On the Product of Weighted Composition Operators and Radial Derivative Operators from the Bloch-Type Space into the Bers-Type Space on the Fourth Loo-Keng Hua Domain

Abstract

Let $H{E}_{IV}$ be the fourth Loo-Keng Hua domain. We study the boundedness of the product of the weighted composition operator and the radial derivative operator from the Bloch-type space ${\mathcal{B}}^{\alpha }\left(H{E}_{IV}\right)$ into the Bers-type space ${\mathcal{A}}^{\beta }\left(H{E}_{IV}\right)$ and provide the necessary and sufficient conditions for their boundedness.

1. Introduction and Preliminaries

Throughout this paper, let $m,n,r,N,N_1,\cdots ,N_r$ be positive integers and $p_1,\cdots ,p_r$ be positive real numbers. Let $z=(z_1,\cdots ,z_n)$ and $w=(w_1,\cdots ,w_n)$ be points in the complex vector space ${\mathbb{C}}^n$. The notation $\overline{z}:=(\overline{z_1},\cdots ,\overline{z_n})$ represents the conjugate of the vector z, and $z^T$ denotes the transpose of z. The inner product of z and w is given by $z{\overline{w}}^T:=\langle z,w\rangle =z_1\overline{w_1}+z_2\overline{w_2}+\cdots +z_n\overline{w_n}$, where $\overline{w_k}$ is the complex conjugate of $w_k$. Additionally, we write the norm of z as

$$|z|=\sqrt{\langle z,z\rangle }=\sqrt{|z_1{|}^2+|z_2{|}^2+\cdots +{|z_n|}^2}.$$

Let $\mathbb{B}=\{z\in {\mathbb{C}}^n:|z|<1\}$ be the open unit ball in ${\mathbb{C}}^n$. As is well known, in the 1930s, Cartan fully characterized the irreducible bounded symmetric domains into six types. The first four types of irreducible domains are called the classical bounded symmetric domains (or the Cartan domain). In this paper, we need the fourth Loo-Keng Hua domain [1], which is realized as the following form:

$$\begin{array}{ccc}& & H{E}_{IV}=H{E}_{IV}(N_1,\cdots ,N_r,N,p_1,\cdots ,p_r)\hfill \\ & & =\left\{(z,\xi ):z\in {\mathfrak{R}}_{IV}\left(N\right),\xi =({\xi }_1,\cdots ,{\xi }_r),{\xi }_j\in {\mathbb{C}}^{N_j},\sum _{j=1}^r{|{\xi }_j|}^{2p_j}<1+{\left|z{z}^T\right|}^2-2z{\overline{z}}^T\right\},\hfill \end{array}$$

where ${\xi }_j=({\xi }_{j_1},\cdots ,{\xi }_{j_{N_j}})$; $j=1,\cdots ,r$; and

$${\mathfrak{R}}_{IV}\left(N\right)=\left\{z\in {\mathbb{C}}^N:1+{\left|z{z}^T\right|}^2-2z{\overline{z}}^T>0,1-{\left|z{z}^T\right|}^2>0\right\}$$

is the fourth Cartan domain. We know that the fourth Loo-Keng Hua domain is not transitive, except in some special cases.

On the fourth Loo-Keng Hua domain, Huang and Jiang studied the generalized Hua’s inequality on the fourth Loo-Keng Hua domain and gave an application ([2]). For more information on the first, second, third, and fourth Loo-Keng Hua domains, please refer to [1,3,4,5,6].

Let $\mathbb{C}$ be the complex plane and $H(H{E}_{IV},\mathbb{C})$ be the space of all holomorphic functions on $H{E}_{IV}$ and $\alpha >0$. We shall consider the Bers-type space on $H{E}_{IV}$, denoted by ${\mathcal{A}}^{\alpha }\left(H{E}_{IV}\right)$, which is given by [2]:

$${\mathcal{A}}^{\alpha }\left(H{E}_{IV}\right)=\left\{f\in H(H{E}_{IV},\mathbb{C}):{\parallel f\parallel }_{{\mathcal{A}}^{\alpha }\left(H{E}_{IV}\right)}<+\infty \right\},$$

where

$${\parallel f\parallel }_{{\mathcal{A}}^{\alpha }\left(H{E}_{IV}\right)}=\underset{(z,\xi )\in H{E}_{IV}}{sup}{\left(1+{\left|z{z}^T\right|}^2-2z{\overline{z}}^T-\sum _{j=1}^r{\left|{\xi }_j\right|}^{2p_j}\right)}^{\alpha }\left|f(z,\xi )\right|.$$

It is not difficult to see that ${\mathcal{A}}^{\alpha }\left(H{E}_{IV}\right)$ is a Banach space under the norm ${\parallel ·\parallel }_{{\mathcal{A}}^{\alpha }\left(H{E}_{IV}\right)}$.

Following Timoney and Su [5,6,7], we shall consider the Bloch-type space on $H{E}_{IV}$, denoted by ${\mathcal{B}}^{\alpha }\left(H{E}_{IV}\right)$, which is given by

$${\mathcal{B}}^{\alpha }\left(H{E}_{IV}\right)=\{f\in H(H{E}_{IV},\mathbb{C}){:\parallel f\parallel }_{{\mathcal{B}}^{\alpha }\left(H{E}_{IV}\right)}<+\infty \},$$

where

$${\parallel f\parallel }_{{\mathcal{B}}^{\alpha }\left(H{E}_{IV}\right)}=\underset{(z,\xi )\in H{E}_{IV}}{sup}{\left(1+{\left|z{z}^T\right|}^2-2z{\overline{z}}^T-\sum _{j=1}^r{\left|{\xi }_j\right|}^{2p_j}\right)}^{\alpha }\left|\nabla f(z,\xi )\right|.$$

It is easy to see that ${\mathcal{B}}^{\alpha }\left(H{E}_{IV}\right)$ is a Banach space under the norm ${\parallel ·\parallel }_{{\mathcal{B}}^{\alpha }\left(H{E}_{IV}\right)}$.

The composition, multiplication, and radial derivative operators on $H(H{E}_{IV},\mathbb{C})$ are defined as follows:

$$\begin{array}{ccc}& & \left(C_{\phi }f\right)(z,\xi )=(f\circ \phi )(z,\xi )=f\left(\phi (z,\xi )\right),\hfill \\ & & \left(M_{\psi }f\right)(z,\xi )=\psi (z,\xi )f(z,\xi ),\hfill \\ & & \mathcal{R}f(z,\xi )=\sum _{j=1}^N{z}_j{\displaystyle \frac{\partial f}{\partial z_j}}(z,\xi )+\sum _{k=1}^r{\xi }_k{\displaystyle \frac{\partial f}{\partial {\xi }_k}}(z,\xi ),(z,\xi )\in H{E}_{IV}.\hfill \end{array}$$

It is well known that

$$\mathcal{R}f(z,\xi )=〈\nabla f(z,\xi ),\overline{(z,\xi )}〉,$$

where

$$\nabla f(z,\xi )=\left({\displaystyle \frac{\partial f}{\partial z_1}}(z,\xi ),\cdots ,{\displaystyle \frac{\partial f}{\partial z_N}}(z,\xi ),{\displaystyle \frac{\partial f}{\partial {\xi }_1}}(z,\xi ),\cdots ,{\displaystyle \frac{\partial f}{\partial {\xi }_r}}(z,\xi )\right)$$

is the complex gradient of function f in $(z,\xi )$ ($N_1=\cdots =N_r=1$).

