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ISRN Mathematical Analysis
Volume 2012 (2012), Article ID 945741, 13 pages
http://dx.doi.org/10.5402/2012/945741
Research Article

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 𝛽={𝛽(𝑛)} be a sequence of positive numbers with 𝛽(0)=1 and 1<𝑝<. We consider the space of sequences 𝑓={𝑓(𝑛)}𝑛=0 such that𝑓𝑝=𝑓𝑝𝛽=𝑛=0||||𝑓(𝑛)𝑝𝛽(𝑛)𝑝<.(1.1) The notation𝑓(𝑧)=𝑛=0𝑓(𝑛)𝑧𝑛(1.2) will be used whether or not the series converges for any value of 𝑧. These are called formal power series and the set of such series is denoted by 𝐻𝑝(𝛽). Let 𝑓𝑘(𝑛)=𝛿𝑘(𝑛). So 𝑓𝑘(𝑧)=𝑧𝑘 and then {𝑓𝑘}𝑘 is a basis such that 𝑓𝑘=𝛽(𝑘). Recall that 𝐻𝑝(𝛽) is a reflexive Banach space with norm 𝛽 and the dual of 𝐻𝑝(𝛽) is 𝐻𝑞(𝛽𝑝/𝑞) where 1/𝑝+1/𝑞=1 and 𝛽𝑝/𝑞={𝛽(𝑛)𝑝/𝑞} [2]. For some other sources on this topic see [112].

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. [1316], 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 [1720]. 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 [2126]. 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 𝐻𝑝(𝛽) and then we will investigate its boundedness and compactness on 𝐻𝑝(𝛽).

Definition 2.1. Let {𝑎𝑛} be a sequence of positive numbers and define 𝐴𝑛=𝑛𝑖=0𝑎𝑖𝛽(𝑖)𝑝.(2.1) The mean operator matrix associated with the sequence {𝑎𝑛} is represented by the matrix 𝐴=[𝑎𝑛𝑘]𝑛,𝑘 and is defined by 𝑎𝑛𝑘=𝑎𝑘𝛽(𝑛)𝑝𝐴𝑛,0𝑘𝑛,0,𝑘>𝑛.(2.2)
From now on, by 𝐴 we denote the mean operator matrix associated with the fixed sequence {𝑎𝑛} as in Definition 2.1.

Theorem 2.2 (see [12, Theorem 1]). If 0<𝑎𝑛𝑎𝑛+1 for all integers 𝑛0, then 𝐴 is a bounded operator on 𝐻𝑝(𝛽).

Theorem 2.3 (see [12, Theorem 2]). Let 1/𝑝+1/𝑞=1 and 𝑏𝑛>0 for 𝑛=0,1, If 𝑀1=sup𝑛𝑛0𝑘=0𝑎𝑘𝛽(𝑛)𝑝+1𝐴𝑛𝑏𝛽(𝑘)𝑘𝑏𝑛1/𝑝𝑀<,2=sup𝑘0𝑛=𝑘𝑎𝑘𝛽(𝑛)𝑝+1𝐴𝑛𝑏𝛽(𝑘)𝑛𝑏𝑘1/𝑞<,(2.3) then 𝐴=[𝑎𝑛𝑘]𝑛,𝑘 is a bounded operator on 𝐻𝑝(𝛽) and 𝐴𝑀11/𝑞𝑀21/𝑝.

Recall that if 𝑎𝑛,𝑏𝑛 are two positive sequences, by 𝑎𝑛𝑏𝑛, we mean that 𝑎𝑛/𝑏𝑛1 whenever 𝑛. Also, we write 𝑎𝑛=𝑜(𝑏𝑛), if 𝑎𝑛/𝑏𝑛0 as 𝑛.

Corollary 2.4. Let lim𝑛𝑛𝑎𝑛/𝐴𝑛 be finite and 1/𝑝+1/𝑞=1. If sup𝑛𝑛0𝑘=0𝑎𝑘𝛽(𝑛)𝑝+1𝑛𝑎𝑛𝑏𝛽(𝑘)𝑘𝑏𝑛1/𝑝<,sup𝑘0𝑛=𝑘𝑎𝑘𝛽(𝑛)𝑝+1𝑛𝑎𝑛𝑏𝛽(𝑘)𝑛𝑏𝑘1/𝑞<,(2.4) then 𝐴 is a bounded operator on 𝐻𝑝(𝛽).

