Bounds for the norm of lower triangular matrices on the Cesàro weighted sequence space

Abstract

This paper is concerned with the problem of finding bounds for the norm of lower triangular matrix operators from $l_{p}(w)$ into $c_{p} (w)$, where $c_{p}(w)$ is the Cesàro weighted sequence space and $(w_{n})$ is a non-negative sequence. Also this problem is considered for lower triangular matrix operators from $c_{p}(w)$ into $l_{p} (w)$, and the norms of certain matrix operators such as Cesàro, Nörlund and weighted mean are computed.

1 Introduction

Let $p\ge1$ and ω denote the set of all real-valued sequences. The space $l_{p}$ is the set of all real sequences $x=(x_{n})\in\omega$ such that

$$\Vert x \Vert _{p}= \Biggl(\sum_{n=1}^{\infty} \vert x_{n} \vert ^{p} \Biggr)^{1/p}< \infty. $$

If $w=(w_{n})\in\omega$ is a non-negative sequence, we define the weighted sequence space $l_{p}(w)$ as follows:

$$l_{p}(w):= \Biggl\{ x=(x_{n})\in\omega : \sum _{n=1}^{\infty}w_{n} \vert x_{n} \vert ^{p}< \infty \Biggr\} , $$

with norm $\Vert \cdot \Vert _{p,w}$, which is defined in the following way:

$$\Vert x \Vert _{p,w}= \Biggl( \sum_{n=1}^{\infty}w_{n} \vert x_{n} \vert ^{p} \Biggr)^{1/p}. $$

Let $C=(c_{n,k})$ denote the Cesàro matrix. We recall that the elements $c_{n,k}$ of the matrix C are given by

$$\begin{aligned} c_{n,k}= \textstyle\begin{cases} \frac{1}{n} & \mbox{for }1\leq k \leq n,\\ 0 & \mbox{for }k>n. \end{cases}\displaystyle \end{aligned}$$

The sequence space defined by

$$\begin{aligned} c_{p}(w) =& \bigl\{ (x_{n})\in\omega : Cx\in l_{p}(w) \bigr\} \\ =& \Biggl\{ (x_{n})\in\omega : \sum_{n=1}^{\infty}w_{n} \Biggl\vert \frac{1}{n}\sum_{i=1}^{n}x_{i} \Biggr\vert ^{p}< \infty \Biggr\} \end{aligned}$$

is called the Cesàro weighted sequence space, and the norm $\Vert \cdot \Vert _{p,w,c}$ of the space is defined by

$$\Vert x \Vert _{p,w,c}= \Biggl( \sum_{n=1}^{\infty}w_{n} \Biggl\vert \frac{1}{n}\sum_{i=1}^{n}x_{i} \Biggr\vert ^{p} \Biggr)^{1/p}. $$

The Cesàro sequence spaces were studied in [1], where $w_{n}=1$ for all n. It is significant that in the special case $w_{n}=1$, we have $l_{p}(w)=l_{p}$ and $c_{p}(w)=c_{p}$.

Let $(w_{n})$ be a non-negative sequence and $A=(a_{n,k})$ be a lower triangular matrix with non-negative entries. In this paper, we shall consider the inequality of the form

$$\Vert Ax \Vert _{p,w,c}\le U \Vert x \Vert _{p,w}, $$

and the inequality of the form

$$\Vert Ax \Vert _{p,w}\le U \Vert x \Vert _{p,w,c}, $$

where $x=(x_{n})$ is a non-negative sequence. The constant U does not depend on x, and we seek the smallest possible value of U. We write $\Vert A \Vert _{p,w,c}$ for the norm of A as an operator from $l_{p}(w)$ into $c_{p}(w)$, $\Vert A \Vert _{p,c}$ for the norm of A as an operator from $l_{p}$ into $c_{p}$, $\Vert A \Vert _{c,p,w}$ for the norm of A as an operator from $c_{p}(w)$ into $l_{p}(w)$, $\Vert A \Vert _{c,p}$ for the norm of A as an operator from $c_{p}$ into $l_{p}$, $\Vert A \Vert _{p,w}$ for the norm of A as an operator from $l_{p}(w)$ into itself and $\Vert A \Vert _{p}$ for the norm of A as an operator from $l_{p}$ into itself.

