THE INTEGRAL FORMULA ON COMPLEX SUBMANIFOLD AND APPLICATION

1998,18(4):466-476

THE INTEGRAL FORMULA ON COMPLEX SUBMANIFQLD AND APPLICATION 1 Chen Shujin ( ~ ;kJ.t ) Department of Mathematics, Xiamen University, Xiamen 361005, China Abstract

First, this paper gives another integral represe;ntation on bounded convex

domain in complex submanifold. Second, using this integral representation, the author easely gets the strengthen consequence of Elgueta.

Key words

1

Complex submanifold, integral formula, convex domain, application.

Introduction

We know that every sttrictly convex domain with Ck-boundary is a strictly pseudoconvex domain with Ck-boundary, where k ~ 1. After an appropriate local change of holomorphic coordinates, the converse is also true.

{z E

en :

en

e

=

be a strictly pseudoconvex domain with k boundary. More precisely D p(z) < O}, where p is a k real valued strictly plurisubharmonic function on a

Let D C

neighborhood liD of

e

b, and grad p i=- 0 on

eo.

Suppose F1 , · · · , Fm are holomorphic functions on liD' set

on transversally.

D = Z(F1 , · · · , Fm ) n D. E aD, then for any E > 0, we have constant 6(0 < fJ < E) and numbers nl,··· ,nn-m-l

We assume Z(F1 , · · · , F m ) meets

Set

Let ( E {l, 2,· .. ,n} so that the map z - t (Znl -(nl'···' Znn_m_l -(nn-m-l' F(z, (*), F1(z),· .. ,Fm(z)) is biholomorphic transformation from the ball B( (*, fJ) onto the neighborhood "c: of point 0 in the space of complex variables w = (Wl' . · . ,wn ) , where

nan

a2

L a:. (C)(Zj - (j) + ~ L az.:z. (C)(Zi - (t)(Zj - (j),. so that the original image G,. of a stirctly convex subdomain 'V(. of lI,. (where 'V(. have Coo F(z,C) =

j=l

boundary and

V(. ell,.)

J

i,j=l

a

J

satisfy

Thus, we can generally transact a problem on strictly pseudoconvex domain as a problem on stirctly convex domain. For instance, by Henkin[l] and Elguete[2], the problem that holomorphic 1 Received

Aug.29,1996; revised Jul.7,1996

Chen: INTEGRAL FORMULA ON COMPLEX SUBMANIFOLD AND APPLICATION

No.4

467

functions in submainfold extends to on strictly pseudoconvex domain is turned to the problem that holomorphic functions in submainfold extends to on strictly convex, by the method of identification of sheaves. Therefore, that we put our researching problem on strictly convex domain is fundamental and important,

First, this paper gives another integral representation on bounded convex

domain in complex submainfold. Second, using this integral representation, we easely get the strengthen consequence of Elgneta[2].

SOl11e Lemmas

2

Definition 1 Let D in en be a bounded convex domain with e 2 boundary. Then there is a e'2 real valued function in l/jj such that D {z : Vjj : <1»(z) < O} and grad <1» =P 0 on every point of aD. Furthermore, if the real Hessian of <1»( z) is strictly positive for z E Vjj, then the

=

domain D is called a stirctly convex domain.

en,

Definition 2 Subset T ~ en is called a real( complex) hyperplane in if for z E T, T - z is real(complex] linear subspace of Lemma 1 Let D be a convex set in = R 2n different from R 2n , nO be the interior set of D, then for any xO E jj - DO, we have hyperplane 7r passing xo, so that D is located in

en. en

one side of

7r, 7r

is called the support hyperplane of D.

Definition 3

Let D ~

= {z E l/jj

en be

an open set with

e2 boundary aD, p be a e2 real function

p(z) < O} and for z E aD, dp(z) the real tangent plane of aD at point (, that is

on

VfJ,

so that D

:

t(

= {w

E en :

2n 0 ( . )

L

:x~

j=l

Xj(w - ()

=1=

o. When (

E aD, we use

= O},

J

where Xj(() is real coordinate of ( E en, so that (j

= Xj(() +iXj+n(().

plane contained in t, is called complex tangent plane of

The maximum complex

aD at point (, and T,

is its expression,

it means

Lemma 2

A complex tangent plane is an analytic plane tangential at ( E aD.

