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Hartogs theorem for forms:

solvability of Cauchy-Riemann operator at critical degree

CHIN-HUEICHANG ANDHSUAN-PEILEE

Abstract. The Hartogs Theorem for holomorphic functions is generalized in two settings: a CR version (Theorem 1.2) and a corresponding theorem based on it forCk ∂-closed forms at the critical degree, 0¯ k≤ ∞(Theorem 1.1). Part of Frenkel’s lemma inCkcategory is also proved.

Mathematics Subject Classification (2000): 32A26 (primary); 32W10 (sec- ondary).

1.

Introduction

Let PN denote the unit polydisc in

C

N,N ≥1. In

C

m+1,m ≥1, set ω= Pm×

zm+1

C

1

2 <|zm+1|<1

, m ≥1.

The classical Hartogs theorem (see [9, p. 55]) states: suppose, for a given holo- morphic function f onω, there is an open setUPm such that f has a holo- morphic extension toU× {zm+1

C

| |zm+1| <1|}, then f can be extended holo- morphically to Pm+1. This phenomenon in higher fiber dimension is suggested by Frenkel’s lemma (see [13, p. 15]).

Letnalways be an integer bigger than 1. Forz

C

m+n we writez=(z,z) withz =(z1, . . . ,zm)andz=(zm+1, . . . ,zm+n). Set

= Pm×

Pn\1 2Pn

where 12Pn = {z

C

n| 2z Pn}. The first part of Frenkel’s lemma says: the Cauchy-Riemann equation

¯u= f, fC(0,q)(), 1≤qm+n, ¯f =0 Received July 29, 2005; accepted in revised form January 5, 2006.

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always has a solutionuC(0,q1)()exceptq =n−1. From now on a(0,n−1) form will be called of the “critical degree”.

Note that in Hartogs theorem the open setU can be replaced by a subset Aof Pmsuch that Ais not contained in a subvariety of codimension one inPm(see [12, p. 16]). The following is our first theorem.

Theorem 1.1. With, A as above. Let f be a Ck ¯-closed(0,n−1)form on, 0≤k ≤ ∞. For zPm, letγz=({z} ×

C

n)be the fiber over zin. For every zA, suppose the Cauchy-Riemann equation onγz,

¯γzv= f, is solvable. Then the Cauchy-Riemann equation on

∂v¯ = f is solvable withvCk().

We explain the notation for the (tangential) Cauchy-Riemann operator used in this paper. In general, if there is no ambiguity, it is denoted by¯; if the ground space X is specified, we use¯X to denote the Cauchy-Riemann (or tangential Cauchy- Riemann) operator on X. Also in integral representations, we usually use ζ for the dummy variable and z for the resulting variable. In this case, the notation¯ζ (respectively,¯z) denotes the (tangential) Cauchy-Riemann operator with respect to ζ (respectively,z) variable.

The proof of Theorem 1.1 depends on Theorem 1.2 which is the CR version of Theorem 1.1. Letρbe aCk (k ≥3) real valued function in

C

m+nwhich is strictly plurisubharmonic in a neighborhood of{ρ≤0}. Letσ be aCkreal valued function in

C

m, strictly plurisubharmonic in a neighborhood of{σ ≤ 0}. We also assume that{σ <0}is connected and relatively compact in

C

m. Set

M = {ρ=0} ∩({σ <0} ×

C

n).

Assume that(respectively,) does not vanish on{ρ =0}(respectively,{σ = 0}) and =0 on∂M. Let f be a(0,q)form onM such that¯Mf =0 in distribution sense. It is proved in [1] that if fL(p0,q)(M)(i.e. f is a(0,q)form with coefficients inLp), 1≤ p≤ ∞, then

¯Mu= f (∗)

is solvable on M withuL(p0,q1)(M) whenever 1 ≤ q < n−1. In the case q =n−1, the above equation is solvable with fC(00,q)(M¯)(i.e. coefficients of f are continuous onM,) if and only if¯ f satisfies the moment condition. Our main result is the following:

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Theorem 1.2. Let M be as above and A be a subset of {σ < 0}not contained in any subvariety of codimension one in{σ <0}. Let f be a∂¯-closed(0,n−1)form with fC0(0,n1)(M¯)Ck(M),0≤kk−3. The equation(∗)is solvable with uC(k0,n2)(M)if and only if all the holomorphic moments of f onz, zA vanish, in other words, for every zA

z

h(z, ζ)f(z, ζ)dζ =0

for every function h holomorphic nearz, wherez= M({z

C

n)is the fiber over zin M.

