Citation: Ruishen Qian, Xiangling Zhu. Invertible weighted composition operators preserve frames on Dirichlet type spaces[J]. AIMS Mathematics, 2020, 5(5): 4285-4296. doi: 10.3934/math.2020273
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Let D be the unit disk in the complex plane C and let H(D) denote the set of functions analytic in D. As usual, let H∞ be the space of bounded analytic functions in D and φa(z)=(a−z)/(1−¯az). Let φ be an analytic self-map of D and ψ∈H(D). The weighted composition operator Wψ,φ, induced by φ and ψ, is defined on H(D) by
(Wψ,φf)(z)=ψ(z)f(φ(z)),f∈H(D). |
When ψ≡1, we get the composition operator, denoted by Cφ. We refer the readers to [1,2] for the theory of composition operators and weighted composition operators.
Let ρ:[0,∞)→[0,∞) be a right continuous and nondecreasing function with ρ(0)=0. We say that an f∈H(D) belongs to the Dirichlet type space Dρ, if
‖f‖2Dρ=|f(0)|2+∫D|f′(z)|2ρ(1−|z|2)dA(z)<∞, |
where dA is the area measure on D normalized so that A(D)=1. When ρ(t)=t, Dρ=H2, the Hardy space. If ρ(t)=tα and α>1, Dρ is the Bergman space A2α−2. When ρ(t)=tα and α<1, it gives the weighted Dirichlet space Dα. Many properties of Dρ spaces were studied by Aleman in [3] and Kerman, Sawyer in [4]. Carleson measure for Dρ spaces was studied by Arcozzi, Rochberg and Sawyer in [5]. For more information about Dρ and Dα, we refer to [6,7,8,9,10,11,12].
Recall that a weight ρ is of upper (resp.lower) type γ (0≤γ<∞) ([13]), if
ρ(st)≤Csγρ(t), s≥1 (resp.s≤1) and 0<t<∞. |
We say that ρ is of upper type less than γ if it is of upper type δ for some δ<γ and ρ is of lower type greater than β if it is of lower type δ for some δ>β. From [13], we see that an increasing function ρ is of finite upper type if and only if ρ(2t)≤Cρ(t).
Let H be a Hilbert space with an inner product ⟨⋅,⋅⟩ and norm ‖⋅‖H. A family {fk}∞k=0 in H is called a frame for H if there exist constants A,B>0 such that
A‖f‖2H≤∞∑k=0|⟨f,fk⟩|2≤B‖f‖2H, for all f∈H. |
A (resp. B) is called the lower (resp. upper) frame bounded. When A=B, the family {fk}∞k=0 is called a tight frame. If A=B=1, we call it a normalized tight frame. The notion of frame was first introduced by Duffin and Schaeffer in [14]. Tight frame is especially popular and now widely used in compressive sensing, image and signal processing, since it provide stable decompositions similar to orthonormal bases (see [15]). We say that a linear operator T on a Hilbert space H preserves frames if {Tfk}∞k=0 is a frame in H for any frame {fk}∞k=0⊆H. Similarly, we call T on H preserves (normalized) tight frames if {Tfk}∞k=0 is a (normalized) tight frame in H for any (normalized) tight frame {fk}∞k=0⊆H. See [16] for more information.
Recently, Manhas, Prajitura and Zhao studied weighted composition operators that preserve frames in [16]. Especially, they build the equivalence between preserve frames and invertible of weighted composition operators on weighted Bergman spaces A2α in the unit ball.
In this paper, we give some characterizations for invertible weighted composition operators on Dirichlet type spaces Dρ when ρ is finite lower type greater than 0 and upper type less than 1. In particular, our result shows that weighted composition operators are invertible if and only if they preserve frames on Dρ. Moreover, we also investigate weighted composition operators that preserve normalized tight frames and tight frames in the weighted Dirichlet space Dα (0<α<1).
Throughout this paper, we say that A≲ if there exists a constant C (independent of A and B ) such that A\leq CB . The symbol A\approx B means that A\lesssim B\lesssim A .