Proof In fact,

t

op~() (Wj -

(j)

j=l Oej

if and only if

t( op«() -

2 j=l

OXj

op(() L --Xj(w j=l OXj n

and

=~

()

t, as

i

op«()

OXj+n

+L n

)(Xj(w - ()

op(()

+ iXj+n(w -

--Xj+n(w - () j=l OXj+n

= 0,

~ op(() ~ op(() L.J --Xj+n(w - () - L.J --Xj(w - () = O. j=l OXj j=l OXj+n

())

=0

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ACTA MATHEMATICA SCIENTIA

Since xj(i(w-()) that

= -Xj+~(w-()

and xj+n(i(w-())

to;~~)(Wj-(j)=O if and only if

j=1

en \

wEt, and (+i(w-()Et,.

L

8cp(() ---a((wj - (j)

j=1

en

Let D C

3[3]

en,

cp(z) < O} is a convex domain in

Vi) :

n

D,

= Xj(w-(),j = 1,···,n, this implies

1

Corollary If D = {z E D,( E D, there has

Lemma

Vol.1S

then for any w E

t- O.

1

be a strictly convex, then for holomorphic function I(z) on

we have (1) '_

= {'l/J1,···, 'l/Jn)(t) = grad~, Fa(() . go = (gOI' · .. ,gon )1 t ) , a = 1,2,· . · ,m; moreover where zED, 'l/J

F

B m (() =

(_1)m(m+1)/2

IV~(()12

1~~

j=1

- Zj )gOj'

) (8F1 8Fm o( ,oo"---a(,d(,oo.,d( ,

=;= (_l)m(n-m)( _1)m(m+1)/2(n _ m)! (

IV~(()12 =

L

8(F1 , · · · , Fm )

8((n-m-1,···, (n)

) -1

d(1/\'" /\ d(n-m,

2

I

O(~ll:::' ~m) t- 0,

1~jl <...
em = (_1)m(n+1)( _1)m(m+1)/2 I(n Lemma 4

n

l: ((j

Fa(z) =

, (1m)

1

m)!(27ri)n-m.

Let M(z) be a holomorphic function of domain D in

en, h(k) is the following

differential operater: n

h(k)(M) = kM

+ L(Zj

- bj )Mj , k > 0; h(O)(¥)

= M.

j=1

Where b is any fixed point in domain D, M j == 8M/8zj , j = 1,2,···, n. If the two holomophic functions M 1 (z) and M 2 (z) of D satisfy h(k)(M1 ) == h(k)(M2 ) , then M 1 (z ) == M 2 (z ) on domain D. Proof We only have to spread M 1 (z) - M 2 (z) to n multiple power series at point b, and note that h(k) is a linear operater, then we get the result. Corollary

If M 1 (z) and M 2 (z) are holomophic on domain D C

en

and satisfy (2)

then we have M 1 (z) == M 2 (z). Lemma 5 Let D C be a strictly convex domain. If
ob, and

en

denote zED,

No.4

Chen: INTEGRAL FORMULA ON COMPLEX

where fJ

~

1 is integer, then for k

= 1,2,· .. , n,

SUBM~NIFOLD AND

469

APPLICATION

we have

(3)

= det(n)( 1/;, Yl, · · · ,Ym, 8,1/;, .. · , 8,1/;) /\ B~ (), and

Proof We set E(()

I

I = F ( Z1, · .. , Zk

+ h, . · . , Zn) -

F (ZI, ... , Zk, ... , Zn) _

h

r

Jan

s1/;k ()
I

(1/;(),( - z)s+l .

We have

Let d

=

inf_ 1(1/;(),(- z)1

(EaD

1(1/;(),( -

> 0, and if Ihl z -

h)\ ~ 1(1/;(),( - z)I-l hll1/;()1 ~ d -

Therefore, we have 1--+0 when h Corollary

is sufficiently small, then there existe

--+

fJ

fJ

> 0 such

that

> O.

0, this implies that (3).

