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ISRN Mathematical Analysis
Volume 2012 (2012), Article ID 945741, 13 pages
doi: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 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 that ‖ 𝑓 ‖ 𝑝 = ‖ 𝑓 ‖ 𝑝 𝛽 = ∞ 𝑛 = 0 | | | | 𝑓 ( 𝑛 ) 𝑝 𝛽 ( 𝑛 ) 𝑝 < ∞ . ( 1 . 1 ) The notation 𝑓 ( 𝑧 ) = ∞ 𝑛 = 0 𝑓 ( 𝑛 ) 𝑧 𝑛 ( 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 𝐻 𝑝 ( 𝛽 ) . Let 𝑓 𝑘 ( 𝑛 ) = 𝛿 𝑘 ( 𝑛 ) . So 𝑓 𝑘 ( 𝑧 ) = 𝑧 𝑘 and then { 𝑓 𝑘 } 𝑘 is a basis such that ‖ 𝑓 𝑘 ‖ = 𝛽 ( 𝑘 ) . Recall that 𝐻 𝑝 ( 𝛽 ) is a reflexive Banach space with norm ‖ ⋅ ‖ 𝛽 and the dual of 𝐻 𝑝 ( 𝛽 ) is 𝐻 𝑞 ( 𝛽 𝑝 / 𝑞 ) where 1 / 𝑝 + 1 / 𝑞 = 1 and 𝛽 𝑝 / 𝑞 = { 𝛽 ( 𝑛 ) 𝑝 / 𝑞 } [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 𝐻 𝑝 ( 𝛽 ) and then we will investigate its boundedness and compactness on 𝐻 𝑝 ( 𝛽 ) .

Definition 2.1. Let { 𝑎 𝑛 } be a sequence of positive numbers and define 𝐴 𝑛 = 𝑛 𝑖 = 0 𝑎 𝑖 𝛽 ( 𝑖 ) 𝑝 . ( 2 . 1 ) The mean operator matrix associated with the sequence { 𝑎 𝑛 } is represented by the matrix 𝐴 = [ 𝑎 𝑛 𝑘 ] 𝑛 , 𝑘 and is defined by 𝑎 𝑛 𝑘 = ⎧ ⎪ ⎨ ⎪ ⎩ 𝑎 𝑘 𝛽 ( 𝑛 ) 𝑝 𝐴 𝑛 , 0 ≤ 𝑘 ≤ 𝑛 , 0 , 𝑘 > 𝑛 . ( 2 . 2 )
From now on, by 𝐴 we denote the mean operator matrix associated with the fixed sequence { 𝑎 𝑛 } as in Definition 2.1.

Theorem 2.2 (see [12, Theorem 1]). If 0 < 𝑎 𝑛 ≤ 𝑎 𝑛 + 1 for all integers 𝑛 ≥ 0 , then 𝐴 is a bounded operator on 𝐻 𝑝 ( 𝛽 ) .

Theorem 2.3 (see [12, Theorem 2]). Let 1 / 𝑝 + 1 / 𝑞 = 1 and 𝑏 𝑛 > 0 for 𝑛 = 0 , 1 , … If 𝑀 1 = s u p 𝑛 𝑛 ≥ 0 𝑘 = 0 𝑎 𝑘 𝛽 ( 𝑛 ) 𝑝 + 1 𝐴 𝑛 𝑏 𝛽 ( 𝑘 ) 𝑘 𝑏 𝑛 1 / 𝑝 𝑀 < ∞ , 2 = s u p ∞ 𝑘 ≥ 0 𝑛 = 𝑘 𝑎 𝑘 𝛽 ( 𝑛 ) 𝑝 + 1 𝐴 𝑛 𝑏 𝛽 ( 𝑘 ) 𝑛 𝑏 𝑘 1 / 𝑞 < ∞ , ( 2 . 3 ) then 𝐴 = [ 𝑎 𝑛 𝑘 ] 𝑛 , 𝑘 is a bounded operator on 𝐻 𝑝 ( 𝛽 ) and ‖ 𝐴 ‖ ≤ 𝑀 1 1 / 𝑞 𝑀 2 1 / 𝑝 .

Recall that if 𝑎 𝑛 , 𝑏 𝑛 are two positive sequences, by 𝑎 𝑛 ∼ 𝑏 𝑛 , we mean that 𝑎 𝑛 / 𝑏 𝑛 → 1 whenever 𝑛 → ∞ . Also, we write 𝑎 𝑛 = 𝑜 ( 𝑏 𝑛 ) , if 𝑎 𝑛 / 𝑏 𝑛 → 0 as 𝑛 → ∞ .

Corollary 2.4. Let l i m 𝑛 → ∞ 𝑛 𝑎 𝑛 / 𝐴 𝑛 be finite and 1 / 𝑝 + 1 / 𝑞 = 1 . If s u p 𝑛 𝑛 ≥ 0 𝑘 = 0 𝑎 𝑘 𝛽 ( 𝑛 ) 𝑝 + 1 𝑛 𝑎 𝑛 𝑏 𝛽 ( 𝑘 ) 𝑘 𝑏 𝑛 1 / 𝑝 < ∞ , s u p ∞ 𝑘 ≥ 0 𝑛 = 𝑘 𝑎 𝑘 𝛽 ( 𝑛 ) 𝑝 + 1 𝑛 𝑎 𝑛 𝑏 𝛽 ( 𝑘 ) 𝑛 𝑏 𝑘 1 / 𝑞 < ∞ , ( 2 . 4 ) then 𝐴 is a bounded operator on 𝐻 𝑝 ( 𝛽 ) .

