of Step 2, we know that O is a domain of holomorphy G. Therefore, G is the envelope of holomorphy of G.

the other hand,

with

Gc

To see that vj and — V1 are plurisubharmonic functions we only have to show that G1 is a domain of holomorphy and then to use Step 1. That G1 is a domain of holomorphy follows from the fact that B is a domain of holomorphy and the strips D S((.) were chosen to be maximal (Exercise).

The result in Proposition 3.2.8 has an analogous formulation in the case of Haxtogs domains.

Corollary 3.2.9. Let B, v,

G :=

(z

=

V.

and V be as in Proposition 3.2.8. Put

E B

xC:

of G is univalent and it is given as

Then the envelope of holomorphy

= Proof Denote G.

Then it

{z

=

Zn) E B

by G the domain

<

<

xC:

that

<

is claimed

to be

IZnI

the envelope of holomorphy of

is clear that G is pseudoconvex.

Put

D :=

{z

E

B xC:

G}

=

{z E B

xC:

<x,, <

Let f E (9(G). Define g E 0(D) as g(z) :=

2,ri) =

In virtue of E.(D). Call this extension Then function g(z),

z

D.

exp(zn)). Observe that Zn + 3.2.8, g extends holomorphically to E(D). Therefore, the + 2iri) = z

Proposition Zn

f(z)

is

Z E

well defined on G. Obviously, f E 0(G) and fIG

Remark. Observe that

the situation discussed in following sense: (a) G is a 1-tubular domain, (b) its basis B is a domain of holomorphy, and (c) for each E B the fiber := E C:

G,

= f.

Hence

Proposition

E G}

=

O.

0

3.2.8 is special in the

is connected.

The remaining part of this section is devoted to the study of envelopes of holomorx Ck, I k < n, which are special k-tubular domains

phy of domains G C

3.2 k-tubular domains

269

in the sense that G + (0, = G 29)• Our goal is to find a Riemann—Stein domain (XG, j3) over and an injective morphism

F: (G, idG) —f (XG

X

J3 x idck)

such that the envelope of holomorphy of G can be given in XG x C" as a 'good' family of convex tube domains in parametrized by the points of XG (cf. [Cas-Tra 19911. [Cas-Tra 1994b], [Mok 1986), and [Sac 1989), see also [Chi-Shc 19971). 3°)

In order to formulate and to prove that result we will need the following two lemmas.

Lemma 3.2.10. Let (X, p) be a Riemann—Stein domain over C" and let G C X be a subdomain. We assume that the envelope of holoinorphy =: G of G is given as a subdomain of X. isan upper semicontinuousfunction. there existsa largest Then, if u: G —÷

plurisubhar,nonic function

with âIG
Proof Fix xo E G and Jo E N such that u(xo) < Jo. We denote by j > Jo. the connected component of {x E G: u(x) < j} that contains xO. Then G1 C G G = UJ>JO G1. Set a3 as the connected component of the interior of the intersection of all subdomains of holomorphy of X containing G3 that contains G3. Then is an increasing sequence of domains of holomorphy. Therefore, O := U G1 is a domain of holomorphy (cf. Corollary 2.2.11 and Theorem 2.5.7) with

GcGcG. Hence,O=O.

Let v: G —+ be a plurisubharmonic function with vIG u. We denote by G1(v) the connected component of (x G: v(x) < j}. J > Jo. that contains G(v) a domain of holomorphy. Therefore, xo. Obviously, C C G is the increasing union of the domains G3, J and any function v E vIG
is locally bounded from above; hence it has a maximal element(cf. Proposition 2.1.16). D

Another proof of Lemma 3.2.10 is given in [Shi 1975].

To prepare the formulation of the second lemma we start describing the situation we will be interested in: Let (S. p) and (X. r) be Riemann domains over and C", respectively. Let F: (X, r) —p (S x Ck, p x idck) be a morphism. Moreover, we assume that (X. r) is a k-tubular domain over C" via aA': X x —p X (cf. Definition 3.2.1).

xck)flG. ERk}. We should note that in ICas-Tra I 994bJ one deals with the problem of the cxplicit description of the envelope of holomorphy of a K-invanant domain C X, X a Stein manifold. K here is a compact Lie group whose complexification Kc acts holomorphically on X. 29)

270

Fix an x

(p X

3 Envelopes of holomorphy for special domains

X. For t

we obtain

E

o F(ax(x, t)) = r(aX(x, t)) = r2(x) + it) = (po F1(x), F2(x) + it) = (p x

172(x) + it),

wherer = (rj.r2) and F = (F1, F2). Since F(ax(x,O)) = F(x), uniqueness of the lifting implies that F o aX(x, t) (F1(x), F2(x) + it) for all t E Therefore, we may think of F: X —÷ S x CA as a k-tubular domain spread over S x Ck wr.l. the map We say that two points x',

x" E X are equivalent

F and we write x'

x",

if Fi(x') = F1(x") =: SE Sandifx',x"belongtothesameconnectedcomponent of As usually, X/ is provided with the quotient topology. We define F: X/— —+ S as F([xJ) := F1(x) and F: X/'-'- —p asF([x]) := here [*J denotes the equivalence class of * w.r.t. '-'-. In general, X/ is not a Hausdorff space. For example, take

X:=CxC\{(O,z2):Rez2=0}, r:=idx, ax: Xx R —÷ X. cix(z,t) :=(z1,z2+it), (S,p) :=(C.id), andF :=r. Observe that X is not a Riemann—Stein domain. On the other side, if X is Riemann— Stein, then we have:

Lemma 3.2.11. Assume the situation as before and, in addition, that (X. r) and (S. p) are Ricmann—Stein domains. Then (X/ —', F) is a Riemann—Stein domain over

F: (X/ -'-. F) —÷ (5, p)

is a morphism between Rieniann domains, and

*:

X —b X/—.

is a holomorphic map with F1 = F o

Proof Obviously, the map

*(x) := [x],

and ri = F o

is continuous. To prove that

is open we have to show

that:

if x', x" X with x' —' x" and if U = U(x") is a sufficiently small neighborhood of x", then there exists a V = V(x') such that

x"-i}. In fact, we may choose U = U(x") such that Flu: U —* U1 x U2 is a homeomorphism, where U1 and U2 are open connected neighborhoods of so := F1 (x') = denote the connected compoFi (x") E S and of F2(x") E Ck. respectively. Let nent of that contains the points x' and x". We recall that is a k-tubular domain over Ct via CIx xRA

—*

a := ax,

3.2 k-tubulardomains

271

and that it is a domain of holomorphy. Therefore, Theorem 3.2.4 implies that F2IXS0

is a classical convex tube domain in Ck.

is injective and that q: F2(Xs0) —+

denote the inverse mapping of

Put ã(r) := rF2(x') + (I — r)F2(x") and a(r) := o &(r), 0 < r < I. Then so' connecting x' and x". a: [0, 1] —÷ X

Fort E

[0. 1], we find a neighborhood V

)< Vi"

denote a connected open neighborhood of so with center&(r), respectively. Since a([0, 1, ]) is compact, we a([0, 1]). Choose an open TN E [0. 1] such that

is a homeomorphism, where V1' and

and an open ball in

can fix points rj connected neighborhood V1 C

fl U1 of so in S. V1'

fl 0 for j Assume that continuously into X. such that

F

If we take a a(ro) =

=F

k. Then

°cok on

V' x

and ipk map V' x (V' fl fl V1).

&(ro)) =

fl then [0, 1] with &(ro) E ã(rO)). Hence, çoj and çok coincide on V x

constructed a global section

V'

V1 x

—+ X, i.e. F o

=

o

So, we have

fl

=

N

j=I

Choosenowanopenconnectedneighborhood V" = V"(&(l)) C U7=i

Tj

V'. Then

x V"). then F1(x) E V' x V") is an open subset of X containing x'. If x E and x can be connected 'over F1(x)' with a point i U. Thus, we have shown that is an open map. The next step consists in proving that X/— is a Hausdorff space: X/—' be two different points. =: s". We choose disjoint neighborhoods Suppose that s' := are disjoint open sets V' = V'(s') and V'1 = V"(s"). Then and over V' and V". Hence ifr(Fj' (V')) and *(F1' (V")) are disjoint neighborhoods of and

i". respectively. =

=: SO we fix representants x' and x" in X of

In the case

= [x') and

= [x"]. In view oft'

and

know thatx', := Fj' (so). Using the fact that x" belong to different connected components of X is a k-tubular domain spread over S x Ck we may assume that u := F2(x') and

s", respectively; i.e.

we

F2(x") belong to + i{O} Wechoose neighborhoods V' = V'(x'), V" = V"(x"), U = U(so), W' = W'(u'), and W" = W"(u") such that

Fly': V' are homeomorphisms.

U

x W',

Fly": V" —*

U x W"

272

3 Envelopes of holomot-phy for special domains

We claim that if U = U(s0) is sufficiently small, then

3iEv': x

(x

X — i}

fl {x E

i.e. iJi(V') fl = 0. Suppose that this is not true, i.e. there are points xj

X with

= 0,

:= Fj(x1) —÷

so

3—400

and

V" such that

x5 E V',

= Fi(x7)

=

—x, x1 '-.-x.

—+

J—400

and F2(x7) are points in

Without loss of generality, we may assume that

+ ilO}-

denote the connected component of Flt (s1) that contains and xi'. Then, as above, F21x, is injective and F2(X1) is a convex tube domain in C', j E N. Put Let

ã1(A) :=

(1 —

A)F2(x) + AF2(x7).

a1(A)

:=

o

(—e. I + e) + i(—e, e) C C, e positive and small. Observe that aj: D —f X is a holomorphic map. If Px denotes the distance function on X w.r.t. the norm I I' then h3 := — log Px° aj are subharmonic functions on D with = h1(ReA). Therefore, hjlEo.ji is a convex function, which implies that h(r) < h1(l)} for aJi r E [0. 11. —* x" we conclude that there is a positive number ro Since x —# x' and where A belongs to D

such that px(x) > ro and px(x7) r

PXfrIj(T)) >

is large.

Combining these facts we obtain for all

N such that

a

• • •

if j

x,

ro/.,ñ) 3 x7. x", 3(&j0(t), J &([0, 1]) U I]). j ? ,jo. where &(r) := (1 — r)F2(x') + tF2(x") forO < r < I. Forany r E [0. l]wechooseaneighborhood U(r) := U(aj0(r)), maps U(r) biholomorphically onto V(r) := x Observe that U(0) V(r) —÷ U(r)the inverse W

mapping of

= roOt"

Then x

on

Moreover, there exists a number

between r' and r" with

11 3(a30(r"),

&j0(f)) = Or"(P(Sj0).

273

3.2 k-tubular domains

Hence, there isaholomorphic mapping 0: W := B(p(s10). = for all r [0. 1]. and 0(ifr(s10), r oO = 1] —* := O(p(sj). &1(t)), r E [0, I.). and define [0, X as Fix aj ? Jo 0 r a1(t) fort E [0, 1] and, since E U(0). we r 0 ã1(t)) = Then = (p(Sj).

=

conclude that

curve

in X with

= a1,

Therefore,

:= 0(p(so). &(r)), r E

Set

[0,

=

j

jo.

1]. Then

=

[0, 1] —* X

= x' and

is a continuous

$(l) =

x". In virtue

of

urn F1(a1(r)) = lim

J-_.o0

J—poO

=

so

=

urn

FoO(r(s3), ä(t)) =

it follows that fi connects x' and x" over so'.

F1

contradicting

rE

[0.

II,

our assumption that

Recall that F 0 = F1. Thus F is continuous and open. Fix a = [xo] E X/ maps U biholomorphiWe choose a neighborhood U := U(x0) c X such that = [x"J with = [x'j cally onto a product domain U1 x U2 C S x Ck.

x',x" E

U.

Then

= F 0 ifr(x') = Fj(x')

F1(x") =

F: (U) —÷ U1 is a topological mapping. So we conclude that F: X/ = F it follows that is a Rieis locally homeomorphic. In virtue Obviously. is holomorphic. mann domain over We want From now on. we use that (S. p) is a Riemann—Stein domain over isRiemann—Stein. This will be done by proving that X/ is to show that (X/ -S-', locally Stein over S via F (cf. Theorem 2.9.2(p4)): Fix ans E SflFi(X). Wechooseaneighborhood V = V(s) C Ssuch that is Stein. (Recall that (X, p) is Riemann—Stein.) Therefore, we have a plurisubharFor monic function u: R which is exhausting. Let V := E V put Hence S

:= inf{u(x) :x E

V'(x) =

It is clear that ü: V —÷ R is an upper semicontinuous exhaustion function.

We choose a neighborhood W = is Stein. W C V such that F: W —÷ W is biholomorphic and so that F) is univalent over W x Ck. Thus, it is a connected We mention that and therefore pseudoconvex. Using the minimum principle of component of C. Kiselman (cf. Theorem 2.3.2) we conclude that ü is plurisubharmonic on W. - Hence ñ E PS.7e(V). and ü exhausts V. Thus, Theorem 2.2.9 yields that V = (V) is a pseudoconvex domain. implies that (X/ —S. 1) is a Finally, since .c was arbitrary, Theorem El Riemann—Stein domain, which finishes the proof of Lemma 3.2.11. Fix a

E V.

F'

Applying Lemma 3.2.11 it turns out that any k-tubular Riemann—Stein domain over has a special form.

274

3 Envelopes of holomorphy for special domains

Corollary 3.2.12. Let (X, r) be a k-tubular Riemann—Stein domain over C". Then there exist a Riemann—Stein domain (Y. q) over Cu1_k, a domain G C Y x Ck, and an isomorphism F: (X. r) —÷ (G. (q x idck)IG) with F oax(-, t) = (F1, F2 + it), tE

Pmof We apply Lemma 3.2.11 with (S. p) =

id)

and F = r. Put (Y, q) :=

F). Then

(X,r) —+ (Y x Ck,q x idck) is a morphism

is the quotient map from Lemma 3.2.11). Fix an xo E X. Observe that r2) is a Riemann—Stein tube domain over Therefore, in virtue of Theorem 3.2.4, r2jX,1(,0)(XO) is injective. Define

F: (Kr) —÷

(Y x

Ck,q x idc&),

F(x) :=

XE X.

Then fr is injective and therefore it is an isomorphism between (X, r) and the k-tubular

0

Riemann—Stein domain (G := F(X),(q x idc*)IG) over C".

In the sequel let G be a domain in Ce_k x Ck satisfying the following property: G —+ G + (0. iiRk) = G. By we denote the projection of G into We define the following equivalence relation on G:

For two points z = (z'. z"), w = (w', w") in G, we say that

az1

n—k-fl<j
Observe that,

• ifz

w, then z' = w'; • if the points z, w belong to the same connected component of then z

We write G/

w.

for the corresponding quotient; the quotient mapping is denoted

by q: G —p G/ G/ is provided with its quotient topology. Cn_k isthemapthatisinducedbyiri,i.e. poq =r1. p:

MoreoveT,

Lemma 3.2.13. In the situation above, the pair (G/ p) is a Riemann domain over p) is holomorphic, and G/ q: (G, idG) —p (G/ is holomorphically separated. Pmof In a first step we will show that the mapping q is open: Take an open subset U C G. We have to verify that the following set {z

is

open. Fix

=

E G : 3wEu: z

= (i'. i") E G and th = (ti", th") E U such that

in particular,

275

3.2 k-tubular domains

Choose r > 0 such that Vt :

Bn_k(z

,

r) x

r)

Now,letf E O(G)with are

G and

C

=OonGforalln —k+1

f(z', z"). z =

(z'. z")

j

r) —f C such

holomorphic functions g, h:

g(z') =

r) x Bk(ti", r) C U.

V2 :=

Vi.

and

n.

Observethatthere

that

h(z') = f(z', z"), z = (z', z") E

=OonGforallcr E Nn_k andn—k+l <j
V2.

So, all r).

open mapping. To prove the Hausdorif property of G/ it suffices to regard two different points of the form and := [(f, := [(f', th")] in the quotient space. Then there exists a function f E (9(G), = 0 on G for the k last j's, with ti,"). i") Choose r > 0 and 1,2, as above such that f(z) f(w), whenever C G, j z E V1 and w E V2. Then q(Vj) are open disjoint subsets of Obviously, p is open and continuous. Arguing as before one easily sees that p is locally homeomorphic, i.e. (G/ p) is a Riemann domain over and that q is holomorphic. Because of the fact that the functions of (9(G/ correspond to those functions with ii — k
j

We continue with the situation above. For abbreviation we set XG := G/ (XG, p) —+ (XG, ft) we denote the envelope of holomorphy of XG. Now, we introduce the following holomorphic map

F=

(F1, F2): G —÷ XG x

Since (j5 o x

o

C',

By

F(z) = F(z'. z") := (x oq(z), z").

q(z), z") = (p o q(z), z") = (z', z") =

z,

we conclude that

F: (G, idG) —b (XG x Ck, 3 x is even a morphism; in particular, F is an open mapping. Moreover, liz = (z'. z"), w = (w', w") E G with F(z) = F(w), then z" = and x o q(z) = x o q(w). Since (9(XG) separates the points of XG, Remark 1.8.5 implies that x is injective. Thus, q(z) = q(w) and therefore z' = w'. Hence the morphism F is injective. Put G := F(G). Then G is a subdomain of XG and

F: (G, idG) —÷ (G, (j'3 x idck)1a) is an isomorphisin.

3 Envelopes of holomorphy for special domains

276

Let (x, z") E G, i.e. (x. z") = F(z', z") with z = (z', z") E G. Since (z'. z" + E IRk we obtain

it) E G, t

, I, F(z,z

/

.

I/

+:t),z +ir) .

= (x oq(z), z" + it) = (F1(z), F2(z) + it) E G (F1. F2). Therefore, the G xlR': —÷ z" + it), is continuous and it satisfies the following relations

:=

with F =: (x,

X

ocô((x. z"). t) = (j, z" + it) = (j5 x

z") + (0, it),

w.r.t. i.e. (G. (j5 X idck)Ia) can be thought as a k-tubular domain over We summarize what has been shown so far in the following lemma.

x be a domain that satisfies G + (0, iIRk) = G. Lemma 3.2.14. Let G c Then there exist a Riemann—Stein domain (XG x Ck, J3 x idck) and an infective x Ck, x idc&) such that with G := F(G) the morphism F: (G, idG) —÷ G x IRk —+ isak-tubular domain over C'1 w.r.t. pair(G,(j x

cô((x, z"). 0 = (x, z" + it). To continue our discussion let a: (G, idG) —+ (G, r) be the envelope of holomorphy of G. Then, in virtue of Theorem 2.12.1, F has a holomorphic extension to Foa = F. G, i.e.thereexistsamorphism F: (G,q) —+ (XG x Ck, j'3 x We know that (G. r) is a k-tubular domain over C" with respect to the following system of mappings G —+ G)IERk, where&, denotes the extension of to O; in particular, we have a, o a = aG(•, 1), t E IR!c.

Our goal is to show that G* is a domain of Put G := F(G). Then a C holomorphy which coincides with the envelope of holomorphy of G. We apply Lemma 3.2.11 to the following situation:

F: (G,r)—+ (XG

ao:

G

—+ G, aa(.,t) := a,.

is a Riemann—Stein domain over C'1!C, where Hence we know that (G/ —, o ifr(x) := r1 (x), x E G. Here i//: G —+ G/ denotes the quotient map of Lemma x Ck. Moreover, we have the morphism 3.2.11 and r =: (r1, r2): O —+

F:

= o 32) XG with In order to prove that F is injective we introduce the following mapping j: XG —p j(q(z)) := oa(z), when z E G. -

12

Observe that we have simplified our notation writing F

instead

of F.

3.2 k-tubular domains

277

We note that the map j is well defined: in fact, if we have points z— (z'. z"), w

= (w', w") E G with j(z)

j(w), then there exists an f E (9(G/ —) with

f(j(z)) f(j(w)). Since Locally oa is independent of the last k coordinates, the = 0, n — k + I < n. Then the fact function g := f 0 0 a E (9(G) satisfies q(w). that g(z) g(w) implies OnGweget: Fojoq = =F1oa = F1 = xoq,whereF =: (Ft, F2). Hence we have the following equality: F o j = x. Since F is a morphism. we even conclude that j E (9(X0, We extend jto a holomorphic map j: XG —* G/ with j o x = j. Then on XG we have the following: F ° J ° x = F o j = x. So the identity theorem yields that F o = idjG; in particular, F is surjective. Moreover, on XG we get Hence Fisinjective. = j,orIofr = Joioj =

j

j

XG is an isomorphism of Riemann Summarizing, we have seen that F: domains. Resolving the equivalence relation '-'-', the injectivity of F yields that all the fibers E XG,are connected. The next step is to prove that the map F: G —* XG x Ck is injective: Fix an x e XG. We recall fort E Rk the identity F oâ1 = (F1. F2 + it). Therefore, the is a tube domain of holomorphy over Ck. So Theorem 3.2.4 fiber '(x), fr2 I

is injective. Moreover, the fiber

implies that the map {z" E

: (x. z")

E

F(G) =: G*}

is a convex tube domain in Ck. Since x was arbitrarily chosen, we conclude that F is injective. We have seen is that F: (G, r) —* (G*. (j3 x and F: (G, idG) —+ (G. (j3 x idck)Ia) x CA is a Riemann—Stein domain and are isomorphisms. In particular. C G* is the envelope of holomorphy of G 'inside of XG x or F: (G, idG) —+ (G*, (j3 x describes the envelope of holomorphy of G.

Hence, the following result has been proven: x CA, I < k < n, be a domain satisfying G + iRk) and an (0, = G. Then there is a Riemann—Stein domain (XG. j3) over X CA, j3 x idck) such that the envelope of injective morphism F: (G, ide) —+ holo,norphy E.(G) of G is given as a subdomain of XG x CA via the morphism F; moreover, for all x E XG the fiber (z" E C' : (x, z") C(G)} is a non-empty lube

Theorem 3.2.15. Let G C

domain in CA.

For this reason we define the

Even more can be said about the envelope

following functions

X(XG) —f

:=

ve

z"): (x, z")

RA E

v

G},

0,

x E X(XG).

278

3 Envelopes of holomorphy for special domains

It is clear that these functions

are upper semicontinuous. In virtue of Lemma 3.2.10 there exists a largest plurisubharmonic function with XG —* Using these functions we have the following precise description of E(G) (Xc). I

Theorem 3.2.16. in the situation described in Theorem 3.2.15, the envelope of holomorphy C(G) is given as

= ((x. z")

E XG

x C":

< Re(v, z"). V E

0).

v

Proof Firstly, we recall the above situation. We know that is a C XG x k-tubular domain w.r.t. the mappings (x, z") —÷ (x, z" + it), t E Rk, with convex fibers. Hence it is given as

= ((x, z") E

XG

x C":

< Re(v, z") for all v E

v

0).

Using Kiselman's theorem (cf. Theorem 2.3.2) we conclude that the functions are plurisubharmonic on XG and, since G c G*, we get ,

'

G*

XG

IX(XG) <

Therefore,

which implies that

j ((x,z")

x Ck:

< Re(v,z"), V E Rk, v #0}=: D*.

On the other hand, D*, as a subdomain of the domain of holomorphy XG x Ck, is given by 'plurisubharmonic inequalities', i.e. D is also a domain of holomorphy, and so D* j G*, which completes the proof of Theorem 3.2.16. 0 LetG beas in Theorem 3.2.15. Put cv := jr1(G), where,r1: G —p denotes the projection of G onto the first (n — k) coordinates. Moreover, assume that G has only connected fibers over cv w.r.t. In. Then = cv. Therefore, Proposition 3.2.8 describes a particular case of the situation handled in Theorem 3.2.15.

Corollary 3.2,17. Let G C be a k-tubular domain, i.e. G + (0, = G, and letw := ,ni(G). Assume that thefibers (iri(z)), z E G, are connected and that C Then the envelope of holomorphy C(G) is univalent and it is given as

x C"

= {(z'. z") where in

< Re(v,

:

is the largest psh function on

z"), v

R", v

0),

satisfying

if k = I, then G=

z,,) E cv x C

< x,, <

:

v < V. and

V

C(G) =

x C:

<x <

(respectively V) is the largest psh (respectively smallest plurisuperha r,nonic) function on with V). v (respectively where

In analogy to Corollary 3.2.9 we obtain the following version for Hartogs domains.

279

3.2 k-tubular domains

Corollary 3.2.18. Let G C D x C fibers

be

a Hartogs domain with connected I-circled

i.e.

G

D

= {(Z,Zn)

xC:

<

where v, —v are upper semiconhinuous on D, v < V. Assume that C(D) C Then is univalent, and his given as

= where

xC

Zn)

<

:

<

and V are defined as above.

Remark 3.2.19. (Cf. the example of Shabat at the beginning of Chapter 1 and Remark 3.1.12(d).) In order to describe the envelope of holomorphy of the example discussed at the very beginning of Chapter I it remains (cf. Remark 3.1.12(d)) to find the envelope of holomorphy of the following domain

D :=

+ iy. w)

E C2

x <0, 0 <

:

y

and

O
ir/2
following domain

:= {(z,

w) E D x C z

b, wi <

whereD:={z=x+iy€C:x
0

is a holomorphic extension of D. In virtue of Corollary 3.2.18 it suffices to determine the largest subhannonic function V: D —÷ Rsuchthat —V <—Von D. Thenwewillobtain

C(D) = {(z, w) E D x C z E b, wi To get

we only have to find a bounded harmonic function h on D such that f 0

( —

if

log2

SecSection3.I.

if

x° <0, y0 = 0 or x0 = 0, 0
(*)

280

3 Envelopes of holomorphy for special domains

That bounded solution of the Dirichlet problem (*) is easily found using the following

biholomorphic mapping c1: D —.

4(z)

sin(—2iz —

ir/2)

— 1,

D,

z

where H4 denotes the upper open halfplane. Observe that E (—00,0)

R,

x=

or

0

0,

0<

y

ifx <0. y=ir/2

C1)(z) E(0,00)

Then h: D —÷

if x <0, y =

h(z) := ((l/ir)Arg(c1(z)) — l)log2, is harmonic on D and

solves the Dirichlet problem (*) from above.

Put V: D —f IL V(4)

log2

if x <0.

—h(z)

if 0 <

y

< y

<5ir/2

<,r/2.

SJf(D) and —V is the largest subharmonic rninorant of —V. Thus

Then —V

I Iwl <2

-

if 0 < y ir/2 F), where F(z, w) =

is the envelope of holomorphy of D. Therefore, (exp z, w), is the envelope of holomorphy of the Shabat example.

Moreover, we also can handle the envelope of holomorphy for Hartogs domains (cf. [Shi 1980]). with connected k-circled fibers z

Let G C D x Ct C C" be a Hartogs domain with connected k-circled fibers E D, where = {w E Cc (z w) E G}.

In a first step we assume that

zED. Following Section 3.1 we know that

is univalent and it is given as the logarithmically convex hull of G:. Therefore it can be written in the form

= {w E C"

r3(z, WI

<

<

(z, WI,..., where r1 (z, .) (respectively

wj E G1(z);

Wi-_I). I

j

k),

(z, .)) is an upper (respectively a lower) semicontinuous

function on G3(z) := ((wi

w3j)

WI

. .

.

< R3(z, w1 SeeSection3.I.

, W.c_i) < 1w31

ws_I), I <S < j — l},

3.2 k-tubular domains I

<

j
Note

r1(z,.) < R3(z,.)

that 0 <

<s<j—

281

and that all

r1(z, •)'s

and R3(z,

•)'s

I.

Put

:=D and Gj:=U{z}xGj(z),

G1

Observe that r3 and — need not will construct domains G c G1, the upper and lower semicontinuous First,

:=

set

such that

>

e

G1

Therefore, we

to be upper semicontinuous on

j =I

k.

and we will denote by r7 and (Ri),

and R3 on G. respectively.

regularization of Tj

= D. Now

let z0

E

D. Then

there are

w°

E Ck and e >

0

w0). E) C G; in particular. we have

and

w0). e) c

{z} x

U

C

(z) x

U

(.0E)

In virtue that

of the

fact that

Iw?I So we get

e)

is a Reinhardt domain it follows for all z E

<

+e < R1(z).

<

—

Hence

E G7

{(z.

xC:

< Iwil <

is open and not empty. Moreover, it is connected and G,t C G2. In particular, and R2 are defined on The next step consists in proving that r < on To do so let (z°. w?) E Then there exist E C such that w° E E(G.o), where w0 such that w° E Using (w° 4). Choose a Reinhardt domain G' that G is open we find > 0 such that e) x G' C G. If e is sufficiently small, then we obtthn w0), e) C Pn—k(Z°, e) x

{z} x

C (zt),,)

In particular, for z E

E C with 41

e) and

—e

< wi

I

< lw?l +

follows that

r2(z.wl) <

<

+r < R2(z).

r(z0, 4) < w?). Since (z°. w?) was arbitrarily chosen, it follows that

Hence

:= ((z. w) 0

E

D x C2

:

(z.

E

< R1(z) s oc are simply numbers.

<

wi) < lw21 <

on

Thus

it

282

3 Envelopes of holomorphy for special domains

is an open connected set with 0 and depend only C G3. Observe that on Iwil. Repeating this construction we obtain the desired domains G C D x and the semicontinuous functions 0 < r7 < on G; moreover, the r7's and the depend only on I <s <j — 1. Put

((z. w) E D x C" : r7(z.

G

uj

<

< 1W31

WI

wi_i) E G. I

wj_i). (z. wi

j

k}.

where

:= {(z, WI

ED x

wi .

:

<

(Zi.

.

.

,

< WS_I) E

WI

G is a Hartogs domain with connected k-circled fibers which contains G. More-

over, if f E (9(G), then for all z E D the function f(z. .) extends holomorphically to the fiber Applying the Hartogs theorem on separate analyticity yields that f itself extends holomorphically to G*. So it remains to find the envelope of holomorphy of

G. Write r7 =: exp(v1) and

=: exp(V3). Then —oc vj < V3 00 and v3, upper semicontinuous. Assume now that D has a univalent envelope of holomorphy, i.e. E.(D) C Then, applying Corollary 3.2.18, has a univalent envelope of holomorphy which is given as —

V3 are

=

((z. WI) E

xC:

< Iwil < exp(VI(z))}.

is the largest psh function on e(D) with v1 and where V1 is the smallest plurisuperharmonic function on e(D) with Vi ID V1. Observe that E is negative on D. Let be an increasing sequence of subdomains D with D = D'.. Then where

= < on In particular, for any z E e(D) there exists a WI E C i.e. such that (z. WI) E isa Hartogs domain with connected 1-circled fibers over C(D). Repeating this argument we obtain that the envelope of holomorphy of G, Therefore,

:= ((z. w) E

x C2: (z, Wj)

E

exp(v2(z,WI)) <

< exp(V2(z.w1))},

3.2 k-tubular domains

283

is univalent and it is given as

=

{(z.

x C

w2)

wi)) < 1w21 <exp(V2(z,

:

(respectively is the largest (respectively smallest) psh (respectively plurisuperharmonic) function on satisfying (respectively V2IG,. where

< V2 on Thus V2). Observe that and V2 depend only on wi I and that is a Hartogs domain with connected 2-circled fibers over C(D) (for z E C(D)

findfirstw1 ECwith(z,w1)E

E Cwith(z,wI,w2) E

Following the same argument successively gives finally that the envelope of holomorphy of G* is univalent and it is given as

=

{(z,

Wk_1, Wk) E

C(G) x C wkI))),

Wk...1)) < IWkI < exp(Vk(z,

exp(i)k(z, WI

where i)k (respectively, 14) is the largest (respectively, smallest) psh (respectively plurisuperharmonic) function on Vk (respectively, VkIG. ? satisfying í3kIG;

Vk. Both functions depend only on Iwil

Iwk_II.

Summarizing we have that C(G) is univalent and that there is an effective description of it. Moreover, E(G) C x Ck is a Hartogs domain with connected k-circled fibers. Now we start treating the general case. Let G C D x Ck C be a Hartogs domain with non-empty coimected k-circled fibers z E D. Without loss of generality, we may assume that if w E 0, and that for any j, z E D, then Wm+1 1 <j
xCk

Go:=((z,w)E

...•wk =0}.

:

Let f E (9(G). Then (cf. Section 3.1) f extends holomorphically to G1

:= {(z.Aiwi

Wk)

ArnWm,

ED

x Ck

l<j<m}. Put G :=

G1 \ Go. Observe that the domain G is of the form we discussed before. Therefore, C(G) is univalent and it is given as

=

{(z, w) E E(D) x CI

:

wi

Wi_I)) <

j

I

k},

where the notions are taken from above, now with respect to G. Note that j ,n. Put G

{(z, w) E €(D) x Ck: IwjI

I < j
w1_i)),

Wj...I)) < IwjI <exp(Vj(z, WI (z. w1

where GT := D.

E

284

3 Envelopes of holomorphy

for special domains

Observe that G is pseudoconvex and that G \ Go = Then every f E 0(G) extends holomorphically first to and then to G, i.e. C(G) = G (recall that extends to G i).

f

Summarizing we obtain the following description of the envelope of holomorphy of Hartogs domains with connected k-circled fibers.

Theorem 3.2.20. Let G C D x fibers. Assume that C univalent and ii is given as

C"'.

be a Hartogs domain with connected k-circled Let m E {O, I k} be as above. Then E(G) is

= {(z. w) E E(D) x w3_f)).

I

j

in.

(z. wi where the notations

k.

Wj_1) E

were taken from above.

V1. and

Observe that E(G) C any point

j

wi—i)). in + I

< IwjI < exp(V3(z. Wi

WI

x Ct is a Hartogs domain with non-empty fibers over

of

3.3

Matrix Reinhardt domains

Let M(n) denote the set of all complex n x n matrices. We identify M(n) with On M(n) we consider the action of the group 9 := 11(n) x 11(n), where 11(n) is the group of all unitary matrices in M(n). given by

M(n) x 9

(Z, (g,

f)) —+ gZf.

where gZf is the matrix multiplication. Analogously, if k E N, we have the following group action

Mk(n) x

((Z1

Observe that in case n =

((zi

Zk), (gi. fi I

fk)) —*

gkZkfk).

this reads as

zk). Of,

36)

37)

Notethatifrn = 0 then = G. For simplicity we write (ga. Il

St. 1k) instead of ((gi ft )

*f ER.

(St. 1k)).

285

3.3 Matrix Reinhardt domains

Definition 3.3.1. A domain D c Mk(n) is said to be a matrix Reinhardt domain (respectively a circular matrix domain) if D is invariant under the group action of D for gkZkfk) (respectively 9), i.e. if (Zi Zk) E D, then (g1Z1f1 k (respectively (gZ1f D for all all unitary matrices gj, f1, 1 E j

(g, f) E 9). In case n = I, D is a matrix Reinhardt domain (respectively a circular matrix domain) if D is a Reinhardt domain (respectively a circular domain) in the usual sense.

Examples 3.3.2. 1) The matrix disc or the generalized disc:

A := {Z where

38)

M(n):

is the n-dimensional unit matrix and'<' means that

—

isa positive

definite Hermitian matrix.

We note that A is a Cartan domain of first type (cf. [Khe 1990] and [Ser-VIa 1994]) and it is holomorphically equivalent to a convex tube domain in Cr; thus A is a domain of holomorphy. Observe that A is a matrix Reinhardt domain, but in case ii > I it is not a scalar Reinhardt domain 2) The matrix polydisc:

P := {(Z1

Zk) E

Z3 E A}.

It is a matrix Reinhardt domain but it is not a scalar Reinhardt domain if,, > 3) The matrix ball:

1.

B:={(Z1 For

min{k,n} >

1,

it is not a matrix Reinhardt domain, but it is a circular matrix

domain.

4) The extended matrix disc:

D := {(AZ1B

AZkB)

Mk(n): (Z,

Zk)

P. A, B

SL(n, C)).

M(n): detM = 1). This example is related to quantum field theory. It is a circular matrix domain but, in general, it is not a matrix Reinhardt domain. Nevertheless, it is shown by P. Heinzner & A. Sergeev (cf. [l-Iei-Ser 1992]) that the extended matrix disc is a domain of holomorphy. Let n k = 2. To see that D is not a matrix Reinhardt domain we present an example due to X. Thou:

where SL(n,C) := (A

38)

Z :=

Z' is the conjugate matrix of Z. A scatar Reinhardi do,nain means simply a Reinhardt domain in the classical sense.

286

3 Envelopes of holomorphy for special domains

Let 0
Observe and denote by Z0 the diagonal matrix [c, 11 E M(2) Z0A0 A if c < 1. Therefore, the pair (Z0, 4) belongs to the extended matrix disc D. I

that [c. ll[I/d, dJ

=:

Fix

/0

1

0

We assume that

(4,

(W1, W2)

= (Z0. [l.cJ) €D.

there exist matrices A, B SL(2, C) and Z1, Z2 E A such that (W1, W2) = Observe that are not singular. It is easy to prove that with Z1, Z2 E A also E A and then E A. Finally, we get

Then

(AZ1

[C2,

1J

= det(W2) W1

E A;

contradiction

Recall that there is a complete characterization for Reinhardt domains to be domains of holomorphy (cf. Proposition 1.9.19). So a natural problem is to characterize the holomorphically convex matrix Reinhardt domains in terms of naturally associated classical Reinhardt domains. Let D C Mk(n) be a matrix Reinhardt domain. We define its diagonal as

d(D)

Dfldiag(n)k,

where diag(n) := Z a diagonal matrix}. We recall from the general theory of matrices that any matrix A E M(n) can be written as A = UAV, U. V E where A = [Xi is the diagonal matrix whose entries ... 0 are the eigenvalues of the positive semidefinite Hermitian matrix Using this representation we conclude that a domain 1)

Mk(n) is matrix Rein-

hardtiff,whenever(Zj Zk) E d(D). Inparticular, Zk) E D,then(Z1 is a Reinhardt open set in the usual sense. d(D) C Observe that d( D) is invariant under arbitrary permutations, i.e.

if ([A11

Am]

[Aki

Ai.ai(fl)]

then

where

Xknj) E d(D).

the a1's denote arbitrary permutations of (1

[Ak.ak(1)

Ak,ty*(n)]) E

d(D),

n}.

Let A bc a positive semidefinite Hermitian matrix and let xj, be an orthonormal k basis of cigenvectors. n. Then there is a uniquely defined positive scmidefinitc = I
287

3.3 Matrix Reinhardt domains

We say that d(D) is piecewise permutational or, jfk = I, permutational. Finally, we point out that D is completely determined by its diagonal, i.e.

D=

Zk) E d(D), U3. V1 E 'U(n),

UkZkVk): (Z1

I

j

k}.

Our aim is to characterize holomorphically convex matrix Reinhardt domains in terms of their diagonals. This has been observed by A. Sergeev (cf. [Serg 1993]) under an additional completeness assumption. The general case was solved independently by X. Thou (cf. [Zho 1990]) and by E. Bedford & J. Dadok (cf. [Bed-Dad 1991]).

Theorem 3.3.3. A inatrir Reinhardi domain D C Mk(n) is a domain of holomorphy and only domain in

its diagonal d(D) is a Reinhardi domain of holomorphy (though: as a

In order to be able to prove the necessary condition of that theorem we need an additional information.

Lemma 33.4. Let

(r) —+ IR be a C2 strictly psh function. Then, after a —* C'1, the second order suitable complex linear coordinate transformation A: terms in the Taylor expansion of ço o A at the origin equals

Zj =Xj where A > 0.

Proof By a standard argument 42)

we

may assume that

is locally given by the

following Taylor expansion at the origin

w(z) = where

S = (S,,) with

11z112

+ Re(zSz') + o(11z112).

(t)

=

In the remaining part of the proof we will identify complex vectors z = x + iy With S =: A + iB, with the real vectors (x, y) E IR'1 x

Zarealmatrix), and v := (x. y) E

near the origin, we rewrite (t) as 'p(z) = p(v) = 11v02 + vTv' + o(11v112),

where T denotes the following matrix

(A 42)

Exercise; cf. Section 2.2.

—B —A

288

if

3 Envelopes of holomorphy for special domains

Observe that if v = (x. y) x + jy, then i(x + iy) (—y, x) =: J(v). So, is an eigenvalue of T with eigenvector v, then is also eigenvalue of T with

eigenvector J(v). Now we choose an orthonormal basis of

such that T

=

vt,

consisting of eigenvectors of T, say

> 0 (this can be achieved by eventually changing the role of

and

Let C: R2" —÷ R2" be the linear mapping such that

j=

= J(v1),

C(e3) =

n,

1

where e3 denotes the j-th standard unit vector in Obviously, C can be thought as a complex linear map from C" to C". We will use C as the coordinate transformation whose existence has been claimed in the lemma. Let w E Ri". Then for the transformed function := q o C we get

= ço(C(w)) = (Cw')'(12,, + T)(Cw') + o(II(Cw')'112)

= w(12,, + C'TC)w' +

o(11w112)

=

1w112

whereDisthediagonalmatrixL) = the claimed second order terms in the Taylor expansion of

+ wDw' +

o(11w112),

at the origin.

0

Let M be a real submanifold of class & of an open set U C C". M is said ii = {0}, where to be totally real if for each z E M we have that denotes the real tangent space to M at the point z and where J defined as J(x, y) := (—y, x), (x. y) E R" x IR" C", means multiplication with i. The definition says that for each z E M none of the complex lines in belong to In particular, dimR M

n.

Theorem 3.3.5. Let be a non-negative C2 strictly psh function in an open set = 0}. Thenforeachpointzo E Z there existsa ball U C C". Put Z := {z E U: I3(zo, r) =: V C U and an & totally real

M of V such that Zn V C M,

i.e. the zero-set of cp is locally contained in a totally real submanifold of class C'. From this result the following remark is obvious.

Corollary 3.3.6. The zero-set of a non-negative C2 strictly psh function on an open set in C" does not contain a real submansfold of class C whose real dimension is larger than n. Proof of Theorem 3.3.5. Without loss of generality we may assume that zo = 0. Moreover, we start working in the coordinate system whose existence was established

in Lemma 3.3.4. Thus, w.l.o.g.

has the following description around the origin:

z=xj+iyj,

289

3.3 Matrix Reinhardt domains

where the A3 's are non-negative. Here we have used that grad colz = 0. Put ax1

Then the form of

implies that

ia

a

a

avi

dxi

dx3

where

=0. 1

(h3

denotes the standard j-th unit vector in

grade —(0) oxI

x W'. In particular, we see that

grads —(0) oxn

linearly independent. Therefore, these vectors remain linearly independent in a small neighborhood V = V(0) C U. So Z fl V C M := M 0 V and M is a real ndimensional submanifold of V of class &. Its tangent space at zero is the orthogonal Thus complement of Re1 + . . . + i.e. it is spanned by the vectors en+1 M is totally real at the origin and therefore also in a neighborhood of 0 Hence the theorem is proven. D are

We will follow a standard procedure to produce a non-negative strictly psh exhaustion function on a holomorphically convex matrix Reinhardt domain D C M1'(n), that is invariant under our group action.

Lemma 3.3.7. On

a

holomorphically convex matrix Reinhardt domain D C Mk (n)

there exists a C°° strictly psh exhaustion function

having the following property:

Zk)=ça(glZifj

EtL(n),j=I

forZ=(Z1

k.

strictly psh exhaustion funcProof Since D is holomorphically convex there is a tion D —+ IL We define a new function on D, namely: ço(Z) :=

I

gk.fk).

J9k

for It is an easy exercise to verify that 0 fulfils all the properties that were claimed in the lemma. where dci denotes the Haar measure

Use that the perturbation.

i

remains an

after a small

Let 9 be a locally compact topological group. Then there is a uniquely defined (up to a positive constant factor) measure on 9, which is invariant under translations from the left. This measure is the so-called Haar measure.

290

3

Envelopes of holomorphy for special domains

Remark 3.3.8. Let D C M(n) be a matrix Reinhardt domain with diagonal d(D). If d(D)containsamatrix[ai then d(D) is connected. with lail = = In fact, the part d* := dfl{[x1 Ed: > 0} of the real part of d is connected. So, if there is the point A := hail E d, then all the connected parts of the real part of d, obtained by permutation from are connected via the point A and its permutations.

Now we are in the position to prove the necessary condition.

Proof of the necessity in Theorem 3.3.3. d(D) is the intersection of D with a certain number of hyperplanes, and therefore it is holomorphically convex. To prove that d(D) is connected we choose a C2 strictly psh exhaustion function 1) —÷ R which is invariant under the action of 9k (cf. Lemma 3.3.7), i.e. is constant on the orbits of Define c := Z E D}. Observe that c E R. Put

M := {Z e D: — c = O}. In virtue of Lemma 3.3.5, M is locally contained in a totally real submanifold. In particular, any orbit Z, Z e M, is locally a subset of a totally real submanifold. Fix a real point Z0 := (hz?1 e d(D) with Zo M; [4

observe such a point always exists. Then the dimension of 9kZ0 is greater than n except for each j = 1 n, have the same k all diagonal elements £= I absolute value. But then using the above remark we conclude that d(D) is connected.

0 Now we turn to prove that a matrix Reinhardt domain D C Mk (n) is holomorphically convex if its diagonal is connected and holomorphically convex as a subdomain in We proceed in proving several preliminary results. We repeat that for a matrix Z M(n) we had introduced the associated diagonal matrix[A1(Z) ? ... the positive semidefinite Hermitian matrix

Lenuna 3.3.9. Let D C M(2) be a matrix Reinhardt domain with diagonal d := d(D). Suppose that d is a logarithmically convex domain in C2. Let P be a real symmetric polynomial in two variables such that the function p: d —÷ R. p([zI,z21) p(zI.z2) := P(1z112, 1z212). ispshond. Then thefunction Q: D —* R. Q(Z) := p(A1(Z), A2(Z)), is psh on D.

