On a product-type operator from Zygmund-type spaces to Bloch-Orlicz spaces

Abstract

The boundedness and compactness of a product-type operator from Zygmund-type spaces to Bloch-Orlicz spaces are investigated in this paper.

1 Introduction

Let $\mathcal{D}$ denote the unit disk in the complex plane $\mathcal {C}$, and let $\mathcal{H}(\mathcal{D})$ be the space of all holomorphic functions on $\mathcal{D}$ with the topology of uniform convergence on compacts of $\mathcal{D}$.

For $0<\alpha<\infty$, the α-Bloch space, denoted by $\mathcal{B}^{\alpha}$, consists of all functions $f\in\mathcal {H}(\mathcal{D})$ such that

$$\sup_{z\in\mathcal{D}}\bigl(1-|z|^{2}\bigr)^{\alpha}\bigl\vert f'(z)\bigr\vert < \infty. $$

By $\mathcal{Z}^{\alpha}$ we denote the Zygmund-type space consisting of those functions $f\in\mathcal{H}(\mathcal{D})$ satisfying

$$\sup_{z\in\mathcal{D}}\bigl(1-|z|^{2}\bigr)^{\alpha}\bigl\vert f''(z)\bigr\vert < \infty. $$

$\mathcal{B}^{\alpha}$ and $\mathcal{Z}^{\alpha}$ are Banach spaces under the norms

$$\begin{aligned}& \|f\|_{\mathcal{B}^{\alpha}}=\bigl\vert f(0)\bigr\vert +\sup_{z\in\mathcal {D}} \bigl(1-|z|^{2}\bigr)^{\alpha}\bigl\vert f'(z)\bigr\vert , \\& \|f\|_{\mathcal{Z}^{\alpha}}=\bigl\vert f(0)\bigr\vert +\bigl\vert f'(0)\bigr\vert +\sup_{z\in\mathcal {D}}\bigl(1-|z|^{2} \bigr)^{\alpha}\bigl\vert f''(z)\bigr\vert , \end{aligned}$$

respectively. For some results on the Zygmund-type spaces on various domains in the complex plane and $\mathcal{C}^{n}$ and operators on them, see, for example, [1–18]. The α-Bloch space is introduced and studied by numerous authors. For the general theory of α-Bloch or Bloch-type spaces and operators of them, see, e.g., [4, 19–41]. Recently, many authors studied different classes of Bloch-type spaces, where the typical weight function, $\omega(z)=1-|z|^{2}$, $z\in\mathcal{D}$, is replaced by a bounded continuous positive function μ defined on $\mathcal{D}$. More precisely, a function $f\in\mathcal{H}(\mathcal{D})$ is called a μ-Bloch function, denoted by $f\in\mathcal{B}^{\mu}$, if $\|f\|_{\mu}=\sup_{z\in\mathcal{D}}\mu(z)|f'(z)|<\infty$. If $\mu(z)=\omega(z)^{\alpha}$, $\alpha>0$, $\mathcal{B}^{\mu}$ is just the α-Bloch space $\mathcal{B}^{\alpha}$. It is readily seen that $\mathcal{B}^{\mu}$ is a Banach space with the norm $\|f\|_{\mathcal{B}^{\mu}}=|f(0)|+\|f\|_{\mu}$.

Recently, Ramos Fernández in [42] used Young’s functions to define the Bloch-Orlicz space. More precisely, let $\varphi: [0,+\infty)\rightarrow[0,+\infty)$ be a strictly increasing convex function such that $\varphi(0)=0$ and note that from these conditions it follows that $\lim_{t\rightarrow+\infty}\varphi(t)=+\infty$. The Bloch-Orlicz space associated with the function φ, denoted by $\mathcal{B}^{\varphi}$, is the class of all analytic functions f in $\mathcal{D}$ such that

$$\sup_{z\in\mathcal{D}}\bigl(1-|z|^{2}\bigr)\varphi\bigl(\lambda \bigl\vert f'(z)\bigr\vert \bigr)< \infty $$

for some $\lambda>0$ depending on f. Also, since φ is convex, it is not hard to see that Minkowski’s functional

$$\|f\|_{\varphi}=\inf \biggl\{ k>0:S_{\varphi} \biggl(\frac{f'}{k} \biggr)\leq1 \biggr\} $$

defines a seminorm for $\mathcal{B}^{\varphi}$, which, in this case, is known as Luxemburg’s seminorm, where

$$S_{\varphi}(f)=\sup_{z\in\mathcal{D}}\bigl(1-|z|^{2}\bigr) \varphi\bigl(\bigl\vert f(z)\bigr\vert \bigr). $$

Moreover, it can be shown that $\mathcal{B}^{\varphi}$ is a Banach space with the norm $\|f\|_{\mathcal{B}^{\varphi}}=|f(0)|+\|f\|_{\varphi}$. We also have that the Bloch-Orlicz space is isometrically equal to a particular μ-Bloch space, where $\mu(z)=\frac{1}{\varphi^{-1}(\frac{1}{1-|z|^{2}})}$ with $z\in \mathcal{D}$. Thus, for any $f\in\mathcal{B}^{\varphi}$, we have

$$\|f\|_{\mathcal{B}^{\varphi}}=\bigl\vert f(0)\bigr\vert +\sup_{z\in\mathcal{D}} \mu(z)\bigl\vert f'(z)\bigr\vert . $$

When φ is the identity map on $[0,+\infty)$, $\mathcal {B}^{\varphi}$ is the so-called Bloch space ℬ.

Let $u\in\mathcal{H}(\mathcal{D})$ and ϕ be an analytic self-map of $\mathcal{D}$. The differentiation operator D, the multiplication operator $M_{u}$ and the composition operator $C_{\phi}$ are defined by

$$(Df) (z)=f'(z),\qquad (M_{u}f) (z)=u(z)f(z),\qquad (C_{\phi}f) (z)=f\bigl(\phi(z)\bigr), \quad f\in\mathcal{H}(\mathcal{D}). $$

There is a considerable interest in studying the above mentioned operators as well as their products (see, e.g., [1–38, 41–56] and the related references therein).

A product-type operator $DM_{u}C_{\phi}$ is defined as follows:

$$(DM_{u}C_{\phi}f) (z)=u'(z)f\bigl(\phi(z) \bigr)+u(z)\phi'(z)f'\bigl(\phi(z)\bigr),\quad u,f\in \mathcal{H}(\mathcal{D}). $$

For $0<\alpha<\infty$ and $\frac{1}{2}<|a|<1$, we define the test functions (see [1])

$$\begin{aligned}& f_{a}(z)=\frac{1}{\overline{a}^{2}} \biggl[\frac{(1-|a|^{2})^{2}}{ (1-\overline{a}z)^{\alpha}}-\frac{1-|a|^{2}}{(1-\overline {a}z)^{\alpha-1}} \biggr], \\& h_{a}(z)=\frac{1}{\overline{a}}\int^{z}_{0} \frac {1-|a|^{2}}{(1-\overline{a}\lambda)^{\alpha}}\, d\lambda,\quad z\in\mathcal{D}. \end{aligned}$$

It is easy to show that $f_{a},h_{a}\in\mathcal{Z}^{\alpha}$ and $f_{a}(a)=0$,

$$f'_{a}(a)=h'_{a}(a)= \frac{1}{\overline {a}}\bigl(1-|a|^{2}\bigr)^{1-\alpha},\qquad f''_{a}(a)=\frac{2\alpha}{(1-|a|^{2})^{\alpha}}, \qquad h''_{a}(a)=\frac {\alpha}{(1-|a|^{2})^{\alpha}}. $$

Esmaeili and Lindström in [1] investigated weighted composition operators between Zygmund-type spaces. Ramos Fernández in [42] studied the boundedness and compactness of composition operators on Bloch-Orlicz spaces. Li and Stević in [5] investigated products of Volterra-type operator and composition operator from $H^{\infty}$ and Bloch spaces to Zygmund spaces, and they in [8] studied products of composition and differentiation operators from Zygmund spaces to Bloch spaces and Bers spaces. Liu and Yu in [25] characterized the boundedness and compactness of products of composition, multiplication and radial derivative operators from logarithmic Bloch spaces to weighted-type spaces on the unit ball. Sharma in [27] studied the boundedness and compactness of products of composition multiplication and differentiation between Bergman and Bloch-type spaces. In [52], Stević investigated the properties of weighted differentiation composition operators from mixed-norm spaces to weighted-type spaces. Stević in [13] studied weighted radial operators from the mixed-norm space to the nth weighted-type space on the unit ball. Stević et al. in [54] characterized the boundedness and compactness of products of multiplication composition and differentiation operators on weighted Bergman spaces. Zhu in [18] studied extended Cesàro operators from mixed-norm spaces to Zygmund-type spaces.

Motivated by the above papers, in this paper, we investigate the boundedness and compactness of the product-type operator $DM_{u}C_{\phi}$ from Zygmund-type spaces to the Bloch-Orlicz space. The paper is organized as follows. In Section 2, we give some necessary and sufficient conditions for the boundedness of the operator $DM_{u}C_{\phi}:\mathcal{Z}^{\alpha}\rightarrow\mathcal {B}^{\varphi}$. In Section 3, we give some necessary and sufficient conditions for the compactness of the operator $DM_{u}C_{\phi}:\mathcal{Z}^{\alpha}\rightarrow\mathcal {B}^{\varphi}$.

Throughout this paper,

$$\mu(z)=\frac{1}{\varphi^{-1}(\frac{1}{1-|z|^{2}})}, $$

and we use letter C to denote a positive constant whose value may change at each occurrence.

2 The boundedness of $DM_{u}C_{\phi}:\mathcal{Z}^{\alpha }\rightarrow\mathcal{B}^{\varphi}$

The following lemma was essentially proved in [3] and [11] (see also [1]).

Lemma 1

For $f\in\mathcal{Z}^{\alpha}$ and $\alpha>0$. Then:

• (i) For $0<\alpha<1$, $|f'(z)|\leq\frac{2}{1-\alpha}\|f\|_{\mathcal{Z}^{\alpha}}$ and $|f(z)|\leq\frac{2}{1-\alpha}\|f\|_{\mathcal{Z}^{\alpha}}$.

