Abstract

Let denote the space of all holomorphic functions on the unit disk of , and let  n be a positive integer, a holomorphic self-map of , and a weight. In this paper, we investigate the boundedness and compactness of a weighted differentiation composition operator , from the logarithmic Bloch spaces to the Zygmund-type spaces.

1. Introduction

Let denote the open unit disk of the complex plane and the space of all analytic functions in .

The logarithmic Bloch space is defined as follows: The space is a Banach space under the norm . Let denote the subspace of consisting of those such that It is obvious that there are unbounded functions. For example, consider the function . There are also bounded functions that do not belong to . In fact, the interpolating Blaschke products do not belong to . It is easily proved that, for , . first appeared in the study of boundedness of the Hankel operators on the Bergman space. Attele in [1] proved that, for , the Hankel operator is bounded if and only if , thus giving one reason, and not the only reason, why log-Bloch-type spaces are of interest. Ye in [2] proved that is a closed subspace of . Galanopoulos in [3] characterized the boundedness and compactness of the composition operator and the boundedness and compactness of the weighted composition operator . Ye in [4] characterized the boundedness and compactness of the weighted composition operator between the logarithmic Bloch space and the -Bloch spaces on the unit disk and the boundedness and compactness of the weighted composition operator between the little logarithmic Bloch space and the little -Bloch spaces on the unit disk. Li in [5] characterized the boundedness and compactness of the weighted composition operator from Bergman spaces into the logarithmic Bloch space on the unit disk. Ye in [6] characterized the boundedness and compactness of the weighted composition operator from the general function space into the logarithmic Bloch space on the unit disk. Colonna and Li in [7] studied the boundedness and compactness of the weighted composition operators from Hardy space into the logarithmic Bloch space and the little logarithmic Bloch space. Petrov in [8] obtains sharp reverse estimates for the logarithmic Bloch spaces on the unit disk. Castillo et al. in [9] characterized the boundedness and compactness of the composition operator from the logarithmic Bloch spaces into weighted Bloch spaces. García Ortiz and Ramos-Fernández in [10] characterized the boundedness and compactness of the composition operators from logarithmic Bloch spaces into Bloch-type spaces.

Let be a weight; that is, is a positive continuous function on . The Zygmund-type space consists of all such that With the norm , it becomes a Banach space. The little Zygmund-type space is a subspace of consisting of those such that When , the Zygmund-type space becomes the Zygmund space [11], while the little Zygmund-type space becomes the little Zygmund space .

Let be the differentiation operator; that is, . If , then the operator is defined by .

The weighted differentiation composition operator, denoted by , is defined as follows [12, 13]: where and is a nonconstant holomorphic self-map of .

If , then becomes the weighted composition operator , defined by which, for , is reduced to the composition operator for some recent articles on weighted composition operators on some -type spaces, for example, [1416] and references therein. If , , then , which was studied in [1721]. When , then , which was studied in [17, 19]. If , , then , that is, the product of differentiation operator and multiplication operator defined by . Zhu in [13] completely characterized the boundedness and compactness of linear operators which are obtained by taking products of differentiation, composition, and multiplication operators from Bergman type spaces to Bers spaces. Stević in [12] studied the boundedness and compactness of the weighted differentiation composition operator from mixed-norm spaces to weighted-type spaces or the little weighted-type space (see also [2224]). Zhu in [25] studied the boundedness and compactness of the generalized weighted composition operator on weighted Bergman spaces. Yang in [21] studied the boundedness and compactness of the operator and from to and spaces. Liu and Yu in [18] studied the boundedness and compactness of the operator between and Zygmund spaces. Ye and Zhou in [26] studied the boundedness and compactness of the weighted composition operators from Hardy to Zygmund type spaces. Stević in [27] studied the boundedness and compactness of the generalized composition operator from mixed-norm space to the Bloch-type space, the little Bloch-type space, the Zygmund space, and the little Zygmund space. For other recently introduced products of operators on spaces of holomorphic functions see [13, 16]. Motivated by the results [12, 18, 23, 24, 27], we consider the boundedness and compactness of the operators from the logarithmic Bloch spaces to the Zygmund-type spaces and the little Zygmund-type spaces. For the proof, we need different test functions and some complex calculations kills.

