On Compactness of Hankel and the $\overline{\partial }$-Neumann Operators on Hartogs Domains in $\mathbb {C}^2$

Abstract

We prove that on smooth bounded pseudoconvex Hartogs domains in $\mathbb {C}^2$ compactness of the $\overline{\partial }$-Neumann operator is equivalent to compactness of all Hankel operators with symbols smooth on the closure of the domain.

Let $\Omega$ be a bounded pseudoconvex domain in $\mathbb {C}^n$ and $L^2_{(0,q)}(\Omega )$ denote the space of square integrable (0, q) forms for $0\le q\le n$. The complex Laplacian $\Box =\overline{\partial }\overline{\partial }^{*}+\overline{\partial }^{*}\overline{\partial }$ is a densely defined, closed, self-adjoint linear operator on $L^2_{(0,q)}(\Omega )$. Hörmander in [7] showed that when $\Omega$ is bounded and pseudoconvex, $\Box$ has a bounded solution operator $N_q$, called the $\overline{\partial }$-Neumann operator, for all q. Kohn in [9] showed that the Bergman projection, denoted by $\mathbf {B}$ below, is connected to the $\overline{\partial }$-Neumann operator via the following formula

$$\begin{aligned} \mathbf {B}=\mathbf {I}-\overline{\partial }^*N_1\overline{\partial }\end{aligned}$$

where $\mathbf {I}$ denotes the identity operator. For more information about the $\overline{\partial }$-Neumann problem we refer the reader to two books [4, 15].

Let $A^2(\Omega )$ denote the space of square integrable holomorphic functions on $\Omega$ and $\phi \in L^{\infty }(\Omega )$. The Hankel operator with symbol $\phi , H_{\phi }: A^2(\Omega )\rightarrow L^2({\Omega })$ is defined by

$$\begin{aligned} H_{\phi }g=[\phi ,\mathbf {B}]g=\left( \mathbf {I}-\mathbf {B}\right) \left( \phi g\right) . \end{aligned}$$

Using Kohn’s formula one can immediately see that

$$\begin{aligned} H_{\phi }g=\overline{\partial }^*N_1( g\overline{\partial }\phi ) \end{aligned}$$

for $\phi \in C^1(\overline{\Omega })$. It is clear that $H_{\phi }$ is a bounded operator; however, its compactness depends on both the function theoretic properties of the symbol $\phi$ as well as the geometry of the boundary of the domain $\Omega$ (see [6]).

The following observation is relevant to our work here. Let $\Omega$ be a bounded pseudoconvex domain in $\mathbb {C}^n$ and $\phi \in C(\overline{\Omega })$. If $\overline{\partial }$-Neumann operator $N_1$ is compact on $L^2_{(0,1)}(\Omega )$ then the Hankel operator $H_{\phi }$ is compact (see [15, Proposition 4.1]).

We are interested in the converse of this observation. Namely,

Assume that $\Omega$ is a bounded pseudoconvex domain in $\mathbb {C}^n$ and $H_{\phi }$ is compact on $A^2(\Omega )$ for all symbols $\phi \in C(\overline{\Omega })$. Then is the $\overline{\partial }$-Neumann operator $N_1$ compact on $L^2_{(0,1)}(\Omega )?$

This is known as D’Angelo’s question and first appeared in [12, Remark 2].

The answer to D’Angelo’s question is still open in general but there are some partial results. Fu and Straube in [13] showed that the answer is yes if $\Omega$ is convex. Çelik and the first author [2, Corollary 1] observed that if $\Omega$ is not pseudoconvex then the answer to D’Angelo’s question may be no. Indeed, they constructed an annulus type domain $\Omega$ where $H_{\phi }$ is compact on $A^2(\Omega )$ for all symbols $\phi \in C(\overline{\Omega })$; yet, the $\overline{\partial }$-Neumann operator $N_1$ is not compact on $L^2_{(0,1)}(\Omega )$.

FormalPara Remark 1

One can extend the definition of Hankel operators from holomorphic functions to the $\overline{\partial }$-closed (0, q)-forms (denoted by $K^2_{(0,q)}(\Omega )$) and ask the analogous problem at the forms level. In this case, an affirmative answer was obtained in [3]. Namely, for $1\le q\le n-1$ if $H^q_{\phi }=[\phi , \mathbf {B}_q]$ is compact on $K^2_{(0,q)}(\Omega )$ for all symbols $\phi \in C^{\infty }(\overline{\Omega })$ then the $\overline{\partial }$-Neumann operator $N_{q+1}$ is compact on $L^2_{(0,q)}(\Omega )$.