• (ii) For $\alpha=1$, $|f'(z)|\leq\log\frac{e}{1-|z|^{2}}\|f\|_{\mathcal{Z}}$ and $|f(z)|\leq\|f\|_{\mathcal{Z}}$.

• (iii) For $\alpha>1$, $|f'(z)|\leq\frac{2}{\alpha-1}\frac{\|f\|_{\mathcal{Z}^{\alpha }}}{(1-|z|^{2})^{\alpha-1}}$. For $\alpha=2$, $|f'(z)|\leq\frac{e}{1-|z|^{2}}\|f\|_{\mathcal{Z}^{2}}$.

• (iv) For $1<\alpha<2$, $|f(z)|\leq\frac{2}{(\alpha-1)(2-\alpha)}\|f\|_{\mathcal {Z}^{\alpha}}$.

• (v) For $\alpha=2$, $|f(z)|\leq2\log\frac{e}{1-|z|^{2}}\|f\|_{\mathcal{Z}^{2}}$.

• (vi) For $\alpha>2$, $|f(z)|\leq\frac{2}{(\alpha-1)(\alpha-2)}\frac{\|f\|_{\mathcal {Z}^{\alpha}}}{(1-|z|^{2})^{\alpha-2}}$.

Lemma 2

If $DM_{u}C_{\phi}:\mathcal{Z}^{\alpha}\rightarrow \mathcal{B}^{\varphi}$ is bounded and $0<\alpha<\infty$, then the following conditions hold:

$$\begin{aligned}& k_{1}=\sup_{z\in\mathcal{D}}\mu(z)\bigl\vert u''(z)\bigr\vert < \infty, \end{aligned}$$

(1)

$$\begin{aligned}& k_{2}=\sup_{z\in\mathcal{D}}\mu(z)\bigl\vert 2u'(z)\phi'(z)+u(z)\phi ''(z) \bigr\vert < \infty, \end{aligned}$$

(2)

$$\begin{aligned}& k_{3}=\sup_{z\in\mathcal{D}}\mu(z)\bigl\vert u(z)\bigr\vert \bigl\vert \phi'(z)\bigr\vert ^{2}< \infty. \end{aligned}$$

(3)

Proof

Suppose that $DM_{u}C_{\phi}:\mathcal{Z}^{\alpha}\rightarrow\mathcal {B}^{\varphi}$ is bounded. Taking the function $f(z)=1\in\mathcal{Z}^{\alpha}$ and using the obvious fact that $\|f\|_{\mathcal{Z}^{\alpha}}=1$, we have that

$$S_{\varphi} \biggl(\frac{(DM_{u}C_{\phi}f)'(z)}{C\|f\|_{\mathcal {Z}^{\alpha}}} \biggr) =S_{\varphi} \biggl( \frac{u''(z)}{C} \biggr)=\sup_{z\in \mathcal{D}}\bigl(1-|z|^{2} \bigr)\varphi \biggl(\frac{|u''(z)|}{C} \biggr)\leq1, $$

from which it follows that (1) holds. Taking the function $f(z)=z\in\mathcal{Z}^{\alpha}$ and using the fact that $\|f\|_{\mathcal{Z}^{\alpha}}=1$, we obtain

$$\begin{aligned}& S_{\varphi} \biggl(\frac{(DM_{u}C_{\phi}f)'(z)}{C\|f\|_{\mathcal {Z}^{\alpha}}} \biggr) \\& \quad = S_{\varphi} \biggl(\frac{u''(z)\phi(z)+2u'(z)\phi'(z)+u(z)\phi ''(z)}{C} \biggr) \\& \quad = \sup_{z\in \mathcal{D}}\bigl(1-|z|^{2}\bigr)\varphi \biggl(\frac{|u''(z)\phi (z)+2u'(z)\phi'(z)+u(z)\phi''(z)|}{C} \biggr)\leq1. \end{aligned}$$

Hence

$$\sup_{z\in\mathcal{D}}\mu(z)\bigl\vert 2u'(z) \phi'(z)+u(z)\phi ''(z)+u''(z) \phi(z)\bigr\vert < \infty. $$

From this, (1) and by the boundedness of $\phi(z)$, condition (2) easily follows. Now taking the function $f(z)=z^{2}\in\mathcal{Z}^{\alpha}$ and using the fact that $\|f\|_{\mathcal{Z}^{\alpha}}=2$, we get

$$ S_{\varphi} \biggl(\frac{u''(z)(\phi(z))^{2}+2\phi (z)(2u'(z)\phi'(z)+u(z)\phi''(z))+2u(z) \phi'(z)^{2}}{2C} \biggr)\leq1. $$

Hence

$$\sup_{z\in\mathcal{D}}\mu(z)\bigl\vert u''(z) \bigl(\phi(z)\bigr)^{2}+2\phi (z) \bigl(2u'(z) \phi'(z)+u(z)\phi''(z)\bigr)+2u(z) \phi'(z)^{2}\bigr\vert < \infty. $$

From this, (1), (2) and the boundedness of $\phi(z)$, we obtain (3). □

Now, we are ready to characterize the boundedness of the product-type operator $DM_{u}C_{\phi}:\mathcal{Z}^{\alpha}\rightarrow\mathcal {B}^{\varphi}$. For this purpose we need to break the problem into five different cases: $0<\alpha<1$, $\alpha=1$, $1<\alpha<2$, $\alpha=2$ and $\alpha>2$.

Theorem 3

Let $u\in\mathcal{H}(\mathcal{D})$, ϕ be an analytic self-map of $\mathcal{D}$ and $0<\alpha<1$. Then $DM_{u}C_{\phi}:\mathcal{Z}^{\alpha}\rightarrow\mathcal {B}^{\varphi}$ is bounded if and only if $k_{1}<\infty$, $k_{2}<\infty$ and

$$ k_{4}=\sup_{z\in\mathcal{D}}\frac{\mu(z)|u(z)||\phi '(z)|^{2}}{(1-|\phi(z)|^{2})^{\alpha}}< \infty. $$

(4)

Proof

Suppose that $DM_{u}C_{\phi}:\mathcal{Z}^{\alpha}\rightarrow\mathcal {B}^{\varphi}$ is bounded, by Lemma 2 we know that $k_{1}, k_{2}, k_{3}<\infty$. Now we will prove (4). Let

$$g_{\phi(\omega)}(z)=f_{\phi(\omega)}(z)-h_{\phi(\omega )}(z)+h_{\phi(\omega)} \bigl(\phi(\omega)\bigr) $$

for all $z\in\mathcal{D}$ and $\omega\in\mathcal{D}$ such that $\frac{1}{2}<|\phi(\omega)|<1$, then $g_{\phi(\omega)}\in\mathcal{Z}^{\alpha}$, and

$$g_{\phi(\omega)}\bigl(\phi(\omega)\bigr)=g'_{\phi(\omega)}\bigl( \phi(\omega)\bigr)=0,\qquad g''_{\phi(\omega)}\bigl( \phi(\omega)\bigr)=\frac{\alpha}{(1-|\phi(\omega )|^{2})^{\alpha}}. $$

By the boundedness of $DM_{u}C_{\phi}:\mathcal{Z}^{\alpha}\rightarrow\mathcal {B}^{\varphi}$, we have $\|DM_{u}C_{\phi}g_{\phi(\omega)}\|_{\mathcal{B}^{\varphi}}\leq C$, then

$$ 1 \geq S_{\varphi} \biggl(\frac{(DM_{u}C_{\phi }g_{\phi(\omega)})'(z)}{C} \biggr)\geq \sup _{\frac{1}{2}< |\phi(\omega)|<1}\bigl(1-|\omega|^{2}\bigr)\varphi \biggl( \frac{\alpha|u(\omega)| |\phi'(\omega)|^{2}}{C(1-|\phi(\omega)|^{2})^{\alpha}} \biggr). $$

It follows that

$$ \sup_{\frac{1}{2}< |\phi(\omega)|<1} \frac{\mu(\omega)|u(\omega)||\phi'(\omega)|^{2}}{ (1-|\phi(\omega)|^{2})^{\alpha}}<\infty. $$

(5)

By $k_{3}<\infty$, we see that

$$ \sup_{|\phi(\omega)|\leq\frac{1}{2}}\frac{\mu (\omega)|u(\omega)||\phi'(\omega)|^{2}}{ (1-|\phi(\omega)|^{2})^{\alpha}}\leq C \sup _{|\phi(\omega)|\leq\frac{1}{2}}\mu(\omega)\bigl\vert u(\omega)\bigr\vert \bigl\vert \phi '(\omega)\bigr\vert ^{2}< \infty. $$

(6)

From (5) and (6), we obtain (4).

Suppose that $k_{1}, k_{2}, k_{4}<\infty$. For each $f\in\mathcal{Z}^{\alpha}\setminus\{0\}$, by Lemma 1(i) we have

$$\begin{aligned}& S_{\varphi} \biggl(\frac{(DM_{u}C_{\phi }f)'(z)}{C\|f\|_{\mathcal{Z}^{\alpha}}} \biggr) \\& \quad \leq \sup_{z\in\mathcal{D}}\bigl(1-|z|^{2}\bigr)\varphi \biggl[\frac{(k_{1}|f(\phi(z))| +k_{2}|f'(\phi(z))|+k_{4}(1-|\phi(z)|^{2})^{\alpha}|f''(\phi (z))|)}{C\mu(z)\|f\|_{\mathcal{Z}^{\alpha}}} \biggr] \\& \quad \leq \sup_{z\in\mathcal{D}}\bigl(1-|z|^{2}\bigr)\varphi \biggl[\frac{k_{1}\frac{2}{1-\alpha}+ k_{2}\frac{2}{1-\alpha}+k_{4}}{C\mu(z)} \biggr] \leq1, \end{aligned}$$

where C is a constant such that $C\geq k_{1}\frac{2}{1-\alpha}+k_{2}\frac{2}{1-\alpha}+k_{4}$. Here we use the fact that

$$ \sup_{z\in\mathcal{D}}\bigl(1-\bigl\vert \phi(z)\bigr\vert ^{2}\bigr)^{\alpha }\bigl\vert f'' \bigl(\phi(z)\bigr)\bigr\vert \leq\|f\|_{\mathcal{Z}^{\alpha}}. $$

