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ISRN Mathematical Analysis

Volume 2012 (2012), Article ID 945741, 13 pages

http://dx.doi.org/10.5402/2012/945741

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

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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