1 Introduction

Let \(\mathbb{D}\) be an open unit disk in the complex plane ℂ and \(H(\mathbb{D})\) be the space of analytic functions on \(\mathbb {D}\). For \(0<\alpha <\infty\), the Bloch-type space (or α-Bloch space) \(\mathcal{B}^{\alpha}\) is the space that consists of all analytic functions f on \(\mathbb{D}\) such that

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

\(\mathcal{B}^{\alpha}\) becomes a Banach space under the norm \(\|f\|_{\mathcal{B}^{\alpha}}=|f(0)|+B_{\alpha }(f)\). When \(\alpha=1\), \(\mathcal{B}^{1}=\mathcal{B}\) is the well-known Bloch space. See [1, 2] for more information on Bloch-type spaces.

Throughout this paper, φ denotes a nonconstant analytic self-map of \(\mathbb{D}\). The composition operator \(C_{\varphi}\) induced by φ is defined by \(C_{\varphi}f = f \circ\varphi\) for \(f \in H(\mathbb{D})\). For a fixed \(u \in H(\mathbb{D})\), define a linear operator \(uC_{\varphi}\) as follows:

$$uC_{\varphi}f =u ( f\circ\varphi) ,\quad f \in H(\mathbb{D}). $$

The operator \(uC_{\varphi}\) is called the weighted composition operator. The weighted composition operator is a generalization of the composition operator and the multiplication operator defined by \(M_{u}f=uf\).

A basic problem concerning composition operators on various Banach function spaces is to relate the operator theoretic properties of \(C_{\varphi}\) to the function theoretic properties of the symbol φ, which attracted a lot of attention recently; the reader can refer to [3].

The differentiation operator D is defined by \(Df=f'\), \(f\in H(\mathbb{D})\). For a nonnegative integer n, we define

$$\bigl(D^{0} f\bigr) (z)=f(z),\qquad \bigl(D^{n} f\bigr) (z)=f^{(n)}(z),\quad n\ge1, f \in H(\mathbb{D}). $$

Let φ be an analytic self-map of \(\mathbb{D}\), \(u \in H(\mathbb {D})\), and let n be a nonnegative integer. Define the linear operator \(D^{n}_{\varphi, u}\), called the generalized weighted composition operator, by (see [4–6])

$$\begin{aligned} \bigl(D^{n}_{\varphi, u} f\bigr) (z) =u(z)\cdot\bigl(D^{n} f\bigr) \bigl(\varphi(z)\bigr) ,\quad f \in H(\mathbb{D}), z\in\mathbb{D}. \end{aligned}$$

When \(n=0\) and \(u(z)=1\), \(D^{n}_{\varphi,u}\) is the composition operator \(C_{\varphi }\). If \(n=0\), then \(D^{n}_{\varphi,u}\) is the weighted composition operator \(uC_{\varphi }\). If \(n=1\), \(u(z)=\varphi'(z)\), then \(D^{n}_{\varphi, u}= DC_{\varphi}\), which was studied in [7–10]. For \(u(z)=1\), \(D^{n}_{\varphi, u}= C_{\varphi}D^{n}\), which was studied in [7, 11–14]. For the study of the generalized weighted composition operator on various function spaces, see, for example, [4–6, 15–19].

It is well known that the composition operator is bounded on the Bloch space by the Schwarz-Pick lemma. Composition operators and weighted composition operators on Bloch-type spaces were studied, for example, in [20–28]. The product-type operators on or into Bloch-type spaces have been studied in many papers recently, see [7–11, 13, 14, 18, 29–36] for example. In [27], Wulan et al. obtained a characterization for the compactness of the composition operators acting on the Bloch space as follows.

Theorem A

Let φ be an analytic self-map of \(\mathbb{D}\). Then \(C_{\varphi}: \mathcal{B}\rightarrow \mathcal{B}\) is compact if and only if

$$\lim_{j\rightarrow\infty}\bigl\| \varphi^{j} \bigr\| _{\mathcal{B}}=0. $$

In [14], Wu and Wulan obtained two characterizations for the compactness of the product of differentiation and composition operators acting on the Bloch space as follows.

Theorem B

Let φ be an analytic self-map of \(\mathbb{D}\), \(n\in \mathbb {N}\). Then the following statements are equivalent.

  1. (a)

    \(C_{\varphi}D^{n}:\mathcal{B}\rightarrow \mathcal{B}\) is compact.

  2. (b)

    \(\lim_{j\rightarrow\infty}\|C_{\varphi}D^{n} I^{j} \|_{\mathcal{B}}=0\), where \(I^{j}(z)=z^{j}\).

  3. (c)

    \(\lim_{|a|\rightarrow1}\|C_{\varphi}D^{n}\sigma_{a}(z)\|_{\mathcal{B}}=0\), where \(\sigma_{a}(z)=(a-z)/(1-\overline{a}z)\) is the Möbius map on \(\mathbb{D}\).

