Weighted Space and Bloch-Type Space on the Unit Ball of an Infinite Dimensional Complex Banach Space

Abstract

Let ${\mathbf {B}}_{{\mathbb {X}}}$ be the open unit ball of a complex Banach space ${\mathbb {X}}$, which may beinfinite dimensional. The weighted composition operator and weighted space defined on ${\mathbf {B}}_{{\mathbb {X}}}$ are introduced. We obtain the boundedness and compactness of the weightedcomposition operator from the Bloch-type spaces to the weighted spaces, and some properties with the Bloch-type spaces are given. Our main results generalize theprevious works on the Euclidean unit ball ${\mathbb {B}}^n$ to the case of ${\mathbf {B}}_{{\mathbb {X}}}$.

1 Introduction

Let ${\mathbb {C}}^n$ be the space of n-dimensional complex variables $z=(z_1, z_2, \ldots , z_n)$. The unit ball

$$\begin{aligned} {\mathbb {B}}^n=\bigg \{z=(z_1,z_2,\ldots ,z_n)\in {\mathbb {C}}^n: \Vert z\Vert ^2=\sum \limits _{k=1}^n|z_k|^2<1\bigg \}, \end{aligned}$$

and ${\mathbb {B}}^1\equiv {\mathbb {U}}$ denotes the unit disk in ${\mathbb {C}}$. Let $H({\mathbb {B}}^n)$ be the family of holomorphic function from ${\mathbb {B}}^n$ to ${\mathbb {C}}$.

A positive continuous function $\mu$ on [0, 1) is called normal if there is $\delta \in [0, 1)$ and $0<a<b<\infty$ such that

$$\begin{aligned}&\frac{\mu (r)}{(1-r)^a} \hbox { is decreasing on }[\delta , 1) \hbox { and } \lim \limits _{r\rightarrow 1}\frac{\mu (r)}{(1-r)^a}=0;\\&\frac{\mu (r)}{(1-r)^b}\hbox { is increasing on }[\delta , 1)\hbox { and } \lim \limits _{r\rightarrow 1}\frac{\mu (r)}{(1-r)^b}=\infty . \end{aligned}$$

Then a normal function $\mu$ is strictly decreasing on $[\delta ,1)$ and $\mu (r)\rightarrow 0$ as $r\rightarrow 1.$ Denote by Aut$({\mathbb {B}}^n)$ the holomorphic automorphism group of ${\mathbb {B}}^n$. If $u\in H({\mathbb {B}}^n)$ and $\varphi \in$ Aut$({\mathbb {B}}^n)$, then the corresponding weighted composition operator is defined by

$$\begin{aligned} uC_{\varphi }(f)(z)=u(z)f(\varphi (z)),\,\,z\in {\mathbb {B}}^n. \end{aligned}$$

(1.1)

The boundedness and the compactness of the operators $uC_{\varphi }$ on Bloch-type spaces, Zygmund spaces, Hardy spaces and the weighted Bergman space attract a lot ofattentions (see, e.g., [6, 8, 12, 14, 16, 18, 19, 22, 23]).

The classical Bloch functions on the open unit disk ${\mathbb {U}}$ have been widelystudied (see, e.g., [1, 17]), and the corresponding notion in higher dimension was firstintroduced by Hahn [11]. Timoney [20] studied in depth the Bloch functions on bounded homogeneous domain in ${\mathbb {C}}^n$ with using Bergman metric (also see, Allen and Colonna [2]). Furthermore, Blasco et al. [3, 4] extended the Bloch space on ${\mathbb {B}}^n$ to the case of unit ball ${\mathbb {B}}_{{\mathbb {H}}}$ of an infinite dimensional complex Hilbert space ${\mathbb {H}}$. The bounded symmetric domains in complex Banach spaces are exactly the open unit balls of JB*-triples which are complex Banach spaces equipped with aJordan triple structure. Moreover, a complex Banach space is a JB*-triple if and only if its open unit ball is homogeneous (see, e.g., Deng and Ouyang [7], Kaup [13]). Recently, Chu et al. [5] generalized the Bloch space on ${\mathbb {B}}^n$ to the case of an infinitedimensional bounded symmetric domain realized as the open unit ball of a JB*-triple ${\mathbb {X}}$ by taking the place of the Bergman metric with the Kobayashi metric (compare with definition in Timoney [20]). In addition, they obtain the criteria for boundedness andcompactness on composition operator between the Bloch spaces on infinitedimensional bounded symmetric domain. By [5], Hamada [10] continued to study the weighted composition operators from the Hardy space $H^{\infty }$ to the Bloch space on bounded symmetric domains. Hamada [9] obtained the boundedness and compactness of the extended Cesàro operators between the Bloch-type spaces, which extended the results in Tang [21] to the case of unit ball of a infinite dimensional complex Banach space.

In this paper, we conform to the definitions of Bloch-type spaces and little Bloch-type spaces as [9], which generalize the corresponding spaces on ${\mathbb {B}}^n$ to the case of the open unit ball ${\mathbf {B}}_{{\mathbb {X}}}$ of an infinite dimensional complex Banach space ${\mathbb {X}}$ with arbitrary norm $\Vert \cdot \Vert$. All the weighted composition operator and weighted space are extended to ${\mathbf {B}}_{{\mathbb {X}}}$ (see, Sect. 2). We study the boundedness and compactness of the weighted composition operator from $\omega$-Bloch space ${\mathcal {B}}_{{\mathcal {R}},\omega }({\mathbf {B}}_{{\mathbb {X}}})$ (resp. little $\omega$-Bloch space ${\mathcal {B}}_{{\mathcal {R}},\omega _0}({\mathbf {B}}_{{\mathbb {X}}})$) to the weighted space on $H_{\mu }^{\infty }({\mathbf {B}}_{{\mathbb {X}}})$ (resp. little weighted space $H_{\mu _0}^{\infty }({\mathbf {B}}_{{\mathbb {X}}}))$ (see, Sect. 4). In Sect. 3, we give the relations between the Bloch-type spaces ${\mathcal {B}}_{{\mathcal {R}},\omega }({\mathbf {B}}_{{\mathbb {X}}})$ and the little Bloch-type spaces ${\mathcal {B}}_{{\mathcal {R}},\omega _0}({\mathbf {B}}_{{\mathbb {X}}})$. Using the work of Hamada [9], we successfully construct two test functions (see, Lemma 2.4), which play a key role in the proof of our main results. Since the Bloch-type spaces ${\mathcal {B}}_{{\mathcal {R}},\omega }({\mathbf {B}}_{{\mathbb {X}}})$ coincide with the Bloch-type spaces ${\mathcal {B}}_{{\mathcal {R}},\omega }({\mathbf {B}}_{{\mathbb {H}}})$ when ${\mathbb {X}}$ is a complex Hilbert space ${\mathbb {H}}$ [see Remark 2.1 in (iii)], our main results extend the corresponding works on ${\mathbb {B}}^n$ (see, e.g., Krantz and Stević [12]) to the case of ${\mathbf {B}}_{{\mathbb {X}}}$ of an infinite dimensional complex Banach space.

2 Preliminaries and Auxiliary Results

Let ${\mathbf {B}}_{{\mathbb {X}}}$ be the unit ball of a complex Banach space ${\mathbb {X}}$ with arbitrary norm $\Vert \cdot \Vert$. Let $H({\mathbf {B}}_{{\mathbb {X}}})$ denote the set of holomorphic mappings from ${\mathbf {B}}_{{\mathbb {X}}}$ into ${\mathbb {C}}.$ For $x\in {\mathbb {X}}{\setminus }\{0\}$, we define

$$\begin{aligned} T(x)=\{l_x\in {\mathbb {X}}^*: l_x(x)=\Vert x\Vert ,\,\,\Vert l_x\Vert =1\}. \end{aligned}$$

Then $T(x)\ne {{\emptyset }}$ in view of the Hahn–Banach theorem.

Let $\omega$ be a normal function on [0, 1), and $\omega$ can be extended to a function on ${\mathbf {B}}_{{\mathbb {X}}}$ by $\omega (z)=\omega (\Vert z\Vert ).$ A function $f\in H({\mathbf {B}}_{{\mathbb {X}}})$ is called a Bloch-type function with respect to $\omega$ if

$$\begin{aligned} \Vert f\Vert _{{\mathcal {B}}_{{\mathcal {R}},\omega }}=\sup \{\omega (z)|{\mathcal {R}}f(z)|:z\in {\mathbf {B}}_{{\mathbb {X}}}\}<+\infty , \end{aligned}$$

(2.1)

where ${\mathcal {R}}f(z)=Df(z)z$ is the radial derivative of f and Df(z) is the Fréchetderivative of f at z.

The class of all Bloch-type functions with respect to $\omega$ on ${\mathbf {B}}_{{\mathbb {X}}}$ is called a Bloch-type space on ${\mathbf {B}}_{{\mathbb {X}}}$ and is denoted by ${\mathcal {B}}_{{\mathcal {R}},\omega }({\mathbf {B}}_{{\mathbb {X}}})$. With the norm

$$\begin{aligned} \Vert f\Vert _{{{\mathcal {R}},\omega }}=|f(0)|+\Vert f\Vert _{{\mathcal {B}}_{{\mathcal {R}},\omega }}, \end{aligned}$$

the Bloch-type space ${\mathcal {B}}_{{\mathcal {R}},\omega }({\mathbf {B}}_{{\mathbb {X}}})$ becomes a Banach space (see, Proposition 2.5 in Hamada [9]).

