1. Introduction Let Ω be a bounded domain in Cn. A natural operator on Ω is the orthogonal projection (1)P:L2Ω→HΩ∩L2Ω,where H(Ω) denotes the holomorphic functions on Ω. There is a corresponding kernel function KΩ(z,w), called the Bergman kernel function on Ω. The nature of the singularity of KΩ(z,w) tells us much about the holomorphic function theory of the domain in question and has been studied extensively since Bergman’s original inquiries [1].

One of the methods for the estimates of the Bergman kernel is to construct maximal size of polydiscs in Ω where we have a plurisubharmonic function with maximal Hessian. For strongly pseudoconvex domains in Cn, these polydiscs are of size δ>0 in normal direction and of size δ1/2 in tangential directions. For weakly pseudoconvex domains, the size of the polydisc in tangential directions depends on the boundary geometry of Ω near z0∈bΩ, and hence we need complete analysis of the boundary geometry near z0.

However these analyses and hence the optimal estimates on the Bergman kernels are done only for special type of pseudoconvex domains of finite type in Cn. These domains are, for example, pseudoconvex domains of finite type in C2 [2–4], decoupled, convex, or uniformly extendable domains of finite type in Cn [5–7], or pseudoconvex domains in Cn with (n-2) positive eigenvalues [8, 9]. For the estimates for weighted Bergman projections, one can also refer to [10–12]. Nevertheless, the optimal estimates for general pseudoconvex domains of finite type in Cn, n>2, are not known, even for n=3 case.

Assume that Ω is a smoothly bounded domain in Cn with smooth defining function r with smooth boundary, bΩ. Regular finite 1-type at z0∈bΩ, denoted by TΩreg(z0), is the maximum order of vanishing of r∘γ for all one complex dimensional regular curve γ, γ(0)=z0, and γ′(0)≠0. Thus TΩreg(z0) satisfies(2)Δn-1z0≤TΩregz0≤Δ1z0,where Δq(z0), 1≤q≤n-1, denotes finite q-type in the sense of D’Angelo [13]. Note that Δn-1(z0)=TBG(z0) where TBG(z0) is the type in the sense of Bloom-Graham.

Remark 1. Consider the domain Ω [13] in C3 defined by (3)rz=Rez3+z12-z232.Then TΩreg(0)=6 and Δ2(0)=4 while Δ1(0)=∞ as the complex analytic curve γ(t)=(t3,t2,0) lies in the boundary. Note that γ(t) is not regular curve.

In the sequel, we let Ω be a smoothly bounded pseudoconvex domain in C3, and assume that TΩreg(z0)=η<∞ where z0∈bΩ. Let Cb(z0,δ0)⊂Ω∪z0 be the “almost tangential curve" connecting a point zδ0∈Ω and z0∈bΩ as defined in (20). Note that dist(zδ,bΩ)≈δ for each zδ∈Cb(z0,δ0). Set τ1=δ1/η, τ2=τ(zδ,δ) where τ(zδ,δ) is defined in (51).

Theorem 2. Let Ω be a smoothly bounded pseudoconvex domain in C3 and assume that TΩreg(z0)<∞ where z0∈bΩ. Then KΩ(zδ,zδ), the Bergman kernel function of Ω at zδ∈Cb(z0,δ0), satisfies(4)KΩzδ,zδ≈δ-2τ1-2τ2-2.

Theorem 3. Let Ω and z0∈bΩ be as in Theorem 2. For each α=(α1,α2,α3), there is a constant Cα>0, independent of δ>0, such that(5)DzαKΩz,zδ≤Cαδ-2-α3τ1-2-α1τ2-2-α2,for z∈Ω and zδ∈Cb(z0,δ0).

Remark 4. (1) In Theorems 2 and 3, we do not assume that Δ1(z0)<∞, but we assume only that TΩreg(z0)<∞ (see Remark 1). With this weaker condition, we get optimal estimates for Bergman kernel function along special “almost tangential” direction, Cb(z0,δ0), but not normal or arbitrary direction.

( 2 ) In [14], Herbort gives an example of a domain ΩH⊂C3 where the Bergman kernel grows logarithmically when z∈Ω approaches to z0∈bΩ in normal direction. Set(6)ΩH=z∈C3∣Rez3+z16+z12z22+z26<0,and for each small δ>0, set zδ=(0,0,-δ). Thus zδ approaches to 0∈bΩH in normal direction as δ→0. In this case, Herbort shows that KΩ(zδ,zδ)≈δ-3(-logδ)-1; that is, the kernel grows logarithmically. For the same domain ΩH in (6), we note that η=TΩreg(z0)=6 and hence τ1=δ1/6, τ2=δ1/3 in (4). Set Cb(z0,δ0)≔δ1/6/2,0,-δ:0≤δ≤δ0. Then zδ≔δ1/6/2,0,-δ∈Cb(z0,δ0) approaches to 0∈bΩ in “almost tangential direction”. In the Appendix of this paper, we will show that (7)KΩHzδ,zδ≈δ-2τ1-2τ2-2=δ-3.

In Section 2, we will construct special coordinates which reflect the regular finite type condition, Δ2(z0)≤TΩreg(z0)=η<∞, and then show that r(z) vanishes to order η in z1-direction. We then consider the slices of Ω by fixing z1. Then the domains become domains in C2, and hence we can handle them. Also, the condition Δ2(z0)<∞ acts like the condition Δ1(z0)<∞ on these slices.

For the estimates of K(z,w), Catlin [2, 15] constructed plurisubharmonic functions with maximal Hessian near each thin δ-strip of bΩ (Section 3 of [2]). In this paper, however, we will construct these functions only on nonisotropic polydiscs Qaδ(zδ)⊂⊂Ω for each zδ∈Cb(z0,δ0) (Proposition 23). This avoids complicated technical parts in Section 3 of [2]. To get estimates of KΩ(z,zδ), z∈Ω, zδ∈Cb(z0,δ0), we consider dilated domains Dδ for each δ>0. Then the polydisc Qaδ(zδ)⊂⊂Ω becomes P(0,1)⊂⊂Dδ, independent of δ>0, where P(0,1) is a polydisc of radius one with center at the origin. Therefore the uniform 1/2-subelliptic estimates for ∂¯-equation hold on P(0,1), and the estimates for KΩ(z,zδ) follow.

