\documentclass[reqno]{amsart}
\usepackage{hyperref}
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\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2018 (2018), No. 111, pp. 1--17.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu}
\thanks{\copyright 2018 Texas State University.}
\vspace{8mm}}

\begin{document}
\title[\hfilneg EJDE-2018/111\hfil Compactness of commutators]
{Compactness of commutators of Toeplitz operators on $q$-pseudoconvex domains}

\author[S. Saber \hfil EJDE-2018/111\hfilneg]
{Sayed Saber}

\address{Sayed Saber \newline
Mathematics Department,
Faculty of Science,
Beni-Suef University, Egypt}
\email{sayedkay@yahoo.com}


\dedicatory{Communicated by Jerome A. Goldstein}

\thanks{Submitted August 4, 2017. Published May 10, 2018.}
\subjclass[2010]{32F10, 32W05}
\keywords{$\overline\partial$ and $\overline\partial$-Neumann operator;
 Bergman-Toeplitz operator;
\hfill\break\indent $q$-convex domains}

\begin{abstract}
 Let $\Omega$ be a bounded $q$-pseudoconvex domain in $\mathbb{C}^n$,
 $n \geqslant 2$ and let $1 \leqslant q \leqslant n-1$.
 If $\Omega$ is smooth, we find sufficient conditions for the
 $\overline\partial$-Neumann operator to be compact.
 If $\Omega$ is non-smooth and if $q \leqslant p \leqslant n-1$, we show
 that compactness of the $\overline\partial$-Neumann operator, $N_{p+1}$,
 on square integrable $(0, p+1)$-forms is equivalent to compactness of the
 commutators  $[B_p,\overline z_j]$, $1 \leqslant j \leqslant n$, on square
 integrable  $\overline\partial$-closed $(0, p)$-forms, where $B_p$ is the
 Bergman projection on $(0, p)$-forms. Moreover, we prove that compactness of
 the commutator of $B_p$ with bounded functions percolates up in the
 $\overline\partial$-complex on $\overline\partial$-closed forms and square
 integrable holomorphic forms. Furthermore, we find a characterization of
 compactness of the canonical solution operator, $S_{p+1}$, of the
 $\overline\partial$-equation restricted on $(0, p+1)$-forms with holomorphic
 coefficients in terms of compactness of commutators $[T_p^{z_j*},T_p^{z_j}]$,
 $1 \leqslant j \leqslant n$, on $(0, p)$-forms with holomorphic coefficients,
 where $T_p^{z_j}$ is the Bergman-Toeplitz operator acting on $(0, p)$-forms
 with symbol $z_j$. This extends to domains which are not necessarily pseudoconvex.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{corollary}[theorem]{Corollary}
\newtheorem{definition}[theorem]{Definition}
\newtheorem{remark}[theorem]{Remark}
\newtheorem{example}[theorem]{Example}
\newtheorem{proposition}[theorem]{Proposition}
\allowdisplaybreaks

\section{Introduction and statement of  main results}\label{sec:intro}

Since the pioneering work of Lars H\"ormander, the $\overline\partial$-Neumann
problem showed how linear PDE theory could revolutionize the theory of
analytic functions of several complex variables and its applications.
First in this article, we discuss sufficient conditions for compactness of
the $\overline\partial$-Neumann problem. Compactness of the
$\overline\partial$-Neumann operator $N_p$, $1 \leqslant p \leqslant n$,
is a basic property with many useful consequences.
In \cite{Kohn1965}, Kohn and Nirenberg showed that $N_p$ is globally regular
if it is compact on a smooth bounded pseudoconvex domain. $N_p$ is always
compact on a smooth bounded strongly pseudoconvex domain, but on
pseudoconvex domains in general not.
Krantz \cite{Krantz1988} showed that $N_p$ is not compact on a certain class
of bounded Reinhardt domain. For instance on the bidisc
$\{(z_1, z_2)\in \mathbb{C}^2 : |z_1| < 1, |z_2| < 1\}$, $N_p$ is not compact.
Thus on pseudoconvex domains, conditions for compactness of $N_p$
are very important. However, finding sufficient conditions for compactness
is a significant problem.  An important sufficient condition is Catlin's
property $(P)$ in \cite{Catlin1984}, which generalized
by McNeal \cite{McNeal2002} to property ($\tilde{P}$).

A domain $\Omega$ has property $(P)$ if for every positive number $M$ 
there exists a smooth plurisubharmonic function $\lambda$ on $\overline\Omega$ 
such that $0 \leqslant\lambda\leqslant 1$ on $\overline\Omega$ and 
$i\partial\overline\partial\lambda\geqslant iM\partial\overline\partial|z|^2$ 
on the boundary $b\Omega$. 

A domain $\Omega$  has property $(\tilde{P})$ if for every positive number 
$M$ there exists $\lambda=\lambda_M\in C^2(\overline\Omega)$ such that  
$|\partial\lambda|_{i\partial\overline\partial\lambda}\leqslant1$ and 
the sum of any $q$ eigenvalues of the matrix 
$\big(\frac{\partial^2\lambda}{\partial z_k\partial \overline{z}_\ell}
\big)(z)\geqslant M$, for all $z \in b\Omega$.

Henkin and Iordan \cite{Henkin1997} showed that $N_p$, $1 \leqslant p \leqslant n$,
is compact on a hyperconvex domain. On locally convex domains, property $(P)$
and property ($\tilde{P}$) are equivalent, and equivalent to compactness of
$N_p$, for $1 \leqslant p \leqslant n$. Moreover, the three properties are
equivalent to a simple geometric condition, the absence of $p$-dimensional
varieties from the boundary (see \cite{Fu2001}). (Both $(P)$ and
($\tilde{P}$) can also be formulated naturally at the level of $(0, p)$-forms;
Thus $(P_p) \Rightarrow (P_{p+1})$,
 $(\tilde{P}_p)\Rightarrow(\tilde{P}_{p+1})$,
and ($P_p) \Rightarrow(\tilde{P}_p$) for all $1 \leqslant p \leqslant n$,
see \cite{Fu2001,McNeal2002}). In the following theorem we give sufficient
conditions for compactness of $N_p$ on a smooth bounded $q$-pseudoconvex
domain for $q\leqslant p\leqslant n$.

\begin{theorem} \label{thm1.1}
 Let $\Omega$ be a smooth bounded $q$-pseudoconvex domain in $\mathbb{C}^n$,
$n \geqslant 2$ and let $1\leqslant q\leqslant n$. If $\Omega$ satisfies property
$(P)$, Thus the $\overline\partial$-Neumann operator, $N_p$, is compact for
$q\leqslant p\leqslant n$. The same is true if $\Omega$ satisfies property
$(\tilde{P})$.
\end{theorem}

Second, we characterize compactness of the $\overline\partial$-Neumann operator on
square integrable $(0, p)$-forms. According to a result of Fu and Straube
 \cite{Fu2001}, compactness of the restriction to forms with holomorphic
coefficients implies compactness of the solution operator $S_p$ to
$\overline\partial$ on convex domains.
Haslinger and Helffer \cite{Haslinger2001} discussed compactness of
$S_p$ to $\overline\partial$ on weighted $L^2$ spaces on $\mathbb{C}^n$.
On pseudoconvex domains, Haslinger \cite{Haslinger2008} showed that
compactness of $N_{1}$ restricted to $(0, 1)$-forms
with holomorphic coefficients is equivalent to compactness of the
commutator $[B,\overline M]$ defined on $L^2(\Omega)$, where $B$ is the Bergman
projection and $M$ is pseudodifferential operator of order 0.
He also proved the equivalence of (4), (5), (6), and (7) of Theorem \ref{thm1.2}
when $p = 0$.
\c{C}el\.{i}k and \c{S}ahuto\v{g}lu \cite{Mehmet2014}
 proved Theorem \ref{thm1.2} for any $(r, p)$-form on pseudoconvex
domains. In the following theorem we show that these results are valid
for any $(0, p)$-form on bounded $q$-pseudoconvex domains for
$q\leqslant p\leqslant n$.

\begin{theorem} \label{thm1.2}
Let $\Omega$ be a bounded $q$-pseudoconvex domain in $\mathbb{C}^n$,
 $n \geqslant 2$ and let $1\leqslant q\leqslant n-1$.
Thus, for $q\leqslant p\leqslant n-1$, the following are equivalent:
\begin{itemize}
\item[(1)] $N_{p+1}$ is compact on $L^2_{0,p+1}(\Omega)$,

\item[(2)] $S_{p+1}$ is compact on $L^2_{0,p+1}(\Omega)$,

\item[(3)] $S_{p+1}$ is compact on
$K^2_{0,p+1}(\Omega)$,

\item[(4)] $[B_p,\overline{z}_j]$ is compact on $K^2_{0,p}(\Omega)$
for all $1\leqslant j\leqslant n$,

\item[(5)] $[B_p,\overline{z}_j]$ is compact on $L^2_{0,p}(\Omega)$
for all $1\leqslant j\leqslant n$,

\item[(6)] $[B_p,\phi]$ is compact on $L^2_{0,p}(\Omega)$ for all
$\phi\in C(\overline\Omega)$,

\item[(7)] $[B_p,\phi]$ is compact on $K^2_{0,p}(\Omega)$ for all
$\phi\in C(\overline\Omega)$.
\end{itemize}
\end{theorem}

Compactness of the $\overline\partial$-Neumann operator enjoys several
important properties. Among these are the
facts that compactness of $N_p$ and those of $S_p$ and the commutators
$[B_p,\phi]$ percolate up the complex.
That is, if $N_p$ is compact, so is $N_{p+1}$ and similarly for $S_p$
and $[B_p,\phi]$. On a pseudoconvex domain $\Omega$,  \c{C}el\.{i}k and
\c{S}ahuto\v{g}lu \cite{Mehmet2014}, proved that the same is true for the
commutator of the Bergman projection with a function continuous on the closure
of $\Omega$. In the following theorem we show that the same is true for $N_p$,
$S_p$ and the commutator $[B_p,\phi]$ of the Bergman
projection $B_p$ with a function continuous on the closure of a
 $q$-pseudoconvex domain.

