C^(inf)-regularity of solutions of the tangential CR-equations on weakly pseudoconvex manifolds

Math. Ann. 311, 147–162 (1998)

Mathematische Annalen c Springer-Verlag 1998

C ∞ -regularity of solutions of the tangential CR-equations on weakly pseudoconvex manifolds Joachim Michel1 , Mei-Chi Shaw2, ? 1 Universit´ e Du Littoral, Centre Universitaire de la Mi-Voix, F-62228 Calais, France (e-mail: [email protected]) 2 Department of Mathematics, University of Notre Dame, Notre Dame, IN 46556, USA (e-mail: [email protected])

Received: 3 August 1997 / Accepted: 17 September 1997

Mathematics Subject Classification (1991): 32F30, 35N10, 32F20 Let Ω be a bounded weakly pseudoconvex domain in Cn with C ∞ boundary bΩ. The existence of C ∞ solutions up to the boundary for the ∂-equation was shown by J. J. Kohn [6]. In the present paper we give an analogous result for the tangential Cauchy-Riemann equations on compact 2n − 1-dimensional submanifolds M in bΩ with boundary. Here the boundary of M is defined by the transversal intersection of bΩ and a hypersurface H = {z |
Partially supported by NSF grant DMS 94-24122

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In this paper we apply the integral formula approach. In order to do this we − = (P1− · · · , Pn− ), depending need two barrier maps W + = (P1+ , · · · , Pn+ ) and W P n on z , ζ ∈ Cn and having values in Cn , such that i =1 Pi± (z , ζ)(ζi − zi ) = 1. One / Ω for W + , z ∈ / Ω, ζ ∈ Ω for W − , and ∂ z W + = 0, ∂ ζ W − = 0. If has z ∈ Ω, ζ ∈ ± ζ approaches bΩ, W blows-up of some finite order in terms of the boundary distance δ(ζ). Weak information about the derivatives of W ± will be enough to show our results. We use an integral formulae method, which was developed for ∂ and later for ∂ b by Peters [17], Lieb/Range [8], Chaumat/Chollet [2], [3], [4], Michel [10], [11], Michel/Perotti [13] and Ma/Michel [9], and which we generalize here. The main idea is the following. If [f ] is a class of a very smooth form on M there exist extensions Er [f ] and Er ∂ b [f ] of [f ] and ∂ b [f ] to differential forms on the ambient space, such that ∂Er [f ] − Er ∂ b [f ] vanishes of a very high order on M . Such a term is needed to compensate the growth of the kernels in the integral formulae. This is done in Sect. 1. Then we use the barriers in order to construct integral kernels Dn,q (z , ζ, λ) depending on the additional parameters λ = (λ1 , λ2 , ...., λk ). For λ = (1, 0, ......., 0) one gets the Bochner-Martinelli-Koppelman kernel. The range of λ can be a single point, the intervall [0, 1] or a triangle. The vertices represent the different barriers we shall use. The starting point is the well-known Bochner-Martinelli-Koppelman integral representation formula for differential forms in a neighborhood of Ω. Here the range of λ is a single point. By passing over to more complicated Stokes chains with higher dimensional ranges for λ by using Stokes formula we get an integral representation formula which can be used to solve the ∂ b equation on M . In our case there are no barriers with good regularity properties up to the boundary in both variables as in the strictly pseudoconvex case. Therefore in order to make the calculations work we have to multiply the kernels Dn,q (z , ζ, λ), which explode if ζ approaches the boundary of Ω, by the regularising factor ∂E [f ] − E (∂ b [f ]). This will lead to differential forms [Tq ([f ])] and [Tq+1 (∂ b [f ])]. With this we obtain in Sect. 2 formulae of the kind

[f ] = ∂ b [Tq ([f ])] + [Tq+1 (∂ b [f ])].

Note that a priori Tq is not a linear operator but can easily be turned into such if it is necessary. Therefore if [f ] is ∂ b -closed and if we choose for Er ∂ b [f ] the trivial extension we obtain solutions for the ∂ b -equation. Derived from these formulae are C k -estimates with loss of regularity. In order to have a C ∞ solution one has to regularize C k solutions. This is done in Sect. 3. We remark here that for the submanifold M we cannot directly apply the translation method of [7]. Fortunately the solution operators can be written as a sum of terms, where each is defined on some wedge in Cn . For these terms one can then apply the translation method to the different wedges.