Proof. Put lim𝑛𝑛𝑎𝑛/𝐴𝑛=𝛽. Then 𝑛𝑎𝑛/𝛽𝐴𝑛 and so 𝑛𝑘=0𝑎𝑘𝛽(𝑛)𝑝+1𝐴𝑛𝑏𝛽(𝑘)𝑘𝑏𝑛1/𝑝𝛽𝑛𝑘=0𝑎𝑘𝛽(𝑛)𝑝+1𝑛𝑎𝑛𝑏𝛽(𝑘)𝑘𝑏𝑛1/𝑝as𝑛,𝑛=𝑘𝑎𝑘𝛽(𝑛)𝑝+1𝐴𝑛𝑏𝛽(𝑘)𝑛𝑏𝑘1/𝑞𝛽𝑛=𝑘𝑎𝑘𝛽(𝑛)𝑝+1𝑛𝑎𝑛𝑏𝛽(𝑘)𝑛𝑏𝑘1/𝑞as𝑘.(2.5) On the other hand sup𝑛𝑛0𝑘=0𝑎𝑘𝛽(𝑛)𝑝+1𝑛𝑎𝑛𝑏𝛽(𝑘)𝑘𝑏𝑛1/𝑝<,sup𝑘0𝑛=𝑘𝑎𝑘𝛽(𝑛)𝑝+1𝑛𝑎𝑛𝑏𝛽(𝑘)𝑛𝑏𝑘1/𝑞<,(2.6) thus Theorem 2.3 implies that 𝐴 is a bounded operator on 𝐻𝑝(𝛽).

Lemma 2.5. Suppose that 𝑛𝑐𝑎𝑛/𝛽(𝑛) is eventually increasing when the constant 𝑐>1𝛾, and eventually decreasing when 𝑐<1𝛾. Let 𝑆11(𝑛)=𝑛𝑛𝑘=1𝑎𝑘𝛽(𝑛)𝑎𝑛𝑘𝛽(𝑘)𝑛1/𝑝,𝑆2(𝑘)=𝑘1/𝑞𝑛=𝑘𝑎𝑘𝛽(𝑛)𝑎𝑛1𝛽(𝑘)𝑛1/(𝑞+1).(2.7) If 𝛾>1/𝑝, then lim𝑛𝑆1(𝑛)=lim𝑘𝑆2(𝑘)=1/(𝛾1/𝑝).

Proof. Let 1/𝑝+1/𝑞=1 and 𝑐2<1𝛾<𝑐1<1. Then in either case there is a positive integer 𝑁 such that 𝑘𝑛𝑐2<𝑎𝑘𝛽(𝑛)𝑎𝑛<𝑘𝛽(𝑘)𝑛𝑐1(2.8) for 𝑁𝑘𝑛. Suppose first that 𝛾>1/𝑝, then lim𝑛𝑛11/𝑝𝑎𝑛𝛽(𝑛)=(2.9) and hence lim𝑛1𝑛𝑁1𝑘=1𝑎𝑘𝛽(𝑛)𝑎𝑛𝛽𝑘(𝑘)𝑛1/𝑝=0.(2.10) Therefore lim𝑛sup𝑆1(𝑛)lim𝑛1𝑛𝑛𝑘=𝑁𝑘𝑛𝑐11/𝑝=10𝑥𝑐11/𝑝𝑑𝑥.(2.11) By calculus integral we get 10𝑥𝑐1/𝑝1𝑑𝑥=11𝑐1/𝑝;𝑐𝑞,(2.12) and so lim𝑛inf𝑆1(𝑛)lim𝑛1𝑛𝑛𝑘=𝑁𝑘𝑛𝑐2𝛿=10𝑥𝑐21/𝑝1𝑑𝑥=1𝑐21/𝑝.(2.13) Letting 𝑐11𝛾 from the right and 𝑐21𝛾 from the left, we have lim𝑛𝑆11(𝑛)=.𝛾1/𝑝(2.14) Also note that lim𝑘sup𝑆2(𝑘)lim𝑘𝑘1/𝑞𝑛=𝑘𝑘𝑛𝑐11𝑛1/𝑞+1,lim𝑘𝑘1/𝑞𝑛=𝑘𝑘𝑛𝑐11𝑛1/𝑞+1=11/𝑞𝑐1.(2.15) If 𝑐11𝛾, then 1/𝑞𝑐1𝛾1/𝑝 and similarly we get lim𝑘inf𝑆2(𝑘)lim𝑘𝑘1/𝑞𝑐2𝑛=𝑘1𝑛11/𝑞𝑐2,lim𝑘𝑘1/𝑞𝑐2𝑛=𝑘1𝑛11/𝑞𝑐2=11/𝑞𝑐2.(2.16) If 𝑐21𝛾, then 1/𝑞𝑐2𝛾1/𝑝. This completes the proof.

