Open Access

A new note on generalized absolute matrix summability

Journal of Inequalities and Applications20122012:166

https://doi.org/10.1186/1029-242X-2012-166

Received: 3 June 2012

Accepted: 13 July 2012

Published: 27 July 2012

Abstract

This paper gives necessary and sufficient conditions in order that a series a n λ n should be summable | B | k , k 1 , whenever a n is summable | A | . Some new results have also been obtained.

MSC:40D25, 40F05, 40G99.

Keywords

summability factorsabsolute matrix summabilityinfinite series

1 Introduction

Let a n be a given infinite series with the partial sums ( s n ) . Let ( p n ) be a sequence of positive numbers such that
P n = v = 0 n p v as ( n ) , ( P i = p i = 0 , i 1 ) .
(1)
The sequence-to-sequence transformation
t n = 1 P n v = 0 n p v s v
(2)
defines the sequence ( t n ) of the Riesz means of the sequence ( s n ) generated by the sequence of coefficients ( p n ) (see [3]). The series a n is said to be summable | R , p n | k k 1 , if (see [1])
n = 1 n k 1 | t n t n 1 | k < .
(3)
Let A = ( a n v ) be a normal matrix, i.e., a lower triangular matrix of nonzero diagonal entries. Then A defines the sequence-to-sequence transformation, mapping the sequence s = ( s n ) to A s = ( A n ( s ) ) , where
A n ( s ) = v = 0 n a n v s v , n = 0 , 1 , .
(4)
The series a n is said to be summable | A | k k 1 , if (see [7])
n = 1 n k 1 | Δ ¯ A n ( s ) | k < ,
(5)
where
Δ ¯ A n ( s ) = A n ( s ) A n 1 ( s ) .

If we take a n v = p v P n , then | A | k summability is the same as | R , p n | k summability.

Before stating the main theorem we must first introduce some further notations.

Given a normal matrix A = ( a n v ) , we associate two lover semimatrices A ¯ = ( a ¯ n v ) and A ˆ = ( a ˆ n v ) as follows:
a ¯ n v = i = v n a n i , n , v = 0 , 1 ,
(6)
and
a ˆ 00 = a ¯ 00 = a 00 , a ˆ n v = a ¯ n v a ¯ n 1 , v , n = 1 , 2 , .
(7)
It may be noted that A ¯ and A ˆ are the well-known matrices of series-to-sequence and series-to-series transformations, respectively. Then, we have
A n ( s ) = v = 0 n a n v s v = v = 0 n a ¯ n v a v
(8)
and
Δ ¯ A n ( s ) = v = 0 n a ˆ n v a v .
(9)

If A is a normal matrix, then A = ( a n v ) will denote the inverse of A. Clearly if A is normal, then A ˆ = ( a ˆ n v ) is normal and has two-sided inverse A ˆ = ( a ˆ n v ) , which is also normal (see [2]).

Sarıgöl [6] has proved the following theorem for | R , p n | k summability method.

Theorem A Suppose that ( p n ) and ( q n ) are positive sequences with P n and Q n as n . Then a n λ n is summable | R , q n | k , k 1 whenever a n is summable | R , p n | , if and only if
(10)
(11)
(12)
provided that
W n = { v = n + 1 v k 1 ( q v Q v Q v 1 ) k } 1 k < .
Theorem B The | R , p n | summability implies the | R , q n | k , k 1 , summability if and only if the following conditions hold:
(13)
(14)
(15)
where
W v = { i = v + 1 i k 1 ( q i Q i Q i 1 ) k } 1 k <

and we regarded that the above series converges for each v and Δ is the forward difference operator.

It may be remarked that the above theorem has been proved by Orhan and Sarıgöl [5].

Lemma ([4])

A = ( a n v ) ( l 1 , l k ) if and only if
sup v n = 1 | a n v | k <
(16)

for the cases 1 k < , where ( l 1 , l k ) denotes the set of all matrices A which map l 1 into l k = { x = ( x n ) : | x n | k < } .

2 Main theorem

The aim of this paper is to generalize Theorem A for the | A | and | B | k summabilities. Therefore we shall prove the following theorem.

Theorem Let k 1 , A = ( a n v ) and B = ( b n v ) be two positive normal matrices such that
(17)
(18)
Then, in order that a n λ n is summable | B | k whenever a n is summable | A | , it is necessary that
(19)
(20)
(21)
Also (19)-(21) and
(22)
(23)
(24)

are sufficient for the consequent to hold.

