#### 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 MacGregorâ€™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 SteviÄ‡ 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 MÃ¶bius invariance. In particular, Shapiroâ€™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 Xiaoâ€™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 , Shapiroâ€™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