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Fredholmness of multiplication of a weighted composition operator with its adjoint on \(H^{2}\) and \(A_{\alpha}^{2}\)

Journal of Inequalities and Applications20182018:23

https://doi.org/10.1186/s13660-018-1615-0

  • Received: 28 August 2017
  • Accepted: 16 January 2018
  • Published:

Abstract

In this paper, we obtain that \(C_{\psi,\varphi}^{\ast}\) is bounded below on \(H^{2}\) or \(A_{\alpha}^{2}\) if and only if \(C_{\psi,\varphi }\) is invertible. Moreover, we investigate the Fredholm operators \(C_{\psi_{1},\varphi_{1}}C_{\psi_{2},\varphi_{2}}^{\ast}\) and \(C_{\psi_{1},\varphi_{1}}^{\ast}C_{\psi_{2},\varphi_{2}}\) on \(H^{2}\) and \(A_{\alpha}^{2}\).

Keywords

  • Hardy space
  • weighted Bergman spaces
  • weighted composition operator
  • Fredholm operator

MSC

  • 47B33
  • 47A53

1 Introduction

Let \(\mathbb{D}\) denote the open unit disk in the complex plane. The Hardy space, denoted \(H^{2}(\mathbb{D})=H^{2}\), is the set of all analytic functions f on \(\mathbb{D}\) satisfying the norm condition
$$\|f\|_{1}^{2}=\lim_{r \rightarrow1} \int_{0}^{2\pi} \bigl\vert f\bigl(re^{i\theta } \bigr) \bigr\vert ^{2}\frac{d\theta}{2\pi}< \infty. $$
The space \(H^{\infty}(\mathbb{D})=H^{\infty}\) consists of all analytic and bounded functions on \(\mathbb{D}\) with supremum norm \(\| f\|_{\infty}=\sup_{z \in\mathbb{D}}|f(z)|\).
For \(\alpha> -1\), the weighted Bergman space \(A^{2}_{\alpha}(\mathbb {D})=A^{2}_{\alpha}\) is the set of functions f analytic in \(\mathbb {D}\) with
$$\|f\|_{\alpha+2}^{2}=(\alpha+1) \int_{\mathbb {D}} \bigl\vert f(z) \bigr\vert ^{2} \bigl(1-|z|^{2}\bigr)^{\alpha}\, dA(z) < \infty, $$
where dA is the normalized area measure in \(\mathbb{D}\). The case where \(\alpha=0\) is known as the (unweighted) Bergman space, often simply denoted by \(A^{2}\).

Let φ be an analytic map from the open unit disk \(\mathbb{D}\) into itself. The operator that takes the analytic map f to \(f \circ \varphi\) is a composition operator and is denoted by \(C_{\varphi}\). A natural generalization of a composition operator is an operator that takes f to \(\psi\cdot f \circ\varphi\), where ψ is a fixed analytic map on \(\mathbb{D}\). This operator is aptly named a weighted composition operator and is usually denoted by \(C_{\psi,\varphi}\). More precisely, if z is in the unit disk, then \((C_{\psi,\varphi}f)(z)=\psi(z)f(\varphi(z))\). For some results on weighted composition and related operators on the weighted Bergman and Hardy spaces, see, for example, [114].

If ψ is a bounded analytic function on the open unit disk, then the multiplication operator \(M_{\psi}\) defined by \(M_{\psi }(f)(z)=\psi(z)f(z)\) is a bounded operator on \(H^{2}\) and \(A_{\alpha }^{2}\) and \(\|M_{\psi}(f)\|_{\gamma} \leq\|\psi\|_{\infty}\|f\| _{\gamma}\) when \(\gamma=1\) for \(H^{2}\) and \(\gamma=\alpha+2\) for \(A_{\alpha}^{2}\). Let P denote the orthogonal projection of \(L^{2}(\partial\mathbb{D})\) onto \(H^{2}\). For each \(b \in{L^{\infty}(\partial\mathbb{D})}\), the Toeplitz operator \(T_{b}\) acts on \(H^{2}\) by \(T_{b}(f)=P(bf)\). Also suppose that \(P_{\alpha}\) is the orthogonal projection of \(L^{2}({\mathbb{D}}, dA_{\alpha})\) onto \(A^{2}_{\alpha}\). For each function \(w \in L^{\infty}(\mathbb{D})\), the Toeplitz operator \(T_{w}\) on \(A^{2}_{\alpha}\) is defined by \(T_{w}(f)=P_{\alpha}(wf)\). Since P and \(P_{\alpha}\) are bounded on \(H^{2}\) and \(A^{2}_{\alpha }\), respectively, the Toeplitz operators are bounded.

Let \(w \in\mathbb{D}\), and let H be a Hilbert space of analytic functions on \(\mathbb{D}\). Let \(e_{w}\) be the point evaluation at w, that is, \(e_{w}(f)=f(w)\) for \(f \in H\). If \(e_{w}\) is a bounded linear functional on H, then the Riesz representation theorem implies that there is a function (usually denoted \(K_{w}\)) in H that induces this linear functional, that is, \(e_{w}(f)=\langle f,K_{w}\rangle\). In this case, the functions \(K_{w}\) are called the reproducing kernels, and the functional Hilbert space is also called a reproducing kernel Hilbert space. Both the weighted Bergman spaces and the Hardy space are reproducing kernel Hilbert spaces, where the reproducing kernel for evaluation at w is given by \(K_{w}(z)=(1-\overline{w}z)^{-\gamma}\) for \(z,w \in\mathbb{D}\), with \(\gamma=1\) for \(H^{2}\) and \(\gamma=\alpha+2\) for \(A_{\alpha }^{2}\). Let \(k_{w}\) denote the normalized reproducing kernel given by \(k_{w}(z)=K_{w}(z)/\|K_{w}\|_{\gamma}\), where \(\|K_{w}\|_{\gamma}^{2}=(1-|w|^{2})^{-\gamma}\).

