Continuous Frames Under the Weighted Composition Operators on the Weighted Bergman Space

Abstract

We investigate the behavior of continuous frames in the weighted Bergman space $A_{\alpha }^2$ over the unit disc under the action of weighted composition operators. Motivated by developments in the discrete frame setting, we provide a comprehensive characterization of those weighted composition operators that preserve continuous frames, including tight and Parseval frames. Additionally, we examine the structure of dual frames in this context and establish necessary and sufficient conditions under which dual frame pairs are preserved by such operators. Explicit constructions of dual pairs induced by weighted composition operators are also presented. The study concludes with an analysis of the scalability of continuous frames and explores its invariance under the action of weighted composition operators.

1. Introduction

Frame theory in Hilbert spaces originated in the work of Duffin and Schaeffer [1] and was subsequently developed further by Daubechies and collaborators [2,3]. In the discrete setting, a frame is a countable indexed family of vectors in a separable Hilbert space that enables every element of the space to be represented in a stable manner, although such representations need not be unique. Inspired by constructions arising in coherent state theory, Antoine and others broadened this concept to encompass families indexed by uncountable sets. In this extended setting, the indexing set is taken to be a locally compact space endowed with a Radon measure, from which the framework of continuous frames naturally arises [4,5,6]. Continuous frames appear naturally in several important analytical tools, including the continuous wavelet transform [5,7] and the short-time Fourier transform [8], where the underlying systems are parametrized continuously.

One of the fundamental problems in frame theory is to characterize the bounded linear operators on Hilbert or Banach spaces that preserve the frame property. In this regard, weighted composition operators have recently become an important object of study. A comprehensive characterization of weighted composition operators preserving the frame structure on the Fock space was established by Han and Wang [9]. Subsequently, Mengestie [10] showed that, within the Fock space, the orbit of any vector under such operators cannot constitute a frame. Furthermore, an exhaustive classification of weighted composition operators that preserve discrete frame structures in weighted Hardy and Bergman spaces was established by Zhao and collaborators [11,12,13].

Continuous frames are particularly relevant for weighted Bergman spaces and weighted composition operators because both the spaces and the operators are governed by continuous analytic and measure-theoretic structures, and continuous frames align naturally with reproducing kernels, integral norms, and the geometric action of composition maps. In this article, we examine weighted composition operators acting on the weighted Bergman space $A_{\alpha }^2$ of the unit disc and characterize those operators that preserve the structure of continuous frames, with particular emphasis on continuous tight and Parseval frames. This study is motivated by the fundamental results in the discrete frame context, particularly those presented in [12]. In addition, we examine the invariance of dual frames under the action of weighted composition operators. We establish necessary and sufficient conditions under which such operators preserve dual frame pairs and further provide explicit formulations for the construction of dual pairs induced by weighted composition operators.

We organized the paper as follows. In Section 2, some necessary preliminary materials are given such that Section 2.1 introduces the concept of continuous frames and summarizes their fundamental properties. Section 2.2 presents the definition of the weighted Bergman space $A_{\alpha }^2$ on the unit disc $\mathbb{D}$ and formulates the weighted composition operator $W_{\psi ,\varphi }$ acting on the space of holomorphic functions $\mathrm{Hol}\left(\mathbb{D}\right)$. Section 3 addresses the behavior of continuous frames under weighted composition operators. Section 3.1 establishes conditions ensuring that the operator $W_{\psi ,\varphi }$ preserves the frame property. In Section 3.2, a characterization of dual frames associated with weighted composition operators acting on weighted Bergman spaces is obtained. The final section examines the scalability of continuous frames, with particular emphasis on its invariance under the action of weighted composition operators.

2. Preliminaries

2.1. Frames

In this section, we give basic definitions related to continuous frames in the literature. For further details, the reader is referred to [4,6,14]. Throughout this work, H is assumed to be a separable Hilbert space over $\mathbb{C}$ and $(X,\mu )$ denotes a positive, $\sigma$-finite measure space.

Definition 1.

A mapping $F:X\to H$ is said to be a continuous frame for H relative to the measure space $(X,\mu )$ if F is weakly measurable and there exist constants $A,B\in (0,\infty )$ with $A\le B$ such that

$${A\left|\right|f\left|\right|}^2\le {\int }_X{\left|\langle f,F\left(w\right)\rangle \right|}^{2}d\mu \left(w\right)\le {B\left|\right|f\left|\right|}^2,\forall f\in H.$$

(1)

The constants A and B are identified as the respective lower and upper frame bounds associated with F. When these constants are equal, that is, when $A=B$, the mapping F is said to be a continuous tight frame. If $A=B=1$, then F is called a continuous Parseval frame. When second inequality in (1) holds, F is called Bessel and B is called the Bessel constant. In the case, where X is countable and μ is taken to be the counting measure, the continuous frame $F:X\to H$ coincides with a discrete frame ${\left(F\left(w\right)\right)}_{w\in X}$ in H.

The operator $T_F:H⟶L^2(X,\mu )$, which is defined by

$$f\mapsto T_F\left(f\right)=\langle f,F\left(w\right)\rangle ,w\in X$$

is called the analysis operator associated with the mapping F. The adjoint operator $T_F^{*}$ is correspondingly called the synthesis operator and weakly defined by $T_F^{*}:L^2(X,\mu )⟶H$ such that

$$T_F^{*}\left(\mathsf{\Phi }\right)={\int }_X\mathsf{\Phi }\left(w\right)F\left(w\right)d\mu \left(w\right),\mathsf{\Phi }\in L^2(X,\mu ).$$

This operator is a well-defined, bounded and linear operator.

