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

Abstract

In this paper, we define the modular space ${\mathcal{Z}}_{\sigma }(s,p)$ 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.

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 [1–9]. Opial [10] introduced the Opial property and proved that the sequence spaces ${\ell }_p$ ($1<p<\mathrm{\infty }$) have this property but $L_p[0,2\pi ]$ ($p\ne 2$, $1<p<\mathrm{\infty }$) 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 $(X,\parallel \cdot \parallel )$ be a real Banach space and let $S(X)$ (resp. $B(X)$) be the unit sphere (resp. the unit ball) of X. A Banach space X has the Opial property if for any weakly null sequence $\{x_n\}$ in X and any x in $X\setminus \{0\}$, 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 $\{x_n\}$ 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 $\beta (A)=inf\{\epsilon >0:A\text{ can be covered by finitely many balls with diameter }\le \epsilon \}$. The function Δ defined by $\mathrm{\Delta }(\epsilon )=inf\{1-inf(\parallel x\parallel :x\in A):A\text{ is closed convex subset of }B(X)\text{ with }\beta (A)\le \epsilon \}$ is called the modulus of noncompact convexity. A Banach space X is said to have property $(L)$, if ${lim}_{\epsilon \to 1^{-}}\mathrm{\Delta }(\epsilon )=1$. This property is an important concept in the fixed point theory and a Banach space X possesses property $(L)$ 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 $T:C\to C(\parallel Tx-Ty\parallel \le \parallel x-y\parallel ,\mathrm{\forall }x,y\in C)$ have a fixed point, that is, there exists $x\in C$ such that $T(x)=x$. Property $(L)$ 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 $(H)$ 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 $x,y\in S(X)$, the inequality $\parallel x-y\parallel >\epsilon$ implies that $\parallel \frac{1}{2}(x+y)\parallel <1-\delta$. For any $x\notin B(X)$, the drop determined by x is the set $D(x,B(X))=conv(\{x\}\cup B(X))$. A Banach space X has the drop property $(D)$ if for every closed set C disjoint with $B(X)$, there exists an element $x\in C$ such that $D(x,B(X))\cap C=\{x\}$. 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 $(H)$. A sequence $\{x_n\}$ is said to be ε-separated sequence for some $\epsilon >0$ if

$$\mathit{sep}(x_n)=inf\{\parallel x_n-x_m\parallel :n\ne m\}>\epsilon .$$

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

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

• (i) $\rho (x)=0$ if and only if $x=0$,

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

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

The modular ρ is called convex if

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

For any modular ρ on X, the space

$$X_{\rho }=\{x\in X:\rho (\sigma x)<\mathrm{\infty }\text{ for some }\sigma >0\}$$

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 $\parallel \cdot \parallel :X\to [0,\mathrm{\infty }]$ defines an F-norm if and only if

• (i) $\parallel x\parallel =0\iff x=0$,

• (ii) $\parallel \alpha x\parallel =\parallel x\parallel$ whenever $|\alpha |=1$,

• (iii) $\parallel x+y\parallel \le \parallel x\parallel +\parallel y\parallel$,

• (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 $d(x,y)=\parallel x-y\parallel$. If the linear metric space $(X,d)$ is complete, then it is called an F-space. The modular space $X_{\rho }$ can be equipped with the following F-norm:

$$\parallel x\parallel =inf\{\alpha >0:\rho \left(\frac{x}{\alpha }\right)\le \alpha \}.$$

If the modular ρ is convex, then the equality $\parallel x\parallel =inf\{\alpha >0:\rho (\frac{x}{\alpha })\le 1\}$ 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 $\rho (2u)\le K\rho (u)+\epsilon$ for all $u\in X_{\rho }$ with $\rho (u)\le a$. 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

$${\mathit{Ces}}_p=\{x\in {\ell }^0:\sum _{n=1}^{\mathrm{\infty }}{(n^{-1}\sum _{i=1}^n|x_i|)}^p<\mathrm{\infty }\},1\le p<\mathrm{\infty },$$

and

$${\mathit{Ces}}_{\mathrm{\infty }}=\{x\in {\ell }^0:\underset{n}{sup}{n}^{-1}\sum _{i=1}^n|x_i|<\mathrm{\infty }\},$$