The weighted composition operator with multiplication symbol $\psi$ and composition symbol $\phi$ is defined by

$$W_{\psi ,\phi }f=\left(M_{\psi }{C}_{\phi }\right)f=\psi (f\circ \phi ),f\in H(H{E}_{IV},\mathbb{C}).$$

Over the past 40 years, the weighted composition operators have been extensively studied on or between the most well-known spaces (see, for example, [2,3,8,9,10,11,12]). Besides the products of the operators $C_{\phi }$ and $M_{\psi }$, there have been some investigations of the products of the operators D, $\mathcal{R}$, $M_{\psi }$, and $C_{\phi }$, where D stands for the differentiation operator. The product-type operators $W_{\psi ,\phi }D$ and $D{W}_{\psi ,\phi }$ were considered in [13,14], respectively. In general, the weighted composition operator is bounded and the radial derivative operator $\mathcal{R}$ is unbounded; one is naturally concerned about when their product $W_{\psi ,\phi }\mathcal{R}$ is bounded. For some products of these and other concrete operators for the case of the unit ball, in [15], Liu and Yu introduced the operator (an extension of Stević–Sharma operator) $T_{{\psi }_1,{\psi }_2,{\psi }_3,\phi }$ and characterized the boundedness and compactness of $T_{{\psi }_1,{\psi }_2,{\psi }_3,\phi }$ from logarithmic Bloch spaces to weighted-type spaces on the unit ball. The product operator ${\mathcal{R}}_{\psi ,\phi }^m=M_{\psi }{C}_{\phi }{\mathcal{R}}^m$ was introduced and studied in [16]. The characterization of the boundedness and compactness of the operators ${\mathcal{R}}_{\psi ,\phi }^m$ from logarithmic Bloch spaces ${\mathrm{B}}_{\mathrm{Log}}$ to weighted-type spaces $H_{\mu }^{\infty }$ on the unit ball was obtained by Stević and Jiang in [17]. The boundedness and compactness of the sum operator ${\mathfrak{G}}_{\overrightarrow{u},\phi }^m={\sum }_{j=0}^m{\mathcal{R}}_{u_j,\phi }^j$ from the mixed-norm space $H(p,q,\varphi )$, where $0<p,q,<+\infty$ and $\varphi$ is normal to the weighted-type space $H_{\mu }^{\infty }$, was given by Huang, Jiang, and Xu in [18]. For some investigations in this direction, see also [19,20,21].

Our work was motivated by the aforementioned research [2,6,17,18]. The primary goal of this paper was to extend the current results on the boundedness of the product-type operator $W_{\psi ,\phi }\mathcal{R}$ from the Bloch spaces ${\mathcal{B}}^{\alpha }\left(H{E}_{IV}\right)$ into the Bers-type space ${\mathcal{A}}^{\beta }\left(H{E}_{IV}\right)$. However, there has been relatively little research on this topic, leaving many open questions, such as the compactness and essential norm of the operator $W_{\psi ,\phi }\mathcal{R}$ from the Bloch spaces ${\mathcal{B}}^{\alpha }\left(H{E}_{IV}\right)$ into the Bers-type space ${\mathcal{A}}^{\beta }\left(H{E}_{IV}\right)$. As a result, the research presented here is of significant importance, and we hope this exposition will inspire further investigation in this area.

From now on, unless specified otherwise, the letter C denotes a positive constant that may vary at each occurrence but is independent of the essential variables.

The following lemmas are used to prove our main results.

Lemma 1

(Minkowski inequality of integration formula [22]). Let $a_k,b_k\ge 0$ for $k=1,2,\cdots ,n$. Then, the following inequality holds:

$$\begin{array}{ccc}& & {\left(\prod _{k=1}^n\left(a_k+b_k\right)\right)}^{1/n}\ge {\left(\prod _{k=1}^n{a}_k\right)}^{1/n}+{\left(\prod _{k=1}^n{b}_k\right)}^{1/n},\hfill \end{array}$$

with equality if and only if $a_k=c{b}_k$ for some constant c and for all $k=1,2,\cdots ,n$.

Lemma 2

([4,23]). Let $Z\in {\mathfrak{R}}_I(m,n)=\{Z\in {\mathbb{C}}^{m\times n}:I_m-Z{\overline{Z}}^T>0\}$, where Z means an $m\times n$ complex matrix, $Z^T$ denotes the transpose of Z, $\overline{Z}$ denotes the conjugate of Z, $I_m-Z{\overline{Z}}^T>0$ denotes $I_m-Z{\overline{Z}}^T$ is a Hermitian positive definite matrix, and m and n are positive integers. Then, there exist two unitary matrices U and V such that

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

where $1>{\lambda }_1\ge {\lambda }_2\ge \cdots \ge {\lambda }_m\ge 0$ and ${\lambda }_1^2,\cdots ,{\lambda }_m^2$ are the characteristic values of $Z{\overline{Z}}^T$.

Lemma 3

([4,23]). Let

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

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

where ${\lambda }_j{\mu }_k<1$ for $j,k=1,2,\cdots ,m$. Then, there exists a permutation matrix P that is $m\times m$ such that

$$\underset{U{\overline{U}}^T=I,V{\overline{V}}^T=I}{inf}\left|det\left(I-{\Lambda }_{1}U{\Lambda }_2{\overline{U}}^{T}V\right)\right|=\left|det\left(I-{\Lambda }_{1}P{\Lambda }_2{P}^T\right)\right|.$$

Lemma 4

([2]). If $z=(z_1,z_2,z_3,z_4)\in {\mathfrak{R}}_{IV}\left(4\right)$, then the following point belongs to ${\mathfrak{R}}_I(2,2)$:

$$Z=\left(\begin{array}{cc}{z}_1+i{z}_2& i{z}_2-z_4\hfill \\ i{z}_3+z_4& z_1-i{z}_2\hfill \end{array}\right).$$

Moreover, it follows that

$$\begin{array}{ccc}& & 1+{\left|z{z}^T\right|}^2-2z{\overline{z}}^T=det\left(I-Z{\overline{Z}}^T\right).\hfill \end{array}$$

In order to prove the main theorems, we need the following propositions.

Proposition 1.

If $A,B\in {\mathfrak{R}}_I(2,2)$, then

$$\begin{array}{ccc}& & {\left[{\left(det\left(I-A{\overline{A}}^T\right)\right)}^{1/2}+{\left(det\left(I-B{\overline{B}}^T\right)\right)}^{1/2}\right]}^2\le 4\left|det\left(I-A{\overline{B}}^T\right)\right|.\hfill \end{array}$$

Moreover,

$$\begin{array}{ccc}& & det\left(I-A{\overline{A}}^T\right)+det\left(I-B{\overline{B}}^T\right)\hfill \\ & & \le 4\left|det\left(I-A{\overline{B}}^T\right)\right|-2{\left(det\left(I-A{\overline{A}}^T\right)\right)}^{1/2}{\left(det\left(I-B{\overline{B}}^T\right)\right)}^{1/2}.\hfill \end{array}$$

(1)

Proof. 