Remark 5. Let Ω be a smoothly bounded pseudoconvex domain in C3, and assume that Δ1(z0)<∞, where z0∈bΩ. Then the conditions of Theorems 2 and 3 are satisfied. Near future, using the results of Theorems 2 and 3, we hope we can prove some function theories on Ω, for example, the existence of peak function for Ω that peaks at z0∈bΩ or necessary conditions for the Hölder estimates for ∂¯-equation.

2. Special Coordinates In the sequel, we let Ω be a smoothly bounded pseudoconvex domain in C3 and assume that m=Δ2(z0)≤η=TΩreg(z0)<∞, z0∈bΩ. Note that m and η are positive integers. Without loss of generality, we may assume that z0=0. In the sequel, we let α=(α1,α2,α3) and β=(β1,β2,β3) be multi-indices and set α′=(α1,α2) and z′=(z1,z2), etc. In Theorem 3.1 in [16], You constructed special coordinates which represent the local geometry of bΩ near z0.

Theorem 6. Let Ω be a smoothly bounded pseudoconvex domain in C3 with smooth defining function r and assume TΩreg(0)=η<∞, 0∈bΩ. Then there is a holomorphic coordinate system z=(z1,z2,z3) about 0 such that(8)1 rz=Rez3+∑α′+β′=mα′,β′>0ηaα′,β′zα′z¯β′+Oz3z+z′η+1,2 rt,0,0≈tη,where(9)aα′,β′≠0 with α2+β2=m for some α2>0, β2>0.

(Idea of the proof) by the standard holomorphic coordinate changes, r(w) has the Taylor series expansion as in (8). Since TΩreg(0)=η, there is a regular curve which we may assume that γ(t)=(t,γ2(t),O(tη)) satisfying rγt≈tη for all sufficiently small t∈C. Set z=(w1,w2+γ2(w1),w3). Then, in z coordinates, r(z) has representation satisfying (8). Also (9) follows from the condition that m=Δ2(0).

Remark 7. (1) The second condition in (8) and property (9) say that r(z) vanishes to order η along z1 axis and order m along z2 axis.

(2) There are much more terms (mixed with z1, z2 and their conjugates), compared to the h-extensible domain cases, in the summation part of (8).

In conjunction with multitype M(0)=(1,m,m3), we need to consider the dominating terms (in size) among the mixed terms in z1 and z2 variables in the summation part of (8). Using the notations of Section 3.2 in [16], set (10)Γ=α′,β′;aα′,β′≠0, m≤α′+β′≤η, and α′,β′>0S=p,q;α1+β1=p, α2+β2=q for some α′,β′∈Γ∪η,0.Then there are (pν,qν)∈S for ν=0,1,…,N and ην,λν>0 for ν=1,…,N, such that (11)1 p0,q0=η,0, pN,qN=0,m, λN=m, η1=η2 p0>p1>…>pN andq0<q1<…<qN3 λ1<λ2<…<λN andη1>η2>…>ηN4 pν-1ην+qν-1λν=1 andpνην+qνλν=15 aα′,β′=0 if α1+β1ην+α2+β2λν<1 for each ν=1,…,N.

Remark 8. (1) Here, pν’s and qν’s are the exponents of z1 and z2, respectively, in the dominating terms in the summation part of (8).

(2) If Δ1(z0)<∞, then the expression in (8) will be similar to that of C2 case in [2], and hence we need not consider the above complicated pairs.

Set t0=η. If 1≤k≤m, then qν-1<k≤qν for some ν=1,…,N. In this case, set(12)tk=ην1-kλν, i.e., tkην+kλν=1.Then (pν-1,qν-1), (tl,l), and (pν,qν) are colinear points in the first quadrant of the plane, and λν (resp., ην) is the intercept of q-axis (resp., p-axis) of this line. Let Lν be the line segment from (pν-1,qν-1) to (pν,qν) for ν=1,…,N, set L=L1∪L2∪…∪LN, ΓL={(α′,β′)∈Γ;(α1+β1,α2+β2)∈L}, and set (13)Λ=α′,β′∈ΓL; α′+β′=pν,qν, α2>0, β2>0, ν=1,…,N.As in Corollary 3.8 and Remark 3.9 in [16], we can rewrite (8) so that(14)rz=Rez3+∑ΓL-Λaα,βz′αz¯′β+∑ν=1N ∑α2+β2=qνα2>0,β2>0Mα2,β2νz1z2α2z¯2β2+Oz3z+∑ν=1N ∑l=qν-1qνz1tl+1z2l+z2m+1,where Mα2,β2ν(z1) is a nontrivial homogeneous polynomial of degree pν given by (15)Mα2,β2νz1=∑α1+β1=pνaα1,β1z1α1z¯1β1,and there are a small constant a0>0, and d∈z1∈C;z1=1 such that(16)Mα2,β2νz1≠0 for z1-d<a0,and Mα2,β2νdδ1/η≈δpν/η,for all α2+β2=qν with all ν=1,…,N-1. Property (16) means that there is α′,β′ with terms mixed in z2 and z¯2 variables for z1-d<a0. Let d=1 be the constant (direction) in (16) and we will fix d in the rest of this paper. In the sequel, we set z^l equal to zl or z¯l, l=1,2,3.

Remark 9. (a) {tk} defined in (12) is strictly decreasing on k.

(b) Each of the summation parts of (14) contains the terms of the form z^1tkz^2k where (tk,k)’s are the pairs, defined in (12), on the polyline L.

(c) Each term of the first summation part in (14) is pure in z2 or z¯2 variables.

(d) Each term of the second summation part in (14) has terms mixed in z2 and z¯2, and it corresponds to the pair of integers (pν,qν), the vertices of the polyline L.

Lemma 10. Let d0(z1)≔r(z1,0,0) be the term containing only z1 or z¯1 variables in the first sum of (14). Then(17)d0z1≈z1η.

Proof. From (8) and (14), we see that rz1,0,0≲z1η. On the other hand, since the regular 1-type at 0∈bΩ is equal to η=TΩreg(0), there is c~1>0 such that rz1,0,0≥c~1z1η.

In the sequel, we let V be a small neighborhood of z0=0∈bΩ where r(z) has expression as in (14). Since (∂r/∂z3)(0)≠0, we may assume that (∂r/∂z3)z≥c0 for all z∈V for a uniform constant c0>0 by shrinking V if necessary. For each fixed δ>0 and for each z=(z1,z2,z3)∈V satisfying z1-dδ1/η<γδ1/η, for a sufficiently small γ>0 to be chosen, we set π(z)=(z1,0,eδ)≔z~∈bΩ, where π(z) is the composition of the projection onto z1z3 plane and then the projection onto bΩ along the Rez3 direction. Using the Taylor series method in z3 variable about eδ, we see that (18)rz1,0,0=2Re∂rz~∂z3-eδ+Oeδ2.Since eδ≪1 and 2Re(∂r/∂z3)=1+Oz≥1/2 on V, it follows from (17) that (19)eδ≈z1η,for z1 near 0.