Now, we can conclude that there exists a constant C such that $\|DM_{u}C_{\phi}f\|_{\mathcal{B}^{\varphi}}\leq C\|f\|_{\mathcal {Z}^{\alpha}}$ for all $f\in\mathcal{Z}^{\alpha}$, so the product-type operator $DM_{u}C_{\phi}:\mathcal{Z}^{\alpha}\rightarrow\mathcal {B}^{\varphi}$ is bounded. □

Theorem 4

Let $u\in\mathcal{H}(\mathcal{D})$ and ϕ be an analytic self-map of $\mathcal{D}$. Then $DM_{u}C_{\phi}:\mathcal{Z}\rightarrow\mathcal {B}^{\varphi}$ is bounded if and only if $k_{1}<\infty$,

$$\begin{aligned}& k_{5}=\sup_{z\in\mathcal{D}}\mu(z)\bigl\vert 2u'(z)\phi'(z)+u(z)\phi ''(z) \bigr\vert \log\frac{e}{1-|\phi(z)|^{2}}< \infty, \end{aligned}$$

(7)

$$\begin{aligned}& k_{6}=\sup_{z\in\mathcal{D}}\frac{\mu(z)|u(z)||\phi '(z)|^{2}}{1-|\phi(z)|^{2}}< \infty. \end{aligned}$$

(8)

Proof

Suppose that $DM_{u}C_{\phi}:\mathcal{Z}\rightarrow \mathcal{B}^{\varphi}$ is bounded, by Lemma 2 we know that $k_{1}, k_{2}, k_{3}<\infty$. Let

$$\begin{aligned}& r(z)=(z-1) \biggl[ \biggl(1+\log\frac{e}{1-z} \biggr)^{2}+1 \biggr], \\& s_{a}(z)=\frac{r(\bar{a}z)}{\bar{a}} \biggl(\log\frac {e}{1-|a|^{2}} \biggr)^{-1} -\int^{z}_{0}\log \frac{e}{1-\bar{a}\lambda}\, d\lambda-c_{1}+c_{2}, \end{aligned}$$

where

$$ c_{1}=\frac{r(|a|^{2})}{\bar{a}} \biggl(\log\frac {e}{1-|a|^{2}} \biggr)^{-1},\qquad c_{2}=\int^{a}_{0} \log\frac{e}{1-\bar{a}\lambda}\, d\lambda $$

for any $a\in\mathcal{D}$ such that $\frac{1}{2}<|a|<1$. Then we have

$$ \bigl\vert s''_{a}(z)\bigr\vert = \frac{2}{1-|z|} \biggl(C+\log\frac{e}{1-|a|} \biggr) \biggl(\log \frac{e}{1-|a|^{2}} \biggr)^{-1} +\frac{1}{1-|z|}\leq\frac{C}{1-|z|} $$

for $\frac{1}{2}<|a|<1$ and $\sup_{\frac{1}{2}<|a|<1}\|s_{a}\|_{\mathcal{Z}}<\infty$.

Now let $a=\phi(\omega)$, $\omega\in\mathcal{D}$ such that $\frac{1}{2}<|\phi(\omega)|<1$, then

$$ s_{\phi(\omega)}\bigl(\phi(\omega)\bigr)=s'_{\phi(\omega)}\bigl( \phi(\omega)\bigr)=0,\qquad s''_{\phi(\omega)}\bigl( \phi(\omega)\bigr)=\frac{\overline{\phi(\omega )}}{1-|\phi(\omega)|^{2}}. $$

By the boundedness of $DM_{u}C_{\phi}:\mathcal{Z}\rightarrow\mathcal {B}^{\varphi}$, we have $\|DM_{u}C_{\phi}s_{\phi(\omega)}\|_{\mathcal{B}^{\varphi }}\leq C$, then

$$ 1 \geq S_{\varphi } \biggl(\frac{(DM_{u}C_{\phi}s_{\phi(\omega)})'(z)}{C} \biggr)\geq \sup _{\frac{1}{2}< |\phi(\omega)|<1}\bigl(1-|\omega|^{2}\bigr)\varphi \biggl( \frac{|u(\omega)||\phi'(\omega)|^{2}|\phi(\omega )|}{C(1-|\phi(\omega)|^{2})} \biggr). $$

From this it follows that

$$ \frac{1}{2}\sup_{\frac{1}{2}< |\phi(\omega)|<1} \frac{\mu(\omega)|u(\omega)||\phi'(\omega)|^{2}}{1-|\phi(\omega )|^{2}} \leq \sup_{\frac{1}{2}<|\phi(\omega)|<1} \frac{\mu(\omega)|u(\omega)||\phi'(\omega)|^{2}|\phi(\omega )|}{1-|\phi(\omega)|^{2}}<\infty. $$

(9)

By $k_{3}<\infty$ we see that

$$ \sup_{|\phi(\omega)|\leq\frac{1}{2}}\frac{\mu (\omega)|u(\omega)||\phi'(\omega)|^{2}}{ 1-|\phi(\omega)|^{2}}\leq \frac{4}{3} \sup_{|\phi(\omega)|\leq\frac{1}{2}}\mu(\omega)\bigl\vert u(\omega) \bigr\vert \bigl\vert \phi '(\omega)\bigr\vert ^{2} < \infty. $$

(10)

From (9) and (10) we obtain $k_{6}<\infty$.

Let

$$ t_{\phi(\omega)}(z)=\frac{r(\overline{\phi(\omega)}z)}{ \overline{\phi(\omega)}} \biggl(\log\frac{e}{1-|\phi(\omega )|^{2}} \biggr)^{-1}-c_{1} $$

for $\omega\in\mathcal{D}$ such that $\frac{1}{2}<|\phi(\omega )|<1$, then, as above, we can get that $t_{\phi(\omega)}\in\mathcal{Z}$ and

$$ t_{\phi(\omega)}\bigl(\phi(\omega)\bigr)=0, \qquad t'_{\phi (\omega)} \bigl(\phi(\omega)\bigr)=\log\frac{e}{1-|\phi(\omega)|^{2}},\qquad t''_{\phi(\omega)} \bigl(\phi(\omega)\bigr)=\frac{2\overline{\phi(\omega )}}{1-|\phi(\omega)|^{2}}. $$

By the boundedness of $DM_{u}C_{\phi}:\mathcal{Z}\rightarrow\mathcal {B}^{\varphi}$, we have $\|DM_{u}C_{\phi}t_{\phi(\omega)}\|_{\mathcal{B}^{\varphi }}\leq C$, then

$$\begin{aligned} 1 \geq& S_{\varphi } \biggl(\frac{(DM_{u}C_{\phi}t_{\phi(\omega)})'(z)}{C} \biggr) \\ \geq& \sup _{\frac{1}{2}< |\phi(\omega)|<1}\bigl(1-|\omega|^{2}\bigr)\varphi \biggl( \frac{|(DM_{u}C_{\phi}t_{\phi(\omega)})'(\omega)|}{C} \biggr) \\ \geq&\sup_{\frac{1}{2}<|\phi(\omega )|<1}\bigl(1-|\omega|^{2}\bigr) \\ &{}\cdot\varphi \biggl(\frac{ |(2u'(\omega)\phi '(\omega)+u(\omega) \phi''(\omega))\log\frac{e}{1-|\phi(\omega)|^{2}}+u(\omega)(\phi '(\omega))^{2}\frac{2\overline{\phi(\omega)}}{1-|\phi(\omega)|^{2}} |}{C} \biggr). \end{aligned}$$

From this and by $k_{6}<\infty$, we get

$$\begin{aligned}& \sup_{\frac{1}{2}< |\phi (\omega)|<1} \mu(\omega)\bigl\vert 2u'(\omega)\phi'(\omega)+u(\omega)\phi''( \omega )\bigr\vert \log\frac{e}{1-|\phi(\omega)|^{2}} \\& \quad \leq C+2Ck_{6}<\infty. \end{aligned}$$

(11)

By $k_{2}<\infty$ we see that

$$\begin{aligned}& \sup_{|\phi(\omega)|\leq\frac {1}{2}}\mu(\omega)\bigl\vert 2u'(\omega)\phi'(\omega)+u(\omega)\phi ''(\omega) \bigr\vert \log\frac{e}{1-|\phi(\omega)|^{2}} \\& \quad \leq C \sup_{|\phi(\omega)|\leq\frac{1}{2}}\mu(\omega)\bigl\vert 2u'(\omega)\phi '(\omega)+u(\omega) \phi''(\omega) \bigr\vert < \infty. \end{aligned}$$

(12)

From (11) and (12) we obtain (7).

Suppose that $k_{1}, k_{5}, k_{6}<\infty$. Then, by Lemma 1(ii) and similar to the proof of Theorem 3, we get that $DM_{u}C_{\phi}:\mathcal{Z}\rightarrow\mathcal{B}^{\varphi}$ is bounded. □

Theorem 5

Let $u\in\mathcal{H}(\mathcal{D})$, ϕ be an analytic self-map of $\mathcal{D}$ and $1<\alpha<2$. Then $DM_{u}C_{\phi}:\mathcal{Z}^{\alpha}\rightarrow\mathcal {B}^{\varphi}$ is bounded if and only if $k_{1}<\infty$,

$$\begin{aligned}& k_{7}=\sup_{z\in\mathcal{D}}\frac{\mu(z)|2u'(z)\phi '(z)+u(z)\phi''(z)|}{(1-|\phi(z)|^{2})^{\alpha-1}}< \infty, \end{aligned}$$

(13)

$$\begin{aligned}& k_{8}=\sup_{z\in\mathcal{D}}\frac{\mu(z)|u(z)||\phi '(z)|^{2}}{(1-|\phi(z)|^{2})^{\alpha}}< \infty. \end{aligned}$$

(14)

Proof

Suppose that $DM_{u}C_{\phi}:\mathcal{Z}^{\alpha}\rightarrow\mathcal {B}^{\varphi}$ is bounded, by Lemma 2 we know that $k_{1}, k_{2}, k_{3}<\infty$. Inequality (14) can be proved as in Theorem 3. Using the test function $f_{\phi(\omega)}(z)$ in Section 1, where $z\in\mathcal{D}$, $\omega\in\mathcal{D}$ such that $\frac{1}{2}<|\phi(\omega)|<1$, then we have that $f_{\phi(\omega)}\in\mathcal{Z}^{\alpha}$, and

$$\begin{aligned}& f_{\phi(\omega)}\bigl(\phi(\omega)\bigr)=0, \\& f'_{\phi(\omega)} \bigl(\phi(\omega)\bigr)=\frac{1}{\overline{\phi(\omega )}(1-|\phi(\omega)|^{2})^{\alpha-1}}, \\& f''_{\phi(\omega)} \bigl(\phi(\omega)\bigr)=\frac{2\alpha}{(1-|\phi(\omega )|^{2})^{\alpha}}. \end{aligned}$$