Proof. Put l i m 𝑛 → ∞ 𝑛 𝑎 𝑛 / 𝐴 𝑛 = 𝛽 . Then 𝑛 𝑎 𝑛 / 𝛽 ∼ 𝐴 𝑛 and so 𝑛 𝑘 = 0 𝑎 𝑘 𝛽 ( 𝑛 ) 𝑝 + 1 𝐴 𝑛 𝑏 𝛽 ( 𝑘 ) 𝑘 𝑏 𝑛 1 / 𝑝 ∼ 𝛽 𝑛 𝑘 = 0 𝑎 𝑘 𝛽 ( 𝑛 ) 𝑝 + 1 𝑛 𝑎 𝑛 𝑏 𝛽 ( 𝑘 ) 𝑘 𝑏 𝑛 1 / 𝑝 a s 𝑛 ⟶ ∞ , ∞ 𝑛 = 𝑘 𝑎 𝑘 𝛽 ( 𝑛 ) 𝑝 + 1 𝐴 𝑛 𝑏 𝛽 ( 𝑘 ) 𝑛 𝑏 𝑘 1 / 𝑞 ∼ 𝛽 ∞ 𝑛 = 𝑘 𝑎 𝑘 𝛽 ( 𝑛 ) 𝑝 + 1 𝑛 𝑎 𝑛 𝑏 𝛽 ( 𝑘 ) 𝑛 𝑏 𝑘 1 / 𝑞 a s 𝑘 ⟶ ∞ . ( 2 . 5 ) On the other hand s u p 𝑛 𝑛 ≥ 0 𝑘 = 0 𝑎 𝑘 𝛽 ( 𝑛 ) 𝑝 + 1 𝑛 𝑎 𝑛 𝑏 𝛽 ( 𝑘 ) 𝑘 𝑏 𝑛 1 / 𝑝 < ∞ , s u p ∞ 𝑘 ≥ 0 𝑛 = 𝑘 𝑎 𝑘 𝛽 ( 𝑛 ) 𝑝 + 1 𝑛 𝑎 𝑛 𝑏 𝛽 ( 𝑘 ) 𝑛 𝑏 𝑘 1 / 𝑞 < ∞ , ( 2 . 6 ) thus Theorem 2.3 implies that 𝐴 is a bounded operator on 𝐻 𝑝 ( 𝛽 ) .

Lemma 2.5. Suppose that 𝑛 𝑐 𝑎 𝑛 / 𝛽 ( 𝑛 ) is eventually increasing when the constant 𝑐 > 1 − 𝛾 , and eventually decreasing when 𝑐 < 1 − 𝛾 . Let 𝑆 1 1 ( 𝑛 ) = 𝑛 𝑛 𝑘 = 1 𝑎 𝑘 𝛽 ( 𝑛 ) 𝑎 𝑛 𝑘 𝛽 ( 𝑘 ) 𝑛 − 1 / 𝑝 , 𝑆 2 ( 𝑘 ) = 𝑘 ∞ 1 / 𝑞 𝑛 = 𝑘 𝑎 𝑘 𝛽 ( 𝑛 ) 𝑎 𝑛 1 𝛽 ( 𝑘 ) 𝑛 1 / ( 𝑞 + 1 ) . ( 2 . 7 ) If 𝛾 > 1 / 𝑝 , then l i m 𝑛 → ∞ 𝑆 1 ( 𝑛 ) = l i m 𝑘 → ∞ 𝑆 2 ( 𝑘 ) = 1 / ( 𝛾 − 1 / 𝑝 ) .

Proof. Let 1 / 𝑝 + 1 / 𝑞 = 1 and 𝑐 2 < 1 − 𝛾 < 𝑐 1 < 1 . Then in either case there is a positive integer 𝑁 such that 𝑘 𝑛 − 𝑐 2 < 𝑎 𝑘 𝛽 ( 𝑛 ) 𝑎 𝑛 < 𝑘 𝛽 ( 𝑘 ) 𝑛 − 𝑐 1 ( 2 . 8 ) for 𝑁 ≤ 𝑘 ≤ 𝑛 . Suppose first that 𝛾 > 1 / 𝑝 , then l i m 𝑛 → ∞ 𝑛 1 − 1 / 𝑝 𝑎 𝑛 𝛽 ( 𝑛 ) = ∞ ( 2 . 9 ) and hence l i m 𝑛 → ∞ 1 𝑛 𝑁 − 1 𝑘 = 1 𝑎 𝑘 𝛽 ( 𝑛 ) 𝑎 𝑛 𝛽 𝑘 ( 𝑘 ) 𝑛 − 1 / 𝑝 = 0 . ( 2 . 1 0 ) Therefore l i m 𝑛 → ∞ s u p 𝑆 1 ( 𝑛 ) ≤ l i m 𝑛 → ∞ 1 𝑛 𝑛 𝑘 = 𝑁 𝑘 𝑛 − 𝑐 1 − 1 / 𝑝 = 1 0 𝑥 − 𝑐 1 − 1 / 𝑝 𝑑 𝑥 . ( 2 . 1 1 ) By calculus integral we get 1 0 𝑥 − 𝑐 − 1 / 𝑝 1 𝑑 𝑥 = 1 1 − 𝑐 − 1 / 𝑝 ; 𝑐 ≠ 𝑞 , ( 2 . 1 2 ) and so l i m 𝑛 → ∞ i n f 𝑆 1 ( 𝑛 ) ≥ l i m 𝑛 → ∞ 1 𝑛 𝑛 𝑘 = 𝑁 𝑘 𝑛 − 𝑐 2 − 𝛿 = 1 0 𝑥 − 𝑐 2 − 1 / 𝑝 1 𝑑 𝑥 = 1 − 𝑐 2 − 1 / 𝑝 . ( 2 . 1 3 ) Letting 𝑐 1 → 1 − 𝛾 from the right and 𝑐 2 → 1 − 𝛾 from the left, we have l i m 𝑛 → ∞ 𝑆 1 1 ( 𝑛 ) = . 𝛾 − 1 / 𝑝 ( 2 . 1 4 ) Also note that l i m 𝑘 → ∞ s u p 𝑆 2 ( 𝑘 ) ≤ l i m 𝑘 → ∞ 𝑘 ∞ 1 / 𝑞 𝑛 = 𝑘 𝑘 𝑛 − 𝑐 1 1 𝑛 1 / 𝑞 + 1 , l i m 𝑘 → ∞ 𝑘 ∞ 1 / 𝑞 𝑛 = 𝑘 𝑘 𝑛 − 𝑐 1 1 𝑛 1 / 𝑞 + 1 = 1 1 / 𝑞 − 𝑐 1 . ( 2 . 1 5 ) If 𝑐 1 → 1 − 𝛾 , then 1 / 𝑞 − 𝑐 1 → 𝛾 − 1 / 𝑝 and similarly we get l i m 𝑘 → ∞ i n f 𝑆 2 ( 𝑘 ) ≥ l i m 𝑘 → ∞ 𝑘 1 / 𝑞 − 𝑐 2 ∞ 𝑛 = 𝑘 1 𝑛 1 − 1 / 𝑞 − 𝑐 2 , l i m 𝑘 → ∞ 𝑘 1 / 𝑞 − 𝑐 2 ∞ 𝑛 = 𝑘 1 𝑛 1 − 1 / 𝑞 − 𝑐 2 = 1 1 / 𝑞 − 𝑐 2 . ( 2 . 1 6 ) If 𝑐 2 → 1 − 𝛾 , then 1 / 𝑞 − 𝑐 2 → 𝛾 − 1 / 𝑝 . This completes the proof.

