Weighted composition operators on $\alpha$-Bloch-Orlicz spaces over the unit polydisc

1.   Introduction

Let ${\mathbb{U}}$ be the unit disk in the complex plane ${\mathbb C}$, ${\mathbb{U}}^{n} = \{z = (z_{1}, z_{2}, \cdots, z_{n}): |z_{i}| < 1, i = 1, 2, \cdots, n\}$ the unit polydisc in the complex vector space ${\mathbb C}^n$, and $\partial{\mathbb{U}}^{n} = \{z = (z_{1}, z_{2}, \cdots, z_{n}): |z_{i}| = 1, \, \, \ i = 1, 2, \cdots, n\}$ the distinguished boundary of ${\mathbb{U}}^{n}$. Let $H({\mathbb{U}}^n)$ be the space of all holomorphic functions on ${\mathbb{U}}^{n}$ and $H^\infty({\mathbb{U}}^n)$ the space of all bounded holomorphic functions on ${\mathbb{U}}^{n}$ (see, for example, ^[12,14,20]).

Let $\psi\in H({\mathbb{U}}^{n})$ and $\varphi = (\varphi_{1}, \varphi_{2}, \cdots, \varphi_{n})$ be a holomorphic self-mapping of ${\mathbb{U}}^{n}$. The weighted composition operator $W_{\psi, \varphi}$ on some subspaces of $H({\mathbb{U}}^n)$ is defined by

If $\psi(z)\equiv1$ on ${\mathbb{U}}^n$, the operator $W_{\psi, {\varphi}}$ is reduced to the composition operator $C_{\varphi}$, while if ${\varphi}(z) = z$, it is reduced to the multiplication operator $M_\psi$. The theory of the (weighted) composition operators on various spaces has quite a long and rich history (we will give some concrete studies. For example, see ^[1,2,3] or^[5] for the studies of composition operators; see ^[7,11,17] from one space to Bloch type spaces; see ^[8,9,24] from one space to weighted spaces or Bloch spaces; see ^[28,29,30] for one on Hardy spaces, Zygmund-Orlicz spaces, or logarithmic Bloch-Orlicz spaces).

Recall that for $\alpha > 0$, the $\alpha$-Bloch space on ${\mathbb{U}}^n$ denoted by ${\mathcal {B}}_{{\alpha}}({\mathbb{U}}^n)$ consists of all $f\in H({\mathbb{U}}^n)$ such that

It is well-known that ${\mathcal {B}}_{\alpha}({\mathbb{U}}^n)$ is a Banach space with the norm $\|f\|_{{\mathcal {B}}_{\alpha}({\mathbb{U}}^n)} = |f(0)|+\|f\|_{\alpha}$. For this and related spaces and operators on them, see, for example, ^[4,19,21] and the references therein.

A positive continuous function on ${\mathbb{U}}$ is called weight. Let $\mu$ be a weight. The $\mu$-Bloch space on ${\mathbb{U}}^n$ denoted by ${\mathcal {B}}_{\mu}({\mathbb{U}}^n)$ consists of all $f\in H({\mathbb{U}}^n)$ such that

It is a Banach space endowed with the norm $\|f\|_{{\mathcal {B}}_{\mu}({\mathbb{U}}^n)} = |f(0)|+\|f\|_{\mu}$. Clearly, the $\mu$-Bloch space is a natural generalization of the ${\alpha}$-Bloch space (see ^{[22,23,25,27]} for other Bloch type spaces, and see ^[10,13] for ${\alpha}$-Bloch space). For some information on this space, also see ^[18].

As an important closed subspace of ${\mathcal {B}}_\mu({\mathbb{U}}^n)$, the little $\mu$-Bloch space ${\mathcal {B}}_{\mu, 0}({\mathbb{U}}^n)$ consists of all $f\in {\mathcal {B}}_{\mu}({\mathbb{U}}^n)$ such that

In order to introduce the ${\alpha}$-Bloch-Orlicz space on ${\mathbb{U}}^n$, we explain here the Bloch-Orlicz space on ${\mathbb{U}}$, which was introduced by Ramos Fernández in ^[18]. More precisely, let $\phi$ be a Young's function, that is, $\phi$ is a strictly increasing convex function on the interval $[0, +\infty)$ such that

Since $\phi(0) = 0$, from the convexity of $\phi$, it clearly follows that $\phi(st)\leq s\phi(t)$ for $0 < s < 1$ and $t > 0$.

