††footnotetext:  File: main.tex, printed: 2024-8-28, 3.17

# Generalized weighted composition operators on the hardy space
$H^{2}(\mathbb{D}^{n})$

Molla Basir Ahamed  Molla Basir Ahamed, Department of Mathematics, Jadavpur
University, Kolkata-700032, West Bengal, India.
<EMAIL_ADDRESS>,  Vasudevarao Allu  Vasudevarao Allu,
School of Basic Science, Indian Institute of Technology Bhubaneswar,
Bhubaneswar-752050, Odisha, India<EMAIL_ADDRESS>and  Taimur Rahman
Taimur Rahman, Department of Mathematics, Jadavpur University, Kolkata-700032,
West Bengal, India<EMAIL_ADDRESS>
###### Abstract.

In this paper, we explore the complex symmetrical characteristics of weighted
composition operators $W_{u,v}$ and weighted composition-differentiation
operators $W_{u,v,k_{1},k_{2},\ldots,k_{n}}$ on the Hardy space
$H^{2}(\mathbb{D}^{n})$ over the Polydisk $\mathbb{D}^{n}$, with respect to
the standard conjugation $\mathcal{J}$. We specify explicit conditions that
confirm the Hermitian characteristics of the operator
$W_{u,v,k_{1},k_{2},\ldots,k_{n}}$ and describe the conditions necessary for
it to exhibit normal behavior. Additionally, we identify the kernels of the
generalized weighted composition-differentiation operators and their
corresponding adjoint operators.

###### Key words and phrases:

Composition Operator, Differentiation Operator, Kernel, Complex symmetry,
Hardy Space

###### AMS Subject Classification: Mathematics Subject Classification:

Primary 47B33, 47B02 Secondary 47A05, 47B15

##  1\. Introduction

Let $\mathbb{T}$ denote the boundary of the unit disk
$\mathbb{D}:=\\{z\in\mathbb{C}:|z|<1\\}$ in the complex plane $\mathbb{C}$.
The polydisk $\mathbb{D}^{n}$ and the torus $\mathbb{T}^{n}$ can be
represented as the Cartesian products of $\mathbb{D}$ and $\mathbb{T}$,
respectively. Let $L^{2}:=L^{2}(\mathbb{T}^{n},\sigma)$ denote the standard
Lebesgue space over $\mathbb{T}^{n}$, where $\sigma$ stands for the normalized
Haar measure on the torus $\mathbb{T}^{n}$. It has been established that the
set
$\\{e_{m_{1},m_{2},\ldots,m_{n}}(z_{1},z_{2},\ldots,z_{n})=\prod_{i=1}^{n}z_{i}^{m_{i}}:m_{1},m_{2},\ldots,m_{n}\in\mathbb{Z}\\}$
forms an orthonormal basis for $L^{2}$. If a function $f$ belongs to $L^{2}$,
it can be expressed by using its Fourier coefficients
$\hat{f}(m_{1},m_{2},\ldots,m_{n})$ as

$\displaystyle
f(z_{1},z_{2},\ldots,z_{n})=\sum_{m_{1},m_{2},\ldots,m_{n}=-\infty}^{\infty}\hat{f}(m_{1},m_{2},\ldots,m_{n})\prod_{i=1}^{n}z^{m_{i}}_{i}.$

Furthermore, we define $L^{\infty}=L^{\infty}(\mathbb{T}^{n})$ as the Banach
space containing all essentially bounded functions defined on
$\mathbb{T}^{n}$.

The Hilbert space denoted as $H^{2}=H^{2}(\mathbb{D}^{n})$, referred to as the
Hardy space, consists of all analytic functions $f$ defined on the polydisk
$\mathbb{D}^{n}$, and satisfies the condition

$\displaystyle||f||^{2}:=\sup_{0<r<1}\int_{\mathbb{T}^{n}}|f(r\zeta)|^{2}\,d\sigma(\zeta)<\infty.$

It is a well-known fact that for a function $f$ belonging to the Hardy space
$H^{2}(\mathbb{D}^{n})$, there exists a boundary function $\lim_{r\to
1}f(r\xi)$, which is an element of $L^{2}$, for almost every
$\xi\in\mathbb{T}^{n}$. The Poisson extension of this boundary function
coincides with $f$ on the polydisk $\mathbb{D}^{n}$. By identifying a function
in $H^{2}(\mathbb{D}^{n})$ with its boundary function, we can consider
$H^{2}(\mathbb{D}^{n})$ as a subset of $L^{2}$. Hence, the norm of $f\in
H^{2}$ can be denoted as

$\displaystyle||f||^{2}=\int_{\mathbb{T}^{n}}|f(\zeta)|^{2}d\sigma(\zeta).$

The reproducing kernel function for the Hardy space $H^{2}$ is defined as

$\displaystyle
K_{\alpha}(z)=K_{\alpha_{1},\alpha_{2},\ldots,\alpha_{n}}(z_{1},z_{2},\ldots,z_{n})=\prod_{i=1}^{n}K_{\alpha_{i}}(z_{i})=\prod_{i=1}^{n}\frac{1}{(1-\bar{\alpha_{i}}z_{i})},$

where
$z=(z_{1},z_{2},\ldots,z_{n}),\alpha=(\alpha_{1},\alpha_{2},\ldots,\alpha_{n})\in\mathbb{D}^{n}$,
and $K_{\alpha_{i}}(z_{i})$’s are the reproducing kernel functions for
$H^{2}(\mathbb{D})$. The normalized reproducing kernel for
$H^{2}(\mathbb{D}^{n})$ is denoted by
$k_{\alpha}(z)=\prod_{i=1}^{n}e_{\alpha_{i}}(z_{i})$, with
$e_{\alpha_{1}}(z_{1}),e_{\alpha_{2}}(z_{2}),\ldots,e_{\alpha_{n}}(z_{n})$
being the normalized reproducing kernel functions for $H^{2}(\mathbb{D})$.

Let $\mathcal{B}(\mathcal{H})$ be the algebra of all bounded linear operators
on a separable complex Hilbert space $\mathcal{H}$. A conjugation on
$\mathcal{H}$ is an antilinear operator
$\mathcal{C}:\mathcal{H}\to\mathcal{H}$ which satisfies
$\langle\mathcal{C}x,\mathcal{C}y\rangle_{\mathcal{H}}=\langle
y,x\rangle_{\mathcal{H}}$ for all $x,y\in\mathcal{H}$ and
$\mathcal{C}^{2}=I_{\mathcal{H}}$, where $I_{\mathcal{H}}$ is the identity
operator on the Hilbert space ${\mathcal{H}}$. An operator
$T\in\mathcal{B}({\mathcal{H}})$ is said to be complex symmetric if there
exists a conjugation $\mathcal{C}$ on $\mathcal{H}$ such that
$T=\mathcal{C}T^{*}\mathcal{C}$. In this case, we say that $T$ is complex
symmetric with conjugation $\mathcal{C}$. The comprehensive examination of
complex symmetric operators was initiated by Garcia, Putinar, and Wogen [17,
18, 19, 20, 21]. The class of complex symmetric operators includes normal
operators, binormal operators, operators with algebraic degree two, Hankel
operators, truncated Toeplitz operators, and the Volterra integration
operator. In the recent times, there has been a considerable amount of
research devoted to investigate the complex symmetric composition operators
that operate on classical Hilbert space of analytic functions (see [8, 16, 29,
31, 34, 38] and references therein).

In this context, $W_{u,v}$ refers to the weighted composition operator, while
$W_{u,v,k_{1},k_{2},\ldots,k_{n}}$ represents the weighted composition-
differentiation operator on $H^{2}(\mathbb{D}^{n}$). The problem of
determining when such an operator acts as a conjugation on the weighted Hardy
space of the unit disk was fully resolved in [31]. Moreover, this finding was
extended to the unit ball in [25, 28, 37]. Our paper will specifically
concentrate on the unit polydisk scenario. However, as we move to higher
dimensions, we confront not just combinatorial complexities but also
unexpected barriers and novel phenomena. For instance, studying the
boundedness of the operator $W_{u,v}$ in $H^{2}(\mathbb{D}^{n})$ becomes more
complicated and challenging.

In general, achieving a complete characterization of all complex symmetric
composition operators is difficult, encouraging researchers to investigate
complex symmetric weighted composition operators with specific conjugation
traits. In this paper, we delve into the analysis of complex symmetric
weighted composition operators using the standard conjugation $\mathcal{J}$.

The organization of this paper is as follows. In section 2, the operators
$W_{u,v}$ and $W_{u,v,k_{1},k_{2},\ldots,k_{n}}$ are shown to possess complex
symmetry on $H^{2}(\mathbb{D}^{n})$ as we explicitly determine the
corresponding forms of $u$ and $v_{j}$ (where $j=1,2,\ldots,n$), taking into
account the standard conjugation $\mathcal{J}$. In Section 3, we establish
conditions that are both necessary and sufficient for
$W_{u,v,k_{1},k_{2},\ldots,k_{n}}$ to exhibit Hermitian characteristics.
Section 4 deals with a sufficient condition for the normality of
$W_{u,v,k_{1},k_{2},\ldots,k_{n}}$. In section 5, we investigate the kernels
of both $W_{u,v,k_{1},k_{2},\ldots,k_{n}}$ and its adjoint operator
$W^{*}_{u,v,k_{1},k_{2},\ldots,k_{n}}$.

##  2\. Complex symmetric weighted composition-differentiation operator on
Hardy space $H^{2}(\mathbb{D}^{n})$

Let $u:\mathbb{D}^{n}\to\mathbb{C}$ be an analytic function and
$v:\mathbb{D}^{n}\to\mathbb{D}^{n}$ be an analytic self-map of the polydisk
$\mathbb{D}^{n}$ given by
$v(z_{1},z_{2},\ldots,z_{n})=(v_{1}(z_{1}),v_{2}(z_{2}),\ldots,v_{n}(z_{n}))$,
where $v_{1},v_{2},\ldots,v_{n}$ are analytic self-maps of the unit disk
$\mathbb{D}$. We now define the weighted composition operator $W_{u,v}$ on the
Hardy space $H^{2}(\mathbb{D}^{n})$ as follows:

$\displaystyle W_{u,v}f(z)=u(z)f(v(z)),\;f\in H^{2}(\mathbb{D}^{n}),$

that is,

$\displaystyle
W_{u,v}f(z_{1},z_{2},\ldots,z_{n})=u(z_{1},z_{2},\ldots,z_{n})f(v_{1}(z_{1}),v_{2}(z_{2}),\ldots,v_{n}(z_{n})),\;f\in
H^{2}(\mathbb{D}^{n}).$

The main aim of this paper is to explore different characteristics of weighted
composition-differentiation operators in the Hardy space
$H^{2}(\mathbb{D}^{n})$, exceeding current research outcomes. Going forward,
we shall refer to the composition-differentiation operator as
$W_{u,v,k_{1},k_{2},\ldots,k_{n}}$, defined as

$\displaystyle
W_{u,v,k_{1},k_{2},\ldots,k_{n}}f(z)=u(z)\frac{\partial^{\sum_{i=1}^{n}k_{i}}f}{\partial
z^{k_{1}}_{1}\partial z^{k_{2}}_{2}\ldots\partial z^{k_{n}}_{n}}(v(z))$

where $f$ is an element of the Hardy space $H^{2}(\mathbb{D}^{n})$. It is
clear and straightforward that $W_{u,v,0,0,\ldots,0}=W_{u,v}$.

