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# Some geometric properties of a new modular space defined by Zweier operator

Fixed Point Theory and Applications20132013:165

https://doi.org/10.1186/1687-1812-2013-165

• Accepted: 5 June 2013
• Published:

## Abstract

In this paper, we define the modular space ${\mathcal{Z}}_{\sigma }\left(s,p\right)$ by using the Zweier operator and a modular. Then, we consider it equipped with the Luxemburg norm and also examine the uniform Opial property and property β. Finally, we show that this space has the fixed point property.

MSC:40A05, 46A45, 46B20.

## Keywords

• Zweier operator
• Luxemburg norm
• modular space
• uniform Opial property
• property $\left(\beta \right)$

## Dedication

Dedicated to Professor Hari M Srivastava

## 1 Introduction

In literature, there are many papers about the geometrical properties of different sequence spaces such as [19]. Opial [10] introduced the Opial property and proved that the sequence spaces ${\ell }_{p}$ ($1) have this property but ${L}_{p}\left[0,2\pi \right]$ ($p\ne 2$, $1) does not have it. Franchetti [11] showed that any infinite dimensional Banach space has an equivalent norm that satisfies the Opial property. Later, Prus [12] introduced and investigated the uniform Opial property for Banach spaces. The Opial property is important because Banach spaces with this property have the weak fixed point property.

## 2 Definition and preliminaries

Let $\left(X,\parallel \cdot \parallel \right)$ be a real Banach space and let $S\left(X\right)$ (resp. $B\left(X\right)$) be the unit sphere (resp. the unit ball) of X. A Banach space X has the Opial property if for any weakly null sequence $\left\{{x}_{n}\right\}$ in X and any x in $X\setminus \left\{0\right\}$, the inequality ${lim}_{n\to \mathrm{\infty }}inf\parallel x\parallel <{lim}_{n\to \mathrm{\infty }}inf\parallel {x}_{n}+x\parallel$ holds. We say that X has the uniform Opial property if for any $\epsilon >0$ there exists $r>0$ such that for any $x\in X$ with $\parallel x\parallel \ge \epsilon$ and any weakly null sequence $\left\{{x}_{n}\right\}$ in the unit sphere of X, the inequality $1+r\le {lim}_{n\to \mathrm{\infty }}inf\parallel {x}_{n}+x\parallel$ holds.

For a bounded set $A\subset X$, the ball-measure of noncompactness was defined by . The function Δ defined by is called the modulus of noncompact convexity. A Banach space X is said to have property $\left(L\right)$, if ${lim}_{\epsilon \to {1}^{-}}\mathrm{\Delta }\left(\epsilon \right)=1$. This property is an important concept in the fixed point theory and a Banach space X possesses property $\left(L\right)$ if and only if it is reflexive and has the uniform Opial property.

A Banach space X is said to satisfy the weak fixed point property if every nonempty weakly compact convex subset C and every nonexpansive mapping $T:C\to C\left(\parallel Tx-Ty\parallel \le \parallel x-y\parallel ,\mathrm{\forall }x,y\in C\right)$ have a fixed point, that is, there exists $x\in C$ such that $T\left(x\right)=x$. Property $\left(L\right)$ and the fixed point property were also studied by Goebel and Kirk [13], Toledano et al. [14], Benavides [15], Benavides and Phothi [16]. A Banach space X is said to have property $\left(H\right)$ if every weakly convergent sequence on the unit sphere is convergent in norm. Clarkson [17] introduced the uniform convexity, and it is known that the uniform convexity implies the reflexivity of Banach spaces. Huff [18] introduced the concept of nearly uniform convexity of Banach spaces. A Banach space X is called uniformly convex (UC) if for each $\epsilon >0$, there is $\delta >0$ such that for $x,y\in S\left(X\right)$, the inequality $\parallel x-y\parallel >\epsilon$ implies that $\parallel \frac{1}{2}\left(x+y\right)\parallel <1-\delta$. For any $x\notin B\left(X\right)$, the drop determined by x is the set $D\left(x,B\left(X\right)\right)=conv\left(\left\{x\right\}\cup B\left(X\right)\right)$. A Banach space X has the drop property $\left(D\right)$ if for every closed set C disjoint with $B\left(X\right)$, there exists an element $x\in C$ such that $D\left(x,B\left(X\right)\right)\cap C=\left\{x\right\}$. Rolewicz [19] showed that the Banach space X is reflexive if X has the drop property. Later, Montesinos [20] extended this result and proved that X has the drop property if and only if X is reflexive and has property $\left(H\right)$. A sequence $\left\{{x}_{n}\right\}$ is said to be ε-separated sequence for some $\epsilon >0$ if
$\mathit{sep}\left({x}_{n}\right)=inf\left\{\parallel {x}_{n}-{x}_{m}\parallel :n\ne m\right\}>\epsilon .$

