Boundedness and Compactness of Weighted Composition Operators from α-Bloch Spaces to Bers-Type Spaces on Generalized Hua Domains of the First Kind

Abstract

We address weighted composition operators $\psi C_{\varphi }$ from $\alpha$-Bloch spaces to Bers-type spaces of bounded holomorphic functions on Y, where Y is a generalized Hua domain of the first kind, and obtain some necessary and sufficient conditions for the boundedness and compactness of those operators.

1. Introduction

Let $\mathsf{\Omega }$ be a bounded domain of ${\mathbb{C}}^n$ and $H(\mathsf{\Omega })$ the class of all holomorphic functions on $\mathsf{\Omega }$. Then, consider a holomorphic self-map $\varphi$ of $\mathsf{\Omega }$ and a function $\psi \in H(\mathsf{\Omega })$. The linear operator

$$(\psi C_{\varphi }f)(z)=\psi (z)f(\varphi (z)),$$

is referred to as a weighted composition operator for $f\in H(\mathsf{\Omega })$. If $\psi (z)\equiv 1$, it reduces to the composition operator, whereas for $\varphi (z)=z$, it becomes the multiplication operator. For any given holomorphic function f, $(\psi C_{\varphi }f)(z)$ represents a generalized composition/multiplication operator. The reader is referred to book [1] for an extensive introduction to the topic.

In this paper, we study the boundedness and the compactness of weighted composition operators from $\alpha$-Bloch spaces ${\mathcal{B}}^{\alpha }$ to Bers-type spaces built on generalized Hua domains of the first kind. On ${\mathrm{GHE}}_{\mathrm{I}}$ the $\alpha$-Bloch space ${\mathcal{B}}^{\alpha }$ consists of all $f\in H({\mathrm{GHE}}_{\mathrm{I}})$, such that

$${\parallel f\parallel }_{{\mathcal{B}}^{\alpha }}:=|f(0,0)|+\underset{(Z,\xi )\in {\mathrm{GHE}}_{\mathrm{I}}}{sup}[det{(I-Z{\overline{Z}}^{\prime })}^k-{\parallel \xi \parallel }_p^2{]}^{\alpha }|\nabla f(Z,\xi )|<\infty ,$$

where

$${\displaystyle \nabla f(Z,\xi )=\left(\frac{\partial f(Z,\xi )}{\partial z_{11}},\frac{\partial f(Z,\xi )}{\partial z_{12}},\cdots ,\frac{\partial f(Z,\xi )}{\partial z_{mn}},\frac{\partial f(Z,\xi )}{\partial {\xi }_1},\cdots ,\frac{\partial f(Z,\xi )}{\partial {\xi }_r}\right).}$$

It is clear that ${\mathcal{B}}^{\alpha }({\mathrm{GHE}}_{\mathrm{I}})$ is a Banach space.

In 1930, Cartan [2] was the the first to characterize the six types of irreducible bounded symmetric domains, which consist of four types of bounded symmetric classical domains, also referred to as Cartan domains, and two exceptional domains, whose complex dimension are 16 and 27, respectively. The Cartan domains are defined as follows:

$$\begin{array}{cc}\hfill & {\Re }_{\mathrm{I}}(m,n):=\left\{Z\in {\mathbb{C}}^{m\times n}:I_m-Z{\overline{Z}}^{\prime }>0\right\},\hfill \\ \hfill & {\Re }_{\mathrm{II}}(p):=\left\{Z\in {\mathbb{C}}^{\frac{p(p+1)}{2}}:I_m-Z{\overline{Z}}^{\prime }>0,Z=Z^{{}^{\prime }}\right\},\hfill \\ \hfill & {\Re }_{\mathrm{III}}(q):=\left\{Z\in {\mathbb{C}}^{\frac{q(q-1)}{2}}:I_m-Z{\overline{Z}}^{\prime }>0,Z=-Z^{{}^{\prime }}\right\},\hfill \\ \hfill & {\Re }_{\mathrm{IV}}(n):=\left\{z\in {\mathbb{C}}^n:1+|z{z}^{{}^{\prime }}{|}^2-2z{\overline{z}}^{\prime }>0,1-{|z{z}^{{}^{\prime }}|}^2>0\right\},\hfill \end{array}$$

where $Z^{\prime }$ denotes the transpose of Z, $\overline{Z}$ denotes the conjugate of Z, and $m,n,p,q$ are positive integers. In 1998, building on the notion of bounded symmetric domains, Yin and Roos constructed a new type of domain called the Cartan–Hartogs domain [3], and Yin introduced the so-called Hua domains [4], which include the Cartan–Hartogs domains, the Cartan–Egg domains, the Hua domains, the generalized Hua domains, and the Hua construction. The generalized Hua domains are defined as follows:

$$\begin{array}{cc}\hfill & {\mathrm{GHE}}_{\mathrm{I}}(N_1,N_2,\cdots ,N_r;m,n;p_1,p_2,\cdots ,p_r;k)\hfill \\ \hfill & =\left\{{\xi }_j\in {\mathbb{C}}^{N_j},Z\in {\Re }_{\mathrm{I}}(m,n):\sum _{j=1}^r{|{\xi }_j|}^{2p_j}<det{(I-Z{\overline{Z}}^{\prime })}^k,j=1,2,\cdots ,r\right\}\hfill \\ \hfill & {\mathrm{GHE}}_{\mathrm{II}}(N_1,N_2,\cdots ,N_r;p;p_1,p_2,\cdots ,p_r;k)\hfill \\ \hfill & =\left\{{\xi }_j\in {\mathbb{C}}^{N_j},Z\in {\Re }_{\mathrm{II}}(p):\sum _{j=1}^r{|{\xi }_j|}^{2p_j}<det{(I-Z\overline{Z})}^k,j=1,2,\cdots ,r\right\}\hfill \\ \hfill & {\mathrm{GHE}}_{\mathrm{III}}(N_1,N_2,\cdots ,N_r;q;p_1,p_2,\cdots ,p_r;k)\hfill \\ \hfill & =\left\{{\xi }_j\in {\mathbb{C}}^{N_j},Z\in {\Re }_{\mathrm{III}}(q):\sum _{j=1}^r{|{\xi }_j|}^{2p_j}<det{(I+Z\overline{Z})}^k,j=1,2,\cdots ,r\right\}\hfill \\ \hfill & {\mathrm{GHE}}_{\mathrm{IV}}(N_1,N_2,\cdots ,N_r;n;p_1,p_2,\cdots ,p_r;k)\hfill \\ \hfill & =\left\{{\xi }_j\in {\mathbb{C}}^{N_j},z\in {\Re }_{\mathrm{IV}}(n):\sum _{j=1}^r|{\xi }_j{|}^{2p_j}<(1+|z{z}^{{}^{\prime }}{{|}^2-2z{\overline{z}}^{\prime })}^k,j=1,2,\cdots ,r\right\}\hfill \end{array}$$

where ${\xi }_j=({\xi }_{j1},\cdots ,{\xi }_{j{N}_j}),j=1,\cdots ,r$, ${\Re }_{\mathrm{I}}(m,n),{\Re }_{\mathrm{II}}(p),{\Re }_{\mathrm{III}}(q),{\Re }_{\mathrm{IV}}(n)$ denote, respectively, the Cartan domains of the first type, second type, third type, and fourth type, $Z^{\prime }$ denotes the transpose of Z, $\overline{Z}$ denotes the conjugate of Z, $N_1,\cdots ,N_r,m,n,p,q$ are positive integers, and $p_1,\cdots ,p_r$ are positive real numbers. For $k=1,m=1,p_1=\cdots =p_r=1$, the generalized Hua domain of the first kind reduces to the unit ball. Without loss of generality, we may assume that $N_j=1$, then ${\xi }_j\in \mathbb{C},j=1,\cdots ,r,\xi =({\xi }_1,\cdots ,{\xi }_r)$ and ${\parallel \xi \parallel }_p^2={\sum }_{j=1}^r{|{\xi }_j|}^{2p_j}$. We define

$${\langle \xi ,t\rangle }_p={\langle {\xi }_1,t_1\rangle }^{p_1}+{\langle {\xi }_2,t_2\rangle }^{p_2}+\cdots +{\langle {\xi }_r,t_r\rangle }^{p_r}.$$

We also write

$$\begin{array}{cc}\hfill {|\langle \xi ,t\rangle }_p|& \le |{\langle {\xi }_1,t_1\rangle }^{p_1}|+|{\langle {\xi }_2,t_2\rangle }^{p_2}|+\cdots +|{\langle {\xi }_r,t_r\rangle }^{p_r}|\hfill \\ \hfill & \le |{\xi }_1{|}^{p_1}|t_1{|}^{p_1}+\cdots +|{\xi }_r{|}^{p_r}{|t_r|}^{p_r}\hfill \\ \hfill & =|\langle \alpha ,\beta \rangle |\le |\alpha ||\beta |={\parallel \xi \parallel }_p{\parallel t\parallel }_p,\hfill \end{array}$$

where $|{\xi }_i{|}^{p_i}={\alpha }_i,{|t_i|}^{p_i}={\beta }_i(i=1,\cdots ,r),\alpha =({\alpha }_1,\cdots ,{\alpha }_r),\beta =({\beta }_1,\cdots ,{\beta }_r).$

For the sake of convenience, the four types of generalized Hua domains will be referred to as ${\mathrm{GHE}}_{\mathrm{I}},{\mathrm{GHE}}_{\mathrm{II}},{\mathrm{GHE}}_{\mathrm{III}}$, and ${\mathrm{GHE}}_{\mathrm{IV}}$.

On ${\mathrm{GHE}}_{\mathrm{I}}$, a Bers-type space ${\mathcal{A}}_{\beta }$ consists of all $f\in H({\mathrm{GHE}}_{\mathrm{I}})$, such that

$${\parallel f\parallel }_{{\mathcal{A}}_{\beta }}:=\underset{(Z,\xi )\in {\mathrm{GHE}}_{\mathrm{I}}}{sup}[det{(I-Z{\overline{Z}}^{\prime })}^k-{\parallel \xi \parallel }_p^2{]}^{\beta }|f(Z,\xi )|<\infty .$$

It is easy to see that ${\mathcal{A}}_{\beta }({\mathrm{GHE}}_{\mathrm{I}})$ is a Banach space with norm $\parallel ·\parallel$.

The boundedness and the compactness of weighted composition operators on (or between) spaces of holomorphic functions on various domains have received considerable attention. Wang and Liu [5] studied the boundedness and the compactness of the weighted composition operators on the Bers-type space on the open unit disc, whereas Zhou and Xu [6] characterized the boundedness and the compactness of the weighted composition operators between $\alpha$-Bloch space and $\beta$-Bloch space, Li [7] investigated the boundedness and the compactness of the weighted composition operators from Hardy space to Bers-type space, and Zhu [8] characterized the boundedness and compactness of $D_{\varphi ,u}^n:\mathcal{B}\to H_{\alpha }^{\infty }$. For the unit poly-disk, Li and Stević [9,10] presented some necessary and sufficient conditions for the boundedness and the compactness of the weighted composition operators between $H^{\infty }$ and $\alpha$-Bloch space, whereas for the open unit ball, Li and Stević [11] studied the boundedness and the compactness of the weighted composition operators between $H^{\infty }$ and Bloch space (see also [12,13,14,15]).

The boundness and compactness of weighted composition have wide applications in differential equations, functional analysis, numerical mathematics, and control theory. For example, in differential equations, the compactness of the operator plays a vital role in proving the global existence of weak/strong solutions of fluid mechanics, see for example, the well-known Aubin–Lions argument [16]; in functional analysis, the compactness of the operator is crucial for the existence of critical points in studying the existence and multiplicity of periodic solutions of nonlinear Dirac equations [17]; in numerical mathematics, the boundness and compactness of the operator are applied in an implicit robust numerical scheme with graded meshes for the modified Burgers model with nonlocal dynamic properties [18], a space-time spectral order sinc-collocation method for the fourth-order nonlocal heat model arising in viscoelasticity [19], and a high-order and efficient numerical technique for the nonlocal neutron diffusion equation representing neutron transport in a nuclear reactor [20].

Jiang [21] has characterized the boundedness and the compactness of the weighted composition operators on the Bers-type space on the Hua domains. Yet, the boundedness and the compactness of the weighted composition operators from $\alpha$-Bloch to ${\mathcal{A}}_{\beta }$ have not been studied in detail. In this paper, we obtain some necessary and sufficient conditions for the boundedness and the compactness of the weighted composition operators from $\alpha$-Bloch to ${\mathcal{A}}_{\beta }$ on generalized Hua domain of the first kind by using a generalization of Hua’s inequalities.

2. Preliminaries

Lemma 1.

Let $\beta >0$, then

$$|f(Z,\xi )|\le \frac{{\parallel f\parallel }_{{\mathcal{A}}_{\beta }}}{[det{(I-Z{\overline{Z}}^{\prime })}^k-{\parallel \xi \parallel }_p^2{]}^{\beta }},$$

(1)

for all $(Z,\xi )\in {\mathrm{GHE}}_{\mathrm{I}}$ and $f\in {\mathcal{A}}_{\beta }({\mathrm{GHE}}_{\mathrm{I}})$.

Proof. 

By the very definition of Bers-type space ${\mathcal{A}}_{\beta }$, we know that

$${\parallel f\parallel }_{{\mathcal{A}}_{\beta }}=\underset{(Z,\xi )\in {\mathrm{GHE}}_{\mathrm{I}}}{sup}[det{(I-Z{\overline{Z}}^{\prime })}^k-{\parallel \xi \parallel }_p^2{]}^{\beta }|f(Z,\xi )|<\infty ,$$

and so,

$$\begin{array}{c}\hfill |f(Z,\xi )|\le \frac{{\parallel f\parallel }_{{\mathcal{A}}_{\beta }}}{[det{(I-Z{\overline{Z}}^{\prime })}^k-{\parallel \xi \parallel }_p^2{]}^{\beta }}.\end{array}$$

□

Lemma 2.

Let $0<a\le 1,0<b\le 1$, and $b\le a$, with q as a positive integer, then

$$a-b\le q(a^{\frac{1}{q}}-b^{\frac{1}{q}}).$$

(2)

Proof. 

$$\begin{array}{cc}\hfill a-b& ={(a^{\frac{1}{q}})}^q-{(b^{\frac{1}{q}})}^q\hfill \\ \hfill & =(a^{\frac{1}{q}}-b^{\frac{1}{q}})(a^{\frac{1}{q}\times (q-1)}+a^{\frac{1}{q}\times (q-2)}{b}^{\frac{1}{q}}+\cdots +b^{\frac{1}{q}\times (q-1)})\hfill \\ \hfill & \le q(a^{\frac{1}{q}}-b^{\frac{1}{q}}).\hfill \end{array}$$

□

Lemma 3

(see [22]). Let $x\ge -1$, if $0<\alpha \le 1$, then

$${(1+x)}^{\alpha }\le 1+\alpha x,$$

(3)

if $\alpha <0$ or $\alpha >1$, then

$${(1+x)}^{\alpha }\ge 1+\alpha x,$$

(4)

and “=” holds if and only if $x=0$ or $\alpha =1$.

Lemma 4

(see [22]). Let $a_k\ge 0,k=1,2,\cdots ,m$, then

$${(a_1·a_2\cdots a_m)}^{\frac{1}{m}}\le \frac{a_1+a_2+\cdots +a_m}{m},$$

(5)

where the equality holds if and only if $a_1=a_2=\cdots =a_m$.

Lemma 5

(see [22]). Let $a_k\in \mathbb{C}$, if $p\ge 1$, then

$$\sum _{k=1}^n|a_k{|}^p\le {\left[\sum _{k=1}^n|a_k|\right]}^p\le n^{p-1}\sum _{k=1}^n{|a_k|}^p.$$

(6)

If $0<p<1$, then

$$\sum _{k=1}^n|a_k{|}^p\ge {\left[\sum _{k=1}^n|a_k|\right]}^p\ge n^{p-1}\sum _{k=1}^n{|a_k|}^p,$$

(7)

where the equality holds if and only if $p>1$, then $|a_1|=\cdots =|a_n|$. If $p=1$, the equality always holds. If $0<p<1$, then at most one of the $a_1,\cdots ,a_n$ is not zero.

Lemma 6

(see [23]). Let

$$Z=\left(\begin{array}{cccc}{z}_{11}& z_{12}& \dots & z_{1n}\\ z_{21}& z_{22}& \dots & z_{2n}\\ ⋮& ⋮& \ddots & ⋮\\ z_{m1}& z_{m2}& \dots & z_{mn}\end{array}\right)$$

be an $m\times n$ matrix $(m\le n)$. Then, there exists an $m\times m$ unitary matrix U and an $n\times n$ unitary matrix V, such that

$$Z=U\left(\begin{array}{ccccccc}{\lambda }_1& 0& \dots & 0& 0& \dots & 0\\ 0& {\lambda }_2& \dots & 0& 0& \dots & 0\\ ⋮& ⋮& & ⋮& ⋮& & ⋮\\ 0& 0& \dots & {\lambda }_m& 0& \dots & 0\end{array}\right)V({\lambda }_1\ge {\lambda }_2\ge \cdots \ge {\lambda }_m\ge 0),$$

and

$$Z{\overline{Z}}^{\prime }=U\left(\begin{array}{cccc}{{\lambda }_1}^2& 0& \dots & 0\\ 0& {{\lambda }_2}^2& \dots & 0\\ ⋮& ⋮& \ddots & ⋮\\ 0& 0& \dots & {{\lambda }_m}^2\end{array}\right){\overline{U}}^{\prime },$$

where ${{\lambda }_1}^2,\cdots ,{{\lambda }_m}^2$ are the characteristic values of $Z{\overline{Z}}^{\prime }$. $I-Z{\overline{Z}}^{\prime }>0⟺{\lambda }_1<1$.

