Weighted Composition-Differentiation Operators on the Hardy and Bergman Spaces

Abstract

We study weighted composition-differentiation operators on the Hardy and Bergman spaces in the unit disk. We first find necessary and sufficient conditions for weighted composition-differentiation operators to be Hilbert–Schmidt. We then characterize convergence of sequences of composition-differentiation operators acting on the Hardy space both in weak operator topology, and in strong operator topology.

1 Introduction

Let ${{{\mathbb {D}}}}$ denote the open unit disk in the complex plane, and let ${\mathcal {H}}$ denote certain functional Hilbert space of analytic functions on the unit disk. For instance, we may assume that ${\mathcal {H}}$ is the classical Hardy space, the Bergman space, the Dirichlet space, and so on. For an analytic self-mapping $\varphi$ on the unit disk, the composition operator $C_\varphi :{\mathcal {H}} \rightarrow {\mathcal {H}}$ is defined by

$$\begin{aligned} C_\varphi (f)=f\circ \varphi . \end{aligned}$$

It is well-known [1, Corollary 3.7] that the composition operator is bounded on the Hardy space $H^2$ (see §2 for definition) and

$$\begin{aligned} \left( \frac{1}{1-|\varphi (0)|^2}\right) ^{1/2}\le \Vert C_\varphi \Vert \le \left( \frac{1+|\varphi (0)|}{1-|\varphi (0)|}\right) ^{1/2}. \end{aligned}$$

For an analytic function $\psi$, the weighted composition operator $C_{\psi ,\varphi }:{\mathcal {H}} \rightarrow {\mathcal {H}}$ is given by

$$\begin{aligned} C_{\psi ,\varphi }(f)=\psi \cdot f\circ \varphi . \end{aligned}$$

In the context of analytic functions, it is easy to verify that the differentiation operator $D(f)=f^\prime$ is not bounded on the Hardy space $H^2$; since $\{z^n\}_{n\ge 1}$ is a sequence of unit vectors in the Hardy space satisfying $\Vert D(z^n)\Vert =n$. Nevertheless, for many analytic self-mappings $\varphi$ on the unit disk, the operator $D_\varphi :H^2\rightarrow H^2$ defined by

$$\begin{aligned} D_\varphi (f)=f^\prime \circ \varphi \end{aligned}$$

is bounded. We follow Fatehi and Hammond [4] to call $D_\varphi$ a composition-differentiation operator. In [9], Shuichi Ohno established a set of sufficient conditions that guarantee when the operator $D_\varphi$ is bounded or compact. In particular, S. Ohno proved that if $\Vert \varphi \Vert _\infty <1$, then $D_\varphi$ is a Hilbert–Schmidt operator; and hence bounded and compact; see [9, Theorem 3.3]. We recall that an operator T on a separable Hilbert space ${\mathcal {X}}$ is said to be Hilbert–Schmidt if for some orthonormal basis ${e_n} \subset \mathcal {X}$ we have

$$\begin{aligned} \sum _{n=0}^\infty \Vert Te_n\Vert ^2<\infty . \end{aligned}$$

According to [9, Corollary 3.2], for a univalent self-map $\varphi$ of the unit disk, the operator $D_\varphi$ on the Hardy space $H^2$ is bounded if and only if

$$\begin{aligned} \sup _{z\in {\mathbb {D}}}\frac{1-|z|}{(1-|\varphi (z)|)^3}<\infty . \end{aligned}$$

Moreover, the operator $D_\varphi$ on $H^2$ is compact if and only if

$$\begin{aligned} \lim _{|z|\rightarrow 1}\frac{1-|z|}{(1-|\varphi (z)|)^3}=0. \end{aligned}$$

Now, let $\psi$ be an analytic function on the unit disk, and define the weighted composition-differentiation operator $D_{\psi ,\varphi }:H^2\rightarrow H^2$ by the following relation:

$$\begin{aligned} D_{\psi ,\varphi }(f)=\psi \cdot f^\prime \circ \varphi . \end{aligned}$$

In Sect. 3, we characterize when the operator $D_{\psi ,\varphi }$ is Hilbert–Schmidt. We prove that the composition-differentiation operator $D_{\psi ,\varphi }$ is Hilbert–Schmidt if and only if

$$\begin{aligned} \sup _{0\le r<1}\left\{ \frac{1}{2\pi }\int _0^{2\pi } \frac{|\psi \left( re^{i\theta }\right) |^2}{(1-\left| \varphi \left( re^{i\theta }\right) \right| ^2)^3}\mathrm{d}\theta \right\} <\infty . \end{aligned}$$

This extends a result of S. Ohno [9, Theorem 3.3] to weighted composition-differentiation operators. We will also discuss the same problem for other types of composition-differentiation operators; in particular, the operators ($k\in {\mathbb {N}}$):

$$\begin{aligned} D^{k}_{\varphi }(f)= & {} f^{(k)}\circ \varphi ,\quad D^{k}_{\psi ,\varphi }(f)=\psi \cdot (f^{(k)}\circ \varphi ), \\ D^{(k)}_{\varphi }(f)= & {} (f^{(k)}\circ \varphi )\varphi ^\prime ,\quad D^{(k)}_{\psi ,\varphi }(f)=\psi \cdot (f^{(k)}\circ \varphi )\varphi ^\prime . \end{aligned}$$

We will find necessary and sufficient conditions for these operators to become Hilbert–Schmidt. In particular, we shall see that the operator $D^{k}_{\psi ,\varphi }$ is Hilbert–Schmidt on $H^2$ if and only if

$$\begin{aligned} \sup _{0\le r<1}\left\{ \frac{1}{2\pi }\int _0^{2\pi } \frac{|\psi \left( re^{i\theta }\right) |^2}{\left( 1-\left| \varphi \left( re^{i\theta }\right) \right| ^2\right) ^{2k+1}}\mathrm{d}\theta \right\} <\infty . \end{aligned}$$

Moreover, the operator $D^{(k)}_{\psi ,\varphi }$ is Hilbert–Schmidt on $H^2$ if and only if

$$\begin{aligned} \sup _{0\le r<1}\left\{ \frac{1}{2\pi }\int _0^{2\pi } \frac{|\psi \left( re^{i\theta }\right) \varphi ^\prime \left( re^{i\theta }\right) |^2}{\left( 1-\left| \varphi \left( re^{i\theta }\right) \right| ^2\right) ^{2k+1}}\mathrm{d}\theta \right\} <\infty . \end{aligned}$$

Similar results are established in the Bergman space (see Sect. 2, for the definition) too. The analogs of the above conditions in the Bergman space are the following; the functions $\psi$ and $\varphi$ have to satisfy

$$\begin{aligned} \int _{{\mathbb {D}}} \frac{|\psi (z)|^2}{(1-\left| \varphi (z)\right| ^2)^{2k+2}}\mathrm{d}A(z)<\infty , \end{aligned}$$

(respectively),

$$\begin{aligned} \int _{{\mathbb {D}}} \frac{\left| \psi (z)\varphi ^\prime (z)\right| ^2}{(1-\left| \varphi (z)\right| ^2)^{2k+2}}\mathrm{d}A(z)<\infty , \end{aligned}$$

where $\mathrm{d}A(z)=\mathrm{d}x\,\mathrm{d}y$ is the area measure on the unit disk.

Our next objective in Sect. 4 is to find the relationships between convergence of the sequence of operators $D_{\psi _n,\varphi _n}$ in operator topologies from one hand, and the convergence of the sequences of functions $\psi _n$ and $\varphi _n$ on the other hand. From historical point of view, Howard Schwarz [11] studied the relationship between the convergence of sequence of composition operators $C_{\varphi _n}$, and the convergence of the sequence of self-maps $\varphi _n$. Valetine Matache [7] extended Schwarz’s work by relating the convergence of $C_{\varphi _n}$ in Hilbert–Schmidt norm to the convergence of the sequence $\varphi _n$. Then appeared Gunatillake’s paper [6] on the relationship between convergence of weighted composition operators $C_{\psi _n,\varphi _n}$, and the convergence of $\{\psi _n\}$ and $\{\varphi _n\}$. This latter was extended by Mehrangiz and Khani-Robati [8] to generalized weighted composition operators on Bloch type spaces. Here we intend to generalize Gunatillake’s result to weighted composition-differentiation operator $D_{\psi ,\varphi }$ in the setting of classical Hardy spaces. More specifically, let ${\mathcal {B}}(H^2)$ denote the Banach algebra of all bounded linear operators on the Hilbert space $H^2$. It is rather well-known that the dual space of ${\mathcal {B}}(H^2)$ is too big, so that the weak and weak-star topology of this space is not so clear. For this reason, it is customary to equip this space with the weak operator topology, the strong operator topology, and the uniform operator topology. We intend to have a characterization of the convergence of $D_{\psi _n,\varphi _n}$ to $D_{\psi ,\varphi }$ with respect to operator topologies in terms of the convergence of $\varphi _n\rightarrow \varphi$ and $\psi _n\rightarrow \psi$ in the weak and strong operator topologies of $H^2$.

