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. .
. U Yq such that
1
•Uj iszeroonb,
<j q. The (1
wi s are entire functions which satisfy the following properties for a 1.0
f
f
•
Wj(a)
•
wj(a)=fwehJ+hiP=fw=0, p<j
=
• • •
=
=
= L(a)
=
= 0.
f
u1wet4'
0.
1
f
j
1
=
0,
p.
k
<j
—(a)= f dço1
is nonsingular. = 0 and the Jacobian matrix For j = I p we extend the function uj to be identically zero on D. Then
Thus
continuous on K and holomorphic on mt K. Therefore the Mergelyan—Bishop theorem implies the existence of a sequence of functions t' E N, holomorphic converge on X and uniformly converging to uj on all of K. In particular, the uniformly to zero on D. The remaining functions u, p < j q, are uniformly approximated on L by functions v1, holomorphic on X. Observe that L is a holomorphically convex compact subset of X and that L' 1) U Yp) is a finite set of points outside of L. Then, fore > 0, it is easy to construct a function he, holomorphic on X, helL' = 0, and — LI < e on L. Therefore we may assume that the functions v3 vanish at the points of L' (multiply vj by he's). u1 is
94
1
Riemann domains
We define a new function v on K by setting v(x) := v1(x), wheneverx D, and := 0, x E K \ D. These functions are continuous on K and holomorphic on mt K. Again applying the Mergelyan—Bishop theorem the v can be approximated uniformly on K by functions holomorphic on X. Thus there exist sequences of functions holomorphic on X and uniformly converging on U .. U Yq to the limit function p q. If /2 E then we put
<j
C" —+ C",
:
+Zqtt'q.p
J
is a holomorphic mapping and. if/2 has been chosen sufficiently large, we conclude from above that j=1
=
det det
q
(J
<j
"+Up
p <j
J
0•
,...,q
Now we introduce the following entire functions
*p.j.v(Z)
j' converges compact uniformly in
Observe that the sequence
to the
function Therefore, for any positive t5 and any sufficiently large v there is a point zo E = 0, 1 q. llzo — all
j
f(x) :=
+
... +
+ zo.P+IwP+i.,L(x) +
... +
satisfies the following two properties
[we' =Oforalli, if v>> I and8 <<1, then lf(x)l
1.11 Riernann surfaces
95
Proof of Theorem 1.11.1. Take a nonvanishing holomorphic 1form w on X. Using Lemma 1.11.3 and Lemma 1.11.10 we find an exhausting sequence of subdoX, and holomorphic functions 1k: X —÷ C, mains (Dk)k€N of X, Dk Dk÷l wefI+"+fA such that = 0 for all closed piecewise & curves y on Dk, IfkI in Dk+l. Putf := 1k. 1 is holomorphic on X and, if y is a closed piecewise C1 curve in X. then we1 0. Finally, we recall that any open Riemann surface is holomorphically convex and that (9(X) separates points in X (ci [Fors 1981]). Hence Theorem 1.11.1 has been verified.
Remark 1.11.11. In general, there is the following result due toO. Forster (cf. [Fors — 1970]): Any ndimensional Stein man can be immersed 62) into The proof of this theorem relies on the above Theorem 1.11.1. For details the reader is asked to contact [Fors 1970]. While any open Riemann surface can be immersed into C, there is a concrete 20} which z2) E P(C2 ) : zo" + z12 + z22 does not admit an immersion into C2, i.e. X is not a Riemann domain over C2 (cf. [Fors 1970]). Thus the class of Riemann—Stein domains over C's, n > I, is a strict subclass of all Stein manifolds of dimension n. dimensional Stein manifold X := {(zo,
62) A holomorphic map 1: X —* Ck, X a complex manifold of dimension n, n k, is called an immersion if its Jacobian is of maximal rank. If n = k, then f is locally biholomorphic.
Chapter 2
Pseudoconvexity
In Chapter I domains of holomorphy were characterized with the help of properties of holomorphic functions (e.g. the holomorphic convexity). Now we want to give a characterization via more geometric objects (e.g. properties of the distance to the boundary).
2.1
Plurisubharmonic functions
Holomorphic convexity may be regarded as a generalization of standard geometric convexity. In the same spirit, subharmonic and plurisubharmonic functions may be considered as a generalization of convex functions. Plurisubharmonic functions are one of the most important tools of modem complex analysis. Therefore and for the reader's convenience, we decided to present an extended list of their properties (although not all of them will be explicitly used in the sequel).
If u: X —+ then for any Let (X, p) be a Riemann domain over a E X and C'1 we define a function Ua,f of one complex variable by the formula
(A €C: DefinItion 2.1.1. A function u: X —p is called plurisubharmonic (shortly if: ps/i) in X (U E • u is upper semicontinuous on X (u Ct(X)) 2), • for every a E X and is subharmonic in a neighborhood C'1 the function of zero: cf. [Rad 1937]. [HayKen 1970], [Ku 1991]. We say that a function u: X —+ IR.). is logarithmically plurisubharmonic (log
psh)iflogu 2)
Recall that for a function U: X —p
the following conditions are eqwvalent:
•i' E slim u(y) = u(x) for any x E X; • there exists a sequence C C(X. R) such that
u pomtwise on X.
97
2.1 Plurisubharmonic functions
The definitions extend in a natural way to Riemann regions 1)
u(X) C !},
E
As usual we put
IC
Remark 2.1.2. Directly from the theory of subharmonic functions (cf. [Via 19661) one gets the following properties of psh functions (Exercise). (a) For an upper semicontinuous function U: x —+ the following conditions are equivalent: (i) (ii)
U
E
IU=l
u(a) = (iii)
<
2ir —
0
dO,
J
< R;
30
f
<
0
< R;
IfJ=l 20
(iv)
f, jf
A —p (u o
is subharmonic; (vii) for any univalent open set U C X the function u o
=
+
(b)
(d) if C In particular, if
(e) If (f) If Ui tion 2.1.16).
and u, C
C
psh on p(U)
= ?SR(X).
R>0
f
(c)
is
(9(X).
\
u pointwise on X, then u E PS3€(X). [—oo, 0]), then u1, ,'PS.Y(X).
u locally uniformly in X, then u
and
UN E .'PSR(X), then max(ui
(g)(Liouvilletypetheorem)IfU E Even to complex manifolds: Let M
uN}
E
u
E
(cf. Proposi
<+oo,thenU
as const.
an ndimensional complex manifold with an atlas is said to be plurisubhannonk on M if for any I E I the function u o co7'is psh on Wi in the classical sense. 4) in particular, the local theory of psb functions on Ricmann domains reduces to domains in C". (Ui. coi)iEI (cf. § 1.1). A function u:
M —.
be
98
2 Pseudoconvexity
I —* IR be convex and increasing. (h) Let I C R be an open interval and let Then o u E .'PSR(X) for every u I). Consequently: If u E .'PSR(X), then e" E .'PSR(X) (in particular, any logpsh function is psh). If u E then 5'SR(X) for every p I. (1) If are logpsh, then + U2 is logpsh. U2 ProposItion 2.1.3 (Maximum principle). Let (X, p) be a Riemann domain and let U .'PSR(X). Ifu
< sup(limsupu(w):
E aD),
Z E D.
Pmof. Let X0 := {x E X: u(x) = u(a)}. Observe that X\
Xo
=
X: u(x)
{x
and therefore Xo is closed. Let xo E Xo. Applying the maximum principle (for subharmonic functions) to each of the functions with = 1, we conclude that lPx(xo) C X0. Thus X0 is open and therefore X = X0. 0
Letu: U —+
Ctm
bedefinedinaneighborhoodUofapointa E X. Wedefine
formal derivatives of u at a au
(p(a)),

aZ3
(d)a Observe that if f
If u: U —÷
(p(a)),
j=
n
1
has all real partial derivatives at p(a)).
(which exist if u o
az1
—(a) :=
o
•o
o
o
az1
••o
a, fi
E
= Dv.Of.
(9(X), then R is such that u 0
is twice IRdifferentiable at p(a), then we
define the Levi form of u at a
j.k=I
Notice that £(11p112)(a;
=
Zj Zk
for any a E X and
E
C". Observe that
Consequently, we have the following Observe that
=
n. Moreover, a function! E
It clear that
J.k =
= 8j.k (the Kronecker
&(X, C) is holomorphic 1ff
Oforj =
1
n.
99
2.1 Plurisubharmonic functions
Proposition 2.1.4. Let u
C2(X, R). Then
£u(a: fl > 0.
u E ?SR(X) Remark 2.1.5. Assume that I C2(I, IR). Then
c
R is an open interval, u E C2(X, I), and
E
n
=
£(ço o u)(a;
+ j=J
a
X,
E C".
'J
Notice that the above formula and Proposition 2.1.4 give a direct proof of Re
are of class C2.
mark 2.1.2(h) for the case where u and
Remark 2.1.6. Let (Y, q) be another Riemann domain over Ctm and let F E 0 ( Y. X). Then for u C2(X, R) we get o
= .Cu(F(b); F'(b)(ij)).
F)(b:
bE
E Ctm,
Y,
where
17(lfl
urn) ECtm
j=I denote the formal derivatives in Y).
F
Consequently, if u D'S.X(X) fl O(Y, X) —cf. Proposition 2.1.26.
&(X, IR), then u o
F
?SR(Y) for any
DefinItion 2.1.7. A function u C(X, R) is called strictly plurisubharmonic if for every domain Y X there exists an e > 0 such that the function u — eIIpfl2 is psh on Y.
Proposition 2.1.8. A function u .Cu(a;
is strictly psh iff
C2 (X,
> 0,
a E X,
E
= £u(a;
Pmof Fore > 0 put VE := u — Note that Fix a domain Y X and put
(observe that e
tion2.l.4, VE
= l}
a E Y,
e
> 0). Then £vf(a;
> 0,
aE
—
Y,
E
C". Hence, by Proposi
E
Fix ana be such that
Lu(a;
XandletYbearelativelycompactneighborhoodofa. Then
=
which finishes the proof.
+
>
>0
Lete >0
0,
U
The following proposition gives a very useful tool for constructing new psh functions.
2 Pseudoconvexity
I 00
Proposition 2.1.9. Let that
C X be open. let
Y
v E
U E .'PSJt°(X).
limsupv(x)
4
Assume
E aY.
Yax—. .
Put
max(v(x), u(x)},

u(x) :=
xE Y
xEX\Y.
Then u E
Proof It clear that ü a vector E C'7 with
Obviously ü is psh on X \ I = 1. and 0
Take a point a
e1(X).
ü(a) = u(a) <
i
i
— 2,rj0I
dO < — I
aY,
üat(re'9) dO
2irj0
and we can apply Remark 2.1.2(a).
U
We will use the following two notations: eZ z
:=
z=
w := (ZIWI
Leta = (al If
z
= (ZI
(Zi
E
W=
Zn).
E C's.
E C", r = (ri
d&P1(a. r) —*
6)
is
bounded from above and measurable 7),
then
we
define
P(u; a,
r: z).
2
(ft
Irje'oj_(zj_aj)12)
+r.
z = (zi I
J(u;a,r):=P(u;a;r;a)= (2r)" If u: IF,, (a, r) —f
A(u; a. r) :=
lfr=(r
(IrrI).. .
E
J
is bounded from above and measurable, then we define
J
P(a,r)
u
=
i
f u(a + r E"
r)wesimplywrite: P(u:a,r:z).J(u;a.r),A(u:a,r).
6 7)
e'9)
That is, the function 10.
x ... x 9 —+ u(a F r . e'°) is Lebesguc measurable.
2.1 Plurisubharmonic functions
101
be upper semiconRemark 2.1.10. Let D be a domain in C" and let 14: D tinuous. Put := {(z. r) D x (IR>0Y': P(z, r) C D}. (a) The functions
—+
c2
A(u;z.r)
are upper semicontinuous (in particular, measurable). Indeed, by Fatou's lemma we have:
J(u; z. r) =
tim sup
urn sup
f
u(z + r
tim sup
u(z + r
(2it)
(z,r)—.(zo,ro)
J
[O.2,rl" (z.r)—.(zo.rn)
I
f
(2ir)" = J(u; ro),
A(u; Z, r) =
lim sup
( u(z + r
lim sup (:.r)—+(Zo,ro)
(z.r)—+(zo.ro)
J
ir
I
Jr" JE"
urn sup
u(z + r
u(zo +
= A(u; zo, ro). (b)
a, r) =
2
. .
r1
I
JO
2
fri I

I
J(u; a, (TI
Tn))ti .
. .
...
t,,
Jo
I J(u; a, (Tin,...,
.
.
dr1
...
JO
PropositIon 2.1.11. Let D be a domain in C" and let u E rh'), 0 < r. r7
a E D, r' = n. Then
j(u;a,r')_<J(u;a,r")forr'_
2 Pseudoconvexity
102
Proof By Remark 2.1.10(b) it is enough to consider only the function J(u: a, •). First, we will prove that
,J(u; a, r') <J(u; The case n
=
I
a,
is well known. Hence
J(u(z', •, z"): a3, r) < j(u(z',
z"): aj. r,"), (z', a3, z") E P(a.
dD(a)). j
= 1, ..
.
,
n.
Consequently, using a finite induction, one can easily get the required inequality. By Fatou's lemma we have
u(a)<
IimJ(u;ar)<
I
f
r—.O
whichprovesthatJ(u;a.r)
Su(a), r+O
\ u(a)whenr \0.
0
2.1.12. Let u2 E .'PSJt9(X). Ifui = U2 Ax almost everywhere in X (cf Section I.2forthe definition of the measure Ax), then U2. ProposItion
Proof WemayassumethatX = DisadomaininC'1. Fixana ED. Sinceui =u2 everywhere, we get A(ui: a. r) = A(u2: a, r) for any 0 < r < d0(a). Hence, by Proposition 2.1.11, 0 = u2(a).
If D C is a domain and L: Aff(C", C'1)), then we define DL := we put UL
a o L: DL
an afline Cisomorphism (L —+ L1(D). Moreover, for any function u: D —+ is
+
Note that U E S'S.Y(D) if uj, E PropositIon
2.1.13. Let D C C" be a domain and let u: 1) —÷ Then the following conditions are equivalent:
semicontinuous.
(i) (ii)
u VrE(OR)n:
VOEDI
UL(Z) < P(UL;
a, r; z),
Z E P(a, r);
V,.E(OR,n:
(iii)
uL(a) <J(uL; a, (iv)
r)
VaED, 30
uL(a)
A(uL; a, r).
be upper
2.1 Plurisubharmonic functions
103
then the inequalities in (ii), (iii), and (iv) hold for such that P(a, r) C DL.
Moreover, every r = (rj
E
Proof
FixL e Aff(C", C??),a E DL,r = (r1,... P(a,r). Theresultiswellknownforn = Zn)
DL,afldZ = (Zi UL(W', zj, w")
P(uL(w', ., w");
r) 1.
C
In particular.
rj; Z1),
j=
n.
I
Hence, after finite induction, we get the required formula. The implication (ii) (iii) is evident. (iii) (iv) follows from Remark 2.1.10(b). (iv) (i): Fix a E D, E C's, = I. It is sufficient to show that
<
I u(a +
irr
JTE
forr > Osufficiently small. LetL E XE C. Now, by Proposition 2.1.11, forO <
u(a) = uL(a)
=
A(uL;a, (rj
urn
I
tim TI
= —i
(
=
I.
(ai an_I,
0
+
Exercise 2.1.14. Let (X. p) be a Riemann domain over C" and u E tPS,lt'(X). Prove
that 1
14(a)
<
in Sketch
f
Ax(Bx(a,r))
u dAx,
a
X,
0
px(a).
Bx(a.r)
u is subharmonic on X in the real sense.
of the proof We may assume that X = 18(a. r) C C", (p = id). Then
r))
=
I
JB(a.r) I
I /
u(a +
I
Jo JP(C"')JO >
I
,
jf o
P(C"')
.
=
where j,t denotes the standard measure in the complex projective space P(C" I)•
2 Pseudoconvexity
104
andu
PropositIon 2.1.15. If u E = 0. lar.
L'(X, bc); inparticu
—00, then u
Proof We may assume that (X, p) = (D. idD), where D is a domain in Suppose that there exists a point a E D such that u = —00 for any neighborhood = —oo for any z P(a, r). U of a. Let 2r := do(a). Observe that J'P(z.r) u Consequently, A(u: z, r) = —00, z E IP(a, r). Hence
u=
—oo
in P(a, r). Let
Do := {z ED: u = —ooinaneighborhoodofz}. We have proved that D0
0. The same method of proof shows that D0 is closed in
0
D. Thus D0 = D — contradiction.
Proposition 2.1.16. Ifafa,nilv
C PSR(X) is locally bounded from above,
then the function
a := (sup IE I
ispsh in X. In particular. max (u
UN }
E IPSJt'(X) for any
UN
denotes the upper regularization of v. v*(x) := urn
Here
.?S.X(X).
v(y). x E
x Proof The problem is local, so assume that (X, p) = (D, idD), where D is a domain
We use Proposition 2.1.13(u). Fix L, a C DL, and r = (ri We have n. Note that UL = dD, (a), j = I in
supP((u1)L; a, r: z) < P(uL; a. r; z), LE1
rn), 0 <Ti <
z e P(a, r).
(El
Since the function z —÷ P(uL; a, r; z) is continuous, we get UL(Z)
Proposition 2.1.17. if a sequence above, then the function
z E P(a, r).
0
is locally bounded front
C
U := (limsupuv)* v. ôc
ispsh on X. Notice that in general u*: X —* [oc. +ocl is upper semicontinuous on X. If v is locally Ct(X, R), v = bounded from above, then v: X —p and v = ç' E C(X. R), v :
2. I
Plunsubharmonic functions
105
Proof We may assume that (X, p) = (D, jdD), where D is a domain in C'1. We use Proposition 2.1.13(u). Fix L Aff(C'1, C'1), a DL, and r = (ri, .
0
Note
that UL =
.
. ,
Then
V++OO
z E
P(a,r).
%'—•
Hence, using the continuity of the function P(UL; a, r;
we get
uL(z)
[1
Psh functions may be considered as a generalization of holomorphic functions in the sense that log If is psh for any holomorphic function f. Consequently, the class where u is psh, extends the class of thin sets. It will of sets of the form turn out that these sets will behave similarly w.r.t. psh functions as thin sets do w.r.t. holomorphic ones. I
DefinItion 2.1.18. A set M C X is called pluripolar connected neighborhood ha and a function Va
if
any point a C M has a
:PSR( Ua) with Va
—00,
Mfl Ua
C
We say that a pluripolar set M C X is complete if any point a E M has a connected neighborhood Ua and a function Va E [PS3C(Ua) with Va (—00).
—00, MflUa =
By Proposition 2.1.15, if M is plunpolar, then Ax(M) = 0. It is clear that any thin set M C X is pluripolar. Remark 2.1.19. (a) We would like to mention the following two results characterizing complete pluripolar sets. A. Zeriahi [Zer 1989]: Let M be a pluripolar subset of a pseudoconvex 10) domain
E D\Mthereexists
DC C'1.
au such that u = —00 Ofl M and u(a) > —oo. Then M is complete pluripolar in D, i.e. there is a v C S'SJ((D) such that M = {z E D: v(z) = —oo}. J. Wiegerinck [Wie 19991: Let D C C be a domain and let S C D be a sequence
in D without accumulation points in D. Let f E \ S). Then M := {(z. w) C (D \ S) x C: w = f(z)} is complete pluripolar in D x C. (b) (Plurithin 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 [LevPol 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 welldefined). 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 nonempty 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 kdimensional Hausdorff measure (cf. § 2.12).
2 Pseudoconvexity
108
[Siu 1974]). The precise result is the following: M is a complex submanifold or if not, then any u E 9SJ((G \ M) extends to a psh function on G. For a sufficient criterion to extend not necessarily bounded psh function through a complex surface see [Kar 1998].
C 5'S3?(X) beasequence locally
Proposition 2.1.23 (Hartogs lemma). Let bounded from above. Assume that for some m C mit. v_p
Then for every compact subset K C X and for every e > 0, there exists a
such that
Proof WemayassumethatX = DisadomaininC". Takean€ >0. itissufficient to show that for every a D there exist 6(a) > 0 and v(a) such that u m + e in v(a). Fix a and 0 < R
<Jim sup sup A(uL,; z, R + 5)
urn sup sup
:EP(a..)
a, R)
Ri"
R)
—
RTh
Ri"
<m+e,
0
provided that 8 is sufficiently small.
Recall that smooth psh functions may be easily described by properties of their Levi forms (Proposition 2.1.4). Thus it is important to be able to approximate (at least locally) a given psh function by smooth psh functions. The required approximation may be given by the following procedure.
Let 4(Zi
Zn)
:=
.
. ..
'1'(z,),
z=
E C", Where 4' E
(zi
is such that:
.supp4' = • 4'(z) = 'l'(IzI), Z E
.1 4' dA2 =
C,
1.
E
e>
0.
109
2.1 Plurisubhaimonic functions
Notice that: ITh • E 'o • supp = P(e), •
=
• •
=
Z
=
(zI
Zn) E
1.
Let
:= (x
X: dx(x) > e},
>0.
For every function u E Lt(X, bc), define
I = j(u
ue(x) : =
The function X
is
—
p(y)) dAx(y)
JPx(x)
+ ew)c1(w)
o
x E X5.
called the eregularization of u. Observe that for a E
E lPx(a, dx(a) — e) we get
I
JPx(a)
=
j(u
+ ew)4(w)
o
Remark 2.1.24 (cf. [Sch 1966]). (a) Ut (b) If E e"(X), then Da,flu€ =
E
(t)
e > 0. for any al +
< k and e > 0
(k E
(c)If K := suppu
In for 0 < e < X, then suppue C C forO < e C < —÷ uniformly onXwhene —÷ Oforany (d)Ifu E then Daflue uniformly on X k. In particular, if u E al + when e —÷ 0 (for any a, fi). Lt(X) (1 < t < +oo) and K := suppu X, then ut E Lt(X) and (e) If u
particular,
E
forO <e < (f)
Moreover,
u when e —÷
0.
)e2 = (Ut,)tj on
fl
e u
pointwise
Pmof It is clear that
in X when e \ E
0.
(Remark 2.1.24(a)). By the second part of (t) we
get 1'I
= (2jr)" JO
p1
... Jo
J(u; a, e(r1
tn)ti .
. .
dt1 ... dry.
110
2 Pseudoconvexity
Consequently, by Proposition 2.1.11,
willapplyRemark2.1.2(a). Fixa E 2ir J —
= 2
+
o
j
f(u
Proposition 2.1.26. Let Then u o F
u. It remains to prove that is psh. We E C", = 1,and0
dO
+
o
f(u o
\
+ ew)
+ ew)D(w)
=
0
ue(a).
(Y, q) be a Riemann domain over Ctm
and let F
(9(Y, X).
any u E
Proof We may assume that u —cc. We already know that the result holds if U E C2(X) (Remark 2.1.6). Let denote the sregularization of u. Put F1(XE). Then o F and u o F E 0
Corollary 2.1.27. Let u: X —+ be upper semicontinuous. Then u ispsh on X ifffor any anah'tic disc ço: E —÷ X the function u o is subharmonic in E.
Lemma 2.1.28. Let D c C" be a domain and let u E logpsh:ffforanya E Candj E (1 n}lhefunction D
z
Iltf). Then U is
Ieozulu(z)
ispsh.
Proof We only need to prove that if v,,,, is psh (for any a and j), then log u is psh. By definition, we have to check that for any zo E D and e the function A log u(zo + is subharmonic (in the region where it is defined), equivalently (cf. [Via 1966]), we have to prove that for any wo C, the function
+
A
0. Put a := Let j be such that 1v0J(zo + A'). Thus ço is subharmonic provided Vaj is psh.
is subharmonic.
Proposition 2.1.29. (a) Any Cnorm q: C" —p (b) Let/i: C" —+ R÷ be such that
h(Az)=IAIh(z), Then
h ispsh
is logpsh.
is logpsh.
AEC, ZEC".
Then
=
0
Ill
2.1 Plurisubhannonic functions
Proof (a) Given arbitrary a,
E
C", the function
C?
A
—÷ q(a
+
is convex, and consequently subharmonic. Thus q E .'PSR (C"). = Now we can apply Lemma 2.1.28 because hand side is psh by Proposition 2.1.26. (b) follows from the proof of (a).
If u
Proposftion 2.1.30. (a)
U
—00, then Lu > 0 in X in the
u
E
and the right
distribution sense, i.e.
dAx(x)
u(x)Cço(x;
0,
E
R÷),
E
(b) If u E '(X, bc) is such that Lu > 0 in X in the distribution sense, then there exists afunction v E ?SJt°(X) such that u = v Axalmost 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 sregularization 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 eregularization 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
Asalmost everywhere on D.
Exploiting the above properties, one can simplify the original description of psh functions, namely:
PropositIon 2.1.31. Afunction u: X —÷ is psh if C X of a and a function (a) for any a E X there exist an open neighborhood V
E
such that u < v in (b) for every a E X and E
the function
is subharmonic in a neighborhood
of zero.
Remark 2.1.32. (1) There is a conjecture that the above proposition remains true without condition (a), i.e. u is psh if for every a E X and E C" the function is
subharmonic in a neighborhood of zero ft]
(2) Proposition 2.1.31 under the assumption that u is locally bounded from above was first proved by Lelong in [Lel 1945].
(3) The following general theorem is true; cf. [Ru 1989]. Let U C Rk, V C Rt be open, k, £ > 2, and let u: U x V —p
be such that
(a)for any (a. b) E U x V there exist an open neighborhood c C U x V of(a, b), such that u
= +oo is due to Avanissian LAva 1961].
The case r = I is due to Arsove EArs 1966]. (4) Notice that the separate subharmonicity (without any additional assumptions) does not imply the global subharmonicity; cf. [WieZei 1991]. Proof of Proposition 2.1.31. We use induction on n. The case n = 1 is trivial. Assume that the result is true for any Riemann domain over C"'. Let (X, p) be a Riemann domain over satisfy (a) and (b). The problem has a local and let u: X —+ character and therefore we may assume that (X, p) = (D, idD) where D = U x V is
a bounded domain in C" =
x C. We may also assume that u
v, where
v E L '(D). Observe that for any k E N the function Uk := max{u, —k) also satisfies u as k / +oo. Consequently, it suffices to prove that (a) and (b). Moreover, for any k E N. Thus we may assume that u —k in D, and finally, uk E taking u + k instead of u, we may assume that u 0. We have to prove that u is upper semicontinuous. We will need the following lemma.
Lemma 2.133. Let A C B U: A x B —* R+ be such that
C R'l be bounded Lebesgue measurable sets and let
2.1 Plurisubharmonic functions
(i)u(x, .)
I 13
is upper semiconrinuous on Bforanyx E A,
(ii)u(.,y) is measurable on Aforanyy E B. Define
ü(y) := I u(x,
y E B.
JA
Then
ü is Lebesgue measurable on B.
Proof Let Uk := min(u, k}, k E N. Observe that Uk satisfies (1) and (ii). Put
yEB,
kEN.
k/
+00. Thus, it suffices to prove that Uk is upper semicontinuous on B for any k. Take a yo B. Since Uk is upper bounded, the Fatou lemma gives:
By the monotone convergence theorem Uk / ü as
limsupük(y)
J
=
A
f
=
0
A
First we prove that u is locally bounded from above. Take (a. b) D and let r > 0 be such that b), 2r) D. By the Fubini theorem, L'(U) for A2a.a. w E V, say for w E B. w) Consequently, by the inductive assumption, u (., w) e PSR(U) for w e B. Notice that u(z, .) is subharmonic in V for any z U. By Lemma 2.1.33, the function B
is
I
irr Jp(,7r)
<
irr
u(z,
E ?(a, r). Consequently, for any
measurable for any
irr
I
w —÷
j
f
(jrr2)"
w)dA2(w) =
( (irr
f
)"
( (,rr—)" v(z,
f
Jrr
I
E
w)dA2(w)
u(z, w)dA2ii_2(z))dA2(w) v(z, w)dA2n_2(z))dA2(w)
P(E.r)
w).
JP((a,b),2r)
Now we prove that u is upper semicontinuous in D. Take (a, b) D and let ro > 0 be such that IPUa, b), ro)
D. Suppose that u(a. b) < M. Since u is locally bounded from above, the inductive assumption yields
2 Pseudoconvexity
114
that u(.. h) is psh. In particular, u(., b) is upper semicontinuous. Hence there exists 0 < ri
A(u(•,b);a,r1) =
1
2
u(z, •) is subharmonic, we get
A(u(z, .); b, r) =
I
irr
u(z, w)dA2(w) \ u(z, b) as r
Jp(b.r)
\ 0,
z E U.
that for each 0 < r < ro the function z —÷ A(u(z, .); b, r) is measurable. Moreover, since u is bounded on IP((a, b), ro), we easily conclude that the integral Recall
I
A(u(z, .); b, r)dA2,,_2(z)
JP(a.r)
is finite for any 0 < r
JP(a,ri)
(lrrI
Finally, for 0
irr2
and
f
P(b,r2)
E IP((a,
b), r), using the mean value theorem (and
the fact that u > 0), we get ,j)
(ir(ri — r) 2 —r
\JT(r2
—
u(z,
r) 2 1
JP(b,r2)u(z,
r2 — r
r)2 M, which shows that u <M in P((a, b), r) provided that r is sufficiently small.
Remark 2.1.34. Let c
be open and let u E
C
R). Then
41u((x + iy); (a + ib)) = .7(u((x, y); (a, b)) + Ru((x, y); (b,
—a)),
where R denotes the real Hessian IfU N
82
x
Q) k
U, Q = (Qi
QN)
RN.
0
2.1 Plurisubharmonic functions
ProposItion 2.1.35.
C C's,
13(x
(if v is convex on U
Then 1)
and let v: U —+
L.e: U be a domain in
U := U +
115
+ iy) := v(x),
Define
x + jy e U.
14)
Proof First consider the case where v is of class e2. By Remark 2.1.34 we get
4.Ci)(x + iy; a + ib) = .lev(x; a) + Jev(x; b), which, of course, implies the required result.
In the general case, assume that i3is psh and let of 13. Observe that + iR" = Hence UE C R". Moreover,
+ it) =
I
13(z
=
denote the sregularization for an open set +
+ it +
J E"
=
1(z
+
=
zE
tE
J E"
=
Hence,
UE —÷
E
IR.
\
v.
By the first part of the proof, u6 is convex in UE for any e > 0. Consequently, v is convex.
v (in R"):
Conversely, assume that v is convex and let v(x +
vE(x) := f
{x E U: B(x, e) C U},
x
JBU)
where 0 is a
function in R" Put Ue : UE + iR" C C", By the first part of the Note that 13. 0 for any e > 0. Consequently, 13 is psh in U.
\
13(x+iy) := v,.(x), x +iy E proof,
is psh in
PropositIon 2.1.36. Let D c C" be an ncircled do,nain and let u: D —f such that Zn) E D. U(ZI, . .. ,Zn) = u(IztI (ZI Put IDI
:= {azil
:=u(rI logD:={xERtt:eXED}, ü(rI
14)
Zn) ED) C
(zi
(ri
IDI.
ü(x):=u(eX),
x€logD.
rn),
Ifv E C2(U, R). then v isconvex 1ff Rv(x; Q) Oforanyx EU and Q ER". I.e. 9 E SUpp9 = B(1). fa(I) 9 = 1, and 9(x) = for any x = (xI e R". It is known that if v is convex in U, then is convex in and e \ v.
be
116
Foranyl=(ii
.
2 Pseudoconvexity
ik)withl
k). the intersection of D with the (n — k)dimensionalsubspace (Zj1 = 0: j Let uj denote the We will identify Di with an (n — k)circled open set in restriction of u to D,. is convex. liz the case where D C we gel: u E zffü (a) (b) In the general case, u E .'PSJt"(D) (i) ü is upper semicontinuous on IDI, ix — land! = (i1 (ii)foranvk =0 i1 < ... <
(iii)for any j =
1
n,
I
if x
{(r? then
x
IDI,
the function
is increasing.
Proof (a) First z E
and
assume that u
D and 9 E R'1.
Consequently, for
E
E C2(D, IR). Recall that u(z) = = £u(e10 z; e'0 . for any z we get: any z E D and 0
E
£u((JziI
? 0.
Hence .Cu(z:
£u(z;
0
One can easily check that 4Lu(z; E)
+
d.jrk
=
Z = (zi
Zn)
E D. r = (rj, ..
.
:= (IziI
,
In particular.
4LuV;
=
arjark
+
j=I
=
(e%)e_x1(al j,k=I
j=I
x€logD, On the other hand,
Rü(x:
(2= j,k=I
j=1
x€logD,
z) for any D, 9 E
2.1 Plurisubharmonic functions
117
Finally,
4Lu(ex;a+ib)=.u(x,e_%.a)+J?n(x,e_X.b),
xElogD,
which implies the required relation.
\
Now, let u be arbitrary and assume that u is psh. Let denote the sregularization is ncircled and for any z E and is psh and u). Observe that of u o W' we get ue(et8 z)
=
z
+ ew)cD(w)
JE"
u(e'0 z + ee'0 .
=
= I u(z +
=
J
Zfl)=UE(IZII of the proof, ü is convex in log(DE). Since
Conversely, assume that v :=
ü
\
UE(z).
Bythefirstpart ü, we conclude that ñ is convex.
is convex in G := log D. Let
denote the
standard eregularization of v:
XE
v(x JB(I)
is convex in
and
\
G€
:=
B(x,e) C G}
{x E G:
v). Define
:= {(Zi := VE(logIzlI
C't: (loglzil log
Ge).
(zi
Zn) E
D and uS \ u. By the first part of the proof, ut E Hence U
E
Then it is clear that (i) is satisfied. Observe
(b) First assume that u
that log D = log(D fl (Ce)"). Hence, by (a), condition (ii) is satisfied for k = 0. n — I, the function uj is psh on D,. Consequently, we can repeat Fork = 1
the above argument and so (ii) is satisfied for any k. To prove (iii) observe that the function
rI_i,A,rI+i is
welldefined, radial, and subharmonic. In particular, the function [0,
is increasing.
rj —÷ J(v; 0,
=
rI_i, rj. rI+i
2 Pseudoconvexity
118
Now, assume that (i), (ii), and (iii) are satisfied. Then obviously u is upper continuous on D. Moreover, by (a), u is psh on D fl and, more generally, each function is psh in D, 11 = (13 1k).
=
with
1.
We want to prove that
u(a) — 2,r j0 1
u(a +
dO
forsuflicientlysmallr >0. Wemayassumethatai•...•at =a,, =0 0 with 0 < £ < n — I. Let 0 < r < dD(a) be so small that ... e a Take! I Recall ispshin for(zt n). that Ui D, fl
I). Hence, using (iii), we get
u(a)< — I 27r Jo 2,r
/ j0
0)dO
+
+
dO
f2iT
0
=— 2ir /
The following results on psh functions will be needed in Chapters 3 and 4 (the proofs of a part of them are beyond the scope of our book).
Theorem 2.1.37 (Choquet's lemma). Let X be a metric space whose topology has I. Assume that the family (uj)JEJ is a countable basis. Let u: X —. locally bounded from above. Then there exists an at most countable subset of! such that
= (supuj)*. lEt
Moreover,
each u is lower semicontinuous, then U,
lElo
Proof We may assume that #1 2: bounded from above, put
S(v) := {(x, t) E X x IR: v(x) > t},
=
iEi
can be found so that
U.
For any function
T(v) := ((x, 1)
v:
X —+
X x R:
locally
v(x)
> t}.
Observe that S(v) is open if v is lower semicontinuous and T(v) is closed if v is upper semicontinuous. Moreover, T(v*) = S(v) 16)• Since v
observe that
is uppersemicontinuous, T(v) is closed. Obviously S(v) C T(v). It remains to C S(v)
2.1 Plurisubharmonic functions
119
For A c I define uA := SUPIEA u,. We want to find a countable set lo C I such that u0 = u. It is equivalent to the equality = T(u). Since S(uA) = we only need to prove that UEA S(u,),
U S(u,) = Usui EEl
IEIO
for a countable set Jo C I. Let } be a countable dense subset of S(uj). For any U E N take an i(i') E I such that Now, it suffices to put E Jo := {i(1), i(2). ... Assume that each u is lower semicontinuous. Consider the open covering (S(u,))jEJ of S(u,). By the Lindelöf theorem, it has a countable subcovering Then S(uj0) S(u,) and consequently, uj. U . . .
Proposition 2.1.38. Let (X, p) be a Riemann domain over C's. (a) Let (uj)jE, C ?S.X(X) be locally bowuled from above. Put u :=
u.
Then the set {x E X: u(x)
=
u1(xo).
In fact the set {x E X: u(x)
is
pluripolar—cf. Theorem 2.1.41.
Proof (a) We may assume that (X. p) = (D, id), where D C is a domain. By Choquet's lemma there exists a countable set Jo C I such that where = u0 Observe that (x E X: u(x)
u(z)
P(u; a, r; z).
z E P(a, r), 0 < r < r.
Hence
zEIP(ar), 0
P(P(u;a, r; .);a, r';z) = P(u:a, r: z).
ZE
r'), 0< r' < r < r.
Thus
P(u*; a, r; z)
In particular, J(u*: a. r) A(u*; a, r).
P(u; a, r, z),
z E IP'(a. r), 0 < r
J(u: a. r), 0 < t < r. Consequently, A(u; a. r) =
2 Pseudoconvexity
120
u and
b) Let Vk := SUP1,>k
(x
E
X: u(x)
E X: vR(x)
< v(x)}
0
and we apply (a).
The problem whether any pluripolar set can be described by one global psh (cf. [Let 1957]) function was open during many years and was finally solved by Josefson in 1978.
Theorem 2.1.39 (Josefson theorem; cf. [Jos 1978]; see also Proposition 4.1.11). If M C C" is pluripolar, then there exists a v v —00, such that MC Notice that a similar result is true for plunpolar subsets of an arbitrary Riemann domain over C", cf. [El 1980]. Proof. We will follow a simplification of the original proof given by J. Siciak in [Sic 19831.
Step 10: Suppose that for each a M there exist r(a) > 0 and Va Va 0 —oo,suchthaiMfllP(a,r(a)) C tç'(—oo). Thenthereexisisav E —00. such that M C v_1(—oo). v Proof of Step 10. By the Lindelöf theorem, there exists a sequence C M such that M C r(a1)). Put V3 := v01, = 1,2 Since for each j the set
j
vJ'(_oo) is of zero measure, there exists a point b IP(I) such that v1(b) > —oo for all j. We may assume that v3 0 on P(j) and —2', j E N 17) 1)efine v := v1. Then v v(b)> —1, and M C 0 5'SR(P(a, r)), Thus fix an a M. By definition there exist r > 0 and u —00, such that M fl IF(a, r) C u'(—oo). u We may assume that a = 0, r = 4, and u <0. Step 2°: (Cf. the theorem of Bremermann, Proposition 4.4.13.) There exists a sequence C (9(P(4)) such that IIft'IIP(2) 1, ii = 1, 2 and u = P(4), where (2.1.1) V
Proof of Step 2°. Define a Hartogs domain
G := ((z, w) E P(4) x C: wi < ft suffices to substitute v1 by a function of the form r,(vj — c3) with Cj := an appropnatc > 0.
Vj
and
2.1 Plurisubharmonic functions
121
It is well known that G is the domain of existence of a function f E (9(G) (cf. [Via 1966]; see also Corollary 2.2.15 and Theorem 2.5.7). Observe that
K := P(2) x E
G.
may assume that ft
I on K. The function f may be represented by a Hartogs (z. w) G,where fv (9(IP>(4)), V =01,2 and the series is locally normally convergent in G. In particular, by the Cauchy I on IP(2), v = 0, 1,2.... Obviously inequalities, u. Finally, since G is 18)• the domain of existence off, we easily conclude that u = 0 Since u —oo, we have 5UPP(l/4) > —oo. Hence there exists a point zo E with > —oo. Consequently, we find an infinite set C N and q N, 2, such that Ifp(zo)I > VE q We
series f(z,w) =
Define
gv(z)
+ zo),
:
z E IP(3),
E
V
Observe that
Step 3°: Foranyq,k,
V
>
1.
Jg1, IIP(
E Nwithi' > >
4q
VE
>8,
(2n+2)!,
(2.1.2)
1,
(2.1.3)
and for any g E f9(P(3)) wit/i IIgIIP(I)
Ig(O)I >
d :=
there exists a polynomial P
E P(l/2): Ig(z)I <
Proof of Step 3°. Put K := k'1,s
g(z) =
such that
<
I< {z
(2.1.4)
c (z
P(1/2): IP(z)I < Write Z
E P(3).
Let ü := O := ((z. w) P(4) x C: wi l'hen G is a Hartogs domain with G C G. Take 0 <(1 < 1 and put G0 := {(z, w) E P(4) x C: <9). Then, by the Hartogs lemma (Proposition 2.1.23). for any compact L n O9 there exists a such that "wi < 9 for (z. w) E L and v Thus the series is locally nonnally convergent in G to a function I E (9(G). Hence G = G and, therefore. = u.
2 Pseudoconvexity
122
By (2.1.3) we get
ae
IaaI < 1,
(2.1.5)
Let A := (Z4. fl [0. vK))", B := (Z÷ fl [0, s))". For any x =
Q(x,z) :=
C C, put
H(x.z) := Q(x,z)g(z) = fl€A
where
and au_p =0 ifa —
:=
The main part of the proof is to show that there exists an x = (XP)fiEA such that
La(X)=0, aEA\B, maxlxpl=1.
(2.1.6) (2.1.7)
>
L(x) := max
(2.1.8)
aEA
Assume for a moment that such an x is already constructed. Define
W(z) := crEA
P(z) := (l/L(x))W(z) = >2(La(X)/L(x))Za,
Z
E C".
aEA
By (2.1.6) P Moreover, by (2.1.6),
By the Cauchy inequalities, we get that IIPIIP(l)
lIP IIPU)
#B
=
.s"
Observe that by (2.1.5) and (2.1.7) < (KvY'. Now, let z such that Ig(z)I Then, by (2.1.6), (2.1.7), and (2.1.8), we get
IP(z)l <
<
z)I
+ IH(x, z) — W(z)l) + \A
=
+ 2" — 2"(I
(Kp)ne
/2(e_KI)
<
(2.1.2)
<
where (*) follows from the Bernoulli inequality. Indeed,
+
/6(e_Av13 +
<
t9)
E A.
I
—
E
P(1/2) be
2.1 Plurisubharmonic functions
123
Thus it remains to construct x (with (2.1.6), (2.1.7), (2.1.8)).
A =: Bo, A,1 := A \ {aI
as"}. A0
Write B = {a'
=
?. The system (of linear equations in the variables x)
La(x)=0,
CXEAI,
has less equations than variables. Consequently, there exists a solution x' = (x),)PEA = I for some B0. such that maxficA = 1. We may assume that Put B1
Now consider the system
Bo \
La(X)O, a€A2. =0. The
system has a solution x2 =
some
such that max$EA
I
for
E Bj.
Put B2 := B0 \
:=
Using this procedure, we get
B€_i :=Bo\{13'
s'1,suchthat
I
La(Xt)0,
x,— £
maxlxfl —1. i
flEA
We only need to show that
max{ILE(xt)I: a E A. Suppose (2.1.9) does not hold,
D(U. where U
=
a
(2.1.9)
A.
£
=
1
s".
(2.1.10)
c Awith#U=#V=:m> l,let =:
V) :=
and V = (SI
{y1
> e_I
1
i.e.
ILa(X')I
ForanysetsU,V
=
£
in the lexicographical order.
By (2.1.5)
I
(Ky)"
v2" <
+ I)!
e".
s"
20)
(2.1.11)
2 Pseudoconvexity
124
We claim that
D(A,, Bt)I >
£ =
0. Sn.
(2.1.12)
Observe that (2.1,12) finishes the proof of Step 3°. Indeed, using (2.1.11) with
£ = s", we get
+s"Ki'/2 < v(Ku)°, which implies that < 2(q + 1)v(KvY'; contradiction 21) We move to the proof of (2.1.12). For £ = 0 the matrix Faa_fl la.flEA is triangular. Hence, by (2.1.4). > ID(Ao, Bo)I =
<s". Put
Suppose that (2.1.12) is true for some £ with 0
if(a.$)
b
if(a, fi) =
—
'I
Then
= 0,
22)
r
flE
Consequently, detlbufilaEAf ,ficHt = 0. On the other hand
det[baflj =
Bt) ±
Be+i).
Now, using (2.1.10). we have
JD(At÷i,
Be+i)I
>
2: ID(At,
which finishes the proof of Step 3°.
Step 4°: Main pmof First we apply Step 3° to and k E N with v E such that with d(v, k) := (2n + 2)!. We get a polynomial E
K
=
k'1
>
such
that <
I
(z E P(1/2):
C {z
(2.1.13)
P(1/2): (2.1.14)
21
Becauseq > 1+! Since
1.
= 1, we have
=
=0.
2.1 Plurisubharmonic functions
Put
— Zo). Then
and by (2.1.13) we have231
E
>
2
IIQu.kIIpi) 2
125
(2.1.15) < .jdR.k)
IIPV,kIIP(5/4) <
(2.1.16)
Moreover, by (2.1.14), we get
<
(z E P(1/4):
C (Z E IP(1/4). gv(z — zo)I
C {z E P(1/4).
—
< e_Kt}
zo)I < <
C {z E P(1/4):
(2.1.17)
4r, r > Observe that by (2.1.16) a locally bounded family of logpsh functions. Put
1.
Thus
is
vk
4r on P(r), r >
Then v is logpsh and Uk We claim that
1.
(2.1.18)
VZEMnP(I/4): v(z)
Vr
Indeed, suppose that there exist T < 0 such that (2.1.18) does not hold. Then there exist k and E M fl P(1/4) such that k'2 > (2n + 2)!, T > and —÷ zo such that 2 eT+I. By Proposition 2.1.38 there exists a sequence > eT/2 for IL 2 Consequently, = —÷ v(zo). Hence for any ji 2 there exists an infinite set C E such that A
> 23)
>
nqv
=
t'j12
vE
We will apply the following elementary inequality:
Ip(:)I
pE
Indeed, fix p, r, and z as above. Let C
:= IIPIIP(r). Consider the function
0< Al c
v(A) := log Observe that v is subharmonic in
z C C". Izi ? r > 0.
K(lzl/r) \ {0) and v(A) 0 for Xl = lzi/r. Thus. v extends to a function subharmonic in K(IzI/r). zl/r. In particular, v(l) <0, which gives OforalIlM
Moreover, < Hence, by the maximum principle, v(A) the required inequality.
126
Pseudoconvexity
2
?
which implies
YE Hence, by (2.1.17), 24) contradiction. and finally u(zo) > e_ Now, we are going to prove that
? 3/10.
(2.1.19)
k —*
< 3/10 on 2(1). Then, by the Hartogs lemma Suppose that urn (cf. Proposition 2.1.23) there exists a ko such that
forzElF(7/8), Thus, urn
<3/9 forz E 2(7/8),
sup
k > k0.
li—.
Using once again the Hartogs lemma, we conclude that
< (3/8)d(v.k) forz
E
2(3/4),
V
V > Lk)(k), k > ko.
E
Consequently, by (2.1.15),
[(3/8)(4/3)I"
<
.k)
for v E
V>
k > /co;
contradiction. Let be as in (2.1.19). By (2.1.18) (with T := > (2n+2)!suchthat <
on
M
— I) we get
fllF(1/4)
E N
with
(2.1.20)
and lim
i1—*+cx)
> 3/10.
v
(2.1.21)
Define v := I
Obviously, by (2.1.20), v
=
—oo
on Mn 2(1/4). Recall that on 2(r), r>
log v which
implies that v
v
—
log 4r
=
is
<4r.
a global psh function
log
and
1, we get
E N,
By (2.1.21),
the series consists of negative psh functions on P(r).
2.1 Plunsubharmonic functions
127
C X be
Proposition 2.1.40. Let (X, p) be a Riemann domain over M1 is j E N. Then M
pluripolar,
Proof We may assume that (X. p) = (D, jdD), where D is a domain in C". By —co, such Josefson's theorem, for each j E N there exists a v1 E
c
that
(—00). Now we argue as in Step 1° of the proof of Josefson's theorem. D
As already mentioned, now we are going to list a few results on psh functions without proofs.
Theorem 2.1.41 (Bedford—Taylor theorem; cf. [Kli 1991], Th. 4.7.6). (a)Assume that u,. Then afamily(u,),E, C 5'SR(X) is Iocallyboundedfivm above. Putu := X: u(x) < u*(x)} the set {x C PSJ€(X) is locally bounded from above.
(b) Assume that a sequence (u
E X:
Putu
Proof (a) is beyond the scope of the book. We show that (a) (b): Let vj :=
u
< v(x)}.
E X:
E
j=l
0
It remains to use (a) and Proposition 2.1.40. Theorem 2.1.42 (cf. [BedTay 1976]). There exists a (nonlinear) operator ?SJ?(X) fl
bc)
—÷
u E
is a nonnegative Borel measure on X}
26)
such that: U E
(b)
bc)
if ,9'SJf(X) fl
u,,
\
fl L°°(X. bc),
u E
then
DefinitIon 2.1.43. The operator the Mon ge—Ampere operator on X.
Definition 2.1.44. A function u E Y
Xandanyfunctionv
As usual Notice that a function u
in the weak sense.
fl
bc)
a —*
is called
called niaxirnal if for any open set uoni)Y,thenv u in Y.
bc) denotes the space of all locally bounded (measurable) functions on X. belongs to L°°(X. bc) if it is locally bounded from below.
2 Pseudoconvexity
128
bc). Then u
Theorem 2.145 (cf. [Ku 199fl, Th. 4.4.2). Let u E = 0.
is maximal
The following class of open sets with 'good' psh exhaustion functions will be useful in the sequel.
Definition 2.1.46. We say that an open set f function u E 1L) such that
X is hyperconve.x if there exists a
t<0. c
Let
be open and let N c
Define
:=sup{u: u The
that 0 < that
(*)
u
u<
is called the plurisubha rmonic measure of N relative to c. Observe I and (cf. Proposition 2.1.16). It is clear hN.c2 = 0 on mt N.
Proposition 2.1.47. Let D C be a bounded domain. Then D is hyperconvex if there exists a continuous function u E S'SR(D, IlL) with (*). Proof (Cf. [Zah 1974]). Let U: D —* [—oo.0) be a psh function with (*). We will construct a continuous psh function vo: D —+ (—oc, 0) with (*). Fix a ball and v 0 in K := B(a. r) C D and let u := hiD. Recall that v
3(a, r). By the Oka theorem (cf. [VIa 19661), for any point b E dB(a. r) we get v(a + t(b — a)) = 0. Thus v = 0 on K. v(b) = The maximum principle implies that v(z) < for any z E D. Fix a t0 > 0 such that u < —zo on K and put uo (l/to)u + 1. Then uo E v. Consequently, the uo < 1, and uo 0 on K. Hence uo hK,D function := v — I satisfies (*). We will show that v is continuous (then satisfies 1
all the required conditions).
Fora E (0,1) let Da := {z E D: v(z)
Da /
D
when a /
= 0 on K. Observe that
I. Moreover,
V
On
Indeed, define h
v}
J
on Da
onD\Da.
1
Then urn sup
D(Z)
a
E
Hence, by Proposition 2.1.9, h e 5'SR(D). Obviously h < I on D and h = 0 on K. s h < v on Da. Thus h hK,D < v. In particular,
2.2 Pseudoconvexity
129
Fix a point zo E D. We want to prove that v is continuous at zo. Let fl(a) := maxb v. Observe thata <$(a) < I. In particular, fl(cr) —+ I when a —÷ 1. Fix
> 0 and a E (0. 1) such that
E Da and fl/a — I < where fi psh functions defined in a neighborhood of Da, C D, such that VE \ v on when e \ 0 (Proposition 2.1.25). One can find e > 0 such that for w := yE E :PS,7€() fl we have w> v on on w K, and w fi + on Consequently,  Let (VF)O
(w —
q)/fl
<
h<.0 on Da.
Hence,
Now, by the continuity of w, there exists a neighborhood U of Zo, U C Da, such that 0 forz E U. Iw(z) — w(zo)I s r?forz E U. Finally, Iv(z) — v(zo)I <
Theorem 2.1.48 (cf. [Kli 1991], Prop. 4.5.12). If is compact, then
c C" is hyperconvex and K C
\ K) = The measure
is
o.
called the equilibrium measure for K.
Proposition 2.1.49 ([Ku 1991], Lemma 5.6.2). Let 2 C C" be hyperconvex, let K
=
(P) = 0. Then
be compact, and let P C K be such that
2.2 Pseudoconvexity Let (X, p) be a Riemann domain over C". Let
For a compact set
C
K C X define
:= {x
E
X:
u(x) 5 supu}. K
Definition 2.2.1. We say that (X, p) is pseudoconvex if for any compact set K C X the set is relatively compact. The definition extends in a natural way to Riemann regions over C".
Remark 2.2.2. (a) Let (X, p) be a Riemann region over C". Then (X, p) is pseudoconvex if for each connected component C of X the domain (C. Plc) Indeed, if (X, p) is pseudoconvex and K C C is compact, then
c
c.
IS
pseudoconvex.
2 Pseudoconvexity
130
Conversely, if any connected component is pseudoconvex and K C X is compact, then K is contained in a finite union of connected components, say K c C1 U. LJCN.
N.Then
c
x.
u. . . u
c
ko(X). Consequently, if (X, p) is holomorphically convex, then (X. p) is pseudoconvex. (b)
Recall (Theorem 1.10.4) that (K, p) isa domain of holomorphy if (9(X) separates points in X and (X, p) is holomorphically convex; cf. Theorem 2.5.7. (c) If X is hyperconvex (Definition 2.1.46), then is pseudoconvex. be such that u(x) < t) for {x E Indeed, let u any z < 0. Take a compact K c c and let o < 0 be such that K C Then
c
a
In view of (b) it is natural to ask whether any pseudoconvex Riemann domain is a domain of holomorphy. This is the famous Levi Problem. The problem, formulated by E. E. Levi in 1910, was solved by Oka only in 1942 for n = 2 and in 1954 by Oka, Norguet, and Bremermann for n > 2. A solution of the Levi problem. via dtechniques, using the theory of holomorphic functions with tempered growth, will be presented in § 2.5.
First we like to discuss pseudoconvex open sets X with smooth boundaries. If u is a function defined in a neighborhood of a point a X, then we put
az1
where
——(a)
denote the formal derivatives; cf. § 2.1.
Let c X be open. Recall that ac is smooth of class Ck (or C(C.smoolh) aS a poini a E ac if there exist an open neighborhood U of a and a function u E Ck(U, IR) such that
c2flU={x€U:u(x)<0}, U\c2={xEU:u(x)>.o},
(')
()
for u is called a local definingfunctionfor means that u is real analytic.
at a; here k E N U {oo) U {co}. where U E
Observethatifu &c(U,R)satisfies(*)andgradu(x) (), i.e. u is a local defining function. We say that is ek.sniooth if is &smooth at any point a E
u satisfies
U,then
2.2 Pseudoconvexity
131
Put
=oJ.
eCu:
XE
Zj
at x; notice that the
is called the complex tangent space to
The space condition
=0 The means that I grad u(x) in the sense of the Hermitian scalar product in is independent of u (Cf. the proposition below). Observe that if definition of n = 1, then (0}.
Proposition 2.2.3. Let c
X be open, a
ac2, and let U be an open neighborhood
of a.
(a) Let uj, u2 E Ck(U, R) be two local defining functions (k E N U {c'o}). Then U2 = vui with v E C"'(U,IR>o). 27) C' (U, (b) The space T5c(ac) is independent of the local defining function u E
un (c)Letui. U2 E C"(U, R), k> 2, be two local defining functions with u2 = R>0) i.c as in (a). Then
where V E
£u2(x;
= v(x)Cu1(x;
x
U fl ac,
E
where £ denotes the Levi form (cf § 2.1).
(d) (Cf. [KraPar 1981]) Letk
E
NU{oo}, k >2, and let cz be Crsmooth. Define —pcz(x),
xE
0,
PX\c(X),
p) with respect to the to the boundary of and a Euclidean norm: cf § 1.1). Then there exist an open neighborhood U of
(recall that pg denotes the
jr: U —* dQ such that:
•S
•
E
II grad
• ,r(x)
S(x) II = 1, x E df2 (consequently, is a local defining function). and is the only point from ac that realizes the Bx(x,
distance to ac, i.e. Bx(x, Iö(x)I) ñ x U.
= {,r(x)) and
= IIp(x) —
(e) The above result does not hold fork = 1. (f) Let k E N U (oo). Then the following conditions are equivalent: 27) The proof will show that also the following result is true: If u2 e Ck(U, R) are such 1(U. R÷). that ti1 is a local defining function andu2 satisfies (*) and (). then u2 = vu withy E
2 Pseudoconvexity
132
is eks,,woth.
(i)
(ii) (here exists afunction U E CJC(X, R) such that
=
{x E
X:
u(x)
={x E X: u(x) >
for u
is called a global defining function for
Proof (a) Using a local C"diffeomorphism we reduce the problem to the case where U = (—1, 1)" with N = 2n, u1(x) = ul(x',xN) = xN, and > Oforany x E U. Then U2(X) = v(x)xN with
(b) Let
(c) Take x E U fl
XNO. be two local defining functions. Then, by (a),
u2 E C8 (U.
where v e C"1(U
XN # 0
XN),
v(x', xN)
=
R>0). In view of the proof of (a), we get gradu2(x) =
and
Then
a
dzi1
—
j.k=I dv
a
=y
duj
d2ui
+ v(x)Luj(x;
=
= v(x)Lui(x:E) (the latter equality follows from the fact that if =0).29) = =0, then 28>
29)
E C' and
E C are such that
2 the fonnula may be obtained by direct differentiation. Notice that the case k 2 3 can be so'ved by direct differentiation: Notice that in the case k
It
j.k=I
8
tiu