Now for each small δ>0, set z(δ)≔(dδ1/η,0,0) and set z~δ=π(z(δ))=(dδ1/η,0,eδ)∈bΩ. For a small constant b>0 to be chosen, set zδ=(dδ1/η,0,eδ-bδ)∈Ω, and for a fixed small δ0>0 satisfying r(zδ0)<0, set (20)Cbz0,δ0≔zδ: zδ=dδ1/η,0,eδ-bδ, 0≤δ≤δ0⊂Ω∪z0,connecting zδ0∈V∩Ω and z0=0∈bΩ.

Following the same arguments as in the proof of Proposition 1.2 in [2], for each fixed z~∈V, we can construct special coordinates about z~ so that, in terms of new coordinates, there is no pure terms in z2 or z-2 variables in the first summation part of r(z) in (14). We will fix z1 variable and consider the coordinate changes only on z′′=(z2,z3) variables.

Proposition 11. For each fixed z~=(z~1,z~2,z~3)∈V, there is a holomorphic coordinate system z′′=Φz~(ζ′′)=(ζ2,Φ3(ζ′′)) such that in the new coordinates ζ′′ defined by (21)Φz~ζ′′=z~2+ζ2,Φ3ζ′′,where (22)Φ3ζ′′=z~3+∂r∂z3z~-1ζ32-∑l=1mclz~ζ2l,and where cl(z~), l=2,3,…,m, depends smoothly on z~, the function given by ρz~1,ζ′′≔rz~1,Φz~ζ′′ satisfies(23)ρz~1,ζ′′=rz~+Reζ3+∑j+k=2j,k>0maj,kz~1ζ2jζ-2k+Oζ3ζ+ζ2m+1.

Proof. For z~∈V, define (24)Φ1w′′=z~2+w2,z~3+∂r∂z3z~-1w32-∂r∂z2z~w2,where w′′=(w2,w3). Then we have (25)ρ2z~1,w′′≔rz~1,Φ1w′′=rz~+Rew3+Ow′′2.

Assume that (22) and (23) hold for l≥2. That is, we have defined Φl-1:C2→C2 so that ρl(z~1,w′′)≔r(z~1,Φl-1(w′′)) can be written as (26)ρlz~1,w′′=Rew3+∑j+k=2j,k>0l-1aj,kl-1z~w2jw-2k+Ow3w′′+w2l.If we define Φl=Φl-1∘ϕl, where (27)ϕlζ′′=ζ2,ζ3-2l!∂lρl∂w2lz~1,0,0ζ2l,then ρl+1(z~1,ζ′′)≔r(z~1,Φl(ζ′′)) satisfies (26) for l replaced by l+1. If we proceed up to l=m and set Φz~=Φm=Φ1∘ϕ2∘⋯∘ϕm, then by setting ρ=ρm+1=rz~,Φz~·, we see that (22) and (23) hold.

In the sequel, we will use the coordinate changes in Proposition 11 only at z~=(z~1,0,z~3)∈V, (in particular at z~δ=(dδ1/η,0,eδ)∈bΩ in Section 3). We want to study the dependence of Φz~ about z~. For each z~=(z~1,0,z~3)∈V, set c0(z~)≔(∂r/∂z3)(z~)=1+O(z~), and we note that (28)clz~=1l!∂lρl∂w2lz~1,0,0, l=1,2,…,m,where ρl is defined in the inductive step of the proof of Proposition 11. Set (29)e0z~=12c0z~-1,e1z~=-c0z~-1∂r∂z2z~ andelz~=-c0z~-1clz~ l=2,…,m,and set ρ0=r. Then ρ1z~1,ζ′′=rz~1,ζ2,z~3+e0ζ3 and (30)ρi+1=ρiz~1,ζ2,ζ3+eiz~ζ2i, i=1,2,…,m.

To study the dependence of Φz~ and hence dependence of aj,k(z~1) about z~1 in (23), we thus need to study the dependence of el(z~) on z~ variable. For a convenience, set z~=(z1,0,z3), i.e., remove tilde’s, and assume that z~ satisfies (31)z1-dδ1/η<γδ1/η,and z3≲z1η,for a sufficiently small γ>0 to be chosen. In view of (19), we see that z~δ=(dδ1/η,0,eδ) satisfies (31). In following we let z be the given coordinates, and we let ζ be the coordinates obtained from holomorphic coordinates changes of z, as in l-th step of coordinate changes in the proof of Proposition 11. Also we let Dks (resp., D~ks), k=1,2,3, denote any partial derivative operator of order s with respect to zk and z¯k (resp., ζk and ζ¯k) variables. According to the coordinate changes in Proposition 11, we note that D1=D~1.

Proposition 12. Assume that z~=(z1,0,z3)∈V satisfies (31). Then for each i=0,1,…,m+1, we have(32)D1l1D~2kρiz~≲z1tk-l1, 0≤k≤m, 0≤l1≤η,and for each α2,β2>0 with α2+β2=qν, for some qν in (14), we have(33)D1l1D~2qνρiz~≈D1l1D2qνrz~≈z1pν-l1, l1≤pν, 2≤i≤m+1.

Proof. We will prove by induction on i. From (14), (17), and (31) one obtains(34)D1l1D2krz~≲z1η-l1+z1tk-l1+z3+z1tk+1-l1≲z1tk-l1,and hence (32) follows for i=0. Since ρ1(z1,ζ′′)=r(z1,ζ2,z3+e0ζ3), it follows, from (31) and chain rule, that (35)D1l1D~2kρ1z~≲D1l1D~2krz~+z3≲z1tk-l1,because we are evaluating at ζ~=(z1,0,0). This proves (32) for i=1.

By induction, assume that (32) holds for i=0,1,…,s. For the ei(z~) defined in (29), it follows, from (34) and induction hypothesis, that(36)D1l1eiz~≲∑j=0l1D1jD2iρiz~≲∑j=0l1z1ti-j≲z1ti-l1,for i=1,…,s. Since we are evaluating at ζ2=0, it follows, for s≥1, that(37)D1l1D~2kρs+1z~=D1l1D~2kρsz~, if k<s, and=D1l1D~2kρsz~+O∑j=0l1D1jesz~, if k≥s.By (30), (36), and (37) and by induction, (32) holds for i=s+1 because tk≤ts if k≥s.