A Banach space X is called nearly uniformly convex (NUC) if for every $\epsilon >0$, there exists $\delta \in \left(0,1\right)$ such that for every sequence $\left({x}_{n}\right)\subseteq B\left(X\right)$ with $\mathit{sep}\left({x}_{n}\right)>\epsilon$, we have $conv\left({x}_{n}\right)\cap \left(\left(1-\delta \right)B\left(X\right)\right)\ne \mathrm{\varnothing }$. Huff [18] proved that every (NUC) Banach space is reflexive and has property $\left(H\right)$. A Banach space X has property $\left(\beta \right)$ if and only if for each $\epsilon >0$, there exists $\delta >0$ such that for each element $x\in B\left(X\right)$ and each sequence $\left({x}_{n}\right)$ in $B\left(X\right)$ with $\mathit{sep}\left({x}_{n}\right)\ge \epsilon$, there is an index k for which $\parallel \frac{x+{x}_{k}}{2}\parallel <1-\delta$.

For a real vector space X, a function $\rho :X\to \left[0,\mathrm{\infty }\right]$ is called a modular if it satisfies the following conditions:
1. (i)

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

2. (ii)

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

3. (iii)

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

The modular ρ is called convex if
1. (iv)

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

For any modular ρ on X, the space
is called a modular space. In general, the modular is not subadditive and thus it does not behave as a norm or a distance. But we can associate the modular with an F-norm. A functional $\parallel \cdot \parallel :X\to \left[0,\mathrm{\infty }\right]$ defines an F-norm if and only if
1. (i)

$\parallel x\parallel =0⇔x=0$,

2. (ii)

$\parallel \alpha x\parallel =\parallel x\parallel$ whenever $|\alpha |=1$,

3. (iii)

$\parallel x+y\parallel \le \parallel x\parallel +\parallel y\parallel$,

4. (iv)

if ${\alpha }_{n}\to \alpha$ and $\parallel {x}_{n}-x\parallel \to 0$, then $\parallel {\alpha }_{n}{x}_{n}-\alpha x\parallel \to 0$.

F-norm defines a distance on X by $d\left(x,y\right)=\parallel x-y\parallel$. If the linear metric space $\left(X,d\right)$ is complete, then it is called an F-space. The modular space ${X}_{\rho }$ can be equipped with the following F-norm:
$\parallel x\parallel =inf\left\{\alpha >0:\rho \left(\frac{x}{\alpha }\right)\le \alpha \right\}.$

If the modular ρ is convex, then the equality $\parallel x\parallel =inf\left\{\alpha >0:\rho \left(\frac{x}{\alpha }\right)\le 1\right\}$ defines a norm which is called the Luxemburg norm.

A modular ρ is said to satisfy the ${\delta }_{2}$-condition if for any $\epsilon >0$, there exist constants $K\ge 2$, $a>0$ such that $\rho \left(2u\right)\le K\rho \left(u\right)+\epsilon$ for all $u\in {X}_{\rho }$ with $\rho \left(u\right)\le a$. If ρ provides the ${\delta }_{2}$-condition for any $a>0$ with $K\ge 2$ dependent on a, then ρ provides the strong ${\delta }_{2}$-condition (briefly $\rho \in {\delta }_{2}^{s}$).

