# 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.

## 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 , 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 . 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].

## 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

, 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

, 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

, 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

, 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.

1. (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 $$c_{p}(w)$$.

2. (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

, 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

, 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. □

## 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

(, 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

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

## 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].

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3. Chen, C-P, Luor, D-C, Ou, Z-Y: Extensions of Hardy inequality. J. Math. Anal. Appl. 273(1), 160-171 (2002)

4. Hardy, GH: An inequality for Hausdorff means. J. Lond. Math. Soc. 1(1), 46-50 (1943)

5. Hardy, GH, Littlewood, JE, Pólya, G: Inequalities, 2nd edn. (1934)

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## Acknowledgements

The second author would like to record his pleasure to Dr. Gholareza Talebi [Department of Mathematics, Faculty of Sciences, Vali-E-Asr University of Rafsanjan] for his valuable conversations during the preparation of the present paper.

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Correspondence to Davoud Foroutannia.

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