1. Introduction
For any open set U⊂Cn, we let Λδ(U) denote the space of functions in Hölder class δ≥0 on U. Let Ω be a smoothly bounded pseudoconvex domain in Cn and z0∈bΩ. Suppose that there exists a neighborhood U of z0 such that, for all ∂¯-closed forms α, with α∈Λδ(Ω), we can solve ∂¯u=α in Ω with a gain of regularity of the solution u; that is,(1)uΛδ+ϵU∩Ω≤CαΛδΩ,for some ϵ>0. In this event, we want to find a necessary condition and determine how large ϵ can be. When z0∈Ω, it is well known that ϵ=1. However, when z0∈bΩ, ϵ>0 depends on the boundary geometry of Ω near z0.
Note that the Hölder estimates of ∂¯-equation are well known when Ω is bounded strongly pseudoconvex domain in Cn. However, for weakly pseudoconvex domains in Cn, Hölder estimates are known only for special pseudoconvex domains, that is, pseudoconvex domains of finite type in C2, convex finite type domains in Cn, and pseudoconvex domains of finite type with diagonal Levi-form in Cn, and so forth. Proving Hölder estimates for general pseudoconvex domains in Cn is one of big questions in several complex variables. Meanwhile, it is of great interest to find a necessary condition or optimal possible gain of the Hölder estimates for ∂¯.
Several authors have obtained necessary conditions for Hölder regularity of ∂¯ on restricted classes of domains [1–4]. Let TBG(z0), the “Bloom-Graham” type, be the maximum order of contact of bΩ with any (n-1)-dimensional complex analytic manifold at z0. If TBG(z0)=N, then Krantz [2] showed that ϵ≤1/N. Krantz’s result is sharp for Ω⊂C2 and when α is a (0,n-1)-form. Also McNeal [3] proved sharp Hölder estimates for (0,1)-form α under the condition that Ω has a holomorphic support function at z0∈Ω. Note that the existence of holomorphic support function is satisfied for restricted domains and it is often the first step to prove the Hölder estimates for ∂¯-equation [4].
Straube [5] proved necessary condition for Hölder regularity gain of Neumann operator N. More specifically, if Neumann operator N has Hölder regularity gain of 2ϵ, then ϵ≤1/η, where η is larger than or equal to order of contact of an analytic variety (possibly singular) V at z0. However, it should be emphasized that there is no natural machinery to pass between necessary conditions for Hölder regularity of ∂¯-Neumann operator and that of ∂¯, in contrast to the case of L2-Sobolev topology.
Let Ω={z:r(z)<0}, where r is a smooth defining function of Ω, and let V be a smooth 1-dimensional analytic variety passing through z0∈bΩ. We say V has order of contact larger than or equal to η with bΩ at z0∈bΩ if there is a positive constant C>0 such that(2)rz≤Cz-z0η,for all z∈V sufficiently close to z0. Here smooth means that γ′(0)≠0 if γ(t) represents a parametrization of V. Recently, the second author, You [6], proved a necessary condition for Hölder estimates for bounded pseudoconvex domains of finite type in C3. That is, if there is a 1-dimensional smooth analytic variety V passing through z0∈bΩ and the order of contact of V with bΩ is larger than or equal to η>0, then the gain of the regularity in Hölder norm should be less than or equal to 1/η. To get a necessary condition for Hölder estimates, we first need a complete analysis of boundary geometry near z0∈bΩ of finite type.
In this paper we prove a necessary condition for the sharp Hölder estimates of ∂¯-equation near z0∈bΩ when Ω is a smoothly bounded pseudoconvex domain in Cn and the Levi-form of bΩ at z0∈bΩ has (n-2)-positive eigenvalues. Our method used to prove the following main theorem will be useful for a study of necessary conditions of Hölder estimates of ∂¯-equation for other kinds of finite type domains.
Theorem 1.