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1. Extension of tangential forms Let Ω ⊂ Cn , n ≥ 2, be a bounded pseudoconvex domain given by a C ∞ defining function ρ : U −→ R, where U ⊃ Ω is an open neighborhood of Ω, such that / 0 and d ρ|bΩ = Ω = {z ∈ U |ρ(z ) < 0}. Let h be a holomorphic function on an open set U1 ⊂ U such that for H := {z ∈ U1 |
n X

hi (z , ζ)(ζi − zi ),

i =1

with hi holomorphic on U1 × U1 . ii) d 0}. Moreover M = {z ∈ bΩ| 0, is a smooth closed hypersurface in Ω. Then we have i) M− ∩bΩ = H ∩bΩ, M− ∩H = bΩ ∩H and these intersections are transversal; ii) Ω is decomposed by M− into two parts R− and Ω \ R− , where R− is the one containing H ∩ Ω. We set finally R+ := {z ∈ U1 \ Ω|ρ(z ) +
f − g = ρ · A + ∂ρ ∧ B . k k If [f ] denotes the class of f we set Cp,q (M ) = {[f ]|f ∈ Cp,q (Cn )}. Because of the expected loss of regularity we are only interested in the case of large k >> 1.

We set ∂ b [f ] := [∂f ]. If |g|k ,U + denotes the C k -norm on U + we set |[f ]|k ,M := inf {|g|k ,U + | g ∼ f }. e : C 0 (U + ) −→ C 0 (U ) be a linear extension operator which preserves Let E k e to differential forms componentwise. C -regularity. We shall apply E

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k Lemma 1. Let [f ] ∈ C0,q (M ) be ∂ b -closed. Then for every 0 ≤ r ≤ k −2r k −2r−1 (U ), gr ∈ C0,q+1 (U ), cr ∈ (k − 2)/2 there exist forms Er [f ] ∈ C0,q k −2r−1 k −2r−2 C0,q+1 (U ), Xr ∈ C0,q (U ), such that

i) Er [f ] has compact support in U1 , [Er [f ]] = [f ], cr = 0 on U + and ∂Er [f ] = ρr gr + cr ; ii) Er+1 [f ] − Er [f ] = ρr+1 Xr ; iii) there exist constants Kr not depending on [f ], with |Er [f ]|k −2r,U + |gr |k −2r−1,U + |cr |k −2r−1,U ≤ Kr |[f ]|k ,M . Proof. We show the result by induction over r. r = 0 is trivial. Let r = 1. k −2 k −2 (U + ), B ∈ C0,q (U + ) Because of [∂f ] = 0 one has with A ∈ C0,q+1 ∂f = ρA + ∂ρ ∧ B . e f −ρE e B , g1 = E e A−∂ E eB, This implies ∂f = ρA+∂(ρB )−ρ∂B . So with E1 [f ] = E e c1 = ∂E1 [f ] − ρg1 , X0 = −E B one is through. Now let the assertion be true for a given r, 1 ≤ r ≤ (k − 4)/2. We only sketch the proof because it is elementary. ∂Er [f ] = ρr gr + cr . 2

Hence 0 = ∂ Er [f ] = ρr−1 (r∂ρr ∧ gr + ρ∂gr + ∂cr ). k −2−2r k −2−2r ∂ρr ∧ gr vanishes on M so there exist A ∈ C0,q+1 (U + ), B ∈ C0,q (U + ), with gr = ρA + ∂ρ ∧ B on U + . e B , gr+1 = E e A − (r + 1)−1 ∂ E e B , cr+1 = We set Er+1 [f ] = Er [f ] − (r + 1)−1 ρr+1 E r+1 −1 e ∂Er+1 [f ] − ρ gr+1 , Xr = −(r + 1) E B .

Then everything goes through. With a little modification one can construct Er [f ] with compact support in U + . k k Lemma 2. Let [f ] ∈ C0,q (M ) with ∂ b [f ] ∈ C0,q+1 (M ). Let for 0 ≤ r ≤ (k − 2)/2 Er ∂ b [f ] be the extension of ∂ b [f ] which was constructed in Lemma 1. Then k −2r k −2r−1 k −2r−1 there exist forms Er [f ] ∈ C0,q (U ), Gr ∈ C0,q+1 (U ), Cr ∈ C0,q+1 (U ), k −2r−2 Yr ∈ C0,q (U ), with

i) Er [f ] has compact support in U1 , [Er [f ]] = [f ], Cr = 0 on U + and ∂Er [f ] − Er ∂ b [f ] = ρr Gr + Cr ; ii) Er+1 [f ] − Er [f ] = ρr+1 Yr ; iii) there exist constants Kr not depending on [f ], with |Er [f ]|k −2r,U + |Gr |k −2r−1,U + |Cr |k −2r−1,U ≤ Kr (|[f ]|k ,M + |∂ b [f ]|k ,M ). Proof. The proof is a straightforward calculation by using the same ideas as in the proof of Lemma 1.