Proof Since P is a symmetric polynomial, there exists a uniquely determined polynomial cD such that

P(xl,x2) = 4)(xi +x2,xIx2), We

x = (xI,x2) ER2

have to verify that the following function

D

Z —+ Q(Z) =

= ispshon D.

= P(IA1(Z)12, 1A2(Z)12)

+ A2(Z)12. IA1(Z)121X2(Z)12) = 4(trace(ZZ*),det(ZZ*))

3.3 Matrix Reinhardt domains

Fix Z0 = (iota, b]Vo

291

D, where [a, b] E d with a, b > 0 and

M(2)

'11(2). Moreover, let W0, W0

with W0 = Uo W0 V0.

where Uo, Vo E What we have to show is

that the following function

—+ Q(Z0 + AW0) = Q(Uo([a, bJ + AW0)Vo) = Q([a, fri + AW0) =: f(A). A

(0) > 0.

C near 0, satisfies that

Calculations lead to the following representation off

AE U :=U(0)cC. where

a + Aw?112 + lb +

gi(A)

:= Al2(lw?212 + g2(A)

+ Aw?1l21h +

a

2 + w?2l2 — 2 Re

g2(A)

with H(A) :=

+Aw?1)(b+Aw2°2). By our assumption we know that the function f,

f(A) := is subharmonic near the origin. In particular, we have (0) ? 0. Looking at the formulas for the we easily obtain that

dA

dA

dA

dAdA

dA

dAdA

—

Thus calculation leads to

d2' d2' —4(0) = —4(0) + ------(a2 dAdA

dAdA

0 it suffices to verify that

To get that

dAdA

dx1

+ b2. a2b2) ? 0.

First assume that our matrix Wo has a special shape, namely w?1 = a. w2°2 = —b. In this case we obtain

gi(A) =a211 +Al2 +b211 — Al2, and, therefore,

0< —

dAdA

= —(a2 + ax1

b2, a2b2)(a2 + b2) +

ax1ax1

(a2 + b2. a2b2)(a2

—

b2)2.

(t)

292

3 Envelopes of holomorphy for special domains

In particular. if a =

h

0. then

+ b2, a2b2) 0. We proceed by contradiction. Assume there is a point [a, b} E d such that + h2, a2b2) <0. Without loss of generality we may assume that a > b > 0, Recall that by assumption d is connected, logarithmically convex, and permutational. Put a Then [a, aJ d and a2 = a!, =: c. We find numbers y and
+

(c/I)2. C2)

> o,

z

—(t2+(c/t)2,c2)<0. y
From (t) we know that

integration leads to

a2i

act) (t,c2)dt= —(52+(c/8)2,c2) <0, 0< Jy2+(c/y)2 I axi dx1dxj

0

which gives the expected contradiction.

In the next step we will prove a similar result for matrix Reinhardt domains in M(n), n > 3.

Lemma 33.10. Let D c M(n), ii > 3, be a matrit Reinhardi domain whose diagonal d = d(D) is a logarithmically convex domain (recall that d is automatically permutational). Suppose P is a real symmetric polynomial in n variables such that the function

p: d —+ IL

= p([zi

p(zj

zn])

P(1z112

is ps/i. Then

Q: D —p IR, is psi:

Q(Z) := p(Ai(Z)

on D.

Pmof The proof is similar to the one of Lemma 3.3.9. First, we write the symmetric polynomial P as a polynomial cD in the elementary symmetric functions sk(X) = sk(Xi

>J1<j<

.

...

P(xj Thus we obtain

Q(Z) = ct)(oi(Z) where (Z) means the sum of all k-th principal minors of ZZ*. Here, the last equality is taken from algebra.

.E

3.3 Matrix Reinhardt domains

d, a3 >

293

matrices (Jo. Vo E 'U(n), and W0 = M(n). We have to prove that the function f Fix

0,

(Jo W0 Vo E

an(Uo([a1

AECnearO, satisfies (0) > 0. Again, we write ak(LaI

a,,] + AW0) = gkGk) +

where

gk(A) :=

+

+

2

2.

I<j,<.<j*
gk(O) = sk(aI

an),

=

gk(O) = 0,

fl

= 0,

aJ

l<ji<

s=I

p.qø(ji.. J&-i} As in the proof of the previous lemma we put

AEC. We get

= where

+

a2 :=

Since the function f is subharmonic, it remains to show that the second term in (1) is non negative. So we claim that for any pair p q

s=I

k=I 3k—I)

294

3 Envelopes of holomorphy for special domains

Observe that for n = 2 this claim already had been verified in the proof of the preceding lemma. So, we proceed by induction and we assume, that claim is true in case of dimension n — 1. Without loss of generality, we may assume that p 1 q. Put d0

:

([z2

and we define a new function

Zn]

:2 :.

[ai, Z2

Zn] E

da

Zn]) = where

:

zn)

(1z212

Evaluating the

elementary symmetric functions we get

Zn)

=

+s2U2)

Observe

and that

that da

whereã2

Using

.

In virtue

of the

induction hypothesis we conclude that

k

2<ji<••<j*i
:= the following equality

—=—+a1 aYk dXk dx*+i we

see

that

>axk k=I ik-l} sn(a2))

k=2

k

cn(a2))

k=I

a?yn_J).

is a Reinhardt domain, logarithmically convex and permutational

is psh on da.

k=I

+Yk

+y2

'l'(v)

k

p.qq(jl

ik—I}

295

3.3 Matrix Reinhardt domains

0, i.e. the lemma is U

which completes the proof of the claim. Thus we have proven. Finally, we obtain the general case.

Lemma 3.3.11. Let D C Mk (n) be a matrix Reinhardt domain such that its diagonal Let P be a piecewise d = d(D) is a pseudoconvex Reinhard: domain in symmetric real polynomial on i.e. P(x11

XkI

Xkuk(I)

Xkyi(fl)

for all variables and all permutations a1. Suppose that the function p: d —* JR. p([zii

ZInJ

Zkn])

[ZkI

IzInI2

:= P(Iziij2

= p(zii

ZIn

ZknI2) = P(zi

IZklI2

Zkn)

ZkI

. Zk)

.

"s'

ispsh on d. Then the function Q: D —* JR.

p(Ai(Z,)

Q(Z)

Zk)ED, ispsh on D.

... >

(We recall that A1(Z1)

> 0 denote the eigenvalues of Z1Z7.)

Proof We proceed in a similar way as before. Let c1 be the real polynomial in nk variables such that

=

P(zi

C1(si(zi

We fix an arbitrary point ([au tL(n), and Vi,, = UXWXVX E

f(A) =

[akl

C

am] + AW1)

+ AW1)

akfl]+AWk)

=

=

An([akl

akn]+AWk))

+ gkn(A)).

CI)(g11(A) +

where, as above,

E

M(n), I <x
= 0, 1 < x

k, I
Moreover, we define

f(A) := 45)

1(gmi(A)

Recallthatforvectorsa. bE

gmñ(A) C" weusethefollowingconvention:

A E C.

a•b = (aibu

3 Envelopes of holomorphy for special domains

296

Then, a tedious calculation leads to

±L(O) x=I k

d2'

= —4(0) + dAdA

—

x=I \dAdA

d2' dAdA

where

and

:=

gkl(O)

virtue of Lemma 3.3.10 we see that the right-hand side of(t) is non negative. Hence fl we have shown that Q is psh. In

Now we start to construct an exhaustion of D by pseudoconvex subdomains. What we will need is the following general approximation result for psh functions.

Lenima 3.3.12. Let G c C" be a Reinhardt domain, G' subdomain, and ii: G —÷ R psh. Suppose that u(z)

= u(IziI

z

= (zi

G

fl

(C,)" a Reuthardr

Zn) E G.

Then there exists a sequence of real polynomials Pk, k E N, in n variables such that pk: G' —+ IR. pk(z) := Pk(1z112 IzflI2),are pshfunctions withpk \

G 1) (C,)" with G' Pmof Fix a Reinhardt subdomain G" G" —÷ IR, Uk \ uIG". Put uk(z)

I

2ir

I j0

The Uk are psh on G" with

for all z E G". On the compact set K := vk(xl

I j0

ük(e"zI

G". We choose psh

.

z E G".

. .

\ uIG'; they satisfy also uk(z) = uk(IzII (x

(IR>0Y':

G'} we set

:=

In virtue of the Weierstrass approximation theorem we find real polynomials Pk such that x K, a al < 2, ID° Pk(x) — D"vt(x)t <

where the positive rA will be chosen later.

3.3 Matrix Reinhardt domains We put

:

Forz

.

E

297

G' we obtain

£pk(Z; X)

E

=

+

•••, IZnl2)IXvI2

(IziI2 v=I

v,/L=I —

eicIIzlIIIXII — eklIXIl 2

= Lvic(z: X) — €k(l + IIzII2)IIXII2, where Vk(Z) Vk(IZI 2 = Uk(Z). = uic(IziI Since Uk is psh we finally reach the following inequality on G'

Lpk(Z; X) With C

sup{l + 1Iz1I2:

z

+ lIzIl2)IIXII2.

E G'} we see that the functions Pic.

pk(z) := jik(z) + ekCIIzll2 + 5k.

zE

are psh on G'. Moreover, pk(z) = Pk(IziI2

where

Pic(x) := Pk(X)+ekC(XI i.e. Pk is a real polynomial in n variables.

Calculations give for z

u(z)

E

G': < vk(IziI2

= Pk(IZI

jznI2)+ekCIIzII2

Pk(Z)

+ ekClJzlI2 + Sj1

2

IznI2)+skCIIzlI2+25k

<

+ skClIzll2 +

Vk_I(1Z112

5 Pk—i(Z) + C11z112(sic

if we choose a sequence

sic—i)

+

2Sk

S Pk—i(Z).

CC'Sic_i/(2 + C2) ,where

sic —* 0, such that

C':=min{11z112: ZEG'}.

0

Hencewehavepk \uonG'.

Proof of the sufficiency in Theorem 3.3.3. Fix an £ E N and put

:= max{— log

ad),

.

—

298

3

on d C

(C's

)k•

Envelopes of holomorphy for special domains

Observe that

psh and invariant under coordinatewise rotations.

is

Therefore,

hi,...,

logd ?

is a convex function. Since log d

dt

itself is convex, it follows

:= (z Ed: ut(z)
logarithmically convex and piecewise

are

that the sublevel sets

d

permutational;

moreover, d1

and

a= By D, we denote that matrix Reinhardt domains whose diagonal d(D1) = dt.

Fix an e > 0 and put

:= (z Edt: Applying Lemma 3.3.12 we

IJ)(Rflk) such that

I
I

find a decreasing

sequence of


real polynomials J-'j E

: d((÷1)5 —÷ IR,

'ij(ZiI

. . .

IZknI2),

, Zkn)

is psh on and \ uj on d(f+I)e. We define new piecewise symmetric polynomials Pj as

Pj(x) :=

Xkok(t)

where summation is over all k-tuples of permutations a1 of (1

xkck(n)), n } and where N

denotes the number of such rearrangements. —÷ Put Pj : pj(Z) :=

Zk

Zk).

Then the p1's are psh and p1 \ ut on be the matrix Reinhardt domain whose diagonal = dfs. From Lemma 3.3.11 we conclude that the functions : Let

Q1(Z1 are

psh and Q3

Zk) :=

\

AI(Zk)

where

Zk) := ue(Ai(Zi) In particular, it turns out that Recall that

is

psh on

AI(Zk)

is given

by dts, i.e.

—p IR.

An(Zk)),

299

3.3 Matrix Reinhardt domains

Observe that

:=

E

Mk(n): fldetZx x=I

is psh on

is continuous on Dt+ Applying the extension theorem for psh functions implies that is even a psh function on Di÷i. Remark that D1 is given as On the other side,

= {Z


Hence D, is pseudoconvex. What remains is to observe that D domains Dt, i.e. D is pseudoconvex.

is

the increasing union of the pseudoconvex U

We like to conclude this section by a few remarks on the so-called extended tube conjecture (cf. [Str-Wig 1964]):

DefinitIon 33.13. (a) For vectors z = (zt. (z, W)m := where

(.,.)

denotes

pmduct. (b)

:=

N

w = (Wi. —

0. (y. Y)m

n>

2,

we put

th),

the Hermitian scalar product. (.,

:= {v E R": y, >

ii)) E

is

called the Minkowski

> 0} is the so-called fufun' light

cone.

:= is the future light tube. := IR" + (d) L(C") := {M E M(n): detM 0, (c)

(MZ,MW)m. Z.W E C").

L(C") is the complex Lorentz gmup. It is easily seen that L(C") consists of two connected components. The one that contains the identity is denoted by L÷(C"). It is called the pmper Lorentz gmup. For n, N E N we put +

.....

TnN .

+ X

+ C X T,,

Then we define TnN := {Mz:

Z E TnN M

where the proper Lorentz transformation M acts as follows on (C")":

Mz = M(zi 47)

ZN) := (Mz1

MZN)

z = (zi

To simplify our notation we write Mz, z E C". instead of

ZN) E

300

3 Envelopes of holomorphy for special domains

The domain is called the extended (n. N)-tube. In theoretical physics. i.e. n = 4, the question, whether the extended tube is a domain of holomorphy, was known for more than thirty years (cf. [Ser-VIa 1994]) as the extended tube conjecture. The answer was found only recently in [Zho 1998], see also [Hei 1998]. The case N = I can be easily answered, namely we have

= Therefore.

is a domain of holomorphy.

Proof of(*). Let z = Lw,

L E L+(C"). Then, applying the Cauchy—

w

[0. 0°).

Schwarz inequality leads immediately to (z. Z)m

If Z

(*)

[0,oo)}.

{z E C'7: (Z.Z)m

[0. oo), then there exist c >

0 and (0. 2,r) such that E C". Then w and (Z,Z)m = (Z.Z)m = ce"°. Put w := (W, W)m. In virtue of the Witt extension theorem we find a Lorentz transformation

E C'7, (Z. Z)m

L such that Lw = z. If L E L+(C"), then z E In the remaining case put := (WI and define L' substituting the last column of L by its Wa_I. —w,,) negative. Then w' and L' E L+(C"). Therefore, z D In general we have the following result (cf. [Hei 19981, [Zho 1998]).

Theorem 3.3.14. For any N We

is a domain of holomorphv

N the extended tube

are not going to present the proof of Theorem 3.3.14, nevertheless we shall

make some remarks to illustrate the main ideas. The proof by Zhou works in the frame

of 2 x 2-matrices. So we will first reformulate the statement in Theorem 3.3.14. In order to do so we have to introduce the following biholomorphic mapping

Z2_iZ3) Z2

We

+

1Z3

Z1 — Z4

z=(zi

z4)€C4.

define the so-called Pauli matrices

/l o\

10

i)'

l\

10

o)

—i\

11

o)'

Then cl is given by

= It is easily seen that the inverse of

=

is given by

trace(Ma4)),

0 —l

M

M(2).

3.3 Matrix Reinhardt domains

301

Then, for vectors z, w E C4, the Minkowski product fulfils the following relations:

det (4(z) + cD(w))

— det

(4(z))

= 2(z, w)m.

— det

g :=

SL(2, C) x SL(2, C). g with the componentwise multiplication is a group. One can check that for any M = (A, B) E g we have Let

(2.2)m=(Z.Z)m,

ZEC4,

where 2 :=

Therefore, the mapping M: C4 —÷ C4. M(z) := 2, is a Lorentz transformation, even more M E Observe that g M —+ M E L + (Ca) is a 2—1 homomorphism. We have

1mM> 0) =: H, 48) H is sometimes called the generalized upper halfpiane. It where Im M : is invariant under the following mappings M —* AMA*. A E SL(2, C). Moreover, the function

u(Z1

(Z1

ZN)EHN,

is strictly psh, weakly exhausting (i.e. u tends to infinity along each boundary sequence in HN), and u(AZIA* AZNA*) = u(Z), A E SL(2, C), Z = (Z1 ZN) E

HN. Observe that (c1 x

x

= {(AZ1Bt

AZNB'): (Zi

ZN)

Therefore, t4N is a domain of holomorphy 1ff It can be seen that

3(A.B)E9

E

11N

(A. B) E g} =:

is a domain of holomorphy.

: Z = (AMIB*

AMNB*)}.

Now the main proof is an application of Kiselman's minimum principle (cf. Sec-

tion 2.3) in a generalized form which is due to J.-J. Loeb (cf. [Loe 1985]) in order to get from u an exhausting psh function on Recall

:=M'.

302

3 Envelopes of holomorphy for special domains

Remark 3.3.15. (a) According to our knowledge there is, so far, no proof for the fact that tflN, ii. N N arbitrarily, is a domain of holomorphy (b) A sufficient criterion is given in [Jar.Pfi 1996b], namely: Let

T(n, N)

{(Mz1

MZN): z = (zi

M E L(C") fl tL(n)}.

ZN)

Then the following is true: Let a: (T(n, N). id) —+ (X, p) be an envelope of holomorphy. Then

domain of holomorphy iffp(X) c Observe that T(n, N) is not a domain of holomorphy for n ? 3 and N

is a 2.

We close this section by quoting the following result on holomorphic extension (cf. [Bro-Eps-Gla 1967]).

Remark 3.3.16. Denote by J :=

fl (R4y" the set of so-called Jost points. It

turns out that


J_—fx_—(xi

Then any function holomorphic in a neighborhood of the set

LI T4N Ui, T4N

extends holomorphically to the extended tube

Hence, T4N is the 'envelope of holomorphy' of

U T4N U J.

3.4 The envelope of holomorphy of X \ M aim of this section is to find a description of the envelope of holomoiphy of X \ M, where M is an arbitrary analytic subset of a Riemann domain X. Recall that Theorem 2.5.9 deals only with such M which are, by assumption, 'extendible' to the envelope of holomorphy of X. The

Singular sets. Let (X. p) be a Riemann domain over C" and let M be a closed subset of X satisfying the following condition

for any domain D C X the set D \ M is connected and dense in D.

(3.4.1)

Notice that mt M = 0. Observe that any closed pluripolar set M C X satisfies (3.4.1). In particular, any proper analytic subset M of X satisfies (3.4.1). Let 0 F C (9(X \ M). We say that a point a E M is non-singular with respect to F (a E if there exists an open neighborhood LI of a such that for each f E F there exists a function

3.4 The envelope of holomorphy of X \ M

303

f E 0(U) with! = f on U \ M. Ifa

then we say that a is M\ singular with respect to F. If Mn5F = 0, i.e. M5F = M, then we say that M is singular with respect to F. If F = (9(X \ M), then we simply say that M is singular and we skip the index F. We would like to point out the essential difference between the notion of 'the singular analytic subset M of X' and 'the singular points Sing(M) of an analytic subset M of X'

Remark 3.4.1. (a) The set M5,, is closed in M and satisfies (3.4.1).

(b) Each function f E F has a holomorphic extension f E (9(X \

(C) M5,F = (Ms,)5?, where respect to F.

F := {f: f E F),

i.e.

is

singular with

let M be Lemma 3.4.2 (cf. Lemma 1.7.5). Let (X, p) be a Riemann domain over a closed subset of X with (3.4.1), and let 0 M). Assume that A is a F c 0(X \ dense subset of

Y := {x E X \ M: dx\M(x)
(i)

M is singular with respect to F;

(ii) dX\M(x)

= inf{d(Lf):

f E F},

X

E A.

(ii): Fix a E A and dx\M(a) r/2> dx\M(xo); contradiction. 0

Proof (i)

;

f

Proposititon 3.4.3 (cf. Proposition 1.7.6). Let (X, p) be a Riemann domain over C", let M be a closed subset of X with (3.4.1). and let F be a natural Fréchet space in

0(X \ M) (cf Section 1.2). (i)

Then the following conditions are equivalent:

M is singular with respect to F;

(ii) there exists an f E F such that M is singular with respect to {f}; (iii) the set of all f F such that M is singular with respect to Baire category in F.

Proof Obviously (iii)

(u)

:

is of the second

(i).

Recall that if M is an analytic subset of a complex manifold 2, then a point a E M is called regular (a E Reg(M)) if there exists an open neighborhood U of a such that M fl U is a complex manifold. If a E Sing(M) := M \ Reg(M). then we say that a is singular — cf. [Chi 19891, § 2.3.

304

3 Envelopes of holomorphy for special domains

(iii): Let Y be as in Lemma 3.4.2 and let A be a countable dense subset of (I) Y. Then, by Lemma 3.4.2, ,•

U

Fx,r,

(x.r)EAxQ dX\M (x)
where

:=

(f E F:

> r}.

F for x E Y and r > dX\M(x). Hence, by the same argument D as in the proof of Proposition 1.7.6, F \ is of the first Baire category. We know that

Remark 3.4.4. (a) If M = with respect to {1/f}.

f'(O), where f E 0(X), f

0, then M is singular

(b) If M is an analytic subset of X, then by [Chi 19891, Appendix I, (a E M is pure M: dima M < n — 2} c M (n — 1)-dimensional. (C) Recall that

if X isa Riemann—Stein domain and M is a pure (n — 1)-dimensional

analytic subset of X, then X \ M is pseudoconvex (cf. Corollary 2.2.17) and therefore M is singular. Consequently, if M is an analytic subset of a Riemann—Stein domain X, then M is singular if M is pure (n — 1)-dimensional. (d) Notice that a pure (n — 1)-dimensional analytic subset M of X need not to be singular if X is not a Riemann—Stein domain. For example, let X

:= E2 \ {(zi. z2): Izil < 2/3, 1/3 < Z21 < 2/3), M := {(l/3,z2): 2/3< 1Z21 < 1).

Then each function holomorphic in X \ M extends to E2.

Proposition 3.4.5. Let (X, p) be a Riemann domain over C". Assume that M is a pure (n — 1)-dimensional analytic subset of X. Let M = UIEI M, be the decomposition of M into irreducible components (cf [Clii 1989], Section 5.4). Then = Uj In

the set Ms.c is also analytic.

Proof Put!0 :={i €1:

!\Io,Nj

M,,j =0,1.

We have to prove that M3,F = N1.

M fl Reg(M). It is clear that B is nonFix an I E jO and let B := empty and open in M fl Reg(M). We will show that B is also relatively closed in M fl Reg(M).

Suppose that b is an accumulation point of B in M fl Reg(M). Using a biholomorphic change of coordinates in a neighborhood of b we reduce the problem to the following elementary remark:

305

3.4 The envelope of holomorphy of X \ M

Given a function f E

:= Then

27r1

that extends to IP(c, r) x E C E", define

x

f

f(z'.

z

— Zn

x

!E. 2

1 extends f to

Since M, fl Reg(M) is connected (cf. [Chi 1989], § 5.3), we conclude that M, fl Reg(M) C Mns,7. Thus (cf. Remark 3.4.1(b)) each function f E F extends to the domain Y := X \ (M \ (N0fl Reg(M))). Observe that Y = (X \ N1) \ (N0 fl Sing(M)). Recall that dim Sing(M) n — 2.

Hence, by [Chi 19891, Appendix I. any function f E F extends to X \ N1. Thus

0 The Dloussky theorem. Let (X. p) be a Riemann domain over Ce?, let

a: (X,p) be

—+ (X.15)

the maximal holomorphic extension, and let M be an analytic subset of X. Let us consider the following condition:

(E) There exists an analytic subset M of X such that C M and aIx\M (X \ M, p) —p (X \ M, 5) is the maximal holomorphic extension. Remark 3.4.6. (a) M must be singular (because X \ M is a domain of holomorphy) .

foranyf E (9(X\M)letf E O(X\M)besuchthatfoalx\M = f. Then C a_I(M). the function isanextension of f to X\a '(M). Hence Indeed,

Suppose that a point a E a'(M) is not singular. Let U be a univalent connected neighborhoodofasuchthateachfunctionf E (9(X\M)extendstof (9(U). Then foreach g E O(X\M) the function o(aIuY' extends toa(U); contradiction. (b) We may always assume that M is singular and a(M) C M. (c) To check (E) it is sufficient to find an analytic subset M of X such that a — '(M) C

M and (X \ M, p) —÷ (X \ M, j3) is a holomorphic extension. Indeed, we only need to take := M,r (cf. Proposition 3.4.5). By virtue of Theorem 2.5.9, we get the following important result.

Theorem 3.4.7 ([Gra-Rem 1956]). Condition (E) holds if there exists an analytic subset M of X such that = Proof Recall that M5 is pure (n — 1)-dimensional (Remark 3.4.4(b)). Since a is locally biholomorphic, a(M5) = a(X) fl M is also pure (n — 1)-dimensional. Since X is Stein, M3 coincides with the union of all pure (n — 1)-dimensional components of M (Remark 3.4.4(b,c) and Proposition 3.4.5). Hence a(M5) C M5. =: M. Now, we Consequently, we may assume that M = M= apply Theorem 2.5.9. 0

The main result of § 3.4 is the following theorem due to Dloussky [Dlo 1977].

306

3 Envelopes of holomorphy for special domains

such that t9(X) separates Theorem 3.4.8. If (X. p) is a Riemann domain over points in X, then (E) holds for any analytic subset M of X. do not know whether Dloussky's theorem is true without the assumption that (9(X) separates points We

Remark 3.4.9. Theorem 3.4.8 remains also true in the category of pluripolar sets, i.e. we assume in (E) that M is a closed plunpolar subset of X (cf. Definition 2.1.18) and we require M to be a closed pluripolar subset of X — cf. [Clii 1993J.

Reduction procedures. The aim of this subsection is to show that the proof of Dloussky's theorem may be reduced to the case where:

n=2,

-

= (K(r) x E)U (E x (E \ K(p))) (for some 0 < r, p < I), X=T= M = H isa singular analytic subset ofT such that H C K(r) x K(p). The reduction procedure will be divided into the following four steps: (n)

• Proposition 3.4.10: It suffices to consider only the case X = T = Tr,p (0 < r,

(n)

p < I), where Tr.p

=

r
zI
:= {(z, w) E C U {(z. w)

0

and

Tr,p

E

< p
• Proposition 3.4.11: It suffices to consider only the case where X = T =

(n)

and M = H is an analytic subset ofT such that for each z E

:={w E

E

C: (z. U')

E

T}).

• Proposition 3.4.13: It suffices to consider only the case where X = T = andM = H is an analytic

(n)

T r.p

x K(p).

• Proposition 3.4.14: It suffices to consider only the case where X

(2)

T

Tr,p

and M = H is an analytic subset of K(r) x K(p). ProposItion 3.4.10.

To

prove Theorem 3.4.8 ills sufficient to consider only the case

(ii)

X=

i.e. to prove that for any domain T = Tr,p and for any analytic subset H of T there exists an analytic subset of E't such that E' \ is the envelope of holomorphy ofT \ H. where

Proof Fix a Riemann domain (X. p) over C" such that (9(X) separates points in X and a singular analytic subset M of X. Let a: (X, p) —+ (X, be the maximaL Recall that

K(a, r) := B1(a, r) C C, K(r) := K(O. r).

3.4 The envelope of holomorphy of X \ M

307

holomorphic extension. Since 0(X) separates points in X. we may assume that X is a subdomain of X, /3 = p on X, and a id. Let (X \ M, p) —p (Y, q) be the maximal holomorphic extension. Then, by (Y, q) —* (X, j3) with o = Id on Proposition 1.9.2, there exists a morphism

X\M.

=

LetY =

(X,j3)beasinPropositionl.5.1l. \ M) is an open subset of Y and that maps biholomorphically W onto V := X \ M C X. Hence, by Proposition 1.5.13, there (Y , q

First observe that W :=

exists

an open set U C Y with W C U such that

Consider(i :=

X —+

Y

.

Then

=

U —* X is biholomorphic.

(X, p) —+ (Y

,

q

) isaholomorphic

extension (recall that Y is connected). Suppose for a moment that (Y =

Then

*Q

(Y

.

q)

.

q ) is a Riemann domain of holomorphy. =

Then M is an

—f (X, /3) is an isomorphism. Put M :=

analytic subset of X and

(Y, q) —* (X \ M, /3) is an isomorphism.

Thus, it remains to show that (Y q ) is a Riemann domain of holomorphy. By Theorem 2.9.1(v) we only need to show that for any T := Tr.p and for any biholo,

morphic

mapping f: T —+ f(T)

c

biholomorphically to 1: E" —÷

=

Y

such that g := o f: T X extends C X, the mapping f extends holomorphi-

callytof: E" —+ Y. Let H :=

fl E). Then H is an analytic subset of T. By virtue of our assumptions there exists an analytic subset H of E'1 such that E" \ H is the envelope of holomorphy of T \ H. By Proposition 1.9.9. f: T \ H —+ Y extends to a locally biholomorphic mapping f: E" \ H —÷ Y. Observe that o = g on T \ H. Hence, by the identity principle, o = on E't \ H. In particular. is biholomorphic. By Proposition 1.5.13 there p: f(E" \ H) —+ \

f

f

exists

an open set U C Y with f(E't\ H) C U such that

biholomorphic. Now we can put / :=

=

U —÷ g(E")

Is

U

0 (n)

Proposition 3.4.11. To prove Theorem 3 .4.8for X = T = T r.p it suffices to consider only those analytic subsets H of T for which each fiber

:= (wE C: (z, w) E H}, as an analytic subset of the whole fiber

:= (wE C: (z,w) E T) is discrete,

z E

For the proof we need the following

3 Envelopes of holomorphy for special domains

308

Lemma 3.4.12. Let H be a pure (n — I )-dimensional analytic subset of a domain and let

D C B

= B(H) := (A

C

xC'

w+

(z + (w +

H}.

Then B has zero measure. ForA

E

C"1 put

1;: C"' x C —÷

w) :=

x C,

+ uA, to).

= is biholomorphic and the analytic subset The lemma says that for A outside a zero measure set in has discrete fibers = (w E := of the domain DA := C: (z. w) HA) = (to E C: (z + Aw, to) e H}. Observe that

isadomain,) > I, Proof of L.emma3.4.I2. ObservethatifD = U then B(H) = B(H fl Di). Thus we may assume that D is a ball. In particular. H = A

cf. [Chi 19891, § 2.8. Let

where f

C x C"': 3e>o:

:= {(z. w, A) e

=

Obviously, B

G

VEEK(F): (z + (to

+ E)A, to +

H}.

(A). Put

xC

:= ((z. to, A) E

x

(z

+ wA, w) E D}.

Observe that

A=((z, u',A)EG: = ((z. to.

A)

G:

w+E.A)EG, f(z +(w+E)A, f(k)(z + wA, w)(A, 1) = 0}

VtEK(f):(z. :

(f(k)(a)(X) denotes the value of the k-th differential of f vector X E

Let r :=

at the point a

D on the

x C). In particular, A is analytic in G. IA• Suppose that for some (Zo, too, A0) E Reg(A) we have

rank r'(Zo, wo, A0) = n — I. Then, by the rank theorem, there exist a neighborhood W0 of A0 and a holomorphic mapping w: Wo —+ C"' x C such that = (zo. wo)

and (p(A),A)E AforallA€ Indeed, by the rank theorem, there exist:

A0) and V C open neighborhoods U C Reg(A)of(zo. —+ U.'V: V biholomorphic mappings 4):

ofA0.

such that

r(U) C V. 4)10.0,0) = (zo.

As), 4'(Ao) = 0, 'V o to CD(a. fi. y) = y for any (a. fi. y) E x C x C"'. The above identity implies that 'V o Write 4) = Cn 4)A): 'VA(O. 0.41(A)) = 41(A) for A E W0 := E"t. Hence 4)x(O. 0. 41(A)) = A for A E W0. We define 0. 41(A))), 42(A) := satisfics all the required conditions.

A

E W0. Then (42(A), A) = 41(0.0.41(A)) and, therefore.

3.4 The envelope of holomorphy of X \ M

:=

+

xC,

W0 xC —÷

x

Wo —+

309

+

+

0) = (zo + w0A0, wO) E D. We know that for any A E W0 there exists an e(A) > 0 such that ifr(A, E H for E r(A)E. Take a neighborhood W C W0o1 A0 and a connected neighborhood V C C of 0 such that *(W x V) C D. Notice that

Consequently, by the identity principle,

( W x V) C H and therefore rank

(A.

n—i forany (A.

E W x V. On the other hand, one can easily prove that

det *'(A,

=

+

+

+ do(A).

0; contradiction. E V such that det We have proved that rank r'(z. w, A) n — 2 for all (z. w, A) E Reg(A). Hence r(A) has dimension < n — 2 and, consequently, r(A) is of zero measure (cf. [Loj

Thus, for any A E W there exists a

1991]) 52)

D

Proof of Proposition 3.4.11. We assume that the extension theorem is true for all analytic sets with discrete fibers. Fix (T, H) with an arbitrary singular subset H ofT. Let I denote the set of all numbers s E Fr, I] such that there exists a singular analytic such that any function from O(T \ H) extends holomorphically to

subset H5 of

Let ro := sup 1. Observe that I is an interval. Moreover, H' fl = H5 for any S. t E I with s t. and any function Put H' := H5. Observe that H' is an analytic subset of T) extends holomorphically to Tr0.p \ H'. Taking HT0 from (9(11 \ we conclude that ro E I. We have to show that = I. Recall that for A E we have defined

PA(z,w):=(z+wA,w). Fix 0 < e << 1. By Lemma 3.4.12 with D := T, there exists a A E

such that the set c1_A(H) has discrete fibers. Observe that D Notice that T' — e) x E is the envelope of holomorphy of T'.

T'.

By the assumption (applied after a rescaling). there exists a singular analytic subset H' of T' such that any function holomorphic in T' \ c1_A(H) extends holomorphically to V \ ,?'. Consequently, each function holomorphic on T \ H extends — 2e) x E C Consequently. to \ 4A(H'). Observe that s := 1 — 2r E I with H5 HU fl

Thus,r0=l.

D

We can argue also directly: first observe that r(Reg(A)) is of zero measure cf. [Fed n — 2 for all

19691. Th. 3.4.3). Next, in the same way as above, we prove that rank r'(z. w, A) (z, w, A) E Reg(Sing(A)). Hence r(Reg(Sing(A)>) is of zero measure, etc. Recall

that 4(1') fl T,,, C H because H is singular.

310

3 Envelopes of holomorphy for special domains

Proposition 3.4.13. To prove Theorem 3 .4 .8 for X = T = ii suffices to consider only those analytic subsets H ofT for which I! C P,7 — (r) x K (p). Proof Suppose that the extension theorem is true for all analytic sets H C x K(p). Fix (T. H) with an arbitrary singular subset H ofT. By Proposition 3.4.11 we may assume that H has discrete fibers. Let ro be as in the proof of Proposition 3.4.11. We have to prove that ro = 1. Suppose that r ro < I. Observe that H fl {IwI > p} (because H is singular). In particular, by the identity principle for analytic sets. HT° has discrete fibers.

The maximality of ro and the compactness of (ro) imply that there exists a (ro) such that we cannot find an open neighborhood LI of zo and a singular zo E analytic subset H of V U x E such that any function from \ HT0) extends holomorphically to V \ I-I. Fix such a Since I! has discrete fibers, there exists p < p' < I such that

{w: wi = p', (zO, w) E H) = 0. Since H is relatively closed in T, there exists an

p'—e
x {p' — e

e>

Iwi

0 (at least so

small that p <

p'+e}) = 0,

(1 — e/ro)zo. Put TI

:= (to. 0) +

C

By our assumption (applied to (T', HTO fl T')) there exists a singular analytic subset

H'

of

x K(p'+r) such that any function holomorphic in

extends holomorphically to

\

Let V : Pn_ i (co. Zr) x E. We have proved that each function from \ Hr0) extends holomorphically to V \ H", where H" H' U (H fl V). Now we take H and we have a contradiction. D

Proposition 3.4.14. It suffices to consider only the case n = 2. Proof We proceed by induction. Suppose that Consider an arbitrary problem (T, H) with a assume that H is

the extension theorem is true in C" —

3. By Proposition 3.4.13 we may

a singular analytic such that H C

the projection

H Notice that H' fl

(z, w) —÷ z E

C H'°.

x K(p). In

particular.

3.4 The envelope of holomorphy of X \ M

311

N, functions an analytic proper subset E of Pn_1 (r) such that

Consequently (cf. Theorem 2.12.2), there exist m

is proper.

(9(Pn_i(r)),

E

•H=

and

where

QH(Z, w) := wm +

• for any := E functions E (9(U), j = I

there exist an open neighborhood U C m, such that:

\E

and

(z)foranyz Uandj #k, j=1 rn), z EU, = {(z,

•foranyzo E E:

<en. Notice that the polynomial Qil is uniquely determined. By Theorem 3.4.7 it is enough to show that the functions 'Pj' J = I holomorphically to

m, extend

xC xC

LetC'

K(r) consider the (n — 1)-dimensional problem (T', 11(q)), where

For any (n—I)

IF'n_2(T) x K(p): r,, w) E H). By the inductive assumption there exists a singular analytic subset 11(q) of such that \ 11(q) is the envelope of holomorphy of T' \ H(q). Moreover, H(r1) fl > p} = 0. In particular, the projection

H07) := ((v, w)

T

E

w) —f

11(q) is

proper. Hence 11(q) =

E

where

w) =

+ j=I

j =I

E

m(q)

rn(q). Notice that 11(q) fl T' C H(q) and therefore

m.

Let G denote the set of all q E K(r) such that there exists a E with Then K(r) \ G is an analytic q) E ci, i.e. G = {q e K(r): ?n...2(r) x {q} subset of K(r) 55)• In particular, for any q E G the projection

w) —'

H(q)

IP'n_2(r)

has fibers consisting of nz different points.

j=I

—"fk =O)forsomef1 k,then

(K(r)\G)flK(m.r) =

fl

tEP.,..2(to,r)

=0. j =

I

k}.

3 Envelopes of holomorphy for special domains

312

Now let Go denote the set of all e G such that the set for Then H(17) fl T' = Go, and consequently,

is singular.

j=l...., m.

,n(rj)—m.

(3.4.2)

We will prove that G \ G0 consists of isolated points (in particular, Go is open).

Take an f (9(T \ H) such that H is singular with respect to (f) (cf. Proposition 3.4.3). We will show that the set S of all e G such that is not singular with respect to lf(. is analytic in G. Suppose that there exists an infinite sequence C S \ qo with —p := G. For any v there existsapoint E H(171) such that •)extends By Proposition 3.4.5 the function f(.. to a neighborhood of .) extends to a neighborhood of any point from fl Reg(H(171)), where denotes any that contains irreducible component of Let

:=

77v)

E},

Q :=

Moreover, since each set

Q is closed in Q V and let a

Zo := (to, '70). Let U, *i 1/fm be as previously with x C. Observe that each point of U connected, U = U' x U" C U" is a regular point of that for any U' the function f(•, with

v

For any v >> I and

(i'

>>

1). Consequently, we may assume to a neighborhood of a point

extends

1.

E U' there exists a j = j(v,

=

ifl,)), Since all the points we may assume that irreducible component of of E. Consequently, there exists a Jo such that j (v) = = I for all v 1. may assume that Consider the expansion ii.

Jo

+ w) = qo, O)withw

q,

m} such that E (I U', must lie in the same v) = j(u) is independent for infinitely many v. We

E

liv) = 0

0.

<0. Hence the function f(.... .) extends to a neighborhood of the point (go, 'in. 1/'i (to. rio)) E H; for all

<0 and

I. Hence, by the identity principle, a11

0 for all

contradiction. m, extend holoBy (3.4.2) and the Hartogs lemma the functions coj, J = I x Go. Observe that they are locally bounded on E"2 x K(r) morphically to (use Viète's formulas). Consequently, by the Riemann theorem on removable singux G and, finally, to m, extend holomorphically first to larities, 'PJ. J = I

x K(r).

3.4 The envelope of holomorphy of X \ M Consequently, we get a singular analytic subset H of 1' := T U

313

x K (r) x E)

such that H fl T = H and each function from O(T \ H) extends holomorphically to > p} = 0 and hence the projection H (z. w) —4 T \ H. Moreover, H fl w E x K(r) is proper. Now we repeat the first part of the proof with respect to the first variable z E E, i.e. m, j=I

if x x K(r)) x E (q, w). Consequently. the functions extend holomorphically to E x E"2.

[1

Oka theorem. Let Z be a connected complex n-dimensional manifold. Recall (cf. § 1.1) that a pair (X, p) is called a Riemann region over Z if Xis a Hausdorif space and p: K —p Z is a local homeomorphism. If X is connected, then we say that (X, p) is a Riemann domain over Z. We say that (X, p) is locally pseudoconvex if for each point zo E Z there exist a such neighborhood V of zo and a biholomorphic mapping Cl,: V —+ cl,(V) C ci) o p) is pseudoconvex, as a Riemann region over C", in the sense that 2.2.

Observe that the above definition is independent of (I) in that sense that if (I): V —+ cD(V), 'I': V —÷ V) are biholomorphic, then Dop) is pseudoconvex (V), o p) is pseudoconvex.

if (p'

Proposition 3.4.15. If(X, p) is a Riemann domain over C", then (X, p) is pseudoconvex in the sense of Section 2.2 (X, p) is locally pseudoconvex.

Proof If (X. p) is pseudoconvex in the standard sense, then we take V := C", c1 := id. Conversely, assume that (X, p) is locally pseudoconvex. We will prove that (X. p) satisfies condition (p4) from Theorem 2.9.2.

Let a E aX. Let zo := p(a) and V be an open neighborhood of zo such that (V), p) is pseudoconvex. Let U be a connected component of p (V) such that U E a (cf. § 1.5). Notice that (U, p) is also pseudoconvex and, therefore, holomorphi-

cally convex (Theorem 2.5.7). Put U := U U {b E ax: U E b}. Then U is a neigh-

borhood ofa in X and X nU = U. Thus (X flU. p) is holomorphically convex. U

If T C Z, then any continuous mapping a : T —÷ X such that p o a = idT is called a section of(X, p) over T (cf. § 1.1). The set of all section overT is denoted by T°(T, X) (sometimes, f°(T. (X, p))). Similarly, as in Proposition 1.1.5, one can easily prove the following identity principlefor sections. It T C Z is connected, ai. E f°(T, X), and ai (Zo) = c2(zo) for some Zo E T, then aI a Riemann domain over To simplify notation put m := n — 1.

x C.

(C) Assume that there exists an R > 0 such that r(Etm x (C \ K(R)). K) let a: iftm x (C \ K(R)) —* X be a fixed section of (X, p).

0 and

314

3 Envelopes of holomorphy for special domains

Consider the family C of all relatively closed subsets L c Em x C such that: • for each z E the set := (wE C: (z, w) E L) is convex and compact, • there exists a section o'L: x C) \ L —+ X such that 0'L = a outside LU(Em K(R)). Obviously, Em x K(R) E C. Put

K:=flL. For z E Em let L(z) denote the family of all convex compact subsets £ C C such that there exists a section (z} x (C \ £) —+ X with az,€(z, ') = a(z, ')outside

tuk(R).

Obviously, if L E C, then

E L(z) for any z E Em. Put

k(z)= fl

e,

ZEEtm.

eEe(z)

Lemma 3.4.16. (a)K EL. = k(z), z Em. (b) Pmof (a) For any L1, L2 the sets C \ ((L1 U (L2)z), z E Em, are connected 56), Hence, by the identity principle for sections, = aL2 in (Em x C) \ (L1 U L2), which shows that O'K ULEe a zo E By the same argument as in (a) we can prove that k(zo) E .C(zo). Obviously. k(zo) C Suppose that \ k(zo) 0. Then there exists an open half-plane Q such that Let r := azo,k(zo). The set B := r(A) c X k(zo) C Q and A := \Q o is compact and univalent. Take an e > 0 such that C := is also univalent and let U = V x W C Em x C be a convex neighborhood of (zo} x A such that U C p(C). Let > 0 be so small that for z E Pm(Z0, C V we have {z} x \ Q) c W 58), 56)

If A, B C C are compact and convex, then U := C \ (A U B) is connected. = 0. then there exists an open half-plane H such that A C H. B c C \ H. We may assume that H = (Rez < O}. Let r > 0 be such that AU B C {IlmzI < r}. In particular, the connected set S := (R + ir) U (i(—r, r)) U (R — ir) is contained in U. Moreover, for any point r] joins zo with S + i[—r. Yoi or the segment + Zo = xo + io U either the segment inside U. In the case where A fl B 0 we may assume that 0 A fl B. Let r > 0 be such that Indeed, if A fl B

A U B C K(r). In particular, the circle S := dK(r) is contained in U. Moreover, for any point 8f) E U the segment [ro. joins zo with S in U. Notice that the result remains true for compact convex subsets of RN. N 2. 57) If S C C is compact and convex, then S = fl Q, where the intersection is taken over all open half-planes Q such that S c Q.

58)

Let

> ObesuchthatPm(zo.

for z C such that K. \ Q C that and let \Q

q)x

C U. ItsufficestoprovethatthereexistsO <S

q

5). Suppose that there exists a sequence —p zo such Recall that K C Em x K(R). In particular, Q) \ the fibers z E E", are unifonnly bounded, and therefore, we may assume that —+ A. On the other hand, since K \ (Em x Q) is relatively closed in x C, we conclude that wo A; contradiction.