Now we prove (33). Assume α2+β2=qν with α2>0, β2>0 where (pν,qν) are the pairs corresponding to the second summation part of (14). Note that the first summation part of (14) will be annihilated by D2qν because it contains the pure terms of z2 or z¯2 mixed with z^1k. Thus it follows from (14), (16), and (31) that(38)D1l1D2qνrz~≈D1l1Mα2,β2νz1+Oz3+z1pν+1-l1≈z1pν-l1.Since ρ1(z1,ζ′′)=r(z1,ζ2,z3+e0ζ3), it follows from (31) and (38) that (39)D1l1D~2qνρ1z~≈D1l1D~2qνrz~+Oz3≈z1pν-l1.Similarly, since ρ2(z1,ζ′′)=ρ1(z1,ζ2,ζ3+e1ζ2), it follows from (36) that (40)D1l1D~2qνρ2z~≈D1l1D~2qνρ1z~+∑j=0l1D1je1≈z1pν-l1,because pν=tqν<t1 for qν≥2. This proves (33) for i=2.

By induction assume that (33) holds for i=2,…,s. If k=qν=αν+βν with αν>0 and βν>0, that is, if D~2k has mixed derivatives of ∂/∂ζ2 and ∂/∂ζ¯2, we note that (37) becomes(41)D1l1D~2kρs+1z~=D1l1D~2kρsz~, if k≤s, and=D1l1D~2kρsz~+O∑j=0l1D1jesz~, if k>s.If k=qν≤s, (33) follows from (41) and induction hypothesis of (33). If qν>s, it follows, from (36), (41), and induction hypothesis of (33), that (42)D1l1D~2qνρs+1z~=D1l1D~2qνρsz~+O∑j=0l1D1jesz~≈z1pν-l1,because ts>pν=tqν for qν>s. Therefore (33) is proved for i=s+1.

Recall the expression of ρ=ρm+1 and coefficient functions aj,k(z~1) in (23).

Corollary 13. Assume that z~=(z1,0,z3) satisfies (31). Then(43)D1l1aj,kz1≲z1tj+k-l1,and if j+k=qν for some qν in (14), then(44)D1l1aj,kz1≈z1pν-l1.

Proof. From (23) we see that (45)D1l1aj,kz1=D1l1D~2j+kρζ~,where ζ~=(z1,0,0) and j,k>0. Hence it follows from (32) that (46)D1l1aj,kz1=D1l1D~2j+kρζ~≲z1tj+k-l1.Assume qν=j+k≤m for some qν. Thus j,k>0 and it follows from (23), (33), and (41) that (47)D1l1aj,kz1=D1l1D~2qνρζ~=D1l1D~2qνρqνζ~≈z1tqν-l1=z1pν-l1,because tqν=pν.

Remark 14. Suppose that qν-1<l≤qν and pν≤tl<pν-1. Then (pν,qν), (tl,l), and (pν-1,qν-1) are colinear points. From the standard interpolation method, we have(48)atlbl≤apνbqν+apν-1bqν-1,for all sufficiently small a,b≥0. Assume that j,k>0 and l=j+k≠qν for any of ν=1,2,…,N. Therefore it follows from (43) and (48) that (49)aj,kz1ζ2jζ¯2k≲z1tj+kζ2j+k≤z1pνζ2qν+z1pν-1ζ2qν-1.Therefore the terms of the form aj,k(z1)ζ2jζ¯2k, with j+k=qν for some qν, in the summation part in (23), are the major terms which bounds the other summation terms from above.

In the sequel, we assume that z~=(z1,0,z3) satisfies (31). As in Section 1 in [2], for each z~=(z1,0,z3), set(50)Alz~=Alz1=maxaj,kz1;j+k=l, l=2,…,m.In view of Remark 14, we will consider Al(z1) only for l=qν, 0≤ν≤N-1. From (9) and (44) we note that(51)Aq0z1=Amz1≈1, andAqνz1≈z1pν, 1≤ν≤N-1,because q0=m. For each sufficiently small δ>0, set(52)τz~,δ=τz1,δ=minδAqνz11/qν;0≤ν≤N-1,and set(53)Tz~,δ=Tz1,δ=minqν;δAqνz11/qν=τz1,δ.From (51) and (52), we see that if δ′<δ, then(54)δ′δ1/2τz~,δ≤τz~,δ′≤δ′δ1/mτz~,δ.

Lemma 15. For each 0<ϵ≤1, T(z~,ϵδ)≤T(z~,δ).

Proof. Set qνϵ=T(z~,ϵδ) and qν=T(z~,δ). Then (55)τz~,ϵδ=ϵδAqνϵz~1/qνϵ=ϵ1/qνϵδAqνϵz~1/qνϵ≥ϵ1/qνϵτz~,δ=ϵ1/qνϵδAqνz~1/qν=ϵ1/qνϵ-1/qνϵδAqνz~1/qν≥ϵ1/qνϵ-1/qντz~,ϵδ.Therefore qνϵ≤qν because 0<ϵ≤1.

Proposition 16. Assume z~=(z1,0,z3) satisfies (31). Then(56)τz~,δ≈τz~δ,δ,where z~δ=(dδ1/η,0,eδ).

Proof. By (31), we note that z1≈δ1/η. Assume that T(z~δ,δ)=qν. Then Aqν(z~δ)≈δpν/η by (51). Therefore it follows, from (50) and (52), that (57)Aqνz~≈z1pν≈δpν/η≈Aqνz~δ,and hence it follows from (52) and (53) that (58)τz~δ,δqν=δAqνz~δ≈δAqνz~≥τz~,δqν.Thus τ(z~δ,δ)≳τ(z~,δ) follows. Similarly, one can show that τ(z~δ,δ)≲τ(z~,δ).