Lemma 7

(see [23]). Let

$${\mathsf{\Lambda }}_1=\left(\begin{array}{cccc}{\lambda }_1& 0& \dots & 0\\ 0& {\lambda }_2& \dots & 0\\ ⋮& ⋮& \ddots & 0\\ 0& 0& \dots & {\lambda }_m\end{array}\right)({\lambda }_1\ge {\lambda }_2\ge \cdots \ge {\lambda }_m\ge 0),$$

$${\mathsf{\Lambda }}_2=\left(\begin{array}{cccc}{\mu }_1& 0& \dots & 0\\ 0& {\mu }_2& \dots & 0\\ ⋮& ⋮& \ddots & 0\\ 0& 0& \dots & {\mu }_m\end{array}\right)({\mu }_1\ge {\mu }_2\ge \cdots \ge {\mu }_m\ge 0),$$

satisfying

$${\lambda }_j{\mu }_k<1(j,k=1,\cdots ,m).$$

Then, there exists a square matrix P, such that

$$\underset{U{\overline{U}}^{\prime }=I,V{\overline{V}}^{\prime }=I}{inf}|det(I-{\mathsf{\Lambda }}_{1}U{\mathsf{\Lambda }}_2{\overline{U}}^{\prime }V)|=|det(I-{\mathsf{\Lambda }}_{1}P{\mathsf{\Lambda }}_2{P}^{\prime })|,$$

and the minimum value is obtained for $U=\mathsf{\Theta }P$ and $V=I$, where

$$\mathsf{\Theta }=\left(\begin{array}{cccc}{e}^{i{\theta }_1}& 0& \dots & 0\\ 0& e^{i{\theta }_2}& \dots & 0\\ ⋮& ⋮& \ddots & 0\\ 0& 0& \dots & e^{i{\theta }_m}\end{array}\right).$$

Lemma 8

(see [22], the Minkowski inequality of integration formula). Let $a_k,b_k\ge 0,k=1,2\cdots ,n$, then

$${\left[\prod _{k=1}^n(a_k+b_k)\right]}^{\frac{1}{n}}\ge {\left(\prod _{k=1}^n{a}_k\right)}^{\frac{1}{n}}+{\left(\prod _{k=1}^n{b}_k\right)}^{\frac{1}{n}},$$

(8)

where the equal sign holds if and only if $a_k=c{b}_k$, $k=1,2,\cdots ,n$.

Lemma 9.

Let $p_i$ $(i=1,2,\cdots ,r)$ be positive integers, $0<km⩽1$, and $t\in [0,1]$, then

$$1-det{(I-t^{2}Z{\overline{Z}}^{\prime })}^k+{\parallel t\xi \parallel }_p^2\le t^2\left[1-det{(I-Z{\overline{Z}}^{\prime })}^k+{\parallel \xi \parallel }_p^2\right],$$

for $(Z,\xi )\in {\mathrm{GHE}}_{\mathrm{I}}$.

Proof. 

Decomposition in polar coordinates gives

$$det{(I-tZ{\overline{Z}}^{\prime })}^k=\prod _{i=1}^m{(1-t{\lambda }_i^2)}^k.$$

Given ${\lambda }_i^2={ħ}_i,i=1,2,\cdots ,m$, we may consider the function

$$f(t)=\prod _{i=1}^m{(1-t{ħ}_i)}^k,t\in [0,1]$$

$$lnf(t)=k\sum _{i=1}^{m}ln(1-t{ħ}_i).$$

Upon differentiating with respect to t, we obtain

$$\begin{array}{cc}\hfill f^{\prime }(t)& =f(t)k\sum _{i=1}^m\frac{-{ħ}_i}{1-t{ħ}_i}\le 0,\hfill \\ \hfill f^{″}(t)& =f^{\prime }(t)k\sum _{i=1}^m\frac{-{ħ}_i}{1-t{ħ}_i}-f(t)k\sum _{i=1}^m\frac{{ħ}_i^2}{{(1-t{ħ}_i)}^2}\hfill \\ \hfill & =f(t)k^2{\left(\sum _{i=1}^m\frac{{ħ}_i}{1-t{ħ}_i}\right)}^2-f(t)k\sum _{i=1}^m\frac{{ħ}_i^2}{{(1-t{ħ}_i)}^2}\hfill \\ \hfill & =f(t)k\left[k{\left(\sum _{i=1}^m\frac{{ħ}_i}{1-t{ħ}_i}\right)}^2-\sum _{i=1}^m\frac{{ħ}_i^2}{{(1-t{ħ}_i)}^2}\right].\hfill \end{array}$$

An application of (6) then gives

$$\begin{array}{cc}\hfill f^{″}(t)& =f(t)k\left[k{\left(\sum _{i=1}^m\frac{{ħ}_i}{1-t{ħ}_i}\right)}^2-\sum _{i=1}^m\frac{{ħ}_i^2}{{(1-t{ħ}_i)}^2}\right]\hfill \\ \hfill & \le f(t)k\left[(km-1)\sum _{i=1}^m\frac{{ħ}_i^2}{{(1-t{ħ}_i)}^2}\right]\hfill \\ \hfill & \le 0.\hfill \end{array}$$

This shows that $f(t)$ is a concave function. It follows that

$$g(t)=1-f(t)=1-\prod _{i=1}^m{(1-t{ħ}_i)}^k,t\in [0,1],$$

is a convex function, and we have

$$1-\prod _{i=1}^m{(1-t{ħ}_i)}^k\le t[1-\prod _{i=1}^m{(1-{ħ}_i)}^k].$$

(9)

The very definition of ${\parallel \xi \parallel }_p^2$ shows that

$$\begin{array}{cc}\hfill {\parallel t\xi \parallel }_p^2& =|t{\xi }_1{|}^{2p_1}+|t{\xi }_2{|}^{2p_2}+\cdots +{|t{\xi }_r|}^{2p_r}\hfill \\ \hfill & \le t^2(|{\xi }_1{|}^{2p_1}+|{\xi }_2{|}^{2p_2}+\cdots +|{\xi }_r{|}^{2p_r})\hfill \\ \hfill & =t^2{\parallel \xi \parallel }_p^2.\hfill \end{array}$$

(10)

Hence, by inequalities (9) and (10), we obtain

$$1-det{(I-t^{2}Z{\overline{Z}}^{\prime })}^k+{\parallel t\xi \parallel }_p^2\le t^2\left[1-det{(I-Z{\overline{Z}}^{\prime })}^k+{\parallel \xi \parallel }_p^2\right].$$

□

Lemma 10.

Let us consider $0<mk\le 1$, some positive intergers $p_j$$(j=1,2,\cdots ,r)$, $t\in [0,1],(Z,\xi )\in {\mathrm{GHE}}_{\mathrm{I}}$, $q=\max \{p_1,p_2,\cdots ,p_r\}$. Then, the following inequality holds

$$|\overline{(Z,\xi )}|\le M\sqrt{1-det{(I-Z{\overline{Z}}^{\prime })}^{\frac{k}{q}}+{\parallel \xi \parallel }_p^{\frac{2}{q}}},$$

where $M=\max \{\sqrt{\frac{q}{k}},\sqrt{r^{1-\frac{1}{q}}}\}$.

Proof. 

If $t\in [0,1]$,$\forall (Z,\xi )\in {\mathrm{GHE}}_{\mathrm{I}}$, then $(tZ,t\xi )\in {\mathrm{GHE}}_{\mathrm{I}}$, ${|Z|}^2=\mathrm{tr}(Z{\overline{Z}}^{\prime })={\lambda }_1^2+{\lambda }_2^2+\cdots +{\lambda }_m^2$. By Lemma 4 and (3), we obtain

$$\begin{array}{cc}\hfill det{(I-Z{\overline{Z}}^{\prime })}^{\frac{k}{q}}& =\prod _{i=1}^m{(1-{\lambda }_i^2)}^{\frac{k}{q}}={\left\{\prod _{i=1}^m{(1-{\lambda }_i^2)}^{\frac{1}{m}}\right\}}^{\frac{mk}{q}}\hfill \\ \hfill & \le {\left\{\frac{1}{m}(m-\sum _{i=1}^m{\lambda }_i^2)\right\}}^{\frac{mk}{q}}\hfill \\ \hfill & ={(1-\frac{1}{m}\sum _{i=1}^m{\lambda }_i^2)}^{\frac{mk}{q}}\hfill \\ \hfill & \le 1-\frac{mk}{q}·\frac{1}{m}{|Z|}^2\hfill \\ \hfill & =1-\frac{k}{q}{|Z|}^2.\hfill \end{array}$$

Then,

$${|Z|}^2\le \frac{q}{k}[1-det{(I-Z{\overline{Z}}^{\prime })}^{\frac{k}{q}}].$$

(11)

Using (7), one has

$$\begin{array}{cc}\hfill {\parallel \xi \parallel }_p^{\frac{2}{q}}& =(|{\xi }_1{|}^{2p_1}+|{\xi }_2{|}^{2p_2}+\cdots +|{\xi }_r{{|}^{2p_r})}^{\frac{1}{q}}\hfill \\ \hfill & \ge r^{\frac{1}{q}-1}(|{\xi }_1{|}^{\frac{2p_1}{q}}+|{\xi }_2{|}^{\frac{2p_2}{q}}+\cdots +|{\xi }_r{|}^{\frac{2p_r}{q}})\hfill \\ \hfill & \ge r^{\frac{1}{q}-1}(|{\xi }_1{|}^2+|{\xi }_2{|}^2+\cdots +|{\xi }_r{|}^2)\hfill \\ \hfill & =r^{\frac{1}{q}-1}{|\xi |}^2.\hfill \end{array}$$

Then,

$${|\xi |}^2\le r^{1-\frac{1}{q}}{\parallel \xi \parallel }_p^{\frac{2}{q}}.$$

(12)

Therefore, by combining (11) and (12), we have

$$\begin{array}{cc}\hfill |\overline{(Z,\xi )}|& =\sqrt{{|Z|}^2+{|\xi |}^2}\hfill \\ \hfill & \le \sqrt{\frac{q}{k}[1-det{(I-Z{\overline{Z}}^{\prime })}^{\frac{k}{q}}]+r^{1-\frac{1}{q}}{\parallel \xi \parallel }_p^{\frac{2}{q}}}\hfill \\ \hfill & \le M\sqrt{1-det{(I-Z{\overline{Z}}^{\prime })}^{\frac{k}{q}}+{\parallel \xi \parallel }_p^{\frac{2}{q}}},\hfill \end{array}$$

(13)

where $M=\max \{\sqrt{\frac{q}{k}},\sqrt{r^{1-\frac{1}{q}}}\}$. □

Lemma 11.

Given $0<km\le 1$, $p_j$ some positive integers $(j=1,2,\cdots ,r)$, $\forall (Z,\xi )\in {\mathrm{GHE}}_{\mathrm{I}}$, $q=\max \{p_1,p_2,\cdots ,p_r\}$, and f a holomorphic function on ${\mathcal{B}}^{\alpha }$$({\mathrm{GHE}}_{\mathrm{I}})$, then there exists a constant C, such that

$$|f(Z,\xi )|\le \left\{\begin{array}{ccc}\hfill & {C\parallel f\parallel }_{{\mathcal{B}}^{\alpha }}\hfill & \hfill 0<\alpha <1\\ \hfill & {C\parallel f\parallel }_{{\mathcal{B}}^{\alpha }}ln\frac{2q}{det{(I-Z{\overline{Z}}^{\prime })}^k-{\parallel \xi \parallel }_p^2}\hfill & \hfill \alpha =1\\ \hfill & {C\parallel f\parallel }_{{\mathcal{B}}^{\alpha }}\frac{1}{[det{(I-Z{\overline{Z}}^{\prime })}^k-{\parallel \xi \parallel }_p^2{]}^{\alpha -1}}\hfill & \hfill \alpha >1\end{array}\right.$$

(14)

where $(Z,\xi )=(z_{11},z_{12},\cdots ,z_{mn},{\xi }_1,{\xi }_2,\cdots ,{\xi }_r)$.

Proof. 

According to Lemmas 2 and 9 and (13),

$$\begin{array}{cc}\hfill |f(Z,\xi )|& =|f(0,0)+{\int }_0^1\langle \nabla f(tZ,t\xi ),\overline{(Z,\xi )}\rangle \mathrm{d}t|\hfill \\ \hfill & \le |f(0,0)|+{\int }_0^1|\nabla f(tZ,t\xi )||\overline{(Z,\xi )}|\mathrm{d}t\hfill \\ \hfill & =|f(0,0)|+|\overline{(Z,\xi )}|{\int }_0^1\frac{[det{(I-t^{2}Z{\overline{Z}}^{\prime })}^k-{\parallel t\xi \parallel }_p^2{]}^{\alpha }|\nabla f(tZ,t\xi )|}{[det{(I-t^{2}Z{\overline{Z}}^{\prime })}^k-{\parallel t\xi \parallel }_p^2{]}^{\alpha }}\mathrm{d}t\hfill \\ \hfill & \le \left[1+{\int }_0^1\frac{|\overline{(Z,\xi )}|}{[det{(I-t^{2}Z{\overline{Z}}^{\prime })}^k-{\parallel t\xi \parallel }_p^2{]}^{\alpha }}\mathrm{d}t\right]{\parallel f\parallel }_{{\mathcal{B}}^{\alpha }}\hfill \\ \hfill & =\left[1+{\int }_0^1\frac{|\overline{(Z,\xi )}|}{[1-(1-det{(I-t^{2}Z{\overline{Z}}^{\prime })}^k+{\parallel t\xi \parallel }_p^2{)]}^{\alpha }}\mathrm{d}t\right]{\parallel f\parallel }_{{\mathcal{B}}^{\alpha }}\hfill \\ \hfill & \le \left[1+M{\int }_0^1\frac{\sqrt{1-det{(I-Z{\overline{Z}}^{\prime })}^{\frac{k}{q}}+{\parallel \xi \parallel }_p^{\frac{2}{q}}}}{[1-t^2(1-det{(I-Z{\overline{Z}}^{\prime })}^k+{\parallel \xi \parallel }_p^2{)]}^{\alpha }}\mathrm{d}t\right]{\parallel f\parallel }_{{\mathcal{B}}^{\alpha }}\hfill \\ \hfill & \le \left[1+M{\int }_0^1\frac{\sqrt{1-\frac{1}{q}(det{(I-Z{\overline{Z}}^{\prime })}^k-{\parallel \xi \parallel }_p^2)}}{[1-t^2(1-\frac{1}{q}(det{(I-Z{\overline{Z}}^{\prime })}^k-{\parallel \xi \parallel }_p^2{))]}^{\alpha }}\mathrm{d}t\right]{\parallel f\parallel }_{{\mathcal{B}}^{\alpha }}\hfill \\ \hfill & =\left[1+M{\int }_0^1\frac{\Im }{{[1-t^2{\Im }^2]}^{\alpha }}\mathrm{d}t\right]{\parallel f\parallel }_{{\mathcal{B}}^{\alpha }}\hfill \\ \hfill & =\left[1+M{\int }_0^1\frac{\Im }{{[(1-t\Im )(1+t\Im )]}^{\alpha }}\mathrm{d}t\right]{\parallel f\parallel }_{{\mathcal{B}}^{\alpha }}\hfill \\ \hfill & \le \left[1+M{\int }_0^1\frac{\Im }{{(1-t\Im )}^{\alpha }}\mathrm{d}t\right]{\parallel f\parallel }_{{\mathcal{B}}^{\alpha }},\hfill \end{array}$$

where $\Im =\sqrt{1-\frac{1}{q}(det{(I-Z{\overline{Z}}^{\prime })}^k-{\parallel \xi \parallel }_p^2)}$.

Case 1: $0<\alpha <1$,

$$\begin{array}{cc}\hfill |f(Z,\xi )|& \le \left\{1+\frac{M}{1-\alpha }[1-{(1-\Im )}^{1-\alpha }]\right\}{\parallel f\parallel }_{{\mathcal{B}}^{\alpha }}\hfill \\ \hfill & \le (1+\frac{M}{1-\alpha }){\parallel f\parallel }_{{\mathcal{B}}^{\alpha }}\hfill \\ \hfill & \le {C\parallel f\parallel }_{{\mathcal{B}}^{\alpha }},\hfill \end{array}$$

(15)

where $C=1+\frac{M}{1-\alpha }$.

Case 2: $\alpha =1$,

$$\begin{array}{cc}\hfill |f(Z,\xi )|& \le \left[1+M{\int }_0^1\frac{\Im }{1-t\Im }\mathrm{d}t\right]{\parallel f\parallel }_{{\mathcal{B}}^{\alpha }}\hfill \\ \hfill & =\left[1+Mln\frac{1}{1-\Im }\right]{\parallel f\parallel }_{{\mathcal{B}}^{\alpha }}\hfill \\ \hfill & =\left[1+Mln\frac{1+\Im }{(1-\Im )(1+\Im )}\right]{\parallel f\parallel }_{{\mathcal{B}}^{\alpha }}\hfill \end{array}$$

(16)

$$\begin{array}{cc}\hfill & \le \left[1+Mln\frac{2}{1-{\Im }^2}\right]{\parallel f\parallel }_{{\mathcal{B}}^{\alpha }}\hfill \\ & \le \left[\frac{1}{ln2}ln\frac{2}{1-{\Im }^2}+Mln\frac{2}{1-{\Im }^2}\right]{\parallel f\parallel }_{{\mathcal{B}}^{\alpha }}\hfill \\ \hfill & \le \left[\frac{1}{ln2}+M\right]ln\frac{2}{1-{\Im }^2}{\parallel f\parallel }_{{\mathcal{B}}^{\alpha }}\hfill \\ \hfill & ={C\parallel f\parallel }_{{\mathcal{B}}^{\alpha }}ln\frac{2q}{det{(I-Z{\overline{Z}}^{\prime })}^k-{\parallel \xi \parallel }_p^2},\hfill \end{array}$$

where $C=\frac{1}{ln2}+M$.

