New Modular Fixed-Point Theorem in the Variable Exponent Spaces ℓ_p(.)

Abstract

In this work, we prove a fixed-point theorem in the variable exponent spaces ${\ell }_{p(.)}$, when $p^{-}=1$ without further conditions. This result is new and adds more information regarding the modular structure of these spaces. To be more precise, our result concerns $\rho$-nonexpansive mappings defined on convex subsets of ${\ell }_{p(.)}$ that satisfy a specific condition which we call “condition of uniform decrease”.

1. Introduction

Variable exponent spaces first appeared in a work of Orlicz in 1931 [1] (see also [2]), where he defined the following space:

$$X=\left\{\left\{x_n\right\}\in {\mathbb{R}}^{\mathbb{N}},\sum _{n=0}^{\infty }{\left|\lambda x_n\right|}^{p\left(n\right)}<\infty ,forsome\lambda >0\right\}.$$

They became very important because of their use in the mathematical modeling of non-Newtonian fluids [3,4]. The typical example of such fluids are electrorheological fluids, the viscosity of which exhibits dramatic and sudden changes when exposed to an electric or magnetic field. The necessity of a clear understanding of the spaces with variable integrability is reinforced by their potential applications.

The properties of this vector space have been extensively studied in [5,6,7]. The norm that was commonly used to investigate the geometrical properties of X is the Minkowski functional associated to the modular unit ball and it is known as the Luxembourg norm. Whereas in the case of classical ${\ell }_p$ spaces, the natural norm is suitable for making calculations, the Luxembourg norm on X is very difficult to manipulate.

In 1950, Nakano [8] introduced for the first time the notion of modular vector space (see also [9,10]). This abstract point of view has been crucial to the development of the research on geometrical and topological properties of the variable exponent spaces ${\ell }_{p(.)}$.

In this work, we will introduce a class of subsets of ${\ell }_{p(.)}$ that have some interesting geometrical properties. This will allow us to prove a new fixed-point theorem concerning ${\ell }_{p(.)}$ spaces. For the study of metric fixed-point theory, we recommend the book [9].

2. Basic Notations and Terminology

For a function $p:\mathbb{N}⟶[1,+\infty )$, define the vector space

$${\ell }_{p(.)}=\left\{\left\{x_n\right\}\in {\mathbb{R}}^{\mathbb{N}},\sum _{n=0}^{\infty }\frac{1}{p\left(n\right)}{\left|\lambda x_n\right|}^{p\left(n\right)}<\infty ,forsome\lambda >0\right\}.$$

Nakano [8,11] introduced the concept of modular vector space.

Proposition 1

([6,9]). Consider the function $\rho :{\ell }_{p(.)}⟶\left[0,+\infty \right]$ defined by

$$\rho \left(x\right)=\rho \left(\left\{x_n\right\}\right)=\sum _{n=0}^{\infty }\frac{1}{p\left(n\right)}{\left|x_n\right|}^{p\left(n\right)}$$

then ρ satisfies the following properties

(1) 

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

(2) 

$\rho \left(\alpha x\right)=\rho \left(x\right)$, if $\left|\alpha \right|=1$,

(3) 

$\rho \left(\alpha x+(1-\alpha )y\right)\le \alpha \rho \left(x\right)+(1-\alpha )\rho \left(y\right),\forall \alpha \in [0,1]$.

for any $x,y\in X$. The function ρ is called a convex modular.

For any subset I of $\mathbb{N}$, we consider the functional

$${\rho }_I\left(x\right)=\sum _{n\in I}{\left|x_n\right|}^{p\left(n\right)}.$$

If $I=\varnothing$, we set ${\rho }_I\left(x\right)=0$. We define on modular spaces a modular topology which is similar to the topology induced by a metric.

Definition 1.

Consider the vector space ${\ell }_{p(.)}$.

(a) 

We say that a sequence $\left\{x_n\right\}\subset {\ell }_{p(.)}$ is ρ-convergent to $x\in {\ell }_{p(.)}$ if and only if $\rho (x_n-x)⟶0$. The ρ-limit is unique if it exists.

(b) 

A sequence $\left\{x_n\right\}\subset {\ell }_{p(.)}$ is called ρ-Cauchy if $\rho (x_n-x_m)⟶0$ as $n,m⟶+\infty$.

(c) 

A nonempty subset $C\subset {\ell }_{p(.)}$ is called ρ-closed if for any sequence $\left\{x_n\right\}\subset C$ which ρ-converges to x implies that $x\in C$.

(d) 

A nonempty subset $C\subset {\ell }_{p(.)}$ is called ρ-bounded if and only if

$${\delta }_{\rho }\left(C\right)=sup\left\{\rho (x-y),x,y\in C\right\}<\infty .$$

Note that $\rho$ satisfies the Fatou property, i.e.,

$$\rho (x-y)\le \underset{n\to +\infty }{lim\; inf}\rho (x-y_n),$$

holds whenever $\left\{y_n\right\}$$\rho$-converges to y, for any $x,y,y_n\in {\ell }_{p(.)}$. Throughout, we will use the notation $B_{\rho }(x,r)$ to denote the $\rho$-ball with radius $r\ge 0$ centered at $x\in {\ell }_{p(.)}$ and defined as

$$B_{\rho }(x,r)=\left\{y\in {\ell }_{p(.)},\rho (x-y)\le r\right\}.$$

Note that Fatou property holds if and only if the $\rho$-balls are $\rho$-closed. That is, all $\rho$-balls are $\rho$-closed in ${\ell }_{p(.)}$.

Definition 2.

Let $C\subset {\ell }_{p(.)}$ be a nonempty subset. A mapping $T:C⟶C$ is called ρ-Lipschitzian if there exists a constant $K\ge 0$ such that

$$\rho \left(T\left(x\right)-T\left(y\right)\right)\le K\rho (x-y),\forall x,y\in C.$$

If $K=1$, T is called ρ-nonexpansive. A point $x\in C$ is called a fixed point of T if $T\left(x\right)=x$.