Suppose that H and \(H'\) are Hilbert spaces and \(A:H \rightarrow H'\) is a bounded operator. The operator A is said to be left semi-Fredholm if there are a bounded operator \(B:H'\rightarrow H\) and a compact operator K on H such that \(BA=I+K\). Analogously, A is right semi-Fredholm if there are a bounded operator \(B':H'\rightarrow H\) and a compact operator \(K'\) on \(H'\) such that \(AB'=I+K'\). An operator A is said to be Fredholm if it is both left and right semi-Fredholm. It is not hard to see that A is left semi-Fredholm if and only if \(A^{\ast}\) is right semi-Fredholm. Hence A is Fredholm if and only if \(A^{\ast}\) is Fredholm. Note that an invertible operator is Fredholm. By using the definition of a Fredholm operator it is not hard to see that if the operators A and B are Fredholm on a Hilbert space H, then AB is also Fredholm on H. The Fredholm composition operators on \(H^{2}\) were first identified by Cima et al. [15] and later by a different and more general method by Bourdon [2]. Cima et al. [15] proved that only the invertible composition operators on \(H^{2}\) are Fredholm. Moreover, MacCluer [16] characterized Fredholm composition operators on a variety of Hilbert spaces of analytic functions in both one and several variables. Recently, Fredholm composition operators on various spaces of analytic functions have been studied (see [13] and [14]).

The automorphisms of \(\mathbb{D}\), that is, the one-to-one analytic maps of the disk onto itself, are just the functions \(\varphi(z)=\lambda\frac{a-z}{1-\overline{a}z}\) with \(|\lambda|=1\) and \(|a| < 1\). We denote the class of automorphisms of \(\mathbb{D}\) by \(\operatorname{Aut}(\mathbb{D})\). Automorphisms of \(\mathbb{D}\) take \(\partial\mathbb{D}\) onto \(\partial\mathbb {D}\). It is known that \(C_{\varphi}\) is Fredholm on the Hardy space if and only if \(\varphi\in\operatorname{Aut}(\mathbb{D})\) (see [2]).

An analytic map φ of the disk to itself is said to have a finite angular derivative at a point ζ on the boundary of the disk if there exists a point η, also on the boundary of the disk, such that the nontangential limit as \(z \rightarrow\zeta\) of the difference quotient \((\eta-\varphi(z))/(\zeta-z)\) exists as a finite complex value. We write \(\varphi'(\zeta)=\angle\lim_{z \rightarrow\zeta} \frac{\eta -\varphi(z)}{\zeta-z}\).

In the second section, we investigate Fredholm and invertible weighted composition operators. In Theorem 2.7, we show that the operator \(C_{\psi,\varphi}^{\ast}\) is bounded below on \(H^{2}\) or \(A^{2}_{\alpha}\) if and only if \(C_{\psi,\varphi}\) is invertible.

In the third section, we investigate the Fredholm operators \(C_{\psi _{1},\varphi_{1}}C_{\psi_{2},\varphi_{2}}^{\ast}\) and \(C_{\psi _{1},\varphi_{1}}^{\ast}C_{\psi_{2},\varphi_{2}}\) on \(H^{2}\) and \(A^{2}_{\alpha}\).

2 Bounded below operators \(C_{\psi,\varphi}^{\ast}\)

Let H be a Hilbert space. The set of all bounded operators from H into itself is denoted by \(B(H)\). We say that an operator \(A \in B(H)\) is bounded below if there is a constant \(c > 0\) such that \(c\| h \| \leq \|A(h)\|\) for all \(h \in H\).

If f is defined on a set V and if there is a positive constant m such that \(|f(z)| \geq m\) for all z in V, then we say that f is bounded away from zero on V. In particular, we say that ψ is bounded away from zero near the unit circle if there are \(\delta> 0\) and \(\epsilon> 0\) such that
$$\bigl\vert \psi(z) \bigr\vert > \epsilon \quad \mbox{for } \delta< |z| < 1. $$

Suppose that T belongs to \(B(H)\). We denote the spectrum of T, the essential spectrum of T, the approximate point spectrum of T, and the point spectrum of T by \(\sigma(T)\), \(\sigma_{e}(T)\), \(\sigma _{\mathrm{ap}}(T)\) and \(\sigma_{p}(T)\), respectively. Moreover, the left essential spectrum of T is denoted by \(\sigma_{\mathrm{le}}(T)\).

Suppose that φ is an analytic self-map of \(\mathbb{D}\). For almost all \(\zeta \in\partial\mathbb{D}\), we define \(\varphi(\zeta )=\lim_{r \rightarrow1}\varphi(r\zeta)\) (the statement of the existence of this limit can be found in [17, Theorem 2.2]). If f is a bounded analytic function on the unit disk such that \(|f(e^{i\theta})|=1\) almost everywhere, then we call f an inner function. We know that if φ is inner, then \(C_{\varphi}\) is bounded below on \(H^{2}\), and therefore \(C_{\varphi}\) has a closed range (see [17, Theorem 3.8]).