Assuming $k_{1},k_{2},\ldots,k_{n}$ are non-negative integers and
$w\in\mathbb{D}^{n}$, we define $K_{w}^{[k_{1},k_{2},\ldots,k_{n}]}(z)$ as the
reproducing kernel for evaluating the $\sum_{i=1}^{n}k_{i}$-th partial
derivative at a specific point $w$, where

$\displaystyle
K_{w}^{[k_{1},k_{2},\ldots,k_{n}]}(z)=\prod_{i=1}^{n}\frac{k_{i}!z^{k_{i}}i}{(1-\bar{w_{i}}z_{i})^{k_{i}+1}},z\in\mathbb{D}^{n}.$

It can be easily verified that
$\bigg{\langle}f(z),K_{w}^{[k_{1},k_{2},\ldots,k_{n}]}(z)\bigg{\rangle}=\frac{\partial^{\sum_{i=1}^{n}k_{i}}f}{\partial
z^{k_{1}}_{1}\partial z^{k_{2}}_{2}\ldots\partial z^{k_{n}}_{n}}(w)$.

Here the weighted composition conjugation on $H^{2}(\mathbb{D}^{n})$ is of the
form

$\displaystyle A_{u,v}f(z)=u(z)\overline{f(\overline{v(z)})},f\in\
H^{2}(\mathbb{D}^{n})$

that is,

$\displaystyle
A_{u,v}f(z_{1},z_{2},\ldots,z_{n})=u(z_{1},z_{2},\ldots,z_{n})\overline{f(\overline{v_{1}(z_{1})},\overline{v_{2}(z_{2})},\ldots,\overline{v_{n}(z_{n})})},f\in\
H^{2}(\mathbb{D}^{n}).$

If the function $u$ is identically equal to $1$, and $v$ is the identity map
of the polydisk $\mathbb{D}^{n}$, the weighted composition conjugation
$A_{u,v}$ simplifies to the standard conjugation $\mathcal{J}$, defined as
$\mathcal{J}f(z):=\overline{f(\bar{z})}$. One can easily see that
$A_{u,v}=W_{u,v}\mathcal{J}$. For a more comprehensive understanding of
weighted composition conjugations, we refer to [31] and references therein.
In the following, we provide two lemmas that are focused on the action of the
adjoint of the composition operator $W_{u,v}$ and composition-differentiation
operator $W_{u,v,k_{1},k_{2},\ldots,k_{n}}$ on the reproducing kernel of
$H^{2}(\mathbb{D}^{n})$. In the entirety of the article, we will use the
notation $m=\sum_{i=1}^{n}{k_{i}}$.

######  Lemma 2.1.

Let $u:\mathbb{D}^{n}\to\mathbb{C}$ be a nonzero analytic function and, let
$v:\mathbb{D}^{n}\to\mathbb{D}^{n}$ with
$v(z_{1},z_{2},\ldots,z_{n})=(v_{1}(z_{1}),v_{2}(z_{2}),\ldots,v_{n}(z_{n}))$
be an analytic self-map of $\mathbb{D}^{n}$ such that $W_{u,v}$ is bounded on
$H^{2}(\mathbb{D}^{n})$. Then for any $w\in\mathbb{D}^{n}$,

$\displaystyle W^{*}_{u,v,}K_{w}(z)=\overline{u(w)}K_{v(w)}(z).$

###### Proof of Lemma 2.1.

Let $f\in H^{2}(\mathbb{D}^{n})$. Then a simple computation shows that

$\displaystyle\langle f,W^{*}_{u,v}K_{w}\rangle=$ $\displaystyle\langle
W_{u,v}f,K_{w}\rangle$ $\displaystyle=$ $\displaystyle\langle
uf(v),K_{w}\rangle$ $\displaystyle=$ $\displaystyle u(w)f(v(w))$
$\displaystyle=$ $\displaystyle\langle f,\overline{u(w)}K_{w}\rangle.$

Therefore, we have

$\displaystyle\langle f,W^{*}_{u,v}K_{w}\rangle=\langle
f,\overline{u(w)}K_{w}\rangle\;\mbox{for all}\;f\in\ H^{2}(\mathbb{D}^{n})$

which leads the following conclusion

$\displaystyle W^{*}_{u,v,}K_{w}(z)=\overline{u(w)}K_{v(w)}(z).$

This completes the proof. ∎

######  Lemma 2.2.

Let $u:\mathbb{D}^{n}\to\mathbb{C}$ be a nonzero analytic function and, let
$v:\mathbb{D}^{n}\to\mathbb{D}^{n}$ with
$v(z_{1},z_{2},\ldots,z_{n})=(v_{1}(z_{1}),v_{2}(z_{2}),\ldots,v_{n}(z_{n}))$
be an analytic self-map of $\mathbb{D}^{n}$ such that
$W_{u,v,k_{1},k_{2},\ldots,k_{n}}$ is bounded on $H^{2}(\mathbb{D}^{n})$. Then
for any $w\in\mathbb{D}^{n}$,

$\displaystyle
W^{*}_{u,v,k_{1},k_{2},\ldots,k_{n}}K_{w}(z)=\overline{u(w)}K^{[k_{1},k_{2},\ldots,k_{n}]}_{v(w)}(z)$

###### Proof of Lemma 2.2.

Let $f\in H^{2}(\mathbb{D}^{n})$. Then a simple computation shows that

$\displaystyle\langle f,W^{*}_{u,v,k_{1},k_{2},\ldots,k_{n}}K_{w}\rangle=$
$\displaystyle\langle W_{u,v,k_{1},k_{2},\ldots,k_{n}}f,K_{w}\rangle$
$\displaystyle=$ $\displaystyle\left\langle u\frac{\partial^{m}f}{\partial
z^{k_{1}}_{1}\partial z^{k_{2}}_{2}\ldots\partial
z^{k_{n}}_{n}}(v),K_{w}\right\rangle$ $\displaystyle=$ $\displaystyle
u(w)\frac{\partial^{m}f}{\partial z^{k_{1}}_{1}\partial
z^{k_{2}}_{2}\ldots\partial z^{k_{n}}_{n}}(v(w))$ $\displaystyle=$
$\displaystyle\left\langle
f,\overline{u(w)}K^{[k_{1},k_{2},\ldots,k_{n}]}_{v(w)}\right\rangle.$

Therefore, we have

$\displaystyle\langle
f,W^{*}_{u,v,k_{1},k_{2},\ldots,k_{n}}K_{w}\rangle=\left\langle
f,\overline{u(w)}K^{[k_{1},k_{2},\ldots,k_{n}]}_{v(w)}\right\rangle\;\mbox{for
all}\;f\in\ H^{2}(\mathbb{D}^{n})$

which leads the following conclusion

$\displaystyle
W^{*}_{u,v,k_{1},k_{2},\ldots,k_{n}}K_{w}(z)=\overline{u(w)}K^{[k_{1},k_{2},\ldots,k_{n}]}_{v(w)}(z).$

This completes the proof. ∎

In the following result, we are aiming to examine the circumstances under
which specific combinations of $u$ and $v_{j}(j=1,2,\ldots,n)$ give rise to
$\mathcal{J}$-symmetric weighted composition-differentiation operators
$W_{u,v}$.

######  Theorem 2.1.

Let $u:\mathbb{D}^{n}\to\mathbb{C}$ be a nonzero analytic function and, let
$v:\mathbb{D}^{n}\to\mathbb{D}^{n}$ with
$v(z_{1},z_{2},\ldots,z_{n})=(v_{1}(z_{1}),v_{2}(z_{2}),\ldots,v_{n}(z_{n}))$
be an analytic self-map of $\mathbb{D}^{n}$ such that $W_{u,v}$ is bounded on
$H^{2}(\mathbb{D}^{n})$. Then $W_{u,v}$ is $\mathcal{J}$-symmetric if, and
only if,

$\displaystyle
u(z_{1},z_{2},\ldots,z_{n})=a\prod_{i=1}^{n}\frac{1}{(1-c_{i}z_{i})}\;\mbox{and}\;v_{j}(z_{j})=\frac{(d_{j}-c_{j}^{2})z_{j}+c_{j}}{1-c_{j}z_{j}}\;\;\mbox{for}\;j=1,2,\ldots,n,$

where $a=u(0,0,\ldots,0)$ and $c_{j}=v_{j}(0)$, $d_{j}=v^{\prime}(0)$ for
$j=1,2,\ldots,n$.

If we substitute the polydisk $\mathbb{D}^{n}$ with the bidisk
$\mathbb{D}^{2}$, then Theorem 2.1 becomes a more comprehensive version of
[27, Theorem 2.1], which Han $et$ $al$ [27] recently established.
Consequently, we can easily derive the following result.

######  Corollary 2.1.

Let $u:\mathbb{D}^{2}\to\mathbb{C}$ be a nonzero analytic function and, let
$v:\mathbb{D}^{2}\to\mathbb{D}^{2}$ with
$v(z_{1},z_{2})=(v_{1}(z_{1}),v_{2}(z_{2}))$ be an analytic self-map of
$\mathbb{D}^{2}$ such that $W_{u,v}$ is bounded on $H^{2}(\mathbb{D}^{2})$.
Then $W_{u,v}$ is $\mathcal{J}$-symmetric if, and only if,

$\displaystyle
u(z_{1},z_{2})=\frac{a}{(1-cz_{1})(1-dz_{2})},\;v_{1}(z_{1})=\frac{(e-c^{2})z_{1}+c}{1-cz_{1}},\;v_{2}(z_{2})=\frac{(f-d^{2})z_{2}+d}{1-dz_{2}},$

where
$a=u(0,0),\;c=v_{1}(0),\;d=v_{2}(0),\;e=v^{\prime}_{1}(0),\;f=v^{\prime}_{2}(0)$.

###### Proof of Theorem 2.1.

First we suppose that $W_{u,v}$ is $\mathcal{J}$-symmetric. Then, we have

(2.1) $\displaystyle
W_{u,v}\mathcal{J}K_{w}(z)=\mathcal{J}W^{*}_{u,v}K_{w}(z)$

for all
$z=(z_{1},z_{2},\ldots,z_{n}),w=(w_{1},w_{2},\ldots,w_{n})\in\mathbb{D}^{n}$.
An easy computation shows that

(2.2) $\displaystyle
W_{u,v}\mathcal{J}K_{w}(z)=u(z)\mathcal{J}K_{w}(v(z))=u(z_{1},z_{2},\ldots,z_{n})\prod_{i=1}^{n}\frac{1}{(1-w_{i}v_{i}(z_{i}))}$

and

(2.3)
$\displaystyle\mathcal{J}W^{*}_{u,v}K_{w}(z)=\mathcal{J}\overline{u(w)}K_{v(w)}(z)=u(w)\overline{K_{v(w)}}(\bar{z})=u(w_{1},w_{2},\ldots,w_{n})\prod_{i=1}^{n}\frac{1}{(1-z_{i}v_{i}(w_{i}))}.$

With the help of equations (2.2) and (2.3), we obtain from (2.1) that

(2.4) $\displaystyle
u(z_{1},z_{2},\ldots,z_{n})\prod_{i=1}^{n}\frac{1}{(1-w_{i}v_{i}(z_{i}))}=u(w_{1},w_{2},\ldots,w_{n})\prod_{i=1}^{n}\frac{1}{(1-z_{i}v_{i}(w_{i}))}$

By substituting $w_{1}=w_{2}=\ldots=w_{n}=0$ in (2.4), we obtain

(2.5) $\displaystyle
u(z_{1},z_{2},\ldots,z_{n})=u(0,0,\ldots,0)\prod_{i=1}^{n}\frac{1}{(1-v_{i}(0)z_{i})}.$