It
=
dui
dv
+
I 8v

= 0.
2.2 Pseudoconvexity
(d) In fact, the problem has situation to the case where:
= M := {(t, f(t)):
•
(—r, r)N c RN
•
f:
t
a
133
local real character. Thus we can easily reduce the
N + I = 2n, i\N(r) :=
C
E
:= (—r', r') C R,
—*
with f(O) = 0,
is a
0, j =
N, We
:= (sgn(y—f(x))dist((x,v),M),(x,y) E AN÷,(s) C want to prove that for small 0 < s
such that: •öE
=
'pN):
—p
IR),
• 1 grad8(x, v)ll = I for any (x, v) E Mfl • 82(x y) = (f(ço(x, y)) — y)2 +
—
x3)2
and co(x, y) is the only
pOint in AN(r) with the above property (for every (x, y) E Recall that
82(x y) = minj(f(t)
Fk(X,y,t) := (f(t) We
—
—
v)2
+
—
y)±() +fk
X)2:t
Xk,
axk
(s)).
k
E
=
N.
I
want to findi = ço(x, y) such that Fk(x, y, t) = 0.
Observe that the functions F1 and I
k=
N.
I
A(r') x
x
FN are of class Jk,t=I
(t)
N
Consequently, by the implicit mapping theorem, there exist small r1 > 0 and a Ck_ I mapping'p: AN+l(5I) —f suchthatt = 'p(x, y) istheonlysolutionof(t)in (TI). Now lets > 0 (s <SI) be so small that for any (x, y) E I (SI) X I (s)
the distance of the point (x, y) to M is realized by a point from x A(r')) fl M (consequently, by the point ('p(x, y), f(ço(x, y)))). In particular, since
(f('p(x, y)) — dXk
y) —
= 0,
k=
I
we get
8(x, y) = (f('p(x, v)) — y)(l +
y))))'.
(x, y) E
N,
134
2 Pseudoconvexity
Hence, using (i), we obtain (Exercise)
—(x, y) =
(I —(x, y) = dy
which shows that 8
1/2'
2
dXk
k=1
N,
) —1 2
(i +
y))) )
12' /
R) and that 1 grad 8fl = I on M
E
fl
(e) It suffices to consider M := {(t, It t E (—1, 1)) (0 < e << 1) and 8(0, y) for small y > 0; Exercise. (f) Assume that (i) is satisfied. Let U and 8 be as in (d). Put := 8 (recall that uo E Ck(U, IR) is a local defining function). Take an e > 0 so small that K := (x EU: )uo(x)I
t<—e,x(0)=0,x(t)=efort>e,andx'(t)>OforItI<e.Define —C.
xE
uE
0
lR) is a global defining function.
DefinItion 2.2.4. Let
X be an open set. We say that
is strongly pseudoconvex
at a point a E acz if there exist an open neighborhood U of a and a local defining R) such that function U E
Lu(x:
> 0,
x EU
n
E
Observe that the definition is independent of u (cf. Proposition 2.2.3(c)). We say that is strongly pseudoconvex if is strongly pseudoconvex at any
pointa E dQ. The notion of the strong pseudoconvexity in invariant under local biholomorphic mappings (cf. Remark 2.1.6). We will see (Proposition 2.2.25(a)) that any strongly pseudoconvex open set is hyperconvex (and, consequently, pseudoconvex). Obviously, if n = 1, then any C2smooth open set X is strongly pseudoconvex.
ProposItion 2.2.5. Let X be open. (a) Assume that ac is C2smooth at a E Let U be an open neighborhood of a and let u E e2(U, R) be strictly psh with () and (). Then u satisfies (v'). In particular. u is a local defining function.
2.2 Pseudoconvexity
135
and let u E C"(U, IR), k 2 2, be a (b) Let U be an open neighborhood of local defining function with (u). Then there exists a c > 0 such that for the function — 1) we have: u,, := > 0,
x
E
E
In particular, u,, is strictly psh in a neighborhood of (notice that u,, is a local function). (c) For k 2, the following conditions are equivalent:
(i)
and strongly pseudoconvex:
is
(ii) there exist an open neighborhood U of ac and a strictly psh function u Ck(U, R) with (*) and (). is strongly pseudoconvex for any boundary point a (d) (Narasimhan)
ac
there exist an open neighborhood and a biholomorphic mapping F: U —÷ F(U) is strongly convex at each point fivm C" such that the open set G := F(U fl In
C" is strongly pseudoconvex.
any strongly convex open set
Recall that an open set G C" is said to be strongly convex at a point 1' E 8G if there exist an open neighborhood V of b and a local defining function v C2(V, R) for G such that
3(v(z; Q) >
E V fl
0,
Q
E
(recall that ,7( denotes the real Hessian; cf. § 2.1), where TZR(aG) denotes the real (2n — 1)dimensional tangent space to at z. It is clear that the definition is independent of v (cf. the proof of Proposition 2.2.3).
Proof (a) If uj C2(U'. R), U' C U, is a local defining function, then u = v E C'(U'. the proof of Proposition 2.2.3(a)). Obviously, grad u(x) = v(x) grad u 1(x), x E U' fl In particular, grad u(x) 0 provided v(x) 0. Suppose that v(b) = 0 for some b U' fl ac2. Consequently, grad v(b) = 0. Then we get 82u 
a
/
azJ
'
8v
8uj
8Zj
dZk
(b)=—(ui—)(b)+—(b)—(b)+v(b) 8Zk'
a
 (b) aZJaZk
= aZJ
(if(p E C',
E C
and
=
(b) It is clear that for any c >
=
0, then
0
the function
=
0); contradiction.
is a local defining function.
Moreover, one can easily check that n
azk
2 ,
xEd,
2 Pseudoconvexity
136
Let K :=
E
= I.
x C'1:
<0}.
Observe that K is compact and
Kfl Now we only need to take a c > 0 such that
K) K) (c) follows from (a) and (b). (d) First observe that if G is strongly convex at each point from M := F(U fl
then analogously as in (b) one can show that for any b E M there exist an open neighborhood V C F(U) and a defining function v E C2(V, R) such that
Rv(z;Q)>0, Then
v is
ZEVflM,
strictly psh in a neighborhood of b. Consequently,
is strongly pseudo
convex at
and let Conversely, assume that c is strongly pseudoconvex. Fix an a E be a local strictly psh defining function (use (c) and (a)). The U E e2(U. 1k), a E U, problem has a local character and therefore we may assume that C's, a = 0, and 0). Write gradu(0) = (1,0
u(z) = 2 Re(zi + B(z)) + Lu(0; z) + where B(z) :=
I
"
j.k=1
Zj
and define
F: C'1 —* C".
F(z) :=(zI +B(z),z2
Zn).
Observe that
• F(0) =
0,
• F is biholomorphic in a neighborhood Uo of 0, •uo = 2Re + Lu(0; w) + o(flwII2). is strongly convex at each point of F(U 0 In particular, G := F(U 0 any small neighborhood U C Uo of 0.
for D
We say that u: X —÷ is an exhaustion function for X if for any I E R the X: u(x) < t} is relatively compact.
set {x
_______
2.2 Pseudoconvexity
137
Proposition 2.2.6. Assume that (X, p) is a holomorphically convex Riemann domain over C's.
(a) Let K C X be compact and let U be an open neighborhood of
Then
there exists a strict!)' psh real analytic exhaustion function u: X —+ IR such that
uOon X \ U. (b) if U: X —+ IR is a strictly psh real analytic exhaustion function (as in (a)), then there exists a sequence tk / +00 such that tt u(lx E X: grad u(x) = k = 1,2 In particular; jfwe put
Xk:={xEX:u(x)
Proof (a) There exists a sequence of holomorphically convex compact sets (K1 such that K1 = *0(X) K1 C mt K1; cf. the proof j 1, and X = ofRemarkl.1O.3. l,suchthatU1 C U,K1 CU1 C 1. Fix a j > I. Any point x E j \ Uj does not belong to K1. Thus IlK,. We may assume E (9(X) such that there exists a function > I for be a neighborhood ofx such that that > I > IlfIlK1. Let is compact, there exists a finite number of points xI Since Kj÷2 \ y tk(j) U I such that KJ+2 \ U Vxk(J). Define fjt := where £(j) is so big that k(j)
k(j)
<
E
If1
Define
> j,
\
xE
oo k(j)
:=
—1
j=1 Observe that the series converges locally uniformly in X. It is clear that v E 9SJt"(X), v is an exhaustion function. To prove that v is real analytic we proceed as follows. Consider the Riemann domain (X, j5) with X := X and
n. Observe
that f
function X x X
E
(9(X) (x, y) —÷
f €(9(X). In particular, for any f
E
(9(X) the
f(x)f(y) is holomorphic. Define s'p: X x X —÷ 00 k(j)
p(x, y) := —1 + J=I v=I
C.
2 Pseudoconvexity
138
The series converges locally uniformly in X x X and therefore q E (9(X x X). Since v(x) = x E X, we conclude that v is real analytic. where e > 0 is so small that u <0 on K. It is clear Now we put u := v + that u satisfies all required conditions. (b) We only need the following aaxiliary lemma (which is a direct consequence of the Sard theorem; cf. [Fed 1969], Th. 3.4.3).
Lemma. Assume that (X, p) is Riemann domain over C" and let u E &"(X, Then the set u((x X: grad u(x) = O}) C R is of Lebesgue measure zero.
IR).
Recall that pseudoconvex domains were defined via properties of psh functions. Now we are going to describe the pseudoconvexity in terms of properties of the boundary
ProposItion 2.2.7. Let (X, p) be a Riensann domain over C. —
log
Then the function
is subharrnonic on X.
Proof First suppose that (X. p) = (D, idD), where D is a domain in C. We may assume that D ç C. Then
_logdo(z)=supf log
1
z€D.
Recall that — logdD is continuous. Thus — logd0 is subharmonic. In the general case fix an a E X and observe that it suffices to prove that there exist a connected neighborhood U C X of a and a domain D C C such that p(U) C D
anddx = dDopOn U(then —logdx = (—logdD)opon U andtherefore, —logdx is subharmonic on U). Define
U := P(a, dx(a)/3),
U P(x),
D
p(Y).
XEU
Observe that
(we have
=
dy on U. Moreover, by Corollary 1.1.6(b), the set Y is univalent 0 E IP(x2) for any xI, x2 E U). Hence dy = dD 0
In the following lemma we collect a few conditions which assure that — log dx,q
is psh on X (where q is a norm on C"). The lemma will be useful in the proofs of Theorems 2.2.9 and 2.9.2.
(X. p) —+ (Y, q) be a morphism of Riemann domains such Lemma 2.2.8. Let that (Y. q) is pseudoconvex. Let q: C" R÷ be a complex norm. Then each of the conditions below implies that — log dx.q E —+ X, where (a) For any sequence of holomorphic mappings CC 1, the following implication is true: is a neighborhood of E, X;
2.2 Pseudoconvexity
(b) (X, p) ispseudoconvex;
139
—
(c) There exists a neighborhood U of dx 30) such that —log
q E .'PS3t(U flX);
o f)(t,.) E
(d) For any continuous mappings f: [0, 1] x E —* X such that 0(E) for any t E [0. 1], the following implication is true:
f(([o, 1) x E) U ({1} x aE)) C X —
:
f((0, 1] x E) C X;
(e) For any continuous mapping f: [0. 1] x E —÷ X for which the mapping o f extends to a holomorphic mapping g: D x E C", where D C C is a
neighborhood of[0, 1J, the following implication is true:
f([0, 1) xE) cx
f({l} xE) c aXorfUl} xE) C
;
X.
(f)Foranydomain T = Tr,p 31), every biholomorphicinapping 1: T —÷ f(T) C X extends to a holomorphic mapping f: E" —÷ X: (g) (Y, q) is a domain of holomorphy (cf Remark 2.2.2(b)) and for any domain T = Tr,p and a biholomorphic mapping f: T —* 1(T) C X such that of extends to a biholomorphic mapping 1: E" —p E") C Y, there exists a holomorphic
mappingf: E" —p Xwithf= fonT. Observe that
in conditions (d)
and (e) the function
.) must be holomorphic on
f(t.
Eforanyt E [0,1). Proof To simplify notation. let (*) denote the condition '— log dx., E :PSJe(X)'. Put moreover u := q• (*): Suppose that (*) does not hold. (a) Then (cf. Remark 2.1.2(a)) there exist: a point a E X, a vector v E C" with q(v)= u(a),anumber0 < < I,andafunctionh1 harmonic inr1Eandcontinuous —Iogu
0,
o
r,E
where xi(A) :=
+
E. Let
A
r1E be such that a =
A1
—loguox,(At) =h2(X1).
Cf. § 1.5. 31)
Recall that(cf. Example 1.8.7) 1',,,
((zi
Zn) E
U {(Zi
forO < r 32)
:=
C":
1z11
Zn) E C":
:= TL), where < r,
j
=
n — I,
I I
n
znl <
— 1, p < Iz11 <
140
2
Pseudoconvexity
LetO
R
X(A) := XL(rA) = Let
+ Arv). A E
E (9(RE) be such that h3 = Re
uox U
Let b
h2(rA), A E RE,
X(AO) and let w E
> IPI
(l/r)E,
Define p
Ao := Aj/r.
exp(—p3). Then
onaE, onE,
(2.2.1)
(2.2.2)
X(AO) = p(Ao).
(2.2.3)
with q(w) = u(b) be such that
+ tw))
0:
(2.2.4)
cf. Proposition 1.1.9. Define
4'(r,A):= poX(A)+f
p(A) p(Ao)
w= p(a)+Arv+t
p(A) p(Ao)
w,
x RE.
=
x and 'P(:, Ao) = p(b) + rw. By (2.2.2) and (2.2.3) we have X(A)) rlp(A)l
q(4'O. A)
S := ([0, 1) x E) U ([0, 1] x
=: Si US2
such that
(t,X)€G.
q('lI(t,A)—pox(A)) < In particular, we define
4(:, A) :=
A)),
(f, A) E U.
where U(A) Bx.q(X(A)). Observe that (by the identity principle) any point A°) E G has a neighborhood W C G such that
(I)(t, A) = (pIu(Ao))'('P(A, t)),
(t, A) E W.
Consequently, 4) is continuous. Moreover, 4)(r,) E 0(E) for any t e [0, 1). Ob+ 1w), t E [0, 1). Since viously (by the identity principle) 4)(:, A0) = X, condition (2.2.4) implies that there exists a sequence [0, 1) /1 = such that u(4)(tk, A0)) —÷ 0. u o 4'(tk,) const on E 33), and infE u o Becawic
u(4(fk.
—4 0 and
U
1flh4(S2) U > 0.
2.2 Pseudoconvexity
141
1. We may assume that Ak —p u o 4)(fk, Ak) for some Ak E E, k := 1. Notice that u(xk) —f 0. Ak), k Now take 1. Then we get a contradiction with (a). .), i'
E. Let
(*): Suppose that (*) does not hold. We keep all the previous notations. we have Observe that, by the maximum principle, for any function a E (b)
sup a = supa o 4) = sup supa o cD(t,.) = sup supa o aE
E
S1
supa. L
where L := 4)(S2). Hence 4)(S) C X (because (X, p) is pseudoconvex). On the other hand u < u(xk) —÷ 0; contradiction. (*): Suppose that (*) does not hold. We keep all the above notations. (c) ; First, similarly as above, the maximum principle implies that q(4)(S)) C Y. with L := *p(cD(S2)). Since (Y, q) is pseudoconvex, we conclude that 4)0 y: [0, 1) —+ X, In particular, we may assume that co(xk) —p Y. Let where y: [0, 1) —÷ S1 denotes the curve obtained by joining the segments
k = 1.2
[(tk, Ak). (tk+i. Ak÷I)i,
k > 1. It is clear that y extends 1. Thus xk = so that y(tk) = (1k, Ak), k continuously to [0. 1] (we put y(l) (1, to)). Hence, by Remark 1.5.4, E
ax.
Recall that, by Proposition 1.5.8 =q'
4):[0,1]x E —+ X
.
1)xE extends continuously to a mapping =", 
Obviously,bycontinuity,cD = 4)onSand p ocD = '1'.
Notice that (D((l, so)) = Let U be a neighborhood of
such that
—
log u E fPSR(U fl X). By continuity
(of 4)), thereexistsane > Osuchthat4)((l —e,1)x
CU. Letk0
Then, applying the be such that (tk, Ak) E (1 — e, 1) x + eE) for all k > maximum principle to the subharmonic function E A —p — logu 0 4)(lk, + eX) with k 2 ko, we conclude that u(4)(tk, + eA)) = u(xk) for any A E and k ko. E such that there exist e > 0 and Ico such that Let Eo denote the set of all E and k > ko (observe that 4)(1, + eE) C + eA)) u(xk) for any A =40
a X). We have proved that
Eo. Obviously, E0 is open. The above argument
shows that Eo is relatively closed in
contradiction (because 4)(l, dE)
q) =
f :=
4)
E and hence 4)(l, E) C
dx;
X).
(*): Suppose that (*) does not hold. We apply the above argument to
(d) (Y.
E. Thus E0 =
= 4)(l, dE)
p up to the moment when 4) is defined. Then, taking (C's, id) and we get a contradiction.
With 10, 11 x E as the topological space and 111 x E as the singular set.
2 Pseudoconvexity
142
(e)
(*): Suppose that (*) does not hold. We argue as above. Observe that the
function of = 'P extends holomorphically to C x RE. Thus we get a contradiction with (e). (f) (*): Suppose that (*) does not hold. We argue as in the first part of the proof. Observe that the vectors v and w must be linearly independent. Suppose they are not. LetXo := Xo —÷ Cbegivenbytherelationp(x) =
p(a)+po(x)v,x E Xo. Then(Xo, po)isaRiemannregionoverC. Clearly,
'x >
OIl d(x.p).q = u on X0 36)• Moreover, x: (l/r)E —÷ Xo and (cf. (2.2.1)). By Proposition 2.2.7 we conclude that — log d(X0,,,0) E S.X(X0). Hence o x > on E. In particular, d(X0p0)(b) > Ip(Ao)I = u(b). On the other hand, property (2.2.4) implies that = d(x.p).q(b) = u(b); contradiction. Fix V1 Vn_2 E C" such that the vectors VI Vn_2, w, v are linearly independent and define
+
%(A) := p(a) + A1v1 + ... + An_2vn_2 + An_i A
=
0. t, A,,) =
Observe that 'Po is injective and
A,,). Moreover,
+ IA,,_ilu(x(A,,)).
IAiIq(vi) + . .. +
q('Po(A) — p0 X(An))
x RE.
E
Let
G0 := {A E C": q(%(A) — P0 x
x (E x aE)) c G0.
x Put
(plu(A)Y'('Po(A)),
A E G0.
By the same method as for 4), one can easily check that
is continuous. Moreover,
0, :, A,,) = CD(t, A,,). Since p ° 4>o = 'Po, we see that 4)o is injective. Consequently, Go —÷ 4)o(Go) C X is biholomorphic. Fix an e > 0 such that
(eE)"2 Take
x
((1 +e)E) x {l —e<
an arbitrary I — <0 < I and let
<1 +e} C Go.
E (O.e] be such that
x (9E) x (1 + B(X.p).q(xo.r)flXo = P(xo.po)(xo,r),xo E X0.
C Go.
9/(1 + e) and
2.2 Pseudoconvexity
143
A(A) (1 + (eAt,..., EAn_2, (1 + T —÷ X. Noticethatf: T —+ f(T)isbiholomorphic. Then,by (f), f extends to a biholomorphic mapping f: —+ f(E") C X. This means that
Define T
f
there exists a holomorphic mapping
4>0: (eE)"2 x (1 +e)E2 —÷ X such that = 4>o on G1 U G2. In particular, the mapping 4>: [0, 1] x E —÷ X defined by 4>(r, := 4>o(O 0. t. A,) gives a continuous extension of 4>; contradiction (because u(4>o(0 0. tj, Ak)) —* 0). (*): Suppose that (*) does not hold. We argue as above. Observe that (g) the mapping r := p o f: T —+ Y is locally biholomorphic. By Proposition 1.9.9, r extends to a locally biholomorphic mapping f: E" —÷ (recall that E" is the envelope of holomorphy of T). Note that
I
q of =qoçoof = pof = 'I'ooA on T. Hence q o = 'Po o A on E" and, therefore. f is injective. Consequently, by (g), f extends to a biholomorphic mapping f: E" —÷ f(E") C X and we conclude the proof as in the previous step.
U
Theorem 2.2.9. Let (X. p) be a Riemann domain over C". Then the following conditions are equivalent: (i)
for any compact K C X the set
(ii)
(X, p) is pseudoconvex;
is compact 37);
for any Cnorm 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 Cnorm q: C" —÷ R+ such that the function — log dx.q on X (cf § 1.1); (iii)
is
there exists an exhaustion function u E .'PS3f(X) fl C(X);
(vi)
(vii) there exists an exhaustion function u E (viii) there exists a strictly psh exhaustion function u E
(X).
Proof (vii)
(ii)
I (I)
(ii)
(iii)
(iv)
(v)
(vi)
I (viii) To simplify notation we put
:=
fl C°°(X).
(i)
psh
144
2
The implications (i)
Pseudoconvexity
(vii)
(ii), (vi)
;
:
(i) are elemen
(ii), and (viii)
The implication (ii) (iii) follows from Lemma 2.2.8(b). The implication (iv) follows from Lemma 1.1.13. The implication (iv) (v) is a direct consequence of Lemma 1.1.12 and Proposition 2.1.16. ',
(iii)
:
(v)
:
.
(vi):
Step 10: Fix a point ao X. For 0 < e < px(ao) let Lie be the connected componentoftheset{x E X: dx(x) > e)thatcontainsao. By Remark 1.1.14(f), for any r E IR the set
:= {x E
<
is the inner Eucidean distance on pIus). We will prove that for any 0 < e << I there exists a strictly psh function E such that for any t R the set
is relatively compact in X, where
(x
is relatively compact in X.
Fix 0 <e << I and let ,j := e/3. Then C {x
P(x, z7) C
Put w(x) := 7u,,2(aO, x) and define
:= f w(p'(p(x) + =
f
w(y)c1(PW —
where ct' is a regularization function in C" (as in § 2.1) and {x E
fl
< t} C {x E U2,,: P(x,
xE It is clear that
01 C
E
X,
t ER.
:= + where C > 0. has the same properties. It remains to prove that there exists a C > 0 such that is strictly psh on It
Observe that the function suffices to show that
2
<
x E (Jr,
E
C",
'1
Zn)
fc 'L'di\2 = 1.
=
...
'1'
C°°(C,R÷), W(z) = 'V(lzI), supp'l' =
2.2 Pseudoconvexity
145
where Co is an absolute constant (then we put C := 2con2/zi). We have
_!
= dZk
+
J
L4, k = I,..., n.
x dzk
Recall that
x") =
lw(x') — w(x")j
x'. x" E lP(x, ii). x E
— p(x")II.
(cf. Remark 1.1.14). Hence a2
(x)
—
aZJaZk
=
! lim!
j
(w(p;' (p(x) +
E" dZk
(p(x) +
+ tei)) —
XEUe, j,k1
dz
E
which finishes the proof of the required result. It is clear that, taking cq (with c > 0) instead of q, we may assume that dx, q (aO) > 1. For v I let denotes the connected component of the set {x X: dx, q (x) > l/v} that contains ao. Observe that for any v 2: 1 there exists an 0 such that C IJ(. Consequently, for any v 2: 1 there exists a strictly psh function vv such that (x0, v N, such that if
:= {x
:= {x E Y1,+ii,_: Vv÷2(X) <
Vv+2(X) < c1,},
then
+ l},
UKvx yEN
Now we construct inductively functions > on \ L,1, = I,..., v — 1, and
Weputui :=
v2.
Supposethatu1
\ Let
such that
on
arealreadyconstructed. Itsufficesto
> v—I onLL\Lt,_I,uy(x)> v—2
E
on
fl
=
and Uy = Uy_I on be defined on by the formula c°L
:=max{—Iogdx,q—log(v—l),vt÷i
= {x
Then >v—
I
on
\
and I
> 0 such that 0}. In particular, there exists a on Now it suffices to put > X
x €
\
2 Pseudoconvexity
146
Finally we put u := exhaustion function. (viii): Let (vi)
ut, and we easily conclude that u is the required
,'
v€Z,
0 for u —1. Let 0 on X. Since u is be a strictly psh function such that u as continuous, u locally uniformly on X. For each u E Z let e(u) > 0 be such that and u < UE(V) < u + I on For simplicity let := Ue(v), cc t' Z. Let x: IR —* [0, I] be an increasing function such that Taking u + const instead of u, we may assume that
e\
uE E
• x(t) = Ofort <—3/2, • x(t) = I fort > 2, • x'(:) >0, x"(t) > Ofor frI <3/2. Define I
vEZ.
LI Hence
•
ifx then + 2—i' > u(x) + 2—u = 2. ± 2— u) = forx in_an open neighborhood of Indeed,ifx E
Hence + 2 — u) = 0. is psh in an open neighborhood of Indeed, if x 2— u< u(x) + I + 2—u 1. Hence, by Remark 2.1.2(h), near —2.
•
• V1 is strictly psh in an open neighborhood of
\
then
I>
U1,(x)
+ 2— v > u(x) + 2— v
\ —1. Thus
then
+
+ 2—i') is psh Indeed, if x E C's,
0,
then
>0 U = x"(U1,(x)+2—u)i grad (recall that x' > > Oin(—3/2,3/2)). We pass to the main part of the proof. We show that there exists a sequence C [I. +oo) such that for each v N the function := cjV1 + ... + We proceed by induction over v. is psh in an open neighborhood of The case v = I is obvious (we may take Cl := 1). Suppose that are already constructed. It is clear that for any 1 is psh in an open neighborhood Thus the function W1,÷1 = + we have to chose such that is psh in an open neighborhood of i\ This is possible because is strictly psh on \ = W Observe that on X. Moreover, {x X: W(x) < v} C X for any v u we get = N. Indeed, if x 2V, then for ci V1(x) + ... + c1, = CI+ V. 112. To get a strictly psh exhaustion function it suffices to take W + D
________ 2.2 Pseudoconvexity
147
Remark 2.2.10. Let (Y, q) be a Riemann domain over and let X be an open subset of Y. Observe that Theorem 2.2.9 remains true for the Riemann region (X, qlx). Corollary 2.2.11. Let (X, p) be a Riemann domain over C". Asswne that X = UVEN
where XL. is a pseudoconvex open subset of X with X is pseudoconvex.
0
Proof dx1, /
Corollary 2.2.12. Let (X, p) be a Riemann domain over C". Assume that Y = mt flLEN X'., where pseudoconver.
is a pseudoconvex open subset of X, V E N.
Then Y is
Proof dy = Corollary 2.2.13. Let (X1, Pj) be a pseudoconvex Riemann domain over C", N. Then X1 x
Proof dx,x...XXN(xI
j
= 1,
x XN 1S pseudoconvex.
xw)=min{dx,(xj):j=l
0
NJ.
Corollary 2.2.14. By Proposition 2.2.7, any Riemann domain over C ispseudoconvex. Corollary 22.15. Let (X, p) be a pseudoconvex Riemann domain over C" and let u be psh on X. Put Y := (x E X: u(x)
Proof First assume that u is additionally continuous. Let K C Y be a compact set and let e > 0 be such that u —e on K. Then
C {x E X: u(x) <—e} fl Hence £J'SJt!(Y) C {x
E X: u(x) < —e} fl
Now, let u be arbitrary. Let denote the eregularization of u. By the first part of the proof (applied to the function — log dx + loge) the set
—logdx(x)+loge<0}
/
is pseudoconvex. Moreover, for each e. the set YE := {x E uc (x) <0} is pseudoY when s \ 0. Now, it remains to use Corollary 2.2.11. convex. Observe that
0
Corollary 2.2.16. Let (X. p) be a pseudoconvex Riernann domain over C" and let Y C X be an open set such thatfor any point a E d Y there exists an open neighborhood Ua such that Y fl Ua ispseudoconvex. Then Y ispseudoconvex; cf Theorem
Proof Let u
be
a continuous psh exhaustion function on X.
Put X,, := lx
E
X: u(s) < v},v €N. Then Xv ispseudoconvex (Corollary2.2.15)and Xv / X. By Corollary 2.2.11 it suffices to show that Y fl X,, is pseudoconvex for each v c N. Fix
avE NandletZ :=
Takeana E
Ifa E Y,thenbyCorollary2.2.12,if
2 Pseudoconvexity
148
:= iPx(a, r) C Y, then Zfl Ua = fl (Ia is pseudoconvex. If a E dY, then by our assumptions, there exists an open neighborhood (Ia such that Y n Ua is pseudoconvex. Hence, by Corollary 2.2.12, Zfl Ua = Y n X, n U0 is pseudoconvex. Observe that for any a E dZ there exists an open neighborhood Va C (Ia of a zn V0. Hence there exists a compact set K C Z such that x = such that — Iogdz E PSR(Z \ K). Take an N > 0 such that N > —logdz on K and put v := max{— logdz, N}. Then v is a continuous psh exhaustion function on Z (recall that Z is relatively compact in X). 0 we take
Corollary 2.2.17. Let (X. p) be a pseudoconvex Riemann domain over C0 and let M be a pure (n — 1)dimensional analytic subset of X. Then X \ M is pseudoconvex Riemann domain. Proof For any a E Mthere exist 0 < r dx(a) and a function f0 E O(18(a, r)), 0, such that M (cf. [Chi 19891, § 2.8). By Remark 1.7.3(g) r) = f0
13(a, r) \ M is a domain of existence and, therefore, pseudoconvex. Now, we apply 0 Corollary 2.2.16.
Corollary 2.2.18. Let (X. p) be a pseudoconvex Riemann domain over C C" be a complex kdimensional affine subspace and let T: H —* C' be an affine isomorphism. Put V := p1(H), q := To ply. Then (Y,q) ispseudoconvex. Proof If u is a continuous psh exhaustion function for X, then uiy is a continuous
0
psh exhaustion function for V. Using the same method of proof, one can easily show the following property.
Corollary 2.2.19. Let (X, p), (V. q) be Riemann domains over C" and Ctm, respectively, and let Z C X x V be a pseudoconve.x domain. Then for any yo E V the fiber Zv() := {x E X: (x, yo) E Z} is apseudoconvex open subset of X. Corollary 2.2.20. Let (X, p), (V. q) be Riemann domains over C" and Ctm, respectivelv and let f: X —* V be holomorphic. Assume that X is pseudoconvex. Then for any pseudoconvex domain Z C Y the region f'(Z) is also pseudoconvex.
Proof Put
:=
Takeacompact K C f2. Then
c It suffices to show that any V E
E f(K)
C
we have: ?SJt(Z
cc Z,
x.
cn
—+ xo. Then for
Take v
v>
v> 1, and, therefore, f(xo) E Z.
Proposition 2.2.21. A domain D C C" is pseudoconvex if the function
D x C" is psh on Dx C".
—÷
1.
Hence
0
2.2 Pseudoconvexity
149
is trivial. Assume that D is pseudoconvex. By Theorem 2.2.9(iv), it suffices to show that
Proof The implication
D
xC",
(2.2.5)
:= (0,0, 1) E C" x C" xC.
where G C Dx C" xC is a pseudoconvex open set and Define
xC
ED x
D x
E
C".
:= f'(D);
note that g(D x C") C G. Since D x C" x C is pseudoconvex, Proposition 2.2.20 implies that G is pseudoconvex. It remains to check (2.2.5): Put G
=
sup(r >
c
0: z + K(r) =sup{r
D}
= sup{r >0: {z) x
x
K(r) c
G}
0 Let D C C" be a balanced domain (i.e. ED C D; cf. Remark 1.9.6(e)). Then there exists a uniquely determined function h : C" —p R÷, with
h(Az)=IAIh(z),
XE C .z E C",
that D = {z E C" : h(z) < l}; observe that h is upper semicontinuous. The function h = hD is called the Minkowski function of D. such
Proposition 2.2.22. (a) Let D C C" be a balanced domain and let h = be its Minkowskifunction. Then D is pseudoconvex Wh E log h E (C") (cf Proposition 2.1.29(b)). (b) Let G C D x Ck be a Hartogs domain with kdimensional 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 nonempty balanced domain (cf Remark 1.9.8(e)). Let H(z, w) = HG(Z. w) := h0.(w), (z. w) E Dx Ck, wherehG. is the Minkowski
function ofthefiber
(notice that H is upper semicontinuous). Then G is pseudoD x C"). In particular, in the special case, let
convex
D is pseudoconvex and log H E
G = ((z. w)
E
D x C": h(w)
where D C C"_' is a domain, u: D —÷ is upper semicontinuous, and h: Ck 0, is an upper setnicontinuous function with h(Aw) = AIh(w), R÷, h A 40) Then G is pseudoconvex if D C, w E C" h PSR(C"), andu E 5'SR(D). Proof (a) If log h E that
then D is pseudoconvex by Corollary 2.2.15. Observe
= l/h(U,
E
C".
11G(Z. w) =
Observe that in the case k = I. h(u')
Ju'I. we get a classical complete Hartogs domain.
150
2 Pseudoconvexity
Consequently, if D is pseudoconvex, then logh is psh by Proposition 2.2.21. (b) If D is pseudoconvex and log H is psh, then G is pseudoconvex by Corollary 2.2.15. Assume that G is pseudoconvex. Then D = (z (z. 0) E G} is pseudoconvex by Corollary 2.2.18. Moreover, 0)
(z, w) E D x Ck.
= 1/H(z, w),
0
Hence, by Proposition 2.2.21, log H is psh. So far, pseudoconvex domains were characterized by the the function — log dx. In the case of smooth open subsets c more, namely:
plurisubharmonicity of X we can say even
Proposition 2.2.23. Let (X, p) be a Riemann domain over C", let X be a C2smooth open set. Then (c2, pIe) is pseudoconvex (iffor any local defining function u E C2(V, R) we have:
£u(x;
> 0,
x E V fl
(Levi condition)
E
Notice that the Levi condition is independent of u (Proposition 2.2.3(c)). Proof Supposethat Since therefore
E
is pseudoconvex and let S and U be as in Proposition 2.2.3(d). we conclude that —log(—8) .'PSJt'(U fl and
£(— log(—8))(x;
—
> 0,
j.k=I (+) Fix a point xo E a vector a vector E C" such that
E
For any x E Q near
=
one can find
0
and E(x) —p when x —p xo. By (+) we conclude that .tS(x; by continuity (recall that S E (32(U)), we get £S(xo; 0.
0. Hence,
Conversely, assume that the Levi condition is satisfied. It suffices to prove that is psh near (cf. Corollary 2.2.16). Let 8, U, and be as in Proposition 2.2.3(e) (note that the Levi condition is satisfied for 8). Suppose that — log
<0 for some x
U fl
and E E C".
2.2 Pseudoconvexity
fl Bx(x, px(x)/2),
ir(x) E
Put
:=
151
:= p(xo) — p(x) (note that
=
Write
IogpQ(s(p(x) + Ce)) =
+ Re(AC + BC2)
+ 211C12 + o(1C12)
ICI<< 1,
(++)
and define
s(p(x) +
+
Observethat = Indeed, by (++), we have:
id << 1.
+ BC2)).
= 0. Moreover,S(11(C))
—8(*(C))
+ BC2)Il
—
+
>
ICI << I.
BC2)I
= lIliiIIexp(Ad + BC2)I(exp(e1C12)
— 1)
>0,
0 < IdI << 1:
contradiction. Hence
0=
= C
:= *'(O) E
(i.e.
0
Moreover, in view of the Taylor formula for 5 o
=Re(a
we get
0>
+
at 0
+°(k1).
')
?'
a recent paper (cf. [ZaiZam 2000]) D. Zaitsev and G. Zampieri present a characterization of pseudoconvexity for domains whose boundaries satisfy a weaker condition than to belong to C2. Their boundaries are allowed to have edges. Such a domain Q C" (or X, (X, p) a Riemann domain over C'1) is pseudoconvex
if: (a) the Levi form is positive semidefinite outside of a set M C Hausdorif measure zero and
with (2n — 2)
(b) the tangent cone (in the sense of Whitney) at any edge point outside M is convex.
2 Pseudoconvexity
152
Proposition 2.2.25. Let X be strongly pseudoconvex. (a) If is smooth (k > 2), then there exist an open neighborhood U of and a strictly ps/i defining Jluwrion u E R). In particular, any strongly pseudoconvex open set is hype rconvex. (b) For any open neighborhood U of c there exists a strongly pseudoconvex C U. C
Proof (a) Let Uo be a neighborhood of
and let uo E ho be a strictly psh defining function (cf. Proposition 2.2.3(c)). Without loss of generality, we may assume that for a sufficiently small e > 0 we have
K :={x EU0:
U0.
Il') be such that x(t) = —r for t Let x E —e, x(O) = 0, x'(t) > 0 for t > —e, and x"(t) > 0 for any (cf. the proof of Proposition 2.2.3(f)). Define U := U U0, —C,
xE E
V
a a'
:= {x
e"(U, IR) is a local defining function, ui is psh, and UI (Jo: uo(x) > —e).
Let u2 be a hR such that
strictly psh exhaustion function for
{x EU: Let1, E
< —€/2}
CL :={x E
I]) be such that irfr = U3(X) :=
I
1
is
strictly psh on
(cf. Proposition 2.2.9). Take
u2(x)
r}.
inaneighborhoodofL. Define xE
10,
Then for any c > 0 the function u = UI + CU3 is strictly psh in a neighborhood of L (because u = Ut +cu2 andu2 is strictly psh) and in (because u = and Uo \ supp c V). Moreover, it is clear that for small c > 0 the function u is strictly psh in a neighborhood of Land < Oon Thus, forO0 such that U. Let E be such that 0 — us m on (cf. Proposition 2.1.25). Put L42 := u1 + '7211p112 > 0); U2 is a strictly psh function on Taking > 0 sufficiently small, we may assume that Th.
e/2
2.3 The Kiselman minimum principle
Remark 2.2.26 (Cf. [For 1976]). For any there exist: G
153
strongly pseudoconvex domain
•N> •a
strongly convex domain D c C"', • an open neighborhood U of G, • a holomorphic mapping F: U —÷ C" such that: • F maps biholomorphically U onto a complex submanifold of CN, • F(G) C
•F(U\G)cC"\D, • F(U) intersects
transversally.
2.3 The Kiselman minimum principle We say that a domain G c is a tube domain if G + = G. Then we have the following necessary shape of pseudoconvex tube domains. Theorem 2.3.1. Any pseudoconve.x lube domain G C C" is of the form G = G1+i R", where G1 is a convex subdomain of R". Proof We may assume that G = to prove that G1 is convex.
G1
+ iR" for a certain domain
c R".
We have
Fix two points aij,ai E G1. Since G1 is connected, there is a continuous curve [0, 1] —÷ G1, y(s) = a•1 for s = 0. I. By S1 we denote the segment Lao, aT] y(r). Put S := {r E [0, II: ST C G1}. It is clear that 0 E S and that with a1 S is an open subset of [0, 1]. Take a ro E [0, 1] such that there exists a sequence c S such that = to. Observe that the function —logdist(., 8G) is plurisubharmonic. Since it is independent of Im z, it is a convex function (cf.
y:
Proposition 2.1.35). Therefore, if x E
then
> c> 0,
dist(x,dGi) > where c is independent off. Hence,
C Gi.
Since [0. 1] is connected, it follows
thatS=[0,l];inpasticular,Sj=[ao.u1JCG1.
0
The main goal of this section is to prove a minimum principle for plurisubharmonic function which has been found by C. Kiselman [Kis 1978]. The proof we will present here is taken from [Kis 1993J. The precise formulation of the result is as follows.
Theorem 2.3.2. Let (X, p) be a Riemann dotnain over C" k and let G be a pseudoconvex subdomain of X x Ck such that for each x E X, the fiber
:={z€Ck: (x,z)€G}
2 Pseudoconvexity
154
is a nonempty (connected) tube domain (i.e. G is a Hartogs domain over X with be such that nonempty co,mected tubular fibers). Let U E
u(x, z) = u(x, z + it),
(x, z) e G, t E
Then the function
g: X—*
g(x):=inf{u(x,z):
x
is plurisubharmonic.
are tubular but not necessarily connected (they must consist of convex components), then the function g is not defined on X but on a Remark. (a) If the fibers
Riemann domain over X. For more details see [Kis 19781. x Ct. Then {0} + iRk may be thought as a closed subgroup of (b) Let G = +). Therefore, one is led to study psh functions on certain Lie groups invariant under a closed subgroup. For a minimum principle in that context see [Boud 1983] and [Loe 1985]. As an immediate consequence of Theorem 2.3.2 we derive the following k x Corollary 2.3.3. Let G C w x Ct C be a pseudoconve.x domain such that each fiber
z' E w, is a nonempty tube domain. Then w is pseudoconvex.
Pmof Observe that the following function U: G —÷ u(z)
= u(z'. z") :=
max{— log dist(z, aG), O} + IIz'112.
z
=
(z'. z") E G,
is plurisubharmonic and independent of Im z". Therefore, the function g: co —÷ isaplurisubharmonicexhaustionfuncfiononw4D, g(z') := inf{u(z', z"): z" i.e. w is pseudoconvex. 0
Remark 2.3.4. Observe that the projection of a pseudoconvex domain is, in general, not pseudoconvex. For example (cf. [Kis 1978]) put
G := {(zI,z2,z3) EC3:
Z2Z31 < l}.
Obviously. G is a pseudoconvex balanced domain in C3. Then
In: C2 x C —+ C2 is the projection onto the first factor. From Theorem 2.2.9 it is clear that n1(G) is not pseudoconvex. Other examples may be found in [Kas where
1980]. [Pfl 1978]. In fact, the situation is much more delicate, namely, it turns out that any domain in , n > 2, is the projection of a pseudoconvex domain in xC (cf. [Shc 1983]).
The proof of Theorem 2.3.2 will be given along a series of lemmas. Some of them will be also of independent interest. Let us start with a result from linear algebra (without giving its proof). IIz'112.
2.3 The Kisehnan minimum principle
155
Lemma 23.5. Let L be a positive semidefinite Hermitian (it x ,:)matrix. Then there exists a Hermitian (n x n)matrLx M with LML = L. We say that such a matrix M is a Hermitian quasiinverse of L.
Lemma 23.6. Let F: C"
R.
F(z) := j=I
i.j=l
bounded from below, where L := <,,<,, is a positive semidefinile Hermizian matrix and b := (b1 E C". If M is a Hermitian quasiinverse of L, then be
LMb' = b'
and
F(z)> —bMb' = F(_(Mb*)r), z E C".
42)
Observe that LML = L implies that LMIj = 1,, j n, where denotes the jth 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 quasiinverse of
L(z, w) := (
 (z. w))
1
.
(z, w) E G.
Then
> bMb' 42)
b'denotesthetransposed(n x
on G,
:=b'.
fl C2(G).
2 Pseudoconvexity
156
where b
(b1
Pmof Fix a point
. :=
( 8zdwi
.
E G. Since u E
d2u
): G
C'1.
it follows that
+ for
all a =
E
C'1.
Thus, in virtue of Lemma 2.3.6, putting a
we obtain
= Now,letU C Cbeanopenset,andlety: U y(z)) = 0,
(z, v(z)) E G,
C" bea&functionsuchthat
j
1
n, z E U.
where u and G are as in Lemma 2.3.7.
Define g: U —b R, g(z) := u(z, y(z)). Differentiation of g with respect to z leads to
= y(z)) +
=
Y(Z))Yjz(Z)
y(z)).
Here we used the assumption on the function y. The equality implies that the function Xz
is still of class (5>1 on U.
Differentiation with respect to
=
yields
+
y(z)) +
With the abbreviations
a :=
b
i,..., (Ylz
Ynz)
the above equation has the following form
=
y(z)) + a(z, y(z)) • fl(z) + b(z, y(z)) • a(z),
z E U.
2.3 The Kiselman minimum principle
157
(z. y(z)) =
To detennine the functions a and b, we differentiate the equation
k=
0,
Forallz cU weobtain o = ak(z, y(z))
k<
1
and
1
H(z) := (Hk3(Z)) := (
82u a Wk a
(z.
y(z))).
a' U__(z. := ( aWkaw1 y(z))).
L(z) :=
Summarizing, the following identities hold for z
z
u.
U:
a(z, y(z)) = —a(z)H(z) — b(z, y(z)) = —$(z)H(z) — &(z)Lt(z) (recall that H =
11'). Keeping the notations we have used so far. we formulate the following criterion for g to be a subharmonic function on U.
Proposition 23.8. Let M be a matrixvalued function on U sue/i that for all z E U the M(z) is a Hermitian quasiinverse of L(z). Then >
—
Z
E U.
In part icula, the function g is subharmonic on U, if the righthand side of this inequality is never negative on U.
Proof In virtue of Lemma 2.3.7 we know that (LMb')(z. y(z)) for y(z)) z E U. Recall from the calculations which were done before Proposition 2.3.8 that a = —aH — and b = — &L' and from Lemma 2.3.6 that LMb' b ='. Then, for z C U, we get
= UZE(Z. y(z)) + a(z, y(z)) • fi(z) + b(z, y(z)) • a(z) > bMb' + + ba' = —1311Mb' — aL'Mb + a$' + ha' = — ab' + a13' + ha' + = fiHMH13' + + aH'13' + aL&' — aHf3' — flLfl' = 13(HMH
= $(HM'H
— —
—
—
L')13t + /3H(ML — L)13 + (—b —
aL')(ML
—
= 13(HM'H
where the last equality uses again the fact that LMW = b'.
—
L)fi' + 0, D
2 Pseudoconvexity
158
Corollary 2.3.9. Let the situation be as in Pmposition 2.3.8. Moreover, assume that the following properties are fulfilled: if z E U and I E IR", then (z. w + it) E G and u(z, w) = u(z, w + it). Then g: U —* R is subharmonic on U. Pivof. Observe that here 11(z) = L(z), z E U.
U
The remaining part of the proof of Theorem 2.3.2 consists of eliminating the smoothness assumptions made so far. Proof of Theorem 2.3.2. We start mentioning that the function g 43) is obviously upper semicontinuous on X. Moreover, for the future we may assume that, without toss of
generality. X = E. i.e. k = n — 1. It suffices to prove that u is subbarmonic on =: E'. In the first step we even assume that u satisfies the following additional condition: (*)
where G(r) := {(z. w) E G: u(z, w) < r}. R÷) as in Section 2.1 and put for
We choose a radial cutoff function '4' E (0, 1]:
e
/(Z,W)\
\ el
I
n—I
(z,w)EC
j,
e
e e
1, yields a plurisubharmomc
ue(z. w) := (u *
w) +
Re wit2
on
:= {(z. u') E G:
Observethalif(z. w)
E
w), e) C G}.
w+it)
andt
E
and u6(z,
w+it) =
ue(z, w). Next, we choose positive numbers a
G(a) +
and fix an
Let e
E (0.
such that G(a) C G(c) C
(0. e21 and Zo
E'. Then we see that
inf{u€(zo, w): (Zo, w) E G(a))
ge(zo) := On
the other hand we mention the following inequalities on w) > u(ZO, w) > c. Thus we conclude that
= and
\
G(c), namely:
w): (zo, w) E
that the infimum is attained at a point (Zo. w) E G(c). Now g is the function from the minimum principle. i.e. from Theorem 2.3.2.
2.4 aoperator
159
From the definition it is clear that the functions
(z, .), z E E', are strongly convex
on
(z, w) E GE} =: By our assumption is connected, z E E'. Hence, in virtue of Theorem 2.3.1, they are convex. Therefore, also the sets
:= (WE
C G}.
z
are convex for sufficiently small e. Observe now that
fl
:'Ez+eE In particular, this implies that the fibers
z
E', are convex domains for all
sufficiently small e. Because of these two remarks we conclude that for any z E E' there is exactly one point Moreover, applying the implicit E such that gt(z) = Ut(Z, function theorem yields that E' —+ is a &function. Thus we are in the situation of Corollary 2.3.9, which gives us that the functions are subharmonic on E'. Observe that g; hence, also g is subharmonic on E'. What remains is to treat the general case. Put Uj(Z, w)
u(z, w)} + (1/j)(max{0, — logdist((z, w), dG)} + 1 Re w112),
G. Then the functions satisfy the assumption (*), so they lead to subharmonic functions gj on E'. Since \ u, it follows that gj \ g. Hence g is a (z, w)
subharmonic function on E', which finishes the proof of Theorem 2.3.2.
0
2.4 3operator The aoperator has been introduced by L. Hörmander ([Hör 1965], [Hör 1990]). We
will mainly follow his approach. Let (K, p) be a Riemann domain over Let A = denote p= X —+ IR be a continuous function. Define the Lebesgue measure on X and let L2(X, q) to be the space of all Ameasurable functions f: X —+ C such that
I
Jx instance, if
i
<+00.
Observe that the space L2(X, may be also defined for more general weights ç, for X —f is upper semicontinuous.
2 Pseudoconvexity
160
with the scalar product
Observe that L2(X,
(f. g)
:= j
(f,
is a complex Hubert space. Put
f E L2(X,
:= (f, It is well known that the space 2)(X) :=
C) is dense in L2(X,
Moreover,
one can easily check that
L2(X,loc)= U We endow the space v
Z+, then
qo
X
U
with the standard Fréchet topology: if 'h E in Coc(X) locally uniformly on it' —+
E
£)(X). v E Z+. We say that Let —÷ Jo in compact set K c X such that supp C K for any v Z.4. and uniformly on X for any a, X) —*
Remark 2.4.1. If
if there exists a
then
for arbitrary a,
(a)
(b) The operator x is
continuous, i.e. if ij,,
(ii, f) —+
E
Jo. then
and
iiofo.
A linear operator C: £)(X) —* C is called a distribution on X (C E .V'(X)) if —k C(fo) whenever Let CR
£)'(X), v e
We say that
Co(f) for any f 46
Shortly.
'70.
1>(X)
Shortly.
—p
Shortly,
—÷ Co.
i)(A)
In.
b
—÷ Co in 2Y(X) 47) if C,,(f) —*
2.4 aoperator
161
Remark 2.4.2. (a) For u E Lt (X, bc) put
[u](f) :=
ufdA,
IE
Moreover, [uJ = 0 in £Y(X) iffu = 0 in L'(X,Ioc). Observe
Then Eu] E that if
I
Jx
L1(X,ioc)
£Y(X)
then
a subspace of (b) For C E .TY(X) and a, fi e
.
tuoj. Thus we can identify Li (X, bc) with
define
f e £)(X).
:= Then
E
.
co, then
Moreover, if
Note that if u E &(X) then
=
lal +
k.
(c) For ,i E C°°(X), C E 2)'(X) define
(,1C)(f) :=
f E £)(X).
E .Z)'(X) (Remark 2.4.1(b)). Moreover, the operator
Then
x £i'(X) is continuous, i.e. if
qo and
(ij. C) —+
E
Co, then
Let C1,C2 E £)'(X)and let U c Xbe open. We say that C1 =
C2
on U if
C1 (1) = C2(f) for any f E
Recall the following localization principle for distributions (cf. [Sch 1966]): C1 = C2 if there exists an open covering (U1)1€1 of X
suchthatC1 =C2 mU, foranyi El. bet supp C denote the supporl of C, i.e. the set of all points a E X f2 E .0(X) 0 on U for any neighborhood U of a. Observe that if are such that f' = f2 in a neighborhood of supp C, then C(f1) = C(f2). denote the space of all differential forms C on X of Fix r, s E Z÷. Let type (r, s) with coefficients in .0'(X) (cf [Nar 1968]), For C E such that C
C
=
>
Cj,jdp'
(2.4.1)
where
{l = C11 E .0'(X),
ik) E Nk: 1
< Ik
:=djij, A...AdpJ,.
2 Pseudoconvexity
162
The representation (2.4.1) is called the canonical representation of C. To simplify notation we will frequently wiite
C= IIi=r. li—s
=
Obviously,
(O}
if r >
n
ors >
n.
Let 'V(X) be a vector subspace of £)'(X). We define V(,5)(X) as the space of all such that C,,j V(X) for all I E! and J In such a forms C way we define
bc) j
C C(r.s)(A') C
Moreover, if
v E Z±, then we say that
E
so).
—* Co in V(,,3)(X)
inthetopologyofV(X)foranyl
—÷
E
we define the support of C:
E
suppC Similarly as in § 2.1, L1(X, bc); UE is defined in
U
suppC,,j.
will denote the rregularization of a function U E := {x X: > e}; cf. Remark 2.1.24. A bc). then we define the eE
If u =
regularization of u as the form
:=
(u,,J)Edpt
E
C > 0.
111=,. IJI=s
Observe that Remark 2.1.24 generalizes in a natural way to the case of differential
forms. From now on we assume that a 'regubarization function' 4' is fixed (cf. § 2.1).
Lemma 2.4.3. If f E L' (X, bc) and f is holomorphic in f(a)foranyO < e <.r.
(a, r), then ft (a) =
Pmoj By the mean value theorem for hobomorphic functions we get
fE(a)
= =
+ ez)4'(z) dA?M(z)
J
f f
10.11"
(Jo
(J
+ rr e'°)dAn(O))4'(r)ri . . ... .
= ro.ir
0
Shortly,
. .
,,(X)
Co.
0
2.4 aoperator
163
Now, for C E £)'(X), we define the eregularizarion of C by the formula Ce(X) :=
.
x
CVE(p(x) — p)),
0,
C 2)(X) is such that Xx = 1 Ofl lFx(x, e). Observe that the definition is independent of a particular choice of Xx• where Xx
Remark2.4.4(cf. [Sch 1966]). (a)Ifu E L'(X,loc),then [uk = [uE],r >0. = for arbitrary a, and e > 0. and (b) E
(c)lf K :=suppC (d)IfK :=suppC
C
X,4d:=dx(K),then = C((xf)€), f
2)(X) is such that x =
where x
foranyo <e <
C
I
0
on
C when e —÷ 0.
K
The definition of the eregularization extends in the natural way to forms from
:=
E
IIl=r, 111=s
X, a E X, and let Lemma 2.4.5. Let C E £)'(X). Assume that K := suppC n. = 0 on Px(a. 3r), j = I 0 < t 4dx(a)}. Suppose that Then
C=
[CC]
on Px(a, t)forO <e
Proof Since
= 0 on Px(a. 2t).
=
j=
1
n, the function
is
holomorphic on Px(a, 2r), 0 < < t. Hence, by Lemma 2.4.3, (C1),, = (C,1)1 =
distribution, i.e.
Fixana X,letO < 7r2r. Hence, by Lemma 2.4.5, C is holomorphic Proof
inlFx(a,r). For u, v
LI
bc) put
:= (u, u) =
(u. v) III=r, IJI=s
Note that (ii, v) C E 0(U)
L1(X, bc), hull E L2(X, bc). 3UE0(U): C = tul on U.
lu,.j12. II=r. IJI=c
164
2 Pseudoconvexity
bc), v E bc); recall that
Remark 2.4.7. Observe that for u E
belongs to
product U A v
(
v,',j'dp" A di")
A dr') A (
>J'
II=r, IJI=s
bc), their wedge
II'I=r'. J'I=s' IKI=r+r', Lj=s+s'
where WKL
J, I
:=
,
J
)u;,Jvf',j'
IlI=r.
J'I=s' IUJ'=K, JUJ'=L
and
J. I', J') E (—I, +l} is the number such that
dp'
A
=
A dp" A
J, 1'. J')dp" A
The notion of the wedge product extends in the natural way to the case where
(cf. Remark 2.4.2(c)).
and v E
UE
Remark 2.4.8 ([Fed 1969], § 1.7). For any r, s, r', s' E such that lu A vii < MlIuIl
lvii,
bc),
uE
there exists a constant M
v E
bc).
Let Mrsrs' denote the minimal constant with the above property. Then
(r ± < I.
r + s < I or r' + s' < 1, then Exercise 2.4.9. Prove that in general Mr.5,r',s' > I. For U, v E
IR)) put
E
(u.
=
f x(u. :=
Note that the space
space and that the space
=
(uj,j, III=r, IJI=s
Jx
= lII=r.
(with the above scalar product) is a complex Hubert is dense in so).
2.4 8operator
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 nonhomogeneous Cauchy—Riemann equation or 8problem. 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 8problem has a solution with the same or similar regularity. Our aim is to prove that the aproblem has a solution if (X, p) is pseudoconvex. The proof will be based on the following result from the Hilbert space theory.
2 Pseudoconvexity
166
Lemma 2.4.12. Assume that
R2 are complex Hubert spaces. Let T
3(I 3 Dom T —+
densely defined operator and let F be a closed subspace 013(2 a linear such that Range T C F. Then the following conditions are equivalent: be
There exists a constant C > 0 such that
(i)
IIfIIx2
CIIT*fII.,(1.
f
E F fl Dom
T*.
(2.4.4)
(ii) For any v E F there e.xistsa u E DomTflRangeT* such thai Tu = vand IIuIIJt
(iii) Range T = F. 3 Dom T be a linear closed densely defined operaRemark 2.4.13. Let tor. Then: (a) Let f E R2. By the Riesz representation theorem we get
f
Dom T*
3M>O VUEDOmT
(b) (cf. [Wei 1980]) The operator
(c) KerT =
(f.
<
is densely defined and T** = T.
(Range
Pmof of Lemma 2.4.12. (i)
(ii): First observe that by Remark 2.4.13(b), (c) we
have:
Ker T* = (Range T**)l = (Range T)1 3 F1.
Consequently,iff =
Ii +f2 E DomT*,wherefi
Fandf2 .1. F,thenf2
E Dom T* and T*f = T*fi. This implies that Range T* = Ker T*. In particular. T*(F fl Dom T*). We will show that Range is a closed subspace of Ri. —* uo 3(1 Then, by Indeed. let C — > 1. Consequently. there exists (2.4.4). 1111'— such that an Jo E —÷ fo. Obviously Jo E F. Let U E Dom T. Then (us.
= lim (T*fv.
=
= (b.
urn (ft,. v*+oo
Thus Jo E Dom T* and T* Jo = uo.
Fixav E FandletL: RangeT*
Cbegivenbytheformula
L(T*f) := (v. f)ae2. 50)
fo =
Tue.
That is. if
DomT x
(us..
f E Dom
—* (uo.fo)
x
then
C DomT and
2.4 aoperator
167
By (2.4.4) L is well defined and
IL(T*f)I
I E Dom
which shows that L is continuous. By the Riesz representation theorem (recall that CIIvII.,e, such that Range T is closed) there exists a u E Range T* with L(T*f) = (u, T*f)jf1, f E Dom Tt. Thus ii Dom T = Dom T and T**U =
=
Tu
v.
(iii) is trivial.
(ii) (iii)
(1):
We only need to prove that the set
A := (f E FflDomT*: IIT*fII.
is bounded. Since F is closed, it suffices to show that A is weakly bounded in F. Take a v E F. Since F = Range T, there exists a u Dom T such that v = Tu. Hence
for any f E A we get:
=
I(v,f)x21 =
U
IuII.x1.
EC(X,R),j = l,2,3,define
j = 1,2,3,
4S+J_I)(X. D0mT:={uE,7t,: auE3t'2},
Tu:=au, u€DomT,
DomS:={f€R2:df€3€3},
Sf:=df, feDomS.
Remark 2.4.14. (a) D(r.s)(X) c Dom T and T and S are densely defined. (b) The operators T and S are (c) Range T C Ker S (cf. Remark 2.4.10(b)).
C Dom S. Consequently,
Now Lemma 2.4.12 (with F = Ker S) gives the following Corollary 2.4.15. Assume that
I EDOmT*flDomS. Then for
any
with av = 0 there exists a u E Dom T fl Range T*
v E
such that du = v and Itu 114?1
Wefixasequenceih
E C2(X,R)such
1]). V E
E
that
112
a (ui,.
Let Dom T X
Consequently.
—+ duo in
Thus fo = auo and finally.
E
eq',
v
— (uo, fo)
(2.4.5)
1,
1lvIK
(2.4.6)
N. X
Then Uv
On the other hand.
Dom T and fo = Tu0.
Uo in
—÷ fo in
2 Pseudoconvexity
168
Exercise 2.4.16. (a) Find a sequence tion (2.4.5).
C
(b) Prove that for any sequence IR) with (2.4.6). a function E
LO,
I]) satisfying condi
[0, 1]) with (2.4.5) there exists
C
=4,.
ForçaE e2(X,R)putq,1
Theorem 2.4.17 (Hormander s Lestimates; [HOr 19651, [Hor 19901). If £4,(x:
>
c E C(X. R>0), then
f(c
I
+ 11Sf
—
(2.4.7)
If + ed'),
c then
If
+
f
11Sf
E
Dom
Tt fl Dom S.
(2.4.8)
Assume for a moment that the above result is true. We will show how it applies to solve aequations 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 aoperator
169
Put
L,:={xEX:w(x)
f
11v112
dA,
u E N.
i
Consequently, we only need to choose x in such a way that
•X
Xo,
•
< +00,
• x'(t) >
+
t
D
R.
Now Corollary 2.4.15, Theorem 2.4.17, and Lemma 2.4.18 give the following
fundamental result.
Theorem 2.4.19. If (X. p) is pseudoconvex, then for any r, s > 0 and bc) with av = 0 there exists a u 6
v
bc) with au = v.
The proof of Theorem 2.4.17 will be divided into a few steps. First, observe that is dense in Dom T* fl Dom S in the sense of the graph norm if the space
DomT*flDomS3f_+ then (2.4.8) is a consequence of (2.4.7). The above mentioned density will be proved
in Propertiesl—7. Next, we will prove (2.4.7) using Properties 8,9.
Property 1. Assume that f E Dom S. Then S(ih,f) —+ Sf in X3
Proof Since d(,7f) = qf E Dom S. For
dq A
17f
E Dom S for any 17
f + qdf, we easily conclude that 8(77f) E R3.
.0(X). Thus
we get (by Remark 2.4.8 and (2.4.6))
<
—
Hence, by the Lebesgue convergence theorem, —* Sf in Finally —+ Sf in
Property 2. Let where
E
f 6 Dom S be such that supp f
—
—4 0 in 3€3. Clearly
0
X. Then
Sf
in
denotes the sregularizarion off.
Proof By Remark 2.1.24, Sf5 = (Sf)5, 0 <s << 1. Hence Sf5 —+ Sf in By virtue of Properties I and 2 we get The sequence
is as in (2.4.5) and (2.4.6).
0
2 Pseudoconvexity
170
Property 3. (a) The space !)(r,s)(X)
is dense
+
U
(b) The space .D(r,s+I)(X)
in Dom T in the norm
in Dom S in the norm
iS
I —+
+ 11Sf
Property 4. Assume that f e Dom T*. Then Moreover.
T*(,hf)
E
1)(X).
in .lti.
Proof By Property I we know that u Dom T we get
= (f.
(ref,
Dom T* for any
Dom T for any u
= (f,
—
(f,
Dom T. Moreover, for
A
=
— (f,
A
Hence
+ If
071.
U E Dom T. x
E Dom Tt. Moreover,
Consequently, by Remark 2.4.13(a), (T*(lif).U)ç1 —
=
f
If lie
for all u E Dom T and, consequently, for all u e
This implies that
Ill
IIT*Oif) — (Exercise). In particular. IIT*OJL,f) —
s
If
and, therefore, by the Lebesgue theorem, T*Olvf) —
—÷ 0 in .7(i. Hence, U
in .Jt'i
Let! E define 11,1K
For! E
(k,
k3)
E (1
n}
:=Oifj E {k1
:= sgn(r)f,.L if j
r(j,ki
=
E .D'(X) by the formulas
k5)=L E
k5}, where r is a permutation such that
2.4 aoperator
Property S. (a)
C Dom T*.
=
fE
II1=r. )KI=s
>
DomT. In particu!a1
e*T*f = Af + Bf, where n
Af :=
(>2[fi.Jx])dp'
>2'
A d15".
j=I
KI=s n
Bf:=(—1)T
fEDomT*.
>2'
III=r. IKI=s
Notice
j=t
that A. B: Dom T —+
Proof Fix an f E
bc).
and let
g :=
AdIK
>2'
E
III=r, KI=sj=I For U
>2'
UJKdp' Ad1öK E
IIJ=r. KI=s
we get
(g,
=
>2(_ly_'
>2'
II=r. IKI=sj=1
Jx
aZJ
n
=
!.jK
>2'
aUIK
dA = (1.
X
By Property 3, the above equality holds for all u E Dom T. This shows that f E
DomTtandg = T*f.
Now, take an arbitrary f E Dom T*. Let
g := T*f =
>2'
III=r, KI=s
gJ,Kdp'
E
2 Pseudoconvexity
172
Take a u E
Then
'°'dA = (T*f,
L I =r
= (f,
= X
III=r.IKI=sj=1
Zj
= I!I=r.
=
(
IIrr,
j=1
Consequently,
(_1)t_1
=
which gives the required formula.
Property 6. Let
D
f E Dom T* be such that supp I
X. Then T*fE
in
Proof Using Property 5 and Remark 2.4.4 we easily conclude that it suffices to prove that —÷ bc). We have in L2
Af + Bf =
=
0
Properties 3, 4, and 6 imply
is dense in Dom T* fl Dom S in the sense of the
Property 7. The space graph norm
DomTtflDomS9f—+ Remark 2.4.20. Let ag
aço
j=1 =——g—. azj Observe that
=
aZk
Property 5 gives
8ZkdZj
g,
gE
n.
2.4 aoperator
173
Property S. n
= (_l)r_1
A
I1I=r. IKI=si=1
+(_l)T_I
>'
fE
dzJ
I!I=r, IKI=sj=I
in
Ix III=r E' j.k1 +
2f
f
f
we have
E
azk
111r.
1I=r,
Pmof Recall that
st= I1I=r.IQI=s+2
71.Q.J.J)——)dP'AdiQ.
(
dZJ
{i.ii
wherea(/, Q, J, j)
E
(—1, +l} is such that A dp1 A
Moreover,weputa(i, Q,
dji' = a(i, Q, J, j)dp1 A
J,j) :=Oif{j,j1 a(i,
Oslo2 =
Q, J
Hence J)
af,.j
—a(i. Q, L,
III=r. tQI=.c+2
j,IE(I (j)uJ=(t)uL=Q 2
=
x
X 'J(l, (j.
£,
aZJ
III=r.
II=r, IJI=s+1 K>).
j)a(1. (j, £, K>), (j. K>), £) sgn(t(e.K)) sgn(r(j.K)) afl.JK
2
= fl=r, Jl=.r+I
Zf
af,j
= fl=r, JI=.c+1
I!I=r, —
U
jfl=r. IKI=sJ.k=1
(JhI,jK afl.kK ôZk
3Zj
174
2 Pseudoconvexity
recallthatQ =(j..e, {j, k1,..., k5J, {ji
=(j,K>)E = (i. k1. .. , Ac3), .
is the permutation mapping (j, Ac1
Property 10
(The
Ac3)
onto
respectively;
.
moreover, r(j,
J.
K)
0
main part of the proof of Theorem 2.4.17). Inequality (2.4.7) is
true.
Proof By
Properties
8 and 9
we get —
I
af,.JK
1I=r. KI=.cj.k=1
Zj
x III=r, IJI=s+lj=1
+2j II!
+ Hence, by Remark 2.4.20, we have
j
IJ=r
>.L' l!I=r j=1
j.k1
IKI=s
+2f
+
(2.4.9)
0
which directly implies (2.4.7).
Proposition 2.4.21. Let (X, p) be a pseudoconvex Riemann domain R>0) are such that that y E C2(X), c E
£y(x; Let v E
y)
be
such that
II
Then there exists a u E
x E X.
Assume
E
dv = 0 and
f I 11v112—dA <+00. C Jx y) such thai du =
v
and (2.4.10)
2!.
L
Proof We may assume that I = 1/2. Let 6 be a CCO strictly psh exhaustion function d for X such that £6(x, > E
X: 8(x)
t ER;
2.4 boperator
175
X, is a pseudoconvex region (cf. Corollary 2.2.15). We will show that: bc) such that 8u1 = (*) For each t R there exists a u, E
I
J xl Suppose
v
and
< 1.
lu1
(2.4.11)
for a moment that (*) is already proven. Fix a sequence
/ +00.
is bounded in By (2.4.11), for any compact K c X the sequence u weakly Consequently, by the Banach—Alaoglu theorem, we may assume that bc). In particular, au = v and with u E
I
< 1,
Jx, which finishes the proof.
a:
lit Suppose that the functions E satisfy (2.4.5), (2.4.6), and, moreover, (notice that such functions do exist). = 1 on v E N, and * = 0 on convex increasing function such Put := y + o (5. where x: IR —÷ IR+ is a Then — 2* > y and that x = 0 on (—oo, z), x oS > 2*, and X'o15 ? ))2 := — := 0 ö)(., E C". Put q'i := q) — 2*, 2: and consider the aoperators We move to the proof of (*). Fix [0, 1]), v E N, and * E
L
D Dom S
wa).
By virtue of Theorem 2.4.17 we get
fE D0mT*flDOmS.
+ 11Sf
Observe that v )(v,
Consequently,
f f I
f Recall that (KerS)1 C
0 = (v.
and v
Dom
fl KerS.
KerS. Thus
f I KerS.
(2.4.12)
2 Pseudoconvexity
176
Finally, by (2.4.12),
< IIT*fIIwi,
f
E
Define L: Range T* C, L(Ttf) := (v, the Riesz theorem, there exists a u, E Range T* C
L(T*f) = (u,,
DomT. Then, L is well defined and, by WI) such that
f E Dom Tt,
and (2.4.13)
< 1. Hence
= Tu, =
v.
0
Condition (2.4.10) follows directly from (2.4.13).
The following auxiliary function will simplify our notation:
+ 11z112
Notice that — log
E
Theorem 2.4.22. Let (X, p) be a pseudoconvex Riemann domain over C", let E and let v E W). ôv = 0. Then there exists a U E bc) such that au = v and
j
o p)4dA
f
(2.4.14)
with av =
In particular, if p(X) is bounded, then for any v E exists a u E W) such that Ôu = v and
0 there
(I whered :=
XE X}.
Pmof Step 1°. Assume that J2y(x:
E
L(—4logSoop)(x:
e2(X). Put y :=
—
4logSo o P. Then we get
=:
x E X,
y) with du =
Consequently, by Proposition 2.4.2 1, there exists a u E
o p)4dA
=
j
2f I$vII2f_dA =
L
E C". v
and
2.5 Solution of the Levi Problem
Step 2°. The general case. Let
:= {x E X:
177
\ ço as e \ 0.
fl
E
is a pseudoconvex region (cf. bc) such that Corollary 2.2.15). Hence, by Step 1° there exists a ur E where
=
v in
> e}. Note that
and
f
o p)4dA
f
(2.4.15)
L
\ 0. Observe that the sequence is locally uniformly bounded from above in X. In particular, by (2.4.15), for each compact K C X the sequence is bounded in Consequently, we may assume that there —+ uweakly. Inparticular,du = vinXand existsau E Fix a sequence
f
o p)4dA =
K
urn fK
o
X, e >0.
K
X
K
which directly implies (2.4.14).
2.5 Solution of the Levi Problem Recall (cf. § 2.2) that the Levi problem is to decide whether any pseudoconvex Riemann domain over is a domain of hobomorphy.
Theorem 2.5.1 ([Jar 1979], [Pfl 1982]). There exists a constant C > • for any pseudoconve.x Riemann domain (X, p) over
•foranyço
E
with
E
0 such that:
L'(X, bc), and
X and 0 < r
• for any a
X
<(I
P(a,r)
1,
f(x) =
Ofor x E 2C
\ (a}, )fl
r S0(p(a))
C'onsequent!v if (X. p) is a pseudoconvex (in particular a holomorphically convex) Riemann domain overC', then 0(X) separates points in X (cf Remark 1.10.5) Proof Fix a function x E C Izi > 1/2. Letc :=
[0, 1]) with x(z) = 1 for IzI S 1/3, x(z) = 0 for Define Xr(Z) := X(z/r), r > 0.
18c.
In fact, we know more: if (X. p) is pseudoconvex, then for any E PSJ?(X) with o L1(X. bc) the space (f E (9(X): < +ac} separates points in If stalks.
2 Pseudoconvexity
178
Fix (X, p),
a,
and r. Let
Xk:={xEXJpJ(x)—pj(a)I
k=0
n},
n.
Note that P(a, r) is the connected component of Xo that contains a (cf. Remark C X,, = X. It is clear that (Xk, p) is a 1.1.2(f)). Obviously C Xo C n (cf. Corollary 2.2.15). n } there exists a function fk E O(Xk) such p'(p(a)) \ (a) and
pseudoconvex region over C", k = 0 We will show that for each k E (0
that fk(a) =
=
1,
0 forx
f
E
2C
<(f
r
P(a.r)
Xk
The case k = n gives the required result. We proceed by induction over k. Put
fo(x) :=
I,
0,
xEP(a,r) xEXO\P(a.r).
It is clear that fo satisfies all the conditions.
Suppose that 1kI is already constructed (for some 1
k
n). Define
pk(a))
j
10 where
XkI: Jpk(x) — pk(a)I < r/3}, := {x Xk: pk(X) — pk(a)I > r/2}.
ilk := (x
E
It is clear that dv =
Then v c
0.
Moreover,
p)6(k_I)dA
<
I
—
dZ
pk(a))
1k—I
p)6(kI)dI\
pk—pk(a)
r
By Theorem 2.4.22 there exists a function u E L2(Xk, bc) such that du = v and
f u E O(Uk). Define 1k
J Xr(Pk
—
pk(a))fk_I
(Pk
—
pk(a))u
on X*i
— (pk
—
pk(a))u
on
—
2.5 Solution of the Levi Problem
dfk = 0 in
Then Uk
/
2/
IXr(Pk
(thus we may assume that Itremains to estimate the integral.
179
and
E
1k
= 1k—I in
o
—
o 2 2
80(p(a))
o Xk
0
)=Mk. Applying the same methods one can get the following result.
Theorem 2.5.2. For any d > 0 there exists a constant C(d) > 0 such that:
•for any pseudoconvex Riemann domain (X, p) over
with
XE X}
•foranyço E S'SR(X) with •foranya E X, 0
E
L'(X, bc),
and
anf E 0(X)suchthatf(a) = 1, f(x) =Oforx E
/ In particular
C(d)
(fP(ar)
X
= 0), the space
__) r
(X) (cf Example 1.2.5(b)) separates points in
stalks.
Let
8x:=min{px,öoop}=min{px,
1
+
11p112
By Theorem 2.2.9, if (X, p) is pseudoconvex, then — bogSx E Recall that for any function 6: X —÷ we put
0(k)(X 8) := {f E 0(X): the function
6'f is bounded on X},
k>
0.
We know (cf. Example 1.2.5(d)) that if 1/8 is locally bounded, then 8) with the norm f —* 116"! lix is a natural Banach space. Observe that if 6 is bounded, then
8) C
8) fork
2 Pseudoconvexity
180
Lemma 2.5.3. Let (X. p) be a Riemann domain over C" and let 8: X —÷ be
such that 8
Px and
x E X, x'
IIp(x') — p(x)II,
18(x') — 8(x)I
Bx(x);
(2.5.1)
cf Remark 1.1.14(e)(.fore.wmple8 = ox). Then
q >0, j E 0(X),
k where (k
c(n.k,q) := In
particular (k = 0,q =
2k+2n
I/q
+ C
2),
Proof Fix k, q, f, and a E X. Since 1j1q inf ml
we get (cf. Exercise 2.1.14)
f
f
—
inf
I
I
f
r)k Já(a.r)
—
—
0<0<1
Ifr'dA}
B(a,r)
—
<
IJq
0)k
I
ix
f
= (c(n. k.
D
Proposition 2.5.4. Let (X, p) be a pseudoconvex Riemann domain over C" and let 8: X —+ ILo be such that •8 • condition (2.5.1) is
•
— logO E .'PSJ((X).
Then:
For any k > 0, a E X there exists a function f {a}, and f(a) = 1, f(x) = Oforx where c(n.k) > Odependsonlyonnandk. (a)
In particular. (b) Ifsup{llp(x)II :x
functionf
0(k+4n)(X 5) such that
8) separates points in X. X)
<
d <+oo, then for any k
0, a E X there exists a
0(X.8)suchlharf(a)= l,f(x)=Oforx p1(p(a))\{a},and
where c(n, k. d) > 0 depends only on n. k, andd. 8) separates points in X.
Proof (a) By Theorem 2.5.1 with çü := —2k logO, r := &8(a), there exists a (a), and function f 0(X) such that f(a) = 1, f(x) = 0 forx
J
X
In2o2k800p6fldA
(j'P(a,r
2c
r 80(p(a))
5
2.5 Solution of the Levi Problem
181
where d(n, k) > 0 depends only on n and k. Hence, by Lemma 2.5.3 (with q = 2) we get [c(n, 2k + 6n, 2)]2 j
If < ?(n, 2k + 6n, 2)d(n,
X2. The case To prove that 8) separates points take xi, E X, xI p(x2) follows from the proposition (with k = 0). If then there exists a) E {l n} with Pj(XI) and p1 E (9m(X. 8x) C
p(xi) =
0(4ht)(X. 8).
0
(b) Use Theorem 2.5.2.
ProposItion 2.5.5. Let (X, p) be a pseudoconvex Riemann domain over C'1 and let 8 be as in Proposition 2.5.4. Then:
8)domain ofholomorphyforanyk > 6n. (a)(X, p) isan (X. p) isan 0(k)(X 8x)domain ofholomorphyforanyk > 6n. (b) If p(X) is bounded, then (X, p) is an
of holomorphy for
anyk> 2n. (X, p) is an
In
8x)do?nain of holomorphy for any k > 2n.
Remark 23.6. In the case (X, p) = (G, id), where G is a domain in C'1,the result may be improved. Namely, any domain of holomorphy G C C'1 is an := min{1, pG}So (cf. § 2.12). Observe of holomorphy for any k > n, where
that if G is bounded, then
SG)
C (9(k)(G 6G) C
=
SG); if G is unbounded, then
8G).
Proof of Proposition 2.5.5. (a) Fix a k > 6n. Let a: (X. p) —p (X, j3) be the 8) separates points in X (Proposition 8)extension. Since maximal 2.5.4(a)), the morphism a is injective and therefore we may assume that X C X, p= a = idx.
PutT := {f
0(X): fIx E
8)). Suppose that there exists an xo E
Let U bearelativelycompactunivalentneighborhoodofxo. Thenthereexistsac0 >
0
such that
If IIu < coIIökfIix,
fE7
(cf. Remark 1.4.5(i)). By Proposition 2.5.4(a) for any a
with f0(a) = I and il8kfa lix
Thus Px
coc(n,
X there exists an fa
E
7
Consequently, a E U fl X.
> 0 on U fl X for some &j > 0; contradiction.
(b) Use Proposition 2.5.4(b).
[]
2 Pseudoconvexity
182
Theorem 2.5.7 (Solution of the Levi Problem). Let (X, p) be a Riemann domain over C'7. Then the following conditions are equivalent: (i)
(X, p) is a domain of holonwrphy;
(ii) (9(X) separates points in X and (X, p) is holomorphically convex;
(iii) (X, p) is holomorphically convex;
(iv) (X, p) is pseudoconvex. Proof By Theorem 1.10.4 we know that (i) (iv) are obvious. The implication (iv)
(ii). The implications (ii) ; (iii) (i) follows from Proposition 2.5.5. 0
From now on the following notions will be used synonymously: • (X. p) is a domain of holomorphy. • (X. p) is holomorphically convex, • (X, p) is a Riemann—Stein domain, • (X. p) is pseudoconvex. Notice that the above result is not true in the category of ramified Riemann domains — cf. [GraRem 1956), [GraRem 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 ([GraRem 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. [GunRos 19651, Th. 0
be such that
C.
First we show that for each I E I the function
f is locally bounded on X.
Indeed, fixi E landa EM. Observethatforanyo < r 3x(a, r/2) \ M
px(a)andx E
we have
PX\M(X)
= dist(p(x). p(13x(a, r) fl M)).
=
Let 0
xE
r) \ M.
It is clear that there exists a c, > 0 such that —
Hence,
forx
.f,(x")I
c, IIp(x') —
x" E Bx(a, r).
Bx(a, r/2) \ M, we have
(x)f(x)I
(ct
If(x)I
inffllp(x) — p(x')II: x' E 8x(a. r) fl
= Now, by the Riemann removable singularity theorem, the function toafunctiong1 E 0(X). E oa
1f extends
By the identity principle, for arbitrary Ii, i2 E I the functions and g,2/J coincide on k \ (M,, U Thus the function f given by formula onX\M1,i E Evidently,foa = f. D The following extension theorem will be useful is Sections 3.1 and 4.5.
ProposItion 2.5.10. Let (X. p) be a Riemann—Stein domain over C" and let H C C" beacomplexaffinesubspacewithl PutM sider (M, PIM) as a Riemann region over Ck with k := dim H (cf corollary 2.2.18). Then 0(M) = 0(X)IM, i.e. any function f E (9(M) extends to afitnction f 0(X). Proof Using induction we reduce the proof to the case H = (Zn = 0} (k = n In fact, one can find I with #1
—
1).
2 Pseudoconvexity
184
Fix a function f E 0(M).
Ua€M Px(a). Write
Put U
xC. Take an a M. Observe that p(M fl Px(a)) = P((p'(a),O),dx(a)) fl = 0). Consequently, we define the local extension fa(x) := 0)), x Px(a). One can easily prove that Ja = Jb on lPx(a) fl Px(b). Thus we get a function
fEO(U)suchthatflM =1.
=
Leti'p E
U. Define

