Abstract and Applied Analysis

Volume 2009 (2009), Article ID 290978, 14 pages

http://dx.doi.org/10.1155/2009/290978

## Weighted Composition Operators from Spaces to Spaces

Department of Mathematics, JiaYing University, Meizhou, GuangDong 514015, China

Received 25 January 2009; Accepted 5 February 2009

Academic Editor: Stevo Stevic

Copyright © 2009 Xiangling Zhu. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

#### Abstract

Let denote the space of all holomorphic functions on the unit ball . Let and be a holomorphic self-map of . In this paper, we investigate the boundedness and compactness of the weighted composition operator from the general function space to the weighted-type space in the unit ball.

#### 1. Introduction

Let be the unit ball of . Let and let be points in , we writeThus . Let be the normalized Lebesgue measure of , that is, Let be the space of all holomorphic functions on . For , letrepresent the radial derivative of . For , , let denote the Möbius transformation of taking to , which is defined bywhere is the orthogonal projection of onto the one dimensional subspace of spanned by , and .

A positive continuous function on is called normal, if there exist positive numbers and and such that (see [1]).

Let be a normal function on . An is said to belong to the weighted-type space , ifwhere is normal on (see, e.g., [2–4]). is a Banach space with the norm . We denote by the subspace of consisting of those such thatWhen , the induced spaces and become the (classical) weighted space and , respectively.

For , recall that the -Bloch space is the space of all for which (see [5])Under the norm , is a Banach space. When , we get the classical Bloch space . For more information of the Bloch space and the -Bloch space (see, e.g., [5–8] and the references therein).

For , the weighted Bergman space is the space of all holomorphic functions on for whichThe Hardy space on the unit ball is defined bywhereand is the normalized surface measure on

For , the space is defined by (see [9])Here denotes the invariant gradient of , and is the invariant Green function defined by , where

Let , . A function is said to belong to (see [10–12]) if is called the general function space since we can get many function spaces, such as Hardy space, Bergman space, Bloch space, space, if we take special parameters of . For example, , , and . If , then is the space of constant functions. For the setting of the unit disk, see [13].

Let and be a holomorphic self-map of . For , the weighted composition operator is defined byThe weighted composition operator is the generalization of a multiplication operator and a composition operator, which is defined by The main subject in the study of composition operators is to describe operator theoretic properties of in terms of function theoretic properties of . The book [14] is a good reference for the theory of composition operators. Recall that a linear operator is said to be bounded if the image of a bounded set is a bounded set, while a linear operator is compact if it takes bounded sets to sets with compact closure.

In the setting of the unit ball, we studied the boundedness and compactness of the weighted composition operator between Bergman-type spaces and in [15]. More general results can be found in [16]. Some necessary and sufficient conditions for the weighted composition operator to be bounded or compact between the Bloch space and are given in [17]. In the setting of the unit polydisk , some necessary and sufficient conditions for a weighted composition operator to be bounded or compact between the Bloch space and are given in [18, 19] (see, also [20] for the case of composition operators). Some related results can be found, for example, in [2, 3, 6, 21–31].

In the present paper, we are mainly concerned about the boundedness and compactness of the weighted composition operator from to the space . Some necessary and sufficient conditions for the weighted composition operator to be bounded and compact are given.

Constants are denoted by in this paper, they are positive and may differ from one occurrence to the other. means that there is a positive constant such that . Moreover, if both and hold, then one says that .

#### 2. Main Results and Proofs

In order to prove our results, we need some auxiliary results which are incorporated in the following lemmas.

Lemma 2.1 (see [12]). * For , , , if , then
and *

The following lemma can be found, for example, in [32].

Lemma 2.2. *
If , then **for some independent of .*

Lemma 2.3. *Assume that* is normal. A
closed set *in**is compact if
and only if it is bounded and satisfies*

The proof of Lemma 2.3 is similar to the proof of Lemma 1 of [33]. We omit the details.

Lemma 2.4. *Assume that*, *is a
holomorphic self-map of*,
is a normal
function on , , , . 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*, one has
*as*

Lemma 2.4 follows by standard arguments similar to those outlined in [14, Proposition 3.11] (see, also, the corresponding lemmas in [20, 34, 35]). We omit the details.

Note that when , the function in is indeed Lipschitz continuous by Lemmas 2.1 and 2.2. By Arzela-Ascoli theorem, similarly to the proof of Lemma 3.6 of [26], we have the following result.

Lemma 2.5. *Let*, , and . Let *be a bounded
sequence in*
which converges
to 0 uniformly on compact subsets of , then

#### 2.1. Case

In this subsection, we consider the case . Our first result is the following theorem.

Theorem 2.6. *Assume that*, *is a
holomorphic self-map of*, *is a normal
function on*, , , , *. Then**is bounded if
and only if**Moreover, when is bounded, the
following relationship**holds.*

*Proof. * Assume that (2.5) holds. For any , by Lemmas 2.1 and 2.2,Therefore, (2.5) implies that is bounded.