Since $u$ is not identically zero on $\mathbb{D}^{n}$, we have that
$u(0,0,\ldots,0)\neq 0$. By the virtue of equations (2.4) and (2.5), it
follows that

(2.6)
$\displaystyle\prod_{i=1}^{n}(1-v_{i}(0)z_{i})\times\prod_{i=1}^{n}(1-w_{i}v_{i}(z_{i}))=\prod_{i=1}^{n}(1-v_{i}(0)w_{i})\times\prod_{i=1}^{n}(1-z_{i}v_{i}(w_{i})).$

By differentiating (2.6) partially with respect to $w_{j}$ (where $1\leq j\leq
n$), we obtain

(2.7) $\displaystyle\prod_{i=1}^{n}$
$\displaystyle(1-v_{i}(0)z_{i})\times\prod_{i\neq
j}^{n}(1-w_{i}v_{i}(z_{i}))\times(-v_{j}(z_{j}))$ $\displaystyle=$
$\displaystyle\prod_{i\neq
j}^{n}(1-v_{i}(0)w_{i})\times\prod_{i=1}^{n}(1-z_{i}v_{i}(w_{i}))\times(-v_{j}(0))$
$\displaystyle+\prod_{i=1}^{n}(1-v_{i}(0)w_{i})\times\prod_{i\neq
j}^{n}(1-z_{i}v_{i}(w_{i}))\times(-z_{j}v^{\prime}_{j}(w_{j})).$

Letting $z_{i}=0$ for $i\neq j$ and $w_{i}=0$ for all $i=1,2,\ldots,n$ in
(2.7), we obtain

$\displaystyle(1-v_{j}(0)z_{j})v_{j}(z_{j})=z_{j}v^{\prime}_{j}(0)+(1-v_{j}(0)z_{j})v_{j}(0).$

This implies that

$\displaystyle
v_{j}(z_{j})=\frac{(v^{\prime}_{j}(0)-v^{2}_{j}(0))z_{j}+v_{j}(0)}{1-v_{j}(0)z_{j}}.$

Thus, we obtain

$\displaystyle
u(z_{1},z_{2},\ldots,z_{n})=a\prod_{i=1}^{n}\frac{1}{(1-c_{i}z_{i})}\;\mbox{and}\;v_{j}(z_{j})=\frac{(d_{j}-c_{j}^{2})z_{j}+c_{j}}{1-c_{j}z_{j}}\;\mbox{for}\;j=1,2,\ldots,n$

with $a=u(0,0,\ldots,0)$ and $c_{j}=v_{j}(0)$, $d_{j}=v^{\prime}(0)$ for
$j=1,2,\ldots,n$.

Conversely, we suppose that

(2.8) $\displaystyle
u(z_{1},z_{2},\ldots,z_{n})=a\prod_{i=1}^{n}\frac{1}{(1-c_{i}z_{i})}\;\mbox{and}\;v_{j}(z_{j})=\frac{(d_{j}-c_{j}^{2})z_{j}+c_{j}}{1-c_{j}z_{j}}\;\mbox{for}\;j=1,2,\ldots,n$

with $a=u(0,0,\ldots,0)$ and $c_{j}=v_{j}(0)$, $d_{j}=v^{\prime}(0)$ for
$j=1,2,\ldots,n$. By using (2.8), we have from equations (2.2) and (2.3) that

(2.9) $\displaystyle W_{u,v}\mathcal{J}K_{w}(z)=$ $\displaystyle
a\prod_{i=1}^{n}\frac{1}{(1-c_{i}z_{i})}\times\prod_{i=1}^{n}\frac{1}{\bigg{(}1-w_{i}\bigg{(}\frac{(d_{i}-c_{i}^{2})z_{i}+c_{i}}{1-c_{i}z_{i}}\bigg{)}\bigg{)}}$
$\displaystyle=$ $\displaystyle
a\prod_{i=1}^{n}\frac{1}{(1-c_{i}z_{i}-w_{i}((d_{i}-c_{i}^{2})z_{i}+c_{i}))}$

and

(2.10) $\displaystyle\mathcal{J}W^{*}_{u,v}K_{w}(z)=$ $\displaystyle
a\prod_{i=1}^{n}\frac{1}{(1-c_{i}w_{i})}\times\prod_{i=1}^{n}\frac{1}{\bigg{(}1-z_{i}\bigg{(}\frac{(d_{i}-c_{i}^{2})w_{i}+c_{i}}{1-c_{i}w_{i}}\bigg{)}\bigg{)}}$
$\displaystyle=$ $\displaystyle
a\prod_{i=1}^{n}\frac{1}{(1-c_{i}w_{i}-z_{i}((d_{i}-c_{i}^{2})w_{i}+c_{i}))}.$

Since

$\displaystyle
1-c_{i}z_{i}-w_{i}((d_{i}-c_{i}^{2})z_{i}+c_{i})=1-c_{i}w_{i}-z_{i}((d_{i}-c_{i}^{2})w_{i}+c_{i}),$

we obtain from equations (2.9) and (2.10) that
$W_{u,v}\mathcal{J}K_{w}(z)=\mathcal{J}W^{*}_{u,v}K_{w}(z)$, which implies
that $W_{u,v}$ is $\mathcal{J}$-symmetric. This completes the proof. ∎

The following result centers on the investigation of which combinations of $u$
and $v$ create complex symmetric weighted composition operators
$W_{u,v,k_{1},k_{2},\ldots,k_{n}}$ with respect to the standard conjugation
$\mathcal{J}$.

######  Theorem 2.2.

Let $u:\mathbb{D}^{n}\to\mathbb{C}$ be a nonzero analytic function and, let
$v:\mathbb{D}^{n}\to\mathbb{D}^{n}$ with
$v(z_{1},z_{2},\ldots,z_{n})=(v_{1}(z_{1}),v_{2}(z_{2}),\ldots,v_{n}(z_{n}))$
be an analytic self-map of $\mathbb{D}^{n}$ such that
$W_{u,v,k_{1},k_{2},\ldots,k_{n}}$ is bounded on $H^{2}(\mathbb{D}^{n})$. Then
$W_{u,v,k_{1},k_{2},\ldots,k_{n}}$ is $\mathcal{J}$-symmetric if, and only if,

$\displaystyle
u(z_{1},z_{2},\ldots,z_{n})=a\prod_{i=1}^{n}\frac{z^{k_{i}}_{i}}{(1-c_{i}z_{i})^{k_{i}+1}}\;\mbox{and}\;v_{j}(z_{j})=\frac{(d_{j}-c_{j}^{2})z_{j}+c_{j}}{1-c_{j}z_{j}}\;\mbox{for}\;j=1,2,\ldots,n,$

where $a=\frac{\partial^{m}}{\partial z^{k_{1}}_{1}\partial
z^{k_{2}}_{2}\ldots\partial z^{k_{n}}_{n}}u(0,0,\ldots,0)$ and
$c_{j}=v_{j}(0)$, $d_{j}=v^{\prime}(0)$ ($j=1,2,\ldots,n$).

######  Remark 2.1.

First, we have to find conditions on $c_{i}$ and $d_{i}$ such that
$v_{i}(\mathbb{D})\subset\mathbb{D}$. A simple computation yields that

$\displaystyle|v_{i}(z_{i})|$
$\displaystyle=\bigg{|}\frac{(d_{i}-c_{i}^{2})z_{i}+c_{i}}{1-c_{i}z_{i}}\bigg{|}$
$\displaystyle=\bigg{|}c_{i}+\frac{d_{i}z_{i}}{1-c_{i}z_{i}}\bigg{|}$
$\displaystyle\geq|c_{i}|-\bigg{|}\frac{d_{i}z_{i}}{1-c_{i}z_{i}}\bigg{|}$
$\displaystyle>|c_{i}|+\bigg{|}\frac{d_{i}z_{i}}{c_{i}z_{i}}\bigg{|}$
$\displaystyle=|c_{i}|+\bigg{|}\frac{d_{i}}{c_{i}}\bigg{|}.$

Therefore, if $|v_{i}(z_{i})|<1$, then we see that
$|c_{i}|+\bigg{|}\frac{d_{i}}{c_{i}}\bigg{|}<1$, which shows that
$|d_{i}|<1-|c_{i}|^{2}$. Thus, it is easy to see that
$v_{i}(\mathbb{D})\subset\mathbb{D}$ if $|d_{i}|<1-|c_{i}|^{2}$.

Next, we have to find the conditions on $c_{i}$ and $d_{i}$ such that
$W_{u,v,k_{1},k_{2},\ldots,k_{n}}$ is bounded. A straightforward computation
shows that

$\displaystyle||W_{u,v,k_{1},k_{2},\ldots,k_{n}}K_{w}(z)||=$
$\displaystyle\bigg{|}\bigg{|}u(z_{1},z_{2},\ldots,z_{n})\prod_{i=1}^{n}\frac{k_{i}!\overline{w_{i}}^{k_{i}}}{(1-\overline{w_{i}}v_{i}(z_{i}))^{k_{i}+1}}\bigg{|}\bigg{|}$
$\displaystyle=$
$\displaystyle\bigg{|}\bigg{|}a\prod_{i=1}^{n}\frac{z^{k_{i}}_{i}}{(1-c_{i}z_{i})^{k_{i}+1}}\times\prod_{i=1}^{n}\frac{k_{i}!\overline{w_{i}}^{k_{i}}}{\bigg{(}1-\overline{w_{i}}\bigg{(}\frac{(d_{i}-c_{i}^{2})z_{i}+c_{i}}{1-c_{i}z_{i}}\bigg{)}\bigg{)}^{k_{i}+1}}\bigg{|}\bigg{|}$
$\displaystyle=$
$\displaystyle\bigg{|}\bigg{|}a\prod_{i=1}^{n}\frac{k_{i}!(\overline{w_{i}}z_{i})^{k_{i}}}{(1-c_{i}z_{i}-\overline{w_{i}}((d_{i}-c_{i}^{2})z_{i}+c_{i}))^{k_{i}+1}}\bigg{|}\bigg{|}$
$\displaystyle\leq$
$\displaystyle|a|\prod_{i=1}^{n}\frac{k_{i}!}{|(1-c_{i}z_{i}-d_{i}\overline{w_{i}}-c_{i}^{2}z_{i}\overline{w_{i}}+c_{i}\overline{w_{i}})|^{k_{i}+1}}$
$\displaystyle\leq$
$\displaystyle|a|\prod_{i=1}^{n}\frac{k_{i}!}{(1-|c_{i}z_{i}|-|d_{i}w_{i}|-|c_{i}^{2}z_{i}w_{i}|-|c_{i}w_{i}|)^{k_{i}+1}}$
$\displaystyle\leq$
$\displaystyle|a|\prod_{i=1}^{n}\frac{k_{i}!}{|(1-c_{i}z_{i}-d_{i}\overline{w_{i}}-c_{i}^{2}z_{i}\overline{w_{i}}+c_{i}\overline{w_{i}})|^{k_{i}+1}}$
$\displaystyle\leq$
$\displaystyle|a|\prod_{i=1}^{n}\frac{k_{i}!}{(1-|c_{i}|-|d_{i}|-|c_{i}^{2}|-|c_{i}|)^{k_{i}+1}}$
$\displaystyle=$
$\displaystyle|a|\prod_{i=1}^{n}\frac{k_{i}!}{(1-2|c_{i}|-|d_{i}|-|c_{i}|^{2})^{k_{i}+1}}.$

It follows that $||W_{u,v,k_{1},k_{2},\ldots,k_{n}}K_{w}(z)||$ is finite if
$1-2|c_{i}|-|d_{i}|-|c_{i}|^{2}>0$. This turns out that
$|d_{i}|<1-2|c_{i}|-|c_{i}|^{2}<(1-|c_{i}|)^{2}$. Since the span of the
reproducing kernel functions is dense in $H^{2}(\mathbb{D}^{n})$, the quantity
$||W_{u,v,k_{1},k_{2},\ldots,k_{n}}f(z)||$ is finite for all $f\in
H^{2}(\mathbb{D}^{n})$ if $|d_{i}|<(1-|c_{i}|)^{2}$. Therefore, the operator
$W_{u,v,k_{1},k_{2},\ldots,k_{n}}$ is bounded on $H^{2}(\mathbb{D}^{n})$ if
$|d_{i}|<(1-|c_{i}|)^{2}$.