3.4 The envelope of holomorphy of X \ M

315

Define

L :=

U

(

{z} x

u

U (z}

(

x (K- fl Q)).

ZEPm(Zø.tS)

that L is relatively closed in Em x C and that the fibers of L are convex and compact. Moreover, w) = w) for Z E Pm(ZO, 8) and This implies that L E C; contradiction. 0 wEW\ Observe

Lemma 3.4.17. Assume that (X, p) is locally pseudoconvex. Let

Y := p1 (Em xC) = X \

x {oo}).

Then (Y, p) is pseudoconvex in the standard sense.

Pmof By Proposition 3.4.15, it suffices to prove that(Y, p) is locally pseudoconvex. Take an arbitrary E and let V c Em xC be such that there exists a biholomorphic mapping 4): V —÷ 4)(V) such that 4) o p) is pseudoconvex. Let V0 C 4)(V) be an arbitrary pseudoconvex neighborhood of 4)(zo). Let W := 4) o p) is also pseudoconvex. In particular, Then, by Corollary 2.2.20, taking V0 sufficiently small, we may assume that V C and 4) = id. It remains to observe that = 0

Lemma 3.4.18. Let (X, p) be as above. Assume that (X, p) is locally pseudoconvex. then K = 0. If k(zo) = Oforapointzo E

PmofLetDo:={z

E

=0}. ThenzoE DoandDoisopen. Let

x C). We know (Lemma 3.4.17) that (Y. p) is pseudoconvex in

Y :=

the standard sense. Let

T := (D0 x C) U (Em x (C \ K(R))). Observe that r := aKIT isa well-defined section of(Y, p) overT. Notice that Em xC is the envelope of holomorphy of T. Hence, since (Y. p) is a domain of holomorphy (Theorem 2.5.7), r extends to a section on f: Em x C —+ Y (Proposition 1.9.2). which implies that K = 0. 0 If k(z)

Oforanyz E Em, then we put D(z) := diam(k(z)), := max{Iw — woI: W E k(z)},

z E Em,

E C;

D is called the diameter function for (X, p) (with respect to R and a). Observe that E £(z) for any z E Em. K(wo.

FirstnotcthatthesetS:=U\Kiscormccted(thefibersS. =

V.arc

connected, the base V is connected, and, therefore, the whole set S must be connected). Consequently, by the identity principle for sections. we only need to observe that tik(zo. = r(zo. u0) = wO) for any point WOE W \

316

3 Envelopes of holomorphy for special domains

Lenuna 3.4.19. (a) D is upper semicontinuous. (b) D(z) = urn +

— 21w1), z E Em.

Proof (a) Let v

1.

vt,, E be such that —÷ Zo and let = IWL, — —* We may assume that E k(zo). Then —+ Wo E k(zo), Iwo — vol < D(zo), which shows that D is upper semicontinuous.

(b) Fix a z and let a. b E k(z) be such that D(z) = Ia — bi. Using the formula

,/fl:7= I +t/2+O(t2)ast +

lim sup

—+ O,weget 21w1]

—

>limsup[Iw—aI+I—w—bI—21W1]

=lim sup [11w12

=llmsuplWl [Ii =lim sup

—

+ al2 +

2Re(tha)

— (2Re(tha)

+ (2Re(thb) + Ib12)/IwI2 _2]

1a12)1Iw12 +

+0

+ Re

Re

(h)]

= la—bl= D(z).

W

—+ oo be such that

Conversely, let

+

urn sup

—

2Re(thb) + 1b12 — 21W1]

E k(z) so that —* ao may assume that as above) we get:

— 21

Take

lim sup

E

WI] = lim

= lWv —* k(z),

+

= lim Re

—

+

bo E

v > 1. We =I— k(z). Then (using the same method

21w1]

((by — WV

—

)

<

— bol

0


Proposition 3.4.20 (Oka theorem). Assume that (X, p) is locally pseudoconvex and K 0 (cf Lemma 3.4.18). Then: (a) log any w E C with IwI > R. In E by Lemma

3.4.19, D E

(b)logD E Proof (a) Fix IWol> R. Let h: C —* C, h(w) h(K(wo, s)) = C\ K(wo, l/s). Let c1: Ctm x —+

+ 1/(w — Ctm

wO).

Observe that

xC, c1(z, w) :=

(z, h(w)).

3.4 The envelope of holomorphy of X \ M

317

Put ,3 := Cl) op = (pi

Pm' h o and consider (X, /5). Obviously, (X. /5) is a pseudoconvex Riemann domain over Em x C. Put a := a o cD Then is a section over Em x (C \ h(K(R))). of (X. /5) We will prove that

=

> 0: 3fEr(z)xK(wo.sL(x,p)• f(z, wO) = a(z.

sup{s

=: ro(z), Indeed, fix a z

wo)}

z E Em.

:=

Etm and let

1ff: {z} x K(wo. s) —÷ X is a section

of (X, /5) with f(z, wo) = &(z, wo), then

r :=

o CD: {z'} x (C \ k(wo, us)) —÷ X

is a section of (X. p) with r(z, oo) = f(z. wo) = ã(z, wo) = a(z, 00). Hence K(wo. us) E £(z) and, consequently. so 5 I/s. Thus I/so ? rO(z). Conversely, let £ := K(wo, so). Recall that £ E £(:). Then

f :=

(z} x (C \ K(wo, i/so)) —* X

o

is a section of (X. /5) with f(z, wo) = a.,(z. 00) = a(z, 00) = ã(z. wo). Thus I/so < ro(z). Let Y :=

x C). Then (Y. /5) is a pseudoconvex Riemann domain over C" (Lemma 3.4.17). Observe (cf. § 2.2) that

ro(z) = Recall that — log WO))

-

wO)),

z

Em

(Theorem 2.2.9), Consequently, log

E

=

£PS.R(Em).

(b)ByLemma2.l.28itissufficienttoshowthatforanya

E

Candj

E

{l

,n}

the function Em

z —p

(3.4.3)

is psh.

FixaandjandletCD: Ctm xC —÷ Ctm xC,cD(z,w) :=(z,exp(az1)w). Define /5 := CD 0 P. Then (X, /5) is a locally pseudoconvex Riemann domain over E" x C. Moreover, r: T —÷ Xis a section of(X, p) 1ff to CD(T) —b Xis a section in particular, a := a o of (X. is a section of (X, /5) over Em x (C \ K(R)), where R := Rexp(IaI). Let K and D be constructed for (X, J3) in the same way as K and D for (X, p) (clearly, K 0). Then D(z) = I Hence, by Lemma 3.4.19 and (a) (applied to (X, /5)), 0 E Consequently, (3.4.3) is plurisubharmonic. 0

318

3 Envelopes of holomorphy for special domains

Hartogs theorem. Let S C E" be relatively closed such that

\ S is pseudoconvex. The aim of this subsection is to describe some analytic properties of Sin terms of fibers z E

Proposition 3.4.21 (Hartogs). Assume that S C E'1 is a relatively closed subset of E" such that G : = E'T \ S is pseudoconvex (i.e. S is pseudoconcave) andfor any z' E'T — I the fiber of S over z' Consists of exactly one point = go(z'). Then go

Remark 3.4.22. The above result may be generalized into the following directions.

Let S C E" be relatively closed such that E" \

S

is pseudoconvex. Then:

(a) (Cf. [Oka 19341, [Nis 1962]) if there exists a non-pluripolar set F C E F the fiber S is analytic. (b)(Cf. [Sad 1984], [Sad 1986])!! is polarfor almost all z' E then S is pluripolar (c) (Cf. [Chi-Sad 1988]) IfS ispluripolai Sn x (0)) = 0, and discrete for almost all z' e then S is analytic. (d) (Cf. [Chi-Sad l9881) if polar for all z' not belonging to some pluripolar set, then S is pluripolar Proof of Proposition 3.4.21. By the Hartogs theorem it suffices to show that cp(aI

foranya1

E 0(E)

n—l.Since

a1_i.a1+i

Gfl({(a1 is

.

xE x

xE)

also pseudoconvex (Corollary 2.2.18), we may assume that n =

2.

First we check that go is continuous. Let E Zo E E and suppose that K(go(zo), r) C F. Then wo go(zo). Let r > 0 be such that K(go(zo), r) for v>> I. Consequently, x K(go(zo), r) C G, v>> 1. By go(z1) the Kontinuitätssatz we get (zo) x K(go(z0), r) C G; contradiction.

Fix a zo E E. We will prove that go is holomorphic near Zo. Using Mobius transforms, we may reduce the problem to the case where zo = go(zo) = 0. By Theorem 2.2.9 we have: — log 8G,e' E Observe that fSG,e2(z, w) = min{jw — I — Iwl), (z, w) E G. Let 0
1—Iwiforizi
psh in a neighborhood of the Compact {(z. w): Izi

r, 1/4

Iwl

have

log wi

1

— I 2,r .'o

log 1w — go(re'°))d9,

1/4 <

3/8.

3/8). We

3.4 The envelope of holomorphy of X \ M

Hence (using the continuity of

2ir logp

J

(1

319

we get

dt

lo:Ipe't —

logipe" _co(reI8)ld:)dO.

=1

3/8.

By the Jensen formula (cf. LRud 1974], Th. 15.18) we get

I

2ir Jo

loglpe"—wIdt=Iogp

independently of w with Iwl
JZr

f

log Ipeul

—

w(rebo)ldt)

dO

= 2jr Iogp.

Consequently,

log Iwi =

— I 2ir .io

and, therefore, for any 1/4 harmonic. Since 1w —

1/4 < Iwi

log 1w —

3/8 the function K(r)

wi

= lw!2 + wi2

z

3/8.

—p log 1w — cp(z)l is

—

we conclude (taking w and —w) that the function thq + is of class Now, taking w and iw with w real, we see that 'p E C°°(K(r)). It remains to show that = 0 in K(r). Take 1/4 < Iwi <3/8 and let u(z) := 1w

—

— —

azaz

—

IaUaU u2 dz

— U

Hence

a2u

w(z)l2. Then

au au

in K(r).

320

3 Envelopes of holomorphy for special domains

After elementary calculations we get -

0=—w 2-w— +

1W— - —— \. az

W

-2

— WU)

...f

+ 2ww —w

/

-

-2

— — —- + W

\

az dz

\

azaz

dz az -

—w

2 aw

+

\

8WdW

2

—- — — —- + cQ —: az az azaz ,

2dz)z

2dWdW

az 8z

dz8z

Oz

Hence

a2w

—

—

—

az

—0

— az

is holomorphic on K (r). It remains to exclude the case where W 0 and the case. Consider the biholomorphic mapping C2 (z. w)

—

In particular. either W or

E

(9(K(r)). Suppose that this (z, z + w) E C2. Then the

domain 0(G) is pseudoconvex. Moreover, for z E we have: (z. w) E 0(S) for z near zero. (z, w — z) E S. Thus fibers of 0(S) consists of exactly one point From the first part of the proof it follows that either or t/i is holomorphic. Observe that = I and, therefore, cannot be antiholomorphic. = z + W(z). Hence On the other hand, if is holomorphic, then w is holomorphic; contradiction. 0

The proof of the case n =

2. We already know that to prove Dloussky's theorem we only need to consider the case where H is a singular analytic subset of T = Tr.p C C2

with H C K(r) x K(p). Letfi: (T\ H, id) —+ (Z, q') be the maximal holomorphic extension. Recall that is injective. Since E2 is the envelope of holomorphy ofT, we see that q'(Z) C E2; cf. Proposition 1.9.2. Put

Q:=

E

x(C\K(p)),

Q':=

E

x(E\K(p)).

3.4 The envelope of holomorphy of X \ M

321

Let (Y, q) denote the Riemann domain over E x C obtained by gluing (Z. q') with Obviously. (Y, q) is locally pseudoconvex. It is evident that (Q, id) along (Y, q) satisfies condition (C) (defined in the subsection 'Oka theorem') with R = p.

:=

U id.

Consider the family U of all pairs (G, M) such that

• G is a planar domain with K(r) C G C E, • M is an analytic, pure one-dimensional subset of G x E with proper projection M

(z, w)

z (in particular, by Remark 3.4.4(c), M must be singular).

•MflT=H.

Obviously (K(r). H) E it. First observe that for any (G, M) E U there exists a section (G

x E)\M

—* Z

= a on (K(r) x E) \ H. Indeed, G x E is the envelope of holomorphy of T fl (G x E) (cf. the proof of Proposition 3.4.5). Hence, by Theorem 3.4.7, (G x E)\ M is the envelope of homomorphy T fl (G x E) \ H. (In particular, M C G x K(p)). Now, we see that (T U (G x E)) \ M is a holomorphic extension ofT \ H. Consequently, there exists such that 0G,M

a morphism

((TU(G x E))\M,id) —* (Z,q') such that aGM = a on T \ H.

For(G1,M1),(G2,M2)€itput(Gi.Mi)-<(G2,M2)ifGi CG2andMi = that if (C1, Mj)IE, is a chain of elements from it, then UIE! M,) E U. Hence,bytheZomtheorem, thereexistsamaximalelement (U,EI (G. M) of U with respect to the relation -<. Fix such a maximal pair (G*, M*). Recall that aG.,M. is an extension of a — to simplify notation we will write a* instead of aG.,M.. Let M2 fl (Gi x E). Observe

M*

z E E.

(z, w)

wm + be the describing Recall that ir is proper. Moreover, let Q polynomial for M* (cf. the proof of Proposition 3.4.14). We are going to prove that G* = E, which will finish the proof.

Lemma 3.4.23. Let y:

[0,1)

be a curve such that

y(t) =

zo E

G' fl E and the set I E [0, 1)}

has exactly m different limit points ((zo, WI) E G*.

Proof (1) First we prove that for any wo E C \

toaneighborhoodof(zo, we).

(zo, Wm)) when t —+ 1. Then Wm} the

section

extends

322

3 Envelopes of holomorphy for special domains

Indeed, let

c2:={w0EC\{wi

.

Wm}

the section

We

extends to a neighborhood of (zo, wo)).

we have: K(w0, s(wo)) C c, where

will prove that for any wo E

j=I

s(wo):= then

=

C

\

Wm

}and the proof will be completed.

Let h: C —÷ C, h(w) := l/(w — wo). Note that = Define'D: CxC —÷ Cx C,D(z, w) :=(z,h(w)). Put4 := Doq and consider the domain (Y, 4). Recall that r is a section of(Y, q) overT ifff := to is a section of (Y. 4) over (l)(T). In particular, there exist S > 0 and R > 0 such that there exists a section a: K(zo, 8) x (C \ K(R)) —-÷ Y of (Y, 4). Observe that fort E LO. 1) such that y(t) E K(zo, 8) we have: Fix a

E

k(yU)) = conv({h(w): (y(t), w) E Ma)), where k (z) denotes the minimal convex compact singular set (cf. Lemma 3.4.16 in the subsection 'Oka theorem'). Let D: K(zo, 8) —p R+ denote the diameter function for (Y. 4) (defined in the subsection 'Oka theorem'). By Proposition 3.4.20(b) and the classical Oka theorem for subharmonic functions (cf. [Vla 19661) we have:

D(zo) = tim sup D(y(t)) = lim sup diam(conv((h(w): (y(t), w) E Ma)))

= lim sup max (0.I)a:—.I

— (

W

:

(y(t), w'), (y(t), w") E Ma

W —

I

WJ—WO

1

WkWO

:j,k=l

ml.

In particular. (Y. 4) has a section f over {zo) x (C \ K(0,

with f(zo, oc) = m}. This —wol: j = 1 directly implies that (Y, q) possesses a section rover {zo} x h'(C \ K(0, = {zo} x K(wo, s(wo)) with r(zo, wo) = a(zo, wo). (2) We pass to the main part of the proof. By (1) we know that there exist numbers sm. 0 < e < minfs1 Sm } such that if we put K0 := K (zo, e), K3

+e).then: • K0

E,

flKk = øforanyj #!c, m,

•thesectiona*extendsonAj := Ko x

a1,j=l

,n.

—e <

<sj +e}toasection

3.4 The envelope of holomorphy of X \ M

the

323

By virtue of the maximality of G*, it remains to show that for each j E (1 set Mt extends to an analytic subset of K0 x Fix a j. Now let (X, ji) denote the Riemann domain obtained by gluing

ni

Ko x (C\ K(wj,s, —e)) andq'' (K0 x K,) <S3 +e}). One can easily prove that (X, j5) is locally pseudoconvex. Let D denote the diameter function for (X, We know that D(y(r)) = 0 for t near I. Recall that log D is subharmonic on K(zo, r) (Proposition 3.4.20(b)). Hence, by the classical Oka theorem for subharmonic functions, we conclude that D 0. This means that there exists a function extends to the K(zo, e) —+ K(w1, Sf) such that x

—S <

complement in Ko x K1 of the graph M1 of

In view of the definition of the envelope

of holomorphy, we conclude that (Ko x is biholomorphic to a connected \ component of and, therefore, it is pseudoconvex. (Ko x Now, by Proposition 3.4.21, must be holomorphic. 0

Lemma 3.4.24. G* is simph' connected.

Proof Let

[0, 1] —+ Gt be an arbitrary closed Jordan curve. We are going to

show that mt

C

:= G* fl

Put

:=

E —+ (I) First, we prove that non-tangential limit of at z. discrete in

Let

{z be the

m}. Notice that is \ universal covering of for a.a. z E ÔE, where denotes the

M*

x E). Observe that is a covering. Let —÷ lifting with F,(0) am}andletFj: E —* Write F3(z)), where Fj: E —+ E, j = I m. Let Aodenotethesetofallz I m,exist. It is well known that (RE) \ Ao is of zero measure. By the identity principle, the set S of all z E Ao such that 17(z) for some j k is of zero measure. t < I, y(l) := Then Fixaz E A := Ao\ Sand let y(:) Indeed, let

= {ai =

y is a curve such that the set (y(t)): t I, namely, 17(z)). = I y(l) = r(z) e Gd'. Observe that

j

[0, 1]) has exactly m limit points when ,n. Hence, by virtue of Lemma 3.4.23,

#,r'(a) <m}.

c

=rnforz c A,weconcludethatE*(z) E

forallz

E A.

(2) Since G* is maximal, to prove that mt C function extends holomotphically to mt Yo. j = I

it suffices to show that each m.

324

3 Envelopes of holomorphy for special domains

In fact, we only need to show that for each j E (1 m } the functions Re Pj and extend harmonically to mt Fix a j and let Ii := Re Let h denote the solution of the Dirichiet problem in mt with boundary data h (cf. ERans 19951). We want to prove that h = h in Im

For, observe that the function h0 := h o — h o (I), it has the radial limits 0 a.e. on 8E. Hence h0

is harmonic in E. Moreover, by 0 and, therefore, h = h on

0

The continuity of h and h implies that h = h on

The main part of the proof Since Gt is simply connected, the exists a biholomorphic

mappingf: E —+ G*. It sufficestoprovethat 11*1=1 Let A denote the set of all z E

such that I t E

exists and the set

[0. l)}

has exactly in limit points when t —+ I. Then, by Lemma 3.4.23 and the maximality

of G*, we conclude that lf*(z)I = l,z E A. It remains to prove that (dE) \ A is of zero measure. o f)*(z), Let A0 denote the set of all z E 8E such that f*(z), exist. It is well known that (SE) \ is of zero measure. Let R(a1 am) denote the resultant of Q and Q' with

j

=

I

Q(w) := wm + yajwm_). 0, the identity principle implies that the setS of all z E A Since 'Pm) such that R((çoi 0 f)*(z) ('Pm o f)4(z)) = 0 is of zero measure. Now, we only

0

need to observe A0 \ S C A.

3.5 Separately holomorphic functions Let

Uc

C Ctm be domains and let 0

A c U, 0

B c V. Define the

cross

X

= X(U. A; V, B) := (U x B) U (A x V).

We say that a function f: X —+ C is separately holomorphic on

V.EA: We may

E (9(V)

&

YWEB:

X (f E 05(X)) if

w) E (9(U).

assume that f(0) = 0. Let f = zg. where g E O(E. Es). The case where

g = consi E 8E is obvious. Thus suppose that g: E —p E1. Put h : log igi. Then h is harmonic on E. Observe that h is bounded. (The only problem is to show that infE > 0. Since f(0) = 0 and f is injective. there exists an e > 0 such that If(z)I > e > 0 for IzI > 1/2. Consequently. Ig(z)I > e for Izi > 1/2.) Since h = 0 a.e. on aE. we conclude that h 0 (cf. [0011969]). which shows that

f is a

contradiction.

325

3.5 Separately holomorphic functions

It is natural to ask whether there is a 'maximal' open set X C C" x cm such that on X fl X (notice that, in general, we do not require that X C X).

for any f E (95(X) there exists an 1€ (9(X) with f =

f

Let U, V. A,Bbe as above.

X = X(U, A: V. B)

{(z, w) E U x V: hAU(Z) +

<

1).

Notice that if U and V are domains of holomorphy, then X is a region of holomorphy (cf. Corollary 2.2.15). Observe that if = 0 on A and = 0 on B, then X C X. In particular, if

A and B are open, then X C X. The main result of this section is the following general answer to the problem of continuation of separately holomorphic functions defined on a cross (ISic 1969], [Zah 1976], ENgu-Zer 1991], see also [Maz 1975], [Kaz 1986], [Shif 1989], [Shif 1994] and the survey article [Ngu 1997]).

V c Ctm be bounded domains and let A C U, B C V be non-pluripolar sets. Assume that U is a domain of holomorphy and A is of type 62) Let X:= X(U, A; V. B), X in X(U, A: V. B). Then for any f 05(X) there exists an f (9(X) such that f = on X fl X. Theorem 3.5.1. Let U c

f

Observe that in the case where U = A = V := E, B K(r), Theorem 3.5.1 reduces to a version of the classical Hartogs lemma. Before the proof we need some auxiliary results.

0
I,

If S is a connected Remark 3.5.2. (a) Let c C C" be an open set and let N C 1 on S or component of c, then = hNnSS on S. Moreover, either <

I

(b)Let c1

foranyz ES. C

c

c

C" be open. If N1 C

1.

Lemma 3.5.3. Let c C C" be a bounded open set, let P C c be N C Then

and let

1*

—

In particular,

1.

Proof We only need to show that Let v E v —oo, be such that P c Since c is bounded, we may assume that v 0 on 61) 62)

Recall that := sup{u: u C u < I on u That is, A is the countable union of closed subsets of c".

(cf. Theorem 2.1.39). 0 on N); cf.

§

21.

326

3 Envelopes of holomorphy for special domains

< Ion 0, we haveu+rv < I oncandu+ev Letting 0onN. Henceu+ev —* 0, we get u hN,c on c \ Consequently, u (cf. the proof of Proposition 2.1.12) and, therefore, hN\p.c

on

e

0

Remark 3.5.4. In the case where is unbounded Lemma 3.5.3 remains true for all plunpolar sets P c f2 such that there exists a function v E PSJe(f2) with v $ —00,

v <0, P C We say that a subset N of an open set C C" is locally non-pluripolar in any connected component S of c the set N fl S is non-pluripolar.

Lemma 3.5.5. If c

C

C" is an open bounded set, N C

if for

then the following

conditions are equivalent: (I)

N is locally non -pluripolar in

< lforanyz E a

(ii)

Pmof We may assume that is connected. The implication (ii) from Lemma 3.5.3. To prove that (i) (ii) suppose that

(I) follows I. The set

P := {z Consequently, there exists a point a E f2 such that = 1. This means that for each k E N there exists a function Uk such that Uk < I on 0 on N, and Uk I —2± Putv := I). Thenv E ?S3((c2)(cf.Remark2.l.2(d)), uk(a) v(a) —I, and v = —oo on N. Thus N is pluripolar; contradiction. 0 Lemma 3.5.6. Let f2, / that

Pmof Let V

E

c be

is bounded. Then

:=

£ E N, N, /

open, and let N, c

\

N. Assume

pointwise on <

Obviously, I. Put

Let v :=

Then


< u(z)}.

:= (z E

It is known that P, is pluripolar (Theorem 2.1.41). Let P := note that P is pluripolar. Observe that v = u, 0 on N \ P. Consequently, by Lemma 3.5.3, V hfr,,\pQ = 0

Lemma 3.5.7. Let N be a non-pluripolar subset of an open bounded set C C".
Pmof Fix ana E (0.1). We have to prove (cf. Lemma 3.5.5) that

< I,

Z E Ga.

Let

No := {z

N:

= 01.

3.5 Separately holomorphic functions

327

Then No c N 11 Ga and therefore it suffices to show that

<

1

for any

Z E Ga. Observe

that N \ N0 C {z E c:

Consequently, the set

<

=

N \ N0 is pluripolar and hence, by Lemma 3.5.3,

P := (z

E

Put

E

notice that P is pluripolar (cf. Proposition 2.1.40). Define

u(z)

on Ga

1

onc2\Ga. (Proposition 2.1.9), u < 1, and u = 0 on N0 \ I

Then u

E

=

u

P.

Thus

(Lemma 3.5.3) and, finally,

< I in Ga.

S

U

Let f C C'7 be hyperconvex (cf. Definition 2.1.46) and let K set. Define /2 = /2K,r2 := (cf. Definition 2.1.43). Recall that rem 2.1.48).

be a compact

\ K) =

is a Borel measure and

0 (Theo-

/

Lemma 3.5.8. Let c C'7, Gt / G c Ctm be open, and let J N C f, G€ j M1 / M C G. Assume that G are bounded, and Nt is locally non-pluripolar in

:=

£ E N. Put

Nt; G1, Mt),

V :=

V1 :=

N; G, M),

N1; G1, M1),

V :=

N,

£

N; G, M):

/ I (Lemma 3.5.6). Let g be such that for each £ E N there exists a E 0(Y1) such that = g on V1 (1 V,. Then = on V1 (t E N)and, therefore, there exists a E 0(Y) such that = g on V fl Y. notice that V1 / Y and

Proof Fix ant Vt

The functions

Nand(zo, wO) E Vt. Recall that

= {(z, w)

(., wo) and

W := {z E c,: with a := 1 —

x G1:

E

< 1}.

wo) are holomorphic in

(z, wo) E Y1} = (z E f21:

<

Obviously. zo E W. Notice that if a =

1,

then W =

(because N1 is locally non-pluripolar in cf. Lemma 3.5.5). We know that w0) = Ie(z, wo) = g(z, wO) for z E N1 fl W. By Lemma 3.5.7, the set N1 fl W is locally non-pluripolar in W. Consequently, wo) on W. (., wo) U

3 Envelopes of holomorphy for special domains

328

Lemma 3.5.9 (cf. [Mit 1961]). Let Ho. be separable Hilbert spaces, dim Ho dim H1 = 00, and let T: Ho —* be a linear infective compact operator 63) such that T(Ho) is dense in Hi. Then there exists an orthogonal basis (hk)kEN C H0 such that is an orthonormal basis in IIbk

=: vk / +00 when k / +00. if there exist a local/v convex nuclear space V and linear continuous op-

V Hi such that T = T2 o T,, then the above basis may erators Ho be chosen in such a way that, the series is convergent for any

e >0. Proof For the convenience of the reader we shall give a rough idea of the proof based on the following textbook on functional analysis [Mei-Vog 19771, see also [Jarc (a) Since T is compact there are orthonormal systems (ek)kcN C Ho. (fk)kEN C = 0. such Hi, and a monotonically decreasing sequence (Sk)kEN C that

Tx =

XE H0,

where the series does converge in the operator norm (cf. [Mei-Vog 19771, PropositiOn 16.3). This representation of T is called the Schmidt representation of the compact operator T. Applying that T(H0) is dense in it follows that (fk)koN an orthonormal basis of (cf. [Mei-Vog 19771, Proposition 12.4). Moreover, since dim = 00 we have Sk > 0, k E N. Finally, the injectivity of T implies that (ek )kEN is an orthonormal basis of Ho. Put bk := 4/sk. k E N. Then Tbk = fk. k E N, and

/ 00.

(b) Before discussing the second assertion of the lemma we like to recall a few general facts from functional analysis, all of them may be found in [Mei-Vog 1977). Although the systems (ek)kEN and (fk)kEN are not uniquely determined the sequence (sk )koN is. It is called the sequence of singular numbers of the compact operator T. For 0 < r
Sr(H'. H") := (A E K(H'. H"):

E

where K(H'. H") denotes the set of all compact linear operators between H' and H" and where (Sk(A))k0N is the sequence of singular numbers associated to the compact

operator A. Sr(H', H") is the so-called Schatten r-class (cf. [Mei-Vog 1977), § 16). We will need the following fact concerning Schatten classes (cf. [Mei-Vog 19771, Lemma 16.7): 63

Recall that a linear operator T: X —* Y. where X and Y are locally convex topological vector spaces, is compact if for any bounded set B C X the set T( B) is relatively compact in Y. We would like to thank Professor K. Floret (Oidenburg University) for the idea of the proof.

3.5 Separately holomorphic functions

329

j = 2,..., 5, be Hubert spaces and let A1: H2J —p H2J+t. j = 1,2, be continuous linear. IfS E Sr(H3. H4), then A2 o S o A1 E Sr(H2, H5),

(*) Let H1,

0 0 such that

V:

(x

e} c U.

Moreover, for any continuous Hubert seminorm p on V there are a Hubert space the completion of V/{x E V: p(x) = 0} with respect to the scalar product, that defines p, and a natural continuous linear operator

XE'V, where [x} denotes the class of .x in V,,,.

Using that T2 is continuous we find a continuous Hilbertseminorm p on V and a —÷ H1 such that T2 = T2 o ii,. continuous linear mapping T2: In virtue of Proposition 28.4 of [Mei-Vog I 977J, for any e > 0 there exists a continp, such that the natural operator ip,q: Vq —+ Vp uous Hubert seminorm q in V, q

(induced by the identity of V) belongs to S€(Vq. Vt). Applying (*) from (b) to

T=

T2

0

oiq o T1 gives that T E

H1). Hence,

.Sic(T)E <00.

0

be a strongly pseudoconvex open and let N C S2 be compact and local!)' non-pluripolar in c. Put Ho := (cf § 1.2), in L2(N. := the closure of Hi

Lemma 3.5.10 ([Ngu-Zer 19911). Let 2 c set

Then the linear operators

are well defined, infective, and continuous Ti is compact. In particular T.=T201\ the operator Ho IN E H1 is compact. Let be as in Lem,na3.5.9. Then for anya E (0, 1)andfor C H0, any compact

I

KC

{z


A Hubert seminorm is a senünorm p that can be defined by a semi-scalar-product. i.e. where (.,.) satisfies all conditions of a Hermitian scalar product except the positive definiteness. &,) Recall that any strongly pseudoconvex open set is hyperconvex; cf. Proposition 2.2.25. 67) Recall that O(c) is a nuclear space.

p(x) =

3 Envelopes of holomorphy for special domains

330

there exists a constant C

C(a, K) > 0 such that

Cvi',

IIbktIK

k E N.

(*)

Proof Step 1°: 7'2 is well defined.

Every function f E (9(7) can be approximated uniformly on N by functions (because strongly pseudoconvex — cf. Proposition 2.7.7). Moreover, uniformly on N, then fsIN C Ho and if Hi. Step 2°: T2 is injective. Suppose that f E (9(c) is such that T2(f) = 0 in H1, i.e. js(P) = 0, where /1 := 0}. By Proposition 2.1.49, we get = and P := lz E N: f(z) Now, since Hence, by Lemma 3.5.5, the set N \ P is locally non-pluripolar in 0 on N \ P. we conclude that f 0. Step 3°: T1 is continuous. It is well known that for any compact L C there exists a constant c(L) > 0 such from

f

f

that

hilL 5j c(L)hIflIL2(Q),

f

(+)

E

which is equivalent to the continuity of T1.

The continuity of T2 is trivial. Step 4°: T1 is compact. Let c H0 be bounded. Then, by (+), it is locally uniformly bounded in Consequently, by the Montel theorem, there exists a subsequence that is locally uniformly convergent.

Step 5°: The estimate (t). Fork Nsuchthatt'k > l,let

uk:= By (+), for any compact L C

log log vk

we get

ilL
k EN.

Thus the sequence (uk)k is locally bounded from above in

Let

u :=limsupuk. 1. Moreover, Since Uk —* +00, we get u P := {z c: u(z) < u*(z)}; P is pluripolar (cf. Theorem 2.1.41). = 0 such that u We will show that there exists a Borel set Q C N with

on N \ Q. Put Qk,p.q

:= {z EN: bk(Z)l

k,

p,q €N

0

3.5 Separately holomorphic functions

(Qk.p.q is compact). Recall that IIbk

=

I2(Qk,p,q)

1.

331

Thus we have

=

IN

Let

Qp,q := U Qk.p,q.

p. q E N.

Then, by Lemma 3.5.9. 00

(d'l

const(q) Vk

—

2

k= I

= 0, where

Consequently,

Qq :=flQp,q, =

Finally,

q €N.

0, where

Q

Qq for every q. Consequently, for each q there exists a p(q) such that zo Qp(q),q. This means that ZO Qk.pq),q for any k. i.e. k E N. Consequently, Ibk(zo)I <

Observe that if zo E N \ Q. then zO

logp(q) Uk(ZO)

Hence u(zO)

+

1

—.

q

l/q. Since q was arbitrary, we conclude that u(zO) = 0.

Thus 0 on N \ (P U Q) and hence = By Proposition 2.1.49 and Lemma 3.5.3 we get = < In particular, SUPK u* < a. Hence, by the Hartogs lemma Thus (Proposition 2.1.23), Uk a on K for k>> I, which implies the required result. 0

PmofofTheore,n 3.5.1. Fix an f Step

10: Reduction to the case where A is compact and B is relatively compact

in V. Since A E

of compact sets such that in C", there exists a sequence is not plunpolar in U fort>> 1. We may assume that Ae / A. In particular, is not pluripolar in U for any 1.

3 Envelopes of holomorphy for special domains

332

of relatively compact and non-pluripolar

/

In a similar way we get a sequence (Bt subsets of V such that B.

I

Suppose that for each £ there exists an fj E (9(X(U, V. Be)) such that Ic = on X(U, Ae; V. fl X(U, Ac; V, Bt). Then, by Lemma 3.5.8, there exists an f e

0(X)suchthatf=fonXflX.

Step 2°: Reduction to the case where U is strongly pseudoconvex, A is compact, B is relatively compact in V. f(., w) E 0(U), w E B, and is continuous and bounded on A x V.

f

By Step 10 we may assume that A is compact and B is relatively compact in V. Since U is a domain of holomorphy, we may find a sequence of relatively compact, strongly pseudoconvex subdomains of U with A C Uk / U. Fix a sequence of relatively compact subdomains of V such that B C V. Fix a k and define Vk

/

Akf := (z E A:

YU,EVA.:

If(z.

< £},

Ie

N.

/

It is clear that Ak., A when I / +oo. In particular, Ak,, is not pluripolar for I 1(k). Fix an I 1(k). Observe that Ak,, is compact and f is continuous on Ak,, x Vk. Wj) —* CE C. lndeed,letAke x Vk (Zj. Wj) —÷ (Zo. wo) E Ax Vk We have to prove that c = f(zo. wo). —* g locally uniformly By the Monte! theorem we may assume that on Vk with g E 0(Vk). For any w E B the function f(., w) is continuous on U and, consequently. w) —÷ f(zo. w) = g(w). Thus f(zo,.) = g on B and, therefore, on the whole 14. (B is non-pluripolar in l'k). Finally, c = g(wo) =

f(zo,

wO).

Suppose that for each I 1(k) we have a function fk,, E c9(X(Uk, Ak,,: Vk, B)) such that Ike = on X(Uk, Ak,: Vk, B) fl X(Uk. Ak,; 14. B). Then, by Lemma 3.5.8,thereexistsanfk E 0(X(Uk, A: Vk, B))suchthatfk = fonX(Uk, A; 14., B)fl

f

X(U, A;Vk, B). Now, once again by Lemma 3.5.8, we conclude that there exists an

f

E

0(X)suchthatf = f on XflX.

Step 3°: The case where U is strongly pseudoconvex, A is compact, B is relatively compact in V. w) E (9(U), w E B, and is continuous and bounded on A x V.

f

We may assume that fl

:=

Ho

:=

I on A x V.

:= the closure of HOIA in L2(A. be the basis from Lemma 3.5.10; := k E N. For any w E B we have f(., w) E Ho and f(., w)IA E H1. Hence Let

p). and let

3.5 Separately holomorphic functions

333

where

ck(w) =

I f(z, w)bk(Z)

=

f f(z,

k E N;

JA

(U) (in particular, locally uniformly in U) and in L2(A, ii). Since f is continuous on A x V, the formula

cf. Lemma 3.5.9. The series is convergent in

ck(w) :=

f f(z. w)bk(Z)

defines a holomorphic function on V. k

converges

w E V,

JA

N. We are going to prove that the series

iocally uniformly in X.

Take a compact K x L c X and let a > maxK h be such > a + fi < I. First, we will prove that there exists a constant C'(L. fi) > 0 such

that

that

k EN.

IICkIIL

(t)

Suppose for a moment that (t) is true. Then, using Lemmas 3.5.10 and 3.5.9. we get

>

IICkIIL IIbkIIK

C'(L,

= C'(L, fi)C(K, which gives the normal convergence on K x L.

We move to the proof of (t). By the Holder inequality, we get w E V. k E N

ICk(W)I

(recall that ICk(W)j

< 1 on A x V). On the other hand, if w

= ICk(W)I =

f f(z,

w)bk(z)

U

Uk

B, then

334

3 Envelopes of holomorphy for special domains

The sequence is bounded from above in V. u := and u —l on B. Let P := (w E V: u(w) < u(w)}; P is u

0,

Thus —l on B \ P. Consequently, I + <

(Lemma 3.5.3). = Hence 1 +ut on L. Now, by the Hartogs lemma, which implies (t).

> 1,

Let

f(z, w) := obviously f is holomorphic. Recall that I =

w) E

f on X fl (U x B). Now, take an

arbitrarypoint(zo.wo) e Xfl(A x V)andleta :=

E (0,1]. Then holomorphic in W := {w E V:
f

3.6 Extension of meromorphic functions The aim of this section is to discuss the problem of continuation of meromorphic functions. The main result (Theorem 3.6.6) says that the envelope of meromorphy of a Riemann domain coincides with its envelope of holomorphy. Let (X, p) be a Riemann domain over C".

Definition 3.6.1. A function f: X \ S —+ C, where S = S(f) is a closed subset of X with mt S = 0, is said to be meromorphic on X (f E if: (a) f E (9(X \ S) and S is singular for I in the sense of § 3.4 (b) for any point a E S there exist an open connected neighborhood U of a and functions (9(U), $ 0, such that = on U \ S (cf. [Rot-Kop 19811). The set

:= X \ 5(f) is called the set of regular points off.

Remark 3.6.2. (a) The following conditions are equivalent:

fe.M(X);

(i)

(ii) foranypointa

XthereexistsanopenneighborhoodUsuchthatfju E .M(U);

c X.

(iii)

(b) Let a, U,

as in Definition 3.6.1(b). We may always assume that the germs Taco and Tat/F are relatively prime in Consequently, we may assume that and are relatively prime for any x 68)

be

That is. f cannot be holomorphically continued across any point of S.

U.

335

3.6 Extension of meromorphic functions

Then S fl U = {x E U:

= O} and, consequently, either S = 0 or S is a pure

(n — 1)-dimensional analytic subset of X. We have the following two possibilities:

Inthiscasewesaythataisapoleoff(a E S'(f)). •

0: In this case we say that a is a point of indeterminacy off (a E 2(f)).

It is clear that 5(f) = ?(f) U 2(f) and ?(f) fl 2(f)

0. Observe that = 0). Thus 2(f) is an analytic subset of X of

2(f) P U = {x E U: ç9(x) = dimension

n

—

2

(cf. [Clii 1989], § 3.7).

is a field.

(c)

(d) (The identity principle for meromorphic functions) If f e and f = 0 on an non-empty open subset of R(f), then f 0. E (9(X), i/i 0, then there exists an f E .M(X)(with S(f) C

such thatf =

on X \ *_1(0) (we shortly writef = Indeed, define f := p/t/.t on X \ Then we extend holomorphically f to is the set of all singular points of (0) in the sense of X \ S, where S := (* § 3.4. We denote the extension again by f. Then f satisfies Definition 3.6.1(a) and, by the identity principle, /,f = in X \ S. (t) Let (Y, q) be another Riemann domain over C

a a C .M(X). P (9(Y \ S). S := 5(g). then f := g o a (9(X \ S)

Indeed, if g E

with S :=

(S). Since a is locally btholomorphic, the function f cannot be holomorphically continued beyond S (cf. § 3.4). Finally, if U is a subdomain of X t,!, such that alu: U —* U is biholomorphic and if i/ig = in U \ S (9(U),

00), then Let 0 F c .M(X). In virtue of the above remarks, the notions of the Fextension and maximal F-extension (F-envelope of meromorphy) from Sections 1.6 and 1.8 may be easily generalized to the case of meromorphic functions. Example 3.63 (The sheaf of I-germs of meromorphic functions; cf. Example 1.6.6). Let l 0 be an arbitraiy set of indices. Let

{(U, (f,),e,): U E For (U,

(f,)IE,). (V,

E

f.

.M(U), I E I),

a

we put

W CU P V, filw = Define

:=

,MW

:= (J

x (a),

(SEC"

—+ C", 69)

Recall that

69)

a)) := a. where f = [(U,

denotes the set of all open neighborhoods of a.

I El.

336

3 Envelopes of holomorphy for special domains

For any I-germ

(as

U

above), consider the set

:= (([(U, (fi)iEI)]L. z): z E U).

The family of such sets U defines a Hausdorff topology on such that each U is univalent (i.e. the projection jrW maps homeomorphically U onto U). Thus is a Riemann region over C's. For any i E I we define a function F: —+ C, F((f, a)) := f,(a). Since o on U, the mapping F1 is meromorphic on F, = f,

Let (X. p) be an arbitrary Riemann domain over

C .M(X).

We define a morphism

a=

(X. p) —*

(I := F). Namely,

a(x) := ([(p(U). (1 °

p(x)),

X

E X.

where U C X is an arbitrary univalent neighborhood of x. Moreover. F1 o a = f for any f F. Consequently, if X denotes the connected component of that contains the set a(X) and := then

a: (X.p) is

(X,13)

an F-extension.

In virtue of the above construction, using the version of the proof of Theorem 1.8.4, we get the following generalization of the Thullen theorem.

Theorem 3.6.4 (Thullen theorem). Let (X, p) be a Riemann domain over C" and let 0 F C .M(X). Then (X, p) has an F-envelope ofmeromorphy. Our next aim is to compare the envelopes of meromorphy and holomorphy. First we consider the following special elementary case.

ProposItion 3.6.5 (Levi) (cf. [Oku-Sak 1957]). Let T = Tr.p (0 < r, p < 1); cf Section 3.4. Then M(T) = .M(E")IT, i.e. any function f E .M(T) extends to a function meromorphic on E".

Proof (Cf. [Siu 1974b)). Fix an f E and let ro := sup(s meromorphically to We may assume that r = ro.

(0,

1): f extends

(I) First consider the case where f E (9(P,,_1 x {p < wi < I)). Then f may be represented by a Hartogs—Laurent series

f(z.

w)

(z. w) E

= wherefk

E

k=—oo

E

Z.

x

< iwi <

I },

3.6 Extension of meromorphic functions

337

denote the set of all z such that f(z. .) is meromorphic For p E Niet in E and has at most p poles counted with multiplicities (i.e. either (9(E) or there exist E N with + we E E, and si p such + that wt}) and the function E \ w —÷ E (9(E \ {wj extends holomorphically to E). In particular, f(z, w)(w — ... (w — for any z E A,, there exist numbers ao(z) Op(Z) E C, not all zero, such that the function w —+ f(z, w) extends holomorphically to E. Thus (*)

kEN. Observe that that dim .1(f)

j

(the projection onto of .1(f)). Recall (Remark 3.6.2(b)). Thus there exists a p such that A,, is not

A,, n

—

2

thin.

Consider the field F

and the vector space IFP+I over IF. Let vk := Takearbitraryk1 1,2 EN,

=

f—k—,,) E

k1 < ...
D:=det Using (*) we get D =

f—k1—!

f—k1—,,

f—k2. f—k2—!

f..k2....p :

Hence, since

0 on

is not thin, we conclude that D

0

Ofl

Consequently, the subspace V C FP+l spanned by the vectors has dimension d p. generate V over F. In particular, for any k E N p. Let vk Vkd, there exist functions ck,o, Ck,t 0, such that ckd E ck.o

(**)

ak,ovk = The columns in the matrix f—,11—,,

f—k2. f—k2—!

.f—k2—p

f—kd. f—kd—1

fkdp

linearly dependent over F. Hence the exist functions b0 all identically zero, such that 0, v = I we get 0, k N, and therefore the function f_k_f are

(z. w)

f(z.

b,, E

not

d. By virtue of (**)

338

3 Envelopes of holomorphy for special domains

extends holomorphically to T and, consequently, to a function g E O(E"). Finally, gives the required meromorphic exthe function (z, w) —÷ g(z. w)/

tension of f to E". (2) Now consider the situation when S := S(f) is such that for each z E ?fl_j the fiber S- := {w E E: (z. w) E S) is discrete in {w E: (z, w) E T); cf. the subsection 'Reduction procedures' in § 3.4. Then, using the method of the proof of Proposition 3.4.13 and (1), we easily conclude that r0 = 1, which gives the required result.

