Abstract

We characterized the boundedness and compactness of weighted differentiation composition operators from BMOA and the Bloch space to Bloch-type spaces. Moreover, we obtain new characterizations of boundedness and compactness of weighted differentiation composition operators.

1. Introduction

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

Let . The -Bloch space on is the space of all holomorphic functions on such that The little -Bloch space consists of all such that Both spaces and are Banach spaces with the norm and is a closed subspace of . If , they become the classical Bloch space and little Bloch space , respectively. For any , the space consists of functions such that For information of such spaces, see, for example, [14].

For , let be the automorphism of that interchanges and . Let the Green function in with logarithmic singularity at be given by The space consists of all in the Hardy space such that is a Banach space under following norm (see, e.g., [5]):

Let and be holomorphic maps on the open unit disk such that . For a nonnegative integer , we define a linear operator as follows: We call it weighted differentiation composition operators, which was defined in [6, 7]. If and , becomes induced by , defined as , . If and , then is the differentiation operator defined as . If , then we get the weighted composition operator defined as . If and , then reduces to . When , then reduces to differentiation composition operator (also named as product of differentiation and composition operator). If we put , then , the product of multiplication and differentiation operator.

The boundedness and compactness of differentiation composition operator between spaces of holomorphic functions have been studied extensively. For example, Hibschweiler; Portnoy and Ohno studied differentiation composition operator on Hardy and Bergman spaces in [8, 9]; Li; Stević and Ohno studied on Bloch type spaces in [1012]; Wu and Wulan gave a new compactness criterion of on the Bloch space in [13]. Recently, the weighted differentiation composition operator between different function spaces has also been investigated by several authors (see, for example, [1421]).

Boundedness, compactness, and essential norm of weighted composition operator between Bloch-type spaces have been studied in [2224]. Recently, Manhas and Zhao [25] and Hyvärinen and Lindström [26] gave a new characterization of boundedness and compactness of in terms of the norm of (for the compactness of composition operator, see [27, 28]).

Motivated by [13, 25, 26], we study the operator from and Bloch space to Bloch-type spaces.

Throughout this paper, constants are denoted by ; they are positive and not necessarily the same at each occurrence. The notation means that there is a positive constant such that . When and , we write .

2. Some Lemmas

It is well known that . From the definition of the norm, we know Indeed, Girela proved that in Corollary 5.2 of [5]. The following lemma is from Lemma 5 in [29] (see also Lemma 4.12 of [4]).

Lemma 1. If , then

The following lemma may be known, but we fail to find its reference; so we give a proof for the completeness of the paper.

Lemma 2. Let . Then,

Proof. Applying Littlewood-Paley identity and Lemma 1, we have It follows from the definitions of Bloch space and space that

By Theorem 6.2 of [5] and the proof of Theorem 1 of [30], we have the following lemma.

Lemma 3. Let be a fixed positive integer and with . If then .

Lemma 4. Suppose that is a fixed positive integer. Let , , and If , then there are two positive constants and , depending only on , such that

Proof. The proof is similar to that of Lemma 2.2 of [13] and is so omitted.

3. Boundedness of

In this section, we characterize the boundedness of from and the Bloch space to Bloch-type spaces.

Theorem 5. Let , , , and a holomorphic self-map of . Then, the following statements are equivalent: (a) is bounded. (b) and are bounded. (c) is bounded. (d) and are bounded. (e) is bounded. (f) and are bounded. (g)  and .(h) and .

Proof. It is obvious that , , , and . Thus, we will prove the theorem according to the following steps. (I): , . (II): , . (III): , . (IV): .
(I): , . Suppose that or holds. We choose the test function . By Lemma 2, we get So Taking and using the fact that , we have We now consider the function It is easy to check that and . Moreover, Thus, and We obtain Thus, for any , we have Using (21) yields Combining (26) with (27), we get We next consider the function Similarly, we get and Moreover, So and . We have, as above,
Thus, for any , Applying (20), we get Combining (34) with (35) yields (II): and . Suppose that is bounded or is bounded. Set If , then for any positive integer , we can find such that If , then choose the test function . It is clear that . From Lemma 2, we have
So If , consider the function where . Let . Then, and It is easy to see that So, by Theorems 5.4 and 5.13 of [4], we have and . By Lemma 1 of [31] and Lemma 3, we get . We have Since is arbitrary, we get . This contradicts the boundedness of and that of .
Now, suppose that is bounded or is bounded. Set If , then for any positive integer , exists such that If , then set . The process as above gives If , consider the function where . Let . Then, and Applying Theorems 5.4 and 5.13 of [4] again yields and . We get and Since is arbitrary, we have . This contradicts the boundedness of .
(III): , . Note that The desired results follow.
(IV): . Suppose that is true. It follows from Proposition 5.1 of [4] that . So,
Conversely, assume that is true. It is easy to see that If , then Hence, is true. From , we obtain that is also true.
From now on, we assume that . For any integer , let Let with be the smallest positive integer such that . Since is not empty for every integer and . By Lemma 4, for , So, is bounded. Similar argument implies Thus, is bounded. Theorem 5 is proved.

4. Compactness of

The following criterion for the compactness is a useful tool and it follows from standard arguments, for example, Proposition 3.11 of [32] or Lemma 2.10 of [33].

Lemma 6. Let , , and , or . Suppose that and are in such that . Then, is compact if and only if for any sequence in with , which converges to zero locally uniformly on ; we have .

We now give the compactness of from and the Bloch space to Bloch-type spaces.

Theorem 7. Let , , , and a holomorphic self-map of . Then, the following statements are equivalent: (a) is compact. (b) is compact and is compact. (c) is compact. (d) is compact and is compact. (e) is compact. (f) is compact and is compact. (g), , (h) and .

Proof. The proof is a modification of that of Theorem 5; so we give a sketch of the proof. We will prove the theorem according to the following steps. (I): , . (II): , . (III): , . (IV): .
(I): , . Suppose that or holds. Then by Theorem 5, we have That is, , .
Let be a sequence in such that as . Now, we consider the function Simple computation shows that and It is also easy to check that uniformly on compact subsets of as . Moreover, We have By Lemma 6, we get We next consider the function Similarly, we get and It is easy to see that converges to zero uniformly on compact subsets of as and Thus, Applying Lemma 6 again, we have Since is arbitrary, we proved that is true.
(II) , . Suppose that or holds. A similar argument to (I) shows that , . Now, suppose that the equations in are not true. Then, there exists a sequence in and such that as and Choose a subsequence of if necessary and suppose that . Let Then, it is easy to check that , , uniformly on compact subsets of and Thus, Those contradict the compactness of and .
(III) , . Let be a norm bounded sequence in that converges to zero uniformly on compact subsets of . Let . For , then there exists such that for , we have Thus, for , we have where and . Since uniformly on compact subsets of as , we have as . It follows from Lemma 6 that is compact.
Similar as above, we know From uniformly on compact subsets of , we have and as . So, , are compact.
(IV): . Suppose that is true. Note that and uniformly on compact subsets of as ; by Lemma 6, we have Conversely, assume that is true. It is easy to see that If , from , we get that is true. If , as in the proof of Theorem 5, let And let with be the smallest positive integer such that . For given , there exists a large enough integer with such that whenever . Let be a norm bounded sequence in that converges to zero uniformly on compact subsets of as . Denote . We get Then, where Since uniformly on compact subsets of , then as . Thus, by Lemma 6, is compact. Similar as above, we can prove that is compact. The proof is complete.

Acknowledgments

This work was supported by NNSF of China (Grant no. 11171203) and NSF of Guangdong Province (Grant nos. 10151503101000025 and S2011010004511).