The Bloch-Orlicz space denoted by ${\mathcal {B}}_\phi({\mathbb{U}})$ consists of all $f\in H({\mathbb{U}})$ such that

for some positive $\lambda$ depending on $f$. The Minkowski's functional

defines a semi-norm for ${\mathcal {B}}_\phi({\mathbb{U}})$, where

${\mathcal {B}}_\phi({\mathbb{U}})$ becomes a Banach space with the norm

Motivated by the Bloch-Orlicz space, the ${\alpha}$-Bloch-Orlicz space on ${\mathbb{U}}$ denoted by ${\mathcal {B}}_{\phi, {\alpha}}({\mathbb{U}})$ was introduced by Liang in ^[15]. Since ${\mathbb{U}}^n$ is an important bounded symmetric domain of ${\mathbb C}^n$, it is natural to define a similar space on the domain and study some concrete operators on it.

The abovementioned facts motivate us to define the ${\alpha}$-Bloch-Orlicz space on ${\mathbb{U}}^n$. The space consists of all $f\in H({\mathbb{U}}^n)$ such that

for some $\lambda > 0$ depending on $f$, and it is denoted by ${\mathcal {B}}_{\phi, {\alpha}}({\mathbb{U}}^n)$. Since $\phi$ is convex, it is easy to see that the Minkowski's functional

defines a semi-norm on ${\mathcal {B}}_{\phi, \alpha}({\mathbb{U}}^n)$, which is known as Luxemburg's semi-norm, where

It can be easily proved that ${\mathcal {B}}_{\phi, \alpha}({\mathbb{U}}^n)$ is a Banach space with the norm

Observe that when $\phi(t) = t$ with $t\geq 0$, we get back the $\alpha$-Bloch space ${\mathcal {B}}_{\alpha}({\mathbb{U}}^n)$. Furthermore, from ^[18] we can suppose that $\phi^{-1}$ is continuously differentiable on $[0, +\infty)$.

In this paper, we mainly study the boundedness and compactness of the operator $W_{\psi, \varphi}$ on ${\mathcal {B}}_{\phi, \alpha}({\mathbb{U}}^n)$. It can be regarded as a continuation of the investigation of concrete operators on these spaces.

Throughout the paper, we will write $W_{\psi, {\varphi}}f$ instead of $W_{\psi, {\varphi}}(f)$. The letter $C$ will denote a positive constant, and the exact value may vary in each case. The notation $a\lesssim b$ means that there is a constant $C > 0$, such that $a\leq Cb$. When $a\lesssim b$ and $b\lesssim a$, we write $a\asymp b$.

2.   Preliminary results

First, we obtain the following result, which is similar to the corresponding result in ^[18].

Lemma 2.1. For each $f\in{\mathcal {B}}_{\phi, \alpha}({\mathbb{U}}^n)\setminus\{0\}$, it follows that

Moreover, for each $f\in{\mathcal {B}}_{\phi, \alpha}({\mathbb{U}}^n)$ and $z\in{\mathbb{U}}^{n}$, the following holds:

Proof. For $f\in{\mathcal {B}}_{\phi, \alpha}({\mathbb{U}}^n)\setminus\{0\}$, by the definition of ${\mathcal {B}}_{\phi, \alpha}({\mathbb{U}}^n)$, there exists a decreasing sequence $\{\lambda_{i}\}\subset {\mathbb R}^{+}$ with $S_{\phi, \alpha}\big(\frac{f}{\lambda_{i}}\big)\leq 1$, such that $\lambda_{i}\rightarrow \|f\|_{\phi, \alpha}$ as $i\rightarrow \infty$. Since the function $\phi$ is increasing, we get

