Product of differentiation and composition operators on the logarithmic Bloch space

Abstract

We obtain a criterion for the boundedness and compactness of the products of differentiation and composition operators $C_{\phi }{D}^m$ on the logarithmic Bloch space in terms of the sequence $\{z^n\}$. An estimate for the essential norm of $C_{\phi }{D}^m$ is given.

MSC:47B38, 30H30.

1 Introduction

Denote by $H(\mathbb{D})$ the space of all analytic functions on the unit disk $\mathbb{D}=\{z:|z|<1\}$ in the complex plane. Let $H^{\mathrm{\infty }}=H^{\mathrm{\infty }}(\mathbb{D})$ denote the space of bounded analytic functions on $\mathbb{A}$. An $f\in H(\mathbb{D})$ is said to belong to the Bloch space ℬ if

$${\parallel f\parallel }_{\mathcal{B}}=\underset{z\in \mathbb{D}}{sup}|f^{\prime }(z)|(1-{|z|}^2)<\mathrm{\infty }.$$

The logarithmic-Bloch space, denoted by $\mathcal{LB}$, consists of all $f\in H(\mathbb{D})$ satisfying

$${\parallel f\parallel }_{log}=\underset{z\in \mathbb{D}}{sup}(1-|z|)|f^{\prime }(z)|log\frac{e}{1-|z|}<\mathrm{\infty }.$$

$\mathcal{LB}$ is a Banach space with the norm ${\parallel f\parallel }_{\mathcal{LB}}=|f(0)|+{\parallel f\parallel }_{log}$. It is well known that $\mathcal{LB}\cap H^{\mathrm{\infty }}$ is the space of multipliers of the Bloch space ℬ (see [1, 2]). For some results on logarithmic-type spaces and operators on them, see, for example, [3–10].

Let φ be an analytic self-map of $\mathbb{D}$. The composition operator $C_{\phi }$ is defined by

$$C_{\phi }(f)=f\circ \phi ,f\in H(\mathbb{D}).$$

The differentiation operator D is defined by $Df=f^{\prime }$, $f\in H(\mathbb{D})$. For a nonnegative integer $m\in \mathbb{N}$, we define

$$D^{m}f=f^{(m)},f\in H(\mathbb{D}).$$

The product of differentiation and composition operators $C_{\phi }{D}^m$ is defined as follows:

$$C_{\phi }{D}^{m}f=f^{(m)}\circ \phi ,f\in H(\mathbb{D}).$$

A basic problem concerning concrete operators on various Banach spaces is to relate the operator theoretic properties of the operators to the function theoretic properties of their symbols, which attracted a lot of attention recently, the reader can refer to [4–37].

It is a well-known consequence of the Schwarz-Pick lemma that the composition operator is bounded on ℬ. See [21–24, 27, 33–35, 37] for the study of composition operators and weighted composition operators on the Bloch space. The product-type operators on or into Bloch type spaces have been studied in many papers recently; see [12–20, 26, 28–32, 34, 36] for example.

Let X and Y be two Banach spaces. Recall that a linear operator $T:X\to Y$ is said to be compact if it takes bounded sets in X to sets in Y which have compact closure. The essential norm of an operator T between X and Y is the distance to the compact operators K, that is, ${\parallel T\parallel }_e^{X\to Y}=inf\{\parallel T-K\parallel :K\text{ is compact}\}$, where $\parallel \cdot \parallel$ is the operator norm. It is easy to see that ${\parallel T\parallel }_e^{X\to Y}=0$ if and only if T is compact. For some results in the topic, see, for example, [11, 20, 22, 24, 26, 28, 37].

In [34], Wu and Wulan obtained a characterization for the compactness of the product of differentiation and composition operators acting on the Bloch space as follows:

Theorem A Let φ be an analytic self-map of $\mathbb{D}$, $m\in \mathbb{N}$. Then $C_{\phi }{D}^m:\mathcal{B}\to \mathcal{B}$ is compact if and only if

$$\underset{n\to \mathrm{\infty }}{lim}{\parallel C_{\phi }{D}^m\left(z^n\right)\parallel }_{\mathcal{B}}=0.$$

The purpose of the paper is to extend Theorem A to the case of $\mathcal{LB}$. We will characterize the boundedness and compactness of $C_{\phi }{D}^m$ in terms of the sequence $\{z^n\}$. Moreover, an estimate for the essential norm of $C_{\phi }{D}^m$ will be given. The main results are given in Sections 3 and 4.

In the paper, we say that a real sequence ${\{a_n\}}_{n\in \mathbb{N}}$ is asymptotic to another real sequence of ${\{b_n\}}_{n\in \mathbb{N}}$ and write ‘$a_n\sim b_n$’ if and only if

$$\underset{n\to \mathrm{\infty }}{lim}\frac{a_n}{b_n}=1.$$

In addition, we say that $A⪯B$ if there exists a constant C such that $A\le CB$. The symbol $A\approx B$ means that $A⪯B⪯A$.

2 Auxiliary lemmas

In this section, we state and prove some auxiliary results which will be used to prove the main results in this paper.

