Sobolev Estimates for the $\overline{\partial }$ and the $\overline{\partial }$-Neumann Operator on Pseudoconvex Manifolds

Abstract

Let D be a relatively compact domain in an n-dimensional Kähler manifold with a $C^2$ smooth boundary that satisfies some “Hartogs-pseudoconvexity” condition. Assume that $\Xi$ is a positive holomorphic line bundle over X whose curvature form $\Theta$ satisfies $\Theta \ge C\omega$, where $C>0$. Then, the $\overline{\partial }$-Neumann operator N and the Bergman projection $\mathcal{P}$ are exactly regular in the Sobolev space $W^{\mathtt{m}}(D,\Xi )$ for some $\mathtt{m}$, as well as the operators $\overline{\partial }N$, ${\overline{\partial }}^{\divideontimes }N$.

1. Introduction

Sobolev estimates are crucial tools in the study of complex analysis on pseudoconvex manifolds. In this paper, we will focus on the Sobolev estimates for the $\overline{\partial }$ operator and the $\overline{\partial }$-Neumann operator on such manifolds. Consider a Hartogs pseudoconvex domain D with a $C^2$ boundary in a Kähler manifold X of complex dimension n, and if $\Xi$ is a positive line bundle over X whose curvature form satisfies $\Theta \ge C\omega$ with constant $C>0$, then the operators N, $\overline{\partial }N$, ${\overline{\partial }}^{\divideontimes }N$ and the Bergman projection $\mathcal{P}$ are regular in the Sobolev space $W^{\mathtt{m}}(D,\Xi )$ for some positive $\mathtt{m}$. This result generalizes the well-known results of Berndtsson–Charpentier [1], Boas–Straube [2], Cao–Shaw–Wang [3], Harrington [4] and Saber [5] and others in the case of the Hartogs pseudoconvex domain in a Kähler manifold for forms with values in a holomorphic line bundle. Indeed, in [1], Berndtsson–Charpentier (see also [6]) obtained the Sobolev regularity for $\mathcal{P}$ for a pseudoconvex domain $\mathsf{\Omega }$. In [2], Boas–Straube proved that the Bergman projection B maps the Sobolev space $W^{\mathtt{m}}\left(\mathsf{\Omega }\right)$ to itself for all $\mathtt{m}>0$ on a smooth pseudoconvex domain in $C^n$ that admits a defining function that is plurisubharmonic on the boundary $b\mathsf{\Omega }$. In [3], Cao–Shaw–Wang obtained the Sobolev regularity of the operators N, $\overline{\partial }N$, ${\overline{\partial }}^{\divideontimes }N$ and $\mathcal{P}$ on a local Stein domain subset of the complex projective space. In [4], Harrington proved this result on a bounded pseudoconvex domain in $C^n$ with a Lipschitz boundary. In [5], Saber proved that the operators N, ${\overline{\partial }}^{\divideontimes }N$ and $\mathcal{P}$ are regular in $W_{r,s}^{\mathtt{m}}\left(D\right)$ for some $\mathtt{m}$ on a smooth weakly q-convex domain in $C^n$. Similar results can be found in [7,8,9,10,11,12,13,14,15,16].

This paper is organized into five sections. The introduction presents an introduction to the subject and contains the history and development of the problem. Section 2 recalls the basic definitions and fundamental results. In Section 3, the basic Bochner–Kodaira–Morrey–Kohn identity is proved on the Kähler manifold. In Section 4, it is proved that the $C^2$ smoothly bounded Hartogs pseudoconvex domains in the Kähler manifold admit bounded plurisubharmonic exhaustion functions. Section 5 deals with the $L^2$ estimates of the $\overline{\partial }$ and $\overline{\partial }$-Neumann operator on the $C^2$ smoothly bounded Hartogs pseudoconvex domains in the Kähler manifold. Section 6 presents the main results.

2. Preliminaries

Assuming that X is a complex manifold of the complex dimension n, $n\ge 2$, $T\left(X\right)$ (resp. $T_x\left(X\right)$) is the holomorphic tangent bundle of X (resp. at $x\in X$) and $\pi :\Xi ⟶X$ is a holomorphic line bundle over X. A system of local complex analytic (holomorphic) coordinates on X is a collection ${\left\{{\gamma }_j\right\}}_{j\in J}$ (for some index set J) of local complex coordinates ${\gamma }_j:{\mathcal{U}}_j⟶C^n$ such that:

(i) ${\displaystyle X={\cup }_{j\in J}{\mathcal{U}}_j}$, i.e., ${\left\{{\mathcal{U}}_j\right\}}_{j\in J}$ is an open covering of X by charts with coordinate mappings ${\mathtt{w}}_j:{\mathcal{U}}_j⟶C^n$ satisfies ${\pi }^{-1}\left({\mathcal{U}}_j\right)={\mathcal{U}}_j\times C$.

(ii) $\left\{f_{ij}\right\}$ is a system of transition functions for $\Xi$; that is, the maps $f_{jk}={\gamma }_j\circ {\gamma }_k^{-1}:{\gamma }_k\left(z\right)⟶{\gamma }_j\left(z\right)$ are biholomorphic for each pair of indices $(j,k)$ with ${\mathcal{U}}_j\cap {\mathcal{U}}_k$ being nonempty (i.e., $f_{jk}$ (resp. $f_{jk}^{-1}={\gamma }_k\circ {\gamma }_j^{-1}$) are holomorphic maps of ${\gamma }_k({\mathcal{U}}_j\cap U_k)$ onto ${\gamma }_j({\mathcal{U}}_j\cap {\mathcal{U}}_k)$ (resp. ${\gamma }_j({\mathcal{U}}_j\cap {\mathcal{U}}_k)$ onto ${\gamma }_k({\mathcal{U}}_j\cap {\mathcal{U}}_k)$)).

Assume that $({\mathtt{w}}_j^1,{\mathtt{w}}_j^2,\cdots ,{\mathtt{w}}_j^n)$ is the local coordinates on ${\mathcal{U}}_j$. A system of functions $\hslash =\left\{{\hslash }_j\right\},j\in J$ is a Hermitian metric along the fibers of $\Xi$ with ${\hslash }_j=\mid f_{ij}{\mid }^2{\hslash }_i$ in ${\mathcal{U}}_i\cap {\mathcal{U}}_j,$ and ${\hslash }_j$ is a $C^{\infty }$ positive function in ${\mathcal{U}}_j$. The (1,0) form of the connection associated with the metric ℏ is given as $\theta =\left\{{\theta }_j\right\}$, ${\theta }_j={\hslash }_j^{-1}\partial {\hslash }_j$. $\Theta =\left\{{\Theta }_j\right\}$ is the curvature form associated with the connection $\theta$ and is given by

$${\Theta }_j=\overline{\partial }{\theta }_j=\partial \overline{\partial }log{\hslash }_j={\displaystyle \sum _{\alpha ,\beta =1}^n}{\Theta }_{j\alpha \overline{\beta }}d{\mathtt{w}}_j^{\alpha }\wedge d{\overline{\mathtt{w}}}_j^{\beta }.$$

Definition 1.

Ξ is positive at $x\in {\mathcal{U}}_j$ if the Hermitian form

$$\sum {\Theta }_{j\alpha \overline{\beta }}{\mu }^{\alpha }{\overline{\mu }}^{\beta },$$

is positive definite on $T_x\left(X\right)$, ∀$\mu \in {\Xi }_x\setminus \left\{0\right\}$.

Along the fibers of $\Xi$, ${\hslash }_0=\left({\hslash }_j^{-1}\right),j\in J$ is a Hermitian metric for which $\Xi$ is positive; i.e., $\partial \overline{\partial }log{\hslash }_j>0$. Then, ${\hslash }_0$ defines a Kähler metric $\mathcal{G}$ on X,

$$\mathcal{G}={\displaystyle \sum _{\alpha ,\beta =1}^n}{\mathtt{g}}_{j\alpha \overline{\beta }}d{\mathtt{w}}_j^{\alpha }d{\overline{\mathtt{w}}}_j^{\beta },{\mathtt{g}}_{j\alpha \overline{\beta }}={\partial }^{2}log{\hslash }_j/\partial {\mathtt{w}}_j^{\alpha }\partial {\overline{\mathtt{w}}}_j^{\beta }.$$

Let ${\mathcal{C}}_{r,s}^{\infty }(X,\Xi )$ (resp. ${\mathcal{D}}_{r,s}^{\infty }(X,\Xi )$) be the space of $C^{\infty }$$(r,s)$ differential forms (resp. with compact support) on X with values in $\Xi$. A form $\psi =\left({\psi }_j\right)\in {\mathcal{C}}_{r,s}^{\infty }(X,\Xi )$ is expressed on ${\mathcal{U}}_j$ as follows:

$${\psi }_j\left(z\right)=\sum _{A_r,B_s}{\psi }_{j{A}_r{B}_s}\left(z\right)d{\mathtt{w}}_j^{A_r}\wedge d{\overline{\mathtt{w}}}_j^{B_s}\otimes s_j,$$

where $A_r=(a_1,\dots ,a_r)$ and $B_s=(b_1,\dots ,b_s)$ are multi-indices and $s_j$ is a section of ${\Xi |}_{{\mathcal{U}}_j}$. Define the inner product

$$(\psi ,\psi )={\hslash }_j\sum _{A_r,B_s}{\psi }_{j{A}_r{\overline{B}}_s}\overline{{\psi }_j^{{\overline{A}}_r{B}_s}},$$

where ${\psi }_j^{{\overline{A}}_r{B}_s}={\sum }_{C_r,D_s}{\mathtt{g}}_j^{c_1{\overline{a}}_1}\cdots {\mathtt{g}}_j^{c_r{\overline{a}}_r}{\mathtt{g}}_j^{b_1{\overline{d}}_1}\cdots {\mathtt{g}}_j^{b_s{\overline{d}}_s}{\psi }_{j{c}_1\cdots c_r{\overline{d}}_1\cdots {\overline{d}}_s}.$ Let

$${\mathcal{C}}_{r,s}^{\infty }(\overline{\mathsf{\Omega }},\Xi )=\{\psi {\mid }_{\overline{\mathsf{\Omega }}};\psi \in {\mathcal{C}}_{r,s}^{\infty }(X,\Xi )\}.$$

Let $\divideontimes :{\mathcal{C}}_{r,s}^{\infty }(X,\Xi )⟶{\mathcal{C}}_{(n-s,n-r)}^{\infty }(X,\Xi )$ be the Hodge star operator, which is a real operator and satisfies

$$\divideontimes \divideontimes \psi ={(-1)}^{r+s}\psi ,$$

For the proof, see Morrow and Kodaira [17]. Set the volume element with respect to $\mathcal{G}$ as $dv$. The inner product $<\psi ,\psi >$ and the norm $\Vert \psi \Vert$ are defined by

$$<\psi ,\psi >={\int }_{\mathsf{\Omega }}(\psi ,\psi )dv={\int }_{\mathsf{\Omega }}{}^t\psi \wedge \divideontimes \overline{\hslash \psi },\mathrm{and}{\Vert \psi \Vert }^2=<\psi ,\psi >.$$

The formal adjoint operator $\psi$ of $\overline{\partial }:{\mathcal{C}}_{r,s-1}^{\infty }(\mathsf{\Omega },\Xi )⟶{\mathcal{C}}_{r,s}^{\infty }(\mathsf{\Omega },\Xi )$ is defined by

$$<\psi \psi ,\psi >=<\psi ,\overline{\partial }\psi >,$$

$\psi \in {\mathcal{C}}_{r,s}^{\infty }(\mathsf{\Omega },\Xi )$ and $\psi \in {\mathcal{D}}_{r,s-1}^{\infty }(\mathsf{\Omega },\Xi )$. Let $\#:{\mathcal{C}}_{r,s}^{\infty }(X,\Xi )⟶{\mathcal{C}}_{s,r}^{\infty }(X,E^{\divideontimes })$ be defined locally as ${(\#\psi )}_j=\overline{{\hslash }_j{\psi }_j}$; the inner product $<\psi ,\psi >$ is given by

$$<\psi ,\psi >={\int }_{\mathsf{\Omega }}^{}{}^t\psi \wedge \divideontimes \#\psi .$$

From Stokes’ theorem, $\psi \in {\mathcal{C}}_{r,s}^{\infty }(\overline{\mathsf{\Omega }},\Xi )$, $\psi \in {\mathcal{C}}_{r,s-1}^{\infty }(\overline{\mathsf{\Omega }},\Xi )$, one obtains

$$<\overline{\partial }\psi ,\psi >=<\psi ,{\overline{\partial }}^{\divideontimes }\psi >+{\int }_{\partial \mathsf{\Omega }}^{}{}^t\psi \wedge \divideontimes \#\psi .$$

Put

$${\mathcal{B}}_{r,s}(\overline{\mathsf{\Omega }},\Xi )=\{\psi \in {\mathcal{C}}_{r,s}^{\infty }(\overline{\mathsf{\Omega }},\Xi );\divideontimes \#\psi {\mid }_{\partial \mathsf{\Omega }}=0\}.$$

As a result,

$$<\overline{\partial }\psi ,\psi >=<\psi ,{\overline{\partial }}^{\divideontimes }\psi >,$$

for $\psi \in {\mathcal{B}}_{r,s}(\overline{\mathsf{\Omega }},\Xi )$.

$L_{r,s}^2(\mathsf{\Omega },\Xi )$ is the Hilbert space of the measurable E-valued $(r,s)$ forms $\psi$, which are square integrable in the sense that ${\parallel \psi \parallel }^2<\infty$. Let $\overline{\partial }:L_{r,s}^2(\mathsf{\Omega },\Xi )⟶L_{r,s+1}^2(\mathsf{\Omega },\Xi )$ and ${\overline{\partial }}^{\divideontimes }:L_{r,s+1}^2(\mathsf{\Omega },\Xi )⟶L_{r,s}^2(\mathsf{\Omega },\Xi )$. In $L_{r,s}^2(\mathsf{\Omega },\Xi )$, the spaces $ker(\overline{\partial },\Xi )$, ${\mathrm{Dom}}_{r,s}(\overline{\partial },\Xi )$ and $\mathrm{Rang}(\overline{\partial },\Xi )$ are the kernel, the domain and the range of $\overline{\partial }$, respectively. A Bergman projection operator $\mathcal{P}:L_{r,s}^2(D,\Xi )⟶L_{r,s}^2(D,\Xi )\cap {ker}_{r,s}\left(E\right)$. Let $\square ={\square }_{r,s}=\overline{\partial }{\overline{\partial }}^{\ast }$$+{\overline{\partial }}^{\ast }\overline{\partial }$ be the unbounded Laplace–Beltrami operator from $L_{r,s}^2(\mathsf{\Omega },\Xi )$ to $L_{r,s}^2(\mathsf{\Omega },\Xi )$ with Dom$({\square }_{r,s},\Xi )=\{\psi \in L_{r,s}^2(\mathsf{\Omega },E)|\psi \in$ Dom$(\overline{\partial },\Xi )\cap$ Dom$({\overline{\partial }}^{\divideontimes },\Xi );\overline{\partial }\psi \in$ Dom$({\overline{\partial }}^{\divideontimes },\Xi )$ and ${\overline{\partial }}^{\divideontimes }\psi \in$ Dom$(\overline{\partial },\Xi )\}$. Let $N_{r,s}$ be the $\overline{\partial }$-Neumann operator on $(r,s)$ forms, solving $N_{r,s}{\square }_{r,s}\psi =\psi$ for any $(r,s)$ form $\psi$ in $L_{r,s}^2(\mathsf{\Omega },\Xi )$. Denote by $\mathcal{P}$ the Bergman operator, mapping a $(r,s)$ form in $L_{r,s}^2(\mathsf{\Omega },\Xi )$ to its orthogonal projection in the closed subspace of $\overline{\partial }$-closed forms.

Let

$${\mathcal{H}}_{r,s}\left(E\right)=ker({\square }_{r,s},\Xi )=\{\psi \in \mathrm{Dom}(\overline{\partial },\Xi )\cap \mathrm{Dom}({\overline{\partial }}^{\divideontimes },\Xi );\overline{\partial }\psi =0\mathrm{and}{\overline{\partial }}^{\divideontimes }\psi =0\}.$$

Let $W_{r,s}^{\mathtt{m}}(\mathsf{\Omega },\Xi )$ be the Sobolev space with $-\frac{1}{2}<\mathtt{m}<\frac{1}{2}$ and let ${\parallel \parallel }_{W_{r,s}^{\mathtt{m}}(\mathsf{\Omega },\Xi )}$ denote its norm. ∀$\psi \in \mathrm{Dom}(\overline{\partial },\Xi )\cap \mathrm{Dom}({\overline{\partial }}^{\divideontimes },\Xi )$, one obtains $\psi \in W_{r,s}^1(\mathsf{\Omega },\mathrm{loc})$. Thus, $\psi$ is an elliptic and $\psi \in W_{r,s}^{\mathtt{m}}(\mathsf{\Omega },\Xi )$ for $-\frac{1}{2}<\mathtt{m}<\frac{1}{2}$ if and only if

$${\parallel \psi \parallel }_{W_{r,s}^{\mathtt{m}}(\mathsf{\Omega },\Xi )}^2={\int }_{\mathsf{\Omega }}{\zeta }^{-2\mathtt{m}}{\left|\psi \right|}^2<\infty .$$

For the proof, see Theorems 4.1 and 4.2 in Jersion and Kenig [18], Lemma 2 in Charpentier [19] and also Theorem C.4 in the Appendix in Chen and Shaw [20].

Proposition 1 ([21,22,23]).

(i) If $\psi \in \mathit{Dom}(\psi ,\Xi )\subset L_{r,s}^2(\mathsf{\Omega },\Xi )$ satisfies supp.$\psi \subseteq \overline{\mathsf{\Omega }}$ and supp.$\psi \psi \subseteq \overline{\mathsf{\Omega }}$, then ${\psi |}_{\mathsf{\Omega }}\in \mathit{Dom}({\overline{\partial }}^{\divideontimes },\Xi )\subset L_{r,s}^2(\mathsf{\Omega },\Xi )$; i.e., ${\psi \psi |}_{\mathsf{\Omega }}={\overline{\partial }}^{\divideontimes }{\psi |}_{\mathsf{\Omega }}$ in $L_{r,s-1}^2(\mathsf{\Omega },\Xi )$. (ii) ${\mathcal{C}}_{r,s}^{\infty }(\overline{\mathsf{\Omega }},\Xi )$ is dense in $\mathit{Dom}(\overline{\partial },\Xi )$ in the sense of ${(\Vert \psi \Vert }^2+\Vert \overline{\partial }\psi {{\Vert }^2)}^{1/2}$. (iii) ${\mathcal{B}}_{r,s}(\overline{\mathsf{\Omega }},\Xi )$ is dense in $\mathit{Dom}({\overline{\partial }}^{\divideontimes },\Xi )$ (resp. $\mathit{Dom}(\overline{\partial },\Xi )\cap \mathit{Dom}({\overline{\partial }}^{\divideontimes },\Xi ))$ in the sense of the norm (${\Vert \psi \Vert }^2+\Vert {\overline{\partial }}^{\divideontimes }\psi {{\Vert }^2)}^{1/2}$ (resp. ${(\Vert \psi \Vert }^2+\Vert \overline{\partial }{\psi \Vert }^2+\Vert {\overline{\partial }}^{\divideontimes }\psi {{\Vert }^2)}^{1/2}$).