In this paper, we provide an affirmative answer to D’Angelo’s question on smooth bounded pseudoconvex Hartogs domains in $\mathbb {C}^2$.

FormalPara Theorem 1

Let $\Omega$ be a smooth bounded pseudoconvex Hartogs domain in $\mathbb {C}^2$. The $\overline{\partial }$-Neumann operator $N_1$ is compact on $L^2_{(0,1)}(\Omega )$ if and only if $H_{\psi }$ is compact on $A^2(\Omega )$ for all $\psi \in C^{\infty }(\overline{\Omega })$.

As mentioned above, compactness of $N_1$ implies that $H_{\psi }$ is compact on any bounded pseudoconvex domain (see [12, 15, Proposition 4.4.1]). The key ingredient of our proof of the converse is the characterization of the compactness of $N_1$ in terms of ground state energies of certain Schrödinger operators as previously explored in [5, 14].

We will need a few lemmas before we prove Theorem 1.

FormalPara Lemma 1

Let $A(a,b)=\{z\in \mathbb {C}:a<|z|<b\}$ for $0{<}a{<}b{<}\infty$ and $d_{ab}(w)$ be the distance from w to the boundary of A(a, b). Then there exists $C{>}0$ such that

$$\begin{aligned} \int _{A(a,b)}(d_{ab}(w))^2|w|^{2n}dV(w)\le \frac{C}{n^2} \int _{A(a,b)}|w|^{2n}dV(w) \end{aligned}$$

for nonzero integer n.

FormalPara Proof

We will use the fact that $d_{ab}(w)=\min \{b-|w|,|w|-a\}$ with polar coordinates to compute the first integral. One can compute that

$$\begin{aligned} \int _{A(a,b)}|w|^{2n}dV(w)= \frac{\pi }{n+1}\big (b^{2n+2}-a^{2n+2}\big ) \end{aligned}$$

for $n\ne -1$. Let $c=\frac{a+b}{2}$. Then

$$\begin{aligned} \int _{A(a,b)}(d_{ab}(w))^2|w|^{2n}dV(w)= & {} \int _{A(a,c)}(|w|-a)^2|w|^{2n}dV(w)\\&+\,\int _{A(c,b)}(b-|w|)^2|w|^{2n}dV(w)\\= & {} 2\pi \int _a^c\big (a^2\rho ^{2n+1}-2a\rho ^{2n+2}+\rho ^{2n+3}\big )d\rho \\&+\,2\pi \int _c^b\big (b^2\rho ^{2n+1}-2b\rho ^{2n+2}+\rho ^{2n+3}\big )d\rho \\= & {} 2\pi \Big (b^{2n+4}-a^{2n+4}\Big )\left( \frac{1}{2n+2}-\frac{2}{2n+3}+\frac{1}{2n+4}\right) \\&+\,2\pi (a^2-b^2)\frac{c^{2n+2}}{2n+2}+4\pi (b-a)\frac{c^{2n+3}}{2n+3}\\= & {} \frac{\pi (b^{2n+4}-a^{2n+4})}{(n+1)(n+2)(2n+3)}-\frac{\pi c^{2n+2}(b^2-a^2)}{(n+1)(2n+3)}. \end{aligned}$$

In the last equality we used the fact that $c=\frac{a+b}{2}$. Then one can show that

$$\begin{aligned} \lim _{n\rightarrow \pm \infty }\frac{n^2\int _{A(a,b)}(d_{ab}(w))^2|w|^{2n}dV(w)}{\int _{A(a,b)}|w|^{2n}dV(w)} =\frac{b^2}{2}. \end{aligned}$$

Therefore, there exists $C{>}0$ such that

$$\begin{aligned} \int _{A(a,b)}(d_{ab}(w))^2|w|^{2n}dV(w)\le \frac{C}{n^2} \int _{A(a,b)}|w|^{2n}dV(w) \end{aligned}$$

for nonzero integer n. $\square$

We note that throughout the paper $\Vert .\Vert _{-1}$ denotes the Sobolev $-1$ norm.

FormalPara Lemma 2

Let $\Omega =\{(z,w)\in \mathbb {C}^2: z\in D \text { and } \phi _1(z)<|w|<\phi _2(z)\}$ be a bounded Hartogs domain. Then there exists $C>0$ such that

$$\begin{aligned} \Vert g(z)w^n\Vert _{-1}\le \frac{C}{n}\Vert g(z)w^n\Vert \end{aligned}$$

for any $g\in L^2(D)$ and nonzero integer n, as long as the right-hand side is finite.