Let Ω be a smoothly bounded pseudoconvex domain in Cn and assume that the Levi-form of bΩ at z0∈bΩ has (n-2)-positive eigenvalues. Assume that there is a smooth holomorphic curve V whose order of contact with bΩ at z0∈bΩ is larger than or equal to η. If there exists a neighborhood U of z0 and a constant C>0 so that, for each α∈L∞0,1(Ω) with ∂¯α=0, there is a u∈Λϵ(U∩Ω¯) such that ∂¯u=α and(3)uΛϵU∩Ω¯≤CαL∞Ω,then ϵ≤1/η.
To prove Theorem 1 we use the analysis of the local geometry near z0∈bΩ in [7] and use the method developed in [6]. In particular Proposition 4 is a key coordinate change which shows that z1 which represents the smooth variety V and the terms mixed with z1 and strongly pseudoconvex directions vanishes up to order m:=[η+1/2], where [x] denotes the largest integer less than or equal to x.
Remark 2.
In general, we note that N:=TBG(z0)≤η. Thus we have ϵ≤1/η≤1/N in (3). We also note that η is a positive integer.
2. Special Coordinates
Let (Ω,z0,η) be as in the statement of Theorem 1 and let r be a smooth defining function of Ω near z0. We may assume that there is a coordinate system z~=(z~1,…,z~n) about z0 such that z0=0 and |∂r/∂z~n|≥c>0, for some constant c>0, in a small neighborhood U of z0. In this section, we construct special coordinates z=(z1,…,zn) near z0∈bΩ which change the given smooth holomorphic curve V into the z1-axis. We will exclude the trivial case, η=2, and hence we assume that η≥3 is a positive integer. Set m:=[η+1/2].
As in the proof of Proposition 2.2 in [7], after a linear change of coordinates followed by standard holomorphic changes of coordinates, we can remove inductively the pure terms such as z~1j, z~¯1k terms as well as z~1jz~α, z~¯1jz~¯α terms, 2≤α≤n-1, in the Taylor series expansion of r(z~) so that r(z~) can be written as(4)rz~=Rez~n+∑j+k≤η, j,k>0a~j,kz~1jz~¯1k+∑α=2n-1z~α2+∑α=2n-1∑j+k≤m, k>0Rea~j,kαz~1jz~¯1kzα+Oz~nz~+z~′′2z~+z~′′z~1m+1+z~1η+1,where z~′′=(z~2,…,z~n-1). Let V be the smooth 1-dimensional variety satisfying (2). Without loss of generality, we may assume that (2) is satisfied in z~-coordinates defined in (4). Let γ:C→V, γ(t)=(γ1(t),…,γn(t)), be a local parametrization of V. We may assume that γ1′(0)≠0, and, hence, after reparametrization, we can write γ(t)=(t,γ2(t),…,γn(t)) and it satisfies(5)rγt≤Ctη.
Lemma 3.
γ
n
(
t
)
vanishes to order at least η.
Proof.
The proof is similar to the proof of Lemma 2.3 in [6]. Since γ(0)=0, γn(t) vanishes to order s>0. Suppose that s<η; that is, γn(t)=asts+O(ts+1) for s<η. In terms of z coordinates in (4), we can write (6)rγt=as2ts+a¯s2t¯s+∑j+k≤η+1, j,k>0cj,ktjt¯k+Ots+1. Since r(γ(t)) vanishes to order at least η, there must be some cancelation between the parenthesis part and summation part. However, this is impossible because parenthesis part consists only of pure terms while summation part consists of mixed power terms.
Proposition 4.
There is a holomorphic coordinate system z with Φ(z)=z~ such that, in terms of z coordinates, r~(z):=r∘Φ(z) can be written as(7)r~z=Rezn+∑j+k=η, j,k>0aj,kz1jz¯1k+∑α=2n-1zα2+∑α=2n-1∑j+k=m, k>0Reaj,kαz1jz¯1kzα+Oznz+z′′2z+z′′z1m+1+z1η+1,and it satisfies(8)r~t,0,…,0,0≲tη.
Proof.