2 Preliminaries

Let f be an analytic function in the unit disk ${\mathbb {D}}$. The function f is said to belong to the Hardy space $H^2$ if

$$\begin{aligned} \Vert f\Vert ^2=\sup _{0\le r<1}\frac{1}{2\pi }\int _0^{2\pi }|f\left( re^{i\theta }\right) |^2\mathrm{d}\theta <\infty . \end{aligned}$$

It is easy to see that for an analytic function $f(z)=\sum _{n=0}^\infty a_nz^n$, the norm of f in $H^2$ is given by

$$\begin{aligned} \Vert f\Vert ^2=\sum _{n=0}^\infty |a_n|^2. \end{aligned}$$

It is well-known (see for instance, [10]) that for $f\in H^2$, the radial limit

$$\begin{aligned} f^*(e^{i\theta }):=\lim _{r\rightarrow 1^-}f\left( re^{i\theta }\right) =\lim _{r\rightarrow 1^-}f_r\left( e^{i\theta }\right) \end{aligned}$$

for almost every $\theta \in [0, 2\pi ]$ exists. The function $f^*$ is known as the radial function of f. The space $H^2$ is a functional Hilbert space, and its inner product is given by

$$\begin{aligned} \langle f,g\rangle =\frac{1}{2\pi }\int _0^{2\pi }f^*\left( e^{i\theta }\right) \overline{g^*\left( e^{i\theta }\right) }\mathrm{d}\theta . \end{aligned}$$

Since the evaluation functionals are bounded, the Hardy space is a reproducing kernel Hilbert space; this means that for each $w\in {\mathbb {D}}$, there is a function

$$\begin{aligned} K_w(z)=\frac{1}{1-{\overline{w}} z}\in H^2 \end{aligned}$$

such that every $f\in H^2$ has the following representation

$$\begin{aligned} f(w)=\langle f, K_w\rangle =\frac{1}{2\pi }\int _0^{2\pi }f^*(e^{i\theta })\overline{K_w^*(e^{i\theta })}\mathrm{d}\theta . \end{aligned}$$

It is also well-known that the functional $w\mapsto f^\prime (w)$ is bounded on $H^2$ ( [1, Theorem 2.16]). It then follows from the Riesz representation theorem that there is a function $K_w^{(1)}\in H^2$ such that

$$\begin{aligned} f^\prime (w)=\langle f, K_w^{(1)}\rangle ,\quad f\in H^2. \end{aligned}$$

It turns out that (see §4)

$$\begin{aligned} K_w^{(1)}(z)=\frac{z}{(1-{\overline{w}} z)^2},\quad (z,w)\in {\mathbb {D}}\times {\mathbb {D}}. \end{aligned}$$

Another functional Hilbert space on the unit disk is the Bergman space $L^2_a$ consisting of all analytic functions f in the unit disk for which the integral

$$\begin{aligned} \frac{1}{\pi }\int _{\mathbb {D}}|f(z)|^2 \mathrm{d}A(z) \end{aligned}$$

is finite; here $\mathrm{d}A(z)=\mathrm{d}x\mathrm{d}y$ is the usual area measure in the complex plane. The norm of f is defined by

$$\begin{aligned} \Vert f\Vert _{L^2_a}=\left( \frac{1}{\pi }\int _{\mathbb {D}}|f(z)|^2 \mathrm{d}A(z)\right) ^{1/2}. \end{aligned}$$

A computation reveals that for $f(z)=\sum _{n=0}^\infty a_nz^n$, we have

$$\begin{aligned} \Vert f\Vert ^2_{L^2_a}=\sum _{n=0}^\infty \frac{|a_n|^2}{n+1}, \end{aligned}$$

from which we conclude that $H^2\subset L^2_a$.

We now turn to recall different notions of convergence in the space of bounded linear operators on a given Hilbert space. Let $\mathcal {B(H)}$ denote the algebra of bounded linear operators on a Hilbert space ${\mathcal {H}}$. We consider a sequence $T_n\in \mathcal {B(H)}$. We say that $T_n$ converges to T in weak operator topology, and write $T_n\rightarrow T$ (WOT) if for each $x\in {\mathcal {H}}$, $T_nx\rightarrow Tx$ weakly. Similarly, we say that $T_n$ converges to T in strong operator topology, and write $T_n\rightarrow T$ (SOT) if for each $x\in {\mathcal {H}}$, $T_nx\rightarrow Tx$ (in norm).

In this paper, we are concerned with bounded operators of the form $D_{\psi ,\varphi }$ on $H^2$. It should be emphasized that if $\psi$ is a bounded analytic function on the open unit disk, and if $\varphi$ is a non-constant self-map of the unit disk such that

$$\begin{aligned} \Vert \varphi \Vert _\infty =\sup \{ |\varphi (z)|: z\in {\mathbb {D}}\} \le r<1, \end{aligned}$$

then the composition-differentiation operator $D_{\psi ,\varphi }$ is bounded; indeed, by a result due to Fatehi and Hammond ( [4, Proposition 4]) we have

$$\begin{aligned} \Vert D_\varphi \Vert \le \left( \frac{r+|\varphi (0)|}{r-|\varphi (0)|}\right) ^{1/2}\left\lfloor \frac{1}{1-r}\right\rfloor r^{\left\lfloor \frac{1}{1-r}\right\rfloor -1}, \end{aligned}$$

(1)

where $\lfloor \cdot \rfloor$ denotes the greatest integer function. On the other hand, if $\psi$ is bounded, then the operator of multiplication by $\psi$, i.e. $M_\psi (f)=\psi f$ is bounded, and its norm does not exceed $\Vert \psi \Vert _\infty$. Therefore,

$$\begin{aligned} D_{\psi ,\varphi }(f)=M_\psi D_\varphi (f),\quad f\in H^2, \end{aligned}$$

is bounded too. We mention that there are instances of bounded operators when $\Vert \varphi \Vert _\infty = 1$. Indeed, if we know that $D_\varphi$ maps $H^2$ into itself, it follows from Closed Graph Theorem that the operator $D_\varphi$ and hence the operator $D_{\psi ,\varphi }$ is bounded (for a bounded analytic function $\psi$). To see this, let $f_n$ be a sequence of analytic self-maps such that $f_n\rightarrow f$ in $H^2$, and $D_{\varphi }(f_n)\rightarrow g$ in $H^2$. But norm convergence implies pointwise convergence, so that $f^\prime _n(\varphi (z))\rightarrow g(z)$ pointwise. Since $f^\prime _n(z)\rightarrow f^\prime (z)$, it follows that $f^\prime _n(\varphi (z))\rightarrow f^\prime (\varphi (z))$. By the uniqueness of limit, we conclude that $f^\prime (\varphi (z))=g(z)$ from which it follows that $D_{\varphi }(f_n)\rightarrow D_{\varphi }(f)$.

3 Hilbert–Schmidt operators

This section is devoted to the study of Hilbert–Schmidt composition-differentiation operators. We start by finding conditions on $\psi$ and $\varphi$ to guarantee that the weighted composition-differentiation operator $D_{\psi ,\varphi }$ is a Hilbert–Schmidt operator. We recall that an operator $T\in \mathcal {B(X)}$ is called Hilbert–Schmidt if for some orthonormal basis ${e_n} \subset \mathcal {X}$ we have $\sum _n\Vert Te_n\Vert ^2<\infty$ (here ${\mathcal {X}}$ is a separable Hilbert space). It is well-known that every Hilbert–Schmidt operator is compact (see [3, page 87]). Assume that $\varphi$ is an analytic self-map of the unit disk, and $\psi$ is an analytic function on the unit disk. In this section, in addition to $D_{\psi ,\varphi }(f)=\psi \cdot f^\prime \circ \varphi$, we shall consider the following weighted composition-differentiation operators on a given functional Hilbert space ${\mathcal {H}}$ (this functional Hilbert space is either the Hardy space $H^2$ or the Bergman space $L^2_a$, moreover $k\ge 1$ is an integer):

$$\begin{aligned} D^{k}_{\varphi }(f)= & {} f^{(k)}\circ \varphi ,\quad D^{k}_{\psi ,\varphi }(f)=\psi \cdot (f^{(k)}\circ \varphi ), \\ D^{(k)}_{\varphi }(f)= & {} (f^{(k)}\circ \varphi )\varphi ^\prime ,\quad D^{(k)}_{\psi ,\varphi }(f)=\psi \cdot (f^{(k)}\circ \varphi )\varphi ^\prime . \end{aligned}$$

Note that in case $k=1$ or $k=2$, it is easier to use the notations

$$\begin{aligned} D^\prime _\varphi (f)= & {} (f^\prime \circ \varphi )\varphi ^\prime ,\quad D^\prime _{\psi ,\varphi }(f)=\psi \cdot (f^\prime \circ \varphi )\varphi ^\prime , \\ D^{\prime \prime }_\varphi (f)= & {} (f^{\prime \prime }\circ \varphi )\varphi ^\prime ,\quad D^{\prime \prime }_{\psi ,\varphi }(f)=\psi \cdot (f^{\prime \prime }\circ \varphi )\varphi ^\prime . \end{aligned}$$

Some authors (see for instance [9]) denote the operator $D^\prime _\varphi (f)$ by

$$\begin{aligned} DC_\varphi f(z)=f^\prime (\varphi (z))\varphi ^\prime (z). \end{aligned}$$

Our first theorem generalizes a similar result proved in [9] for $D_\varphi$.