Given a continuous frame F with upper and lower frames bounds A and B, the operator $S_F=T_F^{*}{T}_F$ is known as the continuous frame operator of F. It is a positive, self-adjoint and invertible operator with bounds $A{I}_H\le S_F\le B{I}_H$.

For any $f\in H$, we have

$$S_F\left(f\right)={\int }_X\langle f,F\left(w\right)\rangle F\left(w\right)d\mu \left(w\right).$$

Definition 2.

Let $F:X\to H$ be a continuous frame for H. A continuous Bessel map $G:X\to H$ is a dual frame of F if

$$f={\int }_X\langle f,F\left(w\right)\rangle G\left(w\right)d\mu \left(w\right),\forall f\in H$$

or

$$\langle f,g\rangle ={\int }_X\langle f,F\left(w\right)\rangle \langle G\left(w\right),g\rangle d\mu \left(w\right),\forall f,g\in H.$$

We call $(F,G)$ a dual pair for H. Let $T_F$ and $T_G$ be the analysis operators of F and G, respectively, then this definition is equivalent to $T_G^{*}{T}_F=I_H$.

condition

$$\langle f,g\rangle ={\int }_X\langle f,F\left(w\right)\rangle \langle G\left(w\right),g\rangle d\mu \left(w\right),\forall f,g\in H$$

is equivalent to

$$\langle f,g\rangle ={\int }_X\langle f,G\left(w\right)\rangle \langle F\left(w\right),g\rangle d\mu \left(w\right),\forall f,g\in H$$

because $T_G^{*}{T}_F=I_H$ if and only if $T_F^{*}{T}_G=I_H$. $S_F^{-1}F$ is called the canonical dual of F.

2.2. Weighted Composition Operators on the Weighted Bergman Space

Let $\mathbb{D}=\{z\in \mathbb{C}:|z|<1\}$ be the open unit disc in $\mathbb{C}$, $\mathrm{Hol}\left(\mathbb{D}\right)$ be the space of holomorphic functions on $\mathbb{D}$, and $dA\left(z\right)={\displaystyle \frac{1}{\pi }}dxdy={\displaystyle \frac{1}{\pi }}rdrd\theta$, $z=x+iy=r{e}^{i\theta }$, be the normalized area measure on $\mathbb{D}$. For $\alpha >-1$, define $d{A}_{\alpha }\left(z\right)={(\alpha +1)(1-|z|}^2{)}^{\alpha }dA\left(z\right)$. Notice that $d{A}_{\alpha }$ is a normalized area measure on $\mathbb{D}$. The weighted Bergman space $A_{\alpha }^2$ of the unit disc $\mathbb{D}$ is defined by;

$$A_{\alpha }^2=\{f\in \mathrm{Hol}\left(\mathbb{D}\right):{\int }_{\mathbb{D}}{\left|f\left(z\right)\right|}^{2}d{A}_{\alpha }\left(z\right)<\infty \}.$$

If $f\in A_{\alpha }^2,$ then ${\left|\right|f\left|\right|}^2={\int }_{\mathbb{D}}{\left|f\left(z\right)\right|}^{2}d{A}_{\alpha }\left(z\right)$ defines a norm on $A_{\alpha }^2$, and the space $A_{\alpha }^2$ is a Banach space. Moreover, it is a Hilbert space with the inner product

$$\langle f,g\rangle ={\int }_{\mathbb{D}}f\left(z\right)\overline{g\left(z\right)}d{A}_{\alpha }\left(z\right).$$

It is a classical fact that the weighted Bergman space $A_{\alpha }^2$ is a reproducing kernel Hilbert space whose reproducing kernel is given by

$$K_{\alpha }(z,w)={\displaystyle \frac{1}{{(1-z\overline{w})}^{2+\alpha }}},z,w\in \mathbb{D}.$$

Let $\psi$ be a holomorphic function on the unit disk $\mathbb{D}$, and let $\varphi :\mathbb{D}\to \mathbb{D}$ be a holomorphic self-map. The associated weighted composition operator $W_{\psi ,\varphi }$ acting on $\mathrm{Hol}\left(\mathbb{D}\right)$ is defined by

$$W_{\psi ,\varphi }\left(f\right)=\psi ·(f\circ \varphi ),f\in \mathrm{Hol}\left(\mathbb{D}\right).$$

Weighted composition operators can be viewed as a natural generalization of composition operators obtained by incorporating multiplication by a holomorphic weight. Such operators have been extensively studied in the context of analytic function spaces; see [15,16] for detailed accounts. The boundedness of composition operators on $A_{\alpha }^2$ follows directly from the Littlewood Subordination Theorem (cf. Section 3.1 in [15]).

The boundedness of the weighted composition operators $W_{\psi ,\varphi }$ on the weighted Bergman spaces $A_{\alpha }^2$ was established in [17] through the use of the generalized Berezin transform. In addition, [18] provided a Carleson measure characterization for the boundedness of these operators in $A_{\alpha }^2$. Throughout the remainder of this paper, we shall assume that $W_{\psi ,\varphi }$ are bounded on the weighted Bergman spaces $A_{\alpha }^2$.

The following proposition is a special case of Proposition 3.5 in [12].

Proposition 1.

