A new note on generalized absolute matrix summability

Abstract

This paper gives necessary and sufficient conditions in order that a series $\sum a_n{\lambda }_n$ should be summable $|B{|}_k$, $k\ge 1$, whenever $\sum a_n$ is summable $|A|$. Some new results have also been obtained.

MSC:40D25, 40F05, 40G99.

1 Introduction

Let $\sum a_n$ be a given infinite series with the partial sums $(s_n)$. Let $(p_n)$ be a sequence of positive numbers such that

$$P_n=\sum _{v=0}^n{p}_v\to \mathrm{\infty }\text{as}(n\to \mathrm{\infty }),(P_{-i}=p_{-i}=0,i\ge 1).$$

(1)

The sequence-to-sequence transformation

$$t_n=\frac{1}{P_n}\sum _{v=0}^n{p}_v{s}_v$$

(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 $\sum a_n$ is said to be summable $|R,p_n{|}_k$$k\ge 1$, if (see [1])

$$\sum _{n=1}^{\mathrm{\infty }}{n}^{k-1}|t_n-t_{n-1}{|}^k<\mathrm{\infty }.$$

(3)

Let $A=(a_{nv})$ 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

$$A_n(s)=\sum _{v=0}^n{a}_{nv}{s}_v,n=0,1,\dots .$$

(4)

The series $\sum a_n$ is said to be summable $|A{|}_k$$k\ge 1$, if (see [7])

$$\sum _{n=1}^{\mathrm{\infty }}{n}^{k-1}|\overline{\mathrm{\Delta }}{A}_n(s){|}^k<\mathrm{\infty },$$

(5)

where

$$\overline{\mathrm{\Delta }}{A}_n(s)=A_n(s)-A_{n-1}(s).$$

If we take $a_{nv}=\frac{p_v}{P_n}$, 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 $A=(a_{nv})$, we associate two lover semimatrices $\overline{A}=({\overline{a}}_{nv})$ and $\hat{A}=({\hat{a}}_{nv})$ as follows:

$${\overline{a}}_{nv}=\sum _{i=v}^n{a}_{ni},n,v=0,1,\dots$$

(6)

and

$${\hat{a}}_{00}={\overline{a}}_{00}=a_{00},{\hat{a}}_{nv}={\overline{a}}_{nv}-{\overline{a}}_{n-1,v},n=1,2,\dots .$$

(7)

It may be noted that $\overline{A}$ and $\hat{A}$ are the well-known matrices of series-to-sequence and series-to-series transformations, respectively. Then, we have

$$A_n(s)=\sum _{v=0}^n{a}_{nv}{s}_v=\sum _{v=0}^n{\overline{a}}_{nv}{a}_v$$

(8)

and

$$\overline{\mathrm{\Delta }}{A}_n(s)=\sum _{v=0}^n{\hat{a}}_{nv}{a}_v.$$

(9)

If A is a normal matrix, then $A^{\prime }=(a_{nv}^{\prime })$ will denote the inverse of A. Clearly if A is normal, then $\hat{A}=({\hat{a}}_{nv})$ is normal and has two-sided inverse ${\hat{A}}^{\prime }=({\hat{a}}_{nv}^{\prime })$, 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$P_n\to \mathrm{\infty }$and$Q_n\to \mathrm{\infty }$as$n\to \mathrm{\infty }$. Then$\sum a_n{\lambda }_n$is summable$|R,q_n{|}_k$, $k\ge 1$whenever$\sum a_n$is summable$|R,p_n|$, if and only if

(10)

(11)

(12)

provided that

$$W_n={\left\{\sum _{v=n+1}^{\mathrm{\infty }}{v}^{k-1}{\left(\frac{q_v}{Q_v{Q}_{v-1}}\right)}^k\right\}}^{\frac{1}{k}}<\mathrm{\infty }.$$

Theorem B The$|R,p_n|$summability implies the$|R,q_n{|}_k$, $k\ge 1$, summability if and only if the following conditions hold:

(13)

(14)

(15)

where

$$W_v={\left\{\sum _{i=v+1}^{\mathrm{\infty }}{i}^{k-1}{\left(\frac{q_i}{Q_i{Q}_{i-1}}\right)}^k\right\}}^{\frac{1}{k}}<\mathrm{\infty }$$

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])

$A=(a_{nv})\in (l_1,l_k)$ if and only if

$$\underset{v}{sup}\sum _{n=1}^{\mathrm{\infty }}{|a_{nv}|}^k<\mathrm{\infty }$$

(16)

for the cases$1\le k<\mathrm{\infty }$, where$(l_1,l_k)$denotes the set of all matrices A which map$l_1$into$l_k=\{x=(x_n):\sum {|x_n|}^k<\mathrm{\infty }\}$.