Motivated by Theorems A and B, in this work we show that \(D^{n}_{\varphi,u}:\mathcal {B}^{\alpha }\to\mathcal{B}^{\beta}\) is bounded (respectively, compact) if and only if the sequence \((j^{\alpha -1}\|D^{n}_{\varphi,u}I^{j}\|_{\mathcal{B}^{\beta}})_{j=n}^{\infty}\) is bounded (respectively, convergent to 0 as \(j\to\infty\)), where \(I^{j}(z)=z^{j}\). Moreover, we use two families of functions to characterize the boundedness and compactness of the operator \(D^{n}_{\varphi, u}\).

Throughout the paper, we denote by C a positive constant which may differ from one occurrence to the next. In addition, we say that \(A\preceq B\) if there exists a constant C such that \(A\leq CB\). The symbol \(A\approx B\) means that \(A \preceq B \preceq A\).

2 Main results and proofs

In this section, we give our main results and proofs. First we characterize the boundedness of the operator \(D^{n}_{\varphi,u}:\mathcal{B}^{\alpha }\to\mathcal{B}^{\beta}\).

Theorem 1

Let n be a positive integer, \(0<\alpha , \beta<\infty\), \(u \in H(\mathbb{D})\) and φ be an analytic self-map of \(\mathbb{D}\). Then the following statements are equivalent.

  1. (a)

    The operator \(D^{n}_{\varphi, u} : \mathcal{B}^{\alpha}\to \mathcal{B}^{\beta}\) is bounded.

  2. (b)

    \(\sup_{j\geq n} j^{ \alpha-1}\|D^{n}_{\varphi, u} I^{j}(z)\|_{\mathcal{B}^{\beta}}<\infty\), where \(I^{j}(z)=z^{j}\).

  3. (c)

    \(u\in\mathcal{B}^{\beta}\), \(\sup_{z\in\mathbb {D}}(1-|z|^{2})^{\beta}|u(z) | |\varphi'(z)|<\infty\) and

    $$\sup_{a\in \mathbb {D}}\bigl\| D^{n}_{\varphi,u}f_{a} \bigr\| _{\mathcal{B}^{\beta}} < \infty,\qquad \sup_{a\in \mathbb {D}}\bigl\| D^{n}_{\varphi,u}h_{a}\bigr\| _{\mathcal{B}^{\beta}} <\infty, $$

    where

    $$f_{a}(z)=\frac{1-|a|^{2}}{(1-\overline{a} z)^{\alpha}} \quad\textit{and} \quad h_{a}(z)= \frac{(1-|a|^{2})^{2}}{(1-\overline{a} z)^{\alpha+1}},\quad z\in \mathbb {D}. $$
  4. (d)
    $$\sup_{z\in\mathbb{D} } \frac{(1-|z |^{2})^{\beta}|u(z)|| \varphi' (z) | }{(1-|\varphi(z)|^{2})^{\alpha +n}} < \infty \quad\textit{and}\quad \sup _{z\in\mathbb{D} } \frac{(1-|z|^{2})^{\beta}|u'(z)|}{(1-|\varphi(z)|^{2})^{\alpha +n-1}}<\infty . $$

Proof

(a) ⇒ (b) This implication is obvious, since for \(j\in\mathbb{N}\), the function \(j^{ \alpha-1} I^{j}\) is bounded in \(\mathcal{B}^{\alpha }\) and \(j^{ \alpha-1}\|I^{j}\|_{\mathcal{B}^{\alpha }} \approx1\).

(b) ⇒ (c) Assume that (b) holds and let \(Q=\sup_{j\ge n}j^{ \alpha-1}\|D^{n}_{\varphi,u}I^{j}\|_{\mathcal{B}^{\beta}} \). For any \(a\in \mathbb {D}\), it is easy to see that \(f_{a}\) and \(h_{a}\) have bounded norms in \(\mathcal{B}^{\alpha}\). It is clear that

$$\begin{aligned}& f_{a}(z)=\bigl(1-|a|^{2}\bigr)\sum _{j=0}^{\infty}\frac{\Gamma(j+\alpha)}{j!\Gamma (\alpha)} \overline{a}^{j}z^{j}, \\& h_{a}(z)=\bigl(1-|a|^{2}\bigr)^{2}\sum _{j=0}^{\infty}\frac{\Gamma(j+1+\alpha )}{j!\Gamma (\alpha+1)}\overline{a}^{j}z^{j}. \end{aligned}$$