(iv) ${\overline{\partial }}^{\divideontimes }=\psi$ on ${\mathcal{B}}_{r,s}(\overline{\mathsf{\Omega }},\Xi )$.

3. The Kähler Identity

As in Takeuchi A. [24,25,26], one can prove the following Kähler identity: Fix the following notation: $C^{\infty }$ sections of $\mathcal{A}\left(T\right(X\left)\right)$, $\mathcal{A}\left(T^{\divideontimes }\left(X\right)\right)$, $\mathcal{A}\left(\overline{T}\left(X\right)\right)$ and $\mathcal{A}\left({\overline{T}}^{\divideontimes }\left(X\right)\right)$ written as ${\sum }_{\alpha =1}^n{\zeta }^{\alpha }\frac{\partial }{\partial z^{\alpha }},{\sum }_{\alpha =1}^n{\psi }_{\alpha }d{z}^{\alpha },{\sum }_{\alpha =1}^n{\eta }^{\overline{\alpha }}\frac{\partial }{\partial z^{\overline{\alpha }}}\mathrm{and}{\sum }_{\alpha =1}^n{\psi }_{\overline{\alpha }}d{z}^{\overline{\alpha }},$ respectively. Use the notation ${\partial }_{\beta }=\frac{\partial }{\partial z^{\beta }}$, ${\overline{\partial }}_{\alpha }=\frac{\partial }{\partial {\overline{z}}^{\alpha }}$. For $\eta ={\sum }_{\alpha =1}^n{\eta }^{\overline{\alpha }}\frac{\partial }{\partial z^{\overline{\alpha }}}\in \mathcal{A}\left(\overline{T}\left(X\right)\right)$, $\psi ={\sum }_{\alpha =1}^n{\psi }_{\overline{\alpha }}d{z}^{\overline{\alpha }}\in \mathcal{A}\left({\overline{T}}^{\divideontimes }\left(X\right)\right)$, define

$${\nabla }_{\beta }{\eta }^{\overline{\alpha }}={\partial }_{\beta }{\eta }^{\overline{\alpha }}\mathrm{and}{\nabla }_{\beta }{\psi }_{\overline{\alpha }}={\partial }_{\beta }{\psi }_{\overline{\alpha }}.$$

A connection $\omega$ for $T\left(X\right)$ is defined as

$$\omega =\left({\omega }_{\alpha }^{\beta }\right),{\omega }_{\alpha }^{\beta }={\displaystyle \sum _{\gamma =1}^n}{\Gamma }_{\gamma \alpha }^{\beta }d{z}^{\gamma },\mathrm{with}{\Gamma }_{\gamma \alpha }^{\beta }={\displaystyle \sum _{\sigma =1}^n}{g}^{\overline{\sigma }\beta }{\partial }_{\gamma }{\mathtt{g}}_{\alpha \overline{\sigma }},$$

and its Riemann curvature tensor

$${\mathcal{R}}_{\overline{\alpha }\beta \overline{\nu }\tau }={\displaystyle \sum _{\mu =1}^n}{\mathtt{g}}_{\mu \overline{\alpha }}{\mathcal{R}}_{\beta \overline{\nu }\tau }^{\mu },{\mathcal{R}}_{\beta \overline{\nu }\tau }^{\alpha }={\partial }_{\overline{\nu }}{\Gamma }_{\tau \beta }^{\alpha }.$$

(1)

One obtains

$${\Gamma }_{\overline{\beta }\overline{\gamma }}^{\overline{\alpha }}=\overline{{\Gamma }_{\beta \gamma }^{\alpha }},{\mathcal{R}}_{\overline{\beta }\nu \overline{\tau }}^{\overline{\alpha }}=\overline{{\mathcal{R}}_{\beta \overline{\nu }\tau }^{\alpha }},\mathrm{and}{\mathcal{R}}_{\alpha \overline{\beta }\nu \overline{\tau }}=\overline{{\mathcal{R}}_{\overline{\alpha }\beta \overline{\nu }\tau }}.$$

(2)

The Ricci curvature is defined by

$${\mathcal{R}}_{\overline{\nu }\tau }={\displaystyle \sum _{\beta =1}^n}{\mathcal{R}}_{\beta \overline{\nu }\tau }^{\beta }.$$

(3)

Following Morrow and Kodaira [13], if $\mathcal{G}$ is a Kähler metric,

$${\Gamma }_{\beta \gamma }^{\alpha }={\Gamma }_{\gamma \beta }^{\alpha },$$

(4)

where

$${\displaystyle \sum _{\tau =1}^n}{\Gamma }_{\tau \alpha }^{\tau }={\partial }_{\alpha }logg,\mathrm{and}{\mathcal{R}}_{\overline{\nu }\tau }={\partial }_{\overline{\nu }}{\partial }_{\tau }logg,\mathrm{where}g=\det \left({\mathtt{g}}_{\alpha \overline{\beta }}\right).$$

For $\zeta ={\sum }_{\alpha =1}^n{\zeta }^{\alpha }\frac{\partial }{\partial z^{\alpha }}\in \mathcal{A}\left(T\left(X\right)\right)$, $\psi ={\sum }_{\alpha =1}^n{\psi }_{\alpha }d{z}^{\alpha }\in \mathcal{A}\left({\overline{T}}^{\divideontimes }\left(X\right)\right)$, one defines

$$\begin{array}{c}{\nabla }_{\beta }{\zeta }^{\alpha }={\partial }_{\beta }{\zeta }^{\alpha }+{\displaystyle \sum _{\gamma =1}^n}{\Gamma }_{\beta \gamma }^{\alpha }{\zeta }^{\gamma },\hfill \\ {\nabla }_{\beta }{\psi }_{\alpha }={\partial }_{\beta }{\psi }_{\alpha }-{\displaystyle \sum _{\gamma =1}^n}{\Gamma }_{\beta \alpha }^{\gamma }{\psi }_{\gamma },\hfill \\ {\nabla }_{\overline{\beta }}{\zeta }^{\alpha }={\partial }_{\overline{\beta }}{\zeta }^{\alpha },\hfill \\ {\nabla }_{\overline{\beta }}{\eta }^{\overline{\alpha }}={\partial }_{\overline{\beta }}{\eta }^{\overline{\alpha }}+{\displaystyle \sum _{\gamma =1}^n}\overline{{\Gamma }_{\beta \gamma }^{\alpha }}{\eta }^{\overline{\gamma }},\hfill \\ {\nabla }_{\overline{\beta }}{\psi }_{\alpha }={\partial }_{\overline{\beta }}{\psi }_{\alpha },\hfill \\ {\nabla }_{\overline{\beta }}{\psi }_{\overline{\alpha }}={\partial }_{\overline{\beta }}{\psi }_{\overline{\alpha }}-{\displaystyle \sum _{\gamma =1}^n}\overline{{\Gamma }_{\beta \alpha }^{\gamma }}{\psi }_{\overline{\gamma }}.\hfill \end{array}$$

(5)

For $\psi \in {\mathcal{C}}_{r,s}^{\infty }(X,\Xi )$, one defines

$$\begin{gathered} {\displaystyle {\nabla }_{\alpha }{\psi }_{{\alpha }_1\dots {\alpha }_r{\overline{\beta }}_1\dots {\overline{\beta }}_s}={\partial }_{\alpha }{\psi }_{{\alpha }_1\dots {\alpha }_r{\overline{\beta }}_1\dots {\overline{\beta }}_s}-{\displaystyle \sum _{j=1}^r}\sum _{\tau }{\Gamma }_{\alpha {\alpha }_j}^{\tau }{\psi }_{{\alpha }_1\dots {\alpha }_{j-1}\tau {\alpha }_{j+1}\dots {\alpha }_r{\overline{\beta }}_1\dots {\overline{\beta }}_s}, \\ {\nabla }_{\alpha }^{(\hslash )}{\psi }_{{\alpha }_1\dots {\alpha }_r{\overline{\beta }}_1\dots {\overline{\beta }}_s}={\nabla }_{\alpha }{\psi }_{{\alpha }_1\dots {\alpha }_r{\overline{\beta }}_1\dots {\overline{\beta }}_s}-{\partial }_{\alpha }log\hslash {\psi }_{{\alpha }_1\dots {\alpha }_r{\overline{\beta }}_1\dots {\overline{\beta }}_s}, \\ {\nabla }_{\overline{\beta }}{\psi }_{{\alpha }_1\dots {\alpha }_r{\overline{\beta }}_1\dots {\overline{\beta }}_s}={\partial }_{\overline{\beta }}{\psi }_{{\alpha }_1\dots {\alpha }_r{\overline{\beta }}_1\dots {\overline{\beta }}_s}-{\displaystyle \sum _{j=1}^s}\sum _{\tau }\overline{{\Gamma }_{\beta {\beta }_j}^{\tau }}{\psi }_{{\alpha }_1\dots {\alpha }_r{\overline{\beta }}_1\dots {\overline{\beta }}_{j-1}\overline{\tau }{\overline{\beta }}_{j+1}\dots {\overline{\beta }}_s}, \\ {\nabla }_{\overline{\beta }}{\psi }^{{\overline{\beta }}_1\dots {\overline{\beta }}_s{\alpha }_1\dots {\alpha }_r}={\partial }_{\overline{\beta }}{\psi }^{{\overline{\beta }}_1\dots {\overline{\beta }}_s{\alpha }_1\dots {\alpha }_r}+{\displaystyle \sum _{j=1}^s}\sum _{\tau }\overline{{\Gamma }_{\beta \tau }^{{\beta }_j}}{\psi }^{{\overline{\beta }}_1\dots {\overline{\beta }}_{j-1}\overline{\tau }{\overline{\beta }}_{j+1}\dots {\overline{\beta }}_s{\alpha }_1\dots {\alpha }_r},} \end{gathered}$$

(6)

For the proof, see Choquet-Bruhat [27], p. 235.

Following Morrow and Kodaira [17], the operators $\overline{\partial }$, $\psi$ are defined as

$$\begin{gathered} \overline{\partial }\psi =\sum _{A_r,B_s}\sum _{\mu }{\nabla }_{\overline{\mu }}{\psi }_{A_r{\overline{B}}_s}d{z}^{\overline{\mu }}\wedge d{z}^{{\alpha }_1}\wedge \dots \wedge d{z}^{{\alpha }_r}\wedge d{z}^{{\overline{\beta }}_1}\wedge \dots \wedge d{z}^{{\overline{\beta }}_s}, \\ {\left({\overline{\partial }}^{\divideontimes }\psi \right)}_{A_r{\overline{B}}_{s-1}}={(-1)}^{r-1}{\displaystyle \sum _{\alpha ,\beta =1}^n}{g}^{\overline{\beta }\alpha }{\nabla }_{\alpha }^{(\hslash )}{\psi }_{\overline{\beta }{A}_r{\overline{B}}_{s-1}}, \end{gathered}$$

(7)

for $\psi \in {\mathcal{C}}_{r,s}^{\infty }(X,\Xi )$.

For a $C^{\infty }$ function $\lambda$ and for a $\psi \in {\mathcal{C}}_{r,s}^{\infty }(X,\Xi )$ at any point of X, one defines

$$\begin{gathered} \mathrm{grad}\lambda =\left(\frac{\partial \lambda }{\partial z^1},\dots ,\frac{\partial \lambda }{\partial z^n},\overline{\frac{\partial \lambda }{\partial z^1}},\dots ,\overline{\frac{\partial \lambda }{\partial z^n}}\right), \\ {\left|\mathrm{grad}\lambda \right|}^2=\left(\mathrm{grad}\lambda \right)\overline{\left(\mathrm{grad}\lambda \right)}={\displaystyle \sum _{\alpha =1}^n}{\left|\frac{\partial \lambda }{\partial z^{\alpha }}\right|}^2+{\displaystyle \sum _{\beta =1}^n}{\left|\overline{\frac{\partial \lambda }{\partial z^{\beta }}}\right|}^2, \\ (\mathcal{L}\left(\lambda \right)\psi ,\psi )=\sum _{B_{s-1}}{\displaystyle \sum _{\beta ,\gamma =1}^n}\frac{{\partial }^2\lambda }{\partial z^{\beta }\partial z^{\overline{\gamma }}}{\psi }_{{\overline{B}}_{s-1}}^{\beta }\overline{{\psi }^{\gamma B_{s-1}}}. \end{gathered}$$

Since $d\lambda \ne 0$ on U, then $\mathrm{grad}\lambda \ne 0$ on U also. Also, set

$$\left(\overline{\nabla },\overline{\nabla }\right)={\hslash }^{-1}\sum _{C_s}\sum _{\mu }{\nabla }_{\overline{\mu }}{\psi }_{{\overline{C}}_s}\overline{{\nabla }^{\mu }{\psi }^{C_s}}.$$

For $\psi \in {\mathcal{C}}_{0,s}^{\infty }(X,\Xi )$, $s\ge 1$, we construct from $\psi$ the two tangent vector fields $\xi$ and $\eta$ to X as follows:

$$\begin{gathered} {\displaystyle \xi =\{{\xi }^{\beta }=\sum _{B_{s-1}}{\displaystyle \sum _{\gamma =1}^n}{\hslash }^{-1}\left({\nabla }_{\overline{\gamma }}{\psi }_{{\overline{B}}_{s-1}}^{\beta }\right)\overline{{\psi }^{\gamma B_{s-1}}},{\xi }^{\overline{\beta }}=0\}, \\ \eta =\{{\eta }^{\gamma }=0,{\eta }^{\overline{\gamma }}=\sum _{B_{s-1}}{\displaystyle \sum _{\beta =1}^n}{\hslash }^{-1}\left({\nabla }_{\beta }^{(\hslash )}{\psi }_{{\overline{B}}_{s-1}}^{\beta }\right)\overline{{\psi }^{\gamma B_{s-1}}}\},} \end{gathered}$$

where $\beta ,\gamma =1,2,\dots ,n.$

Proposition 2 ([24]).

$${\nabla }_{\mu }{\mathtt{g}}_{\alpha \overline{\beta }}=0,{\nabla }_{\overline{\mu }}{\mathtt{g}}_{\alpha \overline{\beta }}=0,and{\nabla }_{\overline{\mu }}{g}^{\overline{\beta }\alpha }=0.$$

Proof. 

Since ${\mathtt{g}}_{\alpha \overline{\beta }}$ is a $C^{\infty }$ section of $T^{\divideontimes }\left(X\right)\otimes {\overline{T}}^{\divideontimes }\left(X\right)$, then Equation (3) gives

$$\begin{array}{cc}{\nabla }_{\mu }{\mathtt{g}}_{\alpha \overline{\beta }}\hfill & ={\partial }_{\mu }{\mathtt{g}}_{\alpha \overline{\beta }}-{\displaystyle \sum _{\tau }}{\Gamma }_{\mu \alpha }^{\tau }{\mathtt{g}}_{\tau \overline{\beta }}\hfill \\ & ={\partial }_{\mu }{\mathtt{g}}_{\alpha \overline{\beta }}-{\displaystyle \sum _{\gamma ,\tau }}{g}^{\overline{\gamma }\tau }\left({\partial }_{\mu }{\mathtt{g}}_{\alpha \overline{\gamma }}\right){\mathtt{g}}_{\tau \overline{\beta }}\hfill \\ & ={\partial }_{\mu }{\mathtt{g}}_{\alpha \overline{\beta }}-{\displaystyle \sum _{\gamma }}{\zeta }_{\beta }^{\gamma }\left({\partial }_{\mu }{\mathtt{g}}_{\alpha \overline{\gamma }}\right)\hfill \\ & ={\partial }_{\mu }{\mathtt{g}}_{\alpha \overline{\beta }}-{\partial }_{\mu }{\mathtt{g}}_{\alpha \overline{\beta }}\hfill \\ & =0.\hfill \\ {\nabla }_{\overline{\mu }}{\mathtt{g}}_{\alpha \overline{\beta }}\hfill & ={\partial }_{\overline{\mu }}{\mathtt{g}}_{\alpha \overline{\beta }}-{\displaystyle \sum _{\tau }}\overline{{\Gamma }_{\mu \beta }^{\tau }}{\mathtt{g}}_{\alpha \overline{\tau }}\hfill \\ & ={\partial }_{\overline{\mu }}{\mathtt{g}}_{\alpha \overline{\beta }}-{\displaystyle \sum _{\gamma ,\tau }}{g}^{\overline{\tau }\gamma }\left({\partial }_{\overline{\mu }}{\mathtt{g}}_{\gamma \overline{\beta }}\right){\mathtt{g}}_{\alpha \overline{\tau }}\hfill \\ & ={\partial }_{\overline{\mu }}{\mathtt{g}}_{\alpha \overline{\beta }}-{\displaystyle \sum _{\gamma }}{\zeta }_{\alpha }^{\gamma }\left({\partial }_{\overline{\mu }}{\mathtt{g}}_{\gamma \overline{\beta }}\right)\hfill \\ & ={\partial }_{\overline{\mu }}{\mathtt{g}}_{\alpha \overline{\beta }}-{\partial }_{\overline{\mu }}{\mathtt{g}}_{\alpha \overline{\beta }}\hfill \\ & =0.\hfill \\ {\nabla }_{\overline{\mu }}{g}^{\overline{\beta }\alpha }\hfill & ={\partial }_{\overline{\mu }}{g}^{\overline{\beta }\alpha }+{\displaystyle \sum _{\tau }}\overline{{\Gamma }_{\mu \tau }^{\beta }}{g}^{\overline{\tau }\alpha }\hfill \\ & ={\partial }_{\overline{\mu }}{g}^{\overline{\beta }\alpha }+{\displaystyle \sum _{\gamma ,\tau }}{g}^{\overline{\beta }\gamma }\left({\partial }_{\overline{\mu }}{\mathtt{g}}_{\gamma \overline{\tau }}\right)g^{\overline{\tau }\alpha }\hfill \\ & ={\partial }_{\overline{\mu }}{g}^{\overline{\beta }\alpha }-{\displaystyle \sum _{\tau ,\gamma }}{\mathtt{g}}_{\gamma \overline{\tau }}\left({\partial }_{\overline{\mu }}{g}^{\overline{\beta }\gamma }\right)g^{\overline{\tau }\alpha }\hfill \\ & ={\partial }_{\overline{\mu }}{g}^{\overline{\beta }\alpha }-{\displaystyle \sum _{\gamma }}{\zeta }_{\gamma }^{\alpha }\left({\partial }_{\overline{\mu }}{g}^{\overline{\beta }\gamma }\right)\hfill \\ & ={\partial }_{\overline{\mu }}{g}^{\overline{\beta }\alpha }-{\partial }_{\overline{\mu }}{g}^{\overline{\beta }\alpha }\hfill \\ & =0.\hfill \end{array}$$

□

Proposition 3 ([24]).

$$\mathrm{div}\xi -\mathrm{div}\eta ={\displaystyle \sum _{\beta =1}^n}{\nabla }_{\beta }{\xi }^{\beta }-{\displaystyle \sum _{\gamma =1}^n}{\nabla }_{\overline{\gamma }}{\eta }^{\overline{\gamma }}.$$

Proof. 