By the boundedness of $DM_{u}C_{\phi}:\mathcal{Z}^{\alpha}\rightarrow\mathcal {B}^{\varphi}$, we have $\|DM_{u}C_{\phi}f_{\phi(\omega)}\|_{\mathcal{B}^{\varphi}}\leq C$, then

$$\begin{aligned} 1 \geq& S_{\varphi} \biggl(\frac{(DM_{u}C_{\phi }f_{\phi(\omega)})'(z)}{C} \biggr) \\ \geq& \sup_{\frac{1}{2}< |\phi(\omega)|<1}\bigl(1-|\omega|^{2}\bigr)\varphi \biggl(\frac{|(DM_{u}C_{\phi} f_{\phi(\omega)})'(\omega)|}{C} \biggr) \\ \geq&\sup_{\frac{1}{2}<|\phi(\omega)|<1}\bigl(1-|\omega |^{2}\bigr)\varphi \biggl(\frac{ | \frac{2u'(\omega)\phi'(\omega)+u(\omega) \phi''(\omega)}{\overline{\phi(\omega)}(1-|\phi(\omega )|^{2})^{\alpha-1}} +\frac{2\alpha u(\omega)\phi'(\omega)^{2}}{(1-|\phi(\omega)|^{2})^{\alpha}} |}{C} \biggr). \end{aligned}$$

From this and by $k_{8}<\infty$, we get

$$ \sup_{\frac{1}{2}< |\phi(\omega)|<1} \frac{\mu(\omega)|2u'(\omega)\phi'(\omega)+u(\omega)\phi ''(\omega)|}{(1-|\phi(\omega)|^{2})^{\alpha-1}} \leq C+2C\alpha k_{8}<\infty. $$

Then, according to the former proof with $k_{2}<\infty$, we can get (13).

Suppose that $k_{1}, k_{7}, k_{8}<\infty$. Then, by Lemma 1(iii) and (iv) and similar to the proof of Theorem 3, we get that $DM_{u}C_{\phi}:\mathcal{Z}^{\alpha}\rightarrow\mathcal {B}^{\varphi}$ is bounded. □

Theorem 6

Let $u\in\mathcal{H}(\mathcal{D})$ and ϕ be an analytic self-map of $\mathcal{D}$. Then $DM_{u}C_{\phi}:\mathcal{Z}^{2}\rightarrow \mathcal{B}^{\varphi}$ is bounded if and only if

$$\begin{aligned}& k_{9}=\sup_{z\in\mathcal{D}}\mu(z)\bigl\vert u''(z)\bigr\vert \log\frac{e}{1-|\phi (z)|^{2}}< \infty, \end{aligned}$$

(15)

$$\begin{aligned}& k_{10}=\sup_{z\in\mathcal{D}}\frac{\mu(z)|2u'(z)\phi '(z)+u(z)\phi''(z)|}{1-|\phi(z)|^{2}}< \infty, \end{aligned}$$

(16)

$$\begin{aligned}& k_{11}=\sup_{z\in\mathcal{D}}\frac{\mu(z)|u(z)||\phi '(z)|^{2}}{(1-|\phi(z)|^{2})^{2}}< \infty. \end{aligned}$$

(17)

Proof

Suppose that $DM_{u}C_{\phi}:\mathcal{Z}^{2}\rightarrow\mathcal{B}^{\varphi}$ is bounded, from Lemma 2 we know that $k_{1}, k_{2}, k_{3}<\infty$. By repeating the arguments in the proof of Theorem 3 and Theorem 5, (16) and (17) can be proved similarly. Hence we only need to show $k_{9}<\infty$. For every $z\in\mathcal{D}$ and $\omega\in\mathcal{D}$ such that $\frac{1}{2}<|\phi(\omega)|<1$, let $p_{\phi(\omega)}(z)=\log\frac{e}{1-\overline{\phi(\omega)}z}$. Clearly $p_{\phi(\omega)}\in\mathcal{Z}^{2}$, and $p_{\phi(\omega)}(\phi(\omega))=\log\frac{e}{1-|\phi(\omega)|^{2}}$,

$$\begin{aligned}& p'_{\phi(\omega)}\bigl(\phi(\omega)\bigr)=\frac{\overline{\phi(\omega )}}{1-|\phi(\omega)|^{2}}, \\& p''_{\phi(\omega)}\bigl(\phi(\omega)\bigr)= \frac{\overline{\phi(\omega)}^{2}}{ (1-|\phi(\omega)|^{2})^{2}}. \end{aligned}$$

By the boundedness of $DM_{u}C_{\phi}:\mathcal{Z}^{2}\rightarrow\mathcal{B}^{\varphi}$, we have $\|DM_{u}C_{\phi}p_{\phi(\omega)}\|_{\mathcal{B}^{\varphi}}\leq C$, then

$$\begin{aligned} 1 \geq& S_{\varphi} \biggl(\frac{(DM_{u}C_{\phi }p_{\phi(\omega)})'(z)}{C} \biggr) \\ \geq& \sup_{\frac{1}{2}< |\phi(\omega)|<1}\bigl(1-|\omega|^{2}\bigr)\varphi \biggl(\frac{|(DM_{u}C_{\phi}p_{\phi(\omega)})'(\omega)|}{C} \biggr) \\ \geq&\sup_{\frac{1}{2}<|\phi(\omega)|<1}\bigl(1-|\omega|^{2}\bigr) \\ &{}\cdot\varphi \biggl(\frac{ |u''(\omega)\log\frac {e}{1-|\phi(\omega)|^{2}} +\frac {(2u'(\omega)\phi'(\omega)+u(\omega)\phi''(\omega))\overline{\phi(\omega)}}{1-|\phi(\omega)|^{2}} + \frac{u(\omega)(\phi'(\omega))^{2}\overline{\phi(\omega )}^{2}}{(1-|\phi(\omega)|^{2})^{2}}|}{C}\biggr). \end{aligned}$$

By $k_{10}, k_{11}<\infty$ we get

$$ \sup_{\frac{1}{2}< |\phi(\omega)|<1}\mu(\omega)\bigl\vert u''(\omega)\bigr\vert \log \frac{e}{1-|\phi(\omega)|^{2}} \leq C+Ck_{10}+Ck_{11}<\infty. $$

(18)

By $k_{1}<\infty$ we see that

$$ \sup_{|\phi(\omega)|\leq\frac{1}{2}}\mu(\omega)\bigl\vert u''(\omega )\bigr\vert \log\frac{e}{1-|\phi(\omega)|^{2}} \leq C \sup_{|\phi(\omega)|\leq\frac{1}{2}}\mu(\omega)\bigl\vert u''( \omega)\bigr\vert < \infty. $$

(19)

From (18) and (19) we obtain (15).

Suppose that $k_{9}, k_{10}, k_{11}<\infty$. Then, by Lemma 1(iii) and (v) and similar to the proof of Theorem 3, we get that $DM_{u}C_{\phi}:\mathcal{Z}^{2}\rightarrow\mathcal{B}^{\varphi}$ is bounded. □

Theorem 7

Let $u\in\mathcal{H}(\mathcal{D})$, ϕ be an analytic self-map of $\mathcal{D}$ and $\alpha>2$. Then $DM_{u}C_{\phi}:\mathcal{Z}^{\alpha}\rightarrow\mathcal {B}^{\varphi}$ is bounded if and only if

$$\begin{aligned}& k_{12}=\sup_{z\in\mathcal{D}}\frac{\mu(z)|u''(z)|}{(1-|\phi (z)|^{2})^{\alpha-2}}< \infty, \end{aligned}$$

(20)

$$\begin{aligned}& k_{13}=\sup_{z\in\mathcal{D}}\frac{\mu(z)|2u'(z)\phi '(z)+u(z)\phi''(z)|}{(1-|\phi(z)|^{2})^{\alpha-1}}< \infty, \end{aligned}$$

(21)

$$\begin{aligned}& k_{14}=\sup_{z\in\mathcal{D}}\frac{\mu(z)|u(z)||\phi '(z)|^{2}}{(1-|\phi(z)|^{2})^{\alpha}}< \infty. \end{aligned}$$

(22)

Proof

Suppose that $DM_{u}C_{\phi}:\mathcal{Z}^{\alpha}\rightarrow\mathcal {B}^{\varphi}$ is bounded, by Lemma 2 we know that $k_{1}, k_{2}, k_{3}<\infty$. With the same argument as in Theorem 5 one can show that (21) and (22) hold.