By Stirling’s formula, we have \(\frac{\Gamma(j+\alpha)}{j!\Gamma(\alpha)}\approx j^{\alpha-1} \) as \(j\rightarrow\infty\). Using linearity we get

$$\begin{aligned}& \bigl\| D^{n}_{\varphi,u}f_{a}\bigr\| _{\mathcal{B}^{\beta}} \le C\bigl(1-|a|^{2}\bigr) \sum_{j=0}^{\infty}|a|^{j} j^{\alpha-1}\bigl\| D^{n}_{\varphi,u}I^{j}\bigr\| _{\mathcal{B}^{\beta}} \preceq Q\quad\mbox{and }\\& \bigl\| D^{n}_{\varphi,u}h_{a}\bigr\| _{\mathcal{B}^{\beta}} \le C\bigl(1-|a|^{2} \bigr)^{2}\sum_{j=0}^{\infty}(j+1)|a|^{j} j^{\alpha-1}\bigl\| D^{n}_{\varphi,u}I^{j}\bigr\| _{\mathcal{B}^{\beta}} \preceq Q. \end{aligned}$$

Therefore, by the arbitrariness of \(a\in \mathbb {D}\),

$$\sup_{a\in \mathbb {D}}\bigl\| D^{n}_{\varphi,u}f_{a} \bigr\| _{\mathcal{B}^{\beta}} < \infty,\qquad \sup_{a\in \mathbb {D}}\bigl\| D^{n}_{\varphi,u}h_{a}\bigr\| _{\mathcal{B}^{\beta}} <\infty. $$

In addition, applying the operator \(D^{n}_{\varphi, u}\) to \(I^{j}\) with \(j=n, n+1\), we obtain

$$\begin{aligned}& \bigl(D^{n}_{\varphi,u}I^{n}\bigr)'(z)=u'(z)n! \quad\mbox{and}\\& \bigl(D^{n}_{\varphi,u}I^{n+1}\bigr)'(z)=u'(z) (n+1)! \varphi (z)+u(z) (n+1)!\varphi '(z), \end{aligned}$$

while for \(j< n\), \((D^{n}_{\varphi,u}I^{j})'(z)=0\). Thus, using the boundedness of the function φ, we have \(u\in\mathcal{B}^{\beta}\) and \(\sup_{z\in \mathbb{D}}(1-|z|^{2})^{\beta}|u(z) | |\varphi'(z)|<\infty\).

(c) ⇒ (d) Assume that (c) holds. Let

$$C_{1}:=\sup_{a\in \mathbb {D}} \bigl\| D^{n}_{\varphi,u}f_{a} \bigr\| _{\mathcal{B}^{\beta}},\qquad C_{2}:= \sup_{a\in \mathbb {D}}\bigl\| D^{n}_{\varphi,u}h_{a}\bigr\| _{\mathcal{B}^{\beta}} . $$

For \(w\in\mathbb{D}\), set

$$g_{w}(z)=\frac{1-|w|^{2}}{(1-\overline{w} z)^{\alpha} } - \frac{\alpha}{\alpha+n}\frac{(1-|w|^{2})^{2}}{(1-\overline{w} z)^{\alpha+1 }} ,\quad w \in \mathbb {D}. $$

It is easy to check that \(g_{w}\in\mathcal{B}^{\alpha }\), \(\|g_{w}\|_{\mathcal{B}^{\alpha }} <\infty\) for every \(w\in\mathbb{D}\). Moreover,

$$\begin{aligned} \sup_{w\in \mathbb {D}}\bigl\| D^{n}_{\varphi,u}g_{w} \bigr\| _{\mathcal{B}^{\beta}} \leq& \sup_{w\in \mathbb {D}}\bigl\| D^{n}_{\varphi,u}f_{w}\bigr\| _{\mathcal{B}^{\beta}}+ \frac{\alpha}{\alpha+n} \sup_{w\in \mathbb {D}}\bigl\| D^{n}_{\varphi,u}h_{w} \bigr\| _{\mathcal{B}^{\beta}}\\ \leq& C_{1}+\frac{\alpha }{\alpha +n}C_{2} < \infty. \end{aligned}$$

In addition,

$$g^{(n)}_{\varphi(\lambda)}\bigl(\varphi(\lambda)\bigr)=0, \qquad \bigl|g^{(n+1)}_{\varphi(\lambda)}\bigl(\varphi(\lambda)\bigr)\bigr|=\alpha (\alpha +1)\cdot \cdot \cdot (\alpha +n-1) \frac{|\varphi(\lambda)|^{n+1}}{(1-|\varphi(\lambda)|^{2})^{\alpha +n}}. $$