Let us denote by ${\ell }^{0}$ the space of all real sequences. The Cesàro sequence spaces
${\mathit{Ces}}_{p}=\left\{x\in {\ell }^{0}:\sum _{n=1}^{\mathrm{\infty }}{\left({n}^{-1}\sum _{i=1}^{n}|{x}_{i}|\right)}^{p}<\mathrm{\infty }\right\},\phantom{\rule{1em}{0ex}}1\le p<\mathrm{\infty },$
and
${\mathit{Ces}}_{\mathrm{\infty }}=\left\{x\in {\ell }^{0}:\underset{n}{sup}{n}^{-1}\sum _{i=1}^{n}|{x}_{i}|<\mathrm{\infty }\right\},$
were introduced by Shiue [21]. Jagers [22] determined the Köthe duals of the sequence space ${\mathit{Ces}}_{p}$ ($1). It can be shown that the inclusion ${\ell }_{p}\subset {\mathit{Ces}}_{p}$ is strict for $1 although it does not hold for $p=1$. Also, Suantai [23] defined the generalized Cesàro sequence space by
where $\rho \left(x\right)={\sum }_{n=1}^{\mathrm{\infty }}{\left(\frac{1}{n}{\sum }_{i=1}^{n}|x\left(i\right)|\right)}^{{p}_{n}}$. If $p=\left({p}_{n}\right)$ is bounded, then
$\mathit{ces}\left(p\right)=\left\{x=\left({x}_{k}\right):\sum _{n=1}^{\mathrm{\infty }}{\left({n}^{-1}\sum _{i=1}^{n}|x\left(i\right)|\right)}^{{p}_{n}}<\mathrm{\infty }\right\}.$
The sequence space $C\left(s,p\right)$ was defined by Bilgin [24] as follows:
$C\left(s,p\right)=\left\{x=\left({x}_{k}\right):\sum _{r=0}^{\mathrm{\infty }}{\left({2}^{-r}\sum _{r}{k}^{-s}|{x}_{k}|\right)}^{{p}_{r}}<\mathrm{\infty },s\ge 0\right\}$
for $p=\left({p}_{r}\right)$ with $inf{p}_{r}>0$, where ${\sum }_{r}$ denotes a sum over the ranges ${2}^{r}\le k<{2}^{r+1}$. The special case of $C\left(s,p\right)$ for $s=0$ is the space
$\mathit{Ces}\left(p\right)=\left\{x=\left({x}_{k}\right):\sum _{r=0}^{\mathrm{\infty }}{\left({2}^{-r}\sum _{r}|{x}_{k}|\right)}^{{p}_{r}}<\mathrm{\infty }\right\}$
which was introduced by Lim [25]. Also, the inclusion $\mathit{Ces}\left(p\right)\subseteq C\left(s,p\right)$ holds. A paranorm on $C\left(s,p\right)$ is given by
$\rho \left(x\right)={\left(\sum _{r=0}^{\mathrm{\infty }}{\left({2}^{-r}\sum _{r}{k}^{-s}|{x}_{k}|\right)}^{{p}_{r}}\right)}^{1/M}$

for $M=max\left(1,H\right)$ and $H=sup{p}_{r}<\mathrm{\infty }$.

The Z-transform of a sequence $x=\left({x}_{k}\right)$ is defined by ${\left(Zx\right)}_{n}={y}_{n}=\alpha {x}_{n}+\left(1-\alpha \right){x}_{n-1}$ by using the Zweier operator

where is the field of all complex or real numbers. The Zweier operator was studied by Şengönül and Kayaduman [26].

Now we introduce a new modular sequence space ${\mathcal{Z}}_{\sigma }\left(s,p\right)$ by
where $\sigma \left(x\right)={\sum }_{r=0}^{\mathrm{\infty }}{\left({2}^{-r}{\sum }_{r}{k}^{-s}|\alpha {x}_{k}+\left(1-\alpha \right){x}_{k-1}|\right)}^{{p}_{r}}<\mathrm{\infty }$ and $s\ge 0$. If we take $\alpha =1$, then ${\mathcal{Z}}_{\sigma }\left(s,p\right)=C\left(s,p\right)$; if $\alpha =1$ and $s=0$, then ${\mathcal{Z}}_{\sigma }\left(s,p\right)=\mathit{Ces}\left(p\right)$. It can be easily seen that $\sigma :{\mathcal{Z}}_{\sigma }\left(s,p\right)\to \left[0,\mathrm{\infty }\right]$ is a modular on ${\mathcal{Z}}_{\sigma }\left(s,p\right)$. We define the Luxemburg norm on the sequence space ${\mathcal{Z}}_{\sigma }\left(s,p\right)$ as follows:
$\parallel x\parallel =inf\left\{t>0:\sigma \left(\frac{x}{t}\right)\le 1\right\},\phantom{\rule{1em}{0ex}}\mathrm{\forall }x\in {\mathcal{Z}}_{\sigma }\left(s,p\right).$

It is easy to see that the space ${\mathcal{Z}}_{\sigma }\left(s,p\right)$ is a Banach space with respect to the Luxemburg norm.

Throughout the paper, suppose that $p=\left({p}_{r}\right)$ is bounded with ${p}_{r}>1$ for all $r\in \mathbb{N}$ and
for $i\in \mathbb{N}$ and $x\in {\ell }^{0}$. In addition, we will require the following inequalities:
$|{a}_{k}+{b}_{k}{|}^{{p}_{k}}\le C\left(|{a}_{k}{|}^{{p}_{k}}+|{b}_{k}{|}^{{p}_{k}}\right),\phantom{\rule{2em}{0ex}}|{a}_{k}+{b}_{k}{|}^{{t}_{k}}\le |{a}_{k}{|}^{{t}_{k}}+|{b}_{k}{|}^{{t}_{k}},$

where ${t}_{k}=\frac{{p}_{k}}{M}\le 1$ and $C=max\left\{1,{2}^{H-1}\right\}$ with $H=sup{p}_{k}$.