The concept of modular uniform convexity was first introduced by Nakano [11], but a weaker definition of modular uniform convexity called $\left(UUC2\right)$ was introduced in [9] and seems to be more suitable to hold in ${\ell }_{p(.)}$ when weaker assumptions on the exponent function $p(·)$ hold. The following definition is given in terms of subsets because of the subsequent results discovered in this work.

Definition 3

([9]). Consider the vector space ${\ell }_{p(.)}$. Let C be a nonempty subset of ${\ell }_{p(.)}$.

1. 

Let $r>0$ and $\epsilon >0$. Define

$$D_2(r,\epsilon )=\left\{(x,y)\in {\ell }_{p(.)}\times {\ell }_{p(.)},\rho \left(x\right)\le r,\rho \left(y\right)\le r,\rho \left(\frac{x-y}{2}\right)\ge \epsilon r\right\}.$$

If $D_2(r,\epsilon )\cap (C\times C)\ne \varnothing$, let

$${\delta }_{2,C}(r,\epsilon )=inf\left\{1-\frac{1}{r}\rho \left(\frac{x+y}{2}\right),(x,y)\in D_2(r,\epsilon )\cap (C\times C)\right\}.$$

If $D_2(r,\epsilon )\cap (C\times C)=\varnothing$, we set ${\delta }_2(r,\epsilon )=1$. We say that ρ satisfies $\left(UC2\right)$ on C if for every $r>0$ and $\epsilon >0$, we have ${\delta }_{2,C}(r,\epsilon )>0$. When $C={\ell }_{p(.)}$, we remark that for every $r>0$, $D_2(r,\epsilon )\ne \varnothing$, for $\epsilon >0$ small enough. In this case, we will use the notation ${\delta }_{2,{\ell }_{p(.)}}={\delta }_2$.

2. 

We say that ρ satisfies $\left(UUC2\right)$ on C if for every $s\ge 0$ and $\epsilon >0$, there exists ${\eta }_2(s,\epsilon )>0$ depending on s and ε such that

$${\delta }_{2,C}(r,\epsilon )\ge {\eta }_2(s,\epsilon )>0forr>s.$$

3. 

We say that ρ is strictly convex on C (in short $\left(SC\right)$), if for every $x,y\in C$ such that

$$\rho \left(x\right)=\rho \left(y\right)and\rho \left(\frac{x+y}{2}\right)=\frac{\rho \left(x\right)+\rho \left(y\right)}{2}implyx=y.$$

In the study of the properties of ${\ell }_{p(.)}$ (see [12]), the following values are very important:

$${\displaystyle p^{+}=\underset{n\in \mathbb{N}}{sup}p\left(n\right)and{p}^{-}=\underset{n\in \mathbb{N}}{inf}p\left(n\right).}$$

In [5], the authors proved that for ${\ell }_{p(.)}$, with $p^{-}>1$, the modular is $\left(UUC2\right)$. This modular geometrical property allows to prove the following fixed-point result:

Theorem 1.

Consider the vector space ${\ell }_{p(.)}$. Assume $p^{-}>1$. Let C be a nonempty ρ-closed convex ρ-bounded subset of ${\ell }_{p(.)}$. Let $T:C⟶C$ be a ρ-nonexpansive mapping. Then T has a fixed point.

In [13], the authors proved a similar fixed-point theorem in the case where $\{n\in \mathbb{N},p(n)=1\}$ has at most one element which is an improvement from $p^{-}>1$.

Before we close this section, we recall the following lemma, of a rather technical nature, which plays a crucial role when dealing with ${\ell }_{p(.)}$ spaces.

Lemma 1.

The following inequalities hold:

(i) 

[14]. If $p\ge 2$, then

$${\left|\frac{a+b}{2}\right|}^p+{\left|\frac{a-b}{2}\right|}^p\le \frac{1}{2}\left({\left|a\right|}^p+{\left|b\right|}^p\right),$$

for any $a,b\in \mathbb{R}$.

(ii) 

[15]. If $1<p\le 2$, then

$${\left|\frac{a+b}{2}\right|}^p+\frac{p(p-1)}{2}{\left|\frac{a-b}{\left|a\right|+\left|b\right|}\right|}^{2-p}{\left|\frac{a-b}{2}\right|}^p\le \frac{1}{2}\left({\left|a\right|}^p+{\left|b\right|}^p\right),$$

for any $a,b\in \mathbb{R}$ such that $\left|a\right|+\left|b\right|\ne 0$.

In this work, using a different approach, we obtain some fixed-point results when $p^{-}=1$ without the known conditions on the function $p(·)$.

3. Uniform Decrease Condition

First, we introduce an interesting class of subsets of ${\ell }_{p(.)}$, which will play an important part in our work. In particular, they enjoy similar modular geometric properties as ${\ell }_{p(.)}$ when $p^{-}>1$. Before, let us introduce the following notations:

$$I_a=\left\{n\in \mathbb{N};p\left(n\right)\ge a\right\}and{J}_a=\mathbb{N}\backslash I_a=\left\{n\in \mathbb{N};p\left(n\right)<a\right\},$$

where $a\in [1,+\infty )$.

Definition 4.

Consider the vector space ${\ell }_{p(.)}$. A nonempty subset C of ${\ell }_{p(.)}$ is said to satisfy the uniform decrease condition (in short $\left(UD\right)$) if for any $\alpha >0$, there exists $a>1$ such that

$$\underset{x\in C}{sup}{\rho }_{J_a}\left(x\right)\le \alpha .$$

Obviously the condition $\left(UD\right)$ passes from a set to its subsets. Moreover, if $p(·)$ is identically equal to 1, then the only $\left(UD\right)$ subset is $C=\left\{0\right\}$. Since this case is not interesting, we will assume throughout that $p(·)$ is not identically equal to 1. Moreover, if $p^{-}>1$, then any nonempty subset of ${\ell }_{p(.)}$ satisfies the condition $\left(UD\right)$. Indeed, let C be a nonempty subset of ${\ell }_{p(.)}$ and $\alpha >0$. Let $a\in (1,p^{-})$. Then $J_a=\varnothing$ which implies

$$\underset{x\in C}{sup}{\rho }_{J_a}\left(x\right)=0\le \alpha .$$

Therefore, the condition $\left(UD\right)$ is interesting to study only when $p^{-}=1$ and $p(·)$ is not identically equal to 1, which will be the case throughout.