Theorem 2.6. Let l i m 𝑛 → ∞ 𝑛 𝑎 𝑛 𝛽 ( 𝑛 ) 𝑝 / 𝐴 𝑛 = 𝛾 , 𝑛 𝑐 𝑎 𝑛 𝛽 ( 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 𝛿 𝑛 = 𝑛 𝑎 𝑛 𝛽 ( 𝑛 ) 𝑝 / 𝐴 𝑛 and suppose first that 0 ≤ 𝛾 < ∞ . Then 𝑛 𝐴 l o g 𝑛 𝐴 − l o g 𝑛 − 1 𝛿 = − 𝑛 l o g 1 − 𝑛 𝑛 ⟶ 𝛾 ( 2 . 1 7 ) as 𝑛 → ∞ , and hence 𝐴 l o g 𝑛 𝐴 − l o g 1 = − 𝑛 𝑛 𝑘 = 2 𝛿 l o g 1 − 𝑘 𝑘 = 𝜖 𝑛 l o g 𝑛 , ( 2 . 1 8 ) where 𝜖 𝑛 → 𝛾 . Consequently 𝐴 𝑛 = 𝐴 1 𝑛 𝜖 𝑛 . Now suppose that 𝛾 = ∞ , then for 𝑛 ≥ 2 , 𝐴 l o g 𝑛 𝐴 − l o g 𝑛 − 1 𝛿 = − l o g 1 − 𝑛 𝑛 ≥ 𝛿 𝑛 𝑛 ( 2 . 1 9 ) since 𝛿 𝑛 → ∞ . If 𝑀 > 0 , then there is 𝑁 1 ∈ ℕ such that 𝛿 𝑛 ≥ 𝑀 + 1 for all 𝑛 ≥ 𝑁 1 .
Without loss of the generality suppose that there is a positive real number 𝑎 > 0 such that 𝛿 𝑛 > 𝑎 for 𝑛 ≤ 𝑁 1 . Note that 𝑁 1 𝑘 = 2 1 𝑘 = l o g 𝑁 1 + 𝑐 + 𝑜 ( 1 ) − 1 . ( 2 . 2 0 ) If 𝑛 > 𝑁 1 , then 𝑛 𝑘 = 2 1 𝑘 = l o g 𝑛 + 𝑐 + 𝑜 ( 1 ) − 1 , 𝑛 𝑘 = 𝑁 1 + 1 1 𝑘 = l o g 𝑛 − l o g 𝑁 1 . ( 2 . 2 1 ) Also, ∑ 𝑛 𝑘 = 2 𝛿 𝑘 / 𝑘 ≥ 𝑎 ∑ l o g 𝑛 𝑁 1 𝑘 = 2 ∑ 1 / 𝑘 + ( 𝑀 + 1 ) 𝑛 𝑘 = 𝑁 1 + 1 1 / 𝑘 , ∑ l o g 𝑛 𝑁 1 𝑘 = 2 ∑ 1 / 𝑘 + 𝑛 𝑘 = 𝑁 1 + 1 1 / 𝑘 ≥ 𝑀 l o g 𝑛 1 + 1 l o g 𝑛 − l o g 𝑁 1 + 𝑎 l o g 𝑁 1 + 𝑐 + 𝑜 ( 1 ) − 1 l o g 𝑛 = 𝑀 1 + 1 + 𝑎 − 𝑀 1 − 1 l o g 𝑁 1 + 𝑎 ( 𝑐 + 𝑜 ( 1 ) − 1 ) , l o g 𝑛 ( 2 . 2 2 ) for large amount of 𝑛 last equality greater than 𝑀 1 . Hence l o g 𝐴 𝑛 ≥ 𝑛 𝑘 = 2 𝛿 𝑘 𝑘 = 𝛾 𝑛 l o g 𝑛 , ( 2 . 2 3 ) where 𝛾 𝑛 → ∞ . It follows that, for any real number 𝑐 , 𝑛 𝑐 𝐴 𝑛 = 𝑛 𝑐 + 𝛾 𝑛 . Since 𝑛 𝑐 − 1 𝐴 𝑛 ∼ 1 𝛾 𝑛 𝑐 𝑎 𝑛 𝛽 ( 𝑛 ) 𝑝 , ( 2 . 2 4 ) thus 𝑛 𝑐 𝑎 𝑛 𝛽 ( 𝑛 ) 𝑝 is eventually increasing for 𝑐 > 1 − 𝛾 , and eventually decreasing for 𝑐 < 1 − 𝛾 . But { 𝛽 ( 𝑛 ) } 𝑛 is bounded, so there are 𝑀 1 , 𝑀 2 > 0 such that 𝑀 1 < 𝛽 ( 𝑛 ) < 𝑀 2 , and 𝑛 𝑐 𝑎 𝑛 = 𝑛 𝛽 ( 𝑛 ) 𝑐 𝑎 𝑛 𝛽 ( 𝑛 ) 𝑝 𝛽 ( 𝑛 ) 𝑝 + 1 , 𝑛 𝑐 𝑎 𝑛 𝛽 ( 𝑛 ) 𝑝 𝛽 ( 𝑛 ) 𝑝 + 1 ≥ 𝑛 𝑐 𝑎 𝑛 𝛽 ( 𝑛 ) 𝑝 𝑀 1 𝑝 + 1 . ( 2 . 2 5 ) This implies that 𝑛 𝑐 𝑎 𝑛 / 𝛽 ( 𝑛 ) is eventually increasing for 𝑐 > 1 − 𝛾 . Similarly 𝑛 𝑐 𝑎 𝑛 / 𝛽 ( 𝑛 ) is eventually decreasing for 𝑐 < 1 − 𝛾 . Thus 𝑛 𝑘 = 1 𝑎 𝑘 𝛽 ( 𝑛 ) 𝑝 + 1 𝐴 𝑛 𝑘 𝛽 ( 𝑘 ) 𝑛 − 1 / 𝑝 ∼ 𝛾 𝑛 𝑛 𝑘 = 1 𝑎 𝑘 𝛽 ( 𝑛 ) 𝑝 + 1 𝑛 𝑎 𝑛 𝑘 𝛽 ( 𝑘 ) 𝑛 − 1 / 𝑝 . ( 2 . 2 6 ) By Lemma 2.5 𝛾 𝑛 𝑛 𝑘 = 1 𝑎 𝑘 𝛽 ( 𝑛 ) 𝑝 + 1 𝑛 𝑎 𝑛 𝑘 𝛽 ( 𝑘 ) 𝑛 − 1 / 𝑝 ( 2 . 2 7 ) is bounded and so 𝑛 𝑘 = 1 𝑎 𝑘 𝛽 ( 𝑛 ) 𝑝 + 1 𝐴 𝑛 𝑘 𝛽 ( 𝑘 ) 𝑛 − 1 / 𝑝 ( 2 . 2 8 ) is bounded. We can see that ∞ 𝑛 = 𝑘 𝑎 𝑘 𝛽 ( 𝑛 ) 𝑝 + 1 𝐴 𝑛 𝑘 𝛽 ( 𝑘 ) 𝑛 1 / 𝑞 ( 2 . 2 9 ) is also bounded. Now by Theorem 2.3, 𝐴 is a bounded operator and so the proof is complete.