FormalPara Proof

We will denote the distance from (z, w) to the boundary of $\Omega$ by $d_{\Omega }(z,w)$. We note that $W^{-1}(\Omega )$ is the dual of $W^1_0(\Omega )$, the closure of $C^{\infty }_0(\Omega )$ in $W^1(\Omega )$. Furthermore,

$$\begin{aligned} \Vert f\Vert _{-1}=\sup \{|\langle f,\phi \rangle |:\phi \in C^{\infty }_0(\Omega ), \Vert \phi \Vert _1\le 1\} \end{aligned}$$

for $f\in W^{-1}(\Omega )$. Then there exists $C_1>0$ such that

$$\begin{aligned} \Vert f\Vert _{-1}\le \Vert d_{\Omega }f\Vert \sup \{\Vert \phi /d_{\Omega }\Vert :\phi \in C^{\infty }_0(\Omega ), \Vert \phi \Vert _1\le 1\} \le C_1\Vert d_{\Omega }f\Vert . \end{aligned}$$

In the second inequality above we used the fact that (see [4, Proof of Theorem C.3]) there exists $C_1>0$ such that $\Vert \phi /d_{\Omega }\Vert \le C_1\Vert \phi \Vert _1$ for all $\phi \in W^1_0(\Omega )$.

Let $d_z(w)$ denote the distance from w to the boundary of $A(\phi _1(z),\phi _2(z))$. Then there exists $C_1>0$ such that

$$\begin{aligned} \Vert g(z)w^n\Vert _{-1}^2\le&C_1\int _{\Omega }(d_{\Omega }(z,w))^2|g(z)|^2|w|^{2n}dV(z,w)\\ \le&C_1\int _D|g(z)|^2\int _{\phi _1(z)<|w|<\phi _2(z)}(d_z(w))^2|w|^{2n}dV(w). \end{aligned}$$

Lemma 1 and the assumption that $\Omega$ is bounded imply that there exists $C_2>0$ such that

$$\begin{aligned} \int _{\phi _1(z)<|w|<\phi _2(z)}(d_z(w))^2|w|^{2n}dV(w)\le \frac{C_2}{n^2}\int _{\phi _1(z)<|w|<\phi _2(z)}|w|^{2n}dV(w). \end{aligned}$$

Then

$$\begin{aligned}&\int _D|g(z)|^2\int _{\phi _1(z)<|w|<\phi _2(z)}(d_z(w))^2|w|^{2n}dV(w)\\&\quad \le \frac{C_2}{n^2}\int _D|g(z)|^2\int _{\phi _1(z)<|w|<\phi _2(z)}|w|^{2n}dV(w)\\&\quad =\frac{C_2}{n^2} \Vert g(z)w^n\Vert ^2. \end{aligned}$$

Therefore, for $C=\sqrt{C_1C_2}$ we have $\Vert g(z)w^n\Vert _{-1}\le \frac{C}{n}\Vert g(z)w^n\Vert$ for nonzero integer n. $\square$

FormalPara Lemma 3

Let $\Omega$ be a bounded pseudoconvex domain in $\mathbb {C}^n$ and $\psi \in C^1(\overline{\Omega })$. Then $H_{\psi }$ is compact if and only if for any $\varepsilon >0$ there exists $C_{\varepsilon }>0$ such that

$$\begin{aligned} \Vert H_{\psi }h\Vert ^2\le \varepsilon \Vert h\overline{\partial }\psi \Vert \Vert h\Vert +C_{\varepsilon }\Vert h\overline{\partial }\psi \Vert _{-1}\Vert h\Vert \end{aligned}$$

(1)

for $h\in A^2(\Omega )$.

FormalPara Proof

First assume that $H_{\psi }$ is compact. Then

$$\begin{aligned} \Vert H_{\psi }h\Vert ^2= \langle H^*_{\psi }H_{\psi } h,h\rangle \le \Vert H^*_{\psi }H_{\psi } h\Vert \Vert h\Vert \end{aligned}$$

for $h\in A^2(\Omega )$. Compactness of $H_{\psi }$ implies that $H^*_{\psi }$ is compact. Now we apply the compactness estimate in [8, Proposition V.2.3] to $H^*_{\psi }$. For $\varepsilon >0$ there exists a compact operator $K_{\varepsilon }$ such that

$$\begin{aligned} \Vert H^*_{\psi }H_{\psi } h\Vert \le&\frac{\varepsilon }{2\Vert \overline{\partial }^*N\Vert } \Vert H_{\psi } h\Vert + \Vert K_{\varepsilon }H_{\psi } h\Vert \\ \le&\frac{\varepsilon }{2} \Vert h\overline{\partial }\psi \Vert + \Vert K_{\varepsilon }H_{\psi } h\Vert . \end{aligned}$$