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$k\ge 1$, $A=(a_{nv})$and$B=(b_{nv})$be two positive normal matrices such that

(17)

(18)

Then, in order that$\sum a_n{\lambda }_n$is summable$|B{|}_k$whenever$\sum a_n$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 $a_{nv}=\frac{p_v}{P_n}$ and $b_{nv}=\frac{q_v}{Q_n}$, then we get Theorem A. Also if we take ${\lambda }_n=1$, 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 $\sum a_n$ and $\sum a_n{\lambda }_n$, respectively. Then, by (8) and (9), we have

$$\overline{\mathrm{\Delta }}{x}_n=\sum _{v=0}^n{\hat{a}}_{nv}{a}_v\text{and}\overline{\mathrm{\Delta }}{y}_n=\sum _{v=0}^n{\hat{b}}_{nv}{a}_v{\lambda }_v.$$

(25)

For $k\ge 1$, we define

$$A=\{(a_i):\sum a_i\text{is summable}|A|\},B=\{(a_i{\lambda }_i):\sum a_i{\lambda }_i\text{is summable}{|B|}_k\}.$$

Then it is routine to verify that these are BK-spaces, if normed by

$$\parallel X\parallel =\{\sum _{n=0}^{\mathrm{\infty }}|\overline{\mathrm{\Delta }}{x}_n|\}$$

(26)

and

$$\parallel Y\parallel ={\left\{\sum _{n=0}^{\mathrm{\infty }}{n}^{k-1}{|\overline{\mathrm{\Delta }}{y}_n|}^k\right\}}^{\frac{1}{k}}$$

(27)

respectively. Since $\sum a_n$ is summable $|A|$ implies $\sum a_n{\lambda }_n$ is summable ${|B|}_k$, by the hypothesis of the theorem,

$$\parallel X\parallel <\mathrm{\infty }\Rightarrow \parallel Y\parallel <\mathrm{\infty }.$$

Now consider the inclusion map $c:A\to B$ 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

$$\parallel Y\parallel \le M\parallel X\parallel .$$

(28)

By applying (25) to $a_v=e_v-e_{v+1}$ ($e_v$ is the v th coordinate vector), we have

$$\overline{\mathrm{\Delta }}{x}_n=\{\begin{array}{cc}0,\hfill & \text{if}n<v,\hfill \\ {\hat{a}}_{nv},\hfill & \text{if}n=v,\hfill \\ {\mathrm{\Delta }}_v{\hat{a}}_{nv},\hfill & \text{if}n>v\hfill \end{array}$$

and

$$\overline{\mathrm{\Delta }}{y}_n=\{\begin{array}{cc}0,\hfill & \text{if}n<v,\hfill \\ {\hat{b}}_{nv}{\lambda }_v,\hfill & \text{if}n=v,\hfill \\ {\mathrm{\Delta }}_v({\hat{b}}_{nv}{\lambda }_v),\hfill & \text{if}n>v.\hfill \end{array}$$

So (26) and (27) give us

$$\parallel X\parallel =\{a_{vv}+\sum _{n=v+1}^{\mathrm{\infty }}|{\mathrm{\Delta }}_v{\hat{a}}_{nv}|\}$$

and

$$\parallel Y\parallel ={\{v^{k-1}{b}_{vv}{|{\lambda }_v|}^k+\sum _{n=v+1}^{\mathrm{\infty }}{n}^{k-1}{|{\mathrm{\Delta }}_v({\hat{b}}_{nv}{\lambda }_v)|}^k\}}^{\frac{1}{k}}.$$

Hence it follows from (28) that

$$v^{k-1}{b}_{vv}{|{\lambda }_v|}^k+\sum _{n=v+1}^{\mathrm{\infty }}{n}^{k-1}{|{\mathrm{\Delta }}_v{\hat{b}}_{nv}{\lambda }_v|}^k\le M^k{a}_{vv}^k+M^k\sum _{n=v+1}^{\mathrm{\infty }}{|{\mathrm{\Delta }}_v{\hat{a}}_{nv}|}^k.$$

Using (17), we can find

$$v^{k-1}{b}_{vv}{|{\lambda }_v|}^k+\sum _{n=v+1}^{\mathrm{\infty }}{n}^{k-1}{|{\mathrm{\Delta }}_v({\hat{b}}_{nv}{\lambda }_v)|}^k=O\left(a_{vv}^k\right).$$