The problem of finding the norm of a lower triangular matrix on the sequence spaces $l_{p}$ and $l_{p}(w)$ has been studied before in [2–8]. In the study, we will expand this problem for matrix operators from $l_{p}(w)$ into $c_{p} (w)$ and matrix operators from $c_{p}(w)$ into $l_{p} (w)$, and we consider certain matrix operators such as Cesàro, Nörlund and weighted mean. The study is an extension of some results obtained by [3, 7].

2 The norm of matrix operators from $l_{p}(w)$ into $c_{p}(w)$

In this section, we tend to compute the bounds for the norm of lower triangular matrix operators from $l_{p}(w)$ into $c_{p} (w)$. In particular, we apply our results for lower triangular matrix operators from $l_{p}$ into $c_{p}$, when $w_{n}=1$ for all n.

Throughout this paper, let $A=(a_{n,k})$ be a matrix with non-negative real entries i.e., $a_{n,k}\ge0$, for all n, k. This implies that $\Vert Ax \Vert _{p,w,c}\le \Vert A \vert x \vert \Vert _{p,w,c}$, and hence the non-negative sequences are sufficient to determine the norm of A. We say that $A=(a_{n,k})$ is lower triangular if $a_{n,k}=0$ for $n< k$. A non-negative lower triangular matrix is called a summability matrix if $\sum_{k=1}^{n}a_{n,k}=1$ for all n.

We first state some lemmas from [3, 7], which are needed for our main result. Set $\xi^{+}=\max(\xi,0)$ and $\xi^{-}=\min(\xi,0)$ and $p^{*}=p/(p-1)$.

Lemma 2.1

[3], Lemma 2.1

Let a and x be two non-negative sequences, then for all n,

$$\sum_{k=1}^{n}a_{k}x_{k} \leq \Biggl\{ \max_{1\le k\le n}\frac{1}{n-k+1}\sum _{j=k}^{n}x_{j} \Biggr\} \sum _{k=1}^{n}(n-k+1) (a_{k}-a_{k-1})^{+}. $$

Lemma 2.2

[3], Lemma 2.2

Let $N\geq1$, and let a and x be two non-negative sequences. If $x_{N}\geq x_{N+1}\geq\cdots\geq0$ and $x_{n}=0$ for $n< N$, then

$$\sum_{k=1}^{n}a_{k}x_{k} \geq \Biggl(\frac{1}{n}\sum_{j=1}^{n}x_{j} \Biggr) \Biggl\{ na_{N}+\frac{n}{n-N+1}\sum _{k=N+1}^{n}(n-k+1) (a_{k}-a_{k-1})^{-} \Biggr\} $$

for all n.

Lemma 2.3

[7], Lemma 1.4

Let $p>1$ and $w=(w_{n})$ be a decreasing sequence with non-negative entries and $\sum_{n=1}^{\infty}\frac{w_{n}}{n}$ be divergent. Let $N\geq1$ and the matrix $C_{N}=(c_{n,k}^{N})$ be with the following entries:

$$\begin{aligned} c_{n,k}^{N}= \textstyle\begin{cases} \frac{1}{n+N-1} & \textit{for }n\geq k, \\ 0 & \textit{for }n< k. \end{cases}\displaystyle \end{aligned}$$

Then $\Vert C_{N} \Vert _{p,w}$ is determined by non-negative decreasing sequences and $\Vert C_{N} \Vert _{p,w}=p^{*}$.

Note that $C_{1}$ is the well-known Cesàro matrix.

Lemma 2.4

[7], Lemma 1.5

If $p>1$ and x and w are two non-negative sequences and also w is decreasing, then

$$\sum_{j=1}^{\infty}w_{j}\max _{1\le i\le j} \Biggl(\frac{1}{j-i+1}\sum _{k=i}^{j}x_{k} \Biggr)^{p}\le \bigl(p^{*}\bigr)^{p}\sum_{k=1}^{\infty}w_{k}x_{k}^{p}. $$

We set $a_{0,0}=0$ and $a_{n,0}=0$ for $n\geq1$ and

$$\begin{aligned} &M_{A}=\sup_{n\geq1} \Biggl\{ \sum _{k=1}^{n}\frac{n-k+1}{n} \Biggl(\sum _{i=k}^{n}a_{i,k}-\sum _{i=k-1}^{n}a_{i,k-1} \Biggr)^{+} \Biggr\} , \\ &m_{A}=\sup_{N\geq1}\inf_{n\geq N} \Biggl\{ \sum_{i=N}^{n}a_{i,N}+ \frac{1}{n-N+1} \sum_{k=N+1}^{n}(n-k+1) \Biggl(\sum_{i=k}^{n}a_{i,k}-\sum _{i=k-1}^{n}a_{i,k-1} \Biggr)^{-} \Biggr\} . \end{aligned}$$

We are now ready to present the main result of this section.