In the second inequality we used the fact that $H_{\psi }h=\overline{\partial }^*N(h\overline{\partial }\psi )$. Since $\Omega$ is bounded pseudoconvex $\overline{\partial }^*N$ is bounded and hence $K_{\varepsilon }\overline{\partial }^*N$ is compact. Now we use the fact that $H_{\psi }h=\overline{\partial }^*N(h\overline{\partial }\psi )$ and [15, Lemma 4.3] for the compact operator $K_{\varepsilon }\overline{\partial }^*N$ to conclude that there exists $C_{\varepsilon }>0$ such that

$$\begin{aligned} \Vert K_{\varepsilon }H_{\psi } h\Vert \le \frac{\varepsilon }{2}\Vert h\overline{\partial }\psi \Vert +C_{\varepsilon }\Vert h\overline{\partial }\psi \Vert _{-1}. \end{aligned}$$

Therefore, for $\varepsilon >0$ there exists $C_{\varepsilon }>0$ such that

$$\begin{aligned} \Vert H_{\psi }h\Vert ^2\le \varepsilon \Vert h\overline{\partial }\psi \Vert \Vert h\Vert +C_{\varepsilon }\Vert h\overline{\partial }\psi \Vert _{-1}\Vert h\Vert \end{aligned}$$

for $h\in A^2(\Omega )$.

To prove the converse assume (1) and choose $\{h_j\}$ a sequence in $A^2(\Omega )$ such that $\{h_j\}$ converges to zero weakly. Then the sequence $\{h_j\}$ is bounded and $\Vert h_j\overline{\partial }\psi \Vert _{-1}$ converges to 0 (as the imbedding from $L^2$ into Sobolev $-1$ is compact). The inequality (1) implies that there exists $C>0$ such that for every $\varepsilon >0$ there exists J such that $\Vert H_{\psi }h_j\Vert ^2 \le C\varepsilon$ for $j\ge J$. That is, $\{H_{\psi }h_j\}$ converges to 0. That is, $H_{\psi }$ is compact. $\square$

The following lemma is contained in [10, Remark 1]. The superscripts on the Hankel operators are used to emphasize the domains.

FormalPara Lemma 4

([10]) Let $\Omega _1$ be a bounded pseudoconvex domain in $\mathbb {C}^n$ and $\Omega _2$ be a bounded strongly pseudoconvex domain in $\mathbb {C}^n$ with $C^2$-smooth boundary. Assume that $U=\Omega _1\cap \Omega _2$ is connected, $\phi \in C^1(\overline{\Omega }_1)$, and $H^{\Omega _1}_{\phi }$ is compact on $A^{2}(\Omega _1)$. Then $H^{U}_{\phi }$ is compact on $A^{2}(U)$.

Now we are ready to prove Theorem 1.

FormalPara Proof of Theorem 1

We present the proof of the nontrivial direction. That is, we assume that $H_{\psi }$ is compact on $A^2(\Omega )$ for all $\psi \in C^{\infty }(\overline{\Omega })$ and prove that $N_1$ is compact. Our proof is along the lines of the proof of [5, Theorem 1.1].

Let $\rho (z,w)$ be a smooth defining function for $\Omega$ that is invariant under rotations in w. That is, $\rho (z,w)=\rho (z,|w|)$,

$$\begin{aligned} \Omega = \{(z, w) \in \mathbb {C}^2: \rho (z, w) < 0\}, \end{aligned}$$

and $\nabla \rho$ is nonvanishing on $b\Omega$. Let $\Gamma _0 = \{(z, w) \in b\Omega : \rho _{|w|}(z, |w|)=0\}$ and

$$\begin{aligned} \Gamma _k = \{(z, w) \in b\Omega : |\rho _{|w|}(z, |w|)|\ge 1/k\} \end{aligned}$$

for $k=1,2,\ldots$. We will show that $\Gamma _k$ is B-regular for $k=0,1,2,\ldots$ by establishing the estimates (2) and (3) below and invoking [5, Lemma 10.2]. Then

$$\begin{aligned} b\Omega =\bigcup _{k=0}^{\infty }\Gamma _k \end{aligned}$$

and [11, Proposition 1.9] implies that $b\Omega$ is B-regular (satisfies Property (P) in Catlin’s terminology). This will be enough to conclude that $N_1$ is compact on $L^2_{(0,1)}(\Omega )$

The proof of the fact that $\Gamma _0$ is B-regular is essentially contained in [5, Lemma 10.1] together with the following fact: Let $\Omega$ be a smooth bounded pseudoconvex domain in $\mathbb {C}^2$. If $H_{\overline{z}}$ and $H_{\overline{w}}$ are compact on $A^2(\Omega )$ then there is no analytic disc in $b\Omega$ (see [6, Corollary 1]).