Let 0<σ<1 be a small constant to be determined (in Remark 22). By Lemma 15, T(z~δ,σδ)≤T(z~δ,δ) for each 0<σ<1, independent of δ>0. Therefore there is a smallest integer s=s(z~δ), 0≤s≤m-1, such that(59)Tz~δ,σs+1δ=Tz~δ,σsδ≔ts.Then ts=qν(s) for some qν(s) by (53). In following, for the fixed integer s=s(z~δ) in (59), set δs=σsδ, τs≔τ(z~δ,δs), and τ1=δ1/η as usual. If we define Φz~(ζ)=(ζ1,ζ2,Φ3(ζ′′)), where Φ3(ζ′′) is defined in (22), we may regard that Φz~(ζ):C3→C3 is a biholomorphism. For each z~δ=(dδ1/η,0,eδ)∈bΩ, set ζ~δ=(dδ1/η,0,0)=Φz~δ-1(z~δ). For each small γ>0, define(60)Rγδsζ~δ≔ζζ1-dδ1/η<γτ1, ζ2<γτs, ζ3<γδs, and,Qγδsz~δ≔Φz~δζ;ζ∈Rγδsζ~δ,and set Rγδ0(ζ~δ)=Rγδ(ζ~δ) and Qγδ0(z~δ)=Qγδ(z~δ) when s=0.

Proposition 17. The function ρ=r∘Φz~δ satisfies(61)D1l1ρζ~δ≲δτ1-l1, and,D1l1D~2kρζ~δ≲δsτ1-l1τs-k, 1≤k≤m.

Proof. Recall that ρ=ρm+1, and z1=δ1/η in (32). When k=0, it follows from (12) (t0=η) and (32) that (62)Dl1ρζ~δ≲δ1/ηη-l1=δτ1-l1.Assume 1≤k≤m. Then by (12), qν-1<k≤qν for some ν, and hence it follows that pν≤tk≤pν-1. Therefore one obtains, from (48)–(52), that(63)z1tkτsk≲z1pν-1τsqν-1+z1pντsqν≲Aqν-1τsqν-1+Aqντsqν≲δs.From (32) and (63), it follows that (64)D1l1D~2kρζ~δ≲z1tk-l1=z1tkτskz1-l1τs-k≲δsτ1-l1τs-k.

Using the z coordinates defined in (14), set (65)L3=∂∂z3 andLk=∂∂zk-∂r∂z3-1∂r∂zk∂∂z3≔∂∂zk+bkz∂∂z3, k=1,2.Then Lk, k=1,2, are tangential holomorphic vector fields and L3r≥c0>0 on V∩Ω for a uniform constant c0>0. For any j,k with j,k>0, define (66)Lj,k∂∂¯rz=L2…L2︸j-1times L¯2…L¯2︸k-1times ∂∂¯rL2,L¯2z.In ζ-coordinates defined by z=z1,Φz~ζ′′≔Φz~(ζ), set Lk′=dΦz~-1Lk, k=1,2,3 and set b~k(ζ)=bk(Φz~(ζ)), k=1,2. If we define (67)Lj,k′∂∂¯ρζ=L2′…L2′︸j-1times L¯2′…L¯2′︸k-1times ∂∂¯ρL2′,L¯2′ζ,then by functoriality, (68)Lj,k∂∂¯rz=Lj,k′∂∂¯ρζ.

Lemma 18. There is a small constant c2>0 such that(69)∂∂¯rzL1,L¯1≥c2δτ1-2, z∈Qγδz~δ,provided γ>0 is sufficiently small.

Proof. Since the level sets of ρ are pseudoconvex, it follows from (61) that(70)∂∂¯ρζL1′,L¯1′=∂2ρ∂ζ1∂ζ¯1ζ+Ob~1≥∂2ρ∂ζ1∂ζ¯1ζ-C~1δτ1-1.Recall that d0(z1)=∑α1+β1=ηaα1,β1z1α1z¯1β1 is the term which contains only z1 or z¯1 variables in the first summation part of (14). Therefore it follows, from (17), (19), and (23), that(71)∂2ρ∂ζ1∂ζ¯1ζ~δ=∂2rz~δ∂z1∂z¯1=∂2d0z1∂z1∂z¯1+Oeδ+z1η-1≈z1η-2=δτ1-2,because z1=dδ1/η=τ1=δ1/η. If ζ∈Rγδ(ζ~δ), it follows from (61) and (71) and by using the Taylor series method that(72)∂2ρ∂ζ1∂ζ¯1ζ≥∂2ρ∂ζ1∂ζ¯1ζ~δ-γC~1δτ1-2≥2c2δτ1-2, ζ∈Rγδζ~δ,provided γ>0 is sufficiently small. Thus (69) follows from (68), (70), and (72).

In the sequel, we let c2 and C2 be the constants which may different from time to time but depend only on the derivatives of r or ρ up to order η. Recall that b~k(ζ)=bk(Φz~(ζ)), k=1,2. By using (61), and by using Taylor series method, one obtains that(73)D1l1D2l2b~kζ≤C2δsτ1-l1τs-l2τ~k-1, ζ∈Rγδsζ~δ,provided γ>0 is sufficiently small, where τ~1=τ1 and τ~2=τs. Note that we can write (74)∂∂¯ρζL2′,L¯2′=∂2ρ∂ζ2∂ζ¯2+R1,where R1=O(b~2). By applying L2′ or L¯2′ successively to ∂∂¯ρ(ζ)(L2′,L¯2′), we obtain that(75)D1l1D~2l2Lj,k′∂∂¯ρζ=D1l1D~2l2∂j+kρ∂ζ2j∂ζ¯2k+D1l1D~2l2Rj+k-1,where, by (73) and by using induction method, Rj+k-1 satisfies(76)D1l1D~2l2Rj+k-1ζ≤C2δsτ1-l1τs-l2-j-k+1, ζ∈Rγδsζ~δ.Combining the estimate in (61), (75), and (76), one obtains that(77)Lj,k′∂∂¯ρζ≤C2δsτs-j-k, ζ∈Rγδsζ~δ.

Assume that (59) holds. Thus ts=qν(s) for some qν(s), and hence it follows from (53) that Aqν(s)(z1)=δsτs-qν(s). Therefore it follows from (23) and (50) that there exist integers j,k>0 with j+k=ts=qν(s), such that(78)∂j+kρ∂ζ2j∂ζ¯2kζ~δ=aj,kz~δ=Aqνsz1=δsτs-qνs=δsτs-j-k.For these j,k>0, it follows from (61), (75), (76), and (78) and by using the Taylor series method that there are constants c2,C2>0 such that (79)c2δsτs-j-k≤Lj,k∂∂¯rz≤C2δsτs-j-k, z∈Qγδsz~δ,provided γ>0 is sufficiently small.

Lemma 19. There is C2>0 such that (80)∂∂¯rzL1,L¯2≤C2γδsτ1-1τs-1, z∈Qγδsz~δ.