Let ψ be a holomorphic function on $\mathbb{D}$ and ϕ be a holomorphic self-map of $\mathbb{D}$ such that $W_{\psi ,\varphi }$ is bounded and $W_{\psi ,\varphi }^{*}$ is bounded below on the weighted Bergman space $A_{\alpha }^2$, where $\alpha >-2$. Then ψ and ϕ satisfies the following conditions.

(1) 

ψ is bounded on $\mathbb{D}$;

(2) 

$1/\psi$ is bounded on $\mathbb{D}$;

(3) 

ϕ is an automorphism of $\mathbb{D}$.

3. Results

3.1. Weighted Composition Operators That Preserves Continuous Frames

Let $U:H\to H$ be a linear operator. U is said to preserve continuous (tight, Parseval) frames if $UF:X\to H$, defined by $w\mapsto UF\left(w\right)$, is a continuous (tight, Parseval) frame for H whenever F is a continuous (tight, Parseval) frame for H with respect to $(X,\mu )$.

We devote this section to characterizing weighted composition operators that preserve the continuous (tight, Parseval) frames. We first establish a result that serves as an analogue to Theorem 1 of [19] and Proposition 2.3 of [20].

Proposition 2.

Suppose U is a bounded linear operator acting on a Hilbert space H. The statements provided below are equivalent:

(1) 

U preserves continuous frames for H relative to $(X,\mu )$;

(2) 

$U^{*}$ is bounded below on H;

(3) 

U is surjective on H.

Proof. 

A well-known fact says that the surjectivity of U and the lower-boundedness of $U^{*}$ are equivalent (see, for example, [21] Theorem 4.13). Hence, (2) and (3) are equivalent. Next, we show that (1) and (2) are equivalent. Suppose F is a continuous frame on $(X,\mu )$ for H, and assume its image under U, denoted by $UF$, also constitutes a continuous frame. It follows from the frame property of $UF$ that there are positive constants $C_1$ and $C_2$ such that for every $f\in H$:

$$C_1{\left|\right|f\left|\right|}^2\le {\int }_X{\left|\langle f,U\left(F\left(w\right)\right)\rangle \right|}^{2}d\mu \left(w\right)={\int }_X|\langle U^{*}f,F\left(w\right)\rangle {|}^{2}d\mu \left(w\right)\le C_2\left|\right|U^{*}f{\left|\right|}^2.$$

The second inequality above is a direct consequence of F being a continuous frame. This relationship immediately implies that $U^{*}$ is bounded from below.

For the converse, assume $U^{*}$ is bounded below. Let $F:(X,\mu )\to H$ be a continuous frame. Notice that the boundedness of U ensures that $U^{*}$ is also bounded. Then there are constants $C_1,C_2,C_3,C_4>0$ such that, for any $f\in H$,

$$\begin{array}{cc}\hfill C_1{C}_2{\left|\right|f\left|\right|}^2\le C_1\left|\right|U^{*}f{\left|\right|}^2\le & {\int }_X|\langle U^{*}f,F\left(w\right)\rangle {|}^{2}d\mu \left(w\right)={\int }_X{\left|\langle f,UF\left(w\right)\rangle \right|}^{2}d\mu \left(w\right)\hfill \\ \hfill \le & C_3\left|\right|U^{*}{f\left|\right|}^2\le C_3{C}_4{\left|\right|f\left|\right|}^2\hfill \end{array}$$

Therefore, $UF$ is a continuous frame for H. □

Proposition 3.

Suppose U is a linear operator on a Hilbert space H. The operator U preserves continuous tight frames with respect to $(X,\mu )$ in H if and only if there exists a positive constant c such that $\left|\right|U^{*}f\left|\right|=c\left|\right|f\left|\right|$ for any $f\in H.$

Proof. 

Assume U is an operator that preserves continuous tight frames, and let $F:(X,\mu )\to H$ be a continuous tight frame. Then there are positive constants A and B such that for any $f\in H$, we have

$${A\left|\right|f\left|\right|}^2={\int }_X{\left|\langle f,UF\left(w\right)\rangle \right|}^{2}d\mu \left(w\right)={\int }_X|\langle U^{*}f,F\left(w\right)\rangle {|}^{2}d\mu \left(w\right)=B\left|\right|U^{*}f{\left|\right|}^2.$$

This means $\left|\right|U^{*}f\left|\right|=c\left|\right|f\left|\right|$ where $c=\sqrt{A/B}>0.$

Conversely, assume that there exists a constant $c>0$ satisfying $\left|\right|U^{*}f\left|\right|=c\left|\right|f\left|\right|$ for all $f\in H.$ Now consider an arbitrary continuous tight frame $F:(X,\mu )\to H$ with frame bound $A>0$. By applying the definition of a tight frame to the vector $U^{*}f$, we obtain the following chain of equalities:

$$A{c}^2{\left|\right|f\left|\right|}^2=A\left|\right|U^{*}{f\left|\right|}^2={\int }_X|\langle U^{*}f,F\left(w\right)\rangle {|}^{2}d\mu \left(w\right)={\int }_X{\left|\langle f,UF\left(w\right)\rangle \right|}^{2}d\mu \left(w\right),$$

which shows that $UF$ satisfies the tight frame condition with frame bound $A{c}^2>0.$ □

The argument used above can be adapted to establish the next proposition.

Proposition 4.

Continuous Parseval frames are preserved by a linear operator U on a Hilbert space H if and only if $U^{*}$ acts as an isometry.

The main result of this work is stated in the following theorem. This theorem shows the relation between weighted composition operators and continuous frames.

Theorem 1.