## 3 Main results

Since ${\ell }_{p}$ is reflexive and convex, $\ell \left(p\right)$-type spaces have many useful applications, and it is natural to consider a geometric structure of these spaces. From this point of view, we generalized the space $C\left(s,p\right)$ by using the Zweier operator and then obtained the equality ${\mathcal{Z}}_{\sigma }\left(s,p\right)=\mathit{Ces}\left(p\right)$, that is, it was seen that the structure of the space $\mathit{Ces}\left(p\right)$ was preserved. In this section, our goal is to investigate a geometric structure of the modular space ${\mathcal{Z}}_{\sigma }\left(s,p\right)$ related to the fixed point theory. For this, we will examine property $\left(\beta \right)$ and the uniform Opial property for ${\mathcal{Z}}_{\sigma }\left(s,p\right)$. Finally, we will give some fixed point results. To do this, we need some results which are important in our opinion.

Lemma 3.1 [2]

If $\sigma \in {\delta }_{2}^{s}$, then for any $L>0$ and $\epsilon >0$, there exists $\delta >0$ such that
$|\sigma \left(u+v\right)-\sigma \left(u\right)|<\epsilon ,$

where $u,v\in {X}_{\sigma }$ with $\sigma \left(u\right)\le L$ and $\sigma \left(v\right)\le \delta$.

Lemma 3.2 [2]

If $\sigma \in {\delta }_{2}^{s}$, convergence in norm and in modular are equivalent in ${X}_{\sigma }$.

Lemma 3.3 [2]

If $\sigma \in {\delta }_{2}^{s}$, then for any $\epsilon >0$, there exists $\delta =\delta \left(\epsilon \right)>0$ such that $\parallel x\parallel \ge 1+\delta$ implies $\sigma \left(x\right)\ge 1+\epsilon$.

Now we give the following two lemmas without proof.

Lemma 3.4 If ${\parallel x\parallel }_{L}<1$ for any $x\in {\mathcal{Z}}_{\sigma }\left(s,p\right)$, then $\sigma \left(x\right)\le {\parallel x\parallel }_{L}$.

Lemma 3.5 For any $x\in {\mathcal{Z}}_{\sigma }\left(s,p\right)$, ${\parallel x\parallel }_{L}=1$ if and only if $\sigma \left(x\right)=1$.

Lemma 3.6 If $liminf{p}_{r}>1$, then for any $x\in {\mathcal{Z}}_{\sigma }\left(s,p\right)$, there exist ${k}_{0}\in \mathbb{N}$ and $\mu \in \left(0,1\right)$ such that
$\sigma \left(\frac{{x}^{k}}{2}\right)\le \frac{1-\mu }{2}\sigma \left({x}^{k}\right)$

for all $k\in \mathbb{N}$ with $k\ge {k}_{0}$, where and ${2}^{r}\le k<{2}^{r+1}$.

Proof Let $k\in \mathbb{N}$ be fixed. Then there exists ${r}_{k}\in \mathbb{N}$ such that $k\in {I}_{{r}_{k}}$. Let γ be a real number $1<\gamma \le liminf{p}_{r}$, and so there exists ${k}_{0}\in \mathbb{N}$ such that $\gamma <{p}_{{r}_{k}}$ for all $k\ge {k}_{0}$. Choose $\mu \in \left(0,1\right)$ such that ${\left(\frac{1}{2}\right)}^{\gamma }\le \frac{1-\mu }{2}$. Therefore, we have
$\begin{array}{rl}\sigma \left(\frac{{x}^{k}}{2}\right)& =\sum _{r=0}^{\mathrm{\infty }}{\left({2}^{-r}\sum _{r}{k}^{-s}|\frac{\alpha x\left(k\right)+\left(1-\alpha \right)x\left(k-1\right)}{2}|\right)}^{{p}_{r}}\\ =\sum _{r=0}^{\mathrm{\infty }}{\left(\frac{1}{2}\right)}^{{p}_{r}}{\left({2}^{-r}\sum _{r}{k}^{-s}|\alpha x\left(k\right)+\left(1-\alpha \right)x\left(k-1\right)|\right)}^{{p}_{r}}\\ \le {\left(\frac{1}{2}\right)}^{\gamma }\sum _{r=0}^{\mathrm{\infty }}{\left({2}^{-r}\sum _{r}{k}^{-s}|\alpha x\left(k\right)+\left(1-\alpha \right)x\left(k-1\right)|\right)}^{{p}_{r}}\\ <\frac{1-\mu }{2}\sigma \left({x}^{k}\right)\end{array}$

for each $x\in {\mathcal{Z}}_{\sigma }\left(s,p\right)$ and $k\ge {k}_{0}$. □

Lemma 3.7 If $\sigma \in {\delta }_{2}^{s}$, then for any $\epsilon \in \left(0,1\right)$, there exists $\delta \in \left(0,1\right)$ such that $\sigma \left(x\right)\le 1-\epsilon$ implies $\parallel x\parallel \le 1-\delta$.

