Abstract and Applied Analysis

Volume 2014, Article ID 892146, 8 pages

http://dx.doi.org/10.1155/2014/892146

## Composition Operators from Certain -Bloch Spaces to Spaces

School of Mathematics and Statistics, Wuhan University, Wuhan 430072, China

Received 26 August 2014; Revised 4 December 2014; Accepted 4 December 2014; Published 22 December 2014

Academic Editor: Narcisa C. Apreutesei

Copyright © 2014 Chunyu Tan and Maofa Wang. 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

Some necessary and sufficient conditions are established for composition operators to be bounded or compact from *μ*-Bloch type spaces to spaces. Moreover, the boundedness, compactness, and Fredholmness of composition operators on little spaces are also characterized.

#### 1. Introduction

Let be the unit disc in the complex plane and the space of all analytic functions on with the topology of uniform convergence on compact subsets of . If , we let , , be the dilation of . The space consists of all functions satisfying . The Bloch space consists of all functions for which equipped with the norm becomes a Banach space (see [1, 2]). For , the -Bloch space consists of all analytic functions on such that Refer to [3] for more details on -Bloch spaces. Recently, many authors have studied different classes of Bloch type spaces, where the typical weight function is replaced by a continuous positive function defined on . More precisely, let be a radial weight function; that is, , , which is decreasing in a neighborhood of , continuous and . The Bloch type space consists of all such that It is easy to check that is a norm on , and is a Banach space equipped with this norm (see, e.g., [4]). Clearly, includes as its special case. Indeed, if with , becomes -Bloch space . When , is just the classical Bloch space . For , is logarithmic Bloch space, which first appeared in characterizing the multipliers of the Bloch spaces. The little Bloch-type space consists of all such that

For , let be the involutive automorphism of the unit disc which interchanges and . We recall that in [5], for , belongs to if is the subclass of consisting of all such that With the norm, is a Banach space, and is the closure of all polynomials in . It is well known that , the space of all analytic functions of the bounded mean oscillation on . is the classical Dirichlet space . For all , is the Bloch space . Also, , the subspace of BMOA consisting of all analytic functions with vanishing mean oscillation, and for , ; see [5, 6] for more details on those spaces.

Let and be two linear subspaces of . If is an analytic self-map of , then induces a composition operator defined by Composition operators have been studied by numerous authors in many subspaces of . Among others, Madigan and Matheson characterized the continuity and compactness of composition operators on the classical Bloch space in [7]. Lou studied composition operators on spaces in [8]. Composition operators between the logarithmic Bloch-type space and are studied in [9–11].

This paper studies composition operators from -Bloch type spaces to spaces. After some necessary background materials, Section 2 gives some function-theoretic characterizations of bounded and compact composition operators by using the Hadamard gap series technique. Section 3 characterizes the continuity, compactness of , and the Fredholmness of on .

Throughout the paper we use the same letter to denote various positive constants which may change at each occurrence. Variables indicating the dependency of constants will be often specified in the parenthesis. We use the notation or for nonnegative quantities and to mean for some inessential constant . Similarly, we use the notation if both and hold.

#### 2. Composition Operators from to

We recall that an analytic function on the unit disk has Hadamard gaps if with for all . The following results are cited from [12].

Theorem A. *Assume that . Then
*

*Theorem B. Assume that is a nonincreasing radial weight satisfying
and such that is a positive nonincreasing absolutely continuous function on the interval satisfying and , where is a sequence such that , . Let . If
then .*

*In the sequel, we always suppose that is as in Theorem B. The next lemma will play a key role in our main results.*

*Lemma 1. There exist two functions such that
*

*Proof. *We consider the function:
where is an appropriately large integer, and is sufficiently small. It follows from Theorem B that . We claim that
for , . Indeed,
For large enough, since
then
Hence
On the other hand, for large enough,
It follows from (19) and (20) that
Since for sufficient large ,
for . That is (15).

Now with a similar argument for , and large enough, we have
where
Now inequality (13) follows immediately from (15) and (23) on the annulus .

On the other hand, in the disc , we have that , , and and have a finite number of zeros in the disc. Hence if and have common zeros in the disc , then one can replace by the function for an appropriate and obtain a pair of functions which satisfy inequality (13).

*We now characterize the boundedness of the composition operator .*

*Theorem 2. Let and be an analytic self-map of the unit disc. Then is bounded if and only if
*

*Proof. *Assume that is bounded; then , for . By Lemma 1, there exist such that . So
which implies (25).

Conversely, for any , it is clear to see that
By (25), . Then is bounded by the closed graph theorem.

*Now, we are going to characterize the compactness of composition operators . In [13], Tjani showed the following result.*

*Lemma 3. Let , be two Banach spaces of analytic functions on . Suppose the following:(1)The point evaluation functions on are continuous.(2)The closed unit ball of is compact subset of in the topology of uniform convergence on compact sets.(3) is continuous when and are given the topology of uniform convergence on compact sets.Then, is a compact operator if and only if, given a bounded sequence in such that uniformly on compact sets, the sequence converges to zero in the norm of .*

*Observe that for any fixed we have
so the point evaluation functionals on are continuous. Thus, as a consequence of Lemma 3, we have the following result.*

*Lemma 4. The composition operator is compact if and only if for every bounded sequence , which converges uniformly to zero on any compact subset of the unit disk, as .*

*We now use Lemma 4 above to give a characterization of compact composition operator .*

*Theorem 5. Let and be an analytic self-map of the unit disc. Then is compact if and only if and
*

*Proof. *We first assume that is compact; then . Since and as , uniformly on any compact subsets of the unit disk, then by Lemma 4, as . So for each and each , there exists such that
If we choose , then
We now consider the functions and for with . Since and uniformly to on any compact subsets of the unit disk, for there exists such that, for all ,
Note that, for ,
Namely, for each and , there is and some constant depending only on such that, for ,
Since is compact, it maps the unit ball of into a relative compact subset of . Thus for each , there exists a finite collection of functions in the unit ball of , such that for each there is a with
If we take , then for
Then
which implies the desired estimate (29) by using Lemma 1 in a similar way as in the proof of Theorem 2.