Let us firstly recall the following construction in [17]. Suppose c_n > 0 for n = 0, 1, \cdots . Define an inner product on H({\mathbb{D}}) by
\langle f, g\rangle = \sum\limits_{n = 0}^\infty \frac{1}{c_n}a_n \bar{b_n}, |
where f(z) = \sum a_nz^n\in H({\mathbb{D}}) and g(z) = \sum b_nz^n\in H({\mathbb{D}}) . Let R(z) = \sum_{n = 0}^\infty c_nz^n. Denote by {\mathcal H}_R for the Hilbert space of analytic functions with
\langle f, f\rangle = \|f\|^2 \lt \infty. |
Let R_\zeta(z) = R(\bar{\zeta} z) , \zeta\in {\mathbb{D}} . Then R_\zeta(z) is the reproducing kernel of {\mathcal H}_R at \zeta \in {\mathbb{D}} , that is, f(\zeta) = \langle f, R_\zeta\rangle for any f\in {\mathcal H}_R .
Lemma 1. Let \rho be of finite upper type less than 1 . Set
R^\rho(z) = 1+\sum\limits_{n = 1}^\infty \frac{1}{n\rho(\frac{1}{n})}z^n, \ z\in {\mathbb{D}}. |
Then {\mathcal H}_{R^\rho} = {\mathcal D}_{\rho} and R^\rho_\zeta(z) = R^\rho(\bar{\zeta} z) is the reproducing kernel for {\mathcal H}_{R^\rho} space at \zeta \in {\mathbb{D}} . Moreover, when f(z) = \sum_{n = 0}^\infty a_nz^n ,
\|f\|_{{\mathcal D}_\rho}^2\approx |a_0|^2+\sum\limits_{n = 1}^\infty n\rho\left(\frac{1}{n}\right)|a_n|^2. |
Proof. Let f(z) = \sum_{n = 0}^\infty a_nz^n . Then
\begin{eqnarray*} \|f\|_{{\mathcal D}_\rho}^2& = &|a_0|^2+\int_{{\mathbb{D}}}|f'(z)|^{2}\rho\left(1-|z|^2\right)dA(z)\\ &\approx&|a_0|^2+\sum\limits_{n = 1}^\infty n^2|a_n|^2\int_0^1 r^{2n-1}\rho(1-r^2)dr. \end{eqnarray*} |
By [9, Lemma 2], for n > 0 , we have
\int_0^1 r^{2n-1}\rho(1-r^2)dr\approx \int_0^1 r^{2n-1}\rho\left(\log\frac{1}{r}\right)dr\approx \frac{1}{n}\rho\left(\frac{1}{n}\right), |
the constants occur here depend only on \rho . Therefore,
\|f\|_{{\mathcal D}_\rho}^2\approx|a_0|^2+\sum\limits_{n = 1}^\infty n\rho\left(\frac{1}{n}\right)|a_n|^2. |
By the definition, f\in {\mathcal H}_{R^\rho} if and only if
\|f\|_{{\mathcal H}_{R^\rho}}^2 = \langle f, f\rangle = |a_0|^2+\sum\limits_{n = 1}^\infty n\rho\left(\frac{1}{n}\right)|a_n|^2 \lt \infty. |
Thus, {\mathcal H}_{R^\rho} = {\mathcal D}_\rho . The proof is complete.
Lemma 2. Let \rho be of finite lower type greater than 0 and upper type less than 1 . Then there exist constants C_1 and C_2 which depending only on \rho such that
C_1 \left(1+\sum\limits_{n = 1}^\infty \frac{1}{n\rho(\frac{1}{n})}t^n\right)\leq \frac{1}{\rho(1-t)}\leq C_2 \left(1+\sum\limits_{n = 1}^\infty \frac{1}{n\rho(\frac{1}{n})}t^n\right) |
for all 0\leq t < 1 .