Under the same assumeption of Lemma 5, we have

h(k)h(k-l) ... h(l)[F(z)]

-1 -

3

ab

h(k)h(k-l) ... h(l) [


A 1\

BF(i) m

~

•

(4)

Integral Formula and Analytic Extension

=

In this part D indicates a strictly convex domain in C"; iJ Z (Fl, · · . , Fm) n D, Z (Fl, ..., F n t ) meets eo transversally. _Theorem 1 If f(z) and j(z) = h(n-m-I).~. h(l)[f(z)] are holomorphic on b. continuous . on iJ (simply denote Z

f,l

E A(D)

E b, and \\7~()1 =1= 0 on

= O(iJ) n CO(D)), where I = 1,2"", n -

eii. we have

m - 1, then when

(5) where C(m , I) = (n-m-l)! (l-I)! Cm, Cm = (-1)' m(n+l)(_1)m(m+I)/2/(27ri)n-m(n _ . - m)! . , G(() !(()/('l/J((),( - b)n-m-l, here bE iJ any fixed point.

=

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ACTA MATHEMATICA SCIENTIA

Proof First assume

1, f

h(n-m-l)···h(l) [

Val.18

E O(D). By computing we get 1

('f/J((),( - z)l

]

= (n-m-1)!(.,p((),(-b}n-m-l

(6)

(1-1)!('f/J((),( - z)n-m

denote the right side of expression (5) as Q(z), and the differential operator

h(n-m-l) ... h(l)

acts on Q(z), then by the corollary of Lemma 5 and expression (6) we obtain

(7)

By Lemma 3 and the right side of expression (7), we have

h(n-m-l) ... h(I)[Q(z)] = !(() = h(n-m-l) ... h(l)[f(z)].

By Lemma 4 we get Q(z) = !(z), that is expression (5). Second, Let f,! E A(D). For D is a convex domain, we have function sequence fj(() E A(v), where v is the neighborhood of b, so that .lim sup I!(() - !j(()1 )-+00

(ED

hence, .lim sup I!(() - !J(()I )-+00

we write

(ED

= 0, = 0,

r

(1- I)! Gj ( ( ) F • [Q(z)] = Cm (n _ 1!1- _ 1)! JaD (.,p((), ( _ z)' 1:;;(.,p, gIl · · ., gm, o,.,p, · · · , o,.,p) t\ B m(0

(1 - I)! +Cm(n _ m -1)!

r

G(() - Gj ( ( ) _ z)'

JaD (.,p((), (

1:; (.,p ,g1, ... ,gm, o,.,p,,,·, o,.,p) -

-.

t\

F

e;(()

(9)

where Gj(() == !J(()/('l/J((),( - b)n-m-l. Therefore, by (8),(9) and the proof of the first part of Theorem 1, we have

Q(z) =,linl !j(z) = f(z). )-+00

Remark When m = 0, in expression (5) C(m, I) = C(O,I) = (~-=-W! = n!w((), we can write (5) as:

(27ri\nn!'

B6(()

r

(1 - I)! . 1 G(() f(z) = (n _ 1)! (27l"i)n Joo (.,p( (), ( _ z)l 1:~(.,p, o,.,p, .. · , o,.,p) t\ w((), where G(()

(10)

= h(n-l) ... h(l)[!(()]/('l/J((),( _b)n-l. Practically, pick 1 = 1 (now the denominator

of the integral kernel is linear about z), theon[4]

f(z)

1

1

r

= (n _ 1)! (27l"i)n Joo

G(() (.,p((), ( _ z) 1:;(.,p, o,.,p, .. · , o,.,p) t\ w((),

( 11)

No.4

Chen: INTEGRAL FORMULA ON COMPLEX SUBMANIFOLD AND APPLICATION

471

where G(() == h(n-l) ... h(l)[f(()]/(-,p((), ( - b)h-l. Moreover, we can write (1) as

f

1

f(()

J

--

(12)

f( z) = (27ri)n eo ('ljJ( (), ( _ z) n 1:;~( 'ljJ, a('ljJ, · · · , a(.,p) 1\ w(().

