Lobachevskii Journal of Mathematics http://ljm.ksu.ru Vol. 27, 2007, 15–29

D. Foroutannia, R. Lashkaripour
LOWER BOUNDS FOR SUMMABILITY MATRICES ON WEIGHTED SEQUENCE SPACES
(submitted by O. E. Tikhonov)

Abstract. The purpose of this paper is ﬁnding a lower bound for summability matrix operators on sequence spaces $l_{p} (w)$ and Lorentz sequence spaces $d (w, p)$ and also the sequence space $e (w, \infty)$ . Also, this study is an extension of some works of Bennett.

1. Introduction

We study the lower bounds of summability matrix operators on $l_{p} (w)$ and Lorentz sequence spaces $d (w, p)$ and also the Banach sequence space $e (w, \infty)$ considered in [1] on $l_{p}$ spaces. The problem of ﬁnding the upper bound and lower bound of certain matrix operators such as Cesaro, Copson and Hausdorff and Hilbert operators are considered in [3], [4], [5], [6] and [7] on weighted sequence spaces.

Let $1 \leq p < \infty$ , $l_{p}$ be the normed linear space of all sequences $(x_{n})$ with ﬁnite norm $∥ x ∥_{p}$ , where

∥ x ∥_{p} = {(\sum_{n = 1}^{\infty} ∣ x_{n} ∣^{p})}^{1 ∕ p} .

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

l_{p} (w) = {(x_{n}) : \sum_{n = 1}^{\infty} w_{n} ∣ x_{n} ∣^{p} < \infty},

with norm, $∥ . ∥_{p, w}$ , which is deﬁned in the following way:

∥ x ∥_{p, w} = {(\sum_{n = 1}^{\infty} w_{n} ∣ x_{n} ∣^{p})}^{1 ∕ p} .

Also, if $(w_{n})$ is a decreasing non-negative sequence such that ${lim}_{n \to \infty} w_{n} = 0$ and $\sum_{n = 1}^{\infty} w_{n} = \infty$ , then the Lorentz sequence space $d (w, p)$ is deﬁned as follows:

d (w, p) = {(x_{n}) : \sum_{n = 1}^{\infty} w_{n} x_{n}^{* p} < \infty},

where $(x_{n}^{*})$ is the decreasing rearrangement of $(∣ x_{n} ∣)$ . Then $d (w, p)$ is the space of null sequences $x$ for which $x^{*}$ is in $l_{p} (w)$ , with norm $∥ x ∥_{d (w, p)} = ∥ x^{*} ∥_{p, w}$ .
Let $X_{k}^{*} = x_{1}^{*} + \dots + x_{k}^{*}$ and $W_{k} = w_{1} + \dots + w_{k}$ . The conjugate space of $d (w, 1)$ is $e (w, \infty)$ , where

e (w, \infty) = {(x_{n}) : {sup}_{k} \frac{X_{k}^{*}}{W_{k}} < \infty},

and its norm is deﬁned by

∥ x ∥_{w, \infty} = {sup}_{k} \frac{X_{k}^{*}}{W_{k}} .

Let $A$ be a matrix with non-negative entries. We consider lower bounds of the form

∥ A x ∥_{p, w} \geq L ∥ x ∥_{p, v}, (∥ A x ∥_{d (w, p)} \geq L ∥ x ∥_{d (v, p)}),

valid for every decreasing non-negative sequence $x$ and $L$ is a constant not depending on $x$ . We seek the largest possible value of $L$ and denote the best lower bound by $L_{p, v, w}$ for matrix operator from $l_{p} (v)$ into $l_{p} (w)$ and also it is denoted by $L_{p, w} (A)$ and $L_{d (w, p)} (A)$ on $l_{p} (w)$ and $d (w, p)$ , respectively.
When $0 < p < 1$ , we use notation $∥ . ∥$ without assuming that it is a norm. We denote transpose matrix of $A$ by $A^{t}$ . Suppose that $A = (a_{n, k})$ is a summability matrix, then $A^{t}$ is quasi-summability matrix and

{(A^{t} x)}_{n} = \sum_{k = n}^{\infty} a_{k, n} x_{k} .

Also, denote by $p^{*}$ the conjugate exponent of $p$ , so that $p^{*} = p ∕ (p - 1)$ .

Throughout this paper, we apply the following lemma and state some statements on weighted sequence space $l_{p} (w)$ and some results on weighted sequence space $d (w, p)$ .