Theorem 3.1

Let $\varphi$ be an analytic self-map of the unit disk, and $\psi$ be an analytic function on the unit disk. Then $D_{\psi ,\varphi }$ is a Hilbert–Schmidt operator on $H^2$ if and only if

$$\begin{aligned} \sup _{0\le r<1}\left\{ \frac{1}{2\pi }\int _0^{2\pi } \frac{|\psi \left( re^{i\theta }\right) |^2}{(1-\left| \varphi \left( re^{i\theta }\right) \right| ^2)^3}\mathrm{d}\theta \right\} <\infty . \end{aligned}$$

Proof

For the orthonormal basis $\{z^n\}$ for $H^2$, we have

$$\begin{aligned} \sum _{n=0}^\infty \Vert D_{\psi ,\varphi }(z^n)\Vert ^2&=\sum _{n=1}^\infty \sup _{0\le r<1}\left\{ \frac{1}{2\pi }\int _0^{2\pi }\left| \psi \left( re^{i\theta }\right) n\varphi \left( re^{i\theta }\right) ^{n-1}\right| ^2\mathrm{d}\theta \right\} \\&= \sup _{0\le r<1}\left\{ \frac{1}{2\pi }\int _0^{2\pi } |\psi \left( re^{i\theta }\right) |^2 \sum _{n=0}^\infty (n+1)^2 \left| \varphi (re^{i\theta })\right| ^{2n}\mathrm{d}\theta \right\} \\&\le \sup _{0\le r<1}\left\{ \frac{1}{2\pi }\int _0^{2\pi } \frac{2 |\psi (re^{i\theta })|^2}{(1-\left| \varphi (re^{i\theta })\right| ^2)^3}\mathrm{d}\theta \right\} . \end{aligned}$$

Therefore, $\sum _{n=0}^\infty \Vert D_{\psi ,\varphi }(z^n)\Vert ^2$ is finite if

$$\begin{aligned} \sup _{0\le r<1}\left\{ \frac{1}{2\pi }\int _0^{2\pi } \frac{|\psi \left( re^{i\theta }\right) |^2}{(1-\left| \varphi \left( re^{i\theta }\right) \right| ^2)^3}\mathrm{d}\theta \right\} <\infty . \end{aligned}$$

For the reverse implication note that

$$\begin{aligned} \infty > \sum _{n=0}^\infty \Vert D_{\psi ,\varphi }(z^n)\Vert ^2&=\sum _{n=1}^\infty \sup _{0\le r<1}\left\{ \frac{1}{2\pi }\int _0^{2\pi }\left| \psi \left( re^{i\theta }\right) n\varphi \left( re^{i\theta }\right) ^{n-1}\right| ^2\mathrm{d}\theta \right\} \\&\ge \sup _{0\le r<1}\left\{ \frac{1}{2\pi }\int _0^{2\pi } \frac{|\psi \left( re^{i\theta }\right) |^2}{\left( 1-\left| \varphi \left( re^{i\theta }\right) \right| ^2\right) ^3}\mathrm{d}\theta \right\} . \end{aligned}$$

$\square$

Corollary 3.2

Let $\varphi$ be an analytic self-map of the unit disk, and $\psi$ be an analytic function on the unit disk. Then $D^\prime _{\psi ,\varphi }$ is a Hilbert–Schmidt operator on $H^2$ if and only if

$$\begin{aligned} \sup _{0\le r<1}\left\{ \frac{1}{2\pi }\int _0^{2\pi } \frac{\left| \psi \left( re^{i\theta }\right) \varphi ^\prime \left( re^{i\theta }\right) \right| ^2}{\left( 1-\left| \varphi \left( re^{i\theta }\right) \right| ^2\right) ^3}\mathrm{d}\theta \right\} <\infty . \end{aligned}$$

Proof

We just note that in this case

$$\begin{aligned} D^\prime _{\psi ,\varphi }(z^n)\left( re^{i\theta }\right) =\psi \left( re^{i\theta }\right) \cdot n\varphi \left( re^{i\theta }\right) ^{n-1}\cdot \varphi ^\prime \left( re^{i\theta }\right) . \end{aligned}$$

The rest calculations remain unchanged. $\square$

Theorem 3.3

Let $\varphi$ be an analytic self-map of the unit disk, and $\psi$ be an analytic function on the unit disk. Then $D^{2}_{\psi ,\varphi }$ is a Hilbert–Schmidt operator on $H^2$ if and only if

$$\begin{aligned} \sup _{0\le r<1}\left\{ \frac{1}{2\pi }\int _0^{2\pi } \frac{|\psi \left( re^{i\theta }\right) |^2}{\left( 1-\left| \varphi \left( re^{i\theta }\right) \right| ^2\right) ^5}\mathrm{d}\theta \right\} <\infty . \end{aligned}$$

(2)

Proof

First, assume that (2) holds true. We note that

$$\begin{aligned} D^{2}_{\psi ,\varphi }(z^n)\left( re^{i\theta }\right) =\psi \left( re^{i\theta }\right) \cdot n(n-1)\varphi \left( re^{i\theta }\right) ^{n-2}. \end{aligned}$$

Using the series expansion

$$\begin{aligned} \sum _{n=0}^\infty (n+1)^2(n+2)^2x^n=\frac{4x^2+16x+4}{(1-x)^5},\quad |x|<1, \end{aligned}$$

we conclude that

$$\begin{aligned} \sum _{n=0}^\infty \Vert&D^{2}_{\psi ,\varphi }(z^n)\Vert ^2=\sum _{n=2}^\infty \sup _{0\le r<1}\left\{ \frac{1}{2\pi }\int _0^{2\pi }\left| \psi \left( re^{i\theta }\right) n(n-1)\varphi \left( re^{i\theta }\right) ^{n-2}\right| ^2\mathrm{d}\theta \right\} \\&=\sup _{0\le r<1}\left\{ \frac{1}{2\pi }\int _0^{2\pi }\left| \psi \left( re^{i\theta }\right) \right| ^2 \sum _{n=0}^\infty (n+2)^2(n+1)^2 \left| \varphi \left( re^{i\theta }\right) \right| ^{2n}\mathrm{d}\theta \right\} \\&=\sup _{0\le r<1}\left\{ \frac{1}{2\pi }\int _0^{2\pi } \frac{|\psi \left( re^{i\theta }\right) |^2\left( 4|\varphi \left( re^{i\theta }\right) |^4+16|\varphi \left( re^{i\theta }\right) |^2+4\right) }{\left( 1-\left| \varphi \left( re^{i\theta }\right) \right| ^2\right) ^5}\mathrm{d}\theta \right\} \\&\le \sup _{0\le r<1}\left\{ \frac{1}{2\pi }\int _0^{2\pi } \frac{24|\psi \left( re^{i\theta }\right) |^2}{\left( 1-\left| \varphi \left( re^{i\theta }\right) \right| ^2\right) ^5}\mathrm{d}\theta \right\} <\infty . \end{aligned}$$