It follows that

$$\begin{aligned} C_{1}+\frac{\alpha }{\alpha +n}C_{2} >& \bigl\| D^{n}_{\varphi ,u}g_{\varphi (\lambda)}\bigr\| _{\mathcal{B}^{\beta}} \\ \geq& \alpha (\alpha +1)\cdot\cdot\cdot(\alpha +n-1) \frac{(1-|\lambda|^{2})^{\beta}|u(\lambda)||\varphi'(\lambda)| |\varphi(\lambda)|^{n+1}}{(1-|\varphi(\lambda)|^{2})^{\alpha +n}} \end{aligned}$$
(2.1)

for any \(\lambda\in \mathbb {D}\). For any fixed \(r\in (0,1)\), from (2.1) we have

$$\begin{aligned} \sup_{|\varphi(\lambda)|>r} \frac{(1-|\lambda|^{2})^{\beta}|u(\lambda)||\varphi'(\lambda)| }{(1-|\varphi(\lambda)|^{2})^{\alpha +n}} \leq& \sup_{|\varphi(\lambda)|>r} \frac{1}{r^{n+1}} \frac{(1-|\lambda|^{2})^{\beta}|u(\lambda)||\varphi'(\lambda)| |\varphi(\lambda)|^{n+1}}{(1-|\varphi(\lambda)|^{2})^{\alpha +n}} \\ \leq& \frac{ C_{1}+\frac{\alpha }{\alpha +n}C_{2} }{r^{n+1}\alpha (\alpha +1)\cdot\cdot\cdot (\alpha +n-1)} < \infty. \end{aligned}$$
(2.2)

From the assumption that \(\sup_{z\in\mathbb{D}}(1-|z|^{2})^{\beta}|u(z) | |\varphi'(z)|<\infty\), we get

$$\begin{aligned} \sup_{|\varphi(\lambda)|\leq r}\frac{(1-|\lambda|^{2})^{\beta}|u(\lambda)||\varphi'(\lambda)| }{(1-|\varphi(\lambda)|^{2})^{\alpha +n}} \leq \frac{ \sup_{|\varphi(\lambda)|\leq r} (1-|\lambda|^{2})^{\beta}|u(\lambda)||\varphi'(\lambda )|}{(1-r^{2})^{\alpha +n}} < \infty. \end{aligned}$$
(2.3)

Therefore, (2.2) and (2.3) yield the first inequality of (d).

Next, note that

$$\begin{aligned} &C_{1}\ge\bigl\| D^{n}_{\varphi,u}f_{\varphi(\lambda)} \bigr\| _{\mathcal {B}^{\beta}} \\ &\quad \geq \alpha (\alpha +1)\cdot\cdot\cdot(\alpha +n-1) \frac{(1-|\lambda|^{2})^{\beta}|u'(\lambda)||\varphi(\lambda )|^{n}}{(1-|\varphi (\lambda)|^{2})^{\alpha +n-1}} \\ &\qquad{}-\alpha (\alpha +1)\cdot\cdot\cdot(\alpha +n)\frac{ (1-|\lambda|^{2})^{\beta}|u(\lambda)||\varphi'(\lambda)| |\varphi(\lambda)|^{n+1}}{(1-|\varphi(\lambda)|^{2})^{\alpha +n}} \end{aligned}$$

for any \(\lambda\in \mathbb {D}\). From (2.1) we get

$$\begin{aligned} &\frac{(1-|\lambda|^{2})^{\beta}|u'(\lambda)||\varphi(\lambda )|^{n}}{(1-|\varphi(\lambda)|^{2})^{\alpha +n-1}}\\ &\quad \leq \frac{ \|D^{n}_{\varphi,u}f_{\varphi(\lambda)}\|_{\mathcal{B}^{\beta}}}{ \alpha (\alpha +1)\cdot\cdot\cdot(\alpha +n-1)} + \frac{(\alpha +n)(1-|\lambda|^{2})^{\beta}|u(\lambda)||\varphi'(\lambda)| |\varphi(\lambda)|^{n+1}}{(1-|\varphi(\lambda)|^{2})^{\alpha +n}}\\ &\quad \leq \frac{ C_{1}}{\alpha (\alpha +1)\cdot\cdot\cdot(\alpha +n-1)} + \frac{(\alpha +n) C_{1}+\alpha C_{2}}{\alpha (\alpha +1)\cdot\cdot\cdot(\alpha +n-1)}\\ &\quad \leq \frac{ (\alpha +n+1)C_{1}+\alpha C_{2}}{\alpha (\alpha +1)\cdot\cdot\cdot(\alpha +n-1)} . \end{aligned}$$

By arbitrary \(\lambda\in\mathbb{D} \), we get

$$\begin{aligned} \sup_{\lambda\in\mathbb{D} }\frac{(1-|\lambda|^{2})^{\beta}|u'(\lambda)||\varphi(\lambda )|^{n}}{(1-|\varphi (\lambda)|^{2})^{\alpha +n-1} } < \infty. \end{aligned}$$
(2.4)

Combining (2.4) with the fact that \(u \in\mathcal{B}^{\beta}\), similarly to the former proof, we get the second inequality of (d).