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2. Homotopy formulae for ∂ b In [12] the first author has constructed a C 2 barrier mapping W + = (P1+ , · · · , Pn+ ) : Ω × (U \ Ω) −→ Cn , which solves ∂ z W + (z , ζ) = 0,

n X

Pj+ (z , ζ)(ζj − zj ) = 1

j =1

and which fulfills the following estimates for all integers S ≥ 0, with constants CS not depending on ζ, |W + (·, ζ)|S ,Ω + |gradζ W + (·, ζ)|S ,Ω ≤

CS . δ(ζ)tS

Here tS is an increasing sequence of positive numbers about which we have no information and δ(ζ) = dist(ζ, bΩ) is the boundary distance of ζ. In [15] we have shown a related result for a barrier W − with ζ ∈ Ω and z ∈ U \ Ω. There much more precise estimates have been obtained. But in the solution operators for ∂ b , which we want to study in this paper, W + and the second barrier W − are mixed. So we cannot use the better information on W − . Therefore we shall not give here all the shown properties of W − and we prefer to treat both barriers in a more streamlined way. Unfortunately there is yet another drawback connected with W − . W − is holomorphic with respect to the variable ζ, acting here as a parameter, but it is only of class C S with respect to z (S can be chosen arbitrarily). So when it is necessary we shall write WS− instead of W − . This fact complicates the construction of C ∞ solutions of ∂ b u = f . More precisely we have shown the following. For any S ≥ 0 there exists a C S map WS− = (P1− , · · · , Pn− ) : (U \ Ω) × Ω −→ Cn which solves ∂ ζ WS− (z , ζ) = 0,

n X

Pj− (z , ζ)(ζj − zj ) = 1,

j =1

and which fulfills the following estimates, with constants CS not depending on ζ and z : CS |DzI WS− (z , ζ)| ≤ δ(ζ)A(S ) for all multi-indices I with |I | ≤ S . (Here DzI denotes a differentiation with respect to z of order |I | and A(S ) is a positive integer). Next we construct integral operators by using the barriers. The domains of integration will be oriented Stokes chains over submanifolds. We shall only integrate forms with compact support in U1 . Therefore in the following calculations we shall ignore all the terms on the boundary parts of U1 . For 2n-dimensional sets we carry over the orientation of Cn . To their boundaries we give the induced orientation. We orient M , M+ and M− in such a way that

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bR+ = −M − M+ , bR− = M − M− , bK = M+ + M− . Set ∆01+− = {(λ0 , λ1 , λ+ , λ− ) ∈ R4 |λν ≥ 0, λ0 + λ1 + λ+ + λ− = 1}. For an ordered subset ∅ = / A ⊂ {0, 1, +, −} we set ∆A = {λ ∈ ∆01+− |λν = 0 for ν ∈ / A}. If ∆A contains only one point we orient it positively. Inductively we orient ∆A , with A = {a1 , · · · , aν }, ν ≥ 2, in such a way that b∆A = ∆a2 ,···,aν − ∆a1 ,a3 ,···,aν + · · · + (−1)ν−1 ∆a1 ,···,aν−1 . For set products of oriented manifolds we choose the product orientation. To guarantee a non-empty domain of definition of the following differential forms we shall never consider sets A with {+, −} ⊂ A. For a = 0, 1, +, − we define the following barrier forms n X Pia (z , ζ)d ζi , ηa = i =1

with Pi0 (z , ζ) =

ζ i −z i 1 |ζ−z |2 , Pi (z , ζ)

=

hi (z ,ζ) h(ζ)−h(z ) .