Lemma 2.7. Let { 𝑎 𝑛 } , { 𝑡 𝑛 } be nonnegative sequences with 𝑡 − 1 = 0 . Then for all 𝑛 ∈ ℕ one has 𝑛 𝑘 = 0 𝑡 𝑘 𝑎 𝑘 ≤ m a x 0 ≤ 𝑘 ≤ 𝑛 1 𝑛 − 𝑘 + 1 𝑛 𝑗 = 𝐾 𝑎 𝑗 𝑛 𝑘 = 1 𝑡 ( 𝑛 − 𝑘 + 1 ) 𝑘 − 𝑡 𝑘 − 1 + + 𝑡 0 ( 𝑛 + 1 ) . ( 2 . 3 0 )

Proof. Employing the summation by parts, we get 𝑛 𝑘 = 0 𝑡 𝑘 𝑎 𝑘 = 𝑛 𝑘 = 0 𝑛 𝑗 = 𝑘 𝑎 𝑗 𝑡 𝑘 − 𝑡 𝑘 − 1 ≤ 𝑛 𝑘 = 0 𝑛 𝑗 = 𝑘 𝑎 𝑗 1 𝑡 𝑛 − 𝑘 + 1 𝑘 − 𝑡 𝑘 − 1 + ( 𝑛 − 𝑘 + 1 ) . ( 2 . 3 1 ) So 𝑛 𝑘 = 0 𝑡 𝑘 𝑎 𝑘 ≤ m a x 0 ≤ 𝑘 ≤ 𝑛 1 𝑛 − 𝑘 + 1 𝑛 𝑗 = 𝑘 𝑎 𝑗 𝑛 𝑘 = 1 𝑡 ( 𝑛 − 𝑘 + 1 ) 𝑘 − 𝑡 𝑘 − 1 + + 𝑡 0 ( 𝑛 + 1 ) , ( 2 . 3 2 ) and at this time the proof is complete.