From the monotonicity of $\phi$ and (2.3), we obtain that $\{S_{i}\}$ is bounded and increasing. Hence, there is a real number $S_{0}$ such that

By (2.3) and $S_i\leq 1$ for each $i\in {\mathbb N}$, we have $S_0\leq S$ and $S_0\leq1$. So, for all $z\in{\mathbb{U}}^{n}$ and $i\in {\mathbb N}$, we have

Letting $i\rightarrow \infty$ in (2.4), we obtain

for all $z\in{\mathbb{U}}^{n}$. Consequently, we obtain that $S = S_0$, and then $S\leq1$. From this, for all $f\in{\mathcal {B}}_{\phi, \alpha}({\mathbb{U}}^n)$ and $z\in{\mathbb{U}}^{n}$, it follows that

Then, for each $k\in\{1, 2, \cdots, n\}$ and all $z\in{\mathbb{U}}^n$, we have

Since $\phi$ is the strictly increasing convex function, it follows that

This completes the proof of the lemma. □

For the convenience, we write

Noting that

from Lemma 2.1 we obtain the following result.

Corollary 2.1. Let ${\alpha} > 0$. If $f\in{\mathcal {B}}_{\phi, {\alpha}}({\mathbb{U}}^n)$, then for all $z\in{\mathbb{U}}^n$, it follows that

We also have the following result, which is similar to that in ^[18].

Lemma 2.2. Let ${\alpha} > 0$. Then, ${\mathcal {B}}_{\phi, \alpha}({\mathbb{U}}^n) = {\mathcal {B}}_{\mu_{\phi, {\alpha}}}({\mathbb{U}}^n)$, where ${\mathcal {B}}_{\mu_{\phi, {\alpha}}}({\mathbb{U}}^n)$ is the special $\mu_{\phi, {\alpha}}$-Bloch space with the weight $\mu_{\phi, {\alpha}}$. Moreover, for each $f\in{\mathcal {B}}_{\phi, {\alpha}}({\mathbb{U}}^n)$, it follows that

Proof. By Lemma 2.1, for all $f\in{\mathcal {B}}_{\phi, \alpha}({\mathbb{U}}^n)$ and $z = (z_{1}, z_{2}, \cdots, z_{n})\in{\mathbb{U}}^{n}$, we have

which implies that ${\mathcal {B}}_{\phi, \alpha}({\mathbb{U}}^n)\subseteq{\mathcal {B}}_{\mu_{\phi, {\alpha}}}({\mathbb{U}}^n)$ and

Conversely, if $f\in{\mathcal {B}}_{\mu_{\phi, {\alpha}}}({\mathbb{U}}^n)$, then we have

for all $z\in{\mathbb{U}}^{n}$, that is,

which implies that

From this, we get

From this, and since $\phi(\frac{s}n)\le\frac1{n}\phi(s)$ for $s\ge 0$ and $n\in {\mathbb N}$, it easily follows that

This shows that $\|f\|_{\phi, \alpha}\leq n\|f\|_{\mu_{\phi, {\alpha}}}$, and then

and ${\mathcal {B}}_{\mu_{\phi, {\alpha}}}({\mathbb{U}}^n)\subseteq{\mathcal {B}}_{\phi, \alpha}({\mathbb{U}}^n)$.

From the above proof, (2.9), and (2.10), we obtain that ${\mathcal {B}}_{\phi, \alpha}({\mathbb{U}}^n) = {\mathcal {B}}_{\mu_{\phi, {\alpha}}}({\mathbb{U}}^n)$ and $\|f\|_{{\mathcal {B}}_{\phi, {\alpha}}({\mathbb{U}}^n)}\asymp\|f\|_{{\mathcal {B}}_{\mu_{\phi, {\alpha}}}({\mathbb{U}}^n)}$. This completes the proof of the lemma. □

The following result is a version of [6, Lemma 3.1]. For the completeness, we give its proof.

