# Some geometric properties of $N\left(\lambda ,p\right)$-spaces

## Abstract

In this paper, we introduce the sequence spaces $N\left(\lambda ,p\right)$ and we show some geometric properties of that spaces. The main purpose of this paper is to show that $N\left(\lambda ,p\right)$ is a Banach space and has the rotund property, the Kadec-Klee property, the uniform Opial property, the $\left(\beta \right)$-property, the k-NUC property and the Banach-Saks property of type p.

MSC:46A45, 40C05, 46B45, 46A35.

## 1 Introduction

By ω, we denote the space of all real valued sequences. Any vector subspace of ω is called a sequence space. We write ${l}_{\mathrm{\infty }}$, c, and ${c}_{0}$ for the spaces of all bounded, convergent and null sequences, respectively. Also by bs, cs, ${l}_{1}$, and ${l}_{\mathrm{\infty }}$, we denote the spaces of all bounded, convergent, absolutely convergent and p-absolutely convergent series, respectively; where $1. Assume here and after that $\left({p}_{k}\right)$ be a bounded sequence of strictly positive real numbers with $sup\left\{{p}_{k}\right\}=H$ and $M=max\left\{1,H\right\}$. Then, the linear space $l\left(p\right)$ was defined by Maddox [1] (see also Simons [2] and Nakano [3]) as follows:

$l\left(p\right)=\left\{x=\left({x}_{k}\right)\in \omega :\sum _{k}{|{x}_{k}|}^{{p}_{k}}<\mathrm{\infty }\right\}\phantom{\rule{1em}{0ex}}\left(0<{p}_{k}\le H<\mathrm{\infty }\right)$

which is complete paranormed space paranormed by

$g\left(x\right)={\left(\sum _{k}{|{x}_{k}|}^{{p}_{k}}\right)}^{\frac{1}{M}}.$

For simplicity in notation, here and in what follows, the summation without limits runs from 1 to ∞.

In [4] was introduced the following numerical sequence $\lambda ={\left({\lambda }_{k}\right)}_{k=0}^{\mathrm{\infty }}$, which is a strictly increasing sequence of positive real numbers tending to infinity, as $k\to \mathrm{\infty }$, that is

We will introduce the following sequence space:

$N\left(\lambda ,p\right)=\left\{x=\left({x}_{n}\right)\in \omega :\sum _{k=1}^{\mathrm{\infty }}{\left(\frac{1}{{\lambda }_{k}}\sum _{i=1}^{k}|\left({\lambda }_{i}-{\lambda }_{i-1}\right){x}_{i}|\right)}^{{p}_{k}}<\mathrm{\infty }\right\}.$

For ${\lambda }_{k}=k$, we obtain the Cesaro sequence space $ces\left(p\right)$ (see [5]). If ${\lambda }_{k}=k$ and ${p}_{k}=p$, then $N\left(\lambda ,p\right)={ces}_{p}$ (see [6]). In case where ${p}_{k}=p$ for all $k\in \mathbb{N}$, then we will denote $N\left(\lambda ,p\right)={N}_{p}$. Some results related to the geometric properties of sequence spaces are given in [79].

## 2 Topological properties

Theorem 2.1 The paranorm on $N\left(\lambda ,p\right)$ is given by the relation

$h\left(x\right)={\left(\sum _{k=1}^{\mathrm{\infty }}{\left(\frac{1}{{\lambda }_{k}}\sum _{i=1}^{k}|\left({\lambda }_{i}-{\lambda }_{i-1}\right){x}_{i}|\right)}^{{p}_{k}}\right)}^{\frac{1}{M}},$

where $M=max\left\{1,H\right\}$ and $H=sup{p}_{k}$.

## 3 Geometrical properties

In this section we will show some geometric properties of the $N\left(\lambda ,p\right)$-spaces, such as the $\left(\beta \right)$-property, the k-NUC property, the Banach-Saks property of type p, and the $\left(H\right)$-property. It is well known that these properties are most important in Banach spaces (see [10, 11] and [1]).

Definition 3.1 A Banach space X is said to be k-nearly uniformly convex (k-NUC) if for any $ϵ>0$, there exists a $\delta >0$ such that for any sequence $\left({x}_{n}\right)\subset B\left(X\right)$ with $sep\left({x}_{n}\right)\ge ϵ$, there are ${n}_{1},{n}_{2},\dots ,{n}_{k}\in \mathbb{N}$ such that

$\parallel \frac{{x}_{{n}_{1}}+{x}_{{n}_{2}}+\cdots +{x}_{{n}_{k}}}{k}\parallel <1-\delta ,$

where $sep\left({x}_{n}\right)=inf\left\{\parallel {x}_{n}-{x}_{m}\parallel :n\ne m\right\}>ϵ$.

Definition 3.2 A Banach space X has property $\left(\beta \right)$ if and only if for each $r>0$ and $ϵ>0$, there exists a $\delta >0$ such that for each element $x\in B\left(X\right)$ and each sequence ${x}_{n}\in B\left(X\right)$ with $sep\left({x}_{n}\right)\ge ϵ$, there is an index k for each

$\parallel \frac{x+{x}_{k}}{2}\parallel \le \delta .$

Definition 3.3 A Banach space X is said to have the Banach-Saks property type p if every weakly null sequence $\left({x}_{k}\right)$ has a subsequence $\left({x}_{kl}\right)$ such that for some $C>0$

$\parallel \sum _{l=0}^{n}{x}_{kl}\parallel

for all $n\in \mathbb{N}$.