Theorem 2.6. Let lim𝑛𝑛𝑎𝑛𝛽(𝑛)𝑝/𝐴𝑛=𝛾, 𝑛𝑐𝑎𝑛𝛽(𝑛)𝑝 be eventually monotonic for any constant 𝑐, and {𝛽(𝑛)} be bounded. Then 𝐴 is a bounded operator if 1/𝛾<𝑝.

Proof. Let 𝛿𝑛=𝑛𝑎𝑛𝛽(𝑛)𝑝/𝐴𝑛 and suppose first that 0𝛾<. Then 𝑛𝐴log𝑛𝐴log𝑛1𝛿=𝑛log1𝑛𝑛𝛾(2.17) as 𝑛, and hence 𝐴log𝑛𝐴log1=𝑛𝑛𝑘=2𝛿log1𝑘𝑘=𝜖𝑛log𝑛,(2.18) where 𝜖𝑛𝛾. Consequently 𝐴𝑛=𝐴1𝑛𝜖𝑛. Now suppose that 𝛾=, then for 𝑛2, 𝐴log𝑛𝐴log𝑛1𝛿=log1𝑛𝑛𝛿𝑛𝑛(2.19) since 𝛿𝑛. If 𝑀>0, then there is 𝑁1 such that 𝛿𝑛𝑀+1 for all 𝑛𝑁1.
Without loss of the generality suppose that there is a positive real number 𝑎>0 such that 𝛿𝑛>𝑎 for 𝑛𝑁1. Note that 𝑁1𝑘=21𝑘=log𝑁1+𝑐+𝑜(1)1.(2.20) If 𝑛>𝑁1, then 𝑛𝑘=21𝑘=log𝑛+𝑐+𝑜(1)1,𝑛𝑘=𝑁1+11𝑘=log𝑛log𝑁1.(2.21) Also, 𝑛𝑘=2𝛿𝑘/𝑘𝑎log𝑛𝑁1𝑘=21/𝑘+(𝑀+1)𝑛𝑘=𝑁1+11/𝑘,log𝑛𝑁1𝑘=21/𝑘+𝑛𝑘=𝑁1+11/𝑘𝑀log𝑛1+1log𝑛log𝑁1+𝑎log𝑁1+𝑐+𝑜(1)1log𝑛=𝑀1+1+𝑎𝑀11log𝑁1+𝑎(𝑐+𝑜(1)1),log𝑛(2.22) for large amount of 𝑛 last equality greater than 𝑀1. Hence log𝐴𝑛𝑛𝑘=2𝛿𝑘𝑘=𝛾𝑛log𝑛,(2.23) where 𝛾𝑛. It follows that, for any real number 𝑐, 𝑛𝑐𝐴𝑛=𝑛𝑐+𝛾𝑛. Since 𝑛𝑐1𝐴𝑛1𝛾𝑛𝑐𝑎𝑛𝛽(𝑛)𝑝,(2.24) thus 𝑛𝑐𝑎𝑛𝛽(𝑛)𝑝 is eventually increasing for 𝑐>1𝛾, and eventually decreasing for 𝑐<1𝛾. But {𝛽(𝑛)}𝑛 is bounded, so there are 𝑀1,𝑀2>0 such that 𝑀1<𝛽(𝑛)<𝑀2, and 𝑛𝑐𝑎𝑛=𝑛𝛽(𝑛)𝑐𝑎𝑛𝛽(𝑛)𝑝𝛽(𝑛)𝑝+1,𝑛𝑐𝑎𝑛𝛽(𝑛)𝑝𝛽(𝑛)𝑝+1𝑛𝑐𝑎𝑛𝛽(𝑛)𝑝𝑀1𝑝+1.(2.25) This implies that 𝑛𝑐𝑎𝑛/𝛽(𝑛) is eventually increasing for 𝑐>1𝛾. Similarly 𝑛𝑐𝑎𝑛/𝛽(𝑛) is eventually decreasing for 𝑐<1𝛾. Thus 𝑛𝑘=1𝑎𝑘𝛽(𝑛)𝑝+1𝐴𝑛𝑘𝛽(𝑘)𝑛1/𝑝𝛾𝑛𝑛𝑘=1𝑎𝑘𝛽(𝑛)𝑝+1𝑛𝑎𝑛𝑘𝛽(𝑘)𝑛1/𝑝.(2.26) By Lemma 2.5𝛾𝑛𝑛𝑘=1𝑎𝑘𝛽(𝑛)𝑝+1𝑛𝑎𝑛𝑘𝛽(𝑘)𝑛1/𝑝(2.27) is bounded and so 𝑛𝑘=1𝑎𝑘𝛽(𝑛)𝑝+1𝐴𝑛𝑘𝛽(𝑘)𝑛1/𝑝(2.28) is bounded. We can see that 𝑛=𝑘𝑎𝑘𝛽(𝑛)𝑝+1𝐴𝑛𝑘𝛽(𝑘)𝑛1/𝑞(2.29) is also bounded. Now by Theorem 2.3, 𝐴 is a bounded operator and so the proof is complete.