Conversely, suppose that is bounded. For , letIt is easy to see thatIf , then obviously
belongs to . From [12], we know that ; moreover, there is a positive constant such that . Therefore,for every , from which we getthat is, (2.5) follows. From (2.7)
and (2.11), we see that (2.6) holds. The proof of this theorem is finished.

Theorem 2.7. *
Assume that , is a
holomorphic self-map of ,
is a normal
function on , , , , . Then
is compact if
and only if
and *

*Proof. * First assume that and the
condition in (2.12) hold. In order to
prove that is compact,
according to Lemma 2.4 it suffices to show that if is bounded in and converges
to uniformly on
compact subsets of as , then as .

Now assume that is a sequence
in such that and uniformly on
compact subsets of as From (2.12), we
have that for every , there is a constant such thatwhen . By Lemmas 2.1 and 2.2, we havewhere . Using the fact that uniformly on
compact subsets of as , we obtainTherefore,Since is an arbitrary
positive number, we have that and, therefore, is compact by
Lemma 2.4.

Conversely, suppose is compact. Let be a sequence
in such that as (if such a
sequence does not exist that condition (2.12) is vacuously satisfied). SetFrom the proof of Theorem 2.6, we
see that for every moreover, . Beside this, converges to 0
uniformly on compact subsets of as . Since is compact, by
Lemma 2.4 we have that as Thusas , from which we obtain (2.12), finishing the proof of the theorem.

Theorem 2.8. *Assume that*, *is a
holomorphic self-map of*, *is a normal
function on*, , , , *. Then**is compact if
and only if*

*Proof. * Suppose that (2.19) holds. From Lemma 2.3, we see that is compact if
and only ifOn the other hand, by Lemmas 2.1
and 2.2, we have thatTaking the supremum in (2.21) over
the the unit ball in the space then letting and applying
(2.19) the result follows.

Conversely, suppose that is compact.
Then is compact and
hence by Theorem 2.7,In addition, is bounded.
Taking , then employing the boundedness of , we getBy (2.22), for every , there exists a ,when By (2.23), for
above chosen , there exists ,when .

Therefore, when and , we have thatIf and , we obtainCombing (2.26) with (2.27), we get
(2.19), as desired.

#### 2.2. Case

Theorem 2.9. *Assume that*, *is a
holomorphic self-map of*, *is a normal
function on*, , , , *. Then**is bounded if
and only if**Moreover, the following
relationship**holds.*

*Proof. * Suppose that (2.28) holds. For any , by Lemmas 2.1 and 2.2, we haveTherefore, (2.28) implies that is a bounded
operator from to .

Conversely, suppose that is bounded. For , letThen by [12] we see that ; moreover, there is a positive constant such that . Hencefor every , that is, we getFrom (2.30) and (2.33), (2.29)
follows. The proof is finished.

Theorem 2.10. *Assume that*, *is a
holomorphic self-map of*, *is a normal
function on*, , , , *. Then**is compact if
and only if**and*

*Proof. * Assume that and
(2.34) hold, and that is bounded in and converges
to 0 uniformly on compact subsets of as . We have that, for every , there is a such thatwhen .

In additionSimilar to the proof of Theorem
2.7, we obtain as Therefore, is compact by
Lemma 2.4.

Conversely, suppose that is compact.
Assume that is a sequence
in such that as (if such a
sequence does not exist that condition (2.34) is vacuously satisfied). SetAfter some calculations or from
[12], we see that for some
positive independent of and converges to 0
uniformly on compact subsets of as . Since is compact, by
Lemma 2.4, we have as . Thusas , which is equivalent to (2.34). The proof of this
theorem is finished.

Similarly to the proof of Theorem 2.8, we can obtain the following results. We omit the details of the proof.

Theorem 2.11. *Assume that*, *is a
holomorphic self-map of*, *is a normal
function on*, , , , *. Then**is compact if
and only if*

#### 2.3. Case

Theorem 2.12. * Assume that, is a
holomorphic self-map of, is a normal
function on, , , , . Then the following statements are equivalent:*(i)

*is bounded;*(ii)

*is compact;*(iii)

*.*

*Proof. * (ii)(i) This implication is obvious.

(i)(iii) Taking , then using the boundedness of the implication
follows.

(iii)(ii) Suppose that . For an , by Lemma 2.1, we see that is continuous
on the closed unit ball and so is bounded in . Therefore,From the above inequality, we
see that is bounded. Let be any bounded
sequence in and uniformly on
compact subsets of as . Employing Lemma 2.5, we haveas . Then the result follows from Lemma 2.3.

Theorem 2.13. * Assume that, is a
holomorphic self-map of, is a normal
function on, , , , . Then the following statements are equivalent:*(i)

*is bounded;*(ii)

*is compact;*(iii)

*.*

*Proof. * (ii)(i) It is obvious.

(i)(iii) Taking , then using the boundedness of , we get the desired result.

(iii)(ii) Suppose that . For any with , we havefrom which we obtainUsing Lemma 2.3, we see that is compact, as
desired.

#### Acknowledgment

The author of this paper is supported by Educational Commission of Guangdong Province, China.

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