Now we state the following simple and well-known lemma, and we frequently use it in this paper.

Lemma 2.1

Let \(C_{\psi,\varphi}\) be a bounded operator on \(H^{2}\) or \(A^{2}_{\alpha}\). Then \(C_{\psi,\varphi}^{\ast}K_{w}=\overline{\psi(w)}K_{\varphi(w)}\) for all \(w \in\mathbb{D}\).

In this paper, for convenience, we assume that \(\gamma=1\) for \(H^{2}\) and \(\gamma=\alpha+2\) for \(A_{\alpha}^{2}\).

Lemma 2.2

Suppose that A and B are two bounded operators on a Hilbert space H. If AB is a Fredholm operator, then B is left semi-Fredholm.

Proof

Suppose that AB is a Fredholm operator. Then there are a bounded operator C and a compact operator K such that \(CAB=I+K\). Therefore B is left semi-Fredholm. □

Zhao [13] characterized Fredholm weighted composition operators on \(H^{2}\). Also, Zhao [14] found necessary conditions of φ and ψ for a weighted composition operator \(C_{\psi ,\varphi}\) on \(A_{\alpha}^{2}\) to be Fredholm. In the following proposition, we obtain a necessary and sufficient condition for \(C_{\psi,\varphi}\) to be Fredholm on \(H^{2}\) and \(A_{\alpha}^{2}\). Then we use it to find when \(C_{\psi_{1},\varphi_{1}}^{\ast}C_{\psi _{2},\varphi_{2}}\) and \(C_{\psi_{1},\varphi_{1}}C_{\psi_{2},\varphi _{2}}^{\ast}\) are Fredholm. The idea of the proof of the next proposition is different from [13] and [14].

Proposition 2.3

The operator \(C^{\ast}_{\psi,\varphi}\) is left semi-Fredholm on \(H^{2}\) or \(A_{\alpha}^{2}\) if and only if \(\varphi\in\operatorname{Aut}(\mathbb{D})\) and \(\psi\in H^{\infty}\) is bounded away from zero near the unit circle. Under these conditions, \(C_{\psi,\varphi}\) is a Fredholm operator.