With z~-coordinates defined in (4), define Φ:Cn→Cn, z~=Φ(z), by(9)Φz=z1,z2+γ2z1,…,zn-1+γn-1z1,zn,and set r~(z)=r∘Φ(z). In terms of z coordinates, r~(z) can be written as(10)r~z=Rezn+∑j+k≤η, j,k>0aj,kz1jz¯1k+∑α=2n-1zα2+∑α=2n-1∑2≤j+k≤m, k>0Reaj,kαz1jz¯1kzα+Oznz+z′′2z+z′′z1m+1+z1η+1.Since γn(t) vanishes to order η, it follows from (5), (9), and (10) that (11)r~t,0,…,0=rt,γ2t,…,γn-1t,0≲tη, and hence (8) is proved. Also we note that (12)r~t,0,…,0=∑j+k≤η, j,k>0aj,ktjt¯k+Otη+1, and hence aj,k=0, for j+k<η, because of (8). This fact together with (10) proves that the first summation part in (7) is homogeneous polynomial of order η.
Now we want to show that aj,kα=0, for j+k<m, in the third summation part in (7). On the contrary, let 0<s<m be the least integer such that aj,kα≠0 for some j+k=s and α. In order to show that this is a contradiction, we use variants of the methods in Lemma 4.1 and Proposition 4.4 in [8]. For t with 0<t<1, define a scaling map(13)z=Htsz≔t1/2sw1,t1/2w2,…,t1/2wn-1,twn,and set ρst=t-1((Hts)∗r~) and then set ρ~=limt→0+ρst. Note that 2s<η, and hence the first summation part in (7) will be disappeared in this limiting process. Also note that ρ~ is the limit in the C∞-topology of ρst which, for each t>0, is a defining function of a pseudoconvex domain Ωt, and hence ρ~ is a defining function of a pseudoconvex domain Ω~ given by(14)ρ~w=Rewn+∑α=2n-1wα2+Re∑α=2n-1Pαw1,w¯1wα,where Pα(w1,w¯1) is a plurisubharmonic, nonholomorphic, polynomial of order s provided it is nontrivial. Therefore the Hessian matrix A:=∂2ρ~/∂wj∂w¯k1≤j,k≤n-1 is semidefinite Hermitian matrix and hence detA≥0. Note that(15)detA=2Re∑α=2n-1∂2Pα∂w1∂w¯1wα-∑α=2n-1∂Pα∂w¯12≥0.Assume Pα is nontrivial for some α; say, α=2. For each |w1|<1, take an appropriate argument of w2 satisfying Re∂2P2/∂w1∂w¯1w2≤0. By (15), it follows that ∂Pα/∂w¯1=0 at w=(w1,w2,0,…,0), and hence Pα is holomorphic function of w1 at w for each 2≤α≤n-1. This is a contradiction proving our proposition.
3. A Construction of Special Functions
Let us take the coordinates z=(z1,…,zn) defined in Proposition 4 near z0∈bΩ. In this section, we construct a family of uniformly bounded holomorphic functions fδδ>0 with large derivatives in zn-direction along some curve Γ⊂Ω defined in (39).
In the sequel, we set z′=(z2,…,zn) and z′′=(z2,…,zn-1). We will consider slices of Ω in z1-direction. From (7), rδ(z′):=r~(dδ1/η,z2,…,zn) can be written as(16)rδz′=Rezn+bηδ+∑k=2n-1zk2+∑α=2n-1Rebαδm/ηzα+Oznδ1/η+znz′+z′′3+z′′δm+1/η+z′′2δ1/η+δ1+1/η,where bη=dη∑j+k=η, j,k>0aj,k and where aj,k’s are fixed constants in (7). Note that bη∈R1. Define (17)w′′=z′′,wn=zn+bηδ, and write w′=z′ for a convenience. Then bηδ term is absorbed in the expression of (16).
