Extension of L^2, di-bar-closed, forms
aa r X i v : . [ m a t h . C V ] M a y EXTENSION OF L , ¯ ∂ -CLOSED, FORMS LUCA BARACCO, STEFANO PINTON AND GIUSEPPE ZAMPIERI
Abstract.
We prove extension of a ¯ ∂ -closed, smooth, form from the intersection of apseudoconvex domain with a complex hyperplane to the whole domain. The extensionform is ¯ ∂ -closed, has harmonic coefficients and its L -norm is estimated by the L -normof the trace. For holomorphic functions this is proved by Ohsawa-Takegoshi [12]. Forforms of higher degree, this is stated in [9]. It seems, however, that the proof contains agap because of the use of a a singular weight and the failure of regularity for the solutionof the related ¯ ∂ -equation.There is a rich literature on the subject (cf. among otheres [7], [14]) but it doesnot seem to contain complete answer to the question. Also, the problem of extendingcohomology classes of ¯ ∂ of higher degree in a compact K¨ahler space is addressed in [8]and [3]. Apart from the formal analogy, this has little in common with our problem inwhich these classes are 0. We also believe, comparing to the preceding literature, thatour approach is original and, somewhat, simpler.MSC: 32F10, 32F20, 32N15, 32T25 We recall
Theorem 1.
Let D ⊂⊂ C n be a bounded smooth pseudoconvex domain of diameter ≤ , α a ¯ ∂ -closed form of degree ≥ such that α J = 0 for / ∈ J and supp α ⊂ { z : | z < δ } .Then there is a solution u = u δ in L to the problem (1) ( ¯ ∂u = α, k z u k ≤ cδ k α k , for c independent of δ , α and D. We refer to [1] for a proof which only relies on the Kohn-H¨ormander-Morrey estimatesin the weighted L space; moreover, “selfboundedness” of the gradient of the weights isnever used. The problem is to extend a ¯ ∂ -closed, smooth, form f from the slice D = D ∩ { z : z = 0 } to the full D with control of the L norm. With the notation D δ forthe δ -disc in the z -plane, this can be achieved by taking a pseudoconvex approximation D ν ր D , choosing δ = δ ν such that the δ -strip D δ × D ν +1 contains D ν ∩ ( D δ × C n − ) takinga family of cut-offs χ δ ( z ) with unitary mass which is 1 for | z | ≤ δ and 0 for | z | > δ , anda family of solutions { u δ } of (1) on D ν for the choice of the form α δ = ¯ ∂χ δ ( z ) f ( z ,...,z n ) z . Eachform f δ := − z u δ + χ δ f is ¯ ∂ -closed in D ν and the family {k f δ k } is uniformly bounded by c k f k D . Then there is a subsequential L -weak limit ˜ f on D , which satisfies ¯ ∂ ˜ f = 0 and(2) k ˜ f k D < ∼ k f k D . If the degree of the forms is 0, that is, if f δ and ˜ f are holomorphic functions, then(3) f δ → ˜ f pointwise . This can be readily checked recalling that, over holomorphic functions, weak convergenceimplies pointwise convergence. On the other hand, ¯ ∂ is elliptic on functions and thus u δ ∈ C ∞ , z u δ | z =0 = 0 and therefore f δ | z =0 = f ; it follows(4) ˜ f | z =0 = f. Thus holomorphic extension with the estimate (2) is proved, for functions. What followsis dedicated to first show that (3) remains true, for an accurate choice of the u δ ’s, ingeneral degree. And next to prove a slightly weaker version of (4), that is,(5) j ∗ ˜ f = f where j : D ֒ → D is the embedding . For the first, we have to take a minimizer with respect to the norm R D ν | z | | · | dV of theaffine space of solutions of (1). The