Now we will prove that $\Gamma _k$ is B-regular for any fixed $k\ge 1$. Let $(z_0,w_0)\in \Gamma _k$, we argue in two cases. The first case is when $\rho _{|w|}(z_0,|w_0|)<0$ and the second case is $\rho _{|w|}(z_0,|w_0|)>0$.

We continue with the first case. Assume that $b\Omega$ near $(z_0,w_0)$ is given by $|w|=e^{-\varphi (z)}$. Let $D(z_0,r)$ denote the disc centered at $z_0$ with radius r and

$$\begin{aligned} U_{a,b}=D(z_0,a)\times \{w\in \mathbb {C}:|w_0|-b<|w|<|w_0|+b\} \end{aligned}$$

for $a,b>0$. Then let us choose $a,a_1,b,b_1>0$ such that $a_1>a,b_1>|w_0|+b$, the open sets

$$\begin{aligned} U=\Omega \cap U_{a,b}= \left\{ (z,w)\in \mathbb {C}^2:z\in D(z_0,a), ~e^{-\varphi (z)}<|w|<|w_0|+b\right\} \end{aligned}$$

and $U_1=\Omega \cap U^{a_1,b_1}$ are connected where

$$\begin{aligned} U^{a_1,b_1}=\left\{ (z,w)\in \mathbb {C}^2:\frac{|z-z_0|^2}{a_1^2}+\frac{|w|^2}{b_1^2}<1\right\} , \end{aligned}$$

and finally $\overline{U}\subset U_1$. Then

$$\begin{aligned} U_1=\left\{ (z,w)\in \mathbb {C}^2:z\in V_1, e^{-\varphi (z)}<|w|<e^{-\alpha (z)}\right\} \end{aligned}$$

where $V_1$ is a domain in $\mathbb {C}$ such that $\overline{D(z_0,a)}\subset V_1\subset D(z_0,a_1)$ and

$$\begin{aligned} \alpha (z)=\log a_1-\log b_1-\frac{1}{2}\log \big (a_1^2-|z-z_0|^2\big ). \end{aligned}$$

One can check that $\alpha$ is subharmonic on $D(z_0,a_1)$, while pseudoconvexity of $\Omega$ implies that the function $\varphi$ is superharmonic on $D(z_0,a_1)$. Furthermore, since B-regularity is invariant under holomorphic change of coordinates, by mapping under $(z,w)\rightarrow (z,\lambda w)$ for some $\lambda >1$, we may assume that

$$\begin{aligned} U_1\subset D(z_0,a_1)\times \{w\in \mathbb {C}:|w|>1\}. \end{aligned}$$

For any $\beta \in C^{\infty }_0(D(z_0,a))$ let us choose $\psi \in C^{\infty }(\overline{V_1})$ such that $\psi _{\overline{z}}=\beta$. Lemma 4 implies that the Hankel operator $H^{U_1}_{\psi }$ (we use the superscript $U_1$ to emphasize the domain) is compact on the Bergman space $A^2(U_1)$.

Let

$$\begin{aligned} \lambda _n(z)=-\log \left( \frac{\pi }{n-1}\left( e^{(2n-2)\varphi (z)} -e^{(2n-2)\alpha (z)}\right) \right) \end{aligned}$$

for $n=2,3,\ldots$. One can check that since $\varphi$ is superharmonic and $\alpha$ is subharmonic, the function $\lambda _n$ is subharmonic. Let $S^{V_1}_{\lambda _n}$ be the canonical solution operator for $\overline{\partial }$ on $L^2(V_1,\lambda _n)$. If $f_n=H^{U_1}_{\psi }w^{-n}$ then we claim that

$$\begin{aligned} f_n(z,w)=g_n(z)w^{-n} \end{aligned}$$

where $g_n=S^{V_1}_{\lambda _n}(\beta d\overline{z})$ and $n=2,3,\ldots$. Clearly $H^{U_1}_{\psi }w^{-n}=f_n\in L^2(U_1)$ and

$$\begin{aligned} \overline{\partial }g_n(z)w^{-n}=\beta (z)w^{-n}d\overline{z}. \end{aligned}$$

To prove the claim we will just need to show that $g_n(z)w^{-n}$ is orthogonal to $A^2(U_1)$. That is, we need to show that $\langle g_n(z)w^{-n}, h(z)w^m\rangle _{U_1}=0$ for any $h(z)\in A^2(V_1)$ and $m\in \mathbb {Z}$. Then

$$\begin{aligned} \langle g_n(z)w^{-n}, h(z)w^m\rangle _{U_1}&=\int _{U_1}g_n(z)w^{-n} \overline{h(z)w^m}dV(z)dV(w)\\&=\int _{V_1}g_n(z)\overline{h(z)}dV(z) \int _{e^{-\varphi (z)}<|w|<e^{-\alpha (z)}}w^{-n}\overline{w^m}dV(w). \end{aligned}$$