were introduced by Shiue [21]. Jagers [22] determined the Köthe duals of the sequence space ${\mathit{Ces}}_p$ ($1<p<\mathrm{\infty }$). It can be shown that the inclusion ${\ell }_p\subset {\mathit{Ces}}_p$ is strict for $1<p<\mathrm{\infty }$ although it does not hold for $p=1$. Also, Suantai [23] defined the generalized Cesàro sequence space by

$$ces(p)=\{x\in {\ell }^0:\rho (\lambda x)<\mathrm{\infty }\text{ for some }\lambda >0\},$$

where $\rho (x)={\sum }_{n=1}^{\mathrm{\infty }}{(\frac{1}{n}{\sum }_{i=1}^n|x(i)|)}^{p_n}$. If $p=(p_n)$ is bounded, then

$$\mathit{ces}(p)=\{x=(x_k):\sum _{n=1}^{\mathrm{\infty }}{\left(n^{-1}\sum _{i=1}^n\right|x(i)\left|\right)}^{p_n}<\mathrm{\infty }\}.$$

The sequence space $C(s,p)$ was defined by Bilgin [24] as follows:

$$C(s,p)=\{x=(x_k):\sum _{r=0}^{\mathrm{\infty }}{(2^{-r}\sum _r{k}^{-s}|x_k|)}^{p_r}<\mathrm{\infty },s\ge 0\}$$

for $p=(p_r)$ 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 $C(s,p)$ for $s=0$ is the space

$$\mathit{Ces}(p)=\{x=(x_k):\sum _{r=0}^{\mathrm{\infty }}{(2^{-r}\sum _r|x_k|)}^{p_r}<\mathrm{\infty }\}$$

which was introduced by Lim [25]. Also, the inclusion $\mathit{Ces}(p)\subseteq C(s,p)$ holds. A paranorm on $C(s,p)$ is given by

$$\rho (x)={\left(\sum _{r=0}^{\mathrm{\infty }}{(2^{-r}\sum _r{k}^{-s}|x_k|)}^{p_r}\right)}^{1/M}$$

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

The Z-transform of a sequence $x=(x_k)$ is defined by ${(Zx)}_n=y_n=\alpha x_n+(1-\alpha )x_{n-1}$ by using the Zweier operator

$$Z=(z_{nk})=\{\begin{array}{cc}\alpha ,\hfill & k=n,\hfill \\ 1-\alpha ,\hfill & k=n-1,\hfill \\ 0,\hfill & \text{otherwise}\hfill \end{array}\text{for }n,k\in \mathbb{N}\text{ and }\alpha \in \mathcal{F}\mathrm{\setminus }\{0\},$$

where ℱ 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 ${\mathcal{Z}}_{\sigma }(s,p)$ by

$${\mathcal{Z}}_{\sigma }(s,p)=\{x\in {\ell }^0:\sigma (tx)<\mathrm{\infty },\text{ for some }t>0\},$$

where $\sigma (x)={\sum }_{r=0}^{\mathrm{\infty }}{(2^{-r}{\sum }_r{k}^{-s}|\alpha x_k+(1-\alpha )x_{k-1}|)}^{p_r}<\mathrm{\infty }$ and $s\ge 0$. If we take $\alpha =1$, then ${\mathcal{Z}}_{\sigma }(s,p)=C(s,p)$; if $\alpha =1$ and $s=0$, then ${\mathcal{Z}}_{\sigma }(s,p)=\mathit{Ces}(p)$. It can be easily seen that $\sigma :{\mathcal{Z}}_{\sigma }(s,p)\to [0,\mathrm{\infty }]$ is a modular on ${\mathcal{Z}}_{\sigma }(s,p)$. We define the Luxemburg norm on the sequence space ${\mathcal{Z}}_{\sigma }(s,p)$ as follows:

$$\parallel x\parallel =inf\{t>0:\sigma \left(\frac{x}{t}\right)\le 1\},\mathrm{\forall }x\in {\mathcal{Z}}_{\sigma }(s,p).$$

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

Throughout the paper, suppose that $p=(p_r)$ 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:

$$|a_k+b_k{|}^{p_k}\le C(|a_k{|}^{p_k}+|b_k{|}^{p_k}),|a_k+b_k{|}^{t_k}\le |a_k{|}^{t_k}+|b_k{|}^{t_k},$$

where $t_k=\frac{p_k}{M}\le 1$ and $C=max\{1,2^{H-1}\}$ with $H=sup{p}_k$.