The little Bloch-type space ${\mathcal {B}}_{{\mathcal {R}},\omega _0}({\mathbf {B}}_{{\mathbb {X}}})$ is a subspace of $\mathcal {B}_{{\mathcal {R}},\omega }({\mathbf {B}}_{{\mathbb {X}}})$ consisting of all f such that

$$\begin{aligned} \lim \limits _{\Vert z\Vert \rightarrow 1}\omega (z)|{\mathcal {R}}f(z)|=0. \end{aligned}$$

(2.2)

Remark 2.1

(i):

  The Bloch-type space ${\mathcal {B}}_{{\mathcal {R}},\omega }({\mathbf {B}}_{{\mathbb {X}}})$ and little Bloch-type space ${\mathcal {B}}_{{\mathcal {R}},\omega _0} ({\mathbf {B}}_{{\mathbb {X}}})$ were first introduced by Hamada [9], which generalize the corresponding spaces defined on the Euclidean unit ball ${\mathbb {B}}^n$ or on the unit disk ${\mathbb {U}}.$

(ii):

 By choosing different functions $\omega$, we have the following special spaces:

• If $\omega (z)=1-\Vert z\Vert ^2$ in (2.1) and (2.2), respectively, then we obtain the Bloch space ${\mathcal {B}}_{{\mathcal {R}}}({\mathbf {B}}_{{\mathbb {X}}})$ and little Bloch space $\mathcal {B}_{{\mathcal {R}},0}({\mathbf {B}}_{{\mathbb {X}}})$ in the unit ball of a complex Banach space (the case in ${\mathbb {B}}^n$, see, e.g., [15]).

• If $\omega (z)=(1-\Vert z\Vert ^2)^{\alpha }$ with $\alpha \in (0,\infty )$ in (2.1) and (2.2), respectively, then we obtain the $\alpha$-Bloch space ${\mathcal {B}}^{\alpha }_{{\mathcal {R}}}({\mathbf {B}}_{{\mathbb {X}}})$ and little $\alpha$-Bloch space ${\mathcal {B}}^{\alpha }_{{\mathcal {R}},0}({\mathbf {B}}_{{\mathbb {X}}})$ in the unit ball of a complex Banach space (the case in ${\mathbb {B}}^n$, see, e.g., [16]).

• If $\omega (z)=(1-\Vert z\Vert ^2)\big (\prod \nolimits _{j=1}^k\ln ^{[j]}\frac{e^k}{1-\Vert z\Vert ^2}\big )$ in (2.1) and (2.2), respectively, then we obtain the iterated logarithmic Bloch space ${\mathcal {B}}_{\log ,{\mathcal {R}}}({\mathbf {B}}_{{\mathbb {X}}})$ and little Bloch space $\mathcal {B}_{\log ,{\mathcal {R}},0}({\mathbf {B}}_{{\mathbb {X}}})$ in the unit ball of a complex Banach space (the case in ${\mathbb {B}}^n$, see, e.g., [12]).

(iii):

  In the case ${\mathbf {B}}_{{\mathbb {X}}}={\mathbf {B}}_{{\mathbb {H}}}$ is the unit ball of a complex Hilbert space ${\mathbb {H}}$, Hamada [9] proved that the condition (2.1) is equal to

$$\begin{aligned} \Vert f\Vert _{{\mathcal {B}}_{{\mathcal {R}},\omega }}=\sup \{\omega (z)\Vert Df(z)\Vert :z\in {\mathbf {B}}_{{\mathbb {X}}}\}<+\infty , \end{aligned}$$

(2.3)

and the condition (2.2) is equal to

$$\begin{aligned} \lim \limits _{\Vert z\Vert \rightarrow 1}\omega (z)\Vert Df(z)\Vert =0. \end{aligned}$$

(2.4)

The same situation holds with (2.3) and (2.4) when ${\mathbf {B}}_{{\mathbb {X}}}={\mathbb {B}}^n$ (see, Tang [21]). In fact, if $f\in H({\mathbf {B}}_{{\mathbb {X}}})$, then the relation $|{\mathcal {R}}f(z)|\leqslant \Vert Df(z)\Vert$ make sure that

$$\begin{aligned} \sup \{\omega (z)\Vert Df(z)\Vert :z\in {\mathbf {B}}_{{\mathbb {X}}}\}<+\infty \Rightarrow f\in \mathcal {B}_{{\mathcal {R}},\omega }({\mathbf {B}}_{{\mathbb {X}}}), \end{aligned}$$

but the converse is not true.

The weighted space $H_{\omega }^{\infty }({\mathbf {B}}_{{\mathbb {X}}})$ consisting of all $f\in H({\mathbf {B}}_{{\mathbb {X}}})$ such that

$$\begin{aligned} \Vert f\Vert _{H_{\omega }^{\infty }}=\sup \{\omega (z)|f(z)|:z\in {\mathbf {B}}_{{\mathbb {X}}}\}<+\infty , \end{aligned}$$

where $\omega$ is normal.

The little weighted space $H_{\omega _0}^{\infty }({\mathbf {B}}_{{\mathbb {X}}})$ is a subspace of $H_{\omega }^{\infty }({\mathbf {B}}_{{\mathbb {X}}})$ consisting of all f such that

$$\begin{aligned} \lim \limits _{\Vert z\Vert \rightarrow 1}\omega (z)|f(z)|=0. \end{aligned}$$

If $u\in H({\mathbf {B}}_{{\mathbb {X}}})$, and $\varphi \in$ Aut$({\mathbf {B}}_{{\mathbb {X}}})$, then the operator $uC_{\varphi }$ is defined by

$$\begin{aligned} uC_{\varphi }(f)(z)=u(z)f(\varphi (z)),\,f\in H({\mathbf {B}}_{{\mathbb {X}}}), z\in {\mathbf {B}}_{{\mathbb {X}}}. \end{aligned}$$

(2.5)

Remark 2.2

We note that (2.5) extends the corresponding weighted compositionoperator in (1.1) on ${\mathbb {B}}^n$ to the case of unit ball ${\mathbf {B}}_{{\mathbb {X}}}$ of a complex Banach space ${\mathbb {X}}$.

Next, we formulate and prove several auxiliary results which are used in the main theorems. Lemma 2.3 was proved by Hamada [9], (the corresponding results in ${\mathbb {B}}^n$, see Tang [21]). Lemma 2.5 is a generalization of the result on ${\mathbb {B}}^n$ (see, Krantz and Stević [12]) to the case of unit ball ${\mathbf {B}}_{{\mathbb {X}}}$ of a complex Banach space ${\mathbb {X}}.$ In Lemma 2.4, we define two test functions, which play a key role in the proof of our main theorems.

Lemma 2.3

([9], Lemma 2.1) Let $\omega$ be a normal function. Denote $k_0=\max (0,[\log _2\frac{1}{\omega (\delta )}])$, $r_k=(\omega |_{[\delta ,1)})^{-1}(\frac{1}{2^k})$ and $n_k=[\frac{1}{1-r_k}]$ for $k> k_0$, where the symbol [x] means the greatest integer not more than x. Let

$$\begin{aligned} G(\zeta )=1+\sum \limits _{k>k_0}^{\infty }2^k\zeta ^{n_k}, \zeta \in {\mathbb {U}}. \end{aligned}$$

Then

(i)  G is a holomorphic function on ${\mathbb {U}}$ such that G(r) is increasing on [0, 1) and

$$\begin{aligned} 0<C_1=\inf \limits _{r\in [0,1)}\omega (r)G(r)\leqslant \sup \limits _{r\in [0,1)}\omega (r)G(r)=C_2<\infty ; \end{aligned}$$

   (ii)  there exists a positive constant $C_3$ such that the inequality

$$\begin{aligned} \int _0^rG(t)\mathrm{{d}}t\leqslant C_3\int _0^{r^2}G(t)\mathrm{{d}}t \end{aligned}$$

holds for all $r\in [r_1,1)$, where $r_1\in (0,1)$ is a constant such that

$$\begin{aligned} \int _0^{r_1}G(t)\mathrm{{d}}t=1. \end{aligned}$$

Lemma 2.4

Let ${\mathbf {B}}_{{\mathbb {X}}}$ be the unit ball of a complex Banach space ${\mathbb {X}}$. For any $\nu \in {\mathbf {B}}_{{\mathbb {X}}}\backslash \{0\}$ and $l_\nu \in T(\nu )$, let

$$\begin{aligned} f_{\nu ,k}(z)=1+k\int _{0}^{\Vert \nu \Vert l_\nu (z)}G(\zeta )\mathrm{{d}}\zeta ,\,\,z\in {\mathbf {B}}_{{\mathbb {X}}}, \end{aligned}$$

where G is the function defined in Lemma 2.3 and $0<k<+\infty$. Then

(a)  $f_{\nu ,k}\in {\mathcal {B}}_{{\mathcal {R}}, \omega _0}({\mathbf {B}}_{{\mathbb {X}}})$ and $\Vert f_{\nu ,k}\Vert _{{\mathcal {R}},\omega }\leqslant 1+kC_2$, where $C_2$ is the constant defined in Lemma 2.3.

(b)   if $\Vert \nu \Vert \geqslant r_1$ and

$$\begin{aligned} F_{\nu ,k}(z)=\frac{1}{f_{\nu ,k}(\nu )}(f_{\nu ,k}(z))^2,\,\,z\in {\mathbf {B}}_{{\mathbb {X}}}, \end{aligned}$$

where $r_1$ is defined in Lemma 2.3, then $F_{\nu ,k}\in {\mathcal {B}}_{{\mathcal {R}},\omega _0}({\mathbf {B}}_{{\mathbb {X}}})$ and

$$\begin{aligned} \Vert F_{\nu ,k}\Vert _{{\mathcal {R}},\omega }\leqslant \frac{1}{k}C_3+ 2(1+k)C_2C_3, \end{aligned}$$

where $C_2,$$C_3$ are the constants defined in Lemma 2.3. Moreover, if $\int _0^1\frac{1}{\omega (t)}\mathrm{{d}}t=\infty$, then $F_{\nu ,k}\rightarrow 0$ uniformly on any closed ball strictly inside ${\mathbf {B}}_{{\mathbb {X}}}$ as $\Vert \nu \Vert \rightarrow 1$.