In 1979, autnor indicated in paper [4] that the two type integral representations of holomorphic functions on n-multiple circular domain in en could be induced by expressions (11) and (12), thereby the integral representation of holomorphic functions on bicircular domain given by Tellllyakov[5] in 1954 is the special case of (11) and (12). In 1985, Gindikin and Henkin[6] extended expression (10) to analytic functionals. Theorem 2 If f(z) and j(z) belong to A(D), then the function defined by

(13)

oi»,

and for any z E ii, f(z) == S(z). is holomorphic for all zED U (aD \ Proof According to Lemma 3, for any ( E aD, we have grad ~(() -I O. By Lemma 2,

E -(()((j j=l,8(j a~

n

Zj) == 0

defines an analytic plane tangent at ( E aD. Thereby for all (E

eii and z E DU(aD\aiJ), we

have (-,p((), ( - z) i- 0 and it is easy to know that the differential form under the integral symbol of the right side of (13) is holomorphic about zED U (aD \ Specially when restraining

z E

b, we can (1)


take proper a : 0

eb:

1 so that:

for any (E ei: we have jtllg~(a()1 f. 0;

(2) we can write z ==

C( m, 1)

1

G(a() •

. ('ljJ( () a

aD

az,

where

zED.

Thus by Theorem 1 we get

-

,a~-z

~

-

)1 det(.,p, gll ... , gm, a(,p,· · · , a(.,p)( a~) (n)

By Lebesgue Theorem, we can take the limit of a

~

1\

F

B m (a() = f(z).

(14)

1, under the integral symbol, Therefore

by (13) and (14) we get: for any zED there has 3(z) == f(z). Theorem 3 Let D == {z E lJjj : ~(z) < O} be a strictly conv~x domain in en with sufficiently smooth boundary. If function f(z) E AS(D) == O(D) n eS(D), s == 0,1,2"", then we have function S(z) E AS(D), so that S(z) == f(z) on

D.

Proof By Theorem 2, we only have to consider the case of z == z" E

fJ

>

Denote k == n - m, let B(z*, 6) == {z E en : Iz - z* I < 6}, for a fixed sufficiently small 0, we can set (if necessary by means of local coordinate transformation)

D n B(z*,6) == {z For

sb.

D n B(z*, 8)

E B(z*,6) C

k

p=l

==

Zk+l

== 0""

,gm

C en is stirctly convex, we have w*(z) == grad~*(z*)

where ~*(z) == ~(z)IDnB(z.,b) == ~(Zl'

sb == I: oii;

en : gl(Z)

so that there has

1/J; ==

Z2,"', Zk,

~~.

i- O.

==

i-

0"",0). Now we express

Zn

== O}

0 for any z" E

an as

As on aiJ we have the equation

p

k

k

p=l

p=l

E w;(()d(p + E w;(()d(p == 0,

eb,

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ACTA MATHEMATICA SCIENTIA

we can write on

Val. 1 8

ob, '11~ J

-

-

-

-

-

d(p = '11! d(j (modd(l' . · . ,d(k, d(l, .. · , [d(j] , ... ,d(k), P

by this, we get where d(

d(U] 1\ d( = (-l)P-j('I1jj'l1 p)d([p] 1\ ac,

= d(l 1\ . · · 1\ d(k, d([p] = d(l 1\ · · . 1\ [d(p] 1\ ... d(k.

Hence on

ab,

we have (15)

where L('I1*) is Levi determinant:

L(\I1*)

=-

0

'11*1

'11!1

'I1~1

'11 k '11: 1

'11!k

'I1~k

\11*kk-

84!*

,'11; = 8 Zj ' 'I1j = y Zj' w:.]

1= 0 on eb. Let D S = 8 Islj8zfl ... 8z~n8z~1 ... 8z~n, where 181

L('1t*)

here

Z+ Z+.

eb, c eb.

that Xl == 1 in B(z*, i6) and Xl == 0 outside B(z*, 6), set X 2

2(z)=Cm

f

,~ -

Z

= ~' z, Zj

=1-

Take Coo function Xl, so Xl. Let

de.tAI\B~(().