The above inequality will be true if and only if each term on the left-hand side is $O(a_{vv}^k)$. Taking the first term,

$$v^{k-1}{b}_{vv}{|{\lambda }_v|}^k=O\left(a_{vv}^k\right)$$

then

$$|{\lambda }_v|=O\left(v^{\frac{1}{k}-1}\frac{a_{vv}}{b_{vv}}\right)$$

which verifies that (19) is necessary. Using the second term, we have

$$\sum _{n=v+1}^{\mathrm{\infty }}{n}^{k-1}{|{\mathrm{\Delta }}_v({\hat{b}}_{nv}{\lambda }_v)|}^k=O\left(a_{vv}^k\right)$$

which is condition (20). Now, if we apply (22) to $a_v=e_{v+1}$, we have

$$\overline{\mathrm{\Delta }}{x}_n=\{\begin{array}{cc}0,\hfill & \text{if}n\le v,\hfill \\ {\hat{a}}_{n,v+1},\hfill & \text{if}n>v\hfill \end{array}$$

and

$$\overline{\mathrm{\Delta }}{y}_n=\{\begin{array}{cc}0,\hfill & \text{if}n\le v,\hfill \\ {\hat{b}}_{n,v+1}{\lambda }_{v+1},\hfill & \text{if}n>v,\hfill \end{array}$$

respectively. Hence

Hence it follows from (28) that

$$\sum _{n=v+1}^{\mathrm{\infty }}{n}^{k-1}{|{\hat{b}}_{n,v+1}{\lambda }_{v+1}|}^k\le M^k{\{\sum _{n=v+1}^{\mathrm{\infty }}|{\hat{a}}_{n,v+1}|\}}^k.$$

Using (18) we can find

$$\sum _{n=v+1}^{\mathrm{\infty }}{n}^{k-1}{|{\hat{b}}_{n,v+1}{\lambda }_{v+1}|}^k=O(1)$$

which is condition (21).

Sufficiency. We use the notations of necessity. Then

$$\overline{\mathrm{\Delta }}{x}_n=\sum _{v=0}^n{\hat{a}}_{nv}{a}_v$$

(29)

which implies

$$a_v=\sum _{r=0}^v{\hat{a}}_{vr}^{\prime }\overline{\mathrm{\Delta }}{x}_r.$$

(30)

In this case

$$\overline{\mathrm{\Delta }}{y}_n=\sum _{v=0}^n{\hat{b}}_{nv}{a}_v{\lambda }_v=\sum _{v=0}^n{\hat{b}}_{nv}{\lambda }_v\sum _{r=0}^v{\hat{a}}_{vr}^{\prime }\overline{\mathrm{\Delta }}{x}_r.$$

On the other hand, since

$${\hat{b}}_{n0}={\overline{b}}_{n0}-{\overline{b}}_{n-1,0}$$

by (22), we have

$$\begin{array}{rcl}\overline{\mathrm{\Delta }}{y}_n& =& \sum _{v=1}^n{\hat{b}}_{nv}{\lambda }_v\left\{\sum _{r=0}^v{\hat{a}}_{vr}^{\prime }\overline{\mathrm{\Delta }}{x}_r\right\}\\ =& \sum _{v=1}^n{\hat{b}}_{nv}{\lambda }_v\{{\hat{a}}_{vv}^{\prime }\overline{\mathrm{\Delta }}{x}_v+{\hat{a}}_{v,v-1}^{\prime }\overline{\mathrm{\Delta }}{x}_{v-1}+\sum _{r=0}^{v-2}{\hat{a}}_{vr}^{\prime }\overline{\mathrm{\Delta }}{x}_r\}\\ =& \sum _{v=1}^n{\hat{b}}_{nv}{\lambda }_v{\hat{a}}_{vv}^{\prime }\overline{\mathrm{\Delta }}{x}_v+\sum _{v=1}^n{\hat{b}}_{nv}{\lambda }_v{\hat{a}}_{v,v-1}^{\prime }\overline{\mathrm{\Delta }}{x}_{v-1}+\sum _{v=1}^n{\hat{b}}_{nv}{\lambda }_v\sum _{r=0}^{v-2}{\hat{a}}_{vr}^{\prime }\overline{\mathrm{\Delta }}{x}_r\\ =& {\hat{b}}_{nn}{\lambda }_n{\hat{a}}_{nn}^{\prime }\overline{\mathrm{\Delta }}{x}_n+\sum _{v=1}^{n-1}({\hat{b}}_{nv}{\lambda }_v{\hat{a}}_{vv}^{\prime }+{\hat{b}}_{n,v+1}{\lambda }_{v+1}{\hat{a}}_{v+1,v}^{\prime })\overline{\mathrm{\Delta }}{x}_v\\ +\sum _{r=0}^{n-2}\overline{\mathrm{\Delta }}{x}_r\sum _{v=r+2}^n{\hat{b}}_{nv}{\lambda }_v{\hat{a}}_{vr}^{\prime }.\end{array}$$