Lemma 1.1. Let $p \geq 0$ and $A = (a_{i, j})$ be a matrix with non-negative entries. The following condition is equivalent to the statement that $A x$ is decreasing for every decreasing, non-negative sequence $x$ in $d (w, p)$ :

(1) $r_{i, n} = \sum_{j = 1}^{n} a_{i, j}$ decreases with $i$ for each $n$ .

Proof. Let $x \in d (w, p)$ be a decreasing, non-negative sequence and $y = A x$ . If (1) holds, by Abel summation, we have

y_{i} = \sum_{j = 1}^{\infty} a_{i, j} x_{j} = \sum_{j = 1}^{\infty} r_{i, j} (x_{j} - x_{j + 1}),

it follows that $A x$ is decreasing. The converse deduce from the fact that $y_{i} = r_{i, n}$ when $x = e_{1} + \dots + e_{n}$ .

Above lemma shows that for the matrix $A$ with condition (1), we have

L_{p, w} (A) = L_{d (w, p)} (A) .

2. Summability matrix operator on $d (w, 1)$ and $e (w, \infty)$

In this section, we consider the lower bound problem for summability matrix operators on $d (w, 1)$ and $e (w, \infty)$ . These are lower triangular matrices with entries of the form:

$a_{n, k} \geq 0$ ;
$a_{n, k} = 0$ if $k > n$ ;
$\sum_{k = 1}^{n} a_{n, k} = 1$ .

We generalize Theorem 1 of [3] for summability matrix operators from $d (v, 1)$ into $d (w, 1)$ . We write $∥ . ∥_{w}$ instead of $∥ . ∥_{d (w, 1)}$ and denote lower bound by $L_{v, w} (A)$ for matrix operator from $d (v, 1)$ into $d (w, 1)$ and it is denoted by $L_{w} (A)$ on $d (w, 1)$ . Moreover, we denote lower bound of matrix operator $A$ from $e (w, \infty)$ into itself with $L_{w, \infty} (A)$ . Throughout this section, we assume $A$ is a summability matrix operator satisfying condition (1) in Lemma 1.1.

Theorem 2.1. Suppose that $A = (a_{i, j})$ is a summability matrix operator from $d (v, 1)$ into $d (w, 1)$ with non-negative entries. We write $S_{n} = W_{n} + \sum_{k = n + 1}^{\infty} w_{k} r_{k, n}$ and $V_{n} = v_{1} + \dots + v_{n}$ . Then

\begin{array}{rcl} L_{v, w} (A) = {inf}_{n} \frac{S_{n}}{V_{n}} . \end{array}

Proof. Denote the stated inﬁmum by M. Let $x$ be in $d (v, 1)$ such that $x_{1} \geq x_{2} \geq \dots \geq 0$ and $y = A x$ . By Abel summation, we have

y_{n} = \sum_{k = 1}^{n - 1} r_{n, k} (x_{k} - x_{k + 1}) + x_{n} .

Hence

\begin{array}{rcl} ∥ A x ∥_{w} & = & \sum_{n = 1}^{\infty} w_{n} (\sum_{k = 1}^{n - 1} r_{n, k} (x_{k} - x_{k + 1}) + x_{n}) \\ = & \sum_{n = 1}^{\infty} S_{n} (x_{n} - x_{n + 1}) \\ \geq & M \sum_{n = 1}^{\infty} V_{n} (x_{n} - x_{n + 1}) \\ = & M ∥ x ∥_{v} . \end{array}

Therefore

L_{v, w} (A) \geq M .

To show that the constant $M$ is the best possible, we take $x_{1} = x_{2} = \dots = x_{n} = 1$ and $x_{k} = 0$ for all $k \geq n + 1$ . Then

∥ x ∥_{v} = V_{n}, ∥ A x ∥_{w} = S_{n} .

Therefore

L_{v, w} (A) = M .

The following lemma is needed in the sequel.

Lemma 2.1. Let $α > 0$ and $X_{(n)} = \sum_{k = n}^{\infty} \frac{1}{k^{1 + α}}$ , then ${(n - 1)}^{α} X_{(n)}$ increases and tends to $\frac{1}{α}$ .

Proof. Let $x_{n} = \frac{1}{n^{1 + α}}$ and $y_{n} = \int_{n - 1}^{n} \frac{d t}{t^{1 + α}}$ , then

Y_{(n + 1)} = \sum_{k = n + 1}^{\infty} y_{k} = \frac{1}{α n^{α}} .