(ljJ«()f.«() )k

} oii ':

8 2 4> *

= lal+If31 = a1+·· ·+a n+f3l +.. ·+f3n,

= (a,f3) E x Without loss of generality we can assume that z" E

8

84!*

-s (n)

Where det(n) A = det(n) ('l/J, g1, · · · ,gm,[),'l/J, · · · ,[),'l/J). We divide as two cases as follow: (1) For 8 = 0,1,·· ., k - 1, we can write

2(z) = C(m, s) where H(()

in (1jJ«()~(~) z)k-.1;~

= h(k-l) ... h(k-.s)[!(()]/ ('l/J((), (

A 1\

B~«(),

(16)

- b)s. Because of the integral kernel of the right

side of (16) is holomorphic about z, then we have a~J~) = 0 for j = 1,2,···, n. So in fact we only have to consider the case of (3 = First, for 0

:s I :s 8 l~()

D .::. z = Where

C=

1,1

C-

:s 8 :s k -

f lan

1, we can write

H(()Iz(() d A F(() ~ [z (1jJ«(), ( _ z)q (;~ 1\ B m := '::'1 z)

C( in, s) (k - 8) · · . (k - s

'l/Jl((),·· ., 'l/Jn((), and

J

o.

+ I-I), 1 ~

q= k- s

+I

~ ( )

+'::'2 Z ,

~ k - 1, and I, (s) consists of

No.4

Chen: INTEGRAL FORMULA ON COMPLEX SUBMANIFOLD AND APPLICATION

473

obviously, 3 2(z) = 3 2(z*).

lim

z-z*

(17)

.z:ED\8D

of Supp Xl C B(z*, 6), and z ~ z*, about 3 l (z ) we only have to consider the case of (, z E jj n B(z*, 8). For a fixed sufficiently small 8 > 0, we can set (if necessary, by means local Becaus~

coordinate transformation)

iJ n B(z*, b) = {z

E B(z*b)

c

en : Fl(z) = Zk+l = 0,·", Fn1(z) = Zn = OJ. k

I L: w;(()((j

For (E afJnB(z*,b) and z E fJnB(z*,b), we have where

"i

j=l

> 0 is a constant, and by. computing we get

-zj)1 ~

k

Tl

L: I(j

j=l

-ZjI2,

k

det( 1/;, gl, ... , gnl' 8(7/;, ... , 8(7/;) = (-1 )n-k (k - I)! (n)

B~(()

,. ,

Thus

l~l(Z)1

< T2

1

L( -1 )j-lw; ./\. 8( '11; . j=l ~tJ

= (_I)(n-k)(n-k+l)/2kId(.

k

_

eE8DnB(z*,b)

(L:

a(d()

I(j - Zj 12 )q

,q

(18)

(19)

= 1,···,k -1,

j=l where

T2

> 0 is a constant. Let

t 2j - l = Re((j - Zj), t 2j

= Im((j -

Zj),j

= 1,2,,··, k, then we

can get k

a(d()~r3dt2···dt2h, and LI(j-zjI2~r4(t~+t~+... +t~k)' j=l

where

T3, T4

> 0 is a constant. Using spherical coordinates, we can get (20)

Taking account of (17) and (20) we obtain lim

z-z*

zED\t)iJ

Hence

n' 3 (z) = n' 3 (z* ).

(21)

n 1'3(z ) E eo(.D)(O ~ 1 ~

Second, for 0

~

1= s

~

s - 1, 1 ~ s ~ h~ - 1). k - 1, we can write

where

Obviously, lim

z-+z·

zED\8D

3 4 ( z)

= '34 (

Z *).

(23)

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ACTA MATHEMATICA SCIENTIA

Vol.IS

By (18) and (19), we get

,. , () == C-1 _

'::'3 Z

8D pnB(z*,b)

==

where C 1

H(()I;(()X 1 ( ( ) d t(llT..* a- llT..* (llT..*(i) i )k e '£' ,Yl,···,Ym, ('£' , '£'

':"

':, -

(n)

Z

==

C( _1)n-k(k - 1)!( _1)(n-k)(n-k+l)/2kL Denote E(()

••• ,

a-('£' liT..* ) B F (( A nl

H(()F,;(()X 1 ( (

).