3 Main results

Since ${\ell }_p$ is reflexive and convex, $\ell (p)$-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 $C(s,p)$ by using the Zweier operator and then obtained the equality ${\mathcal{Z}}_{\sigma }(s,p)=\mathit{Ces}(p)$, that is, it was seen that the structure of the space $\mathit{Ces}(p)$ was preserved. In this section, our goal is to investigate a geometric structure of the modular space ${\mathcal{Z}}_{\sigma }(s,p)$ related to the fixed point theory. For this, we will examine property $(\beta )$ and the uniform Opial property for ${\mathcal{Z}}_{\sigma }(s,p)$. 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

$$|\sigma (u+v)-\sigma (u)|<\epsilon ,$$

where $u,v\in X_{\sigma }$ with $\sigma (u)\le L$ and $\sigma (v)\le \delta$.

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 $\delta =\delta (\epsilon )>0$ such that $\parallel x\parallel \ge 1+\delta$ implies $\sigma (x)\ge 1+\epsilon$.

Now we give the following two lemmas without proof.

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

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

Lemma 3.6 If $liminf{p}_r>1$, then for any $x\in {\mathcal{Z}}_{\sigma }(s,p)$, there exist $k_0\in \mathbb{N}$ and $\mu \in (0,1)$ 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 $\mu \in (0,1)$ such that ${(\frac{1}{2})}^{\gamma }\le \frac{1-\mu }{2}$. Therefore, we have

$$\begin{array}{rl}\sigma \left(\frac{x^k}{2}\right)& =\sum _{r=0}^{\mathrm{\infty }}{\left(2^{-r}\sum _r{k}^{-s}\right|\frac{\alpha x(k)+(1-\alpha )x(k-1)}{2}\left|\right)}^{p_r}\\ =\sum _{r=0}^{\mathrm{\infty }}{\left(\frac{1}{2}\right)}^{p_r}{\left(2^{-r}\sum _r{k}^{-s}\right|\alpha x(k)+(1-\alpha )x(k-1)\left|\right)}^{p_r}\\ \le {\left(\frac{1}{2}\right)}^{\gamma }\sum _{r=0}^{\mathrm{\infty }}{\left(2^{-r}\sum _r{k}^{-s}\right|\alpha x(k)+(1-\alpha )x(k-1)\left|\right)}^{p_r}\\ <\frac{1-\mu }{2}\sigma \left(x^k\right)\end{array}$$

for each $x\in {\mathcal{Z}}_{\sigma }(s,p)$ and $k\ge k_0$. □

Lemma 3.7 If $\sigma \in {\delta }_2^s$, then for any $\epsilon \in (0,1)$, there exists $\delta \in (0,1)$ such that $\sigma (x)\le 1-\epsilon$ implies $\parallel x\parallel \le 1-\delta$.

Proof Suppose that lemma does not hold. So, there exist $\epsilon >0$ and $x_n\in {\mathcal{Z}}_{\sigma }(s,p)$ such that $\sigma (x_n)<1-\epsilon$ 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 $P=sup\{\sigma (2x_n):n\in \mathbb{N}\}$. There exists $D\ge 2$ such that

$$\sigma (2u)\le D\sigma (u)+1$$

(3.2)

for every $u\in {\mathcal{Z}}_{\sigma }(s,p)$ with $\sigma (u)<1$, since $\sigma \in {\delta }_2^s$. We have

$$\sigma (2x_n)\le D\sigma (x_n)+1<D+1$$

for all $n\in \mathbb{N}$ by (3.1). Therefore, $0<P<\mathrm{\infty }$ and from Lemma 3.5 we have

$$\begin{array}{rl}1& =\sigma \left(\frac{x_n}{\parallel x_n\parallel }\right)=\sigma (2s_n{x}_n+(1-s_n)x_n)\\ \le s_n\sigma (2x_n)+(1-s_n)\sigma (x_n)\\ \le s_{n}P+(1-\epsilon )\to (1-\epsilon ).\end{array}$$