The divergence of the vector $\xi$,

$$\mathrm{div}\xi ={\displaystyle \sum _{\beta =1}^n}{\nabla }_{\beta }{\xi }^{\beta }-{\displaystyle \sum _{\beta ,\gamma =1}^n}\left({\Gamma }_{\beta \gamma }^{\beta }-{\Gamma }_{\gamma \beta }^{\beta }\right){\xi }^{\gamma }.$$

Since the metric $\mathcal{G}$ is Kähler, then from Equation (4), $\left({\Gamma }_{\beta \gamma }^{\beta }-{\Gamma }_{\gamma \beta }^{\beta }\right)=0.$ Therefore,

$$\mathrm{div}\xi ={\displaystyle \sum _{\beta =1}^n}{\nabla }_{\beta }{\xi }^{\beta },\mathrm{div}\eta ={\displaystyle \sum _{\gamma =1}^n}{\nabla }_{\overline{\gamma }}{\eta }^{\overline{\gamma }}.$$

Then, the proof is complete. □

Proposition 4 ([24]).

For a $C^{\infty }$ function λ and for $\psi \in {\mathcal{B}}_{0,s}(\overline{\mathsf{\Omega }},\Xi )$, $s\ge 1$,

$$\parallel \overline{\partial }{\psi \parallel }^2+\parallel {\overline{\partial }}^{\divideontimes }{\psi \parallel }^2=\parallel \overline{\nabla }{\psi \parallel }^2+{(\hslash |\mathrm{grad}\lambda \left|\right)}^{-1}{\int }_{\partial \mathsf{\Omega }}(\mathcal{L}\left(\lambda \right)\psi ,\psi )ds+<(\Theta -\mathcal{R})\psi ,\psi >,$$

where $\Theta =\left({\Theta }_{\alpha \overline{\beta }}\right)$ and $\mathcal{R}=\left({\mathcal{R}}_{\alpha \overline{\beta }}\right)$.

Proof. 

Since

$${\nabla }_{\beta }{\xi }^{\beta }={\nabla }_{\beta }\left(\sum _{B_{s-1}}{\displaystyle \sum _{\gamma =1}^n}{\hslash }^{-1}{\nabla }_{\overline{\gamma }}{\psi }_{{\overline{B}}_{s-1}}^{\beta }\overline{{\psi }^{\gamma B_{s-1}}}\right).$$

Since ${\nabla }_{\beta }^{(\hslash )}=({\nabla }_{\beta }-{\partial }_{\beta }log\hslash )$, from Equation (6), then

$$\begin{array}{cc}{\displaystyle {\displaystyle \sum _{\beta =1}^n}}{\nabla }_{\beta }{\xi }^{\beta }=\hfill & {\displaystyle {\displaystyle \sum _{\beta ,\gamma =1}^n}}{\nabla }_{\beta }\left({\hslash }^{-1}{\displaystyle \sum _{B_{s-1}}}{\nabla }_{\overline{\gamma }}{\psi }_{{\overline{B}}_{s-1}}^{\beta }\overline{{\psi }^{\gamma B_{s-1}}}\right)\hfill \\ & ={\hslash }^{-1}{\displaystyle \sum _{\beta ,\gamma }}{\displaystyle \sum _{B_{s-1}}}({\nabla }_{\beta }-{\partial }_{\beta }log\hslash ){\nabla }_{\overline{\gamma }}{\psi }_{{\overline{B}}_{s-1}}^{\beta }\overline{{\psi }^{\gamma B_{s-1}}}+{\hslash }^{-1}{\displaystyle \sum _{\beta ,\gamma }}{\displaystyle \sum _{B_{s-1}}}{\nabla }_{\overline{\gamma }}{\psi }_{{\overline{B}}_{s-1}}^{\beta }\overline{{\nabla }_{\overline{\beta }}{\psi }^{\gamma B_{s-1}}}\hfill \\ & ={\hslash }^{-1}{\displaystyle \sum _{\beta ,\gamma }}{\displaystyle \sum _{B_{s-1}}}{\nabla }_{\overline{\gamma }}{\nabla }_{\beta }^{(\hslash )}{\psi }_{{\overline{B}}_{s-1}}^{\beta }\overline{{\psi }^{\gamma B_{s-1}}}+{\hslash }^{-1}{\displaystyle \sum _{\beta ,\gamma }}{\displaystyle \sum _{B_{s-1}}}{\nabla }_{\overline{\gamma }}{\psi }_{{\overline{B}}_{s-1}}^{\beta }\overline{{\nabla }_{\overline{\beta }}{\psi }^{\gamma B_{s-1}}}\hfill \\ & +{\hslash }^{-1}{\displaystyle \sum _{\beta ,\gamma }}{\displaystyle \sum _{B_{s-1}}}[{\nabla }_{\beta }^{(\hslash )},{\nabla }_{\overline{\gamma }}]{\psi }_{{\overline{B}}_{s-1}}^{\beta }\overline{{\psi }^{\gamma B_{s-1}}}.\hfill \end{array}$$

Then, one obtains the commutator

$$[{\nabla }_{\beta }^{(\hslash )},{\nabla }_{\overline{\gamma }}]{\psi }_{{\overline{B}}_s}=[{\nabla }_{\beta },{\nabla }_{\overline{\gamma }}]{\psi }_{{\overline{B}}_s}+{\Theta }_{\beta \overline{\gamma }}{\psi }_{{\overline{B}}_s}.$$

Using Equation (6), one obtains

$$\begin{array}{c}{\nabla }_{\overline{\gamma }}{\nabla }_{\beta }{\psi }_{{\overline{B}}_s}={\partial }_{\overline{\gamma }}{\partial }_{\beta }{\psi }_{{\overline{B}}_s}-{\displaystyle \sum _{\mu =1}^s}{\displaystyle \sum _{\tau =1}^n}{\Gamma }_{\overline{\gamma }{\overline{\beta }}_{\mu }}^{\overline{\tau }}{\partial }_{\beta }{\psi }_{{\overline{\beta }}_1\dots {\overline{\beta }}_{\mu -1}\overline{\tau }{\overline{\beta }}_{\mu +1}\dots {\overline{\beta }}_s},\hfill \\ {\nabla }_{\beta }{\nabla }_{\overline{\gamma }}{\psi }_{{\overline{B}}_s}={\partial }_{\beta }{\partial }_{\overline{\gamma }}{\psi }_{{\overline{B}}_s}-{\displaystyle \sum _{\mu =1}^s}{\displaystyle \sum _{\tau =1}^n}{\partial }_{\beta }{\Gamma }_{\overline{\gamma }{\overline{\beta }}_{\mu }}^{\overline{\tau }}{\psi }_{{\overline{\beta }}_1\dots {\overline{\beta }}_{\mu -1}\overline{\tau }{\overline{\beta }}_{\mu +1}\dots {\overline{\beta }}_s}-{\displaystyle \sum _{\mu =1}^s}{\displaystyle \sum _{\tau =1}^n}{\Gamma }_{\overline{\gamma }{\overline{\beta }}_{\mu }}^{\overline{\tau }}{\partial }_{\beta }{\psi }_{{\overline{\beta }}_1\dots {\overline{\beta }}_{\mu -1}\overline{\tau }{\overline{\beta }}_{\mu +1}\dots {\overline{\beta }}_s}.\hfill \end{array}$$

Hence, by using Equation (1), one obtains

$$[{\nabla }_{\beta },{\nabla }_{\overline{\gamma }}]{\psi }_{{\overline{B}}_s}=-{\displaystyle \sum _{\mu =1}^s}{\displaystyle \sum _{\tau =1}^n}{\mathcal{R}}_{{\overline{\beta }}_{\mu }\beta \overline{\gamma }}^{\overline{\tau }}{\psi }_{{\overline{\beta }}_1\dots {\overline{\beta }}_{\mu -1}\overline{\tau }{\overline{\beta }}_{\mu +1}\dots {\overline{\beta }}_s}.$$

Therefore, one obtains

$$[{\nabla }_{\beta }^{(\hslash )},{\nabla }_{\overline{\gamma }}]{\psi }_{{\overline{B}}_s}=-{\displaystyle \sum _{B_s}}{\displaystyle {\displaystyle \sum _{\mu =1}^s}}{\displaystyle {\displaystyle \sum _{\tau =1}^n}}{\mathcal{R}}_{{\overline{\beta }}_{\mu }\beta \overline{\gamma }}^{\overline{\tau }}{\psi }_{{\overline{\beta }}_1\dots {\overline{\beta }}_{\mu -1}\overline{\tau }{\overline{\beta }}_{\mu +1}\dots {\overline{\beta }}_s}+{\Theta }_{\beta \overline{\gamma }}{\psi }_{{\overline{B}}_s}.$$

So,

$$\begin{array}{cc}{\hslash }^{-1}\hfill & {\displaystyle \sum _{\beta ,\gamma }}{\displaystyle \sum _{B_{s-1}}}[{\nabla }_{\beta }^{(\hslash )},{\nabla }_{\overline{\gamma }}]{\psi }_{{\overline{B}}_{s-1}}^{\beta }\overline{{\psi }^{\gamma B_{s-1}}}={\hslash }^{-1}{\displaystyle \sum _{\alpha ,\beta ,\gamma }}{g}^{\overline{\alpha }\beta }[{\nabla }_{\beta }^{(\hslash )},{\nabla }_{\overline{\gamma }}]{\psi }_{\overline{\alpha }{\overline{B}}_{s-1}}\overline{{\psi }^{\gamma B_{s-1}}}\hfill \\ & =-{\hslash }^{-1}{\displaystyle \sum _{\alpha ,\beta ,\gamma }}{g}^{\overline{\alpha }\beta }\left({\displaystyle {\displaystyle \sum _{\mu =1}^{s-1}}}{\displaystyle {\displaystyle \sum _{\tau =1}^n}}{\mathcal{R}}_{{\overline{\beta }}_{\mu }\beta \overline{\gamma }}^{\overline{\tau }}{\psi }_{\overline{\alpha }{\overline{\beta }}_1\dots {\overline{\beta }}_{\mu -1}\overline{\tau }{\overline{\beta }}_{\mu +1}\dots {\overline{\beta }}_{s-1}}\right)\overline{{\psi }^{\gamma B_{s-1}}}\hfill \\ & +{\hslash }^{-1}{\displaystyle \sum _{\alpha ,\beta ,\gamma }}{g}^{\overline{\alpha }\beta }{\Theta }_{\beta \overline{\gamma }}{\psi }_{\overline{\alpha }{\overline{B}}_{s-1}}\overline{{\psi }^{\gamma B_{s-1}}}\hfill \\ & =-{\hslash }^{-1}{\displaystyle \sum _{\alpha ,\beta ,\gamma ,\tau }}{g}^{\overline{\alpha }\beta }{\mathcal{R}}_{\overline{\alpha }\beta \overline{\gamma }}^{\overline{\tau }}{\psi }_{\overline{\tau }{\overline{B}}_{s-1}}\overline{{\psi }^{\gamma B_{s-1}}}-{\hslash }^{-1}{\displaystyle \sum _{\alpha ,\beta ,\gamma ,\tau }}{g}^{\overline{\alpha }\beta }{\mathcal{R}}_{{\overline{\beta }}_{\mu }\beta \overline{\gamma }}^{\overline{\tau }}{\psi }_{A_r\overline{\alpha }{\overline{\beta }}_1\dots {\overline{\beta }}_{\mu -1}\overline{\tau }{\overline{\beta }}_{\mu +1}\dots {\overline{\beta }}_{s-1}}\overline{{\psi }^{\gamma B_{s-1}}}\hfill \\ & +{\hslash }^{-1}{\displaystyle \sum _{\alpha ,\gamma }}{\Theta }_{\beta \overline{\gamma }}{\psi }_{{\overline{B}}_{s-1}}^{\beta }\overline{{\psi }^{\gamma B_{s-1}}}.\hfill \end{array}$$

(8)

From the Kähler property of $\mathcal{G}$, Equation (2) gives

$${\mathcal{R}}_{{\overline{\beta }}_{\mu }\overline{\gamma }}^{\overline{\tau }\overline{\alpha }}=\sum _{\beta }{g}^{\overline{\alpha }\beta }{\mathcal{R}}_{{\overline{\beta }}_{\mu }\beta \overline{\gamma }}^{\overline{\tau }}={\mathcal{R}}_{{\overline{\beta }}_{\mu }\overline{\gamma }}^{\overline{\alpha }\overline{\tau }}.$$

Moreover, we remark that ${\psi }_{\overline{\alpha }{\overline{\beta }}_1\dots {\overline{\beta }}_{\mu -1}\overline{\tau }{\overline{\beta }}_{\mu +1}\dots {\overline{\beta }}_{s-1}}=-{\psi }_{\overline{\tau }{\overline{\beta }}_1\dots {\overline{\beta }}_{\mu -1}\overline{\alpha }{\overline{\beta }}_{\mu +1}\dots {\overline{\beta }}_{s-1}}$. Hence, the second term of the right-hand side of Equation (8) is zero, i.e.,

$${\hslash }^{-1}\sum _{\alpha ,\beta ,\gamma ,\tau }{g}^{\overline{\alpha }\beta }{\mathcal{R}}_{{\overline{\beta }}_{\mu }\beta \overline{\gamma }}^{\overline{\tau }}{\psi }_{A_r\overline{\alpha }{\overline{\beta }}_1\dots {\overline{\beta }}_{\mu -1}\overline{\tau }{\overline{\beta }}_{\mu +1}\dots {\overline{\beta }}_{s-1}}\overline{{\psi }^{\gamma B_{s-1}}}=0.$$

As a result, Equation (8) becomes

$${\hslash }^{-1}\sum _{\beta ,\gamma }\sum _{B_{s-1}}[{\nabla }_{\beta }^{(\hslash )},{\nabla }_{\overline{\gamma }}]{\psi }_{{\overline{B}}_{s-1}}^{\beta }\overline{{\psi }^{\gamma B_{s-1}}}=-{\hslash }^{-1}\sum _{\gamma ,\tau }\left(\sum _{\alpha ,\beta }{g}^{\overline{\alpha }\beta }{R}_{\overline{\alpha }\beta \overline{\gamma }}^{\overline{\tau }}\right){\psi }_{\overline{\tau }{\overline{B}}_{s-1}}\overline{{\psi }^{\gamma B_{s-1}}}$$

(9)

On the other hand,

$${\displaystyle \sum _{\alpha ,\beta =1}^n}{g}^{\overline{\alpha }\beta }{\mathcal{R}}_{\overline{\alpha }\beta \overline{\gamma }}^{\overline{\tau }}=\sum _{\alpha ,\beta ,\lambda }{g}^{\overline{\alpha }\beta }{g}^{\overline{\tau }\lambda }{\mathcal{R}}_{\lambda \overline{\alpha }\beta \overline{\gamma }}=\sum _{\alpha ,\lambda }{g}^{\overline{\tau }\lambda }\overline{{\sum }_{\beta }{g}^{\overline{\beta }\alpha }{\mathcal{R}}_{\overline{\beta }\alpha \overline{\lambda }\gamma }}=\sum _{\alpha ,\lambda }{g}^{\overline{\tau }\lambda }\overline{{\mathcal{R}}_{\alpha \overline{\lambda }\gamma }^{\alpha }}=\sum _{\lambda }{g}^{\overline{\tau }\lambda }{\mathcal{R}}_{\overline{\gamma }\lambda }.$$

Hence,

$${\hslash }^{-1}\sum _{\beta ,\gamma }\sum _{B_{s-1}}[{\nabla }_{\beta }^{(\hslash )},{\nabla }_{\overline{\gamma }}]{\psi }_{{\overline{B}}_{s-1}}^{\beta }\overline{{\psi }^{\gamma B_{s-1}}}=(s-1)!{\hslash }^{-1}\sum _{B_{s-1}}{\displaystyle \sum _{\alpha ,\gamma =1}^n}\left({\Theta }_{\beta \overline{\gamma }}-{\mathcal{R}}_{\beta \overline{\gamma }}\right){\psi }_{{\overline{B}}_{s-1}}^{\beta }\overline{{\psi }^{\gamma B_{s-1}}}.$$

We compute the second term of Equation (9). From Equations (1) and (5), one obtains

$$(\overline{\partial }\psi ,\overline{\partial }\psi )={\hslash }^{-1}\sum _{C_s,D_s}{\nabla }_{\overline{\mu }}{\psi }_{{\overline{C}}_s}\overline{{\nabla }^{\tau }{\psi }^{D_s}}{\mathcal{E}}_{\tau D_s}^{\mu C_s},$$

where ${\mathcal{E}}_{\tau D_s}^{\mu C_s}=0$ unless $\mu \notin C_s$, $\tau \notin D_s$ and $\left\{\mu \right\}\cup C_s=\left\{\tau \right\}\cup D_s$, in which case ${\mathcal{E}}_{\tau D_s}^{\mu C_s}$ is the sign of the permutation $\left(\mu C_s\tau D_s\right)$. Consider the terms with $\mu =\tau$. If ${\mathcal{E}}_{\tau D_s}^{\mu C_s}\ne 0$, then we must have $C_s=D_s$ and $\mu \notin C_s$, and hence the sum of these terms is

$${\hslash }^{-1}\sum _{C_s}\sum _{\mu \notin C_s}{\nabla }_{\overline{\mu }}{\psi }_{{\overline{C}}_s}\overline{{\nabla }^{\mu }{\psi }^{C_s}}.$$

Next, we consider the terms with $\mu \ne \tau$. If ${\mathcal{E}}_{\tau D_s}^{\mu C_s}\ne 0$, $\tau \in C_s$, $\mu \in D_s$ with deletion $\tau$ from $C_s$ or $\mu$ from $D_s$ has the same multi-index $B_{s-1}$:

$${\mathcal{E}}_{\tau D_s}^{\mu C_s}={\mathcal{E}}_{\mu \tau B_{s-1}}^{\mu C_s}{\mathcal{E}}_{\tau \mu B_{s-1}}^{\mu \tau B_{s-1}}{\mathcal{E}}_{\tau D_s}^{\tau \mu B_{s-1}}=-{\mathcal{E}}_{\tau B_{s-1}}^{C_s}{\mathcal{E}}_{D_s}^{\mu B_{s-1}},$$

The sum of the terms in question is

$$-{\hslash }^{-1}\sum _{B_{s-1}}\sum _{\mu \ne \tau }{\nabla }_{\overline{\mu }}{\psi }_{\overline{\tau }{\overline{B}}_{s-1}}\overline{{\nabla }^{\tau }{\psi }^{\mu B_{s-1}}}.$$

