Abstract

The boundedness, compactness, and essential norm of weighted composition operators from Dirichlet-Zygmund spaces into Zygmund-type spaces and Bloch-type spaces are investigated in this paper.

1. Introduction

Let denote the space of all analytic functions in the open unit disk . For , the Dirichlet type space is the set of all such that where is the normalized Lebesgue area measure. is a Banach space under the norm . If , we say that belongs to the Dirichlet-Zygmund space, denoted by . To the best of our knowledge, this is the first work to study the Dirichlet-Zygmund space.

Recall that the space , called the minimal Möbius invariant space, is the space of all that admit the representation for some sequence in and . The norm on is defined by

Here, For any , the authors in [1] showed that there exists a constant such that

Therefore, is in fact the space .

We call a weight, if is a continuous, strictly positive and bounded function. is called radial, if for all . Let be a radial weight. Recall that the Zygmund-type space is the space that consists of all such that

is a Banach space under the norm . We say that belongs to the Bloch-type space , if When , is called the Zygmund space, and is called the Bloch space, respectively. In particular, is just the Bloch space when .

The weighted space, denoted by , is the set of all such that When , we denote by . In particular, when , is just the bounded analytic function space.

We denote by the set of all analytic self-maps of for simplicity. Let and . The weighted composition operator is defined as follows. When , is called the composition operator, denoted by . See [2, 3] for more results about the theory of composition operators and weighted composition operators.

For any , by the Schwarz-Pick lemma, we see that is bounded. It was shown in [4] that is compact if and only if . Motivated by [4], Colonna and Li in [5, 6] studied the operators and by and , respectively. Here, is the Lipschtiz space. The composition operator on the space was extensively studied in [1]. In [7], Colonna and Li studied the boundedness and compactness of weighted composition operators from the minimal Möbius invariant space to the Bloch space . In [8], Li studied the boundedness and compactness of the weighted composition operator . See [5, 6, 817] for more results for composition operators, weighted composition operators, and related operators on the Zygmund space and Zygmund-type spaces.

In this paper, we follow the methods of [17] and give some characterizations for the boundedness, compactness, and essential norm of the operator and .

We denoted by a positive constant which may differ from one occurrence to the next. In addition, we will use the following notations throughout this paper: means that there exists a constant such that , while means that .

2. Main Results and Proofs

In this section, we formulate and prove our main results in this paper.

Lemma 1. Suppose . Then, there exists a positive constant such that and for every .

Proof. Suppose and . Then, there exists a constant such that which implies that The inequalities in (8) hold. Here, is the hyperbolic disk (see [3]). From (8), we see that are contained in the disk algebra for . Hence, we get that .

Lemma 2. Let . If , then for all and , there exists a positive constant such that

Proof. Fix . Let and . By Lemma 1, as desired.

Using Lemma 2 and similarly to the proof of Lemma 7 in [18], we get the following lemma.

Lemma 3. Let . Every sequence in bounded in norm has a subsequence which converges uniformly in to a function in .

Lemma 4 (see [5]). 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 an and a subsequence such that uniformly on (iii)The operator is continuous if has the supremum norm and is given by 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 .

The following result is a direct consequence of Lemmas 3 and 4.

Lemma 5. Let and be a weight. If is bounded, then is compact if and only if as for any sequence in bounded in norm which converge to uniformly in .

Theorem 6. Let be a radial, nonincreasing weight tending to zero at the boundary of . Let , , and . Then, the following statements are equivalent. (i)The operator is bounded(ii)and (iii) and

Proof. . For any and , by Lemma 1, we have Hence, Therefore, is bounded.
. Applying the operator to with and using the boundedness of , we get that , , and . Hence, we obtain For any , set It is easy to check that Therefore, by the boundedness of and arbitrary of , we get For , we get From (21) and (22), we obtain From (24), we get On one hand, from (25), we obtain On the other hand, from the fact that , we get From (26) and (27), we see that is finite. Using similar arguments, we see that is also finite.
. From [19], we see that the inequality in is equivalent to the operator is bounded. By [20], the boundedness of is equivalent to From [21], we get , which together with (28) imply that Similarly, the inequality in is equivalent to The proof is complete.

Next, we consider the essential norm of . Recall that the essential norm of is its distance to the set of compact operators , that is,

Here, are Banach spaces, and is a bounded linear operator.

Theorem 7. Let be a radial, nonincreasing weight tending to zero at the boundary of . Let , , and . Suppose that is bounded. Then,

Here,

Proof. First we show that Let be a sequence in the unit disk such that as . Define After a calculation, we get all and belong to and Moreover, and converge to uniformly on as . Hence, for any compact operator , by Lemma 5, we get Hence, as desired.
Next, we show that Let . Define by It is clear that is compact on and . Moreover, uniformly on compact subsets of as . Let such that as . Then, is compact for each . Hence, Thus, we only need to prove that For any with , by the facts that we have where is large enough such that for all . Since , , and uniformly on compact subsets of as , by Lemma 3, we obtain Using Lemma 1 and , we obtain Taking the limit as , we get Similarly, Taking the limit as , we get Hence, by (43), (44), (45), (46), (48), and (50), we get which with (40) implies the desired result.
Finally, we prove that On one hand, by the proof of Theorem 6, we see that the boundedness of is equivalent to the boundedness of and . From [19, 20], we have Hence, On the other hand, from [19, 21], we have