Case 3: $\alpha >1$,

$$\begin{array}{cc}\hfill |f(Z,\xi )|& \le \left[1+\frac{M}{\alpha -1}\left(\frac{1}{{(1-\Im )}^{\alpha -1}}-1\right)\right]{\parallel f\parallel }_{{\mathcal{B}}^{\alpha }}\hfill \\ \hfill & \le \left[C^{\prime }+C^{\prime }\left(\frac{1}{{(1-\Im )}^{\alpha -1}}-1\right)\right]{\parallel f\parallel }_{{\mathcal{B}}^{\alpha }}\hfill \\ \hfill & =C^{\prime }{\parallel f\parallel }_{{\mathcal{B}}^{\alpha }}\frac{1}{{(1-\Im )}^{\alpha -1}}\hfill \\ \hfill & =C^{\prime }{\parallel f\parallel }_{{\mathcal{B}}^{\alpha }}\frac{{(1+\Im )}^{\alpha -1}}{{[(1-\Im )(1+\Im )]}^{\alpha -1}}\hfill \\ \hfill & \le 2^{\alpha -1}{C}^{\prime }{\parallel f\parallel }_{{\mathcal{B}}^{\alpha }}\frac{1}{{(1-{\Im }^2)}^{\alpha -1}}\hfill \\ \hfill & ={C\parallel f\parallel }_{{\mathcal{B}}^{\alpha }}\frac{1}{[det{(I-Z{\overline{Z}}^{\prime })}^k-{\parallel \xi \parallel }_p^2{]}^{\alpha -1}}\hfill \end{array}$$

(17)

where $C={(2q)}^{\alpha -1}{C}^{\prime }$, $C^{\prime }=max\{1,\frac{M}{\alpha -1}\}.$

By combining (15)–(17), the proof of the Lemma is complete. □

Lemma 12.

Let $\varphi =({\varphi }_{11},{\varphi }_{12}\dots {\varphi }_{mn+r})$ be a holomorphic self-map of ${\mathrm{GHE}}_{\mathrm{I}}$ and $\psi \in H({\mathrm{GHE}}_{\mathrm{I}})$. The weighted composition operator $\psi C_{\varphi }:{\mathcal{B}}^{\alpha }({\mathrm{GHE}}_{\mathrm{I}})\to {\mathcal{A}}_{\beta }({\mathrm{GHE}}_{\mathrm{I}})$ is compact if and only if $\psi C_{\varphi }$ is bounded and for any bounded sequence ${\{f_n\}}_{n\ge 1}$ in ${\mathcal{B}}^{\alpha }({\mathrm{GHE}}_{\mathrm{I}})$ converging to 0 uniformly on compact subsets of ${\mathrm{GHE}}_{\mathrm{I}}$, $\parallel \psi C_{\varphi }{f}_n{\parallel }_{{\mathcal{A}}_{\beta }}\to 0$ as $n\to \infty$.

Proof. 

Assume that $\psi C_{\varphi }:{\mathcal{B}}^{\alpha }({\mathrm{GHE}}_{\mathrm{I}})\to {\mathcal{A}}_{\beta }({\mathrm{GHE}}_{\mathrm{I}})$ is compact. Let ${\{f_n\}}_{n\ge 1}$ be a bounded sequence in ${\mathcal{B}}^{\alpha }({\mathrm{GHE}}_{\mathrm{I}})$ and $f_n\to 0$ uniformly on compact subsets of ${\mathrm{GHE}}_{\mathrm{I}}$ as $n\to \infty$.

If $\parallel \psi C_{\varphi }{f}_n{\parallel }_{{\mathcal{A}}_{\beta }}\nrightarrow 0$ as $n\to \infty$, then there exists a subsequence ${\{f_{n_j}\}}_{j\ge 1}$ of ${\{f_n\}}_{n\ge 1}$, such that

$$\underset{j\in N}{inf}{\parallel \psi C_{\varphi }{f}_{n_j}\parallel }_{{\mathcal{A}}_{\beta }}>0.$$

Since $\psi C_{\varphi }$ is compact, there exists a subsequence of the bounded sequence ${\{f_{n_j}\}}_{j\ge 1}$ (without loss of generality, we still write ${\{f_{n_j}\}}_{j\ge 1}$), such that

$$\underset{j\to \infty }{lim}{\parallel \psi C_{\varphi }{f}_{n_j}-f\parallel }_{{\mathcal{A}}_{\beta }}=0,f\in {\mathcal{A}}_{\beta }({\mathrm{GHE}}_{\mathrm{I}}).$$

Let K be a compact subspace of ${\mathrm{GHE}}_{\mathrm{I}}$. From Lemma 1, it follows that

$$|(\psi C_{\varphi }{f}_{n_j}-f)(Z,\xi )|\le \frac{\parallel \psi C_{\varphi }{f}_{n_j}{-f\parallel }_{{\mathcal{A}}_{\beta }}}{[det{(I-Z{\overline{Z}}^{\prime })}^k-{\parallel \xi \parallel }_p^2{]}^{\beta }},j\to \infty$$

for $\forall (Z,\xi )\in K\subset {\mathrm{GHE}}_{\mathrm{I}}.$ Thus, $\psi C_{\varphi }{f}_{n_j}-f\to 0$ uniformly on K. This means that for arbitrary $\epsilon >0,\exists N_1>0$, such that for $j>N_1$, we have

$$|\psi (Z,\xi )f_{n_j}(\varphi (Z,\xi ))-f(Z,\xi )|<\epsilon ,$$

for all $(Z,\xi )\in K$. Since $f_{n_j}\rightrightarrows 0$ on compact subsets of ${\mathrm{GHE}}_{\mathrm{I}}$ as $j\to \infty$, there also exists a positive integer $N_2$, $|f_{n_j}(\varphi (Z,\xi ))|<\epsilon$ for $(Z,\xi )\in K$ whenever $j>N_2$. Let $N=max\{N_1,N_2\}$ and $M={max}_{(Z,\xi )\in K}|\psi (Z,\xi )|$, whenever $j>N$, we have

$$|f(Z,\xi )|⩽|f_{n_j}(\varphi (Z,\xi ))|\underset{(Z,\xi )\in K}{max}|\psi (Z,\xi )|+\epsilon ⩽(M+1)\epsilon ,\forall (Z,\xi )\in K.$$

From the arbitrariness of $\epsilon$, we obtain $f(Z,\xi )\equiv 0$, $\forall (Z,\xi )\in K$. By the uniqueness theorem of analytic functions, we have $f(Z,\xi )\equiv 0,\forall (Z,\xi )\in {\mathrm{GHE}}_{\mathrm{I}}$. This shows that ${lim}_{j\to \infty }{\parallel \psi C_{\varphi }{f}_{n_j}\parallel }_{{\mathcal{A}}_{\beta }}=0$, which contradicts the assumption ${inf}_{j\in N}{\parallel \psi C_{\varphi }{f}_{n_j}\parallel }_{{\mathcal{A}}_{\beta }}>0$.

Conversely, suppose that ${\{f_n\}}_{n\ge 1}$ is a bounded sequence in ${\mathcal{B}}^{\alpha }({\mathrm{GHE}}_{\mathrm{I}})$, then $\parallel f_n{\parallel }_{{\mathcal{B}}^{\alpha }}\le D_1$ for all n. Clearly ${\{f_n\}}_{n\ge 1}$ is uniformly bounded on compact subsets of ${\mathrm{GHE}}_{\mathrm{I}}$. By Montel’s theorem, there exists a subsequence ${\{f_{n_j}\}}_{j\ge 1}$ of ${\{f_n\}}_{n\ge 1}$, such that $f_{n_j}\to f$ uniformly on every compact subset of ${\mathrm{GHE}}_{\mathrm{I}}$ and $f\in {\mathcal{B}}^{\alpha }({\mathrm{GHE}}_{\mathrm{I}})$. For all $(Z_0,{\xi }_0)\in {\mathrm{GHE}}_{\mathrm{I}}$, there exists a compact set $K_{(Z_0,{\xi }_0)}$, such that $(Z_0,{\xi }_0)\in K_{(Z_0,{\xi }_0)}\subset {\mathrm{GHE}}_{\mathrm{I}}$. By Weierstrass’ theorem and because $f_{n_j}\rightrightarrows f$ as $j\to \infty$, for $(Z,\xi )\in K_{(Z_0,{\xi }_0)}$, we obtain $\nabla f_{n_j}\rightrightarrows \nabla f$ as $j\to \infty$. Then, there exists a $J_0>0$, such that for $j>J_0$, we have $|\nabla f_{n_j}(Z,\xi )-\nabla f(Z,\xi )|<1$ for $(Z,\xi )\in K_{(Z_0,{\xi }_0)}$. In addition, $|\nabla f(Z_0,{\xi }_0)|\le |\nabla f(Z_0,{\xi }_0)-\nabla f_{n_j}(Z_0,{\xi }_0)|+|\nabla f_{n_j}(Z_0,{\xi }_0)|$, which suffices to obtain

$$\begin{array}{cc}\hfill & {[det{(I-Z_0{\overline{Z_0}}^{\prime })}^k-{\parallel {\xi }_0{\parallel }_p}^2]}^{\alpha }|\nabla f(Z_0,{\xi }_0)|\hfill \\ \hfill & \le [det{(I-Z_0{\overline{Z_0}}^{\prime })}^k-\parallel {\xi }_0{\parallel }_p^2{]}^{\alpha }|\nabla f(Z_0,{\xi }_0)-\nabla f_{n_j}(Z_0,{\xi }_0)|\hfill \\ \hfill & +[det{(I-Z_0{\overline{Z_0}}^{\prime })}^k-\parallel {\xi }_0{\parallel }_p^2{]}^{\alpha }|\nabla f_{n_j}(Z_0,{\xi }_0)|\hfill \\ \hfill & \le 1+\parallel f_{n_j}{\parallel }_{{\mathcal{B}}^{\alpha }}\hfill \\ \hfill & \le 1+D_1.\hfill \end{array}$$

For all $(Z,\xi )\in {\mathrm{GHE}}_{\mathrm{I}}$, $[det{(I-Z{\overline{Z}}^{\prime })}^k-{\parallel \xi \parallel }_p^2{]}^{\alpha }|\nabla f(Z,\xi )|\le 1+D_1$. We thus have ${\parallel f\parallel }_{{\mathcal{B}}^{\alpha }}\le 1+D_1$ and $\parallel f_{n_j}{-f\parallel }_{{\mathcal{B}}^{\alpha }}\le \parallel f_{n_j}{\parallel }_{{\mathcal{B}}^{\alpha }}+{\parallel f\parallel }_{{\mathcal{B}}^{\alpha }}\le 2D_1+1$ and $f_{n_j}-f\rightrightarrows 0$ on every compact subset of ${\mathrm{GHE}}_{\mathrm{I}}$ as $j\to \infty$. Consequently, we have

$$\underset{j\to \infty }{lim}\parallel \psi C_{\varphi }(f_{n_j}-f){\parallel }_{{\mathcal{A}}_{\beta }}=\underset{j\to \infty }{lim}{\parallel \psi C_{\varphi }{f}_{n_j}-\psi C_{\varphi }f\parallel }_{{\mathcal{A}}_{\beta }}=0,$$

which shows that $\psi C_{\varphi }:{\mathcal{B}}^{\alpha }({\mathrm{GHE}}_{\mathrm{I}})\to {\mathcal{A}}_{\beta }({\mathrm{GHE}}_{\mathrm{I}})$ is compact. □

Lemma 13.

Let $(Z,\xi ),(S,t)\in {\mathrm{GHE}}_{\mathrm{I}}$, if $0<km\le 1$, then

$$det{(I_m-Z{\overline{Z}}^{\prime })}^k+det{(I_m-S{\overline{S}}^{\prime })}^k\le 2|det{(I_m-Z{\overline{S}}^{\prime })}^k|,$$

(18)

and “=” holds if and only if $(Z,\xi )=(S,t)$. If $km>1$, then

$$det{(I_m-Z{\overline{Z}}^{\prime })}^k+det{(I_m-S{\overline{S}}^{\prime })}^k\le 2^{mk}|det{(I_m-Z{\overline{S}}^{\prime })}^k|.$$

(19)

Proof. 

For $m=n$, since $(Z,\xi ),(S,t)\in {\mathrm{GHE}}_{\mathrm{I}}$, there exists two $m\times m$ unitary matrices $U_1,U_2$ and two $n\times n$ unitary matrices $V_1,V_2$ (by Lemma 6), such that

$$Z=U_1\left(\begin{array}{cccc}{\lambda }_1& 0& \dots & 0\\ 0& {\lambda }_2& \dots & 0\\ ⋮& ⋮& \ddots & ⋮\\ 0& 0& \dots & {\lambda }_m\end{array}\right)V_1=U_1{\mathsf{\Lambda }}_1{V}_1(1>{\lambda }_1\ge {\lambda }_2\ge \dots \ge {\lambda }_m\ge 0)$$

$$S=U_2\left(\begin{array}{cccc}{\mu }_1& 0& \dots & 0\\ 0& {\mu }_2& \dots & 0\\ ⋮& ⋮& \ddots & ⋮\\ 0& 0& \dots & {\mu }_m\end{array}\right)V_2=U_2{\mathsf{\Lambda }}_2{V}_2(1>{\mu }_1\ge {\mu }_2\ge \dots \ge {\mu }_m\ge 0).$$

Then, one has

$$\begin{array}{cc}\hfill det(I-Z{\overline{S}}^{\prime })& =det(I-U_1{\mathsf{\Lambda }}_1{V}_1{\overline{V_2}}^{\prime }{\overline{{\mathsf{\Lambda }}_2}}^{\prime }{\overline{U_2}}^{\prime })\hfill \\ \hfill & =det(U_1{\overline{U_1}}^{\prime }-U_1{\mathsf{\Lambda }}_1{V}_1{\overline{V_2}}^{\prime }{\overline{{\mathsf{\Lambda }}_2}}^{\prime }{\overline{U_2}}^{\prime })\hfill \\ \hfill & =det{U}_{1}det({\overline{U_1}}^{\prime }-{\mathsf{\Lambda }}_1{V}_1{\overline{V_2}}^{\prime }{\overline{{\mathsf{\Lambda }}_2}}^{\prime }{\overline{U_2}}^{\prime })\hfill \\ \hfill & =det(I-{\mathsf{\Lambda }}_1{V}_1{\overline{V_2}}^{\prime }{\overline{{\mathsf{\Lambda }}_2}}^{\prime }{V}_2{\overline{V_1}}^{\prime }{V}_1{\overline{V_2}}^{\prime }{\overline{U_2}}^{\prime }{U}_1).\hfill \end{array}$$

By Lemma 7, there exists a square matrix P, such that

$$|det(I-Z{\overline{S}}^{\prime })|\ge |det(I-{\mathsf{\Lambda }}_{1}P{\mathsf{\Lambda }}_2{P}^{\prime })|=\prod _{i=1}^m(1-{\lambda }_i{\mu }_{k_i}),$$

where $k_1,k_2,\cdots ,k_m$ is a permutation of $1,2,\cdots ,m$.

If $0<km\le 1$, and using (7) and Lemma 8, we obtain

$$\begin{array}{cc}\hfill 2|det{(I-Z{\overline{S}}^{\prime })}^k|& =2^{1-mk}·2^{mk}|det{(I-Z{\overline{S}}^{\prime })}^k|\hfill \\ \hfill & =2^{1-mk}[2^m|det(I-Z{\overline{S}}^{\prime }){|]}^k\hfill \\ \hfill & \ge 2^{1-mk}{\left[2^m\prod _{i=1}^m(1-{\lambda }_i{\mu }_{k_i})\right]}^k\hfill \\ \hfill & \ge 2^{1-mk}{\left\{{\{\prod _{i=1}^m[(1-{{\lambda }_i}^2)+(1-{{\mu }_{k_i}}^2)]\}}^{\frac{1}{m}}\right\}}^{mk}\hfill \\ \hfill & \ge 2^{1-mk}{\left\{{\left[\prod _{i=1}^m(1-{{\lambda }_i}^2)\right]}^{\frac{1}{m}}+{\left[\prod _{i=1}^m(1-{{\mu }_{k_i}}^2)\right]}^{\frac{1}{m}}\right\}}^{mk}\hfill \\ \hfill & \ge 2^{1-mk}·2^{mk-1}\left\{{\left[\prod _{i=1}^m(1-{{\lambda }_i}^2)\right]}^{\frac{1}{m}\times mk}+{\left[\prod _{i=1}^m(1-{{\mu }_{k_i}}^2)\right]}^{\frac{1}{m}\times mk}\right\}\hfill \\ \hfill & ={\left[\prod _{i=1}^m(1-{{\lambda }_i}^2)\right]}^k+{\left[\prod _{i=1}^m(1-{{\mu }_{k_i}}^2)\right]}^k\hfill \\ \hfill & =det{(I-Z{\overline{Z}}^{\prime })}^k+det{(I-S{\overline{S}}^{\prime })}^k.\hfill \end{array}$$

(20)

If $km>1$, by using (6) and Lemma 8, we obtain

$$\begin{array}{cc}\hfill 2^{mk}|det{(I-Z{\overline{S}}^{\prime })}^k|& =[2^m|det(I-Z{\overline{S}}^{\prime }){|]}^k\hfill \\ \hfill & \ge {\left[2^m\prod _{i=1}^m(1-{\lambda }_i{\mu }_{k_i})\right]}^k\hfill \\ \hfill & \ge {\left\{{\{\prod _{i=1}^m[(1-{{\lambda }_i}^2)+(1-{{\mu }_{k_i}}^2)]\}}^{\frac{1}{m}}\right\}}^{mk}\hfill \\ \hfill & \ge {\left\{{\left[\prod _{i=1}^m(1-{{\lambda }_i}^2)\right]}^{\frac{1}{m}}+{\left[\prod _{i=1}^m(1-{{\mu }_{k_i}}^2)\right]}^{\frac{1}{m}}\right\}}^{mk}\hfill \\ \hfill & \ge {\left[\prod _{i=1}^m(1-{{\lambda }_i}^2)\right]}^{\frac{1}{m}\times mk}+{\left[\prod _{i=1}^m(1-{{\mu }_{k_i}}^2)\right]}^{\frac{1}{m}\times mk}\hfill \\ \hfill & ={\left[\prod _{i=1}^m(1-{{\lambda }_i}^2)\right]}^k+{\left[\prod _{i=1}^m(1-{{\mu }_{k_i}}^2)\right]}^k\hfill \\ \hfill & =det{(I-Z{\overline{Z}}^{\prime })}^k+det{(I-S{\overline{S}}^{\prime })}^k.\hfill \end{array}$$

(21)

For $m<n$, there exists a unitary matrix $U^{(n)}$, such that

$$Z=({Z_1}^{(m)},0)U,S=({S_1}^{(m)},S_2)U.$$

According to (20), we have

$$\begin{array}{cc}\hfill 2|det{(I-Z{\overline{S}}^{\prime })}^k|& =2|det{(I-Z_1{\overline{S_1}}^{\prime })}^k|\hfill \\ \hfill & \ge det{(I-Z_1{\overline{Z_1}}^{\prime })}^k+det{(I-S_1{\overline{S_1}}^{\prime })}^k\hfill \\ \hfill & \ge det{(I-Z_1{\overline{Z_1}}^{\prime })}^k+det{(I-S_1{\overline{S_1}}^{\prime }-S_2{\overline{S_2}}^{\prime })}^k\hfill \\ \hfill & =det{(I-Z{\overline{Z}}^{\prime })}^k+det{(I-S{\overline{S}}^{\prime })}^k.\hfill \end{array}$$

Thus, the inequality

$$2|det{(I-Z{\overline{S}}^{\prime })}^k|\ge det{(I-Z{\overline{Z}}^{\prime })}^k+det{(I-S{\overline{S}}^{\prime })}^k,$$

(22)

holds when $m\le n$, whereas the equal sign holds if and only if $Z=S$.