Example 1.

Consider the function $p(·)$ defined by

$$p\left(n\right)=1+\frac{1}{n+1},n\in \mathbb{N}.$$

Consider the subset

$$C=\left\{x\in {\ell }_{p(.)};\left|x_n\right|\le \frac{1}{{(n+1)}^2},n\in \mathbb{N}\right\}.$$

C is nonempty, convex and ρ-closed. Let us show that it satisfies the condition $\left(UD\right)$. Indeed, fix $\alpha >0$. Let $N\ge 1$ be such that ${\displaystyle \sum _{k\ge N}\frac{1}{{(k+1)}^2}\le \alpha }$. Set $a=1+\frac{1}{N}$. We have

$$\begin{array}{ccc}{\rho }_{J_a}\left(x\right)\hfill & =\hfill & {\displaystyle \sum _{n\in J_a}\frac{|x_n{|}^{p\left(n\right)}}{p\left(n\right)}}\hfill \\ & \le \hfill & {\displaystyle \sum _{n\ge N}\frac{|x_n{|}^{p\left(n\right)}}{p\left(n\right)}}\hfill \\ & \le \hfill & {\displaystyle \sum _{n\ge N}\frac{1}{p\left(n\right)}\frac{1}{{(n+1)}^{2p\left(n\right)}}}\hfill \\ & \le \hfill & {\displaystyle \sum _{n\ge N}\frac{1}{{(n+1)}^2}}\hfill \\ & \le \hfill & \alpha ,\hfill \end{array}$$

for all $x\in C$, which proves our claim that C is $\left(UD\right)$.

Before we give a characterization of subsets which satisfy the condition $\left(UD\right)$, we need to introduce a new class of subsets of ${\ell }_{p(.)}$.

Definition 5.

Consider the vector space ${\ell }_{p(.)}$ such that $p^{-}=1$ and $p(·)$ not identically equal to 1. Let $f:(0,+\infty )\to (1,2]$ be a nondecreasing function. Define the set $C_f$ to be

$$C_f=\left\{x\in {\ell }_{p(.)};{\rho }_{J_{f\left(\alpha \right)}}\left(x\right)\le \alpha ,forall\alpha >0\right\}.$$

Note that $C_f$ is never empty since $0\in C_f$. Some of the basic properties of $C_f$ are given in the following lemma.

Lemma 2.

Consider the vector space ${\ell }_{p(.)}$ such that $p^{-}=1$ and $p(·)$ not identically equal to 1. Let $f:(0,+\infty )\to (1,2]$ be a non-decreasing function. Then the following properties hold:

1. 

$C_f$ is convex.

2. 

$C_f$ is symmetrical, i.e., $-z\in C_f$ whenever $z\in C_f$.

3. 

The Fatou property implies easily that $C_f$ is ρ-closed as a subset of ${\ell }_{p(.)}$ which in turn implies that $C_f$ is ρ-complete.

Proposition 2.

Consider the vector space ${\ell }_{p(.)}$ such that $p^{-}=1$ and $p(·)$ not identically equal to 1. A subset C of ${\ell }_{p(.)}$ satisfies the condition $\left(UD\right)$ if and only if there exists $f:(0,+\infty )\to (1,2]$ non-decreasing such that $C\subset C_f$.

Proof. 

First, we prove that $C_f$ satisfies the condition $\left(UD\right)$. Fix $\alpha >0$. If we take $a=f\left(\alpha \right)$, we obtain

$$\underset{x\in C_f}{sup}{\rho }_{J_a}\left(x\right)\le \alpha ,$$

which proves our claim. Clearly, any subset C of $C_f$ will also satisfy the condition $\left(UD\right)$. Conversely, let C be a nonempty subset of ${\ell }_{p(.)}$ which satisfies the condition $\left(UD\right)$. For any $\alpha >0$, there exists $a>1$ such that $\underset{x\in C}{sup}{\rho }_{J_a}\left(x\right)\le \alpha$. Set

$$\left[\alpha \right]=\left\{a>1;\underset{x\in C}{sup}{\rho }_{J_a}\left(x\right)\le \alpha \right\}.$$

Define

$$f\left(\alpha \right)=\left\{\begin{array}{cc}2\hfill & if\left[\alpha \right]\subset [2,+\infty ),\hfill \\ sup\left(\left[\alpha \right]\cap (1,2]\right)\hfill & if\left[\alpha \right]\cap (1,2]\ne \varnothing .\hfill \end{array}\right.$$

Clearly, f is well defined and $f\left(\alpha \right)\in (1,2]$, for all $\alpha >0$. Let $\alpha$ and $\beta$ be such that $0<\alpha \le \beta$. We claim that $f\left(\alpha \right)\le f\left(\beta \right)$. Indeed, it is easy to see that $\left[\alpha \right]\subset \left[\beta \right]$. If $\left[\alpha \right]\cap (1,2]\ne \varnothing$, then we have $\left[\beta \right]\cap (1,2]\ne \varnothing$ which easily implies $f\left(\alpha \right)\le f\left(\beta \right)$. Otherwise, assume $\left[\alpha \right]\subset [2,+\infty )$. Let $a\in \left[\alpha \right]$. We have $a\ge 2$ and $a\in \left[\beta \right]$. By definition of the sets J, we have $J_2\subset J_a$. Since ${\rho }_{J_2}\left(x\right)\le J_a\left(x\right)$, for all $x\in {\ell }_{p(.)}$, we obtain

$$\underset{x\in C}{sup}{\rho }_{J_2}\left(x\right)\le \underset{x\in C}{sup}{\rho }_{J_a}\left(x\right)\le \beta ,$$