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

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

Proof Let $\epsilon >0$ and $(x_n)\subset B({\mathcal{Z}}_{\sigma }(s,p))$ with $\mathit{sep}(x_n)\ge \epsilon$ and $x\in B({\mathcal{Z}}_{\sigma }(s,p))$. 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 $i\in \mathbb{N}$, ${(x_n(i))}_{i=1}^{\mathrm{\infty }}$ is bounded, by using the diagonal method, we can find a subsequence $(x_{n_j})$ of $(x_n)$ such that $(x_{n_j}(i))$ 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 $\mathit{sep}({(x_{n_j}^l)}_{j\ge t_l})\ge \epsilon$. Thus, there exists a sequence of positive integers ${(r_l)}_{l=1}^{\mathrm{\infty }}$ 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

$$\sigma \left(x_{r_l}^l\right)\ge \eta \text{for all }l\in \mathbb{N}$$

(3.4)

from Lemma 3.3. However, there exist $k_0\in \mathbb{N}$ and $\mu \in (0,1)$ such that

$$\sigma \left(\frac{v^k}{2}\right)\le \frac{1-\mu }{2}\sigma \left(v^k\right)$$

(3.5)

for all $v\in {\mathcal{Z}}_{\sigma }(s,p)$ and $k\ge k_0$ by Lemma 3.6. There exists $\delta >0$ such that

$$\sigma (y)\le 1-\frac{\mu \eta }{4}\Rightarrow \parallel y\parallel \le 1-\delta$$

(3.6)

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

By Lemma 3.1, there exists ${\delta }_0$ such that

$$|\sigma (u+v)-\sigma (u)|<\frac{\mu \eta }{4},$$

(3.7)

where $\sigma (u)\le 1$ and $\sigma (v)\le {\delta }_0$. Hence, we get that $\sigma (x)\le 1$ since $x\in B({\mathcal{Z}}_{\sigma }(s,p))$. Then there exists $k\ge k_0$ such that $\sigma (x^k)\le {\delta }_0$. Let $u=x_{r_l}^l$ and $v=x^l$. Then

$$\sigma \left(\frac{u}{2}\right)<1\text{and}\sigma \left(\frac{v}{2}\right)<{\delta }_0.$$

We obtain from (3.3) and (3.5) that

$$\sigma \left(\frac{u+v}{2}\right)\le \sigma \left(\frac{u}{2}\right)+\frac{\mu \eta }{4}\le \frac{1-\mu }{2}\sigma (u)+\frac{\mu \eta }{4}.$$

(3.8)

Choose $s_i=r_{l_i}$. By the inequalities (3.2), (3.3), (3.6) and the convexity of the function $f(u)=|u{|}^{p_r}$, we have

