Norm of an integral operator on some analytic function spaces on the unit disk

Abstract

If f is an analytic function in the unit disc , a class of integral operators is defined as follows:

$$I_f(h)(z)={\int }_0^{z}f(w)h^{\prime }(w)dw,h\in H(\mathbb{D}),z\in \mathbb{D}.$$

The norm of $I_f$ on some analytic function spaces is computed in this paper.

MSC:47B38, 32A35.

1 Introduction

Let $\mathbb{D}=\{z:|z|<1\}$ be the unit disk of a complex plane ℂ. Denote by $H(\mathbb{D})$ the class of functions analytic in . Let dσ denote the normalized Lebesgue area measure in and $g(a,z)$ the Green function with logarithmic singularity at a, i.e., $g(a,z)=-log|{\phi }_a(z)|$, where ${\phi }_a(z)=(a-z)/(1-\overline{a}z)$ is the Möbius transformation of .

For $0<p<\mathrm{\infty }$, the $Q_p$ is the space of all functions $f\in H(\mathbb{D})$, for which

$${\parallel f\parallel }_{Q_p}^2={|f(0)|}^2+\underset{a\in \mathbb{D}}{sup}{\int }_{\mathbb{D}}{|f^{\prime }(z)|}^2{(1-{|{\phi }_a(z)|}^2)}^{p}d\sigma (z)<\mathrm{\infty }.$$

(1.1)

We know that $Q_1=BMOA$, the space of all analytic functions of bounded mean oscillation [1, 2]. For all $p>1$, the space $Q_p$ is the same and equal to the Bloch space , consisting of analytic functions f in such that

$${\parallel f\parallel }_{\mathfrak{B}}=|f(0)|+\underset{z\in \mathbb{D}}{sup}|f^{\prime }(z)|(1-{|z|}^2)<\mathrm{\infty }.$$

(1.2)

See [3, 4] for the theory of Bloch functions.

For $\alpha >0$, the α-Bloch space, denoted by ${\mathfrak{B}}^{\alpha }$, is the space of all functions f in , for which

$${\parallel f\parallel }_{{\mathfrak{B}}^{\alpha }}=|f(0)|+\underset{z\in \mathbb{D}}{sup}|f^{\prime }(z)|{(1-{|z|}^2)}^{\alpha }<\mathrm{\infty }.$$

(1.3)

Obviously, ${\mathfrak{B}}^{{\alpha }_1}⫋\mathfrak{B}⫋{\mathfrak{B}}^{{\alpha }_2}$ for $0<{\alpha }_1<1<{\alpha }_2<\mathrm{\infty }$.

For any $f\in H(\mathbb{D})$, the next two integral operators on $H(\mathbb{D})$ are induced as follows:

$$I_f(h)(z)={\int }_0^z{h}^{\prime }(w)f(w)dw\text{and}{J}_f(h)(z)={\int }_0^{z}h(w)f^{\prime }(w)dw(z\in \mathbb{D}).$$

Let $M_f$ denote the multiplication operator, that is, $M_f(h)=fh$.

Let $f\in H(\mathbb{D})$. Then

$$(I_f+J_f)h=fh-f(0)h(0)=M_f(h)-f(0)h(0).$$

If f is a constant, then all results about $I_f$, $J_f$ or $M_f$ are trivial. In general, f is assumed to be non-constant. Both integral operators have been studied by many authors. See [5–21] and the references therein.

Norm of composition operator, weighted composition operator and some integral operators have been studied extensively by many authors, see [22–34] for example. Recently, Liu and Xiong discussed the norm of integral operators $I_f$ and $J_f$ on the Bloch space, Dirichlet space, BMOA space and so on in [35].

In this paper, we study the norm of integral operator $I_f$. The norm of $I_f$ on several analytic function spaces is computed.

2 Main results

In this section, we state and prove our main results. In order to formulate our main results, we need an auxiliary result which is incorporated in the following lemma.