minimizer is a L limit z u µ → z u δ ; in particular, itinherits the estimate, uniform in δ , k z u δ k D < ∼ k f k D . Moreover, u δ does not necessarilysolve ¯ ∂u δ = α δ but it certainly solves ¯ ∂ ( z u δ ) = z α δ . Also, since it is a minimizer, itsatisfies, for every v ∈ C ∞ c | z | u δ , ¯ ∂v ) = ( z u δ , ¯ ∂ ( z v )) = ( ϑ ( z u δ ) , z v ) . Thus z u ∈ ker ϑ for z = 0. But, in fact, also for z = 0. In fact for ψ ∈ C ∞ c , we decompose ψ = ψ + + ψ ǫ where ψ + is supported by | z | > ǫ and ψ ǫ by | z | < ǫ . Then( ϑ ( z u ) , ψ ) = ( ϑ ( z u ) , ψ + + ψ ǫ ) = ( ϑ ( z u ) , ψ ǫ ) = ( z u, ¯ ∂ψ ǫ ) ≤ Cauchy-Schwarz k χ ǫ z u k k ¯ ∂ψ ǫ k → , since z u ∈ L and k ¯ ∂ψ ǫ k is uniformly bounded. Thus ϑ ( z u δ ) = 0 and in particular z u δ ∈ C ∞ . Now, we remark that ˜ f is the limit not only of − z u δ + χ δ f but also of − z u δ .Thus, not only it satisfies ¯ ∂ ˜ f = 0, but also ϑ ˜ f = 0. It follows∆ ˜ f = 0;in particular, ˜ f ∈ C ∞ . As for f δ , we have ¯ ∂f δ = 0 but, we only have ϑf δ = ϑχ δ f , whichyields ϑf δ → L -weakly) and f δ ∈ C ∞ . To carry on our proof, a more subtle analysisis needed. We remark that ϑ ¯ ∂f δ = 0 , XTENSION OF L , ¯ ∂ -CLOSED, FORMS 3 and ¯ ∂ϑf δ = ¯ ∂ϑ ( χ δ f )= (*) ¯ ∂ ( χ δ ϑf )= X i ¯ ∂ i χ δ X j ∂ j f jK + χ δ X i ( X j ¯ ∂ i ∂ j f jK ) , where (*) is explained by the fact that χ δ depends on z and f on z , ..., z n , only. It follows | ∆ f δ | ≤ c | ¯ ∂χ δ | + cχ δ ≤ c χ δ δ . (6)In particular, f δ is harmonic for | z | > δ and f δ → ˜ f uniformly on compact subsets of z = 0. Thus (3) holds for z = 0. As for z = 0, with the notation σ n for the volume ofthe unit ball in C n , we have − Z B r ( z o ) f δ dV − f δ ( z o ) = 1 σ n r n Z r Z b B t (0) [ f δ ( z o + ξ ) − f δ ( z o )] dS ξ dt = 1 σ n r n Z r t Z b B t (0) Z ∇ f δ ( z o + sξ ) · ξt dsdS ξ dt = Divergence Th. σ n r n Z r t Z s Z B t (0) ∆ f δ ( z o + sξ ) dV ξ dsdt = sξ = ζ σ n r n Z r t Z s − n +1 Z B st ( z o ) ∆ f δ dV ζ dsdt. (7)Recall that | ∆ f δ | ≤ c χ δ δ and hence, for z o in the plane z = 0, Z B st ( z o ) | ∆ f δ | dV < ∼ δ Vol(( D δ × B n − st ) ∩ B nst ) ≤ ( st ) n − . (8)Combination of (7) and (8) yields, for z o in the plane z = 0, (cid:12)(cid:12)(cid:12) − Z B r ( z o ) f δ dV − f δ ( z o ) (cid:12)(cid:12)(cid:12) < ∼ r − n Z r t n dt Z ds< ∼ r. (9)We know that we have L weak convergence f δ → ˜ f ; in particular, by the choice of thecharacteristic function of B r ( z o ) as the test function, we have(10) lim δ (cid:12)(cid:12)(cid:12) − Z B r ( z o ) f δ dV − − Z B r ( z o ) ˜ f dV (cid:12)(cid:12)(cid:12) = 0 , r fixed . L. BARACCO, S. PINTON AND G. ZAMPIERI
Finally, by the harmonicity of ˜ f (11) − Z B r ( z o ) ˜ f dV − ˜ f ( z o ) ≡ r .Plugging together (9), (10) and (11), we get, for z o in the plane z = 0(12) lim δ ( f δ ( z o ) − ˜ f ( z o )) = 0 . (In fact, for any ǫ , we first choose r such that (9) is < ǫ ; under this choice of r , we nextchoose δ such that (10) is also < ǫ .) Thus we have proved (3) for a form of general degree;we pass now to (5). This is obtained as an immediate consequence of j ∗ ( z u δ ) = 0, that is(13) ( z u δ ) H | z =0 = 0 for any H which does not contain 1 . We point out that this is by no means evident because we do not know whether u δ issmooth or L . We also point out that our method does not yield z u δ | z =0 = 0 in full, butjust for the “tangential” part (the one which collects multiindices which do not contain1). Thus our program is to prove (13). We decompose(14) u δ = hz + g, for h, g ∈ C ∞ , h + z g ∈ L .We reason by contradiction and assume that for an index J with 1 / ∈ J , that we also write J = iK , we have h iK = 0. We choose a family of test functions(15) ψ ǫ = χ ( | z − ǫ | ǫ ) χ ( | z | ) ...