Proof Suppose that lemma does not hold. So, there exist $\epsilon >0$ and ${x}_{n}\in {\mathcal{Z}}_{\sigma }\left(s,p\right)$ such that $\sigma \left({x}_{n}\right)<1-\epsilon$ and $\frac{1}{2}\le \parallel {x}_{n}\parallel \to 1$. Take ${s}_{n}=\frac{1}{\parallel {x}_{n}\parallel -1}$, and so ${s}_{n}\to 0$ as $n\to \mathrm{\infty }$. Let $P=sup\left\{\sigma \left(2{x}_{n}\right):n\in \mathbb{N}\right\}$. There exists $D\ge 2$ such that
$\sigma \left(2u\right)\le D\sigma \left(u\right)+1$
(3.2)
for every $u\in {\mathcal{Z}}_{\sigma }\left(s,p\right)$ with $\sigma \left(u\right)<1$, since $\sigma \in {\delta }_{2}^{s}$. We have
$\sigma \left(2{x}_{n}\right)\le D\sigma \left({x}_{n}\right)+1
for all $n\in \mathbb{N}$ by (3.1). Therefore, $0 and from Lemma 3.5 we have
$\begin{array}{rl}1& =\sigma \left(\frac{{x}_{n}}{\parallel {x}_{n}\parallel }\right)=\sigma \left(2{s}_{n}{x}_{n}+\left(1-{s}_{n}\right){x}_{n}\right)\\ \le {s}_{n}\sigma \left(2{x}_{n}\right)+\left(1-{s}_{n}\right)\sigma \left({x}_{n}\right)\\ \le {s}_{n}P+\left(1-\epsilon \right)\to \left(1-\epsilon \right).\end{array}$

This is a contradiction. So, the proof is complete. □

Theorem 3.8 The space ${\mathcal{Z}}_{\sigma }\left(s,p\right)$ has property $\left(\beta \right)$.

Proof Let $\epsilon >0$ and $\left({x}_{n}\right)\subset B\left({\mathcal{Z}}_{\sigma }\left(s,p\right)\right)$ with $\mathit{sep}\left({x}_{n}\right)\ge \epsilon$ and $x\in B\left({\mathcal{Z}}_{\sigma }\left(s,p\right)\right)$. For each $l\in \mathbb{N}$, we can find ${r}_{k}\in \mathbb{N}$ such that ${2}^{{r}_{k}}\le l<{2}^{{r}_{k}+1}$. Let
Since for each $i\in \mathbb{N}$, ${\left({x}_{n}\left(i\right)\right)}_{i=1}^{\mathrm{\infty }}$ is bounded, by using the diagonal method, 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 $i\in \mathbb{N}$ with $1\le i\le l$. Therefore, there exists an increasing sequence of positive integers ${t}_{l}$ such that $\mathit{sep}\left({\left({x}_{{n}_{j}}^{l}\right)}_{j\ge {t}_{l}}\right)\ge \epsilon$. Thus, there exists a sequence of positive integers ${\left({r}_{l}\right)}_{l=1}^{\mathrm{\infty }}$ with ${r}_{1}<{r}_{2}<\cdots$ such that $\parallel {x}_{{r}_{l}}^{l}\parallel \ge \frac{\epsilon }{2}$ for all $l\in \mathbb{N}$. Since $\sigma \in {\delta }_{2}^{s}$, there is $\eta >0$ such that
(3.4)
from Lemma 3.3. However, there exist ${k}_{0}\in \mathbb{N}$ and $\mu \in \left(0,1\right)$ such that
$\sigma \left(\frac{{v}^{k}}{2}\right)\le \frac{1-\mu }{2}\sigma \left({v}^{k}\right)$
(3.5)
for all $v\in {\mathcal{Z}}_{\sigma }\left(s,p\right)$ and $k\ge {k}_{0}$ by Lemma 3.6. There exists $\delta >0$ such that
$\sigma \left(y\right)\le 1-\frac{\mu \eta }{4}\phantom{\rule{1em}{0ex}}⇒\phantom{\rule{1em}{0ex}}\parallel y\parallel \le 1-\delta$
(3.6)

for any $y\in {\mathcal{Z}}_{\sigma }\left(s,p\right)$ by Lemma 3.7.