Now we prove that $k_{12}<\infty$. For every $a, z\in\mathcal{D}$, define $q_{a}(z)=\frac{(1-|a|^{2})^{2}}{(1-\bar{a}z)^{\alpha}}$. Then $\sup_{z\in\mathcal{D}}(1-|z|^{2})^{\alpha}|q''_{a}(z)|\leq4\alpha \cdot2^{\alpha}\cdot(\alpha+1)$, which shows that $q_{a}\in\mathcal{Z}^{\alpha}$. Now we let $a=\phi (\omega)$ for every $\omega\in\mathcal{D}$ such that $\frac{1}{2}<|\phi (\omega)|<1$, and we have

$$\begin{aligned}& q_{\phi(\omega)}\bigl(\phi(\omega)\bigr)=\frac{1}{(1-|\phi(\omega )|^{2})^{\alpha-2}}, \\& q'_{\phi(\omega)}\bigl(\phi(\omega)\bigr)=\frac{\alpha\overline{\phi (\omega)}}{(1-|\phi(\omega)|^{2})^{\alpha-1}},\qquad q''_{\phi(\omega)}\bigl(\phi(\omega)\bigr)= \frac{\alpha(\alpha+1)\overline {\phi(\omega)}^{2}}{(1-|\phi(\omega)|^{2})^{\alpha}}. \end{aligned}$$

By the boundedness of $DM_{u}C_{\phi}:\mathcal{Z}^{\alpha}\rightarrow\mathcal {B}^{\varphi}$, we have $\|DM_{u}C_{\phi}q_{\phi(\omega)}\|_{\mathcal{B}^{\varphi}}\leq C$, then

$$\begin{aligned} 1 \geq& S_{\varphi} \biggl(\frac{(DM_{u}C_{\phi}q_{\phi(\omega )})'(z)}{C} \biggr) \\ \geq& \sup_{\frac{1}{2}< |\phi(\omega)|<1}\bigl(1-|\omega|^{2}\bigr) \\ &{}\cdot\varphi \biggl(\frac{ |\frac{u''(\omega )}{(1-|\phi(\omega)|^{2})^{\alpha-2}} +\frac{\alpha\overline{\phi(\omega)}(2u'(\omega)\phi'(\omega )+u(\omega)\phi''(\omega))}{(1-|\phi(\omega)|^{2})^{\alpha-1}} +\frac{\alpha(\alpha+1)\overline{\phi(\omega)}^{2}u(\omega )\phi'(\omega)^{2}}{(1-|\phi(\omega)|^{2})^{\alpha}} |}{C} \biggr). \end{aligned}$$

Then we have

$$\begin{aligned}& \sup_{\frac{1}{2}< |\phi(\omega)|<1}\frac{\mu(\omega )|u''(\omega)|}{(1-|\phi(\omega)|^{2})^{\alpha-2}} \\& \quad \leq C+\sup_{\frac{1}{2}<|\phi(\omega)|<1}\alpha \mu(\omega)\biggl\vert \frac{2u'(\omega)\phi'(\omega)+u(\omega)\phi ''(\omega)}{\overline{\phi(\omega)}(1-|\phi(\omega)|^{2})^{\alpha-1}} +\frac{\alpha+1}{2\alpha}\frac{2\alpha u(\omega)\phi'(\omega)^{2}}{(1-|\phi(\omega)|^{2})^{\alpha}}\biggr\vert \\& \quad \leq C +\sup_{\frac{1}{2}<|\phi(\omega)|<1} \alpha \mu(\omega) \biggl\{ \biggl\vert \frac{2u'(\omega)\phi'(\omega)+u(\omega )\phi''(\omega)}{\overline{\phi(\omega)} (1-|\phi(\omega)|^{2})^{\alpha-1}}\biggr\vert +\frac{\alpha+1}{2\alpha}\biggl\vert \frac{2\alpha u(\omega)\phi'(\omega)^{2}}{(1-|\phi(\omega)|^{2})^{\alpha}}\biggr\vert \biggr\} . \end{aligned}$$

(23)

Since $\frac{\alpha+1}{2\alpha}<1$, then by (21), (22), (23) and according to the former proof with $k_{1}<\infty$ for $|\phi(\omega)|\leq\frac{1}{2}$, then $k_{12}<\infty$. Suppose that $k_{12}, k_{13}, k_{14}<\infty$. Then, by Lemma 1(iii) and (vi) and similar to the proof of Theorem 3, we get that $DM_{u}C_{\phi}:\mathcal{Z}^{\alpha}\rightarrow\mathcal {B}^{\varphi}$ is bounded. □

3 The compactness of $DM_{u}C_{\phi}:\mathcal{Z}^{\alpha }\rightarrow\mathcal{B}^{\varphi}$

In order to prove the compactness of the product-type operator $DM_{u}C_{\phi}:\mathcal{Z}^{\alpha}\rightarrow\mathcal {B}^{\varphi}$, we need the following lemmas. The proof of the following lemma is similar to that of Proposition 3.11 in [43]. The details are omitted.

Lemma 8

Let $u\in\mathcal{H}(\mathcal{D})$, ϕ be an analytic self-map of $\mathcal{D}$ and $0<\alpha<\infty$. Then $DM_{u}C_{\phi}:\mathcal{Z}^{\alpha}\rightarrow\mathcal {B}^{\varphi}$ is compact if and only if $DM_{u}C_{\phi}:\mathcal{Z}^{\alpha}\rightarrow\mathcal {B}^{\varphi}$ is bounded and for any bounded sequence $\{f_{n}\}_{n\in\mathcal{N}}$ in $\mathcal{Z}^{\alpha}$ which converges to zero uniformly on compact subsets of $\mathcal{D}$ as $n\rightarrow\infty$, we have $\|DM_{u}C_{\phi}f_{n}\|_{\mathcal{B}^{\varphi}}\rightarrow0$ as $n\rightarrow\infty$.

The following lemma was essentially proved in paper [11] in Lemma 2.5.

Lemma 9

Fix $0<\alpha<2$ and let $\{f_{n}\}_{n\in\mathcal{N}}$ be a bounded sequence in $\mathcal{Z}^{\alpha}$ which converges to zero uniformly on compact subsets of $\mathcal{D}$ as $n\rightarrow\infty$. Then $\lim_{n\rightarrow\infty}\sup_{z\in\mathcal{D}}|f_{n}(z)|=0$. Moreover, for $0<\alpha<1$, if $\{f_{n}\}_{n\in\mathcal{N}}$ is a bounded sequence in $\mathcal{Z}^{\alpha}$ which converges to zero uniformly on compact subsets of $\mathcal{D}$ as $n\rightarrow\infty$, then $\lim_{n\rightarrow\infty}\sup_{z\in\mathcal{D}}|f'_{n}(z)|=0$.

Theorem 10

Let $u\in\mathcal{H}(\mathcal{D})$, ϕ be an analytic self-map of $\mathcal{D}$ and $0<\alpha<1$. Then $DM_{u}C_{\phi}:\mathcal{Z}^{\alpha}\rightarrow\mathcal {B}^{\varphi}$ is compact if and only if $DM_{u}C_{\phi}:\mathcal{Z}^{\alpha}\rightarrow\mathcal {B}^{\varphi}$ is bounded,

$$ \lim_{|\phi(z)|\rightarrow1}\frac{\mu(z)|u(z)||\phi '(z)|^{2}}{(1-|\phi(z)|^{2})^{\alpha}}=0. $$

(24)

Proof

Suppose that $DM_{u}C_{\phi}:\mathcal{Z}^{\alpha}\rightarrow\mathcal {B}^{\varphi}$ is compact. It is clear that $DM_{u}C_{\phi}:\mathcal{Z}^{\alpha}\rightarrow \mathcal{B}^{\varphi}$ is bounded. By Lemma 2, we have that $k_{1}, k_{2}, k_{3}<\infty$. Let $\{z_{n}\}_{n\in\mathcal{N}}$ be a sequence in $\mathcal{D}$ such that $|\phi(z_{n})|\rightarrow1$ as $n\rightarrow\infty$. Without loss of generality, suppose that $|\phi(z_{n})|>\frac{1}{2}$ for all n. Taking the function

$$g_{n}(z)=\frac{1}{\overline{\phi(z_{n})}^{2}} \biggl[\frac{(1-|\phi (z_{n})|^{2})^{2}}{ (1-\overline{\phi(z_{n})}z)^{\alpha}}- \frac{1-|\phi (z_{n})|^{2}}{(1-\overline{\phi(z_{n})}z)^{\alpha-1}} \biggr] -\frac{1}{\overline{\phi(z_{n})}}\int^{z}_{0} \frac{1-|\phi (z_{n})|^{2}}{(1-\overline{\phi(z_{n})}\lambda)^{\alpha}}\, d\lambda. $$

Then $\sup_{n\in\mathcal{N}}\|g_{n}\|_{\mathcal{Z}^{\alpha }}<\infty$, and $g_{n}\rightarrow0$ uniformly on compact subsets of $\mathcal{D}$. Since $DM_{u}C_{\phi}:\mathcal{Z}^{\alpha}\rightarrow\mathcal {B}^{\varphi}$ is compact, then $\lim_{n\rightarrow\infty}\|DM_{u}C_{\phi}g_{n}\|_{\mathcal {B}^{\varphi}}=0$. Since $\lim_{n\rightarrow\infty}|\phi(z_{n})|=1$, then $\lim_{n\rightarrow\infty}\sup_{z\in\mathcal{D}}|g_{n}(z)|=0$. Moreover, we have

$$ g'_{n}\bigl(\phi(z_{n})\bigr)=0,\qquad g''_{n}\bigl(\phi(z_{n})\bigr)= \frac{\alpha}{(1-|\phi(z_{n})|^{2})^{\alpha}}. $$

Then

$$1\geq S_{\varphi} \biggl(\frac{(DM_{u}C_{\phi}g_{n})'(z_{n})}{\| DM_{u}C_{\phi}g_{n}\|_{\mathcal{B}^{\varphi}}} \biggr)\geq \bigl(1-|z_{n}|^{2}\bigr)\varphi \biggl(\frac{ |u''(z_{n})g_{n}(\phi(z_{n}))+ \frac{\alpha u(z_{n})\phi'(z_{n})^{2}}{(1-|\phi(z_{n})|^{2})^{\alpha}} |}{ \|DM_{u}C_{\phi}g_{n}\|_{\mathcal{B}^{\varphi}}} \biggr). $$

Hence

$$ \biggl\vert \frac{\alpha \mu(z_{n})|u(z_{n})||\phi'(z_{n})|^{2}}{(1-|\phi (z_{n})|^{2})^{\alpha}} -\mu(z_{n})\bigl\vert u''(z_{n})\bigr\vert \bigl\vert g_{n}\bigl(\phi(z_{n})\bigr)\bigr\vert \biggr\vert \leq \|DM_{u}C_{\phi}g_{n}\|_{\mathcal{B}^{\varphi}}. $$

Therefore

$$ \lim_{|\phi(z_{n})|\rightarrow1}\frac{\mu(z_{n})|u(z_{n})||\phi '(z_{n})|^{2}}{(1-|\phi(z_{n})|^{2})^{\alpha}}= \lim_{n\rightarrow\infty} \frac{\alpha \mu(z_{n})|u(z_{n})||\phi'(z_{n})|^{2}}{(1-|\phi (z_{n})|^{2})^{\alpha}}=0. $$