Definition 3.4 Let X be a real vector space. A functional $\sigma :X\to \left[0,\mathrm{\infty }\right)$ is called a modular if

1. (1)

$\sigma \left(x\right)=0$ if and only if $x=\theta$,

2. (2)

$\sigma \left(\alpha x\right)=\sigma \left(x\right)$ for all scalars α with $|\alpha |=1$,

3. (3)

$\sigma \left(\alpha x+\beta y\right)\le \sigma \left(x\right)+\sigma \left(y\right)$ for all $x,y\in X$ and $\alpha ,\beta >0$ with $\alpha +\beta =1$,

4. (4)

the modular σ is called convex if $\sigma \left(\alpha x+\beta y\right)\le \alpha \sigma \left(x\right)+\beta \sigma \left(y\right)$ for all $x,y\in X$ and $\alpha ,\beta >0$, with $\alpha +\beta =1$.

A modular σ is called:

1. (5)

right continuous if ${lim}_{\alpha \to {1}^{+}}\sigma \left(\alpha x\right)=\sigma \left(x\right)$ for all $x\in {X}_{\sigma }$,

2. (6)

left continuous if ${lim}_{\alpha \to {1}^{-}}\sigma \left(\alpha x\right)=\sigma \left(x\right)$ for all $x\in {X}_{\sigma }$,

3. (7)

continuous if it is both right and left continuous,

where ${X}_{\sigma }=\left\{x\in X:{lim}_{\alpha \to {0}^{+}}\sigma \left(\alpha x\right)=0\right\}$. We define ${\sigma }_{p}$ on $N\left(\lambda ,p\right)$ as follows:

${\sigma }_{p}\left(x\right)=\left(\sum _{k=1}^{\mathrm{\infty }}{\left(\frac{1}{{\lambda }_{k}}\sum _{i=0}^{k}|\left({\lambda }_{i}-{\lambda }_{i-1}\right){x}_{i}|\right)}^{{p}_{k}}\right),$

where ${\lambda }_{-1}=0$.

If ${p}_{k}\ge 1$, for all $k\in \mathbb{N}$, by the convexity of the function $t\to {|t|}^{{p}_{k}}$, for all $k\in \mathbb{N}$, ${\sigma }_{p}$ defined above is a modular convex in the $N\left(\lambda ,p\right)$.

Definition 3.5 A modular ${\sigma }_{p}$ is said to satisfy the ${\delta }_{2}$-conditions if for every $ϵ>0$, there exist constant $M>0$ and $m>0$ such that

${\sigma }_{p}\left(2t\right)\le M{\sigma }_{p}\left(t\right)+ϵ$
(3.1)

for all $t\in {X}_{{\sigma }_{p}}$ with ${\sigma }_{p}\left(t\right)\le m$.

Lemma 3.6 ([12])

If ${\sigma }_{p}$ satisfies the ${\delta }_{2}$-conditions, then for any $A>0$ and $ϵ>0$, there exists $\delta >0$ such that

$|{\sigma }_{p}\left(t+w\right)-{\sigma }_{p}\left(t\right)|<ϵ$
(3.2)

whenever $t,w\in {X}_{{\sigma }_{p}}$ with ${\sigma }_{p}\left(t\right)\le A$ and ${\sigma }_{p}\left(w\right)\le \delta$.

Theorem 3.7 ([12])

1. (1)

If ${\sigma }_{p}$ satisfies the ${\delta }_{2}$-conditions, then for any $x\in {X}_{{\sigma }_{p}}$, $\parallel x\parallel =1$ if and only if ${\sigma }_{p}\left(x\right)=1$.

2. (2)

If ${\sigma }_{p}$ satisfies the ${\delta }_{2}$-conditions, then for any sequence $\left({x}_{n}\right)\in {X}_{{\sigma }_{p}}$, $\parallel {x}_{n}\parallel \to 0$ if and only if ${\sigma }_{p}\left({x}_{n}\right)\to 0$.

Theorem 3.8 If ${\sigma }_{p}$ satisfies the ${\delta }_{2}$-conditions, then for any $ϵ\in \left(0,1\right)$, there exists $\delta \in \left(0,1\right)$ such that ${\sigma }_{p}\left(x\right)\le 1-ϵ$ implies $\parallel x\parallel \le 1-\delta$.

Proof The proof of the theorem follows directly from the above two facts. □

Theorem 3.9 For any $x\in N\left(\lambda ,p\right)$ and $ϵ\in \left(0,1\right)$, there exists $\delta \in \left(0,1\right)$, such that ${\sigma }_{p}\left(x\right)\le 1-ϵ$ implies $\parallel x\parallel \le 1-\delta$.

Proof The proof of the theorem follows directly from Theorem 3.8. □

Proposition 3.10 If ${p}_{k}\ge 1$, for all $k\in \mathbb{N}$, then the modular function ${\sigma }_{p}$, on $N\left(\lambda ,p\right)$, satisfies the following conditions:

1. (1)

If $0<\alpha \le 1$, then ${\alpha }^{M}{\sigma }_{p}\left(\frac{x}{\alpha }\right)\le {\sigma }_{p}\left(x\right)$ and ${\sigma }_{p}\left(\alpha x\right)\le \alpha {\sigma }_{p}\left(x\right)$.

2. (2)

If $\alpha \ge 1$, then ${\sigma }_{p}\left(x\right)\le {\alpha }^{M}{\sigma }_{p}\left(\frac{x}{\alpha }\right)$.

3. (3)

If $\alpha \ge 1$, then ${\sigma }_{p}\left(x\right)\le \alpha {\sigma }_{p}\left(\frac{x}{\alpha }\right)$.

4. (4)

The modular function ${\sigma }_{p}\left(x\right)$ is continuous on $N\left(\lambda ,p\right)$.