Conversely, we assume that and (29) holds. Let be a sequence of functions in the unit ball of , such that uniformly on the compact subsets of the unit disc as . We notice that, for ,
Since uniformly on the compact subsets of the unit disc, as , then as uniformly on the compact subsets of the unit disc. So for every , there is such that, for each , , and . Also notice that
By (29) there exists such that, for every , . Thus as , which completes the proof by Lemma 4.

*The following corollary is an immediate result of Theorems 2 and 5.*

*Corollary 6. Let . Then(1) is embedded boundedly into if and only if
(2) is embedded compactly into if and only if
*

*3. Composition Operators from to *

*3. Composition Operators from to**In this section, we investigate composition operators from to . Contrast with the case , here the boundedness and compactness of are equivalent. Last, we also characterize the Fredholmness of composition operators on . We first begin with the following.*

*Lemma 7. Suppose that and is an analytic self-map of . Then is compact if and only if
*

*Proof. *Suppose that is compact; then is relatively compact in , where is the unit ball of . Let ; then there is an -net of . Then for any fixed , there exists such that
Clearly, there is such that
for . So
for . So (42) is proved.

Conversely, suppose that (42) holds and with , converging uniformly to on compact subsets of ; we now prove
For any given , by (42), there is such that, for all ,
that is, . For , and , set
and then , which means that, for each , there exists such that for all . The same as in the proof of Lemma 1.3 in [14], is a continuous function of , so there is a neighbourhood of such that for all . Since and is compact, there exist such that . For , , there exists such that , , . Setting , for all . That is,
On the other hand, since converge to uniformly on compact subsets of , there exists , such that, for all , for . It follows from (42) that . So
It follows from (49) and (50) that for
Combining (47) and (51) implies that as , which completes the proof.

*The following theorem characterizes the equivalence of boundedness and compactness of composition operators from to .*

*Theorem 8. Let and is an analytic self-map of . Then the following are equivalent. (1) is bound.(2) is compact.(3)*

*Proof. *. Assume that is bounded; then by taking . By Lemma 1, there exist such that . So
which implies (52).

Conversely, for any , it is clear that
By (52), . Then is bounded by the closed graph theorem.

. Let be compact. By Lemma 7, for any given , there is such that
for , which implies (52) by Lemma 1.

Conversely, suppose that (52) holds; then for any function ,
as . Hence, is compact by Lemma 7, which completes the proof.

*Finally, we consider the Fredholmness of composition operators on spaces. For a Banach space , recall that a bounded linear operator on is said to be Fredholm if both the dimension of its kernel and the codimension of its image are finite. This occurs if and only if is invertible modulo the compact operators; that is, there is a bounded linear operator such that both and are compact. We also notice that an operator is Fredholm if and only if its dual is Fredholm (see, e.g., [15]).*

*Before giving our result on Fredholmness, we need a useful result due to Wirths and Xiao [16].*

*Lemma 9. Let and with for . Then the following are equivalent. (1).(2).(3) belongs to the closure of the class of the polynomials in the norm .(4)For any there is a such that .*

*Theorem 10. Let be an analytic self-map of the unit disc . Then the following are equivalent.(1) is a Möbius transformation of .(2) is invertible.(3) is Fredholm.*

*Proof. *. If , , then ; that is, . Since is Möbius invariant by [15], we get . If with , we also have . Since any Möbius transformation can be expressed that , is invertible.

is obvious.

. Suppose is Fredholm. Note that cannot be a constant mapping. Otherwise, if , we have and , which contradicts the Fredholmness of .

Assume is not one to one. So there exist , with . Select the neighborhoods , of , , respectively, such that . is a nonempty and open set due to being open by the Open Mapping Theorem, so there exist infinite sequences , such that which are distinct. Hence ; namely, , where is evaluation function, which is a bounded linear functional on . Since contains all polynomials by Lemma 9, we have that each evaluation function is not a linear combination of other evaluation functions, so the sequence is linearly independent in the kernel of the adjoint operator . It is worth pointing out that is also Fredholm. It is a contradiction, so is injective.

We now show is surjective. Assume that is not surjective. Then we can find and such that as . Further, we get, by the Open Mapping Theorem, that as . For arbitrary ,
we get and is bounded uniformly. Again, it is obvious that as . Therefore, . On the other hand, since is also Fredholm, there are operators and on , with compact and bounded, such that . Thus, . Because is compact and is bounded, there exists subsequence such that , , which means . Moreover, is the closure of all polynomials with respect to the norm by Lemma 9, which gets . This implies that . This is a contradiction. So is surjective. Thus is a Möbius transformation, which completes the proof.

*Conflict of Interests*

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

*Acknowledgment*

*Acknowledgment**This research is partially supported by NSFC (11271293).*

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