Conversely, if $D^{2}_{\psi ,\varphi }$ is a Hilbert–Schmidt operator, then

$$\begin{aligned} \infty > \sum _{n=0}^\infty \Vert&D^{2}_{\psi ,\varphi }(z^n)\Vert ^2=\sum _{n=2}^\infty \sup _{0\le r<1}\left\{ \frac{1}{2\pi }\int _0^{2\pi }\left| \psi \left( re^{i\theta }\right) n(n-1)\varphi \left( re^{i\theta }\right) ^{n-2}\right| ^2\mathrm{d}\theta \right\} \\&=\sup _{0\le r<1}\left\{ \frac{1}{2\pi }\int _0^{2\pi } \frac{|\psi \left( re^{i\theta }\right) |^2\left( 4|\varphi \left( re^{i\theta }\right) |^4+16|\varphi \left( re^{i\theta }\right) |^2+4\right) }{(1-\left| \varphi \left( re^{i\theta }\right) \right| ^2)^5}\mathrm{d}\theta \right\} \\&\ge \sup _{0\le r<1}\left\{ \frac{1}{2\pi }\int _0^{2\pi } \frac{4|\psi \left( re^{i\theta }\right) |^2}{\left( 1-\left| \varphi \left( re^{i\theta }\right) \right| ^2\right) ^5}\mathrm{d}\theta \right\} , \end{aligned}$$

which implies that (2) is satisfied. $\square$

Corollary 3.4

Let $\varphi$ be an analytic self-map of the unit disk, and $\psi$ be an analytic function on the unit disk. Then $D^{\prime \prime }_{\psi ,\varphi }$ is a Hilbert–Schmidt operator on $H^2$ if and only if

$$\begin{aligned} \sup _{0\le r<1}\left\{ \frac{1}{2\pi }\int _0^{2\pi } \frac{\left| \psi \left( re^{i\theta }\right) \varphi ^\prime \left( re^{i\theta }\right) \right| ^2}{\left( 1-\left| \varphi \left( re^{i\theta }\right) \right| ^2\right) ^5}\mathrm{d}\theta \right\} <\infty . \end{aligned}$$

Proof

In this case note that

$$\begin{aligned} D^{\prime \prime }_{\psi ,\varphi }(z^n)(re^{i\theta })=\psi (re^{i\theta })\cdot n(n-1)\varphi (re^{i\theta })^{n-2}\varphi ^\prime (re^{i\theta }). \end{aligned}$$

The rest calculations remain unchanged. $\square$

Theorem 3.5

Let $\varphi$ be an analytic self-map of the unit disk, and $\psi$ be an analytic function on the unit disk. Then $D^3_{\psi ,\varphi }$ is a Hilbert–Schmidt operator on $H^2$ if and only if

$$\begin{aligned} \sup _{0\le r<1}\left\{ \frac{1}{2\pi }\int _0^{2\pi } \frac{|\psi (re^{i\theta })|^2}{(1-\left| \varphi (re^{i\theta })\right| ^2)^7}\mathrm{d}\theta \right\} <\infty . \end{aligned}$$

(3)

Proof

For the sake of simplicity, let us assume that $\psi \equiv 1$. We first assume that (3) holds true. Since

$$\begin{aligned} D^3_\varphi (z^n)(re^{i\theta })=n(n-1)(n-2)\varphi (re^{i\theta })^{n-3}, \end{aligned}$$

we can write

$$\begin{aligned} \sum _{n=0}^\infty \Vert D^3_{\varphi }(z^n)\Vert ^2&=\sum _{n=3}^\infty \sup _{0\le r<1}\left\{ \frac{1}{2\pi }\int _0^{2\pi }\left| n(n-1)(n-2)\varphi (re^{i\theta })^{n-3}\right| ^2\mathrm{d}\theta \right\} \\&=\sup _{0\le r<1}\left\{ \frac{1}{2\pi }\int _0^{2\pi } \sum _{n=0}^\infty (n+1)^2(n+2)^2(n+3)^2 \left| \varphi (re^{i\theta })\right| ^{2n}\mathrm{d}\theta \right\} .\\ \end{aligned}$$

Note that

$$\begin{aligned} \sum _{n=0}^\infty (n+1)^2(n+2)^2(n+3)^2x^n=\frac{36(x^3+9x^2+9x+3)}{(1-x)^7},\quad |x|<1, \end{aligned}$$

(4)

so that

$$\begin{aligned} \sum _{n=0}^\infty \Vert D^3_{\varphi }(z^n)\Vert ^2&=\sup _{0\le r<1}\left\{ \frac{1}{2\pi }\int _0^{2\pi } \frac{36\left( |\varphi (re^{i\theta })|^6+9|\varphi (re^{i\theta })|^4+9|\varphi (re^{i\theta })|^2+3\right) }{\left( 1-\left| \varphi (re^{i\theta })\right| ^2\right) ^7}\mathrm{d}\theta \right\} \\&\le \sup _{0\le r<1}\left\{ \frac{1}{2\pi }\int _0^{2\pi } \frac{(36)(22)}{\left( 1-\left| \varphi (re^{i\theta })\right| ^2\right) ^7}\mathrm{d}\theta \right\} <\infty . \end{aligned}$$

On the other hand,

$$\begin{aligned} \infty >\sum _{n=0}^\infty \Vert D^3_{\varphi }(z^n)\Vert ^2&=\sup _{0\le r<1}\left\{ \frac{1}{2\pi }\int _0^{2\pi } \frac{36\left( |\varphi (re^{i\theta })|^6+9|\varphi (re^{i\theta })|^4+9|\varphi (re^{i\theta })|^2+3\right) }{(1-\left| \varphi (re^{i\theta })\right| ^2)^7}\mathrm{d}\theta \right\} \\&\ge \sup _{0\le r<1}\left\{ \frac{1}{2\pi }\int _0^{2\pi } \frac{(36)(3)}{\left( 1-\left| \varphi (re^{i\theta }\right) \right| ^2)^7}\mathrm{d}\theta \right\} . \end{aligned}$$

$\square$

Corollary 3.6

Let $\varphi$ be an analytic self-map of the unit disk, and $\psi$ be an analytic function on the unit disk. Then $D^{(3)}_{\psi ,\varphi }$ is a Hilbert–Schmidt operator on $H^2$ if and only if

$$\begin{aligned} \sup _{0\le r<1}\left\{ \frac{1}{2\pi }\int _0^{2\pi } \frac{|\psi (re^{i\theta })\varphi ^\prime (re^{i\theta })|^2}{(1-\left| \varphi (re^{i\theta })\right| ^2)^7}\mathrm{d}\theta \right\} <\infty . \end{aligned}$$

Proof

We just note that

$$\begin{aligned} D^{(3)}_\varphi (z^n)(re^{i\theta })=n(n-1)(n-2)\varphi (re^{i\theta })^{n-3}\varphi ^\prime (re^{i\theta }). \end{aligned}$$

$\square$

The above pattern can be extended to higher order weighted composition-differentiation operators. We just state the following two results.

Theorem 3.7

Let $\varphi$ be an analytic self-map of the unit disk, $\psi$ be an analytic function on the unit disk, and k be a non-negative integer. Then $D^k_{\psi ,\varphi }$ is a Hilbert–Schmidt operator on $H^2$ if and only if

$$\begin{aligned} \sup _{0\le r<1}\left\{ \frac{1}{2\pi }\int _0^{2\pi } \frac{|\psi (re^{i\theta })|^2}{(1-\left| \varphi (re^{i\theta })\right| ^2)^{2k+1}}\mathrm{d}\theta \right\} <\infty . \end{aligned}$$

Corollary 3.8

Let $\varphi$ be an analytic self-map of the unit disk, $\psi$ be an analytic function on the unit disk, and k be a non-negative integer. Then $D^{(k)}_{\psi ,\varphi }$ is a Hilbert–Schmidt operator on $H^2$ if and only if

$$\begin{aligned} \sup _{0\le r<1}\left\{ \frac{1}{2\pi }\int _0^{2\pi } \frac{|\psi (re^{i\theta })\varphi ^\prime (re^{i\theta })|^2}{(1-\left| \varphi (re^{i\theta })\right| ^2)^{2k+1}}\mathrm{d}\theta \right\} <\infty . \end{aligned}$$

Remark 3.9

If $k=0$ and $\psi$ is the constant function 1, then $D^k_{\psi ,\varphi }=C_\varphi$, so that $C_\varphi$ is a Hilbert–Schmidt operator on $H^2$ if and only if

$$\begin{aligned} \sup _{0\le r<1}\left\{ \frac{1}{2\pi }\int _0^{2\pi } \frac{1}{1-\left| \varphi (re^{i\theta })\right| ^2}\mathrm{d}\theta \right\} <\infty . \end{aligned}$$

This result is of course known (see [12, page 227, Ex. 4]).