For λ ∈ ∆A we set η=

X

λa η a

a∈A

and for q = 0, 1, 2, · · · , n Cn,q = (2πi )−n (−1)q(q−1)/2



 n −1 , q

Dn,q (z , ζ, λ) = Cn,q η ∧ ((∂ ζ + dλ )η)n−q−1 ∧ (∂ z η)q . Here dλ denotes the total differential on ∆A . These double differential forms are called generalized Cauchy-Fantappi`e forms. With Dn,−1 = Dn,n+1 = 0 one has the well known Koppelman formula [18] ∂ z Dn,q−1 = (−1)q (∂ ζ + dλ )Dn,q . Moreover we set X = R+ × ∆0+ + R− × ∆0− + K × ∆01 + M+ × ∆01+ + M− × ∆01− . P \(M × ∆01+− ). Note that for some Then for P z ∈ M Dn,q (z , ·, ·) is defined on terms in we may have a larger domain of definition for Dn,q with respect to z. If A has ν elements dim ∆A = ν − 1. By integrating

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Z Dn,q (z , ζ, λ)

IA = ∆A

we obtain linear combinations of terms ηa1 ∧ · · · ∧ ηaν ∧ (∂ ζ ηa1 )i1 ∧ · · · ∧ (∂ ζ ηaν )iν ∧ (∂ z ηa1 )j1 ∧ · · · ∧ (∂ z ηaν )jν , with i1 + · · · + iν = n − q − ν, j1 + · · · + jν = q. We know (∂ ζ + ∂ z )η1 = 0, ∂ z η+ = 0, ∂ ζ η− = 0. Therefore many of the integrals IA vanish. So a careful but simple analysis gives the following estimates for z ∈ M and ζ an element of the respective domain of definition. Let DzI be a differentiation of order |I | ≤ S . Then there exists a real number MS ≥ 0 and a constant CS with |DzI I0+ (z , ζ)| + |DzI I0− (z , ζ)| ≤

CS , δ(ζ)MS

|DzI I01 (z , ζ)| + |DzI I01+ (z , ζ)| + |DzI I01− (z , ζ)| ≤

CS M S δ(ζ) |
.

Here we have abused a little bit the notation because in the above sums ζ may belong to disjoint domains of definition for different IA . k k (M ), with ∂ b [f ] ∈ C0,q+1 (M ). We choose extensions Er [f ] Now let [f ] ∈ C0,q and Er ∂ b [f ] according to Lemmas 1 and 2. Then for k sufficiently large and r ≤ (k − 2)/2 appropriately chosen with respect to S we get the decomposition

∂Er [f ] − Er ∂ b [f ] ∂Er ∂ b [f ]

= ρMS +1 G + (
with C 1 forms g, h, G, H , such that h and H vanish on U + . If we denote the left-hand sides of the two equations above by L we have in both cases a decomposition L = ρMS +1 A + (
From the construction it follows with a constant C |ρ| ≤ C |
R+

R+ \U +

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Therefore this term is C S on Ω. Analogously Z Z Z Z L(ζ) ∧ I0− , J12 = L(ζ) ∧ I01+ , J22 = L(ζ) ∧ I01− , J3 = L(ζ) ∧ I01 J21 = R−

M−

M+

K

are C S on U \ Ω, Ω ∩ U + , U + \ Ω, U + respectively. Therefore (?) can be written as (?) = J1 + J2 + J3 , where we have set J1 = J11 + J12 , J2 = J21 + J22 , and J1 , J2 , J3 is C S on Ω ∩ U + , U + \ Ω, U + respectively. Moreover there exist constants CS , with |Ji |S ,Di ≤ CS (|Er [f ]|k ,U + |Er ∂ b [f ]|k ,U ), where Di , i = 1, 2, 3 denotes the respective domains of definition. We also have Z Z S S Er [f ] ∧ Dn,q−1 ∈ C0,q−1 (U ), Er ∂ b [f ] ∧ Dn,q ∈ C0,q (U ). U ×∆0

Now we set

U ×∆0

R (∂Er [f ] − Er ∂ b [f ]) ∧ Dn,q−1 + Er [f ] ∧ Dn,q−1 , Σ U ×∆ R R 0 Tq+1 ((∂ b [f ]) := ∂Er ∂ b [f ] ∧ Dn,q + Er ∂ b [f ] ∧ Dn,q . Tq ([f ]) :=

R

Σ

U ×∆0

If ∂ b [f ] = 0 we choose Er ∂ b [f ] = 0. Hence Tq+1 (∂ b [f ]) = 0 in this case. Note that on U × ∆0 Dn,q is the well-known Bochner-Martinelli-Koppelman kernel. Theorem 1. For every integer S ≥ 0 there exist integers A(S ) ≥ 0 and r = R(S ) ≥ 0 with the following properties. A(S ) A(S ) (M ) with ∂ b [f ] ∈ C0,q+1 (M ), q ≥ 1, the forms For every [f ] ∈ C0,q R R = (∂Er [f ] − Er ∂ b [f ]) ∧ Dn,q−1 + Er [f ] ∧ Dn,q−1 , Tq ([f ]) Σ U ×∆ R R 0 Tq+1 (∂ b [f ]) = ∂Er ∂ b [f ] ∧ Dn,q + Er ∂ b [f ] ∧ Dn,q Σ

U ×∆0

are C S on M . If 1 ≤ q ≤ n − 3 one has ∂ b [Tq ([f ])] + [Tq+1 (∂ b [f ])] = [f ]. If q = n − 2, ∂ b [f ] = 0, Er ∂ b [f ] = 0 one has ∂ b [Tn−2 ([f ])] = [f ].