$$\begin{array}{rl}\sigma \left(\frac{x+x_{s_k}}{2}\right)=& \sum _{r=0}^{\mathrm{\infty }}{\left(2^{-r}\sum _r{k}^{-s}\right|\frac{\alpha (x(k)+x_{s_i}(k))+(1-\alpha )(x(k-1)+x_{s_i}(k-1))}{2}\left|\right)}^{p_r}\\ =& \sum _{r=0}^{r_k-1}{\left(2^{-r}\sum _r{k}^{-s}\right|\frac{\alpha (x(k)+x_{s_i}(k))+(1-\alpha )(x(k-1)+x_{s_i}(k-1))}{2}\left|\right)}^{p_r}\\ +\sum _{r=r_k}^{\mathrm{\infty }}{\left(2^{-r}\sum _r{k}^{-s}\right|\frac{\alpha (x(k)+x_{s_i}(k))+(1-\alpha )(x(k-1)+x_{s_i}(k-1))}{2}\left|\right)}^{p_r}\\ \le & \frac{1}{2}\sum _{r=0}^{r_k-1}{\left(2^{-r}\sum _r{k}^{-s}\right|\alpha x(k)+(1-\alpha )x(k-1)\left|\right)}^{p_r}\\ +\frac{1}{2}\sum _{r=0}^{r_k-1}{\left(2^{-r}\sum _r{k}^{-s}\right|\alpha x_{s_i}(k)+(1-\alpha )x_{s_i}(k-1)\left|\right)}^{p_r}\\ +\sum _{r=r_k}^{\mathrm{\infty }}{\left(2^{-r}\sum _r{k}^{-s}\right|\frac{\alpha x_{s_i}(k)+(1-\alpha )x_{s_i}(k-1)}{2}\left|\right)}^{p_r}+\frac{\mu \eta }{4}\\ \le & \frac{1}{2}\sum _{r=0}^{r_k-1}{\left(2^{-r}\sum _r{k}^{-s}\right|\alpha x(k)+(1-\alpha )x(k-1)\left|\right)}^{p_r}\\ +\frac{1}{2}\sum _{r=0}^{r_k-1}{\left(2^{-r}\sum _r{k}^{-s}\right|\alpha x_{s_i}(k)+(1-\alpha )x_{s_i}(k-1)\left|\right)}^{p_r}\\ +\frac{1-\mu }{2}\sum _{r=r_k}^{\mathrm{\infty }}{\left(2^{-r}\sum _r{k}^{-s}\right|\frac{\alpha x_{s_i}(k)+(1-\alpha )x_{s_i}(k-1)}{2}\left|\right)}^{p_r}+\frac{\mu \eta }{4}\\ \le & \frac{1}{2}\sum _{r=0}^{r_k-1}{\left(2^{-r}\sum _r{k}^{-s}\right|\alpha x(k)+(1-\alpha )x(k-1)\left|\right)}^{p_r}\\ +\frac{1}{2}\sum _{r=0}^{\mathrm{\infty }}{\left(2^{-r}\sum _r{k}^{-s}\right|\alpha x_{s_i}(k)+(1-\alpha )x_{s_i}(k-1)\left|\right)}^{p_r}\\ -\frac{\mu }{2}\sum _{r=r_k}^{\mathrm{\infty }}{\left(2^{-r}\sum _r{k}^{-s}\right|\frac{\alpha x_{s_i}(k)+(1-\alpha )x_{s_i}(k-1)}{2}\left|\right)}^{p_r}+\frac{\mu \eta }{4}\\ \le & \frac{1}{2}+\frac{1}{2}-\frac{\mu \eta }{2}+\frac{\mu \eta }{4}\\ =& 1-\frac{\mu \eta }{4}.\end{array}$$

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

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

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

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

Proof Let $\epsilon >0$ and $x\in {\mathcal{Z}}_{\sigma }(s,p)$ be such that $\parallel x\parallel \ge \epsilon$ and $(x_n)$ be a weakly null sequence in $S({\mathcal{Z}}_{\sigma }(s,p))$. By $\sigma \in {\delta }_2^s$, there exists $\zeta \in (0,1)$ independent of x such that $\sigma (x)>\zeta$ by Lemma 3.2. Also since $\sigma \in {\delta }_2^s$, by Lemma 3.1, there is ${\zeta }_1\in (0,\zeta )$ such that

$$|\sigma (y+z)-\sigma (y)|<\frac{\zeta }{4}$$

(3.10)

whenever $\sigma (y)\le 1$ and $\sigma (z)\le {\zeta }_1$. Take $r_0\in \mathbb{N}$ such that

$$\sum _{r=r_0+1}^{\mathrm{\infty }}{\left(2^{-r}\sum _r{k}^{-s}\right|\alpha x(k)+(1-\alpha )x(k-1)\left|\right)}^{p_r}<\frac{{\zeta }_1}{4}.$$

(3.11)

Hence, we have

$$\begin{array}{rl}\zeta <& \sum _{r=1}^{r_0}{\left(2^{-r}\sum _r{k}^{-s}\right|\alpha x(k)+(1-\alpha )x(k-1)\left|\right)}^{p_r}\\ +\sum _{r=r_0+1}^{\mathrm{\infty }}{\left(2^{-r}\sum _r{k}^{-s}\right|\alpha x(k)+(1-\alpha )x(k-1)\left|\right)}^{p_r}\\ \le & \sum _{r=1}^{r_0}{\left(2^{-r}\sum _r{k}^{-s}\right|\alpha x(k)+(1-\alpha )x(k-1)\left|\right)}^{p_r}+\frac{{\zeta }_1}{4}\end{array}$$

