Abstract
The boundedness and compactness of the product of differentiation and composition operators from Bloch spaces into spaces are discussed in this paper.
1. Introduction and Motivation
Let be the open unit disk in the complex plane and let be the class of all analytic functions on . Let be the Euclidean area element on . The Bloch space on is the space of all analytic functions on such that Under the above norm, is a Banach space. Let denote the subspace of consisting of those for which as . This space is called the little Bloch space.
Throughout this paper, we assume that is a nondecreasing and right-continuous function. A function is said to belong to space (see [1]) if where is the Green function with logarithmic singularity at ; that is, ( is a conformal automorphism defined by for ). is a Banach space under the norm From [1], we know that if
Let denote a nonconstant analytic self-map of . Associated with is the composition operator defined by for . The problem of characterizing the boundedness and compactness of composition operators on many Banach spaces of analytic functions has attracted lots of attention recently, for example, [2] and the reference therein.
Let be the differentiation operator on ; then we have . For , the products of differentiation and composition operators and are defined by Operators as well as some other products of linear operators were studied, for example, in [3–9] (see also the references therein).
Recall that a linear operator is said to be bounded if there exists a constant such that for all maps . And is compact if it takes bounded sets in to sets in which have compact closure. For Banach spaces and of , is compact from to if and only if for each sequence in ; the sequence contains a subsequence converging to some limit in .
Considering the definition of spaces and with some conditions, it is difficult to study the operator from Bloch spaces to spaces. In this paper, some sufficient and necessary conditions for the boundedness and compactness of this operator are given.
2. The Boundedness
Lemma 1 (see [10]). If all , then
Theorem 2. Let be an analytic self-map of . Suppose is a nondecreasing and right-continuous function on such that Then the following statements are equivalent:(a) is bounded.(b) is bounded.(c)
Proof.
(c) (a). Suppose that holds. For any and , we have Thus holds.
(a) (b). It is obvious.
(b) (c). Assume holds; that is, there exists a constant such that for all . Conversely, suppose that is bounded. Fix and assume that . Consider the function defined by for ; then for . Sincefor all , . Furthermore, it is clear that , since Thus Note that For this function and this point we have So for all . Then we can implyOn the other hand, we note the functions , which belong to , and we get Then So, we get By the arbitrary of , we have for all .
This completes the proof of this theorem.
3. The Compactness
The following lemma can be proved similarly to [11].
Lemma 3. Let be an analytic self-map of . Then (or ) is compact if and only if (or ) is bounded and for any bounded sequence in which converges to zero uniformly on compact subsets of ; one has .
Lemma 4. Let be an analytic self-map of . Suppose is a nondecreasing and right-continuous function on such that If is compact, then for any there exists a such that, for all in , holds whenever , where is the unit ball of .
Proof. For , let . Then , and uniformly on compact subsets of as . Since is compact, as . That is, for given , there exists such that For , the triangle equality gives Then, we prove that for that given and there exists a such that if , Choose , and we have . Since is compact, . Thus, for given and , there exists an such thatwhenever . Hence, for ,Therefore, for , Thus we have already proved that, for any and , there exists a such that holds whenever .
We finish our proof by showing that the above , in fact, is independent of . Since is compact, is relatively compact in . It means that there is a finite collection of functions in such that, for any and , we can find satisfying On the other hand, if , we have from the previous observation that, for all , By the triangle inequality we obtain that holds whenever . The proof is complete.
Theorem 5. Let be an analytic self-map of . Suppose is a nondecreasing and right-continuous function on such that Then the following statements are equivalent:(a) is compact.(b) is compact.(c) is bounded:
Proof.
(c) (a). Assume holds. Without loss of generality, let be a sequence in which converges to uniformly on compact subsets of , as , where is the unit ball of . By Cauchy’s estimate, we know that also converges to uniformly on compact subsets of . For the sufficiency we will be verifying that converges to in norm. By the assumption, for any , there exists a such that Let ; then we have By is bounded, we know that . It follows that since that as . By Lemma 1, we can obtain that is compact.
(a) (b). It is obvious.
(b) (c). Suppose that is compact. Then it is clear that is bounded. We know that for all . Choose a sequence in which converges to as , and let for . Thus in for all and , where is the unit ball of . By Lemma 4, for any holds for any and . That is, Thus, we obtain by integrating, with respect to , the Fubini theorem, the Poisson formula, and the Fatou’s lemma.
Conflict of Interests
The authors declare that there is no conflict of interests regarding the publication of this paper.
Acknowledgments
This paper was supported in part by NSFC, Approval no. 41174165, the graduate student scientific research innovation project of Jiangsu Province of China, Approval no. KYLX15_0882, the Natural Science Research Program of the Education Department of Jiangsu Province of China, Approval no. 07KJB110069, and a grant of NUIST, Approval no. 20080290.