Since $A,B\in {\mathfrak{R}}_I(2,2)$, by Lemma 2, there exist four $2\times 2$ unitary matrices $U_i$ and $V_i$ ($i=1,2$) such that

$$A=U_1\left(\begin{array}{cc}{\lambda }_1& 0\hfill \\ 0& {\lambda }_2\hfill \end{array}\right)V_1=U_1{\Lambda }_1{V}_1$$

and

$$B=U_2\left(\begin{array}{cc}{\mu }_1& 0\hfill \\ 0& {\mu }_2\hfill \end{array}\right)V_2=U_2{\Lambda }_1{V}_2,$$

$\left(0\le {\lambda }_2\le {\lambda }_1<1,0\le {\mu }_2\le {\mu }_1<1\right)$. We have

$$\begin{array}{ccc}& & det\left(I-A{\overline{A}}^T\right)=(1-{\lambda }_1^2)(1-{\lambda }_2^2),\hfill \\ & & det\left(I-B{\overline{B}}^T\right)=(1-{\mu }_1^2)(1-{\mu }_2^2),\hfill \end{array}$$

and

$$\begin{array}{ccc}& det\left(I-A{\overline{B}}^T\right)& =det\left(I-U_1{\Lambda }_1{V}_1{\overline{U_2{\Lambda }_2{V}_2}}^T\right)\hfill \\ & & =det\left(U_1{\overline{U_1}}^T-U_1{\Lambda }_1{V}_1{\overline{V_2}}^T{\Lambda }_2{\overline{U_2}}^T\right)\hfill \\ & & =det{U}_{1}det\left({\overline{U_1}}^T-{\Lambda }_1{V}_1{\overline{V_2}}^T{\Lambda }_2{\overline{U_2}}^T\right)\hfill \\ & & =det\left({\overline{U_1}}^T-{\Lambda }_1{V}_1{\overline{V_2}}^T{\Lambda }_2{\overline{U_2}}^T\right)det{U}_1\hfill \\ & & =det\left(I-{\Lambda }_1{V}_1{\overline{V_2}}^T{\Lambda }_2{\overline{U_2}}^T{U}_1\right)\hfill \\ & & =det\left(I-{\Lambda }_1{V}_1{\overline{V_2}}^T{\Lambda }_2{V}_2{\overline{V_1}}^T{V}_1{\overline{V_2}}^T{\overline{U_2}}^T{U}_1\right).\hfill \end{array}$$

Set $U=V_1{\overline{V_2}}^T$ and $V=V_1{\overline{V_2}}^T{\overline{U_2}}^T{U}_1$; then, $U{\overline{U}}^T=I$ and $V{\overline{V}}^T=I.$ By Lemmas 1 and 3, there exists a permutation matrix P that is $2\times 2$ such that

$$\begin{array}{ccc}& & 4\left|det\left(I-A{\overline{B}}^T\right)\right|\ge 4\left|det\left(I-{\Lambda }_{1}P{\Lambda }_2{P}^T\right)\right|=4(1-{\lambda }_1{\mu }_{k_1})(1-{\lambda }_2{\mu }_{k_2})\hfill \\ & & =(2-2{\lambda }_1{\mu }_{k_1})(2-2{\lambda }_2{\mu }_{k_2})\hfill \\ & & \ge \left(2-{\lambda }_1^2-{\mu }_{k_1}^2\right)\left(2-{\lambda }_2^2-{\mu }_{k_2}^2\right)\hfill \\ & & =\left(\left(1-{\lambda }_1^2\right)+\left(1-{\mu }_{k_1}^2\right)\right)\left((1-{\lambda }_2^2)+(1-{\mu }_{k_2}^2)\right)\hfill \\ & & \ge {\left[{\left(\left(1-{\lambda }_1^2\right)\left(1-{\lambda }_2^2\right)\right)}^{1/2}+{\left((1-{\mu }_{k_1}^2)(1-{\mu }_{k_2}^2)\right)}^{1/2}\right]}^2\hfill \\ & & \ge {\left[{\left(det\left(I-A{\overline{A}}^T\right)\right)}^{1/2}+{\left(det\left(I-B{\overline{B}}^T\right)\right)}^{1/2}\right]}^2,\hfill \end{array}$$

where $k_1,k_2$ is a permutation of $1,2$, finishing the proof of the proposition. □

Without loss of generality, suppose that $N_1=\cdots =N_r=1$, that is, ${\xi }_j\in \mathbb{C}$; $j=1,\cdots ,r$; $\xi =({\xi }_1,\cdots ,{\xi }_r)$; and ${\parallel \xi \parallel }_p^2={\sum }_{j=1}^r{\left|{\xi }_j\right|}^{2p_j}$$\left(p=(p_1,\cdots ,p_r)\right)$.

Proposition 2.

If $(z,\xi ),(s,\zeta )\in H{E}_{IV}$, then

$$\begin{array}{ccc}& & \left(1+{\left|z{z}^T\right|}^2-2z{\overline{z}}^T-{\parallel \xi \parallel }_p^2\right)+\left(1+{\left|s{s}^T\right|}^2-2s{\overline{s}}^T-{\parallel \zeta \parallel }_p^2\right)\hfill \\ & & \le 4\left(\left|1+z{z}^T\overline{s}{\overline{s}}^T-2z{\overline{s}}^T\right|-{\parallel \xi \parallel }_p{\parallel \zeta \parallel }_p\right).\hfill \end{array}$$

(2)

Moreover, $\left|z{z}^T\right|\le {\left|z\right|}^2$, ${\parallel \xi \parallel }_p+{\left|z\right|}^2<1$, and

$$\begin{array}{ccc}& & \left|(z,\xi )\right|<\sqrt{1+r}.\hfill \end{array}$$

(3)

Proof. 

If $(z,\xi ),(s,\zeta )\in H{E}_{IV}$, then

$$\begin{array}{ccc}& & 1+{\left|z{z}^T\right|}^2-2z{\overline{z}}^T{>\parallel \xi \parallel }_p^2,1+{\left|s{s}^T\right|}^2-2s{\overline{s}}^T>{\parallel \zeta \parallel }_p^2,\hfill \end{array}$$

(4)

and $z,s\in {\mathfrak{R}}_{IV}\left(N\right).$ So there exists a real orthogonal matrix U such that

$$\begin{array}{ccc}& & z=\left(z_1^{*},z_2^{*},z_3^{*},z_4^{*},0,\cdots ,0\right)U,s=\left(w_1^{*},w_2^{*},w_3^{*},w_4^{*},0,\cdots ,0\right)U.\hfill \end{array}$$

Let

$$\begin{array}{ccc}& & z^{*}=(z_1^{*},z_2^{*},z_3^{*},z_4^{*}),s^{*}=(w_1^{*},w_2^{*},w_3^{*},w_4^{*}),\hfill \end{array}$$

$$A=\left(\begin{array}{cc}{z}_1^{*}+i{z}_2^{*}& i{z}_2^{*}-z_4^{*}\hfill \\ i{z}_3^{*}+z_4^{*}& z_1^{*}-i{z}_2^{*}\hfill \end{array}\right),$$

and

$$B=\left(\begin{array}{cc}{w}_1^{*}+i{w}_2^{*}& i{w}_2^{*}-w_4^{*}\hfill \\ i{w}_3^{*}+w_4^{*}& w_1^{*}-i{w}_2^{*}\hfill \end{array}\right).$$

From Lemma 4, we know that $A,B\in {\mathfrak{R}}_I(2,2)$. Moreover, we also have

$$\begin{array}{ccc}& & 1+{\left|z{z}^T\right|}^2-2z{\overline{z}}^T=det\left(I-A{\overline{A}}^T\right),\hfill \\ & & 1+|s{s}^T{|}^2-2s{\overline{s}}^T=det\left(I-B{\overline{B}}^T\right),\hfill \\ & & 1+z{z}^T\overline{s}{\overline{s}}^T-2z{\overline{s}}^T=det\left(I-A{\overline{B}}^T\right).\hfill \end{array}$$

(5)

Using Proposition 1 and conditions (4) and (5), it follows that

$$\begin{array}{ccc}& & \left(1+{\left|z{z}^T\right|}^2-2z{\overline{z}}^T-{\parallel \xi \parallel }_p^2\right)+\left(1+{\left|s{s}^T\right|}^2-2s{\overline{s}}^T-{\parallel \zeta \parallel }_p^2\right)\hfill \\ & & =\left(\det \left(I-A{\overline{A}}^T\right)+det\left(I-B{\overline{B}}^T\right)\right)-\left({\parallel \xi \parallel }_p^2+{\parallel \zeta \parallel }_p^2\right)\hfill \\ & & \le 4\left|\det \left(I-A{\overline{B}}^T\right)\right|-2{\left(\det \left(I-A{\overline{A}}^T\right)\right)}^{1/2}{\left(\det \left(I-B{\overline{B}}^T\right)\right)}^{1/2}-{2\parallel \xi \parallel }_p{\parallel \zeta \parallel }_p\hfill \\ & & =4\left|1+z{z}^T\overline{s}{\overline{s}}^T-2z{\overline{s}}^T\right|\hfill \\ & & -2{\left|1+{\left|z{z}^T\right|}^2-2z{\overline{z}}^T\right|}^{1/2}{\left|1+{\left|s{s}^T\right|}^2-2s{\overline{s}}^T\right|}^{1/2}-{2\parallel \xi \parallel }_p{\parallel \zeta \parallel }_p\hfill \\ & & \le 4\left(\left|1+z{z}^T\overline{s}{\overline{s}}^T-2z{\overline{s}}^T\right|-{\parallel \xi \parallel }_p{\parallel \zeta \parallel }_p\right).\hfill \end{array}$$

That is, (2) holds.