onU
g.1Jcof
Observe that g is well defined, g
C
I
C°°(X), and g =
f on V. Put
onX\M
lo
onV.
Then v is well defined, v and av = 0. By Theorem 2.4.22 there exists a function u L2(X, bc) such that du = v. In particular, du = 0 in V and, consequently, we may assume that u 0(V) (cf. Proposition 2.4.6). Define — Thenf E boc)flO(V), 1= Jon M, anddf = OmX. Using f := g
0(X).
once again Proposition 2.4.6, we conclude that f
D
2.6 Regular solutions Let (X. p) be a Riemann domain over C's. Our aim is to prove that if (X, p) is pseudoconvex, then for any aclosed form V E i.e. av = 0, there exists a U with du = v (cf. Theorem 2.6.4). For k E U {oo} let bc) be the Sobolev space of all u L2(X, bc) such that for any a, k the derivative belongs to with al + flu L2(X, bc), i.e. there exists a Ua8 L2(X, bc) such that in the = sense of distribution (cf. [Nar 1968]). Let further bc) denote the space of all forms of type (r, s) with coefficients in .Wk(X bc).
1968]). In particular,
bc)) C
(b)
k>
bc) C
Remark 2.6.1. (a) (Sobolev lemma)
bc) = bc), k >
0.
Defined:
c',jdp'
aCI.JK
:=
j1
II=r
JI=s+I
KI=s
)d I
0 (cf. [Nar
2.6 Regular solutions
recall that C,,1 K :=
IfJEK
10 I,
o(l, J, K, j)Cj,j
and a(I, J, K, j)d
where J := (j,
if j K, A dp' A df? = dp' A
=0 Remark 2.6.2. (a) Ifs > 1, then Indeed, for C E we have C,k11, =
j,k= I
185
(Ill = r, ILl =
—CJ,JkL
s — 1,
n)andhence
I
k
bc) =
0.
bc)).
bc): Ou E
(u
r
0, k 0.
bc)
(b)
=
bc): au E 'W(rs+2)(X, bc), Oti
E
'Wfr$)(X, boc)},
r,s>0, k>0. bc) is such that
Pmof. (a) The inclusion C' is trivial. Suppose that u E
au = >au, Adp1 E Wfr1)(X,lOC). III=r Then
au,
bc) for any I and, therefore, we may assume that r = 0.
We use induction on k. in
[0, 1]) be such that x = I k = 0: Fix a compact set K c X and let x E X, and neighborhood U of K. Put v := x u. Then v E L2(X), suppv
a
EL2(X),j1 For any w E
0w
—
2
n.
we have:
=1( ———dA=—II' ix ix OwOw
f
x OZ1 dZJ
0Z3
f thdA=—I ix
02w
L'X)
,
j=1
n.

wdA (2.6.1)
2 Pseudoconvexity
186
Let VE denote the eregularization of v, 0 < e << marks 2.1.24, 2.4.4) that
I
(cf. § 2.1). Recall (cf. Re
8v
I aV \ —=1—I
j=l
— whene—+0,
n.
By (2.6.1) we get dUE"
L(X)
dZj
Hence there exists
—÷ 0,j =
a
I
=
dv('
dUE"
,
dZj
dZj
function uj e L2(X) such that n. This implies that
=
U
n.
,
L2(X)
urn E—.O dZJ
=
urn
=
[v1].
—+ Vj in L2(X) when
j=I
n
n. Hence v E W'(X) E L2(X), j = I and, therefore,u E W1(U). Consequently.u C W1(X,loc).  Assume that the result is true fork—I (k > 1). Let u L2(X,loc) be such that be). We already know that u Wk(X, bc). For any al + du E
we get
j=I Applying the case k = Oto the function
n.
we conclude that DCE$u E W1 (X. toe).
Thusu is trivial. Suppose that u e (b) The inclusion bc) and t)u E bc). 814 E We use induction on k. k = 0: Fix a compact set K c X and let x
bc) is such that
[0, 1]) be such that x = 1 in a neighborhood U of K. Put v := x u. Then v E suppv Au = Bu + where B is W(r.5+2)(X) and Ix8u = (8x) a
linear operator with coefficients depending on x 56)• Observe that the main inequality (2.4.9) in the proof of Property lOin § 2.4 is true
for*
=0(noticethatthen
=
Hence,foranyw
we get n
H
1I=r If I=s+ I 56
flu := I
OL2(X)
(X)
+ 118W112(X).
(2.6.2)
2.6 Regular solutions
187
denote the eregularization of v, 0 < r << I. Recall that Moreover, dye —+ dv in and —+ when e —÷ 0. By (2.6.2) (put w = and next e —* 0) we get
=
Let
=
L(X)
IIL2(X)
ii
IlI=r
and in
+
3
JI=s+I Hence, by (a), v E
bc).
Consequently. u E
Assume that the result is true for k — I (k such that du E bc) and Ou E UE bc). For any at + flu k let
1).
bc) be Let u E bc). We already know that
:= I
We have:
=
= Applying the case k = 0 to the form Dcr
we get
E
'W(',.S÷I)(X, bc).
0
bc).
Consequently, u E
Theorem 2.6.4. Assume that (X, p) is pseudoconvex. Then
bc)) =
bc): dv = O}.
{v E
r,
s>
0,
k>
0.
in
= Pmof Fix a v E The case s =
0
{v
dv = 0},
E
r, s >
0.
bc) with dv = 0. is simpler: by Theorem 2.4.19 there exists a u e
bc)
such that u E WJ)(X, bc). Assume that s 1. By Corollary 2.4.15 and the proof of Theorem 2.4.19, there exist functions IR), j = 1, 2, such that if E
T:
thenthereexistsau we have:
Tt(f) =
jDomT v.
ByProperty5fromj 2.4
f E DomT*.
2 Pseudoconvexity
188
= 0, we get
Since u E Range(T*) and
= 0. Hence
= Au,
(2.6.3)
coefficients. is a linear operator with In particular, E bc). Now, by Lemma 2.6.3(b) (with k = 0) we conclude that u E bc). Suppose that we already know that u E bc) and bc) with 0 < f < k — I. Then, by (2.6.3),
where A
hence, again by Lemma 2.6.3(b), u
bc). Induction over I finishes the
E
0
proof.
n > 2, Theorem 2.6.5 ([Ser 19531). Let (X, p) bea Riemann—Siein domain =0. Then there existsau E £)(X)suchthat and Ietv E V
= du.
Remark. Notice that the above result is not true for n =
Indeed.take(X, p) := (C, Suppose that v = for a function u E
1.
v dA2 0. Then, by the Stokes theorem, for
E
sufficiently large R > 0 we have 0
=
f
udz
=
J
A
dz =
B(R)
2i fB(R) v dA2
0;
contradiction.
—* Recall Pmof Consider the Linear operator T = ft: (Theorem 2.6.4) that T is swjective. Observe that Ker T is a closed subspace of
cf. § 2.4). Hence, by
(in the standard Fréchet topology on the Banach theorem, the induced operator
T:
—*
is homeomorphic. Consider A:
Ker T
C,
A([f]) :=
f
VA
f.
Observe that A is well defined.
Indeed, first of all, the integral exists because supp v is compact. Moreover, if (recall that f KerT. then by Theorem 2.6.4 f = 8/i for some h E n > 2 (!)) and, consequently, by the Stokes formula, we get:
JvAf=JvAah=_fa(VAh)=o.
2.6 Regular solutions
189
Put
j=l . n. and define
C: C°°(X) —* C,
C(g) :=
The operator C is linear and continuous and, therefore, it is a distribution with compact support (cf. [Sch 1966]) Let us calculate 8C in the sense of distributions:
=
=
=
= By Theorem 2.6.4 we get v = aw for a w E v
We have proved that
=
= —J —v
in the sense of distributions.
Thus a(w + C) =
0
in the sense of distribution. Consequently, w + C is a holomorphic function (cf. Proposition 2.4.6) and hence C E Finally, we put u := —C. D As an application, we present the following extension theorem.
Theorem 2.6.6 (Hartogs theorem). Let (X, p) be a Riemann domain over C" such that (9(X) separates points in X and let K C X be a compact such that X \ K is connected. Then any function f E (9 (X \ K) extends holomorphically to X. We do not know whether the above result is true without the assumption that
(9(X) separates points
Proof Let
a:
(X.13)
be the maximal holomorphic extension. Since (9(X) separates points, the mapping a is injective (cf. Remark 1.8.5). Consequently, we may assume that X is an open subset of X, p = and a = id. Fixanf E (9(X\K). Let92 E [0, = inaneighborhood U of K. Then the function 1
w :=
onX\K onU
10
is well defined and belongs to C°°(X). Note that w =
Since C is
IC(g)l
f in X \
I
onX
10
onX\suppq.
Put
continuous, there exist a compact K C X. k E and d > for anyg C K.
0
such that
190
2
Pseudoconvexity
Then the form v is well defined, belongs to (X), and = 0. By Theorem 2.6.5, thereexistsau E £)(X)withau = v. Inparticular,u Let Ydenote the union of those connected components of X \ supp that intersect X \ supp u. By the identity principle, we get u = 0 on Y. 
fonXflY. ObservethatXflY 0. \ supp 1' Y and let y: [0, 1] —* X be a curve with y(l) = b. if y([O, 1]) C X \ then y([O, 1]) C Y and hence w—u. Then 7€ (9(X)andf =
PutJ Indeed, y(O) = a,
take a
X
a x flY. Otherwise, let to max{t [0. 1]: y(t) Obviously, to < 1 and y((to, 1]) C Y. Lete > Obesosmallthatc := y(:o+e) e X. ThencE XflY. Since X \ K is connected, the identity principle gives f = in X \ K. 0
f
Let us notice that L. Ehrenpreis [Ehr 1961] was the first who observed that the proof of the above Hartogs theorem may be based on solutions of the dproblem with compact support. See also [For 1998] for an analysis of the proof of the Hartogs theorem.
2.7 Approximation The atheory 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
L2estimates. 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 aoperators
3
3 Dom
42)(X,
Dom S1,
By Theorem 2.4.17 we have:
f
+
> I.
E
C (9(X). Thus
Recall (cf. Proposition 2.4.6) that Ker
I
u> I.
Now, by the proof of Lemma 2.4.12 (with F = Ker E Dom
fl Ker
for each u there exists a
such that
= Let
=
Then by Property 5 from § 2.4 we have:
= in the distribution sense. Define
:= Then V
=—
Moreover, we get (recall that Xv
—i, aZJ
I)
> 1. —
Xv+1)
j = L0
f = =: C,
v> 1.
(2.7.1)
2 Pseudoconvexity
192
Consequently, after selecting a subsequence, we may assume that
—* ho weakly
Obviously,
in
in the distribution sense. Moreover, ho = 0 on X \ Ko (use (2.7.1)). We have:
fi)dA
=
fi3dA
f
=
f
is holomorphic in a neighborhood U of Ko, then we apply the above formula to the function defined as where f on U and 0 on X \ [0, 1]). = near K0. Consequently, fK0 fi3dA = 0. E 0 1
Proposition 2.7.2 (cf. Proposition 2.2.6(a)). Let (X, p) be a pseudoconvex Riemann domain over C", let K c X be a compact set, and let U be an open neighborhood of Then there exists a strictly psh exhaustion function u E such that K C {x E X: u(x) <0) C {x E X: u(x) 0) C U.
Proof Let uo be a continuous psh exhaustion function on X (cf. Theorem 2.2.9). We may assume that uo < 0 on K. Define L := (x E X: uO(x) 2), M := {x X \ U: uQ(x) 0}. The sets L and M are compact. Suppose that M 0. Since M fl = 0, for any point a E M there exists a function Ua such that Ua < 0 on K and ua(a) > 0. Let := {x X: dx(x) > e) and let (Xe) denote the regularization of (cf. § 2.1). It is clear that (U0 there exists an 6(a) > 0 such that L C (ua)E(a) <0 on K and (uQ)E(O)(a) > 0. Put Va := (Ua)E(a). Since v0 is continuous, there exists a neighborhood V0 C X6(0) of a such that Va > 0 in Va. Since M is compact. there exist points ai aN M such that M c Vaj U ... U VaN. Define è := maxfr(a1) e(aN)}, w fl ,..., } E and C on M. Let c := max{ I, maxj. and put f
max{w(x),cuo(x)}
ifuo(x) <2
v is a welldefined continuous psh function on X. Observe that (v0, then v(x) > cuo(x) > 0; ifx E X \ U and uo(x) <0, then x M and therefore v(x) > w(x) > 0. Thus v > 0 on X \ U. If M = 0 we put v := uo. It remains to smooth v. Put
2)
QL.:={xEX:u(x)
2.7 Approximation
Let
193
be the regularization of v and let
E
Then V5 is strictly psh and v5 \ v as e \ 0 on X. Since V is Continuous, V5
V
let r(v) > 0 be such that locally uniformly on X. For each v CC 10, 1]) be such that and V5(v) <0 on K. Let V < < V + I on on Define := = 1 in a neighborhood of and suppq)1 C is a welldefined C°° function on Observe that = 0 on X \ XE(v) and v in a neighborhood of X and vt be a C°' increasing convex function such that X(t) = 0 for Let x: R —÷ R÷). Observe t <0 and x'(') >0 fort >0. Define := x("v + I — v) that:

(a) V,, is psh in a neighborhood of
(cf. Remark 2.1.2(h)).
Indeed,ifx v(x)+l+1—v Indeed, if x E  (c) V,, is strictly psh and> 0 in a neighborhood of c2L \ 0. Hence V,,(x) > 0 and for + I — v > v(x) + I — p > then \ (C" )* we get
LVL(x;
> 0
x'(v,,(x) + 1 —
C JR>o such that for each Now we are going to construct a sequence v and strictly psh in a := N the function We proceed by induction over v. Put W0 := vo and suppose neighborhood of 0 (this condition is empty for are already constructed for some v that Cl is strictly W,, + v = 0). By (a), for any ct÷l > 0, the function is We have_to find such that psh and > v in a neighborhood of V
Fix an A > 0 such strictly psh and> v in a neighborhood of H := \ x E H. C". In virtue of (c) there exists a constant that
B>
C". Hence = E C", which shows > 0 is strictly psh on Recall that >> 0 the function that with >> 0, then W,,÷1 > v on H. on H (cf. (c)). Hence, if > js(use(b)). Thusu = then we get strictly psh function on X. If x KC is a welldefined u (x) = W1 (x) = vO(x) + q x (vj (x)) = vo(x) < 0. Moreover, u > v and therefore E u is an exhaustion function.
>
E
E
(—A
Corollary 2.7.3. Let (X, p) be a pseudoconvex Riemann domain over C" and let 5'SJ? flC In particular. the set K C X be a compact set. Then K is closed.
Take an a Proof Obviously, c X \ {a}. Then by Proposition 2.7.2 there exists a function u such that u <0 on K and u(a) > 0. Consequently, a
and let U := fl C°°(X)
0
Proposition 2.7.4. Let (X, p) be a pseudoconvex Rie,nann domain over C" and let K C X be a compact set. Then for any function fo holomorphic in a neighborhood of there exists a sequence C 0(X) such that lIfe — fouR —+ 0.
2 Pseudoconvexity
194
Proof Take an fo E (9(U), where U is a neighborhood of sition 2.7.2 there exists a strictly psh exhaustion function u
By Propo
C°°(X) such that
Kc{XEX:u(x)
Iv
C
0(X) such that 0.
fo II L2(Ko)
Let 0
0
— fOllL2(K0) —±
— 10 IlK
Proposition 2.7.5. Let (X, p) be a pseudoconvex Riemann domain over C" and let
Y C X be an open set such that (Y, p) is also pseudoconvex. Then the following conditions are equivalent: (i)
the space 0(X)lt',(y) is dense in 0(Y) in the topology of locally uniform convergence, i.e. (Y, X) is a Runge pair;
(ii) for any compact K C Y we have:
=
(iii) for any compact K C V we have: ka(X y = y
y
(iv) for any compact K C Y we have:
Proof Obviously (1) (iii) (ii) and (iii) (iv) by Theorem 2.5.7. Suppose that (iv) is satisfied. Fix a compact K C V. Let Lo := L1 = \ Y. Observe that Lo, L1 are compact and disjoint. In particular, they have disjoint open neighborhoods U0 and U1. Take a function f E 0(Y) and let .'
,'
fo := f on (Jo and fo := I on Ui. Then fo is holomorphic on a neighborhood of L := Ka(x). Obviously, L = Hence, by Proposition 2.7.4, there exists
C 0(X) such that b —f
a sequence
Jo
uniformly on L. In particular,
if we take f 0, f on Lo j K, which shows that (iv) (i). <Moreover, we get a holomorphic function g E 0(X) such that 1/2 on K and > 1/2 on ;
Recall that L1 C
Hence
L1 = 0, which shows that (iv)
(ii).
0
Proposition 2.7.6. Let (X, p) be a pseudoconvex Riemann domain over C" and let = kO(X). K C X be a compact set. Then
c
Proof Obviously, hood of K
=: L. Let U be an arbitrary open neighbor
By Proposition 2.7.2 there exists a strictly psh function u E C°° (X)
X: u(x) < t} X for any t Rand K C Xo C U. By Corollary 2.2.15 Xo is pseudoconvex. Moreover, by Proposition 2.7.4 (applied to compact sets {x E X: u(x) t}, t < 0), any function holomorphic on Xo can be approximated locally uniformly in Xo by functions from 0(X). Thus (Xo, X) is a Runge pair. Hence, by Proposition 2.7.5 we get C Xo C U, which gives the opposite inclusion. 0 such that: X, := {x
Proposition 2.7.7. Let (X, p) be a Riemann domain over C" and let X be strongly pseudoconvex (cf Definition 2.2.3). Then the space is dense in (in the topology of locally uniform convergence).
2.8 The Remmert embedding theorem
195
Proof In virtue of Proposition 2.7.4 it suffices to find a pseudoconvex open set
CX
for any compact K C with C such that C C U (cf. Let a E C2(U, R) be a strictly psh defining function for f2 with Proposition 2.2.25(a) with k = 2). We put := (x E U: u(x) <e} with sufficiently D smallO<e<<1.
2.8 The Remmert embedding theorem FN): X —* and let F = (F1 Let (X, p) be a Riemann region over N n, be holomorphic. Recall (Cf. the proof of Lemma 1.10.12) that F is regular on A
C X if
raFi __Lj(x) = n for any x E A. dZIC
If F
is
regular (on X), injective, and proper, then we say that F
Notice that if F: X —+ dimensional submanifold of CN.
is an embedding, then F(X)
is an embedding. is a complex n
Theorem 2.8.1 (Remmert embedding theorem). Let (X, p) be a Rie,nann—Stein rea holomorphic embedding F: X —* gion over C? 58)• Then there a compact Lemma 2.8.2. Let (X, p) be a Riemann region over C set such that 0(X) separates points in K. Then there exist N E N, N > n, and a holonwrphic mapping F: X —* C's' such that F is infective and regular on K.
Proof Since p is locally biholomorphic, there exists a neighborhood U C X x X of the diagonal {(x,x): x E K} such that if (x', x") E U and p(x') = p(x"), then
= x". 0(X) separates points in K, for any pair (x', x") E K x K \ U there exists C X x X be Let an f(x',x") E 0(X) such that f(X',X")(X) any (y', y") E for a neighborhood of (x', x") such that f(x'.x')(Y') f(x',x")(Y) (x,x') such that U(x'.x"). Since K x K \ U is compact, there exist K x K \ U c U(X'X") U U(X'X",. Li Now we only need to take N := n + £, F := (p, Since
Lemma 2.8.3. Let (X, p) be a Riemann region over C' and let
F = (Ft
FN÷1): X —+
be a holomorphic mapping. For a = (at
Fa := (F1 —aIFN+1 Cf. Remark 110.5.
aN) e
let
FN —aNFN+t): X —* C".
196 FL':
2 Pseudoconvexity
a compact set K C K. Then: (a) If N 2n and F is regular on K, then A2N({a E
Fa
is not regular on K)) =0.
(b) If N > 2n + I and F is injeclive on K, then A2N({a E CN: Fa isnot infective on K)) = 0. Pmof (a) Observe that
CN: Fa is not regular on K}
(a
a(F1
= (a E CN: = (a E CN:
_ai FN+,)
jiaf, j = I
3XEK
= (a E CN: (a, I)
=0, j = I,..., N)
E
x
N + 1),
=: T,
where 1,
X
xC"
j=l
N+l.
k=I
Since N + I > 2n, we get A2(N+I)(4(K x Ca)) = 0. Observe that
(a ECN: (a,t)E 4(K xC")}=:T, Hence the Fubini theorem implies that A2N(T)
t EC1.
= 0.
(b) We have (a E
=
(a
Fa is not injective on K) CN: Fj(x')—aJFN+J(x')= Fj(x")—aJFN÷l(x"),
j=1 = (a E C
N
= (a E CN: (a. I) c
F1(x ) —
xK x C)}=: 7',
)=
j=I
N) N + 1)
2.8 The Remmert embedding theorem
197
where 1,
4):
4)(x', x", t) := t(F(x') — F(x")).
XxXxC
Since N + I >
2,:
+ I, we get A2(N+1)(4)(K x K x C)) =
(a,t)e 4)(K xC")}=tT. Hence the Fubim theorem implies that A2N(T) = 0.
0.
Observe that
tE D
Lemma 2.8.4. Let (X, p) be a Riemann region over C". countable at infinity, such that 0(X) separates points in X. Then: (a) If N 2n, then the set (F
F is no: regular) is of the first Baire
category in 0(X)".
(b)IfN > 2n+1, the,: the set (F€ of the first Baire category in
Fis not infective or not regular) is
Proof (a) Since X is countable at infinity, it suffices to prove that for any compact FisnotregularonK} is nowhere dense in K c X the set E := (F
0(X)N. Fix a K. Observe that E is closed in 0(X)"'. It remains to prove that the set 0(X)" \ E is dense in 0(X)N. Fix A E L X, and e. We are looking for
a mapping F (9(X)N such that F is regular on K and max!. IF — Alt e. By Lemma 2.8.2 there exist M and B E (9(X)M such that B is injective and regular on K. Consider the mapping (A, B): X —÷ CN+M. Then, by Lemma 2.8.3(a) (applied M times), there exists an (M x N)dimensional matrix C with IC Ii such that the mapping F := A + BC: X —p Ct" is regular on K. s/maxL liBli e. Moreover, maxK hF — All = maxL llBCii
:= (F (b) The proof is as in (a). We only need to observe that the set and next proceed 0(X)N: F is not injective or not regular on K) is closed in [J as in (a) (we use M times Lemma 2.8.3(a,b)). Definition 2.8.5. Let (X, p) be a Riemann region over C" and let F: X —* CN be holomorphic. Any relatively compact union of connected components of the set is called an analytic polyhedron of order N. Lemma 2.8.6. Let (X, p) be a Riemann region over C" and let K C X be a holornorphically convex compact set. Then for any open neighborhood U of K there exists an
analytic polyhedron P with K C P
U.
Proof Cf. the proof of Lemma 1.10.12(a). DCII denotes the operator norm of C. i.e. UBCV < IIBIIIICD.
D
2 Pseudoconvexity
198
Lemma 2.8.7. Let (X, p) be a Riemann region over Ce!, let K C X be a compact set, and let P be an analytic polyhedron of order N + 1 such that K C P. If N ? 2", then there exists an analytic polyhedron P of order N such that K C P C P. Proof Assume that P is a relatively compact union of connected components of Q := such
thatK C
FN+,) E O(X)N4* FixO
F'
Consider the Riemann region X := X \ Ff41 (0) and the compact set K := {x E
P: IFN+II 2 c}. Since N 2n, the proof of Lemma 2.8.4(a) (applied to X) implies that there exists an (M x N)dimensional matrix C with arbitrarily small entries such that the mapping F := (FI/FN+I
FN/FN+I) + BC
is regular on K, where B E O(X)M is injective and regular on K. Put
(G,
GN) := FN+IF = (F1
FN)+ FN+IBC E t9(X)N,
GN,FN+I), It is clear that if IICII is small then K C
and U
Qk := (x E X:
I
P. Put N}, k EN.
Let Pk denote the union of those connected components of Qk that intersect K. LetL := {x E 8U: IFN+I(x)I c}. RecallthatLisacompactsetandKflL =
LetO <
0.
Since L c K, rank F'(x) = n for any x E L. Consequently, there exists a constant x > Osuch that for anyx EL > < dk, 2bk < ct, and a k >> I (at least so big that: 2a1' < bk, bk + I/k2
Our aim is to prove that Pk C U for k >> I (then P := Pk will be the required
polyhedron of order N). Suppose that Pk U. Then Pk fl following two possibilities. Xk
L. Then
<
0. Fix an xk E Pk fl
bk + ck < d", j = I
There are the
N. Hence Xk
8U;
contradiction.
E L. Let yk: [0,1] —* be a curve such that yk(O) = xk, E K. fl K = 0, there exists a point Yk 1]) such that (xk, r0) fl Yk ([0, Since IPx (xk, E IPx p(xk)I = I/k2. Notice that Yk Pk and, therefore, E Ip(yk) — xk
N.
I
Moreover, Gk(xk)
IF (xk) — II = I
1/2 < I — (b/c)k < 1F1(xk)Ik
—
(xk ) I
k
< (b/c)k,
I + (b/c)k <2,
j=I
N.
2.8 The Remmert embedding theorem
199
N}besuchthat
.
> We have:
= IFN+l(yk)IkIF(yk)
—
FN÷I(yk) k( Vk
= k
1
+
kf

/l\ /1 1+O(—) (—k \k/ \2
ckfl
x
+
I
k
Fj(xk)
I(' +
F() i
FJ(xk)
fl
i) +
—
FN÷l(Xk) c
—
—
k
+
—
—
— ii)
I
+01—) —I— 'k21
'c
bk
 () xck
k >————i l=———>b 2\8k \cJ / 16k 2
k>>l
0
contradiction.
Proof of Theorem 2.8.1. Recall that (9(X) separates points in X (cf. Theorem 2.5.1). Hence by Lemma 2.8.4(b), there exists a holomorphic, injective, and regular mapping
G: X —+ Assume
that we have already constructed a holomorphic mapping F: X —+
such that {x E
X: IF(x)I
X,
kEN.
Then, by Lemma 2.8.3, there exists an (2n + 1) x (2n + 1)dimensional matrix C with ICGI s IGI such that the mapping H := F + CG is injective and regular on X 60)• 60) Indeed, let (K1 be a sequence of compact sets with = X. Consider the injective and regular mapping If := (F, G) and N = k := (2n + I) + 2n. By Lemma 2.8.3 for each j there exists a zero measure set C Ce" such that the mapping
(H1+aIHk÷I
Hk+akHk÷1)
(*)
is injective and regular on for anya 1). Taking T := 1) we get azero measure set such that the mapping is injective and regular on X for any a T. Fix an a T with 101 e. Now we repeat the same procedure with respect to the mapping (*) and N := k — 1. We find b.c E Ck_l with e such that the mapping Ibi, Id
(H1+blHk+cIHk÷1 is regular and injective on X. Finally, finite induction finishes the proof.
200
2 Pseudoconvexity
Moreover,
(XE X: IH(x)I
{x E X: IF(x)I
kEN,
consequently. H is the required mapping.
We move to the construction of F. Recall that X is holomorphically convex and therefore there exists a sequence of holomorphically convex compact subsets of X such that K3 C mt and X = U'°=, K3. By Lemmas 2.8.6, 2.8.7, there exists a sequence of analytic polyhedra of order < 2n such that K3 C P3 C j 1. Assume that we have found a holomorphic mapping F: X —+ such that
IF(x)I > k + Mk+I,
XE Pk+1
\
k EN,
(t)
where Mk := SUPPk GI. k N. Then for any k E N we get {x
X: IF(x)I
X.
Thus we turn to the construction of a mapping F with (t). Assume that we have constructed a holomorphic mapping
F: X —+
such
that
IF(x)I>k+M&+i,
XEdPk, k€N.
(j)
Then put
Ak := {X Pk+I \ Bk := {x Pk: IF(x)I
F(x)I
Condition (1) shows that for any k the sets Ak, Bk are compact (and disjoint). Directly from the definition we conclude that fl = Ak U Bk. By the approximation theorem (Proposition 2.7.4) there exists a function 1k E 0(X) such that
flXIfkI
1/2".
XEAk,kEN. Define FN+1
2.9 The Docquier—Qrauert criteria
Observe that
mt Bk
= X.
201
Hence the above series is locally uniformly convergent
on X and defines a holomorphic function. It is easy to check that F4v+i(x)I > k + Mk+l for x Ak, which directly implies (t). Thus we move to the construction of a mapping F with (i). Suppose that Pk is a union of connected components of Qk H1 where Since Pk...1 CC Pk, we get maxp5_1 IHkI < I and JHk(x)I = I for Hk E x aPk. Consequently, there exist ak > I and m,, N such that if we put
then
maxlllkl <
xEapk. kEN. Define
F It is clear that the series is locally uniformly convergent in X and that F satisfies (i).
0
2.9
The DocquierGrauert criteria
The aim of this section is to localize the description of the pseudoconvexity. 61) Theorem 2.9.1. Let (X. p) be a Riemann domain over C's. Then the following conditions are equivalent:
(i)
(X, p) is a Riemann—Stein domain;
(ii) for any continuous mappmgs f: [0. 1] x E —+ X (cf Section 1.5) with o f)(t,•) e (9(E)for any t [0, 1), the following implication is true: 1) x E) U ({1} x SE)) C X, then f([0, 1] x E) C X;
(iii) for any continuous mapping 1: [0, 1] x E
Xfor which the mapping
oI
extends to a holomorphic mapping g: D x E —÷ C", where D C C is neighborhood of [0, II, the following implication is true: 1) x E)c X,thenf({1} xE) C dXorf({l) xE) C X. 61)
The reader is advised to recall the main results from § 1.5.
a
2 Pseudoconvexity
202
(iv)for any domain T =
T
a holomorphic mapping f: a
Riemann—S rein domain
(Y, q) over C" and a morphism
and a biholomorphic (X, p) —* (Y, q) such that for any domain T = mapping f: T —* f(T) C X such that o f extends to a biholornorphic mapping E" —÷ 1(E") C Y, there exists a holomorphic mapping f: E" —÷ X such that f = I on T:
(vi) there exist a Riemann—Stejn domain (Y, q) over C" and a morphism
w: (X,p) —+ (Y,q) such that there exists a neighborhood U of d X with — log
E
?S3t'(U fl X).
Remark. There is a domain of holomorphy G C C3 and a continuous curve a [0, II —÷ C2suchthatUo<,<1((a(t)}xE) c Gand(a(1),0) E G,but(a(1)}xE G (cf. LFav 1979]).
Proof of Theorem 2.9.1. By Lemma 2.2.8 and Theorem 2.5.7, each of the conditions (ii), (iii), (iv), (v), (vi) implies (i).
(ii): Put K := f([0, 1] x dE). Then K is compact subset of X. Let
(i)
For each t E [0, 1) the function v(f(t, .)) therefore, by the maximum principle, supg v(f(t, .)) vE
Hence
(i)
f([0, I) x E) C :
is
subharmonic in E and
v(f(t, .))
V.
X.
(iii): Suppose that the theorem does not hold, i.e. that there are points X,,
j = l.2,withf(l,Aj) E X,butf(1,A2) E ax. Observe that dx(A) = 0, where A := {f(l — 1/k, A2): k E N). Because of Theorem 1.10.4 we find a holomorphic function h E c9(X) such that
lim Ih o f(l — 1/k, A2)I = +00.
=10,1] x Eand B' := ([0, 1)x E)U((1,.A1)}. Forevery point (r, A) B' we choose a neighborhood U(t, A) := U(f(t, A)) C X such that U(r, A) —÷ p(U(r, A)) is biholomorphic. Since o = there exists ForabbreviationweputB
a bidisc V(r. A) c D x E, centered at (r, A), such that
g(V(r,A)) C p(U(r,A)) 62)
Recall that
:=
and
f
f(V(r,A)flB) C U(r,A).
where
n—l.IznI
Zn) E
C":
n—I, p <
< flu.
203
2.9 The Docquier—Grauert criteria
Put
f(r.A) :=
V(r. A)
o
X.
f on
Obviously, the function
V(r, A) fl B. Fix two such parameters (t', A'), (r". A") E B' such that V0 := 0; in particular, this intersection is connected. Observe V(t". A") fl V(v', A') that f(t',A')IVo = 81v0 =
f(r'.A')(To, A0) = f(ro. where (To, A0)
:=
+ r", A' + A") E
lifting it follows that
=
Put V := U(r.A)EB'
A0)
= f(r".x")(to, A0),
fl B'. In virtue of the uniqueness of the
V0
on V0.
A). Then V is an open subset of C2 with B' c V. Define the function 7: V —÷ x, := f(r.A). It is clear that fIB' = f a'. Chooser with max{JA1I, A21} < r < I. Then there is an open subset W' of C V
such that [0, 1) C W' and W' x K(r) C V. E —* K(r) with t(0) = Ai and choose Take a biholomorphic mapping neighborhoods W1 = W1(l) C D and U1 := U1(0) C C such that W1 x £(Ui) C V. We introduce the following function
F: (W' x E)U(W1 x U1)—÷ C,
F(r,A) :=h(f(r,e(A))).
Put W := W' U Wi. In what follows we study the Taylor expansion of F(r, .), r A = 0:
E W,
at the point
F(r, A)
= 1 for Let R: W —+ (0, +oo] denote the Hartogs radius of this series, then R(r) any r E W'. Since —log R is a psh function on W, Oka's theorem (cf. [VIa 1966])
leads to the following additional information:
—logR(l) = lim(—logR(r))
0:
r/1
1. in particular, we havethat R(l) We choose r' E (0, 1) in such a way that
a neighborhood
W
and uniformly on
< r'
x K(r')
1.
Then there exists
x K (r'). Thus we obtain a holomorphic
function (W' U W)
<
that
W
C,
F(r, A) :=
204
2 Pseudoconvexity
The fact that
= tim
IF(l.
IF(l — 1/k,
r' (A2))I = urn
k*+oo
leads then to a contradiction. The implications (i) (iv) and (I) The implication (i)
:
Ihof(1 — 1/k.
= +oo
(v) follow from Proposition 1.9.1.
(vi) is obvious ((Y, q)
(X. p). p := id. U := X).
0
Let
x 8E.
x E,
The main local criteria for the pseudoconvexity are contained in the following theorem.
Theorem 2.9.2 ([DocGra 1960]). Let (X, p) be a Riemann domain over C". Then the following conditions are equivalent: (I)
(X. p) is a Riemann—Stein domain (cf Theorem 2.5.7);
(p1) there exists afunction u e .'PSJt°(X) such that (x E X: u(x)
Xfor any
Xfor any compact K C X;
(ps) X =
where
is a is St rongly pseudoconvex domain with real analytic
boundary, and
v
1;
(pa) there exist a Riemann—Stein domain (Y, q) over C" and a morphism
(X. p) —* (Y,q) =q'
such that any point a d x has a neighborhood U C X such that the region (X fl U. p) is holomorphically convex;
—* X, (ps) (Kontinuitätssatz) for any sequence of holomorphic mappings where D,, C C is a neighbo rhood of E, v 1, we have thefollowing implication:
'f (p6)
X, then
for any biholomorphic mapping f: W —÷ f(W) C X. where W C C" is a neighborhood of & if
X, then
X;
(p7) there exist a Riemann—Siein domain (Y, q) over C" and a morphism
p: (X.p) —+ (Y,q) such that there is no continuous mapping f: E" —÷ X with the following properties: The notations (pi) —
(p) are taken from IDocGra 1960].
205
2.9 The Docquier—Grauert criteria
(ti) (t2)
(t4)

C X,
nax
0,
o f extends to a biholornorphic mapping in a neighborhood of
there exist a Riemann—Stein domain (Y. q) over C" and a morphism
q: (X,p) —÷ (Y.q) =9'
=9'
such that any point a E d X has an open neighborhood U C X such that there is no continuous mapping f: E" —÷ U with the above properties (ti —t4).
Proposition 2.9.3. if f is as in
then there exists a bi/wiomnorphic mapping
f: W —* f(W)c X, where Wisan open neighborhood of&suchthatf = I on
Proof. We know that the mapping
o f extends to a biholomorphic mapping In particular, f is injective on E" and is injective on the compact f(E"). Since is locally biholomorphic on f(A), for any 0
g: V —+ g(V) C Y, where V is an open neighborhood of
Proof of Theorem 2.9.2.
(1)
(3) (i) (4) (i)
(ps)
(i)
(ps)
(p4)
(7)
*
(12)
)
(1)
(pi) follows from Theorem 2.2.9.
(1) (i) (2) (i)
(4
(i)
.'
(p7) follows from Theorem 2.2.9.
(p3) follows from Proposition 2.2.6(b) and Corollary 2.2.11.
(pd) is trivial: we take (Y, q) = (X, p) and U := X. —÷
X,v > l,beasin(p5)withA :=
X. Put K := A. It suffices to show that
c
Take a
206
2
Pseudoconvexity
and fix a v. Then v o vE the maximum principle we get:
is subharmonic in
maxv<maxv.
aE
E
and, consequently, by
K
(ps): Let f: W —+ f(W) c X be as in (p6) with cc X. (6)(p2) Take K := Let D := X. It suffices to We know that D show that f(E") c D. Since f(E'1) is open (f is btholomorphic), we only need to :
show that Take an arbitrary function v E c the maximum principle, we have:
sup v = f(E")
sup v o
E'
f
sup v o
f = sup v o f =
sup
Then, by
v <sup v. K
Thus f(E'1) c =w
Let a E ax and let U C X be such that (X fl U,p) (7) (p4) is holomorphically convex. We already know that (i) (p6). Thus the region (X fl U, p) satisfies (ps). Suppose that f: E" —+ U is a continuous mapping satisfying (ti—ti) of (p7). By Proposition 2.9.3 there exists a biholomorphic mapping f: W —p f(W) C X, where W is an open neighborhood of such that I = I on
By
(ti) we get
X. Now condition (p6) implies that
f(E") = f(s) f(s) X which contradicts (t3). (8) (ps) : (p6): Let f: W —* f(W) be a biholomorphic mapping of an open neighborhood of such that A := X. Take an arbitrary sequence = f(z,) t'> 1. Then
f(s),
=
Zv,n) E
v>
C A and, therefore, we can select a convergent subsequence. (9) (p6)
1,
and define cov(A) C
A),
X. Now
(p7): Use (7) with U := X =c',
(10) (p7)
is tnvial: we take U := X (i): It suffices to check condition (v) from Theorem 2.9.1. Let
(11) (p7) ço: (X. p) —+ (Y, q), T = Tr,p, f: T X, and be as in(v) from Theorem 2.9.1. < I such that f extends holomorphically to We may assume that there is no r p'E. Since extends ço o fIg. Fix p < p' < and let Q := 1
Proposition 1.5.8 implies that f I extends continuously to a mapping f: Q —+ X x p'E). to the mapping f substituted by Now we can apply Note that extends 4 c f to a neighborhood of Q. Consequently, K := f(Q) C X. Since c f = is injective on K. Hence there exists a on Q, the mapping neighborhood U of K such that c'Iu is also injective. Now, the mapping (coIuY' 01 gives a holomorphic extension off to W := Finally, f must extend to some
with r'> r; contradiction.
207
2.9 The Docquier—Grauert criteria
(l2)(p)
=c (i): Take an a E aX and U as in (p). Let yo := e(a, V). where V is a connected neighborhood of
and suppose (and e(a, V)  '(V) that belongs to a). Fix 0 < r
•,
that U fl X = denotes the connected component of such that Py(yo, r) V and let := e(a, lPy(yo, r)). =* Let and let T: —÷ X be as in Remark 1.5.10. satisfies (p7) with respect to the morphism At first we show that
Letf: 
beacontinuousmappingsuchthatf(Si\)
E't —+
=
=IIfr
f extends to a biholomorphic mapping in an open
neighborhood of E".