Proof. Without loss of generality, we assume that 1/2 < t < 1 . Then,
\begin{eqnarray*} \sum\limits_{n = 1}^\infty \frac{1}{n\rho(\frac{1}{n})}t^n &\approx& \sum\limits_{n = 1}^\infty \int_{\frac{1}{n+1}}^{\frac{1}{n}}\frac{t^{\frac{1}{x}}}{x\rho(x)}dx \approx \int_0^1\frac{t^{\frac{1}{x}}}{x\rho(x)}dx\\ &\approx&\int_1^\infty\frac{t^y}{y\rho(\frac{1}{y})}dy \ \ \left(y = \frac{x}{- \ln t}\right)\\ &\approx&\int_{- \ln t}^\infty\frac{e^{-x}}{x\rho(\frac{1}{x}\ln\frac{1}{t})}dx\\ &\approx&\frac{1}{\rho(\ln\frac{1}{t})}\int_{- \ln t}^\infty\frac{e^{-x}\rho(\ln\frac{1}{t})}{x\rho(\frac{1}{x}\ln\frac{1}{t})}dx. \end{eqnarray*} |
Since \rho is of finite lower type greater than 0 and upper type less than 1 , there exist \gamma and \delta , satisfied 0 < \gamma < \delta < 1 , such that
\rho(st)\lesssim s^{\gamma}\rho(t), \ \ s\leq 1, | (1) |
and
\rho(st)\lesssim s^{\delta}\rho(t), \ \ s\geq 1, | (2) |
where 0 < t < \infty . Therefore,
\begin{eqnarray*} \sum\limits_{n = 1}^\infty \frac{1}{n\rho(\frac{1}{n})}t^n &\lesssim&\frac{1}{\rho(\ln\frac{1}{t})}\left(\int_0^\infty e^{-x}x^{\gamma-1}dx+\int_0^\infty e^{-x}x^{\delta-1}dx\right)\\ &\approx&\frac{1}{\rho(1-t)}\left(\Gamma(\gamma)+\Gamma(\delta)\right), \end{eqnarray*} |
where \Gamma(.) is the Gamma function. Hence
1+\sum\limits_{n = 1}^\infty \frac{1}{n\rho(\frac{1}{n})}t^n\lesssim \frac{1}{\rho(1-t)}. |
On the other hand, since \rho is nondecreasing, we have
\begin{eqnarray*} \sum\limits_{n = 1}^\infty \frac{1}{n\rho(\frac{1}{n})}t^n &\approx&\frac{1}{\rho(\ln\frac{1}{t})}\int_{- \ln t}^\infty\frac{e^{-x}\rho(\ln\frac{1}{t})}{x\rho(\frac{1}{x}\ln\frac{1}{t})}dx\\ &\gtrsim& \frac{1}{\rho(\ln\frac{1}{t})}\int_{ \ln2}^\infty\frac{e^{-x}\rho(\ln\frac{1}{t})}{x\rho(\frac{1}{x}\ln\frac{1}{t})}dx\\ &\gtrsim& \frac{1}{\rho(1-t)}\int_{ \ln2}^\infty e^{-x}x^{-1}dx \approx \frac{1}{\rho(1-t)}. \end{eqnarray*} |
The proof is complete.
Lemma 3. Let \rho be of finite lower type greater than 0 and upper type less than 1 . Let r_z^\rho = \frac{R_{\xi}^\rho(z)}{\|R_{\xi}^\rho(z)\|_{D_\rho}} denote the normalized reproducing kernel for {\mathcal H}_{R^\rho} . Then r_z^\rho\rightarrow 0 weakly in {\mathcal D}_\rho as |z|\rightarrow 1 .
Proof. By Lemmas 1 and 2, we have
\|R_z^\rho\|_{{\mathcal D}_\rho}\approx\frac{1}{\sqrt{\rho(1-|z|^2)}}. |
Take any complex polynomial p , we deduce that
\langle p, r_z^\rho\rangle = \langle p, \frac{R_z^\rho}{\|R_z^\rho\|_{{\mathcal D}_\rho}}\rangle = \frac{p(z)}{\|R_z^\rho\|_{{\mathcal D}_\rho}}\approx p(z)\sqrt{\rho(1-|z|^2)}. |
Since a polynomial is bounded on {\mathbb{D}} , we obtain
\lim\limits_{|z|\rightarrow1}\langle p, r_z^\rho\rangle = 0. |
It is well known that polynomials are dense in {\mathcal D}_\rho . So r_z^\rho\rightarrow 0 weakly in {\mathcal D}_\rho as |z|\rightarrow 1 . The proof is complete.
Let \mu be a finite positive Borel measure on {\mathbb{D}} . Recall that \mu is a {\mathcal D}_{\rho} -Carleson measure if the inclusion map i: {\mathcal D}_{\rho}\rightarrow L^2(\mu) is bounded, that is
\int_{{\mathbb{D}}}|f(z)|^2d\mu(z)\leq C\|f\|_{{\mathcal D}_{\rho}}^2 |
for all f\in {\mathcal D}_{\rho} . The best constant C , denoted by \|\mu\|_{\rho} , is said to be the norm of \mu . The following lemma can be found in [9, Lemma 7].
Lemma 4. Let \rho be of finite lower type greater than 0 and upper type less than 1 . Then g\in H({\mathbb{D}}) is a multiplier of {\mathcal D}_\rho if and only if g\in H^{\infty} and |g(z)|^2\rho (1-|z|^2)dA(z) is {\mathcal D}_\rho -Carleson measure.