According to (21), we see that

$$\begin{array}{cc}\hfill 2^{mk}|det{(I-Z{\overline{S}}^{\prime })}^k|& =2^{mk}|det{(I-Z_1{\overline{S_1}}^{\prime })}^k|\hfill \\ \hfill & \ge det{(I-Z_1{\overline{Z_1}}^{\prime })}^k+det{(I-S_1{\overline{S_1}}^{\prime })}^k\hfill \\ \hfill & \ge det{(I-Z_1{\overline{Z_1}}^{\prime })}^k+det{(I-S_1{\overline{S_1}}^{\prime }-S_2{\overline{S_2}}^{\prime })}^k\hfill \\ \hfill & =det{(I-Z{\overline{Z}}^{\prime })}^k+det{(I-S{\overline{S}}^{\prime })}^k.\hfill \end{array}$$

Thus, the inequality

$$2^{mk}|det{(I-Z{\overline{S}}^{\prime })}^k|\ge det{(I-Z{\overline{Z}}^{\prime })}^k+det{(I-S{\overline{S}}^{\prime })}^k,$$

(23)

holds when $m\le n$, with the equality holding if and only if $Z=S$ and $mk=1$. □

Lemma 14.

Assume $(Z,\xi ),(S,t)\in {\mathrm{GHE}}_{\mathrm{I}}$ and $0<km\le 1$, then

$$[det{(I_m-Z{\overline{Z}}^{\prime })}^k-{\parallel \xi \parallel }_p^2]+[det{(I_m-S{\overline{S}}^{\prime })}^k-{\parallel t\parallel }_p^2]\le 2||det{(I_m-Z{\overline{S}}^{\prime })}^k{|-\parallel \xi \parallel }_p{\parallel t\parallel }_p|,$$

(24)

with equality that holds if and only if $(Z,\xi )=(S,t)$.

Proof. 

Starting from the inequality $a^2+b^2\ge 2ab$, we obtain

$${\parallel \xi \parallel }_p^2+{\parallel t\parallel }_p^2\ge {2\parallel \xi \parallel }_p{\parallel t\parallel }_p.$$

Then, by (18), we have

$$\begin{array}{cc}\hfill 2||det{(I-Z{\overline{S}}^{\prime })}^k{|-\parallel \xi \parallel }_p{\parallel t\parallel }_p|& =2|det{(I-Z{\overline{S}}^{\prime })}^k{|-2\parallel \xi \parallel }_p{\parallel t\parallel }_p\hfill \\ \hfill & \ge det{(I-Z{\overline{Z}}^{\prime })}^k+det{(I-S{\overline{S}}^{\prime })}^k-{\parallel \xi \parallel }_p^2-{\parallel t\parallel }_p^2\hfill \\ \hfill & =[det{(I-Z{\overline{Z}}^{\prime })}^k-{\parallel \xi \parallel }_p^2]+[det{(I-S{\overline{S}}^{\prime })}^k-{\parallel t\parallel }_p^2].\hfill \end{array}$$

This completes the proof. □

Lemma 15.

Assume $(Z,\xi ),(S,t)\in {\mathrm{GHE}}_{\mathrm{I}}$ and $0<km\le 1$, then

$$\left[det{(I_m-Z{\overline{Z}}^{\prime })}^k-{\parallel \xi \parallel }_p^2\right]\left[det{(I_m-S{\overline{S}}^{\prime })}^k-{\parallel t\parallel }_p^2\right]\le ||det{(I_m-Z{\overline{S}}^{\prime })}^k{|-\parallel \xi \parallel }_p{\parallel t\parallel }_p{|}^2,$$

(25)

Proof. 

By the elementary inequality $\frac{a+b}{2}\ge \sqrt{ab}$ and Lemma 14, we have

$$\begin{array}{cc}\hfill & \left[det{(I_m-Z{\overline{Z}}^{\prime })}^k-{\parallel \xi \parallel }_p^2\right]\left[det{(I_m-S{\overline{S}}^{\prime })}^k-{\parallel t\parallel }_p^2\right]\hfill \\ \hfill & \le {\left\{\frac{[det{(I_m-Z{\overline{Z}}^{\prime })}^k-{\parallel \xi \parallel }_p^2]+[det{(I_m-S{\overline{S}}^{\prime })}^k-{\parallel t\parallel }_p^2]}{2}\right\}}^2\hfill \\ \hfill & \le ||det{(I_m-Z{\overline{S}}^{\prime })}^k{|-\parallel \xi \parallel }_p{\parallel t\parallel }_p{|}^2\hfill \end{array}$$

□

Lemma 16

(see [24]). Assume $Z,S\in {\Re }_I(m,n)$, then there exists a constant C, such that

$$|det(I_m-Z{\overline{S}}^{\prime })|{\left\{\sum _{\genfrac{}{}{0pt}{}{1\le g\le m\hfill }{1\le l\le n\hfill }}{|\mathrm{tr}[{(I_m-Z{\overline{S}}^{\prime })}^{-1}{I}_{gl}{\overline{S}}^{\prime }]|}^2\right\}}^{\frac{1}{2}}\le C,$$

(26)

where $I_{gl}$ is an $m\times n$ matrix, where the elements of the gth row and lth column are one and the other elements are zero.

3. Boundedness of $\mathit{\psi }{\mathit{C}}_{\mathit{\varphi }}\mathbf{:}{\mathcal{B}}^{\mathit{\alpha }}\mathbf{\to }{\mathcal{A}}_{\mathit{\beta }}$

Theorem 1.

Assume that $\alpha =1,\beta >0$, $0<km\le 1$, and that $p_j$ ($j=1,2,\cdots ,r$) are positive integers. Let $\varphi =({\varphi }_{11},{\varphi }_{12}\cdots {\varphi }_{mn+r})$ be a holomorphic self-map of ${\mathrm{GHE}}_{\mathrm{I}}$, with $\psi \in H({\mathrm{GHE}}_{\mathrm{I}})$ and $(Z_{\varphi },{\xi }_{\varphi })=\varphi (Z,\xi )$. If

$$K_1:=\underset{(Z,\xi )\in {\mathrm{GHE}}_{\mathrm{I}}}{sup}|\psi (Z,\xi )|[det{(I-Z{\overline{Z}}^{\prime })}^k-{\parallel \xi \parallel }_p^2{]}^{\beta }ln\frac{2q}{det{(I-Z_{\varphi }{\overline{Z_{\varphi }}}^{\prime })}^k-{\parallel {\xi }_{\varphi }\parallel }_p^2}<\infty .$$

(27)

Then, the weighted composition operator $\psi C_{\varphi }:{\mathcal{B}}^{\alpha }({\mathrm{GHE}}_{\mathrm{I}})\to {\mathcal{A}}_{\beta }({\mathrm{GHE}}_{\mathrm{I}})$ is bounded.

Conversely, if the weighted composition operator $\psi C_{\varphi }:{\mathcal{B}}^{\alpha }({\mathrm{GHE}}_{\mathrm{I}})\to {\mathcal{A}}_{\beta }({\mathrm{GHE}}_{\mathrm{I}})$ is bounded, then

$$\begin{array}{c}\hfill K_2:=\underset{(Z,\xi )\in {\mathrm{GHE}}_{\mathrm{I}}}{sup}|\psi (Z,\xi )|[det{(I-Z{\overline{Z}}^{\prime })}^k-{\parallel \xi \parallel }_p^2{]}^{\beta }det{(I-Z_{\varphi }{\overline{Z_{\varphi }}}^{\prime })}^{1-k}\\ \hfill \times ln\frac{2}{det{(I-Z_{\varphi }{\overline{Z_{\varphi }}}^{\prime })}^k-{\parallel {\xi }_{\varphi }\parallel }_p^2}<\infty .\end{array}$$

(28)

Proof. 

Assume that (27) holds. By Lemma 11 and for $f\in {\mathcal{B}}^{\alpha }({\mathrm{GHE}}_{\mathrm{I}})$, we know that

$$\begin{array}{cc}\hfill & [det{(I-Z{\overline{Z}}^{\prime })}^k-{\parallel \xi \parallel }_p^2{]}^{\beta }|(\psi C_{\varphi }f)(Z,\xi )|\hfill \\ \hfill & =[det{(I-Z{\overline{Z}}^{\prime })}^k-{\parallel \xi \parallel }_p^2{]}^{\beta }|\psi (Z,\xi )||f(\varphi (Z,\xi ))|\hfill \\ \hfill & \le C|\psi (Z,\xi )|[det{(I-Z{\overline{Z}}^{\prime })}^k-{\parallel \xi \parallel }_p^2{]}^{\beta }\times ln\frac{2q}{det{(I-Z_{\varphi }{\overline{Z_{\varphi }}}^{\prime })}^k-{\parallel {\xi }_{\varphi }\parallel }_p^2}{\parallel f\parallel }_{{\mathcal{B}}^{\alpha }}\hfill \\ \hfill & \le C{K}_1{\parallel f\parallel }_{{\mathcal{B}}^{\alpha }}.\hfill \end{array}$$

For all $(Z,\xi )\in {\mathrm{GHE}}_{\mathrm{I}}$, we have

$$\parallel \psi C_{\varphi }{f\parallel }_{{\mathcal{A}}_{\beta }}=\underset{(Z,\xi )\in {\mathrm{GHE}}_{\mathrm{I}}}{sup}[det{(I-Z{\overline{Z}}^{\prime })}^k-{\parallel \xi \parallel }_p^2{]}^{\beta }|(\psi C_{\varphi }f)(Z,\xi )|\le C{K}_1{\parallel f\parallel }_{{\mathcal{B}}^{\alpha }},$$

which implies that $\psi C_{\varphi }:{\mathcal{B}}^{\alpha }({\mathrm{GHE}}_{\mathrm{I}})\to {\mathcal{A}}_{\beta }({\mathrm{GHE}}_{\mathrm{I}})$ is bounded.

Conversely, assume that $\psi C_{\varphi }:{\mathcal{B}}^{\alpha }({\mathrm{GHE}}_{\mathrm{I}})\to {\mathcal{A}}_{\beta }({\mathrm{GHE}}_{\mathrm{I}})$ is bounded. For any $(S,t)\in {\mathrm{GHE}}_{\mathrm{I}}$, let us introduce a test function $f_{(S,t)}\in H({\mathrm{GHE}}_{\mathrm{I}})$, such that

$$f_{(S,t)}(Z,\xi ):=det{(I-S{\overline{S}}^{\prime })}^{1-k}ln\frac{2}{det{(I-Z{\overline{S}}^{\prime })}^k-{\langle \xi ,t\rangle }_p}.$$

This means that

$$\begin{array}{ccc}\hfill \frac{\partial f_{(S,t)}}{\partial z_{gl}}& =& \frac{k·det{(I-Z{\overline{S}}^{\prime })}^{k-1}·det{(I-S{\overline{S}}^{\prime })}^{1-k}}{det{(I-Z{\overline{S}}^{\prime })}^k-{\langle \xi ,t\rangle }_p}\times det(I-Z{\overline{S}}^{\prime })\mathrm{tr}[{(I-Z{\overline{S}}^{\prime })}^{-1}{I}_{gl}{\overline{S}}^{\prime }],\hfill \\ & & 1\le g\le m,1\le l\le n,\hfill \\ \hfill \frac{\partial f_{(S,t)}}{\partial {\xi }_j}& =& \frac{det{(I-S{\overline{S}}^{\prime })}^{1-k}·p_j{\xi }_j^{p_j-1}{\overline{t_j}}^{p_j}}{det{(I-Z{\overline{S}}^{\prime })}^k-{\langle \xi ,t\rangle }_p},j=1,\cdots ,r.\hfill \end{array}$$

In view of (18), it follows that

$$2|det(I-Z{\overline{S}}^{\prime }){|}^{\frac{1}{m}}\ge det{(I-Z{\overline{Z}}^{\prime })}^{\frac{1}{m}}+det{(I-S{\overline{S}}^{\prime })}^{\frac{1}{m}}.$$

Then,

$$\begin{array}{cc}\hfill 2^{m(1-k)}[|det(I-Z{\overline{S}}^{\prime }){{|}^{\frac{1}{m}}]}^{m(1-k)}& \ge {[det{(I-Z{\overline{Z}}^{\prime })}^{\frac{1}{m}}+det{(I-S{\overline{S}}^{\prime })}^{\frac{1}{m}}]}^{m(1-k)}\hfill \\ \hfill & \ge {[det{(I-S{\overline{S}}^{\prime })}^{\frac{1}{m}}]}^{m(1-k)},\hfill \end{array}$$

which means that

$$2^{m(1-k)}{|det(I-Z{\overline{S}}^{\prime })|}^{1-k}\ge det{(I-S{\overline{S}}^{\prime })}^{1-k}.$$

(29)

According to (29) and Lemmas 14 and 16, there exists a constant $C_1>0$, such that

$$\begin{array}{cc}\hfill & [det{(I-Z{\overline{Z}}^{\prime })}^k-{\parallel \xi \parallel }_p^2]|\nabla f_{(S,t)}(Z,\xi )|\hfill \\ \hfill & =\frac{det{(I-Z{\overline{Z}}^{\prime })}^k-{\parallel \xi \parallel }_p^2}{|det{(I-Z\overline{S})}^k-{\langle \xi ,t\rangle }_p|}\times det{(I-S{\overline{S}}^{\prime })}^{1-k}\times \{k^2{|det{(I-Z{\overline{S}}^{\prime })}^{k-1}|}^2\hfill \\ \hfill & \times \sum _{\genfrac{}{}{0pt}{}{1\le g\le m\hfill }{1\le l\le n\hfill }}|det(I-Z{\overline{S}}^{\prime })\mathrm{tr}[{(I-Z{\overline{S}}^{\prime })}^{-1}{I}_{gl}{\overline{S}}^{\prime }]{|}^2+\sum _{j=1}^r{|p_j{\xi }_j^{p_j-1}{\overline{t_j}}^{p_j}|}^2{\}}^{\frac{1}{2}}\hfill \\ \hfill & \le \frac{[det{(I-Z{\overline{Z}}^{\prime })}^k-{\parallel \xi \parallel }_p^2]\times det{(I-S{\overline{S}}^{\prime })}^{1-k}}{||det{(I-Z{\overline{S}}^{\prime })}^k|-|{\langle \xi ,t\rangle }_p||}\times \{k{|det(I-Z{\overline{S}}^{\prime })|}^{k-1}\hfill \\ \hfill & \times {\left[\sum _{\genfrac{}{}{0pt}{}{1\le g\le m\hfill }{1\le l\le n\hfill }}|det(I-Z{\overline{S}}^{\prime }){|}^2{|\mathrm{tr}[{(I-Z{\overline{S}}^{\prime })}^{-1}{I}_{gl}{\overline{S}}^{\prime }]|}^2\right]}^{\frac{1}{2}}+{\left[\sum _{j=1}^r{|p_j{\xi }_j^{p_j-1}{\overline{t_j}}^{p_j}|}^2\right]}^{\frac{1}{2}}\}\hfill \\ \hfill & \le \frac{[det{(I-Z{\overline{Z}}^{\prime })}^k-\parallel \xi {\parallel }_p^2]}{||det{(I-Z{\overline{S}}^{\prime })}^k{|-\parallel \xi \parallel }_p{\parallel t\parallel }_p|}\hfill \\ \hfill & \times \left\{k{C}_1{|det(I-Z{\overline{S}}^{\prime })|}^{k-1}\times det{(I-S{\overline{S}}^{\prime })}^{1-k}+{\left[\sum _{j=1}^r{|p_j{\xi }_j^{p_j-1}{\overline{t_j}}^{p_j}|}^2\right]}^{\frac{1}{2}}\times det{(I-S{\overline{S}}^{\prime })}^{1-k}\right\}\hfill \\ \hfill & \le \frac{2[det{(I-Z{\overline{Z}}^{\prime })}^k-\parallel \xi {\parallel }_p^2]}{[det{(I-Z{\overline{Z}}^{\prime })}^k-{\parallel \xi \parallel }_p^2]+[det{(I-S{\overline{S}}^{\prime })}^k-{\parallel t\parallel }_p^2]}\times \left\{2^{m(1-k)}k{C}_1+{\left[\sum _{j=1}^r{|p_j|}^2\right]}^{\frac{1}{2}}\right\}\hfill \\ \hfill & \le \frac{2[det{(I-Z{\overline{Z}}^{\prime })}^k-\parallel \xi {\parallel }_p^2]}{det{(I-Z{\overline{Z}}^{\prime })}^k-{\parallel \xi \parallel }_p^2}\times \left\{2^{m(1-k)}k{C}_1+{\left[\sum _{j=1}^r{|p_j|}^2\right]}^{\frac{1}{2}}\right\}\hfill \\ \hfill & \le 2\times \left\{2^{m(1-k)}k{C}_1+{\left[\sum _{j=1}^r{|p_j|}^2\right]}^{\frac{1}{2}}\right\}\hfill \\ \hfill & \le C.\hfill \end{array}$$