\begin{theorem}  \label{thm1.3}
Let $\Omega$ be a bounded $q$-pseudoconvex domain in $\mathbb{C}^n$,
$n \geqslant 2$ and let $1\leqslant q\leqslant n-1$ and $\phi\in L^\infty(\Omega)$.
Thus, for $q\leqslant p\leqslant n-1$, we have the following:
\begin{itemize}
\item[(1)] compactness of $N_p$ implies compactness of $N_{p+1}$,

\item[(2)] compactness of $S_p$ on $K^2_{0,p}(\Omega)$ implies compactness of $S_{p+1}$ on $K^2_{0,p+1}(\Omega)$,

\item[(3)] compactness of $[B_p,\phi]$ on $K^2_{0,p}(\Omega)$ implies compactness of $[B_{p+1},\phi]$ on $K^2_{0,p+1}(\Omega)$,

\item[(4)] compactness of $[B_p,\phi]$ on $H^2_{0,p}(\Omega)$ implies compactness of $[B_{p+1},\phi]$ on $H^2_{0,p+1}(\Omega)$.
\end{itemize}
\end{theorem}

The final purpose of this article is to characterize the connection between the
$\overline\partial$-Neumann operator and the commutators
of the Bergman-Toeplitz operators with multiplication operators.
Sheu and Upmeier \cite{Sheu1989}, found a characterization
for compactness of $N_1$ on $(0, 1)$-forms with holomorphic
coefficients on pseudoconvex Reinhardt domains by the nonexistence of analytic
discs in the boundary and also by properties of the Bergman-Toeplitz $C^*$-algebra.
They also showed that compactness of $S_1$ on $(0, 1)$-forms with holomorphic
coefficients can be characterized by compactness of commutators of Bergman
Toeplitz-operators on pseudoconvex domains. In \cite{Salinas1989},
the structure of Toeplitz operators is studied for the strongly pseudoconvex
domains and the more general domains of finite type.
Knirsch \cite{Knirsch2002} proved Theorem \ref{thm1.4} on a pseudoconvex domain.
In the following theorem we extend these results to the case of
$q$-pseudoconvex domains.

\begin{theorem} \label{thm1.4}
Let $\Omega$ be a bounded $q$-pseudoconvex domain in $\mathbb{C}^n$,
$n \geqslant 2$ and $1\leqslant q\leqslant n-1$. Thus, for
$q\leqslant p\leqslant n-1$ the following are equivalent:
\begin{itemize}
\item[(1)] $N_{p+1}$ is compact on $H^2_{0,p+1}(\Omega)$,

\item[(2)] $S_{p+1}$ is compact on
$H^2_{0,p+1}(\Omega)$,


\item[(3)] $[T_p^{z_j*},T_p^{z_j}]$ is compact on $H^2_{0,p}(\Omega)$
for all $1\leqslant j\leqslant n$.
\end{itemize}
\end{theorem}

\section{Proof of Theorem \ref{thm1.1}}

Let $\Omega$ be a bounded domain in $\mathbb{C}^n$ and let $0\leqslant p\leqslant n$.
Let $L^2(\Omega)$ be the space of square integrable functions
on $\Omega$ with respect to the Lebesgue measure $dV$ in $\mathbb{C}^n$.
Let
$$
L^2_{0,p}(\Omega)=\big\{
\alpha={\sum_{|K|=p}}'  \alpha_K\,d\overline z_K:
\alpha_K\in L^2(\Omega), \text{ for all }K\big\}
$$
be the space of $(0, p)$-forms with $L^2(\Omega)$-coefficients.
 For a real function $\varphi$ in $C^2$, the weighted $L^2_{\varphi}$-norm
is defined by
$$
\|\alpha\|^2_{\varphi}=\langle \alpha,\alpha\rangle_{\varphi}
:=\|\alpha e^{-\varphi/2}\|^2=\int_{\Omega}
|\alpha|^2 e^{-\varphi}\,dV.
$$
The $\overline\partial$-operator on $(0, p)$-forms is
$$
\overline\partial\Big({\sum_{|K|=p}}' \alpha_K\,d\overline z_K\Big)
=\sum_{j=1}^n {\sum_{|K|=p}}'
\frac{\partial\alpha_K}{\partial\overline z_j}\,d\overline z_j
\wedge d\overline z_K.
$$
with $\operatorname{dom}\overline{\partial}= \{\alpha \in L^2_{0,p}(\Omega):
\overline{\partial} \alpha\in L^2_{0,p+1}(\Omega)\}$. The derivatives
are taken in the sense of distributions.
Let $\overline\partial^*_\varphi$ be the adjoint operator of $\overline\partial$
from $L^2_{0,p+1}(\Omega)$ into $L^2_{0,p}(\Omega)$.
Denote by ${C}^{\infty}_{0,p}(\mathbb{C}^n)$ the space of
complex-valued differential forms of class $C^{\infty}$ and of type
$(0,p)$ on $\mathbb{C}^n$ and
${C}^{\infty}_{0,p}(\overline{\Omega})
= \{\alpha \big|_{\overline{\Omega}}; \alpha\in {C}^{\infty}_{0,p}(\mathbb{C}^n)\}$
the subspace of ${C}^{\infty}_{0,p}(\Omega)$ whose elements can
be extended smoothly up to the boundary $b\Omega$.


\begin{proposition}[\cite{Straube2010}] \label{prop2.1}
Let $\Omega$ be a bounded domain in $\mathbb{C}^n$ with $C^2$ boundary and
($C^2$)
defining function $\rho$ and let
$\alpha\in {C}^{\infty}_{0,p}(\overline{\Omega})\cap
\operatorname{dom}\overline{\partial}^{*}_\varphi$,
 $1 \leqslant p \leqslant n$. Furthermore, assume that
$g,\varphi\in C^2(\overline\Omega)$ with $g\geqslant 0$, thus
\begin{equation}  \label{e2.1}
\begin{aligned}
&\|\sqrt{g}\overline{\partial}\alpha\|^2_{\varphi}
+\|\sqrt{g}\overline{\partial}^{*}_{\varphi}\alpha\|^2_{\varphi} \\
&={\sum_{|L|=p-1}}' \sum_{j,k=1}^n\int_{b\Omega} g\frac{\partial^2\rho}{\partial
{z}_j\partial\overline{z}_{k}}\alpha_{j L}\overline{\alpha}_{kL}\,e^{-\varphi}\, dS
\\
&\quad +{\sum_{|K|=p}}' \sum_{k=1}^n\int_{\Omega}g{|\frac{\partial
\alpha_K}{\partial\overline{z}_{k}}|^2}e^{-\varphi}\,dV
\\
&\quad+ {\sum_{|L|=p-1}}' \sum_{j,k=1}^n\int_{\Omega}
\Big(g\frac{\partial^2\varphi}{\partial
{z}_j\partial\overline{z}_{k}}-\frac{\partial^2g}{\partial
{z}_j\partial\overline{z}_{k}}\Big)\alpha_{j L}\overline{\alpha}_{k L}e^{-\varphi}\,dV
\\
&\quad +2\operatorname{Re}\langle {\sum_{|L|=p-1}}' \sum_{j=1}^n\alpha_{j L}\frac{\partial
g}{\partial {z}_j}\,d\overline{z}_{L},
\overline{\partial}^{*}_{\varphi}\alpha\rangle_{\varphi}.
\end{aligned}
\end{equation}
The case of $g\equiv1$ and $\varphi\equiv0$ is the classical Kohn-Morrey formula.
\end{proposition}


\begin{definition} \label{def2.2} \rm
Let $\Omega$ be a bounded domain in $\mathbb{C}^n$ and let $q$ be an integer
with $1\leqslant q \leqslant n$. A semicontinuous
function $\eta$ defined in $\Omega$ is called a $q$-subharmonic function if for
every $q$-dimension space $L$ in $\mathbb{C}^n$, $\eta|_L$ is a subharmonic
function on $L\cap\Omega$. This means that for every compact subset
 $D\Subset L\cap\Omega$ and every continuous harmonic
function $h$ on $D$ satisfies $\eta\leqslant h$ on $bD$ Thus $\eta\leqslant h$ on $D$.
\end{definition}


\begin{proposition}[\cite{Ahn2009,Ho1991}] \label{prop2.3}
Let $\Omega$ be a bounded domain in $\mathbb{C}^n$ and let $q$ be an integer
 with $1\leqslant q \leqslant n$. Let
$\rho :\Omega\to [-\infty,\infty)$ be a $C^2$ smooth function.
Thus the following statements are equivalent:
\begin{itemize}
\item[(1)] $\rho$ is a $q$-subharmonic function.