i.e., $2\in \left[\beta \right]$. This fact, will force $f\left(\beta \right)=2$. In all cases, we have $f\left(\alpha \right)\le f\left(\beta \right)$. In other words, the function $f:(0,+\infty )\to (1,2]$ is non-decreasing. Finally, let us show that $C\subset {\ell }_g$, where $g\left(\alpha \right)=(1+f(\alpha \left)\right)/2$, for all $\alpha >0$. Since $1<f\left(\alpha \right)$, then we have $1<g\left(\alpha \right)<f\left(\alpha \right)$, for all $\alpha >0$. If $\left[\alpha \right]\subset [2,+\infty )$, pick $a\in \left[\alpha \right]$. Then $g\left(\alpha \right)=3/2<a$ which implies $J_{g\left(\alpha \right)}\subset J_a$. Hence

$${\rho }_{J_{g\left(\alpha \right)}}\left(x\right)\le {\rho }_{J_a}\left(x\right),forallx\in C,$$

which implies $\underset{x\in C}{sup}{\rho }_{J_{g\left(\alpha \right)}}\left(x\right)\le \underset{x\in C}{sup}{\rho }_{J_a}\left(x\right)\le \alpha$. Otherwise, assume $\left[\alpha \right]\cap (1,2]\ne \varnothing$, then $f\left(\alpha \right)=sup\left(\left[\alpha \right]\cap (1,2]\right)$. Since $g\left(\alpha \right)<f\left(\alpha \right)$, there exists $a\in \left[\alpha \right]$ such that $g\left(\alpha \right)<a\le f\left(\alpha \right)$. Similar argument will show that

$$\underset{x\in C}{sup}{\rho }_{J_{g\left(\alpha \right)}}\left(x\right)\le \underset{x\in C}{sup}{\rho }_{J_a}\left(x\right)\le \alpha .$$

In both cases, we showed that $\underset{x\in C}{sup}{\rho }_{J_{g\left(\alpha \right)}}\left(x\right)\le \alpha$, for all $\alpha >0$, i.e., $C\subset C_g$ as claimed. □

Proposition 2 allows us to focus on the subsets $C_f$ instead of subsets which satisfy the condition $\left(UD\right)$. The next result is amazing and surprising since it tells us that the subsets $C_f$ enjoy nice modular geometric properties despite the fact that $p^{-}=1$.

Theorem 2.

Consider the vector space ${\ell }_{p(.)}$ such that $p^{-}=1$ and $p(·)$ not identically equal to 1. Let $f:(0,+\infty )\to (1,2]$ be a non-decreasing function. Then, ρ is $\left(UUC2\right)$ on $C_f$.

Proof. 

Let $r>0$ and $\epsilon >0$. Let $x,y\in C_f$ such that $\rho \left(x\right)\le r$, $\rho \left(y\right)\le r$ and $\rho \left({\displaystyle \frac{x-y}{2}}\right)\ge r\epsilon$. Since $\rho$ is convex, we have

$$r\epsilon \le \rho \left({\displaystyle \frac{x-y}{2}}\right)\le {\displaystyle \frac{\rho \left(x\right)+\rho \left(y\right)}{2}}\le r,$$

which implies $\epsilon \le 1$. Set ${\displaystyle \alpha =\frac{r\epsilon }{2}}$. The properties of $C_f$ imply ${\displaystyle \frac{x-y}{2}\in C_f}$. So

$${\rho }_{J_{f\left(\alpha \right)}}\left({\displaystyle \frac{x-y}{2}}\right)\le \alpha ,$$

which implies

$${\rho }_{I_{f\left(\alpha \right)}}\left({\displaystyle \frac{x-y}{2}}\right)=\rho \left({\displaystyle \frac{x-y}{2}}\right)-{\rho }_{J_{f\left(\alpha \right)}}\left({\displaystyle \frac{x-y}{2}}\right)\ge r\epsilon -\alpha =\frac{r\epsilon }{2}.$$

Next, set

$$K=I_{f\left(\alpha \right)}\cap \left\{n,p\left(n\right)\ge 2\right\}andL=I_{f\left(\alpha \right)}\cap \left\{n,p\left(n\right)<2\right\}.$$

Since $I_{f\left(\alpha \right)}=K\cup L$, we obtain ${\rho }_{I_{f\left(\alpha \right)}}\left(z\right)={\rho }_K\left(z\right)+{\rho }_L\left(z\right)$, for all $z\in C_f$. From our assumptions, we have

$${\rho }_K\left({\displaystyle \frac{x-y}{2}}\right)\ge \frac{r\epsilon }{4}or{\rho }_L\left({\displaystyle \frac{x-y}{2}}\right)\ge \frac{r\epsilon }{4}.$$

Assume first that

$${\rho }_K\left({\displaystyle \frac{x-y}{2}}\right)\ge \frac{r\epsilon }{4}.$$

Using Lemma 1, we obtain

$${\rho }_K\left(\frac{x+y}{2}\right)+{\rho }_K\left(\frac{x-y}{2}\right)\le \frac{{\rho }_K\left(x\right)+{\rho }_K\left(y\right)}{2},$$

which implies

$${\rho }_K\left(\frac{x+y}{2}\right)\le \frac{{\rho }_K\left(x\right)+{\rho }_K\left(y\right)}{2}-\frac{r\epsilon }{4}.$$

Using the convexity of the modular, we have

$${\rho }_{L\cup J_{f\left(\alpha \right)}}\left(\frac{x+y}{2}\right)\le \frac{{\rho }_{L\cup J_{f\left(\alpha \right)}}\left(x\right)+{\rho }_{L\cup J_{f\left(\alpha \right)}}\left(y\right)}{2},$$

which implies

$$\rho \left(\frac{x+y}{2}\right)\le \frac{\rho \left(x\right)+\rho \left(y\right)}{2}-\frac{\epsilon r}{4}\le r\left(1-\frac{\epsilon }{4}\right).$$

