1. Introduction

Let and be Banach spaces of analytic functions on a domain in , an analytic function on , and let be an analytic function mapping into itself. The weighted composition operator with symbols and from to is the operator with range in defined by where denotes the multiplication operator with symbol , and denotes the composition operator with symbol .

Let be the set of analytic functions on . For the Hardy space is the space consisting of all such that Let denote the space of all for which .

The Bloch space on the open unit disk is the Banach space consisting of the analytic functions on such that The Bloch norm is given by . Using the Schwarz-Pick lemma, it is easy to see that the Hardy space is contained in and . The inclusion is proper, as the function shows.

The little Bloch space, denoted by , is defined as the set of the analytic functions on such that . It is well known that is a closed separable subspace of . The interested reader is referred to [1] for more information on the Bloch space.

The logarithmic Bloch space is defined as the set of functions on such that It is a Banach space under the norm defined by . Clearly, if , then , so is a subset of the little Bloch space.

The little logarithmic Bloch space, denoted by , is defined as the subspace of whose elements satisfy the condition

The space arises in connection to the study of certain operators with symbol. Arazy [2] proved that the multiplication operator is bounded on the Bloch space if and only if . In [3], Brown and Shields extended this result to the little Bloch space.

The space also arises in the study of Hankel operators on the Bergman one space. The Bergman space on is defined to be the set of analytic functions on whose modulus is Lebesgue integrable over .

The Hankel operator on is defined as , where is the identity operator, and is the standard Bergman projection from into . In [4], Attele showed that is bounded on if and only if .

The study of operators with symbol on the logarithmic Bloch space began with the characterizations of the bounded and the compact composition operators given in [5] by Yoneda. In [6], Galanopoulos extended these results to the weighted composition operators on . He also introduced a class of Banach spaces () closely related to and studied the Taylor coefficients of the functions in . In [7], Ye characterized the bounded and the compact weighted composition operators on the little logarithmic Bloch space . See [8, 9] for the study of the weighted composition operators on Bloch spaces and weighted Bloch spaces.

In this paper, we characterize the bounded and the compact weighted composition operators from the Hardy space (with ) to the logarithmic Bloch space as well as to its subspace . The paper consists of five sections. Specifically, in Section 2, we consider the bounded weighted composition operators mapping into and . In particular, we show that where the notation stands for , for some positive constants and . In Section 3, we look at the issue of compactness of such operators.

In Section 4, we characterize the bounded and the compact weighted composition operators mapping into in the case when . Finally, in Section 5, we study the operators mapping into .

2. Boundedness of from into and

In the following theorem, we give two characterizations of boundedness when the operator maps into .

Theorem 2.1. Let be an analytic function on , and let be an analytic self-map of . The following statements are equivalent. (a)The operator is bounded.(b)(c) and .

Proof. (a) (b). For , the function is bounded and . Therefore, if is bounded, then .
(b) (c) Let be an upper bound for , . Taking , we deduce that , so .
For and , define the sets
Fix an integer , and . For , by the product rule, we have
In the proof of Theorem 2 of [10], it was shown that
For , there exists such that . So From (2.2) and (2.4), we deduce that is finite.
(c) (a) Let with and pick . Then and, since , Thus, by (2.5) and (2.6), we deduce that , completing the proof.

We next turn our attention to the weighted composition operators mapping into the little logarithmic Bloch space.

Theorem 2.2. Let be an analytic function on , and let be an analytic self-map of . The following statements are equivalent. (a)The operator is bounded.(b)For each integer and (c) and .

Proof. (a) (b) is proved as in the case of the operator mapping into .
(b) (c) Suppose that (b) holds. If for some integer , then for all , we have If is properly contained in , then for , arguing as in the proof of (b) (c) in Theorem 2.1, there exists such that , so that where and .
By the assumption of , we have as . On the other hand, since , as . Therefore,
(c) (a) Assume that (c) holds. To prove that is bounded, it suffices to show that for each , since the boundedness of the operator can be shown as in the proof of Theorem 2.1. Since , for and , we have as . On the other hand, by (2.6), as . Hence, as , completing the proof.

In Section 3, we shall prove that all bounded weighted composition operators from into are compact.

3. Compactness of from into and

The following criterion for compactness follows by a standard argument similar, for example, to that outlined in Proposition 3.11 of [11].

Lemma 3.1. Let be analytic on , an analytic self-map of , . The operator is compact if and only if for any bounded sequence in which converges to zero uniformly on compact subsets of , we have as .

The proof of the following result is similar to the proof of Lemma 1 of [12]. Hence we omit it.

Lemma 3.2. A closed set K in is compact if and only if it is bounded and satisfies the following:

We now introduce two one-parameter families of functions which will be used to characterize the compactness of the operators under consideration.

Fix and, for , define

Theorem 3.3. Let be analytic on , an analytic self-map of , and assume that is bounded. Then the following conditions are equivalent: (a) is compact.(b) and (c) and (d) and

Proof. We begin by showing that (a), (b), and (c) are equivalent.
Suppose that is compact and that is a sequence in such that as . Since the sequences and are bounded in and converge to 0 uniformly on compact subsets of , by Lemma 3.1, it follows that and as .
Assume that (b) holds. Fix . A straightforward calculation shows that Eliminating , we obtain that Thus, for , Taking the limit as , we deduce that On the other hand, using (3.3), we obtain that Taking the limit as , we obtain that proving (c).
Suppose that (c) holds. Let be a bounded sequence in converging to 0 uniformly on compact subsets of . Set . Then, given , there exists such that for , Then, for , noting that , we obtain that Thus, for , we have On the other hand, for , Thus, by the uniform convergence to 0 of and on compact sets, we see that (3.11) holds also in this case for sufficiently large. Hence, for sufficiently large. Since , we deduce that as . Therefore, is compact.
(a) (d) Suppose that is compact. Since the sequence defined by , is bounded in and converges to 0 uniformly on compact subsets, by Lemma 3.1, it follows that as . The second condition in (d) follows from the equivalence of (a) and (c).
(d) (c) Fix and choose such that for all and for .
For , there exists such that . Using the product rule, we may write the following so that The left-hand side of (3.14) can be written as which, by (2.3), is bounded below by Thus, from (3.14), we deduce that proving (c). The equivalence of statements (a)–(d) is now established.

Next, we characterize the compact weighted composition operators from into .

Theorem 3.4. Let be an analytic function on , and let be an analytic self-map of . The following statements are equivalent. (a)The operator is compact.(b)For each integer , and .(c) and

Proof. (a) (b) is immediate.
(b) (c) It suffices to show that if (b) holds, then as . Fix and let be an integer such that for all . Observe that, since for , and, by assumption, the functions and are in , we have Therefore, there is , such that for , Thus, if , and , then On the other hand, if , then there exists such that , so, as shown in the proof of (d) implies (c) of Theorem 3.3, we have as . Since is arbitrary, the result follows.
(c) (a) Let be a bounded sequence in converging to 0 uniformly on compact subsets, and let . We wish to show that and as . As shown in the proof of (c) implies (a) of Theorem 3.3, for and , as . Thus, . The convergence to 0 of is proved as in the case of the operator mapping into .