Lemma 2.1 Let $0<p<1$. For any $z_0\in \mathbb{D}$, the function

$$g_{z_0}(z)=\frac{z_0-z}{1-{\overline{z}}_{0}z}-z_0$$

(2.1)

is analytic in and ${\parallel g_{z_0}\parallel }_{Q_p}=1/{(p+1)}^{1/2}$.

Proof By (1.1) and [[1], Proposition 1, p.109], we have

$$\begin{array}{rl}{\parallel g_{z_0}\parallel }_{Q_p}^2& =\underset{a\in \mathbb{D}}{sup}{\int }_{\mathbb{D}}{|g_{z_0}^{\prime }(z)|}^2{(1-{|{\phi }_a(z)|}^2)}^{p}d\sigma (z)\\ =\underset{b\in \mathbb{D}}{sup}{\int }_{\mathbb{D}}{(1-{|{\phi }_b(z)|}^2)}^{p}d\sigma (z),\end{array}$$

where $b={\phi }_{z_0}(a)$. Taking $w={\phi }_b(z)$, we have

$${\parallel g_{z_0}\parallel }_{Q_p}^2=\underset{b\in \mathbb{D}}{sup}{(1-{|b|}^2)}^2{\int }_{\mathbb{D}}\frac{{(1-{|w|}^2)}^p}{{|1-\overline{b}w|}^4}d\sigma (w).$$

Since

$$\frac{1}{{(1-\overline{b}w)}^2}=\sum _{n=0}^{\mathrm{\infty }}\frac{\mathrm{\Gamma }(n+2)}{n!\mathrm{\Gamma }(2)}{\overline{b}}^n{w}^n=\sum _{n=0}^{\mathrm{\infty }}\frac{\mathrm{\Gamma }(n+2)}{n!}{\overline{b}}^n{w}^n,$$

we have

$$\begin{array}{rl}{\int }_{\mathbb{D}}\frac{{(1-{|w|}^2)}^p}{{|1-\overline{b}w|}^4}d\sigma (w)& =\sum _{n=0}^{+\mathrm{\infty }}\frac{\mathrm{\Gamma }{(n+2)}^2}{{(n!)}^2}{|b|}^{2n}{\int }_{\mathbb{D}}{(1-{|w|}^2)}^p{|w|}^{2n}d\sigma (w)\\ =\sum _{n=0}^{+\mathrm{\infty }}\frac{\mathrm{\Gamma }{(n+2)}^2}{{(n!)}^2}{|b|}^{2n}{\int }_0^1{(1-r)}^p{r}^{n}dr\\ =\sum _{n=0}^{+\mathrm{\infty }}\frac{\mathrm{\Gamma }{(n+2)}^2}{{(n!)}^2}\frac{\mathrm{\Gamma }(p+1)\mathrm{\Gamma }(n+1)}{\mathrm{\Gamma }(n+p+2)}{|b|}^{2n}\\ =\sum _{n=0}^{+\mathrm{\infty }}\frac{\mathrm{\Gamma }(p+1)\mathrm{\Gamma }{(n+2)}^2}{n!\mathrm{\Gamma }(n+p+2)}{|b|}^{2n}.\end{array}$$

A simple computation shows

$$\frac{\mathrm{\Gamma }(p+1)\mathrm{\Gamma }{(n+2)}^2}{n!\mathrm{\Gamma }(n+p+2)}=\frac{(n+1)!(n+1)}{(p+1)(p+2)\cdots (p+n+1)}.$$

Also, it is easy to see

$$\frac{1}{p+1}⩽\frac{n+1}{p+n+1}⩽\frac{(n+1)!(n+1)}{(p+1)(p+2)\cdots (p+n+1)}⩽\frac{n+1}{p+1}.$$