:= Tof. Then ettherg(E")fl a X 0. 3!Py(y0, r) Putg
øorg(E'1)
C X and
On the other hand, since lFy(y0.
r) is pseudoconvex,
Py(yo, r);
contradiction.
Thus by (ii) — neighborhood
E
such that dx =
C
Finally, by Proposition 1.5.9, there exists a on Wand therefore — tog dx =c"
In particular, there exists a neighborhood Z C X of a x such that — log dx
fl X). Now, it remains to use Theorem 2.9.1 (vi).
E
0
The Docquier—Grauert criteria remain even true in the following much more general
context (which, unfortunately, is beyond the scope of our book).
Theorem 2.9.4 ([DocGra 1960]). Let M be an ndimensional connected Stein manifold (cf Section 1.10) and let (X, p) be a Riemann domain over M (cf Section 1.1). Then the following conditions are equivalent: (1)
X is a Stein manifold;
(X) such that {x E X: u (x) < t I
(ps) there exists a function u E
X for any
tE
Xfor any compact K C X;
(p2)
(ps) X =
where
boundary, and
is a is strongly pseudoconvex domain with real analytic v
I;
(p4) any point a e ax (the boundary ax is taken with respect to the mapping p: X —* M analogously as in Section 1.5) has a neighborhood U C X such that the X fl U is Stein;
—* (ps) (Kontinuitatssatz) for any sequence of holomorphic mappings where C C is a neighborhood of E, v> 1, we have thefollowing implication:
if
X, then
2 Pseudoconvexity
208
(p6) for any biholomorphic mapping f: W neighborhood of X, then
f(W) C X, where W C
f(s)
is a
X;
there is no continuous mapping f: E" —÷ X with the following properties: X,
(t2)
f extends to a biholomorphic mapping in a neighborhood of E?z; (p) any point a E ax has an open neighborhood U C X such that there is no continuous mapping f: E'1 —÷ U with the above properties (t I—t4). As an application of Theorem 2.9.2, we get:
Theorem 2.9.5 (Kajiwara theorem, IKai 1965]). Let
:= liin((X1. p),
((X, p),
(cf Proposition 1.6.8). Suppose that all the domains (X,, pi), i E I, are pseudoconvex. Then (X, p) is pseudoconvex.
Pmof To prove that X is pseudoconvex we use the DocquierGrauert criterion (cf. Theorem 2.92(p6)). Assume that f: W —p f(W) C X is a biholomorphic mapping,
X. Wehavetoprovethat
where W
f(s)
X. Let V be a relatively compact neighborhood of
where
U
x E. By Proposition 1.6.8 there exist Jo and a relatively compact V0 —÷ V is biholomorphic. Observe that for open set V0 c X10 such that any 1/2 < t < 1 there exists an e = e(t) > 0 such that f(G,) C V. where G, G, —* V0.
Let*,
holomorphic mapping '1',: G, —* X30, where G, := (I E)" — x < I + e(t) } (O, o is the envelope of holomorphy of G,). By the identity principle we get =f on G,' fl G,'. Thus, there on G,. In particular. '1', is injective on C, and = exists a biholomorphic mapping 4': Uo —f C Xj0, where Uo C U is an neighborhood of such that Since o '1' = f on Uo. Moreover, X10 is pseudoconvex, we get 'P(s) = X. 0 X10. Finally, {
I
2.10 The division theorem F1 and let 0(X) be given. FN Roughly speaking, the division problem is to characterize those cases where there exist functionsuj (9(X) such that UN
Let (X, p) be a Riemann domain over
U°
=
... + UNFN.
209
2.10 The division theorem
In the case where the functions u0, F1,..., FN satisfy some additional regularity conditions (e.g. they are bounded), one may ask whether the functions ui. be found in the same class of regularity; Corollary 2.10.5, Lemma 4.4.6.
.
.
.
, UN
can
It is clear that to solve the division problem one must at least assume that {x E F1(x) .... = FN(x) = 0) C E X: u0(x) = 0}. Below we will present a general division method based on the socalled Koszul complex introduced in [Hör 1967] and used, for instance, in [KelTay 1971], [Cno X:
1972], [Sko 1972]).
We will need the following extension theorem for forms.
ProposItion 2.10.1. Let (X, p) be a Riemann domain over C" and let S be an analytic subset of X with dim S n — 1. Suppose thatu E bc), v E bc)
•,fau=OonX\S,thenau=oonX,
• if!
(9(X
\
5) fl
bc), then f exterwLc holomorphically to X.
Pmof First consider the special case where (K, p) = id), S = Ek x {0}fl_k (0 k n — 1), u E v E We have to prove that for any çø E
f Fix such a ço. Let Obviously
çø.
= I on
[0. 1]) be such that
E
= (z', z") E Ck x
z
17E(z)
VA
e > 0, and let
Define
:= (I —
\ 5). Hence
E
(_1)r+s+1
A
JE"
j
VA
Observe that
/
/
JE"
/
JE"
VAp
e—.O JE"
by the Lebesgue theorem. On the other hand Ud(p€
JE"
=
j(l
By the Lebesgue theorem
—
11ç)U
A
U —
U
—÷ 0 as e —* 0. Put
j(I)
—
when e —+ 0. It remains to show that Note that
A
:= Ek x
1(2)1 < CIIUIJL2(A
A drk A
JE"
) .
210
2 Pseudoconvexity
a constant independent of e). Since IIu IIL2(AE) —+ 0 as e —÷ 0, we only need to show that is bounded when e —÷ 0. Indeed, we have: (C denotes
<
<
Now we move to the main proof. Let d := dim S.
If d = 0, then we apply the special case locally (after a biholomorphic change of coordinates) with k := 0 and we get the assertion.
Suppose that I sd Sfl — I. Then inaneighborhood of any regularpomt ofS we apply the special case with k := d. This shows that there exists an analytic set S' C X with dim S' cd — I such that du = v on X \ S'. Finite induction finishes the proof. Let (X. p) be a Riemann—Stein domain over C"
and let
F=(F1 be
a holomorphic mapping. F
=
0. For r, s E Z+, define
bc): au
(r.s)(X) := {u
loc)}.
E
Fort ENputM(N,t):=(Nfl[i,N]Yanddefine := {u = (UI)IEM(N,)
=
= sgn(J)uI},
V!EM(N,1): Ui E A?rs), where
E, denotes the group of all permutations of t elements 65)• Observe that
=
(0)
ifr > n ors —p
> n on
> N. Define := du1,
I E M(N, t).
Let S
:= (x E X: Fj(x) = ...= F,v(x) = 0}.
Notethat S isananalytic subsetofX withdim S sn—i. lathe sequelthefollowing
direct consequence of Proposition 2.10.1 will be frequently used. Notice that the first part of our construction of the Koszul complex may be performed also for an arbitrajy Ricmann domain. 65) The definition of says that the system (u()JEM(pj,) IS SkeWsymmetriC 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 beskewsvmmetricsvs: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.slI)
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 skewsymmetric
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 skewsymmetric 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 skewsymmetric system of forms from (X, bc) and dh° should be considered in the sense of distributions. Observe that by (2.10.6) and (b)
Ph° = u0.
(2.10.7)
Moreover, by (2.10.4), using the inequality XF H lihOlle
II
L2(X)
C
II
2 log II Fli, we get
u0 (II Fll
H
L2(X)
The main idea of the construction is to find an element g1
u :=
—
such that
Pg'
solves our division problem. Notice that we have Pu = u0 independently of g1. The problem is to get (ii) and (iii).
=0. Theneithers =norr = N—I. We = 0, i.e. h0 itself solves our problem. Ifs = n, then obviously dh° E = (0k.
will show that in these cases
If r = N — 1. then AN
U ence
i
= 8Ph° =
= 0. Recall that P is injective on
—
Thus we may assume that
s
2.10 The division theorem
215
On X \ S we define formally the following sequence
j = 1,2
h' Using (2.10.4), (2.10.5), and (a) we get < IIah'II
CIIu°JJ
I
j = 0. 1,2
U8FIV4'
CIIu°II
In particular, by (d),
j=
(2.10.8)
.
for j =
E
0
— I
and
for
E
Moreover,onX\Swehave
E
1.
(2.10.9)
= ah'—',
j=
N— t
1
(2.10.10)
— 1.
Indeed, by (2.10.6), (2.10.2), (2.10.7), and (a), we get inductively:
Ph' =
PQdh°
=
=
dh°
—
QPah° = ah°
—
= aW—' — QPah'—' Qa
aW—2 = dh'',
QdPh° = ah°
= ah'—' j=2
—
Qau° = ah°,
—
N—i—i.
Now we will show that
=0. Consider the following two cases: = n — s: Then ahL E
=
(2.10.11)
{0}.
= N —: — 1: Then and therefore it suffices to show that E Pahu = 0. Indeed, by (2.10.10) we get Pah,L = = = 0. By (2.10.8) we get II
II
L2(X)
FII_(2i
< C IIu°Il
0L2(X) < CL
By virtue of Theorem 2.4.22, using (2.10.11), we conclude that there exists a g'2 E such that agIL = hIL and
° Suppose
that for some j E (1
such
CR. —
1}
we have already constructed gIL,
that g
k
E
At+1+k
iir.s+k—1'
= hk —
op)2 11L2(X) < Cu,
(2.10.12)
k=
j +I
/1,
(2.10.13)
216
2 Pseudoconvexity
where := 0. We will construct g1. Observe that by (2.10.10) and (2.10.12) we have: —
=
=
—
pgJ+2) =
—
such that
Hence, by Theorem 2.4.22, there exists a and H JIg3
II
ag' =
0. h
(2.l0.l4) —
Jib' — PgJ+I II exp(—fp —
o p)211
—
and
E
—
It remains to estimate the last term. By (2.10.8) and (2.10.13), using the inequalities
obtain
h' — pg f+'
L2(X)
Cl llu°jI . ilaFil'
.
Ii F
1—(2j+
+ Cl II
U FII H
Cu + cli
lie
—2(p—j)
. I!dFIi
L2(X)
L2(X)
P)2 HL2(x)
CL (2.10.15)
Consequently, after finite induction, we construct g1. Put u := h° — Pg1 E
Then, by (2.10.7), (2.10.14), and (2.10.15)
Pu = Ph° =
u0,
= 0,
HIIuIICHL2(X)
(2.10.16)
Thus we have proved the following
Theorem 2.10.2 (Division theorem). Let (X, p) be a Riemann—S:ein domain over FN): X —+ Cw be a holomorphic mapping, F 0. Let Let F = (F1 r.s nand00(depending only 0 on n, N, r, s, 1) such that
for any
E ,9'SJ?(X) and for any element u0
Pu0 = 0. du0 = 0. and 11
ltllu°it
< +00.
.
such that Pu = u0, au = 0, and
there erists a u E
< CL where ii := min{n
—
s, N
—
In the case z = 0 we get:
i — l}.
With
2.10 The division theorem
217
Corollary 2.103. Let (X, p) and F = (F1,..., FN) be as in Theorem 2.10.2. Let < r, s ii. Then there exists a constant C > 0 (depending only on n, N, r, s) such thatforanyço E and for any aclosed form u0 E with
0
I := there exists a
u = (uj,
.
< +00,
. IIFII .. ,
such that au1 = ... = dUN = 0
E
and
5CR.
where ji := min{n
s, N — I).
—
In particular if r =
s
=
then we get the following division theorem for holo
0.
morphic functions. .p E
There exists a constant C > 0 (depending only on n and N) such that for any and for any u0 E 0(X) with II
:= IIIu°I
there exists a u = (Ui
UN)
< +00,
.
E
0(X)" such that
u0 =FIUI+•••+FNUN. Cli,
II
whereji :=min{n,N —1). The following lemma will be useful in the proof of Corollary 2.10.5.
Lemma 2.10.4. Let 8: X —+ R>0 be such that 18(x') — 8(x)I S IIp(x')
Then (a) where
lix
X
—
d(n, k, v)ll8kfllx,
d(n, k, v) :=
In particuku the operator d": anyk Oandv E L'(X), then
k >
E X, x'
0, v e
f
E
O(k)(X 8),
(k +IvI)k+IvI k
8)
8)
is continuous for
k >0, q >0, f E
218
2 Pseudoconvexity
Proof (a) Fix k. v, f, and a E X. By the Cauchy inequalities, and by (c2) we get <
8
I
<
I 15k111
e)k }
I
0
(b)is obvious.
Corollary 2.10.5 (Division theorem for functions with tempered growth). Let (X, p) be a Riemann—Siein domain over C'T and let
8: X —÷ (0.11 the following conditions: := min{px. So o p}. • 1S(x') — 8(x)I lp(x') — p(x)II, x E X. x' E — logS (cf Proposition 2.5.4).
.8
•5ae Let x > 0, N E
N,
forsonzeb>0,
and F = 0, u0
5)N 67) F
FN) E 0(X) we have (F1
0. Assume that
onX. Then there existuj v
:=
S)withu0 = u1Fj +
UN E
+ fi + js(2x + 3) + n, 1tL := min{n.
Proof Put
:=
k
I
:=
+
N .—
+UNFN on X, where
II.
Then 21i+l)5k11
<
Hence, by the division theorem (with ço := —k log 8), there exist functions UI
Notice that if (X. p) = (G. Id). where G is a domain in C". then > 0. E L2(C"), Recall that 8) < +oo}. (1 E 9(X):
and
.5
67)
UN€0(X) E
L2(X) (because
219
2.11 Spectrum
suchthatu°=uiFi
and —ILXF) ii L2(X)
CL
where C > 0 depends only on n and N. By Lemma 2.10.4 XF = log UFII + max{log (IIaFIRI + 11p112)), log IIFII}
where A depends only on x and H
IF liii
A — (2x
+ 3) logS,
Consequently.
HIIUIISIIL2(x =I
Finally, by Lemma 2.5.3, U3 E
N.
D
Corollary 2.10.6 (Division theorem). Let (X, p) be a Riemann—Stein domain over UN E (9(X) = 0. Then there exist and let F E (9(X)N be such that such that 1 = F1 + ... + UN FN. Proof By Corollary 2.10.3 we only need to find a function ço E ?S,1e(X) such that
< +00. Since
E L2(X, bc), wecanproceedas intheproofofLemma2.4.18. U
Exercise 2.10.7. Let (X, p) and F be as in Theorem 2.10.2. Assume additionally thatd:=sup(IIp(x)II: XE X} <+00. Prove that in this case the division theorems 2.10.2 and 2.10.3 remain true with XF
:= log IIFII + max{log 118F11, log IIFII}
and with a constant C depending only on n, N, r, 5, t, and d.
As a corollary prove that the division theorem 2.10.5 remains true with v :=
l)+n.
2.11 Spectrum and let A be a closed astable subalgebra Notice that I A. of (9(X) with p E (e.g. A = clO(X) Let 6 = (5(A) denote the spectrum of the algebra A, that is the set of all continuous A —÷ C (in characters on A, i.e. nonzero continuous algebra homomorphisms = 1). Let (X. p) be a Riemann domain over
220
2 Pseudoconvexity
Remark. In fact, any homomorphism .4 —÷ C is continuous; cf. Theorem 2.11.5. This is a particular case of the socalled Michaelconjecture.
Forf E Adefinef:6 For F = (f, 6—. C't.
IN) E
{f:f
Let,4 := E A). IN). In particular, we define
—÷
.4N
put F :=
.4 —. C be the evaluation at a, i.e. := f(a). = Observethatfoa = f, f Ed.. 6. Puta: X —+ 6,a(x)
For a E X let E
In particular, j%oa = p.
Proposition 2.11.1 (ERos 1963]). There exists a topology on 6 such that: • (6, is a Riemann region over C", • a: (X, p) (6, j5) is a morphism,
• A C 0(6), • a: (X, p) —* (X, j3) is the maximal .4extension, where X denotes the connected component of 6 thai contains a(X). Proof First, we endow (5 with a topology. Take a
set K =
C X such that
IIfIIK,
inequalities, we get
f EA. V :=
Let
f
E (5. Then there exists a compact Hence, by the Cauchy
0
(2.11.1)
Put
f€A.
>
Note that by (2.11.1) the series is convergent. Observe that: • f, ) is holomorphic in P(zo, dx(K));
•çp(4. l,z)= 1; • •
1, z)I f + g. z)
r < dx(K); =
I, z) +
Thus for each z E iP(zo, r), 0 < r belongs to 6 and
f,
g, z); the mapping .4
f
.4. Hence
we may assume that c
E .4, k
= I; ef. Remark 1.4.5(k).
N. Consequently,
1, z)
A.
ByconunuitythcrcexistacompactK C Xandaconstantc >
f
f —÷
cUfUp, E A. Thus
221
2.11 Spectrum
r):=
z): z E 1?(zO, r)). We endow €5 with the topology O0. — r, then we get Indeed, if z E ?(w, p) C IP(zo. r) andO < p
Let U =
•,
in which the system
(z—w) I)
=
—
=
Zo)L)(z — w)t
—
=
zoYL(z
—
(z—zoY (2.11.2)
Recall that z)) = z. Hence the mapping j)Iu is a homeomorphism onto P(zo, r) and , z), z E IP(zo, r). = The topology of €5 is Hausdorff. r) fl r) 0 Indeed, let If then 6, for small r > 0. Assume that = zO and let r > 0 be such that = r) = 0. Suppose r) fl 2r) is defined, j = 1,2. We will show that •, w) = w) for some w E IP(zo, r). Then, by (2.11.2), for f A that we have: w), 1, zo)
1, zo) =
(1)
=
u'). f, zo)
=
f, zo) =
contradiction.
We have proved that (6,
is a Riemann region over C's. Moreover, for f
we get:
z€P(zo.r). which shows that A C
0(6).
Now, we will show that a is continuous. Indeed, if x E IPx(a), then
f, p(x)) = Hencea(x)
—
.,
p(x)).
f(x) =
= a(x)(f).
A
222
2 Pseudoconvexity
Thusa:(X.p)—+(15,j5)isamorphism. is
an Aextension. Let (X. p) —* (Y, q) be any other holomorphic extension. Define a: Y —÷
a(y) :=
Then, by Remark 1.4.5(i), a is well defined. It is clear
o
anda = a. Moreover,a is continuous. Indeed, forb E y E Py(b, r) (r > 0 small) we have:
that ,5oa =
q
f q(y)) =
Y
and
(q(y) —
—q(b))"
=
=
I
= a(y)(f). Hence a(Y) C X and so a is a morphism of (Y, q) into (X. j3).
0
Let
Hom(A, C) :=
A —+ C:
is an algebra homomorphism),
C) := Hom(.4. C) \
10).
Proposition 2.11.2(Igusa theorem, cf. [Igu 1952]). Let (X. p) be a Riemann—Stein domain over C's. Then the mapping
X3 x
C)
E
any algebra homomorphism C).
is bijective. In
0(X) —+ C is continuous
and, consequently, (5(0(X)) =
Proof Since 0(X) separates points, r is injective. Fix a
E
C) and
let
:=
{f E 0(X):
We want to show that there exists an First observe that
E
=0).
X such that f(xo) = 0 for any f E
0. Otherwise, by Corollary 2.10.6, there exist UI
Hence 1
contradiction.
UN E 0(X) with
('ii)
223
2.11 Spectrum
Let zo :=
By virtue of (*) with F := p — zo we conclude that
)
p(X). Let Jo (9(X) be such that foIP'(ZO) is injective (cf. Remark 1.8.12). Take an f E 0(X). Then by (*) with F := (f, fo. p — Zo) we see that there exists a such that fo(xi) = f(x1) = 0. Since lois injective on pomtxf E the point xj is independent of f. zo
Proposition 2.11.3. Let (X, p) be a Riemann—Stein domain over C",
E X, and let
—+ f(xo)for
E X. t E T, be a Moore—Smith sequence. Then x, —+
anyf E0(X). Proof Suppose that f(x,) —p f(xo) for any f
E
0(X). Let F: X
be a Remmert embedding (cf. Theorem 2.8.1). Put M := F(X), zo := F(xo), g(zo)foranyg E (9(M). Inparticular,takingas z, := F(x,),t E T. Theng(z,) we easily conclude that z, —p zo. Finally, since g the coordinate functions in D F: X —* M is homeomorphic, we conclude that x, —÷ xo.
Remark 2.11.4. The above proof may be essentially simplified for ordinary sequences: Let
f
such that E (9(X). Let P E N}. Then A is relatively compact. Otherwise, by Theorem 1.10.4, A := v E N} = +00. On the other hand there exists an fo E 0(X) with fo(xv) —+ fo(xo): contradiction. be an accumulation point of A. Suppose that Let xO. = Then there exists an fo E 0(X) with = # lo(xo). On the other hand, = fo(xv) = fo(xo); contradiction. E
X, v
be
Theorem 2.11.5. Let (X, p) be a Riemann domain over C" and let .4 be a closed dstable subalgebra of 0(X) with p E .4". Then: (a) 6(A) = C). with the weak topology fmm A' 69) then (b) If we endow (6(A), fr) is a Riemann domain over C",
a: (X,p) —p (6(A).j)isanwrphism.
A = 0(6(A)), a: (X, p)
(6(A), j3) is the maximal Aextension.
Proof. (a) Let X be as in Proposition 2.11.1. First, observe that A is a closed i3is E .4 (Remark 1.4.5). Moreover, (X, stable subalgebra of 0(6(A)) with separates points in X domain of holomorphy (Remark 1.8.2(a)) and an = 0(X). (Proposition 1.8.10). Hence, by Theorem 1.10.4,
2 Pseudoconvexity
224
Since ax: (9(X) —*
is an algebra isomorphism, the mapping —÷
is well defined. Moreover, since ax is a topological isomorphism, A maps bijectively tS(.4) onto G(9(Xfl. Thus, assertion (a) follows from Proposition 2.11.2. (b) We already know that = (9(X). Moreover, by Proposition 2.11.2 we know that the mapping X is bijective. Hence, for E y —÷ r(y) = = any E 6(A) there exists a yo E X with = This simply means that = yo. i.e. X = 6(A).
Thus, it remains to use Proposition 2.11.3 to prove that the topology of 6M) (introduced in Proposition 2.11.1) coincides with the weak topology from A'.
Theorem 2.11.6. Let (X, p), (Y, q) be Riemann Stein domains over
and
0
Ctm.
respectively, and let
(9(Y) —+0(X) be an algebra homomorphism with mapping
(1)
F: such that
=
X
1.
Then there thsts a holomorphic
Y
=
Proof Take an x
X and let
0(Y)
g —L
E C.
= 1. By Proposition 2.11.2 there = g(y) for any g E (9(Y). Thus we
Then is an algebra homomorphism with exists (exactly one) point y Y such that
have defined a mapping F: X —+ Y such that E = F F = m, we only need to verify j =I that F is continuous. Let X —+ xo E X and g E 0(Y). Then, from the detinition of F. we get = = g(F(xo)). Hence, by Proposition 2.11.3,
0
F(xo).
2.12 Liftings of holomorphic mappings H Theorem 2.11.6 allows us to lift arbitrary holomorphic mappings to envelopes of holomorphy. The case of proper holomorphic mappings will be studied in Theorem 2.12.5.
Theorem 2.12.1. Let (X, p). (V. q) be Riemann domains over lively, and let
a: (X,p) be the maximal holomorphic
(X.13),
fi: (Y.q) —+
Then:
and Ctm. respec
2.12 Liftings of holomorphic mappings II
225
(a) For any holomorphic mapping t: X Y there exists a (exactly one) holomorphic mapping 1: X Y such that i o a = o t. if(Y, q) is a Rienzann—Stein domain over Cm, then any holomorphic In
mapping r: X —* Y extends to a holomorphic mapping i: X —k Y such that 1
70)
oa=
(b) For any topological space B and for any continuous t: X x B —+ Y such that
for each b E B the mapping Tb := r(., b) : X —* V is holomorphic, the mapping
1: X x B
b) :=
V given by the formula
is continuous.
Proof (a) By the identity principle i is uniquely determined. Let
:=(a*)_l
(9(Y) —÷ (9(X),
is an algebra homomorphism and = 1. By Theorem 2.11.6 there exists a holomorphic mapping 1: X —* Y such that = f Consequently, = a* 01*, which implies that fi or = I oa (because (9(Y) separates points). (b)LetX x B (x,,l,,) —* (xo,bo) E Xx BbeaMoore—Smith sequence. By Proposition 2.11.3 we only need to show that f(I(x,. b,)) —+ f(I(xo. b0)) for any function f (9(Y). Fix an f. Recall that Obviously,
f(I(x,b)) = (a*)_I(fo.florh)(x), r is continuous, we get f
Since
0
o Tb,
—+ f 0
(x,b) E Xx B. 0
locally uniformly in X.
Hence, by Remark 1.4.5(i) weget uniformly in k, which completes the proof.
(f
locally
Now we turn to discuss proper mappings. Let (X, p), (Y, q) be Riemann domains over C" and let F: X —+ phic. Define ,
Y
be holomor
xEX.
Theorem 2.12.2. Let (X. p), (Y, q) be Rienzann domains over C" and let
F:
X
Y
be a proper holomorphic mapping7U. Put S := (iF) — (0). Then: • S X,
•
the set
:= F(S) is analytic in V and E
70)
In other words, ifs: (X, p) —p (X. is the maximal holomorphic extension, then for any Riemann—Stein domain Y over Cm (with arbitrary n,), the morphism a gives also a holornorphic extension w.r.t. all holomorphic mappings X —+ V. 71)
Recall that F is proper if F1(K) is compact for every compact K
V.
226
2 Pseudoconvexity
• the mapping
Y\E
FIx\p—I(Z):
an analytic covering. i.e. there exists a number k E N (called the multiplicity of F) such that for any b E Y \ E there exists an open neighborhood V C Y \ E such that F 1(V) = U1 U•   U where U1 Uk are pairwise disjoint open sets, and —+ V is biholomorphic, j = I k, F'u,: is
will
follow [Rud 19801. We need a
Hausdorif measures. The fined by the
pth
few
auxiliary
results.
Hausdorff measure on RN is the outer measure de
formula
:=
sup
= lim
3€,
where
A=(JA,.
diamA1
ACRN
with respect to the Eucidean norm in RN; diam 0 := 0). It is that = #A. Notice that our definition differs from the standard one by a constant factor (ci. [Fed 1969]). (the
diameter is taken
clear
Property
(HI). If A C R" and F: A —÷ RN' satisfies the Lipschitz conditions with < p > 0.
constant L, then
Property(H2). if A cW"
and
< +00, then
> p.
Proof
3t(A)
Hence .1?(A) Property
—p 0 when 8 —+
0
0.
(H3). if A C RN then x [0, 1])
Proof LetA = number such that k1 diam
p > 0.
<8< 1,)
EN. Letk1betheminimalnatural A < 2). We find intervals
227
2.12 Liftings of holomorphic mappings II
1.
= 1,.
Observe that diam(A1 x
x [0, 1])
Then
x j=1 v=1
x [0, 1])
Hence
0
2P+23ft(A), which implies the required estimate.
Property (114). Let Q C RN' be an Ndimensional cube 72)• Then
forp
—
N,
forp>
N.
Proof By (H2) it suffices to prove only that J€N(Q) E (0, +oo). By (HI) and (H3) we may assume that N' = N and Q = [0, 1]". Fork €N divide Q intokN cubes with edge 1/k. Then
= Hence
RN(Q)
j
A with diam A3 N. Let B1 be a closed Euclidean ball Let Q = with radius diam such that A C B3 (B3 0 if A3 = 0), j E N. Let B denote the unit Euclidean ball in R" and let a = aN AN(B). Then
!AN(U A3)>
=
AN(Q)
0 Property (HS). Let (X, p) be a Riemann domain over fl
subset of X such that connected.
and let A be a closed
= Ofor any a E X. Then the set X \
A is
Proof It suffices to prove the following claim. be a convex domain and let A be a relatively closed subset of D (*) Let D C such that RN_I (A) = 0. Then D \ A is connected. 72)
That is. Q = L([O.
11N
x (o}N'_N) where L: RN'
is an affinc isomorphism.
228
2 Pseudoconvexity
Indeed, suppose that (*) is true, and let 00, b X \ A. Since X is connected, there exist £ N and points aj at X such that such that ao lP(ai), P(a1) fl
j
0, = I £ — I, andb E P(at). Observe that intA = 0 (by(H4)). in particular, there exist points £ — I. By (*) E (115(a3) 1) \ A, j = 1 the set \ A is connected for j = I £. Thus we can join ao with b1 by a curve in IP(a1) \ A. next b, with in IP(a2) \ A, and finally with bin \ A. We move to the proof of (*). Suppose that D \ A is disconnected and let a and b lie in two different connected components of D \ A. Let Q C D \ A be an Observe that (N — 1)dimensional compact cube containing a such that [a, b] I Q A fl[b, x I 0 for any x Q (because a and bare in different connected components). Let ir denote the conic projection with center at b onto the (N — 1)dimensional affine A fir plane containing Q. Observe that ir is Lipschitz on Ao let L be the Lipschitz constant ofJrIA0. Then, by (HI) and(H4), we get
0<
= RN_l (,r(Ao))
=0; D
contradiction.
Property (H6). Let D be a domain in C",
R2''(Z)= 0.
f
0(D), 1
0, Z
:=
Then
Proof By (H2) it suffices to prove that for any a D there exists a neighborhood K1, where is compact and U c D such that Zn U = <+oo,
j
N.
We apply induction on n. The case n = 1 is trivial. Assume the result is true for
n — 1.
Then, by the Weierstrass Preparation Theorem, we may assume that D
=
E" and
that f is an irreducible Weierstrass polynomial f(z', Zn)
=4
Let
z1
:= Z
:={z€Z:
\ Z1 =
{z E
1(z) =
1(z) = 0).
We will prove the result separately for Z1 and Zo. If a = (a',a,,) Z1, then in a neighborhood of a the set Z1 is the graph of a holomorphic function Zn = Consequently, by (Hi) and(H4), Zi = KJ'),
where K)" is compact and
< +oo, j
N.
That is. Q is an (N — 1)dimensional cube containing a and contained in the (N — 1)dimensional auline hyperplane perpendicular to the segment [a. fri.
229
2.12 Liftings of holomorphic mappings II
On the other hand, 4 C Z' x E, where Z' := R1 (0) and
—f C
R: is
the resultant of the polynomials f K1. where K is compact and
we can write Zo =
and
a—.
By the inductive assumption, Z' =
<
+00, j
E N.
Now, by (H3),
< +00,
is compact and
KJ°>, where
JEN. C] Proper mappings. Let (X, p), (Y, q) be Riemann domains over C" and let F: X —÷
I
be a proper holomorphic mapping. First observe that F is closed.
Property (P1).
is finite for any b E Y.
Proof The set M := is a compact analytic subset of X. Let M0 be a connected component of M. By the maximum principle the mapping p is constant on Mo. Hence M0 must consist of one point. Consequently. M Consists of isolated points. Since M is compact, M is finite. U
Property (P2). LetS := (JF)1(0). Then S
X.
Proof. Suppose that S = X. Let 1
Xo:=(xEX:rankl L
3Zk
1 (x)I=r}:
J
Xo is open. Then, by the rank theorem, the set (FIx0Y' (b) is infinite for b which contradicts (P1).
E
F(X0).
0
Property (P3). Let E := F(S). Then E is nowhere dense, Y \ E is connected, and F is surjective.
Proof Since F is closed, is closed. By (P2) and (H6) 3&''(p(S fl IP(a))) = 0 for any a X. Hence fl IP'(b))) = 0 for any b E Y. Consequently, is nowhere dense and I \ E is connected (by (H5)). By the implicit mapping theorem the mapping is open. The set F(X) is closed in V. Suppose that F(X) Y. Since F is proper and FIx\s is open, we easily conclude that aF(X) C E. Therefore, by (H5). Y \ is connected. On the other hand, we have the decomposition Y
\ 8F(X) =
contradiction. Consequently, we get
(mt F(X)) U (Y
\ F(X)); C]
2 Pseudoconvexity
230
Property (P4). FIx\Fl(E): X \
—*
Y
\E
is an analytic covering. More
(b) = kfor any b E
over, there exists a k E N such that
I \ E.
Property (PS). F is open.
Proof Fix an a E X and let Uo
X be a neighborhood ofa such that
fl On = {a}.
Let V be an open neighborhood of F(a) such that V fl F(aU0) = 0. Put U := V is proper (for any compact K C V we have Uo fl F'(V). Then Flu: U = fl Uo) and hence, by (P3), F(U) = V. 0
= £, F'(b) = {ai and letU u; Property(P6). Leib E ag, respectively. Then there exists an be arbitrary open neighborhoods of ai open neighborhood V of b such that F' (V) has exactly £ connected components t. In particular, by (P4), Ui U, wi:ha1 E U1 C UandF(U1) = V. j = 1 £
= k, then b
Moreover,
Proof Let Qi (2 C U,j =
E.
Qe be neighborhoods of t,andQ,LflQV = Oforv
at, respectively, such that
PutA := A is closed. Since F is closed and b F(A), there exists an open neighborhood V of I. b such that V fl F(A) = 0. Let := (2 fl F'(V), j = I and let T be a connected component of U := Uj. Recall Fix a j (I that V fl F(8U) = 0. Now we can easily prove (like in (P5)) that FIT: T —÷ V is 1
0.
proper. In particular, F(T) = V and soT fl
Consequently, the number of all connected components of all sets t, are connected and must be equal C. Hence j =I
j
=1
=
V,
j =
C.
Assume now that C = k. Then
U1 —÷ V must be bijective and hence
0 Proof of Theorem 2.12.2. By (P3), (P4), and (P6) we only need to prove that E is analytic.
ak}andletUl
UkandVbeasin
(P6). Define g
notice that g is holomorphic on V. Thus we have defined a holomorphic function
g: Y \
We will prove that limy\E?Y...+b g(y) = 0 for any b E E (then, by the Radó theorem (cf. [Nar 1971]), g E 0(Y) and E = Fix a b E E. Let V0 be a relatively compact neighborhood of b. Put E —÷
C := max(IJF(x)I: XE
F'(Vo)}.
.
2.12 Liftings of holomorphic mappings!!
Vc JgJ
231
.
be as in(P6). We may assume that ai Sand IJFI 5 on Ui. Then, 0 e on V \ E, and consequently, by continuity, <e on V \ E.
V0
Proposition 2.12.3 ([Ker 1960]). Let (X, p), (Y, q) be Riemann domains over
F: X —+
let
Y
be proper and holomorphic, and let k denote the multiplicity of F. Then for any function f E (9(X) there exists a manic polynomial
(y,z)€
xC,
Ck E (9(Y) such that
with coefficients Cl {z
C: P(y, z) = 0} =
(t)
YE Y.
In particular,
fk +
o
F)fk_i
o.
Proof LetEbeasinTheorem2.12.2.Takeapointb€ Y\EandletV,Uj Uk be as in Theorem 2.12.2. Put çoj := (FIu,Y': V k, and define U3, j = 1 P(y,z) :=
(z
...
—
(z
—
f(Wk(y))
=
zk
(y, z)
V x C:
notice that c1(y) is the value of a symmetric polynomial on and, therefore, cj(y) depends only on the set V. j = I k. y The functions Cl ck can be defined on Y \ E. Obviously, they are holomorphic. Moreover, (t) holds evidently on Y \ E. Let K C Y be compact. Then is compact, and consequently. the set C C is compact. In particular, the functions cl ck are bounded on K \ E. Since E is analytic, the Riemann removable singularity theorem implies that Ck extend holomorphically to Y (we denote the extensions by the same letters). Cl Observe that, by continuity, . .
k1
holds on Y.
j'
isclear. = Itremainstocheck(t)forb e E. Theinclusion and let Y \ E —f b. Then there exists numbers = {al, E Nsuchthatkj +kt =kandforv>> I weget .
j=l
232
2 Pseudoconvexity
where #Pv.j
(j =
aLjk,}, and P1..1 —* Pv.k = (cf. (P6)). Now, by continuity, we have
as
v —f +00
e
z) =
fl fl(z —
—*
j=Iu=l
fl(z
—
f(aj))k) = P(b, z),
j=1
0
which implies the inclusion c' in (t).
Proposition 2.12.4. Let (X, p), (Y. q) be Riemann domains over
F: X —.
and let
V
be a holomorphic mapping such that for any function f E (9(X) there exist k and a monic
P(y, z) =
+
(y, z)
E
N
V x C,
k E (9(Y) such that
wit/i
k
+
o
(1)
Q
I
Assume
that X is Stein. Then F is pmper.
Pmof Take a compact K C V. To prove that the set L := F' (K) is compact it suffices to show that every function f E (9(X) is bounded on L (cf. Theorem I.1O.4(vii)). Take an f E 0(X) and let k and P be as in the assumptions. Let k}.Then f(L) C {z E C: aYEK. P(y, z) = O} C (Izi
C + 1).
C]
Theorem 2.12.5 ([Ker 1960]). Let (X. p), (V. q) be Riemann domains over C'1 and let
F: X —÷
V
be pmper and holomorphic. Let
a: (X,p)—÷ (X,13),
fi: (Y,q)—÷
the maximal holomorphic extension. Let F: X —* Theorem 2.12.1). Then: be
(a)
F: X —÷
V
Y
be the lifting of F (cf
233
2.12 Liftings of holomorphic mappings II
=
(b) F(a(X)) = 13(Y), (c) FIk\a(x): X \ a(X) —÷ Y \ (d) FIa(x): a(X) —* 13(Y)
13(Y) is proper and surjecnve,
Recently, the above result have been regained in a special case in [BerPin 1995]. (a) We will apply Proposition 2.12.4. Take an / tion 2.12.3 there exist 0(Y) such that Ck
Proof
(f oa)k +
0(X). By Proposi
0 F)(f oa)k_J =0. =C3,J = 1
k. Then,bytheidentityprinciple.
we get
+
F)fk_J =0.
(b) Since F is the lifting of F, we get F(a(X)) = = 13(Y). Y) and let I 0(X) be a holomorphic function Now take $'o = 13(Yo) (vo E that separates points of the set T := F(90) (cf. Remark 1.8.12). Let k, and
j=
I
k,beasin(a)andsuchthat
=0) =
(z €C: zk
yE
Y
(cf. Proposition 2.12.3). Then
/(T) C
{z E
C:
zk
+
= 0) =
{z
e C: zk +
(f
=
= 0} C
f(Tfla(X)).
Consequently, C a(X). (c) For any compact K C Y \ 13(Y) we get
(X which, by (a), implies that
\ a(X)),
is compact. Moreover, by (a) and (b),
weget fr(k\a(X))= Y\13(Y). (d) Follows from (a) and (b).
D
Using the above results, we prove that the property to be a Stein domain is stable under proper mappings.
234
2 Pseudoconvexity
Proposition 2.12.6 ([Ker 1960]). Let (X, p), (Y, q) be Riemann domains over C" and let
F: X —÷
Y
be proper and holoinorphic. Then X is Stein iffY is Stein.
Proof The implication follows from Proposition 1.10.13 and Theorem 2.5.7. Now, assume that X is Stein. By Theorem I.10.4(vii), we only need to show that (9(Y) separates points in Y and that for any infinite set B C Y with no limit points in Y there exists a function g E 0(Y) such that sup8 Igi = +oo. Take b1,!'2 Y with b1 b2. Then there exists a function f (9(X) that separates points in the set U F'(b2). By Proposition 2.12.3 there exists a monic polynomial
(y,z)E YxC. with coefficients cj
ck
{z
(9(Y) such that
E C: P(y, z) = 0} =
fk + In particular, by (*)
o
(*)
y E Y,
F)fk_)
(**)
0.
P(b2, .). Hence there exists a Jo such that c10(bi) #
cj0(b2).
Let B C Y be an infinite set with no limit points in Y. Then A := F'(B) is an infinite subset of X with no limit points. Let I E 0(X) be such that SUPA Ill = +oo. Let P be as above. Suppose that all the function k are bounded on B. Then 0 F o F are bounded on A. Hence, by (**), the functions would be bounded on A. Consequently, there exists a Jo such that the function c10 is unbounded
f
onB.
0
Chapter 3
Envelopes of holomorphy for special domains
3.1 Univalent enveLopes of holomorphy LetG cC't
c (9(G). Themainaimofthissectionistodiscuss
conditions which assure that the
of holomorphy of (G, Id) is univalent, i.e. there exists a domain G C C G, such that idGO: (G, Id) —÷ (G, Id) is the maximal 5extension. In such a situation we will shortly write C C't. The problem has been studied, for instance, in [CasTra 1991], [CasTra 1994], [CasTra I 994b].
It is useful to consider also the following more general question. Given a Riemann—
Stein domain (X0, P0) over C", we like to characterize those domains X C X0 and families C (9(X) for which the ienvelope of holomorphy of (X, may be realized as a subdomain of Xo, i.e. there exists a domain X c Xo, X c X, such that idXk: (X, polx) —* (X, Polk) is the maximal iextension; we will shortly write C(X, C Xo. The case (X0. P0) = (C". id) reduces to the previous problem of univalent envelopes of holomorphy.
Remark 3.1.1. (a) Let G C C". D C Ctm be domains,
V :=
{f E (9(G x D):
VU,ED:
C (9(G). Put
f(., w) E
By virtue of Proposition 1.8. 15 we get the following two results: If C C" and D is a domain of holomorphy, then and x D, V) = e(G, x D. If C(G) C C" and e(D) C Ctm then x D) C x
x D, V) c and
C(G x D) =
(b)IfE(X) C Xo, then (by Corollary 2.2.12)
fl
C(X)=int
z.
XcZcX0 Z is a domain of holomosphy
(c)
Let
F: X0 —÷
X0 be biholomorphic and let X. I c Xo be domains such that Then F(C(X)) c In particular, if
C X0, C(Y) C X0, and F(X) Recall that
:= £(G, 0(G)).
c I.
236
3 Envelopes of holomorphy for special domains
F(X)= X,then
and
fl
Z.
x czcx0 Z isadomain of holomorphy
Remark 3.1.2. (a) (Cf. Remark 1.9.6) Let G C C" be starlike and let 4 C 0(G) be such that for any f 4 and t E (0, 1] the function G z —p f(tz) belongs to 4. Then 4) is a starlike domain in C". Notice that if (0, l]G = G but 0 G, then £(G, 8) need not be univalent.
For example. G := C \ R_, 8
{L,: t E (0, lj}, where L,(z) := Log(rz),
z E C \ IR_, and Log denotes the principal branch of the logarithm.
(b) (Cf. Remark 1.9.6) Let G C C" be balanced and let 4 C 0(G) be such that 4 and E E \ 1O} the function G z —÷ belongs to 4. Then 4) is a balanced domain in C". In particular, if G is balanced, then
for any f
fl
DJG D is a balanced
domain of holomorphy
(C) (Cf. Corollary 1.9.18) Let G C C" be ncircled and let 4 C 0(G) be such that (8E)" the function G z —+ 4 and z) belongs to 4. Then is an ucircled domain in C".
for any f E
In particular, if G C C" is ncircled, then
fl
D.
DDG D is an ncircled domain of holomorphy
(d) Let G C C" be a complete ncircled domain (cf. Proposition 1.9.19) and let be such that for any f E 4 and E (E \ {0})" the function G z —+ G, belongs to 4. Then E(G, 4) is a complete ncircled domain in C". z), z Notice that in (a)—(d) one can take 4 :=
4 C (9(G) 
(e) Recall that if G C C" be an ncircled domain such that G fl x {0} x $ 0 for some E (1 n}, then any function holomorphic on G extends holomorphically to the domain C"—1)
j
z:
E
x E x {1}"', z e
cf. the proof of Proposition 1.9.19. Consequently, if G C C" is a ncircled domain with 0 E G, then any function holomorphic on G extends to the domain
G :=
E E", z€G};
237
3.1 Univalent envelopes of holomorphy
G is complete ncircled. In particular, if 0 E G, then
E(G)=int
fl
D.
DJG
D isa complete ncircled domain of bolomorphy
(f) (Cf. Remark 1.9.8) Let G C D x C" be a Hartogs domain with nonempty starlike fibers. Let 7 C (9(D x Ck) be such that D x is a 7domain of holomorphy, E (0, 1]thefunction andleU c TlGandforanyf E G Then is univalent, (z. w) —+ f(z. 1w) belongs to C D x Ct, and E(G, is a Hartogs domain with nonempty starlike fibers. (g) (Cf. Remark 1.9.8) Let G C D x C" be a Hartogs domain with nonempty balanced fibers. Let 7 C (9(D x Ck) be such that D x is a 7domain of bobmorphy, and let C (9(G) be such that J and for any f E and E E \ (0) the function G Then C(G, S) is univalent, (z. w) —f f(z, çw) belongs to C D x Ck, and C(G, is a Hartogs domain with nonempty balanced E(G,
j
fibers.
(h) Let G C D x be a Hartogs domain with nonempty connected kcircled x fibressuchthatGfl(D x C"—') øforsomej (1 k). Then any function holomorphic on G extends hobomorphically to the Hartogs domain
:=
.
xEx
w): ç
Consequently, if G c D x
(z, w)
G}.
Ck
is a Hartogs domain with nonempty connected kcircled fibers such that G fl (D x 101) 0, then any function holomorphic on G extends to the following Hartogs domain with complete kcircled fibers
a
:= {(z.
. w):
E
Ek
(z, w)
Indeed, since the fibers are connected, any function f
G}.
(9(G) may be represented
where byaHartogs—Laurentseriesf(z, w) = Z") x and the series is locally normally convergent. Now, the condition Gfl (D x (0) x Ck_J) x 0 implies that Consequently, 0 for fi Z'—1 x the series is locally normally convergent on GW. (i) Let G c D x C" be a Hartogs domain with nonempty kcircled fibers (we do not assume that the fibers are connected) such that D x (0} C G. Then any function
C''
holomorphic on G extends hobomorphically to the Hartogs domain G (defined in (h)). Indeed, any function f E (9(G) maybe represented by the Harlogs—Laurent series (9(G),foranyz E Dthefunctionfp(z..) f(z. w) = E$EZk ffl(z, w)wfi, where and the series is is constant on any connected component of the fiber E locally normally convergent. On the other side, let
Go:={(z.w)EG:VcEEA:
•W)EG)
3 Envelopes of holomorphy for special domains
238
x (0) C G, the set GOC D x Ck is a Hartogs domain with nonempty and fp(z. w) 0 on G0 for fi complete kcircled fibers. Thus fp (z, w) E G, and for (z. w) G0 and E Hence f(z, w) = Since D
consequently, the series is convergent in G. Now we will present two criteria for the univalence of the envelope of holomorphy.
PropositIon 3.1.3. Let (X, p), (Y, q) be Riemann domains over
and let
r: (X,p) —* (Y,q) be a morphism. Let 7 be a aswble vector subspace of 0(Y) and put := C1O(X)
Assume that
a: (X,p) are maximal
fi: (Y,q) —÷
and 7extensions, respectively. Let
f:
(X,j3) —÷
and I is infective. be a lifting oft (Proposition 1.9.2). Then = clo(k) X C Y C Xo are if (Xo. P0) is a Riemann—Stein domain and in domains such that C Xo and(X, Y) isa Rungepair(cf Proposition 2.7.5), then and L(Y)) is a Runge pair. C C(X) c Xo.
Pmof We know that is a astable closed subspace of 0(k); cf. Remark 1.4.5(j). Since as: is a topological isomorphism (in the topologies of locally uniform is a is a dense subset of convergence), we conclude that dense subset of It remains to observe that separates points in stalks (ProposiTo prove that I is injective, recall that separates points tion 1.8.10). By the first part of the proof we conclude that in stalks. k be such that 1(n) = f(x2). In particular, Let = points xl, x2 are obviously not separated by functions from 15(0(Y)). Thus
The
=
X2.
cJ
Proposition 3.1.4. Let D C
be a domain of holomorphy and let M be a pure (n — 1)dimensional analytic subset of D. Then for any domain C C D we have: is univalent
if C(G) is univalent, then
Proof If quently,
\ M) is univalent.
\ M) =
\ M.
is univalent, then by Theorem 2.5.9, C(G \ M) = \ M) is univalent.
\ M.
Conse
239
3.1 Univalent envelopes of holomorphy
Now assume that E(G \ M) is univalent. Let a: (G, Id) —÷ (G, j3) be the maximal holomorphic extension. By Proposition 1.9.2 there exists a morphism
t:
that r o a = idG.D (obviously z
such
rem
(G,13) —÷
(D,id)
fr). Put M :=
Hence, by Theo
2.5.9,
aIG\M:
(G \ M, id) —+ (G \ M,
the maximal holomorphic extension. Hence \ M). Id) is isomorphic to Consequently, G \ M must be univalent. Suppose that EM (G \ M, j3). are such that j3(ai) = p(a2) =: zo and Letr := min{dO(aI),dO(a2)}. We easily conclude that r) U r)) \ M) 115(zo, r). Hence, there exist points b1,!,2 G \ M, b1 contradiction. !2, such that 13(b1) = Thus G is univalent. U is
Using Proposition 3.1.3, one can characterize several classes of domains with univalent envelopes of holomorphy, for instance:
Example 3.1.5 (Envelopes of holomorphy of circular domains). Let G C C" be a circular domain E G for any ( E dE and z G). Put
zeG}: C(G) is a Ctcone. Assume that for any z E G the section := G} is connected 2) and E(C(G)) is univalent Then is univalent and circled (and (E(G), E(C(G))) is a Runge pair). Indeed, by Proposition 3.1.3, it suffices to prove that (G, C(G)) is a Runge pair. Since the sections of G are connected, any f E (9(G) may be represented by a series
f(z) = where I
2nz
z E G,
I'
I
ZEG.
"
Clearly. (9(G). Moreover, the series converges locally uniformly in G. One can easily show that
= Consequently, 2)
That is,
e G,
extends to a holomorphic function on C(G). is an annulus.
Observe that 0 a G connected.
z
C(G) = C". Moreover, if G is balanced, then each section
is
240
3 Envelopes of holomorphy for special domains
Remark 3.1.6. Let G C
be an ncircled domain. Recall that the logarithmic image of G is the set
logG := {(xI
ER'1: (eX1
E G}.
c")
For a domain U C W1 put
expU := ((zi
(log Izil
Zn) E
U}.
log
Notice that exp log G = G and exp U is an ncircled domain with log exp U = U. By Proposition 1.9.19, G is a domain of holomorphy if log G is convex. As a simple corollary we get the following formula:
E(G) = exp(convlogG). where, as usually, cony A denotes the convex hull of A.
Let U C R" be a convex domain and let U1 denote the projection of U onto the
first) axes(l < j2,thedomain
can
be written in the form U3
= ((xi
x R: (1(xi
<
where the functions £,. —L1: U1_1 —*
are upper semicontinuous and
forany(xi
<
(which will be
This simple observation leads to the following description of used in the next section).
= ((zi
Zn)
z1,)
r3(zi
(zi where
0< ri <
R1
<
< Zj—i)
R1(zi
G1,
j
< +oo and if) >2, then:
rj:
G —* [0. +00) is upper semicontinuous. R3: G1 —÷ (0. +00] is lower semicontinuous, z1_i) < R3(zi
z1—t)
forany
(zi
Zj—l) E G1.
is a (j — 1)circled domain of holomorphy, C r1, R3 depend only on (Izil In fact, if U := cony log G and t.
are as above, then
:=
Zj_i) :=exp(L1(loglziI
log
=1
3.1 Univalent envelopes of holomorphy
241
Proposilion 3.1.7 (Envelopes of holomorphy of Hartogs domains). (a) Let G C D x Ck be a Hartogs domain with nonempty balanced fibers. Let that for any f E and E \ {O} the function G
IC
C
0(G) be such
f(z.
(z. w) —÷
belongs to Then there exist: • a Riemann domain (D, over •amorphisma: (D.id) (D,co).and with nonempty balanced fibers • a Hartogs domain G C D x such that
•a x idck(G) C G, • for any f E there exists an f