Proof

(a) Using Lemma 2.3, we obtain

$$\begin{aligned} \omega (z)|{\mathcal {R}}f_{\nu ,k}(z)|=k\omega (z)|G(\Vert \nu \Vert l_\nu (z))|\Vert \nu \Vert |l_\nu (z)|\leqslant k\omega (\Vert z\Vert )G(\Vert z\Vert )\leqslant kC_2. \end{aligned}$$

Therefore, $f_{\nu ,k}\in {\mathcal {B}}_{{\mathcal {R}},\omega }({\mathbf {B}}_{{\mathbb {X}}})$ and $\Vert f_{\nu ,k}\Vert _{{\mathcal {R}},\omega }\leqslant 1+kC_2.$ Moreover, since ${\mathcal {R}}f_{\nu ,k}$ is bounded on ${\mathbf {B}}_{{\mathbb {X}}}$, we have $\lim \nolimits _{\Vert z\Vert \rightarrow 1}\omega (z)|{\mathcal {R}}f_{\nu ,k}(z)|=0,$ which implies $f_{\nu ,k}\in \mathcal {B}_{{\mathcal {R}},\omega _0}({\mathbf {B}}_{{\mathbb {X}}})$.

(b) By some simple estimates and using the Lemma 2.3, we have

$$\begin{aligned} \omega (z)|{\mathcal {R}}F_{\nu ,k}(z)|&=2k\cdot \omega (z)\frac{1}{f_{\nu ,k}(\nu )}|f_{\nu ,k}(z)|G(\Vert \nu \Vert l_\nu (z))|\Vert \nu \Vert |l_\nu (z)|\nonumber \\&\leqslant 2k\cdot \omega (\Vert z\Vert )\frac{1}{f_{\nu ,k}(\nu )}|f_{\nu ,k}(z)|G(\Vert z\Vert )\nonumber \\&\leqslant 2kC_2\frac{1+k\int _{0}^{\Vert \nu \Vert }G(\zeta )\mathrm{{d}}\zeta }{1+k\int _{0}^{\Vert \nu \Vert ^2}G(\zeta )\mathrm{{d}}\zeta }\nonumber \\&\leqslant 2kC_2\frac{1+kC_3\int _{0}^{\Vert \nu \Vert ^2}G(\zeta )\mathrm{{d}}\zeta }{k\int _{0}^{\Vert \nu \Vert ^2}G(\zeta )\mathrm{{d}}\zeta }\nonumber \\&= 2kC_2\Bigg (\frac{1}{k}\frac{1}{\int _{0}^{\Vert \nu \Vert ^2}G(\zeta )\mathrm{{d}}\zeta }+C_3\Bigg )\nonumber \\&= 2kC_2\Bigg (\frac{1}{k}\frac{\int _{0}^{r_1}G(\zeta )\mathrm{{d}}\zeta }{\int _{0}^{\Vert \nu \Vert ^2}G(\zeta )\mathrm{{d}}\zeta }+C_3\Bigg )\nonumber \\&\leqslant 2kC_2\Bigg (\frac{1}{k}\frac{\int _{0}^{\Vert \nu \Vert }G(\zeta )\mathrm{{d}}\zeta }{\int _{0}^{\Vert \nu \Vert ^2}G(\zeta )\mathrm{{d}}\zeta }+C_3\Bigg )\nonumber \\&\leqslant 2kC_2\Bigg (\frac{1}{k}C_3+C_3\Bigg )=2(1+k)C_2C_3. \end{aligned}$$

(2.6)

By (2.6), we obtain $F_{\nu ,k}\in {\mathcal {B}}_{{\mathcal {R}},\omega }({\mathbf {B}}_{{\mathbb {X}}})$ and

$$\begin{aligned} \Vert F_{\nu ,k}\Vert _{{\mathcal {R}},\omega }&\leqslant |F_{\nu ,k}(0)|+ 2(1+k)C_2C_3\nonumber \\&=\big |\frac{1}{f_{\nu ,k}(\nu )}\big |+ 2(1+k)C_2C_3\nonumber \\&=\frac{1}{1+k\int _{0}^{\Vert \nu \Vert ^2}G(\zeta )\mathrm{{d}}\zeta }+ 2(1+k)C_2C_3\nonumber \\&\leqslant \frac{\int _{0}^{\Vert \nu \Vert }G(\zeta )\mathrm{{d}}\zeta }{k\int _{0}^{\Vert \nu \Vert ^2}G(\zeta )\mathrm{{d}}\zeta }+ 2(1+k)C_2C_3\nonumber \\&=\frac{1}{k}C_3+ 2(1+k)C_2C_3. \end{aligned}$$

(2.7)

Moreover, since ${\mathcal {R}}F_{\nu ,k}$ is bounded on ${\mathbf {B}}_{{\mathbb {X}}}$, we have $\lim \nolimits _{\Vert z\Vert \rightarrow 1}\omega (z)|{\mathcal {R}}F_{\nu ,k}(z)|=0,$ which implies $F_{\nu ,k}\in \mathcal {B}_{{\mathcal {R}},\omega _0}({\mathbf {B}}_{{\mathbb {X}}})$.

Next, if $\int _0^1\frac{1}{\omega (t)}\mathrm{{d}}t=\infty$, then we can prove that $F_{\nu ,k}\rightarrow 0$ uniformly for $\Vert z\Vert \leqslant r\,\,(0<r<1)$ as $\Vert \nu \Vert \rightarrow 1$ by a similar way in ([9], Lemma 2.7). The proof is finished. $\square$

Lemma 2.5

Suppose that $u\in H({\mathbf {B}}_{{\mathbb {X}}}),$$\mu$ is normal and $\varphi \in Aut({\mathbf {B}}_{{\mathbb {X}}})$. Then theoperator $uC_\varphi$ : ${\mathcal {B}}_{{\mathcal {R}},\omega }({\mathbf {B}}_{{\mathbb {X}}})\rightarrow H_{\mu }^{\infty }({\mathbf {B}}_{{\mathbb {X}}})$ is compact if and only if $uC_\varphi$: ${\mathcal {B}}_{{\mathcal {R}},\omega }({\mathbf {B}}_{{\mathbb {X}}})\rightarrow H_{\mu }^{\infty }({\mathbf {B}}_{{\mathbb {X}}})$ is bounded, and for any bounded sequence $\{f_k\}_{k\in {\mathbb {N}}}$ in $\mathcal {B}_{{\mathcal {R}},\omega }({\mathbf {B}}_{{\mathbb {X}}})$ converging to zero uniformly on compact subset of ${\mathbf {B}}_{{\mathbb {X}}}$ as $k\rightarrow \infty$, we have that $\lim \nolimits _{k\rightarrow \infty }\Vert uC_\varphi (f_k)\Vert _{H_{\mu }^{\infty }}=0$.

Proof

It is similar to the proofs of the corresponding results [cf. ([4], Lemma 4.4) and ([12], Lemma 5)]. We omit the proof here.

Throughout this paper, the notation $A\asymp B$ means that there is a positive constant C such that $\frac{A}{C}\leqslant B\leqslant CA$. $\square$

3 Some Properties with Spaces ${\mathcal {B}}_{{\mathcal {R}},\mu }({\mathbf {B}}_{{\mathbb {X}}})$ and ${\mathcal {B}}_{{\mathcal {R}},\mu _0}({\mathbf {B}}_{{\mathbb {X}}})$

In this section, we prove ${\mathcal {B}}_{{\mathcal {R}},\mu _0}({\mathbf {B}}_{{\mathbb {X}}})$ is a closed subset of $\mathcal {B}_{{\mathcal {R}},\mu }({\mathbf {B}}_{{\mathbb {X}}})$ in Theorem 3.1, and the transform relationship between of them is given by delay function in Theorem 3.2, which generalize the previous works on ${\mathbb {B}}^n$ to the case of unit ball ${\mathbf {B}}_{{\mathbb {X}}}$ (see, Theorem 2 and Theorem 3 in Krantz and Stević [12]).

Theorem 3.1

Let ${\mathbf {B}}_{{\mathbb {X}}}$ be the unit ball of a complex Banach space ${\mathbb {X}}$. Then ${\mathcal {B}}_{{\mathcal {R}},\mu _0}({\mathbf {B}}_{{\mathbb {X}}})$ is a closed subset of $\mathcal {B}_{{\mathcal {R}},\mu }({\mathbf {B}}_{{\mathbb {X}}})$, where $\mu$ is normal on [0, 1).

Proof

Let $\{f_j\}_{j\in {\mathbb {N}}}$ be a sequence in $\mathcal {B}_{{\mathcal {R}},\mu _0}({\mathbf {B}}_{{\mathbb {X}}})$ such that

$$\begin{aligned} \lim \limits _{j\rightarrow +\infty }\Vert f_j-f\Vert _{{{\mathcal {R}},\mu }}=0, \,\,f\in {\mathcal {B}}_{{\mathcal {R}},\mu }({\mathbf {B}}_{{\mathbb {X}}}). \end{aligned}$$

(3.1)

Using the (3.1), we have that, for every $\varepsilon >0$, there is an $j_0\in {\mathbb {N}}$ such that

$$\begin{aligned} \Vert f_j-f\Vert _{{{\mathcal {R}},\mu }}<\varepsilon \end{aligned}$$

(3.2)

for $j\geqslant j_0$. In particular, taking $j=j_0$ in (3.2), it gives that

$$\begin{aligned} \sup \limits _{z\in {\mathbf {B}}_{{\mathbb {X}}}}\mu (\Vert z\Vert )|{\mathcal {R}}f_{j_0}(z)&-{\mathcal {R}}f(z)| \leqslant |f_{j_0}(z)-f(0)|\nonumber \\&+\sup \limits _{z\in {\mathbf {B}}_{{\mathbb {X}}}}\mu (\Vert z\Vert )|{\mathcal {R}}f_{j_0}(z)-{\mathcal {R}}f(z)|<\varepsilon . \end{aligned}$$

(3.3)

