COMPACT AND HILBERT–SCHMIDT WEIGHTED COMPOSITION OPERATORS ON WEIGHTED BERGMAN SPACES

Abstract

Let u and $\varphi $ be two analytic functions on the unit disk D such that $\varphi (D) \subset D$ . A weighted composition operator $uC_{\varphi }$ induced by u and $\varphi $ is defined on $A^2_{\alpha }$ , the weighted Bergman space of D, by $uC_{\varphi }f := u \cdot f \circ \varphi $ for every $f \in A^2_{\alpha }$ . We obtain sufficient conditions for the compactness of $uC_{\varphi }$ in terms of function-theoretic properties of u and $\varphi $ . We also characterize when $uC_{\varphi }$ on $A^2_{\alpha }$ is Hilbert–Schmidt. In particular, the characterization is independent of $\alpha $ when $\varphi $ is an automorphism of D. Furthermore, we investigate the Hilbert–Schmidt difference of two weighted composition operators on $A^2_{\alpha }$ .

1 Introduction

Let D be the unit disk $\{z \in {\mathbb C} : |z| < 1\}$ in the complex plane ${\mathbb C}$ and T be the unit circle $\{z \in {\mathbb C} : |z| = 1\}$ . For $0 < p < \infty $ and $\alpha> -1$ , the weighted Bergman space $A^p_{\alpha }$ of D consists of all analytic functions f in $L^p(D, dA_{\alpha })$ , that is,

$$ \begin{align*}\|f\|_{A^p_{\alpha}}^p := \int_D |f(z)|^p \,dA_{\alpha}(z) < \infty, \end{align*} $$

where $dA_{\alpha }(z) := (\alpha + 1)(1 - |z|^2)^{\alpha } dA(z)$ and $dA(z) := ({1}/{\pi })\,dx\,dy$ is the normalized area measure on D. It is known that $A^2_{\alpha }$ is a closed subspace of $L^2(D, dA_{\alpha })$ and is thus a Hilbert space with the inner product $\langle \cdot ,\cdot \rangle $ given by

$$ \begin{align*}\langle f, g\rangle := \int_D f(z)\overline{g(z)} \, dA_{\alpha}(z) \quad \text{for every } f, g \in A^2_{\alpha}. \end{align*} $$

In what follows, we denote the norm on $A^2_{\alpha }$ by $\|\cdot \|$ for brevity. By writing $f(z) = \sum _{k = 0}^{\infty } a_k z^k$ , we have

$$ \begin{align*}\|f\|^2 = \sum_{k = 0}^{\infty} \frac{k! \, \Gamma(\alpha + 2)}{\Gamma(\alpha + 2 + k)}|a_k|^2, \end{align*} $$

where $\Gamma $ is the usual gamma function. If we let

(1-1) $$ \begin{align} e_k(z) = \sqrt{\frac{\Gamma(\alpha + 2 + k)}{k!\,\Gamma(\alpha + 2)}}\, z^k \quad {\textrm{for}} \ k = 0, 1, \ldots, \end{align} $$

then $\{e_k\}_{k=0}^{\infty }$ is the standard orthonormal basis for $A^2_{\alpha }$ . Furthermore, if w is an arbitrary point in D, then $\langle f, k_w\rangle = f(w)$ for all $f \in A^2_{\alpha }$ , where $k_w(z) := 1/(1 - \overline {w}z)^{\alpha + 2}$ is the reproducing kernel representing the point evaluation functional on $A^2_{\alpha }$ at $z = w$ . Moreover, $\|k_w\|^2 = 1/(1 - |w|^2)^{\alpha + 2}$ .

Let u and $\varphi $ be two analytic functions on D such that $\varphi (D) \subset D$ . They induce a weighted composition operator $uC_{\varphi }$ from $A^2_{\alpha }$ into the linear space of all analytic functions on D by

$$ \begin{align*}uC_{\varphi}(f)(z) := u(z)f(\varphi(z)) \quad \text{for \ every } f \in A^2_{\alpha} \textrm{ and} \ z \in D. \end{align*} $$

When $u \equiv 1$ , the corresponding operator, denoted by $C_{\varphi }$ , is known as a composition operator. From exercise 3.1.3 in [3, page 127], $C_{\varphi }$ is always bounded. However, this is not necessarily true for weighted composition operators. When $uC_{\varphi }$ maps $A^2_{\alpha }$ into itself, we say $uC_{\varphi }$ is a weighted composition operator on $A^2_{\alpha }$ . In this case, $u = uC_{\varphi }1 \in A^2_{\alpha }$ . An appeal to the closed graph theorem shows that every operator $uC_{\varphi }$ on $A^2_{\alpha }$ is bounded. Furthermore, if $g \in A^2_{\alpha }$ and $w \in D$ , then

$$ \begin{align*}\langle (uC_{\varphi})^*k_w, g\rangle = \langle k_w, uC_{\varphi}g\rangle = \overline{u(w)g(\varphi(w))} = \langle \overline{u(w)}k_{\varphi(w)}, g\rangle. \end{align*} $$

Thus,

$$ \begin{align*}(uC_{\varphi})^*k_w = \overline{u(w)}k_{\varphi(w)}. \end{align*} $$

During the past two decades, several authors have studied the properties of (weighted) composition operators on $A^p_{\alpha }$ with Berezin transforms and Carleson-type measures (see for example [4, 5, 11, 13]). In Section 2, we obtain sufficient conditions for the compactness of $uC_{\varphi }$ in terms of function-theoretic properties of u and $\varphi $ . In Section 3, we characterize Hilbert–Schmidt weighted composition operators and the Hilbert–Schmidt difference of two weighted composition operators on $A^2_{\alpha }$ .

2 Compact weighted composition operators

A bounded linear operator T from a Banach space $B_1$ to a Banach space $B_2$ is said to be compact if it maps bounded subsets of $B_1$ into relatively compact subsets of $B_2$ . Equivalently, T is compact if and only if it maps every bounded sequence $\{x_n\}_{n = 1}^{\infty }$ in $B_1$ onto a sequence $\{Tx_n\}_{n = 1}^{\infty }$ in $B_2$ which has a convergent subsequence. It was shown in [13, Theorem 4.3] that $uC_{\varphi }$ is compact on $A^p_{\alpha }$ if and only if

$$ \begin{align*}\lim_{\delta \to 0^+} \sup_{\zeta \in T} \frac{\mu_{\alpha,\, p} \circ \varphi^{-1}(S(\zeta, \delta))}{\delta^{\alpha + 2}} = 0, \end{align*} $$

where $S(\zeta , \delta ) := \{z \in D : |z - \zeta | < \delta \}$ and $\mu _{\alpha ,\, p} \circ \varphi ^{-1}$ is the measure such that $\|uC_{\varphi }f\|_{A^p_{\alpha }}^p = \int _D |f|^p d\mu _{\alpha ,\, p} \circ \varphi ^{-1}$ for all $f \in A^p_{\alpha }$ . Later, Čučković and Zhao estimated the essential norm of $uC_{\varphi }$ and deduced that $uC_{\varphi }$ is compact on $A^2_0$ if and only if

$$ \begin{align*}\lim_{|a| \to 1^-} \int_D \frac{(1 - |a|^2)^2 |u(z)|^2}{|1 - \overline{a}\varphi(z)|^4} \, dA(z) = 0 \end{align*} $$

[4, Corollary 2]. These characterizations, however, are rather implicit and less tractable. In this section, we provide more explicit sufficient conditions that guarantee $uC_{\varphi }$ is compact on $A^2_{\alpha }$ . To this end, we first state a useful result to the study of compact weighted composition operators on $A^2_{\alpha }$ .

Lemma 2.1. Let $uC_{\varphi }$ be a weighted composition operator on $A^2_{\alpha }$ . The following two statements are equivalent:

1. (i) $uC_{\varphi }$ is compact on $A^2_{\alpha }$ ;

2. (ii) if $\{f_n\}_{n = 1}^{\infty }$ is a bounded sequence in $A^2_{\alpha }$ and $f_n \to 0$ uniformly on compact subsets of D, then $\|uC_{\varphi }f_n\| \to 0$ .