Let $W_{\psi ,\varphi }$ be bounded on $A_{\alpha }^2$. Then the following statements are equivalent:

(1) 

$W_{\psi ,\varphi }$ maps continuous frames in $A_{\alpha }^2$ to continuous frames.

(2) 

The adjoint operator $W_{\psi ,\varphi }^{*}$ is bounded below on $A_{\alpha }^2$.

(3) 

$W_{\psi ,\varphi }$ is a surjective operator on $A_{\alpha }^2$.

(4) 

$W_{\psi ,\varphi }$ is an invertible operator on $A_{\alpha }^2$.

(5) 

ψ and $1/\psi$ are bounded, and ϕ is an automorphism of $\mathbb{D}$.

Proof. 

The equivalence of conditions (1), (2), and (3) is established directly by Proposition 2. Furthermore, Proposition 1 provides the implication from (2) to (5). To connect the remaining conditions, we refer to Theorem 2.0.1 in [22], which demonstrates that (4) and (5) are equivalent. Although the cited theorem specifically addresses the Hardy space, the underlying arguments remain valid for the weighted Bergman space $A_{\alpha }^2$ on the unit disc. To complete the proof, it is sufficient to verify that (5) implies (2). Given the equivalence of (4) and (5), we may take $W_{\psi ,\varphi }$ to be an invertible operator on $A_{\alpha }^2$. Since invertibility of an operator is preserved by its adjoint, $W_{\psi ,\varphi }^{*}$ must also be invertible. An invertible operator is necessarily bounded below, which confirms condition (2) and concludes the argument. □

The next result gives a criterion for weighted composition operators to preserve continuous Parseval frames.

Theorem 2.

Suppose $W_{\psi ,\varphi }$ is a bounded operator on $A_{\alpha }^2$. Then the following conditions are equivalent:

(1) 

$W_{\psi ,\varphi }$ preserves continuous Parseval frames on $A_{\alpha }^2$.

(2) 

$W_{\psi ,\varphi }^{*}$ is an isometry on $A_{\alpha }^2$.

(3) 

$W_{\psi ,\varphi }$ is a unitary operator on $A_{\alpha }^2$.

(4) 

There exists a complex number λ, with $\left|\lambda \right|=1$, such that $\psi =\lambda k_{\alpha }(·,{\varphi }^{-1}\left(0\right))$, where

$$k_{\alpha }(z,w)={\displaystyle \frac{K_{\alpha }(z,w)}{\left|\right|K_{\alpha }(z,w)\left|\right|}}={\left[{\displaystyle \frac{{(1-|w|}^2{)}^{1/2}}{(1-z\overline{w})}}\right]}^{2+\alpha }$$

is the normalized reproducing kernel, and ϕ is an automorphism of the unit disc $\mathbb{D}$.

Proof. 

The fact that (1) and (2) are equivalent is a direct consequence of Proposition 4. In addition, the equivalence of (2), (3), and (4) has been established in [23] (see Corollary 3.6). □

The preceding result allows us to draw the following conclusion.

Corollary 1.

Assume $W_{\psi ,\varphi }$ is a bounded operator in $A_{\alpha }^2$. Then the following assertions are equivalent:

(1) 

$W_{\psi ,\varphi }$ preserves continuous tight frames on $A_{\alpha }^2$.

(2) 

There is a positive constant c such that $\left|\right|W_{\psi ,\varphi }^{*}f\left|\right|=c\left|\right|f\left|\right|$ for any $f\in A_{\alpha }^2$.

(3) 

ϕ is an automorphism of the unit disc $\mathbb{D}$, and a complex number β exists such that $\psi =\beta k_{\alpha }(·,{\varphi }^{-1}\left(0\right))$. In fact, the constants c and β in (2) and (3) satisfy $c=\left|\beta \right|$.

Proof. 

By Proposition 3, (1) and (2) are equivalent. Now, we want to show that (2) and (3) are equivalent. Suppose (2) is true. Note that

$${\displaystyle \frac{1}{c}}{W}_{\psi ,\varphi }\left(f\right)={\displaystyle \frac{1}{c}}\psi ·(f\circ \varphi )=W_{\psi /c,\varphi }\left(f\right).$$

Hence, we have ${\displaystyle \frac{1}{c}}{W}_{\psi ,\varphi }^{*}=W_{\psi /c,\varphi }^{*}$. By the assumption, this implies that

$$\left|\right|W_{\psi /c,\varphi }^{*}f\left|\right|={\displaystyle \frac{1}{c}}\left|\right|W_{\psi ,\varphi }^{*}f\left|\right|={\displaystyle \frac{c}{c}}\left|\right|f\left|\right|=\left|\right|f\left|\right|.$$

This means $W_{\psi /c,\varphi }^{*}$ is an isometry on $A_{\alpha }^2$. Hence, by Theorem 2, $\varphi$ is an automorphism on $\mathbb{D}$ and ${\displaystyle \frac{\psi }{c}}=\lambda k_{\alpha }(·,{\varphi }^{-1}\left(0\right))$, where $\left|\lambda \right|=1$. Hence, we get (3) with $\beta =c\lambda$.