Lemma 5. ([1]) Let \varphi be an analytic self-map of {\mathbb{D}} . Then, for any z\in{\mathbb{D}} ,
\frac{1-|\varphi(0)|}{1+|\varphi(0)|}\leq\frac{1-|\varphi(z)|}{1-|z|}. |
Lemma 6. Let \rho be of finite lower type greater than 0 and upper type less than 1 . Suppose that \varphi is an automorphism on {\mathbb{D}} . Then f\circ\varphi\in{\mathcal D}_\rho for all f\in{\mathcal D}_\rho.
Proof. Suppose that \varphi(z) = \eta\frac{a-z}{1-\overline{a}z} , where a, z\in{\mathbb{D}} and |\eta| = 1 . Then
\begin{equation} \nonumber \begin{split} \|C_{\varphi}f\|_{{\mathcal D}_\rho}^2 = &|f\circ\varphi(0)|^2+\int_{{\mathbb{D}}}|(f\circ\varphi)'(z)|^2\rho(1-|z|^2)dA(z) \\ = &|\langle f, R_{\varphi(0)}^{\rho}\rangle|^2+\int_{\varphi({\mathbb{D}})}|f'(w)|^2\rho(1-|\varphi^{-1}(w)|^2)dA(w) \\ = &|\langle f, R_{\varphi(0)}^{\rho}\rangle|^2+\int_{\varphi({\mathbb{D}})}|f'(w)|^2\rho(1-|w|^2)\frac{\rho(1-|\varphi^{-1}(w)|^2)}{\rho(1-|w|^2)}dA(w), \end{split} \end{equation} |
where z = \varphi^{-1}(w) . Noting that
\|R_{\varphi(0)}^{\rho}\|_{{\mathcal D}_\rho}\approx\frac{1}{\sqrt{\rho(1-|a|^2)}}, |
we get
|\langle f, R_{\varphi(0)}^{\rho}\rangle|^2\lesssim\|f\|_{{\mathcal D}_\rho}^2\|R_{\varphi(0)}^{\rho}\|_{{\mathcal D}_\rho}^2\approx\frac{\|f\|_{{\mathcal D}_\rho}^2}{\rho(1-|a|^2)} \lt \infty. |
Since \rho is of finite lower type greater than 0 and upper type less than 1 , similar to Lemma 2, we can deduce that
\begin{equation} \nonumber \begin{split} \frac{\rho(1-|\varphi^{-1}(w)|^2)}{\rho(1-|w|^2)}\lesssim&\left(\frac{1-|\varphi^{-1}(w)|^2}{1-|w|^2}\right)^{\gamma}+\left(\frac{1-|\varphi^{-1}(w)|^2}{1-|w|^2}\right)^{\delta} \\\lesssim&\left(\frac{1-|\varphi^{-1}(w)|}{1-|w|}\right)^{\gamma}+\left(\frac{1-|\varphi^{-1}(w)|}{1-|w|}\right)^{\delta}. \end{split} \end{equation} |
Combined with Lemma 5 again, we obtain
\frac{1-|\varphi^{-1}(w)|}{1-|w|} = \frac{1-|z|}{1-|\varphi(z)|}\leq\frac{1+|a|}{1-|a|}. |
That is,
\frac{\rho(1-|\varphi^{-1}(w)|^2)}{\rho(1-|w|^2)}\lesssim\left(\frac{1+|a|}{1-|a|}\right)^\gamma+\left(\frac{1+|a|}{1-|a|}\right)^\delta. |
Noting the fact that \varphi({\mathbb{D}})\subseteq{\mathbb{D}} , we get the desired result. The proof is complete.
Lemma 7. [16] Suppose that T is a bounded linear operator on a Hilbert space \mathcal{H} . Then the following statements are equivalent.
(i) T preserves frames on \mathcal{H} .
(ii) T is surjective on \mathcal{H} .
(iii) T is bounded below on \mathcal{H} .
Lemma 8. [16] Suppose that T is a bounded linear operator on a Hilbert space \mathcal{H} . Then T preserves tight frames on \mathcal{H} if and only if there is constant \lambda > 0 such that \|T^*f\|_{\mathcal{H}} = \lambda\|f\|_{\mathcal{H}} for any f\in\mathcal{H} .
Lemma 9. [16] Suppose that T is a bounded linear operator on a Hilbert space \mathcal{H} . Then T preserves normalized tight frames on \mathcal{H} if and only if \|T^*f\|_{\mathcal{H}} = \|f\|_{\mathcal{H}} for any f\in\mathcal{H} .