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In all these cases there exist constants CS , which are independent of [f ], with |[Tq ([f ])]|S ,M + |[Tq+1 (∂ b [f ])]|S ,M ≤ CS (|[f ]|A(S ),M + |∂ b [f ]|A(S ),M ). Proof of Theorem 1. In the following we can neglect terms which lead to integrals over M and which contain ∂Er [f ] − Er ∂ b [f ] or ∂Er ∂ b [f ]. With this in mind we have in the sense of Stokes chains X b = −M+ × ∆0+ + R+ × ∆+ − R+ × ∆0 − M− × ∆0− + R− × ∆− − R− × ∆0 +M− × ∆01 + M+ × ∆01 + K × ∆1 − K × ∆0 − M+ × ∆1+ + M+ × ∆0+ −M+ × ∆01 − M− × ∆1− + M− × ∆0− − M− × ∆01 = R+ × ∆+ + R− × ∆− + K × ∆1 − (R+ + R− + K ) × ∆0 − M+ × ∆1+ − M− × ∆1− . In the following we want to differentiate Tq ([f ]) with respect to z , but Tq ([f ]) is only defined on M . This is possible because when we apply for example ∂ z to Tq ([f ]) we apply ∂ z at first to every term in the sum defining Tq ([f ]). These terms are defined on open sets so this will be possible. Then we restrict to M . The starting point is the Bochner-Martinelli-Koppelman formula [18] which reads, because of supp(Er [f ]) ⊂ U , as Z Z ∂Er [f ] ∧ Dn,q + ∂ z Er [f ] ∧ Dn,q−1 . Er [f ] = U ×∆0

U ×∆0

But knowing the result it is much more convenient to proceed in reverse order. Let 1 ≤ q ≤ n − 2 and ∂ b [f ] = 0 for q = n − 2. Applying Stokes theorem we obtain with Θ = ∂Er [f ] − Er ∂ b [f ] R R ∂Tq ([f ]) = Θ ∧ ∂ z Dn,q−1 + ∂ z Er [f ] ∧ Dn,q−1 Σ U ×∆ 0 R R = (−1)q Θ ∧ (∂ ζ + dλ )Dn,q + ∂ z Er [f ] ∧ Dn,q−1 ΣR U ×∆ R R 0 =− Θ ∧ Dn,q − ∂Er ∂ b [f ] ∧ Dn,q + ∂ z Er [f ] ∧ Dn,q−1 b(Σ) Σ U ×∆ R R 0 = −Tq+1 (∂ b [f ]) − Θ ∧ Dn,q + Er ∂ b [f ] ∧ Dn,q b(Σ) UR ×∆0 +∂ z Er [f ] ∧ Dn,q−1 . U ×∆0

P
vanish because of q ≥ 1 and ∂ z η+ = Integrals over ∆+ , ∆1 and ∆1+ in b P
∂ z η1 = 0. The integral over ∆− in b vanishes because of n − q − 1 ≥ 1 and ∂ ζ η− = 0. This yields by using the Bochner-Martinelli formula Z ∂Tq ([f ]) + Tq+1 (∂ b [f ]) = Er [f ] + M− ×∆1−

Θ ∧ Dn,q .

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If q ≤ n − 3 the integral on the right-hand side vanishes. Let q = n − 2 and ∂ b [f ] = 0. Then Θ = ∂Er [f ]. 1 2 2 Set H˜ = H ∩ Ω. R− is decomposed by H˜ in R− and R− , with M− ⊂ R− . Fix z with 0. When we integrate with respect to λ Z Θ(ζ) ∧ Dn,n−2 (z , ζ, λ) ∆1−

can be written as

1 g(z , ζ), h(ζ) − h(z )

where g(z , ζ) is a ∂-closed C S (n, n − 1)-form (with respect to ζ) on R− . 1 has piecewise smooth pseudoconMoreover g(z , ζ) vanishes on M . Now R− vex boundary. It was shown in [12] that there exists for S sufficiently large a C 1 1 (n, n − 2)-form v on R− such that ∂v = g. Then we have Z M−

g(z , ζ) − h(ζ) − h(z )