Lemma 2.3. For a fixed $a\in{\mathbb{U}}$, there exists a function $f_{a, \alpha}\in H({\mathbb{U}})$ such that

Proof. We set

Therefore, $u$ is a real and continuously differentiable function in the sense that its partial derivatives exist and are continuous on ${\mathbb{U}}$. Furthermore, for all $z\in{\mathbb{U}}$, the function $u$ satisfies

Now, we let $f_{a, \alpha}(z) = u(z)e^{iv(z)}$, where $v$ is a real function defined on ${\mathbb{U}}$. In the order that $f_{a, \alpha}$ is a holomorphic function on ${\mathbb{U}}$, then its real part and its imaginary part must satisfy the Cauchy-Riemann equations, that is,

It is easy to see that if

then (2.12) holds. To find a function $v$ that satisfies (2.13), we define

where $h$ is a real function that satisfies

Then, by a computation, we see that $v$ satisfies (2.13). This completes the proof. □

Using the function $f_{a, {\alpha}}$, we can construct some special functions in the space ${\mathcal {B}}_{\phi, {\alpha}}({\mathbb{U}}^n)$.

Lemma 2.4. For the fixed $a\in{\mathbb{U}}$, the function

belongs to ${\mathcal {B}}_{\phi, \alpha}({\mathbb{U}}^n)$ for $l\in\{1, 2, \cdots, n\}$. Moreover, $\|g_{a, \alpha}\|_{\phi, \alpha} = 1$.

Proof. It is clear that $g_{a, {\alpha}}$ is holomorphic on ${\mathbb{U}}^n$. The result holds due to the following equality:

where

is the automorphism of ${\mathbb{U}}$. From (2.15), we obtain that $\|g_{a, \alpha}\|_{\phi, \alpha} = 1$. This completes the proof. □

The following result characterizes the compactness of the operator $W_{\psi, \varphi}$. The proof is standard, so it is omitted (see Proposition 3.11 in ^[5] or Theorem 3.1 in ^[16]).

Lemma 2.5. Let ${\alpha} > 0$. Then, the operator $W_{\psi, \varphi}$ is compact on ${\mathcal {B}}_{\phi, {\alpha}}({\mathbb{U}}^n)$ if, and only if, for each bounded sequence $\{f_{i}\}\subset{\mathcal {B}}_{\phi, \alpha}({\mathbb{U}}^n)$ such that $\{f_{i}\}\to0$ uniformly on every compact subset of ${\mathbb{U}}^{n}$ as $i\to\infty$, it follows that $\lim_{i\to\infty}\|W_{\psi, \varphi}f_{i}\|_{\phi, \alpha} = 0$.

3.   Main results and proofs

In this section, we assume that the function $\mu_{\phi, {\alpha}}$ satisfies the following condition:

In this case, we will give some examples that satisfy the condition (3.1). Moreover, we will characterize the boundedness and compactness of the operator $W_{\psi, \varphi}$ on ${\mathcal {B}}_{\phi, {\alpha}}({\mathbb{U}}^n)$. As the applications, the characterizations of the boundedness and compactness of $C_{\varphi}$ and $M_\psi$ on ${\mathcal {B}}_{\phi, {\alpha}}({\mathbb{U}}^n)$ are obtained.

Theorem 3.1. Let $\alpha > 0$, $\psi\in H({\mathbb{U}}^{n})$, and ${\varphi}$ be a holomorphic self-mapping of ${\mathbb{U}}^{n}$. Then, the operator $W_{\psi, \varphi}$ is bounded on ${\mathcal {B}}_{\phi, \alpha}({\mathbb{U}}^n)$ if, and only if,

and

Proof. For each fixed $k\in\{1, 2, \cdots, n\}$, we write

and

Suppose that $L$, $M < +\infty$. Then, we clearly have $L_k\leq L$ and $M_k\leq M$. For every $f\in{\mathcal {B}}_{\phi, \alpha}({\mathbb{U}}^n)\setminus\{0\}$, from the convexity of $\phi$, Lemmas 2.1 and 2.2, it follows that

Hence, from (3.4), we obtain

On the other hand, it follows from Corollary 2.1 that

From (3.5) and (3.6), we obtain that

which shows that the operator $W_{\psi, \varphi}$ is bounded on ${\mathcal {B}}_{\phi, \alpha}({\mathbb{U}}^n)$.