Since $f_{(S,t)}(0,0)\le ln2$, one has

$$\begin{array}{cc}\hfill \parallel f_{(S,t)}{\parallel }_{{\mathcal{B}}^{\alpha }}& =|f_{(S,t)}(0,0)|+\underset{(Z,\xi )\in {\mathrm{GHE}}_{\mathrm{I}}}{sup}[det{(I-Z{\overline{Z}}^{\prime })}^k-{\parallel \xi \parallel }_p^2{]}^{\alpha }|\nabla f_{(S,t)}(Z,\xi )|\hfill \\ \hfill & \le C+ln2.\hfill \end{array}$$

Therefore, we have

$$\begin{array}{cc}\hfill & \infty >(C+ln2)\parallel \psi C_{\varphi }{\parallel }_{{\mathcal{B}}^{\alpha }\to {\mathcal{A}}_{\beta }}\hfill \\ \hfill & \ge \parallel \psi C_{\varphi }{f}_{(S,t)}{\parallel }_{{\mathcal{A}}_{\beta }}\hfill \\ \hfill & =\underset{(Z,\xi )\in {\mathrm{GHE}}_{\mathrm{I}}}{sup}[det{(I-Z{\overline{Z}}^{\prime })}^k-{\parallel \xi \parallel }_p^2{]}^{\beta }|\psi (Z,\xi )f_{(S,t)}(\varphi (Z,\xi ))|\hfill \\ \hfill & \ge |\psi (Z,\xi )|[det{(I-Z{\overline{Z}}^{\prime })}^k-{\parallel \xi \parallel }_p^2{]}^{\beta }det{(I-S{\overline{S}}^{\prime })}^{1-k}\times |ln\frac{2}{det{(I-Z_{\varphi }{\overline{S}}^{\prime })}^k-{\langle {\xi }_{\varphi },t\rangle }_p}|.\hfill \end{array}$$

Let us now consider

$$(S,t)=(Z_{\varphi },{\xi }_{\varphi })=\varphi (Z,\xi ),$$

so that

$$\begin{array}{c}\hfill \underset{(Z,\xi )\in {\mathrm{GHE}}_{\mathrm{I}}}{sup}|\psi (Z,\xi )|[det{(I-Z{\overline{Z}}^{\prime })}^k-{\parallel \xi \parallel }_p^2{]}^{\beta }det{(I-Z_{\varphi }{\overline{Z_{\varphi }}}^{\prime })}^{1-k}ln\frac{2}{det{(I-Z_{\varphi }{\overline{Z_{\varphi }}}^{\prime })}^k-{\parallel {\xi }_{\varphi }\parallel }_p^2}<\infty .\end{array}$$

The proof is thus completed. □

Theorem 2.

Assume that $\alpha >1,\beta >0$, $0<km\le 1$, and that $p_j$ are positive integers $(j=1,2,\cdots ,r)$. Let $\varphi =({\varphi }_{11},{\varphi }_{12}\cdots {\varphi }_{mn+r})$ be a holomorphic self-map of ${\mathrm{GHE}}_{\mathrm{I}}$, with $\psi \in H({\mathrm{GHE}}_{\mathrm{I}})$ and $(Z_{\varphi },{\xi }_{\varphi })=\varphi (Z,\xi )$. If

$$K_3:=\underset{(Z,\xi )\in {\mathrm{GHE}}_{\mathrm{I}}}{sup}|\psi (Z,\xi )|\frac{[det{(I-Z{\overline{Z}}^{\prime })}^k-{\parallel \xi \parallel }_p^2{]}^{\beta }}{[det{(I-Z_{\varphi }{\overline{Z_{\varphi }}}^{\prime })}^k-\parallel {\xi }_{\varphi }{{\parallel }_p^2]}^{\alpha -1}}<\infty ,$$

(30)

then, the weighted composition operator $\psi C_{\varphi }:{\mathcal{B}}^{\alpha }({\mathrm{GHE}}_{\mathrm{I}})\to {\mathcal{A}}_{\beta }({\mathrm{GHE}}_{\mathrm{I}})$ is bounded.

Conversely, if the weighted composition operator $\psi C_{\varphi }:{\mathcal{B}}^{\alpha }({\mathrm{GHE}}_{\mathrm{I}})\to {\mathcal{A}}_{\beta }({\mathrm{GHE}}_{\mathrm{I}})$ is bounded, then,

$$K_4:=\underset{(Z,\xi )\in {\mathrm{GHE}}_{\mathrm{I}}}{sup}|\psi (Z,\xi )|\frac{[det{(I-Z{\overline{Z}}^{\prime })}^k-{\parallel \xi \parallel }_p^2{]}^{\beta }det{(I-Z_{\varphi }{\overline{Z_{\varphi }}}^{\prime })}^{1-k}}{[det{(I-Z_{\varphi }{\overline{Z_{\varphi }}}^{\prime })}^k-\parallel {\xi }_{\varphi }{{\parallel }_p^2]}^{\alpha -1}}<\infty .$$

(31)

Proof. 

Assume that (30) holds. By Lemma 11 and for $f\in {\mathcal{B}}^{\alpha }({\mathrm{GHE}}_{\mathrm{I}})$, we have

$$\begin{array}{cc}\hfill [det{(I-Z{\overline{Z}}^{\prime })}^k-{\parallel \xi \parallel }_p^2{]}^{\beta }|(\psi C_{\varphi }f)(Z,\xi )|& =[det{(I-Z{\overline{Z}}^{\prime })}^k-{\parallel \xi \parallel }_p^2{]}^{\beta }|\psi (Z,\xi )·(C_{\varphi }f)(Z,\xi )|\hfill \\ \hfill & =[det{(I-Z{\overline{Z}}^{\prime })}^k-{\parallel \xi \parallel }_p^2{]}^{\beta }|\psi (Z,\xi )||f(\varphi (Z,\xi ))|\hfill \\ \hfill & \le C|\psi (Z,\xi )|\frac{[det{(I-Z{\overline{Z}}^{\prime })}^k-{\parallel \xi \parallel }_p^2{]}^{\beta }}{[det{(I-Z_{\varphi }{\overline{Z_{\varphi }}}^{\prime })}^k-\parallel {\xi }_{\varphi }{{\parallel }_p^2]}^{\alpha -1}}{\parallel f\parallel }_{{\mathcal{B}}^{\alpha }}\hfill \\ \hfill & \le C{K}_3{\parallel f\parallel }_{{\mathcal{B}}^{\alpha }}.\hfill \end{array}$$

For all $(Z,\xi )\in {\mathrm{GHE}}_{\mathrm{I}}$, we obtain

$$\parallel \psi C_{\varphi }{f\parallel }_{{\mathcal{A}}_{\beta }}=\underset{(Z,\xi )\in {\mathrm{GHE}}_{\mathrm{I}}}{sup}[det{(I-Z{\overline{Z}}^{\prime })}^k-{\parallel \xi \parallel }_p^2{]}^{\beta }|(\psi C_{\varphi }f)(Z,\xi )|\le C{K}_3{\parallel f\parallel }_{{\mathcal{B}}^{\alpha }}.$$

This implies that $\psi C_{\varphi }:{\mathcal{B}}^{\alpha }({\mathrm{GHE}}_{\mathrm{I}})\to {\mathcal{A}}_{\beta }({\mathrm{GHE}}_{\mathrm{I}})$ is bounded.

Conversely, assume that $\psi C_{\varphi }:{\mathcal{B}}^{\alpha }({\mathrm{GHE}}_{\mathrm{I}})\to {\mathcal{A}}_{\beta }({\mathrm{GHE}}_{\mathrm{I}})$ is bounded. For $(S,t)\in {\mathrm{GHE}}_{\mathrm{I}}$, define a test function $f_{(S,t)}\in H({\mathrm{GHE}}_{\mathrm{I}})$, such that

$$f_{(S,t)}(Z,\xi ):=\frac{det{(I-S{\overline{S}}^{\prime })}^{1-k}}{{[det{(I-Z{\overline{S}}^{\prime })}^k-{\langle \xi ,t\rangle }_p]}^{\alpha -1}}.$$

For the test function f, we have

$$\begin{array}{ccc}\hfill \frac{\partial f_{(S,t)}}{\partial z_{gl}}& =& \frac{k(\alpha -1)·det{(I-Z{\overline{S}}^{\prime })}^{k-1}·det{(I-S{\overline{S}}^{\prime })}^{1-k}}{{[det{(I-Z{\overline{S}}^{\prime })}^k-{\langle \xi ,t\rangle }_p]}^{\alpha }}\hfill \\ & & \times det(I-Z{\overline{S}}^{\prime })\mathrm{tr}[{(I-Z{\overline{S}}^{\prime })}^{-1}{I}_{gl}{\overline{S}}^{\prime }],1\le g\le m,1\le l\le n,\hfill \\ \hfill \frac{\partial f_{(S,t)}}{\partial {\xi }_j}& =& \frac{(\alpha -1)p_j{\xi }_j^{p_j-1}{\overline{t_j}}^{p_j}·det{(I-S{\overline{S}}^{\prime })}^{1-k}}{{[det{(I-Z{\overline{S}}^{\prime })}^k-{\langle \xi ,t\rangle }_p]}^{\alpha }},j=1,\cdots ,r.\hfill \end{array}$$

From (29) and Lemmas 14 and 16, there exists a constant $C_1>0$, such that

$$\begin{array}{cc}\hfill & [det{(I-Z{\overline{Z}}^{\prime })}^k-{\parallel \xi \parallel }_p^2{]}^{\alpha }|\nabla f_{(S,t)}(Z,\xi )|\hfill \\ \hfill & =\frac{[det{(I-Z{\overline{Z}}^{\prime })}^k-{\parallel \xi \parallel }_p^2{]}^{\alpha }}{|det{(I-Z{\overline{S}}^{\prime })}^k-{\langle \xi ,t\rangle }_p{|}^{\alpha }}\times det{(I-S{\overline{S}}^{\prime })}^{1-k}\times (\alpha -1)\times \{k^2{|det{(I-Z{\overline{S}}^{\prime })}^{k-1}|}^2\hfill \\ & \times \sum _{\genfrac{}{}{0pt}{}{1\le g\le m\hfill }{1\le l\le n\hfill }}|det(I-Z{\overline{S}}^{\prime })\mathrm{tr}[{(I-Z{\overline{S}}^{\prime })}^{-1}{I}_{gl}{\overline{S}}^{\prime }]{|}^2+\sum _{j=1}^r{|p_j{\xi }_j^{p_j-1}{\overline{t_j}}^{p_j}|}^2{\}}^{\frac{1}{2}}\hfill \\ \hfill & \le \frac{[det(I-Z{\overline{Z}}^{\prime }){|}^k-{\parallel \xi \parallel }_p^2{]}^{\alpha }\times det{(I-S{\overline{S}}^{\prime })}^{1-k}}{||det(I-Z{\overline{S}}^{\prime }){|}^k-|{\langle \xi ,t\rangle }_p{||}^{\alpha }}\times (\alpha -1)\times \{k{|det(I-Z{\overline{S}}^{\prime })|}^{k-1}\hfill \\ & \times {\left[\sum _{\genfrac{}{}{0pt}{}{1\le g\le m\hfill }{1\le l\le n\hfill }}|det(I-Z{\overline{S}}^{\prime }){|}^2{|\mathrm{tr}[{(I-Z{\overline{S}}^{\prime })}^{-1}{I}_{gl}{\overline{S}}^{\prime }]|}^2\right]}^{\frac{1}{2}}+{\left[\sum _{j=1}^r{|p_j{\xi }_j^{p_j-1}{\overline{t_j}}^{p_j}|}^2\right]}^{\frac{1}{2}}\}\hfill \\ \hfill & \le \frac{[det{(I-Z{\overline{Z}}^{\prime })}^k-{\parallel \xi \parallel }_p^2{]}^{\alpha }}{||det(I-Z{\overline{S}}^{\prime }){|}^k-{\parallel \xi \parallel }_p{\parallel t\parallel }_p{|}^{\alpha }}\times (\alpha -1)\times \{k{C}_{1}det{(I-S{\overline{S}}^{\prime })}^{1-k}{|det(I-Z{\overline{S}}^{\prime })|}^{k-1}\hfill \\ & +{\left[\sum _{j=1}^r{|p_j{\xi }_j^{p_j-1}{\overline{t_j}}^{p_j}|}^2\right]}^{\frac{1}{2}}det{(I-S{\overline{S}}^{\prime })}^{1-k}\}\hfill \\ \hfill & \le {\left\{\frac{2[det{(I-Z{\overline{Z}}^{\prime })}^k-\parallel \xi {\parallel }_p^2]}{[det{(I-Z{\overline{Z}}^{\prime })}^k-{\parallel \xi \parallel }_p^2]+[det{(I-S{\overline{S}}^{\prime })}^k-{\parallel \xi \parallel }_p^2]}\right\}}^{\alpha }(\alpha -1)\times (k{C}_1{2}^{m(1-k)}+C_2)\hfill \\ \hfill & \le {\left\{\frac{2[det{(I-Z{\overline{Z}}^{\prime })}^k-\parallel \xi {\parallel }_p^2]}{det{(I-Z{\overline{Z}}^{\prime })}^p-{\parallel \xi \parallel }_p^2}\right\}}^{\alpha }(\alpha -1)\times (k{C}_1{2}^{m(1-k)}+C_2)\hfill \\ \hfill & \le 2^{\alpha }(\alpha -1)C_3\hfill \\ \hfill & =C_4.\hfill \end{array}$$

Since $f_{(S,t)}(0,0)\le 1$, we obtain

$$\begin{array}{cc}\hfill \parallel f_{(S,t)}{\parallel }_{{\mathcal{B}}^{\alpha }}& = |f_{(S,t)}(0,0)|+\underset{(Z,\xi )\in {\mathrm{GHE}}_{\mathrm{I}}}{sup}[det{(I-Z{\overline{Z}}^{\prime })}^k-{\parallel \xi \parallel }_p^2{]}^{\alpha }|\nabla f_{(S,t)}(Z,\xi )|\hfill \\ \hfill & \le C_4+1.\hfill \end{array}$$

It follows that

$$\begin{array}{cc}\hfill \infty >(C_4+1){\parallel \psi C_{\varphi }\parallel }_{{\mathcal{B}}^{\alpha }\to {\mathcal{A}}_{\beta }}& \ge \parallel \psi C_{\varphi }{f}_{(S,t)}{\parallel }_{{\mathcal{A}}_{\beta }}\hfill \\ \hfill & =\underset{(Z,\xi )\in {\mathrm{GHE}}_{\mathrm{I}}}{sup}[det{(I-Z{\overline{Z}}^{\prime })}^k-{\parallel \xi \parallel }_p^2{]}^{\beta }|\psi (Z,\xi )f_{(S,t)}(\varphi (Z,\xi ))|\hfill \\ \hfill & \ge |\psi (Z,\xi )|\frac{[det{(I-Z{\overline{Z}}^{\prime })}^k-{\parallel \xi \parallel }_p^2{]}^{\beta }det{(I-S{\overline{S}}^{\prime })}^{1-k}}{|det{(I-Z_{\varphi }{\overline{S}}^{\prime })}^k-{\langle {\xi }_{\varphi },t\rangle }_p{|}^{\alpha -1}}.\hfill \end{array}$$

We write $(S,t)=(Z_{\varphi },{\xi }_{\varphi })=\varphi (Z,\xi )$, then

$$\begin{array}{c}\hfill \underset{(Z,\xi )\in {\mathrm{GHE}}_{\mathrm{I}}}{sup}|\psi (Z,\xi )|\frac{[det{(I-Z{\overline{Z}}^{\prime })}^k-{\parallel \xi \parallel }_p^2{]}^{\beta }det{(I-Z_{\varphi }{\overline{Z_{\varphi }}}^{\prime })}^{1-k}}{[det{(I-Z_{\varphi }{\overline{Z_{\varphi }}}^{\prime })}^k-\parallel {\xi }_{\varphi }{{|}_p^2]}^{\alpha -1}}<\infty .\end{array}$$

This completes the proof of the theorem. □

Corollary 1.