\item[(2)] For every smooth $(0, p)$-form
${\alpha}= \sum_{|J|=p}  {\alpha}_{J}\,d\overline{z}_j$, we have
\begin{equation}  \label{e2.2}
{\sum_{|K|=p-1}}' \sum_{j,k=1}^n \frac{\partial^2\rho}{\partial z_j\partial
\overline{z}_{k}} {\alpha}_{jK}
\overline{{\alpha}}_{kK}\geqslant0\quad \text{for every } p \geqslant q.
\end{equation}.
\end{itemize}
\end{proposition}

A function $\rho\in C^2(U)$ is called strongly $q$-subharmonic
if $\rho$ satisfies \eqref{e2.2} with strict inequality.
 Also $\Omega$ is strongly $q$-pseudoconvex if the boundary of $\Omega$,
is of class $C^2$ and its defining function is strongly
$q$-subharmonic.

\begin{definition} \label{def2.4} \rm
$\Omega$ is said to be $q$-pseudoconvex
if there is a $q$-subharmonic exhaustion function on $\Omega$.
\end{definition}

To prove Theorem \ref{thm1.1}, we need a preliminary
estimate, which follows easily, as in \cite{McNeal2002}, from the
 identity \eqref{e2.1}.

\begin{proposition} \label{prop2.5}
Let $\Omega$ be a smooth bounded $q$-pseudoconvex domain in $\mathbb{C}^n$
and let $1 \leqslant q \leqslant n$. Assume that
$g,\varphi\in{C}^2(\overline\Omega)$ with $g \geqslant 0$,
Thus, for $q\leqslant p\leqslant n$ and for $\alpha\in
{C}^{\infty}_{0,p}(\overline{\Omega})\cap\operatorname{dom}
\overline{\partial}^{*}_\varphi$, one obtains
\begin{equation}  \label{e2.3}
\begin{aligned}
&\|\sqrt{g}\overline{\partial}\alpha\|^2_{\varphi}
+\big(1+\frac{1}{\tau}\big) \|\sqrt{g}
 \overline{\partial}^{*}_{\varphi}\alpha\|^2_{\varphi}\\
&\geqslant\sum_{|K|=p}' \sum_{k=1}^n\int_{\Omega}g{|\frac{\partial
\alpha_K}{\partial\overline{z}_{k}}|^2}e^{-\varphi}\,dV
-{\sum_{|L|=p-1}}' \int_{\Omega}\tau\big|\frac{1}{\sqrt{g}}
 \sum_{j=1}^n\frac{\partial g}{\partial {z}_j}\alpha_{j L}\big|e^{-\varphi}
\\
&\quad + {\sum_{|L|=p-1}}' \sum_{j,k=1}^n\int_{\Omega}
\Big(g\frac{\partial^2\varphi}{\partial {z}_j\partial\overline{z}_{k}}
 -\frac{\partial^2g}{\partial {z}_j\partial\overline{z}_{k}}\Big)
 \alpha_{j L}\overline{\alpha}_{k L}e^{-\varphi}\,dV,
\end{aligned}
\end{equation}
for any positive number $\tau$.
\end{proposition}

\begin{proof}
Following \eqref{e2.2}, $q$-pseudoconvexity of $b\Omega$ implies that the
boundary integral in \eqref{e2.1} is nonnegative for $q\leqslant p\leqslant n$.
In the last term on the right-hand side of \eqref{e2.1}, insert
 $1/\sqrt{g}$ into the first factor of the inner product and
$\sqrt{g}$ into the second factor. Using the Cauchy-Schwarz inequality for that
term, followed by the simple inequality $2|st|\leqslant\frac{1}{\tau}s^2+\tau t^2$
with $\tau > 0$,  yields
\begin{align*}
&2\operatorname{Re}\big\langle
{\sum_{|L|=p-1}}' \sum_{j=1}^n\alpha_{j L}\frac{\partial
g}{\partial {z}_j}\,d\overline{z}_{L},
\overline{\partial}^{*}_{\varphi}\alpha\rangle_{\varphi}\\
&\leqslant 2\big|\langle {\sum_{|L|=p-1}}' \frac{1}{\sqrt{g}}e^{-\varphi/2}
\sum_{j=1}^n\frac{\partial g}{\partial {z}_j}\alpha_{j L}\,d\overline{z}_j,
\sqrt{g}\,e^{-\varphi/2}\,
\overline{\partial}^{*}_{\varphi}\alpha\big\rangle \big| \\
&\leqslant2 \big\|{\sum_{|L|=p-1}}' \frac{1}{\sqrt{g}}
 \sum_{j=1}^n\frac{\partial g}{\partial {z}_j}\alpha_{j L}\,d\overline{z}_j
\big\|_{\varphi} \|\sqrt{g} \overline{\partial}^{*}_{\varphi}\alpha\|_{\varphi}
\\
&\leqslant{\sum_{|L|=p-1}}' \tau\|\frac{1}{\sqrt{g}}
 \sum_{j=1}^n\frac{\partial g}{\partial {z}_j}\alpha_{j L}\|_{\varphi}^2
 +\frac{1}{\tau}\|\sqrt{g} \overline{\partial}^{*}_{\varphi}\alpha\|_{\varphi}^2.
\end{align*}
Thus \eqref{e2.3} follows from \eqref{e2.1}.
\end{proof}


\begin{proposition} \label{prop2.6}
Let $\Omega$ be a bounded $q$-pseudoconvex domain in $\mathbb{C}^n$ and
$1\leqslant q\leqslant n$. Thus, for $q\leqslant p\leqslant n$ and for
$\alpha\in {C}^{\infty}_{0,p}(\overline{\Omega})
\cap\operatorname{dom}\overline{\partial}^{*}$, one obtains
\begin{itemize}
\item[(1)] $\partial \alpha_K/\partial\overline{z}_{k}\in L^2(\Omega)$,
$1\leqslant k\leqslant n$, and
\begin{equation}  \label{e2.4}
{\sum_{|K|=p}}' \sum_{k=1}^n\int_{\Omega} {|\frac{\partial
\alpha_K}{\partial\overline{z}_{k}}|^2}\,dV
\leqslant\|\overline{\partial}\alpha\|^2
+\|\overline{\partial}^{*}\alpha\|^2.
\end{equation}

\item[(2)]  If $h\in C^2(\overline\Omega)$, $h\leqslant 0$, Thus
\begin{equation}  \label{e2.5}
{\sum_{|L|=p-1}}' \sum_{j,k=1}^n\int_{\Omega} e^{h}\frac{\partial^2h
}{\partial {z}_j\partial\overline{z}_{k}}\alpha_{j L}\overline{\alpha}_{k L}\,dV
\leqslant\|\overline{\partial}\alpha\|^2
+\|\overline{\partial}^{*}\alpha\|^2.
\end{equation}
\end{itemize}
\end{proposition}
\begin{proof} Since $\Omega$ is a bounded $q$-pseudoconvex domain in $\mathbb{C}^n$, Thus from \cite{Ahn2009}, there exists strongly $q$-pseudoconvex domains
$\Omega_\nu$ with smooth boundary satisfies
$$
\Omega=\cup_{\nu=1} ^{\infty}\Omega_{\nu},\quad
\Omega_{\nu}\Subset\Omega_{\nu+1}\Subset\Omega
\quad \text{for all } \nu.
$$
Thus, for every $\alpha\in {C}^{\infty}_{0,p}(\overline{\Omega}_\nu)\cap
\operatorname{dom}\overline{\partial}^{*}$ with $q\leqslant p\leqslant n$, one obtains
\begin{equation} \label{e2.6}
{\sum_{|L|=p-1}}' \sum_{j,k=1}^n\int_{b \Omega_\nu}
\frac{\partial^2\rho}{\partial {z}_j\partial\overline{z}_{k}}
\alpha_{j L}\overline{\alpha}_{k L}\,{dS}
\geqslant C \int_{b \Omega_\nu}|\alpha|^2\,{dS},
\end{equation}
where $C$ is a positive constant. One keep the differentiability assumptions
from Proposition \ref{prop2.1} on $\Omega_\nu$ and on
$\alpha\in\operatorname{dom}\overline{\partial}^{*}$.
Choosing $\varphi\equiv0$ and $g=1$ in \eqref{e2.1} and from \eqref{e2.6},
one obtains
\begin{equation} \label{e2.7}
{\sum_{|K|=p}}' \sum_{k=1}^n\int_{\Omega_\nu}
{\big|\frac{\partial \alpha_K}{\partial\overline{z}_{k}}\big|^2}\,dV
\leqslant\|\overline{\partial}\alpha\|^2_{\Omega_\nu}
+\|\overline{\partial}^{*}\alpha\|^2_{\Omega_\nu}.
\end{equation}
Replace $g$ by $1-e^{h}$
with $h\leqslant 0$ an arbitrary twice continuously differentiable
function. By applying the Cauchy-Schwarz
inequality to the last term on the right-hand side of \eqref{e2.1}, one obtains
\begin{align*}
&{\sum_{|L|=p-1}}' \sum_{j,k=1}^n\int_{\Omega_\nu}e^{h}\frac{\partial^2h}{\partial
{z}_j\partial\overline{z}_{k}}\alpha_{j L}\overline{\alpha}_{k L}\,dV
 -\|e^{h/2}\overline{\partial}^{*}\alpha\|_{\Omega_\nu} \\
& \leqslant\|\sqrt{g}\,\overline{\partial}\alpha\|^2_{\Omega_\nu}
+\|\sqrt{g}\overline{\partial}^{*}\alpha\|^2_{\Omega_\nu}.
\end{align*}
Since $g+ e^{h} = 1$ and $g\geqslant1$, it follows that
\begin{equation} \label{e2.8}
{\sum_{|L|=p-1}}' \sum_{j,k=1}^n\int_{\Omega_\nu}e^{h}\frac{\partial^2h}{\partial
{z}_j\partial\overline{z}_{k}}\alpha_{j L}\overline{\alpha}_{k L}\,dV
\leqslant\|\overline{\partial}\alpha\|^2_{\Omega_\nu}
+\|\overline{\partial}^{*}\alpha\|^2_{\Omega_\nu},
\end{equation}
for all $\alpha\in {C}^{\infty}_{0,p}(\overline{\Omega}_{\nu})\cap
\operatorname{dom}\overline{\partial}^{\ast}$, $p\geqslant q$.
Estimates \eqref{e2.7} and \eqref{e2.8} were derived under the assumption
that $\alpha$ is continuously differentiable on $\overline\Omega_\nu$,
it holds by density for all square-integrable forms
$\alpha\in\operatorname{dom}\overline{\partial}\cap$
$\operatorname{dom}\overline{\partial}^{*}$. The latter property carries over
to arbitrary bounded $q$-pseudoconvex domains by
exhausting a nonsmooth by smooth ones, and thus so does inequality \eqref{e2.4}
and \eqref{e2.5}.
\end{proof}