Conversely, assume that the operator $W_{\psi, \varphi}$ is bounded on ${\mathcal {B}}_{\phi, \alpha}({\mathbb{U}}^n)$. By setting $\hat{f}(z)\equiv1$ (clearly, $1\in{\mathcal {B}}_{\phi, \alpha}({\mathbb{U}}^n)$), we obtain $\psi = W_{\psi, \varphi}\hat{f}\in{\mathcal {B}}_{\phi, \alpha}({\mathbb{U}}^n)$. Then,

From (3.7), for each $k\in\{1, 2, \cdots, n\}$ and all $z\in{\mathbb{U}}^n$, it follows that

which shows

By Lemma 2.4, we see that the following function belongs to ${\mathcal {B}}_{\phi, {\alpha}}({\mathbb{U}}^n)$,

Moreover, $\|g_{\varphi_{l}(w), \alpha}\|_{\phi, \alpha} = 1$. Then, we have

From (3.9), we obtain

which leads to

By summing in (3.10) from $1$ to $n$, we obtain

Then, by Corollary 2.1 and the fact $\|g_{\varphi_{l}(w), {\alpha}}\|_{\phi, {\alpha}} = 1$, (3.11) becomes

Since

by Lemma 2.3 we get

Now, letting $z = w$ in (3.13), by (3.8) and $c_0 = \int_{0}^{1}\frac{1}{\mu_{\phi, \alpha}(t)}dt < +\infty$, we have

which shows

The proof is finished. □

By directly applying Theorem 3.1, we derive the following two results.

Corollary 3.1. Let $\alpha > 0$ and $\psi\in H({\mathbb{U}}^{n})$. Then, the operator $M_\psi$ is bounded on ${\mathcal {B}}_{\phi, \alpha}({\mathbb{U}}^n)$ if, and only if, $\psi\in H^\infty({\mathbb{U}}^n)\cap{\mathcal {B}}_{\mu_{\phi, {\alpha}}}({\mathbb{U}}^n)$.

Corollary 3.2. Let $\alpha > 0$ and ${\varphi}$ be a holomorphic self-mapping of ${\mathbb{U}}^{n}$. Then, the operator $C_{\varphi}$ is bounded on ${\mathcal {B}}_{\phi, \alpha}({\mathbb{U}}^n)$ if, and only if,

The compactness of the operator $W_{\psi, \varphi}$ has been characterized from $p$-Bloch space ${\mathcal {B}}_{p}({\mathbb{U}}^n)$ to $q$-Bloch space ${\mathcal {B}}_{q}({\mathbb{U}}^n)$ in ^[26], which motivates us to consider the same problem on ${\mathcal {B}}_{\phi, \alpha}({\mathbb{U}}^{n})$.

Theorem 3.2. Let $\alpha > 0$, $\psi\in H({\mathbb{U}}^{n})$, and ${\varphi}$ be a holomorphic self-mapping of ${\mathbb{U}}^{n}$. Then, the operator $W_{\psi, \varphi}$ is compact on ${\mathcal {B}}_{\phi, \alpha}({\mathbb{U}}^{n})$ if, and only if, $W_{\psi, \varphi}$ is bounded on ${\mathcal {B}}_{\phi, \alpha}({\mathbb{U}}^{n})$,