Lemma 2.1 For $m,n\in \mathbb{N}$, define the function $H_{m,n}:[0,1)\to [0,\mathrm{\infty })$ by

$$H_{m,n}(x)=\frac{n!}{(n-m-1)!}{x}^{n-m-1}{(1-x)}^{m+1}log\frac{e}{1-x}.$$

(2.1)

Then the following statements hold:

• (i) For $n,m\in \mathbb{N}$ and $n\ge m+1$, there is a unique $x_{m,n}\in [0,1)$ such that $H_{m,n}(x_{m,n})$ is the absolute maximum of $H_{m,n}$.

• (ii)
  $$\underset{n\to \mathrm{\infty }}{lim}{x}_{m,n}=1$$

  (2.2)

and

$$\underset{n\to \mathrm{\infty }}{lim}[n(1-x_{m,n})]=m+1.$$

(2.3)

• (iii)
  $$\underset{n\to \mathrm{\infty }}{lim}\frac{{max}_{0<t<1}{H}_{m,n}(t)}{log(n+1)}={\left(\frac{m+1}{e}\right)}^{m+1}.$$

  (2.4)

Proof Directly computing we have

$$H_{m,n}^{\prime }(x)=\frac{n!}{(n-m-1)!}{x}^{n-m-2}{(1-x)}^m((n-m-1-nx)log\frac{e}{1-x}+x).$$

Define

$$g_{m,n}(x)=(n-m-1-nx)log\frac{e}{1-x}+x,x\in [0,1).$$

(2.5)

It is easy to see that $g_{m,n}$ is continuous on [0,1) and $g_{m,n}(0)=n-m-1\ge 0$, ${lim}_{x\to 1^{-}}{g}_{m,n}(x)=-\mathrm{\infty }$. Furthermore,

$$g_{m,n}^{\prime }(x)=-nlog\frac{e}{1-x}+n-\frac{m+1}{1-x}+1<0,x\in [0,1).$$

Then $g_{m,n}$ is decreasing on $[0,1)$. When $n=m+1$, we get ${max}_{0\le x<1}{H}_{m,n}(x)=H_{m,n}(0)$. When $n>m+1$, the intermediate value theorem of continuous function gives the result that there exists a unique $x_{m,n}\in (0,1)$ such that $g_{m,n}(x_{m,n})=0$. So we have

$$\underset{0<t<1}{max}{H}_{m,n}(x)=H_{m,n}(x_{m,n}).$$

(i) has been proved. By (2.5), we have $g_{m,n}(x_{m,n})=0$. Thus

$$(\frac{n-m-1}{n}-x_{m,n})log\frac{e}{1-x_{m,n}}=-\frac{x_{m,n}}{n}.$$

It follows from ${lim}_{n\to \mathrm{\infty }}\frac{x_{m,n}}{n}=0$ and $log\frac{e}{1-x_{m,n}}\ge 1$ that (2.2) holds. Also, $g_{m,n}(x_{m,n})=0$ gives the result that

$$\frac{n-m-1}{n}-x_{m,n}=-\frac{x_{m,n}}{nlog\frac{e}{1-x_{m,n}}}.$$

So we have

$$n(1-x_{m,n})-m-1=-\frac{x_{m,n}}{log\frac{e}{1-x_{m,n}}}.$$

This gives the result (2.3). The proof of (ii) is complete.

Note that

$$nlog{x}_{m,n}\sim nlog[1+(x_{m,n}-1)]\sim n(x_{m,n}-1)\to -m-1\text{as }n\to \mathrm{\infty }.$$

This and (2.2) give

$$\underset{n\to \mathrm{\infty }}{lim}{x}_{m,n}^{n-m-1}=e^{-m-1}.$$

(2.6)

By (2.3) and (2.6) we obtain

$$\begin{array}{rcl}\underset{n\to \mathrm{\infty }}{lim}\frac{H_{m,n}(x_{m,n})}{log(n+1)}& =& \underset{n\to \mathrm{\infty }}{lim}\frac{n!x_{m,n}^{n-m-1}{(1-x_{m,n})}^{m+1}log\frac{e}{1-x_{m,n}}}{(n-m-1)!log(n+1)}\\ =& e^{-m-1}\underset{n\to \mathrm{\infty }}{lim}\frac{n!{((m+1)/n)}^{m+1}log\frac{en}{m+1}}{(n-m-1)!log(n+1)}={\left(\frac{m+1}{e}\right)}^{m+1},\end{array}$$

which shows that (iii) hold. The proof is complete. □

Lemma 2.2 Let $m,n\in \mathbb{N}$ and $n-m-1>0$. Let $r_{m,n}=(n-m-1)/n$. Then $H_{m,n}$ is increasing on $[r_{m,n-m},r_{m,n}]$ and

$$\underset{r_{m,n-m}\le x\le r_{m,n}}{min}{H}_{m,n}(x)=H_{m,n}(r_{m,n-m})\sim {\left(\frac{m+1}{e}\right)}^{m+1}log(n+1)\mathit{\text{as }}n\to \mathrm{\infty }.$$