Theorem 2.8 (see [26, Theorem 8]). Let 1 / 𝑝 + 1 / 𝑞 = 1 , { 𝑥 𝑛 } be a positive sequence, then ∞ 𝑗 = 0 m a x 0 ≤ 𝑖 ≤ 𝑗 1 𝑗 − 𝑖 + 1 𝑗 𝑘 = 𝑖 𝑥 𝑘 𝑝 ≤ 𝑞 𝑝 ∞ 𝑘 = 0 𝑥 𝑝 𝑘 . ( 2 . 3 3 )

Theorem 2.9. Let { 𝑎 𝑛 } be a positive sequence and 𝑀 3 = s u p 𝑛 ≥ 0 𝑛 𝑘 = 1 𝑛 − 𝑘 + 1 𝐴 𝑛 𝑎 𝑘 − 𝑎 𝛽 ( 𝑘 ) 𝑘 − 1 𝛽 ( 𝑘 − 1 ) + 𝛽 ( 𝑛 ) 𝑝 + 1 + ( 𝑛 + 1 ) 𝑎 0 𝐴 𝑛 𝛽 ( 0 ) 𝛽 ( 𝑛 ) 𝑝 + 1 ( 2 . 3 4 ) be finite. Then 𝐴 is bounded and ‖ 𝐴 ‖ ≤ 𝑀 3 𝑞 .

Proof. Let ∑ 𝑓 ( 𝑧 ) = ∞ 𝑛 = 0 𝑓 ( 𝑛 ) 𝑧 𝑛 ∈ 𝐻 𝑝 ( 𝛽 ) , thus 𝐴 ( 𝑓 ) ( 𝑧 ) = ∞ 𝑛 = 0 𝑛 𝑘 = 0 𝑎 𝑘 𝛽 ( 𝑛 ) 𝑝 𝐴 𝑛 𝑧 𝑓 ( 𝑘 ) 𝑛 . ( 2 . 3 5 ) By definition of ‖ ⋅ ‖ 𝛽 , we have 𝑛 ≥ 0 𝛽 ( 𝑛 ) 𝑝 | | | | | 𝑛 𝑘 = 0 𝑎 𝑘 𝛽 ( 𝑛 ) 𝑝 𝐴 𝑛 𝑓 | | | | | ( 𝑘 ) 𝑝 ≤ 𝑛 ≥ 0 𝑛 𝑘 = 0 𝑎 𝑘 𝛽 ( 𝑛 ) 𝑝 + 1 𝐴 𝑛 | | | | 𝛽 ( 𝑘 ) 𝑓 ( 𝑘 ) 𝛽 ( 𝑘 ) 𝑝 . ( 2 . 3 6 ) In Lemma 2.7, consider 𝑡 𝑘 = 𝑎 𝑘 / 𝛽 ( 𝑘 ) and 𝑎 𝑗 = | 𝑓 ( 𝑗 ) | 𝛽 ( 𝑗 ) . Then 𝑛 ≥ 0 𝑛 𝑘 = 0 𝑎 𝑘 𝛽 ( 𝑛 ) 𝑝 + 1 𝐴 𝑛 | | | | 𝛽 ( 𝑘 ) 𝑓 ( 𝑘 ) 𝛽 ( 𝑘 ) 𝑝 ≤ 𝑛 ≥ 0 m a x 0 ≤ 𝑘 ≤ 𝑛 1 𝑛 − 𝑘 + 1 𝑛 𝑗 = 𝑘 | | | | 𝑓 ( 𝑗 ) 𝛽 ( 𝑗 ) 𝑝 × 𝑛 𝑘 = 1 𝑛 − 𝑘 + 1 𝐴 𝑛 𝑎 𝑘 − 𝑎 𝛽 ( 𝑘 ) 𝑘 − 1 𝛽 ( 𝑘 − 1 ) + 𝛽 ( 𝑛 ) 𝑝 + 1 + ( 𝑛 + 1 ) 𝑎 0 𝐴 𝑛 𝛽 ( 0 ) 𝛽 ( 𝑛 ) 𝑝 + 1 𝑝 . ( 2 . 3 7 ) Now, Theorem 2.8 implies that 𝑛 ≥ 0 m a x 0 ≤ 𝑘 ≤ 𝑛 1 𝑛 − 𝑘 + 1 𝑛 𝑗 = 𝑘 | | | | 𝑓 ( 𝑗 ) 𝛽 ( 𝑗 ) 𝑝 𝑀 𝑝 3 ≤ 𝑀 𝑝 3 𝑞 𝑝 ∞ 𝑘 = 1 | | | | 𝑓 ( 𝑘 ) 𝑝 𝛽 ( 𝑘 ) 𝑝 , ( 2 . 3 8 ) and so we get ‖ 𝐴 𝑓 ‖ ≤ 𝑀 3 𝑞 ‖ 𝑓 ‖ 𝛽 for all 𝑓 ∈ 𝐻 𝑝 ( 𝛽 ) . Thus 𝐴 ∈ 𝐵 ( 𝐻 𝑝 ( 𝛽 ) ) and indeed ‖ 𝐴 ‖ ≤ 𝑀 3 𝑞 . This completes the proof.

Corollary 2.10. Let 1 / 𝑝 + 1 / 𝑞 = 1 , 𝑎 𝑘 / 𝛽 ( 𝑘 ) ≥ 𝑎 𝑘 − 1 / 𝛽 ( 𝑘 − 1 ) and 𝑀 4 = s u p 𝑛 𝑛 ≥ 0 𝑘 = 0 𝑎 𝑘 𝛽 ( 𝑛 ) 𝑝 + 1 𝛽 ( 𝑘 ) 𝐴 𝑛 < ∞ . ( 2 . 3 9 ) Then 𝐴 is a bounded operator on 𝐻 𝑝 ( 𝛽 ) and ‖ 𝐴 ‖ ≤ 𝑀 4 .

