de Gruyter Expositions in Mathematics 34
Editors
0. H. Kegel. Albert-Ludwigs-Universitat, Freiburg V. P. Maslov, Acad...
123 downloads
1062 Views
9MB Size
Report
This content was uploaded by our users and we assume good faith they have the permission to share this book. If you own the copyright to this book and it is wrongfully on our website, we offer a simple DMCA procedure to remove your content from our site. Start by pressing the button below!
Report copyright / DMCA form
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,IwI0, 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 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' 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)) 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 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). 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. =
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)
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.
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 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
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.
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 dx(a) there exists an f E .8 such that d(T0f)
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
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
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 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 0. fi,,
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 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, < 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}
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
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 > I and8 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 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
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
—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 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 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 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)
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
l,let =:
V) :=
and V = (SI
{y1
> e_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
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)
/co;
contradiction. Let be as in (2.1.19). By (2.1.18) (with T := > (2n+2)!suchthat
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
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).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)
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 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 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
=
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 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) 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)
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 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
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
Ofor frI > 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) 0 be such that u —e on K. Then
C {x E X: u(x)
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
—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))
10. = 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
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 O VUEDOmT
(b) (cf. [Wei 1980]) The operator
(c) KerT =
(f.
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
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)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 E
X: 8(x) 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
0.
K
• for any pseudoconve.x Riemann domain (X, p) over
•foranyço
E
with
E
0 such that:
L'(X, bc), and
X and 0 < 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
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, 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)
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 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 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 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. 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. 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) 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) 0 be such that locally uniformly on X. For each v CC 10, 1]) be such that and V5(v) 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. 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) 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'
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 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
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
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
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} 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.
=
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
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) 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
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
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
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 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 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 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) 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 o,
z
—(t2+(c/t)2,c2)
(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) =: 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
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 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 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
Iwi
0 (at least so
small that p
> 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.
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
—
)
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 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
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. 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 +
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: 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 —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 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
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
? 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 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
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
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
0
1a1
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
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
Oforj=s+l
assumethata1 0
361
properly chosen) and
in the
/
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
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,
0, b: E —f iR+ such
that
D=int
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) O,andc > 0 such that E Z",j = I
:=flz E
DC C
D1 :=fl(z
—( —
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 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 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
o sufficiently
+ 1 >2. DelIneg:
:=
—* Ras
4.
I
+ 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
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.
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
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
+
0 and = 0, such that c (0, 1),
ED, (z',O)
ifz =
DnH.
0 such that
I
If(z',zn)I2dA2n(z)
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 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 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
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 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
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
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 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. Then we get for z E D 11 V1 and arbitrary X We
Fix 0
—
£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
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 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 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)
>
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 —
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
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 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 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 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:ao := (a E A: a > Ao;
:= 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;
:=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
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