While the above lemma is a generalization of [3, Proposition 3.11], it can also be obtained by a Hilbert space argument. From exercise 4.7.1 in [17, page 97], a sequence of functions in $A^2_{\alpha }$ is weakly convergent to zero if and only if this sequence is norm bounded and converges to zero uniformly on compact subsets of D. Lemma 2.1 now follows from this fact and [17, Theorem 1.14].

One simple sufficient condition for the compactness of $uC_{\varphi }$ , which is analogous to [7, Theorem 2], is given below.

Theorem 2.2. Suppose that $uC_{\varphi }$ is a weighted composition operator on $A^2_{\alpha }$ . If $\overline {\varphi (D)} \subset D$ , then $uC_{\varphi }$ is compact.

Proof. Since $\overline {\varphi (D)} \subset D$ , there is a constant M such that $0 < M < 1$ and $|\varphi (z)| \leq M$ for all $z \in D$ . Let $\{f_n\}_{n = 1}^{\infty }$ be a bounded sequence in $A^2_{\alpha }$ such that $f_n \to 0$ uniformly on compact subsets of D. In particular, this sequence converges to zero uniformly on $\overline {S(0, M)}$ . Then there exists some $N \in {\mathbb N}$ for which $|f_n(\varphi (z))| < \epsilon $ whenever $n> N$ and $z \in D$ . With $u \in A^2_{\alpha }$ , it follows that $\|uC_{\varphi } f_n\| \leq \epsilon \|u\|$ for all $n> N$ . By Lemma 2.1, $uC_{\varphi }$ is compact.

We remark that the condition $\overline {\varphi (D)} \subset D$ in Theorem 2.2 is sufficient, but not necessary for the compactness of $uC_{\varphi }$ . This is shown below.

Example 2.3. Let $u(z) = z-1$ and $\varphi (z) = (z + 1) / 2$ . Note that $1 \in \overline {\varphi (D)}$ . Choose any $\varepsilon> 0$ . With $u(1) = 0$ and the continuity of u at $z = 1$ , there is a sufficiently small $\delta> 0$ such that $|u|^2 < \varepsilon $ on $S(1, \delta )$ . We show that $uC_{\varphi }$ is compact by using Lemma 2.1.

Let $\{f_n\}_{n = 1}^{\infty }$ be a sequence in $A^2_{\alpha }$ such that $\|f_n\| \leq 1$ for all $n \in {\mathbb N}$ and $f_n \to 0$ uniformly on compact subsets of D. Since $\varphi $ is continuous on the compact set $\overline {D \setminus S(1, \delta )}$ , the set $\varphi (\overline {D \setminus S(1, \delta )})$ is compact in D. Then there exists some $N \in {\mathbb N}$ for which if $n> N$ and $z \in D \setminus S(1, \delta )$ , we have

$$ \begin{align*}|f_n(\varphi(z))|^2 < \varepsilon. \end{align*} $$

These, together with the fact that $C_{\varphi }$ is bounded on $A^2_{\alpha }$ , imply

$$ \begin{align*} \|uC_{\varphi} f_n\|^2 & = \int_{S(1, \delta)} |u(z)|^2 |f_n(\varphi(z))|^2\, dA_{\alpha}(z) + \int_{D \setminus S(1, \delta)} |u(z)|^2 |f_n(\varphi(z))|^2\, dA_{\alpha}(z) \\[3pt] & \leq \varepsilon\int_{S(1, \delta)} |f_n(\varphi(z))|^2\, dA_{\alpha}(z) + \varepsilon\int_{D \setminus S(1, \delta)} |u(z)|^2\, dA_{\alpha}(z) \\[3pt] & \leq \varepsilon \|C_{\varphi} f_n\|^2 + \varepsilon \int_D |u(z)|^2\, dA_{\alpha}(z) \\[3pt] & \leq (\|C_{\varphi}\|^2 + 4)\varepsilon \end{align*} $$

whenever $n> N$ .

In this example, $\varphi $ has an angular derivative at $z = 1$ because $(1 - \varphi (z)) / (1 - z) = 1 / 2$ . Then it follows from [3, Corollary 3.14] that $C_{\varphi }$ is not compact on $A^2_{\alpha }$ . However, $uC_{\varphi }$ is compact.

There is another question of interest: does the compactness of $C_{\varphi }$ guarantee that of $uC_{\varphi }$ ? The answer to this question is generally no, at least when u is unbounded on D. To see this, we first state a necessary condition for $uC_{\varphi }$ to be compact.

Theorem 2.4. If $uC_{\varphi }$ is a compact weighted composition operator on $A^2_{\alpha }$ , then

(2-1) $$ \begin{align} \lim_{|z| \to 1^-} |u(z)|\bigg(\frac{1 - |z|^2}{1 - |\varphi(z)|^2}\bigg)^{{\alpha}/{2} + 1} = 0. \end{align} $$

This theorem is a simple generalization of [4, Proposition 1]: since

$$ \begin{align*}\|(uC_{\varphi})^*K_z\| = (1 - |z|^2)^{{\alpha}/{2} + 1}|u(z)|\|k_{\varphi(z)}\| = |u(z)|\bigg(\frac{1 - |z|^2}{1 - |\varphi(z)|^2}\bigg)^{{\alpha}/{2} + 1}, \end{align*} $$

where $K_z$ is the normalized reproducing kernel corresponding to the point evaluation functional on $A^2_{\alpha }$ at z, the condition in Equation (2-1) follows from the compactness of $(uC_{\varphi })^*$ and the result that $K_z \to 0$ weakly in $A^2_{\alpha }$ as $|z| \to 1^-$ .

While the validity of the converse of Theorem 2.4 awaits further investigation, the condition in Equation (2-1) actually is equivalent to the compactness of composition operators on $A^2_{\alpha }$ [17, Theorem 11.8]. Under additional assumptions on u and $\varphi $ , however, the condition in Equation (2-1) does characterize the compactness of $uC_{\varphi }$ . This will be shown in Theorem 2.8.

Example 2.5. Let $u(z) = 1 / (1 - z)^{1/2 + {\alpha }/{4}}$ and $\varphi (z) = 1 - (1 - z)^{1/2}$ . From [10, Example 3.4], $\varphi $ has no finite angular derivative at any point of T. Thus, $C_{\varphi }$ is compact by [3, Theorem 3.22]. However,

$$ \begin{align*}\lim_{r \to 1^-} u(r)\bigg[\frac{1 - r^2}{1 - (\varphi(r))^2}\bigg]^{{\alpha}/{2} + 1} = \lim_{r \to 1^-} \bigg[\frac{1 + r}{2- (1 - r)^{{1}/{2}}}\bigg]^{{\alpha}/{2} + 1} = 1 \ (\ne 0). \end{align*} $$

According to Theorem 2.4, $uC_{\varphi }$ is not compact on $A^2_{\alpha }$ .

When $C_{\varphi }$ is compact, how can we choose u such that $uC_{\varphi }$ is compact? The next result provides one criterion. Its statement and proof are similar to those of [10, Theorem 4.1].

Theorem 2.6. Suppose $u \in A^2_{\alpha }$ and $C_{\varphi }$ is compact on $A^2_{\alpha }$ . If there is a constant c with $0 < c < 1$ such that u is bounded on the set $\{z \in D : |\varphi (z)|> c\}$ , then $uC_{\varphi }$ is compact on $A^2_{\alpha }$ .

We prove a ‘converse’ of Theorem 2.4 with extra assumptions on u and $\varphi $ . While Moorhouse showed that the condition in Equation (2-1) characterizes the compactness of $uC_{\varphi }$ when u is bounded on D [14, Corollary 1], the validity of our result does not require the boundedness of u. The following lemma is needed.