Lemma 2.7. Let {𝑎𝑛},{𝑡𝑛} be nonnegative sequences with 𝑡1=0. Then for all 𝑛 one has 𝑛𝑘=0𝑡𝑘𝑎𝑘max0𝑘𝑛1𝑛𝑘+1𝑛𝑗=𝐾𝑎𝑗𝑛𝑘=1𝑡(𝑛𝑘+1)𝑘𝑡𝑘1++𝑡0(𝑛+1).(2.30)

Proof. Employing the summation by parts, we get 𝑛𝑘=0𝑡𝑘𝑎𝑘=𝑛𝑘=0𝑛𝑗=𝑘𝑎𝑗𝑡𝑘𝑡𝑘1𝑛𝑘=0𝑛𝑗=𝑘𝑎𝑗1𝑡𝑛𝑘+1𝑘𝑡𝑘1+(𝑛𝑘+1).(2.31) So 𝑛𝑘=0𝑡𝑘𝑎𝑘max0𝑘𝑛1𝑛𝑘+1𝑛𝑗=𝑘𝑎𝑗𝑛𝑘=1𝑡(𝑛𝑘+1)𝑘𝑡𝑘1++𝑡0(𝑛+1),(2.32) and at this time the proof is complete.

Theorem 2.8 (see [26, Theorem 8]). Let 1/𝑝+1/𝑞=1, {𝑥𝑛} be a positive sequence, then 𝑗=0max0𝑖𝑗1𝑗𝑖+1𝑗𝑘=𝑖𝑥𝑘𝑝𝑞𝑝𝑘=0𝑥𝑝𝑘.(2.33)

Theorem 2.9. Let {𝑎𝑛} be a positive sequence and 𝑀3=sup𝑛0𝑛𝑘=1𝑛𝑘+1𝐴𝑛𝑎𝑘𝑎𝛽(𝑘)𝑘1𝛽(𝑘1)+𝛽(𝑛)𝑝+1+(𝑛+1)𝑎0𝐴𝑛𝛽(0)𝛽(𝑛)𝑝+1(2.34) be finite. Then 𝐴 is bounded and 𝐴𝑀3𝑞.