By the usual integral comparison,

\frac{1}{α n^{α}} \leq X_{(n)} \leq \frac{1}{α {(n - 1)}^{α}},

which implies the stated limit. Write $z_{n} = n^{1 + α} y_{n}$ , then

z_{n + 1} = {(n + 1)}^{1 + α} \int_{n}^{n + 1} \frac{d t}{t^{1 + α}} = {(n + 1)}^{1 + α} \int_{n - 1}^{n} \frac{d t}{{(t + 1)}^{1 + α}} .

For $n - 1 \leq t \leq n$ , we have $\frac{n + 1}{t + 1} \leq \frac{n}{t}$ , hence $\frac{{(n + 1)}^{1 + α}}{{(t + 1)}^{1 + α}} \leq \frac{n^{1 + α}}{t^{1 + α}}$ and $z_{n + 1} \leq z_{n}$ . Therefore $(\frac{x_{n}}{y_{n}})$ is increasing, then so is $(\frac{X_{(n)}}{Y_{(n)}})$ .

Theorem 2.2. Let $A$ be a summability matrix operator from $d (w, 1)$ into itself. If for $k > n$

r_{k, n} \leq \frac{n}{k},

and $w_{n} \leq \frac{1}{n}$ for all $n$ , then

\begin{array}{rcl} L_{w} (A) = 1 . \end{array}

Proof. We have $S_{n} = W_{n} + \sum_{k = n + 1}^{\infty} w_{k} r_{k, n}$ , hence $S_{n} \geq W_{n}$ for all $n$ , and so

L_{1, w} (A) \geq 1 .

Since $r_{k, n} \leq \frac{n}{k}$ for $k > n$ ,

S_{n} \leq W_{n} + n \sum_{k = n + 1}^{\infty} \frac{1}{k^{2}} .

Applying Lemma 2.1, $n \sum_{k = n + 1}^{\infty} \frac{1}{k^{2}} \leq 1$ . Therefore

\frac{S_{n}}{W_{n}} \leq 1 + \frac{1}{W_{n}} .

Since $W_{n} \to \infty$ as $n \to \infty$ , hence

L_{1, w} (A) \leq 1 .

This deduces the statement.

We note that the Hausdorff matrix, Nörlund mean matrix, weighted mean matrix, and in particular Cesaro matrix are summability matrix operators.

We denote the Cesaro matrix by $C$ , with entries:

\begin{array}{rcl} c_{i, j} = \{\begin{matrix} \frac{1}{i} & if i \geq j \\ 0 & if i < j . \end{matrix} \end{array}

The lower bound problem of $C$ is discussed in [3].

Corollary 2.1. If $N \geq 0$ and $w_{n} = \frac{1}{n + N}$ , then

L_{w} (C) = 1 .

Theorem 2.3. Let $A$ be a summability matrix operator from $e (w, \infty)$ into itself. If $w_{1} = 1$ and $w_{n} \geq a_{n, 1}$ for all $n$ , then

\begin{array}{rcl} L_{w, \infty} (A) = 1 . \end{array}

Proof. Let $x$ be in $e (w, \infty)$ such that $x_{1} \geq x_{2} \geq \dots \geq 0$ and $y = A x$ . Then for all $n$

y_{n} = \sum_{k = 1}^{n} a_{n, k} x_{k} \geq x_{n} \sum_{k = 1}^{n} a_{n, k} = x_{n} .

Hence

\frac{X_{n}}{W_{n}} \leq \frac{Y_{n}}{W_{n}}, (\forall n) .

Therefore $∥ x ∥_{w, \infty} \leq ∥ A x ∥_{w, \infty}$ and so $L_{w, \infty} (A) \geq 1$ . To show that the constant is the best possible, we take $x_{1} = 1$ and $x_{n} = 0$ for all $n \geq 2$ . Then $∥ x ∥_{w, \infty} = 1$ and $y_{n} = a_{n, 1}$ . Thus $y_{n} \leq w_{n}$ and

Y_{n} \leq W_{n}, (\forall n) .

Therefore $∥ A x ∥_{w, \infty} \leq 1$ and $L_{w, \infty} (A) \leq 1$ . This establishes the proof of the theorem.