Proof. Note that 𝑛 ≥ 0 𝑛 𝑘 = 0 𝑎 𝑘 𝛽 ( 𝑛 ) 𝑝 + 1 𝐴 𝑛 | | | | 𝛽 ( 𝑘 ) 𝑓 ( 𝑘 ) 𝛽 ( 𝑘 ) 𝑝 ≤ 𝑛 ≥ 0 m a x 0 ≤ 𝑘 ≤ 𝑛 1 𝑛 − 𝑘 + 1 𝑛 𝑗 = 𝑘 | | | | 𝑓 ( 𝑗 ) 𝛽 ( 𝑗 ) 𝑝 𝑛 𝑘 = 0 𝑎 𝑘 𝛽 ( 𝑛 ) 𝑝 + 1 𝛽 ( 𝑘 ) 𝐴 𝑛 𝑝 . ( 2 . 4 0 ) Theorem 2.8 implies that 𝑛 ≥ 0 m a x 0 ≤ 𝑘 ≤ 𝑛 1 𝑛 − 𝑘 + 1 𝑛 𝑗 = 𝑘 | | | | 𝑓 ( 𝑗 ) 𝛽 ( 𝑗 ) 𝑝 𝑀 𝑝 4 ≤ 𝑀 𝑝 4 𝑞 𝑝 ∞ 𝑘 = 1 | | | | 𝑓 ( 𝑘 ) 𝑝 𝛽 ( 𝑘 ) 𝑝 , ( 2 . 4 1 ) and so by Theorem 2.9 we obtain ‖ 𝐴 𝑓 ‖ ≤ 𝑞 𝑀 4 ‖ 𝑓 ‖ 𝛽 for all 𝑓 ∈ 𝐻 𝑝 ( 𝛽 ) . Thus 𝐴 ∈ 𝐵 ( 𝐻 𝑝 ( 𝛽 ) ) and indeed ‖ 𝐴 ‖ ≤ 𝑀 4 𝑞 . This completes the proof.

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

Theorem 2.11. Let 𝑆 be a nonempty subset of 𝐻 𝑝 ( 𝛽 ) . Then 𝑆 is relatively compact if and only if the following hold:(i)there exists 𝑀 > 0 , such that for all ∑ ∞ 𝑛 = 0 𝑓 ( 𝑛 ) 𝑧 𝑛 ∈ 𝑆 , | 𝑓 ( 𝑖 ) 𝛽 ( 𝑖 ) | ≤ 𝑀 for all 𝑖 ∈ ℕ ∪ { 0 } ;(ii)given 𝜖 > 0 , there is 𝑛 0 ∈ ℕ such that ∑ ∞ 𝑛 = 𝑛 0 | 𝑓 ( 𝑛 ) | 𝑝 𝛽 ( 𝑛 ) 𝑝 < 𝜖 𝑝 for all ∑ ∞ 𝑛 = 0 𝑓 ( 𝑛 ) 𝑧 𝑛 ∈ 𝑆 .