Let π be the projection onto bΩ along zn-direction. Set zδ=(dδ1/η,0,…,0) and set z~δ=π(zδ):=(dδ1/η,0,…,0,z~n). Note that |z~n|≲δ. Define a biholomorphism Φδ:Cn-1→Cn-1, Φδ(ζ′)=z′=(ζ′′,Φn(ζ′)), by(18)ζ′′=z′′,Φnζ′=ζn+z~n-∑α=2n-1bαδm/ηζα,and set ρδ(ζ′):=rδ∘Φδ(ζ′). Then ρδ(0′)=0, and, in terms of ζ′ coordinates, ρδ(ζ′) can be written as (19)ρδζ′=Reζn+∑k=2n-1ζk2+Oζnζ′+ζ′′3+Oζnδ1/η+ζ′′δm+1/η+ζ′′2δ1/η+δ1+1/η.
Set Ω~δ:=Ω∩{(dδ1/η,z2,…,zn)}, the z1 slice of Ω, and set U~δ=U∩{(dδ1/η,z2,…,zn)}. Also set Ωδ=Φδ-1(Ω~δ), and set Uδ={(dδ1/η,ζ′);Φ(dδ1/η,ζ′)∈U~δ}. Then Ωδ is pseudoconvex domain in Cn-1 and bΩδ∩Uδ is uniformly strongly pseudoconvex, independent of δ>0, provided U is sufficiently small. In the same manner as in Proposition 4.1 in [9] or Proposition 2.5 in [10] (our case is much simpler because bΩδ∩Uδ is uniformly strongly pseudoconvex independent of δ), we can push out bΩδ near z~δ∈bΩδ∩Uδ uniformly independent of δ>0: For each small γ>0, set Bγ={ζ′:|ζ′|<γ}. Set (20)Jδζ′=δ2+ζn2+∑k=2n-1ζk41/2, and for each small σ>0 we set (21)Wδ,a,σ=ζ′:ρδζ′<σJδζ′∩Ba, where a>0 is chosen so that B2a⊂Uδ. Then Wδ,a,σ is the maximally pushed out domain of Ωδ near z~δ reflecting strong pseudoconvexity.
To connect the pushed out part Wδ,a,σ and Ωδ, we use a bumping family Ωδt0≤t≤τ⊂Cn-1 with front Ba as in Theorem 2.3 in [11] or Theorem 2.6 in [10] (again the construction of a bumping family is much simpler because Ωδ is uniformly strongly pseudoconvex). Set (22)Dδ,σt=Ωδt∖Ba∪Wδ,a,σ∩Ωδt. Then Dδ,σt becomes a pseudoconvex domain in Cn-1 which is pushed out near the origin provided t>0 and σ>0 are sufficiently small. In the sequel, we fix these t0 and σ0 and we note that these choices of t0 and σ0>0 are independent of δ>0. Set Dδ:=Dδ,σ0t0⊂Cn-1.
According to Section 3 of [10], or by a method similar to dimension two case of [9], there exists L2(Dδ) holomorphic function fδ satisfying(23)∂fδ∂ζn0,…,0,-bδ2≥1δ,for some b∈R independent of δ where b is taken so that (0,…,0,-bδ/2)∈Ωδ⊂Cn-1. Note that fδ is independent of z1.
Recall that the domains Ωδ or Dδ are the domains in Cn-1 obtained by fixing ζ1=dδ1/η. Define a biholomorphism Ψ:Cn→Cn by(24)Ψζ1,ζ′=ζ1,Φδζ′,and set ρ(ζ)=r~∘Ψ(ζ). For a small constant 0<c<d to be determined, set (25)Pδ,c≔ζ:ζ1-dδ1/η<cδ1/η, ζk<a1, k=2,…,n, where a1=a/2n. In terms of ζ coordinates, for each 0<σ≤σ0, and for each 0<c<d, set (26)Ωδ,cσ=Pδ,c∩ζ1,ζ′:ρdδ1/η,ζ′<σJδζ′, which is obtained by moving Wδ,a,σ along ζ1 direction, and set (27)Ωδ,c=Pδ,c∩ζ:ρζ<0. Note that Ωδ,cσ and Ωδ,c are small neighborhoods of zδ including ζ1 direction.
Lemma 5.
For sufficiently small c>0, we have Ωδ,c⋐Ωδ,cσ/2, or, equivalently,(28)ρdδ1/η,ζ′-ρζ<σ2Jδζ′, for ζ=ζ1,ζ′∈Pδ,c.