It should be noted that if we take a n v = p v P n and b n v = q v Q n , then we get Theorem A. Also if we take λ n = 1 , then we get Theorem B.

Proof of the Theorem Necessity. Let ( x n ) and ( y n ) denote A-transform and B-transform of the series a n and a n λ n , respectively. Then, by (8) and (9), we have
Δ ¯ x n = v = 0 n a ˆ n v a v and Δ ¯ y n = v = 0 n b ˆ n v a v λ v .
(25)
For k 1 , we define
A = { ( a i ) : a i is summable | A | } , B = { ( a i λ i ) : a i λ i is summable | B | k } .
Then it is routine to verify that these are BK-spaces, if normed by
X = { n = 0 | Δ ¯ x n | }
(26)
and
Y = { n = 0 n k 1 | Δ ¯ y n | k } 1 k
(27)
respectively. Since a n is summable | A | implies a n λ n is summable | B | k , by the hypothesis of the theorem,
X < Y < .
Now consider the inclusion map c : A B defined by c ( x ) = x . This is continuous, which is immediate as A and B are BK-spaces. Thus there exists a constant M such that
Y M X .
(28)
By applying (25) to a v = e v e v + 1 ( e v is the v th coordinate vector), we have
Δ ¯ x n = { 0 , if n < v , a ˆ n v , if n = v , Δ v a ˆ n v , if n > v
and
Δ ¯ y n = { 0 , if n < v , b ˆ n v λ v , if n = v , Δ v ( b ˆ n v λ v ) , if n > v .
So (26) and (27) give us
X = { a v v + n = v + 1 | Δ v a ˆ n v | }
and
Y = { v k 1 b v v | λ v | k + n = v + 1 n k 1 | Δ v ( b ˆ n v λ v ) | k } 1 k .
Hence it follows from (28) that
v k 1 b v v | λ v | k + n = v + 1 n k 1 | Δ v b ˆ n v λ v | k M k a v v k + M k n = v + 1 | Δ v a ˆ n v | k .
Using (17), we can find
v k 1 b v v | λ v | k + n = v + 1 n k 1 | Δ v ( b ˆ n v λ v ) | k = O ( a v v k ) .
The above inequality will be true if and only if each term on the left-hand side is O ( a v v k ) . Taking the first term,
v k 1 b v v | λ v | k = O ( a v v k )
then
| λ v | = O ( v 1 k 1 a v v b v v )
which verifies that (19) is necessary. Using the second term, we have
n = v + 1 n k 1 | Δ v ( b ˆ n v λ v ) | k = O ( a v v k )
which is condition (20). Now, if we apply (22) to a v = e v + 1 , we have
Δ ¯ x n = { 0 , if n v , a ˆ n , v + 1 , if n > v
and
Δ ¯ y n = { 0 , if n v , b ˆ n , v + 1 λ v + 1 , if n > v ,
respectively. Hence
Hence it follows from (28) that
n = v + 1 n k 1 | b ˆ n , v + 1 λ v + 1 | k M k { n = v + 1 | a ˆ n , v + 1 | } k .
Using (18) we can find
n = v + 1 n k 1 | b ˆ n , v + 1 λ v + 1 | k = O ( 1 )

which is condition (21).

