Abstract

In this note we express the norm of composition followed by differentiation from the logarithmic Bloch and the little logarithmic Bloch spaces to the weighted space on the unit disk and give an upper and a lower bound for the essential norm of this operator from the logarithmic Bloch space to .

1. Introduction

Let be the open unit disk in the complex plane , be the space of all analytic functions on , and be the space of bounded analytic functions on with the norm .

An analytic function is said to belong to the logarithmic Bloch space if and to the little logarithmic Bloch space if It can be easily proved that is a Banach space, under the norm and that is a closed subspace of . Some sources for results and references about the logarithmic Bloch functions are the papers of Yoneda [1], Stević [2], and the authors of [3–8].

Let be a weight, that is, a positive continuous function on . The weighted space consists of all such that where is a weight.

Let be a holomorphic self-map of . The composition operator is defined by Let be the differentiation operator. The product is defined by The operator is probably studied for the first time by Hibschweiler and Portnoy in [9], where the boundedness and compactness of between Bergman and Hardy spaces are investigated. In [10], Stević calculated the norm of the operator from the classical Bloch space to . Recently there has been some interest in calculating operator norms and essential norms of composition and related operators (see, e.g., [11–18] and the references therein). Motivated by the papers [10, 19], we continue here this line of research by calculating .

Suppose that and are Banach spaces and is a bounded linear operator. The essential norm of is its distance to the compact operators. More precisely, where denotes the operator norm. If , it is simply denoted by . Since the set of all compact operators is a closed subset of the set of bounded operators, it follows that an operator is compact if and only if .

Essential norm formulas for composition operators are known in various settings. When acts from the Hardy space to itself, Shapiro [20] gives a formula for in terms of the Nevanlinna counting function for . In [21], Donaway gives upper and lower estimates for when maps the Bloch, Dirichlet, or a Besov type space to itself. The essential norm of the operator from -Bloch spaces to space was estimated recently by Stević in [10]. In this note we give upper and lower estimates for .

2. The Operator Norm of or

In this section we prove a nice formula. Namely, we calculate the norm of the operator .

Theorem 1. Assume is a weight on . Then the following are equivalent:(a) is a bounded operator;(b) is a bounded operator;(c)
Moreover, one has

Proof. . By the fact and the definition of operator norm, we easily obtain that is a bounded operator and . Suppose that is a bounded operator from to . Taking the test function , we easily have for every . It implies that (c) holds when .
Fixing , we consider the function
Since is increasing on and , we have Moreover, since as , it follows that for every . Thus, for each we obtain that for every . Letting , we obtain that for every . It implies that (c) also holds when .
. For every , we easily obtain that Hence is a bounded operator. Also, we obtain Moreover, from (9), (10), (15), and (17), we obtain

3. Estimates of Essential Norm of or

In this section we will estimate the essential norm of . For this purpose we need some lemmas.

Lemma 2. If , then .

This can be done in exactly the same way as in the proof of [3, Lemma 2.1].

Lemma 3. Let be an analytic self-map of and be a weight on . Assume that is a bounded operator from to ; then is compact if and only if for any bounded sequence in , which converges to uniformly on compact subsets of , one has as .

Proof. Necessity. Suppose that is compact. Let be a bounded sequence in with uniformly on compact subsets of . Assume that there is a subsequence and an such that for all . Since is compact, we can find a further subsequence and a function such that . Then we obtain that, for , Hence uniformly on compact subsets of . Also, since uniformly on compact subsets of , uniformly on compact subsets of . It follows that and hence , contradicting the fact that for all . Therefore we must have that .
Sufficiency. Let be a bounded sequence in . Then Lemma 2 and Montel’s Theorem tell us that forms a normal family, and hence there exists a subsequence converging uniformly on compact sets to some function . It is easy to see that must be in . Then is a bounded sequence in converging to uniformly on compact subsets of and by the hypothesis guarantees that in . Thus is compact.

Lemma 4. Let be a weight on and be an analytic self-map of with . Suppose that is bounded. Then is compact.

Proof. Suppose that is a bounded sequence in which converges to uniformly on compact subsets of . By Cauchy’s inequality we easily obtain that also converges to uniformly on compact subsets of . Since is bounded, one can take the test function to see that . Then we obtain that as , since is contained in the disk , which is a compact subset of . Hence, by Lemma 3, the operator is compact.

Lemma 5. Let . Then , where .

Since is decreasing on , one may easily prove the result.

Theorem 6. Let be a weight on and be an analytic self-map of . Suppose that is bounded. Then

Proof. If , by Lemma 4, it follows that is compact which is equivalent to . On the other hand, it is clear that in this case the condition is vacuous, so that it is understood that
Now suppose that . Assume that is a sequence in such that as . Let where . Then we have , Clearly uniformly on compact subsets of as . It follows that Thus, . Let . Then and uniformly on compact subsets of as . Since , then it follows that converges to weakly in . Thus, for any compact operator , . Therefore Hence Thus the first inequality in (21) follows. The second inequality in (21) is obvious. Now we prove the third one.
Let be fixed and . By Lemma 4 we obtain that the operator is compact for every . It follows that By Cauchy’s inequality, we obtain that On the other hand, by Lemma 5, we obtain that where . Hence, for for all and all , we have Letting and then letting , we obtain that The proof of the theorem is finished.