Theorem 2.4. Suppose that $A = (a_{n, k})$ is a summability matrix operator from $e (w, \infty)$ into itself, then

\begin{array}{rcl} L_{w, \infty} (A^{t}) = 1 . \end{array}

Proof. Let $x$ be in $e (w, \infty)$ such that $x_{1} \geq x_{2} \geq \dots \geq 0$ and $y = A^{t} x$ . Then for all $n$

\begin{array}{rcl} Y_{n} = \sum_{k = 1}^{n} y_{k} & = & \sum_{k = 1}^{n} (\sum_{i = k}^{\infty} a_{i, k} x_{i}) \\ = & X_{n} + \sum_{i = n + 1}^{\infty} (\sum_{k = 1}^{n} a_{i, k}) x_{i} . \end{array}

Hence $X_{n} \leq Y_{n}$ for all $n$ and so $∥ x ∥_{w, \infty} \leq ∥ A^{t} x ∥_{w, \infty}$ . Therefore

L_{w, \infty} (A) \geq 1 .

To show that the constant is the best possible, we take $x_{1} = 1$ and $x_{n} = 0$ for all $n \geq 2$ . We have $A^{t} x = x$ , hence $∥ A^{t} x ∥_{w, \infty} = ∥ x ∥_{w, \infty}$ and so we have the statement.

3. Summability matrix on $l_{p} (w)$

In this section, we consider summability matrix operator and its transpose on $l_{p} (w)$ . First, we shall give some lemmas which will be useful in the sequel.

Lemma 3.1. Let $(x_{n})$ , $(y_{n})$ and $(w_{n})$ be non-negative sequences. If $(w_{n})$ is decreasing and

\sum_{i = 1}^{n} x_{i} \leq \sum_{i = 1}^{n} y_{i} (n = 1, 2, \dots),

then

\sum_{i = 1}^{n} w_{i} x_{i} \leq \sum_{i = 1}^{n} w_{i} y_{i} (n = 1, 2, \dots) .

Proof. Suppose that $X_{i} = \sum_{k = 1}^{i} x_{k}$ and $Y_{i} = \sum_{k = 1}^{i} y_{k}$ . Fixing $n$ and summing by parts, we have

\begin{array}{rcl} \sum_{i = 1}^{n} w_{i} x_{i} & = & \sum_{i = 1}^{n - 1} (w_{i} - w_{i + 1}) X_{i} + w_{n} X_{n} \\ \leq & \sum_{i = 1}^{n - 1} (w_{i} - w_{i + 1}) Y_{i} + w_{n} Y_{n} \\ = & \sum_{i = 1}^{n} w_{i} y_{i} . \end{array}

Lemma 3.2. Let $p \geq 1$ and $(x_{n})$ , $(y_{n})$ and $(w_{n})$ be non-negative sequences. If $(x_{n})$ and $(w_{n})$ are decreasing and

\sum_{i = 1}^{n} x_{i} \leq \sum_{i = 1}^{n} y_{i} (n = 1, 2, \dots),

then

\sum_{i = 1}^{n} w_{i} x_{i}^{p} \leq \sum_{i = 1}^{n} w_{i} y_{i}^{p} (n = 1, 2, \dots) .

Proof. Applying Lemma 3.1 and then Hölder’s inequality, we have

\begin{array}{rcl} \sum_{i = 1}^{n} x_{i}^{p} & = & \sum_{i = 1}^{n} x_{i} x_{i}^{p - 1} \leq \sum_{i = 1}^{n} y_{i} x_{i}^{p - 1} \\ \leq & {(\sum_{i = 1}^{n} y_{i}^{p})}^{1 ∕ p} {(\sum_{i = 1}^{n} x_{i}^{p})}^{1 ∕ p^{*}} . \end{array}

Therefore $\sum_{i = 1}^{n} x_{i}^{p} \leq \sum_{i = 1}^{n} y_{i}^{p}$ and again by Lemma 3.1, we deduce

\sum_{i = 1}^{n} w_{i} x_{i}^{p} \leq \sum_{i = 1}^{n} w_{i} y_{i}^{p} .

In the following, we consider the lower bound problem for quasi-summability matrix operators.

Theorem 3.1. Suppose that $p \geq 1$ and $A = (a_{n, k})$ is a summability matrix operator from $l_{p} (v)$ into $l_{p} (w)$ . Also, let $v = (v_{n})$ , $w = (w_{n})$ be non-negative decreasing sequences. If $v_{1} = w_{1}$ and

\sum_{i = 1}^{n} v_{i} \leq \sum_{i = 1}^{n} w_{i},

for all $n$ , then

\begin{array}{rcl} L_{p, v, w} (A^{t}) = 1 . \end{array}

Proof. Let $x = (x_{n})$ be a non-negative decreasing sequence and $y = A^{t} x$ . Then

\begin{array}{rcl} \sum_{i = 1}^{n} y_{i} & = & \sum_{i = 1}^{n} \sum_{k = i}^{\infty} a_{k, i} x_{k} \\ \geq & \sum_{i = 1}^{n} \sum_{k = i}^{n} a_{k, i} x_{k} \\ = & \sum_{i = 1}^{n} x_{i} . \end{array}