Therefore, one obtains

$$\begin{array}{cc}\left(\overline{\partial }\psi ,\overline{\partial }\psi \right)\hfill & ={\hslash }^{-1}{\displaystyle \sum _{C_s}}{\displaystyle \sum _{\mu \notin C_s}}{\nabla }_{\overline{\mu }}{\psi }_{{\overline{C}}_s}\overline{{\nabla }^{\mu }{\psi }^{C_s}}-{\hslash }^{-1}{\displaystyle \sum _{B_{s-1}}}{\displaystyle \sum _{\mu \ne \tau }}{\nabla }_{\overline{\mu }}{\psi }_{\overline{\tau }{\overline{B}}_{s-1}}\overline{{\nabla }^{\tau }{\psi }^{\mu B_{s-1}}}\hfill \\ & ={\hslash }^{-1}{\displaystyle \sum _{C_s}}{\displaystyle \sum _{\mu }}{\nabla }_{\overline{\mu }}{\psi }_{{\overline{C}}_s}\overline{{\nabla }^{\mu }{\psi }^{C_s}}-{\hslash }^{-1}{\displaystyle \sum _{B_{s-1}}}{\displaystyle \sum _{\mu \ne \tau }}{\nabla }_{\overline{\mu }}{\psi }_{\overline{\tau }{\overline{B}}_{s-1}}\overline{\left({\displaystyle \sum _{\gamma }}{g}^{\overline{\gamma }\tau }{\nabla }_{\overline{\gamma }}\right){\psi }^{\mu B_{s-1}}}\hfill \\ & ={\hslash }^{-1}{\displaystyle \sum _{C_s}}{\displaystyle \sum _{\mu }}{\nabla }_{\overline{\mu }}{\psi }_{{\overline{C}}_s}\overline{{\nabla }^{\mu }{\psi }^{C_s}}-{\hslash }^{-1}{\displaystyle \sum _{B_{s-1}}}{\displaystyle \sum _{\mu \ne \tau }}{\displaystyle \sum _{\gamma }}{g}^{\overline{\tau }\gamma }{\nabla }_{\overline{\mu }}{\psi }_{\overline{\tau }{\overline{B}}_{s-1}}\overline{{\nabla }_{\overline{\gamma }}{\psi }^{\mu B_{s-1}}}.\hfill \end{array}$$

(10)

Since ${\nabla }_{\overline{\tau }}{g}^{\overline{\beta }\alpha }=0$, then by using Proposition 2, one obtains

$$\left(\overline{\partial }\psi ,\overline{\partial }\psi \right)=\left(\overline{\nabla },\overline{\nabla }\right)-{\hslash }^{-1}\sum _{B_{s-1}}\sum _{\mu ,\gamma }{\nabla }_{\overline{\mu }}{\psi }_{{\overline{B}}_{s-1}}^{\gamma }\overline{{\nabla }_{\overline{\gamma }}{\psi }^{\mu B_{s-1}}}.$$

Then, one obtains

$${\hslash }^{-1}\sum _{B_{s-1}}\sum _{\beta ,\gamma }{\nabla }_{\overline{\beta }}{\psi }_{{\overline{B}}_{s-1}}^{\gamma }\overline{{\nabla }_{\overline{\gamma }}{\psi }^{\beta B_{s-1}}}=(s-1)!\left[\left(\overline{\nabla },\overline{\nabla }\right)-\left(\overline{\partial }\psi ,\overline{\partial }\psi \right)\right].$$

Therefore,

$${\displaystyle \sum _{\beta =1}^n}{\nabla }_{\beta }{\xi }^{\beta }={\hslash }^{-1}\sum {\nabla }_{\overline{\gamma }}{\nabla }_{\beta }^{(\hslash )}{\psi }_{{\overline{B}}_{s-1}}^{\beta }\overline{{\psi }^{{\overline{A}}_r\gamma B_{s-1}}}+(s-1)!\left\{((\Theta -\mathcal{R})\psi ,\psi )+\left(\overline{\nabla },\overline{\nabla }\right)-\left(\overline{\partial }\psi ,\overline{\partial }\psi \right)\right\}.$$

(11)

Using Equation (7),

$$\begin{array}{cc}{\displaystyle \sum _{\gamma =1}^n}{\nabla }_{\overline{\gamma }}{\eta }^{\overline{\gamma }}\hfill & ={\hslash }^{-1}{\displaystyle \sum _{\beta ,\gamma =1}^n}{\displaystyle \sum _{B_{s-1}}}({\nabla }_{\overline{\gamma }}-{\partial }_{\overline{\gamma }}log\hslash ){\nabla }_{\beta }^{(\hslash )}{\psi }_{{\overline{B}}_{s-1}}^{\beta }\overline{{\psi }^{\gamma B_{s-1}}}\hfill \\ & +{\hslash }^{-1}{\displaystyle \sum _{\beta ,\gamma }}{\displaystyle \sum _{B_{s-1}}}{\nabla }_{\beta }^{(\hslash )}{\psi }_{{\overline{B}}_{s-1}}^{\beta }\overline{{\nabla }_{\gamma }{\psi }^{\gamma B_{s-1}}}\hfill \\ & ={\hslash }^{-1}{\displaystyle \sum _{\beta ,\gamma }}{\displaystyle \sum _{B_{s-1}}}{\nabla }_{\overline{\gamma }}{\nabla }_{\beta }^{(\hslash )}{\psi }_{{\overline{B}}_{s-1}}^{\beta }\overline{{\psi }^{\gamma B_{s-1}}}+{\hslash }^{-1}{\displaystyle \sum _{\beta ,\gamma }}{\displaystyle \sum _{B_{s-1}}}{\nabla }_{\beta }^{(\hslash )}{\psi }_{{\overline{B}}_{s-1}}^{\beta }\overline{{\nabla }_{\gamma }^{(\hslash )}{\psi }^{\gamma B_{s-1}}}.\hfill \end{array}$$

(12)

Hence, by using Equation (11), one obtains

$${\displaystyle \sum _{\gamma =1}^n}{\nabla }_{\overline{\gamma }}{\eta }^{\overline{\gamma }}=(s-1)!\left({\overline{\partial }}^{\divideontimes }\psi ,{\overline{\partial }}^{\divideontimes }\psi \right)+{\hslash }^{-1}\sum _{\beta ,\gamma }\sum _{B_{s-1}}{\nabla }_{\overline{\gamma }}{\nabla }_{\beta }^{(\hslash )}{\psi }_{{\overline{B}}_{s-1}}^{\beta }\overline{{\psi }^{\gamma B_{s-1}}}.$$

(13)

Subtracting Equation (13) from Equation (12) and from Proposition 2, one obtains

$$\frac{1}{(s-1)!}\left(\mathrm{div}\xi -\mathrm{div}\eta \right)=\left(\overline{\nabla },\overline{\nabla }\right)-\left(\overline{\partial }\psi ,\overline{\partial }\psi \right)-\left({\overline{\partial }}^{\divideontimes }\psi ,{\overline{\partial }}^{\divideontimes }\psi \right)+((\Theta -\mathcal{R})\psi ,\psi ).$$

By integrating this identity over $\mathsf{\Omega }$ and by applying the divergence theorem, one obtains

$$\frac{1}{(s-1)!}{\int }_{\partial \mathsf{\Omega }}\left(\xi -\eta \right).\mathrm{n}\mathrm{ds}=\parallel \overline{\nabla }{\psi \parallel }^2-\parallel \overline{\partial }{\psi \parallel }^2-{\parallel {\overline{\partial }}^{\divideontimes }\psi \parallel }^2+<(\Theta -\mathcal{R})\psi ,\psi >,$$

(14)

with the outer unit normal vector n to $\partial \mathsf{\Omega }$, which is given at each point $x\in \partial \mathsf{\Omega }$ by $n=\frac{\mathrm{grad}\lambda }{\left|\mathrm{grad}\lambda \right|}$, and the projection of the vector $\left(\xi -\eta \right)$ on the vector n is $\left(\xi -\eta \right).n$. Now, we compute $\eta .\mathrm{grad}\lambda$. Since

$$\eta .\mathrm{grad}\lambda ={\displaystyle \sum _{\gamma =1}^n}{\eta }^{\overline{\gamma }}{\partial }_{\overline{\gamma }}\lambda ={\hslash }^{-1}{\displaystyle \sum _{\beta =1}^n}\sum _{B_{s-1}}{\nabla }_{\beta }^{(\hslash )}{\psi }_{{\overline{B}}_{s-1}}^{\beta }\overline{\left({\displaystyle \sum _{\gamma =1}^n}{\psi }^{\gamma B_{s-1}}{\partial }_{\gamma }\lambda \right)},$$

at any point of X, then for $\psi \in {\mathcal{B}}_{0,s}(\overline{\mathsf{\Omega }},\Xi )$, $s\ge 1$, one obtains

$$\eta .\mathrm{grad}\lambda =0\mathrm{on}\partial \mathsf{\Omega }.$$

Hence,

$$\eta .n=0,\mathrm{on}\partial \mathsf{\Omega }.$$

(15)

Now we compute $\xi .n$. from Equation (5); one obtains

$$\begin{array}{cc}\xi .n\hfill & =\frac{1}{\left|\mathrm{grad}\lambda \right|}\left(\xi .\mathrm{grad}\lambda \right)=\frac{1}{\left|\mathrm{grad}\lambda \right|}{\displaystyle \sum _{\beta =1}^n}{\xi }^{\beta }{\partial }_{\beta }\lambda \hfill \\ & =\frac{1}{\hslash \left|\mathrm{grad}\lambda \right|}{\displaystyle \sum _{\gamma =1}^n}\sum _{B_{s-1}}\left({\displaystyle \sum _{\beta =1}^n}{\nabla }_{\overline{\gamma }}{\psi }_{{\overline{B}}_{s-1}}^{\beta }{\partial }_{\beta }\lambda \right)\overline{{\psi }^{\gamma B_{s-1}}}.\hfill \end{array}$$

(16)

Again, for $\psi \in {\mathcal{B}}_{0,s}(\overline{\mathsf{\Omega }},\Xi )$, $s\ge 1$, one obtains

$${\displaystyle \sum _{\beta =1}^n}{\psi }_{{\overline{B}}_{s-1}}^{\beta }{\partial }_{\beta }\lambda =0\mathrm{on}\partial \mathsf{\Omega }.$$

Since $\lambda \equiv 0$ on $\partial \mathsf{\Omega }$, then we can write

$${\displaystyle \sum _{\beta =1}^n}{\psi }_{{\overline{B}}_{s-1}}^{\beta }{\partial }_{\beta }\lambda =\lambda {\varphi }_{{\overline{B}}_{s-1}},$$

on the neighborhood U of $\partial \mathsf{\Omega }$, where ${\varphi }_{{\overline{B}}_{s-1}}$ is a $C^{\infty }$ section of ${\bigwedge }^{s-1}{\overline{T}}^{\divideontimes }\left(X\right)\otimes \Xi$. So,

$${\displaystyle \sum _{\beta =1}^n}{\nabla }_{\overline{\gamma }}{\psi }_{{\overline{B}}_{s-1}}^{\beta }{\partial }_{\beta }\lambda +{\displaystyle \sum _{\beta =1}^n}{\psi }_{{\overline{B}}_{s-1}}^{\beta }{\partial }_{\beta }{\partial }_{\overline{\gamma }}\lambda ={\varphi }_{{\overline{B}}_{s-1}}{\partial }_{\overline{\gamma }}\lambda +\lambda {\nabla }_{\overline{\gamma }}{\varphi }_{{\overline{B}}_{s-1}}\mathrm{on}U.$$

Then, we multiply this equation by ${\hslash }^{-1}\overline{{\psi }^{\gamma B_{s-1}}}$ and sum it with respect to $\gamma$. Since $\psi \in {\mathcal{B}}_{0,s}(\overline{\mathsf{\Omega }},\Xi )$, one obtains

$$\begin{array}{cc}{\hslash }^{-1}{\displaystyle \sum _{B_{s-1}}}\hfill & {\displaystyle {\displaystyle \sum _{\beta ,\gamma =1}^n}}{\nabla }_{\overline{\gamma }}{\psi }_{{\overline{B}}_{s-1}}^{\beta }{\partial }_{\beta }\lambda \overline{{\psi }^{\gamma B_{s-1}}}+{\hslash }^{-1}{\displaystyle \sum _{B_{s-1}}}{\displaystyle {\displaystyle \sum _{\beta ,\gamma =1}^n}}{\partial }_{\beta }{\partial }_{\overline{\gamma }}\lambda {\psi }_{{\overline{B}}_{s-1}}^{\beta }\overline{{\psi }^{\gamma B_{s-1}}}\hfill \\ & ={\hslash }^{-1}{\displaystyle \sum _{B_{s-1}}}{\displaystyle {\displaystyle \sum _{\gamma =1}^n}}{\varphi }_{{\overline{B}}_{s-1}}{\partial }_{\overline{\gamma }}\lambda \overline{{\psi }^{\gamma B_{s-1}}}+{\hslash }^{-1}{\displaystyle \sum _{B_{s-1}}}{\displaystyle {\displaystyle \sum _{\gamma =1}^n}}\lambda {\nabla }_{\overline{\gamma }}{\varphi }_{{\overline{B}}_{s-1}}\overline{{\psi }^{\gamma B_{s-1}}}\hfill \\ & =0,\hfill \end{array}$$

on $\partial \mathsf{\Omega }$. Therefore, by dividing by $\left|\mathrm{grad}\lambda \right|$, (16) becomes

$$\xi .n=-\frac{1}{\hslash \left|\mathrm{grad}\lambda \right|}{\displaystyle \sum _{B_{s-1}}}{\displaystyle {\displaystyle \sum _{\beta ,\gamma =1}^n}}{\partial }_{\beta }{\partial }_{\overline{\gamma }}\lambda {\psi }_{{\overline{B}}_{s-1}}^{\beta }\overline{{\psi }^{\gamma B_{s-1}}}\mathrm{on}\partial \mathsf{\Omega }.$$

Then,

$$\xi .n=-\frac{1}{\hslash \left|\mathrm{grad}\lambda \right|}(\mathcal{L}\left(\lambda \right)\psi ,\psi ),$$

(17)

on $\partial \mathsf{\Omega }$. Thus, the proposition is proved by substituting Equations (15) and (17) in Equation (14). □

4. Bounded P.S.H. Functions and Hartogs Pseudoconvexity in Kähler Manifolds

Definition 2 ([28]).

Ω is the smooth local Stein domain if ∀ point $z\in \partial \mathsf{\Omega }$, and ∃ is a neighborhood U if z satisfies $U\cap \mathsf{\Omega }$, which is Stein.

Definition 3 ([29]).

We say that Ω is Hartogs pseudoconvex if there exists a smooth bounded function h on Ω such that

$$i\partial \overline{\partial }(-log\delta +h)\ge C\omega \mathrm{in}\mathsf{\Omega },$$

(18)

for some $C>0$, where ω is the Kähler form associated with the Kähler metric.

In particular, every Hartogs pseudoconvex domain admits a strictly plurisubharmonic exhaustion function and is thus a Stein manifold.

Next, we will examine several examples of Hartogs pseudoconvex domains.

Example 1.

Suppose X is a complex manifold with a continuous strongly plurisubharmonic function and $\mathsf{\Omega }⋐X$ is a Stein domain. According to [30], there exists a Kähler metric on X such that Ω is Hartogs pseudoconvex.

Example 2 ([29]).

All the local Stein-domain subsets of a Stein manifold are in the Hartogs pseudoconvex domain.

Example 3 ([29]).

Every $C^2$ pseudoconvex domain in the ${\mathcal{C}}^n$ subset of a Stein manifold is a Hartogs pseudoconvex domain.

Example 4 ([30]).

Any local Stein domain subset of a Kähler manifold with positive holomorphic bisectional curvature satisfies Equation (18) on $U\cap \mathsf{\Omega }$.

Example 5 ([30]).

If Ω is a local Stein domain of the complex projective space ${\mathcal{P}}^n$, then Ω satisfies Equation (18).

The canonical line bundle K of X is defined by transition functions $\left(k_{ij}\right)$

$$k_{ij}=\frac{\partial ({\mathtt{w}}_j^1,{\mathtt{w}}_j^2,\dots ,{\mathtt{w}}_j^n)}{\partial ({\mathtt{w}}_i^1,{\mathtt{w}}_i^2,\dots ,{\mathtt{w}}_i^n)}\mathrm{on}{U}_i\cap U_j,$$

with

$${\mathtt{g}}_i={\left|k_{ij}\right|}^2{\mathtt{g}}_j\mathrm{on}{U}_i\cap U_j.$$

Hence, $g=\left\{{\mathtt{g}}_i\right\}$ determines a metric of K. Let $h=\left\{h_i\right\}$ be a Hermitian metric of $\Xi$ and $\partial \overline{\partial }logh$ its curvature tensor. So, $\{\hslash =g.h\}$ determines a Hermitian metric of $\Xi \otimes K$ and

$$\partial \overline{\partial }log\hslash =\partial \overline{\partial }logh+\partial \overline{\partial }logg.$$

Then, from Proposition 4,

$$\parallel \overline{\partial }{\psi \parallel }^2+\parallel {\overline{\partial }}^{\divideontimes }{\psi \parallel }^2={\parallel \overline{\nabla }\psi \parallel }^2+<\Theta \psi ,\psi >+\frac{1}{\hslash \left|\mathrm{grad}\lambda \right|}{\int }_{\partial D}(\mathcal{L}\left(\lambda \right)\psi ,\psi )ds,$$

(19)

for $\psi \in {\mathcal{B}}_{0,s}(\overline{D},\Xi \otimes K)$, $s\ge 1$. Using $h=h_{\mathtt{m}}={\zeta }^{\mathtt{m}}h$, one obtains

$${\Theta }_{\mathtt{m}}=\Theta -\mathtt{m}\partial \overline{\partial }(-log\zeta ).$$

With respect to the $\mathcal{G}$ and $h_{\mathtt{m}}$, and for $\psi ,\psi \in {\mathcal{C}}_{n,s}^{\infty }(\overline{D},\Xi )$, we define the global inner product $<\psi ,\psi {>}_{\mathtt{m}}$ and the norm ${\parallel \psi \parallel }_{W_{n,s}^{\mathtt{m}}(D,\Xi )}$ by

$$<\psi ,\psi {>}_{\mathtt{m}}={\int }_D{(\psi ,\psi )}_{\mathtt{m}}dv\mathrm{and}{\parallel \psi \parallel }_{W_{n,s}^{\mathtt{m}}(D,\Xi )}^2=<\psi ,\psi {>}_{\mathtt{m}}.$$

Then, (19) becomes

$$\begin{array}{cc}\parallel \overline{\partial }{\psi \parallel }_{W_{n,s}^{\mathtt{m}}(D,\Xi )}^2+\parallel {\overline{\partial }}_{\mathtt{m}}^{\divideontimes }{\psi \parallel }_{W_{n,s}^{\mathtt{m}}(D,\Xi )}^2\hfill & ={\parallel \overline{\nabla }\psi \parallel }_{W_{n,s}^{\mathtt{m}}(D,\Xi )}^2+\frac{1}{\hslash \left|\mathrm{grad}\lambda \right|}{\int }_{\partial D}{(\mathcal{L}\left(\lambda \right)\psi ,\psi )}_{\mathtt{m}}ds+<{\Theta }_{\mathtt{m}}\psi ,\psi {>}_{\mathtt{m}}\hfill \\ & +<\mathtt{m}\partial \overline{\partial }(-log\zeta )\psi ,\psi {>}_{\mathtt{m}}.\hfill \end{array}$$

(20)

As Theorem 1.1 in [31], one obtains

Theorem 1.