and

Proof. Assume that the operator $W_{\psi, \varphi}$ is bounded on ${\mathcal {B}}_{\phi, \alpha}({\mathbb{U}}^{n})$ and (3.14), (3.15) holds, respectively. To show that the operator $W_{\psi, {\varphi}}$ is compact on ${\mathcal {B}}_{\phi, {\alpha}}({\mathbb{U}}^n)$, using Lemma 2.5, we just need to prove that for each bounded sequence $\{f_j\}$ in ${\mathcal {B}}_{\phi, {\alpha}}({\mathbb{U}}^n)$ such that $f_j\to0$ uniformly on any compact subset of ${\mathbb{U}}^n$ as $j\to\infty$, we have that $\lim_{j\to\infty}\|W_{\psi, {\varphi}}f_j\|_{\phi, {\alpha}} = 0$. Let $\{f_{j}\}$ be a such sequence in ${\mathcal {B}}_{\phi, {\alpha}}({\mathbb{U}}^n)$. Set $C_{0} = \sup_{j\in {\mathbb N}}\|f_{j}\|_{\phi, \alpha}$. For $\varepsilon > 0$, by (3.14) and (3.15), there exists a $\delta\in(0, 1)$ such that

and

for all $z\in E = \{z\in{\mathbb{U}}^n: dist(\varphi(z), \partial{\mathbb{U}}^{n}) < \delta\}$. From this, for all $z\in E$ we have

which shows

On the other hand, ${\mathbb{U}}^n\setminus E = \{z\in{\mathbb{U}}^n:dist(\varphi(z), \partial{\mathbb{U}}^{n})\geq\delta\}$ is a compact subset of ${\mathbb{U}}^{n}$. Hence, $f_{j}\to0$ uniformly on ${\mathbb{U}}^n\setminus E$ as $j\to\infty$. From Cauchy's estimate, it follows that $\frac{\partial f_{j}}{\partial z_{l}}\rightarrow 0$ uniformly on ${\mathbb{U}}^n\setminus E$ as $j\to\infty$ for each $l\in\{1, 2, \cdots, n\}$. Since $W_{\psi, \varphi}$ is bounded on ${\mathcal {B}}_{\phi, {\alpha}}({\mathbb{U}}^n)$, from Theorem 3.1 we have that $M < +\infty$.

Set $f_l(z) = z_{l}, \, \, z\in{\mathbb{U}}^n$. Then, $f_l\in{\mathcal {B}}_{\phi, \alpha}({\mathbb{U}}^{n})$. Since $W_{\psi, {\varphi}}$ is bounded on ${\mathcal {B}}_{\phi, {\alpha}}({\mathbb{U}}^n)$, $W_{\psi, {\varphi}}f_l = \psi\varphi_{l}\in{\mathcal {B}}_{\phi, {\alpha}}({\mathbb{U}}^n)$. From the definition of ${\mathcal {B}}_{\phi, {\alpha}}({\mathbb{U}}^n)$, we have

Since $\mu_{\phi, \alpha}(z_{k}) = \frac{1}{\phi^{-1}(\frac{1}{(1-|z_{k}|^{2})^{\alpha}})}$, from (3.16) we obtain

Therefore,

Since $|\varphi_{l}(z)| < 1, \, \ l\in\{1, 2, \cdots, n\}$, we get

For the convenience, we write

Then, $C_l < +\infty$ for each $l\in\{1, 2, \cdots, n\}$. Hence, for each $l\in\{1, 2, \cdots, n\}$, we have

as $j\to\infty$. Moreover, it follows by (3.17) that the operator $W_{\psi, \varphi}$ is compact on ${\mathcal {B}}_{\phi, \alpha}({\mathbb{U}}^{n})$.

Now, suppose that the operator $W_{\psi, \varphi}$ is compact on ${\mathcal {B}}_{\phi, {\alpha}}({\mathbb{U}}^n)$. It is clear that $W_{\psi, \varphi}$ is bounded on ${\mathcal {B}}_{\phi, {\alpha}}({\mathbb{U}}^n)$. First, we prove that condition (3.14) holds. If the condition (3.14) is not right, then there is a constant $\varepsilon_0 > 0$ and a sequence $\{z^{m}\}$ in ${\mathbb{U}}^{n}$ with $w^{m} = \varphi(z^{m})\to\partial{\mathbb{U}}^{n}$ as $m\rightarrow \infty$, such that

for all $m\in {\mathbb N}.$ Since $W_{\psi, \varphi}$ is bounded on ${\mathcal {B}}_{\phi, {\alpha}}({\mathbb{U}}^n)$, by the condition (3.2) in Theorem 3.1, we see that for all $l\in\{1, 2, \cdots, n\}$, the sequence

is bounded. Hence, there is a subsequence of $\{z^{m}\}$ (for simplicity, here we assume that it is the sequence $\{z^{m}\}$) such that the following limit exists:

for every $l\in\{1, 2, \cdots, n\}.$ Also, we may assume that for every $l\in\{1, 2, \cdots, n\}$, the following limit exists:

