Products of Composition and Differentiation between the Fractional Cauchy Spaces and the Bloch-Type Spaces

Abstract

The operators and are defined by and where is an analytic self-map of the unit disc and is analytic in the disc. A characterization is provided for boundedness and compactness of the products of composition and differentiation from the spaces of fractional Cauchy transforms to the Bloch-type spaces , where and . In the case , the operator is compact is bounded and . For , is compact is bounded and .

1. Introduction

Let and let denote the family of functions analytic on . Let denote the Banach space of complex Borel measures on , endowed with the total variation norm. For , the space of fractional Cauchy transforms is the family of functions of the form where . The principal branch of the logarithm is used here. The space is a Banach space, with norm where varies over all measures in for which (1) holds. The families have been studied extensively [1, 2]. Interest in these spaces was first established in connection with the classical family of normalized univalent functions. It is known that for any [2]. The reference [2] also includes MacGregor’s construction of a function with .

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

The relation (1) implies that , and there is a constant depending only on such that for all .

Let be an analytic self-map of . The composition operator is defined by for . The differentiation operator is defined by . In this paper, the products and are studied. Conditions on are given, necessary and sufficient to imply boundedness or compactness of and .

Products of composition and differentiation on the Bloch space were studied by Ohno in [3]. In [4], Li and Stević studied and acting between the weighted Bergman spaces and the Bloch-type spaces. In [5], Hibschweiler and Portnoy studied these operators between Bergman and Hardy spaces.

2. Preliminary Results

Fix . For fixed and for , the relation (1) yields a constant depending only on such that [2].

For each , [2].

We follow the convention that denotes a positive constant, the precise value of which will differ from one appearance to the next.

Lemma 1 and Lemma 2 will be used to develop test functions for . Proofs appear in [6].

Lemma 1. Fix . For , define Then, , and there is a constant such that for all .

Lemma 2. Fix . For , define Then, , and there is a constant such that for all .

3. The Operator

In [7], Shapiro proved that the condition is necessary for to be compact, for Banach spaces obeying boundary regularity and Möbius invariance. In particular, Shapiro’s result applies to the Lipschitz spaces and thus, to the space when [8].

Theorem 3. Fix and . Let be an analytic self-map of .

Proof. First, assume that is bounded, that is, there is a constant such that for all . It is clear that and . Thus, and for all . It follows that and thus, .
Let and define By Lemma 1 and the preliminary results, there is a constant independent of such that , and thus, . It follows that for all . Calculations yield and The substitution in (11) now yields and thus, By the relation (9), Thus, It follows that By Xiao’s result [9], is bounded. Furthermore, (16) yields as . Thus, is compact [9], and it follows as in [7] that . It has been established that the conditions , and are necessary if is bounded.
Next, assume that , , and . To show that is compact, let be a bounded sequence in with uniformly on compact subsets of as . It is enough to prove that as . First, note that as . For , (9) yields Since and uniformly on compact subsets as , the argument shows that as . Thus, as , and is compact, as required.
The remaining implication is clear, and the proof is complete.

Theorem 4. Fix and . Let be an analytic self-map of . Then, and

Proof. Fix and as described.
First, assume (21) and (22). Let . By (21) and the introductory remarks in Section 2, A similar argument using (22) yields for all . Thus, . Since , it now follows that , as required.
For the converse, assume that for a constant independent of . In particular, .
The argument leading to (16) remains valid for . Thus, (22) holds. It remains to prove (21). First, note that For , define for . By Lemma 1 and Lemma 2, there is a constant independent of such that . Thus, for all . It follows that for all . An argument using and yields The relations (25) and (28) establish relation (21), and the proof is complete.

Theorem 5. Fix and assume . Let be a self-map of for which is bounded. and

Proof. First, assume that is bounded and relations (30) and (31) hold. Let be a bounded sequence in such that uniformly on compact subsets of . As previously noted, there is a constant depending only on such that for and . Relation (31) now implies that given , there exists , such that for all .
Since is bounded, relation (9) holds, and thus, for all . Since uniformly on , there exists such that for all . The relations (33) and (35) yield for .
A similar argument using and (30) yields such that for . The relations (36) and (37) yield as .
Since as , the argument yields as for any sequence as described, and therefore, is compact.
For the converse, assume that is compact. We may assume that . Let be any sequence in with as . For , define By the lemmas above, . Also, uniformly on compact subsets. Therefore, and as . Calculations yield and Substitution into (40) yields as . Since is a generic sequence with as , this yields the relation (30).
A similar argument using the functions yields the relation (31). The details are omitted.

Theorem 3 implies that if is bounded for fixed with , then is compact for all . The next corollary gives a related result when .

Corollary 6. Fix and . Let be a self-map of and assume that is bounded. Then, is compact for any .

Proof. By assumption, there is a constant such that for all . Fix with and let . Then, and [2]. Therefore, is bounded and Theorem 5 applies.
Since is bounded, (21) yields and therefore, A similar argument using (22) yields Theorem 5 now yields is compact.

4. The Operator

In this section, characterizations are given for self-maps for which is bounded or compact. The proofs are similar to those in Section 3, so details are kept to a minimum.

Theorem 7. Fix and .

Proof. First, assume that there is a constant independent of such that . In particular, . For , define There is a constant independent of such that , and it follows that for all . The substitution yields for all . Therefore, Since , It follows that and therefore By [9], is bounded. A further argument as in the proof of Theorem 3 yields that is compact. Since , Shapiro’s result [7] applies and yields . Thus, the conditions and are necessary in order for to be bounded.
Next, assume and . Let be a bounded sequence in with uniformly on compact subsets of . First, note that as . For , Since uniformly on compact subsets, the argument yields and is compact.
The remaining implication is trivial, and the proof is complete.

Theorem 8. Fix and . Let be a self-map of .

Proof. First, assume that the supremum is finite.
Let . By previous remarks, . By an argument as in the proof of Theorem 4, and thus, as required.
To complete the proof, assume that for a constant independent of . The argument leading to (53) remains valid for . This proves the opposite implication, and the proof is complete.

Theorem 9. Fix and . Let be a self-map of and assume that is bounded.

Proof. First, assume that is bounded and the limit condition holds. Let be a bounded sequence in with uniformly on compact subsets as . Clearly, as . As in previous arguments, for all . The hypothesis now implies that, given , there exists , such that for all . Since and since uniformly on compact subsets, as . By (60) and (61), as . The argument yields as for any sequence as described above. Thus, is compact.
Now, assume that is compact. We may assume that . Let be any sequence in with as . For , define for . By Lemma 1, for all . Also, uniformly on compact subsets. Therefore, as . Given , there exists such that for all . In particular, for . Calculations yield for . Since is a generic sequence with , it follows that The proof is complete.

Assume that is bounded for fixed and . By Theorem 7, and . It follows that is compact for any . Corollary 10 gives a related result in the case .

Corollary 10. Fix and assume that is bounded. Then, is compact for any .

Proof. Fix and let . Then, and [2]. Therefore, is bounded and Theorem 9 applies.
By Theorem 8, there is a constant with for all . Therefore, as. By Theorem 9, is compact.

Data Availability

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Conflicts of Interest

The author declares that there is no conflict of interest regarding the publication of this paper.

Acknowledgments

This research was conducted under the employ of the Department of Mathematics and Statistics at the University of New Hampshire, Durham, NH, USA.