(3) It remains to reduce the proof to the situation as in (2). Let S := S(f). For put

E

4'A(z,w):=(z+wA,w). is biholomorphic and 4'j' = 4'A; cf. § 3.4. Fix 0 < e << I. By Lemma 3.4.12, there exists a E such that the set S' := has discrete fibers. Observe that (I)_A(T) D =: T'. The function g := f o 4'A is meromorphic on T' and holomorphic on T' \ S'. Hence, Recall that

by (2), the function g extends to extends meromorphically f to 4'x(Pn_,(1 — e) e —* 0 finishes the proof.

e)

x E). Consequently, o — 2e) x E. Letting

x E) j

0

Theorem 3.6.6. Let (X, p) be a Riemann domain over C", let 0

7 c .M (X), and

let

a: (X. p) —÷ (X, j5) be the maximal 7-extension (cf Theorem 3.6.4). Then (X, j3) is Stein. In the envelope of meromorphy of (X, p) coincides with the envelope

of holomorphy of(X,

70)

Proof We may assume that the maximal 7-extension is realized as the connected component of as in Proposition 3.6.4. To prove that (X, is Stein it suffices to verify condition (v) from Theorem 2.9.1 with (Y, q) = (C". Id) and = p. Thus letT = < r,p < I and let ci): T —÷ 4'(T) C Xbeabiholomorphic mapping such that ji o 4' extends to a biholomorphic mapping 4': E" —+ 4' ( E") C C". We want to prove that ci) extends to a holomorphic mapping Ci): E" —p X. Forany f F let F1 E .M(X)be as in Example 3.6.3. Then Fjo4' is meromorphic

on T and, consequently, by Proposition 3.6.5, it extends to a function I E .M(E"). Now define

:= ([(4'(E"), (fo

)fEc

cf. Example 3.6.3. One can easily check that 4'; E" —÷

z E E"; is a holomorphic

extension of ci). Obviously. (i)(E") C X. 70)

In the case of Reinhardt domains a simpler proof may be found in ISak 19701.

0

3.6 Extension of meromorphic functions

339

Corollary 3.6.7 ([Kaj-Sak 1967]). Let (X, p) be a Riemann domain over C" and let .M(X). Then there exist 0(X), 0, such that f = co/i/i.

f

The above corollary justifies to introduce meromorphic functions on Riemann domains over C" as global quotients of holomorphic functions; cf. Remark 3.6.2(e).

Proof The result is well-known if X is Stein — cf. [Hor 1990] follows via Theorem 3.6.6: if

The general case

f

(cf. Reis the envelope of meromorphy and f E is such that f o a = 0, such that i/if = on mark 3.6.2(1)), then first we find i/i E 0(X), iii (cf. X \ S(f). Consequently, we get (i/i o a)f = o a on X \ S(f). Thus f =

9

Remark 3.6.2(e)).

In [Kaj-Kaz 1971] the problem of continuation and quotient representation of meromorphic functions on a domain over X x C", where X is a product of complex projective spaces, is discussed. Below we present an alternative proof of the most important part of Theorem 3.6.6 based on the Dloussky Theorem 3.4.8. Theorem 3.6.8. Let (X, p) be a Riemann domain over C" such that 0(X) separates points in X. Let a: (X. p) —÷ (X, j5) be the maximal holomorphic extension. Then each function memmorphic on X extends meromorphically to X.

Proof (1) First observe that if M c X is a singular analytic set and if M denotes its extension obtained via the Dloussky theorem, then any irreducible component of M intersects

a(X).

Indeed, suppose that B is an irreducible component of M such that B fla(X)

= 0.

Let A denote the union of all other components of M. Then A is a pure (n — 1)dimensional subset of X such that M = (A). Consequently, by Theorem 2.5.9, X \ A is the envelope of holomorphy of X \ M. Since X \ M is also the envelope of holomorphy of X \ M, we get M = A: contradiction.

(2) Take a function f E .M(X) and let S := S(f). We may assume that S 0. Recall (cf. Definition 3.6.1(a)) that S is a pure (n — 1)-dimensional singular analytic subset of X. Let S denote the extension of S obtained from the Dloussky theorem. Then there exists an f 0(X \ S) such that 10 a = Jon X \ S. It remains to prove that f is meromorphic on X.

f

71) The proof is based on the sheaf theory: Assume that (X, 0. p) is Stein and f E .M(X). Let t9, denote the sheaves of germs of holomorphic and meromorphic functions on X. respectively. .4(. Since f(9 is isomorphic Then the multiplication by f defines a sheaf homomorphism 1: .M is also coherent. By Theorem A there is a non-trivial to 0 and (9 is coherent, the sheaf F := (9 fl global e (9(X). X —÷ F.

3 Envelopes of holomorphy for special domains

340

(3) Let A be an irreducible component of S. Then Ao := A flReg(S) is a connected

complex manifold. Let A denote the set of all points a E A0 such that f is meromorphic in a neighborhood of a. It is clear that A is open in Ao. 0. Let b be an accumulation point of A in A0. The problem of Moreover, by (I), A meromorphic continuation of f to a neighborhood of b reduces, via a biholomorphic transformation, to the question whether a function g holomorphic in E" \ M, where x E, extends meromorphically to M := x (0), and meromorphic in E". The positive answer to this question follows directly from Proposition 3.6.5 with T := (p arbitrary). Thus A = Ao and, consequently, f is meromorphic in X \ Sing(S). In the next step we will prove that f extends meromorphically to a neighborhood of any point from Reg(Sing(S)). Take a b E Reg(Sing(S)). The problem of meromorphic continuation of f to a neighborhood of b reduces, via a biholomorphic transformation, to the question whether a function g meromorphic in E" \ M, where We use once again x {l/2} x (1/2}, extends meromorphically to M := (n — 1)-dimensional

Proposition 3.6.5 with T := T113213.

Thus f is meromorphic in X \ Sing(Sing(S)). Now, finite induction finishes the proof.

0

Chapter 4

Existence domains of special

families of holomorphic functions

4.1 Special domains In this section we study pseudoconvex domains with many symmetries and give characterizations for them to be s-domains of holomorphy, where C (9. t'°°.domalns in the plane. In the plane there is the following description of analogous to the Cartan—Thullen domains of holomorphy in terms of theorem (cf. [Ahe-Sch 1975]).

Theorem 4.1.1. Let D c C be a bounded domain. if D is J(>°-convex. Then, D is an .le°-domain of

Let K C D

Pmof

be

K. Letf E Je°°(D).

compact and fix a

We claim that

iE

5

S := dist(z°, K). In fact, (*) is clear for j = for some j > 0. Set

(*)

where

0.

We proceed with induction. Assume that (*) holds

I

C.

Then

g(z) := (1(z)

—

(k)_O (z

—

—

g E 3e°°(D) and (j + l)!g(z0) =

Ifz e K, we get Ig(z)I

Therefore,

S If

+

+ I)! = lg(z°)I s

S

i.e. (*) is verified.

8/2). Using that D is an of (*) it follows that T,of converges on R°°-domain of holomorphy we conclude that 8/2 pD(z0). Because

342

4 Existence domains of special families of holomorphic functions

Fix a b E K with Jz° — bl =

and c E

with pD(z°) = lz°

—

ci. Then,

PD(K) — PD(Z0). Therefore, PD(Z0) which implies that D is .1(°°-convex. 2: Here we proceed via contradiction. So, let's assume that D is not an domain of holomorphy. Then there exist a E D and 0 < r < R such that 2pD(Z°) Hence,

lb — ci — Ic — z°I

.]3(a,r)C DandB(a,r)flaD={b}; • for every f E

the Taylor expansion Ta! off converges on B(a, R). Choose c E dB(a, r) fl D and e > 0 such that B(c, e) D. Moreover, fix numbers

r


0. PutK :=]3(a,ri)\D. Observe

that C \ K is connected and contains D and that also B(a, r2) \ K is connected. Then any f E \ K) extends to an entire function. Thus, \ K) = C. Using

the Riemann mapping theorem implies that K is totally disconnected. Thus we find an open set U with b E U C B(a, and C D. In virtue of the above conditions., we see that b is an accumulation point of contradiction.

0 The proof of the implication

was

based on the fact that D is a bounded

It seems to be unknown whether this implication remains true without

domain.

assuming that D is bounded

Remark 4.1.2. According to [Rud 1955] there are sequences (ak)k€N C IR>o, \ O,and(ck)kEN,ak > > O.k E N,suchthat Ic—.

ck

k=I

a,, — Ck

<00,

K(ak,ck)flK(aj,cJ)=O,

Put

G :=C\(UK(ak,ck)U{O)). kEN

G is an .1(°°-convex domain in C satisfying the following property: if f E 3(°°(G),

then lf(x)l <

x < 0. In other words, G is X°°-convex but not Caratheodory complete. Observe that G is biholomorphically equivalent to a of holomorphy, but bounded domain G' c C. So G' is an R°°-convex G' is not Carathéodory complete. Conversely, any Carathéodory complete domain is f(oo) =

0,

3C°°-convex (cf. [Jar-Pfl 1993] for more details).

Balanced domains. First recall that a domain D C C" is balanced if ED C D, or equivalently, if there is an upper semicontinuous function h: C" —÷ [0, oo), homogeneous (i.e. h(Az)

IA$h(z), A E C, z E C"), such that

D= Dh =(z E

C": h(z) <1).

343

4.1 Special domains

Observe that Ii = hD is uniquely defined; h is the Minkowski function of D. Pseudoconvexity of D = Dh can be characterized in properties of h. More precisely, we know that (cf. Proposition 2.2.22(a))

if

D is pseudoconvex

if

hE

log/i E

Introduce the following notation:

:=

u

{U E

homogeneous }.

In order to discuss characterizations of balanced domains that are R°°-domains of holomorphy we need the following function. Let D = be a balanced domain. —+ [0.00). Put/i = hD:

h(z) : =

=

QE

deg Q? 1.

Q

degQ

Obviously. ii is homogeneous and 0

1,

I on D) IQI 1Q1lIdesQ < Il). ')

h.

Ii

Starting from D = Dh we define h*

=

as

the upper semicontinuous regularization of ); h*(Z) < l}.

fz E

Observe that h* is a homogeneous psh function with 0
Dh

be as above. Then every f E

extends as a of holomorphy, and D* is fat.

D* is an

Proof Let f

We may assume that f is not identically constant and that on D. Then we have the following expansion off in a series of homogeneous polynomials

Ill <

I

with

IM

1)
Then, using the definition of/i we get

< h(z) < h*(z),

z E

j > I.

h* < I on D*, we see that the series is locally uniformly convergent on D*; in particular, it gives the holomorphic extension of 1. It remains to see that D* is an Je°°-domain of holomorphy. Assume this is not the case. Then there are open sets U1, U2 with 0 U1 C D* flU2 c U2, U2 connected. such that, if f then flu1 = flui for a suitable f E (9(U2). Since

Using the fact that is homogeneous, we find a point a E U2 \ D* with Therefore, there is Q 1. with QI 1 on 1. deg Q

h(a) > '>

degree

Observe that the zero polynomial 1.

is counted as a homogeneous polynomial of any

344

4 Existence domains of special families of holomorphic functions

D and Q(a)

11 deg

>

I

f

1.

Hence

I

<

QI

I on D*,

but Q(a) I > I

I.

Taking

0

(Q — Q(a)y' leads to a contradiction.

Corollary 4.1.4. Let D = Dh. Then the following pmperties are equivalent: (i)

of holomorphy;

D is an

(ii) there is a sequence (Qj)1EN C 1Q31 < I, j E Z+, on D (iii) h = h*.

Q = j, such that

=

In particular, if D = Dh is an R°° -domain of holonwrphy, then D is fat.

Proof It remains to prove (i) (ii): We know that there exists an E Je°°(D), ft < 1, such that f cannot be holomorphically extended through Let Q3 be the expansion of f in a series of homogeneous polynomials, i.e. Q3 E = 0 or deg = j. Then, as above, < I on D. Hence,
f

I

I

Suppose now that there is an a E C'7 with


Proposition 4.1.5. Let D = Dh be a bounded balanced pseudoconvex domain with continuous h. Then h = h. In particular h* = h and D is an .7(°° -domain of holomorphy.

Proof Obviously, h

Ii. To establish the inverse inequality it suffices to prove that

h(a) = I impliesh(a)? I. Fix an a E C" with h(a) = I. Recall that we know that D i5 8%om(C")cot'vex 2) Let 0 < t < I. Then the continuity of h gives that K, := {z E C": h(z) t} 2)

For the convenience of the reader we give the pmof: Let K C D be compact. Suppose C D that converges to a boundary point of D such that IQ(z*)I It QIItc. E Observe that there is an e > 0 such that h(z) (I + z E K. Put Q K, := tAz: Al i + e, 2 E K). Then K1 C D is compact. Since D is pseudoconvex. we find

there is a sequence (zk

f e 0(D) with

hulk, = I.

lf(zk)I —' oo.

k-. Developing I as a series of homogeneous polynomials, i.e. I = inequalities yields to hQ1(z)I <(I +rY',z E K. Therefore, lQj(zk)h hf(zk)I < oc;contradiction.

and applying Cauchy

(I +e)1.k EN.

Hence,

4.1 Special domains

345

is a compact subset of D. Thus, its hull is compact, too. Thereforewechoosenumbersr(:) (t, 1),O <1< 1,suchthatr(t)a Then we find Q,

I, with

deg Qi

E

= II Q1 IlK, < PQ,(r(t)a)I.

Now, the left equality leads to IQ,(tz/h(z))I virtue of the definition of h this gives In particular for z = r(t)a we obtain

h(a) = urn 1(r(f)a)

1,

z

0, or h on

< ,i in

t

urn

t-.i

=

1,

which completes the proof.

EJ

Remark. From Proposition 4.1.5 one easily obtains that any bounded complete pseudoconvex Reinhardt domain is an of holomorphy. (Exercise: show that in this case the Minkowski function is always continuous.) On the other side, the following example D := {z E C2: < I)

shows that something like the boundedness condition is necessary. (Exercise: show that .7e°°(D) = C.)

c OE

Example 4.1.6 [Ceg 1993]. Let

and choose

be dense in

C

= I. Wedefine u: C2 —÷ [0, oc).

u(z) :=

log Izi — aJz21.

j=i Then u ,'PSR(C°) (Exercise). Moreover, the function h: C2 —+ [0, oo), h1 exp(u) is homogeneous and psh. So it gives a bounded balanced pseudoconvex domain

D := {z

C2: h(z) := hj(z) + max{IziI,

1Z21}

< 1}.

Observe that

N(h) : = {z

C2: h is not continuous at z}

= (z E C2:

Izi I = 1Z21 and

is of Lebesgue measure zero. Therefore, D is fat cise).

u(z) > —oo}

and N(h) is not pluripolar (Exer-

Observe that any pseudoconvex D = Dh, for which N (h) is fat (Exercise).

is

a set of Lebesgue measure zero,

346

4 Existence domains of special families of holomorphic functions

To calculate h fix a point z0 = (z?,z20) E x Remember that z0 is a limit point of a sequence ((aj(k)z?, along which u = —00. Therefore, if I along that sequence. Hence, I Q(z°) 1. (C2), I Q I < I on D, then QI QE The maximum principle finally gives that IQI I on E x E. Hence, < I idc'u I. is obvious. Hence, D* = E x E. That h1 Summarizing, there are fat bounded balanced pseudoconvex domains which are not J(°°-domains of holomorphy, although, in contrast to the Cartan—ThuIlen Theorem 1.8.4, they are Hartogs domains of that type have been given by I

I

I

N. Sibony [Sib 19751. We conclude this discussion mentioning that the Hartogs triangle is an of holoniorphy but not (Exercise). Thus there is no analogous result like the Cartan—Thullen Theorem for of holomorphy. Combining Proposition 4.1.5 and the preceding example one might think that the size of N(h) is responsible for being an J(°°-domain of holomorphy. In fact, we have the following result due to J. Siciak [Sic 19841, [Sic 1985].

Theorem 4.1.7. Let D = Dh be a balanced domain. Then the following properties are equivalent:

(i)

D is an

(ii)

Ii

-domain of holomorphy;

=

(iii) D = mt (iv) D is an R°° fl A°°-domain of holomorphy. if D is bounded, then the above properties are equivalent to

(v) h

and N(h) ispluripolar.

E

In order to be able to prove Theorem 4.1.7 we need some special facts from the theory of psh functions, in particular, the generalization of Josefson's theorem due to J. Siciak [Sic 1982]. Before proving this generalization we provide a few preparations, all of them can be found in the work of J. Siciak.

Lemma 4.1.8. Let K C C" be compact. Then sup{exp(u): u

whereC(C")

E

£(C"). uIK 5 01 = sup{IPI11k: kEN, P 3c>o: u(z)

{u E

IPIxI < I),

C+log(l + IIzII).

z

(t)

E C"}.

Proof Obviously, the right hand side in (t) is majorized by the left part. Now, let v = exp(u), u E £(C") with uIK <0. We denote by C E (0, 1). the e-regularization of v (cf. section 2.1). Recall that v when e \ 0. Put (e Recall that

+

+

+ 1z112)'12,

ZE

C",

(C") means the set of all polynomials of degree less or equal k.

347

4.1 Special domains

and —

w) :=

where (z, w)

E

I

+

IwI(e +

if w 0 ifw=O,

+ 1w12)'12,

i sIIzII.

C" x C.

Then

1)

Moreover, the family

is

=

= v(z).

and urn

locally unifonnly bounded from above. Set

i)iij. j ? 2.

V3

< u(z), z E C".

Observe that

=

{z E

an ij > 0 and put

Fix

C": v(z) < I K, and

Obviously, K c G7. Therefore, the Hartogs lemma (applied to K, j gives an index j7 such that Vj(Z) 1+ z In virtue of Proposition 4.1.5 we have

Vu, = sup(IQI

IQI <

QE

<

I Ofl

1}}.

Then any competitor Q in that supremum has the property that

IQ(.

Ion K.

1)11(1

Hence,

kEN, P E

Vj <(1 Letting first j —p 00 and then In

all: E

IPI <

f

0

—÷ 0 gives (t).

the case that the compact set in Lemma 4.1.9 is circled, i.e. exp(it)K = K for R, more can be shown, namely

Lemma 4.1.9. Let K C C" be a circled and

compact. Then

sup{exp(u): u E .C(C"). uiK where

Ion K),

:=

Moreover,

QE

=

0)

= max{l,

'l'K},

Ion

degq 2: 1, iQi

K)5).

where

:= sup(h: h

?S3€hom(C").

his

1),

S cC".

Proof Observe that it suffices to show '<' in (i). In virtue of Lemma 4.1.8 the left expression is majorized by the following one k €N, P E .'Pk(C"), IPixi

Observethatmtheca.seofD= D5wehaveh =

l} =:

0,

Z

E C".

4 Existence domains of special families of holomorphic functions

348

Recall that K is circled. Now, let P E .'P(C't) be a competitor of o. We write whereq := degP and E P= = j. Fix az E K and regard the holomorphic function E

P(Az).

C,

By the maximum principle and the Cauchy inequalities we conclude that I Q(z)I z

Fq+l,

•

K —

s I,

Usingthe representation ofPwe have

s 'l'x onC".

K. Hence,

—(

(q + 1)44(z),

if'4'K(Z) 1.

instead of P the analogous estimate leads to

Dealing with

IPIs 5 (sq + 1)max{1. or p11/q

sEN.

(sq +

Hence, I P1' 7k <max{ 1.

N, k > q. Since P and deg P }, k max{l, 'l'K}. Using the compactness of K and the homogeneity of 4K and is an easy consequence from (1).

k were arbitrary,

weobtaino

Corollary 4.1.10. Let S C sup(exp(u):

the last equality D

be circled and bounded. Then

u E £(C"),

<max{l, c1s}.

(*)

Proof For abbreviation we denote the left-hand side of (*) by First we will show that

L5 = sup(LQ: f a balanced domain, S C Obviously, L5 >

=:

Assume now, there is no equality. Then Ls(a) > A >

some a E C". Choose u E £(C't) with uls

for 0 such that exp(u(a)) > A. Observe

that

Sc Exploiting that Thus

S

:=(Z EC": exp(u(z)) < l+e},

e>0.

is circled one easily finds a balanced domain S C C .'R. Ife —÷ O.thenA < exp(u(a)) 5

exp(u —log(1 +e))

contradiction. Fix now a balanced domain f S. Choose an increasing exhausting sequence of circled compact sets K1 C with U mt K1 = c. We know that ?

4.1 Special domains

j

ThenP(a.r) C

EN. Fora

._lonP(a,r),jj0.

Then,

< I on IF'(a,

and CI

E

with u <

E

349

r), J >

? Ja,and Therefore, u

I on ?(a, r). Since a was arbitrary, we obtain

I on Hence inf{CIK: K C K compact and circled } = Taking into account that = max{ I, for circled compact sets (cf. Lemma 0 4.1.9), the proof is finished. u

}

After all these preparations we present now the announced generalization of Josefson's result.

PropositIon 4.1.11. Let S C C' be pluripolar and circled. Then there is an h E h

0, With S

C

Proof The proof will need several steps. Step I: It suffices to prove the proposition in case when S is bounded. := Sn 3(0, J). are Assume S to be arbitrary. Then S = where bounded, circled, and pluripolar. Assume now that there are U) E 11) 0, Without with Si C loss of generality, SUp{u1(Z): z 1. Then, using the Hartogs lemma we find a point a E and a subsequence (uJ(k))kEN with gives a function in uJ(k)(a) > 1/2. Then, u := Step 2: Let S be bounded, circled, and pluripolar. Recall that

= sup{h(z): h Then it suffices to know

=

his S 1).

00.

In fact, let us assume that D = oo. For any m that 4s(zm) m. Therefore there are u1 sup{u1(z): z E

as wanted.

=

I

and

E

N there is a point Zm E with

lim sup{u1(z): z E S} =

with uj(k)(a) As above, we pass to a subsequence a Further, we choose a new subsequence := B,,.

< l/e on S. Then u :=

such

0.

1/2 for a certain such that

is the function we like to have.

Step 3: We claim that = 00. Suppose the claim is not true. Then exp(23), j E N. We set vj := u1(sup(u1(z): z E E,, with Observe that v3 Again, as above, the Hartogs lemma allows to find a E B,, and a subsequence with v)(k)(a) ? 1/2. Finally, we put V :=

1-_I

k=I

1/2&

Vj(k)

350

4 Existence domains of special families of holomorphic functions

It is clear that 0

v

6 ,I'S340m. Moreover, for z E V we obtain

=

fl

v(z)

0.

Therefore, V is pluripolar: contradiction. Applying (*) we conclude that Now, choose r > 0 such that 6 Sc B(O,r) and put M := z 6 B(O,r)}}. Since 2 IIzII/r we find another number R > 2r such that M(M + I) if R/2 IIzII < R. Recall that S is pluripolar, i.e. there exists a u 6 u < 0 on B(0, R), such

thatS c

Then, fore> Owe define I

max{eu(z) + log(M + I).

Izil


1k

> R;

well defined, it belongs to £(C"), and u1 0 on S. In virtue of Corollary 4.1.10 we conclude that max{O, log c1s). In particular, if IzIl < R, then UE is

eu(z) + log(M + I) Finally, e —+ 0 contradiction.

max(O. log

max(O. log

leads toM + I

max{l, M} for

IIzII

< r; D

Now we give the proof of Theorem 4.1.7. Proof of Theorem 4.1.7. Let D of(i) and (ii).

be

as in the theorem. We already know the equivalence

If I <

Assume condition (i). Let f E constant function. We write f(z) = and we recall that lQ3l <

I

Q3(z),

on D. Then

<

I

I. f(O) = 0, and let f be not a z 6 D. where Q1 E

on the domain (Exercise) G

mt b 0. The maximum principle gives that even

< I on G. Therefore, the series is locally uniformly convergent on G and gives a holomorphic extension of f to G. Hence, G C 0: (iii) is true. (iv): Suppose that (iv) is not true. We find, as usual, open sets U1, (iii) with the standard properties. Then there is a 6 U2 \ Choose p 6

with p(a)I > sup(Ip(z)I: z 6 0) =:

c.

Then f := (p

—

U2

cY' 6 R°° fl

Thus f is not extendible to U2; contradiction.

That (iv) implies (i) is obvious. What remains is to prove that (I) 4=> (v) when D is bounded. So, starting from now, D is assumed to be bounded. (i) => (v): Recall that h (respectively h) is semicontinuous from above (respectively from below). Hence h is continuous at all points, where h and h coincide. The remaining set is given by S := {z E D: I:(z) < h(z) = h(z)). Using the result of E. Bedford & B.A. Taylor (cf. Theorem 2.1.41), 5 is pluripolar. Hence, (v) is proven.

4.1 Special domains

351

(v) (i): We have to show that h = k". Since h is homogeneous it is clear that N(h) is a circled piuripolar set. So Siciak's result (cf. Proposition 4.1.11) can be Since D is 0, with N(h) C applied. Therefore we find u E .9'S340rn, u I on D. Recalling that u and h are logarithmically bounded we may assume that u 0 < t < 1. Moreover, v1 < I on D (use psh it follows that v, := hi_lu: E that N(h) C u'(O)). Then, in virtue of Lemma 4.1.9 we have

v,(z) =

h(z),

(*)

z E C".

is a set h(z). Since 0, then letting t —+ 0 leads to h(z) If u(z) of Lebesgue measure zero, we finally get, using Proposition 2.1.12, that h = h

0

everywhere.

C (0, oc) be as in

Example 4.1.12. Let the sequences (aj)ri C 13E and Example 4.1.6. We define u: C3 —÷ [0.00) as

u(z) :=

max{log Izi

—

log 1Z31}.

z E

C3.

j=I u

is psh on C3. Then the following function

h: C3 —p

h(z) := exp(u(z)) + Izi,

belongs to ll'S34om(C3). h(z) > 0 if z 0. Therefore. D := Dh := (z E C3: h(z) < l} is a bounded pseudoconvex balanced domain. Observe that N(h) C {z E C3: Z3 = 01: in particular. N(h) is pturipolar. Hence, D is a fat 7('°-dornain of

holomorphy. Moreover, using Example 4.1.6 we easily see that E x E x (0) C

= D, i.e. the of hotomorphy D that mt We know for an difference of the polynomially convex hull of D and D is 'thin', but it is, as the example showed, in general not empty. The example is a 3-dimensional one. In fact. it turns out that such an example is not possible in C2 (cf. [Sic 1985]).

Theorem 4.1.13. Let D = Dh be a bounded balanced domain in C2. Then the following pmperlies are equivalent: of holomorphy; (1) D is an (ii) N(h) is pluripolar and ii = lie, where h1 denotes the lower semicontinuous regularization of h; (iii) D is polynomially convex and D is fat. For the proof the following result of J. Siciak [Sic 1985] will be used.

Lemma 4.1.14. Letu

E

S3e(C). Assume that N(u) is polar and that

a+tog(l +IidI) where a, b E R.

+IidD.

4 Existence domains of special families of holomorphic functions

352 Then

kEN, P E

= log

(l/k)logIPI

u}.

Pmof The proof will be based on some classical facts on systems of extremal points due to J. Siciak (cf. LSic 19641). First, we define the function b : C x C —+ [0, co), Iz

— wI

if (z, w)

—

E

C2

ifz=oo, WEC

—

if z E C, w = 00 ifz=w=oo.

0,

Using the lower estimate for u shows that b is upper semicontinuous on C x C. Then, for j E N, the function

:=

[0,00),

:

b(Zk,

is upper semicontinuous, too. So we find

=

zt),

z

=

(zo

Z3).

E

with

Put

p1IIk _<exp(u)}. In that situation Siciak's result tells us that

•

is harmonic outside of F;

log

• log

F \ N(u).

that

extended maximum principle 6) for = subharmonic functions implies that u $ log outside of F; in particular, log outside of F. which completes the proof of the lemma. 0 log

u

on

Therefore,

the

Pmof of Theorem 4.1.13. (i)

I)

Obviously, (iii) (i). (ii): Observe that it suffices to prove the following two inequalities:
sh function satisfying the condition of Lemma 4.1.14 (recall that D is assumed to be bounded and cf. Theorem 4.1.7). Therefore, log

I)

kEN, P E .'Pk(C), (1/k)logIPI 6)

See

(Rans 1995], Theorem 3.6.9.

u}).

353

4.1 Special domains

Takeapolynomialp E ?(C)with(l/k)logIpI

u,degp k. Put Q(zI.z2) := Then Q is a homogeneous polynomial

when p(z) = withIQIIIk h(., I). (ii) (iii): We have

D=

{z

h(z) < i} =

E C2:

Moreover, since N(h) is pluripolar, h = h*. Hence, D is fat.

We conclude the discussion of balanced domains with the following example

Example. Fors E N let Q3 s}

E

)%om(C"), 1 cf <s, be given such that = 0.

30'EC'

Moreover, letd1 := deg Q1 and D := {Z E

fl

0.

=

E (0, oo), I <j <s. with

1.. Then.

< 1)

IQ

is an unbounded pseudoconvex balanced domain with continuous Mmkowski function.

Lemma 4.1.15. For D the following properties are equivalent: (i)

(ii)

D is an

-domain of holomorphy;

eQ for all j;

(iii) (iv) There exists a Q

deg Q > 0, such that

E

< I on D.

Proof. The implications (ii) (i), (1) (iii), and (iii) ; (iv) are trivial. So it remains to verify (iv) (ii): Using homogeneity it follows from the assumption on Q that 1Q1l/d < Q p.t/d1 where d := deg Q. Therefore, the function lJd Q11Pu'di is psh and bounded from above outside an analytic set, so it extends as a psh function to C". In particular, it is constant. So there is a c > 0 with

1Q11/d

= cIQ1IP/d

.

.

354

4 Existence domains of special families of holomorphic functions

a certain unit vector b E C" we

Now, fix a j and choose can write

+Ab)= +

#0,

XE C.

Ab) = Amq(A).

AE

C, with q(0)

0.

Therefore, Amq(A) =

A E C, where gj is holomorphic near 0 0 and g1 (0) # 0. From this representation immediately follows that E Q.

Finally, we mention the characterization of those balanced domains that are

0(N) (D,

of holomorphy 7) (cf. LJar-Pfl 1987]).

Proposition 4.1.16. Let D = Dh be a balanced domain in C" and let N ? 0. Then the following properties are equivalent: (i)

D is an

SD)-domain of holomorphy;

=

(ii) There is a sequence (QJ)JEZ+ C <

Observe that, in case N = Proof. The proof of (1)

,'

0,

0 or deg

= j, such that

this proposition is the same as Corollary 4.1.4.

(ii) is left to the reader (compare the proof of Corol-

lary 4.1.4). (ii)

'.

U2flD ç with f I

(i): Suppose that (i) is not true. Then there are open sets 0

U1 C E 0(U2) U2 connected, suchthatforeveryf E = flu. Now, in virtue of the open mapping theorem we conclude that

U2,

VaEU, 3c(a)>0

1a1 <

fE

c(a). Hence, (ii) gives h 1 on U2. Then, applying the In particular, 1Q1(a)l 0 maximum principle, it follows that h = 1 on U2; contradiction. Remark. The following question seems to be open: What are characterizations of bounded pseudoconvex circled domains that are .N°°-domains of holomorphy

Reinhardt domains. We already know that any pseudoconvex bounded complete Reinhardt domain is an R°°-domain of holomorphy because it is balanced with continuous Minkowski function. Here we will study arbitrary pseudoconvex Reinhardt domains. For further need we fix the following notations.

V1:=((z1

Recall that &N)(D 8D) := (1 E 0(D):

j=l bounded on D).

4.1 Special domains

Moreover, if a = (al .

.

.

, as,) E

355

then we put

is well defined. C" (a) then the product Iz 11a1 ... Let us repeat the characterization of pseudoconvex Reinhardt domains (cf. Proposition 1.9.19):

A Reinhardi domain D C C" is pseudoconvex

log

n}then(zi where log D := {x image of D.

E

IR": exp(x) := (exp(xi)

D C R" is convex and, if

exp(xn))

E D} is the logarithmic

From that description it is clear that

intD\DC i.e. if D is not

V0.

fat, then the difference between

mt D and D is

contained in some

set. In particular, a holomorphic function on D with 'slow growth' near that set extends holomorphically to mt D. We are led to study relations between geometric properties of log D and the stnictare of, for example, analytic

DefinitIon 4.1.17. Let X C R" be a convex domain by E(X) the vector subspace of R", that satisfies (a)

(think of X as log D).

We denote

X + E(X) =

for any vector subspace F C R" with X + F = X holds that dim F dim E(X). (b)

It is easily seen that E(X) is

of E(X). Let X ç

IR"

(aa)aEA C R"

uniquely defined and

that any F as in (b) is a subsp

be a convex domain. Then, using convexity, we find A C that

and

such

X=

mt

fl

aEA

:= {x E W (x — 0a. a) < 0), i.e. X is given as the 'intersection' of halfspaces. In particular, if a Reinhardt domain is given as where

D :=

mt

fl Da,

aEA

where

:=

{z

C"(a):

.

< Ca),

Ca > 0.

356

4 Existence domains of special families of holomorphic functions

then the associated vector subspace E(X) of X

:= logD = mt

fl{x ER": (x,a) aEA

can

be explicitly calculated as

E(X)={x ER": (x,ct)=OifaEA}. The Da 's are so-called elementary Reinhardi domains. Using the convex domain log D of a pseudoconvex Reinhardt domain we define

Definition 4.1.18. (a) A vector subspace F of R" is called to be of rational type, if F is generated by F fl Q"; otherwise, we say that F is of irrational type. (b) A pseudoconvex Reinhardt domain D C C" is called to be of rational or (irrational) type if E(log D) is of rational (irrational) type. We shall see that the type of a Reinhardt domain is responsible for being an domain of holomorphy. Before we formulate that result we have to introduce a new subfamily of holomorphic functions, namely:

=

fl

SD) j

N>O

Then the following description of Reinhardt domains, that are .7e°°-domains of holo-

morphy. is true (cf. [Jar-Pfl 1985] and [Jar-NI 1987]).

Theorem 4.1.19. Let D c C" be a Reinhardt domain of holomorphy. Then the following conditions are equivalent: (i)

D is an

-domain of holonwrphy;

(ii) D is an

of holomorphy:

(iii) D is fat and of rational type:

(iv) D =

mt

Da for some A C

where Da are elementary Reinhardt

domains as above. In any boundedfal Reinhardi domain of holomorphy is an ,7t'°° -domain of holomorphy.

Proof (ii)

(iii): It is clear (cf. the remarks at the beginning of this subsection) that D is fat. Put X := log D and F := [E(X)1 fl Assume that E(X) is not of rational type. Then it is clear that dim F > dim E(X). and let be the expansion of f into its Laurent series. Let f E (*) We will show that a,, = 0 whenever v E(X)'.

357

4.1 Special domains

Suppose for the moment that (*) is true. Then for x E X,

+ v, v)) =

>

E F we get

V

avexp((x,

> vEE(X)1flZ"

VEZ"

Consequently, the series is convergent in

logz := (log IziI

:= {z E

E

X + [E(X)' flQ"]'}.

C D. Or in the real picture, X + F

Since D is an

X which contradicts dim F > dim E(X). Now we are going to prove (*), Fix a u

such that (w. v) >

0.

C

E(X)-1-. Choose w E E(X), 11w II = 1. and some x° E X. Put x(1)

Fix 0 < N < (w, t')

exp(x° + 1w), t E lit Since x(t) E D, the Cauchy inequalities imply <

sup{If(z)I: 1z31 = x1(t). j = I

n}

II080(x(t))_N(x(t)t)I v)/N))_N,

I ER,

independent oft. where M := sup(öD(x(l))exp(t(w, v)/N): t E IR}. In order to get av = 0 it suffices to show that M = oo. Suppose that M (I + lIx(z)112r"2}. Ift E T, then

where c := exp ( — (x°, v)) is

Thus,

exp ( — 21(w,

v)/N) +

exp

+ tWj — r(w, v)/N))

= pD(x(t)) for I >

Consequently, T is bounded. Therefore,

Now, we

will estimate PD. Let d := dist(x0 + Rw, aX) = dist(x0, DX). Fix a t E R and let Z E i)D, 0, such that PD(X(t)) ? 0.5hz — x(1)II. Write logz = x0 + 1w + u with Z1 " 'Zn Then we obtain that uE lull > d. Fix a j = j(t) such that 1u11 > PD(X(t)) where

II

d0 :=

— 1)/2. choose Jo such that there is a sequence (1k for all k and tk = oo. Then

Finally,

M > pD(x(lk))exp(tk(w, v)/N)

contradiction.

+ tkWj(

E R,

Ik

to with j (1k) = Jo

+ 1k(w, v)/N) —*

oo;

358

4 Existence domains of special families of holomorphic functions

(iii)

(iv): We only have to prove that

a #0, such that

\ Dthereexistsa E

foranya E

Suppose for a moment that (*) holds. Fix a b e mt D and a

b

(*)

< IaaI} =:

C {z E

1), a E

such that

D. where

:= Choose V

{z

C"(a): IzaI < luau.

:=B(b,r) C Dandtakeaz E V. Wemayassumethatzi =

o< t

=zj =0

EN, converging 0, 1 to z. Since l(Zk)al < laal, k E N, it follows that aj £. Thus za is = laaI. Then, in virtue of defined and lzaI laal, z e V. That b D* yields the maximum principle, we get Iza I = laa I on V; contradiction. Therefore, using the fact that 1) is tat, we get and

. .

D=intD C

n. Choose points

j

C DU(ChI\(C.)?!).

fl

mt

E

aE(C.Y'\D

Then, taking the interior on the right-hand side gives D = mt flaE(c )"\D D. So it remains to prove (*). Fix an a (C.)" \ D. Put := loga logD. Recall that log D is convex and of rational type, i.e. up to an 'rational' orthonormal isomorphism we can write log 1) in the following form

logD

x

=

where is a convex domain in with E(D1) = 0. with vertex at By D2 we denote the open convex cone in

that is spanned by D1. Since E(D1) = 0, does not contain any real affine line. l'herefore, '12)). where '12 runs through an non-empty open D2 C (x2 E Rn_k (x2, '12) < We choose an '12 subset of 0 such that its coordinates are all integers. Then the inverse rational isomorphism leads to an non-zero rational vector a with log 1) C E R": (x, a) < a)). Without loss of generality, a may be chosen in V'. Taking the exponential leads to the claim (*). The remaining implications are trivial; hence the theorem has been proven. D

Recall that the inclusion map p00(1)) —÷ (9(O+)(1)) is continuous, when is equipped with the following sequence of norms (IIISNIID)N>o. Then, in view of the Banach theorem, either = or .l?oc(D) is of first In the case of Reinhardi domains, that Baire category in the Fréchet space of holomorphy, always the second case holds, i.e. there are a lot of are unbounded functions belonging to

359

4.1 Special domains

Proposition 4.1.20. Let D ç C"

be

a Reinhardi domain that is an 3€°° -domain of

ç

holonwrphy. Then

Proof Accordingtomeorem4.l.I9therearea

Z",a

such that DC Da =: {z E C"(a): ZaI < Now, we are going to construct f E (9(Da) with urn If(z)I = oo when z approaches a within D. Put E —* C,

Q(A)

log(l/(l — A)).

It is clear that = oo. Moreover, for any e > 0 we have that (I — A)IQ(A)Ih/E remains bounded on E. Now, we introduce the following holomorphic function on Da:

1(z) := It is clear that

—+

Z E Da.

oo and that (1 —

is bounded on Da.

ZED

What remains to show is that

isboundedonD.

(*)

In fact, let N > 0 be arbitrary. Choosing r := N/IaI gives that f E The fact (*) is a special case of the following result we quote without giving its proof (cf. [Jar-Pfl 1987] and [Nar 1967)).

:P(C"), P there exists a constant c = c(P, Lemma. Let P iJt

0, with d := deg P. Then, for every > 0, such that for any domain G C C" and for any

E :PSR(G) the following inequality holds:

To apply this result weonly have tochoose P(z) := (I

:= 111ft

0 Observe that the proof of Theorem 4.1.19 implicitly has shown that any n-circled (9(N)(D SD)-domaln of holomorphy, 0 < N < I, has to be a fat domain. Conversely, the following is true (cf. [Jar-Pfl 1987]). ProposItion 4.1.21. Let D be a fat pseudoconvex Reinhardt domain in C" and let SD)-domain of holomorphy. N > 0. Then D is an

Proof Let D

C" and suppose that D is not an (D, )-dornain of holomorphy. Since D is fat, there are open sets U1, U2, U2 connected, with 0 U1 C D flU2 ç there is an f E 0(U2) with = U2 C (Ce)", such that for any f (9(N)(D

f I

4 Existence domains of special families of holomorphic functions

360

a

on U1. Choose a U2 \ 1). Exploiting that log D is convex we find a and c > 0 such that

DC

G

C"(a):

(z

...

Iz1,aI


0,

G.

f

(9(NJ(G SG) there is f (9(U2) with f = on Ui (*) Our aim is to find a contradiction to (*). Without loss of generality we may Moreover, by assumption, for any f

s,a1>Oforj=s+l

assumethata1
remaining a's. Furthermore, c = I. Then G is given as

G= where u(z) :=

log

u(z) <0),

x

{z E

— logc, z

=: G. Observe that

x

E

u

For a moment let us assume that we have the following information for u:

= 0,

urn v—+Qo

? u, Applying the Banach theorem, for any b

I on D.

(**)

U2 we find c(b) > 0 such that

f c(b) connectivity of U2 delivers that

E

=

E (.9(G) and on U2. Therefore, for each b

on U1 the U2 we obtain

0. In particular, we see that u <0 on U2 which contradicts the maximum principle (recall that u(a) 0). and So it remains to prove the property (**). Put R(z) := zi ... z choosee > Owith3ke
C N such that

—f

00

—

I

I

<j S k.

Put

Then

zr',

...

:=

z E G.

0(G). Moreover, onG.

lim V —+

In virtue of the lemma which was quoted at the end of the proof of Proposition 4.1.20 it remains to find 0 such that

ê on D.

v E N.

4.1 Special domains

(Then use the sequences property (**)) Fix a z E D. Then

1g1(z)I

(ë > 0

361

properly chosen) and

in the

/

<

1

exp(q1,u(z))

11

.

E N.

z31

Consequently,

fl

IR(z)I . where

izji)2

0

> 0 is properly chosen.

So far we have a complete characterization of those Reinhardt domains that are of holomorphy. Even more is true. There is an effective way to describe the for a given Reinhardt domain. Definition 4.1.22. Let X C R" be a convex domain. Put inductively.

K1(X) := X + (E(x)-'K(X)

nQ")'.

U

K1(X),

:=

j

I;

M(X) := E(K(X)).

1='

It is an easy exercise from Linear Algebra to verify for a vector subspace E C IR" the equivalence of the following facts: • E is of rational type; • E-'- is of rational type;

•dim(E'flQ")' =dimE. Using these properties we obtain

if

Remark. (a) E(X) is of rational type K1 (X) = X. (b)LetX IR". Then there isaq n—dim E(X)with K(X) = Kq(X). (c) M(X) is of rational type and K(X) =X + M(X). (d) K(X) = mt fl Y. where the intersection is taken over all convex domains Y CR" with X C Y such that E(Y) is of rational type. Theorem 4.1.23. Let D c C" be a pseudoconvex Reinhardt domain. Then D mt exp K(log D)) is the .R°° -envelope of holornorphv of D. in particular,

(a)R°°(D)=C

K(logD)=IR":

(b) the Laurent series of any f E

Proof Put

G :=

D)'

atzt.

intexp(K(logL))). Then G is a fat Reinhardt domain with of i.e. G is of rational type. Hence G is an

E(log G) = M(log D), holomorphy.

is of the form

362

4 Existence domains of special families of holomorphic functions

C (9(G). Fix an f

What remains to show is that expand f in its Laurent series

f(z) =

Then we

where X := log D.

Looking at the definition of K1 (X) and the form of the series it is clear that f extends to

a bounded holomorphic function on exp(K, (X)). Repeating this argument leads to an bounded extension off on exp(K(X)). Finally, the Riemann theorem on removable singularities gives the extension to G. 0

In more concrete, situations the description of the .X°°-envelope is even more precise. Let & = E and letc1 >0, j = I...., N. Put

G:=flGj. :=

C"(&):

(z

andr:=rank[a'

where

j < N),

I

aNJ

Moreover, we introduce the following notation:

J :=

I < ii <

<1,
a"] =r};

A, :=Z'tfl(R+ahl M := fl M1. IEJ

With the help of these notations we have the following result.

Proposition 4.1.24. In the above case the following is true: M(X) X := log G, and, consequentlv K(X) = X + M. In particular, A1 = 1 E J.

N,and

Proof

fl Fix an I

=

I = (ii

ir) E J. Then there is an

= {x

E

E

R" such that

x',aIk) <0),

k

=

I

r.

= M, where = C

4.1 Special domains

363

Let

:=

mt

fl {x E R":

a)
(x —

aEA1

So E(Yj) is Observe that X c X, c Y, and E(Yj) = M,. Observe that a E of rational type and, therefore, M(X) C M(Y1) = Mi. Since I E J was arbitrarily chosen, we conclude that M(X) C M. If M(X) = R", nothing is to prove. Now, let M(X) ç R". Assume that

K(X)=intflllfl, E

E B. Then there is an! =

Fix a

Xi C

E B}.