From (3.18), there must be an $l_0\in\{1, 2, \cdots, n\}$ (here, we can assume that $l_0 = 1$) such that

In order to obtain a contradiction, we divide into the following two cases for consideration.

Case 1. Assume that $|w_{1}^{m}|\rightarrow 1$ as $m\rightarrow \infty$. Set

Then, $h_m\in{\mathcal {B}}_{\phi, {\alpha}}({\mathbb{U}}^n)$ and $h_{m}\to0$ uniformly on compact subsets of ${\mathbb{U}}^{n}$ as $m\rightarrow \infty$. Moreover, by an easy computation,

Then, from (3.19) and (3.20), we get

as $m\to\infty$, which is a contradiction, since $\|W_{\psi, {\varphi}} h_m\|_{\mu_{\phi, \alpha}}\to0$ as $m\to\infty$.

Case 2. Assume that $|w_{1}^{m}|\rightarrow \rho < 1$ as $m\rightarrow \infty$. Since $w^{m}\rightarrow \partial{\mathbb{U}}^{n}$, there is an $l\in\{2, \cdots, n\}$ such that $|w_{l}^{m}|\rightarrow 1$ as $m\rightarrow \infty$. If there is a $\varepsilon_{2} > 0$ such that

we can also assume that

Similar to Case 1, we obtain a contradiction by using the functions

Now, for the $l$ chosen above, assume that

Set $\widetilde{h}_{m} = h_{m}+\widehat{h}_{m}$. Then, $\{\widetilde{h}_{m}\}\subset{\mathcal {B}}_{\phi, \alpha}({\mathbb{U}}^{n})$ and $\widetilde{h}_{m}\rightarrow 0$ uniformly on compact subsets of ${\mathbb{U}}^{n}$ as $m\rightarrow \infty$. We have

as $m\to\infty$, which also is a contradiction, since $\|W_{\psi, {\varphi}}\widetilde{h}_m\|_{\mu_{\phi, {\alpha}}}\to0$ as $m\to\infty$.

Now, we begin to prove that condition (3.15) holds. Assume that condition (3.15) is not right. Then, there exist a positive constant $\varepsilon_{4}$ and a sequence $\{z^{m}\}$ in ${\mathbb{U}}^{n}$ with $w^{m} = \varphi(z^{m})\rightarrow \partial{\mathbb{U}}^{n}$ as $m\to\infty$, such that

for all $m\in {\mathbb N}.$ Since $W_{\psi, \varphi}$ is bounded on ${\mathcal {B}}_{\mu_{\phi, {\alpha}}}({\mathbb{U}}^n)$, from (3.3) in Theorem 3.1, we know that the sequence

is bounded. Hence, there is a subsequence of $\{z^{m}\}$ such that

Also, we may assume that for every $l\in\{1, 2, \cdots, n\}$, there is a finite limit

Case 3. Assume that $|w_{1}^{m}|\rightarrow 1$ as $m\rightarrow \infty$. Set

Then, $\{S_{m}\}\subset{\mathcal {B}}_{\phi, \alpha}({\mathbb{U}}^{n})$ and $S_{m}\rightarrow 0$ uniformly on compact subsets of ${\mathbb{U}}^{n}$ as $m\rightarrow \infty$. Moreover, we also have

So, we get

as $m\rightarrow \infty$, which is a contradiction.

Case 4. Assume that $|w_{1}^{m}|\rightarrow \rho < 1$ as $m\rightarrow \infty$. Since $w^{m}\rightarrow \partial{\mathbb{U}}^{n}$, there is an $l\in\{2, \cdots, n\}$ such that $|w_{l}^{m}|\rightarrow 1$ as $m\rightarrow \infty$.