(d) ⇒ (a) For any \(f \in\mathcal{B}^{\alpha }\), we have

$$\begin{aligned} &\bigl(1-|z |^{2}\bigr)^{\beta}\bigl| \bigl(D^{n}_{\varphi,u} f\bigr)'(z) \bigr| \\ &\quad=\bigl(1-|z |^{2}\bigr)^{\beta}\bigl| \bigl(f^{(n)}(\varphi)u \bigr)'(z) \bigr| \\ &\quad\leq \bigl(1-|z |^{2}\bigr)^{\beta}\bigl|u(z)\bigr|\bigl| \varphi' (z) \bigr| \bigl|f^{(n+1)}\bigl(\varphi(z)\bigr)\bigr|+ \bigl(1-|z |^{2}\bigr)^{\beta}\bigl| u' (z) \bigr| \bigl|f^{(n)} \bigl(\varphi(z)\bigr)\bigr| \\ &\quad \leq C\frac{(1-|z |^{2})^{\beta}|u(z)|| \varphi' (z) | }{(1-|\varphi(z)|^{2})^{\alpha +n}}\|f\|_{\mathcal{B}^{\alpha }} +C \frac{(1-|z|^{2})^{\beta}|u'(z)|}{(1-|\varphi(z)|^{2})^{\alpha +n-1}}\|f\| _{\mathcal {B}^{\alpha }}, \end{aligned}$$
(2.5)

where in the last inequality we used the fact that for \(f \in \mathcal{B}^{\alpha }\) (see [2])

$$\sup_{z\in \mathbb {D}}\bigl(1-|z|^{2}\bigr)^{\alpha }\bigl|f'(z)\bigr| \asymp \bigl|f'(0)\bigr|+\cdots+\bigl|f^{(n)}(0)\bigr|+\sup _{z\in \mathbb {D}}\bigl(1-|z|^{2}\bigr)^{\alpha +n}\bigl|f^{(n+1)}(z)\bigr|. $$

Moreover

$$\bigl|\bigl(D^{n}_{\varphi,u} f\bigr) (0)\bigr|=\bigl|f^{(n)}\bigl( \varphi(0) \bigr)u(0) \bigr|\leq\frac{|u(0) |}{(1-|\varphi(0)|^{2})^{\alpha +n-1}}\|f\|_{\mathcal{B}^{\alpha }}. $$

From (d) we see that

$$\bigl\| D^{n}_{\varphi,u} f\bigr\| _{\mathcal{B}^{\beta}}=\bigl|\bigl(D^{n}_{\varphi,u} f\bigr) (0)\bigr|+ \sup_{z\in \mathbb {D}}\bigl(1-|z |^{2} \bigr)^{\beta}\bigl| \bigl(D^{n}_{\varphi,u} f\bigr)'(z) \bigr|< \infty. $$

Therefore the operator \(D^{n}_{\varphi,u}:\mathcal{B}^{\alpha }\rightarrow\mathcal{B}^{\beta}\) is bounded. The proof is complete. □

For the study of the compactness of \(D^{n}_{\varphi,u}:\mathcal{B}^{\alpha}\rightarrow\mathcal{B}^{\beta}\), we need the following lemma, which can be proved in a standard way; see, for example, Proposition 3.11 in [3].

Lemma 2

Let n be a positive integer, \(0<\alpha , \beta<\infty\), \(u \in H(\mathbb{D})\) and φ be an analytic self-map of \(\mathbb{D}\). Then \(D^{n}_{\varphi,u}:\mathcal{B}^{\alpha}\rightarrow\mathcal{B}^{\beta}\) is compact if and only if \(D^{n}_{\varphi,u}:\mathcal{B}^{\alpha}\rightarrow\mathcal{B}^{\beta}\) is bounded and for any bounded sequence \((f_{j})_{j\in{ \mathbb {N}}}\) in \(\mathcal{B}^{\alpha}\) which converges to zero uniformly on compact subsets of \(\mathbb{D}\), \(\|D^{n}_{\varphi,u} f_{j} \|_{\mathcal {B}^{\beta}}\to0\) as \(j\to\infty\).

Theorem 3

Let n be a positive integer, \(0<\alpha , \beta<\infty\), \(u \in H(\mathbb{D})\) and φ be an analytic self-map of \(\mathbb{D}\) such that \(D^{n}_{\varphi,u}:\mathcal{B}^{\alpha}\to\mathcal{B}^{\beta}\) is bounded. Then the following statements are equivalent.

  1. (a)

    \(D^{n}_{\varphi,u}:\mathcal{B}^{\alpha}\to\mathcal{B}^{\beta}\) is compact.

  2. (b)

    \(\lim_{j\rightarrow\infty} j^{\alpha -1}\|D^{n}_{\varphi, u} I^{j} \|_{\mathcal{B}^{\beta}}=0\), where \(I^{j}(z)=z^{j}\).

  3. (c)

    \(\lim_{|\varphi (a)|\to1}\|D^{n}_{\varphi,u}f_{\varphi(a)}\|_{\mathcal{B}^{\beta}}=0\) and \(\lim_{|\varphi (a)|\to1}\|D^{n}_{\varphi,u}h_{\varphi(a)}\|_{\mathcal{B}^{\beta}}=0\).