Suppose X is an n-dimensional complex manifold and $D⋐X$ is a Hartogs pseudoconvex. $\zeta \left(z\right)=\mathrm{dist}(z,\partial D)=-\rho \left(z\right)$, where ζ is the Kähler metric ω on X. If $\mathtt{m}>0$, then

$$i\partial \overline{\partial }(-{\zeta }^{\mathtt{m}})\ge c_{\mathtt{m}}\left|{\zeta }^{\mathtt{m}}\right|\omega ,$$

(21)

for some constant $c_{\mathtt{m}}>0$.

Proof. 

Using Equation (18) and if $\rho =-\zeta$,

$$-\rho i\partial \overline{\partial }\rho +i\partial \rho \wedge \overline{\partial }\rho \ge C{\rho }^2\omega .$$

(22)

Let $\left(e_i\right)$ be an orthonormal basis for $T(\partial D)$ near p. In this case, near $p\in \partial D$, choose local coordinates that satisfy $x_{2n}=\rho$, $e_i\left(p\right)=0$, $i=1,2,\dots ,n-1$. The Hermitian form for $i\partial \overline{\partial }\rho$ is denoted by $\left(a_{ij}\right)$. The inequality (22) gives the coordinates

$$-\rho {\displaystyle \sum _{i,j=1}^n}{a}_{ij}{\eta }_i{\overline{\eta }}_j+{|\partial \rho |}^2|{\eta }_n{|}^2\ge C{\rho }^2{\displaystyle \sum _{j=1}^n}{\left|{\eta }_j\right|}^2.$$

(23)

If ${\eta }_n=0$,

$${\displaystyle \sum _{i,j=1}^{n-1}}{a}_{ij}{\eta }_i{\overline{\eta }}_j\ge C\left|\rho \right|{\displaystyle \sum _{j=1}^{n-1}}{\left|{\eta }_j\right|}^2.$$

Expanding (23), one obtains

$$-\rho {\displaystyle \sum _{i,j=1}^{n-1}}{a}_{ij}{\eta }_i{\overline{\eta }}_j+2\mathtt{Re}(-\rho ){\displaystyle \sum _{k=1}^{n-1}}{a}_{nk}{\eta }_n{\overline{\eta }}_k-\rho a_{nn}|{\eta }_n{|}^2+{|\partial \rho |}^2|{\eta }_n{|}^2\ge {C\left|\rho \right|}^2{\displaystyle \sum _{j=1}^{n-1}}{\left|{\eta }_j\right|}^2.$$

for $j\le n-1$, replacing v by ${\eta }_j/(-\rho )$,

$${\displaystyle \sum _{i,j=1}^{n-1}}\left(\frac{a_{ij}}{-\rho }\right){\eta }_i{\overline{\eta }}_j+2\mathtt{Re}(-\rho ){\displaystyle \sum _{k=1}^{n-1}}{a}_{nk}{\eta }_n{\overline{\eta }}_k-\rho a_{nn}|{\eta }_n{|}^2+{|\partial \rho |}^2|{\eta }_n{|}^2\ge C^{\prime }{\displaystyle \sum _{j=1}^{n-1}}{\left|{\eta }_j\right|}^2.$$

(24)

The inequality’s left side can be expressed as follows:

$$Q(z,\eta )+{|\partial \rho |}^2{\left|{\eta }_n\right|}^2.$$

For $z\in D$, we assume that

$$\tilde{Q}(\varsigma ,\eta )=\underset{z\to \varsigma }{lim\; inf}Q(z,\eta )=\underset{t\to 0}{lim}\underset{|z-\varsigma |<t}{inf}Q(z,\eta ).$$

From Equation (24), one obtains

$$\tilde{Q}(\varsigma ,\eta )+{|\partial \rho |}^2\left(\varsigma \right)|{\eta }_n{|}^2\ge C^{\prime }{\displaystyle \sum _{j=1}^{n-1}}{\left|{\eta }_j\right|}^2.$$

(25)

Take a look at $\tilde{Q}(p,(0,{\eta }_n)\ge 0$; for a small enough $C^{\prime }$,

$$\tilde{Q}(\varsigma ,v)+{|\partial \rho |}^2\left(\varsigma \right)|{\eta }_n{|}^2\ge C^{\prime }{\left|{\eta }_n\right|}^2,$$

in a neighborhood of p. On the sphere $\left|\eta \right|=1$, inequality (25) still holds for $|{\eta }^{\prime }|\le \sigma$ in a neighborhood of ${\eta }^{\prime }=0$, where $\eta =({\eta }^{\prime },{\eta }_n)$. This gives us

$$Q(z,\eta )+{|\partial \rho |}^2\left(z\right)|{\eta }_n{|}^2\ge \frac{C^{\prime }}{2}{\left|{\eta }_n\right|}^2,$$

for $\zeta \left(z\right)<{\sigma }^{\prime }$, $|{\eta }^{\prime }|\le \sigma$. But, when $|{\eta }^{\prime }|>\sigma$ and $\left|\eta \right|=1$, one obtains $|{\eta }^{\prime }{|}^2\ge {\sigma }_0{\left|{\eta }_n\right|}^2$, where ${\sigma }_0={\sigma }^2{(1-{\sigma }^2)}^{-1}$. So, by using (25),

$$Q(z,\eta )+{|\partial \rho |}^2|{\eta }_n{|}^2\ge {\sigma }^{\ast \ast }{\left|{\eta }_n\right|}^2\mathrm{for}\mathrm{some}{\sigma }^{\ast \ast }>0$$

and for $\zeta \left(z\right)<{\sigma }^{\ast }$,

$$Q(z,\eta )+{|\partial \rho |}^2|{\eta }_n{|}^2\ge \frac{{\sigma }^{\ast \ast }}{2}|{\eta }_n{|}^2+\frac{C^{\prime }}{2}{\displaystyle \sum _{j=1}^{n-1}}{\left|{\eta }_j\right|}^2.$$

Recalling this one yields

$$-\rho {\displaystyle \sum _{i,j=1}^n}{a}_{ij}{\eta }_i{\overline{\eta }}_j+{|\partial \rho |}^2|{\eta }_n{|}^2\ge \frac{\sigma }{2}|{\eta }_n{|}^2+\frac{C^{\prime }}{2}{\displaystyle \sum _{j=1}^{n-1}}{\left|{\eta }_j\right|}^2$$

Which means

$$\begin{array}{cc}-i\partial \overline{\partial }{(-\rho )}^{\mathtt{m}}\hfill & =i\mathtt{m}{(-\rho )}^{\mathtt{m}}\left(\frac{\partial \overline{\partial }\rho }{-\rho }+(1-\mathtt{m})\frac{\partial \rho \wedge \overline{\partial }\rho }{{\rho }^2}\right)\hfill \\ & \ge \frac{C^{\prime }}{2}\mathtt{m}{\left|\rho \right|}^{\mathtt{m}}\omega .\hfill \end{array}$$

□

Lemma 1.

Let $D⋐X$ be a $C^2$ Hartogs pseudoconvex in an n-dimensional complex manifold X. Suppose ${\mathtt{m}}_0={\mathtt{m}}_0\left(D\right)>0$ is the order of plurisubharmonicity for $\zeta \left(z\right)=d(z,\partial D)$:

$${\mathtt{m}}_0\left(D\right)=sup\{0<\epsilon \le 1|i\partial \overline{\partial }(-{\zeta }^{\epsilon })\ge \mathrm{on}D\}.$$

Then, ∀$0<\mathtt{m}<{\mathtt{m}}_0$ and $\varphi =-\mathtt{m}log\zeta$; there exists

$$it\partial \overline{\partial }\varphi \ge i\partial \varphi \wedge \overline{\partial }\varphi ,$$

(26)

with $0<t=\frac{\mathtt{m}}{{\mathtt{m}}_0}<1$. Also, there exists $C_{\mathtt{m}}>0$, which satisfies

$$i\partial \overline{\partial }(-{\zeta }^{\mathtt{m}})\ge C_{\mathtt{m}}{\zeta }^{\mathtt{m}}\left(\frac{i\partial \zeta \wedge \overline{\partial }\zeta }{{\zeta }^2}+\omega \right).$$

(27)

Proof. 

By Equation (21), ∃${\mathtt{m}}_0>0$ satisfies $i\partial \overline{\partial }(-{\zeta }^{{\mathtt{m}}_0})\ge 0$ on D. Since

$$\begin{array}{cc}i\partial \overline{\partial }(-{\zeta }^{{\mathtt{m}}_0})\hfill & =-i\partial \left({\mathtt{m}}_0{\zeta }^{{\mathtt{m}}_0-1}\overline{\partial }\zeta \right)=-i{\mathtt{m}}_0({\mathtt{m}}_0-1){\zeta }^{{\mathtt{m}}_0-2}\partial \zeta \wedge \overline{\partial }\zeta -i{\mathtt{m}}_0{\zeta }^{{\mathtt{m}}_0-1}\partial \overline{\partial }\zeta \hfill \\ & ={\mathtt{m}}_0{\zeta }^{{\mathtt{m}}_0}\left((1-{\mathtt{m}}_0)\frac{\partial \zeta \wedge \overline{\partial }\zeta }{{\zeta }^2}+i\frac{\partial \overline{\partial }(-\zeta )}{\zeta }\right).\hfill \end{array}$$

(28)

Then

$$(1-{\mathtt{m}}_0)\frac{i\partial \zeta \wedge \overline{\partial }\zeta }{{\zeta }^2}+i\frac{\partial \overline{\partial }(-\zeta )}{\zeta }\ge 0.$$

(29)

Also, by using Equation (18), one obtains

$$i\partial \overline{\partial }(-log\zeta )=\frac{i\partial \zeta \wedge \overline{\partial }\zeta }{{\zeta }^2}+i\frac{\partial \overline{\partial }(-\zeta )}{\zeta }\ge \omega .$$

(30)

Therefore, from Equations (29) and (30), one obtains

$$i\partial \overline{\partial }(-log\zeta )\ge {\mathtt{m}}_0\frac{i\partial \zeta \wedge \overline{\partial }\zeta }{{\zeta }^2}.$$

(31)

Since $\partial \varphi =-\mathtt{m}\frac{\partial \zeta }{\zeta }$ and $\overline{\partial }\varphi =-\mathtt{m}\frac{\overline{\partial }\zeta }{\zeta }$, then

$$i\partial \varphi \wedge \overline{\partial }\varphi ={\mathtt{m}}^2\frac{i\partial \zeta \wedge \overline{\partial }\zeta }{{\zeta }^2}.$$

(32)

Then, from Equations (31) and (32), one obtains

$$i\partial \overline{\partial }(-log\zeta )\ge {\mathtt{m}}_0\frac{i\partial \varphi \wedge \overline{\partial }\varphi }{{\mathtt{m}}^2}.$$

Then, Equation (26) is proved.

To prove Equation (27), choose $0<\kappa <\min \{1,\frac{{\mathtt{m}}_0-\mathtt{m}}{{\mathtt{m}}_0}\}$, and by using Equation (28), one obtains

$$\begin{array}{cc}i\partial \overline{\partial }(-{\zeta }^{\mathtt{m}})\hfill & =-i\mathtt{m}(\mathtt{m}-1){\zeta }^{\mathtt{m}-2}\partial \zeta \wedge \overline{\partial }\zeta -i\mathtt{m}{\zeta }^{\mathtt{m}-1}\partial \overline{\partial }\zeta =\mathtt{m}{\zeta }^{\mathtt{m}}\left((1-\mathtt{m})\frac{i\partial \zeta \wedge \overline{\partial }\zeta }{{\zeta }^2}+\frac{i\partial \overline{\partial }(-\zeta )}{\zeta }\right)\hfill \\ & =\mathtt{m}{\zeta }^{\mathtt{m}}\left(({\mathtt{m}}_0-\mathtt{m}-\kappa {\mathtt{m}}_0)\frac{i\partial \zeta \wedge \overline{\partial }\zeta }{{\zeta }^2}+(1-\kappa )\frac{i\partial \overline{\partial }(-{\zeta }^{{\mathtt{m}}_0})}{{\mathtt{m}}_0{\zeta }^{{\mathtt{m}}_0}}+\kappa i\partial \overline{\partial }(-log\zeta )\right)\hfill \\ & \ge C_{\mathtt{m}}\mathtt{m}{\zeta }^{\mathtt{m}}\left(\frac{i\partial \zeta \wedge \overline{\partial }\zeta }{{\zeta }^2}+\omega \right).\hfill \end{array}$$

Then, Equation (27) is proved. □

5. The ${\mathbf{L}}^{\mathbf{2}}$ Estimates of $\overline{\partial }$

As in [21,22,23,32,33], one proves the following results:

Theorem 2.

Let $D⋐X$ be a $C^2$ Hartogs pseudoconvex in an n-dimensional complex manifold X. Let Ξ be a positive line bundle over X whose curvature form Θ satisfies $\Theta \ge C\omega$, where $C>0$. Let $\psi \in L_{n,s}^2(D,{\zeta }^{\mathtt{m}},\Xi )$, $1\le s\le n$, a $\overline{\partial }$-closed form. Then, for $0<\mathtt{m}<{\mathtt{m}}_0$, there exists $\psi \in L_{n,s-1}^2(D,{\zeta }^{\mathtt{m}},\Xi )$, which satisfies $\overline{\partial }\psi =\psi$ and

$${\int }_{\mathsf{\Omega }}{\left|\psi \right|}^2{\zeta }^{\mathtt{m}}dv\le C{\int }_{\mathsf{\Omega }}{\left|\psi \right|}^2{\zeta }^{\mathtt{m}}dv.$$

(33)

Proof. 

The boundary term in Equation (20) vanishes since $\mathtt{m}>0$. For $u\in {\mathcal{B}}_{n,s}(\overline{D},\Xi )$, $s\ge 1$, and since the curvature form $\Theta$ of $\Xi$ satisfies

$$\Theta \ge C_D\omega \mathrm{on}D\mathrm{with}{C}_D>0.$$

then by using Equation (18), one obtains

$$<{\Theta }_{\mathtt{m}}\psi ,\psi {>}_{\mathtt{m}}\ge C_{\mathtt{m}}<\psi ,\psi {>}_{\mathtt{m}}.$$

(34)

Also, from the assumption of pseudoconvex on D, one obtains

$$\parallel \overline{\partial }{u\parallel }_{W_{n,s}^{\mathtt{m}}(D,\Xi )}^2+\parallel {\overline{\partial }}_{\mathtt{m}}^{\divideontimes }{u\parallel }_{W_{n,s}^{\mathtt{m}}(D,\Xi )}^2\ge C_{\mathsf{\Omega }}{\parallel u\parallel }_{W_{n,s}^{\mathtt{m}}(D,\Xi )}^2,$$

for all $u\in {\mathcal{B}}_{n,s}(D,\Xi )$. Let $u\in {\mathcal{D}}_{n,s}^{\infty }(D,E)$, with $u=u_1+u_2$, $u_1\in ker(\overline{\partial },\Xi )$ and $u_2\in ker{(\overline{\partial },E)}^{\perp }=\overline{\mathrm{Im}({\overline{\partial }}_{\mathtt{m}}^{\divideontimes },E)}\subset ker({\overline{\partial }}_{\mathtt{m}}^{\divideontimes },\Xi )$. Then, for every $(n,s)$ form u with compact support, one obtains

$$\begin{array}{cc}|<u,\psi {>}_{\mathtt{m}}|\hfill & =|<u_1+u_2,\psi {>}_{\mathtt{m}}|\hfill \\ & =|<u_1,\psi {>}_{\mathtt{m}}|+|<u_2,\psi {>}_{\mathtt{m}}|\hfill \\ & =|<u_1,\psi {>}_{\mathtt{m}}|\le \parallel u_1{\parallel }_{W_{n,s}^{\mathtt{m}}(D,\Xi )}{\parallel \psi \parallel }_{W_{n,s}^{\mathtt{m}}(D,\Xi )}\hfill \\ & \le \frac{1}{\sqrt{C}}\parallel {\overline{\partial }}_{\mathtt{m}}^{\divideontimes }{u}_1{\parallel }_{W_{n,s}^{\mathtt{m}}(D,\Xi )}{\parallel \psi \parallel }_{W_{n,s}^{\mathtt{m}}(D,\Xi )}\hfill \\ & =\frac{1}{\sqrt{C}}\parallel {\overline{\partial }}_{\mathtt{m}}^{\divideontimes }{u\parallel }_{W_{n,s}^{\mathtt{m}}(D,\Xi )}{\parallel \psi \parallel }_{W_{n,s}^{\mathtt{m}}(D,\Xi )}.\hfill \end{array}$$

Using the Riesz representation theorem, the linear form

$${\overline{\partial }}_{\mathtt{m}}^{\divideontimes }u⟼<u,\psi {>}_{\mathtt{m}},$$

is continuous on Rang$({\overline{\partial }}^{\divideontimes },\Xi )$ in the $L^2$ norm and has norm $\le C$, with

$${\parallel \psi \parallel }_{W_{n,s}^{\mathtt{m}}(D,\Xi )}=C.$$

Following Hahn–Banach theorem, ∃ is an element that is E valued $(n,s-1)$ from u on D (with a smooth boundary) perpendicular to $ker(\overline{\partial },E)$ with ${\parallel \psi \parallel }_{W_{n,s}^{\mathtt{m}}(D,\Xi )}\le C$,

$$<{\overline{\partial }}_{\mathtt{m}}^{\divideontimes }u,\psi {>}_{\mathtt{m}}=<u,\psi {>}_{\mathtt{m}},$$

for all $L^{2}u$ with both $\overline{\partial }u$ and ${\overline{\partial }}_{\mathtt{m}}^{\divideontimes }u$ and also $L^2$. Hence,

$$\overline{\partial }\psi =\psi ,$$

and

$${\parallel \psi \parallel }_{W_{n,s}^{\mathtt{m}}(D,\Xi )}\le C{\parallel \psi \parallel }_{W_{n,s}^{\mathtt{m}}(D,\Xi )}.$$

Exhaust a general pseudoconvex domain D by a sequence $D_{\mu }$ of $C^{\infty }$ pseudoconvex domains:

$$D={\cup }_{\mu =1}^{\infty }D,$$

with $D_{\mu }\subset D_{\mu +1}\subset D$ for each $\mu$. On each $D_{\mu }$, ∃ a ${\psi }_{\mu }\in L_{n,s-1}^2(D_{\mu },{\zeta }^{\mathtt{m}},\Xi )$ satisfies

$$\overline{\partial }{\psi }_{\mu }=\psi \mathrm{in}{D}_{\mu },$$

and

$${\int }_{D_{\mu }}|{\psi }_{\mu }{|}^2{\zeta }^{\mathtt{m}}dv\le C{\int }_{D_{\mu }}{\left|\psi \right|}^2{\zeta }^{\mathtt{m}}dv\le c{\int }_D{\left|\psi \right|}^2{\zeta }^{\mathtt{m}}dv.$$

Choose a subsequent ${\psi }_{\mu }$ of ${\psi }_{\mu }$, satisfying

$${\psi }_{\mu }⟶\psi ,$$

in $L_{n,s-1}^2(D,{\zeta }^{\mathtt{m}},\Xi )$ weakly. Moreover,

$${\int }_D{\left|\psi \right|}^2{\zeta }^{\mathtt{m}}dv\le lim\; infC{\int }_D{\left|\psi \right|}^2{\zeta }^{\mathtt{m}}dv\le c{\int }_D{\left|\psi \right|}^2{\zeta }^{\mathtt{m}}dv.$$

□

Theorem 3.