(2.7)

Consequently,

$$\underset{r_{m,n-m}\le x\le r_{m,n}}{min}\frac{H_{m,n}(x)}{{\parallel z^n\parallel }_{\mathcal{LB}}}=\frac{H_{m,n}(r_{m,n-m})}{{\parallel z^n\parallel }_{\mathcal{LB}}}\sim \frac{{(m+1)}^{m+1}}{e^m}\mathit{\text{as }}n\to \mathrm{\infty }.$$

(2.8)

Proof Since $n-m-1>0$, we have

$$H_{m,n}^{\prime }(r_{m,n})=\frac{n!}{(n-m-1)!}{\left(\frac{n-m-1}{n}\right)}^{n-m-2}{\left(\frac{m+1}{n}\right)}^m\left(\frac{n-m-1}{n}\right)>0.$$

By Lemma 2.1, we have $r_{m,n}<x_{m,n}$, where $x_{m,n}$ is given as in Lemma 2.1. Since $H_{m,n}^{\prime }(x)>0$ for $x\in (0,x_{m,n})$, we see that $H_{m,n}$ is increasing on $[r_{m,n-m},r_{m,n}]$. Thus

$$\begin{array}{rcl}\underset{r_{m,n-m}\le x\le r_{m,n}}{min}{H}_{m,n}(x)& =& H_{m,n}(r_{m,n-m})\\ =& \frac{n!}{(n-m-1)!}{\left(\frac{n-2m-1}{n-m}\right)}^{n-m-1}{\left(\frac{m+1}{n-m}\right)}^{m+1}log\frac{e(n-m)}{m+1}.\end{array}$$

Applying the important limit ${lim}_{n\to \mathrm{\infty }}{(1+\frac{1}{n})}^n=e$ we obtain the result that (2.7) holds.

By Lemma 2.1 we have

$${\parallel z^n\parallel }_{\mathcal{LB}}=\underset{|z|<1}{sup}n{|z|}^{n-1}(1-|z|)log\frac{e}{1-|z|}=H_{0,n}(x_{0,n}),$$

(2.9)

where $x_{0,n}$ is given in Lemma 2.1. By Lemma 2.1 we have

$$\begin{array}{rl}\underset{n\to \mathrm{\infty }}{lim}\frac{H_{m,n}(r_{m,n-m})}{{\parallel z^n\parallel }_{\mathcal{LB}}}& =\underset{n\to \mathrm{\infty }}{lim}\frac{H_{m,n}(r_{m,n-m})}{log(n+1)}\frac{log(n+1)}{{\parallel z^n\parallel }_{\mathcal{LB}}}\\ =\underset{n\to \mathrm{\infty }}{lim}\frac{H_{m,n}(r_{m,n-m})}{log(n+1)}\underset{n\to \mathrm{\infty }}{lim}\frac{log(n+1)}{{\parallel z^n\parallel }_{\mathcal{LB}}}=\frac{{(m+1)}^{m+1}}{e^m}.\end{array}$$

This gives (2.8). The proof is complete. □

Lemma 2.3 [3]

For $m\in \mathbb{N}$. Then $f\in \mathcal{LB}$ if and only if

$$\underset{z\in \mathbb{D}}{sup}{(1-|z|)}^m|f^{(m)}(z)|log\frac{e}{1-|z|}<\mathrm{\infty }.$$

Moreover,

$${\parallel f\parallel }_{\mathcal{LB}}\approx \sum _{j=0}^{m-1}|f^{(j)}(0)|+\underset{z\in \mathbb{D}}{sup}{(1-|z|)}^m|f^{(m)}(z)|log\frac{e}{1-|z|}.$$

3 The boundedness of $C_{\phi }{D}^m$ on $\mathcal{LB}$

In this section, we will state the boundedness criterion for the operator $C_{\phi }{D}^m$ on $\mathcal{LB}$. Since the boundedness of $C_{\phi }{D}^m$ on $\mathcal{LB}$ gives $\phi \in \mathcal{LB}$, we may always assume that $\phi \in \mathcal{LB}$. The main result of this section is stated as follows.

Theorem 3.1 Let $m\in \mathbb{N}$ and φ be an analytic self-map of $\mathbb{D}$ such that $\phi \in \mathcal{LB}$. Then $C_{\phi }{D}^m$ is bounded on $\mathcal{LB}$ if and only if

$$\underset{n\in \mathbb{N}}{sup}\frac{{\parallel C_{\phi }{D}^m(z^n)\parallel }_{\mathcal{LB}}}{{\parallel z^n\parallel }_{\mathcal{LB}}}<\mathrm{\infty }.$$

(3.1)