For $\alpha >1,k=m=1,p_1=\cdots =p_r=1$, we have the case of the unit ball $\mathbf{B}=\{z\in {\mathbb{C}}^{n+r}:|z{|}^2<1\}$ and $\psi C_{\varphi }:{\mathcal{B}}^{\alpha }(\mathbf{B})\to {\mathcal{A}}_{\beta }(\mathbf{B})$ is bounded if and only if

$$\underset{z\in B}{sup}\frac{|\psi (z)|(1-{|z|}^2{)}^{\beta }}{{(1-|\varphi (z)|}^2{)}^{\alpha -1}}<\infty ,$$

when $\beta =0$. This result is equivalent to that obtained by Li and Stević in [11].

4. Compactness of $\mathit{\psi }{\mathit{C}}_{\mathit{\varphi }}\mathbf{:}{\mathcal{B}}^{\mathit{\alpha }}\mathbf{\to }{\mathcal{A}}_{\mathit{\beta }}$

Theorem 3.

Assume that $\alpha =1,\beta >0$, $0<km\le 1$, and that $p_j$$(j=1,2,\cdots ,r)$ are positive integers. Let $\varphi =({\varphi }_{11},{\varphi }_{12}\cdots {\varphi }_{mn+r})$ be a holomorphic self-map of ${\mathrm{GHE}}_{\mathrm{I}}$, with $\psi \in H({\mathrm{GHE}}_{\mathrm{I}})$ and $(Z_{\varphi },{\xi }_{\varphi })=\varphi (Z,\xi )$. If $\psi \in {\mathcal{A}}_{\beta }$ and

$$\underset{\varphi (Z,\xi )\to \partial {\mathrm{GHE}}_{\mathrm{I}}}{lim}|\psi (Z,\xi )|[det{(I-Z{\overline{Z}}^{\prime })}^k-{\parallel \xi \parallel }_p^2{]}^{\beta }ln\frac{2q}{det{(I-Z_{\varphi }{\overline{Z_{\varphi }}}^{\prime })}^k-{\parallel {\xi }_{\varphi }\parallel }_p^2}=0,$$

(32)

then the weighted composition operator $\psi C_{\varphi }:{\mathcal{B}}^{\alpha }({\mathrm{GHE}}_{\mathrm{I}})\to {\mathcal{A}}_{\beta }({\mathrm{GHE}}_{\mathrm{I}})$ is compact.

Conversely, if the weighted composition operator $\psi C_{\varphi }:{\mathcal{B}}^{\alpha }({\mathrm{GHE}}_{\mathrm{I}})\to {\mathcal{A}}_{\beta }({\mathrm{GHE}}_{\mathrm{I}})$ is compact, then $\psi \in {\mathcal{A}}_{\beta }$ and

$$\underset{\varphi (Z,\xi )\to \partial {\mathrm{GHE}}_{\mathrm{I}}}{lim}|\psi (Z,\xi )|[det{(I-Z{\overline{Z}}^{\prime })}^k-{\parallel \xi \parallel }_p^2{]}^{\beta }det{(I-Z_{\varphi }{\overline{Z_{\varphi }}}^{\prime })}^{1-k}\times ln\frac{2}{det{(I-Z_{\varphi }{\overline{Z_{\varphi }}}^{\prime })}^k-{\parallel {\xi }_{\varphi }\parallel }_p^2}=0.$$

(33)

Proof. 

Assume that (32) holds. We have

$$\underset{(Z,\xi )\in {\mathrm{GHE}}_{\mathrm{I}}}{sup}|\psi (Z,\xi )|[det{(I-Z{\overline{Z}}^{\prime })}^k-{\parallel \xi \parallel }_p^2{]}^{\beta }ln\frac{2q}{det{(I-Z_{\varphi }{\overline{Z_{\varphi }}}^{\prime })}^k-{\parallel {\xi }_{\varphi }\parallel }_p^2}<\infty .$$

If $\psi C_{\varphi }$ is bounded, consider the bounded sequence ${\{f_k\}}_{k\ge 1}$ in ${\mathcal{B}}^{\alpha }({\mathrm{GHE}}_{\mathrm{I}})$, which converges to 0 uniformly on compact subsets of ${\mathrm{GHE}}_{\mathrm{I}}$. Hence, there exists $M_1>0$, such that $\parallel f_k{\parallel }_{{\mathcal{B}}^{\alpha }}\le M_1,k=1,2,\cdots$. By (32), this means that $\forall \epsilon >0$, $\exists \delta \in (0,1)$, such that for $\mathrm{dist}(\varphi (Z,\xi ),\partial {\mathrm{GHE}}_{\mathrm{I}})<\delta$, we have

$$|\psi (Z,\xi )|[det{(I-Z{\overline{Z}}^{\prime })}^k-{\parallel \xi \parallel }_p^2{]}^{\beta }ln\frac{2q}{det{(I-Z_{\varphi }{\overline{Z_{\varphi }}}^{\prime })}^k-{\parallel {\xi }_{\varphi }\parallel }_p^2}<\epsilon .$$

(34)

According to Lemma 11, we obtain

$$\begin{array}{cc}\hfill & [det{(I-Z{\overline{Z}}^{\prime })}^k-{\parallel \xi \parallel }_p^2{]}^{\beta }|(\psi C_{\varphi }{f}_k)(Z,\xi )|\hfill \\ & =[det{(I-Z{\overline{Z}}^{\prime })}^k-{\parallel \xi \parallel }_p^2{]}^{\beta }|\psi (Z,\xi )·(C_{\varphi }{f}_k)(Z,\xi )|\hfill \\ \hfill & =[det{(I-Z{\overline{Z}}^{\prime })}^k-{\parallel \xi \parallel }_p^2{]}^{\beta }|\psi (Z,\xi )||f_k(\varphi (Z,\xi ))|\hfill \\ \hfill & \le C|\psi (Z,\xi )|[det{(I-Z{\overline{Z}}^{\prime })}^k-{\parallel \xi \parallel }_p^2{]}^{\beta }{\parallel f_k\parallel }_{{\mathcal{B}}^{\alpha }}\hfill \\ \hfill & \times ln\frac{2q}{det{(I-Z_{\varphi }{\overline{Z_{\varphi }}}^{\prime })}^k-{\parallel {\xi }_{\varphi }\parallel }_p^2}\hfill \\ \hfill & \le C{M}_1\epsilon .\hfill \end{array}$$

(35)

On the other hand, let us introduce the set

$$E_{\delta }:=\{(Z,\xi )\in {\mathrm{GHE}}_{\mathrm{I}}:\mathrm{dist}(\varphi (Z,\xi ),\partial {\mathrm{GHE}}_{\mathrm{I}})\ge \delta \},$$

which is a compact subset of ${\mathrm{GHE}}_{\mathrm{I}}$. By the assumptions, $f_k$ converges to 0 uniformly on any compact subset of ${\mathrm{GHE}}_{\mathrm{I}}$. From this, and since $\psi \in {\mathcal{A}}_{\beta }$, for such $\epsilon$, we have

$$\begin{array}{cc}\hfill & [det{(I-Z{\overline{Z}}^{\prime })}^k-{\parallel \xi \parallel }_p^2{]}^{\beta }|(\psi C_{\varphi }{f}_k)(Z,\xi )|\hfill \\ \hfill & =[det{(I-Z{\overline{Z}}^{\prime })}^k-{\parallel \xi \parallel }_p^2{]}^{\beta }|\psi (Z,\xi )·(C_{\varphi }{f}_k)(Z,\xi )|\hfill \\ \hfill & =[det{(I-Z{\overline{Z}}^{\prime })}^k-{\parallel \xi \parallel }_p^2{]}^{\beta }|\psi (Z,\xi )||f_k(\varphi (Z,\xi ))|\hfill \\ \hfill & \le {\parallel \psi \parallel }_{{\mathcal{A}}_{\beta }}\epsilon .\hfill \end{array}$$

(36)

Combining (35) and (36), we have

$$\parallel \psi C_{\varphi }{f}_k{\parallel }_{{\mathcal{A}}_{\beta }}=\underset{(Z,\xi )\in {\mathrm{GHE}}_{\mathrm{I}}}{sup}[det{(I-Z{\overline{Z}}^{\prime })}^k-{\parallel \xi \parallel }_p^2{]}^{\beta }|(\psi C_{\varphi }{f}_k)(Z,\xi )|\to 0,k\to \infty .$$

Consequently, making use of Lemma 12, we finally have that $\psi C_{\varphi }:{\mathcal{B}}^{\alpha }({\mathrm{GHE}}_{\mathrm{I}})\to {\mathcal{A}}_{\beta }({\mathrm{GHE}}_{\mathrm{I}})$ is compact.

Conversely, suppose $\psi C_{\varphi }:{\mathcal{B}}^{\alpha }({\mathrm{GHE}}_{\mathrm{I}})\to {\mathcal{A}}_{\beta }({\mathrm{GHE}}_{\mathrm{I}})$ is compact. Let $f\equiv 1$, we have

$$[det{(I-Z{\overline{Z}}^{\prime })}^k-{\parallel \xi \parallel }_p^2{]}^{\beta }|\psi (Z,\xi )|=[det{(I-Z{\overline{Z}}^{\prime })}^k-{\parallel \xi \parallel }_p^2{]}^{\beta }|(\psi C_{\varphi }f)(Z,\xi )|<\infty .$$

This shows that $\psi \in {\mathcal{A}}_{\beta }.$ Consider now a sequence $(S^i,t^i)=\varphi (Z^i,{\xi }^i)$ in ${\mathrm{GHE}}_{\mathrm{I}}$, such that $\varphi (Z^i,{\xi }^i)\to \partial {\mathrm{GHE}}_{\mathrm{I}}$ as $i\to \infty$. If such a sequence does not exist, then condition (33) obviously holds. Moreover, let us introduce the following sequence of test functions ${\{f_i\}}_{i\ge 1}$:

$$\begin{array}{cc}\hfill f_i(Z,\xi )& ={\left\{ln\frac{2}{det{(I-S^i{\overline{S^i}}^{\prime })}^k-{\parallel t^i\parallel }_p^2}\right\}}^{-1}\times {\left\{ln\frac{2}{det{(I-Z{\overline{S^i}}^{\prime })}^k-{\langle \xi ,t^i\rangle }_p}\right\}}^2\hfill \\ \hfill & \times det{(I-S^i{\overline{S^i}}^{\prime })}^{1-k}.\hfill \end{array}$$

Differentiation gives

$$\begin{array}{ccc}\hfill \frac{\partial f_i}{\partial z_{gl}}& =& \frac{2k\times det{(I-Z{\overline{S^i}}^{\prime })}^{k-1}det{(I-S^i{\overline{S^i}}^{\prime })}^{1-k}}{det{(I-Z{\overline{S^i}}^{\prime })}^k-{\langle \xi ,t^i\rangle }_p}\times \frac{ln\frac{2}{det{(I-Z{\overline{S^i}}^{\prime })}^k-{\langle \xi ,t^i\rangle }_p}}{ln\frac{2}{det{(I-S^i{\overline{S^i}}^{\prime })}^k-{\parallel t^i\parallel }_p^2}}\hfill \\ & & \times det(I-Z{\overline{S^i}}^{\prime })\times \mathrm{tr}[{(I-Z{\overline{S^i}}^{\prime })}^{-1}{I}_{gl}{\overline{S^i}}^{\prime }],1\le g\le m,1\le l\le n,\hfill \\ \hfill \frac{\partial f_i}{\partial {\xi }_j}& =& \frac{2p_j{\xi }_j^{p_j-1}{\overline{t_j}}^{p_j}\times det{(I-S^i{\overline{S^i}}^{\prime })}^{1-k}}{det{(I-Z{\overline{S^i}}^{\prime })}^k-{\langle \xi ,t^i\rangle }_p}\times \frac{ln\frac{2}{det{(I-Z{\overline{S^i}}^{\prime })}^k-{\langle \xi ,t^i\rangle }_p}}{ln\frac{2}{det{(I-S^i{\overline{S^i}}^{\prime })}^k-{\parallel t^i\parallel }_p^2}},j=1,\cdots ,r.i=1,2,\cdots \hfill \end{array}$$

From (29) and Lemmas 14 and 16, there exists a constant $C_5>0$, such that

$$\begin{array}{cc}\hfill & [det{(I-Z{\overline{Z}}^{\prime })}^k-{\parallel \xi |}_p^2]|\nabla f_i(Z,\xi )|\hfill \\ \hfill & =\frac{[det{(I-Z{\overline{Z}}^{\prime })}^k-{\parallel \xi \parallel }_p^2]det{(I-S^i{\overline{S^i}}^{\prime })}^{1-k}}{|det{(I-Z{\overline{S^i}}^{\prime })}^k-{\langle \xi ,t^i\rangle }_p|}\times |\frac{ln\frac{2}{det{(I-Z{\overline{S^i}}^{\prime })}^k-{\langle \xi ,t^i\rangle }_p}}{ln\frac{2}{det{(I-S^i{\overline{S^i}}^{\prime })}^k-{\parallel t^i\parallel }_p^2}}|\hfill \\ \hfill & \times \{4k^2{|det{(I-Z{\overline{S^i}}^{\prime })}^{k-1}|}^2\hfill \\ \hfill & \times \sum _{\genfrac{}{}{0pt}{}{1\le g\le m\hfill }{1\le l\le n\hfill }}|det(I-Z{\overline{S^i}}^{\prime })\mathrm{tr}[{(I-Z{\overline{S^i}}^{\prime })}^{-1}{I}_{gl}{\overline{S^i}}^{\prime }]{|}^2+4\sum _{j=1}^r{|p_j{\xi }_j^{p_j-1}{\overline{t_j}}^{p_j}|}^2{\}}^{\frac{1}{2}}\hfill \\ \hfill & \le \frac{[det{(I-Z{\overline{Z}}^{\prime })}^k-{\parallel \xi \parallel }_p^2][det{(I-S^i{\overline{S^i}}^{\prime })}^{1-k}}{||det{(I-Z{\overline{S^i}}^{\prime })}^k|-|{\langle \xi ,t^i\rangle }_p||}\times \frac{|ln\frac{2}{|det{(I-Z{\overline{S^i}}^{\prime })}^k-{\langle \xi ,t^i\rangle }_p|}|+\pi }{ln\frac{2}{det{(I-S^i{\overline{S^i}}^{\prime })}^k-{\parallel t^i\parallel }_p^2}}\hfill \\ \hfill & \times \{2k|det(I-Z{\overline{S^i}}^{\prime }){|}^{k-1}\hfill \\ \hfill & \times {\left[\sum _{\genfrac{}{}{0pt}{}{1\le g\le m\hfill }{1\le l\le n\hfill }}{|det(I-Z{\overline{S^i}}^{\prime })\mathrm{tr}[{(I-Z{\overline{S^i}}^{\prime })}^{-1}{I}_{gl}{\overline{S^i}}^{\prime }]|}^2\right]}^{\frac{1}{2}}+2{\left[\sum _{j=1}^r{|p_j{\xi }_j^{p_j-1}{\overline{t_j}}^{p_j}|}^2\right]}^{\frac{1}{2}}\}\hfill \\ \hfill & \le \frac{det{(I-Z{\overline{Z}}^{\prime })}^k-{\parallel \xi \parallel }_p^2}{||det{(I-Z{\overline{S^i}}^{\prime })}^k{|-\parallel \xi \parallel }_p\parallel t^i{\parallel }_p|}\times \frac{|ln\frac{2}{|det{(I-Z{\overline{S^i}}^{\prime })}^k-{\langle \xi ,t^i\rangle }_p|}|+\pi }{ln\frac{2}{det{(I-S^i{\overline{S^i}}^{\prime })}^k-{\parallel t^i\parallel }_p^2}}\hfill \\ \hfill & \times \left\{2k{C}_5{|det(I-Z{\overline{S^i}}^{\prime })|}^{k-1}det{(I-S^i{\overline{S^i}}^{\prime })}^{1-k}+2{\left[\sum _{j=1}^r{|p_j{\xi }_j^{p_j-1}{\overline{t_j}}^{p_j}|}^2\right]}^{\frac{1}{2}}det{(I-S^i{\overline{S^i}}^{\prime })}^{1-k}\right\}\hfill \\ \hfill & \le \frac{2[det{(I-Z{\overline{Z}}^{\prime })}^k-\parallel \xi {\parallel }_p^2]}{[det{(I-Z{\overline{Z}}^{\prime })}^k-{\parallel \xi \parallel }_p^2]+[det{(I-S^i{\overline{S^i}}^{\prime })}^k-{\parallel t\parallel }_p^2]}\times \frac{|ln\frac{2}{|det{(I-Z{\overline{S^i}}^{\prime })}^k-{\langle \xi ,t^i\rangle }_p|}|+\pi }{ln\frac{2}{det{(I-S^i{\overline{S^i}}^{\prime })}^k-{\parallel t^i\parallel }_p^2}}\hfill \\ \hfill & \times \left\{2k{C}_5{|det(I-Z{\overline{S^i}}^{\prime })|}^{k-1}det{(I-S^i{\overline{S^i}}^{\prime })}^{1-k}+C_6\right\}\hfill \\ \hfill & \le \frac{2[det{(I-Z{\overline{Z}}^{\prime })}^k-\parallel \xi {\parallel }_p^2]}{det{(I-Z{\overline{Z}}^{\prime })}^k-{\parallel \xi \parallel }_p^2}\times \frac{|ln\frac{2}{|det{(I-Z{\overline{S^i}}^{\prime })}^k-{\langle \xi ,t^i\rangle }_p|}|+\pi }{ln\frac{2}{det{(I-S^i{\overline{S^i}}^{\prime })}^k-{\parallel t^i\parallel }_p^2}}\times (2k{C}_5·2^{m(1-k)}+C_6)\hfill \\ \hfill & \le 2\times (2k{C}_5·2^{m(1-k)}+C_6)\times \frac{|ln\frac{2}{|det{(I-Z{\overline{S^i}}^{\prime })}^k-{\langle \xi ,t^i\rangle }_p|}|+\pi }{ln\frac{2}{det{(I-S^i{\overline{S^i}}^{\prime })}^k-{\parallel t^i\parallel }_p^2}}\hfill \\ \hfill & \le C_7\times \frac{|ln\frac{2}{|det{(I-Z{\overline{S^i}}^{\prime })}^k-{\langle \xi ,t^i\rangle }_p|}|+\pi }{ln\frac{2}{det{(I-S^i{\overline{S^i}}^{\prime })}^k-{\parallel t^i\parallel }_p^2}}.\hfill \end{array}$$

We now have two cases.