Thus,

$${\parallel g_{z_0}\parallel }_{Q_p}^2⩽\underset{b\in \mathbb{D}}{sup}\frac{{(1-{|b|}^2)}^2}{p+1}\sum _{n=0}^{+\mathrm{\infty }}(n+1){|b|}^{2n}=\underset{b\in \mathbb{D}}{sup}\frac{{(1-{|b|}^2)}^2}{p+1}\frac{1}{{(1-{|b|}^2)}^2}=\frac{1}{p+1},$$

and

$${\parallel g_{z_0}\parallel }_{Q_p}^2⩾\underset{b\in \mathbb{D}}{sup}\frac{{(1-{|b|}^2)}^2}{p+1}\sum _{n=0}^{+\mathrm{\infty }}{|b|}^{2n}=\underset{b\in \mathbb{D}}{sup}\frac{{(1-{|b|}^2)}^2}{p+1}\frac{1}{1-{|b|}^2}=\frac{1}{p+1}.$$

Then the proof is complete. □

First, we consider the norm of $I_f$ on $Q_p$, $0<p<1$.

Theorem 2.2 Let $0<p<1$. If $f\in H(\mathbb{D})$, then $I_f$ is bounded on $Q_p$ if and only if $f\in H^{\mathrm{\infty }}$. Moreover,

$$\parallel I_f\parallel ={\parallel f\parallel }_{H^{\mathrm{\infty }}}.$$

Proof For any $h\in Q_p$ with ${\parallel h\parallel }_{Q_p}=1$, it is trivial that $\parallel I_f\parallel ⩽{\parallel f\parallel }_{H^{\mathrm{\infty }}}$. To prove the converse, define $c={sup}_{z\in \mathbb{D}}|f(z)|$. Given any $ϵ>0$, there exists $z_1\in \mathbb{D}$ such that $|f(z_1)|>c-ϵ$. Let $h(z)=g_{z_1}(z)/{\parallel g_{z_1}\parallel }_{Q_p}$, where

$$g_{z_1}(z)=\frac{z_1-z}{1-{\overline{z}}_{1}z}-z_1.$$

It is easy to see that

$${\parallel h\parallel }_{Q_p}=1,|h^{\prime }(z_1)|(1-{|z_1|}^2)=1/{\parallel g_{z_1}\parallel }_{Q_p}.$$

Henceforth,

$$\begin{array}{rl}{\parallel I_f\parallel }^2& ⩾{\parallel I_{f}h\parallel }_{Q_p}^2=\underset{a\in \mathbb{D}}{sup}{\int }_{\mathbb{D}}{|h^{\prime }(z)f(z)|}^2{(1-{|{\phi }_a(z)|}^2)}^{p}d\sigma (z)\\ =\underset{a\in \mathbb{D}}{sup}{\int }_{\mathbb{D}}{|h^{\prime }({\phi }_a(w))f({\phi }_a(w)){\phi }_a^{\prime }(w)|}^2{(1-{|w|}^2)}^{p}d\sigma (w).\end{array}$$

Taking $w={\mathit{re}}^{i\theta }$ and by the subharmonicity of ${|h^{\prime }({\phi }_a(w))f({\phi }_a(w)){\phi }_a^{\prime }(w)|}^2$, we obtain

$$\begin{array}{rl}{\parallel I_f\parallel }^2& ⩾\underset{a\in \mathbb{D}}{sup}{\int }_{\mathbb{D}}{|h^{\prime }(z)f(z)|}^2{(1-{|{\phi }_a(z)|}^2)}^{p}d\sigma (z)\\ =\underset{a\in \mathbb{D}}{sup}{\int }_0^1\frac{1}{\pi }{\int }_0^{2\pi }{\left|h^{\prime }\left({\phi }_a\left({\mathit{re}}^{i\theta }\right)\right)f\left({\phi }_a\left({\mathit{re}}^{i\theta }\right)\right){\phi }_a^{\prime }\left({\mathit{re}}^{i\theta }\right)\right|}^2{(1-r^2)}^{p}rdrd\theta \\ ⩾\underset{a\in \mathbb{D}}{sup}{|h^{\prime }(a)f(a)|}^2{(1-{|a|}^2)}^{2}2{\int }_0^1{(1-r^2)}^{p}rdr\\ =\frac{1}{p+1}\underset{a\in \mathbb{D}}{sup}{|h^{\prime }(a)f(a)|}^2{(1-{|a|}^2)}^2⩾\frac{1}{p+1}{|h^{\prime }(z_1)f(z_1)|}^2{(1-{|z_1|}^2)}^2\\ ⩾\frac{1}{p+1}\frac{{|f(z_1)|}^2}{{\parallel g_{z_1}\parallel }_{Q_p}^2}.\end{array}$$