Proof ⇒) Assume that $C_{\phi }{D}^m$ is bounded on $\mathcal{LB}$, that is, ${\parallel C_{\phi }{D}^m\parallel }_{\mathcal{LB}\to \mathcal{LB}}<\mathrm{\infty }$. Since the sequence $\{z^n/{\parallel z^n\parallel }_{\mathcal{LB}}\}$ is bounded in the logarithmic Bloch space $\mathcal{LB}$, we have

$$\frac{{\parallel C_{\phi }{D}^m(z^n)\parallel }_{\mathcal{LB}}}{{\parallel z^n\parallel }_{\mathcal{LB}}}\le {\parallel C_{\phi }{D}^m\parallel }_{\mathcal{LB}\to \mathcal{LB}}{\parallel \frac{z^n}{{\parallel z^n\parallel }_{\mathcal{LB}}}\parallel }_{\mathcal{LB}}\le {\parallel C_{\phi }{D}^m\parallel }_{\mathcal{LB}\to \mathcal{LB}}<\mathrm{\infty },$$

for any $n\in \mathbb{N}$, from which the implication follows.

⇐) We now assume that the condition (3.1) holds. On the one hand, for the case ${sup}_{z\in \mathbb{D}}|\phi (z)|<1$, there is an $r\in (0,1)$ such that $|\phi (z)|<r$. By (3.1), for any given $f\in \mathcal{LB}$, we have

$$\begin{array}{rcl}{\parallel C_{\phi }{D}^{m}f\parallel }_{\mathcal{LB}}& =& \underset{z\in \mathbb{D}}{sup}(1-|z|)log\frac{e}{1-|z|}|f^{(m+1)}(\phi (z)){\phi }^{\prime }(z)|\\ \le & \underset{z\in \mathbb{D}}{sup}{\parallel \phi \parallel }_{\mathcal{LB}}\frac{|f^{(m+1)}(\phi (z))|{(1-|\phi (z)|)}^{m+1}log\frac{e}{1-|\phi (z)|}}{{(1-|\phi (z)|)}^{m+1}log\frac{e}{1-|\phi (z)|}}\\ ⪯& \underset{z\in \mathbb{D}}{sup}\frac{{\parallel \phi \parallel }_{\mathcal{LB}}{\parallel f\parallel }_{\mathcal{LB}}}{{(1-r)}^{m+1}ln\frac{e}{1-r}}<\mathrm{\infty }.\end{array}$$

The last estimate shows that the operator $C_{\phi }$ is bounded on $\mathcal{LB}$.

On the other hand, for the case ${sup}_{z\in \mathbb{D}}|\phi (z)|=1$. Let N be the smallest positive integer such that ${\mathbb{D}}_N$ is not empty, where

$${\mathbb{D}}_n=\{z\in \mathbb{D}:r_{m,n-m}\le |\phi (z)|\le r_{m,n}\}$$

and $r_{m,n}$ is given in Lemma 2.2. Note that $H_{m,n}(|\phi (z)|)>0$, when $z\in {\mathbb{D}}_n$, $n\ge N$, by (2.8) we obtain

$$ϵ:=\underset{z\in {\mathbb{D}}_n}{inf}\frac{H_{m,n}(|\phi (z)|)}{{\parallel z^n\parallel }_{\mathcal{LB}}}>0.$$

For any given $f\in \mathcal{LB}$, by Lemma 2.3 we have

$$\begin{array}{rl}{\parallel C_{\phi }{D}^{m}f\parallel }_{\mathcal{LB}}& =\underset{z\in \mathbb{D}}{sup}(1-|z|)log\frac{e}{1-|z|}|f^{(m+1)}(\phi (z)){\phi }^{\prime }(z)|\\ =\underset{n\ge N}{sup}\underset{z\in {\mathbb{D}}_n}{sup}(1-|z|)log\frac{e}{1-|z|}|f^{(m+1)}(\phi (z)){\phi }^{\prime }(z)|\\ =\underset{n\ge N}{sup}\underset{z\in {\mathbb{D}}_n}{sup}(1-|z|)log\frac{e}{1-|z|}|f^{(m+1)}(\phi (z)){\phi }^{\prime }(z)|\frac{{\parallel z^n\parallel }_{\mathcal{LB}}}{H_{m,n}(|\phi (z)|)}\frac{H_{m,n}(|\phi (z)|)}{{\parallel z^n\parallel }_{\mathcal{LB}}}\\ ⪯\frac{{\parallel f\parallel }_{\mathcal{LB}}}{ϵ}\underset{n\ge N}{sup}\underset{z\in {\mathbb{D}}_n}{sup}\frac{n!}{(n-m-1)!}(1-|z|)log\frac{e}{1-|z|}|{\phi }^{\prime }(z)|\frac{{|\phi (z)|}^{n-m-1}}{{\parallel z^n\parallel }_{\mathcal{LB}}}\\ \le \frac{{\parallel f\parallel }_{\mathcal{LB}}}{ϵ}\underset{n\ge N}{sup}\frac{{\parallel C_{\phi }{D}^m(z^n)\parallel }_{\mathcal{LB}}}{{\parallel z^n\parallel }_{\mathcal{LB}}}.\end{array}$$

The proof is complete. □

4 The essential norm of $C_{\phi }{D}^m$ on $\mathcal{LB}$

Denote $K_{r}f(z)=f(rz)$ for $r\in (0,1)$. Then $K_r$ is a compact operator on the space $\mathcal{LB}$. It is easy to see that $\parallel K_r\parallel \le 1$. We denote by I the identity operator.