###### Proof of Theorem 2.2.

Let $W_{u,v,k_{1},k_{2},\ldots,k_{n}}$ be $\mathcal{J}$-symmetric. Then, we
have

(2.11) $\displaystyle
W_{u,v,k_{1},k_{2},\ldots,k_{n}}\mathcal{J}K_{w}(z)=\mathcal{J}W^{*}_{u,v,k_{1},k_{2},\ldots,k_{n}}K_{w}(z)$

for all
$z=(z_{1},z_{2},\ldots,z_{n}),w=(w_{1},w_{2},\ldots,w_{n})\in\mathbb{D}^{n}$.
Moreover, it can be easily seen through computation that

(2.12) $\displaystyle W_{u,v,k_{1},k_{2},\ldots,k_{n}}\mathcal{J}K_{w}(z)=$
$\displaystyle u(z)\bigg{(}\frac{\partial^{m}}{\partial z^{k_{1}}_{1}\partial
z^{k_{2}}_{2}\ldots\partial
z^{k_{n}}_{n}}\prod_{i=1}^{n}\frac{1}{(1-w_{i}z_{i})}\bigg{)}(v(z))$
$\displaystyle=$ $\displaystyle
u(z_{1},z_{2},\ldots,z_{n})\prod_{i=1}^{n}\frac{k_{i}!w^{k_{i}}_{i}}{(1-w_{i}v_{i}(z_{i}))^{k_{i}+1}}$

and

(2.13) $\displaystyle\mathcal{J}W^{*}_{u,v,k_{1},k_{2},\ldots,k_{n}}K_{w}(z)=$
$\displaystyle\mathcal{J}\overline{u(w)}K^{[k_{1},k_{2},\ldots,k_{n}]}_{v(w)}(z)$
$\displaystyle=$ $\displaystyle
u(w)\overline{K^{[k_{1},k_{2},\ldots,k_{n}]}_{v(w)}(\bar{z})}$
$\displaystyle=$ $\displaystyle
u(w_{1},w_{2},\ldots,w_{n})\prod_{i=1}^{n}\frac{k_{i}!z^{k_{i}}_{i}}{(1-z_{i}v_{i}(w_{i}))^{k_{i}+1}}.$

Thus, the equation (2.11) is equivalent to

(2.14) $\displaystyle
u(z_{1},z_{2},\ldots,z_{n})\prod_{i=1}^{n}\frac{k_{i}!w^{k_{i}}_{i}}{(1-w_{i}v_{i}(z_{i}))^{k_{i}+1}}=u(w_{1},w_{2},\ldots,w_{n})\prod_{i=1}^{n}\frac{k_{i}!z^{k_{i}}_{i}}{(1-z_{i}v_{i}(w_{i}))^{k_{i}+1}}.$

Substituting $z_{1}=0,z_{2}=0,\ldots,z_{n}=0$ in (2.14), we obtain
$u(0,z_{2},\ldots,z_{n})=0,u(z_{1},0,\ldots,z_{n})=0,\ldots,u(z_{1},z_{2},\ldots,z_{n-1},0)=0$
and $u(0,0,\ldots,0)=0$ respectively, which leads us to the conclusion that
$u$ contains factors of $z_{1},z_{2},\ldots,z_{n}$.

Let
$u(z_{1},z_{2},\ldots,z_{n})=h(z_{1},z_{2},\ldots,z_{n})\prod_{i=1}^{n}z^{t_{i}}_{i}$,
where $t_{i}(i=1,2,\ldots,n)\in\mathbb{N}$ and $h$ is analytic with
$h(0,z_{2},\ldots,z_{n})\neq 0,h(z_{1},0,\ldots,z_{n})\neq
0,\ldots,h(z_{1},z_{2},\ldots,z_{n-1},0)\neq 0$ and $h(0,0,\ldots,0)\neq 0$.
Then, we obtain from (2.14) that

(2.15) $\displaystyle
h(z_{1},z_{2},\ldots,z_{n})\prod_{i=1}^{n}\frac{k_{i}!w^{k_{i}}_{i}z^{t_{i}}_{i}}{(1-w_{i}v_{i}(z_{i}))^{k_{i}+1}}=h(w_{1},w_{2},\ldots,w_{n})\prod_{i=1}^{n}\frac{k_{i}!w^{t_{i}}_{i}z^{k_{i}}_{i}}{(1-z_{i}v_{i}(w_{i}))^{k_{i}+1}}.$

Now, we claim that $k_{i}=t_{i}$ for all $i=1,2,\ldots,n$. If $k_{i}>t_{i}$,
it follows from (2.15) that

(2.16) $\displaystyle
h(z_{1},z_{2},\ldots,z_{n})\prod_{i=1}^{n}\frac{k_{i}!w^{k_{i}-t_{i}}_{i}}{(1-w_{i}v_{i}(z_{i}))^{k_{i}+1}}=h(w_{1},w_{2},\ldots,w_{n})\prod_{i=1}^{n}\frac{k_{i}!z^{k_{i}-t_{i}}_{i}}{(1-z_{i}v_{i}(w_{i}))^{k_{i}+1}}$

Setting $z_{1}=0$ in (2.16), we obtain $h(0,z_{2},\ldots,z_{n})=0$, which is a
contradiction because $h(0,z_{2},\ldots,z_{n})\neq 0$. If $k_{i}<t_{i}$, we
obtain from (2.15) that

(2.17) $\displaystyle
h(z_{1},z_{2},\ldots,z_{n})\prod_{i=1}^{n}\frac{k_{i}!z^{t_{i}-k_{i}}_{i}}{(1-w_{i}v_{i}(z_{i}))^{k_{i}+1}}=h(w_{1},w_{2},\ldots,w_{n})\prod_{i=1}^{n}\frac{k_{i}!w^{t_{i}-k_{i}}_{i}}{(1-z_{i}v_{i}(w_{i}))^{k_{i}+1}}.$

Putting $w_{1}=0$ in (2.17), we get $h(z_{1},z_{2},\ldots,z_{n})\equiv 0$ on
$\mathbb{D}^{n}$, which is a contradiction. Therefore $k_{i}=t_{i}$ for all
$i=1,2,\ldots,n$ and hence the equation (2.15) reduces to

(2.18) $\displaystyle
h(z_{1},z_{2},\ldots,z_{n})\prod_{i=1}^{n}\frac{1}{(1-w_{i}v_{i}(z_{i}))^{k_{i}+1}}=h(w_{1},w_{2},\ldots,w_{n})\prod_{i=1}^{n}\frac{1}{(1-z_{i}v_{i}(w_{i}))^{k_{i}+1}}.$

Letting $w_{i}=0$ for all $i=1,2,\ldots,n$ in (2.18), we obtain

$\displaystyle
h(z_{1},z_{2},\ldots,z_{n})=h(0,0,\ldots,0)\prod_{i=1}^{n}\frac{1}{(1-v_{i}(0)z_{i})^{k_{i}+1}}$

and hence

(2.19) $\displaystyle
u(z_{1},z_{2},\ldots,z_{n})=h(0,0,\ldots,0)\prod_{i=1}^{n}\frac{z^{k_{i}}_{i}}{(1-v_{i}(0)z_{i})^{k_{i}+1}}.$

Since $h(0,0,\ldots,0)\neq 0$, it follows from equations (2.14) and(2.19) that

(2.20)
$\displaystyle\prod_{i=1}^{n}(1-v_{i}(0)z_{i})^{k_{i}+1}\times\prod_{i=1}^{n}(1-w_{i}v_{i}(z_{i}))^{k_{i}+1}=\prod_{i=1}^{n}(1-v_{i}(0)w_{i})^{k_{i}+1}\times\prod_{i=1}^{n}(1-z_{i}v_{i}(w_{i}))^{k_{i}+1}.$

Taking the partial derivative with respect to the variable $w_{j}(1\leq j\leq
n)$ in (2.20), we obtain

(2.21) $\displaystyle(k_{j}$
$\displaystyle+1)\prod_{i=1}^{n}(1-v_{i}(0)z_{i})^{k_{i}+1}\times\prod_{i\neq
j}^{n}(1-w_{i}v_{i}(z_{i}))^{k_{i}+1}\times(1-w_{j}v_{j}(z_{j}))^{k_{j}}(-v_{j}(z_{j}))$
$\displaystyle=$ $\displaystyle(k_{j}+1)\prod_{i\neq
j}^{n}(1-v_{i}(0)w_{i})^{k_{i}+1}\times\prod_{i=1}^{n}(1-z_{i}v_{i}(w_{i}))^{k_{i}+1}\times(1-v_{j}(0)w_{j})^{k_{j}}(-v_{j}(0))$
$\displaystyle+(k_{j}+1)\prod_{i=1}^{n}(1-v_{i}(0)w_{i})^{k_{i}+1}\times\prod_{i\neq
j}^{n}(1-z_{i}v_{i}(w_{i}))^{k_{i}+1}\times(1-z_{j}v_{j}(w_{j}))^{k_{j}}(-z_{j}v^{\prime}_{j}(w_{j})).$

Letting $w_{i}=0$ for all $i=1,2,\ldots,n$ and $z_{i}=0$ for $i\neq j$ in
(2.21), we obtain

$\displaystyle(1-v_{j}(0)z_{j})^{k_{j}+1}(v_{j}(z_{j}))=(1-v_{j}(0)z_{j})^{k_{j}+1}(v_{j}(0))+(1-v_{j}(0)z_{j})^{k_{j}}(z_{j}v^{\prime}_{j}(0)),$

which implies that

$\displaystyle
v_{j}(z_{j})=\frac{(v^{\prime}_{j}(0)-v^{2}_{j}(0))z_{j}+v_{j}(0)}{1-v_{j}(0)z_{j}}.$

Thus, we easily obtain the following

$\displaystyle
u(z_{1},z_{2},\ldots,z_{n})=a\prod_{i=1}^{n}\frac{z^{k_{i}}_{i}}{(1-c_{i}z_{i})}\;\mbox{and}\;v_{j}(z_{j})=\frac{(d_{j}-c_{j}^{2})z_{j}+c_{j}}{1-c_{j}z_{j}}\;\;\mbox{for}\;j=1,2,\ldots,n,$

where $a=\frac{\partial^{m}}{\partial z^{k_{1}}_{1}\partial
z^{k_{2}}_{2}\ldots\partial z^{k_{n}}_{n}}u(0,0,\ldots,0)$ and
$c_{j}=v_{j}(0)$, $d_{j}=v^{\prime}(0)$( $j=1,2,\ldots,n$). Conversely,
suppose that