Proof. Let 𝑆 be relatively compact, thus there exist 𝑔 1 , … , 𝑔 𝑘 ∈ 𝐻 𝑝 ( 𝛽 ) such that 𝑆 ⊆ 𝑘 𝑖 = 1 𝐵 𝑔 𝑖 . , 1 ( 2 . 4 2 ) For every ∑ 𝑓 ( 𝑧 ) = ∞ 𝑛 = 0 𝑓 ( 𝑛 ) 𝑧 𝑛 ∈ 𝑆 , there is 𝑔 𝑖 such that 𝑓 ∈ 𝐵 ( 𝑔 𝑖 , 1 ) . By Minkowski inequality we get ∞ 𝑛 = 0 | | | | 𝑓 ( 𝑛 ) 𝑝 𝛽 ( 𝑛 ) 𝑝 ≤ ⎡ ⎢ ⎢ ⎣ ∞ 𝑛 = 0 | | 𝑓 ( 𝑛 ) − ̂ 𝑔 𝑖 ( | | 𝑛 ) 𝑝 𝛽 ( 𝑛 ) 𝑝 1 / 𝑝 + ∞ 𝑛 = 0 | | ̂ 𝑔 𝑖 ( | | 𝑛 ) 𝑝 𝛽 ( 𝑛 ) 𝑝 1 / 𝑝 ⎤ ⎥ ⎥ ⎦ 𝑝 ≤ ‖ ‖ 𝑓 − 𝑔 𝑖 ‖ ‖ + ‖ ‖ 𝑔 𝑖 ‖ ‖ 𝑝 ≤ ‖ ‖ 𝑔 1 + 𝑖 ‖ ‖ 𝑝 ≤ ‖ ‖ 𝑔 1 + m a x 𝑖 ‖ ‖ ∶ 𝑖 = 1 , … , 𝑘 𝑝 . ( 2 . 4 3 ) Thus for every 𝑓 ∈ 𝑆 and 𝑛 ∈ ℕ ∪ { 0 } , we get | | 𝑓 | | ‖ ‖ 𝑔 ( 𝑛 ) 𝛽 ( 𝑛 ) ≤ 1 + m a x 𝑖 ‖ ‖ . ∶ 𝑖 = 1 , … , 𝑘 ( 2 . 4 4 ) So (i) holds. Now suppose that 𝜖 is an arbitrary positive number. Since 𝑆 is relatively compact, thus there exist ℎ 1 , … , ℎ 𝑘 ∈ 𝐻 𝑝 ( 𝛽 ) such that 𝑆 ⊆ 𝑘 𝑖 = 1 𝐵 ℎ 𝑖 , 𝜖 2 . ( 2 . 4 5 ) Since ℎ 𝑖 ∈ 𝐻 𝑝 ( 𝛽 ) , there exists 𝑁 𝑖 ∈ ℕ such that ∞ 𝑛 = 𝑁 𝑖 | | | ℎ 𝑖 ( | | | 𝑛 ) 𝑝 𝛽 ( 𝑛 ) 𝑝 < 𝜖 𝑝 2 𝑝 ( 2 . 4 6 ) for 𝑖 = 1 , … , 𝑘 . Put 𝑁 0 𝑁 = m a x 𝑖 ∶ 𝑖 = 1 , … , 𝑘 , ( 2 . 4 7 ) and consider 𝑓 ∈ 𝑆 . Then there exists 𝑖 ∈ { 1 , … , 𝑘 } , such that 𝑓 ∈ 𝐵 ( ℎ 𝑖 , 𝜖 / 2 ) . Hence we get ∞ 𝑛 = 𝑁 0 | | | | 𝑓 ( 𝑛 ) 𝑝 𝛽 ( 𝑛 ) 𝑝 ≤ ⎡ ⎢ ⎢ ⎣ ⎛ ⎜ ⎜ ⎝ ∞ 𝑛 = 𝑁 0 | | | ℎ 𝑓 ( 𝑛 ) − 𝑖 | | | ( 𝑛 ) 𝑝 𝛽 ( 𝑛 ) 𝑝 ⎞ ⎟ ⎟ ⎠ 1 / 𝑝 + ⎛ ⎜ ⎜ ⎝ ∞ 𝑛 = 𝑁 0 | | | ℎ 𝑖 | | | ( 𝑛 ) 𝑝 𝛽 ( 𝑛 ) 𝑝 ⎞ ⎟ ⎟ ⎠ 1 / 𝑝 ⎤ ⎥ ⎥ ⎦ 𝑝 ≤ ‖ ‖ 𝑓 − ℎ 𝑖 ‖ ‖ + 𝜖 2 𝑝 ≤ 𝜖 𝑝 . ( 2 . 4 8 ) So (ii) holds.
Conversely, assume that 𝜖 > 0 be given and let (i) and (ii) hold. By condition (ii), there exists 𝑛 0 ∈ ℕ such that ∞ 𝑛 = 𝑛 0 | | | | 𝑓 ( 𝑛 ) 𝑝 𝛽 ( 𝑛 ) 𝑝 < 𝜖 𝑝 2 , ( 2 . 4 9 ) for all 𝑓 ∈ 𝑆 . Let 𝑀 𝑛 0 be the closed linear span of the set { 1 , 𝑧 , … , 𝑧 𝑛 0 − 1 } in 𝐻 𝑝 ( 𝛽 ) . Consider ℂ 𝑛 0 and 𝑀 𝑛 0 with norms ‖ ‖ 𝑧 1 , … , 𝑧 𝑛 0 ‖ ‖ = 𝑛 0 𝑛 = 1 | | 𝑧 𝑛 | | 𝑝 𝛽 ( 𝑛 ) 𝑝 1 / 𝑝 , ( 2 . 5 0 ) for all ( 𝑧 𝑖 ) 𝑛 0 𝑖 = 1 ∈ ℂ 𝑛 0 , and ‖ ‖ ‖ ‖ 𝑛 0 − 1 𝑖 = 0 𝑎 𝑖 𝑧 𝑖 ‖ ‖ ‖ ‖ = ⎛ ⎜ ⎜ ⎝ 𝑛 0 − 1 𝑖 = 0 | | 𝑎 𝑖 | | 𝑝 𝛽 ( 𝑖 ) 𝑝 ⎞ ⎟ ⎟ ⎠ 1 / 𝑝 ( 2 . 5 1 ) for all ∑ 𝑛 0 − 1 𝑖 = 0 𝑎 𝑖 𝑧 𝑖 ∈ 𝑀 𝑛 0 . Define 𝐿 ∶ 𝑀 𝑛 0 → ℂ 𝑛 0 , by 𝐿 ⎛ ⎜ ⎜ ⎝ 𝑛 0 − 1 𝑖 = 0 𝑎 𝑖 𝑧 𝑖 ⎞ ⎟ ⎟ ⎠ = 𝑎 0 , … , 𝑎 𝑛 0 − 1 . ( 2 . 5 2 ) Clearly, we can see that 𝐿 is a bounded linear operator. Now, consider the compact subset 𝑧 𝑖 𝑛 0 𝑖 = 1 ∶ 𝑛 0 𝑖 = 1 | | 𝑧 𝑖 | | 𝑝 𝛽 ( 𝑖 ) 𝑝 ≤ 𝑛 0 𝑀 𝑝 ( 2 . 5 3 ) in ℂ 𝑛 0 . Then we have ⎧ ⎪ ⎨ ⎪ ⎩ 𝑛 0 − 1 𝑖 = 0 𝑓 ( 𝑖 ) 𝑧 𝑖 ∶ ∞ 𝑛 = 0 𝑓 ( 𝑛 ) 𝑧 𝑛 ⎫ ⎪ ⎬ ⎪ ⎭ ∈ 𝑆 ⊆ 𝐿 − 1 𝑧 𝑖 𝑛 0 𝑖 = 1 ∶ 𝑛 0 𝑖 = 1 | | 𝑧 𝑖 | | 𝑝 𝛽 ( 𝑖 ) 𝑝 ≤ 𝑛 0 𝑀 𝑝 . ( 2 . 5 4 ) Since 𝐿 − 1 𝑧 𝑖 𝑛 0 𝑖 = 1 ∶ 𝑛 0 𝑖 = 1 | | 𝑧 𝑖 | | 𝑝 𝛽 ( 𝑖 ) 𝑝 ≤ 𝑛 0 𝑀 𝑝 ( 2 . 5 5 ) is a compact subspace of 𝑀 𝑛 0 , so there exist 𝑔 1 , … , 𝑔 𝑘 ∈ 𝑀 𝑛 0 such that 𝐿 − 1 𝑧 𝑖 𝑛 0 𝑖 = 1 ∶ 𝑛 0 𝑖 = 1 | | 𝑧 𝑖 | | 𝑝 𝛽 ( 𝑖 ) 𝑝 ≤ 𝑛 0 𝑀 𝑝 ∈ 𝑘 𝑖 = 1 𝐵 𝑔 𝑖 , 𝜖 2 1 / 𝑝 . ( 2 . 5 6 ) Hence for every ⎧ ⎪ ⎨ ⎪ ⎩ 𝑓 ∈ 𝑛 0 − 1 𝑖 = 0 𝑓 ( 𝑖 ) 𝑧 𝑖 ∶ 𝑓 ( 𝑧 ) = ∞ 𝑛 = 0 𝑓 ( 𝑛 ) 𝑧 𝑛 ⎫ ⎪ ⎬ ⎪ ⎭ ∈ 𝑆 ( 2 . 5 7 ) there is 𝑖 ∈ { 1 , … , 𝑘 } satisfying 𝑛 0 − 1 𝑛 = 0 | | 𝑓 ( 𝑛 ) − ̂ 𝑔 𝑖 | | ( 𝑛 ) 𝑝 𝛽 ( 𝑛 ) 𝑝 ≤ 𝜖 𝑝 2 . ( 2 . 5 8 ) Also, we have ‖ ‖ 𝑓 − 𝑔 𝑖 ‖ ‖ 𝛽 𝑝 ≤ 𝑛 0 − 1 𝑛 = 0 | | 𝑓 ( 𝑛 ) − ̂ 𝑔 𝑖 | | ( 𝑛 ) 𝑝 𝛽 ( 𝑛 ) 𝑝 + ∞ 𝑛 = 𝑛 0 | | | | 𝑓 ( 𝑛 ) 𝑝 𝛽 ( 𝑛 ) 𝑝 ≤ 𝜖 𝑝 2 + 𝜖 𝑝 2 ≤ 𝜖 𝑝 . ( 2 . 5 9 ) Thus, 𝑆 is relatively compact and so the proof is complete.