For the second case, assume

$${\rho }_L\left(\frac{x-y}{2}\right)\ge \frac{\epsilon r}{4}.$$

Set

$$c=\frac{\epsilon }{8},L_1=\left\{n,|x_n-y_n|\le c\left(|x_n|+|y_n|\right)\right\}and{L}_2=L\backslash L_1.$$

Since $c<1$, we obtain

$${\rho }_{L_1}\left(\frac{x-y}{2}\right)\le \sum _{n\in L_1}\frac{c^{p\left(n\right)}}{p\left(n\right)}{\left(\frac{|x_n|+|y_n|}{2}\right)}^{p\left(n\right)}\le \frac{c}{2}\sum _{n\in L_1}\frac{|x_n{|}^{p\left(n\right)}+{\left|y_n\right|}^{p\left(n\right)}}{p\left(n\right)}.$$

Hence

$${\rho }_{L_1}\left(\frac{x-y}{2}\right)\le \frac{c}{2}\left({\rho }_{L_1}\left(x\right)+{\rho }_{L_1}\left(y\right)\right)\le \frac{c}{2}\left(\rho \left(x\right)+\rho \left(y\right)\right)\le \frac{c}{2}r.$$

Our assumption on ${\displaystyle {\rho }_L\left(\frac{x-y}{2}\right)}$ implies

$${\rho }_{L_2}\left(\frac{x-y}{2}\right)={\rho }_L\left(\frac{x-y}{2}\right)-{\rho }_{L_1}\left(\frac{x-y}{2}\right)\ge r\frac{\epsilon }{4}-\frac{c}{2}r\ge r\frac{\epsilon }{8}.$$

For any $n\in L_2$, we have

$$f\left(\frac{r\epsilon }{2}\right)-1=f\left(\alpha \right)-1\le p\left(n\right)-1\le p\left(n\right)(p\left(n\right)-1)$$

$$c\le c^{2-p\left(n\right)}\le {\left(\frac{|x_n-y_n|}{|x_n|+|y_n|}\right)}^{2-p\left(n\right)}.$$

Using Lemma 1, we obtain

$${\left|\frac{x_n+y_n}{2}\right|}^{p\left(n\right)}+\frac{\left(f\right(\alpha )-1)}{2}c{\left|\frac{x_n-y_n}{2}\right|}^{p\left(n\right)}\le \frac{1}{2}\left(|x_n{|}^{p\left(n\right)}+{\left|y_n\right|}^{p\left(n\right)}\right),$$

for any $n\in L_2$. Hence

$${\rho }_{L_2}\left(\frac{x+y}{2}\right)\le \frac{{\rho }_{L_2}\left(x\right)+{\rho }_{L_2}\left(y\right)}{2}-\frac{r\left(f\left(\alpha \right)-1\right){\epsilon }^2}{128},$$

which implies

$$\rho \left(\frac{x+y}{2}\right)\le r\left(1-\frac{\left(f\left(\alpha \right)-1\right){\epsilon }^2}{128}\right).$$

Both cases imply that $\rho$ is $\left(UC2\right)$ on $C_f$ with

$${\delta }_{2,C_f}(r,\epsilon )\ge min\left(\frac{\epsilon }{4},\frac{{\displaystyle \left(f\left(\frac{r\epsilon }{2}\right)-1\right){\epsilon }^2}}{128}\right)>0,$$

since $f\left(a\right)>1$, for any $a>0$. Since $f(·)$ is nondecreasing, we may set

$${\eta }_2(r,\epsilon )=min\left(\frac{\epsilon }{4},\frac{{\displaystyle \left(f\left(\frac{r\epsilon }{2}\right)-1\right){\epsilon }^2}}{128}\right)$$

to see that in fact $\rho$ is $\left(UUC2\right)$ on $C_f$ which completes the proof of Theorem 2. □

The following lemma will be useful:

Lemma 3.

Consider the vector space ${\ell }_{p(.)}$ such that $p^{-}=1$ and $p(·)$ not identically equal to 1. Let $f:(0,+\infty )\to (1,2]$ be a non-decreasing function. Set ${\displaystyle g\left(\alpha \right)=f\left(\frac{\alpha }{4}\right)}$, for $\alpha >0$. We have

$$C_f+C_f=\left\{x+y;x,y\in C_f\right\}\subset C_g.$$

Proof. 

Let $x,y\in C_f$. For any $n\in J_{g\left(\alpha \right)}=\left\{{\displaystyle n;p\left(n\right)\le f\left(\frac{\alpha }{4}\right)}\right\}$, we have

$${\left|\frac{x_n+y_n}{2}\right|}^{p\left(n\right)}\le \frac{1}{2}\left(|x_n{|}^{p\left(n\right)}+{\left|y_n\right|}^{p\left(n\right)}\right),$$

which implies

$$\frac{1}{p\left(n\right)}{\left|x_n+y_n\right|}^{p\left(n\right)}\le \frac{2^{p\left(n\right)-1}}{p\left(n\right)}\left(|x_n{|}^{p\left(n\right)}+{\left|y_n\right|}^{p\left(n\right)}\right).$$

Hence

$$\begin{array}{ccc}{\displaystyle {\rho }_{J_{f\left(\frac{\alpha }{4}\right)}}(x+y)}\hfill & \le \hfill & {\displaystyle 2^{f\left(\frac{\alpha }{4}\right)-1}\left({\rho }_{J_{f\left(\frac{\alpha }{4}\right)}}\left(x\right)+{\rho }_{J_{f\left(\frac{\alpha }{4}\right)}}\left(y\right)\right)}\hfill \\ & \le \hfill & {\displaystyle 2\left(\frac{\alpha }{4}+\frac{\alpha }{4}\right)}\hfill \\ & =\hfill & \alpha .\hfill \end{array}$$

Therefore ${\rho }_{J_{g\left(\alpha \right)}}(x+y)\le \alpha$, that is $x+y\in C_g$, which completes the proof of Lemma 3. □

In the next section, we will prove a fixed-point theorem for modular nonexpansive mappings.

4. Application

As an application to Theorem 2, we will prove a fixed-point result for modular nonexpansive mappings. The classical ingredients will be needed. First, we prove the proximinality of $\rho$-closed convex subsets which satisfies the condition $\left(UD\right)$.