The following results can be deduced by [18, Corollary 3.6].
Lemma 10. Suppose that \psi\in H({\mathbb{D}}) and \varphi is an analytic self-map of {\mathbb{D}} such that W_{\psi, \varphi} is bounded on a Hilbert space \mathcal{H}_\gamma ( 0 < \gamma < \infty ) with reproducing kernel functions \frac{1}{(1-\overline{w}z)^{\gamma}}, \ \ w, z\in{\mathbb{D}} . Then the following statements are equivalent.
(i) {W_{\psi, \varphi}} is co-isometry on \mathcal{H}_\gamma , that is, {W_{\psi, \varphi}}{W_{\psi, \varphi}}^* = I .
(ii) {W_{\psi, \varphi}} is an unitary operator on \mathcal{H}_\gamma .
(iii) \varphi is an automorphism on {\mathbb{D}} and \psi = \xi r_{\varphi^{-1}(0)}^\alpha , where |\xi| = 1 .
In this section, we state and prove the main result in this paper.
Theorem 1. Let \rho be of finite lower type greater than 0 and upper type less than 1 . Suppose that \psi\in H({\mathbb{D}}) and \varphi is an analytic self-map of {\mathbb{D}} such that W_{\psi, \varphi} is bounded on {\mathcal D}_\rho . Then the following statements are equivalent.
(i) {W_{\psi, \varphi}} preserves frames on {\mathcal D}_\rho .
(ii) {W_{\psi, \varphi}} is surjective on {\mathcal D}_\rho .
(iii) {W_{\psi, \varphi}}^{*} is bounded below on {\mathcal D}_\rho .
(iv) \psi and \frac{1}{\psi} are multipliers of {\mathcal D}_\rho and \varphi is an automorphism of {\mathbb{D}} .
(v) {W_{\psi, \varphi}} is invertible on {\mathcal D}_\rho .
Proof. (i)\Leftrightarrow(ii)\Leftrightarrow(iii) . These implications can be deduced by Lemma 7.
(iii)\Rightarrow(iv) . Let w\in{\mathbb{D}} and R_z^\rho be the reproducing kernel function in {\mathcal D}_\rho . After a calculation,
\begin{equation} \nonumber \begin{split} W_{\psi, \varphi}^*(R_z^\rho)(w) = &\langle W_{\psi, \varphi}^*(R_z^\rho), R_w^\rho\rangle = \langle R_z^\rho, W_{\psi, \varphi}(R_w^\rho)\rangle \\ = &\langle R_z^\rho, \psi\cdot R_w^\rho\circ\varphi\rangle = \overline{\psi(z)R_w^\rho(\varphi(z))} \\ = &\overline{\psi(z)}R_{\varphi(z)}^\rho(w). \end{split} \end{equation} |
By the assumption that W_{\psi, \varphi} is bounded and W_{\psi, \varphi}^* is bounded below on {\mathcal D}_\rho , we see that there is a constant C > 0 such that
\|W_{\psi, \varphi}^*f\|_{{\mathcal D}_\rho}\geq C\|f\|_{{\mathcal D}_\rho}, \ \ \ \ f\in {\mathcal D}_\rho. |
Thus,
\|W_{\psi, \varphi}^*(R_z^\rho)\|_{{\mathcal D}_\rho}\geq C\|R_z^\rho\|_{{\mathcal D}_\rho}, |
that is,
|\psi(z)|\|R_{\varphi(z)}^\rho\|_{{\mathcal D}_\rho}\geq C\|R_z^\rho\|_{{\mathcal D}_\rho}. |
Therefore,
|\psi(z)|\geq C\frac{\|R_z^\rho\|_{{\mathcal D}_\rho}}{\|R_{\varphi(z)}^\rho\|_{{\mathcal D}_\rho}}\gtrsim\frac{\sqrt{\rho(1-|\varphi(z)|)}}{\sqrt{\rho(1-|z|)}}. |
Since
\frac{\rho(1-|z|)}{\rho(1-|\varphi(z)|)}\lesssim\left(\frac{ 1-|z| }{ 1-|\varphi(z)|}\right)^\gamma+\left(\frac{ 1-|z| }{ 1-|\varphi(z)|}\right)^{\delta}, |
we have
\frac{\sqrt{\rho(1-|\varphi(z)|)}}{\sqrt{\rho(1-|z|)}}\gtrsim\frac{1}{\sqrt{\left(\frac{ 1-|z| }{ 1-|\varphi(z)|}\right)^{\gamma}+\left(\frac{ 1-|z| }{ 1-|\varphi(z)|}\right)^{\delta}}}. |
By Lemma 5, we obtain
\frac{1}{\left(\frac{ 1-|z| }{ 1-|\varphi(z)|}\right)^{\gamma}+\left(\frac{ 1-|z| }{ 1-|\varphi(z)|}\right)^{\delta}}\geq\frac{1}{\left(\frac{1+|\varphi(0)|}{1-|\varphi(0)|}\right)^\gamma+\left(\frac{1+|\varphi(0)|}{1-|\varphi(0)|}\right)^{\delta}}. |
So,
|\psi(z)|\gtrsim\frac{\rho(1-|z|)}{\rho(1-|\varphi(z)|)}\gtrsim\frac{1}{\sqrt{\left(\frac{1+|\varphi(0)|}{1-|\varphi(0)|}\right)^\gamma+\left(\frac{1+|\varphi(0)|}{1-|\varphi(0)|}\right)^{\delta}}} \gt 0. |
Hence, \frac{1}{\psi} is bounded.