(9(G) such that f(a(z). w) = f(z. w),
(z, w)€ G.
:= {f: f
•G isthe domain of existence of thefamily
In the case where D is univalent, the envelope E (G.
= 0(G) and a: (D, id) —÷
is also univalent.
(D, ço) is the maxi,nal holomorphic
extension of D, then: • (D, satisfies all the above conditions, • a x idCk: (G. Id) —+ (G, ço X idck) is the maximal holomorphic extension. (b) Let G
fibers. Let
CDx
Ck be a Hartogs domain with nonempty connected kcircled
C 0(G) be such that for any f G
andç E (aE)r the function
E
IC
(z, w) —+ f(z.
z)
belongs to Then there exist: • a Riemann domain (D. 60) over
•amorphisma: (D,id) —+ (D,ço),and • a Harrogs domain G C D x Ck with nonempty connected kcircled fibers such that
•a x idck(G) •for any f
C G,
there exists an f E 0(G) such that f(a(z). w) = f(z. w),
(z.w)€G, • G is the domain of existence of the family

:= {f: f
In the case where b is univalent, the envelope e(G,
is also univalent.
= 0(G) and a: (D. id) —i' (D, 60) is the maximal holomorphic extension of D, then: • (D, 60) satisfies all the above conditions, • a x idck: (G. Id) —+ (G. ço x idc&) is the maximal holomorphic extension. Remark 3.1.8. Notice that in general, if G C D x C is an arbitrary Hartogs domain with nonempty circular fibers such that is univalent, then €(G) need not be univalent.
242
3 Envelopes of holomorphy for special domains
For example, let G be the domain from the introduction to Chapter I. Then x C is a Hartogs domain with nonempty circular fibers such that E(G) is GC not univalent (see also Remark 3.1.12(d)).
Proof (a) Any function f z, w) =
E
may be expanded into a series of the form f(z.
where fp E 0(D)
j/3=v
w) =
E
the series is locally normally convergent in G. Put F := E fE Let cr: (D, id) —÷ (D. çp) denote the maximal Fextension. For f E and v E where Z+, define E 0(D) denotes the 1131=v extension of
(ffl ° a =
f .)
let D1 denote the region of normal convergence of the series It is clear that bj c 15 x Ck is a Hartogs domain with balanced
fibers (they may be empty: cf. Remark 3.1.9) and a x idc*(G) C D1.
Let D,
be the connected component of D1 that contains a x idck(G). Obviously, the sum of the series .) defines a holomorphic extension / of f to D1. i.e. /(a(z), w) = f(z, w). Notice that = fc for any E E \ (0) 6), Let G denote (G). Then G C /5 x C' the connected component of mt Dj that contains ax is a Hartogs domain with balanced fibers such that a x idc*: (G, id) —÷ (G, x id) is an iextension. Let /5 be the connected component of the set {x E iS: 01 that contains cr(D). We want to show that 0 is the sdomain of existence. Suppose that (xo. wo) E G Define a new Hartogs is such that d(T(r0 u'O)f) > R > u'o) for any f E domain
R) x
E E \ (0), w E IP'k(wO, R)}.
We will prove that for any f E
exists a function f
w) = f(x. w) for (x, U') E be such that
u's). Take (z. w) E
E
and let
such that E E \ (01
E Pk(WO, R). Define
f(z. u') := T(x()u,o)(fc)(z, observe that f is well defined. holomorphic on c, and (x. w) E wo)
u') = f(x. w) for
That is. the set of those (x. u') E D x Ck for which there exists an open neighborhood U C D x Ck such that the series is convergent. That is, w) = f(x. for any (x. u') Dj.
i(z.
R) x
(z. w) e
j = 1.2.
=
j = 1.2. We have to show that gi = X2 on follows directly from the identity principle. If
) fl
fl
Pk(wO. R)):
Xj
is holomorphic.
fl 0. then the equality = 0. then we may easily find points N—l,and
fl
0. Consequently, we may successively apply the previous case.
243
3.1 Univalent envelopes of holornorphy
One can easily check that f(z. w) = R)),
E
and the series is convergent locally normally in
E
= fflnearxo(f of existence, we get R
where
Since(D,q)istheFdomain
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 Fextension. 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 kcircled 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 kcircled fibers such a x idck(G). Then G C D x that a x (G, id) —+ (G, x id) is an sextension. Let D be the connected ø} that contains a(D). component of the set (x ED: We have to show that G is the idomain of existence. Suppose that (xO, wo) G Define a new Hartogs wo) for any f is such that d(T(x0w0)f) > R > domain
R) x
Fixanf E
.
w:
(aE)k, w
Pk(wO. R)}.
Take(z,w)
Pk(WO. R). Define
!(z, w) := Use the fact that
f(x. u') for (x. w) E 9)
w) =
1) x Ck for which there exists an open neighborhood l'hat is. the set of those (x, w) <0 for some) E (I n}. is well defined in U (if is convergent. cCi' x x
U C Dx Ck such that
thenU
is a Hartogs domain with connected circled fibers and wo).
244
3
Envelopes of holomorphy for special domains
observe that f is well defined, holomorphic on and w) = J(x, w) for where fp is holo(x, w) E wo). Moreover, f(z, w) = EPEZk morphic in IP(Q(xo), R), E Zk, and the series is convergent locally normally in Hence, R and R) x Pk(wO, R) C G; contradiction. It remains to prove that the fibers are connected for any x E D. Consider the set c of all (xo, wO) E D x Ck for which there exists an open such that for any f E there exists a constant M(f) > 0 neighborhood U C D x such that the function U (x. w) —p is welldefined and
M(f),
U, for any
W := {(e'°' lw'1 I
7L'. Observe that
c Dx
has kcircled fibers. Directly from the definition of G we conclude that G C c. First observe that and let U be as above. We may assume Indeed, take (xo, wo) QC that U = V x W, where V is a neighborhood of xO and W is a kcircled neighborhood of wo. Now, let A = A1 x x Ak c W be a compact neighborhood of wo such that each is a closed disc or a closed annulus. Then there exists a 9 E (0, 1) such < for (x, w) E V x A and E Zk, which implies that is normally convergent on V x A. that the series Thus we only need to prove that the fibers of are connected. For, take (xO, We are going to find a domain W C with W. We may E (xo, assume that w', w" (R>o)lc. By definition, there exist a neighborhood V of xo and r > 0 with r) U Pk(wg, r) C (C5)k, such that for any f 4S there exists an M(f) > 0 such that r) U r)) < M(f) for (x, w) V x and,fl E Zk. Define (x, w)
etOk
Pk(w0,r), (wi' R"
Then W C (Ce)" is a domain containing w', w", and
t
10, 1]).
<M(f) for (x, w)
E
Vx In the case = 0(G), similarly as in (a), we have to show that D is Stein. Define := {(x, w) D x Ck: (x, e'°) G} By Corollary 2.2.20, is t ERk}andthatcisaHartogs Stein Noticethatcz={(x,w+ir): (x,w)E
domain over D with connected fibers
Hence, by Corollary 2.3.3, D is Stein.
E
Remark 3.1.9. Notice that in general Dj need not be connected and D may be smaller than D. For example, let Recall that
e" =
We apply Corollary 2.2.20 to the mapping D x Ck (z, w) —' (z, E D x Ck. isconnectedkcircled, isconnectedkcircled. Consequently, the c (R>o)* isconnected. Thisimpliesthat((Ieu't I e'°* I): wkI): w E set ((Iwj I u' E cZX) is connected, which finally shows that the whole fiber is connected. l
245
3.1 Univalent envelopes of holomorphy
D:={x+iy€C: < G := {(z. w) E D x C: {f E 0(G): 3OER: f(z, w) =
(z. w) E G}.
lyl> lxi. Iwl be a Hartogs domain with nonempty 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 nonempty balanced fibers. (b) Let G C D x Ck be a Hartogs domain with nonempty connected kcircled 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 kcircled fibers.
(c) Let G c D x Ck be a Hartogs domain with nonempty connected kcircled 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 nonempty kcircled fibers (cf Example 3.1.16). In particular, (fG c D x Ck is a Hartogs domain of holomorphy with nonempty balanced (resp. connected kcircled) 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 nonempty and balanced (resp. connected kcircled) 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 nonempty and balanced (resp. connected kcircled). Assume that we know that C(G) C D x Ck is a univalent Hartogs domain with nonempty 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 nonempty complete kcircled 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 nonempty kcircled 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 nonempty kcircled 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 nonempty connected kcircled 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 Fgerms 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 kcircled fibers such that /3: (G, idG) —* (G, x idc*) is an iextension. Let D be the connected component Then
f
oftheset{x ED: The proof that the fibers
x E D, are connected is exactly the same as in
Proposition 3.1.7(b).
The univalence of the envelope of holomorphy is not biholomorphically invariant (like the Runge property), namely: is univalent. Assume that Remark 3.1.12. Let D c C" be a domain such that F: D —÷ G is biholomorphic. By Proposition 1.9.9, F has a locally biholomorphic extension F: E(D) —+ C's. Moreover, we know (Remark 1.9.10) that
F1: (G, id) —p
F)
is the maximal holomorphic extension. In particular,
C(G) is univalent if F is infective. Consider the following four examples:

The case where K := C(D) \ D is compact: Put G := F(C(D)) 15) Observe that G \ G C F(K) is compact. Consequently, by the Hartogs Theorem 2.6.6, the mapping F1: G —* D extends holomorphically to G; let F' denote the extension. (a)
Now, one can easily check that F: Thus C(G) is univalent.
(b) The case where
—* G is biholomorphic and
\ D) = 0:
Suppose that a, b
= (F)1. a
are such that F(a) = F(b) =: c. Since F is locally biholomorphic. there exist of a, b, c, respectively, such that U0 fl Ub = and open neighborhoods U0. U,,, are biholomorphic. Let V : F(Ua 11 D) fl and Flub: Ub —± U0 —+ F(Ub fl D). Observe that V 0. Now, it is clear that there exist points a' E U0 fl D, b' E tjh fl D such that F(a') = F(b') e V; contradiction. Note that G is open.
248
3 Envelopes of holomorphy for special domains
Thus in this case
is also univalent.
(c) Let U C R2 be a domain such that:
—x'I < for some (XI. xi), (xi. x') cony U. Put D := U + iIR2. Recall that E(D) = (cony U) + (Bochner theorem). Let z2). Then F is injective on D but is not injective on F(zi, z2) := In particular. E.(G) is not univalent. (d) Let
U'): 0< r < I,
D :=
E
cPE(0,lr/21 E L2jr,
5jr/2)
I
< ti'I < 2
The mapping F(z, w) := (exp(z). w) is injective on D and G := F(D) is the domain from the example in the introduction to Chapter 1. Observe that D is a Hartogs domain over R := {x + iy E C: x < 0. 0 < v < 5ir/2} with connected circular fibers. Hence D) is univalent (cf. Corollary 3.1.10(b), see also Remark 3.2.19) and C(D) J R x E. Thus F is not injective on e(D) and therefore E(G) is not univalent.
Forr,pE(0,.1)put x K(r))
H = Hr.p := Observe that C(H) =
F: H —p
U
p<
E
< l} x E).
(cf. Example 1.8.7).
c
be a bihotomorphic mapping. As above, F denotes the holomorphic extension of F to P,,. Then there is the following criterion for F to be injective (cf. [Kas 1974]).
Let
G
PropositIon 3.1.13. F is infective Prnof
fl
\ H) = 0.
'==*' is obvious. So it remains to prove
S :=
sup{s
Put
e (0, II:
injective}.
It suffices to show that S = I. Suppose S < I. In virtue of the assumption then r S < I. In the first step we shall see that FI{ZEpn: is injective. To prove this we put
F1 (F(z0)) fl (z E Obviously,
the assumption
I
= 5) is exactly one pointj.
F(H) Ii FOP,, \ H) = 0 implies that 0
T
p.
3.! Univalent envelopes of holomorphy
Suppose now that T > 0. Then there are sequences
S=
w1,j
p, and
I.
1VI / T,
249 C
with
E N, such that
fr(z') = F(w1), j
N.
may assume that both sequences converge. Put
We
urn
=:
z0,
=:
urn
w0.
i—+oo
Since F is locally biholomorphic, it follows that z0 w°. We choose disjoint neighborhoods U = U(w0) IF,, and V = V(z°) := IF,, with F(U) = F(V) such that Flu and Fly are injective. DeBn(Z°, r) fine CD := (FJuY'. In virtue of the definition ofT we know that, if z E V, lznl = S. > Moreover, using that F is locally biholomorphic and the property of S we conclude
that
lcD,,(F(z))l > S lcD,,(F(z))l
>S
if z E V.
=
5, and 1±1 > T,
(*)
if z E V. IZ,,l <S.
With this information in mind we look at the irreducible analytic subset A :=
in V. Then B := CD(fr(A)) is an irreducible analytic subset {z E V: Zn = of U, w0 E B. Moreover, the function g: C" —÷ C, g(z) := z,,, satisfies >S and Ig(w°)i = S. Then the maximum principle gives that IZ,,l = S for any z B. which contradicts (*). Hence we get T = 0. Following the recent argument we even have that F is injective on (z E IF,,
IZnIS}. \ H) = 0 and the local biholornorphicity such that F is also injective on (z IF,,: lz,,l < S +e}:
Then the assumption that F(H) fl F(IF',,
of F leads to an
0
0
contradiction to the definition of S.
So far we don't know any injective holornorphic mapping F: H —b C", such that F is not injective on IF,,
A similar statement (as in Proposition 3.1.13) remains true in the unbounded case. Here we only give the precise formulation. (The reader is asked for the proof as an exercise.)
Fixr,p E
(0, 1) and put
H:
Let F: H
x K(r)) U
p<
< l} xC).
G be a biholornorphic mapping and denote by F its holornorphic extension to C(H) = P,,_i x C =: H. Then the following is true:
250
3 Envelopes of holomorphy for special domams
if F(H)
and
IF(z')I —+ oo for any sequence (z1)jeN C H
with
I —÷ 00. then fr is infective on Whether this statement remains true if we only assume that F is injective on
H seems to be unknown Example 3.1.14 ([CasTra 1991]). Let
:= G1
((re', w) E C2: I <
r
<e. 0
<
<
is a Hartogs domain over (I < Izi <e) with circular fibers which are in general is not univalent; cf. Remark 3.1.8 for another example
disconnected 16j Then of that type.
Indeed, we can argue as in Remark 3.1.12(d). Let R := (x + iy e C: 0 < x < 0 < y <4ir},
:= 1(x + iy, w) E R x C: y — D1
2
1,
< IwI
is a Hartogs domain over R with connected circular fibers. Then
E(Di)=1(x +iy,w)E
R
xC:
IwI
Observe that the mapping F: C2 —* C2, F(z, w) (eZ, w), maps biholomorphically D, onto G1. It is clear that F is not injective on e(D1). Consequently, by Remark 3.1.12(a), C(G1) is not univalent.
Remark 3.1.15. Envelopes of holomorphy of arbitrary Hartogs domains 0 C C2 are characterized by the following result by E. Ligocka [Lig 1998]. G C D x C be an arbitrary Jiartogs domain with nonempty circular fibers. Then there exist: x C with nonenlptv connected circular • a Hartogs domain of holomorphy G C fibers, where either = E or = C, and • a surf ective holomorphic mapping CD: G —* such that CD: 0 —+ E(G) is a covering.
Example 3.1.16 ([CasTra 19911). Let
02 := {(re't,
Z2,
w) E C3: I < r <e.
0 <
r
<4jr, 1z21
< IwI <e'}.
Then 02 is a Hartogs domain of holomorphy over the domain
G2:={(re",z2)EC2: I
17)
(I. e). we havc = K(r/2) U {5x/2 —2 < wi < 5ir/2). an f e (9(D1). Since D1 fl (R x (0)) 0. the Hartogs—Laurent series of I is
For cxamplc. fors e Take
convergent in the domain D1 defined by the right hand side. Since the function (z. w) —* Iwi — is psh. D1 is a domain of holomorphy (Corollary 2.2.15). Thus D1.
z
3.1 Univalent envelopes of holomorphy
251
G2 has circular fibers which are in general disconnected 18) The projection G2 is not a domain of holomorphy — cf. Remark 2.3.4.
Indeed,letR:={x+iyEC:0<x
D2 := {(x + iy, Z2. w) E R x C x C5: 1z2t < y.
Observe that the mapping F(zj, Z2. w) := (en, Z2, w) transforms biholomorphically
onto G2. Since the functions
RxCxC53(zI,z2,w)—*1z21—ImzI, R xC x (z,,z2. w) —÷ Imzj —1 R xC x C5 (zi, Z2, w) —* log wi —!mzi are psh, D2 is a domain of holomorphy. Consequently, G2 is a domain of holomorphy. G2 is a Hartogs domain over (1 < zl <e} with complete circular fibers. Thus, by Proposition 2.2.22(b). to prove that G2 is not a domain of holomorphy, it suffices to observe that the function u defined on (1 < Izi < e} by the formula u(re") := I < r < e, 0< t < 2,r, cannot be subharmonic (because it has a —Iog(t global maximum). I
Example 3.1.17 ([CasTra 1991]). Let
:= {(re", Z2, a;)
C3:
I
< IwI
<e').
Then G3 is a Hartogs domain over G1 (cf. Example 3.1.14) with connected circular fibers. Moreover, C(G3) = G2 (Example 3.1.16) and therefore C(G3) has disconnected fibers. In particular, by Corollary 3.1.10, cannot be univalent (which was already directly shown in Example 3.1.14). Indeed, recall that D2 is biholomorphic toG2 via F(zi, Z2, w) := (eZi, Z2, w). In particular, F1 maps btholomorphically G3 onto
D3 :={(x +iy,z2,w) ER xC x C4: y —2<1Z21< Y,
<
<eV},
where R := {x + iy C: 0 <x < 1, 0 < y <4jr}. Hence, we only need to prove that any function I (9(D3) extends holomorphically to D2. By the Cauchy integral formula, the extension f E (9(D2) may be given by the formula 
f(zi, 18>
Z2, w) :=
For example,
I
2,r:
j(kl=r
ifs E (I, e),
then
< r,
—
—2 < r < v.
Z2
= {e"12' < wi < c"12) U
< wi <
252
3 Envelopes of holomorphy for special domains
Remark 3.1.18. Let V,, := C'7' x {O} and consider the biholomorphism C'7 \
_!*
(z'. Zn)
E
C C't \ Vn be a circular domain. Then G := 4(G) is a Hartogs domain over (Z'/Zn: (z'. E G} with nonempty circular fibers. Observe that for any gives an isomorphism 3 W —+ WZn E (z'. Zn) E G. the mapping between the section G x G has connected fibers. It is clear that P(C(G)) = D
By virtue of Proposition 3.1.4, for any circular domain G C C'7, we have: is univalent
E.(G \
\
is univalent
is univalent.
Remark 3.1.19. (a) Let G C C'7 be a circular domain with connected sections such that E(C(G)) = C'7. Then is univalent and 0 E We already know that is univalent (Remark 3.1.5) and that (E(G), C'7) is a Runge pair. Suppose that 0 Fix a point a E G and consider the holomorphic
function E(G) fl (Ca) 3 (fo is well defined because 0 Proposition2.5.l0,thereexistsanf E =
E
By E(G)0.
Consequently, since C'7) is a Runge pair, the function E may be approximated, locally uniformly in by complex polynomials. Since is a circular open subset of C, we get a contradiction. Notice that if the sections are not connected, then E(G) need not be univalent (cf. Example 3.1.20).
(b) (Cf. [CasTra 1991]) Let G C C'7 be a circular domain of holomorphy with connected sections. Then C(G) is a domain of holomorphy. Indeed, by Remark 3.1.18 and Corollary 3.1.10, we get the following sequence of implications: G is a circular domain of holomorphy with connected sections G
\
is a circular domain of holomorphy with connected sections
G := (D(G \ V1,) C D x
isaHartogsdomainofholomorphy
with nonempty connected circular fibers D is a domain of holomorphy is a domain of holomorphy C(G \ Vn) = cj'(D' x C(G) \ V1, is a domain of holomorphy. In particular, by Proposition 3.1.4, Cecone.
is
univalent. Note that E(C(G))
is a
The same argument shows that C(G) \ E is a domain of holomorphy for any complex (n — 1)dimensional subspace of C". Thus
\
(0}
= C(G) \
{0J.
253
Univalent envelopes of holomorphy
3.1
IfO then = C's, and This completes the proof if 0 therefore, by (a), 0 E = G. Thus C(G) = C". which completes the proof. Notice that if the sections are not connected, then C(G) need not be a domain of holomorphy — cf. [CasTra 1991], where it is shown that there exists a circular domain of holomorphy G C C2 (whose sections are in general disconnected) such that C(G) = (C) Let G C C" be a circular domain with 0 E G (we do not assume that sections are connected). Then C(G) is univalent. Indeed, by Remarks 3.1.18, 3.1.2(i) and Corollary 3.1.10, we get the following sequence of implications:
G is a circular domain with 0 E G
4(G \
U
C
C is a Hartogs domain
with nonempty circular fibers such that
C G'
C(G) is univalent C(G
C(G \
=>
\
V,,) is univalent Vi.,) is
univalent
is univalent.
(d) Let G C 0
be a circular domain with connected sections such that G. Then E(G) is univalent and has connected sections if C(C(G)) is univa
lent. Moreover, if C(C(G)) is univalent, then = C(C(G)). Indeed, we already know that if C(C(G)) is univalent, then (Remark 3.1.5). Assume that C(G) is univalent and has connected sections. Then:
E(G \
is
is univalent
univalent and has connected sections
G := 4(G \ V,,) C D x
is a Hartogs domain with nonempty
connected fibers such that e(G) is univalent and has connected fibers C(D) is univalent (Corollary 3.1.10) t(C(G) \ V,,) is univalent C(C(G)) is univalent. Moreover, \ V,,) = C(t(G \ V,,)) = C = C(C(G))\ V,,. Consequently, C(C(G))\(0} = = t(C(G)\ and C(C(G)) \ (0}. ff0 C(C(G)), then C(C(G)) = C". Hence, by (a), 0 E hence 0 E C(t(G)).
Obviously, (C(D) x
Example 3.1.20 ([CasTra 19911). There exists a circular domain G C C2 (with disconnected sections) such that C(G) = and is not univalent (Cf. Remark 3.1.19(a)).
254
3 Envelopes of holomorphy for special domains
Indeed, put
1/2
Let X := Xo U X1 U X2. Endow X with the topology induced from C x JR. Define X
A E C.
(A, 1)
Then (X, p) is a simply connected Riemann domain over C 19), Therefore, in virtue of the uniformization theorem (cf. LFors 1981], [Spr 1957]), there exists a injective holomorphic mapping F0: X —÷ C. Let 6,r
Z2
Finally define
G := {z E
q(z) = < 1/25).
< 1/25 or IVzJI — I +
Ilizil — I —
One can prove that G is a circular domain such that: = (A E • sections Az E G}, z E G, may be disconnected (they may have at most five disjoint components),
• C(G) = • for any z E G the element by z. say = • the mapping
E
G is
X from the definition of G is uniquely determined
(.3
z
EX
holomorphic (and p o C) = q). is not univalent. Define It remains to show that
D := {z E
IIq(z)I
I
—
(z
ED: (0.6ir/25)flargq(z)
<
0).
Observe that D is a circular domain with connected sections, 8D C G, and Do C G.
For f E (9(G) define
f(z :=2n':1(f
ICI=c+
—f
—
Z E D, cj :=
l±t . 0 E (0, 3ir/2) fl argq(z). IIzJI
19)
The details of the construction are left for the reader.
255
3.1 Univalent envelopes of holomorphy
=f
on D0. Then f 0(D) and f Letfo 0(G). Fixaz
= F0 F0 o Pa.) o q(z) = const. Hence
we obtain fo(crz) =
Fo 0
8fo
(1—c, l+e)with0 <e << 1,
G. Thenforct o
o
p
=
o
8fo az2
zI—(z)+z2———(z) =0,
Fo 0
0 q(az)
=
z E G,
which, by the identity principle, implies that
zi—(z)
dfo
+ z2—(z) = 0,
8Z1
z E D.
dz2
* ( )
1/4/i), b := (13/15)a. Note thatq(a) = q(b) = —1, a. bE G, and the segment [a. b] is contained in 1). Thus, by (*), f is constant along [a, b]. In particular, fo(a) = fo(b). (—1,,r) = e(b). Hence fo(a) f0(b), On the other hand,®(a) = (—1,0) which implies that E(G) cannot be univalent. Take a :=
The following result generalizes Remark 1.9.6(a).
Proposition 3.1.21. Let (X0, be a 7domain of holomorphy (7 C 0(Xo)). Let X be a univalent subdomain of Xo. Assume that 8 c 0(X) is such that TIx C 8. Let H: Xo x [0, 1] —* Xo be a continuous mapping such that:
H(.,t) E 0(Xo,Xo) for anyt E [0,1], 0) = idx0,
H(Xo, 1) C X,
H(X,[0, Ij) cx, a locally connected B C [0, 1] and a continuous Pnapping
there
T: C" x B —f such that
f€
T(.,t)
8,
t E B},
Aut(C")foranyt B, po(H(x, r)) = T(po(x), t)forany(t,x) E B x X, H(Xo,:) C Xfor anyt e [0, 1] \ B. Then
8) C Xo.
Proof (Cf. [Tra
phy. Note
that
1988]) Let a: (X, poix) —÷ (X, /) be the a is injecuve. Since (X0, po) is a 7domain of holomorphy, there
exists a morphism
t: (X,Th —* (X0.p0) (r 0 a = idx,x0). We want to show that r is injective. It is clear that r is injective on
a(X).
256
3 Envelopes of holomorphy for special domains
Let H: X x B — E
[0.
X denote the lifting of HIXXB (cf. Proposition 1.9.4). For 1] \ B, put H(x, to) := a(H(r(x), to)), X E X. Observe that
H(cr(x'), 1) = a(H(x', 1)), (x', 1) e X x [0, p(H(x, 1)) = po(H(r(x). 1)). (x, 1) E X x [0, 1). '0) In particular, H is continuous on a(X) x [0, lJ, and consequently, by the monodromy theorem, H is globally continuous on X x [0, lJ. Notice that H(•. 0) = Recall that (9(X) separates points in X. Thus to prove that r is injective, it suffices to show that if r(bi) = r(b2) for some b1,h2 E X, then f(bj) = f(b2) for any
f
E (9(X).
X such that a point a = a(a') E a(X). Assume moreover that b1, b2 Eare Let f In particular, 13(b1) = j5(b2) =: (9(X). Fix = r(h2) curves = 1.2, with X, [0,11 —+ j = 1,2. j
Fix
two Let
Yf := jo
J=
1,2. Consequently. we have two curves af: [0,11 —÷ (9, yj(1)), E 10, II, j = 1,2 (cf. Example 1.3.1). We like to prove
c1(t) :=
i
=(12(I). Define
that
O
H(a,3t).
1/3<1 <2/3
I), 1).
—
2/3
H(b3, 3(1 — 1)).
=
Note that
a,
=
h3,
j=
j=
:=
1,2.
1,
1
1,2. Thus the curves
j=
1,2. induce two
newcurvesp3: [0,11 —p E [0, lJ,j = 1,2. := Note that po(X) for 0 < t < 2/3. Indeed, if 0 < t < 1/3, then t5j(') = t 2/3, 15(H(a, 31)) = po(H(r(a(a')). 31)) = po(H(a'. 3r)) E po(X); if 1/3 po(X) (because then = — I), I)) = — I)), 1))
H(Xo. I) c X). In particular, c5i(2/3) = 62(2/3) = po(H(co. I)). Recall that a(X) is univalent and jS(a(X)) = po(X). Consequently, by the identity principle for liftings. we get = (151a(x)r'(6j(t)) forO < : < 2/3, j = 1,2. Thus 6i(2/3) = 62(2/3) and
= F2(2/3).
therefore
Moreover,61(t) =82(1) =
1.
Let
0
a,
— 2u)/(3
—
4u)),
0
u
1/2, 2u/3
0
l)t), I),2u — I),
H(a, 3(2u — —
—
20
If x = a(x').
po(H(r(x).
1)):
l)(l
—
1)),
I
I — 2u/3
1—2u/3
1/2
then by the previous equality it remains to use the identity principle.
we have j(H(x, f)) =
1)))
=
3.1 Univalent envelopes of holomorphy
257
1,2. Hence po(a'), —
p(4(u t))=
2u)/(3
—
414))).
0
0 <
z
<2u/3 I — 214/3
214/3
O
Zo,
po(H(a'.
3(2u — 1)1)),
1/2
1, 0
u
I)), 2u — I)), 1/2 u — 1)(1 — t))), 1/2< u <
I
< 1/3
po(H(r(yj(3t —
1,
1/3
t
po(H(co, 3(2u
1,
2/3 <
t
is a homotopy joining yj with 1,2.
2/3 <
I
By virtue of the monodromy theorem, we get
0
Proposition 3.1.22 ([Tra 19881). Let G C D C C" be domains and let
H: D x [0,11 —+ C" be a continuous mapping
such
that:
H(',O)=idD, H(D x {1}) C G, H(.,:) C"), t E [0, 1], H(G x [0, 1]) C G. Assume that D is a domain of holomorphy Then:  (a) The envelope of holomorphy G of G is univalent, G C D, and H (G x [0, 11) C
G.
(b) if moreover
H(D x [0, 1]) C D, H(., t) is mi ectivefor each t
[0, 1),
(H(D x (t}), D) isaRunge pairfor each t
[0, 1),
then (G, D) is a Proof. (a) We argue as in the proof of Proposition 3.1.21 of the lifting follows from Theorem 2.12.1(b)). (b)
— Exercise (the existence
Fix a compact K C G. We want to show that
C G. We will prove
more, namely that x [0, 1]) c G (then
=
x {0}) C G). Let Jo := {t
[0. 1]:
x {t}) C G).
Obviously 1 Moreover, the continuity of H implies that is open. To prove that Jo is closed (and consequently, that Jo = [0, I]) it suffices to show that
\ (1) C {t e [0, 1): H(KO(Th x {t}) C L}.
258
3 Envelopes of holomorphy for special domains
where