Remark 1.3.

(a) Whenzis strictly pseudoconvex in{z} ×

C

n, it is well-known that the solv- ability of the tangential Cauchy-Riemann equation¯zu = f with f of top degree is equivalent to the vanishing of all holomorphic moments of f onz. So ifzis strictly pseudoconvex in{z} ×

C

nfor everyz A, the statement of Theorem 1.2 can be phrased as:

The equation(∗)is solvable withuC0(M)if and only if¯zv= f is solvable for everyzA.

(b) The CR function version of Hartogs theorem was proved in [2] where the con- dition for Ais stronger but no convexity condition or boundary regularity is required forM.

We recall the representaion formula for¯b-closed form f onM (cf. [1]):

(−1)qf(z)= ¯

M

f(ζ )q1(

r

,

r

)(ζ,z)+(−1)q

M

f(ζ )∧(

r

,

r

,

s

)(ζ,z)

+

M

f(ζ )q(

r

,

s

)(ζ,z), zM (2.7) where(

r

,

r

), (

r

,

r

,

s

), (

r

,

s

)are defined in Section 2. It is clear from this representation that the obstruction to the solvability of(∗)is the last integral which is null whenq <n−1 by type consideration. We therefore in Section 2 define for

fC(00,n1)(M¯),¯Mf =0 in distribution sense, the following transform:

T f(z)=(−1)n1

M

f(ζ )(

r

,

s

)(ζ,z), (2.8) which may be taken as the global moment of f (cf. ((3.1))). Section 3 consists of the properties of the operatorT needed in this paper. For f in the domain ofT

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we show that T f is defined in a set containing M and is¯-closed there, so (2.7) becomes a “jump formula” for f (see Remark 3.3(a) for more details). Next, we show thatT can be defined locally over the base set{σ <0}. Finally, the operatorT is proved to be just defined fiberwise over every pointz ∈ {σ <0}. This enables us to defineT onMwith arbitrary base setBin

C

m. In section 4 we define the moment operator

M

h forT f (or f) with respect to a holomorphic functionh. It turns out that

M

h(f)is a holomorphic function in the base set. Using this property we prove a more general Theorem 4.4 which implies Theorem 1.2 immediately. Section 5 contains the proof of Theorem 1.1. The interesting thing here is a procedure which improves the method in [11] to produce aCk solution for 0≤k≤ ∞. The results of this paper hold for(p,n−1)forms, 0 ≤ pn. For simplicity, we only deal with the case p=0.

Finally, closely related to the topics in this paper, there is another Hartogs theorem (see [9, p. 56, 63]) whose higher dimensional analogue is written in a forthcoming paper.

ACKNOWLEDGEMENTS. We thank the referee for comments that helped to im- prove the presentation of this paper.

2.

Preliminaries

We write ζ

C

m+n as , ζ)where ζ

C

m and ζ

C

n. Similarly, for differential forms we writed f =(df,df), and∂f =(∂f, ∂f), whered·,d· denote respectively the differentials with repect to the first 2mvariables and those with respect to the last 2nvariables; likewise for·(∂¯·)and·(∂¯·). Also we use =1. . .m+n, =1. . .mand =m+1. . .m+n; similarly for¯,¯,¯, etc..

The following notations and exterior calculus developed by Harvey and Polk- ing [5] will be used in construting kernels needed in this paper:

LetE1, . . . ,Eα (which are called sections) be a collection ofN-tuples ofC2 functions in(ζ,z)

C

N×

C

N. Following Harvey-Polking [5] we use

(E1, . . . ,Eα) = E1,dζ

E1, ζz∧ · · · ∧ Eα,dζ

Eα, ζz (2.1)

λ1+...+λα=N−α

¯ζ,zE1,dζ E1, ζz

λ1

∧ · · · ∧

¯ζ,zEα,dζ Eα, ζz

λα

where x,y =

xiyi for vectors x,y in

C

N and dζ here is understood to be the N-vector (dζ1, . . . ,dζN). Then is C1 away from the singular set

Z = α

1{(ζ,z)| Ej, ζz = 0}. We can rewrite as (E1, . . . ,Eα) = N1

0 q(E1, . . . ,Eα), whereq is the sum of components ofwhich are of

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degreeq ind¯zj, j = 1, . . . ,N. Outside the singular set Z we have the following identity:

¯ζ,z(E1, . . . ,Eα)=α

j=1

(−1)j(E1, . . . ,Ej, . . . ,Eα). (2.2)

To construct the sections we use the results of Fornæss [3] which we briefly outline in the following and refer to [3] for details:

We first observe that for any strongly convex domainG

C

N withCk,k2 boundary, there exist aCk functionµwith positive real Hessian, a constantc> 0 such thatG ={µ <0}, and =0 in a neighborhood of∂G. Furthermore, if we define

H

(ξ, η)= N

1

∂µ

∂ξj(ξ)(ξjηj), then it satisfies

H

(ξ, η)µ(ξ)µ(η)+c|ξ−η|2

for all ξ,η in a small neighborhood of G. The section¯ (∂ξ∂µ1(ξ),· · ·,∂ξ∂µN(ξ))will serve the purpose of this paper in case the given domain is strongly convex.

On the other hand, Fornæss proved in [3] that any strongly pseudoconvex do- main X

C

N with Ck, k 2 boundary, admits an embedding into a bounded strongly convex domain Y

C

N withCk boundary for some N, such that the boundary of X is mapped into the boundary ofY, and the map is 1-1 holomophic in a neighborhood of X¯ (cf. [3, Theorem 9] for explicit statements).

Now for any strongly pseudoconvex domain X

C

N withCk,k ≥2 bound- ary, the above oberservation and Fornæss’ embedding theroem together imply the existence of

H

and the section forY

C

N. Their pull-backs to

C

N then give the following resuts (cf. [3, Theorem 16]):

There exist a Ck function ν which is strictly plurisubharmonic in a neigh- borhood of X¯ with X = {ν < 0}, a constant > 0 and a function H(ξ, η)Ck1(X×X), where X = {η∈

C

N, ν(η) < }satisfying

H(ξ,·) is holomorphic in X, (2.3)

nj(ξ, η)Ck1(X×X), j =1, . . . ,N, holomorphic in η, such that (2.4)

H(ξ, η)= N

1

nj(ξ, η)(ξjηj),

c>0, such that∀η∈ ¯X,ξ ∈ ¯X (2.5) 2Re H(ξ, η)ν(ξ)ν(η)+c|ξ−η|2,

dξH(ξ, η)|ξ=η =∂ν(ξ). (2.6)

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Letρ,σ be as in Section 1. In view of the above discussion, for the strongly pseudoconvex domain{ρ < 0} ⊂

C

m+n there exist

r

j’s which correspond to the nj’s in (2.4), and we define the section

r

(ζ,z) as (

r

1, . . . ,

r

m+n). Similarly for the domain{σ < 0} ⊂

C

m we define the section

s

,z) =(

s

1, . . . ,

s

m). We then use

s

(ζ,z)for the section(

s

1,z), . . . ,

s

m,z),0,· · ·,0

n

). Let

r

(ζ,z)= (

r

1(ζ,z), . . . ,

r

m+n(ζ,z)), where

r

j(ζ,z)= −

r

j(z, ζ ). Thus

r

,

r

and

s

areCk1 in a neighborhood ofM¯ × ¯M.

The kernels(

r

,

r

), (

r

,

r

,

s

), (

r

,

s

)etc., are defined according to for- mula (2.1).

For fC(00,q)(M¯), satisfying ¯Mf = 0 in distribution sense on M with 1≤qn+m−1, we recall the following basic representation formula from [1, p. 543]: forzM

(−1)qf(z)= ¯

M

f(ζ )∧q1(

r

,

r

)(ζ,z)+(−1)q

M

f(ζ )∧(

r

,

r

,

s

)(ζ,z)

+

M

f(ζ)q(

r

,

s

)(ζ,z). (2.7)

The last integral in (2.7) is null whenq < n−1 by type consideration. For fC(00,n1)(M¯),¯M f =0 in distribution sense, we define the following transform:

T f(z)=(−1)n1

M

f(ζ )(

r

,

s

)(ζ,z). (2.8) Remark 2.1. A(0,n−1)form f defined in a subset of

C

m+ncan be decomposed as follows:

f = s

j=1

fj where fj =

|α|+|β|=n1

|α|=j1

fjαdz¯αdz¯βands=min(m+1,n). (2.9)

Moreover, when f is¯-closed we have:

¯zf1=0, ¯zfj = −¯zfj+1, j =1, . . . ,s−1, and¯zfs =0. (2.10) The definition ofT immediately gives

T f =T f1=(T f)1. (2.11)

3.