Applying Lemma 3.2, we deduce that $∥ y ∥_{p, w} \geq ∥ x ∥_{p, w}$ . Since $(x_{i}^{p})$ is a non-negative decreasing sequence and $\sum_{i = 1}^{n} v_{i} \leq \sum_{i = 1}^{n} w_{i}$ , we have $∥ x ∥_{p, v} \leq ∥ x ∥_{p, w}$ . Therefore $∥ y ∥_{p, w} \geq ∥ x ∥_{p, v}$ and $L_{p, v, w} (A^{t}) \geq 1$ . Further $a_{1, 1} = 1$ and $A^{t} e_{1} = e_{1}$ , hence $∥ A^{t} e_{1} ∥_{p, w} = ∥ e_{1} ∥_{p, v}$ . This completes the proof of the theorem.

We establish a lower bound for summability matrices with increasing rows.

Lemma 3.3. Suppose $p \geq 1$ and $A = (a_{i, j})$ , $B = (b_{i, j})$ are summability matrices. If

\begin{array}{rcl} \sum_{j = 1}^{n} a_{i, j} \leq \sum_{j = 1}^{n} b_{i, j}, (i, n = 1, 2, \dots) (I) \end{array}

then

L_{p, w} (A) \leq L_{p, w} (B) .

Proof. Let $x$ be a decreasing non-negative sequence. Applying Lemma 3.1 for $(I)$ , we have

\sum_{j = 1}^{i} a_{i, j} x_{j} \leq \sum_{j = 1}^{i} b_{i, j} x_{j} .

Hence $∥ A x ∥_{p, w} \leq ∥ B x ∥_{p, w}$ , and so

L_{p, w} (A) \leq L_{p, w} (B) .

In the following statement, we compare lower bound of summability matrix with Cesaro matrix.

Theorem 3.2. Suppose $p \geq 1$ and $A = (a_{i, j})$ is a summability matrix. If the rows of $A$ are increasing, then

L_{p, w} (A) \leq L_{p, w} (C) .

Proof. We show that

\sum_{j = 1}^{n} a_{i, j} \leq \sum_{j = 1}^{n} c_{i, j} (i, n = 1, 2, \dots) .

It is clear that for $n \geq i$ , we have

\sum_{j = 1}^{n} a_{i, j} = \sum_{j = 1}^{n} c_{i, j} = 1 .

When $n < i$ , since the rows of $A$ are increasing, we have

\begin{array}{rcl} i \sum_{j = 1}^{n} a_{i, j} & = & n \sum_{j = 1}^{n} a_{i, j} + (i - n) \sum_{j = 1}^{n} a_{i, j} \\ \leq & n \sum_{j = 1}^{n} a_{i, j} + n \sum_{j = n + 1}^{i} a_{i, j} \\ = & n \sum_{j = 1}^{i} a_{i, j} \\ = & n . \end{array}

Hence

\sum_{j = 1}^{n} a_{i, j} \leq \frac{n}{i} = \sum_{j = 1}^{n} c_{i, j} .

We now apply Lemma 3.3 for $A$ and $C$ to establish the theorem.

In the following we state some result of Theorem 3.2 showing the exact value of the lower bound for summability matrix, where $w_{n} = \frac{1}{n}$ .

Corollary 3.1. Suppose $p \geq 1$ and $A = (a_{i, j})$ is a summability matrix with increasing rows. If $w$ is deﬁned by $w_{n} = \frac{1}{n}$ , then

L_{p, w} (A) = 1 .

Proof. Let $x$ be a decreasing, non-negative sequence in $l_{p} (w)$ . We have

\begin{array}{rcl} ∥ A x ∥_{p, w}^{p} & = & \sum_{n = 1}^{\infty} w_{n} {(\sum_{k = 1}^{n} a_{n, k} x_{k})}^{p} \\ \geq & \sum_{n = 1}^{\infty} w_{n} x_{n}^{p} {(\sum_{k = 1}^{n} a_{n, k})}^{p} \\ = & ∥ x ∥_{p, w}^{p}, \end{array}

hence $L_{p, w} (A) \geq 1$ . Applying Theorem 3.1 and Corollary 2.1, we deduce that

L_{p, w} (A) \leq 1 .