Proof. By functoriality, we have (81)∂∂¯rzL1,L¯2=∂∂¯ρζL1′,L¯2′=∂2ρ∂ζ1∂ζ¯2ζ+Ob~1ζ+b~2ζ.From (23), we see that (82)D1∂2ρ∂ζ1∂ζ¯2ζ~δ=Oeδ=Oδ=∂2ρ∂ζ1∂ζ¯2ζ~δ,and it follows from (61) that (83)D~2∂2ρ∂ζ1∂ζ¯2ζ~δ≲δsτ1-1τs-2.Therefore (80) follows from (73), (81), and (83) and by using Taylor series method.

Note that T(z~δ,σsδ)≔ts=qν(s), for some qν(s), and hence there exist j>0, k>0 with j+k=ts. In view of (79), we may assume that (84)L2ReLj-1,k∂∂¯rz≈δsτs-ts, z∈Qγδsz~δ,is valid (when j=1, we replace Lj-1,k by Lj,k-1). Set (85)Gz=ReLj-1,k∂∂¯rz.By using the estimates (73)–(76), one obtains that (86)L1G≤D1∂ts-1ρ∂ζ2j-1∂ζ-2k+D1Rj+k-1+b1G≤C2δsτ1-1τs-ts+1,and similarly, (87)∂∂¯GLj,L¯kz≤C2δsτ~j-1τ~k-1τs-ts+1, j,k=1,2,for z∈Qγδs(z~δ), where τ~1=τ1 and τ~2=τs.

Lemma 20. Assume that (59) holds. Then(88)Gz≤C2σ1/tsδsτs-ts+1, z∈Qγδsz~δ.

Proof. Suppose z∈Qγδs(z~δ). In view of (51)–(53), (56), and (59), we see that (89)σs+1δAts-1z1/ts-1≥τz,σs+1δ=σs+1δAtsz1/ts=σ1/tsτz,σsδ≈σ1/tsτs,and hence it follows that (90)Ats-1z≲σ1/tsδsτs-ts+1.This together with (73)–(78) implies the estimate (88).

In the sequel, we write (91)L=a1L1+a2L2+a3L3.

Lemma 21. There is a positive number σ>0, independent of z~δ and δ, such that if z∈Qγδs(z~δ) and if (59) holds, then there are constants c2>0 and C2>0, independent of z~, δ and σ>0, such that(92)∂∂¯G2L,L¯z≥c2δs2τs-2tsa22-C2δs2τ1-2τs-2ts+2a12-C2a32.

Proof. Suppose z∈Qγδs(z~δ). From (87) and (88), we note that(93)Gz∂∂¯GzLj,L¯k≲σ1/tsδs2τs-2ts+2τ~j-1τ~k-1,for j,k=1,2 where τ~1=τ1 and τ~2=τs. Using (84)–(88) and (93) and by using small (large) constant method, one obtains that (94)∂∂¯Gz2L,L¯=2LGz2+2Gz∂∂¯GL,L¯z≥2c2δs2τs-2tsa22+4Re∑1≤j<k≤3LjGL¯kGaja¯k+2∑1≤j≤k≤3Gz∂∂¯GzLj,L¯kaja¯k≥c2δs2τs-2tsa22-C2δs2τ1-2τs-2ts+2a12-C2a32,for some c2>0 and C2>0 provided σ>0 is sufficiently small.

Remark 22. From now on, we fix constants c2>0 and C2>0, which depend only on the derivatives of r or ρ of order up to η on V, satisfying (69), (73), (80), and (86)–(92), and set C2=c2-1 for a convenience. Now we choose and fix γ>0 and then fix σ>0 so that(95)40C22γ1/2≤1, and420C24γ-7/2σ2/m≤116.

3. Estimates on the Bergman Kernels Recall that z~δ=π(z(δ))=(dδ1/η,0,eδ)∈bΩ where z(δ)=(dδ1/η,0,-δ) and where π is the projection defined before (19). Also note that Φz~δ(ζ~δ)=z~δ where ζ~δ=(dδ1/η,0,0) and where Φz~δ is the holomorphic coordinate function defined in Proposition 11 about z~=z~δ. Also recall Cb(z0,δ0) defined in (20). In this section we estimate the Bergman kernel function KΩ(z,zδ), for z∈Ω and zδ∈Cb(z0,δ0).

To get optimal estimates of the Bergman kernel, we need to construct a plurisubharmonic function which has maximal Hessian near each thin neighborhood of bΩ as in [2, 15]. It contains complicated estimates depending on the type conditions of each boundary points. In this paper, however, we will construct such functions only at z~δ∈bΩ. This will make the estimates much simpler than those in [2, 15] but still contain many complicated estimates.

Note that σ>0 and γ>0 are fixed in Remark 22 and hence the type ts and the integer s defined in (59) depend only on z~δ∈bΩ. Recall that δs=σsδ, τ1=δ1/η, τ2=τ(z~δ,δ), and τs=τ(z~δ,δs). From (54) we have(96)σs/2τ2≤τs≤σs/mτ2.Let us write L=a1L1+a2L2+a3L3.

Proposition 23. There exist a smooth plurisubharmonic function gz~δ on Ω¯ that satisfies the following:

(i) gz~δz≤1, for z∈Ω¯, and gz~δ is supported in Qγδs(z~δ)∩Ω¯.

(ii) There exist a small constant b>0 such that if z∈Q2bδ(z~δ)∩Ω¯, then (97)∂∂¯gz~δL,L¯z≈τ1-2a12+τ2-2a22+δ-2a32.

(iii) If Φz~δ(ζ)=(z1,z2,Φ3(ζ)) where Φ3 is defined in (22), then (98)D~αgz~δ∘Φζ≤Cατ1-α1τ2-α2δ-α3.holds for all ζ∈Rγδs(ζ~δ) where D~α=D~1α1D~2α2D~3α3.

Proof. For each fixed z~δ, we note that the integers s=s(z~δ) and ts, defined in (59), will be fixed. Set τ~1=τ1 and τ~2=τs. Note that γ-2σ-4sτ~i2≤1 provided δ>0 is sufficiently small. Since δs=σsδ, it follows from (80) that(99)∂∂¯rzL1,L¯2a1a¯2≤C2γδτ1-2a12+σ2sτs-2a22, and∂∂¯rzLi,L¯3aia¯3≤C2γσ2sδτ~i-2ai2+C2γ-1σ-2sδ-1τ~i2a32≤C2γσ2sδτ~i-2ai2+δ-1a32, i=1,2,for z∈Qγδs(z~δ). From now on, we fix λ=420C22γ-9/2 and set λs=σ-2sλ.