L
:= (H(K x [0.
(since G is a domain of holornorphy, L is a compact subset of G). SE [0, 1) and suppose that x {t}) c G. Since H(•, t): D —÷
Thus, fix a H(D x {:})
is biholomorphic, we get x {t})
= (H(K x
O(H(Dx(tl))
Since (H(D x {t}), 1)) is a Runge pair, we get
x (t}) = (H(K x {1}))0(D) Finally, since H(KO(Th x (t}) C G, we have x {t})
= (H(K
x
C L.
Corollary 3.1.23 (cf. [AIm 19221). The envelope of holomorphy of any starlike domain G C C" is univalent and Runge. Proof
TakeD := C" and H(z,r) :=(l—t)z,t E[0, l],z
C".
0
3.2 ktubular domains A domain G c C" is called to be a tube domain (cf. Section 2.3) if G is of the form G = B + iR". where B is a domain in R". We say that B is the basis of G. Observe that G is invariant under all mappings z —÷ z + it, t R". It is well known that the envelope of holomorphy of G can be realized as a domain in C" and, even more, it is explicitly given as = conv(B) + iR". This result was proven by K. Stein (cf. [Stei 1937]) in the case n = 2, and later by S. Bochner (cf. [Boc 1938]) in the general case (see also S. Hitotumatu [Hit 1950]). In this section we intent to generalize the definition of a tube domain and to discuss the envelopes of holomorphy of such general tube domains (cf. [Abe 1985], [CoeLoe 1986], and [Sac 1989]). We shall see, that the results we get during this discussion also contain the classical ones.
DefinItion 3.2.1. Let k, n E N, I k
(a)ax(•,O) = (b) p(ax(x, 1)) = (pi(x), p2(x) + it), x E X, t 21)
Compare Example 1.9.11 with g = Rk.
3.2 ktubular domains
259
We note that if X = G is a domain in C", then G is a tube domain (in the classical sense) if and only if it is an ntubular domain over C" w.r.t. to aG: G x R" —* G, aQ(z, t) := z + it. in the sense of this new definition. Often, instead of using the term ntubular domain, we will say that (X, p) is a lube domain over C". Moreover, we point out that the map ax is uniquely determined (cf. Proposition 1.9.12).
Remark. Let (X, p) be a ktubular domain over C". Assume that it is isomorphic to a Riemann domain (X', p'). Then (X', p') is, in a natural way, again a ktubular domain over C" (Exercise).
Let (X. p) be a ktubular domain over C" and denote by ax the corresponding map. Wefixapointxo E Xandputzo := pI(xo) E =: (pi. p2). Then Ck. := isspreadoverC" viap2: denotes the connected component of the fiber p2(x) := p2(x). Moreover, if that contains the point then X(x0) := is a ktubular domain over Ck w.r.t. the map x
—*
ô(x,t) := ax(x. t).
Therefore, a ktubular domain may be thought as a family of fibers consisting of unions
of tube domains over Ck parameterized by points in an open set in Cn_k.
Example. Let (X, p) be a Riemann domain over Cn_k, and let c C X x Ck be then a subdomain satisfying the following property: if (x, z) E c and if t E Put aç2: c x —+ z), I) := (x, z + it). Then (x. z + it) E (p x idck)k2) is a ktubular domain over C" w.r.t. ac2. For further use let us recall Proposition 1.9.12: Let (X, p) and ax be as above. Then ax gives a group homomorphism IR" —÷ Aut(X), ã(t) := ax(•, I); in particular. if t', t" E Rk then ax (ax (x, I'), t") = cX(x, t' + t"), x E X.
The aim of our discussion on ntubular domains is to see the univalence of their envelopes of holomorphy. In analogy to a Riemann domain over C" we introduce the following notation: a where B is a connected Hausdorif space and where w: B —+ R" is a pair (B, local homeomorphism, is called a domain over R". Having such a domain (B, (p) over III." there is an associated ntubular domain over C".
Lemma 3.2.2. Let (B, be a domain overlR". Put X := B x and p: X —* C", p(x. y) : = + iy. Then (X. p) is an ntubular domain over C". Proof
Take
andtElR".
ax: X x R" —+
X as ax((x, y), t) := (x, y + 1), where (x. Y) E X U
We say that such (X, p) is a tube domain over C" with basis (B. Inversely, any ntubular domain can be thought as a tube domain over C" with some basis (B. cQ); namely, we have the following lemma.
260
3 Envelopes of holomorphy for special domains
Lemma 3.2.3. Let (X, p) be an ntubular domain over C" via ax: X x —÷ X. Put B := p'(p(X) fl (IR" + i(O})) and p: B —* IR", p := ir o where ir: C" =R" +iR" —+ R", ir(x+iy) :=x. Then(B,ço)isadomainoverR",and (X. p) is isomorphic to the tube domain over C" with basis (B,
Proof Set B' := p(X) fl (R" + i{O}); then B =
We
claim that p(X) =
B' + iR".
if p(x) = E + iii. then p(ax(x, = p(x) — = R" + i{0), i.e. B' or p(x) E B' + ilR". Conversely, E B' + ilR", then = p(x) for + some x E X. Hence p(ax(x, = p(x) + iij, i.e. + E p(X). In fact,
Next, we introduce the following two mappings:
F: B x R" —p X, G: X —p B xlR",
:=
'i).
G(x):=(ax(x,—Lmp(x)),Imp(x)).
It is clear that F and G are continuous mappings and that F o G = jdx and G o F = id8 Thus B x R" is homeomorphic to X. In particular, B is a connected Hausdorif space.
Let x E
B
C X. Then there is an open neighborhood U = U(x), such that
pIu U —* p(U) is a topological map. In particular, the set p(U) fl (IR" + i{O}) = p(B fl U) is open in R"
R" + i{O}. Thus (PIBn(J:
BflU
—+ ço(B flU)
is homeomorphic; i.e. (B, ço) is a domain over R'1. Therefore, B x R" —÷ C", j3(x, v) := ço(x) + iy, gives a tube domain over C" with basis (B, Observe that p o F = i.e. (B x IR", ,3) is isomorphic to (X, p) via F: B x
0 In virtue of the classical result of S. Bochner, (conv(B) + iR", is the envelope of holomorphy of (B + iR", idB+jRn), i.e. the classical tube domains have schlicht envelopes of holomorphy. We shall prove the same result for ntubular domains over C". Here we will follow a paper of M. Abe (cf. [Abe 1985]). We mention that there is another approach to this result due to G. Coeuré & J. J. Loeb (cf. [CoeLoe 19861).
Theorem 3.2.4. Let (X, p) be an n tubular Riemann—Stein domain over C". Then p is infective and p(X) fl (R" + i{O}) is a convex domain in IR" R" + i{O}; i.e. (X, p) is isomorphic to the classical tube domain (p(X) fl (R" + i{0})) + ilR". 22) Proof Let B' := p(X)fl (IR" + i{0}) and B :=
Then, in virtue of Lemma 3.2.3, (X, p) is isomorphic to the tube domain (Y, q) := (B x IR", ç' + i . with basis (B, q), where := p18: B —÷ R". 22)
Here, and later on, p( X) fl (R" + 1(01) is identified with a subset of P".
3.2 ktubular domains
261
We have to prove that q is injective. Let Then there exist a continuous curve y: [0, 1] —÷ B xI E B with xo y(l) = such that: and N e N with y(O) =
• •
is injective in a neighborhood of o is the line segment connecting
N:
j
I
:=
with
o
1 <j
o
[0, 1] —* B. = x0, We will show that there are a continuous curve ii) with the following properties: = and an open neighborhood U of
• •
with
is the line segment connecting
o
in particular, q(xo) C Obviously, we may assume that N ? 2. Fix a r E (0, 1] and put
: = {Yi + s(yo — yi) + A : = {r E (0, 1]: C B,
—
yl): 0 s.f and s + t < r}. compact, y(l/N) E K1,
such that 'PIKr injective and
= A1).
< I such that (0, To] C A. is a local homeomorphism, there exists 0 < Moreover, we point out that if is injective on a compact subset L C B, then remains to be injective on an open neighborhood of L. In particular, this implies that for some A is an open subset of (0, 1]. Therefore, either A = (0. I] or A = (0, rj E (0,1]. Suppose now that A = (0,ri). < < Then, using the uniqueness of the lifting, we obtain Let 0 < Put D K1. Then RIo: D —÷ A1, is a topological map. C U1<1, Now, we consider the following holomorphic map g: C2 —+ C", Since
:= y' + Observe that
(i
—
—
A) =
Yi) +
A) + (i/2)77A(y2 —
—
yo) for
VI).
= + irj
E
C.
C, A E C, and
E [—1,1] x
€A11
Set x [0, ri)
—+ Y,
r)),
+ Ui. t) :
—
yo)).
Then f is continuous and the equation q o f = g holds on E x [0, ti). Because of Proposition 1.5.8 there exists a continuous extension
f: E x [0.rjj
—÷ Y := YU.dY,
of Y w.r.t. q. Since where (Y, is the leads to the same equation on E x [0. TI].
of = g on E
x 10, TI), continuity
262
3 Envelopes of holomorphy for special domains
Observe that g(—l, r) = y' + r(yo y([O. 1/Nj) x {0} C Y forO < r <
f(—l, therefore, f(E, r1)
r
/
— yi). 0
r
1, and therefore f(—l, r) E
Hence
r)
y((O, 1/NJ) x
{0} C Y;
dY. In virtue of Theorem 2.9.1(11) we get f(E, r1) C X.
Sincef(1,ri)EYandf(—l,r1)EY,itfollowsthatf([—l,IJx[0,r11)CY.Put K :=jriof([—l, l]x[O,r1]),whereir1: Y = BxW1 —÷ Bdenotestheprojection onto B. Then ço(K) =
Hence we have seen that TI
E K, and çc'Ix is injective. A. Therefore, A = (0. 1]. In particular, there exists
acontinuouscurvey2: [0,1] —k Bwithy2(0) =xoandy2(l) = y(2/N),suchthat is injective on a neighborhood of y2([O. I]) and Yo
is the line segment joining
o
and V2.
Repeating the above argument we finally see that there is a continuous curve yN: [0, 1] —* B joining xo with xi, such that is injective in a neighborhood of to w(xi). In particular, YNUO. I]) and so that ço o YN is the line segment from p(xo) W(xI) and, moreover, Lw(xo),co(xi)] c w(B) = B'. So is globally El injective, and B' = w(B) = p(X) fl (R't + i{0}) is convex.
Corollary 3.2.5. Let (X, p) be a ktubular Riemann—Stein domain over C't, and let Xo E X. Then X(xo) is a ktubular domain of holomorphy over Ck, B(x0) := fl (Rk + i{0}) C Rk is convex, and X(xo) is isomorphic to (B(xo) + iRk id).
Remark. Already in IBre l954], H. J. Bremermann had mentioned the following extended domain' over the shell example: Let (B, p) be a
0
:= B x R2 over C2 with basis
(B,çp)isgivenas(z E C2: < example of a Riemann domain over C2, whose holomorphic functions do not separate points. In the last step of our discussion on ntubular domains we shall determine their envelopes of holomorphy. For this reason we recall the following result (cf. Proposiw.r.t. ax. Then its envelope of tion 1.9.15): Let (X, p) be a ktubular domain over X x Rk X, where holomorphy (X, is also a ktubular domain over C't via Rk. t): X —÷ X,t X —+ Xis the holomorphic extension
Now we are able to prove the main result on tube domains over C".
Theorem 3.2.6. Let (X, p) be a tube domain over C" with B
:= p(X) fl (IR" + i(0}) C R".
Recall our convention that we identify JR" + i 10) with R".
23)
3.2 ktubular domains
263
:=convBandX := B+iR'7. Thenp: (X,p) —+ (X.id*) is an envelope of holomorphy of(X, p).
Proof Let a: (X, p) —+ (X, j5) be an envelope of holomorphy of X. In view of the above remark we know that (X,
is a tube domain over C'1. Put X' :=
and
B' := X' fl and p(X) C X'. + i(O}) c R'1. Then X' = B' + is a morphism between Riemann domains over Thus p :(X. p) —+ (X', C'1. Since (X, j3) is a Riemann—Stein domain, we know that is injective and that xI is convex, i.e. X' is a domain of holomorphy.  Let (9(X). Thenthereexistsan f E (9(X) with ba = f. Put F X' C. Fisholomorphicon X'with Fop = on X. Hence (X', idx') is an envelope of holomorphy of (X, p) via p: X —÷ X'.
f€
f
Since X' is convex, it follows that p(X) C X C X'. (9(X). So there exists an h (9(p(X)); then g o p Let g O(X') with h o p = g o p, i.e. h = g on p(X). Hence every function g (9(p(X)) is the restriction of a function h (9(X'), which implies, since X is convex, that X = X'.
0 So far we have seen that every tube domain (X, p) over C" has a schlicht envelope of holomorphy, namely, conv(B) + ilR'1. where B := p(X) fl (IR" + i{O}) C R". In
particular, if X = B + iIR" is a classical tube domain in C", then conv(B) + iR" is its envelope of holomorphy; i.e. the classical result of K. Stein and S. Bochner is a special case of what has been proven here. One might ask how to describe the envelope of holomorphy for a generalized tube domain B1 + iB2 c C", where B1, B2 are domains in IR" (cf. [Kaj 1963]).
Theorem 3.2.7. A domain B1 + i B2 is a domain of holomorphy ([and only ([the B3 are convex. Proof We only discuss the nontrivial implication, i.e. we start with the assumption that B1 + i is pseudoconvex. The proof proceeds via several steps. Claim 1. Let
X (af, bj)
C
C" be a nonpseudoconvex domain and
Then X2 := B +i((ai.di) x X(aj,bj))isnotpseudoconvex.
let d1 E Otherwise,
K2 andthenalso K3 := B+i((ai +bi —dp,bi) xX(aj. bj))would
be pseudoconvex. Since (X2 \ X3) fl (X3 \ X2) = 0, it follows that K1 = X2 U X3 is pseudoconvex 24); contradiction.
Claim 2. Let X1 (as above) be a nonpseudoconvex domain and let e E (0, 24)
Exercise.
Fora proof see lPfl I975b1.
3 Envelopes of holomorphy for special domains
264
+ e) x X(aj. bj)) is not pseudoconvex.
Then B + i((O. Use
Claim I and induction over k.
Claim 3. Let X1 (as above) be not pseudoconvex and let 0 <
—
<
— ai.
x X(aj. bj)) is not pseudoconvex.
Then B +
Choosek EN such that If k > 1, observe that
Ifk = l,use directly Claim I.
<
(bI._aI fore < min{b'
—(h1 _ai)/2k. (b1
Then Claim2together with Claim I
gives the desired result.
Applying Claim 3 step by step we arrive at
Claim 4. Let B+i j
=I
bj)benotpseudoconvex andletO <
bj —ai,
n. Then B + i X(a. b) is not pseudoconvex.
Claim 5. For a nonconvex domain B C R'1 there exists a positive number r0 such
that for all r >
the domain B + i X (—r, r) is not pseudoconvex. 1=' < < Otherwise we would find real numbers r1 < —÷ oo, such n
that the domains Xk := B + i X (—rk, rk) are pseudoconvex. Observe that Xk C 1='
Xk+I. Then, in virtue of Corollary 2.2.11, it follows that B + iR'1 =
is
pseudoconvex. Hence, by Theorem 2.3.1. B is convex; contradiction.
Claim 6. If the domain B + i X (aj.
is pseudoconvex, then B is convex.
Otherwise, using Claim 5 there is a positive number r with the following properties: n
• B + i X (—r, r) is not pseudoconvex,
j=l
it.
Then Claim 4 gives the contradiction. Now we are in the situation to conclude our proof. Since —i(B1 + iB2) = B2 + i(—B1) is pseudoconvex, it suffices to show that B1 is convex. Choose numbers a3.
3.2 ktubular domains b3 so
265
that X (aj. b3) c B2. Then the domain
B1
+i X(aj,bj)=(Bi +iB2)fl(R"+iX(aj.bj))
is pseudoconvex; hence, by Claim 6, we conclude that Bi is convex.
0
Remark. It would be interesting to know whether the above result remains true in the following general situation: Let (Gd, 1, 2, be domains over R". Define X := G1 x G2 and p: X —÷ p(x) = p(xI.x2) := 9i(xj) + içp2(x2). Then
i
(X, p) is a Riemann domain over C". Assume that X is pseudoconvex. What does this imply for (G1, In [Kaj 1963), J. Kajiwara claimed that the envelope of holomorphy of a domain Bi +i B2 is given by cony Bi +1 cony B2. In [Pfl 1974] it was mentioned that, in general,
this statement is false. Instead of presenting the rather complicated counterexample of [Pfl 1974] we give here a much simpler one. we learnt from B. V. Shabat (cf. [Sha 1976]). Put B1
: = (x ER2: 1/2< lxii < I),
82: = {y = (yi. Y2) ER2:
< 1/4.
y3i
j=
1,2).
that the function f: G := + iB2 —÷ C, 1(z) := (zf + is holomorphic on G, but there is no holomorphic function F: cony Bi + lB2 —÷ C
Observe
with FIG = 1. We point out that
it is not clear to us whether G has a schlicht envelope of
holomorphy There is another counterexample due to J. Siciak (communication) which is. in addition, starlike. i.e. its envelope of holomorphy is univalent (cf. Proposition 1.9.6(c)). Namely:
LetD:=(z=x+iyeC: lxi X
<
< l}andput
:= (D x (—1, l))L1((—l, I) x D) C
We denote by h the solution of the Dirichlet problem for D \ (—1, 1) with boundary
valuesOon (—i, I)and I on 8D\(—1, 1). Observethath issubharmonicon D. Then X := (z E
Dx D:
±h(z2) < 1}
is a domain of holomorphy containing X. The function z = x + iy —+ F'(z) — y is harmonic on
and h(z) > y, but h(z) $
y
:= (z E D: y > 0) on dD+. Therefore, h(iy) > y if 0 < y < 1. As a
consequence we get that the point (1/2, 1/2)
X.
266
3 Envelopes of hotomorphy for special domains
Wechoosee E (0. l)andp > Ososmallthat(i(l—e)/2.i(1—e)/2) with B1 := {x ER2: 1x11 < I — e) and B2 := {y E 1R2: we
Xandthat, < 1—rh
have that G := B1 + lB2 C X. Then the envelope of holomorphy ç cony G.
of 6
exists as a subset in C2 and it is contained in X; i.e.
Before we proceed with the general discussion on ktubular domains, we recall the following classical result (cf. [VIa 1966].
PropositIon 3.2.8. Let B c
n
be a domain of holoniorphy. and let
—V: B —+
v, be
2.
upper semicontinuous functions with v < V on B. Set
G :=
{z
=
+iYn) E B x C:
Then the envelope of holornorphy
E(G) = {z =
<
<
of G is univalent, and it is given as
B xC:
+i'vn) E
<Xn <
where C' (respectively V) is the largest plurLsubhannonic minorant (respectively the
smallest plurisuperharmonic 25) majorant) of v (respectively of V) on B.
Remark. Proposition 3.2.8 shows that a 1tubular domain G need not to be a domain of holomorphy, although each of its fibers := {z,, E C: < x,, < B, is a convex domain. In order that G is pseudoconvex, it is necessary that the boundaries of G2 (as a function on B) fulfil an additional 'convexity'—condition. Proof of Proposition 3.2.8. The proof will be given in several steps. Step I: Assume that G of Proposition 3.2.8 is a domain of holomorphy. We shall show that then v and — V are necessarily plurisubharmonic functions on B: By our assumption we know that G is pseudoconvex, i.e. ço := — log dist(., is plurisubharmonic on G. Observe that is independent of the coordinate y1,. i.e.
+
=
=
for all
z
Zn) E
Bx
+
G.
Define
D :=
(z
=
<
<
C:
F: G D, F(z) := (E. exp(zn)), is a locally biholomorphic surjection. Set 1k: D —+ := Observe that 1k is well defined and that it is a plurisubharmonic exhaustion function on D.
Then
Therefore, D is a pseudoconvex domain. E B and choose a ball U = Fix a such that v(f) < < < that a function U: G —p (—oc,
r)
B and a real numbers rj
z E U.
oc] isplurisuperharmonic if —u € ?S3€(G).
3.2 ktubular domains
267
Di := U x K(r2) and 1)2 := (U x C) fl 1) are pseudoconvex domains with 0 and (Di \ D2)fl (D2 \ D1) = 0. Therefore,
Then
Di fl 1)2
13
:=
EU xC: IznI
UD2 =
Applying Theorem 2.2.9, we conclude that for
is pseudoconvex
= —logexp(V(.)) = is psh on U, i.e. — V is locally psh on B. Hence — V Z
:= (0,... ,O, 1)
—V(S)
E
To see that also v is a plurisubharmonic on B we only mention that the mapping provides a biholomorphism from G onto the pseudoconvex = (Z. Zn)
domain {z e B xC: —
V(Tz)
<
Step 2: Conversely, assume that v and — V are plurisubharmonic functions on B.
Since B x C is pseudoconvex it immediately follows that
Step 3: Now we discuss the general case described in the hypothesis of Proposition 3.2.8: Let f be a holomorphic function on G. For E B we choose the largest strip
:= {z E C:
<x <
such that f(E..) is the restriction of an Let v1 and —V1 be the upper semicontinuous regularization of v[ and of _v(, respectively. Then v[ < vj t' < V < V1 < V(, and f7 is holomorphic on the strip :=
lz E C: vf(z) <x < V1(z)}.
Hartogs' theorem on separate analyticity (cf. Lemma 1.8.14) it is easily seen that the function Using
f: G1 :=
(z
BxC:
< Xn <
—+ C,
1(z) :
is holomorphic on Gj. So far, we have seen that any f E (9(G) is the restriction of a function f
GcG1. 27)
28)
Compare the footnote to the proof of Theorem 3.2.7. Observe that z —' — and z — — belong Observe that such a strip always exists.
to PSR(B)
(9(Gj),
268
3 Envelopes of holomorphy for special domains
Assuming that v1 and — where
V1
are piurisubharmonic on B it is clear that G C B x C:
G :={z On
<
in virtue of Step 2, we know that O is a domain of holomorphy G. Therefore, G is the envelope of holomorphy of G.
the other hand,
with
Gc
To see that vj and — V1 are plurisubharmonic functions we only have to show that G1 is a domain of holomorphy and then to use Step 1. That G1 is a domain of holomorphy follows from the fact that B is a domain of holomorphy and the strips D S((.) were chosen to be maximal (Exercise).
The result in Proposition 3.2.8 has an analogous formulation in the case of Haxtogs domains.
Corollary 3.2.9. Let B, v,
G :=
(z
=
V.
and V be as in Proposition 3.2.8. Put
E B
xC:
of G is univalent and it is given as
Then the envelope of holomorphy
= Proof Denote G.
Then it
{z
=
Zn) E B
by G the domain
<
<
xC:
that
<
is claimed
to be
IZnI
the envelope of holomorphy of
is clear that G is pseudoconvex.
Put
D :=
{z
E
B xC:
G}
=
{z E B
xC:
<x,, <
Let f E (9(G). Define g E 0(D) as g(z) :=
2,ri) =
In virtue of E.(D). Call this extension Then function g(z),
z
D.
exp(zn)). Observe that Zn + 3.2.8, g extends holomorphically to E(D). Therefore, the + 2iri) = z
Proposition Zn
f(z)
is
Z E
well defined on G. Obviously, f E 0(G) and fIG
Remark. Observe that
the situation discussed in following sense: (a) G is a 1tubular domain, (b) its basis B is a domain of holomorphy, and (c) for each E B the fiber := E C:
G,
= f.
Hence
Proposition
E G}
=
O.
0
3.2.8 is special in the
is connected.
The remaining part of this section is devoted to the study of envelopes of holomorx Ck, I k < n, which are special ktubular domains
phy of domains G C
3.2 ktubular domains
269
in the sense that G + (0, = G 29)• Our goal is to find a Riemann—Stein domain (XG, j3) over and an injective morphism
F: (G, idG) —f (XG
X
J3 x idck)
such that the envelope of holomorphy of G can be given in XG x C" as a 'good' family of convex tube domains in parametrized by the points of XG (cf. [CasTra 19911. [CasTra 1994b], [Mok 1986), and [Sac 1989), see also [ChiShc 19971). 3°)
In order to formulate and to prove that result we will need the following two lemmas.
Lemma 3.2.10. Let (X, p) be a Riemann—Stein domain over C" and let G C X be a subdomain. We assume that the envelope of holoinorphy =: G of G is given as a subdomain of X. isan upper semicontinuousfunction. there existsa largest Then, if u: G —÷
plurisubhar,nonic function
with âIG
Proof Fix xo E G and Jo E N such that u(xo) < Jo. We denote by j > Jo. the connected component of {x E G: u(x) < j} that contains xO. Then G1 C G G = UJ>JO G1. Set a3 as the connected component of the interior of the intersection of all subdomains of holomorphy of X containing G3 that contains G3. Then is an increasing sequence of domains of holomorphy. Therefore, O := U G1 is a domain of holomorphy (cf. Corollary 2.2.11 and Theorem 2.5.7) with
GcGcG. Hence,O=O.
Let v: G —+ be a plurisubharmonic function with vIG u. We denote by G1(v) the connected component of (x G: v(x) < j}. J > Jo. that contains G(v) a domain of holomorphy. Therefore, xo. Obviously, C C G is the increasing union of the domains G3, J and any function v E vIG
is locally bounded from above; hence it has a maximal element(cf. Proposition 2.1.16). D
Another proof of Lemma 3.2.10 is given in [Shi 1975].
To prepare the formulation of the second lemma we start describing the situation we will be interested in: Let (S. p) and (X. r) be Riemann domains over and C", respectively. Let F: (X, r) —p (S x Ck, p x idck) be a morphism. Moreover, we assume that (X. r) is a ktubular domain over C" via aA': X x —p X (cf. Definition 3.2.1).
xck)flG. ERk}. We should note that in ICasTra I 994bJ one deals with the problem of the cxplicit description of the envelope of holomorphy of a Kinvanant domain C X, X a Stein manifold. K here is a compact Lie group whose complexification Kc acts holomorphically on X. 29)
270
Fix an x
(p X
3 Envelopes of holomorphy for special domains
X. For t
we obtain
E
o F(ax(x, t)) = r(aX(x, t)) = r2(x) + it) = (po F1(x), F2(x) + it) = (p x
172(x) + it),
wherer = (rj.r2) and F = (F1, F2). Since F(ax(x,O)) = F(x), uniqueness of the lifting implies that F o aX(x, t) (F1(x), F2(x) + it) for all t E Therefore, we may think of F: X —÷ S x CA as a ktubular domain spread over S x Ck wr.l. the map We say that two points x',
x" E X are equivalent
F and we write x'
x",
if Fi(x') = F1(x") =: SE Sandifx',x"belongtothesameconnectedcomponent of As usually, X/ is provided with the quotient topology. We define F: X/— —+ S as F([xJ) := F1(x) and F: X/'' —p asF([x]) := here [*J denotes the equivalence class of * w.r.t. ''. In general, X/ is not a Hausdorff space. For example, take
X:=CxC\{(O,z2):Rez2=0}, r:=idx, ax: Xx R —÷ X. cix(z,t) :=(z1,z2+it), (S,p) :=(C.id), andF :=r. Observe that X is not a Riemann—Stein domain. On the other side, if X is Riemann— Stein, then we have:
Lemma 3.2.11. Assume the situation as before and, in addition, that (X. r) and (S. p) are Ricmann—Stein domains. Then (X/ —', F) is a Riemann—Stein domain over
F: (X/ '. F) —÷ (5, p)
is a morphism between Rieniann domains, and
*:
X —b X/—.
is a holomorphic map with F1 = F o
Proof Obviously, the map
*(x) := [x],
and ri = F o
is continuous. To prove that
is open we have to show
that:
if x', x" X with x' —' x" and if U = U(x") is a sufficiently small neighborhood of x", then there exists a V = V(x') such that
x"i}. In fact, we may choose U = U(x") such that Flu: U —* U1 x U2 is a homeomorphism, where U1 and U2 are open connected neighborhoods of so := F1 (x') = denote the connected compoFi (x") E S and of F2(x") E Ck. respectively. Let nent of that contains the points x' and x". We recall that is a ktubular domain over Ct via CIx xRA
—*
a := ax,
3.2 ktubulardomains
271
and that it is a domain of holomorphy. Therefore, Theorem 3.2.4 implies that F2IXS0
is a classical convex tube domain in Ck.
is injective and that q: F2(Xs0) —+
denote the inverse mapping of
Put ã(r) := rF2(x') + (I — r)F2(x") and a(r) := o &(r), 0 < r < I. Then so' connecting x' and x". a: [0, 1] —÷ X
Fort E
[0. 1], we find a neighborhood V
)< Vi"
denote a connected open neighborhood of so with center&(r), respectively. Since a([0, 1, ]) is compact, we a([0, 1]). Choose an open TN E [0. 1] such that
is a homeomorphism, where V1' and
and an open ball in
can fix points rj connected neighborhood V1 C
fl U1 of so in S. V1'
fl 0 for j Assume that continuously into X. such that
F
If we take a a(ro) =
=F
k. Then
°cok on
V' x
and ipk map V' x (V' fl fl V1).
&(ro)) =
fl then [0, 1] with &(ro) E ã(rO)). Hence, çoj and çok coincide on V x
constructed a global section
V'
V1 x
—+ X, i.e. F o
=
o
So, we have
fl
=
N
j=I
Choosenowanopenconnectedneighborhood V" = V"(&(l)) C U7=i
Tj
V'. Then
x V"). then F1(x) E V' x V") is an open subset of X containing x'. If x E and x can be connected 'over F1(x)' with a point i U. Thus, we have shown that is an open map. The next step consists in proving that X/— is a Hausdorff space: X/—' be two different points. =: s". We choose disjoint neighborhoods Suppose that s' := are disjoint open sets V' = V'(s') and V'1 = V"(s"). Then and over V' and V". Hence ifr(Fj' (V')) and *(F1' (V")) are disjoint neighborhoods of and
i". respectively. =
=: SO we fix representants x' and x" in X of
In the case
= [x') and
= [x"]. In view oft'
and
know thatx', := Fj' (so). Using the fact that x" belong to different connected components of X is a ktubular domain spread over S x Ck we may assume that u := F2(x') and
s", respectively; i.e.
we
F2(x") belong to + i{O} Wechoose neighborhoods V' = V'(x'), V" = V"(x"), U = U(so), W' = W'(u'), and W" = W"(u") such that
Fly': V' are homeomorphisms.
U
x W',
Fly": V" —*
U x W"
272
3 Envelopes of holomotphy for special domains
We claim that if U = U(s0) is sufficiently small, then
3iEv': x
(x
X — i}
fl {x E
i.e. iJi(V') fl = 0. Suppose that this is not true, i.e. there are points xj
X with
= 0,
:= Fj(x1) —÷
so
3—400
and
V" such that
x5 E V',
= Fi(x7)
=
—x, x1 '.x.
—+
J—400
and F2(x7) are points in
Without loss of generality, we may assume that
+ ilO}
denote the connected component of Flt (s1) that contains and xi'. Then, as above, F21x, is injective and F2(X1) is a convex tube domain in C', j E N. Put Let
ã1(A) :=
(1 —
A)F2(x) + AF2(x7).
a1(A)
:=
o
(—e. I + e) + i(—e, e) C C, e positive and small. Observe that aj: D —f X is a holomorphic map. If Px denotes the distance function on X w.r.t. the norm I I' then h3 := — log Px° aj are subharmonic functions on D with = h1(ReA). Therefore, hjlEo.ji is a convex function, which implies that h(r) < h1(l)} for aJi r E [0. 11. —* x" we conclude that there is a positive number ro Since x —# x' and where A belongs to D
such that px(x) > ro and px(x7) r
PXfrIj(T)) >
is large.
Combining these facts we obtain for all
N such that
a
• • •
if j
x,
ro/.,ñ) 3 x7. x", 3(&j0(t), J &([0, 1]) U I]). j ? ,jo. where &(r) := (1 — r)F2(x') + tF2(x") forO < r < I. Forany r E [0. l]wechooseaneighborhood U(r) := U(aj0(r)), maps U(r) biholomorphically onto V(r) := x Observe that U(0) V(r) —÷ U(r)the inverse W
mapping of
= roOt"
Then x
on
Moreover, there exists a number
between r' and r" with
11 3(a30(r"),
&j0(f)) = Or"(P(Sj0).
273
3.2 ktubular domains
Hence, there isaholomorphic mapping 0: W := B(p(s10). = for all r [0. 1]. and 0(ifr(s10), r oO = 1] —* := O(p(sj). &1(t)), r E [0, I.). and define [0, X as Fix aj ? Jo 0 r a1(t) fort E [0, 1] and, since E U(0). we r 0 ã1(t)) = Then = (p(Sj).
=
conclude that
curve
in X with
= a1,
Therefore,
:= 0(p(so). &(r)), r E
Set
[0,
=
j
jo.
1]. Then
=
[0, 1] —* X
= x' and
is a continuous
$(l) =
x". In virtue
of
urn F1(a1(r)) = lim
J_.o0
J—poO
=
so
=
urn
FoO(r(s3), ä(t)) =
it follows that fi connects x' and x" over so'.
F1
contradicting
rE
[0.
II,
our assumption that
Recall that F 0 = F1. Thus F is continuous and open. Fix a = [xo] E X/ maps U biholomorphiWe choose a neighborhood U := U(x0) c X such that = [x"J with = [x'j cally onto a product domain U1 x U2 C S x Ck.
x',x" E
U.
Then
= F 0 ifr(x') = Fj(x')
F1(x") =
F: (U) —÷ U1 is a topological mapping. So we conclude that F: X/ = F it follows that is a Rieis locally homeomorphic. In virtue Obviously. is holomorphic. mann domain over We want From now on. we use that (S. p) is a Riemann—Stein domain over isRiemann—Stein. This will be done by proving that X/ is to show that (X/ S', locally Stein over S via F (cf. Theorem 2.9.2(p4)): Fix ans E SflFi(X). Wechooseaneighborhood V = V(s) C Ssuch that is Stein. (Recall that (X, p) is Riemann—Stein.) Therefore, we have a plurisubharFor monic function u: R which is exhausting. Let V := E V put Hence S
:= inf{u(x) :x E
V'(x) =
It is clear that ü: V —÷ R is an upper semicontinuous exhaustion function.
We choose a neighborhood W = is Stein. W C V such that F: W —÷ W is biholomorphic and so that F) is univalent over W x Ck. Thus, it is a connected We mention that and therefore pseudoconvex. Using the minimum principle of component of C. Kiselman (cf. Theorem 2.3.2) we conclude that ü is plurisubharmonic on W.  Hence ñ E PS.7e(V). and ü exhausts V. Thus, Theorem 2.2.9 yields that V = (V) is a pseudoconvex domain. implies that (X/ —S. 1) is a Finally, since .c was arbitrary, Theorem El Riemann—Stein domain, which finishes the proof of Lemma 3.2.11. Fix a
E V.
F'
Applying Lemma 3.2.11 it turns out that any ktubular Riemann—Stein domain over has a special form.
274
3 Envelopes of holomorphy for special domains
Corollary 3.2.12. Let (X, r) be a ktubular Riemann—Stein domain over C". Then there exist a Riemann—Stein domain (Y. q) over Cu1_k, a domain G C Y x Ck, and an isomorphism F: (X. r) —÷ (G. (q x idck)IG) with F oax(, t) = (F1, F2 + it), tE
Pmof We apply Lemma 3.2.11 with (S. p) =
id)
and F = r. Put (Y, q) :=
F). Then
(X,r) —+ (Y x Ck,q x idck) is a morphism
is the quotient map from Lemma 3.2.11). Fix an xo E X. Observe that r2) is a Riemann—Stein tube domain over Therefore, in virtue of Theorem 3.2.4, r2jX,1(,0)(XO) is injective. Define
F: (Kr) —÷
(Y x
Ck,q x idc&),
F(x) :=
XE X.
Then fr is injective and therefore it is an isomorphism between (X, r) and the ktubular
0
Riemann—Stein domain (G := F(X),(q x idc*)IG) over C".
In the sequel let G be a domain in Ce_k x Ck satisfying the following property: G —+ G + (0. iiRk) = G. By we denote the projection of G into We define the following equivalence relation on G:
For two points z = (z'. z"), w = (w', w") in G, we say that
az1
n—kfl<j
Observe that,
• ifz
w, then z' = w'; • if the points z, w belong to the same connected component of then z
We write G/
w.
for the corresponding quotient; the quotient mapping is denoted
by q: G —p G/ G/ is provided with its quotient topology. Cn_k isthemapthatisinducedbyiri,i.e. poq =r1. p:
MoreoveT,
Lemma 3.2.13. In the situation above, the pair (G/ p) is a Riemann domain over p) is holomorphic, and G/ q: (G, idG) —p (G/ is holomorphically separated. Pmof In a first step we will show that the mapping q is open: Take an open subset U C G. We have to verify that the following set {z
is
open. Fix
=
E G : 3wEu: z
= (i'. i") E G and th = (ti", th") E U such that
in particular,
275
3.2 ktubular domains
Choose r > 0 such that Vt :
Bn_k(z
,
r) x
r)
Now,letf E O(G)with are
G and
C
=OonGforalln —k+1
f(z', z"). z =
(z'. z")
j
r) —f C such
holomorphic functions g, h:
g(z') =
r) x Bk(ti", r) C U.
V2 :=
Vi.
and
n.
Observethatthere
that
h(z') = f(z', z"), z = (z', z") E
=OonGforallcr E Nn_k andn—k+l <j
V2.
So, all r).
open mapping. To prove the Hausdorif property of G/ it suffices to regard two different points of the form and := [(f, := [(f', th")] in the quotient space. Then there exists a function f E (9(G), = 0 on G for the k last j's, with ti,"). i") Choose r > 0 and 1,2, as above such that f(z) f(w), whenever C G, j z E V1 and w E V2. Then q(Vj) are open disjoint subsets of Obviously, p is open and continuous. Arguing as before one easily sees that p is locally homeomorphic, i.e. (G/ p) is a Riemann domain over and that q is holomorphic. Because of the fact that the functions of (9(G/ correspond to those functions with ii — k
j
We continue with the situation above. For abbreviation we set XG := G/ (XG, p) —+ (XG, ft) we denote the envelope of holomorphy of XG. Now, we introduce the following holomorphic map
F=
(F1, F2): G —÷ XG x
Since (j5 o x
o
C',
By
F(z) = F(z'. z") := (x oq(z), z").
q(z), z") = (p o q(z), z") = (z', z") =
z,
we conclude that
F: (G, idG) —b (XG x Ck, 3 x is even a morphism; in particular, F is an open mapping. Moreover, liz = (z'. z"), w = (w', w") E G with F(z) = F(w), then z" = and x o q(z) = x o q(w). Since (9(XG) separates the points of XG, Remark 1.8.5 implies that x is injective. Thus, q(z) = q(w) and therefore z' = w'. Hence the morphism F is injective. Put G := F(G). Then G is a subdomain of XG and
F: (G, idG) —÷ (G, (j'3 x idck)1a) is an isomorphisin.
3 Envelopes of holomorphy for special domains
276
Let (x, z") E G, i.e. (x. z") = F(z', z") with z = (z', z") E G. Since (z'. z" + E IRk we obtain
it) E G, t
, I, F(z,z
/
.
I/
+:t),z +ir) .
= (x oq(z), z" + it) = (F1(z), F2(z) + it) E G (F1. F2). Therefore, the G xlR': —÷ z" + it), is continuous and it satisfies the following relations
:=
with F =: (x,
X
ocô((x. z"). t) = (j, z" + it) = (j5 x
z") + (0, it),
w.r.t. i.e. (G. (j5 X idck)Ia) can be thought as a ktubular domain over We summarize what has been shown so far in the following lemma.
x be a domain that satisfies G + (0, iIRk) = G. Lemma 3.2.14. Let G c Then there exist a Riemann—Stein domain (XG x Ck, J3 x idck) and an infective x Ck, x idc&) such that with G := F(G) the morphism F: (G, idG) —÷ G x IRk —+ isaktubular domain over C'1 w.r.t. pair(G,(j x
cô((x, z"). 0 = (x, z" + it). To continue our discussion let a: (G, idG) —+ (G, r) be the envelope of holomorphy of G. Then, in virtue of Theorem 2.12.1, F has a holomorphic extension to Foa = F. G, i.e.thereexistsamorphism F: (G,q) —+ (XG x Ck, j'3 x We know that (G. r) is a ktubular domain over C" with respect to the following system of mappings G —+ G)IERk, where&, denotes the extension of to O; in particular, we have a, o a = aG(•, 1), t E IR!c.
Our goal is to show that G* is a domain of Put G := F(G). Then a C holomorphy which coincides with the envelope of holomorphy of G. We apply Lemma 3.2.11 to the following situation:
F: (G,r)—+ (XG
ao:
G
—+ G, aa(.,t) := a,.
is a Riemann—Stein domain over C'1!C, where Hence we know that (G/ —, o ifr(x) := r1 (x), x E G. Here i//: G —+ G/ denotes the quotient map of Lemma x Ck. Moreover, we have the morphism 3.2.11 and r =: (r1, r2): O —+
F:
= o 32) XG with In order to prove that F is injective we introduce the following mapping j: XG —p j(q(z)) := oa(z), when z E G. 
12
Observe that we have simplified our notation writing F
instead
of F.
3.2 ktubular domains
277
We note that the map j is well defined: in fact, if we have points z— (z'. z"), w
= (w', w") E G with j(z)
j(w), then there exists an f E (9(G/ —) with
f(j(z)) f(j(w)). Since Locally oa is independent of the last k coordinates, the = 0, n — k + I < n. Then the fact function g := f 0 0 a E (9(G) satisfies q(w). that g(z) g(w) implies OnGweget: Fojoq = =F1oa = F1 = xoq,whereF =: (Ft, F2). Hence we have the following equality: F o j = x. Since F is a morphism. we even conclude that j E (9(X0, We extend jto a holomorphic map j: XG —* G/ with j o x = j. Then on XG we have the following: F ° J ° x = F o j = x. So the identity theorem yields that F o = idjG; in particular, F is surjective. Moreover, on XG we get Hence Fisinjective. = j,orIofr = Joioj =
j
j
XG is an isomorphism of Riemann Summarizing, we have seen that F: domains. Resolving the equivalence relation ''', the injectivity of F yields that all the fibers E XG,are connected. The next step is to prove that the map F: G —* XG x Ck is injective: Fix an x e XG. We recall fort E Rk the identity F oâ1 = (F1. F2 + it). Therefore, the is a tube domain of holomorphy over Ck. So Theorem 3.2.4 fiber '(x), fr2 I
is injective. Moreover, the fiber
implies that the map {z" E
: (x. z")
E
F(G) =: G*}
is a convex tube domain in Ck. Since x was arbitrarily chosen, we conclude that F is injective. We have seen is that F: (G, r) —* (G*. (j3 x and F: (G, idG) —+ (G. (j3 x idck)Ia) x CA is a Riemann—Stein domain and are isomorphisms. In particular. C G* is the envelope of holomorphy of G 'inside of XG x or F: (G, idG) —+ (G*, (j3 x describes the envelope of holomorphy of G.
Hence, the following result has been proven: x CA, I < k < n, be a domain satisfying G + iRk) and an (0, = G. Then there is a Riemann—Stein domain (XG. j3) over X CA, j3 x idck) such that the envelope of injective morphism F: (G, ide) —+ holo,norphy E.(G) of G is given as a subdomain of XG x CA via the morphism F; moreover, for all x E XG the fiber (z" E C' : (x, z") C(G)} is a nonempty lube
Theorem 3.2.15. Let G C
domain in CA.
For this reason we define the
Even more can be said about the envelope
following functions
X(XG) —f
:=
ve
z"): (x, z")
RA E
v
G},
0,
x E X(XG).
278
3 Envelopes of holomorphy for special domains
It is clear that these functions
are upper semicontinuous. In virtue of Lemma 3.2.10 there exists a largest plurisubharmonic function with XG —* Using these functions we have the following precise description of E(G) (Xc). I
Theorem 3.2.16. in the situation described in Theorem 3.2.15, the envelope of holomorphy C(G) is given as
= ((x. z")
E XG
x C":
< Re(v, z"). V E
0).
v
Proof Firstly, we recall the above situation. We know that is a C XG x ktubular domain w.r.t. the mappings (x, z") —÷ (x, z" + it), t E Rk, with convex fibers. Hence it is given as
= ((x, z") E
XG
x C":
< Re(v, z") for all v E
v
0).
Using Kiselman's theorem (cf. Theorem 2.3.2) we conclude that the functions are plurisubharmonic on XG and, since G c G*, we get ,
'
G*
XG
IX(XG) <
Therefore,
which implies that
j ((x,z")
x Ck:
< Re(v,z"), V E Rk, v #0}=: D*.
On the other hand, D*, as a subdomain of the domain of holomorphy XG x Ck, is given by 'plurisubharmonic inequalities', i.e. D is also a domain of holomorphy, and so D* j G*, which completes the proof of Theorem 3.2.16. 0 LetG beas in Theorem 3.2.15. Put cv := jr1(G), where,r1: G —p denotes the projection of G onto the first (n — k) coordinates. Moreover, assume that G has only connected fibers over cv w.r.t. In. Then = cv. Therefore, Proposition 3.2.8 describes a particular case of the situation handled in Theorem 3.2.15.
Corollary 3.2,17. Let G C be a ktubular domain, i.e. G + (0, = G, and letw := ,ni(G). Assume that thefibers (iri(z)), z E G, are connected and that C Then the envelope of holomorphy C(G) is univalent and it is given as
x C"
= {(z'. z") where in
< Re(v,
:
is the largest psh function on
z"), v
R", v
0),
satisfying
if k = I, then G=
z,,) E cv x C
< x,, <
:
v < V. and
V
C(G) =
x C:
<x <
(respectively V) is the largest psh (respectively smallest plurisuperha r,nonic) function on with V). v (respectively where
In analogy to Corollary 3.2.9 we obtain the following version for Hartogs domains.
279
3.2 ktubular domains
Corollary 3.2.18. Let G C D x C fibers
be
a Hartogs domain with connected Icircled
i.e.
G
D
= {(Z,Zn)
xC:
<
where v, —v are upper semiconhinuous on D, v < V. Assume that C(D) C Then is univalent, and his given as
= where
xC
Zn)
<
:
<
and V are defined as above.
Remark 3.2.19. (Cf. the example of Shabat at the beginning of Chapter 1 and Remark 3.1.12(d).) In order to describe the envelope of holomorphy of the example discussed at the very beginning of Chapter I it remains (cf. Remark 3.1.12(d)) to find the envelope of holomorphy of the following domain
D :=
+ iy. w)
E C2
x <0, 0 <
:
y
and
O
ir/2
following domain
:= {(z,
w) E D x C z
b, wi <
whereD:={z=x+iy€C:x
0
is a holomorphic extension of D. In virtue of Corollary 3.2.18 it suffices to determine the largest subhannonic function V: D —÷ Rsuchthat —V <—Von D. Thenwewillobtain
C(D) = {(z, w) E D x C z E b, wi To get
we only have to find a bounded harmonic function h on D such that f 0
( —
if
log2
SecSection3.I.
if
x° <0, y0 = 0 or x0 = 0, 0
(*)
280
3 Envelopes of holomorphy for special domains
That bounded solution of the Dirichlet problem (*) is easily found using the following
biholomorphic mapping c1: D —.
4(z)
sin(—2iz —
ir/2)
— 1,
D,
z
where H4 denotes the upper open halfplane. Observe that E (—00,0)
R,
x=
or
0
0,
0<
y
ifx <0. y=ir/2
C1)(z) E(0,00)
Then h: D —÷
if x <0, y =
h(z) := ((l/ir)Arg(c1(z)) — l)log2, is harmonic on D and
solves the Dirichlet problem (*) from above.
Put V: D —f IL V(4)
log2
if x <0.
—h(z)
if 0 <
y
< y
<5ir/2
<,r/2.
SJf(D) and —V is the largest subharmonic rninorant of —V. Thus
Then —V
I Iwl <2

if 0 < y ir/2 F), where F(z, w) =
is the envelope of holomorphy of D. Therefore, (exp z, w), is the envelope of holomorphy of the Shabat example.
Moreover, we also can handle the envelope of holomorphy for Hartogs domains (cf. [Shi 1980]). with connected kcircled fibers z
Let G C D x Ct C C" be a Hartogs domain with connected kcircled fibers E D, where = {w E Cc (z w) E G}.
In a first step we assume that
zED. Following Section 3.1 we know that
is univalent and it is given as the logarithmically convex hull of G:. Therefore it can be written in the form
= {w E C"
r3(z, WI
<
<
(z, WI,..., where r1 (z, .) (respectively
wj E G1(z);
Wi_I). I
j
k),
(z, .)) is an upper (respectively a lower) semicontinuous
function on G3(z) := ((wi
w3j)
WI
. .
.
< R3(z, w1 SeeSection3.I.
, W.c_i) < 1w31
ws_I), I <S < j — l},
3.2 ktubular domains I
<
j
Note
r1(z,.) < R3(z,.)
that 0 <
<s<j—
281
and that all
r1(z, •)'s
and R3(z,
•)'s
I.
Put
:=D and Gj:=U{z}xGj(z),
G1
Observe that r3 and — need not will construct domains G c G1, the upper and lower semicontinuous First,
:=
set
such that
>
e
G1
Therefore, we
to be upper semicontinuous on
j =I
k.
and we will denote by r7 and (Ri),
and R3 on G. respectively.
regularization of Tj
= D. Now
let z0
E
D. Then
there are
w°
E Ck and e >
0
w0). E) C G; in particular. we have
and
w0). e) c
{z} x
U
C
(z) x
U
(.0E)
In virtue that
of the
fact that
Iw?I So we get
e)
is a Reinhardt domain it follows for all z E
<
+e < R1(z).
<
—
Hence
E G7
{(z.
xC:
< Iwil <
is open and not empty. Moreover, it is connected and G,t C G2. In particular, and R2 are defined on The next step consists in proving that r < on To do so let (z°. w?) E Then there exist E C such that w° E E(G.o), where w0 such that w° E Using (w° 4). Choose a Reinhardt domain G' that G is open we find > 0 such that e) x G' C G. If e is sufficiently small, then we obtthn w0), e) C Pn—k(Z°, e) x
{z} x
C (zt),,)
In particular, for z E
E C with 41
e) and
—e
< wi
I
< lw?l +
follows that
r2(z.wl) <
<
+r < R2(z).
r(z0, 4) < w?). Since (z°. w?) was arbitrarily chosen, it follows that
Hence
:= ((z. w) 0
E
D x C2
:
(z.
E
< R1(z) s oc are simply numbers.
<
wi) < lw21 <
on
Thus
it
282
3 Envelopes of holomorphy for special domains
is an open connected set with 0 and depend only C G3. Observe that on Iwil. Repeating this construction we obtain the desired domains G C D x and the semicontinuous functions 0 < r7 < on G; moreover, the r7's and the depend only on I <s <j — 1. Put
((z. w) E D x C" : r7(z.
G
uj
<
< 1W31
WI
wi_i) E G. I
wj_i). (z. wi
j
k}.
where
:= {(z, WI
ED x
wi .
:
<
(Zi.
.
.
,
< WS_I) E
WI
G is a Hartogs domain with connected kcircled fibers which contains G. More
over, if f E (9(G), then for all z E D the function f(z. .) extends holomorphically to the fiber Applying the Hartogs theorem on separate analyticity yields that f itself extends holomorphically to G*. So it remains to find the envelope of holomorphy of
G. Write r7 =: exp(v1) and
=: exp(V3). Then —oc vj < V3 00 and v3, upper semicontinuous. Assume now that D has a univalent envelope of holomorphy, i.e. E.(D) C Then, applying Corollary 3.2.18, has a univalent envelope of holomorphy which is given as —
V3 are
=
((z. WI) E
xC:
< Iwil < exp(VI(z))}.
is the largest psh function on e(D) with v1 and where V1 is the smallest plurisuperharmonic function on e(D) with Vi ID V1. Observe that E is negative on D. Let be an increasing sequence of subdomains D with D = D'.. Then where
= < on In particular, for any z E e(D) there exists a WI E C i.e. such that (z. WI) E isa Hartogs domain with connected 1circled fibers over C(D). Repeating this argument we obtain that the envelope of holomorphy of G, Therefore,
:= ((z. w) E
x C2: (z, Wj)
E
exp(v2(z,WI)) <
< exp(V2(z.w1))},
3.2 ktubular domains
283
is univalent and it is given as
=
{(z.
x C
w2)
wi)) < 1w21 <exp(V2(z,
:
(respectively is the largest (respectively smallest) psh (respectively plurisuperharmonic) function on satisfying (respectively V2IG,. where
< V2 on Thus V2). Observe that and V2 depend only on wi I and that is a Hartogs domain with connected 2circled fibers over C(D) (for z E C(D)
findfirstw1 ECwith(z,w1)E
E Cwith(z,wI,w2) E
Following the same argument successively gives finally that the envelope of holomorphy of G* is univalent and it is given as
=
{(z,
Wk_1, Wk) E
C(G) x C wkI))),
Wk...1)) < IWkI < exp(Vk(z,
exp(i)k(z, WI
where i)k (respectively, 14) is the largest (respectively, smallest) psh (respectively plurisuperharmonic) function on Vk (respectively, VkIG. ? satisfying í3kIG;
Vk. Both functions depend only on Iwil
Iwk_II.
Summarizing we have that C(G) is univalent and that there is an effective description of it. Moreover, E(G) C x Ck is a Hartogs domain with connected kcircled fibers. Now we start treating the general case. Let G C D x Ck C be a Hartogs domain with nonempty coimected kcircled fibers z E D. Without loss of generality, we may assume that if w E 0, and that for any j, z E D, then Wm+1 1 <j
xCk
Go:=((z,w)E
...•wk =0}.
:
Let f E (9(G). Then (cf. Section 3.1) f extends holomorphically to G1
:= {(z.Aiwi
Wk)
ArnWm,
ED
x Ck
l<j<m}. Put G :=
G1 \ Go. Observe that the domain G is of the form we discussed before. Therefore, C(G) is univalent and it is given as
=
{(z, w) E E(D) x CI
:
wi
Wi_I)) <
j
I
k},
where the notions are taken from above, now with respect to G. Note that j ,n. Put G
{(z, w) E €(D) x Ck: IwjI
I < j
w1_i)),
Wj...I)) < IwjI <exp(Vj(z, WI (z. w1
where GT := D.
E
284
3 Envelopes of holomorphy
for special domains
Observe that G is pseudoconvex and that G \ Go = Then every f E 0(G) extends holomorphically first to and then to G, i.e. C(G) = G (recall that extends to G i).
f
Summarizing we obtain the following description of the envelope of holomorphy of Hartogs domains with connected kcircled fibers.
Theorem 3.2.20. Let G C D x fibers. Assume that C univalent and ii is given as
C"'.
be a Hartogs domain with connected kcircled Let m E {O, I k} be as above. Then E(G) is
= {(z. w) E E(D) x w3_f)).
I
j
in.
(z. wi where the notations
k.
Wj_1) E
were taken from above.
V1. and
Observe that E(G) C any point
j
wi—i)). in + I
< IwjI < exp(V3(z. Wi
WI
x Ct is a Hartogs domain with nonempty fibers over
of
3.3
Matrix Reinhardt domains
Let M(n) denote the set of all complex n x n matrices. We identify M(n) with On M(n) we consider the action of the group 9 := 11(n) x 11(n), where 11(n) is the group of all unitary matrices in M(n). given by
M(n) x 9
(Z, (g,
f)) —+ gZf.
where gZf is the matrix multiplication. Analogously, if k E N, we have the following group action
Mk(n) x
((Z1
Observe that in case n =
((zi
Zk), (gi. fi I
fk)) —*
gkZkfk).
this reads as
zk). Of,
36)
37)
Notethatifrn = 0 then = G. For simplicity we write (ga. Il
St. 1k) instead of ((gi ft )
*f ER.
(St. 1k)).
285
3.3 Matrix Reinhardt domains
Definition 3.3.1. A domain D c Mk(n) is said to be a matrix Reinhardt domain (respectively a circular matrix domain) if D is invariant under the group action of D for gkZkfk) (respectively 9), i.e. if (Zi Zk) E D, then (g1Z1f1 k (respectively (gZ1f D for all all unitary matrices gj, f1, 1 E j
(g, f) E 9). In case n = I, D is a matrix Reinhardt domain (respectively a circular matrix domain) if D is a Reinhardt domain (respectively a circular domain) in the usual sense.
Examples 3.3.2. 1) The matrix disc or the generalized disc:
A := {Z where
38)
M(n):
is the ndimensional unit matrix and'<' means that
—
isa positive
definite Hermitian matrix.
We note that A is a Cartan domain of first type (cf. [Khe 1990] and [SerVIa 1994]) and it is holomorphically equivalent to a convex tube domain in Cr; thus A is a domain of holomorphy. Observe that A is a matrix Reinhardt domain, but in case ii > I it is not a scalar Reinhardt domain 2) The matrix polydisc:
P := {(Z1
Zk) E
Z3 E A}.
It is a matrix Reinhardt domain but it is not a scalar Reinhardt domain if,, > 3) The matrix ball:
1.
B:={(Z1 For
min{k,n} >
1,
it is not a matrix Reinhardt domain, but it is a circular matrix
domain.
4) The extended matrix disc:
D := {(AZ1B
AZkB)
Mk(n): (Z,
Zk)
P. A, B
SL(n, C)).
M(n): detM = 1). This example is related to quantum field theory. It is a circular matrix domain but, in general, it is not a matrix Reinhardt domain. Nevertheless, it is shown by P. Heinzner & A. Sergeev (cf. [lIeiSer 1992]) that the extended matrix disc is a domain of holomorphy. Let n k = 2. To see that D is not a matrix Reinhardt domain we present an example due to X. Thou:
where SL(n,C) := (A
38)
Z :=
Z' is the conjugate matrix of Z. A scatar Reinhardi do,nain means simply a Reinhardt domain in the classical sense.
286
3 Envelopes of holomorphy for special domains
Let 0
Observe and denote by Z0 the diagonal matrix [c, 11 E M(2) Z0A0 A if c < 1. Therefore, the pair (Z0, 4) belongs to the extended matrix disc D. I
that [c. ll[I/d, dJ
=:
Fix
/0
1
0
We assume that
(4,
(W1, W2)
= (Z0. [l.cJ) €D.
there exist matrices A, B SL(2, C) and Z1, Z2 E A such that (W1, W2) = Observe that are not singular. It is easy to prove that with Z1, Z2 E A also E A and then E A. Finally, we get
Then
(AZ1
[C2,
1J
= det(W2) W1
E A;
contradiction
Recall that there is a complete characterization for Reinhardt domains to be domains of holomorphy (cf. Proposition 1.9.19). So a natural problem is to characterize the holomorphically convex matrix Reinhardt domains in terms of naturally associated classical Reinhardt domains. Let D C Mk(n) be a matrix Reinhardt domain. We define its diagonal as
d(D)
Dfldiag(n)k,
where diag(n) := Z a diagonal matrix}. We recall from the general theory of matrices that any matrix A E M(n) can be written as A = UAV, U. V E where A = [Xi is the diagonal matrix whose entries ... 0 are the eigenvalues of the positive semidefinite Hermitian matrix Using this representation we conclude that a domain 1)
Mk(n) is matrix Rein
hardtiff,whenever(Zj Zk) E d(D). Inparticular, Zk) E D,then(Z1 is a Reinhardt open set in the usual sense. d(D) C Observe that d( D) is invariant under arbitrary permutations, i.e.
if ([A11
Am]
[Aki
Ai.ai(fl)]
then
where
Xknj) E d(D).
the a1's denote arbitrary permutations of (1
[Ak.ak(1)
Ak,ty*(n)]) E
d(D),
n}.
Let A bc a positive semidefinite Hermitian matrix and let xj, be an orthonormal k basis of cigenvectors. n. Then there is a uniquely defined positive scmidefinitc = I
287
3.3 Matrix Reinhardt domains
We say that d(D) is piecewise permutational or, jfk = I, permutational. Finally, we point out that D is completely determined by its diagonal, i.e.
D=
Zk) E d(D), U3. V1 E 'U(n),
UkZkVk): (Z1
I
j
k}.
Our aim is to characterize holomorphically convex matrix Reinhardt domains in terms of their diagonals. This has been observed by A. Sergeev (cf. [Serg 1993]) under an additional completeness assumption. The general case was solved independently by X. Thou (cf. [Zho 1990]) and by E. Bedford & J. Dadok (cf. [BedDad 1991]).
Theorem 3.3.3. A inatrir Reinhardi domain D C Mk(n) is a domain of holomorphy and only domain in
its diagonal d(D) is a Reinhardi domain of holomorphy (though: as a
In order to be able to prove the necessary condition of that theorem we need an additional information.
Lemma 33.4. Let
(r) —+ IR be a C2 strictly psh function. Then, after a —* C'1, the second order suitable complex linear coordinate transformation A: terms in the Taylor expansion of ço o A at the origin equals
Zj =Xj where A > 0.
Proof By a standard argument 42)
we
may assume that
is locally given by the
following Taylor expansion at the origin
w(z) = where
S = (S,,) with
11z112
+ Re(zSz') + o(11z112).
(t)
=
In the remaining part of the proof we will identify complex vectors z = x + iy With S =: A + iB, with the real vectors (x, y) E IR'1 x
Zarealmatrix), and v := (x. y) E
near the origin, we rewrite (t) as 'p(z) = p(v) = 11v02 + vTv' + o(11v112),
where T denotes the following matrix
(A 42)
Exercise; cf. Section 2.2.
—B —A
288
if
3 Envelopes of holomorphy for special domains
Observe that if v = (x. y) x + jy, then i(x + iy) (—y, x) =: J(v). So, is an eigenvalue of T with eigenvector v, then is also eigenvalue of T with
eigenvector J(v). Now we choose an orthonormal basis of
such that T
=
vt,
consisting of eigenvectors of T, say
> 0 (this can be achieved by eventually changing the role of
and
Let C: R2" —÷ R2" be the linear mapping such that
j=
= J(v1),
C(e3) =
n,
1
where e3 denotes the jth standard unit vector in Obviously, C can be thought as a complex linear map from C" to C". We will use C as the coordinate transformation whose existence has been claimed in the lemma. Let w E Ri". Then for the transformed function := q o C we get
= ço(C(w)) = (Cw')'(12,, + T)(Cw') + o(II(Cw')'112)
= w(12,, + C'TC)w' +
o(11w112)
=
1w112
whereDisthediagonalmatrixL) = the claimed second order terms in the Taylor expansion of
+ wDw' +
o(11w112),
at the origin.
0
Let M be a real submanifold of class & of an open set U C C". M is said ii = {0}, where to be totally real if for each z E M we have that denotes the real tangent space to M at the point z and where J defined as J(x, y) := (—y, x), (x. y) E R" x IR" C", means multiplication with i. The definition says that for each z E M none of the complex lines in belong to In particular, dimR M
n.
Theorem 3.3.5. Let be a nonnegative C2 strictly psh function in an open set = 0}. Thenforeachpointzo E Z there existsa ball U C C". Put Z := {z E U: I3(zo, r) =: V C U and an & totally real
M of V such that Zn V C M,
i.e. the zeroset of cp is locally contained in a totally real submanifold of class C'. From this result the following remark is obvious.
Corollary 3.3.6. The zeroset of a nonnegative C2 strictly psh function on an open set in C" does not contain a real submansfold of class C whose real dimension is larger than n. Proof of Theorem 3.3.5. Without loss of generality we may assume that zo = 0. Moreover, we start working in the coordinate system whose existence was established
in Lemma 3.3.4. Thus, w.l.o.g.
has the following description around the origin:
z=xj+iyj,
289
3.3 Matrix Reinhardt domains
where the A3 's are nonnegative. Here we have used that grad colz = 0. Put ax1
Then the form of
implies that
ia
a
a
avi
dxi
dx3
where
=0. 1
(h3
denotes the standard jth unit vector in
grade —(0) oxI
x W'. In particular, we see that
grads —(0) oxn
linearly independent. Therefore, these vectors remain linearly independent in a small neighborhood V = V(0) C U. So Z fl V C M := M 0 V and M is a real ndimensional submanifold of V of class &. Its tangent space at zero is the orthogonal Thus complement of Re1 + . . . + i.e. it is spanned by the vectors en+1 M is totally real at the origin and therefore also in a neighborhood of 0 Hence the theorem is proven. D are
We will follow a standard procedure to produce a nonnegative strictly psh exhaustion function on a holomorphically convex matrix Reinhardt domain D C M1'(n), that is invariant under our group action.
Lemma 3.3.7. On
a
holomorphically convex matrix Reinhardt domain D C Mk (n)
there exists a C°° strictly psh exhaustion function
having the following property:
Zk)=ça(glZifj
EtL(n),j=I
forZ=(Z1
k.
strictly psh exhaustion funcProof Since D is holomorphically convex there is a tion D —+ IL We define a new function on D, namely: ço(Z) :=
I
gk.fk).
J9k
for It is an easy exercise to verify that 0 fulfils all the properties that were claimed in the lemma. where dci denotes the Haar measure
Use that the perturbation.
i
remains an
after a small
Let 9 be a locally compact topological group. Then there is a uniquely defined (up to a positive constant factor) measure on 9, which is invariant under translations from the left. This measure is the socalled Haar measure.
290
3
Envelopes of holomorphy for special domains
Remark 3.3.8. Let D C M(n) be a matrix Reinhardt domain with diagonal d(D). If d(D)containsamatrix[ai then d(D) is connected. with lail = = In fact, the part d* := dfl{[x1 Ed: > 0} of the real part of d is connected. So, if there is the point A := hail E d, then all the connected parts of the real part of d, obtained by permutation from are connected via the point A and its permutations.
Now we are in the position to prove the necessary condition.
Proof of the necessity in Theorem 3.3.3. d(D) is the intersection of D with a certain number of hyperplanes, and therefore it is holomorphically convex. To prove that d(D) is connected we choose a C2 strictly psh exhaustion function 1) —÷ R which is invariant under the action of 9k (cf. Lemma 3.3.7), i.e. is constant on the orbits of Define c := Z E D}. Observe that c E R. Put
M := {Z e D: — c = O}. In virtue of Lemma 3.3.5, M is locally contained in a totally real submanifold. In particular, any orbit Z, Z e M, is locally a subset of a totally real submanifold. Fix a real point Z0 := (hz?1 e d(D) with Zo M; [4
observe such a point always exists. Then the dimension of 9kZ0 is greater than n except for each j = 1 n, have the same k all diagonal elements £= I absolute value. But then using the above remark we conclude that d(D) is connected.
0 Now we turn to prove that a matrix Reinhardt domain D C Mk (n) is holomorphically convex if its diagonal is connected and holomorphically convex as a subdomain in We proceed in proving several preliminary results. We repeat that for a matrix Z M(n) we had introduced the associated diagonal matrix[A1(Z) ? ... the positive semidefinite Hermitian matrix
Lenuna 3.3.9. Let D C M(2) be a matrix Reinhardt domain with diagonal d := d(D). Suppose that d is a logarithmically convex domain in C2. Let P be a real symmetric polynomial in two variables such that the function p: d —÷ R. p([zI,z21) p(zI.z2) := P(1z112, 1z212). ispshond. Then thefunction Q: D —* R. Q(Z) := p(A1(Z), A2(Z)), is psh on D.
Proof Since P is a symmetric polynomial, there exists a uniquely determined polynomial cD such that
P(xl,x2) = 4)(xi +x2,xIx2), We
x = (xI,x2) ER2
have to verify that the following function
D
Z —+ Q(Z) =
= ispshon D.
= P(IA1(Z)12, 1A2(Z)12)
+ A2(Z)12. IA1(Z)121X2(Z)12) = 4(trace(ZZ*),det(ZZ*))
3.3 Matrix Reinhardt domains
Fix Z0 = (iota, b]Vo
291
D, where [a, b] E d with a, b > 0 and
M(2)
'11(2). Moreover, let W0, W0
with W0 = Uo W0 V0.
where Uo, Vo E What we have to show is
that the following function
—+ Q(Z0 + AW0) = Q(Uo([a, bJ + AW0)Vo) = Q([a, fri + AW0) =: f(A). A
(0) > 0.
C near 0, satisfies that
Calculations lead to the following representation off
AE U :=U(0)cC. where
a + Aw?112 + lb +
gi(A)
:= Al2(lw?212 + g2(A)
+ Aw?1l21h +
a
2 + w?2l2 — 2 Re
g2(A)
with H(A) :=
+Aw?1)(b+Aw2°2). By our assumption we know that the function f,
f(A) := is subharmonic near the origin. In particular, we have (0) ? 0. Looking at the formulas for the we easily obtain that
dA
dA
dA
dAdA
dA
dAdA
—
Thus calculation leads to
d2' d2' —4(0) = —4(0) + (a2 dAdA
dAdA
0 it suffices to verify that
To get that
dAdA
dx1
+ b2. a2b2) ? 0.
First assume that our matrix Wo has a special shape, namely w?1 = a. w2°2 = —b. In this case we obtain
gi(A) =a211 +Al2 +b211 — Al2, and, therefore,
0< —
dAdA
= —(a2 + ax1
b2, a2b2)(a2 + b2) +
ax1ax1
(a2 + b2. a2b2)(a2
—
b2)2.
(t)
292
3 Envelopes of holomorphy for special domains
In particular. if a =
h
0. then
+ b2, a2b2) 0. We proceed by contradiction. Assume there is a point [a, b} E d such that + h2, a2b2) <0. Without loss of generality we may assume that a > b > 0, Recall that by assumption d is connected, logarithmically convex, and permutational. Put a Then [a, aJ d and a2 = a!, =: c. We find numbers y and
+
(c/I)2. C2)
> o,
z
—(t2+(c/t)2,c2)<0. y
From (t) we know that
integration leads to
a2i
act) (t,c2)dt= —(52+(c/8)2,c2) <0, 0< Jy2+(c/y)2 I axi dx1dxj
0
which gives the expected contradiction.
In the next step we will prove a similar result for matrix Reinhardt domains in M(n), n > 3.
Lemma 33.10. Let D c M(n), ii > 3, be a matrit Reinhardi domain whose diagonal d = d(D) is a logarithmically convex domain (recall that d is automatically permutational). Suppose P is a real symmetric polynomial in n variables such that the function
p: d —+ IL
= p([zi
p(zj
zn])
P(1z112
is ps/i. Then
Q: D —p IR, is psi:
Q(Z) := p(Ai(Z)
on D.
Pmof The proof is similar to the one of Lemma 3.3.9. First, we write the symmetric polynomial P as a polynomial cD in the elementary symmetric functions sk(X) = sk(Xi
>J1<j<
.
...
P(xj Thus we obtain
Q(Z) = ct)(oi(Z) where (Z) means the sum of all kth principal minors of ZZ*. Here, the last equality is taken from algebra.
.E
3.3 Matrix Reinhardt domains
d, a3 >
293
matrices (Jo. Vo E 'U(n), and W0 = M(n). We have to prove that the function f Fix
0,
(Jo W0 Vo E
an(Uo([a1
AECnearO, satisfies (0) > 0. Again, we write ak(LaI
a,,] + AW0) = gkGk) +
where
gk(A) :=
+
+
2
2.
I<j,<.<j*
gk(O) = sk(aI
an),
=
gk(O) = 0,
fl
= 0,
aJ
l<ji<
s=I
p.qø(ji.. J&i} As in the proof of the previous lemma we put
AEC. We get
= where
+
a2 :=
Since the function f is subharmonic, it remains to show that the second term in (1) is non negative. So we claim that for any pair p q
s=I
k=I 3k—I)
294
3 Envelopes of holomorphy for special domains
Observe that for n = 2 this claim already had been verified in the proof of the preceding lemma. So, we proceed by induction and we assume, that claim is true in case of dimension n — 1. Without loss of generality, we may assume that p 1 q. Put d0
:
([z2
and we define a new function
Zn]
:2 :.
[ai, Z2
Zn] E
da
Zn]) = where
:
zn)
(1z212
Evaluating the
elementary symmetric functions we get
Zn)
=
+s2U2)
Observe
and that
that da
whereã2
Using
.
In virtue
of the
induction hypothesis we conclude that
k
2<ji<••<j*i
:= the following equality
—=—+a1 aYk dXk dx*+i we
see
that
>axk k=I ikl} sn(a2))
k=2
k
cn(a2))
k=I
a?yn_J).
is a Reinhardt domain, logarithmically convex and permutational
is psh on da.
k=I
+Yk
+y2
'l'(v)
k
p.qq(jl
ik—I}
295
3.3 Matrix Reinhardt domains
0, i.e. the lemma is U
which completes the proof of the claim. Thus we have proven. Finally, we obtain the general case.
Lemma 3.3.11. Let D C Mk (n) be a matrix Reinhardt domain such that its diagonal Let P be a piecewise d = d(D) is a pseudoconvex Reinhard: domain in symmetric real polynomial on i.e. P(x11
XkI
Xkuk(I)
Xkyi(fl)
for all variables and all permutations a1. Suppose that the function p: d —* JR. p([zii
ZInJ
Zkn])
[ZkI
IzInI2
:= P(Iziij2
= p(zii
ZIn
ZknI2) = P(zi
IZklI2
Zkn)
ZkI
. Zk)
.
"s'
ispsh on d. Then the function Q: D —* JR.
p(Ai(Z,)
Q(Z)
Zk)ED, ispsh on D.
... >
(We recall that A1(Z1)
> 0 denote the eigenvalues of Z1Z7.)
Proof We proceed in a similar way as before. Let c1 be the real polynomial in nk variables such that
=
P(zi
C1(si(zi
We fix an arbitrary point ([au tL(n), and Vi,, = UXWXVX E
f(A) =
[akl
C
am] + AW1)
+ AW1)
akfl]+AWk)
=
=
An([akl
akn]+AWk))
+ gkn(A)).
CI)(g11(A) +
where, as above,
E
M(n), I <x
= 0, 1 < x
k, I
Moreover, we define
f(A) := 45)
1(gmi(A)
Recallthatforvectorsa. bE
gmñ(A) C" weusethefollowingconvention:
A E C.
a•b = (aibu
3 Envelopes of holomorphy for special domains
296
Then, a tedious calculation leads to
±L(O) x=I k
d2'
= —4(0) + dAdA
—
x=I \dAdA
d2' dAdA
where
and
:=
gkl(O)
virtue of Lemma 3.3.10 we see that the righthand side of(t) is non negative. Hence fl we have shown that Q is psh. In
Now we start to construct an exhaustion of D by pseudoconvex subdomains. What we will need is the following general approximation result for psh functions.
Lenima 3.3.12. Let G c C" be a Reinhardt domain, G' subdomain, and ii: G —÷ R psh. Suppose that u(z)
= u(IziI
z
= (zi
G
fl
(C,)" a Reuthardr
Zn) E G.
Then there exists a sequence of real polynomials Pk, k E N, in n variables such that pk: G' —+ IR. pk(z) := Pk(1z112 IzflI2),are pshfunctions withpk \
G 1) (C,)" with G' Pmof Fix a Reinhardt subdomain G" G" —÷ IR, Uk \ uIG". Put uk(z)
I
2ir
I j0
The Uk are psh on G" with
for all z E G". On the compact set K := vk(xl
I j0
ük(e"zI
G". We choose psh
.
z E G".
. .
\ uIG'; they satisfy also uk(z) = uk(IzII (x
(IR>0Y':
G'} we set
:=
In virtue of the Weierstrass approximation theorem we find real polynomials Pk such that x K, a al < 2, ID° Pk(x) — D"vt(x)t <
where the positive rA will be chosen later.
3.3 Matrix Reinhardt domains We put
:
Forz
.
E
297
G' we obtain
£pk(Z; X)
E
=
+
•••, IZnl2)IXvI2
(IziI2 v=I
v,/L=I —
eicIIzlIIIXII — eklIXIl 2
= Lvic(z: X) — €k(l + IIzII2)IIXII2, where Vk(Z) Vk(IZI 2 = Uk(Z). = uic(IziI Since Uk is psh we finally reach the following inequality on G'
Lpk(Z; X) With C
sup{l + 1Iz1I2:
z
+ lIzIl2)IIXII2.
E G'} we see that the functions Pic.
pk(z) := jik(z) + ekCIIzll2 + 5k.
zE
are psh on G'. Moreover, pk(z) = Pk(IziI2
where
Pic(x) := Pk(X)+ekC(XI i.e. Pk is a real polynomial in n variables.
Calculations give for z
u(z)
E
G': < vk(IziI2
= Pk(IZI
jznI2)+ekCIIzII2
Pk(Z)
+ ekClJzlI2 + Sj1
2
IznI2)+skCIIzlI2+25k
<
+ skClIzll2 +
Vk_I(1Z112
5 Pk—i(Z) + C11z112(sic
if we choose a sequence
sic—i)
+
2Sk
S Pk—i(Z).
CC'Sic_i/(2 + C2) ,where
sic —* 0, such that
C':=min{11z112: ZEG'}.
0
Hencewehavepk \uonG'.
Proof of the sufficiency in Theorem 3.3.3. Fix an £ E N and put
:= max{— log
ad),
.
—
298
3
on d C
(C's
)k•
Envelopes of holomorphy for special domains
Observe that
psh and invariant under coordinatewise rotations.
is
Therefore,
hi,...,
logd ?
is a convex function. Since log d
dt
itself is convex, it follows
:= (z Ed: ut(z)
logarithmically convex and piecewise
are
that the sublevel sets
d
permutational;
moreover, d1
and
a= By D, we denote that matrix Reinhardt domains whose diagonal d(D1) = dt.
Fix an e > 0 and put
:= (z Edt: Applying Lemma 3.3.12 we
IJ)(Rflk) such that
I
I
find a decreasing
sequence of
real polynomials J'j E
: d((÷1)5 —÷ IR,
'ij(ZiI
. . .
IZknI2),
, Zkn)
is psh on and \ uj on d(f+I)e. We define new piecewise symmetric polynomials Pj as
Pj(x) :=
Xkok(t)
where summation is over all ktuples of permutations a1 of (1
xkck(n)), n } and where N
denotes the number of such rearrangements. —÷ Put Pj : pj(Z) :=
Zk
Zk).
Then the p1's are psh and p1 \ ut on be the matrix Reinhardt domain whose diagonal = dfs. From Lemma 3.3.11 we conclude that the functions : Let
Q1(Z1 are
psh and Q3
Zk) :=
\
AI(Zk)
where
Zk) := ue(Ai(Zi) In particular, it turns out that Recall that
is
psh on
AI(Zk)
is given
by dts, i.e.
—p IR.
An(Zk)),
299
3.3 Matrix Reinhardt domains
Observe that
:=
E
Mk(n): fldetZx x=I
is psh on
is continuous on Dt+ Applying the extension theorem for psh functions implies that is even a psh function on Di÷i. Remark that D1 is given as On the other side,
= {Z
Hence D, is pseudoconvex. What remains is to observe that D domains Dt, i.e. D is pseudoconvex.
is
the increasing union of the pseudoconvex U
We like to conclude this section by a few remarks on the socalled extended tube conjecture (cf. [StrWig 1964]):
DefinitIon 33.13. (a) For vectors z = (zt. (z, W)m := where
(.,.)
denotes
pmduct. (b)
:=
N
w = (Wi. —
0. (y. Y)m
n>
2,
we put
th),
the Hermitian scalar product. (.,
:= {v E R": y, >
ii)) E
is
called the Minkowski
> 0} is the socalled fufun' light
cone.
:= is the future light tube. := IR" + (d) L(C") := {M E M(n): detM 0, (c)
(MZ,MW)m. Z.W E C").
L(C") is the complex Lorentz gmup. It is easily seen that L(C") consists of two connected components. The one that contains the identity is denoted by L÷(C"). It is called the pmper Lorentz gmup. For n, N E N we put +
.....
TnN .
+ X
+ C X T,,
Then we define TnN := {Mz:
Z E TnN M
where the proper Lorentz transformation M acts as follows on (C")":
Mz = M(zi 47)
ZN) := (Mz1
MZN)
z = (zi
To simplify our notation we write Mz, z E C". instead of
ZN) E
300
3 Envelopes of holomorphy for special domains
The domain is called the extended (n. N)tube. In theoretical physics. i.e. n = 4, the question, whether the extended tube is a domain of holomorphy, was known for more than thirty years (cf. [SerVIa 1994]) as the extended tube conjecture. The answer was found only recently in [Zho 1998], see also [Hei 1998]. The case N = I can be easily answered, namely we have
= Therefore.
is a domain of holomorphy.
Proof of(*). Let z = Lw,
L E L+(C"). Then, applying the Cauchy—
w
[0. 0°).
Schwarz inequality leads immediately to (z. Z)m
If Z
(*)
[0,oo)}.
{z E C'7: (Z.Z)m
[0. oo), then there exist c >
0 and (0. 2,r) such that E C". Then w and (Z,Z)m = (Z.Z)m = ce"°. Put w := (W, W)m. In virtue of the Witt extension theorem we find a Lorentz transformation
E C'7, (Z. Z)m
L such that Lw = z. If L E L+(C"), then z E In the remaining case put := (WI and define L' substituting the last column of L by its Wa_I. —w,,) negative. Then w' and L' E L+(C"). Therefore, z D In general we have the following result (cf. [Hei 19981, [Zho 1998]).
Theorem 3.3.14. For any N We
is a domain of holomorphv
N the extended tube
are not going to present the proof of Theorem 3.3.14, nevertheless we shall
make some remarks to illustrate the main ideas. The proof by Zhou works in the frame
of 2 x 2matrices. So we will first reformulate the statement in Theorem 3.3.14. In order to do so we have to introduce the following biholomorphic mapping
Z2_iZ3) Z2
We
+
1Z3
Z1 — Z4
z=(zi
z4)€C4.
define the socalled Pauli matrices
/l o\
10
i)'
l\
10
o)
—i\
11
o)'
Then cl is given by
= It is easily seen that the inverse of
=
is given by
trace(Ma4)),
0 —l
M
M(2).
3.3 Matrix Reinhardt domains
301
Then, for vectors z, w E C4, the Minkowski product fulfils the following relations:
det (4(z) + cD(w))
— det
(4(z))
= 2(z, w)m.
— det
g :=
SL(2, C) x SL(2, C). g with the componentwise multiplication is a group. One can check that for any M = (A, B) E g we have Let
(2.2)m=(Z.Z)m,
ZEC4,
where 2 :=
Therefore, the mapping M: C4 —÷ C4. M(z) := 2, is a Lorentz transformation, even more M E Observe that g M —+ M E L + (Ca) is a 2—1 homomorphism. We have
1mM> 0) =: H, 48) H is sometimes called the generalized upper halfpiane. It where Im M : is invariant under the following mappings M —* AMA*. A E SL(2, C). Moreover, the function
u(Z1
(Z1
ZN)EHN,
is strictly psh, weakly exhausting (i.e. u tends to infinity along each boundary sequence in HN), and u(AZIA* AZNA*) = u(Z), A E SL(2, C), Z = (Z1 ZN) E
HN. Observe that (c1 x
x
= {(AZ1Bt
AZNB'): (Zi
ZN)
Therefore, t4N is a domain of holomorphy 1ff It can be seen that
3(A.B)E9
E
11N
(A. B) E g} =:
is a domain of holomorphy.
: Z = (AMIB*
AMNB*)}.
Now the main proof is an application of Kiselman's minimum principle (cf. Sec
tion 2.3) in a generalized form which is due to J.J. Loeb (cf. [Loe 1985]) in order to get from u an exhausting psh function on Recall
:=M'.
302
3 Envelopes of holomorphy for special domains
Remark 3.3.15. (a) According to our knowledge there is, so far, no proof for the fact that tflN, ii. N N arbitrarily, is a domain of holomorphy (b) A sufficient criterion is given in [Jar.Pfi 1996b], namely: Let
T(n, N)
{(Mz1
MZN): z = (zi
M E L(C") fl tL(n)}.
ZN)
Then the following is true: Let a: (T(n, N). id) —+ (X, p) be an envelope of holomorphy. Then
domain of holomorphy iffp(X) c Observe that T(n, N) is not a domain of holomorphy for n ? 3 and N
is a 2.
We close this section by quoting the following result on holomorphic extension (cf. [BroEpsGla 1967]).
Remark 3.3.16. Denote by J :=
fl (R4y" the set of socalled Jost points. It
turns out that
J_—fx_—(xi
Then any function holomorphic in a neighborhood of the set
LI T4N Ui, T4N
extends holomorphically to the extended tube
Hence, T4N is the 'envelope of holomorphy' of
U T4N U J.
3.4 The envelope of holomorphy of X \ M aim of this section is to find a description of the envelope of holomoiphy of X \ M, where M is an arbitrary analytic subset of a Riemann domain X. Recall that Theorem 2.5.9 deals only with such M which are, by assumption, 'extendible' to the envelope of holomorphy of X. The
Singular sets. Let (X. p) be a Riemann domain over C" and let M be a closed subset of X satisfying the following condition
for any domain D C X the set D \ M is connected and dense in D.
(3.4.1)
Notice that mt M = 0. Observe that any closed pluripolar set M C X satisfies (3.4.1). In particular, any proper analytic subset M of X satisfies (3.4.1). Let 0 F C (9(X \ M). We say that a point a E M is nonsingular with respect to F (a E if there exists an open neighborhood LI of a such that for each f E F there exists a function
3.4 The envelope of holomorphy of X \ M
303
f E 0(U) with! = f on U \ M. Ifa
then we say that a is M\ singular with respect to F. If Mn5F = 0, i.e. M5F = M, then we say that M is singular with respect to F. If F = (9(X \ M), then we simply say that M is singular and we skip the index F. We would like to point out the essential difference between the notion of 'the singular analytic subset M of X' and 'the singular points Sing(M) of an analytic subset M of X'
Remark 3.4.1. (a) The set M5,, is closed in M and satisfies (3.4.1).
(b) Each function f E F has a holomorphic extension f E (9(X \
(C) M5,F = (Ms,)5?, where respect to F.
F := {f: f E F),
i.e.
is
singular with
let M be Lemma 3.4.2 (cf. Lemma 1.7.5). Let (X, p) be a Riemann domain over a closed subset of X with (3.4.1), and let 0 M). Assume that A is a F c 0(X \ dense subset of
Y := {x E X \ M: dx\M(x)
(i)
M is singular with respect to F;
(ii) dX\M(x)
= inf{d(Lf):
f E F},
X
E A.
(ii): Fix a E A and dx\M(a)r/2> dx\M(xo); contradiction. 0
Proof (i)
;
f
Proposititon 3.4.3 (cf. Proposition 1.7.6). Let (X, p) be a Riemann domain over C", let M be a closed subset of X with (3.4.1). and let F be a natural Fréchet space in
0(X \ M) (cf Section 1.2). (i)
Then the following conditions are equivalent:
M is singular with respect to F;
(ii) there exists an f E F such that M is singular with respect to {f}; (iii) the set of all f F such that M is singular with respect to Baire category in F.
Proof Obviously (iii)
(u)
:
is of the second
(i).
Recall that if M is an analytic subset of a complex manifold 2, then a point a E M is called regular (a E Reg(M)) if there exists an open neighborhood U of a such that M fl U is a complex manifold. If a E Sing(M) := M \ Reg(M). then we say that a is singular — cf. [Chi 19891, § 2.3.
304
3 Envelopes of holomorphy for special domains
(iii): Let Y be as in Lemma 3.4.2 and let A be a countable dense subset of (I) Y. Then, by Lemma 3.4.2, ,•
U
Fx,r,
(x.r)EAxQ dX\M (x)
where
:=
(f E F:
> r}.
F for x E Y and r > dX\M(x). Hence, by the same argument D as in the proof of Proposition 1.7.6, F \ is of the first Baire category. We know that
Remark 3.4.4. (a) If M = with respect to {1/f}.
f'(O), where f E 0(X), f
0, then M is singular
(b) If M is an analytic subset of X, then by [Chi 19891, Appendix I, (a E M is pure M: dima M < n — 2} c M (n — 1)dimensional. (C) Recall that
if X isa Riemann—Stein domain and M is a pure (n — 1)dimensional
analytic subset of X, then X \ M is pseudoconvex (cf. Corollary 2.2.17) and therefore M is singular. Consequently, if M is an analytic subset of a Riemann—Stein domain X, then M is singular if M is pure (n — 1)dimensional. (d) Notice that a pure (n — 1)dimensional analytic subset M of X need not to be singular if X is not a Riemann—Stein domain. For example, let X
:= E2 \ {(zi. z2): Izil < 2/3, 1/3 < Z21 < 2/3), M := {(l/3,z2): 2/3< 1Z21 < 1).
Then each function holomorphic in X \ M extends to E2.
Proposition 3.4.5. Let (X, p) be a Riemann domain over C". Assume that M is a pure (n — 1)dimensional analytic subset of X. Let M = UIEI M, be the decomposition of M into irreducible components (cf [Clii 1989], Section 5.4). Then = Uj In
the set Ms.c is also analytic.
Proof Put!0 :={i €1:
!\Io,Nj
M,,j =0,1.
We have to prove that M3,F = N1.
M fl Reg(M). It is clear that B is nonFix an I E jO and let B := empty and open in M fl Reg(M). We will show that B is also relatively closed in M fl Reg(M).
Suppose that b is an accumulation point of B in M fl Reg(M). Using a biholomorphic change of coordinates in a neighborhood of b we reduce the problem to the following elementary remark:
305
3.4 The envelope of holomorphy of X \ M
Given a function f E
:= Then
27r1
that extends to IP(c, r) x E C E", define
x
f
f(z'.
z
— Zn
x
!E. 2
1 extends f to
Since M, fl Reg(M) is connected (cf. [Chi 1989], § 5.3), we conclude that M, fl Reg(M) C Mns,7. Thus (cf. Remark 3.4.1(b)) each function f E F extends to the domain Y := X \ (M \ (N0fl Reg(M))). Observe that Y = (X \ N1) \ (N0 fl Sing(M)). Recall that dim Sing(M) n — 2.
Hence, by [Chi 19891, Appendix I. any function f E F extends to X \ N1. Thus
0 The Dloussky theorem. Let (X. p) be a Riemann domain over Ce?, let
a: (X,p) be
—+ (X.15)
the maximal holomorphic extension, and let M be an analytic subset of X. Let us consider the following condition:
(E) There exists an analytic subset M of X such that C M and aIx\M (X \ M, p) —p (X \ M, 5) is the maximal holomorphic extension. Remark 3.4.6. (a) M must be singular (because X \ M is a domain of holomorphy) .
foranyf E (9(X\M)letf E O(X\M)besuchthatfoalx\M = f. Then C a_I(M). the function isanextension of f to X\a '(M). Hence Indeed,
Suppose that a point a E a'(M) is not singular. Let U be a univalent connected neighborhoodofasuchthateachfunctionf E (9(X\M)extendstof (9(U). Then foreach g E O(X\M) the function o(aIuY' extends toa(U); contradiction. (b) We may always assume that M is singular and a(M) C M. (c) To check (E) it is sufficient to find an analytic subset M of X such that a — '(M) C
M and (X \ M, p) —÷ (X \ M, j3) is a holomorphic extension. Indeed, we only need to take := M,r (cf. Proposition 3.4.5). By virtue of Theorem 2.5.9, we get the following important result.
Theorem 3.4.7 ([GraRem 1956]). Condition (E) holds if there exists an analytic subset M of X such that = Proof Recall that M5 is pure (n — 1)dimensional (Remark 3.4.4(b)). Since a is locally biholomorphic, a(M5) = a(X) fl M is also pure (n — 1)dimensional. Since X is Stein, M3 coincides with the union of all pure (n — 1)dimensional components of M (Remark 3.4.4(b,c) and Proposition 3.4.5). Hence a(M5) C M5. =: M. Now, we Consequently, we may assume that M = M= apply Theorem 2.5.9. 0
The main result of § 3.4 is the following theorem due to Dloussky [Dlo 1977].
306
3 Envelopes of holomorphy for special domains
such that t9(X) separates Theorem 3.4.8. If (X. p) is a Riemann domain over points in X, then (E) holds for any analytic subset M of X. do not know whether Dloussky's theorem is true without the assumption that (9(X) separates points We
Remark 3.4.9. Theorem 3.4.8 remains also true in the category of pluripolar sets, i.e. we assume in (E) that M is a closed plunpolar subset of X (cf. Definition 2.1.18) and we require M to be a closed pluripolar subset of X — cf. [Clii 1993J.
Reduction procedures. The aim of this subsection is to show that the proof of Dloussky's theorem may be reduced to the case where:
n=2,