By Lemma 3.1, there exists ${\delta }_{0}$ such that
$|\sigma \left(u+v\right)-\sigma \left(u\right)|<\frac{\mu \eta }{4},$
(3.7)
where $\sigma \left(u\right)\le 1$ and $\sigma \left(v\right)\le {\delta }_{0}$. Hence, we get that $\sigma \left(x\right)\le 1$ since $x\in B\left({\mathcal{Z}}_{\sigma }\left(s,p\right)\right)$. Then there exists $k\ge {k}_{0}$ such that $\sigma \left({x}^{k}\right)\le {\delta }_{0}$. Let $u={x}_{{r}_{l}}^{l}$ and $v={x}^{l}$. Then
$\sigma \left(\frac{u}{2}\right)<1\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\sigma \left(\frac{v}{2}\right)<{\delta }_{0}.$
We obtain from (3.3) and (3.5) that
$\sigma \left(\frac{u+v}{2}\right)\le \sigma \left(\frac{u}{2}\right)+\frac{\mu \eta }{4}\le \frac{1-\mu }{2}\sigma \left(u\right)+\frac{\mu \eta }{4}.$
(3.8)
Choose ${s}_{i}={r}_{{l}_{i}}$. By the inequalities (3.2), (3.3), (3.6) and the convexity of the function $f\left(u\right)=|u{|}^{{p}_{r}}$, we have
$\begin{array}{rl}\sigma \left(\frac{x+{x}_{{s}_{k}}}{2}\right)=& \sum _{r=0}^{\mathrm{\infty }}{\left({2}^{-r}\sum _{r}{k}^{-s}|\frac{\alpha \left(x\left(k\right)+{x}_{{s}_{i}}\left(k\right)\right)+\left(1-\alpha \right)\left(x\left(k-1\right)+{x}_{{s}_{i}}\left(k-1\right)\right)}{2}|\right)}^{{p}_{r}}\\ =& \sum _{r=0}^{{r}_{k}-1}{\left({2}^{-r}\sum _{r}{k}^{-s}|\frac{\alpha \left(x\left(k\right)+{x}_{{s}_{i}}\left(k\right)\right)+\left(1-\alpha \right)\left(x\left(k-1\right)+{x}_{{s}_{i}}\left(k-1\right)\right)}{2}|\right)}^{{p}_{r}}\\ +\sum _{r={r}_{k}}^{\mathrm{\infty }}{\left({2}^{-r}\sum _{r}{k}^{-s}|\frac{\alpha \left(x\left(k\right)+{x}_{{s}_{i}}\left(k\right)\right)+\left(1-\alpha \right)\left(x\left(k-1\right)+{x}_{{s}_{i}}\left(k-1\right)\right)}{2}|\right)}^{{p}_{r}}\\ \le & \frac{1}{2}\sum _{r=0}^{{r}_{k}-1}{\left({2}^{-r}\sum _{r}{k}^{-s}|\alpha x\left(k\right)+\left(1-\alpha \right)x\left(k-1\right)|\right)}^{{p}_{r}}\\ +\frac{1}{2}\sum _{r=0}^{{r}_{k}-1}{\left({2}^{-r}\sum _{r}{k}^{-s}|\alpha {x}_{{s}_{i}}\left(k\right)+\left(1-\alpha \right){x}_{{s}_{i}}\left(k-1\right)|\right)}^{{p}_{r}}\\ +\sum _{r={r}_{k}}^{\mathrm{\infty }}{\left({2}^{-r}\sum _{r}{k}^{-s}|\frac{\alpha {x}_{{s}_{i}}\left(k\right)+\left(1-\alpha \right){x}_{{s}_{i}}\left(k-1\right)}{2}|\right)}^{{p}_{r}}+\frac{\mu \eta }{4}\\ \le & \frac{1}{2}\sum _{r=0}^{{r}_{k}-1}{\left({2}^{-r}\sum _{r}{k}^{-s}|\alpha x\left(k\right)+\left(1-\alpha \right)x\left(k-1\right)|\right)}^{{p}_{r}}\\ +\frac{1}{2}\sum _{r=0}^{{r}_{k}-1}{\left({2}^{-r}\sum _{r}{k}^{-s}|\alpha {x}_{{s}_{i}}\left(k\right)+\left(1-\alpha \right){x}_{{s}_{i}}\left(k-1\right)|\right)}^{{p}_{r}}\\ +\frac{1-\mu }{2}\sum _{r={r}_{k}}^{\mathrm{\infty }}{\left({2}^{-r}\sum _{r}{k}^{-s}|\frac{\alpha {x}_{{s}_{i}}\left(k\right)+\left(1-\alpha \right){x}_{{s}_{i}}\left(k-1\right)}{2}|\right)}^{{p}_{r}}+\frac{\mu \eta }{4}\\ \le & \frac{1}{2}\sum _{r=0}^{{r}_{k}-1}{\left({2}^{-r}\sum _{r}{k}^{-s}|\alpha x\left(k\right)+\left(1-\alpha \right)x\left(k-1\right)|\right)}^{{p}_{r}}\\ +\frac{1}{2}\sum _{r=0}^{\mathrm{\infty }}{\left({2}^{-r}\sum _{r}{k}^{-s}|\alpha {x}_{{s}_{i}}\left(k\right)+\left(1-\alpha \right){x}_{{s}_{i}}\left(k-1\right)|\right)}^{{p}_{r}}\\ -\frac{\mu }{2}\sum _{r={r}_{k}}^{\mathrm{\infty }}{\left({2}^{-r}\sum _{r}{k}^{-s}|\frac{\alpha {x}_{{s}_{i}}\left(k\right)+\left(1-\alpha \right){x}_{{s}_{i}}\left(k-1\right)}{2}|\right)}^{{p}_{r}}+\frac{\mu \eta }{4}\\ \le & \frac{1}{2}+\frac{1}{2}-\frac{\mu \eta }{2}+\frac{\mu \eta }{4}\\ =& 1-\frac{\mu \eta }{4}.\end{array}$