  4. (d)
    $$\lim_{|\varphi(z)|\rightarrow1}\frac{(1-|z |^{2})^{\beta}|u(z)||\varphi'(z)|}{(1-|\varphi(z)|^{2})^{n+\alpha }}=0 \quad \textit{and}\quad \lim _{ |\varphi (z)|\rightarrow1}\frac{(1-|z|^{2})^{\beta}|u'(z)|}{(1-|\varphi(z)|^{2})^{n+\alpha -1}}=0. $$

Proof

(a) ⇒ (b) Assume that \(D^{n}_{\varphi,u}:\mathcal {B}^{\alpha}\to\mathcal{B}^{\beta}\) is compact. Since the sequence \(\{j^{\alpha -1}I^{j}\}\) is bounded in \(\mathcal{B}^{\alpha}\) and converges to 0 uniformly on compact subsets, by Lemma 2 it follows that \(j^{\alpha -1}\|D^{n}_{\varphi, u} I^{j}\| _{\mathcal{B}^{\beta}} \to0\) as \(j\to\infty\).

(b) ⇒ (c) Suppose that (b) holds. Fix \(\varepsilon >0\) and choose \(N\in \mathbb {N}\) such that \(j^{\alpha -1}\|D^{n}_{\varphi,u}I^{j}\|_{\mathcal {B}^{\beta}} <\varepsilon \) for all \(j\ge N\). Let \(z_{k} \in \mathbb{D}\) such that \(|\varphi (z_{k})|\to1\) as \(k\to\infty\). Arguing as in the proof of Theorem 1, we have

$$\begin{aligned} &\bigl\| D^{n}_{\varphi,u}f_{\varphi(z_{k})}\bigr\| _{\mathcal{B}^{\beta}} \\ &\quad\le C \bigl(1-\bigl|\varphi (z_{k})\bigr|^{2} \bigr)\sum_{j=0}^{\infty}\bigl|\varphi (z_{k})\bigr|^{j}j^{\alpha -1} \bigl\| D^{n}_{\varphi,u}I^{j}\bigr\| _{\mathcal {B}^{\beta}}\\ &\quad=C \bigl(1-\bigl|\varphi (z_{k})\bigr|^{2}\bigr) \Biggl(\sum _{j=0}^{N-1} \bigl|\varphi (z_{k})\bigr|^{j}j^{\alpha -1} \bigl\| D^{n}_{\varphi,u}I^{j}\bigr\| _{\mathcal{B}^{\beta}} + \sum_{j=N}^{\infty} \bigl|\varphi (z_{k})\bigr|^{j}j^{\alpha -1}\bigl\| D^{n}_{\varphi,u}I^{j} \bigr\| _{\mathcal{B}^{\beta}} \Biggr)\\ &\quad\le CQ\bigl(1-\bigl|\varphi (z_{k})\bigr|^{N}\bigr) + C\varepsilon , \end{aligned}$$

where \(Q=\sup_{j\ge n}j^{\alpha -1}\|D^{n}_{\varphi,u}I^{j}\|_{\mathcal{B}^{\beta}} \). Since \(|\varphi (z_{k})|\to1\) as \(k\to\infty\), from the last inequality and the arbitrariness of ε, we get \(\lim_{k\rightarrow\infty}\|D^{n}_{\varphi,u}f_{\varphi(z_{k})}\|_{\mathcal{B}^{\beta}} =0\), i.e., \(\lim_{|\varphi (a)|\to 1}\|D^{n}_{\varphi,u}f_{\varphi(a)}\|_{\mathcal{B}^{\beta}} =0\).

Notice that

$$\sum_{j=0}^{N-1}(j+1)r^{j}= \frac{1-r^{N}-Nr^{N}(1-r)}{(1-r)^{2}},\quad 0\le r< 1, $$

arguing as in the proof of Theorem 1, we get

$$\begin{aligned} \bigl\| D^{n}_{\varphi,u}h_{\varphi (z_{k})}\bigr\| _{\mathcal{B}^{\beta}} \le& C\bigl(1-\bigl|\varphi (z_{k})\bigr|^{2} \bigr)^{2}\sum_{j=0}^{\infty}\bigl| \varphi (z_{k})\bigr|^{j}j^{\alpha }\bigl\| D^{n}_{\varphi,u}I^{j} \bigr\| _{\mathcal {B}^{\beta}}\\ \leq& C\bigl(1-\bigl|\varphi (z_{k})\bigr|^{2}\bigr)^{2}\sum _{j=0}^{N-1} (j+1)\bigl|\varphi (z_{k})\bigr|^{j}j^{\alpha -1} \bigl\| D^{n}_{\varphi,u}I^{j}\bigr\| _{\mathcal{B}^{\beta}}\\ &{} + C\bigl(1-\bigl|\varphi (z_{k})\bigr|^{2}\bigr)^{2}\sum _{j=N}^{\infty}(j+1)\bigl|\varphi (z_{k})\bigr|^{j}j^{\alpha -1} \bigl\| D^{n}_{\varphi,u}I^{j}\bigr\| _{\mathcal{B}^{\beta}}\\ \le& C (1-\bigl|\varphi (z_{k})\bigr|^{N}-N\bigl|\varphi (z_{k})\bigr|^{N} \bigl(1-\bigl|\varphi (z_{k})\bigr| \bigr)+ C\varepsilon . \end{aligned}$$