E J such that

(i1

={x ER":

c E(X,)1 =

E

R'1: (v, fi) = 0,

<0}.

(x

=A1&'

IRa"

E (RT) Ara", A := (A1 It remains to show that A E (R÷)T. Let L: R" —÷ W be given by (x, asr)), x E R". L(x) := ((x, a")

Then <

(x',a'k),

k=

r}

I

and

<0}. Since L(X) c

we see that A E (IR+Y. Therefore, fi E A,. was arbitrarily A1, which immediately leads to M C M(X). U chosen, so we obtain B C Let

us discuss a concrete example.

Example 4.1.25. Let

< '.

G := (z E C3: where 0 <

<

1,

j = 1,2,3,

(aI,a2,a3) and a2 :=

(I—an,

cr1

(cr1 —

a3)/(aj —

cr2)

< l},

Q. Set

a1 :=

I —cr2.1 —cr3).

Itiseasytoseethatrank[a1.a2] Z(l,

IZIII_a1z211_a2lZ3II_03

= 2andthatA11 2) = Z3fl(R÷a' +R+a2)=

1,1). Therefore,

M(IogG)= {v ER3: vI+v2+v3 =0} and K(logG)= {x ER3: xI+x2+x3 <0). Hence, the

of holomorphy of G is given by

G={ZEC3:Iz,I•1z21.1z31<1). 8)

the following fact from Linear Algebra (cf. IJar-Pft 19871): Let X0. X1 XN be =: r and 0 X1 fl a"Jand C Xo. Thenthereexistl

Use

open halfspaces of R" and let a' J.dX1, a1 e (W')0, with rank(a'

Xjifl•••flXjrCXO•

364

4 Existence domains of special families of holomorphic functions

Remark 4.1.26. The following question was posed by 1. Ohsawa in 1994. Let be pseudoconvex and let H be a one-codimensional hyperplane in C". c c Given a bounded holomorphic function f on fl H: does f extend to a bounded holomorphic function on There are positive answers to that question in case that is strongly pseudoconvex (cf. for example LHen 1972]). In the general case, there are various counterexamples

published by M. Tsuji (cf. [Tsu 1995], [Tsu 1996], [Tsu 1996b1), and [Ham-Tsu 1996]). Using the example before there is another counterexample.

=

Repeat that

Define

H,:=(z}xC2.

A€C.

and

< I. Moreover, p(A) E G implies that t"l < I Choosing := 0.5, t > I. and at > yields that =: 10 < I. Take f(A) 1/(A — s), where 10 < S < I.

Observe that çp(A) E G, if and < I.

<

Then f is a bounded holomorphic function on the one dimensional linear subman= t. = I/t} of the two dimensional pseudoconvex domain ifold {z E G: G2 := (z E G: zi = rJ. Observe that f is singular at the points. Suppose that any bounded holomorphic function on the hyperplane intersection of a pseudoconvex domain extends as a bounded holomorphic function, then f extends first to G2 and then to G as a bounded holomorphic function f. Then, according to our example. f extends to G. Therefore, f extends to the unit disc; contradiction.

Recall that .7t°°(D) j Ak(D), k E whenever D is a bounded domain. In case that D is unbounded this inclusion is, in general, no longer true. Moreover, we have the following information on .4k.domains of holomorphy for Reinhardt domains (cf. [Jar-Pfl 1997]). Theorem 4.1.27. Let D c

be a fat Reinhardt domain of holomorphy and let

domain of holomorphy.

Then D is an

k E

a Reinhardt domain, is an

In

holomorphv. k

of

Z÷.

The proof is based on the following geometric lemma.

Lemma 4.1.28. Let a

:= Fix an

(Re)" and e > 0. Put (z

C"(a): lzijaI

,,.

.1a,,

< I +e}.

> 0. Then there exists a neighborhood U of the set (d Do) \

? dD(((Z)

Z E U fl D0.

? IZI ' ...

Proof We may assume that at a5 > 0, as÷I =0 Fix an a E (aD0) \ Note that at . ...

such that

=

< 0, where 0 < s < n. ...

4.1 Special domains

365

We have to show that there is a neighborhood U of a such that

1z21) C

: z3=0}and2:=(2

2).

Let U be a neighborhood of a such that < ft(I

j=I

1,

<1 +e0,

+ Izl_eJl)ai ft (1—

z EU,

j=I where

(0

e3

0),

0, 1, 0

the I

is standing at the

fl(lzjl +

I + EO.

j=I Therefore,

IP(z,

Z E Un Do.

j=s-4-I 1z21)

Proof of Theorem is a family A

Then

z E U

C

4.1.27.

c

fl D0 \

C". Since 0 is

We may always assume that D

and a function C:

D

0

V0.

mt 11(z

A —÷ R>0

C"(a):

.

. .

fat, there

such that

< c(a)}.

.

aEA

Consequently, it suffices to consider the case

D

= {z E C"(a): lz1Ia1 ...

<

where c > 0 and a

Moreover, we may assume that all a3's are not zero (otherwise we can pass to lower dimensions) and that c = 1. -domain of holomorphy. Then there are open sets Now, suppose that D is not an connected, with 0 Uj, U2, Ui C 0 r) U2 U2 such that for each f there is an f E (9(1/2) with flu1 = flu1. Moreover, we may also assume that U2 fl V0 = 0 and 0 fl U2 c (12 (recall that D is fat). here we use the notation from Lemma 4.1.28. Choose eo > 0 such that U2 Oandg E SD,0). on

such that is existence domain of g. Take an £ N, £ > 2N + 3k + 1. Suppose we already knew that I := zt1gID ,4k(fl) (*). Then (9(U2)

extends g to 1/2; contradiction. What remains to prove is (*): obviously it suffices to verify that

= 0,

a C (SD) \

t, a E

In + laP

k.

and recall that some of thea1 's do vanish. According to Lemma Fix an a E 0) \ 2.10.4(a) we know that
366

4 Existence domains of special families of holomorphic functions

where the positive number co depends only on n, N, and k. Then, in virtue of Lemma

4.1.28 we conclude forz E D flU, z near a, that

ciIztl_tl_2(N+k)1I

.__÷

0

where c1 is independent of z. This proves (*).

Remark. Directly from Theorem 4.1.27 it follows that the so-called Hartogs triangle [Sib 19751. (z Izil < 1z21 < 1) is an As-domain of holomorphy,k E

In this paper N. Sibony also showed that the Hartogs triangle is not an of holomorphy. Here is the full characterization of of holomorphy inside the class of Reinhardt domains (cf. [Jar-Pfl 1997]).

Theorem 4.1.29. Let D c C" be a Reinhardt domain of holomorphv. Then the following conditions are equivalent:

(i)

Disfatand VJE(I

n)

:

where

:= ((ZI

C": z =

Az1

(ii) D is an

(Zi

Zn)

of holomorphy, where

=1! E (9(D): (iii) D is

an

D, A E E};

: l)°f E

of holomorphy;

(iv) D is an (9(D)-domain of holomorphy. Observe that the Hartogs triangle does not satisfy condition (I) and so it is not an

of holomorphy. Before we can proceed proving this theorem we have to verify the following lemma.

Lemma 4.1.30. Let D ç C" be a Reinhardt domain that is an holomorphy. E C

of

where

:= (f

(9(D): VaEE

:

E

The condition (*) in (i) was introduced by S. Fu (cf. [Fu 1994]), when he studied the completeness of Reinhardt domains.

367

4.1 Special domains

Z÷ and assume that

Let k0

:={a Then there exist

Ac

Ioi=ko}c E.

E

and functions a: A —* R>0, b: E —f iR+ such

that

D=int


fl

Moreover, if E =

(t)

then D satisfies (*) of(i) in Theorem 4.1.29.

Proof Observe that sequently, there is f E

(D) carries the natural structure of a Fréchet space. Consuch that D is the domain of existence off. Write

zED, t)EA

where

Ac

is such that

0 if v E A. Note that for z

D always

0

whenever vj <0 and that

aaf(z) =

aE

E

E

E. Using the Cauchy

inequalities we obtain

D, (v, a)

(LI)

A x E,

C b, where D denotes the domain defined on the right-hand side in (t). is convergent in D. Suppose now It is clear that for each a E the series of that D ç 1). Since D is connected, there exist d E D, r > 0 such that 1F1(d, r) C D Thus D

but

P(d,r)

f

D.

extends to g3 E 0(D) is such that each derivative Observe that if g (9(P(d, r)), j = 1 n, then the function g itself extends holomorphically to P(d, r). Indeed, the extension may be given by the following formula

= g(d) +

— dj) J

+ t(z

—

d))dt,

z

P(d, r).

j=1 The above property and the fact that

contradiction. Hence, D = D.

E imply that f extends to IP'(d, r);

368

4 Existence domains of special families of holomorphic functions

Now, we assume that E = Let 3D fl V10 0, 1 <Jo n. Since D = = D it suffices to show that vj0 > 0 for each v E A. So, fix such := (ai := max{0,vj}, I j n. It is clearthat 0. Therefore

to prove that a E A. Puta

— —

j. is bounded on D. Then tYj0

L',

1.—I

Zj

lJj
0 because of 3D fl

Vj0

$ 0.

El

Now we will be able to verify Theorem 4.1.29.

Pmof of Theorem 4.1.29. We may assume that D C". The implications (iv) (iii) (ii) are evident. (ii) (i): Recall that mt D \ D C V0. In particular, if D is an holomorphy, •

:

:=

{f

E

(9(D):

domain.

: fIDflG E

of

fl G)),

then D is necessarily fat.

n}. Then for any r > 0 the Suppose that 3D fl $ 0 for some Jo E {l domain Dr := DflIP(0, r) is an with E := Hence, by Lemma 4.1.30, if Dr fl Vj0 0, then = Dr. Consequently, b(M = D. Hence, condition (*) in (i) is true. (iv): Suppose that D is not an (9(D)-domain of holomorphy. As usual there (i) are open set U1 and U2 with the well known properties such that for each f E 0(D) there is f (9(U2) with f = Jon Ui. Recall that D can be written as ',

D=

mt

ritZ E C"(a):

.

. .

<

aEA

where A C (IR"),, and c: A —* R>0. Fix a E A, c > 0, and e > 0 such that

DC {z E C"(a):

..

< c)

.

C De

:= (z E C"(a): IziIal . ... .

<(1

and U2 - Observe that

is a domain of holomorphy. Therefore, it remains to prove that to get the desired contradiction. To do so, we first remark that obviously D \ Vo C Now, let Jo be such that 3D fl 0. Then the condition (*) in (i) yields

Dc

... Consequently,

>

0.

Thus (3D) fl V0 C

Z E D, A E E.

0

4.1 Special domains

369

Remark. Analyzing the proof of Theorem 4.1.29, (i)

(iv), it is easily seen that also the following is true (Exercise): Assume that the Reinhardt domain D C C" is an -domain of holomorphy. Then any condition in Theorem 4.1.29 is equivalent to the following one: ,'

(v) D is an .lf.Oo(D) fl (9(D)-do,nain of holo,norphy. So far we have discussed holomorphic functions with certain bounded derivatives or with derivatives that extend continuously to the closure of the domain. One relation between such functions will be described in the next lemma.

Lemma 4.1.31. Let D cC" beaReinhardi domain. Then

C A°(D). In

k E N, where

C

:= {f E 0(D):

:

If D is assumed to be bounded, then Ak(D) C

LetD := {z E C2: zil < 1z2U. ()bservethatthefunctionf(z) := zr/4z €D, belongs to \ ,4k(D) i.e. the jump of regularity in the lemma is in some sense necessary. Proof. Note that D has a univalent envelope of holomorphy. So we may assume that D is a domain of holomorphy and that D C". Fix an f E Obviously,

jf(z') — f(z")I

z

E D,

= l}XD(Z', z"),

z'. z" E D,

where XE) denote the arc-length distance on D. To show that f extends continuously to D it suffices to prove that for any a E dD there are a constant c > 0 and a neighborhood U = U(a) such that

XD(Z. z")

c(IIz' — z"II + IIpj(z')lI + IIpi(z")II).

J := (ii

=

.•a,,

#0. Putw' :=

z'. z" E D fl U,

= = 0 and the other a1's different from 0. z1,). and p0 =: id. p1: C" —p C3, pj(z) := (Zj Now fix such apointa. We may assume that) = (1 s), i.e. al = . .a5 = 0 where

.

SinceDisReinhardt,

W' E D and

XD(Z, it;') < 2x(fz'11 +

.

+

Let w" be defined in the same way as w'. Thus it remains to show that there is a constant c' > 0 and a neighborhood U = U(a) such that

xD(w, W") < c'IIz'

— z"II.

z'. z" E D fl U.

Using continuity it suffices to consider only the case where 0 0, # 1z71 = I n. Let L1 = .. . := Log := the principle branch of the logarithm. j Furthermore, if j > s + I, then L3 may denote a branch of the logarithm. I

370

4 Existence domains of special families of holomorphic functions

We define y: [0. II —÷ y3(t) := exp((l — + tLj(w7)), j = n. Since D is logarithmically convex, y([O, 1]) C D. What is left to be

1

shown is that for any j there is a > 0 such that the length provided that z', z" are near the point a. — z71

1ff <s,thenl

of yj is estimated by

11z1 — 1z711

If j > s + 1, then a neighborhood U3 ofa3 is chosen such that Izj for all z1 E U1. Consequently, for points

=

j

< IajI/2

—a31

E U3 we obtain

— Lj(z7)Idt

— z71 = 41z — z71

0

Hence, the lemma has been proven.

Now we pass to the characterization of of holomorphy inside the class of Reinhardt domains. There is the following result (cf. [Jar-Pfl 1987] and [Jar-Pfl 1997]).

Theorem 4.1.32. Let D C C" be a Reinhardt domain of holomorphy. following conditions are equivalent:

D

(i)

C" andforeachk E

the

Then the

domain D iscm

where

Lr(D) := {f

(9(D):

p E [1, ,o], k E

E

:

:=

:=

fl

p€(I,00J

Disfatandthereexistsap

(ii)

[l,oo) such that

{0};

(iii) D is fat and E(log D) = {0}; (iv)

D D

C" and,foreachk

E

D is fat, and for every a (Ce)" \ D there exist sequences pmvided that D is balanced) such that (C IICJZ'1IL2(D)

1, j EN,

Pmof The implications (v) (ii) and (iv) follows directly from Lemma 4.1.31. (iii) (v): Fix a E (Ce)" \ D and j :

:= loga. Notethatx0

= 00.

lim

:

X. Chooseabasisa1

(i)

:

(ii) are obvious. (i)

N. Put

a"

,'

(iv)

as above X := logD and E Z" ofE(logD)-'- =

Special domains

4.1

(E Z'4.

371

if D is balanced) such that

xc fix ER": By A we denote the matrix A := (a1 assume that IdetAl J2ir"lai ... 1 = (I 1) N". Then we get .

j

IIL2(D) = (2,r)"

(x _xO,ak)
Xo.

a"]. Without loss of generality we may Put v1 =: a1 + + a" — I where

.

exp((x, 2(v +

(2ir)" f exp((x, 2(v + +

exp(2(x°,

= IdetAl where X1 := €R": < Letc1 := Then

j

= = j

=

n}.

I

I = i.

I

which gives the property in (v). (ii) (iii): Put X := log D and suppose that E(X) (0). By assumption, there is an f L(D) \ (0). We write f as its Laurent series Then 1(z) = :

JD

f

=

f f

DI

ff If IDI := ((Izil z = (Zi Z" such that Recall that we may assume that X = Then we obtain

E D}. Consequently, there exists a

vo

ID

= (2jrY

= (27r)"f exp ((x'. where

j

exp

Y

x F. where F := E(X) and Y C X'.

((x. pvo + 2))dAn(x)

+ 2'))dAn_ni(x'))f

((x",

x := (x', x"), etc, and m := dim F; contradiction.

+ 2"))dAm(X") = 00,

4 Existence domains of special families of holomorphic functions

372

(iii) (i): Fix a k E and suppose that D is not of holomorphy. Then we find open sets U1, U2. U2 connected, with 0 U1 c U2 n D ç U2, such that foranyf E there existf E 0(U2) with f = f on U1. We may assume DandDflV0=0. thatU2 Since E(log D) = (0) and D is fat, there are JR-linearly independent vectors > O,andc > 0 such that E Z",j = I

:=flz E

DC C

D1 :=fl(z

<(I

E

Using the fact that the al's are linearly independent we may assume

and U2

thatc= I. Now, we fix a E U2 \ D1 and then Jo such that Ia&O I > (I + e). Moreover, we set a := For N E N we define ., Na

fN(z):=

.

ZED1.

—

Obviously, E 0(D1). It suffices to verify that (+) fN that

—

Ua)O) =

Assuming this for a moment it follows

E

on U2; contradiction.

So, what remains is the proof of(+): observe that aafN, a E

al <

finite sum of terms of the form —a

d

E

7L,

k). Thus it suffices to find N such that

E (0

I

< p < 00.

Let A

:=

and

B

A'.

Put

Tj(x)

:=

J =I

n, x = (xi

E

JRfl•

k, is a

4.1 Special domains

If p E [I, oo) and U

J

E

Z", then we have

f

=

exp ((x. pi' +

log D0

= (2ir)"

Ii T(v)>—( —

exp ((Be, pv +

J

2) > 0,

then

373

j =I 2jr

p\IdetAi"'

< 1. Hence, if v = Na

n. In particular, if

—T(2)). / —

j=l

n,

a and if

p E [I, oo) and al k. Moreover. No > 7)(a) for all j and therefore, Na — a E R+a' which shows that I foraillal
+ R+a", D

Remark. There are also other classes of holomorphic functions C 0(D), D a Reinhardt domain of holomorphy, for which there is a characterization of D to be a i-domain of holomorphy. For more details we refer to [Jar-Pfl 1997).

Hartogs domains. Here we discuss J?°°-domains of holomorphy in the class of Hartogs domains. First we recall what is meant here by a Hartogs domain. x Ctm —* Let c C C" be any domain and let H: be an upper semicontinuous function with H(z, Aw) = AIH(z. w), z E c, w E Ctm, and A E C. Then the domain D := H) := ((z, w)) E xCm H(z, w) < 1} is a Hartogs domain with ,n-dimensional balanced fibers over (cf. Remark 1.9.8(e)). In the case when H(z, w) := wi exp(u(z)). u: —+ R upper semicontinuous, the domain D := D(c, u) := {(z, w) E C"41: wi < exp(—u(z))}

simply called a complete Hartogs domain with basis c. Observe that the class of Hartogs domains with balanced fibers contains the class of balanced domains and the class of complete Hartogs domains. Recall (Proposition 2.2.22(b)) that H) is pseudoconvex if and only if is pseudoconvex and log H E x Cm). In the case of a complete Hartogs domain we have: u) is pseudoconvex if is pseudoconvex and u E The discussion of conditions under which a pseudoconvex complete Hartogs domain is an of holomorphy was initiated by N. Sibony in his paper [Sib 1975]. The following example is given there. is

374

4 Existence domains of special families of holomorphic functions

Example 4.133. Let (aJ)JEN C E be a sequence of points in E such that any b is non-tangential limit of a certain subsequence of )JEN. Put

ü: E—+

E

R_00,

where the a1 's are positive and so chosen that ü —oo. Observe that the summands are negative sh functions on E. thus ü E SR(E) and = —00, j E N. Then, u := exp(ü) E SR(E) with u < I and = 0, j E N. We set as above

D := D(E, u) := 1(z, w) E E x C: wi <exp(—u(z))}. D is pseudoconvex.

Proposition 4.1.34. The domain D defined in Example 4.1.33 is fat, R00 ( D) -convex, and D ç E2. Moreover, [0, 1) —÷ (0,00), E C, then any f E 0(D) with *(iwi) on D extends holomorphicall to the bidisc E2. If(z, w)i

= ,Y00(E2) (i.e. D is not an 3? 00 -domain of holomorphy).

in

Proof That D is fat is an easy consequence of the fact that u is continuous (Exercise). Obviously. D ç E2 (observe that u 0). Moreover, E x (l/e)E C D. Now, fix and f E 3?°°(D) as in the theorem. We develop f in its Hartogs series, i.e.

f(z, w) =

where

E

0(E).

j=O

Then, the maximum principle and the Cauchy inequalities lead to the following esti-

mate

ifr(l/e)exp(j),

Z

E E, z E E, 0 < r < I.

exp(ju(z)),

<

(*)

The first line in (*) means that the functions h1 are bounded on E. The second one gives that

r < 1.

ihJ(ak)I < Now, choose r3 := sup(r E (0, 1): i,fr(r) (I of Fatou yields for j E N with rj < I that

I. Applying the theorem

l!i11 S rJ'tJi(rj)

E

and 1h1(z)i

ifr1 = I.

r51(l

z

e E,

E,

375

4.1 Special domains

Since rj —÷ 1, the Hartogs series is locally uniformly convergent on E2, i.e. it J—400

the holomorphic extension f of f on E2. To see that D is it suffices to observe that any fe 0(D) has a locally uniformly convergent Hartogs series f(z, u') = h(z)w1. In particular, f can be locally uniformly approximated by polynomials in 0 gives

We like to stress the fact that D is but not an of holomorphy, an observation which does not fit with the Cartan—Thullen theorem. Now we turn to conditions for a Hartogs domain H) with balanced fibers to be an X°°-domain of holomorphy. Here we will need the following psh function associated with H.

DefinitIon 4.1.35. Let

H(z, w)

c

and H E

sup {IQ(z,

w) w) E

We

x Ctm) be as above. We set

x C".

o(c2),

H on

x C"}.

remind the reader that a similar construction has been used in the case of

balanced domains.

ProposItion 4.1.36. Let I) = D(c, H) be a Hartogs domain with rn-dimensional fibers and let f E R°°(D). Then f extends holomorphically to 0* := D(cZ, H), where H* is, as usual, the upper semicontinuous regularization of H. Moreove,

(t) Proof Without loss of generality let If I < I. We write f as its Hartogs series, i.e.

f(z. w) =

w).

w) =

aa(z)w".

e

j=0 Using the Cauchy inequalities we get 1Q1(z. w)I

H(z. w)

w) E 2 x Ctm.

Hence, the series converges locally uniformly on D*, which gives the holomorphic

extension off to What remains is the proof of the fatness-property (t). Assuming the contrary we find a point (a, b) E mt D* fl x Ctm) \ D" There exists an open connected neighborhood U = U1 x U2 of(a, b) with U C b. Observe that any bounded Runge domain is

376

4 Existence domains of special families of holomorphic functions

Fix a point (z'.

w') E U and a sequence ((z". w'))t,EN C U fl D* such that

w") = (z'. w'). Take a homogeneous polynomial

j

Q(z. w) =

N, Ua


With

Ofl S7 X Ctm.

I we conclude that IQ(z', < 1. Therefore, H(z', u") Since (z'. w') was arbitrarily chosen in U, we get H* 1 on U. In particular, H*(a, b) < I. Since (a. h) D* it follows that H*(a, b) = I. So we have H = on U. Therefore, U fl D* = 0; contradiction. LI

H = H is a necessary condition of holomorphy. Moreover, we have the

Observe that the Proposition 4.1.36 implies that

for D = D(c2, H) to be an following result.

Theorem 4.1.37. Let D = H) be a pseudocom'ex Hartogs domain with rn-dimensional fibers. Assume that is an R°° -domain of holomorphv or that 00, whenever lwlI = I. H(z. w) 3:

Then the following conditions are equivalent:

H=H*;

(i)

(ii) D is an 3t°°° -domain of Proof First, let us assume that c is an of holomorphy. It remains to prove the implication (i) (ii). Let us start mentioning that the : of holomorphy D of D is again a Hartogs domain with rn-dimensional fibers, i.e. D

=

H). AssumenowthatD ç D. Wefind(a,b)

D\DwithH'(a,b) > I

(recall the fatness-property of D*). Then there is another point (ã,h) E D with H(ä. b) > 1. So we can choose a competitor from the definition of H. say Q. with > I. Since (Q — Q(ñ, it extends holomorphically to lQ(ã, D; contradiction. Now we turn to the second case, i.e. we assume that H(z. w) tends to infinity for ac and lwll = I. Suppose that D is not an of holomorphy. z Then there are open sets 0 U1 c U2 fl D ç U2, U2 connected, such that, if then there is an f E (9(U2) with flu1 = flu. According to the fE proof of the first part (observe that, in fact, there we do not need the knowledge on the x Cm. shape of the of holomorphy of D) we may assume that U2 c Thus we find (z', w') U2 with z1 Q. Fix a point (z°. w0) U1. We connect these two points with a curve y = (yi. [0, II —÷ U2. Put to (0, I) such that

y(to)

dD and y(t)

D, 0 < t < to. In virtue to the remark above we may also

The condition on the boundary behaviour of H even implies that y2('o) = 0. Take an e > 0 such the tubular neighborhood assume that yl (to) E

V := ((yl (1) x

e): 0

t

to)

4.1 Special domains

377

belongs to U2. Thus V contains a boundary point of D

contradiction.

0

In the special case of a complete 1-lartogs domain D = D(2, U), we will reformulate the result stated in Proposition 4.1.37. Here, H(z, u') = Iwiexp(u(z)). (z. w) E x C. Thus, H(z. w)

= wi sup{If(z)I": j

E N,

fE

< exp(u)}.

Therefore

Corollary 4.138. Let D = basis

is an

Then

u) be a complete Hartogs domain. Assume thai its °-domain of holomorphy or that urn u(z) = 00.

the following two conditions are equivalent:

D is an

(i)

of holomorphy:

j EN.

(ii) u =

Remark

II)

f E (9(d), tog Ifi

da

domain in C's. Define

0,

f E (9(d),

ñ(z) := sup((1/j) log If(z)I:

j EN, f E

4.139.

ü(z)

Let u

:= sup{ctoglf(z)I: C>

and

clog fl

u}.

log Ill

z

d.

ju}.

Then <ü. In [Sib 1975]. the set := (U E u = u*} played the fundamental role in the formulation of Corollary 4.1.38. Let us summarize a few properties of • In virtue of the Choquet lemma (cf. Theorem 2.1.37) one can describe ü using a denumerable set of functions, i.e. there are sequences (Cj)JEN C R>o and (fj)JEN C (9(d) with Cj log u, j N, such that

= •

In the case that u

sequences(kJ)JEN

(sup(c1

log

E 9S,7('(d2), u >

j E

0, even more is true, namely we find

CNand (fJ)1EN C(9(d2)with(l/k1)loglf1I

= (sup((l/k1)logIf1I:

j in particular, u =

j

E

EN) =lim(l/k1)loglf1l; J—.oc

=

II) This result can be partially found in [Sib 19751: moreover, the proof there is more complicated than the one presented here.

378

4 Existence domains of special families of holomorphic functions

In fact, the second statement is an easy consequence from the first one. To get the

'urn' simplytakeas anewsequencethe followingone: (I/k1) log Ifi (I/k2) log

V

= So it suffices to prove the first claim. By assumption we have u = where > u. jE For each cj we choose a sequence of positive rational numbers such that

qj.&

k

N}. Since u > 0 we have

log

u

for all

j,k. Fix a

E

If v(z°) > 0, then it is easily seen that

u(z0). In the case that v(z°) z

and implies that

0 it follows that ä(z°)

Then there is a sequence = v*(z) = u(z) > 0. Therefore, u(z). Hence u =

The last step consists only in observing that

log

0 C

= v(z°) u(z0). Thus,

=z

with

=

zi(z). which

= .L log f7& I.

• The same conclusion as before remains true whenever the function u E is bounded from below (or from above).

• Let u, v one side. Then u + v

and assume that both functions are bounded from at least 12)

In fact, from the remarks before we know that

= (sup((I/k)loglfl: k €N. f€ O(c2), (I/k) log 1.11 v= (sup((I/k)logIfI: keN, f E 9(c2), (I/k)logIfI <

u

Then,

observing that (l/k)logIfl + (l/!)IogIgI =

proves the

claim.

• Let u C Then the remarks before and the Bedford—Taylor theorem (cf. Theorem 2.1.41) leads to the fact that u is continuous outside of a pluripolar set. • Let c be pseudoconvex such that allows a neighborhood basis of pseudoconvex belongs domains. Then any u C(U) fl U an open neighborhood of to (compare the theorem of Bremerrnann, Proposition 4.4. I 3).

• C(E) fl

C

fact, letu (5'(E)flSR(E). Wechoosev e with tlIaF = v is the solution of the Dirichlet problem with boundary values given by

In i.e.

12)

It is not clear to the authors

on the boundedness

whether the statement remains true without the assumption

4.1 Special domains

379

Then there is an f E (9(E) with Re f = VIE. Since v = log I exp(f)I on E, we have VIE E SR(E). Put := u — v. Then 13 E C(E) fl SJt'(E) and VISE = 0. Define 13j := max{13, I/f + j(IzI — I )}. Then E SJt"(U), where U is an open neighborhood of E. Observe that (13j)jEN converges uniformly on E to 13. Therefore, there are positive numbers such that 13 = sup{ — SJ: j E N} on E. According to the remarks before we know that v1IE — E S.7€(E). Hence, 13IE E SJ?(E). Finally, since u = + v, we get UIE E SJt°(E). 13

• If u :=

=

Izi, then

u

responsible for D(u2. u) to be an

(Exercise). Thus

instead

of ü' is

of holomorphy.

Looking more carefully at the construction in Example 4.1.33 one even can construct a complete Hartogs domain D that is an .Y°°-domain of holomorphy but not an A(D)-domain of holomorphy (cf. [Sib 1975]). Recall that such an example does not exist within the class of Reinhardt domains.

Example 4.1.40. Fix sequences c1
Observe /, < U

E C(E) fl

0

dense in

C

and (Cj)jEN C R>0,

E .7€(E)fl S.7t'(K(3/2)). Put u := exp(i/i). Then and that Corollary 4.1.38 implies that hence, u E

D := D(E. u) := {(z. w) E C2:

wi <exp(—u(z))}

is an .7€°°-domain of holomorphy with E x {w E C

:

intD=D.

ç

E2

(l/e)} C D and

Iwi

Now, let f E .4(D), Ii! IIt = I. We write f as its I-lartogs series

f(z, w) = >Jhj(z)w1. j=O

=

I

j

'1

C(E) and < exp(fu(z)), z E, and then z E. Therefore, the Hartogs series off is locally uniformly convergent on E x E, i.e. it gives an extension

Then

-

-

Moreover, one can show that f is continuous on E2. We skip that argument.

Semitubular domains. Let us recall what we mean by a semitubular domain.

DefinItion 4.1.41. A domainD C

Dc

is called sernitubular if there are a domain

andfunctionsr1: D —÷ R,j = l,2,suchthat

380

4 Existence domains of special families of holomorphic functions

(a)

rI is upper semicontinuous, r2 is lower semicontinuous. E D xC: < r2) := <

(b) D = D(D,

DisthebasisofD. Semirubular domains have been studied in [Mok 1986]. Here we only quote the

main result from [Mok 1986] without giving the proof.

Proposition 4.1.42. Let D = D(D, r2) be a semitubular domain of holomorphy in C's. Then r1 and are psh on D. Moreover:

extends holomorphically to the semitubular domain 0* :=

(1) Any

D(D, Fj, —(-—r,) ). (ii)if 0 is an

of holomorphv, then (iii) if D is an .Jt"°° -domain of holomorphy and if

an

E

9S.1((D).

—r2 E S'SJt'(D), then D is

of holonzorphv.

For the proof the reader is asked to study Mokri's paper.

Nonschlicht envelopes of holomorphy. Here we are going to modify the example of N. Sibony in order to obtain a bounded pseudoconvex domain in C2, whose is not univalent. This construction is due to [Bac-Fäl 1991). Even moze is true, there is a bounded domain of holomorphy whose J€°°- envelope of holomorphy has infinite volume (cf. [Bac 1992]). The construction of the second example follows the ideas of the first one but it is more involved. That's why we only present the first example. The construction of the example will use the following gluing theorem for domains

of holomorphy. Theorem 4.1.43. Let G1, G2 be pseudoconvex domains in C" with G1 fl G2 = 0,

8G,, j = I, 2. Assu,ne that G1 is strongly pseudoconvex at z' with

and let z' C2-boundary such that

let y: [0, 1] —÷ C" be a piecewise linear Jordan curve

•y(f) EC" \(G1 UG2).0 < r 0 such that for any e E (0, eo) there exist a pseudoconvex domain with

G1 U

U y([0, I]) C

C

U

U (y([0,

fl

\3(y(0),e) =

fl

\ B(y( I), e) = G2 \ B(y(l), e).

G1

\B(y(0),e),

The proof is divided into two steps both interesting for themselves.

4.1 Special domains

381

Lemma 4.1.44 (Barbell lemma). Let B = Then there is E (0, 1) such that for with any E (0, so) there exists a pseudoconvex domain G5 C

—2 6 <2€, (ZECn ImzI+IIzII :

Rezi>1+e}=Gefl{ZEC

E C x C"'.

where z = (Zi,

x What this lemma is saying is that B U L, where L := has a basis of pseudoconvex neighborhoods all of them are looking as a slightly larger ball or a tube neighborhood of L, respectively, outside of a small ball around the point z0 := 0). Applying the lemma one can build a domain of holomorphy that looks (1, 0 like a barbell or even like an atomium. We should mention that this result can be found also in LShc 1983] where a different proof is given.

Proof Let 0 <

E C x C"t. We define the

< 1/2 and write z = (xi +

e

following function

C" —+ < 1—c,

I

2e6,

I +e, —1 —e)5

+ (—2 — 5€ + 15e5)(xj — I

+ (—8€ — +256,

+ 20s6)(xj —

— 1

—

e)3)

is the uniquely defined polynomial of degree 5, which exto the whole real line. Moreover, a simple calculation shows that for sufficiently small e we have <0 on (I — e. I + e). Thus is monotonically decreasingon this interval. Therefore, Wt(xl) 2€, E (I —e. I +e). we define Using this function Observe that

tends WE IR\( I —e. I +e) to a C2-function

Cn —+ IL

ur(z)

'

-2 + lizil

—

Our aim now is to show that for sufficiently small e the domain

:=

{z E

< 0)

is pseudoconvex. 13) 13)

One can show that zI

is not possible.

— toE(xI) s not subharmonic. Therefore, a short direct proof

382

4

domains of special families of holomorphic functions

Existence

Put D := (z i

C:

(xi) <0). Observe that

—

is a Hartogs domain with < 1),

(n — 1)-dimensional balanced fibers, i.e. = {z E C": E where H : is upper semicontinuous. Recall that

such a domain is C is pseudoconvex (which is obvious in our situation) C and log H ff'SR(D51 ). Therefore we may assume that n = 2. First we calculate pseudoconvex 1ff the basis

grad u5(z) If

+ iyj, Z2).

= (—

(xi) = 0. Thus xj = — i/i + and therefore D5 has a C2-boundary. #0 for each z E It remains to discuss the Levi form of at boundary points z of D5 into the

zE

D5 and yi = Z2 = # 0. Hence grad

d

0,

then

direction of tangential vectors X, i.e. it suffices to deal with the following vector

x= We have to show that

£up(z: X)

+

+

= (—

0,

or

+

+ 2)1z212 +

0.

In order to get the last inequality it suffices to verify the following one — 2)


(t)

z E

2 Oforz

R\(l —e, I

E

dD€,(t)holds

forz E 8D5 with (—i/l +e6,oo)\(1 —e, 1 +e). is a second To discuss the remaining part of the boundary we first observe that order polynomial with negative second derivative, if e is sufficiently small. Calculation shows that

• > 0 and — e). — e is sufficiently small. In particular, ,,, W5 I(i—€,I—

> 0,

+e—

'P€ 1(I+5t_

1-1-s)

+

<0, if

<

We discuss now three cases. Case 10: Letx1 E

(I —

s. 1 —

=: 15,1:

From the remark before we know that Thus — s

is monotomcally increasing on (1 —

—

— I.

4.1 Special domains

383

0. So it suffices to see that

(t) is obviously fulfilled when

—2) <

3) <

—

> 0.

E

X1

According to the behaviour of the second derivative we get

s which gives that (t) is true on

Case 2°: Let

E (1

i

+e—

<—v.

—

for sufficiently small e. 1

+ e) =:

According to the remark on the

+ e). < 2, e sufficiently small. Hence (t) is

is monotonically decreasing on (1 +

third derivative we know that So we conclude that

+e—

I

clear.

Case 3°: Letx1 E (I

—

I

+e —

:= 16e3p(t + e + 1), It is easily seen that for t E

=: 1(3. PUt 3€

1

€2

E (_ -i-,

=: .1(3.

e sufficiently small, we have

(1) *E(t) <

+ 8€ + C1e2) + C2e9, _12(8t + 24€ — C3e2), < —:(24t + 48€ + C4e2), where C1, j = 1, 2, 3, 4, are positive constants. Let t E with —2 0. Exploiting (l)-(3) we get (2) (3)

+ 24e)2 —

1448e2

+ t4D1e3 + t4(8t + 24e)2

—

t4C5e3

is sufficiently small and where C5, D1, 1)2 are positive constants. Hence, (t) is also true on I(,3 for sufficiently small We have proved that DE, e small, is Levi pseudoconvex. Hence, D( is pseudocon-

where

0

vex.

As an immediate consequence (use a btholomorphic rescaling) we get the following result.

Corollary 4.1.45. Let B and €o be as in the barbell lemma. Furthermore, for r (€)

+

we put

c:={z€C":Rez1<(l—e)r(r)},

384

4 Existence domains of special families of holomorphic functions

If t E (0. to), then there is a pseudoconvex domain Ge with

{z E C'1: Im

wherez =(zi,E) E

+

)r(e)2 } fl

<

fl

=

Cx C'1'.

The second lemma is an observation of H. Boas (cf. [Boa 1996]), that is based on a result due to Y. Eliashberg (cf. [Eli 1990]).

Lemma 4.1.46. Let D be a domain in C" and let z0 be a strongly pseudoconve.x C2-boundarv point. Then there exist CO > 0 and pseudoconvex domains DE C D, 0 <


D\B(z°,e) = such that the boundary of

\]3(z°,e)

near z° looks like a ball with z0 as a boundary point.

Proof Without loss of generality we may assume that z° = 0. Fix an > 0 such that there are a strictly psh defining function p E C2(l3(2eo)) of D and a ball B(a, r) C D flB(e0) with z0 E 8B(a, r). Denote by E C2(C'1) the global defining function of the ball B(a, r). Observe that p and coincide at z0 together with their first derivatives. Thus there is a positive constant c such that — p(z)I < cIIzlI2, z E B(eo).

Fix an e E (0, to). Our aim is to find good' cutoff functions a, E such that max{*, p + ca — fi} equals near 0 and becomes p + ca away from 0. In afirst stepputa*(z) := where& Owillbe chosen in s) and ã(t) = t2 ti < a, a small. Since the function p + ca* does not coincide with p outside of a neighborhood of 0, we have to modify a* to get the wanted a. Choose a non-negative function such that 'l)(z) = I on 13(1) and CD(z) = cb(DzII) everywhere. Put Cl) E Cl)o(z) :=

and

a := a*cl)o. Observe that p + ca =

Now, we want to find a <

and

\

p on

& such that p + ca

is

strictly psh on

Therefore we calculate its Levi form

£(p + ca)(z: X) = L(p + ca*cl)o)(z; X) = £p(z; X) +

c

+ ccl)o(z) +2cRe j.k=I

(&"IZjI

+

&'(Izj) )ix1i 2 IzI) 3

+ ca*(z)Lcl)o(z; X)

2iZji 8Zk

4.1 Special domains

385

(4d +

+

1z1

—

c(dta*(z)e2 d > 0. and d1,d2

£p(z; X) > dIIXII2 on 13(eo) x small. We have to find a and & such that

where

> d, a
4d + cã"(t) + Construction of &: FixO < y <min{2,

IR>o sufficiently

+ 1 >2. DelIneg:

:=

—* Ras

4.

I
g(t) :=

Ii.

t

— IL),

Finally, put

xi: [0. Then Xi

—÷ R,

C2 and Xi(r) =

x(t) Therefore,

=

t

xi(t)

:=f '1 f (—

xg(x)dx)ds.

[0. 1]. Moreover, observe that Xi > 0 and that

x'(r) =

xg(x)dx,

xg(x)dx + gO).

=

=

gO)

t E [0. ii]. Next, we choose

= jç(0) = o.

1) with X2(O) = xlOL) and

E

:=

FixanN E NsuchthatthenewfunctionX2: [0. N] —÷ the following inequality

x'(t)

N

+

1

—*

JR.

sup

x(t) =1 X2(t/1),

y.

——

386 Let

4 Existence domains of special families of holomorphic functions

x(') := 0, z >

+ N. Then x E C2(R+).

Now, leta > Put

a:

R÷

&(r) :=

R,

a2X(t)

Then this gives a C2-function on IR+ satisfying

a(f) = 0.

f

> F.

,,

a (1) +

?

Therefore we get

£(p+ca)(z; X) >

>0

1X112+cd IZ,l>(7

for X

0. Hence, the function p +ur is strictly psh on and fulfils the following inequality p + ca* > on 18(a). Finally, we set := In virtue of the strict plurisubharmonicity of p + ca we find a > 0 such that also the function p + ca —

=: p + ca —

=:

strictly psh on 18(eo). Observe that Pi = p outside of strictly less than i/i near 0. Moreover, pi(z) = p(z) + cIlzII2 when Thus there. Put remains

-

p(z) =

I

Uzil

1 pi(z),

a < lizil <

Then the modified pseudoconvex domain

:= (D \ 18(ro)) U Hence the theorem is proven.

and that P1 is a. <

is given by (z E 18(eo):

< 0}.

0

Remark. One can even show if D has a then there exists a modified domain having also a e2-boundary. To do this one has to apply the approximation theorem for plurisubharmonic functions of R. Richberg (cf. tRic 1968J) to smooth the maximum we got at the end of the proof.

Proof of Theorem 4.1.43. Put around each corner of y a small ball and then apply the previous two lemmas. 0

387

4.1 Special domains

Now we are able to present the announced example.

Example 4.1.47. First, we repeat the construction of the Sibony example. Let is > 0.7. j E N, and so that any b E E E be given such that non-tangential limit of a certain subsequence of (a1

>ct1

22, XE

E,

Then we put

u

:= exp(ü),

where the positive a's are chosen such that the series converges uniformly on K (1/2). Furthermore, we set D := D(E, u) := {(z, w) E E x C: wi <exp(—u(z))}.

ç E2 is pseudoconvex and each f E 3t4'°°(D) extends holomorphically to E2. Moreover, the same conclusion remains true for small perturbations of D 'over' D

K(l/2). defining function of D is given by p(z, w) := iwlexp(u(z)) — I. < 0.5, wi 0. A few calculations show that the boundary of D 'over' K(l/2) contains a lot of points a E E2 at which p is strictly psh. Fix one of them, e.g. A local

izi

a

piecewise linear Jordan curve y: [0, I] —*

C2

having the

following properties

• y(O) =

a and y'(O) points into the direction of the outer normal vector of D

at a, I,

• starting from the first corner y(ti) = (yj(t1). y2('I)), y2(1I)

0. the curve y

continues into the direction of(O, y2(tI)) upto the next corner y(12) =

y2(12))

with )'z(12)

= • the curve y continues piecewise linearly in the plane z =

up to the moment it reaches its last corner y(IN)

= (yi(ti),

Yl (Ii)

with 1y2(t)i > I

y(t2)).

•the final piece ofy consistsofthe segmentfrom y(tN)to (y1(t1), —y2(tI)) E E2. Applying Theorem 4.1.43 there is a pseudoconvex domain L) containing DUy([0, l])andcloseto DUy([0, I]) lJ) does not intersect D. Therefore, if f E R°°(D), then there is an f E (9(E2), liD = liD. Ontheotherside,observethatthefunctiong(z, w) := belongs to .7€°°(D). So g extends holomorphically to E2; in particular, the value

at the point y(l) is different from the value of g at this point. Hence the R°°-envelope of holomorphy of D cannot be univalent.

388

4 Existence domains of special families of holomorphic functions

4.2 The Ohsawa—Takegoshi extension theorem The aim of this section is to present an extension theorem for L2-holomorphic functions on D fl H to the whole of D, where D is a bounded pseudoconvex domain in C" and where H is an I -codimensional linear subspace of C". This theorem has been proved first by Ohsawa—Takegoshi (cf. [Ohs-Tak 1987]). After that starting paper T. Ohsawa has given different generalizations (cf. [Ohs 19881, [Ohs 1995], [Ohs 1994]). The original proof of Ohsawa—Takegoshi was based on a d-theorem involving complete Kähler metrics, inspired by a theorem of Donelly and Fefferman. In 1995, Siu (cf. [Slu 19961) found a simpler proof which avoids the use of general metrics and the somehow complicated commutator identities used in the orig nal proof. Siu's proof is based only on the Hormander—Kohn—Morrey formalism for the in domains

in C". In this spirit, B. Berndtsson and Diederich—Herbort also gave another proofs of the extension theorem (cf. [Ber 19961 and [Die-Her 1999]). Here we will follow the proof of Siu. The extension theorem reads as follows:

Theorem 4.2.1. Let D C C" be a bounded pseudoconvex domain, let H := {z E C" z,, = O}, :

and let

Then there is a constant C = C(D), depending only on the E diameter of D and the dimension n, such that any f E O( D fl H) 14) with

I

JDnH

If(z', O)12e

extends to a holomorphic function F E 0(D) (i.e. F(z'. 0)

= f(z', 0), (z', 0) E D)

with

ID

C

If(z',

Observe that the assumption about the special shape of H does not mean any restriction of generality in the category of one codimensional subspaces (use an affine transformation). Moreover, in [Ohs 1988] it is shown that Theorem 4.2.1 remains also true if H is thought as a pure dimensional closed complex submanifold of C". Even more,

it is claimed that the result is also true if D is a relatively compact pseudoconvex subdomain of a Stein manifold.