χ ( | z n | );we also use the notation ψ ( z ′ ) := χ ( | z | ) ...χ ( | z n | ). We arrange coordinates so thatsupp ψ ǫ ⊂⊂ D ; we notice that supp ψ ǫ ⊂⊂ { z : z = 0 } . Let z u µ → z u δ be the approxi-mation of the minimizer u δ in R | z | | · | dV norm. We have, for any ψ ∈ Dom ( ¯ ∂ ∗ ) suchthat ¯ ∂ ∗ ψ | z =0 = 0, | ( ¯ ∂ ( u µ − u δ ) , ψ ) | ≤ (cid:18)Z | z ( u µ − u δ | dV (cid:19) (cid:18)Z (cid:12)(cid:12)(cid:12) ¯ ∂ ∗ ψz (cid:12)(cid:12)(cid:12) dV (cid:19) → . (16)Remember also that ¯ ∂u µ ≡ ¯ ∂χ δ fz ∈ C ∞ (independent of µ ) that we also call α . Thus, by(16) applied for ψ = ψ ǫ ¯ ω H , with H = 1 iK , we get(17) lim ǫ Z ( ¯ ∂u δ ) H ψ ǫ dV = lim ǫ Z α H ψ ǫ dV = 0 . XTENSION OF L , ¯ ∂ -CLOSED, FORMS 5 Now, Z ∂ ¯ z ( u δ ) iK ψ ǫ = − Z ( u δ ) iK ¯ z − ǫǫ ˙ χ ( | z − ǫ | ǫ ) ψ ( z ′ ) dV = (14) − Z h iK ( z ′ ) ψ ( z ′ ) dV z ′ (cid:16) ǫ Z ∆ ǫ (3 ǫ ) ˙ χ ( | z − ǫ | ǫ ) ¯ z − ǫz dx dy (cid:17) − Z g iK (0 , z ′ ) ψ ( z ′ ) dV z ′ (cid:16) ǫ Z ∆ ǫ (3 ǫ ) ˙ χ ( | z − ǫ | ǫ )(¯ z − ǫ ) dx dy (cid:17) + O ( ǫ )= − Z h iK ( z ′ ) ψ ( z ′ ) dV z ′ (cid:16) ǫ Z ∆ ǫ (3 ǫ ) ˙ χ ( | z − ǫ | ǫ ) ¯ z − ǫz dx dy (cid:17) + O ( ǫ ) + O ( ǫ ) . (18)We show that the integral in the last line is a constant different from 0. Under thecoordinate change in C , z − ǫ = ρζ , ρ ∈ [0 , ǫ ] , ζ ∈ ∂ ∆, we have Z ∆ ǫ (3 ǫ ) ˙ χ ( | z − ǫ | ǫ ) ¯ z − ǫz dx dy = Z ǫ ˙ χ ( ρ ǫ ) Z ∂ ∆ ρ ¯ ζρζ + 3 ǫ ρiζ dζ dρ = Z ǫ ˙ χ ( ρ ǫ ) ρ iǫ Z ∂ ∆ ζ (1 + ρζ ǫ ) dζ dρ = Residue Th. Z ǫ ˙ χ ( ρ ǫ ) ρ iǫ πi (cid:18) − ρ ǫ (cid:19) dρ = − π ǫ Z ǫ ˙ χ ( ρ ǫ ) ρ dρ = by the change t = ρ ǫ − π ǫ Z ˙ χ ( t ) t ǫ dt = − πǫ (cid:12)(cid:12)(cid:12) χ ( t ) t − Z χ ( t ) dt ]= πǫ . (19) L. BARACCO, S. PINTON AND G. ZAMPIERI
Thus, if h iK = 0, then the limit for ǫ → h K is, we have, when i = 1, Z ∂ ¯ z i ( u δ ) K ψ ǫ dV = − Z ( u δ ) K χ (cid:18) | z − ǫ | ǫ (cid:19) ∂ ¯ z i χ ( z ′ ) dV = − Z h K ( z ′ ) ∂ ¯ z i χ ( z ′ ) Z ∆ ǫ (3 ǫ ) χ (cid:16) | z − ǫ | ǫ (cid:17) z dx dy dV z ′ + O ( ǫ )= − Z h K ( z ′ ) ˙ χ ( z ′ ) O ( ǫ ) + O ( ǫ )= O ( ǫ ) . (20)(Clearly, in case ( u δ ) K is bounded, that is h K = 0, we have that (20) is indeed O ( ǫ ).)In any case, for iK not containing 1 and with H = 1 iK , from the identity ( ¯ ∂u δ ) H = ∂ ¯ z (( u δ ) iK ) − ∂ ¯ z i (( u δ ) K ), we get, combining (18) and (20), Z ( ¯ ∂u δ ) iK ψ ǫ dV = c + 0( ǫ ) , c = 0 , which contradicts (17). Summarizing up, we have proved that in the decomposition( u δ ) iK = h iK z + g iK , we cannot have h iK = 0 when 1 / ∈ iK . Thus, ( u δ ) iK is smoothand in particular z ( u δ ) iK | z =0 = 0, and hence ( f δ ) iK | z =0 = f iK . If we plug into (12), weget ˜ f iK | z =0 = f iK . All in all, we have got Theorem 2.
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Dipartimento di Matematica, Universit`a di Padova, via Trieste 63, 35121 Padova, Italy
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