$\displaystyle
u(z_{1},z_{2},\ldots,z_{n})=a\prod_{i=1}^{n}\frac{z^{k_{i}}_{i}}{(1-c_{i}z_{i})^{k_{i}+1}}\;\mbox{and}\;v_{j}(z_{j})=\frac{(d_{j}-c_{j}^{2})z_{j}+c_{j}}{1-c_{j}z_{j}}\;\mbox{for}\;j=1,2,\ldots,n,$

where

$\displaystyle a=\frac{\partial^{m}}{\partial z^{k_{1}}_{1}\partial
z^{k_{2}}_{2}\ldots\partial
z^{k_{n}}_{n}}u(0,0,\ldots,0)\;\;\mbox{and}\;c_{j}=v_{j}(0),\;d_{j}=v^{\prime}(0)\;\;\mbox{for}\;\;j=1,2,\ldots,n).$

In view of (2.12) and (2.13), a simple computation shows that

(2.22) $\displaystyle W_{u,v,k_{1},k_{2},\ldots,k_{n}}\mathcal{J}K_{w}(z)=$
$\displaystyle
u(z_{1},z_{2},\ldots,z_{n})\prod_{i=1}^{n}\frac{k_{i}!w^{k_{i}}_{i}}{(1-w_{i}v_{i}(z_{i}))^{k_{i}+1}}$
$\displaystyle=$ $\displaystyle
a\prod_{i=1}^{n}\frac{z^{k_{i}}_{i}}{(1-c_{i}z_{i})^{k_{i}+1}}\times\prod_{i=1}^{n}\frac{k_{i}!w^{k_{i}}_{i}}{\bigg{(}1-w_{i}\bigg{(}\frac{(d_{i}-c_{i}^{2})z_{i}+c_{i}}{1-c_{i}z_{i}}\bigg{)}\bigg{)}^{k_{i}+1}}$
$\displaystyle=$ $\displaystyle
a\prod_{i=1}^{n}\frac{k_{i}!(w_{i}z_{i})^{k}_{i}}{(1-c_{i}z_{i}-w_{i}((d_{i}-c_{i}^{2})z_{i}+c_{i}))^{k_{i}+1}}$

and

(2.23) $\displaystyle\mathcal{J}W^{*}_{u,v,k_{1},k_{2},\ldots,k_{n}}K_{w}(z)=$
$\displaystyle
u(w_{1},w_{2},\ldots,w_{n})\prod_{i=1}^{n}\frac{k_{i}!z^{k_{i}}_{i}}{(1-z_{i}v_{i}(w_{i}))^{k_{i}+1}}$
$\displaystyle=$ $\displaystyle
a\prod_{i=1}^{n}\frac{w^{k_{i}}_{i}}{(1-c_{i}w_{i})^{k_{i}+1}}\times\prod_{i=1}^{n}\frac{k_{i}!z^{k_{i}}_{i}}{\bigg{(}1-z_{i}\bigg{(}\frac{(d_{i}-c_{i}^{2})w_{i}+c_{i}}{1-c_{i}w_{i}}\bigg{)}\bigg{)}^{k_{i}+1}}$
$\displaystyle=$ $\displaystyle
a\prod_{i=1}^{n}\frac{k_{i}!(w_{i}z_{i})^{k}_{i}}{(1-c_{i}w_{i}-z_{i}((d_{i}-c_{i}^{2})w_{i}+c_{i}))^{k_{i}+1}}$

for any
$z=(z_{1},z_{2},\ldots,z_{n}),w=(w_{1},w_{2},\ldots,w_{n})\in\mathbb{D}^{n}$.
Since

$\displaystyle
1-c_{i}z_{i}-w_{i}((d_{i}-c_{i}^{2})z_{i}+c_{i})=1-c_{i}w_{i}-z_{i}((d_{i}-c_{i}^{2})w_{i}+c_{i}),$

we obtain from (2.22) and (2.23) that
$W_{u,v,k_{1},k_{2},\ldots,k_{n}}\mathcal{J}K_{w}(z)=\mathcal{J}W^{*}_{u,v,k_{1},k_{2},\ldots,k_{n}}K_{w}(z)$,
which implies that $W_{u,v,k_{1},k_{2},\ldots,k_{n}}$ is
$\mathcal{J}$-symmetric. This completes the proof. ∎

##  3\. Hermitian property of composition-differentiation operators on the
Hardy space $H^{2}(\mathbb{D}^{n})$

A bounded linear operator $T$ is Hermitian (self-adjoint) if $T^{*}=T$. Recent
research has been focused on investigating the self-adjointness of weighted
composition-differentiation operators in the reproducing kernel Hilbert space
(see [15, 31, 33] and other related works). In 2021, Fatehi and Hummond [14]
presented a complete characterization of the self-adjointness of weighted
composition-differentiation operators in the Hardy space over the unit disk
$\mathbb{D}$. In 2023, Liu _et al_ [32] discussed the self-adjointness of the
weighted composition-differentiation operator in the Bergman space
$\mathcal{A}^{2}_{\alpha}(\mathbb{D})$. In 2012, Le [30] derived
characterizations of self-adjoint weighted composition operators on Hilbert
spaces of analytic functions over the unit ball
$\mathbb{B}_{n}:=\\{z\in\mathbb{C}^{n}:|z|<1\\}$. On a similar note, in [25],
Han and Wang investigated the self-adjointness of the composition operator on
Hilbert spaces of analytic functions over the same unit ball $\mathbb{B}_{n}$.
No research has been carried out on the self-adjointness of the composition-
differentiation operator in relation to the Hardy spaces over the polydisk
$\mathbb{D}^{n}$.

Our aim is to investigate the self-adjoint properties over the polydisk
$\mathbb{D}^{n}$. In the following result, we establish a condition that is
both necessary and sufficient for the bounded composition-differentiation
operator $W_{u,v,k_{1},k_{2},\ldots,k_{n}}$ to satisfy Hermitian properties on
$H^{2}(\mathbb{D}^{n})$.

######  Theorem 3.1.

Let $u:\mathbb{D}^{n}\to\mathbb{C}$ be a nonzero analytic function and, let
$v:\mathbb{D}^{n}\to\mathbb{D}^{n}$ with
$v(z_{1},z_{2},\ldots,z_{n})=(v_{1}(z_{1}),v_{2}(z_{2}),\ldots,v_{n}(z_{n}))$
be an analytic self-map of $\mathbb{D}^{n}$ such that
$W_{u,v,k_{1},k_{2},\ldots,k_{n}}$ is bounded on $H^{2}(\mathbb{D}^{n})$. Then
$W_{u,v,k_{1},k_{2},\ldots,k_{n}}$ is Hermitian if, and only if,

$\displaystyle
u(z_{1},z_{2},\ldots,z_{n})=a\prod_{i=1}^{n}\frac{z^{k_{i}}_{i}}{(1-\bar{c_{i}}z_{i})^{k_{i}+1}}\;\mbox{and}\;v_{j}(z_{j})=\frac{(d_{j}-|c_{j}|^{2})z_{j}+c_{j}}{1-\bar{c_{j}}z_{j}}\;\mbox{for}\;j=1,2,\ldots,n,$

where $a,d_{j}(j=1,2,\ldots,n)\in\mathbb{R}$ and
$c_{j}(j=1,2,\ldots,n)\in\mathbb{C}$.

The defining formulas of $W_{u,v}$ and $W_{u,v,k_{1},k_{2},\ldots,k_{n}}$ make
it evident that $W_{u,v,0,0,\ldots,0}=W_{u,v}$. Moreover, the implications of
Theorem 3.1 reveal that when $k_{1}=k_{2}=\ldots=k_{n}=0$,
$W_{u,v,k_{1},k_{2},\ldots,k_{n}}$ coincides with $W_{u,v}$. Consequently, we
can readily derive the following result from Theorem 3.1.

######  Corollary 3.1.

Let $u:\mathbb{D}^{n}\to\mathbb{C}$ be a nonzero analytic function and, let
$v:\mathbb{D}^{n}\to\mathbb{D}^{n}$ with
$v(z_{1},z_{2},\ldots,z_{n})=(v_{1}(z_{1}),v_{2}(z_{2}),\ldots,v_{n}(z_{n}))$
be an analytic self-map of $\mathbb{D}^{n}$ such that $W_{u,v}$ is bounded on
$H^{2}(\mathbb{D}^{n})$. Then $W_{u,v}$ is Hermitian if, and only if,

$\displaystyle
u(z_{1},z_{2},\ldots,z_{n})=a\prod_{i=1}^{n}\frac{1}{(1-\bar{c_{i}}z_{i})^{k_{i}+1}}\;\mbox{and}\;v_{j}(z_{j})=\frac{(d_{j}-|c_{j}|^{2})z_{j}+c_{j}}{1-\bar{c_{j}}z_{j}}\;\mbox{for}\;j=1,2,\ldots,n,$

where $a,d_{j}(j=1,2,\ldots,n)\in\mathbb{R}$ and
$c_{j}(j=1,2,\ldots,n)\in\mathbb{C}$.

###### Proof of Theorem 3.1.

Let $W_{u,v,k_{1},k_{2},\ldots,k_{n}}$ be Hermitian. Then
$W^{*}_{u,v,k_{1},k_{2},\ldots,k_{n}}=W_{u,v,k_{1},k_{2},\ldots,k_{n}}$, which
is equivalent to

(3.1) $\displaystyle
W^{*}_{u,v,k_{1},k_{2},\ldots,k_{n}}K_{w}(z)=W_{u,v,k_{1},k_{2},\ldots,k_{n}}K_{w}(z)$

for all $w,z\in\mathbb{D}^{n}$. A simple computation shows that

(3.2) $\displaystyle W_{u,v,k_{1},k_{2},\ldots,k_{n}}K_{w}(z)=$ $\displaystyle
u(z)\bigg{(}\frac{\partial^{m}}{\partial z^{k_{1}}_{1}\partial
z^{k_{2}}_{2}\ldots\partial
z^{k_{n}}_{n}}\prod_{i=1}^{n}\frac{1}{(1-\bar{w_{i}}z_{i})}\bigg{)}(v(z))$
$\displaystyle=$ $\displaystyle
u(z_{1},z_{2},\ldots,z_{n})\prod_{i=1}^{n}\frac{k_{i}!\bar{w_{i}}^{k_{i}}}{(1-\bar{w_{i}}v_{i}(z_{i}))^{k_{i}+1}}$

and

(3.3) $\displaystyle W^{*}_{u,v,k_{1},k_{2},\ldots,k_{n}}K_{w}(z)=$
$\displaystyle\overline{u(w)}K^{[k_{1},k_{2},\ldots,k_{n}]}_{v(w)}(z)$
$\displaystyle=$
$\displaystyle\overline{u(w_{1},w_{2},\ldots,w_{n})}\prod_{i=1}^{n}\frac{k_{i}!z^{k_{i}}_{i}}{(1-z_{i}\overline{v_{i}(w_{i})})^{k_{i}+1}}.$

By using the equations (3.2) and (3.3), we obtain from (3.1) that

(3.4) $\displaystyle
u(z_{1},z_{2},\ldots,z_{n})\prod_{i=1}^{n}\frac{k_{i}!\bar{w_{i}}^{k_{i}}}{(1-\bar{w_{i}}v_{i}(z_{i}))^{k_{i}+1}}=\overline{u(w_{1},w_{2},\ldots,w_{n})}\prod_{i=1}^{n}\frac{k_{i}!z^{k_{i}}_{i}}{(1-z_{i}\overline{v_{i}(w_{i})})^{k_{i}+1}}.$