Theorem 2.5

Suppose that $p>1$ and $w=(w_{n})$ is a decreasing sequence with non-negative entries. If $A=(a_{n,k})$ is a lower triangular matrix with non-negative entries, then we have the following statements.

• (i) $\Vert A \Vert _{p,w,c}\leq p^{*}M_{A}$. Moreover, if $M_{A}<\infty$, then A is a bounded matrix operator from $l_{p}(w)$ into $c_{p}(w)$.

• (ii) If $\sum_{n=1}^{\infty}\frac{w_{n}}{n}$ is divergent and $(\frac{w_{n}}{w_{n+1}})$ is decreasing, then $\Vert A \Vert _{p,w,c}\geq p^{*}m_{A}$.

Therefore if $w=(w_{n})$ is a decreasing sequence with non-negative entries and $(\frac{w_{n}}{w_{n+1}})$ is decreasing and $\sum_{n=1}^{\infty}\frac{w_{n}}{n}=\infty$, then

$$p^{*}m_{A}\leq \Vert A \Vert _{p,w,c}\leq p^{*}M_{A}. $$

In particular, if $w_{n}=1$ for all n and if $M_{A}<\infty$, then A is a bounded matrix operator from $l_{p}$ into $c_{p}$ and $p^{*}m_{A}\leq \Vert A \Vert _{p,c}\leq p^{*}M_{A}$.

Proof

(i) Let $(x_{n})$ be a non-negative sequence. By using Lemma 2.1, we get

$$\begin{aligned} &\sum_{k=1}^{n} \Biggl(\frac{1}{n}\sum _{i=k}^{n}a_{i,k} \Biggr)x_{k}\\ &\quad\leq\Biggl\{ \max_{1\le k\le n}\frac{1}{n-k+1} \sum_{j=k}^{n}x_{j} \Biggr\} \sum _{k=1}^{n}\frac {n-k+1}{n} \Biggl(\sum _{i=k}^{n}a_{i,k}-\sum _{i=k-1}^{n}a_{i,k-1} \Biggr)^{+} \\ &\quad \leq M_{A}\max_{1\le k\le n} \Biggl\{ \frac{1}{n-k+1}\sum _{j=k}^{n}x_{j} \Biggr\} . \end{aligned}$$

By applying Lemma 2.4, we deduce that

$$\begin{aligned} \sum_{n=1}^{\infty}w_{n} \Biggl(\sum _{k=1}^{n} \Biggl(\frac{1}{n}\sum _{i=k}^{n}a_{i,k} \Biggr)x_{k} \Biggr)^{p} \leq&M_{A}^{p} \sum_{n=1}^{\infty} w_{n}\max _{1\leq k\le n} \Biggl(\frac{1}{n-k+1}\sum _{j=k}^{n}x_{j} \Biggr)^{p} \\ \leq&\bigl(p^{*}M_{A}\bigr)^{p}\sum _{k=1}^{\infty}w_{k}x_{k}^{p}. \end{aligned}$$

(ii) We have $m_{A}=\sup_{N\geq1}\beta_{N}$, where

$$\beta_{N}=\inf_{n\geq N} \Biggl\{ \sum _{i=N}^{n}a_{i,N}+\frac{1}{n-N+1} \sum _{k=N+1}^{n}(n-k+1) \Biggl(\sum _{i=k}^{n}a_{i,k}-\sum _{i=k-1}^{n}a_{i,k-1} \Biggr)^{-} \Biggr\} . $$