Proof The proof of the proposition is similar to Proposition 2.1 in [13]. □

Now we will define the following two norms (the first is known as the Luxemburg norm and the second as the Amemiya norm) in $N\left(\lambda ,p\right)$:

${\parallel x\parallel }_{L}=inf\left\{\alpha >0:{\sigma }_{p}\left(\frac{x}{\alpha }\right)\le 1\right\}$
(3.3)

and

${\parallel x\parallel }_{A}=\underset{\alpha >0}{inf}\frac{1}{\alpha }\left\{1+{\sigma }_{p}\left(\alpha \cdot x\right)\right\}.$
(3.4)

Proposition 3.11 Let $x\in N\left(\lambda ,p\right)$. Then the following relations are satisfied:

1. (1)

If ${\parallel x\parallel }_{L}<1$, then ${\sigma }_{p}\left(x\right)\le {\parallel x\parallel }_{L}$.

2. (2)

If ${\parallel x\parallel }_{L}>1$, then ${\sigma }_{p}\left(x\right)\ge {\parallel x\parallel }_{L}$.

3. (3)

${\parallel x\parallel }_{L}=1$ if and only if ${\sigma }_{p}\left(x\right)=1$.

4. (4)

${\parallel x\parallel }_{L}<1$ if and only if ${\sigma }_{p}\left(x\right)<1$.

5. (5)

${\parallel x\parallel }_{L}>1$ if and only if ${\sigma }_{p}\left(x\right)>1$.

Proof (1) Let $x\in N\left(\lambda ,p\right)$ and ${\parallel x\parallel }_{L}<1$. Let also $ϵ>0$ such that $0<ϵ<1-{\parallel x\parallel }_{L}$. On the other hand from the definition of the norm by relation (3.3) we find that there exists a $\alpha >0$ such that ${\parallel x\parallel }_{L}+ϵ>\alpha$ and ${\sigma }_{p}\left(\frac{x}{\alpha }\right)\le 1$. From the above relations and property (1) of Proposition 3.10, we obtain

$\frac{{\parallel x\parallel }_{L}+ϵ}{\alpha }>1$

and

${\sigma }_{p}\left(x\right)\le \frac{{\parallel x\parallel }_{L}+ϵ}{\alpha }{\sigma }_{p}\left(x\right)=\frac{{\parallel x\parallel }_{L}+ϵ}{\alpha }{\sigma }_{p}\left(\alpha \cdot \frac{x}{\alpha }\right)\le \frac{{\parallel x\parallel }_{L}+ϵ}{\alpha }\cdot \alpha \cdot {\sigma }_{p}\left(\frac{x}{\alpha }\right)\le {\parallel x\parallel }_{L}+ϵ.$

The previous statement is valid for every $ϵ>0$, from which it follows that ${\sigma }_{p}\left(x\right)\le {\parallel x\parallel }_{L}$.

1. (2)

In this case we will choose $ϵ>0$ such that $0<ϵ<1-\frac{1}{{\parallel x\parallel }_{L}}$, and we obtain $1<\left(1-ϵ\right){\parallel x\parallel }_{L}<{\parallel x\parallel }_{L}$. Now using into consideration definition of the norm (3.3) and relation (1) of Proposition 3.10, we get

$1<{\sigma }_{p}\left(\frac{x}{\left(1-ϵ\right){\parallel x\parallel }_{L}}\right)\le \frac{1}{\left(1-ϵ\right){\parallel x\parallel }_{L}}{\sigma }_{p}\left(x\right)\phantom{\rule{1em}{0ex}}⇒\phantom{\rule{1em}{0ex}}\left(1-ϵ\right){\parallel x\parallel }_{L}\le {\sigma }_{p}\left(x\right)$

for every $ϵ\in \left(0,1-\frac{1}{{\parallel x\parallel }_{L}}\right)$. Finally we have proved that ${\parallel x\parallel }_{L}\le {\sigma }_{p}\left(x\right)$.

1. (3)

Since ${\sigma }_{p}\left(x\right)$ is continuous function (see [14]), this property follows immediately.

2. (4)

Follows from properties (1) and (3).

3. (5)

Follows from properties (2) and (3). □

Theorem 3.12 $N\left(\lambda ,p\right)$ is a Banach space under the Luxemburg and Amemiya norms.

Proof We will prove that $N\left(\lambda ,p\right)$ is a Banach space under the Luxemburg norm. In a similar way we can prove that $N\left(\lambda ,p\right)$ is a Banach space under the Amemiya norm. In what follows we need to show that every Cauchy sequence in $N\left(\lambda ,p\right)$ is convergent according to the Luxemburg norm. Let $\left\{{x}_{k}^{n}\right\}$ be any Cauchy sequence in $N\left(\lambda ,p\right)$ and $ϵ\in \left(0,1\right)$. Thus there exists a natural number ${n}_{0}$, such that for any $n,m\ge {n}_{0}$ we get ${\parallel {x}^{\left(n\right)}-{x}^{\left(m\right)}\parallel }_{L}<ϵ$. From Proposition 3.11 we get

${\sigma }_{p}\left({x}^{\left(n\right)}-{x}^{\left(m\right)}\right)\le {\parallel {x}^{\left(n\right)}-{x}^{\left(m\right)}\parallel }_{L}<ϵ$
(3.5)

for all $n,m\ge {n}_{0}$. This implies that

$\sum _{k=1}^{\mathrm{\infty }}{\left(\frac{1}{{\lambda }_{k}}\sum _{i=1}^{k}|\left({\lambda }_{i}-{\lambda }_{i-1}\right)\left({x}_{i}^{\left(n\right)}-{x}_{i}^{\left(m\right)}\right)|\right)}^{{p}_{k}}<ϵ.$
(3.6)