Unless $m=-n$ the integral $\int _{e^{-\varphi (z)}<|w|<e^{-\alpha (z)}}w^{-n}\overline{w^m}dV(w)=0$. So let us assume that $m=-n$. In that case we get

$$\begin{aligned} \int _{V_1}g_n(z)\overline{h(z)}dV(z) \int _{e^{-\varphi (z)}<|w|<e^{-\alpha (z)}}w^{-n}\overline{w^m}dV(w) =\int _{V_1}g_n(z)\overline{h(z)}e^{-\lambda _n(z)}dV(z). \end{aligned}$$

The integral on the right-hand side above is zero because $g_n$ is orthogonal to $A^2(V_1,\lambda _n)$. Therefore,

$$\begin{aligned} g_n(z)w^{-n}=H^{U_1}_{\psi }w^{-n}. \end{aligned}$$

The equality above implies that $\frac{\partial g_n}{\partial \overline{z}}=\frac{\partial \psi }{\partial \overline{z}}=\beta$. Then the compactness estimate (1) implies that

$$\begin{aligned} \int _{D(z_0,a)}|g_n(z)|^2e^{-\lambda _n(z)}dV(z)\le & {} \Vert g_n(z)w^{-n}\Vert _{U_1}^2\\\le & {} \varepsilon \Vert \beta (z)w^{-n}\Vert _{U_1}\Vert w^{-n}\Vert _{U_1} \\&+\,C_{\varepsilon }\Vert \beta (z)w^{-n}\Vert _{W^{-1}(U_1)}\Vert w^{-n}\Vert _{U_1} \\= & {} \varepsilon \left( \int _{D(z_0,a)}|\beta (z)|^2e^{-\lambda _n(z)}dV(z)\right) ^{1/2}\\&\quad \times \, \left( \int _{V_1} e^{-\lambda _n(z)}dV(z)\right) ^{1/2} \\&+\, C_{\varepsilon } \Vert \beta (z)w^{-n}\Vert _{W^{-1}(U_1)} \left( \int _{V_1} e^{-\lambda _n(z)}\right) ^{1/2}. \end{aligned}$$

Then by Lemma 2 there exists $C>0$ such that

$$\begin{aligned} \Vert \beta (z)w^{-n}\Vert _{W^{-1}({U_1})}\le \frac{C}{n}\Vert \beta (z)w^{-n}\Vert _{U_1} = \frac{C}{n}\Vert \beta \Vert _{L^2(D(z_0,a),\lambda _n)}. \end{aligned}$$

We note that to get the equality above we used the fact that $\beta$ is supported in $D(z_0,a)$. Hence we get

$$\begin{aligned} \Vert g_n\Vert ^2_{L^2(D(z_0,a),\lambda _n)} \le \left( \varepsilon +\frac{CC_{\varepsilon }}{n} \right) \Vert \beta \Vert _{L^2(D(z_0,a),\lambda _n)}\Vert 1\Vert _{L^2(V_1,\lambda _n)}. \end{aligned}$$

For any $\varepsilon >0$ there exists an integer $n_{\varepsilon }$ such that

$$\begin{aligned} \frac{CC_{\varepsilon }}{n}+ \frac{\pi a_1}{\sqrt{n-1}}\le \varepsilon \end{aligned}$$

for $n\ge n_{\varepsilon }$. Then

$$\begin{aligned} \Vert g_n\Vert _{L^2(D(z_0,a),\lambda _n)}^2\le 2\varepsilon \Vert \beta \Vert _{L^2(D(z_0,a),\lambda _n)} \Vert 1\Vert _{L^2(V_1,\lambda _n)} \le 2\varepsilon ^2 \Vert \beta \Vert _{L^2(D(z_0,a),\lambda _n)} \end{aligned}$$

for $n\ge n_{\varepsilon }$ because $U\subset D(z_0,a)\times \{w\in \mathbb {C}:|w|>1\}$ and

$$\begin{aligned} \Vert 1\Vert _{L^2(V_1,\lambda _n)}&\le \Vert 1\Vert _{L^2(D(z_0,a_1),\lambda _n)}\\&=\left( \int _{D(z_0,a_1)}\frac{\pi }{n-1}\left( e^{(2n-2)\varphi (z)} -e^{(2n-2)\alpha (z)}\right) dV(z)\right) ^{1/2}\\&\le \left( \int _{D(z_0,a_1)}\frac{\pi }{n-1}dV(z)\right) ^{1/2}\\&= \frac{\pi a_1}{\sqrt{n-1}}. \end{aligned}$$