Z H˜

g(z , ζ) = h(ζ) − h(z )

and hence Z Z Θ(ζ)∧Dn,n−2 (z , ζ) = M− ×∆1−

H˜



Z ∂ζ 2 R−

g(z , ζ) = h(ζ) − h(z )

Z H˜

g(z , ζ) h(ζ) − h(z )

∂v(ζ) = h(ζ) − h(z )



Z bM

=0

v(ζ) . h(ζ) − h(z )

1 If we could approximate h(ζ)−h(z ) uniformly on bM by a sequence {hn }n of 1 smooth functions in R− , which are holomorphic in the interior, we would get Z Z Z Θ(ζ) ∧ Dn,n−2 (z , ζ) = lim hn (ζ)v(ζ) = − lim hn (ζ)g(z , ζ) = 0. M− ×∆1−

n→∞ bM

n→∞

M

That this is possible is shown by the following Lemma. Let for a compact set L A∞ (L) denote the space of C ∞ functions on L which are holomorphic in the interior. Lemma 3. Every holomorphic function in a neighborhood of H ∩ Ω can be 1 ). uniformly approximated on H ∩ Ω by functions of A∞ (R− Remark. This result is new even in the strictly pseudoconvex case. 1 1 Proof. We have H ∩ Ω = {z ∈ R− ||e h(z ) | ≤ 1}. Let µ : H ∩ Ω −→ R− × {w ∈ h(z ) C||w| ≤ 1}, z → (z , e ),be the Oka map. Let V be a neighborhood of H ∩ Ω and b holomorphic in V . We choose a χ ∈ C0∞ (Cn ) with support in V which is identically 1 in a smaller neighborhood of H ∩ Ω. Set

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α(z , w) =

∂χ(z ) ∧ b(z ) . e h(z ) − w

∞ 1 1 (R− × {w ∈ C||w| ≤ 1}), with ∂α = 0. R− × {w ∈ C||w| ≤ 1} Then α ∈ C0,1 has a piecewise smooth pseudoconvex boundary so again by [12] there exists 1 × {w ∈ C||w| ≤ 1}), with ∂v = α. Then B (z , w) = χ(z )b(z ) − v ∈ C ∞ (R− h(z ) 1 − w)v(z ) is in A∞ (R− × {w ∈ C||w| ≤ 1}). Obviously one has (e

B (µ(z )) = b(z ). Let 0 < ε < 1 and Bε (z , w) = B (z , (1 − ε)w). Then Bε → B uniformly on 1 × {w ∈ C||w| ≤ 1} when ε → 0. It is easy to see that there are functions R− 1 ) with aν ∈ A∞ (R− ∞ X B (z , w) = aν (z )wν . ν=0 1 × {w ∈ C||w| ≤ 1}. The corresponding series for Bε converges uniformly on R− 1 × {w ∈ C||w| ≤ 1} for Therefore we can approximate Bε uniformly on R− sufficiently large M by

B˜ (z , w) =

M X

aν (z )((1 − ε)w)ν .

ν=0 1 Then B˜ (z , e h(z ) ) defines a function in A∞ (R− ) which approximates b uniformly on H ∩ Ω.

We set

X

0

= R+ × ∆+ − M+ × ∆1+ .

An analogous calculation as for Theorem 1 shows, if the starting point is T10 ([f ]) (T10 will be defined in Theorem 2), the following theorem: Theorem 2. Let n ≥ 2. For every integer S ≥ 0 there exist integers A(S ) ≥ 0 and r = R(S ) ≥ 0 with the following properties. For every function [f ] ∈ C A(S ) (M ) with ∂ b [f ] ∈ C A(S ) (M ) the forms R T00 ([f ]) = (∂Er [f ] − Er ∂ b [f ]) ∧ Dn,0 , Σ0 R R Er ∂ b [f ] ∧ Dn,0 T10 (∂ b [f ]) = ∂Er ∂ b [f ] ∧ Dn,0 + Σ

U ×∆0

are C S on M . T00 ([f ]) is holomorphic on U + ∩Ω and C S on U + ∩Ω. If ∂ b [f ] = 0 we can assume that T10 (∂ b [f ]) = 0. If n ≥ 3 we have

[T00 ([f ])] + [T10 (∂ b [f ])] = [f ].