For each fixed k and for all $n,m\ge {n}_{0}$,

$\frac{1}{{\lambda }_{k}}\sum _{i=1}^{k}|\left({\lambda }_{i}-{\lambda }_{i-1}\right)\left({x}_{i}^{\left(n\right)}-{x}_{i}^{\left(m\right)}\right)|<ϵ.$

Hence ${\left({y}_{k}^{\left(n\right)}\right)}_{k}={\left(\frac{1}{{\lambda }_{k}}{\sum }_{i=1}^{k}|\left({\lambda }_{i}-{\lambda }_{i-1}\right){x}_{i}^{\left(n\right)}|\right)}_{k}$ is a Cauchy sequence in . Since is a complete normed space, there exists a ${\left({y}_{k}\right)}_{k}={\left(\frac{1}{{\lambda }_{k}}{\sum }_{i=1}^{k}|\left({\lambda }_{i}-{\lambda }_{i-1}\right){x}_{i}|\right)}_{k}\in \mathbb{R}$ such that $\left({y}_{k}^{\left(n\right)}\right)\to {y}_{k}$ as $n\to \mathrm{\infty }$. Therefore, as $n\to \mathrm{\infty }$, by relation (3.6) we have

$\sum _{k=1}^{\mathrm{\infty }}{\left(\frac{1}{{\lambda }_{k}}\sum _{i=1}^{k}|\left({\lambda }_{i}-{\lambda }_{i-1}\right)\left({x}_{i}-{x}_{i}^{\left(m\right)}\right)|\right)}^{{p}_{k}}<ϵ$

for all $m\ge {n}_{0}$. In the sequel we will show that $\left({y}_{k}\right)$ is a sequence form $N\left(\lambda ,p\right)$. From Proposition 3.10 and relation (3.5) we have

$\underset{n\to \mathrm{\infty }}{lim}{\sigma }_{p}\left({x}^{\left(n\right)}-{x}^{\left(m\right)}\right)={\sigma }_{p}\left(x-{x}^{\left(m\right)}\right)\le {\parallel x-{x}^{\left(m\right)}\parallel }_{L}<ϵ$

for all $m\ge {n}_{0}$. This implies that $\left({x}^{\left(n\right)}\right)\to x$ as $m\to \mathrm{\infty }$. So we have $x={x}^{\left(n\right)}-\left({x}^{\left(n\right)}-x\right)\in N\left(\lambda ,p\right)$. This proves that $N\left(\lambda ,p\right)$ is a complete normed space under the Luxemburg norm. □

In what follows we will show results related to the Luxemburg norm, and for this reason we will denote it just $\parallel \cdot \parallel$.

Theorem 3.13 The space $N\left(\lambda ,p\right)$ is rotund if and only if ${p}_{k}>1$ for all $k\in \mathbb{N}$.

Proof Let $N\left(\lambda ,p\right)$ be rotund and choose $k\in \mathbb{N}$ such that ${p}_{k}=1$. Consider the two sequences given by

$x=\left(0,0,\dots ,0,\frac{{\lambda }_{k}}{{2}^{k}\cdot |{\lambda }_{k}-{\lambda }_{k-1}|},0,0,\dots \right)$

and

$y=\left(0,0,\dots ,0,\frac{2{\lambda }_{k}}{{3}^{k}\cdot |{\lambda }_{k}-{\lambda }_{k-1}|},0,0,\dots \right).$

Then obviously $x\ne y$ and

${\sigma }_{p}\left(x\right)={\sigma }_{p}\left(y\right)={\sigma }_{p}\left(\frac{x+y}{2}\right)=1.$

Then from Proposition 3.11, property (3), it follows that $x,y,\frac{x+y}{2}\in S\left[N\left(\lambda ,p\right)\right]$, which leads to the contradiction that the sequence space $N\left(\lambda ,p\right)$ is not rotund. Hence ${p}_{k}>1$, for all $k\in \mathbb{N}$.

Conversely, let $x\in S\left[N\left(\lambda ,p\right)\right]$ and $y,z\in S\left[N\left(\lambda ,p\right)\right]$ such that $x=\frac{y+z}{2}$. By the convexity of ${\sigma }_{p}$ and property (3) from Proposition 3.11, we have

$1={\sigma }_{p}\left(x\right)\le \frac{{\sigma }_{p}\left(y\right)+{\sigma }_{p}\left(z\right)}{2}\le \frac{1}{2}+\frac{1}{2}=1,$

which gives ${\sigma }_{p}\left(y\right)={\sigma }_{p}\left(z\right)=1$ and

${\sigma }_{p}\left(x\right)=\frac{{\sigma }_{p}\left(y\right)+{\sigma }_{p}\left(z\right)}{2}.$
(3.7)

From the previous relation we obtain

$\begin{array}{c}\sum _{k=1}^{\mathrm{\infty }}{\left(\frac{1}{{\lambda }_{k}}\sum _{i=1}^{k}|\left({\lambda }_{i}-{\lambda }_{i-1}\right){x}_{i}|\right)}^{{p}_{k}}\hfill \\ \phantom{\rule{1em}{0ex}}=\frac{1}{2}\left(\sum _{k=1}^{\mathrm{\infty }}{\left(\frac{1}{{\lambda }_{k}}\sum _{i=1}^{k}|\left({\lambda }_{i}-{\lambda }_{i-1}\right){y}_{i}|\right)}^{{p}_{k}}+\sum _{k=1}^{\mathrm{\infty }}{\left(\frac{1}{{\lambda }_{k}}\sum _{i=1}^{k}|\left({\lambda }_{i}-{\lambda }_{i-1}\right){z}_{i}|\right)}^{{p}_{k}}\right).\hfill \end{array}$