We may assume that the level sets of r are pseudoconvex on V and L3r2≥c02>0 on V∩Ω, where we may assume that c02≥4c2. Also 4C2γ1/2≤c2/10 by (95). Therefore it follows from (69) and (99) that(100)λsδ-1∂∂¯rL,L¯+λsδ-12Lr2=λsδ-1∑k=13∂∂¯rLk,L¯kak2+2λsδ-1Re∑1≤j<k≤3∂∂¯rLj,L¯kaja¯k+λs2δ-2a32L3r2≥λsδ-14c25δτ1-2a12+∂∂¯rzL2,L¯2-c210γ1/2σ2sδτs-2a22+3c2λs2δ-2a32,for z∈Qγδs(z~δ).

Let ψ(ζ) be defined by (101)ψζ=χτ1-2ζ1-dδ1/η2+τs-2ζ22+δs-2ζ32,where χ is a smooth function such that χ(t)=1 for t<γ2/9 and χ(t)=0 for t≥γ2, satisfying Dkχ≤Ckγ-2k. Set Ψ(z)=ψ((Φz~δ)-1(z)). Note that Φz~δ-1(z) has similar expression as in (22). Thus it follows, from (22), (29), (30), and chain rule, that(102)DαΨz≤Cαγ-2ατ1-α1τs-α2δs-α3 z∈Qγδsz~δ.Here α=(α1,α2,α3) and α=α1+α2+α3. Since C2=c2-1, one obtains(103)∂∂¯ΨzL,L¯≤C2γ-4τ1-2a12+τs-2a22+δs-2a32,λsδ-1LiΨzaia3≤10C2γ-4τ~i-2ai2+c210λs2δ-2a32, i=1,2,where τ~1=τ1 and τ~2=τs.

Suppose that z satisfies Ψ(z)≥1/4. Using the fact that Lkr=0, k=1,2, and the fact that 84C2γ-9/2=(c2/5)λ≤(c2/5)λs, it follows from (100)–(103) that(104)∂∂¯Ψeλsδ-1rL,L¯=eλsδ-1r∂∂¯ΨL,L¯+2λsδ-1∑i=13ReLiΨL¯3raia¯3+eλsδ-1rλsδ-1Ψ∂∂¯rL,L¯+λs2δ-2ΨLr2≥14eλsδ-1r3c25λsτ1-2a12+c2λs2δ-2a32+14eλsδ-1rλsδ-1∂∂¯rzL2,L¯2-2c25γ1/2λτs-2a22.We note that the negative part in (104) contains γ1/2λ instead of γ1/2λs.

Let h be a smooth convex function such that h(t)=0 for t≤1/2 and h(t)>0 for t>1/2 and h(9/8)≤1. Set Gz~δ(z)=Ψ(z)eλsδ-1r(z) and set gz~δ(z)=h(Gz~δ(z)). Suppose T(z~δ,δ)=2. Then s=0, and hence (79) holds for δs=δ with j=k=1; that is,(105)c2δτ2-2≤∂∂¯rzL2,L¯2≤C2δτ2-2, z∈Qγδz~δ.For those z with Ψ(z)≥1/4, it follows from (104) (with λs=λ) and (105) that(106)∂∂¯Gz~δz≥3c2λ20eλδ-1rτ1-2a12+τ2-2a22+δ-2a32.If Ψ(z)≤1/4, then Gz~δ(z)≤1/4 and hence gz~δ(z)=0. Hence gz~δ is a smooth plurisubharmonic function supported on Qγδ(z~δ), and gz~δ≤1.

Now assume T(z~δ,δ)>2 and assume that (59) holds. Then (79) holds for some positive integers j,k with j+k=ts. Let G(z) be the function defined in (85). From (88) and (95), we see that(107)λγ1/2δs-2τs2ts-2Gz2≤λγ1/2C22σ2/ts≤420C24γ-7/2σ2/m≤116,z∈Qγδs(z~δ), because λ=420C22γ-4 and ts≤m. Set (108)gz~δz=hΨzeλsδ-1rz+ϕλγ1/2δs-2τs2ts-2Gz2,where ϕ(t) is a smooth function that satisfies ϕ(t)=t, for t≤1/16, ϕ(t)=0 for t≥1, and ϕ(t)≤1/8 for all t. Thus gz~δ∈C0∞(Qγδs(z~δ)) and gz~δ≤1 because h(9/8)≤1. By (107) we note that ϕ(z)=z on Qγδs(z~δ), and we also note that gz~δ=0 if Ψ(z)eλsδ-1r(z)≤3/8, in particular, gz~δ=0 outside Qγδs(z~δ). From (92), we obtain that(109)λγ1/2δs-2τs2ts-2∂∂¯Gz2L,L¯≥γ1/2c2λτs-2a22-C2λτ1-2a12-C2λδs-2a32≥c2γ1/2λτs-2a22-c240λτ1-2a12-c240λδs-2a32,for z∈Qγδs(z~δ), because γ1/2C2≤c2/40.

Assuming that Ψ(z)eλsδ-1r(z)≥3/8, we note that the negative coefficient part of a22 of the Hessian of Ψ(z)eλsδ-1r(z) in (104) is controlled by the first term in the third line of (109), and the error terms of the coefficients of a12 and a32 in the third line of (109) are controlled by the corresponding coefficients of the Hessian of Ψ(z)eλsδ-1r(z) in (104). In either T(z~δ,δ)=2 or T(z~δ,δ)>2 cases, it follows from (104), (106), and (109) that(110)∂∂¯gz~δL,L¯≥c232eλsδ-1rzλsτ1-2a12+γ1/2λτs-2a22+λs-2δ-2a32,for z∈Qγδs(z~δ).

Note that parameters, c2, C2, γ, σ, and λ, are fixed in Remark 22, independent of δ>0. Therefore the upper bound of gz~δ follows from (84)–(88), (96), (99), (102), and (103). Note that eλsδ-1r(z)>e-1/4>3/4, if r(z)>-δ/4λs=-δσ2s/4λ, and this property holds on Q2bδ(z~δ) if we take b>0 sufficiently small; say, 0<2b<σ2m/λ2. Also note that Ψ=1 on Q2bδ(z~δ). This fact together with (96) and (110) proves properties (i) and (ii). Property (iii) follows from (22), (30), (32), and (96).