Properties of Tf

In this section we always assume that fC(00,n1)(M¯)and¯Mf =0 in distribution sense.

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Lemma 3.1. T f =0if there exists uC(00,n2)(M¯)such that∂¯Mu= f on M.

Proof. Suppose there existsuC(00,n2)(M¯)satisfying¯Mu = f. By the defini- tion ofT f, we have

T f =(−1)n1

M

¯ζu(ζ )(

r

,

s

)(ζ,z)=

M

u∧ ¯ζ(

r

,

s

)(ζ,z) by Stokes’ theorem. Invoking (2.2) the last integral becomes

M

u((

r

)(

s

)− ¯z(

r

,

s

))=0 by type considerations. This proves the lemma.

Lemma 3.2. There is an open neighborhood

N

of M in{σ <0} ×

C

n, depending on ρ only, such that T fC1(

N

). On the set U =

N

∩ {ρ ≥ 0} we have

∂(¯ T f)=0.

Proof. Observe that in the definition ofT the integration is just over∂M. By (2.3)- (2.6), we see thatT f is well-defined forzin an open set

N

, depending onρonly, containing Min{σ <0} ×

C

nand isCk2there.

We show that ∂(¯ T f) = 0 in the interior of U. Consider m > 1 first. The identity (2.2) and type considerations imply¯z(

r

,

s

)= −¯ζ(

r

,

s

)in this case.

Thus

∂(¯ T f) =(−1)n1

M

f ∧ ¯z(

r

,

s

)=(−1)n

M

f ∧ ¯ζ(

r

,

s

)

= −

M

¯ζ(f(

r

,

s

))= −

M

dζ(f(

r

,

s

))=0

by Stokes’ theorem. Form = 1, the section

s

is the Cauchy kernel which is holo- morphic in bothζandz. Hence (2.2) and type consideration give

¯z(T f)=(−1)n1

M

f(

r

).

ForzU\Mwe can apply Stokes’ theorem to the above integral and get¯z(T f)= 0 as¯f = ¯ζ

r

=0.

SinceT fCk2(

N

), we conclude that∂(¯ T f)=0 onU by continuity. The lemma is proved.

Remark 3.3.

(a) In (2.7) the form inside the parenthesis after¯Mis inC1(M)provided that f is inC1(M), and so it can be extended to aC1form in{σ <0} ×

C

n. By Lemma 3.2 the last form in (2.7) is actually inCk2(U)and is¯-closed. Therefore, in this case, (2.7) becomes a “jump formula” for f. A more general jump formula can be found in [2], but we don’t need it here.

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(b) In view of (2.10),(2.11) and Lemma 3.2, we see that all the coefficients ofT f are holomorphic inz.

Letσ˜ be aC1function defined in a neighborhood of{σ ≤0}such that{ ˜σ <0} ⊂ {σ < 0}. SetM˜ = {ρ = 0} ∩({ ˜σ < 0} ×

C

n). Supposedρdσ˜ = 0 on M.˜ Denote by b =(b˜,0,· · ·,0

n

), whereb˜ is the section for the Bochner-Martinelli kernel in

C

m. For f in the domain ofT, define

Tf(z)=(−1)n1

M˜

f(

r

,b)(ζ,z).

Lemma 3.1 and the proof of Lemma 3.2 still hold forTand consequentlyTf is a

¯-closed form inU˜ =

N

∩ {ρ≥0} ∩({ ˜σ <0} ×

C

n). The next lemma shows that T is “locally” defined over

C

m.

Lemma 3.4. Let f be in the domain of T . Then T f =Tf onU .˜ Proof. Forz ∈ ˜U, apply (2.2) to get

T f =(−1)n1

M

f(

r

,

s

)(ζ,z)

=(−1)n1

M

f((

r

,b)(

s

,b)− ¯ζ,z(

r

,

s

,b))

=(−1)n1

M

f(

r

,b) by Stokes’ theorem and type considerations,

=(−1)n1

M˜

f(

r

,b) +(−1)n1

M\ ˜M

dζ(f(

r

,b)) by Stokes’ theorem.