Case A${}_1$. If $|det{(I-Z{\overline{S^i}}^{\prime })}^k-{\langle \xi ,t^i\rangle }_p|\le 2$, then

$$\begin{array}{cc}\hfill \frac{|ln\frac{2}{|det{(I-Z{\overline{S^i}}^{\prime })}^k-{\langle \xi ,t^i\rangle }_p|}|+\pi }{ln\frac{2}{det{(I-S^i{\overline{S^i}}^{\prime })}^k-{\parallel t^i\parallel }_p^2}}& \le \frac{ln\frac{2}{|det{(I-Z{\overline{S^i}}^{\prime })}^k|-|{\langle \xi ,t^i\rangle }_p|}+\pi }{ln\frac{2}{det{(I-S^i{\overline{S^i}}^{\prime })}^k-{\parallel t^i\parallel }_p^2}}\hfill \\ \hfill & \le \frac{ln\frac{2}{|det{(I-Z{\overline{S^i}}^{\prime })}^k{|-\parallel \xi \parallel }_p{\parallel t^i\parallel }_p}+\pi }{ln\frac{2}{det{(I-S^i{\overline{S^i}}^{\prime })}^k-{\parallel t^i\parallel }_p^2}}\hfill \\ \hfill & \le \frac{ln\frac{4}{[det{(I-Z{\overline{Z}}^{\prime })}^k|-{\parallel \xi \parallel }_p^2]+[det{(I-S^i{\overline{S^i}}^{\prime })}^k-\parallel t^i{{\parallel }^2]}_p}+\pi }{ln\frac{2}{det{(I-S^i{\overline{S^i}}^{\prime })}^k-{\parallel t^i\parallel }_p^2}}\hfill \\ \hfill & \le \frac{ln\frac{4}{det{(I-S^i{\overline{S^i}}^{\prime })}^k-{\parallel t^i\parallel }_p^2}+\pi }{ln\frac{2}{det{(I-S^i{\overline{S^i}}^{\prime })}^k-{\parallel t^i\parallel }_p^2}}\hfill \\ \hfill & \le 2+\frac{\pi }{ln\frac{2}{det{(I-S^i{\overline{S^i}}^{\prime })}^k-{\parallel t^i\parallel }_p^2}}\hfill \\ \hfill & \le C_8,\hfill \end{array}$$

(37)

where $C_8=2+\frac{\pi }{ln2}$.

Case A${}_2$. If $|det{(I-Z{\overline{S^i}}^{\prime })}^k-{\langle \xi ,t^i\rangle }_p|>2$, then

$$\begin{array}{cc}\hfill \frac{|ln\frac{2}{|det{(I-Z{\overline{S^i}}^{\prime })}^k-{\langle \xi ,t^i\rangle }_p|}|+\pi }{ln\frac{2}{det{(I-S^i{\overline{S^i}}^{\prime })}^k-{\parallel t^i\parallel }_p^2}}& =\frac{|ln2-ln|det{(I-Z{\overline{S^i}}^{\prime })}^k-{\langle \xi ,t^i\rangle }_p||+\pi }{ln\frac{2}{det{(I-S^i{\overline{S^i}}^{\prime })}^k-{\parallel t^i\parallel }_p^2}}\hfill \end{array}$$

(38)

$$\begin{array}{cc}\hfill & \le \frac{ln|det{(I-Z{\overline{S^i}}^{\prime })}^k-{\langle \xi ,t^i\rangle }_p|+\pi }{ln\frac{2}{det{(I-S^i{\overline{S^i}}^{\prime })}^k-{\parallel t^i\parallel }_p^2}}\hfill \\ \hfill & \le \frac{ln(|det{(I-Z{\overline{S^i}}^{\prime })}^k|+|{\langle \xi ,t^i\rangle }_p|)+\pi }{ln2}.\hfill \end{array}$$

Since $Z{\overline{S^i}}^{\prime }=C={(c_{ij})}_{m\times m},$ $c_{ij}={\sum }_{g=1}^n{z}_{ig}{s}_{jg}(i,j=1,\cdots ,m)$, we may write $I-C=D={(d_{ij})}_{m\times m}$ with

$$d_{ij}=\left\{\begin{array}{ccc}\hfill & 1-(\sum _{g=1}^n{z}_{ig}{s}_{jg})\hfill & \hfill i=j\\ \hfill & -\sum _{g=1}^n{z}_{ig}{s}_{jg}\hfill & \hfill i\ne j\end{array}\right.$$

(39)

Using (39), we have $det(I-C)={\sum }_{j_1{j}_2\cdots j_m}{(-1)}^{\tau (j_1{j}_2\cdots j_m)}{d}_{1j_1}{d}_{2j_2}\cdots d_{m{j}_m}$ and

$$|det(I-C)|\le m!{(n+1)}^m=G$$

Hence,

$$\frac{|ln\frac{2}{|det{(I-Z{\overline{S^i}}^{\prime })}^k-{\langle \xi ,t^i\rangle }_p|}|+\pi }{ln\frac{2}{det{(I-S^i{\overline{S^i}}^{\prime })}^k-{\parallel t^i\parallel }_p^2}}\le \frac{ln(G^k+\parallel \xi {\parallel }_p\parallel t^i{\parallel }_p)+\pi }{ln2}\le \frac{ln(G^k+1)+\pi }{ln2}\le C_9.$$

By using both cases $A_1$ and $A_2$, we have $[det(I-Z{\overline{Z}}^{\prime })-{\parallel \xi \parallel }_p^2]|\nabla f_i(Z,\xi )|\le Q{C}_7$ and then $\parallel f_i{\parallel }_{{\mathcal{B}}^{\alpha }}\le Q{C}_7$, which means that $\parallel f_i{\parallel }_{{\mathcal{B}}^{\alpha }}$ is bounded, where $Q=max\{C_8,C_9\}.$ It follows that ${\{f_i\}}_{i\ge 1}\in {\mathcal{B}}^{\alpha }({\mathrm{GHE}}_{\mathrm{I}})$ and

$$\begin{array}{cc}\hfill |f_i(Z,\xi )|& ={\left\{ln\frac{2}{det{(I-S^i{\overline{S^i}}^{\prime })}^k-{\parallel t^i{\parallel }_p}^2}\right\}}^{-1}\times |ln\frac{2}{det{(I-Z{\overline{S^i}}^{\prime })}^k-{\langle \xi ,t^i\rangle }_p}{|}^2\hfill \\ \hfill & \times det{(I-S^i{\overline{S^i}}^{\prime })}^{1-k}\hfill \\ \hfill & \le {\left\{ln\frac{2}{det{(I-S^i{\overline{S^i}}^{\prime })}^k-{\parallel t^i\parallel }_p^2}\right\}}^{-1}\times det{(I-S^i{\overline{S^i}}^{\prime })}^{1-k}\hfill \\ \hfill & \times {\left\{|ln\frac{2}{|det{(I-Z{\overline{S^i}}^{\prime })}^k-{\langle \xi ,t^i\rangle }_p|}|+\pi \right\}}^2.\hfill \end{array}$$

If $|det{(I-Z{\overline{S^i}}^{\prime })}^k-{\langle \xi ,t^i\rangle }_p|\le 2$, then

$$\begin{array}{cc}\hfill |f_i(Z,\xi )|& \le {\left\{ln\frac{2}{det{(I-S^i{\overline{S^i}}^{\prime })}^k-{\parallel t^i\parallel }_p^2}\right\}}^{-1}\times det{(I-S^i{\overline{S^i}}^{\prime })}^{1-k}\hfill \\ \hfill & \times {\left\{ln\frac{2}{|det{(I-Z{\overline{S^i}}^{\prime })}^k|-|{\langle \xi ,t^i\rangle }_p|}+\pi \right\}}^2\hfill \\ \hfill & \le {\left\{ln\frac{2}{det{(I-S^i{\overline{S^i}}^{\prime })}^k-{\parallel t^i\parallel }_p^2}\right\}}^{-1}\times det{(I-S^i{\overline{S^i}}^{\prime })}^{1-k}\hfill \\ \hfill & \times {\left\{ln\frac{2}{|det{(I-Z{\overline{S^i}}^{\prime })}^k{|-\parallel \xi \parallel }_p{\parallel t^i\parallel }_p}+\pi \right\}}^2\hfill \\ \hfill & \le {\left\{ln\frac{2}{det{(I-S^i{\overline{S^i}}^{\prime })}^k-{\parallel t^i\parallel }_p^2}\right\}}^{-1}\times det{(I-S^i{\overline{S^i}}^{\prime })}^{1-k}\hfill \\ \hfill & \times {\left\{ln\frac{4}{[det{(I-Z{\overline{Z}}^{\prime })}^k-{\parallel \xi \parallel }_p^2]+[det{(I-S^i{\overline{S^i}}^{\prime })}^k-{\parallel t^i\parallel }_p^2]}+\pi \right\}}^2\hfill \\ \hfill & \le {\left\{ln\frac{2}{det{(I-S^i{\overline{S^i}}^{\prime })}^k-{\parallel t^i\parallel }_p^2}\right\}}^{-1}\times det{(I-S^i{\overline{S^i}}^{\prime })}^{1-k}\hfill \\ \hfill & \times {\left\{ln\frac{4}{det{(I-Z{\overline{Z}}^{\prime })}^k-{\parallel \xi \parallel }_p^2}+\pi \right\}}^2\hfill \end{array}$$

Since $0<det{(I-S^i{\overline{S^i}}^{\prime })}^{1-k}\le 1$, we take $i\to \infty$ and obtain $(S^i,t^i)\to \partial {\mathrm{GHE}}_{\mathrm{I}}$. This implies $det{(I-S^i{\overline{S^i}}^{\prime })}^k-{\parallel t^i\parallel }_p^2\to 0$, then ${\left\{ln\frac{2}{det{(I-S^i{\overline{S^i}}^{\prime })}^k-{\parallel t^i\parallel }_p^2}\right\}}^{-1}\to 0$. Let us now consider a compact subset E of ${\mathrm{GHE}}_{\mathrm{I}}$. For $(Z,\xi )\in E$, it is easy to see that $det{(I-Z{\overline{Z}}^{\prime })}^k-{\parallel \xi \parallel }_p^2$ has a positive lower bound. Thus, we have $f_i(Z,\xi )\rightrightarrows 0,i\to \infty$ on all compact subsets of ${\mathrm{GHE}}_{\mathrm{I}}$. If $|det{(I-Z{\overline{S^i}}^{\prime })}^k-{\langle \xi ,t^i\rangle }_p|>2$, then

$$\begin{array}{cc}\hfill |f_i(Z,\xi )|& \le {\left\{ln\frac{2}{det{(I-S^i{\overline{S^i}}^{\prime })}^k-{\parallel t^i\parallel }_p^2}\right\}}^{-1}\times det{(I-S^i{\overline{S^i}}^{\prime })}^{1-k}\hfill \\ \hfill & \times {\left\{|ln2-ln(|det{(I-Z{\overline{S^i}}^{\prime })}^k-{\langle \xi ,t^i\rangle }_p|)|+\pi \right\}}^2\hfill \\ \hfill & \le {\left\{ln\frac{2}{det{(I-S^i{\overline{S^i}}^{\prime })}^k-{\parallel t^i\parallel }_p^2}\right\}}^{-1}\times det{(I-S^i{\overline{S^i}}^{\prime })}^{1-k}\hfill \\ \hfill & \times {\left\{ln(|det{(I-Z{\overline{S^i}}^{\prime })}^k|+|{\langle \xi ,t^i\rangle }_p|)+\pi \right\}}^2\hfill \\ \hfill & \le {\left\{ln\frac{2}{det{(I-S^i{\overline{S^i}}^{\prime })}^k-{\parallel t^i\parallel }_p^2}\right\}}^{-1}\times det{(I-S^i{\overline{S^i}}^{\prime })}^{1-k}\hfill \\ \hfill & \times {\{ln(G^k+1)+\pi \}}^2\hfill \end{array}$$

From $0<det{(I-S^i{\overline{S^i}}^{\prime })}^{1-k}\le 1$ and ${\left\{ln\frac{2}{det{(I-S^i{\overline{S^i}}^{\prime })}^k-{\parallel t^i\parallel }_p^2}\right\}}^{-1}\to 0$ as $i\to \infty$, one concludes that $f_i(Z,\xi )\to 0,i\to \infty$.

The above proof shows that $f_i(Z,\xi )\rightrightarrows 0,i\to \infty$ on all compact subsets of ${\mathrm{GHE}}_{\mathrm{I}}$. By Lemma 12, this implies that $\parallel \psi C_{\varphi }{f}_i{\parallel }_{{\mathcal{A}}_{\beta }}\to 0$. Therefore, we conclude that

$$\begin{array}{cc}\hfill 0\leftarrow & \parallel \psi C_{\varphi }{f}_i{\parallel }_{{\mathcal{A}}_{\beta }}\hfill \\ \hfill & =\underset{\varphi (Z,\xi )\in {\mathrm{GHE}}_{\mathrm{I}}}{sup}{[det{(I-Z{\overline{Z}}^{\prime })}^k-{\parallel \xi \parallel }_p^2]}^{\beta }|\psi (Z,\xi )|{\left\{ln\frac{2}{det{(I-S^i{\overline{S^i}}^{\prime })}^k-{\parallel t^i\parallel }_p^2}\right\}}^{-1}\hfill \\ \hfill & \times |ln\frac{2}{det{(I-Z_{\varphi }{\overline{S^i}}^{\prime })}^k-{\langle {\xi }_{\varphi },t^i\rangle }_p}{|}^2\times det{(I-S^i{\overline{S^i}}^{\prime })}^{1-k}\hfill \\ \hfill & \ge {[det{(I-Z^i{\overline{Z^i}}^{\prime })}^k-{\parallel {\xi }^i\parallel }_p^2]}^{\beta }|\psi (Z^i,{\xi }^i)|{\left\{ln\frac{2}{det{(I-S^i{\overline{S^i}}^{\prime })}^k-{\parallel t^i\parallel }_p^2}\right\}}^{-1}\hfill \\ \hfill & \times {\left\{ln\frac{2}{det{(I-S^i{\overline{S^i}}^{\prime })}^k-{\parallel t^i\parallel }_p^2}\right\}}^2\times det{(I-S^i{\overline{S^i}}^{\prime })}^{1-k}\hfill \\ \hfill & =|\psi (Z^i,{\xi }^i)|{[det{(I-Z^i{\overline{Z^i}}^{\prime })}^k-{\parallel {\xi }^i\parallel }_p^2]}^{\beta }det{(I-S^i{\overline{S^i}}^{\prime })}^{1-k}\times ln\frac{2}{det{(I-S^i{\overline{S^i}}^{\prime })}^k-{\parallel t^i\parallel }_p^2}.\hfill \end{array}$$

□

Theorem 4.

Assume that $\alpha >1,\beta >0$, $0<km\le 1$, and that $p_j$ are some positive integers $(j=1,2,\cdots ,r)$. Let $\varphi =({\varphi }_{11},{\varphi }_{12}\cdots {\varphi }_{mn+r})$ be a holomorphic self-map of ${\mathrm{GHE}}_{\mathrm{I}}$, with $\psi \in H({\mathrm{GHE}}_{\mathrm{I}})$ and $(Z_{\varphi },{\xi }_{\varphi })=\varphi (Z,\xi )$. If $\psi \in {\mathcal{A}}_{\beta }$ and

$$\underset{\varphi (Z,\xi )\to \partial {\mathrm{GHE}}_{\mathrm{I}}}{lim}\frac{|\psi (Z,\xi )|{[det{(I-Z{\overline{Z}}^{\prime })}^k-{\parallel \xi \parallel }_p^2]}^{\beta }}{[det{(I-Z_{\varphi }{\overline{Z_{\varphi }}}^{\prime })}^k-\parallel {\xi }_{\varphi }{{\parallel }_p^2]}^{\alpha -1}}=0,$$

(40)

then the weighted composition operator $\psi C_{\varphi }:{\mathcal{B}}^{\alpha }({\mathrm{GHE}}_{\mathrm{I}})\to {\mathcal{A}}_{\beta }({\mathrm{GHE}}_{\mathrm{I}})$ is compact.

Conversely, if the weighted composition operator $\psi C_{\varphi }:{\mathcal{B}}^{\alpha }({\mathrm{GHE}}_{\mathrm{I}})\to {\mathcal{A}}_{\beta }({\mathrm{GHE}}_{\mathrm{I}})$ is compact, then $\psi \in {\mathcal{A}}_{\beta }$ and

$$\underset{\varphi (Z,\xi )\to \partial {\mathrm{GHE}}_{\mathrm{I}}}{lim}\frac{|\psi (Z,\xi )|{[det{(I-Z{\overline{Z}}^{\prime })}^k-{\parallel \xi \parallel }_p^2]}^{\beta }}{[det{(I-Z_{\varphi }{\overline{Z_{\varphi }}}^{\prime })}^k-\parallel {\xi }_{\varphi }{{\parallel }_p^2]}^{\alpha -\frac{1}{k}}}=0.$$

(41)

Proof. 