Throughout this paper, we will use the letter to denote a positive constant that can change its value at each occurrence.

2. Auxiliary Results

Here we prove and quote some auxiliary results which will be used in the proofs of the main results in this paper.

Lemma 1. Let be a positive integer. Suppose ; there exists a constant such that

Proof. We use induction on . Using the definition of the logarithmic Bloch spaces we have the case holds for . Assume the case holds; since let ; then we have , so By the Cauchy integral formula we obtain Note that we have for every . Hence the case holds. The desired result follows. The proof of this lemma is complete.

Lemma 2 (see [4, 28]). Let then .

The following criterion for the compactness is a useful tool and it follows from standard arguments (e.g., [29, Proposition 3.11] or [30, Lemma 2.10]).

Lemma 3. Let , and let n be a nonnegative integer, a holomorphic self-map of , and a weight. Then is compact if and only if is bounded and, for any bounded sequence in which converges to zero uniformly on compact subsets of as , we have as .

Lemma 4. A closed set in is compact if and only if is bounded and satisfies

The proof is similar to that of Lemma 1 in [31]; hence we omit it.

3. Boundedness and Compactness of from to Spaces

In this section, we study the boundedness and compactness of .

Theorem 5. Let , and let n be a nonnegative integer, a holomorphic self-map of , and a weight. Then the following statements are equivalent:(1) is bounded;(2) is bounded;(3)

Proof. . Suppose that (17), (18), and (19) hold. Then, for every and , by Lemma 1, we have On the other hand, we have Applying conditions (20) and (21), we deduce that the operator is bounded.
. This implication is clear.
. Assume that is bounded; that is, there exists a constant , such that for all . For , we have that Taking ; we have that By (23), (24), and the boundedness of the function , we get In the same way, taking , we have that By (23), (25), (26), and the boundedness of the function , we have that For a fixed , set We get that By Lemma 2 we have Hence, and .
On the other hand for each fix , by (30), we obtain that it follows that for each fix . From (29), we have and Hence By (33), we obtain that And from (23), we have Thus combining (35) with (34) we get the condition (17).
For a fixed , set It is easy to see that Using Lemma 2, we easily get that and with a direct calculation. From (37), we have , Hence From (39), we obtain that By (25), we have Thus combining (40) with (41) we get the condition (18).
Next, we prove (19). To see this, for a fixed , put It is easy to see that From Lemma 2 we obtain that and with a direct calculation. From (43), we have , Hence By (45), we obtain that By (27), we have Thus combining (47) with (46) we get the condition (19), finishing the proof of the theorem.

Theorem 6. Let , and let n be a nonnegative integer, a holomorphic self-map of , and a weight. Then the following statements are equivalent:(1) is compact;(2) is compact;(3) is bounded and

Proof. . Assume that is bounded and that conditions (48), (49), and (50) hold. For any bounded sequence in which converges to zero uniformly on compact subsets of . To establish the assertion, it suffices, in view of Lemma 3, to show that We assume that . From (48), (49), and (50) we have that, for any , there exists ; when , we have From the boundedness of by Theorem 5, we see that (23), (25), and (27) hold. Since uniformly on compact subsets of , Cauchy's estimate gives that , , and converge to uniformly on compact subsets of ; there exists a such that implies that From (52) and (53) and Lemma 1 we have when . It follows that the operator is compact.
. It is obvious.
. Assume that is compact. Then it is clear that is bounded. By Theorem 5 we get that is bounded. Let be a sequence in such that as . We can use the test functions Note that for . We see that converges to uniformly on ; hence, converges to uniformly on compact subsets of and from (30) and (33) we have , . Then is a bounded sequence in which converges to 0 uniformly on compact subsets of . By Lemma 3, we have Note that From (33) and using the compactness of we obtain From (59) and , it follows that and consequently (48) holds.
Next, let By a direct calculation, we obtain that converges to uniformly on compact subsets of , , and . By Lemma 3, we have Note that From (39) and using the compactness of we obtain From (64) and , we have it implies that (49) holds.
In order to prove (50), choose By a direct calculation, we may easily prove that converges to uniformly on compact subsets of , , and . By Lemma 3, we have Note that From (45) and using the compactness of we obtain From (69) and , it follows that and consequently (50) holds, finishing the proof of the theorem.