In the last integral we use (2.2) again to get

dζ(

r

,b)= ¯ζ(

r

,b)= −¯z(

r

,b)+(

r

)(b) and the integral vanishes by type considerations. So

T f =(−1)n1

M˜

f(

r

,b)=Tf. This completes the proof.

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Lemma 3.4 has two implications. First, in Lemma 3.1 the assumption onucan be weakened byuC(00,n2)(M). Next, it suggests that the strong pseudoconvexity of the base set can be relaxed when one definesT f. This is seen more precisely in Lemma 3.5.

Fort ≥0, set

Mt = {ρ=t} ∩({σ <0} ×

C

n) M˜t = {ρ=t} ∩({ ˜σ <0} ×

C

n) and

z,t = {ρ=t} ∩({z} ×

C

n) forz ∈ {σ <0}. Whent =0 we havez=z,0,M˜ = ˜M0, andM =M0.

Now fixz0 ∈ {σ <0}. As in Lemma 3.4, letσ˜ = |zz0|22, where >0 is chosen so that{ ˜σ <0} ⊂ {σ <0}and˜ =0 on∂M. For˜ small enough there existt >0 small such thatz0,t is strongly pseudoconvex in{z0} ×

C

nand

{z| |zz0|< } × {z|(z0,z)z0,t} ⊂ ˜U \M. Fix suchandt. By Lemma 3.4 we have forz0∈ ˜U satisfyingρ(z0) >t

T f(z0)=(−1)n1

M˜

f(ζ)(

r

,b)(ζ,z0)

=(−1)n1

M˜

(T f)(ζ )(

r

,b)(ζ,z0) where the last equality follows from (2.7) and Lemma 3.1 forT.

Since in the last integral

(

r

,b)= R(ζ,z)0(b)

where R(ζ,z)is a form holomorphic in ζ. Applying Stokes’ theorem to the last integral in the above formula, we have

T f(z0)=(−1)n1

S

(T f)(ζ )(

r

,b)(ζ,z0)

by type consideration and the fact that¯ζ0(b)=0, where S = {ζ| |ζz0| = } × {ζ|(z0, ζ)z0,t}.

Lettend to zero. It follows from Lemma 1.14 of [Ky] that T f(z0)=(−1)n1cm

ζz 0,t

(T f)(z0, ζ)

r

z

0,t)(ζ,z0) wherecm= ((m2πi1)m)! and

r

˜z

0,t is the section constructed fromρ(˜ z)=ρ(z0,z)t.

We thus have the following lemma:

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Lemma 3.5. For any z0 ∈ {σ <0}, for any z0 =(z0,z0)∈ ˜U and for any t > 0 such thatρ(z0) >t and{z|ρ(z0,z) <t}is strictly pseudoconvex inU˜ ∩({z0

C

n), we have

T f(z0)=(−1)n1cm

ζz 0,t

(T f)(z0, ζ)

r

z

0,t)(ζ,z0). (3.1) In particular, if{z|ρ(z0,z) <0}is strictly pseudoconvex which holds for z0in a dense open set in{σ <0}, we have

T f(z0)=(−1)n1cm

ζz 0

f(z0, ζ)

r

z

0)(ζ,z0). (3.1) Proof. It remains to prove(3.1). The denseness of suchz0follows from Sard’s the- orem. In view of the fact that (3.1) holds for any 0<t<ρ(z0)with{z|ρ(z0,z)<

t}strictly pseudoconvex inU˜ ∩ ({z0

C

n),(3.1)is obtained by taking limit as sucht goes to 0 in (3.1) and by (2.7) and Stokes’theorem.

Remark 3.6. Formula (3.1) (or((3.1))) is of fundamental importance. It shows thatT f can be defined byρonly: the connectivity and the strong pseudoconvexity of the base set{σ <0}can be relaxed. Moreover, it shows thatT f has a continuous extension to the setU1=

N

∩ {ρ≥0} ∩({σ ≤0} ×

C

n).

Now instead of{σ < 0}we take the base set to be an arbitrary open set Bin

C

m. It is easy to see that all the preceding results about T f still holds. Indeed, through (3.1) and((3.1))the properties ofT f become even more transparent.

4.

The moment operator M

h(g)

and the proof of Theorem 1.2

Letρbe the same as before andBbe an arbitrary relatively compact open subset in

C

m. Set

M˜ = {ρ=0} ∩(B×

C

n).

For an open neighborhoodO ofM, we set˜

U˜ =O∩ {ρ≥0} ∩(B×

C

n).