On the other hand, $f_{j_0}\in {\mathcal {B}}_{{\mathcal {R}},\mu _0}({\mathbf {B}}_{{\mathbb {X}}})$ makes sure that, for every $\varepsilon >0$, there is a $\delta >0$ such that

$$\begin{aligned} \mu (\Vert z\Vert )|{\mathcal {R}}f_{j_0}(z)|<\varepsilon \end{aligned}$$

(3.4)

for $\delta<\Vert z\Vert <1$. Thus, from (3.3) and (3.4), we have

$$\begin{aligned} \mu (\Vert z\Vert )|{\mathcal {R}}f(z)|&= \mu (\Vert z\Vert )|{\mathcal {R}}f(z)-{\mathcal {R}}f_{j_0}(z)+{\mathcal {R}}f_{j_0}(z)|\\&\leqslant \mu (\Vert z\Vert )|{\mathcal {R}}f(z)-{\mathcal {R}}f_{j_0}(z)|+\mu (\Vert z\Vert )|{\mathcal {R}}f_{j_0}(z)|\\&\leqslant 2\varepsilon \end{aligned}$$

for $\delta<\Vert z\Vert <1$, which implies that $f\in {\mathcal {B}}_{{\mathcal {R}},\mu _0}({\mathbf {B}}_{{\mathbb {X}}})$. The proof of this theorem is completed. $\square$

Theorem 3.2

Let ${\mathbf {B}}_{{\mathbb {X}}}$ be the unit ball of a complex Banach space ${\mathbb {X}}$. Assume that $f\in {\mathcal {B}}_{{\mathcal {R}},\mu }({\mathbf {B}}_{{\mathbb {X}}})$ and $f_r(z)=f(rz),\,\, r\in [0,1), \,\,z\in {\mathbf {B}}_{{\mathbb {X}}}.$ Then $f\in {\mathcal {B}}_{{\mathcal {R}},\mu _0}({\mathbf {B}}_{{\mathbb {X}}})$ if and only if

$$\begin{aligned} \lim \limits _{r\rightarrow 1^{-}}\Vert f-f_r\Vert _{{\mathcal {R}},\mu }=0, \end{aligned}$$

(3.5)

where $\mu$ is normal decreasing function of $\Vert z\Vert$ with $\mu (0)<+\infty .$

Proof

After some simple computations, it is easy to see that

$$\begin{aligned} {\mathcal {R}}f_r(z)={\mathcal {R}}f(rz), z\in {\mathbf {B}}_{{\mathbb {X}}}. \end{aligned}$$

(3.6)

Since $\mu$ is normal decreasing function of $\Vert z\Vert$, by (3.6), it follows that

$$\begin{aligned} \mu (\Vert z\Vert )|{\mathcal {R}}f_r(z)|&= \mu (\Vert z\Vert )|{\mathcal {R}}f(rz)|\leqslant \Vert f\Vert _{{\mathcal {B}}_{{\mathcal {R}},\mu }}\frac{\mu (\Vert z\Vert )}{\mu (\Vert rz\Vert )} \leqslant \Vert f\Vert _{\mathcal {B}_{{\mathcal {R}},\mu }}\frac{\mu (\Vert z\Vert )}{\mu (r)}. \end{aligned}$$

(3.7)

Assume that (3.5) holds. Let $f\in {\mathcal {B}}_{{\mathcal {R}},\mu }({\mathbf {B}}_{{\mathbb {X}}})$, then by (3.7), we have

$$\begin{aligned} \lim \limits _{\Vert z\Vert \rightarrow 1^{-}}\mu (\Vert z\Vert )|{\mathcal {R}}f_r(z)|=0. \end{aligned}$$

This implies that $f_r\in {\mathcal {B}}_{{\mathcal {R}},\mu _0}({\mathbf {B}}_{{\mathbb {X}}})$. Furthermore, $\mathcal {B}_{{\mathcal {R}},\mu _0}({\mathbf {B}}_{{\mathbb {X}}})$ is a closed subset of ${\mathcal {B}}_{{\mathcal {R}},\mu }({\mathbf {B}}_{{\mathbb {X}}})$ (see, Theorem 3.1), then (3.5) implies that $f\in {\mathcal {B}}_{{\mathcal {R}},\mu _0}({\mathbf {B}}_{{\mathbb {X}}})$.

Now assume $f\in {\mathcal {B}}_{{\mathcal {R}},\mu _0}({\mathbf {B}}_{{\mathbb {X}}})$. Then, for every $\varepsilon >0$, there is a $\delta \in (0,1)$ such that

$$\begin{aligned} \mu (z)|{\mathcal {R}}f(z)|<\varepsilon , \end{aligned}$$

(3.8)

as $\delta ^2<\Vert z\Vert <1$. By (3.6), we have

$$\begin{aligned} \Vert f-f_r\Vert _{{\mathcal {R}},\mu }&=\sup \limits _{z\in {\mathbf {B}}_{{\mathbb {X}}}}\mu (z)|{\mathcal {R}}f(rz)-{\mathcal {R}}f(z)|\nonumber \\&\leqslant \sup \limits _{\Vert z\Vert \leqslant \delta }\mu (z)|{\mathcal {R}}f(rz)-{\mathcal {R}}f(z)|\nonumber \\&\quad +\sup \limits _{\Vert z\Vert >\delta }\mu (z)|{\mathcal {R}}f(rz)-{\mathcal {R}}f(z)|. \end{aligned}$$

(3.9)

It is obviously

$$\begin{aligned} \lim \limits _{r\rightarrow 1^{-}}\sup \limits _{\Vert z\Vert \leqslant \delta }|{\mathcal {R}}f(rz)-{\mathcal {R}}f(z)|=0 \end{aligned}$$

(3.10)

and

$$\begin{aligned} \sup \limits _{\Vert z\Vert \leqslant \delta }\mu (z)=\sup \limits _{\Vert z\Vert \leqslant \delta }\mu (\Vert z\Vert )\leqslant \mu (0)<+\infty . \end{aligned}$$

(3.11)

Thus, using (3.10) and (3.11), it follows that

$$\begin{aligned} \lim \limits _{r\rightarrow 1^{-}}\sup \limits _{\Vert z\Vert \leqslant \delta }\mu (z)|{\mathcal {R}}f(rz)-{\mathcal {R}}f(z)|=0. \end{aligned}$$

(3.12)

On the other hand, by (3.8), for all $z\in {\mathbf {B}}_{{\mathbb {X}}}$ and $r\in [0,1)$ such that $\delta<\Vert z\Vert<1, \delta<r<1$, we have

$$\begin{aligned} \mu (z)|{\mathcal {R}}f(rz)|<\mu (rz)|{\mathcal {R}}f(rz)|<\varepsilon . \end{aligned}$$

(3.13)

Hence, by (3.8) and (3.13), it follows that

$$\begin{aligned} \sup \limits _{\delta<\Vert z\Vert<1}\mu (z)|{\mathcal {R}}f(rz)-{\mathcal {R}}f(z)|&\leqslant \sup \limits _{\delta<\Vert z\Vert<1}\mu (z)|{\mathcal {R}}f(rz)|\nonumber \\&+\sup \limits _{\delta<\Vert z\Vert<1}\mu (z)|{\mathcal {R}}f(z)|<2\varepsilon \end{aligned}$$

(3.14)

for every $r\in (\delta ,1)$. Using (3.12) and (3.14) in (3.9), we obtain (3.5). The proof is finished. $\square$

4 $uC_{\varphi }:{\mathcal {B}}_{{\mathcal {R}},\omega }({\mathbf {B}}_{{\mathbb {X}}})$(or ${\mathcal {B}}_{{\mathcal {R}},{\omega _0}}({\mathbf {B}}_{{\mathbb {X}}}))\rightarrow H_{\mu }^{\infty }({\mathbf {B}}_{{\mathbb {X}}})$(or $H_{\mu _0}^{\infty }({\mathbf {B}}_{{\mathbb {X}}}))$

In this section, we study the boundedness and compactness of operator $uC_{\varphi }: {\mathcal {B}}_{{\mathcal {R}},\omega }({\mathbf {B}}_{{\mathbb {X}}})\rightarrow H_{\mu }^{\infty }({\mathbf {B}}_{{\mathbb {X}}})$ (resp. $\mathcal {B}_{{\mathcal {R}},\omega _0}({\mathbf {B}}_{{\mathbb {X}}})\rightarrow H_{\mu _0}^{\infty }({\mathbf {B}}_{{\mathbb {X}}}))$, which generalize thecorresponding results on ${\mathbb {B}}^n$ to the case of unit ball ${\mathbf {B}}_{{\mathbb {X}}}$ of a infinite dimensional complex Banach space ${\mathbb {X}}$ (see, Theorems 16–19 in Krantz and Stević [12]). Thefollowing set $E_{\varepsilon ,\rho }$ is needed when consider the compactness of the operator $uC_{\varphi }$. For $\forall \varepsilon >0$ and $\rho \in (0,1)$, we define

$$\begin{aligned} E_{\varepsilon ,\rho }=\bigg \{z\in {\mathbf {B}}_{{\mathbb {X}}}: \Vert z\Vert \leqslant \rho , \exists s\in \bigg [1,\frac{1}{\rho }\bigg ], \mathrm{{s.t.}}\,\, \mu (sz)|u(sz)|\geqslant \varepsilon \bigg \}. \end{aligned}$$

(4.1)

Theorem 4.1

Let ${\mathbf {B}}_{{\mathbb {X}}}$ be the unit ball of a complex Banach space. Assume that $u\in H({\mathbf {B}})$, $\varphi \in$ Aut$({\mathbf {B}}_{{\mathbb {X}}})$, $\omega$ and $\mu$ are normal on [0, 1). Then $uC_{\varphi }: {\mathcal {B}}_{{\mathcal {R}},\omega }({\mathbf {B}}_{{\mathbb {X}}})\rightarrow H_{\mu }^{\infty }({\mathbf {B}}_{{\mathbb {X}}})$ is bounded if and only if

$$\begin{aligned} \sup \limits _{z\in {\mathbf {B}}_{{\mathbb {X}}}}\mu (z)|u(z)|\Bigg (1+\int _0^{\Vert \varphi (z)\Vert }\frac{1}{\omega (t)}\mathrm{{d}}t\Bigg )<\infty . \end{aligned}$$