Therefore $L_{p, w} (A) = 1$ . Bennett considered summability matrices with increasing or decreasing rows in [1]. For example

\begin{array}{rcl} Γ (2) = [\begin{matrix} 1 & 0 \\ 1 ∕ 3 & 2 ∕ 3 & 0 \\ 1 ∕ 6 & 2 ∕ 6 & 3 ∕ 6 & 0 \\ 1 ∕ 10 & 2 ∕ 10 & 3 ∕ 10 & 4 ∕ 10 & 0 \\ . & . & . & . & . & . \end{matrix}] \end{array}

$Γ (2)$ is the summability matrix with increasing rows and for $w_{n} = \frac{1}{n}$ , we have

L_{p, w} (Γ (2)) = 1 .

4. Quasi-summability matrix

In this section, we establish a lower bound for quasi-summability matrix on $l_{p} (w)$ for $0 < p \leq 1$ , where sequences are non-negative. Note that we shall use the norm only as a notation and do not use norm’s properties. First, we compare lower bound of quasi-summability matrix with Copson matrix.

Lemma 4.1. Suppose $0 < p \leq 1$ and $u$ , $v$ and $w$ are N-tuples with non-negative entries. If $u$ , $w$ are decreasing, and

\sum_{i = n}^{N} u_{i} \geq \sum_{i = n}^{N} v_{i} (n = 1, \dots, N),

then

\sum_{i = n}^{N} w_{i} u_{i}^{p} \geq \sum_{i = n}^{N} w_{i} v_{i}^{p}, (n = 1, \dots, N) .

Proof. Set $x_{i} = u_{N - i + 1}$ and $y_{i} = v_{N - i + 1}$ for $1 \leq i \leq N - n + 1$ , we have $\sum_{i = 1}^{N - n + 1} x_{i} = \sum_{i = n}^{N} u_{i}$ and $\sum_{i = 1}^{N - n + 1} y_{i} = \sum_{i = n}^{N} v_{i}$ . Applying Lemma 3.1 and Hölder’s inequality, we deduce the statement.

Lemma 4.2. Suppose $0 < p \leq 1$ and $A = (a_{i, j})$ , $B = (b_{i, j})$ are summability matrices. If the rows of $B$ are decreasing and

\begin{array}{rcl} \sum_{i = 1}^{n} a_{j, i} \geq \sum_{i = 1}^{n} b_{j, i}, (j, n = 1, 2, \dots) (I) \end{array}

then

L_{p, w} (A^{t}) \leq L_{p, w} (B^{t}) .

Proof. Let $N$ be ﬁxed and $x$ be a sequence with non-negative entries. We deﬁne $u$ and $v$ by

u_{i} = \sum_{j = i}^{N} b_{j, i} x_{j}, v_{i} = \sum_{j = i}^{N} a_{j, i} x_{j}, (i = 1, 2, \dots) .

It is clear that $u_{i}$ decreases with $i$ . Deﬁnition of summability matrix and $(I)$ follow that

\sum_{i = n}^{N} u_{i} \geq \sum_{i = n}^{N} v_{i}, (n = 1, 2, \dots, N),

hence applying Lemma 4.1, we deduce that

\sum_{i = n}^{N} w_{i} u_{i}^{p} \geq \sum_{i = n}^{N} w_{i} v_{i}^{p}, (n = 1, 2, \dots, N) .

Therefore $∥ B^{t} x ∥_{p, w} \geq ∥ A^{t} x ∥_{p, w}$ , and so

L_{p, w} (A^{t}) \leq L_{p, w} (B^{t}) .

The transpose of Cesaro matrix is called the Copson matrix and we denote it with $C^{t}$ . Applying Theorem 2.1 of [6], we have

L_{p, w} (C^{t}) = p .

Theorem 4.1. Suppose $A = (a_{i, j})$ is a summability matrix. If the rows of $A$ are decreasing, then

L_{p, w} (A^{t}) \leq p,

where $0 < p \leq 1$ .

Proof. It is clear that the rows of $C$ are decreasing. For $n \geq j$ , we have

\sum_{i = 1}^{n} a_{j, i} = \sum_{i = 1}^{n} c_{j, i} = 1 .

Also, when $n < j$ , since the $j^{t h}$ row of $A$ is decreasing, the average

\frac{1}{n} \sum_{i = 1}^{n} a_{j, i}

decreases with $n$ . Hence

\sum_{i = 1}^{n} a_{j, i} \geq \frac{n}{j} (j = 1, 2, \dots),

and so we have $(I)$ . We now apply Lemma 4.2 for $A^{t}$ and $C^{t}$ to establish the theorem.