By applying a similar argument as in the proof of Theorem 2.2, we can assume
that
$u(z_{1},z_{2},\ldots,z_{n})=h(z_{1},z_{2},\ldots,z_{n})\prod_{i=1}^{n}z^{k_{i}}_{i}$
and $h$ is analytic with $h(0,z_{2},\ldots,z_{n})\neq
0,h(z_{1},0,\ldots,z_{n})\neq 0,\ldots,h(z_{1},z_{2},\ldots,z_{n-1},0)\neq 0$
and $h(0,0,\ldots,0)\neq 0$. Then, we obtain from (3.4) that

(3.5) $\displaystyle
h(z_{1},z_{2},\ldots,z_{n})\prod_{i=1}^{n}\frac{1}{(1-\bar{w_{i}}v_{i}(z_{i}))^{k_{i}+1}}=\overline{h(w_{1},w_{2},\ldots,w_{n})}\prod_{i=1}^{n}\frac{1}{(1-z_{i}\overline{v_{i}(w_{i})})^{k_{i}+1}}.$

Putting $w_{i}=0$ for all $i=1,2,\ldots,n$ in (3.5), we obtain

$\displaystyle
h(z_{1},z_{2},\ldots,z_{n})=\overline{h(0,0,\ldots,0)}\prod_{i=1}^{n}\frac{1}{(1-\overline{v_{i}(0)}z_{i})^{k_{i}+1}}$

and hence

$\displaystyle
u(z_{1},z_{2},\ldots,z_{n})=a\prod_{i=1}^{n}\frac{z^{k_{i}}_{i}}{(1-\bar{c_{i}}z_{i})^{k_{i}+1}}$

with $a=\overline{h(0,0,\ldots,0)}$ and $c_{i}=v_{i}(0)$ for $i=1,2,\ldots,n$.
Substituting this in (3.4), we obtain

(3.6)
$\displaystyle\bar{a}\prod_{i=1}^{n}(1-\bar{c_{i}}z_{i})^{k_{i}+1}\times\prod_{i=1}^{n}(1-\bar{w_{i}}v_{i}(z_{i}))^{k_{i}+1}=a\prod_{i=1}^{n}(1-c_{i}\bar{w_{i}})^{k_{i}+1}\times\prod_{i=1}^{n}(1-z_{i}\overline{v_{i}(w_{i})})^{k_{i}+1}$

for any
$z=(z_{1},z_{2},\ldots,z_{n}),w=(w_{1},w_{2},\ldots,w_{n})\in\mathbb{D}^{n}$.
If we set $w_{1}=w_{2}=\ldots=w_{n}=0$, then it is easy to see that
$\bar{a}=a$, and hence $a\in\mathbb{R}$. Taking the derivative partially with
respect to $\bar{w_{j}}(1\leq j\leq n)$ in (3.6), we obtain

(3.7) $\displaystyle(k_{j}+1)$
$\displaystyle\prod_{i=1}^{n}(1-\bar{c_{i}}z_{i})^{k_{i}+1}\times\prod_{i\neq
j}^{n}(1-\bar{w_{i}}v_{i}(z_{i}))^{k_{i}+1}\times(1-\bar{w_{j}}v_{j}(z_{j}))^{k_{j}}(-v_{j}(z_{j}))$
$\displaystyle=$ $\displaystyle(k_{j}+1)\prod_{i\neq
j}^{n}(1-c_{i}\bar{w_{i}})^{k_{i}+1}\times\prod_{i=1}^{n}(1-z_{i}\overline{v_{i}(w_{i})})^{k_{i}+1}\times(1-c_{j}\bar{w_{j}})^{k_{j}}(-c_{j})$
$\displaystyle+(k_{j}+1)\prod_{i=1}^{n}(1-c_{i}\bar{w_{i}})^{k_{i}+1}\times\prod_{i\neq
j}^{n}(1-z_{i}\overline{v_{i}(w_{i})})^{k_{i}+1}\times(1-z_{j}\overline{v_{j}(w_{j})})^{k_{j}}(-z_{j}\overline{v^{\prime}_{j}(w_{j})})$

for any
$z=(z_{1},z_{2},\ldots,z_{n}),w=(w_{1},w_{2},\ldots,w_{n})\in\mathbb{D}^{n}$.
Setting $w_{1}=w_{2}=\cdots=w_{n}=0$ in (3.7), we have

$\displaystyle(1-\bar{c_{j}}z_{j})^{k_{j}+1}v_{j}(z_{j})=(1-\bar{c_{j}}z_{j})^{k_{j}+1}c_{j}+(1-\bar{c_{j}}z_{j})^{k_{j}}(z_{j}\overline{v^{\prime}_{j}(0)})$

which implies that

(3.8) $\displaystyle
v_{j}(z_{j})=\frac{(1-\bar{c_{j}}z_{j})c_{j}+z_{j}\overline{v^{\prime}_{j}(0)}}{1-\bar{c_{j}}z_{j}}=\frac{(d_{j}-|c_{j}|^{2})z_{j}+c_{j}}{1-\bar{c_{j}}z_{j}}\;\mbox{for
all}\;j=1,2,\ldots,n$

with $d_{j}=\overline{v^{\prime}_{j}(0)}$ and $c_{j}=v_{j}(0)$.

Differentiating (3.8) partially with respect to $z_{j}$, we obtain
$v^{\prime}_{j}(0)=d_{j}$. Therefore, we see that $\bar{d_{j}}=d_{j}$ which
implies that $d_{j}(1\leq j\leq n)$ are real numbers. Conversely, assume that

$\displaystyle
u(z_{1},z_{2},\ldots,z_{n})=a\prod_{i=1}^{n}\frac{z^{k_{i}}_{i}}{(1-\bar{c_{i}}z_{i})^{k_{i}+1}}\;\mbox{and}\;v_{j}(z_{j})=\frac{(d_{j}-|c_{j}|^{2})z_{j}+c_{j}}{1-\bar{c_{j}}z_{j}}\;\mbox{for}\;j=1,2,\ldots,n,$

where $a,d_{j}(j=1,2,\ldots,n)\in\mathbb{R}$ and
$c_{j}(j=1,2,\ldots,n)\in\mathbb{C}$. In view of (3.2) and (3.3), we have

(3.9) $\displaystyle W_{u,v,k_{1},k_{2},\ldots,k_{n}}K_{w}(z)=$ $\displaystyle
u(z_{1},z_{2},\ldots,z_{n})\prod_{i=1}^{n}\frac{k_{i}!\bar{w_{i}}^{k_{i}}}{(1-\bar{w_{i}}v_{i}(z_{i}))^{k_{i}+1}}$
$\displaystyle=$ $\displaystyle
a\prod_{i=1}^{n}\frac{z^{k_{i}}_{i}}{(1-\bar{c_{i}}z_{i})^{k_{i}+1}}\times\prod_{i=1}^{n}\frac{k_{i}!\bar{w_{i}}^{k_{i}}}{\bigg{(}1-\bar{w_{i}}\bigg{(}\frac{(d_{i}-|c_{i}|^{2})z_{i}+c_{i}}{1-\bar{c_{i}}z_{i}}\bigg{)}\bigg{)}^{k_{i}+1}}$
$\displaystyle=$ $\displaystyle
a\prod_{i=1}^{n}\frac{k_{i}!(\bar{w_{i}}z_{i})^{k_{i}}}{(1-\bar{c_{i}}z_{i}-\bar{w_{i}}((d_{i}-|c_{i}|^{2})z_{i}+c_{i}))^{k_{i}+1}}$

and

(3.10) $\displaystyle W^{*}_{u,v,k_{1},k_{2},\ldots,k_{n}}K_{w}(z)=$
$\displaystyle\overline{u(w_{1},w_{2},\ldots,w_{n})}\prod_{i=1}^{n}\frac{k_{i}!z^{k_{i}}_{i}}{(1-z_{i}\overline{v_{i}(w_{i})})^{k_{i}+1}}$
$\displaystyle=$
$\displaystyle\overline{a\prod_{i=1}^{n}\frac{w^{k_{i}}_{i}}{(1-\bar{c_{i}}w_{i})^{k_{i}+1}}}\times\prod_{i=1}^{n}\frac{k_{i}!z^{k_{i}}_{i}}{\bigg{(}1-z_{i}(\overline{\bigg{(}\frac{(d_{i}-|c_{i}|^{2})w_{i}+c_{i}}{1-\bar{c_{i}}w_{i}}\bigg{)}}\bigg{)}^{k_{i}+1}}$
$\displaystyle=$
$\displaystyle\bar{a}\prod_{i=1}^{n}\frac{k_{i}!(\bar{w_{i}}z_{i})^{k_{i}}}{(1-c_{i}\bar{w_{i}}-z_{i}((\bar{d_{i}}-|c_{i}|^{2})\bar{w_{i}}+\bar{c_{i}}))^{k_{i}+1}}$

for any
$z=(z_{1},z_{2},\ldots,z_{n}),w=(w_{1},w_{2},\ldots,w_{n})\in\mathbb{D}^{n}$.
Since

$\displaystyle
1-\bar{c_{i}}z_{i}-\bar{w_{i}}((d_{i}-|c_{i}|^{2})z_{i}+c_{i})=$
$\displaystyle
1-\bar{c_{i}}z_{i}-\bar{w_{i}}(d_{i}-|c_{i}|^{2})z_{i}-c_{i}\bar{w_{i}}$
$\displaystyle=$ $\displaystyle
1-c_{i}\bar{w_{i}}-z_{i}((d_{i}-|c_{i}|^{2})\bar{w_{i}}+\bar{c_{i}})$

and $a,d_{j}(j=1,2,\ldots,n)\in\mathbb{R}$, it follows from (3.9) and (3.10)
that

$\displaystyle
W^{*}_{u,v,k_{1},k_{2},\ldots,k_{n}}K_{w}(z)=W_{u,v,k_{1},k_{2},\ldots,k_{n}}K_{w}(z),$

which implies that $W_{u,v,k_{1},k_{2},\ldots,k_{n}}$ is Hermitian on
$H^{2}(\mathbb{D}^{n})$. This completes the proof. ∎

##  4\. Sufficient condition for the normality of composition-differentiation
operators on the Hardy space $H^{2}(\mathbb{D}^{n})$

Normal operators are those that commute with their own adjoint. In other
words, a bounded linear operator T is classified as normal if, and only if,
$T^{*}T=TT^{*}$. In 2023, Liu _et al_ [32] investigated the normality of
composition-differentiation operators on the Bergman space, presenting
necessary and sufficient conditions in their study. Lo and Loh [33]
investigated the class of normal generalized weighted composition operators on
$H_{\gamma}$ and their relevance to the property of complex symmetry. In the
last few years, substantial research has been dedicated to investigate the
normality of weighted composition-differentiation operators in the reproducing
kernel Hilbert space (see [15, 14, 29, 7] and relevant references). Le [30]
has conducted a study on the normality of weighted composition operators on
Hilbert spaces of analytic functions over the unit ball $\mathbb{B}_{n}$.
Similarly, Han and Wang [25] have explored the normality of the composition
operator on Hilbert spaces of analytic functions over the same unit ball
$\mathbb{B}_{n}$.

Until now, the entirety of normality research has been restricted to the unit
disk and the unit ball, and it seems that no investigations have been carried
on the polydisk. Our main aim is to fill this gap by exploring the normality
properties over the polydisk and we obtain Theorem 4.1. However, the
significance of Theorem 4.1 lies in its limited scope, as it does not cover
all normal weighted composition-differentiation operators on
$H^{2}(\mathbb{D}^{n})$. Therefore, we establish the precise conditions for an
operator $W_{u,v,k_{1},k_{2},\ldots,k_{n}}$ to be normal, assuming that $u$
and $v_{j}(z_{j})$ (where $j=1,2,\ldots,n$) adopt a specific form that
generalizes the symbols explained in Theorem 4.1. Despite our efforts, we have
not yet determined whether an operator $W_{u,v,k_{1},k_{2},\ldots,k_{n}}$,
with $v_{j}(0)(0)\neq 0$ for $j=1,2,\ldots,n$, can be normal for different
choices of $u$ and $v_{j}(z_{j})$. We now establish a sufficient condition
that guarantees the bounded operator $W_{u,v,k_{1},k_{2},\ldots,k_{n}}$ to be
normal, given the complex symmetry under the standard conjugation
$\mathcal{J}$.