Let $u\in C^{\infty }_0(D(z_0,a))$ and $n\ge n_{\varepsilon }$. Then

$$\begin{aligned}&\int _{D(z_0,a)}|u(z)|^2e^{\lambda _n(z)}dV(z)\\ {}&\quad = \sup \left\{ |\langle u,\beta \rangle _{D(z_0,a)}|^2: \beta \in C^{\infty }_0(D(z_0,a)), \Vert \beta \Vert _{L^2(D(z_0,a),\lambda _n)}^2\le 1 \right\} \\&\quad \le \sup \left\{ |\langle u,(g_n)_{\overline{z}}\rangle _{D(z_0,a)}|^2: \Vert g_n\Vert _{L^2(D(z_0,a),\lambda _n)}^2\le 2\varepsilon ^2 \right\} \\&\quad =\sup \left\{ |\langle u_z,g_n \rangle _{D(z_0,a)}|^2: \Vert g_n\Vert _{L^2(D(z_0,a),\lambda _n)}^2\le 2\varepsilon ^2 \right\} \\&\quad \le 2\varepsilon ^2 \int _{D(z_0,a)}|u_z(z)|^2e^{\lambda _n(z)}dV(z). \end{aligned}$$

There exists $0<c<1$ such that $e^{-\varphi (z)}<c e^{-\alpha (z)}$ for $z\in D(z_0,a)$. Then

$$\begin{aligned} \frac{\pi }{n-1} e^{(2n-2)\varphi (z)}\big (1-c^{2n-2}\big )<e^{-\lambda _n(z)}<\frac{\pi }{n-1} e^{(2n-2)\varphi (z)}. \end{aligned}$$

So for large n we have

$$\begin{aligned} \frac{\pi }{2(n-1)} e^{(2n-2)\varphi (z)}<e^{-\lambda _n(z)} <\frac{\pi }{n-1} e^{(2n-2)\varphi (z)} \end{aligned}$$

and

$$\begin{aligned}&\frac{n-1}{\pi }\int _{D(z_0,a)}|u(z)|^2e^{(2-2n)\varphi (z)}dV(z)\\&\quad <\int _{D(z_0,a)}|u(z)|^2e^{\lambda _n(z)}dV(z) \\&\quad \le 2\varepsilon ^2 \int _{D(z_0,a)}|u_z(z)|^2e^{\lambda _n(z)}dV(z)\\&\quad \le \frac{4\varepsilon ^2(n-1)}{\pi } \int _{D(z_0,a)}|u_z(z)|^2e^{(2-2n)\varphi (z)}dV(z). \end{aligned}$$

That is, for any $\varepsilon >0$ and $u\in C^{\infty }_0(D(z_0,a))$

$$\begin{aligned} \int _{D(z_0,a)}|u(z)|^2e^{(2-2n)\varphi (z)}dV(z)\le 4\varepsilon ^2 \int _{D(z_0,a)}|u_z(z)|^2e^{(2-2n)\varphi (z)}dV(z) \end{aligned}$$

(2)

for large n.

The estimate in (2) is identical to the one in [5, p. 38, proof of Lemma 10.2]. That is $\lambda ^m_{n\varphi }(D(z_0,a))\rightarrow \infty$ as $n\rightarrow \infty$ (see [5, Definition 2.3]). Since $\varphi$ is smooth and subharmonic, [5, Theorem 1.5] implies that $\lambda ^e_{n\varphi }(D(z_0,a))\rightarrow \infty$ as $n\rightarrow \infty$. We note that [5, Theorem 1.5] implies that if $\lambda ^m_{n\varphi }(D(z_0,a))\rightarrow \infty$ as $n\rightarrow \infty$ then $\lambda ^e_{n\varphi }(D(z_0,a))\rightarrow \infty$ as $n\rightarrow \infty$. This is enough to conclude that $\Gamma _k$ is B-regular. This argument is contained in the proof of Proposition 9.1 converse of (1) in [5, p. 33]. We repeat the argument here for the convenience of the reader.