So, the inequality (3.4) implies that $\parallel \frac{x+{x}_{{s}_{k}}}{2}\parallel \le 1-\delta$. Consequently, the space ${\mathcal{Z}}_{\sigma }\left(s,p\right)$ possesses property $\left(\beta \right)$. □

Since property $\left(\beta \right)$ implies NUC, NUC implies property $\left(D\right)$ and property $\left(D\right)$ implies reflexivity, we can give the following result from Theorem 3.8.

Corollary 3.9 The space ${\mathcal{Z}}_{\sigma }\left(s,p\right)$ is nearly uniform convex, reflexive and also it has property $\left(D\right)$.

Theorem 3.10 The space ${\mathcal{Z}}_{\sigma }\left(s,p\right)$ has the uniform Opial property.

Proof Let $\epsilon >0$ and $x\in {\mathcal{Z}}_{\sigma }\left(s,p\right)$ be such that $\parallel x\parallel \ge \epsilon$ and $\left({x}_{n}\right)$ be a weakly null sequence in $S\left({\mathcal{Z}}_{\sigma }\left(s,p\right)\right)$. By $\sigma \in {\delta }_{2}^{s}$, there exists $\zeta \in \left(0,1\right)$ independent of x such that $\sigma \left(x\right)>\zeta$ by Lemma 3.2. Also since $\sigma \in {\delta }_{2}^{s}$, by Lemma 3.1, there is ${\zeta }_{1}\in \left(0,\zeta \right)$ such that
$|\sigma \left(y+z\right)-\sigma \left(y\right)|<\frac{\zeta }{4}$
(3.10)
whenever $\sigma \left(y\right)\le 1$ and $\sigma \left(z\right)\le {\zeta }_{1}$. Take ${r}_{0}\in \mathbb{N}$ such that
$\sum _{r={r}_{0}+1}^{\mathrm{\infty }}{\left({2}^{-r}\sum _{r}{k}^{-s}|\alpha x\left(k\right)+\left(1-\alpha \right)x\left(k-1\right)|\right)}^{{p}_{r}}<\frac{{\zeta }_{1}}{4}.$
(3.11)
Hence, we have
$\begin{array}{rl}\zeta <& \sum _{r=1}^{{r}_{0}}{\left({2}^{-r}\sum _{r}{k}^{-s}|\alpha x\left(k\right)+\left(1-\alpha \right)x\left(k-1\right)|\right)}^{{p}_{r}}\\ +\sum _{r={r}_{0}+1}^{\mathrm{\infty }}{\left({2}^{-r}\sum _{r}{k}^{-s}|\alpha x\left(k\right)+\left(1-\alpha \right)x\left(k-1\right)|\right)}^{{p}_{r}}\\ \le & \sum _{r=1}^{{r}_{0}}{\left({2}^{-r}\sum _{r}{k}^{-s}|\alpha x\left(k\right)+\left(1-\alpha \right)x\left(k-1\right)|\right)}^{{p}_{r}}+\frac{{\zeta }_{1}}{4}\end{array}$
(3.12)
and this implies that
$\begin{array}{rl}\sum _{r=1}^{{r}_{0}}{\left({2}^{-r}\sum _{r}{k}^{-s}|\alpha x\left(k\right)+\left(1-\alpha \right)x\left(k-1\right)|\right)}^{{p}_{r}}& >\zeta -\frac{{\zeta }_{1}}{4}\\ >\zeta -\frac{\zeta }{4}\\ =\frac{3\zeta }{4}.\end{array}$
(3.13)
Since ${x}_{n}{\to }^{w}0$, by the inequality (3.10), there exists ${r}_{0}\in \mathbb{N}$ such that
$\sum _{r=1}^{{r}_{0}}{\left({2}^{-r}\sum _{r}{k}^{-s}|\alpha \left({x}_{n}\left(k\right)+x\left(k\right)\right)+\left(1-\alpha \right)\left({x}_{n}\left(k-1\right)+x\left(k-1\right)\right)|\right)}^{{p}_{r}}>\frac{3\zeta }{4}.$
(3.14)
Again, by ${x}_{n}{\to }^{w}0$, there is ${r}_{1}>{r}_{0}$ such that for all $r>{r}_{1}$