Suppose that $DM_{u}C_{\phi}:\mathcal{Z}^{\alpha}\rightarrow \mathcal{B}^{\varphi}$ is bounded and (24) holds. Then $k_{1}, k_{2}, k_{3}<\infty$ by Lemma 2 and for every $\epsilon>0$, there is $\delta\in(0,1)$ such that

$$ \frac{\mu(z)|u(z)||\phi'(z)|^{2}}{(1-|\phi(z)|^{2})^{\alpha }}< \epsilon $$

(25)

whenever $\delta<|\phi(z)|<1$. Assume that $\{f_{n}\}_{n\in\mathcal {N}}$ is a sequence in $\mathcal{Z}^{\alpha}$ such that $\sup_{n\in\mathcal{N}} \|f_{n}\|_{\mathcal{Z}^{\alpha}}\leq L$, and $f_{n}$ converges to 0 uniformly on compact subsets of $\mathcal{D}$ as $n\rightarrow \infty$. Let $K=\{z\in\mathcal{D}:|\phi(z)|\leq\delta\}$. Then by $k_{1}, k_{2}, k_{3}<\infty$ and (25) it follows that

$$\begin{aligned}& \sup_{z\in\mathcal{D}}\mu(z)\bigl\vert (DM_{u}C_{\phi }f_{n})'(z) \bigr\vert \\& \quad \leq \sup_{z\in\mathcal{D}}\mu(z)\bigl\vert u''(z) \bigr\vert \bigl\vert f_{n}\bigl(\phi(z)\bigr)\bigr\vert + \sup _{z\in\mathcal{D}}\mu(z)\bigl\vert 2u'(z) \phi'(z)+u(z)\phi ''(z)\bigr\vert \bigl\vert f'_{n}\bigl(\phi(z)\bigr)\bigr\vert \\& \qquad {}+\sup_{z\in K}\mu(z)\bigl\vert u(z)\bigr\vert \bigl\vert \phi'(z)\bigr\vert ^{2}\bigl\vert f''_{n}\bigl(\phi(z)\bigr)\bigr\vert + \sup _{z\in\mathcal{D}\setminus K}\mu(z)\bigl\vert u(z)\bigr\vert \bigl\vert \phi '(z)\bigr\vert ^{2}\bigl\vert f''_{n} \bigl(\phi(z)\bigr)\bigr\vert \\& \quad \leq k_{1}\sup_{z\in\mathcal{D}}\bigl\vert f_{n}\bigl(\phi(z)\bigr)\bigr\vert +k_{2}\sup _{z\in \mathcal{D}}\bigl\vert f'_{n}\bigl(\phi(z) \bigr)\bigr\vert +k_{3}\sup_{z\in K}\bigl\vert f''_{n}\bigl(\phi(z)\bigr)\bigr\vert \\& \qquad {} +\sup_{z\in\mathcal{D}\setminus K}\frac{\mu(z)\vert u(z)\vert \vert \phi'(z)\vert ^{2}(1-\vert \phi(z)\vert ^{2})^{\alpha }\vert f''_{n}(\phi(z))\vert }{ (1-\vert \phi(z)\vert ^{2})^{\alpha}} \\& \quad \leq k_{1}\sup_{\omega\in\mathcal{D}}\bigl\vert f_{n}(\omega)\bigr\vert +k_{2}\sup_{\omega \in\mathcal{D}} \bigl\vert f'_{n}(\omega)\bigr\vert + k_{3} \sup_{\vert \omega \vert \leq\delta}\bigl\vert f''_{n}( \omega)\bigr\vert +L\epsilon. \end{aligned}$$

Here we use the fact that $\sup_{z\in\mathcal{D}}(1-|\phi(z)|^{2})^{\alpha}|f''_{n}(\phi (z))|\leq\|f_{n}\|_{\mathcal{Z}^{\alpha}}\leq L$. So we obtain

$$\begin{aligned} \begin{aligned}[b] &\|DM_{u}C_{\phi}f_{n} \|_{B^{\varphi}} \\ &\quad = \bigl\vert u'(0)f_{n}\bigl(\phi(0)\bigr)+u(0) \phi'(0)f'_{n}\bigl(\phi(0)\bigr)\bigr\vert + \sup_{z\in\mathcal{D}}\mu(z)\bigl\vert (DM_{u}C_{\phi}f_{n})'(z) \bigr\vert \\ &\quad \leq \bigl\vert u'(0)\bigr\vert \bigl\vert f_{n}\bigl(\phi(0)\bigr)\bigr\vert +\bigl\vert u(0)\bigr\vert \bigl\vert \phi'(0)\bigr\vert \bigl\vert f'_{n} \bigl(\phi (0)\bigr)\bigr\vert \\ &\qquad {} +k_{1}\sup_{\omega\in\mathcal{D}}\bigl\vert f_{n}(\omega)\bigr\vert +k_{2}\sup_{\omega\in\mathcal{D}} \bigl\vert f'_{n}(\omega)\bigr\vert + k_{3} \sup_{\vert \omega \vert \leq\delta}\bigl\vert f''_{n}( \omega)\bigr\vert +L\epsilon. \end{aligned} \end{aligned}$$

(26)

Since $f_{n}$ converges to 0 uniformly on compact subsets of $\mathcal {D}$ as $n\rightarrow\infty$, Cauchy’s estimation gives that $f'_{n}$, $f''_{n}$ also do as $n\rightarrow\infty$. In particular, since $\{\omega:|\omega|\leq\delta\}$ and $\{\phi(0)\}$ are compact, it follows that

$$\lim_{n\rightarrow\infty}\bigl\{ \bigl\vert u'(0)\bigr\vert \bigl\vert f_{n}\bigl(\phi(0)\bigr)\bigr\vert +\bigl\vert u(0) \bigr\vert \bigl\vert \phi '(0)\bigr\vert \bigl\vert f'_{n}\bigl(\phi(0)\bigr)\bigr\vert \bigr\} =0\quad \mbox{and}\quad \lim_{n\rightarrow\infty}k_{3}\sup _{\vert \omega \vert \leq\delta }\bigl\vert f''_{n}( \omega)\bigr\vert =0. $$

Moreover, since $0<\alpha<1$, by Lemma 9 we have

$$\lim_{n\rightarrow\infty}\sup_{\omega\in\mathcal {D}}\bigl\vert f_{n}(\omega)\bigr\vert =0,\qquad \lim_{n\rightarrow\infty}\sup _{\omega\in\mathcal {D}}\bigl\vert f'_{n}(\omega)\bigr\vert =0. $$

Hence, letting $n\rightarrow\infty$ in (26), we get

$$\lim_{n\rightarrow\infty}\|DM_{u}C_{\phi}f_{n} \|_{B^{\varphi}}=0. $$

Employing Lemma 8 the implication follows. □

Theorem 11

Let $u\in\mathcal{H}(\mathcal{D})$ and ϕ be an analytic self-map of $\mathcal{D}$. Then $DM_{u}C_{\phi}:\mathcal{Z}\rightarrow\mathcal {B}^{\varphi}$ is compact if and only if $DM_{u}C_{\phi}:\mathcal{Z}\rightarrow\mathcal {B}^{\varphi}$ is bounded,

$$\begin{aligned}& \lim_{|\phi(z)|\rightarrow1}\mu(z)\bigl\vert 2u'(z) \phi'(z)+u(z)\phi ''(z)\bigr\vert \log \frac{e}{1-|\phi(z)|^{2}} =0, \end{aligned}$$

(27)

$$\begin{aligned}& \lim_{|\phi(z)|\rightarrow1}\frac{\mu(z)|u(z)||\phi '(z)|^{2}}{1-|\phi(z)|^{2}}=0. \end{aligned}$$

(28)

Proof

Suppose that $DM_{u}C_{\phi}:\mathcal{Z}\rightarrow \mathcal{B}^{\varphi}$ is compact. It is clear that $DM_{u}C_{\phi}:\mathcal{Z}\rightarrow\mathcal{B}^{\varphi}$ is bounded. By Lemma 2, we have that $k_{1}, k_{2}, k_{3}<\infty$. Let $\{z_{n}\}_{n\in \mathcal{N}}$ be a sequence in $\mathcal{D}$ such that $|\phi(z_{n})|\rightarrow1$ as $n\rightarrow\infty$. Without loss of generality, we may suppose that $|\phi(z_{n})|>\frac{1}{2}$ for all n. Taking the function

$$s_{n}(z)=\frac{r(\overline{\phi(z_{n})}z)}{\overline{\phi(z_{n})}} \biggl(\log\frac{e}{1-|\phi(z_{n})|^{2}} \biggr)^{-1}- \biggl(\log\frac{e}{1-|\phi(z_{n})|^{2}} \biggr)^{-2} \int ^{z}_{0}\log^{3}\frac{e}{1-\overline{\phi(z_{n})}\lambda }\, d \lambda. $$

Then $\sup_{n\in\mathcal{N}}\|s_{n}\|_{\mathcal{Z}}<\infty$ by the proof of Theorem 4, and $s_{n}\rightarrow0$ uniformly on compact subsets of $\mathcal{D}$ by a direct calculation. Consequently, $\lim_{n\rightarrow\infty}\sup_{z\in\mathcal{D}}|s_{n}(z)|=0$ by Lemma 9. Since $DM_{u}C_{\phi}:\mathcal{Z}\rightarrow\mathcal{B}^{\varphi}$ is compact, then $\lim_{n\rightarrow\infty}\|DM_{u}C_{\phi}s_{n}\|_{\mathcal {B}^{\varphi}}=0$. Moreover, we have

$$ s'_{n}\bigl(\phi(z_{n})\bigr)=0,\qquad s''_{n}\bigl(\phi(z_{n})\bigr)=- \frac{\overline {\phi(z_{n})}}{1-|\phi(z_{n})|^{2}}. $$

Then

$$\begin{aligned} 1 \geq& S_{\varphi} \biggl(\frac{(DM_{u}C_{\phi}s_{n})'(z_{n})}{\| DM_{u}C_{\phi}s_{n}\|_{\mathcal{B}^{\varphi}}} \biggr) \\ \geq&\bigl(1-|z_{n}|^{2}\bigr)\varphi \biggl( \frac{ |u''(z_{n})s_{n}(\phi(z_{n}))+ \frac{-\overline{\phi(z_{n})}u(z_{n})\phi'(z_{n})^{2}}{1-|\phi (z_{n})|^{2}} |}{ \|DM_{u}C_{\phi}g_{n}\|_{\mathcal{B}^{\varphi}}} \biggr). \end{aligned}$$