Since $x=\frac{y+z}{2}$, we get

$\begin{array}{c}\sum _{k=1}^{\mathrm{\infty }}{\left(\frac{1}{{\lambda }_{k}}\sum _{i=1}^{k}|\left({\lambda }_{i}-{\lambda }_{i-1}\right)\left({y}_{i}+{z}_{i}\right)|\right)}^{{p}_{k}}\hfill \\ \phantom{\rule{1em}{0ex}}=\frac{1}{2}\left(\sum _{k=1}^{\mathrm{\infty }}{\left(\frac{1}{{\lambda }_{k}}\sum _{i=1}^{k}|\left({\lambda }_{i}-{\lambda }_{i-1}\right){y}_{i}|\right)}^{{p}_{k}}+\sum _{k=1}^{\mathrm{\infty }}{\left(\frac{1}{{\lambda }_{k}}\sum _{i=1}^{k}|\left({\lambda }_{i}-{\lambda }_{i-1}\right){z}_{i}|\right)}^{{p}_{k}}\right).\hfill \end{array}$

This implies that

$\begin{array}{c}{\left(\frac{1}{{\lambda }_{k}}\sum _{i=1}^{k}|\left({\lambda }_{i}-{\lambda }_{i-1}\right)\left({y}_{i}+{z}_{i}\right)|\right)}^{{p}_{k}}\hfill \\ \phantom{\rule{1em}{0ex}}=\frac{1}{2}\left({\left(\frac{1}{{\lambda }_{k}}\sum _{i=1}^{k}|\left({\lambda }_{i}-{\lambda }_{i-1}\right){y}_{i}|\right)}^{{p}_{k}}+{\left(\frac{1}{{\lambda }_{k}}\sum _{i=1}^{k}|\left({\lambda }_{i}-{\lambda }_{i-1}\right){z}_{i}|\right)}^{{p}_{k}}\right).\hfill \end{array}$
(3.8)

From the previous relation we get ${y}_{i}={z}_{i}$ for all $i\in \mathbb{N}$, respectively, $z=y$. That is, the sequence space $N\left(\lambda ,p\right)$ is rotund. □

In what follows we will give two facts without proof because their proofs follow directly from Proposition 3.10 and Proposition 3.11.

Theorem 3.14 Let $x\in N\left(\lambda ,p\right)$. Then the following statements hold:

1. (i)

For $0<\alpha <1$ and $\parallel x\parallel >\alpha$ we get ${\sigma }_{p}\left(x\right)>{\alpha }^{M}$.

2. (ii)

If $\alpha \ge 1$ and $\parallel x\parallel <\alpha$, then we have ${\sigma }_{p}\left(x\right)<{\alpha }^{M}$.

Theorem 3.15 Let $\left({x}_{n}\right)$ be a sequence in $N\left(\lambda ,p\right)$. Then the following statements hold:

1. (i)

${lim}_{n\to \mathrm{\infty }}\parallel {x}_{n}\parallel =1$ implies ${lim}_{n\to \mathrm{\infty }}{\sigma }_{p}\left({x}_{n}\right)=1$.

2. (ii)

${lim}_{n\to \mathrm{\infty }}{\sigma }_{p}\left({x}_{n}\right)=0$ implies ${lim}_{n\to \mathrm{\infty }}\parallel {x}_{n}\parallel =0$.

Theorem 3.16 Let $x\in N\left(\lambda ,p\right)$ and $\left({x}^{\left(n\right)}\right)\subset N\left(\lambda ,p\right)$. If ${\sigma }_{p}\left({x}^{\left(n\right)}\right)\to {\sigma }_{p}\left(x\right)$ as $n\to \mathrm{\infty }$ and ${x}_{k}^{\left(n\right)}\to {x}_{k}$ as $n\to \mathrm{\infty }$ for all $k\in \mathbb{N}$, then ${x}^{\left(n\right)}\to x$ as $n\to \mathrm{\infty }$.

Proof The proof of the theorem is similar to Theorem 2.9 in [13]. □

Theorem 3.17 The sequence space $N\left(\lambda ,p\right)$ has the Kadec-Klee property.

Proof It is enough to prove that every weakly convergent sequence on the unit sphere is convergent in norm. Let $x\in N\left(\lambda ,p\right)$ and $\left({x}^{\left(n\right)}\right)\in N\left(\lambda ,p\right)$ such that $\parallel {x}^{\left(n\right)}\parallel \to 1$ and ${x}^{\left(n\right)}\stackrel{w}{\to }x$ be given. From the properties of Theorem 3.15 it follows that ${\sigma }_{p}\left({x}^{\left(n\right)}\right)\to 1$ as $n\to \mathrm{\infty }$. On the other hand, from Proposition 3.11, we get ${\sigma }_{p}\left(x\right)=1$. Therefore we have ${\sigma }_{p}\left({x}^{\left(n\right)}\right)\to {\sigma }_{p}\left(x\right)$, as $n\to \mathrm{\infty }$. Since ${x}^{\left(n\right)}\stackrel{w}{\to }x$ and ${p}_{k}\left(x\right)={x}_{k}$ is a continuous functional, ${x}_{k}^{\left(n\right)}\to {x}_{k}$ as $n\to \mathrm{\infty }$ and for $k\in \mathbb{N}$. Now the proof of the theorem follows from Theorem 3.16. □

Theorem 3.18 For any $1, the space ${N}_{p}$ has the uniform Opial property.

We omit this proof.

To prove the following theorem we will use the same technique given in [15] and will consider that ${lim}_{n}inf{p}_{n}>1$.