= (K(r) x E)U (E x (E \ K(p))) (for some 0 < r, p < I), X=T= M = H isa singular analytic subset ofT such that H C K(r) x K(p). The reduction procedure will be divided into the following four steps: (n)
• Proposition 3.4.10: It suffices to consider only the case X = T = Tr,p (0 < r,
(n)
p < I), where Tr.p
=
r
zI
:= {(z, w) E C U {(z. w)
0
and
Tr,p
E
< p
• Proposition 3.4.11: It suffices to consider only the case where X = T =
(n)
and M = H is an analytic subset ofT such that for each z E
:={w E
E
C: (z. U')
E
T}).
• Proposition 3.4.13: It suffices to consider only the case where X = T = andM = H is an analytic
(n)
T r.p
x K(p).
• Proposition 3.4.14: It suffices to consider only the case where X
(2)
T
Tr,p
and M = H is an analytic subset of K(r) x K(p). ProposItion 3.4.10.
To
prove Theorem 3.4.8 ills sufficient to consider only the case
(ii)
X=
i.e. to prove that for any domain T = Tr,p and for any analytic subset H of T there exists an analytic subset of E't such that E' \ is the envelope of holomorphy ofT \ H. where
Proof Fix a Riemann domain (X. p) over C" such that (9(X) separates points in X and a singular analytic subset M of X. Let a: (X, p) —+ (X, be the maximaL Recall that
K(a, r) := B1(a, r) C C, K(r) := K(O. r).
3.4 The envelope of holomorphy of X \ M
307
holomorphic extension. Since 0(X) separates points in X. we may assume that X is a subdomain of X, /3 = p on X, and a id. Let (X \ M, p) —p (Y, q) be the maximal holomorphic extension. Then, by (Y, q) —* (X, j3) with o = Id on Proposition 1.9.2, there exists a morphism
X\M.
=
LetY =
(X,j3)beasinPropositionl.5.1l. \ M) is an open subset of Y and that maps biholomorphically W onto V := X \ M C X. Hence, by Proposition 1.5.13, there (Y , q
First observe that W :=
exists
an open set U C Y with W C U such that
Consider(i :=
X —+
Y
.
Then
=
U —* X is biholomorphic.
(X, p) —+ (Y
,
q
) isaholomorphic
extension (recall that Y is connected). Suppose for a moment that (Y =
Then
*Q
(Y
.
q)
.
q ) is a Riemann domain of holomorphy. =
Then M is an
—f (X, /3) is an isomorphism. Put M :=
analytic subset of X and
(Y, q) —* (X \ M, /3) is an isomorphism.
Thus, it remains to show that (Y q ) is a Riemann domain of holomorphy. By Theorem 2.9.1(v) we only need to show that for any T := Tr.p and for any biholo,
morphic
mapping f: T —+ f(T)
c
biholomorphically to 1: E" —÷
=
Y
such that g := o f: T X extends C X, the mapping f extends holomorphi
callytof: E" —+ Y. Let H :=
fl E). Then H is an analytic subset of T. By virtue of our assumptions there exists an analytic subset H of E'1 such that E" \ H is the envelope of holomorphy of T \ H. By Proposition 1.9.9. f: T \ H —+ Y extends to a locally biholomorphic mapping f: E" \ H —÷ Y. Observe that o = g on T \ H. Hence, by the identity principle, o = on E't \ H. In particular. is biholomorphic. By Proposition 1.5.13 there p: f(E" \ H) —+ \
f
f
exists
an open set U C Y with f(E't\ H) C U such that
biholomorphic. Now we can put / :=
=
U —÷ g(E")
Is
U
0 (n)
Proposition 3.4.11. To prove Theorem 3 .4.8for X = T = T r.p it suffices to consider only those analytic subsets H of T for which each fiber
:= (wE C: (z, w) E H}, as an analytic subset of the whole fiber
:= (wE C: (z,w) E T) is discrete,
z E
For the proof we need the following
3 Envelopes of holomorphy for special domains
308
Lemma 3.4.12. Let H be a pure (n — I )dimensional analytic subset of a domain and let
D C B
= B(H) := (A
C
xC'
w+
(z + (w +
H}.
Then B has zero measure. ForA
E
C"1 put
1;: C"' x C —÷
w) :=
x C,
+ uA, to).
= is biholomorphic and the analytic subset The lemma says that for A outside a zero measure set in has discrete fibers = (w E := of the domain DA := C: (z. w) HA) = (to E C: (z + Aw, to) e H}. Observe that
isadomain,) > I, Proof of L.emma3.4.I2. ObservethatifD = U then B(H) = B(H fl Di). Thus we may assume that D is a ball. In particular. H = A
cf. [Chi 19891, § 2.8. Let
where f
C x C"': 3e>o:
:= {(z. w, A) e
=
Obviously, B
G
VEEK(F): (z + (to
+ E)A, to +
H}.
(A). Put
xC
:= ((z. to, A) E
x
(z
+ wA, w) E D}.
Observe that
A=((z, u',A)EG: = ((z. to.
A)
G:
w+E.A)EG, f(z +(w+E)A, f(k)(z + wA, w)(A, 1) = 0}
VtEK(f):(z. :
(f(k)(a)(X) denotes the value of the kth differential of f vector X E
Let r :=
at the point a
D on the
x C). In particular, A is analytic in G. IA• Suppose that for some (Zo, too, A0) E Reg(A) we have
rank r'(Zo, wo, A0) = n — I. Then, by the rank theorem, there exist a neighborhood W0 of A0 and a holomorphic mapping w: Wo —+ C"' x C such that = (zo. wo)
and (p(A),A)E AforallA€ Indeed, by the rank theorem, there exist:
A0) and V C open neighborhoods U C Reg(A)of(zo. —+ U.'V: V biholomorphic mappings 4):
ofA0.
such that
r(U) C V. 4)10.0,0) = (zo.
As), 4'(Ao) = 0, 'V o to CD(a. fi. y) = y for any (a. fi. y) E x C x C"'. The above identity implies that 'V o Write 4) = Cn 4)A): 'VA(O. 0.41(A)) = 41(A) for A E W0 := E"t. Hence 4)x(O. 0. 41(A)) = A for A E W0. We define 0. 41(A))), 42(A) := satisfics all the required conditions.
A
E W0. Then (42(A), A) = 41(0.0.41(A)) and, therefore.
3.4 The envelope of holomorphy of X \ M
:=
+
xC,
W0 xC —÷
x
Wo —+
309
+
+
0) = (zo + w0A0, wO) E D. We know that for any A E W0 there exists an e(A) > 0 such that ifr(A, E H for E r(A)E. Take a neighborhood W C W0o1 A0 and a connected neighborhood V C C of 0 such that *(W x V) C D. Notice that
Consequently, by the identity principle,
( W x V) C H and therefore rank
(A.
n—i forany (A.
E W x V. On the other hand, one can easily prove that
det *'(A,
=
+
+
+ do(A).
0; contradiction. E V such that det We have proved that rank r'(z. w, A) n — 2 for all (z. w, A) E Reg(A). Hence r(A) has dimension < n — 2 and, consequently, r(A) is of zero measure (cf. [Loj
Thus, for any A E W there exists a
1991]) 52)
D
Proof of Proposition 3.4.11. We assume that the extension theorem is true for all analytic sets with discrete fibers. Fix (T, H) with an arbitrary singular subset H ofT. Let I denote the set of all numbers s E Fr, I] such that there exists a singular analytic such that any function from O(T \ H) extends holomorphically to
subset H5 of
Let ro := sup 1. Observe that I is an interval. Moreover, H' fl = H5 for any S. t E I with s t. and any function Put H' := H5. Observe that H' is an analytic subset of T) extends holomorphically to Tr0.p \ H'. Taking HT0 from (9(11 \ we conclude that ro E I. We have to show that = I. Recall that for A E we have defined
PA(z,w):=(z+wA,w). Fix 0 < e << 1. By Lemma 3.4.12 with D := T, there exists a A E
such that the set c1_A(H) has discrete fibers. Observe that D Notice that T' — e) x E is the envelope of holomorphy of T'.
T'.
By the assumption (applied after a rescaling). there exists a singular analytic subset H' of T' such that any function holomorphic in T' \ c1_A(H) extends holomorphically to V \ ,?'. Consequently, each function holomorphic on T \ H extends — 2e) x E C Consequently. to \ 4A(H'). Observe that s := 1 — 2r E I with H5 HU fl
Thus,r0=l.
D
We can argue also directly: first observe that r(Reg(A)) is of zero measure cf. [Fed n — 2 for all
19691. Th. 3.4.3). Next, in the same way as above, we prove that rank r'(z. w, A) (z, w, A) E Reg(Sing(A)). Hence r(Reg(Sing(A)>) is of zero measure, etc. Recall
that 4(1') fl T,,, C H because H is singular.
310
3 Envelopes of holomorphy for special domains
Proposition 3.4.13. To prove Theorem 3 .4 .8 for X = T = ii suffices to consider only those analytic subsets H ofT for which I! C P,7 — (r) x K (p). Proof Suppose that the extension theorem is true for all analytic sets H C x K(p). Fix (T. H) with an arbitrary singular subset H ofT. By Proposition 3.4.11 we may assume that H has discrete fibers. Let ro be as in the proof of Proposition 3.4.11. We have to prove that ro = 1. Suppose that r ro < I. Observe that H fl {IwI > p} (because H is singular). In particular, by the identity principle for analytic sets. HT° has discrete fibers.
The maximality of ro and the compactness of (ro) imply that there exists a (ro) such that we cannot find an open neighborhood LI of zo and a singular zo E analytic subset H of V U x E such that any function from \ HT0) extends holomorphically to V \ II. Fix such a Since I! has discrete fibers, there exists p < p' < I such that
{w: wi = p', (zO, w) E H) = 0. Since H is relatively closed in T, there exists an
p'—e
x {p' — e
e>
Iwi
0 (at least so
small that p <
p'+e}) = 0,
(1 — e/ro)zo. Put TI
:= (to. 0) +
C
By our assumption (applied to (T', HTO fl T')) there exists a singular analytic subset
H'
of
x K(p'+r) such that any function holomorphic in
extends holomorphically to
\
Let V : Pn_ i (co. Zr) x E. We have proved that each function from \ Hr0) extends holomorphically to V \ H", where H" H' U (H fl V). Now we take H and we have a contradiction. D
Proposition 3.4.14. It suffices to consider only the case n = 2. Proof We proceed by induction. Suppose that Consider an arbitrary problem (T, H) with a assume that H is
the extension theorem is true in C" —
3. By Proposition 3.4.13 we may
a singular analytic such that H C
the projection
H Notice that H' fl
(z, w) —÷ z E
C H'°.
x K(p). In
particular.
3.4 The envelope of holomorphy of X \ M
311
N, functions an analytic proper subset E of Pn_1 (r) such that
Consequently (cf. Theorem 2.12.2), there exist m
is proper.
(9(Pn_i(r)),
E
•H=
and
where
QH(Z, w) := wm +
• for any := E functions E (9(U), j = I
there exist an open neighborhood U C m, such that:
\E
and
(z)foranyz Uandj #k, j=1 rn), z EU, = {(z,
•foranyzo E E:
<en. Notice that the polynomial Qil is uniquely determined. By Theorem 3.4.7 it is enough to show that the functions 'Pj' J = I holomorphically to
m, extend
xC xC
LetC'
K(r) consider the (n — 1)dimensional problem (T', 11(q)), where
For any (n—I)
IF'n_2(T) x K(p): r,, w) E H). By the inductive assumption there exists a singular analytic subset 11(q) of such that \ 11(q) is the envelope of holomorphy of T' \ H(q). Moreover, H(r1) fl > p} = 0. In particular, the projection
H07) := ((v, w)
T
E
w) —f
11(q) is
proper. Hence 11(q) =
E
where
w) =
+ j=I
j =I
E
m(q)
rn(q). Notice that 11(q) fl T' C H(q) and therefore
m.
Let G denote the set of all q E K(r) such that there exists a E with Then K(r) \ G is an analytic q) E ci, i.e. G = {q e K(r): ?n...2(r) x {q} subset of K(r) 55)• In particular, for any q E G the projection
w) —'
H(q)
IP'n_2(r)
has fibers consisting of nz different points.
j=I
—"fk =O)forsomef1 k,then
(K(r)\G)flK(m.r) =
fl
tEP.,..2(to,r)
=0. j =
I
k}.
3 Envelopes of holomorphy for special domains
312
Now let Go denote the set of all e G such that the set for Then H(17) fl T' = Go, and consequently,
is singular.
j=l...., m.
,n(rj)—m.
(3.4.2)
We will prove that G \ G0 consists of isolated points (in particular, Go is open).
Take an f (9(T \ H) such that H is singular with respect to (f) (cf. Proposition 3.4.3). We will show that the set S of all e G such that is not singular with respect to lf(. is analytic in G. Suppose that there exists an infinite sequence C S \ qo with —p := G. For any v there existsapoint E H(171) such that •)extends By Proposition 3.4.5 the function f(.. to a neighborhood of .) extends to a neighborhood of any point from fl Reg(H(171)), where denotes any that contains irreducible component of Let
:=
77v)
E},
Q :=
Moreover, since each set
Q is closed in Q V and let a
Zo := (to, '70). Let U, *i 1/fm be as previously with x C. Observe that each point of U connected, U = U' x U" C U" is a regular point of that for any U' the function f(•, with
v
For any v >> I and
(i'
>>
1). Consequently, we may assume to a neighborhood of a point
extends
1.
E U' there exists a j = j(v,
=
ifl,)), Since all the points we may assume that irreducible component of of E. Consequently, there exists a Jo such that j (v) = = I for all v 1. may assume that Consider the expansion ii.
Jo
+ w) = qo, O)withw
q,
m} such that E (I U', must lie in the same v) = j(u) is independent for infinitely many v. We
E
liv) = 0
0.
<0. Hence the function f(.... .) extends to a neighborhood of the point (go, 'in. 1/'i (to. rio)) E H; for all
<0 and
I. Hence, by the identity principle, a11
0 for all
contradiction. m, extend holoBy (3.4.2) and the Hartogs lemma the functions coj, J = I x Go. Observe that they are locally bounded on E"2 x K(r) morphically to (use Viète's formulas). Consequently, by the Riemann theorem on removable singux G and, finally, to m, extend holomorphically first to larities, 'PJ. J = I
x K(r).
3.4 The envelope of holomorphy of X \ M Consequently, we get a singular analytic subset H of 1' := T U
313
x K (r) x E)
such that H fl T = H and each function from O(T \ H) extends holomorphically to > p} = 0 and hence the projection H (z. w) —4 T \ H. Moreover, H fl w E x K(r) is proper. Now we repeat the first part of the proof with respect to the first variable z E E, i.e. m, j=I
if x x K(r)) x E (q, w). Consequently. the functions extend holomorphically to E x E"2.
[1
Oka theorem. Let Z be a connected complex ndimensional manifold. Recall (cf. § 1.1) that a pair (X, p) is called a Riemann region over Z if Xis a Hausdorif space and p: K —p Z is a local homeomorphism. If X is connected, then we say that (X, p) is a Riemann domain over Z. We say that (X, p) is locally pseudoconvex if for each point zo E Z there exist a such neighborhood V of zo and a biholomorphic mapping Cl,: V —+ cl,(V) C ci) o p) is pseudoconvex, as a Riemann region over C", in the sense that 2.2.
Observe that the above definition is independent of (I) in that sense that if (I): V —+ cD(V), 'I': V —÷ V) are biholomorphic, then Dop) is pseudoconvex (V), o p) is pseudoconvex.
if (p'
Proposition 3.4.15. If(X, p) is a Riemann domain over C", then (X, p) is pseudoconvex in the sense of Section 2.2 (X, p) is locally pseudoconvex.
Proof If (X. p) is pseudoconvex in the standard sense, then we take V := C", c1 := id. Conversely, assume that (X, p) is locally pseudoconvex. We will prove that (X. p) satisfies condition (p4) from Theorem 2.9.2.
Let a E aX. Let zo := p(a) and V be an open neighborhood of zo such that (V), p) is pseudoconvex. Let U be a connected component of p (V) such that U E a (cf. § 1.5). Notice that (U, p) is also pseudoconvex and, therefore, holomorphi
cally convex (Theorem 2.5.7). Put U := U U {b E ax: U E b}. Then U is a neigh
borhood ofa in X and X nU = U. Thus (X flU. p) is holomorphically convex. U
If T C Z, then any continuous mapping a : T —÷ X such that p o a = idT is called a section of(X, p) over T (cf. § 1.1). The set of all section overT is denoted by T°(T, X) (sometimes, f°(T. (X, p))). Similarly, as in Proposition 1.1.5, one can easily prove the following identity principlefor sections. It T C Z is connected, ai. E f°(T, X), and ai (Zo) = c2(zo) for some Zo E T, then aI a Riemann domain over To simplify notation put m := n — 1.
x C.
(C) Assume that there exists an R > 0 such that r(Etm x (C \ K(R)). K) let a: iftm x (C \ K(R)) —* X be a fixed section of (X, p).
0 and
314
3 Envelopes of holomorphy for special domains
Consider the family C of all relatively closed subsets L c Em x C such that: • for each z E the set := (wE C: (z, w) E L) is convex and compact, • there exists a section o'L: x C) \ L —+ X such that 0'L = a outside LU(Em K(R)). Obviously, Em x K(R) E C. Put
K:=flL. For z E Em let L(z) denote the family of all convex compact subsets £ C C such that there exists a section (z} x (C \ £) —+ X with az,€(z, ') = a(z, ')outside
tuk(R).
Obviously, if L E C, then
E L(z) for any z E Em. Put
k(z)= fl
e,
ZEEtm.
eEe(z)
Lemma 3.4.16. (a)K EL. = k(z), z Em. (b) Pmof (a) For any L1, L2 the sets C \ ((L1 U (L2)z), z E Em, are connected 56), Hence, by the identity principle for sections, = aL2 in (Em x C) \ (L1 U L2), which shows that O'K ULEe a zo E By the same argument as in (a) we can prove that k(zo) E .C(zo). Obviously. k(zo) C Suppose that \ k(zo) 0. Then there exists an open halfplane Q such that Let r := azo,k(zo). The set B := r(A) c X k(zo) C Q and A := \Q o is compact and univalent. Take an e > 0 such that C := is also univalent and let U = V x W C Em x C be a convex neighborhood of (zo} x A such that U C p(C). Let > 0 be so small that for z E Pm(Z0, C V we have {z} x \ Q) c W 58), 56)
If A, B C C are compact and convex, then U := C \ (A U B) is connected. = 0. then there exists an open halfplane H such that A C H. B c C \ H. We may assume that H = (Rez < O}. Let r > 0 be such that AU B C {IlmzI < r}. In particular, the connected set S := (R + ir) U (i(—r, r)) U (R — ir) is contained in U. Moreover, for any point r] joins zo with S + i[—r. Yoi or the segment + Zo = xo + io U either the segment inside U. In the case where A fl B 0 we may assume that 0 A fl B. Let r > 0 be such that Indeed, if A fl B
A U B C K(r). In particular, the circle S := dK(r) is contained in U. Moreover, for any point 8f) E U the segment [ro. joins zo with S in U. Notice that the result remains true for compact convex subsets of RN. N 2. 57) If S C C is compact and convex, then S = fl Q, where the intersection is taken over all open halfplanes Q such that S c Q.
58)
Let
> ObesuchthatPm(zo.
for z C such that K. \ Q C that and let \Q
q)x
C U. ItsufficestoprovethatthereexistsO <S
q
5). Suppose that there exists a sequence —p zo such Recall that K C Em x K(R). In particular, Q) \ the fibers z E E", are unifonnly bounded, and therefore, we may assume that —+ A. On the other hand, since K \ (Em x Q) is relatively closed in x C, we conclude that wo A; contradiction.
3.4 The envelope of holomorphy of X \ M
315
Define
L :=
U
(
{z} x
u
U (z}
(
x (K fl Q)).
ZEPm(Zø.tS)
that L is relatively closed in Em x C and that the fibers of L are convex and compact. Moreover, w) = w) for Z E Pm(ZO, 8) and This implies that L E C; contradiction. 0 wEW\ Observe
Lemma 3.4.17. Assume that (X, p) is locally pseudoconvex. Let
Y := p1 (Em xC) = X \
x {oo}).
Then (Y, p) is pseudoconvex in the standard sense.
Pmof By Proposition 3.4.15, it suffices to prove that(Y, p) is locally pseudoconvex. Take an arbitrary E and let V c Em xC be such that there exists a biholomorphic mapping 4): V —÷ 4)(V) such that 4) o p) is pseudoconvex. Let V0 C 4)(V) be an arbitrary pseudoconvex neighborhood of 4)(zo). Let W := 4) o p) is also pseudoconvex. In particular, Then, by Corollary 2.2.20, taking V0 sufficiently small, we may assume that V C and 4) = id. It remains to observe that = 0
Lemma 3.4.18. Let (X, p) be as above. Assume that (X, p) is locally pseudoconvex. then K = 0. If k(zo) = Oforapointzo E
PmofLetDo:={z
E
=0}. ThenzoE DoandDoisopen. Let
x C). We know (Lemma 3.4.17) that (Y. p) is pseudoconvex in
Y :=
the standard sense. Let
T := (D0 x C) U (Em x (C \ K(R))). Observe that r := aKIT isa welldefined section of(Y, p) overT. Notice that Em xC is the envelope of holomorphy of T. Hence, since (Y. p) is a domain of holomorphy (Theorem 2.5.7), r extends to a section on f: Em x C —+ Y (Proposition 1.9.2). which implies that K = 0. 0 If k(z)
Oforanyz E Em, then we put D(z) := diam(k(z)), := max{Iw — woI: W E k(z)},
z E Em,
E C;
D is called the diameter function for (X, p) (with respect to R and a). Observe that E £(z) for any z E Em. K(wo.
FirstnotcthatthesetS:=U\Kiscormccted(thefibersS. =
V.arc
connected, the base V is connected, and, therefore, the whole set S must be connected). Consequently, by the identity principle for sections. we only need to observe that tik(zo. = r(zo. u0) = wO) for any point WOE W \
316
3 Envelopes of holomorphy for special domains
Lenuna 3.4.19. (a) D is upper semicontinuous. (b) D(z) = urn +
— 21w1), z E Em.
Proof (a) Let v
1.
vt,, E be such that —÷ Zo and let = IWL, — —* We may assume that E k(zo). Then —+ Wo E k(zo), Iwo — vol < D(zo), which shows that D is upper semicontinuous.
(b) Fix a z and let a. b E k(z) be such that D(z) = Ia — bi. Using the formula
,/fl:7= I +t/2+O(t2)ast +
lim sup
—+ O,weget 21w1]
—
>limsup[Iw—aI+I—w—bI—21W1]
=lim sup [11w12
=llmsuplWl [Ii =lim sup
—
+ al2 +
2Re(tha)
— (2Re(tha)
+ (2Re(thb) + Ib12)/IwI2 _2]
1a12)1Iw12 +
+0
+ Re
Re
(h)]
= la—bl= D(z).
W
—+ oo be such that
Conversely, let
+
urn sup
—
2Re(thb) + 1b12 — 21W1]
E k(z) so that —* ao may assume that as above) we get:
— 21
Take
lim sup
E
WI] = lim
= lWv —* k(z),
+
= lim Re
—
+
bo E
v > 1. We =I— k(z). Then (using the same method
21w1]
((by — WV
—
)
<
— bol
0
Proposition 3.4.20 (Oka theorem). Assume that (X, p) is locally pseudoconvex and K 0 (cf Lemma 3.4.18). Then: (a) log any w E C with IwI > R. In E by Lemma
3.4.19, D E
(b)logD E Proof (a) Fix IWol> R. Let h: C —* C, h(w) h(K(wo, s)) = C\ K(wo, l/s). Let c1: Ctm x —+
+ 1/(w — Ctm
wO).
Observe that
xC, c1(z, w) :=
(z, h(w)).
3.4 The envelope of holomorphy of X \ M
317
Put ,3 := Cl) op = (pi
Pm' h o and consider (X, /5). Obviously, (X. /5) is a pseudoconvex Riemann domain over Em x C. Put a := a o cD Then is a section over Em x (C \ h(K(R))). of (X. /5) We will prove that
=
> 0: 3fEr(z)xK(wo.sL(x,p)• f(z, wO) = a(z.
sup{s
=: ro(z), Indeed, fix a z
wo)}
z E Em.
:=
Etm and let
1ff: {z} x K(wo. s) —÷ X is a section
of (X, /5) with f(z, wo) = &(z, wo), then
r :=
o CD: {z'} x (C \ k(wo, us)) —÷ X
is a section of (X. p) with r(z, oo) = f(z. wo) = ã(z, wo) = a(z, 00). Hence K(wo. us) E £(z) and, consequently. so 5 I/s. Thus I/so ? rO(z). Conversely, let £ := K(wo, so). Recall that £ E £(:). Then
f :=
(z} x (C \ K(wo, i/so)) —* X
o
is a section of (X. /5) with f(z, wo) = a.,(z. 00) = a(z, 00) = ã(z. wo). Thus I/so < ro(z). Let Y :=
x C). Then (Y. /5) is a pseudoconvex Riemann domain over C" (Lemma 3.4.17). Observe (cf. § 2.2) that
ro(z) = Recall that — log WO))

wO)),
z
Em
(Theorem 2.2.9), Consequently, log
E
=
£PS.R(Em).
(b)ByLemma2.l.28itissufficienttoshowthatforanya
E
Candj
E
{l
,n}
the function Em
z —p
(3.4.3)
is psh.
FixaandjandletCD: Ctm xC —÷ Ctm xC,cD(z,w) :=(z,exp(az1)w). Define /5 := CD 0 P. Then (X, /5) is a locally pseudoconvex Riemann domain over E" x C. Moreover, r: T —÷ Xis a section of(X, p) 1ff to CD(T) —b Xis a section in particular, a := a o of (X. is a section of (X, /5) over Em x (C \ K(R)), where R := Rexp(IaI). Let K and D be constructed for (X, J3) in the same way as K and D for (X, p) (clearly, K 0). Then D(z) = I Hence, by Lemma 3.4.19 and (a) (applied to (X, /5)), 0 E Consequently, (3.4.3) is plurisubharmonic. 0
318
3 Envelopes of holomorphy for special domains
Hartogs theorem. Let S C E" be relatively closed such that
\ S is pseudoconvex. The aim of this subsection is to describe some analytic properties of Sin terms of fibers z E
Proposition 3.4.21 (Hartogs). Assume that S C E'1 is a relatively closed subset of E" such that G : = E'T \ S is pseudoconvex (i.e. S is pseudoconcave) andfor any z' E'T — I the fiber of S over z' Consists of exactly one point = go(z'). Then go
Remark 3.4.22. The above result may be generalized into the following directions.
Let S C E" be relatively closed such that E" \
S
is pseudoconvex. Then:
(a) (Cf. [Oka 19341, [Nis 1962]) if there exists a nonpluripolar 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. [ChiSad 1988]) IfS ispluripolai Sn x (0)) = 0, and discrete for almost all z' e then S is analytic. (d) (Cf. [ChiSad l9881) if polar for all z' not belonging to some pluripolar set, then S is pluripolar Proof of Proposition 3.4.21. By the Hartogs theorem it suffices to show that cp(aI
foranya1
E 0(E)
n—l.Since
a1_i.a1+i
Gfl({(a1 is
.
xE x
xE)
also pseudoconvex (Corollary 2.2.18), we may assume that n =
2.
First we check that go is continuous. Let E Zo E E and suppose that K(go(zo), r) C F. Then wo go(zo). Let r > 0 be such that K(go(zo), r) for v>> I. Consequently, x K(go(zo), r) C G, v>> 1. By go(z1) the Kontinuitätssatz we get (zo) x K(go(z0), r) C G; contradiction.
Fix a zo E E. We will prove that go is holomorphic near Zo. Using Mobius transforms, we may reduce the problem to the case where zo = go(zo) = 0. By Theorem 2.2.9 we have: — log 8G,e' E Observe that fSG,e2(z, w) = min{jw — I — Iwl), (z, w) E G. Let 0
1—Iwiforizi
psh in a neighborhood of the Compact {(z. w): Izi
r, 1/4
Iwl
have
log wi
1
— I 2,r .'o
log 1w — go(re'°))d9,
1/4 <
3/8.
3/8). We
3.4 The envelope of holomorphy of X \ M
Hence (using the continuity of
2ir logp
J
(1
319
we get
dt
lo:Ipe't —
logipe" _co(reI8)ld:)dO.
=1
3/8.
By the Jensen formula (cf. LRud 1974], Th. 15.18) we get
I
2ir Jo
loglpe"—wIdt=Iogp
independently of w with Iwl
JZr
f
log Ipeul
—
w(rebo)ldt)
dO
= 2jr Iogp.
Consequently,
log Iwi =
— I 2ir .io
and, therefore, for any 1/4 harmonic. Since 1w —
1/4 < Iwi
log 1w —
3/8 the function K(r)
wi
= lw!2 + wi2
z
3/8.
—p log 1w — cp(z)l is
—
we conclude (taking w and —w) that the function thq + is of class Now, taking w and iw with w real, we see that 'p E C°°(K(r)). It remains to show that = 0 in K(r). Take 1/4 < Iwi <3/8 and let u(z) := 1w
—
— —
azaz
—
IaUaU u2 dz
— U
Hence
a2u
w(z)l2. Then
au au
in K(r).
320
3 Envelopes of holomorphy for special domains
After elementary calculations we get 
0=—w 2w— +
1W—  —— \. az
W
2
— WU)
...f
+ 2ww —w
/