Lemma 2.7. If $f \in A^2_{\alpha }$ , then

$$ \begin{align*}c\,\|f\|^2 \leq |f(0)|^2 + \int_D |f'(z)|^2 (1 - |z|^2)^2 \, dA_{\alpha}(z) \leq d\,\|f\|^2, \end{align*} $$

where $c := \min \{1, [(\alpha + 1)(\alpha + 2)] / (\alpha + 3)\}$ and $d := \max \{1, (\alpha + 1)(\alpha + 2)\}$ .

The proof of this lemma is direct and follows from a straightforward computation of the integral $\int _D |f'(z)|^2 (1 - |z|^2)^2 \,dA_{\alpha }(z)$ in terms of the Taylor coefficients of f. An immediate consequence of Lemma 2.7 is that $f \in A^2_{\alpha }$ if and only if $f' \in L^2(D, dA_{\alpha + 2})$ . Indeed, this is a particular case of a more general result in [8, Proposition 1.11]. Moreover, the lemma implies that $\|f\|$ is equivalent to $\|f'\|_{A_{\alpha + 2}^2}$ if $f \in A^2_{\alpha }$ and $f(0)=0$ .

Theorem 2.8. Let $uC_{\varphi }$ be a weighted composition operator on $A^2_{\alpha }$ . If

1. (i) $\varphi $ is univalent on D;

2. (ii) $ \lim _{|z| \to 1^-} |u'(z)|(1 - |z|^2) = 0$ ; and

3. (iii) $ \lim _{|z| \to 1^-} |u(z)|(({1 - |z|^2})/({1 - |\varphi (z)|^2}))^{{\alpha }/{2} + 1} = 0$ ;

then $uC_{\varphi }$ is compact on $A^2_{\alpha }$ .

Proof. Fix any $\varepsilon> 0$ . By conditions (ii) and (iii), there is a constant r with ${1 / 2 < r < 1}$ such that

$$ \begin{align*}|u'(z)|^2(1 - |z|^2)^2 < \varepsilon \quad\textrm{and}\quad |u(z)|^2(1 - |z|^2)^{\alpha + 2} < \varepsilon(1 - |\varphi(z)|^2)^{\alpha + 2} \end{align*} $$

whenever $r < |z| < 1$ . Let $\{f_n\}_{n = 1}^{\infty }$ be a sequence in $A^2_{\alpha }$ with $\|f_n\| \leq 1$ for all $n \in {\mathbb N}$ and $f_n \to 0$ uniformly on compact subsets of D. By Lemma 2.7,

(2-2) $$ \begin{align} \|uC_{\varphi}f_n\|^2 \leq \frac{1}{c}\bigg[|u(0) f_n(\varphi(0))|^2 + \int_D |(u \cdot f_n \circ \varphi)'(z)|^2 (1 - |z|^2)^2\, dA_{\alpha}(z)\bigg], \end{align} $$

where c is the constant defined in Lemma 2.7. Then

$$ \begin{align*}|(u \cdot f_n \circ \varphi)'(z)|^2 \leq 2(|(u(z)f_n'(\varphi(z))\varphi'(z)|^2 + |u'(z)f_n(\varphi(z))|^2), \end{align*} $$

so that

$$ \begin{align*}\int_D |(u \cdot f_n \circ \varphi)'(z)|^2 (1 - |z|^2)^2 \, dA_{\alpha}(z) \leq 2(A_n + B_n + C_n + D_n), \end{align*} $$

where

$$ \begin{align*}A_n := \int_{\overline{S(0, r)}} |u(z)|^2 |f_n'(\varphi(z))|^2 |\varphi'(z)|^2 (1 - |z|^2)^2\, dA_{\alpha}(z), \end{align*} $$

$$ \begin{align*}B_n := \int_{D \setminus \overline{S(0, r)}} |u(z)|^2 |f_n'(\varphi(z))|^2 |\varphi'(z)|^2 (1 - |z|^2)^2\, dA_{\alpha}(z), \end{align*} $$

$$ \begin{align*}C_n := \int_{\overline{S(0, r)}} |u'(z)|^2 |f_n(\varphi(z))|^2 (1 - |z|^2)^2\, dA_{\alpha}(z), \end{align*} $$

and

$$ \begin{align*}D_n := \int_{D \setminus \overline{S(0, r)}} |u'(z)|^2 |f_n(\varphi(z))|^2 (1 - |z|^2)^2\, dA_{\alpha}(z). \end{align*} $$

Both sets $\{\varphi (0)\}$ and $\varphi (\overline {S(0, r)})$ are compact in D. Thus, there exists some $N \in {\mathbb N}$ for which if $n> N$ and $z \in \overline {S(0, r)}$ , then

(2-3) $$ \begin{align} |f_n(\varphi(0))|^2, \ |f_n(\varphi(z))|^2, \ |f_n'(\varphi(z))|^2 < \varepsilon. \end{align} $$

From the continuity of $u\varphi '$ and $u'$ on the compact set $\overline {S(0, r)}$ , there is a positive constant M such that

$$ \begin{align*}|u(z)\varphi'(z)|^2, \ |u'(z)|^2 \leq M \end{align*} $$

for all $z \in \overline {S(0, r)}$ . Therefore, if $n> N$ , we have

(2-4) $$\begin{align} A_n + C_n \leq 2M\varepsilon \int_{\overline{S(0, r)}}\, dA_{\alpha}(z) \leq 2M\varepsilon \int_D\, dA_{\alpha}(z) = 2M\varepsilon. \end{align} $$

The boundedness of $C_{\varphi }$ on $A^2_{\alpha }$ implies that

(2-5) $$\begin{align} D_n \leq \varepsilon\int_{D \setminus \overline{S(0, r)}} |f_n(\varphi(z))|^2\, dA_{\alpha}(z) \leq \varepsilon\|C_{\varphi}f_n\|^2 \leq \|C_{\varphi}\|^2 \varepsilon. \end{align} $$

It remains to estimate $B_n$ . Note that

$$ \begin{align*} B_n & = (\alpha + 1)\int_{D \setminus \overline{S(0, r)}} |u(z)|^2 |f_n'(\varphi(z))|^2 |\varphi'(z)|^2 (1 - |z|^2)^{\alpha + 2}\, dA(z) \\[3pt] & \leq (\alpha + 1)\varepsilon\int_{D \setminus \overline{S(0, r)}} |f_n'(\varphi(z))|^2 |\varphi'(z)|^2 (1 - |\varphi(z)|^2)^{\alpha + 2}\, dA(z). \end{align*} $$

Put $w = \varphi (z)$ . By the change-of-variable formula in [11, page 891] and the univalence of $\varphi $ ,

(2-6) $$ \begin{align} B_n & \leq (\alpha + 1)\varepsilon\int_D |f_n'(w)|^2 (1 - |w|^2)^{\alpha + 2}\, dA(w) \nonumber \\[3pt] & \leq \varepsilon\int_D |f_n'(w)|^2 (1 - |w|^2)^2\, dA_{\alpha}(w) \nonumber \\[3pt] & \leq \varepsilon d\|f_n\|^2 \nonumber \\[3pt] & \leq \varepsilon d , \end{align} $$

where d is the constant defined in Lemma 2.7. From Equations (2-2)–(2-6), it now follows that

$$ \begin{align*}\|uC_{\varphi}f_n\|^2 \leq \frac{\varepsilon}{c}(|u(0)|^2 + 4M + 2\|C_{\varphi}\|^2 + 2d) \end{align*} $$

for all $n> N$ . Hence, $\|uC_{\varphi }f_n\| \to 0$ as $n \to \infty $ .