Let X, D and Ξ be the same as Theorem 2. Let $\psi \in L_{n,s}^2(D,\Xi )$, $1\le s\le n$, with $\overline{\partial }\psi =0$. Thus, ∃$\psi \in L_{n,s-1}^2(D,\Xi )$ satisfies $\overline{\partial }\psi =\psi$ and

$$\parallel \psi \parallel \le \parallel \psi \parallel .$$

Proof. 

Since

$$\frac{1}{\hslash \left|\mathrm{grad}\lambda \right|}{\int }_{\partial \mathsf{\Omega }}(\mathcal{L}\left(\lambda \right)u,u)ds\ge C{\int }_{\partial \mathsf{\Omega }}{\left|u\right|}^{2}ds,$$

and from Equation (18), one obtains

$$\parallel \overline{\partial }{u\parallel }^2+{\parallel \psi u\parallel }^2\ge C_D{\parallel u\parallel }^2,$$

∀$u\in {\mathcal{B}}_{n,s}(D,\Xi )$. This completes the proof of Theorem 3. □

Following Theorem 3, as in [34,35], one can prove the following:

Theorem 4.

Let X, D and Ξ be the same as Theorem 2. Then, □ has a closed range and ${ker}_{n,s}(\square ,E)=\left\{0\right\}$. For each $1\le s\le n$, there exists a bounded linear operator

$$N_{n,s}:L_{n,s}^2(D,\Xi )⟶L_{n,s}^2(D,\Xi ),$$

which satisfies

(i) Rang$(N_{n,s},\Xi )\subset$ Dom$({\square }_{n,s},\Xi )$ and ${\square }_{n,s}{N}_{n,s}$$=N_{n,s}{\square }_{n,s}=I$ on Dom$({\square }_{n,s},\Xi )$.

(ii) ∀$\psi \in L_{n,s}^2(D,\Xi )$, $\psi =\overline{\partial }{\overline{\partial }}^{\divideontimes }{N}_{n,s}\psi +{\overline{\partial }}^{\divideontimes }\overline{\partial }{N}_{n,s}\psi .$

(iii) For $\psi \in L_{n,s}^2(D,\Xi )$, one obtains

$$\begin{gathered} \parallel N_{n,s}\psi \parallel \le c_0\parallel \psi \parallel , \\ \parallel \overline{\partial }{N}_{n,s}\psi \parallel \le c_0\parallel \psi \parallel , \\ \parallel {\overline{\partial }}^{\divideontimes }{N}_{n,s}\psi \parallel \le c_0\parallel \psi \parallel . \end{gathered}$$

(iv)

$$\begin{gathered} N_{(n,s+1)}\overline{\partial }=\overline{\partial }{N}_{n,s}\mathit{on}\mathit{Dom}(\overline{\partial },\Xi ),1\le s\le n-1, \\ {\overline{\partial }}^{\divideontimes }{N}_{n,s}=N_{n,s-1}{\overline{\partial }}^{\divideontimes }\mathit{on}\mathit{Dom}({\overline{\partial }}^{\divideontimes },E),2\le s\le n. \end{gathered}$$

(v) If $\psi \in L_{n,s}^2(D,\Xi )$ and $\overline{\partial }\psi =0$, then $\psi =\overline{\partial }{\overline{\partial }}^{\divideontimes }{N}_{n,s}\psi$ and $u={\overline{\partial }}^{\divideontimes }{N}_{n,s}\psi$.

Proof. 

$$L_{n,s}^2(D,\Xi )=\overline{\mathrm{Rang}({\square }_{n,s},\Xi )}\oplus ker({\square }_{n,s},\Xi ).$$

We need to show that

$$ker({\square }_{n,s},\Xi )=ker(\overline{\partial },\Xi )\cap ker({\overline{\partial }}^{\divideontimes },\Xi )=\left\{0\right\}.$$

(35)

To show that

$$ker(\overline{\partial },\Xi )\cap ker({\overline{\partial }}^{\divideontimes },\Xi )=\left\{0\right\}.$$

(36)

We note that if $\psi \in L_{n,s}^2(\overline{\partial },\Xi ),$ then by using Theorem 4, ∃ a $\psi \in L_{n,s-1}^2(\mathsf{\Omega },\Xi )$ satisfies $\psi =\overline{\partial }\psi .$ If $\psi$ is also in $ker({\overline{\partial }}^{\divideontimes },\Xi )$, one obtains

$$0=<{\overline{\partial }}^{\divideontimes }\overline{\partial }\psi ,\psi >={\parallel \overline{\partial }\psi \parallel }^2.$$

Thus, $\psi =0$ and Equation (35) is proved. We shall show that $\mathrm{Rang}({\square }_{n,s},\Xi )$ is closed. Following Theorem 4, ∀$\psi \in L_{n,s}^2(D,\Xi ),s>0$ with $\overline{\partial }\psi =0$ and ∃ a $\psi \in L_{n,s-1}^2(D,\Xi )$ satisfies $\psi =\overline{\partial }\psi$ and

$${\parallel \psi \parallel }^2\le c_0{\parallel \psi \parallel }^2,$$

where $c_0=c_0\left(D\right)>0$. Thus, $\mathrm{Rang}(\overline{\partial },\Xi )$ is closed in every degree. Thus,

$${\parallel \psi \parallel }^2\le c_0(\parallel \overline{\partial }{\psi \parallel }^2+\parallel {\overline{\partial }}^{\divideontimes }\psi {\parallel }^2),$$

for $\psi \in \mathrm{Dom}(\overline{\partial },E)\cap \mathrm{Dom}({\overline{\partial }}^{\divideontimes },\Xi )$ and $\psi \perp ker(\overline{\partial },\Xi )\cap$$ker({\overline{\partial }}^{\divideontimes },\Xi )$. Thus, from (36),

$${\parallel \psi \parallel }^2\le c_0(\parallel \overline{\partial }{\psi \parallel }^2+\parallel {\overline{\partial }}^{\divideontimes }\psi {\parallel }^2),$$

for $\psi \in$ Dom$(\overline{\partial },\Xi )\cap$ Dom$({\overline{\partial }}^{\divideontimes },\Xi )$. Thus, ∀$\psi \in$ Dom$({\square }_{n,s},\Xi )$,

$$\begin{array}{cc}{\parallel \psi \parallel }^2\hfill & \le c_0[<\overline{\partial }\psi ,\overline{\partial }\psi >+<{\overline{\partial }}^{\divideontimes }\psi ,{\overline{\partial }}^{\divideontimes }\psi >]\hfill \\ & =c_0[<{\overline{\partial }}^{\divideontimes }\overline{\partial }\psi ,\psi >+<\overline{\partial }{\overline{\partial }}^{\divideontimes }\psi ,\psi >]\hfill \\ & =c_0<\square \psi ,\psi >\hfill \\ & \le c_0\parallel \square \psi \parallel \parallel \psi \parallel .\hfill \end{array}$$

Thus,

$$\parallel \psi \parallel \le c_0\parallel \square \psi \parallel ,$$

(37)

i.e., $\mathrm{Rang}({\square }_{n,s},\Xi )$ is closed. Therefore,

$$L_{n,s}^2(D,\Xi )=\mathrm{Rang}(\square ,\Xi )=\overline{\partial }{\overline{\partial }}^{\divideontimes }\mathrm{Dom}({\square }_{n,s},E)\oplus {\overline{\partial }}^{\divideontimes }\overline{\partial }\mathrm{Dom}({\square }_{n,s},\Xi ).$$

Also, from Equation (37), ${\square }_{n,s}$ is 1-1 and $\mathrm{Rang}({\square }_{n,s},\Xi )$ is the whole space $L_{n,s}^2(D,E)$. Thus, there exists a unique inverse

$$N_{n,s}:L_{n,s}^2(D,\Xi )⟶L_{n,s}^2(D,\Xi ),$$

which satisfies $\square N=N\square =I$ and

$$\parallel N_{n,s}\psi \parallel \le c_0\parallel \psi \parallel .$$

∀$\psi \in L_{n,s}^2(D,\Xi )$. Also, by (ii),

$$\begin{array}{cc}<{\overline{\partial }}^{\divideontimes }{N}_{n,s}\psi ,{\overline{\partial }}^{\divideontimes }{N}_{n,s}\psi >+<\overline{\partial }{N}_{n,s}\psi ,\overline{\partial }{N}_{n,s}\psi >\hfill & =<(\overline{\partial }{\overline{\partial }}^{\divideontimes }+{\overline{\partial }}^{\divideontimes }\overline{\partial })N_{n,s}\psi ,N_{n,s}\psi >\hfill \\ & =<{\square }_{n,s}{N}_{n,s}\psi ,N_{n,s}\psi >\hfill \\ & \le \parallel \psi \parallel \parallel N_{n,s}\psi \parallel \hfill \\ & \le c_0{\parallel \psi \parallel }^2.\hfill \end{array}$$

Then

$$\parallel {\overline{\partial }}^{\divideontimes }{N}_{n,s}{\psi \parallel }^2\le c_0{\parallel \psi \parallel }^2,$$

and

$$\parallel \overline{\partial }{N}_{n,s}{\psi \parallel }^2\le c_0{\parallel \psi \parallel }^2.$$

Now, we show that ${\overline{\partial }}^{\divideontimes }{N}_{n,s}=N_{n,s}{\overline{\partial }}^{\divideontimes }$ on Dom$({\overline{\partial }}^{\divideontimes },\Xi )$. Using (ii), ${\overline{\partial }}^{\divideontimes }u={\overline{\partial }}^{\divideontimes }\overline{\partial }{\overline{\partial }}^{\divideontimes }{N}_{n,s}u.$ Then,

$$N_{n,s}{\overline{\partial }}^{\divideontimes }u=N_{n,s}{\overline{\partial }}^{\divideontimes }\overline{\partial }{\overline{\partial }}^{\divideontimes }{N}_{n,s}u=N_{n,s}({\overline{\partial }}^{\divideontimes }\overline{\partial }+\overline{\partial }{\overline{\partial }}^{\divideontimes }){\overline{\partial }}^{\divideontimes }{N}_{n,s}u={\overline{\partial }}^{\divideontimes }{N}_{n,s}u.$$

Similarly, one can prove $\overline{\partial }{N}_{n,s}=N_{n,s}\overline{\partial }$ on Dom$(\overline{\partial },\Xi )$. From (ii),

$$\psi =\overline{\partial }{\overline{\partial }}^{\divideontimes }{N}_{n,s}\psi +\overline{\partial }{\overline{\partial }}^{\divideontimes }\overline{\partial }{N}_{n,s}\psi .$$

Thus, $\overline{\partial }\psi =0$ implies $\overline{\partial }{\overline{\partial }}^{\divideontimes }\overline{\partial }{N}_{n,s}\psi =0$ and

$$<\overline{\partial }{\overline{\partial }}^{\divideontimes }\overline{\partial }{N}_{n,s}\psi ,\overline{\partial }{N}_{n,s}\psi >={\parallel {\overline{\partial }}^{\divideontimes }\overline{\partial }{N}_{n,s}\psi \parallel }^2=0.$$

Since $\overline{\partial }{N}_{n,s}\psi \in \mathrm{Dom}\left({\overline{\partial }}^{\divideontimes }\right).$ Thus, $\psi =\overline{\partial }{\overline{\partial }}^{\divideontimes }{N}_{n,s}\psi$ and $u={\overline{\partial }}^{\divideontimes }{N}_{n,s}\psi$ is the solution which is unique and orthogonal to $ker(\overline{\partial },\Xi )$. □

Corollary 1.

Let X, D and Ξ be the same as Theorem 2. Then, for all $\psi \in L_{n,s}^2(D,\Xi )$ that satisfies $\overline{\partial }\psi =0$, the canonical solution $u={\overline{\partial }}^{\divideontimes }{N}_{n,s}\psi$ satisfies the estimate

$${\parallel u\parallel }^2\le C{\parallel \psi \parallel }^2.$$

Proof. 

From (iv), one obtains $\overline{\partial }{N}_{n,s}\psi =N_{(n,s+1)}\overline{\partial }\psi =0$. Since

$$\parallel N_{n,s}\psi \parallel \le c_0\parallel \psi \parallel .$$

Thus,

$$\begin{array}{cc}{\parallel u\parallel }^2\hfill & =<{\overline{\partial }}^{\divideontimes }{N}_{n,s}\psi ,{\overline{\partial }}^{\divideontimes }{N}_{n,s}\psi >\hfill \\ & =<\overline{\partial }{\overline{\partial }}^{\divideontimes }{N}_{n,s}\psi ,N_{n,s}\psi >\hfill \\ & =<(\overline{\partial }{\overline{\partial }}^{\divideontimes }+{\overline{\partial }}^{\divideontimes }\overline{\partial })N_{n,s}\psi ,N_{n,s}\psi >\hfill \\ & =<\psi ,N_{n,s}\psi >\hfill \\ & \le \parallel \psi \parallel \parallel N_{n,s}\psi \parallel \hfill \\ & \le c{\parallel \psi \parallel }^2.\hfill \end{array}$$

Thus, the proof follows. □

Let ${\square }_{n,0}={\overline{\partial }}^{\divideontimes }\overline{\partial }\mathrm{on}{L}_{n,0}^2(D,\Xi ).$ Set

$${\mathcal{H}}_{n,0}(\mathsf{\Omega },\Xi )=ker({\square }_{n,0},E)=\{\psi \in L_{n,0}^2(D,\Xi )|\overline{\partial }\psi =0\}.$$

Since $\overline{\partial }\psi =0$, then ${\mathcal{H}}_{n,0}(D,\Xi )$ is a closed subspace of $L_{n,0}^2(D,\Xi )$. Let

$$\mathcal{P}:L_{n,0}^2(D,\Xi )⟶{\mathcal{H}}_{n,0}(D,\Xi ),$$

be the Bergman projection operator.

Lemma 2 ([16]).

Let X, D and Ξ be the same as Theorem 2. Then,

$$N_{n,0}:L_{n,0}^2(D,\Xi )⟶L_{n,0}^2(D,\Xi ),$$

satisfies

(i) $\mathit{Rang}(N_{n,0},\Xi )\subset \mathit{Dom}({\square }_{n,0},\Xi )$, ${\square }_{n,0}{N}_{n,0}=N_{n,0}{\square }_{n,0}=I-{\mathcal{P}}_{n,0}$.

(ii) ∀$\psi \in L_{n,0}^2(D,\Xi )$; one obtains $\psi ={\overline{\partial }}^{\divideontimes }\overline{\partial }{N}_{n,0}\psi \oplus {\mathcal{P}}_{n,0}\psi .$

(iii) $N_{n,1}\overline{\partial }=\overline{\partial }{N}_{n,0}$ on Dom$(\overline{\partial },\Xi )$, ${\overline{\partial }}^{\divideontimes }{N}_{n,1}=N_{n,0}{\overline{\partial }}^{\divideontimes }$ on Dom$({\overline{\partial }}^{\divideontimes },\Xi )$.

(iv) $N_{n,0}\psi ={\overline{\partial }}^{\divideontimes }{N}_{n,1}^2\overline{\partial }\psi$ if $\psi \in \mathit{Dom}(\overline{\partial },\Xi )$.

(v) ∀$\psi \in L_{n,0}^2(D,\Xi )$,

$$\parallel N_{n,0}\psi \parallel \le C\parallel \psi \parallel ,$$

$$\parallel \overline{\partial }{N}_{n,0}\psi \parallel \le \sqrt{C}\parallel \psi \parallel .$$

Proof. 

Let $\psi \in \mathrm{Dom}({\square }_{n,0},\Xi )\cap {\left({\mathcal{H}}_{n,0}\left(E\right)\right)}^{\perp }$. Since $\mathrm{Rang}(\overline{\partial },\Xi )$ is closed in every degree, $\mathrm{Rang}({\overline{\partial }}^{\divideontimes },\Xi )$ is closed. Thus, $\psi \perp ker(\overline{\partial },\Xi )$ and $\psi \in \mathrm{Rang}({\overline{\partial }}^{\divideontimes },\Xi )$. Let $\psi =\overline{\partial }u$; then, $\psi \in L_{n,1}^2(D,E)$ since $u\in \mathrm{Dom}({\square }_{n,0},\Xi )$. Using (v) in Theorem 5, $v\equiv {\overline{\partial }}^{\divideontimes }{N}_{n,0}\psi$ is the solution of $\overline{\partial }v=\psi$, which is unique and $v\perp ker(\overline{\partial },\Xi )$. Thus, $v=u$. By using Equation (36), one obtains

$${\parallel u\parallel }^2\le {c\parallel \psi \parallel }^2=c\parallel \overline{\partial }{u\parallel }^2=c<{\square }_{n,0}u,u>\le c\parallel {\square }_{n,0}u\parallel \parallel u\parallel .$$

Thus, ${\square }_{n,0}$ is bounded below on $\mathrm{Dom}({\square }_{n,0},\Xi )\cap {\left({\mathcal{H}}_{n,0}\left(E\right)\right)}^{\perp }$ and ${\square }_{n,0}$ has a closed range and (i) and (ii) is proved. Then, from the strong Hodge decomposition,

$$L_{(n,0}^2(\mathsf{\Omega },\Xi )=\mathrm{Rang}({\square }_{n,0},E)\oplus {\mathcal{H}}_{n,0}(\mathsf{\Omega },E)={\overline{\partial }}^{\divideontimes }\overline{\partial }\left(\mathrm{Dom}({\square }_{n,0},E)\right)\oplus {\mathcal{H}}_{n,0}(\mathsf{\Omega },\Xi ),$$

for all $\psi \in \mathrm{Rang}({\square }_{n,0},\Xi )$, there is a unique $N_{n,0}\psi \perp {\mathcal{H}}_{n,0}(D,\Xi )$ that satisfies ${\square }_{n,0}{N}_{n,0}\psi =\psi$. Extending $N_{n,0}$ to $L_{n,0}^2(D,\Xi )$ by requiring $N_{n,0}{\mathcal{P}}_{n,0}=0$, $N_{n,0}$ satisfies (i) and (ii). (iii) is proved as before. If $\psi \in \mathrm{Dom}(\overline{\partial },\Xi )$,

$$N_{n,0}u=(I-{\mathcal{P}}_{n,0})N_{n,0}u=N_{n,0}\left({\overline{\partial }}^{\divideontimes }\overline{\partial }\right)N_{n,0}u={\overline{\partial }}^{\divideontimes }{N}_{n,0}^2\overline{\partial }u.$$

Thus, (iv) holds on $\mathrm{Dom}(\overline{\partial },\Xi )$. From (iii) in Theorem 5,

$$\parallel N_{n,1}\psi \parallel \le C\parallel \psi \parallel .$$

for all $\psi \in {\mathcal{C}}_{n,0}^{\infty }(\overline{\mathsf{\Omega }},E)$,

$$\parallel N_{n,0}{\psi \parallel }^2=<\overline{\partial }{\overline{\partial }}^{\divideontimes }{N}_{n,1}^2\overline{\partial }\psi ,N_{n,1}^2\overline{\partial }\psi >=<N_{n,1}\overline{\partial }\psi ,N_{n,1}^2\overline{\partial }\psi >=\parallel N_{n,1}\overline{\partial }\psi \parallel \parallel N_{n,1}^2\overline{\partial }\psi \parallel \le C\parallel N_{n,1}\overline{\partial }{\psi \parallel }^2.$$

(38)

On the other hand, one obtains

$$\parallel N_{n,1}\overline{\partial }{\psi \parallel }^2=<N_{n,1}\overline{\partial }\psi ,N_{n,1}\overline{\partial }\psi >=<N_{n,1}^2\overline{\partial }\psi ,\overline{\partial }\psi >=<{\overline{\partial }}^{\divideontimes }{N}_{n,1}^2\overline{\partial }\psi ,\psi >\le \parallel N_{n,0}\psi \parallel \parallel \psi \parallel .$$

(39)

Combining Equation (38) and Equation (39), one obtains

$$\parallel N_{n,0}\psi \parallel \le C\parallel \psi \parallel ,$$

and

$$\begin{array}{cc}\parallel \overline{\partial }{N}_{n,0}{\psi \parallel }^2\hfill & =<{\overline{\partial }}^{\divideontimes }\overline{\partial }{N}_{n,0}\psi ,N_{n,0}^2\psi >\hfill \\ & =<(I-{\mathcal{P}}_{n,0})\psi ,N_{n,0}\psi >\hfill \\ & \le \parallel \psi \parallel \parallel N_{n,0}\psi \parallel \hfill \\ & \le C{\parallel \psi \parallel }^2.\hfill \end{array}$$

Then, the proof follows. □

6. Sobolev Estimates

As in Cao–Shaw–Wang [3,35], one prove the following results:

Proposition 5.

$$\begin{array}{c}{\overline{\partial }}^{\divideontimes }\psi =-{\#}^{-1}\divideontimes \overline{\partial }\divideontimes \#\psi ,\hfill \\ {\overline{\partial }}_{\mathtt{m}}^{\divideontimes }\psi ={\overline{\partial }}^{\divideontimes }\psi +\mathtt{m}\divideontimes \left(\frac{\partial \zeta }{\zeta }\wedge \divideontimes \psi \right).\hfill \end{array}$$

(40)

Proof. 