Proposition 3.

Consider the vector space ${\ell }_{p(.)}$ such that $p^{-}=1$ and $p(·)$ not identically equal to 1. Let $f:(0,+\infty )\to (1,2]$ non-decreasing. Any nonempty ρ-closed convex subset C of $C_f$ is proximinal, i.e., for any $x\in C_f$ such that

$$d_{\rho }(x,C)=inf\left\{\rho (x-y);y\in C\right\}<\infty ,$$

there exists a unique $c\in C$ such that $d_{\rho }(x,C)=\rho (x-c).$

Proof. 

Without loss of generality, we assume that $x\notin C$. Since C is $\rho$-closed we have, $R=d_{\rho }(x,C)>0$. For any $n\ge 1$, there exists $y_n\in C$ such that $\rho (x-y_n)<R(1+1/n)$. We claim that $\{y_n/2\}$ is $\rho$-Cauchy. Assume not. Then there exists a subsequence $\left\{y_{\varphi \left(n\right)}\right\}$ of $\left\{y_n\right\}$ and ${\epsilon }_0>0$ such that

$$\rho \left(\frac{y_{\varphi \left(n\right)}-y_{\varphi \left(m\right)}}{2}\right)\ge {\epsilon }_0,$$

for any $n>m\ge 1$. According to Lemma 3, $\{x-y_{\varphi \left(n\right)}\}$ is in $C_g$, where $g\left(\alpha \right)=f(\alpha /4)$, for any $\alpha >0$. Fix $n>m\ge 1$. We have

$$max\left\{\rho \left(x-y_{\varphi \left(n\right)}\right),\rho \left(x-y_{\varphi \left(m\right)}\right)\right\}\le R\left(1+\frac{1}{\varphi \left(m\right)}\right).$$

Since

$${\epsilon }_0=R\left(1+\frac{1}{\varphi \left(m\right)}\right)\frac{{\epsilon }_0}{{\displaystyle R\left(1+\frac{1}{\varphi \left(m\right)}\right)}}\ge R\left(1+\frac{1}{\varphi \left(m\right)}\right){\epsilon }_1,$$

with ${\displaystyle {\epsilon }_1=\frac{{\epsilon }_0}{2R}}$, and using Theorem 2, we obtain

$$\begin{array}{ccc}{\displaystyle \rho \left(x-\frac{y_{\varphi \left(n\right)}+y_{\varphi \left(m\right)}}{2}\right)}\hfill & \le \hfill & {\displaystyle R(1+1/\varphi \left(m\right))\left(1-{\delta }_{2,C_g}\left(R\left(1+\frac{1}{\varphi \left(m\right)}\right),{\epsilon }_1\right)\right)}\hfill \\ & & \\ & \le \hfill & {\displaystyle R(1+1/\varphi \left(m\right))\left(1-{\eta }_2(R,{\epsilon }_1)\right),}\hfill \end{array}$$

where

$${\eta }_2(R,{\epsilon }_1)=min\left(\frac{{\epsilon }_1}{4},\frac{{\displaystyle \left(g\left(\frac{R{\epsilon }_1}{2}\right)-1\right){\epsilon }_1^2}}{128}\right).$$

Since $y_{\varphi \left(n\right)}$ and $y_{\varphi \left(m\right)}$ are in C and C is convex, we obtain

$$R=d_{\rho }(x,C)\le \rho \left(x-\frac{y_{\varphi \left(n\right)}+y_{\varphi \left(m\right)}}{2}\right)\le R(1+1/\varphi \left(m\right))\left(1-{\eta }_2(R,{\epsilon }_1)\right).$$

If we let $m\to +\infty$, we obtain

$$R\le R(\left(1-{\eta }_2(R,{\epsilon }_1)\right)<R.$$

This contradiction implies that $\{y_n/2\}$ is $\rho$-Cauchy. Since ${\ell }_{p(·)}$ is $\rho$-complete, there exists $y\in {\ell }_{p(·)}$ such that $\{y_n/2\}$$\rho$-converges to y. Since C is convex and $\rho$-closed, we conclude that $2y\in C$. Using the Fatou property, we have

$$\begin{array}{ccc}R=d_{\rho }(x,C)\hfill & \le \hfill & \rho (x-2y)\hfill \\ & \le \hfill & {\displaystyle \underset{m\to +\infty }{lim\; inf}\rho \left(x-\left(y+\frac{y_m}{2}\right)\right)}\hfill \\ & \le \hfill & {\displaystyle \underset{m\to +\infty }{lim\; inf}\underset{n\to +\infty }{lim\; inf}\rho \left(x-\frac{y_n+y_m}{2}\right)}\hfill \\ & \le \hfill & {\displaystyle \underset{m\to +\infty }{lim\; inf}\underset{n\to +\infty }{lim\; inf}\frac{\rho (x-y_n)+\rho (x-y_m)}{2}=R=d_{\rho }(x,C).}\hfill \end{array}$$

If we set $c=2y$, we obtain $d(x,C)=\rho (x-c)$. The uniqueness of the point c comes from the fact that $\rho$ is strictly convex on $C_g$ since it is $\left(UUC2\right)$. □

The next result discusses an intersection property known as the property $\left(R\right)$ [9]. Recall that a nonempty $\rho$-closed convex subset C of ${\ell }_{p(·)}$ is said to satisfy the property $\left(R\right)$ if for any decreasing sequence of nonempty $\rho$-closed $\rho$-bounded convex subsets of C have a nonempty intersection.

Proposition 4.

Consider the vector space ${\ell }_{p(.)}$ such that $p^{-}=1$ and $p(·)$ not identically equal to 1. Let $f:(0,+\infty )\to (1,2]$ be a non-decreasing function. Then $C_f$ satisfies the property $\left(R\right)$.

Proof. 