It is well known that a univalent inner function is an automorphism ([1, Corollary 3.8]). To prove that \varphi is an automorphism, we only need to prove that \varphi is an inner function and \varphi is univalent.
First, we prove that \varphi is an inner function. Since W_{\psi, \varphi} is bounded on {\mathcal D}_\rho , applying {W_{\psi, \varphi}} on the constant function 1 , we have \psi\in{\mathcal D}_\rho . By Lemma 3, r_z^\rho \rightarrow 0 weakly in {\mathcal D}_\rho . Thus,
\lim\limits_{|z|\rightarrow1}\frac{\psi(z)}{\|R_z^\rho\|_{{\mathcal D}_\rho}} = \lim\limits_{|z|\rightarrow1}\left\langle \psi, r_z^\rho\right\rangle = 0. |
Noting that
\frac{|\psi(z)|}{\|R_z^\rho\|_{{\mathcal D}_\rho}} = |\psi(z)|\sqrt{\rho(1-|z|)}\gtrsim\sqrt{\rho(1-|\varphi(z)|)}, |
we get
\lim\limits_{|z|\rightarrow1}\rho(1-|\varphi(z)|) = 0, |
which implies that \lim_{|z|\rightarrow1}|\varphi(z)| = 1. In other word, \varphi is an inner function.
Next we prove that \varphi is univalent. Suppose \varphi(z) = \varphi(w) , where z, w\in{\mathbb{D}} . Then clearly R_{\varphi(z)}^\rho = R_{\varphi(w)}^\rho . Since |\psi| > 0 and {W_{\psi, \varphi}}^{*}R_{z}^\rho = \overline{\psi(z)} R_{\varphi(z)}^\rho , we obtain
{W_{\psi, \varphi}}^{*}\left(\frac{R_{z}^\rho}{\overline{\psi(z)}}\right) = {W_{\psi, \varphi}}^{*}\left(\frac{R_{w}^\rho}{\overline{\psi(w)}}\right). |
So \frac{R_{z}^\rho}{\overline{\psi(z)}} = \frac{R_{w}^\rho}{\overline{\psi(w)}}. Let f = 1 . Then
\langle f, \frac{R_{z}^\rho}{\overline{\psi(z)}}\rangle = \langle f, \frac{R_{w}^\rho}{\overline{\psi(w)}}\rangle, |
which implies that \psi(z) = \psi(w) . Hence, R_{z}^\rho = R_{w}^\rho . From \langle \xi, R_{z}^\rho\rangle = \langle\xi, R_{w}^\rho\rangle, we deduce that z = w , that is, \varphi is univalent. Hence, \varphi is an automorphism.