Let $N\geq1$, so that $\beta_{N}\geq0$. If $y=(y_{n})$ is a decreasing sequence with non-negative entries and $\Vert y \Vert _{p,w}=1$, we set $x_{1}=x_{2}=\cdots=x_{N-1}=0$ and

$$x_{n+N-1}=\biggl(\frac{w_{n}}{w_{n+N-1}}\biggr)^{1/p}y_{n} $$

for all $n\geq1$. So $\Vert x \Vert _{p,w}= \Vert y \Vert _{p,w}=1$, and from Lemma 2.2 it follows that

$$\begin{aligned} \Vert A \Vert _{p,w,c}^{p}&\geq\sum _{n=1}^{\infty}w_{n} \Biggl(\sum _{k=1}^{n} \Biggl(\frac{1}{n}\sum _{i=k}^{n}a_{i,k} \Biggr)x_{k} \Biggr)^{p} \\ &\geq \beta_{N}^{p}\sum_{n=1}^{\infty}w_{n} \Biggl(\frac{1}{n}\sum_{j=1}^{n}x_{j} \Biggr)^{p} \\ &=\beta_{N}^{p}\sum_{n=1}^{\infty}w_{n+N-1} \Biggl(\frac{1}{n+N-1}\sum_{j=1}^{n}x_{j+N-1} \Biggr)^{p} \\ &=\beta_{N}^{p}\sum_{n=1}^{\infty}w_{n+N-1} \Biggl(\frac{1}{n+N-1}\sum_{j=1}^{n} \biggl(\frac{w_{j}}{w_{j+N-1}}\biggr)^{1/p}y_{j} \Biggr)^{p} \\ &\geq\beta_{N}^{p} \Vert C_{N}y \Vert _{p,w,}^{p}. \end{aligned}$$

By Lemma 2.3, we conclude that $\Vert A \Vert _{p,w,c}\geq p^{*}\beta_{N}$, so

$$\Vert A \Vert _{p,w,c}\geq p^{*}m_{A}. $$

 □

In what follows we assume that $w=(w_{n})$ is a decreasing sequence with non-negative entries and $(\frac{w_{n}}{w_{n+1}})$ is decreasing and $\sum_{n=1}^{\infty}\frac{w_{n}}{n}=\infty$.

At first we bring a corollary of Theorem 2.5 for a lower triangular matrix $A=(a_{n,k})$. The rows of $C_{1}A$ are increasing, where $C_{1}$ is the Cesàro matrix and

$$\begin{aligned} (C_{1}A)_{n,k}=\sum_{i=1}^{\infty}c^{1}_{n,i}a_{i,k}= \frac{1}{n}\sum_{i=k}^{n}a_{i,k},\quad (n, k=1,2,\ldots). \end{aligned}$$

Corollary 2.6

Suppose that $p>1$ and $A=(a_{n,k})$ is a non-negative lower triangular matrix that $\sum_{i=k-1}^{n}a_{i,k-1}\le \sum_{i=k}^{n}a_{i,k}$ for $1< k \leq n$. Then

$$\Vert A \Vert _{p,w,c}=p^{*}\sup_{n\geq1} a_{n,n}. $$

In particular, $\Vert I \Vert _{p,w,c}=p^{*}$, where I is the identity matrix.

Proof

Since the finite sequence $(\sum_{i=k}^{n}a_{i,k} )_{k=1}^{n}$ is increasing for each n, we have

$$\begin{aligned} \Biggl(\sum_{i=k}^{n}a_{i,k}-\sum _{i=k-1}^{n}a_{i,k-1} \Biggr)^{+} =\sum_{i=k}^{n}a_{i,k}- \sum_{i=k-1}^{n}a_{i,k-1} \end{aligned}$$

for $1\leq k \le n$. Hence

$$\begin{aligned} M_{A} =&\sup_{n\geq1} \Biggl\{ \sum _{k=1}^{n}\frac{n-k+1}{n} \Biggl(\sum _{i=k}^{n}a_{i,k}-\sum _{i=k-1}^{n}a_{i,k-1} \Biggr) \Biggr\} \\ =&\sup_{n\geq1}\frac{1}{n}\sum _{k=1}^{n}\sum_{i=k}^{n}a_{i,k} \le\sup_{n\geq1} a_{n,n}. \end{aligned}$$

Moreover,

$$\begin{aligned} \Biggl(\sum_{i=k}^{n}a_{i,k}-\sum _{i=k-1}^{n}a_{i,k-1} \Biggr)^{-}=0\quad (1\leq k \le n) \end{aligned}$$

and

$$\begin{aligned} m_{A}=\sup_{N\geq1}\inf_{n\geq N}\sum _{i=N}^{n}a_{i,N}=\sup _{n\geq1} a_{n,n}. \end{aligned}$$