Sufficiency. We use the notations of necessity. Then
Δ ¯ x n = v = 0 n a ˆ n v a v
(29)
which implies
a v = r = 0 v a ˆ v r Δ ¯ x r .
(30)
In this case
Δ ¯ y n = v = 0 n b ˆ n v a v λ v = v = 0 n b ˆ n v λ v r = 0 v a ˆ v r Δ ¯ x r .
On the other hand, since
b ˆ n 0 = b ¯ n 0 b ¯ n 1 , 0
by (22), we have
Δ ¯ y n = v = 1 n b ˆ n v λ v { r = 0 v a ˆ v r Δ ¯ x r } = v = 1 n b ˆ n v λ v { a ˆ v v Δ ¯ x v + a ˆ v , v 1 Δ ¯ x v 1 + r = 0 v 2 a ˆ v r Δ ¯ x r } = v = 1 n b ˆ n v λ v a ˆ v v Δ ¯ x v + v = 1 n b ˆ n v λ v a ˆ v , v 1 Δ ¯ x v 1 + v = 1 n b ˆ n v λ v r = 0 v 2 a ˆ v r Δ ¯ x r = b ˆ n n λ n a ˆ n n Δ ¯ x n + v = 1 n 1 ( b ˆ n v λ v a ˆ v v + b ˆ n , v + 1 λ v + 1 a ˆ v + 1 , v ) Δ ¯ x v + r = 0 n 2 Δ ¯ x r v = r + 2 n b ˆ n v λ v a ˆ v r .
(31)
By considering the equality
k = v n a ˆ n k a ˆ k v = δ n v ,
where δ n v is the Kronecker delta, we have that
b ˆ n v λ v a ˆ v v + b ˆ n , v + 1 λ v + 1 a ˆ v + 1 , v = b ˆ n v λ v a ˆ v v + b ˆ n , v + 1 λ v + 1 ( a ˆ v + 1 , v a ˆ v v a ˆ v + 1 , v + 1 ) = b ˆ n v λ v a v v b ˆ n , v + 1 λ v + 1 ( a ¯ v + 1 , v a ¯ v , v ) a v v a v + 1 , v + 1 = b ˆ n v λ v a v v b ˆ n , v + 1 λ v + 1 ( a v + 1 , v + 1 + a v + 1 , v a v v ) a v v a v + 1 , v + 1 = Δ v ( b ˆ n v λ v ) a v v + b ˆ n , v + 1 λ v + 1 a v v a v + 1 , v a v v a v + 1 , v + 1
and so
Δ ¯ y n = b n n λ n a n n Δ ¯ x n + v = 1 n 1 Δ v ( b ˆ n v λ v ) a v v Δ ¯ x v + v = 1 n 1 b ˆ n , v + 1 λ v + 1 a v v a v + 1 , v a v v a v + 1 , v + 1 Δ ¯ x v + r = 0 n 2 Δ ¯ x r v = r + 2 n b ˆ n v λ v a ˆ v r .
Let
Since
| T n ( 1 ) + T n ( 2 ) | k 2 k ( | T n ( 1 ) | k + | T n ( 2 ) | k )
to complete the proof of Theorem, it is sufficient to show that
n = 1 n k 1 | T n ( i ) | k < for i = 1 , 2 .
Then
T n ( 1 ) ¯ = n 1 1 k T n ( 1 ) = n 1 1 k b n n λ n a n n Δ ¯ x n + n 1 1 k v = 1 n 1 Δ v ( b ˆ n v λ v ) a v v Δ ¯ x v + n 1 1 k v = 1 n 1 b ˆ n , v + 1 λ v + 1 a v v a v + 1 , v a v v a v + 1 , v + 1 Δ ¯ x v = v = 1 c n v Δ ¯ x v ,
where
c n v = { n 1 1 k ( Δ v ( b n v λ v ) a v v + b ˆ n , v + 1 λ v + 1 a v v a v + 1 , v a v v a v + 1 , v + 1 ) , if 1 v n 1 , n 1 1 k b n n λ n a n n , if v = n , 0 , if v > n .
Now
| T n ( 1 ) ¯ | k < whenever | Δ ¯ x n | <
is equivalently
sup v n = 1 | c n v | k <
(32)
by Lemma. But it follows from conditions (20), (21) and (23) that
n = v | c n v | k = O ( 1 ) { n k 1 | b n n λ n a n n | k + n = v + 1 n k 1 | Δ v ( b ˆ n v λ v ) a v v + b ˆ n , v + 1 λ v + 1 a v v a v + 1 , v a v v a v + 1 , v + 1 | k } = O ( 1 ) as v .
Finally,
T n ( 2 ) ¯ = n 1 1 k T n ( 2 ) = n 1 1 k r = 0 n 2 Δ ¯ x r v = r + 2 n b ˆ n v a ˆ v r λ v = r = 0 d n r Δ ¯ x r ,
where
d n r = { n 1 1 k v = r + 2 n b ˆ n v a ˆ v r λ v , if 0 r n 2 , 0 , if r > n 2 .
Now
| T n ( 2 ) ¯ | k < whenever | Δ ¯ x n | <
is equivalently
sup r n = 1 | d n r | k <
(33)
by Lemma. But it follows from conditions (21) and (24) that
n = r + 2 | d n r | k = O ( 1 ) n = r + 2 n k 1 { v = r + 2 | b ˆ n v a ˆ v r λ v | } k = O ( 1 ) n = r + 2 n k 1 | b ˆ n , r + 1 λ r + 1 | k = O ( 1 ) as r .
Therefore, we have
n = 1 n k 1 | T n ( i ) | k < for i = 1 , 2 .

This completes the proof of the Theorem. □

Authors’ Affiliations

(1)
Department of Mathematics, Erciyes University

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Copyright

© Özarslan and Ari; licensee Springer 2012

This article is published under license to BioMed Central Ltd. This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/2.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.