The complex Laplacian
$\square_p= \overline\partial\overline\partial^*
+ \overline\partial^*\overline\partial$ acts as an unbounded selfadjoint
operator on $L^2_{0,p}(\Omega)$, $1 \leqslant p \leqslant n$,
it is surjective and thus has a continuous inverse,
the $\overline\partial$-Neumann operator $N_p$. The space
$$
K^2_{0,p}(\Omega)=\{\alpha\in L^2_{0,p}(\Omega):\overline\partial\alpha=0\}
$$
is a closed subspace of $L^2_{0,p}(\Omega)$ because $\overline\partial$
is a closed and densely defined operator. A bounded, linear operator
$$
S_{p+1}: L^2_{0,p+1}(\Omega)\cap K^2_{0,p}(\Omega)\to L^2_{0,p}(\Omega)
$$
is called a canonical solution operator for $\overline\partial$ if
$\overline\partial S_{p+1}\alpha=\alpha$ for all
$\alpha\in L^2_{0,p+1}(\Omega)\cap K^2_{0,p}(\Omega)$ and
$S_{p+1}\alpha\perp K^2_{0,p}(\Omega)$.
The Bergman projection $B_p : L^2_{0,p}(\Omega) \to K^2_{0,p}(\Omega)$
is the orthogonal projection  from $L^2_{0,p}(\Omega)$ onto $K^2_{0,p}(\Omega)$
and $B_0$ is the classical Bergman projection.

\begin{proposition} \label{prop2.7}
Let $\Omega$ be a bounded $q$-pseudoconvex domain in $\mathbb{C}^n$ for
$1\leqslant q\leqslant n$. Thus, for $q\leqslant p\leqslant n$,
there exists a bounded linear operator $N_p: L^2_{0,p}(\Omega)\to
L^2_{0,p}(\Omega)$ which has the following properties:
\begin{itemize}
\item[(1)] $\mathcal{R}ang(N_p)\subset\operatorname{dom}\square_p$,
$N_p \square_p= I$ on $\operatorname{dom}\square_p$,

\item[(2)] for $\alpha\in L^2_{0,p}(\Omega)$, we have $\alpha
=\overline\partial\overline\partial^{*} N_p\alpha\,
\oplus\overline\partial^{*}\overline\partial N_p\alpha$,

\item[(3)] $\overline{\partial} N_p=N_p\overline{\partial}$ on
$\operatorname{dom}\overline\partial$,
$q\leqslant p \leqslant n$,
$\overline{\partial}^{*} N_p=N_p\overline{\partial}^{*}$ on
$\operatorname{dom}\overline\partial^{*}$, $q+1\leqslant p \leqslant n$,

\item[(4)] if $\overline\partial \alpha=0$,
Thus $u=\overline\partial^{*}N_p\alpha$
solves the equation $\overline\partial u=\alpha$,

\item[(5)] $N_p$, $\overline\partial N_p$ and
 $\overline\partial^{*} N_p$ are bounded
operators with respect to the $L^2$-norms,

\item[(6)] the Bergmann projection $B_p$ is given by
\begin{equation} \label{e2.9}
B_p=Id-S_{p+1}\overline\partial.
\end{equation}
\end{itemize}
\end{proposition}

\begin{proof} If $z_0$ is a point of $\Omega_{\nu}$, and
$h(z)=-1+|z-z_0|^2/d^2$, where $d=\sup_{z,z'\in \Omega_{\nu}}|z-z'|$
is the diameter of $\Omega_{\nu}$,  Thus \eqref{e2.8} implies the
fundamental estimate
$$
\|\alpha\|^2_{\Omega_{\nu}}
\leqslant\Big(\frac{d^2e}{p}\Big)\Big(\|\overline\partial
\alpha\|^2_{\Omega_{\nu}}+\|\overline\partial^{\ast}
\alpha\|^2_{\Omega_{\nu}}\Big).
$$
This estimate was derived under the assumption that
$\alpha$ is continuously differentiable on $\overline\Omega_{\nu}$,
it holds by density for all square-integrable
forms  $\alpha\in$ $\operatorname{dom}\overline{\partial}\cap
\operatorname{dom}\overline{\partial}^{*}$.
Thus, by exhausting as in Proposition \ref{prop2.6} and for $p\geqslant q$, one
obtains
\begin{equation} \label{e2.10}
\|\alpha\|_{\Omega}\leqslant \Big(\frac{d^2e}{p}\Big)\|\square_p
\alpha\|_{\Omega}.
\end{equation}
Since $\square_p$ is a linear closed densely defined operator, Thus,
from \cite[Theorem 1.1.1]{H65}; $\operatorname{Rang}(\square_p)$ is closed.
Thus, from \cite[(1.1.1)]{H65} and the fact that $\square_p$ is self adjoint, one
obtains the Hodge decomposition
$$
L^2_{0,p}(\Omega) = \overline\partial\overline{\partial}^{\ast}
\operatorname{dom} \square_p \oplus\overline{\partial}^{\ast} \overline
\partial \operatorname{dom} \square_p .
$$
Since $\square_p:\operatorname{dom} \square_p \to
\operatorname{Rang}(\square_p)=L^2_{0,p}(\Omega)$ is one to one on
$\operatorname{dom}\square_p$ from \eqref{e2.10},
there exists a unique bounded inverse
operator $N_p:\operatorname{Rang}(\square_p)\to \operatorname{dom} \square_p$
satisfies $N_p \square_p \alpha=\alpha$ on $\operatorname{dom} \square_p $
and satisfies $\square_p N_p=I$ on
$L^2_{0,p}(\Omega)$. Thus (1) and (2) follow.

To show that $\overline{\partial}^{\ast} N_p=N_p\overline{\partial}^{\ast}$
on $\operatorname{dom} \overline{\partial}^{\ast}$, by using (2),
 $\overline{\partial}^{\ast}\alpha=\overline{\partial}^{\ast}
\overline{\partial}\overline{\partial}^{\ast}N_p\alpha$, for $\alpha \in
\operatorname{dom} \overline{\partial}^{\ast} $. Thus
$$
N_p\overline{\partial}^{\ast}\alpha
=N_p\overline{\partial}^{\ast}\overline{\partial}
\overline{\partial}^{\ast}N_p\alpha=N_p(\overline{\partial}^{\ast}
\overline{\partial}+
\overline{\partial}\overline{\partial}^{\ast})\overline{\partial}^{\ast}N_p
\alpha=\overline{\partial}^{\ast}N_p\alpha.
$$
A similar argument shows that $\overline\partial N_p=N_p\overline\partial$ on
$\operatorname{dom}\overline\partial$. By using (3) and since
$\overline\partial\alpha=0$, one obtains $\overline\partial
N_p\alpha=N_p\overline\partial\alpha=0$. Thus, by using (2), we obtains
$\alpha=\overline\partial\overline{\partial}^{\ast} N_p\alpha$.
Thus $u=\overline{\partial}^{\ast}N_p\alpha$ satisfies the equation
$\overline\partial u=\alpha$. Since
$\operatorname{Rang}N_p\subset \operatorname{dom}\square_p$, Thus by
applying \eqref{e2.10} to $N_p\alpha$ instead of $\alpha$, (5) follows.
To prove (6), we consider the two complementary cases,
$\alpha \in K^2_{0,p}(\Omega)$ and $\alpha \perp K^2_{0,p}(\Omega)$,
and prove this expression for each. First, if $\alpha \in K^2_{0,p}(\Omega)$,
then $(Id-S_{p+1}\overline\partial) \alpha = \alpha$ as expected.
If $\alpha \perp K^2_{0,p}(\Omega)$, then
$S_{p+1}\overline\partial \alpha = \alpha$ since the ranges of
$\overline\partial$ and $\overline\partial^{\ast}$ are closed.
Thus, $S_{p+1}\alpha = 0$ and the proof follows.
\end{proof}

\begin{proof}[Proof of Theorem \ref{thm1.1}]
 The first part of Theorem \ref{thm1.1} follows by using \eqref{e2.5},
as in \cite[Propositions 4.2 and 4.8]{Straube2010}.
And by applying Proposition \ref{prop2.5}, the second part of Theorem \ref{thm1.1}
follows as in \cite[Theorem 4.1 and Proposition 4.2]{McNeal2002}
and \cite{Straube2010} respectively.
\end{proof}

\section{Proof of Theorem \ref{thm1.2}}

Let $\Omega$ be a bounded domain in $\mathbb{C}^n$ and let
$0 \leqslant p \leqslant n$ and $\phi\in L^\infty(\Omega)$. Denote by
$$
H^2(\Omega)=\{\alpha\in L^2(\Omega):\alpha\text{ is holomorphic on }\Omega\}
$$
the Bergman space which is a closed subspace of $L^2(\Omega)$. The space
$$
H^2_{0,p}(\Omega)=\big\{\alpha={\sum_{|K|=p}}'  \alpha_K\,d\overline z_K:
\alpha_K\in H^2(\Omega), \text{ for all } K\big\}
$$
is the space of $(0, p)$-forms with holomorphic coefficients, is equipped with the
induced norm from $L^2_{0,p}(\Omega)$ and so $H^2_{0,p}(\Omega)$ is a closed
subspace of $L^2_{0,p}(\Omega)$.
For $p = 0$, $H^2_{0,0}(\Omega)=K^2_{0,0}(\Omega)$ is called the Bergman space,
but for $p > 1$ only one obtains $H^2_{0,p}(\Omega)\subsetneqq K^2_{0,p}(\Omega)$.