Hence, according to Theorem 2.5, we obtain the desired result. □

Example 2.7

Let $A=(a_{n,k})$ be defined by

$$\begin{aligned} a_{n,k}= \textstyle\begin{cases} \frac{1}{n^{2}} & \mbox{for } k< n, \\ \frac{2n-1}{n} & \mbox{for } k=n, \\ 0 & \mbox{for } k>n. \end{cases}\displaystyle \end{aligned}$$

Since the finite sequence $(\sum_{i=k}^{n}a_{i,k} )_{k=1}^{n}$ is increasing for each n and $\sup_{n\geq1} a_{n,n}=2$, by Corollary 2.6, we have $\Vert A \Vert _{p,w,c}=2p^{*}$.

Now, in the second case, we state some corollaries of Theorem 2.5 for a lower triangular matrix A, where the rows of $C_{1}A$ are decreasing.

Corollary 2.8

Suppose that $p>1$ and $A=(a_{n,k})$ is a lower triangular matrix with $\sum_{i=k-1}^{n}a_{i,k-1}\ge\sum_{i=k}^{n}a_{i,k}$ for $1< k \leq n$. Then

$$p^{*} \Biggl(\inf_{n\geq 1}\frac{1}{n}\sum _{k=1}^{n}\sum_{i=k}^{n}a_{i,k} \Biggr)\leq \Vert A \Vert _{p,w,c}\leq p^{*} \Biggl(\sup _{n\geq1}\sum_{i=1}^{n}a_{i,1} \Biggr). $$

In particular, for summability matrices the left-hand side of the above inequality reduces to $p^{*}$.

Moreover, if the right-hand side of the above inequality is finite, then A is a bounded matrix operator from $l_{p}(w)$ into $c_{p}(w)$.

Proof

Since the finite sequence $(\sum_{i=k}^{n}a_{i,k} )_{k=1}^{n}$ is decreasing for each n, we have

$$\begin{aligned} \Biggl(\sum_{i=k}^{n}a_{i,k}-\sum _{i=k-1}^{n}a_{i,k-1} \Biggr)^{+}=0\quad (1< k \le n), \end{aligned}$$

and $(\sum_{i=1}^{n}a_{i,1}-\sum_{i=0}^{n}a_{i,0} )^{+} =\sum_{i=1}^{n}a_{i,1}$. Hence $M_{A}=\sup_{n\geq1}\sum_{i=1}^{n}a_{i,1}$. Moreover,

$$\begin{aligned} \Biggl(\sum_{i=k}^{n}a_{i,k}-\sum _{i=k-1}^{n}a_{i,k-1} \Biggr)^{-} =\sum_{i=k}^{n}a_{i,k}- \sum_{i=k-1}^{n}a_{i,k-1}, \end{aligned}$$

for $1< k \leq n$, so

$$\begin{aligned} m_{A} =&\sup_{N\geq1}\inf_{n\geq N} \Biggl\{ \sum_{i=N}^{n}a_{i,N}+ \frac{1}{n-N+1} \sum_{k=N+1}^{n}(n-k+1) \Biggl(\sum_{i=k}^{n}a_{i,k}-\sum _{i=k-1}^{n}a_{i,k-1} \Biggr) \Biggr\} \\ =&\sup_{N\geq1}\inf_{n\geq N}\frac{1}{n-N+1}\sum _{k=N}^{n}\sum_{i=k}^{n}a_{i,k} \\ \ge&\inf_{n\geq1}\frac{1}{n}\sum _{k=1}^{n}\sum_{i=k}^{n}a_{i,k}. \end{aligned}$$

Therefore, by Theorem 2.5, we prove the desired result. □

The two examples of Corollary 2.8 are given as follows.

Example 2.9

Suppose that $\alpha\ge2$ and the matrix $A=(a_{n,k})$ is defined by

$$\begin{aligned} a_{n,k}= \textstyle\begin{cases} \frac{1}{n^{\alpha}} & \mbox{for } n\geq k, \\ 0 & \mbox{for } n< k. \end{cases}\displaystyle \end{aligned}$$

Since $\sum_{i=k}^{n}a_{i,k}=\sum_{i=k}^{n} \frac{1}{i^{\alpha}}$ and $\sum_{i=k-1}^{n}a_{i,k-1}\ge\sum_{i=k}^{n}a_{i,k}$ for $1< k \leq n$, we have $0\le \Vert A \Vert _{p,w,c}\leq p^{*}\zeta(\alpha)$, where $\zeta(\alpha )=\sum_{n=1}^{\infty}\frac{1}{n^{\alpha}}$.