(2.2)

By Lemma 2.1 we have

$$\parallel I_f\parallel ⩾|f(z_1)|>c-ϵ.$$

Since ϵ is arbitrary, we have $\parallel I_f\parallel ⩾{sup}_{z\in \mathbb{D}}|f(z)|$ and the proof is complete. □

Next, we consider the norm of $I_f$ from $Q_p$ ($0<p<1$) to .

Theorem 2.3 Let $0<p<1$. If $f\in H(\mathbb{D})$, then $I_f$ is bounded from $Q_p$ space to space if and only if $f\in H^{\mathrm{\infty }}$. Moreover, we have

$$\parallel I_f\parallel ={(p+1)}^{1/2}{\parallel f\parallel }_{H^{\mathrm{\infty }}}.$$

Proof If $f\in H^{\mathrm{\infty }}$, then (1.2) gives

$${\parallel I_{f}h\parallel }_{\mathfrak{B}}=\underset{z\in \mathbb{D}}{sup}|f(z)h^{\prime }(z)|(1-{|z|}^2)⩽{\parallel f\parallel }_{H^{\mathrm{\infty }}}\underset{z\in \mathbb{D}}{sup}|h^{\prime }(z)|(1-{|z|}^2).$$

From a part of the proof of estimate (2.2) for $f\equiv 1$, we see that

$$\underset{z\in \mathbb{D}}{sup}|h^{\prime }(z)|(1-{|z|}^2)⩽{(p+1)}^{1/2}{\parallel h\parallel }_{Q_p},$$

and so

$${\parallel I_{f}h\parallel }_{\mathfrak{B}}⩽{\parallel f\parallel }_{H^{\mathrm{\infty }}}{(p+1)}^{1/2}{\parallel h\parallel }_{Q_p}.$$

This leads to

$$\parallel I_f\parallel ⩽{(p+1)}^{1/2}{\parallel f\parallel }_{H^{\mathrm{\infty }}}.$$

On the other hand, define $c={sup}_{z\in \mathbb{D}}|f(z)|$. Given any $ϵ>0$, there exists $z_1\in \mathbb{D}$ such that $|f(z_1)|>c-ϵ$. Let $h(z)=g_{z_1}(z)/{\parallel g_{z_1}\parallel }_{Q_p}$, where

$$g_{z_1}(z)=\frac{z_1-z}{1-{\overline{z}}_{1}z}-z_1.$$

This together with Lemma 2.1 gives the following:

$$\begin{array}{rl}\parallel I_f\parallel & ⩾{\parallel I_{f}h\parallel }_{\mathfrak{B}}=\underset{z\in \mathbb{D}}{sup}|f(z)h^{\prime }(z)|(1-{|z|}^2)⩾|f(z_1)h^{\prime }(z_1)|(1-{|z_1|}^2)\\ =|f(z_1)|/{\parallel g_{z_1}\parallel }_{Q_p}>{(p+1)}^{1/2}(c-ϵ).\end{array}$$

Since ϵ is arbitrary, we have

$$\parallel I_f\parallel ⩾{(p+1)}^{1/2}\underset{z\in \mathbb{D}}{sup}|f(z)|={(p+1)}^{1/2}{\parallel f\parallel }_{H^{\mathrm{\infty }}}.$$

The proof is complete. □

Finally, we consider the norm of the integral operator $I_f$ on ${\mathfrak{B}}^{\alpha }$, $0<\alpha <1$.