Conversely, assume (3) is true. Let $\tilde{\psi }={\displaystyle \frac{\psi }{\left|\beta \right|}}$. Then, $\varphi$ and $\tilde{\psi }$ satisfy condition (4) in Theorem 2. Hence, $W_{\tilde{\psi },\varphi }^{*}$ is an isometry on $A_{\alpha }^2$ by the Theorem 2, that is, $\left|\right|W_{\tilde{\psi },\varphi }^{*}f\left|\right|=\left|\right|f\left|\right|,f\in A_{\alpha }^2$. But, we have $\left|\right|W_{\tilde{\psi },\varphi }^{*}f\left|\right|=\left|\right|{\displaystyle \frac{\psi }{\left|\beta \right|}}·f\circ \varphi \left|\right|={\displaystyle \frac{\left|\right|W_{\psi ,\varphi }^{*}f\left|\right|}{\left|\beta \right|}}=\left|\right|f\left|\right|$. This shows that $\left|\right|W_{\psi ,\varphi }^{*}f\left|\right|=\left|\beta \right|\left|\right|f\left|\right|,f\in A_{\alpha }^2$. Thus, the result follows. □

3.2. Dual Frames Associated with Weighted Composition Operators

In this part, we investigate the behavior of dual frames and how they are transformed by weighted composition operators. Recall that a dual pair (F, G) is characterized by the reconstruction formula

$$\langle f,g\rangle ={\int }_X\langle f,F\left(w\right)\rangle \langle G\left(w\right),g\rangle d\mu \left(w\right)$$

which must hold for every $f,g$ in the Hilbert space $H.$

The theorem below provides a characterization of the conditions under which a weighted composition operator preserves the image of a dual pair.

Theorem 3.

Assume $W_{\psi ,\varphi }$ is a bounded operator on $A_{\alpha }^2$. For any dual pair $(F,G)$ for $A_{\alpha }^2$, the following properties are equivalent:

(1) 

$(W_{\psi ,\varphi }F,W_{\psi ,\varphi }G)$ is a dual pair for $A_{\alpha }^2$.

(2) 

$W_{\psi ,\varphi }$ is a unitary operator on $A_{\alpha }^2$.

(3) 

$W_{\psi ,\varphi }^{*}$ is an isometry on $A_{\alpha }^2$.

(4) 

ϕ is an automorphism on $\mathbb{D}$ and $\psi =\lambda k_{\alpha }(·,{\varphi }^{-1}\left(0\right))$ for some $\lambda \in \mathbb{C}$ with $\left|\lambda \right|=1$.

Proof. 

We already know that (2), (3), and (4) are equivalent. By corollary of 3.9 of [24], (2) implies (1). Thus, the proof is complete once we establish that (1) implies (3). For any $f,g\in A_{\alpha }^2$, we have

$$\begin{array}{cc}\hfill \langle f,g\rangle & ={\int }_X\langle f,W_{\psi ,\varphi }F\left(w\right)\rangle \langle W_{\psi ,\varphi }G\left(w\right),g\rangle d\mu \left(w\right)\hfill \\ \hfill & ={\int }_X\langle W_{\psi ,\varphi }^{*}f,F\left(w\right)\rangle \langle G\left(w\right),W_{\psi ,\varphi }^{*}g\rangle d\mu \left(w\right)\hfill \\ \hfill & =\langle W_{\psi ,\varphi }^{*}f,W_{\psi ,\varphi }^{*}g\rangle \hfill \\ \hfill & =\langle W_{\psi ,\varphi }{W}_{\psi ,\varphi }^{*}f,g\rangle .\hfill \end{array}$$

Since $\langle f,g\rangle =\langle W_{\psi ,\varphi }{W}_{\psi ,\varphi }^{*}f,g\rangle$ for any $f,g\in A_{\alpha }^2$, we have $W_{\psi ,\varphi }{W}_{\psi ,\varphi }^{*}=I$. Hence, $W_{\psi ,\varphi }^{*}$ is an isometry on $A_{\alpha }^2$. □

The invertibility of the weighted composition operator $W_{\psi ,\varphi }$ ensures that the image of any continuous frame F remains a continuous frame, as established in Theorem 1. However, this does not ensure that $(F,W_{\psi ,\varphi }F)$ forms a dual pair. The following corollary provides a characterization of this scenario using the frame operator $S_F$ associated with $F.$

Corollary 2.

Suppose F is a continuous frame for $A_{\alpha }^2$ and $W_{\psi ,\varphi }$ is a bounded, invertible operator on $A_{\alpha }^2$. Then, $(F,W_{\psi ,\varphi }F)$ is a dual pair if and only if $S_F=W_{\psi ,\varphi }^{-1}$.

Proof. 

Suppose F is a continuous frame. Then by Theorem 1, $W_{\psi ,\varphi }F$ is a continuous frame on $A_{\alpha }^2$.

Now suppose that $(F,W_{\psi ,\varphi }F)$ is a dual pair. For any $f,g\in A_{\alpha }^2$, we have

$$\begin{array}{cc}\hfill \langle f,g\rangle & ={\int }_X\langle f,F\left(w\right)\rangle \langle W_{\psi ,\varphi }F\left(w\right),g\rangle d\mu \left(w\right)\hfill \\ \hfill & ={\int }_X\langle f,F\left(w\right)\rangle \langle F\left(w\right),W_{\psi ,\varphi }^{*}g\rangle d\mu \left(w\right)\hfill \\ \hfill & =\langle S_{F}f,W_{\psi ,\varphi }^{*}g\rangle \hfill \\ \hfill & =\langle W_{\psi ,\varphi }{S}_{F}f,g\rangle .\hfill \end{array}$$

Hence, $W_{\psi ,\varphi }{S}_F=I\iff S_F=W_{\psi ,\varphi }^{-1}$. The converse is clear since $S_F^{-1}F=W_{\psi ,\varphi }F$ defines the canonical dual of F. □

In essence, the following two results are analogous to Theorems 2.1 and 2.3 of [25]. Specifically, they establish that a weighted composition operator can generate a sequence of dual frames from an initial dual frame. Furthermore, they show that the discrepancy between any general dual frame and the canonical dual can be represented as the image of a Bessel mapping through the weighted composition operator.