2
— — — + W
\
az dz
\
azaz
dz az 
—w
2 aw
+
\
8WdW
2
— — — — + cQ —: az az azaz ,
2dz)z
2dWdW
az 8z
dz8z
Oz
Hence
a2w
—
—
—
az
—0
— az
is holomorphic on K (r). It remains to exclude the case where W 0 and the case. Consider the biholomorphic mapping C2 (z. w)
—
In particular. either W or
E
(9(K(r)). Suppose that this (z, z + w) E C2. Then the
domain 0(G) is pseudoconvex. Moreover, for z E we have: (z. w) E 0(S) for z near zero. (z, w — z) E S. Thus fibers of 0(S) consists of exactly one point From the first part of the proof it follows that either or t/i is holomorphic. Observe that = I and, therefore, cannot be antiholomorphic. = z + W(z). Hence On the other hand, if is holomorphic, then w is holomorphic; contradiction. 0
The proof of the case n =
2. We already know that to prove Dloussky's theorem we only need to consider the case where H is a singular analytic subset of T = Tr.p C C2
with H C K(r) x K(p). Letfi: (T\ H, id) —+ (Z, q') be the maximal holomorphic extension. Recall that is injective. Since E2 is the envelope of holomorphy ofT, we see that q'(Z) C E2; cf. Proposition 1.9.2. Put
Q:=
E
x(C\K(p)),
Q':=
E
x(E\K(p)).
3.4 The envelope of holomorphy of X \ M
321
Let (Y, q) denote the Riemann domain over E x C obtained by gluing (Z. q') with Obviously. (Y, q) is locally pseudoconvex. It is evident that (Q, id) along (Y, q) satisfies condition (C) (defined in the subsection 'Oka theorem') with R = p.
:=
U id.
Consider the family U of all pairs (G, M) such that
• G is a planar domain with K(r) C G C E, • M is an analytic, pure onedimensional subset of G x E with proper projection M
(z, w)
z (in particular, by Remark 3.4.4(c), M must be singular).
•MflT=H.
Obviously (K(r). H) E it. First observe that for any (G, M) E U there exists a section (G
x E)\M
—* Z
= a on (K(r) x E) \ H. Indeed, G x E is the envelope of holomorphy of T fl (G x E) (cf. the proof of Proposition 3.4.5). Hence, by Theorem 3.4.7, (G x E)\ M is the envelope of homomorphy T fl (G x E) \ H. (In particular, M C G x K(p)). Now, we see that (T U (G x E)) \ M is a holomorphic extension ofT \ H. Consequently, there exists such that 0G,M
a morphism
((TU(G x E))\M,id) —* (Z,q') such that aGM = a on T \ H.
For(G1,M1),(G2,M2)€itput(Gi.Mi)<(G2,M2)ifGi CG2andMi = that if (C1, Mj)IE, is a chain of elements from it, then UIE! M,) E U. Hence,bytheZomtheorem, thereexistsamaximalelement (U,EI (G. M) of U with respect to the relation <. Fix such a maximal pair (G*, M*). Recall that aG.,M. is an extension of a — to simplify notation we will write a* instead of aG.,M.. Let M2 fl (Gi x E). Observe
M*
z E E.
(z, w)
wm + be the describing Recall that ir is proper. Moreover, let Q polynomial for M* (cf. the proof of Proposition 3.4.14). We are going to prove that G* = E, which will finish the proof.
Lemma 3.4.23. Let y:
[0,1)
be a curve such that
y(t) =
zo E
G' fl E and the set I E [0, 1)}
has exactly m different limit points ((zo, WI) E G*.
Proof (1) First we prove that for any wo E C \
toaneighborhoodof(zo, we).
(zo, Wm)) when t —+ 1. Then Wm} the
section
extends
322
3 Envelopes of holomorphy for special domains
Indeed, let
c2:={w0EC\{wi
.
Wm}
the section
We
extends to a neighborhood of (zo, wo)).
we have: K(w0, s(wo)) C c, where
will prove that for any wo E
j=I
s(wo):= then
=
C
\
Wm
}and the proof will be completed.
Let h: C —÷ C, h(w) := l/(w — wo). Note that = Define'D: CxC —÷ Cx C,D(z, w) :=(z,h(w)). Put4 := Doq and consider the domain (Y, 4). Recall that r is a section of(Y, q) overT ifff := to is a section of (Y. 4) over (l)(T). In particular, there exist S > 0 and R > 0 such that there exists a section a: K(zo, 8) x (C \ K(R)) —÷ Y of (Y, 4). Observe that fort E LO. 1) such that y(t) E K(zo, 8) we have: Fix a
E
k(yU)) = conv({h(w): (y(t), w) E Ma)), where k (z) denotes the minimal convex compact singular set (cf. Lemma 3.4.16 in the subsection 'Oka theorem'). Let D: K(zo, 8) —p R+ denote the diameter function for (Y. 4) (defined in the subsection 'Oka theorem'). By Proposition 3.4.20(b) and the classical Oka theorem for subharmonic functions (cf. [Vla 19661) we have:
D(zo) = tim sup D(y(t)) = lim sup diam(conv((h(w): (y(t), w) E Ma)))
= lim sup max (0.I)a:—.I
— (
W
:
(y(t), w'), (y(t), w") E Ma
W —
I
WJ—WO
1
WkWO
:j,k=l
ml.
In particular. (Y. 4) has a section f over {zo) x (C \ K(0,
with f(zo, oc) = m}. This —wol: j = 1 directly implies that (Y, q) possesses a section rover {zo} x h'(C \ K(0, = {zo} x K(wo, s(wo)) with r(zo, wo) = a(zo, wo). (2) We pass to the main part of the proof. By (1) we know that there exist numbers sm. 0 < e < minfs1 Sm } such that if we put K0 := K (zo, e), K3
+e).then: • K0
E,
flKk = øforanyj #!c, m,
•thesectiona*extendsonAj := Ko x
a1,j=l
,n.
—e <
<sj +e}toasection
3.4 The envelope of holomorphy of X \ M
the
323
By virtue of the maximality of G*, it remains to show that for each j E (1 set Mt extends to an analytic subset of K0 x Fix a j. Now let (X, ji) denote the Riemann domain obtained by gluing
ni
Ko x (C\ K(wj,s, —e)) andq'' (K0 x K,) <S3 +e}). One can easily prove that (X, j5) is locally pseudoconvex. Let D denote the diameter function for (X, We know that D(y(r)) = 0 for t near I. Recall that log D is subharmonic on K(zo, r) (Proposition 3.4.20(b)). Hence, by the classical Oka theorem for subharmonic functions, we conclude that D 0. This means that there exists a function extends to the K(zo, e) —+ K(w1, Sf) such that x
—S <
complement in Ko x K1 of the graph M1 of
In view of the definition of the envelope
of holomorphy, we conclude that (Ko x is biholomorphic to a connected \ component of and, therefore, it is pseudoconvex. (Ko x Now, by Proposition 3.4.21, must be holomorphic. 0
Lemma 3.4.24. G* is simph' connected.
Proof Let
[0, 1] —+ Gt be an arbitrary closed Jordan curve. We are going to
show that mt
C
:= G* fl
Put
:=
E —+ (I) First, we prove that nontangential limit of at z. discrete in
Let
{z be the
m}. Notice that is \ universal covering of for a.a. z E ÔE, where denotes the
M*
x E). Observe that is a covering. Let —÷ lifting with F,(0) am}andletFj: E —* Write F3(z)), where Fj: E —+ E, j = I m. Let Aodenotethesetofallz I m,exist. It is well known that (RE) \ Ao is of zero measure. By the identity principle, the set S of all z E Ao such that 17(z) for some j k is of zero measure. t < I, y(l) := Then Fixaz E A := Ao\ Sand let y(:) Indeed, let
= {ai =
y is a curve such that the set (y(t)): t I, namely, 17(z)). = I y(l) = r(z) e Gd'. Observe that
j
[0, 1]) has exactly m limit points when ,n. Hence, by virtue of Lemma 3.4.23,
#,r'(a) <m}.
c
=rnforz c A,weconcludethatE*(z) E
forallz
E A.
(2) Since G* is maximal, to prove that mt C function extends holomotphically to mt Yo. j = I
it suffices to show that each m.
324
3 Envelopes of holomorphy for special domains
In fact, we only need to show that for each j E (1 m } the functions Re Pj and extend harmonically to mt Fix a j and let Ii := Re Let h denote the solution of the Dirichiet problem in mt with boundary data h (cf. ERans 19951). We want to prove that h = h in Im
For, observe that the function h0 := h o — h o (I), it has the radial limits 0 a.e. on 8E. Hence h0
is harmonic in E. Moreover, by 0 and, therefore, h = h on
0
The continuity of h and h implies that h = h on
The main part of the proof Since Gt is simply connected, the exists a biholomorphic
mappingf: E —+ G*. It sufficestoprovethat 11*1=1 Let A denote the set of all z E
such that I t E
exists and the set
[0. l)}
has exactly in limit points when t —+ I. Then, by Lemma 3.4.23 and the maximality
of G*, we conclude that lf*(z)I = l,z E A. It remains to prove that (dE) \ A is of zero measure. o f)*(z), Let A0 denote the set of all z E 8E such that f*(z), exist. It is well known that (SE) \ is of zero measure. Let R(a1 am) denote the resultant of Q and Q' with
j
=
I
Q(w) := wm + yajwm_). 0, the identity principle implies that the setS of all z E A Since 'Pm) such that R((çoi 0 f)*(z) ('Pm o f)4(z)) = 0 is of zero measure. Now, we only
0
need to observe A0 \ S C A.
3.5 Separately holomorphic functions Let
Uc
C Ctm be domains and let 0
A c U, 0
B c V. Define the
cross
X
= X(U. A; V, B) := (U x B) U (A x V).
We say that a function f: X —+ C is separately holomorphic on
V.EA: We may
E (9(V)
&
YWEB:
X (f E 05(X)) if
w) E (9(U).
assume that f(0) = 0. Let f = zg. where g E O(E. Es). The case where
g = consi E 8E is obvious. Thus suppose that g: E —p E1. Put h : log igi. Then h is harmonic on E. Observe that h is bounded. (The only problem is to show that infE > 0. Since f(0) = 0 and f is injective. there exists an e > 0 such that If(z)I > e > 0 for IzI > 1/2. Consequently. Ig(z)I > e for Izi > 1/2.) Since h = 0 a.e. on aE. we conclude that h 0 (cf. [0011969]). which shows that
f is a
contradiction.
325
3.5 Separately holomorphic functions
It is natural to ask whether there is a 'maximal' open set X C C" x cm such that on X fl X (notice that, in general, we do not require that X C X).
for any f E (95(X) there exists an 1€ (9(X) with f =
f
Let U, V. A,Bbe as above.
X = X(U, A: V. B)
{(z, w) E U x V: hAU(Z) +
<
1).
Notice that if U and V are domains of holomorphy, then X is a region of holomorphy (cf. Corollary 2.2.15). Observe that if = 0 on A and = 0 on B, then X C X. In particular, if
A and B are open, then X C X. The main result of this section is the following general answer to the problem of continuation of separately holomorphic functions defined on a cross (ISic 1969], [Zah 1976], ENguZer 1991], see also [Maz 1975], [Kaz 1986], [Shif 1989], [Shif 1994] and the survey article [Ngu 1997]).
V c Ctm be bounded domains and let A C U, B C V be nonpluripolar sets. Assume that U is a domain of holomorphy and A is of type 62) Let X:= X(U, A; V. B), X in X(U, A: V. B). Then for any f 05(X) there exists an f (9(X) such that f = on X fl X. Theorem 3.5.1. Let U c
f
Observe that in the case where U = A = V := E, B K(r), Theorem 3.5.1 reduces to a version of the classical Hartogs lemma. Before the proof we need some auxiliary results.
0
I,
If S is a connected Remark 3.5.2. (a) Let c C C" be an open set and let N C 1 on S or component of c, then = hNnSS on S. Moreover, either <
I
(b)Let c1
foranyz ES. C
c
c
C" be open. If N1 C
1.
Lemma 3.5.3. Let c C C" be a bounded open set, let P C c be N C Then
and let
1*
—
In particular,
1.
Proof We only need to show that Let v E v —oo, be such that P c Since c is bounded, we may assume that v 0 on 61) 62)
Recall that := sup{u: u C u < I on u That is, A is the countable union of closed subsets of c".
(cf. Theorem 2.1.39). 0 on N); cf.
§
21.
326
3 Envelopes of holomorphy for special domains
< Ion0, we haveu+rv < I oncandu+ev Letting 0onN. Henceu+ev —* 0, we get u hN,c on c \ Consequently, u (cf. the proof of Proposition 2.1.12) and, therefore, hN\p.c
on
e
0
Remark 3.5.4. In the case where is unbounded Lemma 3.5.3 remains true for all plunpolar sets P c f2 such that there exists a function v E PSJe(f2) with v $ —00,
v <0, P C We say that a subset N of an open set C C" is locally nonpluripolar in any connected component S of c the set N fl S is nonpluripolar.
Lemma 3.5.5. If c
C
C" is an open bounded set, N C
if for
then the following
conditions are equivalent: (I)
N is locally non pluripolar in
< lforanyz E a
(ii)
Pmof We may assume that is connected. The implication (ii) from Lemma 3.5.3. To prove that (i) (ii) suppose that
(I) follows I. The set
P := {z Consequently, there exists a point a E f2 such that = 1. This means that for each k E N there exists a function Uk such that Uk < I on 0 on N, and Uk I —2± Putv := I). Thenv E ?S3((c2)(cf.Remark2.l.2(d)), uk(a) v(a) —I, and v = —oo on N. Thus N is pluripolar; contradiction. 0 Lemma 3.5.6. Let f2, / that
Pmof Let V
E
c be
is bounded. Then
:=
£ E N, N, /
open, and let N, c
\
N. Assume
pointwise on <
Obviously, I. Put
Let v :=
Then
< u(z)}.
:= (z E
It is known that P, is pluripolar (Theorem 2.1.41). Let P := note that P is pluripolar. Observe that v = u, 0 on N \ P. Consequently, by Lemma 3.5.3, V hfr,,\pQ = 0
Lemma 3.5.7. Let N be a nonpluripolar subset of an open bounded set C C".
Pmof Fix ana E (0.1). We have to prove (cf. Lemma 3.5.5) that
< I,
Z E Ga.
Let
No := {z
N:
= 01.
3.5 Separately holomorphic functions
327
Then No c N 11 Ga and therefore it suffices to show that
<
1
for any
Z E Ga. Observe
that N \ N0 C {z E c:
Consequently, the set
<
=
N \ N0 is pluripolar and hence, by Lemma 3.5.3,
P := (z
E
Put
E
notice that P is pluripolar (cf. Proposition 2.1.40). Define
u(z)
on Ga
1
onc2\Ga. (Proposition 2.1.9), u < 1, and u = 0 on N0 \ I
Then u
E
=
u
P.
Thus
(Lemma 3.5.3) and, finally,
< I in Ga.
S
U
Let f C C'7 be hyperconvex (cf. Definition 2.1.46) and let K set. Define /2 = /2K,r2 := (cf. Definition 2.1.43). Recall that rem 2.1.48).
be a compact
\ K) =
is a Borel measure and
0 (Theo
/
Lemma 3.5.8. Let c C'7, Gt / G c Ctm be open, and let J N C f, G€ j M1 / M C G. Assume that G are bounded, and Nt is locally nonpluripolar in
:=
£ E N. Put
Nt; G1, Mt),
V :=
V1 :=
N; G, M),
N1; G1, M1),
V :=
N,
£
N; G, M):
/ I (Lemma 3.5.6). Let g be such that for each £ E N there exists a E 0(Y1) such that = g on V1 (1 V,. Then = on V1 (t E N)and, therefore, there exists a E 0(Y) such that = g on V fl Y. notice that V1 / Y and
Proof Fix ant Vt
The functions
Nand(zo, wO) E Vt. Recall that
= {(z, w)
(., wo) and
W := {z E c,: with a := 1 —
x G1:
E
< 1}.
wo) are holomorphic in
(z, wo) E Y1} = (z E f21:
<
Obviously. zo E W. Notice that if a =
1,
then W =
(because N1 is locally nonpluripolar in cf. Lemma 3.5.5). We know that w0) = Ie(z, wo) = g(z, wO) for z E N1 fl W. By Lemma 3.5.7, the set N1 fl W is locally nonpluripolar in W. Consequently, wo) on W. (., wo) U
3 Envelopes of holomorphy for special domains
328
Lemma 3.5.9 (cf. [Mit 1961]). Let Ho. be separable Hilbert spaces, dim Ho dim H1 = 00, and let T: Ho —* be a linear infective compact operator 63) such that T(Ho) is dense in Hi. Then there exists an orthogonal basis (hk)kEN C H0 such that is an orthonormal basis in IIbk
=: vk / +00 when k / +00. if there exist a local/v convex nuclear space V and linear continuous op
V Hi such that T = T2 o T,, then the above basis may erators Ho be chosen in such a way that, the series is convergent for any
e >0. Proof For the convenience of the reader we shall give a rough idea of the proof based on the following textbook on functional analysis [MeiVog 19771, see also [Jarc (a) Since T is compact there are orthonormal systems (ek)kcN C Ho. (fk)kEN C = 0. such Hi, and a monotonically decreasing sequence (Sk)kEN C that
Tx =
XE H0,
where the series does converge in the operator norm (cf. [MeiVog 19771, PropositiOn 16.3). This representation of T is called the Schmidt representation of the compact operator T. Applying that T(H0) is dense in it follows that (fk)koN an orthonormal basis of (cf. [MeiVog 19771, Proposition 12.4). Moreover, since dim = 00 we have Sk > 0, k E N. Finally, the injectivity of T implies that (ek )kEN is an orthonormal basis of Ho. Put bk := 4/sk. k E N. Then Tbk = fk. k E N, and
/ 00.
(b) Before discussing the second assertion of the lemma we like to recall a few general facts from functional analysis, all of them may be found in [MeiVog 1977). Although the systems (ek)kEN and (fk)kEN are not uniquely determined the sequence (sk )koN is. It is called the sequence of singular numbers of the compact operator T. For 0 < r
Sr(H'. H") := (A E K(H'. H"):
E
where K(H'. H") denotes the set of all compact linear operators between H' and H" and where (Sk(A))k0N is the sequence of singular numbers associated to the compact
operator A. Sr(H', H") is the socalled Schatten rclass (cf. [MeiVog 1977), § 16). We will need the following fact concerning Schatten classes (cf. [MeiVog 19771, Lemma 16.7): 63
Recall that a linear operator T: X —* Y. where X and Y are locally convex topological vector spaces, is compact if for any bounded set B C X the set T( B) is relatively compact in Y. We would like to thank Professor K. Floret (Oidenburg University) for the idea of the proof.
3.5 Separately holomorphic functions
329
j = 2,..., 5, be Hubert spaces and let A1: H2J —p H2J+t. j = 1,2, be continuous linear. IfS E Sr(H3. H4), then A2 o S o A1 E Sr(H2, H5),
(*) Let H1,
00 such that
V:
(x
e} c U.
Moreover, for any continuous Hubert seminorm p on V there are a Hubert space the completion of V/{x E V: p(x) = 0} with respect to the scalar product, that defines p, and a natural continuous linear operator
XE'V, where [x} denotes the class of .x in V,,,.
Using that T2 is continuous we find a continuous Hilbertseminorm p on V and a —÷ H1 such that T2 = T2 o ii,. continuous linear mapping T2: In virtue of Proposition 28.4 of [MeiVog I 977J, for any e > 0 there exists a continp, such that the natural operator ip,q: Vq —+ Vp uous Hubert seminorm q in V, q
(induced by the identity of V) belongs to S€(Vq. Vt). Applying (*) from (b) to
T=
T2
0
oiq o T1 gives that T E
H1). Hence,
.Sic(T)E <00.
0
be a strongly pseudoconvex open and let N C S2 be compact and local!)' nonpluripolar in c. Put Ho := (cf § 1.2), in L2(N. := the closure of Hi
Lemma 3.5.10 ([NguZer 19911). Let 2 c set
Then the linear operators
are well defined, infective, and continuous Ti is compact. In particular T.=T201\ the operator Ho IN E H1 is compact. Let be as in Lem,na3.5.9. Then for anya E (0, 1)andfor C H0, any compact
I
KC
{z
A Hubert seminorm is a senünorm p that can be defined by a semiscalarproduct. i.e. where (.,.) satisfies all conditions of a Hermitian scalar product except the positive definiteness. &,) Recall that any strongly pseudoconvex open set is hyperconvex; cf. Proposition 2.2.25. 67) Recall that O(c) is a nuclear space.
p(x) =
3 Envelopes of holomorphy for special domains
330
there exists a constant C
C(a, K) > 0 such that
Cvi',
IIbktIK
k E N.
(*)
Proof Step 1°: 7'2 is well defined.
Every function f E (9(7) can be approximated uniformly on N by functions (because strongly pseudoconvex — cf. Proposition 2.7.7). Moreover, uniformly on N, then fsIN C Ho and if Hi. Step 2°: T2 is injective. Suppose that f E (9(c) is such that T2(f) = 0 in H1, i.e. js(P) = 0, where /1 := 0}. By Proposition 2.1.49, we get = and P := lz E N: f(z) Now, since Hence, by Lemma 3.5.5, the set N \ P is locally nonpluripolar in 0 on N \ P. we conclude that f 0. Step 3°: T1 is continuous. It is well known that for any compact L C there exists a constant c(L) > 0 such from
f
f
that
hilL 5j c(L)hIflIL2(Q),
f
(+)
E
which is equivalent to the continuity of T1.
The continuity of T2 is trivial. Step 4°: T1 is compact. Let c H0 be bounded. Then, by (+), it is locally uniformly bounded in Consequently, by the Montel theorem, there exists a subsequence that is locally uniformly convergent.
Step 5°: The estimate (t). Fork Nsuchthatt'k > l,let
uk:= By (+), for any compact L C
log log vk
we get
ilL
k EN.
Thus the sequence (uk)k is locally bounded from above in
Let
u :=limsupuk. 1. Moreover, Since Uk —* +00, we get u P := {z c: u(z) < u*(z)}; P is pluripolar (cf. Theorem 2.1.41). = 0 such that u We will show that there exists a Borel set Q C N with
on N \ Q. Put Qk,p.q
:= {z EN: bk(Z)l
k,
p,q €N
0
3.5 Separately holomorphic functions
(Qk.p.q is compact). Recall that IIbk
=
I2(Qk,p,q)
1.
331
Thus we have
=
IN
Let
Qp,q := U Qk.p,q.
p. q E N.
Then, by Lemma 3.5.9. 00
(d'l
const(q) Vk
—
2
k= I
= 0, where
Consequently,
Qq :=flQp,q, =
Finally,
q €N.
0, where
Q
Qq for every q. Consequently, for each q there exists a p(q) such that zo Qp(q),q. This means that ZO Qk.pq),q for any k. i.e. k E N. Consequently, Ibk(zo)I <
Observe that if zo E N \ Q. then zO
logp(q) Uk(ZO)
Hence u(zO)
+
1
—.
q
l/q. Since q was arbitrary, we conclude that u(zO) = 0.
Thus 0 on N \ (P U Q) and hence = By Proposition 2.1.49 and Lemma 3.5.3 we get = < In particular, SUPK u* < a. Hence, by the Hartogs lemma Thus (Proposition 2.1.23), Uk a on K for k>> I, which implies the required result. 0
PmofofTheore,n 3.5.1. Fix an f Step
10: Reduction to the case where A is compact and B is relatively compact
in V. Since A E
of compact sets such that in C", there exists a sequence is not plunpolar in U fort>> 1. We may assume that Ae / A. In particular, is not pluripolar in U for any 1.
3 Envelopes of holomorphy for special domains
332
of relatively compact and nonpluripolar
/
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 nonpluripolar in l'k). Finally, c = g(wo) =
f(zo,
wO).
Suppose that for each I 1(k) we have a function fk,, E c9(X(Uk, Ak,,: Vk, B)) such that Ike = on X(Uk, Ak,: Vk, B) fl X(Uk. Ak,; 14. B). Then, by Lemma 3.5.8,thereexistsanfk E 0(X(Uk, A: Vk, B))suchthatfk = fonX(Uk, A; 14., B)fl
f
X(U, A;Vk, B). Now, once again by Lemma 3.5.8, we conclude that there exists an
f
E
0(X)suchthatf = f on XflX.
Step 3°: The case where U is strongly pseudoconvex, A is compact, B is relatively compact in V. w) E (9(U), w E B, and is continuous and bounded on A x V.
f
We may assume that fl
:=
Ho
:=
I on A x V.
:= the closure of HOIA in L2(A. be the basis from Lemma 3.5.10; := k E N. For any w E B we have f(., w) E Ho and f(., w)IA E H1. Hence Let
p). and let
3.5 Separately holomorphic functions
333
where
ck(w) =
I f(z, w)bk(Z)
=
f f(z,
k E N;
JA
(U) (in particular, locally uniformly in U) and in L2(A, ii). Since f is continuous on A x V, the formula
cf. Lemma 3.5.9. The series is convergent in
ck(w) :=
f f(z. w)bk(Z)
defines a holomorphic function on V. k
converges
w E V,
JA
N. We are going to prove that the series
iocally uniformly in X.
Take a compact K x L c X and let a > maxK h be such > a + fi < I. First, we will prove that there exists a constant C'(L. fi) > 0 such
that
that
k EN.
IICkIIL
(t)
Suppose for a moment that (t) is true. Then, using Lemmas 3.5.10 and 3.5.9. we get
>
IICkIIL IIbkIIK
C'(L,
= C'(L, fi)C(K, which gives the normal convergence on K x L.
We move to the proof of (t). By the Holder inequality, we get w E V. k E N
ICk(W)I
(recall that ICk(W)j
< 1 on A x V). On the other hand, if w
= ICk(W)I =
f f(z,
w)bk(z)
U
Uk
B, then
334
3 Envelopes of holomorphy for special domains
The sequence is bounded from above in V. u := and u —l on B. Let P := (w E V: u(w) < u(w)}; P is u
0,
Thus —l on B \ P. Consequently, I + <
(Lemma 3.5.3). = Hence 1 +ut on L. Now, by the Hartogs lemma, which implies (t).
> 1,
Let
f(z, w) := obviously f is holomorphic. Recall that I =
w) E
f on X fl (U x B). Now, take an
arbitrarypoint(zo.wo) e Xfl(A x V)andleta :=
E (0,1]. Then holomorphic in W := {w E V:
f
3.6 Extension of meromorphic functions The aim of this section is to discuss the problem of continuation of meromorphic functions. The main result (Theorem 3.6.6) says that the envelope of meromorphy of a Riemann domain coincides with its envelope of holomorphy. Let (X, p) be a Riemann domain over C".
Definition 3.6.1. A function f: X \ S —+ C, where S = S(f) is a closed subset of X with mt S = 0, is said to be meromorphic on X (f E if: (a) f E (9(X \ S) and S is singular for I in the sense of § 3.4 (b) for any point a E S there exist an open connected neighborhood U of a and functions (9(U), $ 0, such that = on U \ S (cf. [RotKop 19811). The set
:= X \ 5(f) is called the set of regular points off.
Remark 3.6.2. (a) The following conditions are equivalent:
fe.M(X);
(i)
(ii) foranypointa
XthereexistsanopenneighborhoodUsuchthatfju E .M(U);
c X.
(iii)
(b) Let a, U,
as in Definition 3.6.1(b). We may always assume that the germs Taco and Tat/F are relatively prime in Consequently, we may assume that and are relatively prime for any x 68)
be
That is. f cannot be holomorphically continued across any point of S.
U.
335
3.6 Extension of meromorphic functions
Then S fl U = {x E U:
= O} and, consequently, either S = 0 or S is a pure
(n — 1)dimensional analytic subset of X. We have the following two possibilities:
Inthiscasewesaythataisapoleoff(a E S'(f)). •
0: In this case we say that a is a point of indeterminacy off (a E 2(f)).
It is clear that 5(f) = ?(f) U 2(f) and ?(f) fl 2(f)
0. Observe that = 0). Thus 2(f) is an analytic subset of X of
2(f) P U = {x E U: ç9(x) = dimension
n
—
2
(cf. [Clii 1989], § 3.7).
is a field.
(c)
(d) (The identity principle for meromorphic functions) If f e and f = 0 on an nonempty open subset of R(f), then f 0. E (9(X), i/i 0, then there exists an f E .M(X)(with S(f) C
such thatf =
on X \ *_1(0) (we shortly writef = Indeed, define f := p/t/.t on X \ Then we extend holomorphically f to is the set of all singular points of (0) in the sense of X \ S, where S := (* § 3.4. We denote the extension again by f. Then f satisfies Definition 3.6.1(a) and, by the identity principle, /,f = in X \ S. (t) Let (Y, q) be another Riemann domain over C
a a C .M(X). P (9(Y \ S). S := 5(g). then f := g o a (9(X \ S)
Indeed, if g E
with S :=
(S). Since a is locally btholomorphic, the function f cannot be holomorphically continued beyond S (cf. § 3.4). Finally, if U is a subdomain of X t,!, such that alu: U —* U is biholomorphic and if i/ig = in U \ S (9(U),
00), then Let 0 F c .M(X). In virtue of the above remarks, the notions of the Fextension and maximal Fextension (Fenvelope of meromorphy) from Sections 1.6 and 1.8 may be easily generalized to the case of meromorphic functions. Example 3.63 (The sheaf of Igerms of meromorphic functions; cf. Example 1.6.6). Let l 0 be an arbitraiy set of indices. Let
{(U, (f,),e,): U E For (U,
(f,)IE,). (V,
E
f.
.M(U), I E I),
a
we put
W CU P V, filw = Define
:=
,MW
:= (J
x (a),
(SEC"
—+ C", 69)
Recall that
69)
a)) := a. where f = [(U,
denotes the set of all open neighborhoods of a.
I El.
336
3 Envelopes of holomorphy for special domains
For any Igerm
(as
U
above), consider the set
:= (([(U, (fi)iEI)]L. z): z E U).
The family of such sets U defines a Hausdorff topology on such that each U is univalent (i.e. the projection jrW maps homeomorphically U onto U). Thus is a Riemann region over C's. For any i E I we define a function F: —+ C, F((f, a)) := f,(a). Since o on U, the mapping F1 is meromorphic on F, = f,
Let (X. p) be an arbitrary Riemann domain over
C .M(X).
We define a morphism
a=
(X. p) —*
(I := F). Namely,
a(x) := ([(p(U). (1 °
p(x)),
X
E X.
where U C X is an arbitrary univalent neighborhood of x. Moreover. F1 o a = f for any f F. Consequently, if X denotes the connected component of that contains the set a(X) and := then
a: (X.p) is
(X,13)
an Fextension.
In virtue of the above construction, using the version of the proof of Theorem 1.8.4, we get the following generalization of the Thullen theorem.
Theorem 3.6.4 (Thullen theorem). Let (X, p) be a Riemann domain over C" and let 0 F C .M(X). Then (X, p) has an Fenvelope ofmeromorphy. Our next aim is to compare the envelopes of meromorphy and holomorphy. First we consider the following special elementary case.
ProposItion 3.6.5 (Levi) (cf. [OkuSak 1957]). Let T = Tr.p (0 < r, p < 1); cf Section 3.4. Then M(T) = .M(E")IT, i.e. any function f E .M(T) extends to a function meromorphic on E".
Proof (Cf. [Siu 1974b)). Fix an f E and let ro := sup(s meromorphically to We may assume that r = ro.
(0,
1): f extends
(I) First consider the case where f E (9(P,,_1 x {p < wi < I)). Then f may be represented by a Hartogs—Laurent series
f(z.
w)
(z. w) E
= wherefk
E
k=—oo
E
Z.
x
< iwi <
I },
3.6 Extension of meromorphic functions
337
denote the set of all z such that f(z. .) is meromorphic For p E Niet in E and has at most p poles counted with multiplicities (i.e. either (9(E) or there exist E N with + we E E, and si p such + that wt}) and the function E \ w —÷ E (9(E \ {wj extends holomorphically to E). In particular, f(z, w)(w — ... (w — for any z E A,, there exist numbers ao(z) Op(Z) E C, not all zero, such that the function w —+ f(z, w) extends holomorphically to E. Thus (*)
kEN. Observe that that dim .1(f)
j
(the projection onto of .1(f)). Recall (Remark 3.6.2(b)). Thus there exists a p such that A,, is not
A,, n
—
2
thin.
Consider the field F
and the vector space IFP+I over IF. Let vk := Takearbitraryk1 1,2 EN,
=
f—k—,,) E
k1 < ...
D:=det Using (*) we get D =
f—k1—!
f—k1—,,
f—k2. f—k2—!
f..k2....p :
Hence, since
0 on
is not thin, we conclude that D
0
Ofl
Consequently, the subspace V C FP+l spanned by the vectors has dimension d p. generate V over F. In particular, for any k E N p. Let vk Vkd, there exist functions ck,o, Ck,t 0, such that ckd E ck.o
(**)
ak,ovk = The columns in the matrix f—,11—,,
f—k2. f—k2—!
.f—k2—p
f—kd. f—kd—1
fkdp
linearly dependent over F. Hence the exist functions b0 all identically zero, such that 0, v = I we get 0, k N, and therefore the function f_k_f are
(z. w)
f(z.
b,, E
not
d. By virtue of (**)
338
3 Envelopes of holomorphy for special domains
extends holomorphically to T and, consequently, to a function g E O(E"). Finally, gives the required meromorphic exthe function (z, w) —÷ g(z. w)/
tension of f to E". (2) Now consider the situation when S := S(f) is such that for each z E ?fl_j the fiber S := {w E E: (z. w) E S) is discrete in {w E: (z, w) E T); cf. the subsection 'Reduction procedures' in § 3.4. Then, using the method of the proof of Proposition 3.4.13 and (1), we easily conclude that r0 = 1, which gives the required result.
(3) It remains to reduce the proof to the situation as in (2). Let S := S(f). For put
E
4'A(z,w):=(z+wA,w). is biholomorphic and 4'j' = 4'A; cf. § 3.4. Fix 0 < e << I. By Lemma 3.4.12, there exists a E such that the set S' := has discrete fibers. Observe that (I)_A(T) D =: T'. The function g := f o 4'A is meromorphic on T' and holomorphic on T' \ S'. Hence, Recall that
by (2), the function g extends to extends meromorphically f to 4'x(Pn_,(1 — e) e —* 0 finishes the proof.
e)
x E). Consequently, o — 2e) x E. Letting
x E) j
0
Theorem 3.6.6. Let (X, p) be a Riemann domain over C", let 0
7 c .M (X), and
let
a: (X. p) —÷ (X, j5) be the maximal 7extension (cf Theorem 3.6.4). Then (X, j3) is Stein. In the envelope of meromorphy of (X, p) coincides with the envelope
of holomorphy of(X,
70)
Proof We may assume that the maximal 7extension is realized as the connected component of as in Proposition 3.6.4. To prove that (X, is Stein it suffices to verify condition (v) from Theorem 2.9.1 with (Y, q) = (C". Id) and = p. Thus letT = < r,p < I and let ci): T —÷ 4'(T) C Xbeabiholomorphic mapping such that ji o 4' extends to a biholomorphic mapping 4': E" —+ 4' ( E") C C". We want to prove that ci) extends to a holomorphic mapping Ci): E" —p X. Forany f F let F1 E .M(X)be as in Example 3.6.3. Then Fjo4' is meromorphic
on T and, consequently, by Proposition 3.6.5, it extends to a function I E .M(E"). Now define
:= ([(4'(E"), (fo
)fEc
cf. Example 3.6.3. One can easily check that 4'; E" —÷
z E E"; is a holomorphic
extension of ci). Obviously. (i)(E") C X. 70)
In the case of Reinhardt domains a simpler proof may be found in ISak 19701.
0
3.6 Extension of meromorphic functions
339
Corollary 3.6.7 ([KajSak 1967]). Let (X, p) be a Riemann domain over C" and let .M(X). Then there exist 0(X), 0, such that f = co/i/i.
f
The above corollary justifies to introduce meromorphic functions on Riemann domains over C" as global quotients of holomorphic functions; cf. Remark 3.6.2(e).
Proof The result is wellknown if X is Stein — cf. [Hor 1990] follows via Theorem 3.6.6: if
The general case
f
(cf. Reis the envelope of meromorphy and f E is such that f o a = 0, such that i/if = on mark 3.6.2(1)), then first we find i/i E 0(X), iii (cf. X \ S(f). Consequently, we get (i/i o a)f = o a on X \ S(f). Thus f =
9
Remark 3.6.2(e)).
In [KajKaz 1971] the problem of continuation and quotient representation of meromorphic functions on a domain over X x C", where X is a product of complex projective spaces, is discussed. Below we present an alternative proof of the most important part of Theorem 3.6.6 based on the Dloussky Theorem 3.4.8. Theorem 3.6.8. Let (X, p) be a Riemann domain over C" such that 0(X) separates points in X. Let a: (X. p) —÷ (X, j5) be the maximal holomorphic extension. Then each function memmorphic on X extends meromorphically to X.
Proof (1) First observe that if M c X is a singular analytic set and if M denotes its extension obtained via the Dloussky theorem, then any irreducible component of M intersects
a(X).
Indeed, suppose that B is an irreducible component of M such that B fla(X)
= 0.
Let A denote the union of all other components of M. Then A is a pure (n — 1)dimensional subset of X such that M = (A). Consequently, by Theorem 2.5.9, X \ A is the envelope of holomorphy of X \ M. Since X \ M is also the envelope of holomorphy of X \ M, we get M = A: contradiction.
(2) Take a function f E .M(X) and let S := S(f). We may assume that S 0. Recall (cf. Definition 3.6.1(a)) that S is a pure (n — 1)dimensional singular analytic subset of X. Let S denote the extension of S obtained from the Dloussky theorem. Then there exists an f 0(X \ S) such that 10 a = Jon X \ S. It remains to prove that f is meromorphic on X.
f
71) The proof is based on the sheaf theory: Assume that (X, 0. p) is Stein and f E .M(X). Let t9, denote the sheaves of germs of holomorphic and meromorphic functions on X. respectively. .4(. Since f(9 is isomorphic Then the multiplication by f defines a sheaf homomorphism 1: .M is also coherent. By Theorem A there is a nontrivial to 0 and (9 is coherent, the sheaf F := (9 fl global e (9(X). X —÷ F.
3 Envelopes of holomorphy for special domains
340
(3) Let A be an irreducible component of S. Then Ao := A flReg(S) is a connected
complex manifold. Let A denote the set of all points a E A0 such that f is meromorphic in a neighborhood of a. It is clear that A is open in Ao. 0. Let b be an accumulation point of A in A0. The problem of Moreover, by (I), A meromorphic continuation of f to a neighborhood of b reduces, via a biholomorphic transformation, to the question whether a function g holomorphic in E" \ M, where x E, extends meromorphically to M := x (0), and meromorphic in E". The positive answer to this question follows directly from Proposition 3.6.5 with T := (p arbitrary). Thus A = Ao and, consequently, f is meromorphic in X \ Sing(S). In the next step we will prove that f extends meromorphically to a neighborhood of any point from Reg(Sing(S)). Take a b E Reg(Sing(S)). The problem of meromorphic continuation of f to a neighborhood of b reduces, via a biholomorphic transformation, to the question whether a function g meromorphic in E" \ M, where We use once again x {l/2} x (1/2}, extends meromorphically to M := (n — 1)dimensional
Proposition 3.6.5 with T := T113213.
Thus f is meromorphic in X \ Sing(Sing(S)). Now, finite induction finishes the proof.
0
Chapter 4
Existence domains of special
families of holomorphic functions
4.1 Special domains In this section we study pseudoconvex domains with many symmetries and give characterizations for them to be sdomains of holomorphy, where C (9. t'°°.domalns in the plane. In the plane there is the following description of analogous to the Cartan—Thullen domains of holomorphy in terms of theorem (cf. [AheSch 1975]).
Theorem 4.1.1. Let D c C be a bounded domain. if D is J(>°convex. Then, D is an .le°domain of
Let K C D
Pmof
be
K. Letf E Je°°(D).
compact and fix a
We claim that
iE
5
S := dist(z°, K). In fact, (*) is clear for j = for some j > 0. Set
(*)
where
0.
We proceed with induction. Assume that (*) holds
I
C.
Then
g(z) := (1(z)
—
(k)_O (z
—
—
g E 3e°°(D) and (j + l)!g(z0) =
Ifz e K, we get Ig(z)I
Therefore,
S If
+
+ I)! = lg(z°)I s
S
i.e. (*) is verified.
8/2). Using that D is an of (*) it follows that T,of converges on R°°domain of holomorphy we conclude that 8/2 pD(z0). Because
342
4 Existence domains of special families of holomorphic functions
Fix a b E K with Jz° — bl =
and c E
with pD(z°) = lz°
—
ci. Then,
PD(K) — PD(Z0). Therefore, PD(Z0) which implies that D is .1(°°convex. 2: Here we proceed via contradiction. So, let's assume that D is not an domain of holomorphy. Then there exist a E D and 0 < r < R such that 2pD(Z°) Hence,
lb — ci — Ic — z°I
.]3(a,r)C DandB(a,r)flaD={b}; • for every f E
the Taylor expansion Ta! off converges on B(a, R). Choose c E dB(a, r) fl D and e > 0 such that B(c, e) D. Moreover, fix numbers
r
0. PutK :=]3(a,ri)\D. Observe
that C \ K is connected and contains D and that also B(a, r2) \ K is connected. Then any f E \ K) extends to an entire function. Thus, \ K) = C. Using
the Riemann mapping theorem implies that K is totally disconnected. Thus we find an open set U with b E U C B(a, and C D. In virtue of the above conditions., we see that b is an accumulation point of contradiction.
0 The proof of the implication
was
based on the fact that D is a bounded
It seems to be unknown whether this implication remains true without
domain.
assuming that D is bounded
Remark 4.1.2. According to [Rud 1955] there are sequences (ak)k€N C IR>o, \ O,and(ck)kEN,ak > > O.k E N,suchthat Ic—.
ck
k=I
a,, — Ck
<00,
K(ak,ck)flK(aj,cJ)=O,
Put
G :=C\(UK(ak,ck)U{O)). kEN
G is an .1(°°convex domain in C satisfying the following property: if f E 3(°°(G),
then lf(x)l <
x < 0. In other words, G is X°°convex but not Caratheodory complete. Observe that G is biholomorphically equivalent to a of holomorphy, but bounded domain G' c C. So G' is an R°°convex G' is not Carathéodory complete. Conversely, any Carathéodory complete domain is f(oo) =
0,
3C°°convex (cf. [JarPfl 1993] for more details).
Balanced domains. First recall that a domain D C C" is balanced if ED C D, or equivalently, if there is an upper semicontinuous function h: C" —÷ [0, oo), homogeneous (i.e. h(Az)
IA$h(z), A E C, z E C"), such that
D= Dh =(z E
C": h(z) <1).
343
4.1 Special domains
Observe that Ii = hD is uniquely defined; h is the Minkowski function of D. Pseudoconvexity of D = Dh can be characterized in properties of h. More precisely, we know that (cf. Proposition 2.2.22(a))
if
D is pseudoconvex
if
hE
log/i E
Introduce the following notation:
:=
u
{U E
homogeneous }.
In order to discuss characterizations of balanced domains that are R°°domains of holomorphy we need the following function. Let D = be a balanced domain. —+ [0.00). Put/i = hD:
h(z) : =
=
QE
deg Q? 1.
Q
degQ
Obviously. ii is homogeneous and 0
1,
I on D) IQI 1Q1lIdesQ < Il). ')
h.
Ii
Starting from D = Dh we define h*
=
as
the upper semicontinuous regularization of ); h*(Z) < l}.
fz E
Observe that h* is a homogeneous psh function with 0
Dh
be as above. Then every f E
extends as a of holomorphy, and D* is fat.
D* is an
Proof Let f
We may assume that f is not identically constant and that on D. Then we have the following expansion off in a series of homogeneous polynomials
Ill <
I
with
IM
1)
Then, using the definition of/i we get
< h(z) < h*(z),
z E
j > I.
h* < I on D*, we see that the series is locally uniformly convergent on D*; in particular, it gives the holomorphic extension of 1. It remains to see that D* is an Je°°domain of holomorphy. Assume this is not the case. Then there are open sets U1, U2 with 0 U1 C D* flU2 c U2, U2 connected. such that, if f then flu1 = flui for a suitable f E (9(U2). Since
Using the fact that is homogeneous, we find a point a E U2 \ D* with Therefore, there is Q 1. with QI 1 on 1. deg Q
h(a) > '>
degree
Observe that the zero polynomial 1.
is counted as a homogeneous polynomial of any
344
4 Existence domains of special families of holomorphic functions
D and Q(a)
11 deg
>
I
f
1.
Hence
I
<
QI
I on D*,
but Q(a) I > I
I.
Taking
0
(Q — Q(a)y' leads to a contradiction.
Corollary 4.1.4. Let D = Dh. Then the following pmperties are equivalent: (i)
of holomorphy;
D is an
(ii) there is a sequence (Qj)1EN C 1Q31 < I, j E Z+, on D (iii) h = h*.
Q = j, such that
=
In particular, if D = Dh is an R°° domain of holonwrphy, then D is fat.
Proof It remains to prove (i) (ii): We know that there exists an E Je°°(D), ft < 1, such that f cannot be holomorphically extended through Let Q3 be the expansion of f in a series of homogeneous polynomials, i.e. Q3 E = 0 or deg = j. Then, as above, < I on D. Hence,
f
I
I
Suppose now that there is an a E C'7 with
Proposition 4.1.5. Let D = Dh be a bounded balanced pseudoconvex domain with continuous h. Then h = h. In particular h* = h and D is an .7(°° domain of holomorphy.
Proof Obviously, h
Ii. To establish the inverse inequality it suffices to prove that
h(a) = I impliesh(a)? I. Fix an a E C" with h(a) = I. Recall that we know that D i5 8%om(C")cot'vex 2) Let 0 < t < I. Then the continuity of h gives that K, := {z E C": h(z) t} 2)
For the convenience of the reader we give the pmof: Let K C D be compact. Suppose C D that converges to a boundary point of D such that IQ(z*)I It QIItc. E Observe that there is an e > 0 such that h(z) (I + z E K. Put Q K, := tAz: Al i + e, 2 E K). Then K1 C D is compact. Since D is pseudoconvex. we find
there is a sequence (zk
f e 0(D) with
hulk, = I.
lf(zk)I —' oo.
k. Developing I as a series of homogeneous polynomials, i.e. I = inequalities yields to hQ1(z)I <(I +rY',z E K. Therefore, lQj(zk)h hf(zk)I < oc;contradiction.
and applying Cauchy
(I +e)1.k EN.
Hence,
4.1 Special domains
345
is a compact subset of D. Thus, its hull is compact, too. Thereforewechoosenumbersr(:) (t, 1),O <1< 1,suchthatr(t)a Then we find Q,
I, with
deg Qi
E
= II Q1 IlK, < PQ,(r(t)a)I.
Now, the left equality leads to IQ,(tz/h(z))I virtue of the definition of h this gives In particular for z = r(t)a we obtain
h(a) = urn 1(r(f)a)
1,
z
0, or h on
< ,i in
t
urn
t.i
=
1,
which completes the proof.
EJ
Remark. From Proposition 4.1.5 one easily obtains that any bounded complete pseudoconvex Reinhardt domain is an of holomorphy. (Exercise: show that in this case the Minkowski function is always continuous.) On the other side, the following example D := {z E C2: < I)
shows that something like the boundedness condition is necessary. (Exercise: show that .7e°°(D) = C.)
c OE
Example 4.1.6 [Ceg 1993]. Let
and choose
be dense in
C
= I. Wedefine u: C2 —÷ [0, oc).
u(z) :=
log Izi — aJz21.
j=i Then u ,'PSR(C°) (Exercise). Moreover, the function h: C2 —+ [0, oo), h1 exp(u) is homogeneous and psh. So it gives a bounded balanced pseudoconvex domain
D := {z
C2: h(z) := hj(z) + max{IziI,
1Z21}
< 1}.
Observe that
N(h) : = {z
C2: h is not continuous at z}
= (z E C2:
Izi I = 1Z21 and
is of Lebesgue measure zero. Therefore, D is fat cise).
u(z) > —oo}
and N(h) is not pluripolar (Exer
Observe that any pseudoconvex D = Dh, for which N (h) is fat (Exercise).
is
a set of Lebesgue measure zero,
346
4 Existence domains of special families of holomorphic functions
To calculate h fix a point z0 = (z?,z20) E x Remember that z0 is a limit point of a sequence ((aj(k)z?, along which u = —00. Therefore, if I along that sequence. Hence, I Q(z°) 1. (C2), I Q I < I on D, then QI QE The maximum principle finally gives that IQI I on E x E. Hence, < I idc'u I. is obvious. Hence, D* = E x E. That h1 Summarizing, there are fat bounded balanced pseudoconvex domains which are not J(°°domains of holomorphy, although, in contrast to the Cartan—ThuIlen Theorem 1.8.4, they are Hartogs domains of that type have been given by I
I
I
N. Sibony [Sib 19751. We conclude this discussion mentioning that the Hartogs triangle is an of holoniorphy but not (Exercise). Thus there is no analogous result like the Cartan—Thullen Theorem for of holomorphy. Combining Proposition 4.1.5 and the preceding example one might think that the size of N(h) is responsible for being an J(°°domain of holomorphy. In fact, we have the following result due to J. Siciak [Sic 19841, [Sic 1985].
Theorem 4.1.7. Let D = Dh be a balanced domain. Then the following properties are equivalent:
(i)
D is an
(ii)
Ii
domain of holomorphy;
=
(iii) D = mt (iv) D is an R°° fl A°°domain of holomorphy. if D is bounded, then the above properties are equivalent to
(v) h
and N(h) ispluripolar.
E
In order to be able to prove Theorem 4.1.7 we need some special facts from the theory of psh functions, in particular, the generalization of Josefson's theorem due to J. Siciak [Sic 1982]. Before proving this generalization we provide a few preparations, all of them can be found in the work of J. Siciak.
Lemma 4.1.8. Let K C C" be compact. Then sup{exp(u): u
whereC(C")
E
£(C"). uIK 5 01 = sup{IPI11k: kEN, P 3c>o: u(z)
{u E
IPIxI < I),
C+log(l + IIzII).
z
(t)
E C"}.
Proof Obviously, the right hand side in (t) is majorized by the left part. Now, let v = exp(u), u E £(C") with uIK <0. We denote by C E (0, 1). the eregularization of v (cf. section 2.1). Recall that v when e \ 0. Put (e Recall that
+
+
+ 1z112)'12,
ZE
C",
(C") means the set of all polynomials of degree less or equal k.
347
4.1 Special domains
and —
w) :=
where (z, w)
E
I
+
IwI(e +
if w 0 ifw=O,
+ 1w12)'12,
i sIIzII.
C" x C.
Then
1)
Moreover, the family
is
=
= v(z).
and urn
locally unifonnly bounded from above. Set
i)iij. j ? 2.
V3
< u(z), z E C".
Observe that
=
{z E
an ij > 0 and put
Fix
C": v(z) < I K, and
Obviously, K c G7. Therefore, the Hartogs lemma (applied to K, j gives an index j7 such that Vj(Z) 1+ z In virtue of Proposition 4.1.5 we have
Vu, = sup(IQI
IQI <
QE
<
I Ofl
1}}.
Then any competitor Q in that supremum has the property that
IQ(.
Ion K.
1)11(1
Hence,
kEN, P E
Vj <(1 Letting first j —p 00 and then In
all: E
IPI <
f
0
—÷ 0 gives (t).
the case that the compact set in Lemma 4.1.9 is circled, i.e. exp(it)K = K for R, more can be shown, namely
Lemma 4.1.9. Let K C C" be a circled and
compact. Then
sup{exp(u): u E .C(C"). uiK where
Ion K),
:=
Moreover,
QE
=
0)
= max{l,
'l'K},
Ion
degq 2: 1, iQi
K)5).
where
:= sup(h: h
?S3€hom(C").
his
1),
S cC".
Proof Observe that it suffices to show '<' in (i). In virtue of Lemma 4.1.8 the left expression is majorized by the following one k €N, P E .'Pk(C"), IPixi
Observethatmtheca.seofD= D5wehaveh =
l} =:
0,
Z
E C".
4 Existence domains of special families of holomorphic functions
348
Recall that K is circled. Now, let P E .'P(C't) be a competitor of o. We write whereq := degP and E P= = j. Fix az E K and regard the holomorphic function E
P(Az).
C,
By the maximum principle and the Cauchy inequalities we conclude that I Q(z)I z
Fq+l,
•
K —
s I,
Usingthe representation ofPwe have
s 'l'x onC".
K. Hence,
—(
(q + 1)44(z),
if'4'K(Z)1.
instead of P the analogous estimate leads to
Dealing with
IPIs 5 (sq + 1)max{1. or p11/q
sEN.
(sq +
Hence, I P1' 7k <max{ 1.
N, k > q. Since P and deg P }, k max{l, 'l'K}. Using the compactness of K and the homogeneity of 4K and is an easy consequence from (1).
k were arbitrary,
weobtaino
Corollary 4.1.10. Let S C sup(exp(u):
the last equality D
be circled and bounded. Then
u E £(C"),
<max{l, c1s}.
(*)
Proof For abbreviation we denote the lefthand side of (*) by First we will show that
L5 = sup(LQ: f a balanced domain, S C Obviously, L5 >
=:
Assume now, there is no equality. Then Ls(a) > A >
some a E C". Choose u E £(C't) with uls
for 0 such that exp(u(a)) > A. Observe
that
Sc Exploiting that Thus
S
:=(Z EC": exp(u(z)) < l+e},
e>0.
is circled one easily finds a balanced domain S C C .'R. Ife —÷ O.thenA < exp(u(a)) 5
exp(u —log(1 +e))
contradiction. Fix now a balanced domain f S. Choose an increasing exhausting sequence of circled compact sets K1 C with U mt K1 = c. We know that ?
4.1 Special domains
j
ThenP(a.r) C
EN. Fora
._lonP(a,r),jj0.
Then,
< I on IF'(a,
and CI
E
with u <
E
349
r), J >
? Ja,and Therefore, u
I on ?(a, r). Since a was arbitrary, we obtain
I on Hence inf{CIK: K C K compact and circled } = Taking into account that = max{ I, for circled compact sets (cf. Lemma 0 4.1.9), the proof is finished. u
}
After all these preparations we present now the announced generalization of Josefson's result.
PropositIon 4.1.11. Let S C C' be pluripolar and circled. Then there is an h E h
0, With S
C
Proof The proof will need several steps. Step I: It suffices to prove the proposition in case when S is bounded. := Sn 3(0, J). are Assume S to be arbitrary. Then S = where bounded, circled, and pluripolar. Assume now that there are U) E 11) 0, Without with Si C loss of generality, SUp{u1(Z): z 1. Then, using the Hartogs lemma we find a point a E and a subsequence (uJ(k))kEN with gives a function in uJ(k)(a) > 1/2. Then, u := Step 2: Let S be bounded, circled, and pluripolar. Recall that
= sup{h(z): h Then it suffices to know
=
his S 1).
00.
In fact, let us assume that D = oo. For any m that 4s(zm) m. Therefore there are u1 sup{u1(z): z E
as wanted.
=
I
and
E
N there is a point Zm E with
lim sup{u1(z): z E S} =
with uj(k)(a) As above, we pass to a subsequence a Further, we choose a new subsequence := B,,.
< l/e on S. Then u :=
such
0.
1/2 for a certain such that
is the function we like to have.
Step 3: We claim that = 00. Suppose the claim is not true. Thenexp(23), j E N. We set vj := u1(sup(u1(z): z E E,, with Observe that v3 Again, as above, the Hartogs lemma allows to find a E B,, and a subsequence with v)(k)(a) ? 1/2. Finally, we put V :=
1_I
k=I
1/2&
Vj(k)
350
4 Existence domains of special families of holomorphic functions
It is clear that 0
v
6 ,I'S340m. Moreover, for z E V we obtain
=
fl
v(z)
0.
Therefore, V is pluripolar: contradiction. Applying (*) we conclude that Now, choose r > 0 such that 6 Sc B(O,r) and put M := z 6 B(O,r)}}. Since 2 IIzII/r we find another number R > 2r such that M(M + I) if R/2 IIzII < R. Recall that S is pluripolar, i.e. there exists a u 6 u < 0 on B(0, R), such
thatS c
Then, fore> Owe define I
max{eu(z) + log(M + I).
Izil
1k
> R;
well defined, it belongs to £(C"), and u1 0 on S. In virtue of Corollary 4.1.10 we conclude that max{O, log c1s). In particular, if IzIl < R, then UE is
eu(z) + log(M + I) Finally, e —+ 0 contradiction.
max(O. log
max(O. log
leads toM + I
max{l, M} for
IIzII
< r; D
Now we give the proof of Theorem 4.1.7. Proof of Theorem 4.1.7. Let D of(i) and (ii).
be
as in the theorem. We already know the equivalence
If I <
Assume condition (i). Let f E constant function. We write f(z) = and we recall that lQ3l <
I
Q3(z),
on D. Then
<
I
I. f(O) = 0, and let f be not a z 6 D. where Q1 E
on the domain (Exercise) G
mt b 0. The maximum principle gives that even
< I on G. Therefore, the series is locally uniformly convergent on G and gives a holomorphic extension of f to G. Hence, G C 0: (iii) is true. (iv): Suppose that (iv) is not true. We find, as usual, open sets U1, (iii) with the standard properties. Then there is a 6 U2 \ Choose p 6
with p(a)I > sup(Ip(z)I: z 6 0) =:
c.
Then f := (p
—
U2
cY' 6 R°° fl
Thus f is not extendible to U2; contradiction.
That (iv) implies (i) is obvious. What remains is to prove that (I) 4=> (v) when D is bounded. So, starting from now, D is assumed to be bounded. (i) => (v): Recall that h (respectively h) is semicontinuous from above (respectively from below). Hence h is continuous at all points, where h and h coincide. The remaining set is given by S := {z E D: I:(z) < h(z) = h(z)). Using the result of E. Bedford & B.A. Taylor (cf. Theorem 2.1.41), 5 is pluripolar. Hence, (v) is proven.
4.1 Special domains
351
(v) (i): We have to show that h = k". Since h is homogeneous it is clear that N(h) is a circled piuripolar set. So Siciak's result (cf. Proposition 4.1.11) can be Since D is 0, with N(h) C applied. Therefore we find u E .9'S340rn, u I on D. Recalling that u and h are logarithmically bounded we may assume that u 0 < t < 1. Moreover, v1 < I on D (use psh it follows that v, := hi_lu: E that N(h) C u'(O)). Then, in virtue of Lemma 4.1.9 we have
v,(z) =
h(z),
(*)
z E C".
is a set h(z). Since 0, then letting t —+ 0 leads to h(z) If u(z) of Lebesgue measure zero, we finally get, using Proposition 2.1.12, that h = h
0
everywhere.
C (0, oc) be as in
Example 4.1.12. Let the sequences (aj)ri C 13E and Example 4.1.6. We define u: C3 —÷ [0.00) as
u(z) :=
max{log Izi
—
log 1Z31}.
z E
C3.
j=I u
is psh on C3. Then the following function
h: C3 —p
h(z) := exp(u(z)) + Izi,
belongs to ll'S34om(C3). h(z) > 0 if z 0. Therefore. D := Dh := (z E C3: h(z) < l} is a bounded pseudoconvex balanced domain. Observe that N(h) C {z E C3: Z3 = 01: in particular. N(h) is pturipolar. Hence, D is a fat 7('°dornain of
holomorphy. Moreover, using Example 4.1.6 we easily see that E x E x (0) C
= D, i.e. the of hotomorphy D that mt We know for an difference of the polynomially convex hull of D and D is 'thin', but it is, as the example showed, in general not empty. The example is a 3dimensional one. In fact. it turns out that such an example is not possible in C2 (cf. [Sic 1985]).
Theorem 4.1.13. Let D = Dh be a bounded balanced domain in C2. Then the following pmperlies are equivalent: of holomorphy; (1) D is an (ii) N(h) is pluripolar and ii = lie, where h1 denotes the lower semicontinuous regularization of h; (iii) D is polynomially convex and D is fat. For the proof the following result of J. Siciak [Sic 1985] will be used.
Lemma 4.1.14. Letu
E
S3e(C). Assume that N(u) is polar and that
a+tog(l +IidI) where a, b E R.
+IidD.
4 Existence domains of special families of holomorphic functions
352 Then
kEN, P E
= log
(l/k)logIPI
u}.
Pmof The proof will be based on some classical facts on systems of extremal points due to J. Siciak (cf. LSic 19641). First, we define the function b : C x C —+ [0, co), Iz
— wI
if (z, w)
—
E
C2
ifz=oo, WEC
—
if z E C, w = 00 ifz=w=oo.
0,
Using the lower estimate for u shows that b is upper semicontinuous on C x C. Then, for j E N, the function
:=
[0,00),
:
b(Zk,
is upper semicontinuous, too. So we find
=
zt),
z
=
(zo
Z3).
E
with
Put
p1IIk _<exp(u)}. In that situation Siciak's result tells us that
•
is harmonic outside of F;
log
• log
F \ N(u).
that
extended maximum principle 6) for = subharmonic functions implies that u $ log outside of F; in particular, log outside of F. which completes the proof of the lemma. 0 log
u
on
Therefore,
the
Pmof of Theorem 4.1.13. (i)
I)
Obviously, (iii) (i). (ii): Observe that it suffices to prove the following two inequalities:
sh function satisfying the condition of Lemma 4.1.14 (recall that D is assumed to be bounded and cf. Theorem 4.1.7). Therefore, log
I)
kEN, P E .'Pk(C), (1/k)logIPI 6)
See
(Rans 1995], Theorem 3.6.9.
u}).
353
4.1 Special domains
Takeapolynomialp E ?(C)with(l/k)logIpI
u,degp k. Put Q(zI.z2) := Then Q is a homogeneous polynomial
when p(z) = withIQIIIk h(., I). (ii) (iii): We have
D=
{z
h(z) < i} =
E C2:
Moreover, since N(h) is pluripolar, h = h*. Hence, D is fat.
We conclude the discussion of balanced domains with the following example
Example. Fors E N let Q3 s}
E
)%om(C"), 1 cf <s, be given such that = 0.
30'EC'
Moreover, letd1 := deg Q1 and D := {Z E
fl
0.
=
E (0, oo), I <j <s. with
1.. Then.
< 1)
IQ
is an unbounded pseudoconvex balanced domain with continuous Mmkowski function.
Lemma 4.1.15. For D the following properties are equivalent: (i)
(ii)
D is an
domain of holomorphy;
eQ for all j;
(iii) (iv) There exists a Q
deg Q > 0, such that
E
< I on D.
Proof. The implications (ii) (i), (1) (iii), and (iii) ; (iv) are trivial. So it remains to verify (iv) (ii): Using homogeneity it follows from the assumption on Q that 1Q1l/d < Q p.t/d1 where d := deg Q. Therefore, the function lJd Q11Pu'di is psh and bounded from above outside an analytic set, so it extends as a psh function to C". In particular, it is constant. So there is a c > 0 with
1Q11/d
= cIQ1IP/d
.
.
354
4 Existence domains of special families of holomorphic functions
a certain unit vector b E C" we
Now, fix a j and choose can write
+Ab)= +
#0,
XE C.
Ab) = Amq(A).
AE
C, with q(0)
0.
Therefore, Amq(A) =
A E C, where gj is holomorphic near 0 0 and g1 (0) # 0. From this representation immediately follows that E Q.
Finally, we mention the characterization of those balanced domains that are
0(N) (D,
of holomorphy 7) (cf. LJarPfl 1987]).
Proposition 4.1.16. Let D = Dh be a balanced domain in C" and let N ? 0. Then the following properties are equivalent: (i)
D is an
SD)domain of holomorphy;
=
(ii) There is a sequence (QJ)JEZ+ C <
Observe that, in case N = Proof. The proof of (1)
,'
0,
0 or deg
= j, such that
this proposition is the same as Corollary 4.1.4.
(ii) is left to the reader (compare the proof of Corol
lary 4.1.4). (ii)
'.
U2flD ç with f I
(i): Suppose that (i) is not true. Then there are open sets 0
U1 C E 0(U2) U2 connected, suchthatforeveryf E = flu. Now, in virtue of the open mapping theorem we conclude that
U2,
VaEU, 3c(a)>0
1a1 <
fE
c(a). Hence, (ii) gives h 1 on U2. Then, applying the In particular, 1Q1(a)l 0 maximum principle, it follows that h = 1 on U2; contradiction. Remark. The following question seems to be open: What are characterizations of bounded pseudoconvex circled domains that are .N°°domains of holomorphy
Reinhardt domains. We already know that any pseudoconvex bounded complete Reinhardt domain is an R°°domain of holomorphy because it is balanced with continuous Minkowski function. Here we will study arbitrary pseudoconvex Reinhardt domains. For further need we fix the following notations.
V1:=((z1
Recall that &N)(D 8D) := (1 E 0(D):
j=l bounded on D).
4.1 Special domains
Moreover, if a = (al .
.
.
, as,) E
355
then we put
is well defined. C" (a) then the product Iz 11a1 ... Let us repeat the characterization of pseudoconvex Reinhardt domains (cf. Proposition 1.9.19):
A Reinhardi domain D C C" is pseudoconvex
log
n}then(zi where log D := {x image of D.
E
IR": exp(x) := (exp(xi)
D C R" is convex and, if
exp(xn))
E D} is the logarithmic
From that description it is clear that
intD\DC i.e. if D is not
V0.
fat, then the difference between
mt D and D is
contained in some
set. In particular, a holomorphic function on D with 'slow growth' near that set extends holomorphically to mt D. We are led to study relations between geometric properties of log D and the stnictare of, for example, analytic
DefinitIon 4.1.17. Let X C R" be a convex domain by E(X) the vector subspace of R", that satisfies (a)
(think of X as log D).
We denote
X + E(X) =
for any vector subspace F C R" with X + F = X holds that dim F dim E(X). (b)
It is easily seen that E(X) is
of E(X). Let X ç
IR"
(aa)aEA C R"
uniquely defined and
that any F as in (b) is a subsp
be a convex domain. Then, using convexity, we find A C that
and
such
X=
mt
fl
aEA
:= {x E W (x — 0a. a) < 0), i.e. X is given as the 'intersection' of halfspaces. In particular, if a Reinhardt domain is given as where
D :=
mt
fl Da,
aEA
where
:=
{z
C"(a):
.
< Ca),
Ca > 0.
356
4 Existence domains of special families of holomorphic functions
then the associated vector subspace E(X) of X
:= logD = mt
fl{x ER": (x,a) aEA
can
be explicitly calculated as
E(X)={x ER": (x,ct)=OifaEA}. The Da 's are socalled 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. [JarPfl 1985] and [JarNI 1987]).
Theorem 4.1.19. Let D c C" be a Reinhardt domain of holomorphy. Then the following conditions are equivalent: (i)
D is an
domain of holonwrphy;
(ii) D is an
of holomorphy:
(iii) D is fat and of rational type:
(iv) D =
mt
Da for some A C
where Da are elementary Reinhardt
domains as above. In any boundedfal Reinhardi domain of holomorphy is an ,7t'°° domain of holomorphy.
Proof (ii)
(iii): It is clear (cf. the remarks at the beginning of this subsection) that D is fat. Put X := log D and F := [E(X)1 fl Assume that E(X) is not of rational type. Then it is clear that dim F > dim E(X). and let be the expansion of f into its Laurent series. Let f E (*) We will show that a,, = 0 whenever v E(X)'.
357
4.1 Special domains
Suppose for the moment that (*) is true. Then for x E X,
+ v, v)) =
>
E F we get
V
avexp((x,
> vEE(X)1flZ"
VEZ"
Consequently, the series is convergent in
logz := (log IziI
:= {z E
E
X + [E(X)' flQ"]'}.
C D. Or in the real picture, X + F
Since D is an
X which contradicts dim F > dim E(X). Now we are going to prove (*), Fix a u
such that (w. v) >
0.
C
E(X)1. Choose w E E(X), 11w II = 1. and some x° E X. Put x(1)
Fix 0 < N < (w, t')
exp(x° + 1w), t E lit Since x(t) E D, the Cauchy inequalities imply <
sup{If(z)I: 1z31 = x1(t). j = I
n}
II080(x(t))_N(x(t)t)I v)/N))_N,
I ER,
independent oft. where M := sup(öD(x(l))exp(t(w, v)/N): t E IR}. In order to get av = 0 it suffices to show that M = oo. Suppose that M(I + lIx(z)112r"2}. Ift E T, then
where c := exp ( — (x°, v)) is
Thus,
exp ( — 21(w,
v)/N) +
exp
+ tWj — r(w, v)/N))
= pD(x(t)) for I >
Consequently, T is bounded. Therefore,
Now, we
will estimate PD. Let d := dist(x0 + Rw, aX) = dist(x0, DX). Fix a t E R and let Z E i)D, 0, such that PD(X(t)) ? 0.5hz — x(1)II. Write logz = x0 + 1w + u with Z1 " 'Zn Then we obtain that uE lull > d. Fix a j = j(t) such that 1u11 > PD(X(t)) where
II
d0 :=
— 1)/2. choose Jo such that there is a sequence (1k for all k and tk = oo. Then
Finally,
M > pD(x(lk))exp(tk(w, v)/N)
contradiction.
+ tkWj(
E R,
Ik
to with j (1k) = Jo
+ 1k(w, v)/N) —*
oo;
358
4 Existence domains of special families of holomorphic functions
(iii)
(iv): We only have to prove that
a #0, such that
\ Dthereexistsa E
foranya E
Suppose for a moment that (*) holds. Fix a b e mt D and a
b
(*)
< IaaI} =:
C {z E
1), a E
such that
D. where
:= Choose V
{z
C"(a): IzaI < luau.
:=B(b,r) C Dandtakeaz E V. Wemayassumethatzi =
o< t
=zj =0
EN, converging 0, 1 to z. Since l(Zk)al < laal, k E N, it follows that aj £. Thus za is = laaI. Then, in virtue of defined and lzaI laal, z e V. That b D* yields the maximum principle, we get Iza I = laa I on V; contradiction. Therefore, using the fact that 1) is tat, we get and
. .
D=intD C
n. Choose points
j
C DU(ChI\(C.)?!).
fl
mt
E
aE(C.Y'\D
Then, taking the interior on the righthand side gives D = mt flaE(c )"\D D. So it remains to prove (*). Fix an a (C.)" \ D. Put := loga logD. Recall that log D is convex and of rational type, i.e. up to an 'rational' orthonormal isomorphism we can write log 1) in the following form
logD
x
=
where is a convex domain in with E(D1) = 0. with vertex at By D2 we denote the open convex cone in
that is spanned by D1. Since E(D1) = 0, does not contain any real affine line. l'herefore, '12)). where '12 runs through an nonempty open D2 C (x2 E Rn_k (x2, '12) < We choose an '12 subset of 0 such that its coordinates are all integers. Then the inverse rational isomorphism leads to an nonzero rational vector a with log 1) C E R": (x, a) < a)). Without loss of generality, a may be chosen in V'. Taking the exponential leads to the claim (*). The remaining implications are trivial; hence the theorem has been proven. D
Recall that the inclusion map p00(1)) —÷ (9(O+)(1)) is continuous, when is equipped with the following sequence of norms (IIISNIID)N>o. Then, in view of the Banach theorem, either = or .l?oc(D) is of first In the case of Reinhardi domains, that Baire category in the Fréchet space of holomorphy, always the second case holds, i.e. there are a lot of are unbounded functions belonging to
359
4.1 Special domains
Proposition 4.1.20. Let D ç C"
be
a Reinhardi domain that is an 3€°° domain of
ç
holonwrphy. Then
Proof Accordingtomeorem4.l.I9therearea
Z",a
such that DC Da =: {z E C"(a): ZaI < Now, we are going to construct f E (9(Da) with urn If(z)I = oo when z approaches a within D. Put E —* C,
Q(A)
log(l/(l — A)).
It is clear that = oo. Moreover, for any e > 0 we have that (I — A)IQ(A)Ih/E remains bounded on E. Now, we introduce the following holomorphic function on Da:
1(z) := It is clear that
—+
Z E Da.
oo and that (1 —
is bounded on Da.
ZED
What remains to show is that
isboundedonD.
(*)
In fact, let N > 0 be arbitrary. Choosing r := N/IaI gives that f E The fact (*) is a special case of the following result we quote without giving its proof (cf. [JarPfl 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 ncircled (9(N)(D SD)domaln of holomorphy, 0 < N < I, has to be a fat domain. Conversely, the following is true (cf. [JarPfl 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