\begin{example}[\cite{Knirsch2002}] \label{exam3.1} \rm
Let $\alpha=\sum_{j=1}^n\alpha_j \, d\overline z_j\in L^2_{0,1}(\Omega)$. Thus
$$
\alpha\in K^2_{0,1}(\Omega)\Leftrightarrow
\frac{\partial\alpha_j}{\partial\overline z_{k}}
=\frac{\partial\alpha_{k}}{\partial\overline z_j},\quad 1\leqslant j<k\leqslant n,
$$
which can be seen from
$$
\overline\partial\alpha
=\sum_{j,k=1}^n\frac{\partial\alpha_j}{\partial\overline z_{k}}\, d\overline
z_{k}\wedge \, d\overline z_j
=\sum_{1\leqslant j<k\leqslant n}\Big[\frac{\partial\alpha_j}
{\partial\overline z_{k}}-\frac{\partial\alpha_{k}}{\partial\overline z_j}
\Big]\, d\overline z_{k}\wedge \, d\overline z_j.
$$
Now let $\alpha=\sum_{j=1}^3 \alpha_j\, d\overline z_j\in L^2_{0,1}(\Omega)$ and
$g=\sum_{j=1}^3 g_j\, d\overline z_j\in L^2_{0,1}(\Omega)$ with
 $\alpha_1 := z_1+\overline{z}_2+\overline{z}_3$,
$\alpha_2 := \overline{z}_1 + z_2 + \overline{z}_3$,
$\alpha_3 := \overline{z}_1 + \overline{z}_2$ and $g_1:= \alpha_3$,
$g_2 = \alpha_1$, $g_3 := \alpha_2$. Using the above
equivalence we have $\alpha\in K^2_{0,1}(\Omega)$, but $g\notin K^2_{0,1}(\Omega)$.
\end{example}


\begin{remark}[\cite{Knirsch2002}] \label{rmk3.2} \rm
The structure of $H^2_{0,p}(\Omega)$ is less complicated than $K^2_{0,p}(\Omega)$.
If $\alpha\in H^2_{0,p}(\Omega)$, Thus every form with the same coefficients,
 but with different indicates is also in $H^2_{0,p}(\Omega)$.
But for $\alpha\in K^2_{0,p}(\Omega)$, $p > 0$.
\end{remark}

\begin{lemma} \label{lem3.3}
Let $\Omega$ be a bounded $q$-pseudoconvex domain in $\mathbb{C}^n$, $n\geqslant2$.
\begin{itemize}
\item[(i)] Let $\alpha\in K^2_{0,p+1}(\Omega)$, where $1 \leqslant q \leqslant n-1$.
Thus, for $q\leqslant p\leqslant n-1$, there exist $\alpha_j\in K^2_{0,p}(\Omega)$,
$1 \leqslant j \leqslant n$, satisfies
\begin{equation} \label{e3.1}
\alpha=\sum_{j=1}^n \alpha_j\wedge d\overline z_j\quad\text{and}\quad
\sum_{j=1}^n \|\alpha_j\|\lesssim\|\alpha\|.
\end{equation}

\item[(ii)] Let $\alpha\in H^2_{0,p+1}(\Omega)$, where
$1 \leqslant q \leqslant n-1$. Thus, for $q\leqslant p\leqslant n-1$,
there exist $\alpha_j\in H^2_{0,p}(\Omega)$, $1 \leqslant j \leqslant n$,
satisfies
\begin{equation} \label{e3.2}
\alpha=\sum_{j=1}^n \alpha_j\wedge d\overline z_j\quad \text{and}\quad
\|\alpha\|^2=\sum_{j=1}^n \|\alpha_j\|^2.
\end{equation}

\item[(iii)] Let $\alpha\in K^2_{0,p+1}(\Omega)$, where
$1 \leqslant p \leqslant n-1$. Thus, for $q\leqslant p\leqslant n-1$,
there exist $\alpha_j\in K^2_{0,p}(\Omega)$, $1 \leqslant j \leqslant n$, satisfies
\begin{equation} \label{e3.3}
[B_{p+1},\phi]\alpha=(Id-B_{p+1})\Big(\sum_{j=1}^n\left([B_p,\phi]\alpha_j\right)
\wedge d\overline z_j\Big).
\end{equation}
\end{itemize}
\end{lemma}

\begin{proof}
(i) As in \cite[Lemma 1]{Mehmet2014}, for
\[
f={\sum_{|K|=p}}'  f_K\,d\overline z_K= S_{p+1}\alpha,
\]
 one can write 
\[
f=\sum_{j=1}^n f_j\,\wedge d\overline z_j,
\]
where the $f_j$'s are square integrable $(0, p-1)$-forms satisfies there are
no common terms between $f_j\wedge d\overline{z}_j$ and
$f_{k}\wedge d\overline{z}_{k}$ if $j\neq k$. This can be done as follows:
Let $\vee$ denote the adjoint of the exterior
multiplication. That is, if $f$ is a $(0, p)$-form $d\overline z_j\vee f$
is a $(0, p-1)$-form satisfies
$$
\langle h\wedge d\overline z_j,f\rangle =\langle h, d\overline z_j\vee\,f\rangle
\quad \text{for all }h\in C^{\infty}_{0,p-1}(\mathbb{C}^n).
$$
Thus, one can define
\begin{equation} \label{e3.4}
\begin{gathered}
f_1 = d\overline z_{1}\vee f, \\
f_j = d\overline z_j\vee\left(f-\sum_{k=1}^{j-1} f_{k}\wedge d\overline z_{k}\right),
\quad \text{for } j = 2, 3, \dots , n.
\end{gathered}
\end{equation}
Namely, $f_1$ is defined by collecting all terms that contain $d\overline z_{1}$
and writing their sum as $f_{1}\wedge d\overline z_{1}$.
Thus, one defines $f_1$ by collecting the terms in $f-f_1\wedge d\overline z_{1}$
with $d\overline z_{1}$ and writing their sum as $f_2\wedge d\overline z_{2}$ etc.
Since $\overline\partial \alpha = 0$ and $f$ is in the range of
$\overline\partial^{\ast}$, we have
$\overline\partial f=\alpha$ and $\overline\partial^{\ast}f=0$.
So  $f\in \operatorname{dom}\overline{\partial}\cap
\operatorname{dom}\overline{\partial}^{*}$. Also since $f_j$ consists of terms
$f_K$ for some $|K| = p$, ``bar'' derivatives of $f_j$'s come
from ``bar'' derivatives of $f$. Thus
$$
\sum_{j,k=1}^n \|\frac{\partial f_j}{\partial\overline{z}_k}\|
\lesssim{\sum_{|K|=p}}' \sum_{k=1}^n \|\frac{\partial f_K}{\partial\overline{z}_k}\|.
$$
And by using \eqref{e2.4}, one obtains
$$
\sum_{j,k=1}^n \|\frac{\partial f_j}{\partial\overline{z}_{k}}\|
\lesssim\|\overline{\partial}f\|
+\|\overline{\partial}^{*}f\|=\|\alpha\|.
$$
Hence, $\|\overline{\partial}f_j\|\lesssim\|\alpha\|$ for every $j$, and
$$
\alpha = \overline{\partial}f =
\sum_{j=1}^n \overline{\partial}f_j \wedge d\overline{z}_j.
$$
Thus \eqref{e3.1} follows by defining $\alpha_j = \overline{\partial}f_j$.


(ii) By defining $\alpha_1$ and $\alpha_j$ as $f_1$ and $f_j$ in \eqref{e3.4}.
Namely, $\alpha_1$ is defined by collecting all terms that contain
$d\overline z_{1}$ and writing their sum as $\alpha_{1}\wedge d\overline z_{1}$.
Thus, one defines $\alpha_1$ by collecting the terms in
$\alpha-\alpha_1\wedge d\overline z_{1}$ with $d\overline z_{1}$
 and writing their sum as $\alpha_2\wedge d\overline z_{2}$ etc.
The proof is completed as in (i).

(iii) Since both sides of \eqref{e3.3} are orthogonal to $K^2_{0,p+1}(\Omega)$
we need only to show that for any $h\in L^2_{0,p+1}(\Omega)$ that is orthogonal
to $K^2_{0,p+1}(\Omega)$, one obtains
$$
\big\langle [B_{p+1},\phi]\alpha-(Id-B_{p+1})\Big(\sum_{j=1}^n
\left([B_p,\phi]\alpha_j\right)\wedge d\overline z_j\Big),h>=0.
$$
One can computes that
\begin{align*}
&\big\langle [B_{p+1},\phi]\alpha-(Id-B_{p+1})\Big(\sum_{j=1}^n
 \left([B_p,\phi]\alpha_j\right)\wedge d\overline z_j\Big),h\big\rangle \\
&=-\langle \phi\alpha,h\rangle -\langle \sum_{j=1}^nB_p(\phi\alpha_j)
 \wedge d\overline z_j,h\rangle
 +\langle \sum_{j=1}^n\phi\alpha_j\wedge d\overline z_j,h\rangle \\
&=-\langle \sum_{j=1}^nB_p(\phi\alpha_j)\wedge d\overline z_j,h\rangle.
\end{align*}
The fact that
$\overline\partial(f\wedge d\overline z_j)
=\overline\partial f\wedge d\overline z_j$
 implies that the $(0, p+1)$-forms $B_p(\phi\alpha_j)\wedge d\overline z_j$ are
$\overline\partial$-closed for $j = 1, \dots , n$. Thus
$$
\langle \sum_{j=1}^nB_p(\phi\alpha_j)\wedge d\overline z_j,h\rangle = 0
$$
and \eqref{e3.3} follows.
\end{proof}

\begin{remark} \label{rmk3.4} \rm
Indeed, any $(0, p+1)$-form $\alpha={\sum_{|J|=p+1}}' \alpha_{J}\,d
\overline z_{J}$ can be written in the form
$$
\alpha=\frac{1}{(p+1)}\sum_{j=1}^n{\sum_{|L|=p-1}}'
\alpha_{jL}d\overline z_j\wedge\,d\overline z_{L},
$$
and if $\alpha$ has holomorphic coefficients,
the $\alpha_{jL}$ are holomorphic. Let $\alpha\in K^2_{0,p+1}(\Omega)$, one
obtains
$$
[B_p,\phi]\alpha=B_p\phi \alpha-\phi B_p \alpha
=\phi  \alpha-S_{p+1}\overline\partial(\phi \alpha)
-\phi \alpha=-S_{p+1}(\overline\partial \phi \wedge \alpha),
$$
for all
$\phi\in C(\overline\Omega)$. Thus, by taking $\phi = \overline{z}_j$, one obtains
\begin{equation} \label{e3.5}
\begin{aligned}
-\sum_{j=1}^n[B_p,\overline{z}_j]\Big({\sum_{|L|=p-1}}'
 \alpha_{jL} \,d\overline z_{L}\Big)
&=\sum_{j=1}^nS_{p+1}\Big(d\overline z_j\wedge{\sum_{|L|=p-1}}'
 \alpha_{jL} \,d\overline z_{L}\Big)\\
&=S_{p+1}\Big(\sum_{j=1}^n{\sum_{|L|=p-1}}'  \alpha_{jL} \,d\overline z_j
 \wedge\,d\overline z_{L}\Big)\\
&=(p+1)\,S_{p+1}\alpha.
\end{aligned}
\end{equation}
\end{remark}

\begin{lemma} \label{lem3.5}
Let $\Omega$ be a bounded $q$-pseudoconvex domain in $\mathbb{C}^n$,
$n \geqslant 2$ and $1\leqslant q\leqslant n-1$.
Thus, for $q\leqslant p\leqslant n-1$, the following are equivalent:
\begin{itemize}
\item[(1)] $S_{p+1}$ is compact on $K^2_{0,p+1}(\Omega)$,