(4.2)

Moreover, if $uC_{\varphi }: {\mathcal {B}}_{{\mathcal {R}},\omega }({\mathbf {B}}_{{\mathbb {X}}})\rightarrow H_{\mu }^{\infty }({\mathbf {B}}_{{\mathbb {X}}})$ is bounded, then

$$\begin{aligned} \Vert uC_{\varphi }\Vert _{{\mathcal {R}},\mu }\asymp \sup \limits _{z\in {\mathbf {B}}_{{\mathbb {X}}}}\mu (z)|u(z)|\Bigg (1+\int _0^{\Vert \varphi (z)\Vert }\frac{1}{\omega (t)}\mathrm{{d}}t\Bigg ). \end{aligned}$$

(4.3)

Proof

Assume that (4.2) holds and $f\in {\mathcal {B}}_{{\mathcal {R}},\omega }({\mathbf {B}}_{{\mathbb {X}}})$, $z\in {\mathbf {B}}_{{\mathbb {X}}}$. Using Proposition 2.4 in [9], we have

$$\begin{aligned} \Vert uC_{\varphi }(f)(z)\Vert _{{\mathcal {B}}_{{\mathcal {R}},\omega }\rightarrow H_{\mu }^{\infty }}&=\sup \limits _{z\in {\mathbf {B}}_{{\mathbb {X}}}}\mu (z)|uC_{\varphi }(f)(z)|=\sup \limits _{z\in {\mathbf {B}}_{{\mathbb {X}}}}\mu (z)|f(\varphi (z))u(z)|\nonumber \\&=\sup \limits _{z\in {\mathbf {B}}_{{\mathbb {X}}}}\mu (z)|u(z)||f(\varphi (z))| \nonumber \\&\leqslant C_4\Vert f\Vert _{{\mathcal {R}}, \omega }\sup \limits _{z\in {\mathbf {B}}_{{\mathbb {X}}}}\mu (z)|u(z)|\Bigg (1+\int _0^{\Vert \varphi (z)\Vert }\frac{1}{\omega (t)}\mathrm{{d}}t\Bigg ). \end{aligned}$$

(4.4)

From (4.2) and (4.4), it follows that

$$\begin{aligned} \Vert uC_{\varphi }\Vert _{H_{\mu }^{\infty }}\leqslant C_4\sup \limits _{z\in {\mathbf {B}}_{{\mathbb {X}}}}\mu (z)|u(z)|\Bigg (1+\int _0^{\Vert \varphi (z)\Vert }\frac{1}{\omega (t)}\mathrm{{d}}t\Bigg )<\infty \end{aligned}$$

(4.5)

and $uC_{\varphi }: {\mathcal {B}}_{{\mathcal {R}},\omega }({\mathbf {B}}_{{\mathbb {X}}})\rightarrow H_{\mu }^{\infty }({\mathbf {B}}_{{\mathbb {X}}})$ is bounded.

Conversely, assume that $uC_{\varphi }: {\mathcal {B}}_{{\mathcal {R}},\omega }({\mathbf {B}}_{{\mathbb {X}}})\rightarrow H_{\mu }^{\infty }({\mathbf {B}}_{{\mathbb {X}}})$ is bounded. For $\nu \in {\mathbf {B}}_{{\mathbb {X}}}, k\geqslant 0$ and $l_\nu \in T(\nu )$, we give the test function

$$\begin{aligned} f_{\nu ,k}(z)=1+k\int _0^{\Vert \nu \Vert l_\nu (z)}G(\zeta )\mathrm{{d}}\zeta , z\in {\mathbf {B}}_{{\mathbb {X}}}, \end{aligned}$$

(4.6)

where G is defined as Lemma 2.3. By Lemma 2.4, then $f_{\nu ,k}\in {\mathcal {B}}_{{\mathcal {R}},\omega _0}({\mathbf {B}}_{{\mathbb {X}}})\subset {\mathcal {B}}_{{\mathcal {R}},\omega }({\mathbf {B}}_{{\mathbb {X}}})$ and $\Vert f_{\nu ,k}\Vert _{{\mathcal {R}},\omega }\leqslant C_2$. Let $r_1$ be the constant in Lemma 2.3. Thus, for $\nu \in {\mathbf {B}}_{{\mathbb {X}}}$ with $\Vert \varphi (\nu )\Vert \geqslant r_1$, we get

$$\begin{aligned}&\mu (\nu )|u(\nu )|\Bigg (1+\int _0^{\Vert \varphi (\nu )\Vert }\frac{1}{\omega (t)}\mathrm{{d}}t\Bigg )\leqslant \mu (\nu )|u(\nu )|\Bigg (1+\int _0^{\Vert \varphi (\nu )\Vert }\frac{G(t)}{C_1}\mathrm{{d}}t\Bigg )\nonumber \\&\quad \leqslant \mu (\nu )|u(\nu )|\Bigg (1+\frac{C_3}{C_1}\int _0^{\Vert \varphi (\nu )\Vert ^2}G(t)\mathrm{{d}}t\Bigg ) \nonumber \\&\quad \leqslant \mu (\nu )|u(\nu )|f_{\varphi (\nu ),\frac{C_3}{C_1}}(\varphi (\nu ))|\leqslant \sup \limits _{\nu \in {\mathbf {B}}_{{\mathbb {X}}}}\mu (\nu )|u(\nu )||f_{\varphi (\nu ),\frac{C_3}{C_1}}(\varphi (\nu ))|\nonumber \\&\quad =\sup \limits _{\nu \in {\mathbf {B}}_{{\mathbb {X}}}}\mu (\nu )|uC_{\varphi }[f_{\varphi (\nu ),\frac{C_3}{C_1}}](\nu )| = \Vert uC_{\varphi }[f_{\varphi (\nu ),\frac{C_3}{C_1}}]\Vert _{\mathcal {B}_{{\mathcal {R}},\omega }\rightarrow H_{\mu }^{\infty }}\nonumber \\&\quad \leqslant \Vert uC_{\varphi }\Vert \Vert f_{\varphi (\nu ), \frac{C_3}{C_1}}\Vert _{{\mathcal {R}},\omega }\leqslant C_2\Vert uC_{\varphi }\Vert <\infty . \end{aligned}$$

(4.7)

If $\Vert \varphi (\nu )\Vert <r_1$, then by Lemma 2.3, we have

$$\begin{aligned}&\mu (\nu )|u(\nu )|\Bigg (1+\int _0^{\Vert \varphi (\nu )\Vert }\frac{1}{\omega (t)}\mathrm{{d}}t\Bigg )\leqslant \mu (\nu )|u(\nu )|\Bigg (1+\int _0^{\Vert \varphi (\nu )\Vert }\frac{G(t)}{C_1}\mathrm{{d}}t\Bigg )\nonumber \\&\quad \leqslant \mu (\nu )|u(\nu )|\Bigg (1+\int _0^{r_1}\frac{G(t)}{C_1}\mathrm{{d}}t\Bigg )=\left( 1+\frac{1}{C_1}\right) \mu (\nu )|u(\nu )|\nonumber \\&\quad \leqslant \left( 1+\frac{1}{C_1}\right) \sup \limits _{\nu \in {\mathbf {B}}_{{\mathbb {X}}}}\mu (\nu )|u(\nu )|\leqslant \left( 1+\frac{1}{C_1}\right) \Vert uC_{\varphi }(1)\Vert _{{\mathcal {B}}_{{\mathcal {R}},\omega }\rightarrow H_{\mu }^{\infty }}\nonumber \\&\quad \leqslant (1+\frac{1}{C_1})\Vert uC_{\varphi }\Vert <\infty . \end{aligned}$$

(4.8)

The inequalities (4.7) and (4.8) yield (4.2), as desired. Moreover, from (4.4), (4.7) and (4.8), we obtain (4.3). This completes the proof. $\square$

Theorem 4.2

Let ${\mathbf {B}}_{{\mathbb {X}}}$ be the unit ball of a complex Banach space ${\mathbb {X}}$. Assume $u\in H({\mathbf {B}})$, $\varphi \in$ Aut$({\mathbf {B}}_{{\mathbb {X}}})$, $\omega$ and $\mu$ are normal functions on [0, 1). Let the set $E_{\varepsilon ,\rho }$ is relatively compact in ${\mathbf {B}}_{{\mathbb {X}}}$ for any $\varepsilon >0$ and $\rho \in (0,1)$. The following statements are true.

(a):

  If $\int _{0}^{1}\frac{1}{\omega (t)}\mathrm{{d}}t<\infty$, then $uC_{\varphi }: {\mathcal {B}}_{{\mathcal {R}},\omega }({\mathbf {B}}_{{\mathbb {X}}})\rightarrow H_{\mu }^{\infty }({\mathbf {B}}_{{\mathbb {X}}})$ is compact if and only if $u\in H_{\mu }^\infty ({\mathbf {B}}_{{\mathbb {X}}})$.