Example 2.10

Suppose that the matrix $A=(a_{n,k})$ is defined by

$$\begin{aligned} a_{n,k}= \textstyle\begin{cases} \frac{1}{n(n+1)} & \mbox{for }n\geq k, \\ 0 & \mbox{for } n< k. \end{cases}\displaystyle \end{aligned}$$

Since $\sum_{i=k}^{n}a_{i,k}=\sum_{i=k}^{n}\frac{1}{i(i+1)}$, by Corollary 2.8, we have $0\le \Vert A \Vert _{p,w,c}\leq p^{*}$.

We apply the above corollary to the following two special cases.

Let $(a_{n})$ be a non-negative sequence with $a_{1}>0$, and $A_{n}=a_{1}+\cdots+a_{n}$. The Nörlund matrix $N_{a}=(a_{n,k})$ is defined as follows:

$$\begin{aligned} a_{n,k}= \textstyle\begin{cases} \frac{a_{n-k+1}}{A_{n}}& \mbox{for }1\le k\le n,\\ 0& \mbox{for } k>n. \end{cases}\displaystyle \end{aligned}$$

Also the weighted mean matrix $M_{a}=(a_{n,k})$ is defined by

$$\begin{aligned} a_{n,k}= \textstyle\begin{cases} \frac{a_{k}}{A_{n}}& \mbox{for }1\le k\le n,\\ 0& \mbox{for } k>n. \end{cases}\displaystyle \end{aligned}$$

Corollary 2.11

Suppose that $p>1$ and $N_{a}=(a_{n,k})$ is the Nörlund matrix and $(a_{n})$ is an increasing sequence. Then

$$p^{*}\leq \Vert N_{a} \Vert _{p,w,c}\leq p^{*} \Biggl(\sup _{n\geq1}\sum_{i=1}^{n} \frac{a_{i}}{A_{i}} \Biggr). $$

Proof

Since $N_{a}$ is a summability matrix and $\sum_{i=1}^{n}a_{i,1}=\sum_{i=1}^{n}\frac{a_{i}}{A_{i}}$, by applying Corollary 2.8, we have the desired result. □

Corollary 2.12

Suppose that $p>1$ and $M_{a}=(a_{n,k})$ is the weighted mean matrix and $(a_{n})$ is a decreasing sequence. Then

$$p^{*}\leq \Vert M_{a} \Vert _{p,w,c}\leq p^{*}a_{1} \Biggl(\sup_{n\geq1}\sum_{i=1}^{n} \frac{1}{A_{i}} \Biggr). $$

Proof

Since $M_{a}$ is a summability matrix and $\sum_{i=1}^{n}a_{i,1}=\sum_{i=1}^{n}\frac{a_{1}}{A_{i}}$, by Corollary 2.8, the proof is obvious. □

Finally, in the third case, if the rows of $C_{1}A$ are neither increasing nor decreasing, we present the following theorem.

Theorem 2.13

Suppose that $p>1$ and $A=(a_{n,k})$ is a non-negative lower triangular matrix. If A is a bounded matrix operator from $l_{p}(w)$ into itself, then A is a bounded matrix operator from $l_{p}(w)$ into $c_{p}(w)$ and

$$\Vert A \Vert _{p,w,c}\leq p^{*} \Vert A \Vert _{p,w}. $$

Proof

We have

$$\begin{aligned} \Vert Ax \Vert _{p,w,c}^{p} =&\sum _{n=1}^{\infty}w_{n} \Biggl\vert \frac{1}{n}\sum_{k=1}^{n}\sum _{j=1}^{k} a_{k,j}x_{j} \Biggr\vert ^{p} \\ =&\sum_{n=1}^{\infty}w_{n} \Biggl\vert \sum_{j=1}^{n} (C_{1}A)_{n,j}x_{j} \Biggr\vert ^{p}= \bigl\Vert (C_{1}A)x \bigr\Vert _{p,w}^{p}. \end{aligned}$$

Hence, by Lemma 2.3, we conclude that $\Vert A \Vert _{p,w,c}= \Vert C_{1}A \Vert _{p,w}\le p^{*} \Vert A \Vert _{p,w}$. □

We apply the above theorem to the following two Nörlund and weighted mean matrices.