Set

Similar to Case 3, we obtain

which also is a contradiction. Now, for the $l$ chosen above, assume that

Let

Then, $\{\widetilde{S}_{m}\}\subset{\mathcal {B}}_{\phi, \alpha}({\mathbb{U}}^{n})$ and $\widetilde{S}_{m}\rightarrow 0$ uniformly on compact subsets of ${\mathbb{U}}^{n}$ as $m\rightarrow \infty$. For such sequence, we have

as $m\rightarrow \infty$, from which we obtain a contradiction. Hence, the proof is finished. □

According to Theorem 3.2, we get the following two results.

Corollary 3.3. Let $\alpha > 0$ and $\psi\in H({\mathbb{U}}^{n})$. Then, the operator $M_\psi$ is compact on ${\mathcal {B}}_{\phi, \alpha}({\mathbb{U}}^n)$ if, and only if, $\psi\in H_0^\infty({\mathbb{U}}^n)\cap{\mathcal {B}}_{\mu_{\phi, {\alpha}}, 0}({\mathbb{U}}^n)$.

Corollary 3.4. Let $\alpha > 0$ and ${\varphi}$ be a holomorphic self-mapping of ${\mathbb{U}}^{n}$. Then, the operator $C_{\varphi}$ is compact on ${\mathcal {B}}_{\phi, \alpha}({\mathbb{U}}^n)$ if, and only if, $C_\varphi$ is bounded on ${\mathcal {B}}_{\phi, \alpha}({\mathbb{U}}^n)$ and

In the final of the paper, we give the following example in order to show that there exists $\phi$ such that $\mu_{\phi, {\alpha}}(t)$ satisfies the condition (3.1).

Example 3.1. Let $\phi(t) = t^{p}$ and $p > \max\{1, {\alpha}\}$. Then, $\mu_{\phi, {\alpha}}(t)$ satisfies the condition (3.1).

Proof. From the condition $p > 1$, it follows that $\phi$ is a strictly increasing convex function in $[0, +\infty)$. Now, we prove that

It is not hard to see that $\phi^{-1}(t) = t^{\frac{1}{p}}$, and then $\mu_{\phi, {\alpha}}(t) = (1-t^{2})^{\frac{\alpha}{p}}$. Since the function $\mu_{\phi, {\alpha}}(t)$ is unbounded in the point $t = 1$, we just need to show that the following integral is convergent,

In fact, we have

Since $0 < \alpha/p < 1$ and $0 < (\frac{1}{2})^{\frac{\alpha}{p}} < +\infty$, by the comparison rule, the integral (3.22) is convergent. □

4.   Conclusions

In this paper, we define the $\alpha$-Bloch-Orlicz space on ${\mathbb{U}}^n$ by using Young's function and show that its norm is equivalent with a special $\mu$-Bloch space. We completely characterize the boundedness and compactness of the weighted composition operator $W_{\psi, \varphi}$ on the $\alpha$-Bloch-Orlicz space in terms of the behaviors of the symbols $\psi$ and $\varphi$. In addition, we give an example that satisfies the condition (3.1), which shows the rationality of this condition. As some applications, the corresponding results of the operators $M_\psi$ and $C_{\varphi}$ are obtained. This paper can be viewed as a continuation and extension of our previous studies.

Author contributions

Fuya Hu: Writing and editing, formal analysis and methodology; Chengshi Huang: Commenting and reviewing; Zhijie Jiang: Writing-original draft and investigation. All authors have read and approved the final version of the manuscript for publication.

Use of Generative-AI tools declaration

We declare we have not used Artificial Intelligence (AI) tools in the creation of this article.

Acknowledgments

The authors would like to express their sincere thanks to the anonymous referees for valuable comments that are very helpful to improve this paper. This work was supported by Sichuan Science and Technology Program (No. 2024NSFSC0416).

Conflict of interest

The authors declare that they have no competing interests.