(b):

  If $\int _{0}^{1}\frac{1}{\omega (t)}\mathrm{{d}}t=\infty$, then $uC_{\varphi }: {\mathcal {B}}_{{\mathcal {R}},\omega }({\mathbf {B}}_{{\mathbb {X}}})\rightarrow H_{\mu }^{\infty }({\mathbf {B}}_{{\mathbb {X}}})$ is compact if and only if $uC_{\varphi }: {\mathcal {B}}_{{\mathcal {R}},\omega }({\mathbf {B}}_{{\mathbb {X}}})\rightarrow H_{\mu }^{\infty }({\mathbf {B}}_{{\mathbb {X}}})$ is bounded and

$$\begin{aligned} \lim \limits _{\Vert \varphi (z)\Vert \rightarrow 1}\mu (z)|u(z)|\Bigg (1+\int _0^{\Vert \varphi (z)\Vert }\frac{1}{\omega (t)}\mathrm{{d}}t\Bigg )=0. \end{aligned}$$

(4.9)

Proof

(a) Suppose that $uC_{\varphi }: {\mathcal {B}}_{{\mathcal {R}},\omega }({\mathbf {B}}_{{\mathbb {X}}})\rightarrow H_{\mu }^{\infty }({\mathbf {B}}_{{\mathbb {X}}})$ is compact. Then it is clear that $uC_{\varphi }$ is bounded. Take the test function ${\widehat{f}}(z)\equiv 1, z\in {\mathbf {B}}_{{\mathbb {X}}}.$ It is easy to know that ${\widehat{f}}\in {\mathcal {B}}_{{\mathcal {R}},\omega }({\mathbf {B}}_{{\mathbb {X}}})$. We have

$$\begin{aligned} \sup \limits _{z\in {\mathbf {B}}_{{\mathbb {X}}}}\mu (z)uC_{\varphi }[{\widehat{f}}](z)= \sup \limits _{z\in {\mathbf {B}}_{{\mathbb {X}}}}\mu (z)|u(z)|<\infty , \end{aligned}$$

which implies $u\in H_{\mu }^\infty ({\mathbf {B}}_{{\mathbb {X}}}).$

Conversely, if $u\in H_{\mu }^\infty ({\mathbf {B}}_{{\mathbb {X}}})$, then $\int _{0}^{1}\frac{1}{\omega (t)}\mathrm{{d}}t<\infty$ makes sure that (4.2) holds. Thus, $uC_{\varphi }: {\mathcal {B}}_{{\mathcal {R}},\omega }({\mathbf {B}}_{{\mathbb {X}}})\rightarrow H_{\mu }^{\infty }({\mathbf {B}}_{{\mathbb {X}}})$ is bounded by Theorem 4.1. For any $\varepsilon >0$, there exists $\rho \in (\frac{1}{2},1)$ such that

$$\begin{aligned} \mu (z)|u(z)|\Bigg (1+\int _\rho ^{\Vert \varphi (z)\Vert }\frac{1}{\omega (t)}\mathrm{{d}}t\Bigg )<\frac{\varepsilon }{3}, \end{aligned}$$

(4.10)

when $z\in {\mathbf {B}}_{{\mathbb {X}}}$ with $\rho<\Vert \varphi (z)\Vert <1.$ Let $\{f_j\}_{j\in {\mathbb {N}}}$ be a bounded sequence in $\mathcal {B}_{{\mathcal {R}},\omega }({\mathbf {B}}_{{\mathbb {X}}})$ which converges to 0 uniformly on any compact subset of ${\mathbf {B}}_{{\mathbb {X}}}$. We may assume that $\Vert f_j\Vert _{{\mathcal {R}},\omega }\leqslant 1$. Then $|f_j|\leqslant C_\rho$ for all j and $\Vert z\Vert \leqslant \rho$ by Proposition 2.4 in [9], where

$$\begin{aligned} C_\rho =C_4\Big (1+\int _{0}^{\rho }\frac{1}{\omega (t)}\mathrm{{d}}t\Big ). \end{aligned}$$

There exists a positive N such that

$$\begin{aligned} |f_j(w)|\leqslant \frac{\varepsilon }{3\Vert u\Vert _{H_\mu ^\infty }+1}, j>N, \,\,w\in E_{\varepsilon /(3C_\rho ),\rho }. \end{aligned}$$

Therefore, for $\Vert \varphi (z)\Vert \leqslant \rho$ and $t=1$ or for $\rho<\Vert \varphi (z)\Vert <1$ and $t=\frac{\rho }{\Vert \varphi (z)\Vert }$, we have

$$\begin{aligned} \mu (z)|u(z)||f_j(t\varphi (z))|<\frac{\varepsilon }{3},\,\,j>N. \end{aligned}$$

(4.11)

Next, we will finish the proof through a case by case check:

Case I:  if $z\in {\mathbf {B}}_{{\mathbb {X}}}$ with $\Vert \varphi (z)\Vert \leqslant \rho$, by (4.11), then we have

$$\begin{aligned} \mu (z)|u(z)||f_j(\varphi (z))|<\frac{\varepsilon }{3},\,\, j>N \end{aligned}$$

(4.12)

holds.

Case II:  if $z\in {\mathbf {B}}_{{\mathbb {X}}}$ with $\rho<\Vert \varphi (z)\Vert <1$, by (4.10), (4.11) and some simple calculation, we have

$$\begin{aligned}&\mu (z)|u(z)||f_j(\varphi (z))|\nonumber \\&\quad \leqslant \mu (z)|u(z)|\Big |f_j(\varphi (z))-f_j\Big (\rho \frac{\varphi (z)}{\Vert \varphi (z)\Vert }\Big )\Big |+ \mu (z)|u(z)|\Big |f_j\Big (\rho \frac{\varphi (z)}{\Vert \varphi (z)\Vert }\Big )\Big |\nonumber \\&\quad \leqslant \mu (z)|u(z)\int _{\frac{\rho }{\Vert \varphi (z)\Vert }}^1|{\mathcal {R}}f_j(t\varphi (z))|\frac{1}{t}\mathrm{{d}}t+\mu (z)|u(z)|\Big |f_j\Big (\rho \frac{\varphi (z)}{\Vert \varphi (z)\Vert }\Big )\Big | \nonumber \\&\quad \leqslant \mu (z)|u(z)|\frac{\Vert \varphi (z)\Vert }{\rho }\int _{\frac{\rho }{\Vert \varphi (z)\Vert }}^1|{\mathcal {R}}f_j(t\varphi (z))|\mathrm{{d}}t+\mu (z)|u(z)|\Big |f_j\Big (\rho \frac{\varphi (z)}{\Vert \varphi (z)\Vert }\Big )\Big |\nonumber \\&\quad \leqslant 2\mu (z)|u(z)|\Vert \varphi (z)\Vert \int _{\frac{\rho }{\Vert \varphi (z)\Vert }}^1|{\mathcal {R}}f_j(t\varphi (z))|\mathrm{{d}}t+\mu (z)|u(z)|\Big |f_j\Big (\rho \frac{\varphi (z)}{\Vert \varphi (z)\Vert }\Big )\Big |\nonumber \\&\quad \leqslant 2\mu (z)|u(z)|\Vert \varphi (z)\Vert \int _{\frac{\rho }{\Vert \varphi (z)\Vert }}^1\frac{1}{\omega (t\Vert \varphi (z)\Vert )}\mathrm{{d}}t+\mu (z)|u(z)|\Big |f_j\Big (\rho \frac{\varphi (z)}{\Vert \varphi (z)\Vert }\Big )\Big | \nonumber \\&\quad \leqslant 2\mu (z)|u(z)|\int _{\rho }^{\Vert \varphi (z)\Vert }\frac{1}{\omega (t)}\mathrm{{d}}t+\mu (z)|u(z)|\Big |f_j\Big (\rho \frac{\varphi (z)}{\Vert \varphi (z)\Vert }\Big )\Big |\nonumber \\&\quad < \frac{2}{3}\varepsilon +\frac{1}{3}\varepsilon =\varepsilon \end{aligned}$$

(4.13)

for $j>N.$

For any $\varepsilon >0$, employing the (4.12) and (4.13), we obtain

$$\begin{aligned} \Vert uC_{\varphi }f_j\Vert _{H_{\mu }^{\infty }}<\varepsilon ,\,\, j>N. \end{aligned}$$

(4.14)

Using Lemma 2.5 and (4.14), $uC_{\varphi }: \mathcal {B}_{{\mathcal {R}},\omega }({\mathbf {B}}_{{\mathbb {X}}})\rightarrow H_{\mu }^{\infty }({\mathbf {B}}_{{\mathbb {X}}})$ is compact.

(b) If $uC_{\varphi }: {\mathcal {B}}_{{\mathcal {R}},\omega }({\mathbf {B}}_{{\mathbb {X}}})\rightarrow H_{\mu }^{\infty }({\mathbf {B}}_{{\mathbb {X}}})$ is compact, then $uC_{\varphi }: {\mathcal {B}}_{{\mathcal {R}},\omega }({\mathbf {B}}_{{\mathbb {X}}})\rightarrow H_{\mu }^{\infty }({\mathbf {B}}_{{\mathbb {X}}})$ is bounded. Now, we assume that conditions (4.9) dose not hold, then there exist $\varepsilon >0$ and a sequence $\{\varphi (z_k)\}_{k\in {\mathbb {N}}}\subset {\mathbf {B}}_{{\mathbb {X}}}$ such that

$$\begin{aligned} \lim \limits _{k\rightarrow \infty } \Vert \varphi (z_k)\Vert =1 \end{aligned}$$

(4.15)

and

$$\begin{aligned} \mu (z_k)|u(z_k)|\Bigg (1+\int _0^{\Vert \varphi (z_k)\Vert }\frac{1}{\omega (t)}\mathrm{{d}}t\Bigg )\geqslant \varepsilon , \,\,k=1,2,3,\ldots \end{aligned}$$

(4.16)

For $\nu \in {\mathbf {B}}_{{\mathbb {X}}}$ and $l_\nu \in T(\nu )$, we give the test function

$$\begin{aligned} F_{\varphi (z_k),\frac{C_3}{C_1}}(z)=\frac{1}{1+\frac{C_3}{C_1}\int _0^{\Vert \varphi (z_k)\Vert ^2}G(\zeta )\mathrm{{d}}\zeta } \Bigg (1+\frac{C_3}{C_1}\int _0^{\Vert \varphi (z_k)\Vert l_{\varphi (z_k)}(z)}G(\zeta )\mathrm{{d}}\zeta \Bigg )^2, \end{aligned}$$