3 Hilbert–Schmidt weighted composition operators

An important class of compact operators is the Hilbert–Schmidt operators. Let $H_1$ and $H_2$ be separable Hilbert spaces and $T : H_1 \to H_2$ be a bounded linear operator. Then T is said to be Hilbert–Schmidt if $\sum _{k=0}^{\infty } \|T e_k\|_{H_2}^2 < \infty $ for some orthonormal basis $\{e_k\}_{k=0}^{\infty }$ of $H_1$ . The value of this sum is independent of the choice of an orthonormal basis. It is well known that every Hilbert–Schmidt operator is compact, but the converse is not necessarily true. In what follows, we take $\{e_k\}_{k=0}^{\infty }$ to be the standard orthonormal basis for $A^2_{\alpha }$ , as given by Equation (1-1) in Section 1. We also recall a few identities for useful reference:

1. (a) $ {1}/{(1 - x)^{\alpha + 2}} = \sum _{k=0}^{\infty } ({\Gamma (\alpha + 2 + k)}/{k! \,\Gamma (\alpha + 2)}) x^k$ for $|x| < 1$ ;

2. (b) $ 1 - |({w - z})/({1 - \overline {w}z})|^2 = {(1 - |w|^2)(1 - |z|^2)}/{|1 - \overline {w}z|^2}$ and $ 1 - \overline {w}(({w - z})/ ({1 - \overline {w}z})) = ({1 - |w|^2})/({1 - \overline {w}z})$ for every $w, z \in D$ .

Using the criterion for $uC_{\varphi }$ to belong to the Schatten class, Čučković and Zhao obtained a characterization for Hilbert–Schmidt weighted composition maps on $A^2_0$ [4, Corollary 3]. We first generalize this result to the weighted Bergman space and provide a direct proof.

Theorem 3.1. Let $uC_{\varphi }$ be a weighted composition operator on $A^2_{\alpha }$ . Then $uC_{\varphi }$ is Hilbert–Schmidt if and only if

(3-1) $$ \begin{align} \int_D \frac{|u(z)|^2}{(1 - |\varphi(z)|^2)^{\alpha + 2}} \, dA_{\alpha}(z) < \infty. \end{align} $$

Proof. Direct computation gives

$$ \begin{align*} \sum_{k=0}^{\infty} \|uC_{\varphi} e_k\|^2 & = \sum_{k=0}^{\infty} \frac{\Gamma(\alpha + 2 + k)}{k! \,\Gamma(\alpha + 2)} \int_D |u(z)|^2 |\varphi(z)|^{2k}\, dA_{\alpha}(z) \\[3pt] & = \int_D |u(z)|^2 \sum_{k=0}^{\infty} \frac{\Gamma(\alpha + 2 + k)}{k! \,\Gamma(\alpha + 2)} |\varphi(z)|^{2k}\, dA_{\alpha}(z) \\[3pt] & = \int_D \frac{|u(z)|^2}{(1 - |\varphi(z)|^2)^{\alpha + 2}} \, dA_{\alpha}(z). \end{align*} $$

Interchanging the summation and integral sums in the second equality is legitimate because the terms are all non-negative. The assertion now follows.

It is shown in Theorem 2.2 that if $\overline {\varphi (D)} \subset D$ , then $uC_{\varphi }$ is compact. By Theorem 3.1, $uC_{\varphi }$ is also Hilbert–Schmidt. The next result shows that when $\varphi $ is an automorphism of D, the characterization of when a weighted composition operator is Hilbert–Schmidt becomes simpler.

Corollary 3.2. Let $\varphi $ be an automorphism of D. Then the weighted composition operator $uC_{\varphi }$ is Hilbert–Schmidt on $A^2_{\alpha }$ if and only if

(3-2) $$ \begin{align} \int_D \frac{|u(z)|^2}{(1 - |z|^2)^2} \, dA(z) < \infty. \end{align} $$

Proof. By the Schwarz–Pick theorem [3, page 48], we have

$$ \begin{align*}\frac{1 - |\varphi(z)|}{1 - |z|} \geq \frac{1 - |\varphi(0)|}{1 + |\varphi(0)|}. \end{align*} $$

Thus,

(3-3) $$ \begin{align} \frac{|u(z)|^2}{(1 - |z|^2)^{\alpha + 2}} \geq \bigg(\frac{1}{2} \cdot \frac{1 - |\varphi(0)|}{1 + |\varphi(0)|}\bigg)^{\alpha + 2}\frac{|u(z)|^2}{(1 - |\varphi(z)|^2)^{\alpha + 2}}. \end{align} $$

Write $\varphi (z) = c(a - z) / (1 - \overline {a}z)$ , where $a \in D$ and $|c| = 1$ . Since

$$ \begin{align*}1 - |\varphi(z)|^2 = \frac{(1 - |a|^2)(1 - |z|^2)}{|1 - \overline{a}z|^2} \quad\mathrm{and}\quad |1 -\overline{a}z| \geq 1 - |a| \end{align*} $$

for every $z \in D$ , it follows that

(3-4) $$ \begin{align} \frac{|u(z)|^2}{(1 - |z|^2)^{\alpha + 2}} \leq \bigg(\frac{1 + |a|}{1 - |a|}\bigg)^{\alpha + 2}\frac{|u(z)|^2}{(1 - |\varphi(z)|^2)^{\alpha + 2}}. \end{align} $$

We obtain the desired result by combining Equations (3-3) and (3-4), Theorem 3.1, and the fact that ${dA_{\alpha }(z) \kern1.2pt{=}\kern1.2pt (\alpha + 1) (1 - |z|^2)^{\alpha } dA(z)}$ .

The condition in Equation (3-2) is independent of the parameter $\alpha $ and can be expressed as ‘ $u \in L^2(D, d\tau )$ ’, where $\tau $ is the Möbius invariant measure on D defined by

(3-5) $$ \begin{align} d\tau(z) = \frac{1}{(1 - |z|^2)^2}\, dA(z). \end{align} $$

Here the term ‘invariant measure’ is justified by the fact that if $\varphi $ is an automorphism of D, then

$$ \begin{align*}\int_D |(f \circ \varphi)'(z)|^2 (1 - |z|^2)^2\, d\tau(z) = \int_D |f'(z)|^2 (1 - |z|^2)^2\, d\tau(z) \end{align*} $$

for all analytic functions f on D [17, Section 5.3.1].

Corollary 3.2 is also in contrast to the corresponding result for the Hardy space $H^2$ of D: if $\varphi $ is an automorphism, then it follows from [12, Theorem 9] that the only Hilbert–Schmidt weighted composition operator on $H^2$ is the zero operator.

Example 3.3. Let $u(z) = 1 / (1 - z)^{1/4}$ and $\varphi $ be any automorphism of D. Since, for all $z \in D$ we have $1 - |z|^2 \leq 2|1-z|$ , it follows that

$$ \begin{align*} \int_D \frac{|u(z)|^2}{(1 - |z|^2)^2} \, dA(z) & = \int_D \frac{1}{|1-z|^{{1}/{2}}(1 - |z|^2)^2} \, dA(z) \\[3pt] & \geq \frac{1}{4}\int_D \frac{1}{|1-z|^{{5}/{2}}} \, dA(z). \end{align*} $$

By [3, Lemma 7.3], $\int _D ({1}/{|1-z|^{5 / 2}}) \, dA(z) = \infty $ . According to Corollary 3.2, $uC_{\varphi }$ is not Hilbert–Schmidt on $A^2_{\alpha }$ .

The inequality in Equation (3-3) in fact holds for all analytic self-maps $\varphi $ of D. Thus, Equation (3-2) provides a sufficient condition for $uC_{\varphi }$ to be Hilbert–Schmidt on $A^2_{\alpha }$ . However, this condition is not necessary, as shown by the following example.