In fact, for $\psi \in {\mathcal{C}}_{r,s-1}^{\infty }(D,\Xi )$ and $\psi \in {\mathcal{C}}_{r,s-1}^{\infty }(\overline{D},\Xi )$, one obtains

$$\begin{array}{cc}\overline{\partial }({}^t\psi \wedge \divideontimes \#\psi )\hfill & ={}^t\overline{\partial }\psi \wedge \divideontimes \#\psi +{(-1)}^{r+s}{}^t\psi \wedge \overline{\partial }\divideontimes \#\psi =\hfill \\ & {}^t\overline{\partial }\psi \wedge \divideontimes \#\psi +{}^t\psi \wedge \divideontimes \divideontimes \overline{\partial }\divideontimes \#\psi =\hfill \\ & {}^t\overline{\partial }\psi \wedge \divideontimes \#\psi +{}^t\psi \wedge \divideontimes \#({\#}^{-1}\divideontimes \overline{\partial }\divideontimes \#)\psi .\hfill \end{array}$$

Since ${}^t\psi \wedge \divideontimes \#\psi$ is of type $(n,n-1)$, then

$$\begin{array}{c}\partial ({}^t\psi \wedge \divideontimes \#\psi )=0,\hfill \\ \overline{\partial }({}^t\psi \wedge \divideontimes \#\psi )=d({}^t\psi \wedge \divideontimes \#\psi ).\hfill \end{array}$$

Then, by Stokes theorem, one obtains

$$\begin{array}{cc}0={\int }_{\mathsf{\Omega }}d({}^t\psi \wedge \divideontimes \#\psi )\hfill & ={\int }_{\mathsf{\Omega }}\overline{\partial }({}^t\psi \wedge \divideontimes \#\psi )\hfill \\ & ={\int }_{\mathsf{\Omega }}{}^t\overline{\partial }\psi \wedge \divideontimes \#\psi +{\int }_{\mathsf{\Omega }}{}^t\psi \wedge \divideontimes \#({\#}^{-1}\divideontimes \overline{\partial }\divideontimes \#)\psi ,\hfill \end{array}$$

i.e.,

$${\int }_D{}^t\overline{\partial }\psi \wedge \divideontimes \#\psi =-{\int }_D{}^t\psi \wedge \divideontimes \#({\#}^{-1}\divideontimes \overline{\partial }\divideontimes \#)\psi ,$$

i.e.,

$${\int }_D{}^t\overline{\partial }\psi \wedge \divideontimes \#\psi ={\int }_D{}^t\psi \wedge \divideontimes \#{\overline{\partial }}^{\divideontimes }\psi .$$

Therefore,

$$\psi \psi =-{\#}^{-1}\divideontimes \overline{\partial }\divideontimes \#\psi .$$

Then,

$$\begin{array}{cc}{\overline{\partial }}_{\mathtt{m}}^{\divideontimes }\psi \hfill & =-{\zeta }^{\mathtt{m}}{\#}^{-1}\divideontimes \overline{\partial }\divideontimes {\zeta }^{-\mathtt{m}}\#\psi \hfill \\ & =-{\zeta }^{\mathtt{m}}{\zeta }^{-\mathtt{m}}{\#}^{-1}\divideontimes \overline{\partial }\divideontimes \#\psi -{\zeta }^{\mathtt{m}}{\#}^{-1}\divideontimes \left(-\mathtt{m}{\zeta }^{-\mathtt{m}-1}\overline{\partial }\zeta \wedge \divideontimes \#\psi \right)\hfill \\ & ={\overline{\partial }}^{\divideontimes }\psi +\mathtt{m}{\#}^{-1}\divideontimes \left(\frac{\overline{\partial }\zeta }{\zeta }\wedge \divideontimes \#\psi \right).\hfill \end{array}$$

But,

$$\frac{\overline{\partial }\zeta }{\zeta }\wedge \divideontimes \#\psi =\frac{\overline{\partial }\zeta }{\zeta }\wedge \divideontimes \overline{b}\overline{\psi }=\overline{b}\frac{\overline{\partial }\zeta }{\zeta }\wedge \divideontimes \overline{\psi },$$

and

$$\#\left(\frac{\partial \zeta }{\zeta }\wedge \divideontimes \psi \right)=\overline{b}\overline{\frac{\partial \zeta }{\zeta }\wedge \divideontimes \psi }=\overline{b}\frac{\overline{\partial }\zeta }{\zeta }\wedge \divideontimes \overline{\psi }.$$

Then,

$$\frac{\overline{\partial }\zeta }{\zeta }\wedge \divideontimes \#\psi =\#\left(\frac{\partial \zeta }{\zeta }\wedge \divideontimes \psi \right),$$

i.e.,

$${\#}^{-1}\divideontimes \frac{\overline{\partial }\zeta }{\zeta }\wedge \divideontimes \#\psi =\divideontimes \left(\frac{\partial \zeta }{\zeta }\wedge \divideontimes \psi \right).$$

Then,

$${\overline{\partial }}_{\mathtt{m}}^{\divideontimes }\psi ={\overline{\partial }}^{\divideontimes }\psi +\mathtt{m}\divideontimes \left(\frac{\partial \zeta }{\zeta }\wedge \divideontimes \psi \right),$$

where ${\psi }_0=\psi$. □

Theorem 5.

Let X, D and Ξ be the same as Theorem 2. Let $\psi \in L_{n,s}^2(D,\Xi )\cap \mathrm{Dom}(\overline{\partial },E)\cap \mathrm{Dom}({\overline{\partial }}^{\divideontimes },\Xi )$, $1\le s\le n$. Then,

$$\parallel \overline{\partial }{\psi \parallel }_{W_{n,s}^{\mathtt{m}}(D,\Xi )}^2+{\parallel {\overline{\partial }}^{\divideontimes }\psi \parallel }_{W_{n,s}^{\mathtt{m}}(D,\Xi )}^2\ge C_{\mathtt{m}}\left({\parallel \partial \varphi \wedge \divideontimes \psi \parallel }_{W_{n,s}^{\mathtt{m}}(D,\Xi )}^2+\parallel \overline{\nabla }{\psi \parallel }_{W_{n,s}^{\mathtt{m}}(D,\Xi )}^2+{\int }_D(-\hslash ){\left|\psi \right|}^{2}dv\right),$$

(41)

where $C_{\mathtt{m}}>0$ is an independent constant of ψ.

Proof. 

As Lemma 1, one obtains

$$\begin{gathered} \partial \overline{\partial }\hslash =\mathtt{m}{\zeta }^{\mathtt{m}}\left((1-\mathtt{m})\frac{\partial \zeta \wedge \overline{\partial }\zeta }{{\zeta }^2}-\frac{\partial \overline{\partial }\zeta }{\zeta }\right), \\ \partial \overline{\partial }(-log{\zeta }^{\mathtt{m}})=\mathtt{m}\left(\frac{\partial \zeta \wedge \overline{\partial }\zeta }{{\zeta }^2}-\frac{\partial \overline{\partial }\zeta }{\zeta }\right). \end{gathered}$$

Then

$${\zeta }^{\mathtt{m}}\partial \overline{\partial }(-log{\zeta }^{\mathtt{m}})=\partial \overline{\partial }\hslash +{\mathtt{m}}^2\left(\frac{\partial \zeta \wedge \overline{\partial }\zeta }{{\zeta }^2}\right).$$

Therefore, for $\psi \in {\mathcal{C}}_{n,s}^1(\overline{D},E)\cap \mathrm{Dom}({\overline{\partial }}^{\divideontimes },\Xi )$, and by using Equation (18), one obtains

$$\parallel \overline{\partial }{\psi \parallel }_{W_{n,s}^{\mathtt{m}}(D,\Xi )}^2+\parallel {\overline{\partial }}_{\mathtt{m}}^{\divideontimes }{\psi \parallel }_{W_{n,s}^{\mathtt{m}}(D,\Xi )}^2={\parallel \overline{\nabla }\psi \parallel }_{W_{n,s}^{\mathtt{m}}(D,\Xi )}^2+<{\Theta }_{\mathtt{m}}\psi ,\psi {>}_{\mathtt{m}}+<(\partial \overline{\partial }\hslash )\psi ,\psi >+{\mathtt{m}}^2<\left(\frac{\partial \zeta \wedge \overline{\partial }\zeta }{{\zeta }^2}\right)\psi ,\psi >.$$

(42)

Also, by using Equation (40), one obtains

$$\begin{array}{cc}\parallel {\overline{\partial }}_{\mathtt{m}}^{\divideontimes }{\psi \parallel }_{\mathtt{m}}^2\hfill & =\parallel {\overline{\partial }}^{\divideontimes }{\psi \parallel }_{\mathtt{m}}^2+2\mathtt{Re}<{\overline{\partial }}^{\divideontimes }\psi ,\mathtt{m}\divideontimes \left(\frac{\partial \zeta }{\zeta }\wedge \divideontimes \psi \right){>}_{\mathtt{m}}+\parallel \mathtt{m}\divideontimes \left(\frac{\partial \zeta }{\zeta }\wedge \divideontimes \psi \right){\parallel }_{\mathtt{m}}^2,\hfill \\ & =\parallel {\overline{\partial }}^{\divideontimes }{\psi \parallel }_{\mathtt{m}}^2+2\mathtt{Re}<{\overline{\partial }}^{\divideontimes }\psi ,\mathtt{m}\divideontimes \left(\frac{\partial \zeta }{\zeta }\wedge \divideontimes \psi \right){>}_{\mathtt{m}}+{\mathtt{m}}^2{\parallel \frac{\partial \zeta }{\zeta }\wedge \divideontimes \psi \parallel }_{\mathtt{m}}^2,\hfill \end{array}$$

(43)

and since for all $\kappa >0$,

$$2\mathrm{Re}<{\overline{\partial }}^{\divideontimes }\psi ,\mathtt{m}\divideontimes \left(\frac{\partial \zeta }{\zeta }\wedge \divideontimes \psi \right){>}_{\mathtt{m}}|\le \frac{\mathtt{m}}{\kappa }{\int }_{\mathsf{\Omega }}(-\hslash )|{\overline{\partial }}^{\divideontimes }{\psi |}^{2}dv+\kappa \mathtt{m}{\int }_D(-\hslash ){|\mathtt{m}\divideontimes \left(\frac{\partial \zeta }{\zeta }\wedge \divideontimes \psi \right)|}^{2}dv,$$

(44)

and since

$$<(\partial \overline{\partial }\hslash )\psi ,\psi >\ge C_0\left({\int }_D{(-\hslash )\left|\psi \right|}^{2}dv+{\int }_D(-\hslash ){|\frac{\partial \zeta }{\zeta }\wedge \divideontimes \psi |}^{2}dv\right).$$

(45)

Then, by using Equations (43)–(45), the identity (42) becomes

$$\parallel \overline{\partial }{\psi \parallel }_{W_{n,s}^{\mathtt{m}}(D,\Xi )}^2+{\parallel {\overline{\partial }}^{\divideontimes }\psi \parallel }_{W_{n,s}^{\mathtt{m}}(D,\Xi )}^2\ge C_{\mathtt{m}}\left({\parallel \partial \varphi \wedge \divideontimes \psi \parallel }_{W_{n,s}^{\mathtt{m}}(D,\Xi )}^2+\parallel \overline{\nabla }{\psi \parallel }_{W_{n,s}^{\mathtt{m}}(D,\Xi )}^2+{\int }_D(-\hslash ){\left|\psi \right|}^{2}dv\right).$$

Then the proof follows from the density of ${\mathcal{C}}_{n,s}^1(\overline{D},E)\cap \mathrm{Dom}({\overline{\partial }}^{\divideontimes },\Xi )$ in $\mathrm{Dom}(\overline{\partial },\Xi )\cap$$\mathrm{Dom}({\overline{\partial }}^{\divideontimes },E)$ in the sense of ${\left({\parallel \psi \parallel }_{W_{n,s}^{\mathtt{m}}(D,\Xi )}^2+\parallel \overline{\partial }{\psi \parallel }_{W_{n,s}^{\mathtt{m}}(D,\Xi )}^2+{\parallel {\overline{\partial }}^{\divideontimes }\psi \parallel }_{W_{n,s}^{\mathtt{m}}(D,\Xi )}^2\right)}^2.$ □

Corollary 2.

Let X, D and Ξ be the same as Theorem 2. Then,

$$\begin{gathered} \parallel \overline{\partial }{N}_{n,s}{\psi \parallel }_{W_{n,s}^{\mathtt{m}}(D,\Xi )}\le {C\parallel \psi \parallel }_{W_{n,s}^{\mathtt{m}}(D,\Xi )},\psi \in ker({\overline{\partial }}^{\divideontimes },E),0\le s\le n-1. \\ \parallel {\overline{\partial }}^{\divideontimes }{N}_{n,s}{\psi \parallel }_{W_{n,s}^{\mathtt{m}}(D,\Xi )}\le C{\parallel \psi \parallel }_{W_{n,s}^{\mathtt{m}}(D,\Xi )},\psi \in ker(\overline{\partial },\Xi ),2\le s\le n. \end{gathered}$$

(46)

Proof. 

Since $\overline{\partial }{N}_{n,s}\psi \in \mathrm{Dom}(\overline{\partial },E)\cap \mathrm{Dom}({\overline{\partial }}^{\divideontimes },\Xi )$, $0\le s\le n-1$. Then, substituting $\overline{\partial }{N}_{n,s}\psi$ into Equation (41), for $\psi \in ker({\overline{\partial }}^{\divideontimes },\Xi )$, one obtains

$$\parallel \overline{\partial }\overline{\partial }{N}_{n,s}{\psi \parallel }_{W_{n,s}^{\mathtt{m}}(D,\Xi )}^2+{\parallel \psi \overline{\partial }{N}_{n,s}\psi \parallel }_{W_{n,s}^{\mathtt{m}}(D,\Xi )}^2\ge C_{\mathtt{m}}\left(\parallel \partial \varphi \wedge \divideontimes \overline{\partial }{N}_{n,s}{\psi \parallel }_{W_{n,s}^{\mathtt{m}}(D,\Xi )}^2+\parallel \overline{\nabla }\overline{\partial }{N}_{n,s}{\psi \parallel }_{W_{n,s}^{\mathtt{m}}(D,\Xi )}^2+{\int }_D(-\hslash ){\left|\overline{\partial }{N}_{n,s}\psi \right|}^{2}dv\right).$$

Then, by using the fact that $\psi =\left({\overline{\partial }}^{\divideontimes }\overline{\partial }+\overline{\partial }{\overline{\partial }}^{\divideontimes }\right)N\psi$, $\overline{\partial }\overline{\partial }=0$ and ${\overline{\partial }}^{\divideontimes }N\psi =N{\overline{\partial }}^{\divideontimes }\psi =0$, one obtains

$${\parallel \psi \parallel }_{W_{n,s}^{\mathtt{m}}(D,\Xi )}^2\ge C_{\mathtt{m}}{\int }_D(-\hslash ){\left|\overline{\partial }N\psi \right|}^{2}dv.$$

Then, the first equation of Equation (46) is proved by choosing $C=\frac{1}{C_{\mathtt{m}}}$. Similarly, for $2\le s\le n$, ${\overline{\partial }}^{\divideontimes }N\psi \in \mathrm{Dom}(\overline{\partial },E)\cap \mathrm{Dom}({\overline{\partial }}^{\divideontimes },\Xi )$. Then, substituting ${\overline{\partial }}^{\divideontimes }{N}_{n,s}\psi$ into Equation (41), for $\psi \in ker(\overline{\partial },\Xi )$, one obtains

$$\parallel \overline{\partial }{\overline{\partial }}^{\divideontimes }{N\psi \parallel }_{W_{n,s}^{\mathtt{m}}(D,\Xi )}^2+{\parallel \psi {\overline{\partial }}^{\divideontimes }N\psi \parallel }_{W_{n,s}^{\mathtt{m}}(D,\Xi )}^2\ge C_{\mathtt{m}}\left(\parallel \partial \varphi \wedge \divideontimes {\overline{\partial }}^{\divideontimes }{N\psi \parallel }_{W_{n,s}^{\mathtt{m}}(D,\Xi )}^2+\parallel \overline{\nabla }{\overline{\partial }}^{\divideontimes }{N\psi \parallel }_{W_{n,s}^{\mathtt{m}}(D,\Xi )}^2+{\int }_D(-\hslash ){\left|{\overline{\partial }}^{\divideontimes }N\psi \right|}^{2}dv\right).$$

Then, by using the fact that $\psi =\left({\overline{\partial }}^{\divideontimes }\overline{\partial }+\overline{\partial }{\overline{\partial }}^{\divideontimes }\right)N\psi$, ${\overline{\partial }}^{\divideontimes }{\overline{\partial }}^{\divideontimes }=0$ and $\overline{\partial }N\psi =N\overline{\partial }\psi =0$, one obtains

$${\parallel \psi \parallel }_{W_{n,s}^{\mathtt{m}}(D,\Xi )}^2\ge C_{\mathtt{m}}{\int }_D(-\hslash ){\left|{\overline{\partial }}^{\divideontimes }N\psi \right|}^{2}dv.$$

Then, Equation (48) is proved by choosing $C=\frac{1}{C_{\mathtt{m}}}$. □

Theorem 6.