Theorem 2.12. Let the mean matrix operator 𝐴 be bounded on 𝐻 𝑝 ( 𝛽 ) , and l i m 𝑚 → ∞ ∞ 𝑛 = 𝑚 𝛽 ( 𝑛 ) 𝑝 2 + 𝑝 𝐴 𝑝 𝑛 1 / 𝑝 𝑚 𝑘 = 0 𝑎 𝑘 𝛽 ( 𝑘 ) 𝑞 1 / 𝑞 = 0 , ( 2 . 6 0 ) where 1 / 𝑝 + 1 / 𝑞 = 1 . Then 𝐴 is a compact operator on 𝐻 𝑝 ( 𝛽 ) .

Proof. Let 𝐵 𝐻 𝑝 ( 𝛽 ) be the closed unit ball of 𝐻 𝑝 ( 𝛽 ) . Define 𝑆 = 𝐴 ( 𝐵 𝐻 𝑝 ( 𝛽 ) ) and note that 𝑆 is a bounded subset of 𝐻 𝑝 ( 𝛽 ) . Put 𝑟 𝑛 = | 𝑓 ( 𝑛 ) | 𝑎 𝑛 , 𝑢 𝑛 = 𝛽 ( 𝑛 ) 𝑝 2 + 𝑝 / 𝐴 𝑝 𝑛 , 𝑣 𝑘 = ( 𝛽 ( 𝑘 ) / 𝑎 𝑘 ) 𝑝 , and 𝐸 𝑚 = ∞ 𝑛 = 𝑚 𝑢 𝑛 1 / 𝑝 𝑚 𝑘 = 0 𝑣 𝑘 1 − 𝑞 1 / 𝑞 . ( 2 . 6 1 ) Note that l i m 𝑚 → ∞ 𝐸 𝑚 = 0 . So for every 𝜖 > 0 , there exists 𝑚 0 ∈ ℕ such that 𝐸 𝑚 < 𝜖 / ( 𝑞 𝑝 − 1 𝑝 ) 1 / 𝑝 for all 𝑚 ≥ 𝑚 0 . Note that if 𝑓 ( 𝑧 ) = ∞ 𝑘 = 0 𝑓 ( 𝑘 ) 𝑧 𝑘 ∈ 𝐵 𝐻 𝑝 ( 𝛽 ) , ( 2 . 6 2 ) then 𝐴 𝑓 ( 𝑧 ) = ∞ 𝑛 = 0 𝑛 𝑘 = 0 𝑎 𝑘 𝛽 ( 𝑛 ) 𝑝 𝑓 ( 𝑘 ) 𝐴 𝑛 𝑧 𝑛 ∈ 𝑆 . ( 2 . 6 3 ) Since ‖ 𝑓 ‖ 𝑝 𝛽 ≤ 1 , we have ∞ 𝑛 = 𝑚 | | | | 𝐴 𝑓 ( 𝑛 ) 𝑛 𝛽 ( 𝑛 ) 𝑝 ≤ ∞ 𝑛 = 𝑚 𝛽 ( 𝑛 ) 𝑝 2 + 𝑝 𝐴 𝑝 𝑛 𝑛 𝑘 = 0 𝑎 𝑘 | | | | 𝑓 ( 𝑘 ) 𝑝 = ∞ 𝑛 = 𝑚 𝑢 𝑛 𝑛 𝑘 = 0 𝑟 𝑘 𝑝 ≤ 𝜖 𝑝 ∞ 𝑘 = 0 𝑟 𝑘 𝑝 𝑣 𝑘 ≤ 𝜖 𝑝 ∞ 𝑘 = 0 | | | | 𝑓 ( 𝑘 ) 𝑝 𝛽 ( 𝑘 ) 𝑝 ≤ 𝜖 𝑝 . ( 2 . 6 4 ) Thus by Theorem 2.11, 𝑆 is compact and so the proof is complete.

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