The proof of this extension result consists of different steps.

space

—

In this section, D fl H will be understood as a subset of C" The correct interpretation is easily seen from the context.

as

well

as

its projection to the

389

4.2 The Ohsawa—Takegoshi extension theorem

The proof follows similar ideas from functional analysis already used in the solution j = 0, 1, 2. be given Hubert spaces and let

of the a-equation in Section 2.4. Let

T

-+

H1

iDomS

S

be linear, closed, and densely defined operators with ST = 0. Assume moreover that L : H1 —÷ H1 is a linear semi-positive 15) bijective operator. Then the following statement holds.

Lemma 4.2.2. Asswne that (Lu, u)H1

+

I

Then, for any g E Ker

V E Dom T4 fl Dom S.

S, there exists a u Tu

=g

Dom T fl Range T* with

and

Proof Using the property that always (Lx,

R it follows from evaluating

E

(L(x + y), x + y)111 = (x + v, L(x + y))111. (L(x + iy), x + iy)111 = (x + iy, L(x + iy))111 that (Lx,y)111 =(x,Ly)H1, Then

(L(x + zy), x + ty)111 > 0, t

C, immediately leads to

< (Lx,

(Lx,

(Ly, y)H1,

Therefore we conclude that with g = (v.

2 I

= (v. <

-

I

Lg)H1 I-

x,yEHi.

E

x,

Y

E H1.

the following inequalities hold

(Lv, v)H1 (Lg,

+

= t' E Dorn

fl Ker S.

Then, following the argument in Section 2.4 we find an element u E Range T*flDom T with the claimed properties. D

To be able to apply Lemma 4.2.2 we describe the situation of the main theorem in terms of Hilbert spaces. L is called semi-positive if (Lx, x)ff1

0. x E H1.

4 Existence domains of special families of holomorphic functions

390

Let D C C° be a bounded pseudoconvex domain given as

D={z E where p C2(b) where

: p(z)

fl 13

<0}withDflH :p(z)>0},

I grad p(z)I

=

1

lift < £, j(i) =Oift >1,

C°°(R, R) with

2/(1 —1).

and

Fix an f E (9(b) with J'DflH If(z'. For sufficiently small e E (0, 1/2) we put

:= j(1rI2/e2), t where XE(Z)

e ?S3((D) fl

and let

on

c

We choose ( E (0, 1) and a cut-off function j

j(:) =

0,

:=

zE

E

< 00.

:=

C,

is a s-closed

Observe that

satisfying

= (l/e)4f

ID where DE := {z

<00,

D : £e2 <

Let us introduce the following functions

:=

+

and

z

x (A/2)E; in particular, where A > is taken such that D C ? log 2, Z E D.e E (0. 1/2). Then 0 we have 1

X) >

11X112,

z

Fore> Olet

D, X E

0 < e < 1/2.

ife=0 ife>0.

11

Similar as in the proof of the classical d-problem we define a sequence of operators, namely:

Ho :=

j Dom

H2

:=

u'),

where

:= 16)

See Section 2.2.

u

Dom TE,

and

:= ..j&

e

> 0,

4.2 The Ohsawa—Takegoshi extension theorem

=

Dom

are linear, closed, and densely defined operators satisfying = Dom S. and Dom = Dom T*, where T := To and

and

Obviously,

391

Dom T, Dom

S:=So. Finally, we put n—I

fl—I

2

+

H1 —÷

+ (7 +IZflI2)2

Observe that is a semi-positive bijective operator. What we like to do now is to prove the estimate in Lemma 4.2.2. In order to realize

the necessary calculations we need the following lemma (cf. [Hör 1965], [Pfl I 975b]).

Lemma 4.2.3. Under the above notations we have that in Dom T' fl Dom with respect to the graph-norm. Moreover, aform v

I) (D) fl Dom TE* is dense 0.

Proof In order not to interrupt too much the flow of the proof we postpone the proof of the density result to the end of this section. So it remains to prove the second statement in the lemma. In fact, we do only that part we will need in further argument. Let

= Then

E C(Q1)(D) fl Dom

(u, T*v)110 = (Tu,

In particular, using u E

fl Dom

= U E

Dom T.

and Stokes theorem we get

T*v = j=1

0 for some point Z0 E 3D. Without loss of generality, let Re 13 > 0 near z0. Choosing a function 0 < ü E with u(z°) > 0 and suppü C tz E : Rei3(z) > O}. Then we have UID E Dom T and Put 13 :=

Suppose that 13(z°)

(a, T*V)HI = (Tu.

= J03._I

dZj

17) =

17)

Recall the following version of the Stokes theorem.

J ,

,

where ds denotes the surface measure.

az1

ao az3

f,gE&(D).

392

4 Existence domains of special families of holomorphic functions

= (U, T*v)111 + Lb i.e. the boundary integral is zero, contradicting the above assumption. Hence the second condition is verified.

C]

Calculation to get the estimate in Lemma 4.2.2. According to Lemma 4.2.3 we may

startwitha(O, 1)-formu =

<e<

E

Then

j=I We

like to verify the following inequality (compare Lemma 4.2.2): (L€u,

+

u)111

Starting with the right-hand side we have

Su)jq, = (yeT*u, T*U)HO +

(aETu, T*u)flo +

SU)H2

+ (*).

We proceed by transforming the term (*):

+f

(*) =fD

D

—

aUi)(auk dZj

dZk

j
dzk

Observe that u E Dom T* implies that this integration by parts did not lead boundary term. (*)

=

f

(d(fliT*u),

+ JD 'k

D

-f =

f

D

j
dZJ

- dZk +f

k

8Zj

8Zk

aZk —

dUi)auke_*dA dZk

j.k

dZJ

Using Stokes theorem (recall our assumptions on the boundary function p of D) we proceed as follows:

(*) =

T*u),

+ (S*O7eTU), u)y1

jk

aZJ

-

aZk

dZj

4.2 The Ohsawa—Takegoshi extension theorem

393

where we used the following abbreviation

:=

—

In the next step we treat the last boundary integral. Since u

Dom

it follows

= 0. Therefore there is a function 0 near such that there the following equality = Op holds. Differentiation yields on that

u1

Multiplying with Uk and summing up the last equality leads to

= on dD. Thus we end with the following form of (*):

(*) = (dOkT*u).

u)H1

+ (S*(,,tTu), u)111

+f

+f

>

Observe that the last term is non-negative. Moreover, the third term can be rewritten

in the following way:

f

I

JaD

rkL,—--—uke az1

ds

j,k

= ID

+

1k

>:

where the first term is again non-negative. Thus we obtain the following inequality

for(*) (*) > (a(,,5Ttu), u)H1 + (S*(ii€Tu), u)H1

+

JD.

dz1

aZf

= E1 + (rmTTtu, u)j.j1 + E2 + (,,(S*Su, u)H1

=:

394

4 Existence domains of special families of holomorphic functions

where

E1 :=

u)H1 —

(7kTT*u, u)H1

=

f

Zj

D

Observe that IIET*u is a &-function on D, i.e. the operator T is nothing than a. —

E2

where, as above,

S*Su

—

Thus

E2=—

I

Jo

dZf

j.k

dZk

Now we are going to analyze a part of (*)j, namely —

E2

=

u)H1

+

u)111

e_*)e_*dlt2n

+J

+

Uj

= ID 'h:

—

f

I

d2u3

— Jo

+1

auk

—

+

—

a* au1 -

azjazk)

a2u* az1

i,,,

Thus we obtain

(*)> E1 + E2 + J >

aZJaZk

j,k

= Et + f j,k

+

aZfaZk

az1

+

azk

:= (*)2.

395

4.2 The Ohsawa—Takegoshi extension theorem

We apply Stokes theorem to the last term in (*)2:

f

D j,k 8z1 azk

=

( az;kujuke

ID

+

+

1

ujuke*ds

f

aZfr

liD

= where the last equality follows again from the fact that u E Dom T*, i.e.

a,kap —--—UJUkIaD = 0. j.k

aZJ dZk

Using the term of E1 we have the following inequality

(*)

+

E1

f

u) —

+ 2ReJ

u) —

In virtue of the form of

(*)

we obtain

ID —

—i

e

JD e2 +

L.et

'j

D.

D

JD

+ IznI2

us repeat what we have got so far:

+

f

+

+

f

—

JD6+IZnI

,,e

dAb—

JDe +IZnI

e

396

4 Existence domains of special families of holomorphic functions

and 'k leads to the following estimate:

Applying the definition of the functions IEU

2

i

c

(62 Ye —

ID (lIü

12

ID

=

+

u)jj1, (4.2.1)

:=

where lull2

+ ID

Therefore, using Lemma 4.2.2, we end up with the

following conclusion

Lemma 4.2.4. Under the above conditions there exi,cts a

=

E Ho, 0 <e < 1/2, with

and

f

26 J (l—t)r

D

(62

+

We define Xe! — s/Ye + qeZnUe on D;

is a holomorphic extension of f IDnII which satisfies the following estimate: 1/2

(IL, 2

(j.

fl

2

2

1/2

+ Iznl

)

Ill

where

:=

sup

{lznls/y€ +

23 It + A2(1 := (Z E D :

2

=:

e}.

Since f is square-integrable on D, the first term on the right-hand side tends to 0 for e —p 0. The second term gives

ID(

+

f sup

{lf(z',

zn) +

397

4.2 The Ohsawa—Takegoshi extension theorem

+ <

I

J

\(H xC)

If(z'.

Zn))

SUp {lf(z'.

J

Dr1H

+

J

xC)

If(z',

Zn).

In virtue of uniform continuity of f the first integral tends to

I

JDOH

If(z',

Recall that D is assumed to have a smooth boundary. Therefore, we find so > 0 and = 0, such that c (0, 1),

ED, (z',O)

ifz =

DnH.

<6, then dist(z'. Dfl H) <

Thus, (1/62) x the second integral tends toO ifs —÷ 0. Therefore, using standard methods we find a sequence (Ff1

0(D) with F =

f on IL) fl H such that

(I + A2)a f f where a denotes the supremum of eC on D. 2

D

DflH

converging to F

If(z',

Summarizing we have verified the following result

Lemma 4.23. Let D C C

be a bounded pseudoconvex domain with smooth bound-

f 0(D), and let C2(D) fl fPSJt4'(D). Then there is an effective positive and a holoniorphic function F E (9(D) satisfying

or),, let

E

ID


From here a standard argument leads to the proof of Theorem 4.2.1.

Proof of Theorem 4.2.1. In virtue of Proposition 2.5.9 we may assume that f E 0(D). We write D as the union of an increasing sequence of bounded pseudoconvex domains D1, b1 c D, with smooth boundaries. Moreover, we approximate 1t, by fl

a decreasing sequence of functions 'Pj,k E

In virtue of Lemma 4.2.5, there are functions Fj.k E

with f =

FJ.A

on

fl H such that

I

cf

DflH

If(z',

Observe that we may use the same constant C in all steps because of the effective form of our constant from above. Finally, the usual diagonal process and Montel's theorem finishes the proof of the 0 extension theorem.

398

4 Existence domains of special families of holomorphic functions

Corollary 4.2.6. Let D and .p as in Theorem 4.2.1 and let H be a complex affine 0. Then there is a constant C linear subspace of C'1. Assume that D' := D fl H depending on the diameter of D such that any function f E 0(D') fl L2(D',

C

ID

ID'

Proof W.I.o.g., we may assume that

H={ZEC'1 wherek
Remark. One may try to generalize the extension theorem of Ohsawa—Takegoshi for functions belonging to U' instead to L2. But what turns out is that then the situation

is drastically different from the one discussed in Theorem 4.2.1. As it is shown in [Die-Maz 1997] the following happens: Let n 3, 2 < p < 00, and e > 0. Then there exist a bounded pseudoconvex domain D C C'1 with anda holomorphicfunclion f E 0(D fl H) wit!:

I

D fl H

0.

<00,

J Dflhi

which does not have a holomorphic extension to D in the space

18)•

For example, D can be given as an algebraically biholomorphic image of a complex JzjI2Pi < I). ellipsoid & := {z C'1 :

As a simple consequence of Corollary 4.2.6, we get a great class of pseudoconve domains being of holomorphy.

Theorem 4.2.7. An)' boundedfat pseudoconvex domain is an

-domain of holomor-

phy.

Proof Assume there is a bounded fat pseudoconvex domain D C C'1 that is not B2 with the same center domain of holomorphy. Hence there are open balls Bi Recall that

=

fl (9(D); cf. § 1.2.

399

4.2 The Ohsawa—Takegoshi extension theorem

E D, B1 c D but B1 fl dD an F E (9(B2) with Fl81 = b

Since

0, such that for any function f

there exists

E

D is fat there is a point a E B2 \ D. Set if as that affine line defined by a

= Zn =Oandb = 0.

andthecenterb. Wemayassumethatll isgivenbyz2 =

is square-integrable on D' := H fl D. Then the function 1, := (Zi — But F itself extends holomorphically as F to B2. Therefore it extends to F E which contradicts In virtue of the identity theorem we conclude FIB.,nH =

0

withthesingularityoffatai. Recall that

5D): cf. Lemma 2.5.3. Therefore we conclude

C

Corollary 4.2.8. Any fat bounded pseudoconvex domain is an

So)-domain

of holonwrphy.

on Looking at the punctured disc in the plane it is clear that any extends holomorphically to the whole disc. Thus it is natural to assume some fatness in Theorem 4.2.7. More general, the following observation is true:

Theorem 4.2.9. Let G be any domain in

and let A be an analytic subset of G, \ A) = In particular, if G ispseudoconvex and if the G. Then -domain of holomorphy although again codimension of A is 1, G \ A is never an A

pseudoconve.x.

Proof 19) It suffices to prove the theorem in a neighborhood of a regular point zo of

A. Therefore we may assume that zo = 0 and that A is defined by Zn = 0 near the origin.

Then, in a neighborhood of the origin, we write f

f(z', Zn) =

as

a Laurent series,

where z = (z', Zn).

In virtue of the assumption there is r > 0 such that

I

If(z',zn)I2dA2n(z) <00.

JlI:'I
I

If(z', Zn)I2dA2(Zn)

<00.

J{0
Evaluating that integral leads, for almost all z', to the following equation

f

tf(z', =

—

+ ir

ak(z)12 I

Cf. the proof of Proposition 2.10.1.

k+1

400

4 Existence domains of special families of holomorphic functions

Since e E (0. r) is arbitrary it follows that ak(z') = 0 for all k < 0 and for almost all z'. and thus for all z'. Hence f extends holomorphically to a full neighborhood of the

0

origin.

In LJar-Pfl-Zwo l999J (see also LJar-Pfl 1989]) it is shown that any bounded balanced pseudoconvex domain is Bergman complete. In particular this implies that all such domains are -domains of holomorphy. Here we give another more direct proof (cf. [Jar-Pfl 19961) based also on the extension theorem of Ohsawa—Takegoshi.

Theorem 4.2.10. Any bounded pseudoconvex balanced domain 1}

is an

-domain of holomorphy.

Pmof The proof is by induction on the dimension n. If n = 1, it is trivial. Now let n > 1. Suppose that G is not an of holomorphy. Since G is balanced, we may assume that there are domains 0 V C U fl G ç U such that: (a) for any f E there exists an f E 0(U) with ft v = fly; (b) there is a continuous curve y : [0, 1] —÷ U with y(0) E V, y(l) =: b E G, C G;

(c)b=(b1,0

0).

To see that (b) is always possible we proceed as follows. Let U, V be as in (a) I] —+ U be an arbitrary curve with y(O) E V. y(l) =: bE and let y : and y([O, 1)) C G. Observethatthereisr0 E (0, ljsuchthat[O, r0)b C Gandr0b E It suffices to show that we can replace U, V. y by TOU, r0V, Toy. Take an arbitrary and put fro(Z) := f(roz), z E G. Then fT0 E and function f E therefore there exists an Define 1(z) := E (9(U) with = fr0lv. z E r0U. Then f E (9(r0U) with flr0v = flr0v. We chooser e (0. 1) and a domain W with ry([0, I]) C WC GnU, ry(0) E V, and [r, ljb c U. Take now the(n — 1)-dimensional domains U and V and an 0 such that Er. 1]b c U x (rE) C U and rb E V x {O} C (U x {0}) fl W. We define G := G fl x (0)) and consider G as a bounded balanced domain of holomorphy

Let f

Using the Ohsawa—Takegoshi Theorem we find a function F E

with Flã = f. By our construction there exists an F E 0(U) such that Fly = Fly. Then the identity theorem implies that F and F coincide on W; in Thus fj extends as a holomorphic function to U with particular. F (., 0)1 = f I U

G, which contradicts our assumption for the (n — 1)-dimensional case.

0

Remark. We do not know how to characterize unbounded balanced -domains of holomorphy. Even more, there is no description of unbounded balanced domains The remaining part of this section is devoted to the proof of the density result of Lemma 4.2.3.

401

4.2 The Ohsawa—Takegoshi extension theorem

Proof of Lemma 4.2.3. Let

I

= DomTt flDomS

E

be given. We have to find a sequence

c

=

fvf E C2(D)

such that aD

(1)

(2) Ill

= 0, — Sf11112 + ITs!1,

+

—

T5f11110

0,

fr). J = 0, 1,2. Observe that (1) implies that E Dom Tt. Since e2(D), can be read as the result of the differential operator T5 directly applied to Using the special form of the operators TF and SF, the above properties remain true if we substitute T5 (respectively S) by T (respectively by Sf), i.e. we obtain the density result claimed in Lemma 4.2.3. Recall that the operators T5 and S are differential operators. Thus we may look at the following system of differential equations: where H1 :=

/T5\ (

s )w

JT5f\ = ( sj )

,

w E

First, we rewrite (*) in its real form. Let

For the operator S we obtain

where the

x n)-matrices

=

are given as

—I.

j=t+t. I

1,

0,

otherwise.

Observe that the empty sum has to be read as 0.

(*)

402

4 Existence domains of special families of holomorphic functions

For

we get

where the (2 x 2,z)-matrices At, At are given by the following entities I Uk'-'

j=n+t

I.

ifk=l.

0,

otherwise,

I,

ifk=l. j=

0,

£

otherwise.

Moreover, the entities of the (2 x 2n)-matrix B, belong to C0(D), D D. Recall that i/i E C2(D) Summarizing, the left-hand side of the system (*) has the following real form At ,' \ o o

1

fu

/

c1)(u)(E)(U) 1' := and where denotes the (2n x 2n)-unit matrix. Now we are going to prove that, if(a, b) = (ai,..., b1

where B

rank

A'ae

and

+>

o

0

Ct

0, then

_co)be]

0

t=i

(i)

Cl

(ii)

0

are maximal. The case (i) is simple, so details are left for the reader. In the remaining part we will concentrate on (ii). We may assume that at = = aj_i = 0 and a3 0. Observe that the matrix under discussion has 2n rows. Therefore we consider the following (2n x 2n)submatrix OI.i—1

—b1_,

—a,L_,

a,

a(i)

b,•••b1,

—b1

—b(i)

°I.i'-I

a(i)

01_1.1

Oi—i,n—,

b1L..1

Oi—,,n---,

—a(i)' —b111_, On—ti—i

O,,,

b(i)'

°i—IM—i

b

b(i)

O7?_g,I_I

b(i)'

—b,I,1....1

—a,I,_,

°j—1,1

Oi—lji—i

°n—Li—I

—a(i)'

(ui

.

4.2 The Ohsawa—Takegoshi extension theorem

where a(i) := (a,+i . an), b(i) := (b1+i

403

ba), 'k denotes the k-dimensional unit-matrix, Ojk is the zero matrix with j rows and k columns, and' means the matrix transpose. Because of a 0 we find real numbers n, such yj. and 8j, i + I that

j

(aj a

a

J

Write the i-tb column, respectively the (n+i )-th column, of(iii) as the first, respectively the second, column. Then, adding the a1- multiple of the (2 + (i — I) + j)-th column and the p1-multiple of the (2 + (i — I) + (n — i) + (1 — I) + j)-th column to the first column and adding the of the (2 + (1 — I) + j)-th column and the -multiple of the (2+ (i — I) + (n — i) + (i — I) +j)-th column to the second column, we end up with a matrix of the following shape (02n—2,2

kF

*'

M

where after a short calculation the (2 x 2)-matrix F has the following form:

1+

(a2+b2)( a,

I

>

b a,

The remaining ((2n — 2) x (2n — 2))-matrix M looks like

/

°i—I.n—i

M—

I °ni.t1

—

—b111_1

b,11_1

0i—I.n—i

0n—i,i—i

°i—I.n—i

Adding the p-multiple of the ((n — I) + f)-th column to the f-tb column we get

detM = Hence the matrix (ii) has maximal rank along dD.

Recall the shape of the operator T:

T'f =

—

1=1

Then, for

E

we get

4 Existence domains of special families of holomorphic functions

404

where Ii

:=

E Ho,

tively h) to C", and (

I (respectively h) denotes the trivial extension off (respec-

,

is the distributional pairing on D.

Put N := 2n, K := 2 +

+ 2n, K0 = 2, J :=

and

A A I

o

/

\12n

Im f,,), and let g Moreover, let u = (Ui (Re fi Im fi be the real form of (T*f. Sf, f). In particular, u and g are tuples of L2(G)-functions, where G := D. Then At, I < ( < N, (respectively B) are (K x J) -matrices with &(G) entities where G := D. Moreover, the rank (respectively with C°(G) entities of the following matrices

is locally constant near

and

Here At(Ko) denotes the matrix consisting of the first Ko rows of A'. With these notations, u and g fulfil the following differential equations (in the sense of distributions) N

j on

ifl
fIkK

G

(a

(t)

Summarizing, we are led to prove the following lemma

Lemma 1. Let G c IR", N > I, be a bounded domain with a C2-boundai-y and a G. Moreover let At, defining function p, and let G be another domain with G I

< t! < N (respectively B) be (K x J ) -matrices with C' (G) (respectively with

C°(G)) entities. Assu,ne that the rank of the matrices is locally constant near dG.

Then,foranytupleofL2(G)-functionsu = (UI

I

uj)andg =

satisfying (t). there exists a sequence of C2(G)-functions that

+ Bus'

—

=0 e=i

I

on3G.

= (u

At (K0)

gK) such

405

4.2 The Ohsawa—Takegoshi extension theorem

The rank condition above leads to the idea to reduce the proof of Lemma 1 to a similar local result (cp. Lemma 3) via coordinate transformations and partition of unity. As a preparation we discuss first the case that the matrix A" is of a particular simple form (cp. (1:)). Lemma 2. Put

Q:={xEQ:xN
Q:={xER":lxJI
:

IxjI
{x E Qi

XN

<0),



natural numbers and let At, I N (respectively B) be (K x J)-matrices with '(Q)-entities (respectively with C°( Q)-enrities bkJ). Moreover we assume that

if I
(1,

(1)

otherwise.

0,

Letu = (uj, .. ., uj)andg =

g1 E L2(Q),fulfihthefollowing

differential equations (in the sense of distributions):

t=ij=l

4

+

Ui =

gk

{

j=I

Q.

and 1k denote the trivial extensions to RN. Then there exists a sequence of

—

—

where

•

+ H

•

=

0,

1

j

5

Proof Choose a cut-off function

L2(Q)

:=

such that

1 5k 5K;

I

> E

1,

Put

—* 0,

—

i=t

£=ij=1

=

for suitable positive numbers 17,., v E N. (RN -

with the following properties:

and

I

e > 0. Then there are linear operators

I.

: L2(Q) —÷

defined by

:=fRN-I

g E L2(Q), X = (i,XN)

E

Q,

406

4 Existence domains of special families of holomorphic functions

where i denotes the trivial extension of g to RN. Observe that maps L2(Q) into itself. It is clear that r8(g), g E L2(Q), allows partial derivatives in directions I (this equation is j
gj(x)

:=

Fix.r0 := (R

—

—

[

x E Q.

—

R1)/2 and letO <S

Leta E

Put

I ThenaE Therefore,

I

=(fk,a6)Q-. (=1

where

(,

K,

j=I

j=1

) means the distributional pairing. Calculations lead to the following equa-

tions:

•


I

=

• •

=

= (u3,

=

(j I

C < N.

Hence we may rewrite our differential equations on Q1

—

as:

—

j=I I

We only mention that these equations hold also on Qi. if 1 functionsfl instead a E

j

Ko, i.e. for aU test

407

4.2 The Ohsawa—Takegoshi extension theorem

j ro (respectively J — (r — E £)'( Q j) (respectively DXN E )-functions), that we (respectively by L2 (

In virtue of the assumption we conclude that, if I

<j

ro)

£)'( Q

J), then the distributions are given by L2( Q

call g,

i.e.

=

fi E

1

J

aE

(gJ,a)Q— =

<

j < K0;

(r — rO) <1

—

J.

From here we get

+ II

t=I j=I

+ j=I

—

j=t

N-I J > t=I j=I

—

+

0.

+ IITe(fk) —

—

Therefore it remains to approximate the following functions For that reason we choose new

g.

and

4JDX(TE(uJ),

functions

such that

O),

I

and

JRN

If

Then, we define the following C2(Qj)-

S > 0, then we put functions 1

*

if ro <

*

if

I

f< j<

J

It is clear that

ur(x) = 0

I < j
if

*

U

*

—

XN

for suitable positive

> in

+ 0 ifl

I<

£


Moreover, since TeUj allows a weak xN-denvative on Qi for I <j
on Qj

for I — (r

Then applying the usual diagonal process concludes the proof of the lemma.

The next step consists in discussing a general matiü

0

408

4 Existence

domains of special families of holomorphic functions

Lemma 3. Assume the situation

of the

above lemma except condition (i). Instead of

we assume that rank AN = r (respectively rank AN (Ko) of Q fl {x RN : XN = O}. RN : IxaI < R'}. Then there is a cube Q' := {x

= TO) lfl a neighborhood Q'

Q, such that the

conclusion of Lemma 2 remains true.

Proof We may assume that there is a cube Qi = {x that

rank(afl:l<j,k
-

u :=

I

AN(r0)

Thenü (L2(QT))J. Writing the matrix AN(ro)

as

we get det U = det U1

< R1}

:

Q such

onQi.

1)UIQ_ =: UuIQ-.

AN(ro) =: U = (Ui,

U1 :=

RN

E

:

U2) with

k, j
I

Then the following matrix

0 on

the inverse to U and its coefficients belong to In virtue of the assumptions and the definition of ü, we conclude that the following differential equations are satisfied: is

D,.,

U3

t=Ij=I

+

djkUJ

=

1k on

+ >djküj = fk on Qi, t=i

j1

where

+>sAjbkA

dk) := t=I

Put

1

< k < K,

j=I

A=i

I
409

4.2 The Ohsawa—Takegoshi extension theorem

and

:1
D := Observe that the matrix

+ By.

is of the following form

/Iro

°

\u5

U4 U6

rows. U4 have Ko — where 'r0 is the ro-dimensional unit-matrix and implies that U4 = 0 Since V is not singular on rank condition for

on Qi. Assume for a moment that the statement of Lemma 2 is fulfilled for the system (i)

with respect to a cube Q2 := (x E RN : Ix,I < R2) V E N, such that

(L2(Q-))

ü(x) = Then, defining

0.

:=

1<

j
Qi. i.e. there exist functions

—÷o,

> —rfl. for suitable positive

leads to

+

+ BVÜL —

f=

+

—

f.

and u(x) = 0 as above. Therefore, only the system (i) has to be studied. If ro < K0 we look at the following part of the differential equations:

..f

—

+

(=1 j=I

a E .V(Q1) andr0
ÜjdvjctdApj

(ii)

=

j=1

where

we obtain

N-I I —

t=i

>J

—fQi

ÜVDXN(c,a)dAN

=

f

(iii)

410

4 Existence domains of special families of holomorphic functions

Then taking the sum of (iii) over v =

I,.... ro and subtracting it from the former

formula (ii) gives

j=ro+I

f

j=1 I

= JQ1

—

N\

fvc;Lv)adAN.

Recall that the equations in (iii) are true also for all a E and ro < i.t S K. Therefore we obtain a new system of equations of the following form satisfying the assumption of Lemma 2

(iv)

where Ft are )-matrices and H is a C°( Q with F'(ro) = Ct (rO) and H(r0) D(ro), and where g = consists of L2(Q1)-functions. Observe that now it suffices to prove the statement of Lemma 2 for the new system (iv) on a small cube. Recall that rank F" = rank A" = r near the origin and that F" can be written as F1

where is the (J ro)-zero-matrix rank Fi = — ro near the origin. Transformations like the ones from above finally lead to the situation of Lemma 2

on a smaller cube Q2 Q i. Hence, using Lemma 2 and the transformations back gives the proof of Lemma 3. 0 As already mentioned before starting with Lemma 2, partition of unity and coordinate transformations then complete the proof of Lemma 1. Details are left to the reader. So the proof of the density result is finished. 0

4.3 The Skoda division theorem In Section 4.2 we saw that any fat bounded pseudoconvex domain in C" is an domain of holomorphy. The aim here is show that for a certain class of bounded pseudoconvex domains there exist even functions that are unbounded near the boundary. In particular, such domains are The main tool to derive these results is a division theorem due to H. Skoda (cf. [Sko 1972], [Pfi l975b]).

411

4.3 The Skoda division theorem

Theorem 4.3.1. Let D C E \ D. Assume that

be

apseudoconvex domain,

a> 1, and

E


liz — zOii

ID

Then there exist functions h1,..., h,, E (9(D) satisfying the following properties:

z

—

D,

ID

ID g(z) := liz

z°i12, andq(a) :=

—

Remark. It should be pointed out that what we have formulated in Theorem 4.3.1 is only a particular case of Skoda's original result.

The proof of Theorem 4.3.1 is based on the following Hubert space lemma and the general inequality (4.2.1) in Section 4.2.

be Hubert spaces, F a subspace of Lemma 4.3.2. Let Rj, j = 1, 2, 3, and be dense!)' defined closed linear operators and let T : Jtj .Y2, S: .7?2 —* let A : be a continuous linear operator with Range T C Ker S. satisfying A(Ker T) C F and C a positive constant such that

+

IIA'y +

X

E Dom

fl Dom S.

E F there exists an xo e Ker T with

Then, for any

Proof First DomT*.

)' E F,

>

we show that iIA*y

+

T*xIi.#tj

Ciiyfl.y for any y E F and any x

Fix y e F and x e Dom T*. Since S is a closed operator, we write x in the form x

+x2withxl E KerSandx2

(KerS)'. ThenO=(x2.Tub,1,u E DomT.

Therefore, X2 E Dom T* and T*X2 = 0 and then Hence IIA*y

Now, fix

a

T*xI

+

Dom T* with T*x = T*xi.

+ iISxi

>

E F. We define the following well defined continuous linear

functional CD:

ASF + Range

—* C.

CD(A*y + T*x) :=

412

4 Existence domains of special families of holomorphic functions

with

+ T*x)i

D(A* i.e.

1(1)11

IIyoII.YIIA*y

<

+

Applying the Hahn—Banach theorem and the Riesz represen-

C

with

tation theorem we find xo E

and

then (T*x. xo)J(1 = 0 = (x, for all x Dom T*. In particular, if y = 0 Therefore Dom T** = Dom T and T**xo = Txo = 0, i.e. Ker T. (A*y. xO)J(1 = (y, Axo)J(; therefore Moreover, if y E F, then (y. — Ax0 E F, we conclude that Axo = = 0. Since 0 (V. Axo — In order to apply Lemma 4.3.2 we introduce the following Hilbert spaces.

:=

:= i/i +a(n — l)logg,

where where

By where g(z) := liz — z 02 ii

T

S

—+ .7€3 we denote the densely defined closed operators that are given as usual via the d-operator. Moreover, .

JC2 and

A(hi

is a continuous linear operator: gj(z) := — As a first step towards the proof of Theorem 4.3.1 we consider the following special

configuration.

Lemma 4.3.3. Let D be a bounded pseudoconve.x domain with & boundary, let i/i DC D',andletz° D. Then the claim in Theorem4.3.l is true. Pmof Using the notions introduced before the lemma we shall discuss the following setting with respect to Lemma 4.3.2: where

T

= (T

T).

= (S

S).

The operator A is as above and F := (9(D) fl .7(. Observe that the constant function I belongs to F.

It remains to prove the inequality of Lemma 4.3.2. So fix v = (vi E Dom T* fl Dom S F. Remark that v1 :=

Dom T* fl Dom Sand u

andT*v=(T*vi

T*vn)andSv=(SvI

413

4.3 The Skoda division theorem

Now we start our estimate X := IIA*u +

+ = IIA*u

+

+2 Re(A*u,

+

It is easy to see that

Au = where gj(z) = z3

—

E

and g(z) = liz

z°ll2. Moreover, since

—

z0

D, we have

= E Dom T and Hence we obtain the following estimate: 2Re(Atu,

=

2 Re

f ID

'kJ

Here we used the following simple inequality 2ab

Together with the estimate (4.2.1) of Section 4.2 we get X

(I _a

—

a

0, a. b E

1k.

414

4 Existence domains of special families of holomorphic functions

What remains is the comparison of the last two terms in the above formula. Recall that = + a(n — l)logg and that fl &(D). Therefore it suffices E to estimate a(n —

v1)

= a(n — l)g2

—

>

j=I k.t=I

j=I k
?a(n— l)g2>>IgkvJJ —gJvfkI2 j=I Il

(ss)

Pt

2

>

k=I j=I fl

11

(**)

2

? =

—

k

where (*) is a consequence of the Binet—Cauchy multiplication theorem for the determinant of a 2 x n-matrix multiplied with its transposed conjugate (often also called Lagrange identity) and (**) is simply the Schwarz inequality. Hence we are finished with the following inequality

IIA*u +

+

U E F, v E Dom T fl Dom S.

? (1 —

which allows to apply Lemma 4.3.2. Hence the lemma is proven.

9

Now we are able to present the complete proof of Theorem 4.3.1.

Pmof of Theorem 4.3.1. We exhaust D by a sequence of bounded pseudoconvex domains j E N, with &-boundaries such that D1 D and we approximate the plurisubharmonic function * hi11, by a monotonically decreasing sequence

of

E

EN.

415

4.3 The Skoda division theorem

In virtue of Lemma 4.3.3 we find functions hJki following properties

—

hJkn E (9(D1) having the

= I,

z2)

ZE

ID3

where hfk := (hJki Observe that the assumption on z0 and liz

leads to

—

ID3

ID liz — zOii

using the mean value inequality for psh functions we see that the sequences are locally bounded. Thus the theorem of Montel provides us with holomorphic functions h1 E (9(D3) satisfying

Then,

—

Moreover,

=

1.

Z E D1.

we obtain ilh

(z)112g(z1'""

ID1 where

here

:= (h11

E

(9(D1)".

The final step in the proof consists (as usual) in using again Montel's theorem and

0

a diagonal process. Details are left for the reader.

DefinitIon 43.4. Let D C C" be a domain and z° E dD. We say that D general cone condition at z°, if there exist a sequence C C". positive numbers fi I, r I such that =z0

fulfils

the

D, and

and

Examples of a domain satisfying at each of its boundary point the general cone condition are domains with k > 2 (Exercise). In [Pfl 19751 the following application of Skoda's division theorem is given.

416

4 Existence domains of special families of holomorphic functions

Theorem 4.3.5. I.et D be a bounded pseudoconvex domain in C" satisfying the genera! cone condition at the boundary point z0. Then for any sequence in D —÷ 1—' 0O

there exists an f

= 00.

with SUPJEN

E

Pmof Suppose that there is a sequence (zJ)Ti C D, z' —÷ z0, that does not satisfy the claim in the theorem, i.e. all sequences are bounded. fE Denoting by we obtain —÷ C the evaluation map 81(f) := a pointwise bounded sequence of linear continuous functionals on the Hubert space (D). Then the Banach—Steinhaus theorem implies the existence of a constant C > 0 with :

= Let that

and

<2a(n

r be the data of the general cone condition. We choose a > 1 such — 1) + 2 =: e <2n + fl—1. In virtue of Theorem 4.3.1 (take * = 0)

we find functions

such that

E

—

WI) = I,

thatq(a) = a(a

Put such that

z ED, V E N.

liz —

—

ID Recall

fE

CIifiIL2(D).

—

rllz° — and R > 0 such < I. Then we obtain

C(a,

D)f

that D c

liz

R) for all v. Let v>>

—

C(a, D) f B(O. R)\B(O.r,)

< C(a, < C(a, D, r)llz° — observe that the exponent 2n —

e

<0.

Now exploiting the equation for the hr/s we conclude

I

<

— w)'i <

< nCC(a. D, r)IIz° — wv

—

wtiI —

IihvjOL2(D)

z°iI + liz0

—

1

417

4.3 The Skoda division theorem < nCC(a, D, r)(IIz° —

+ liz0 —

—

z°IJ)

< 0.25 + nCC(a, D, r)iIz° —

v then ji —÷

e)

I

oo

leads to a

contradiction.

U

Remark. Theorem 4.3.5 has been used to prove a necessary condition for the hypoellipticity of the a-problem (cf. [Cat 1981], [Die-Pfl 1981]): the precise result reads as follows: Let D C be a bounded pseudoconvex domain with C2-boundary and assume hypoellipticity for (p. q)-forms (i.e. every equation aa = ft with a a-closed form ft on D, I ,)-form a with singsuppa C q < n, can be solved by an Then does not contain a complex analytic variety of dimension singsupp ft larger or equal q. Sufficient conditions for the hypoellipticity can be found in the work of J. J. Kohn [Koh 1979].

A more direct consequence is the following corollary.

Corollary 4.3.6. Let D C C" be a bounded pseudoconvex domain. Assume that D satisfies the general cone condition at all of its boundary points. Then D is

f with lirnif(z)I =

and it carries an

-convex,

E

Proof The first claim is a direct consequence of Theorem 4.3.5. So it remains to construct f. First, we choose a sequence and UK1 = D and a dense sequence C D with C reorder the points of the sequence as follows

of compact sets

C D. We

The new sequence will be denoted by Let Bk := 8(wk, pD(wk)). Using Theorem 4.3.5 we construct inductively sequences

C D.

C N,

c IR,

C

with the following properties:

fl Kj(,+1),

\

E

= I,

singsuppa := D \ (z E

E

N,

i' E N,

+

+ 20)

V

D:

>

V.

I) E N.

—

alUflb extends to a

on U n b}.

418

4 Existence domains of special families of holomorphic functions

Then

f : D —p

f(z) :=

C,

z

gives a square-integrable holomorphic function on D with

ED, ?

C + v —+

00.

0

Therefore, f satisfies the property claimed in the corollary.

The following example (cf. [Jar-Pfi-Zwo 19991) shows that Theorem 4.3.5 becomes

false if D is only assumed to be a fat domain.

Example43.7. Forj E Nleta1 :=

j E N, satisfying the following properties: •K(aJ,sJ)flK(ak,sk)= ø,k • forD :=

E

\

<

EN;

K(a1, s1) U {O}) there is an M > 0 such that Ill

Mon

for all f E with IfIIL2(D) = 1. Thus, D is a fat bounded pseudoconvex domain, it has no cone condition at 0, an it does not fulfil the claim in Theorem 4.3.5. Construction of the sf's: Step 1: We claim that there are M > 0 and (rj)jEN C IR>o such that for DN := E \ K (aj. r1) the Bergman kernel KDN of Dpi can be estimated by

KDN(Z)<M, Recall that the Bergman

of a domain G is given as

KG(Z) := sup{Jf(z)12 : f E and that KG is continuous on G. In

Ill

=

11.

-

fact, since D,(z) := E \ K(a1, 1)

/ E \ {ai}, we conclude applying

Ramadanov's theorem (cf. [Jar-Pfi 19931, Th. 6.1.15) that KD1(z)

Thus, we can chooser1 E (0,

\ KE\(a)} = KEIE\(ai). and M > 0 such that

K(ai,r1), j >2, SetD1 21)

and

KE\k(aIrI)(Z) < M, Z E K.

:=E\K(a1,r1). For more details on the Bergman kernel contact [Jar-NI 1993].

419

4.3 The Skoda division theorem

Now assume that we have already constructed positive numbers ri

r,v such

that

1<j,kkI

ae E Dk :=

N;

KDk(Z)<M. ZEK, Since DN(t) 1= DN \ K(aN+1, t) / DN \ {aN÷1}, we find, as before, a positive :—.0

number rN+I such that

K(aN+I, rN+I) n (U k(aj, ri)) =

£>N+I, K,

KDN+,(z) < M, z where DN+J

:= E \

K(a1, r1).

Hence, Step I is verified. Step 2: There are positive numbers Sj

(0, rj), j

N, such that for

0= E\ the

following is true: KD(Z) <M. z Fix an increasing sequence of compact sets LN, N N, of with To prove the statement of Step 2 it suffices to find a family LN =

of positive numbers (4J)NEN,N<JEN such that

SNN
GN:=E\(

£(aj,4)UUK(uj,SJ)U{0}). j=N-+-I

j=1

a moment that the sequence (4) is already constructed. Puts1 := and set D := E \ U (0}). Then GN C 0. Therefore, KD on GN; in particular, K0(z) < M, z E LN. Since N is arbitrary, K0 <M on K. Suppose for

420

4 Existence domains of special families of holomorphic functions

What remains is the construction of (4). This will be done using induction with respect to N. Observe that if '>0.

:= E

\

U

R(a1,

U {0}).

/

then

t-+O

Therefore,

KD.(t) \

KJy

=

K01. If t1 <

I

is sufficiently small, then

0

< M, Put

z E L1.

j > 2.

r1 and

Assume that the construction has been successfully done for the Nth step, i.e. is already defined with the claimed properties. Repeating the argument (4 from above we put for t > 0

E\( U j=N+2

j=I

:=E\( (J

/

Then

j=N+I

j=1

Therefore K0.(t) \

with the Bergman kernel of E \ particular, K0. KDN+I. Hence we find

:=

and

completes Step 2.

Remark.

:=

j>

and K0.

coincides

j DN+l.

U

In

< I sufficiently small such that if

N + 2, then KGN+I < M on LN+I, which

It seems to be unknown whether there is a general weaker condition

than the cone condition for bounded pseudoconvex domains from that a statement like the one in Theorem 4.3.5 follows According to Corollary 4.2.8, any bounded fat pseudoconvex domain D C C" is an

t50)-domain of holomorphy, where

:= min{p0(z), (1 +

z E D.

4.3 The Skoda division theorem

421

Moreover, we know that any Riemann—Stein domain (X, p) over C" is an 6x)-domain of holomorphy, if k > 6n. Skoda's division theorem can be used to improve these kind of results for univalent Riemann—Stein domains, i.e. for pseudoconvex domains in C". Let D c C" be a pseudoconvex domain. Put

:D

SD

—p

R,

SD(Z)

:= min{l.

+ 11z1121'12. z E D.

that 8j) < and that, if D is bounded, then '5D < c8r) (for certain < of c> 0). Thus, in virtue of Proposition 2.5.5, any such D is an holomorphy, k > 6n. In fact, more is true. Observe

ProposItion 43.8. Let D c C" be a pseudoconvex domain and let z0 E 8 D. Then. for any C E (0, 1) there exist functions E (9(n+e)(D, 'SD) z

—

z

Proof We apply Theorem 4.3.1 in the

:= max{—2elogpD,0} Thus we find functions h1

E

D} < C(n, e)(l

+ Iz°IIY'.

following situation:

and

2a(n

— 1)

+ 2—

2e

<2,z.

(9(D) with the following properties:

ZED. and —

ID

min{l,

ID Iz — z°II


ID liz — z°tl

E SD) is an easy consequence of the mean value theorem for psh functions. In fact, fix an a E D. Then 2pb(z) ? pD(a) and 211z — z°ll ta — if z E La, p(a)/2). Therefore, settings := we get

That

2min{l, 1h1

(a) I Pa —

ITS'

JP(a,s)

—

min{ I, < Pn (a)

e)2
422

4 Existence domains of special families of holomorphic functions

Thus Ihj(z)ISD(z)fl+E

c(n.

+

(1

11z112)

Corollary 43.9. Any pseudoconvex domain D C is there exists anf E Ol:D.SD) with limlf(z)I = 00,

If D is bounded,

+ IIz°11",

z

E D.

0

So)-convex, and E dD, e E (0,1).

the same is true after substituting 8D by

Proof The proof is similar to the one of Corollary 4.3.6, therefore it is left for the

0

reader.

Remark. Observe that the corollary remains true for

instead of We don't know whether the claims of the corollary are true for any pseudoconvex domain D C C" if we substitute by 5D. Moreover, it would be interesting to know the infimum of k's such that any D has the properties of Corollary 4.3.9 for (9(k)(D or (9(k)(D SD), respectively

4.4

The Catlin-Hakim-Sibony theorem

According to the example of N. Sibony we know that there are bounded fat pseudoconvex domains that are not 3e°°-domains of holomorphy. On the other hand, if a bounded pseudoconvex domain D c C" can be written as

D=

mt fl{G

c C" G domain of holomorphy with D

C G},

(*)

then D is an A°°-domain of holomorphy. Observe, that D is pseudoconvex and fat, i.e. D = mt D. Sometimes the set on the right side in (*) is called the Nebenhülle of D. In fact, otherwise there would exist open sets U1, U2 with U2 connected and 0 U1 c D fl U2 ç U2, such that 3FEO(u2):

Flu1 = flu1.