The Bergman space

As is well-known, the Bergman spaces are special cases of weighted Hardy spaces; we mean by a weighted Hardy space, a space of analytic functions on the unit disk such that the monomials $\{z^n\}_{n\ge 0}$ constitute an orthonormal basis for the space. If $\Vert z^n\Vert =\beta (n)$, then

$$\begin{aligned} \Vert f\Vert ^2=\left\| \sum _{n=0}^\infty a_nz^n\right\| ^2=\sum _{n=0}^\infty |a_n|^2\beta (n)^2. \end{aligned}$$

If $\beta (n)=1$ for each n, we get the Hardy space, and if $\beta (n)=(n+1)^{-1/2}$ we get the Bergman space consisting of functions f that are analytic in the unit disk and satisfy

$$\begin{aligned} \Vert f\Vert ^2_{L^2_a}=\frac{1}{\pi }\int _{\mathbb {D}}|f(z)|^2 \mathrm{d}A(z)<\infty , \end{aligned}$$

where $\mathrm{d}A(z)=\mathrm{d}x\mathrm{d}y$ is the usual area measure. It is easy to see that for $f(z)=\sum _{n=0}^\infty a_nz^n$, we have

$$\begin{aligned} \Vert f\Vert ^2_{L^2_a}=\sum _{n=0}^\infty \frac{|a_n|^2}{n+1}. \end{aligned}$$

An easy computation shows that the sequence $\{\sqrt{n+1}\,z^n\}_{n=0}^\infty$ constitutes an orthonormal basis for the Bergman space $L^2_a$. It should be mentioned that the operator $C_\varphi (f)=f\circ \varphi$ is bounded on $L^2_a$ and its norm satisfies (see [12, Theorem 10.3.2])

$$\begin{aligned} \Vert C_\varphi \Vert _{L^2_a} \le \left( \frac{1+|\varphi (0)|}{1-|\varphi (0)|}\right) ^{1/2}. \end{aligned}$$

The subscript $L^2_a$ reflects the norm of operator acting on $L^2_a$. Therefore, for $\psi \in H^\infty$, the operator $D_{\psi ,\varphi }$ is bounded as long as $D_\varphi$ maps the Bergman space into itself; however there are instances in which this operator is unbounded.

We are now in a position to characterize weighted composition-differentiation operators on the Bergman space that are Hilbert–Schmidt.

Theorem 3.10

Let $\varphi$ be an analytic self-map of the unit disk, and $\psi$ be an analytic function on the unit disk. Then $D_{\psi ,\varphi }$ is a Hilbert–Schmidt operator on $L_a^2$ if and only if

$$\begin{aligned} \int _{\mathbb {D}} \frac{|\psi (z)|^2}{(1-\left| \varphi (z)\right| ^2)^4}\mathrm{d}A(z)<\infty . \end{aligned}$$

Proof

Let $\varphi$ satisfy the above condition, and $\{\sqrt{n+1}\,z^n\}$ be an orthonormal basis for $L_a^2$. Using the elementary expansion

$$\begin{aligned} \sum _{n=0}^\infty (n+1)^2(n+2)x^n=\frac{4x+2}{(1-x)^4},\quad |x|<1, \end{aligned}$$

we have

$$\begin{aligned} \sum _{n=0}^\infty \Vert D_{\psi ,\varphi }(\sqrt{n+1}\, z^n)\Vert ^2_{L^2_a}&= \sum _{n=1}^\infty \int _{\mathbb {D}}\left| \psi (z) n\sqrt{n+1}\,\varphi (z)^{n-1}\right| ^2\mathrm{d}A(z)\\&=\int _{\mathbb {D}}|\psi (z)|^2 \sum _{n=0}^\infty (n+1)^2(n+2) \left| \varphi (z)\right| ^{2n}\mathrm{d}A(z)\\&=\int _{\mathbb {D}} \frac{|\psi (z)|^2\left( 4|\varphi (z)|^2+2\right) }{(1-\left| \varphi (z)\right| ^2)^4}\mathrm{d}A(z)\\&\le 6\int _{\mathbb {D}} \frac{|\psi (z)|^2}{(1-\left| \varphi (z)\right| ^2)^4}\mathrm{d}A(z)<\infty . \end{aligned}$$

On the other hand, if $D_{\psi ,\varphi }$ is Hilbert–Schmidt on $L_a^2$, then we have

$$\begin{aligned} \infty >\sum _{n=0}^\infty \left\| D_{\psi ,\varphi }(\sqrt{n+1}\,z^n)\right\| ^2_{L^2_a}&= \sum _{n=1}^\infty \int _{\mathbb {D}}\left| \psi (z) n\sqrt{n+1}\,\varphi (z)^{n-1}\right| ^2\mathrm{d}A(z)\\&=\int _{\mathbb {D}} \frac{|\psi (z)|^2\left( 4|\varphi (z)|^2+2\right) }{(1-\left| \varphi (z)\right| ^2)^4}\mathrm{d}A(z)\\&\ge 2\int _{\mathbb {D}} \frac{|\psi (z)|^2}{(1-\left| \varphi (z)\right| ^2)^4}\mathrm{d}A(z). \end{aligned}$$

$\square$

Corollary 3.11

Let $\varphi$ be an analytic self-map of the unit disk, and $\psi$ be an analytic function on the unit disk. Then $D^\prime _{\psi ,\varphi }$ is a Hilbert–Schmidt operator on $L_a^2$ if and only if

$$\begin{aligned} \int _{\mathbb {D}} \frac{|\psi (z)\varphi ^\prime (z)|^2}{(1-\left| \varphi (z)\right| ^2)^4}\mathrm{d}A(z)<\infty . \end{aligned}$$

Proof

It is enough to notice that

$$\begin{aligned} D^\prime _{\psi ,\varphi }(\sqrt{n+1}\, z^n)=\psi (z) n\sqrt{n+1}\,\varphi (z)^{n-1}\varphi ^\prime (z). \end{aligned}$$

$\square$

Theorem 3.12

Let $\varphi$ be an analytic self-map of the unit disk, and $\psi$ be an analytic function on the unit disk. Then $D^{2}_{\psi ,\varphi }$ is a Hilbert–Schmidt operator on $L_a^2$ if and only if

$$\begin{aligned} \int _{{\mathbb {D}}} \frac{|\psi (z)|^2}{(1-\left| \varphi (z)\right| ^2)^6}\mathrm{d}A(z)<\infty . \end{aligned}$$

(5)

Proof

Let $\psi$ and $\varphi$ satisfy the condition (5). Then, using

$$\begin{aligned} \sum _{n=0}^\infty (n+1)^2(n+2)^2(n+3)x^n=\frac{36x^2+72x+12}{(1-x)^6},\quad |x|<1, \end{aligned}$$

(6)

we have

$$\begin{aligned} \sum _{n=0}^\infty \left\| D^2_{\psi ,\varphi }(\sqrt{n+1}\, z^n)\right\| ^2&=\sum _{n=2}^\infty \int _{{\mathbb {D}}}\left| \psi (z) n(n-1)\sqrt{n+1}\,\varphi (z)^{n-2}\right| ^2\\&=\int _{{\mathbb {D}}}|\psi (z)|^2\sum _{n=0}^\infty (n+1)^2(n+2)^2(n+3)|\varphi (z)|^{2n}\\&=\int _{{\mathbb {D}}}|\psi (z)|^2\frac{36|\varphi (z)|^4+72|\varphi (z)|^2+12}{(1-|\varphi (z)|^2)^6}. \end{aligned}$$

This implies that

$$\begin{aligned} \int _{\mathbb {D}}\frac{12 |\psi (z)|^2}{(1-|\varphi (z)|^2)^6}\mathrm{d}A(z)&\le \sum _{n=0}^\infty \left\| D^2_{\psi ,\varphi }(\sqrt{n+1}\, z^n)\right\| ^2 \\&\le \int _{\mathbb {D}}\frac{120 |\psi (z)|^2}{(1-|\varphi (z)|^2)^6}\mathrm{d}A(z), \end{aligned}$$

from which the result follows. $\square$

Corollary 3.13

Let $\varphi$ be an analytic self-map of the unit disk, and $\psi$ be an analytic function on the unit disk. Then $D^{(2)}_{\psi ,\varphi }$ is a Hilbert–Schmidt operator on $L_a^2$ if and only if

$$\begin{aligned} \int _{{\mathbb {D}}} \frac{|\psi (z)\varphi ^\prime (z)|^2}{(1-\left| \varphi (z)\right| ^2)^6}\mathrm{d}A(z)<\infty . \end{aligned}$$

Theorem 3.14

Let $\varphi$ be an analytic self-map of the unit disk, and $\psi$ be an analytic function on the unit disk. Then $D^{3}_{\psi ,\varphi }$ is a Hilbert–Schmidt operator on $L_a^2$ if and only if

$$\begin{aligned} \int _{{\mathbb {D}}} \frac{|\psi (z)|^2}{(1-\left| \varphi (z)\right| ^2)^8}\mathrm{d}A(z)<\infty . \end{aligned}$$

(7)

Proof

We write

$$\begin{aligned} \sum _{n=0}^\infty \left\| D^3_{\psi ,\varphi }(\sqrt{n+1}\, z^n)\right\| ^2&=\sum _{n=3}^\infty \int _{{\mathbb {D}}}\left| \psi (z) n(n-1)(n-2)\sqrt{n+1}\,\varphi (z)^{n-3}\right| ^2\\&=\int _{{\mathbb {D}}}|\psi (z)|^2\sum _{n=0}^\infty (n+1)^2(n+2)^2(n+3)^2(n+4)|\varphi (z)|^{2n}. \end{aligned}$$