It follows that

$$ \biggl\vert \frac{\mu(z_{n})|\phi(z_{n})||u(z_{n})||\phi '(z_{n})|^{2}}{1-|\phi(z_{n})|^{2}} -\mu(z_{n})\bigl\vert u''(z_{n})\bigr\vert \bigl\vert s_{n}\bigl(\phi(z_{n})\bigr)\bigr\vert \biggr\vert \leq \|DM_{u}C_{\phi}s_{n}\|_{\mathcal{B}^{\varphi}}. $$

Therefore

$$ \lim_{|\phi(z_{n})|\rightarrow1}\frac{\mu(z_{n})|u(z_{n})||\phi '(z_{n})|^{2}}{1-|\phi(z_{n})|^{2}}= \lim _{n\rightarrow\infty}\frac{\mu(z_{n})|\phi (z_{n})||u(z_{n})||\phi'(z_{n})|^{2}}{1-|\phi(z_{n})|^{2}}=0. $$

(29)

On the other hand, let

$$ t_{n}(z)=\frac{\overline{\phi(z_{n})}z-1}{\overline {\phi(z_{n})}} \biggl[ \biggl(1+\log\frac{e}{1-\overline{\phi(z_{n})}z} \biggr)^{2}+1 \biggr] \biggl(\log\frac{e}{1-|\phi(z_{n})|^{2}} \biggr)^{-1}-c_{n}, $$

where

$$ c_{n}=\frac{|\phi(z_{n})|^{2}-1}{\overline{\phi(z_{n})}} \biggl[ \biggl(1+\log\frac{e}{1-|\phi(z_{n})|^{2}} \biggr)^{2}+1 \biggr] \biggl(\log\frac{e}{1-|\phi(z_{n})|^{2}} \biggr)^{-1} $$

such that $\lim_{n\rightarrow\infty}c_{n}=0$. By a direct calculation, we may easily prove that $t_{n}\rightarrow0$ uniformly on compact subsets of $\mathcal{D}$, and $\sup_{n\in\mathcal{N}}\| t_{n}\|_{\mathcal{Z}}<\infty$ by the proof of Theorem 4. Since $DM_{u}C_{\phi}:\mathcal{Z}\rightarrow\mathcal{B}^{\varphi}$ is compact, then $\lim_{n\rightarrow\infty}\|DM_{u}C_{\phi}t_{n}\|_{\mathcal {B}^{\varphi}}=0$. Moreover, we have

$$ t_{n}\bigl(\phi(z_{n})\bigr)=0,\qquad t'_{n} \bigl(\phi(z_{n})\bigr)=\log\frac{e}{1-|\phi(z_{n})|^{2}},\qquad t''_{n}\bigl(\phi(z_{n})\bigr)= \frac{2\overline{\phi(z_{n})}}{1-|\phi(z_{n})|^{2}}. $$

Then

$$\begin{aligned} 1 \geq& S_{\varphi} \biggl(\frac{(DM_{u}C_{\phi}t_{n})'(z_{n})}{ \|DM_{u}C_{\phi}t_{n}\|_{\mathcal{B}^{\varphi}}} \biggr) \\ \geq& \bigl(1-|z_{n}|^{2}\bigr) \\ &{}\cdot\varphi \biggl(\frac{ |(2u'(z_{n})\phi'(z_{n})+u(z_{n})\phi ''(z_{n}))\log\frac{e}{1-|\phi(z_{n})|^{2}}+ \frac{2\overline{\phi(z_{n})}u(z_{n})\phi'(z_{n})^{2}}{1-|\phi (z_{n})|^{2}} |}{ \|DM_{u}C_{\phi}t_{n}\|_{\mathcal{B}^{\varphi}}} \biggr). \end{aligned}$$

It follows that

$$\begin{aligned}& \mu(z_{n})\bigl\vert 2u'(z_{n}) \phi'(z_{n})+u(z_{n})\phi ''(z_{n}) \bigr\vert \log\frac{e}{1-|\phi(z_{n})|^{2}} \\& \quad \leq\|DM_{u}C_{\phi}t_{n}\|_{\mathcal{B}^{\varphi}} + \frac{2\mu(z_{n})|\phi(z_{n})||u(z_{n})||\phi '(z_{n})|^{2}}{1-|\phi(z_{n})|^{2}}. \end{aligned}$$

(30)

Letting $n\rightarrow\infty$ in (30) and combining with (29), we can get

$$ \lim_{|\phi(z_{n})|\rightarrow1}\mu(z_{n})\bigl\vert 2u'(z_{n})\phi '(z_{n})+u(z_{n}) \phi''(z_{n}) \bigr\vert \log \frac{e}{1-|\phi(z_{n})|^{2}} =0. $$

(31)

The implication follows from (29) and (31).

Conversely, by Lemma 1(ii), Lemma 2, Lemma 8 and Lemma 9, we can prove the converse implication similar to Theorem 10, so we omit the details. □

Theorem 12

Let $u\in\mathcal{H}(\mathcal{D})$, ϕ be an analytic self-map of $\mathcal{D}$ and $1<\alpha<2$. Then $DM_{u}C_{\phi}:\mathcal{Z}^{\alpha}\rightarrow\mathcal {B}^{\varphi}$ is compact if and only if $DM_{u}C_{\phi}:\mathcal{Z}^{\alpha}\rightarrow\mathcal {B}^{\varphi}$ is bounded,

$$\begin{aligned}& \lim_{|\phi(z)|\rightarrow1}\frac{\mu(z)|2u'(z)\phi '(z)+u(z)\phi''(z)|}{(1-|\phi(z)|^{2})^{\alpha-1}}=0, \end{aligned}$$

(32)

$$\begin{aligned}& \lim_{|\phi(z)|\rightarrow1}\frac{\mu(z)|u(z)||\phi '(z)|^{2}}{(1-|\phi(z)|^{2})^{\alpha}}=0. \end{aligned}$$

(33)

Proof

Suppose that $DM_{u}C_{\phi}:\mathcal{Z}^{\alpha}\rightarrow\mathcal {B}^{\varphi}$ is compact. It is clear that $DM_{u}C_{\phi}:\mathcal{Z}^{\alpha}\rightarrow \mathcal{B}^{\varphi}$ is bounded. By Lemma 2, we have that $k_{1}, k_{2}, k_{3}<\infty$. Let $\{z_{n}\}_{n\in\mathcal{N}}$ be a sequence in $\mathcal{D}$ such that $|\phi(z_{n})|\rightarrow1$ as $n\rightarrow\infty$. Without loss of generality, we may suppose that $|\phi(z_{n})|>\frac{1}{2}$ for all n. Then (33) can be proved as the method of (24) in Theorem 10, so we only need to show that (32) holds. Taking the function

$$ f_{n}(z)=\frac{1}{\overline{\phi(z_{n})}^{2}} \biggl[\frac{(1-|\phi (z_{n})|^{2})^{2}}{ (1-\overline{\phi(z_{n})}z)^{\alpha}}- \frac{1-|\phi(z_{n})|^{2}}{ (1-\overline{\phi(z_{n})}z)^{\alpha-1}} \biggr]. $$

Then $\sup_{n\in\mathcal{N}}\|f_{n}\|_{\mathcal{Z}^{\alpha }}<\infty$, and $f_{n}\rightarrow0$ uniformly on compact subsets of $\mathcal{D}$. Since $DM_{u}C_{\phi}:\mathcal{Z}^{\alpha}\rightarrow\mathcal {B}^{\varphi}$ is compact, it gives $\lim_{n\rightarrow\infty}\|DM_{u}C_{\phi}f_{n}\|_{\mathcal {B}^{\varphi}}=0$. Moreover, we have

$$f_{n}\bigl(\phi(z_{n})\bigr)=0,\qquad f'_{n} \bigl(\phi(z_{n})\bigr)=\frac{1}{\overline{\phi (z_{n})}(1-|\phi(z_{n})|^{2})^{\alpha-1}},\qquad f''_{n} \bigl(\phi(z_{n})\bigr)=\frac{2\alpha}{(1-|\phi(z_{n})|^{2})^{\alpha}}. $$

Then

$$\begin{aligned} 1 \geq& S_{\varphi} \biggl(\frac{(DM_{u}C_{\phi}f_{n})'(z_{n})}{\| DM_{u}C_{\phi}f_{n}\|_{\mathcal{B}^{\varphi}}} \biggr) \\ \geq&\bigl(1-|z_{n}|^{2}\bigr)\varphi \biggl( \frac{ |\frac{2u'(z_{n})\phi '(z_{n})+u(z_{n})\phi''(z_{n})}{\overline{\phi(z_{n})}(1-|\phi(z_{n})|^{2})^{\alpha-1}}+ \frac{2\alpha u(z_{n})\phi'(z_{n})^{2}}{(1-|\phi(z_{n})|^{2})^{\alpha}} |}{ \|DM_{u}C_{\phi}f_{n}\|_{\mathcal{B}^{\varphi}}} \biggr). \end{aligned}$$

It follows that

$$\biggl\vert \frac{\mu(z_{n})|2u'(z_{n})\phi'(z_{n})+u(z_{n})\phi''(z_{n})|}{ |\phi(z_{n})|(1-|\phi(z_{n})|^{2})^{\alpha-1}} -\frac{2\alpha\mu(z_{n})|u(z_{n})||\phi'(z_{n})|^{2}}{(1-|\phi (z_{n})|^{2})^{\alpha}}\biggr\vert \leq \|DM_{u}C_{\phi}f_{n}\|_{\mathcal{B}^{\varphi}}. $$

Therefore

$$\lim_{|\phi(z_{n})|\rightarrow1}\frac{\mu(z_{n})|2u'(z_{n})\phi '(z_{n})+u(z_{n})\phi''(z_{n})|}{ (1-|\phi(z_{n})|^{2})^{\alpha-1}}=0. $$

By Lemma 1(iii), Lemma 2, Lemma 8 and Lemma 9, we can prove the converse implication similar to Theorem 10, so we omit the details. □