Proof

Let \(C_{\psi,\varphi}\) be Fredholm on \(H^{2}\) or \(A_{\alpha}^{2}\). Assume that ψ is not bounded away from zero near the unit circle. Then for each positive integer n, there is \(x_{n} \in\mathbb{D}\) such that \(1-1/n < |x_{n}| < 1\) and \(|\psi (x_{n})| < 1/n\). Then there exist a subsequence \(\{x_{n_{m}}\}\) and \(\zeta\in\partial\mathbb{D}\) such that \(x_{n_{m}} \rightarrow\zeta \) as \(m \rightarrow\infty\). Since \(\psi(x_{n_{m}})\rightarrow0\) as \(m\rightarrow\infty\), by Lemma 2.1 we see that
$$\begin{aligned} \lim_{m \rightarrow\infty} \bigl\Vert C_{\psi,\varphi}^{\ast }k_{x_{n_{m}}} \bigr\Vert _{\gamma} =&\lim_{m \rightarrow\infty} \bigl\vert \psi (x_{n_{m}}) \bigr\vert \frac{\|K_{\varphi(x_{n_{m}})}\|_{\gamma}}{\| K_{x_{n_{m}}}\|_{\gamma}} \\ \leq&\limsup_{m \rightarrow\infty} \bigl\vert \psi(x_{n_{m}}) \bigr\vert \biggl(\frac {(1+|x_{n_{m}}|)(1-|x_{n_{m}}|)}{(1+|\varphi(x_{n_{m}})|)(1-|\varphi (x_{n_{m}})|)} \biggr)^{\gamma/2} \\ \leq&2^{\gamma/2}\lim_{m \rightarrow\infty} \bigl\vert \psi (x_{n_{m}}) \bigr\vert \limsup_{m \rightarrow\infty} \biggl( \frac {1-|x_{n_{m}}|}{1-|\varphi(x_{n_{m}})|} \biggr)^{\gamma/2} \\ =&0, \end{aligned}$$
where the last equality follows from the fact that \(\liminf\frac {1-|\varphi(x_{n_{k}})|}{1-|x_{n_{k}}|} \neq0\) (see [17, Corollary 2.40]). Since \(k_{x_{n_{m}}}\) tends to zero weakly as \(m \rightarrow\infty\) (see [17, Theorem 2.17]), by [18, Theorem 2.3, p. 350], \(C_{\psi,\varphi}^{\ast}\) is not left semi-Fredholm. This is a contradiction. Hence ψ is bounded away from zero near the unit circle. Denote the inner product on \(H^{2}\) or \(A_{\alpha}^{2}\) by by \(\langle\cdot,\cdot\rangle_{\gamma}\), where \(\gamma=1\) for \(H^{2}\) and \(\gamma=\alpha+2\) for \(A_{\alpha }^{2}\). Define the bounded linear functional \(F_{\psi}\) by \(F_{\psi }(f)=\langle f,\psi\rangle_{\gamma}\) for each f that belongs to \(H^{2}\) or \(A_{\alpha}^{2}\), where \(\gamma=1\) for \(H^{2}\) and \(\gamma =\alpha+2\) for \(A_{\alpha}^{2}\). We know that, for each \(\zeta\in \partial\mathbb{D}\), \(K_{r\zeta}/\|K_{r\zeta}\|_{\gamma}\) tends to zero weakly as \(r \rightarrow1\). Then
$$\lim_{r \rightarrow1}F_{\psi} \biggl(\frac{K_{r\zeta}}{\|K_{r\zeta }\|_{\gamma}} \biggr)= \lim_{r \rightarrow1} \biggl\langle \frac{K_{r\zeta }}{\|K_{r\zeta}\|_{\gamma}},\psi \biggr\rangle _{\gamma}=0, $$
and so \(|\psi(r\zeta)|/\|K_{r\zeta}\|_{\gamma} \rightarrow0\) as \(r \rightarrow1\). Now, we show that φ is inner. For each \(\zeta\in\partial \mathbb{D}\) such that \(\varphi(\zeta):=\lim_{r\rightarrow1}\varphi (r\zeta)\) exists, by Lemma 2.1 we have
$$\begin{aligned} \lim_{r \rightarrow1} \bigl\Vert C_{\psi,\varphi}^{\ast}k_{r\zeta} \bigr\Vert _{\gamma} =&\lim_{r \rightarrow1}\frac{|\psi(r\zeta)|}{\| K_{r\zeta}\|_{\gamma}} \biggl(\frac{1}{1-|\varphi(r\zeta )|^{2}} \biggr)^{\gamma/2} \\ \leq&\lim_{r \rightarrow1}\frac{|\psi(r\zeta)|}{\|K_{r\zeta}\| _{\gamma}} \biggl(\frac{1}{1-|\varphi(r\zeta)|} \biggr)^{\gamma/2}. \end{aligned}$$
Since \(|\psi(r\zeta)|/\|K_{r\zeta}\|_{\gamma} \rightarrow0\) as \(r \rightarrow1\), if \(\varphi(\zeta) \notin\partial\mathbb{D}\), then \(\lim_{r \rightarrow1}\|C_{\psi,\varphi}^{\ast}k_{r\zeta}\| _{\gamma}=0\), which is a contradiction (see [18, Theorem 2.3, p. 350]). Hence φ is inner. Since \(C_{\psi,\varphi}^{\ast }\) is left semi-Fredholm, by Lemma 2.2, \(T_{\psi}^{\ast}\) is left semi-Fredholm. Then \(\operatorname{dim} \operatorname{ker} T_{\psi}^{\ast}\) is finite. It follows from Lemma 2.1 that ψ has only finite zeroes in \(\mathbb {D}\). If φ is constant on \(\mathbb{D}\), then \(\operatorname{dim} \operatorname{ker} C_{\psi,\varphi}^{\ast}= \operatorname{dim} (\operatorname{ran} C_{\psi,\varphi})^{\bot}=\infty\), a contradiction. If \(\varphi(a)=\varphi(b)\) for some \(a,b \in\mathbb{D}\) with \(a \neq b\), then by using the idea similar to that used in [2, Lemma] there exist infinite sets \(\{a_{n}\}\) and \(\{b_{n}\}\) in \(\mathbb{D}\) which are disjoint and such that \(\varphi(a_{n})= \varphi(b_{n})\). We can assume that \(\psi(a_{n})\psi(b_{n}) \neq0\) because ψ has only finite zeroes in \(\mathbb{D}\). By Lemma 2.1 we see that
$$C_{\psi,\varphi}^{\ast} \biggl(\frac{K_{a_{n}}}{\overline{\psi (a_{n})}}-\frac{K_{b_{n}}}{\overline{\psi(b_{n})}} \biggr)=K_{\varphi(a_{n})}-K_{\varphi(b_{n})} \equiv0. $$
Therefore, \(K_{a_{n}}/\overline{\psi(a_{n})}-K_{b_{n}}/\overline {\psi(b_{n})} \in\operatorname{ker} C_{\psi,\varphi}^{\ast}\). It is not hard to see that \(\{K_{a_{n}}/\overline{\psi (a_{n})}-K_{b_{n}}/\overline{\psi(b_{n})}\}\) is a linearly independent set in the kernel of \(C_{\psi,\varphi}^{\ast}\), and so we have our desired contradiction. Hence φ must be univalent. Then [17, Corollary 3.28] implies that φ is an automorphism of \(\mathbb{D}\). Since \(C_{\psi,\varphi}C_{\varphi }^{-1}=M_{\psi}\) is a bounded multiplication operator on \(H^{2}\) and \(A^{2}_{\alpha}\), by [19, p. 215], \(\psi\in H^{\infty}\).
Conversely, suppose that \(\varphi\in\operatorname{Aut}(\mathbb{D})\) and \(\psi\in H^{\infty}\) is bounded away from zero near the unit circle. Since \(C_{\varphi}\) is invertible, \(C_{\varphi}\) has a closed range. Since \(\psi\not\equiv0\), \(\operatorname{ker} T_{\psi}=(0)\). We infer that \(T_{\psi}\) has a closed range by [18, Corollary 2.4, p. 352], [20, Theorem 3], and [12, Theorem 8], so by [18, Proposition 6.4, p. 208], \(T_{\psi}\) is bounded below. We claim that \(C_{\psi,\varphi}\) has a closed range. This can be seen as follows. Suppose that \(\{h_{n}\}\) is a sequence such that \(\{C_{\psi ,\varphi}(h_{n})\}\) converges to f as \(n\rightarrow\infty\). Since \(T_{\psi}\) has a closed range, \(\{C_{\psi,\varphi}(h_{n})\}\) converges to \(T_{\psi}g\) for some g as \(n\rightarrow\infty\). Since \(T_{\psi}\) is bounded below, there is a constant \(c > 0\) such that \(\|T_{\psi}(C_{\varphi}(h_{n})-g)\| \geq c\|C_{\varphi }(h_{n})-g\|\). Therefore \(C_{\varphi}(h_{n}) \rightarrow g\) as \(n \rightarrow\infty\). There exists h such that \(C_{\varphi}(h)=g\) because \(C_{\varphi}\) has a closed range. Hence \(f=C_{\psi,\varphi }(h)\), as desired. Hence \(\operatorname{ran} C_{\psi,\varphi}\) is closed and \(\operatorname{ker} C_{\psi,\varphi}=(0)\). [20, Theorem 3] and [12, Theorem 10] imply that \(T_{\psi}\) is Fredholm, and so \(\operatorname{ker} T_{\psi}^{\ast }\) is finite dimensional. Since \(\varphi\in \operatorname{Aut}(\mathbb{D})\), it is not hard to see that
$$\operatorname{ker} C_{\psi,\varphi}^{\ast}=(\operatorname{ran} C_{\psi,\varphi })^{\bot}=(\operatorname{ran} T_{\psi})^{\bot}= \operatorname{ker} T_{\psi }^{\ast}. $$
Therefore, \(\operatorname{dim} \operatorname{ker} C_{\psi,\varphi}^{\ast} < \infty\), and the conclusion follows from [18, Corollary 2.4, p. 352]. □