Proof.
Assume ζ∈Ωδ,c. Then(29)ρζ-ρdδ1/η,ζ′≤cδ1/ηmaxζ~1-dδ1/η<cδ1/ηD1ρζ~1,ζ′.Note that Φδ is independent of ζ1. Since ρ(ζ)=r~∘Ψ(ζ), it follows from (7) and (24) that(30)D1ρζ~1,ζ′≲δ1-1/η+ζn+ζ′′2+δm-1/ηζ′′≲δ-1/ηJδζ′,because δ(m-1)/η|ζ′′|≤δ-1/η(δ2m/η+ζ′′2) and 2m≥η. Combining (29) and (30), we obtain (28) provided c>0 is sufficiently small.
For each σ>0 and a2>0, set Uδ,a2σ:=Ω¯δ,a2σ. Since fδ is independent of ζ1, we see that fδ is holomorphic on Ωδ,cσ. We will show that fδ is bounded uniformly on Uδ,a2σ/8 for some 0<a2<c≤a1 to be determined. For each q=(q1,q′)∈Uδ,a2σ/8, set τ1=a2δ1/η, τk=a2Jδ(q′)1/2, 2≤k≤n-1, and τn=a2Jδ(q′), and define a nonisotropic polydisc Qa2(q) by (31)Qa2q≔ζ:ζk-qk<τk, 1≤k≤n. In order to proceed as in Section 7 of [9], we first show the following lemma which is similar to Lemma 4.3 in [9].
Lemma 6.
There is an independent constant 0<a2<c such that(32)Qa2q⊂Uδ,a23σ/4⊂Ωδ,cσ, for q=q1,q′∈Uδ,a2σ/8.
Proof.
Assume ζ=(ζ1,ζ′)∈Qa2(q). Then we have (33)Jδq′2≤δ2+2ζn2+2ζn-qn2+8∑k=2n-1ζk4+ζk-qk4≤8Jδζ′2+2a2Jδq′2+8∑k=2n-1a2Jδq′2=8Jδζ′2+8n-14a22Jδq′2. If we take a2>0 so that (8n-14)a22≤1/2, we obtain that (1/4)Jδ(q′)≤Jδ(ζ′). This shows that q∈Qa~2(ζ), where a~2=4a2. By the same argument, we have Jδ(ζ′)≤4Jδ(q′) provided (8n-14)a~22≤1/2. Therefore, if 0<a2≤1/8·1/4n-7, we obtain that(34)14Jδq′≤Jδζ′≤4Jδq′, for ζ∈Qa2q.Since δm/η≤δ1/2≤Jδ(ζ′)1/2, it follows from (7) that(35)Dkρζ≲Jδζ′1/2, ζ=ζ1,ζ′∈Qa2q, 2≤k≤n-1.Combining (34) and (35), one obtains(36)∇′ρdδ1/η,ζ~′·ζ′-q′≤C2a21/2Jδq′,for each ζ,ζ~∈Qa2(q), for some C2>0, where ∇′ denotes the gradient of ζ′=(ζ2,…,ζn) variables.
Now we prove (32). Assume q∈Uδ,a2σ/8 and ζ∈Qa2(q). Since ρ(dδ1/η,q′)≤σ/8Jδ(q′), we can write(37)ρdδ1/η,ζ′≤σ8Jδq′+∇′ρdδ1/η,ζ~′ζ′-q′,for some ζ~=(ζ~1,ζ~′)∈Qa2(q). Combining (34), (36), and (37), we obtain that (38)ρdδ1/η,ζ′≤σ2Jδζ′+C2a21/2Jδq′<3σ4Jδζ′, provided 16C2a21/2<σ. This proves (32).
Let bη=dη∑j+k=η, j,k>0aj,k be the number in (16), and define(39)Γ=z:z=dδ1/η,0,…,0,-bδ2-bηδ, δ>0.Then Γ=Ψ(Γ~), where Γ~={ζ:ζ=(dδ1/η,0,…,0,-bδ/2)} and where Ψ is defined in (24), and b>0 is the number in (23). Note that Γ⊂Ω for all sufficiently small δ>0 provided d>0 is sufficiently small.