Proof. Let 𝑓(𝑧)=𝑛=0𝑓(𝑛)𝑧𝑛𝐻𝑝(𝛽), thus 𝐴(𝑓)(𝑧)=𝑛=0𝑛𝑘=0𝑎𝑘𝛽(𝑛)𝑝𝐴𝑛𝑧𝑓(𝑘)𝑛.(2.35) By definition of 𝛽, we have 𝑛0𝛽(𝑛)𝑝|||||𝑛𝑘=0𝑎𝑘𝛽(𝑛)𝑝𝐴𝑛𝑓|||||(𝑘)𝑝𝑛0𝑛𝑘=0𝑎𝑘𝛽(𝑛)𝑝+1𝐴𝑛||||𝛽(𝑘)𝑓(𝑘)𝛽(𝑘)𝑝.(2.36) In Lemma 2.7, consider 𝑡𝑘=𝑎𝑘/𝛽(𝑘) and 𝑎𝑗=|𝑓(𝑗)|𝛽(𝑗). Then 𝑛0𝑛𝑘=0𝑎𝑘𝛽(𝑛)𝑝+1𝐴𝑛||||𝛽(𝑘)𝑓(𝑘)𝛽(𝑘)𝑝𝑛0max0𝑘𝑛1𝑛𝑘+1𝑛𝑗=𝑘||||𝑓(𝑗)𝛽(𝑗)𝑝×𝑛𝑘=1𝑛𝑘+1𝐴𝑛𝑎𝑘𝑎𝛽(𝑘)𝑘1𝛽(𝑘1)+𝛽(𝑛)𝑝+1+(𝑛+1)𝑎0𝐴𝑛𝛽(0)𝛽(𝑛)𝑝+1𝑝.(2.37) Now, Theorem 2.8 implies that 𝑛0max0𝑘𝑛1𝑛𝑘+1𝑛𝑗=𝑘||||𝑓(𝑗)𝛽(𝑗)𝑝𝑀𝑝3𝑀𝑝3𝑞𝑝𝑘=1||||𝑓(𝑘)𝑝𝛽(𝑘)𝑝,(2.38) and so we get 𝐴𝑓𝑀3𝑞𝑓𝛽 for all 𝑓𝐻𝑝(𝛽). Thus 𝐴𝐵(𝐻𝑝(𝛽)) and indeed 𝐴𝑀3𝑞. This completes the proof.

Corollary 2.10. Let 1/𝑝+1/𝑞=1, 𝑎𝑘/𝛽(𝑘)𝑎𝑘1/𝛽(𝑘1) and 𝑀4=sup𝑛𝑛0𝑘=0𝑎𝑘𝛽(𝑛)𝑝+1𝛽(𝑘)𝐴𝑛<.(2.39) Then 𝐴 is a bounded operator on 𝐻𝑝(𝛽) and 𝐴𝑀4.

Proof. Note that 𝑛0𝑛𝑘=0𝑎𝑘𝛽(𝑛)𝑝+1𝐴𝑛||||𝛽(𝑘)𝑓(𝑘)𝛽(𝑘)𝑝𝑛0max0𝑘𝑛1𝑛𝑘+1𝑛𝑗=𝑘||||𝑓(𝑗)𝛽(𝑗)𝑝𝑛𝑘=0𝑎𝑘𝛽(𝑛)𝑝+1𝛽(𝑘)𝐴𝑛𝑝.(2.40) Theorem 2.8 implies that 𝑛0max0𝑘𝑛1𝑛𝑘+1𝑛𝑗=𝑘||||𝑓(𝑗)𝛽(𝑗)𝑝𝑀𝑝4𝑀𝑝4𝑞𝑝𝑘=1||||𝑓(𝑘)𝑝𝛽(𝑘)𝑝,(2.41) and so by Theorem 2.9 we obtain 𝐴𝑓𝑞𝑀4𝑓𝛽 for all 𝑓𝐻𝑝(𝛽). Thus 𝐴𝐵(𝐻𝑝(𝛽)) and indeed 𝐴𝑀4𝑞. This completes the proof.

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

Theorem 2.11. Let 𝑆 be a nonempty subset of 𝐻𝑝(𝛽). Then 𝑆 is relatively compact if and only if the following hold:(i)there exists 𝑀>0, such that for all 𝑛=0𝑓(𝑛)𝑧𝑛𝑆, |𝑓(𝑖)𝛽(𝑖)|𝑀 for all 𝑖{0};(ii)given 𝜖>0, there is 𝑛0 such that 𝑛=𝑛0|𝑓(𝑛)|𝑝𝛽(𝑛)𝑝<𝜖𝑝 for all 𝑛=0𝑓(𝑛)𝑧𝑛𝑆.