In order to give the lower and upper estimate for the essential norm of $C_{\phi }{D}^m$ on $\mathcal{LB}$, we need the following result.

Lemma 4.1 There is a sequence $\{r_k\}$, with $0<r_k<1$ tending to 1, such that the compact operator

$$L_n=\frac{1}{n}\sum _{k=1}^n{K}_{r_k}$$

on $\mathcal{LB}$ satisfies:

• (i) for any $t\in (0,1)$, ${lim}_{n\to \mathrm{\infty }}{sup}_{{\parallel f\parallel }_{\mathcal{LB}}\le t}{sup}_{|z|\le t}|{((I-L_n)f)}^{\prime }(z)|=0$,

(iia) ${lim}_{n\to \mathrm{\infty }}{sup}_{{\parallel f\parallel }_{\mathcal{LB}}\le 1}{sup}_{|z|<1}|(I-L_n)f(z)|\le 1$,

(iib) ${lim}_{n\to \mathrm{\infty }}{sup}_{{\parallel f\parallel }_{\mathcal{LB}}\le 1}{sup}_{|z|<s}|(I-L_n)f(z)|=0$, for any $s\in (0,1)$,

• (iii) ${lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}\parallel I-L_n\parallel \le 1$.

Proof (i) follows from (iib) by Cauchy’s formula. The proof of (iii) is similar to the proof of Proposition 8 in [25]. Hence we omit it. Next we prove (iia) and (iib). The argument is much like that given in the proof of Proposition 2.1 of [25] or Lemmas 1 and 2 in [22]. For any $0<s<1$, we choose an increasing sequence $r_k$ tending to 1 such that $r_k\ge 1-\frac{1-s}{k^2}$. For any given $z\in \mathbb{D}$ and $r_k$, $k=1,2,3,\dots$ , there exists an $s_k\in (r_k,1)$ such that

$$|f(z)-f_{r_k}(z)|=z{f}^{\prime }(s_{k}z)(z-r_{k}z).$$

(4.1)

For any $f\in \mathcal{LB}$ with ${\parallel f\parallel }_{\mathcal{LB}}\le 1$, we have

$$\begin{array}{rcl}|(I-L_n)f(z)|& \le & \frac{1}{n}\sum _{k=1}^n|f(z)-f_{r_k}(z)|\le \frac{1}{n}\sum _{k=1}^n|f^{\prime }(s_{k}z)|(1-r_k)\\ \le & \frac{1}{n}\sum _{k=1}^n\frac{1-r_k}{(1-|r_{k}z|)log\frac{e}{1-|r_{k}z|}}\le \frac{1}{n}\sum _{k=1}^{n}1=1.\end{array}$$

Thus

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\underset{{\parallel f\parallel }_{\mathcal{LB}}\le 1}{sup}\underset{|z|<1}{sup}|(I-L_n)f(z)|\le 1.$$

This shows that (iia) holds.

If $|z|\le s$, by the equality (4.1), we have

$$\begin{array}{rcl}|(I-L_n)f(z)|& \le & \frac{1}{n}\sum _{k=1}^n\frac{1-r_k}{(1-|sz|)log\frac{e}{1-|sz|}}\\ \le & \frac{1}{n}\sum _{k=1}^n\frac{1-r_k}{(1-s)}=\frac{1}{n}\sum _{k=1}^n\frac{1}{k^2}\le \frac{{\pi }^2}{6n}.\end{array}$$

The above estimate gives (iib). The proof is complete. □

The following lemma can be proved in a standard way; see, for example Proposition 3.11 in [11].

Lemma 4.2 Let $m\in \mathbb{N}$ and φ be an analytic self-map of $\mathbb{D}$. Then $C_{\phi }{D}^m$ is compact on $\mathcal{LB}$ if and only if $C_{\phi }{D}^m$ is bounded on $\mathcal{LB}$ and for any bounded sequence $\{f_n\}$ in $\mathcal{LB}$ which converges to zero uniformly on compact subsets of $\mathbb{D}$, then ${\parallel C_{\phi }{D}^m{f}_n\parallel }_{\mathcal{LB}}\to 0$ as $n\to \mathrm{\infty }$.