Theorem 4.

Suppose $F:X\to A_{\alpha }^2$ is a continuous frame with associated frame operator $S_F$, and let $W_{\psi ,\varphi }$ denote a bounded linear operator on $A_{\alpha }^2$. If $G_1:X\to A_{\alpha }^2$ is a dual of F, then

$$G_{i+1}\left(w\right)=S_F^{-1}F\left(w\right)+W_{\psi ,\varphi }{S}_F{G}_i\left(w\right)-W_{\psi ,\varphi }F\left(w\right)$$

is a sequence of duals of F, where $i\in \mathbb{N}$.

Proof. 

One can easily verify that $G_i$ defines a continuous Bessel mapping for $A_{\alpha }^2$ for each $i\in \mathbb{N}$. To show that $G_i$ is a dual of F, we use induction. Let $f,g\in A_{\alpha }^2$. If $i=1$, then

$$\begin{array}{cc}\hfill {\int }_X\langle f,F\left(w\right)\rangle & \langle G_2\left(w\right),g\rangle d\mu \left(w\right)={\int }_X\langle f,F\left(w\right)\rangle \langle S_F^{-1}F\left(w\right),g\rangle d\mu \left(w\right)+\hfill \\ & {\int }_X\langle f,F\left(w\right)\rangle \langle W_{\psi ,\varphi }{S}_F{G}_1\left(w\right),g\rangle d\mu \left(w\right)-{\int }_X\langle f,F\left(w\right)\rangle \langle W_{\psi ,\varphi }F\left(w\right),g\rangle d\mu \left(w\right)\hfill \\ & =\langle f,g\rangle +\langle f,S_F{W}_{\psi ,\varphi }^{*}g\rangle -\langle S_{F}f,W_{\psi ,\varphi }^{*}g\rangle \hfill \\ & =\langle f,g\rangle .\hfill \end{array}$$

Now, assume $G_n$ is a dual of F. We need to show that $G_{n+1}$ is also a dual of F. But one can easily show, using a similar computation as above, that

$${\int }_X\langle f,F\left(w\right)\rangle \langle G_{n+1}\left(w\right),g\rangle d\mu \left(w\right)=\langle f,g\rangle ,$$

for any $f,g\in A_{\alpha }^2$. Hence, the proof is complete. □

Theorem 5.

Suppose $F:X\to A_{\alpha }^2$ is a continuous frame with frame operator $S_F$, and $W_{\psi ,\varphi }$ is a bounded invertible on $A_{\alpha }^2$. A continuous Bessel mapping $G:X\to A_{\alpha }^2$ serves as a dual of F if and only if it can be expressed as

$$G\left(w\right)=S_F^{-1}F\left(w\right)+W_{\psi ,\varphi }L\left(w\right),w\in X,$$

for some continuous Bessel mapping $L:X\to A_{\alpha }^2$ that satisfies ${\int }_X\langle f,F\left(w\right)\rangle L\left(w\right)d\mu \left(w\right)=0$ for every $f\in A_{\alpha }^2.$

Proof. 

Suppose G is a dual of F. Set $L\left(w\right)=W_{\psi ,\varphi }^{-1}G\left(w\right)-W_{\psi ,\varphi }^{-1}{S}_F^{-1}F\left(w\right)$. It is evident that L defines a continuous Bessel mapping for $A_{\alpha }^2$. Then

$$\begin{array}{cc}\hfill {\int }_X\langle f,F\left(w\right)\rangle \langle L\left(w\right),g\rangle d\mu \left(w\right)=& {\int }_X\langle f,F\left(w\right)\rangle \langle W_{\psi ,\varphi }^{-1}G\left(w\right),g\rangle d\mu \left(w\right)-\hfill \\ & {\int }_X\langle f,F\left(w\right)\rangle \langle W_{\psi ,\varphi }^{-1}{S}_F^{-1}F\left(w\right),g\rangle d\mu \left(w\right)\hfill \\ \hfill =& {\int }_X\langle f,F\left(w\right)\rangle \langle G\left(w\right),{\left(W_{\psi ,\varphi }^{-1}\right)}^{*}g\rangle d\mu \left(w\right)-\hfill \\ & {\int }_X\langle f,F\left(w\right)\rangle \langle S_F^{-1}F\left(w\right),{\left(W_{\psi ,\varphi }^{-1}\right)}^{*}g\rangle d\mu \left(w\right)\hfill \\ \hfill =& \langle f,{\left(W_{\psi ,\varphi }^{-1}\right)}^{*}g\rangle -\langle f,{\left(W_{\psi ,\varphi }^{-1}\right)}^{*}g\rangle \hfill \\ \hfill =& 0.\hfill \end{array}$$

This means ${\left\{\langle f,F\left(w\right)\rangle \right\}}_{w\in X}\in \mathrm{Ran}{\left(T_L\right)}^{\perp }=\mathrm{Ker}\left(T_L^{*}\right)$, where $T_L^{*}$ is the synthesis operator of L. Thus

$$0=T_L^{*}\left(\langle f,F\left(w\right)\rangle \right)={\int }_X\langle f,F\left(w\right)\rangle L\left(w\right)d\mu \left(w\right).$$

