Abstract

We study Fredholm (weighted) composition operators between general weighted Banach spaces of analytic functions on the open unit disc with sup-norms.

1. Introduction

Let be Banach spaces (infinite dimensional spaces). We denote by the space of bounded linear operators from to . A bounded linear operator is said to be Fredholm if the spaces and are finite dimensional. Every Fredholm operator has closed range. It is known that is Fredholm if and only if is invertible modulo the compact operators and if and only if its dual map is Fredholm. For background on Fredholm operators we refer to [1].

Let and be analytic functions on the open unit disk of the complex plane such that . These maps define, on the space of analytic functions on , the so-called weighted composition operator by . It combines the classical composition operator with the pointwise multiplication operator . These operators have been studied by different authors on various types of function spaces. For more information about composition operators, we refer the reader to the excellent monographs of Cowen and MacCluer [2] and Shapiro [3]. In his thesis Schwartz [4] showed that a composition operator on the Hardy space is invertible if and only if is a conformal automorphism of the unit disk. Cima et al. [5] proved that only the invertible operators on are Fredholm. Various authors have considered the composition operator on several Banach spaces of analytic functions and have studied when is Fredholm. See, for example, [6] for the Hardy and Bergman spaces in , [7] for various spaces on domain in , [8] for a variety of Hilbert spaces of analytic functions on domains in , [9] for the space of bounded analytic functions on the open unit ball of a complex Banach space , and [10] for Banach spaces of analytic functions on satisfying certain conditions. Recently, Zhao has given a characterization of bounded Fredholm weighted composition operator on Dirichlet spaces [11], on Hardy spaces [12], and on a class of weighted Hardy spaces [13].

In this paper we consider the weighted composition operator defined on the weighted Banach spaces of holomorphic functions and on the smaller spaces (see Section 2 for the definition). In this framework, Contreras and Hernández-Díaz [14] and Montes-Rodríguez [15], continuing work by [16], characterized the boundedness and compactness of between weighted Banach spaces of analytic functions and . We are interested in finding a characterization of Fredholm weighted composition operators in terms of properties of the symbol and of the multiplier function .

Our paper is motivated mainly by the works [10, 17]. In the first one, Bonet et al. characterized when is a Fredholm operator if it is considered between or . Recently, Galindo and Lindström considered in [10] composition operators on a class of Banach spaces of analytic functions defined on and , satisfying certain conditions, and they proved that the composition operator associated with the analytic self-map is Fredholm if and only if is a biholomorphic map and if and only if is invertible. We will present a characterization of Fredholm composition operators and Fredholm weighted composition operators when they are considered between or , for a general typical weight function (see Theorem 10). Similar results to the ones presented here for more concrete weights are already established in Section 3 of [18]. Although some of the techniques are known and have already been used in the mentioned paper, we present here proofs obtained independently in a more general setting with some applications. Some particular cases of Banach spaces considered in [10, 18] are the standard weighted Bergman spaces with and , the little Bloch space, and the weighted Banach spaces , , where . For any weight, does not satisfy in general condition (C1) of [10] or conditions (C1) and of [18]. Hence, we complement recent work by Contreras, Galindo, Hernández-Daz, Hyvärinen, Lindström, Nieminen, and Saukko, among others.

2. Notation and Preliminaries

Let be the open unit disk in the complex plane and let us denote by the set of analytic functions from into . A weight function is a radial bounded continuous function that is nonincreasing with respect to such that . In the literature these types of weight functions are called typical weight functions. The weighted Banach spaces of analytic functions are defined as follows: endowed with the norm . The function is not a weight function according to our assumptions. In this case and .

Many results on weighted spaces of analytic functions are formulated in terms of the so-called associated weight function which is defined by the formulawhere denotes the point evaluation of . By [19], we know that is also a weight function; we have that , and for each we can find , , such that . It is well known that is isometrically isomorphic to and is isometric to , where the inclusion map is the canonical injection from a Banach space into its bidual. We refer to [19–21] for more information about these spaces.

For with and or , we consider the composition operator and the weighted composition operator is defined as follows: provided it is well defined. We always assume that given a weighted composition operator there exists a point such that .

In case the composition operator is bounded and is a bounded analytic function on the unit disk, , the multiplication operator , defined as , is bounded ([17, Proposition  2.1]), and the weighted composition operator is also bounded and can be decomposed as , where and .

By Schwarz Lemma, if , then for every and therefore is bounded on every space . If satisfies the Lusky condition then Theorem  2.3 in [16] ensures that is bounded on and . For instance, the standard weights , are weight functions which have .

The point evaluations play an important role in our results. Our first two lemmas are known. We give a short proof for the reader’s convenience.

Lemma 1. For every , the point evaluation , , is continuous and .