Theorem 4.3 Let $m\in \mathbb{N}$ and φ be an analytic self-map of $\mathbb{D}$. Suppose that $C_{\phi }{D}^m$ is bounded on $\mathcal{LB}$. Then the estimate for the essential norm of $C_{\phi }{D}^m$ on $\mathcal{LB}$ is

$${\parallel C_{\phi }{D}^m\parallel }_e^{\mathcal{LB}\to \mathcal{LB}}\approx \underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\frac{{\parallel C_{\phi }{D}^m(z^n)\parallel }_{\mathcal{LB}}}{{\parallel z^n\parallel }_{\mathcal{LB}}}.$$

(4.2)

Proof We first give the lower estimate for the essential norm. Without loss of generality, we assume that $n\ge m+1$. Choose the sequence of function $f_n(z)=z^n/{\parallel z^n\parallel }_{\mathcal{LB}}$, $n\in \mathbb{N}$. Then ${\parallel f_n\parallel }_{\mathcal{LB}}=1$, and $\{f_n\}$ converges to zero weakly on $\mathcal{LB}$ as $n\to \mathrm{\infty }$. Thus we have

$$\underset{n\to \mathrm{\infty }}{lim}{\parallel K{f}_n\parallel }_{\mathcal{LB}}=0$$

for any given compact operator K on $\mathcal{LB}$. The basic inequality gives

$${\parallel C_{\phi }{D}^m-K\parallel }^{\mathcal{LB}\to \mathcal{LB}}\ge {\parallel (C_{\phi }{D}^m-K)f_n\parallel }_{\mathcal{LB}}\ge {\parallel C_{\phi }{D}^m{f}_n\parallel }_{\mathcal{LB}}-{\parallel K{f}_n\parallel }_{\mathcal{LB}}.$$

Thus we obtain

$${\parallel C_{\phi }{D}^m-K\parallel }^{\mathcal{LB}\to \mathcal{LB}}\ge \underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}{\parallel C_{\phi }{D}^m{f}_n\parallel }_{\mathcal{LB}}\ge \underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}{\parallel C_{\phi }{D}^m{f}_n\parallel }_{\mathcal{LB}}.$$

So we have

$${\parallel C_{\phi }{D}^m\parallel }_e^{\mathcal{LB}\to \mathcal{LB}}=\underset{K}{inf}\parallel C_{\phi }{D}^m-K\parallel \ge \underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\frac{{\parallel C_{\phi }{D}^m(z^n)\parallel }_{\mathcal{LB}}}{{\parallel z^n\parallel }_{\mathcal{LB}}}.$$

Now we give the upper estimate for the essential norm. For the case of ${sup}_{z\in \mathbb{D}}|\phi (z)|<1$, there is a number $\delta \in (0,1)$ such that ${sup}_{z\in \mathbb{D}}|\phi (z)|<\delta$. In this case, the operator $C_{\phi }{D}^m$ is compact on $\mathcal{LB}$. In fact, choose a bounded sequence ${\{f_n\}}_{n\in \mathbb{N}}$ in $\mathcal{LB}$ which converges to zero uniformly on compact subset of $\mathbb{D}$. From Cauchy’s integral formula, $\{f_n^{(m+1)}\}$ converges to zero on compact subsets of $\mathbb{D}$ as $n\to \mathrm{\infty }$. It follows that

$$\begin{array}{rcl}\underset{n\to \mathrm{\infty }}{lim}{\parallel C_{\phi }{D}^m{f}_n\parallel }_{\mathcal{LB}}& =& \underset{n\to \mathrm{\infty }}{lim}(\left|f_n^{(m)}(\phi (0))\right|+{\parallel C_{\phi }{D}^m{f}_n\parallel }_{log})\\ =& \underset{n\to \mathrm{\infty }}{lim}\underset{z\in \mathbb{D}}{sup}(1-|z|)log\frac{e}{1-|z|}|f_n^{(m+1)}(\phi (z)){\phi }^{\prime }(z)|\\ \le & {\parallel \phi \parallel }_{\mathcal{LB}}\underset{n\to \mathrm{\infty }}{lim}\underset{z\in \mathbb{D}}{sup}\left|f_n^{(m+1)}(\phi (z))\right|\\ =& {\parallel \phi \parallel }_{\mathcal{LB}}\underset{n\to \mathrm{\infty }}{lim}\underset{|w|\le \delta }{sup}|f_n^{(m+1)}(w)|=0.\end{array}$$

Then the operator $C_{\phi }{D}^m$ is compact on $\mathcal{LB}$ by Lemma 4.2. This gives

$${\parallel C_{\phi }{D}^m\parallel }_e^{\mathcal{LB}\to \mathcal{LB}}=0.$$

(4.3)

On the other hand, by Lemma 2.1 and (2.9) we obtain

$${\parallel z^n\parallel }_{\mathcal{LB}}=H_{0,n}(x_{0,n})\ge H_{0,n}(r_{0,n})\ge \frac{1}{2}log(en),$$

which implies that

$$\begin{array}{r}\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\frac{{\parallel C_{\phi }{D}^m(z^n)\parallel }_{\mathcal{LB}}}{{\parallel z^n\parallel }_{\mathcal{LB}}}\\ \le e\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\underset{z\in \mathbb{D}}{sup}(1-|z|)log\frac{e}{1-|z|}\frac{n!}{(n-m-1)!}{|\phi (z)|}^{n-m-1}|{\phi }^{\prime }(z)|\\ \le e{\parallel \phi \parallel }_{\mathcal{LB}}\underset{n\to \mathrm{\infty }}{lim}{n}^m{\delta }^{n-m-1}=0.\end{array}$$

Combining the last inequality with (4.3), we get the desired result.