\item[(2)] $[B_p,\overline{z}_j]$ is compact on $K^2_{0,p}(\Omega)$
for all $1\leqslant j\leqslant n$.
\end{itemize}
\end{lemma}

\begin{proof}
The implication $(1) \Rightarrow (2)$ follows from \eqref{e3.5}.
The implication $(2) \Rightarrow (1)$ follows from \eqref{e3.1} as follows.
Assume that $\{\alpha^k\}$ is a bounded sequence in $K^2_{0,p+1}(\Omega)$, Thus
for each $k$ there exist $\overline\partial$-closed $(0, p)$-forms $\alpha^k_j$
for $1\leqslant j\leqslant n$ satisfies
$$
\alpha^k=\sum_{j=1}^n \alpha^k_j\wedge d\overline z_j\quad\text{and}\quad
\sum_{j=1}^n \|\alpha^k_j\|\leqslant\|\alpha^k\|.
$$
Thus, from \eqref{e3.5}, one obtains
$$
S_{p+1}(\alpha^k)=(-1)^{p+1}\sum_{j=1}^n[B_p,\overline{z}_j](\alpha^k_j).
$$
Furthermore, if $[B_p,\overline{z}_j]$ is compact on $\overline\partial$-closed
$(0, p)$-forms for $1\leqslant j\leqslant n$, the sequences
$\{[B_p,\overline{z}_j](\alpha_j^k)\}$ have convergent subsequences for each $j$.
Hence $S_{p+1}$ is compact on
$K^2_{0,p+1}(\Omega)$. Thus the proof follows.
\end{proof}

\begin{lemma} \label{lem3.6}
Let $\Omega$ be a bounded $q$-pseudoconvex domain in $\mathbb{C}^n$,
$n \geqslant 2$ and let $1\leqslant q\leqslant n-1$. Thus, for
$q\leqslant p\leqslant n-1$, the following are equivalent:
\begin{itemize}
\item[(1)] $N_{p+1}$ is compact on $L^2_{0,p+1}(\Omega)$,

\item[(2)] $[B_p,\phi]$ is compact on $K^2_{0,p}(\Omega)$ for all
$\phi\in C(\overline\Omega)$.
\end{itemize}
\end{lemma}

\begin{proof}
The implication $(1) \Rightarrow (2)$ follows as
\cite[Proposition 4.1]{Straube2010}.
We prove only the implication $(2) \Rightarrow (1)$.
Assume that $[B_p,\phi]$ is compact for all
$\phi\in C(\overline\Omega)$ and $f\in K^2_{0,p+2}(\Omega)$. Thus,
by Lemma \ref{lem3.5},  $S_{p+1}$ is compact. Moreover, by \eqref{e3.1},
there exist $f_j\in K^2_{0,p+1}(\Omega)$ with
$$
f=\sum_{j=1}^n f_j\wedge d\overline z_j\quad \text{and}\quad
\sum_{j=1}^n \|f_j\|\lesssim\|f\|.
$$
Thus
$$
S(f)=\sum_{j=1}^n S_{p+1}(f_j)\wedge d\overline z_j
$$
 solves $\overline\partial u = f$ and $S$ is compact. Thus
$S_{p+2} = (Id - B_{p+1})S$ is compact on  $K^2_{0,p+2}(\Omega)$
and $N_{p+1}$ is compact by the Range's formula \cite{Range1984}:
\begin{equation} \label{e3.6}
N_p = (S_p)^*S_p + S_{p+1}(S_{p+1})^*.
\end{equation}
Thus the proof follows.
\end{proof}


\begin{proof}[Proof of Theorem \ref{thm1.2}]
 The equivalence $(1) \Leftrightarrow (2)$ follows from \eqref{e3.6}.
The equivalence $(2) \Leftrightarrow (3)$ follows from the compactness of
 $S_{p+1}$ on $\overline\partial$-closed forms
is equivalent to compactness of $S_{p+1}$ on $L^2_{0,p+1}(\Omega)$ as $S_{p+1}$
vanishes on the orthogonal complement of $K^2_{0,p+1}(\Omega)$.
The equivalence $(3) \Leftrightarrow (4)$  follows from Lemma \ref{lem3.5}.
The equivalence $(5) \Leftrightarrow (6)$  follows from
\cite[Corollary 1]{Mehmet2014}. The equivalence of (1) and (7) follows from
Lemma \ref{lem3.6}. The implication $(2) \Rightarrow (4)$ is easy;
$(2) \Rightarrow (5)$ follows from \cite[Proposition 4.1]{Straube2010}.
The implications $(6) \Rightarrow (7)$ and $(7) \Rightarrow (4)$ are obvious.
\end{proof}


\begin{corollary} \label{coro3.7}
For $q\leqslant p\leqslant n-1$, the following are equivalent:
\begin{itemize}
\item[(1)] $[B_p,\phi]$ is compact on $H^2_{0,p}(\Omega)$, for all
$\phi\in C(\overline\Omega)$,


\item[(2)] $N_{p+1}$ is compact on $H^2_{0,p+1}(\Omega)$.
\end{itemize}
\end{corollary}

The proof of the above corollary follows by using \eqref{e3.2} as in
Lemma \ref{lem3.6}.



\section{Proof of Theorem \ref{thm1.3}}

The proof will be based on several steps.
\smallskip

\noindent\textbf{Step 1.}
(1) follows, by using \eqref{e2.4}  with \eqref{e3.6}, as
in \cite[Proposition 4.5]{Straube2010}.
Let $\alpha={\sum_{|J|=p+1}}' \alpha_{J}\, d\overline z_{J}\in \operatorname{dom}
\overline\partial\cap\operatorname{dom}\overline\partial^*$.
 For $k = 1, \dots  , n$, one defines $(0, p)$-forms
$\alpha_k={\sum_{|K|=p}}'\alpha_{kK}\, d\overline z_K$.
Thus  $\alpha_k \in\operatorname{dom}\overline\partial\cap\operatorname{dom}
\overline\partial^*$. For $\operatorname{dom}\overline\partial$, this holds
because the components of $\overline\partial\alpha_k$ are linear combinations
of terms $\partial\alpha_J/\partial\overline z_j$, and their
$L^2$-norm is controlled by
$\|\overline\partial\alpha\|+\|\overline\partial^*\alpha\|$ from \eqref{e2.4}.
To see that $\alpha_k \in\operatorname{dom}\overline\partial^*$,
note first that inner products with $\alpha_k$ are closely related to inner
products with $u$: if
$u={\sum_{|K|=p}}' a_K\, d\overline z_K\in L^2_{0,p}(\Omega)$,
Thus for $k$ fixed,
\begin{equation} \label{e4.1}
\begin{aligned}
\langle d\overline z_{k}\wedge u,\alpha\rangle
&=\langle {\sum_{|K|=p}}' a_K (d\overline z_{k}\wedge d\overline z_K),
{\sum_{|J|=p+1}}'\alpha_{J}\, d\overline z_{J}\rangle\\
&={\sum_{|K|=p}}' a_K\overline{\alpha_{kK}}
=\langle u,\alpha_k\rangle .
\end{aligned}
\end{equation}
The inner products are in $L^2_{0,p+1}(\Omega)$ and $L^2_{0,p}(\Omega)$,
respectively. Thus, for
$\beta \in\operatorname{dom}\overline\partial$,
\begin{equation} \label{e4.2}
\begin{aligned}
\langle \overline\partial\beta,\alpha_k\rangle
&=\langle d\overline z_{k}\wedge\overline\partial\beta,\alpha\rangle
 =-\langle \overline\partial(d\overline z_{k}\wedge \beta),\alpha \rangle \\
&=-\langle d\overline z_{k}\wedge \beta,\overline\partial^*\alpha\rangle
&=-\langle \beta,\gamma_k\rangle ,
\end{aligned}
\end{equation}
where $\gamma_k={\sum_{|L|=p-1}}' (\overline\partial^*\alpha)_{kL}\, d\overline z_{L}$.
The last equality follows as in \eqref{e4.1}. \eqref{e4.2}
shows that $\alpha_k \in\operatorname{dom}\overline\partial^*$, and that
\begin{equation} \label{e4.3}
\overline\partial^*\alpha_k=-\gamma_k.
\end{equation}
Now fix $\varepsilon > 0$. The compactness estimate for the $p$-forms
$\alpha_k$ gives
\begin{equation} \label{e4.4}
\begin{aligned}
\|\alpha\|^2&=\frac{1}{(p+1)}\sum_{k=1}^n\|\alpha_k\|^2 \\
&\leqslant\frac{1}{(p+1)}\sum_{k=1}^n
 \Big(\varepsilon(\|\overline\partial\alpha_k\|^2+\|\overline\partial^*\alpha_k\|^2)
+C_\varepsilon\|\alpha_k\|^2_{W^{-1}(\Omega)}\Big),
\end{aligned}
\end{equation}
where the first equality follows from the definition of $\alpha_k$ and the
observation that in the sum on the right-hand side of this equality,
$\|\alpha_J\|^2$ occurs precisely $(p+1)$ times for each strictly
increasing multi-index $J$ of length $p + 1$.
Both $\|\overline\partial\alpha_k\|^2$ and
$\|\overline\partial^*\alpha_k\|^2$ are dominated by
$\|\overline\partial\alpha\|^2+\|\overline\partial^*\alpha\|^2$,
independently of $\varepsilon$. For $\|\overline\partial\alpha_k\|^2$ this
was noted at the beginning of the proof, for $\|\overline\partial^*\alpha_k\|^2$,
this follows from \eqref{e4.3}.
Since $\|\alpha_k\|^2_{W^{-1}(\Omega)}\lesssim\|\alpha\|^2_{W^{-1}(\Omega)}$,
by the definition of $\alpha_k$, \eqref{e4.4} implies a compactness estimate
for $\alpha$.
\smallskip


