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A new note on generalized absolute matrix summability

Abstract

This paper gives necessary and sufficient conditions in order that a series a n λ n should be summable |B | k , k1, whenever a n is summable |A|. Some new results have also been obtained.

MSC:40D25, 40F05, 40G99.

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,i1).
(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 k1, 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 As=( 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 k1, 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 asn. Then a n λ n is summable|R, q n | k , k1whenever 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 , k1, 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 cases1k<, 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 Letk1, A=( a n v )andB=( 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 k1, 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:AB 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

YMX.
(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 <fori=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 <fori=1,2.

This completes the proof of the Theorem. □

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Özarslan, H., Ari, T. A new note on generalized absolute matrix summability. J Inequal Appl 2012, 166 (2012). https://doi.org/10.1186/1029-242X-2012-166

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Keywords

  • summability factors
  • absolute matrix summability
  • infinite series