Theorem 7. Let , and let n be a nonnegative integer, a holomorphic self-map of , and a weight. Then is a bounded operator provided that the following conditions are satisfied:

Proof. Suppose that (71), (72), and (73) hold. It is clear that (17), (18), and (19) hold. By Theorem 5 we have that is bounded. In order to prove is bounded, it is enough to show that, for any , . Using (71), (72), and (73) we have that, for any , there is a constant , such that implies Then, for any , from Lemma 1 we obtain that when . Hence for all , completing the proof of the theorem.

Theorem 8. Let , and let n be a nonnegative integer, a holomorphic self-map of , and a weight. If is a bounded operator, then (17), (18), and (19) hold and the following conditions are satisfied:

Proof. Assume that is bounded; it is clear that is bounded. By Theorem 5 we have that (17), (18), and (19) hold. On the other hand, for all , . Take ; we have that then (76) holds. Let ; we have that By (80), (76), and the boundedness of the function , we get Hence, (77) holds. In the same way, let ; we have that By (76), (77), (82), and the boundedness of the function , we have that That is, (78) holds. The proof is completed.

Theorem 9. Let , and let n be a nonnegative integer, a holomorphic self-map of , and a weight. Then is a bounded operator if and only if is a bounded operator and (76), (77), and (78) hold.

Proof. Assume that is a bounded operator; it is clear that is a bounded operator. On the other hand, for all , . Taking , we have that then (76) holds. Let ; we have that By (85), (76), and the boundedness of the function , we get Hence, (77) holds. In the same way, take ; we have that By (76), (77), (87), and the boundedness of the function , we have that That is, (78) holds.
Conversely, suppose that is a bounded operator and (76), (77), and (78) hold. For each polynomial we get that Hence, . On the other hand, since polynomials are dense in , thus, for each , there is a sequence of polynomials such that Since is a bounded operator, by Theorem 5 we have is a bounded operator. Since and is the closed subset of , we see that ; thus is a bounded operator. The proof is completed.

Theorem 10. Let , and let n be a nonnegative integer, a holomorphic self-map of , and a weight. Then the following statements are equivalent:(1) is compact;(2) is compact;(3)(71), (72), and (73) hold.

Proof. . Suppose that (71) (72), and (73) hold. By Theorem 7 we know that is bounded. Taking the supremum in inequality (20) over all such that and letting yields Hence, by Lemma 4, we see that the operator is compact.
. This implication is clear.
. Assume that is compact. Firstly, it is obvious is bounded. By Theorem 9 we have that (76), (77), and (78) hold. On the other hand, we have that is compact. By Theorem 6 we have that (48), (49), and (50) hold. We prove that (76) and (48) imply (71). The proof of (72) and (73) is similar; hence, it will be omitted. From (48), it follows that, for every , there exists such that when . Using (76) we see that there exists such that when . Therefore when and , by (93), we have On the other hand, when and , by (94), we obtain From (95) and (96) we have we obtain that (71) holds, as desired. The proof is completed.

Conflict of Interests

The authors declare that there is no conflict of interests regarding the publication of this paper.

Acknowledgments

The authors thank the referees for their thoughtful comments and helpful suggestions. This research was supported by the Natural Science Foundation of China (11171285) and the Priority Academic Program Development of Jiangsu Higher Education Institutions.