(31)

By considering the equality

$$\sum _{k=v}^n{\hat{a}}_{nk}^{\prime }{\hat{a}}_{kv}={\delta }_{nv},$$

where ${\delta }_{nv}$ is the Kronecker delta, we have that

$$\begin{array}{rcl}{\hat{b}}_{nv}{\lambda }_v{\hat{a}}_{vv}^{\prime }+{\hat{b}}_{n,v+1}{\lambda }_{v+1}{\hat{a}}_{v+1,v}^{\prime }& =& \frac{{\hat{b}}_{nv}{\lambda }_v}{{\hat{a}}_{vv}}+{\hat{b}}_{n,v+1}{\lambda }_{v+1}(-\frac{{\hat{a}}_{v+1,v}}{{\hat{a}}_{vv}{\hat{a}}_{v+1,v+1}})\\ =& \frac{{\hat{b}}_{nv}{\lambda }_v}{a_{vv}}-\frac{{\hat{b}}_{n,v+1}{\lambda }_{v+1}({\overline{a}}_{v+1,v}-{\overline{a}}_{v,v})}{a_{vv}{a}_{v+1,v+1}}\\ =& \frac{{\hat{b}}_{nv}{\lambda }_v}{a_{vv}}-\frac{{\hat{b}}_{n,v+1}{\lambda }_{v+1}(a_{v+1,v+1}+a_{v+1,v}-a_{vv})}{a_{vv}{a}_{v+1,v+1}}\\ =& \frac{{\mathrm{\Delta }}_v({\hat{b}}_{nv}{\lambda }_v)}{a_{vv}}+{\hat{b}}_{n,v+1}{\lambda }_{v+1}\frac{a_{vv}-a_{v+1,v}}{a_{vv}{a}_{v+1,v+1}}\end{array}$$

and so

$$\begin{array}{rcl}\overline{\mathrm{\Delta }}{y}_n& =& \frac{b_{nn}{\lambda }_n}{a_{nn}}\overline{\mathrm{\Delta }}{x}_n+\sum _{v=1}^{n-1}\frac{{\mathrm{\Delta }}_v({\hat{b}}_{nv}{\lambda }_v)}{a_{vv}}\overline{\mathrm{\Delta }}{x}_v+\sum _{v=1}^{n-1}{\hat{b}}_{n,v+1}{\lambda }_{v+1}\frac{a_{vv}-a_{v+1,v}}{a_{vv}{a}_{v+1,v+1}}\overline{\mathrm{\Delta }}{x}_v\\ +\sum _{r=0}^{n-2}\overline{\mathrm{\Delta }}{x}_r\sum _{v=r+2}^n{\hat{b}}_{nv}{\lambda }_v{\hat{a}}_{vr}^{\prime }.\end{array}$$

Let

Since

$$|T_n(1)+T_n(2){|}^k\le 2^k\left(\right|T_n(1){|}^k+|T_n(2){|}^k)$$

to complete the proof of Theorem, it is sufficient to show that

$$\sum _{n=1}^{\mathrm{\infty }}{n}^{k-1}|T_n(i){|}^k<\mathrm{\infty }\text{for}i=1,2.$$

Then

$$\begin{array}{rcl}\overline{T_n(1)}& =& n^{1-\frac{1}{k}}{T}_n(1)\\ =& n^{1-\frac{1}{k}}\frac{b_{nn}{\lambda }_n}{a_{nn}}\overline{\mathrm{\Delta }}{x}_n+n^{1-\frac{1}{k}}\sum _{v=1}^{n-1}\frac{{\mathrm{\Delta }}_v({\hat{b}}_{nv}{\lambda }_v)}{a_{vv}}\overline{\mathrm{\Delta }}{x}_v+n^{1-\frac{1}{k}}\sum _{v=1}^{n-1}{\hat{b}}_{n,v+1}{\lambda }_{v+1}\frac{a_{vv}-a_{v+1,v}}{a_{vv}{a}_{v+1,v+1}}\overline{\mathrm{\Delta }}{x}_v\\ =& \sum _{v=1}^{\mathrm{\infty }}{c}_{nv}\overline{\mathrm{\Delta }}{x}_v,\end{array}$$