(3.12)

and this implies that

$$\begin{array}{rl}\sum _{r=1}^{r_0}{\left(2^{-r}\sum _r{k}^{-s}\right|\alpha x(k)+(1-\alpha )x(k-1)\left|\right)}^{p_r}& >\zeta -\frac{{\zeta }_1}{4}\\ >\zeta -\frac{\zeta }{4}\\ =\frac{3\zeta }{4}.\end{array}$$

(3.13)

Since $x_n{\to }^{w}0$, by the inequality (3.10), there exists $r_0\in \mathbb{N}$ such that

$$\sum _{r=1}^{r_0}{\left(2^{-r}\sum _r{k}^{-s}\right|\alpha (x_n(k)+x(k))+(1-\alpha )(x_n(k-1)+x(k-1))\left|\right)}^{p_r}>\frac{3\zeta }{4}.$$

(3.14)

Again, by $x_n{\to }^{w}0$, there is $r_1>r_0$ such that for all $r>r_1$

$$\parallel x_{n_{|r_0}}\parallel <1-{(1-\frac{\zeta }{4})}^{1/M},$$

(3.15)

where $p_r\le M\in \mathbb{N}$ for all $r\in \mathbb{N}$. Therefore, we obtain that

$$\parallel x_{n_{|\mathbb{N}-r_0}}\parallel >{(1-\frac{\zeta }{4})}^{1/M}$$

(3.16)

by the triangle inequality of the norm. It follows from the definition of the Luxemburg norm that

$$\begin{array}{rl}1& \le \sigma \left(\frac{x_{n_{|\mathbb{N}-r_0}}}{{(1-\frac{\zeta }{4})}^{1/M}}\right)\\ =\sum _{r=r_0+1}^{\mathrm{\infty }}{\left(\frac{2^{-r}{\sum }_r{k}^{-s}|\alpha x_n(k)+(1-\alpha )x_n(k-1)|}{{(1-\frac{\zeta }{4})}^{1/M}}\right)}^{p_r}\\ \le {\left(\frac{1}{{(1-\frac{\zeta }{4})}^{1/M}}\right)}^M\sum _{r=r_0+1}^{\mathrm{\infty }}{\left(2^{-r}\sum _r{k}^{-s}\right|\alpha x_n(k)+(1-\alpha )x_n(k-1)\left|\right)}^{p_r}\end{array}$$

(3.17)

and this implies that

$$\sum _{r=r_0+1}^{\mathrm{\infty }}{\left(2^{-r}\sum _r{k}^{-s}\right|\alpha x_n(k)+(1-\alpha )x_n(k-1)\left|\right)}^{p_r}\ge 1-\frac{\zeta }{4}.$$

(3.18)

By (3.7), (3.8), (3.11), (3.15) and since $x_n{\to }^{w}0\Rightarrow x_n\to 0$ (coordinatewise), we have for any $r>r_1$ that

$$\begin{array}{rl}\sigma (x_n+x)=& \sum _{r=1}^{r_0}{\left(2^{-r}\sum _r{k}^{-s}\right|\alpha (x_n(k)+x(k))+(1-\alpha )(x_n(k-1)+x(k-1))\left|\right)}^{p_r}\\ +\sum _{r=r_0+1}^{\mathrm{\infty }}{\left(2^{-r}\sum _r{k}^{-s}\right|\alpha (x_n(k)+x(k))+(1-\alpha )(x_n(k-1)+x(k-1))\left|\right)}^{p_r}\\ \ge & \sum _{r=r_0+1}^{\mathrm{\infty }}{\left(2^{-r}\sum _r{k}^{-s}\right|\alpha (x_n(k)+x(k))+(1-\alpha )(x_n(k-1)+x(k-1))\left|\right)}^{p_r}\\ -\frac{\zeta }{4}+\frac{3\zeta }{4}\\ \ge & \frac{3\zeta }{4}+(1-\frac{\zeta }{4})-\frac{\zeta }{4}\\ =& 1+\frac{\zeta }{4}.\end{array}$$

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 ${\mathcal{Z}}_{\sigma }(s,p)$ has property $(L)$ and the fixed point property.