)

Using

(12), by (24) we get

3 3 (z ) =

c,

f_

k E(()L('1T*) . d(;: A d(. }8D nB(z*,lJ ) ['E '1T~(()((j - Zj)]k . p(()

(25)

p

j=l

Because of for ( E

1

ob, n B(z*, ~), z E B(z*,~) n (D \

oi», we have (26)

~; =

where w;(z*) have

=f

t

('1T;)() -

1=1

:*p~~g'1TPj((») ((j

- Zj). Moreover because of '1T p(z*) -:j:. 0, then

0, therefore we can find 61 , c > 0, so that for all ((, z) E B(z*, 61 ) x B(z*. 61 ) , we

Iw;(()1 > 2c > 0 and L(w*(()) =1= 0 as E(()L(W*)d([p] 1\ d( k

[~

j=l

w;(()((j - Zj)]kW p(( )

well as

Iw;(() + ~;I > c > o.

So by (26) we have

(-I)k+ pE(()L(w*)

==-------(k - I)W;(()(w;(() + q,;)

(27)

Denote

then when z ---. z", we get

=

c,

1-

8D

(-l)k+P J(()d(

k 1 [~ W;(()((j _ Zj)]k-l

j=l

A d([p] A d([p]

No.4

475

Chen: INTEGRAL FORMULA ON COMPLEX SUBMANIFOLD AND APPLICATION

k

[2: W;()((j -

Zj)]k-1

j=1

(28)

where a = min(<5, <5 1 ) . More, applying (15),(18) and (19), we get k

k

L(w*)[ 2: '11; (()((j - Zj )]k-1

L( -1)j-1 w; 1\ 8('11: 1\ d( j=1

j=1

~p«) ( 2L

= C2

_ !L.£L)

a(p r_ laD [2: 'l/Jj(()((j L(~)

k

j=1

~ p a(p

(29)

Zj)]k-l

where C2 is a constant. About (29), applying the method of getting (20), similarly we have lim

z-z·

D"S(z)

= D"S(z*), 1 ~ s ~ k -

(30)

1,

zED\8D

Hence D S 3 ( z ) E CO(D), 1 ~ s ~ k - 1, (2) For s 2 k, denote s

n'3(z) where H 1 (()

=C

= k - 1 + v, v = 1, 2, · .., we can write

liJ (1/J~)~~~';f~+v 1:;~

= h(k-1) · · · h(1)[/( ()]/ ("p((), ( '= ( )

~5

Z

=6

r

A A B.;'(() := 3 5(z) + 3 6(z),

(32)

b)k-1, and

H 1(()F,,(()X1 ( ( ) d t A 1\ B F ( (

JaiJ ("p((), ( _ z)k+v (:)

)

m'

obviously,

(33) Using the same method as (29), we can obtain

(34)

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ACTA MATHEMATICA SCIENTIA

where C3 is a constant. Repeeating the method of getting (34)

Vol.18

l/

times, then we obtain (35)

where C4 is a constant, W(() is a continuously differentiable function. About (35), applying the method of getting (20), sirmilarly we have lim

,Z-,Z*

Ss(z)

= o.

(36)

'zED\8D

By (33) and (36) we obtain

lim

z-+z·

D"S(z)

= DSS(z*), s ~ k.

(37)

zED\8D

Hence DSS(z) E CO(D), s ~ k. Summing up the above, for f(z) E A"(D),S(z) E AS(D) have been obtained, where s = 0,1,2,···. By Theorem 2, for z E b. we have S(z) = f(z). References 1 Henkin G M. Continuation of bounded holomorphic functions from submanifolds in general position to strictly pseudoconvex domains. Math USSR Izvestija, 1972,6: 536-562 2 Elgueta M. Extension to strictly pseudoconvex domains of holomorphic functions in a submanifold in general position and up to the boundary. ILLINOIS Journal of Math, 1980,24(1): 1-17 3 Chen Shujin. General integral representation of the holomorphic functions on the analytic subvariety. Publ RIMS Kyto Univ, 1993,29: 511-533 4 Chen Shujin. The integral representation of analytic functions on the bounded convex domain in en. Acta Math Sinica, 1979,22(6): 743-750 5 Temlyakov A A. The integral representation of analytic functions of two complex variables. V chen Zap Mos ObI In-Ta, 1954,20:7-16 6 Gindikin S G. Henkin G M. The Cauchy-Fantappie formula in projective space. In: Multidimensional complex analysis. 50-63 K vasnoyarsk Inst Fiziki, 1985[Russian]

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