Let $V=\{z\in D(z_0,a):\Delta \varphi (z)>0\}$ and $K_0=\overline{D(z_0,a/2)}\setminus V$. Then V is open and $K_0$ is a compact subset of $D(z_0,a)$. Furthermore, $\Delta \varphi =0$ on $K_0$. If $K_0$ has non-trivial fine interior then it supports a nonzero function $f\in W^1(\mathbb {C})$ (see [15, Proposition 4.17]). Then

$$\begin{aligned} \lambda ^e_{n\varphi }(D(z_0,a))\le \frac{\Vert \nabla f\Vert ^2}{\Vert f\Vert ^2} <\infty \quad \text { for all }\quad n. \end{aligned}$$

Which is a contradiction. Hence $K_0$ has empty fine interior which implies that $K_0$ satisfies property (P) (see [15, Proposition 4.17] or [11, Proposition 1.11]). Therefore, for $M>0$ there exists an open neighborhood $O_M$ of $K_0$ and $b_M\in C^{\infty }_0(O_M)$ such that $|b_M|\le 1/2$ on $O_M$ and $\Delta b_M>M$ on $K_0$. Furthermore, using the assumption that $|w|>0$ on $\Gamma _k$ one can choose $M_1$ such that the function $g_{M_1}(z,w)=M_1(|w|^2e^{\varphi (z)}-1)+b_M(z)$ has the following properties: $|g_{M_1}| \le 1$ and the complex Hessian $H_{g_{M_1}}(W)\ge M\Vert W\Vert ^2$ on $\Gamma _k\cap \overline{D(z_0,a)}$ where W is complex tangential direction. Then [1, Proposition 3.1.7] implies that $\Gamma _k\cap \overline{D(z_0,a/2)}$ satisfies property (P) (hence it is B-regular). Therefore, [15, Corollary 4.13] implies that $\Gamma _k$ is B-regular.

The computations in the second case (that is $\rho _{|w|}(z_0,|w_0|)>0$) are very similar. So we will just highlight the differences between the two cases. We define

$$\begin{aligned} U^{a_1,b_1}=\left\{ (z,w)\in \mathbb {C}^2: |w|>b_1|z-z_0|^2+a_1\right\} \end{aligned}$$

and

$$\begin{aligned} U_1=\Omega \cap U^{a_1,b_1} =\left\{ (z,w)\in \mathbb {C}^2:z\in V_1,e^{-\alpha (z)}<|w|<e^{-\varphi (z)}\right\} \end{aligned}$$

where $V_1$ is a domain in $\mathbb {C}$ and where $\alpha (z)=-\log (b_1|z-z_0|^2+a_1)$ is a strictly superharmonic function. One can show that $bU^{a_1,b_1}$ is strongly pseudoconvex. We choose $a,a_1,b,b_1>0$ such that such that $\overline{D(z_0,a)}\subset V_1$ and U is given by

$$\begin{aligned} U=\Omega \cap U_{a,b} = \left\{ (z,w)\in \mathbb {C}^2:z\in D(z_0,a), e^{-\alpha (z)}<|w|<e^{-\varphi (z)}\right\} \end{aligned}$$

where $U_{a,b}=D(z_0,a)\times \{w\in \mathbb {C}:|w_0|-b<|w|<|w_0|+b\}$. Furthermore, we define

$$\begin{aligned} \lambda _n(z)=-\log \left( \frac{\pi }{n+1}\left( e^{-(2n+2)\varphi (z)} -e^{-(2n+2)\alpha (z)}\right) \right) \end{aligned}$$

for $n=0,1,2,\ldots$ and by scaling $U_1$ in w variable if necessary, we will assume that $U_1\subset D(z_0,a_1)\times \{w\in \mathbb {C}:|w|<1\}$ so that $\Vert 1\Vert _{L^2(D(z_0,a_1),\lambda _n)}$ goes to zero as $n\rightarrow \infty$. One can check that $\lambda _n$ is subharmonic for all n.

We take functions $\beta \in C^{\infty }_0(D(z_0,a))$ and consider symbols $\psi \in C^{\infty }(\overline{V_1})$ such that $\psi _{\overline{z}}=\beta$. Then we consider the functions $H_{\psi }w^n$ for $n=0,1,2,\ldots$. Calculations similar to the ones in the previous case reveal that $g_n(z)w^n=H_{\psi }w^n$ where $g_n=S^{V_1}_{\lambda _n}(\beta d\overline{z})$. Using similar manipulations and again the compactness estimate (1) we conclude that for any $\varepsilon >0$ there exists an integer $n_{\varepsilon }$ such that for $u\in C^{\infty }_0(D(z_0,a))$ and $n\ge n_{\varepsilon }$ we have

$$\begin{aligned} \int _{D(z_0,a)}|u(z)|^2e^{(2n+2)\varphi (z)}dV(z)\le \varepsilon \int _{D(z_0,a)}|u_z(z)|^2e^{(2n+2)\varphi (z)}dV(z). \end{aligned}$$

(3)

Finally, an argument similar to the one right after (2) implies that $\Gamma _k$ is B-regular. $\square$