Research Article | Open Access
Xiangling Zhu, Nanhui Hu, "Weighted Composition Operators from Besov Zygmund-Type Spaces into Zygmund-Type Spaces", Journal of Function Spaces, vol. 2020, Article ID 2384971, 7 pages, 2020. https://doi.org/10.1155/2020/2384971
Weighted Composition Operators from Besov Zygmund-Type Spaces into Zygmund-Type Spaces
The boundedness, compactness, and essential norm of weighted composition operators from Besov Zygmund-type spaces into Zygmund-type spaces are investigated in this paper.
Let denote the open unit disk in the complex plane and the space of all analytic functions in . For an analytic self-map of and , the weighted composition operator is defined as follows:
When , is just the composition operator, denoted by . In the past several decades, composition operators and weighted composition operators have received much attention and appear in various settings in the literature (see, for example, [2–5, 8, 10, 13, 15, 16, 19]).
Let . The Bloch type space consists of those functions for which
is a Banach space under the above norm. It is known that when , is the classical Bloch space.
For , an is said to be in the Zygmund-type space , if
It is easy to check that is a Banach space under the norm . When , is the Zygmund space. When , is just the Bloch type space . In particular, when , is just the Bloch space . Hence, the Zygmund space is the space of all such that with norm
Let be the normalized area measure on . For , the Besov space, denoted by , is the space of all such that
This space is a Banach space with the following norm In particular, is the classical Dirichlet space. Besov spaces are Möbius invariant in the sense that for all and , the set of all Möbius maps of (see [1, 19]).
In , Colonna and Tjani introduced a new class type space , called the Besov Zygmund-type space, which consists of all such that . Since the Besov space is contained in the Bloch space, it follows that the Besov Zygmund-type space is a subset of the Zygmund space, and hence, it is contained in the disk algebra.
Colonna and Li studied the boundedness and compactness of the operator and in [2, 3], respectively. Here, is the space of bounded analytic functions. (See [2, 3, 6–14, 16, 17] for more results of composition operators, weighted composition operators, and related operators on the Zygmund space and Zygmund-type spaces.) Colonna and Tjani characterized the boundedness and compactness of in .
In this work, we give some characterizations for the boundedness, compactness, and the essential norm of the operator .
Throughout the paper, we denote by a positive constant which may differ from one occurrence to the next. In addition, we say that if there exists a constant such that . The symbol means that .
2. Main Results and Proofs
In this section, we formulate and prove our main results in this paper. For this purpose, we need the following lemmas.
Lemma 1. Suppose . Then, there exists a positive constant such that for every .
Proof. For , it is well known that Then, the inequalities in (6) follow from the definition of the Besov Zygmund-type space. Since the Zygmund space is continuously embedded into , as shown in Lemma 2.1 of , we get that . The proof is complete.
Lemma 2 (see ). Let . Every sequence in bounded in norm has a subsequence which converges uniformly in to a function in .
Lemma 3 (see ). Let be a Banach space that is continuously contained in the disk algebra, and let be any Banach space of analytic functions on . Suppose that (i)the point evaluation functionals on are continuous(ii)for every sequence in the unit ball of that exists and a subsequence such that uniformly on (iii)the operator is continuous if has the supremum norm and is given the topology of uniform convergence on compact setsThen, is a compact operator if and only if given a bounded sequence in such that uniformly on , then the sequence as .
Lemma 4. Let and . If is bounded, then is compact if and only if as for any sequence in bounded in norm which converge to uniformly in .
The following estimates are fundamental in operator theory and function spaces on the unit disk (see (, Lemma 3.10)).
Lemma 5 (see ). Suppose that is real, and (i)If then as a function of is bounded on (ii)If then (iii)If then Now, we are in a position to give the following characterization of bound composition operators from to .
Theorem 6. Let , , , and be an analytic self-map of . Then, is bounded if and only if ,
Proof. First, suppose that , (11) and (12) hold. For arbitrary and , by Lemma 1, we have
Therefore, is bounded.
Conversely, suppose that is bounded. Applying the operator to with , and using the boundedness of , we get that , , and . Hence, For such that , set Then, Thus, for , we have and by Lemma 5, So, Therefore, by the boundedness of , we get Since , for any such that , we have which implies that By (14), we get From (23) and (24), we see that (11) holds.
For , define So, By Lemma 5, we see that and . By the boundedness of , we get After a calculation, Hence, for any , On the one hand, from (28), we obtain On the other hand, by (15), we get From (29) and (30), we see that (12) holds. The proof is complete.
Next, we estimate the essential norm of . Recall that the essential norm of is defined as the distance from to the set of compact operators , that is,
Theorem 7. Let , , , and be an analytic self-map of such that is bounded. Then, Here,
Proof. First we prove that
Let be a sequence in such that as . Define
From the proof of Theorem 6, we see that and belong to . Moreover, and converge to uniformly on as . Hence, for any compact operator , by Lemma 4, we obtain
Here, we used the fact that since and converges to uniformly on as .
Therefore, we get as desired.
Next, we prove that Let . Define by It is clear that is compact on and ; moreover, uniformly on compact subsets of as . Let such that as . Then, for each , is compact. Hence, Thus, we only need to show that For any with , from the facts that we have where is large enough such that for all . Since as , by Lemma 2, we have Since and uniformly on compact subsets of as , we have Now, we estimate . Using Lemma 1 and the fact that , we have Taking the limit as , we get Similarly, again by Lemma 1, Taking the limit as , we get Hence, by (43), (44), (45), (46), (48), and (50), we get which with (40) implies the desired result. The proof is complete.
From Theorem 7 and the result that if and only if is compact, we get the following corollary.
Corollary 8. Let , , , and be an analytic self-map of such that is bounded. Then, is compact if and only if
No data were used to support this study.
Conflicts of Interest
The authors declare that they have no conflicts of interest.
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