In the next proposition, we give a necessary condition of ψ for an operator \(C_{\psi,\varphi}^{\ast}\) to be bounded below on \(H^{2}\) and \(A^{2}_{\alpha}\). Then we use Proposition 2.4 to obtain all invertible weighted composition operators on \(H^{2}\) and \(A^{2}_{\alpha}\).

Proposition 2.4

Let ψ be an analytic map of \(\mathbb {D}\), and let φ be an analytic self-map of \(\mathbb{D}\). If \(C_{\psi,\varphi}^{\ast}\) is bounded below on \(H^{2}\) or \(A^{2}_{\alpha}\), then \(\psi\in H^{\infty}\) is bounded away from zero on \(\mathbb{D}\), and \(\varphi\in\operatorname{Aut}(\mathbb{D})\).

Proof

Let \(\varphi\equiv d\) for some \(d \in\mathbb{D}\). Since \(C_{\psi,\varphi}^{\ast}\) is bounded below, there is a constant \(c > 0\) such that \(\|C_{\psi,\varphi}^{\ast} f\|_{\gamma} \geq c\|f\| _{\gamma}\) for all f. Then for each \(w \in\mathbb{D}\), by Lemma 2.1, \(\|C_{\psi,\varphi}^{\ast}K_{w}\|_{\gamma}=|\psi(w)|\|K_{d}\| _{\gamma} \geq c\|K_{w}\|_{\gamma}\). Therefore, for each \(w \in\mathbb{D}\),
$$\bigl\vert \psi(w) \bigr\vert \geq\frac{c}{\|K_{d}\|_{\gamma}}\frac {1}{(1-|w|^{2})^{\gamma/2}}. $$
It is easy to see that ψ is bounded away from zero on \(\mathbb {D}\). Now assume that φ is not a constant function. Suppose that ψ is not bounded away from zero on \(\mathbb{D}\). Therefore, there exist a sequence \(\{x_{n}\}\) in \(\mathbb{D}\) and \(a \in\overline{\mathbb{D}}\) such that \(x_{n} \rightarrow a\) and \(|\psi (x_{n})| \rightarrow0\) as \(n \rightarrow\infty\). First, suppose that \(a \in\mathbb{D}\). By Lemma 2.1 we have
$$\bigl\Vert C_{\psi,\varphi}^{\ast}k_{a} \bigr\Vert _{\gamma}= \bigl\vert \psi(a) \bigr\vert \biggl(\frac {1-|a|^{2}}{1-|\varphi(a)|^{2}} \biggr)^{\gamma/2}=0. $$
Since \(C_{\psi,\varphi}^{\ast}\) is bounded below, \(0 \geq c\|k_{a}\| _{\gamma}=c\), a contradiction. Now assume that \(a \in\partial\mathbb {D}\). It is not hard to see that there is a subsequence \(\{x_{n_{m}}\}\) such that \(\{\varphi(x_{n_{m}})\}\) converges. By Lemma 2.1 we see that
$$ \limsup_{m \rightarrow\infty} \bigl\Vert C_{\psi,\varphi}^{\ast }k_{x_{n_{m}}} \bigr\Vert _{\gamma}=\limsup_{m\rightarrow\infty} \bigl\vert \psi (x_{n_{m}}) \bigr\vert \biggl(\frac{1-|x_{n_{m}}|^{2}}{1-|\varphi (x_{n_{m}})|^{2}} \biggr)^{\gamma/2}. $$
(1)
If \(\{\varphi(x_{n_{m}})\}\) converges to a point in \(\mathbb{D}\), then (1) is equal to zero. Now assume that \(\{\varphi(x_{n_{m}})\}\) converges to a point in \(\partial\mathbb{D}\). If φ has a finite angular derivative at a, then by the Julia-Carathéodory theorem we have
$$\limsup_{m\rightarrow\infty}\frac{1-|x_{n_{m}}|^{2}}{1-|\varphi (x_{n_{m}})|^{2}}=\frac{1}{|\varphi'(a)|}, $$
which shows that (1) is equal to zero. If φ does not have a finite angular derivative at a, then
$$\limsup_{m\rightarrow\infty}\frac{1-|x_{n_{m}}|}{1-|\varphi(x_{n_{m}})|}=0, $$
so again (1) is equal to zero. Since \(C_{\psi,\varphi}^{\ast}\) is bounded below and \(\|k_{x_{n_{m}}}\|_{\gamma}=1\), we have \(c=0\), is a contradiction. Therefore, ψ is bounded away from zero on \(\mathbb {D}\). Since by [18, Proposition 6.4, p. 208], \(0 \notin\sigma _{\mathrm{ap}}(C_{\psi,\varphi}^{\ast})\), we have that \(\lim_{r \rightarrow 1}\|C_{\psi,\varphi}^{\ast}k_{r\zeta}\|_{\gamma} \neq0\) for all \(\zeta\in\partial\mathbb{D}\). We employ the idea of the proof of Proposition 2.3 to see that φ is a univalent inner function. Thus \(\varphi\in\operatorname{Aut}(\mathbb{D})\) (see [17, Corollary 3.28]). Moreover, since \(C_{\psi,\varphi}\) is a bounded operator, as we saw in the proof of Proposition 2.3, we conclude that \(\psi\in H^{\infty}\), and the proposition follows. □