Assume that (40) holds. We have

$$\underset{(Z,\xi )\in {\mathrm{GHE}}_{\mathrm{I}}}{sup}\frac{|\psi (Z,\xi )|{[det{(I-Z{\overline{Z}}^{\prime })}^k-{\parallel \xi \parallel }_p^2]}^{\beta }}{[det{(I-Z_{\varphi }{\overline{Z_{\varphi }}}^{\prime })}^k-\parallel {\xi }_{\varphi }{{\parallel }_p^2]}^{\alpha -1}}<\infty .$$

From Theorem 2, it follows that $\psi C_{\varphi }:{\mathcal{B}}^{\alpha }({\mathrm{GHE}}_{\mathrm{I}})\to {\mathcal{A}}_{\beta }({\mathrm{GHE}}_{\mathrm{I}})$ is bounded. Let ${\{f_k\}}_{k\ge 1}$ be a bounded sequence in ${\mathcal{B}}^{\alpha }({\mathrm{GHE}}_{\mathrm{I}})$ with $f_k$ that converges to 0 uniformly on compact subsets of ${\mathrm{GHE}}_{\mathrm{I}}$. There exists $M_2>0$, such that $\parallel f_k{\parallel }_{{\mathcal{B}}^{\alpha }}\le M_2,k=1,2,\cdots$. By (40), for any $\epsilon >0$, there is a constant $\delta \in (0,1)$, such that

$$\frac{|\psi (Z,\xi )|{[det{(I-Z{\overline{Z}}^{\prime })}^k-{\parallel \xi \parallel }_p^2]}^{\beta }}{[det{(I-Z_{\varphi }{\overline{Z_{\varphi }}}^{\prime })}^k-\parallel {\xi }_{\varphi }{{\parallel }_p^2]}^{\alpha -1}}<\epsilon ,$$

(42)

for $\mathrm{dist}(\varphi (Z,\xi ),\partial {\mathrm{GHE}}_{\mathrm{I}})<\delta$. Using Lemma 11, we have

$$\begin{array}{cc}\hfill & {[det{(I-Z{\overline{Z}}^{\prime })}^k-{\parallel \xi \parallel }_p^2]}^{\beta }|(\psi C_{\varphi }{f}_k)(Z,\xi )|\hfill \\ \hfill & ={[det{(I-Z{\overline{Z}}^{\prime })}^k-{\parallel \xi \parallel }_p^2]}^{\beta }|\psi (Z,\xi )·(C_{\varphi }{f}_k)(Z,\xi )|\hfill \\ \hfill & ={[det{(I-Z{\overline{Z}}^{\prime })}^k-{\parallel \xi \parallel }_p^2]}^{\beta }|\psi (Z,\xi )||f_k(\varphi (Z,\xi ))|\hfill \\ \hfill & \le C|\psi (Z,\xi )|\parallel f_k{\parallel }_{{\mathcal{B}}^{\alpha }}\frac{{[det{(I-Z{\overline{Z}}^{\prime })}^k-{\parallel \xi \parallel }_p^2]}^{\beta }}{[det{(I-Z_{\varphi }{\overline{Z_{\varphi }}}^{\prime })}^k-\parallel {\xi }_{\varphi }{{\parallel }_p^2]}^{\alpha -1}}\hfill \\ \hfill & \le C{M}_2\epsilon .\hfill \end{array}$$

(43)

On the other hand, if we set

$$E_{\delta }:=\{(Z,\xi )\in {\mathrm{GHE}}_{\mathrm{I}}:\mathrm{dist}(\varphi (Z,\xi ),\partial {\mathrm{GHE}}_{\mathrm{I}})\ge \delta \},$$

we have that $E_{\delta }$ is a compact subset of ${\mathrm{GHE}}_{\mathrm{I}}$. For $\epsilon$ defined in (42), $f_k$ converges to 0 uniformly on any compact subset of ${\mathrm{GHE}}_{\mathrm{I}}$. For $\psi \in {\mathcal{A}}_{\beta }$, we have

$$\begin{array}{cc}\hfill & {[det{(I-Z{\overline{Z}}^{\prime })}^k-{\parallel \xi \parallel }_p^2]}^{\beta }|(\psi C_{\varphi }{f}_k)(Z,\xi )|\hfill \\ \hfill & ={[det{(I-Z{\overline{Z}}^{\prime })}^k-{\parallel \xi \parallel }_p^2]}^{\beta }|\psi (Z,\xi )·(C_{\varphi }{f}_k)(Z,\xi )|\hfill \\ \hfill & ={[det{(I-Z{\overline{Z}}^{\prime })}^k-{\parallel \xi \parallel }_p^2]}^{\beta }|\psi (Z,\xi )||f_k(\varphi (Z,\xi ))|\hfill \\ \hfill & \le {\parallel \psi \parallel }_{{\mathcal{A}}_{\beta }}\epsilon .\hfill \end{array}$$

(44)

According to inequalities (43) and (44), we see that

$$\parallel \psi C_{\varphi }{f}_k{\parallel }_{{\mathcal{A}}_{\beta }}=\underset{(Z,\xi )\in {\mathrm{GHE}}_{\mathrm{I}}}{sup}{[det{(I-Z{\overline{Z}}^{\prime })}^k-{\parallel \xi \parallel }_p^2]}^{\beta }|(\psi C_{\varphi }{f}_k)(Z,\xi )|\to 0,k\to \infty .$$

Consequently, making use of Lemma 12, we have that $\psi C_{\varphi }:{\mathcal{B}}^{\alpha }({\mathrm{GHE}}_{\mathrm{I}})\to {\mathcal{A}}_{\beta }({\mathrm{GHE}}_{\mathrm{I}})$ is compact.

Conversely, suppose that $\psi C_{\varphi }:{\mathcal{B}}^{\alpha }({\mathrm{GHE}}_{\mathrm{I}})\to {\mathcal{A}}_{\beta }({\mathrm{GHE}}_{\mathrm{I}})$ is compact. Then, $\psi C_{\varphi }:{\mathcal{B}}^{\alpha }({\mathrm{GHE}}_{\mathrm{I}})\to {\mathcal{A}}_{\beta }({\mathrm{GHE}}_{\mathrm{I}})$ is bounded. Let $f\equiv 1$, we obtain

$$[det{(I-Z{\overline{Z}}^{\prime })}^k-{\parallel \xi \parallel }_p^2{]}^{\beta }|\psi (Z,\xi )|=[det{(I-Z{\overline{Z}}^{\prime })}^k-{\parallel \xi \parallel }_p^2{]}^{\beta }|(\psi C_{\varphi }f)(Z,\xi )|<\infty .$$

This shows that $\psi \in {\mathcal{A}}_{\beta }.$ Consider now a sequence $(S^i,t^i)=\varphi (Z^i,{\xi }^i)$ in ${\mathrm{GHE}}_{\mathrm{I}}$ such that $\varphi (Z^i,{\xi }^i)\to \partial {\mathrm{GHE}}_{\mathrm{I}}$ as $i\to \infty$. If such a sequence does not exist, then condition (41) obviously holds.

Moreover, let us introduce a sequence of test functions ${\{f_i\}}_{i\ge 1}$:

$$f_i(Z,\xi ):=\frac{{[det{(I-S^i{\overline{S^i}}^{\prime })}^k-{\parallel t^i\parallel }_p^2]}^{\frac{1}{k}-1+\alpha }}{{[det{(I-Z{\overline{S^i}}^{\prime })}^k-{\langle \xi ,t^i\rangle }_p]}^{2\alpha -1}}.$$

Differentiation gives

$$\begin{array}{ccc}\hfill \frac{\partial f_i}{\partial z_{gl}}& =& \frac{(2\alpha -1)k·det{(I-Z{\overline{S^i}}^{\prime })}^{k-1}{[det{(I-S^i{\overline{S^i}}^{\prime })}^k-{\parallel t^i\parallel }_p^2]}^{\frac{1}{k}-1+\alpha }}{{[det{(I-Z{\overline{S^i}}^{\prime })}^k-{\langle \xi ,t^i\rangle }_p]}^{2\alpha }}\hfill \\ & & \times det(I-Z{\overline{S^i}}^{\prime })\mathrm{tr}[{(I-Z{\overline{S^i}}^{\prime })}^{-1}{I}_{gl}{\overline{S^i}}^{\prime }],\hfill \\ \hfill \frac{\partial f_i}{\partial {\xi }_j}& =& \frac{(2\alpha -1)p_j{\xi }_j^{p_j-1}{\overline{t_j}}^{p_j}{[det{(I-S^i{\overline{S^i}}^{\prime })}^k-{\parallel t^i\parallel }_p^2]}^{\frac{1}{k}-1+\alpha }}{{[det{(I-Z{\overline{S^i}}^{\prime })}^k-{\langle \xi ,t^i\rangle }_p]}^{2\alpha }}.\hfill \end{array}$$

From (29) and Lemma 15, it follows that there exists a constant $C_{10}>0$, such that

$$\begin{array}{cc}\hfill & [det{(I-Z{\overline{Z}}^{\prime })}^k-{\parallel \xi \parallel }_p^2{]}^{\alpha }|\nabla f_i(Z,\xi )|\hfill \\ \hfill & =\frac{(2\alpha -1)[det{(I-Z{\overline{Z}}^{\prime })}^k-{\parallel \xi \parallel }_p^2{]}^{\alpha }{[det{(I-S^i{\overline{S^i}}^{\prime })}^k-{\parallel t^i\parallel }_p^2]}^{\frac{1}{k}-1+\alpha }}{|det{(I-Z{\overline{S^i}}^{\prime })}^k-{\langle \xi ,t^i\rangle }_p{|}^{2\alpha }}\times \{k^2{|det{(I-Z{\overline{S^i}}^{\prime })}^{k-1}|}^2\hfill \\ & \times \sum _{\genfrac{}{}{0pt}{}{1\le g\le m\hfill }{1\le l\le n\hfill }}|det(I-Z{\overline{S^i}}^{\prime })\mathrm{tr}[{(I-Z{\overline{S^i}}^{\prime })}^{-1}{I}_{gl}{\overline{S^i}}^{\prime }]{|}^2+\sum _{j=1}^r{|p_j{\xi }_j^{p_j-1}{\overline{t_j}}^{p_j}|}^2{\}}^{\frac{1}{2}}\hfill \\ \hfill & \le \frac{(2\alpha -1)[det{(I-Z{\overline{Z}}^{\prime })}^k-{\parallel \xi \parallel }_p^2{]}^{\alpha }{[det{(I-S^i{\overline{S^i}}^{\prime })}^k-{\parallel t^i\parallel }_p^2]}^{\alpha }}{|det{(I-Z{\overline{S^i}}^{\prime })}^k-{\langle \xi ,t^i\rangle }_p{|}^{2\alpha }}\hfill \\ \hfill & \times \{k|det(I-Z{\overline{S^i}}^{\prime }){|}^{k-1}\times {[det{(I-S^i{\overline{S^i}}^{\prime })}^k-{\parallel t^i\parallel }_p^2]}^{\frac{1}{k}-1}\hfill \\ \hfill & \times {\left[\sum _{\genfrac{}{}{0pt}{}{1\le g\le m\hfill }{1\le l\le n\hfill }}{|det(I-Z{\overline{S^i}}^{\prime })\mathrm{tr}[{(I-Z{\overline{S^i}}^{\prime })}^{-1}{I}_{gl}{\overline{S^i}}^{\prime }]|}^2\right]}^{\frac{1}{2}}\hfill \\ \hfill & +{\left[\sum _{j=1}^r{|p_j{\xi }_j^{p_j-1}{\overline{t_j}}^{\prime p_j}|}^2\right]}^{\frac{1}{2}}\times {[det{(I-S^i{\overline{S^i}}^{\prime })}^k-{\parallel t^i\parallel }_p^2]}^{\frac{1}{k}-1}\}\hfill \\ \hfill & \le (2\alpha -1)\times \left\{k·C_1{|det(I-Z{\overline{S^i}}^{\prime })|}^{k-1}{[det{(I-S^i{\overline{S^i}}^{\prime })}^k]}^{\frac{1}{k}-1}+C_{10}\right\}\hfill \\ \hfill & \le (2\alpha -1)\times \left\{k·C_1{|det(I-Z{\overline{S^i}}^{\prime })|}^{k-1}det{(I-S^i{\overline{S^i}}^{\prime })}^{1-k}+C_{10}\right\}\hfill \\ \hfill & \le (2\alpha -1)\times \left\{C_1·k·2^{m(1-k)}+C_{10}\right\}\hfill \\ \hfill & \le C^{″}.\hfill \end{array}$$

This shows that $f_i\in {\mathcal{B}}^{\alpha }({\mathrm{GHE}}_{\mathrm{I}})$, $i=1,2,\cdots$ and

$$\begin{array}{cc}\hfill |f_i(Z,\xi )|& =\frac{{[det{(I-S^i{\overline{S^i}}^{\prime })}^k-{\parallel t^i\parallel }_p^2]}^{\frac{1}{k}-1+\alpha }}{|det{(I-Z{\overline{S^i}}^{\prime })}^k-{\langle \xi ,t^i\rangle }_p{|}^{2\alpha -1}}\hfill \\ \hfill & \le \frac{{[det{(I-S^i{\overline{S^i}}^{\prime })}^k-{\parallel t^i\parallel }_p^2]}^{\frac{1}{k}-1+\alpha }}{||det{(I-Z{\overline{S^i}}^{\prime })}^k|-|{\langle \xi ,t^i\rangle }_p{||}^{2\alpha -1}}\hfill \\ \hfill & \le \frac{{[det{(I-S^i{\overline{S^i}}^{\prime })}^k-{\parallel t^i\parallel }_p^2]}^{\frac{1}{k}-1+\alpha }}{|det{(I-Z{\overline{S^i}}^{\prime })}^k{|-\parallel \xi \parallel }_p\parallel t^i{{\parallel }_p|}^{2\alpha -1}}\hfill \\ \hfill & \le \frac{2^{2\alpha -1}{[det{(I-S^i{\overline{S^i}}^{\prime })}^k-{\parallel t^i\parallel }_p^2]}^{\frac{1}{k}-1+\alpha }}{[det{(I-Z{\overline{Z}}^{\prime })}^k-{\parallel \xi \parallel }_p^2+det{(I-S^i{\overline{S^i}}^{\prime })}^k-\parallel t^i{{\parallel }_p^2]}^{2\alpha -1}}\hfill \\ \hfill & \le \frac{2^{2\alpha -1}{[det{(I-S^i{\overline{S^i}}^{\prime })}^k-{\parallel t^i\parallel }_p^2]}^{\frac{1}{k}-1+\alpha }}{[det{(I-Z{\overline{Z}}^{\prime })}^k-{\parallel \xi \parallel }_p^2{]}^{2\alpha -1}}.\hfill \end{array}$$

Taking $i\to \infty$, we have $(S^i,t^i)\to \partial {\mathrm{GHE}}_{\mathrm{I}}$. This implies that $det{(I-S^i{\overline{S^i}}^{\prime })}^k-{\parallel t^i\parallel }_p^2\to 0$. If E is a compact subset of ${\mathrm{GHE}}_{\mathrm{I}}$, for $(Z,\xi )\in E$, we have that $det{(I-Z{\overline{Z}}^{\prime })}^k-{\parallel \xi \parallel }_p^2$ has a positive lower bound. Thus, we have $f_i(Z,\xi )\rightrightarrows 0,i\to \infty$ on all compact subsets of ${\mathrm{GHE}}_{\mathrm{I}}$. According to Lemma 12, we have that $\parallel \psi C_{\varphi }{f}_i{\parallel }_{{\mathcal{A}}_{\beta }}\to 0$. Hence,

$$\begin{array}{cc}\hfill 0\leftarrow \parallel \psi C_{\varphi }{f}_i{\parallel }_{{\mathcal{A}}_{\beta }}& =\underset{\varphi (Z,\xi )\in {\mathrm{GHE}}_{\mathrm{I}}}{sup}[det{(I-Z{\overline{Z}}^{\prime })}^k-{\parallel \xi \parallel }_p^2{]}^{\beta }|\psi (Z,\xi )|\frac{{[det{(I-S^i{\overline{S^i}}^{\prime })}^k-{\parallel t^i\parallel }_p^2]}^{\frac{1}{k}-1+\alpha }}{|det{(I-Z_{\varphi }{\overline{S^i}}^{\prime })}^k-{\langle {\xi }_{\varphi },t^i\rangle }_p{|}^{2\alpha -1}}\hfill \\ \hfill & \ge [det{(I-Z^i{\overline{Z^i}}^{\prime })}^k-\parallel {\xi }^i{{\parallel }_p^2]}^{\beta }\frac{|\psi (Z^i,{\xi }^i)|}{[det{(I-Z_{\varphi }^i{\overline{Z_{\varphi }^i}}^{\prime })}^k-\parallel {\xi }_{\varphi }^i{{\parallel }_p^2]}^{\alpha -\frac{1}{k}}}.\hfill \end{array}$$

□

Corollary 2.

For $\alpha >1,k=m=1,p_1=\cdots =p_r=1$, we are back to the case of the unit ball $\mathbf{B}=\{z\in {\mathbb{C}}^{n+r}:|z{|}^2<1\}$, and $\psi C_{\varphi }:{\mathcal{B}}^{\alpha }(\mathbf{B})\to {\mathcal{A}}_{\beta }(\mathbf{B})$ is compact if and only if $\psi \in {\mathcal{A}}_{\beta }$ and

$$\underset{\varphi (z)\to \partial B}{lim}\frac{|\psi (z)|(1-{|z|}^2{)}^{\beta }}{{(1-|\varphi (z)|}^2{)}^{\alpha -1}}=0,$$

when $\beta =0$. Also, in this case, the result is analog to that obtained by Li and Stević in [11].