Corollary 2.14

[7], Corollary 1.3

Suppose that $p>1$ and $N_{a}=(a_{n,k})$ is the Nörlund matrix and $(a_{n})$ is a decreasing sequence with $a_{n}\downarrow\alpha$ and $\alpha>0$. Then

$$\Vert N_{a} \Vert _{p,w}=p^{*}. $$

Corollary 2.15

Suppose that $p>1$ and $N_{a}=(a_{n,k})$ is the Nörlund matrix and $(a_{n})$ is a decreasing sequence with $a_{n}\downarrow\alpha$ and $\alpha>0$. Then

$$\Vert N_{a} \Vert _{p,w,c}\le\bigl(p^{*} \bigr)^{2}. $$

Proof

By applying Theorem 2.13 and Corollary 2.14, we have the desired result. □

Corollary 2.16

[7], Corollary 1.4

Suppose that $p>1$ and $M_{a}=(a_{n,k})$ is the weighted mean matrix and $(a_{n})$ is an increasing sequence with $a_{n}\uparrow\alpha$ and $\alpha<\infty$. Then

$$\Vert M_{a} \Vert _{p,w}=p^{*}. $$

Corollary 2.17

Suppose that $p>1$ and $M_{a}=(a_{n,k})$ is the weighted mean matrix and $(a_{n})$ is an increasing sequence with $a_{n}\uparrow\alpha$ and $\alpha<\infty$. Then

$$\Vert M_{a} \Vert _{p,w,c}\le\bigl(p^{*} \bigr)^{2}. $$

Proof

By using Theorem 2.13 and Corollary 2.16, the proof is clear. □

3 The norm of matrix operators from $c_{p}(w)$ into $l_{p}(w)$

In this section, we compute the bounds for the norm of lower triangular matrix operators from $c_{p}(w)$ into $l_{p}(w)$. In particular, when $w_{n}=1$ for all n, the bounds for the norm of lower triangular matrix operators from $c_{p}$ into $l_{p}$ are deduced. Moreover, we apply our results for Cesàro, Nörlund and weighted mean matrices.

We begin with a proposition which is needed to prove the main theorem of this section.

Proposition 3.1

([6], Proposition 5.1). Let $p>1$ and $w=(w_{n})$ be a decreasing sequence with non-negative entries, and let $C_{1}$ be the Cesàro matrix. Then we have $\Vert C_{1} \Vert _{p,w}\leq p^{*}$.

Theorem 3.2

Suppose that $p>1$ and $w=(w_{n})$ is a sequence with non-negative entries and $A=(a_{n,k})$ is a lower triangular matrix with non-negative entries. We have

$$\frac{1}{p^{*}} \Vert A \Vert _{p,w}\leq \Vert A \Vert _{c,p,w}\leq\sup_{n\ge1} \Bigl(n\sup_{1\le k\le n} a_{n,k} \Bigr). $$

Moreover, if the right-hand side of the above inequality is finite, then A is a bounded matrix operator from $c_{p}(w)$ into $l_{p}(w)$. In particular, if $w_{n}=1$ for all n, then we have

$$\frac{1}{p^{*}} \Vert A \Vert _{p}\leq \Vert A \Vert _{c,p}\leq\sup_{n\ge1} \Bigl(n\sup_{1\le k\le n} a_{n,k} \Bigr). $$

Proof

Suppose that $x\in c_{p}(w)$

$$\begin{aligned} \Vert Ax \Vert _{p,w}^{p} =&\sum _{n=1}^{\infty}w_{n} \Biggl\vert \sum _{k=1}^{n}a_{n,k}x_{k} \Biggr\vert ^{p} \\ \leq&\sum_{n=1}^{\infty}w_{n} \Biggl( \sup_{{1\le k\le n}} a_{n,k}\sum_{k=1}^{n}x_{k} \Biggr)^{p} \\ \leq& \sup_{n\ge1} \Bigl(n\sup_{1\le k\le n} a_{n,k} \Bigr)^{p} \sum_{n=1}^{\infty}w_{n} \Biggl(\frac{1}{n}\sum_{k=1}^{n}x_{k} \Biggr)^{p} \\ =& \sup_{n\ge1} \Bigl(n\sup_{1\le k\le n} a_{n,k} \Bigr)^{p} \Vert x \Vert _{p,w,c}^{p}. \end{aligned}$$