U2 contains a point b D and therefore, in virtue of (*) a point a that is not in the intersection. So we can choose a pseudoconvex domain G, D C G, with a G. G is pseudoconvex, thus there is Jo E (9(G) such that flu1 is never restriction of a holomorphic function on U2. Observe that fob E which contradicts our assumption.

Definition 4.4.1. Let D C C" be a bounded pseudoconvex domain.

423

4.4 The Catlin-Hakim-Sibony theorem

(a) We say that D allows a weak Stein neighborhood basis if the property (*) is fulfilled. (b) D is said to have a Stein neighborhood basis, if

D=fl(GcCtt:Gpseudoconvex. DCG}.

D=intD.

It is clear that condition (b) always implies condition (a).

RecallthatthesocalledHartogstriangle D := (z E C2 1z21 < Izil < l} isa bounded pseudoconvex domain without a weak Stein neighborhood basis; we have :

15

ç fl{G cC2 : G domain of holomorphy, DC G} =

E2.

While the Hartogs triangle has no regular boundary, there are also pseudoconvex domains with the famous worm domains, that do not satisfy (*) (cf. [Die-For 1997J). In order to describe the worm domains choose a A IR —+ such :

that: (i)

A(z)=0,

1<0,

(ii)

A(t) > 1,

t > 1,

(iii)

A"(t) > lOOA'(t),

(iv)

A'(t) > 0,

£

(v)

A'(t) > 100,

if A(t) >

Forr> I weputp,. =

p

:C x

p(z) := Izi +

—÷

t

> 0,

IR,

—I+

— 1) + A(1z212 — r2),

and then we define

I+r2}andthat

ObservethatDc{ZEC2:IziI<2, Dfl {z E C2:

I

1Z21

r} =

(z E

C2: Izi +exp(i log 1z212)I < I}.

ProposItion 4.4.2. For fixed r> 1, the domain Dr = D is a bounded pseudoconvex domain with C°°-boundary, a D fails to be strongly pseudoconvex precisely on the set r}. {z E C2 : Zi = 0, 1z21 1

424

4 Existence domains of special families of holomorphic functions

Proof It is easy to see that D is a domain. Calculations give the following derivatives of p: aP

-

—(z) =

2 zi + exp(—i log 1z21 ),

az1

I

exp(ilogIz2I 2 exp(—i log 1Z212) — (Z

—

—

1)

+ z2A'(1z212 — r2),

and

a2p

(z)=l.

-

a2

-

"

aziaz2 az2az2

(z)

log 1z212),

(z) =

exp(i log 1Z212) + Zi exp(—i log lZ2l2)) + A'(1z212 — r2)

—1Z21

+

6A"(1z21

1Z21

2

— I) + lZ214A'(1Z212 — I) + 1Z2I2A'(1Z2I2 — r2).

To prove that D has a C°° -boundary we assume for a z0 E d D that both derivatives vanish. Thus log 1412)

=0.

Since p(z°) = Owe have I = A(14L2 — I) + A(1412 — r2). Therefore, 181 < 1 or 41 > r. On the other hand, o

=

=

—1

+

—

r2).

If 41 < I or 141 > r, then exactly one term in this sum is zero; contradiction. To prove that D is pseudoconvex it suffices (cf. Proposition 2.2.23) to verify the positive semi-definiteness of the Levi form of p along the complex tangent plane, i.e.

for X(z) :=

£p(z; X(z))

z E aD.

Calculations lead to the following formula:

£p(z; X(z))

=

1z21

l)+A(1z212—r2)+1z112

2Izi

+1z214A(1Z212

I )+Iz2I2A'(1z212— I )+1z214A"(1z212—r2)+1z212A'(1z212—r2))

+ 1Z21 6(A(1Z21 —

2Re

2 —

I))2 + 1z212(A'(1z212 — r2))2

exp(i log 1Z212))(1Z21 4A'(1z21

2

—

1) —

—

r2)).

425

4.4 The Catlin—Hakim—Sibony theorem

Here we used that z E and that A'(1z212 — l)A'(1z212 There are three cases for z to discuss:

—

r2)

=

0.

Case I): LetI < 1Z21 r. Then £p(z: X(z)) = log and 0 = p(z) = (zi + exp(i log — I. Thus £p(z; X(z)) = 1Z11/IZ2h. So the Levi form here is positive definite except when zI = 0.

Case 2): Let

S

1z21 < I. Then an inspection of the formula for the Levi

form shows that it contains only one term that might be negative, namely the last one. Therefore, if Izi + exp(i log 1z212)12 ? 1/2 we obtain using properties (i) and (iii) of A that

£p(z; X(z))

2

— I)

16A'(1z2r2

— 1)

Izi + exp(i log 1z212)12A"(1z2i

?

—

I)

—

2lzi 11Z21 4A(1z21

34A'(1z2[2

—

2

I) >

— I) 0.

+ exp(i log 1z212)12 < 1/2, then 1/2 < A(1z212 — I) (observe that z EdD). Therefore, property (v) of A leads to A'(1z212 — 1) > 100 and then we get

If

Lp(z; X(z)) > Case

3): If r 2 <

6(A'(1z21

2 —

l))2

—

1) >

l6A'(1z212

100 . 84

> 0.

2 I + r and Izi +exp(z log 1z21 22 )j > 1/2, then .

1Z21

X(z))>

—

r2)

—

4A'(1z212

—

r2) > 0.

In the remaining case we have as above that A(1z212 — r2) > 1/2, and therefore in

virtue of (v) we get

£p(z;

X(z)) > r2(A'(1z2(2 — r2))2 — 4A'(1z212 — r2) > 100. 96 > 0.

0

Hence Proposition 4.4.2 is proven.

Theorem 4.4.3.

If r >

then

= D (as above) has no weak Stein neighborhood

basis.

Pmof Itisclearthat 1z21=lOrër}CD.

{zEC2:zi=0, Applying the has to contain

it follows that any pseudoconvex domain G, L) C G, {z E C2 :

Izi + II

1,

I

Hence D has no weak Stein neighborhood basis.

1z21 <

0

We should mention that there are bounded pseudoconvex domains without a Stein neighborhood basis, but with a weak Stein neighborhood basis (cf. [Ste 1987]).

426

4 Existence domains of special families of holomorphic functions

Remark In fact, Theorem 4.4.3 remains true as long as r

exp(jr/2) while, if

1 cr <exp(ir/2), then D has a Stein neighborhood basis. Furthermore, Theorem 4.4.3 shows that Dr. r > e'T, has no global defining psh function. Even more is true, namely has no global defining e2-thnction with a positive semi-definite Levi form on More information with respect to this question may be found in [Beh 1984], [For 1979]. We also recall that any bounded pseudoconvex domain with C'°-boundary allows a Stein neighborhood basis (cf. [Die-For 1977]).

The discussion so far shows that it is not clear whether bounded pseudoconvex domains with C°°-boundaries are of holomorphy. That the answer to that question is positive was independently established by D. Catlin [Cat 1980] and M. Hakim & N. Sibony [Hak-Sib 1980]. Here we will follow the argument of M. Hakim & N. Sibony.

Theorem 4.4.4. Any bounded pseudoconvex domain D C an -domain of holomorphy. -

The

with C°°-boundary is

proof of Theorem 4.4.4 is based on the following regularity result of the

a-problem due to J. J. Kohn (cf. [Koh 1973], [Koh 1977]).

Theorem 4.4.5. Let G c C" be a bounded pseudoconve,x domain with Then any a-closed (0. s)-form a E C(0S)(G). 0 cs
The proof of this result is beyond of the scope of our book, therefore the reader is asked to contact Kohn's papers. We only mention that in [Koh 1973] a weaker form of Theorem 4.4.5 was given; the proof was completed in [Koh 1977] along an idea due to L. Hormander. A sketch of the proof is also given in the nice survey article [Koh 1982]. Now, we start with the preparation to the proof of Theorem 4.4.4. be as in Theorem 4.4.4 and let F1 .4°°(D) FN have no common zeros on D. Then there exist functions UN E .40c(D) such that u1F1 = Ion D.

Lemma 4.4.6. Let D C

Proof The proof is similar to that of the division theorem in Section 2.10. except that here we use Kohn's solution of the 8-problem. Therefore we only sketch the main steps using the analogous notation like in Section 2.10. Define

:=

0 <

s

:= {u = (UI)IEM(N.,) : Uj E A(OS)(D),

aEE,, IEM(N,t)),

= sgn(a)u,,

427

4.4 The Catlin—Hakim—Sibony theorem

where

=

M(N, r) :=

(i1

.

i1)

E N': 1 < it. i, < N),

Observe that a gives linear operators 8 : recall the following linear operators

P:

Moreover, we

—+

(Pu)j :=

(D)

1 < : < N.

Fju(J,J),

I e M(N, t),

and

Q: A(05)(D) —+ 1+1

(Qu)1 :=

I E M'(N, t + 1),

We know that

pd=8p, PoP=O, PoQ+QoP=id.

(*)

Put then = 1. It remains to modify := (Fj/11F112)jEM(N1) E that it becomes an N-tuple of functions in A°°(D). Let

so

U1

Observe that Pu =

0.

.— uU0I E Al

Then we define inductively:

:= QuJT

and

uj := auJ_1 E

2
Observe that because of(*) we have = and = 0. In the case that N n, the fact that PuZ = 0 implies that = 0; recall, that the F1's have no common zero. Therefore the form in uZi are 8-closed. If N > n. then is 8-closed. We discuss the case N n in detail, the other case is left as an simple exercise to the reader. Using the Kohn result there is E i N_2)(D) with = uZ_i. Put

-N-I

.

UN_2 . 22)

N-I

N

UN_2 —' VN_2

ANI

For the correct interpretation of P and Q for i = 0 sec Section 2.10.

428

4 Existence domains of special families of holomorphic functions

Then

N-2

—

UN_2 —

— —

UUN2

Therefore we can repeat the argument solving by

uZI

—

PvZI,

=

and

etc. In the final step we reach

modifying uZl

with

0

which gives the desired N functions in Pmof of Theorem 4.4.4. Otherwise there are as usual open sets U1, U2 with

0

U1 c D fl

ç

U2

U2,

is surjective. U2 connected, such that the restriction map 0(U2) —p 1.°°(D), Since D has a C°°-boundary, there is a point z0 E U2 \ D. Defining Fj (z) := Zj

—

4. Theorem 4.4.6 gives functions

UI

= 1,

E .4O0(D) with

z

€D.

According to our assumption, the u3 lu 's extend to Üj E 0(U2). Therefore,

—4) = I

0

on U2; contradiction.

The division theorem can be also used to get the following result on continuous non-zero algebra homomorphism from .4°°(D) to C.

Theorem 4,4.7. Any continuous character x of the Fréchet algebra as an evaluation map f —+ f(a) with a E D.

Pmof Let

be a such continuous character. We claim that it is continuous with

respect to the supremum norm. Otherwise, there is a function f — x(f) has an inverse g E

I

(D) is given

x(f)) = l;contradiction.

E

such that Ix(f)l

> If 1Ir. But then

in particular, this leads to (I =

x(g)x(f —

-

We introduce the Banach algebra A := as its norm. Then, with II. in virtue of the above claim, x extends to a character j on the Banach algebra A. We claim now that is a point evaluation. Otherwise, for any E D we find a function E A with

429

4.4 The Catlin—Hakim—Sibony theorem

Putting

:= fi —

E A we have but

with an appropriate constant we may even assume that in the next step we take new functions with IIhç — E

Multiplying

Putting

> 1.

lb <

E

—

it follows that

x(hc)=O Since D is

h1'... ,hN E

and

compact we can choose a finite number of the functions hc, say having no common zero on D. Then, in virtue of Theo-

rem 4.4.6 we have the following representation of 1:

I=

gj E

Then applying x to this equation leads to a contradiction.

U

Remark. Let D be as above. Then even the following is true (cf. LHak-Sib 1980]): • any continuous character on .4(D) is given by an evaluation map. of holomorphy which are not In Recall that there are the case of bounded pseudoconvex domains with a e°°-boundary we will have always In fact, we shall even prove a stronger result. Let D be as above and let K C D be compact. We introduce the following notion.

Definition 4.4.8. Let D and K be as before. (a)

:=

(b)

:=

where

(z E

D: If(z)l

(z E D

Ill IlK. f

E

u E ?SRk(D)}

: u(z) < 00.

Remark. The S'SJ?0(D)-convex hulls of the compact sets in b are responsible for the hypoellipticity of the a-problem (cf. LSib 1983]). More precisely: Let D be a pseudoconvex domain with C°°-boundary. Then D is hypoelliptic for the (0. 1)-forms if and only if dD = aD fl K, K c D compact.

The aim of the remaining part of this section is to provide a proof of the following result (cf. [Hak-Sib 1980] and (Cat 1980]).

430

4 Existence domains of special families of holomorphic functions

Theorem 4.49. Let D C C" be a bounded pseudoconvex domain with C°°-boundar and let K C D be compact. Then

= fl In

(*)

=

n aD.

D is .4°° -convex.

Before we are able to present the proof of Theorem 4.4.9 we have to establish some more results.

Recall that the worm domain is an example of a bounded pseudoconvex domain with C°°-boundary that does not allow a global psh defining function. Nevertheless, any such domain allows a nice psh exhaustion function; in particular, it is hyperconvex (cf. [Die-For 1977b] and [Ran 1981]).

Theorem 4.4.10. Let D C C" be a bounded pseudoconvex domain with C3-boundary and let r : U = U(D) —÷ R be a global defining function for D. Pu: *(z) := 11z112.

—

Then there are £ > 0 and

E (0, I) such that the function

: D —÷ JR.

is continuous on D and stmngly plurisubharmonic

(z) := — (—r(z) on D, 0 <

U andp: V —÷ aDsuchthatforallz E V we have Pmof Choose V = V(dD) IIp(z) — zil = PD(z). For z E V we write every vector X E C" as X = r(X; z) + v(X; z), where v(X; z) :=

2

(re, (z)

(z)).

r(X;

z) := X — v(X; z).

Z)j = 0, i.e. r(X; z) is a holomorphic tangent vector at the level set (r = r(z)}. Since D is assumed to be pseudoconvex, it follows that for 0. Therefore, z E dD and X E C" the following inequality holds: £r(z: r(X: z) Observe that

for z E V,

£r(z: r(X; z)) — Lr(p(z); r(X; p(z))) = O(Ir(Z)I)11X112. = V'(aD) V and C1 > 0 such that £r(z; r(X; z)) > —CiIr(z)111X112, z E V1, X E C".

Thus we find V' Then

£r(z; X) "

= £r(z; r(X: z)) + £r(z; v(X; z)) + 2Re > _C2(Ir(z)111X112 + IIXII where C2 > C1.

.

a2

-

j,k=I 8ZJ8Zk

r(z)r(X;

z)

431

4.4 The Catlin—Hakim—Sibony theorem

In the next step we calculate the Levi form of

on D

X) =

X),

where

D(z; X)

er2(z)(11x112

+(l

—r(z)(Lr(z;X)

have to show that D(z; X) > 0, z E D fl V1 ift>> 1 and << 1. < 1/2, z E D, and > 1. Then we get for z E D 11 V1 and arbitrary X We

Fix 0 <

D(z; X) > —

£r2(z)11X112(1 — 1/2)

r(z)(r(z)C20X112

Using the formula 2ab

—

IIXII(C2

+

+

a1a2 + ab2 we get 2

Ir(z)I Xfl (C2 + 2iitllz II) where C2 + 2iitIIzII

(z)X1 <

+

flX 112,

C3 on D. Therefore we finally obtain

D(z; X) > r2(z)11X112(f/2 + C2 —

> 0,

z E D fl

X E C" \ (Oh

whenever e + > and 0 < ij Choose a compact subset K of D such that D \ Vi C K. Then, on K we have the

following estimate for D(z; X) D(z; X)

— C4)11X112,

? e > 0, z E K, and C4 sufficiently large to give an upper estimate for the remaining summands on K. Hence, if £ is sufficiently large, then is strongly psh on 1) for sufficiently small D where

Remark. In (Ker-Ros 1981] it is shown that even a relatively compact pseudoconvex subdomain with e' -boundary in a Riemann—Stein domain (X. p) over is hyperconvex.

Besides the existence of good psh exhaustion functions as stated in the previous theorem the other ingredient for the proof of Theorem 4.4.9(**) is the Jensen measure (cf. [Gam 1978]).

432

4 Existence domains of special families of holomorphic functions

C" be a domain and let K C D be compact. Then, there exists a probability measure p. suppp C K, such that

Proposition 4.4.11. Let D

u(z0)

uE

p is called Jensen measure for the point z°.

Proof We will construct a positive linear functional L : C(K, IR) —p R with the following properties: IILII = I, L(g)

0. if g E C(K, R), g

u

—

u(z0) fora suitable u E

Assume that L has been constructed. Then, in virtue of the Riesz representation theorem, there is a regular Borel measure p on K, p > 0, p(K) = such that 1

L(g) = I gdp,

g E C(K, R).

JK

Now, fix a u E

Then L(u — u(z0)) = fK(u — u(z°))dp > 0. Hence, udp. :5 JK What remains is to find the functional L: Set B := C(K. IR). Then B with the sup-norm is a real Banach space. We define the following two subsets 14(z0)

C1 := {g E B : 2UE.9'SJ(°(D)

C2 := {g

ii(z°)

:

B

? 0. g

? u on K},

: g <0}.

The C3 's are non-empty and convex cones of B. Moreover, C2 is open. Let g E C1.

Then g > u on K for a suitable u E u(z°) ? 0. Choose z1 E K with u(z1) >0. Hence. g(z') 2 u(z') >0, i.e. u C2. Applying the Hahn—Banach separation theorem we find a continuous linear functional L : B —+ R, DLII = 1, and a ER such that

Lie2
b:=inf{L(g) lfa <0.then2a L(g/2)= L(g)/2; contradiction. A similar argument leads to b <0. Hence, a = a = b = 0. a :=sup{L(g) :g EC2),

Now, ifg E B. g 0. then -(g + r) E C2, and therefore —L(g) — L(e) 0. When 0, continuity of L gives L(g) 2 0, i.e. L is positive. Hence the existence of

r —÷

the functional L, we have used at the beginning, is shown.

0

Thus we have the following result from which Theorem 4.4.9(**) immediately follows.

433

4.4 The Catlin-Hakim-Sibony theorem

be a bounded pseudoconvex domain with C3Proposition 4.4.12. Let D c Then for any compact set K C D the following equality holds

fl 8D.

fl

1i 8D c fl is obvious. Thereand choose a Jensen measure for z0 with

Proof The fact that (KflaD)

fore, we fix a z° E C K. Take

fl

from Theorem 4.4.10 with uIaD = 0 and u <0 on D. Then

uE

u(z0) = 0 Thus !2({z

K : u(z) <0)) =

f

0.

C aD.

0. Hence,

Let now v E 9S3(0(D) be arbitraiy. Then v(z°)

f

=

f

<

KnaD

sup leKfldD

fl 8D.

in particular. z° E

What remains is the proof of (*) in Theorem 4.4.9. To present that proof we need several preparations. The first result due to Bremermann [Bre 1954] describes locally the continuous

psh functions in terms of holomorphic functions. The proof given here is due to N. Sibony [Sib 1975].

ProposItion 4.4.13. Let D

C" be a domain and let K C D be compact. Then, if

e>OanduEfPSR(D)flC(D),therearef1

NE

(0, co) such that

u(z) —e

:

j=1

N)
z E K.

Proof Fix u and as in the proposition. Put

Di :=

x

((z. w)

wi <exp(—u(z)), z

D}.

Then Dj isa pseudoconvex domain in Let z° K. We define a function f = 120 by means of the following power series.

f(w) :=

434

4 Existence domains of special families of holomorphic functions

The radius of convergence of f is at least exp(—u(z°)). Thus f can be thought as a holomorphic function on {(z, w) E D1 : z = z°). Hence there is a holomorphic

extension f Dj —+

C

off.

Then, for any z E D we develop -

into a power series

00

f(z, w) =

0)

with a1(z) := J

i=l Since f is holomorphic on D1, we know that

lim! log 1a1(z)I

u(z),

z E D.

Applying the Hartogs Lemma (cf. Proposition 2.1.8) we find Jo such that


z E K, J

Jo.

Observe, here the continuity of u has been used. Taking into account that log a3 (z°)l = ju(z°), j E N, there is an open neighborhood U (z°) c D on which log aj0 I j0u (z) — e/2 holds. Hence, a standard covering argument leads to functions gN E 0(D) and positive numbers Cl, .. . , cjj with u(z) — e/2 < max{c3 log 1g3(z)I

:j =1

u(z) + e/2,

N}

z E K.

0

Then final claim follows if we set f1 := g1

Lemma 4.4.14. Any bounded domain D C C" with C2-boundary has the following property: D U is strictly starlike w.r.t. a point a E D n U. 23)

Proof The reader is asked to give the proof of this geometric fact by himself as an exercise or to contact [Kra-Par 1999]. 0

Corollary 4.4.15. Let D be as in Lemma 4.4.14, z0 E OD, e > 0, and let u E ?SR0(D). Then there are a neighborhood U = U(z0), functions Ii IN E

0 (C"), and positive u(z)—e

numbers

, Cj.q

such that

z E DflU.

N)
max{c1logIf1(z)I :j = 1

Proof Let z0 E dD, 1 > s > 0, and u be given as in the corollary. In virtue of Lemma 4.4.14 we choose a neighborhood U = U(z0) and a point z' E D flU such that D fl U is pseudoconvex and starlike with respect to z1. Observe that D1 23)

z> I.

{z' + t(z

—

z') : z E

That is. 0 fl U is starlike w.r.t. a and a +

U fl D}. —

a)

I

>

1,

0 fl U for any

E

fl D) and

435

4.4 The Catlin—Hakim—Sibony theorem

is also pseudoconvex and starlike with respect to z1. Moreover, we define the continuous psh functions

u,:D,—÷R,

ZED,,t>l.

that u is continuous and psh on D, Now, we choose > I such that

Recall

lu,, On

in

particular, u

s + I

for a suitable real s.

<e/2.

—

the other side, in virtue of Proposition 4.4.13 we find functions 1' and positive numbers ci N with the following property

f,v E

c9(D,,)

U,,(Z) —e/4 Observe that D,,

log is

llPj

N}

I

u(z) — e

=

z E DflU.

starlike. Thus, using Corollary 3.1.23 there exist polynomials N,with

=: rj.

—

Summarizing, we have the following inequality on D s

u,,(z),

< u,,(z) — e/2

log

:j =

max(c1 log

N}

I

log 1p10(z)I +

cj0

log 1p3(z)I

fl U:

:j=I

N}

+e/4

log

+ e/4 cj1 log fit (z)l + c11r11 /lf.i1 (z)l + e/4 (z)l + u,,(z) + cj1rj1 (z)l + e/4 u(z) + e/2 + e/2

which ends the proof of the corollary.

D

While Corollary 4.4.15 says that any u E can be locally uniformly approximated by finite maxima of functions of the type clog If!. c > 0, we need a global version of this fact in order to prove Theorem 4.4.9(*) (cf. [Gam-Sib

f

I 980J).

Proposition 4.4.16. Let D C boundary and let u E Postponing the proof of

be a bounded pseudoconvex domain with C°°Then u is the limit offunctions of the type

Proposition 4.4.16

we first complete

the proof of Theo-

rem 4.4.9.

Pmof of Theorem 4.4.9(*)• It is obvious E

and u E

PSR°(D).

that

Assume

Therefore, fix

c

that u(z0) >

Then, in virtue

CEK

436

4 Existence domains of special families of holomorphic functions

of Proposition 4.4.16, we find functions ft. f,v E N

and

positive numbers

such that

max{c1logIf1(z0)I

:j =

N}=ckloglfk(z°)i >

I

CEK

for a suitable k. Thus lfk(z°)I >

0

contradiction. CEK

The proof of Proposition 4.4.16 either needs deep results from the a-theory (cf. [Cat 1980]) or a detour to function algebra (cf. [Hak-Sib 1980]). Here we shall follow the second approach.

PmofofPmposition 4.4.16. Denote by A the closure of .4°°(D) in e(D). Then A is a Banach algebra, and we know that its spectrum is given by D (cf. the proof of Theorem 4.4.7). Fix a u E

-

Without loss of generality, we may assume that u

ü(z):=sup{clog(f(z)I:c>0. fEA, Fact 1. 1: suffices to

0. Put

zED.

that u = ü on D.

Pmof of Fact I. Fixe > Oanda ED. Assumingu = ü > Oandfa E A with log lfaI u and u(a) — e/2
that fa(a)

0. Using the definition of A we choose fa E

with

lila — fallb < min{caifa(a)I, elfa(a)l, EI}(2Ca)' =: Then we get

ca log Ifa(a)i

Ca

log(Ifa(a)l

—

Moreover, if z E D with lfo(z)l >

Ca log lfa(Z)I $ c0 log(Ifa(z)I +

Ca

log

lfa(a)I

—

u(a) —

then Ca

log Ifa(Z)I +

zi(z) + €.

0

Then, a compactness argument concludes the proof.

In order to show that u =

ü

we need the following form of a maximum principle.

The generalized maximum principle. Let a E D. let U = U (a) be a neighborhood ofa. Assumeforafunction h E C(U fl D) that h
: c >0, f E A,clogigl

h}, Z E U fl D.

437

4.4 The Catlin—Hakim—Sibony theorem

Fact 2. The generalized maximum principle implies that u = ü. ü. Then

Proof of Fact 2. We will proceed via contradiction. Thus, assume that u

a := sup{u(z) — ü(z) z e D} > 0. Put

&:=(zE D:u(z)—ü(z)=a). 0 and 6 is compact. Since ü is lower semicontinuous, & We introduce the following Banach algebra on 6 Ale

:= (lie : f E

We choose a point a S that belongs to the Shilov boundary of Ale 24)• Recall that according to our assumption there is a neighborhood U = U(a) such that

u(z)=sup{clogjf(z)i:c> 0, fE A, clogifi

_
zE UnD.

Using the fact that the Shilov boundary of AI€ is the closure of the set of strongly boundary points for Ale (cf. [Sto 1971], Corollary 7.24) we choose g e A with

=2.

ilgIie\u <

<0 near dU fl 6 fl D. Therefore, there is c>>

Observe that log

I

such that

z€aUflD, —ca + log i1gIl13

0.

From here it follows that c(u

—a)+loglgl

c(u — a) + log Igi

cuon D, fl D.

cü on

Put h := c(u — a) + log IgI. Then the generalized maximum principle applies to h. Thus, we have c(u — a) + log Igi cü on Un D. In particular, we get log ig(z)i 0, 0 Z E U fl 6, which contradicts the choice of g. What remains is the proof of the generalized maximum principle. This needs a small detour in considering Banach algebras generated by Hartogs series. Let i, be semicontinuous from below on D. Put Y

Y is a compact set in B

:=

:=

The functions in

:=

:= {(z,

E

Dx

C

:

< exp(—v(z))}.

On Y we consider the following Banach algebra

where

?(A) :=

: N E No,

E

A).

are called Hariogs polynomials.

24) Let (X, A) be a uniform algebra. i.e. X is compact and A is a subalgebra of C(X) with the y,there is followingproperties: C C A, A isclosedw.r.t. lIx,and foranypair(x, y) E X x X, x E A, 0 f(x) f(y). Then there is a unique minimal closed set F' C X such that liflix = Ill hr. E A. r is called the Shilov boundary of(X. A) (cf. (Sto 1971], Th. 7.4).

f f

438

4 Existence domains of special families of holomorphic functions

Fact 3. Spec B = {(z, C) E b x C: ICI

exp(—ii(z))} =:

:= sup{cloglf(z)I : c> 0, f Pmof of Fact 3. Let x

E

E

A,

where

S v},

z ED.

Spec B. Then x is determined by its action on ?(A).

Therefore,

x IA is given by evaluation at the point a E Dand where Co := x (C). Therefore, Spec B can be thought as a compact subset S( B) of D xC via identifying the characters with the points at which the characters evaluate. Let (z°, Co) E S(B). Then IF(z°, Co)I IFOy, FE ?(A). PutCi := Coexp(ico). LetF= Then

IF(z°,Ci)I =

where

= IIFIIy,

= IF(z°,Co)I

and FE

F(z,C) :=

Hence. S(B)is invariant

under rotation in the C-variable. Moreover, using the classical maximum principle for

F(z,) E .'P(C), we obtain

S(B) = ((z, C) E D x C: Ri < R(z)},

where R : D —÷ R÷. Since Spec B is compact, the function R is upper semicontinuous.

What remains to see is that R = exp(—i). Fix (z°,Co) ED x Cwith Col > exp(—i'(z°)). Then there arec> Oandf E A such that -

—logICoI I. Hence the evaluation at (z°, Co) is not continuous, i.e. it cannot be identified with a character on B. Now, we are going to prove the converse inclusion. Fix (z°, Co) E Y. Since o x {0} C Spec B, we may suppose that Co 0. In particular, i3(z0) <00. Since Spec B is compact, we may assume that ICol <exp(—i3(z°)). Let F E Without

with Fl
This equality is. as before, understood in the way that any character on B can be identified with an evaluation map .?(A) F —* F(z. Y. (z.

439

4.4 The Catlin—Hakim—Sibony theorem

disc K(exp(—v(z'))). If F = f1 E A,theCauchyinequalitiesgivethat Hence, (1/j)loglf3l vonD. Therefore, (1/j)logIf1I < lion 1). Plugging this information into F we obtain 17(1 —

F —÷ F(z°.

The last estimate is independent of F, i.e. the evaluation

extends continuously to B. Hence (z°,

E Spec

B. Fact 3 is proven.

0

The last step in the proof of Proposition 4.4.16 consists in proving the generalized maximum principle. In this proof we need a part of a result due to Edwards (cf. [Gam

l978],Th. 1.2).

Fact 4. Let M be a compact space, Z0 E M, and let .'R be a family of continuous functions fmm M to 1—00. 00). that contains the constant functions, that separates the points of M, and that is closed under taking sums and multiplying with 1/k, k E N. Then, for each lower semi-continuous function q : D —÷ (—00. 00] the following inequality is true:

sup(w(z0):wER,w
Proof of Fact 4. For abbreviation, let L. respectively R, denote the left, respectively the right, side of the inequality in Fact 4. First, we consider the case that q is continuous. Assume that L < b < R for a certain real b. We define similarly as in the proof of Proposition 4.4.11 the convex cone

Ci :={g E C(M.R):

h(z0)=Oandh

g}.

C1. Applying the separation theorem for convex cones leads to a non-zero measure r on M such that Then q — b

r is non-negative on C, and

I

(q

—

h)dr

0:

JM

Defining a := r/r(M) gives a Jensen measure for z0 w.r.t. b < L; contradiction. Hence, fM qda Now, we discuss the general case, i.e. q is lower semi-continuous. Choose an increasing sequence of continuous functions q1 : M —p R with q1 = q in particular, h(z0)

fM hdr for h E .R.

440

4 Existence domains of special families of holomorphic functions

on M. Then we know that = R1, j e N, where and R denote the left and right side from the inequality for q1, respectively. Because of the weak-star compactness we can find Jensen measures with J'M Moreover, take the Jensen = measure a as a weak-star accumulation point of (cj)JEN. Fix a j and let j k. Then < L. For = Rk = k —÷ oo we obtain fM q3da < L. Hence, if j —+ oo, then fM qda < L; in particular. the claimed inequality has been proven. El Fact 5. The generalized maximum principle is true. Pmof of Fact 5. Define the following lower semi-continuous function on U fl D:

w(z) :=

u(z),

ifzEUflD

u(z),

if z E 1W fl D.

-

-

Put

th(z):=supfrloglf(z)l:c>0. fE A, cloglfl < wonUflD}, First, we shall prove that ii' following inclusion ((z. C) E (U fl D) x C

:

ICI

on U

zE

Un!).

fl D. To do this it suffices to establish the

exp(—ü(z)))

C {(z. C) E (U fl D) x C

:

exp(—th(z))}.

IC!

Define

Q := ((z. C): z E Un D.

exp(—u(z)) orz

= ((z. C) : z E U fl D.

i)U fl D,

CI

exp(—ü(z))}

exp(—w(z))}

ICI

and

K := ((z. C) : z E U fl D,

exp(—ü(z))}.

Applying the local maximum principle for Banach algebras (cf. (Sto 1971], Corollary 9.13). we obtain

F(BIK) C (['(B) fl K)

U

c

Q,

where BIK is the uniform closure of B on K, where ['denotes the Shilov boundary, where Y = and where means the relative boundary w.r.t. Y.

Fix a point (z. C) E Y, z e Un D, and a Hartogs polynomial F E

F(z, C)l Therefore, (z. C)

Then

IIFIlr(BIK)

Q, which gives the desired inclusion. Hence, th < ü on U fl D.

4.5 Structure of envelopes of holomorphy By definition, we have h

w on U fl D. Let

((i/k) log If and fix a z° E U fl

: k E N, f

E

A},

D. If a is any Jensen measure for z0 w.r.t.

h(z0) = sup{h(z°)

:

441

i

h on U fl

15)

f

then

hdcr

Unb

JUnb wdor.

In virtue of Fact 4, taking the mflmum over all such a's, leads to h(z0)

Hence, the generalized maximum principle holds. Hence, the proof of Proposition 4.4.16 is completed.

0 0

4.5 Structure of envelopes of holomorphy We know that there are domains in C" whose envelopes of holomorphy are not longer

domains in C" but Riemann domains over C". That was one reason to enlarge the category of domains in C" to the one of Riemann domains over C". Conversely, the following observation shows that not every Riemann domain of holomorphy over C" can be obtained as the envelope of holomorphy of a domain in C". Theorem 4.5.1. Let G c C", n 2, be a bounded C°° -smooth pseudoconvex domain (or a bounded C2-smooth pseudoconvex do,nain with a Stein neighborhood basis). Then there exists a locally biholonwrphic Pnapping p: G such that the Riemann domain (G, p) is not the envelope of holomorphy of any domain in C".

Proof Choose points z', z" is strictly such that liz' — z"II = diam G. Then pseudoconvex near z' and near z". Without loss of generality we may assume that z' = (1,0 0). z" = —z', and G C B(z', 2) fl B(z", 2). Now choose S E (0, 1/4) sufficiently small such that with

w':=z'+(2—S,0

0)

the intersection flB(w', 2) consists of strongly pseudoconvex boundary points and is a graph over the tangential hyperplane of l)G at z'. With A := 2ir/(2 — c5) it is obvious that the mapping

p: G —+ C",

p(z)

locally biholomorphic, Ic. (G, p) is a pseudoconvex Riemann domain over C". Assume that the statement in the theorem is false for p. Then there exists a subdomain Go c G satisfying the following two properties: is

442

4 Existence domains of special families of holomorphic functions

• PIG0 iS injective, i.e. Go is a univalent subdomain of G, • every holomorphic function on G0 extends holomorphically to the whole of G. Observe that the strongly pseudoconvex boundary points of G are peak-points for the algebra C(G) fl (9(G) [Hak-Sib 1975], [Pfi 1974] and [Epe 19631, [Ros 1961]). Thus they belong to Go (cf. Remark 1.4.5(h)). Otherwise there exist such a

z0E aG\Go,2€ G\Go,2nearz°,andfE e(G)fl(9(G)suchthat

f(z°) =

1

>

>

the function g := I /(f — is holomorphic on G0 and, therefore, it extends E (9(G). Taking into account that g(f — = I on G0, the identity theorem leads to I = = 0; contradiction. — Then

to a function Put

G':=G—(2—ö,0

0).

Since z'. z" are strongly pseudoconvex boundary points of G and since G0 is connected. there exists a point E Go fl i)G' fl B(z', 2). Choose a neighborhood V C Go and put th

0)E

8G is strongly pseudoconvex in th Thus we find a point

ZEGofl(V+(2—&O

0)).

:= — (2 — 0 0). Then are different points in Go with which contradicts the assumption that Go is univalent. Put

=

0

Remark 4.5.2. The above example relies on the fact that G has a lot of peak points. On the other hand, by a result of J. E. and E. L. Stout [For-Sto 1977]. we can obtain a surjective, locally biholomorphic mapping q: E'7 —÷ Then (E0, p oq), where p is the mapping of Theorem 4.5.1, is not the envelope of holomorphy of any univalent subdomain Go C E". The proof follows the same ideas as above, namely, if Go would be an univalent subdomain of E" such that (E", q o p) is its envelope of holomorphy, then C q(Go). But then PIqIG& could not be injective. Instead of following [For-Sto 19771 we will use here another approach in order to reach the same conclusion (cf. [For-Sto l977b]).

Lemma 4.5.3. Let s' < s and r + s' + s < t be positive real numbers. Put X := K(0. r) U K(r + s, s') U [0,t] C C. Let E C(X) satisfy • lint x is holornorphic, • lint x)' is continuously extendible to the closed discs K (0. r) U K (r + s, s'). Recall

there exists an f

that a point z0 e aG is called to be a peak-point for the algebra C(G) fl 0(G) if C(G) fl 0(G) such that I = f(zO) > If(z)I, z G \ {z°).

4.5 Structure of envelopes of holomorphy

443

is a &-function.

If e > max{

0 is given, then there exists a complex polynomial p E ?(C') such that

<e, p(t) =

lip' —

—

must be read as the continuous extension of

(Remark:

usual derivative of

Proof Fix an e >

on 0.

P(C1) with

is

x)'

and as the

the real axis, respectively.)

Using the classical theorem of Mergelyan, we find a polynomial — P1 lx <es, where e1 := e/(4(t + r + s')). Put

P2(Z) := pi(z) Then p2

and p'(t) =

+ *'(t)

a polynomial with p2(t) =

Z E C.

—

and ilp2 —

< 2ei.

Finally, we set

p(z) :=

f

+ *(t),

z E

C,

where the integration is along an arbitrary piecewise C1-curve connecting t and z in C. Obviously, we have p(t) = and p'(z) = p2(z), z E C. It remains toestimate lip— Letx XflR. Then we get

lp(x) — fr(x)l

=

f

—

p2(r)dr +

(jX

f p2(t) — *'(t)idt lfz

E

*'(r)dr + 2(t —

<2(t +

<e.

K(0,r)UK(r+s,s'),then = (ftm

m denotes 0 (respectively r + s). if z E K(0, r) (respectively z E K(r + s. As above it follows that

here

lp(z) — *(z)I

f

P2(t) — *'(r)ldt + iz — mlllp2 —

Hence, by continuity, we end up with

< 2tci + 2(r + s')ei < E.

<e.

—

D

As a consequence we will derive the following approximation result in the complex plane.

Lemma 4.5.4. Let N be an integer greater than

I

and let

O=to
444

4 Existence domains of special families of holomorphic functions

be positive numbers. Put X := [0, t] U U7=0 K(t1, r1).

• •

lint x is holomorphic, x)' extends continuously to U7=0

let

C(X) with

rj),

• 'I'

belongs to the class C'. Then,for any e > 0, there is a polynomial p

max{IIp

< r,

lip' —

—

Proof Observe that the case N

=

1

and

max{lIpi where X :

1.

Thus, ifl

(tN_I+rN_I, IN—rN),

with

< e/2,

—

—

[—ro,

p(t) = *(t), p'(t) = %fr'(t).

is the one treated in Lemma 4.5.3. Now, suppose

thatLemma4.5.4istrueforanarbitrary N—I > we find a polynomial pi

such that

i]

p1(l) =

p'(l) =

K(tj, rj).

Put X := K(0, IN—I + rN_I) U K(IN, rN) U [0, t]. If we substitute the function by the new function X —+ C,

J pi(z) 1

ifz E K(O,tN_I +rN_I)UEO,t] ifz [1,1] U K(tN, rN),

then the assumptions of Lemma 4.5.3 27) are fulfilled. Therefore, there exists a poly-

nomial p E .1'(C) satisfying all conditions of that lemma with respect to Hence p is the desired polynomial for X and e.

and e/2.

0

Remark. It should be noted that the claim in the preceding lemma remains true even if we deal with a sequence instead of a finite number of pairwise disjoint closed discs attached to the real line and an arbitrary positive continuous function on the real axis, i.e. there is a Carleman type approximation result for this infinite geometric configuration (cf. [For-Sto 1977]). Now we turn to study the higher dimensional situation. First we recall the following fact which will be used later.

Lemma 4.5.5. The union of two disjoint convex compact sets in C" is a polynomially convex set. in polynomially convex.

the union of two disjoint bounded closed polydiscs in C" is

c C", j = 1,2, be compact convex sets with K1 fl K2 = 0. Put K := K1 U K2. Then there are a point = (il + ii,, + ii,,) E C", real numbers a functional L: C" —÷ IR, Proof Let K3

z=x+iy=(xi+iyi 27)

Taker=tN_I +rN_1.r+s=:N.ands'=rN.

445

4.5 Structure of envelopes of holomorphy such that

LIK1 <0 =


Re P1K2. We may assume that IlpIIx = 1. Then the compact convex sets disjoint and therefore, Y p(Ki) U p(K2) is a polynomially convex set in C.

are

Now suppose that K is not polynomially convex, i.e. there exists Zo E K \ K. Then Ip(zol < 1. Assume that p(zo) Y. Then there exists a q E .'P(Ct) such that Iq(p(zo))I > I > flq fly > flq o P11K which contradicts that zo E K. Hence we may assume that p(zo) E p(K1); in particular, p(zo) 0. In virtue of the polynomial convexity of K1 we find P E such that

P(z0) =

>

IIPIIKI

1,

IPflK2 >0.

Applying the fact that Y is polynomially convex and that the function g(A)

10.

:=!

A,

AEp(K2) A p(Ki)

is holomorphic on a neighborhood of Y, we get another polynomial g E P(C') satisfying A E

p(K2),

2IIPIIK,

Put f

Ig(A) — Al

< IpLo)l,

AE

—

(g o p)P. Then f E If(z)I

and

Z

E K,

such that and

If(zo)l = g(p(zoflflP(zo)l

contradiction.

U

Observe that according to a result of E. Kaffin (cf. [Kal 1965]) there are three mutually disjoint closed polydiscs Q in Ci', n > 1, such that their union is not polynomially convex. 29) Nevertheless, the following approximation Q3 result holds.

u

Lemma 4.5.6. put X := 28)

29)

:=

0

0), ri), where r1

(0, 1/2), 0 < j
Cf. [Rud 1974b1. Th. 3.4.

Notice that the union of three mutually disjoint closed balls in C" is always polynomially convex (cf. lKal 1965]).

446

4 Existence domains of special families of holomorphic functions

Then any funct ion f that is holomorphic on a neighborhood of X can be on X by polynomials.

uniformly

approximated

Proof Fix an arbitrary e >

0. Since

with

polynomial P1 E

U

is polynomially convex, there is a


ilPi —

=0.

c

< N.

In virtue of the polynomial convexity of Y1 U

we find a new polynomial P2 with

lIP2 — lilA,)
max(11P2

Continuing this line of construction we end up with polynomials Pj max{11P3+i —

Hence IIPN —

Pilhy

<s/N,

liPj+I —

J=

I

PN such that

N — 1.

I lix <E.

0

Next we study the following geometric situation in C" (cf. Lemma 4.5.4):

Let X := L U

where

:= P((j, 0

0), r3),

0 < r, < 1/2,

Lemma4.5.7. Leif E C(X). AssumethatfisofclassC1 onLandholomorphicon a neighborhood of U7=0 Then, given a positive number e, there exists an entire function F such that

IIF—fllx <e, 8ZI

0)—

<e, ax1

(zi,O

0) E

X.

Pivof Set N

Xi :=Ukij.rju[_rl,N+rN]cC j=O

0). E X1. Then the new function! satisfies the required f(z := f(zi, 0 conditions in Lemma 4.5.4. Therefore, we find an entire function h: C —+ C with and

max{llh — flIx,. Iih' Defineg(z1

—

!'llx,} <s/2.

Zn) := f(z)—f(zi),z E X. Thefunctiongisholomorphiconallthe 0) e Thusgmaybewrittenasg =

4.5 Structure of envelopes of holomorphy where

...

,

are holomorphic functions on with

j

0

4.5.6, there are polynomials P2,...,

<s/(2n). F(z) Zk Pk(Z) + h(z1), z E satisfies the following estimates: Izkllgk(z) —

—

virtue of Lemma

In

2
Setting

IF(z) — f(z)j

N.

447

Pk(z)I

entire function F which

+ ifzi — h(zi)I <e.
axi

X.

z

X1,

which completes the proof.

D

j

N. Moreover, We keep the notation X and L from (*) with Tj := 1/4, 0 we assume that there is an open neighborhood U of X, an open subset V in C". and a CLdiffeomorphism CI,: U —+ V. that is holomorphic on neighborhoods of < N. Under these conditions the following lemma is true.

j

PropositIon 4.5.8. Assume the situation just described. Then there exists a biholomorphic mapping *: P,, —÷ i/j(IP',,) such that U7=0 C V, i.e. a C is biholomorphic to the unit polydisc.

certain neighborhood of ( u7=o

Proof Choose positive numbers s" < s" < s' <

< 1/4 such that the following

s

conditions hold:

• := • U' := L +

U, U, • c1 is holomorphic on

j

0

N.