Let $\left\{C_n\right\}$ be a decreasing sequence of nonempty $\rho$-closed $\rho$-bounded convex subsets of $C_f$. Let $x\in C_1$. We have

$$d_{\rho }(x,C_n)=inf\{\rho (x-x_n);x_n\in C_n\}\le sup\left\{\rho (x-y),x,y\in C_1\right\}={\delta }_{\rho }\left(C_1\right)<\infty .$$

Since $\left\{C_n\right\}$ is decreasing, the sequence $\left\{d_{\rho }(x,C_n)\right\}$ is increasing bounded above by ${\delta }_{\rho }\left(C_1\right)$. Set $R=\underset{n\to +\infty }{lim}{d}_{\rho }(x,C_n)=\underset{n}{sup}{d}_{\rho }(x,C_n)$. If $R=0$, then $x\in C_n$ for any $n\ge 1$, which will imply ${\displaystyle \bigcap _{n\ge 1}}{C}_n\ne \varnothing$. Otherwise, assume $R>0$. Using Proposition 3, there exists $c_n\in C_n$ such that $d_{\rho }(x,C_n)=\rho (x-c_n)$, for any $n\ge 1$. Similar argument as the one used in the proof of Proposition 3 will show that $\{c_n/2\}$ is $\rho$-Cauchy and converges to $c\in {\ell }_{p(·)}$. Since $\left\{C_n\right\}$ is a decreasing sequence of $\rho$-closed subsets, we conclude that $2c\in {\displaystyle \bigcap _{n\ge 1}}{C}_n$. Again this will show that ${\displaystyle \bigcap _{n\ge 1}}{C}_n\ne \varnothing$ which completes the proof of Proposition 4. Moreover, using Fatou property, we note that

$$\rho (x-2c)\le \underset{m\to +\infty }{lim\; inf}\underset{n\to +\infty }{lim\; inf}\rho \left(x-\frac{c_n+c_m}{2}\right),$$

which will imply

$$d_{\rho }\left(x,\bigcap _{n\ge 1}{C}_n\right)=\underset{n\to +\infty }{lim}{d}_{\rho }(x,C_n).$$

□

Remark 1.

Let us note that under the assumptions of Proposition 4, the conclusion still holds when we consider any family ${\left\{C_{\alpha }\right\}}_{\alpha \in \Gamma }$ of nonempty, convex, ρ-closed subsets of C, where $(\Gamma ,\prec )$ is upward directed, such that there exists $x\in C$ which satisfies ${\displaystyle \underset{\alpha \in \Gamma }{sup}{d}_{\rho }(x,C_{\alpha })<\infty }$. Indeed, set ${\displaystyle d=\underset{\alpha \in \Gamma }{sup}{d}_{\rho }(x,C_{\alpha })}$. Without loss of generality, we may assume $d>0$. For any $n\ge 1$, there exists ${\alpha }_n\in \Gamma$ such that

$$d\left(1-\frac{1}{n}\right)<d_{\rho }(x,C_{{\alpha }_n})\le d.$$

Since $(\Gamma ,\prec )$ is upward directed, we may assume ${\alpha }_n\prec {\alpha }_{n+1}$ which implies $C_{{\alpha }_{n+1}}\subset C_{{\alpha }_n}$. Proposition 4 implies $C_0={\displaystyle \bigcap _{n\ge 1}}{C}_{{\alpha }_n}\ne \varnothing$. Clearly $C_0$ is ρ-closed and using the last noted point in the proof of Proposition 4, we obtain

$$d_{\rho }(x,C_0)=\underset{n\to +\infty }{lim}{d}_{\rho }(x,C_{{\alpha }_n})=\underset{n\ge 1}{sup}{d}_{\rho }(x,C_{{\alpha }_n})=d.$$

Let $c_0\in C_0$ such that $d_{\rho }(x,C_0)=\rho (x-c_0)$. We claim that $c_0\in C_{\alpha }$, for any $\alpha \in \Gamma$. Indeed, fix $\alpha \in \Gamma$. If for some $n\ge 1$ we have $\alpha \prec {\alpha }_n$, then obviously we have $c_0\in C_{{\alpha }_n}\subset C_{\alpha }$. Therefore let us assume that $\alpha \nprec {\alpha }_n$, for any $n\ge 1$. Since $\Gamma$ is upward directed, there exists ${\beta }_n\in \Gamma$ such that ${\alpha }_n\prec {\beta }_n$ and $\alpha \prec {\beta }_n$, for any $n\ge 1$. We can also assume that ${\beta }_n\prec {\beta }_{n+1}$ for any $n\ge 1$. Again we have $C_1={\displaystyle \bigcap _{n\ge 1}}{C}_{{\beta }_n}\ne \varnothing$. Since $C_{{\beta }_n}\subset C_{{\alpha }_n}$, for any $n\ge 1$, we obtain $C_1\subset C_0$. Moreover we have

$$d=d_{\rho }(x,C_0)\le d_{\rho }(x,C_1)=\underset{n\ge 1}{sup}{d}_{\rho }(x,C_{{\beta }_n})\le d.$$

Hence, $d_{\rho }(x,C_1)=d$ which implies the existence of a unique point $c_1\in C_1$ such that $d_{\rho }(x,C_1)=\rho (x-c_1)=d$. Since ρ is $\left(SC\right)$ on $C_f$, we obtain $c_0=c_1$. In particular, we have $c_0\in C_{{\beta }_n}$, for any $n\ge 1$. Since $\alpha \prec {\beta }_n$, we conclude that $C_{{\beta }_n}\subset C_{\alpha }$, for any $n\ge 1$, which implies $c_0\in C_{\alpha }$. Since α was taking arbitrary in $\Gamma$, we obtain $c_0\in {\displaystyle \bigcap _{\alpha \in \Gamma }}{C}_{\alpha }$, which implies ${\displaystyle \bigcap _{\alpha \in \Gamma }}{C}_{\alpha }\ne \varnothing$ as claimed.

The next result is necessary to obtain the fixed-point theorem sought for $\rho$-nonexpansive mappings.

Proposition 5.