Proof. Let 𝑆 be relatively compact, thus there exist 𝑔1,,𝑔𝑘𝐻𝑝(𝛽) such that 𝑆𝑘𝑖=1𝐵𝑔𝑖.,1(2.42) For every 𝑓(𝑧)=𝑛=0𝑓(𝑛)𝑧𝑛𝑆, there is 𝑔𝑖 such that 𝑓𝐵(𝑔𝑖,1). By Minkowski inequality we get 𝑛=0||||𝑓(𝑛)𝑝𝛽(𝑛)𝑝𝑛=0||𝑓(𝑛)̂𝑔𝑖(||𝑛)𝑝𝛽(𝑛)𝑝1/𝑝+𝑛=0||̂𝑔𝑖(||𝑛)𝑝𝛽(𝑛)𝑝1/𝑝𝑝𝑓𝑔𝑖+𝑔𝑖𝑝𝑔1+𝑖𝑝𝑔1+max𝑖𝑖=1,,𝑘𝑝.(2.43) Thus for every 𝑓𝑆 and 𝑛{0}, we get ||𝑓||𝑔(𝑛)𝛽(𝑛)1+max𝑖.𝑖=1,,𝑘(2.44) So (i) holds. Now suppose that 𝜖 is an arbitrary positive number. Since 𝑆 is relatively compact, thus there exist 1,,𝑘𝐻𝑝(𝛽) such that 𝑆𝑘𝑖=1𝐵𝑖,𝜖2.(2.45) Since 𝑖𝐻𝑝(𝛽), there exists 𝑁𝑖 such that 𝑛=𝑁𝑖|||𝑖(|||𝑛)𝑝𝛽(𝑛)𝑝<𝜖𝑝2𝑝(2.46) for 𝑖=1,,𝑘. Put 𝑁0𝑁=max𝑖𝑖=1,,𝑘,(2.47) and consider 𝑓𝑆. Then there exists 𝑖{1,,𝑘}, such that 𝑓𝐵(𝑖,𝜖/2). Hence we get 𝑛=𝑁0||||𝑓(𝑛)𝑝𝛽(𝑛)𝑝𝑛=𝑁0|||𝑓(𝑛)𝑖|||(𝑛)𝑝𝛽(𝑛)𝑝1/𝑝+𝑛=𝑁0|||𝑖|||(𝑛)𝑝𝛽(𝑛)𝑝1/𝑝𝑝𝑓𝑖+𝜖2𝑝𝜖𝑝.(2.48) So (ii) holds.
Conversely, assume that 𝜖>0 be given and let (i) and (ii) hold. By condition (ii), there exists 𝑛0 such that 𝑛=𝑛0||||𝑓(𝑛)𝑝𝛽(𝑛)𝑝<𝜖𝑝2,(2.49) for all 𝑓𝑆. Let 𝑀𝑛0 be the closed linear span of the set {1,𝑧,,𝑧𝑛01} in 𝐻𝑝(𝛽). Consider 𝑛0 and 𝑀𝑛0 with norms 𝑧1,,𝑧𝑛0=𝑛0𝑛=1||𝑧𝑛||𝑝𝛽(𝑛)𝑝1/𝑝,(2.50) for all (𝑧𝑖)𝑛0𝑖=1𝑛0, and 𝑛01𝑖=0𝑎𝑖𝑧𝑖=𝑛01𝑖=0||𝑎𝑖||𝑝𝛽(𝑖)𝑝1/𝑝(2.51) for all 𝑛01𝑖=0𝑎𝑖𝑧𝑖𝑀𝑛0. Define 𝐿𝑀𝑛0𝑛0, by 𝐿𝑛01𝑖=0𝑎𝑖𝑧𝑖=𝑎0,,𝑎𝑛01.(2.52) Clearly, we can see that 𝐿 is a bounded linear operator. Now, consider the compact subset 𝑧𝑖𝑛0𝑖=1𝑛0𝑖=1||𝑧𝑖||𝑝𝛽(𝑖)𝑝𝑛0𝑀𝑝(2.53) in 𝑛0. Then we have 𝑛01𝑖=0𝑓(𝑖)𝑧𝑖𝑛=0𝑓(𝑛)𝑧𝑛𝑆𝐿1𝑧𝑖𝑛0𝑖=1𝑛0𝑖=1||𝑧𝑖||𝑝𝛽(𝑖)𝑝𝑛0𝑀𝑝.(2.54) Since 𝐿1𝑧𝑖𝑛0𝑖=1𝑛0𝑖=1||𝑧𝑖||𝑝𝛽(𝑖)𝑝𝑛0𝑀𝑝(2.55) is a compact subspace of 𝑀𝑛0, so there exist 𝑔1,,𝑔𝑘𝑀𝑛0 such that 𝐿1𝑧𝑖𝑛0𝑖=1𝑛0𝑖=1||𝑧𝑖||𝑝𝛽(𝑖)𝑝𝑛0𝑀𝑝𝑘𝑖=1𝐵𝑔𝑖,𝜖21/𝑝.(2.56) Hence for every 𝑓𝑛01𝑖=0𝑓(𝑖)𝑧𝑖𝑓(𝑧)=𝑛=0𝑓(𝑛)𝑧𝑛𝑆(2.57) there is 𝑖{1,,𝑘} satisfying 𝑛01𝑛=0||𝑓(𝑛)̂𝑔𝑖||(𝑛)𝑝𝛽(𝑛)𝑝𝜖𝑝2.(2.58) Also, we have 𝑓𝑔𝑖𝛽𝑝𝑛01𝑛=0||𝑓(𝑛)̂𝑔𝑖||(𝑛)𝑝𝛽(𝑛)𝑝+𝑛=𝑛0||||𝑓(𝑛)𝑝𝛽(𝑛)𝑝𝜖𝑝2+𝜖𝑝2𝜖𝑝.(2.59) Thus, 𝑆 is relatively compact and so the proof is complete.