Bourdon [21, Theorem 3.4] obtained the following corollary; we give another proof (see also [22, Theorem 2.0.1]).

Corollary 2.5

Let ψ be an analytic map of \(\mathbb {D}\), and let φ be an analytic self-map of \(\mathbb{D}\). The weighted composition operator \(C_{\psi,\varphi}\) is invertible on \(H^{2}\) or \(A_{\alpha}^{2}\) if and only if \(\varphi\in\operatorname{Aut}(\mathbb{D})\) and \(\psi\in H^{\infty}\) is bounded away from zero on \(\mathbb{D}\).

Proof

Let \(C_{\psi,\varphi}\) be invertible. Then \(C_{\psi ,\varphi}^{\ast}\) is bounded below. The conclusion follows from Proposition 2.4. The reverse direction is trivial since \(C_{\varphi}\) and \(T_{\psi}\) are invertible. □

Note that if \(C_{\psi,\varphi}\) is invertible, then \(C_{\psi,\varphi }^{\ast}\) is bounded below. Hence by Proposition 2.4 and Corollary 2.5 we can see that \(C_{\psi,\varphi}^{\ast}\) is bounded below if and only if \(C_{\psi,\varphi}\) is invertible.

The algebra \(A(\mathbb{D})\) consists of all continuous functions on the closure of \(\mathbb{D}\) that are analytic on \({\mathbb{D}}\). In the next corollary, we find some Fredholm weighted composition operators that are not invertible.

Corollary 2.6

Suppose that \(\varphi\in\operatorname{Aut}(\mathbb {D})\) and \(\psi\in A(\mathbb{D})\). Assume that \(\{z \in\mathbb{D}: \psi(z)=0\}\) is a nonempty finite set and \(\psi(z) \neq0\) for all \(z \in\partial \mathbb{D}\). Then \(C_{\psi,\varphi}\) is Fredholm, but it is not invertible.

Proof

It is easy to see that ψ is bounded away from zero near the unit circle. Therefore the result follows from Proposition 2.3 and Corollary 2.5. □

Theorem 2.7

Suppose that ψ is an analytic map of \(\mathbb{D}\) and φ is an analytic self-map of \(\mathbb{D}\). The operator \(C_{\psi,\varphi}^{\ast}\) is bounded below on \(H^{2}\) or \(A^{2}_{\alpha}\) if and only if \(\varphi\in\operatorname{Aut}(\mathbb {D})\) and \(\psi\in H^{\infty}\) is bounded away from zero on \(\mathbb {D}\).

3 The operators \(C_{\psi_{1},\varphi_{1}}C_{\psi_{2},\varphi _{2}}^{\ast}\) and \(C_{\psi_{1},\varphi_{1}}^{\ast}C_{\psi _{2},\varphi_{2}}\)

In this section, we find all Fredholm operators \(C_{\psi_{1},\varphi _{1}}C_{\psi_{2},\varphi_{2}}^{\ast}\) and \(C_{\psi_{1},\varphi _{1}}^{\ast}C_{\psi_{2},\varphi_{2}}\).