Since

$$\begin{array}{ccc}& & 1+{\left|z{z}^T\right|}^2-2z{\overline{z}}^T=det\left(I-A{\overline{A}}^T\right)\hfill \\ & & =(1-{\lambda }_1^2)(1-{\lambda }_2^2)\hfill \\ & & \le {\displaystyle \frac{1}{4}}{\left(\left(1-{\lambda }_1^2\right)+\left(1-{\lambda }_2^2\right)\right)}^2\hfill \\ & & ={\left(1-{\displaystyle \frac{1}{2}}\left({\lambda }_1^2+{\lambda }_2^2\right)\right)}^2\hfill \\ & & ={\left(1-{\displaystyle \frac{{\left|A\right|}^2}{2}}\right)}^2,\hfill \end{array}$$

and

$$\begin{array}{ccc}& {|A|}^2& =|z_1^{*}+i{z}_2^{*}{|}^2+|i{z}_2^{*}-z_4^{*}{|}^2+|i{z}_3^{*}+z_4^{*}{|}^2+{|z_1^{*}-i{z}_2^{*}|}^2\hfill \\ & & =2\left(|z_1^{*}{|}^2+|z_2^{*}{|}^2+|z_3^{*}{|}^2+{|z_4^{*}|}^2\right)\hfill \\ & & ={2|z|}^2,\hfill \end{array}$$

then,

$$\begin{array}{ccc}& & \left|z{z}^T\right|\le {\left|z\right|}^2,\hfill \end{array}$$

(6)

which improves the estimation ${\left|z{z}^T\right|}^2\le 2z{\overline{z}}^T=2{\left|z\right|}^2$ in [2] [Lemma 2.5] and [11] [Lemma 19]. It follows that

$$\begin{array}{ccc}& & 1+{\left|z{z}^T\right|}^2-2z{\overline{z}}^T\le 1+{\left|z\right|}^4-2{\left|z\right|}^2={\left(1-{\left|z\right|}^2\right)}^2.\hfill \end{array}$$

Thus

$${\parallel \xi \parallel }_p^2<1+{\left|z{z}^T\right|}^2-2z{\overline{z}}^T\le {\left(1-{\left|z\right|}^2\right)}^2,$$

that is

$${\parallel \xi \parallel }_p+{\left|z\right|}^2<1.$$

Since

$$\begin{array}{ccc}& |{\xi }_j{|}^{2p_j}& \le \sum _{j=1}^r|{\xi }_j{|}^{2p_j}={\parallel \xi \parallel }_p^2<1,(j=1,2,\cdots ,r),\hfill \end{array}$$

we have

$$\begin{array}{ccc}& \left|\right(z,\xi \left)\right|& =\sqrt{{\left|z\right|}^2+{\left|\xi \right|}^2}\hfill \\ & & <\sqrt{1+|{\xi }_1{|}^2+|{\xi }_2{|}^2+\cdots +{\left|{\xi }_r\right|}^2}\hfill \\ & & <\sqrt{1+r}.\hfill \end{array}$$

The proof of Proposition 2 is then complete.

Remark 1.

By direct computations, using the triangle inequality, we have

$$\begin{array}{ccc}& {\left|z{z}^T\right|}^2& ={\left|z_1^2+z_2^2+\cdots +z_N^2\right|}^2\hfill \\ & & \le {\left(|z_1{|}^2+|z_2{|}^2+\cdots +{|z_N|}^2\right)}^2\hfill \\ & & ={\left|z\right|}^4,\hfill \end{array}$$

and we can also obtain (6).

□

Next, we give the following generalized Hua’s inequality on $H{E}_{IV}$.

Proposition 3.

If $(z,\xi ),(s,\zeta )\in H{E}_{IV}$, then

$$\begin{array}{ccc}& & \left(1+{\left|z{z}^T\right|}^2-2z{\overline{z}}^T-{\parallel \xi \parallel }_p^2\right)\left(1+{\left|s{s}^T\right|}^2-2s{\overline{s}}^T-{\parallel \zeta \parallel }_p^2\right)\hfill \\ & & \le 4{\left(\left|1+z{z}^T\overline{s}{\overline{s}}^T-2z{\overline{s}}^T\right|-{\parallel \xi \parallel }_p{\parallel \zeta \parallel }_p\right)}^2.\hfill \end{array}$$

Proof. 

If $(z,\xi ),(s,\zeta )\in H{E}_{IV}$, then by the elementary inequality $\sqrt{ab}\le {\displaystyle \frac{a+b}{2}}$ ($a\ge 0,b\ge 0)$ and Proposition 2:

$$\begin{array}{ccc}& & \left(1+{\left|z{z}^T\right|}^2-2z{\overline{z}}^T-{\parallel \xi \parallel }_p^2\right)\left(1+{\left|s{s}^T\right|}^2-2s{\overline{s}}^T-{\parallel \zeta \parallel }_p^2\right)\hfill \\ & & \le {\displaystyle \frac{1}{4}}{\left(\left(1+{\left|z{z}^T\right|}^2-2z{\overline{z}}^T-{\parallel \xi \parallel }_p^2\right)+\left(1+{\left|s{s}^T\right|}^2-2s{\overline{s}}^T-{\parallel \zeta \parallel }_p^2\right)\right)}^2.\hfill \\ & & \le 4{\left(\left|1+z{z}^T\overline{s}{\overline{s}}^T-2z{\overline{s}}^T\right|-{\parallel \xi \parallel }_p{\parallel \zeta \parallel }_p\right)}^2.\hfill \end{array}$$

□

2. The Boundedness of the Operator ${\mathit{W}}_{\mathit{\psi },\mathit{\phi }}\mathcal{R}:{\mathcal{B}}^{\mathit{\alpha }}\left({\mathit{HE}}_{\mathit{IV}}\right)\to {\mathcal{A}}^{\mathit{\beta }}\left({\mathit{HE}}_{\mathit{IV}}\right)$

In this section, we formulate and prove results regarding the boundedness of the operator $W_{\psi ,\phi }\mathcal{R}:{\mathcal{B}}^{\alpha }\left(H{E}_{IV}\right)\to {\mathcal{A}}^{\beta }\left(H{E}_{IV}\right)$.

For $(z,\xi )\in H{E}_{IV}$, we have $z\in {\mathfrak{R}}_{IV}\left(N\right)$; thus, by Proposition 2, $\left|z\right|<1$ and ${\parallel \xi \parallel }_p^2={\sum }_{j=1}^r{\left|{\xi }_j\right|}^{2p_j}<1$. Set $q=min\{p_1,p_2,\cdots ,p_r\}$ and $Q=max\{p_1,p_2,\cdots ,p_r\}.$

Theorem 1.

Suppose $\psi \in H(H{E}_{IV},\mathbb{C})$, $\phi \in H(H{E}_{IV},H{E}_{IV})$, and $\alpha ,\beta >0$.