Multiplying (4) through $x^4$, differentiating with respect to x, and dividing both sides by $x^3$, we obtain

$$\begin{aligned} \sum _{n=0}^\infty (n+1)^2(n+2)^2(n+3)^2(n+4)x^n=\frac{36(110x^3+72x^2+54x+12)}{(1-x)^8},\quad |x|<1. \end{aligned}$$

This implies that

$$\begin{aligned}&\sum _{n=0}^\infty \left\| D^3_{\psi ,\varphi }(\sqrt{n+1}\, z^n)\right\| ^2\\&=\int _{{\mathbb {D}}} |\psi (z)|^2\sum _{n=0}^\infty \frac{36(110|\varphi (z)|^6+72|\varphi (z)|^4+54|\varphi (z)|^2+12)}{(1-|\varphi (z)|^2)^8}. \end{aligned}$$

Since $0\le |\varphi (z)|\le 1$, it follows that the operator $D^3_{\psi ,\varphi }$ is Hilbert–Schmidt if and only if (7) holds true. $\square$

Corollary 3.15

Let $\varphi$ be an analytic self-map of the unit disk, and $\psi$ be an analytic function on the unit disk. Then $D^{(3)}_{\psi ,\varphi }$ is a Hilbert–Schmidt operator on $L_a^2$ if and only if

$$\begin{aligned} \int _{{\mathbb {D}}} \frac{\left| \psi (z)\varphi ^\prime (z)\right| ^2}{(1-\left| \varphi (z)\right| ^2)^8}\mathrm{d}A(z)<\infty . \end{aligned}$$

The above pattern can be used repeatedly to establish the following results.

Theorem 3.16

Let $\varphi$ be an analytic self-map of the unit disk, $\psi$ be an analytic function on the unit disk, and k be a non-negative integer. Then $D^{k}_{\psi ,\varphi }$ is a Hilbert–Schmidt operator on $L_a^2$ if and only if

$$\begin{aligned} \int _{{\mathbb {D}}} \frac{|\psi (z)|^2}{(1-\left| \varphi (z)\right| ^2)^{2k+2}}\mathrm{d}A(z)<\infty . \end{aligned}$$

Corollary 3.17

Let $\varphi$ be an analytic self-map of the unit disk, $\psi$ be an analytic function on the unit disk, and k be a non-negative integer. Then $D^{(k)}_{\psi ,\varphi }$ is a Hilbert–Schmidt operator on $L_a^2$ if and only if

$$\begin{aligned} \int _{{\mathbb {D}}} \frac{\left| \psi (z)\varphi ^\prime (z)\right| ^2}{(1-\left| \varphi (z)\right| ^2)^{2k+2}}\mathrm{d}A(z)<\infty . \end{aligned}$$

Remark 3.18

If $k=0$ and $\psi$ is the constant function 1, then $D^k_{\psi ,\varphi }=C_\varphi$, so that $C_\varphi$ is a Hilbert–Schmidt operator on $L_a^2$ if and only if

$$\begin{aligned} \int _{{\mathbb {D}}} \frac{1}{(1-\left| \varphi (z)\right| ^2)^{2}}\mathrm{d}A(z)<\infty . \end{aligned}$$

This result is of course known (see [12, page 227, Ex. 5]).

4 Convergence in operator topologies

We begin by computing the adjoint of the composition-differentiation operator $D_{\psi ,\varphi }$. The following lemma is known; but for the convenience of reader, we include a proof here.

Lemma 4.1

[5]. Let $D_{\psi ,\varphi }$ be a bounded operator on $H^2$. Then $D^*_{\psi ,\varphi }(K_w)=\overline{\psi (w)}K^{(1)}_{\varphi (w)}$, where $K^{(1)}_{w}(z)$ is the reproducing kernel corresponding to differentiation functional $w\mapsto f^\prime (w)$ on $H^2$.

Proof

Let $w\in {\mathbb {D}}$ be fixed. It is well-known that the functional $w\mapsto f^\prime (w)$ is bounded on $H^2$ (see [1, Theorem 2.16]). It then follows from Riesz representation theorem that there is a function $K_w^{(1)}\in H^2$ such that

$$\begin{aligned} f^\prime (w)=\langle f, K_w^{(1)}\rangle ,\quad f\in H^2. \end{aligned}$$

On the other hand, we have already seen that

$$\begin{aligned} f(w)=\langle f, K_w\rangle =\frac{1}{2\pi }\int _0^{2\pi }f^*(e^{i\theta })\frac{1}{1-we^{-i\theta }}\mathrm{d}\theta . \end{aligned}$$

This implies that

$$\begin{aligned} f^\prime (w)=\frac{1}{2\pi }\int _0^{2\pi }f^*(e^{i\theta })\frac{e^{-i\theta }}{(1-we^{-i\theta })^2}\mathrm{d}\theta , \end{aligned}$$

from which, by the uniqueness of kernel function, we obtain

$$\begin{aligned} K_w^{(1)}(z)=\frac{z}{(1-{\overline{w}} z)^2},\quad (z,w)\in {\mathbb {D}}\times {\mathbb {D}}.\end{aligned}$$

(8)

We now assume that $f\in H^2$, and write

$$\begin{aligned} \langle f, D^*_{\psi ,\varphi }(K_w)\rangle&=\langle D_{\psi ,\varphi }(f), K_w\rangle \\&=\langle \psi \cdot f^\prime \circ \varphi , K_w\rangle \\&=\psi (w)\langle f^\prime \circ \varphi , K_w\rangle \\&=\psi (w)\langle f, K^{(1)}_{\varphi (w)}\rangle \\&=\langle f, \overline{\psi (w)}K^{(1)}_{\varphi (w)}\rangle , \end{aligned}$$

from which the result follows. $\square$

Assume that a sequence of self-maps $\varphi _n$ on the unit disk satisfies $\Vert \varphi _n\Vert _\infty<r <1$, and that $\psi _n$ is a sequence of bounded analytic functions on the unit disk satisfying

$$\begin{aligned} \lim _{n\rightarrow \infty }\frac{\Vert \psi _n\Vert _\infty }{(r-|\varphi _n(0)|)^{1/2}}=0. \end{aligned}$$

It then follows from $\Vert D_{\psi _n,\varphi _n}\Vert \le \Vert \psi _n\Vert _\infty \Vert D_{\varphi _n}\Vert$ and (1) that

$$\begin{aligned} \Vert D_{\psi _n,\varphi _n}\Vert \rightarrow 0,\quad n\rightarrow \infty , \end{aligned}$$

while $\varphi _n$ may not be convergent. To avoid this, we often consider sequences of operators $D_{\psi _n,\varphi _n}$ with nonzero limits.

Lemma 4.2

Let $\varphi _n$ be a sequence of analytic self-mappings of the unit disk such that for each n we have $\Vert \varphi _n\Vert _\infty <1$, and let $\psi _n$ be a sequence in $H^2$ such that $D_{\psi _n,\varphi _n}$ is bounded. Assume that $D_{\psi _n,\varphi _n}$ converges in the weak operator topology to a nonzero operator T. Then we have

(a) $\psi _n$ converges weakly to a nonzero function $\psi$,

(b) $\varphi _n$ converges weakly to an analytic self-mapping of ${\mathbb {D}}$,

(c) T is a weighted composition-differentiation operator on $H^2$.

Proof

For part (a), consider $g(z)=z$. By assumption, $D_{\psi _n,\varphi _n}(g)\rightarrow T(g)$ weakly (in $H^2$). This means that $\psi _n g^\prime (\varphi _n)\rightarrow T(g)$ weakly; or $\psi _n \rightarrow T(g)$ weakly. To prove that T(g) is a nonzero function, assume on the contrary that $T(g)=0$. Let f be a polynomial, then $D_{\psi _n,\varphi _n}(f)\rightarrow T(f)$ weakly, which implies that for every $w\in {\mathbb {D}}$,

$$\begin{aligned} D_{\psi _n,\varphi _n}(f)(w)\rightarrow T(f)(w), \end{aligned}$$

or equivalently,

$$\begin{aligned} \lim _{n\rightarrow \infty } \psi _n(w)f^\prime (\varphi _n(w))=T(f)(w). \end{aligned}$$

Since $\psi _n(w)\rightarrow 0$, and the sequence $f^\prime (\varphi _n(w))$ is bounded, it follows that $T(f)(w)=0$. Since the polynomials are dense in $H^2$, we conclude that for every $h\in H^2$, $T(h)=0$ which contradicts our assumption on T. Thus (a) is proved.