where

$$c_{nv}=\{\begin{array}{cc}{n}^{1-\frac{1}{k}}(\frac{{\mathrm{\Delta }}_v(b_{nv}{\lambda }_v)}{a_{vv}}+{\hat{b}}_{n,v+1}{\lambda }_{v+1}\frac{a_{vv}-a_{v+1,v}}{a_{vv}{a}_{v+1,v+1}}),\hfill & \text{if}1\le v\le n-1\text{,}\hfill \\ n^{1-\frac{1}{k}}\frac{b_{nn}{\lambda }_n}{a_{nn}},\hfill & \text{if}v=n,\hfill \\ 0,\hfill & \text{if}v>n\text{.}\hfill \end{array}$$

Now

$$\sum {\left|\overline{T_n(1)}\right|}^k<\mathrm{\infty }\text{whenever}\sum |\overline{\mathrm{\Delta }}{x}_n|<\mathrm{\infty }$$

is equivalently

$$\underset{v}{sup}\sum _{n=1}^{\mathrm{\infty }}{|c_{nv}|}^k<\mathrm{\infty }$$

(32)

by Lemma. But it follows from conditions (20), (21) and (23) that

$$\begin{array}{rcl}\sum _{n=v}^{\mathrm{\infty }}{|c_{nv}|}^k& =& O(1)\{n^{k-1}{\left|\frac{b_{nn}{\lambda }_n}{a_{nn}}\right|}^k+\sum _{n=v+1}^{\mathrm{\infty }}{n}^{k-1}|\frac{{\mathrm{\Delta }}_v({\hat{b}}_{nv}{\lambda }_v)}{a_{vv}}+{\hat{b}}_{n,v+1}{\lambda }_{v+1}\frac{a_{vv}-a_{v+1,v}}{a_{vv}{a}_{v+1,v+1}}{|}^k\}\\ =& O(1)\text{as}v\to \mathrm{\infty }.\end{array}$$

Finally,

$$\overline{T_n(2)}=n^{1-\frac{1}{k}}{T}_n(2)=n^{1-\frac{1}{k}}\sum _{r=0}^{n-2}\overline{\mathrm{\Delta }}{x}_r\sum _{v=r+2}^n{\hat{b}}_{nv}{\hat{a}}_{vr}^{\prime }{\lambda }_v=\sum _{r=0}^{\mathrm{\infty }}{d}_{nr}\overline{\mathrm{\Delta }}{x}_r,$$

where

$$d_{nr}=\{\begin{array}{cc}{n}^{1-\frac{1}{k}}{\sum }_{v=r+2}^n{\hat{b}}_{nv}{\hat{a}}_{vr}^{\prime }{\lambda }_v,\hfill & \text{if}0\le r\le n-2,\hfill \\ 0,\hfill & \text{if}r>n-2\text{.}\hfill \end{array}$$

Now

$$\sum {\left|\overline{T_n(2)}\right|}^k<\mathrm{\infty }\text{whenever}\sum |\overline{\mathrm{\Delta }}{x}_n|<\mathrm{\infty }$$

is equivalently

$$\underset{r}{sup}\sum _{n=1}^{\mathrm{\infty }}{|d_{nr}|}^k<\mathrm{\infty }$$

(33)

by Lemma. But it follows from conditions (21) and (24) that

$$\begin{array}{rcl}\sum _{n=r+2}^{\mathrm{\infty }}{|d_{nr}|}^k& =& O(1)\sum _{n=r+2}^{\mathrm{\infty }}{n}^{k-1}{\left\{\sum _{v=r+2}^{\mathrm{\infty }}\left|{\hat{b}}_{nv}{\hat{a}}_{vr}^{\prime }{\lambda }_v\right|\right\}}^k\\ =& O(1)\sum _{n=r+2}^{\mathrm{\infty }}{n}^{k-1}{|{\hat{b}}_{n,r+1}{\lambda }_{r+1}|}^k\\ =& O(1)\text{as}r\to \mathrm{\infty }.\end{array}$$

Therefore, we have

$$\sum _{n=1}^{\mathrm{\infty }}{n}^{k-1}|T_n(i){|}^k<\mathrm{\infty }\text{for}i=1,2.$$

This completes the proof of the Theorem. □