Theorem 3.19 The Banach space $N\left(\lambda ,p\right)$ has the k-NUC property for every $k\ge 2$.

Proof Let $ϵ>0$ and $\left({x}_{n}\right)\subset B\left(N\left(\lambda ,p\right)\right)$ with $sep\left({x}_{n}\right)\ge ϵ$. For each $m\in \mathbb{N}$, let

${x}_{n}^{m}=\left(\stackrel{m-1}{\stackrel{⏞}{0,0,\dots ,0}},{x}_{n}\left(m\right),x\left(m+1\right),\dots \right).$
(3.9)

Since for each $i\in \mathbb{N}$, ${\left({x}_{n}\left(i\right)\right)}_{n=1}^{\mathrm{\infty }}$ is bounded, by the diagonal method (see [16]), we find that for each $m\in \mathbb{N}$, we can find a subsequence $\left({x}_{{n}_{j}}\right)$ of $\left({x}_{n}\right)$ such that $\left({x}_{{n}_{j}}\left(i\right)\right)$ converges for each $i\in \mathbb{N}$, $1\le i\le m$. Therefore, there exists an increasing sequence of positive integers $\left({t}_{m}\right)$ such that $sep\left({\left({x}_{{n}_{j}}^{m}\right)}_{j>{t}_{m}}\right)\ge ϵ$. Hence, there is a sequence of positive integers ${\left({r}_{m}\right)}_{m\in \mathbb{N}}$ with ${r}_{1}<{r}_{2}<{r}_{3}<\cdots$ such that $\parallel {x}_{{r}_{m}}^{m}\parallel \ge \frac{ϵ}{2}$ for all $m\in \mathbb{N}$. Then by Theorem 3.15, we may assume that there exists $\eta >0$ such that

(3.10)

Let $\alpha >0$ be such that $1<\alpha <{lim inf}_{n}{p}_{n}$. For fixed integer $k\ge 2$, let ${ϵ}_{1}=\left(\frac{\left({k}^{\alpha -1}-1\right)}{\left(k-1\right){k}^{\alpha }}\right)\cdot \frac{\eta }{2}$. Then by Lemma 3.6, there is a $\delta >0$ such that

$|{\sigma }_{p}\left(u+v\right)-{\sigma }_{p}\left(u\right)|\le {ϵ}_{1},$
(3.11)

whenever ${\sigma }_{p}\left(u\right)\le 1$ and ${\sigma }_{p}\left(v\right)\le \delta$. Since by Proposition 3.11, property (1), we get ${\sigma }_{p}\left({x}_{n}\right)\le 1$, $\mathrm{\forall }n\in \mathbb{N}$. Then there exist positive integers ${m}_{i}$ ($i=1,2,\dots ,k-1$) with ${m}_{1}<{m}_{2}<\cdots <{m}_{k-1}$ such that ${\sigma }_{p}\left({x}_{{p}_{i}}^{{m}_{i}}\right)\le \delta$ and $\alpha \le {p}_{j}$ for all $j\ge {m}_{k-1}$. Define ${m}_{k}={m}_{k-1}+1$. By (3.10), we have ${\sigma }_{p}\left({x}_{{r}_{{m}_{k}}}^{{m}_{k}}\right)\ge \eta$. Let ${s}_{i}=i$ for $1\le i\le k-1$ and ${s}_{k}={r}_{{m}_{k}}$. From relations (3.10), (3.11), and the convexity of the function ${f}_{i}\left(u\right)={|u|}^{{p}_{i}}$ ($i\in \mathbb{N}$), we have

$\begin{array}{c}{\sigma }_{p}\left(\frac{{x}_{{s}_{1}}+{x}_{{s}_{2}}+\cdots +{x}_{{s}_{k}}}{k}\right)\hfill \\ \phantom{\rule{1em}{0ex}}=\sum _{n=1}^{\mathrm{\infty }}{\left(\frac{1}{{\lambda }_{n}}\sum _{i=1}^{n}|\left({\lambda }_{i}-{\lambda }_{i-1}\right)\frac{{x}_{{s}_{1}}\left(i\right)+{x}_{{s}_{2}}\left(i\right)+\cdots +{x}_{{s}_{k}}\left(i\right)}{k}|\right)}^{{p}_{n}}\hfill \\ \phantom{\rule{1em}{0ex}}=\sum _{n=1}^{{m}_{1}}{\left(\frac{1}{{\lambda }_{n}}\sum _{i=1}^{n}|\left({\lambda }_{i}-{\lambda }_{i-1}\right)\frac{{x}_{{s}_{1}}\left(i\right)+{x}_{{s}_{2}}\left(i\right)+\cdots +{x}_{{s}_{k}}\left(i\right)}{k}|\right)}^{{p}_{n}}\hfill \\ \phantom{\rule{2em}{0ex}}+\sum _{n={m}_{1}+1}^{\mathrm{\infty }}{\left(\frac{1}{{\lambda }_{n}}\sum _{i=1}^{n}|\left({\lambda }_{i}-{\lambda }_{i-1}\right)\frac{{x}_{{s}_{1}}\left(i\right)+{x}_{{s}_{2}}\left(i\right)+\cdots +{x}_{{s}_{k}}\left(i\right)}{k}|\right)}^{{p}_{n}};\hfill \end{array}$

from (3.11) we get

$\begin{array}{c}\sum _{n=1}^{{m}_{1}}{\left(\frac{1}{{\lambda }_{n}}\sum _{i=1}^{n}|\left({\lambda }_{i}-{\lambda }_{i-1}\right)\frac{{x}_{{s}_{1}}\left(i\right)+{x}_{{s}_{2}}\left(i\right)+\cdots +{x}_{{s}_{k}}\left(i\right)}{k}|\right)}^{{p}_{n}}\hfill \\ \phantom{\rule{2em}{0ex}}+\sum _{n={m}_{1}+1}^{\mathrm{\infty }}{\left(\frac{1}{{\lambda }_{n}}\sum _{i=1}^{n}|\left({\lambda }_{i}-{\lambda }_{i-1}\right)\frac{{x}_{{s}_{2}}\left(i\right)+\cdots +{x}_{{s}_{k}}\left(i\right)}{k}|\right)}^{{p}_{n}}+{ϵ}_{1}\le \hfill \end{array}$

from the convexity of ${f}_{i}\left(u\right)={|u|}^{{p}_{i}}$ ($i\in \mathbb{N}$), it follows that