As for part (b), we note that for each nonzero z and each w in the unit disk we have

$$\begin{aligned} \langle K_z, T^*(K_w)\rangle =\lim _{n\rightarrow \infty }\langle K_z, D_{\psi _n,\varphi _n}^*(K_w)\rangle . \end{aligned}$$

We now use Lemma 4.1 together with () to obtain

$$\begin{aligned} T^*(K_w)(z)=\lim _{n\rightarrow \infty }\overline{\psi _n(w)}K^{(1)}_{\varphi _n(w)}(z)=\lim _{n\rightarrow \infty }\overline{\psi (w)}\frac{z}{(1-\overline{\varphi _n(w)}z)^2}. \end{aligned}$$

(9)

Let $E=\{\zeta \in {\mathbb {D}}: \psi (\zeta )\ne 0\}$. Since $\psi$ is a nonzero analytic function, it follows that E is topologically "big" in ${\mathbb {D}}$; meaning that it has accumulation points in the unit disk. Since the sequence

$$\begin{aligned} \overline{\psi (w)}\frac{z}{(1-\overline{\varphi _n(w)}z)^2} \end{aligned}$$

is convergent, it follows that for each $w\in E$, the sequence $\varphi _n(w)$ converges, say to $\varphi$. On the other hand, $\Vert \varphi _n\Vert _\infty <1$, so that $\varphi _n$ forms a normal family. This implies that a subsequence of $\varphi _n$ converges to an analytic function h on every compact subset of the unit disk. Thus h and $\varphi$ must agree on the whole disk, that is, $\varphi _n\rightarrow \varphi$ uniformly on compact subsets of ${\mathbb {D}}$. It is clear that $|\varphi (w)|\le 1$ for each $w\in {\mathbb {D}}$. Note also that uniform boundedness of $\varphi _n$ implies that this sequence is norm bounded in $H^2$. Therefore, $\varphi _n\rightarrow \varphi$ weakly in $H^2$. The last thing to be proved is that $\varphi$ is a self mapping of the unit disk. To this end, let $w\in E$. Since we have from (9)

$$\begin{aligned} T^*(K_w)(z)=\overline{\psi (w)}\frac{z}{(1-\overline{\varphi (w)}z)^2}, \end{aligned}$$

(10)

and $T^*(K_w)\in H^2$, we get $|\varphi (w)|<1$. In case that $w\notin E$, the function $\psi$ does not vanish in a punctured closed disk around w, so that on the boundary of this small disk we have again $|\varphi |<1$, and by the maximum principle $|\varphi (w)|<1$.

Finally, for part (c), it follows from (10) that

$$\begin{aligned} \langle T^*(K_w), K_z\rangle =\overline{\psi (w)}\langle K^{(1)}_{\varphi (w)}, K_z\rangle , \end{aligned}$$

and hence

$$\begin{aligned} T(K_z)(w)=\langle T(K_z), K_w\rangle =\langle \psi (w)K_z, K^{(1)}_{\varphi (w)}\rangle =\psi (w)K^\prime _z(\varphi (w)). \end{aligned}$$

Since the span of the reproducing kernel functions is dense in $H^2$, we have $T(f)(w)=\psi (w)f^\prime (\varphi (w))$, or $T(f)=D_{\psi ,\varphi }(f)$, which is the desired result. $\square$

The following theorem describes the conditions under which the sequence of bounded operators $D_{\psi _n,\varphi _n}$ converges to $D_{\psi ,\varphi }$ in the weak operator topology.

Theorem 4.3

Let $\{\varphi _n\}_{n\ge 1}$ and $\varphi$ be analytic self-maps of the unit disk such that $\Vert \varphi _n\Vert _\infty <1$, and let $\{\psi _n\}_{n\ge 1}$ and $\psi$ be elements in $H^2$. Assume that each $D_{\psi _n,\varphi _n}$ is bounded, and that $D_{\psi ,\varphi }$ is a bounded nonzero operator on $H^2$. Then $D_{\psi _n,\varphi _n}$ converges to $D_{\psi ,\varphi }$ in weak operator topology if and only if

(a) $\psi _n$ converges weakly to $\psi$ in $H^2$,

(b) $\varphi _n$ converges weakly to $\varphi$ in $H^2$,

(c) $\sup _n\Vert D_{\psi _n,\varphi _n}\Vert <\infty .$

Proof

Assume that $D_{\psi _n,\varphi _n}$ converges to $D_{\psi ,\varphi }$ in weak operator topology. For (a), put $f(z)=z$. It follows that $D_{\psi _n,\varphi _n}(f)\rightarrow D_{\psi ,\varphi }(f)$ weakly; or $\psi _n\rightarrow \psi$ weakly.

For (b), let $E\subset {\mathbb {D}}$ be the set on which $\psi$ does not vanish. Consider $g(z)=z^2/2$. Since $D_{\psi _n,\varphi _n}(g)\rightarrow D_{\psi ,\varphi }(g)$ weakly, it follows that $\psi _n\varphi _n\rightarrow \psi \varphi$ weakly. In particular, this convergence is pointwise. Therefore, on E where $\psi$ is nonzero, $\varphi _n\rightarrow \varphi$ pointwise. On the other hand, by part (b) of the preceding lemma, $\varphi _n\rightarrow h$ weakly, where h is an analytic self-map of the unit disk. Therefore, $\varphi _n(z)\rightarrow h(z)$ for each $z\in {\mathbb {D}}$. This implies that $h=\varphi$ on E, and hence on the whole disk (E is a big set!).

The last part is clear from the Banach-Steinhaus theorem; for each $f\in H^2,\,D_{\psi _n,\varphi _n}(f)$ is weakly convergent, hence pointwise bounded (the bound depends on f). Banach-Steinhaus theorem now implies that $D_{\psi _n,\varphi _n}$ is uniformly bounded.

To prove the converse statement, assume that parts (a), (b), and (c) hold true. Clearly, $\varphi _n\rightarrow \varphi$ pointwise, and $\psi _n\rightarrow \psi$ pointwise. For $z,w\in {\mathbb {D}}$, we have

$$\begin{aligned} \overline{\psi (w)}\frac{z}{(1-\overline{\varphi (w)}z)^2}=\lim _{n\rightarrow \infty }\overline{\psi _n(w)}\frac{z}{(1-\overline{\varphi _n(w)}z)^2}, \end{aligned}$$

and hence

$$\begin{aligned} \langle D_{\psi ,\varphi }(K_z), K_w\rangle =\lim _{n\rightarrow \infty }\langle D_{\psi _n,\varphi _n}(K_z), K_w\rangle . \end{aligned}$$

We recall that a sequence $T_n\in \mathcal {B(H)}$ converges to T in weak operator topology if $\sup _n \Vert T_n\Vert <\infty$ and for every x, y in a dense subset of ${\mathcal {H}}$ we have $\langle T_nx, y\rangle \rightarrow \langle Tx, y\rangle$ (see [2, Chap. IX, Proposition 1.3(b)]). Invoking this statement, and noting that the reproducing kernel functions $\{K_z: z\in {\mathbb {D}}\}$ is dense in $H^2$, and the sequence of operators $D_{\psi _n,\varphi _n}$ is bounded, by part (c), we conclude that $D_{\psi _n,\varphi _n}\rightarrow D_{\psi ,\varphi }$ in weak operator topology. $\square$

Remark 4.4

If we consider the sequence $D_{\varphi _n}$ in which $\sup _n \Vert \varphi _n\Vert _\infty \le r<1$, and if we assume that $\varphi _n\rightarrow \varphi$ weakly, then from the estimate (1)

$$\begin{aligned} \Vert D_{\varphi _n}\Vert \le \left( \frac{r+|\varphi _n(0)|}{r-|\varphi _n(0)|}\right) ^{1/2}\lfloor \frac{1}{1-r}\rfloor r^{\lfloor \frac{1}{1-r}\rfloor -1}, \end{aligned}$$

we conclude that $\sup _n\Vert D_{\varphi _n}\Vert$ is bounded, so that the condition (c) is redundant.

In the following we turn to the convergence of $D_{\psi _n,\varphi _n}$ in strong operator topology. Note that the weak convergence of $\varphi _n\rightarrow \varphi$ and the weak convergence of $\psi _n\rightarrow \psi$ are not sufficient for the convergence of $D_{\psi _n,\varphi _n}\rightarrow D_{\psi ,\varphi }$ in the strong operator topology. For example, let $\psi _n=\varphi _n=z^n/2$. It is clear that both of this sequences converge to the zero function weakly (elements of an orthonormal basis in a Hilbert space converge to zero weakly). Therefore, by Theorem 4.3, $D_{\psi ,\varphi }\rightarrow 0$ (WOT). Now let $f(z)=z$, then for each $m\ne n$, we have

$$\begin{aligned} \Vert D_{\psi _n,\varphi _n}(f)-D_{\psi _m,\varphi _m}(f)\Vert =\sqrt{2}/2, \end{aligned}$$

which means that $D_{\psi _n,\varphi _n}$ is not convergent in strong operator topology.