(4.17)

where G is defined as Lemma 2.3 and $z\in {\mathbf {B}}_{{\mathbb {X}}}$. In fact, (4.15) implies that we can assume that $\Vert \varphi (z_k)\Vert >r_1$, where $r_1$ is the constant in Lemma 2.3. Furthermore, if we take $f_k(z)=F_{\varphi (z_k),\frac{C_3}{C_1}}(z)$ for $z\in {\mathbf {B}}_{{\mathbb {X}}}$, from Lemma 2.4, then $\{f_k\}_{k\in N}$ is a bounded sequence in ${\mathcal {B}}_{{\mathcal {R}},\omega _0}({\mathbf {B}}_{{\mathbb {X}}})$ and $f_k\rightarrow 0$ uniformly on any compact subset of ${\mathbf {B}}_{{\mathbb {X}}}$. Therefore, we have

$$\begin{aligned} \lim \limits _{k\rightarrow \infty }\Vert uC_{\varphi }(f_k)\Vert _{H_{\mu }^{\infty }}=0. \end{aligned}$$

(4.18)

On the other hand, using (4.16) and Lemma 2.3 with $\Vert \varphi (z_k)\Vert >r_1$, we have

$$\begin{aligned} \Vert uC_{\varphi }(f_k)\Vert _{H_{\mu }^{\infty }}&=\sup \limits _{z\in {\mathbf {B}}_{{\mathbb {X}}}}\mu (z)|u(z)||f_k(\varphi (z))|\geqslant \mu (z_k)|u(z_k)||f_k(\varphi (z_k))|\nonumber \\&=\mu (z_k)|u(z_k)|\Bigg (1+\frac{C_3}{C_1}\int _0^{\Vert \varphi (z_k)\Vert ^2}G(t)\mathrm{{d}}t\Bigg )\nonumber \\&\geqslant \mu (z_k)|u(z_k)|\Bigg (1+\frac{1}{C_1}\int _0^{\Vert \varphi (z_k)\Vert }G(t)\mathrm{{d}}t\Bigg )\nonumber \\&\geqslant \mu (z_k)|u(z_k)|\Bigg (1+\int _0^{\Vert \varphi (z_k)\Vert }\frac{1}{\omega (t)}\mathrm{{d}}t\Bigg )\geqslant \varepsilon , \end{aligned}$$

(4.19)

(4.19) is a contradiction compare with (4.18). Thus, we obtain (4.9).

Conversely, assume that $uC_{\varphi }: {\mathcal {B}}_{{\mathcal {R}},\omega }({\mathbf {B}}_{{\mathbb {X}}})\rightarrow H_{\mu }^{\infty }({\mathbf {B}}_{{\mathbb {X}}})$ is bounded and (4.9) holds. By Theorem 4.1, we know that condition (4.2) holds. Furthermore, since $\int _{0}^{1}\frac{1}{\omega (t)}\mathrm{{d}}t=\infty$, we get (4.2) implies $u\in H_\mu ^\infty ({\mathbf {B}}_{{\mathbb {X}}})$. Moreover, according to (4.9), there exists $\rho \in (\frac{1}{2},1)$ such that

$$\begin{aligned} \mu (z)|u(z)|\Bigg (1+\int _0^{\Vert \varphi (z)\Vert }G(t)\mathrm{{d}}t\Bigg )<\frac{\varepsilon }{3},\,\,\rho<\Vert \varphi (z)\Vert <1, \end{aligned}$$

where $\varepsilon$ is any small positive number. The rest of the proof is similar to the case (a), we omit it. This proof is completed. $\square$

Theorem 4.3

Let ${\mathbf {B}}_{{\mathbb {X}}}$ be the unit ball of a complex Banach space. Assume $u\in H({\mathbf {B}}_{{\mathbb {X}}})$, $\varphi \in$ Aut$({\mathbf {B}}_{{\mathbb {X}}})$, $\omega$ and $\mu$ are normal functions on [0, 1). Then $uC_{\varphi }: {\mathcal {B}}_{{\mathcal {R}},{\omega _0}}({\mathbf {B}}_{{\mathbb {X}}})\rightarrow H_{\mu ,0}^{\infty }({\mathbf {B}}_{{\mathbb {X}}})$ is bounded if and only if $u\in H_{\mu _0}^\infty ({\mathbf {B}}_{{\mathbb {X}}})$ and

$$\begin{aligned} M=:\sup \limits _{z\in {\mathbf {B}}_{{\mathbb {X}}}}\mu (z)|u(z)|\Bigg (1+\int _0^{\Vert \varphi (z)\Vert }\frac{1}{\omega (t)}\mathrm{{d}}t\Bigg )<\infty . \end{aligned}$$

(4.20)

Proof

First assume that $u\in H_{\mu ,0}^\infty ({\mathbf {B}}_{{\mathbb {X}}})$ and (4.20) hold. By Theorem 4.1, we know that $uC_{\varphi }: {\mathcal {B}}_{{\mathcal {R}},\omega }({\mathbf {B}}_{{\mathbb {X}}})\rightarrow H_{\mu }^{\infty }({\mathbf {B}}_{{\mathbb {X}}})$ is bounded. Therefore, we only need to prove that $uC_{\varphi }(f)\in H_{\mu ,0}^\infty ({\mathbf {B}}_{{\mathbb {X}}})$ for any $f\in \mathcal {B}_{{\mathcal {R}},{\omega _0}}({\mathbf {B}}_{{\mathbb {X}}})$. Now, let $f\in {\mathcal {B}}_{{\mathcal {R}},{\omega _0}}({\mathbf {B}}_{{\mathbb {X}}})$. Applying (4.20), then there exists $\rho _0\in (\frac{1}{2},1)$ such that

$$\begin{aligned} \omega (z)|{\mathcal {R}}f(z)|<\frac{\varepsilon }{4M},\,\, \rho _0\leqslant \Vert z\Vert <1, \end{aligned}$$

(4.21)

where $\varepsilon$ is an arbitrarily small positive number.

Taking ${\widetilde{z}}=\rho _0\frac{\varphi (z)}{\Vert \varphi (z)\Vert }$, then $\Vert {\widetilde{z}}\Vert =\rho _0<1$. For any $z\in {\mathbf {B}}_{{\mathbb {X}}}$ with $\rho _0<\Vert z\Vert <1$, using (4.21), we have

$$\begin{aligned} |f(\varphi (z))-f({\widetilde{z}})|&=\Big |\int _{\frac{\rho _0}{\Vert \varphi (z)\Vert }}^1\frac{{\mathcal {R}}f(t\varphi (z))}{t}\mathrm{{d}}t\Big |\nonumber \\&\leqslant \frac{\Vert \varphi (z)\Vert }{\rho _0}\int _{\frac{\rho _0}{\Vert \varphi (z)\Vert }}^1|{\mathcal {R}}f(t\varphi (z))|\mathrm{{d}}t \nonumber \\&\leqslant \frac{\varepsilon }{4M}\frac{\Vert \varphi (z)\Vert }{\rho _0}\int _{\frac{\rho _0}{\Vert \varphi (z)\Vert }}^1\frac{1}{\omega (t\Vert \varphi (z)\Vert )}\mathrm{{d}}t\nonumber \\&\leqslant \frac{\varepsilon }{4\rho _0M}\int _{\rho _0}^{\Vert \varphi (z)\Vert }\frac{1}{\omega ({\tilde{t}})}\mathrm{{d}}{\tilde{t}}\nonumber \\&\leqslant \frac{\varepsilon }{2M}\int _{\rho _0}^{\Vert \varphi (z)\Vert }\frac{1}{\omega ({\tilde{t}})}\mathrm{{d}}{\tilde{t}}. \end{aligned}$$

(4.22)

In the above equality (4.22), we use the fact that

$$\begin{aligned} \rho _0=\frac{\rho _0}{\Vert \varphi (z)\Vert }\Vert \varphi (z)\Vert<\Vert t\varphi (z)\Vert<t<1. \end{aligned}$$

Set $K=\sup \nolimits _{\Vert z\Vert \leqslant \rho _0}|f(z)|$. By Proposition 2.4 in [9], $K<\infty .$ Since $u\in H_{\mu ,0}^\infty ({\mathbf {B}}_{{\mathbb {X}}})$, we have, for any $\varepsilon >0$, there is $\rho _1\in (\rho _0,1)$ such that

$$\begin{aligned} \mu (z)|u(z)|<\frac{\varepsilon }{2K}, \end{aligned}$$

(4.23)

whenever $\rho _1<\Vert z\Vert <1$. Combining (4.22) and (4.23), for $\rho _1<\Vert z\Vert <1$, we obtain

$$\begin{aligned} \mu (z)|uC_{\varphi }(f)(z)|&=\mu (z)|u(z)||f(\varphi (z))|\nonumber \\&=\mu (z)|u(z)||f(\varphi (z))-f({\widetilde{z}})+f({\widetilde{z}})|\nonumber \\&\leqslant \mu (z)|u(z)||f(\varphi (z))-f({\widetilde{z}})|+\mu (z)|u(z)||f({\widetilde{z}})|\nonumber \\&\leqslant \frac{\varepsilon }{2M}\mu (z)|u(z)|\int _{\rho _0}^{{\Vert \varphi (z)\Vert }}\frac{1}{\omega ({\tilde{t}})}\mathrm{{d}}{\tilde{t}}+\mu (z)|u(z)||f({\widetilde{z}})|\nonumber \\&\leqslant \frac{\varepsilon }{2M}M+\frac{\varepsilon }{2K}K=\varepsilon . \end{aligned}$$

(4.24)

Hence, (4.24) implies that $uC_{\varphi }(f)\in H_{\mu ,0}^\infty ({\mathbf {B}}_{{\mathbb {X}}})$.