Since \varphi is an automorphism, \varphi^{-1} is also an automorphism. By Lemma 6, C_{\varphi^{-1}} is also bounded on {\mathcal D}_\rho . Therefore, {W_{\psi, \varphi}}\circ C_{\varphi^{-1}} is bounded. For any f\in{\mathcal D}_\rho , since
{W_{\psi, \varphi}}\circ C_{\varphi^{-1}}f = \psi\cdot(f\circ\varphi^{-1}\circ\varphi) = \psi f, |
we see that \psi is multipliers of {\mathcal D}_\rho . By Lemma 4, we known that \psi\in H^{\infty} . Moreover, noting that \frac{1}{\psi}\in H^{\infty} , by Lemma 4 again we have
\begin{equation} \nonumber \begin{split} &\int_{{\mathbb{D}}}\left|\left(\frac{f(z)}{\psi(z)}\right)'\right|^2\rho(1-|z|^2)dA(z) \\ = &\int_{{\mathbb{D}}}\left|\frac{f'(z)\psi(z)-f(z)\psi'(z)}{\psi^2(z)}\right|^2\rho(1-|z|^2)dA(z) \\\lesssim&\int_{{\mathbb{D}}}\left|f'(z)\right|^2\rho(1-|z|^2)dA(z)+\int_{{\mathbb{D}}}\left|f(z)\right|^2\left|\psi'(z)\right|^2\rho(1-|z|^2)dA(z) \\\lesssim&\int_{{\mathbb{D}}}\left|f'(z)\right|^2\rho(1-|z|^2)dA(z), \end{split} \end{equation} |
which implies that \frac{1}{\psi} is also a multiplier of {\mathcal D}_\rho .
(iv)\Rightarrow(v) . From [19, Theorem 3.3] and Lemmas 1 and 6, we only need to verified that
\liminf\limits_{n\rightarrow\infty} \sqrt[n]{n\rho(\frac{1}{n})} = 1. |
From (1) , we have
\rho(1-|z|^2)\lesssim(1-|z|^2)^{\gamma}\rho(1). | (3) |
Since \gamma < \delta < 1 , there exist \alpha satisfy \delta < \alpha < 1 . From [13, Lemma 4], we known that \rho is of finite upper type less than \alpha < 1 if and only if
\int_{t}^{\infty}\rho(s)s^{-\alpha-1}ds\lesssim\rho(t)t^{-\alpha}, \ \ 0 \lt t \lt \infty. |
Noted that
\int_{t}^{\infty}\rho(s)s^{-\alpha-1}ds\geq\rho(t)\int_{t}^{\infty}s^{-\alpha-1}ds\approx\frac{\rho(t)}{t^{\alpha}}. |
That is,
\rho_1(t): = \int_{t}^{\infty}\rho(s)s^{-\alpha-1}ds\approx\rho(t)t^{-\alpha}, \ \ 0 \lt t \lt \infty, |
and \rho_1(t) is nonincreasing. Therefore,
\rho_1(1)\leq\rho_1(1-|z|^2)\approx\rho(1-|z|^2)(1-|z|^2)^{-\alpha}. |
Thus,
\rho(1-|z|^2)\gtrsim(1-|z|^2)^{\alpha}. | (4) |
Combine with (3) and (4), there exist positive constants C_1 and C_2 such that
\left(C_1\sqrt{n\left(\frac{1}{n}\right)^{\alpha}}\right)^{\frac{1}{n}}\leq\left(\sqrt{n\rho\left(\frac{1}{n}\right)}\right)^{\frac{1}{n}}\leq\left(C_2\sqrt{n\left(\frac{1}{n}\right)^{\gamma}}\right)^{\frac{1}{n}}. |
Noting that
\lim\limits_{n\rightarrow\infty}\sqrt[n]{C_1} = \lim\limits_{n\rightarrow\infty}\sqrt[n]{C_2} = 1 |
and
\lim\limits_{n\rightarrow\infty}\left(\sqrt{n\left(\frac{1}{n}\right)^{\alpha}}\right)^{\frac{1}{n}} = \lim\limits_{n\rightarrow\infty}\left(\sqrt{n\left(\frac{1}{n}\right)^{\gamma}}\right)^{\frac{1}{n}} = 1, |
we get the desired result.
(v)\Rightarrow (iii) . Since {W_{\psi, \varphi}} is invertible on {\mathcal D}_\rho , we see that {W_{\psi, \varphi}}^{*} is also invertible on {\mathcal D}_\rho , which implies that {W_{\psi, \varphi}}^{*} is bounded below on {\mathcal D}_\rho . The proof is complete.
Next we investigate equivalent characterizations of weighted composition operators {W_{\psi, \varphi}} preserves normalized tight frames and tight frames on {\mathcal D}_\rho . However, we have to restrict ourself on the space {\mathcal D}_\alpha when 0 < \alpha < 1 . Then, we also give the similar results like [16, Theorem 3.7 and Corollary 3.8].