In the following, we evaluate lower bound of summability matrix with increasing rows.

Lemma 4.3([1], Lemma 3.13). Let $0 < p \leq 1$ , and $u$ be N-tuple with positive entries. Then

\begin{array}{rcl} p \sum_{k = n}^{N} u_{k} {(\sum_{j = k}^{N} u_{j})}^{p - 1} \leq {(\sum_{j = n}^{N} u_{j})}^{p}, (n = 1, \dots, N), (I I) \end{array}

and if $p \geq 1$ , the inequality in $(I I)$ is reversed. The constant $p$ is best possible in either version of $(I I)$ and there is strict inequality unless $p = 1$ or $u = 0$ .

Theorem 4.2. Let $0 < p \leq 1$ and $A = (a_{i, j})$ be the summability matrix. If the rows of $A$ are increasing, then

L_{p, w} (A^{t}) \geq p .

Proof. Suppose $x$ is a sequence with non-negative terms. Applying Lemma 4.3, we have

\begin{array}{rcl} ∥ A^{t} x ∥_{p, w}^{p} = \sum_{n = 1}^{\infty} w_{n} {(\sum_{k = n}^{\infty} a_{k, n} x_{k})}^{p} & \geq & p \sum_{n = 1}^{\infty} w_{n} \sum_{k = n}^{\infty} a_{k, n} x_{k} {(\sum_{j = k}^{\infty} a_{j, n} x_{j})}^{p - 1} \\ = & p \sum_{k = 1}^{\infty} x_{k} \sum_{n = 1}^{k} w_{n} a_{k, n} {(\sum_{j = k}^{\infty} a_{j, n} x_{j})}^{p - 1} \\ \geq & p \sum_{k = 1}^{\infty} w_{k} x_{k} \sum_{n = 1}^{k} a_{k, n} {(\sum_{j = k}^{\infty} a_{j, n} x_{j})}^{p - 1} . \end{array}

Since $A$ has increasing rows,

\sum_{j = k}^{\infty} a_{j, n} x_{j}, (n = 1, \dots, k),

is increasing with n, hence

\sum_{n = 1}^{k} a_{k, n} {(\sum_{j = k}^{\infty} a_{j, n} x_{j})}^{p - 1} \geq {(\sum_{j = k}^{\infty} a_{j, n} x_{j})}^{p - 1} .

Thus applying Hölder’s inequality, it follows that

\begin{array}{rcl} ∥ A^{t} x ∥_{p, w}^{p} & \geq & p \sum_{k = 1}^{\infty} w_{k} x_{k} {(\sum_{j = k}^{\infty} a_{j, k} x_{j})}^{p - 1} \\ \geq & p \sum_{k = 1}^{\infty} w_{k}^{1 ∕ p} x_{k} {(w_{k}^{1 ∕ p^{*} (p - 1)} \sum_{j = k}^{\infty} a_{j, k} x_{j})}^{p - 1} \\ \geq & p {(\sum_{k = 1}^{\infty} w_{k} x_{k}^{p})}^{1 ∕ p} {(\sum_{k = 1}^{\infty} {(w_{k}^{1 ∕ p} \sum_{j = k}^{\infty} a_{j, k} x_{j})}^{p})}^{1 ∕ p^{*}} \\ = & p ∥ x ∥_{p, w} ∥ A^{t} x ∥_{p, w}^{p - 1} . \end{array}

Therefore $∥ A^{t} x ∥_{p, w} \geq p ∥ x ∥_{p, w}$ , and so $L_{p, w} (A^{t}) \geq p$ .

Proposition 4.1([6], Proposition 2.2). Let $0 < p, q < 1$ , and $A$ be a matrix with non-negative entries. Then

\begin{array}{rcl} ∥ A x ∥_{q, w} \geq L ∥ x ∥_{p, w}, \end{array}

for all non-negative $x$ if and only if

\begin{array}{rcl} ∥ A^{t} y ∥_{p^{*}, w} \geq L ∥ y ∥_{q^{*}, w} . \end{array}

for all non-negative $y$ .

Theorem 4.3. Let $p < 0$ and $A = (a_{i, j})$ be the summability matrix. If $A$ has increasing rows, then

\sum_{j = 1}^{\infty} w_{j} {(\sum_{k = 1}^{j} a_{j, k} x_{k})}^{p} \leq {(p^{*})}^{p} \sum_{k = 1}^{\infty} w_{k} x_{k}^{p},

for any sequence $x$ with positive terms. Proof. Since $0 < p^{*} < 1$ , Theorem 4.2 follows that

L_{p^{*}, w} (A^{t}) \geq p^{*} .