For each zδ=(dδ1/η,0,eδ-bδ)∈Cb(z0,δ0), set ζδ≔Φz~δ-1(zδ)=(dδ1/η,0,-bδ).

Proposition 24. There is a small constant a>0 such that R2aδ(ζδ)⊂⊂Ω for all sufficiently small δ>0.

Proof. From (22)–(29), we obtain that(111)ρζδ=rzδ=-bδ+Oδ1+1/η<-bδ2,for all sufficiently small δ>0. Assume ζ∈R2aδ(ζδ) and write (112)ρζ=ρζ-ρdδ1/η,ζ2,ζ3+ρdδ1/η,ζ2,ζ3-ρζδ+ρζδ≔E1+E2+ρζδ.From (61), and by using Taylor series method, one obtains that (113)E1≤amaxζ~1-dδ1/η<2aδ1/ηD1ρζ~1,ζ2,ζ3δ1/η≤2aC2δ,for a uniform constant C2>0. Similarly, we obtain E2≤4aC2δ. Combining these estimates and (111) and if we set a=b/24C2, then we obtain that (114)ρζ<6aC2δ-bδ2=-bδ4, ζ∈Raδζδ.

Remark 25. (1) Set g~δ(ζ)≔gz~δ∘Φz~δ(ζ). Then, by functoriality, Proposition 23 holds, where gz~δ is replaced by g~δ, and Qγδ(z~δ) is replaced by Rγδ(ζ~δ).

For each fixed δ>0, and for each fixed z~δ=(dδ1/η,0,eδ)∈bΩ, set Ωz~δ=Φz~δ-1(Ω). Note that detJCΦz~δ-1z=2(∂r/∂z3)z~δ≥2c0>0 on V. Thus it follows, from transformation formula, that(115)KΩz,zδ=4∂r∂z3z~δ2KΩz~δζ,ζδ.In view of Propositions 23 and 24, there is a smooth plurisubharmonic weight function gz~δ which has maximal Hessian on Qaδ(zδ)⊂⊂Ω. We also note that τ(z~δ,δ)≈τ(zδ,δ) by (56). If we use these properties and (115), we get the following estimates for the Bergman kernel function KΩ(zδ,zδ) at zδ∈Cb(z0,δ0) as in Theorem 6.1 in [2]:(116)KΩzδ,zδ≈δ-2δ-2/ητzδ,δ-2, zδ∈Cbz0,δ0.This proves Theorem 2.

Now we want to get derivative estimates of K(z,zδ) for z∈Ω and zδ∈Cb(z0,δ0). In view of (115), we will estimate KΩz~δ(ζ,ζδ) where z=Φz~δ(ζ) and zδ=Φz~δ(ζδ). We will follow the methods in [3, 9] which use dilated coordinates. For each fixed δ>0, we recall that τ1=δ1/η, τ2=τ(zδ,δ) and τ3=δ. Define a dilation map Dδ given by(117)Dδζ=ζ1-dδ1/ηaτ1,ζ2aτ2,ζ3+bδaτ3≔w1,w2,w3=w,set (118)ρδw≔δ-1ρ∘Dδ-1w,Ωδ=w∈C3;ρδw<0,and set (119)λδw≔g~δ∘Dδ-1w,where g~δ(ζ)≔gz~δ∘Φz~δ(ζ) and where gz~δ is defined in Proposition 23.

Set (120)L3δ=∂∂w3,Lkδ=∂∂wk-∂ρδ∂w3-1∂ρδ∂wk∂∂w3, k=1,2,and write Lδ=b1L1δ+b2L2δ+b3L3δ. The properties of λδ(w), which follow from Propositions 23 and 24 and Remark 25, are summarized in the following proposition.

Proposition 26. For each δ>0 there is λδ(w), defined on Ωδ, such that

(1) λδ(w) is smooth plurisubharmonic in Ωδ, and λδ≤1;

(2) suppλδ(w)⊂P(0,C~), for some C~=2a-1γ>1;

(3) ∂∂¯λδ(Lδ,L¯δ)(w)≈b12+b22+b32 if w∈P(0,1);

(4) Dwαλδw≤Cα.

The weight function with the properties in Proposition 26 is the key ingredient for the derivative estimates of the Bergman kernel function off the diagonal. Set P=P(0,C~) and let Nδ be the Neumann operator on Ωδ. Then we have the following L2 estimates of Nδ (Proposition 3.14 in [3]).

Proposition 27. Let h∈L2 be a (0,1) form and supph⊂P. Then there is C>0, independent of δ>0, so that(121)∫Ωδ∩PNδh2≤Ch2.

Note that Dδ(ζδ)=0. Set (122)P0,r≔w=w1,w2,w3:wk≤r, k=1,2,3.From (117) and Proposition 24, we note that(123)DδRaδζδ=P0,1⊂⊂P0,2⊂⊂Ωδ,independent of δ>0. Let ξ1,ξ2∈C0∞(P(0,1)) with ξ1=1 in a neighborhood of 0 and ξ2=1 on suppξ1. From (123), we see that suppξ2⊂P(0,1)⊂⊂P(0,2)⊂⊂Ωδ, independent of δ>0. Therefore we have the following elliptic estimates:(124)ξ1fs+22≤Csξ2□δfs2+f2, s≥0, f∈Dom∂¯∩Dom∂¯∗,where □δ is the complex Laplacian on Ωδ.

Remark 28. The estimates in (124) are on the polydisc P(0,1)⊂⊂P(0,2)⊂⊂Ωδ, strictly inside of Ωδ, independent of δ>0. Therefore we gain two derivatives in (124) and it is stable; that is, Cs is independent of δ>0. Also we note that we do not require that Δ1(z0)<∞. Since P(0,2)⊂⊂P=P(0,C~) where C~=2a-1γ>2, we can also apply the estimate (121) on P(0,2).

Let ϕ∈C0∞(P(0,1)), ∫ϕ=1, and ϕ be polyradial. In terms of w-coordinates in (117), we have the following well known representation of Bergman kernel function on Ωδ. (125)KΩδw,0=ϕw-∂¯∗Nδ∂¯ϕw.Let χ∈C∞(Ωδ) with χ=1 outside P(0,1) and χ=0 on suppϕ. Combining (121)–(125), we can prove the following lemma as in the proof of Theorem 4.2 in [3].

Lemma 29. For each s≥0 there is Cs>0 such that (126)χNδ∂¯ϕs≤Cs.

Now, if we use the estimate (126) with s=α+3, we can prove Theorem 3 as in the proof of Theorem 4.2 in [3].