\noindent\textbf{Step 2.}
(2) follows by using \eqref{e3.1}, as in \cite[Lemma 3]{Mehmet2014}.
In fact, we assume that $\{\alpha^k\}$ is a bounded sequence of
$\overline\partial$-closed $(0, p+1)$-forms. Thus, by \eqref{e3.1}, there exist
$\overline\partial$-closed $(0, p)$-forms $\{\alpha^k_j\}$'s satisfies
$$
\alpha^k=\sum_{j=1}^n \alpha^k_j\wedge d\overline z_j\quad\text{and}\quad
\sum_{j=1}^n \|\alpha^k_j\|\lesssim\|\alpha^k\|.
$$
Let us define
$f^k=\sum_{j=1}^n S_p(\alpha^k_j)\wedge d\overline z_j$.
Thus $\overline\partial f^k = \alpha^k$ and compactness of $S_p$ implies that
$\{f^k\}$ has a convergent subsequence. Thus, $\overline\partial$ has a compact
solution operator on $(0, p + 1)$-forms. Hence, the canonical solution operator,
$S_{p+1}$, is compact on $K^2_{0,p+1}(\Omega)$.
\smallskip

\noindent\textbf{Step 3.}
(3) follows by using \eqref{e3.1}, as in \cite[Theorem 2]{Mehmet2014}.
If $p = n - 1$ Thus $K^2_{0,p+1}(\Omega)=L^2_{0,n}(\Omega)$ and
$[B_{p+1},\phi]$ is the zero operator,
hence compact. So, we need to prove this part for $n \geqslant 3$ and
$q\leqslant p\leqslant n-2$ and for $\alpha\in K^2_{0,p+1}(\Omega)$.
Thus, from \eqref{e3.1}, there exist
$\alpha_j\in K^2_{0,p}(\Omega)$, $1 \leqslant j \leqslant n$, satisfies
$$
\alpha=\sum_{j=1}^n \alpha_j\wedge d\overline z_j\quad \text{and}\quad
\sum_{j=1}^n \|\alpha_j\|\lesssim\|\alpha\|.
$$
Let $\{u^k\}\in K^2_{0,p+1}(\Omega)$ be a bounded sequence. Thus \eqref{e3.1}
implies that for each $k$ and $1\leqslant j\leqslant n$ there exists
$\{u^k_j\}\in K^2_{0,p}(\Omega)$ satisfies
$$
u^k=\sum_{j=1}^n u^k_j\wedge d\overline z_j\quad \text{and}\quad
\sum_{j=1}^n \|u^k_j\|\lesssim\|u^k\|.
$$
Moreover, compactness of $[B_p,\phi]$ on $K^2_{0,p}(\Omega)$ implies that
$\{[B_p,\phi]u^k_j\}$, $1\leqslant j\leqslant n$, has a convergent subsequence.
By using \eqref{e3.3}, $[B_{p+1},\phi]$ is compact on $K^2_{0,p+1}(\Omega)$.
\smallskip

\noindent\textbf{Step 4.}
(4) follows by using \eqref{e3.2}, as in the part (3).


\begin{corollary} \label{coro4.1}
Compactness of $[B_p,\phi]$ on $K^2_{0,p}(\Omega)$, for a fixed $\phi$,
does not necessarily imply compactness
of $[B_p,\phi]$ on $L^2_{0,p}(\Omega)$.
\end{corollary}

 The proof of the above corollary follows from \cite{Mehmet2014}.

\section{Proof of Theorem \ref{thm1.4}}

We identify
$\phi\in L^\infty(\Omega)$ with the multiplication operator
$\phi: L^2_{0,p}(\Omega) \to L^2_{0,p}(\Omega)$ acting by
$$
\phi:\alpha={\sum_{|K|=p}}'  \alpha_K\,d\overline z_K\mapsto
{\sum_{|K|=p}}'  (\phi\alpha_K)\,d\overline z_K.
$$
It follows that $\phi$ is a bounded operator with
$\|\phi\|\leqslant\|\phi\|_{\infty}$ and $\phi^* =\overline\phi $.
The composition
$$
T^\phi_p=B_p\phi B_p: L^2_{0,p}(\Omega) \to L^2_{0,p}(\Omega)
$$
is called the Bergman-Toeplitz operator acting on $(0, p)$-forms
with symbol $\phi$. The Bergman-Toeplitz operators are bounded operators
with $\|T^\phi_p\|\leqslant\|\phi\|_\infty$ and
$T^{\phi^*}_p=T^{\overline\phi}_p$. Clearly
$T^{\phi}_p \alpha=B_p\phi(\alpha)$, for all $\alpha \in  K^2_{0,p}(\Omega)$.

\begin{lemma}[\cite{Knirsch2002}] \label{lem5.1}
 The selfcommutator $[T_p^{z_j*},T_p^{z_j}]$ of $T_p^{z_j}$, is compact if
and only if  the operator $(Id-B_p)\overline z_j$ is compact.
\end{lemma}

\begin{proof}  Given a Toeplitz operator with
symbol $\phi\in H^\infty(\Omega)$, the selfcommutator of $T^{\phi}_p$, is given by
\begin{equation} \label{e5.1}
[T^{\phi^*}_p,T^{\phi}_p]f=B_p \overline\phi \phi
  f-B_p\phi B_p\overline\phi f=B_p\phi  (Id-B_p)\overline\phi  f,\quad
f\in H^2_{0,p}(\Omega).
\end{equation}
Thus, $[T^{\phi^*}_p,T^{\phi}_p]$ is compact if and only if the operator
$f \to (Id-B_p)\overline\phi  f$ from $H^2_{0,p}(\Omega)$ into
$L^2_{0,p}(\Omega)$ is compact. Using the $i^{th}$ coordinate function $z_i$
in place of $\phi$ in \eqref{e5.1}, one obtains
$$
[T_p^{z_j*},T_p^{z_j}]f=B_p\, z_j  (Id-B_p)\overline z_j  f,\quad
f\in H^2_{0,p}(\Omega).
$$
Thus, $[T_p^{z_j*},T_p^{z_j}]$ is compact if and only if the operator
$(Id-B_p)\overline z_j$ is compact.
\end{proof}

To prove a formula for $S_{p+1}$
restricted on $(0, p+1)$-forms with holomorphic coefficients,
the following definitions are needed
\begin{gather*}
\mathcal{J}^p :=\{(j_1,\dots ,j_p)\in\{1,\dots ,n\}^p: j_1<\dots <j_n\},\\
\mathcal{M}^p :=\{(j_1,\dots ,j_p)\in\{1,\dots ,n\}^p: j_1\neq\dots \neq j_n\}.
\end{gather*}
Let $K, M \in \mathcal{M}^p=\cup_{J\in \mathcal{J}^p}\mathcal{M}^p_J$.
If $K$ and $M$ have the same components, one can write $K \sim M$
and $\mathcal{M}^p_K:=\{M\in\mathcal{M}^p: K \sim M\}$ is the
equivalence class of $K$. So one can writes
$$
\alpha={\sum_{|K|=p}}'  \alpha_K\,d\overline z_K
=\sum_{J\in \mathcal{J}^p} \alpha_K\,d\overline z_K
$$
for $(0, p)$-forms $\alpha$ with strongly increasing $p$-tuple $J$ and
$$
\sum_{K\in\mathcal{M}^p} \alpha_K\,d\overline z_K
=\sum_{J\in \mathcal{J}^p} \sum_{K\in\mathcal{M}_J^p }  \alpha_K\,d\overline z_K
$$
for $(0, p)$-forms with non strongly increasing $p$-tuple $K$.
It is clear that $\mathcal{M}_J^p\cap \mathcal{J}^p=\{J\}$, for all
$J\in \mathcal{J}^p$. Thus the mapping
$\mathcal{M}^p \ni M \longmapsto J(M)\in \mathcal{M}_{M}^p\cap\mathcal{J}^p$
is unique, which we need essentially in the proof of
Lemma \ref{lem5.2} with the facts
$|\mathcal{M}_{M}^p|= p!$ and
 $|\mathcal{M}^p|= p!\,|\mathcal{J}^p|$.
As in \cite[Theorem 3.1]{Knirsch2002}, we prove the following lemma.

\begin{lemma} \label{lem5.2}
Let $\Omega$ be a bounded $q$-pseudoconvex domain in $\mathbb{C}^n$ and
let $1 \leqslant q \leqslant n - 1$, $n \geqslant 2$.
Let $\alpha={\sum_{|K|=p+1}}' \alpha_K\, d\overline z_{J}\in H^2_{0,p+1}(\Omega)$,
Thus, for $q\leqslant p\leqslant n-1$, one obtains
\begin{equation} \label{e5.2}
S_{p+1}\alpha=\frac{1}{(p+1)}\sum_{j=1}^n[(Id-B_p)\overline z_j]
\Big({\sum_{|K|=p,\,j \notin K}}' \alpha_{J(j, K)}
\varepsilon^{j, K}_{J(j, K)} d\overline z_K\Big),
\end{equation}
where ${J(j, K)}$ denotes the strongly increasing $(p+ 1)$-tuple with
the same components as the $(p + 1)$-tuple $(j, K)$ and
$\varepsilon^{j, K}_{J(j, K)}$ is the sign of the permutation
$\begin{pmatrix}
j, K
\\
{J(j, K)}
\end{pmatrix}$.
\end{lemma}