Remark 7.
In the above discussion, σ>0 is any number such that 0<σ≤σ0. Thus, in particular, we can fix σ=σ0.
Theorem 8.
f
δ
is bounded holomorphic function in Ωδ,a2σ/8 and, along Γ, fδ satisfies(40)∂fδ∂ζndδ1/η,0,…,0,-bδ2≥1δ,for some b∈R independent of δ.
Proof.
By (23) and (24), we already know that there is a L2 holomorphic function fδ on Ωδ,cσ satisfying estimate (40). We only need to show that fδ is bounded in Ωδ,a2σ/8. Assume q∈Uδ,a2σ/8=Ω¯δ,a2σ/8⋐Ωδ,cσ. Then Qa2(q)⊂Ωδ,cσ by Lemma 6. Now if we use the mean value theorem on polydisc Qa2(q)⊂Ωδ,cσ and the fact that fδ∈L2(Ωδ,cσ) is holomorphic we will get the boundedness of fδ on Ω¯δ,a2σ/8.
4. Proof of Theorem 1
Without loss of generality, we may assume that Ω={ζ∈Cn;ρ(ζ)<0}, where ρ(ζ)=ρ∘Ψ(ζ) and where Ψ is given in (24). Let f=fδ be the bounded holomorphic function in Ωδ,a2σ/8 defined in Theorem 8, and set α=∂¯gδ, where (41)gδ=ϕζ1-dδ1/ηcδ1/ηϕζ2a2ϕζ3a2⋯ϕζna2f0,ζ2,…,ζn and where
(42)
ϕ
t
=
1
,
t
≤
1
2
,
0
,
t
≥
3
4
.
Note that(43)αL∞≲δ-1/η.Now set (44)hζ1,…,ζn=uζ1,ζ2,…,ζn-gδ, where u∈Λϵ(U∩Ω) solves ∂¯u=α as in the statement of Theorem 1, and hence h is holomorphic. Set q1δ(θ)=(dδ1/η+4/5cδ1/ηeiθ,0,…,0,-bδ/2) and q2δ(θ)=(dδ1/η+4/5cδ1/ηeiθ,0,…,0,-bδ), where θ∈R. Let us estimate the lower and upper bounds of the integral (45)Hδ=12π∫02πhq1δθ-hq2δθdθ. From the definition of ϕ we have gδ(q1δ(θ))=gδ(q2δ(θ))=0, and it follows from (3) and (43) that(46)Hδ=12π∫02πuq1δθ-uq2δθdθ≲δϵαL∞≲δϵ-1/η.
For the lower bound estimate, we start with an estimate of the holomorphic function f=fδ with a large nontangential derivative constructed in Theorem 8. For each sufficiently small δ>0, set ζδ′=(0,…,0,-bδ/2) and ζ~δ′=(0,…,0,-bδ), and set ζδ=(dδ1/η,ζδ′) and ζ~δ=(dδ1/η,ζ~δ′). Then Taylor’s theorem of f in ζn variable shows that (47)f0,…,0,ζn=fζδ′+∂f∂ζnζδ′ζn+bδ2+Oζn+bδ22. Now we take ζn=-bδ. Since ∂f/∂ζnζδ′≥1/δ, it follows that(48)fζ~δ′-fζδ′=∂f∂ζnζδ′-bδ2+Oδ2≳1,for all sufficiently small δ>0. Returning to the lower bound estimate of Hδ, the mean value property, (3), (43), and (48) give us(49)Hδ=12π∫02πhq1δθ-hq2δθdθ=hζδ-hζ~δ≥fζ~δ′-fζδ′-uζ~δ-uζδ≳1-δϵ-1/η,because gδ(ζδ)=f(ζδ′) and gδ(ζ~δ)=f(ζ~δ′). If we combine (46) and (49), we obtain that(50)1≲δϵ-1/η.If we assume ϵ>1/η and δ→0, (50) will be a contradiction. Therefore, ϵ≤1/η.