Let g be aC1 ¯-closed(0,n−1) form inU˜. Let V be an open set in B andh be a function holomorphic in a neighborhood of{ρ = 0} ∩(V ×

C

n). Define the moment operator ofgwith respect tohonV by

M

h(g)(z)=

ζz

h(z, ζ)g(z, ζ)dζ, where=m+1∧ · · · ∧m+n. (4.1) Lemma 4.1.

M

h(g)(z)is a holomorphic function in V .

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Proof. First observe that

M

h(g)(z)=

ζz

h(z, ζ)g1(z, ζ)dζ

where g1 is defined by (2.9). Fix z0V. Choose a domain D in

C

n with C1 boundary and a neighborhoodW ofz0 inV such that

z0⊂ {z0} ×D, {ρ≤0} ∩(W ×

C

n)W ×D andW ×D ⊂ ˜U. For any function h holomorphic in W ×D it follows from Stokes’ theorem and (2.9), (2.10) that

M

h(g)(z)=

z

h(z, ζ)g1(z, ζ)dζ =

D

h(z, ζ)g1(z, ζ)dζ wheneverzW. Now by (2.10)

¯z

M

h(g)(z)=

D

h(z, ζ)∂¯zg1(z, ζ)dζ= −

D

h(z, ζ)∂¯ζg2(z, ζ)dζ wheneverzW and so

¯

M

h(g)(z)= −

D

dζ(h(z, ζ)g2(z, ζ))dζ=0. This completes the proof.

Remark 4.2. Suppose f is a¯-closed(0,n−1)form inC0(M¯˜). In view of Remark 3.6, T f is well-defined inU˜ =

N

∩ {ρ 0} ∩(B×

C

n)where

N

is given by Lemma 3.2. Locally, for eachzB there is a small open neighborhoodW ofz contained in Bsuch that forz∈ ˜U(W ×

C

n) f can be represented as in (2.7), we see immediately that

M

h(f)=

M

h(T f)for any holomorphic functionh. In other words, the moment of f is well-defined and is holomorphic inz.

With Remark 4.2 we have:

Corollary 4.3. Fix zB. All the holomorphic moments of f onz vanish is equivalent to T f(z) = 0 onU˜∩ ({z} ×

C

n). Moreover, supposez is strictly pseudoconvex, equation∂¯zu = f is solvable onz if and only if T f(z)= 0for all z∈ ˜U({z} ×

C

n).

Proof. To prove the first statement, we need only show thatMh(f)(z)=0 for allh holomorphic nearzimpliesT f(z)=0 onU˜ ∩({z} ×

C

n). The proof of Lemma 4.1 and Remark 4.2 give

0=

M

h(f)(z)=

z

h(z, ζ)(T f)1(z, ζ)dζ

=

z,t

h(z, ζ)T f(z, ζ)dζ, t >0,

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wheneverz,t ⊂ ˜Uandhis holomorphic nearz,t. The last equality follows from (2.10), (2.11) and Stokes’ theorem. Now suppose z,t is strictly pseudoconvex.

Let h(z, ζ)dζ =

r

z). By Lemma 3.5 we have T f(z) = 0 for z ∈ ˜U({z} ×

C

n),ρ(z) t. By Sard’s theorem 0 is a limit point of sucht so the first statement is proved. The second statement follows from the first statement and Remark 1.3(a).

Theorem 4.4. Letρbe as in Theorem 1.2. Let B be a connected relatively compact open subset in

C

m. Set

M˜ = {ρ=0} ∩(B×

C

n).

Let f be a(0,n−1)form in C0(M¯˜)with∂¯ f = 0onM in distribution sense. Let˜ A be a subset of B not contained in any subvariety of codimension one in B. If for every zA all the holomorphic moments of f onz vanish, then T f vanishes identically onU˜ =

N

∩ {ρ≥0} ∩(B×

C

n)where

N

is defined in Lemma 3.2.

Proof. Step 1. The assumption on the set Aimplies that there is a point p ∈ ¯B such that the intersection of Awith any neighborhood of p is not contained in a subvariety of codimension one in B. LetV be any connected neighborhood of p in B. Let¯ h be any function holomorphic near M˜ ∩ (V ×

C

n). By Lemma 4.1

M

h(f)(z)is a holomorphic function inVB. Remark 4.2 and the assumption give that

M

h(f)(z)=

M

h(T f)(z)= 0 for allz V A. By the choice of p, we must have

M

h(f)(z)identically equal to zero onV.