Theorem 13

Let $u\in\mathcal{H}(\mathcal{D})$ and ϕ be an analytic self-map of $\mathcal{D}$. Then $DM_{u}C_{\phi}:\mathcal{Z}^{2}\rightarrow \mathcal{B}^{\varphi}$ is compact if and only if $DM_{u}C_{\phi}:\mathcal{Z}^{2}\rightarrow\mathcal {B}^{\varphi}$ is bounded,

$$\begin{aligned}& \lim_{|\phi(z)|\rightarrow1}\mu(z)\bigl\vert u''(z) \bigr\vert \log\frac{e}{1-|\phi (z)|^{2}}=0, \end{aligned}$$

(34)

$$\begin{aligned}& \lim_{|\phi(z)|\rightarrow1}\frac{\mu(z)|2u'(z)\phi '(z)+u(z)\phi''(z)|}{1-|\phi(z)|^{2}}=0, \end{aligned}$$

(35)

$$\begin{aligned}& \lim_{|\phi(z)|\rightarrow1}\frac{\mu(z)|u(z)||\phi '(z)|^{2}}{(1-|\phi(z)|^{2})^{2}}=0. \end{aligned}$$

(36)

Proof

Suppose that $DM_{u}C_{\phi}:\mathcal{Z}^{2}\rightarrow\mathcal{B}^{\varphi}$ is compact. It is clear that $DM_{u}C_{\phi}:\mathcal{Z}^{2}\rightarrow\mathcal {B}^{\varphi}$ is bounded. By Lemma 2, we have that $k_{1}, k_{2}, k_{3}<\infty$. Let $\{z_{n}\}_{n\in\mathcal{N}}$ be a sequence in $\mathcal{D}$ such that $|\phi(z_{n})|\rightarrow1$ as $n\rightarrow\infty$. Without loss of generality, we may suppose that $|\phi(z_{n})|>\frac{1}{2}$ for all n. Then, by repeating the arguments in the proof of Theorem 10 and Theorem 12, (35) and (36) can be proved similarly, so we only need to show that (34) holds. Taking the function

$$ p_{n}(z)= \biggl(1+ \biggl(\log\frac{e}{1-\overline{\phi (z_{n})}z} \biggr)^{2} \biggr) \biggl(\log\frac{e}{1-|\phi (z_{n})|^{2}} \biggr)^{-1}. $$

(37)

Then we have

$$\begin{aligned}& p'_{n}(z)=\frac{2\overline{\phi(z_{n})}}{1-\overline{\phi (z_{n})}z} \biggl(\log \frac{e}{ 1-\overline{\phi(z_{n})}z} \biggr) \biggl(\log\frac{e}{1-|\phi (z_{n})|^{2}} \biggr)^{-1}, \end{aligned}$$

(38)

$$\begin{aligned}& p''_{n}(z)=\frac{2\overline{\phi(z_{n})}^{2}}{(1-\overline{\phi (z_{n})}z)^{2}} \biggl(\log \frac{e}{ 1-\overline{\phi(z_{n})}z}+1 \biggr) \biggl(\log\frac{e}{1-|\phi (z_{n})|^{2}} \biggr)^{-1}. \end{aligned}$$

(39)

It is easy to show that $\{p_{n}\}_{n\in\mathcal{N}}$ is a bounded sequence in $\mathcal{Z}^{2}$, and $p_{n}\rightarrow0$ uniformly on compact subsets of $\mathcal{D}$. Since $DM_{u}C_{\phi}:\mathcal{Z}^{2}\rightarrow \mathcal{B}^{\varphi}$ is compact, then $\lim_{n\rightarrow\infty}\|DM_{u}C_{\phi}p_{n}\|_{\mathcal {B}^{\varphi}}=0$.

From (37), (38) and (39), we can get that

$$\begin{aligned}& \mu(z_{n})\bigl\vert u''(z_{n}) \bigr\vert \biggl[\log\frac{e}{1-|\phi(z_{n})|^{2}}+ \biggl(\log\frac{e}{1-|\phi(z_{n})|^{2}} \biggr)^{-1} \biggr] \\& \qquad {}- \frac{2\mu(z_{n})|\phi(z_{n})||2u'(z_{n})\phi'(z_{n})+u(z_{n})\phi ''(z_{n})|}{1-|\phi(z_{n})|^{2}} \\& \qquad {}-\frac{2\mu(z_{n})|\phi(z_{n})|^{2}|u(z_{n})||\phi '(z_{n})|^{2} [1+ (\log\frac{e}{1-|\phi(z_{n})|^{2}} )^{-1} ]}{ (1-|\phi(z_{n})|^{2})^{2}} \\& \quad \leq\|DM_{u}C_{\phi}p_{n}\|_{\mathcal{B}^{\varphi}}. \end{aligned}$$

Since $\lim_{n\rightarrow\infty} (\log\frac{e}{1-|\phi (z_{n})|^{2}} )^{-1}=0$, and by (35) and (36), we can get (34).

By Lemma 1(iii) and (v), Lemma 2 and Lemma 8, we can prove the converse implication similar to Theorem 10, so we omit the details. □

Theorem 14

Let $u\in\mathcal{H}(\mathcal{D})$, ϕ be an analytic self-map of $\mathcal{D}$ and $\alpha>2$. Then $DM_{u}C_{\phi}:\mathcal{Z}^{\alpha}\rightarrow\mathcal {B}^{\varphi}$ is compact if and only if $DM_{u}C_{\phi}:\mathcal{Z}^{\alpha}\rightarrow\mathcal {B}^{\varphi}$ is bounded,

$$\begin{aligned}& \lim_{|\phi(z)|\rightarrow1}\frac{\mu(z)|u''(z)|}{(1-|\phi (z)|^{2})^{\alpha-2}}=0, \end{aligned}$$

(40)

$$\begin{aligned}& \lim_{|\phi(z)|\rightarrow1}\frac{\mu(z)|2u'(z)\phi '(z)+u(z)\phi''(z)|}{(1-|\phi(z)|^{2})^{\alpha-1}}=0, \end{aligned}$$

(41)

$$\begin{aligned}& \lim_{|\phi(z)|\rightarrow1}\frac{\mu(z)|u(z)||\phi '(z)|^{2}}{(1-|\phi(z)|^{2})^{\alpha}}=0. \end{aligned}$$

(42)

Proof

Suppose that $DM_{u}C_{\phi}:\mathcal{Z}^{\alpha}\rightarrow\mathcal {B}^{\varphi}$ is compact. It is clear that $DM_{u}C_{\phi}:\mathcal{Z}^{\alpha}\rightarrow \mathcal{B}^{\varphi}$ is bounded. By Lemma 2, we have that $k_{1}, k_{2}, k_{3}<\infty$. Let $\{z_{n}\}_{n\in\mathcal{N}}$ be a sequence in $\mathcal{D}$ such that $|\phi(z_{n})|\rightarrow1$ as $n\rightarrow\infty$. Without loss of generality, we may suppose that $|\phi(z_{n})|>\frac{1}{2}$ for all n. Then, by repeating the arguments in the proof of Theorem 10 and Theorem 12, (41) and (42) can be proved similarly, so we only need to show that (40) holds.

Now let $q_{n}(z)=\frac{(1-|\phi(z_{n})|^{2})^{2}}{(1-\overline{\phi (z_{n})}z)^{\alpha}}$, then $\sup_{n\in\mathcal{N}}\|q_{n}\|_{\mathcal{Z}^{\alpha }}<\infty$, $q_{n}\rightarrow0$ uniformly on compact subsets of $\mathcal{D}$, and

$$\begin{aligned}& q_{n}\bigl(\phi(z_{n})\bigr)=\frac{1}{(1-|\phi (z_{n})|^{2})^{\alpha-2}}, \\& q'_{n}\bigl(\phi(z_{n})\bigr)= \frac{\alpha\overline{\phi(z_{n})}}{(1-|\phi (z_{n})|^{2})^{\alpha-1}},\qquad q''_{n}\bigl( \phi(z_{n})\bigr)=\frac{\alpha(\alpha +1)\overline{\phi(z_{n})}^{2}}{(1-|\phi(z_{n})|^{2})^{\alpha}}. \end{aligned}$$

Then we have

$$\begin{aligned}& \frac{\mu(z_{n})|u''(z_{n})|}{(1-|\phi (z_{n})|^{2})^{\alpha-2}} \\& \quad \leq\|DM_{u}C_{\phi}q_{n}\|_{\mathcal {B}^{\varphi}} \\& \qquad {}+\alpha\bigl\vert \phi(z_{n})\bigr\vert \mu(z_{n})\biggl\vert \frac{2u'(z_{n})\phi'(z_{n})+u(z_{n})\phi ''(z_{n})}{(1-|\phi(z_{n})|^{2})^{\alpha-1}}+ (\alpha+1)\overline{ \phi(z_{n})}\frac{u(z_{n})\phi '(z_{n})^{2}}{(1-|\phi(z_{n})|^{2})^{\alpha}}\biggr\vert \\& \quad \leq\|DM_{u}C_{\phi}q_{n}\|_{\mathcal {B}^{\varphi}}+ \alpha \mu(z_{n})\biggl\vert \frac{2u'(z_{n})\phi'(z_{n})+u(z_{n})\phi''(z_{n})}{ \overline{\phi(z_{n})}(1-|\phi(z_{n})|^{2})^{\alpha-1}} + \frac{\alpha+1}{2\alpha}\frac{2\alpha u(z_{n})\phi'(z_{n})^{2}}{ (1-|\phi(z_{n})|^{2})^{\alpha}}\biggr\vert \\& \quad \leq\|DM_{u}C_{\phi}q_{n}\|_{\mathcal {B}^{\varphi}} \\& \qquad {}+\alpha \mu(z_{n}) \biggl\{ \biggl\vert \frac{2u'(z_{n})\phi'(z_{n})+u(z_{n})\phi''(z_{n})}{ \overline{\phi(z_{n})}(1-|\phi(z_{n})|^{2})^{\alpha-1}} \biggr\vert + \frac{\alpha+1}{2\alpha}\biggl\vert \frac{2\alpha u(z_{n})\phi'(z_{n})^{2}}{ (1-|\phi(z_{n})|^{2})^{\alpha}}\biggr\vert \biggr\} . \end{aligned}$$

(43)

Since $\frac{\alpha+1}{2\alpha}<1$, then by (41), (42) and letting $n\rightarrow\infty$ in (43), we can get (40).

For the converse, by Lemma 1(iii) and (vi), Lemma 2 and Lemma 8, we can prove the converse implication similar to Theorem 10, so we omit the details. □