(1) 

If

$$\begin{array}{c}\hfill L:=\underset{(z,\xi )\in H{E}_{IV}}{sup}{\displaystyle \frac{{\left(1+{\left|z{z}^T\right|}^2-2z{\overline{z}}^T-{\parallel \xi \parallel }_p^2\right)}^{\beta }}{{\left(1+{\left|w{w}^T\right|}^2-2w{\overline{w}}^T-{\parallel \eta \parallel }_p^2\right)}^{\alpha }}}\left|\psi (z,\xi )\right|\left|(w,\eta )\right|<\infty ,\end{array}$$

(7)

where $(w,\eta ):=\phi (z,\xi )$, then $W_{\psi ,\phi }\mathcal{R}:{\mathcal{B}}^{\alpha }\left(H{E}_{IV}\right)\to {\mathcal{A}}^{\beta }\left(H{E}_{IV}\right)$ is bounded.

(2) 

If $W_{\psi ,\phi }\mathcal{R}:{\mathcal{B}}^{\alpha }\left(H{E}_{IV}\right)\to {\mathcal{A}}^{\beta }\left(H{E}_{IV}\right)$ is bounded, then

$$\begin{array}{ccc}& & \underset{(z,\xi )\in H{E}_{IV}}{sup}{\displaystyle \frac{{\left(1+{\left|z{z}^T\right|}^2-2z{\overline{z}}^T-{\parallel \xi \parallel }_p^2\right)}^{\beta }}{{\left|1+w{w}^T\overline{w}{\overline{w}}^T-2w{\overline{w}}^T-{\parallel \eta \parallel }_p^2\right|}^{\alpha }}}{\left|\psi (z,\xi )\right|\parallel \eta \parallel }_p^2<\infty .\hfill \end{array}$$

(8)

Proof. 

First, suppose that (7) holds. For $f\in {\mathcal{B}}^{\alpha }\left(H{E}_{IV}\right)$, $\forall (z,\xi )\in H{E}_{IV}$, we have that

$$\begin{array}{ccc}& & {\left(1+{\left|z{z}^T\right|}^2-2z{\overline{z}}^T-\sum _{j=1}^r{\left|{\xi }_j\right|}^{2p_j}\right)}^{\beta }\left|\left(W_{\psi ,\phi }\mathcal{R}f\right)(z,\xi )\right|\hfill \\ & & ={\left(1+{\left|z{z}^T\right|}^2-2z{\overline{z}}^T-{\parallel \xi \parallel }_p^2\right)}^{\beta }\left|\psi (z,\xi )\mathcal{R}f\left(\phi (z,\xi )\right)\right|\hfill \\ & & \le {\left(1+{\left|z{z}^T\right|}^2-2z{\overline{z}}^T-{\parallel \xi \parallel }_p^2\right)}^{\beta }\left|\psi (z,\xi )\right||\nabla f\left(\phi (z,\xi )\right)|\left|\phi (z,\xi )\right|\hfill \\ & & \le {\displaystyle \frac{{\left(1+{\left|z{z}^T\right|}^2-2z{\overline{z}}^T-{\parallel \xi \parallel }_p^2\right)}^{\beta }}{{\left(1+{\left|w{w}^T\right|}^2-2w{\overline{w}}^T-{\parallel \eta \parallel }_p^2\right)}^{\alpha }}}{\left|\psi (z,\xi )\right|\left|(w,\eta )\right|\parallel f\parallel }_{{\mathcal{B}}^{\alpha }\left(H{E}_{IV}\right)}\hfill \\ & & \le {M\parallel f\parallel }_{{\mathcal{B}}^{\alpha }\left(H{E}_{IV}\right)}.\hfill \end{array}$$

(9)

Then, from (7) and (9), we obtain

$$\begin{array}{ccc}& & {∥W_{\psi ,\phi }\mathcal{R}f∥}_{{\mathcal{A}}^{\beta }\left(H{E}_{IV}\right)}\le M{\parallel f\parallel }_{{\mathcal{B}}^{\alpha }\left(H{E}_{IV}\right)},\hfill \end{array}$$

from which the boundedness of the operator $W_{\psi ,\phi }\mathcal{R}:{\mathcal{B}}^{\alpha }\left(H{E}_{IV}\right)\to {\mathcal{A}}^{\beta }\left(H{E}_{IV}\right)$ follows.

Now, assume that $W_{\psi ,\phi }\mathcal{R}:{\mathcal{B}}^{\alpha }\left(H{E}_{IV}\right)\to {\mathcal{A}}^{\beta }\left(H{E}_{IV}\right)$ is bounded. Then, there is a positive constant C such that

$$\parallel W_{\psi ,\phi }{\mathcal{R}f\parallel }_{{\mathcal{A}}^{\beta }\left(H{E}_{IV}\right)}\le C{\parallel f\parallel }_{{\mathcal{B}}^{\alpha }\left(H{E}_{IV}\right)},$$

for all $f\in {\mathcal{B}}^{\alpha }\left(H{E}_{IV}\right)$.

To prove (8) holds for $\alpha =1$, fix $(s,\zeta )\in H{E}_{IV}$. By Proposition 2, we consider the test function $f_{(s,\zeta )}:H{E}_{IV}\to H{E}_{IV}$ given by

$$\begin{array}{ccc}& & f_{(s,\zeta )}(z,\xi )=log{\displaystyle \frac{1}{1+z{z}^T\overline{s}{\overline{s}}^T-2z{\overline{s}}^T-{\langle \xi ,\zeta \rangle }_p}},\hfill \end{array}$$

(10)

for $(z,\xi )\in H{E}_{IV},$ where ${\langle \xi ,\zeta \rangle }_p={\langle {\xi }_1,{\zeta }_1\rangle }^{p_1}+\cdots +{\langle {\xi }_r,{\zeta }_r\rangle }^{p_r}$ and ${|\langle \xi ,\zeta \rangle }_p{|\le \parallel \xi \parallel }_p{\parallel \zeta \parallel }_p$. Then, for $j=1,\cdots ,N$ and $k=1,\cdots ,r,$ we can check that

$$\begin{array}{c}\hfill {\displaystyle \frac{\partial f_{(s,\zeta )}}{\partial z_j}}(z,\xi )={\displaystyle \frac{2\overline{s_j}-2z_j\overline{s}{\overline{s}}^T}{1+z{z}^T\overline{s}{\overline{s}}^T-2z{\overline{s}}^T-{\langle \xi ,\zeta \rangle }_p}},\end{array}$$

(11)

and

$$\begin{array}{c}\hfill {\displaystyle \frac{\partial f_{(s,\zeta )}}{\partial {\xi }_k}}(z,\xi )={\displaystyle \frac{p_k{\xi }_k^{p_k-1}{\overline{{\zeta }_k}}^{p_k}}{1+z{z}^T\overline{s}{\overline{s}}^T-2z{\overline{s}}^T-{\langle \xi ,\zeta \rangle }_p}}.\end{array}$$

(12)