Theorem 2.12. Let the mean matrix operator 𝐴 be bounded on 𝐻𝑝(𝛽), and lim𝑚𝑛=𝑚𝛽(𝑛)𝑝2+𝑝𝐴𝑝𝑛1/𝑝𝑚𝑘=0𝑎𝑘𝛽(𝑘)𝑞1/𝑞=0,(2.60) where 1/𝑝+1/𝑞=1. Then 𝐴 is a compact operator on 𝐻𝑝(𝛽).

Proof. Let 𝐵𝐻𝑝(𝛽) be the closed unit ball of 𝐻𝑝(𝛽). Define 𝑆=𝐴(𝐵𝐻𝑝(𝛽)) and note that 𝑆 is a bounded subset of 𝐻𝑝(𝛽). Put 𝑟𝑛=|𝑓(𝑛)|𝑎𝑛, 𝑢𝑛=𝛽(𝑛)𝑝2+𝑝/𝐴𝑝𝑛, 𝑣𝑘=(𝛽(𝑘)/𝑎𝑘)𝑝, and 𝐸𝑚=𝑛=𝑚𝑢𝑛1/𝑝𝑚𝑘=0𝑣𝑘1𝑞1/𝑞.(2.61) Note that lim𝑚𝐸𝑚=0. So for every 𝜖>0, there exists 𝑚0 such that 𝐸𝑚<𝜖/(𝑞𝑝1𝑝)1/𝑝 for all 𝑚𝑚0. Note that if 𝑓(𝑧)=𝑘=0𝑓(𝑘)𝑧𝑘𝐵𝐻𝑝(𝛽),(2.62) then 𝐴𝑓(𝑧)=𝑛=0𝑛𝑘=0𝑎𝑘𝛽(𝑛)𝑝𝑓(𝑘)𝐴𝑛𝑧𝑛𝑆.(2.63) Since 𝑓𝑝𝛽1, we have 𝑛=𝑚||||𝐴𝑓(𝑛)𝑛𝛽(𝑛)𝑝𝑛=𝑚𝛽(𝑛)𝑝2+𝑝𝐴𝑝𝑛𝑛𝑘=0𝑎𝑘||||𝑓(𝑘)𝑝=𝑛=𝑚𝑢𝑛𝑛𝑘=0𝑟𝑘𝑝𝜖𝑝𝑘=0𝑟𝑘𝑝𝑣𝑘𝜖𝑝𝑘=0||||𝑓(𝑘)𝑝𝛽(𝑘)𝑝𝜖𝑝.(2.64) Thus by Theorem 2.11, 𝑆 is compact and so the proof is complete.

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