Conversely, let $G\left(w\right)=S_F^{-1}F\left(w\right)+W_{\psi ,\varphi }L\left(w\right)$ and ${\int }_X\langle f,F\left(w\right)\rangle L\left(w\right)d\mu \left(w\right)=0$ for each $f\in A_{\alpha }^2$. Then

$$\begin{array}{cc}\hfill {\int }_X\langle f,F\left(w\right)\rangle \langle G\left(w\right),g\rangle d\mu \left(w\right)=& {\int }_X\langle f,F\left(w\right)\rangle \langle S_F^{-1}F\left(w\right),g\rangle d\mu \left(w\right)+\hfill \\ & {\int }_X\langle f,F\left(w\right)\rangle \langle W_{\psi ,\varphi }L\left(w\right),g\rangle d\mu \left(w\right)\hfill \\ \hfill =& \langle f,g\rangle +{\int }_X\langle f,F\left(w\right)\rangle \langle L\left(w\right),W_{\psi ,\varphi }^{*}g\rangle d\mu \left(w\right)\hfill \\ \hfill =& \langle f,g\rangle +\langle {\int }_X\langle f,F\left(w\right)\rangle L\left(w\right)d\mu \left(w\right),W_{\psi ,\varphi }^{*}g\rangle \hfill \\ \hfill =& \langle f,g\rangle \hfill \end{array}$$

for each $f,g\in A_{\alpha }^2.$ Thus G is a dual of F. □

3.3. Scalability of Continuous Frames Under the Weighted Composition Operators

This part of the paper first introduces the concept of scalability for continuous frames and examines its relationship with weighted composition operators acting on the weighted Bergman space $A_{\alpha }^2$ of the unit disc $\mathbb{D}$. We then establish the existence of scalable dual frames and provide criteria characterizing when a weighted composition operator preserves scalability. For more detail about scalability, we refer reader to [26,27,28].

Definition 3.

Let H be a complex Hilbert space, and let $(X,\mu )$ be a measure space equipped with a positive $\sigma$-finite measure μ. Suppose $m:X⟶{\mathbb{R}}^{+}$ is a measurable function. A mapping $F:X⟶H$ is said to be a scalable continuous frame if the function $\left(mF\right):X⟶H$, given by $w\mapsto m\left(w\right). F\left(w\right)$ for each $w\in X$, constitutes a continuous Parseval frame for H.

Theorem 6.

Every scalable continuous frame has a scalable dual frame.

Proof. 

Let $F:X\to H$ be a scalable continuous frame with the scaling function $m:X\to {\mathbb{R}}^{+}.$ Since the $mF$ forms a Parseval continuous frame, its frame operator coincides with the identity operator on H. Consequently, for any $f\in H$,

$$f={\int }_X\langle f,m\left(w\right)F\left(w\right)\rangle m\left(w\right)F\left(w\right)d\mu \left(w\right)={\int }_X\langle f,F\left(w\right)\rangle m{\left(w\right)}^{2}F\left(w\right)d\mu \left(w\right).$$

So, by definition, this means $m{\left(w\right)}^{2}F\left(w\right)$ is a dual frame of $F\left(w\right)$. Further, if we scale the dual frame $m^{2}F$ by ${\displaystyle \frac{1}{m}}$, we get $mF$, which is Parseval. □

Theorem 7.

Let $F:X\to H$ be a continuous frame, and let T be an invertible operator on $H.$ Then the following statements are equivalent:

(1) 

There exists a measurable function $m:X\to {\mathbb{R}}^{+}$ such that the mapping $mTF:X\to H$, defined by $\left(mTF\right)\left(w\right)=m\left(w\right)TF\left(w\right)$ for all $w\in X,$ forms a continuous Parseval frame for $H.$

(2) 

There is a measurable mapping $m:X\to {\mathbb{R}}^{+}$ such that the frame operator of $mF$ is ${\left(T^{*}T\right)}^{-1}$.

Proof. 

Suppose (1) is true. Then the frame operator $S_{mTF}$ of $mTF$ is the identity operator $I_H$ on H. Then

$$\begin{array}{cc}\hfill f=& {\int }_X\langle f,m\left(w\right)TF\left(w\right)\rangle m\left(w\right)TF\left(w\right)d\mu \left(w\right)=T\left({\int }_X\langle T^{*}f,m\left(w\right)F\left(w\right)\rangle m\left(w\right)F\left(w\right)d\mu \left(w\right)\right)\hfill \\ \hfill =& T{S}_{mF}{T}^{*}f,\hfill \end{array}$$

where $S_{mF}$ is the frame operator of $mF.$ This implies $T{S}_{mF}{T}^{*}=I_H$, or, $S_{mF}={\left(T^{*}T\right)}^{-1}$.

Conversely, suppose (2) holds. Then $S_{mF}={\left(T^{*}T\right)}^{-1}$ implies that $T{S}_{mF}{T}^{*}=I_H$. Then

$$\begin{array}{cc}\hfill f=T{S}_{mF}{T}^{*}f=& T\left({\int }_X\langle T^{*}f,m\left(w\right)F\left(w\right)\rangle m\left(w\right)F\left(w\right)d\mu \left(w\right)\right)\hfill \\ \hfill =& {\int }_X\langle f,m\left(w\right)TF\left(w\right)\rangle m\left(w\right)TF\left(w\right)d\mu \left(w\right).\hfill \end{array}$$

This implies that the frame operator $S_{mTF}$ of $mTF$ is the identity operator $I_H$ on H. Therefore, $mTF$ is a continuous Parseval frame. □

Corollary 3.