A linear-fractional self-map of \(\mathbb{D}\) is a mapping of the form \(\varphi(z)=(az+b)/(cz+d)\) with \(ad-bc \neq0\) such that \(\varphi (\mathbb{D}) \subseteq\mathbb{D}\). We denote the set of those maps by \(\operatorname{LFT}(\mathbb{D})\). Suppose \(\varphi(z)=(az+b)/(cz+d)\). It is well known that the adjoint of \(C_{\varphi}\) acting on \(H^{2}\) and \(A^{2}_{\alpha}\) is given by \(C_{\varphi}^{\ast}=T_{g}C_{\sigma}T_{h}^{\ast}\), where \(\sigma(z)=({\overline{a}z-\overline{c}})/({-\overline {b}z+\overline{d}})\) is a self-map of \(\mathbb{D}\), \(g(z)=(-\overline{b}z+\overline {d})^{-\gamma}\), and \(h(z)=(cz+d)^{\gamma}\). Note that g and h are in \(H^{\infty}\) ([17, Theorem 9.2]). If \(\varphi(\zeta)=\eta\) for \(\zeta,\eta\in\partial\mathbb{D}\), then \(\sigma(\eta)=\zeta\). We know that φ is an automorphism if and only if σ is, and in this case, \(\sigma=\varphi^{-1}\). The map σ is called the Krein adjoint of φ. We denote by \(F(\varphi)\) the set of all points in \(\partial\mathbb{D}\) at which φ has a finite angular derivative.

Example 3.1

Suppose that \(\varphi\in\operatorname{LFT}(\mathbb{D})\) is not an automorphism of \(\mathbb{D}\). Assume that \(\psi\in H^{\infty}\) is continuously extendable to \(\mathbb{D} \cup F(\varphi )\). Assume that \(C_{\psi,\varphi}C_{\psi,\varphi}^{\ast}\) is considered as an operator on \(H^{2}\) or \(A_{\alpha}^{2}\). Since φ is not an automorphism of \(\mathbb{D}\), \(\overline {\varphi(\mathbb{D})} \subseteq\mathbb{D}\) or there is only one point \(\zeta\in\partial\mathbb{D}\) such that \(\varphi(\zeta) \in \partial\mathbb{D}\). If \(\overline{\varphi(\mathbb{D})} \subseteq \mathbb{D}\), then by [17, p. 129], \(C_{\varphi}\) is compact. It is easy to see that \(C_{\psi,\varphi}C_{\psi,\varphi}^{\ast}\) is a compact operator. Since compact operators are not Fredholm, we can see that \(C_{\psi,\varphi}C_{\psi,\varphi}^{\ast}\) is not Fredholm.

In the other case, assume that \(F(\varphi)=\{\zeta\}\). Because for each \(w \in\partial\mathbb{D}\) such that \(w \neq\zeta\), \(\sigma (\varphi(w)) \notin\partial\mathbb{D}\), we obtain \(\sigma \circ \varphi\notin\operatorname{Aut}(\mathbb{D})\). Since \(C_{\sigma\circ \varphi}\) is not Fredholm (see e.g. [2] and [16]), \(0 \in \sigma_{e}(C_{\sigma\circ\varphi})\). By [23, Corollary 2.2] and [4, Proposition 2.3] there is a compact operator K such that
$$C_{\psi,\varphi}C_{\psi,\varphi}^{\ast}= \bigl\vert \psi(\zeta ) \bigr\vert ^{2}C_{\varphi}C_{\varphi}^{\ast}+K. $$
Also, [23, Theorem 3.1], [23, Proposition 3.6], and [24, Theorem 3.2] imply that there is a compact operator \(K'\) such that
$$ C_{\psi,\varphi}C_{\psi,\varphi}^{\ast}= \bigl\vert \psi(\zeta ) \bigr\vert ^{2} \bigl\vert \varphi'(\zeta) \bigr\vert ^{-\gamma}C_{\sigma\circ\varphi}+K'. $$
(2)
From the fact that \(0 \in\sigma_{e}(C_{\sigma\circ\varphi})\) and equation (2) we can infer that \(0 \in\sigma_{e}(C_{\psi,\varphi }C_{\psi,\varphi}^{\ast})\). Then \(C_{\psi,\varphi}C_{\psi,\varphi }^{\ast}\) is not Fredholm.

By the preceding example it seems natural to conjecture that if \(C_{\psi,\varphi}C_{\psi,\varphi}^{\ast}\) is Fredholm, then \(\varphi\in\operatorname{Aut}(\mathbb{D})\). We will prove our conjecture in Theorem 3.2 and show that if \(C_{\psi_{1},\varphi_{1}}C_{\psi _{2},\varphi_{2}}^{\ast}\) is Fredholm on \(H^{2}\) or \(A_{\alpha }^{2}\), then \(C_{\psi_{1},\varphi_{1}}\) and \(C_{\psi_{2},\varphi _{2}}\) are Fredholm.

Theorem 3.2

The operator \(C_{\psi_{1},\varphi_{1}}C_{\psi _{2},\varphi_{2}}^{\ast}\) is Fredholm on \(H^{2}\) or \(A_{\alpha}^{2}\) if and only if \(\varphi_{1},\varphi_{2} \in\operatorname{Aut}(\mathbb {D})\), \(\psi_{1},\psi_{2} \in H^{\infty}\), and \(\psi_{1}\) and \(\psi _{2}\) are bounded away from zero near the unit circle.

Proof

Let \(C_{\psi_{1},\varphi_{1}}C_{\psi_{2},\varphi _{2}}^{\ast}\) be Fredholm. Therefore \(C_{\psi_{2},\varphi_{2}}^{\ast }\) is left semi-Fredholm. By Proposition 2.3 we see that \(\varphi_{2} \in\operatorname{Aut}(\mathbb{D})\) and \(\psi_{2} \in H^{\infty}\) is bounded away from zero near the unit circle. Since \(C_{\psi _{2},\varphi_{2}}C_{\psi_{1},\varphi_{1}}^{\ast}\) is Fredholm, again we can see that \(\varphi_{1}\) is an automorphism of \(\mathbb {D}\) and \(\psi_{1} \in H^{\infty}\) is bounded away from zero near the unit circle.