If n = 2 and ∂ b [f ] = 0 we have

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[T00 ([f ])] = [f ]. In all these cases there exists a constant CS , which is independent of [f ], with
|[T00 ([f ])]|S ,U + ∩Ω + |[T10 (∂ b [f ])]|S ,M ≤ CS |[f ]|A(S ),M + |∂ b [f ]|A(S ),M . If [f ] ∈ C ∞ (M ), with ∂ b [f ] = 0, there exists an extension E∞ [f ] ∈ C ∞ (U ) with compact support in U1 , such that ∂E∞ [f ] vanishes of infinitely high order on M . W + is C ∞ on Ω with respect to z . Therefore we obtain with a small modification of the above proof: Corollary. Under the same conditions as in Theorem 2 there exists for every CR function f ∈ C ∞ (M ) an extension Z e T0 ([f ]) = ∂E∞ f ∧ Dn,0 Σ0

of f , which is C ∞ on U + ∩ Ω and holomorphic on U + ∩ Ω.

3. Regular solutions for ∂ b A(S ) Let [g] ∈ C0,q (M ), 1 ≤ q ≤ n − 2, with ∂ b [g] = 0. Then a solution of the ∂ b -equation is given by (with ΘS = ∂Er [g], r = R(S )) Z Z vS = ∂Er [g] ∧ Dn,q−1 + Er [g] ∧ Dn,q−1 . Σ

U ×∆0

The second integral, which we denote by J4S (g), is C S on U . The first integral can be decomposed into J1S (g) + J2S (g) + J3S (g), with S S S (U + ∩ Ω), J2S (g) ∈ C0,q−1 (U + \ Ω), J3S (g) ∈ C0,q−1 (U + ). J1S (g) ∈ C0,q−1

Let {Vµ }1≤µ≤N be an open covering of U + ∩ Ω with the following property. For any µ there exist 2 non-tangential vectors tµ+ and tµ− such that for sufficiently + − (z ) = z + tµ+ , Tµ, (z ) = z + tµ− , small  > 0 one has, with Tµ, + (Vµ ∩ U + ∩ Ω) ⊂ (U + ∩ Ω)o , Tµ, − (Vµ ∩ (U + \ Ω)) ⊂ (U + \ Ω)o . Tµ,

Moreover the distance of the set on the left-hand side to the boundary of the set on the right-hand side should be comparable to . This is possible because b(U + ∩ Ω) is piecewise smooth. Let {ρµ }1≤µ≤N be a subordinated partition of unity. We set with  > 0 sufficiently small

Regularity of the tangential CR-equations

159

a JiS,,µ (z ) = JiS (Tµ, (z )), with a = “ + ” for i = 1, 3, a = “ − ” for i = 2

and JiS, (g) =

N X µ=1

ρµ JiS,,µ , for i = 1, 2, 3.

S S (g) is C ∞ in a neighborhood of U + ∩ Ω, J3, (g) is C ∞ in a neighborThen J1, + S S hood of U ∩ Ω and J2, (g) is even better than C in a neighborhood of U + \ Ω with respect to U + . From the construction it also follows that JiS, (g) converges in C S -norms for  −→ 0 to JiS (g), i = 1, 2, 3, on U + ∩ Ω, U + ∩ Ω, U + \ Ω, respectively. ∞ (M ) with ∂ b [f ] = 0, 1 ≤ q ≤ n − 2. Then there exists Theorem 3. Let [f ] ∈ C0,q ∞ [u] ∈ C0,q−1 (M ) with ∂ b [u] = [f ].

Proof. Case 1, when 2 ≤ q ≤ n − 2. Let ϕ be a radial symmetric test function with ϕ ≥ 0, Z supp(ϕ) ⊂ {z ∈ Cn ||z | < 1}, ϕ(z )dV = 1. For δ > 0 we set ϕδ (z ) =

1 z ϕ( ). 2n δ δ

A(A(S )) (M ) with For every integer S ≥ 0 there exists [uS ] ∈ C0,q−1

∂ b [uS ] = [f ]. A(S ) Now we want to construct a modified sequence of solutions [u˜ S ] ∈ C0,q−1 (M ) with |[u˜ S +1 ] − [u˜ S ]|S −1,M ≤ 2−S .

Then

 [u∞ ] := lim

k −→∞

will be a C

∞

[u˜ k ] +

∞ X

 [u˜ i +1 − u˜ i ]

i =k

solution.

With [u˜ 0 ] = [u0 ] we shall construct [u˜ S ] by induction. We suppose that u˜ 0 , · · · , u˜ S have already been constructed. Then A(S ) (M ) [gS ] := [uS +1 − u˜ S ] ∈ C0,q−1

with ∂ b [gS ] = 0.