Applying Proposition 4.1, we deduce that

L_{p, w} (A) \geq p^{*} .

Hence for any sequence $x$ with positive terms, we have

∥ A x ∥_{p, w} \geq p^{*} ∥ x ∥_{p, w} .

This completes the proof of theorem.

Corollary 4.1. Let $p > 0$ , and $x$ be a positive sequence. Then

\begin{array}{rcl} \sum_{j = 1}^{\infty} w_{j} {max}_{1 \leq i \leq j} {(\frac{j - i + 1}{\frac{1}{x_{i}} + \dots + \frac{1}{x_{j}}})}^{p} \leq {(\frac{p + 1}{p})}^{p} \sum_{k = 1}^{\infty} w_{k} x_{k}^{p} . \end{array}

Proof. We apply Theorem 4.3, by replacing $p$ with $- p$ and $x_{k}$ with $\frac{1}{x_{k}}$ . The left hand side of inequality is $∥ A x ∥_{p, w}^{p}$ , where $A = (a_{i, j})$ is a summability matrix operator with

\begin{array}{rcl} a_{j, k} = \{\begin{matrix} \frac{1}{j - i_{j} + 1} & k = i_{j}, \dots, j \\ 0 & otherwise \end{matrix} \end{array}

where $i_{j}$ is chosen to be ﬁrst value of i at which the maximum

{max}_{1 \leq i \leq j} (\frac{1}{j - i + 1} \sum_{k = i}^{j} x_{k})

is attained( $j = 1, 2, \dots$ ). The rows of $A$ are increasing and its entries depend on $x$ , but this is not damaging, because applying Theorem 4.3, for any positive sequence $x$ , we have

∥ A x ∥_{p, w} \leq p^{*} ∥ x ∥_{p, w}

and this establishes the statement.

The following statement is an extension of famous inequality due to Carlemann([2], Theorem 334).

Corollary 4.2. If $(x_{k})$ is a sequence with non-negative terms, then

\sum_{i = 1}^{\infty} w_{j} {max}_{1 \leq i \leq j} {(x_{i} \dots x_{j})}^{\frac{1}{j - i + 1}} \leq e \sum_{k = 1}^{\infty} w_{k} x_{k} .

Proof. We apply Corollary 4.1, by replacing $x_{k}$ with $x_{k}^{1 ∕ p}$ and tending $p \to 0$ , we have the statement.

References

[1] G. Bennett, Inequalities complimentary to Hardy, Quart. J. Math. Oxford(2), 49 (1998), 395–432.

[2] G. H. Hardy, J. E. Littlewood and G. Polya, Inequalities, 2nd edition, Cambridge University press, Cambridge 2001.

[3] G. J. O. Jameson and R. Lashkaripour, Lower bounds of operators on weighted $l_{p}$ spaces and Lorentz sequence spaces, Glasgow Math. J. 42(2000), 211–223.

[4] G. J. O. Jameson and R. Lashkaripour, Norms of certain operators on weighted $l_{p}$ spaces and Lorentz sequence spaces, J. Inequalities in Pure and Applied Mathematics. Volume 3, Issue 1, Article 6(2002).

[5] R. Lashkaripour and D. Foroutannia, Inequalities involving upper bounds for certain matrix operators, Proc. Indian Acad. Sci.(Math. Sci) Volume 116, Agust(2006), 325–336.

[6] R. Lashkaripour and D. Foroutannia, Lower bounds for matrices on weighted sequence spaces, Journal of Sciences, Islamic Republic of Iran. Vol. 18, No. 1(2007), 49–56.

[7] R. Lashkaripour and D. Foroutannia, Some inequalities involving upper bounds for some matrix operators I, Czech. Math. J., to appear.

Department of Mathematics, Sistan and Baluchestan University, Zahedan, Iran

E-mail address: $d_{-}$ foroutan@math.com

E-mail address: lashkari@hamoon.usb.ac.ir

Received March 21, 2007

1. Introduction

2. Summability matrix operator on d(w, 1) and e(w,∞)

3. Summability matrix on lp(w)

4. Quasi-summability matrix

References

2. Summability matrix operator on $d (w, 1)$ and $e (w, \infty)$

3. Summability matrix on $l_{p} (w)$