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.

  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

[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].