160

J. Michel, M.-C. Shaw

We have [gS ] = ∂ b (J1S (gS ) + · · · + J4S (gS )) with an appropriate barrier WS− and an extension ER(S ) [gS ]. Now we set for  > 0 and z ∈ M S S S (gS ) + J3, (gS ) + ϕδ ? (J2, (gS ) + J4S (gS ))]. [gS,δ ] = ∂ b [J1, ∞ (M ). Then for  > 0 sufficiently small we have [gS,δ ] ∈ C0,q

Now

S S (gS ) − J1S (gS )] + ∂ b [J3, (gS ) − J3S (gS )] [gS,δ − gS ] = ∂ b [J1, S (gS ) + J4S (gS )) − J2S (gS ) − J4S (gS )]. +∂ b [ϕb ∗ (J2,

Therefore it is easy to see that for δ = 2 and  small |[gS,δ ] − [gS ]|S −1,M ≤ 2−S . We set

[u˜ S +1 ] = [uS +1 − gS,δ ].

A(S +1) Then [u˜ S +1 ] ∈ C0,q−1 (M ), ∂ b [u˜ S +1 ] = [f ] and

|[u˜ S +1 ] − [u˜ S ]|S −1,M = |[gS ] − [gS,δ ]|S −1,M ≤ 2−S . Case 2, when q = 1. ∞ (U + ∩ Ω) be ∂-closed. There is a We need a preliminary result. Let g ∈ C0,1 Seeley extension operator, cf. [21], ∞ ∞ (U + ∩ Ω) −→ E0,1 (U ), E : C0,1

with supp E g ⊂ U1 and |E g|k ,U ≤ Ck |g|k ,U + ∩Ω for all k ≥ 0, with Ck independent of g. By an analogous calculation as for Theorem 1 one can show that Z Z ∂E g ∧ Dn,0 + E g ∧ Dn,0 , T (g) = Σ∗

U ×∆0

P∗

= R+ × ∆0+ + ((K + R− ) \ U + ) × ∆01 − M+ × ∆01+ , is a linear solution with operator for the ∂-equation on U + ∩Ω, cf. [12]. Moreover T (g) ∈ C ∞ (U + ∩Ω) and for all S ≥ 0 there exists an integer N (S ) ≥ 0 with |T (g)|S ,U + ∩Ω ≤ CS |g|N (S ),U + ∩Ω . Now for every S ≥ 0 there exists a [uS ] ∈ C A(N (S )) (M ), with ∂ b [uS ] = [f ]. In the first steps we proceed in the same way as in case q ≥ 2 and then we set

Regularity of the tangential CR-equations

161

[gS ] = [uS +1 − u˜ S ]. Then [gS ] ∈ C A(N (S )) (M ) and ∂ b [gS ] = 0. By Theorem 2 we have with an appropriate r Z  ∂Er [gS ] ∧ Dn,0 . [gS ] = Σ0

Z

Set

∂Er [gS ] ∧ Dn,0 .

geS := Σ0

Then ∂e gS = 0 and geS ∈ C

N (S )

(U + ∩ Ω).

We set for  ≥ 0, geϕ (z )

N X

=

+ ρµ (z )e gS (Tµ, (z )).

µ=1

Then for  > 0 geS ∈ C ∞ (U + ∩ Ω) and ∂e gS =

N X

+ ∂ρµ ∧ (e gS ◦ Tµ, − geS ).

µ=1

gS converges to ∂e gS = 0 in C N (S ) -norms if  tends to 0. Therefore ∂e  gS |N (S ),U + ∩Ω will be small if  is small. So |∂e Set

gS ). w = T (∂e

Then |w|S ,U + ∩Ω ≤ CS |∂e gS |N (S ),U + ∩Ω , w ∈ C ∞ (U + ∩ Ω) and ∂w = ∂e gS . This implies that for  sufficiently small gS ] − [e gS ]|S −1,M ≤ 2−S . |[w]|S −1,M + |[e We set

[u˜ S +1 ] = [uS +1 − geS + w].

Then ∂ b [u˜ S +1 ] = [f ], [u˜ S +1 ] ∈ C A(N (S )) (M ) and |[u˜ S +1 ] − [u˜ S ]|S −1,M ≤ 2−S . Consequently

 lim

k −→∞

is a C

∞

[u˜ k ] +

∞ X i =k

solution. This proves Theorem 3.

 [u˜ i +1 − u˜ i ]

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J. Michel, M.-C. Shaw

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