Put U := U7=0 U U'. Observe that,if Therefore, we may write

z,,) — 4)(O, Z2 where the mapping

g :=

Zn)

U, then also (0,

z

= zig(zi

Zn) E U.

Z2

Zn),

Z

U,

g,1) is of class e' on U and holomorphic on neigh-

that, ifz E =0,thenz Thus, for Now we apply Lemma 4.5.7 (with rj = 1/4 + s) to the functions any positive e we find an entire mapping C" —p C" that satisfies the following properties:

< j
Observe

IIg(z) — g€(z)II <e/(2(N

8xi

8x1

+

1)),

z

L

:

tE[O,N+l/4+s].

448

4

Existence domains of special families of holomorphic functions

:= zige(z) + 4)(0.

Put

Zn),

Z2

z

E U.

14E(z) — 4)(z)II < a(1)

—(t,O

0) —

dx1

—(t,O

Z

0)

an

It is easy to see that then E

<e,

I E [0.

N + 1/4].

Thus we may assume, if e is sufficiently small, that 4)E is injective on the

C

's,

C

Recall that w(') := 4)0', 0 0). 1 E [0. N + 1/41, is a regular curve. Therefore we may assume that WF(t) := 4)((I, 0 0) is a regular curve on the interval [0, N + 1/4] that is also injective. We fix au t, for which all the former statements hold, and we write (I) := Observe, that 4) need not to be regular at every point of L. On the other hand, we know that det 4)'O, 0 0) is a non-constant holomorphic function on the complex plane. Thus there is only a finite number of singular points of 4) on the intervals

[j+l/4+s'.j+l—l/4—s'J, j=0

N—I.

So we find a C°°-function a: C —÷ C satisfying the following properties:

•a(A)=A,ifA E K(j, l/4+s"), • a(1) = r + ifi0'), t R, where is a sufficiently small real valued function with sufficiently small first derivative, • det 4'(a(t), 0 0) 0,1 E [0. N]. • 4) is injective on U ((a(t). 0 0): 1 E 10, NI). + Then, applying Lemma 4.5.4, we choose an entire function &: C —+ C with: • a is injective and regular on U7=0 K(j, 1/4 + s") U [0, NJ,

•a(0) =0, • K(j, 1/4)

ã(K(j, 1/4 + s")) C K(j. 1/4 + s'),

• 4) is injective and regular in a small neighborhood of

K(j, 1/4) U ((&(t), 0

Then 4)o(z) :=

0): 0 < t
in an injective holomorphic mapping in a

Z2

with 4)0(U0) C V and

neighborhood Uo of u7

C

0 < j
C.

We

choose S in such a way that 5

1/4 + s" on a neighborhood of K(j, 1/4 + s"), and

{z E C"

:

Zi E [0,

0 <j

NI. IziI < S(zi).

2

< N, j

,z}

C Uo.

449

4.5 Structure of envelopes of holomorphy

Using Lemma 4.5.4 we approximate by entire functions. Thus we find an entire function S: C —* C and a simply connected domain US,

[0. N] UU K(j, 1/4 + s") c

U5

such that S is without zeros on U5 and

E C" : Zi E U*. IzjI <

C

If

E —f

2

< j < n}).

U5 denotes the biholomorphic Riemann mapping, we put

E" —÷ V,

C iJ'(E") C V. Hence the

Then is a bihotomorphic mapping with lemma is proven.

0

After all the preceding preparations we now formulate the announced result due to J. E. and E. L. Stout (cf. [For-Sto 1977b]).

Theorem 4.5.9. Let G be an bounded domain in C" (for example G = B,,). Then there is an open subset M c G that is biholomorphically equivalent to the unit polvdisc E" and that contains almost all of G, i.e. G \ M has Lebesgue measure zero.

Remark. What the theorem says is that any bounded domain G has almost global coordinates in E". We point out that this theorem remains true for complex manifolds.

Proof of Theorem 4.5.9. Using Vitali's covering theorem we know that there is a countable family of mutually disjoint polydiscs P3 := e3) C G, j E N, such that Pj covers almost all of G. We can extract a finite number of those polydiscs, say

such that k(1)

/i

I

j=I Moreover, we shrink the polydiscs k(l) (I) P, )> I

(1)\

I

)> to concentric polydiscs

with

Using Proposition 4.5.8 we are able to find an open subset Di of G satisfying the following two properties: k(l)

j=l b1 is biholomorphically equivalent to E".

450

4 Existence domains of special families of holomorphic functions

Obviously there is a domain Di with k(i)

u j=i is

D1

biholomorphically equivalent to E".

Therefore, A217(G \ D1) < Repeating the above argument we find polydiscs wise disjoint, such that

p2)

CG\

k(2)

and

with

concentric polydiscs k2)

3(2))>

\

Thus we get

u p2))

+

\ D1)

Again applying Proposition 4.5.8 leads to a domain 1)2 with k(2)

U

U D1

j=i

D2 is biholomorphically equivalent to E'1. Finally we end up with a domain D2 such that k (2)

U

p]fl

U D1

D2

D2.

j=I D2

is biholomorphically equivalent to E",

Am(aD2) = 0,

pair-

4.5 Structure of envelopes of holomorphy

451

Repeating this procedure we obtain a monotonically increasing sequence of domains

D1 is

biholomorphically equivalent to E".

What remains is to prove that D :=

is

also biholomorphically equivalent to

Fix a point xo E D1 and denote by E" —+ D1, j E N, a biholomorphic mapping with (0) = xO. In virtue of Montet's theorem we may assume that the sequence converges uniformly on compact subsets of E" to a holomorphic mapping E" —p C" with ifr(E") C D. ToshowthatD C E N, the inverse mapping of Then, if we consider a subsequence that converges on D10 compact-uniformly to a mapping E (c°Jk)kEN of = E (9(D10, E"). (Use CE". the maximum principle for each component function of çQ.) Therefore, = wo (observe that converges uniformly on a neighborhood W := W(ço(wo)) E"),

i.e.

E *(E").

In virtue of the compact-uniform convergence of the sequence (fri )j EN we know that det 3 —+ det compact-uniformly on E". Moreover, det is without I

zeros on E". (Recall that the are biholomorphic.) Therefore, applying Hurwitz theorem, either det does nowhere vanish on or it is identically zero there. In the latter case Sard's theorem immediately leads to a contradiction to the fact that Hence, is locally biholomorphic on E". Dc Now, we shall prove that = D. If this is not true, then there would exist a point zo E E" with wo Fix neighborhoods V V' = *(zo) E V(zo) E" such that fr is injective on V'. V'(zo) We claim that there existaball B(wo, r) and an index j1 with r) = 0 for all j Otherwise we could find sequences (wj)JEN C C". w1 = and such that w1 = *k(j)(Zj). where (mfrk(j))jEN is a suitable c subsequence of Without loss of generality we may assume that z1 —* dV. Then mjrJ(k)(Zj) = = WO *(zo) which contradicts the injectivity of on V'. 13(wo, r) fl fl Observe that Choose Jo ? j' such that B(wo, r) is open in r) and that fl13(u0, r) is closed in r). To see the last claim, let (Zk)AEN C V such that C B(wo, r) and =

B(wo,r). Again, we may assume that ifrj0(zk) —÷ does not intersect r), we conclude that E V, i.e. Hence B(w0, r) fl r), i.e. = E contradiction.

= wj.

Since

E B(wo,

C

r) fl D;

452

4

Existence domains of special families of holomorphic functions

Finally, we will verify that is injective. In fact, let z'. z" be two points in E" We choose a compact neighborhood W of z'. z" with W C E". i/i(W). Then K C for a certain index Jo. Using the same notation and Put K argument as before we see that with

=

(z'))

(z')) +

—

(z'))

+ Here we used that

)kEN is

(z"))

—

(z"))

— çoft

(z")) —÷

0.

uniformly convergent on a small neighborhood of

is continuous at = E" —p D is a biholomorphic mapping.

tji(z")} and that Hence

D

Remark 4.5.10. In particular. the theorem gives an open subset M of B covering almost all of 18 that is biholomorphically to the polydisc P,,. But recall that the ball and the polydisc themselves are not biholomorphically equivalent. Whether IP,, contains a biholomorphic copy of 18,, with full measure seems to be unknown Using a similar but more elaborated technique as before J. E. and E. L. Stout (cf. [For-Sto 19771) even proved that any domain G C C" is the projection of a Riemann—Stein domain (P,,, p), where the fiber z E G, consists of not more than (2n + I )4" + 2 points. It seems to be unknown whether this upper bound

could be given as Cn with an universal positive constant C They also proved that domain' (IP,,, p) can be substituted by a 'ball-Riemann domain the (3,,, p)' (cf. [For-Sto 1982]). Now we are able to show that also the polydisc with a certain biholomorpic mapping is never the envelope of a domain in C"

Theorem 4.5.11. There is a locally biholomorphic mapping q: E" —÷ C" such that the Riemann domain (E". q) is not the envelope of holomorphy of any univalent domain in C"

Proof According Theorem 4.5.9 we choose a subdomain G' C 18,, with A2,,(G') = A2,,(13,,) and a biholomorphic mapping j3: E" —+ G'. Moreover, let p: B,, —÷ C" denote the locally biholomorphic map from Theorem 4.5.1. Put q := poj3. Obviously, q is locally biholomorphic, i.e. (E". q) is a Riemann domain over C". Assume that E" contains a subdomain Go such that q is injective on G0 and any holomorphic function on Go extends holomorphically to E". ft(Go). Otherwise, one would find a point Z() E p13,, \ /5(Go). We claim that d]3,, Then take a function f, continuous on 18,, and holomorphic on 18,,, such that

f(zo) =

I

>

If(z)I.

A2,,(G') = A2,,(B). we can choose Obviously, we have that a :=

z E 18,, \ (zo}.

C E" with lim

Since zo.

I

and lirnf o

=

= I.

If

4.5 Structure of envelopes of holomorphy

we fIx

a

large vo with

l/(f o

—

fo

on

453

o j5(z'°l > a, then we get a holomorphic function g := Go that does not extend to a holomorphic function on the

whole of E'7; contradiction. Now arguing as in the proof of Theorem 4.5.1 concludes the proof.

9

is the projection of the On the other hand we will show that any domain G C envelope of holomorphy of some proper subdomain G' C G. i.e. if p) denotes the envelope of holomorphy of G', then p(C(G')) = G. In other words, there exists a proper subdomain G' C G such that any function f E (9(G') has a many-valued' holomorphic extension to the whole of G. FirstwehavetointroduceafewauxiliarysetsinC",n > 1. C" —÷ Cbe a complex linear functional and let a E IR, r > 0, and a E C". A set of the form

X = (z E C" :

liz — all

= r, Ree(z)

a)

is called a closed spherical cap (with center a and radius r) if it contains more than OB(a, r). Analogously. one point and X

X := (z E C"

:

liz — all

= r, Ref(z)

> a}

X C X dB(a,r). By a punctured spherical cap (with center a E C") we mean a set of the form X), where X denotes a closed spherical cap with center a and where the (X \ X3's are open spherical caps, centered at a, with X3 C X and X1 fl Xk = 0, j k.

isanopen spherical capif 0

Lemma 4.5.12. Anyfunction f that is in a neighborhood of a punctured spherical cap X extends holomorphically to a neighborhood of conv(X). Proof By the Hartogs' theorem (cf. Theorem 2.6.6) it suffices to study only a special case, namely, X = {z E C" llzll = 1, Rez1 = a), where —l
X := {z E C": I —e < where

Ilzll <

I

+e. IRezi —cr1 <

e e (0, I — al) is properly chosen. For

denote by

E

the middle point of the following interval

:= Recail thatn >

(I — e)2 — 2.

h(E) :=

30

Rez1

:= + e)2 —

— cr1

<

e, we where

454

4 Existence domains of special families of holomorphic functions

Define

F(z) :=

f

I

z

2iri

— Zn

-

= (z. Zn) E conv(X)

fl Bn.

Observe that for E is holomorphic on the circle IRezi — cr1 < e, i.e. the function F is well defined. Moreover, if E conv(X) fl 13,,, aK(O. then for z. sufficiently near to z°, it follows that E I (i). Applying the Cauchy theorem we conclude that for such Z we have

F(z)

=

—i-,

=

f

_L

— Zn

f

— Zn

i.e. F is holomorphic near z0. Hence, F is holomorphic on conv(X) fl 13,,. > I — e. Then f(L.) Zn) E conv(X) with I — e < IIZII < I and Fix Z = < — is holomorphic in E C : (I — < (I + e)2 — 11z112}. i.e. in the full disc K(O, 1(1 + e)2 — 11z112). Therefore, we get = Zn). Finally, the identity theorem guaranties that f and F coincide on their common domain of 0 definition conv(X) fl (B,, \ (I E)13n), i.e. F extends f holomorphically.

Before we continue, we mention the following simple observation: whenever Y1. Y2 denote two disjoint concentric punctured spherical caps with conv(Y1) fl

c cn be a domain, let (K1

j be a local/v finite family of

co?npact subsets of G. and let (U1 )jEJ be afamily of open subsets with K1 C U1 C G. k(j)) of closed Then there exists a local! finite family { J. I < k spherical caps XJ.k with center zero such that: :

j

•conv(X1k) C U3foreachj,k,

• XJk fl Xj'k' = 0, • K1 C

k)

(j', k'),

k))for each j.

Proof First observe that the family (Kj)JEJ is necessarily denumerable, i.e. we may take J = N. Moreover, let us fix an exhaustion of G. i.e. a sequence of relatively compact subdomains G1 C G with

The compact set

meets at most a finite number of the Kg's. say K1 K1)

finiteness of (ICy )jEN. Putting ej :=

ii ifj > m(l). Usethelocal

/2 it follows that the

-neighborhoods of the

K,j > m(I), does not meet G1. Um(I) to be open neighborhoods of

We define

G,

C U1,

I <j

ni(l).

Kn,(1).

-

Km(I) with

455

4.5 Structure of envelopes of holomorphy

Then

there exists a j(2) such that

K1

C G(2). We choose m(2) > m(1)

such that G1(2) does not intersect j > m(2). Then we find a positive < ei with Moreover, we choose neighborhoods of > m(2). dist(K1. GJ(2)) > C j
<j

C U3 with K, C

Repeating this process we obtain a sequence of open sets U3

LI3

G such that (UJ)JEN is locally finite. Summarizing, we may assume that our family (U1 )JEN is already locally finite with

G. j

E

N.

Now fix j and a point zo E

0: We choose an open cap Z, center at 0 and radius Izoll. with zo E Z Case zo and conv(Z) C U1. Ifs > 0 is sufficiently small, then the (I + e)-dilation Y = (I + e)Z of Z is a closed cap with the following properties: zo E int(conv(Y)) C conv(Y) C LII. Case zo = 0: Here we choose a small ball

Y :=

r)

U1 and a closed cap

> —l/2}

= 81B(0,r)fl{z :

such that Zo = 0 int(conv(Y)) C conv(Y) C U1. Yj.k(J) of the closed caps Y2, z E K1, is compact, a finite number Y31 Since constructed before have the property that k(j)

K1 C U int(conv(YJk)). Recall that the caps YJ.k are subsets of concentric spheres around zero with radii Pj.k. We define Xj.k := (I + Sj,k)Yj.k with sufficiently small Ej.kS such that k(j)

conv(X1k) C

K1 C (I + 6j,k)Pj.k

(I + 6j'k')Pjk'

if (j, k)

(j', k').

Hence, the family (Xj,k) j.k of closed spherical caps is a locally finite one and it satisfies

all the conditions desired. We have already mentioned that any domain in

D is the projection of a Riemann—

Our next aim is to show that it seems there Stein domain (X, p) with X = is no general relation between a domain G C C" and its envelope of holomorphy

a: (G,id) —÷

(G,13G).

Theorem 4.5.14. Any domain of holomorphy G C C", n ? 2, is the project ion of the there is envelope of holomorphy of a simple subdomain G' C G, or more

456

4 Existence domains of special families of holomorphic functions

a subdomain G' c G, that is topologically equivalent to an open cube, such that, = G. a: (G', Id) —÷ (G', j3) is its envelope of holonwrphy, then

Pmof We choose a locally finite sequence of compact sets C G, j E N, with K1 = G. Taking U3 := G we can apply Lemma 4.5.13 to obtain a system of closed spherical caps j E N, I k k(j), with the following properties: Xjk fl X3'k' = 0 C G, j

if (i, J) E

(j', k'),

I
N,

U int(conv(X3k)) = j.k

Now let D be an arbitrary subdomain of G that contains the union of all the XJ,k. Denote by aD: (D, id) —÷ (D, PD) its envelope of holornorphy. Then the inclusion i: D —÷ G lifts to the envelopes of holomorphy as i: (D, PD) —* (G. id) so that i is a morphism. Therefore, pD(D) C G. Recall that X.k may be written as XJ.k = {z E C" : ReAJk(z) > a,

lizil

= RJk},

where Ajk are linear functionals and the R3,k are certain positive real numbers. If Cj,k are sufficiently small, then XJk := {z E

: ReA3.k(z) > a, RJ.k

—

Cj,k

< IIzII < RJ,k}

c

D.

In virtue of Lemma 4.5.12 we know that conv(X3,k) = int(conv(XJ.k)) is the envelope of holomorphy of X3,k. Therefore, as just before, we conclude that the inclusion lj.k : XJk —+ D lifts to a morphism

ijk : (conv(X3k), id) —+ (I). PD). Thus, we obtain PD 0 i3k(conv(X3k)) G

= U K3 C

= int(conv(XJk)) and, therefore, C

Pt,(D).

j.k j=I Hence it is shown that G is covered by the projection of the envelope of holomorphy of D. What remains to show is that D can be chosen as a subdomain that is topologically equivalent to an open cube: We choose smoothly bounded domains Vjk containing XJ.k that are topologically is locally finite and its sets are equivalent to open cubes such that the system ( mutually disjoint. Rename the Xj.k (respectively Vj.k) as Yt (respectively W1). Then := 1]) we connect W,with Wt4.i via a Jordan arc ye: [0, 1] —+ G such that intersects the W1's only in ye(O) E W1 and in ye(I) E We+i. Using small pairwise disjoint tube neighborhoods of I', we end up with a domain G' := Ue(We U T1) 0 that is topologically equivalent to a cube.

4.5 Structure of envelopes of holomorphy

457

Remark 4.5.15. 1) It is known that if a: (G, id) —÷ (G.

) is the envelope of holomorphy of a univalent domain G C C? then the natural map ir(a): irj (G) —p between the first homotopy groups induced by a is surjective (cf. [Ker 1961]). Or more heuristically, the envelope of holomorphy of G has no 'new one-dimensional holes'. With the methods used above it is not difficult to see that there is no higherdimensional analogue. The following is even true: any domain of holomorphy G c C" contains a subdomain Go with Jrk(GO) = Hk(GO, Z) = Hk(G0, Z) = 0, k > 2 such that G is the envelope of holomorphy of Go (cf. [For-Zam 1983]).

2) So far, we do not know any characterization of a Riemann—Stein domain (X, p) over C" that is the envelope of holomorphy of some univalent domain. According to and W. R. Zame (cf. [For-Zam 1983]) this seems to be a quite difficult J. E. problem.

In Theorem 4.5.14 it was assumed that G is a domain of holomorphy. But, with some effort, it turns out that, in fact, this assumption is superfluous. Here we give only

the formulation of the result. For a proof the reader is asked to contact the original paper [For-Zam 19831.

Theorem 4.5.16. If G C C" is a domain, then there is a subdomain G' of G, topologi= G, wherea: (G', id) —+ (G'.

j3)

is its envelope of holomorphy. We

know that not every Riemann—Stein domain (X, p) over C" is envelope of

holomorphy of an univalent domain. Nevertheless, there is a large class of Riemann— by univalent domains, as the following theorem Stein domains that are will show. Before we state this result we have to prove a preliminary geometric fact that will be used later.

Lemma 4.5.17. Let (X, p) be a Riemann domain over C" and let (Kj)JEN be a sequence of compact subsets of X. Put K := (J3 K1. If the family (p(K1 ))JEN is locally finite in p(K) 31) and tf P1K is infective, then there is an open set U C X, K C U, such that already Ply jS infective. Proof Observe that p(K1) is a compact subset of C". Since the family

is locally finite in p(K), there exists an open set V1. p(K1) C Vi, and an index fl V1 = 0, if) > nii. Now we choose an open neighborhood X with p(W1) C V1. Put L1 K1 C K. Then L1 is a Wi = Wi(Ki)

nii such that

compact subset of X and, by assumption, the projection p is injective on L1. Using that p is locally biholomorphic we may assume that PI*1uL, is injective (eventually after shnnking W1). Since p( W1) fl p(K3) = 0, f > m it follows that p is injective on W1 U K. In the next step we replace Ki, K2 by the new compact set W1 UK2. Itis clear that

the new sequence p(W, U 31)

p(K3)..... is again Locally finite in

Local finiteness is here meant with respect to the relative topology of p(K).

U p(K).

458

4 Existence domains of special families of holomorphic functions

Thus using the same reasoning as before gives an open neighborhood V2 of p( W1 UK2) and an open set W2 such that W1 U K2 C W2 X, p(W2) C V2. and V2 meets only is injective. a finite number of the p(K1)'s, and Then we can construct open sets W,j j K1, W1 X such that is injective and

Then

WJUKk+I C

W :=

W1

isthedesiredopenneighborhood

of K, on which p is injective.

D

Now we formulate and prove what we have announced before.

Theorem 4.5.18. Let (X,

p)

be a Riemann—Stein domain over

C" which is a cover

Then there is a domain G C C", whose envelope of holonwrphy (G, Id) —÷ (X, p), i.e. X is the envelope of holomorphy of G.

32),

is given by a

Pmof We only discuss the case where the fibers p'(z), z E p(X) =: D, contain infinitely many points. The remaining case, which is much simpler, is left for the reader.

We start by choosing locally finite sequences (K1 respectively of compact subsets K3 C D, respectively of domains U3, with K3 C U3 C D such that U3 = • U3 K = and • —÷ U3 biholomorphic, =

•

if

fl (if

0, then £ =

The aim of the following construction is to establish the situation of Lemma 4.5.17.

of closed spher-

So, in virtue of Lemma 4.5.13, we select a family (Xj.k)JEN, ical caps with center at zero satisfying the following properties: fl = 0 if (j, k) (j', k'), • • (Xjk : j N, I k k(j)} is locally finite, i

•conv(X3k)C

U3 foreachj ENandeachl • D = U.k int(conv(X3k)).

Fix a pair (j,

X3'k'

k).

We recall that, ifconv(XJk )flccrnv(X)'k')

connecting X,k and X3'.k'

that

0, then conv(X3,k)fl

0. Thus we choose an arc r(j,

0 or XJ.k fl COflv(Xj'k')

lies entirely in conv(X3,k )U

Xj',k'

k, j', k')

or XJ,k Uconv(XJ',k').

The pair (j, k) was arbitrarily chosen, thus we have constructed a family of arcs

F := {F(j, k, j', k') : conv(X3k) fl

0).

Since (COflv(X3,k))3.k is locally finite, so the system F. In the next step we try to separate a little bit the XJ.k and the arcs F(j, k, j', k') to obtain a sequence of compact sets in X, namely their corresponding preimages under plu, such that the projection map p is injective on their union. Fix X3.k C U3. Since F is locally finite, we may replace this spherical cap by a punctured spherical cap X3,k in such a way that XJJC does not meet any of the arcs 32

Recall that (X. p) is a cover if for any z E

U = U(z) and open sets homeomorphic.

C X.

j

E

J,

such

p(X)

that

there

exist a connected open neighborhood = Uj€1 U3 and pltj1: Uj —' U is

459

4.5 Structure of envelopes of holomorphy

in F. Moreover, we choose small dilations (1 + TJk,,)X),k =: XJk. 0 if £ —* oo, of X3,k such that • C C j, e N, int(conv(Xk)) c fl = 0 if (j, k, £) (j', k', £'). • Finally, for (j, k. j'. k') with conv(XJ.k) fl 0 we choose pairwise disjoint arcs k, j', k') connecting XJk and such that r'(j, k, j', k') lies in U or Xk U Conv(X;k,), that k, j', k') does not intersect any F(r, s, r', s') and other and that rt(j. k, j', k') ÷ than XJk,

r(j, k, j', k'), if £ Now we transfer the geometric situation, constructed so far in D, to X itself. Put

f't(j, k, j', k') := (pIu1,)_l(rt(j,

k,

j', k')).

Thus we obtain a sequence of compact subsets of X such that p is injective on their union K

U ZkU U ft(j,k,j',k').

j,k.(

In virtue of the convergence

j,k.j',k'.f

k, j', k') —÷

—+

r(j, k, J', k'), it

is clear that the family {p(Zk), k, j', k'))} is locally finite in p(K). Thus, according to Lemma 4.5.17, we find an open set Y, K C Y, on which the map p is injective.

At the end it suffices to show that any holomorphic function f: Y —* C extends holomorphically to X. In fact, using Lemma 4.5.12, f extends holomorphically to (pIu).kY'(conv(Xk)). If two such 'extension-sets' intersect, then, for the corresponding indices, we have conv(XJk) fl 0. Since r1(j. k, j', k') connects Xk and the two extensions coincide on their common domain of definition, i.e. all those local extensions lead to a global holomorphic function F: X —* C

withFly=f.

C]

Remark 4.5.19. We mention another topological problem connected with Riemann domains over C'1. By the uniformization result of Koebe (cf. [Fors 1981]) it is known that any Riemann surface, which is topologically equivalent to a domain in the complex

plane, is already holomorphically equivalent to a domain in C. So the question was raised whether a parallel result to Koebe's is valid for Riemann—Stein domains over C's. There is the example given byE. L. Stout (cf. [Sto 1986]) of the following Stein manifold X

:=

{z E C8

:

= l},

which is real-analytically equivalent to a domain in C7 but not holomorphically to any

460

4 Existence domains of special families of holomorphic functions

domain in C7. Stein domain

Up to our knowledge there is no such an example of a Riemann—

We conclude this section mentioning two open questions posed in [For-Zam 1983]: such that for any choice of a locally biholomoiphic a) ftJ Is there a domain G C mapping p: G —. C" the pair (X, p) is the envelope of holomorphy of a univalent domain in C"

Let (X. p)be aRiemann—Stein domain over C". Is there always anotherprojec-

tion p': X —÷ C" such that (X, p') is the envelope of holomorphy of some univalent domain in C"

List of symbols

General symbols the set of natural numbers, 0

N

N;

N0 :=NU{0}; Z := the ring of integer numbers; Q := the field of rational numbers; R the field of real numbers; C := the field of complex numbers; C U {oo} = the Riemann sphere; C Rez := the real part ofz E C, Imz := the imaginary part ofz E C; := x — iy = the conjugate of z = x + iy; Izi

:=

y2 =

the module of a complex number z = x

+ iy;

:= the Cartesian product of n-copies of the set A, e.g.

:= A \ (O}, e.g. := Ia E A: a ? 0}, e.g.

R÷;

Z÷ =

No,

=

=:

A_ :={aEA:a<0}; A>o := (a E A: a > A
OJ,

:= (—oo} U IL

e.g. R>o;

:= RU {+oo}; A,b E B}, A, BC X, Xisavectorspace;

A+ B := (a +b: a E A.B:_—{a.b:aEA,b€B}, AcIK.BCX,XisaIK-vectorspace: Dom T : = the domain of a linear operator T;

Range T := the range of a linear operator T; Ker T := the kernel of a linear operator T; := the canonical basis in C", (el 0

:=

1:=(l

=

1)EN";

1

if) =k;

462

List of symbols

= the

(z, w) :=

Hermitian

scalar product in C";

th,,), W=(WI. z

.w

zjwj; ZnWn),Z(Z1 = (zi

:= (z.

zw:=(zIwl eZ

:=

lizil :=

E C";

=the Eucidean norm in C";

=

lz,,i} = the maximum norm in C";

Izi := max{Izil

jdAX: A —+ X, idA,X(X) := x, x E A; idA := idA,X if A = X or it is clear what the outer space X is; := the number of elements of A; diam A := the diameter of the set A C C" with respect to the Euclidean distance; cony A := the convex hull of the set A;

Bq(a,r):={zEC":q(z.a)0,whereq:C" —*R+isa C-norm; Bq(a, +oo) = C";

Bq(a. r) := Bq(a, r): liz—all
C" and radius r > 0;

:=B,,(0,r), r >0;

3(r)

:=

B,,

a E

B,, (I)

the unit Euclidean ball in C";

K(a,r) :=Bi(a.r): K(r) := K(0, r);

E := K(l) =

C: Xl < 1) =the unit disc (Einheitskreis); IF(a. r) = ?,,(a. r) := B1 [(a, r) = {z e C" lz — ai
:

a

C" and radius IP(r) = iP,,(r) := E

r>

0;

:=P,,(l)=theunitpolydiscinC"; noticethatP,, = E"; x K(a,,,r,,) = the polydisc with center P(a.r) = := K(ai,ri) x I?>,,

a

E

C" and multiradius (polyradius)

IP(a,r) IP(r) =

r=

(ri

r,,)

E

(R>0)"; notice that

1);

:=

IID(0, r);

=the distingui shed boundary of P(a.r); T,a/ =

(")

:= ((z. w)

(n)

Tr.p

=

:

Tr.'p;

E

C"' x C: Izi <

r,

Iwl

<

p
463

List of symbols top T := the topology of T;

mtxA := the interiorofA A

in the topology of X, AC X; A := the closure of A in the topology of X. A c X; X A is relatively compact in X;

:= zr ...

Z = (Zi. .. .,Zn)

aE

E

a

= (ai

a,1) E Z'7;

A —÷ C;

FIA := {fIA : 1€ F), F is a family of mappings X —+ Y and A c X;

(W(k) holds fork>> I) (there exists Ico such that W(k) holds fork > (W(e) holds for 0 < e 4Z I) (there exists > 0 such that W(e) holds for 0<6 supp f := {x : f(x) 0) = the support off; K

Ktm) := the space of all polynomial mappings F: K E {IR. C};

:=

the

spaceof all

FE

:= the space of all F E

d: such that for any j

E

{l

n} the

degree of F w.r.t. the j—th variable is d1; := the space of all homogeneous polynomials from

Ct(X) := the set of all upper semicontinuous functions

u

: X —*

:= d(uOPa)(p(a)) j = 1,... ,n, u: X —+ C, := (X, p) is a Riemann domain over C", a E X; := n, the — + j= formal derivatives of u at a, u: X —p C, (X, p) is a Riemann domain over C", 1

a E X; gradu(a) :=

= the gradient of u ata;

:= a) = the real gradient of u at a, u: X —÷ IR, (X, p) is a Riemann domain over C". Ea X; notice that after the

(xI,yI C", we have

+ (Xl

+iyi

u(a) = grad u(a);

Da.P

Ck(X, Y) := the space of all Ck(X) := Ck(X C);

F: X —*

Y, k E

U foe) U {cv};

E

464

List of symbols

AN := Lebesgue measure in RN; := Lebesgue measure in X, (X, p) is a Riemann domain over C"; := the space of all p-integrable functions on X (w.r.t. Ax); LP(X, bc) := the space of all locally p-integrable functions on X (w.r.t. Ax); (9(X. Y) := the space of all hobomorphic mappings F: X —p (9(X) := (9(X, C) = the space of all holomorphic functions f: X —* C; := (9(X) fl = the space of all p-integrable holomorphic functions on X;

:= {f E (9(A) :=

lix < +oo};

(9(X):

(9(U)IA;

U U JA U open in C"

:= the space of all bounded holomorphic functions on X;

tf"'(p(a))

= the j-th partial complex derivative of f at a, j = n, f E (9(X),(X,p)isaRiemanndomainoverC",a E X; o o := = v-th partial complex deriv=

:= (a'f)(a)

ative off ata,

E

Taf(Z) :=

=thelaylorseriesof

=

at a, f

(9(X), (X, p) is a Riemann domain over C", a X: convergent for all z P(p(a), r)} = the radius of d(Taf) := sup{r > 0: Ta 1(Z) convergence of the Taylor series Ta!, f E (9(X), (X. p) is a Riemann domain over C": f

:=

{f E (9(12)

:

:

c°a

= aai in

Tx := the topology of locally uniform convergence in X; co* (9(Y) —+ X —* Y;

(rp*y1(f)

V

:= {flP:

f

I

E 4},

E

C

Aut(X) := the group of all automorphisms of X; AUta(X) := {h Aut(X): h(a) = a}; := subharmonic; the set of all subharmonic functions on 12, 12 C C; psh:plurisubharmonic; := the set of all plurisubharmonic functions on X; sh

£u(a:

:=

=

the Levi form of u at a.

kE

U {oo};

465

List of symbols

Symbols in Chapter 1

f'(T, (X, p)) := the set of all sections of a Riemann domain (K. p) over a set T C the open univalent neighborhood U C X of a such that Bxq(a,r) = Bq(a,r) p(U) = Bq(a, r), (X, p) is a Riemann domain over C";

dx.q(a) := sup{r (0, +co]: Bx,q(a. r) exists}; dx,q(A) := inf{dx,q(x): x A}, A C X;

Bx,q(a, r), 0
PIâq(a);

X: dx,q(a) = +oo};

:= (a

Bx(a, r) = 3(a, r) 1(a); px(a) := 1(A); px(A)

Bxi 1(a, r);

Bxii 1(a);

Bx(a) = B(a)

lFx(a, r) = IF(a, r) :=

(a, r);

(a); dx(a) := : dx,i (A); dx(A) I). A(r) := Px(a) = := BX! (a); Pa

Pa,I

r)

;

z

+

z E C".

r>

0;

A and p(A) =

A

y(11_j)II: N E N, 0 = L(y) length ofacurvey: [0,1] —÷ C"; [0, 1) —* X is a curve ax(x', x") := inf{L(p o

<

r); < 1N = 1) =

the

= x', = x") := the inner Eucidean distance on (X, p); (0. jr) := the sheaf of germs of holomorphic functions (of n complex variables); := (x E X: If(x)I < IIfIIK. f family

* :=

C

the

i-hull of a compact K c X w.r.t. a

(9(X); = the polynomial hull of a compact K C C";

=(p

aX

:=

çQ:

the

p—boundary of a Riemann domain (X, p) with respect to a morphism

(X,p)—÷ (Y,q);

=

dX:= dX;

466 =4'

List of symbols =

=4'

X

:=XLJaX;

=

=4'

X:=X;

:=

q o

=p

=

X —+

p

the extension of p;

SQ

*

—4'

Sf)

X=X;

(a: (X,p) —÷ a,: (X. p) —+ (Y,,q1),i El;

(X,p) —÷

((X.p),(Q,),EI) = ((X, p), (PI)IEI) =

:=

the kernel

Pi)IEI'

ofafamily (X,, ps)IEI; := the direct limit of a family

((X1. Pi)IE1, (aj,i)i4j);

:=

sheaf of I-germs of holomorphic functions: C(X,.8) = C((X, p)..8) := the .8-envelope of (X, p); E(X) = E((X, p)) := X, p). (9(X)). the

Symbols in Chapter 2 log—psh := logarithmically plurisubharmonic;

:= (u

?SR(X): u(X) C J}. I C u(a + r e'0) dO;

P(u: a, r: z) := P(u; a, r; z) := P(u; a, (r,..., r); z); J(u: a, r) := u(a + r e'6) do;

J(u; a, r) := J(u: a, (r A(u; a. r) :=

r)); fP(a,r)

u dAb;

A(u; a, r) := A(u; a, (r Aff(C", C"') : = the set of all affine C-isomorphisms L: C" —÷ Cm; Jt4'v(x; Q) :=

Qk = the Hessian of v at x;

:= u

L1(X, bc),

:= (XE X: dx(x) >

XE

:= that supp '1' = E.

(x E X:

Zn) := ... = 'V(IzI), z E C, f'4' dA2 = 1;

u(x)

SUPK

u}

=

the

E

R÷) is such

.8-hull of a compact K C X w.r.t. a

family .8 C TaC (

E

:=

:=

the

C":

= 0} = the complex tangent space to

real tangent space to ac at a:

at a;

List of symbols

467

L2(X.p) := the space of all A-measurable functions f: <+oo; Ix If fv

X

—' C

there exists a compact set K C X such that supp

Jo

—*

V E Z÷ and

uniformly on X for any a.

such that

C K for any

E

the space of all distributions on X; —+ Co(f) for any f

Co

Ix ufdA,

[u)(f)

f

E

2)(X);

u E L'(X, bc);

E

supp C := the support of C, C E the space of all differential forms on X of type (r. s) with coefficients in

A...Adpa, Adpj1

C=

=

= iI=s

the canonical representation ofaforni. where

I

E

:= the space of all forms C E and J E

E

:= min{px,

Nk: I

<

such that C1j E 11(X) for all

p};

8x :=min{l,px}6oop. Symbols in Chapter 3

Reg(M)

the set of all regular points of an analytic set M:

Sing(M) := the set of all singular points of an analytic set M;

.M(X)

the field of meromorphic functions on X. Symbols in Chapter 4

?SRhom = :=

the set of all absolutely homogeneous functions from

u(z)

{u E

= flN>o {u E

8D)

:u=

:= {z E D: If(z)I < < {z E 0 :

:= ek(D) fl

Iflix. f E uE

0


C + log(l + IzII)};

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E. M. Chirka, Complex Analytic Sets, Kiuwer Academic Publishers, 1989.

[Per 1973]

J.-P. Ferrier, Spectral Theory and Complex Analysis, North—Holland, 1973.

[Fuk 1963]

[Gra-Fri 19761

B. A. Fuks, Introduction to the Theory of Analytic Functions of Several Complex Variables, Amer. Math. Soc., 1963. H. Grauert & K. Fritzsche, Several Complex Variables, Springer-Verlag, 1976.

(Gun 1990]

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[Gun-Ros 1965] R. Gunning & H. Rossi, Analytic Functions of Several Complex Variables. Prentice-Hall, Englewood Cliffs, NJ., 1965. [Hay-Ken 1970] W. K. Hayman & P. B. Kennedy, Subharmonic Functions, Academic Press, 1970.

[Hen-Lei 1984] G. M. Henkin & J. Leiterer, Theory of Functions on Complex Manifolds, Akademie-Verlag. Berlin 1984. [Hor 19901

L. HOrmander, An Introduction to Complex Analysis in Several Variables. North Holland, 1990.

[Kli 1991]

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LKra 19821

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LNar 1968]

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470

BibLiography

[Sha 1976]

B. V. Shabat, An Introduction to Complex Analysis, vol. 1, II (in Russian), Nauka, Moscow 1976.

(VIa 1966]

V. S. Vladiniirov, Methods of the Theory of Functions of Many Complex Variables, The M.I.T. Press, 1966.

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[Ahe-Sch 1975]

P. R. Ahem & R. Schneider, Isometrics of X°°, Duke Math. 1.42(1975), 321—326.

[Alni 19221

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EArs 1966]

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[Bac 1992]

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[Bac-Fal 1991]

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[Bed-Dad 19911

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Index

A

accumulation point of a filter action of a Lie group analytic covering analytic polyhedron arbitrarily continuable domain arcwise Euclidean distance atlas

29 68 226 197 16 11 3

B

balanced domain barbell lemma basis of a fitter

65 321

Bedford—Taylor theorem boundary

127

of a Riemann domain of class Ck point of a Riemann domain bounded holomorphic function

29

30 130

30 21)

C

canonical representation Cartan-Thulten theorem

162 25

Catlin—Hakim—Sibony theorem

422

Cauchy inequalities character Choquet's lemma circular domain circular matrix domain function manifold mapping smooth boundary classical monodromy theorem complete Hartogs domain

19

219 118 65

285 3 3 3 1311

16

373

complete n-circled domain complete pluripolar set complex derivative Lorentz group manifold tangent space condition (S) cone condition continuation along a curve convergence of a filter of a filter basis cover

73 1115

19

299 3 131

40 415 23 29 29 16

D

defining function 132 3-closed form 124 3-problem 165 3-stability 26 S-tempered holomorphic function 21 diameter function 315 worm domain 423 direct limit of Riemaim domains 42 directed set 41 distance to the boundary 5 to the boundary in a direction 10 distribution 1611 division problem 208 theorem 219 for functions with tempered growth 218 domain

of existence of I

484

Index

of holomorphy E

elementary Reinhardt domains 356 embedding 195 envelope of holomorphy 55 equilibrium measure 129 exhaustion function L36 extended (n, N)-tube 299 extended tube conjecture 299. 300

with k-circled fibers with starlike fibers Hartogs polynomials Hausdorff measure Hermitian quasi-inverse matrix Hessian holomorphic convexity extension

F

envelope of meromorphy extension fat domain

51

66 432 226 155

114 25

26

function hull

3. 1

mapping

3. ig

homogeneous psh function

-

ff1

Hörmander's L2-estimates hyperconvexity hypoellipticity

343 168 128.

417, 429

cp-boundary

of a Riernann domain point of a Riemann domain

311 311

filter filter basis

formal derivative function of class Ck future light cone tube

299 299

G

general cone condition geometrically Runge germ global defining function graph norm

86 22 132 169

45. 336 identity principle for holomorphic functions for liftings for sections Igusa theorem inner Euclidean distance irrational type isomorphism of Riemann domains of Riemann regions

19

4 313 222 11

356 25 25

J

Jensen measure Josefson theorem

432 120

K

H Haar measure Hartogs lemma for psh functions theorem type lemma Hartogs domain with balanced fibers with circular fibers

289 108 3-1-8

k-tubular domain kernel of Riemann domains Kiselman minimum principle Kontinuitätssatz Koszul complex

60

258.

43 153

204 209

L

6L 373 66

Lebesgue measure length of a curve

20 11

485

index

multiplicity of a proper mapping 226

Levi 151) condition 98 form 130 Problem 336 theorem 68 Lie group lifting 4 224 of a holomorphic mapping Liouville theorem for psh functions 92 local defining function 130 313 local pseudoconvexity 161 localization principle 88 localization theorem 3 locally homeomorphic mapping locally non-pluripolar Set 326 IS locally uniform convergence logarithmic image Th 355 logarithmic plurisubharmonicity 299 Lorentz transformation

M

manifold of class mapping of class matrix ball polydisc

3

3 285 285

Reinhardt domain 6 maximal ball 54 maximal holomorphic extension maximal meromorphic extension 335 122 maximal psh function maximum principle for psh functions 98 20 measurable set Mergelyan—Bishop theorem 82 meromorphic function 334 149. 343 Minkowski function 299 Minkowski product 122 Monge-Ampère operator monodromy theorem 15. 16 morphism 25 of Riemann domains 25 of Riemann regions

N

,z-circled

domain in C" 65 20 Riemann domain over C" 20 natural Fréchet space non-homogeneous Cauchy—Riemann equation 165

0 Ohsawa—Takegoshi extension theorem

Oka theorem order of an analytic polyhedron

388

316 192

P

p-integrable holomorphic function 20 partial complex derivative 300 Pauli matrix 282 piecewise permutational [06 pluri-thin set I05 pluripolar set 96 plurisubharmonic function plurisubharmonic measure [28 Poincaré—Volterra theorem

point of indeterminacy of a meromorphic function pole of a meromorphic function projection proper Lorentz group proper mapping pseudoconcave set pseudoconvexity psh exhaustion function

14 335 335

299

84 318 129

430

R

radius of convergence of a Taylor series rank theorem rational type real analytic manifold real tangent space

[9 308

356 3

135

486

Index

regular mapping

[95

regular point of a meromorphic function

334

regular point of an analytic set regularization of a distribution of a form of a function Reinhardt domain Remmert embedding theorem removable singulanties of psh functions Riemann domain ove r C" over M over R"

303

region over C" over M surface

surface of f Riemann Stein domain over C" region over C" ring of germs Runge approximation theorem

L63

163

separately holomorphic functions 324 separation of points 40 sheaf of germs 21 sheaf of 1-germs 44. 3.35 Sibony example 374 singular analytic sets 302 singular point of an analytic set 303 Skoda division theorem 410 smooth boundary of class Ck 130 space 184 2.19

3

3

259

3

23

starlike domam Stem manifold Stein neighborhood basis

strong convexity strong pseudoconvexity subharmomc function

hull

schlicht envelope of holomorphy set section

semitubular domains

135

134 103

SUI)I)OIt

of a distribution of a form

163

76 22 86

S

domain of existence domain of holomorphy envelope of holomorphy extension

76 423

strictly plurisubhannonic function 99 strictly starlike set 43.4

T

7 Taylor series theorem Bedlord-Thylor theorem Cartan-Thullen theorem

convexity

M

25 50 55 55

26 2.5

260 3

4. 313 380

121 75

Cathn-Hakim-Sibony theorem 422 classical monodromy theorem 16 Hartogs theorem Igusa theorem 222 Josefson theorem 120 Levi theorem 336 Liouville theorem for psh functions 91 localization theorem 88 Mergelyan—Bishop theorem 81 monodromy theorem 15

487

Index

Ohsawa—Takegoshi extension

theorem

Oka theorem Poincaré—Volterra theorem

rank theorem Remmert embedding theorem Runge approximation theorem Skoda division theorem Thullen theorem thin set Thullen theorem topology of locally uniform convergence

3R8

totally real submanifold tube domain

3.th 14

308 195

86

U

union of Riemann domains univalent set upper regularization

40 3 11)4

410 336

36 336 18

W weak separation of points

weak Stein neighborhood basis

423

ESSN 093M-6572 ISBN

'I—