Let X, D and Ξ be the same as Theorem 2. Let $\psi \in L_{n,s}^2(D,{\zeta }^{-\mathtt{m}},\Xi )$, $1\le s\le n$, a $\overline{\partial }$-closed form. Then, for $0\le \mathtt{m}<{\mathtt{m}}_0$, ∃$\psi ={\overline{\partial }}_{\mathtt{m}}^{\divideontimes }N\psi \in L_{n,s-1}^2(D,{\zeta }^{-\mathtt{m}},\Xi )$ satisfies $\overline{\partial }\psi =\psi$ and

$${\int }_D{\left|\psi \right|}^2{\zeta }^{-\mathtt{m}}dv\le C{\int }_D{\left|\psi \right|}^2{\zeta }^{-\mathtt{m}}dv.$$

(47)

Proof. 

Let $\chi =\psi e^{\varphi }=\psi {\zeta }^{-\mathtt{m}}$, $\varphi =-\mathtt{m}log\zeta$. Then, $\chi$ is orthogonal to all $\overline{\partial }$-closed forms of $L_{n,s-1}^2(D,{\zeta }^{-\mathtt{m}},\Xi )$. Equation (33) gives

$${\int }_D{\left|\chi \right|}^2{\zeta }^{\mathtt{m}}dv\le C{\int }_D{\left|\overline{\partial }\chi \right|}^2{\zeta }^{\mathtt{m}}dv.$$

For $\varphi =-\mathtt{m}log\zeta$, one obtains

$$\overline{\partial }\chi =e^{\varphi }\overline{\partial }\psi +e^{\varphi }\overline{\partial }\varphi \wedge \psi ={\zeta }^{-\mathtt{m}}\overline{\partial }\psi +{\zeta }^{-\mathtt{m}}\overline{\partial }\varphi \wedge \psi .$$

Then,

$${\int }_D{\left|\psi \right|}^2{\zeta }^{-\mathtt{m}}dv={\int }_D{\left|\chi \right|}^2{\zeta }^{\mathtt{m}}dv\le C{\int }_D{\left|\overline{\partial }\chi \right|}^2{\zeta }^{\mathtt{m}}dv.$$

Then,

$$\begin{array}{cc}{\int }_D{\left|\psi \right|}^2{\zeta }^{-\mathtt{m}}dv\hfill & \le C{\int }_D{|\overline{\partial }\psi +\overline{\partial }\varphi \wedge \psi |}^2{\zeta }^{-\mathtt{m}}dv\hfill \\ & \le C\left(\left(1+\frac{1}{\tau }\right){\int }_D\left|\psi \right|{\zeta }^{-\mathtt{m}}dv+\left(1+\tau \right){\int }_D{|\overline{\partial }\varphi \wedge \psi |}^2{\zeta }^{-\mathtt{m}}dv\right),\hfill \end{array}$$

for every $\tau >0$. Since

$$|\overline{\partial }{\varphi \wedge \psi |}^2\le \mid \psi {\mid }^2|\overline{\partial }{\varphi |}^2\le t^2{\left|\psi \right|}^2,$$

by choosing $\tau$, which satisfies $(1+\tau )t^2<1,$ (i.e., $0<\tau <{\left(\frac{{\mathtt{m}}_0}{\mathtt{m}}\right)}^2-1$),

$${\int }_D{\left|\psi \right|}^2{\zeta }^{-\mathtt{m}}dv\le C\frac{\left(1+\frac{1}{\tau }\right)}{[1-(1+\tau )t^2]}{\int }_D{\left|\psi \right|}^2{\zeta }^{-\mathtt{m}}dv.$$

It follows that $\overline{\partial }\psi =\psi$ and

$${\int }_D{\left|\psi \right|}^2{\zeta }^{-\mathtt{m}}dv\le \tilde{C}{\int }_D{\left|\psi \right|}^2{\zeta }^{-\mathtt{m}}dv.$$

□

Theorem 7.

Let X, D and Ξ be the same as Theorem 2. The Bergman projection $\mathcal{P}:L_{n,s}^2(D,\Xi )⟶L_{n,s}^2(D,E)\cap ker(\overline{\partial },\Xi )$ is bounded from $W_{n,s}^{\mathtt{m}/2}(D,\Xi )$ to $W_{n,s}^{\mathtt{m}/2}(D,\Xi )$, where $0\le s\le n-1$.

Proof. 

From Lemma 2, $\mathcal{P}=I-{\overline{\partial }}^{\divideontimes }{N}_{r,s+1}\overline{\partial }$. Then, by using Equation (47), ${\overline{\partial }}^{\divideontimes }N$ is bounded on $ker(\overline{\partial },\Xi )$ with

$$\parallel {\overline{\partial }}^{\divideontimes }{N\psi \parallel }_{-\mathtt{m}}\le C{\parallel \psi \parallel }_{-\mathtt{m}},$$

(48)

for $\psi \in ker(\overline{\partial },\Xi )$, $1\le s\le n-1$. The Bergman projection with respect to the weighted space $L^2(D,{\zeta }^{\mathtt{m}},\Xi )$ is denoted by ${\mathcal{P}}_{\mathtt{m}}$. ∀ $\psi ,\phi \in L_{n,0}^2(D,\Xi )$ with $\overline{\partial }\psi =0$, and one obtains

$$<\mathcal{P}\phi ,\psi >=<\phi ,\psi >=<{\zeta }^{-\mathtt{m}}\phi ,\psi {>}_{\mathtt{m}}=<{\mathcal{P}}_{\mathtt{m}}{\zeta }^{-\mathtt{m}}\phi ,\psi {>}_{\mathtt{m}}=<{\zeta }^{\mathtt{m}}{\mathcal{P}}_{\mathtt{m}}{\zeta }^{-\mathtt{m}}\phi ,\psi >.$$

This implies that

$$\mathcal{P}={\mathcal{P}}^2=\mathcal{P}{\zeta }^{\mathtt{m}}{\mathcal{P}}_{\mathtt{m}}{\zeta }^{-\mathtt{m}}=(I-{\overline{\partial }}^{\divideontimes }N\overline{\partial }){\zeta }^{\mathtt{m}}{\mathcal{P}}_{\mathtt{m}}{\zeta }^{-\mathtt{m}}={\zeta }^{\mathtt{m}}{\mathcal{P}}_{\mathtt{m}}{\zeta }^{-\mathtt{m}}-{\overline{\partial }}^{\divideontimes }N(\overline{\partial }{\zeta }^{\mathtt{m}}\wedge {\mathcal{P}}_{\mathtt{m}}{\zeta }^{-\mathtt{m}}),$$

(49)

because $\overline{\partial }{\mathcal{P}}_{\mathtt{m}}=0$. ∀$\psi \in L^2(D,\Xi )$,

$$\parallel {\zeta }^{\mathtt{m}}{\mathcal{P}}_{\mathtt{m}}{\zeta }^{-\mathtt{m}}{\psi \parallel }_{-\mathtt{m}}^2\le \parallel {\mathcal{P}}_{\mathtt{m}}{\zeta }^{-\mathtt{m}}{\psi \parallel }_{W_{n,s}^{\mathtt{m}}(D,\Xi )}^2\le \parallel {\zeta }^{-\mathtt{m}}{\psi \parallel }_{W_{n,s}^{\mathtt{m}}(D,\Xi )}^2={\parallel \psi \parallel }_{-\mathtt{m}}^2.$$

(50)

With (46), one obtains

$$\begin{array}{cc}\parallel {\overline{\partial }}^{\divideontimes }N(\overline{\partial }{\zeta }^{\mathtt{m}}\wedge {\mathcal{P}}_{\mathtt{m}}{\zeta }^{-\mathtt{m}}\psi ){\parallel }_{-\mathtt{m}}^2\hfill & \le C\parallel \overline{\partial }{\zeta }^{\mathtt{m}}\wedge {\mathcal{P}}_{\mathtt{m}}{\zeta }^{-\mathtt{m}}{\psi \parallel }_{-\mathtt{m}}^2\hfill \\ & \le C\parallel {\zeta }^{\mathtt{m}/2}{\mathcal{P}}_{\mathtt{m}}{\zeta }^{-\mathtt{m}}{\psi \parallel }^2\hfill \\ & =C\parallel {\mathcal{P}}_{\mathtt{m}}{\zeta }^{-\mathtt{m}}{\psi \parallel }_{W_{n,s}^{\mathtt{m}}(D,\Xi )}^2\hfill \\ & \le C\parallel {\zeta }^{-\mathtt{m}}{\psi \parallel }_{W_{n,s}^{\mathtt{m}}(D,\Xi )}^2=\hfill \\ & C{\parallel \psi \parallel }_{-\mathtt{m}}^2.\hfill \end{array}$$

(51)

With Equations (49) to (51), one obtains

$${\parallel \mathcal{P}\psi \parallel }_{-\mathtt{m}}^2\le C{\parallel \psi \parallel }_{-\mathtt{m}}^2.$$

(52)

We note that $W^{\mathtt{m}/2}(D,\Xi )\subset L^2(D,{\zeta }^{-\mathtt{m}},\Xi )$. From Equation (52), one obtains

$$\parallel {\mathcal{P}}_{\mathtt{m}}{\psi \parallel }_{-\mathtt{m}}^2\le {C\parallel \psi \parallel }_{-\mathtt{m}}^2\le C_1{\parallel \psi \parallel }_{\mathtt{m}/2}^2.$$

(53)

Using Equation (52), one obtains that the Bergman projection satisfies

$${\parallel \mathcal{P}\psi \parallel }_{\mathtt{m}/2}\le C_2{\parallel \psi \parallel }_{\mathtt{m}/2}^2.$$

(54)

Then, the Theorem is proved. □

In the following, the Sobolev boundary regularity for N, $\overline{\partial }N$ and ${\overline{\partial }}^{\divideontimes }N$ is studied.

Theorem 8.

Let X, D and Ξ be the same as Theorem 2. Then, ∀$0<\mathtt{m}<{\mathtt{m}}_0$, N is bounded from $W_{n,s}^{\mathtt{m}/2}(D,\Xi )$ to $W_{n,s}^{\mathtt{m}/2}(D,\Xi )$ and $0\le s\le n-1$. Also, ∀$\psi \in W_{n,s}^{\mathtt{m}}(D,\Xi )$, and one obtains the following estimates:

$$\begin{gathered} {\parallel N\psi \parallel }_{W_{n,s}^{\mathtt{m}/2}(D,\Xi )}\le 2C^2{\parallel \psi \parallel }_{W_{n,s}^{\mathtt{m}/2}(D,\Xi )}^2, \\ \parallel \overline{\partial }{N\psi \parallel }_{W_{n,s}^{\mathtt{m}/2}(D,\Xi )}\le {C\parallel \psi \parallel }_{W_{n,s}^{\mathtt{m}/2}(D,\Xi )}^2, \\ \parallel {\overline{\partial }}^{\divideontimes }{N\psi \parallel }_{W_{n,s}^{\mathtt{m}/2}(D,\Xi )}\le C{\parallel \psi \parallel }_{W_{n,s}^{\mathtt{m}/2}(D,\Xi )}^2, \end{gathered}$$

(55)

where C depends only on $\mathtt{m}$.

Proof. 

Since $\mathcal{P}=I-{\overline{\partial }}^{\divideontimes }N\overline{\partial }$, then ${\overline{\partial }}^{\divideontimes }N\psi ={\overline{\partial }}^{\divideontimes }N\mathcal{P}\psi$. Let ${\mathcal{P}}^{\prime }={\overline{\partial }}^{\divideontimes }N\overline{\partial }$ be another projection operator into $ker({\overline{\partial }}^{\divideontimes },\Xi )$. Then, $\mathcal{P}=I-{\mathcal{P}}^{\prime }$. It follows that $\overline{\partial }N\psi =\overline{\partial }N{\mathcal{P}}^{\prime }\psi$. The self-adjoint property of $\mathcal{P}$ and ${\mathcal{P}}^{\prime }$ gives

$${\parallel \mathcal{P}\psi \parallel }_{W_{n,s}^{\mathtt{m}}(D,\Xi )}+\parallel {\mathcal{P}}^{\prime }{\psi \parallel }_{W_{n,s}^{\mathtt{m}}(D,\Xi )}\le C_3{\parallel \psi \parallel }_{W_{n,s}^{\mathtt{m}}(D,\Xi )}.$$

Thus, by using Equation (54), and for $s\ge 0$, one obtains

$$\parallel \overline{\partial }{N\psi \parallel }_{W_{n,s}^{\mathtt{m}}(D,\Xi )}=\parallel \overline{\partial }N{\mathcal{P}}^{\prime }{\psi \parallel }_{W_{n,s}^{\mathtt{m}}(D,\Xi )}\le C_4\parallel {\mathcal{P}}^{\prime }{\psi \parallel }_{W_{n,s}^{\mathtt{m}}(D,\Xi )}\le C_4{C}_3{\parallel \psi \parallel }_{W_{n,s}^{\mathtt{m}}(D,\Xi )},$$

(56)

and for $s\ge 2$, one obtains

$$\parallel {\overline{\partial }}^{\divideontimes }{N\psi \parallel }_{W_{n,s}^{\mathtt{m}}(D,\Xi )}=\parallel {\overline{\partial }}^{\divideontimes }{N\mathcal{P}\psi \parallel }_{W_{n,s}^{\mathtt{m}}(D,\Xi )}\le C_4{\parallel \mathcal{P}\psi \parallel }_{W_{n,s}^{\mathtt{m}}(D,\Xi )}\le C_4{C}_3{\parallel \psi \parallel }_{W_{n,s}^{\mathtt{m}}(D,\Xi )}.$$

Since for all $\psi \in ker(\overline{\partial },\Xi )$, one obtains

$${\overline{\partial }}^{\divideontimes }N\psi ={\overline{\partial }}_{\mathtt{m}}^{\divideontimes }{N}_{\mathtt{m}}\psi -{\mathcal{P}}_{\mathtt{m}}{\overline{\partial }}_{\mathtt{m}}^{\divideontimes }{N}_{\mathtt{m}}\psi .$$

Thus, for all $\psi \in L_{n,1}^2(D,\Xi )$, one obtains

$$\begin{array}{cc}\parallel {\overline{\partial }}^{\divideontimes }{N\psi \parallel }_{W_{n,s}^{\mathtt{m}}(D,\Xi )}\hfill & =\parallel {\overline{\partial }}^{\divideontimes }{N\mathcal{P}\psi \parallel }_{W_{n,s}^{\mathtt{m}}(D,\Xi )}=\parallel {\overline{\partial }}_{\mathtt{m}}^{\divideontimes }{N}_{\mathtt{m}}\mathcal{P}\psi -{\mathcal{P}}_{\mathtt{m}}{\overline{\partial }}_{\mathtt{m}}^{\divideontimes }{N}_{\mathtt{m}}{\mathcal{P}\psi \parallel }_{W_{n,s}^{\mathtt{m}}(D,\Xi )}\hfill \\ & \le C_5{\parallel \mathcal{P}\psi \parallel }_{W_{n,s}^{\mathtt{m}}(D,\Xi )}\le C_5{C}_3{\parallel \psi \parallel }_{W_{n,s}^{\mathtt{m}}(D,\Xi )}.\hfill \end{array}$$

(57)

Since ${\left(\overline{\partial }N\right)}^{\divideontimes }=N{\overline{\partial }}^{\divideontimes }={\overline{\partial }}^{\divideontimes }N$ and ${\left({\overline{\partial }}^{\divideontimes }N\right)}^{\divideontimes }=N\overline{\partial }=\overline{\partial }N$. Use Equations (56) and (57), and by choosing $C=max\{C_3{C}_4,C_3{C}_5\}$, the second and third inequality of Equation (55) follows. Since

$$N=\overline{\partial }{\overline{\partial }}^{\divideontimes }{N}^2+{\overline{\partial }}^{\divideontimes }\overline{\partial }{N}^2=\overline{\partial }N{\overline{\partial }}^{\divideontimes }N+{\overline{\partial }}^{\divideontimes }N\overline{\partial }N.$$

Equations (56) and (57) give

$${\parallel N\psi \parallel }_{\mathtt{m}/2\left(D\right)}\le 2C^2{\parallel \psi \parallel }_{\mathtt{m}/2\left(D\right)}^2.$$

□

Theorem 9.

Let X, D and Ξ be the same as Theorem 2. Then, ∀ $0<\mathtt{m}<{\mathtt{m}}_0$ and N is bounded from $W_{n,s}^{\mathtt{m}/2}(D,\Xi )$ to $W_{n,s}^{\mathtt{m}/2}(D,\Xi )$, where $0\le s\le n-1.$ Also, ∀ $\psi \in W_{n,s}^{\mathtt{m}/2}(D,\Xi )$, and one obtains the following estimates:

$$\begin{gathered} {\parallel N\psi \parallel }_{W_{n,s}^{-\mathtt{m}/2}(D,\Xi )}\le {C\parallel \psi \parallel }_{W_{n,s}^{-\mathtt{m}/2}(D,\Xi )}, \\ \parallel \overline{\partial }{N\psi \parallel }_{W_{n,s}^{-\mathtt{m}/2}(D,\Xi )}\le {C\parallel \psi \parallel }_{W_{n,s}^{-\mathtt{m}/2}(D,\Xi )}, \\ \parallel {\overline{\partial }}^{\divideontimes }{N\psi \parallel }_{W_{n,s}^{-\mathtt{m}/2}(D,\Xi )}\le C{\parallel \psi \parallel }_{W_{n,s}^{-\mathtt{m}/2}(D,\Xi )}. \end{gathered}$$

Proof. 

With respect to the $L^2$ norm, if ${\mathcal{S}}^{\divideontimes }$ is the adjoint map of $\mathcal{S}$, one obtains

$$\begin{array}{cc}{\parallel \mathcal{S}f\parallel }_{W_{n,s}^{\mathtt{m}/2}(D,\Xi )}\hfill & =\underset{g\in L^2}{sup}\frac{<\mathcal{S}f,g{>}_{L^2}}{{\parallel g\parallel }_{W_{n,s}^{\mathtt{m}/2}(D,\Xi )}}\hfill \\ & =\underset{g\in L^2}{sup}\frac{<f,{\mathcal{S}}^{\divideontimes }g{>}_{L^2}}{{\parallel g\parallel }_{W_{n,s}^{-\mathtt{m}/2}(D,\Xi )}}\hfill \\ & \le \parallel {\mathcal{S}}^{\divideontimes }{\parallel }_{W_{n,s}^{-\mathtt{m}/2}(D,\Xi )}{\parallel g\parallel }_{W_{n,s}^{\mathtt{m}/2}(D,\Xi )}.\hfill \end{array}$$

(58)

Then, by using Theorem 9 and Equation (58), the proof follows. □

7. Conclusions

Sobolev estimates for the $\overline{\partial }$ and the $\overline{\partial }$-Neumann operator on pseudoconvex manifolds are fundamental results in complex analysis. They allow us to understand the behavior of holomorphic functions and provide important tools for solving the $\overline{\partial }$ equation. These estimates have applications in various areas of mathematics, such as the study of complex geometry and partial differential equations on pseudoconvex manifolds.