$\parallel {x}_{{n}_{|{r}_{0}}}\parallel <1-{\left(1-\frac{\zeta }{4}\right)}^{1/M},$
(3.15)
where ${p}_{r}\le M\in \mathbb{N}$ for all $r\in \mathbb{N}$. Therefore, we obtain that
$\parallel {x}_{{n}_{|\mathbb{N}-{r}_{0}}}\parallel >{\left(1-\frac{\zeta }{4}\right)}^{1/M}$
(3.16)
by the triangle inequality of the norm. It follows from the definition of the Luxemburg norm that
$\begin{array}{rl}1& \le \sigma \left(\frac{{x}_{{n}_{|\mathbb{N}-{r}_{0}}}}{{\left(1-\frac{\zeta }{4}\right)}^{1/M}}\right)\\ =\sum _{r={r}_{0}+1}^{\mathrm{\infty }}{\left(\frac{{2}^{-r}{\sum }_{r}{k}^{-s}|\alpha {x}_{n}\left(k\right)+\left(1-\alpha \right){x}_{n}\left(k-1\right)|}{{\left(1-\frac{\zeta }{4}\right)}^{1/M}}\right)}^{{p}_{r}}\\ \le {\left(\frac{1}{{\left(1-\frac{\zeta }{4}\right)}^{1/M}}\right)}^{M}\sum _{r={r}_{0}+1}^{\mathrm{\infty }}{\left({2}^{-r}\sum _{r}{k}^{-s}|\alpha {x}_{n}\left(k\right)+\left(1-\alpha \right){x}_{n}\left(k-1\right)|\right)}^{{p}_{r}}\end{array}$
(3.17)
and this implies that
$\sum _{r={r}_{0}+1}^{\mathrm{\infty }}{\left({2}^{-r}\sum _{r}{k}^{-s}|\alpha {x}_{n}\left(k\right)+\left(1-\alpha \right){x}_{n}\left(k-1\right)|\right)}^{{p}_{r}}\ge 1-\frac{\zeta }{4}.$
(3.18)
By (3.7), (3.8), (3.11), (3.15) and since ${x}_{n}{\to }^{w}0⇒{x}_{n}\to 0$ (coordinatewise), we have for any $r>{r}_{1}$ that
$\begin{array}{rl}\sigma \left({x}_{n}+x\right)=& \sum _{r=1}^{{r}_{0}}{\left({2}^{-r}\sum _{r}{k}^{-s}|\alpha \left({x}_{n}\left(k\right)+x\left(k\right)\right)+\left(1-\alpha \right)\left({x}_{n}\left(k-1\right)+x\left(k-1\right)\right)|\right)}^{{p}_{r}}\\ +\sum _{r={r}_{0}+1}^{\mathrm{\infty }}{\left({2}^{-r}\sum _{r}{k}^{-s}|\alpha \left({x}_{n}\left(k\right)+x\left(k\right)\right)+\left(1-\alpha \right)\left({x}_{n}\left(k-1\right)+x\left(k-1\right)\right)|\right)}^{{p}_{r}}\\ \ge & \sum _{r={r}_{0}+1}^{\mathrm{\infty }}{\left({2}^{-r}\sum _{r}{k}^{-s}|\alpha \left({x}_{n}\left(k\right)+x\left(k\right)\right)+\left(1-\alpha \right)\left({x}_{n}\left(k-1\right)+x\left(k-1\right)\right)|\right)}^{{p}_{r}}\\ -\frac{\zeta }{4}+\frac{3\zeta }{4}\\ \ge & \frac{3\zeta }{4}+\left(1-\frac{\zeta }{4}\right)-\frac{\zeta }{4}\\ =& 1+\frac{\zeta }{4}.\end{array}$

Since $\sigma \in {\delta }_{2}^{s}$, it follows from Lemma 3.3 that there is τ depending on ζ only such that $\parallel {x}_{n}+x\parallel \ge 1+\tau$. □

Corollary 3.11 The space ${\mathcal{Z}}_{\sigma }\left(s,p\right)$ has property $\left(L\right)$ and the fixed point property.

## Authors’ Affiliations

(1)
Department of Mathematics, Firat University, Elaızğ, 23119, Turkey
(2)
Department of Statistics, Bitlis Eren University, Bitlis, 13000, Turkey
(3)
Department of Mathematics, Muş Alparslan University, Muş, 49100, Turkey

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