$\begin{array}{c}\phantom{\rule{1em}{0ex}}\le \sum _{n=1}^{{m}_{1}}\frac{1}{k}\sum _{j=1}^{k}{\left(\frac{1}{{\lambda }_{k}}\sum _{i=1}^{n}|\left({\lambda }_{i}-{\lambda }_{i-1}\right){x}_{{s}_{j}}\left(i\right)|\right)}^{{p}_{n}}\hfill \\ \phantom{\rule{2em}{0ex}}+\sum _{n={m}_{1}+1}^{{m}_{2}}{\left(\frac{1}{{\lambda }_{n}}\sum _{i=1}^{n}|\left({\lambda }_{i}-{\lambda }_{i-1}\right)\frac{{x}_{{s}_{2}}\left(i\right)+{x}_{{s}_{3}}\left(i\right)+\cdots +{x}_{{s}_{k}}\left(i\right)}{k}|\right)}^{{p}_{n}}\hfill \\ \phantom{\rule{2em}{0ex}}+\sum _{n={m}_{2}+1}^{\mathrm{\infty }}{\left(\frac{1}{{\lambda }_{n}}\sum _{i=1}^{n}|\left({\lambda }_{i}-{\lambda }_{i-1}\right)\frac{{x}_{{s}_{2}}\left(i\right)+{x}_{{s}_{3}}\left(i\right)+\cdots +{x}_{{s}_{k}}\left(i\right)}{k}|\right)}^{{p}_{n}}+{ϵ}_{1}\hfill \\ \phantom{\rule{1em}{0ex}}\le \sum _{n=1}^{{m}_{1}}\frac{1}{k}\sum _{j=1}^{k}{\left(\frac{1}{{\lambda }_{k}}\sum _{i=1}^{n}|\left({\lambda }_{i}-{\lambda }_{i-1}\right){x}_{{s}_{j}}\left(i\right)|\right)}^{{p}_{n}}\hfill \\ \phantom{\rule{2em}{0ex}}+\sum _{n={m}_{1}+1}^{{m}_{2}}{\left(\frac{1}{{\lambda }_{n}}\sum _{i=1}^{n}|\left({\lambda }_{i}-{\lambda }_{i-1}\right)\frac{{x}_{{s}_{2}}\left(i\right)+{x}_{{s}_{3}}\left(i\right)+\cdots +{x}_{{s}_{k}}\left(i\right)}{k}|\right)}^{{p}_{n}}\hfill \\ \phantom{\rule{2em}{0ex}}+\sum _{n={m}_{2}+1}^{\mathrm{\infty }}{\left(\frac{1}{{\lambda }_{n}}\sum _{i=1}^{n}|\left({\lambda }_{i}-{\lambda }_{i-1}\right)\frac{{x}_{{s}_{3}}\left(i\right)+{x}_{{s}_{4}}\left(i\right)+\cdots +{x}_{{s}_{k}}\left(i\right)}{k}|\right)}^{{p}_{n}}+2{ϵ}_{1}\hfill \\ \phantom{\rule{1em}{0ex}}\le \sum _{n=1}^{{m}_{1}}\frac{1}{k}\sum _{j=1}^{k}{\left(\frac{1}{{\lambda }_{k}}\sum _{i=1}^{n}|\left({\lambda }_{i}-{\lambda }_{i-1}\right){x}_{{s}_{j}}\left(i\right)|\right)}^{{p}_{n}}\hfill \\ \phantom{\rule{2em}{0ex}}+\sum _{n={m}_{1}+1}^{{m}_{2}}\frac{1}{k}\sum _{j=2}^{k}{\left(\frac{1}{{\lambda }_{k}}\sum _{i=1}^{n}|\left({\lambda }_{i}-{\lambda }_{i-1}\right){x}_{{s}_{j}}\left(i\right)|\right)}^{{p}_{n}}\hfill \\ \phantom{\rule{2em}{0ex}}+\sum _{n={m}_{2}+1}^{{m}_{3}}\frac{1}{k}\sum _{j=3}^{k}{\left(\frac{1}{{\lambda }_{k}}\sum _{i=1}^{n}|\left({\lambda }_{i}-{\lambda }_{i-1}\right){x}_{{s}_{j}}\left(i\right)|\right)}^{{p}_{n}}\hfill \\ \phantom{\rule{2em}{0ex}}+\sum _{n={m}_{3}+1}^{{m}_{4}}\frac{1}{k}\sum _{j=4}^{k}{\left(\frac{1}{{\lambda }_{k}}\sum _{i=1}^{n}|\left({\lambda }_{i}-{\lambda }_{i-1}\right){x}_{{s}_{j}}\left(i\right)|\right)}^{{p}_{n}}+\cdots \hfill \\ \phantom{\rule{2em}{0ex}}+\sum _{n={m}_{k-1}+1}^{{m}_{k}}\frac{1}{k}\sum _{j=k-1}^{k}{\left(\frac{1}{{\lambda }_{k}}\sum _{i=1}^{n}|\left({\lambda }_{i}-{\lambda }_{i-1}\right){x}_{{s}_{j}}\left(i\right)|\right)}^{{p}_{n}}\hfill \\ \phantom{\rule{2em}{0ex}}+\sum _{n={m}_{k}+1}^{\mathrm{\infty }}{\left(\frac{1}{{\lambda }_{k}}\sum _{i=1}^{n}|\left({\lambda }_{i}-{\lambda }_{i-1}\right)\frac{{x}_{{s}_{k}}\left(i\right)}{k}|\right)}^{{p}_{n}}+\left(k-1\right){ϵ}_{1}\hfill \\ \phantom{\rule{1em}{0ex}}\le \frac{{\sigma }_{p}\left({x}_{{s}_{1}}\right)+{\sigma }_{p}\left({x}_{{s}_{2}}\right)+\cdots +{\sigma }_{p}\left({x}_{{s}_{k-1}}\right)}{k}+\frac{1}{k}\sum _{n=1}^{{m}_{k}}{\left(\frac{1}{{\lambda }_{k}}\sum _{i=1}^{n}|\left({\lambda }_{i}-{\lambda }_{i-1}\right){x}_{{s}_{k}}\left(i\right)|\right)}^{{p}_{n}}\hfill \\ \phantom{\rule{2em}{0ex}}+\sum _{n={m}_{k}+1}^{\mathrm{\infty }}{\left(\frac{1}{{\lambda }_{k}}\sum _{i=1}^{n}|\left({\lambda }_{i}-{\lambda }_{i-1}\right)\frac{{x}_{{s}_{k}}\left(i\right)}{k}|\right)}^{{p}_{n}}+\left(k-1\right){ϵ}_{1}\hfill \\ \phantom{\rule{1em}{0ex}}\le \frac{k-1}{k}+\frac{1}{k}\sum _{n=1}^{{m}_{k}}{\left(\frac{1}{{\lambda }_{k}}\sum _{i=1}^{n}|\left({\lambda }_{i}-{\lambda }_{i-1}\right){x}_{{s}_{k}}\left(i\right)|\right)}^{{p}_{n}}\hfill \\ \phantom{\rule{2em}{0ex}}+\frac{1}{{k}^{\alpha }}\cdot \sum _{n={m}_{k}+1}^{\mathrm{\infty }}{\left(\frac{1}{{\lambda }_{k}}\sum _{i=1}^{n}|\left({\lambda }_{i}-{\lambda }_{i-1}\right){x}_{{s}_{k}}\left(i\right)|\right)}^{{p}_{n}}+\left(k-1\right){ϵ}_{1}\hfill \\ \phantom{\rule{1em}{0ex}}\le 1-\frac{1}{k}+\frac{1}{k}\left[1-\sum _{n={m}_{k}+1}^{\mathrm{\infty }}{\left(\frac{1}{{\lambda }_{k}}\sum _{i=1}^{n}|\left({\lambda }_{i}-{\lambda }_{i-1}\right)\frac{{x}_{{s}_{k}}\left(i\right)}{k}|\right)}^{{p}_{n}}\right]\hfill \\ \phantom{\rule{2em}{0ex}}+\frac{1}{{k}^{\alpha }}\cdot \sum _{n={m}_{k}+1}^{\mathrm{\infty }}{\left(\frac{1}{{\lambda }_{k}}\sum _{i=1}^{n}|\left({\lambda }_{i}-{\lambda }_{i-1}\right)\frac{{x}_{{s}_{k}}\left(i\right)}{k}|\right)}^{{p}_{n}}+\left(k-1\right){ϵ}_{1}\hfill \\ \phantom{\rule{1em}{0ex}}\le 1+\left(k-1\right){ϵ}_{1}-\left(\frac{{k}^{\alpha -1}-1}{{k}^{\alpha }}\right)\sum _{n={m}_{k}+1}^{\mathrm{\infty }}{\left(\frac{1}{{\lambda }_{k}}\sum _{i=1}^{n}|\left({\lambda }_{i}-{\lambda }_{i-1}\right){x}_{{s}_{k}}\left(i\right)|\right)}^{{p}_{n}}\hfill \\ \phantom{\rule{1em}{0ex}}\le 1+\left(k-1\right){ϵ}_{1}-\left(\frac{{k}^{\alpha -1}-1}{{k}^{\alpha }}\right)\eta \hfill \\ \phantom{\rule{1em}{0ex}}=1-\left(\frac{{k}^{\alpha -1}-1}{{k}^{\alpha }}\right)\frac{\eta }{2}.\hfill \end{array}$

Now from Theorem 3.9, there exists a $\vartheta >0$ such that

$\parallel \frac{{x}_{{s}_{1}}+{x}_{{s}_{2}}+\cdots +{x}_{{s}_{k}}}{k}\parallel <1-\vartheta .$

□

The proof of the following results we omit.

Theorem 3.20 The Banach space $N\left(\lambda ,p\right)$ has the $\left(\beta \right)$-property.

Theorem 3.21 The Banach space $N\left(\lambda ,p\right)$ has the Banach-Saks property of type p.

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Correspondence to Naim L Braha.

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Braha, N.L. Some geometric properties of $N\left(\lambda ,p\right)$-spaces. J Inequal Appl 2014, 112 (2014). https://doi.org/10.1186/1029-242X-2014-112

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• DOI: https://doi.org/10.1186/1029-242X-2014-112

### Keywords

• normed sequence spaces
• rotund property
• $\left(\beta \right)$-property