Conversely, let \(\varphi_{1},\varphi_{2} \in\operatorname{Aut}(\mathbb {D})\) and \(\psi_{1},\psi_{2} \in H^{\infty}\) be bounded away from zero near the unit circle. By Proposition 2.3, \(C_{\psi_{1},\varphi _{1}}\) and \(C_{\psi_{2},\varphi_{2}}^{\ast}\) are Fredholm, so the result follows. □

In the following theorem for functions \(\psi_{1},\psi_{2} \in A(\mathbb{D})\), we find all Fredholm operators \(C_{\psi_{1},\varphi _{1}}^{\ast}C_{\psi_{2},\varphi_{2}}\) when \(\varphi_{1}\) and \(\varphi_{2}\) are univalent self-maps of \(\mathbb{D}\).

Theorem 3.3

Suppose that \(\psi_{1},\psi_{2} \in A(\mathbb{D})\). Let \(\varphi_{1}\) and \(\varphi_{2}\) be univalent self-maps of \(\mathbb{D}\). The operator \(C_{\psi_{1},\varphi _{1}}^{\ast}C_{\psi_{2},\varphi_{2}}\) is Fredholm on \(H^{2}\) or \(A_{\alpha}^{2}\) if and only if \(C_{\psi_{1},\varphi_{1}}\) and \(C_{\psi_{2},\varphi_{2}}\) are Fredholm on \(H^{2}\) or \(A_{\alpha }^{2}\), respectively.

Proof

Let \(C_{\psi_{1},\varphi_{1}}^{\ast}C_{\psi _{2},\varphi_{2}}\) be Fredholm on \(H^{2}\) or \(A_{\alpha}^{2}\). Then \(C_{\psi_{2},\varphi_{2}}^{\ast}C_{\psi_{1},\varphi_{1}}\) is also Fredholm. It is easy to see that \(C_{\varphi_{2}}\) and \(C_{\varphi _{1}}\) are left semi-Fredholm. Therefore, \(0 \notin\sigma _{\mathrm{le}}(C_{\varphi_{1}})\) and \(0 \notin\sigma_{\mathrm{le}}(C_{\varphi _{2}})\). Since \(\operatorname{dim} \operatorname{ker} C_{\psi_{1},\varphi _{1}}^{\ast}C_{\psi_{2},\varphi_{2}} < \infty\) and \(\operatorname{dim} \operatorname{ker} C_{\psi_{2},\varphi_{2}}^{\ast}C_{\psi _{1},\varphi_{1}} < \infty\), \(\psi_{1} \not\equiv0\), \(\psi_{2} \not\equiv0\), and \(\varphi _{1}\) and \(\varphi_{2}\) are not constant functions. By the open mapping theorem we know that \(0 \notin\sigma_{p}(C_{\varphi_{1}})\) and \(0 \notin\sigma_{p}(C_{\varphi_{2}})\). Now [18, Proposition 4.4, p. 359] implies that \(0 \notin\sigma_{\mathrm{ap}}(C_{\varphi _{1}})\) and \(0 \notin\sigma_{\mathrm{ap}}(C_{\varphi_{2}})\). Hence by [18, Proposition 6.4, p. 208], \(\operatorname{ran} C_{\varphi_{1}}\) and \(\operatorname{ran} C_{\varphi_{2}}\) are closed. By [1, Theorem 5.1], \(\varphi_{1},\varphi_{2} \in\operatorname{Aut}(\mathbb{D})\). Since \(C_{\varphi_{1}}^{\ast}\) and \(C_{\varphi_{2}}\) are invertible, \((C_{\varphi_{1}}^{\ast})^{-1}\) and \(C_{\varphi_{2}}^{-1}\) are Fredholm. Then \(T_{\psi_{1}}^{\ast}T_{\psi_{2}}\) is Fredholm, and so \(0 \notin\sigma_{e}(T_{\overline{\psi_{1}}\psi_{2}})\). We get from [25] and [20, Theorem 2] that \(\psi_{1}\) and \(\psi _{2}\) never vanish on \(\partial\mathbb{D}\). Since \(\psi_{1} \not \equiv0\) and \(\psi_{2} \not\equiv0\), \(\psi_{1}\) and \(\psi_{2}\) have only finite zeroes on \(\mathbb{D}\). This implies that there is \(r < 1\) such that for each w with \(r < |w| < 1\), \(\psi_{1}(w) \neq0\) and \(\psi_{2}(w) \neq0\). Therefore, \(\psi_{1}\) and \(\psi_{2}\) are bounded away from zero near the unit circle. Therefore, by Proposition 2.3, \(C_{\psi_{1},\varphi_{1}}\) and \(C_{\psi_{2},\varphi_{2}}\) are Fredholm on \(H^{2}\) or \(A_{\alpha}^{2}\). The reverse implication follows from the fact stated before Theorem 3.2. □

Declarations

Acknowledgements

The author would like to thank the referees for their valuable comments and suggestions.

Authors’ contributions

The author read and approved the final manuscript.

Competing interests

The author declares that he has no competing interests.

Open Access This article is distributed under the terms of the Creative Commons Attribution 4.0 International License (http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made.

Authors’ Affiliations

(1)
Department of Mathematics, Shiraz University of Technology, Shiraz, Iran

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