######  Theorem 4.1.

Let $u:\mathbb{D}^{n}\to\mathbb{C}$ be a nonzero analytic function and, let
$v:\mathbb{D}^{n}\to\mathbb{D}^{n}$ with
$v(z_{1},z_{2},\ldots,z_{n})=(v_{1}(z_{1}),v_{2}(z_{2}),\ldots,v_{n}(z_{n}))$
be an analytic self-map of $\mathbb{D}^{n}$ such that
$W_{u,v,k_{1},k_{2},\ldots,k_{n}}$ is bounded on $H^{2}(\mathbb{D}^{n})$.
Suppose $W_{u,v,k_{1},k_{2},\ldots,k_{n}}$ is $\mathcal{J}$-symmetric and
$v_{j}(0)=0$ for all $j=1,2,\ldots,n$. Then $W_{u,v,k_{1},k_{2},\ldots,k_{n}}$
is normal.

###### Proof of Theorem 4.1.

Since $W_{u,v,k_{1},k_{2},\ldots,k_{n}}$ is $\mathcal{J}$-symmetric and
$v_{j}(0)=0$ for all $j=1,2,\ldots,n$, it follows from Theorem 2.2 that

$\displaystyle
u(z_{1},z_{2},\ldots,z_{n})=a\prod_{i=1}^{n}z^{k_{i}}_{i}\;\mbox{and}\;v_{j}(z_{j})=d_{j}z_{j}\;\mbox{for}\;j=1,2,\ldots,n,$

where $a,d_{j}(j=1,2,\ldots,n)\in\mathbb{C}$. Then for any non-negative
integers $m_{1},m_{2},\ldots,m_{n}$, we have

(4.1) $\displaystyle||$ $\displaystyle
W_{u,v,k_{1},k_{2},\ldots,k_{n}}e_{m_{1},m_{2},\ldots,m_{n}}||^{2}$
$\displaystyle=$
$\displaystyle\sum_{j_{1},j_{2},\ldots,j_{n}=0}^{\infty}|\langle
W_{u,v,k_{1},k_{2},\ldots,k_{n}}e_{m_{1},m_{2},\ldots,m_{n}},e_{j_{1},j_{2},\ldots,j_{n}}\rangle|^{2}$
$\displaystyle=$
$\displaystyle\sum_{j_{1},j_{2},\ldots,j_{n}=0}^{\infty}\bigg{|}\bigg{\langle}u\bigg{(}\frac{\partial^{m}}{\partial
z^{k_{1}}_{1}\partial z^{k_{2}}_{2}\ldots\partial
z^{k_{n}}_{n}}\prod_{i}^{n}z^{m_{i}}_{i}\bigg{)}(v),e_{j_{1},j_{2},\ldots,j_{n}}\bigg{\rangle}\bigg{|}^{2}$
$\displaystyle=$
$\displaystyle\sum_{j_{1},j_{2},\ldots,j_{n}=0}^{\infty}\bigg{|}\bigg{\langle}a\prod_{i=1}^{n}z^{k_{i}}_{i}\times\bigg{(}\prod_{i}^{n}\frac{m_{i}!}{(m_{i}-k_{i})!}(d_{i}z_{i})^{m_{i}-k_{i}}\bigg{)},e_{j_{1},j_{2},\ldots,j_{n}}\bigg{\rangle}\bigg{|}^{2}$
$\displaystyle=$
$\displaystyle\sum_{j_{1},j_{2},\ldots,j_{n}=0}^{\infty}\bigg{|}\bigg{\langle}a\prod_{i}^{n}\frac{m_{i}!d^{m_{i}-k_{i}}_{i}}{(m_{i}-k_{i})!}\times\prod_{i=1}^{n}z^{m_{i}}_{i},e_{j_{1},j_{2},\ldots,j_{n}}\bigg{\rangle}\bigg{|}^{2}$
$\displaystyle=$
$\displaystyle\sum_{j_{1},j_{2},\ldots,j_{n}=0}^{\infty}\bigg{|}a\prod_{i}^{n}\frac{m_{i}!d^{m_{i}-k_{i}}_{i}}{(m_{i}-k_{i})!}\bigg{|}^{2}\bigg{|}\bigg{\langle}e_{m_{1},m_{2},\ldots,m_{n}},e_{j_{1},j_{2},\ldots,j_{n}}\bigg{\rangle}\bigg{|}^{2}$
$\displaystyle=$
$\displaystyle\bigg{|}a\prod_{i}^{n}\frac{m_{i}!d^{m_{i}-k_{i}}_{i}}{(m_{i}-k_{i})!}\bigg{|}^{2}$

and

(4.2) $\displaystyle||$ $\displaystyle
W^{*}_{u,v,k_{1},k_{2},\ldots,k_{n}}e_{m_{1},m_{2},\ldots,m_{n}}||^{2}$
$\displaystyle=$
$\displaystyle\sum_{j_{1},j_{2},\ldots,j_{n}=0}^{\infty}|\langle
W^{*}_{u,v,k_{1},k_{2},\ldots,k_{n}}e_{m_{1},m_{2},\ldots,m_{n}},e_{j_{1},j_{2},\ldots,j_{n}}\rangle|^{2}$
$\displaystyle=$
$\displaystyle\sum_{j_{1},j_{2},\ldots,j_{n}=0}^{\infty}|\langle
e_{m_{1},m_{2},\ldots,m_{n}},W_{u,v,k_{1},k_{2},\ldots,k_{n}}e_{j_{1},j_{2},\ldots,j_{n}}\rangle|^{2}$
$\displaystyle=$
$\displaystyle\sum_{j_{1},j_{2},\ldots,j_{n}=0}^{\infty}\bigg{|}\bigg{\langle}e_{m_{1},m_{2},\ldots,m_{n}},u\bigg{(}\frac{\partial^{m}}{\partial
z^{k_{1}}_{1}\partial z^{k_{2}}_{2}\ldots\partial
z^{k_{n}}_{n}}\prod_{i}^{n}z^{j_{i}}_{i}\bigg{)}(v)\bigg{\rangle}\bigg{|}^{2}$
$\displaystyle=$
$\displaystyle\sum_{j_{1},j_{2},\ldots,j_{n}=0}^{\infty}\bigg{|}\bigg{\langle}e_{m_{1},m_{2},\ldots,m_{n}},a\prod_{i=1}^{n}z^{k_{i}}_{i}\times\bigg{(}\prod_{i}^{n}\frac{j_{i}!}{(j_{i}-k_{i})!}(d_{i}z_{i})^{j_{i}-k_{i}}\bigg{)}\bigg{\rangle}\bigg{|}^{2}$
$\displaystyle=$
$\displaystyle\sum_{j_{1},j_{2},\ldots,j_{n}=0}^{\infty}\bigg{|}\bigg{\langle}e_{m_{1},m_{2},\ldots,m_{n}},a\prod_{i}^{n}\frac{j_{i}!d^{j_{i}-k_{i}}_{i}}{(j_{i}-k_{i})!}\times\prod_{i=1}^{n}z^{j_{i}}_{i}\bigg{)}\bigg{\rangle}\bigg{|}^{2}$
$\displaystyle=$
$\displaystyle\sum_{j_{1},j_{2},\ldots,j_{n}=0}^{\infty}\bigg{|}a\prod_{i}^{n}\frac{j_{i}!d^{j_{i}-k_{i}}_{i}}{(j_{i}-k_{i})!}\bigg{|}^{2}\bigg{|}\bigg{\langle}e_{m_{1},m_{2},\ldots,m_{n}},e_{j_{1},j_{2},\ldots,j_{n}}\bigg{\rangle}\bigg{|}^{2}$
$\displaystyle=$
$\displaystyle\bigg{|}a\prod_{i}^{n}\frac{m_{i}!d^{m_{i}-k_{i}}_{i}}{(m_{i}-k_{i})!}\bigg{|}^{2}$

as the set $\\{e_{m_{1},m_{2},\ldots,m_{n}}:m_{1},m_{2},\ldots,m_{n}\geq 0\\}$
forms an orthonormal basis for $H(\mathbb{D}^{n})$. Thus, it follows from
(4.1) and (4.2) that for any non-negative integer $m_{1},m_{2},\ldots,m_{n}$,
we have

(4.3)
$\displaystyle||W_{u,v,k_{1},k_{2},\ldots,k_{n}}e_{m_{1},m_{2},\ldots,m_{n}}||^{2}=\bigg{|}a\prod_{i}^{n}\frac{m_{i}!d^{m_{i}-k_{i}}_{i}}{(m_{i}-k_{i})!}\bigg{|}^{2}=||W^{*}_{u,v,k_{1},k_{2},\ldots,k_{n}}e_{m_{1},m_{2},\ldots,m_{n}}||^{2}.$

Consequently, (4.3) leads to the conclusion that

$\displaystyle||W_{u,v,k_{1},k_{2},\ldots,k_{n}}f||=||W^{*}_{u,v,k_{1},k_{2},\ldots,k_{n}}f||$

for any $f\in H^{2}(\mathbb{D}^{n})$, and hence
$W_{u,v,k_{1},k_{2},\ldots,k_{n}}$ is a normal operator. This completes the
proof. ∎

From Theorem 4.1, it is clear that when $k_{1}=k_{2}=\ldots=k_{n}=0$,
$W_{u,v,k_{1},k_{2},\ldots,k_{n}}$ coincides with $W_{u,v}$. Consequently, we
can readily derive the following result from Theorem 4.1.

######  Corollary 4.1.

Let $u:\mathbb{D}^{n}\to\mathbb{C}$ be a nonzero analytic function and, let
$v:\mathbb{D}^{n}\to\mathbb{D}^{n}$ with
$v(z_{1},z_{2},\ldots,z_{n})=(v_{1}(z_{1}),v_{2}(z_{2}),\ldots,v_{n}(z_{n}))$
be an analytic self-map of $\mathbb{D}^{n}$ such that $W_{u,v}$ is bounded on
$H^{2}(\mathbb{D}^{n})$. Suppose $W_{u,v}$ is $\mathcal{J}$-symmetric and
$v_{j}(0)=0$ for all $j=1,2,\ldots,n$. Then $W_{u,v}$ is normal.