By Proposition 3, we have

$$\begin{array}{ccc}& {\left|\nabla f_{(s,\zeta )}(z,\xi )\right|}^2& =\sum _{j=1}^N{\left|{\displaystyle \frac{\partial f(s,\zeta )}{\partial z_j}}(z,\xi )\right|}^2+\sum _{k=1}^r{\left|{\displaystyle \frac{\partial f(s,\zeta )}{\partial {\xi }_k}}(z,\xi )\right|}^2\hfill \\ & & ={\displaystyle \frac{4{\sum }_{j=1}^N{\left|\overline{s_j}-z_j\overline{s}{\overline{s}}^T\right|}^2+{\sum }_{k=1}^r{\left|p_k{\xi }_k^{p_k}{\overline{{\zeta }_k}}^{p_k}\right|}^2}{{\left|1+z{z}^T\overline{s}{\overline{s}}^T-2z{\overline{s}}^T-{\langle \xi ,\zeta \rangle }_p\right|}^2}}\hfill \\ & & \le {\displaystyle \frac{4{\sum }_{j=1}^N{\left|\overline{s_j}-z_j\overline{s}{\overline{s}}^T\right|}^2+Q^2{\sum }_{k=1}^r|{\xi }_k{|}^{2p_k}{\left|{\zeta }_k\right|}^{2p_k}}{{\left(\left|1+z{z}^T\overline{s}{\overline{s}}^T-2z{\overline{s}}^T\right|-\left|{\langle \xi ,\zeta \rangle }_p\right|\right)}^2}}\hfill \\ & & \le {\displaystyle \frac{4{\left|\overline{s}-z\overline{s}{\overline{s}}^T\right|}^2+r{Q}^2}{{\left(\left|1+z{z}^T\overline{s}{\overline{s}}^T-2z{\overline{s}}^T\right|-{\parallel \xi \parallel }_p{\parallel \zeta \parallel }_p\right)}^2}}\hfill \\ & & \le 16{\displaystyle \frac{4{\left|\overline{s}-z\overline{s}{\overline{s}}^T\right|}^2+r{Q}^2}{{\left[\left(1+{\left|z{z}^T\right|}^2-2z{\overline{z}}^T-{\parallel \xi \parallel }_p^2\right)+\left(1+{\left|s{s}^T\right|}^2-2s{\overline{s}}^T-{\parallel \zeta \parallel }_p^2\right)\right]}^2}}\hfill \\ & & \le {\displaystyle \frac{64{\left|\overline{s}-z\overline{s}{\overline{s}}^T\right|}^2+16r{Q}^2}{{\left(1+{\left|z{z}^T\right|}^2-2z{\overline{z}}^T-{\parallel \xi \parallel }_p^2\right)}^2}}\hfill \\ & & \le {\displaystyle \frac{C}{{\left(1+{\left|z{z}^T\right|}^2-2z{\overline{z}}^T-{\sum }_{j=1}^r{\left|{\xi }_j\right|}^{2p_j}\right)}^2}}.\hfill \end{array}$$

It can be easily seen that $f_{(s,\zeta )}\in \mathcal{B}\left(H{E}_{IV}\right)$ with

$$\begin{array}{ccc}& & \underset{(s,\zeta )\in H{E}_{IV}}{sup}{\parallel f_{(s,\zeta )}\parallel }_{\mathcal{B}\left(H{E}_{IV}\right)}\le C.\hfill \end{array}$$

(13)

Since

$$\begin{array}{ccc}& & |\mathcal{R}{f}_{(s,\zeta )}(z,\xi )|\hfill \\ & & =\left|\sum _{j=1}^N{z}_j{\displaystyle \frac{\partial f(s,\zeta )}{\partial z_j}}(z,\xi )+\sum _{k=1}^r{\xi }_k{\displaystyle \frac{\partial f(s,\zeta )}{\partial {\xi }_k}}(z,\xi )\right|\hfill \\ & & =\left|\sum _{j=1}^N{z}_j{\displaystyle \frac{2{\overline{s_j}}^T-2z_j\overline{s}{\overline{s}}^T}{1+z{z}^T\overline{s}{\overline{s}}^T-2z{\overline{s}}^T-{\langle \xi ,\zeta \rangle }_p}}+\sum _{k=1}^r{\xi }_k{\displaystyle \frac{p_k{\xi }_k^{p_k-1}{\overline{{\zeta }_k}}^{p_k}}{1+z{z}^T\overline{s}{\overline{s}}^T-2z{\overline{s}}^T-{\langle \xi ,\zeta \rangle }_p}}\right|\hfill \\ & & ={\displaystyle \frac{\left|2{\sum }_{j=1}^N\left(z_j{\overline{s_j}}^T-z_j^2\overline{s}{\overline{s}}^T\right)+{\sum }_{k=1}^r\left(p_k{\xi }_k^{p_k}{\overline{{\zeta }_k}}^{p_k}\right)\right|}{\left|1+z{z}^T\overline{s}{\overline{s}}^T-2z{\overline{s}}^T-{\langle \xi ,\zeta \rangle }_p\right|}},\hfill \end{array}$$

so taking $(z,\xi )=(w,\eta ):=\phi (z,\xi )$, then taking $(s,\zeta )=(w,\eta )$, and by (6), we obtain

$$\begin{array}{ccc}& & {\left|w\right|}^2>{\left|w\right|}^4\ge {\left|w{w}^T\right|}^2\hfill \end{array}$$

(14)

and

$$\begin{array}{ccc}& & |\mathcal{R}{f}_{(s,\zeta )}(z,\xi )|=|\mathcal{R}{f}_{(w,\eta )}(w,\eta )|\hfill \\ & & ={\displaystyle \frac{\left|2{\sum }_{j=1}^N\left(w_j{\overline{w_j}}^T-w_j^2\overline{w}{\overline{w}}^T\right)+{\sum }_{k=1}^r\left(p_k{\eta }_k^{p_k}{\overline{{\eta }_k}}^{p_k}\right)\right|}{\left|1+w{w}^T\overline{w}{\overline{w}}^T-2w{\overline{w}}^T-{\langle \eta ,\eta \rangle }_p\right|}}\hfill \\ & & ={\displaystyle \frac{\left|2{\sum }_{j=1}^N\left(w_j{\overline{w_j}}^T-w_j^2\overline{w}{\overline{w}}^T\right)+{\sum }_{k=1}^r\left(p_k{\left|{\eta }_k\right|}^{2p_k}\right)\right|}{\left|1+w{w}^T\overline{w}{\overline{w}}^T-2w{\overline{w}}^T-{\parallel \eta \parallel }_p^2\right|}}\hfill \\ & & ={\displaystyle \frac{\left|{2\left(\right|w|}^2-|w{w}^T{|}^2)+{\sum }_{k=1}^r\left(p_k{\left|{\zeta }_k\right|}^{2p_k}\right)\right|}{\left|1+w{w}^T\overline{w}{\overline{w}}^T-2w{\overline{w}}^T-{\parallel \eta \parallel }_p^2\right|}}\hfill \\ & & \ge {\displaystyle \frac{{q\parallel \eta \parallel }_p^2}{\left|1+w{w}^T\overline{w}{\overline{w}}^T-2w{\overline{w}}^T-{\parallel \eta \parallel }_p^2\right|}}.\hfill \end{array}$$

(15)

Hence, inequalities (13) and (15) imply that

$$\begin{array}{ccc}& & C∥W_{\psi ,\phi }\mathcal{R}∥\ge {\parallel f_{(s,\zeta )}\parallel }_{\mathcal{B}\left(H{E}_{IV}\right)}∥W_{\psi ,\phi }\mathcal{R}∥\ge \hfill \\ & & \ge {∥W_{\psi ,\phi }\mathcal{R}{f}_{(s,\zeta )}∥}_{{\mathcal{A}}^{\beta }\left(H{E}_{IV}\right)}\hfill \\ & & =\underset{(z,\xi )\in H{E}_{IV}}{sup}{\left(1+{\left|z{z}^T\right|}^2-2z{\overline{z}}^T-\sum _{j=1}^r{\left|{\xi }_j\right|}^{2p_j}\right)}^{\beta }\left|\psi (z,\xi )\mathcal{R}{f}_{(s,\zeta )}\left(\phi (z,\xi )\right)\right|\hfill \\ & & \ge {\left(1+{\left|z{z}^T\right|}^2-2z{\overline{z}}^T-\sum _{j=1}^r{\left|{\xi }_j\right|}^{2p_j}\right)}^{\beta }\left|\psi (z,\xi )\mathcal{R}{f}_{(s,\zeta )}\left(\phi (z,\xi )\right)\right|\hfill \\ & & \ge {\left(1+{\left|z{z}^T\right|}^2-2z{\overline{z}}^T-\sum _{j=1}^r{\left|{\xi }_j\right|}^{2p_j}\right)}^{\beta }\left|\psi (z,\xi )\mathcal{R}{f}_{(w,\eta )}(w,\eta )\right|\hfill \\ & & \ge q{\displaystyle \frac{{\left(1+{\left|z{z}^T\right|}^2-2z{\overline{z}}^T-{\parallel \xi \parallel }_p^2\right)}^{\beta }}{1+|w{w}^T{|}^2-2w{\overline{w}}^T-{\parallel \eta \parallel }_p^2}}{\left|\psi (z,\xi )\right|\parallel \eta \parallel }_p^2.\hfill \end{array}$$

Thus,

$$\begin{array}{ccc}& & \underset{(z,\xi )\in H{E}_{IV}}{sup}{\displaystyle \frac{{\left(1+{\left|z{z}^T\right|}^2-2z{\overline{z}}^T-{\parallel \xi \parallel }_p^2\right)}^{\beta }}{1+{\left|w{w}^T\right|}^2-2w{\overline{w}}^T-{\parallel \eta \parallel }_p^2}}{\left|\psi (z,\xi )\right|\parallel \eta \parallel }_p^2<\infty ,\hfill \end{array}$$

which means that (8) holds for $\alpha =1$.