Consider the vector space ${\ell }_{p(.)}$ such that $p^{-}=1$ and $p(·)$ are not identically equal to 1. Let $f:(0,+\infty )\to (1,2]$ be a nondecreasing function. Then $C_f$ has the ρ-normal structure property, i.e., for any nonempty ρ-closed convex ρ-bounded subset C of ${\ell }_f$ not reduced to one point, there exists $x\in C$ such that

$${\displaystyle \underset{y\in C}{sup}\rho (x-y)<{\delta }_{\rho }\left(C\right).}$$

Proof. 

Let C be a $\rho$-closed convex $\rho$-bounded subset C of $C_f$ not reduced to one point. Since C is not reduced to one point, we have ${\delta }_{\rho }\left(C\right)>0$. Let $x,y\in C$ such that $x\ne y$. Set

$${\epsilon }_0=\frac{1}{{\delta }_{\rho }\left(C\right)}\rho \left(\frac{x-y}{2}\right)>0.$$

Fix $c\in C$. Using Lemma 3, we have $x-c$ and $y-c$ are in $C_f-C_f\subset C_g$, where $g\left(\alpha \right)=f(\alpha /4)$, for any $\alpha >0$. So far we have

$$max\left\{\rho (x-c),\rho (y-c)\right\}\le {\delta }_{\rho }\left(C\right)and\rho \left(\frac{x-y}{2}\right)\ge {\delta }_{\rho }\left(C\right){\epsilon }_0.$$

Theorem 2 implies

$$\rho \left(c-\frac{x+y}{2}\right)\le {\delta }_{\rho }\left(C\right)\left(1-{\delta }_{2,C_g}\left(R,{\epsilon }_0\right)\right).$$

Since c was taken arbitrary in C, we conclude that

$$\underset{c\in C}{sup}\rho \left(c-\frac{x+y}{2}\right)\le {\delta }_{\rho }\left(C\right)\left(1-{\delta }_{2,C_g}\left({\delta }_{\rho }\left(C\right),{\epsilon }_0\right)\right)<{\delta }_{\rho }\left(C\right)>0.$$

Therefore the proof of Proposition 5 is complete. □

Putting all this together, we are ready to prove the main fixed-point result of our work.

Theorem 3.

Consider the vector space ${\ell }_{p(.)}$ such that $p^{-}=1$ and $p(·)$ are not identically equal to 1. Let C be a nonempty ρ-closed convex ρ-bounded subset of ${\ell }_{p(.)}$, which satisfies the condition $\left(UD\right)$. Any ρ-nonexpansive mapping $T:C\to C$ has a fixed point.

Proof. 

Since C satisfies the condition $\left(UD\right)$, Proposition 2 secures the existence of a nondecreasing function $f:(0,+\infty )\to (1,2]$ such that C is a subset of $C_f$. The conclusion is trivial if C is reduced to one point. Therefore, we will assume that C is not reduced to one point, i.e., ${\delta }_{\rho }\left(C\right)>0$. Consider the family

$$\mathcal{F}=\left\{K\subset C,K\ne \varnothing ,\rho -closedconvexandT\left(K\right)\subset K\right\}$$

The family $\mathcal{F}$ is not empty since $C\in \mathcal{F}$. Since C is bounded, we use Remark 1 to be able to use Zorn’s lemma and conclude that $\mathcal{F}$ contains a minimal element $K_0$. Let us show that $K_0$ is reduced to one point. Assume not, i.e., $K_0$ contains more than one point. Set $co\left(T\left(K_0\right)\right)$ to be the intersection of all $\rho$-closed convex subset of C containing $T\left(K_0\right)$. Hence $co\left(T\left(K_0\right)\right)\subset K_0$ since $K_0\in \mathcal{F}$. Moreover, we have

$$T\left(co\left(T\left(K_0\right)\right)\right)\subset T\left(K_0\right)\subset co\left(T\left(K_0\right)\right),$$

which implies that $co\left(T\left(K_0\right)\right)\in \mathcal{F}$. $K_0$ being a minimal element of $\mathcal{F}$ we deduce that $K_0=co\left(T\left(K_0\right)\right)$. Using Proposition 5, we deduce the existence of $x_0\in K_0$ such that

$$r_0=\underset{y\in K_0}{sup}\rho (x_0-y)<{\delta }_{\rho }\left(K_0\right).$$

Define the subset $K=\left\{x\in K_0,\underset{y\in K_0}{sup}\rho (x-y)\le r_0\right\}$. K is not empty since $x_0\in K$. Note that we have $K={\displaystyle \bigcap _{y\in K_0}}{B}_{\rho }(y,r_0)\cap K_0$. Using the properties of modular balls, K is a $\rho$-closed and convex subset of $K_0$. Next, we prove that $T\left(K\right)\subset K$. Indeed, let $x\in K$. Since T is $\rho$-nonexpansive, we have

$$\rho (T\left(x\right)-T\left(y\right))\le \rho (x-y)\le r_0,$$

for all $y\in K_0$. So we have $T\left(y\right)\in B_{\rho }(T\left(x\right),r_0)\cap K_0$, which implies $T\left(K_0\right)\subset B_{\rho }(T\left(x\right),r_0)$. Since $K_0=co\left(T\left(K_0\right)\right)$, we conclude that $K_0\subset B_{\rho }(T\left(x\right),r_0)$, which implies

$$\rho (T\left(x\right)-y)\le r_0,$$

for all $y\in K_0$. Hence $T\left(x\right)\in K$. Since x was taken as arbitrary in K, we obtain $T\left(K\right)\subset K$. The minimality of $K_0$ will force $K=K_0$. Hence

$$r_0<{\delta }_{\rho }\left(K_0\right)={\delta }_{\rho }\left(K\right)\le r_0.$$

This is a contradiction. Therefore, $K_0$ is reduced to one point and it is a fixed point of T because $T\left(K_0\right)\subset K_0$. □

Remark 2.

In Theorem 3, the condition $\left(UD\right)$ can be replaced by the following condition which is slightly more general:

$$there exists x_0\in {\ell }_{p(.)} such that x_0+C satisfies the condition \left(UD\right).$$