Step 2. Claim: There is a neighborhoodW of pin B¯ such thatT f vanishes identi- cally onU˜ ∩(W×

C

n).

Proof of the Claim. By Corollary 4.3 it suffices to show that for every zW,

M

h(f)(z)= 0 for allhholomorphic nearz. If there is no such neighborhood, then there exists a sequence{zj}1Bsuch thatzjpas j → ∞andT f(zj,·) does not vanish identically for all j =1,2, . . ..

Chooset > 0 so that p,t is aCk strictly pseudoconvex real hypersuface in U˜ ∩({p} ×

C

n). LetW be a connected open neighborhood of pin Bsuch thatW

× {z|(p,z)p,t} ⊂ ˜U\ ˜M.

For every j = 1,2, . . . there is a functionhj holomorphic nearzj such that

M

hj(f)(zj) =0. For simplicity we may assume thatzj is strictly pseudoconvex in{zj} ×

C

n (or we do as in the proof of Corollary 4.3).

Fix jso thatzjW. SetD1= {z|ρ(p,z) <t}andD2= {z|ρ(zj,z) <

0}. By our choice both D1, D2 are strictly pseudoconvex domains in

C

n and D2

D1 for every j = 1,2, . . .. Since ρ(zj,·) is plurisubharmonic in D1 if ρ is plurisubharmonic inU˜ and which can be assumed, it follows from Corollary 5.4.3 of [8] that D1,D2form a Runge pair.

Thushj(zj,·)can be uniformly approximated onD¯2by functions holomorphic onD1. Let{gk}k=1be such a sequence of holomorphic functions onD1. By Step 1,

(13)

for allk≥1

M

gk(f)(z)=

M

gk(T f)(z)=0 for allzW. Therefore we must have

M

hj(f)(zj)=0, contradicting our assumption onzj andhj. This completes the proof of the claim.

Step 3. Set

S

≡ {z| zB, T f(z,·) ≡ 0 onU˜ ∩({z} ×

C

n)}. By Step 2

S

has non-empty interior. In fact, the argument in Step 2 shows that the interior of

S

is both open and closed in B. Since B is connected, we conclude that

S

= B.

Theorem 4.4 is proved.

From the proof of Theorem 4.4 we have the following:

Corollary 4.5. Under the assumptions of Theorem4.4, the following statements are equivalent:

(a) For all zA, every holomorphic moment of f onzvanishes.

(b) T f(z)≡0onU .˜

(c) T f(z)≡0for all zz, zA.

(d) ¯z,tu = T f(z,·)is solvable on z,t ⊂ ˜U for every zA and for every t ≥0, provided thatz,t is strictly pseudoconvex in{z} ×

C

n.

Corollary 4.6. Under the assumptions of Theorem 4.4, if the set{z|z= ∅, zB}is not contained in a subvariety of codimension one in B, then T f ≡0onM.˜ In particular, if B = {σ <0}, then∂¯M is solvable at q=n−1.

On the other hand, we have

Remark 4.7. LetM = {ρ=0} ∩({ ˜σ <0} ×

C

n)whereρis as in the assumption of Theorem 4.4 andσ˜ is anyC1function satisfying˜ =0 on∂M. Suppose M and{ ˜σ < 0} are connected and m < n, then for fC(00,q)(M¯)satisfying

¯Mf = 0 in distribution sense onM withmqn+m −1, considering type, the following representation formula holds forzM

(−1)qf(z)= ¯

M

f(ζ )∧q1(

r

,

r

)(ζ,z)+(−1)q

M

f(ζ )∧(

r

,

r

,b)(ζ,z)

+

M

f(ζ)q(

r

,b)(ζ,z).

Note that the last integral vanishes formqn−2 by type consideration. In other words,¯M is always solvable formqn−2 in this case. Thus the tangential Cauchy-Riemann equation(∗)is solvable atq = n−1 iffTf(z)= T f(z)= 0.

In view of Corollary 4.6, if{ ˜σ <0}is not contained inπ({ρ =0}), the projection of{ρ=0}to the

C

mplane, then(∗)is solvable atq =n−1.

Proof of Theorem 1.2. The solutionuis obtained from Corollary 4.5 and formula (2.7). The regularity ofucan be proved by routine procedure, see e.g. [14], and we omit the details here.

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