Finally, if $\alpha \ne 1$, we consider the test function $g_{(s,\zeta )}:H{E}_{IV}\to H{E}_{IV}$ given by

$$\begin{array}{ccc}& & g_{(s,\zeta )}(z,\xi )={\displaystyle \frac{1}{{\left(1+z{z}^T\overline{s}{\overline{s}}^T-2z{\overline{s}}^T-{\langle \xi ,\zeta \rangle }_p\right)}^{\alpha -1}}},\hfill \end{array}$$

(16)

for $(z,\xi )\in H{E}_{IV}.$ Then, we can check that for $j=1,\cdots ,N$ and $k=1,\cdots ,r,$

$$\begin{array}{c}\hfill {\displaystyle \frac{\partial g_{(s,\zeta )}}{\partial z_j}}(z,\xi )={\displaystyle \frac{2(1-\alpha )(z_j\overline{s}{\overline{s}}^T-\overline{s_j})}{{\left(1+z{z}^T\overline{s}{\overline{s}}^T-2z{\overline{s}}^T-{\langle \xi ,\zeta \rangle }_p\right)}^{\alpha }}},\end{array}$$

(17)

and

$$\begin{array}{c}\hfill {\displaystyle \frac{\partial g_{(s,\zeta )}}{\partial {\xi }_k}}(z,\xi )={\displaystyle \frac{-(1-\alpha )p_k{\xi }_k^{p_k-1}{\overline{{\zeta }_k}}^{p_k}}{{\left(1+z{z}^T\overline{s}{\overline{s}}^T-2z{\overline{s}}^T-{\langle \xi ,\zeta \rangle }_p\right)}^{\alpha }}}.\end{array}$$

(18)

By Proposition 2, we have

$$\begin{array}{ccc}& & {\left|\nabla g_{(s,\zeta )}(z,\xi )\right|}^2\hfill \\ & & =\sum _{j=1}^N{\left|{\displaystyle \frac{\partial g(s,\zeta )}{\partial z_j}}(z,\xi )\right|}^2+\sum _{k=1}^r{\left|{\displaystyle \frac{\partial g(s,\zeta )}{\partial {\xi }_k}}(z,\xi )\right|}^2\hfill \\ & & ={\displaystyle \frac{{(1-\alpha )}^2\left(4{\sum }_{j=1}^N{\left|z_j\overline{s}{\overline{s}}^T-\overline{s_j}\right|}^2+{\sum }_{k=1}^r{\left|p_k{\xi }_k^{p_k}{\overline{{\zeta }_k}}^{p_k})\right|}^2\right)}{{\left|1+z{z}^T\overline{s}{\overline{s}}^T-2z{\overline{s}}^T-{\langle \xi ,\zeta \rangle }_p\right|}^{2\alpha }}}\hfill \\ & & \le {\displaystyle \frac{{(1-\alpha )}^2\left(4{\left|z\overline{s}{\overline{s}}^T-\overline{s}\right|}^2+r{Q}^2\right)}{{\left|1+z{z}^T\overline{s}{\overline{s}}^T-2z{\overline{s}}^T-{\langle \xi ,\zeta \rangle }_p\right|}^{2\alpha }}}\hfill \\ & & \le {\displaystyle \frac{16{(1-\alpha )}^2\left(4{\left|z\overline{s}{\overline{s}}^T-\overline{s}\right|}^2+r{Q}^2\right)}{{\left[\left(1+{\left|z{z}^T\right|}^2-2z{\overline{z}}^T-{\parallel \xi \parallel }_p^2\right)+\left(1+{\left|s{s}^T\right|}^2-2s{\overline{s}}^T-{\parallel \zeta \parallel }_p^2\right)\right]}^{2\alpha }}}\hfill \\ & & \le {\displaystyle \frac{C}{{\left(1+{\left|z{z}^T\right|}^2-2z{\overline{z}}^T-{\sum }_{j=1}^r{\left|{\xi }_j\right|}^{2p_j}\right)}^{2\alpha }}},\hfill \end{array}$$

which implies that $g_{(s,\zeta )}\in {\mathcal{B}}^{\alpha }\left(H{E}_{IV}\right)$ with

$$\begin{array}{ccc}& & \underset{(s,\zeta )\in H{E}_{IV}}{sup}{\parallel g_{(s,\zeta )}\parallel }_{{\mathcal{B}}^{\alpha }\left(H{E}_{IV}\right)}\le C.\hfill \end{array}$$

(19)

From (17) and (18), we obtain

$$\begin{array}{ccc}& & |\mathcal{R}{g}_{(s,\zeta )}(z,\xi )|=\left|\sum _{j=1}^N{z}_j{\displaystyle \frac{\partial g(s,\zeta )}{\partial z_j}}(z,\xi )+\sum _{k=1}^r{\xi }_k{\displaystyle \frac{\partial g(s,\zeta )}{\partial {\xi }_k}}(z,\xi )\right|\hfill \\ & =& \left|\sum _{j=1}^N{z}_j{\displaystyle \frac{2(1-\alpha )(z_j\overline{s}{\overline{s}}^T-\overline{s_j})}{{\left(1+z{z}^T\overline{s}{\overline{s}}^T-2z{\overline{s}}^T-{\langle \xi ,\zeta \rangle }_p\right)}^{\alpha }}}-\sum _{k=1}^r{\xi }_k{\displaystyle \frac{(1-\alpha )p_k{\xi }_k^{p_k-1}{\overline{{\zeta }_k}}^{p_k}}{{\left(1+z{z}^T\overline{s}{\overline{s}}^T-2z{\overline{s}}^T-{\langle \xi ,\zeta \rangle }_p\right)}^{\alpha }}}\right|\hfill \\ & =& {\displaystyle \frac{|\alpha -1|\left|2{\sum }_{j=1}^N\left(z_j^2\overline{s}{\overline{s}}^T-z_j\overline{s_j}\right)-{\sum }_{k=1}^r\left(p_k{\xi }_k^{p_k}{\overline{{\zeta }_k}}^{p_k}\right)\right|}{{\left|1+z{z}^T\overline{s}{\overline{s}}^T-2z{\overline{s}}^T-{\langle \xi ,\zeta \rangle }_p\right|}^{\alpha }}},\hfill \end{array}$$

so taking $(z,\xi )=(w,\eta ):=\phi (z,\xi )$, then taking $(s,\zeta )=(w,\eta )$, and using (14), we obtain