\begin{proof}  First we show that
\begin{equation} \label{e5.3}
\alpha={\sum_{|J|=p+1}}' \alpha_{J}\, d\overline z_{J}
=\frac{1}{(p+1)}\sum_{j=1}^n {\sum_{|K|=p}}' \alpha_{J(j, K)}
\varepsilon^{j, K}_{J(j, K)} d\overline z_j\wedge d\overline z_K.
\end{equation}
For this we consider the equivalence class
 $\mathcal{M}_J^{p+1}=\{M_1,\dots ,M_{(p+1)!}\}$ and we get
\begin{align*}
(p+1)!\alpha
&={\underset{J\in \mathcal{J}^{p+1}}\sum}\alpha_K
  \big[d\overline z_{J}+\underset{(p+1)!}{\dots }+d\overline z_{J}\big]\\
&=\sum_{J\in \mathcal{J}^{p+1}} \alpha_K
  \big[\varepsilon_{J}^{M_1}d\overline z_{M_1}+\underset{(p+1)!}{\dots }
  +\varepsilon_{J}^{M_{(p+1)!}}d\overline z_{M_{(p+1)!}}\big]\\
&=\sum_{J\in \mathcal{J}^{p+1}} \alpha_K \sum_{M\in \mathcal{M}_J^{p+1}}
 \varepsilon_{J}^{M}d\overline z_{M}\\
&=\sum_{M\in \mathcal{M}^{p+1}} \alpha_{J(M)} \varepsilon_{J(M)}^{M}d\overline z_{M}\\
&=\sum_{j=1}^n \sum_{L\in \mathcal{M}^{p}} \alpha_{J(j, L)}
  \varepsilon_{J(j, L)}^{j, L}d\overline z_j\wedge d\overline z_{L}\\
&=\sum_{j=1}^n {\sum_{|K|=p}}' \sum_{L\in \mathcal{M}^{p}_K} \alpha_{J(j, L)}
 \varepsilon_{J(j, L)}^{j, L}\varepsilon_L^K d\overline z_j\wedge d\overline z_K.
\end{align*}
Consider the inner sum
\begin{align*}
\sum_{L\in \mathcal{M}^{p}_K} \alpha_{J(j, L)}
  \varepsilon_{J(j, L)}^{j, L}\varepsilon_L^K
&=\sum_{L\in \mathcal{M}^{p}_K} \alpha_{J(j, K)}
  \varepsilon_{J(j, K)}^{j, L}\varepsilon_L^K \\
&=\sum_{L\in \mathcal{M}^{p}_K} \alpha_{J(j, K)}
  \varepsilon_{J(j, K)}^{j, L}\varepsilon_{j,L}^{j,K}\\
&=\sum_{L\in \mathcal{M}^{p}_K} \alpha_{J(j, K)} \varepsilon_{J(j, K)}^{j,K}\\
&=p!\alpha_{J(j, K)} \varepsilon_{J(j, K)}^{j,K}
\end{align*}
and so we get
$$
(p+1)!\alpha=p!\sum_{j=1}^n{\sum_{|K|=p}}' \alpha_{J(j, K)}
 \varepsilon_{J(j, K)}^{j,K}d\overline z_j\wedge d\overline z_K.
$$
Thus \eqref{e5.3} follows. Let
$h={\sum_{|K|=p}}' h_K\, d\overline z_K\in\, K^2_{0,p}(\Omega)$, fix $j$ and consider
$$
\overline z_j\,h={\sum_{|K|=p}}' \overline z_j\,h_K\, d\overline z_K.
$$
Thus
$$
\overline\partial(\overline z_jh)={\sum_{|K|=p}}'  h_K\, d\overline z_j
\wedge d\overline z_K
$$
and with the projection formula \eqref{e2.9} one obtains
\begin{equation} \label{e5.4}
[(Id-B_p)\overline z_j](h)
=[S_{p+1}\overline\partial\overline z_j](h)=S_{p+1}
\Big({\sum_{|K|=p}}'  h_K\, d\overline z_j\wedge d\overline z_K\Big).
\end{equation}
Using the assumption that $\alpha\in H^2_{0,p+1}(\Omega)$ and fix $j$, one defines
\[
h_j={\sum_{|K|=p,\,j \notin K}}' \alpha_{J(j, K)}
 \varepsilon_{J(j, K)}^{j,K} d\overline z_K
\]
it follows that $h_j\in H^2_{0,p}(\Omega)\subset K^2_{0,p}(\Omega)$.
Thus, one can use \eqref{e5.4} for $h_j$ and obtains
\begin{align*}
&[(Id-B_p)\overline z_j]
\Big({\sum_{|K|=p,\,j \notin K}}'
\alpha_{J(j, K)} \varepsilon_{J(j, K)}^{j,K} d\overline z_K\Big) \\
&=S_{p+1}\Big({\sum_{|K|=p}}' \alpha_{J(j, K)} \varepsilon_{J(j, K)}^{j,K}
\, d\overline z_j\wedge d\overline z_K\Big).
\end{align*}
From \eqref{e5.3}, one obtains
\begin{align*}
S_{p+1}(\alpha)
&=\frac{1}{(p+1)}\sum_{j=1}^n S_{p+1}
 \Big({\sum_{|K|=p}}' \alpha_{J(j, K)}\, \varepsilon_{J(j, K)}^{j,K}
 \, d\overline z_j\wedge d\overline z_K\Big) \\
&=\frac{1}{(p+1)}\sum_{j=1}^n[(Id-B_p)\overline z_j]
 \Big({\sum_{|K|=p,\,j \notin K}}' \alpha_{J(j, K)}
  \varepsilon_{J(j, K)}^{j,K} d\overline z_K\Big).
\end{align*}
Thus \eqref{e5.2} follows.
\end{proof}

\begin{proof}[Proof of Theorem \ref{thm1.4}]
 The equivalence $(1) \Leftrightarrow (2)$ follows from \eqref{e3.6}.
Here, we prove only the equivalence of $(2)$ and $(3)$ as
in \cite[Theorem 4.4]{Knirsch2002}.

First we prove $(2) \Rightarrow (3)$. Let $S_{p+1}$ be a compact on
$H^2_{0,p}(\Omega)$. With Lemma \ref{lem5.1} it is enough to show compactness
of $(Id-B_p)\overline z_j$ for all $j = 1, \dots , n$.
Let $f^m={\sum_{|K|=p}}' f^m_K\, d\overline z_K$ be
a bounded sequence in $H^2_{0,p}(\Omega)$. It is clear that for every $j$
$$
u_j^m=\overline\partial(\overline z_jf^m)
={\sum_{|K|=p}}' f^m_K\, d\overline z_j\wedge d\overline z_K
$$
is a bounded sequence in $H^2_{0,p+1}(\Omega)$. By our assumption there exists
a subsequence $u_j^{m_k}$ satisfies $S_{p+1}(u_k^{m_k})$ converges in
$L^2_{0,p}(\Omega)$. Thus, from \eqref{e5.1} and \eqref{e2.9}, one obtains
convergence of
$[(Id-B_p)\overline z_j](f^{m_k})=S_{p+1}\overline\partial(\overline z_jf^{m_k})
=S_{p+1}(u_j^{m_k})$
in $L^2_{0,p}(\Omega)$ for every $j$.

Second, we prove $(3) \Rightarrow (2)$.
Let $f^m={\sum_{|J|=p+1}}' f^m_{J}\, d\overline z_{J}$ be a bounded sequence in
$H^2_{0,p+1}(\Omega)$. We have to show the existence of a subsequence $f^{m_l}$
satisfies $S_{p+1}(f^{m_l})$
converges in $L^2_{0,p}(\Omega)$. With the equivalence of Lemma \ref{lem5.1}, one
can assume that $(Id-B_p)\overline z_j$ is
compact on $H^2_{0,p}(\Omega)$ for all $j = 1, \dots , n$.
For a fix $j$, one defines
$$
h_j^m={\sum_{{|K|=p}\, j \notin K}}' f^m_{J(j, K)} \varepsilon^{j, K}_{J(j, K)}\,
d\overline z_K.
$$
By using formula \eqref{e5.2}, one obtains
\begin{align*}
S_{p+1}(f^m)
&=\frac{1}{(p+1)}\sum_{j=1}^n[(Id-B_p)\overline z_j]
\Big({\sum_{|K|=p\, j \notin K}}' f^m_{J(j, K)}\varepsilon^{j, K}_{J(j, K)}\,
d\overline z_K\Big)\\
&=\frac{1}{(p+1)}\sum_{j=1}^n[(Id-B_p)\overline z_j] (h_j^m).
\end{align*}
Clearly $h^m_j$ is a bounded sequence in $H^2_{0,p}(\Omega)$ for every $j$.
 Since $(Id-B_p)\overline z_j$ is compact on $H^2_{0,p}(\Omega)$ for every $j$,
there exists a subsequence $(h^{m_{l(1)}}_1)$ satisfies
$((Id-B_p)\overline z_{1})(h^{m_{l(1)}}_1)$ converges in $L^2_{0,p}(\Omega)$.
 Clearly $(h^{m_{l(1)}}_2)$ is also a bounded sequence in $H^2_{0,p}(\Omega)$.
Thus it exists a further subsequence $(h^{m_{l(2)}}_2)$ satisfies the
sequences $((Id-B_p)\overline z_{1})(h^{m_{l(2)}}_2)$ converge in
$L^2_{0,p}(\Omega)$. By continuing this process, one obtains a subsequence
$(h^{m_{l(j)}}_j)$ satisfies $((Id-B_p)\overline z_{1})(h^{m_{l(j)}}_j)$
converges in $L^2_{0,p}(\Omega)$ for all $j = 1, \dots , n$.
By defining $m_l = m_{l(n)}$ and with formula \eqref{e5.2} one obtains
convergence of
$$
S_{p+1}(f^{m_l})=\frac{1}{(p+1)}\sum_{j=1}^n((Id-B_p)\overline z_j)\,(h_j^{m_l}).
$$
in $L^2_{0,p}(\Omega)$ for every $j$.
\end{proof}

\begin{corollary} \label{coro5.3}
 For $q\leqslant p\leqslant n-1$, compactness of $S_{p+1}$ on
$K^2_{0,p+1}(\Omega)$ implies compactness of $[T_p^{z_j*},T_p^{z_j}]$ on
$K^2_{0,p}(\Omega)$ for all $j$ with $1\leqslant j\leqslant n$.
\end{corollary}

The proof of the above corollary follows by repeating the implication
 $(2) \Rightarrow (3)$ of Theorem \ref{thm1.4} for $f\in K^2_{0,p}(\Omega)$.

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\end{document}