Lemma 4.5

Let $\{\varphi _n\}_{n\ge 1}$ and $\varphi$ be analytic self-maps of the unit disk such that $\Vert \varphi _n\Vert _\infty <1$, and let $\{\psi _n\}_{n\ge 1}$ and $\psi$ be elements in $H^2$ where $\psi$ is nonzero. Assume that each $D_{\psi _n,\varphi _n}$ and $D_{\psi ,\varphi }$ are bounded operators on $H^2$. If $D_{\psi _n,\varphi _n}$ converges to $D_{\psi ,\varphi }$ in strong operator topology, then we have

(a) $\psi _n$ converges to $\psi$ in $H^2$,

(b) $\varphi _n$ converges to $\varphi$ in $H^2$.

Proof

For part (a), set $f(z)=z$, then by assumption

$$\begin{aligned} \Vert D_{\psi _n,\varphi _n}(f)- D_{\psi ,\varphi }(f)\Vert \rightarrow 0, \end{aligned}$$

which is the same as saying $\psi _n\rightarrow \psi$ in $H^2$.

For (b), we have to prove that $\Vert \varphi _n-\varphi \Vert \rightarrow 0$ as $n\rightarrow \infty$. Let $g(z)=z^2/2$, it then follows from the assumption that

$$\begin{aligned} \Vert \psi _n\varphi _n-\psi \varphi \Vert \rightarrow 0. \end{aligned}$$

From part (a) and the fact that $\Vert \varphi _n\Vert _\infty <1$ we conclude that

$$\begin{aligned} \Vert (\psi _n-\psi )\varphi _n\Vert \rightarrow 0. \end{aligned}$$

Since

$$\begin{aligned} \Vert (\varphi _n-\varphi )\psi \Vert \le \Vert (\psi -\psi _n)\varphi _n\Vert +\Vert (\psi _n\varphi _n-\psi \varphi )\Vert , \end{aligned}$$

it follows that

$$\begin{aligned} \Vert (\varphi _n-\varphi )\psi \Vert \rightarrow 0,\quad n\rightarrow \infty . \end{aligned}$$

(11)

Put $x_n:=\Vert \varphi _n-\varphi \Vert$, and let $x_{n_k}$ be an arbitrary subsequence of $x_n$. By (11) we have

$$\begin{aligned} \Vert (\varphi _{n_k}-\varphi )\psi \Vert \rightarrow 0, \end{aligned}$$

from which it follows that a subsequence of $(\varphi _{n_k}-\varphi )\psi$ converges to zero at almost every point of the unit disk:

$$\begin{aligned} (\varphi _{n_{k_j}}(z)-\varphi (z))\psi (z)\rightarrow 0,\quad \text {a.e.} \end{aligned}$$

Since the analytic function $\psi$ is not identically zero, it can not vanish on a set with accumulation point in ${\mathbb {D}}$, so that it does not vanish almost everywhere in ${\mathbb {D}}$, implying that

$$\begin{aligned} \varphi _{n_{k_j}}(z)-\varphi (z)\rightarrow 0,\quad \text {a.e.} \end{aligned}$$

Since $\varphi _n$, and its limit function, are bounded, the Bounded Convergence Theorem applies;

$$\begin{aligned} x_{n_{k_j}}=\Vert \varphi _{n_{k_j}}-\varphi \Vert \rightarrow 0,\quad j\rightarrow \infty . \end{aligned}$$

We have proved that every subsequence of $x_n$ (here $x_{n_k}$), in its turn has a subsequence that converges to zero, thus $x_n$ converges to zero. $\square$

Theorem 4.6

Let $\{\varphi _n\}_{n\ge 1}$ and $\varphi$ be analytic self-maps of the unit disk such that $\Vert \varphi _n\Vert _\infty <1$, and let $\{\psi _n\}_{n\ge 1}$ and $\psi$ be elements in $H^2$ where $\psi$ is nonzero. Assume that each $D_{\psi _n,\varphi _n}$ and $D_{\psi ,\varphi }$ are bounded operators on $H^2$ where $D_{\psi ,\varphi }$ is nonzero. Then $D_{\psi _n,\varphi _n}$ converges to $D_{\psi ,\varphi }$ in strong operator topology if and only if

(a) $\psi _n$ converges to $\psi$ in $H^2$,

(b) $\varphi _n$ converges to $\varphi$ in $H^2$,

(c) $\sup _n\Vert D_{\psi _n,\varphi _n}\Vert <\infty .$

Proof

Assume that $D_{\psi _n,\varphi _n}$ converges to $D_{\psi ,\varphi }$ in strong operator topology. Parts (a) and (b) follow from Lemma 4.5. Part (c) is a consequence of Theorem 4.3.

For the converse, assume that parts (a), (b) and (c) hold true. We first verify that for every polynomial f,

$$\begin{aligned} \Vert D_{\psi _n,\varphi _n}(f)-D_{\psi ,\varphi }(f)\Vert \rightarrow 0,\quad n\rightarrow \infty . \end{aligned}$$

(12)

Note that on the line segment joining $\varphi _n(z)$ to $\varphi (z)$ we have

$$\begin{aligned} f^\prime (\varphi _n(z))-f^\prime (\varphi (z))=\int f^{\prime \prime } (w)dw, \end{aligned}$$

from which it follows that for each $z\in {\mathbb {D}}$,

$$\begin{aligned} |f^\prime (\varphi _n(z))-f^\prime (\varphi (z))|\le \Vert f^{\prime \prime }\Vert _\infty |\varphi _n(z)-\varphi (z)|, \end{aligned}$$

or

$$\begin{aligned} \Vert f^\prime \circ \varphi _n-f^\prime \circ \varphi \Vert \le \Vert f^{\prime \prime }\Vert _\infty \Vert \varphi _n-\varphi \Vert . \end{aligned}$$

(13)

Again we set

$$\begin{aligned} x_n=\Vert D_{\psi _n,\varphi _n}(f)-D_{\psi ,\varphi }(f)\Vert , \end{aligned}$$

and assume that

$$\begin{aligned} x_{n_k}=\Vert D_{\psi _{n_k},\varphi _{n_k}}(f)-D_{\psi ,\varphi }(f)\Vert \end{aligned}$$

is an arbitrary subsequence of $x_n$. We prove that there is a subsequence $x_{n_{k_j}}$ that converges to 0. To this end, we see from part (b),

$$\begin{aligned} \Vert \varphi _{n_k}-\varphi \Vert \rightarrow 0, \end{aligned}$$

so that (13) implies

$$\begin{aligned} \Vert f^\prime \circ \varphi _{n_k}-f^\prime \circ \varphi \Vert \rightarrow 0. \end{aligned}$$

Thus we can find a subsequence, say $f^\prime \circ \varphi _{n_{k_j}}-f^\prime \circ \varphi$, that converges to zero at almost every point of the unit disk. Moreover, we have

$$\begin{aligned} \left| \psi (z)\big (f^\prime \circ \varphi _{n_{k_j}}(z)-f^\prime \circ \varphi (z)\big )\right| \le 2|\psi (z)|\Vert f^{\prime }\Vert _\infty . \end{aligned}$$

The Dominated Convergence Theorem now implies that

$$\begin{aligned} \Vert \psi (f^\prime \circ \varphi _{n_{k_j}}-f^\prime \circ \varphi )\Vert \rightarrow 0,\quad j\rightarrow \infty . \end{aligned}$$

(14)

On the other hand, by part (a),

$$\begin{aligned} \Vert (\psi _{n_{k_j}}-\psi )f^\prime \circ \varphi _{n_{k_j}}\Vert ^2\le \Vert f^\prime \Vert ^2_\infty \Vert \psi _{n_{k_j}}-\psi \Vert ^2\rightarrow 0,\quad j\rightarrow \infty . \end{aligned}$$

(15)

Finally, we write

$$\begin{aligned} \Vert D_{\psi _n,\varphi _n}(f)-D_{\psi ,\varphi }(f)\Vert&=\Vert \psi _n f^\prime \circ \varphi _n -\psi f^\prime \circ \varphi \Vert \\&\le \Vert (\psi _n-\psi )f^\prime \circ \varphi _n\Vert +\Vert \psi (f^\prime \circ \varphi _n -f^\prime \circ \varphi )\Vert , \end{aligned}$$

from which, using (14) and (15) we conclude that

$$\begin{aligned} \Vert D_{\psi _{n_{k_j}},\varphi _{n_{k_j}}}(f)-D_{\psi ,\varphi }(f)\Vert \rightarrow 0,\quad j\rightarrow \infty . \end{aligned}$$

We recall that by [2, Chap. IX, Proposition 1.3(d)]), a sequence $T_n\in \mathcal {B(H)}$ converges to T in strong operator topology if on a dense subset of ${\mathcal {H}}$ we have $\Vert T_nx-Tx\Vert \rightarrow 0$. This fact together with (12) proves that $D_{\psi _n,\varphi _n}\rightarrow D_{\psi ,\varphi }$ in strong operator topology. $\square$