Let $F:X\to A_{\alpha }^2$ be a continuous frame and assume that $W_{\psi ,\varphi }$ is invertible on $A_{\alpha }^2$. If there is a measurable mapping $m:X\to {\mathbb{R}}^{+}$ such that $mF$ and $m{W}_{\psi ,\varphi }F$ are Parseval frames for $A_{\alpha }^2$, then $W_{\psi ,\varphi }$ is unitary.

Proof. 

The proof follows from Theorem 7. □

The following result is an application of Theorem 2.

Theorem 8.

Let $F:X\to A_{\alpha }^2$ be a scalable continuous frame with scaling function m. Let $W_{\psi ,\varphi }$ be bounded on $A_{\alpha }^2$. Then $W_{\psi ,\varphi }F$ is scalable continuous frame with scaling function m if and only if $W_{\psi ,\varphi }^{*}$ is an isometry.

Proof. 

Suppose $W_{\psi ,\varphi }F$ is a scalable continuous frame with scaling function m, that is, $m\left(w\right)W_{\psi ,\varphi }F\left(w\right)$ is a continuous Parseval frame. Notice that

$$m\left(w\right)W_{\psi ,\varphi }F\left(w\right)=W_{\psi ,\varphi }m\left(w\right)F\left(w\right).$$

(2)

This means $W_{\psi ,\varphi }$ preserves continuous Parseval frames on $A_{\alpha }^2.$ Thus by Theorem 2, $W_{\psi ,\varphi }^{*}$ is an isometry.

Conversely, suppose $W_{\psi ,\varphi }^{*}$ is an isometry. Then by Theorem 2, $W_{\psi ,\varphi }$ preserves continuous Parseval frames. By (2) this means $W_{\psi ,\varphi }F$ is a scalable continuous frame with scaling function m. □

Example 1.

Now we will construct an example of a scalable continuous frame for the weighted Bergman space $A_{\alpha }^2$. Recall that the reproducing kernel $K_{\alpha }(z,w)$ for $A_{\alpha }^2$ is given by

$$K_{\alpha }(z,w)={\displaystyle \frac{1}{{(1-z\overline{w})}^{2+\alpha }}},z,w\in \mathbb{D},$$

and the normalized reproducing kernel $k_{\alpha }(z,w)$ at $w\in \mathbb{D}$ is

$$k_{\alpha }(z,w)={\displaystyle \frac{K_{\alpha }(z,w)}{\left|\right|K_{\alpha }(·,w)\left|\right|}}={\displaystyle \frac{{(1-|w|}^2{)}^{\frac{\alpha +2}{2}}}{{(1-\overline{w}z)}^{\alpha +2}}}.$$

Let $(X,\mu )=(\mathbb{D},d{A}_{\alpha }\left(w\right)),$ and define $F\left(w\right)=k_{\alpha }(·,w)$. Then F is a mapping from $\mathbb{D}$ to the weighted Bergman space $A_{\alpha }^2.$ In general F is not a continuous frame for $A_{\alpha }^2$ over the unit disc $\mathbb{D}$ because lower frame bound fails to exist. However, the following simple computation shows that F is Bessel. Let $f\in A_{\alpha }^2$. Then

$$\langle f,F\left(w\right)\rangle =\langle f,k_{\alpha }(·,w)\rangle ={\displaystyle \frac{1}{\left|\right|K_{\alpha }(·,w)\left|\right|}}\langle f,K_{\alpha }(·,w)\rangle ={(1-|w|}^2{)}^{{\displaystyle \frac{\alpha +2}{2}}}f\left(w\right).$$

Hence

$${\int }_{\mathbb{D}}{\left|\langle f,F\left(w\right)\rangle \right|}^{2}d{A}_{\alpha }\left(w\right)={\int }_{\mathbb{D}}{\left|f\left(w\right)\right|}^2{(1-|w|}^2{)}^{\alpha +2}d{A}_{\alpha }\left(w\right)\le {\int }_{\mathbb{D}}{\left|f\left(w\right)\right|}^{2}d{A}_{\alpha }\left(w\right)={\left|\right|f\left|\right|}^2.$$

In fact one can say more, that is, F is a scalable continuous frame with scaling function $m\left(w\right)=\left|\right|K_{\alpha }(·,w)\left|\right|$. Indeed,

$${\int }_{\mathbb{D}}{\left|\langle f,m\left(w\right)F\left(w\right)\rangle \right|}^{2}d{A}_{\alpha }\left(w\right)={\int }_{\mathbb{D}}|\langle f,K_{\alpha }(·,w)\rangle {|}^{2}d{A}_{\alpha }\left(w\right)={\int }_{\mathbb{D}}{\left|f\left(w\right)\right|}^{2}d{A}_{\alpha }\left(w\right)={\left|\right|f\left|\right|}^2.$$

Moreover, using the elementary relation

$${\int }_{\mathbb{D}}\langle f,K_{\alpha }(·,w)\rangle \langle K_{\alpha }(·,w),g\rangle d{A}_{\alpha }\left(w\right)={\int }_{\mathbb{D}}f\left(w\right)\overline{g\left(w\right)}d{A}_{\alpha }\left(w\right)=\langle f,g\rangle ,$$

valid for all $f,g\in A_{\alpha }^2$, we conclude that the continuos Parseval frame $mF$ serves as its own dual frame.