Conversely, assume that $uC_{\varphi }: {\mathcal {B}}_{{\mathcal {R}},{\omega _0}}({\mathbf {B}}_{{\mathbb {X}}})\rightarrow H_{\mu ,0}^{\infty }({\mathbf {B}}_{{\mathbb {X}}})$ is bounded. Taking $f(z)\equiv 1$, then $f\in {\mathcal {B}}_{{\mathcal {R}},{\omega _0}}({\mathbf {B}}_{{\mathbb {X}}}).$ Thus,

$$\begin{aligned} \lim \limits _{\Vert z\Vert \rightarrow 1}\mu (z)|uC_{\varphi }(1)|= \lim \limits _{\Vert z\Vert \rightarrow 1}\mu (z)|u(z)|=0, \end{aligned}$$

which implies that $u\in H_{\mu ,0}^\infty ({\mathbf {B}}_{{\mathbb {X}}})$. On the other hand, For $\nu \in {\mathbf {B}}_{{\mathbb {X}}}, k\geqslant 0$ and $l_\nu \in T(\nu )$, we give the test function

$$\begin{aligned} f_{\nu ,k}(z)=1+k\int _0^{\Vert \nu \Vert l_\nu (z)}G(\zeta )\mathrm{{d}}\zeta , \,\,z\in {\mathbf {B}}_{{\mathbb {X}}}, \end{aligned}$$

(4.25)

where G is defined as Lemma 2.3. Applying the facts ${\mathcal {B}}_{{\mathcal {R}},\omega _0}({\mathbf {B}}_{{\mathbb {X}}})\subset {\mathcal {B}}_{{\mathcal {R}},\omega }({\mathbf {B}}_{{\mathbb {X}}})$ and $H_{\mu ,0}^\infty ({\mathbf {B}}_{{\mathbb {X}}})\subset H_{\mu }^\infty ({\mathbf {B}}_{{\mathbb {X}}})$, we can obtain (4.20) by (4.7) and (4.8). The proof is finished. $\square$

Theorem 4.4

Let ${\mathbf {B}}_{{\mathbb {X}}}$ be the unit ball of a complex Banach space ${\mathbb {X}}$. Assume $u\in H({\mathbf {B}}_{{\mathbb {X}}}),$$\varphi \in$ Aut$({\mathbf {B}}_{{\mathbb {X}}})$, $\omega$ and $\mu$ are normal functions on [0, 1). The set $E_{\varepsilon ,\rho }$ is relatively compact in ${\mathbf {B}}_{{\mathbb {X}}}$ for any $\varepsilon >0$ and $\rho \in (0,1)$. Then $uC_{\varphi }: \mathcal {B}_{{\mathcal {R}},{\omega _0}}({\mathbf {B}}_{{\mathbb {X}}})\rightarrow H_{\mu _0}^{\infty }({\mathbf {B}}_{{\mathbb {X}}})$ is compact if and only if

$$\begin{aligned} \lim \limits _{\Vert z\Vert \rightarrow 1}\mu (z)|u(z)|\Bigg (1+\int _0^{\Vert \varphi (z)\Vert }\frac{1}{\omega (t)}\mathrm{{d}}t\Bigg )=0. \end{aligned}$$

(4.26)

Proof

Assume that (4.26) holds. Taking any $f\in \mathcal {B}_{{\mathcal {R}},{\omega _0}}({\mathbf {B}}_{{\mathbb {X}}})$, by Proposition 2.4 in [9], we have

$$\begin{aligned}&\mu (z)|uC_{\varphi }(f)(z)|=\mu (z)|u(z)||f(\varphi (z))|\nonumber \\&\quad \leqslant C_4\mu (z)|u(z)|\Bigg (1+\int _{0}^{\Vert \varphi (z)\Vert }\frac{1}{\omega (t)}\mathrm{{d}}t\Bigg )\Vert f\Vert _{{\mathcal {R}},\omega }, \end{aligned}$$

(4.27)

which implies

$$\begin{aligned} \lim \limits _{\Vert z\Vert \rightarrow 1}\sup \{\mu (z)|uC_{\varphi }(f)(z)|: \,\,\Vert f\Vert _{{\mathcal {R}},\omega }\leqslant 1\}=0. \end{aligned}$$

(4.28)

Therefore, $uC_{\varphi }: {\mathcal {B}}_{{\mathcal {R}},{\omega _0}}({\mathbf {B}}_{{\mathbb {X}}})\rightarrow H_{\mu _0}^{\infty }({\mathbf {B}}_{{\mathbb {X}}})$ is compact.

Conversely, assume that $uC_{\varphi }: {\mathcal {B}}_{{\mathcal {R}},{\omega _0}}({\mathbf {B}}_{{\mathbb {X}}})\rightarrow H_{\mu _0}^{\infty }({\mathbf {B}}_{{\mathbb {X}}})$ is compact, then $uC_{\varphi }$ is bounded. Taking the test function ${\widehat{f}}(z)=1$, we have

$$\begin{aligned} \lim \limits _{\Vert z\Vert \rightarrow 1}\mu (z)uC_{\varphi }(f)(z)=\lim \limits _{\Vert z\Vert \rightarrow 1}\mu (z)|u(z)|=0. \end{aligned}$$

(4.29)

Now we prove

$$\begin{aligned} \lim \limits _{\Vert \varphi (z)\Vert \rightarrow 1}\mu (z)|u(z)|\Bigg (1+\int _0^{\Vert \varphi (z)\Vert }\frac{1}{\omega (t)}\mathrm{{d}}t\Bigg )=0 \end{aligned}$$

(4.30)

holds. In fact, if (4.30) does not hold, then there exist $\varepsilon >0$ and a sequence $\{\varphi (z_j)\}\subset {\mathbf {B}}_{{\mathbb {X}}}$ such that $\lim \nolimits _{j\rightarrow \infty }\Vert \varphi (z_j)\Vert =1$ and

$$\begin{aligned} \mu (z_j)|u(z_j)|\Bigg (1+\int _0^{\Vert \varphi (z_j)\Vert }\frac{1}{\omega (t)}\mathrm{{d}}t\Bigg )\geqslant \varepsilon , \,\,j=1,2,3,\ldots \end{aligned}$$

(4.31)

Taking the same test functions $f_k(z)$ as (4.17) and using the similar ways inTheorem 4.2, we note that $\{f_k\}$ is a bounded sequence in ${\mathcal {B}}_{{\mathcal {R}},{\omega _0}}({\mathbf {B}}_{{\mathbb {X}}})$ and $f_k\rightarrow 0$uniformly on any compact subset of ${\mathbf {B}}_{{\mathbb {X}}}$. Since $uC_{\varphi }: {\mathcal {B}}_{{\mathcal {R}},{\omega _0}}({\mathbf {B}}_{{\mathbb {X}}})\rightarrow H_{\mu _0}^{\infty }({\mathbf {B}}_{{\mathbb {X}}})$ is compact, we may assume that there exists some $h\in H_{\mu _0}^{\infty }({\mathbf {B}}_{{\mathbb {X}}})$ such that $\Vert uC_{\varphi }(f_k)-h\Vert _{H_{\mu ,0}^{\infty }}\rightarrow 0$ as $k\rightarrow \infty$. Then for each $z\in {\mathbf {B}}_{{\mathbb {X}}}$, we have

$$\begin{aligned} h(z)=\lim \limits _{k\rightarrow \infty }uC_{\varphi }(f_k)(z)=uC_{\varphi }(\lim \limits _{k\rightarrow \infty }f_k)(z)= uC_{\varphi }(0)(z)=0. \end{aligned}$$

Thus, we have $\Vert uC_{\varphi }(f_k)\Vert _{H_{\mu ,0}^{\infty }}\rightarrow 0$ as $k\rightarrow \infty$. This contradicts with (4.31). Thus, (4.30) holds. By (4.30) we have that for any $\varepsilon >0$ there exists $r\in (0,1)$ such that

$$\begin{aligned} \mu (z)|u(z)|\Bigg (1+\int _0^{\Vert \varphi (z)\Vert }\frac{1}{\omega (t)}\mathrm{{d}}t\Bigg )<\varepsilon \end{aligned}$$

(4.32)

with $r<\Vert \varphi (z)\Vert <1.$ On the other hand, using (4.29), there exists $\rho \in (0,1)$ such that

$$\begin{aligned} \mu (z)|u(z)|<\frac{\varepsilon }{1+\int _0^{r}\frac{1}{\omega (t)}\mathrm{{d}}t} \end{aligned}$$

(4.33)

with $\rho<\Vert z\Vert <1.$ Therefore, when $\rho<\Vert z\Vert <1$ and $r<\Vert \varphi (z)\Vert <1$, from (4.32) and (4.33), we have

$$\begin{aligned} \mu (z)|u(z)|\Bigg (1+\int _0^{\Vert \varphi (z)\Vert }\frac{1}{\omega (t)}\mathrm{{d}}t\Bigg )<\varepsilon . \end{aligned}$$

(4.34)

If $\rho<\Vert z\Vert <1$ and $\Vert \varphi (z)\Vert \leqslant r$, from (4.33), then

$$\begin{aligned} \mu (z)|u(z)|\Bigg (1+\int _0^{\Vert \varphi (z)\Vert }\frac{1}{\omega (t)}\mathrm{{d}}t\Bigg )< \mu (z)|u(z)|\int _0^{r}\frac{1}{\omega (t)}\mathrm{{d}}t<\varepsilon . \end{aligned}$$

(4.35)

Combining (4.34) and (4.35), we obtain

$$\begin{aligned} \lim \limits _{\Vert z\Vert \rightarrow 1}\mu (z)|u(z)|\int _0^{\Vert \varphi (z)\Vert }\frac{1}{\omega (t)}\mathrm{{d}}t=0. \end{aligned}$$

(4.36)

This proof is completed. $\square$

Remark 4.5

This assumption that the set $E_{\varepsilon ,\rho }$ is relatively compact in ${\mathbf {B}}_{{\mathbb {X}}}$ isautomatically satisfied when dim(${\mathbb {X}})<+\infty$ (see Theorems 4.2, 4.4).