ISRN Mathematical AnalysisVolume 2012 (2012), Article ID 945741, 13 pagesdoi:10.5402/2012/945741

Research Article

Boundedness and Compactness of the Mean Operator Matrix on Weighted Hardy Spaces

Bahmann Yousefi and Ebrahim Pazouki

Department of Mathematics, Payame Noor University, P.O. Box 19395-3697, Tehran, Iran

Received 5 January 2012; Accepted 16 February 2012

Academic Editors: E. Beretta and V. Kravchenko

Copyright © 2012 Bahmann Yousefi and Ebrahim Pazouki. 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.

We investigate the boundedness and the compactness of the mean operator matrix acting on the weighted Hardy spaces.

1. Introduction

First in the following, we generalize the definitions coming in [1]. Let be a sequence of positive numbers with and . We consider the space of sequences such that The notation will be used whether or not the series converges for any value of . These are called formal power series and the set of such series is denoted by . Let . So and then is a basis such that . Recall that is a reflexive Banach space with norm and the dual of is where and [2]. For some other sources on this topic see [1–12].

The study of weighted Hardy spaces lies at the interface of analytic function theory and operator theory. As a part of operator theory, research on weighted Hardy spaces is of fairly recent origin, dating back to valuable work of Allen Shields [1] in the mid- 1970s. The mean operator matrix has been the focus of attention for several decades and many of its properties have been studied. Some of basic and useful works in this area are due to Browein et al. [13–16], which are pretty large works that contain a number of interesting results and indeed they are mainly of auxiliary nature. Also, some properties of mean operator matrices have been studied recently by Lashkaripour on weighted sequence spaces [17–20]. In this paper, we have given conditions under which the mean operator matrix is bounded and compact as an operator acting on weighted Hardy spaces. More details of our works are as follows: the idea of Theorem 2.6 comes from [16]. In Theorem 2.9, we extend the method used in [20, Theorem 1.2] to show the boundedness of the mean operator matrix acting on the weighted Hardy spaces. Some inequalities are useful to find a bound for the mean operator matrix acting on weighted Hardy spaces [21–26]. For example the inequality proved in [26, Theorem 8] is used in the proof of Theorem 2.11.

2. Main Results

In this section we define an operator acting on and then we will investigate its boundedness and compactness on .

Definition 2.1. Let be a sequence of positive numbers and define The mean operator matrix associated with the sequence is represented by the matrix and is defined by From now on, by we denote the mean operator matrix associated with the fixed sequence as in Definition 2.1.

Theorem 2.2 (see [12, Theorem 1]). If for all integers , then is a bounded operator on .

Theorem 2.3 (see [12, Theorem 2]). Let and for If then is a bounded operator on and .

Recall that if are two positive sequences, by , we mean that whenever . Also, we write , if as .

Corollary 2.4. Let be finite and . If then is a bounded operator on .

Proof. Put . Then and so On the other hand thus Theorem 2.3 implies that is a bounded operator on .

Lemma 2.5. Suppose that is eventually increasing when the constant , and eventually decreasing when . Let If , then .

Proof. Let and . Then in either case there is a positive integer such that for . Suppose first that , then and hence Therefore By calculus integral we get and so Letting from the right and from the left, we have Also note that If , then and similarly we get If , then . This completes the proof.

Theorem 2.6. Let , be eventually monotonic for any constant , and be bounded. Then is a bounded operator if .

Proof. Let and suppose first that . Then as , and hence where . Consequently . Now suppose that , then for , since . If , then there is such that for all . Without loss of the generality suppose that there is a positive real number such that for . Note that If , then Also, for large amount of last equality greater than . Hence where . It follows that, for any real number , . Since thus is eventually increasing for , and eventually decreasing for . But is bounded, so there are such that , and This implies that is eventually increasing for . Similarly is eventually decreasing for . Thus By Lemma 2.5 is bounded and so is bounded. We can see that is also bounded. Now by Theorem 2.3, is a bounded operator and so the proof is complete.

Lemma 2.7. Let be nonnegative sequences with . Then for all one has

Proof. Employing the summation by parts, we get So and at this time the proof is complete.

Theorem 2.8 (see [26, Theorem 8]). Let , be a positive sequence, then

Theorem 2.9. Let be a positive sequence and be finite. Then is bounded and .

Proof. Let , thus By definition of , we have In Lemma 2.7, consider and . Then Now, Theorem 2.8 implies that and so we get for all . Thus and indeed . This completes the proof.

Corollary 2.10. Let , and Then is a bounded operator on and .

Proof. Note that Theorem 2.8 implies that and so by Theorem 2.9 we obtain for all . Thus and indeed . This completes the proof.

Now, we characterize compactness of subsets of and then we will investigate compactness of the mean operator matrix on .

Theorem 2.11. Let be a nonempty subset of . Then is relatively compact if and only if the following hold: (i)there exists , such that for all , for all ;(ii)given , there is such that for all .

Proof. Let be relatively compact, thus there exist such that For every , there is such that . By Minkowski inequality we get Thus for every and , we get So (i) holds. Now suppose that is an arbitrary positive number. Since is relatively compact, thus there exist such that Since , there exists such that for . Put and consider . Then there exists , such that . Hence we get So (ii) holds.Conversely, assume that be given and let (i) and (ii) hold. By condition (ii), there exists such that for all . Let be the closed linear span of the set in . Consider and with norms for all , and for all . Define , by Clearly, we can see that is a bounded linear operator. Now, consider the compact subset in . Then we have Since is a compact subspace of , so there exist such that Hence for every there is satisfying Also, we have Thus, is relatively compact and so the proof is complete.

Theorem 2.12. Let the mean matrix operator be bounded on , and where . Then is a compact operator on .

Proof. Let be the closed unit ball of . Define and note that is a bounded subset of . Put , , , and Note that . So for every , there exists such that for all . Note that if then Since , we have Thus by Theorem 2.11, is compact and so the proof is complete.

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