Therefore, \(\lim_{k\to\infty}\|D^{n}_{\varphi,u}h_{\varphi (z_{k})}\|_{\mathcal{B}^{\beta}} \le C\varepsilon \). By the arbitrariness of ε, we obtain the desired result.

(c) ⇒ (d) To prove (d) we only need to show that if \((z_{k})_{k\in \mathbb {N}}\) is a sequence in \(\mathbb{D}\) such that \(|\varphi(z_{k})| \rightarrow1\) as \(k\rightarrow\infty\), then

$$\lim_{k\to\infty}\frac{(1-|z_{k} |^{2})^{\beta}|u(z_{k})||\varphi'(z_{k})|}{(1-|\varphi(z_{k})|^{2})^{\alpha +n}} =0, \qquad \lim_{k\to\infty} \frac{(1-|z_{k}|^{2})^{\beta}|u'(z_{k})|}{(1-|\varphi (z_{k})|^{2})^{\alpha +n-1}}=0. $$

Let \((z_{k})_{k\in \mathbb {N}}\) be such a sequence that \(|\varphi(z_{k})| \rightarrow1\) as \(k\rightarrow\infty\). Arguing as in the proof of Theorem 1, we obtain

$$\lim_{k\to\infty}\bigl\| D^{n}_{\varphi,u}g_{\varphi (z_{k})}\bigr\| _{\mathcal{B}^{\beta}} \le \lim _{k\to\infty}\bigl\| D^{n}_{\varphi,u}f_{\varphi (z_{k})}\bigr\| _{\mathcal{B}^{\beta}} + \frac {\alpha }{n+\alpha }\lim_{k\to\infty}\bigl\| D^{n}_{\varphi,u}h_{\varphi (z_{k})} \bigr\| _{\mathcal{B}^{\beta}} =0. $$

Hence \(\lim_{k\to\infty}\|D^{n}_{\varphi,u}g_{\varphi (z_{k})}\|_{\mathcal{B}^{\beta}} = 0\). Similarly to the proof of Theorem 1, we have

$$\frac{n! (1-|z_{k}|^{2})^{\beta}|u(z_{k})||\varphi'(z_{k})||\varphi(z_{k})|^{n+1} }{(1-|\varphi(z_{k})|^{2})^{\alpha +n}}\leq\bigl\| D^{n}_{\varphi,u} g_{\varphi (z_{k})} \bigr\| _{{\mathcal{B}^{\beta}}}\rightarrow0 \quad\mbox{as } k\rightarrow\infty, $$

which implies

$$\begin{aligned} \lim_{k\to\infty}\frac{(1-|z_{k}|^{2})^{\beta}|u(z_{k})||\varphi'(z_{k})| }{(1-|\varphi(z_{k})|^{2})^{\alpha +n}} = \lim_{k\to\infty}\frac{(1-|z_{k}|^{2})^{\beta}|u(z_{k})||\varphi'(z_{k})||\varphi(z_{k})|^{n+1} }{(1-|\varphi(z_{k})|^{2})^{\alpha +n}}=0. \end{aligned}$$
(2.6)

In addition,

$$\begin{aligned} &\bigl\| D^{n}_{\varphi,u} f_{\varphi (z_{k})} \bigr\| _{{\mathcal {B}^{\beta}} }+ \frac{ (n+1)!(1-|z_{k}|^{2})^{\beta}|u(z_{k})||\varphi'(z_{k})| |\varphi(z_{k})|^{n+1}}{(1-|\varphi(z_{k})|^{2})^{\alpha +n}}\\ &\quad\geq \frac{ n!(1-|z_{k}|^{2})^{\beta}|u'(z_{k})||\varphi(z_{k})|^{n}}{(1-|\varphi (z_{k})|^{2})^{\alpha +n-1}}. \end{aligned}$$

From (2.6) and the assumption that \(\|D^{n}_{\varphi,u} f_{\varphi (z_{k})} \|_{{\mathcal{B}^{\beta}} }\to0\) as \(k\to\infty\), we have

$$\lim_{k\to\infty} \frac{(1-|z_{k}|^{2})^{\beta}|u'(z_{k})|}{(1-|\varphi(z_{k})|^{2})^{n} } =\lim_{k\to\infty} \frac{(1-|z_{k}|^{2})^{\beta}|u'(z_{k})||\varphi(z_{k})|^{n}}{(1-|\varphi(z_{k})|^{2})^{\alpha +n-1}}=0, $$

as desired.