Proof. If , fix : Then .
For all , it is clear that . On the other hand, Under the hypothesis that is a typical weight we know that the closed unit ball of is dense in the closed unit ball of for the compact open topology (see [20]). Then if , , there exists such that in the compact open topology . Then . Hence . Since , we obtain

Lemma 2. If is a weight function, then (i),(ii) for all .

Proof. (i) By [16, Proposition 1.1], we have . From formula (2) and Lemma 1 we obtain the conclusion.
(ii) Let be a polynomial; then Since is equicontinuous in and the polynomials are dense in , we have for all .

For Banach spaces (infinite dimensional spaces) a bounded linear operator is said to be Fredholm if the spaces and are finite dimensional. Note that a consequence of the Fredholmness is that the condition implies that the range of is closed. It is known that is Fredholm if and only if is Fredholm and if and only if there are ,, and such that ,  , where denote the set of compact operators on . As a consequence, invertible operators are Fredholm and compact operators cannot be Fredholm. We refer to [1, Chapter III] for the proofs of these results.

3. Fredholm Weighted Composition Operators

Recall that the weighted composition operator is bounded if and only is finite ([14, Proposition  3.1]) and the operator is bounded if and only if and is finite ([14, Proposition  3.2]).

Lemma 3. Let and let satisfy . The operator is Fredholm if and only if is Fredholm.

Proof. It is a consequence of the fact that and that coincides with the biadjoint map of whenever both operators are well defined.

Lemma 4. Let , , and . If is Fredholm, then there can be at most finitely many points at which vanishes.

Proof. We first prove that for every zero of . To do this suppose that and . Then there exists such that . Hence .
Now, suppose that there exists an infinite sequence such that . The sequence is linearly independent in (see [17] pp. 178) and . Then .
On the other hand, since is Fredholm, is also Fredholm; hence , which is a contradiction.

Proposition 5. Let be analytic function and . If is Fredholm, then is injective.

Proof. Our proof starts with the observation that for every point in there is a neighborhood where is injective. Otherwise, there exists and there are disjoint infinite sequences and such that and . By Lemma 4, has only a finite number of zeros in ; then we can assume . For every , we may define . Then . Indeed, Moreover, is an infinite linearly independent sequence in . Indeed, if we suppose then for all polynomial we obtain For every it is sufficient to take a polynomial such that to conclude .
By assuming that is Fredholm, then is Fredholm and is finite. This fact contradicts the fact that is an infinite linearly independent sequence in .
Note that cannot be constant. We are now in position to show that is injective. Assume is not injective. Then there are , , such that . By our observation above there are and , , such that and are injective. Then and are open in . Hence is open in and it is an open neighborhood of . Therefore As in the first part of the proof, we can assume that . We define as above. It is an infinite linearly independent sequence in . A contradiction since is Fredholm.

Remark 6. For with () the operator is compact for every . Then is not Fredholm. In fact, the associated weight is also radial and decreasing. Hence for all and From [14, Proposition  3.1], is bounded. Since , with , by [16, Corollary  3.2] we obtain that the composition operator is compact and . Then the operator can be written as , where , . Since and , we deduce that is compact.

Proposition 7. Let be an analytic function and . If or is Fredholm, then is an automorphism.

Proof. By Lemma 3, it is enough to consider only the case . If is Fredholm then is also a Fredholm operator. From [1, Chapter III, Theorem  13] there are a bounded operator and a compact operator , such that . For , we consider . By Lemma 2, weakly converges to 0 in as goes to 1. Since is a compact operator, as . Hence there is , , such that for all with . Sincewe obtain that For with , we haveThen for with , From Lemma 2, for , we know that . Hence, by (2), It follows that and then is an inner function.
By [2, Corollary  3.28] it suffices to prove that is univalent to show that is an automorphism. Since is Fredholm, from Lemma 3 we obtain that is Fredholm. By Proposition 5, we get that is injective and the proof is complete.

Lemma 8. If is an automorphism on and is bounded, then is bounded and .

Proof. Since is an automorphism, it is known that there are and such that , where . Then . The Möbius transformation satisfies that . Now applying that is radial, we obtain As is bounded, by [16, Proposition  2.1] we obtain that is bounded. An easy computation shows that .

We observe that Lemma 8 is also true if we replace by .

Proposition 9. Let be an analytic function and . If and are bounded on or and is also Fredholm, then and are Fredholm.

Proof. By Lemma 3 we only need to consider the case . Proposition 7 implies that is an automorphism. Hence by the lemma above is bounded and is Fredholm. Thus the operator is well defined, bounded, and Fredholm.

We obtain a characterization of Fredholm weighted composition operators by applying the results above.

Theorem 10. Let be an analytic function and . If is bounded on or , then the following assertions are equivalent:(1) is Fredholm.(2) is Fredholm.(3) and are Fredholm.(4) and are Fredholm.(5), , and is an automorphism.