Theorem 2.4 Let $0<\alpha <1$ and $f\in H(\mathbb{D})$. Then the integral operator $I_f$ is bounded on ${\mathfrak{B}}^{\alpha }$ if and only if $f\in H^{\mathrm{\infty }}$. Moreover,

$$\parallel I_f\parallel ={\parallel f\parallel }_{H^{\mathrm{\infty }}}.$$

Proof For any $h\in {\mathfrak{B}}^{\alpha }$ with ${\parallel h\parallel }_{{\mathfrak{B}}^{\alpha }}=1$, by (1.3) we have

$${\parallel I_{f}h\parallel }_{{\mathfrak{B}}^{\alpha }}=\underset{z\in \mathbb{D}}{sup}{(1-{|z|}^2)}^{\alpha }|f(z)||h^{\prime }(z)|⩽{\parallel h\parallel }_{{\mathfrak{B}}^{\alpha }}\cdot {\parallel f\parallel }_{H^{\mathrm{\infty }}}.$$

This implies $\parallel I_f\parallel ⩽{\parallel f\parallel }_{H^{\mathrm{\infty }}}$.

Now we need to show the reverse inequality. Define $c={sup}_{z\in \mathbb{D}}|f(z)|$. Given any $ϵ>0$, there exists $z_1\in \mathbb{D}$ such that $|f(z_1)|>c-ϵ$. Put

$$h(z)={\int }_{\mathrm{\Gamma }(z)}\frac{{(1-{|z_1|}^2)}^{\alpha }}{{(1-{\overline{z}}_1\zeta )}^{2\alpha }}d\zeta ,$$

(2.3)

where $\mathrm{\Gamma }(z)$ is any path in from 0 to z, and a single-valued analytic branch is specified. By Theorem 13.11 in [[36], p.274], we know h is an analytic function in and $h^{\prime }(z)={(1-{|z_1|}^2)}^{\alpha }/{(1-{\overline{z}}_{1}z)}^{2\alpha }$. Also, it is easy to check ${\parallel h\parallel }_{{\mathfrak{B}}^{\alpha }}=1$. In fact,

$$\begin{array}{rl}{\parallel h\parallel }_{{\mathfrak{B}}^{\alpha }}& =\underset{z\in \mathbb{D}}{sup}|h^{\prime }(z)|{(1-{|z|}^2)}^{\alpha }=\underset{z\in \mathbb{D}}{sup}\frac{{(1-{|z_1|}^2)}^{\alpha }}{{|1-{\overline{z}}_{1}z|}^{2\alpha }}{(1-{|z|}^2)}^{\alpha }\\ ⩽\underset{z\in \mathbb{D}}{sup}\frac{{(1-{|z_1|}^2)}^{\alpha }{(1-{|z|}^2)}^{\alpha }}{{(1-|z_1||z|)}^{2\alpha }}⩽1.\end{array}$$

(2.4)

On the other hand, we have

$$\begin{array}{rl}{\parallel h\parallel }_{{\mathfrak{B}}^{\alpha }}& =\underset{z\in \mathbb{D}}{sup}|h^{\prime }(z)|{(1-{|z|}^2)}^{\alpha }=\underset{z\in \mathbb{D}}{sup}\frac{{(1-{|z_1|}^2)}^{\alpha }}{{|1-{\overline{z}}_{1}z|}^{2\alpha }}{(1-{|z|}^2)}^{\alpha }\\ ⩾\frac{{(1-{|z_1|}^2)}^{\alpha }}{{(1-{|z_1|}^2)}^{2\alpha }}{(1-{|z_1|}^2)}^{\alpha }=1.\end{array}$$

(2.5)

Hence, the assertion follows by (2.4) and (2.5). Thus

$$\begin{array}{rl}\parallel I_f\parallel & ⩾{\parallel I_{f}h\parallel }_{{\mathfrak{B}}^{\alpha }}=\underset{z\in \mathbb{D}}{sup}|f(z)h^{\prime }(z)|{(1-{|z|}^2)}^{\alpha }⩾|f(z_1)h^{\prime }(z_1)|{(1-{|z_1|}^2)}^{\alpha }\\ ⩾|f(z_1)|\frac{{(1-{|z_1|}^2)}^{\alpha }}{{(1-{|z_1|}^2)}^{2\alpha }}{(1-{|z_1|}^2)}^{\alpha }=|f(z_1)|>c-ϵ.\end{array}$$

Since the ϵ is arbitrary, the proof is complete. □