Next, we assume that ${sup}_{z\in \mathbb{D}}|\phi (z)|=1$. Let $L_n$ be the sequence of operators given in Lemma 4.1. Since $L_n$ is compact on $\mathcal{LB}$ and $C_{\phi }{D}^m$ is bounded on $\mathcal{LB}$, then $C_{\phi }{D}^m{L}_n$ is also compact on $\mathcal{LB}$. Hence

$$\begin{array}{rcl}{\parallel C_{\phi }{D}^m\parallel }_e^{\mathcal{LB}\to \mathcal{LB}}& \le & \underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}{\parallel C_{\phi }{D}^m-C_{\phi }{D}^m{L}_n\parallel }_{\mathcal{LB}\to \mathcal{LB}}\\ =& \underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}{\parallel C_{\phi }{D}^m(I-L_n)\parallel }_{\mathcal{LB}\to \mathcal{LB}}\\ =& \underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\underset{{\parallel f\parallel }_{\mathcal{LB}}\le 1}{sup}{\parallel C_{\phi }{D}^m(I-L_n)f\parallel }_{\mathcal{LB}}\\ =& \underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\underset{{\parallel f\parallel }_{\mathcal{LB}}\le 1}{sup}{\parallel {((I-L_n)f)}^{(m)}\circ \phi \parallel }_{\mathcal{LB}}\\ \le & I_1+I_2,\end{array}$$

where

$$I_1=\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\underset{{\parallel f\parallel }_{\mathcal{LB}}\le 1}{sup}\left|{((I-L_n)f)}^{(m)}(\phi (0))\right|$$

and

$$I_2=\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\underset{{\parallel f\parallel }_{\mathcal{LB}}\le 1}{sup}\underset{z\in \mathbb{D}}{sup}|{((I-L_n)f)}^{(m+1)}(\phi (z)){\phi }^{\prime }(z)|(1-|z|)log\frac{e}{1-|z|}.$$

It follows from Lemma 4.1(iib) and Cauchy’s integral formula that $I_1=0$.

For each positive integer $n\ge m+1$, we define

$${\mathbb{D}}_n=\{z\in \mathbb{D}:r_{m,n-m}\le |\phi (z)|<r_{m,n}\},$$

where $r_{m,n}$ is given in Lemma 2.1. Let k be the smallest positive integer such that ${\mathbb{D}}_k\ne 0$. Since ${sup}_{z\in \mathbb{D}}|\phi (z)|=1$, ${\mathbb{D}}_n$ is not empty for every integer $n\ge k$ and $\mathbb{D}={\bigcup }_{n=k}^{\mathrm{\infty }}{\mathbb{D}}_n$, we have

$$\underset{{\parallel f\parallel }_{\mathcal{LB}}\le 1}{sup}\underset{z\in \mathbb{D}}{sup}|{((I-L_n)f)}^{(m+1)}(\phi (z)){\phi }^{\prime }(z)|(1-|z|)log\frac{e}{1-|z|}=I_{21}+I_{22},$$

where

$$I_{21}=\underset{{\parallel f\parallel }_{\mathcal{LB}}\le 1}{sup}\underset{k\le i\le N-1}{sup}\underset{z\in {\mathbb{D}}_i}{sup}|{((I-L_n)f)}^{(m+1)}(\phi (z)){\phi }^{\prime }(z)|(1-|z|)log\frac{e}{1-|z|}$$

and

$$I_{22}=\underset{{\parallel f\parallel }_{\mathcal{LB}}\le 1}{sup}\underset{N\le i}{sup}\underset{z\in {\mathbb{D}}_i}{sup}|{((I-L_n)f)}^{(m+1)}(\phi (z)){\phi }^{\prime }(z)|(1-|z|)log\frac{e}{1-|z|}.$$

Here N is a positive integer determined as follows.

By (2.8),

$$\underset{i\to \mathrm{\infty }}{lim}\frac{{\parallel z^i\parallel }_{\mathcal{LB}}}{H_{m,i}(r_{m,i-m})}=\frac{e^m}{{(m+1)}^{m+1}}.$$

Hence, for any given $\epsilon >0$, there exists an N such that

$$\frac{{\parallel z^i\parallel }_{\mathcal{LB}}}{H_{m,i}(r_{m,i-m})}\le \frac{e^m}{{(m+1)}^{m+1}}+\epsilon$$