(d) ⇒ (a) Assume that \((f_{k})_{k\in \mathbb {N}}\) is a bounded sequence in \(\mathcal{B}^{\alpha }\) converging to 0 uniformly on compact subsets of \(\mathbb{D}\). By the assumption, for any \(\varepsilon>0\), there exists \(\delta\in(0,1)\) such that

$$\begin{aligned} \frac{(1-|z|^{2})^{\beta}|\varphi '(z)||u(z)|}{(1-|\varphi (z)|^{2})^{\alpha +n}}< \varepsilon \quad \mbox{and}\quad \frac{(1-|z|^{2})^{\beta}|u'(z)|}{(1-|\varphi(z)|^{2})^{\alpha +n-1}}< \varepsilon \end{aligned}$$
(2.7)

when \(\delta<|\varphi(z)|<1\). Suppose that \(D^{n}_{\varphi,u}:\mathcal{B}^{\alpha }\to\mathcal{B}^{\beta}\) is bounded, by Theorem 1, we have

$$\begin{aligned} C_{3}=\sup_{z \in\mathbb {D}}\bigl(1-|z|^{2} \bigr)^{\beta}\bigl|u'(z)\bigr| < \infty \end{aligned}$$
(2.8)

and

$$\begin{aligned} C_{4}=\sup_{z \in\mathbb{D}}\bigl(1-|z |^{2} \bigr)^{\beta} \bigl|u(z)\bigr|\bigl|\varphi'(z)\bigr| < \infty. \end{aligned}$$
(2.9)

Let \(K=\{ z\in\mathbb{D}:|\varphi(z)| \leq\delta\}\). Then by (2.8) and (2.9) we have that

$$\begin{aligned} & \sup_{z\in\mathbb{D}} \bigl(1-|z|^{2}\bigr)^{\beta}\bigl| \bigl(D^{n}_{\varphi ,u}f_{k}\bigr)'(z)\bigr|\\ &\quad\leq\sup_{z\in K}\bigl(1-|z|^{2}\bigr)^{\beta}\bigl|u(z)\bigr|\bigl| \varphi'(z)\bigr| \bigl|f_{k}^{(n+1)}\bigl(\varphi(z)\bigr)\bigr|+ \sup_{z\in K} \bigl(1-|z |^{2}\bigr)^{\beta}\bigl|u' (z) \bigr| \bigl|f_{k}^{(n)}\bigl(\varphi(z)\bigr)\bigr|\\ &\qquad{} +C\sup_{z\in\mathbb{D}\setminus K} \frac{(1-|z|^{2})^{\beta}|u(z)||\varphi'(z)|}{(1-|\varphi(z)|^{2})^{\alpha +n}} \|f_{k} \|_{\mathcal{B}^{\alpha }}+C\sup_{z\in\mathbb{D}\setminus K} \frac{(1-|z|^{2})^{\beta}|u'(z)|}{(1-|\varphi(z)|^{2})^{\alpha +n-1}} \|f_{k}\| _{\mathcal{B}^{\alpha }}\\ &\quad\leq C_{4} \sup_{z\in K} \bigl|f_{k}^{(n+1)} \bigl(\varphi(z)\bigr)\bigr|+C_{3}\sup_{z\in K} \bigl|f_{k}^{(n)}\bigl(\varphi(z)\bigr)\bigr| +C\varepsilon \|f_{k}\|_{{\mathcal{B}}^{\alpha }}, \end{aligned}$$

i.e., we get

$$\begin{aligned} \bigl\| D^{n}_{\varphi,u}f_{k}\bigr\| _{\mathcal{B}^{\beta}} =&C_{4} \sup_{|w| \leq\delta} \bigl|f_{k}^{(n+1)}(w)\bigr|+C_{3} \sup_{|w| \leq\delta} \bigl|f_{k}^{(n)}(w)\bigr| \\ &{} +C\varepsilon\|f_{k}\|_{{\mathcal{B}}^{\alpha }}+\bigl|u(0)\bigr|\bigl|f^{(n)}_{k} \bigl(\varphi (0)\bigr)\bigr|. \end{aligned}$$
(2.10)

Since \(f_{k}\) converges to 0 uniformly on compact subsets of \(\mathbb{D}\) as \(k\to\infty\), Cauchy’s estimate gives that \(f^{(n)}_{k} \to0\) as \(k\to\infty\) on compact subsets of \(\mathbb{D}\). Hence, letting \(k\to\infty\) in (2.10) and using the fact that ε is an arbitrary positive number, we obtain \(\|D^{n}_{\varphi,u} f_{k}\|_{\mathcal{B}^{\beta}}\rightarrow0\) as \(k\to\infty\). Applying Lemma 2 the result follows. □