$$\begin{array}{ccc}& & |\mathcal{R}{g}_{(s,\zeta )}(z,\xi )|=|\mathcal{R}{g}_{(w,\eta )}(w,\eta )|\hfill \\ & & ={\displaystyle \frac{|\alpha -1|\left|2{\sum }_{j=1}^N\left(z_j^2\overline{s}{\overline{s}}^T-z_j\overline{s_j}\right)-{\sum }_{k=1}^r\left(p_k{\xi }_k^{p_k}{\overline{{\zeta }_k}}^{p_k}\right)\right|}{{\left|1+w{w}^T\overline{w}{\overline{w}}^T-2w{\overline{w}}^T-{\parallel \eta \parallel }_p^2\right|}^{\alpha }}}\hfill \\ & & ={\displaystyle \frac{|\alpha -1|\left|2{\sum }_{j=1}^N\left(w_j^2\overline{w}{\overline{w}}^T-w_j\overline{w_j}\right)-{\sum }_{k=1}^r\left(p_k{\left|{\zeta }_k\right|}^{2p_k}\right)\right|}{{\left|1+w{w}^T\overline{w}{\overline{w}}^T-2w{\overline{w}}^T-{\parallel \eta \parallel }_p^2\right|}^{\alpha }}}\hfill \\ & & ={\displaystyle \frac{|\alpha -1|\left|{2\left(\right|w|}^2-|w{w}^T{|}^2)+{\sum }_{k=1}^r\left(p_k{\left|{\zeta }_k\right|}^{2p_k}\right)\right|}{{\left|1+w{w}^T\overline{w}{\overline{w}}^T-2w{\overline{w}}^T-{\parallel \eta \parallel }_p^2\right|}^{\alpha }}}\hfill \\ & & \ge {\displaystyle \frac{{q|\alpha -1|\parallel \eta \parallel }_p^2}{{\left|1+w{w}^T\overline{w}{\overline{w}}^T-2w{\overline{w}}^T-{\parallel \eta \parallel }_p^2\right|}^{\alpha }}}.\hfill \end{array}$$

(20)

Hence, using (19) and (20), we have that

$$\begin{array}{ccc}& & C∥W_{\psi ,\phi }\mathcal{R}∥\ge {\parallel g_{(s,\zeta )}\parallel }_{{\mathcal{B}}^{\alpha }\left(H{E}_{IV}\right)}∥W_{\psi ,\phi }\mathcal{R}∥\ge \hfill \\ & & \ge {∥W_{\psi ,\phi }\mathcal{R}{g}_{(s,\zeta )}∥}_{{\mathcal{A}}^{\beta }\left(H{E}_{IV}\right)}\hfill \\ & & =\underset{(z,\xi )\in H{E}_{IV}}{sup}{\left(1+{\left|z{z}^T\right|}^2-2z{\overline{z}}^T-\sum _{j=1}^r{\left|{\xi }_j\right|}^{2p_j}\right)}^{\beta }\left|\psi (z,\xi )\mathcal{R}{g}_{(s,\zeta )}\left(\phi (z,\xi )\right)\right|\hfill \\ & & \ge {\left(1+{\left|z{z}^T\right|}^2-2z{\overline{z}}^T-\sum _{j=1}^r{\left|{\xi }_j\right|}^{2p_j}\right)}^{\beta }\left|\psi (z,\xi )\mathcal{R}{g}_{(s,\zeta )}\left(\phi (z,\xi )\right)\right|\hfill \\ & & \ge {\left(1+{\left|z{z}^T\right|}^2-2z{\overline{z}}^T-\sum _{j=1}^r{\left|{\xi }_j\right|}^{2p_j}\right)}^{\beta }\left|\psi (z,\xi )\mathcal{R}{g}_{(w,\eta )}(w,\eta )\right|\hfill \\ & & \ge q|\alpha -1|{\displaystyle \frac{{\left(1+{\left|z{z}^T\right|}^2-2z{\overline{z}}^T-{\parallel \xi \parallel }_p^2\right)}^{\beta }}{{\left|1+w{w}^T\overline{w}{\overline{w}}^T-2w{\overline{w}}^T-{\parallel \eta \parallel }_p^2\right|}^{\alpha }}}{\left|\psi (z,\xi )\right|\parallel \eta \parallel }_p^2.\hfill \end{array}$$

Taking the supremum over all $(z,\xi )\in H{E}_{IV}$, we obtain

$$\begin{array}{ccc}& & \underset{(z,\xi )\in H{E}_{IV}}{sup}{\displaystyle \frac{{\left(1+{\left|z{z}^T\right|}^2-2z{\overline{z}}^T-{\parallel \xi \parallel }_p^2\right)}^{\beta }}{{\left|1+w{w}^T\overline{w}{\overline{w}}^T-2w{\overline{w}}^T-{\parallel \eta \parallel }_p^2\right|}^{\alpha }}}{\left|\psi (z,\xi )\right|\parallel \eta \parallel }_p^2<\infty ,\hfill \end{array}$$

that is, (8) holds for $\alpha \ne 1$. Hence, the proof is complete. □

Let $\phi \in H(H{E}_{IV},H{E}_{IV})$, and for all $(z,\xi )\in H{E}_{IV}$, we write $(w,\eta )=\phi (z,\xi )$. By Proposition 2, ${\parallel \eta \parallel }_p<1$. We say that $\phi$ satisfies the $\rho$-condition, if there exists $\rho \in (0,1)$ such that ${\parallel \eta \parallel }_p\ge \rho$. We have the following theorem.

Theorem 2.

Suppose $\psi \in H(H{E}_{IV},\mathbb{C})$; φ satisfies the ρ-condition; and $\alpha ,\beta >0$. Then, $W_{\psi ,\phi }\mathcal{R}:{\mathcal{B}}^{\alpha }\left(H{E}_{IV}\right)\to {\mathcal{A}}^{\beta }\left(H{E}_{IV}\right)$ is bounded if and only if

$$\begin{array}{c}\hfill M:=\underset{(z,\xi )\in H{E}_{IV}}{sup}{\displaystyle \frac{{\left(1+{\left|z{z}^T\right|}^2-2z{\overline{z}}^T-{\parallel \xi \parallel }_p^2\right)}^{\beta }}{{\left(1+{\left|w{w}^T\right|}^2-2w{\overline{w}}^T-{\parallel \eta \parallel }_p^2\right)}^{\alpha }}}\left|\psi (z,\xi )\right|<\infty ,\end{array}$$

(21)

where $(w,\eta ):=\phi (z,\xi )$.

Proof. 

If (21) holds, using (3), we obtain that (7) holds; so Theorem 1 implies that $W_{\psi ,\phi }\mathcal{R}:{\mathcal{B}}^{\alpha }\left(H{E}_{IV}\right)\to {\mathcal{A}}^{\beta }\left(H{E}_{IV}\right)$ is bounded. On the other hand, we can easily prove it; this is left to interested readers. □

3. Conclusions

Let $\Omega$ be a bounded domain in ${\mathbb{C}}^n$. There has been significant interest in operators acting on the subspaces of $H(\Omega ,\mathbb{C})$. However, there are still relatively few results concerning the product of the weighted composition operator and the radial derivative operator from Bloch spaces ${\mathcal{B}}^{\alpha }\left(H{E}_{IV}\right)$ into the Bers-type space ${\mathcal{A}}^{\beta }\left(H{E}_{IV}\right)$. Our hope is that this exposition will stimulate further research in this area.

In this study, we aimed to investigate the boundedness of the operator $W_{\psi ,\phi }\mathcal{R}$ from the Bloch-type spaces ${\mathcal{B}}^{\alpha }\left(H{E}_{IV}\right)$ into the Bers-type space ${\mathcal{A}}^{\beta }\left(H{E}_{IV}\right)$. This work serves as a solid starting point for discussion and further inquiry. It is worth noting that working with operators on the fourth Loo-Keng Hua domain $H{E}_{IV}$ presents challenges compared with operators on the subspace of all holomorphic functions on the unit ball. This difficulty primarily arises from the complexity of obtaining test functions in the Bloch-type space ${\mathcal{B}}^{\alpha }\left(H{E}_{IV}\right)$.