when $i\ge N$. For such N it follows that

$$\begin{array}{rcl}{I}_{22}& =& \underset{{\parallel f\parallel }_{\mathcal{LB}}\le 1}{sup}\underset{N\le i}{sup}\underset{z\in {\mathbb{D}}_i}{sup}|{((I-L_n)f)}^{(m+1)}(\phi (z)){\phi }^{\prime }(z)|(1-|z|)log\frac{e}{1-|z|}\\ =& \underset{{\parallel f\parallel }_{\mathcal{LB}}\le 1}{sup}\underset{N\le i}{sup}\underset{z\in {\mathbb{D}}_i}{sup}|{((I-L_n)f)}^{(m+1)}(\phi (z)){\phi }^{\prime }(z)|\\ \cdot (1-|z|)log\frac{e}{1-|z|}\frac{H_{m,i}(|\phi (z)|)}{{\parallel z^i\parallel }_{\mathcal{LB}}}\frac{{\parallel z^i\parallel }_{\mathcal{LB}}}{H_{m,i}(|\phi (z)|)}\\ ⪯& (\frac{e^m}{{(m+1)}^{m+1}}+\epsilon )\underset{{\parallel f\parallel }_{\mathcal{LB}}\le 1}{sup}{\parallel (I-L_n)f\parallel }_{\mathcal{LB}}\underset{N\le i}{sup}\underset{z\in {\mathbb{D}}_i}{sup}|{\phi }^{\prime }(z)|\\ \cdot (1-|z|)log\frac{e}{1-|z|}\frac{i!}{(i-m-1)!}\frac{{|\phi (z)|}^{i-m-1}}{{\parallel z^i\parallel }_{\mathcal{LB}}}\\ \le & (\frac{e^m}{{(m+1)}^{m+1}}+\epsilon )\parallel I-L_n\parallel \underset{N\le i}{sup}\frac{{\parallel C_{\phi }{D}^m(z^i)\parallel }_{\mathcal{LB}}}{{\parallel z^i\parallel }_{\mathcal{LB}}}.\end{array}$$

Thus

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}{I}_{22}⪯(\frac{e^m}{{(m+1)}^{m+1}}+\epsilon )\underset{N\le i}{sup}\frac{{\parallel C_{\phi }{D}^m(z^i)\parallel }_{\mathcal{LB}}}{{\parallel z^i\parallel }_{\mathcal{LB}}}.$$

(4.4)

By (ii) of Lemma 4.1 and Cauchy’s integral formula, we have

$$\begin{array}{r}\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}{I}_{21}\\ =\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\underset{{\parallel f\parallel }_{\mathcal{LB}}\le 1}{sup}\underset{k\le i<N-1}{sup}\underset{z\in {\mathbb{D}}_i}{sup}\left|{((I-L_n)f)}^{(m+1)}(\phi (z))\right||{\phi }^{\prime }(z)|(1-|z|)log\frac{e}{1-|z|}\\ \le {\parallel \phi \parallel }_{\mathcal{LB}}\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\underset{{\parallel f\parallel }_{\mathcal{LB}}\le 1}{sup}\underset{|\phi (z)|<r_{m,N-1}}{sup}\left|{((I-L_n)f)}^{(m+1)}(\phi (z))\right|\\ =0,\end{array}$$

which together with (4.4) implies that

$$I_2⪯(\frac{e^m}{{(m+1)}^{m+1}}+\epsilon )\underset{N\le i}{sup}\frac{{\parallel C_{\phi }{D}^m(z^i)\parallel }_{\mathcal{LB}}}{{\parallel z^i\parallel }_{\mathcal{LB}}}.$$

(4.5)

From (4.5) we obtain

$${\parallel C_{\phi }{D}^m\parallel }_e^{\mathcal{LB}\to \mathcal{LB}}\le I_1+I_2⪯(\frac{e^m}{{(m+1)}^{m+1}}+\epsilon )\underset{N\le i}{sup}\frac{{\parallel C_{\phi }{D}^m(z^i)\parallel }_{\mathcal{LB}}}{{\parallel z^i\parallel }_{\mathcal{LB}}}.$$

By the arbitrariness of ε, we get

$${\parallel C_{\phi }{D}^m\parallel }_e^{\mathcal{LB}\to \mathcal{LB}}⪯\frac{e^m}{{(m+1)}^{m+1}}\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\frac{{\parallel C_{\phi }{D}^m(z^n)\parallel }_{\mathcal{LB}}}{{\parallel z^n\parallel }_{\mathcal{LB}}}.$$

The proof is complete. □

From Theorem 4.3, we obtain the following result.

Corollary 4.4 Let $m\in \mathbb{N}$ and φ be an analytic self-map of $\mathbb{D}$ such that $C_{\phi }{D}^m$ is bounded on $\mathcal{LB}$. Then $C_{\phi }{D}^m$ is compact on $\mathcal{LB}$ if and only if

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\frac{{\parallel C_{\phi }{D}^m(z^n)\parallel }_{\mathcal{LB}}}{{\parallel z^n\parallel }_{\mathcal{LB}}}=0.$$

Especially, when $m=0$, from the proof of Theorem 4.3, we get the exact formula for essential norm of composition operator on $\mathcal{LB}$.

Corollary 4.5 Let φ be an analytic self-map of $\mathbb{D}$. Suppose that $C_{\phi }$ is bounded on $\mathcal{LB}$; then

$${\parallel C_{\phi }\parallel }_e^{\mathcal{LB}\to \mathcal{LB}}=\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\frac{{\parallel {\phi }^n\parallel }_{\mathcal{LB}}}{{\parallel z^n\parallel }_{\mathcal{LB}}}.$$