Example 3.4. Let $u(z) = (1 - z)^{(\alpha + 1) / 4}$ and $\varphi (z) = 1 - (1 - z)^{1/2}$ . Then $u \in A^2_{\alpha }$ . We claim that $uC_{\varphi }$ is Hilbert–Schmidt on $A^2_{\alpha }$ . Since $\varphi $ takes D into a polygonal region inscribed in T, there exist positive constants $c, \delta $ such that $\delta < 1/2$ , and $1 - |\varphi (z)| \geq c|1 - z|^{1/2}$ on $S(1, \delta )$ . Write

$$ \begin{align*} & \int_D \frac{|u(z)|^2}{(1 - |\varphi(z)|^2)^{\alpha + 2}} \, dA_{\alpha}(z) \\[3pt] &\quad = \int_{S(1, \delta)} \frac{|u(z)|^2}{(1 - |\varphi(z)|^2)^{\alpha + 2}} \, dA_{\alpha}(z) + \int_{D \setminus S(1, \delta)} \frac{|u(z)|^2}{(1 - |\varphi(z)|^2)^{\alpha + 2}} \, dA_{\alpha}(z). \end{align*} $$

By choosing $1 + {\alpha }/{2} < \beta < \tfrac 32 + \alpha $ , we have

$$ \begin{align*}(1 - |\varphi(z)|^2)^{\alpha + 2} \geq (1 - |\varphi(z)|)^{\alpha + 2} \geq c^{\alpha + 2}|1 - z|^{1 + {\alpha}/{2}} \geq c^{\alpha + 2}|1 - z|^{\beta} \end{align*} $$

for $z \in S(1, \delta )$ . Thus,

$$ \begin{align*} \int_{S(1, \delta)} \frac{|u(z)|^2}{(1 - |\varphi(z)|^2)^{\alpha + 2}} \, dA_{\alpha}(z) & \leq \frac{1}{c^{\alpha + 2}}\int_{S(1, \delta)} \frac{1}{|1 - z|^{\beta - ({\alpha + 1})/{2}}} \, dA_{\alpha}(z) \\[3pt] & \leq \frac{1}{c^{\alpha + 2}}\int_D \frac{1}{|1 - z|^{\beta - ({\alpha + 1})/{2}}} \, dA_{\alpha}(z) \\[3pt] & < \infty, \end{align*} $$

since $|\,\beta - ({\alpha + 1})/{2}| < 1 + {\alpha }/{2}$ . On $D \setminus S(1, \delta )$ , the continuity of $\varphi $ ensures that $|\varphi (z)| \leq d$ for a constant d with $0 < d < 1$ . Then

$$ \begin{align*} \int_{D \setminus S(1, \delta)} \frac{|u(z)|^2}{(1 - |\varphi(z)|^2)^{\alpha + 2}} \, dA_{\alpha}(z) & \leq \frac{1}{(1 - d^2)^{\alpha + 2}}\int_{D \setminus S(1, \delta)} |u(z)|^2\, dA_{\alpha}(z) \\[3pt] & \leq \frac{1}{(1 - d^2)^{\alpha + 2}}\|u\|^2 \\[3pt] & < \infty. \end{align*} $$

From Theorem 3.1, $uC_{\varphi }$ is Hilbert–Schmidt. However, since $1 - |z|^2 \leq 2|1 - z|$ on D, we have

$$ \begin{align*}\int_D \frac{|u(z)|^2}{(1 - |z|^2)^2} \, dA(z) \geq \frac{1}{4}\int_D \frac{1}{|1 - z|^{({3 - \alpha})/{2}}} \, dA(z) = \infty, \end{align*} $$

provided that $|(3 - \alpha ) / 2| \geq 1$ , that is, $-1 < \alpha \leq 1$ or $\alpha \geq 5$ .

The rest of this section is devoted to characterizing when $uC_{\varphi } - vC_{\psi }$ on $A^2_{\alpha }$ is Hilbert–Schmidt, where v and $\psi $ are two analytic functions on D such that $\psi (D) \subset D$ . This problem originates from the study of the topological structure of the space of (weighted) composition operators on $A^2_{\alpha }$ . There has been extensive investigation about differences of composition operators on the Hardy space $H^2$ of D (see for example [1, 6, 16]). The compact difference of two composition operators between weighted Bergman spaces was completely characterized in [9, 14, 15].

In [2], Choe et al. topologized the space of composition operators on $A^2_{\alpha }$ and described its components. By putting

$$ \begin{align*}\phi(z) = \frac{\psi(z) - \varphi(z)}{1 - \overline{\psi(z)}\varphi(z)} \end{align*} $$

for $z \in D$ , they also characterized the Hilbert–Schmidt difference of two composition operators $C_{\varphi }$ and $C_{\psi }$ in terms of $|\phi |$ , which is known as the pseudo-hyperbolic distance between $\varphi $ and $\psi $ . We generalize such characterization to the weighted case and construct an example to illustrate the result.

Theorem 3.5. Let $uC_{\varphi }$ and $vC_{\psi }$ be two weighted composition operators on $A^2_{\alpha }$ . Then the following statements are equivalent.

1. (i) The operator $uC_{\varphi } - vC_{\psi }$ is Hilbert–Schmidt on $A^2_{\alpha }$ .

2. (ii) $ {|\phi | u}/{(1 - |\varphi |^2)^{1 + {\alpha }/{2}}}, \, {v}/{(1 - |\psi |^2)^{1 + {\alpha }/{2}}} - {u(1 - |\psi |^2)^{1 + {\alpha }/{2}}}/{(1 - \overline {\psi }{\varphi })^{\alpha + 2}} \in L^2 (D, dA_{\alpha })$ .

3. (iii) $ {|\phi | v}/{(1 - |\psi |^2)^{1 + {\alpha }/{2}}}, \, {u}/{(1 - |\varphi |^2)^{1 + {\alpha }/{2}}} - {v(1 - |\varphi |^2)^{1 + {\alpha }/{2}}}/{(1 - \overline {\varphi }{\psi })^{\alpha + 2}} \in L^2 (D, dA_{\alpha })$ .

Proof. We first compute $\sum _{k=0}^{\infty } \|(uC_{\varphi } - vC_{\psi })e_k\|^2$ :

$$ \begin{align*} & \sum_{k=0}^{\infty} \|(uC_{\varphi} - vC_{\psi})e_k\|^2 \\[6pt] &\quad = \sum_{k=0}^{\infty} \frac{\Gamma(\alpha + 2 + k)}{k! \,\Gamma(\alpha + 2)} \|u\varphi^k - v\psi^k\|^2 \\[6pt] &\quad = \sum_{k=0}^{\infty} \frac{\Gamma(\alpha + 2 + k)}{k! \,\Gamma(\alpha + 2)} \int_D |u\varphi^k - v\psi^k|^2\, dA_{\alpha} \\[6pt] &\quad = \sum_{k=0}^{\infty} \frac{\Gamma(\alpha + 2 + k)}{k! \,\Gamma(\alpha + 2)} \int_D [|u|^2|\varphi|^{2k} + |v|^2|\psi|^{2k} - 2\,{\textrm{Re}}(u\overline{v}(\varphi\overline{\psi})^k)] \, dA_{\alpha}. \end{align*} $$

Interchanging the summation and integral signs in the last equality is valid because all the terms ${\Gamma (\alpha + 2 + k)}/{k! \,\Gamma (\alpha + 2)}|u\varphi ^k - v\psi ^k|^2$ are nonnegative. Then

$$ \begin{align*} & \sum_{k=0}^{\infty} \|(uC_{\varphi} - vC_{\psi})e_k\|^2 \\[6pt] &\quad = \int_D \sum_{k=0}^{\infty} \frac{\Gamma(\alpha + 2 + k)}{k! \,\Gamma(\alpha + 2)} [|u|^2|\varphi|^{2k} + |v|^2|\psi|^{2k} - 2\,{\textrm{Re}}(u\overline{v}(\varphi\overline{\psi})^k)] \, dA_{\alpha} \\[6pt] &\quad = \int_D \bigg[|u|^2 \sum_{k=0}^{\infty} \frac{\Gamma(\alpha + 2 + k)}{k! \,\Gamma(\alpha + 2)} |\varphi|^{2k} + |v|^2 \sum_{k=0}^{\infty} \frac{\Gamma(\alpha + 2 + k)}{k! \,\Gamma(\alpha + 2)} |\psi|^{2k} \\[6pt] & \quad\quad -2\,{\textrm{Re}}\bigg(u\overline{v}\sum_{k=0}^{\infty} \frac{\Gamma(\alpha + 2 + k)}{k! \,\Gamma(\alpha + 2)}(\varphi\overline{\psi})^k\bigg)\bigg]\, dA_{\alpha} \\[6pt] &\quad = \int_D \bigg[\frac{|u|^2}{(1 - |\varphi|^2)^{\alpha + 2}} + \frac{|v|^2}{(1 - |\psi|^2)^{\alpha + 2}} - 2\,{\textrm{Re}}\bigg(\frac{u\overline{v}}{(1 - \varphi\overline{\psi})^{\alpha + 2}}\bigg)\bigg]\, dA_{\alpha}. \end{align*} $$