Theorem 2. Let 0 < \alpha < 1 , \psi\in H({\mathbb{D}}) and \varphi be an analytic self-map of {\mathbb{D}} . Suppose that W_{\psi, \varphi} is bounded on {\mathcal D}_\alpha . Then the following statements are equivalent.
(i) {W_{\psi, \varphi}} preserves normalized tight frames on {\mathcal D}_\alpha .
(ii) {W_{\psi, \varphi}}^{*} is an isometry on {\mathcal D}_\alpha .
(iii) {W_{\psi, \varphi}} is an unitary operator on {\mathcal D}_\alpha .
(iv) \varphi is an automorphism on {\mathbb{D}} and \psi = \xi r_{\varphi^{-1}(0)}^\alpha , where |\xi| = 1 .
Proof. (i)\Leftrightarrow(ii) . It follows from Lemma 9.
(ii)\Leftrightarrow(iv)\Leftrightarrow(iii) . Noting the fact that the {\mathcal D}_\alpha space is a Hilbert space with the following reproducing kernel:
R_w(z) = \frac{1}{(1-\overline{w}z)^\alpha}, |
then the result can be deduced by Lemma 10.
Theorem 3. Let 0 < \alpha < 1 , \psi\in H({\mathbb{D}}) and \varphi be an analytic self-map of {\mathbb{D}} . Suppose that W_{\psi, \varphi} is bounded on {\mathcal D}_\alpha . Then the following statements are equivalent.
(i) {W_{\psi, \varphi}} preserves tight frames on {\mathcal D}_\alpha .
(ii) There is a constant c > 0 such that \|{W_{\psi, \varphi}}^{*}f\|_{{\mathcal D}_\alpha} = c\|f\|_{{\mathcal D}_\alpha} for any f\in{\mathcal D}_\alpha .
(iii) \varphi is an automorphism on {\mathbb{D}} and there is a complex number s such that \psi = s r_{\varphi^{-1}(0)}^\alpha .
Proof. \it (i)\Leftrightarrow(ii) . It follows by Lemma 8.
(ii)\Rightarrow(iii) . Since \frac{1}{c}{W_{\psi, \varphi}} = W_{\frac{\psi}{c}, \varphi} , from \|{W_{\psi, \varphi}}^{*}f\|_{{\mathcal D}_\alpha} = c\|f\|_{{\mathcal D}_\alpha} , we deduce that
\|W_{\frac{\psi}{c}, \varphi}^{*}f\|_{{\mathcal D}_\alpha} = \|f\|_{{\mathcal D}_\alpha}. |
In other word, W_{\frac{\psi}{c}, \varphi}^{*} is an isometry on {\mathcal D}_\alpha . By Theorem 2, we get that \varphi is an automorphism on {\mathbb{D}} and there exists a complex number \xi with |\xi| = 1 such that \frac{\psi}{c} = \xi r_{\varphi^{-1}(0)}^\alpha . Setting s = c\xi , we get the desired result.
(iii)\Rightarrow(ii) . Let \phi = \frac{\psi}{|s|} . Then by Theorem 2, W_{\phi, \varphi}^* is an isometry on {\mathcal D}_\alpha , which implies that
\|W_{\phi, \varphi}^*f\|_{{\mathcal D}_\alpha} = \|f\|_{{\mathcal D}_\alpha}, \ \ f\in{\mathcal D}_\alpha. |
This is clearly the same as
\|W_{\psi, \varphi}^*f\|_{{\mathcal D}_\alpha} = |s|\|f\|_{{\mathcal D}_\alpha}, \ \ f\in{\mathcal D}_\alpha. |
The proof is complete.
In this paper, we mainly prove that the weighted composition operator W_{\psi, \varphi} is invertible on Dirichlet type spaces {\mathcal D}_\rho if and only if it preserve frames. We also show that W_{\psi, \varphi} is an unitary operator if and only if W_{\psi, \varphi} preserve normalized tight frames on \mathcal{D}_{\alpha} ( 0 < \alpha < 1 ). Weighted composition operators preserve tight frames on \mathcal{D}_{\alpha} ( 0 < \alpha < 1 ) are also investigated.
The authors thank the referee for useful remarks and comments that led to the improvement of this paper. This work was supported by NNSF of China (No. 11801250), Overseas Scholarship Program for Elite Young and Middle-aged Teachers of Lingnan Normal University, the Key Program of Lingnan Normal University (No. LZ1905), and Department of Education of Guangdong Province (No. 2018KTSCX133).
We declare that we have no conflict of interest.
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