###### Proof of Corollary 4.1.

Since $W_{u,v}$ is $\mathcal{J}$-symmetric and $v_{j}(0)=0$ for all
$j=1,2,\ldots,n$, it follows from Theorem 2.1 that

$\displaystyle
u(z_{1},z_{2},\ldots,z_{n})=a\;\mbox{and}\;v_{j}(z_{j})=d_{j}z_{j}\;\mbox{for}\;j=1,2,\ldots,n,$

where $a,d_{j}(j=1,2,\ldots,n)\in\mathbb{C}$.
Now, we can proceed with the proof in a manner analogous to the proof of
Theorem 4.1. By setting $k_{i}=0$ for all $i=1,2,\ldots,n$ in equation (4.3),
we obtain

$\displaystyle||W_{u,v}e_{m_{1},m_{2},\ldots,m_{n}}||^{2}=\bigg{|}a\prod_{i}^{n}d^{m_{i}}_{i}\bigg{|}^{2}=||W^{*}_{u,v}e_{m_{1},m_{2},\ldots,m_{n}}||^{2},$

which leads to the conclusion that $||W_{u,v}f||=||W^{*}_{u,v}f||$ for any
$f\in H^{2}(\mathbb{D}^{n})$, and hence $W_{u,v}$ is a normal operator. This
completes the proof. ∎

##  5\. Kernel of generalized composition-differentiation operators on Hardy
space $H^{2}(\mathbb{D}^{n})$

In the study of composition operators for several spaces of functions, one of
the significant findings is to explore the kernel of the operators. In this
section, we investigate the kernel of generalized composition-differentiation
operators $W_{u,v,k_{1},k_{2},\ldots,k_{n}}$ on the Hardy space
$H^{2}(\mathbb{D}^{n})$. The kernel of an operator refers to the collection of
vectors that map to the zero vector. To explore the kernel of generalized
composition-differentiation operators on $H^{2}(\mathbb{D}^{n})$, we must
introduce the space containing all polynomials in $\mathbb{C}^{n}$ of degree
$\leq m$.

Let $\mathcal{P}^{n}_{m}(\mathbb{C})$ be the space containing all polynomials
in $\mathbb{C}^{n}$ of degree $\leq m$ over the complex plane $\mathbb{C}$.
The explicit definition of this space is as follows:

$\displaystyle\mathcal{P}^{n}_{m}(\mathbb{C})=\bigg{\\{}\sum_{a_{1}+a_{2}+\cdots+a_{n}\leq
m}A_{a_{1},a_{2},\ldots,a_{n}}\prod_{i=1}^{n}z_{i}^{a_{i}}:A_{a_{1},a_{2},\ldots,a_{n}}\in\mathbb{C}\bigg{\\}}.$

In 2014, Jung _et al_ [29, Proposition 3.2] have proved that the kernel of the
adjoint of a complex symmetric weighted composition operator on
${H}^{2}(\mathbb{D})$ is simply $\\{0\\}$. In 2021, Han and Wang [26,
Proposition 2.9] demonstrated that the kernel of the adjoint of a nonzero
$J$-symmetric weighted composition-differentiation operator on the Hardy
Hilbert space ${H}^{2}(\mathbb{D})$ is the whole complex plane $\mathbb{C}$.

It is natural to raise the following question.

######  Question 5.1.

Can we explore the kernel of $W_{u,v,k_{1},k_{2},\ldots,k_{n}}$ on
$H^{2}(\mathbb{D}^{n})$ explicitly?

In this section, our primary aim is to answer the Question 5.1. It has been
studied extensively the kernel of the composition-differentiation operator in
different spaces defined over the unit disk $\mathbb{D}$. We will explore the
kernel of the composition-differentiation operator
$W_{u,v,k_{1},k_{2},\ldots,k_{n}}$ on $H^{2}(\mathbb{D}^{n})$ in the Hardy
space $H^{2}(\mathbb{D}^{n})$ over the polydisk $\mathbb{D}^{n}$.

By establishing the following result, we will show that both the kernels of
the generalized composition-differentiation operators
$W_{u,v,k_{1},k_{2},\ldots,k_{n}}$ and their adjoint are equal to
$\mathcal{P}^{n}_{m-1}(\mathbb{C})$. Proving the following result is greatly
influenced by the role of the Identity theorem in several complex variables.

######  Theorem 5.1.

Let $u:\mathbb{D}^{n}\to\mathbb{C}$ be a nonzero analytic function and, let
$v:\mathbb{D}^{n}\to\mathbb{D}^{n}$ with
$v(z_{1},z_{2},\ldots,z_{n})=(v_{1}(z_{1}),v_{2}(z_{2}),\ldots,v_{n}(z_{n}))$
be an analytic self-map of $\mathbb{D}^{n}$ such that
$W_{u,v,k_{1},k_{2},\ldots,k_{n}}$ is bounded on $H^{2}(\mathbb{D}^{n})$. If
$W_{u,v,k_{1},k_{2},\ldots,k_{n}}$ is complex symmetric with standard
conjugation $\mathcal{J}$, then
$ker(W_{u,v,k_{1},k_{2},\ldots,k_{n}})=ker(W^{*}_{u,v,k_{1},k_{2},\ldots,k_{n}})=\mathcal{P}^{n}_{m-1}(\mathbb{C})$.

###### Proof of Theorem 5.1.

Suppose that $W_{u,v,k_{1},k_{2},\ldots,k_{n}}$ is complex symmetric on
$H^{2}(\mathbb{D}^{n})$ with the standard conjugation $\mathcal{J}$. Then

(5.1) $\displaystyle
W_{u,v,k_{1},k_{2},\ldots,k_{n}}\mathcal{J}f(z)=\mathcal{J}W^{*}_{u,v,k_{1},k_{2},\ldots,k_{n}}f(z),\;f\in\
H^{2}(\mathbb{D}^{n}).$

If $f\in\ ker(W_{u,v,k_{1},k_{2},\ldots,k_{n}})$, then

$\displaystyle W_{u,v,k_{1},k_{2},\ldots,k_{n}}(f(z))\equiv
0\;\mbox{on}\;\mathbb{D}^{n}.$

That is,

$\displaystyle u(z)\frac{\partial^{m}f}{\partial z^{k_{1}}_{1}\partial
z^{k_{2}}_{2}\ldots\partial z^{k_{n}}_{n}}(v(z))\equiv
0\;\mbox{on}\;\mathbb{D}^{n},$

equivalently, we have

$\displaystyle\frac{\partial^{m}f}{\partial z^{k_{1}}_{1}\partial
z^{k_{2}}_{2}\ldots\partial z^{k_{n}}_{n}}(v(z))\equiv
0\;\mbox{on}\;\mathbb{D}^{n}.$

Since $v(\mathbb{D}^{n})$ is open, so by Identity Theorem in several
variables, we have

$\displaystyle\frac{\partial^{m}}{\partial z^{k_{1}}_{1}\partial
z^{k_{2}}_{2}\ldots\partial z^{k_{n}}_{n}}(f(z))\equiv
0\;\mbox{on}\;\mathbb{D}^{n}$

which implies that

$\displaystyle
f(z)=\sum_{a_{1}+a_{2}+\cdots+a_{n}\leq(m-1)}A_{a_{1},a_{2},\ldots
a_{n}}\prod_{i=1}^{n}z^{a_{i}}_{i}$

for some $A_{a_{1},a_{2},\dots
a_{n}}(a_{1}+a_{2}+\cdots+a_{n}\leq(m-1))\in\mathbb{C}$. Clearly,

(5.2) $\displaystyle
ker(W_{u,v,k_{1},k_{2},\ldots,k_{n}})\subseteq\mathcal{P}^{n}_{m-1}(\mathbb{C}).$

Let $f$ be an element of $\mathcal{P}^{n}_{m-1}(\mathbb{C})$. It is clear that
$f$ is a polynomial of degree $\leq(m-1)$ in $n$ variables. Thus, by applying
the Identity Theorem in several variables, we have

$\displaystyle W_{u,v,k_{1},k_{2},\ldots,k_{n}}(f(z))$
$\displaystyle=u(z)\frac{\partial^{m}f}{\partial z^{k_{1}}_{1}\partial
z^{k_{2}}_{2}\ldots\partial z^{k_{n}}_{n}}(v(z))$ $\displaystyle\equiv
u(z)\frac{\partial^{m}}{\partial z^{k_{1}}_{1}\partial
z^{k_{2}}_{2}\ldots\partial z^{k_{n}}_{n}}(f(z))\;\mbox{on}\;\mathbb{D}^{n}$
$\displaystyle\equiv 0\;\mbox{on}\;\mathbb{D}^{n}$

which turns out that $f\in\ ker(W_{u,v,k_{1},k_{2},\ldots,k_{n}})$. Thus, we
have

(5.3)
$\displaystyle\mathcal{P}^{n}_{m-1}(\mathbb{C})\subseteq\ker(W_{u,v,k_{1},k_{2},\ldots,k_{n}}).$

Hence, based on (5.2) and (5.3), we conclude that
$ker(W_{u,v,k_{1},k_{2},\ldots,k_{n}})=\mathcal{P}^{n}_{m-1}(\mathbb{C})$.
Furthermore, for $f\in\ ker(W^{*}_{u,v,k_{1},k_{2},\ldots,k_{n}})$, we have
$W^{*}_{u,v,k_{1},k_{2},\ldots,k_{n}}(f)=0$. Since $D_{n,\psi,\phi}$ is
complex symmetric on $H^{2}(\mathbb{D}^{n})$ with conjugation $\mathcal{J}$,
we have from (5.1) that

$\displaystyle
W_{u,v,k_{1},k_{2},\ldots,k_{n}}\mathcal{J}f=\mathcal{J}W^{*}_{u,v,k_{1},k_{2},\ldots,k_{n}}f=0.$

That is,

$\displaystyle\mathcal{J}f\in\ker(W_{u,v,k_{1},k_{2},\ldots,k_{n}})=\mathcal{P}^{n}_{m-1}(\mathbb{C}),$

equivalently, we have

$\displaystyle f\in\mathcal{P}^{n}_{m-1}(\mathbb{C}).$

Hence,

(5.4) $\displaystyle
ker(W^{*}_{u,v,k_{1},k_{2},\ldots,k_{n}})\subseteq\mathcal{P}^{n}_{m-1}(\mathbb{C}).$

Let $g\in\mathcal{P}^{n}_{m-1}(\mathbb{C})$. Then, we have
$W^{*}_{u,v,k_{1},k_{2},\ldots,k_{n}}g=\mathcal{J}W_{u,v,k_{1},k_{2},\ldots,k_{n}}\mathcal{J}g=0$
which implies that $g\in\ker(W^{*}_{u,v,k_{1},k_{2},\ldots,k_{n}})$. Hence,

(5.5) $\displaystyle\mathcal{P}^{n}_{m-1}(\mathbb{C})\subseteq
ker(W^{*}_{u,v,k_{1},k_{2},\ldots,k_{n}}).$

Consequently, from (5.4) and (5.5), we obtain
$ker(W^{*}_{u,v,k_{1},k_{2},\ldots,k_{n}})=\mathcal{P}^{n}_{m-1}(\mathbb{C})$.
Therefore,

$\displaystyle
ker(W_{u,v,k_{1},k_{2},\ldots,k_{n}})=ker(W^{*}_{u,v,k_{1},k_{2},\ldots,k_{n}})=\mathcal{P}^{n}_{m-1}(\mathbb{C}).$

This completes the proof. ∎

##  6\. Concluding remarks

In the article, we have discussed the $\mathcal{J}$-symmetric exhibited by the
partial composition-differentiation operator
$W_{u,v,k_{1},k_{2},\ldots,k_{n}}$. Our research has involved investigating
normality and Hermitian characteristics of these operators. Additionally, we
have determined the kernels of the operators
$W_{u,v,k_{1},k_{2},\ldots,k_{n}}$and their adjoints. furthermore, we have
observed that the discussion on normality can be extended to encompass
scenarios where $v_{j}(0)\neq 0(j=1,2,\ldots,n)$. Moreover, the normality and
symmetry of these operators can be discussed independently using different
conjugation methods.

##  7\. Declaration

Compliance of Ethical Standards

Conflict of interest The authors declare that there is no conflict of interest
regarding the publication of this paper.

Data availability statement Data sharing not applicable to this article as no
datasets were generated or analysed during the current study.

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