Hence

$$\begin{aligned} \frac{ \Vert Ax \Vert _{p,w}}{ \Vert x \Vert _{p,w,c}}\le\sup_{n\ge1} \Bigl(n\sup _{1\le k\le n} a_{n,k} \Bigr) \end{aligned}$$

and

$$\begin{aligned} \Vert A \Vert _{c,p,w}\leq\sup_{n\ge1} \Bigl(n\sup _{1\le k\le n} a_{n,k} \Bigr). \end{aligned}$$

On the other hand, Proposition 3.1 concludes that $\Vert x \Vert _{p,w,c}= \Vert C_{1}x \Vert _{p,w}\leq p^{*} \Vert x \Vert _{p,w}$, so

$$\begin{aligned} \frac{ \Vert Ax \Vert _{p,w}}{ \Vert x \Vert _{p,w,c}}\ge\frac{1}{p^{*}}\frac{ \Vert Ax \Vert _{p,w}}{ \Vert x \Vert _{p,w}}. \end{aligned}$$

Therefore $\frac{1}{p^{*}} \Vert A \Vert _{p,w}\leq \Vert A \Vert _{c,p,w}$, and the proof is complete. □

Corollary 3.3

If $p>1$, then the generalized Cesàro matrix $C_{N}$ is bounded from $c_{p}(w)$ into $l_{p}(w)$ and

$$\Vert C_{N} \Vert _{c,p,w}=1. $$

Proof

Since

$$\begin{aligned} \sup_{n\ge1} \Bigl(n\sup_{1\le k\le n} a_{n,k} \Bigr)=\sup_{n\ge 1}\frac{n}{n+N-1}=1, \end{aligned}$$

by using Lemma 2.3 and Theorem 3.2, the proof is obvious. □

We apply the above theorem to the following two special cases.

Corollary 3.4

Suppose that $p>1$ and $N_{a}=(a_{n,k})$ is the Nörlund matrix and $(a_{n})$ is a decreasing sequence with $a_{n}\downarrow\alpha$ and $\alpha>0$. Then

$$1\leq \Vert N_{a} \Vert _{c,p,w}\leq a_{1}\sup _{n\ge1}\frac{n}{A_{n}}. $$

Proof

By Theorem 3.2 and Corollary 2.14, the proof is clear. □

Example 3.5

Let $\alpha>0$ and $a_{n}=\alpha+\frac{1}{n^{\alpha+1}}$ for all n. The sequence $(a_{n})$ is decreasing and $a_{n}\downarrow\alpha$ and also $a_{1}\sup_{n\ge1}\frac{n}{A_{n}}=1+\frac{1}{\alpha}$. So

$$1\leq \Vert N_{a} \Vert _{c,p,w}\leq1+\frac{1}{\alpha}. $$

Specially $\Vert N_{a} \Vert _{c,p,w}\rightarrow1$, when $\alpha\rightarrow\infty$.

Corollary 3.6

Suppose that $p>1$ and $M_{a}=(a_{n,k})$ is the weighted mean matrix and $(a_{n})$ is an increasing sequence with $a_{n}\uparrow\alpha$ and $\alpha<\infty$. Then

$$1\leq \Vert M_{a} \Vert _{c,p,w}\leq\sup _{n\ge1}\frac{na_{n}}{A_{n}}. $$

Proof

By using Theorem 3.2 and Corollary 2.16, the proof is obvious. □

Example 3.7

Let $a_{n}=1-\frac{1}{(n+1)^{2}}$ for all n. The sequence $(a_{n})$ is increasing and $a_{n}\uparrow1$ and also

$$\sup_{n\ge1}\frac{na_{n}}{A_{n}}=\frac{3a_{3}}{A_{3}}\simeq1.091. $$

So

$$1\leq \Vert M_{a} \Vert _{c,p,w}\leq1.091. $$

4 Conclusions

In the present study, we considered the problem of finding bounds for the norm of lower triangular matrix operators from $l_{p}(w)$ into $c_{p} (w)$ and from $c_{p}(w)$ into $l_{p} (w)$. Moreover, we computed the norms of certain matrix operators such as Cesàro, Nörlund and weighted mean, and we extended some results of [3, 7].