Abstract

We characterize boundedness and compactness of weighted composition operators mapping the families of fractional Cauchy transforms into the Bloch-type spaces. Corollaries are obtained about composition operators and multiplication operators.

1. Introduction

Let denote the open unit disc in the complex plane and let denote the space of functions analytic in . Let denote the Banach space of complex-valued Borel measures on , endowed with the total variation norm. For , the space of fractional Cauchy transforms is the collection of functions of the form where . The principal branch of the logarithm is used here. The space is a Banach space, with norm given by where varies over all measures in for which (1) holds. The families have been studied extensively [1, 2].

Let . The Bloch-type space is the Banach space of functions analytic in such that , with norm

The integral representation (1) implies that and there is a constant depending only on such that for .

It is known that any univalent belongs to for any . MacGregor [3] constructed a univalent function such that . Let denote the normalized function . Then . Since , the classical family of schlicht functions, the Distortion Theorem [4] yields .

Let be an analytic self-map of and let . The weighted composition operator is defined for by If , then the operator reduces to the composition operator defined by . If is the identity function, then is the multiplication operator defined by .

This paper characterizes and for which is bounded or compact. Corollaries are obtained for the operators and .

2. Boundedness

We follow the convention that denotes a positive constant, which may vary from one appearance to the next.

Theorem 1. Fix and . Let and let be an analytic self-map of . Ifthen is bounded.

Proof. Assume the hypotheses. By a theorem due to Ohno et al., is bounded [5]. Thus there is a constant such that for all . Relation (4) now yields for all .

The focus of this paper is to prove the converse of Theorem 1. Several lemmas are needed. Proofs of Lemmas 2 and 3 appear in [2].

Lemma 2. Fix and let . Let for . Then and .

Lemma 3. Fix and let . If , then and there is a positive constant independent of such that

The first statement in Lemma 4 is due to MacGregor [3]. The norm inequality is due to Hibschweiler and Nordgren [6].

Lemma 4. Let . If and , then and

Lemma 5 will be used to develop test functions needed for the proof of the converse.

Lemma 5. Fix . Let and define Then and there is a constant independent of such that for all .

Proof. First assume and fix generic . A calculation shows that is in the Hardy space and . Since the inclusion is bounded, this case is complete.
Fix . Then By the case for and Lemma 2, is the product of a function in and a function in . By Lemma 4, and there is a constant independent of such that for all .
Finally fix . By the previous case, and for all . By Lemma 3, and . The proof is complete.

We now prove the converse of Theorem 1. The test functions used in the proof first appeared in [5], in the context of the spaces .

Theorem 6. Fix and . Let and let be an analytic self-map of . Assume that is bounded. Then

Proof. Fix , , , and as described. By assumption there is a constant independent of such that for all .
The argument will first establish that . Let and define By Lemmas 2 and 5, there is a constant such that for all . Therefore for all . Since it follows that In particular, (17) yields .
To obtain the second condition in the theorem, let and define By Lemma 5, there is a constant independent of such that . Relation (13) yields for all . Since it follows that for all .
First consider with . By the triangle inequality, relation (21) yieldsfor such . By relation (17), it follows thatFinally consider with . Let in relation (13). Thus and for all . Therefore for all . ThereforeRelations (23) and (26) yield and the proof is complete.

Let . Ohno et al. [5] characterized and for which is bounded. Theorems 1 and 6 and their result yield the following corollary.

Corollary 7. Fix . Let and let be an analytic self-map.

Xiao [7] characterized the self-maps for which is bounded for .

Corollary 8. Fix , and as above.

Proof. The equivalence of the first two conditions follows from Corollary 7. The equivalence of the second and third conditions is due to Xiao.

Let and let . The function is a multiplier of into if for every . By the Closed Graph Theorem, it follows that is bounded. The collection of all such multipliers is denoted . In [5], Ohno et al. characterized .

Let denote the set of analytic functions for which is bounded. Corollary 9 follows from Corollary 7 and the characterization in [5] for the case .

Corollary 9. Fix and let .

3. Compactness

A characterization is given for functions for which is compact.

Lemma 10. Fix and let . Define by Then and there is a constant such that for all .

Proof. First fix and let . A particular case of Lemma 5 provides a constant independent of such that for all . Since Lemma 4 now implies that and for all .
When , By the previous case and Lemma 2, is the product of a function in and a function in . By Lemma 4, and for all .
Fix . By the previous cases and for all . Lemma 3 shows that and . A similar argument applies when . The proof is complete.

Lemma 11 is the standard sequential criterion for compactness.

Lemma 11. Fix . The operator is compact if and only if as for any sequence in with and uniformly on compact subsets of as .

Theorem 12. Fix . Assume that is bounded. The operator is compact if and only if

Proof. Fix and assume that is bounded.
First assume the limit conditions (35) and (36). Corollary 7 implies that is bounded and it now follows as in [5] that is compact. Suppose that is a sequence in such that for all and uniformly on compact subsets. By relation (4), and thus as . By Lemma 11, is compact.
Now assume that is compact. We may assume that . Let be any sequence in with as . For define By Lemma 5, for all . Also uniformly on compact subsets of as . Thus as and as . Calculations yield as .
The argument will first establish that as . As in [5], define the test functions where and . Then uniformly on compact subsets as . By Lemmas 10 and 5, there is a constant with for all . It now follows thatIn particular, as . relation (40) is established. Since is a generic sequence with as , relation (35) holds.
To complete the proof note that relations (39) and (40) yield as . Since , as . Condition (36) follows and the proof is complete.