Since $\varphi = (\psi - \phi ) / (1 - \overline {\psi }\phi )$ , it follows that

$$ \begin{align*} & \frac{|u|^2}{(1 - |\varphi|^2)^{\alpha + 2}} + \frac{|v|^2}{(1 - |\psi|^2)^{\alpha + 2}} - 2\,{\textrm{Re}}\bigg(\frac{u\overline{v}}{(1 - \varphi\overline{\psi})^{\alpha + 2}}\bigg) \\[3pt] &\quad = |u|^2\bigg[\frac{|1 - \overline{\psi}\phi|^2}{(1 - |\psi|^2)(1 - |\phi|^2)}\bigg]^{\alpha + 2} + \frac{|v|^2}{(1 - |\psi|^2)^{\alpha + 2}} - 2\,{\textrm{Re}}\bigg[u\overline{v}\bigg(\frac{1 - \overline{\psi}\phi}{1 - |\psi|^2}\bigg)^{\alpha + 2}\bigg] \\[3pt] &\quad = \frac{1}{(1 - |\psi|^2)^{\alpha + 2}}\bigg[|u|^2\bigg(\frac{|1 - \overline{\psi}\phi|^2}{1 - |\phi|^2}\bigg)^{\alpha + 2} + |v|^2 - 2\,{\textrm{Re}}(u\overline{v}(1 - \overline{\psi}\phi)^{\alpha + 2})\bigg] \\[3pt] &\quad = \frac{1}{(1 - |\psi|^2)^{\alpha + 2}}\bigg[|u|^2\bigg(\frac{|1 - \overline{\psi}\phi|^2}{1 - |\phi|^2}\bigg)^{\alpha + 2} + |v - u(1 - \overline{\psi}\phi)^{\alpha + 2}|^2 \ - |u|^2 |1 - \overline{\psi}\phi|^{2\alpha + 4}\bigg] \\[3pt] &\quad = \frac{1}{(1 - |\psi|^2)^{\alpha + 2}}\bigg[|u|^2\bigg(\frac{|1 - \overline{\psi}\phi|^2}{1 - |\phi|^2}\bigg)^{\alpha + 2}(1 - (1 - |\phi|^2)^{\alpha + 2}) \ + |v - u(1 - \overline{\psi}\phi)^{\alpha + 2}|^2\bigg] \\[3pt] &\quad = \frac{[1 - (1 - |\phi|^2)^{\alpha + 2}] |u|^2}{(1 - |\varphi|^2)^{\alpha + 2}} + \bigg|\frac{v}{(1 - |\psi|^2)^{1 + {\alpha}/{2}}} - \frac{u(1 - |\psi|^2)^{1 + {\alpha}/{2}}}{(1 - \overline{\psi}{\varphi})^{\alpha + 2}}\bigg|^2. \end{align*} $$

Therefore,

$$ \begin{align*} & \sum_{k=0}^{\infty} \|(uC_{\varphi} - vC_{\psi})e_k\|^2 \\[3pt] &\quad = \int_D \bigg[\frac{(1 - (1 - |\phi|^2)^{\alpha + 2}) |u|^2}{(1 - |\varphi|^2)^{\alpha + 2}} + \bigg|\frac{v}{(1 - |\psi|^2)^{1 + {\alpha}/{2}}} - \frac{u(1 - |\psi|^2)^{1 + {\alpha}/{2}}}{(1 - \overline{\psi}{\varphi})^{\alpha + 2}}\bigg|^2\bigg]\, dA_{\alpha}. \end{align*} $$

The operator $uC_{\varphi } - vC_{\psi }$ is Hilbert–Schmidt on $A^2_{\alpha }$ if and only if

$$ \begin{align*}\frac{\sqrt{1 - (1 - |\phi|^2)^{\alpha + 2}} \, u}{(1 - |\varphi|^2)^{1 + {\alpha}/{2}}}, \, \frac{v}{(1 - |\psi|^2)^{1 + {\alpha}/{2}}} - \frac{u(1 - |\psi|^2)^{1 + {\alpha}/{2}}}{(1 - \overline{\psi}{\varphi})^{\alpha + 2}} \in L^2(D, dA_{\alpha}). \end{align*} $$

Moreover, write

$$ \begin{align*}\frac{\sqrt{1 - (1 - |\phi|^2)^{\alpha + 2}} \, u}{(1 - |\varphi|^2)^{1 + {\alpha}/{2}}} = \sqrt{\frac{1 - (1 - |\phi|^2)^{\alpha + 2}}{|\phi|^2}} \cdot \frac{|\phi| u}{(1 - |\varphi|^2)^{1 + {\alpha}/{2}}}. \end{align*} $$

Note that the function $f(x) = [1 - (1 - x^2)^{\alpha + 2}] / x^2$ is continuous and positive on $(0, 1]$ . This, in conjunction with the fact $\lim _{x \to 0^+} f(x) = \alpha + 2> 0$ , implies that f is bounded above and away from zero on $(0, 1)$ . Thus,

$$ \begin{align*}\frac{\sqrt{1 - (1 - |\phi|^2)^{\alpha + 2}} \, u}{(1 - |\varphi|^2)^{1 + {\alpha}/{2}}} \in L^2(D, dA_{\alpha}) \end{align*} $$

if and only if $|\phi | u / (1 - |\varphi |^2)^{1 + {\alpha }/{2}} \in L^2(D, dA_{\alpha })$ . This establishes the equivalence of statements (i) and (ii).

Furthermore, upon switching the roles of $u, v$ and the roles of $\varphi , \psi $ in the preceding calculations, we obtain

$$ \begin{align*} & \sum_{k=0}^{\infty} \|(vC_{\psi} - uC_{\varphi})e_k\|^2 \\[3pt] &\quad = \int_D \bigg[\frac{(1 - (1 - |\phi|^2)^{\alpha + 2}) |v|^2}{(1 - |\psi|^2)^{\alpha + 2}} + \bigg|\frac{u}{(1 - |\varphi|^2)^{1 + {\alpha}/{2}}} - \frac{v(1 - |\varphi|^2)^{1 + {\alpha}/{2}}}{(1 - \overline{\varphi}{\psi})^{\alpha + 2}}\bigg|^2\bigg]\, dA_{\alpha}. \end{align*} $$

By a similar argument, statements (i) and (iii) are also equivalent.

Taking $v = 0$ and $\varphi = \psi $ in the above theorem, we obtain the characterization in Equation (3-1) for a single Hilbert–Schmidt weighted composition operator. There are also two nontrivial consequences of Theorem 3.5. The first one characterizes the Hilbert–Schmidt difference of two composition operators on $A^2_{\alpha }$ . The second one, which generalizes [2, Corollary 3.8], states that the Hilbert–Schmidt property of the difference of weighted composition operators on a smaller space extends to larger spaces.

Corollary 3.6 [2, Corollary 3.7].

The operator $C_{\varphi } - C_{\psi }$ is Hilbert–Schmidt on $A^2_{\alpha }$ if and only if

$$ \begin{align*}\frac{|\phi|}{(1 - |\varphi|^2)^{1 + {\alpha}/{2}}}, \, \frac{|\phi|}{(1 - |\psi|^2)^{1 + {\alpha}/{2}}} \in L^2(D, dA_{\alpha}). \end{align*} $$

Proof. The ‘only if’ part is evident by taking $u = v = 1$ in Theorem 3.5. To prove the ‘if’ part, assume that both $|\phi | / (1 - |\varphi |^2)^{1 + {\alpha }/{2}}$ and $|\phi | / (1 - |\psi |^2)^{1 + {\alpha }/{2}}$ are in $L^2(D, dA_{\alpha })$ . Write

$$ \begin{align*} &\frac{1}{(1 - |\psi|^2)^{1 + {\alpha}/{2}}} - \frac{(1 - |\psi|^2)^{1 + {\alpha}/{2}}}{(1 - \overline{\psi}{\varphi})^{\alpha + 2}} = \frac{1 - (1 - \overline{\psi}{\phi})^{\alpha + 2}}{(1 - |\psi|^2)^{1 + {\alpha}/{2}}} \\[3pt] &\quad = \frac{1 - (1 - \overline{\psi}{\phi})^{\alpha + 2}}{\phi} \cdot \frac{\phi}{(1 - |\psi|^2)^{1 + {\alpha}/{2}}}. \end{align*} $$

By the continuity of the function $g(z) = [1 - (1 - \overline {w}{z})^{\alpha + 2}] / z$ ( $w \in D$ ) on $D \setminus \{0\}$ and the fact that $\lim _{z \to 0} g(z)$ exists (and equals $(\alpha + 2)\overline {w}$ ), the expression $[1 - (1 - \overline {\psi }{\phi })^{\alpha + 2}] / \phi $ is bounded on the set $\{z \in D : \phi (z) \ne 0\}$ . Thus,

$$ \begin{align*}\frac{1}{(1 - |\psi|^2)^{1 + {\alpha}/{2}}} - \frac{(1 - |\psi|^2)^{1 + {\alpha}/{2}}}{(1 - \overline{\psi}{\varphi})^{\alpha + 2}} \in L^2(D, dA_{\alpha}) \end{align*} $$

as well. In light of Theorem 3.5, $C_{\varphi } - C_{\psi }$ is Hilbert–Schmidt on $A^2_{\alpha }$ .

Corollary 3.7. Let $uC_{\varphi }$ and $vC_{\psi }$ be two weighted composition operators on $A^2_{\alpha }$ . If $uC_{\varphi } - vC_{\psi }$ is Hilbert–Schmidt on $A^2_{\alpha }$ , then $uC_{\varphi } - vC_{\psi }$ is also Hilbert–Schmidt on $A^2_{\beta }$ for every $\beta> \alpha > -1$ .

Proof. Since $uC_{\varphi } - vC_{\psi }$ is Hilbert–Schmidt on $A^2_{\alpha }$ , the following functions are all in $L^2(D, d\tau )$ , where $\tau $ is the measure defined in Equation (3-5):

1. (i) $ |\phi | u (({1 - |z|^2})/({1 - |\varphi |^2}))^{1 + {\alpha }/{2}}$ ;

2. (ii) $ |\phi | v (({1 - |z|^2})/({1 - |\psi |^2}))^{1 + {\alpha }/{2}}$ ;

3. (iii) $ [{v}/{(1 - |\psi |^2)^{1 + {\alpha }/{2}}} - {u(1 - |\psi |^2)^{1 + {\alpha }/{2}}}/{(1 - \overline {\psi }{\varphi })^{\alpha + 2}}](1 - |z|^2)^{1 + {\alpha }/{2}}$ ;

since

(3-6) $$ \begin{align} \bigg(\frac{1 - |z|}{1 - |\varphi|}\bigg)^{\beta - \alpha} \leq \bigg(\frac{1 + |\varphi(0)|}{1 - |\varphi(0)|}\bigg)^{\beta - \alpha} \end{align} $$

[3, page 48] and

(3-7) $$ \begin{align} \frac{1}{2}\bigg(\frac{1 - |z|}{1 - |\varphi|}\bigg) \leq \frac{1 - |z|^2}{1 - |\varphi|^2} \leq 2\bigg(\frac{1 - |z|}{1 - |\varphi|}\bigg) \end{align} $$

(both Equations (3-6) and (3-7) hold if $\varphi $ is replaced by $\psi $ ), we have

$$ \begin{align*} & \int_D |\phi|^2 |u|^2 \bigg(\frac{1 - |z|^2}{1 - |\varphi|^2}\bigg)^{\,\beta + 2}\, d\tau(z) \\[3pt] &\quad \leq 2^{\alpha + \beta + 4}\bigg(\frac{1 + |\varphi(0)|}{1 - |\varphi(0)|}\bigg)^{\beta - \alpha}\int_D |\phi|^2 |u|^2 \bigg(\frac{1 - |z|^2}{1 - |\varphi|^2}\bigg)^{\alpha + 2}\, d\tau(z)\\[3pt] &\quad < \infty. \end{align*} $$

From the proof of Corollary 3.6,

$$ \begin{align*} & \bigg|\frac{v}{(1 - |\psi|^2)^{1 + {\beta}/{2}}} - \frac{u(1 - |\psi|^2)^{1 + {\beta}/{2}}}{(1 - \overline{\psi}{\varphi})^{\,\beta + 2}}\bigg|^2 \\[3pt] &\quad = \bigg|v\bigg[\frac{1}{(1 - |\psi|^2)^{1 + {\beta}/{2}}} - \frac{(1 - |\psi|^2)^{1 + {\beta}/{2}}}{(1 - \overline{\psi}{\varphi})^{\,\beta + 2}}\bigg] + (v - u)\frac{(1 - |\psi|^2)^{1 + {\beta}/{2}}}{(1 - \overline{\psi}{\varphi})^{\,\beta + 2}}\bigg|^2 \\[3pt] &\quad \leq 2\bigg[\bigg|\frac{1 - (1 - \overline{\psi}{\phi})^{\,\beta + 2}}{\phi}\bigg|^2 \frac{|\phi|^2 |v|^2}{(1 - |\psi|^2)^{\,\beta + 2}} + |v-u|^2 \frac{(1 - |\psi|^2)^{\,\beta + 2}}{|1 - \overline{\psi}{\varphi}|^{2\beta + 4}}\bigg] \\[3pt] &\quad \leq 2\bigg[M_{\beta} \frac{|\phi|^2 |v|^2}{(1 - |\psi|^2)^{\,\beta + 2}} + |v - u|^2 \bigg(\frac{1 - |\phi|^2}{1 - |\varphi|^2}\bigg)^{\,\beta + 2}\bigg], \end{align*} $$

where $M_{\beta }$ is a constant depending on $\beta $ only. Then

$$ \begin{align*} & \bigg|\frac{v}{(1 - |\psi|^2)^{1 + {\beta}/{2}}} - \frac{u(1 - |\psi|^2)^{1 + {\beta}/{2}}}{(1 - \overline{\psi}{\varphi})^{\,\beta + 2}}\bigg|^2 (1 - |z|^2)^{\,\beta + 2} \\[3pt] &\quad \leq 2\bigg[M_{\beta} |\phi|^2 |v|^2 \bigg(\frac{1 - |z|^2}{1 - |\psi|^2}\bigg)^{\,\beta + 2} + (1 - |\phi|^2)^{\,\beta + 2}|v - u|^2\bigg(\frac{1 - |z|^2}{1 - |\varphi|^2}\bigg)^{\,\beta + 2}\bigg]. \end{align*} $$

Note that

(3-8) $$ \begin{align} & \int_D |\phi|^2 |v|^2 \bigg(\frac{1 - |z|^2}{1 - |\psi|^2}\bigg)^{\,\beta + 2}\, d\tau(z) \nonumber \\[3pt] &\quad \leq 2^{\alpha + \beta + 4}\bigg(\frac{1 + |\psi(0)|}{1 - |\psi(0)|}\bigg)^{\beta - \alpha}\int_D |\phi|^2 |v|^2 \bigg(\frac{1 - |z|^2}{1 - |\psi|^2}\bigg)^{\alpha + 2}\, d\tau(z) \nonumber \\[3pt] &\quad < \infty. \end{align} $$

Moreover, if we put $c = 2^{\alpha + \beta + 4}(({1 + |\varphi (0)|})/({1 - |\varphi (0)|}))^{\beta - \alpha }$ , then appealing to the proof of Corollary 3.6 again gives

$$ \begin{align*} & (1 - |\phi|^2)^{\,\beta + 2}|v - u|^2\bigg(\frac{1 - |z|^2}{1 - |\varphi|^2}\bigg)^{\,\beta + 2} \\[3pt] &\quad \leq c(1 - |\phi|^2)^{\alpha + 2}|v - u|^2\bigg(\frac{1 - |z|^2}{1 - |\varphi|^2}\bigg)^{\alpha + 2} \\[3pt] &\quad = c\bigg|v\bigg[\frac{1}{(1 - |\psi|^2)^{1 + {\alpha}/{2}}} - \frac{(1 - |\psi|^2)^{1 + {\alpha}/{2}}}{(1 - \overline{\psi}{\varphi})^{\alpha + 2}}\bigg] + (v - u)\frac{(1 - |\psi|^2)^{1 + {\alpha}/{2}}}{(1 - \overline{\psi}{\varphi})^{\alpha + 2}} \\[3pt] & \quad\quad -v\bigg[\frac{1}{(1 - |\psi|^2)^{1 + {\alpha}/{2}}} - \frac{(1 - |\psi|^2)^{1 + {\alpha}/{2}}}{(1 - \overline{\psi}{\varphi})^{\alpha + 2}}\bigg]\bigg|^2 (1 - |z|^2)^{\alpha + 2} \\[3pt] &\quad \leq 2c\bigg[\bigg|\frac{v}{(1 - |\psi|^2)^{1 + {\alpha}/{2}}} - \frac{u(1 - |\psi|^2)^{1 + {\alpha}/{2}}}{(1 - \overline{\psi}{\varphi})^{\alpha + 2}}\bigg|^2 \\[3pt] & \quad\quad +\bigg|\frac{1 - (1 - \overline{\psi}{\phi})^{\alpha + 2}}{\phi}\bigg|^2 \frac{|\phi|^2 |v|^2}{(1 - |\psi|^2)^{\alpha + 2}}\bigg](1 - |z|^2)^{\alpha + 2} \\[3pt] &\quad \leq 2c\bigg[\bigg|\frac{v}{(1 - |\psi|^2)^{1 + {\alpha}/{2}}} - \frac{u(1 - |\psi|^2)^{1 + {\alpha}/{2}}}{(1 - \overline{\psi}{\varphi})^{\alpha + 2}}\bigg|^2(1 - |z|^2)^{\alpha + 2} \\[3pt] & \quad\quad + M_{\alpha} |\phi|^2 |v|^2 \bigg(\frac{1 - |z|^2}{1 - |\psi|^2}\bigg)^{\alpha + 2}\bigg], \end{align*} $$

where $M_{\alpha }$ is a constant depending on $\alpha $ only. Thus,

$$ \begin{align*}\int_D (1 - |\phi|^2)^{\,\beta + 2}|v - u|^2\bigg(\frac{1 - |z|^2}{1 - |\varphi|^2}\bigg)^{\,\beta + 2}\, d\tau(z) < \infty. \end{align*} $$

This, in conjunction with Equation (3-8), implies that

$$ \begin{align*}\bigg[\frac{v}{(1 - |\psi|^2)^{1 + {\beta}/{2}}} - \frac{u(1 - |\psi|^2)^{1 + {\beta}/{2}}}{(1 - \overline{\psi}{\varphi})^{\,\beta + 2}}\bigg](1 - |z|^2)^{1 + {\beta}/{2}}\in L^2(D, d\tau). \end{align*} $$

According to Theorem 3.5, $uC_{\varphi } - vC_{\psi }$ is also Hilbert–Schmidt on $A^2_{\beta }$ .

Example 3.8. Let $u(z) = v(z) = z$ , $\varphi (z) = az + 1 - a$ and $\psi (z) = \varphi (z) + \varepsilon (1 - \varphi (z))^b$ , where a, b, $\varepsilon $ are positive constants such that $a \leq 1/2$ , $b> 2$ , and $\varepsilon $ is to be determined. Since ${\textrm{Re}} (z) < 1$ for $z \in D$ , we have

$$ \begin{align*} 1 - |\varphi(z)|^2 - |1 - \varphi(z)|^2 & = 2a - 2a^2 - 2a^2|z|^2 + 2a(2a - 1){\textrm{Re}}(z) \\[3pt] & \geq 2a - 2a^2 - 2a^2|z|^2 + 2a(2a - 1) \\[3pt] & = 2a^2(1 - |z|^2) \\[3pt] &> 0, \end{align*} $$

that is,

$$ \begin{align*}1 - |\varphi(z)|^2> |1 - \varphi(z)|^2 = a^2|1 - z|^2. \end{align*} $$

Note that $0 < |1 - \varphi (z)| = a|1-z| < 1$ on D. In what follows, we choose $\varepsilon < 1/4$ . Then

$$ \begin{align*} 1 - |\psi(z)|^2 &> 1 - |\varphi(z)|^2 - 2\varepsilon |1 - \varphi(z)|^b - \varepsilon^2 |1 - \varphi(z)|^{2b} \\[3pt] & > (1 - 2\varepsilon - \varepsilon^2)|1 - \varphi(z)|^2 \\[3pt] & > \tfrac{7}{16}a^2|1 - z|^2 \\[3pt] & > 0, \end{align*} $$

or $\psi (D) \subset D$ . We claim that $uC_{\varphi } - vC_{\psi }$ is Hilbert–Schmidt on $A^2_{\alpha }$ if ${3\alpha }/{4} + \tfrac 72 < b < {5\alpha }/{4} + \tfrac 92$ . Since

$$ \begin{align*} |1 - \overline{\psi(z)}\varphi(z)| & = |1 - |\psi(z)|^2 + \overline{\psi(z)}(\psi(z) - \varphi(z))| \\[3pt] & \geq 1 - |\psi(z)|^2 - \varepsilon|1 - \varphi(z)|^b \\[3pt] &> \tfrac{7}{16}a^2|1 - z|^2 - \tfrac{1}{4}a^2|1 - z|^2 \\[3pt] & = \tfrac{3}{16}a^2|1 - z|^2, \end{align*} $$

we have

$$ \begin{align*}|\phi(z)| = \bigg|\frac{\psi(z) - \varphi(z)}{1 - \overline{\psi(z)}\varphi(z)}\bigg| < \frac{16\varepsilon|1 - \varphi(z)|^b}{3a^2|1 - z|^2} < \frac{4}{3}a^{b-2}|1 - z|^{b-2}. \end{align*} $$

Thus,

$$ \begin{align*}\int_D \frac{|\phi(z)|^2 |u(z)|^2}{(1 - |\varphi(z)|^2)^{\alpha + 2}} \, dA_{\alpha}(z) \leq M_1 \int_D \frac{1}{|1 - z|^{8 - 2b + 2\alpha}} \, dA_{\alpha}(z); \end{align*} $$

and from the proof of Corollary 3.6,

$$ \begin{align*} & \int_D \bigg|\frac{v(z)}{(1 - |\psi(z)|^2)^{1 + {\alpha}/{2}}} - \frac{u(z)(1 - |\psi(z)|^2)^{1 + {\alpha}/{2}}}{(1 - \overline{\psi(z)}{\varphi(z)})^{\alpha + 2}}\bigg|^2\, dA_{\alpha}(z) \\[3pt] &\quad = \int_D \bigg|\frac{1 - (1 - \overline{\psi(z)}\phi(z))^{\alpha + 2}}{\phi(z)}\bigg|^2 \frac{|\phi(z)|^2 |z|^2}{(1 - |\psi(z)|^2)^{\alpha + 2}}\, dA_{\alpha}(z) \\[3pt] &\quad \leq M_2\int_D \frac{1}{|1 - z|^{8 - 2b + 2\alpha}} \, dA_{\alpha}(z), \end{align*} $$

where $M_1$ and $M_2$ are positive constants depending on a, b, and $\alpha $ . The integral $\int _D ({1}/{|1 - z|^{8 - 2b + 2\alpha }}) \, dA_{\alpha }(z)$ is finite if and only if $|8 - 2b + 2\alpha | < 1 + {\alpha }/{2}$ , that is, ${3\alpha }/{4} + \tfrac 72 < b < {5\alpha }/{4} + \tfrac 92$ . The claim now follows from Theorem 3.5.